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XtraEdge for IIT-JEE 1 SEPTEMBER 2010







Dear Students,

Three Things you Must Know to Attract SuccessEveryone wants success. Some people spend their every waking moment pursuing it,to the detriment of all else. For others, attaining success seems impossible. Theyconclude that it is destined for a select few. The rest of us are to remain "contentwith 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 adestination." Jim Rohn describes it as " .... a condition that must be attractednot pursued." Success is something you must work hard and long to earn, for yourself. It has aprice, 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 atemporary illusion. Success that will remain with you, and bring you joy ratherthan sorrow, requires a learning process, a time to grow out of old habits andinto new ones, a time to learn what works and what doesn't. And you mustpay 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 successmean to you ? Identify, in specific terms, what you regard as success. Whattraits or skills will you need to achieve this goal? Find 2 or 3 people who havewhat you want. Write down the habits that have made them successuf andresolve to copy them. This is called mentoring learning from others who havearrived 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 beensaid that, "At some point in time, the pursuit of your goals becomes secondaryand what you have become in the process .... is what is most important. It's notthe distance you go .... so much as the going itself" (Les Brown).

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

When you are ready for success you attract it, with little effort. Whenyou 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 eludespeople.Now that you know how to attract success, why not get started on the journey thatwill take you where you want to go? Any one can succeed, but unfortunately notevery one will. Fate does not foist it upon you. You can have anything you want inlife, 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 startattracting success. It's literally yours for the taking!Presenting forever positive ideas to your success.Yours truly

Pramod Maheshwari,

B.Tech., IIT Delhi

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

one, of any kind, in any manner, there from.

• No Portion of the magazine can bepublished/ reproduced without the writtenpermission of the publisher

•

All disputes are subject to the exclusiveurisdiction of the Kota Courts onl .

Owned & Published by Pramod Maheshwari,112, Shakti Nagar, Dadabari, Kota & Printedby Naval Maheshwari, Published & Printed at112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari

Impatience never commanded success.

Volume - 6 Issue - 3

September, 2010 (Monthly Magazine)

Editorial / Mailing Office :

112-B, Shakti Nagar, Kota (Raj.) 324009

Tel. : 0744-2500492, 2500692, 3040000

e-mail : [email protected]

Editor :

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[B.Tech. IIT-Delhi]

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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 didnot realize how close they were to successwhen 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 6Mr. 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 ConceptAliphatic 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, formsubmission 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 ofour 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 andtechnologists, 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 thecountry 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 DoctoralProgrammes 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 betweenpolicy 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 theywould 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 politicalinitiative 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 theoverwhelming 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 FormLightweight 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 heatand 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 tohigher-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 containedin 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 materialfunctions 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 haveimproved 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 thearmor 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 designsare 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 credibleprofessional 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 amajor 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 Facts1. 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

roduce.




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 Careand 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 programencourages 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 recognizesthe outstanding contributions made by him as a CorporateLeader. Through his achievements, Mr. Vineet Agrawalhas 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 ElectricalEngineering, 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 StorySuccess Story

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

Dr. Alok AggarwalB.Tech, IIT Delhi in 1980

Ph.D. in Hopkins University, (1984)




PHYSICS

1. A transverse harmonic disturbance is produced in astring. The maximum transverse velocity is 3 m/s andmaximum transverse acceleration is 90 m/s2. If thewave velocity is 20 m/s then find the waveform.

[IIT-2005]

Sol. The wave form of a transverse harmonic disturbancey = 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)

ωωa

a 2

=3

90 ⇒ ω = 30 rad/s ...(iv)

Substituting the value of ω in (i) we get

a =30

3 = 0.1 m ...(v)

Now

k =λπ2

=v/v

2π =

v

v2π =

v

ω =

20

30 =

2

3 ...(vi)

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

y = 0.1 sin φ±± x23t30

2. A 5m long cylindrical steel wire with radius 2 × 10 –3 m is suspended vertically from a rigid support andcarries a bob of mass 100 kg at the other end. If the

bob gets snapped, calculate the change in temperatureof the wire ignoring radiation losses. (For the steelwire : Young's modulus = 2.1 × 1011 Pa; Density= 7860 kg/m3; Specific heat = 420 J/kg-K).

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

under tension and hence in the deformed state.

Therefore it has potential energy (U) which is given by the formula.

U =2

1 × stress × stain × volume

=2

1 ×

Y

)Stress( 2

× πr 2l

=2

1

Y

)r /Mg( 22π× πr 2l =

2

1

Yr

gM2

22

π

l ...(i)

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

Q = mc ∆T ...(iii)From (i) and (iii) Since U = Q Therefore

∴ mc∆T =Yr

gM

2

12

22

π

l

∴ ∆T =Ycmr

gM

2

12

22

π

l

Herem = mass of string = density × volume of string

= ρ × πr 2l

∴ ∆T =

ρπ Yc)r (

gM

2

122

22

=2

1 ×

7860420101.2)10214.3(

)10100(1123

2

××××××

×−

= 0.00457ºC

3. The x – y plane is the boundary between twotransparent 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 inmedium –2.

Sol.Y

^^ j38i36 +

^ j38

^i36

M'

M

XO

Z M'

XO

–10^K

^^^k 10 – j38i36A +=

→

^^ j38i36 +

Fig(1) Fig(2)

Figure 1 shows vector i36 +^ j38

Figure 2 shows vector

→

A =^

i36 +^

j38 –^

k 10 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 ).(– k 10 – j38i36(

++

+ = cos i

KNOW IIT-JEEBy Previous Exam Questions




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

2/122

^^^

])38()36[(

j38i36n

+

+=

^^^ j5

4i

8

3n +=

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

2

3 =

r sin

isin =

r sin

º60sin ⇒ r = 45º

From the triangle ORS

^r = (sin r) ^n – ( cos r)^k

= (sin 45º)

+

^^ j

5

4i

5

3 – (cos 45º)

^k

= ]k 5 – j4i3[25

1 ^^^+

4. Along horizontal wire AB, which is free to move in avertical plane and carries a steady current of 20A, isin 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 shownin figure. Show that when AB is slightly depressed, itexecutes simple harmonic motion. Find the period ofoscillations.

A B

DC

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

weight per unit length =πµ40

r II2 21 ...(i)

when the rod is depressed by a distance x, then theforce acting on the upper wire increases and behaveas a restoring force

A

Fmag

I1 = 20A x

B

B'mgr = 0.01 mA'

C I2 = 30A D

Restoring force/length =

π

µ

4

0

x – r

II2 21 –

π

µ

4

0

r

II2 21

=π

µ

40 2I1I2

r

1 –

x – r

1

⇒ Restoring force/length =π

µ4

0 2I1I2

r )x – r (

)x – r ( – r

=π

µ

40

)x – r (r

xII2 21

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

Restoring force/length F =π

µ4

0 2

21

r

II2x

Since F ∝ x ∴ The motion is simple harmonic

∴π

µ

40

221

r

II2 = (mass per unit length) ω2 ...(ii)

From (i) (Mass per unit length) × g =

π

µ

4

0

r

II2 21

Mass per unit length =π

µ

40

rg

II2 21 ...(iii)

From (ii) and (iii)

π

µ

40

221

r

II2 =

π

µ

40

rg

II2 21 × ω2

⇒ ω =r

g ⇒

T

2π =

r

gt

⇒ T = 2π g

r = 2p

8.9

01.0 = 0.2 sec

5. In the figure both cells A and B are of equal emf.Find R for which potential difference across batteryA will be zero, long time after the switch is closed.Internal resistance of batteries A and B are r 1 and r 2 respectively (r 1 > r 2).s

CR

LR

R

R

R

R

S

BA

r 1 r 2

Sol. After a long time capacitor will be fully charged,

hence no current will flow through capacitor and allthe current will flow from inductor. Since current isD.C., resistance of L is zero.

∴ R eg =

+ R

2

R ×

2

1 + r 1 + r 2

=4

R 3 + r 1 + r 2

I =eqR

ε+ε ⇒ I =

egR

2ε =

21 r r 4/R 3

3

++ε

Potential drop across A isε – I r 1 = 0

⇒ ε =21 r r 4/R 3

2

++ε

r 1

⇒ r 1 = r 2 + 3R/4

or R =3

4(r 1 – r 2)




CHEMISTRY

6. An organic compound A, C8H4O3, in dry benzene inthe presence of anhydrous AlCl3 gives compound B.The compound B on treatment with PCl5 followed byreaction with H2/Pd(BaSO4) gives compound C,which on reaction with hydrazine gives a cyclised

compound D(C14H10 N2). 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

OH2 NNH2

C6H5

N

N

The formation of D from C may be explained asfollows.

C6H5

C6H5

O

O

NH2

NH2

O –

NH2

NH2

O –

+

+ C6H5O –

N – H

N – H

OH

C6H5

N

N

7. An alkyl halide X, of formula C6H

13Cl on treatment

with potassium t-butoxide gives two isomeric alkenesY and Z(C6H12). Both alkenes on hydrogenation give2, 3-dimethyl butane. Predict the structures of X, Yand Z. [IIT-1996]

Sol. The alkyl halide X, on dehydrohalogenation givestwo isomeric alkenes.

X136 ClHC

HCl – ;

butoxidetK

∆

−− → 126HCZY +

Both, Y and Z have the same molecular formulaC6H12(CnH2n). Since, both Y and Z absorb one mol ofH2 to give same alkane 2, 3-dimethyl butane, hencethey should have the skeleton of this alkane.

Y and Z (C6H12) NiH2 → CH3 – CH – CH – CH3

CH3 CH3

2,3-dimethyl butane

The above alkane can be prepared from two alkenesCH3 – C = C – CH3

CH3 CH3 2,3-dimethyl

butene-2

(Y)

and CH3 – CH – C = CH2

CH3 CH3

2,3-dimethyl butene-1(Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3

CH3 CH3

(Y)

H2

NiCH3 – CH – CH – CH3

CH3 CH3

CH3 – CH – C = CH2

CH3 CH3

(Z)

H2

NiCH3 – CH – CH – CH3

CH3 CH3

Both, Y and Z can be obtained from following alkylhalide :

CH3 – C – CH – CH3

CH3 CH3

2-chloro-2,3-dimethyl butane

(X)

K-t- utoxide

∆; –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 2750as it vaporizes at 20ºC and that of liquid toluene(density = 0.867 g ml –1) increases by a factor of 7720at 20ºC. A solution of benzene and toluene at 20ºChas a vapour pressure of 46.0 torr. Find the molefraction of benzene in vapour above the solution. [IIT-1996]

Sol. Given that,Density of benzene = 0.877 g ml –1 Molecular mass of benzene (C6H6)

= 6 × 12 + 6 × 1 = 78

∴ Molar volume of benzene in liquid form =

877.0

78ml

=877.0

78 ×

1000

1L = 244.58 L

And molar volume of benzene in vapour phse

=877.0

78 ×

1000

2750L = 244.58 L

Density of toluene = 0.867 g ml –1 Molecular mass of toluene (C6H5CH3)

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




=867.0

92ml =

867.0

92 ×

1000

1 L

And molar volume of toluene in vapour phase

=867.0

92 ×

1000

7720L = 819.19 L

Using the ideal gas equation,PV = nRT

At T = 20ºC = 293 K

For benzene, P = 0BP =

V

nRT

=58.244

293082.01 ×× = 0.098 atm

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

P = 0TP =

V

nRT

=19.819

293082.01 ×× = 0.029 atm

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

PB = 0BP 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 = 0.457According to Dalton's law,

PB = PM 'Bx (in vapour phase)

or mole fraction of benzene in vapour form,

'Bx =

M

B

P

P =

0.46

457.048.74 × = 0.74

9. The values of Λ∞ for HCl, NaCl and NaAc (sodiumacetate) are 420, 126 and 91 Ω –1 cm2 mol –1,respectively. The resistance of a conductivity cell is520 Ω when filled with 0.1 M acetic acid and dropsto 122 Ω when enough NaCl is added to make thesolution 0.1 M in NaCl as well. Calculate the cellconstant and hydrogen-ion concentration of thesolution. 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 solutionk = 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 =G

k =

)00627.0(

)cm0126.0(1

11

−

−−

Ω

Ω = 2.01 cm –1

Conductivity of 0.1 M HAc solution

k =R

K =

Ω

−

520

cm01.2 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 isgiven as

α =∞

Λ

Λ

m

m =

)molcm385(

)molcm65.38(121

121

−−

−−

Ω

Ω ≈ 0.1

and the hydrogen-ion concentration of 0.1 M HAcsolution is

[H+] = cα = (0.1 M)(0.1) = 0.01 MThus, its pH is pH = – log[H+]/M = – log(0.01) = 2

10. An organic compound A, C8H4O3, in dry benzene inthe presence of anhydrous AlCl3 gives compound B.The compound B on treatment with PCl5 followed byreaction with H2/Pd(BaSO4) gives compound C,which on reaction with hydrazine gives a cyclisedcompound D(C14H10 N2). 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

OH2 NNH2

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 – H

OH

C6H5

N

N




MATHEMATICS

11. With usual notation, if in a triangle ABC

11

c b +=

12

ac + =

13

ba +, then prove that

7

Acos

= 19

Bcos

= 25

Ccos

[IIT-1984]

Sol. Let11

c b +=

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

ac b 222 −+

= 2

222

)30(2492536

λλ−λ+λ =

51

cos B =ac2

bca 222 −+

=2

222

70

364925

λ

λ−λ+λ =

35

19

cos C =ab2

c ba 222 −+

=2

222

84

253649

λ

λ−λ+λ =

7

5

∴ 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 fromD 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 = ACLet 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,

a

x +

h

y = 1 ...(1)

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

h

x –

a

y = 0 ⇒ x =

a

hy ...(2)

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

as follows :

2a

hy +

h

y = 1 ⇒ y =

22

2

ha

ha

+

∴ E =

++ 22

2

22

2

ha

ha,

ha

ah

Since F is mid-point of DE,

∴ F

++ )ha(2

ha,

)ha(2

ah22

2

22

2

∴ slope of AF =

)ha(2

ah0

)ha(2

hah

22

2

22

2

+−

+−

=2

222

ah

ha)ha(h2

−

−+

⇒ m1 =ah

)h2a( 22 +− ...(3)

and slope of BE =

aha

ah

0ha

ha

22

22

2

++

−+ =

2322

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 realnumbers. Prove that if f(x) is an integer whenever x isan 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 integerwhenever x is an integer. [IIT-1998]

Sol. Suppose f(x) = Ax2 + Bx + C is an integer wheneverx 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. Also2A, 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 thearch) is in the form of a rectangle surrounded by asemi-circle. The semi-circular portion is fitted withcoloured glass while the rectangular part is fitted withclear glass. The clear glass transmits three times asmuch light per square meter as the coloured glass does.What is the ratio for the sides of the rectangle so thatthe 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 + π b = K (say) ...(1) Now, let the light transmission rate (per squaremetre) of the coloured glass be L and Q be the totalamount of transmitted light.

