xtraedge_2009_10

XtraEdge for IIT-JEE 1 OCTOBER 2009

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

Every effort has been made to avoid errors oromission in this publication. In spite of this, errorsare possible. Any mistake, error or discrepancynoted may be brought to our notice which shall betaken care of in the forthcoming edition, hence anysuggestion is welcome. It is notified that neither thepublisher nor the author or seller will beresponsible for any damage or loss of action to anyone, of any kind, in any manner, there from.

• 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.

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

"Faliure is Success if we learn from it" Volume - 5 Issue - 4

October, 2009 (Monthly Magazine) Editorial / Mailing Office :

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

NEXT MONTHS ATTRACTIONS

Key Concepts & Problem Solving strategy for IIT-JEE.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Much more IIT-JEE News.

Xtra Edge Test Series for JEE-2010 & 2011

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."

• "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.

• Along with success comes a reputation for wisdom.

• They can because they think they can.

• Nothing can stop the man with the right Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.

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

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 Class 12 marks may become critical for IIT admission IIT-K postpones launch of its dream satellite

IITian ON THE PATH OF SUCCESS 7 Dr. Rai Mahesh Kumar Sinha

KNOW IIT-JEE 8 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 45

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

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 6] Students’ Forum

Physics Fundamentals Magnetic Field Gravitation

CATALYST CHEMISTRY 27

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

DICEY MATHS 36

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

Study Time........

Test Time ..........


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.

"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


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

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

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


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 non-PhDs 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 non-PhDs, 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 non-PhDs 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".

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 IIT-Rajasthan 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


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 co-ordinating 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

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.

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.

He concluded by saying that the projects are being funded through two MoUs signed between IIT-Kanpur 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

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.

Robots play soccer at IIT-KGP 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.

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 IIT-KGP 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.


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.

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.

Dr. Rai Mahesh Kumar Sinha Ph.D. (I.I.T. Kanpur)

M.Tech (I.I.T. Kharagpur)

M.Sc.Tech (Allabhad Univ.)

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


PHYSICS

1. 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]

m v

M

O

Sol. We know that →τ = dtLd→

⇒ →τ × dt = →Ld

When angular impulse ( →τ ×

→td ) 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 ⇒ mv × L = I0 × ω ...(i) where I0 is the moment of inertia of the wooden log-

bullet system after collision about O I0 = Iwooden log + Ibullet

= 31 ML2 + ML2 ...(ii)

From (i) and (ii)

ω =

+

×

22 mLML31

Lmv

⇒ ω =

+ mL

3ML

mv = L)m3M(

mv3+

2. (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 ?

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

u = 21

ε0E2 (Energy/Volume)

(i) Energy stored within the sphere (U1) Electric field at a distance r is

E = 04

1πε

. 3RQ . r

U = 21

∈0E2 = 20ε

2

30

rRQ.

41

πε

Volume of element dV = (4πr2)dr Energy stored in this volume dU = U(dr)

dU = (4πr2dr) 2

30

0 rRQ.

41

2

πεε

dU = 6

2

0 RQ.

401πε

.r4dr

∴ U1 = ∫R

0dU = ∫πε

R

0

46

0

2drr

R8Q

= R0

56

0

2]r[

R8Q

πε

U1 = R

Q.40

1 2

0πε ...(1)

(ii) Energy stored outside the sphere (U2) Electric field at a distance r is

E = 20 R

Q.4

1πε

KNOW IIT-JEE By Previous Exam Questions


∴ U = 21

ε0E2 = 2

20

0

RQ.

41

2

πεε

∴ dU = µ . dV = (4πr2dr)

πε

ε2

20

0

RQ.

41

2

dU = 0

2

8Qπε 2r

dr

∴ U2 = ∫α

RdU =

0

2

8Qπε

. ∫α

R 2rdr =

R8Q

0

2

πε ...(2)

Therefore, total energy of the system is

U = U1 + U2 = R40

Q

0

2

πε+

R8Q

0

2

πε

or U = 203

RQ

0

2

πε

(b) Comparing this with gravitational forces, the gravitational potential energy of earth will be

U = –53

RGM2

by replacing Q2 by M2 and 04

1πε

by G.

g = R

GM2

∴ G = M

gR 2

U = 53− MgR

Therefore, energy needed to completely disassemble the earth against gravitational pull amongst its constituent particle will be given by

E = |U| = 53 MgR

Substituting the values, we get

E = 53 (10m/s2) (2.5 × 1031 kg-m)

E = 1.5 × 1032 J (c) This is the case of a charged spherical conductor

of radius R, energy of which is given by = C

Q21 2

or U = R4

Q.21

0

2

πε =

R8Q

0

2

πε

3. 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 semicircles as shown in figure. [IIT-2000]

KI N

ML

I

z

x

y

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

a velocity →v = –v0 i . Find the instantaneous force

→F

on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B0 j is

applied determine the force 1F→

and 2F→

on the semicircles KLM and KNM due to the field and the

net force →F on the loop.

Sol. (a) Magnetic field (→B ) at the origin = magnetic field

due to semicircle KLM + Magnetic field due to other semicircle KNM

∴ →B =

R4Iµ0 (– i ) +

R4Iµ0 ( j )

→B = –

R4Iµ0 i +

R4Iµ0 j

= R4Iµ0 (– i + j )

∴ Magnetic force acting on the particle

→F = q(

→v ×

→B )

= q(–v0 i ) × (– i + j )R4Iµ0

→F = – k

R4Iqvµ 00

(b) →F KLM =

→F KNM =

→F KM

And →F KM = BI(2R) i = 2BIR i

→1F =

→2F = 2BIR i

Total force on the loop,

→F =

→

1F + →

2F

or →F = 4BIR i

Note : If a current carrying wire ADC (of any shape)

is placed in a uniform magnetic field →B .

Then, →F ADC =

→F AC

or |→F ADC| = i (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.


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

µ1 = 2

µ2 = 2

µ3 = 3

Sol. For total internal reflection on interface AB

sin i = µ

112

= µµ

2

1 = 22 =

21 ; i = 45º

for total internal reflection on interface CD

sin i = µ

132

= µµ

2

3 = 23

⇒ i = 60º ⇒ The minimum angle for total internal reflection

for both the interface is 60º. 5. 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

2Å = n

λ

2 ...(1)

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

2Å

N N N N N N n loops

2.5Å

N N N N N N (n+1) loops λ/2

N

and 2.5 Å = (n + 1)2λ ...(2)

∴ 25.2 =

n1n + ,

45 =

n1n + or n = 4

Hence, from equation (1),

2Å = 42λ i.e. λ = 1Å

d will be minimum, when

n = 1, dmin = 2λ =

2Å1 = 0.5 Å

Now, de broglie wavelength is given be

λ = mK2h or K =

m2.h

2

2

λ

∴ K = 1931210

234

106.1101.92)101()1063.6(

−−−

−

××××××× eV

= 6.11.98

)63.6( 2

×× × 102 eV = 151 eV

CHEMISTRY

6. A solution of 0.2 g of a compound containing Cu2+ and C2O4

2– 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 C2O4

2– 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 C2O4

2– with MnO4– are :

MnO4– + 8H+ + 5e– → Mn2+ + 4H2O] × 2

C2O42– → 2CO2 + 2e–] × 5

2MnO4– + 5C2O4

2– + 16H+ → 2Mn2+ + 10CO2 + 8H2O From this equation, we conclude 2 mol MnO4

– ≡ 5 mol C2O4

–. Hence,

Amount of C2O42– in the solution =

25 (22.6 mL)

(0.02 M) =

25

L1000

6.22 (0.02 mol L–1) = 0.00113 mol.

The chemical equations involved during the treatment of KI and the titration with Na2S2O3 are

2Cu2+ + 4I– → Cu2I2 + I2 and I2 + 2S2O3

2– → 2I– + S4O62–

From these equations, we conclude 2 mol Cu2+ ≡ 4 mol I– ≡ 1 mol I2 and 1 mol I2 ≡ 2 mol S2O3

2– Now, Amount of S2O3

2– consumed = (11.3 mL)

(0.05 M) =

L1000

3.11 (0.05 mol L–1)

=

1000

3.11 (0.05) mol = 0.000565 mol

Amount of Cu2+ equivalent to the above amount of S2O3

2– = 0.000565 mol

Hence, −

+

242

2

OC of AmountCu ofAmount =

00113.0000565.0 =

21


7. 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 ∆fHº(CO2, g) = –393.3 kJ mol–1 [IIT-1990] Sol. From the enthalpy of combustion of ethane and

propane, we write

(1) C2H6(g) + 27 O2(g) → 2CO2(g) + 3H2O(1) :

∆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

8. 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 teq 0.7 × 0.06 mol 0.3 × 0.06 mol 0.3 × 0.06 mol = 0.018 mol = 0.018 mol (i) The equilibrium constant KC is

KC = [NH3][H2S] =

L2mol018.0

L2mol018.0

= 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.

9. 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 →gas) smelling (foul

isocyanide alkylRNC + 3KCl

+ 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.

Amount of gas liberated = 1molmL22400mL112

− =2001

mol

From the above equation, it is obvious that

Amount of compound RNH2 = 2001 mol

If M is the molar mass of RNH2, then

M

g295.0 = 2001 mol

or M = 0.295 × 200 g mol–1 = 59 g mol–1.


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 CH3 – C

OH

group.

Hence, it is concluded that ROH is CH3 – CH – CH3

OH

.

Thus, the original compound is CH3 – CH – CH3

NH2 Isopropylamine

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

C : H : : 12

7.85 : 1

3.14 : : 7.14 : 14.3 : : 1 : 2

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

= 100

5 × 38.05 g

Mass of hydrocarbon absorbing 160 g (= 1 mol) of

Br2 = )100/05.385(

0.1×

× 160 g = 84.1 g.

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

)A(126HC → 4KMnO.conc

)C(84 OHC + CH3COOH

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

CH3C ≡ CCH3 +H2O

HgSO4/H2SO4 CH3C = CHCH3

OH

→ CH3CCH2CH3

O 2-butanone

(C)

2-butyne

Finally, the structure of A would be

C = O + HOOCCH3 CH3

CH3CH2

[O]C = CHCH3

CH3

CH3CH2

(C) 3-methylpent-2-ene(A)

The structure of B is

CH3CH2C = CHCH3

CH3

H2 CH3CH2CHCH2CH3

CH33-methyl pentane

(B)

Hence, Molecular formula of A is C6H12

Structure of A is CH3CH2C = CHCH3

CH3 3-methyl pent-2-ene

Structure of B is CH3CH2CHCH2CH3

CH33-methyl pentane

Structure of C is CH3CCH2CH3

O2-butanone

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 = n3

6nn

6C)32.33( ×+−

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]


Sol. Let the equation of straight line L be y = mx

P ≡

++ 1mm,

1m1 ; Q ≡

++ 1mm3,

1m3

Now equation of L1 : y – 2x = 1m2m

+− ...(1)

equation of L2 : y + 3x = 1m9m3

++ ...(2)

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

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

A(–a, 0)

C

O

B y

D

L

E

πx =

– a

/2

x = a

x

If 2

ma

=

2a 2

(m2 + 1)

⇒ 2m1 =

21 (m2 + 1) ⇒ 2 = m4 + m2

⇒ 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

= 2 22 OKOB −

= 24

a2

a 22− = 2

4a 2

= a

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

Thus area of the trapezium

BCDE = 21 (BC + DE) (KL)

= 21 (a + 4a)

2a3 =

4a15 2

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

the vectors

→a = a1 i + a2 j + a3 k ;

→b = b1 i + b2 j + b3 k

→c = c1 i + c2 j + c3 k

If ar, br, cr, where r = 1, 2, 3 are non-negative real

numbers and ∑=

++3

1rrrr )cba( = 3L. Show that

V ≤ L3. [IIT-2002]

Sol. V = |)cb.(a|→→→

× ≤ 23

22

21 aaa ++

23

22

21 bbb ++ 2

322

21 ccc ++ ...(1)

Now, L = 3

)ccc()bbb()aaa( 321321321 ++++++++

[(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 = 2

1a + 22a + 2

3a + 2a1a2 + 2a1a3 + 2a2a3 ≥ 21a + 2

2a + 23a

⇒ (a1 + a2 + a3) ≥ 23

22

21 aaa ++

Similarly, (b1 + b2 + b3) ≥ 23

22

21 bbb ++

and (c1 + c2 + c3) ≥ 23

22

21 ccc ++

∴ from (1) and (2) L3 ≥ [( 2

1a + 22a + 2

3a )( 23

22

21 bbb ++ )( 2

322

21 ccc ++ )]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.

⇒ 222 cba

|dcba|

++

+γ+β+α = 90%. h

⇒ aα + bβ + cγ + d = ± 0.9h 222 cba ++

⇒ locus is, ax + by + cz + d = ±0.9h 222 cba ++ ⇒ locus of A´ is a plane parallel to the plane ABCD


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

PV

φ

U

(0, a)

1. What is the atomocity and the shape of the Gaseous

molecule if b = 3 and a = 2. 2. Write the relation of adiabatic index of the gas in

terms of a or b or in terms of both a and b. 3. 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.

4. If b start varying with respect to time as b(t) = c0 + c1t2 where c0 and c1 are positive constants

then find the slope of dtdf v/s t graph where f is the

degrees of freedom for the gas. 5. A particle enters in the given magnetic field

→B = B0 k where B0 is a constant with the velocity of

jbiav +=→

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)

1. Circular path with radius of the circular path

is r = 0

22

B.qbam +

2. Helix and the pitch of the Helix is 0B.q

m2π .a

3. Helix and the pitch of the Helix is 0B.q

m2π .b

4. Same path as followed by circulating electrons which is responsible for the unstable Rutherford atomic model, means spiral path of decreasing radius.

