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No success is possible unless you believe that you can succeed.

Volume - 5 Issue - 11 May, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Cover Design Govind Saini, Om Gocher Layout : Mohammed Rafiq Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

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

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

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

Pramod Maheshwari, B.Tech., IIT Delhi

1

MAY 2010

XtraEdge for IIT-JEE

2

MAY 2010

Volume-5 Issue-11 May, 2010 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

Regulars ..........

Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Xtra Edge Test Series for JEE – 2011 & 2012 AIEEE-2010 Examination Paper

PAGE

NEWS ARTICLE

4

IITian ON THE PATH OF SUCCESS

8

IIT-Develops technology to produce stealth aircraft Urine-processing technologies yield rich cash flow potential Mr. Sujal Patel

KNOW IIT-JEE

10

Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS S Success Tips for the Months • "All of us are born for a reason, but all of us don't discover why. Success in life has nothing to do with what you gain in life or accomplish for yourself. It's what you do for others." • "Don't confuse fame with success. Madonna is one; Helen Keller is the other." • "Success is not the result of spontaneous combustion. You must first set yourself on fire." • "Success does not consist in never making mistakes but in never making the same one a second time."

8-Challenging Problems [Set# 1] Students’ Forum Physics Fundamentals Electrostatics-I 1- D Motion, Projectile Motion

CATALYSE CHEMISTRY

DICEY MATHS

41

Mathematical Challenges Students’ Forum Key Concept Complex Number Matrices & Determinants

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

• "Failure is success if we learn from it."

XTRAEDGE TEST SERIES

XtraEdge for IIT-JEE

33

Key Concept Gaseous State & Real Gases General organic Chemistry Understanding : Physical Chemistry

• "A strong, positive self-image is the best possible preparation for success." • "The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself."

17

49

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper IIT - 2010 Examination Paper with Solution

3

66

MAY 2010

IIT develops technology

called because of their very small

convert it into fertiliser. The Delhi

to

size - are known to exhibit unique

government is willing to consider

physical and chemical properties.

a

produce

stealth

aircraft

revenue-share

commercial

The IIT team found that crystals of

venture selling the phosphates and

Materials scientists at the Indian

“barium hexaferrite” with particle

nitrates in urine.

Institute

size of 10-15 nanometres have the

of

Technology

in

Roorkee (IIT-R) have developed

ability

microwave absorbing nanocom-

(Human hair, for comparison, is

posite coatings that could make

100,000 nanometres thick). They

aircraft almost invisible to radar.

developed special processes for

The

synthesizing the nanopowder and

technology

for

building

to

absorb

invisible, or stealth aircraft, is a

formulating

closely

Sharma

guarded

secret

of

it

microwaves.

as

a

said

coating.

that

the

developed countries and a handful

nanocomposite coating on the

of laboratories in India are doing

aluminium

research

area.

percent of incident microwaves at

of

Radars

in that

this emit

pulses

sheet

absorbed

89

On the outskirts of Delhi, a littleknown non-government organisation, Fountain for Development Research and Action, is laying the ground for the first urine bank. It has diverted urine from two schools, where it has installed odour-free urinals, into a tank and transferred the run-off to a village nearby for use as fertiliser.

15 giga hertz - the frequency

Director Madhab Nayak says the

microwave radiation identify flying

normally

foundation is working towards

aircraft by detecting the radiation

reflecting only 11 percent. A

making

reflected by the aircraft’s metallic

stealth

potential

body. The nanocomposite coatings

absorb all the incident radiation

developed by Rahul Sharma, R.C.

and reflect nothing.

used

by

aircraft

radars

should

—

ideally

Agarwala and Vijaya Agarwala at IIT-R absorb most of the incident

Urine-processing

radiation and reflect very little.

technologies yield rich

Sharma, who revealed his team’s

farmers as

aware

of

replacement

its for

expensive urea. "There is no such thing as waste," says Vijayaraghavan M Chariar, assistant professor at the Centre for

Rural

Development

and

cash flow potential

Technology at IIT. "Urine consists

held

The stink is out of urine, literally

of a lot of inorganic salts, which

recently at the Indian Institute of

and metaphorically, with a growing

produce gases only when mixed

Science

believes

number of researchers spotting

with water. It is, in fact, pure

their nano-product is a significant

commercial and ecological value in

fertiliser," he added.

step in developing a technology to

a liquid most people consider

enable

waste.

work

at

an

nanomaterials in

international

conference Bangalore,

aircraft

escape

surveillance

and

equipment

from

radar

protect

“jamming”.Nanoparticles

XtraEdge for IIT-JEE

IIT has come up with a cheap, odour-free, urinal which it has

its

The Indian Institute of Technology

successfully tested on campus. The

electronic

(IIT) Delhi, for instance, is working

odour-free

to harvest this human waste and

technology with simple science to

-

so

4

urinal

combines

MAY 2010

translate into a significant water-

vandalism,

saving initiative (urine smells only

working perfectly."

when mixed with water, which this technology eliminates).

In

these

urinals

are

Chariar, explaining how it could be diverted for use as a nutrient

preparation

for

the

by a simple plumbing.

Commonwealth Games, the Delhi

The urine tank could deliver the

Urine is collected through a tank

government is planning to install

liquid nutrient directly to plants

placed

1,000 such urinals at a nominal

about two to three metres below

cost of Rs 3 lakh.

the soil, he says.

Chariar is already working on the

The Centre for Banana Research

second phase of his project, which

in Trichy is already using it for

was

banana

underground,

harvested

and used as liquid fertiliser two to three metres below the ground on a five-acre field on campus, said Chariar, who can talk animatedly about this human waste and how its poor treatment alone has led to sanitation problems.

initiated

by

Unicef

and

plantations

University of agriculture Sciences,

for setting up a small reactor to

Bangalore, too is looking at its

extract nitrates and phosphates

varied uses.

from urine. "This could become a

simple technology, called Zerodor,

micro-enterprise from the urinal,"

IIT

developed by Chariar, that fits

says Chariar.

electricity

The Delhi government is also

water

looking

Kolkata:

and diverts the urine through a drain where it is collected and

the

Stockholm Environment Institute,

The public urinal at IIT uses a

into the waste coupler in the pan

and

at

installing

Chariar’s

student

produces

from

waste

Waste

water

technology at a few parks in the

management is a big issue world

city, while harvesting urine in

wide and specially in India where

those places.

there is acute shortage of the

Chariar has even designed similar

precious resource in many places

urinals for women. "We have filed

but a 23-year-old student of IIT

for trademark registration and we

Kharagpur claims he has found a

are in discussion with companies

solution.

for marketing it," he says. With a

Apart

little more investment, he says, a

management of waste water he

hydrophobic

pans

has also demonstrated producing

could make it water resistant and

electricity from it, which could go

completely drain the urine, leaving

a long way in protecting the

Meanwhile, the Delhi government,

no room for any oxidisation,

earth's resources.

which has already installed 200

which can also cause odour.

harvested.

The idea is not to

allow it to mix with water at any stage. Chariar has already transferred this technology to Good Yield Environmental

Technologies,

a

Kolkata firm, and filed for a patent. Chariar claims that Zerodor is a low cost product and would need replacement in only about two years.

such odourless urinals in different parts of the city, uses a different and

perhaps

technology.

more

expensive

Amiya

Chandra,

deputy commissioner of the city’s municipal "Other

corporation, than

problems

XtraEdge for IIT-JEE

says, of

In

the

coating

developed

on

world,

communities have been quick to realise

the

huge

economic

potential of urine. "Communities in Germany are exporting urine to neighbouring

countriesthat

are

using it on their farms, says 5

from

finding

solutions

A look at reservoirs used for water supply Manoj Mandelia, who is pursuing integrated

MTech

at

IIT

Kharagpur, there was no policy in the

country

which

examined

MAY 2010

waste as part of a cycle of

Kharagpur, West Bengal,PIN -

challenges. The consortium's plan

production-consumption-ecovery.

721302, India or by email at

relies on IIT experts in electrical

[email protected].

and

"Waste

management

still

The

computer

engineering;

constituted a linear system of

nomination form is also available

mechanical,

collection

which

on the official award website

aerospace

creates health and environmental

The objective behind the award is

architecture;

hazards," he said.

to commemorate the spirit and

members of the Wanger Institute

drive of Dr. Nina Saxena, who

for Sustainable Energy Research to

personified technical excellence.

tackle these challenges.

and

disposal

"I developed a product which uses the concept of microbial fuel cell (MFC is a bio-electrochemical system that drives a current by mimicking bacterial interactions found in nature), which could not only treat waste water but also produce

electricity

in

the

process," explains Mandelia who heads a team of five people in the project.

The fourth edition will continue the

tradition

pioneering

of

rewarding

innovations

betterment

of

for society.

A distinguished committee is being formed

by

IIT

Kharagpur

to

adjudge the nominations for this award. The award committee is chaired

by

Director

of

IIT

Kharagpur and is comprised of

The Water Diviner

Deans

The project, named LOCUS which

members of IIT Kharagpur and

stands for Localised Operation of Bio-cells

Using

Sewage,

can

and

selected

faculty

well known alumnus, based in India and in the US.

water to about 60-80 per cent.

Nominations

for

announces

of

the

and

university's

departments and research centers will also work together to offer wind energy courses addressing the technical, operational, social, and environmental aspects of wind energy

in

industry.

consultation

To

involvement

in

with

ensure

student

the

project,

fellowships will be offered annually to undergraduate and graduate students

in

wind

energy

engineering fields of study. Faculty

will also be invited to IIT to attend

Wind

workshops and to share ideas with

Turbine Green

to Jobs,

Research, and Education

the

their American counterparts. The wind energy consortium will work with small wind turbine manufacturer Viryd Technologies to procure and install an 8KW

in Technology Award Kharagpur

business;

IIT’s New On-Campus

Nina Saxena Excellence IIT

engineering;

university consortium members

Support IIT Kharagpur Calls for

and

and students from international

achieve chemical oxygen demand (COD) reduction levels in waste

Many

materials,

Viryd wind turbine on IIT's Main the

Campus, and to deliver a second

fourth edition of Nina Saxena

turbine to one of IIT's engineering

Excellence in Technology Award

The consortium members will

laboratories to perform turbine

to invite entries of this year in

research

energy

reliability studies. The consortium

areas of technical innovations.

challenges identified in the U.S.

will also work with wind energy

Nominations for entries to the

Department

"20%

developer Invenergy to install a

award are open until April 30,

Wind Energy by 2030" report,

1.5MW GE wind turbine adjacent

2010. Entries can be submitted

including wind technology, grid

to a wind farm in Marseilles, Ill.

either by post to the Director, IIT

system integration, and workforce

The

XtraEdge for IIT-JEE

the

wind

of

Energy's

6

close

proximity

of

IIT's

MAY 2010

Marseilles turbine to an existing

pharmacy,

wind

institutes like IITs and NITs.

admission into fresh semester in

Many students from the state

January

made

academic record"

farm

provides

an

ideal

opportunity to study turbine-toturbine wake interaction, wind farm interaction, and wind energy efficiencies in addition to turbine reliability studies.

science

it

to

in

premier

the

top-100.

graduate

levels, this

were

year

denied

for

"bad

Srujana (JNTU, Kukatpally) ranked

.Prof V N Pal, who is an alumni of

22,

the

Karthik

Mufaquam

Nagarjuna

Ali

51

and

44, Pavan

institute,

has

taken

the

students' issue to President Patil,

Kishore got the 102nd rank in

who is the Visitor of the institute.

Hyderabad boy tops in

ECE stream.

Prof Pal met the President on

GATE 2010

Srinivas Reddy got the 68th rank

HYDERABAD

in EEE, and V. Suryanarayana 31 in

termination of students of IIT

Harikrishna, a final year computer

Mechanical.

Kanpur at length.

science engineering student from

GATE 2010 score is valid for two

He explained the socio-economic

the city, has topped the national

years

condition of these students.

level Graduate Aptitude Test in

announcement

Engineering

according to the details published

Vidya Balan to address

on GATE website

IIT

:

Malladi

(GATE-2010).

He

achieved the feat in his first

from

the of

date the

of

GATE.

March 15 by IIT-Guwahati, which

discussed

the

issue

of

She has been invited by one of the

Terminated IIT students seek

The results were announced on

and

results,

attempt, scoring 99.99 percentile by scoring 83.55 per cent in

April one at Rashtrapati Bhawan

intervention

of

Prez,PM

Indian

Institutes

of

Technology to be a keynote speaker at an upcoming seminar. In an interview to a leading daily,

conducted

Terminated

"bad

she confirmed her invitation from

GATE

this

performance", 38 students of IIT

one of the IITs. She said, “I have

year. Mallad,

Kanpur have taken the issue to

been approached by one of the

who topped

President Pratibha Patil and Prime

IITs for being a keynote speaker at

the national-

Minister Manmohan Singh with a

a seminar they are holding. The

level

plea that the institute reconsider

topic is ‘The changing face of the

the decision.

Indian heroine’. I am excited about

said,

The students have written to

it, but am figuring out dates and

“My aim is to join ME at the Indian

Prime Minister and the PMO has

prior

Institute of Science, Bengaluru. I

forwarded the matter to the HRD

moment. But, I hope this comes

want to become a scientist,”

Ministry for "appropriate action".

through.”

Harikrishna, 21, said.

The Ministry has again forwarded

A MA in sociology, Vidya Balan

Students who clear GATE are

it to the institute for action, a

believes

eligible for admission to masters’

ministry official said.

groomed her and given her the

degree courses in engineering,

The 38 students, including 24 from

confidence to address seminars at

technology,

under-graduate

these prestigious universities.

Graduate Engineering

Aptitude

Test

(GATE-2010),

in

architecture,

XtraEdge for IIT-JEE

for

and 7

14

post-

commitments

her

at

education

the

has

MAY 2010

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

MR. SUJAL PATEL To his competitors, Sujal Patel is now a name to reckon

most complex back-end operational challenges. That

with. His company Isilon Systems, in the clustered storage

experience gave him the insight for a new type of solution,

space, has not only earned its position as the fastest

a type of virtualized storage optimized for media. His

growing technology company in North America, but in

experience gave him the insight to a real customer need,

the five years of selling its products, Patel has transformed

and his deep technical knowledge gave him the ability to

the company from zero sales, into a company with a $100

spot a solution.

million run rate, $80 million in cash, and no debt.

I did not want to wait 10-15 years, treading cautiously at

“Slightly better is not a good term,” says Patel, the CEO

every step before taking the plunge. So when I got the

of Isilon Systems and a pioneer in the clustered storage

chance to found Isilon, I jumped at the opportunity,”

space. “For a technology to be adopted in an existing

beams Patel.

market, you really need to have a technology that is

That’s also a reason which led venture capitalists to take

substantially better — 10 (times) better than what is in

this 26 year old lad seriously. A few months after Patel

the marketplace today,” adds Patel. “It has to be so much

founded Isilon in 2001, the NASDAQ came down

better that it is overwhelming for people who buy that

crashing, bringing the ‘Dot Com Burst’. The venture

product and service.” When Patel says so, he is not being

capital market was in disarray. With the existing

merely theoretical. The unsatiated desire to bring out the

companies dropping their revenues, there was not much

best led this 35 year old entrepreneur to steer his

hope for new companies to find potential investment. To

company successfully, in a market space, which was

make matters worse, there were about 250 other

crowded by biggies like EMC and NetApp; not to mention

startups in the storage space. Patel was undeterred. Sure

the 250 odd startups in the storage space.

about his ideas, he approached close to 40 venture

Patel founded Isilon at the age of 26. Prior to this, Patel

capitalists, and with perseverance, eventually he managed

spent his initial career days at Real Networks. As an

to gain their confidence. Five months after the company’s

engineer, Patel used to solve some of Real Networks’

beginning, Patel had managed to raise $8.4 million venture

XtraEdge for IIT-JEE

8

MAY 2010

capital in what was one of the toughest markets to raise

Next, Patel began revamping the business strategy by

money.

reaching out to the broader enterprise segments. It was not easy. The segment he wanted to target comprised of

But all was not rosy yet. The road ahead proved to be

Fortune 50 companies who did not have much

more challenging than raising the funds. The next three

expectation from a startup like Isilon. With the

years were spent in building the products. Developing the

organizational re-structuring, he completely overhauled

company’s IP not only took a tremendous amount of

the company’s services and products to meet the

money, but also ate in to the time to get into the market.

expectations

By the time the product was ready to debut in

of

the

large

enterprise

customers.

Eventually, more and more Fortune 50 companies began

the markets in 2003, the company’s debt was near

to adopt Isilon’s products.

$20 million.

