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"Faliure is Success if we learn from it" Volume - 5 Issue - 4 October, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009


Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain [B.E] Cover Design & Layout Niranjan Jain 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

Dear Students, Find a mentor who can be your role model and your friend ! A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right. Some basic rules to know mentors : • The best mentors are successful people in their own field. Their behaviors are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them. It is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control. • Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field. Be cautious while searching for a mentor : • Select people to be your mentors who have the highest ethical standards and a genuine willingness to help others. • Choose mentors who have and will share superb personal development habits with you and will encourage you to follow suit. • Incorporate activities into your mentor relationship that will enable your mentor to introduce you to people of influence or helpfulness. • Insist that your mentor be diligent about monitoring your progress with accountability functions. • Encourage your mentor to make you an independent, competent, fully functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.) Getting benefited from a role-mode : Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors : • What would they do in my situation? • What do they do every day to encourage growth and to move closer to a goal ? • How do they think in general ? in specific situations ? • Do they have other facts of life in balance ? What effect does that have on their well-being ? • How do their traits apply to me ? • Which traits are worth working on first ? Later ? A final word : Under the right circumstances mentors make excellent role models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them . Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi



Volume-5 Issue-4 October, 2009 (Monthly Magazine)




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







Class 12 marks may become critical for IIT admission IIT-K postpones launch of its dream satellite

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


Dr. Rai Mahesh Kumar Sinha

Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS S Success Tips for the Months • "The way to succeed is to double your error rate." • "Success is the ability to go from failure to failure without losing your enthusiasm."

8-Challenging Problems [Set# 6] Students’ Forum Physics Fundamentals Magnetic Field Gravitation


• Along with success comes a reputation for wisdom. • They can because they think they can. • Nothing can stop the man with the Nothing can stop the man with the mental attitude from achieving his nothing on earth can help the man the wrong mental attitude.

right right goal; with

• Keep steadily before you the fact that all true success depends at last upon yourself.

XtraEdge for IIT-JEE


Key Concept Halogen Derivatives Halogen & Noble Gases Family Understanding: Organic Chemistry

• "Success is the maximum utilization of the ability that you have." • We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.




Mathematical Challenges Students’ Forum Key Concept Limit, Continuity & Differentiability Parabola, ElLipse & Hyperbola

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


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



Class 12 marks may become critical for IIT admission CHENNAI: Marks scored in the Class XII board examinations are likely to become a key determining factor in addition to the performance in the nerve-wracking Joint Entrance Examination (JEE) for admission into the prestigious Indian Institutes of Technology (IITs) by 2011. In a couple of months, a pan-IIT committee constituted by the union human resources development (HRD) ministry to suggest reforms to the JEE is expected to submit its report recommending ways and means to factor in the marks scored by students in higher secondary examinations while preparing the IIT merit list. A meeting of all IIT directors and JEE representatives held in Chennai over the weekend discussed the proposed changes. The proposal comes amidst widespread concern among top academicians over the current IIT admission system which is entirely dependant on JEE scores and ignores academic performances in board exams. The inherent weakness of such a system is that the IITs have been able to largely attract only students who have been "conditioned for the JEE" by high profile coaching centres in Kota and Hyderabad. Such students who lack "raw intelligence", as described by IIT Madras director M S Ananth in the past, are sometimes at sea after entering the campus.

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"We hope to devise a methodology to compute a normalised class XII cut-off eligibility score for each educational board (CBSE, ICSE, and State Boards). If it is approved, only students who have scored this cut-off mark would become eligible to appear for the JEE and consecutively for admission," IIT Madras deputy director V G Idichandy, who is heading the committee, said on Monday. The present eligibility norm of an aggregate score of 60% in Class XII determined by the IIT standing council, as opposed to 85% recommended by a JEE review committee four years ago, is considered too low a benchmark. "We are collecting data on Class XII results of the past four to five years from different boards in all states to base our recommendation on. Much will depend on how we compute an acceptable method to normalise the marks scored in different boards. You have nearly 40 boards of education in India," Professor Idichandy said. However, the more difficult part will be to convince authorities of all the boards to declare Class XII results within a specified timeframe every academic year. "This will be crucial for us as we have to base the JEE on the Class XII results. I personally think that this is where a common school board, at least at the level of higher secondary education, which has been proposed by the HRD minister Kapil Sibal, will be of help in determining any all-India merit list," he contended.


Idichandy acknowledged that the JEE cannot be abolished "but we want to give as much importance as possible, for the performance of students at the school level" in the IIT admissions.

Single-digit cutoffs continue to dog IIT NEW DELHI: In an unforeseen effect of RTI, globally respected IITs have been stuck in a spiral of low cut-offs in their joint entrance examinations (JEE) for the last three years even for general candidates. Despite all their efforts to pull out of the single-digit cut-offs they had fallen into in 2007 and 2008 (1,4 & 3 and 5,0 & 3 in Maths, Physics and Chemistry, respectively), IITs could improve only marginally this year, as evident from the marks announced earlier this month. Out of the maximum possible marks of 160 in each subject in 2009, the cut-offs in Maths and Chemistry barely broke into double digits (11 marks each) while it remained a single-digit score in Physics (8 marks). This is even after IITs abandoned the cut-off formula they had adopted in 2007 and 2008 (20 percentile or the best of the bottom 20 per cent of the candidates) and tried a new one in 2009 (average or mean of the marks of all the candidates). Such ridiculously low cut-offs have been dogging IITs ever since they found themselves at a loss to explain to the Central Information Commission the basis on which they had arrived at the respectably high cut-offs of 37, 48 and 55 in the 2006 JEE, which was the first OCTOBER 2009

to be held after RTI came into force in November 2005. In their third and latest attempt to explain the 2006 cut-offs, they set up a committee last month consisting of directors of IIT Guwahati and IIT Bombay, Gautam Barua and Dewang Khakhar, to submit a report to the Calcutta high court showing the exact calculations. The calculations contained in the 11-page report reveal that, in a major departure from the norms of fair selection, IITs had in the 2006 JEE excluded hundreds of high-aggregate scoring candidates even before arriving at the subject cut-offs, which was meant to be the first level of screening. It is because of this serious flaw in the implementation of the 2006 formula that IITs, in their two earlier attempts before the CIC and high court, could not account for the major mismatch between the stated cut-offs (37, 48 and 55) and those yielded by the two different formulas claimed by them (while the first formula produced cut-offs of -8, -3 & -6, the second resulted in 7,4 and 6). In a bid to bridge this wide gap, the Barua-Khakhar committee took recourse to the "iterative process", which is used to increase the cut-offs "with every iteration" to get the desired number of candidates. But while determining the cut-off of one subject through the iterative process, the committee eliminated the candidates who had high marks in the other two subjects. Thus, although they were supposed to be calculated separately through the iterative process, the cut-off of one subject affected the cut-offs of the other two subjects. The committee did not however admit this paradox anywhere in its report. Had the IITs implemented their belatedly-disclosed iterative procedure in a fair and

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transparent manner, the cut-offs would have actually been 42, 44 and 51, thereby reducing the deviation among the three subject cut-offs to 9 marks instead of 18 marks. This would have very significantly changed the composition of the merit list in 2006. And had they applied the iterative process in the JEE of the past three years as well, the IITs would have been able to take their low cut-offs to a more respectable level and spared themselves the embarrassment of admitting general candidates who got, for instance, 5% in Physics in last year's JEE.

Now, IIT counselling system goes online MUMBAI: If you've made it to an Indian Institute of Technology, you no longer need to travel to the campus to book your seat. The tech schools have decided to take the counselling process online, thus allowing students to submit their preferences a mix of streams and IITs from home. Currently, students from across the country travel to the closest IIT after they make their mark in the Joint Entrance Exam. "Now, all general category students will be allowed to submit their preferences online. However, all other candidates will have to travel to the nearest IIT campus for the same as they have to submit their certificates to us,'' said IIT-Guwahati director Gautam Barua. The decision to conduct the counselling online was taken when the directors recently met in Chennai to discuss plans for the upcoming JEE in April 2010. In another key decision, the IIT directors agreed to centrally conduct two or more rounds of seat allocation, to ensure that seats don't go abegging. While this year, the IITs for the first time conducted a second 4

round of seat allotments, it was held at the institute level. Students who took admission were offered internal betterment before the second allotment had taken place. So, if a student with a ranking of 1,104 in JEE-2009 did not take the seat allotted to him in IIT-B, another candidate with a lower ranking got his place (if he had opted for that subject and IIT-B in his preference form). Also, if a candidate signed up at IIT-Delhi in the first round, s/he were not allowed to move to say IIT-Madras or IIT-Bombay even if a slot opened there and these institutes were listed in his/her choices. "Now, we want to remove that barrier. A student will be allowed to move out of one IIT and join another, if he prefers to do so in the later rounds of seat allotment,'' added Barua. In another relief to students, the IITs have decided to put out the answer key of the entrance exam, soon after the exam ends.

HRD allows IITs to take non-PhDs as lecturers NEW DELHI: Close to three decades ago, the Indian Institutes of Technology (IITs) upped the bar for selecting faculty: only PhDs were allowed to take classes. Diluting that lofty standard, the HRD ministry has now allowed non-PhDs to join as lecturers. What's more shocking is that at least 10% jobs have been reserved at the lecturer's level, an obsolete term that has been scrapped from academia around the world. Making it tough for IITs to attract talent at the level of assistant professor is another clause that mandates the tech schools to take only those with three years' experience. IIT directors fear it might result in bright students preferring to take up posts at foreign universities where a fresher begins his career as an OCTOBER 2009

assistant professor and not as a lecturer. Earlier, the IITs too were taking fresh, bright PhDs at assistant professor level. While the directive on taking nonPhDs as lecturers is optional, the directors are clueless why it was inserted. "We don't need it. The four-tier recruitment concept is regressive and I don't understand why the government needs to disturb something that is in good equilibrium," asked an IIT director, who refused to be named. Currently, none of the IITs has faculty members who are nonPhDs, barring a few of them who joined the tech schools in the 70's when the country did not have too many PhDs. But the ministry says the decision to take nonPhDs has not been thrust upon IITs. "There is no coercion involved. Faculty crunch is a fact," one official said. "That clause was fine at the development stage. In the early years of the IITs, when we advertised for two posts, we used to get five applications. Now we get about 40 to 50, all of who are PhDs. But even now there are vacant posts for faculty merely because we are extremely choosy about who we pick," said a dean from IIT-Bombay. But some see no harm in this optional clause. "Allowing us to take non-PhDs is just an enabling clause. But what worries most of us is the provision that does not allow us to take bright PhDs fellows as assistant professors," said Gautam Barua, director of IIT-Guwahati. Several directors are seeing red over the fact that drawing up a rule to take 10% faculty as lecturers puts them in a "peculiar not-very-good position". Whether to take a candidate as a lecturer or as an assistant professor, said another director, "must be left to the good judgment of the selection panel".

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The same rules apply to other central technical institutes like Indian Institutes of Management, National Institutes of Technology and the Indian Institutes of Science Education and Research.

IIT-G ranks 10th among top tech institutes The Indian Institute of Technology, Guwahati, an outcome of the Assam Accord, has earned the distinction of being ranked 10th in a list of 67 top engineering and technology institutes in India. The honour attests to the relatively young institution’s impeccable academic and research credentials. The coveted top spot has been taken by Bangalore-based Indian Institute of Science, followed by IIT Kanpur in the second place and IIT Mumbai in the third slot. According to media reports the ranking has been made taking into account citations, publications and research record available between 1999 and 2008 in the Scopus International bibliographical database. Published in the Current Science the list has been prepared by G Pratap and BM Gupta of the National Institute of Science Communication and Information Resources and National Institute of Science, Technology and Development Studies, reports stated.

Central team to study IIT & IIM sites A Central team will visit the desert state next week to study the sites proposed to set up IITRajasthan and IIM. The final decision about the location of the institutes is likely only after the team's visit to the sites suggested by the state-constituted Vyas committee," Vipin Chandra Sharma, principal secretary, higher education, told TOI on Tuesday.


The state government had recently sent the committee report to the Union HRD ministry, which decided to see the proposed sites before taking a final call. "An HRD expert team would visit the state next week," Sharma said. He had gone to Delhi to discuss the setting up of the institutes in the state. "The Centre wants to expedite the site selection process as the project has already been delayed," Sharma pointed out. The previous BJP government had proposed Kota as the location for the IIT. However, the HRD ministry rejected this on the ground that Kota is not connected by air and also cited the presence of tutorials as another deterrent. After assuming power, chief minister Ashok Gehlot constituted the Vyas committee, which recommended IIT at Jodhpur and IIM at Udaipur. Interestingly, Rajasthan is the only one among seven states where the issue of IIT location is still dragging. This, despite the fact that the state encompasses 11% of the country's land and apparently it possesses the largest land bank. One of the tiniest, Himachal Pradesh has, however, identified the land near Mandi. The other states such as Andhra Pradesh, Bihar, Madhya Pradesh, Gujarat and Punjab too have sealed land for the new IITs. IIT-R at present is functional at IIT-Kanpur, which is overburdened with the presence of the two batches of IIT-R.