Coloured

glass

Clear glassa a

Then, Q = 2ab(3L) +2

1π b2(L)

Q =2

Lπ b2 + 12ab

Q =2

Lπ b2 + 6b (K – 4b – π b)

Q =2

L6Kb – 24b2 – 5π b2

db

dQ =

2

L6K – 48b – 10π b = 0

⇒ b =π+1048

K 6 ...(2)

and2

2

db

Qd =2L –48 + 10πLa

Thus, Q is maximum and from (1) and (2),

(48 + 10π) b = 6K and K = 2a + 4b + π b

⇒ (48 + 10π) b = 62a + 4b + π b

Thus, the ratio =a

b2 =

π+6

6

15. If f : [–1, 1] → R and f´(0) =

∞→ n

1nf lim

n and f(0) = 0.

Find the value of :

π∞→

2limn

(n + 1)cos –1

n

1 – n given that

0 <

−

∞→ n

1coslim 1

n <

2

π [IIT-2004]

Sol. Hereπ∞→

2limn

(n + 1)cos –1

n

1 – n

=∞→n

lim n

−

+

π− 1

n

1cos

n

11

2 1

=∞→n

lim nf

n

1

Where f

n

1 =

+

π n

11

2cos –1

n

1 – 1 = f´(0)

= ∞→ n

1nf lim)0´(f given n

∴ π∞→

2limn

(n + 1)cos –1

n

1 – n = f´(0) ...(1)

where f(x) =π2

(1 + x) cos –1x – 1, f(0) = 0

⇒ f´(x) =

+−

−+

π− xcos

x1

1)x1(

2 1

2

⇒ f´(0) =

π

+−π 2

12

= 1 –π2

...(2)

∴ from equation (1) and (2)

π∞→

2limn

(n + 1) cos –1

n

1 – n = 1 –

π2

bility

• 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 degreesof freedom f 1 and f 2 and molar heat capacities at

constant volume1VC and

2VC respectively, the

lnP versus lnV graph is plotted for adiabatic process as shown, then

yx

ln V

Ln P

(A) f 1 > f 2 (B) f 2 > f 1

(C)12 VV CC < (D)

21 VV CC >

Q. 3 A particle of charge q and mass m movesrectilinearly 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 atβα

=x

(D) the maximum acceleration of the particle

ism

qα

Q. 4 Speed of a body moving in a circular pathchanges 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 areaconnected to a battery of internal resistance zero.If the length of the resistor is doubled bystretching 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 halfQ. 6 A mosquito with 8 legs stands on water surface

and each leg makes depression of radius ‘a’. If thesurface tension and angle of contact are ‘T’ andzero 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 Iin different direction. Four loops enclosing thewires 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, Jodh ur Branch

Physics Challenging Problems

Solut ions wi l l be ubl i shed in next i ssue

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 bythe

magnetic force tomove a unit chargealong the loop is zero

Q. 8 An ideal monoatomic gas undergoes differenttypes 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 willalso increase

(B) PV2 = constant (Q) If volume increasesthen temperature willdecrease

(C) C = Cv + 2R (R) For expansion, heatwill have to besupplied to the gas

(D) C = Cv – 2R (S) If temperatureincreases then workdone 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

NMe b

volt2016

120VV da =×=−

volt402

6

120VV e b =×=−

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

20

E2

E =∈σ+ ………..(1)

10

EE =−∈σ

………..(2)

Adding (1) & (2)

120

EE +=∈σ

012 A)EE(q ∈+=

5

96

120 107.8

10941032.0

)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]

K

b bd

AC 0K

+−

ε=

We set b = 0

Cd

AC 0

K =ε

= if CK = 2C

Then,d b2

b2K

d

A2

K

b bd

A 00

−=⇒

ε=

+−

ε

d b&0K ≤>∴

0d b2&

d b2

b2K >−

−

=∴

d b2

d≤<∴

2

d b >∴

6. [A,B,C,D]

For option (A)

1 2 3

10V

20Vi

i = 4

10 – 20 = 4

10 = 2.5 A

For option (B)

1 2 3

20V

i =3

20A

olution

Physics Challenging Problems

Set # 4

8ue s ti ons wer e Publ i s he d i n Au us t I ss ue




For Option (C)

1 2 3

10V

20Vi

A35

610

3211020i ==++

−=

For Option (D)

1 2 3

20V

Amp201

20i ==

7. [B,D]

Rxθ

θ

C

→v

m521

101

qB

mvR =

××

==

Coordinates of center ⇒

3

5

35sinR x +=×+=θ+=

45

45cosR y −=×−=θ−=

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

π=×

×π×=

π=

21

12

qB

m2T

8. [B,C]P1 P2

9Ω

Q1 Q2

2Ω 2Ω

504.01v b ××==ε l

volt2.0=ε

A02.0i10

2.0

R i

eq

=⇒=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 Earthdoes, 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 onthe axis of a circular ring of radius R = 30 cm.Distance of the particle from centre of the ring isa = 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 ofradius x and radial thickness de as shown in Figurer.Its area is dS = 2πx dxDistance of every point of this ring from point

charge is r = 22 xa +

∴ Electric field strength at circumference of this

ring is E =2

0 r

q

4

1

πε

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

ring considered is given by cosθ =r

a Flux passing

through this ring is dφ =→→dsE

dφ = d dS cos θ

=

+π

+πε 22220 xa

a)dxx2(

)xa(

q

4

1

Hence, total flux passing through the given ring is

φ = ∫=

= +ε

R x

0x 2/3220 )xa(

xdx

2

aq =

+ε 220 xa

1 –

q

1

2

qa

= 105 NC –1 m2

2. A parallel plate capacitor is filled by a di-electricwhose di-electric constant varies with potentialdifference V according to law K = aV, wherea = 2 volt –1. An air capacitor having samedimensions charged to a potential difference of

V0 = 28 volt is connected in parallel to theuncharged capacitor filled with above mentioneddi-electric. Calculate ratio of charge on capacitorfilled by aforesaid di-electric to charge in aircapacitor in steady state.

Sol. Let area of plates of each capacitor be A and letseparation between them be d. Capacitance of air

capacitor, C0 =d

A0ε

And capacitance of capacitor, filled with di-electric,C =

d

KA0ε = aVC0 = 2VC0

Initial charge in air capacitor, q0 = C0V0

When air capacitor is connected across the othercapacitor, some charge flows from air-capacitor tothe other capacitor so that potential differencesacross two capacitors in steady state becomes equal.Let the potential difference be V.Charge of capacitor filled with di-electric will beequal to

q1 = CV = 2C0V2

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

But q1 + q2 = q0

or 2C0V2 + C0V = C0V0

From above equation V = – 4 or 3.5 Negative value is absurd. Therefore, V = 3.5 volts.

∴ 2

1

q

q =

VC

VC2

0

20 = 2V = 7 Ans.

3. A stationary circular loop of radius a is located in amagnetic field which varies with time from t = 0 tot = T according to law B = B0 . t (T – t). If plane ofloop is normal to the direction of field and resistanceof the loop is R, calculate(i) amount of heat generated in the loop during thisinterval, and(ii) magnitude of charge flown through the loopfrom instant t = 0 to the instant when currentreverses 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, andemf is induced. consequently, and emf is induced in

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

Students' Forum PHYSICS




the loop and a current starts to flow through theloop. 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 = –dt

dφ = – πa2 B0(T – 2t)

Induced current i =R e = 202 )T – t2(

R Baπ ...(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

42

dt.)T – t2(R

Badt.P

=R

TBa3o

20

42

π Ans. (1)

The current reverses its sign when its magnitudereduces to zero. Let this happen at instant t = t 0.Substituting t by t0 in equation (1),

∴ R

Ba 02π

(2t0 – T) = 0 or t0 =2

T

Substituting i bydt

dq 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( = –

R 4

TBa 20

2π

or magnitude of charge that flows =R 4

TBa 20

2π

Ans. (2)

4. Oxygen is used as working substance in an engineworking on the cycle shown in Figure

P

V

1

4

32

Processes 1-2, 2-3, 3-4 and 4-1 are isothermal,isobaric, adiabatic and isochoric, respectively. Ifratio of maximum to minimum volume of oxygenduring the cycle is 5 and that of maximum tominimum absolute temperature is 2, assumingoxygen to be an ideal gas, calculate efficiency of theengine.Given, (0.4)0.4 = 0.693 and loge 5 1.6094

Sol. Volume of gas is minimum at state 2 during thecycle. Let it be V0. Then maximum volume of gasduring the cycle will be equal to 5V0 which is atstates 4 and 1. Therefore, V4 = V1 = 5V0.Temperature during the cycle is maximum at the endof isobaric process 2 → 3 i.e. state 3 and minimumat the end of isochoric cooling process 4 → 1 i.e.state 1. Let minimum absolute temperature be T0.Then T1 = T0 and T3 = 2T0.Since gas is Oxygen which is di-atomic, therefore,

Cv =2

5R, C p =

2

7 R and γ =

5

7

Since, process 1 → 2 is isothermal, therefore,temperature during the process remains constant.Hence temperature T2 is also equal to T0.Considering n mole of the gas,Work done by the gas during isothermal process1 → 2,

W12 = nRT1.log1

2

V

V = – nRT0 loge 5

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

Now considering isobaric process 2 → 3

2

3

V

V =

2

3

T

T = 2 or V3 = 2V0

Heat supplied to gas during the process,

Q23 = nCP(T3 – T2) =2

7 nRT0

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

Now considering adiabatic process 3 → 4,

V3 = 2V0 , T3 = 2T0V4 = 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 bythe 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)0.4Heat supplied to the gas during heating process,

QS = Q23 =2

7 ( 6 – loge 5 – 5 × 0.40.4) = 0.2642

or η =sQ

W =

7

2 ( 6 – loge 5 – 5 × 0.40.4) = 0.2642

or η = 26.42 % Ans.




5. A thin convex lens of focal length f and havingaperture diameter d is used to focus sun rays on ascreen. If speed of light in vacuum is c, absolutetemperature of sun TS and rays fall on the lensnormally, calculate pressure experienced by thesurface on which rays are converged.

Sol. When light rays incident on the screen, momentum

of rays reduces to zero. Due to change in momentumof incident rays, the surface experiences a force.Let radius of sun be R s 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 = 2sR 4π

∴ Rate of radiation from sun, E = 2s

4s R 4T πσ

Intensity of sun rays at earth,

I = 2r 4

E

π = 2

2s

4s

r

R Tσ Wm

–2

Area of convex lens, A =2

2

d

π =

4

1πd2m2

∴ Power incident on the lens,

P = AI =2

22s

4s

r 4

d.R T πσW

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

screen =c

P

Just after incidence, momentum of rays reducesreduces to zero, therefore, magnitude of rate ofchange of momentum of rays,

dt

dp =

0 –

c

P =

cr 4

dR T2

22s

4s πσ

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

Hence, force on screen, F =c4

dR T2

22s

4s

ρ

πσ

But for a lens,u

v

O

I= Where I is the size of image.

∴ Radius of image circle formed on the screen

=u

v. O =

r

f . R s

∴ Area of image circle, a =2

sR r

f

π

It means force F acts on area a

∴ Pressure =a,area

F,force =

cf 4

dT2

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 changedinto 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 chargethrough a certain surface.