Finally he decided to increase the company’s focus on Cruising over Obstacles

R&D. Innovation has always been an essential part of his life. A very innovative and inquisitive person, Patel is

Not only confident and unshakable, Patel is also a man of

known to get at least six ideas every day. Even as a

clear vision, with no illusions about his capabilities. At 26,

kid he was known for asking around 500 questions

as the Founder of the company, he served as the CEO for

about everything under the Sun. Thus, Patel diverted

the initial three years. But he knew that, if he wanted to

about

make Isilon the next big thing in storage space, he needed

25

percent

of

the

company’s

revenue

towards R&D and fostered innovation within the

someone who knew the dynamics of running a large

organization.

organization. He promptly stepped down and appointed as a new CEO, who knew what it takes to grow into the

So what drives this confident and zealous man? It is the

larger league. “I floated the company but knew my

self-belief, passion and the creative way of looking at

limitations in the business front. We had three products

problems and coming up with solutions, says Patel. Even

ready to debut in the market and if we wanted then to

during his college days, while working on the Internet, he

succeed, we needed an expert who knew the right strings

looked for opportunities to think about new ways, solve

to pull,” says Patel. His focus and timely decisions were

problems, or to bring innovative techniques/technologies

fruitful in the subsequent years when Isilon was on a

to the market place.

dream run, literally growing at 200 percent year-on-year. By the time the company went public, it was a $60 million

To run a business successfully, it is also important that

company.

one should be honest with oneself, the team and the stakeholders. The foremost thing that Patel did after

“My goal was to see Isilon become a $100 million company by 2007 and become a player to be reckoned in

taking over the company was to communicate with

the $4 billion global storage market for its technology.

customers and partners and update them about the

And we were still $40 million short. I knew that despite

happenings within the organization, reinstating their faith

the economic instability, the company I had founded had

in him and the company. “I went and talked to each of our

great potential if one could maximize it,” says Patel.

investors and customers, telling what we planned

To start with, he re-structured the entire organization,

and how much earnings we expected through the

replacing every executive in the management, including

quarters. I believe that apart from your technology

CTO, CFO, Head of engineering, Head of operations and

offerings, one reason companies want to do business

others. Such an action is quite unheard of, especially

with you is the goodwill you develop in tough times,”

immediately after a company had gone public.

says Patel.

XtraEdge for IIT-JEE

9

MAY 2010

KNOW IIT-JEE By Previous Exam Questions

To find the vertical velocity at the time of collision let us consider the motion of P in vertical and horizontal directions. 49 49 m/s m/s 2 2 P 49m/s Q

PHYSICS 1.

In Searle's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 100 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). Find maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free from error. [IIT-2004] Sol. Maximum percentage error in Y is given by L W Y= × 2 X πD 4  ∆D  ∆x ∆L  ∆Y  = 2 + +   L  D  x  Y  max .

49m/s

245m

Horizontal direction Sx = 122.5 Tx = ? vx = 49 2 ∴ ⇒

49 2

=

displacement time

122.5 tx

tx =

(122.5) 2 49 ay = – 9.8 m/s2

ty =

∴

135º

A B Both particle travel in the same vertical plane and undergo a collision After the collision, P retraces its path, Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground? Take g = 9.8 m/s2. Sol. mp = 20 g mQ = 40 g The horizontal velocities of both the particles are same and since both are projected simultaneously, these particle will meet exactly in the middle of AB (horizontally).

XtraEdge for IIT-JEE

velocity =

(122.5) 2 49 Vertical Direction vy = ? 49 uy = 2

∴

Particles P and Q of mass 20 gm and 40 gm respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q make 45º and 135º angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49 m/s. The separation AB is 245 m. [IIT-1982] Q P

45º

49 / 2 m / s B

A 49 / 2 m / s

 0.001   0.001   0.1  = 2  = 0.0489 + +  0.05   0.125   110  So maximum percentage error = 4.89%.

2.

135º

45º

vy = uy + aty =

mpVp

mQVQ

49 2

– 9.8 ×

(122.5) 2 =0 49

–

+

Before collision mpVp

mQVQ′

After collision Also, v2 – u2 = 2as 2

10

∴

 49   = 2 (– 9.8) × s –   3

⇒

s = 61.25

MAY 2010

⇒ The collision takes place at the maximum height where the velocities of both the particles will be in the horizontal direction. Applying conservation of linear momentum in the horizontal direction with the information that P retraces its path therefore its momentum will be same in magnitude but different in direction. Momentum of system after collision ∴ mpmp – mQvQ = – mPvP + mQv′Q 2m P v P − m Q v Q ∴ v′Q = mQ

Let v be the velocity given. The centripetal force is provided by the resultant gravitational attraction of the two masses

∴ ⇒

2 × 0.02 × (49 / 2 ) − 0.040(49 / 2 ) = 0.040 49  0.04  49 = ×0=0  0.040 −1 = 2   2 New Path of Q after Collision Considering vertical Motion of Q uy = 0 sy = – 61.25 ay = – 9.8 ty = ? 1 1 S = ut + at2 = × (– 9.8) × t2 = (–61.25) 2 2 ∴ t = 3.53 sec Considering Horizontal motion of Q : Since the VQ' = 0, therefore the particle Q falls down

FR = F 2 + F 2 + 2F 2 cos 60º m×m = 3F= 3G a2 2 2 m mv 3G 2 = r a

v2 =

3G ma a2 × 3

Gm a Time period of circular motion

⇒

v=

T=

2πr 2π a 3 a3 = = 2π v 3Gm Gm

a 4. Two fixed charges –2Q and Q are located at the points with coordinates (–3a, 0) and (+3a, 0) respectively in the x-y plane. [IIT-1991] (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Give the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole x-axis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantative argument that the particle eventually crosses the circle. Find its speed when it does so. Sol. (a) Let P be a point in the X-Y plane with coordinates (x, y) at which the potential due to charges –2Q and +Q placed at A and B respectively be zero. Y

vertically so it falls down on the mid point of AB. Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time [IIT-1988] period of the circular motion. Sol. The radius of the circle 3.

P(x,y) y

2

r= =

2 2 a a − 3 4 a

A B +Q C X′ (–3a,0) O x (5a,0) (–3a,0) (3a-x) (3a+x) Y′ K (2Q) K ( + Q) = ∴ 2 2 (3a + x ) + y (3a − x ) 2 + y 2

3 m

F FR

m

F

⇒ 2 (3a − x ) 2 + y 2 =

a/ 3 a

⇒ ⇒ ⇒ ⇒ ⇒

m

v

XtraEdge for IIT-JEE

11

2

X

(3a + x ) 2 + y 2

2

4[(3a – x) + y ] = [(3a + x)2 + y2] 4[6a2 + x2 – 12ax + y2] = [6a2 + x2 + 12ax + y2] 3x2 + 3y2 – 30ax + 27a2 = 0 x2 + y2 – 10ax + 9a2 = 0 (x – 5a)2 + (y – 0)2 = (4a)2 MAY 2010

This is the equation of a circle with centre at (5a, 0) and radius 4a. Thus c (5a, 0) is the centre of the circle. (b) For x > 3a To find V(x) at any point on X-axis let us consider a point (arbitrary) M at a distance x from the origin. +Q –2Q (–3a,0)

(3a,x) x

O

X

C B A D (a) Are there positive and negative terminals on the galvanometer ? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. (c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X. Sol. (a) No. There are no positive and negative terminals on the galvanometer. R 0.6 ρ 12 Ω (b) & (c) Q Bridge is balanced AJ = = R JB 0.4 ρ X

M

The potential at M will be V(x) =

K (−2Q) K (+ Q) + x + 3a ( x − 3a )

where k =

1 4πε 0

2   1 − For |x| > 3a ∴ V(x) = KQ  x − 3 a x + 3a   Similarly, for 2   1 0 < |x| < 3a V(x) = KQ  −   3a − x 3a + x 

⇒ X=8Ω where ρ is the resistance per unit length.

Since circle of zero potential cuts the x-axis at (a, 0) and (9a, 0) hence V(x) = 0 at x = a at x = 9a • From the above expressions V(x) → ∞

at x → 3a

and

V(x) → – ∞ and

A

x → – 3a. •

V(x) → 0 as x → + ∞

•

V(x) varies at

1 in general. x

a

12Ω

D

CHEMISTRY The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and K2O in the mixture ? [IIT-1979] Sol. Mass of sample of feldspar containing Na2O and K2O = 0.5 g. According to the question, Na2O + 2HCl → 2NaCl + H2O ..(1) 2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g + 2HCl → 2KCl + H2O ...(2) K2O 2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g Let, Mass of NaCl = x g ∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate, NaCl + AgNO3 → AgCl + NaNO3 ...(3) 23 + 35.5 = 58.5g 108 + 35.5 = 143.5g ...(4) KCl + AgNO3 → AgCl + KNO3 39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3)

X

3a

Applying Energy Conservation (K.E. + P.E.)centre = (KE. + P.E.)circumference  Qq 2Qq  1  Qq 2Qq  0 +K = mv2 + K  − −   8a  2  2a  6a 12a 

5.

J

6.

–3a

⇒

X C G

V

(c)

12Ω

KQq 1 mv-2 = 4a 2

⇒v=

1  Qq  KQq =   2ma 4πε 0  2ma 

A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12Ω are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [IIT-2002]

XtraEdge for IIT-JEE

12

MAY 2010

58.5 g of NaCl yields = 143.5 g AgCl 143.5 ∴ x g of NaCl yields = x g AgCl 58.5 And from eq. (4), 74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields 143.5 = (0.1180 – x)g AgCl 74.5 Total mass of AgCl 143.5 143.5 x+ (0.1180 – x) = 0.2451 58.5 74.5 which gives, x = 0.0342 Hence, Mass of NaCl = x = 0.0342 g And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1), 117 g of NaCl is obtained from = 62 g Na2O ∴ 0.0342 g NaCl is obtained from 62 = × 0.032 = 0.018 g Na2O 117 From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from 94 = × 0.0838 = 0.053 g K2O 149 0.018 Step 3. % of Na2O in feldspar = × 100 = 3.6% 0.5 0.053 % of K2O in feldspar = × 100 = 10.6 % 0.5 7. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ..... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ? [IIT-1996] Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in fig. Three such cells form one hcp unit. For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside, hence 8 Number of atoms per unit cell = +1=2 8 O

But

c=

2 2 3

a

c β α

b

a γ

∴ Volume of the hexagonal cell 3 2 2 2 a . a = a3 2 = 2 3 and radius of the atom, r = a/2 Hence, fraction of total volume of atomic packing Volume of 2 atoms factor = Volume of the hexagonal cell 3

4 a 4 2 × π  2 × πr 3 π 3 2 3 = = = 3 3 a 2 3 2 a 2 = 0.74 = 74% ∴ The percentage of void space = 100 – 74 = 26%

A basic nitrogen compound gave a foul smelling gas when treated with CHCl3 and alc. KOH. A 0.295 g sample of the substance, dissolved in aq. HCl and treated with NaNO2 solution at 0ºC liberated a colourless, odourless gas whose volume corresponds to 112 ml at STP. After the evolution of gas was completed, the aq. solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and I2 gave a yellow precipitate. Identify the original sustance. Assume that it contains one N atom per molecule. [IIT-1993] Sol. As the compound on heating with CHCl3 and alc. KOH gives foul smelling gas, it should be any primary amine. 8.

∆ RNH2 + CHCl3 + 3KOH →

a 60º N

b

= Area of the base × height =

XtraEdge for IIT-JEE

C

+ 3KCl + 3H2O

3 2 a 2 ( Q b = a)

Area of the base = b × ON = b × a sin 60º = Volume of the hexagonal cell

RN

(Alkyl isocyanide (foul smelling gas)

Since the compound on treating with NaNO2 and HCl at 0ºC produces a colourless gas, the compound must be a p-aliphatic amine, because if it was aromatic diazonium salt might have been produced

3 2 a.c 2

13

MAY 2010

CH3 – CH – CH3 + I2 + 2NaOH →

NaNO2 + HCl → NaCl + HNO2 RNH2 + HNO2 → ROH + N2 + H2O Thus, the gas liberated is N2. 112 ml 1 = ml Amount of gas produced = 22,400 ml 200

OH CH3 – C – CH3 + 2NaI + 2H2O O

Acetone

CH3 – C – CH3 + 3I2 + 3NaOH →

From the above equation, it is obvious that amount of

O

1 mol. 200 If M is the molar mass of RNH2, then

compound RNH2 =

CI3 – C – CH3 O

∆ CI3 – C – CH3 + NaOH → CH3COONa + CHI3

1 0.295 g mol = 200 Molar mass (M ) –1

∴ M = 0.295 × 200 g mol = 59 g mol Thus, the molar mass of alkyl group (R—) will be, 59 – 16 = 43 g mol–1. Hence, R = C3H7 —, i.e., CH3CH2CH2 – CH3 or CH– CH3

An organic compound (A) C8H6, on treatment with dil. H2SO4 containing HgSO4 gives a compound (B), which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous AlCl3. The compound (B) when treated with iodine in aq. KOH, yields (C) and a yellow compound (D). Identify (A), (B), (C) and (D) with justification. Show, how (B) is formed from (A). [IIT-1994] Sol. The given reactions may be formulated as follows : 9.

The original compound may be either CH3CH2CH2NH2 or CH3 – CH – CH3 NH2

From equation, it is clear that the liquid obtained after distillation is ROH. Since it gives yellow ppt.

C8H6 Dil H2SO4 HgSO4 (A)

with NaOH and I2, it must have CH3 — C — group. Hence, it is concluded that ROH is CH3 – CH – CH3. OH Thus, the original compound is CH3 – CH – CH3. NH2 ∆

Isopropyl amine

CH3

CH – N

C + 3KCl +

CH3

CH3

(A)

Acetophenone (B)

CH – OH + N2 + H2O

CH3COCl AlCl3; – HCl

Isopropyl alcohol

XtraEdge for IIT-JEE

COCH3

dil H2SO4 HgSO4; H2O

CH – NH2 + HNO2 → CH3

C6H6 + Acid chloride

C = CH2

C≡CH

Isopropyl isocyanide

CH3

AlCl3 ∆ ∆ I2 + KOH

(B)

(C) + (D) The reaction of compound (B) with I2 in KOH is iodoform reaction. The compound (B) must have a –COCH3 group so as to exhibit iodoform reaction. Since (B) is obtained from benzene by Friedal-Crafts reaction, it is an aromatic ketone (C6H5COCH3). The compound (C) must be potassium salt of an acid. The compound (A) may be represented as C6H5C2H. Since it gives C6H5COCH3 on treating with dil. H2SO4 and HgSO4, it must contain a triple bond (–C ≡ CH) in the side chain. Here, the given reactions may be formulated as follows : OH

OH

CH3

Yellow ppt.

O

–1

The different equations are : CH3 CH– NH2 + CHCl3 + 3KOH CH3

+ 3NaI + 3H2O

Benzene

14

MAY 2010

O + 3I2 + 4KOH

∆

(X)

K 2 CO 3 + BaCl2 → 2KCl + BaCO 3

COOK

∆ BaCO 3 → BaO + CO 2 ↑

(Y)

+ CHI3

Hence, (A) is CO2 (B) is H2O (X) is KHCO3 (Y) is K2CO3 and (Z) is BaCO3.

(B) Phenyl acetylene

MATHEMATICS

Acetophenone

COOK

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

(D) CHI 3

Idoform

Potassium benzoate

10. When 20.02 g of a white solid (X) is heated, 4.4 g of an acid gas (A) and 1.8g of a neutral gas (B) are evolved leaving behind a solid residue (Y) of weight 13.8 g. Gas (A) turns lime water milky and (B) condenses into a liquid which changes anhydrous CuSO4 blue. The aqueous solution of (Y) is alkaline to litmus and gives 19.7 g of a white precipitate (Z) with BaCl2 solution. (Z) gives CO2 with an acid. [IIT-1989] Identify (A), (B), (X), (Y) and (Z). Sol. (i) Since acidic gas (A) turns lime water milky hence it is CO2 or SO2, both of which form white insoluble compound with Ca(OH)2 (ii) Neutral gas (B) condenses into a liquid which turns anhydrous CuSO4 (white) into blue (CuSO4.5H2O), hence (B) is H2O. (iii) (Y) gives alkaline solution and its solution forms white precipitate (Z) with BaCl2 solution. (Z) on heating gives the acid gas CO2, hence (Z) is BaCO3 and therefore (Y) is a metal carbonate. (iv) Since (Y) and (A) are produced from (X), thus (X) is a metal bicarbonate. (X) → (A) + (B) + (Y) 4.4 g

1.80 g

y B

A(–a, 0)

⇒

x

a2 a (m2 + 1)   = 2 m

1 2

=

1 2 (m + 1) ⇒ 2 = m4 + m2 2

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

(Y)

Q 4.4g CO2 is obtained from 20.02 g of MHCO3 ∴ 4g CO2 is obtained from 200.2 g of MHCO3 200.2 Molecular weight of MHCO3 = = 100.1 2 Atomic weight of M = 39 Thus, the metal M is potassium and then (X) is KHCO3. The equations are :

XtraEdge for IIT-JEE

L

2

If

13.8 g

( B)

x=a

D

∆ 2MHCO 3 → CO 2 + H2 O + M 2 CO 3 (A)

O

E

C

From the above values we may write a general equation for a bicarbonate. (X)

(A)

COCH3

C≡CH

20.02 g

( Z)

( Z)

(D)

(C) Potassium benzoate

(C)

( B)

–3KI;–3H2O

(B)

Hence. (A)

(A)

(Y)

πx = – a/2

– C – CH3

∆ 2KHCO 3 → K 2 CO 3 + CO 2 + H 2 O

15

MAY 2010

= 2 OB 2 − OK 2 2

⇒

2

2

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

=2

⇒ locus of A´ is a plane parallel to the plane ABCD 14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001] Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws. Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements in the range. Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0

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

a = a1 ˆi + a2 ˆj + a3 kˆ ;

→

b = b1 ˆi + b2 ˆj + b3 kˆ

→

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

r

+ b r + c r ) = 3L. Show that

r =1

V ≤ L3.