Two IIT-K profs part of Chandrayan-II mission Banking upon the rich expertise the IIT-Kanpur has in field of research and technology, the Indian Space Research Organisation (ISRO), which plans to send a lunar rover as a part of Chandrayan-II mission to the moon in the year 2012, has handed over the responsibility of OCTOBER 2009

development and testing of computer vision based autonomous 3D map generation and development and validation of kinematic traction control models (a sub-controller which will correct the path of the rover due to slip and slide) to the two professors of this prestigious institute. Dr Ashish Dutta, Associate Professor, Dept of Mechanical Engineering at IIT-K who is working on the development and validation of kinematic traction control models said, "In 2012 ISRO plans to send a lunar rover as a part of Chandrayan-II mission to the moon. The landing module would carry a mobile robot (rover) that would emerge out of the lander to explore the surface and also perform scientific experiments." The IIT-K is involved in the following two aspects of the Chandrayan-II mission, first is the development and testing of computer vision based autonomous 3D terrain map generation and obstacle detection algorithms for path planning and second is development and validation of kinematic traction control models (a sub-controller which will correct the path of the rover due to slip and slide) for coordinating the six wheels of the rover based on wheels and surface interaction, said Dr Dutta "The lunar terrain consists of loose sand, dust, craters, ash etc. It is expected that due to slip, sinkage of the wheels the rover may not function as desired and drift from its desired path or may even overturn. Hence, terrain properties strongly influence rover mobility and eventually the success of the mission", he added. Dr KS Venkatesh, Associate Professor, Department of Electrical Engineering, IIT-Kanpur is working on the visual navigation of the lunar surface. Dr Dutta further elucidated that the vision

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based map generation using a single stationary camera and structured light has already been completed. The system is capable of functioning in real time with reasonable computing resources. This method is now being extended to mobile platforms where the cameras would be mounted on a prototype rover moving on an uneven terrain.

hope that our project would not get delayed further," Dhande said. The satellite weighing around five kilograms is 34 cm long and 10 cm broad and has been designed to collect information regarding flood and drought situations in the country.

In order to identify the wheel and surface interaction parameters, a one-wheel test set up has been developed to study the variation of slip, friction etc for different types of lunar like terrain conditions. A kinematic control model of a six wheel rover with a rocker-bogey mechanism has also been developed. Finally, the vision based system would give us the 3D map of the terrain based on which the traction control algorithm would give the safest path for the rover, said he.

On Sunday students, researchers and scientists of IIT Kharagpore (IIT-KGP) had gathered to watch a game of football. As they rooted for their teams in the five-a-side match, all the players have been built by them and by students from three other technical institutes.

He concluded by saying that the projects are being funded through two MoUs signed between IITKanpur and VSSC (Vikram Sarabhai Space Centre). VSSC is a centre of the Department of Space, Government of India.

IIT-K postpones launch of its dream satellite The launch of IIT-K's ISRO funded dream project, nano satellite 'Jugnu' has been postponed to next year, Director IIT-Kanpur said. "The project designed by the students and the scientists of the institution was scheduled to be launched by the end of this year but now it has been rescheduled for some time between Jan-March next year," Sanjay Govind Dhande, Director, IIT-Kanpur said today. Ruling out any link between the satellite's schedule with ISRO's Chandrayaan moon mission, Dhande said the institute will complete the project on time. "The students engaged in the project are a bit dejected by the jolt faced by Chandrayaan but we 6

Robots play soccer at IIT-KGP

The techies were witness to the prestigious RoboCup, being held in India for the first time. The RoboCup Challenge @ India 2009 was held from August 28 till 30 at the IIT campus, around 120 km southwest from Kolkata. The host of this event, IIT Kharagpur, has become the first institute in the country to obtain an approval from the International RoboCup Federation. “This is the first time that a RoboCup Challenge is being held in South-East Asia. It is adding yet another crown to IIT’s achievements in the field of robotics,” said its coordinator Mithilesh Gurujala. Each team spent around Rs 80,000 to built their robots. Although IITKGP and Hyderabad-based institutes International Institute of Information Technology were finalists, the match could not be completed on Sunday due to some technical snag. “There was also some problem with the batteries. We’re working on it and hopefully by 1 am on Monday we’ll be able to hold the final rounds,” said, Gurujala, also the referee for the matches.


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

Dr. Rai Mahesh Kumar Sinha Ph.D. (I.I.T. Kanpur) M.Tech (I.I.T. Kharagpur) M.Sc.Tech (Allabhad Univ.)

Dr. Rai Mahesh Kumar Sinha completed his graduation Allahabad University, and after that he was awarded by master degree as Master of Technology in Industrial Electronics from I.I.T., Kharagpur in 1969 then achieved Ph.D. in Computer Science, Indian Institute of Technology, Kanpur in 1973. Presently, he is working as Professor in I.I.T., Kanpur and related to various research works Areas of Interest: • Artificial Intelligence • Natural Language Processing, Machine Translation, Speech to Speech Translation, Indian Language Technology • Vision, Pattern Recognition, OCR, Document Processing • Computer Architecture Research & Projects R.M.K. Sinha works primarily in the area of Applied Artificial Intelligence. He applies AI techniques to document processing, text recognition, computer vision, speech processing, natural language processing and in design of knowledge based systems. Intercommunicating layers of knowledge and their integration is key to his design approach. R.M.K. Sinha also applies artificial neural networks and fuzzy computing techniques in pattern recognition. In natural language processing, one of the primary aims is to design machine aids for translation from English to Indian languages & vice-versa and among Indian languages. R.M.K. Sinha's approach is based on a new concept of using Pseudo-Interlingua, word expert model utilizing Karak theory, pattern directed rule base and hybrid example base. His investigations also include exploring design and development of special parallel architectures for computer vision and natural language processing.

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R.M.K. Sinha has been working on R & D for Indian Language Technology for the last three decades and his research has touched and provided direction to almost all facets of providing technological solution to the problem of overcoming the language barrier in the country. The multi-lingual GIST technology and several other packages for Indian language processing have been developed under his supervision. Some of the major projects that have been initiated and executed / currently being executed under his supervision are the following: • Machine Translation • Speech to Speech Translation • Optical Character Recognition • Vision Course Projects • Spell-checker design Honours & Recognition • Associate of UNESCO Chair in Communication: ORBICOM, Quebec, Canada. • Senior Member Institution of Electrical and Electronic Engineers (IEEE), USA. • Member Technical Advisory Committee of Centre for Development of Advanced Computing (CDAC), India. • Founder President, Society for Machine Aids for Translation and Communication (SMATAC) India. • Adjudged Best CS Teacher, Asian Institute of Technology, Bangkok, 1983. • Invited Expert on occasion of release of CD for Hindi fonts and Web-site by Smt. Sonia Gandhi, Vigyan Bhavan, June 20, 2005. • Member Selection Committees for IITs, Universities and Ministries.



KNOW IIT-JEE By Previous Exam Questions

Assume the earth to be a sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be 2.5 × 1031 kg-m. (c) If the same charge of Q as in part (a) above is given to a spherical conductor of the same radius R, what will be the energy of the system ? [IIT-1992] Sol. (a) In this case the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field with energy density dr


A wooden log of mass M and length L is hinged by a frictionless nail at O. A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision about O. [IIT-2005] O

M m v Sol.

We know that τ =

→ dL → ⇒ τ × dt = d L dt

1 ε0E2 (Energy/Volume) 2 (i) Energy stored within the sphere (U1) Electric field at a distance r is 1 Q E= . .r 4πε 0 R 3 u=

When angular impulse ( τ × d t ) is zero, the angular momentum is constant. In this case for the wooden log-bullet system, the angular impulse about O is constant. Therefore, [angular momentum of the system]initial = angular momentum of the system]final ...(i) ⇒ mv × L = I0 × ω where I0 is the moment of inertia of the wooden logbullet system after collision about O I0 = Iwooden log + Ibullet 1 = ML2 + ML2 ...(ii) 3 From (i) and (ii) mv × L ω= 1 2 2  3 ML + mL    3mv mv ⇒ ω= =  ML  (M + 3m)L  3 + mL   2.

ε  1 1 Q  ∈0E2 = 0  . r 2 2  4πε 0 R 3  Volume of element dV = (4πr2)dr Energy stored in this volume dU = U(dr)

ε  1 Q  dU = (4πr dr) 0  . 3 r 2  4πε 0 R 



dU =

1 Q2 . 6 .r4dr 40πε 0 R

U1 =




dU = Q2

8πε 0 R 6

Q2 8πε0 R 6

R 4 0

r dr

[r 5 ]0R

1 Q2 . 40πε 0 R (ii) Energy stored outside the sphere (U2) Electric field at a distance r is 1 Q E= . 2 4πε 0 R

(a) A charge of Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system. (b) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull amongst its constituent particles ?

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U1 =




ε  1 1 Q ∴ U = ε0E2 = 0  . 2 2 2  4πε 0 R  ε ∴ dU = µ . dV = (4πr2dr)  0 2  dU =


 1 Q   4πε . R 2 0 

   

  



Q Q + 40πε 0 R 8πε 0 R

Sol. (a) Magnetic field ( B ) at the origin = magnetic field due to semicircle KLM + Magnetic field due to other semicircle KNM → µ I µ I ∴ B = 0 (– ˆi ) + 0 ( ˆj ) 4R 4R → µ0 I ˆ µ0 I ˆ B =– i + j 4R 4R µ I = 0 (– ˆi + ˆj ) 4R ∴ Magnetic force acting on the particle

1 by G. 4πε 0

GM 2 R gR 2 ∴ G= M −3 MgR U= 5 Therefore, energy needed to completely disassemble the earth against gravitational pull amongst its constituent particle will be given by 3 E = |U| = MgR 5 Substituting the values, we get 3 E = (10m/s2) (2.5 × 1031 kg-m) 5 E = 1.5 × 1032 J (c) This is the case of a charged spherical conductor g=


F = q( v × B )

µ I = q{(–v0 ˆi ) × (– ˆi + ˆj )} 0 4R → µ 0 qv 0 I ˆ k F =– 4R →

(b) F KLM = F KNM = F KM → And F KM = BI(2R) ˆi = 2BIR ˆi →

F1 = F2 = 2BIR ˆi Total force on the loop, →

F = F1 + F2

or F = 4BIR ˆi Note : If a current carrying wire ADC (of any shape)

1 Q2 2 C

is placed in a uniform magnetic field B .


Q 1 Q = . 8πε 0 R 2 4πε 0 R

Then, F ADC = F AC →

or | F ADC| = ˆi (AC)B From this we can conclude that net force on a current carrying loop in uniform magnetic field is zero. In the question, segments KLM and KNM also form a loop and they are also placed in a uniform magnetic field but in this case net force on the loop will not be zero. It would had been zero if the current in any of the segments was in opposite direction.

A circular loop of radius R is bent along a diameter and given a shape as shown in figure. One of the semicircles (KNM) lies in the x-z plane and the other one (KLM) in the y-z plane with their centres at origin. Current I is flowing through each of the [IIT-2000] semicircles as shown in figure.

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net force F on the loop.

3 GM 2 5 R

of radius R, energy of which is given by =

applied determine the force F 1 and F 2 on the semicircles KLM and KNM due to the field and the


by replacing Q2 by M2 and

or U =


a velocity v = –v0 ˆi . Find the instantaneous force F on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B0 ˆj is

3 Q2 or U = 20 πε 0 R (b) Comparing this with gravitational forces, the gravitational potential energy of earth will be



K →



x N

(a) a particle of charge q is released at the origin with

dU =

U = U1 + U2 =





Q 2 dr 8πε 0 r 2

α dr Q2 Q2 . = R 8πε 0 R r 2 8πε 0 R Therefore, total energy of the system is

∴ U2 =





What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces? [IIT-2005]

µ1 =

λ 1Å = = 0.5 Å 2 2 Now, de broglie wavelength is given be n = 1, dmin =



µ2 = 2

∴ K=

µ3 = 3

= Sol. For total internal reflection on interface AB


1 2µ


Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above can form. [IIT-1997] Sol. As nodes are formed at each of the atomic sites, hence λ 2Å = n   ...(1) 2







N N (n+1) loops


N n loops

N λ/2

λ ...(2) 2 2.5 n +1 5 n +1 = , = or n = 4 ∴ 2 n 4 n Hence, from equation (1), λ i.e. λ = 1Å 2Å = 4 2 d will be minimum, when and

2.5 Å = (n + 1)

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(6.63 × 10 −34 ) 2 (1× 10

−10 2

) × 2 × 9.1× 10 −31 × 1.6 × 10 −19


(6.63) 2 × 102 eV = 151 eV 8 × 9.1× 1.6

 5   22.6  (0.02 M) =    L  (0.02 mol L–1) = 0.00113 mol. 2 1000    The chemical equations involved during the treatment of KI and the titration with Na2S2O3 are 2Cu2+ + 4I– → Cu2I2 + I2 and I2 + 2S2O32– → 2I– + S4O62– From these equations, we conclude 2 mol Cu2+ ≡ 4 mol I– ≡ 1 mol I2 and 1 mol I2 ≡ 2 mol S2O32– Now, Amount of S2O32– consumed = (11.3 mL)  11.3  (0.05 M) =  L  (0.05 mol L–1)  1000 

[Q Distance between successive nodes = λ/2] Hence from the figure 2Å N

λ2 .2m

A solution of 0.2 g of a compound containing Cu2+ and C2O42– ions on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 M Na2S2O3 solution for complete reduction. Find out the mole ratio of Cu2+ to C2O42– in the compound. Write down the balanced redox reactions involved in the above titration. [IIT-1991] Sol. The chemical equations involved in the titration of C2O42– with MnO4– are : MnO4– + 8H+ + 5e– → Mn2+ + 4H2O] × 2 C2O42– → 2CO2 + 2e–] × 5 2MnO4– + 5C2O42– + 16H+ → 2Mn2+ + 10CO2 + 8H2O From this equation, we conclude 2 mol MnO4– ≡ 5 mol C2O4–. Hence, 5 Amount of C2O42– in the solution =   (22.6 mL) 2






⇒ i = 60º ⇒ The minimum angle for total internal reflection for both the interface is 60º.


or K =


1 2 = = ; i = 45º µ 2 2 2 for total internal reflection on interface CD µ 3 1 sin i = 3 = 3 = 2 2µ 2µ sin i =


 11.3  =  (0.05) mol = 0.000565 mol  1000  Amount of Cu2+ equivalent to the above amount of S2O32– = 0.000565 mol Hence,


