XtraEdge for IIT-JEE SEPTEMBER 2010 - Career · PDF fileXtraEdge for IIT-JEE 4 SEPTEMBER 2010 IIT-K To Coordinate 2011 JEE The IIT-K will be coordinating the IIT Joint Entrance Exam

XtraEdge for IIT-JEE 1 SEPTEMBER 2010



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

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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 4 IIT-K To Coordinate 2011 JEE Vice president addresses IIT Delhi

IITian ON THE PATH OF SUCCESS 6 Mr. Vineet Agrawal & Dr. Alok Aggarwal

KNOW IIT-JEE 8 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 59

Class XII – IIT-JEE 2011 Paper

Class XI – IIT-JEE 2012 Paper

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 5] Students’ Forum Physics Fundamentals Current Electricity Circular Motion, Rotational Motion

CATALYSE CHEMISTRY 35

Key Concept Aliphatic Hydrocarbon Oxygen Family & Hydrogen Family Understanding : Inorganic Chemistry

DICEY MATHS 48

Mathematical Challenges Students’ Forum Key Concept Probability Binomial Theorem

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.


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 produce.


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 Ph.D. in Hopkins University, (1984)


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

ωa

a 2 =

390 ⇒ ω = 30 rad/s ...(iv)

Substituting the value of ω in (i) we get

a = 303 = 0.1 m ...(v)

Now

k = λπ2 =

v/v2π =

vv2π =

vω =

2030 =

23 ...(vi)

From (iv), (v) and (vi) the wave form is

y = 0.1 sin

φ±± x

23t30

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 = 21 × stress × stain × volume

= 21 ×

Y)Stress( 2

× πr2l

= 21

Y)r/Mg( 22π × πr2l =

21

YrgM2

22

πl ...(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 = Yr

gM21

2

22

πl

∴ ∆T = YcmrgM

21

2

22

πl

Here m = mass of string = density × volume of string = ρ × πr2l

∴ ∆T = ρπ Yc)r(

gM21

22

22

= 21 ×

7860420101.2)10214.3()10100(

1123

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 = i36 + ^j38 – 10

^k is

incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium –2.

Sol.

Y

^^ j38i36 + ^j38

^i36

M'

M

XO

Z M'

XO

–10 ^K

^^^ k10–j38i36A +=→

^^ j38i36 +

Fig(1) Fig(2)

Figure 1 shows vector i36 + ^j38

Figure 2 shows vector →A = ^i36 + ^j38 – ^k10

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

2222

^^^^

)1(–)10(–)38()36(

)k).(–k10–j38i36(

++

+ = cos i

KNOW IIT-JEE By Previous Exam Questions


⇒ i = 60 Unit vector in the direction MOM' from figure (1) is

2/122

^^^

])38()36[(j38i36n

+

+=

^^^ j54i

83n +=

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

23 =

rsinisin =

rsinº60sin ⇒ r = 45º

From the triangle ORS

^r = (sin r) ^n – ( cos r) ^k

= (sin 45º)

+ ^^ j

54i

53 – (cos 45º)

^k

= ]k5–j4i3[25

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

DC Sol. When AB is steady, Weight per unit length = Force per unit length

weight per unit length = π

µ4

0 rII2 21 ...(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

µ4

0 x–rII2 21 –

πµ4

0 rII2 21

= π

µ4

0 2I1I2

r1–

x–r1

⇒ Restoring force/length = π

µ4

0 2I1I2

r)x–r()x–r(–r

= π

µ4

0 )x–r(r

xII2 21

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

Restoring force/length F = π

µ4

0 221

rII2 x

Since F ∝ x ∴ The motion is simple harmonic

∴ π

µ4

0 221

rII2 = (mass per unit length) ω2 ...(ii)

From (i) (Mass per unit length) × g = π

µ4

0 rII2 21

Mass per unit length = π

µ4

0

rgII2 21 ...(iii)

From (ii) and (iii)

π

µ4

0 221

rII2 =

πµ4

0

rgII2 21 × ω2

⇒ ω = rg ⇒

T2π =

rgt

⇒ T = 2π gr = 2p

8.901.0 = 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 RR

R

RR

S

BA

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

2R ×

21 + r1 + r2

= 4R3 + r1 + r2

I = eqRε+ε ⇒ I =

egR2ε =

21 rr4/R33

++ε

Potential drop across A is ε – I r1 = 0

⇒ ε = 21 rr4/R3

2++

ε r1

⇒ r1 = r2 + 3R/4

or R = 34 (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–

+

+ C6H5O–

N – H

N – HOH

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.

X136 ClHC

HCl–;

butoxidetK

∆

−− → 126HCZY +

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→ CH3 – 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 CH32,3-dimethyl butene-1

(Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3

CH3 CH3(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

CH3 CH32-chloro-2,3-dimethyl butane

(X)

K-t-butoxide

∆; –HCl

CH2 = C — CH – CH3

CH3 CH3

Cl

+ CH3 – C = C – CH3

CH3 CH3

(Z) 20% (Y) 80%

Hence, X,

CH3 – C – CH – CH3

CH3 CH3

Cl

Y, CH3 – C = C – CH3

CH3 CH3

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 =877.078 ml

= 877.078 ×

10001 L = 244.58 L

And molar volume of benzene in vapour phse

= 877.078 ×

10002750 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


= 867.092 ml =

867.092 ×

10001 L

And molar volume of toluene in vapour phase

= 867.092 ×

10007720 L = 819.19 L

Using the ideal gas equation, PV = nRT At T = 20ºC = 293 K

For benzene, P = 0BP =

VnRT

= 58.244

293082.01 ×× = 0.098 atm

= 74.48 torr (Q 1 atm = 760 torr) Similarly, for toluene,

P = 0TP =

VnRT

= 19.819

293082.01 ×× = 0.029 atm

= 22.04 torr (Q 1 atm = 760 torr) According to Raoult's law, PB = 0

BP xB = 74.48 xB

PT = 0TP xT = 22.04 (1 – xB)

And PM = 0BP xB + 0

TP xT or 46.0 = 74.48 xB + 22.04 (1 – xB) Solving, xB = 0.457 According to Dalton's law, PB = PM '

Bx (in vapour phase) or mole fraction of benzene in vapour form,

'Bx =

M

B

PP =

0.46457.048.74 × = 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 :

∞Λm (HCl) = 420 Ω–1 cm2 mol–1,

∞Λm (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 = Ω122

1 – Ω520

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–1cm2 dm–3 = 12.6 Ω–1 cm2(10 cm)–3 = 0.0126 Ω–1 cm–1 Cell constant,

K = Gk =

)00627.0()cm0126.0(

1

11

−

−−

ΩΩ = 2.01 cm–1

Conductivity of 0.1 M HAc solution

k = RK =

Ω

−

520cm01.2 1

Molar conductivity of 0.1 M HAc solution

Λm(HAc) = ck =

)dmmol1.0(cm)520/01.2(3

11

−

−−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 ∞Λm (HAc) = ∞Λm (HCl) + ∞Λm (NaAc) – ∞Λm (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 = )molcm385(

)molcm65.38(121

121

−−

−−

ΩΩ ≈ 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.

C6H5C6H5

OO

NH2

NH2

O–

NH2

NH2 O–

+

+ C6H5O–

N – H

N – H

OH

C6H5

N

N


MATHEMATICS

11. With usual notation, if in a triangle ABC

11

cb + = 12

ac + = 13

ba + , then prove that

7

Acos = 19

Bcos = 25

Ccos [IIT-1984]

Sol. Let 11

cb + = 12

ac + = 13

ba + = 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 = bc2

acb 222 −+

= 2

222

)30(2492536

λ

λ−λ+λ = 51

cos B = ac2

bca 222 −+

= 2

222

70364925

λλ−λ+λ =

3519

cos C = ab2

cba 222 −+

= 2

222

84253649

λλ−λ+λ =

75

∴ cos A : cos B : cos C = 51 :

3519 :

75

= 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 CDB

Then, equation of AC is,

ax +

hy = 1 ...(1)

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

hx –

ay = 0 ⇒ x =

ahy ...(2)

Solving (1) and (2) we get the coordinates of point E as follows :

2ahy +

hy = 1 ⇒ y = 22

2

haha

+

∴ E =

++ 22

2

22

2

haha,

haah

Since F is mid-point of DE,

∴ F

++ )ha(2ha,

)ha(2ah

22

2

22

2

∴ slope of AF =

)ha(2ah0

)ha(2hah

22

2

22

2

+−

+−

= 2

222

ahha)ha(h2

−−+

⇒ m1 = ah

)h2a( 22 +− ...(3)

and slope of BE = a

haah

0ha

ha

22

2

2

2

++

−+ = 232

2

ahaaha++

⇒ m2 = 22 h2a

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. ⇒ C, A + B, (A + B) – (A – B) = 2A 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) + 21

πb2(L)

Q = 2L πb2 + 12ab

Q = 2L πb2 + 6b (K – 4b – πb)

Q = 2L 6Kb – 24b2 – 5πb2

dbdQ =

2L 6K – 48b – 10πb = 0

⇒ b = π+1048

K6 ...(2)

and 2

2

dbQd =

2L –48 + 10πLa

Thus, Q is maximum and from (1) and (2), (48 + 10π) b = 6K and K = 2a + 4b + πb ⇒ (48 + 10π) b = 62a + 4b + πb

Thus, the ratio = ab2 =

π+66

15. If f : [–1, 1] → R and f´(0) =

∞→ n1nflim

n and f(0) = 0.

Find the value of :

π∞→

2limn

(n + 1)cos–1

n1 – n given that

0 <

−

∞→ n1coslim 1

n <

2π [IIT-2004]

Sol. Here π∞→

2limn

(n + 1)cos–1

n1 – n

= ∞→n

lim n

−

+

π− 1

n1cos

n112 1

= ∞→n

lim nf

n1

Where f

n1 =

+

π n112 cos–1

n1 – 1 = f´(0)

=

∞→ n1nflim)0´(fgiven

n

∴ π∞→

2limn

(n + 1)cos–1

n1 – n = f´(0) ...(1)

where f(x) = π2 (1 + x) cos–1x – 1, f(0) = 0

⇒ f´(x) =

+−

−+

π− xcos

x1

1)x1(2 12

⇒ f´(0) =

π

+−π 2

12 = 1 – π2 ...(2)

∴ from equation (1) and (2)

π∞→

2limn

(n + 1) cos–1

n1 – n = 1 –

π2

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 ablest navigator.


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

1VC and 2VC respectively, the

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)

12 VV CC < (D) 21 VV CC >

Q. 3 A particle of charge q and mass m moves

rectilinearly under the action of an electric field .xE β−α= 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 βα /

(C) mean position of the particle is at βα

=x

(D) the maximum acceleration of the particle

is mqα

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) Ta16π

(C) 8

Ta (D) π16

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 →ld 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

Solut ions wil l be published in next issue

Set # 5


Column – I Column – II

(A) Along closed loop 1 (P) ∫ µ=→→

id.B 0l

(B) Along closed loop 2 (Q) ∫ µ−=→→

id.B 0l

(C) Along closed loop 3 (R) ∫ =→→

0d.B l

(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

CROSSWORD PUZZLE NO.-12


1. [B,D]

C1 C2

2Ω 1Ω 3Ω

120 V

1Ω 2Ω 3Ω a d G H

N M e b

volt2016

120VV da =×=−

volt4026

120VV eb =×=−

volt20VV ed =−∴ Charge on C40202C1 µ=×=

volt6036

120VV HG =×=−

volt6036

120VV NM =×=−

0VV MG =−∴ Charge on C2 is zero

2. [B] N08.0)EE(q 21 =−

Columb10258

10105.21008.0q 6

25−×=

××

×=

C1032.0 6−×= C32.0 µ=

3. [C]

02 ∈σ

– σ

E 2

0E

2E =

∈σ

+ ………..(1)

10

EE =−∈σ ………..(2)

Adding (1) & (2)

120

EE +=∈σ

012 A)EE(q ∈+=

5

96

120 107.810941032.0

)EE(qA

×

××π××=

+∈=

−

4. [B] 20

E2

E =∈σ

+ 21 EEE2 −=

10

E2

E =∈σ

+− 2

EEE 21 −

=

20

105.2 5×=

= 12.5 ×104V/m

5. [B]

Kbbd

AC 0

K+−

ε=

We set b = 0

CdA

C 0K =

ε= if CK = 2C

Then, db2

b2Kd

A2

Kbbd

A 00

−=⇒

ε=

+−

ε

db&0K ≤>∴

0db2&db2

b2K >−−

=∴

db2d

≤<∴ 2db >∴

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

8 Questions were Published in August Issue


For Option (C)

1 2 3

10V

20V i

A35

610

3211020i ==++

−=

For Option (D)

1 2 3

20V

Amp20120i ==

7. [B,D]

y

R x θ

θ

C

→v

m521

101qBmvR =

××

==

Coordinates of center ⇒

3535sinRx +=×+=θ+=

4545cosRy −=×−=θ−=

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

π=×

×π×=

π=

2112

qBm2T

8. [B,C]

P1 P2

9Ω

Q1 Q2

2Ω 2Ω

504.01vb ××==ε l volt2.0=ε

A02.0i10

2.0R

ieq

=⇒=ε

=

mA20A102i 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

qx

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 = 22 xa + ∴ Electric field strength at circumference of this

ring is E = 20 r

q4

1πε

Inclination θ of →E with the normal to surface of the

ring considered is given by cosθ = ra Flux passing

through this ring is dφ = →→dsE

dφ = d dS cos θ

=

+π

+πε 22220 xa

a)dxx2()xa(

q4

1

Hence, total flux passing through the given ring is

φ = ∫=

= +ε

Rx

0x 2/3220 )xa(

xdx2aq =

+ε 220 xa

1–q1

2qa

= 105 NC–1 m2

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 = dA0ε

And capacitance of capacitor, filled with di-electric,

C = dKA0ε

= 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

qq =

VCVC2

0

20 = 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' ForumPHYSICS


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 B0 (tT – t2)

Induced emf, e = – dtdφ = – πa2 B0(T – 2t)

Induced current i = Re = 20

2)T–t2(

RBaπ ...(1)

Thernal power generated at this instant, P = i2R

or P = R

Ba 30

42π (2t – T)2

During an elemental time interval dt, heat generated

= P.dt = R

Ba 30

42π (2t –T)2 .dt

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

Q = ∫∫π

=T

0

220

42dt.)T–t2(

RBadt.P

= R

TBa 3o

20

42π Ans. (1)

The current reverses its sign when its magnitude reduces to zero. Let this happen at instant t = t0. Substituting t by t0 in equation (1),

∴ R

Ba 02π

(2t0 – T) = 0 or t0 = 2T

Substituting i by dtdq in equation (1),

dq = dt)T–t2(R

Ba 02π

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

q = R

Ba 02π

∫2/T

0dt)T–t2( = –

R4TBa 2

02π

or magnitude of charge that flows = R4

TBa 20

2π

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 loge 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 = 25 R, Cp =

27 R and γ =

57

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.log1

2

VV = – nRT0 loge 5

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

Now considering isobaric process 2 → 3

2

3

VV

= 2

3

TT

= 2 or V3 = 2V0

Heat supplied to gas during the process,

Q23 = nCP(T3 – T2) = 27 nRT0

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–

VP–VP 4433

γ =

1–)T–T(nR 43

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 = 27 ( 6 – loge 5 – 5 × 0.40.4) = 0.2642

or η = sQ

W = 72 ( 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

= 4sTσ Wm–2

Surface area of sun = 2sR4π

∴ Rate of radiation from sun, E = 2s

4s R4T πσ

Intensity of sun rays at earth,

I = 2r4Eπ

= 2

2s

4s

rRTσ

Wm–2

Area of convex lens, A = 2

2d

π =

41

πd2m2

∴ Power incident on the lens,

P = AI = 2

22s

4s

r4d.RT πσ

W

But these rays are converged on a screen by the lens, therefore, rate of incidence of momentum on the

screen = cP

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

dtdp =

0–

cP =

cr4dRT

2

22s

4s πσ

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

Hence, force on screen, F = c4

dRT2

22s

4s

ρ

πσ

But for a lens, uv

OI

= Where I is the size of image.

∴ Radius of image circle formed on the screen

= uv . O =

rf . Rs

∴ Area of image circle, a = 2

sRrf

π

It means force F acts on area a

∴ Pressure = a,areaF,force =

cf4dT

2

24sσ

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 = tqlim

0t ∆∆

→∆ ⇒

dtdq

Electric charge passing a surface in time

t = q = ∫t

0dtI

Average current I =

∫∫

t

0

t

0

dt

dtI

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 = Tq ; T =

ωπ2 ⇒ I =

πω

2q

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 J

∆S

or J = θ∆

∆cosSI

Its unit A/m2 Electric current can be defined as flux of current

density vector.

i.e., i = ∫→→dS.j

Relation between drift velocity and current

density dv = – enj

→

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 = Alρ

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

∫→→

cdS.j = –

dtdq

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 PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


electric field is established inside the metallic conductor. Due to this, electron modify their random motion and starts to drift slowly in the opposite direction of electric field. The average velocity of drifting possessed by electron is known as drift velocity.

→

dv = →

τ E

me

where, dv→

= drift velocity, e = electron, τ = relaxation time, m = mass of electron

→E = electric field

Variation of Resistance with Temperature : Let a metallic conductor of length l and cross-sectional area A.

Rt = R0(1 + αt) where, Rt = resistance of conductor at temperature tºC, R0 = resistance of conductor at 0ºC, α = temperature coefficient.

A

i

Some Important Points : (a) 'α' is proportionality constant known as

temperature coefficient of resistance variation. (b) The value of α does not depend upon initial and

final resistance of the conductor. (c) The value of α depends upon the unit which is

chosen. (d) The value of α may by negative. Electric Conductance (G) :

It is reciprocal of resistance, G = R1

Its unit is per ohm.

Electric conductivity σ = ρ1

Ohm's law in vector form :

E = ρ→i

where, ρ = τ2ne

m = receptivity of material

According to ohm's law, electric current passing through a conductor is proportional to the potential difference between end of the conductor

i.e., V = IR In case of ohm´s law, V-I graph is straight line.

I

V

α

Ohm's Law fails in tube, crystal diodes, thyristors

etc. EMF and PD of a Cell : A device which supplies

electric energy is called a seat of emf. The seat of emf is also called a cell.