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

a

b

YR

x

d

v

c l

l

Part-2Part-1

C/R-1

C/R-2

2v

C/R-3

B⊗

B

C/R – Conducting Rail R = 10 Ω vBl = 30 volts 6. Current passing through resistance R and it's

direction. 7. P/d across terminals a and c. 8. Energy of deutron accelerated by potential difference

across b and c.

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

Physics Challenging Problems

Solut ions wi l l be publ ished in next issue

Set # 6


1. 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.

2. 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.

3. 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.

4. 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.

5. As hydrogen atom HA1 Ground state

–13.6 eV

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

–3.4 eV

n = 1 Nucleus

n = 2

Accepts energy photon of 1.9eV so the final energy

= –3.4 + 1.9 = –1.5 eV corresponds to n = 3

electronn = 3

n = 2

n = 1

–1.5 eV

–3.4 eV

–13.6 eV

Position of electron in hydrogen atom HA1 and HA2 & hydrogen spectra

No. of spectral lines emitted = 2

)1n(n − =2

)13(3 − = 3

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

7. Energy of photons from hydrogen spectrum tube are-

1.9 eV

Ist line of Balmer series

10.2 eV

Ist line of Lymen series

12.1 eV

IInd line ofLymen 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. 8. 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.

Solution Physics Challenging Problems

Set # 5

8 Quest ions were Publ i shed in September Issue


1. 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

cmvr =

21

21

mmm.0m.0

++ = 0 ...(1)

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 :

m1v1 = m2v2 ...(2) The gravitational force between the masses is

conservative. Calculating the potential energy between the two masses at the moment of release, we arrive at :

Ep = – ∫∞

LdrF = – ∫∞

L

221

rmGm dr = –

LmGm 21 ...(3)

When the masses arrive at distance l we have :

–L

mGm 21 = –l

21mGm + 21 m1v1

2 + 21 m2v2

2 ..(4)

Using this result along with Eq. (2), we obtain

−

+=

−

+=

Ll1m

mmG2v

Ll1m

mmG2v

21

21

22

22

21

21

l

l ..(5)

We are also interested in the relative velocity, which can be expressed :

12vr = 1v

r – 2vr = (v1 + v2) r

We can solve this equation simply by plugging the calculated v1 and v2 into it, or as following :

| 12vr |2 = | 1v

r – 2vr |2 = ( 1v

r – 2vr ). ( 1v

r – 2vr )

= 21v

r + 22v

r – 2 1vr . 2v

r

= 21v + 2

2v + 2v1v2 ...(6) Notice that the directions of the velocities are

opposite. Using Eq. (2) and Eq. (5), we have :

| 12vr |2 = 2

1v + 22v + 2

2

1

mm 2

1v

= 21 mm

G2+

−

L11

l( 2

1m + 22m + 2m1m2)

= 2G

−

L11

l(m1 + m2) ...(7)

and therefore,

| 12vr

| = )mm(L11G2 21 +

−l

..(8)

Another method of finding v12 is to use one of the masses as the frame of reference;

a0

m1 m2

for example, m1 (fig. 1). The frame of m1 is not

inertial. Its acceleration, a0, is :

a0 = 1m

F = 221

1 rmGm

m1 = 2

2

rGm ...(9)

The force exerted on m2 in this frame, F´, is the sum of the gravity and D'alembert's force,

F´ = – 221

rmGm – m2

2

2

rGm

= – Gm2(m1 + m2) 2r1

The difference between the initial and the final kinetic energies, ∆Ek, must equal the work done by the force F´, ∆Ek = W,

where ∆Ek = 21 m2

22v .

Therefore,

W = rd.F2

1

r

r

rrr

r∫ = ∫ +−l

L 2212 r1)mm(Gm dr

Hence, v12 = )mm(L11G2 21 +

−l

as expected

2. 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.

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

Students' ForumPHYSICS


(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.

m1

k

m2

g

Sol. (i) Denoting the position of the center of mass at t = 0

by x0, we can write: x(t) = x0 + 21 gt2

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 :

k = 0

1

LLgm

− ...(1)

The equations of motion are :

m1

L0

m2

x1 x

m1 1••x = m1g – k(x1 – x2 – L0) ...(2)

m2 2••x = 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,

21

2211

mm)t(´xm)t(´xm

++ = x0(t) = x0(0) = 0 ...(4)

This leads us to:

x´t(t) = – 1

2

mm x2(t) ...(5)

The distance between the two masses is given by : R(t) = R´(t) = x´1(t) – x´2(t)

= x´1(t) + 2

1

mm x´1(t)

= 2

21

mmm + x´1(t) ...(6)

Now in the current frame of reference, the equations of motion are defined as:

11 ´xm && = – k(x´1 – x´2 – L0) ...(7) 22 ´xm && = + k(x´1 – x´2 – L0) ...(8) We define r(t) ≡ R(t) – L0 as the displacement from

equilibrium. From Eq. (6), we have :

+=

+=

+=

)t(´xm

mm)t(r

)t(´xm

mm)t(r

L–)t(´xm

mm)t(r

12

21

12

21

012

21

&&&&

&& ...(9)

From Eq. (5) we have

=

=

−=

)t(xmm–´x

)t(xmm–´x

)t(xmm´x

21

21

21

21

21

21

&&&&

&& ...(10)

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,

rmm

mm

21

21 &&+

= µ r&& ...(11)

where µ = 21

21

mmmm+

is the reduced mass of the

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 :

1x&& – 2x&& = – k

+

21 m1

m1 (x1 – x2 – L0)

= – µk (x1 – x2 – L0) ...(12)

The solution to these equation is:

r(t) = A cos

φ+t

µk ...(13)

where A and φ are determined from the initial conditions, R(0) = L0, and R(0) = 0. Hence,

R(t) = r(t) + L0

= L0 + (L – L0) cos

t

µk ..(14)

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 = 21

21

mmmm

+

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

v t = t0 ω

v´ µ = 0ω´

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 torque relative to this point vanishes ( N

r= 0); so,

dtJdr

= 0.

+ω=++==

RvMIpRJJRMvJ

cm,infinal

0initial

&rr ...(1)

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,

Jf = IRv + MvR ...(2)

Applying the principle of conservation of momentum, Ji = Jf, we obtain:

v =

2

0

MRI1

v

+ ...(3)

(ii) For a spherical rigid body we know that

I = 52 MR2. Therefore,

v = 5/7

v0 = 75 v0 ...(4)

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,

W = ∫ rd.frr

= 0

for rdr

= 0. (iii) On a frictionless surface, the linear and angular

momentum are conserved. Therefore, ω and v are unchanged, or :

ω=ω=

´v´v

4. 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.

∴ volume of sphere at 20ºC is V0 = 0

210ρ

where ρ0 is density of water at 20ºC. volume of sphere at 25º C becomes equal to V = V0(1 + 3αs∆θ)

= 0

210ρ

[1 + 3 × 10–5(25 – 20)] = 0

0315.210ρ

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. 5. Suppose a nucleus X, initially at rest, undergoes α-

decay according to the equation, X225

92 → 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 α-


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

Mass of α-particle = 32 × 10–26 kg

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,

R = qBsinmv θ

or v sin θ = m

RqB = 1.2 × 107 ms–1

where q (charge of α-particle) = 3.2 × 10–19 coulomb.

and pitch, p = qB

cosmv2 θπ

or v cos θ = m2

pqBπ

= 9 × 106 ms–1.

∴ velocity of emitted α-particle,

v = 22 )sinv()cosv( θ+θ = 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,

my = 003.4

03.221 × 32 × 10–26 kg

∴ 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

or E = 21 myV2 +

21 mαv2 = 4.77 MeV

Hence, mass lost during α-decay,

∆m = 931E u = 0.005 u

∴ 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

= 225

931md × MeV

= 7.84 MeV Ans.

Regents Physics You Should Know

Nuclear Physics : • Alpha particles are the same as helium nuclei

and have the symbol .

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

• Deuterium ( ) is an isotope of hydrogen

( )

• 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).


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

∆I

θ B

v

θ B

+q 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

∆ Fr

= I(∆ lr

× Br

) or q( vr

× Br

) 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 = 20

rsinI

4µ θ∆

πl

P

rθ

I

∆l

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 rotated from ∆ l

r towards r

r through the smaller

angle.

Magnetic field at the centre of a Circular Coil :

B = a2NIµ0 tesla (T)

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 = 2/322

20

)xa(2NIaµ+

tesla (T)

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

Point :

B = dI

4µ0

π(sin θ1 + sin θ2)

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.

When the conductor is long B = dI2

4µ0

π

Magnetic Moment of a Loop : Magnetic moment of a current loop(m) = IS (current × area)

or m = IS ampere metre2 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 right-

hand rule allows you to determine the magnetic force on a moving charged particle.

Step 2 : Set up the problem using the following steps :

Draw the velocity vector vr

and magnetic field Br

with their tails together so that you can visualize the plane in which these two vector lie.

Identify the angle φ between the two vectors.

Magnetic Field

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


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

r or B

r.

Step 3 : Execute the solution as follows :

Express the magnetic force using Eq. BvqFrrr

×=

The magnitude of the force is given by Eq. F = qvB sin φ.

Remember that Fr

is perpendicular to the plane of the vectors v

r and B

r. The direction of Bv

rr× is

determined by the right-hand rule; keep referring to until you're sure you understand this rule. If q is negative, the force is opposite to Bv

rr× .

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 of motion, Σ F

r= m a

r, with the net force given by

Fr

Σ = q )BvE(rrr

×+ . 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 express all vector quantities (including E

r, B

r, v

r,

Fr

and ar

) 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.

R = B|q|

mv and

ω = Rv = v

mvB|q| =

mB|q|

If your calculation involves a more complex trajectory, use F

rΣ = m a

r 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.

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).

Draw the current element lr

d , being careful to ensure that it points in the direction of the current.

Draw the unit vector r . 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 Br

. Step 3 : Execute the solution as follows :

Use eq. dB = 20

rsindI

4µ φ

πl or Bd

r= 2

0

rrdI

4µ ×

πlr

to express the magnetic field Bdr

at P from the representative current element.

Add up all the Bdr

's to find the total field at point P. In some situations the Bd

r's at point P have the

same direction for all the current elements; then the magnitude of the total B

r field is the sum of

the magnitudes of the Bdr

's. But often the Bdr

's have different direction for different current elements. Then you have to set up a coordinate system and represent each Bd

r in terms of its

components. The integrals for the total Br

is then expressed in terms of an integral for each component.

Sometimes you can use the symmetry of the situation to prove that one component of B

r 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 4 : Evaluate your answer : Often your answer will be a mathematical expression for B

r as a

function of the position of the field point. Check the answer by examining its behavior in as many limits as you can.

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

DP 0

B

2d

Sol. The point is at a distance 22 dD + from each wire.

∴ magnitude of field due each = 22

0

dD

i24µ

+π

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.

A

D P0

B

0

B sinθ B

θ

θ B cosθ B cosθ

B B sinθ

d

d

∴ B´ (field) at P = 2B cos θ along OP

= 2 × π4

µ022 dD

i2

+ 22 dD

d

+ =

)dD(idµ

220

+π

2. 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

or v = m

Bqr = 27

19

1064.645.0102.32.1

−

−

××××

= 2.6 × 107 ms–1 (b) Bqv = mv2/r

⇒ Bq = mv/r = rmE2

= 2mv

21EQ

or E = m2

rqB 222= 27

22192

1064.6245.0)102.3(2.1

−

−

×××××

= 2.25 × 10–12 J (c) E(energy acquired) = Vq

or V = qE = 19

12

102.31025.2

−

−

×× ⇒ V = 7.0 × 106 V

3. Use Biot –Savart law to calculate the magnetic field

B at the common centre of the following circuits.

r2 r1

b

c

0

i

i

a

d

θ

r1

r2

i

i

d a 0 b c

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.

b

c

0

i

a

d

α

dα

dl i

Then dB1 = 21

0

rº90sinid

4µ l

π perpendicular into the

plane of the paper

⇒ dB1 = 21

10

r)dr(i

4µ α

π (Q dl = r1dα)

⇒ B1 = ∫θ

απ 01

0 dri

4µ

= 1

0

r4iµ

πθ

Similarly, B2, field at O due to cd = 2

0

r4iµ

πθ

out of the

plane of the paper.

∴ B, field due to the loop abcd =

−

πθ

21

0

r1

r1

4iµ

perpendicular into the plane. The second circuit is a special case of the above

when θ = π

∴ B =

−

210 r

1r1iµ

41


4. 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.

θ 0

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. 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.

a I2

b I1

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.

dF (force of attraction) = x2Iµ 10

π×

bdxI2

= b2IIµ 210

π×

xdx

∴ F = b2IIµ 210

π ∫+ba

x xdx

= b2IIµ 210

πln

bba +

WHAT ARE EARTHQUAKES?

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.

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.

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.


Newton's Law of Gravitation : Two point masses m1 and m2, separated by' a distance

r, attract each other with a force

F = G 221

rmm

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

m1 F F m2

r 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 g

r r Any mass sets up a gravitational field around it. The

gravitational field strength at a distance r from a point mass m is

g = –G 2rm

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

V = – Grm

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.

Gravitational field (g) and potential (V) due to a spherical shell and a solid sphere :

R

M

rR

M

r

(i) Outside (i) Outside

g = – G 2rM g = – G 2r

M

gr

= – G 2rM r g

r = – G r

rM

2

V = – Gr

M V = – Gr

M

(ii) Inside (ii) Inside g = 0

g = – G 3RM r

where R is the radius of the sphere

V = – GRM

where R is the radius of the shell

V = – G 3R2M (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

2emv

21 = Vm = G

RM m

where M is the mass of the planet and R is its radius.

or ve = RGM2

Gravitation PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


If g is the gravitational field intensity, then

mg = 2RGM2 or GM = gR2

∴ ve = gR2

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

Fattraction = r

mv20

where r is the radius of the orbit, measured from the centre of the planet.