Amount of Cu 2 + Amount of

C 2 O 24 −


1 0.000565 = 2 0.00113


Using the data given below, calculate the bond enthalpy of C–C and C–H bonds. ∆CHº(ethane) = –1556.5 kJ mol–1 ∆CHº (propane) = –2117.5 kJ mol–1 C(graphite) → C(g); ∆H = 719.7 kJ mol–1 Bond enthalpy of H–H = 435.1 kJ mol–1 ∆fHº(H2O, 1) = –284.5 kJ mol–1 [IIT-1990] ∆fHº(CO2, g) = –393.3 kJ mol–1 Sol. From the enthalpy of combustion of ethane and propane, we write 7 (1) C2H6(g) + O2(g) → 2CO2(g) + 3H2O(1) : 2 ∆CH = 3∆fH(H2O, 1) + 2∆fH(CO2, g) – ∆fH(C2H6, g) Thus, ∆fH(C2H6,g) = – ∆CH + 3∆fH(H2O, 1)+ 2∆fH(CO2, g) = (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol–1 = – 83.6 kJ mol–1 (2) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(1) ∆CH = 3∆f H(CO2, g)+ 4∆fH(H2O), 1) – ∆fH(C3H8, g) Thus ∆fH(C3H8, g) = –∆CH + 3∆fH(CO2, g) + 4∆fH(H2O, 1) = (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol–1 = –101.0 kJ mol–1 To calculate the εC–H and εC–C, we carry out the following manipulations. (i) 2C(graphite) + 3H2(g) → C2H6(g) ∆H = – 83.6 kJ mol–1 2C(g) → 2C (graphite) ∆H = –2 × 719.7 kJ mol–1 6H(g) → 3H2(g) ∆H = –3 × 435.1 kJ mol–1 Add 2C(g) + 6H(g) → C2H6(g) ∆H(i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol–1 = – 2828.3 kJ mol–1 (ii) 3C(graphite) + 4H2(g) → C3H8(g) ∆H = –101.0 kJ mol–1 3C(g) → 3C (graphite) ∆H = –3 × 719.7 kJ mol–1 8H(g) → 4H2(g) ∆H = – 4 × 435.1 kJ mol–1 Add 3C(g) + 8H(g) → C3H8(g) ∆H(ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol–1 = – 4000.5 kJ mol–1 Now, ∆H(i) = εC–C – 6εC–H = –2828.3 kJ mol–1 ∆H(ii) = –2εC–C – 8εC–H = –4000.5 kJ mol–1 Solving for εC–C and εC–H, we get εC–H = 414.0 kJ mol–1 and εC–H = 344.3 kJ mol–1

When 3.06 g of solid NH4HS is introduced into a two-litre evacuated flask at 27ºC, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate Kc and Kp for the reaction at 27ºC. (ii) What would happen to equilibrium when more solid NH4HS is introduced into the flask ? Sol. The reaction along with the given data is NH4HS(s) NH3(g) + H2S(g) t = 0 3.06g (= 0.06mol) 0 0 0.3 × 0.06 mol 0.3 × 0.06 mol teq 0.7 × 0.06 mol = 0.018 mol = 0.018 mol (i) The equilibrium constant KC is  0.018 mol   0.018 mol  KC = [NH3][H2S] =    2L 2L    = 8.1 × 10–5 (mol/L)2 The equilibrium constant Kp is Kp = Kc(RT)∆vg = (8.1 × 10–5 mol2/L2) (0.082 atm L mol–1 K–1) (300 K)2 = 4.90 × 10–2 atm2 (ii) There will not be any effect on the equilibrium by introducing more of solid NH4HS as the equilibrium constant is independent of the quantity of solid. 8.


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A basic volatile nitrogen compound gave a foul smelling gas when treated with choroform and alcoholic potash. A 0.295 g sample of the substance, dissolved in aqueous HCl, and treated with NaNO2 solution at 0 ºC liberated a colourless, odourless gas whose volume corresponded to 112 mL at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N atom per molecule. [IIT-1993] Sol. Since the compound gives a foul smellings gas on treating with CHCl3 and alcoholic KOH, the compound must be a primary amine. RNH2 + CHCl3 + 3KOH → RNC + 3KCl 9.

alkyl isocyanide (foul smelling gas)

+ 3H2O ...(1) Since the compound on treating with NaNO2/HCl at 0 ºC produce a colourless gas, the compound must be an aliphatic primary amine. RNH2 + HNO2 → ROH + N2 + H2O Thus, the gas produced is nitrogen. 1 112 mL = mol Amount of gas liberated = −1 22400 mL mol 200 From the above equation, it is obvious that 1 Amount of compound RNH2 = mol 200 If M is the molar mass of RNH2, then 0.295 g 1 = mol M 200 or M = 0.295 × 200 g mol–1 = 59 g mol–1. 11


Thus, the molar mass of alkyl group R is (59 – 16)g, i.e. 43 g mol–1. Hence, R must be C3H7. From Eq. (2), it is obvious that the liquid obtained after distillation is ROH. Since this gives yellow precipitates with alkali and iodine (iodoform test), it must contain CH – C group.

Finally, the structure of A would be [O] CH3 CH3 C = CHCH3 C = O + HOOCCH3 CH3CH2 CH3CH2 (A)

The structure of B is


Hence, it is CH3 – CH – CH3 .

OH concluded






3-methyl pentane (B)

Hence, Molecular formula of A is C6H12 Structure of A is CH3CH2C = CHCH3




3-methyl pent-2-ene

10. A certain hydrocarbon A was found to contain 85.7 per cent carbon and 14.3 per cent hydrogen. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon B. 1.0 g of hydrocarbon A just decolourized 38.05 g of a 5 percent solution (by mass) of Br2 in CCl4. Compound A, on oxidation with concentrated KMnO4, gave compound C (molecular formula C4H8O) and compound C could easily be prepared by the action of acidic aqueous mercuric sulphate on 2-butyne. Determine the molecular formula of A and deduce the structures A, B and C. [IIT-1984] Sol. The ratio of atoms in the compound A is 85.7 14.3 C:H:: : : : 7.14 : 14.3 : : 1 : 2 12 1 Thus, Empirical formula of A is CH2. Since the compound A consumes 1 mol of hydrogen, the molecule of A contains only one carbon-carbon double bond. From the data on the absorption of bromine, we can calculate the molar mass of A as shown in the following. Mass of bromine absorbed by 1.0 g of hydrocarbon 5 = × 38.05 g 100 Mass of hydrocarbon absorbing 160 g (= 1 mol) of 1.0 × 160 g = 84.1 g. Br2 = (5 × 38.05 / 100) Hence, Molar mass of A is 84.1 g mol–1 The number of repeating CH2 group in one molecule of A will be 6(= 84.1/14). Hence, Molecular formula of A is C6H12. Now, it is given that

Structure of B is CH3CH2CHCH2CH3 CH3

3-methyl pentane

Structure of C is CH3CCH2CH3 O


MATHEMATICS 11. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001] Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws. Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements in the range. Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0 ⇒ M = 3n – 3.2n + 3 Hence the total number of favourable cases = (3n – 3.2n + 3). 6C3

⇒ required probability =

conc. KMnO

4 C 6 H12     → C 4 H 8 O + CH3COOH

(3n − 3.2 n + 3) × 6 C3 6n


(A )

12. A straight line L through the origin meets the line x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L2 are drawn, parallel to 2x – y = 5 and 3x + y = 5 respectively. Lines L1 and L2 intersect at R, shown that the locus of R as L varies, is a straight line. [IIT-2002]

The compound C is obtained by the hydration of 2-butyne. Hence, its structure obtained from the reaction is 2-butyne




OH Thus, the original compound is CH3 – CH – CH3








2-butanone (C)

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Sol. Let the equation of straight line L be y = mx m  3m   1  3 , , P≡  ; Q≡    m +1 m +1  m +1 m +1

Thus area of the trapezium 1 BCDE = (BC + DE) (KL) 2 2 1  3a  15a = (a + 4a)   = 2 4  2 

m−2 ...(1) m +1 3m + 9 equation of L2 : y + 3x = ...(2) m +1 By eliminating 'm' from equation (1) and (2), we get locus of R as x – 3y + 5 = 0, which represents a straight line. Now equation of L1 : y – 2x =

14. Let V be the volume of the parallelopiped formed by the vectors →

a = a1 ˆi + a2 ˆj + a3 kˆ ;

b = b1 ˆi + b2 ˆj + b3 kˆ

13. From a point A common tangents are drawn to the circle x2 + y2 = a2/2 and parabola y2 = 4ax. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. [IIT-1996] Sol. Equation of any tangent to the parabola, y2 = 4ax is y = mx + a/m. This line will touch the circle x2 + y2 = a2/2

y πx = – a/2


A(–a, 0)

c = c1 ˆi + c2 ˆj + c3 kˆ If ar, br, cr, where r = 1, 2, 3 are non-negative real 3

numbers and

→ → →

a 12 + a 22 + a 32

b12 + b 22 + b 32


c12 + c 22 + c 32


(a1 + a 2 + a 3 ) + (b1 + b 2 + b 3 ) + (c1 + c 2 + c 3 ) 3 [(a1 + a2 + a3) (b1 + b2 + b3) (c1 + c2 + c3)]1/3 ∴ L3 ≥ [(a1 + a2 + a3)(b1 + b2 + b3)(c1 + c2 + c3)] ..(2) Now, (a1 + a2 + a3)2 = a12 + a 22 + a 32 + 2a1a2 + 2a1a3 + 2a2a3 ≥ a12 + a 22 + a 32 Now, L =





⇒ (a1 + a2 + a3) ≥


a a (m2 + 1)   = 2 m 1 1 ⇒ = (m2 + 1) ⇒ 2 = m4 + m2 2 2 m ⇒ m4 + m2 – 2 = 0 ⇒ (m2 – 1)(m2 + 2) = 0 ⇒ m2 – 1 = 0, m2 = – 2 (which is not possible). ⇒ m=±1 Therefore, two common tangents are y = x + a and y = –x – a These two intersect at A(–a, 0) The chord of contact of A(–a, 0) for the circle x2 + y2 = a2/2 is (–a)x + 0.y = a2/2 or x = – a/2 and chord of contact of A(–a, 0) for the parabola y2 = 4ax is 0.y = 2a(x – a) or x = a Again length of BC = 2BK If

Similarly, and

a 12 + a 22 + a 32

(b1 + b2 + b3) ≥

(c1 + c2 + c3) ≥

b12 + b 22 + b 32

c12 + c 22 + c 32

∴ from (1) and (2) L3 ≥ [( a12 + a 22 + a 32 )( b12 + b 22 + b 32 )( c12 + c 22 + c 32 )]1/3 ≥ V 15. T is a prallelopiped in which A, B, C and D are vertices of one face and the just above it has corresponding vertices A´, B´, C´, D´, T is now compressed to S with face ABCD remaining same and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A´´ is a plane. [IIT-2004] Sol. Let the equation of the plane ABCD be ax + by + cz + d = 0, the point A´´ be (α, β, γ) and the height of the parallelopiped ABCD be h. | aα + b β + cγ + d | ⇒ = 90%. h a 2 + b2 + c2

= 2 OB2 − OK 2

⇒ aα + bβ + cγ + d = ± 0.9h a 2 + b 2 + c 2

a2 a2 a2 =2 =2 =a − 2 4 4 and we know that DE is the latus rectum of the parabola so its length is 4a.

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+ b r + c r ) = 3L. Show that [IIT-2002]

Sol. V = | a .( b × c ) | ≤



r =1

V ≤ L3.

C 2

∑ (a

⇒ locus is, ax + by + cz + d = ±0.9h a 2 + b 2 + c 2 ⇒ locus of A´ is a plane parallel to the plane ABCD



Physics Challenging Problems

Set # 6

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Dev Sharma Director Academics, Jodhpur Branch

So l ut i ons wi l l b e pub l i s he d i n ne x t i s s ue

1. Circular path with radius of the circular path

Passage # 1 (Ques. 1 to 4) The internal energy 'U' v/s PV graph where P is the pressure and V is the volume of an ideal gas filled up in a piston cylinder system is shown below If tan φ = b then

is r =

m a 2 + b2 q.B 0

2. Helix and the pitch of the Helix is


2πm .a q.B0

2πm .b q.B 0 4. Same path as followed by circulating electrons which is responsible for the unstable Rutherford atomic model, means spiral path of decreasing radius. 3. Helix and the pitch of the Helix is

(0, a) φ PV 1.

What is the atomocity and the shape of the Gaseous molecule if b = 3 and a = 2.


Write the relation of adiabatic index of the gas in terms of a or b or in terms of both a and b.




Passage # 2 (Ques. 6 to 8) Two conducting wires are sliding in two separate portions, the details of motion are given along with the figure. If terminals a and d are grounded then


If 'a' start varying with respect to time as a(t) = 2(3 + t) and b remains constant then draw the graph of CV v/s a where CV is the molar specific heat at constant volume.



Part-2 c vl



l C/R-3

If b start varying with respect to time as b(t) = c0 + c1t2 where c0 and c1 are positive constants df v/s t graph where f is the then find the slope of dt degrees of freedom for the gas.




x C/R – Conducting Rail R = 10 Ω vBl = 30 volts

A particle enters in the given magnetic field →

B = B0 kˆ where B0 is a constant with the velocity of


Current passing through resistance R and it's direction.


P/d across terminals a and c.


Energy of deutron accelerated by potential difference across b and c.

v = aˆi + bˆj where a and b are the positive constants. The place where the magnetic field exists and the particle moves is filled with the resistive medium then path followed by the particle is(Charge on particle q and mass m)

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B B ⊗











Physics Challenging Problems Que s t i o ns we r e Publ i sh e d i n Se p t e m be r I s s ue

As energies of light photons falling on metallic surface because of trichromatic light are 2eV, 2.8eV and 3eV The work function of the metal is 2.5eV so light waves/photons corresponding to frequency ω2 and ω3 are able to have photo electric effect but not of frequency ω1. so KEmax. due to light of frequency ω2 is (KEmax.)2 = E2 – W = 2.8 – 2.5 = 0.3eV KEmax. due to light of frequency ω3 is (KEmax.)3 = E3 – W = 3 – 2.5 = 0.5eV As (KEmax.)3 > (KEmax.)2 so fastest photo electron is related with (KEmax.)3 and the stopping potential will be 0.5 volt.

