XtraEdge_2010_09

Dear Students,

Three Things you Must Know to Attract Success

Everyone wants success. Some people spend their every waking moment pursuing it, to the detriment of all else. For others, attaining success seems impossible. They conclude that it is destined for a select few. The rest of us are to remain "content with such things as we have". Having it all is not "in our stars".

When you strive for success with the wrong assumptions, you will never reach it. It's like traveling somewhere with the wrong map.

Zig Ziglar says that, "Success is a process, not an event," "a journey, not a

destination." Jim Rohn describes it as " .... a condition that must be attracted not pursued."

Success is something you must work hard and long to earn, for yourself. It has a price, sometimes a very high one. And most people are n't really and truly ready to pay that price, to do what success demands. If success has eluded you so far, perhaps you should try changing your assumptions. You need to accept that : • You must go through a growing process, which will require time and patience,

in order to achieve success. There are no short cuts. Anything else is a temporary illusion. Success that will remain with you, and bring you joy rather than sorrow, requires a learning process, a time to grow out of old habits and into new ones, a time to learn what works and what doesn't. And you must pay your dues, in full, in advance! so don't be in a hurry.

• You will need to acquire traits and skills that attract it. What does success mean to you ? Identify, in specific terms, what you regard as success. What traits or skills will you need to achieve this goal? Find 2 or 3 people who have what you want. Write down the habits that have made them successuf and resolve to copy them. This is called mentoring learning from others who have arrived where you want to go. Once you learn to do what it takes, you qualify. And when you qualify, success comes looking for you. You just can't be denied! • You must be ready to travel the road to success, oftentimes alone. It's been

said that, "At some point in time, the pursuit of your goals becomes secondary and what you have become in the process .... is what is most important. It's not the distance you go .... so much as the going itself" (Les Brown).

Remember, when parents try to teach their children to crawl, what they do? They put their favorite toy in front of them and teased them forward, inch by inch. They were after the toy, which kept them motivated. When they became good at reaching the toy, they had learned to crawl. After that, they could reach any destination they wanted. The DESTINATION was less important. They became champion crawlers in the PROCESS!

When you are ready for success you attract it, with little effort. When you are not, it runs from you, no matter how hard you chase. In other words, you repel it! Most likely, this is the reason that success eludes people.

Now that you know how to attract success, why not get started on the journey that will take you where you want to go? Any one can succeed, but unfortunately not every one will. Fate does not foist it upon you. You can have anything you want in life, if you're ready to pay the price. But if you consider the process too hard, too slow, or too long and lonely, you have qualified your self as a looser; painful but true. So don't short change yourself with short-cuts. Go out there today and start attracting success. It's literally yours for the taking!

Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

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Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari

Impatience never commanded success.

Volume - 6 Issue - 3 September, 2010 (Monthly Magazine)

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Volume-6 Issue-3 September, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics,, Chemistry & Maths Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2011 & 2012

S

Success Tips for the Months

• What matters is not what you have, but what you can do.

• It is not about what you can't do. It is about what you can do.

• Never mind what others do; do better than yourself, beat your own record from day to day, and you are a success.

• If you don't know where you're going, who will follow you?

• If you want to be smart, find friends who are smarter than you are.

• Don't be irreplaceable. If you can't be replaced, you can't be promoted.

• Never test the depth of the water with both feet.

• Many of life's failures are people who did not realize how close they were to success when they gave up.

CONTENTS

INDEX PAGE

NEWS ARTICLE

Vice president addresses IIT Delhi

IITian ON THE PATH OF SUCCESS

KNOW IIT-JEE

XTRAEDGE TEST SERIES

Class XI – IIT-JEE 2012 Paper

Regulars ...

DYNAMIC PHYSICS

Students’ Forum Physics Fundamentals Current Electricity

Circular Motion, Rotational Motion

CATALYSE CHEMISTRY

Aliphatic Hydrocarbon

Oxygen Family & Hydrogen Family Understanding : Inorganic Chemistry

DICEY MATHS

Study Time...

Test Time ...

IIT-K To Coordinate 2011 JEE The IIT-K will be coordinating the IIT Joint Entrance Exam (IIT-JEE) for 2011. The members of the IIT Joint Admission Board (JAB) will be attending a meeting at IIT-K on August 21 to finalise the schedule. Representatives of all 15 IITs are expected to attend the meeting. The 2010 exam was coordinated by IIT-Chennai. IIT-K Director Sanjay Govind Dhande said: “Not many changes are likely to be made in the examination format. We will decide the dates for the form distribution, form submission and examination . A final decision will be taken in the meeting by the JAB members.” The name of the IIT-JEE chairman will also be announced soon after the meeting. The IIT-JEE is divided into seven zones headed by prominent IITs. Each year, one of these seven IITs coordinates the exam.

Vice President addresses IIT Delhi convocation

New Delhi: Vice President of India M. Hamid Ansari has said that there has not been sufficient appreciation of engineering education being a key enabler of India's growth and a vital element in shaping of our national destiny.

Addressing the audiences at the forty-first convocation ceremony of the Indian Institute of Technology (IIT)-Delhi on Saturday, Ansari said that questions about the ability of the present framework of engineering education to respond to national requirements in adequate measure remain unanswered.

He further said that the quality of teaching and employability of graduates is one aspect of it; the dearth of qualified and motivated faculty is another. It is for this reason that the National Knowledge Commi-ssion called for "a new paradigm in regulation, accredi-tation, governance and faculty development" across the engineering education spectrum.

Ansari further stated, "It would seem that an essential concomitant of technological advance is the effort by society, including its professional segment of engineers and technologists, to ensure that it sustains and promotes social cohesiveness through necessary correctives."

"Technological, scientific or digital divides in societies cannot further the larger human cause. Today's professionals cannot function in isolation of the social and political context nor can they remain in ivory towers or professional silos," he said. He drew attention to some data that makes for disturbing reading viz. less than 1 per cent of IIT undergraduates in the country pursue Masters or Ph D courses within the IIT system; less than 15 per cent of those graduating from IITs move towards teaching or research, whether in India or abroad; the IIT system produces less than 1.5 per cent of the total engineering graduates in the country but accounts for over 70 per cent of those pursuing Doctoral programmes in engineering and technology. Also, in terms of international grading of academic output based on publications, citations of faculty, and patents applied for and granted, India fares poorly in comparison to even some developing countries.

Only IIT-Mumbai and IIT-Delhi find a place in the 2009 Times Higher Education ranking of 50 engineering and information technology institutions. However, no Indian university, not even an IIT, figures in the top 100 of the Shanghai Jiao Tong

University Institute of Higher Education's Academic Ranking of World Universities, or in the top 100 of the 2009 Times Higher Education World University Rankings. Ansari informed the gathering, "Students from India and those of Indian origin and numbering 35,300 accounted for over one-third of all foreign engineering students in the United States in 2009. Out of these, around 26,000 students were enrolled in Masters Programmes constituting over 65 per cent of all foreign masters students, and 5690 were enrolled in Doctoral Programmes constituting around one fifth of all foreign doctoral students."

"These figures shed light on the opportunity loss for our academic institutions, and eventually to the nation, to benefit from the research potential and effort of the best and brightest graduating from our engineering institutions, including the IITs," he said. Emphasizing on the need to focus on accessible, affordable and applicable learning, the Vice President said that "We need to close the gap between policy intent and actual delivery. The requirement to up-skill or re-skill 500 million people by 2020 in order to meet growth requirements underlines the need for undertaking this on a war footing. Curricular reforms, faculty development and promotion of a spirit of entrepreneurship and innovation are imperative and compelling."