G 2rMm =

rmv2

0

or v0 = r

GM

If a satellite is projected with velocity v < v0 the path is a small ellipse with point of projection as apogee and if v0 < v < 2 v0 the path is a bigger ellipse with point of projection as perigee. If v = 2 v0 the path is parabolic and if v > 2 v0 the path is hyperbolic.

1. 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.

∴ 0 = 21 M1v1

2 + 21 M2v2

2 – sMGM 21

(It is zero because in the beginning, both kinetic energy and potential energy are zero.)

0 = M1v1 + M2v2

Solving the equations v12 =

)MM(sGM2

21

22

+

and v22 =

)MM(sGM2

21

22

+

V(velocity of approach) = v1 – (–v2) = v1 + v2

= s

)MM(G2 21 +

2. 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)

mdv2

= Gm 2dm ⇒ v2 =

dGms

⇒ v = ωd

∴ v2 = v

Gmsω or ω = s

3

Gmv

or T = 3s

vGm2π

= 33

3011

)109.34(1097.11067.62

×××××π −

= 225 days

Taking the equation in terms of d and ω

2s

dGmm

= mω2d ⇒ T2 = s

2

Gm4π d3

∴ T2 ∝ d3 This is Kepler's third law. 3. 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.

Sol. E (energy of the satellite) = – r

GMm + 21 mv2

By the dynamics of circular motion

2rGMm =

rmv2

⇒ v = r

GM

∴ E = –r

GMm + 21

rGMm = –

21

rGMm

⇒ dE = 2rGMm

21 dr

Also – dE = power × dt = Fv dt = av3dt

∴ 2rGMm

21 dr = –a

2/3

rGM

dt

⇒ dt = –GMa2

m r–1/2 dr

∴ t = – GMa2

m∫ −R

nR

2/1 drr = GMa

Rm ( n – 1)

Q GM = gR2, t = gRa

m ( n – 1)

Solved Examples


4. 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.

M

v v orbit

Now vescape = RGM2

∆v = vfinal – vinitial = vorbit – vescape

⇒ ∆v = R

GM – RGM2 = –

RGM ( 2 – 1)

The negative sign means the speed has to be decreased.

∴ required change in speed = R

GM ( 2 – 1)

5. 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

2RGMm = mω2r or ω2 = 3

2

RgR (Q GM= gRe

2)

⇒ ω = 3

2e

RgR

∴ ωrel = 3

2e

RgR

+ ωe

⇒ ωrel = 213

122

102104.610

××× + 7.27 × 10–5

(Q ωe = 86400

2π = 7.27 × 10–5)

⇒ ωrel = 22.6 × 10–5 + 7.27 × 10–5 = 30 × 10–5 rad s–1

∴ τ = rel

2ω

π = 510302

−×π

= 2.09 × 104s = 5 hr 48 min

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

2. X-ray unit : It is the unit of length.

1 X-ray unit = 10–13 m

3. Slug : It is the unit of mass.

1 slug = 14.59 kg

4. Chandra Shekhar limit : It is the largest practical unit of mass.

1 Chandra Shekhar limit = 1.4 × Solar mass

5. Shake : It is the unit of time.

1 Shake = 10–6 second

6. Barn : It is the unit of area.

1 barn = 10–28 m2

7. Cusec : It is the unit of water flow.

1 cusec = 1 cubic foot per second flow

8. Match No. : This unit is used to express velocity of supersonic jets.

1 match no. = velocity of sound

= 332 m/sec.

9. Knot : This unit is used to express velocity of ships in water.

1 knot = 1.852 km/hour

10. Rutherford : It is the unit of radioactivity.

1 rutherford (rd) = 1 × 106 disintegrations/sec

11. Dalton : It is the unit of mass.

1 dalton = 121 mass of C12 = 931 MeV

= 1 a.m.u.

12. Curie : It is the unit of radioactivity.

1 curie = 3.7 × 1010 disintegration / sec


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.

– C : X

δ+ δ–

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 CH3F 139 452 1.847 CH3Cl 178 351 1.860 CH3Br 193 293 1.830 CH3I 214 234 1.636

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.

(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 :

I

:Cl:

II

:Cl:

III

+Cl:

IV

+Cl:

V

+Cl::–

–

–:

(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 sp2-hybridized 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.

Organic Chemistry

Fundamentals

HALOGEN DERIVATIVES

KEY CONCEPT


C

Cl sp2 169 pm

Cl

H H

H sp3

177 pm

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.

— C — X

X δ+

δ–

δ–

δ+

Haloarene or aryl

halide Haloalkane or alkyl

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

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

(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 + + Cl–

Chlorobenzene Phenyl cation

CH2 = CH – Cl CH2 = CH + Cl–+

Vinyl chloride Vinyl cation

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

Aryl halide

NH2–

NH3

Benzyne

NH2–

NH3

Aniline

NH2

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

H

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.

(1) X

+ NH2–

I

X+ NH3

:–

(2) X

I

+ X–

:–Benzyne

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.

(3)

Benzyne

+ NH2–

II

NH2

:–

(4)

NH2

II

+ NH2–

:– Aniline

+ NH3

NH2

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 14C and in half it was held by an adjacent carbon.

NH2– *

(47%)NH3

Cl

* NH2

* NH2+

(53%) 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.

NH2– *

NH3

Cl

*

*NH2

*

NH2–

NH3

NH2–

NH3

* NH2

*H2N

(b) Fact. Compounds containing two groups ortho to

halogen like 2-bromo-3-methyl anisole, do not react at all.

CH3 No reaction

NH2–

BrCH3O

NH3

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

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.


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)

F 1681 – 333 – 513

Cl 1256 – 349 – 370

Br 1143 – 325 – 339

I 1009 – 296 – 274

At – – 270 –

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

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 Bond energy (free

energy of dissociation) (kJ mol–1)

Bond length X2 (Å)

F 126 1.43

Cl 210 1.99

Br 158 2.28

I 118 2.66

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.

Inorganic Chemistry

Fundamentals

HALOGEN & NOBLE GASES FAMILY

KEY CONCEPT


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 Acid Dimer

CN– : cyanide ion HCN hydrogen cyanide (CN)2 : cyanogen SCN– : thiocynate ion HSCN : thiocyanic acid (SCN)2 :

selenocyanogen SeCN– : selenocyanate ion (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, 4-dihydroxybenzene) 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 β-quinol 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.

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 5p 5d

(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.

Xe

F

F F

F

(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 5p 5d

(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


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

F

F

FXeF

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 = 6(3σbp + 2lp + 1πbp)

F Xe F

O 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)

Xe

F

F F

F

O

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 = 7(4σbp + 1lp + 2πbp)

F

Xe

F

O

O

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

F

Xe

F

O

O

F

FF

F O

O Xe

O XeO3F2 XeO2F4

(Trigonal bipyramidal) (Octahedral)

MEMORABLE POINTS

• Parsec is the unit of Distance

• Estimated radius of universe is 1025 m

• Estimated age of Sun is 1018 s

• 18/5 km h–1 equal to 1 ms–1

• 1 femtometre (1 fm) is equal to 10–15 m

• Dot product of force and velocity is Power

• Moment of momentum is equal to

Angular momentum

• Rocket propulision is based on the principle of

Conservation of linear momentum

• The largest of astronomical unit, light year and i


1. 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.

CxHy +

+

4yx O2 → xCO2 +

2y H2O

10 ml 10

+

4yx ml → 10 x ml

According to above equation 10 ml of hydrocarbon

will require 10

+

4yx ml of oxygen for complete

combustion and 10x ml CO2 will be produced. According to the question, Volume of CO2 produced = 20 = 10x

x = 2 ...(i)

∴ Volume of oxygen used = 25 = 10

+

4yx ml

...(ii) Substituting the value of x in Eqn. (ii)

25 = 10

+

4y2

25 = 20 + 2y5

or 25 – 20 = 2y5

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.

2. 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.

Sol. A(C8H10) OH)ii(

O)i(

2

3 → )B(

264 OHC

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.

H – C ≡ C – H 106HC

H2

+

− → C3H5 – C ≡ C – C3H5

the C3H5 – correspond to cyclopropyl (∆) radical hence compound (A) is

CH – C ≡ C – CH CH2

CH2

CH2

CH2 1,2-dicyclopropyl ethane

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2).

CH – C ≡ C – CH CH2

CH2

CH2

CH2 (A)

(i) O3

CH – C – C – CH CH2

CH2

CH2

CH2

H2O O

O – O

(B)

CH – C – C – CHCH2

CH2

CH2

CH2 +H2O2

O O

CH – COOH CH2

CH2 2

warm

Compound (B) is prepared from cyclopropyl bromide as follows :

CH – BrCH2

CH2 (C)

Mgeither

CH.MgBrCH2

CH2

C = O

∆

O

Cyclopropyl magnesium bromide

CH – COOH CH2

CH2

CH.COOMgBr CH2

CH2

Addition compound

HOH

dil. HCl; –MgBrOH

UNDERSTANDINGOrganic Chemistry


Hence, A, –C≡C–

B, –COOH C, –Br 3. 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).

C CH3

CH3

C = O + HCN CH3

CH3

2H2O

OH

CN

H+ C CH3

CH3

OH

COOH (D)

The conversion of (B) to (C) and then (C) to (D) indicates that the reaction (B) to (C) is Hell-Volhard-Zelinsky reaction.

Thus we have :

C CH3

CH3

BrCOOH

H+

C CH3

CH3

OH

COOH (D)

C CH3

CH3

H

COOH (B)

Br2/P

(C)

Hence, we conclude that the aldehyde from which

compound (B) is obtained by oxidation has the structure :

CH3

CH3 – CH – CHO

2-methyl propanal Now, the ozonolysis of the compound (A) gives

acetone and isobutyraldehyde propanal), i.e.,

C = O + O = HC – CH – CH3 CH3

CH3

CH3 –2[O]

C – CH – CH – CH3 CH3

CH3

O

O – O

C = O + CH3 – CH – CHO CH3

CH3

C = CH – CH CH3

CH3

CH3

CH3

O3

CH3

H2O Zn

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

follows :

CH3

CH3 – C = CH – CH – CH3

2,4-dimethyl pentene-2

A,

CH3

CH3

CH3 – CH – COOH

2-methyl propanoic acid

B,

CH3

CH3 – C – COOH

2-bromo-2-methyl propanoic acid

C,

Br

CH3

CH3 – C – COOH

2-hydroxy-2-methyl propanoic acid

D,

OH

4. 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.

CH2

COOHCOOH

2NaHCO3 CH2 COONa COONa

+ 2CO2 +2H2O

∆ CH3COOH + CO2 Soda-lime

∆CH4 + 2CO2

(B)

Acid (B) can be prepared from (A), C3H6Cl2, which

should be 1,3-dichloro propane.

CH2

CH2ClCH2Cl

2NaOH(aq.)CH2

CH2OH CH2OH (–2NaCl)

Propane 1,3-diol

3[O]CH2

COOH COOH

(B)

+ H2O

(A)

Esterification of (B) with ethanol gives malonic ester

which is a synthetic reagent of high importance. CH2

COOHCOOH

∆ CH2 COOC2H5

COOC2H5– 2H2OMalonic ester(B)

+ C2H5OH


LiAlH4 CH2CH2OH CH2OH

CH2 COOH COOH

(B)–2H2O

2SOCl2 CH2 CH2ClCH2Cl–2SO2; –2HCl

(A)

Hence, (A) CH2

CH2ClCH2Cl

(B) CH2

COOHCOOH

5. 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).

Element % Relative no. of atoms

Simplest ratio

C 76.6 12

6.76 = 6.38 06.138.6 = 6

H 6.38 138.6 = 6.38

06.138.6 = 6

O 17.02 16

02.17 = 1.06 06.106.1 = 1

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 :

OH

(A)

CO2 + NaOH OH

(B)

COONa HCl

–NaCl

OH

(C)

COOH

OH

(E)

COOCH3 CH3OHConc. H2SO4

Methyl salicylate Oil of winter green

(Hair tonic)

OH

(D)

COOC6H5

Antiseptic (Salol)

C6H5OH/POCl3

–HCl

140ºC

Elements Named for Places 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.

• 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)

• Germanium : Germany

• Hafnium : Hafnia, Latin for Copenhagen

• 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

• Terbium : Ytterby, Sweden

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

• Ytterbium : Ytterby, Sweden

• Yttrium : Ytterby, Sweden


1. 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?

2. 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

x.g (x) < ∫x

0

)x(g dx; for ∀ x > 0.

3. 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.

4. 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

OA = →a and OB = →b , find OC .

5. 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.

6. 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). 7. 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 8. Let A & B be the matrices such that AAT = I and

AB= BA. Prove that ABT = ATB. 9. If a2 + b2 + c2 = 1, b + ic = (1 + a) z, prove that

c1iba

++ =

iz1iz1

−+ , where a, b, c are real numbers and z

is a complex number.

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

`tàxÅtà|vtÄ VtÄÄxÇzxá 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 Joint Director Academics, Career Point, KotaSolut ions wi l l be publ ished in next i ssue

6Set

Maths Facts • 40 when written "forty" is the only number

with letters in alphabetical order, while "one" is the only one with letters in reverse order.

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

• 111 111 111 x 111 111 111

= 12345678 9 87654321

• 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.

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

• 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.