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

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

Its unit is C/s or ampere.

Electric current has direction as well asmagnitude 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 : Thistype 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 tomechanical transfer of charged particle is called

convection current. Convection current in differentsituation.

Case I : If a point charge is rotating with constantangular velocity ω.

I =T

q; T =

ω

π2 ⇒ I =

π

ω

2

q

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

I = R λω

θ

J

∆S

or J =θ∆

∆cosS

I

Its unit A/m2

Electric current can be defined as flux of currentdensity vector.

i.e., i = ∫ →→dS. j

Relation between drift velocity and current

density dv = –en

j→

Here, negative sign indicates that drifting of electrontakes place in the opposite direction of currentdensity.

The average thermal velocity of electron is zero.

Electric resistance : Electric resistance (R) isdefined as the opposition to the flow of electric

charge through the material.It is a microscopic quantity.

Its symbol is

Its unit is ohm.

(a)

(b)

R =A

lρ

where, R = resistance,

ρ = resistivity of the material,

l = length of the conductor,

A = area of cross section

Continuity Equation :

∫→→

cdS. j = –

dt

dq

The continuity equation is based on conservation principle of charge.

Drift Velocity (vd) : When a potential differenceis applied between ends of metallic conductor, an

Current Electricity

PHYSICS FUNDAMENTAL FOR IIT- EE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY




electric field is established inside the metallicconductor. Due to this, electron modify theirrandom motion and starts to drift slowly in theopposite direction of electric field. The averagevelocity of drifting possessed by electron isknown as drift velocity.

→dv =

→

τE

m

e

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.

R t = R 0(1 + αt)

where,

R t = resistance of conductor at temperature tºC,

R 0 = resistance of conductor at 0ºC,

α = temperature coefficient.

A

i

Some Important Points :

(a) 'α' is proportionality constant known astemperature coefficient of resistance variation.

(b) The value of α does not depend upon initial andfinal resistance of the conductor.

(c) The value of α depends upon the unit which ischosen.

(d) The value of α may by negative.

Electric Conductance (G) :

It is reciprocal of resistance, G =R

1

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 passingthrough 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, thyristorsetc.

EMF and PD of a Cell : A device which supplieselectric energy is called a seat of emf. The seat ofemf is also called a cell.

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

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

When the terminals of a cell are connected to anexternal resistance, the cell is said to be in closedcircuit.

E.M.F. has no electrostatic origin.Internal Resistance of a Cell (r) : Internalresistance of a cell is the resistance of itselectrolyte.

The internal resistance of cell :

(a) Varies directly as concentration of the solution ofthe cell.

(b) Varies directly as the separation betweenelectrodes i.e., length of solution betweenelectrodes.

(c) Varies inversely as the area of immersedelectrodes.

(d) is independent of the material of electrodes.Potential difference across the cell :

Potential difference across the first cell

V1 = E1 + Ir 1 (discharging of cell)

R

i

E1r 1 E2r 2

Potential difference across the second cellV2 = E2 – Ir 2 (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

Ri R i

When a battery being charged, the terminalvoltage is greater than its emf V = E + Ir.

Kirchhoff's Law : Kirchhoff's law is able tosolve complicated circuit problems.

(i) First Law : Incoming current = Outgoing current

I1 + I2 = I3 + I4 + I5

I2

I5

I1I3

I4

This law is based upon conservation principle ofcharge.

(ii) Second Law : (Loop rule or voltage law.) Thislaw is based upon conservation principle ofenergy.

Grouping of resistors :

Case I : Resistors in series

R MN = R eq = R 1 + R 2

M R 1 R 2 N

In general,

R eq = R 1 + R 2 + ... + R n

Case II : Resistors in parallel

MNR

1 =

eqR

1 =

1R

1 +

2R

1

M

R 1

R 2 N

In general,

MNR

1 =

1R

1 +

2R

1+ ... +

nR

1

Problem solving strategy. : Power and Energy in circuits

Step 1 Identify the relevant concepts :

The ideas of electric power input and output can beapplied to any electric circuit. In most cases you’llknow when these concepts are needed, because the

problem will ask you explicitly to consider power orenergy.

Step 2 Set up the problem using the following steps :

Make a drawing of the circuit.

Identify the circuit elements, including sources ofemf and resistors.

Determine the target variables. Typically theywill be the power input or output for each circuitelement, or the total amount of energy put into ortaken 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 circuitwhen the current I runs through the source from –to +. The energy is converted from chemicalenergy in a battery, from mechanical energy in agenerator, or whatever. In this case the source hasa positive power output to the circuit or,equivalently, a negative power input to thesource.

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 passesthrough the source in the direction from + to – .This occurs in charging a storage battery, whenelectrical energy is converted back to chemicalenergy. 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 currentthrough a resistor, It removes energy from acircuit at a rate given by VI = I2R = V2/R, whereV is the potential difference across the resistor.

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

You may need to calculated the total energydelivered to or extracted from a circuit element ina 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. Thisconservation can be expressed in either of two forms:

“net power input = net power output” or “thealgebraic sum of the power inputs to the circuitelements is zero.”

Problem solving strategy : Series and Parallel

Step 1 Identify the relevant concepts : Many resistornetworks are made up of resistors in series, in

parallel, or a combination of the two. The keyconcept is such a network can be replaced by a singleequivalent resistor.




Step 2 Set up the problem using the following steps:

Make a drawing of the resistor network.

Determine whether the resistors are connected inseries or parallel. Note that you can often considernetworks such as combinations of series and

parallel arrangements.

a R 1 x R 2 y R 3 b

I I(a) R 1, R 2, and R 3 in series

R 1

I

(b) R 1, R 2, and R 3 in parallel

R 2

R 3 I

a b

Determine what the target variables are. Theycould include the equivalent resistance of thenetwork, the potential difference across eachresistor, or the current through each resistor.


Use Eq. R eq = R 1 + R 2 + R 3 (resistors in series)

oreqR

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 toseries and parallel combinations.

When calculating potential differences, rememberthat when resistors are connected in series, thetotal potential differences across the combinationequals the sum of the individual potentialdifferences. When they are connected in parallel,the potential difference across the parallelcombination.

Keep in mind the analogous statements forcurrent. 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 totalcurrent through the combination equals the sumof the currents through the individual resistors.

Step 4 Evaluate your answer : Check whether yourresults are consistent. If resistors are connected inseries, the equivalent resistance should be greaterthan that of any individual resistor; if they areconnected 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’srules are important tools for analyzing any circuitmore complicated than a single loop.

Step 2 Set up the problem using the following steps :

Draw a large circuit diagram so you have plentyof room for labels. Label all quantities, known

and unknown, including an assumed direction foreach unknown current and emf. Often you willnot know in advance the actual direction of anunknown current or emf, but this does not matter.If the the actual direction of a particular quantityis opposite to your assumption, the result willcome out with a negative sign. If you areKirchhoff’s rules correctly, they will give you thedirections as well as the magnitudes of unknowncurrents and emfs.

When you label currents, it is usually best to usethe junction rule immediately to express thecurrents of as few quantities as possible. For

example, fig (a) shows a circuit correctly labeled;fig. (b) shows the same circuit, relabeled byapplying the junction rule to point a to eliminate I3.

r 1 ε1r 2 ε2

I1

I1

I3 R 3

I2

I2

R 1 a R 2

+ +

(a)

r 1 ε1 r 2 ε2

I1

I1

I1 + I2 R 3

I2

I2

R 1 a R 2

+ +

(b)

Determine which quantities are the targetvariables.


Choose any closed loop in the network anddesignate a direction (clockwise orcounterclockwise) to travel around the loop whenapplying the loop rule. The direction does nothave to be the same as any assumed currentdirection.




Travel around the loop in the designateddirection, adding potential differences as youcross them. Remember that a positive potentialand a negative potential difference corresponds toa decrease in potential. An emf is counted as

positive when you traverse it from (–) to (+), andnegative when you go from (+) to (–). An IR termis 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. Ineach part of the figure “travel” is the directionthat we imagine going around a loop while usingKirchhoff’s loop law, not necessary the directionof current.

Equate the sum is Step 2 to zero.

If necessary, choose another loop to get adifferent relation among the unknowns, andcontinue until you have as many independentequations as unknowns or until every circuitelement has been included in a at least one of thechosen loops.

Solve the equations simultaneously to determinethe unknowns. This step involves algebra, not

physics, but it can be fairly complex. Be carefulwith algebraic manipulations; one sign error will

prove fatal to the entire solution.

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

– V b.

Step 4 Evaluate your answer : Check all the step inyour algebra. A useful strategy is to consider a loopother than the ones you used to solve the problem; ifthe sum of potential drops around this loop is notzero, you made an error somewhere in yourcalculations. As always, ask yourself whether isanswer make sense.

Travel

+ –+ ε

Travel

–+ – ε

Travelε ε

Travel

–+ –IR

Travel

R

I

Travel

+ – +IR

Travel

R

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 removedfrom 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 asthe bridge is balanced. So the 3Ω resistance in BDarm is ineffective and can be omitted from the circuit.The resistance of ABC branch is 2Ω + 4Ω = 6Ω asAB and BC are in series. Similarly the resistance ofA 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

R

1 =

6

1 +

3

1 =

2

1 ∴ R = 2Ω

(i) The current drawn from 2 volt accumulator is

i =R

E =

2

2 = I amp.

(ii) The current through 3Ω resistor is zero.(iii) When the 3Ω resistor is removed from thecircuit, there will be no change.

2. A battery of e.m.f. 5 volt and internal resistance 20Ω is connected with a resistance R 1 = 50 Ω and aresistance R 2 = 40Ω. A voltmeter of resistance 1000 Ω is used to measure the potential difference across R 1.What percentage error is made in the reading ?




Sol. The circuit is shown in fig.

V

1000Ω

20Ω

R 2 = 50Ω

5V

R 1 = 50Ω

When voltmeter is not connected

current in the circuit i =21 R R r

E

++

∴ i =405020

5

++ =

110

5 =

22

1A

Potential difference across R 1 = i × R 1

=22

1 × 50 = 2.27 volt.

When the voltmeter is connected across R 1.In this case the galvanometer resistance is in parallelwith R 1. Hence

Equivalent resistance =501000

501000

+

× = 47.62 ohm

Current in the circuit

=62.474020

5

++ =

62.107

5A

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 isconnected across 400 ohm resistance. Calculate whatthe same voltmeter will read when it is connectedacross 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 equallyshared.

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

´R

1 =

R

1 +

400

1 =

R 400

R 400 + or

R 400

R 400

+

But R´ should be equal to 300 ohm. Hence

R 400

R 400

+ = 300 ∴ R = 1200 ohm

Thus, voltmeter resistance is 1200 ohm.When the voltmeter is connected across 300 ohm, theeffective 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 240ohm and 400 ohm.Potential diff. across 240 ohm : Potential differenceacross 400 ohm

= 240 : 400 = 3 : 5As total potential is 60 V, hence potential differenceacross 240 ohm, i.e., across resistance 300 ohm will

be

8

3 × 60 = 22.5 V.

4. In the circuit shown in fig. E, F, G and H are cells ofe.m.f. 2, 1, 3 and 1 volt and their internal resistancesare 2, 1, 3 and 1 ohm respectively. Calculate

E –+

+ –

–

– +

+

A B

D C

G

F H2Ω

(i) The potential difference between B and D and(ii) the potential difference across the terminals ofeach 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 andADBA, we have

2i + 1i + 2i1 = 2 – 1 = 1or 3i + 2i1 = 1 ...(2)

2VA B

D C

3V

1V

1V

2Ω

2Ω

3Ω

1Ω

i

i2

1Ωi1




Applying Kirchhoff's second law to mesh DCBD, weget

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

i1 =13

1 amp., i2 =

13

6amp. and i =

13

5amp.

(i) Potential difference between B and D= 2i1 = 2

13

1 =

13

2 volt.

(ii) Potential difference across G

= E – i2R = 3 –13

36× = 1.61 V

Potential difference across H

= 1 –

−

13

6(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 oppositecorner. Find the joint resistance between the corners.

Sol. The skeleton ABCDEFGH, is shown in fig.

A

E Fi/6

B

GH

i/3i/3

i/6

i/6i/3

CD i/6

i/3

i/6

ii/6

i/3

i/3

This skeleton consists of twelve wires. Let theresistance of each wire be r. Here the current i entersat corner A and leaves at corner G. The current i atcorner 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, Fand H, the currents again combine to give currents,each of magnitude i/3 along CG, FG and HGrespectively. At corner G, all these currents combineso that the current leaving at G is i.Let R be the equivalent resistance between thecorners A and G. Taking any one of the paths sayABCG, we have

VAG = VAB + VBC + VCG

iR =3

ir +

6

ir +

3

ir

or R =6

5r

According to given problem r = 6 ohm

∴ R =6

5 × 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 withuniform speed, its is said to execute a uniformcircular motion.