[IIT-2002]

→ → →

Sol. V = | a .( b × c ) | ≤

a 12

+ a 22

b12 + b 22 + b 32

+ a 32 c12 + c 22 + c 32

⇒ M = 3n – 3.2n + 3 Hence the total number of favourable cases = (3n – 3.2n + 3). 6C3

...(1)

(a 1 + a 2 + a 3 ) + (b1 + b 2 + b 3 ) + (c1 + c 2 + c 3 ) 3 [(a1 + a2 + a3) (b1 + b2 + b3) (c1 + c2 + c3)]1/3 ∴ L3 ≥ [(a1 + a2 + a3)(b1 + b2 + b3)(c1 + c2 + c3)] ..(2) Now, (a1 + a2 + a3)2 = a 12 + a 22 + a 32 + 2a1a2 + 2a1a3 + 2a2a3 ≥ a 12 + a 22 + a 32

Now, L =

⇒ (a1 + a2 + a3) ≥ Similarly, and

⇒ required probability =

(c1 + c2 + c3) ≥

b12 + b 22 + b 32

c12 + c 22 + c 32

∴ from (1) and (2) L3 ≥ [( a 12 + a 22 + a 32 )( b12 + b 22 + b 32 )( c12 + c 22 + c 32 )]1/3 ≥ V

m   1 , P≡  ; Q≡ m + 1 m +1 

13. T is a prallelopiped in which A, B, C and D are vertices of one face and the just above it has corresponding vertices A´, B´, C´, D´, T is now compressed to S with face ABCD remaining same and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A´´ is a plane. [IIT-2004] Sol. Let the equation of the plane ABCD be ax + by + cz + d = 0, the point A´´ be (α, β, γ) and the height of the parallelopiped ABCD be h.

XtraEdge for IIT-JEE

(3n − 3.2 n + 3) ×6 C 3 6n

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

a 12 + a 22 + a 32

(b1 + b2 + b3) ≥

= 90%. h

⇒ locus is, ax + by + cz + d = ±0.9h a 2 + b 2 + c 2

1 (BC + DE) (KL) 2 2 1  3a  15a = (a + 4a)   = 4 2  2

∑ (a

a 2 + b2 + c2

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

BCDE =

numbers and

| aα + bβ + cγ + d |

Now equation of L1 : y – 2x =

3m   3 ,   m + 1 m +1  m−2 m +1

...(1)

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

equation of L2 : y + 3x =

16

MAY 2010

Physics Challenging Problems

Set # 1

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

So lutions will b e p ub lished in n ex t issue 1.

Two capacitors C1 and C2, can be charged to a

3.

potential V/2 each by having

V

varying magnetic field B = [(4T/s) t + 0.3T] in a

C2

C1

S1

cylindrical region of radius 4m. An equilateral R

R

In the figure shown there exists a uniform time

triangular conducting loop is placed in the

O

magnetic field with its centroide on the axis of the

S2

field and its plane perpendicular to the field. A

(A) S1 closed and S2 open + + + +

(B) S1 open and S2 closed (C) S1 and S2 both closed (D) cannot be charged at V/2 2.

B

+ + + + + + +

+ + + + + + + + + + + + + + + + + + C

(A) e.m.f. induced in any one rod is 16V

Energy liberated in the de-excitation of hydrogen

(B) e.m.f. induced in the complete ∆ABC is

atom from 3rd level to 1st level falls on a photo-

48 3V

cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown

(C) e.m.f. induced in the complete ∆ABC is 48V

hydrogen like gas, excited to 2nd energy level, it is

(D) e.m.f. induced in any one rod is 16 3V

found that the de-Broglie wavelength of the 4.

fastest photoelectrons, now ejected has decreased

6 parallel plates are arranged as shown. Each

by a factor of 3. For this new gas, difference of

plate has an area A and Distance between them is

energies of 2nd Lyman line and 1st Balmer line if

as shown. Plate 1-4 and plates 3-6 are connected

found to be 3 times the ionization potential of the

equivalent capacitance across 2 and 5 can be

hydrogen atom. Select the correct statement(s)

writted as

(A) The gas is lithium

nA ∈0 . Find min value of n. (n, d are d

natural numbers)

(B) The gas is helium (C) The work function of photo-cathode is 8.5eV (D) The work function of photo-cathode is 5.5eV

d d

XtraEdge for IIT-JEE

17

4 5 6

1 2 3 2d

d d

MAY 2010

5.

Match the following

6.

Assuming the potential at infinity to be zero, the

Column – I

Column – II

potential at a point located at a distance a/2 from

(A) A light conducting

(P) Magnetic field B

the centre of the sphere will be

circular flexible

is doubled.

loop of wire of

2 1 a − b  

(A)

Q 4πε 0

(C)

Q 1 1  − 4πε 0  a b 

(B)

Q 4πε 0

 11  8a − 

1 b 

radius r carrying current I is placed in uniform magnetic

(D) None of these

field B, the tension in the loop is doubled

7.

if (B) Magnetic field at a

slowly from inner surface of the shell to surface

(Q) Inductance is

point due to a long

increased by four

straight current

times.

of the sphericalball will be

carrying wire at a point near the wire is doubled if (C) The energy stored

Work done by external agent in taking a charge q

1 1 (A) kQq  −  a c

1 1 (B) kQq  −  b a 

1 1  (C) kQq  −  a b

1 1  (D) kQq  −  c a 

(R) Current I is

in the inductor will

doubled

8.

become four times

surface is fixed to be zero. Now the charge on the

(D) The force acting on a (S) Radius r is moving charge,

Now the outer shell is grounded, i.e., the outer inner ball will be

doubled

(A) zero

moving in a constant magnetic field will be

(C)

doubled if

(B) Q

Q1 1 1  + −  Ca c b

(D)

Q1 1 1  + −  b a c b

(T) Velocity v is Doubled Passage # (Q. No. 6 to Q. No. 8 )

A solid, insulating ball of radius ‘a’ is surrounded

Cartoon Law of Physics

by a conducting spherical shell of inner radius ‘b’

Any body passing through solid matter will leave a

and outer radius ‘c’ as shown in the figure. The

perforation conforming to its perimeter.

inner ball has a charge Q which is uniformly

Also

distribute throughout is volume. The conducting

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silhouette

of

passage,

this

pressure explosions and of reckless cowards who

Answer the following questions.

Q

the

phenomenon is the specialty of victims of directed-

spherical shell has a charge –Q.

are so eager to escape that they exit directly

–Q c

b

called

through the wall of a house, leaving a cookiecutout-perfect hole. The threat of skunks or matrimony often catalyzes this reaction.

a

18

MAY 2010

Students' Forum PHYSICS 1.

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

(v + v s − v t ) 1 =r ∆t v 1300  hits  = 40 × = 43.33   … (5) 1200  second  (b) Using the same reasoning for this case, we obtain: v − vs + v t r′ = r v 1100  hits  = 40 × = 36.66  … (6)  1200  second  (c) In this case: v + vs + v t r′ = r v 1700  hits  = 40 × = 56.67  … (7)  1200  second  (d) Here, v − vs − v t r′ = r v 700  hits  = 40 × = 23.33  … (8)  1200  second 

m A fighter plane flies at a velocity of 300   . On  sec  the fighter plane there is a gun which shoots at a rate of 40 rounds per second with a muzzle velocity of m 1200   . The shots are aimed at another fighter  sec 

r′ =

m plane flying at a velocity of 200   . Find the rate  sec  at which the projectiles hit the target plane : (a) When the two planes move in the same direction, and the target plane is in front of the shooting plane. (b) The same as (a), when the target plane is in the rear of the shooting plane. (c) When the two planes move towards one another. (d) When the two planes move away from one another. Sol. Denote by vs the velocity of the plane from which the shots are fired, by vt the velocity of the target plane and by L the distance between them at the certain moment of time when the shooting plane starts to shoot. Denote by r the rate of fire of the gun and by v the muzzle velocity. (a) The time it takes for the first projectile to reach the target plane is: L …(1) t1 = v + vs − v t 1 After a time of the second projectile is shot, and r the distance between the planes at this time is given by: v − vt …(2) L' = L – s r Thus, the time it takes the second projectile to arrive at the target plane is: v − vt L− s r …(3) t2 = v + vs − v t which is 1 ∆t = t2 + – t1 r vs − v t 1 v = …(4) = – r r(v + vs − v t ) r(v + v s − v t ) after the first shot. Naturally, the time increment does not depend on the initial distance; thus the rate of hitting is:

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2.

Consider the system described in figure.

m1 m2

(a) Use the equations of energy conservation to find the velocities of masses m1 and m2 after they are released from rest and pass a distance y (assume m2 > m1). (b) Use the expression obtained in the first section to find the acceleration of the masses. Sol. (a) We write the change in the potential energy of the masses : m1 : = ∆Ep = m1gy ...(1) m2 : ∆Ep = – m2gy ...(2) The change in the kinetic energy is 1 1 ...(3) ∆Ek = m1v12 + m2v22 2 2

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MAY 2010

r r Because | v1 | = | v 2 | ≡ v, we obtain

(b) The maximal value is obtained when N = 0 and ay´ = 0. (The block is still upon the inclined surface, as stipulated.) Therefore, a = g cot α (c) The equation of motion with a constant acceleration 'a' is : 1 x(t) = x0 + v0t + at2 2 1 In our case, we have L = a x´ t2. Therefore, 2 2L t= g sin α + a cos α

1 (m1 + m2)v2 ...(4) 2 Since there are no external forces except gravity, which is a conservative force, we know, using energy conservation that : ∆Ep + ∆Ek = 0 ...(5) By substituting 1 (m1 – m2)gy + (m1 + m2)v2 = 0 ...(6) 2

∆Ek =

1/ 2

 2g(m 2 − m1 )   Hence, v(y) =   m1 + m 2  (b) By definition,

1/ 2

a=

 2g(m 2 − m1 )  dv  =  dt  m1 + m 2 

...(7)

y

.

(d) Because 'a' is changed to (–a), the equation of motion in the xˆ´ direction becomes : a x´ = g sin α – a cos α In order for the mass to slide up the incline, the condition a x´ > 0 must be met. Thus, a > g tan α

d y dt

1/ 2

 2g (m 2 − m1 )  dy 1  ...(8) . =   m1 + m 2  2 y dt dy and because = v we substitute Eq. (7) into Eq. dt (8) and obtain. m − m1 a= 2 g m 2 + m1

3.

Two long wires are placed on a smooth horizontal table. Wires have equal but opposite charges. Magnitude of linear charge density on each wire is λ. Calculate (for unit length of wires) work required to increase the separation between the wires from a to 2a. Sol. Since, wires have opposite charge, therefore, they attract each other. To increase separation between the wires, work is to be done against this force of attraction. Let at some instant separation between the wires be x as shown in Fig. To calculate force of attraction between the wires, first electric field due to charge on one wire at position of the second wire is to be calculated. Therefore, considering a cylindrical surface of radius x and of unit length, co-axial with positively charged wire, + – 4.

A smooth incline of lift angle α is accelerated at a rate α. A block of mass m is placed on the incline. At t = 0 the block is released, and begins moving (see figure.) N m

y´

a mg

x´ α

(a) Write the equations of motion for the block. (b) What is the maximal value of 'a' for which the block will remain attached to the incline ? (c) How much time is required for the block to slide a distance L along the incline ? (d) If we accelerate the incline in the opposite direction, what is the minimal value of a necessary for the block to slide up the incline ? Sol. (a) D'alembert's force exists in the acclerated system, on the plane of the incline. Therefore, r r m a = mg xˆ ´ sin α – mg yˆ ´ cos α + N yˆ ´ – m a r a is expressed by the unit vectors of the accelerated system r as : a = – a xˆ ´ cos α – a yˆ ´ sin α and in component form : a x´ = g sin α + a cos α  a = N − g cos α + a sin α ( N ≥ 0)  y´ m

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+

–

+

–

+

x

–

Its area = 2π x × 1 Charge enclosed within it = λ ∴ Flux passing through the cylindrical surface =

λ ε0

Electric field,

E = Flux passing per unit area. (λ / ε 0 ) λ = = 2π x 2πε 0 x Magnitude of charge on unit length of second wire = λ ∴ Force of attraction per unit length is F = λ E

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MAY 2010

λ2 2πε 0 x To increase the separation, wires are to be pulled apart by applying an infinitesimally greater force (F + dF). ∴ Work done to increase separation from x to (x + dx), dW = (F + dF) dx ≈ F. dx x = 2a x = 2 a λ2 F dx = dx or Total work done = x = a 2πε 0 x x =a

or

F=

∫

=

∫

λ2 loge 2 2πε 0

Two short electric dipoles having dipole moment p1 and p2 are placed co-axially and uni-directionally, at a distance r apart. Calculate nature and magnitude of force between them. Sol. Let second dipole having dipole moment p2 consist of charges (+ q2) and (– q2) which are separated by an elemental distance 2dr as shown in Fig. Then p2 = q2 2dr …(1) 5.

→ p1

2.dr r

– q2

+ q2

Since, dipoles are separated by a distance r, it means distance between their centres is r. Distance of charges (– q2) & (+ q2) from centre of first dipole is (r – dr) & (r + dr), respectively. If electric field strength due to and at distance r from dipole having dipole moment p1 is E, then electric field strength at position of two charges will be (E – dE) & (E + dE), respectively. 1 2p1 (rightward) Where E= 4π ε 0 r 3

The typical size of a meteor is about one cubic centimeter, which is equivalent to the size of a sugar cube.

2.

Each day, Earth accumulate 10 to 100 tons of material.

3.

There are over 100 billion galaxies in the universe.

4.

The largest galaxies contain nearly 400 billion stars.

5.

The risk of a falling meteorite striking a human occurs once every 9,300 years.

6.

A piece of a neutron star the size of a pin point would way 1 million tons.

7.

Europa, Jupiter’s covered in ice.

8.

Light reflecting off the moon takes 1.2822 seconds to reach Earth.

9.

There has only been one satellite destroyed by a meteor, it was the European Space Agency’s Olympus in 1993.

moon,

is

completely

10. The International Space Station orbits at 248 miles above the Earth. 11. The Earth orbits the Sun at 66,700mph. 12. Venus spins in the opposite direction compared to the Earth and most other planets. This means that the Sun rises in the West and sets in the East.

2p 1 . 41 dr ∴ dE = – 3 . 4π ε 0 r Force on charge (– q2) is F1 = (E – dE) q2 (leftward) and that on charge (+ q2) is F2 = (E + dE) q2 (rightwards) Hence, net force on second dipole is F = F2 – F1 (rightward) 2p 1 or F = dE 2q2 = – 3 . 1 dr 2q2 4π ε 0 r 4 But 2q2 dr = p2 1 6p1p 2 ∴ Net force = – 4πε 0 r 4 (–ve) sign indicates that actual direction of force on second dipole is leftward or force between two dipoles is of attraction and its magnitude is 6p p 1 . 14 2 F= 4πε 0 r

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1.

13. The Moon is moving away from the Earth at about 34cm per year. 14. The Sun, composed mostly of helium and hydrogen, has a surface temperature of 6000 degrees Celsius. 15. A manned rocket reaches the moon in less time than it took a stagecoach to travel the length of England.