The Vice President stated that the evolutionary context of any technology determines the purposes to which they would be deployed. "Where such technologies evolve as societal products, they carry the ability to serve larger social purposes. Increasingly, in this era of globalization transforming technologies are emerging in corporate contexts and are being deployed to primarily serve narrow corporate interests and stakeholders."

"Thus, the shrinking base of stakeholders in the development and deployment of technologies is fast eroding their political and social legitimacy. It is increasingly felt that these technologies are widening societal inequalities and deepening political conflict. The situation has also been compounded by the lack of political initiative and social impetus by national leaderships and community elders," he stated.

Quoting American scientist Bill Hubbard, the Vice President said that "progress of biology, neuroscience and computer science will make possible in the foreseeable future technolo-gies of mind and life that will invalidate the working social assumptions of societies."

"The graduating students today represent the young citizenry that constitutes the overwhelming majority of our population. It is for you to question if the technologies that you have imbibed and would develop in future are being co-opted in the massive social and political projects that our nation has undertaken since independence - of ameliorating the condition of each of our citizens so that they have access to opportunity to lead better lives and utilize their potential," Hamid Ansari said. Congratulating IIT students who had been conferred awards and medals at the occasion, the Vice President wished them success for their professional and personal endeavors.

"I am confident that the graduating students would live up to the oath that you have undertaken to be honest in the discharge of your duties, to uphold the dignity of the individual and integrity of the profession, and to utilize your knowledge for the service of the country and of mankind," he said.

Nano-Clay Used to Form Lightweight Composite Ballistic Armor with Superior Strength and Blast Resistance

MKP Structural Design Associates, Inc. (Ann Arbor, MI) garnered U.S. Patent 7,694,621 for lightweight composite ballistic armor made with nano-clay. The armor is intended for use in military and tactical vehicles and armored civilian

vehicles as well as buildings protecting people, machinery, supplies and fuel, according to inventor Zheng-Dong Ma. Features of the composite armor system include ultra-light-weight, flexibility, superior ballistic and blast resistance, superior strength and durability for structural integrity, capability to resist heat and flame, ease of manufacture, maintenance and repair. Damage to the armor is restricted to a limited range due to the fact that long cracks in the polymer matrix can be stopped from further propagation due to the presence of nano-clay particles in the matrix.

The terrorist attacks of Sep. 11, 2001 in New York City and Washington, D.C., and the current war in Iraq, have heightened the need for ballistic armor. Military vehicles, in particular, are vulnerable to higher-potency weapons such as rocket-launched grenades and other projectiles. Military personnel want lightweight, fast and maneuverable vehicles, but they also want vehicle occupants to be fully protected.

Ballistic steel armor plates, while relatively inexpensive, add thousands of pounds to a vehicle, many of which were not designed to carry such loads. This has resulted in numerous engine and transmission failures as well as problems with vehicle suspensions and brakes. The additional weight reduces fuel efficiency and makes it impossible to carry additional personnel in the vehicle in case of emergency. For these reasons, designers are beginning to adopt more lightweight composite armor across the board for military and tactical vehicles.

MKP’s front plate is preferably composed of ceramic pellets arranged in a periodic pattern designed for improving the ballistic resistance, especially in the presence of multiple hits. The pellets may be contained in a single-layered or three-dimensional metal or fiber network filled by thermoset or thermoplastic polymer material. The polymer may be further improved by use of nano clay to improve resistance to crack propagation. The ceramic pellet will have an optimally designed shape, which enhances the transferring of impact load onto surrounding pellets. This feature

results in desired compress stress among the pellets, which reduces the crack propagation and improves the out-of-plane impact resistance performance.

The ceramic pellets in the tile are seated in a fabric network, and are molded with the selected thermoset or thermoplastic polymer material. The polymer material functions as impact absorber and position keeper of the pellets and may have nano-clay particles molded in to further improve resistance to crack propagation. The fabric network in the ceramic layer has two major functions: one is to keep the pellets in a desired arrangement and the other is to reinforce the ceramic layer during the ballistic impact.

The back plate features ultra-light weight and outstanding out-of-plane stiffness/strength. It is designed to have improved bending stiffness and strength for optimizing the armor performance. The back plate, combined with one or more face plates, is referred to herein as an Armor Tile.

MKP Structural Design Associates Armor Tile The fabric net is designed to hold the armor tiles (ceramic layer and back plate) in place and form an integrated armor kit that can fill into various vehicle contours. The optimally designed supporting structure also provides the advanced features of low cost and ease of installation, replacement, and repair. Since its establishment in 2001, MKP Structural Design Associates has been dedicated to the development of new technologies for simulating, designing, and manufacturing innovative structural and material concepts. These can be used for a wide range of applications, including next-generation air and ground vehicle systems.

XtraEdge for IIT-JEE 6 SEPTEMBER 2010 At least two alternative supporting

structures are possible. The first is a net structure to which the armor kits are attached. The benefit of this design is it is lightweight and easy to install on different kinds of surfaces. The second one is made of fabric cloths, such as a para-aramid fiber, which has arrays of pockets that the armor tiles can be inserted in. This concept is similar to the body armor except a large number of armor inserts will be used. In terms of materials, different kinds of materials are combined to defeat the projectile effectively. Ceramic pellets or cylinders function to damage and to rotate the projectiles. Optimized cable network provides reinforcement when tension and bending loads exist on the armor plate. Matrix material functions to absorb shock waves and to keep the structural integrityVarious lightweight armor designs are now becoming commercially available. Cellular Materials International, Inc. of Charlottesville, Va. offers a product called Microtrussm, a periodic cellular material designed to absorb a larger amount of energy than solid material of equal mass. When a blast hits the face of the sandwich panel, the face plate will stretch and wrinkle followed by the propagation of the impulse force into the core. The core will then buckle and collapse, absorbing the maximum kinetic energy of the blast. The back face plate takes the remaining blast pressure towards the end of the blast event where the intensity of the impulse force is considerably reduced. Thus, the periodic structure maximizes the absorption of the impulse energy created by the blast and distributes or diffuses the intensity of the force, leading to protection of the assets behind the sandwich structures.

Engineering college of IIT Bombay is one of the most credible professional college in India

Germany and Australia have joined a growing number of developed nations keen to tie up with the new Indian Institutes of Technology that have opened in the last two years. Both have formally told the Indian government their universities would like to collaborate with the new IITs. The reason is clear: the developed

world is looking at India both for trained technical manpower and as a potential research hub.

Germany wants to collaborate with IIT-Mandi that started in 2009 while Australia is interested in IIT-Patna, started in 2008, top government sources have told HT. They join Japan, France and the United Kingdom, which are already in talks with the government to collaborate with the new IITs in Hyderabad, Jodhpur and Ropar, respectively.

The proposed collaboration involves the foreign partner providing technical knowhow and assistance to the IITs, and engaging in exchange programmes and joint research, sources said.

The talks so far with Japan, France and the UK suggest that the foreign partners are keen to tap Indian talent - both in terms of trained engineers and research - through their collaboration with the IITs, the sources said.

Japan, for instance, wants IIT-Hyderabad to incorporate the Japanese language and the country's management practices in its course structure - a move that would ease the integration of the institute's graduates into Japanese firms. Top Japanese companies are also expected to help train students at this IIT.

The early IITs too were hand-held and assisted - financially and technically - by foreign countries when they were started half a century ago, though that was largely to help a newly independent, struggling nation find its educational feet.

IIT-Bombay was helped by the erstwhile Soviet Union and UNESCO, IIT-Kanpur and Madras by the US and Germany and IIT-Delhi by Britain.

IIT-Kanpur plans presence in Bangalore, US

Bangalore: The Indian Institute of Technology-Kanpur (IIT-K) plans a presence in India's IT hub here, the US and Malaysia, institute director S.G. Dhande said here on Saturday.