–3.4 eV n=2 n=1 Nucleus Accepts energy photon of 1.9eV so the final energy = –3.4 + 1.9 = –1.5 eV corresponds to n = 3 electron n=3 –1.5 eV Position of electron in hydrogen atom n=2 HA1 and HA2 & –3.4 eV hydrogen spectra

As light waves/photons corresponding to frequency ω1 are not able eject the photo electrons so there is no effect on stopping potential and photo current.


–13.6 eV n ( n − 1) 3(3 − 1) = =3 No. of spectral lines emitted = 2 2

As light waves/photons corresponding to frequency ω2 are able to have the photo electric effect but not ejecting the fastest moving photoelectron so, Photo current decreases but there is no effect on stopping potential.


First and second spectral lines of Lymean series and first spectral line of Balmer series.


Energy of photons from hydrogen spectrum tube are1.9 eV 10.2 eV 12.1 eV

As light corresponding to frequency ω3 is able to have photo electric effect and responsible to eject fastest photoelectron so, - Photo current decreases - As no photo electric effect takes place due to this light intensity = 0 So stopping potential is not 0.5 volt instead of that it is 0.3 volt because now the fastest electron is due to light of frequency ω2.

Ist line of Ist line of IInd line of Balmer series Lymen series Lymen series If 2nd line of Lymen series get aabsorbed by absorption column then KEmax. of fastest elect. = E – W = 10.2 – 2 = 8.2 eV (No photo electric emission due to 1st line of Balmer series) So stopping potential is 8.2 volt.

As hydrogen atom HA1 Ground state –13.6 eV

8. n=1 Nucleus Accepts energy photon of 12.1eV so the final energy = –13.6 + 12.1 = –1.5 eV corresponds to n = 3 As hydrogen atom HA2 First excited state

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Set # 5


If absorption column get removed then photon of energies 1.9eV, 10.2eV and 12.1eV falls on to metal. Now, KEmax of fastest elect. KEmax. = E – W = 12.1-2 = 10.1eV and stopping potential = 10.1volt so stopping potential increases from 8.2 volt to 10.1volt and as no. of photoelectrons ejected will be more because of simultaneous presence of 10.2eV and 12.1eV photons so photocurrent also increases.


Students' Forum PHYSICS

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

Two particles of masses m1 and m2 separated a distance L from each other are released from their initial rest state. What will their velocity be when the distance between them is l ? Sol. Notice that when the masses were released, the velocity of the center of mass was r 0.m1 + 0.m 2 =0 ...(1) v cm = m1 + m 2 Because both of the initial velocities are zero. Thus, the total momentum of the system is zero. We denote the velocities of the masses m1 and m2 as v1 and v2, respectively. Using the law of conservation of linear momentum, derived from the absence of external forces, we obtain : ...(2) m1v1 = m2v2 The gravitational force between the masses is conservative. Calculating the potential energy between the two masses at the moment of release, we arrive at : L L Gm m Gm1m 2 1 2 ...(3) Ep = – F dr = – dr = – 2 ∞ ∞ L r When the masses arrive at distance l we have : Gm1m 2 1 1 Gm1m 2 – =– + m1v12 + m2v22 ..(4) L 2 2 l Using this result along with Eq. (2), we obtain  2 2G 1 l  m 22  −  v1 = m1 + m 2  l L ..(5)  2 G 1 l  2 2 v 2 = m1  −   m1 + m 2 l L



1 1  = 2G  −  (m1 + m2) l L and therefore,


1 1  ..(8) 2 G  −  ( m1 + m 2 ) l L Another method of finding v12 is to use one of the masses as the frame of reference; m2 m1

r | v12 | =

a0 for example, m1 (fig. 1). The frame of m1 is not inertial. Its acceleration, a0, is : F 1 Gm1m 2 Gm 2 = = ...(9) a0 = m1 m1 r2 r2 The force exerted on m2 in this frame, F´, is the sum of the gravity and D'alembert's force, Gm1m 2  Gm  F´ = – – m2  2 2  2 r  r  1 = – Gm2(m1 + m2) 2 r The difference between the initial and the final kinetic energies, ∆Ek, must equal the work done by the force F´, ∆Ek = W, 1 where ∆Ek = m2 v 22 . 2 Therefore, r r2 r r l 1 W = r F.d r = − Gm 2 ( m1 + m 2 ) 2 dr r1 L r

We are also interested in the relative velocity, which can be expressed : r r r v12 = v1 – v 2 = (v1 + v2) rˆ We can solve this equation simply by plugging the calculated v1 and v2 into it, or as following : r r r r r r r | v12 |2 = | v1 – v 2 |2 = ( v1 – v 2 ). ( v1 – v 2 ) r r r r = v12 + v 22 – 2 v1 . v 2

Hence, v12 = 2.

...(6) = v12 + v 22 + 2v1v2 Notice that the directions of the velocities are opposite. Using Eq. (2) and Eq. (5), we have : m r | v12 |2 = v12 + v 22 + 2 1 v12 m2

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2G 1 1  2 2  −  ( m1 + m 2 + 2m1m2) m1 + m 2  l L 


1 1  2G − (m1 + m 2 ) as expected l L

A mass m1 is hung on an ideal (massless) spring. Another mass m2 is connected to the other end of the spring (see figure). The whole system is held at rest. At t = 0, m2 is released and the system falls freely due to gravity. Assume that the natural length of the spring is L0, its initial stretched length (before t = 0) is L and the acceleration due to gravity is g. (i) Find the position of the centre of mass as a function of time. OCTOBER 2009

Now in the current frame of reference, the equations of motion are defined as: m1&x&´1 = – k(x´1 – x´2 – L0) ...(7)

(ii) Write the equations of motion for the two masses in the frame of the laboratory. (iii) Find the distance between m1 and m2 as a function of time. m2 g

m 2 &x&´2 = + k(x´1 – x´2 – L0)


m1 Sol. (i) Denoting the position of the center of mass at t = 0 1 by x0, we can write: x(t) = x0 + gt2 2 where the downwards direction is defined as positive. (ii) Let us first find the force constant of the spring from the force equation of the initial state : m1g k= ...(1) L − L0 The equations of motion are : m2



Using the above six equations we can rewrite Eqs. (7-8) using only terms of r(t). The two equations will be identical, since using eq. (5) leaves us with only one degree of freedom. Therefore, m1m 2 &r& = µ &r& ...(11) m1 + m 2


m1 ••

m1 x1 = m1g – k(x1 – x2 – L0)

where µ =


m1 + m 2 x´1(t) m2

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m1m 2 is the reduced mass of the m1 + m 2

system. Another way to write the equation of motion is by dividing Eq. (2) by m1, dividing Eq. (3) by m2 and subtracting Eq. (3) from Eq. (2). The resulting equation is :

m2 x•• 2 = m2g + k(x1 – x2 – L0) ...(3) Note that (x1 – x2 – L0) > 0 implies positive acceleration (downwards) for m2, and negative acceleration (upwards) for m1. (iii) An easy way to solve this problem is by transforming to frame which accelerates with our system at g. In this frame, D' alernbert's force exists, balancing gravity so that we are left only with the force applied by the spring. Also, in this frame, the center of mass x´0 is at rest, so we choose x´0 is at rest, so we choose x´0 = 0. Therefore, m1x´1 ( t ) + m 2 x´2 ( t ) = x0(t) = x0(0) = 0 ...(4) m1 + m 2 This leads us to: m ...(5) x´t(t) = – 2 x2(t) m1 The distance between the two masses is given by : R(t) = R´(t) = x´1(t) – x´2(t) m = x´1(t) + 1 x´1(t) m2 =


We define r(t) ≡ R(t) – L0 as the displacement from equilibrium. From Eq. (6), we have :  m1 + m 2 x´1 ( t ) – L 0 r ( t ) = m2  m1 + m 2 & ...(9) x& ´1 ( t ) r ( t ) = m2  m1 + m 2 && &x&´1 ( t ) r ( t ) = m 2  From Eq. (5) we have  m2 x 2 (t ) x´1 = − m1  m2  ...(10) x& 2 ( t ) x& ´1 = – m1  m2 && x´1 = – m &x& 2 ( t ) 1 

 1 1   (x1 – x2 – L0) &x&1 – &x& 2 = – k  + m m 2   1 k = – (x1 – x2 – L0) ...(12) µ The solution to these equation is:  k  ...(13) r(t) = A cos  t + φ  µ    where A and φ are determined from the initial conditions, R(0) = L0, and R(0) = 0. Hence, R(t) = r(t) + L0  k  = L0 + (L – L0) cos  ..(14) t  µ    Note : the transformation to the frame of the centre of mass used here is quite common in two-body problems. In general, we replaced:




m1, m2 → µ =

v0 5 = v0 ...(4) 7/5 7 Note that this problem cannot be solved easily by using the principle of conservation of energy. The reason is the existence of the force of friction. When the mass stops sliding, the friction does not vanish, but it does not do any work, because the point of contact between the body and the plane, A, is temporarily at rest. Therefore, r r W = f .d r = 0 r for d r = 0. (iii) On a frictionless surface, the linear and angular momentum are conserved. Therefore, ω and v are unchanged, or :  v´= v  ω´= ω

m1m 2 m1 + m 2


x1, x2 → r = x2 – r1 This reduces the number of equations and variables. 3.

A cylindrical rigid body of mass M, radius R and moment of inertia I about its center of mass is thrown horizontaly onto a plane at a velocity v0. Initially, the body slides and does not roll. Gradually, it starts to roll as a result of the friction of the plane, until it finally rolls at a velocity v without sliding. (i) Compute the final velocity v. (ii) Find v for a spherical body. (iii) Later, the sphere reaches a perfectly smooth area of the plane. Find the angular velocity and the velocity of the center of mass on the smooth area. v0 t=0

A hollow steel sphere, weighing 200 kg is floating on water. A weight of 10 kg is to be placed on it in order to submerge when the temperature is 20ºC. How much less weight is to be placed when temperature increases to 25ºC ? Given γwater = 1.5 × 10–4 / ºC, αsteel = 1 × 10–5/ ºC Sol. At the instant of submergence, Total mass of sphere and weight placed on it = mass of water displaced ∴ mass of water displaced at 20ºC = (200 + 10)kg = 210 kg and volume of the sphere = volume of water displaced by it. 210 ∴ volume of sphere at 20ºC is V0 = ρ0 4.



t = t0


µ =0

Sol. (i) We use the principal of conservation of angular momentum about the point of contact between the body and the plane A to solve the problem. The r torque relative to this point vanishes ( N = 0); so, r dJ = 0. dt J initial = Mv 0 R r r ...(1)  J final = J in ,cm + R + p = Iω + Mv& R

where ρ0 is density of water at 20ºC. volume of sphere at 25º C becomes equal to V = V0(1 + 3αs∆θ) 210 210.0315 [1 + 3 × 10–5(25 – 20)] = = ρ0 ρ0 Density of water at 25ºC becomes equal to ρ = ρ0(1 – γw∆θ) = 0.99925 ρ0 Mass of water to be displaced at 25ºC in order to submerge the sphere = V.ρ = 209.847 kg ∴ Required difference of weight to be placed on it = 210 – 209.87 = 0.126 kg Ans.

We arrive at Eq. (1) by relying upon the fact that the final angular momentum about the point A equals the angular momentum in the center of mass frame, Iω, plus the angular momentum of the center of mass point about the point A. At t = 0 the only motion of the mass is the rolling; therefore, v = ωR. Hence, v + MvR ...(2) Jf = I R Applying the principle of conservation of momentum, Ji = Jf, we obtain: v0 ...(3) v= I 1+ MR 2 (ii) For a spherical rigid body we know that 2 I = MR2. Therefore, 5

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Suppose a nucleus X, initially at rest, undergoes αdecay according to the equation, 225 92 X

→ Y + α

The emitted α-particle is found to move along a helical path in a uniform magnetic field of induction B = 5T. Radius and pitch of the helix traced by the α18


particle are R = 5 cm and p = 7.5 π cm, respectively. Calculate binding energy per nucleon of nucleus X. Given that, m(Y) = 221.03 u m(α) = 4.003 u m(n) = 1.009 u m(p) = 1.008 u 2 Mass of α-particle = × 10–26 kg 3 1 u = 931 MeV/c2 Sol. Let velocity of emitted α-particle be v at angle θ with the direction of magnetic field. Then radius of helical path traced by the α-particle, mv sin θ R= qB

You Should Know Nuclear Physics : •

Alpha particles are the same as helium nuclei and have the symbol


RqB = 1.2 × 107 ms–1 or v sin θ = m where q (charge of α-particle) = 3.2 × 10–19 coulomb. 2πmv cos θ and pitch, p = qB

The atomic number is equal to the number of protons (2 for alpha)

Deuterium (

pqB = 9 × 106 ms–1. or v cos θ = 2πm ∴ velocity of emitted α-particle,

The number of nucleons is equal to protons + neutrons (4 for alpha)

Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator.

Natural radiation is alpha ( ), beta ( and gamma (high energy x-rays)

A loss of a beta particle results in an increase in atomic number.

All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2)

Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers).

Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.

Rutherford discovered the positive nucleus using his famous gold-foil experiment.

Fusion requires that hydrogen be combined to make helium.

Fission requires that a neutron causes uranium to be split into middle size atoms and produce extra neutrons.

Radioactive half-lives can not be changed by heat or pressure.

One AMU of mass is equal to 931 meV of energy (E = mc2).



Regents Physics


( v cos θ) 2 + ( v sin θ) 2

= 1.5 × 107 ms–1. When an α-particle is emitted with velocity v from a stationary nucleus X, decay product (nucleus Y) recoils. According to law of conservation of momentum, that recoil velocity V of Y is given by myV = mav ...(1) where mass of nucleus Y, 221.03 2 my = × × 10–26 kg 4.003 3 ∴ V = 2.715 × 105 ms–1 ∴ Total energy released during α-decay of nucleus X is E = kinetic energy of nucleus Y + kinetic energy of α-particle 1 1 E = myV2 + mαv2 = 4.77 MeV 2 2 Hence, mass lost during α-decay, E u = 0.005 u ∆m = 931 ∴ mass of nucleus, X, mx = my + mα + ∆m = 225.038 u mass defect in nucleus X, md = [92mp + (225 – 92) mn] – mx ∴ md = 1.895 u ∴ Binding energy per nucleon in nucleus X m × 931 MeV = d 225 = 7.84 MeV Ans.