A battery is a device which manages a potential difference between its two terminals.

e = EMF of the battery is the work done by the force per unit charge.

When the terminals of a cell are connected to an external resistance, the cell is said to be in closed circuit.

E.M.F. has no electrostatic origin. Internal Resistance of a Cell (r) : Internal

resistance of a cell is the resistance of its electrolyte.

The internal resistance of cell : (a) Varies directly as concentration of the solution of

the cell. (b) Varies directly as the separation between

electrodes i.e., length of solution between electrodes.

(c) Varies inversely as the area of immersed electrodes.

(d) is independent of the material of electrodes. Potential difference across the cell : Potential difference across the first cell V1 = E1 + Ir1 (discharging of cell)

R

i

E1r1 E2r2

Potential difference across the second cell V2 = E2 – Ir2 (charging of cells)

Concept of Rise up and Drop up of voltage: (a) Ideal cell

Rise up

+E

Drop up

–E


(b) Real cell

Rise up

E – ir

Drop up

–E – ir

r,E

i r,E i

(c) Electric resistance

Drop up

–IR

Rise up

+IR

R i R i

When a battery being charged, the terminal

voltage is greater than its emf V = E + Ir. Kirchhoff's Law : Kirchhoff's law is able to

solve complicated circuit problems. (i) First Law : Incoming current = Outgoing current I1 + I2 = I3 + I4 + I5

I2 I5

I1 I3

I4

This law is based upon conservation principle of

charge. (ii) Second Law : (Loop rule or voltage law.) This

law is based upon conservation principle of energy.

Grouping of resistors : Case I : Resistors in series RMN = Req = R1 + R2

M R1 R2 N

In general, Req = R1 + R2 + ... + Rn Case II : Resistors in parallel

MNR1 =

eqR1 =

1R1 +

2R1

M

R1

R2

N

In general,

MNR1 =

1R1 +

2R1 + ... +

nR1

Problem solving strategy. : Power and Energy in circuits

Step 1 Identify the relevant concepts : The ideas of electric power input and output can be

applied to any electric circuit. In most cases you’ll know when these concepts are needed, because the

problem will ask you explicitly to consider power or energy.

Step 2 Set up the problem using the following steps : Make a drawing of the circuit. Identify the circuit elements, including sources of

emf and resistors. Determine the target variables. Typically they

will be the power input or output for each circuit element, or the total amount of energy put into or taken out of a circuit element in a given time.

Step 3 Execute the solution as follows : A source of emf ε delivers power εI into a circuit

when the current I runs through the source from – to +. The energy is converted from chemical energy in a battery, from mechanical energy in a generator, or whatever. In this case the source has a positive power output to the circuit or, equivalently, a negative power input to the source.

A source of emf power εI from a circuit – that is, it has a negative power output, or, equivalently, a positive power input–when currents passes through the source in the direction from + to –.This occurs in charging a storage battery, when electrical energy is converted back to chemical energy. In this case the source has a negative power output to the circuit or, equivalently, a positive power input to the source.

No matter what the direction of the current through a resistor, It removes energy from a circuit at a rate given by VI = I2R = V2/R, where V is the potential difference across the resistor.

There is also a positive power input to the internal resistance r of a source, irrespective of the direction of the current. The internal resistance always removes energy from the circuit, converting it into heat at a rate I2r.

You may need to calculated the total energy delivered to or extracted from a circuit element in a given amount of time. If integral is just the product of power and elapsed time.

Step 4 Evaluate your answer : Check your results, including a check that energy is conserved. This conservation can be expressed in either of two forms: “net power input = net power output” or “the algebraic sum of the power inputs to the circuit elements is zero.”

Problem solving strategy : Series and Parallel

Step 1 Identify the relevant concepts : Many resistor networks are made up of resistors in series, in parallel, or a combination of the two. The key concept is such a network can be replaced by a single equivalent resistor.


Step 2 Set up the problem using the following steps: Make a drawing of the resistor network. Determine whether the resistors are connected in

series or parallel. Note that you can often consider networks such as combinations of series and parallel arrangements.

a R1 x R2 y R3 b

I I(a) R1, R2, and R3 in series

R1

I

(b) R1, R2, and R3 in parallel

R2

R3 I

a b

Determine what the target variables are. They could include the equivalent resistance of the network, the potential difference across each resistor, or the current through each resistor.

Step 3 Execute the solution as follows : Use Eq. Req = R1 + R2 + R3 (resistors in series)

or eqR

1 =1R

1 +2R

1 +3R

1 +...(resistors in parallel)

to find the equivalent resistance for a series or a parallel combination, respectively.

If the network is more complex, try reducing in to series and parallel combinations.

When calculating potential differences, remember that when resistors are connected in series, the total potential differences across the combination equals the sum of the individual potential differences. When they are connected in parallel, the potential difference across the parallel combination.

Keep in mind the analogous statements for current. When resistors are connected in series, the current is the same through every resistor and equals the current through the series combination. When resistors are connected in parallel the total current through the combination equals the sum of the currents through the individual resistors.

Step 4 Evaluate your answer : Check whether your results are consistent. If resistors are connected in series, the equivalent resistance should be greater than that of any individual resistor; if they are connected in parallel, the equivalent resistance should be less than that of any individual resistor.

Problem solving st. : Kirchhoff’s Rules : Step 1 Identify the relevant concepts : Kirchhoff’s

rules are important tools for analyzing any circuit more complicated than a single loop.

Step 2 Set up the problem using the following steps : Draw a large circuit diagram so you have plenty

of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. Often you will not know in advance the actual direction of an unknown current or emf, but this does not matter. If the the actual direction of a particular quantity is opposite to your assumption, the result will come out with a negative sign. If you are Kirchhoff’s rules correctly, they will give you the directions as well as the magnitudes of unknown currents and emfs.

When you label currents, it is usually best to use the junction rule immediately to express the currents of as few quantities as possible. For example, fig (a) shows a circuit correctly labeled; fig. (b) shows the same circuit, relabeled by applying the junction rule to point a to eliminate I3.

r1 ε1 r2 ε2

I1

I1

I3 R3

I2

I2

R1 a R2

+ +

(a)

r1 ε1 r2 ε2

I1

I1

I1 + I2 R3

I2

I2

R1 a R2

+ +

(b) Determine which quantities are the target

variables. Step 3 Execute the solution as follows : Choose any closed loop in the network and

designate a direction (clockwise or counterclockwise) to travel around the loop when applying the loop rule. The direction does not have to be the same as any assumed current direction.


Travel around the loop in the designated direction, adding potential differences as you cross them. Remember that a positive potential and a negative potential difference corresponds to a decrease in potential. An emf is counted as positive when you traverse it from (–) to (+), and negative when you go from (+) to (–). An IR term is negative if you travel through the resistor in the same direction as the assumed current and positive if you pass it in the opposite direction. Figure. summarizes these sign conventions. In each part of the figure “travel” is the direction that we imagine going around a loop while using Kirchhoff’s loop law, not necessary the direction of current.

Equate the sum is Step 2 to zero. If necessary, choose another loop to get a

different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in a at least one of the chosen loops.

Solve the equations simultaneously to determine the unknowns. This step involves algebra, not physics, but it can be fairly complex. Be careful with algebraic manipulations; one sign error will prove fatal to the entire solution.

You can use this same bookkeeping system to find the potential Vab of any point a with respect to any other point b. Start at b and add the potential changes you encounter in going from b to a, using the same sign rules as in Step 2. The algebraic sum of the these changes is vab = Va

– Vb. Step 4 Evaluate your answer : Check all the step in

your algebra. A useful strategy is to consider a loop other than the ones you used to solve the problem; if the sum of potential drops around this loop is not zero, you made an error somewhere in your calculations. As always, ask yourself whether is answer make sense.

Travel

+ – + ε

Travel

– + – ε

Travelε ε

Travel

– + –IR

TravelR

I

Travel

+ – +IR

TravelR

I When using Kirchhoff’s rules, follow these sign

conventions as you travel around a circuit loop.

Solved Examples

1. In a circuit shown in fig. (i) find the current drawn from the accumulator. (ii) find the current through the 3 ohm resistor, (iii) What happens when 3 ohm resistor is removed

from the circuit ? 2V

1Ω

3Ω2Ω 2Ω

4Ω

A

C B

D

Sol. The equivalent Wheatstone's bridge network of the

given circuit is shown in fig. B

2Ω 4Ω

3Ω

1Ω 2Ω

D

2 Volt

A C

Here the points B and D are at the same potential as

the bridge is balanced. So the 3Ω resistance in BD arm is ineffective and can be omitted from the circuit. The resistance of ABC branch is 2Ω + 4Ω = 6Ω as AB and BC are in series. Similarly the resistance of A D C branch is 1Ω + 2Ω = 3Ω.

The two resistances, i.e., 6 ohm and 3 ohm are in parallel. The equivalent resistance R is given by

R1 =

61 +

31 =

21 ∴ R = 2Ω

(i) The current drawn from 2 volt accumulator is

i = RE =

22 = I amp.

(ii) The current through 3Ω resistor is zero. (iii) When the 3Ω resistor is removed from the

circuit, there will be no change. 2. A battery of e.m.f. 5 volt and internal resistance 20Ω

is connected with a resistance R1 = 50 Ω and a resistance R2 = 40Ω. A voltmeter of resistance 1000 Ω is used to measure the potential difference across R1. What percentage error is made in the reading ?


Sol. The circuit is shown in fig.

V 1000Ω

20Ω

R2 = 50Ω

5V

R1 = 50Ω

When voltmeter is not connected

current in the circuit i = 21 RRr

E++

∴ i = 405020

5++

= 110

5 = 221 A

Potential difference across R1 = i × R1

= 221 × 50 = 2.27 volt.

When the voltmeter is connected across R1. In this case the galvanometer resistance is in parallel

with R1. Hence

Equivalent resistance = 501000501000

+× = 47.62 ohm

Current in the circuit

= 62.474020

5++

= 62.107

5 A

Potential difference measured by voltmeter

= 62.107

5 × 47.62 = 2.21 volt.

Percentage error = 27.2

21.227.2 − × 100 = 2.6%

3. In the circuit fig. a voltmeter reads 30 V when it is

connected across 400 ohm resistance. Calculate what the same voltmeter will read when it is connected across the 300 Ω resistance ?

V

30 V

300 Ω

400 Ω

60 V Sol. Potential difference across 400 ohm = 30 V Potential difference across 300 ohm = (60 – 30) = 30 V This shows that the potential difference is equally

shared.

Let R be the voltmeter resistance. The resistance 400 and voltmeter resistance R are in parallel. Their equivalent resistance R´ is given by

´R

1 = R1 +

4001 =

R400R400 + or

R400R400

+

But R´ should be equal to 300 ohm. Hence

R400

R400+

= 300 ∴ R = 1200 ohm

Thus, voltmeter resistance is 1200 ohm. When the voltmeter is connected across 300 ohm, the

effective resistance R" is given by

"R

1 = 1200

1 + 300

1 = 1200

41+ = 1200

5

∴ R´´ = 5

1200 = 240 ohm.

Now the potential difference is shared between 240 ohm and 400 ohm.

Potential diff. across 240 ohm : Potential difference across 400 ohm

= 240 : 400 = 3 : 5 As total potential is 60 V, hence potential difference

across 240 ohm, i.e., across resistance 300 ohm will be

83 × 60 = 22.5 V.

4. In the circuit shown in fig. E, F, G and H are cells of

e.m.f. 2, 1, 3 and 1 volt and their internal resistances are 2, 1, 3 and 1 ohm respectively. Calculate

E – +

+–

–

–+

+

A B

D CG

F H2Ω

(i) The potential difference between B and D and (ii) the potential difference across the terminals of

each of the cells G and H. Sol. Fig. shows the current distribution. Applying Kirchhoff's first law at point D, we have i = i1 + i2 ...(1) Applying Kirchhoff's second law to mesh and

ADBA, we have 2i + 1i + 2i1 = 2 – 1 = 1 or 3i + 2i1 = 1 ...(2)

2V A B

D C3V

1V

1V

2Ω

2Ω

3Ω

1Ω

ii2

1Ωi1


Applying Kirchhoff's second law to mesh DCBD, we get

3i2 – 1i2 – 2i1 = 3 – 1 or 4i2 – 2i1 = 2 ...(3) Solving eqs. (1), (2) and (3), we get

i1 = 131 amp., i2 =

136 amp. and i =

135 amp.

(i) Potential difference between B and D

= 2i1 = 2

131 =

132 volt.

(ii) Potential difference across G

= E – i2R = 3 – 13

36× = 1.61 V

Potential difference across H

= 1 –

−

136 (1) = 1.46 V.

5. Twelve equal wires, each of resistance 6 ohm are

joined up to form a skeleton cube. A current enters at one corner and leaves at the diagonally opposite corner. Find the joint resistance between the corners.

Sol. The skeleton ABCDEFGH, is shown in fig.

A

E F i/6

B

GH

i/3 i/3 i/6

i/6i/3

C D i/6

i/3

i/6

i i/6 i/3

i/3

This skeleton consists of twelve wires. Let the

resistance of each wire be r. Here the current i enters at corner A and leaves at corner G. The current i at corner A is divided into three equal parts (i/3) because the resistance of each wire is the same. At B, D and E, the current i/3 is divided into two equal parts each having magnitude i/6. At the corners C, F and H, the currents again combine to give currents, each of magnitude i/3 along CG, FG and HG respectively. At corner G, all these currents combine so that the current leaving at G is i.

Let R be the equivalent resistance between the corners A and G. Taking any one of the paths say ABCG, we have

VAG = VAB + VBC + VCG

iR = 3i r +

6i r +

3i r

or R = 65 r

According to given problem r = 6 ohm

∴ R = 65 × 6 = 5 ohm.

FRACTIONAL DISTILLATION OF AIR

Did you know that the air we breathe isn’t just oxygen, infact it’s made up of a number of different gases such as nitrogen, oxygen, carbon dioxide, argon, neon and many others. Each of these gases carry useful properties so separating them from the air around us is extremely beneficial.

The process is called fractional distillation and consists of two steps, the first relies on cooling the air to a very low temperature (i.e. converting it into a liquid), the second involves heating it up thus allowing each gas within the mixture to evaporate at its own boiling point. The key to success here is that every element within air has its own unique boiling temperature. As long as we know these boiling temperatures we know when to collect each gas.

So what are the real world benefits of separating and extracting these gases? Well liquid oxygen is used to power rockets, oxygen gas is used in breathing apparatus, nitrogen is used to make fertilizers, the nitric acid component of nitrogen is used in explosives.

The other gases all have their own uses too, for example argon is used to fill up the empty space in most light bulbs (thanks to its unreactive nature). Carbon dioxide is used in fire extinguishers and is great for putting out fires in burning liquids and electrical fires. There really are too many uses to list but suffice it to say that fractional distillation is an extremely useful process for humans the world over.


Circular Motion : When a particle moves on a circular path with

uniform speed, its is said to execute a uniform circular motion.

Angular Velocity : It is the rate of change of angular displacements of the body. If the radial line in the adjoining figure rotates through an angle θ(radian) in time t (seconds) then its angular velocity.

O θ

ω = tθ radian / second

If it takes the radial line a time T to complete one revolution, then

ω = T2π

and if n revolutions are made in 1s then

n = T1 and ω = 2πn

The angular acceleration of the particle is given by

α = t∆ω∆

Linear Velocity : Linear velocity = angular velocity × radius

v = ω × r linear acceleration of particle (a) = a × r Centripetal Acceleration : When a particle moves

with uniform speed v in a circle of radius r it is acted upon by an acceleration v2/r in the direction of centre. It is called centripetal acceleration. The acceleration has a fixed magnitude but its direction is continuously changing. It is always directed towards the centre of the circle.

Centripetal Forces : If the particle of mass m moves with uniform velocity v in circle of radius r, then

force acting on it towards the centre is r

mv2. This is

called centripetal force. It has a fixed magnitude and is always directed towards the centre.

Without centripetal force, a body can not move on a circular path. Earth gets this force from the gravitational attraction between earth and sun; electron moves in circular path due to electrostatic attraction between it and nucleus. A cyclic or car while taking turn, gets the centripetal force from the friction between road and type. To create this force, the vehicle tilts itself towards the centre. If it makes angle θ with the vertical in tilted position then than θ = v2/rg. where v is its velocity and r is the radius of the path. In order to avoid skidding (or slipping), the angle of tilt θ with vertical should be less than angle of friction λ. i.e. tan θ < tan λ

or rgv2

< µ (since coefficient of friction µ = tan λ)

In limiting condition rgv2

= µ or v = g.r.µ

This is the maximum safe speed at the turn. Since centripetal force is provided by the friction, it

can never be more than the maximum value µR = (µmg) or frictional force.

Motion in a vertical circle : When a body tied at one end of a string is revolved in a vertical circle, it has different speed at different points of the circular path. Therefore, the centripetal force and tension in the string change continuously. At the highest point A of motion.

mg Ta

r

Tbvb B

mg

A va

C

Ta + mg = r

mv2a or Ta =

rmv2

a – mg

This tension, at highest point will be zero, for a minimum velocity vc given by

0 = r

mv2c – mg or vc = gr

Circular Motion, Rotational Motion

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


This minimum speed is called critical speed (vc). If the speed at A is less than this value, the particle will not reach up to the highest point. To reach with this speed at A, the body should have speed at B given by the conservation laws viz.

Decrease in kinetic energy = increase in potential energy

21 mvb

2 – 21 mva

2 = mg.2r

vb2 = va

2 + 4gr

for critical speed va = vc = gr

∴ vb2gr + 4gr or vb = gr5

Therefore, the body should have speed at B at least gr5 , so that it can just move in vertical circle.

Tension in string at B is given by.

Tb – mg = r

mv2b or Tb = mg +

rvgr5m = 6mg

This means that the string should be able to stand to a tension, equal to six times the weight of the body otherwise the string will break.

At any other point P making angle θ with the vertical, from the figure.

A

C

B mg

mg cos θ Q

P

vp T θ

T – mg cos θ = r

mv2p or T = m

θ+ cosg

rv2

r

At point A, θ = 180º; Ta = m

− g

rv2

a

At point B, θ = 0º; Tb = m

+ g

rv2

b

Conical pendulum : A conical pendulum consists of a string AB (fig.)

whose upper end is fixed at A and other and B is tied with a bob. When the bob is drawn aside and is given a horizontal push. Let it describe a horizontal circle with constant angular speed ω in such a way that AB makes a constant angle θ with the vertical. As the string traces the surface of a cone, it is known as conic pendulum.