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

Oθ

ω =t

θ radian / second

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

ω =T

2π

and if n revolutions are made in 1s then

n = T

1 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 moveswith uniform speed v in a circle of radius r it is actedupon by an acceleration v2/r in the direction of centre.

It is called centripetal acceleration. The accelerationhas a fixed magnitude but its direction iscontinuously changing. It is always directed towardsthe centre of the circle.

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

force acting on it towards the centre isr

mv2

. This is

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

Without centripetal force, a body can not move on acircular path. Earth gets this force from thegravitational attraction between earth and sun;electron moves in circular path due to electrostaticattraction between it and nucleus. A cyclic or carwhile taking turn, gets the centripetal force from thefriction between road and type. To create this force,the vehicle tilts itself towards the centre. If it makesangle θ with the vertical in tilted position then thanθ = v2/rg. where v is its velocity and r is the radius ofthe path. In order to avoid skidding (or slipping), the

angle of tilt θ with vertical should be less than angleof friction λ. i.e. tan θ < tan λ

orrg

v2

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

In limiting conditionrg

v2

= µ or v = g.r .µ

This is the maximum safe speed at the turn.

Since centripetal force is provided by the friction, itcan never be more than the maximum valueµR = (µmg) or frictional force.

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

mg

Ta

r

T

v bB

mg

Ava

C

Ta + mg =r

mv2a or Ta =

r

mv2a – mg

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

0 =r

mv2c – mg or vc = gr

Circular Motion, Rotational Motion

PHYSICS FUNDAMENTAL FOR IIT- EE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY




This minimum speed is called critical speed (vc). Ifthe speed at A is less than this value, the particle willnot reach up to the highest point. To reach with thisspeed at A, the body should have speed at B given bythe conservation laws viz.

Decrease in kinetic energy = increase in potentialenergy

21 mv b

2 –21 mva

2 = mg.2r

v b2 = va

2 + 4gr

for critical speed va = vc = gr

∴ v b2gr + 4gr or v b = gr 5

Therefore, the body should have speed at B at least

gr 5 , so that it can just move in vertical circle.

Tension in string at B is given by.

T b – mg =

r

mv2 b or T b = mg +

r

vgr 5m = 6mg

This means that the string should be able to stand to atension, equal to six times the weight of the bodyotherwise the string will break.

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

A

C

Bmg

mg cos θQ

P

v p Tθ

T – mg cos θ =r

mv2 p

or T = m

θ+ cosg

r

v2r

At point A, θ = 180º; Ta = m

− g

r

v2a

At point B, θ = 0º; T b = m

+ g

r

v2 b

Conical pendulum :

A conical pendulum consists of a string AB (fig.)whose upper end is fixed at A and other and B is tiedwith a bob. When the bob is drawn aside and is givena horizontal push. Let it describe a horizontal circlewith constant angular speed ω in such a way that ABmakes a constant angle θ with the vertical. As thestring traces the surface of a cone, it is known asconic pendulum.

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

T cos θ = mg

The horizontal component is equal to the centripetalforce 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 ofmass.

Consider the case of a body of an arbitrary shape of nXY plane as shown in fig. Let the body consist ofnumber of

P1(x1, y1)(x2, y2)

P2(x, y )

P3(x3, y3)

XO

Y

particles P1, P2, P3, .... of masses m1, m2, m3, ..... andcoordinates (x1, y1), (x2, y2), (x3, y3), ..... If )y,x( be

the coordinates of centre of mass, then

x =.....mmm

....xmxmxm

321

332211

+++

+++ =

n

nn

m

xm

Σ

Σ

and y =....mmm

...ymymym

321

332211

+++

+++ =

n

nn

m

ym

Σ

Σ

When there is a continuous distribution of massinstead 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 byintegration 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 r 1 and r 2 respectively with respectto origin. Now the position vector r of the centre ofmass is given by

(m1 + m2)r = m1r 1 + m2r 2 ...(1)

Thus, the product of the total mass of the system and position vector of the centre of mass is equal to thesum of the products of the individual masses andtheir respective position vectors. Hence

r =21

2211

mm

r mr m

+

+ ...(2)

Now the velocity of centre of mass of the system is

given by v = dt

dr

The acceleration of the centre of mass is given by

a =dt

dv =

dt

d

dt

d =

2

2

dt

xd

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

f(total) = Mdt

dv

When no external force acts on the system, then

0 = M dt

dv

or dt

dv

= 0

∴ v = constant

Therefore, when no external force acts on the system,the centre of mass of an isolated system move withuniform 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 constitutingthe body and the square of their respective distancefrom the axis.

Radius of Gyration : If we consider that the wholemass of the body is concentrated at a distance K fromthe axis of rotation, then moment of inertia I can beexpressed as

I = MK 2

where M is the total mass of the body and K is theradius of gyration. Thus the quantity whose squarewhen multiplied by the total mass of the body givesthe moment of inertia of the body about that axis isknown as radius of gyration.

Theorems on moment of inertia :

Theorem of parallel axes : According to thistheorem, the moment of inertia I of a body about anyaxis is equal to its moment of inertia about a parallelaxis through centre of mass IG plus Ma2 where M isthe mass of the body and a is the perpendiculardistance between the axes, i.e., I = IG + Ma2

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

I = Ix + Iy

Table of moment of inertia :

Body AxisMoment of

inertia

1. Thin uniform rod

of length l

Through itscentre and

perpendicular toits length

12

M 2l

2. Thin rectangularsheet of sides aand b.


perpendicular toits plane

M

+

12

b

12

a 22

3. Thickrectangular barof length l,

breadth b andthickness t.

Through itsmidpoint and

perpendicular toits length

M

+

12

b

12

22l

4. Uniform solidsphere of radiusR

About a diameter52 MR 2

5. Circular ring ofradius R.



MR 2

6. Disc of radius R. Through itscentre and


2

1MR 2

7. Solid cylinder of

length l andradius R.

(i) Through its

centre and parallel to itslength

(ii) Through itscentre and

perpendicular toits length.

2

1MR 2

M

+

124

R 22 l

Angular momentum of a rotating body :

In case of rotating body about an axis, the sum of themomentum of the linear momentum of all the




particles about the axis of rotation is called angularmomentum about the axis.

Q Also the angular momentum of rigid body aboutan axis is the product of moment of inertia and theangular 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 = 2

1mv

2

Rotational K.E. = 2

1Iω

2

Kinematics equation of a rotating rigid body :

The angular velocity of a rotating rigid body isdefined as the rate of change of angular displacement,

i.e.,→ω = )dt/d(

→θ

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

→α =

dt

d→ω

=2

2

dt

d→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 +2

1αt2

and ω2 = ω02 + 2 αθ

These equations are similar to usual kinematics

equation of translatory motion.

v = u + at, s = ut +2

1at2

and v2 = u2 + 2as

Problem Solving Strategy : Rotational Dynamics for

Rigid Bodies :

Our strategy for solving problems in rotationaldynamics is very similar to the strategy for solving

problems that in involve Newton’s second law.

Step-1 : Identify the relevant concepts : The equationΣτ = Iαz is useful whenever torques act on a rigid

body–that is, whenever forces act on a rigid body insuch a way as to change the state of the body’srotation.

In some cases you may be able to use an energyapproach instead. However, if the target variable is a

force, a torque, an acceleration, an angularacceleration, or an elapsed time, the approach usingΣτ = 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 bodyor bodies to be analyzed.

For each body, draw a free-body diagramisolating the body and including all the forces(and only those forces) that act on the body,including its weight. Label unknown quantitieswith algebraic symbols. A new consideration isthat you must show the shape of the bodyaccurately, including all dimensions and angles

you will need for torque calculations.Choose coordinate axes for each body andindicate a positive sense of rotation for eachrotating body. If there is a linear acceleration, it’susually simplest to pick a positive axis in itsdirection. If you know the sense of αz in advance,

picking it as the positive sense of rotationsimplifies the calculations. When you represent aforce in terms of its components, cross out theoriginal 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, rotationalmotion, or both. Depending on the behavior of the body in question, apply ΣF = ma

r, Στz = Iαz, or

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

There may be geometrical relations between themotions of two or more bodies, as with a stringthat unwinds from a pulley while turning it or awheel that rolls without slipping. Express theserelations in algebraic form, usually as relations

between two linear accelerations or between alinear acceleration and an angular acceleration.

Check that the number of equations matches the

number of unknown quantities. Then solve theequations to find the target variable(s).

Step-4 : Evaluate your answer : Check that thealgebraic signs of your results make sense. As anexample, suppose the problem is about a spool ofthread. If you are pulling thread off the spool, youranswers should not tell you that the spool is turningin the direction the results for special cases orintuitive expectations. Ask yourself : Does this resultmake sense ?”




Problem Solving Strategy: Equilibrium of a Rigid Body

Step-1 : Identify the relevant concepts : The first andsecond conditions for equilibrium are usefulwhenever there is a rigid body that is not rotating andnot accelerating in space.Step-2 : Set up the problem using the following steps:

Draw a sketch of the physical situation, includingdimensions, and select the body in equilibrium to

be analyzed.Draw a free-body diagram showing the forcesacting on the selected body and no others. Do notinclude forces exerted by this body on other

bodies. Be careful to show correctly the point atwhich each force acts; this is crucial for correcttorque calculations. You can't represent a rigid

body as a point.Choose coordinate axes and specify a positivesense of rotation for torques. Represent forces interms of their components with respect to the axesyou have chosen; when you do this, cross out theoriginal force so that you don't included it twice.

In choosing a point to compute torques, note thatif a force has a line of action that goes through a

particular point, the torque of the force withrespect to that point is zero. You can ofteneliminate unknown forces or components fromthe torque equation by a clever choice of point foryour calculation. The body doesn't actually haveto be pivoted about an axis through the chosen

point.Step-3 : Execute the solution as follows :

Write equations expressing the equilibriumconditions. Remember that ΣFx = 0, ΣFy = 0, andΣτz = 0 are always separate equations; never add

x-and y-components in a single equation. Alsoremember that when a force is represented in termof its components, you can compute the torque ofthat force by finding the torque of eachcomponent separately, each with its appropriatelever arm and sign, and adding the results. This isoften easier than determining the lever arm of theoriginal force.You always need as many equations as you haveunknowns. Depending on the number ofunknowns, you may need to compute torqueswith respect to two or more axes to obtain enoughequations. Often, there are several equally goodsets of force and torque equations for a particular

problem; there is usually no single "right"combination of equations. When you have asmany independent equations as unknowns, youcan solve the equations simultaneously.

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

Solved Examples

1. A particle a moves along a circle of radius R = 50 cmso that its radius vector r relative to point O (fig.)rotates with the

A

R

CO

r

A

R

CO

r

Y

Xθ

θ

(a) (b) constant angular velocity ω = 0.40 rad/sec. Find themodulus of the velocity of the particle and modulusand direction of its total acceleration.

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

)2sin(

r

θ−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, vdt

dr = – 4R cos θ sin θ

dt

dθi + 2R cos 2θ

dt

dθ j

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

a =dt

dv = 4 R cos 2 θ

dt

dθi – 4 R ω sin 2 θ

dt

dθ j

= – 4 R ω2 cos 2 θ i – 4R ω2 sin 2θ j

|a| = 4 R ω2

2. A particle describes a horizontal circle on the smooth

inner surface of a conical funnel as shown in fig. Ifthe height of the plane of the circle above the vertex9.8 mark cm, find the speed of the particle.

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

R sin α

R

αr

h=9.8 cmα

mg

mv2/r

R

c o s 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) actingradially 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 =

2v

r g ...(3)

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

h

r =

2v

r g 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 4mradius with a constant speed of 5m/sec. What is itsangular momentum about (i) the centre of circle (ii) a

point on the axis of the circle and 3 m distant from itscentre ? Which of these will always be in samedirection ?

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/secand 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 secondcase the direction changes.

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

Sol. (a) As the body is rotating about its axis ofsymmetry, the angular momentum vector coincideswith 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 =2

1 × 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., 4rev/sec or 8π radians/sec, the kinetic energy ofrotation 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 andlength l can rotate freely about a vertical axis passingthrough its end A. At a certain moment the end Bstarts experiencing a constant force F which is always

perpendicular to the original position of the stationaryrod and directed in the horizontal plane. Find the

angular velocity of the rod as a function of itsrotation 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)

AB

X

F

FY

θl

→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 =3

1m l2 (for rod) and α = ω (dω/dφ)

∴ l F cos φ =3

1 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 =3

1m l2 ∫

ωωω

0d

l F sin φ =3

1m l2(ω2/2) ∴ ω =

φlm

sinF6




Addition of hydrogen halides to Alkenes :

Markovnikov’s Rule

Hydrogen halides (HI, HBr, HCl, and HF) add to thedouble bond of alkenes :

C = C + HX → – C – C –

H H

These additions are sometimes carried out bydissolving the hydrogen halide in a solvent, such asacetic acid or CH2Cl2, or by bubbling the gaseous

hydrogen halide directly into the alkene and using thealkene itself as the solvent. HF is prepared as polyhydrogen fluoride in pyridine. The order ofreactivity of the hydrogen halides is HI > HBr > HCl> HF, and unless the alkene is highly substituted, HClreacts so slowly that the reaction is not one that isuseful as a preparative method. HBr adds readily, thereaction may follow an alternate course. However,adding silica gel or alumina to the mixture of thealkene and HCl or HBr in CH2Cl2 increases the rateof addition dramatically and makes the reaction aneasy 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. Theaddition of HBr to propene, for example, couldconceivably lead to either 1-bromopropane or2-bromopropane. The main product, however is2-bromopropane :

CH2 = CHCH3 + HBr → CH3CHCH3

Br 2-Bromopropane

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

C = CH2 + HBr → CH3 – C – CH3

Br tert-Butyl bromide

CH3

H3C H3C

2-Methylpropene

(isobutylene)

Consideration of many examples like this led theRussian chemist Vladimir Markovnikov in 1870 toformulate 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 anillustration :

CH2 = CHCH3 → CH2 – CHCH3

H Br

Markovnikov addition

product H Br

Carbon atom

with thegreater

number of

hydrogen atoms

Reactions that illustrate Markovnikov’s rule are saidto 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 thisis the step that occurs first (before the addition of thenucleophilic portion of the adding reagent), it is thestep that determines the overall orientation of thereaction.