16. The nearest known black hole is 1,600 light years (10 quadrillion miles/16 quadrillion kilometers) away.

21

MAY 2010

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22

MAY 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Electrostatics-I KEY CONCEPTS & PROBLEM SOLVING STRATEGY

•

The unit of electric field is Newton per coulomb or volt per metre. The electric field strength at a distance r from a point charge q in a medium of permittivity ε is given by 1 q E= 4πε r 2 → 1 q rˆ Vectorially E = 4πε r 2 With reference to any origin

Coulomb's Law : 1 q1q 2 F0 = (in vacuum) 4πε 0 r 2 →

Vectorially F =

1 q1q 2 rˆ 4πε 0 r 2

In any material medium F =

1 q1q 2 4πε 0 ε r r 2

where εr is a constant of the material medium called its relative permittivity, and ε0 is a universal constant, called the permittivity of free space. 1 ε0 = 8.85 × 10–12 or = 9 × 109 4πε 0

→

Where R is the position vector of the field point and →

r , the position vector of q. Due to a number of discrete charges

Where ε is called the absolute permittivity of the medium. Obviously, εr = F0/F. Remember εr = ∞ for conductors. Conductors and insulators Each body contains enormous amounts of equal and opposite charges. A 'charged' body contains an excess of either positive or negative charge. In a conductor, some of the negative charges are free to move around. In an insulator (also called a dielectric), the charges cannot move. They can only undergto small localized displacements, causing polarization. Induction When a charged body A is brought near another body B, unlike charges are induced on the near surface of B (called bound charges) and like charges appear on the far surface of B (called free charges) If B is a conductor, the free charges can be removed by earthing B, e.g., by touching it. If B is an insulator, separation of like and unlike charges will still occur due to induction. However, the like charges cannot then be removed by earthing B. Electric Field And Potential Electric Field An electric field of strength E is said to exist at a point if a test charge ∆q at that point experiences a force given by →

→

∆ F = ∆q F

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or

→

→

q R− r E= 4πε → → 3 R− r

The unit of ε0 is C2N m–2 or farad per metre. 1 q1q 2 Also F = 4πε 0 ε r r 2

•

→

→

→

E=

i= N

∑ i =1

→

→

q1 R − ri 4πε → → 3 R − ri

Electric Potential The electric potential at a point is the work done by an external agent in bringing a unit positive charge from infinity up to that point along any arbitrary path. ∆W∞ →P (by an external agent ) VP = volt(V) or JC–1 ∆q

The potential difference between two points P and Q is given by ∆WQ→P (by agent ) volt (V) VP – VQ = ∆q The potential at a distance r from a point charge q in a medium of permittivity ε is 1 q 1 q ϕ or V = = → → 4πε r 4πε R− r with reference to any arbitrary origin. Due to a number of charges i= N

ϕ or V =

→

∆F E= ∆q

i =1

23

1

∑ 4πε

q1 →

→

R − ri

MAY 2010

In a conductor, all points have the same potential. If charge q (coulomb) is placed at a point where the potential is V (volt), the potential energy of the system is qV (joule). It follows that if charges q1, q2 are separated by distance r, the mutual potential qq energy of the system is 1 2 . 4πεr • Relation Between Field (E) And Potential (V) The negative of the rate of change of potential along a given direction is equal to the component of the field that direction. ∂V Er = – along r ∂r ∂V perpendicular to r and E⊥ = r∂θ When two points have different potentials, an electric field will exist between them, directed from the higher to the lower potential. • Lines of Force A line of force in an electric field is such a curve that the tangent to it at any point gives the direction of the field at that point. Lines of force cannot intersect each other because it is physically impossible for an electric field to have two directions simultaneously. • Equipotential Surfaces The locus of points of equal potential is called an equipotential surface. Equipotential surfaces lie at right angles to the electric field. Like lines of force, they can never intersect. Note: For solving problems involving electrostatic units, remember the following conversion factors: 3 × 109 esu of charge = 1 C 1 esu of potential = 300 V • Electric Flux The electric flux over a surface is the product of its surface area and the normal component of the electric field strength on that surface. Thus, →

Problems in electrostatics can be greatly simplified by the use of Gaussian surfaces. These are imaginary surfaces in which the electric intensity is either parallel to or perpendicular to the surface everywhere. There are no restrictions in constructing a Gaussian surface. The following results follow from Gauss's law 1. In a charged conductor, the entire charge resides only on the outer surface. (It must always be remembered that the electric field is zero inside a conductor.) 2. Near a large plane conductor with a charge density σ (i.e., charge per unit area), the electric intensity is E = σ/ε0 along the normal to the plane 3. Near an infinite plane sheet of charge with a charge density σ, the electric intensity is E = σ/2ε0 along the normal to the plane 4. The electric intensity at a distance r from the axis of a long cylinder with λ charge per unit length (called the linear density of charge), is → 1 λ E= along r 2πε 0 r Problem solving strategy: Coulomb's Law : Step 1 : The relevant concepts : Coulomb's law comes into play whenever you need to know the electric force acting between charged particles. Step 2 : The problem using the following steps : Make a drawing showing the locations of the charged particles and label each particle with its charge. This step is particularly important if more than two charged particles are present. If three or more charges are present and they do not all lie on the same line, set up an xycoordinate system. Often you will need to find the electric force on just one particle. If so, identify that particle. Step 3 : The solution as follows : For each particle that exerts a force on the particle of interest, calculate the magnitude of that force 1 | q1q 2 | using equation F = 4πε 0 r 2

→

dϕ = (E cos θ) ds = En ds = E . ds E

ds

N

O

Sketch the electric force vectors acting on the particle(s) of interest due to each of the other particles (that is, make a free-body diagram). Remember that the force exerted by particle 1 on particle 2 points from particle 2 toward particle 1 if the two charges have opposite signs, but points from particle 2 directly away from particle 1 if the charges have the same sign. Calculate the total electric force on the particle(s) of interest. Remember that the electric force, like any force, is a vector. When the forces acting on a charge are caused by two or more other charges, the total force on the charge is the vector sum of

The total electric flux over a surface is obtained by summing :

ϕE =

→

→

∑ E .∆ s

∫

or

→

→

E .d s

Gauss's Theorem The total electric flux across a 1 closed surface is equal to times the total charge ε0 inside the surface.

Mathematically

→

→

∑ E .∆ s

= q/ε0

where q is the total charge enclosed by the surface.

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24

MAY 2010

the indivual forces. It's often helpful to use components in an xy-coordinate system. Be sure to use correct vector notation; if a symbol represents a vector quantity, put an arrow over it. If you get sloppy with your notation, you will also get sloppy with your thinking. As always, using consistent units is essential. With the value of k = 1/4πε0 given above, distances must be in meters, charge in coulombs, and force in newtons. If you are given distance in centimeters, inches, or furlongs, donot forget to convert ! When a charge is given in microcoulombs (µC) or nanocoulombs (nC), remember that 1µC = 10–6C and 1nC = 10–9C. Some example and problems in this and later chapters involve a continuous distribution of charge along a line or over a surface. In these cases the vector sum described in Step 3 becomes a vector integral, usually carried out by use of components. We divide the total charge distribution into infinitesimal pieces, use Coulomb's law for each piece, and then integrate to find the vector sum. Sometimes this process can be done without explicit use of integration. In many situations the charge distribution will be symmetrical. For example, you might be asked to find the force on a charge Q in the presence of two other identical charges q, one above and to the left of Q and the other below and to the left of Q. If the distance from Q to each of the other charges are the same, the force on Q from each charge has the same magnitude; if each force vector makes the same angle with the horizontal axis, adding these vectors to find the net force is particularly easy. Whenever possible, exploit any symmetries to simplify the problem-solving process. Step 4 : your answer : Check whether your numerical results are reasonable, and confirm that the like charges repel opposite charges attract. Problem solving strategy : Electric-field calculations Step 1: the relevant concepts : Use the principle of superposition whenever you need to calculate the electric field due to a charge distribution (two or more point charges, a distribution over a line, surface, or volume or a combination of these). Step 2: The problem using the following steps : Make a drawing that clearly shows the locations of the charges and your choice of coordinate axes. On your drawing, indicate the position of the field point (the point at which you want to calculate the r electric field E ). Sometimes the field point will be at some arbitrary position along a line. For r example, you may be asked to find E at point on the x-axis. Step 3 : The solution as follows :

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Be sure to use a consistent set of units. Distances must be in meters and charge must be in coulombs. If you are given centimeters or nanocoulombs, do not forget to convert. When adding up the electric fields caused by different parts of the charge distribution, remember that electric field is a vector, so you must use vector addition. Don't simply add together the magnitude of the individual fields: the directions are important, too. Take advantage of any symmetries in the charge distribution. For example, if a positive charge and a negative charge of equal magnitude are placed symmetrically with respect to the field point, they produce electric fields of the same magnitude but with mirror-image directions. Exploiting these symmetries will simplify your calculations. Must often you will use components to compute vector sums. Use proper vector notation; distinguish carefully between scalars, vectors, and components of vectors. Be certain the components are consistent with your choice of coordinate axes. r In working out the directions of E vectors, be careful to distinguish between the source point and the field point. The field produced by a point charge always points from source point to field point if the charge is positive; it points in the opposite direction if the charge is negative. In some situations you will have a continuous distribution of charge along a line, over a surface, or through a volume. Then you must define a small element of charge that can be considered as a point, finds of all charge elements. Usually it is r easiest to do this for each component of E separately, and often you will need to evaluate one or more integrals. Make certain the limits on your integrals are correct; especially when the situation has symmetry, make sure you don't count the charge twice. r Step 4 : your answer : Check that the direction of E is reason able. If your result for the electric-field magnitude E is a function of position (say, the coordinate x), check your result in any limits for which you know what the magnitude should be. When possible, check your answer by calculating it in a different way. Problem solving strategy : Gauss's Law Step 1 : Identify the relevant concepts : Gauss's law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniform over a plane. In these situations r we determine the direction of E from the symmetry of the charge distribution. If we are given the charge distribution. we can use Gauss's law to find the the r magnitude of E . Alternatively, if we are given the field, we can use Gauss's law to determine the details 25

MAY 2010

of the charge distribution. In either case, begin your analysis by asking the question, "What is the symmetry ?" Step 2 : Set up the problem using the following steps Select the surface that you will use with Gauss's law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface. The Gaussian surface does not have to be a real physical surface, such as a surface of a solid body. Often the appropriate surface is an imaginary geometric surface; it may be in empty space, embedded in a solid body, or both. Usually you can evaluate the integral in Gauss's law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively. Step 3 : Execute the solution as follows : Carry out the integral in Eq. r r Q ΦE = E cos φ dA = E dA = E.dA = encl ε0

∫

∫

surface is not necessarily zero. In that case, however, the integral over the Gaussian surface – is always zero. Once you have evaluated the integral, use eq. to solve for your target variable. Step 4 : Evaluate your answer : Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it make sense.

Solved Examples 1.

Supposing that the earth has a charge surface density of 1 electron/metre2, calculate (i) earth's potential, (ii) electric field just outside earths surface. The electronic charge is – 1.6 × 10–19 coulomb and earth's radius is 6.4×106 metre (ε0 = 8.9 × 10–12 coul2/nt–m2). Sol. Let R and σ be the radius and charge surface density of earth respectively. The total charge, q on the earth surface is given by q = 4 p R2 σ (i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated at its centre. Thus, 1 q . V= 4πε 0 R

∫

(various forms of Gauss's law) This may look like a daunting task, but the symmetry of the charge distribution and your careful choice of a Gaussian surface makes it straightforward. Often you can think of the closed surface as being made up of several separate surfaces, such as the side and ends of a cylinder. The integral

∫

=

Substituting the given values V=

E dA

over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. Some of these integrals may be zero, as in points 4 and 5 below. r If E is perpendicular (normal) at every point to a surface with area A, if points outward from the interior of the surface, and if it equal to EA. If r instead E is perpendicular and inward, then E⊥ =

(ii) E =

=

⊥ dA

nt − m joule = – 0.115 volt. = – 0.115 coul coul

q 4πR 2 σ σ 1 1 = . = 2 4πε 0 R 4πε 0 ε0 R2 − 1.6 × 10 −19 coul / metre 2 −12

2

2

= – 1.8 × 10–8 nt/coul.

Determine the electric field strength vector if the potential of this field depends on x, y co-ordinates as (a) V = a(x2 – y2) and (b) V = axy. Sol. (a) V = a(x2 – y2) ∂V ∂V Hence, Ex = – = – 2ax, Ey = – = + 2ay ∂x ∂y 2.

, E⊥ is always the

perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian

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(8.9 × 10 −12 coul 2 / nt − m 2 )

8.9 × 10 coul / nt − m The negative sign shows that E is radially inward.

∫

∫E

(6.4 × 10 −6 metre) × (−1.6 × 10 −19 coul / metre 2 )

= – 0.115

– E and E ⊥ dA = – EA. r If E is tangent to a surface at every point, then E⊥ = 0 and the integral over that surface is zero. r If E = 0 at every point on a surface, the integral is zero. In the integral

1 4πR 2 σ R.σ . = 4πε 0 R ε0

∴ E = – 2axi + 2ayj or E = – 2a(xi – yj) (b) V = a x y

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MAY 2010

Hence, Ex = –

∴ 3.

∂V ∂V = –ay, Ey = – = – ax ∂x ∂y

Φ1 =

E = – ayi – axj = – a[yi + xj]

∴

O

∴

R

Sol. Let q and q′ be the charges on inner and outer sphere. Then q + q′ = Q …(1) As the surface densities are equal, hence q q' = 2 4πr 4πR 2 (∴ Surface density = charge/area) ∴ q R2 = q′ r2 …(2) From eq. (1) q′ = (Q – q), hence q R2 = (Q – q)r2 q(R2 + r2) = Q r2

and q′ = Q – q =

R2 + r2 Now potential at O is given by 1 q 1 q' V= + 4πε 0 r 4πε 0 r

= =

Φ′ =

∫

q 5ε 0

A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of a radius 5 cm. (a) Find the electric field at a point 2 cm away from the centre. (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere. Sol. (a) See fig. (a) Let P be a point where we have to calculate the electric field. We draw a Gaussian surface (shown dotted) through point P. The flux through this surface is q = 6 × 10–8 C

QR2 R2 + r2

5cm

P

2cm

Fig. (a)

Q (r + R ) 4πε 0 (R 2 + r 2 )

Φ=

Fig. (b)

∫ E.dS = E ∫ dS = 4π(2 ×10

−2 2

) E

According to Gauss's law, Φ = q/ε0 ∴ 4π × (2 × 10–2)2 E = q/ε0

S1 and S2 are two parallel concentric spheres enclosing charges q and 2q respectively as shown in fig. (a) What is the ratio of electric flux through S1 and S2 ? (b) How will the electric flux through the sphere S1 change, if a medium of dielectric constant 5 is introduced in the space inside S1 in place of air ? 4.

or E =

q −2 2

4πε 0 × (2 × 10 )

=

(9 × 10 9 ) × (4 × 10 −8 ) 4 × 10 − 4

= 9 × 105 N/C (b) See fig. (b) We draw a Gaussian surface (shown dotted) through the material of hollow sphere. We know that the electric field in a conducting material is zero, therefore the flux through this Gaussian surface is zero. Using Gauss's law, the total charge enclosed must be zero. So, the charge on the inner surface of hollow sphere is 6 × 10–8 C. So, the charge on the outer surface will be 10 × 10–8 C.

2q q S1

Sol. (a) Let Φ1 and Φ2 be the electric flux through spheres S1 and S2 respectively.

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q + 2q 3q = ε0 ε0

5.

Q r2 Q r2 1 1 + 4πε 0 (R 2 + r 2 )r 4πε 0 (R 2 + r 2 )r

S2

Φ2 =

q / ε0 1 Φ1 = = Φ 2 3q / ε 0 3

∫

r

Q r2

and

(b) Let E be the electric field intensity on the surface of sphere S1 due to charge q placed inside the sphere. When dielectric medium of dielectric constant K is introduced inside sphere S1, then electric field intensity E′ is given by E′ = E/K Now the flux Φ′ through S1 becomes 1 q E.dS = Φ′ = E ' .dS = K Kε 0

A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such the surface densities are equal. Find the potential at the common centre. q q′

∴ q=

q ε0

27

MAY 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

1-D Motion, Projectile Motion KEY CONCEPTS & PROBLEM SOLVING STRATEGY

Graphs During analysis of a graph, the first thing is see the physical quantities drawn along x-axis and y-axis. If y = mx, the graph is a straight line passing through the origin with slope = m. [see fig. (a)]

Kinematics : Velocity (in a particular direction) Displacement (in that direction ) = Time taken r r r r dx (VAB ) x = VAx – VBx and (VAB ) x = t Where dx is the displacement in the x direction in time t. Swimmer crossing a river

m = tanθ Y θ is acute and m is positive

θ

θ

d

vs

θ

Time taken to cross the river =

fig.(a)

d Vs cos θ

For minimum time, θ should be zero. x

X (ii)

c is positive m is negative

d . vs

Also x = vr × t For reaching a point just opposite the horizontal component of velocity should be zero. v sinθ = Vr | Displacement | |Average Velocity| = time r r r r r v–u v + (– u ) a = = t t

X (ii) fig(c)

For y = kx2, where k is a constant, we get parabola [see fig (d)] Y

v 2 + u 2 – 2uv cos θ t Where θ is the angle between v and u. The direction of acceleration is along the resultant of r r v and (– u ).