Let l be the length of string AB. The forces acting on the bob are (i) weight mg acting downwards, (ii) tension T along the sting (horizontal) component is T sin θ and vertical component is T cos θ).

T cos θ = mg The horizontal component is equal to the centripetal

force i.e.,

A

h

r T sin θ T

T cosθ

B

mg

O

Rotational Motion : Centre of mass of a system of particles : The point at which the whole mass of the body may

be supposed to be concentrated is called the centre of mass.

Consider the case of a body of an arbitrary shape of n XY plane as shown in fig. Let the body consist of number of

P1(x1, y1)(x2, y2)

P2(x, y )

P3(x3, y3)

X O

Y

particles P1, P2, P3, .... of masses m1, m2, m3, ..... and

coordinates (x1, y1), (x2, y2), (x3, y3), ..... If )y,x( be the coordinates of centre of mass, then

x = .....mmm

....xmxmxm

321

332211

++++++

= n

nn

mxm

ΣΣ

and y = ....mmm

...ymymym

321

332211

++++++

= n

nn

mym

ΣΣ

When there is a continuous distribution of mass instead of being discrete, we treat an infinitesimal element of the body of mass dm whose position is (x, y, z). In such a case, we replace summation by integration in above equations. Now we have,

x = ∫∫

dm

dmx =

M

dmx∫

y = ∫∫

dm

dmy =

M

dmy∫


z = ∫∫

dm

dmz =

M

dmz∫

where M is the total mass. Motion of centre of mass : Consider two particles of masses m1 and m2 located

at position vectors r1 and r2 respectively with respect to origin. Now the position vector r of the centre of mass is given by

(m1 + m2)r = m1r1 + m2r2 ...(1) Thus, the product of the total mass of the system and

position vector of the centre of mass is equal to the sum of the products of the individual masses and their respective position vectors. Hence

r = 21

2211

mmrmrm

++ ...(2)

Now the velocity of centre of mass of the system is

given by v = dtdr

The acceleration of the centre of mass is given by

a = dtdv =

dtd

dtd = 2

2

dtxd

The equation describing the motion of the centre of mass may be written as

f(total) = Mdtdv

When no external force acts on the system, then

0 = Mdtdv or

dtdv = 0

∴ v = constant Therefore, when no external force acts on the system,

the centre of mass of an isolated system move with uniform velocity.

Moment of inertia and radius of gyration : Moment of Inertia : The moment of inertia of a

body about an axis is defined as the sum of the products of the masses of the particles constituting the body and the square of their respective distance from the axis.

Radius of Gyration : If we consider that the whole mass of the body is concentrated at a distance K from the axis of rotation, then moment of inertia I can be expressed as

I = MK2 where M is the total mass of the body and K is the

radius of gyration. Thus the quantity whose square when multiplied by the total mass of the body gives the moment of inertia of the body about that axis is known as radius of gyration.

Theorems on moment of inertia : Theorem of parallel axes : According to this

theorem, the moment of inertia I of a body about any axis is equal to its moment of inertia about a parallel axis through centre of mass IG plus Ma2 where M is the mass of the body and a is the perpendicular distance between the axes, i.e., I = IG + Ma2

Theorem of perpendicular axes : According to this theorem, the moment of inertia I of the body about a perpendicular axis is equal to the sum of moment of inertia of the body about two axes right angles to each other in the plane of the body and intersecting at a point where the perpendicular axis passes, i.e.,

I = Ix + Iy Table of moment of inertia :

Body Axis Moment of inertia

1. Thin uniform rod of length l

Through its centre and perpendicular to its length

12M 2l

2. Thin rectangular sheet of sides a and b.

Through its centre and perpendicular to its plane

M

+

12b

12a 22

3. Thick rectangular bar of length l, breadth b and thickness t.

Through its midpoint and perpendicular to its length

M

+

12b

12

22l

4. Uniform solid sphere of radius R

About a diameter 52 MR2

5. Circular ring of radius R.

Through its centre and perpendicular to its plane

MR2

6. Disc of radius R. Through its centre and perpendicular to its plane

21 MR2

7. Solid cylinder of length l and radius R.

(i) Through its centre and parallel to its length (ii) Through its centre and perpendicular to its length.

21 MR2

M

+

124R 22 l

Angular momentum of a rotating body : In case of rotating body about an axis, the sum of the

momentum of the linear momentum of all the


particles about the axis of rotation is called angular momentum about the axis.

Q Also the angular momentum of rigid body about an axis is the product of moment of inertia and the angular velocity of the body about that axis.

L = r × p = Iω Translational and rotational quantities :

Translational Motion Rotational Motion Displacement = s Angular displacement = θ

Velocity = v Angular velocity = ω

Acceleration = a Angular acceleration = α

Inertia = m Moment of inertia = I

Force = F Torque = τ

Momentum = mv Angular momentum = Iω

Power = Fv Rotational power = τω

Kinetic energy = 21 mv2 Rotational K.E. =

21 Iω2

Kinematics equation of a rotating rigid body : The angular velocity of a rotating rigid body is

defined as the rate of change of angular displacement,

i.e., →ω = )dt/d(

→θ

Similarly, the angular acceleration is defined as the rate of change of angular velocity, i.e.,

→α =

dtd

→ω = 2

2

dtd

→θ

Let a body be rotating with constant angular

acceleration →α with initial angular velocity

→ω0 . If θ

is the initial angular displacement, then its angular

velocity →ω and angular displacement θ at any time is

given by the following equations ω = ω0 + αt

0 = ω0t + 21

αt2

and ω2 = ω02 + 2 αθ

These equations are similar to usual kinematics equation of translatory motion.

v = u + at, s = ut + 21 at2

and v2 = u2 + 2as Problem Solving Strategy : Rotational Dynamics for Rigid Bodies : Our strategy for solving problems in rotational

dynamics is very similar to the strategy for solving problems that in involve Newton’s second law.

Step-1 : Identify the relevant concepts : The equation Στ = Iαz is useful whenever torques act on a rigid body–that is, whenever forces act on a rigid body in such a way as to change the state of the body’s rotation.

In some cases you may be able to use an energy approach instead. However, if the target variable is a force, a torque, an acceleration, an angular acceleration, or an elapsed time, the approach using Στ = Iα2 is almost always the most efficient one.

Step-2 : Setup the problem using the following steps: Draw a sketch of the situation and select the body

or bodies to be analyzed. For each body, draw a free-body diagram

isolating the body and including all the forces (and only those forces) that act on the body, including its weight. Label unknown quantities with algebraic symbols. A new consideration is that you must show the shape of the body accurately, including all dimensions and angles you will need for torque calculations.

Choose coordinate axes for each body and indicate a positive sense of rotation for each rotating body. If there is a linear acceleration, it’s usually simplest to pick a positive axis in its direction. If you know the sense of αz in advance, picking it as the positive sense of rotation simplifies the calculations. When you represent a force in terms of its components, cross out the original force to avoid including it twice.

Step-3 : Execute the solution as follows : For each body in the problem, decide whether it

under goes translational motion, rotational motion, or both. Depending on the behavior of the body in question, apply ΣF = m a

r, Στz = Iαz, or

both to the body. Be careful to write separate equations of motion for each body.

There may be geometrical relations between the motions of two or more bodies, as with a string that unwinds from a pulley while turning it or a wheel that rolls without slipping. Express these relations in algebraic form, usually as relations between two linear accelerations or between a linear acceleration and an angular acceleration.

Check that the number of equations matches the number of unknown quantities. Then solve the equations to find the target variable(s).

Step-4 : Evaluate your answer : Check that the algebraic signs of your results make sense. As an example, suppose the problem is about a spool of thread. If you are pulling thread off the spool, your answers should not tell you that the spool is turning in the direction the results for special cases or intuitive expectations. Ask yourself : Does this result make sense ?”


Problem Solving Strategy: Equilibrium of a Rigid Body Step-1 : Identify the relevant concepts : The first and

second conditions for equilibrium are useful whenever there is a rigid body that is not rotating and not accelerating in space.

Step-2 : Set up the problem using the following steps: Draw a sketch of the physical situation, including

dimensions, and select the body in equilibrium to be analyzed.

Draw a free-body diagram showing the forces acting on the selected body and no others. Do not include forces exerted by this body on other bodies. Be careful to show correctly the point at which each force acts; this is crucial for correct torque calculations. You can't represent a rigid body as a point.

Choose coordinate axes and specify a positive sense of rotation for torques. Represent forces in terms of their components with respect to the axes you have chosen; when you do this, cross out the original force so that you don't included it twice.

In choosing a point to compute torques, note that if a force has a line of action that goes through a particular point, the torque of the force with respect to that point is zero. You can often eliminate unknown forces or components from the torque equation by a clever choice of point for your calculation. The body doesn't actually have to be pivoted about an axis through the chosen point.

Step-3 : Execute the solution as follows : Write equations expressing the equilibrium

conditions. Remember that ΣFx = 0, ΣFy = 0, and Στz = 0 are always separate equations; never add x-and y-components in a single equation. Also remember that when a force is represented in term of its components, you can compute the torque of that force by finding the torque of each component separately, each with its appropriate lever arm and sign, and adding the results. This is often easier than determining the lever arm of the original force.

You always need as many equations as you have unknowns. Depending on the number of unknowns, you may need to compute torques with respect to two or more axes to obtain enough equations. Often, there are several equally good sets of force and torque equations for a particular problem; there is usually no single "right" combination of equations. When you have as many independent equations as unknowns, you can solve the equations simultaneously.

Step-4 : Evaluate your answer : A useful way to check your results is to rewrite the second condition for equilibrium, Στz = 0, using a different choice of origin. If you've done everything correctly, you'll get the same answers using this new choice of origin as you did with your original choice

Solved Examples

1. A particle a moves along a circle of radius R = 50 cm

so that its radius vector r relative to point O (fig.) rotates with the

A

R

CO

r

A

R

C O

r

Y

Xθ

θ

(a) (b) constant angular velocity ω = 0.40 rad/sec. Find the

modulus of the velocity of the particle and modulus and direction of its total acceleration.

Sol. Consider X and Y axes as shown in fig. Using sine law in triangle CAO, we get

)2sin(

rθ−π

= θsin

R or θθcossin2

r = θsin

R

∴ r = 2 R cos θ Now r = r cos θ i + r sin θ j = 2 R cos2θ i + 2R cos θ sin θ j

Now, vdtdr = – 4R cos θ sin θ

dtdθ i + 2R cos 2θ

dtdθ j

= – 2 R sin 2 θ ω i + 2 R cos 2θ ω j ∴ |v| = 2 ω R Further

a = dtdv = 4 R cos 2 θ

dtdθ i – 4 R ω sin 2 θ

dtdθ j

= –4 R ω2 cos 2 θ i – 4R ω2 sin 2θ j |a| = 4 Rω2

2. A particle describes a horizontal circle on the smooth inner surface of a conical funnel as shown in fig. If the height of the plane of the circle above the vertex 9.8 mark cm, find the speed of the particle.

Sol. The forces acting on the particle are shown in fig. They are

R sin α

Rα

r h=9.8 cmα

mg

mv2/r

R c

os α

(i) weight m g acting vertically downwards. (ii) normal reaction R of smooth surface of the cone. (iii) reaction of the centripetal force (mv2/r) acting

radially outward. Hence, R sin α = m g ...(1) and R cos α = (mv2/r) ...(2)


Dividing eq. (1) by eq. (2), we get

tan α =

2vrg ...(3)

From figure, tan α = (r/h) ...(4) From eqs. (3) and (4), we get

hr = 2v

rg or v = )hg(

∴ v = [9.8 × (9.8 × 10–2)]1/2 = 9.98 m/s

3. A particle of 10 kg mass is moving in a circle of 4m radius with a constant speed of 5m/sec. What is its angular momentum about (i) the centre of circle (ii) a point on the axis of the circle and 3 m distant from its centre ? Which of these will always be in same direction ?

Sol. The situation is shown in fig. L1

L2

3 0 4

5

(a) We know that L = r × mv

L = m v r sin θ Here m = 10 kg, r = 4 m, v = 5 m/sec and θ = 90º ∴ L = 10 × 5 × 4 × 1 = 200 kg-m2/sec.

(b) In this case r = )34( 22 + = 5m. ∴ L = 10 × 5 × 5 = 250 kg-m2/sec. From figure it is obvious that angular momentum in

first case always has same direction but in second case the direction changes.

4. A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 1 kg-m2 and its rate of rotation 2 rev./sec. (a) what is its angular momentum ? (b) what additional work will have to be done to double its rate of rotation ?

Sol. (a) As the body is rotating about its axis of symmetry, the angular momentum vector coincides with the axis of rotation.

∴ Angular momentum L = Iω ...(1)

Kinetic energy of rotation E = 21 Iω2

or 2E = Iω2 ∴ I2ω2 = 2IE or Iω = )IE2( ...(2)

From eqs. (1) and (2), L = )IE2( ...(3) ω = 2 rev/sec = 2 × 2π or 4π radian/sec.

∴ E = 21 × 1 × (4π)2 = 8π2 joule

Now L = )IE2( = )812( 2π××

= )16( 2π = 4π = 12.57 kg.m2/sec. (b) When the rate of rotation is doubled, i.e., 4

rev/sec or 8π radians/sec, the kinetic energy of rotation is given by

E = 21 × 1 × (8π)2 = 32π2 joule

Additional work required = Final K.E. of rotation – Initial K.E. of rotation = 32π2 – 8π2 = 24 π2 = 236.8 joule

5. A thin horizontal uniform rod AB of mass m and length l can rotate freely about a vertical axis passing through its end A. At a certain moment the end B starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed in the horizontal plane. Find the angular velocity of the rod as a function of its rotation angle φ counted relative to the initial position.

Sol. The situation of the rod at an angle φ is shown in fig. Here

r = i l cos φ + + j l sin φ and F = j F (Force is always perpendicular to rod)

A B X

F

F Y

θ l

→τ = r × F = (i l cos φ + j l sin φ) × (j F)

= l F cos φ k

| →τ | = l F cos φ

We know that τ = 1 α

Here I = 31 m l2 (for rod) and α = ω (dω/dφ)

∴ l F cos φ = 31 m l2 . ω (dω/dφ)

or l F cos φ dφ = 31 m l2 . ω dω

Integrating within proper limits, we have

l F ∫φ

φφ0

dcos = 31 m l2 ∫

ωωω

0d

l F sin φ = 31 m l2(ω2/2) ∴ ω =

φ

lmsinF6


Addition of hydrogen halides to Alkenes : Markovnikov’s Rule Hydrogen halides (HI, HBr, HCl, and HF) add to the

double bond of alkenes :

C = C + HX → – C – C –

H H These additions are sometimes carried out by

dissolving the hydrogen halide in a solvent, such as acetic acid or CH2Cl2, or by bubbling the gaseous hydrogen halide directly into the alkene and using the alkene itself as the solvent. HF is prepared as polyhydrogen fluoride in pyridine. The order of reactivity of the hydrogen halides is HI > HBr > HCl > HF, and unless the alkene is highly substituted, HCl reacts so slowly that the reaction is not one that is useful as a preparative method. HBr adds readily, the reaction may follow an alternate course. However, adding silica gel or alumina to the mixture of the alkene and HCl or HBr in CH2Cl2 increases the rate of addition dramatically and makes the reaction an easy one to carry out.

The addition of HX to an unsymmetrical alkene could conceivably occur in two ways. In practice, however, one product usually predominates. The addition of HBr to propene, for example, could conceivably lead to either 1-bromopropane or 2-bromopropane. The main product, however is 2-bromopropane :

CH2 = CHCH3 + HBr → CH3CHCH3

Br2-Bromopropane

When 2-methylpropene reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide :

C = CH2 + HBr → CH3 – C – CH3

Brtert-Butyl bromide

CH3 H3C

H3C

2-Methylpropene (isobutylene)

Consideration of many examples like this led the Russian chemist Vladimir Markovnikov in 1870 to formulate what is now known as Markovnikov’s rule. One way to state this rule is to say that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that

already has the greater number of hydrogen atoms. The addition of HBr to propene is an illustration :

CH2 = CHCH3 → CH2 – CHCH3

H BrMarkovnikov addition

product H Br

Carbon atom with the greater number of hydrogen atoms

Reactions that illustrate Markovnikov’s rule are said

to be Markovnikov additions. A mechanism for addition of a hydrogen halide to an

alkene involves the following two steps : Step 1 :

C = C + H – X → +C – C – + X

H

– slow

The π electrons of the alkene form a bond with a proton from HX to form a carbocation and a halide ion

Step 2 :

X + +C – C – → – C – C –

H

fast

The halide ion reacts with the carbocation by donating an electron pair; the result is an alkyl halide

–H

X

Modern Statement of Markovniov’s Rule : According to Modern statement of Markovnikov’s

rule, In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the adding reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as an intermediate. Because this is the step that occurs first (before the addition of the nucleophilic portion of the adding reagent), it is the step that determines the overall orientation of the reaction.

Notice that this formulation of Markovnikov’s rule allows us to predict the outcome of the addition of a

such as ICl. Because of the greater electro negativity of chlorine, the positive portion of this molecule is iodine. The addition of ICl to 2-methylpropene takes place in the following way and produces 2-chloro-1-iodo-2-methylpropane :

Organic Chemistry

Fundamentals

ALIPHATIC HYDROCARBON

KEY CONCEPT


C = CH2 + I – Cl →

H3C

H3C

2-Methylpropene

C – CH2 – I H3C

H3C +

Cl–

δ+ δ–

→ CH3 – C – CH2 – I

CH3

2-Chloro-1-iodo-2-methylpropane

Cl

An Exception to Markovnikov’s Rule : This rule exception concerns the addition of HBr to

alkenes when the addition is carried out in the presence of peroxides (i.e., compounds with the general formula ROOR). When alkenes are treated with HBr in the presence of peroxides, an anti-Markovnikov addition occurs in the sense that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms. With propene, for example, the addition takes place as follows :

CH3CH = CH2 + HBr →ROOR CH3CH2CH2Br This addition occurs by a radical mechanism, and not

by the ionic mechanism. This anti-Markovnikov addition occurs only when HBr is used in the presence of peroxides and does not occur significantly with HF, HCl, and HI even when peroxides are present.