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) is an isotope of hydrogen





Magnetic Field KEY CONCEPTS & PROBLEM SOLVING STRATEGY Magnetic field : A magnetic field of strength B is said to exist at a point if a current element or a moving charged particle passing through the point experiences a sideways force equal in magnitude to ∆F = I∆ lB sin θ or qvB sin θ ∆

Magnetic field at the centre of a Circular Coil :

µ 0 NI tesla (T) 2a where, a = radius of the coil, N = its number of turns, I = current. The direction of B is along the axis of the coil. Magnetic Field a Point on the Axis of a Coil : B=


I θ






where ∆l is the length of the current element, q is the charge moving with velocity v, and θ is the angle between the direction of B and the current element, or between B and v, 0 < θ < π. The direction of the force ∆F is always perpendicular to the plane containing ∆l and B, or v and B. In the figures, this would mean the plane of the paper. The sense of ∆F is that in which a screw would move if rotated from ∆l or v to B through θ. In this case, this would mean a clockwise rotation, causing ∆F to be directed into the paper. In vector notation, this is summarized as r r r r r q( v × B ) ∆ F = I(∆ l × B ) or The unit of B is tesla (T) or newton per ampere metre or weber per square metre. B is called the magnetic induction. Biot-Savart Law : A moving charge or any current element give rise to a magnetic field. This is given by (∆B)p =


µ0 I (sin θ1 + sin θ2) 4π d where d is the perpendicular distance of the point from the conductor, θ1 and θ2 are the angles subtended by the upper and lower portions of the conductor at the point. µ 0 2I 4π d Magnetic Moment of a Loop : Magnetic moment of a current loop(m) = IS (current × area) or m = IS ampere metre2 When the conductor is long


Torque on a current Loop τ = m B sin θ where θ is the angle between normal to the loop and the magnetic field. Energy of a Current Loop in a Magnetic Field U = Uθ = 0 + mB(1 – cos θ) Work Done in Turning a Current Loop


W = mB(1 – cos θ)

Problem Solving Strategy : Magnetic Forces : Step 1 : Identify the relevant concepts : The righthand rule allows you to determine the magnetic force on a moving charged particle. Step 2 : Set up the problem using the following steps : r r Draw the velocity vector v and magnetic field B with their tails together so that you can visualize the plane in which these two vector lie.

where (∆B)p is the contribution of ∆l to the magnetic field at P, while µ0 is a universal magnetic constant with the value µ0 = 4π × 10–7 weber/ampere metre or henry per metre. The direction of (∆B)P is perpendicular to the plane containing ∆l and r, with the same sense as the motion of a screw which is r r rotated from ∆ l towards r through the smaller angle.

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tesla (T)


P θ

2( a 2 + x 2 ) 3 / 2

where x is the distance of the point from its centre. Magnetic Field due to a Straight Conductor at a Point :

µ 0 I∆l sin θ 4π r2 I

µ 0 NIa 2

Identify the angle φ between the two vectors.



Identity the target variables. This may be the magnitude and direction of the force, or it may be r r the magnitude or direction of v or B .

Problem Solving Strategy : Magnetic Field Calculations : Step 1 : Identify the relevant concepts : The law of Biot and Savart allows you to calculate the magnetic field due to a current –carrying wire of any shape. The idea is to calculate the field due to a representative current element in the wire, then combine the contributions from all such elements to find the total field. Step 2 : Setup the problem using the following steps : Make a diagram showing a representative current element and the point P at which the field is to be determined (the field point). r Draw the current element d l , being careful to ensure that it points in the direction of the current. Draw the unit vector rˆ . Note that it is always directed from the current element (the source point) to the field point P. Identify the target variables. Usually they will be the magnitude and direction of the magnetic field r B. Step 3 : Execute the solution as follows : r r µ 0 I d l × rˆ µ 0 I dl sin φ Use eq. dB = or dB = 4π 4π r 2 r2 r to express the magnetic field dB at P from the representative current element. r Add up all the dB 's to find the total field at point r P. In some situations the dB 's at point P have the same direction for all the current elements; then r the magnitude of the total B field is the sum of r r the magnitudes of the dB 's. But often the dB 's have different direction for different current elements. Then you have to set up a coordinate r system and represent each dB in terms of its r components. The integrals for the total B is then expressed in terms of an integral for each component. Sometimes you can use the symmetry of the r situation to prove that one component of B must vanish. Always be alert for ways to use symmetry to simplify the problem. Look for ways to use the principle of superposition of magnetic fields. Later in this chapter we'll determine the fields produced by certain simple conductor shapes; if you encounter a conductor of a complex shape that can be represented as a combination of these simple shapes, you can use superposition to find the field of the complex shape. Examples include a rectangular loop and a semicircle with straight line segments on both sides.

Step 3 : Execute the solution as follows :

r r r Express the magnetic force using Eq. F = q v × B The magnitude of the force is given by

Eq. F = qvB sin φ. r Remember that F is perpendicular to the plane of r r r r the vectors v and B . The direction of v × B is determined by the right-hand rule; keep referring to until you're sure you understand this rule. If q r r is negative, the force is opposite to v × B .

Step 4 : Evaluate your answer : Whenever you can, solve the problem in two ways. Do it directly from the geometric definition of the vector product. Then find the components of the vectors in some convenient axis system and calculate the vector product algebraically from the components. Verify that the results agree. Problem Solving Strategy : Motion in Magnetic Fields : Step 1 : Identify the relevant concepts : In analyzing the motion of a charged particle in electric and magnetic fields, you will apply Newton's second law r r of motion, Σ F = m a , with the net force given by r r r r ΣF = q (E + v × B) . Often, other forces such as gravity can be neglected. Step 2 : Setup the problem using the following steps : Determine the target variable(s). Often the use of components is the most efficient approach. Choose a coordinate system and then r r r express all vector quantities (including E , B , v , r r F and a ) in terms of their components in this system.

Step 3 : Execute the solutions as follows : If the particle moves perpendicular to a uniform magnetic field, the trajectory is circle with a radius and angular speed given by Eqs. mv R= and |q|B


|q|B v |q|B =v = R m mv

If your calculation involves a more complex r r trajectory, use ΣF = m a in component form: ΣFx = max, and so forth. This approach is particularly useful when both electric and magnetic fields are present.

Step 4 : Evaluate your answer : Check whether your results are reasonable.

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Step 4 : Evaluate your answer : Often your answer r will be a mathematical expression for B as a function of the position of the field point. Check the answer by examining its behavior in as many limits as you can.

B 2 q 2 r 2 1.2 2 (×3.2 × 10 −19 ) 2 × 0.45 2 = 2m 2 × 6.64 × 10 − 27 –12 = 2.25 × 10 J (c) E(energy acquired) = Vq or

E =

or V =

Solved Examples 1.


Two long wires a distance 2d apart carry equal antiparallel currents i, as shown in the figure. Calculate the magnetic induction at a point P equidistant from the wires at a distance D from a point midway between the wires. A

E 2.25 × 10 −12 = ⇒ V = 7.0 × 106 V q 3.2 × 10 −19

Use Biot –Savart law to calculate the magnetic field B at the common centre of the following circuits. i





D 2 + d 2 from each wire. µ 2i ∴ magnitude of field due each = 0 2 4π D + d 2

Sol. The point is at a distance

The direction of the field due to A is at right angles to AP and that due to B is at right angles to BP. Resolving the field along OP and perpendicular to it, the normal components cancel out and the components along OP are added. B sinθ A B d θ B cosθ 0 0 D P θ B cosθ d B B B sinθ ∴ B´ (field) at P = 2B cos θ along OP µ µ 0 id 2i d = =2× 0 4π D 2 + d 2 π( D 2 + d 2 ) D2 + d2






d dl

i b

a α

dα 0 µ idl sin 90º Then dB1 = 0 perpendicular into the 4π r12 plane of the paper µ i(r dα) ⇒ dB1 = 0 1 2 (Q dl = r1dα) 4π r1

⇒ B1 =


An alpha particle travels in a circular path of radius 0.45 m in a magnetic field with B = 1.2 Wb m–2. Calculate (a) its speed, (b) its kinetic energy, and (c) the potential difference through which if would have to be associated to achieve this energy. Mass of alpha particle = 6.64 × 10–27 kg. Sol. (a) Bqv = mr2/r ⇒ v = Bqr/m

µ0 i 4π r1



dα =

µ 0 iθ 4πr1

Similarly, B2, field at O due to cd =

µ 0 iθ out of the 4πr2

plane of the paper.

∴ B, field due to the loop abcd =

Bqr 1.2 × 3.2 × 10 −19 × 0.45 = m 6.64 × 10 − 27 7 –1 = 2.6 × 10 ms (b) Bqv = mv2/r

µ 0 iθ  1 1  − 4π  r1 r2

  

perpendicular into the plane. The second circuit is a special case of the above when θ = π 1 1 1 ∴ B = µ 0 i −  4  r1 r2 

or v =

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0 Sol. The field due to the straight portions da and bc is zero as the centre O is at end-on position relative to them. The field due to the curved parts are opposite as can seen by the screw rule. To find the magnitude due to either conductor, consider an element of width dl at angular distance α, from the radius Od. i


⇒ Bq = mv/r =












2mE  1 2 Q E = mv  r 2   22



A wire ring whose radius is 4 cm is at right angles to the general direction of a radically symmetrical diverging magnetic field as shown in the figure. The flux density in the region occupied by the wire itself is 0.1 Wb m–2 and the direction of the field everywhere is at an angle of 60º with the plane of the ring. Find the magnitude and direction of the force on the ring when the current in it is 15.9 A.




Earthquakes like hurricanes are not only super destructive forces but continue to remain a mystery in terms of how to predict and anticipate them. To understand the level of destruction associated with earthquakes you really need to look at some examples of the past.

Sol. Let us resolve the field along and perpendicular to the axis of the ring. The resolved parts are B sin θ and B cos θ. The forces on the elements of the ring due to the 'Bsinθ' component are in the plane of the ring and are distributed symmetrically towards the centre all along the ring, so they sum up to zero. But the forces on the elements due to 'Bcosθ' component are along the normal to the ring, hence they sum up to a resultant along that direction. ∴ F = ΣI∆lB cos θ = BIl cos θ = BI2πR cos θ (∴ l = 2πR) or F = 2πBIR cos θ = 2π × 0.1 × 15.9 × (4 × 10–2) cos 60º = 0.2N 5.

If we go back to the 27th July 1976 in Tangshan, China, a huge earthquake racked up an official death toll of 255,000 people. In addition to this an estimated 690,000 were also injured, whole families, industries and areas were wiped out in the blink of a second. The scale of destruction is hard to imagine but earthquakes of all scales continue to happen all the time. So what exactly are they ? Well the earths outer layer is made up of a thin crust divided into a number of plates. The edges of these plates are referred to as boundaries and it’s at these boundaries that the plates collide, slide and rub against each other. Over time when the pressure at the plate edges gets too much, something has to give which results in the sudden and often violent tremblings we know as earthquakes.

A long straight conductor carrying I1, is placed in the plane of a ribbon carrying current I2 parallel to the previous one. The width of the ribbon is b and the straight conductor is at a distance 'a' from the near edge. Find the force of attraction between the two. b

I1 a


The strength of an earthquake is measured using a machine called a seismograph. It records the trembling of the ground and scientists are able to measure the exact power of the quake via a scale known as the richter scale. The numbers range from 1-10 with 1 being a minor earthquake (happen multiple times per day and in most case we don’t even feel them) and 7-10 being the stronger quakes (happen around once every 10-20 years). There’s a lot to learn about earthquakes so hopefully we’ll release some more cool facts in the coming months.

Sol. Consider a thin strip at a distance x and of thickness dx. It is equivalent to a long straight conductor carrying (I2dx/b) current. µ I I dx dF (force of attraction) = 0 1 × 2 2πx b µ 0 I1I 2 dx = × 2πb x µ 0 I1I 2 a + b dx ∴ F= 2πb x x µ 0 I1I 2 a+b = ln 2πb b

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Gravitation KEY CONCEPTS & PROBLEM SOLVING STRATEGY Gravitational field (g) and potential (V) due to a spherical shell and a solid sphere : M M

Newton's Law of Gravitation : Two point masses m1 and m2, separated by' a distance r, attract each other with a force F=G

m1m 2

r2 where G = 6.67 × 10–11 Nm2 kg–2 = universal constant of gravitation. m1 m2 F F

(i) Outside g=–G


This force between two masses acts equally on both masses, acts though in opposite directions. It does not depend on the medium present between the two masses. Gravitation Field : This is a region in space where any mass will experience a force. The gravitational field strength (g) at a point is the force acting on a unit mass placed at that point. It is a vector. m




M r (ii) Inside g=0

M r (ii) Inside V=–G



r R3 where R is the radius of the sphere

M R where R is the radius of the shell


M 2R 3

(3R2 – r2)

Escape Velocity : The minimum velocity to be imparted to a body on the surface of a planet, so that it is carried beyond the gravitational field of that planet, is called the escape velocity of that planet. Obviously to carry the body beyond the gravitational field, the amount of energy needed is that which is required to bring it from infinity up to the surface of the planet. This is exactly the potential energy of the body. Potential energy per unit mass is equal to the potential of the field. So if m is the mass of the body and vc is the escape velocity, then


r2 The negative sign indicates that the gravitational field is always attractive. Gravitational Potential : The gravitational potential (V) at a points is the work that has to be done to bring a unit mass from infinity to that point. It is a scalar. The gravitational potential at a distance r from a point mass m is m r The negative sign arises because in bringing the unit mass from infinity, work is done by the system, so that its potential energy decreases. The potential at a point does not depend on the actual path followed in bringing the unit mass from infinity. Thus, gravitational force is a conservative force. V=–G