1. Let S1 ≡ a1x2 + 2h1xy + b1y2 + 2g1x + 2f1y + c1 = 0 & S2 ≡ a2x2 + 2h2xy + b2y2 + 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.

2. c = 2R; ∆4

abc = R; s∆ = r

A

BC

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

s∆ – 2.

s∆ .R; 2r

∆ = 1 + rR2

∆ = r2 + 2Rr

3. m1(OQ) = hk & m2(PR) = –

1m1 =

hk+β−

so m1m2 = –1

⇒ hk .

hk+β−

= – 1

k2 = +βh – h2; so βh = h2 + k2

β = h

kh 22 + so point R is

+ 0,h

kh 22

O

(h,k) rR

(b,0)A(a,0)

Q(a,a)

Now (PR)2 = (RA)2

k2 + 222

hkhh

+− =

222

hkha

+−

⇒ k2 + h2 + 222

hkh

+ – 2(h2 + k2)

= a2 + 222

hkh

+ – 2ah

kh 22 +

⇒ k2 + h2 – a2 = 2(h2 + k2) – 2ah

kh 22 +

–a2 = (h – 2a) h

kh 22 +

(x2 + y2) (x – 2a) + a2x = 0 4. 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.

5. L.H.S. xa =

ay ⇒ x =

ya 2

∫∞

+

0

ya

ayf

y/a)y/a(n

2

2l

−2

2

ydya

I = ∫∞

+

0ya

ayf

y1 (ln a2 – lny) dy

MATHEMATICAL CHALLENGES SOLUTION FOR SEPTEMBER ISSUE (SET # 5)


I = ∫∞

+

0xa

axf (2ln a – lny)

ydy

2I = ∫∞

+

0xa

axf (lna)

xdx ⇒ I = ∫

∞

+

0xa

axf

xnal dx

6. Let the fixed points be P(α, 0) & Q (– α, 0) and

variable line be

y = mx + c as given 2m1

|cm|

+

+α .2m1

|cm|

+

+α− = a; where

'a' is a constant. so |c2 – m2α2| = a (1 + m2) ...(1) Now let foot of the perpendicular from (α, 0) be (h, k)

then c = k – mh & – m1 =

α−hk ⇒ m = –

kh α−

so c = k + k

)h(h α− = k

hhk 22 α−+

use these in (1)

22

2

2

222

k)h(

k)hhk(

αα−

−α−+ = a 2

2

k)h(1 α−

+

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

7. y = f(x) = ∫ −x

0

zzx 2e dz = ∫ −

x

0

zzx 2e.e dz

y´ = ∫ −x

0

zzx 2e.ze dz + 1 = – ∫ −−

x

0

zzx )ze2(e21 2

dz + 1

= –

− ∫ −−

x

0

zxzx0

zxz dze.xe)e.e(21 22

+ 1

= 21 xy + 1 ⇒

dxdy –

21 xy = 1

I.F. = ∫− dx2x

e = 4/x2e−

solution is

y . 4/x2e− = ∫ − 4/x2

e dx = ∫ −x

0

4/z2e dz

y = ∫ −x

0

4/z4/x 22ee dz proved.

8. Since 1−α

α + 1−β

β = 4a7a

2

2

−−

)(1

)(2β+α−+αβ

β+α−αβ = 4a7a

2

2

−−

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

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

and –a2

β ∈ (1, 4) ⇒ 1 < 2

1a 2 + < 4

⇒ a ∈ (– 7 , –1) ∪ (1, 7 ) ...(4)

Hence a ∈ (–2, – 3 ] ∪ [ 3 , 2)

9. f´(x) = 35 x2/3 –

310 x –1/3 = 3/1x

)2x(35 −

sign. dia of f´(x)

(2, 0)(5, 0)

(0,–3(4)1/3)

20+ +

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

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 ∑∈xp

minp = 1.99C49 + 2.98C49 + 3.97C49 + .... + 51.49C49

= Coeff. of x49 in [(1 + x)99 + 2(1 + x)98 + ..... + 51(1+x)49]

= Coeff. of (1 + x)99

x111

)x1(51

x111

)x1(11 4851

+−

+−

+−

+−

= Coeff. of x

)x1(51

)x1(x

)x1()x1( 49

2

2

4899

−+

−

+

+−+

= coeff. of x51 in [(1 + x)101 – (1 + x)50] + coeff. of x50 in 51(1 + x)49 = 101C51 – 0 + 0 = 101C51


1. Find all possible negative real values of 'a' such that :

∫ −− −0

a

tt2 )9.29( dt ≥ 0

Sol. Here ∫ −− −0

a

tt2 )9.29( dt ≥ 0

⇒0

a

tt2

9log9.2

9log29

−−

−

−−

≥ 0 ⇒ ( )0att2 )49(9 −− +− ≥ 0

⇒ 9–2a – 4.9–a + 3 ≥ 0 ⇒ t2 – 4t + 3 ≥ 0; where t = 9–a and t ∈(1, ∞ ) ⇒ (t – 1) (t – 3) ≥ 0 ⇒ t ≤ 1 or t ≥ 3 is possible as t > 1.

∴ 9–a ≥ 3 ⇒ a ≤ – 21 ; |3–a ≥ 3 ⇒ a ≤ –1

2. Let ABCD be any arbitrary plane quadrilateral in the space having E as the point of intersection of its diagonals. If ∆1 and ∆2 be the areas of triangles DEC and AEB, using vector method prove that

∆ ≥ 1∆ + 2∆ , where ∆ is the area of the quadrilateral ABCD. Also discuss the case when the equality holds.

Sol. Let the position vector of the points A, B, C and and

D with respect to E be →a , →b , – λ1.

→a and –λ2→b ;

where λ, λ2 ∈ R+

Now, ∆1 = 21 |DECE|

→→× = |ba|

221 →→

×λλ

)0(E

)a,(C 1λ−

)b(B)a(A

)b,(D 2λ−

⇒ 1∆ = |2||ba|21

1λλ×→→

=2

|ba|→→

×21λλ ...(i)

and ∆2 = |EABE|21 →→

× = 21 |ba|

→→×

⇒ 2∆ = 2

|ba|→→

× ...(ii)

also ∆ = |DBCA|21 →→

× = 2

)1)(1( 21 λ+λ+|ba|

→→×

∴ ∆ = 2

|ba|→→

× )1)(1( 21 λ+λ+

∆ = 2

|ba|→→

×21211 λλ+λ+λ+

where, 2

21 λ+λ ≥ 21λλ

∴ ∆ = 2

|ba|→→

×212121 λλ+λλ+

≥ |ba|21 →→

× + 221 )1( λλ+

= 2

|ba|→→

× + 2

|||ba| 21λλ×→→

∴ ∆ ≥ 2∆ + 1∆ using (i) and (ii) It is clear that equality holds if λ1 = λ2 and in this

case side AB and DC will become parallel.

3. Let S be the coefficients of x49 in given expression f(x) and if P be product of roots of the equation

f(x) = 0, then find the value of PS , given that :

f(x) = (x – 1)2

− 2

2x

−

21x

− 3

3x

−

31x ,

.........

− 25

25x

−

251x

Sol. Here we can write f(x) as :

f(x) =

−

−

−− 25

25x...3

3x2

2x)1x(

×

−

−

−−

251x...

31x

21x)1x(

Now roots of f(x) = 0 are;

12, 22, 32, ..... , 252 and 1, 21 ,

31 , .....,

251

Now f(x) is the polynomial of degree 50, So coefficient of x49 will be : S = – (sum of roots)

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

Students' Forum

MATHS


= – (12 + 22 + ... + 252) –

++++

251....

31

211

= –

+

×× K6

512625 where, K = ∑=

25

1n n1

⇒ S = –(K + 5525). Product of roots :

12 . 22 . 32 .... 252 . 1 . 21 .

31 ....

251 = 1 . 2 . 3 ...25

∴ P = 25 !

Hence PS =

!25)5525K( +− , where K = ∑

=

25

1n n1

4. Find the point inside a triangle from which the sum of the squares of distance to the three side is minimum. Find also the minimum value of the sum of squares of distance.

Sol. If a, b, c are the lengths of the sides of the ∆ and x, y, z are length of perpendicular from the points on the sides BC, CA and AB respectively, we have to minimise : ∆ = x2 + y2 + z2

we have, 21 ax +

21 by +

21 cz = ∆

⇒ ax + by + cz = 2∆ A

B C x

y z

where ∆ is the area of ∆ABC. We have the identity : ⇒ (x2 + y2 + z2) (a2 + b2 + c2) – (ax + by + cz)2 = (ax – by)2 + (by – cz)2 + (cz – ax)2 ⇒ (x2 + y2 + z2)(a2 + b2 + c2) ≥ (ax + by + cz)2 ⇒ (x2 + y2 + z2) (a2 + b2 + c2 ≥ 4∆2

⇒ x2 + y2 + z-2 ≥ 222

2

cba4

++∆

Equality holds only when

ax =

by =

cz = 222 cba

czbyax++++ = 222 cba

2++

∆

∴ The minimum value of ∆ is ;

222

2

cba4

++∆ = 222 cba

s)cs)(bs)(as(4++

−−−

5. From an external point P(α, 2) a variable line is

drawn to meet the ellipse 4

y9

x 22+ = 1 at the points

A and D. Same line meets the x-axis and y-axis at the points B and C respectively. Find the range of values of 'α' such that PA. PD = PB.PC.

Sol. We have been given,

xB

D

CO

A

yP(α,2)

PA.PD = PB.PC Equation of any line through point 'P' is :

θα−

cosx =

θ−

sin2y = r

or x = α + r cos θ, y = 2 + r sin θ Putting this point in the equation of given ellipse, we get 4(r cos θ + α)2 + 9(2 + r sin θ)2 = 36 ⇒ r2 (4 cos2 θ + 9 sin2θ) + 4r (9 sin θ + 2 α cos θ) + 4 α2 = 0 Since PA and PD are the roots of this quadratic in r,

we get

PA.PD = )sin9cos4(

422

2

θ+θα ...(i)

Similarly, putting x = r cos θ + α, y = r sin θ + 2 in the equation of coordinate axis i.e. xy = 0

(r cos θ + α). (r sin θ + 2) = 0 ⇒ r2 sin θ cos θ + r (2 cos θ + α sin θ) + 2α = 0 Since PB and PC and the roots of this quadratic in 'r',

we get, PB.PC = θθ

αcossin

2 = α

α2sin

4 ...(ii)

Thus, we get

θ

α2sin

4 = θ+θ

α22

2

sin9cos44 from (i) and (ii)

⇒ θ

α2sin

4 = )2cos1(9)2cos1(4

8 2

θ−+θ+α

⇒ θ2sin

1 = θ−

α2cos513

2

⇒ 13 = 5 cos 2θ + 2α sin 2θ

where; 5 cos 2θ + 2α sin 2θ ≤ 2425 α+

∴ 254 2 +α ≥ 13 ⇒ α2 ≥ 4

25169 − = 36

⇒ α ∈ (–∞, – 6] ∪ [6, ∞) 6. Let f(x) be a polynomial with integral coefficients

suppose that both f(1) and f(2) are odd. Then, prove that for any integer n, f(n) ≠ 0.

Sol. Suppose f(x) = 0 for some integer n. Then (x – n) divides f(x) So; f(x) = (x – n) g (x) Now, f(1) = (1 – n) . g(1) and f(2) = (2 – n) g (2) Now g (1) and g(2) are both integers, and one of

(1 – n) or (2 – n) is even. So one of f(1) or f(2) is even, which is contradictory,

so there is no integer n, for which f(n) = 0


Limits : Theorems of Limits : If f(x) and g(x) are two functions, then (i) )]x(g)x(f[lim

ax±

→ = )x(flim

ax→ ± )x(glim

ax→

(ii) )]x(g).x(f[limax→

= )x(flimax→

. )x(glimax→

(iii)

→ )x(g

)x(flimax

= )x(glim

)x(flim

ax

ax

→

→ if )x(glimax→

≠ 0

(iv) )]x(kf[limax→

= k )x(flimax→

, where k is constant.

(v) )x(flimax→

= )x(flimax→

(vi) q/p

ax)x(flim

→ =

q/p

ax)x(flim

→ , where p and q are

integers. Some important expansions :

(i) sin x =

+−+− ....!7

x!5

x!3

xx753

(ii) cos x =

+−+− ....!6

x!4

x!2

x1642

(iii) sin h x =

∞+++ ....!5

x!3

xx53

(iv) cos h x =

∞+++ ....!4

x!2

x142

(v) tan x =

+++ ....15x2

3xx

53

(vi) log(1 + x) =

+−+− ....4

x3

x2

xx432

(vii) ex =

++++ ....!3

x!2

xx132

(viii) ax =

+++ ....)a(log!2

xalogx1 22

(ix) (1 – x)–1 = 1 + x + x2 + x3 + ......

(x) sin–1x =

++++ ......7

x.65.

43.

21

5x.

43.

21

3x.