Alcohols from Alkenes through Oxymercuration-Demercuration Markovnikov Addition : A useful laboratory procedure for synthesizing

alcohols from alkenes that avoids rearrangement is a two-step method called oxymercuration -demercuration.

Alkenes react with mercuric acetate in a mixture of tetrahydrofurane (THF) and water to produce (hydroxyalkyl) mercury compounds. These (hydroxyalkyl) mercury compounds can be reduced to alcohols with sodium borohydride :

Step 1 : Oxymercuration

C = C THF + H2O + Hg 23OCCH

||O

– C – C –

HO Hg – OCCH3

+ CH3COH

O

O

Step 2 : Demercuration

– C – C –

HO Hg – OCCH3

+ OH– + NaBH4 O

– C – C – + Hg + CH3CO–

HO H

O

In the first step, oxymercuration, water and mercuric

acetate add to the double bond; in the second step, demercuration, sodium borohydride reduces the acetoxymercury group and replaces it with hydrogen. (The acetate group is often abbreviated – OAc.)

Both steps can be carried out in the same vessel, and both reactions take place very rapidly at room temperature or below. The first step–oxymercuration–usually goes to completion within a period of 20s – 10 min. The second step – demercuration – normally requires less than an hour. The overall reaction gives alcohols in very high yields, usually greater than 90%.

Oxymercuration–demercuration is also highly regioselective. The net orientation of the addition of the elements of water, H – and –OH, is in accordance with Markovnikov’s rule. The H– becomes attached to the carbon atom of the double bond with the greater number of hydrogen atoms :

R– C – C – H

HO H

(1) Hg(OAc)2/THF–H2O C = C

(2) NaBH4, OH–

H H H H

H R

+ HO – H

The following are specific examples :

Pentene1

CHCH)CH(CH 2223−

=)s15(

OHTHF

)OAc(Hg

2

2

− →

CH3(CH2)2

HgOAcOH||CHCH 2−

)h1(OH

NaBH4− →

CH3(CH2)2

OH|CHCH3 + Hg

2-Pentanol (93%)

CH3

Hg(OAc)2

THF-H2O (20 s)

1-Methylcyclopentanol

OH H3C HgOAc

H

NaBH4

OH–

(6 min)

OHH3C

+ Hg

1-Methylcyclopentene






Rearrangements of the carbon skeleton seldom occur in oxymercuration - demercuration. The oxymercuration - demercuration of 3, 3-dimethyl-1-butene is a striking example illustrating this feature. It is in direct contrast to the hydration of 3, 3-dimethyl-1-butene.

3

23

3

CH|

CHCHCCH|

CH

=− OH,NaBH)2(

OHTHF/)OAc(Hg)1(

4

22 →−

3, 3-Dimethyl-1-butene

OHCH||CHCH—CCH

|CH

3

33

3

3,3-Dimethyl-2-butanol (94%) Analysis of the mixture of products by gas

chromatography failed to reveal the presence of any 2, 3-dimethyl-2-butanol. The acid-catalyzed hydration of 3, 3-dimethyl-1-butene, by contrast, gives 2, 3-dimethyl-2-butanol as the major product.

A mechanism that accounts for the orientation of addition in the oxymercuration stage, and one that also explains the general lack of accompanying rearrangements. Central to this mechanism is an

electrophilic attack by the mercury species, gOAcH+

, at the less substituted carbon of the double bond (i.e., at the carbon atom that bears the greater number of hydrogen atoms), and the formation of a bridged intermediate.

Hydroboration : Synthesis of Alkylboranes Hydroboration of an alkene is the starting point for a

number of useful synthetic procedures, including the anti-Markovnikov syn hydration procedure. Hydroboration was discovered by Herbert C. Brown, and it can be represented in its simplest terms as follows :

C = C + H — B hydroboration — C — C —

H BAlkene Boron hydride

Alkylborane Hydroboration can be accomplished with diborane

(B2H6), which is a gaseous dimer of borane (BH3), or more conveniently with a reagent prepared by dissolving diborane in THF. When diborane is introduced to THF, it reacts to form a Lewis acid–base complex of borane (the Lewis acid) and THF. The complex is represented as BH3 : THF.

B2H6 + 2 O 2H – B – O

H

H

– +

Diborane THF (tetrahydrofuran)

BH3 : THF

Solutions containing the BH3 : THF complex can be obtained commercially. Hydroboration reactions are usually carried out in ethers : either in diethyl ether (CH3CH2)2O, or in some higher molecular weight ether such as “diglyme” [(CH3OCH2CH2)2O, diethylene glycol dimethyl ether]. Great care must be used in handling diborane and alkylboranes because they ignite spontaneously in air (with a green flame). The solution of BH3: THF must be used in an inert atmosphere (e.g., argon or nitrogen) and with care.

Stereochemistry of Hydroboration :

H B

HH

H — B

+

We can see the results of a syn addition in our examples involving the hydroboration of 1-methylcyclopentene ring :

CH3

H B

H

CH3

HH

syn additionanti-Markovnikov

+ enantiomer

+H

H – BH

Science Jokes A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"

1. 1 couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!".

2. Q. What is volume of a person who lost all his memory ? A. 1/3 π r2h

Because he keeps on saying, “main CONE hu!"


Oxygen : Oxygen occurs as two non-metallic forms, dioxygen

O2 and ozone O3. Dioxygen O2 is stable as a diatomic molecule, which accounts for it being a gas. The bonding in the O2 molecules is not as simple as it might at first appear. If the molecule had two covalent bonds, then all electron would be paired and the molecule should be diamagnetic.

O + O → O O or O = O Dioxygen is paramaginetic and therefore contains

unpaired electrons. The explanation of this phenomenon was one of the early successes of the molecular orbital theory.

Liquid dioxygen is pale blue in colour, and the solid is also blue. The colour arises from electronic transitions which excite the ground state (a triplet state) to a singlet state. This transition is 'forbidden' in gaseous dioxygen. In liquid or solid dioxygen a single photon may collide with two molecules simultaneously and promote both to excited states, absorbing red – yellow – green light, so O2 appears blue. The origin of the excited singlet states in O2 lies in the arrangement of electrons in the antibonding π*2py and π*2pz molecular orbitals, and is shown below.

Second excited state (electrons have opposite

spins

π*pyπ*pz

singlet

State 1Σg

+ Energy /kJ

157

First excited state (electrons

paired )

singlet 1∆g 92

Ground state (electrons have parallel spins)

triplet 3Σg– 0

Singlet O2 is excited, and is much more reactive than normal ground state triplet dioxygen. Singlet dioxygen can be generated photochemically by irradiating normal dioxygen in the presence of a sensitizer such as fluorescein, methylence blue or some polycyclic hydrocarbons. Singlet dioxygen can also be made chemically :

H2O2 + OCl– →EtOH O2(1∆g) + Η2Ο + Cl–

Singlet dioxygen can add to a diene molecule in the 1, 4 positions, rather like a Diels–Alder reaction. It may add 1, 2 to an alkene which can be cleaved into two carbonyl compounds.

CH – CH

CH2 CH2 + singlet O2 O – O

Singlet dioxygen may be involved in biological

oxidations. Ozone O3 is the triatomic allotrope of oxygen. It is

unstable, and decomposes to O2. The structure of O3 is angular, with an O – O – O bond angle of 116º48´. Both O – O bond lengths are 1.28 Å, which is intermediate between a single bond (1.48 Å in H2O2) and a double bond (1.21 Å in O2). The older valence bond representation as resonance hybrid now seldom used. The structure is described as the central O atom using sp2 hybrid orbitals to bond to the terminal O atoms. The central atom has one lone pair, and the terminal O atoms have two lone pairs. This leaves four electrons for π bonding. The pz atomic orbitals from the three atoms form three delocalized molecular orbitals covering all three atoms. One MO is bonding, one non-bonding, and one antibonding. The four π electron fill the bonding and non-bonding MOs and thus contribute one delocalized π bond to the molecule in addition to the two σ bonds. Thus the bond order is 1.5, and the π system is described as a four-electron three-centre bond.

Trioxides of Sulphur : The only important trioxide in this group, SO3, is

obtained by reaction of sulfur dioxide with molecular oxygen, a reaction that is thermodynamically very favorable but extremely slow in the absence of a catalyst. Platinum sponge, V2O5, and NO serve as catalysts under various conditions. Sulfur trioxide reacts vigorously with water to form sulfuric acid. Commercially, for practical reasons, SO3 is absorbed in concentrated sulfuric acid, to give oleum, which is then diluted. Sulfur trioxide is used as such for preparing sulfonated oils and alkyl arenesulfonate detergents. It is also a powerful but generally indiscriminate oxidizing agent; however, it will selectively oxidize pentachlorotoluene and similar compounds to the alcohol.

The free molecule, in the gas phase, has a planar, triangular structure that may be considered to be a

Inorganic Chemistry

Fundamentals

OXYGEN FAMILY & HYDROGEN FAMILY

KEY CONCEPT


resonance hybrid involving pπ – pπ S – O bonding, with additional π bonding via overlap of filled oxygen pπ orbitals with empty sulfur dπ orbitals, to account for the very short S – O distance of 1.41 Å

S

:O:

O O S

:O:

O S

:O:

O O O In view of this affinity of S in SO3 for electrons, it is

not surprising that SO3 functions as a fairly strong Lewis acid toward the bases that it does not preferentially oxidize. Thus the trioxide gives crystalline complexes with pyridine, trimethylamine, or dioxane, which can be used, like SO3 itself, as sulfonating agents for organic compounds.

The structure of solid SO3 is complex. At least three well-defined phase are known. γ-Sulfur trioxide, formed by condensation of vapors at – 80ºC or below, is an icelike solid containing cyclic trimers with structure.

S S

S O

O

O

O

O

O

O

O

O

A more stable, asbestos-like phase (β-SO3) has

infinite helical chains of linked SO4 tetrahedra and the most stable form, α-SO3, which also has an asbestos-like appearance, presumably has similar chains crosslinked into layers.

– S – O – S – O – S – O –

O

O

O

O

O

O Liquid γ-SO3, which is a monomer-trimer mixture,

can be stabilized by the addition of boric acid. In the pure state it is readily polymerized by traces of water.

Compounds of Sulphur and Nitrogen : A number of ring and chain compounds containing S

and N exist. The elements N and S are diagonally related in the periodic table, and have similar charge densities. Their electronegativities are close (N 3.0, S 2.5) so covalent bonding is expected. The compounds formed have unusual structures which cannot be explained by the usual bonding theories.

Attempting to work out oxidation states is unhelpful or misleading.

The best known is tetrasulphur tetranitride S4N4, and this is starting point for many other S – N compounds. S4N4 may be made as follows :

6SCl2 + 16NH3 → S4N4 + 2S + 14NH4Cl

6S2Cl2 + 16NH3 → 4CCl S4N4 + 8S + 12NH4Cl

6S2Cl2 + 4NH4Cl → S4N4 + 8S + 16HCl S S

NN NN

S S S4N4 is a solid, m.p. 178ºC, It is thermochromic, that

is it changes colour with temperature. At liquid nitorgen temperatures it is almost colourless, but at room temperature it is orange-yellow, and at 100ºC it is red. It is stable in air, but may detonate with shock, grinding or sudden heating. The structure is a heterocylic ring. This is cradle shaped and differs structurally from the S8 ring, which is crown shaped. The X-ray structure shows that the average S – N bond length is 1.62 Å. Since the sum of the covalent radii for S and N is 1.78 Å, the S – N bonds seem to have some double bond character. The fact that the bonds are of equal length suggest that this is delocalized. The S.....S distances at the top and bottom of the cradle are 2.58 Å. The van der waals (non-bonded) distance S ... S is 3.30 Å, and the single bond distance S – S is 2.08 Å.

This indicates weak S – S bonding, and S4N4 is thus a cage structure.

Many different sizes of rings exist, for example cyclo-S2N2, cyclo-S4N2, cyclo-S4N3Cl, cyclo-S3N3Cl3. In addition bicyclo compounds S11N2, S15N2, S16N2,S17N2 and S19N2 are known. The last four may be regarded as two heterocyclic S7N ring, with the N atoms joined through a chain of 1 – 5S atoms.

S4N4 is very slowly hydrolysed by water, but reacts rapidly with warm NaOH with the break-up of the ring:

S4N4 + 6NaOH + 3H2O → Na2S2O3 + 2Na2SO3 + 4NH3 If S4N4 is treated with Ag2F in CCl4 solution then

S4N4F4 is formed. This has an eight-membered S – N ring, with the F atoms bonded to S. This results from breaking the S – S bonds across the ring. Similarly the formation of adducts such as S4N4.BF3 or S4N4.SbF5 (in which the extra group is bonded to N) breaks the S – S bonds and increases the mean S – N distance from 1.62 Å to 1.68 Å. This is presumably because the electron attracting power of BF3 or SbF5 withdraws some of the π electron density.

Reduction of S4N4 with SNCl2 in MeOH gives tetrasulphur tetraimide S4(NH4). Several imides can be made by reacting S4N4 with S, or S2Cl2 with NH3. These imides are related to an S8 ring in which one or more S atoms have been replaced by imide NH groups, for example in S7NH, S6(NH)2, S5(NH)3 and S4(NH)4.


If S4N4 is vaporized under reduced pressure and passed through silver wool, then disulphur dinitrogen S2N2 is formed.

S4N4 + 4Ag → S2N2 + 2Ag2S + N2 S2N2 is a crystalline solid, which is insoluble in water

but soluble in many organic solvents. It explodes with shock or heat. The structure is cyclic and the four atoms are very nearly square planar.

The most important reaction of S2N2 is the slow polymerization of the solid or vapour to form polythiazyl (SN)x. This is a bronze coloured shiny solid that looks like a metal. It conducts electricity and conductivity increases as the temperature decreases, which is typical of a metal. It becomes a superconductor at 0.26 K. The crystal structure shows that the four-membered rings in S2N2 have opened and polymerized into a long chain polymer. The atoms have a zig-zag arrangement, and the chain is almost flat. Conductivity is much greater along the chains than in other directions, and so the polymer behaves as a one-dimensional metal. The resistivity is quite high at room temperature.

Ortho and Para Hydrogen : The hydrogen molecule H2 exists in two different

forms known as ortho and para hydrogen. The nucleus of an atom has nuclear spin, in a similar way to electrons having a spin. In the H2 molecule, the two nuclei may be spinning in either the same direction, or in opposite directions. This gives rise to spin isomerism, that is two different forms of H2 may exist. These are called ortho and para hydrogen. Spin isomerism is also found in other symmetrical molecules whose nuclei have spin momenta, e.g. D2, N2, F2, Cl2. There are considerable differences between the physical properties (e.g. boiling points, specific heats and thermal conductivities) of the ortho and para forms, because of differences in their internal energy. There are also difference in the band spectra of the ortho and para forms of H2.

The para form has the lower energy, and at absolute zero the gas contains 100% of the para form. As the temperature is raised, some of the para form changes into the ortho form. At high temperatures the gas contains about 75% ortho hydrogen.

Para hydrogen is usually prepared by passing a mixture of the two forms of hydrogen through a tube packed with charcoal cooled to liquid air temperature. Para hydrogen prepared in this way can be kept for weeks at room temperature in a glass vessel, because the ortho-para conversion is slow in the absence of catalysts. Suitable catalysts include activated charcoal, atomic hydrogen, metals such as Fe, Ni, Pt and W and paramagnetic substances or ions (which contain unpaired electrons) such as O2, NO, NO2, Co2+ and Cr2O3.

Brief description: nickel is found as a constituent in most meteorites and often serves as one of the criteria for distinguishing a meteorite from other minerals. Iron meteorites, or siderites, may contain iron alloyed with from 5 to nearly 20% nickel. The USA 5-cent coin (whose nickname is "nickel") contains just 25% nickel. Nickel is a silvery white metal that takes on a high polish. It is hard, malleable, ductile, somewhat ferromagnetic, and a fair conductor of heat and electricity. Nickel carbonyl, [Ni(CO)4], is an extremely toxic gas and exposure should not exceed 0.007 mg M-3.

Basic information Name: Nickel Symbol: Ni Atomic number: 28 Atomic weight: 58.6934 (2) Standard state: solid at 298 K Group in periodic table: 10 Group name: (none) Period in periodic table: 4 Block in periodic table: d-block Colour: lustrous, metallic, silvery tinge Classification: Metallic Small and large samples of nickel foil like this, as well as sheet, wire, mesh and rod (and nickel alloys in foil, sheet, wire, insulated wire and rod form) can be purchased from Advent Research Materials via their web catalogue. ISOLATION : Isolation: it is not normally necessary to make nickel in the laboratory as it is available readily commercially. Small amounts of pure nickel can be islated in the laborotory through the purification of crude nickel with carbon monoxide. The intermediate in this process is the highly toxic nickel tetracarbonyl, Ni(CO)4. The carbonyl decomposes on heating to about 250°C to form pure nickel powder.


1. A colourless salt (A), soluble in water, gives a mixture of three gases (B), (C) and (D) along with water vapours. Gas (B) is blue towards litmus paper, gas (C) red and gas (D) is neutral. Gas (B) is also obtained when (A) is heated with NaOH and gives brown ppt. with K2HgI4. Solution thus obtained gives white ppt. (E) with CaCl2 solution in presence of CH3COOH. Precipatete (E) decolorises −

4MnO /H+. Gas (C) turns lime water milky while gas (D) burns with blue flame and is fatal when inhaled. Identify (A) to (D) and explain chemical reactions.

Sol. Gas (B) gives brown ppt. with K2HgI4

⇒ gas (B) is NH3 ⇒ gas (A) has NH4+

(C) turns lime water milky ⇒ gas (C) can be SO2 or CO2 Gas (D) is also obtained along with (C). Gas (D)

burns with blue flame and is fatal when inhaled ⇒ gas (D) is CO ⇒ gas (C) is CO2 ⇒ (A) has C2O4

2–

It is confirmed by the fact that CaCl2 gives white ppt. CaC2O4(E) which decolourises MnO4

–/H+ ⇒ (A) is (NH4)2C2O4 Explanation :

(NH4)2 C2O4 →∆ 2NH3 + CO2 + CO + H2O (A) (B) (C) (D) (B) is blue towards litmus (basic) (C) is red toward litmus (acidic) (D) is neutral (NH4)2C2O4+2NaOH →∆ Na2C2O4+

)B(NH2 3 +2H2O

Na2C2O4+ CaCl2 → )E(.pptWhite

OCaC 42 ↓ + 2NaCl

NH3 + K2HgI4 →

O Hg

Hg NH2I

brown ppt (Iodide of Million’s base)

2MnO4– +16H++5C2O4

2– → 10CO2+2Mn2++ H2O violet colourless

2. (i) A blue compound (A) when heated, loses its water of crystallization and becomes white (B), When (B) absorbs moisture again becomes blue.