IIT-K intends to set up research centres in the US and Malaysia as part of a plan to compete with top universities, he said. The presence in Bangalore, the US and Malaysia will be preceded by establishing a centre at Noida, near the national capital, the institute's first footprint outside Kanpur in Uttar Pradesh, Dhande told an innovation convention here.

The convention was organized by IIT-K alumni as part of the golden jubilee celebrations of the institute. Dhande did not elaborate on the kind of presence the institute planned in Bangalore, which has transformed into a major hub of new economy sectors such information technology and biotechnology. In Malaysia, the IIT-K plans a research centre in Penang while the facility in the US may come up either in Silicon Valley, Boston or Washington, Dhande said. Opening up of centres abroad was essential to compete internationally, he said.

"If we want to be internationally recognized, we must have presence in different parts of the world," Dhande said. Karnataka has been pleading with the central government that an IIT be set up in the state. The central government has promised to do so but no time frame has been indicated. IANS

Global suitors woo new IITs

Space Quick Facts

1. Saturn’s rings are made up of particles of ice, dust and rock. Some particles are as small as grains of sand while others are much larger than skyscrapers. 2. Jupiter is larger than 1,000 Earths. 3. The Great Red Spot on Jupiter is a

hurricane-like storm system that was first detected in the early 1600’s.

4. Comet Hale-Bopp is putting out approximately 250 tons of gas and dust per second. This is about 50 times more than most comets

Mr. Vineet Agrawal received his Bachelor’s Degree in Mechanical Engineering with Distinction from IIT Delhi in 1983. He obtained his MBA degree from Bajaj Institute of Management Studies, Mumbai in 1985.

Mr. Vineet Agrawal is presently the President of the Wipro Consumer Care and Lighting business since 2002.

Mr. Vineet Agrawal joined Wipro as a Management Trainee in 1985. He was rotated through various positions before being appointed as the Chief Executive of Wipro Peripherals Business in 1999. Subsequently, he was appointed as the Corporate Executive Vice President handling the Six Sigma Quality function, Innovation foray and the CSR initiative for the Wipro Corporation in 2000. He has been a member of the Chief Executive Council member of Wipro since 2000.

Mr. Vineet Agrawal has led the business of Wipro Consumer Care and Lighting from 300 Cr to 2000 Cr in 6 years. In 2007, in a bold move, his business acquired the 700 Cr Unza – a South Asian Co. – this being the largest overseas personal care acquisition by an Indian Co. Mr. Agrawal’s responsibility includes running the FMCG Indian business of Wipro, both in India and abroad. He also handles the Office Interior business which includes the Office Modular furniture business and the Lighting business.

Mr. Vineet Agrawal pioneered Wipro’s Social Responsibility with Wipro Applying Thought in Schools program. This was an initiative that he conceptualized and initiated in 2001 with a belief that this was something Wipro needs to get into. This initiative has grown to impact 800 schools and 700,000 children. The program encourages creativity in children by improving the teaching method in schools.

Mr. Vineet Agrawal’s strength lies in Strategy development and executing it on ground. He was chosen to lead the complex Wipro Repositioning exercise during 1996-99.

In honouring Mr. Vineet Agrawal, IIT Delhi recognizes the outstanding contributions made by him as a Corporate Leader. Through his achievements, Mr. Vineet Agrawal has brought glory to the name of the Institute..

Dr. Alok Aggarwal received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained his Ph.D. in Electrical Engineering and Computer Science from Johns Hopkins University in 1984.

Dr. Alok Aggarwal joined IBM Research Division in Yorktown Heights New York in 1984. During the fall of 1987 and 1989, he was on sabbatical from IBM and taught two courses (in two terms) at the Massachusetts Institute of Technology (MIT) and also supervised two Ph.D. students. During 1991 and 1996, along with other colleagues from IBM, he created and sold a "Supply Chain Management Solution" for paper mills, steel mills and other related industries. In July 1997, Dr. Aggarwal "Founded" the IBM India Research Laboratory that he set-up inside the Indian Institute of Technology Delhi. Dr. Aggarwal started this Laboratory from "ground zero" and by July 2000, he had built it into a 60-member team (with 30 PhDs and 30 Masters in Electrical Engineering, Computer Science, and in Business Administration). In August 2000, Dr. Aggarwal became the Director of Emerging Business Opportunities for IBM Research Division worldwide. Dr. Alok Aggarwal has published 86 Research papers and he has also been granted 8 patents from the US Patents and Trademark Office. Along with his colleagues at Evalueserve, in 2003, he has pioneered the concept of “Knowledge Process Outsourcing (KPO)” and wrote the first article in this regard. Dr. Aggarwal has served as a Chairperson of the IEEE Computer Society's Technical Committee on Mathematical Foundations of Computing and has been on the editorial boards of SIAM Journal of Computing, Algorithmica, and Journal of Symbolic Computation. During 1998-2000, Dr. Aggarwal was a member of Executive Committee on Information Technology of the Confederation of the Indian Industry (CII) and also of the Telecom Committee of Federation of Indian Chamber of Commerce and Industry (FICCI). He is currently a Chartered Member of The Indus Entrepreneur (TiE) organization. In honouring Dr. Alok Aggarwal, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Researcher. Through his achievements, Dr. Alok Aggarwal has brought glory to the name of the Institute.

Success Story

Success Story

This articles contains stories of persons who have succeed after graduation from different IIT's

Dr. Alok Aggarwal

B.Tech, IIT Delhi in 1980

PHYSICS

1. A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s and maximum transverse acceleration is 90 m/s2_{. If the}

wave velocity is 20 m/s then find the waveform. [IIT-2005] Sol. The wave form of a transverse harmonic disturbance

y = a sin (ωt ± kx ± φ)

Given vmax = aω = 3 m/s ...(i)

Amax = aω2 = 90 m/s2 ....(ii)

Velocity of wave v = 20 m/s ...(iii) Dividing (ii) by (i)

ω ω a a 2 ₌ 3 90 _{⇒ ω = 30 rad/s ...(iv)} Substituting the value of ω in (i) we get

a = 30 3 = 0.1 m ...(v) Now k = λ π 2 = v / v 2π = v v 2π = v ω = 20 30 = 2 3 ...(vi) From (iv), (v) and (vi) the wave form is y = 0.1 sin_ ± x±φ_

2 3 t 30

2. A 5m long cylindrical steel wire with radius 2 × 10–3

m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire : Young's modulus = 2.1 × 1011 _{Pa; Density}

= 7860 kg/m3_{; Specific heat = 420 J/kg-K).}

[IIT-2001] Sol. When the mass of 100 kg is attached, the string is

under tension and hence in the deformed state. Therefore it has potential energy (U) which is given by the formula.