21x

753

(xi) tan–1x =

−+− .....x

51x

31x 53

Some important Limits : (i) xsinlim

0x→ = 0

(ii) xcoslim0x→

= 1

(iii) x

xsinlim0x→

= 1 = xsin

xlim0x→

(iv) x

xtanlim0x→

= 1 = xtan

xlim0x→

(v) x

)x1log(lim0x

+→

= 1

(vi) x

0xelim

→ = 1

(vii) x

1elimx

0x

−→

= 1

(viii) x

1alimx

0x

−→

= logea

(ix) axaxlim

nn

ax −−

→ = nan–1

(x) x

x x11lim

+

∞→ = e =

x

x x11lim

+

−∞→

(xi) x/10x

)x1(lim +→

= e

(xii) x

x xa1lim

+

∞→ = ea

(xiii) nx

alim∞→

=

<>∞

1aif,01aif,

i.e. a∞ = ∞, if a > 1 and a∞ = 0, if a < 1

(xiv) x

1)x1(limn

0x

−+→

= n

(xv) x

xsinlim1

0x

−

→ = 1 =

xxtanlim

1

0x

−

→

LIMIT, CONTINUITY & DIFFERENTIABILITY

Mathematics Fundamentals

MATH


(xvi) xsinlim 1ax

−

→ = sin–1a, |a| ≤ 1

(xvii) xcoslim 1ax

−

→= cos–1a, |a| ≤ 1

(xviii) xtanlim 1ax

−

→ = tan–1a, – ∞ < a < ∞

(xix) xloglim eex→ = 1

(xx) 20x xxcos1lim −

→ =

21

Let )x(flimax→

= l and )x(glimax→

= m, then

(xxi) )x(gax

))x(f(lim→

= lm

(xxii)If f(x) ≤ g(x) for every x in the deleted neighbourhood (nbd) of a, then )x(flim

ax→ ≤ )x(glim

ax→.

(xxiii) If f(x) ≤ g(x) ≤ h(x) for every x in the deleted nbd of a and )x(flim

ax→ = l = )x(hlim

ax→, then )x(glim

ax→ = l.

(xxiv) )x(foglimax→

= f

→)x(glim

ax = f(m)

In particular (a) )x(floglimax→

= log

→)x(flim

ax= log l

(b) )x(faxelim

→ =

)x(flimaxe → = el

(xxv) If )x(flimax→

= + ∞ or – ∞, then )x(f

1limax→

= 0.

Evaluation of Limits (Working Rules) :

By factorisation : To evaluate )x()x(lim

ax ψφ

→, factorise

both φ(x) and ψ(x), if possible, then cancel the common factor involving a from the numerator and the denominator. In the last obtain the limit by substituting a for x.

Evaluation by substitution : To evaluate )x(flimax→

,

put x = a + h and simplify the numerator and denominator, then cancel the common factor involving h in the numerator and denominator. In the last obtain the limit by substituting h = 0.

By L – Hospital's rule : Apply L-Hospital's rule to

the form 00 or

∞∞ .

)x(g)x(flim

ax→=

)x´(g)x´(flim

ax→ =

)x(g)x(flim n

n

ax→

By using expansion formulae : The expansion formulae can also be used with advantage in simplification and evaluation of limits.

By rationalisation : In case if numerator or denominator (or both) are irrational functions,

rationalisation of numerator or denominator (or both) helps to obtain the limit of the function.

Continuity : f(x) is continuous at x = a if )x(flim

ax→exists and is

equal to f(a) i.e. if )x(flim–ax→

= f(a) = )x(flimax +→

.

Discontinuous functions : A function f is said to be discontinuous at a point a of its domain D if is not continuous there at. The point a is then called a point of discontinuity of the function. The discontinuity may arise due to any of the following situations:

(a) )x(flimax +→

or )x(flimax −→

of both may not exist.

(b) )x(flimax +→

as well as )x(flimax −→

may exist but are

unequal. (c) )x(flim

ax +→ as well as )x(flim

ax −→both may exist but

either of the two or both may not be equal to f(a). We classify the point of discontinuity according to

various situations discussed above. Removable discontinuity : A function f is said to

have removable discontinuity at x = a if )x(flim

ax −→= )x(flim

ax +→but their common value is not

equal to f(a). Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.

Discontinuity of the first kind : A function f is said to have a discontinuity of the first kind at x = a if

)x(flimax −→

and )x(flimax +→

both exist but are not equal.

f is said to have a discontinuity of the first kind from the left at x = a if )x(flim

ax −→exists but not equal to

f(a). Discontinuity of the first kind from the right is similarly defined.

Discontinuity of second kind : A function f is said to have a discontinuity of the second kind at x = a if neither )x(flim

ax −→nor )x(flim

ax +→exists.

f if said to have discontinuity of the second kind from the left at x = a if )x(flim

ax −→does not exist.

Similarly, if )x(flimax +→

does not exist, then f is said to

have discontinuity of the second kind from the right at x = a.

Differentiability : f(x) is said to be differentiable at x = a if R´ = L´

i.e. h

)a(f)ha(fLt0h

−+→

= h

)a(f)ha(fLt0h −

−−→

Note : We discuss R, L or R´, L´ at x = a when the function is defined differently for x > a or x < a and at x = a.


Parabola : The locus of a point which moves such that its

distance from a fixed point is equal to its distance from a fixed straight line, i.e. e = 1 is called a parabola.

X´ Z O S(a, 0)

M Y

Y

L

L´ P´

P

N X

Its equation in standard form is y2 = 4ax (i) Focus S (a, 0) (ii) Equation of directrix ZM is x + a = 0 (iii) Vertex is O (0, 0) (iv) Axis of parabola is X´OX Some definitions : Focal distance : The distance of a point on parabola

from focus is called focal distance. If P(x1, y1) is on the parabola, then focal distance is x1 + a.

Focal chord : The chord of parabola which passes through focus is called focal chord of parabola.

Latus rectum : The chord of parabola which passes through focus and perpendicular to axis of parabola is called latus rectum of parabola. Its length is 4a and end points are L(a, 2a) and L´(a, – 2a).

Double ordinate : Any chord which is perpendicular to the axis of the parabola is called its double ordinate.

Equation of tangent at P(x1, y1) is yy1 = 2a(x + x1) and equation of tangent in slope form is

y = mx + ma

Here point of contact is

ma2,

ma

2

Equation of normal at P (x1, y1) is

y – y1 = a2y1− (x – x1)

and equation of normal in slope form is y = mx – 2am – am3

Here foot of normal is (am2, –2am) The line y = mx + c may be tangent to the parabola if

c = a/m and may be normal to the parabola if c = –2am – am3.

Chord of contact at point (x1, y1) is yy1 = 2a (x + x1) Ellipse : If a point moves in a plane in such a way that ratio of

its distances from a fixed point (focus) and a fixed straight line (directrix) is always less than 1, i.e. e < 1 called an ellipse

Standard equation of an ellipse is 2

2

ax + 2

2

by = 1

where b2 = a2 (1 – e2) Now, When a > b

S´ (–ae,0)A´

(–a,0)A´

(–a,0)

O (0,0)

(ae,0) S XZ

M

L1 B(0,b)

Y

L

Z´X´

M´

L´

Y´

B´(0,– b)

In this position, (i) Major axis 2a and minor axis 2b (ii) Foci, S´(–ae, 0) and S(ae, 0) and centre O(0, 0) (iii) Vertices A´ (–a, 0) and A(a, 0) (iv) Equation of directries ZM and Z´M´ are

x ± ea = 0, Z

0,

ea and Z´

− 0,

ea

(v) Length of latus rectum is ab2 2

= LL´ = L1L1´

The coordinates of points of intersection of line y = mx + c and the ellipse are given by

++

−222

2

222

2

mab

b,mab

ma

PARABOLA, ELLIPSE & HYPERBOLA

Mathematics Fundamentals

MATH


Equation of tangents of ellipse in term of m is

y = mx ± 222 mab + and the line y = mx + c is a tangent of the ellipse, if

c = ± 222 mab + The length of chord cuts off by the ellipse from the

line y = mx + c is

222

22222

mabcbma.m1ab2

+−++

The equation of tangent at any point (x1, y1) on the ellipse is

21

axx + 2

1

byy = 1

and at the point (a cos φ, b sin φ) on the ellipse, the tangents is

a

cosx φ + b

siny φ = 1

Parametric equations of the ellipse are x = a cos θ and y = b sin θ.

The equation of normal at any point (x1, y1) on the ellipse is

1

21

xa)xx( − =

1

21

yb)yy( −

also at the point (a cos φ, b sin φ) on the ellipse, the equation of normal is

ax sec φ – by cosec φ = a2 – b2 Focal distance of a point P(x1, y1) are a ± ex1 Chord of contact at point (x1, y1) is

21

axx + 2

1

byy = 1

Chord whose mid-point is (h, k) is

2ahx + 2b

ky = 2

2

ah + 2

2

bk i.e. T = S1

The locus of point of intersection of two perpendicular tangents drawn on the ellipse is x2 + y2 = a2 + b2. This locus is a circle whose centre is the centre of the ellipse and radius is length of line joining the vertices of major and minor axis. This circle is called "director circle".

The eccentric angle of point P on the ellipse is made by the major axis with the line PO, where O is centre of the ellipse.

(a) The sum of the focal distance of any point on an ellipse is equal to the major axis of the ellipse.

(b) The point (x1, y1) lies outside, on or inside the ellipse f(x, y) = 0 according as f(x1, y1) > = or < 0.

The locus of mid-point of parallel chords of an ellipse

is called its diameter and its equation is y = ma

xb2

2−

which is passes through centre of the ellipse.

The two diameter of an ellipse each of which bisect the parallel chords of others are called conjugate diameters. Therefore, the two diameters y = m1x and

y = m2x will be conjugate diameter if m1m2 = – 2

2

ab .

Hyperbola : When the ratio (defined in parabola and ellipse) is

greater than 1, i.e. e > 1, then the conic is said to be hyperbola.

Since the equation of the hyperbola 2

2

ax – 2

2

by = 1

differs from that of the ellipse 2

2

ax + 2

2

by = 1 in

having –b2, most of the results proved for the ellipse are true for the hyperbola, if we replace b2 by – b2 in their proofs. We therefore, give below the list of corresponding results applicable in case of hyperbola.

Standard equation of hyperbola is 2

2

ax – 2

2

by = 1

where b2 = a2 (e2 – 1)

M´ Y M L

L´

XX´

L1

L1´ B´

Y´

A´Z´ O Z (–ae,0)S´

(–a,0)(0

,b)

(0,0

) A (a,0)S(ae,0)

(0,b

) B

In this case,

Foci are S (ae, 0) and S´(–ae, 0). Equation of directrices ZM and Z´M´ are

x m ea = 0, Z

0,

ea and Z´

− 0,

ea

Transverse axis AA´ = 2a, conjugate axis BB´ = 2b. Centre O (0, 0).

Length of latus rectum LL´ = L1L1´ = ab2 2

The difference of focal distance from any point P(x1, y1) on hyperbola remains constant and is equal to the length of transverse axis. i.e.

S´P ~ SP = (ex1 + a) – (ex1 – a) = 2a The equation of rectangular hyperbola

x2 – y2 = a2 = b2 i.e. in standard form of hyperbola put a = b. Hence e = 2 for rectangular hyperbola.

XtraEdge for IIT-JEE OCTOBER 2009 45

a

PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. According to Debye's T3 Law, the specific heat of many solids at low temperature T varies according to the relation c = αT3, where α is a constant. The heat energy required to raise the temperature of 4 kg mass from T = 1 K to T = 3K is

(A) 208 α (B) 20 α (C) 80 α (D) 8 α

2. A system is taken from state A to B along two different paths 1 and 2. The work done on the system along these two paths is W1 and W2 respectively. The heat absorbed by the system along these two paths is Q1 and Q2 respectively. The internal energy at A and B is UA and UB respectively

(A) W1 = W2 = UB – UA (B) Q1 = Q2 = UA – UB (C) Q1 + W1 = Q2 + W2 = UA + UB

(D) Q1 + W2 = Q2 + W1 = UB – UA

3. The indicator diagram for two processes 1 and 2 carried on an ideal gas is shown in figure. If m1 and m2 be the slopes (dP/dV) for Process 1 and Process 2 respectively, then

Process1

Process2 V O

A

(A) m1 = m2 (B) m1 > m2 (C) m1 < m2 (D) None of these

4. A wave equation is given by y = A cos(ωt – kx), where symbols have their usual meanings. If vp is the maximum particle velocity and v is the wave velocity of the wave then

(A) vp can never be equal to v (B) vp = v for λ = 2πA (C) vp = v for λ = A/2π

(D) vp = v for λ = πA

5. A string of length 2L, obeying Hooke's Law, is stretched so that its extension is L. The speed of the transverse wave tavelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave traveling on the string will be

(A) 2

1 v (B) 2 v (C) 21 v (D) 22 v

IIT-JEE 2010

XtraEdge Test Series # 6

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect, Atomic Structure, Radioactivity, X-ray, Nuclear Physics, Matter Waves, Photoelectric Effect, Practical Physics. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements, Nitrogen Family, Oxygen Family, Halogen Family & Noble Gas, Salt Analysis, Metallurgy, Co-ordination Compounds, Transitional Elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D, Probability, Determinants, Matrices. Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type questions with only one correct answer in each. +3 marks will be

awarded for correct answer and -1 mark for wrong answer. • Question 14 to 19 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.

Section - II • Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.


6. The potential difference across the Coolidge tube is 20 kV and 10 mA current flows through the voltage supply. Only 0.5% of the energy carried by the electrons striking the target is converted into X-rays. The power carried by X-ray beam is P.

(A) P = 0.1 W (B) P = 1 W (C) P = 2 W (D) P = 10 W 7. A radioactive sample consists of two distinct species

is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represents the form of this plot ?

(A)

tτ

N

(B)

tτ

N

(C)

tτ

N

(D)

tτ

N

8. A star initially has 1040 deutrons. It produces energy

via the processes 1H2 + 1H2 → 1H3 + p and 1H2 + 1H3 → 2He4 + n. If the average power radiated by the star is 1016 W, the deutron supply of the star is exhausted in a time of the order of

(The masses of nuclei are : m(H2) = 2.014 u, m(p) = 1.007 u, m(n) = 1.008 u, m(He4) = 4.001 u)

(A) 106 s (B) 108 s (C) 1012 s (D) 1016 s 9. In an excited state of hydrogen like atom an electron

has total energy of – 3.4 eV. If the kinetic energy of the electron is E and its de Broglie wavelength is λ, then

(A) E = 6.8 eV, λ ~ 6.6 × 10–10 m (B) E = 3.4 eV, λ ~ 6.6 × 10–10 m (C) E = 3.4 eV, λ ~ 6.6 × 10–11 m (D) E = 6.8 eV, λ ~ 6.6 × 10–11 m This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Use the following Key to choose the appropriate

answer. (A) If both (A) and (R) are true, and (R) is the

correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the

correct explanation of (A). (C) If (A) is true, but (R) is false. (D) If (A) is false, but (R) is true.