(ii) Aqueous solution of (A) gives white ppt. with BaCl2 solution and on reaction with K4[Fe(CN)6] gives a brown chocolate colour (C).

(iii) A reacts with KI to give I2 and a white ppt. (D) which dissolves in excess of KI to give a brown coloured complex salt (E).

(iv) (A) reacts with KCN to give a white ppt. (F) which dissolves in excess of KCN to give soluble (G).

(v) (A) reacts with NH4OH to give a pale blue ppt. (H) which dissolves in excess of NH4OH in presence of (NH4)2SO4 to give a deep blue colour (I).

(vi) (A) reacts with NaOH to give an insoluble ppt. which on boiling gives a black ppt (J). The black ppt. reacts with glucose to give a red ppt. (K).

What are (A) to (K) ? Give balanced equations for all the observations.

Sol. Observation (ii) shows that (A) contains sulphate (SO4

2–) ions because it gives white ppt. of BaSO4. Since (A) gives chocolate colour with K4[Fe(CN)6], hence it also contains Cu2+ ions, hence (A) is CuSO4.5H2O.

(i))A(Blue

24 OH5.CuSO →∆

White)B(4CuSO

MoistureOH5 2 →

Blue)A(24 OH5.CuSO

(ii) CuSO4 + BaCl2 → .pptWhite

4BaSO ↓ + CuCl2

2CuSO4 + K4[Fe(CN)6] → colour Chocolate )C(

62 ])CN(Fe[Cu + 2K2SO4

(iii) (A)

4CuSO + 2KI → Green

2CuI + K2SO4

2CuI2 → ppt.(D)

22ICu ↓ + I2

(D)

22ICu + 2KI → iodide cuprous Pot. (E)

2 ]CuI[K2

(iv) CuSO4 + 2KCN → unstable Yellow

2)CN(Cu + K2SO4

2Cu(CN)2 → ppt. white(F)

22 )CN(Cu ↓ + (CN)2 ↑

Cu2(CN)2 + 6KCN → Soluble (G)

43 ])CN(Cu[K2

(v) 2CuSO4 + 2NH4OH → ↓blue Pale (H)

24 )OH(Cu.CuSO

+ (NH4)2SO4 CuSO4.Cu(OH)2 + (NH4)2SO4 + 6NH4OH → 4

blue Deep (I)43 SO])NH(Cu[2 + 8H2O

(vi) CuSO4 + 2NaOH → Cu(OH)2 ↓ + Na2SO4 Cu(OH)2 →∆

.pptBlack)J(CuO + H2O

CH2OH(CHOH)4CHO + 2CuO →∆ CH2OH(CHOH)4COOH +

.pptdRe)K(2OCu ↓

UNDERSTANDINGInorganic Chemistry


3. A hydrated metallic salt A, light green in colour, gives a white anhydrous residue B after being heated gradually. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue and a mixture of two gases E and F. The gaseous mixture, when passed through acidified permanganate, discharge the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Identify A, B, C, D, E and F.

[IIT-1988] Sol. The given observations are as follows. (i)

)A(saltmetallicHydrated →heat

)B(residueanhydrouswhite

(ii) Aqueous solution of B →NO

)C(compoundbrowndark

(iii) Salt B heatingstrong →

)D(residueBrown +

)F()E(gasesTwo+

Gaseous mixture (E) + (F)

acidified KMnO4

BaCl2 solution

Pink colour is discharged

White precipitate(iv)

The observation (ii) shows that B must be ferrous sulphate since with NO, it gives dark brown compound according to the reaction

[Fe(H2O)6]2+ + NO → browndark

252 )]NO()OH(Fe[ + + H2O

Hence, the salt A must be FeSO4.7H2O The observation (iii) is 2FeSO4 →

brown)D(

32OFe + 43421

)F()E(32 SOSO

+

+

The gaseous mixture of SO2 and SO3 explains the observation (iv), namely,

colourpink

4MnO2 − + 5SO2 + 2H2O → colourno

2Mn2 + −+ 24SO5 + 4H+

2H2O + SO2 + SO3 4H+ + SO32– + SO4

2– Ba2+ + SO3

2– → pptwhite

3BaSO

Ba2+ + SO42– →

pptwhite4BaSO

Hence, the various compounds are (A) FeSO4.7H2O (B) FeSO4 (C) [Fe(H2O)5NO]SO4 (D) Fe2O3 (E) and (F) SO2 and SO3

4. Compound (A) is a light green crystalline solid. It gives the following tests :

(i) It dissolves in dilute H2SO4 without evolving any gas.

(ii) A drop of KMnO4 is added to the above solution. The pink colour disappears.

(iii) Compound (A) is heated strongly. Gases (B) and (C) with pungent smell came out. A brown residue (D) is left behind.

(iv) The gas mixture (B) and (C) is passed into dichromate solution. The solution turns green.

(v) The green solution from step (iv) gives a white ppt. (E) with a solution of Ba(NO3)2.

(vi) Residue (D) from (v) is heated on charcoal in reducing flame. It gives a magnetic substance. Identify compounds (A) to (E) and predict all the equations. [IIT-1980]

Sol. The fore said observations may be briefly summarised as follows :

(a) solidgreen Light

A → 42SOH.Dil )A(

ofSolution

→ 4KMnO Pink colour disappears

(b) A →∆ gas

smelling PungentCB + +

residueBrownD

(c) Solution

722 OCrK → + CB Green solution

→ 23 )NO(Ba .pptWhite

E

(d) residueBrown

D∆

→ coalChar a magnetic substance

From the last step, one may conclude that brown residue (D) (hence also compound (A)) must be a salt of iron. Since (A) decolourises KMnO4 solution hence it should be a salt of Fe (II). The reactions involved are given below.

MnO4– + 8H+ + 5e– → Mn2+ + 4H2O

[Fe2+ → Fe3+ + e–] × 5

Pink4MnO− +

Green

2Fe5 + + 8H+ → Colourless

2Mn + + 5Fe3+ + 4H2O

From observations of (b) and (c), one concludes that compound (A) should be FeSO4 as on heating, it gives pungent gases SO2 and SO3.

)A(

4FeSO2 →∆

)Brown()D(

32OFe + )B(2SO +

)C(3SO

SO2, gas turns dichromate solution green due to formation of green coloured sulphate of chromium (III), the different equations are,

Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O

[SO2 + 2H2O → 4H+ + SO42– + 2e–] × 3

Cr2O72– + 3SO2 + 2H+ → 2Cr3+ + 3SO4

2– + H2O White ppt. (E) is of BaSO4

Green342 )SO(Cr + 3Ba(NO3)2 →

)E(4BaSO3 ↓ + 2Cr(NO3)3

Hence, (A) is FeSO4 (B) is SO2 (C) is SO3 (D) is Fe2O3 and (E) is BaSO4


5. A substance (X) is soluble in conc. HCl. When to this solution NaOH solution is added, a white precipitate is produced. This precipitate dissolves in excess of NaOH solution giving a strongly reducing solution. Heating of (X) with sulphur gives a brown powder (Y) which is soluble in warm yellow ammonium sulphide solution. When HCl is added to the latter, a grey precipitate is produced. Heating of (X) in air gives a water soluble compound gives white gelatinous precipitate. Identify the compounds giving the reactions involved.

Sol. The reaction sequence is as follows :

(1) X → HCl.Conc Dissolves →NaOH White ppt.

NaOHExcess →

Dissolves and the solution is strongly reducing.

(2) X ∆

→S

BrownY → x24 S)NH( Dissolves

→HCl Grey ppt.

(3) X HeatO2→

42SOH conc.in Soluble

ZFused

NaOH → Soluble →+H

White gelatinous ppt. (a) According to step (2), (X) appears to be tin.

)X(Sn + 2S →

)Y(2SnS

SnS2 + (NH4)2Sx → tethiostanna.Amm324 SnS)NH(

(NH4)2SnS3 + 2HCl → .pptGrey

2SnS ↓ + 2NH4Cl + H2S

(b) Step (1) and (3) can also be explained, if (X) is tin

)X(Sn + 2HCl → SnCl2 + H2

SnCl2 + 2NaOH → Sn(OH)2 + 2NaCl SnCl2 + 2NaOH →

reducingStrongly 22SnONa + 2HCl

(c) Sn + O2 →Fuse (Z)

2SnO

SnO2 + 2NaOH →Fuse Na2SnO3 + H2O

Na2SnO3 + 2HCl → ppt. gelatinous White

32SnOH ↓ + 2NaCl

Brief description: cobalt is a brittle, hard, transition metal with magnetic properties similar to those of iron. Cobalt is present in meteorites. Ore deposits are found in Zaire, Morocco and Canada. Cobalt-60 (60Co) is an artificially produced isotope used as a source of γ rays (high energy radiation). Cobalt salts colour glass a beautiful deep blue colour.

Basic information about and classifications of cobalt :

Name : Cobalt Symbol : Co Atomic number : 27 Atomic weight : 58.933195 (5) Standard state : solid at 298 K Group in periodic table : 9 Group name : (none) Period in periodic table : 4 Block in periodic table: d-block Colour : lustrous, metallic, greyish tinge Classification : Metallic Marmite, which we all eat here in England and which is what makes us English, is a source of vitamin B12, actually a compound containing cobalt. The equivalent, but altogether blander, in Australia is Vegemite. Marmite is available in the USA. Try mixing it with peanut butter.

This sample is from The Elements Collection, an attractive and safely packaged collection of the 92 naturally occurring elements that is available for sale.

ISOLATION :

Isolation: it is not normally necessary to make cobalt in the laboratory as it is available readily commercially. Many ores contain cobalt but not many are of economic importance. These include the sulphides and arsenides linnaeite, Co3S4, cobaltite, CoAsS, and smaltite, CoAs2. Industrially, however, it is normally produced as a byproduct from the produstion of copper, nickel, and lead.


1. Show that the conics through the intersection of two

rectangular hyperbolas are also rectangular hyperbolas. If A, B, C & D be the four points of intersection of these two rectangular hyperbolas, then find the orthocentre of the triangle ABC.

2. Find the area of a right angle triangle if it is known that the radius of circle inscribed in the triangle is r and that of the circumscribed circle is R.

3. Q is any point on the line x = a. If A is the point (a, 0) and QR, the bisector of the angle OQA, meets OX in R, then prove that the locus of the foot of the perpendicular from R to OQ has the equation

(x – 2a) (x2 + y2) + a2x = 0

4. Show that the equation z4 + 2z3 + 3z2 + 4z + 5 = 0 with (z ∈ C) have no

purely real as well as purely imaginary root.

5. Prove that

x

xnax

xaf

0

l

+∫

∞

dx = lnax

dxax

xaf

0

+∫

∞

6. A straight line moves so that the product of the perpendiculars on it form two fixed points is a constant. Prove that the locus of the foot of the perpendiculars from each of these points upon the straight line is a circle, the same for each.

7. Prove the identity :

∫ −x

0

zzx 2e dz = ∫ −

x

0

4/z4x

22

ee dz, deriving for the

function f(x) = ∫ −x

0

zzx 2e dz a differential equation and

solving it. 8. Let α, β be the roots of a quadratic equation, such

that αβ = 4 and 1−α

α + 1−β

β = 4a7a

2

2

−− Find the set

of values of a for which α, β ∈ (1, 4)

9. Investigate the function f(x) = x5/3 – 5x2/3 for points of extremum and find the values of k such that the equation x5/3 – 5x2/3 = k has exactly one positive root.

10. Let A = 1, 2, 3, ....., 100. If X is a subset of A containing exactly 50 elements then show that

∑∈xp

minp = 101C51.

`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 wil l be published in next issue

5Set

Interesting Facts

• Fish have no eyelids. They can't blink, wink or close their eyes to sleep.

• For centuries salt makers used conical wicker basket moulds called barrows to make their 'lumps' (like bricks only bigger!), which then had to be crushed before the salt could be used. The word 'lump' has passed into the English language. Workers had to 'lump' the salt and their job was known as 'lumping'.

• The first spacecraft to visit Venus was Mariner 2 in 1962. It was subsequently visited by many others (more than 20 in all). Including Pioneer Venus and the Soviet Venera 7 - the first spacecraft to land on another planet - and Venera 9 which returned the first photographs of the surface.

• Adults lose nearly one percent of their natural ability to mend genetic damage with each year that passes. The older you are, then, the less able your system is to fix the cell errors that lead to cancer. It has also been found that young people with skin cancer have the repair capacity of people 30 years older.

• The tail section of an airplane gives the bumpiest ride.

• The plague in Zurich killed 3,700 of the cities 6,000 inhabitants in 1567.


1. Let x2 + y2 = a2 ....(1) and x2 + y2 + 2gx + c = 0 ....(2) are two circles they cut orthogonally. Hence c – a2 = 0 so from (2) x2 + y2 + 2gx + a2 = 0 ...(3) Let P (a cos α, a sin α) be any point on 1st circle. It’s

polar w.r.t. 2nd circle is a x cos α + a y sin α + g(x + a cos α) + a2 = 0 ....(4) other end of diameter of 1st circle through P is Q(– a cos α, – a sin α) This satisfies eqn. ..(4) Hence proved. 2. Let A be at origin & position vectors of B, C, D are ,

br

, cr

& dr

respectively. Perpendicular from B and C to the faces ACD and ABD meet at H with position vector h

r,

A

B

C

D

dc b

so BH ⊥ AC ⇒ c).bh(

rrr− = 0 ⇒ c.h

rr = c.b

rr

BH ⊥ AD ⇒ d).bh(rrr

− = 0 ⇒ d.hrr

= d.brr

CH ⊥ AB ⇒ b).ch(rrr

− = 0 ⇒ b.hrr

= b.crr

and CH ⊥ AD ⇒ d).ch(rrr

− = 0 ⇒ d.hrr

= b.crr

so d.brr

= d.crr

⇒ )cb(rr

− . dr

= 0 ...(1) so BC ⊥ AD , proved. Now, let any point M (with position vector m ) be

on BCD such that AM ⊥ plane BCD then . m . )bc(

rr− = 0 = m . )cd(

rr−

so m . br

= m . cr

= m . dr

...(2) Now, let P be any point on AM with position vector

t m such that DP is perpendicular to ABC, then (t m – d

r). b

r= 0 = (t m – d

r). c

r so t m . b

r= b

r. d

r &

t m . cr

= dr

. cr

which are same equations using (1) and (2) in them. So such a scalar t can be obtained as

t = b.m

d.br

rr

3. I (u) = ∫π

+−0

2 dx)uxcosu21(nl ;

I (– u) = ∫π

++0

2 dx)uxcosu21(nl

Use ∫a

0

dx)x(f = ∫ −a

0

dx)xa(f

I (u) = I (–u) I (u) + I (– u)

= ∫π

+++−0

22 dx)uxcosu21()uxcosu21(nl

= ∫π

−−0

2222 dx]xcosu4)u1[(nl

= ]xcosu4u2u1[n 2224

0

−++∫π

l dx

= ]ux2cosu21[n 42

0

+−∫π

l dx

Now let 2x = y

⇒ Ι (u) + I (– u) = ]uycosu21[n21 42

2

0

+−∫π

l dy

=21 I (u2) + ]uycosu21[n

21 42

2

0

+−∫π

l dy

Now let y = 2π – t

I (u) + I (– u) =21 I (u2) + ]utcosu21[n

21 42

0

+−∫π

l (– dt)

=21 I (u2) +

21 I (u2)

2I (u) = I (u2) as I(u) = I (– u) (or using f (2a– x) = f (x) Prop).

so I (u) =21 I(u2)

similarly find ]uxcosu21[n3 22

0

+−∫π

l dx

& show I (u) = I (– u) = 21 I (u2) = )u(I

21 n2n

MATHEMATICAL CHALLENGES SOLUTION FOR AUGUST ISSUE (SET # 4)


4. Let f(x) = (x – α) (x – β) so f(n) f (n + 1) = (n – α) (n – β) (n + 1 – α)(n + 1– β) = (n – α) (n + 1 – β) (n – β) (n + 1 – α) = [n (n + 1) – n(α + β) – α + αβ] [n(n + 1) – n (α + β) – β + αβ] = [n (n + 1) + na + b – α] [n (n + 1) + an + b – β] = (m – α) (m – β) ; let m = n(n + 1) + an + b = f (m) 5. Let xy = c2 be rectangular hyperbola. Let A (ct1, c/t1) and b (ct2, c/t2) be two fixed points on

it. and P (ct, c/t) be any variable point. Line AP : x + y t1t = c (t1 + t) Line BP : x + y t2 t = c (t2 + t) These lines intersect with x- axis at M (c(t1 + t), 0)

and N(c(t2 + t), 0). Length MN = |c (t1 – t2) | which is a constant.

Similarly intercept on y- axis can be obtained as

−

21 t1

t1c . Hence proved.

6. xy

dyxdxy − + 2

22

)yx(dxydyx

−− = 0

⇒ 2yx)dyxdxy(y − +

2

22

)yx(xydyxydydxydxxydxxydyx

−+−−−+

⇒ xy . d(x/y) +

2)yx()dydx(xy)dyxydx(y)dxydyx(x

−−−+−+ = 0

⇒ y/x

)y/x(d + 2)yx()yx(dxy)xy(dy)xy(d.x

−−−− = 0

⇒ y/x

)y/x(d + 2)yx()yx(dxy)xy(d)yx(

−−−− = 0

⇒ y/x

)y/x(d + d

− yxyx

= 0

⇒ ln (x/y) + yx

yx−

= c

7. Let 2

2

ax + 2

2

by = 1 be the ellipse and

y = m1 (x – ae) and y = m2 (x – ae) are two chords through its focus (ae, 0). Any conic through the extremities of these chords can be defined as y – m1 (x – ae) y – m2 (x – ae) +

λ

−+ 1

by

ax

2

2

2

2 = 0 ...(1)

It it passes through origin, then m1m2 a2 e2 – λ = 0 ...(2) Solving (1) with x- axis

m1m2 (x – a e)2 + λ

−1

ax

2

2= 0

using (2) in it (x – ae)2 + e2 (x2 – a2 ) = 0 (1 + e2) x2 – 2ae x = 0

x = 0 & x = 2e1ae2

+

so other point on x- axis through which this conic

passes is

+0,

e1ae2

2 which is a fixed point.