U = 2 1

× stress × stain × volume

= 2 1 _× Y ) Stress ( 2 _{× πr}2 l = 2 1 Y ) r / Mg ( _π 2 2 × πr2 l = 2 1 Y r g M 2 2 2 π l _...(i)

This energy is released in the form of heat, thereby raising the temperature of the wire

Q = mc ∆T ...(iii)

From (i) and (iii) Since U = Q Therefore ∴ mc∆T = Y r g M 2 1 2 2 2 π l ∴ ∆T = Ycm r g M 2 1 2 2 2 π l Here

m = mass of string = density × volume of string = ρ × πr2 l ∴ ∆T = ρ πr ) Yc ( g M 2 1 2 2 2 2 = 2 1 × 7860 420 10 1 . 2 ) 10 2 14 . 3 ( ) 10 100 ( 11 2 3 2 × × × × × × × − = 0.00457ºC

3. The x – y plane is the boundary between two transparent media. Medium –1 with z ≥ 0 has a refractive index 2 and medium –2 with z ≤ 0 has a refractive index 3. A ray of light in medium –1 given by the vector A = _{6 3}^_{i + 8 3}^_{j – 10}^_kis incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium –2. Sol. Y ^ ^ j 3 8 i 3 6 + ^ j 3 8 ^ i 3 6 M' M X O Z M' X O –10K^ ^ ^ ^_{8 3j–10k} i 3 6 A= + → ^ ^ j 3 8 i 3 6 + Fig(1) Fig(2) Figure 1 shows vector 6 3i + 8 3^j

Figure 2 shows vector A→ = 6 3^i + 8 3^j – 10^k The perpendicular to line MOM' is Z-Axis which has a unit vector of ^k. Angle between vector IO→ and

→

ZO can be found by dot product →

IO. ZO→ = (IO) (ZO) cos i 2 2 2 2 ^ ^ ^ ^ ) 1 (– ) 10 (– ) 3 8 ( ) 3 6 ( ) k ).(– k 10 – j 3 8 i 3 6 ( + + + = cos i

KNOW IIT-JEE

⇒ i = 60

Unit vector in the direction MOM' from figure (1) is

2 / 1 2 2 ^ ^ ^ ] ) 3 8 ( ) 3 6 [( j 3 8 i 3 6 n + + = ^ ^ ^ _j 5 4 i 8 3 n= +

To find the angle of refraction, we use snell's law

2 3 ₌ r sin i sin ₌ r sin º 60 sin _{⇒ r = 45º}

From the triangle ORS ^ r = (sin r) n^ – ( cos r) ^k = (sin 45º) _      + ^ ^ j 5 4 i 5 3 – (cos 45º) ^k = [3i 4j–5k] 2 5 1 ^ ^ ^ +

4. Along horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30A, as shown in figure. Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations.

A B D C

Sol. When AB is steady,

Weight per unit length = Force per unit length weight per unit length =

π µ 4 0 r I I 2 _{1 2} ...(i)

when the rod is depressed by a distance x, then the force acting on the upper wire increases and behave as a restoring force A Fmag I1= 20A _x B B' mg r = 0.01 m A' C _{I2 = 30A} D Restoring force/length = π µ 4 0 x – r I I 2_{1 2} – π µ 4 0 r I I 2 _{1 2} = π µ 4 0 _2I 1I2   r 1 – x – r 1 ⇒ Restoring force/length = π µ 4 0 _2I 1I2       r ) x – r ( ) x – r ( – r = π µ 4 0 ) x – r ( r x I I 2_{1 2}

when x is small i.e., x <<r then r – x ≈ r

Restoring force/length F = π µ 4 0 22 1 r I I 2 x Since F ∝ x ∴ The motion is simple harmonic ∴ π µ 4 0 2 2 1 r I I

2 _{= (mass per unit length) ω}2 _...(ii)

From (i) (Mass per unit length) × g = π µ 4 0 r I I 2_{1 2}

Mass per unit length = π µ 4 0 rg I I 2_{1 2} ...(iii) From (ii) and (iii)

π µ 4 0 2 2 1 r I I 2 = π µ 4 0 rg I I 2_{1 2 × ω}2 ⇒ ω = r g _⇒ T 2π = r g t ⇒ T = 2π g r _{= 2p} 8 . 9 01 . 0 _{= 0.2 sec}

5. In the figure both cells A and B are of equal emf. Find R for which potential difference across battery A will be zero, long time after the switch is closed. Internal resistance of batteries A and B are r1 and r2

respectively (r1 > r2).s C R L R R R R R S B A r1 r2

Sol. After a long time capacitor will be fully charged, hence no current will flow through capacitor and all the current will flow from inductor. Since current is D.C., resistance of L is zero. ∴ Reg =       + R 2 R × 2 1 + r1 + r2 = 4 R 3 + r1 + r2 I = eq R ε + ε _{⇒ I =} eg R 2ε = 2 1 r r 4 / R 3 3 + + ε

Potential drop across A is ε – I r1 = 0 ⇒ ε = 2 1 r r 4 / R 3 2 + + ε r1 ⇒ r1 = r2 + 3R/4 or R = 3 4 (r1 – r2)

CHEMISTRY

6. An organic compound A, C8H4O3, in dry benzene in

the presence of anhydrous AlCl3 gives compound B.

The compound B on treatment with PCl5 followed by

reaction with H2/Pd(BaSO4) gives compound C,

which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D.

Explain the formation of D from C. [IIT-2000] Sol. The given reactions are as follows.

O O + O AlCl3 O O OH PCl5 H2/Pd (BaSO4) C6H5 H C C O O H2NNH2 C6H5 N N

The formation of D from C may be explained as follows. C6H5 C6H5 O O NH2 NH2 O– NH2 NH2 O– + + C6H5 O – N – H N – H OH C6H5 N N

7. An alkyl halide X, of formula C6H13Cl on treatment

with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give

2, 3-dimethyl butane. Predict the structures of X, Y

and Z. [IIT-1996]

Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes.

X13 6H Cl C HCl – ; butoxide t K ∆ − −_  →  12 6H C Z Y+

Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of

H2 to give same alkane 2, 3-dimethyl butane, hence

they should have the skeleton of this alkane. Y and Z (C6H12) Ni H2 →  _CH 3 – CH – CH – CH3 CH3 CH3 2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3 CH3 CH3 2,3-dimethyl butene-2 (Y) and CH3 – CH – C = CH2 CH3 CH3 2,3-dimethyl _butene-1 (Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3 CH3CH3 (Y) H2 Ni CH3 – CH – CH – CH3 CH3 CH3 CH3 – CH – C = CH2 CH3 CH3 (Z) H2 Ni CH3 – CH – CH – CH3 CH3 CH3

Both, Y and Z can be obtained from following alkyl halide : CH3 – C – CH – CH3 CH3CH3 2-chloro-2,3-dimethyl butane (X) K-t-butoxide ∆; –HCl CH2 = C — CH – CH3 CH3 CH3 Cl + CH3 – C = C – CH3 CH3CH3 (Z) 20% (Y) 80% Hence, X, CH3 – C – CH – CH3 CH3CH3 Cl Y, CH3 – C = C – CH3 CH3CH3 Z, CH3 – CH – C = CH2 CH3 CH3

8. The molar volume of liquid benzene

(density = 0.877 g ml–1_{) increases by a factor of 2750}

as it vaporizes at 20ºC and that of liquid toluene (density = 0.867 g ml–1_{) increases by a factor of 7720}

at 20ºC. A solution of benzene and toluene at 20ºC has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in vapour above the solution.