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r M g = – G 2 rˆ r


Any mass sets up a gravitational field around it. The gravitational field strength at a distance r from a point mass m is


r M g = – G 2 rˆ r V=–G




(i) Outside

g r

g = –G



M 1 mv e2 = Vm = G m 2 R where M is the mass of the planet and R is its radius. or


ve =



If g is the gravitational field intensity, then mg = ∴

ve =


A planet of mass m moves along a circle around the sun of mass ms with velocity v = 34.9 kms–1 with respect to the heliocentric frame of reference, that is, with the sun at the centre of the frame. Find the period of revolution of this planet around the sun and show that Kepler's third law, that is, the cube of the orbital radius is proportional to the square of the time revolution of planets, Given that ms = 1.97 × 1030 kg, G = 6.67 × 10–11 units. Sol. From the dynamics of circular motion (assuming circular orbit)

or GM = gR2

R2 2gR

Satellites and Orbital speed (V0) : A satellite is a small body revolving around a larger body under the gravitational attraction of the latter. The force of gravitational attraction provides the necessary centripetal force so that the satellite may be in rotational equilibrium. The speed at which rotational equilibrium is attained is called the orbital speed. Let it be v0. Then for rotational equilibrium

Gm s m v2 = Gm 2 ⇒ v2 = d d d ⇒ v = ωd m

mv 02 r where r is the radius of the orbit, measured from the centre of the planet. Fattraction =


Mm r2


∴ v2 = or T =

mv 02

v0 =

and if v0 < v <

2 v0 the path is a bigger ellipse with

point of projection as perigee. If v = parabolic and if v >


Two masses M1 and M2 at an infinite distance from each other and initially at rest, start interacting gravitationally. Find their velocity of approach when they are a distance s apart. Sol. Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved. 1 1 GM1M 2 ∴ 0 = M1v12 + M2v22 – 2 2 s (It is zero because in the beginning, both kinetic energy and potential energy are zero.) 0 = M1v1 + M2v2

mv 2 GM ⇒v= 2 r r r 1 GMm 1 GMm GMm ∴ E=– + =– r 2 r 2 r 1 GMm dr ⇒ dE = 2 r2 Also – dE = power × dt = Fv dt = av3dt GMm

1 GMm  GM  dr = –a   2 r2  r  m r–1/2 dr ⇒ dt = – 2a GM

2GM 22 s ( M1 + M 2 )

2GM 22 s ( M1 + M 2 ) V(velocity of approach) = v1 – (–v2) = v1 + v2

2G (M1 + M 2 ) s

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v22 =


v3 2π × 6.67 × 10 −11 × 1.97 × 10 30

An artificial satellite (of mass m) of the earth (radius R and mass M) moves in an orbit whose radius is n times the radius of the earth. Assuming resistance to the motion to be proportional to the square of velocity, that is F = av2, find how long the satellite will take to fall on to the earth. 1 GMm Sol. E (energy of the satellite) = – + mv2 r 2 By the dynamics of circular motion

Solved Examples


2πGm s


2 v0 the path is

2 v0 the path is hyperbolic.

Solving the equations v12 =

v3 Gm s



= 225 days (34.9 × 10 3 ) 3 Taking the equation in terms of d and ω Gmm s 4π 2 3 2 2 d = mω d ⇒ T = Gm s d2 ∴ T2 ∝ d3 This is Kepler's third law. =


GM r If a satellite is projected with velocity v < v0 the path is a small ellipse with point of projection as apogee or

Gm s ω v


m 2a GM

Q GM = gR2, t =




3/ 2


r −1/ 2 dr =

m a gR

m R a GM

( n – 1)

( n – 1)



A spaceship approaches the moon (mass = M and radius = R) along a parabolic path which is almost tangential to its surface. At the moment of maximum approach, the brake rocket is fired to convert the spaceship into a satellite of the moon. Find the change in speed. Sol. If v is the velocity at the vertex of the parabola, then v is also the escape velocity because if it is thrown with this velocity it will follow the parabolic path never to return to the moon.

At a Glance

Some Important Practical Units 1.

Par sec : It is the largest practical unit of distance. 1 par sec = 3.26 light year



X-ray unit : It is the unit of length. 1 X-ray unit = 10–13 m

v v orbit


2GM R ∆v = vfinal – vinitial = vorbit – vescape

1 slug = 14.59 kg

Now vescape =


GM 2GM GM – =– ( 2 – 1) R R R The negative sign means the speed has to be decreased.

⇒ ∆v =

∴ required change in speed =


R2 ⇒


⇒ ωrel =

= mω r

gR e2 R3


or ω


∴ ωrel =

10 × 6.4 2 × 1012 23 × 10 21

gR 2 R3 gR e2 R3


Barn : It is the unit of area. 1 barn = 10–28 m2


Cusec : It is the unit of water flow. 1 cusec = 1 cubic foot per second flow


Match No. : This unit is used to express velocity of supersonic jets. 1 match no. = velocity of sound = 332 m/sec.



(Q GM= gRe )

Knot : This unit is used to express velocity of ships in water. 1 knot = 1.852 km/hour

+ ωe

10. Rutherford : It is the unit of radioactivity. 1 rutherford (rd) = 1 × 106 disintegrations/sec


+ 7.27 × 10

11. Dalton : It is the unit of mass.

2π (Q ωe = = 7.27 × 10–5) 86400 ⇒ ωrel = 22.6 × 10–5 + 7.27 × 10–5 = 30 × 10–5 rad s–1 2π 2π ∴ τ= = ωrel 30 × 10 −5 = 2.09 × 104s = 5 hr 48 min

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Shake : It is the unit of time. 1 Shake = 10–6 second

GM ( 2 – 1) R

A satellite is revolving in a circular equatorial orbit of radius R = 2 × 104 km from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth = 6400 km and g (acceleration due to gravity) = 10 ms–2. Sol. Let ω be the actual angular velocity of the satellite from east to west and ωc be the angular speed of the earth (west to east). Then ωrelative = ω –(–ωe) = ω + ωe ⇒ ω = ωrel – ωe By the dynamics of circular motion 2

Chandra Shekhar limit : It is the largest practical unit of mass. 1 Chandra Shekhar limit = 1.4 × Solar mass



Slug : It is the unit of mass.

1 dalton =

1 mass of C12 = 931 MeV 12

= 1 a.m.u.

12. Curie : It is the unit of radioactivity. 1 curie = 3.7 × 1010 disintegration / sec





Organic Chemistry Fundamentals

(i) Resonance effect : In haloarenes (e.g., chlorobenzene), the lone pairs of electrons on the halogen atom are delocalized on the benzene ring as shown below :

Nature of the C–X Bond : Due to electronegativity difference between the carbon and the halogen, the shared pair of electron lies closer to the halogen atom. δ+ δ– –C : X


As a result, the halogen carries a small negative charge, i.e., δ – while the carbon carries a small positive charge, i.e., δ+. Consequently C–X bond is a polar covalent bond. Since the size of halogen atom increases as we move down the group in the periodic table, fluorine atom is the smallest and iodine the largest. Consequently, the carbon-halogen bond length increases and bond enthalpy decreases from C – F to C – I. Further, as we move from F to I, the electronegativity of the halogen decreases, therefore, the polarity of the C–X bond and hence the dipole moment of the haloalkane should also decrease accordingly. But the dipole moment of CH3F is slightly lower than that of CH3Cl. The reason being that although the magnitude of –ve charge on the F atom is much higher than that on the Cl atom but due to small size of F as compared to Cl the C – F bond distance is so small that the product of charge and distance, i.e., dipole moment of CH3F turns out to be slightly lower than that of CH3Cl. The bond lengths, bond enthalpies and dipole moments of halomethanes are given in table. Some Physical Data of Halomethanes (CH3–X) Halo methane

C–X bond length /pm

C–X bond enthalpy/kJ mol–1

Dipole moment / Debye


139 178

452 351

1.847 1.860












+Cl: –


V I II III IV (a) As a result, C – Cl bond acquires some double bond character, i.e., Cl is attached to C by little more than a single pair of electrons. On the other hand, in case of alkyl halides (say methyl chloride) carbon is attached to chloring by a pure single bond. Consequently, C – X bond in aryl halides is little stronger than in alkyl halides, and hence cannot be easily broken.

Like aryl, vinyl halides such as vinyl chloride can be represented as a resonance hybrid of the following structures : CH2 = CH – Cl:

+ :CH2 – CH = Cl:

As a result, C – X bond is vinyl halides, like in haloarenes, is little more stronger than in alkyl halides and hence cannot be easily broken.

(b) As discussed above, aryl halides are stabilised by resonance out alkyl halides are not. Consequently, the energy of activation for the displacement of halogen from aryl halides is much greater than that from alkyl halides. Thus, aryl halides are much less reactive than alkyl halides towards nucleophilic substitution reactions. (ii) Difference in hybridization of carbon atom in C – X bond. In haloalkanes (e.g., methyl chloride), the halogen is attached to sp3-hybridized carbon while in halogens or vinyl halides, the halogen is attached to sp2-hybridized carbon. Since a sp2hybridized orbital is smaller in size as compared to sp3-orbital of carbon, therefore, the C – Cl bond in chlorobenzene or vinyl chloride should be shorter and hence stronger than in methyl chloride. This has been confirmed by the X-ray analysis which shows that the C – Cl bond in chlorobenzene is 169 pm whereas in methyl chloride, it is 177 pm.

Reactivity of Haloarenes : Both haloalkanes (alkyl halides) and haloarenes (aryl halides) or vinyl halides contain a C – X bond but haloarenes and vinyl halides are extremely less reactive than haloalkanes towards nucleophilic substitution reactions. The following reasons can be given to account for the low reactivity of aryl and vinyl halides.

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Elimination-addition mechanism for nucleophilic aromatic substitution. Benzyne : When an aryl halide like chlorobenzene is treated with the very strong basic amide ion, NH2–, in liquid ammonia, it is converted into aniline. This is not the simple displacement that, on the surface, it appears to be. Instead, the reaction involves two stages : elimination and then addition. The intermediate is the molecule called benzyne (or dehydrobenzene). X NH2 NH2– NH2– NH3 NH3

Cl 177 pm sp3 C

Cl 169 pm H



Thus, in chlorobenzene, C – Cl bond is stronger than in methyl chloride and hence difficult to break.

(iii) Polarity (or Nature) of the carbon-halogen bond. Another reason for the low reactivity of aryl halides over alkyl halides is their lesser polar character. The sp2-hybrid carbon due to greater s-character is more electronegative than a sp3-hybrid carbon Therefore, the sp2-hybrid carbon of C – X bond in aryl halides or vinyl halides has less tendency to release electrons to the halogen than a sp3-hybrid carbon in alkyl halides. As a result , the C – X bond in aryl halides or vinyl halides is less polar than in alkyl halides. This is supported by the observation that the dipole moment of chlorobenzene is just 1.7 D as compared to the dipole moment of methyl cholride, i.e, 1.86 D. Consequently, the halogen atom present in aryl haldides cannot be easily displaced by nucleophiles. δ– X δ+


Aryl halide

Benzyne has the structure shown in fig. in which an additional bond is formed between two carbons (the one originally holding the halogen and the one originally holding the hydrogen) by sideways overlap of sp2 orbitals. This new bond orbital lies along the side of the ring, and has little interaction with the π cloud lying above and below the ring. The sideways overlap is not very good, the new bond is a weak one, and benzyne is a highly reactive molecule. H H

δ– δ+ —C—X

Haloarene or aryl halide (C – X bond is less polar and hence X cannot be displaced easily by nucleophiles.)


Haloalkane or alkyl halide (C – X bond is more polar than in aryl halides and hence X can be easily displaced by nucleophiles)

H Benzyne molecule. The sideways overlap of sp2 orbitals form a π bond out of the plane of the aromatic π cloud.

The elimination stage, in which benzyne is formed, involves two steps : abstraction of a hydrogen ion (step 1) by the amide ion to form ammonia and carbanion I, which then loses halide ion (step 2) to form benzyne.

(iv) Instability of phenyl cation : In haloarenes and vinyl halides, the phenyl cation or the vinyl cation formed as a result of self-ionization is not stabilized by resonance because the sp2-hybridized orbital of carbon having the +ve charge is perpendicular to the p-orbitals of the phenyl ring or the vinyl group. Therefore, these cations are not formed hence aryl and vinyl halide do not undergo nucleophilic substitution reactions (SN1 mechanism). +

– Cl Chlorobenzene

Phenyl cation

CH2 = CH – Cl Vinyl chloride

+ CH2 = CH + Cl– Vinyl cation

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X + NH2–

+ NH3 I


+ X–


+ Cl–





Elimination The addition stage, in which benzyne is consumed, may also involve two steps : attachment of the amide ion (step 3) to form carbanion II, which then reacts with an acid, ammmonia, to abstract a hydrogen ion (step 4). It may be that step (3) and step (4) are



concerted, and addition involves a single step; if this is so, the transition state is probably one in which attachment of nitrogen has proceeded to a greater extent attachment of hydrogen, so that it has considerable carbanion character. NH2

+ NH2–



NH2 + NH2–

NH2 + NH3 :–


Chemistry Facts




After firing 5 billion billion zinc ions at a speed of 18,460 miles per second (30,000 kilometers per second) at lead, the German scientists at Darmstadt, Germany created a single atom of 112 protons (ununbium) that survived for one third (1/3) of a millisecond.

If an electric current is passed through a solution or molten salt (the electrolyte), ions will migrate to the electrodes: positive ions (cations) to the negative electrode (cathode) and negative ions (anions) to the positive electrodes (anions).

The positron was discovered in 1932 by the U.S. physicist Carl Anderson at California Institute of Technology, United States.

Fritz Haber developed chlorine gas for use by the Germans in World War I. (Unable to live with his, his wife commited suicide in 1915).

The flatulence of a single sheep could power a small truck for 25 miles (40 kilometers) a day. The digestive process produces methane gas, which can be burned as fuel.

Cesiums has a diameter of 0.0000002 inches (0.0000005 millimeter).

Hydrogen atoms with no neutrons make up 99.985% percent of all hydrogen atoms. The remaining 0.015% percent contain one neutron.