Hence proved. 8. Let x = c ∈ R

f ′ (c±) = 0h

lim→ h

)c(f)hc(f±

−±

= 0h

lim→ h

)c(fch1cf

±

−

±

; c ≠ 0

= 0h

lim→ h

)c(f)c/h1(f2

)c2(f

±

−+

= 0h

lim→ h2

)c(f2ch1f)c2(f

±

−

±

= 0h

lim→ h2

)1(f)c2(fch1f)c2(f

±

−

±

;

(using x = 2c & y = 1 in

2xyf =

2)y(f).x(f )

= f (2c) 0h

lim→ c.

ch2

)1(fch1f

±

−

±

= c2

)1('f)c2(f

= c2

)1('f)c2(f ; as given f′(1) = f(1)

f ′ (c ±) = c2

)c(fx

So f(x) is differentiable for ∀ x ∈ R except x = 0

Now f′(x) = x

)x(f

⇒ )x(f)x('f =

x1

so ln f(x) = ln x + ln c ⇒ f(x) = cx

Now as f

2xy =

2)y(f)x(f


let y = 1 in it 2. f(x/2) = f(x) (f(1)

22xc =cx f(1)

so f(1) = 1 as c ≠ 0 so f(x) = x 9. Let i lines are there, no two of which are parallel and

no three of which are coincident. Introduction of (i + 1)th line will introduce (i + 1) new parts. Let Pi denotes the number of parts in which plane is being divided by i lines, then

Pi + 1 = Pi + (i + 1)

Pi + 1–Pi = i + 1 using i = 1, 2, 3, ......, n – 1 P2 – P1 = 2, P1 – P2 = 3 M M Pn – Pn–1 = n Add these equation Pn – Pi = 2 + 3 +.....+ n Pn = Pi + 2 + 3 + ....+ n

= 2 + 2 + 3+ .....+ n = 1 + 2

)1n(n +

Pn = 21 (n2 + n + 2)

10. 21 tan x +

x3cosxsin +

x9cosx3sin +

x27cosx9sin =

21 tan 27 x

L.H.S. : Consider on

21 tan x +

x3cosxsin =

21

xcosxsin +

x3cosxsin

= x3cosxcos2

xcosxsin2x3cosxsin +

= x3cosxcos4

x2sin2x3cosxsin2 +

= x3cosxcos4

x2sin2x2sinx4sin +−

= x3cosxcos4x2sinx4sin +

= x3cosxcos4xcosx3sin2 =

21 tan 3x

similarly 21 tan 3x +

x9cosx3sin =

21 tan 9x

and 21 tan 9x +

x27cosx9sin =

21 tan 27

on adding all these we get

x3cosx3sin +

x9cosx3sin +

x27cosx9sin =

21 (tan 27x – tan x)

Proved.

Brief Description : pure vanadium is a greyish silvery metal, and is soft and ductile. It has good corrosion resistance to alkalis, sulphuric acid, hydrochloric acid, and salt waters. The metal oxidizes readily above 660°C to form V2O5. Industrially, most vanadium produced is used as an additive to improve steels.

Table : basic information about and classifications of vanadium

• Name : Vanadium

• Symbol : V

• Atomic number : 23

• Atomic weight : 50.9415 (1)

• Standard state : solid at 298 K

• up in periodic table : 5

• Group name : (none)

• Period in periodic table : 4

• Block in periodic table : d-block

• Colour : silvery grey metallic

• Classification : Metallic

Isolation: vanadium is available commercially and production of a sample in the laboratory is not normally required. Commercially, routes leading to matallic vanadium as main product are not usually required as enough is produced as byproduct in other processes.

In industry, heating of vanadium ore or residues from other processes with salt, NaCl, or sodium carbonate, Na2CO3, at about 850°C gives sodium vanadate, NaVO3. This is dissolved in water and acidified to give a red solid which in turn is melted to form a crude form of vanadium pentoxide, "V2O5". Reduction of vanadium pentoxide with calcium, Ca, gives pure vanadium. An alternative suitable for small scales is the reduction of vanadium pentachloride, VCl5, with hydrogen, H2, or magnesium, Mg. Many other methodsare also in use.

Industrially, most vanadium is used as an additive to improve steels. Rather than proceed via pure vanadium metal it is often sufficient to react the crude of vanadium pentoxide, "V2O5", with crude iron. This produces ferrovanadium suitable for further work.



1. Let f(x) satisfies the differential equation

x . dx

)x(df + f(x) = g(x) where f(x) and g(x) are

continuous functions and f(x) is a decreasing function

∀ x > 0. Prove that x . g(x) < ∫x

0)x(g dx ∀ x > 0.

Sol. Given that x . dx

)x(df + f(x) = g(x)

⇒ dx

)x(df = x

)x(f)x(g − < 0

⇒ g(x) < f(x) ∀ x > 0 ...(1) Again x . df(x) + f(x) . dx = g(x) dx

⇒ ∫ +x

0)x(fx(d ) = ∫

x

0)x(g dx

⇒ x . f(x) = ∫x

0)x(g dx

f(x) = ∫x

0)x(g

x1 dx

⇒ g(x) < ∫x

0)x(g

x1 dx (using eqn(1))

⇒ x . g(x) < ∫x

0)x(g dx ∀ x > 0

2. Let P(x) be a polynomial of degree n such that

P(i) = 1i

i+

for i = 0, 1, 2 ..... n. If n is odd than find

the value of P(n + 1). Sol. Let Q(x) = (x + 1) P(x) – x clearly Q(x) is polynomial of degree n + 1. Also

Q(i) = (i + 1) 1i

i+

– i = 0 for i = 1, 2, 3 .....n

Thus we can assume Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a

constant. Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n) 1 = (–1)n + 1 k(n + 1) !

⇒ k = !)1n(

1+

(Q n is odd)

Thus P(x) =

+

+−−−

+x

!)1x()nx)....(2x)(1x(x

1x1 ,

where n is odd ∴ P(n + 1) = 1

3. Evaluate : ∑=

n

0rr

n C (r – nx)2 xr (1 – x)n – r;

x ≠ 0, 1, n ∈ N > 2

Sol. ∑=

n

0rr

n C (r – nx)2 xr (1 – x)n – r

= ∑=

n

0rr

n C (r2 + n2x2 – 2nxr)r

x1x

−(1 – x)n

= (1 – x)n

−∑=

rn

0rr

n2

x1xCr

+ n2x2 rn

0rr

n

x1xC

−∑=

– 2nx

−∑=

rn

0rr

n

x1xCr

we know that ∑=

n

0r

rr

n yC = (1 + y)r ...(i)

⇒ ∑=

−

n

0r

r

rn

x1xC =

n

x1x1

−+ = (1 – x)–n

Differentiating (i) w.r.t. y we get

⇒ ∑=

n

0rr

n C.r yr–1 = n(1 + y)n – 1

∴ ∑=

n

0rr

n C.r yr = ny(1 + y)n–1 ...(ii)

⇒ ∑=

n

0rr

n C.rr

x1x

−= n

1n

x1x1

x1x −

−+

−

= nx(1 – x)–n Differentiating (ii) w.r.t. y we get

⇒ ∑=

n

0rr

n2 C.r yr–1 = n(1 + y)n – 2 y(n – 1) + (1 + y

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

Students' ForumMATHS


= ny(1 + y)n – 2(ny + 1)

⇒ rn

0rr

n2

x1xC.r

−∑=

= nx1

x−

2n

x1x1

−

−+

+

−1

x1nx

= nx(nx + 1 – x) (1 – x)–n Given sum is equal to (1 – x)n nx(nx + 1 – x) (1 – x)–n + n2x2(1 – x)–n – 2nx . nx (1 – x)–n = nx(nx + 1 – x) + n2x2 – 2n2x2 = nx(1 – x)

4. For any complex no. z, Prove that following inequalities.

(i) 1|z|

z− ≤ arg (z)

(ii) |z – 1| ≤ ||z| – 1| + |z| |arg (z)| Sol. (i) Ist method : Let z = |z| (cos θ + i sin θ) or |z|eiθ where θ = arg (z) L.H.S. = |cos θ + i sin θ – 1|

= 2

cos.2

sin2i2

sin2 2 θθ+

θ−

= 2

sin2 θ2

cosi2

sin θ−

θ ≤ 22θ = θ = arg (z)

(Q sin x ≤ x for x > 0)

IInd method : |z|

z complex no. lies on a circle with

unit radius i.e. x2 + y2 = 1

θ Q (1, 0)

P

|z|

z

QP = |z|

z – 1

| QP | ≤ length of the arc QP

1|z|

z− ≤ 1 arg (z)

Hence proved (ii) Ist method : L.H.S. = |z – 1| = |z – |z| + |z| – 1| By using ∆ inequality |z – 1| ≤ |z – |z|| + | |z| – 1|

≤ | |z| – 1| + |z| 1|z|

z−

≤ | |z| – 1| + |z| arg (z) [by using (i)] IInd method : Let p(z) is any point on the circle with

radius |z|.

θ R Q (1, 0)

P(z)

O

In ∆PQR, use ∆ PQR, use ∆ inequality PQ ≤ PR + QR |z – 1| ≤ |z| arg (z) + ||z| – 1| Note : you can take Q point out side the circle, too.

5. Find the least value of n for which (n – 2)x2 + 8x + n + 4 > sin–1(sin 12)

+ cos–1(cos 12) ∀ x ∈R when n∈ N. Sol. we have sin–1(sin 12) + cos–1(cos 12)

= sin–1(sin (4π – (4π – 12)))

+ cos–1(cos (4π – (4π – 12)))

= –(4π – 12) + 4π – 12 = 0

So that, (n – 2)x2 + 8x + n + 4 > 0 ∀ x ∈ R

⇒ n – 2 > 0 ⇒ n ≥ 3 and 82 – 4(n – 2) (n + 4) < 0

or n2 + 2n – 24 > 0 ⇒ n > 4 ⇒ n ≥ 5 ⇒ n = 5

6. Suppose a function f(x) satisfies the following

conditions f(x + y) = )y(f).x(f1

)y(f)x(f+

+ ∀ x, y and

f´(0) = 1. Also – 1 < f(x) < 1 for all x ∈R, then find the set of values of x where f(x) is differentiable and also find the value of

∞→xlim [f(x)]x.

Sol. First put x = 0, y = 0 ⇒ f(0) = 0

Now, f´(x) = h

)x(f)hx(flim0x

−+→

= h

)x(f)h(f).x(f1

)h(f)x(f

lim0x

−+

+

→

=

+−−

−−

→ )h(f).x(f1)x(f1

0h)0(f)h(flim

2

0x = 1 – f2(x)

integrating we get 21 ln

−+

)x(f1)x(f1 = x + c

⇒ f(x) = xx

xx

eeee

−

−

+−

clearly f(x) is differentiable for all x ∈R.

x

x)]x(f[lim

∞→ =

x

xx

xx

x eeeelim

+

−−

−

∞→

=

+

−−

−

∞→ xx

xx

x eeeelim

e x = 1


Some Definitions : Experiment : A operation which can produce some

well defined outcomes is known as an experiment. Random experiment : If in each trail of an

experiment conducted under identical conditions, the outcome is not unique, then such an experiment is called a random experiment.

Sample space : The set of all possible outcomes in an experiment is called a sample space. For example, in a throw of dice, the sample space is 1, 2, 3, 4, 5, 6. Each element of a sample space is called a sample point.

Event : An event is a subset of a sample space. Simple event : An event containing only a single

sample point is called an elementary or simple event. Events other than elementary are called composite or compound or mixed events.

For example, in a single toss of coin, the event of getting a head is a simple event.

Here S = H, T and E = H In a simultaneous toss of two coins, the event of

getting at least one head is a compound event. Here S = HH, HT, TH, TT and E = HH, HT, TH Equally likely events : The given events are said to

be equally likely, if none of them is expected to occur in preference to the other.

Mutually exclusive events : If two or more events have no point in common, the events are said to be mutually exclusive. Thus E1 and E2 are mutually exclusive in E1 ∩ E2 = φ.

The events which are not mutually exclusive are known as compatible events.

Exhaustive events : A set of events is said to be totally exhaustive (simply exhaustive), if no event out side this set occurs and at least one of these event must happen as a result of an experiment.

Independent and dependent events : If there are events in which the occurrence of one does not depend upon the occurrence of the other, such events are known as independent events. On the other hand, if occurrence of one depend upon other, such events are known as dependent events.

Probability : In a random experiment, let S be the sample space

and E ⊆ S, then E is an event. The probability of occurrence of event E is defined as

P(E) = Sin element distinct ofnumber Ein elementsdistinct ofnumber =

n(S)n(E)

= outcomes possible all ofnumber

E of occurrence tofavourable outocomes ofnumber

Notations : Let A and B be two events, then A ∪ B or A + B stands for the occurrence of at

least one of A and B. A ∩ B or AB stands for the simultaneous

occurrence of A and B. A´ ∩ B´ stands for the non-occurrence of both A

and B. A ⊆ B stands for "the occurrence of A implies

occurrence of B". Random variable : A random variable is a real valued function whose

domain is the sample space of a random experiment. Bay’s rule : Let (Hj) be mutually exclusive events such that

P(Hj) > 0 for j = 1, 2, ..... n and S = Un

1jjH

=. Let A be

an events with P(A) > 0, then for j = 1, 2, .... , n

P

AH j =

∑=

n

1kkk

jj

)H/A(P)H(P

)H/A(P)H(P

Binomial Distribution : If the probability of happening of an event in a single

trial of an experiment be p, then the probability of happening of that event r times in n trials will be nCr pr (1 – p)n – r.

Some important results :

(A) P(A) = cases ofnumber Total

Aevent tofavourable cases ofNumber

= n(S)n(A)

PROBABILITY Mathematics Fundamentals M

ATHS


)AP( = cases ofnumber Total

Aevent tofavourablenot cases ofNumber

= n(S)

)An(

(B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E.

Thus odds in favour of an event E

= cases leunfavourab ofNumber

cases favourable ofNumber = ba

Similarly, odds against an event E

= cases favorable ofNumber

cases leunfavourab ofNumber = ab

Note : If odds in favour of an event are a : b, then the

probability of the occurrence of that event is

baa+

and the probability of non-occurrence of

that event is ba

b+

.

If odds against an event are a : b, then the probability of the occurrence of that event is

bab+

and the probability of non-occurrence of

that event is ba

a+

.

(C) P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1 P(φ) = 0 P(S) = 1 If S = A1, A2, ..... An, then P(A1) + P(A2) + .... + P(An) = 1 If the probability of happening of an event in one

trial be p, then the probability of successive happening of that event in r trials is pr.

(D) If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or

P(A + B) = P(A) + P(B) If A and B are any two events, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B) If the probabilities of happening of n independent

events be p1, p2, ...... , pn respectively, then

(i) Probability of happening none of them = (1 – p1) (1 – p2) ........ (1 – pn) (ii) Probability of happening at least one of them = 1 – (1 – p1) (1 – p2) ....... (1 – pn) (iii) Probability of happening of first event and not

happening of the remaining = p1(1 – p2) (1 – p3) ....... (1 – pn) If A and B are any two events, then

P(A ∩ B) = P(A) . P

AB or

P(AB) = P(A) . P

AB

Where P

AB is known as conditional probability

means probability of B when A has occured. Difference between mutually exclusiveness and

independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments.

(E) P(A A ) = 0

P(AB) + P( AB ) = 1

P( A B) = P(B) – P(AB)

P(A B ) = P(A) – P(AB)

P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards : Coins : A coin has a head side and a tail side. If

an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated.

Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated.

Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards.

In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack.


Binomial Theorem (For a positive Integral Index) : If n is a positive integer and x, a are two real or

complex quantities, then (x + a)n = nC0 xn + nC1xn – 1 a + nC2 xn – 2 a2 + ... + nCrxn–r ar + .... + nCn – 1x an–1 + nCnan ..(1) The coefficient nC0, nC1, ......, nCn are called binomial

coefficients. Properties of Binomial Expansion : There are (n + 1) terms in the expansion of

(x + a)n, n being a positive integer. In any term of expansion (1), the sum of the

exponents of x and a is always constant = n. The binomial coefficients of term equidistant

from the beginning and the end are equal, i.e. nCr = nCn – r (0 ≤ r ≤ n).

The general term of the expansion is (r + 1)th term usually denoted by Tr + 1 = nCr xn – r ar (0 ≤ r ≤ n).

The middle term in the expansion of (x + a)n (a) If n is even then there is just one middle term, i.e.

th

12n

+ term.

(b) if n is odd, then there are two middle terms, i.e. th

12n

+ term and

th

23n

+ term.

The greatest term in the expansion of (x + a)n, x, a ∈ R and x, a > 0 can be obtained as below :

∴ r

1r

TT + =

xa

r1rn +−

or r

1r

TT + – 1 =

rx)xa(ra)1n( +−+

=

−

+++ r

xaa)1n(

rx)xa( =

rxxa + |k – r|,

where k = xa

a)1n(++

Now, suppose that

(i) k = xa

a)1n(++ is an integer. We have

Tr + 1 > Tr ⇔ r

1r

TT + > 1 ⇔ r < k (i.e. 1 ≤ r < k)

Along , Tr + 1 = Tr ⇔ r

1r

TT + = 1 ⇔ r = k,

i.e. Tk + 1 = Tk > Tk–1 > .... > T3 > T2 > T1 In this case there are two greatest terms Tk and

Tk+1.

(ii) k = xa

a)1n(++ is not an integer. Let [k] be the

greatest integer in k. We have

Tr+1 > Tr ⇔ r

1r

TT + ⇔ r < k = [k] + (fraction)

⇔ r ≤ [k] i.e. T1 < T2 < T3 < ..... < T[k] – 1 < T[k] < T[k] + 1 In this case there is exactly one greatest term viz.

([k] + 1)th term. Term independent of x in the expansion of

(x + a)n – Let Tr + 1 be the term independent of x. Equate to zero the index of x and you will find the value of r.