[IIT-1996] Sol. Given that,

Density of benzene = 0.877 g ml–1

Molecular mass of benzene (C6H6)

= 6 × 12 + 6 × 1 = 78 ∴ Molar volume of benzene in liquid form =

877 . 0 78 ml = 877 . 0 78 × 1000 1 L = 244.58 L

And molar volume of benzene in vapour phse

= 877 . 0 78 _× 1000 2750_{L = 244.58 L} Density of toluene = 0.867 g ml–1

Molecular mass of toluene (C6H5CH3)

= 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92 ∴ Molar volume of toluene in liquid form

= 867 . 0 92 ml = 867 . 0 92 × 1000 1 L

And molar volume of toluene in vapour phase

= 867 . 0 92 × 1000 7720 L = 819.19 L Using the ideal gas equation,

PV = nRT At T = 20ºC = 293 K For benzene, P = PB0 = nRT_V = 58 . 244 293 082 . 0 1× × = 0.098 atm = 74.48 torr (Q 1 atm = 760 torr) Similarly, for toluene,

P = 0 T P = V nRT = 19 . 819 293 082 . 0 1× × = 0.029 atm = 22.04 torr (Q 1 atm = 760 torr) According to Raoult's law,

PB = PB0xB = 74.48 xB

PT = PT0xT = 22.04 (1 – xB)

And PM = PB0xB + PT0xT

or 46.0 = 74.48 xB + 22.04 (1 – xB)

Solving, xB = 0.457

According to Dalton's law,

PB = PM x'B (in vapour phase) or mole fraction of benzene in vapour form,

' B x = M B P P = 0 . 46 457 . 0 48 . 74 × = 0.74

9. The values of Λ∞ _{for HCl, NaCl and NaAc (sodium}

acetate) are 420, 126 and 91 Ω–1 _cm2 _mol–1_,

respectively. The resistance of a conductivity cell is 520 Ω when filled with 0.1 M acetic acid and drops to 122 Ω when enough NaCl is added to make the solution 0.1 M in NaCl as well. Calculate the cell constant and hydrogen-ion concentration of the solution. Given :

∞

Λ_m(HCl) = 420 Ω–1 _cm2 _mol–1_,

∞

Λ_m(NaCl) = 126 Ω–1 _cm2 _mol–1_,

and Λm(NaAc) = 91 Ω–1 cm2 mol–1

Sol. Resistance of 0.1 M HAc = 520 Ω

Resistance of 0.1 M HAc + 0.1 M NaCl = 122 Ω Conductance due to 0.1 M NaCl,

G = Ω 122 1 _– Ω 520 1 _{= 0.00627 Ω}–1

Conductivity of 0.1 M NaCl solution k = Λmc = (126 Ω–1 cm2 mol–1)(0.1 mol dm–3) = 12.6 Ω–1_cm2 _dm–3 _{= 12.6 Ω}–1 _cm2_{(10 cm)}–3 = 0.0126 Ω–1 _cm–1 Cell constant, K = G k = ) 00627 . 0 ( ) cm 0126 . 0 ( 1 1 1 − − − Ω Ω = 2.01 cm–1

Conductivity of 0.1 M HAc solution

k = R K ₌ Ω − 520 cm 01 . 2 1

Molar conductivity of 0.1 M HAc solution Λm(HAc) = c k ₌ ) dm mol 1 . 0 ( cm ) 520 / 01 . 2 ( 3 1 1 − − − Ω = 0.038 65 Ω–1 _cm–1 _dm3 _mol–1 = 38.65 Ω–1 _cm2 _mol–1

According to Kohlrausch law, Λ∞_{(HAc) is given by}

∞

Λ_m(HAc) = Λ∞_m(HCl) + Λ∞_m(NaAc) – Λ∞_m(NaCl) = (420 + 91 – 126) Ω–1 _cm2 _mol–1

= 385 Ω–1 _cm2 _mol–1

Therefore, the degree of dissociation of acetic acid is given as α = _∞ Λ Λ m m ₌ ) mol cm 385 ( ) mol cm 65 . 38 ( 1 2 1 1 2 1 − − − − Ω Ω _{≈ 0.1}

and the hydrogen-ion concentration of 0.1 M HAc solution is

[H+_{] = cα = (0.1 M)(0.1) = 0.01 M}

Thus, its pH is pH = – log{[H+_{]/M} = – log(0.01) = 2}

10. An organic compound A, C8H4O3, in dry benzene in

the presence of anhydrous AlCl3 gives compound B.

The compound B on treatment with PCl5 followed by

reaction with H2/Pd(BaSO4) gives compound C,

which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D.

Explain the formation of D from C. [IIT-2000] Sol. The given reactions are as follows.

O O + O AlCl3 O O OH PCl5 H2/Pd (BaSO4) C6H5 H C C O O H2NNH2 C6H5 N N

The formation of D from C may be explained as follows. C6H5 C6H5 O O NH2 NH2 O– NH2 NH2 O– + + C6H5 O – N – H N – H OH C6H5 N N

MATHEMATICS

11. With usual notation, if in a triangle ABC 11 c b+ = 12 a c+ = 13 b a+

, then prove that

7 A cos ₌ 19 B cos ₌ 25 C cos _[IIT-1984] Sol. Let 11 c b+ = 12 a c+ = 13 b a+ _{= λ} ⇒ (b + c) = 11λ, c + a = 12λ, a + b = 13λ ⇒ 2(a + b + c) = 36λ or a + b + c = 18λ Now, b + c = 11λ and a + b + c = 18λ ⇒ a = 7λ c + a = 12, and a + b + c = 18λ ⇒ b = 6λ a + b = 13λ and a + b + c = 18λ ⇒ c = 5λ ∴ cos A = bc 2 a c b2+ 2− 2 = 2 2₂ 2 ) 30 ( 2 49 25 36 λ λ − λ + λ = 5 1 cos B = ac 2 b c a2+ 2− 2 = ₂ 2 2 2 70 36 49 25 λ λ − λ + λ = 35 19 cos C = ab 2 c b a2+ 2− 2 = ₂ 2 2 2 84 25 36 49 λ λ − λ + λ = 7 5

∴ cos A : cos B : cos C = 5 1 : 35 19 : 7 5 = 7 : 19 : 25 12. Let ABC be a triangle with AB = AC. If D is

mid-point of BC, the foot of the perpendicular drawn from D to AC and F and mid-point of DE. Prove that AF is perpendicular to BE. [IIT-1989] Sol. Let BC be taken as x-axis with ortigin at D, the

mid-point of BC, and DA will be y-axis AB = AC

Let BC = 2a, then the coordinates of B and C are (–a, 0) and (a , 0) let A(0, h)

y A E F C D B

Then, equation of AC is,

a x ₊

h

y _{= 1 ...(1)}

and equation of DE ⊥ AC and passing through origin is,

h x – a y _{= 0 ⇒ x =} a hy ...(2)

Solving (1) and (2) we get the coordinates of point E as follows : 2 a hy + h y = 1 ⇒ y = ₂ 2 ₂ h a h a + ∴ E = _       + + 2 2 2 2 2 2 h a h a , h a ah

Since F is mid-point of DE,

∴ F _       + + 2(a h ) h a , ) h a ( 2 ah 2 2 2 2 2 2 ∴ slope of AF = ) h a ( 2 ah 0 ) h a ( 2 h a h 2 2 2 2 2 2 + − + − = 2 2₂ 2 ah h a ) h a ( h 2 − − + ⇒ m1 = ah ) h 2 a ( 2+ 2 − ...(3) and slope of BE = a h a ah 0 h a h a 2 2 2 2 2 + + − + ₌ 2 3 2 2 ah a a h a + + ⇒ m2 = _{2 2} h 2 a ah + ...(4) from (3) and (4), m1m2 = – 1 ⇒ AF ⊥ BE.

13. Let f(x) = Ax2 _{+ Bx + C where, A, B, C are real}

numbers. Prove that if f(x) is an integer whenever x is an integer, then the numbers 2A, A + B and C are all integers. Conversely, prove that if the numbers 2A, A + B and C are all integers, then f(x) is an integer whenever x is an integer. [IIT-1998] Sol. Suppose f(x) = Ax2 _{+ Bx + C is an integer whenever}

x is an integer

∴ f(0), f(1), f(–1) are integers.

⇒ C, A + B + C, A – B + C are integers ⇒ C, A + B, A – B are integers.

Conversely suppose 2A, A + B and C are integers. Let n be any integer. We have

f(n) = An2 _{+ Bn + C = 2A}     2 ) 1 – n ( n + (A + B)n + C Since n is an integer, n(n – 1)/2 is an integers. Also 2A, A + B and C are integers.