⇒ |a| =

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fig(b) Y

in this case resultant velocity Vs2 + v 2r and t =

X (i)

vR vr

VR =

c

c

vs

(ii)

if y = mx + c, the graph is a straight line not passing through the origin and having an intercept c which may be positive or negative [see fig. (b,) (c)]] Y c is negative Y m is '+' ve m is '+' ve

vr

vssinθ

X

X

(i)

vscosθ

Y m = tanθ θ is abtuse and m is positive

Parabola Fig.(d)

X

x2 + y2 = r2 is equation of a circle with centre at origin and radius r. 28

MAY 2010

For (x – a)2 + (y – b)2 = r2, the motion is in a circular path with centre at (a, b) and radius r

point and it is u cos θ directed in the horizontal direction. The mechanical energy of a projectile remain constant throughout the path. the following approach should be adopted for solving problems in two-dimensional motion : Resolve the 2-D motion in two 1-D motions in two mutually perpendicular directions (x and y direction) Resolve the vector quantitative along these directions. Now use equations of motion separately for x-direction and y-directions. If you do not resolve a 2-D motions in two 1-D motions in two 1-D motion then use equations of motion in vector form r r r r r 1 r 2 r r r v = u + at ; s = ut + a t ; v.v– u .u = 2a s 2 r r 1 ( u + v )t s= 2 When y = f(x) and we are interested to find (a) The values of x for which y is maximum for minimum (b) The maximum/minimum values of y then we may use the concept of maxima and minima. Problem solving strategy : Motion with constant Acceleration : Step 1: Identify the relevant concepts : In most straight-line motion problems, you can use the constant-acceleration equations. Occasionally, however, you will encounter a situation in which the acceleration isn't constant. In such a case, you'll need a different approach

x 2 y2 + = 1 is equation of an ellipse a 2 b2 x × y = constant give a rectangular hyperbola.

Note : To decide the path of motion of a body, a relationship between x and y is required. Area under-t graph represents change in velocity. Calculus method is used for all types of motion (a = 0 or a = constt or a = variable) Differentiate w.r.t Differentiate w.r.t time time v = f(t) a = f(t)

s = f(t)

integrate w.r.t. time

integrate w.r.t. time

S stand for displacement a=v

dv dv d 2s = = ds dt dt 2

Also vx =

dx dy ⇒ vy = dt dt

dv y dv x d2x d2y = = and ay = dt dt dt 2 dt 2 The same concept can be applied for z-co-ordintae. Projectile motion : P ucosθ u ucosθ ucosθ usinθ θ ucosθ ucosθ Q θ u usinθ

ax =

g g F F

g F

dυ x d  dx  d 2 x =  = dt  dt  dt 2 dt Step 2: Set up the problem using the following steps: You must decide at the beginning of a problem where the origin of coordinates are usually a matter of convenience. If is often easiest to place the particle at the origin at time t = 0; then x0 = 0. It is always helpful to make a motion diagram showing these choices and some later positions of the particle. Remember that your choice of the positive axis direction automatically determines the positive directions for velocity and acceleration. If x is positive to the right of the origin, the vx and ax are also positive toward the right. Restate the problem in words first, and then translate this description into symbols and equations. When does the particle arrive at a certain point (that is, what is the value of t)? where is the particle when its velocity has a specified value (that is, what is the value of x when vx has the specified value)? "where is the motorcyclist when his velocity is 25m/s?"

ax =

g F

Projectile motion is a uniformly accelerated motion. For a projectile motion, the horizontal component of velocity does not change during the path because there is no force in the horizontal direction. The vertical component of velocity goes on decreasing with time from O to P. At he highest point it becomes zero. From P to Q again. the vertical component of velocity increases but in downwards direction. Therefore the minimum velocity is at the topmost

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29

MAY 2010

velocity (either in terms of components or in terms of magnitude and direction) and asked to find the coordinates and velocity components as some later time. In other problems you might be given two points on the trajectory and asked to find the initial velocity. In any case, you'll be using equations

Translated into symbols, this becomes "What is the value of x when vx = 25 m/s?" Make a list of quantities such as x, x0,vx,v0x,ax and t. In general, some of the them will be known quantities, and decide which of the unknowns are the target variables. Be on the lookout for implicit information. For example. "A are sits at a stoplight" Usually means v0x = 0. Step 3 : Execute the solution : Choose an equation from Equation vx = v0x + axt 1 x = x0 + v0xt + axt2 (constant acceleration only) 2

x = (v0 cosα0)t (projectile motion) through ...(1) vy = v0 sin α0 – gt (projectile motion)

make sure that you have as many equations as there are target variables to be found. It often helps to state the problem in words and then translate those words into symbols. For example, when does the particle arrive at a certain point ? (That is at what value of t?) Where is the particle when its velocity has a certain value? (That is, what are the values of x and y when vx or vy has the specified value ?) At the highest point in a trajectory, vy = 0. so the question "When does the particle reach its highest points ?" translates into "When does the projectile return to its initial elevation?" translates into "What is the value of t when y = y0 ?"

v 2x = v 02 x + 2ax(x – x0) (constant accelerations only)

 v + vx  x – x0 =  0 x  t (constant acceleration only) 2   that contains only one of the target variables. Solve this equation for the equation for the target variable, using symbols only. then substitute the known values and compute the value of the target variable. sometimes you will have to solve two simultaneous equations for two unknown quantities. Step 4 : Evaluate your answer : Take a herd look at your results to see whether they make sense. Are they within the general range of values you expected?

Step 3 : Execute the solution use equation (1) & (2)

to find the target variables. As you do so, resist the temptation to break the trajectory into segments and analyze each segment separately. You don't have to start all over, with a new axis and a new time scale, when the projectile reaches its highest point ! It's almost always easier to set up equation (1) & (2)

Problem solving strategy : Projectile Motion : Step 1 : Identify the relevant concepts : The key concept to remember is the throughout projectile motion, the acceleration is downward and has a constant magnitude g. Be on the lookout for aspects of the problem that do not involve projectile motion. For example, the projectile-motion equations don't apply to throwing a ball, because during the throw the ball is acted on by both the thrower's hand and gravity. These equations come into play only after the ball leaves the thrower's hand.

at the starts and continue to use the same axes and time scale throughout the problem. Step 4 : Evaluate your answer : As always, look at your results to see whether they make sense and whether the numerical values seem reasonable. Relative Velocity : Step 1 : Identify the relevant concepts : Whenever you see the phrase "velocity relative to" or "velocity with respect to", it's likely that the concepts of relative will be helpful.

Step 2 : Set up the problem using the following steps

Define your coordinate system and make a sketch showing axes. Usually it's easiest to place the origin to place the origin at the initial (t = 0) position of the projectile. (If the projectile is a thrown ball or a dart shot from a gun, the thrower's hand or exits the muzzle of the gun.) Also, it's usually best to take the x-axis as being horizontal and the y-axis as being upward. Then the initial position is x0 = 0 and y0 = 0, and the components of the (constant) acceleration are ax = 0, ay = – g.

Step 2 : Set up the problem : Label each frame of reference in the problem. Each moving object has its own frame of reference; in addition, you'll almost always have to include the frame of reference of the earth's surface. (Statements such as "The car is traveling north at 90 km/h" implicitly refer to the car's velocity relative to the surface of the earth.) Use the labels to help identify the target variable. For example, if you want to find the velocity of a car (C) with respect to a bus (B), your target variable is vC/B.

List the unknown and known quantities, and decide which unknowns are your target variables. In some problems you'll be given the initial

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...(2)

Step 3 : Execute the solution : Solve for the target variable using equation

vP/A = vP/B + vB/A (relative velocity along a line) ...(1) 30

MAY 2010

(As the ball returns to its initial position, the change in position, the change in position vector of the ball, that is the net displacement will be zero).

(If the velocities are not along the same direction, you'll need to use the vector from of this equations, derived later in this section.) It's important to note the order of the double subscripts in equation (1) vA/B always means "velocity of A relative to B." These subscripts obey an interesting kind of algebra, as equation (1) shown. If regard each one as a fraction, then the fraction on the left side is the product of the fractions on the right sides : P/A = (P/B) (B/A). This is a handy rule you can use when applying Equation (1) to any number of frames of reference. For example, if there are three different frames of reference A, B, and C, we can write immediately. vP/A = vP/C + vC/B + VB/A Step 4 : Evaluate your answer : Be on the lookout for stray minus signs in your answer. If the target variable is the velocity of a car relative to a bus (vV/B), make sure that you haven't accidentally calculated the velocity of the bus relative of the car (vB/C). If you have made this mistake, you can recover using equation. vA/B = – vB/A

→

∴ | V aV | = 0. 2.

(a) (b) (c)

Sol. (a)

Solved Examples 1.

(b)

A small glass ball is pushed with a speed V from A. It moves on a smooth surface and collides with the wall at B. If it loses half of its speed during the collision, find the distance, average speed and velocity of the ball till it reaches at its initial position. A

(c)

B

0.5V

V

d Sol. The ball moves from A to B with a constant speed V. Since it loses half of its speed on collision, it returns from B to A with a constant speed V/2. ∴ V1 = V and V2 = V/2 d1 + d 2 Using the formula, VaV = (d1 / V1 ) + (d 2 / V2 )

If the motion is observed by one of parents, answer to case (a) case (b) gets altred. It is because the speed of the child w.r.t. either of mother or father is 9 Km/hour.

Putting d1 = d2 = d; V1 = V and V2 = V/2 d1 + d 2 2V = We obtain, VaV = (dV) + (d / 0.5V ) 3

A particle is projected with velocity v0 = 100 m/s at an angle θ = 30º with the horizontal. Find : (a) velocity of the particle after 2 sec. (b) angle between initial velocity and the velocity after 2 sec. (c) the maximum height reached by the projectile (d) horizontal range of the projectile 3.

From the formula, →

→

average velocity = V aV =

→

→

→

| s1 + s 2 | | s net | = s1 s 2 t net + V1 V2 →

Since s1 = s2 = d and snet = | s1 + s 2 | = 0

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A long belt is moving horizontally with a speed of 4 Km/hour. A child runs on this belt to and fro with a speed of 9 Km/hour (with respect to the belt) between his father and mother located 50 m apart on the moving belt. For an observer on a stationary platform outside, what is the speed of the child running in the direction of motion of the belt, speed of the child running opposite to the direction of motion of the belt and time taken by the child in case (a) and (b) ? Which of the answers change, if motion is viewed by one of the parents ? Let us consider positive direction of x-axis from left to right Here, vB = + 4 Km/hour Speed of child w.r.t. belt, vC = = 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = 9 + 4 = 13 Km/hour Here, vB = + 4 Km/hour, vC = – 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = – 9 + 4 = –5 Km/hour The negative sign shows that the child appears to run in a direction opposite to the direction of motion of the belt. Distance between the parents, s = 50 m = 0.05 Km Since parents and child are located on the same belt, the speed of the child as observe by stationary observer in either direction (either father to mother or from mother to father) will be 9 Km/hour. Time taken by the child in case (a) and (b), 0.50 km = 20 sec. t= 9 km / hour

31

MAY 2010

→

→

As the ball collides elastically and the inclined plane is fixed, the ball follows the law of reflection. Now along the incline, velocity component after impact is v sin α and acceleration is g sin α. Perpendicular to the incline, velocity component is vcos α and acceleration (– g cos α). Hence, if we measure x and y-coordinate along the incline and perpendicular to the incline, then x = (v sin α) t + ½ (g sin α)t2 and y = (v cos α) t – ½ (g cos α)t2 When the ball hits the plane for a second time, y = 0, (v cos α)t – ½(g cos α)t2 or t = (2v/g) Putting this value of t in x,

→

Sol. (a) v t = v xt ˆi + v yt ˆj

where ˆi and ˆj are the unit vectors along +ve x and +ve y-axis respectively →

v t =(ux + axt) ˆi + (uy + ayt) ˆj

Here,

ux = v0 cos θ = 50 3 m/s, ax = 0 uy = v0 sin θ = 50 m/s, ay = – g (Q g acts downwards) →

v t = 50 3 ˆi + (50 – 10 × 2) ˆj

=[50 3 ˆi + 30 ˆj ] m/s ∴

→

| v 2 | = ( v 2x + v 2y ) = (50 3 ) 2 + (30) 2

x=

→

v 0 = 50 3 ˆi + 50 ˆj

(b)

5.

→

v 2 = 50 3 ˆi + 30 ˆj

∴

→

→

v 0 . v 2 = 7500 + 1500 = 9000 →

→

If α is the angle between v 0 and v 2 →

Then, cos α =

→

v0 . v2

→

→

| v0 | × | v2 |

(c)

9000 = 100 × 91.65

∴

ymax =

v 02 sin 2θ g

v 02 =

or,

Rg = Rg sin 2θ

as θ = 45º

v0 45º

v 02 sin 2 θ = 125 m 2g

1.22m A

u 2 sin 2θ (d) R = = 1732 m g

4.

A batsman hits a ball at a height of 1.22m above the ground so that ball leaves the bat at an angle 45º with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position of the batsman. Will the ball clear the wall if its range is 106.68 m. Take g = 10 m/s2

Sol. R(range) =

α = cos–1 (0.98) = 10.8º 2 2 vy – u y = 2ayy At y = ymax, vy = 0 ∴ 0 – v02 sin2 θ = 2 (–g)ymax

4 v 2 sin α = 8h sin α g

B 106.68m

or,

A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle α with the horizontal. Having fallen the distance 'h', the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time ? α

v0 = (Rg )

…(1)

Equation of trajectory y = x tan 45º – or,

α

y=x–

gx 2 2 v 02 cos 2 45º

gx 2 gx 2 =x– 2Rg.½ Rg

Putting x = 97.53, we get y = 97.53 –

10 × (97.53) 2 = 8.35 cm 106.68 × 10

Hence, height of the ball from the ground level is h = 8.35 + 1.22 = 9.577 m As height of the wall is 7.31 m so the ball will clear the wall.

α

Sol. Just before impact magnitude of velocity of the ball,

v = (2gh )

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32

MAY 2010

KEY CONCEPT

GASEOUS STATE & REAL GASES

Physical Chemistry Fundamentals

increases continuously with pressure in the higher pressure region. (3) For CO2, there is a large dip in the beginning. In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low pressure region.

Real Gases : Deviation from Ideal Behaviour : Real gases do not obey the ideal gas laws exactly under all conditions of temperature and pressure. Experiments show that at low pressures and moderately high temperatures, gases obey the laws of Boyle, Charles and Avogadro approximately, but as the pressure is increased or the temperature is decreased, a marked departure from ideal behaviour is observed. Ideal gas

0

100 200 300 p/101.325 bar Plots of Z versus p of a few gases This graph gives an impression that the nature of the deviations depend upon the nature of the gas. In fact, it is not so. The determining factor is the temperature relative to the critical temperature of the particular gas; near the critical temperature, the pV curves are like those for CO2, but when far away, the curves are like those for H2 (below fig.)

V

Plot of p versus V of hydrogen, as compared to that of an ideal gas The curve for the real gas has a tendency to coincide with that of an ideal gas at low pressures when the volume is large. At higher pressures, however, deviations are observed. Compressibility Factor : The deviations can be displayed more clearly, by plotting the ratio of the observed molar volume Vm to the ideal molar volume Vm,ideal (= RT/p) as a function of pressure at constant temperature. This ratio is called the compressibility factor Z and can be expressed as

Vm ,ideal

T1>T2>T3>T4

1.0

T4 T3 T2 T1 ideal gas

Z 0

p = Vm RT

200 400 p/101.325 kPa

600

Plots of Z versus p of a single gas at various temperatures Provided the pressure is of the order of 1 bar or less, and the temperature is not too near the point of liquefaction, the observed deviations from the ideal gas laws are not more than a few percent. Under these conditions, therefore, the equation pV = nRT and related expressions may be used. Van der Waals Equation of state for a Real gas Causes of Deviations from Ideal Behaviour : The ideal gas laws can be derived from the kinetic theory of gases which is based on the following two important assumptions:

Plots of Compressibility Factor versus Pressure : For an ideal gas Z = 1 and is independent of pressure and temperature. For a real gas, Z = f(T, p), a function of both temperature and pressure. A graph between Z and p for some gases at 273.15 K, the pressure range in this graph is very large. It can be noted that: (1) Z is always greater than 1 for H2. (2) For N2, Z < 1 in the lower pressure range and is greater than 1 at higher pressures. It decreases with increase of pressure in the lower pressure region, passes through a minimum at some pressure and then

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CO2

Z

p

Z=

CH4

ideal gas

1.0

H2

Vm

H2 N2

t = 0ºC

33

MAY 2010

(i) The volume occupied by the molecules is negligible in comparison to the total volume of the gas. (ii) The molecules exert no forces of attraction upon one another. Derivation of van der Waals Equation : Van der Waals was the first to introduce systematically the correction terms due to the above two invalid assumptions in the ideal gas equation piVi = nRT. His corrections are given below. Correction for volume : Vi in the ideal gas equation represents an ideal volume where the molecules can move freely. In real gases, a part of the total volume is, however, occupied by the molecules of the gas. Hence, the free volume Vi is the total volume V minus the volume occupied by the molecules. If b represents the effective volume occupied by the molecules of 1 mole of a gas, then for the amount n of the gas Vi is given by ...(1) Vi = V – nb Where b is called the excluded volume or co-volume. The numerical value of b is four times the actual volume occupied by the gas molecules. This can be shown as follows. If we consider only bimolecular collisions, then the volume occupied by the sphere of radius 2r represents the excluded volume per pair of molecules as shown in below Fig.