10. Assertion : In a standing wave formed in a stretched wire the energy of each element of wire remains constant.

Reason : The net energy transfer in a standing wave is zero.

11. Assertion : On a T-V graph (T on y-axis), the curve

for adiabatic expansion would be a monotonically decreasing curve.

Reason : The slope of an adiabatic process represented on T-V graph is always + ve.

12. Assertion : Energy is released in nuclear fission. Reason : Total binding energy of the fission

fragments is larger than the total binding energy of the parents nucleus.

13. Assertion : If the accelerating potential in an X-ray

tube in increased, the wavelength of the characteristic X-ray do not change.

Reason : When an electron beam strikes the target in an X-ray tube, part of kinetic energy is converted into X-ray energy.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 14 to 16) In the following figure the spring is at its natural

length. In both the chambers 'n' moles of monoatomic gas is filled at temperature T. Heat is supplied to the right chamber. Piston is non-conducting and vessel is non-conducting. Temperature of the left chamber does not change. Piston is displaced by L/4.

n, T

K

n, T

14. Change in temperature of right chamber is –

(A) 2T + nR

LK165 2

(B) T/3 + nR16

KL2

(C) 2T/3 + nR16

KL5 2 (D) 2T/3 +

nRKL5 2

15. Change in the internal energy of the right chamber is

(A) nRT + 3215 KL2 (B) nRT +

32KL2

(C) 2nRT + 15 KL2 (D) 3nRT + 3215 KL2


16. Heat transferred into the system is

(A) 2nRT + 2

KL2 (B) nRT +

2KL2

(C) 3nRT + 32

KL2 (D) nRT +

32KL2

Passage : II (No. 17 to 19) Two hydrogen like atoms A and B are of different

masses and each atom contains equal number of protons and neutrons. The energy difference between the radiation corresponding to first Balmer lines emitted A and B is 5.667 eV when A & B moving with the same velocity, strikes a heavy target they rebound back with the same velocity. In this process the atom B imparts twice the momentum to the target the A imparts.

17. Ionization energy of Atom B is – (A) 27.2 eV (B) 13.6 eV (C) 10.2 eV (D) 54.4 eV

18. Atomic number of atom A is (A) 1 (B) 2 (C) 3 (D) 4 19. Mass number of atom B & Atom A (A) 2, 4 (B) 4, 2 (C) 2, 1 (D) 4, 1 The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q Q

R R R R

S S S S Q P R S

20. Consider a situation (i) that two sound waves, y1 = (0.2m) sin 504π (t – x/300) and y2 = (0.6 m) sin 490π (t – x/300) are superimposed. Consider another situation (ii) that two sound waves, y1 = (0.2m) sin 504π(t – x/300)

y2 = (0.4 m) sin 504π(t + x/300), are superimposed. Match the Column-I with Column-II :

Column -I Column-II (A) In situation (i) (P) Stationary waves

are formed (B) In situation (ii) (Q) There will be the

phenomenon of 'Beats'

(C) When two waves of same frequency and amplitude and travelling in opposite directions superimpose

(R) Amplitude of the resultant wave will vary periodically with position

(D) If the intensity of sound alternately increases and decreases periodically as a result of superposition of waves of slightly different frequencies

(S) Amplitude of the resultant wave will vary periodically with time

21. The specific heat capacity of a material is given as C = AT, where A is a constant, T is temperature. The substance is heated from 27ºC to 127ºC. Unit of A is J/kg/K2. Then match quantities in column I to that in column II.

Column -I Column-II (A) Mean specific heat in the

range 27ºC to 127ºC is (P) 400 A

(B) Actual specific heat at 127ºC is

(Q) 350 A

(C) Graph of specific heat versus temperature is

(R)

(D) Graph of amount of heat

transferred versus temperature is

(S)

22. Match the Column-I with Column-II

Column -I Column-II (A) An electron moves in

an orbit in a Bohr atom

(P) Total Energy

= 2

EnergyPotential

(B) As a satellite moves in a circular orbit around total earth

(Q) Kinetic Energy = Magnitude of Energy

(C) In Rutherford's α-scattering experiment, as an α-particle moves in the electric field of a nucleus

(R) Motion in under a central force

(D) As an object, release from some height above ground, falls towards earth, assuming negligible air resistance

(S) Mechanical energy is coserved


CHEMISTRY

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. For the reaction : [Ag(CN)2]– Ag+ + 2CN– The equilibrium constant at 25º C is 4 × 10–19. If a

solution is 0.1 M in KCN and 0.03 M in AgNO3 originally, at equilibrium, the concentration of Ag+ is

(A) 7.5 × 10–16 M (B) 7.5 × 10–18 M (C) 1.25 × 10–19 M (D) 1.25 × 10–17 2. At 25º C, solubility product of Zn(OH)2 is 10–14. If

NH4OH is 50% dissociated, then the concentration of Zn2+ in its 0.1 M solution is

(A) 2 × 10–12 (B) 1 × × 10–14 (C) 10–12 (D) 4 × 10–12 3. Which of the following pairs of compounds will have

identical B.P ?

(A)

Me

Cl H

Et

&

Me

MeH

CH2Cl

(B)

CH3

Br

H

CH3

OH

H

&

OH

H

H

CH3

CH3

Br

(C)

Me

H

Me

OH H OH

OH H

&

Me

H

Me

OHH OH

MeHO

(D) None of these 4. Which of the following statements are not correct ? (A) Me – CH = C = CH – Me is optically active

(B) Me

MeC

H

Me is optically inactive

(C) All the compounds having chiral centre with L.P. as one of the group, are non-resolvable.

(D) All geometrical isomers are diastereomers

5. Identify most acidic hydrogen in given compound.

OH

O – H

S – H

OH

O O

H

db

c

a

(A) a (B) b (C) c (D) d 6. A yellow powder reacted with F2 to form a colourless

gas X which is used as gaseous insulator in high power generators. It does not get hydrolysed. Another compound is obtained by reaction of sulphur dichloride with NaF. It can be easily hydrolysed and has see-saw shape. X and Y respectively are

(A) AgI, AgBr (B) SF6, SF4 (C) SF4, SF6 (D) SCl4,SCl6 7. Which of the following statement is true ? (A) Si–O bond is stronger than C–O bond (B) Dimethyl ether acts as better lewis base but not

disilyl ether (C) (CH3)3SiO– is more stable than (CH3)3CO–

(D) All of these 8. Here A, B, C and D are respectively

(A) fused Na2CO3

+ air (B) H2SO4 + H2O Evaporation

(C)

Yellow solution

Pb (CH3COO)2

Yellow ppt.

(D)

Orange Coloured

Green Solid

(A) FeSO4, FeCl3, Fe(OH)3, PbCl2 (B) FeCl2, FeSO4, Fe(OH)3, PbSO4 (C) Cr2O3, Na2CrO4, Na2Cr2O7, PbCrO4

(D) FeSO4, Fe2(CO3)2, Fe(OH)3, PbCO3 9. When sulphur is dissolved in oleum, a deep blue

coloured solution containing polyatomic sulphur cation is obtained. The formula of cation present is

(A) +24S (B) +2

8S

(C) +219S (D) +2

16S This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the



10. Assertion : KMnO4 is a coloured compound Reason : Colour of KMnO4 is due to charge transfer. 11. Assertion : [Co(NH3)5Cl]Cl2 reacts with excess of

AgNO3 to form 2 moles of AgCl (white ppt) Reason : [Co(NH3)6]Cl3 gives 2 moles of Cl– which

react with AgNO3 to forms 2 moles of AgCl. 12. Assertion : AlF3 is almost insouble in anhydrous HF

but dissolves in NaF. Reason : NaF produces free F– 13. Assertion : Cu+ ion does not exist in solution. Reason : Cu+ ion undergoes disproportionation in

aqueous solution. This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 14 to 16) The stability of complexes depend upon stability

constant. Higher the value of stability constant, more will be stability of complex. It can also be determined with the help of dissociation constant. Higher the value of dissociation constant, lesser will be stability. Smaller cation, with higher charge can form more stable complex. Stronger the ligand, more stable will be complex. Polydentate ligands form more stable complex than unidentate ligand. If multidentate ligand is cyclic, it further increases the stability, it is called macrocyclic effect.

14. Formation of complex involves (A) exothermic, decrease in entropy (B) endothermic, decrease in entropy (C) exothermic, increase in entropy (D) endothermic, increase in entropy 15. Which of the following is most stable (A) [Co(en)3]3+ (B) [Co(NH3)6]3+ (C) [Co(H2O)6]3+ (D) [CoF6]3–

16. Which of the following exist as dimer (A) Al(CH3)3 (B) CH3Li (C) Si(CH3)4 (D) Be(CH3)2 Passage : II (No. 17 to 19) A black coloured compound (A) on reaction with dil

H2SO4 form a gas 'B' and a solution of compound (C). When gas B is passed through solution of compound (C), a black coloured compound 'A' is obtained which is soluble in 50% HNO3 and forms blue coloured complex 'D' with excess of NH4OH and chocolate brown ppt. 'E' with K4[Fe(CN)6]

17. 'A' is (A) CuS (B) FeS (C) PbS (D) HgS 18. 'D' is (A) Cu(OH)2 (B) [Cu(NH3)2]SO4 (C) [Cu(NH3)4](NO.3)2 (D) [Cu(NH3)6]SO4 19. 'E' is (A) Cu2[Fe(CN)6] (B) [Cu4[Fe(CN)6] (C) Cu3[Fe(CN)6]2 (D) None of these The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

20. Match the following :

Column -I Column-II (A) Compound show

Geometrical isomerism

(P)

Me Me

(B) Compound is chiral (Q) = C

MeH

MeH

(C) Compound having plane of symmetry

(R) C = C

Me

H MeH

(D) Compound having centre of symmetry

(S)

MeH Me

H


21. Match the following : Column -I Column-II

(A) Two electron three centre bond

(P) (BN)x

(B) Four electron three centre bond

(Q) B2H6

(C) sp3 hybrid orbitals (R) AlCl3

(D) Inorganic graphite (S) B4H10


Column -I Column-II (A) KHCO3 (P) Exists in solid state (B) NaHCO3 (Q) Soluble in water (C) LiHCO3 (R) Hydrogen bonding (D) NH4HCO3 (S) Dimeric anion

MATHEMATICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The line joining A(b cos α, b sin α) and B (a cos β,

a sin β) is produced to the point M(x, y) so that

AM : MB = b : a, then x cos2

β+α + y sin2

β+α =

(A) –1 (B) 0 (C) 1 (D) a2 + b2 2. A line meets the coordinate axes in A and B. A circle

is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is

(A) m(m + n) (B) m + n (C) n(m + n) (D) (1/2) (m + n) 3. Let P (a sec θ, b tan θ) and Q (a sec φ, b tan φ) where

θ + φ = π/2, be two points on the hyperbola

2

2

ax – 2

2

by = 1. If (h, k) is the point of intersection of

normals at P and Q, then k is equal to

(A) a

ba 22 + (B) –

+a

ba 22

(C) b

ba 22 + (D) –

+b

ba 22

4. The coordinates of the end point of the latus rectum

of the parabola (y – 1)2 = 2(x + 2), which does not lie on the line 2x + y + 3 = 0 are

(A) (–2, 1) (B) (– 3/2, 1) (C) (–3/2, 2) (D) (–3/2, 0)

5. If q is the angle between the line r = 2i + 3j – k + (i + j + k) t and the plane

r.(3i – 4j + 5k) = q, then

(A) cos θ = 15

62 (B) sin θ = 15

62

(C) cos θ = –70

711 (D) sin θ = – 70

711

6. If a, b are nonzero vectors and a is perpendicular to b then a nonzero vector r satisfying r.a = α, for some scalar α, a × r = b is

(A) 2||)(

abaa ++α (B) 2|| b

baa ×+α

(C) 2||(

ab)aa ×−α (D) 2||

(b

b)aa ×−α

7. The number of values of k for which the system of equations

(k + 1)x + 8y = 4k, kx + (k + 3)y = 3k – 1 has no solution is

(A) 0 (B) 3 (C) 2 (D) infinite 8. If 2

1l + 21m + 2

1n = 1, and l1 l2 + m1m2 + n1n2 = 0, and

∆ =

333

222

111

nmnmnm

l

l

l

then

(A) |∆| = 3 (B) |∆| = 2 (C) |∆| = 1 (D) ∆ = 0 9. If the probability of choosing an integer k out of 2m

integers 1, 2, 3, ...., 2m is inversely proportional to k4(1 ≤ k ≤ 2m), then the probability that chosen number is odd, is

(A) equal to 1/2 (B) less than ½ (C) greater than 1/2 (D) less than 1/3 This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Use the following Key to choose the appropriate





10. Assertion : If A and B are two 3 × 3 matrices such that AB = O, then A = O or B = O.

Reason : If A, B and X are three 3 × 3 matrices such that AX = B, |A| ≠ 0, then X = A–1 B.

11. Assertion : Let p < 0 and α1, α2, ... α9 be the nine

roots of x9 = p, then

∆ =

987

654

321

ααααααααα

= 0

Reason : If two rows of a determinant are identical, then determinant equals zero.