We get f(n) is an integer for all integer n

14. A window of perimeter (including the base of the arch) is in the form of a rectangle surrounded by a semi-circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light?[IIT-1991] Sol. Let '2b' be the diameter of the circular portion and 'a'

be the lengths of the other sides of the rectangle. Total perimeter = 2a + 4b + πb = K (say) ...(1) Now, let the light transmission rate (per square metre) of the coloured glass be L and Q be the total amount of transmitted light.

Coloured glass Clear glass a a Then, Q = 2ab(3L) + 2 1_πb2_(L) Q = 2 L_{πb2 _{+ 12ab}} Q = 2 L_{πb2 _{+ 6b (K – 4b – πb)}} Q = 2 L {6Kb – 24b2 _{– 5πb}2_} db dQ = 2 L_{{6K – 48b – 10πb} = 0} ⇒ b = π +10 48 K 6 ...(2) and 2₂ db Q d ₌ 2 L_{{–48 + 10π}La}

Thus, Q is maximum and from (1) and (2), (48 + 10π) b = 6K and K = 2a + 4b + πb ⇒ (48 + 10π) b = 6{2a + 4b + πb}

Thus, the ratio = a b 2 = π + 6 6 15. If f : [–1, 1] → R and f´(0) =       ∞ → _n 1 nf lim n and f(0) = 0. Find the value of :

π ∞ → 2 lim n (n + 1)cos –1 _      n 1 – n given that 0 <       − ∞ → _n 1 cos lim 1 n < ₂ π [IIT-2004] Sol. Here π ∞ → 2 lim n (n + 1)cos –1 _      n 1 – n = ∞ → nlimn _     −             + π − ₁ n 1 cos n 1 1 2 1 = ∞ → nlimnf _     n 1 Where f       n 1 _{= }      + π n 1 1 2 _cos–1 _      n 1 _{– 1 = f´(0)}             = ∞ → n 1 nf lim ) 0 ´( f given n ∴ π ∞ → 2 lim n (n + 1)cos –1 n 1 – n = f´(0) ...(1) where f(x) = π 2 (1 + x) cos–1_{x – 1, f(0) = 0} ⇒ f´(x) =         + − − + π − _x cos x 1 1 ) x 1 ( 2 1 2 ⇒ f´(0) =       π + − π 1 2 2 _{= 1 –} π 2 _...(2)

∴ from equation (1) and (2)

π ∞ → 2 lim n (n + 1) cos –1 _      n 1 – n = 1 – π 2

Ability

• We can accomplish almost anything win tin our ability if we but think we can.

• He is the best sailor who can steer within fewest points of the wind, and exact a motive power out of the greatest obstacles.

• Our work is the presentation of our capabilities. • The wind and the waves are always on the side of the

Q. 1 Which of the following graphs are hyperbolic (A) P – V diagram for an isothermal process (B) Current density vs area of cross section in a current carrying wire

(C) Velocity of incompressible fluid vs area of cross-section for steady flow of fluid through a pipe

(D) Wavelength corresponding to which emissive power is maximum vs temperature of blackbody

Q. 2 For two different gases x and y, having degrees of freedom f1 and f2 and molar heat capacities at

constant volume C and _V1 C respectively, the _V2 lnP versus lnV graph is plotted for adiabatic process as shown, then

y x ln V Ln P (A) f1 > f2 (B) f2 > f1 (C) CV₂ <CV₁ (D) CV1 >CV2 Q. 3 A particle of charge q and mass m moves

rectilinearly under the action of an electric field

. x

E=α−β Here α and β are positive constants and x is the distance from the point where the particle was initially at rest then –

(A) motion of particle is oscillatory (B) amplitude of the particle is α /β

(C) mean position of the particle is at

β α = x

(D) the maximum acceleration of the particle is

m qα

Q. 4 Speed of a body moving in a circular path changes with time as v = 2t, then –

(A) Magnitude of acceleration remains constant (B) Magnitude of acceleration increases (C) Angle between velocity and acceleration remains constant

(D) Angle between velocity and acceleration increases

Q. 5 Consider a resistor of uniform cross section area connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it then

(A) current will become four times

(B) the electric field in the wire will become half (C) the thermal power produced by the resistor will become one fourth

(D) the product of the current density and conductance will become half

Q. 6 A mosquito with 8 legs stands on water surface and each leg makes depression of radius ‘a’. If the surface tension and angle of contact are ‘T’ and zero respectively then the weight of mosquito is

(A) 8T.a (B) 16πTa (C) 8 Ta (D) π 16 Ta

Q. 7 Three wires are carrying same constant current I in different direction. Four loops enclosing the wires in different manner are shown. The direction of d→_l is shown in the figure

Loop-1 Loop-2 Loop-3 Loop-4

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

Solutions will be published in next issue

Column – I Column – II (A) Along closed loop 1 (P)

∫

(B) Along closed loop 2 (Q)

∫

(C) Along closed loop 3 (R)

∫

(D) Along closed loop 4 (S) net work done by the

magnetic force to move a unit charge along the loop is zero Q. 8 An ideal monoatomic gas undergoes different

types of processes which are described in column – I. Match the corresponding effect in column – II. The letters have usual meaning.

Column – I Column – II (A) P = 2V2 _{(P) If volume increases}

then temperature will also increase

(B) PV2 _{= constant (Q) If volume increases}

then temperature will decrease

(C) C = Cv + 2R (R) For expansion, heat

will have to be supplied to the gas (D) C = Cv – 2R (S) If temperature

increases then work done by gas is positive

Across:

1. The particles are far apart and moving fast. (3) 6. When a liquid changes to a gas fast for a cup of tea!(7) 7. Ice thawing is an example. (7)

9. The partices in a liquid are all ? up! (7) 11. The 'bits' of solids, liquids and gases. (9)

13. One way in which solids are different than gases or liquids. Its an easy question really, its not ? (4)

14. The particles of a solid are under going this without causing a sound! and more so on heating! (9)

15. The 'pattern' of particles in a solid is very? (7)

16. What a solid does on heating (without melting) as the atoms get more excited! (7)

Down:

2. The particles are closest together in this state. (5) 3. Very difficult to do with a solid, not much spare space! (8) 4. The particles are close together but they can still move

around quite freely in this state. (6)

5. A word that means particles spreading in liquids and gases because of their random movement. (9)

8. This is happening to a gas when it is cooled to form a liquid. (10)

10. When a liquid changes to a solid it is ? (8)

11. This is caused by gas particles hitting the side of a container millions of times a second! (8)

12. You must obtain this to check out a theory in a scientific court! (8) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

1. [B,D] C1 C2 2Ω 1Ω 3Ω 120 V 1Ω 2Ω 3Ω a d G H N M e b volt 20 1 6 120 V Va − d = × = volt 40 2 6 120 V Vb− e = × = volt 20 V Vd − e = ∴ Charge on C1=2×20=40µC volt 60 3 6 120 V V_G − _H = × = volt 60 3 6 120 V V_M− _N = × = 0 V V_G− _M = ∴ Charge on C2 is zero 2. [B] q(E₁−E₂)=0.08N Columb 10 25 8 10 10 5 . 2 10 08 . 0 q 6 2 5 − × = × × × = ₌0.32_×10−6C =0.32µC 3. [C] 0 2∈ σ – σ E 2 0 E 2 E = ∈ σ + ………..(1) 1 0 E E= − ∈ σ _{………..(2)} Adding (1) & (2) 1 2 0 E E + = ∈ σ 0 1 2 E )A E ( q= + ∈ 5 9 6 1 2 0 8.7 10 10 9 4 10 32 . 0 ) E E ( q A × × × π × × = + ∈ = − 4. [B] ₂ 0 E 2 E = ∈ σ + 2E=E₁−E₂ 1 0 E 2 E = ∈ σ + − 2 E E E₌ 1− 2 20 10 5 . 2 × 5 = = 12.5 ×104_V/m 5. [B] K b b d A C 0 K + − ε = We set b = 0 C d A C_K=ε0 = if CK = 2C Then, d b 2 b 2 K d A 2 K b b d A ₀ 0 − = ⇒ ε = + − ε d b & 0 K> ≤ ∴ 0 d b 2 & d b 2 b 2 K − > − = ∴ d b 2 d _{< ≤} ∴ 2 d b> ∴ 6. [A,B,C,D]