Correction for Forces of Attraction : Consider a molecule A in the bulk of a vessel as shown in Fig. This molecule is surrounded by other molecules in a symmetrical manner, with the result that this molecule on the whole experiences no net force of attraction. A B

(i) 2r excluded volume

Excluded volume per pair of molecules

N' =

Thus, excluded volume per pair of molecules

or

N' ∝

n V

Correction term ∝ n/V ...( 2a) (ii) The number of molecules striking the side of the vessel per unit time Larger this number, larger will be the decrease in the rate of change of momentum. Consequently, the correction term also has a larger value,. Now, the number of molecules striking the side of vessel in a unit time also depends upon the number of molecules present in unit volume of the container, and hence in the present case:

Excluded volume per molecule 1   4 3  4 3 8 πr  = 4  πr  2 3  3 

= 4 (volume occupied by a molecule) Since b represents excluded volume per mole of the gas, it is obvious that

Correction term ∝ n / V ...(2b) Taking both these factors together, we have

 4  b = N A 4 πr 3    3

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nN A V

Thus, the correction term is given as :

4 4  = π(2r)3 = 8  πr 3  3 3 

=

Arrangement of molecules within and near the surface of a vessel Now, consider a molecule B near the side of the vessel, which is about to strike one of its sides, thus contributing towards the total pressure of the gas. There are molecules only on one side of the vessel, i.e. towards its centre, with the result that this molecule experiences a net force of attraction towards the centre of the vessel. This results in decreasing the velocity of the molecule, and hence its momentum. Thus, the molecule does not contribute as much force as it would have, had there been no force of attraction. Thus, the pressure of a real gas would be smaller than the corresponding pressure of an ideal gas, i.e. pi = p + correction term ...(2) This correction term depends upon two factors: The number of molecules per unit volume of the vessel Large this number, larger will be the net force of attraction with which the molecule B is dragged behind. This results in a greater decrease in the velocity of the molecule B and hence a greater decrease in the rate of change of momentum. Consequently, the correction term also has a large value. If n is the amount of the gas present in the volume V of the container, the number of molecules per unit volume of the container is given as

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MAY 2010

n Correction term ∝   V

der Waals equation is not a complete solution of the behaviour of real gases. Applicability of the Van Der Waals Equation : Since the van der Waals equation is applicable to real gases, it is worth considering how far this equation can explain the experimental behaviours of real gases. The van der Waals equation for 1 mole of a gas is

n   V

n2

or Correction term = a

...( 3) V2 Where a is the proportionality constant and is a measure of the forces of attraction between the molecules. Thus pi = p + a

n2

 p + a  Vm2 

...(4)

V2 The unit of the term an2/V2 will be the same as that of the pressure. Thus, the SI unit of a will be Pa m6 mol–2. It may be conveniently expressed in kPa dm6 mol–2. When the expressions as given by Eqs (1) and (4) are substituted in the ideal gas equation piVi = nRT, we get 2    p + n a  (V – nb) = nRT  V 2  

 p + a  Vm2 

or

...(5)

6

kPa dm mol

H2 He N2 O2 Cl2 NO NO2 H2O CH4 C2H6 C3H8 C4H10(n) C4H10(iso) C5H12(n) CO CO2

21.764 3.457 140.842 137.802 657.903 135.776 535.401 553.639 228.285 556.173 877.880 1466.173 1304.053 1926.188 150.468 363.959

–2

3

dm mol –1

Z=1–

pVm +

a Vm RT

a =RT Vm

...(ii)

pb ...(iiii) RT Here Z is greater than 1 and increases linearly with pressure. This explains the nature of the graph in the high pressure region. A high temperature and low pressure If temperature is high, Vm will also be sufficiently large and thus the term a / Vm2 will be negligibly small. At this stage, b may also be negligible in comparison to Vm. Under these conditions, Eq. (i) reduces to an ideal gas equation of state: pVm = RT Hydrogen and helium The value of a is extremely small for these gases as they are difficult to liquefy. Thus, we have the equation of state as p(Vm – b) = RT, obtained from the van der Waals equation by ignoring the term a / Vm2 . Hence, Z is always greater than 1 and it increases with increase of p. The van dar Waals equation is a distinct improvement over the ideal gas law in that it gives qualitative reasons for the deviations from ideal behaviour. However, the generality of the equation is lost as it contains two constants, the values of which depend upon the nature of the gas.

p(Vm – b) = RT or

0.026 61 0.023 70 0.039 13 0.031 83 0.056 22 0.027 89 0.044 24 0.030 49 0.042 78 0.063 80 0.084 45 0.122 6 0.114 2 0.146 0 0.039 85 0.042 67

are characteristics of the gas. The values of these constants are determined by the critical constants of the gas. Actually, the so-called constant vary to some extent with temperature and this shows that the van

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  Vm = RT or  

From the above equation it is clear that in the low pressure region, Z is less than 1. On increasing the pressure in this region, the value of the term (a/VmRT) increase as V is inversely proportional to p. Consequently, Z decreases with increase of p. At high pressure When p is large , Vm will be small and one cannot ignore b in comparison to Vm. may be considered However, the term a / Vm2 negligible in comparison to p in Eq. (i) Thus,

b

a

..(i)

At low pressure When pressure is low, the volume is sufficiently large and b can be ignored in comparison to Vm in Eq. (i). Thus, we have

This equation is applicable to real gases and is known as the van der Waals equation. Values of van der Waals Constants : The constants a and b in van der Waals equation are called van der Waals constants and their values depend upon the nature of the gas. They Van Der Waals Constants Gas

  (Vm – b) = RT  

35

Z=1+

MAY 2010

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36

MAY 2010

KEY CONCEPT

GENERAL ORGANIC CHEMISTRY

Organic Chemistry Fundamentals Stability of different decreasing order : ⊕ > C

types

of

⊕ CH

carbocations

in

Θ

⊕ ⊕

⊕

⇒ Steric bulk example : CH3

⊕

(Ph)3 C > (Ph)2 CH > Ph – C H2 ≥ ⊕ ⊕ ⊕ CH2 = CH – CH2 ≥ R – C – R > R – CH – R

Θ

H 3C – C – O

⊕ ⊕ > R – CH2 > CH2 = CH

weaker nucleophile Stronger base

A special stability is associated with cycloproyl methyl cations and this stability increases with every additional cyclopropyl group. This is undoubtedly because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p-orbital of the cation carbon. H H H H H Cyclopropyl methyl cation orbital representation conjugation with the p-like orbital of the ring

Nucleophilicity versus basicity : If the nucleophilic atoms are from the same period of the periodic table, strength as a nucleophile parallels strength as a base. For example : H2O < NH3

CH3OH ≈ H2O < CH3CO2Θ < CH3OΘ ≈ OHΘ Increasing base strength Increasing nucleophile strength Nucleophile strength increases down a column of the periodic table (in solvents that can have hydrogen bond, such as water and alcohols). For example : Θ


Br– > Cl– > F– The order of basicity is opposite : F– > Cl– > Br– > I–, the reason that alkyl fluorides are ineffective substrates in SN2 reactions is related, to the relatively low acidity of HF (pKa = 3). Sulfonic acids, R SO2OH are similar to sulfuric acid in acidity and the sulfonate ion RSO3– is a very good leaving group. Alky benzenesulfonates, alkyl p-toluenesulfonates are therefore, very good substrates in SN2 reactions. The triflate ion (CF3SO3–) is one of the best leaving groups known, it is the anion of CF3SO3H which is a strong acid much stronger than sulfuric acid.

C

⇒

decreases

CH3

R

H

Θ

increasing nucleophilic strength decreasing base strength

>

>

Θ

Θ

F < Cl < Br < I

Θ

RO < RS

R3N < R3P

XtraEdge for IIT-JEE

37

MAY 2010

The substitution of tritium for hydrogen gives isotope effects which are numerically larger (kH/kT = 16). E2 elimination like SN2 process takes place in one step (without the formation of any intermediates). As the attacking base begins to abstract a proton from a carbon next to the leaving group, the C – H bond begins to break, a new carbon-carbon double bond begins to form and leaving group begins to depart. In confirmation with this mechanism, the base induced elimination of HBr from (I) proceeds 7.11 times faster than the elimination of DBr from (II). Thus C–H or C – D bond is broken in the rate determining step. If it was not so there would not have been any rate difference. H

Kinetic Isotope Effects : The kinetic isotope effect is a change of rate that occurs upon isotopic substitution and is generally expressed as a ratio of the rate constants, k light/k heavy. A normal isotope effect is one where the ratio of k light to k heavy is greater than 1. In an inverse isotope effect, the ratio is less than 1. A primary isotope effect is one which results from the making or breaking of a bond to an isotopically substituted atom and this must occur in the rate determining step. A secondary isotope effect is attributable to isotopic substitution of an atom not involved in bond making or breaking in the rate determining step. Thus when a hydrogen in a substrate is replaced by deuterium, there is often a change in the rate. Such changes are known as deuterium, isotope effects and are expressed by the ratio kH/kD, the typical value for this ratio is 7. The ground state vibrational energy (the zero-point vibrational energy) of a bond depends on the mass of the atoms and is lower when the reduced mass is higher. Consequently, D – C, D – O, D – N bonds, etc., have lower energies in the ground state than the corresponding H – C, H – O, H – N bonds, etc. Thus, complete dissociation of deuterium bond would require more energy than that for a corresponding hydrogen bond in the same environment. In case a H – C, H –O, or H – N bond is not broken at all in a reaction or is broken in a non-rate-determining step, substitution of deuterium for hydrogen generally does not lead to a change in the rate, however, if the bond is broken in the rate-determining step, the rate must be lowered by the substitution. This helps in determination of mechanism. In the bromination of acetone, the rate determining step is the tautomerization of acetone which involves cleavage of a C–H bond. In case this mechanistic assignment is correct, one should observe a substantial isotop effect on the bromination of deuterated acetone. Indeed kH/kD was found to be around 7.

– C – CH2Br H

Faster reaction

D – C – CH2Br D

Base

– CD = CH2

Slower reaction

1-Bromo-2,2-dideuterio-2-phenylethane (II)

No deuterium isotope effect is found in E1 reactions since the rupture of C – H (or C – D) bond occurs after the rate determing step, rather than during it. Thus no rate difference can be measured between a deuterated and a non deuterated substrate. Mechanism Review : Substitution versus Elimination SN2 Primary substrate Back-side attack of Nu : with respect to LG Strong/polarizable unhindered nucleophile Bimolecular in ratedetermining step Concerted bond forming/bond breaking Inverse of stereochemistry Favored by polar aprotic solvent.

Bromoacetone rate-determining step OH CH3C = CH2

Several mechanisms get support from kinetic isotope effect. Some of these are, oxidation of alcohols with chromic acid and electrophilic aromatic substitution. An example of a secondary isotope effect, where it is sure that the C – H bond does not break at all in the reaction. Secondary isotope effects for kH/kD are generally between 0.6 and 2.0.

SN2 and E2 Secondary or primary substrate Strong unhindered base/nucleophile leads to SN2 Strong hindered base/nucleophile leads to E2 Low temperatrue (SN2)/high temperature (E2)

(CZ3)2CHBr + H2O → (CZ3)2CHOH + HBr the solvolysis of isopropyl bromide where Z = H or D, kH/kD is 1.34 Secondary isotope effect.

XtraEdge for IIT-JEE

– CH = CH2

1-Bromo-2-phenylethane (I)

CH3COCH3 + Br2 → CH3COCH2Br

CH3COCH3

Base

38

SN1 and E1 Tertiary substrate Carbocation intermediate Weak nucleophile/base (e.g., solvent) Unimolecular in ratedetermining step Racemization if SN1 Removal of β-hydrogen if E1 Protic solvent assists ionization of LG Low temperature (SN1)/high temperature (E2) E2 Tertiary or secondary substrate Concerted anti-coplanar TS Bimolecular in ratedetermining step Strong hindered base High temperature

MAY 2010

UNDERSTANDING

Physical Chemistry

1.

What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10–10 M2 K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103 M–1 and K2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3] = 7.14 × 103 M–1 Sol. If x be the concentration of AgCl in the solution, then [Cl–] = x From the Ksp for AgCl, we derive K sp 1.7 × 10 −10 M 2 = [Ag+] = x [Cl − ] If we assume that the majority of the dissolved Ag+ goes into solution as Ag(NH3)2+ then [Ag(NH3)2+] = x Since two molecules of NH3 are required for every Ag(NH3)2+ ion formed, we have [NH3] = 0.20 M – 2x Therefore,  1.7 × 10 −10 M 2  (0.20M − 2x ) 2    x [Ag + ][ NH 3 ]2   Kinst = = x [Ag( NH 3 ) +2 ]

k (0.0126 Ω −1cm −1 ) = = 2.01 cm–1 G (0.00627 Ω −1 ) Conductivity of 0.1 M HAc solution K 2.01 cm −1 k= = R 520 Ω Molar conductivity of 0.1 M HAc solution k (2.01 / 520)Ω −1 cm −1 Λm(HAc) = = c (0.1 mol dm −3 )

K=

= 0.038 65 Ω–1 cm–1 dm3 mol–1 = 38.65 Ω–1 cm2 mol–1 According to Kohlrausch law, Λ∞(HAc) is given by Λ∞m (HAc) = Λ∞m (HCl) + Λ∞m (NaAc) – Λ∞m (NaCl) = (420 + 91 – 126) Ω–1 cm2 mol–1 = 385 Ω–1 cm2 mol–1 Therefore, the degree of dissociation of acetic acid is given as Λ (38.65 Ω −1 cm 2 mol −1 ) α = ∞m = ≈ 0.1 Λm (385 Ω −1 cm 2 mol −1 ) and the hydrogen-ion concentration of 0.1 M HAc solution is [H+] = cα = (0.1 M)(0.1) = 0.01 M Thus, its pH is pH = – log{[H+]/M} = – log(0.01) = 2

= 6.0 × 10–8 M2 From which we derive (0.20M − 2 x ) 2 6.0 × 10 −8 M 2 = = 3.5 × 102 x2 1.7 × 10 −10 M 2 which gives x = [Ag(NH3)2+] = 9.6 × 10–3 M, which is the solubility of AgCl in 0.20 M NH3 2.

The values of Λ∞ for HCl, NaCl and NaAc (sodium acetate) are 420, 126 and 91 Ω–1 cm2 mol–1, respectively. The resistance of a conductivity cell is 520 Ω when filled with 0.1 M acetic acid and drops to 122 Ω when enough NaCl is added to make the solution 0.1 M in NaCl as well. Calculate the cell constant and hydrogen-ion concentration of the solution. Given : Λ∞m (HCl) = 420 Ω–1 cm2 mol–1,

Potassium alum is KA1(SO4)2.12H2O. As a strong electrolyte, it is considered to be 100% dissociated into K+, Al3+, and SO42–. The solution is acidic because of the hydrolysis of Al3+, but not so acidic as might be expected, because the SO42– can sponge up some of the H3O+ by forming HSO4–. Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm3 of solution, calculate its [H3O+] : (a) Considering the hydrolysis Al3+ + 2H2O Al(OH)2+ + H3O+ –5 with Kh = 1.4 × 10 M (b) Allowing also for the equilibrium H3O+ + SO42– HSO4– + H2O –2 with K2 = 1.26 × 10 M 11.4 g Sol. (a) Amount of alum = = 0.024 mol 474.38 g mol −1 3.