The number of term in the expansion of

(x + y + z)n is 2

)2n)(1n( ++ , where n is a positive

integer. Pascal Triangle In(x + a)n when expanded the various coefficients

which occur are nC0, nC1, nC2, .... The Pascal triangle gives the values of these coefficients for n = 0, 1, 2, 3, 4, 5, ....

n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 n = 6 1 6 15 20 15 6 1 n = 7 1 7 21 35 35 21 7 1

n = 8 1 8 28 56 70 56 28 8 1

BINOMIAL THEOREM Mathematics Fundamentals M

ATHS


Rule : It is to be noted that the first and least terms in each row is 1. The terms equidistant from the beginning and end are equal. Any number in any row is obtained by adding the two numbers in the preceding row which are just at the left and just at the right of the given number, e.g. the number 21 in the row for n = 7 is the sum of 6 (left) and 15 (right) which occur in the preceding row for n = 6.

Important Cases of Binomial Expansion : If we put x = 1 in (1), we get (1 + a)n = nC0 + nC1a + nC2a2 + .................. + nCrar + ........... nCnan ...(2) If we put x = 1 and replace a by – a, we get (1 – a)n = nC0 – nC1a + nC2a2 – .................. + (–1)r nCrar + .... + (–1)n nCnan ...(3) Adding and subtracting (2) and (3), and then

dividing by 2, we get

21 (1 + a)n + (1 – a)n = nC0 + nC2a2

+ nC4a4 + .... ...(4)

21 (1 + a)n – (1 – a)n = nC1a + nC3a3

+ nC5a5 + ...... ...(5) Properties of Binomial Coefficients : If we put a = 1 in (2) and (3), we get 2n = nC0 + nC2 + ...... + nCr + ....+ nCr + ..... nCn–1 + nCn and 0 = nC0 – nC1 + nC2 – ...... + ...... + (–1)n nCn

∴ nC0 + nC2 + .... = nC1 + nC3 + .... = 21 [2n ± 0]

= 2n – 1 ...(6) Due to convenience usually written as C0 + C2 + C4 + ... = C1 + C3 + C5 + .... = 2n – 1 and C0 + C1 + C2 + C3 + ...... + Cn = 2n

Where nCr ≡ Cr = !)rn(!r

!n−

= !r

)1rn)...(2n)(1n(n +−−−

Some other properties to remember : C1 + 2C2 + 3C3 + ... + nCn = n . 2n – 1 C1 – 2C2 + 3C3 – .... = 0 C0 + 2C1 + 3C2 + ... + (n + 1)Cn = (n + 2) 2n –1

C0Cr + C1Cr+1 + ... + Cn – rCn = !)rn(.!)rn(

!)n2(+−

C02 + C1

2 + C22 + .... + Cn

2 = 2)!n(!)n2(

C02 – C1

2 + C22 – C3

2 + ...

=

evenisnif,C.(–1)odd isn if,0

2/nnn/2

Binomial Theorem for Any Index : The binomial theorem for any index states that

(1 + x)n = 1 + !1

nx + !2

)1n(n − x2

+ !3

)2n)(1n(n −− x3 +..... ...(7)

Where n is any index (positive or negative) The general term in expansion (7) is

Tr + 1 = !r

)1rn)......(1n(n +−− xr

In this expansion there are infinitely many terms. This expansion is valid for |x| < 1 and first term

unity. When x is small compared with 1, we see that the

terms finally get smaller and smaller. If x is very small compared with 1, we take 1 as a first approximation to the value of (1 + x)n or 1 + nx as a second approximation.

Replacing n by – n in the above expansion, we get

(1 + x)–n = 1 – nx + !2

)1n(n + x2 – !3

)2n)(1n(n ++ x3

+ ... + (–1)r

!r)1rn)...(2n)(1n(n −+++ xr + ...

Replacing x by – x in this expansion, we get

(1 – x)–n = 1 + nx + !2

)1n(n + x2 + !3

)2n)(1n(n ++ x3 +

....... + !r

)1rn)...(2n)(1n(n −+++ xr + ....

Important expansions for n = –1, –2 are : (1 + x)–1 = 1 – x + x2 – x3 + ...+ (–1)rxr + .... to ∞ (1 – x)–1 = 1 + x + x2 + x3 + ..... + xr + .... to ∞ (1 + x)–2 = 1 – 2x + 3x2 – .... + (–1)r(r + 1)xr +…. (1 – x)–2 = 1 + 2x + 3x2 + ..... + (r + 1)xr + .... to ∞ (1 + x)–3 = 1 – 3x + 6x2 – 10x3 + ...

+ (–1)r

!2)2r)(1r( ++ xr + .....

(1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ...

...... + !2

)2r)(1r( ++ xr + ....

XtraEdge for IIT-JEE SEPTEMBER 2010 59

PHYSICS

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. Two plane mirror which are perpendicular are forming two sides of a vessel filled with liquid of refractive index = 1.5. After all possible refraction & reflection, find the deviation (δ) in ray –

Incident ray 30º

µ = 1.5

α β

(A) δ = 0º (B) δ = 180º (C) δ = 90º

(D) we can't find out deviation as other two angles in

figure is not given

2. Two identical simple pendulums A and B are fixed at same point. They are displaced by very small angles α and β (β > α) and released from rest. Find the time after which B reaches its initial position for the first time. Collisions are elastic and length of strings is l.

A B

α β

(A) gl

π (B) g

2 lπ

(C) gl

απβ (D)

g2 l

απβ

3. A block of mass 2 kg is kept at origin at t = 0 and is

having velocity 54 m/s in positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U = – x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t = 0 is -

(A) 8 m/s (B) 4 m/s (C) 2410 m/s (D) 20 m/s 4. Power applied to a particle varies with time as

P = (3t2 – 2t + 1)watts, where t is time in seconds. Then the change in kinetic energy between time t = 2s to t = 4s is -

(A) 46 J (B) 52 J (C) 92 J

(D) 104 J

IIT-JEE 2011

XtraEdge Test Series # 5

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity, Refl. At Plane surface, Ref. at Curved surface, Refraction at Plane surface, Prism (Deviation & Dispersion), Refraction at Curved surface, Wave Nature of Light: Interference. Chemistry : Gaseous state, Chemical Energetics, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis, Reaction Mechanism, Alkane, Alkene, Alkyne, Alcohol, Ether & Phenol, Practical Organic Chemistry, Aromatic Hydrocarbons, Halogen Derivatives, Carboxylic Acid & Its Derivatives, Nitrogen Compounds, Amines, Carbohydrates, Amino Acid, Protein & Polymers. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number, Indefinite Integration, Definite Integration, Area Under the Curve, Differential Equations. Instructions : Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 5 to 9 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - III • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly

matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.. Section - IV • Question 12 to 19 are Numerical type question. +4 marks will be awarded for correct answer and –1 mark for wrong answer.


Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

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

(A) 1.6 3 cm (B) 1.6 cm (C) 4.8 cm

(D) none of these 6. As shown in figure

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

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

179 N (D) force applied by 20 kg block on inclined plane is

223 N 7. Two converging lens have focal

length 20 cm & 30 cm. Optical axis of both lens coincide. This lens system is used to form an image of an object. It turn out that size of the image does not depend on the distance between the lens system & the object. If L is distance between lens & M is magnification after all possible refraction -

(A) L = 10 cm (B) L = 50 cm

(C) |M| = 23 (D) |M| =

32

8. A research vessel has a round glass window in

bottom for observing the seabed. The diameter of the glass window is 60 cm, the thickness of the glass is 20 mm and the index of refraction of water is 1.33 and that of the glass is 1.55. The seabed is 6.0 m beneath the window. A man in interior of vessel can see seabed through window.

Interior of vessel air

Glass

waterwindow Sea bed`

6 m

(A) Area of seabed that can be seen through the window is 40 m2 (Approx)]

(B) Viewable area of seabed from interior of vessel will increases as thickness of glass of window increases

(C) Viewable area of seabed from interior of vessel is independent of thickness of glass of window

(D) Area of seabed that can be seen through window is 160 m2 (Approx)

9. A SHM is given by y = (sin ωt + cos ωt). Which of

the following statement are true - (A) The amplitude is 1m (B) The amplitude is 2 m (C) When t = 0, the amplitude is 0 m (D) When t = 0, the amplitude is 1 m This section contains 2 questions (Questions 10, 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

10. In the arrangement shown below force F is just sufficient to keep equilibrium of 100 N block, T1, T2 and T3 are tension, in strings AB, CD and EF and T4 is total force of all tensions on block 100 N

Match the following

θ θ

m1

m2

µ = 0.5θ = 37º

θ = 37º

θ = 37º m

20 kg

L

O

f = 20 cm f = 30 cm

FBDF

T2T1

T3

E

100 N

A

C


Column-I Column-II

(A) T1 (P) 7

400 N

(B) T2 (Q) 7

100 N

(C) T3 (R) 7

200

(D) T4 (S) 100 N (T) None of these 11. A block of mass m = 1 kg is at

rest with respect to a rough wedge as shown. The wedge starts moving up from rest with an acceleration of a = 2 m/s2 and the block remains at rest with respect to wedge. Then in 4 sec of motion (if θ = 60º & g = 10 m/s2) work done on block.

Column-I Column-II (A) By gravity (in magnitude) (P) 144 J (B) By normal reaction (Q) 32 J (C) By frictional force (R) 160 (D) By all forces (S) 48 J (T) None of these

This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. A disc of radius 1 m is rolling on

a plane horizontal mirror with constant angular speed ω = 3 rad/s (sec figure). Calculate velocity of image of point Q with respect to point Q itself. (given θ = 30º)

13. In the figure shown the lens has focal length f and velocity v as shown. If the final image velocity is kv. Find k.

14. Light travelling in air falls at an angle of incidence 2º one face of a thin prism of angle 4º and refractive index 1.5. The medium on the other side is water (R.I = 4/3). Find the deviation produce by prism in degrees.

15. The ratio of sizes of two images, obtained on a fixed screen of a candle for two positions of a thin lens (focal length f) is 4. If distance between candle and screen is 36 units (> 4f). Then find f.

16. IN YDSE one of the two

slits is covered with a glass slab. At P and Q ; 8th and 12th maximum are observed. Find the order of maxima at O if thickness of slab is minimum for this to happen.

17. In the arrangement

shown the rod is freely pivoted at point O and is in contact with the equilateral triangular block which can moves on the horizontal frictionless ground. As the block is given a speed v forward, the rod rotates about point O. Find the angular velocity of rod in rad/s at the instant when θ = 30º. [Take v = 20 m/s, a = 1 m]

18. A cylinder of mass M radius R is resting on a

horizontal platform (which is parallel to x – y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in X-direction given by X = A cos ωt. There is sufficient friction present in the surface of contact that can prevent the slipping between the cylinder and platform. The minimum torque in N-m acting on the cylinder during its motion is.

[take M = 4 kg, R = 1 m, A = 2m, ω = 1 rad/s] 19. A body of mass m = 4 kg starts moving with velocity

v0 in a straight line is such a way that on the body work is being done at the rate which is proportional to the square of velocity as given by P = βv2 where β

= 2693.0 . Find the time elapsed in seconds before

velocity of body is doubled.

m

µ

θ

a

Qω

θ

P

v

O

f/2 f/2

Q

O

PPO = OQ

v

3a2O

θ


CHEMISTRY


1. An alkene (A), C16H16 on ozonolysis gives only one product B(C8H8O). Compound (B) on reaction with NH2OH, H2SO4, ∆ gives N-methyl benzamide. The compound 'A', is –

(A) CH3 C = C

H CH3 H

(B)

C = C CH3 CH3

(C) CH2–CH = CH–CH2

(D) CH = CH

CH3

CH3

2. As per Boyle's law V ∝ 1/P at constant temperature,

As per charles law V ∝ T at constant pressure. Therefore, by combining, one concluded that T ∝ 1/P hence, PT = constant

(A) PT = constant is correct, because volume remain same in both the laws

(B) PT = constant is incorrect, because volume remain same at the constant temperature and at the constant pressure

(C) PT = constant is correct, because volume at constant temperaute and volume at constant pressure are not same

(D) PT = constant is incorrect, because volume at constant temperature and volume at constant pressure for the same amount of gas are different

3. Among the following statements on the nitration of aromatic compounds, the false one is -

(A) The rate of nitration of benzene is almost the same as that of hexadeuterobenzene

(B) The rate of nitration of toluene is greater than that of benzene

(C) The rate of nitration of benzene is greater than that of hexadeuterobenzene

(D) Nitration is an electrophilic substitution reaction

4. A compression of an ideal gas is represented by curve AB, which of the following is wrong

B(vB)

A(vA)

log V

log P

(A) number of collision increases B

A

VV times

(B) number of moles in this process is constant (C) it is isothermal process (D) it is possible for ideal gas Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

5. Which is correct for ∆Gº (A) ∆G = ∆H – T∆S (B) at equilibrium ∆Gº = 0 (C) at equilibrium ∆G = – RT log K

(D) ∆G = ∆Gº + RT log K

6. Which of the following statement(s) is (are) correct about friction ?

(A) The coefficient of friction between two bodies is largely independent of area of contact

(B) The frictional force can never exceed the reaction force on a body from the supporting surface

(C) Rolling friction is generally smaller than sliding friction

(D) Friction is due to irregularities of the surfaces in contact.

7. (Consider the reaction

C – OH

O

EtOH

)(NH,Na 3 → l A 22

23

ClCH

SMe,O → B + C

D∆

Identify the correct representation of structure of the products -


(A) A is

COOH

(B)The intermediate formed in the conversion of B to D is enol

(C) The structure of C is

O O

(D) A can also be formed from the reaction + COOH

8. Reduction of But-2-yne with Na and liquid NH3 gives an alkene which upon catalytic hydrogenation with D2 / Pt gives an alkane. The alkene and alkane formed respectively are -

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

9. Identify the compounds which do not give positive iodoform test in the following sequence of the reaction

(i) Hydrolysis (ii) Heating (iii) I2 + NaOH

(A) CH3 – CH – C – Et

O

CO2Et

(B)

CO2Et

O

(C)

O

O–Et

O

(D) Me – CH – CO2Et

CO2Et

This section contains 2 questions (Questions 10, 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the following Column I Column II (A) HCOOH (P) Decarboxylation on heating (B) CH3COOH (Q) Reation with Br2

(C) OHCOOH

(R) Cu2+(alkaline)→Cu2O

(D) PhCH2COOH (S) Decarbonylation or decarboxylation on treatment with conc. H2SO4 (T) None of these

11. Match the following Column –I Column II (A) Cyclopropenyl carbocation (P) Hyperconjugation (B) Cyclopentadienyl anion (Q) All carbon atoms are sp2 hybridized (C) Benzyne (R) Aromatic nature (D) t-Butyl carbocation (S) Diamagnetic (T) None of these


012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Which of the following is the strongest nucleophilic site in the following species ?

O SO O

O••

•••• –

••O••

••–

••–4

•• – 3

2

1

O O


13. Which nitrogen is protonated readily in guanidine ?

HN = C

2 NH2

NH2 3

1

14. CH3

OH H3C OH–

H

2→

+ [F]

4

2

CCl

Br→ 43421

possibleareproductssuch5

284 BrHC

How many structures of [F] are possible ?

15. A nanopeptide contains…………..peptide linkages. 16. 0.45 g of an acid of mol wt 90 is neutralised by 20

mL 0.5 N NaOH. The basicity of acid will be 17. A Gaseous mixture of 2 moles of A, 3 moles of B, 5

moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is 1.5 atm. The total pressure is

18. A solution containing 4 m mole of An+ ions requires

1.6 m mole of MnO4– for oxidation of An+ to AO3

– in acidic medium. The value of n is

19. The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80. The atomic weight of the metal is

MATHEMATICS


1. If k > 0, |z| = |w| = k, and α = wzk

w–z2 +

, then Re (α)

equals (A) 0 (B) k/2 (C) k

(D) None of these

2. If n∈N, then value of S = )C(

)C()1(–r

2rr

nn

0r

r+

=∑ is

(A) 2n

1+

(B) 2n

2+

(C) n + 2

(D) n + 1 3. The equation of the curve satisfying the differential

equation y2 (x2 + 1) = 2xy1 passing through the point, (0, 1) and having slope of tangent at x = 0 as 3 is

(A) y = x2 + 3x + 2 (B) y2 = x2 + 3x + 1 (C) y = x3 + 3x + 1 (D) none of these

4. If I = ∫ ++ dx)]xsecx(tanxtan21[ 2/1 then I equals

(A) – log |sec x + tan x| + log |cos x| + C

(B) xcos

)4/2/xtan(log π+ + C

(C) log |tan (x/2 + π/4) cos x| + C (D) log |cot (π/4 – x/2) cos x| + C Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. Solution of )x2–x(loglog 2

3x2x28x6x 22 ++++ = 0 is

(A) a natural number (B) a negative integer (C) – 1

(D) none of these 6. Sum to n terms of the series S = 12 + 2 (2)2 + 32 + 2(42) + 52 + 2(62) + … is

(A) 21 n (n + 1)2 when n is even

(B) 21 n2 (n + 1) when n is odd

(C) 41 n2 (n + 2) when n is odd

(D)

41 n (n + 2)2 when n is even

7. The number of ways in which we can arrange the 2n

letters x1, x2,…,xn, y1,y2,…yn in a line so that the suffixes of letters x and those of the letters y are respectively in ascending order of magnitude is -

(A) (nC0)2 + (nC1)2 + … + (nCn)2 (B) 2nCn (C) 2n[1.3.5 … (2n – 1]/n!

(D) 2nCn – 1 8. The values of α which satisfy

∫α

π 2/

xsin dx = sin 2α (α ∈ [0, 2π]) are equal to

(A) π/2 (B) 3π/2 (C) 7π/6

(D) 11π/6

9. If ∫ + x5cos21x7cos–x8cos dx is expressed as

K sin 3x + M sin 2x + C then (A) K = – 1/3 (B) K = 1/3 (C) M = – 1/2

(D) M = 1/2



A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. The number of real roots of Column-I Column-II

(A) x9x ++ = 2.7 (P) 2

(B) |1x|

x–13 2

+ = 1 (Q) 1

(C) x–5x + = 1 (R) infinite

(D) 1–x2–x–1–x2x + = 2 (S) 0 (T) None 11. Match the Column Column-I Column-II

(A)

∫ ∫ +→

a

y

a

yx

tsintsin

0xdte–dte

x1lim

22 = (P) 1

(B)

∫

∫+

+

→

yx

0

t2

2yx

0

t

0x dte

dtelim

2

2

= (Q) ye2sin

(C) 3

x

y

0x x

dxxsinlim

2

∫→

= (R) 0

(D) x

dttcoslim

x

y

2

0x

∫→

= (S) 2/3

(T) None


012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. The value of ( )8y – 31 if (1 + x2)

dxdy = x(1 – y),

y(0) = 4/3 is ………… . 13. The area bounded by the curves x = y2 and

x = 3 – 2y2 is ………. .