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

such as ICl. Because of the greater electro negativityof chlorine, the positive portion of this molecule isiodine. 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 – IH3C

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 toalkenes when the addition is carried out in the

presence of peroxides (i.e., compounds with thegeneral formula ROOR). When alkenes are treatedwith HBr in the presence of peroxides, an anti-Markovnikov addition occurs in the sense that thehydrogen atom becomes attached to the carbon atomwith the fewer hydrogen atoms. With propene, forexample, 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 occursignificantly with HF, HCl, and HI even when

peroxides are present.

Alcohols from Alkenes through Oxymercuration-

Demercuration Markovnikov Addition :

A useful laboratory procedure for synthesizingalcohols from alkenes that avoids rearrangement is a

two-step method called oxymercuration -demercuration.

Alkenes react with mercuric acetate in a mixture oftetrahydrofurane (THF) and water to produce(hydroxyalkyl) mercury compounds. These(hydroxyalkyl) mercury compounds can be reducedto 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 mercuricacetate add to the double bond; in the second step,

demercuration, sodium borohydride reduces theacetoxymercury 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 roomtemperature 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 highyields, usually greater than 90%.

Oxymercuration–demercuration is also highly

regioselective. The net orientation of the addition ofthe elements of water, H – and –OH, is in accordancewith Markovnikov’s rule. The H– becomes attachedto the carbon atom of the double bond with thegreater 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 :

Pentene1CHCH)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 occurin oxymercuration - demercuration. Theoxymercuration - 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 gaschromatography failed to reveal the presence of any2, 3-dimethyl-2-butanol. The acid-catalyzedhydration 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 ofaddition in the oxymercuration stage, and one thatalso explains the general lack of accompanyingrearrangements. 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 ofhydrogen atoms), and the formation of a bridgedintermediate.

Hydroboration : Synthesis of Alkylboranes

Hydroboration of an alkene is the starting point for anumber of useful synthetic procedures, including theanti-Markovnikov syn hydration procedure.Hydroboration was discovered by Herbert C. Brown,and it can be represented in its simplest terms asfollows :

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), ormore conveniently with a reagent prepared bydissolving diborane in THF. When diborane isintroduced 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 areusually carried out in ethers : either in diethyl ether(CH3CH2)2O, or in some higher molecular weightether such as “diglyme” [(CH3OCH2CH2)2O,diethylene glycol dimethyl ether]. Great care must beused in handling diborane and alkylboranes becausethey ignite spontaneously in air (with a green flame).The solution of BH3: THF must be used in an inertatmosphere (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 ourexamples involving the hydroboration of1-methylcyclopentene ring :

CH3

H B

H

CH3

HH

syn addition

anti-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, dioxygenO2 and ozone O3. Dioxygen O2 is stable as a diatomicmolecule, which accounts for it being a gas. The

bonding in the O2 molecules is not as simple as itmight at first appear. If the molecule had twocovalent bonds, then all electron would be paired andthe 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 themolecular orbital theory.

Liquid dioxygen is pale blue in colour, and the solidis also blue. The colour arises from electronictransitions which excite the ground state (a tripletstate) to a singlet state. This transition is 'forbidden'in gaseous dioxygen. In liquid or solid dioxygen asingle photon may collide with two moleculessimultaneously and promote both to excited states,absorbing red – yellow – green light, so O2 appears

blue. The origin of the excited singlet states in O2 liesin the arrangement of electrons in the antibonding

π*2py and π*2pz molecular orbitals, and is shown below.

Second excitedstate (electronshave opposite

spins

π*pyπ*pz

singlet

State1Σg

+

Energy /kJ

157

First excitedstate (electrons

paired )singlet 1∆g 92

Ground state(electrons have parallel spins)

triplet 3Σg – 0

Singlet O2 is excited, and is much more reactive thannormal ground state triplet dioxygen. Singletdioxygen can be generated photochemically byirradiating normal dioxygen in the presence of asensitizer such as fluorescein, methylence blue orsome polycyclic hydrocarbons. Singlet dioxygen canalso be made chemically :

H2O2 + OCl – → EtOH O2(1∆g) + Η2Ο + Cl –

Singlet dioxygen can add to a diene molecule in the1, 4 positions, rather like a Diels–Alder reaction. Itmay add 1, 2 to an alkene which can be cleaved intotwo carbonyl compounds.

CH – CH

CH2 CH2 + singlet O2 O – O

Singlet dioxygen may be involved in biologicaloxidations.

Ozone O3 is the triatomic allotrope of oxygen. It isunstable, 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 isintermediate 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 seldomused. The structure is described as the central O atomusing sp2 hybrid orbitals to bond to the terminal Oatoms. The central atom has one lone pair, and theterminal O atoms have two lone pairs. This leavesfour electrons for π bonding. The pz atomic orbitalsfrom the three atoms form three delocalizedmolecular orbitals covering all three atoms. One MOis bonding, one non-bonding, and one antibonding.

The four π electron fill the bonding and non-bondingMOs and thus contribute one delocalized π bond tothe molecule in addition to the two σ bonds. Thus the

bond order is 1.5, and the π system is described as afour-electron three-centre bond.

Trioxides of Sulphur :

The only important trioxide in this group, SO3, isobtained by reaction of sulfur dioxide with molecularoxygen, a reaction that is thermodynamically veryfavorable but extremely slow in the absence of acatalyst. Platinum sponge, V2O5, and NO serve ascatalysts under various conditions. Sulfur trioxidereacts vigorously with water to form sulfuric acid.Commercially, for practical reasons, SO3 is absorbedin concentrated sulfuric acid, to give oleum, which isthen diluted. Sulfur trioxide is used as such for

preparing sulfonated oils and alkyl arenesulfonatedetergents. It is also a powerful but generallyindiscriminate oxidizing agent; however, it willselectively oxidize pentachlorotoluene and similarcompounds 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 filledoxygen pπ orbitals with empty sulfur dπ orbitals, toaccount for the very short S – O distance of 1.41 Å

S

:O:

O O

S

:O:

O

S

:O:

OO O In view of this affinity of S in SO3 for electrons, it isnot surprising that SO3 functions as a fairly strongLewis acid toward the bases that it does not

preferentially oxidize. Thus the trioxide givescrystalline complexes with pyridine, trimethylamine,or dioxane, which can be used, like SO3 itself, assulfonating agents for organic compounds.

The structure of solid SO3 is complex. At least threewell-defined phase are known. γ-Sulfur trioxide,formed by condensation of vapors at – 80ºC or

below, is an icelike solid containing cyclic trimerswith structure.

S S

SO

O

O

O

O

O

O

O

O

A more stable, asbestos-like phase (β-SO3) hasinfinite helical chains of linked SO4 tetrahedra andthe most stable form, α-SO3, which also has anasbestos-like appearance, presumably has similarchains 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 Sand N exist. The elements N and S are diagonallyrelated in the periodic table, and have similar chargedensities. Their electronegativities are close (N 3.0, S2.5) so covalent bonding is expected. The compoundsformed have unusual structures which cannot beexplained by the usual bonding theories.

Attempting to work out oxidation states is unhelpfulor misleading.

The best known is tetrasulphur tetranitride S4 N4, andthis is starting point for many other S – Ncompounds. S4 N4 may be made as follows :

6SCl2 + 16NH3 → S4 N4 + 2S + 14NH4Cl

6S2Cl2 + 16NH3 → 4CCl S4 N4 + 8S + 12NH4Cl

6S2Cl2 + 4NH4Cl → S4 N4 + 8S + 16HCl

S S

N N N N

S SS4 N4 is a solid, m.p. 178ºC, It is thermochromic, thatis it changes colour with temperature. At liquidnitorgen temperatures it is almost colourless, but atroom temperature it is orange-yellow, and at 100ºC itis red. It is stable in air, but may detonate with shock,grinding or sudden heating. The structure is aheterocylic ring. This is cradle shaped and differsstructurally 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 covalentradii for S and N is 1.78 Å, the S – N bonds seem tohave some double bond character. The fact that the

bonds are of equal length suggest that this isdelocalized. 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 S4 N4 is thus acage structure.

Many different sizes of rings exist, for examplecyclo-S2 N2, cyclo-S4 N2, cyclo-S4 N3Cl, cyclo-S3 N3Cl3. In addition bicyclo compounds S11 N2,S15 N2, S16 N2,S17 N2 and S19 N2 are known. The lastfour may be regarded as two heterocyclic S7 N ring,with the N atoms joined through a chain of 1 – 5S

atoms.S4 N4 is very slowly hydrolysed by water, but reactsrapidly with warm NaOH with the break-up of thering:

S4 N4 + 6NaOH + 3H2O → Na2S2O3 + 2Na2SO3 + 4NH3

If S4 N4 is treated with Ag2F in CCl4 solution thenS4 N4F4 is formed. This has an eight-membered S – Nring, with the F atoms bonded to S. This results from

breaking the S – S bonds across the ring. Similarlythe formation of adducts such as S4 N4.BF3 orS4 N4.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 S4 N4 with SNCl2 in MeOH givestetrasulphur tetraimide S4(NH4). Several imides can

be made by reacting S4 N4 with S, or S2Cl2 with NH3.These imides are related to an S8 ring in which one ormore S atoms have been replaced by imide NHgroups, for example in S7 NH, S6(NH)2, S5(NH)3 andS4(NH)4.




If S4 N4 is vaporized under reduced pressure and passed through silver wool, then disulphur dinitrogenS2 N2 is formed.

S4 N4 + 4Ag → S2 N2 + 2Ag2S + N2

S2 N2 is a crystalline solid, which is insoluble in water but soluble in many organic solvents. It explodeswith shock or heat. The structure is cyclic and the

four atoms are very nearly square planar.The most important reaction of S2 N2 is the slow

polymerization of the solid or vapour to form polythiazyl (SN)x. This is a bronze coloured shinysolid that looks like a metal. It conducts electricityand conductivity increases as the temperaturedecreases, which is typical of a metal. It becomes asuperconductor at 0.26 K. The crystal structure showsthat the four-membered rings in S2 N2 have openedand polymerized into a long chain polymer. Theatoms have a zig-zag arrangement, and the chain isalmost flat. Conductivity is much greater along thechains than in other directions, and so the polymer

behaves as a one-dimensional metal. The resistivity isquite high at room temperature.

Ortho and Para Hydrogen :

The hydrogen molecule H2 exists in two differentforms known as ortho and para hydrogen. Thenucleus of an atom has nuclear spin, in a similar wayto electrons having a spin. In the H2 molecule, thetwo nuclei may be spinning in either the samedirection, or in opposite directions. This gives rise tospin isomerism, that is two different forms of H2 mayexist. These are called ortho and para hydrogen. Spinisomerism is also found in other symmetricalmolecules 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 orthoand para forms, because of differences in theirinternal energy. There are also difference in the bandspectra of the ortho and para forms of H2.

The para form has the lower energy, and at absolutezero the gas contains 100% of the para form. As thetemperature is raised, some of the para form changesinto the ortho form. At high temperatures the gascontains about 75% ortho hydrogen.

Para hydrogen is usually prepared by passing amixture of the two forms of hydrogen through a tube

packed with charcoal cooled to liquid airtemperature. Para hydrogen prepared in this way can

be kept for weeks at room temperature in a glassvessel, because the ortho-para conversion is slow inthe absence of catalysts. Suitable catalysts includeactivated charcoal, atomic hydrogen, metals such asFe, Ni, Pt and W and paramagnetic substances or ions(which contain unpaired electrons) such as O2, NO,

NO2, Co2+ and Cr 2O3.

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 (whosenickname 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. Smallamounts 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 amixture of three gases (B), (C) and (D) along withwater vapours. Gas (B) is blue towards litmus paper,gas (C) red and gas (D) is neutral. Gas (B) is alsoobtained when (A) is heated with NaOH and gives

brown ppt. with K 2HgI4. Solution thus obtained giveswhite ppt. (E) with CaCl2 solution in presence of

CH3COOH. Precipatete (E) decolorises −4MnO /H+.