For option (A)

1 2 3 10V 20V i i = 4 10 – 20 = 4 10 = 2.5 A For option (B) 1 2 3 20V i = 3 20_A

Solution

Physics Challenging Problems

Set # 4

For Option (C) 1 2 3 10V 20V i A 3 5 6 10 3 2 1 10 20 i = = + + − = For Option (D) 1 2 3 20V Amp 20 1 20 i= = 7. [B,D] y R x θ θ C → v m 5 2 1 10 1 qB mv R = × × = = Coordinates of center ⇒ 3 5 3 5 sin R x=+ θ=+ × =+ 4 5 4 5 cos R y=− θ=− × =−

Position of particle not given so R = 5m Center may be anywhere

π = × × π × = π = 2 1 1 2 qB m 2 T 8. [B,C] P1 P2 9Ω Q1 Q2 2Ω 2Ω 5 04 . 0 1 v b = × × = ε _l volt 2 . 0 = ε A 02 . 0 i 10 2 . 0 R i eq = ⇒ = ε = mA 20 A 10 2 i= × −2 =

1. Barium compounds are the source for the different greens in fireworks.

2. There are 60,000 miles (97,000 km) in blood vessels in every human.

3. The average person produces about 400 to 500 ml of cerebrospinal fluid every day.

4. Ernest Rutherford discovered that the atom had a nucleus in 1911.

5. Impacts by comets or asteroids can also generate giant tsunamis.

6. Basic surgery would cure 80% of the over 45 million blind people in the world. Sixty percent of whom live in sub-Saharan Africa, China and India.

7. Studies have confirmed that ginkgo increases blood flow to the retina, and can slow retinal deterioration resulting in an increase of visual acuity. In clinical tests ginkgo has improved hearing loss in the elderly. It also improves circulation in the extremities relieving cold hands and feet, swelling in the limbs and chronic arterial blockage.

8. Venus may well once have had water like Earth does, but because of the scorching surface temperature of 482 degrees C (900 degrees F). Any sign of it has long ago evaporated.

9. About 95 percent of every edible fat or oil consists of fatty acids. Fatty acids all are based on carbon chains - carbon atoms linked together one after another in a single molecule. Different fatty acids are defined as saturated, monounsaturated, or polyunsaturated depending on how effectively hydrogen atoms have linked onto those carbon chains.

10. On average women cry 5.3 times a month. Men only 1.4.

11. The Medal of Honor is the highest award for valour in action against an enemy force which can be bestowed upon an individual serving in the Armed Services of the United States.

1. A particle having charge q = 8.85 µC is placed on the axis of a circular ring of radius R = 30 cm. Distance of the particle from centre of the ring is a = 40 cm. Calculate electrical flux passing through the ring

.

Sol. Electric field strength at point in plane of ring depends upon its distance from centre of the ring. Magnitude of electric field field is same at all those points which are equidistant from the centre and co-planer with the ring.

r θ q x R E a θ

Therefore, consider a copaner and concentric ring of radius x and radial thickness de as shown in Figurer. Its area is dS = 2πx dx

Distance of every point of this ring from point charge is r = a2+x2

∴ Electric field strength at circumference of this ring is E = ₂ 0 r q 4 1 πε

Inclination θ of →E with the normal to surface of the ring considered is given by cosθ =

r a

Flux passing through this ring is dφ = →Eds→

dφ = d dS cos θ = _       + π         + πε₀ 2 2 _a2 _x2 a ) dx x 2 ( ) x a ( q 4 1 Hence, total flux passing through the given ring is φ =

∫

2. A parallel plate capacitor is filled by a di-electric whose di-electric constant varies with potential difference V according to law K = aV, where a = 2 volt–1_{. An air capacitor having same}

dimensions charged to a potential difference of

V0 = 28 volt is connected in parallel to the

uncharged capacitor filled with above mentioned di-electric. Calculate ratio of charge on capacitor filled by aforesaid di-electric to charge in air capacitor in steady state.

Sol. Let area of plates of each capacitor be A and let separation between them be d. Capacitance of air capacitor, C0 =

d A 0 ε

And capacitance of capacitor, filled with di-electric, C = d KA 0 ε = aVC0 = 2VC0

Initial charge in air capacitor, q0 = C0V0

When air capacitor is connected across the other capacitor, some charge flows from air-capacitor to the other capacitor so that potential differences across two capacitors in steady state becomes equal. Let the potential difference be V.

Charge of capacitor filled with di-electric will be equal to

q1 = CV = 2C0V2

and charge on air capacitor will be equal to q2 = C0V

But q1 + q2 = q0

or 2C0V2 + C0V = C0V0

From above equation V = – 4 or 3.5

Negative value is absurd. Therefore, V = 3.5 volts. ∴ 2 1 q q = V C V C 2 0 2 0 _{= 2V = 7 Ans.}

3. A stationary circular loop of radius a is located in a magnetic field which varies with time from t = 0 to t = T according to law B = B0 . t (T – t). If plane of

loop is normal to the direction of field and resistance of the loop is R, calculate

(i) amount of heat generated in the loop during this interval, and

(ii) magnitude of charge flown through the loop from instant t = 0 to the instant when current reverses its direction.

Neglect self inductance of the loop.

Sol. Since, magnetic field strength B varies with time, therefore, flux linked wit the loop varies with time. But whenever flux linked wit a circuit changes, and emf is induced. consequently, and emf is induced in

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

Students' Forum

the loop and a current starts to flow through the loop. Due to flow of current heat is generated. A an instant t, flux linked with the loop.

φ = B × area of the loop ∴ φ = πa2 _B 0 (tT – t2) Induced emf, e = – dt dφ _{= – πa}2 _B 0(T – 2t) Induced current i = R e = 2 0_(2t–T)2 R B a π ...(1) Thernal power generated at this instant, P = i2_R

or P = R B a4 3₀ 2 π (2t – T)2

During an elemental time interval dt, heat generated = P.dt = R B a4 30 2 π (2t –T)2 _.dt

∴ Total heat generated from t = 0 to t = T,

Q =

∫

∫

Substituting t by t0 in equation (1), ∴ R B a2 ₀ π (2t0 – T) = 0 or t0 = 2 T Substituting i by dt dq in equation (1), dq = (2t–T)dt R B a2 0 π

∴ Charge that flows from t = 0 to t = T/2,

q = R B a2 0 π

∫

or magnitude of charge that flows =

R 4 T B a2 ₀ 2 π Ans. (2) 4. Oxygen is used as working substance in an engine

working on the cycle shown in Figure

P V 1 4 3 2

Processes 1-2, 2-3, 3-4 and 4-1 are isothermal, isobaric, adiabatic and isochoric, respectively. If ratio of maximum to minimum volume of oxygen during the cycle is 5 and that of maximum to minimum absolute temperature is 2, assuming oxygen to be an ideal gas, calculate efficiency of the engine.