Λ∞m (NaCl) = 126 Ω–1 cm2 mol–1,

and Λm(NaAc) = 91 Ω–1 cm2 mol–1 Sol. Resistance of 0.1 M HAc = 520 Ω Resistance of 0.1 M HAc + 0.1 M NaCl = 122 Ω Conductance due to 0.1 M NaCl, 1 1 – = 0.00627 Ω–1 G= 122 Ω 520 Ω Conductivity of 0.1 M NaCl solution k = Λmc = (126 Ω–1 cm2 mol–1)(0.1 mol dm–3) = 12.6 Ω–1cm2 dm–3 = 12.6 Ω–1 cm2(10 cm)–3 = 0.0126 Ω–1 cm–1 Cell constant,

XtraEdge for IIT-JEE

Molarity of the prepared solution = Hydrolysis of Al3+ is 39

0.024 mol

0.1 dm 3 = 0.24 M

MAY 2010

Al3+ + 2H2O

Al(OH)2+ + H3O+ 2+

Kh =

Thus, (177 K )(8.314 MPa cm 3 K –1mol −1 ) (8)(6.485 MPa ) 3 = 28.36 cm mol–1

+

[Al(OH) ][H 3O ]

b(NO) =

[Al3+ ] If x is the concentration of Al3+ that has hydrolyzed, we have ( x )( x ) = 1.4 × 10–5 M Kh = 0.24 M − x Solving for x, we get [H3O+] = x = 1.82 × 10–3 M (b) We will have to consider the following equilibria. Al3+ + 2H2O Al(OH)2+ + H3O+ 2– + H3O + SO4 HSO4– + H2O Let z be the concentration of SO42– that combines with H3O+ and y be the net concentration of H3O+ that is present in the solution. Since the concentration z of SO42– combines with the concentration z of H3O+, it is obvious that the net concentration of H3O+ produced in the hydrolysis reaction of Al3+ is (y + z). Thus, the concentration (y + z) of Al3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are Al3+ + 2H2O Al(OH) 2+ + H 3O + 0.24 M − y − z

H 3O y

+

+

y+ z

SO 24 − 0.48 M − z

Thus, Kh =

HSO 4− z

and 550 K )(8.314 MPa cm 3 K −1mol −1 ) (8)(4.56 MPa ) 3 = 125.35 cm mol–1 Hence b(NO) < b(CCl4) (ii) Since a = 27pcb2 therefore a(NO) = (27) (6.485 MPa) (28.36 cm3 mol–1)2 = 140827 MPa cm6mol–2 ≡ 140.827 kPa dm6 mol–2 a(CCl4) = (27) (4.56 MPa) (125.35 cm3 mol–1)2 = 1934538 MPa cm6 mol–2 ≡ 1934.538 KPa dm6mol–2 Hence a(NO) < a(CCl4) (iii) Since Vc = 3b therefore, Vc(NO) = 3 × (28.36 cm3 mol–1) = 85.08 cm3 mol–1 Vc(CCl4) = 3 × (125.35 cm3 mol–1) = 376.05 cm3 mol–1 Hence Vc(NO) < Vc(CCl4) (iv) NO is more ideal in behaviour at 300 K and 1.013 MPa, because its critical temperature is less than 300 K, whereas for CCl4 the corresponding critical temperature is greater than 300 K.

b(CCl4) =

y

+ H2O

( y + z)( y) = 1.4 × 10–5 M (0.24M − y − z)

...(i)

5. At 298 K, the emf of the cell -3 -3 Hg 0.01 mol dm KCl 1 mol dm–3 KNO3 0.01 mol dm KOH Hg

1 z = ...(ii) y(0.48M − z) 1.26 × 10 − 2 M From Eq. (ii), we get (0.48 M ) y z= (1.26 × 10 − 2 M ) + y Substituting this in Eq. (i), we get   (0.48M ) y y+ y −2  (1.26 × 10 M ) + y   = 1.4 × 10–5   (0.48M ) y  0.24 − y −   ( 1 . 26 × 10 − 2 M ) + y   Making an assumption that y 1 such that |f '(x)| < | f (x) | for all x ∈ (α, ∞ ) (D) there exists β > 0 such that |f '(x)| + | f '(x) | ≤ β for all x ∈ (0, ∞ ) Sol.[B,C]

2 105 71 3π − (D) 15 2

22 (A) −π 7

1

∫ 0

x 4 (1 − x ) 4

0

Let f be a real valued function defined on the interval (0, ∞ ) by f(x) = ln x +

2 r 1

39.

41.

z

y2 = 1 intersect at the points A and B 4 Equation of a common tangent with positive slope to the circle as well as to the hyperbola is -

(A) 2x –

z2

x2 – 9

5 y – 20 = 0

(B) 2x – 5 y + 4 = 0 (C) 3x – 4y + 8 = 0 (D) 4x – 3y + 4 = 0

So option A, C, D are true. Ans.[B]

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73

MAY 2010

y = m(x – 4) ± 4 1 + m 2

Sol.

y = mx ±

Sol.

|A | = a2 – bc if A is symmetric b = c then | A | = (a + b) (a – b) So a & b can attain 2(p – 1) solution It A is skew symmetric then a = b = c = 0 So total no. of solution = 2p – 2 + 1 = 2p – 1

45.

The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is [Note : The trace of a matrix is the sum of its diagonal entries.] (B) p 3 – (p – 1)2 (A) (p – 1) (p2 – p + 1) 2 (C) (p –1) (D) (p –1) (p2 – 2)

9m 2 − 4

– 4m ± 4 1 + m 2 = ±

9m 2 − 4

16m2 + 16 + 16m2 m 32 m 1 + m 2 = 9m2 – 4 m 32m 1 + m 2 = – 23m2 – 20 1024m2 + 1024 m4 = 529m4 + 400 + 920 m2 495 m4 + 104 m2 – 400 = 0 (5m2 – 4) (99m2 + 100) = 0 2 4 ∴ m2 = ∴m=± 5 5

Ans.[C]

So tangent with positive slope 2 4 y= x± 5 5 2x –

46.

5y±4=0

Ans.

Equation of the circle with AB as its diameter is (A) x2 + y2 – 12 x + 24 = 0 (B) x2 + y2 + 12 x + 24 = 0 (C) x2 + y2 + 24 x – 12 = 0 (D) x2 + y2 – 24x – 12 = 0

43.

SECTION – IV Integer type This section contains TEN paragraphs. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

Ans.[A] x2 + y2 – 8x = 0 Sol. 4x2 – 9y2 = 36

47.

 4x 2 − 36   – 8x = 0 x +   9   2 13x – 72x – 36 = 0 (x – 6) (13x + 6) = 0 −6 x = 6, 13 2

Sol.

(x – 6)2 + (y – 12 ) (y + 12 ) = 0 x2 + y2 – 12x + 24 = 0

dy ( X – x) dx dy y–x = x3 dx dy – y = – x3 x dx

Y–y=

T1 y – = – x2 T2 x

Paragraph for Question No. 44 to 46

Let p be an odd prime number and Tp be the following set of 2 × 2 matrices :   a b  Tp = A =  : a , b, c ∈ {0,1,2,......, p − 1}    c a 

I.F. = y = x

1 x

∫ − xdx + C

y − x2 = +C x 2 1 1= − +C 2

The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is (B) 2(p – 1) (A) (p – 1)2 2 (D) 2p –1 (C) (p –1) + 1

∴y=

Ans.[D]

XtraEdge for IIT-JEE

Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to –

Ans.[9]

x = 6, y = ± 12 ∴ Equation of required circle

44.

The number of A in Tp such that det (A) is not divisible by p is (B) p3 – 5p (A) 2p2 3 (C) p –3p (D) p3 – p2 [D]

74

∴C=

3 2

x (− x 2 + 3) 2

MAY 2010

48.

51.

The number of values of θ in the interval nπ  π π for n = 0 , ± 1, ± 2  − ,  such that θ ≠ 5 2 2  

Ans.[2] Sol. 1 = 4a2 – b2 ... (1) 2a =1 e e a= ... (2) 2 ... (3) also b2 = a2 (e2 – 1) (1) & (3) 1 = 4a2 – a2e2 + a2 ⇒ 1 = 5a2 – a2e2 5 e2 e4 – ⇒1= 4 4 ⇒ e4 – 5e2 + 4 = 0 ⇒ (e2 – 4) (e2 – 1) = 0 ∴ e=2

and tan θ = cot 5 θ as well as sin 2 θ = cos 4 θ is – Ans.[3] Sol.

π 12 π 5π , ± 4 12

tan θ = cot 5θ ⇒ θ = (2n + 1) So θ = ±

π , ± 12

∴ Sin 2θ = cos 4θ ⇒ sin 2θ =

1 2

or – 1

π 5π –π , , satisfies the given 12 12 4 conditions So total number of solution = 3

only θ =

49.

The

maximum 1

value

of

sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ

f (θ)

Sol.[2]

= =

the

52.

expression

is

6 , then | d | is –

1 2

ˆi

2

sin θ + 3 sin θ cos θ + 5 cos θ 1

r r If a and b r ˆi − 2ˆj and a= 5 r r r r 2a + b . a × b

(

Sol.

Plane is normal to vector ˆi − 2ˆj + kˆ 1(X – 1) – 2 (Y –2) + 1(Z – 3) = 0 X – 2Y + Z = 0 |d| 6= ⇒|d|=6 6 53.

)]

For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by if [ x ] is odd  x − [x] f (x) =  1 + [ x ] − x if [ x ] is even Then the value of

π 2 10 f ( x ) cos πx dx is 10 −10

∫

Ans.[4]

| a |=| b |= 1 & a ⋅ b = 0

Sol.

(2 a + b ) ⋅ [(a × b ) × (a – 2 b )]

= ( 2 a + b ) ⋅ [ b + 2a ] = | b | 2 + 4 | a | 2 = 5

XtraEdge for IIT-JEE

2 3 4

= ˆi (−1) − ˆj(−2) + kˆ (−1)

Ans.[5] Sol.

ˆj kˆ

3 4 5

are vector is space given by 2ˆi + ˆj + 3kˆ , then the value of bˆ = 14 r r × a − 2b is

) [( ) (

If the distance between the plane Ax – 2y + z = d and the plane containing the lines x −1 y − 2 z − 3 x −2 y−3 z−4 = = = = and is 2 3 4 3 4 5

Ans.[6]

1 + 4 cos 2 θ + 3 sin θ cos θ 1 = 3 1 + 2(1 + cos 2θ) + sin 2θ 2 1 = 3 3 + 2 cos 2θ + sin 2θ 2 1 So f(θ)max = = 2 9 3– 4+ 4

50.

The line 2x + y = 1 is tangent to the hyperbola x 2 y2 =1. If this line passes through the point − a 2 b2 of intersection of the nearest directrix and the xaxis, then the eccentricity of the hyperbola is

x (− x 2 + 3) 2 −3(−9 + 3) =9 f(–3) = 2

∴ f(x) =

75

V[ x ] is odd  {x} f(x) =  1 = {x} V[ x ] is even graph of y = f(x) is MAY 2010

56. 0 1

–5 –4 –3 –2 –1

2 3

Q f(x) & cos πx both are even functions 2

π 10

So, I =

=

x sin 3θ =

10

∫ f (x) cos πx dx

∫ f (x) cos(π x)dx

Ans.[3] Sol. (xyz) sin 3θ + y (– cos 3θ) + z (– cos 3θ) = 0 (xyz) sin 3θ + y (–2 sin 3θ) + z (–2 cos 3θ) = 0 (xyz) sin 3θ + y (– cos 3θ − sin 3θ ) + z (– 2cos 3θ) = 0 For y0z0 ≠ 0 ⇒ Nontrivial solution

0

∴ f(x) & cos πx both are periodic then 2

I=π

2

∫ f (x) cos(πx) dx 0

2  1 = π 2  (1 − x ) cos x (πx )dx + ( x − 1) cos(πx )dx   0 1   2 2 + 2 = π  2  =4  π 

∫

54.

∫

ω

ω2

ω

z + ω2

1

1

z+ω

ω

2

(

= 0 is equal to =

)

100 2 100 2 + ∑ k − 3k + 1 Sk is – 100! k =1

∑

⇒θ=

π 2π , 3 3

⇒ cos 3θ = 0

⇒θ=

π π 5π 7π , , , 6 2 6 6

These two donot satisfy system of equations

π 5π 9π , , 4 4 4

π 5π 3π , , = 3 12 12 4

No. of solutions = 3

|(k2 – 3k + 1)Sk|

PHYSICS

K =1 100

=1+1+

∑

K =3

=2+

1

⇒ sin 3θ = 0

⇒θ=

K Sk = K 100

cos 3θ

⇒ sin 3θ = cos 3θ ⇒ 3θ =

Ans.[4] Sol.

− cos 3θ − 2 cos 3θ = 0

sin3θ cos3θ [(4sin 3θ – 2 cos 3θ – 2sin 3θ) – (2cos 3θ – cos 3θ – sin 3θ) + 2 cos 3θ – 2 sin 3θ] =0 ⇒ (sin3θ cos3θ) [2 sin3θ – 2cos 3θ – cos 3θ + sin 3θ + 2cos 3θ – 2sin 3θ] = 0 ⇒ (sin3θ cos3θ) (sin3θ – cos 3θ) = 0

Let Sk, k = 1, 2, ……, 100, denote the sum of the infinite geometric series whose first term is k k −1 1 and the common ratio is . Then the value k! k of

− 2 sin 3θ

sin 3θ cos3θ 1 2 sin 3θ 2 =0 1 cos 3θ + sin 3θ 2

Ans.[1] Sol. On solving the determinant It become z3 = 0 So no. of solutions = 1 55.

− cos 3θ

sin 3θ

1

2π 2π + i sin . 3 3 Then the number of distinct complex number z z +1

sin 3θ

sin 3θ − cos 3θ − sin 3θ − 2 cos 3θ

Let ω be the complex number cos

satisfying

2 cos 3θ 2 sin 3θ + y z

(xyz) sin3θ = (y + 2z) cos 3θ + y sin 3θ) have a solution (x0, y0, z0) with y0z0 ≠ 0, is

−10

10

π2 5

The number of all possible values of θ, where 0 < θ < π, for which the system of equations (y + z) cos 3 θ = (xyz) sin 3θ

∑

(k 2 – 3k + 1) k –1

SECTION – I

Single Correct Choice Type

k –1 k – k–2 k –1

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

100 =4 =2+2– 99

XtraEdge for IIT-JEE

76

MAY 2010

57.

A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity isP

Ans. Sol.

[A] In the given range block is in equilibrium so P – mg sin θ + f = 0 f = mg sinθ – P Equation of straight line with negative slope.

59.

A real gas behaves like an ideal gas if its (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high [D] Reason : PV = nRT holds true in case of low pressure and high temperature conditions.

4R 3R

4R

Ans. Sol.

2GM 2GM (B) − (4 2 − 5) (4 2 − 5) 7R 7R GM 2GM (C) (D) ( 2 − 1) 5R 4R [A] ∆Wext = U2 – U1 for unit +ve mass U1 = V1 and U2 = V2 = 0 Gdm V1 = dV = − 2 (r + 16R 2 )1 / 2

(A)

Ans. Sol.

∫

60.

∫

t L

GM 2πrdr = − 7πR 2 (r 2 + 16R 2 )1 / 2 Put r2 + 16R2 = t2

∫

V1 = −

2GM 7R 2

4 2R

∫ dt = −

5R

2GM (4 2 − 5) 7R

Ans. Sol.

2GM (4 2 − 5) 7R A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is µ and tan θ > µ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sinθ – µ cosθ) to P2 = mg(sinθ + µ cosθ), the frictional force f versus P graph will look like –

∆Wext = U2 – U1 =

58.

61.

P θ f P2 P1

P

(B)

f

(C)

P1

P2

P

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Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is1 1 1 = + (B) R100 = R40 + R60 (A) R 100 R 40 R 60

P2 P

(D)

Ans.

[D]

Sol.

Rated power = R∝

f P1

(A) directly proportional to L (B) directly proportional to t (C) independent of L (D) independent of t [C] ρL ρL R= = Lt A ρ R= t R is independent of L

(C) R100 > R60 > R40

f

(A)

Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is-

∴ P1

P2

(D)

1 1 1 > + R 100 R 60 R 40

V2 R

1 Rated power

P1 > P2 > P3

1 1 1 > > R1 R 2 R 3

1 1 1 > > R 100 R 60 R 40

P

77

MAY 2010

62.

To verify Ohm's law, student is provided with a test resistor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct to carry out the experiment is-

63.

G1

R2

G2

RT

R1

(A)

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current i in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into plane of the paper, the wire takes the shape of a circle. The tension in the wire isx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

V

R1

(C)

G2

RT

R2

Ans. Sol.

(B)

IBL 2π

Fmagnetic = iB×Rdθ T cos

dθ 2

T cos

T G2

RT

(C)

2T sin

i

iB × R dθ = 2T sin

V

2 πR = L L R= 2π

G2

RT

64.

(D)

R1 V

[C] R2

Converted ammeter = G

G

Ans.

dθ 2

∴T=

iBL 2π

An AC voltage source of variable angular frequency ω and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased (A) the bulb glows dimmer (B) the bulb glows brighter (C) total impedance of the circuit is unchanged (D) total impedance of the circuit increases [B] C R

R1

Sol.