The very first shell of an atom (innermost) can hold only up to two electrons.

Aniline Addition

Some facts on which the above mechanism is based. (a) Fact. Labeled chlorobenzene in which 14C held the chlorine atom was allowed to react with amide ion. In half the aniline obtained the amino group was held by 14 C and in half it was held by an adjacent carbon. Cl *


NH2 *

* +





Interpretation : In benzyne the labeled carbon and the ones next to it become equivalent, and NH2– adds randomly (except for a small isotope effect) to one or the other. * * Cl *

NH2– NH2 *



NH3 *

NH2– NH3



(b) Fact. Compounds containing two groups ortho to halogen like 2-bromo-3-methyl anisole, do not react at all. Br CH3O


NH2– NH3

No reaction

Interpretation : With no ortho hydrogen to be lost, benzyne cannot form.

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Inorganic Chemistry Fundamentals

HALOGEN & NOBLE GASES FAMILY halogen atom gains an electron, and X → X–. Thus, the halogen all form halide ions. Bond energy in X2 Molecule : The elements all form diatomic molecules. It would be expected that the bond energy in the X2 molecules would decrease as the atoms become larger, since increased size results in less effective overlap of orbitals. Cl2, Br2 and I2 show the expected trend (table) but the bond energy for F2 does not fit the expected trend. Bond energy and bond lengths of X2

Ionization Energy : The ionization energies of the halogens show the usual trend to smaller values as the atoms increase in size. The values are very high, and there is little tendency for the atoms to lose electrons and form positive ions. Ionization and hydration energies, electron affinity First ionization energy (kJ mol–1)

Electron affinity (kJ mol–1)

Hydration energy X– (kJ mol–1)



– 333

– 513



– 349

– 370



– 325

– 339



– 296

– 274


– 270

Bond energy (free energy of dissociation) (kJ mol–1)

The ionization energy for F is appreciably higher than for the others because of its small size. F always has an oxidation state of (–1) except in F2. It forms compounds either by gaining an electron to form F–, or by sharing an electron to form a covalent bond. Hydrogen has an ionization energy of 1311 kJ mol–1, and it forms H+ ions. It is at first surprising that the halogens Cl, Br and I have lower ionization energies than H, yet they do not form simple X+ ions. The ionization energy is the energy required to produce an ion from a single isolated gaseous atom. Usually we have a crystalline solid, or a solution, so the lattice energy or hydration energy must also be considered. Because H+ is very small , crystals containing H+ have a high lattice energy, and in solution the hydration energy is also very high (1091 kJ mol–1). The negative ions also have a hydration energy. Thus H+ ions are formed because the lattice energy, or the hydration energy, exceeds the ionization energy. In contrast the halide ions X+ would be large and thus have low hydration and lattice energies. Since the ionization energy would be larger than the lattice energy or hydration energy, these ions are not normally formed. However, a few compounds are know where I+ is stabilized by forming a complex with a Lewis base. for example [I(pyridine)2]+ NO3–. The electron affinities for the halogens are all negative. This shows that energy is evolved when a

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Bond length X2 (Å)













The bond energy in F2 is abnormally low (126 kJmol–1), and this is largely responsible for its very high reactivity. (Other elements in the first row of the periodic table also have weaker bonds than the elements which follow in their respective groups. For example in Group 15 the N – N bond in hydrazine is weaker than P – P, and in Group 16 the O – O bond in peroxides is weaker than S – S.) Two different explanation have been suggested for the low bond energy : (1) Mulliken postulated that in Cl2, Br2 and I2 some pd hybridization occurred, allowing some multiple bonding. This would make the bonds stronger than in F2 in which there are no d orbitals available. (2) Coulson suggested that since fluorine atoms are small, the F – F distance is also small (1.48 Å), and hence internuclear repulsion is appreciable. The larger electron –electron repulsions between the lone pairs of electrons on the two fluorine atoms weaken the bond. Pseudohalogens and Pseudohalides : A few ions are known, consisting of two or more atoms of which at least one is N, that have properties similar to those of the halide ions. They are therefore called pseudohalide ions. Pseudohalide ions are univalent, and these form salts resembling the halide salts. 30


For example, the sodium salts are soluble in water, but the silver salts are insoluble. The hydrogen compounds are acids like the halogen acids HX. Some of the pseudohalide ions combine to form dimers comparable with the halogen molecules X2. These include cyanogen (CN)2, thiocyanogen (SCN)2 and selenocyanogen (SeCN)2. The important pseudohalogens Anion –



CN : cyanide ion

HCN hydrogen cyanide (CN)2 : cyanogen

SCN– : thiocynate ion

HSCN : thiocyanic acid (SCN)2 : selenocyanogen

SeCN– : selenocyanate ion

The gases Ar, Kr and Xe may be trapped in cavities in a similar way when water is frozen under a high pressure of the gas. These are clathrate compounds, but are more commonly called 'the noble gas hydrates'. They have formulae approximating to 6H2O : 1 gas atom. He and Ne are not trapped because they are too small. The heavier noble gases can also be trapped in cavities in synthetic zeolites, and samples have been obtained containing up to 20% of Ar by weight. Clathrates provide a convenient means of storing radioactive isotopes of Kr and Xe produced in nuclear reactors.

Structure and bonding in Xenon compounds : (i) Structure and bonding in XeF4 : The structure of XeF4 is square planar, with Xe–F distances of 1.95 Å. The valence bond theory explains this by promoting two electrons as shown : 5s 5d 5p

(SeCN)2 : selenocyanogen

OCN– :cyanate ion

HOCN : cyanic acid

NCN2– : cyanamide ion

H2NCN : cyanamide

ONC– : fulminate ion

HONC : fulminic acid

N3– : azide ion

HN3 : hydrogen azide

The best known pseudohalide is CN–. This resembles Cl–, Br– and I– in the following respects : 1. It forms an acid HCN. 2. It can be oxidized to form a molecule cyanogen (CN)2. 3. It forms insoluble salts with Ag+, Pb2+ and Hg+. 4. Interpseudohalogen compounds ClCN, BrCN and ICN can be formed. 5. AgCN is insoluble in water but soluble in ammonia, as is AgCl. 6. It forms a large number of complexes similar to halide complexes. e.g. [Cu(CN)4]2– and [CuCl4]2–, and [Co(CN)6]3– and [CoCl6]3–. Clathrate Compounds : Clathrate compounds of the noble gases are well known. Normal chemical compounds have ionic or covalent bonds. However, in the clathrates atoms or molecules of the appropriate size are trapped in cavities in the crystal lattice of other compounds. Though the gases are trapped, they do not form bonds. If an aqueous solution of quinol (1, 4dihydroxybenzene) is crystallized under a pressure of 10 – 40 atmospheres of Ar, Kr or Xe, the gas becomes trapped in cavities of about 4Å diameter in the β-quinol structure. When the clathrate is dissolved, the hydrogen bonded arrangement of βquinol breaks down and the noble gas escapes. Other small molecules such as O2, SO2, H2S, MeCN and CH3OH form clathrates as well as Ar, Kr and Xe. The smaller noble gases He and Ne do not form clathrate compounds because the gas atoms are small enough to escape from the cavities. The composition of these clathrate compounds corresponds to 3 quinol : 1 trapped molecule, through normally all the cavities are not filled.

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(Electronic Structure of Xe-excited state) (four unpaired electrons form bonds to four fluorine atoms six electron pairs form octahedral structure with two positions occupied by lone pairs) The Xe atom bonds to four F atoms. The xenon 5px orbital forms a three-centre MO with 2p orbitals from two F atoms just as in XeF2. The 5py orbital forms another three-centre MO involving two more F atoms. The two three-centre obitals are at right angles to each other, thus giving a square planar molecule.


F Xe



(ii) Structure and bonding in XeF6 : The structure of XeF6 is a distorted octahedron. The bonding in XeF6 has caused considerable controversy which is not completely resolved. The structure may be explained in valence bond terms by promoting three electrons in Xe : 5s 5d 5p

(Electronic structure of Xenon-exicted state)

The six unpaired electrons form bonds with fluorine atoms. The distribution of seven orbitals gives either a capped octahedron or a pentagonal bipyramid (as in IF7). (A capped octahedron has a lone pair pointing through one of the faces of the octahedron) Since there are six bonds and one lone pair, a capped 31


octahedron would give a distorted octahedral molecule. The molecular orbital approach fails with XeF6, since three three-centre molecular orbitals systems mutually at right angles would give a regular octahedral shape. F

Total number of electron pairs = 7(5σbp + 1lp + 1dπ-pπbp)

Hybridization : sp3d2 (to accommodate 5σbp and 1lp) Geometry : Square pyramidal Structure of XeO2F2 : Total number of electron in valence shell of Xe = 14 (8 from Xe + 2 from F + 4 from O) Total number of electron pairs

F Xe



= 7(4σbp + 1lp + 2πbp) F

F The vibrational spectrum of gaseous XeF6 indicates C3v, symmetry, i.e. an octahedron distorted by the lone pair at the centre of one triangular face. The structure of the molecule rapidly fluctuates between structures where the lone pair occupies each of the eight triangular faces. In various non-aqueous solvents, xenon hexafluoride forms a tetramer Xe4F24. Solid xenon hexafluoride is polymorphic. Except at very low temperatures it contains tetramers, where four square pyramidal XeF5+ ions are joined to two similar ions by means of two bridging F–ions. The XeF distances are 1.84 Å on the square pyramidal units and 2.23 Å and 2.60 Å in the bridging groups. Xenon Oxyfluorides : Structure of XeOF2 : Total number of electrons in valence shell of Xe:12 (8 from Xe + 2 from O and 2 from F) Total number of electrons pairs

O Xe O F

Hybridization: sp3d (to accommodate 4bp + 1lp) Geometry : Trigonal bipyramidal or Sea-saw. Similarly : Structure of XeO3F2 and XeO2F4 O F F F O

Xe O




F XeO3F2


(Trigonal bipyramidal)

= 6(3σbp + 2lp + 1πbp)







• Parsec is the unit of O

• Estimated radius of universe is

Hybridisation = sp3d (to accommodate 3bp and 2lp) Geometry = T-shaped Structure of XeOF4 : Total number of electron in valence shell of Xe : 14 (8 from Xe + 2 from O + 4 from F)

• 18/5 km h–1 equal to

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1 ms–1

• 1 femtometre (1 fm) is equal to • Dot product of force and velocity is

10–15 m Power

• Moment of momentum is equal to


Angular momentum Xe


1025 m 1018 s

• Estimated age of Sun is



• Rocket propulision is based on the principle of

Conservation of linear momentum


• The largest of astronomical unit, light year and i 32



Organic Chemistry



10 ml. of gaseous hydrocarbon were mixed with 100 ml of oxygen and the mixture was exploded. On cooling, the volume was reduced to 95 ml On adding KOH, the volume was further reduced to 75 ml. The residual gas was found to be oxygen. All volumes were measure under the same condition of temperature and pressure. Calculate the molecular formula of hydrocarbon. Sol. Volume of hydrocarbon = 10 ml Volume of mixed oxygen = 100 ml Volume of the mixture after explosion and cooling (Vol. of CO2 + unused O2) = 95 ml ∴ Unused oxygen = 75 ml (given) ∴ Volume of used oxygen = 100 – 75 = 25 ml Contraction in volume on treatment with KOH, i.e., volume of CO2 produced = 95 – 75 = 20 ml If the molecular formula of hydrocarbon is CxHy, its combustion will take place according to the equation. y y  CxHy +  x +  O2 → xCO2 + H2O 4 2 

(i ) O

Sol. A(C8H10) 3 → C 4 H 6 O 2 ( ii ) H 2O

2H H – C ≡ C – H − → C3H5 – C ≡ C – C3H5 + C 6 H10

the C3H5 – correspond to cyclopropyl (∆) radical hence compound (A) is CH2 CH2 CH – C ≡ C – CH CH2 CH2 1,2-dicyclopropyl ethane

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2). CH2 (i) O3 CH2 CH – C ≡ C – CH CH2 CH2 (A)


CH – C – C – CH





H2O warm


CH – C – C – CH





CH – COOH CH2 (B) Compound (B) is prepared from cyclopropyl bromide as follows : O Mg CH2 CH2 CH.MgBr C = O CH – Br either ∆ CH2 CH2 (C) Cyclopropyl 2

5y 2

5y 2 or y=2 Now on substitution of the value of x and y in the formula CxHy, the molecular formula of hydrocarbon come to beC2H2.

or 25 – 20 =

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( B)

Since compound (A) adds one mol of O3, hence it should have either a C = C or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n) and if it was alkyne it should have the formula C8H14; it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne.

y  10 ml 10  x +  ml → 10 x ml 4  According to above equation 10 ml of hydrocarbon y  will require 10  x +  ml of oxygen for complete 4   combustion and 10x ml CO2 will be produced. According to the question, Volume of CO2 produced = 20 = 10x x=2 ...(i) y  ∴ Volume of oxygen used = 25 = 10  x +  ml 4   ...(ii) Substituting the value of x in Eqn. (ii) y  25 = 10  2 +  4   25 = 20 +

A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also give equations for the reactions.