Gas (C) turns lime water milky while gas (D) burnswith blue flame and is fatal when inhaled. Identify(A) to (D) and explain chemical reactions.

Sol. Gas (B) gives brown ppt. with K 2HgI4

⇒ gas (B) is NH

3⇒ gas (A) has NH

4

+ (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 + K 2HgI4 → O

Hg

Hg NH2I

brown ppt

(Iodide of Million’s base)

2MnO4

–

+16H+

+5C2O4

2–

→ 10CO2+2Mn2+

+ H2Oviolet colourless

2. (i) A blue compound (A) when heated, loses its waterof crystallization and becomes white (B), When (B)absorbs moisture again becomes blue.(ii) Aqueous solution of (A) gives white ppt. withBaCl2 solution and on reaction with K 4[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 browncoloured 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 presenceof (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 allthe 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 K 4[Fe(CN)6],hence it also contains Cu2+ ions, hence (A) isCuSO4.5H2O.

(i))A(Blue

24 OH5.CuSO →∆

White)B(4CuSO

Moisture

OH5 2 → Blue)A(

24 OH5.CuSO

(ii) CuSO4 + BaCl2 → . pptWhite

4BaSO ↓ + CuCl2

2CuSO4 + K 4[Fe(CN)6] → colour Chocolate)C(

62 ])CN(Fe[Cu + 2K 2SO4

(iii)(A)

4CuSO + 2KI → Green

2CuI + K 2SO4

2CuI2 → ppt.(D)

22ICu ↓ + I2

(D)22ICu + 2KI →

iodidecuprousPot.(E)2 ]CuI[K 2

(iv) CuSO4 + 2KCN → unstableYellow

2)CN(Cu + K 2SO4

2Cu(CN)2 → ppt.white(F)

22 )CN(Cu ↓ + (CN)2 ↑

Cu2(CN)2 + 6KCN → Soluble(G)

43 ])CN(Cu[K 2

(v) 2CuSO4 + 2NH4OH → ↓ bluePale(H)

24 )OH(Cu.CuSO

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

→ 4 blueDeep(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 heatedgradually. B is soluble in water and its aqueoussolution reacts with NO to give a dark browncompound C. B on strong heating gives a brownresidue and a mixture of two gases E and F. Thegaseous mixture, when passed through acidified

permanganate, discharge the pink colour and when passed through acidified BaCl2 solution, gives awhite 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(

compound browndark

(iii) Salt Bheating

strong → )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 ferroussulphate since with NO, it gives dark browncompound according to the reaction

[Fe(H2O)6]2+ + NO →

browndark

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

Hence, the salt A must be FeSO4.7H2OThe observation (iii) is

2FeSO4 →

brown

)D(32OFe +

43421

)F()E(

32 SOSO

+

+

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

colour pink 4MnO2 − + 5SO2 + 2H2O →

colour no

2Mn2 + −+ 24SO5 + 4H+

2H2O + SO2 + SO3 4H+ + SO32– + SO4

2–

Ba2+ + SO32– →

pptwhite3BaSO

Ba2+ + SO42– →

pptwhite4BaSO

Hence, the various compounds are(A) FeSO4.7H2O (B) FeSO4

(C) [Fe(H2O)5 NO]SO4 (D) Fe2O3 (E) and (F) SO2 and SO3

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

(i) It dissolves in dilute H2SO4 without evolving anygas.

(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 intodichromate 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 inreducing flame. It gives a magnetic substance.Identify compounds (A) to (E) and predict all theequations. [IIT-1980]

Sol. The fore said observations may be brieflysummarised as follows :

(a)solidgreenLight

A → 42SOH.Dil )A(

of Solution

→ 4KMnO Pink colour disappears

(b) A → ∆

gassmellingPungent

CB + +residueBrown

D

(c)Solution

722 OCr K → + CB Green solution

→ 23 ) NO(Ba . pptWhite

E

(d)residueBrown

D∆ → coalChar a magnetic substance

From the last step, one may conclude that brownresidue (D) (hence also compound (A)) must be a saltof iron. Since (A) decolourises KMnO4 solutionhence it should be a salt of Fe (II). The reactionsinvolved are given below.

MnO4 – + 8H+ + 5e – → Mn2+ + 4H2O[Fe2+ → Fe3+ + e – ] × 5

Pink 4MnO− +

Green

2Fe5 + + 8H+ → Colourless

2Mn + + 5Fe3+ + 4H2O

From observations of (b) and (c), one concludes thatcompound (A) should be FeSO4 as on heating, itgives pungent gases SO2 and SO3.

)A(4FeSO2 → ∆

)Brown()D(

32OFe +)B(2SO +

)C(3SO

SO2, gas turns dichromate solution green due toformation of green coloured sulphate of chromium

(III), the different equations are,Cr 2O7

2– + 14H+ + 6e – → 2Cr 3+ + 7H2O

[SO2 + 2H2O → 4H+ + SO42– + 2e – ] × 3

Cr 2O72– + 3SO2 + 2H+ → 2Cr 3+ + 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) isFe2O3 and (E) is BaSO4




5. A substance (X) is soluble in conc. HCl. When to thissolution NaOH solution is added, a white precipitateis 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 ammoniumsulphide solution. When HCl is added to the latter, a

grey precipitate is produced. Heating of (X) in airgives a water soluble compound gives whitegelatinous precipitate. Identify the compounds givingthe reactions involved.

Sol. The reaction sequence is as follows :

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

NaOH

Excess →

Dissolves and the solution is strongly reducing.

(2) X∆

→ SBrown

Y → x24 S) NH( Dissolves

→ HCl Grey ppt.

(3) XHeat

O2 →

42SOHconc.inSoluble

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

22SnO Na + 2HCl

(c) Sn + O2 → Fuse (Z)

2SnO

SnO2 + 2NaOH → Fuse Na2SnO3 + H2O

Na2SnO3 + 2HCl → ppt.gelatinousWhite32SnOH ↓ + 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 (60

Co) is an artificiallyproduced 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 B 12, 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 tworectangular hyperbolas are also rectangularhyperbolas. If A, B, C & D be the four points ofintersection of these two rectangular hyperbolas, thenfind the orthocentre of the triangle ABC.

2. Find the area of a right angle triangle if it is knownthat the radius of circle inscribed in the triangle is rand 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 inR, 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 equationz4 + 2z3 + 3z2 + 4z + 5 = 0 with (z ∈ C) have no

purely real as well as purely imaginary root.

5. Prove that

xxn

ax

xaf

0

l

+∫∞ dx = lna

xdx

ax

xaf

0

+∫∞

6. A straight line moves so that the product of the perpendiculars on it form two fixed points is aconstant. Prove that the locus of the foot of the

perpendiculars from each of these points upon thestraight line is a circle, the same for each.

7. Prove the identity :

∫ −x

0

zzx 2

e dz = ∫ −x

0

4/z4

x2

2

ee dz, deriving for the

function f(x) = ∫ −x

0

zzx 2

e dz a differential equation and

solving it.

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

that αβ = 4 and1−α

α +

1−ββ

=4a

7a2

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 pointsof extremum and find the values of k such that theequation x5/3 – 5x2/3 = k has exactly one positive root.

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

∑∈x p

min p = 101C51.

`tàxÅtà|vtÄ tÄÄxÇzxá

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety

of possible twists and turns of problems in mathematics that would be very helpful in facing

IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and

we hope that this section would prove a rich resource for practicing challenging problems and

enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari

Joint Director Academics, Career Point, Kota Solut ions wi l l be publ ished in next i ssue

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' haspassed 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 = 0so 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 isa x cos α + a y sin α + g(x + a cos α) + a2 = 0 ....(4)other end of diameter of 1st circle through P isQ(– 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 Cto the faces ACD and ABD meet at H with position

vector hr

,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

⇒ )c b( rr− . d

r = 0 ...(1)

so BC ⊥ AD , proved.

Now, let any point M (with position vector m ) beon 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 vectort m such that DP is perpendicular to ABC, then

(t m – dr

). br

= 0 = (t m – dr

). cr

so t m . br

= br

. dr

&

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[n2142

2

0

+−∫π

l dy

=2

1I (u2) + ]uycosu21[n

2

1 422

0

+−∫π

l dy

Now let y = 2π – t

I (u) + I (– u) =2

1I (u2) + ]utcosu21[n

2

1 420

+−∫π

l (– dt)

=2

1I (u2) +

2

1I (u2)

2I (u) = I (u2) as I(u) = I (– u)

(or using f (2a– x) = f (x) Prop).

so I (u) =2

1I(u2)

similarly find ]uxcosu21[n3 22

0

+−∫π

l dx

& show I (u) = I (– u) =2

1I (u2) = )u(I

2

1 n2n

MATHEMATICAL CHALLENGESSOLUTION 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 (ct

1, c/t

1) and b (ct

2, c/t

2) 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 (t

2 + t)

These lines intersect with x- axis at M (c(t1 + t), 0)

and N(c(t2 + t), 0). Length MN = |c (t

1 – t

2) | which is

a constant.Similarly intercept on y- axis can be obtained as

−

21 t

1

t

1c . Hence proved.

6. xy

dyxdxy − +

2

22

)yx(

dxydyx

−

− = 0

⇒ 2yx

)dyxdxy(y −+

2

22

)yx(

xydyxydydxydxxydxxydyx

−

+−−−+

⇒ x

y. 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

− yx

yx= 0

⇒ ln (x/y) +yx

yx

− = c

7. Let2

2

a

x+

2

2

b

y= 1 be the ellipse and

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

λ

−+ 1

b

y

a

x2

2

2

2

= 0 ...(1)

It it passes through origin, thenm1m2 a

2 e2 – λ = 0 ...(2)Solving (1) with x- axis

m1m

2 (x – a e)2 + λ

−1

a

x2

2

= 0

using (2) in it(x –

ae)2 + e2 (x2 – a2 ) = 0

(1 + e2) x2 – 2ae x = 0

x = 0 & x =2e1

ae2

+

so other point on x- axis through which this conic

passes is

+0,

e1

ae22

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(f ch1cf

±

−

±

; c ≠ 0

=0h

lim→ h

)c(f )c/h1(f 2

)c2(f

±

−+

=0h

lim→ h2

)c(f 2c

h1f )c2(f

±

−

±

=0h

lim

→ h2

)1(f )c2(f c

h1f )c2(f

±

−

±;

(using x = 2c & y = 1 in

2

xyf =

2

)y(f ).x(f )

= f (2c)0h

lim→

c.c

h2

)1(f c

h1f

±

−

± =

c2

)1('f )c2(f

=c2

)1('f )c2(f ; as given f ′(1) = f(1)

f ′ (c ±) =c2

)c(f x

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

Now f ′(x) =x

)x(f

⇒ )x(f

)x('f =

x

1

so ln f(x) = ln x + ln c⇒ f(x) = cx

Now as f

2

xy=

2

)y(f )x(f




let y = 1 in it2. f(x/2) = f(x) (f(1)

22

xc=cx f(1)

so f(1) = 1 as c ≠ 0so f(x) = x

9. Let i lines are there, no two of which are parallel andno 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 beingdivided by i lines, then

Pi + 1 = Pi + (i + 1)

Pi + 1 –Pi = i + 1

using i = 1, 2, 3, ......, n – 1P2 – P1

= 2,P1 – P2 = 3

M M

Pn – Pn–1 = n

Add these equationPn – Pi = 2 + 3 +.....+ nPn = Pi + 2 + 3 + ....+ n

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

)1n(n +

Pn =2

1 (n2 + n + 2)

10. 2

1tan x +

x3cos

xsin+

x9cos

x3sin +

x27cos

x9sin =

2

1tan 27 x

L.H.S. : Consider on

2

1tan x +

x3cos

xsin=

2

1

xcos

xsin +

x3cos

xsin

=x3cosxcos2

xcosxsin2x3cosxsin +

=x3cosxcos4

x2sin2x3cosxsin2 +

=x3cosxcos4

x2sin2x2sinx4sin +−

=x3cosxcos4

x2sinx4sin +

=

x3cosxcos4

xcosx3sin2=

2

1 tan 3x

similarly2

1tan 3x +

x9cos

x3sin=

2

1tan 9x

and2

1tan 9x +

x27cos

x9sin=

2

1tan 27

on adding all these we get

x3cos

x3sin+

x9cos

x3sin +

x27cos

x9sin=

2

1 (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, withhydrogen, 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(f x(d ) = ∫

x

0)x(g dx

⇒ x . f(x) = ∫x

0)x(g dx

f(x) = ∫

x

0 )x(gx

1

dx

⇒ g(x) < ∫x

0)x(g

x

1dx (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 aconstant.