12. Assertion : If A and B are two events such that

P(B) = 1, then A and B are independent. Reason : A and B are independent if and only if

P(A ∩ B) = P(A) P(B) 13. Assertion : The lines a1x + b1y + c1 = 0, a2x + b2y +

c2 = 0 and a3x + b3y + c3 = 0 are concurrent if

333

222

111

aaacbacba

= 0

Reason : The area of the triangle formed by three concurrent lines is zero.


Passage : I (No. 14 to 16) Let k be the length of any edge of a regular tetrahedron. (A tetrahedron whose edges are all equal in length is called a regular tetrahedron.) The angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane whereas the angle between two planes is equal to the angle between the normals. Let O be the origin of reference and A, B and C vertices with position vectors a, b and c respectively of the regular tetrahedron. 14. The angle between any edge and a face not

containing the edge is (A) cos–1(1/2) (B) cos–1 (1/4)

(C) cos–1 (1/ 3 ) (D) π/3 15. The angle between any two faces is

(A) cos–1 (1/ 3 ) (B) cos–1 (1/4) (C) π/3 (D) cos–1 (1/2)

16. The value of [a b c]2 is (A) k2 (B) (1/2)k2 (C) (1/3)k2 (D) k3

Passage : II (No. 17 to 19) Let A(x1, y1), B(x2, y2) and C(x3, y3) be three points.

Area of triangle with vertices A, B and C is given by

21 |∆|

where ∆ = 1yx1yx1yx

33

22

11

17. Points A, B, C are collinear if and only if (A) ∆ = 0 (B) ∆ > 0 (C) ∆ < 0 (D) ∆ ≤ 0

18. If zk = xk + iyk for k = 1, 2, 3 and ∆1 = 1zz1zz1zz

33

22

11

then area of ∆ABC is

(A) 41 |∆1| (B)

i41 |∆1|

(C) 21 |∆1| (D)

i21 |∆1|

19. If P(x, y) is such that

1yx1yx1yx

22

11 + 1yx1yx1yx

33

11 = 0

then the line through A and P is (A) median of ∆ABC (B) bisector of ∠A (C) altitude through vertex A (D) perpendicular bisector of the side BC The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S


20. Column –I Column-II (A) Centroid of the triangle with

vertices A(2, 3, 7), B(6, 7, 5), C(1, 2, 3)

(P) (1, 6, 5)

(B) Mid-point of the line joining the points A(7, 9, 11) and B(–5, 3, –1)

(Q) (3, 4, 5)

(C) A point on the line

2x =

3y =

5z , at a distance 2

from the origin.

(R) (3, 3, 2)

(D) Coordinates of the point dividing the join of (5, 5, 0) and (0, 0, 5) in the ratio 2:3.

(S) (4/ 38 ,

6/ 38 ,

10/ 38 )

21. Let ak = nCk for 0 ≤ k ≤ n and Ak =

−

k

1k

a00a

and

B = ∑−

=+

1n

1k1kk A.A =

b00a

,

Column -I Column-II

(A)

a

(P) 1nn2+

(2nCn)

(B) a – b (Q) 0 (C) a + b (R) 2nCn+1

(D) a/b (S) 1 22. Cards are dealt one by one from wellshuffled pack of

52 playing cards until r (1 ≤ r ≤ 4) aces are obtained. If pr denotes the probability of drawing r aces for the first time at the nth draw (with n ≥ 4), and Pr = (52Cn)Pr, then

Column -I Column-II (A) P1 (P) 52 – nC3 (B) P2 (Q) (n – 1) (52 – nC2) (C) P3 (R) (n – 1C2) (52 – nC1) (D) P4 (S) n –1C3

How does GPS Work? GPS or Global Positioning System is a satellite-based navigation system that consists of network of 18 to 24 satellites that are placed in the earth’s orbit. It was originally intended for certain military applications, but in the late 70’s, the government came up with a system that was also meant for civilian use. GPS works anywhere in the world, in all weather conditions, 24/7. There are however no costs for the use. So, how does GPS work?

Global Positioning System satellites circle the earth in the same orbit, twice a day and transmit signals down to the stations located on mother earth. The information retrieved from these signals is taken and then triangulation is used to accurately calculate the user’s precise location. The GPS receiver will then compare the time when the signal was transmitted by the satellite to the time the signal was received. This time difference will then tell the receiver how far away the satellite actually is. With these distance measurements from a couple of the satellites, the receiver will then be able to determine the user’s exact position and will then display it on an electronic map in the unit.

It is important to note that these GPS receivers only receive information and don’t transmit signals in any way. For unobstructed transmittance of signals, receivers are required to have an unobstructed view of the sky, so units are generally placed outdoors and tend to perform very poorly if placed near tall buildings or within forests. GPS operations are highly dependent on accurate time references that are generally provided by atomic clocks at the U.S. Naval Observatory. Each and every GPS satellite will have an atomic clock placed on board.

So, we already know that these satellites transmit information that indicates the current time and its current location. All these GPS satellites synchronize operations so that any repeating signals can be transmitted at the very same instant. These signals move at the speed of light and arrive at the receivers end at slightly varied times, as some satellites are farther away than the others. the distance to the satellites is calculated by estimating the time taken for the signals to reach the GPS receiver. Once the receiver has estimated the distance of at least 4 of these satellites, it can then calculate their positions in three dimensions (latitude and longitude and even the altitude). Once the receiver is locked on to the signals of at least three of these satellites, it can then calculate a 2 d position (longitude and latitude) and can also track movements. Once the position is determined, the unit can then calculate other factors like the speed, trip distance, track, distance to the destinations, sunrise time and sunset time, and so on.

Today, there are at least 24 functional satellites doing the rounds at all times. The GPS satellites that are operated by the U.S. Air Force orbit the earth with a period of 12 hours. Ground stations can also precisely track each satellite’s orbit.


PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A spherical black body of radius r0 radiates a power P

at temperature T0. Another spherical black body of radius r0/2 and at temperature 2T0 emits a power

(A) P (B) 2P (C) 4P (D) 8P 2. The average translational kinetic energy of O2

(relative molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (relative molar mass 28) molecules in eV at the same temperature is

(A) 0.0015 (B) 0.003 (C) 0.048 (D) 0.768 3. The internal energy of an ideal gas is

(A) zero (B) PV (C) 1

PV−γ

(D) 1

PV−γ

γ

4. 1 g of water on evaporation at atmospheric pressure

forms 1671 cm3 of steam. Heat of vaporisation at this pressure is 540 calg–1. The increase in internal energy is

(A) 250 cal (B) 500 cal (C) 1000 cal (D) 1500 cal 5. A motion is described by y = 3ex.e–3t where y, x are

in metre and t is in second. (A) This represents equation of progressive wave

propagating along –x direction with 3 ms–1. (B) This represents equation of progressive wave

propagating along +x direction with 3 ms–1 (C) This does not represent a progressive wave

equation (D) Data is insufficient to arrive at any conclusion of

this sort. 6. A tuning fork A of frequency as given by the

manufacture is 512 Hz is being tested using an accurate oscillator. It is found that they produce 2 beats per second when the oscillator reads 514 Hz and 6 beats per second when it reads 510 Hz. The actual frequency of the fork in Hz is

(A) 508 (B) 512 (C) 516 (D) 518

7. The speed of longitudinal wave is 100 times the speed of transverse wave in a taut brass wire. If the Young's modulus of wire is 1011 Nm–2, then the stress in the wire is

(A) 105 Nm–2 (B) 106 Nm–2 (C) 107 Nm–2 (D) 108 Nm–2

IIT-JEE 2011

XtraEdge Test Series # 6

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism, Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D

Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type questions with only one correct answer in each. +3 marks will be

awarded for correct answer and -1 mark for wrong answer. • Question 14 to 19 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.

Section - II • Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.


8. Two sounding bodies are producing progressive waves given by y1 = 4 sin(400πt) and y2 = 3 sin(404πt), where t is in second which superpose near the ears of a person. The person will hear.

(A) 2 beats per second with intensity ratio 4/3 between maxima and minima

(B) 2 beats per second with intensity ratio 49 between maxima and minima

(C) 4 beats per second with intensity ratio 7 between maxima and minima

(D) 4 beats per second with intensity ratio 4/3 between maxima and minima.

9. An air bubble doubles in radius on rising from the

bottom of a lake to its surface. Assuming that the bubble rises slowly and the atmospheric pressure to be equal to a column of water of height H, the depth of the lake is

(A) 4H (B) 5H (C) 7H (D) 14H This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Use the following Key to choose the appropriate




10. Assertion : Sound produced by an open organ pipe is

more melodious than that produced by a closed organ pipe.

Reason : Air can flow in a better way in an open organ pipe.

11. Assertion : Two tuning forks having frequencies 410

Hz and 524 Hz are kept close and made to vibrate. Beats will not be heard.

Reason : Sound waves superimpose only when the frequencies of superposing waves are equal or nearly equal.

12. Assertion : A blue star is hotter than a red star. Reason : According to Stefan's law, a black body at a

higher temperature radiates more power per unit area. 13. Assertion : A hot body is kept in some surrounding.

As it cools, its temperature falls from 80ºC to 78ºC in a time duration t1 and from 50ºC to 48ºC in time duration t2. The temperature of surrounding is constantly 20ºC, then t1 > t2.

Reason : According to Newton's law of cooling, rate of cooling depends on the difference of temperature of the body and the surrounding.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 14 to 16) Many waveforms are described in terms of

combinations of travelling waves. Superposition principle is used to analyse such wave combinations. Two pulses travelling on same string are described by

y1 = 2)t4x3(

52 +−

, y2 = 2)6t4x3(

52 +−+

−

14. The direction in which pulse is travelling is (A) y1 is in positive x-axis, y2 is in positive x-axis (B) y1 is in negative x-axis, y2 is in negative x-axis (C) y1 is in positive x-axis, y2 is in negative x-axis (D) y1 is in negative x-axis, y2 is in positive x-axis 15. The time when the two waves cancel everywhere – (A) 1 sec (B) 0.5 sec (C) 0.25 sec (D) 0.75 sec 16. The point where two waves always cancel (A) 0.25 m (B) 0.5 m (C) 0.75 m (D) 1 m Passage : II (No. 17 to 19) One mole of diatomic gas is taken through following

cyclic process. The process CA is P = (Constant)V. Temperature at C is 100 K.

B A

P

3P0

P0

V C

17. Temperature at A is (A) 300 K (B) 900 K (C) 600 K (D) 1200 L 18. Molar heat capacity for process CA is (A) R (B) 2R (C) 3R (D) 4R 19. Work done in the cycle is - (A) 200 R (B) – 200 R (C) 400 R (D) – 400 R


The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q Q

R R R R

S S S S Q P R S

20. Match the standing waves formed in column-II due to

plane progressive waves in Column-I and also with conditions in column-I.

Column -I Column-II

(A) Incident wave is y = A sin(kx – ωt)

(P) y = 2A cos kx sin ωt

(B) Incident wave is y = A cos(kx – ωt)

(Q) y = 2A sin kx ωt

(C) x = 0 is rigid support

(R) y = 2A sin kx cos ωt

(D) x = 0 is flexible support

(S) y = 2A cos kx cos ωt

21. Match Columns-I and II

Column –I Column-II

(A) Wien's displacement law explains

(P) Why days are hot and night cold in deserts

(B) Planck's law explains

(Q) Why a blackened platinum wire, when gradually heated, appears first dull red and then blue

(C) Kirchhoff's law explains

(R) The distribution of energy in black body spectrum at shorter as well as longer wavelengths

(D) Newton's second law explains

(S) Why some stars are hotter than others

22. The figures given below depict different processes for

a given amount of an ideal gas.

1/V

P

V

PAdiabatic

(i) (ii)

V

P

V

P

(iii) (iv) Column -I Column-II

(A) In fig. (i) (P) Heat is absorbed by the system

(B) In fig (ii) (Q) Work is done on the system

(C) In fig (iii) (R) Heat is rejected by the system

(D) In fig. (iv) (S) Work is done by the system

CHEMISTRY

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. At constant pressure P, A dissociates on heating

according to the equation A(g) B(g) + C(g) The equilibrium partial pressure of A at T K is 1/9 P,

the equilibrium Kp at TK is

(A) 98 P (B)

964 P (C)

916 P (D) 9 P

2. Calculate the pH of 6.66 × 10–3 M solution of

Al(OH)3. Its first dissociation is 100% where as second dissociation is 50% and third dissociation is negligible.

(A) 2 (B) 12 (C) 11 (D) 13 3. pH of the blood in the body is maintained by buffer

solution of (A) glucose and salt concentration (B) protein and salt concentration (C) CO3

3– and HCO3–

(D) Salt and carbonate ion

4. IUPAC name of the following compound is : OH

CH3

(A) 2-methyl-3-cyclohexenol (B) 3-methyl-1-cyclohexen-4-ol (C) 4-hydroxy-3-methyl-1-cyclohexene (D) 2-hydroxy-1-methylcyclohexene


5. Which will form geometrical isomers ?

(A)

Cl

Cl

(B) CH3CH = NOH

(C)

(D) All

6. The dissolution of Al(OH)3 by a solution of NaOH

results in the formation of (A) [Al(H2O)4(OH)2]+ (B) [Al(H2O)3(OH)3] (C) [Al(H2O)2(OH)4]– (D) [Al(H2O)6](OH)3 7. Helium-oxygen mixture is used by deep sea divers in

preference to nitrogen-oxygen mixture because (A) helium is much less soluble in blood than

nitrogen (B) nitrogen is highly soluble in water (C) helium is insoluble in water (D) nitrogen is less soluble in blood than helium 8. SF4 + BF3 → (A). The compound 'A' is (A) [SF5]–[BF2]+ (B)[SF3]+[BF4]– (C) SF6 (D) S2F4 9. Red lead on reaction with dil. HNO3 forms (A) PbO (B) PbO2 (C) PbO + Pb(NO3)2 (D) PbO2 + Pb(NO3)2 This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Use the following Key to choose the appropriate




10. Assertion : For a reaction at equilibrium, the free

energy for the reaction is minimum. Reason : The free energy for both reactants and

products decreases and become equal. 11. Assertion : Tropylium cation is more stable than

(CH3)3C⊕ Reason : It is stabilized by both resonance effect and

inductive effect.