Given, (0.4)0.4 _{= 0.693 and log}

e 5 1.6094

Sol. Volume of gas is minimum at state 2 during the cycle. Let it be V0. Then maximum volume of gas

during the cycle will be equal to 5V0 which is at

states 4 and 1. Therefore, V4 = V1 = 5V0.

Temperature during the cycle is maximum at the end of isobaric process 2 → 3 i.e. state 3 and minimum at the end of isochoric cooling process 4 → 1 i.e. state 1. Let minimum absolute temperature be T0.

Then T1 = T0 and T3 = 2T0.

Since gas is Oxygen which is di-atomic, therefore, Cv = 2 5_{R, C} p = 2 7 _{R and γ =} 5 7

Since, process 1 → 2 is isothermal, therefore, temperature during the process remains constant. Hence temperature T2 is also equal to T0.

Considering n mole of the gas,

Work done by the gas during isothermal process 1 → 2, W12 = nRT1.log 1 2 V V = – nRT0 loge 5

But for isothermal process Q = W, therefore, Q12 = – nRT0 loge 5

Now considering isobaric process 2 → 3

2 3 V V = 2 3 T T = 2 or V3 = 2V0

Heat supplied to gas during the process, Q23 = nCP(T3 – T2) =

2 7 _nRT

0

Work done by gas during the process, W23 = nR(T3 – T2) = nRT0

Now considering adiabatic process 3 → 4, V3 = 2V0 , T3 = 2T0

V4 = 5V0 , T4 = ?

Using T.Vγ – 1 = constant

(2T0) (2V0)γ – 1 = T4(5V0)γ – 1

or T4 = 2 (0.4)0.4 T0

Work done by the gas during the process, W24 = 1 – V P – V P_{3 3 4 4} γ = –1 ) T – T ( nR _{3 4} γ = 5 nRT0 (1 – (0.4)0.4) T0

During isochoric process 4 → 1, no work is done by the gas and heat is rejected from the gas.

Hence, W41 = 0 and Q41 is negative

∴ Net work done by the gas during the cycle, W = W12 + W23 + W34 + W41 =

nRT0 {6 – loge 5 – 5 × (0.4)0.4}

Heat supplied to the gas during heating process, QS = Q23 = 2 7 _{( 6 – log} e 5 – 5 × 0.40.4) = 0.2642 or η = s Q W = 7 2 ( 6 – loge 5 – 5 × 0.40.4) = 0.2642 or η = 26.42 % Ans.

5. A thin convex lens of focal length f and having aperture diameter d is used to focus sun rays on a screen. If speed of light in vacuum is c, absolute temperature of sun TS and rays fall on the lens

normally, calculate pressure experienced by the surface on which rays are converged.

Sol. When light rays incident on the screen, momentum of rays reduces to zero. Due to change in momentum of incident rays, the surface experiences a force. Let radius of sun be Rs and its distance from earth

be. r.

According to Stefan's law,

rate of radiation from per unit area of surface of sun = σ WmTs4 –2

Surface area of sun = 4πR2_s

∴ Rate of radiation from sun, E = 2 s 4 s4 R

T π

σ

Intensity of sun rays at earth,

I = ₂ r 4 E π = 2 2 s 4 s r R T σ Wm–2

Area of convex lens, A = 2 2 d       π = 4 1 _πd2_m2

∴ Power incident on the lens,

P = AI = ₂ 2 2 s 4 s r 4 d . R T π σ W But these rays are converged on a screen by the lens, therefore, rate of incidence of momentum on the screen =

c P

Just after incidence, momentum of rays reduces reduces to zero, therefore, magnitude of rate of change of momentum of rays,

dt dp _{= }      _–0 c P ₌ c r 4 d R T 2 2 2 s 4 s π σ

But the magnitude of rate of change of momentum = force experience by the screen.

Hence, force on screen, F =

c 4 d R T 2 2 2 s 4 s ρ π σ

But for a lens,

u v

OI = Where I is the size of image.

∴ Radius of image circle formed on the screen = u v_{. O =} r f _{. R} s

∴ Area of image circle, a =

2 s R r f       π

It means force F acts on area a ∴ Pressure = a , area F , force ₌ c f 4 d T 2 2 4 s σ Ans.

What is mercury poisoning?

CHEMICAL DANGER Too much mercury can make you sick, but sometimes the symptoms are hard to distinguish from other illnesses.

What's mercury?

There are three kinds of mercury. Depending on what the exposure is, you could have different symptoms and disease states.

Elemental, or metal mercury, is found in thermometers. The problem with that is the inhalation of fumes that come off that mercury. Playing with it and ingesting it is not as toxic. That kind of mercury causes significant amounts of neurological damage. As the exposure gets longer, there may be additional changes in the bone marrow that affect the ability to produce blood cells, infertility and problems with heart rhythm.

Mercury salts, which are basically industrial, if you breathe in or ingest them, gravitate more toward the kidney and not so much the nervous system.

• The organic mercury is what gets into the food chain. It's put into the water by chemical plants that are manufacturing things and they get into shellfish and fish, or elemental mercury that gets into the water is changed into organic mercury by sea life; we eat fish or shellfish and we get mercury exposure. That organic mercury acts very similarly to the elemental form. It affects a lot of nervous system damage. If a woman is pregnant, this can also cause birth defects and loss of the fetus if the levels get high enough.

Is mercury something we need in our diets, or is no amount nutritionally safe or necessary?

No level is normal. Zero is normal. It doesn’t have a specific reason to be in our body. As long as we live on this Earth, because it's in Earth's crust and in the atmosphere, we're going to be exposed. But there is no specific function for that metal in our body.

The issue is one of looking at the total body burden: How much mercury is in the body and what's known to be a normal background? Theoretically, there's going to be a baseline level, a general population average, but depending on where you live, that level may be higher or lower. If you live near a coast, you're more up to eating seafood. Or you may be in an industrial area where mercury is put into the water or the air.

Review of Concepts :

Electric current is the rate of transfer of charge through a certain surface.

The direction of electric current is as that of flow of positive charge.

If a charge ∆q cross an area in time ∆t, then the average current = ∆q/∆t

Its unit is C/s or ampere.

Electric current has direction as well as magnitude but it is a scalar quantity.

Electric current obeys simple law of algebra. i.e., I = I1 + I2 α I1 I2 I1 Types of Current :

Steady state current or constant current : This type of current is not function of time.

Transient or variable current : This type of current passing through a surface depends upon time.

i.e., I = f(t) or I = t q lim 0 t ∆ ∆ → ∆ ⇒ dt dq

Electric charge passing a surface in time t = q =

∫

∫

∫

Convection Current : The electric due to mechanical transfer of charged particle is called convection current. Convection current in different situation.

Case I : If a point charge is rotating with constant angular velocity ω. I = T q ; T = ω π 2 _{⇒ I} = π ω 2 q

Case II : If a non-conducting ring having λ charge per unit length is rotating with constant angular velocity ω about an axis passing through centre of ring and perpendicular to the plane of ring.

I = R λω θ Jˆ ∆S or J = θ ∆ ∆ cos S I Its unit A/m2

Electric current can be defined as flux of current density vector.

i.e., i =

∫

Relation between drift velocity and current density v_d = –

en j

→

Here, negative sign indicates that drifting of electron takes place in the opposite direction of current density.

The average thermal velocity of electron is zero. Electric resistance : Electric resistance (R) is

defined as the opposition to the flow of electric charge through the material.

It is a microscopic quantity. Its symbol is

Its unit is ohm.

(a) (b) R = A l ρ where, R = resistance, ρ = resistivity of the material, l = length of the conductor, A = area of cross section Continuity Equation :

∫

The continuity equation is based on conservation principle of charge.

Drift Velocity (vd) : When a potential difference

is applied between ends of metallic conductor, an

Current Electricity

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