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

1x

1,

where n is odd

∴ P(n + 1) = 1

3. Evaluate :

∑=

n

0r

r n C (r – nx)2 xr (1 – x)n – r ;

x ≠ 0, 1, n ∈ N > 2

Sol. ∑=

n

0r

r n C (r – nx)2 xr (1 – x)n – r

= ∑=

n

0r

r n C (r 2 + n2x2 – 2nxr)

r

x1

x

−

(1 – x)n

= (1 – x)n

−∑

=

r n

0r

r n2

x1

xCr

+ n2

x2

r n

0r r

n

x1

xC

−∑=

– 2nx

−∑=

r n

0r r

n

x1

xCr

we know that ∑=

n

0r

r r

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

⇒ ∑=

−

n

0r

r

r n

x1

xC =

n

x1

x1

−

+ = (1 – x) –n

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

⇒ ∑=

n

0r

r n C.r yr–1 = n(1 + y)n – 1

∴ ∑=

n

0r

r n C.r yr = ny(1 + y)n–1 ...(ii)

⇒ ∑=

n

0r

r n C.r

r

x1

x

−

= n1n

x1

x1

x1

x−

−

+−

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

⇒ ∑=

n

0r

r 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' Forum MATHS




= ny(1 + y)n – 2(ny + 1)

⇒ r n

0r

r n2

x1

xC.r

−∑

=

= nx1

x

−

2n

x1

x1

−

−

+

+

−1

x1

nx

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

+ n2

x2

(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 followinginequalities.

(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

sin2θ

2cosi

2sin

θ−

θ ≤ 2

2

θ = 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 withradius |z|.

θ RQ (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 = 0So 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(f 1

)y(f )x(f

++

∀ x, y and

f´(0) = 1. Also – 1 < f(x) < 1 for all x ∈R, then findthe set of values of x where f(x) is differentiable andalso 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(f lim

0x

−+→

=h

)x(f )h(f ).x(f 1

)h(f )x(f

lim0x

−+

+

→

=

+−

−

−

−→ )h(f ).x(f 1

)x(f 1

0h

)0(f )h(f lim

2

0x = 1 – f 2(x)

integrating we get

2

1ln

−

+

)x(f 1

)x(f 1 = x + c

⇒ f(x) =xx

xx

ee

ee−

−

+

−

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

x

x)]x(f [lim

∞→ =

x

xx

xx

x ee

eelim

+

−−

−

∞→

=

+

−−

−

∞→ xx

xx

x ee

eelim

e x = 1




Some Definitions :

Experiment : A operation which can produce somewell defined outcomes is known as an experiment.

Random experiment : If in each trail of anexperiment conducted under identical conditions, theoutcome is not unique, then such an experiment iscalled a random experiment.

Sample space : The set of all possible outcomes inan 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 singlesample point is called an elementary or simple event.Events other than elementary are called composite orcompound or mixed events.

For example, in a single toss of coin, the event ofgetting 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 occurin preference to the other.

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

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

Exhaustive events : A set of events is said to be

totally exhaustive (simply exhaustive), if no event outside this set occurs and at least one of these eventmust happen as a result of an experiment.

Independent and dependent events : If there areevents in which the occurrence of one does notdepend upon the occurrence of the other, such eventsare known as independent events. On the other hand,if occurrence of one depend upon other, such eventsare known as dependent events.

Probability :

In a random experiment, let S be the sample spaceand E ⊆ S, then E is an event.

The probability of occurrence of event E is defined as

P(E) =Sinelementdistinctof number

Einelementsdistinctof number=

n(S)

n(E)

=outcomes possibleallof number

Eof occurrencetofavourableoutocomesof number

Notations :Let A and B be two events, then

A ∪ B or A + B stands for the occurrence of atleast one of A and B.

A ∩ B or AB stands for the simultaneousoccurrence of A and B.

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

A ⊆ B stands for "the occurrence of A impliesoccurrence of B".

Random variable :

A random variable is a real valued function whosedomain is the sample space of a random experiment.

Bay’s rule :

Let (H j) be mutually exclusive events such that

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

1 j jH

=. Let A be

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

P

A

H j =

∑=

n

1k k k

j j

)H/A(P)H(P

)H/A(P)H(P

Binomial Distribution :

If the probability of happening of an event in a singletrial of an experiment be p, then the probability ofhappening of that event r times in n trials will benCr p

r (1 – p)n – r .

Some important results :

(A) P(A) =casesof numberTotal

Aeventtofavourablecasesof Number

=n(S)

n(A)

PROBABILITY

Mathematics Fundamentals

M A T H S




)AP( =casesof numberTotal

Aeventtofavourablenotcasesof Number

=n(S)

)An(

(B) Odd in favour and odds against an event : As aresult of an experiment if “a” of the outcomes are

favourable to an event E and b of the outcomes areagainst it, then we say that odds are a to b in favourof E or odds are b to a against E.

Thus odds in favour of an event E

=casesleunfavourabof Number

casesfavourableof Number=

b

a

Similarly, odds against an event E

=casesfavorableof Number

casesleunfavourabof Number=

a

b

Note :

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

probability of the occurrence of that event is

ba

a

+ 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

ba

b

+ and the probability of non-occurrence of

that event is ba

a

+.

(C) P(A) + P( A ) = 10 ≤ 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 onetrial be p, then the probability of successivehappening of that event in r trials is pr .

(D) If A and B are mutually exclusive events, thenP(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 independentevents 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 nothappening 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

A

B or

P(AB) = P(A) . P

A

B

Where P

A

B is known as conditional probability

means probability of B when A has occured.

Difference between mutually exclusiveness and

independence : Mutually exclusiveness is usedwhen the events are taken from the sameexperiment and independence is used when theevents 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. Ifan experiment consists of more than a coin, thencoins are considered to be distinct if not otherwisestated.

Dice : A die (cubical) has six faces marked 1, 2,3, 4, 5, 6. We may have tetrahedral (having fourfaces 1, 2, 3, 4,) or pentagonal (having five faces1, 2, 3, 4, 5) die. As in the case of coins, If wehave more than one die, then all dice areconsidered 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. Thereare 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 orcoart card namely king, queen and jack. So there arein all 12 face cards (4 kings, 4 queens and 4 jacks).Also there are 16 honour cards, 4 of each suit namelyace, king, queen and jack.




Binomial Theorem (For a positive Integral Index) :

If n is a positive integer and x, a are two real orcomplex quantities, then

(x + a)n = nC0 xn + nC1x

n – 1 a + nC2 xn – 2 a2 +

... + nCr xn–r ar + .... + nCn – 1x an–1 + nCna

n ..(1)

The coefficient nC0,nC1, ......,

nCn are called binomialcoefficients.

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 theexponents of x and a is always constant = n.

The binomial coefficients of term equidistantfrom 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 termusually 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

12

n

+ term.

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

12

n

+ term and

th

2

3n

+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

T

T + =x

a

r

1r n +−

or

r

1r

T

T + – 1 =

rx

)xa(r a)1n( +−+

=

−+

++r

xa

a)1n(

rx

)xa( =

rx

xa +|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

T

T + > 1 ⇔ r < k (i.e. 1 ≤ r < k)

Along , Tr + 1 = Tr ⇔ r

1r

T

T + = 1 ⇔ r = k,

i.e. Tk + 1 = Tk > Tk–1 > .... > T3 > T2 > T1

In this case there are two greatest terms Tk andTk+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

T

T + ⇔ 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 is2

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

integer.

Pascal Triangle

In(x + a)n when expanded the various coefficientswhich 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 1n = 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 A T H S




Rule : It is to be noted that the first and least terms ineach row is 1. The terms equidistant from the

beginning and end are equal. Any number in any rowis obtained by adding the two numbers in the

preceding row which are just at the left and just at theright of the given number, e.g. the number 21 in therow 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 + nC2a

2 + ..................

+ nCr ar + ........... nCna

n ...(2)

If we put x = 1 and replace a by – a, we get

(1 – a)n = nC0 –nC1a + nC2a

2 – ..................

+ (–1)r nCr ar + .... + (–1)n nCna

n ...(3)

Adding and subtracting (2) and (3), and thendividing by 2, we get

2

1

(1 + a)n

+ (1 – a)n

=n

C0 +n

C2a2

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

2

1(1 + a)n – (1 – a)n = nC1a + nC3a

3

+ 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

∴ n

C0 +n

C2 + .... =n

C1 +n

C3 + .... = 2

1[2

n

± 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 =!)r n(!r

!n

−

=!r

)1r n)...(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 – r Cn =!)r n(.!)r n(

!)n2(

+−

C02 + C1

2 + C22 + .... + Cn

2 =2)!n(

!)n2(

C02 – C1

2 + C22 – C3

2 + ...

=

evenisnif ,C.(–1)oddisnif ,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 −x

2

+!3

)2n)(1n(n −− x3 +..... ...(7)

Where n is any index (positive or negative)

The general term in expansion (7) is

Tr + 1 =!r

)1r n)......(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 theterms finally get smaller and smaller. If x is verysmall compared with 1, we take 1 as a firstapproximation to the value of (1 + x)n or 1 + nx asa 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

)1r n)...(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

)1r n)...(2n)(1n(n −+++xr + ....

Important expansions for n = –1, –2 are :

(1 + x) –1 = 1 – x + x2 – x3 + ...+ (–1)r xr + .... 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 areforming two sides of a vessel filled with liquid ofrefractive index = 1.5. After all possible refraction &reflection, find the deviation (δ) in ray –

Incident ray30º

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 atsame point. They are displaced by very small anglesα and β (β > α) and released from rest. Find the timeafter which B reaches its initial position for the firsttime. Collisions are elastic and length of strings is l.

A B

α β

(A)g

lπ (B)

g2 l

π

(C)g

l

α

πβ (D)

g

2 l

α

πβ

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 (SIunits). 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 asP = (3t2 – 2t + 1)watts, where t is time in seconds.Then the change in kinetic energy between timet = 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 Curvedsurface, 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.






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 ofspring constant 1000 N/m asconnected by an ideal pulleyas shown and system isarranged in vertical plane. Atequilibrium θ is 60º andmasses m1 and m2 are 2 kgand 3 kg respectively. Thenelongation in each springwhen 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 andstrings are massless.If mass m of hanging

block is the minimummass to set theequilibrium of systemthen – (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 focallength 20 cm & 30 cm. Opticalaxis of both lens coincide. Thislens system is used to form animage of an object. It turn outthat size of the image does notdepend on the distance betweenthe lens system & the object. If L is distance betweenlens & M is magnification after all possible refraction -(A) L = 10 cm (B) L = 50 cm

(C) |M| =2

3 (D) |M| =

3

2

8. A research vessel has a round glass window in bottom for observing the seabed. The diameter of theglass window is 60 cm, the thickness of the glass is20 mm and the index of refraction of water is 1.33and that of the glass is 1.55. The seabed is 6.0 m

beneath the window. A man in interior of vessel cansee seabed through window.

Interior of vessel air

Glass

water window

Sea bed`6 m

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

(B) Viewable area of seabed from interior of vesselwill increases as thickness of glass of windowincreases

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

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

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

A

B

C

D

P

Q R S T

TS

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 justsufficient to keepequilibrium of 100 N block,T1, T2 and T3 are tension, instrings AB, CD and EF andT4 is total force of alltensions on block 100 N

Match the following

θ θ

m1

m2

µ = 0.5

θ = 37º

θ = 37º

θ = 37º m

20 kg

L

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 atrest with respect to a roughwedge as shown. The wedgestarts moving up from rest withan acceleration of a = 2 m/s2 and the block remains at restwith 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

89

0

1

2

3

4

5

6

7

89

0

1

2

3

4

5

6

7

89

0

1

2

3

4

5

6

7

89

X Y Z W

12. A disc of radius 1 m is rolling ona plane horizontal mirror with

constant angular speed ω = 3

rad/s (sec figure). Calculatevelocity of image of point Qwith respect to point Q itself.(given θ = 30º)

13. In the figure shown the lenshas focal length f andvelocity v as shown. If thefinal image velocity is kv.Find k.

14. Light travelling in air falls at an angle of incidence 2ºone 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 indegrees.

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

16. IN YDSE one of the twoslits is covered with aglass slab. At P and Q ;8th and 12th maximum are

observed. Find the orderof maxima at O ifthickness of slab is minimum for this to happen.