Addition compound

magnesium bromide HOH dil. HCl; –MgBrOH







Hence, A, B,




CH3 – C = CH – CH – CH3 CH3




An alkene (A) on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid (B). When (B) is treated with Br2 in presence of P, it yields a compound (C) which on hydrolysis gives a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrolysis. Identify the compounds (A), (B), (C) and (D). Sol. The structure of compound can be easily obtained by the fact that it is synthesized from acetone as follows (Streeker synthesis). OH CH3 CH3 C = O + HCN C CN CH3 CH3 CH3

2H2O H+


2-methyl propanoic acid

Br C,





Compound (A), C3H6Cl2, on reduction with LiAlH4 gives propane. Treatment of (A) with aqueous alkali followed by oxidation gives (B) C3H4O4 which gives effervescence with NaHCO3. Esterification of (B) with ethanol gives (C), C7H12O4, which is well known synthetic reagent. When (B) is heated alone, the product is ethanoic acid, but while heating with soda-lime it gives methane. Compound (B) on reduction with LiAlH4 gives a diol which on reaction with SOCl2 gives back compound (A). Identify all the compounds and give balanced equation of the reactions. Sol. Compound (B) gives effervescence with NaHCO3 solution. Hence it is a dicarboxylic acid, since it on heating alone gives acetic acid and with soda-lime CH4, it means two –COOH in it are at different carbon atoms. COONa COOH 2NaHCO3 CH2 CH2 + 2CO2 +2H2O COONa COOH








C (D)

CH3 – C – COOH 2-hydroxy-2-methyl propanoic acid



CH3 – C – COOH 2-bromo-2-methyl propanoic acid

The conversion of (B) to (C) and then (C) to (D) indicates that the reaction (B) to (C) is Hell-VolhardZelinsky reaction. Thus we have : Br Br2/P CH3 H CH3 C C COOH CH3 COOH CH 3


Hence, we conclude that the aldehyde from which compound (B) is obtained by oxidation has the structure : CH3 – CH – CHO




2-methyl propanal

Now, the ozonolysis of the compound (A) gives acetone and isobutyraldehyde propanal), i.e., CH3




O O3


CH3 H2O Zn



CH4 + 2CO2



Propane 1,3-diol 3[O]

C – CH – CH – CH3 O–O



+ H2O


Esterification of (B) with ethanol gives malonic ester which is a synthetic reagent of high importance. COOH COOC2H5 ∆ CH2 CH2 + C2H5OH COOH COOC2H5 – 2H2O

C = O + CH3 – CH – CHO

CH3 Hence the compounds (A), (B), (C) and (D) are as follows :

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Acid (B) can be prepared from (A), C3H6Cl2, which should be 1,3-dichloro propane. CH2Cl 2NaOH(aq.) CH2OH CH2 CH2 CH2Cl (–2NaCl) CH2OH


C = O + O = HC – CH – CH3

C = CH – CH


2,4-dimethyl pentene-2



Malonic ester



CH2 (B)


Elements Named for Places


2SOCl2 CH2 –2SO2; –2HCl

This is an alphabetical list of element toponyms or elements named for places or regions. Ytterby in Sweden has given its name to four elements: Erbium, Terbium, Ytterbium and Yttrium.

CH2Cl (A)

Hence, (A) CH2


(B) CH2



An organic compound (A) has 76.6% C and 6.38% H. Its vapour density is 47. It gives characteristic colour with FeCl3 solution. Compound (A) when treated with CO2 and NaOH at 140º C under pressure gives (B), which on acidification gives (C). (C) reacts with phenol in presence of POCl3 to give (D), which is a well known antiseptic. (C) also reacts with methanol in presence of H2SO4 to give (E), which is used as a hair tonic. What are (A) to (E) ? Explain the reaction involved. Sol. (i) Calculation of empirical and molecular formula of (A).

Americium : America, the Americas

Berkelium : University of California at Berkeley

Californium : State of California and University of California at Berkeley

Copper : probably named for Cyprus

Darmstadtium : Darmstadt, Germany

Dubnium : Dubna, Russia

Erbium : Ytterby, a town in Sweden

Europium : Europe

Francium : France

Gallium : Gallia, Latin for France. Also named for Lecoq de Boisbaudran, the element's discoverer (Lecoq in Latin is gallus)



Relative no. of atoms

Simplest ratio



76.6 = 6.38 12

6.38 =6 1.06



6.38 = 6.38 1

6.38 =6 1.06

Germanium : Germany

1.06 =1 1.06

Hafnium : Hafnia, Latin for Copenhagen


17.02 = 1.06 16


Hassium : Hesse, Germany

Holmium : Holmia, Latin for Stockholm

Lutetium : Lutecia, ancient name for Paris

Magnesium : Magnesia prefecture in Thessaly, Greece

Polonium : Poland

Rhenium : Rhenus, Latin for Rhine, a German province

Ruthenium : Ruthenia, Latin for Russia

Scandium : Scandia, Latin for Scandinavia


Strontium : Strontian, a town in Scotland

Conc. H2SO4

Terbium : Ytterby, Sweden

Thulium : Thule, a mythical island in the far north (Scandinavia?)

Ytterbium : Ytterby, Sweden

Yttrium : Ytterby, Sweden

Hence, Empirical formula of (A) = C6H6O Empirical formula wt. = 94 Molecular wt. = V.D. × 2 = 47 × 2 = 94 So, molecular formula of (A) is C6H6O (ii) Since (A) gives colour with FeCl3, hence it is phenol. (iii) All the reactions are :



CO2 + NaOH 140ºC








Methyl salicylate Oil of winter green (Hair tonic)



Antiseptic (Salol)

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`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 S ol ut i o ns wi l l be p ubl i s he d i n ne x t i s s ue Joint Director Academics, Career Point, Kota


Find the greatest value of the expression (a–x) (b–y) (c– z) (ax + by + cz), where a, b, c are known positive quantities and a – x, b – y, c – z are also positive?


Let f(x) satisfies the differential equation xf’ (x) + f(x) = g(x), where f(x) and g(x) are continuous functions. If f(x) is decreasing function for all x ∈ R+, then prove that

10. Let n is an odd positive integer, show that (without using mathematical induction) (n2 – 1)n is divisible by 24. Here n > 1.


x.g (x) <

∫ g(x) dx; for ∀ x > 0.

Maths Facts



If a chord of the circle x2 + y2 = 32 makes equal intercepts of length p on the coordinate axes, then find the range of p.

40 when written "forty" is the only number with letters in alphabetical order, while "one" is the only one with letters in reverse order.


The arc AC of a circle subtends a right angle at the centre O. B divides the arc AC in the ratio 2 : 1. If

1 googol = 10100; 1 googolplex = 10 googol = 1010100 .

111 111 111 x 111 111 111

OA = a and OB = b , find OC . 5.

= 12345678 9 87654321

Out of 20 consecutive numbers 4 are chosen at random. Prove that the chance of their sum being even is greater than that of their sum being odd.


Find a point P on the line 3x + 2y + 10 = 0 such that |PA – PB| is maximum when A is (4, 2) and B is (2, 4).


Secants are drawn from a given point A to cut a given circle at the pairs of points P1, Q1; P2, Q2; ...., Pn, Qn. Show that AP1 . AQ1 = AP2 . AQ2 = .... = APn . AQn

Pi (3.14159...) is a number that cannot be written as a fraction.

If you add up the numbers 1-100 consecutively (1+2+3+4+5...) the total is 5050.

The billionth digit of Pi is 9.

1 and 2 are the only numbers where they are the values of the numbers of factors they have.

2 and 5 are the only primes that end in 2 or 5.


Let A & B be the matrices such that AAT = I and AB= BA. Prove that ABT = ATB.

The largest prime number is 9,808,358 digits long; more than the number of atoms in the universe.


If a2 + b2 + c2 = 1, b + ic = (1 + a) z, prove that 1 + iz a + ib = , where a, b, c are real numbers and z 1 − iz 1+ c is a complex number.

The digits to the right of the Pi's (3.141...) decimal point can keep going forever, and there is no pattern to these digits at all.

XtraEdge for IIT-JEE







Let S1 ≡ a1x2 + 2h1xy + b1y2 + 2g1x + 2f1y + c1 = 0 & S2 ≡ a2x2 + 2h2xy + b2 y2 + 2g2x + 2f2y + c2 = 0 be the rectangular hyperbolas. So a1 + b1 = 0 & a2 + b2 = 0 Now S1 + λS2 = 0 represents the conics through their points of intersection i.e. A, B, C and D. The sum of coefficient of x2 & y2 in it is (a1 + λa2) + (b1 + λb2) = (a1 + b1) + λ (a2 + b2) = 0 Hence, it will also be rectangular hyperbola. Now for λ when it represents pair of straight lines then also sum of coeff. of x2 & y2 will be zero. Hence those lines will be perpendicular. So AD & BC will be perpendicular. Similarly BD & AC and CD & AB will also pairs of perpendicular lines. Hence D will be orthocentre of triangle ABC. In fact orthocentre of triangle forms by any of 3 of these points will be the fourth point.


Now (PR)2 = (RA)2

 h2 + k2 k2 +  h −  h 

 h2 + k2 = a2 +   h 

∆2 2

– 2.

s ∆ = r2 + 2Rr 3.




2 2   – 2a h + k  h 

h2 + k2 h

h2 + k2 h (x2 + y2) (x – 2a) + a2x = 0 –a2 = (h – 2a)


Let z = x be the purely real root then f(x) = x4 + 2x3 + 3x2 + 4x + 5 = 0 f ´(x) = 4x3 + 6x2 + 6x + 4 = 0 ⇒ 4(x + 1) (4x2 + 4 – 4x + 6x) = 0 ⇒ (x + 1) (4x2 + 2x + 4) = 0 ⇒ x = –1 is only real root & f(–1) = 1 – 2 + 3 – 4 + 5 = 3 > 0 no real root of f(x). Now let z = iy be the purely imaginary roots then y4 – 2iy3 – 3y2 + 4iy + 5 = 0 so y4 – 3y2 + 5 = 0 and 2y3 + 3y2 = 0 must have simultaneous solution which is not possible. as y = 0, y = – 3/2 are the roots of 2nd but they do not satisfy.




2R ∆ ∆ .R; 2 = 1 + r s r

k 1 k & m2(PR) = – = h m1 −β+ h so m1m2 = –1 k k ⇒ . =–1 h −β+ h so βh = h2 + k2 k2 = +βh – h2;  h2 + k2  h2 + k2 β= so point R is  , 0  h  h   m1(OQ) =

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   

  – 2(h2 + k2)  

⇒ k2 + h2 – a2 = 2(h2 + k2) – 2a

a2 + b2 = c2 a + b + c = 2s a + b = 2(s – R) as c = 2R a2 + b2 + 2ab = 4(s – R)2 = 4R2 + 2ab ab = 2(s – R)2 – 2R2 = 2s2 – 4sR 2∆ = 2s2 – 4sR



2 2    = a − h + k   h  

 h2 + k2 ⇒ k2 + h2 +   h 

∆ =r s

abc = R; 4∆ A

c = 2R;

(h,k) r O R A(a,0) (b,0)

a2 a y = ⇒x= x y a


 y a  ln (a 2 / y) f  +  a y  a2 / y ∞ 



a 1

∫ f  a + y  y (ln a

 − a 2 dy     y2    2

– lny) dy




Let f(x) = x2 – (a2 + 1)x + 4 Since both roots of f(x) = 0 lie in (1, 4), hence D = (a2 + 1)2 – 16 ≥ 0

dy x a I = f  +  (2ln a – lny) a x y   0

dx x a ⇒I= 2I = f  +  (lna) a x x   0


⇒ a ∈(–∞, – 3 ) ∪ ( 3 , ∞) and f(1) > 0 ⇒ 1 – (a2 + 1) 4 > 0 ⇒ a ∈ (–2, 2) and f(4) > 0 ⇒ 16 – (a2 + 1) 4 + 4 > 0 ⇒ a ∈ (–2, 2)

 x a  lna f +  dx a x x 0

Let the fixed points be P(α, 0) & Q (– α, 0) and variable line be | mα + c | | − mα + c | . = a; where y = mx + c as given 1+ m2 1+ m2 'a' is a constant. so |c2 – m2α2| = a (1 + m2) ...(1) Now let foot of the perpendicular from (α, 0) be (h, k) h −α 1 k then c = k – mh & – = ⇒m=– m k h −α

and –


(h − α) 2 k2

α2 = a 1+


y = f(x) =



e zx − z dz =




ze zx .e − z dz + 1 = –


(2, 0) (5, 0) 1/3

(0,–3(4) ) x = 2 is local min. x = 0 is local max. f(x) is non diff. at x = 0 f(0) = 0 f(2) = 25/3 – 5.22/3.(2 – 5) = –3.22/3 = – 3(4)1/3 f(x) = x2/3 (x – 5) f(x) passes through (0, 0), (5, 0) If x5/3 – 5x2/3 = k has exactly one positive root then from sketch. k>0


.e − z dz

2 1 zx e (−2ze − z ) dz + 1 2

∫ 0

x  2 2 1 = – (e − z . e zx ) 0x − xe − z . e zx dz  + 1 2  0   1 1 dy – xy = 1 = xy + 1 ⇒ 2 2 dx


I.F. = e 2 solution is y . e −x y = ex




= x




= e−x


−z 2 / 4



10. There will be 99C44 subsets in which 1 will be least element similarly there will be 98C49 subsets in which 2 will be least element so p min = 1.99C49 + 2.98C49 + 3.97C49 + .... + 51.49C49





dx =


−z2 / 4

= Coeff. of x49 in [(1 + x)99 + 2(1 + x)98 + ..... + 51(1+x)49] 1 1− (1 + x ) 51 51(1 + x ) 48 = Coeff. of (1 + x)99 − 1 1   1− 1 −   1 + x 1 + x  



dz proved.



β α a2 − 7 Since + = 2 β −1 α −1 a −4

= Coeff. of

(1 + x ) 99 − (1 + x ) 48

51(1 + x ) 49 −x

x2 (1 + x ) 2 51 = coeff. of x in [(1 + x)101 – (1 + x)50] + coeff. of x50 in 51(1 + x)49 101 = C51 – 0 + 0 = 101C51

2αβ − (α + β) a2 − 7 = 2 αβ + 1 − (α + β) a −4 as given αβ = 4, so α + β = a2 + 1 Hence the equation is x2 – (a2 + 1) x + 4 = 0

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5 2/3 10 –1/3 5 ( x − 2) x – x = 3 x1 / 3 3 3 sign. dia of f´(x) + + 2 0





f´(x) =

(h − α) 2



y´ =



Hence a ∈ (–2, – 3 ] ∪ [ 3 , 2)

|k2 + h2 – hα + hα – α2| |k2 + h2 – hα – hα + α2 | = a(k2 + (h – α)2) so x2 + y2 = (α2 + a) 7.


a +1 β ∈ (1, 4) ⇒ 1 <
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