14. If I = 3 ∫π

θ0

3sin (1 + 2cosθ) ( 1 + cos θ)2 dθ then

the value of I is ……… . 15. Number of divisors of the form 4n + 2 (n ≥ 0) of

integers 240 is………. 16. If |z| < 1/2 and 4|(1 + i)z3 + iz| < k, find the least

integral value of k. 17. If x1 + x2 + x3 = 1, x2x3 + x3x1 + x1x2 = 1 and x1x2x3 = 1,

find value of |x1| + |x2| + |x3|.

18. Let an = 121–2

3

3

+.

131–3

3

3

+.

141–4

3

3

+...

1n1–n

3

3

+, find

∞→nlim 3an.

19. If x = (7 + 34 )2n = [x] + f, then x(1 – f) is equal to……..



PHYSICS


1. A container has a liquid filled upto the height H. There is a hole at height H/4. The area of hole is 'a'. Density of liquid is ρ. Torque due to the efflux coming out about an axis passing through 'A' and perpendicular to the plane of figure is –

H/4

H –→–→ –→

A

(A) 4

agH2ρ (B) 4

agH3 2ρ

(C) 2agH83

ρ (D) ρagH2

2. A container has a hole at a height of 2m. If the time taken by the efflux to strike the inclined plane perpendicularly is 1 sec. Then the height of liquid level initially is (Take g = 10 m/s2) –

45º2m.

(A) 2 m (B) 5 m (C) 7 m

(D) 3 m

3. A block of mass 2 kg is kept at origin at t = 0 and is having velocity 54 m/s in positive x direction. Its potential energy is defined as U = –x3 + 6x2 + 15 (SI units). Its velocity when the applied force is minimum (after the time t = 0) is –

(A) 8 m/s (B) 4 m/s (C) 2410 m/s (D) None of these

4. A particle of mass m is released from point A on smooth fixed circular track as shown. If the particle is released from rest at t = 0, then variation of normal reaction N with (θ) angular displacement from initial position is –

IIT-JEE 2012

XtraEdge Test Series # 5

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity. Chemistry : Gaseous state, Chemical Energetic, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number.

Instructions : Section - I • Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 5 to 9 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - III • Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly

matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.. Section - IV • Question 12 to 19 are Numerical type question. +4 marks will be awarded for correct answer and –1 mark for wrong answer.


A m

R O

(A)

N

3mg

θ

(B)

N

θ

3mg

(C)

N

3mg

θ

(D)

N

3mg

θ

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. A vessel filled with liquid is resting on the rough

horizontal surface. A hole is made in the vessel as shown. Then –

µ

h • H

•

(A) The torque due to friction about the centre of gravity is of into the plane of the paper

(B) The torque due to normal reaction force between container and ground about center of gravity is out of the plane of paper

(C) Torque due to friction about center of gravity is zero

(D) Torque due to normal reaction force between container and ground about center of gravity is zero

6. A trolley of mass m1 is to be moved such as to keep

block A of mass m2 at rest with respect to it. A bucket of mass m3 (with water) in it is placed on trolley. Coefficient of friction between the block A and trolley is µ. The trolley is moved with acceleration so that block does not slip –

BA

(A) The minimum coefficient of friction between bucket and trolley is µ/2

(B) The acceleration of trolley is g/µ (C) The inclination of water surface in the bucket

with horizontal in absence of any slipping is tan–

1 1/µ (D) Force on trolley is (m1 + m2 + m3) g/µ 7. As shown is figure the string BC is 10 cm long and

has a linear mass density of 10 kg/m while the string ED is massless. If both strings are inextensible and pulley is ideal then when the system is released from rest the ratio of tension in the string.

D B

2kg 4kg E C

(A) at points E and C is 45

(B) at points E and C is 54

(C) at points D and E is 1

(D) at points D and E is 21

8. A wedge of mass m1 and a block of mass m2 is in equilibrium as shown. Inclined surface of the wedge has an inclination α with the horizontal and all contacts are smooth. The normal reaction on the wedge may be –

m2

m1

α

(A) m2g cos α (B) m2g sin α cos α (C) m1g + m2g cos2 α (D) m1g+m2g sin α cos α

9. A rough L-shaped rod is located in a horizontal plane and a sleeve of mass m is inserted in the rod. The rod is rotated with a constant angular velocity ω in the horizontal plane. The lengths l1 and l2 are shown in figure. The normal reaction and frictional force acting on the sleeve when it just starts slipping are (µ = coefficient of friction between rod and sleeve) –


ω

L-shaped rod

l1

sleeve

l2

(A) N = mω2l1 (B) f = mω2l2

(C) N = m 21

42g lω+ (D) f = µN


A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. As shown block C of mass 5 kg is pulled by a force F and its acceleration is found to be 3 m/s2. The masses of blocks A and B are 10 kg and 5 kg respectively while the string passing over ideal pullies is ideal and is under tension T. If acceleration of blocks A and B are a1 and a2 respectively then if all surfaces are smooth and g = 10 m/s2 –

F

B

aC = 3 m/s2

T A

C

Column-I Column-II (A) F (P) 2 (B) T (Q) 1 (C) a1 (R) 55 (D) 2 a2 (S) 70 (T) None of these

11. A single conservative force acts on a body of mass 1 kg that moves along the x-axis. The potential energy U(x) is given by U(x) = 20 + (x–2)2 where x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J then –

Column-I Column-II (A) minimum value (P) 29 of x in meters (B) maximum value (Q) 7.38 of x in meters (C) maximum potential (R) 49 energy in joules (D) maximum kinetic (S) – 3.38 Energy in joules (T) None of these


012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. A particle moves in a straight line with its retardation

proportional to its displacement 'x'. Change in kinetic energy is proportional to nth power of x, where n is -

13. A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a force F(x) = – K/(2x2) where K = 10 Nm2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Find its velocity when it reaches x = 0.50 m.

14. An artillery gun is mounted on a railway truck standing on straight horizontal rails. The total mass of the truck with gun shells and crew is M = 50 tons and the mass of each shell is m = 25 kg. The gun fires in a horizontal direction along the railway. The initial velocity of the shells is V0 = 1000 m/s. What will the speed of truck after the second shot? Disregard friction and air resistance.


15. A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each will same K.E. E0. The energy of explosion will be K times of E0, then the value of K is -

16. A cubical block of mass 6kg and side 16.1 cm is

placed on frictionless horizontal surface. If is hit by a cue at the top as to impart-impulse in horizontal direction. Minimum impulse imparted to topple the block must be greater than.

17. A disc is rotating freely its axis. Percentage change in

angular velocity of disc if temperature decreases by 20ºC is (coefficient of linear expansion of material of disc is 5 × 10–4/ºC )

18. A glass capillary sealed at the upper end is of length

0.11 m and internal diameter 2 × 10–5 m. This tube is immersed vertically into a liquid of surface tension 5.06 × 10–2 N/m. When the length x × 10–2. m of the tube is immersed in liquid then the liquid level inside and outside the capillary tube becomes the same, then the value of x is : (Assume atmospheric pressure is

1.01 × 105 2m

N )

19. A wire of length '2m' is clamped horizontally

between two fixed support. A mass m = 5 kg is hanged from middle of wire. The vertical and depression in wire (in cm) in equilibrium is (Young modulus of wire = 2.4 × 109 N/m2, cross-sectional area = 1 cm2)

CHEMISTRY


1. Given that H2O (l) → H2O(g) ; ∆H = + 43.7 kJ H2O (s) → H2O (l) ; ∆H = + 6.05 kJ ∆Hsublimation of ice is - (A) 49.75 kJ mol–1 (B) 37.65 kJ mol–1 (C) 43.7 kJ mol–1 (D) – 43.67 kJ mol–1

2. A real gas of molar mass 60 g mol–1 has density at critical point equal to 0.80 g/cm3 and its critical

temperature is given by Tc = 821

104 5× K. Then the

van der Waal's constant 'a' (atm L2 mol–2) will be (A) 0.025 (B) 0.325 (C) 3.375 (D) 33.750

3. Adiabatic reversible expansion of a monoatomic gas (M) and a diatomic gas (D) at an initial temperatrue Ti has been carried out independently from initial volume V1 to final volume V2. The final temperature (TM for monoatomic gas and TD for diatomic gas) attained will be :

(A) TM = TD > Ti (B) TM < TD < Ti (C) TM > TD > Ti (D) TM = TD = Ti

4. What volume of 2N K2Cr2O7 solution is require to oxidise 0.81 g of H2S in acid medium -

(A) 47.8 (B) 23.8 (C) 40 ml

(D) 72 ml

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. 20 volume of H2O2 is equal to - (A) 20 % H2O2 by mass (B) 6 % H2O2 by mass (C) 1.764 N (D) 3.528 N 6. Which of the following is/are state function ? (A) q (B) q – w (C) q + w (D) q / w 7. According to kinetic theory of gases :

(A) the pressure exerted by a gas is proportional to mean square velocity of the molecules

(B) the pressure exerted by the gas is proportional to the root mean square velocity of the molecules

(C) the root mean square velocity is inversely proportional to the temperature

(D) the mean translational K.E. of the molecule is directly proportional to the absolute temperature

8. According to Charles' law :

(A) V ∝ T1 (B)

PdTdV

= K

(C) PdV

dT

= K (D)

P2T

V–T1

= 0

9. Consider in Redox reaction 2S2O3

2– + I2 → S4O62– + 2I–

(A) S2O32– gets reduced to S4O6

2– (B) S2O3

2– gets oxidised to S4O62–

(C) I2 gets reduced to I– (D) I2 gets oxidised to I–



A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Column-I Column-II

(A) If force of attraction (P)

+ 2V

aP (V–b)= RT

among the gas molecules be negligible (B) If the volume of the (Q) PV = RT –a/V gas molecules be negligible (C) At STP (for real gas) (R) PV = RT + Pb (D) At low pressure and (S) PV = RT at high temperature (T) PV/RT = 1–a/VRT 11. Match of the following : Column-I Column-II (A) A process carried (P) Adiabatic out infinitesimally slowly (B) A process in which (Q) ∆G = 0 no heat enters or leaves the system (C) A process carried (R) Sublimation out at constant temperature (D) A process in (S) Reversible equilibrium (T) Isothermal


012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Oxidation number of Fe in Na2[Fe(CN)5NO] is

13. The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is.

14. How many mole of electrons are needed to convert one mole of nitrate ion to hydrazine.

15. Calculate the volume occupied by 8.8 g of CO2 at 31.1ºC and 1 bar pressure. (R = 0.083 bar litre K–1 mol–1)

16. A compound exists in the gaseous phase both as monomer (A) and dimer (A2). The molecular weight of A is 48. In an experiment 96 g of the compound was confined in a vessel of volume 33.6 litre and heated to 273º C. Calculate the pressure developed if the compound exists as dimer to the extent of 50% by weight under these conditions.

17. The haemoglobin from the red blood corpuscles of most mammals contains approximately 0.33% of iron by weight. The molecular weight of haemoglobin as 67,200. The number of iron atoms in each molecule of haemoglobin is (atomic weight of iron = 56) :

18. 0.7 g of Na2CO3.xH2O were dissolved in water and the volume was made to 100 mL, 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralisation. The value of x is

19. The temperature of a 5 mL of strong acid increases by 5ºC when 5 ml of a strong base is added to it. If 10 mL of each are mixed, temperature should increase by


MATHEMATICS


1. The set of x : log1/3 log4(x2 – 5) > 0 is equal to (A) (3, ∞) (B) ( 6 , 3)

(C) (–3, – 6 ) ∪ ( 6 , 3) (D) none of these 2. The coefficient of xk (0 ≤ k ≤ n) in the expansion of

E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is (A) nCk (B) n+1Ck (C) n+1Ck+1

(D) none of these 3. Let a > 0, b > 0 and c > 0. Then both the roots of the

equation ax2 + bx + c = 0 (A) are real and negative (B) have negative real parts (C) are rational numbers (D) none of these

4. If three positive real numbers a, b, c are in A.P. such that abc = 4, then the minimum possible value of b is

(A) 23/2 (B) 22/3 (C) 21/3

(D) 25/2 Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

5. Let z = 7i3–5i5

i353–i2–1i5–i211

++

, then

(A) z is purely real (B) z is purely imaginary

(C) (z – –z )i is purely real

(D) z + –z = 0 5. Let S denote the set of all real values of a for which

the roots of equation x2 – 2ax + a2 – 1 = 0 lie between 5 and 10, then S equals

(A) (–1, 2) (B) (2, 9) (C) (4, 9)

(D) (6, 9)

7. If 2 and 31 appear as two terms in an A.P., then (A) common difference of the A.P. is a rational

number (B) all the terms of the A.P. must be rational (C) all the terms of the A.P. must be integers

(D) sum to any number of terms of the A.P. must be

rational 8. Given that the 4th term in the expansion of

10

8x32

+ has the maximum numerical value, then x

lies in the interval

(A)

2164,2 (B)

2,–

2360–

(C)

2,–

2164– (D)

2360,–2

9. If log x2 – log 2x = 3 log 3 – log 6 then x equals (A) 9 (B) 3 (C) 4

(D) 5 This section contains 2 questions (Questions 10, 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. z lies on ……… if Column-I Column-II (A) |z – 3| + |z – i| = 10 (P) circle

(B) i–z3–z2 = 2 (Q) hyperbola

(C) z2 + 2–z = 5 (R) straight line

(D) i2–z

6–z = 3 (S) ellipse

(T) parabola


11. The value of Column-I Column-II (A) 1.1! + 2.2! + 3.3! (P) (n + 2)2n–1 – (n + 1) + … + n.n! (B) n.nC1 + (n – 1).nC2 (Q) 2nCn

+ (n – 2). nC3 + … +1.nCn (C) 2C2 + 3C2 + 4C2 + …+ nC2 (R) (n + 1)! – 1

(D) ∑=

n

0r

2r

n )C( (S) n+1C3

(T) 0 This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. The remainder when 22003 is divided by 17 is… 13. The number of real values of x which satisfy the

equation 7x6x7x6–x

2

2

++

+ = 1 are…

14. Let x1, x2, ... ∈ (0, π) denote the values of x satisfying

the equation

)upto....|x|cosxcos|x|cos1( 3227 ∞++++ = 93,

find the value of π1 (x1 + x2 + ...)

15. If x satisfies log1–x(3) – log1–x (2) =1/2, find – 4x. 16. Rakshit is allowed to select (n + 1) or more books out

of (2n + 1) distinct books. If the number of ways in which he may not select all of them is 255, then value of n is…

17. If 1, x1, x2, x3 are the roots of x4 – 1 = 0 and ω is a complex cube root of unity, find the value of

)x–)(x–)(x–()x–)(x–)(x–(

321

32

22

12

ωωωωωω

18. If the lengths of the sides of a right triangle ABC

right angled at C are in A.P., find 5 (sin A + sin B). 19. A class contains 4 boys and g girls. Every Sunday

five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is…

• William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

• The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period. (It is actually 24 hours and 37 minutes)

• Space debris travels through space at over 18,000 mph.

• The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system.

• A car travelling at a constant speed of 60 miles per hour would take longer than 48 million years to reach the nearest star (other than our Sun), Proxima Centauri. This is about 685,000 average human lifetimes

• Scientists estimate that the contents of our universe consists of 4 percent ordinary atoms (baryons) in stars, nebulae and diffuse intergalactic gas. Dark Matter provides about 30 percent; and Dark Energy provides the rest of about 66% percent.

• One parsec is equal to 19.2 million million miles.

• The coldest known star is an unnamed star about 160 light years from Earth. Its surface temperature is only 2600F which is 7400F cooler than the Sun!

• Venus is the second closest planet to the Sun, and the sixth largest overall.

• The first manned space flight happened on the 12th April 1961, when Yuri Gagarin made a complete orbit of the Earth before landing safely back in Russia.


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 Ans B B A A A A,C B,D C,D B,D

10 A → Q B → R C → P D → S Column Matching 11 A → R B → P C → Q D → S

12 13 14 15 16 17 18 19 Numerica l Type 3 3 1 8 2 5 4 8

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 Ans B D C A A,D C B,C,D C A,C,D

10 A → S B → R C → Q D → P Column Matching 11 A → Q,R,S B → R C → Q D → P

12 13 14 15 16 17 18 19 Numerica l Type 4 1 3 8 2 6 3 9

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 Ans A B C B B,C A,B A,B,C A,B,C,D B,C

10 A → S B → P C → Q D → R Column Matching 11 A → Q B → R C → S D → P

12 13 14 15 16 17 18 19 Numerica l Type 3 4 8 4 3 3 2 1

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 Ans B C A A A,B B,C,D A,C B,C B,C,D

10 A → S B → R C → Q D → P Column Matching 11 A → S B → Q C → R D → P

12 13 14 15 16 17 18 19 Numerica l Type 2 1 1 6 4 2 1 5

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 Ans A C B B B,D C B,D B,C,D B,C

10 A → R B → Q,T C → P D → S Column Matching 11 A → S B → P C → T D → Q

12 13 14 15 16 17 18 19 Numerica l Type 2 2 7 5 2 4 2 5

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 Ans C C B B A,C D A,B,D A,C A

10 A → S B → R C → Q D → P Column Matching 11 A → R B → P C → S D → Q

12 13 14 15 16 17 18 19 Numerica l Type 8 1 1 5 4 1 7 5

IIT- JEE 2011 (September issue)

IIT- JEE 2012 (September issue)


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IIT JEE becoming more competitive examination day by day. Regular change in pattern making it more challenging.

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Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.

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Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.

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XtraEdge for IIT-JEE SEPTEMBER 2010 - Career · PDF fileXtraEdge for IIT-JEE 4 SEPTEMBER 2010 IIT-K To Coordinate 2011 JEE The IIT-K will be coordinating the IIT Joint Entrance Exam