17. In the arrangementshown the rod isfreely pivoted at

point O and is incontact with theequilateraltriangular blockwhich can moves on the horizontal frictionlessground. As the block is given a speed v forward, the

rod rotates about point O. Find the angular velocityof rod in rad/s at the instant whenθ = 30º. [Take v = 20 m/s, a = 1 m]

18. A cylinder of mass M radius R is resting on ahorizontal platform (which is parallel to x – y plane)with its axis fixed along the y-axis and free to rotateabout its axis. The platform is given a motion in X-direction given by X = A cos ωt. There is sufficientfriction present in the surface of contact that can

prevent the slipping between the cylinder and platform. The minimum torque in N-m acting on thecylinder 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 velocityv0 in a straight line is such a way that on the bodywork is being done at the rate which is proportionalto the square of velocity as given by P = βv2 where β

=2

693.0. Find the time elapsed in seconds before

velocity of body is doubled.

mµ

θ

a

Q

ω

θ

P

v

O

f/2 f/2

Q

O

P

PO = OQ

v

3a2

Oθ




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. Thecompound 'A', is –

(A)CH3

C = CH

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/Phence, 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 atthe constant pressure

(C) PT = constant is correct, because volume atconstant 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 thatof benzene

(C) The rate of nitration of benzene is greater thanthat of hexadeuterobenzene

(D) Nitration is an electrophilic substitution reaction

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

B(vB)

A(vA)log V

log P

(A) number of collision increasesB

A

V

V times

(B) number of moles in this process is constant

(C) it is isothermal process

(D) it is possible for ideal gas




your response in OMR sheet against the questionnumber of that question. + 4 marks will be given for


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 islargely independent of area of contact

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

(C) Rolling friction is generally smaller than slidingfriction

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

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 toD is enol

(C) The structure of C isOO

(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 catalytichydrogenation with D2 / Pt gives an alkane. Thealkene 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 andmeso-2, 3-dideuterobutane

9. Identify the compounds which do not give positiveiodoform test in the following sequence of thereaction(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









S, D-T then the correctly bubbled 4 × 5 matrix should

be as follows :

A

B

C

D

P

Q R S T

TS

P

P

P Q R

R

R

Q

Q

S

S T

T

P Q R S T





wrong answer.

10. Match the following

Column I Column II

(A) HCOOH (P) Decarboxylation

on heating

(B) CH3COOH (Q) Reation with Br 2

(C) OH

COOH (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








question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,

respectively, then the correct darkening of bubbles will


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. Which of the following is the strongest nucleophilicsite 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

OHH3C OH –

H

2

→ +

[F]4

2

CCl

Br → 43421

possibleare productssuch5

284 Br HC

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 20mL 0.5 N NaOH. The basicity of acid will be

17. A Gaseous mixture of 2 moles of A, 3 moles of B, 5moles of C and 10 moles of D is contained in avessel. 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 requires1.6 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 themolecular weight of its chloride is 80. The atomicweight of the metal is

MATHEMATICS



which ONLY ONE is correct. Mark your response inOMR sheet against the question number of that



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 2r

r nn

0r

r +

=∑ is

(A)2n

1+

(B)2n

2+

(C) n + 2

(D) n + 1

3. The equation of the curve satisfying the differentialequation y2 (x

2 + 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







answer.

5. Solution of )x2 – x(loglog 23x2x28x6x 22 ++++

= 0 is

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

(D) none of these

6. Sum to n terms of the seriesS = 12 + 2 (2)2 + 32 + 2(42) + 52 + 2(62) + … is

(A)2

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

(B)2

1 n2 (n + 1) when n is odd

(C)4

1 n2 (n + 2) when n is odd

(D)4

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

7. The number of ways in which we can arrange the 2nletters x1, x2,…,xn, y1,y2,…yn in a line so that thesuffixes of letters x and those of the letters y arerespectively 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 ∫ + x5cos21

x7cos – 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











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

TS

P

P

P Q R

R

R

Q

Q

S

S T

T

P Q R S T





wrong answer.

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

x

1lim

22 = (P) 1

(B)

∫

∫+

+

→

yx

0

t2

2yx

0

t

0xdte

dte

lim2

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








question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,respectively, then the correct darkening of bubbles will


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 value of ( )8y –3

1 if (1 + x2)

dx

dy = x(1 – y),

y(0) = 4/3 is ………… .

13. The area bounded by the curves x = y2 andx = 3 – 2y2 is ………. .

14. If I = 3 ∫π

θ03sin (1 + 2cosθ) ( 1 + cos θ)2 dθ then

the value of I is ……… .

15. Number of divisors of the form 4n + 2 (n ≥ 0) ofintegers 240 is……….

16. If |z| < 1/2 and 4|(1 + i)z3 + iz| < k, find the leastintegral 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 =12

1 – 23

3

+.

13

1 – 33

3

+.

14

1 – 43

3

+...

1n

1 – n3

3

+, find

∞→nlim 3an.

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

to……..







PHYSICS



which ONLY ONE is correct. Mark your response inOMR sheet against the question number of that



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 effluxcoming 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) 2agH8

3ρ (D) ρagH2

2. A container has a hole at a height of 2m. If the timetaken by the efflux to strike the inclined plane

perpendicularly is 1 sec. Then the height of liquidlevel 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 isminimum (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 onsmooth fixed circular track as shown. If the particle isreleased from rest at t = 0, then variation of normalreaction 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 ofMomentum, 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.




Am

R O

(A)

N

3mg

θ

(B)

N

θ

3mg

(C)

N

3mg

θ

(D)

N

3mg

θ







answer.

5. A vessel filled with liquid is resting on the roughhorizontal surface. A hole is made in the vessel asshown. Then –

µ

h• H

•

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

(B) The torque due to normal reaction force betweencontainer and ground about center of gravity isout of the plane of paper

(C) Torque due to friction about center of gravity iszero

(D) Torque due to normal reaction force between

container and ground about center of gravity iszero

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 ontrolley. Coefficient of friction between the block Aand trolley is µ. The trolley is moved withacceleration so that block does not slip –

B

A

(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 andhas a linear mass density of 10 kg/m while the stringED is massless. If both strings are inextensible and

pulley is ideal then when the system is released fromrest the ratio of tension in the string.

D B

2kg 4kg

E C

(A) at points E and C is4

5

(B) at points E and C is5

4

(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 inequilibrium as shown. Inclined surface of the wedgehas an inclination α with the horizontal and allcontacts are smooth. The normal reaction on thewedge 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 planeand a sleeve of mass m is inserted in the rod. The rodis rotated with a constant angular velocity ω in thehorizontal plane. The lengths l1 and l2 are shown infigure. The normal reaction and frictional forceacting 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









S, D-T then the correctly bubbled 4 × 5 matrix shouldbe as follows :

A

B

C

D

P

Q R S T

TS

P

P

P Q R

R

R

Q

Q

S

S T

T

P Q R S T





wrong answer.

10. As shown block C of mass 5 kg is pulled by a force Fand its acceleration is found to be 3 m/s2. The massesof blocks A and B are 10 kg and 5 kg respectivelywhile the string passing over ideal pullies is ideal andis under tension T. If acceleration of blocks A and Bare a1 and a2 respectively then if all surfaces aresmooth 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 bodyof mass 1 kg that moves along the x-axis.The potential energy U(x) is given byU(x) = 20 + (x–2)2 where x is in meters. At x = 5.0 mthe particle has a kinetic energy of 20 J then –Column-I Column-II

(A) minimum value (P) 29of x in meters

(B) maximum value (Q) 7.38of x in meters

(C) maximum potential (R) 49energy in joules

(D) maximum kinetic (S) – 3.38Energy in joules

(T) None of theseThis section contains 8 questions (Q.12 to 19).






darkened. For example, if the correct answers toquestion numbers X, Y, Z and W (say) are 6, 0, 9 and 2,

respectively, then the correct darkening of bubbles will


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 particle moves in a straight line with its retardation proportional to its displacement 'x'. Change in kineticenergy 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 forceF(x) = – K/(2x2) where K = 10 Nm2. At time t = 0 itis at x = 1.0 m and its velocity is v = 0. Find itsvelocity when it reaches x = 0.50 m.

14. An artillery gun is mounted on a railway truckstanding on straight horizontal rails. The total mass ofthe truck with gun shells and crew is M = 50 tons andthe mass of each shell is m = 25 kg. The gun fires in ahorizontal direction along the railway. The initialvelocity of the shells is V0 = 1000 m/s. What will thespeed of truck after the second shot? Disregardfriction and air resistance.




15. A stationary body explodes into four identicalfragments 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, thenthe 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 horizontaldirection. Minimum impulse imparted to topple the

block must be greater than.

17. A disc is rotating freely its axis. Percentage change inangular velocity of disc if temperature decreases by20ºC is (coefficient of linear expansion of material ofdisc is 5 × 10 –4/ºC )

18. A glass capillary sealed at the upper end is of length0.11 m and internal diameter 2 × 10 –5 m. This tube isimmersed vertically into a liquid of surface tension5.06 × 10 –2 N/m. When the length x × 10 –2. m of the

tube is immersed in liquid then the liquid level insideand outside the capillary tube becomes the same, thenthe 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 ishanged from middle of wire. The vertical anddepression in wire (in cm) in equilibrium is (Youngmodulus of wire = 2.4 × 109 N/m2, cross-sectionalarea = 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 atcritical 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 temperatrueTi has been carried out independently from initialvolume 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 K 2Cr 2O7 solution is require tooxidise 0.81 g of H2S in acid medium -(A) 47.8 (B) 23.8(C) 40 ml (D) 72 ml







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 ∝ T

1 (B)

PdT

dV

= K

(C)PdV

dT

= K (D)

P2T

V – T

1

= 0

9. Consider in Redox reaction

2S2O32– + I2 → S4O6

2– + 2I – (A) S2O3

2– gets reduced to S4O62–

(B) S2O32– gets oxidised to S4O6

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












S, D-T then the correctly bubbled 4 × 5 matrix shouldbe as follows :

A

B

C

D

P

Q R S T

TS

P

P

P Q R

R

R

Q

Q

S

S T

T

P Q R S T





wrong answer.

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








question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,

respectively, then the correct darkening of bubbles willlook 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. Oxidation number of Fe in Na2[Fe(CN)5 NO] is

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

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

15. Calculate the volume occupied by 8.8 g of CO2 at31.1ºC and 1 bar pressure. (R = 0.083 bar litre K –1 mol –1)

16. A compound exists in the gaseous phase both asmonomer (A) and dimer (A2). The molecular weightof A is 48. In an experiment 96 g of the compoundwas confined in a vessel of volume 33.6 litre andheated to 273º C. Calculate the pressure developed ifthe compound exists as dimer to the extent of 50% byweight under these conditions.

17. The haemoglobin from the red blood corpuscles ofmost mammals contains approximately 0.33% of iron

by weight. The molecular weight of haemoglobin as67,200. 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 thissolution required 19.8 mL of N/10 HCl for completeneutralisation. 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 10mL 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 ofE = 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 theequation 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. suchthat 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. Eachquestion has four choices (A), (B), (C) and (D), out of





answer.

5. Let z =

7i3 – 5i5

i353 – i2 – 1

i5 – 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 whichthe roots of equation x2 – 2ax + a2 – 1 = 0 lie between5 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 of10

8

x32

+ has the maximum numerical value, then x

lies in the interval

(A)

21

64,2 (B)

2,–

23

60 –

(C)

2,–

21

64 – (D)

23

60,– 2

9. If log x2 – log 2x = 3 log 3 – log 6 then x equals

(A) 9 (B) 3(C) 4

(D) 5








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

TS

P

P

P Q R

R

R

Q

Q

S

S T

T

P Q R S T





wrong answer.

10. z lies on ……… if

Column-I Column-II(A) |z – 3| + |z – i| = 10 (P) circle

(B)i – z

3 – 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 ofColumn-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



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 bubbleswill 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

equation7x6x

7x6 – x2

2

++

+ = 1 are…

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

)upto....|x|cosxcos|x|cos1( 32

27

∞++++

= 9

3

,find the value of

π1

(x1 + x2 + ...)

15. If x satisfieslog1–x(3) – log1–x (2) =1/2, find – 4x.

16. Rakshit is allowed to select (n + 1) or more books outof (2n + 1) distinct books. If the number of ways inwhich he may not select all of them is 255, then valueof n is…

17. If 1, x1, x2, x3 are the roots of x4 – 1 = 0 and ω is acomplex 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 ABCright angled at C are in A.P., find 5 (sin A + sin B).

19. A class contains 4 boys and g girls. Every Sundayfive students, including at least three boys go for a

picnic to Appu Ghar, a different group being sentevery week. During, the picnic, the class teachergives each girl in the group a doll. If the total numberof 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 mill ion 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,D10 A→ Q B→ R C→ P D→ SColumn

Matching 11 A→ R B→ P C→ Q D→ S12 13 14 15 16 17 18 19Numerica 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,D10 A→ S B→ R C→ Q D→ PColumn

Matching 11 A→ Q,R,S B→ R C→ Q D→ P12 13 14 15 16 17 18 19Numerica l

Type 4 1 3 8 2 6 3 9

M THEM TICS

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,C10 A→ S B→ P C→ Q D→ RColumn

Matching 11 A→ Q B→ R C→ S D→ P12 13 14 15 16 17 18 19Numerica 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,D10 A→ S B→ R C→ Q D→ PColumn

Matching 11 A→ S B→ Q C→ R D→ P12 13 14 15 16 17 18 19Numerica 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,C10 A

→ R B→ Q,T C

→ P D→ SColumn

Matching 11 A→ S B→ P C→ T D→ Q

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

M THEM TICS

Ques 1 2 3 4 5 6 7 8 9

Ans C C B B A,C D A,B,D A,C A10 A→ S B→ R C→ Q D→ PColumn

Matching 11 A→ R B

→ P C→ S D

→ Q12 13 14 15 16 17 18 19Numerica l

Type 8 1 1 5 4 1 7 5

IIT- JEE 2011 (September issue)

IIT- JEE 2012 (September issue)




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