12. Assertion : K2CO3 can be prepared by Solvay process like Na2CO3.

Reason : KHCO3 is highly water soluble. 13. Assertion : PbI4 does not exist although PbCl4 exist. Reason : Both Pb4+ and I–1 are strong oxidant and

strong reductant respectively.


Passage : I (No. 14 to 16) Lithium only forms monoxide when heated in

oxygen. Sodium forms monoxide and peroxide in excess of oxygen. Other alkali metals form super oxide with oxygen i.e., MO2. The abnormal behaviour of lithium is due to small size. The larger size of nearer alkali metals also decides the role in formation of superoxides. The three ions related to each other as follows :

ionOxide

2O − → 2O2/1 ionPeroxide

22O − → 2O

ionSuperoxide2O2 −

All the three ions abstract proton from water. 14. Consider the following reaction : M + O2 →

oxide)per su(2MO (M = alkali metal)

Select the correct statement : (A) M can not be Li and Na (B) M can not be Cs and Rb (C) M can not be Li and Rb (D) None of these 15. Lithium does not form stable peroxide because : (A) of its small size (B) d-orbital are absent in it (C) it is highly reactive and form superoxide in place

of peroxide (D) covalent nature of peroxide

16. Which compound will liberate oxygen when react with water :

(A) Na2O2 (B) KO2 (C) Na2O (D) Cs2O2

Passage : II (No. 17 to 19) All the boron trihalides except BI3 may be prepared

by direct reaction between the elements. Boron trihalides consist of trigonalplanar BX3 molecules. Unlike the halides of the other elements in the group they are monomeric in the gas, liquid and solid states, BF3 and BCl3 are gases, BBr3 is a volatile liquid and BI3 is a solid. Boron trihalides are Lewis acids because they form Lewis complexes with suitable bases.


BF3(g) + : NH3(g) → F3B – NH3(g) However, boron chlorides, bromides and iodides are

susceptible (sensitive) to protolysis by mild proton sources such as water, alcohols and even amines for example BCl3 undergoes rapid hydrolysis.

BCl3(g) + 3H2O(l) → B(OH)3(aq.) + 3HCl(aq.) It is supposed that the first step in the above reaction

is the formation of the complex Cl3B ← OH2 which then eliminates HCl and reacts further with water.

17. Which of the following is the best order of Lewis

acid strength of BF3, BCl3 and BBr3 ? (A) BF3 > BCl3 > BBr3 (B) BF3 = BCl3 = BBr3 (C) BF3 < BCl3 <BBr3 (D) BBr3 > BF3 > BCl3 18. Which of the following reaction is incorrect ?

(A) BF3(g) + −.)aq(F → [ ]− .)aq(4BF

(B) BCl3(g) + 3EtOH(l) → B(OEt)3(l) + 3HCl(g)

(C) BBr3(l) + F3BN(CH3)3(g) → BF3(g) +Br3BN(CH3)3(g) (D) BCl3(g) +

Excess)(55 NHC2 l → Cl3B(C5H5N)2(s)

19. Which of the following is correct ? (A) B(OCH3)3 is much weaker Lewis acid than BBr3 (B) B(OH)3(aq.) behave as a triprotic acid (C) [H2BO3](aq.)

– is a conjugate base of H3BO3(aq.)

(D) BF3 does not react with ethers. The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q Q

R R R R

S S S S Q P R S


Column -I Column-II (A) Na + Liq. NH3 (P) Paramagnetic (B) Li (Q) Blue coloured

(C) KO2 (R) Strongest reducing

agent

(D) [e(NH3)x]– (S) Highest ionisation energy

21. Match the following : Column -I Column-II

(A) Enantiomers (P) meso-Tartaric acid (B) Enantiomerism (Q) CH3CH = C = CHC2H5

(C) Diastereomers (R) Possess identical

physical and chemical properties

(D) Diastereomerism (S) Possess different physical properties


Column -I Column-II

(A) Saturated solution of AgCl

(P) Common ion effect

(B) Unsaturated solition of AgCl

(Q) IP = SP

(C) Supersaturated solution of AgCl

(R) IP > SP

(D) Solution of AgCl in presence of NaCl

(S) IP < SP

MATHEMATICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Q, R and S are the points on the line joining the

points P (a, x) and T (b, y) such that PQ = QR = RS =

ST, then

++

8y3x5,

8b3a5 is the mid point of the

segment (A) PQ (B) QR (C)RS (D) ST 2. The line x + y = 1 meets x-axis at A and y-axis at B.P

is the mid-point of AB (fig.) P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA; M2 is that from P2 to OP; P3 is that from M2 to OA and so on. If Pn denotes the nth foot of the perpendicular on OA from Mn–1, then OPn =

A x

P M1

M2

O P3 P2 P1

B

y

(A) 1/2 (B) 1/2n

(C) 1/2n/2 (D) 1/ 2


3. The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x – 6y + 9 sin2α + 13 cos2α = 0 is 2α. The equation of the locus of the point P is

(A) x2 + y2 + 4x – 6y + 4 = 0 (B) x2 + y2 + 4x – 6y – 9 = 0 (C) x2 + y2 + 4x – 6y – 4 = 0 (D) x2 + y2 + 4x – 6y + 9 = 0 4. The directrix of the parabola y2 + 4x + 3 = 0 is (A) x –3/4 = 0 (B) x + 1/4 = 0 (C) x – 1/4 = 0 (D) x – 4/3 = 0 5. Equation of the locus of the pole with respect of the

ellipse 2

2

ax + 2

2

by = 1, of any tangent line to the

auxiliary circle is the curve

4

2

ax + 4

2

by = λ2 where

(A) λ2 = a2 (B) λ2 = 1/a2 (C) λ2 = b2 (D) λ2 = 1/b2 6. If PQ is a double ordinate of the hyperbola

2

2

ax – 2

2

by = 1 such that OPQ is an equilateral

triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola, satisfies

(A) 1 < e < 2/ 3 (B) e = 2/ 3

(C) e = 3 /2 (D) e > 2/ 3 7. Equation of the plane through (3, 4, – 1) which is

parallel to the plane r.(2i – 3j + 5k) = 0 is (A) r.(2i – 3j + 5k) + 11 = 0 (B) r.(3i – 4j + k) + 11 = 0 (C) r.(3i + 4j – k) + 7 = 0 (D) r.(2i – 3j + 5k) + 11 = 0 8. If a = λ(i + j – k), b = µ(i – j + k), and c are unit

vectors perpendicular to the vector a and coplanar with a and b, then a unit vector d perpendicular to both a and c is

(A) 6

1 (2i – j + k) (B) 2

1 (j + k)

(C) 6

1 (i – 2j + k) (D) 2

1 (j – k)

9. If a, b, c are three non-coplanar vectors such that

a + b + c = αd and b + c + d = βa then j + 4k then a + b + c + d is equal to

(A) 0 (B) αa (C) βb (D) (α + β)c

This section contains 4 questions numbered 10 to 13, (Assertion and Reason type question). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Use the following Key to choose the appropriate




10. Assertion : The points A(3, 4), B(2, 7), C(4, 4) and

D(3, 5) are such that one of them lies inside the triangle formed by the other points.

Reason : Centroid of a triangle lies inside the triangle.

11. Assertion : The line 9x + y – 28 = 0 is a chord of

contact of a point P with respect to the circle 2x2 + 2y2 – 3x + 5y – 7 = 0.

Reason : The line joining the points of contacts of the tangents drawn from a point P outside a circle to the circle is the chord of contact of P with respect to the circle.

12. Assertion : The line bx – ay = 0 will not meet the

hyperbola 2

2

ax – 2

2

by = 1 (a > b > 0)

Reason : The line y = mx + c does not meet the

hyperbola 2

2

ax – 2

2

by = 1if c2 = a2m2 – b2

13. Assertion : The lines 1

1x − = 1

y−

= 1

1z + and

21x − =

21y + =

3z are coplanar and equation of the

plane containing them is 5x + 2y – 3z – 8 = 0

Reason : the line 1

2x − = 2

1y + = 3z is

perpendicular to the plane 3x + 6y + 9z – 8 = 0 and parallel to the plane x + y – z = 0.


Passage : I (No. 14 to 16) A(3, 7) and B(6, 5) are two points. C : x2 + y2 – 4x – 6y – 3 = 0 is a circle.


14. The chords in which the circle C cuts the members of the family S of circles through A and B are concurrent at

(A) (2, 3) (B) (2, 23/3) (C) (3, 23/2) (D) (3, 2) 15. Equation of the member of the family S which bisects

the circumference of C is (A) x2 + y2 – 5x – 1 = 0 (B) x2 + y2 – 5x + 6y – 1 = 0 (C) x2 + y2 – 5x – 6y – 1 = 0 (D) x2 + y2 + 5x – 6y – 1= 0 16. If O is the origin and P is the centre of C, then

difference of the squares of the lengths of the tangents from A and B to the circle C is equal to

(A) (AB)2 (B) (OP)2 (C) |(AP)2 – (BP)2| (D) None of these Passage :II (No. 17 to 19) a = 6i + 7j + 7k, b = 3i + 2j – 2k, P(1, 2, 3). 17. The position vector of L, the foot of the

perpendicular from P on the line r = a + λb is (A) 6i + 7j + 7k (B) 3i + 2j – 2k (C) 3i + 5j + 9k (D) 9i + 9j + 5k 18. The image of the point P in the line r = a + λb is (A) (11, 12, 11) (B) (5, 2, –7) (C) (5, 8, 15) (D) (17, 16, 7) 19. If A is the point with position vector a

r then Area of

the ∆PLA in sq. units is equal to

(A) 63 (B) 2/177

(C) 17 (D) 7/2

The section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

D C B A

P P P P

Q Q Q Q

R R R R

S S S S Q P R S

20. If P(x, y) is a point in the coordinate plane such that

Column -I Column-II (A) P is equidistant

from (a + b, a – b) and (a – b), (a + b)

(P) 3x2 + 3y2 – 2xy = 0

(B) P is at a distance a + b, from (a, b)

(Q) y – 2x = 0

(C) distance of P from x-axis is twice its distance from y-axis

(R) x2 + y2 – 2ax – 2by – 2ab = 0

(D) distance of P from the origin is the mean of the its distances from the coordinate axes.

(S) x = y

21. Match the column.

Column -I Column-II

(A) y = mx + )ba( 22 + 2m1+ (P) 2

2

ax + 2

2

by = 1

(B) y = mx + 222 bma + (Q) 2

2

ax – 2

2

by = 1

(C) y = mx + 222 bma − (R) b2y2 = 4ax (D) y = mx + a/b2x m (S) x2 + y2 = a2 + b2

22. If a and b are two units vectors inclined at angle α to

each other then Column –I Column-II

(A) |a + b| < 1 if (P) 3

2π < α < π

(B) |a – b| = |a + b| if (Q) π/2 < θ ≤ π (C) |a + b| < 2 (R) α = π/2

(D) |a – b| < 2 (S) 0 ≤ θ < π/2

Honesty

• To be persuasive, You must be believable.

To be believable, You must be credible.

To be credible, You must be truthful.

• An honest man is the noblest work of God.

• If I am honesty in all my dealings, I can never experience fear.

• Prefer a loss to a dishonest gain; one brings pain for the moment, the other for all time.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans C D C B D B D C B D Ques 11 12 13 14 15 16 17 18 19 Ans D A B C A B D A B

20 A → Q, S B → P, R C → P, R D → Q, S 21 A → Q B → P C → S D → R 22 A → P,Q,R,S B → P,Q,R,S C → R, S D → R, S

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans B D B C A B D C B A Ques 11 12 13 14 15 16 17 18 19 Ans C A A A A A A C A

20 A → P,R,S B → Q C → P,R,S D → S 21 A → Q, R B → S C → Q, R D → P 22 A → P,Q,R,S B → P,Q,R C → Q D → P

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B B D C B C B C C D Ques 11 12 13 14 15 16 17 18 19 Ans D A B C D B A A A

20 A → Q B → P C → S D → R 21 A → R B → Q C → P D → S 22 A → P B → Q C → R D → S

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 Ans C C C B B C C B C C Ques 11 12 13 14 15 16 17 18 19 Ans C B D C D D B C B

20 A → P, R B → Q, S C → Q, R D → P, S 21 A → Q, S B → R C → P D → Not match 22 A → P, S B → S C → P, S D → Q, R

CHEMISTRY Ques 1 2 3 4 5 6 7 8 9 10 Ans C B C A D C A B D A Ques 11 12 13 14 15 16 17 18 19 Ans A D A A A B C D A

20 A → P, Q, R B → S C → P D → P,Q,R 21 A → R B → Q C → S D → P, Q 22 A → Q B → S C → R D → P, R

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B B D C B D A B A A Ques 11 12 13 14 15 16 17 18 19 Ans D C B B C C C C B

20 A → Q B → R C → Q D → P 21 A → S B → P C → Q D → R 22 A → P B →R C → Q D → S

IIT- JEE 2010 (October issue)

IIT- JEE 2011 (October issue)

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