Voltmeter should be connected in parallel to RT and Ammeter should be connected in series with RT.

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T

T = iBR

R2

Converted ammeter =

dθ 2

T = tension

R2

G1

dθ 2

dθ

R1

G1

Sol.

(B)

[C]

V

Ans.

IBL π IBL (D) 4π

(A) IBL

G1

78

V0, ω ~

MAY 2010

Z=

 1  R2 +   ωC 

2

Q

ω increased z decreased ∴ current in circuit increase ∴ Bulb glow brighter.

SECTION – II Multiple Correct Choice Type This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 65.

Ans. Sol.

Ans. Sol.

A student uses a simple pendulum 1 m length to determine g, the acceleration due to gravity. He uses a stop watch the least count of 1 sec for this records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ? (A) Error ∆T in measuring T, the time period, is 0.05 seconds (B) Error ∆T in measuring T, the time period, is 1 second (C) Percentage error in the determination of g is 5% (D) Percentage error in the determination of g is 2.5 % [A,C] Total time( t ) Time period (T) = no. of oscillations So

67.

Ans. Sol.

1 × 2 = 0.05 sec ∆T = 40

T = 2π

l ; g

T2 =

4π 2 l ; g

∆g ∆T ∆l = + 2 ; g l T

g =

4π 2 l ; T2

68.

A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

XtraEdge for IIT-JEE

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, As shown in figure. Its pressure at A is P0. Choose the correct option(s) from the following V B 4V C

A

T T (A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0 ln 4 (C) Pressure at C is P0/4 T (D) Temperature at C is 0 4 [A,B] From figure AB → isothermal process So TA = TB ⇒ Internal energies will be same. V  WAB = nRT0 ln  2  = P0V0 ln 4  V1 

It is not given that line BC passes through origin. So we can't find pressure or temperature at point C.

∆l =0 l

∆T = 0.05 sec ; T = 2 sec. ∆l 2 × 0.05 Putting we get = = 0.05 l 2 ∆g × 100 = 5 % g 66.

(A) |Q1| > |Q2| (B) |Q1| < |Q2| (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero [A,D] Number of field lines emitting from Q1 is more than number of field lines reaching at Q2 So | Q 1 | > | Q2 | r and if so E at a point which is right to Q2 will be zero.

V

∆t ∆T 1sec = = t T 40 sec

Q

79

A Point mass 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are) correct for the system of these two masses ? (A) Total momentum of the system is 3 kg ms–1 (B) Momentum of 5 kg mass after collision is 4 kg ms–1 (C) Kinetic energy of the centre of mass is 0.75 J (D) Total kinetic energy of the system is 4 J MAY 2010

Ans. Sol.

B

[A,C] O 1kg

v1

5kg

2m/sec

1kg

5kg

60º 135º 30º = r

60º

v2

r1

After collision

Before collision

r2

90º A

D

60º 60º 30º Final Ray

A rat OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60º (see figure). If the refractive

Incident Ray

SECTION – III Paragraph Type

3 , which

This section contains 2 paragraphs.. Based upon the first paragraph 2 multiple choice question and based upon the second paragraph 3 multiple choice question have to be answered. Each of these questions has four choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

60º C

60º

45º

it hits at E By geometry angle r1 = 45º, r2 = 45º 1 1 and sin 45º = We know sinθC = 3 2 1 1 > 2 3 so 45º > θC So total internal reflection occurs. After reflection angle of incidence at AD will be 30º so ray comes out making an angle 60º with the normal at AD.

Momentum of 5 kg = 5 × v2 = 5 kg-m/sec 1× 3 + 5 × 0 = 0.5 m/sec vCM = 6 1 2 kCM = × (M1 + M 2 ) v CM = 0.75 joule 2 1 ksystem = × 1 × 9 = 4.5 joule. 2

index of the material of the prism is of the following is (are) correct ? B

E

60º

Collision is elastic so v2 + 2 = v1 ......(i) Conservation of momentum, 1 × v1 + 0 = – 2 × 1 + 5 × v2 ....(ii) Solving v1 = 3 m/sec v2 = 1 m/sec r Total momentum p system = 1 × v1 = 3 kg-m/sec

69.

C

135º

Paragraph for Question No. 70 to 71 90º

Electrical resistance of certain materials, knows as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC (0) if they are placed in a magnetic field, i.e. the critical temperature TC(B) is a function of the magnetic field strength B. The dependence of TC (B) on B is shown in the figure. TC(B)

75º

A

Ans.

D (A) The ray gets totally internally reflected at face CD (B) The ray comes out through face AD (C) The angle between the incident ray and the emergent ray is 90º (D) The angle between the incident ray and the emergent ray is 120º [A,B,C]

Sol.

Refraction at first surface AB :

sin 60º = sin r

TC(0)

3 1

r = 30º

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O

80

B

MAY 2010

70.

In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correct variation of R with T in these fields ?

Paragraph for Question No. 72 to 74 When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time m , as can be seen period is proportional to k easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = αx4 (α > 0) for | x | near the origin and becomes a constant equal to V0 for | x | ≥ X0 (see figure)

R

B2

(A)

B1 T

O R

B2

V(x)

B1

(B)

V0

T

O R

72. B1

(C)

B2 T

O R

Ans. Sol.

B1

73.

B2

(D) O

Ans. Sol.

x

X0

T

[A] B2 > B1 TC(B1) so TC(B2) < Dashed Solid line line Resistance ∝ Temperature above critical .

For periodic motion of small amplitude A, the time period T of this particle is proportional tom 1 m (B) (A) A α A α (C) A

A superconductor has TC(0) = 100 K. When a magnetic field of 7.5. Tesla is applied, its TC decreases to 75 K. For this material one can definitely say that when (A) B = 5 Tesla, TC (B) = 80 K (B) B = 5 Tesla, 75 K < TC (B) < 100 K (C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = Tesla, TC(B) = 70 K Ans. [B] Sol. TC(0) = 100 K, B = 0 B = 7.5 tesla TC (B) = 75 If B = 5 Tesla, TC(B) should be greater than 75 K

α m

1 A

α m

[B] V(x) = α x4 [V( x )] [ML2 T −2 ] = = [ML–2T–2] [α] = [ x ]4 [L4 ] Time period ∝ (Amplitude)x (α)y (Mass)z [T] = [L]x [ML–2T–2]y [M]2 1 1 Solving x = –1, y = − , z = 2 2 1 M T = A–1 α–1/2 M1/2 = A α

74.

The acceleration of this particle for | x | > X0 is (A) proportional to V0 V0 (B) proportional to mX 0 (C) proportional to

81

(D)

Ans Sol.

71.

XtraEdge for IIT-JEE

If the total energy of the particle is E, it will perform periodic motion only if(A) E < 0 (B) E > 0 (C) V0 > E > 0 (D) E > V0 [C] Energy Total should be less than maximum potential energy so E < V0 and E > 0.

V0 mX 0

(D) zero MAY 2010

Ans. Sol.

76.

[D] for |x| > x0 U = constant dU =0 F=– dx acceleration = zero.

Ans. Sol.

SECTION – IV Integer Type This section contains TEN questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

77. 75.

Ans.

Sol.

A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the car in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m/s. [7] v1 v2

Ans. Sol.

f0 * S

The focal length of thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its m image changes from m25 to m50. The ratio 25 m 50 is ? [6] f m= f +u 20 20 2 m25 = = –4 ; m50 = =– 20 − 25 3 20 − 50 m 25 =6 m 50 An α-particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are λα and λp λp respectively. The ratio , to the nearest integer λα is ? [3] After accelerating through V0 KE of a particle becomes = qV0 evolts so KEα = 200 eV KEp = 100 eV h λdebroglied = 2MKE

 v + v1  f 1 = f0    v 

λP = λα

 v  f ( v + v1 )  = 0 f2 = f1  ( v − v1 )  v − v1  f (v + v 2 ) f 2' = 0 (v − v 2 )

= 2 × 1.414 78.

f ' 2 −f 2 ( v + v 2 ) ( v + v1 ) = – = 0.012 f0 ( v − v 2 ) ( v − v1 )

Ans. Sol.

( v + v 2 )( v − v1 ) − ( v + v1 )( v − v 2 ) = 0.012 ( v − v 2 )( v − v1 ) 2

2

i1

2v( v 2 − v1 ) = 0.012 ( v − v1 )( v − v 2 )

=~ 3

R

E

1Ω

1Ω

1Ω

E

E

i1 =

i2 R

1Ω

2 ( v 2 − v1 ) = 0.012 v 18 (v2 – v1) = 0.006 × 330 × 5 = 7.128

4 × 200 = 2 2 1× 100

When two identical batteries of internal resistance 1Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 then the value of R is Ω is ? [4]

v + v( v 2 − v1 ) − v1v 2 − v + v1v 2 − v( v1 − v 2 ) = 0.012 ( v − v 2 )( v − v1 )

XtraEdge for IIT-JEE

M α KE α = M p KE p

E

2E R+2

Eeq = E 2

 2E  J1 =   .R R+2

82

req = 0.5Ω

MAY 2010

i2 =

E R + 0.5

Ares = Ares =

2

 E   R J2 =   R + 0.5 

From J1 = 2.25 J2 2

 2E    R = 2.25  R +2

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad/s. If the Young's modulus of the material of the wire is n × 109 Nm–2, the value of n is ?

Ans.

[4]

2

1.5 2 = R + 2 R + 0 .5

R=

2 =4Ω 0.5

Ans.

Two spherical bodies A(radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B ? [9]

Sol.

λA = 500 nm

79.

x

stress = Yx

a= … (i)

ω=

;

AY m

4.9 × 10 –7 × n × 10 9 0.1

140 = 70 n n=4

2

EA 1 = =   × (3)4 = 9 EB σ(4grB 2 )(T24 )  3 

82.

A binary star consists of two stars (mass 2.2 MS) and B(mass 11 MS), where MS is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is ?

Ans. Sol.

[6]

When two progressive waves y1 = 4sin(2x – 6t)

π  and y2 = 3 sin  2 x − 6 t −  are superimposed, 2  the amplitude of the resultant wave is ? [5] y1 = 4 sin (2x –6t) y2 = 3 sin (2x –6t –π/2)

rA

π φ= 2

XtraEdge for IIT-JEE

AYx m

140 =

From Stefan's law

Ans. Sol.

(l = 1 m)

F = Yx A F = Ayx Ayx = ma

rA 1 = 3 rB

80.

x l

stress =Y strain

λATA = λBTB

σ(4πrA 2 )(T14 )

strain =

Sol.

λB = 1500 nm T1 T = A =3 T2 TB

32 + 4 2 + 0 = 5

81.

 E    R  R + 0.5 

2R + 1 = 1.5 R + 3 0.5 R = 2

A12 + A 22 + 2A1A 2 cos φ

83

d

rB

MAY 2010

LA = mAω rA2 LB = mBωrB2 Ratio (K) =

Ans. Sol.

m r 2 LA + LB L = A + 1 = A A2 + 1 LB LB m B rB

rA m B 11 = = =5 rB m A 2.2

Ratio (K) =

83.

1 × 52 + 1 = 6 5

Gravitational acceleration on the surface of a

[8] The amount of heat required to raise the temp from –5°C to 0°C. Q1 = m × 2100 × 10–3 × 5 = 10.5 m Joule The amount of heat required to melt 1 gm ice = 10–3 × 3.36 × 105 = 336 J 420 = 336 + 10.5 m 10.5 = 84 m = 8 gm.00

ATTITUDE

6 g, where g is the gravitational 11 acceleration on the surface of the earth. The

planet is

2 times 3 that of the earth. If the escape speed on the surface of the earth is taken to be 11 km/s, the escape speed on the surface of the planet in km/s will be ? [3]

average mass density of the planet is

Ans. Sol.

υp υc

=

2g p R p 2g e R e

=

gp ge

×

Rp Re

•

Great effort springs naturally from a great attitude.

•

Like success, failure is many things to many people. With Positive Mental Attitude, failure is a learning experience, a rung on the ladder, a plateau at which to get your thoughts in order and prepare to try again.

•

Your attitude, not your determine your altitude.

•

Develop an attitude of gratitude, and give thanks for everything that happens to you, knowing that every step forward is a step toward achieving something bigger and better than your current situation.

•

You can adopt the attitude there is nothing you can do, or you can see the challenge as your call to action.

•

"An optimist is a person who sees a green light everywhere, while the pessimist sees only the red stoplight... The truly wise person is colorblind."

•

Positive thinking will let you do everything better than negative thinking will.

•

You cannot control what happens to you, but you can control your attitude toward what happens to you, and in that, you will be mastering change rather than allowing it to master you.

•

You can do it if you believe you can!

…

(1)

⇒

Mp 4 πR 3 p 3

GM p R 2p

=

3 M 2 Me 2 Rp ⇒ P = … (2) 3 4 πR 3 3 R 3e Me e 3

2 Mp 6 GM e 6 Rp = ⇒ = 11 R e 2 11 R e2 Mc

…(3)

from (2) and (3) 3 2 Rp 2 Rp 6 Rp 3 6 = ⇒ = 3 2 3 Re 11 R e Re 22

… (4)

from (1) and (4) υp υe

=

6 3 6 = × 11 22

18 3 = 242 11

vp = 3 km /sec. 84.

A piece of ice (heat capacity = 2100 Jkg–1ºC–1 and latent heat = 3.36 × 108 J kg–1) of mass m grams is at –5ºC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is ?

XtraEdge for IIT-JEE

84

aptitude,

will

MAY 2010

XtraEdge for IIT-JEE

85

MAY 2010

IIT-JEE 2010 PAPER-II (PAPER & SOLUTION) Time : 3 Hours

Total Marks : 237

Instructions : • The question paper consists of 3 Parts (Chemistry, Mathematics and Physics). And each part consists of four • • • •

Sections. For each question in Section I: you will be awarded 5 marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus two (–2) mark will be awarded. For each question in Section II: you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles is darkened. No negative marks will be awarded for incorrect answers in this section. For each question in Section III: you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. For each question in Section IV: you will be awarded 2 marks for each row in which you have darkened the bubbles(s) corresponding to the correct answer. Thus, each question in this section carries a maximums of 8 marks. There is no negative marks awarded for incorrect answer(s) in this section.

CHEMISTRY

were separately subjected to nitration using

SECTION – I Single Correct Choice Type

HNO3/H2SO4 mixture. The major product formed in each case respectively, is

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is (A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic Ans. [A] Sol. σ1s 2 σ*1s 2 σ 2s 2 σ* 2s 2 π 22 p x = π 2 p y 1 (6 – 4) = 1 2 Diamagnetic The compounds P, Q and S COOH HO

P

H3C

H3C

HO

NO2

NO2

O || C

O

O 2N

OCH3

COOH

B. Ο =

2.

OCH3

COOH

(A)

(B) HO

NO2

H 3C NO2

OCH3

O || C

Q O || C

O

NO2 O

S

XtraEdge for IIT-JEE

86

MAY 2010

OCH3

COOH

(C) H 3C

HO

NO2

Ans. Sol.

NO2 O || C

O

OCH3

(D) H3C

& 4R =

NO2

NO2

R= O || C

Ans.

L 2 2

2× π R2 L2

COOH

Sol.

=

NO2 OH OCH3

4. NO2

HNO3/H2SO4

(D) OSF2

[D]

CH3

Sol. O || C

S F

HNO3/H2SO4

O

5.

O

F

In the reaction O

O || C

3.

L2 4 × 2 × 100 = 78. 5% L2

The species having pyramidal shape is (B) BrF3 (A) SO3 (C) SiO 32 –

Ans. CH3

× 100

2× π

HNO3/H2SO4 OH OCH3

area of circle × 100 area of square

% packing efficiency (η) =

NO2

O

2 L

=

[C] COOH

(B) 68.02% (D) 78.54%

NO2

COOH HO

(A) 39.27 % (C) 74.05% [D] Area of square = L2

H3 C NO2

C NH2

O

(1) NaOH/Br2 O (2)

T

C Cl

the structure of the product T is O O C (A) H3C O– C

The packing efficiency of the two-dimensional square unit cell shown below is

(B)

NH C O

CH3

L

XtraEdge for IIT-JEE

87

MAY 2010

(C) H3C

NH C O O O C NH–C

(D) H3C

Ans. Sol.

CH3

C

NaOH/Br2

CH3 Hoffmann's NH2 degradation Cl O C

NH2 +

Ans. Sol.

  10.5 NA  Ag =    108

8.

–HCl

Ans. Sol. 9.

– – Cl– Cl Cl

8 = 2.82 BM

SECTION – II Integer Type

Ans. Sol.

Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is [7] m d= ⇒ 10.5 g/cc V 10.5 Number of atoms of Ag in 1cc ⇒ × NA 108

XtraEdge for IIT-JEE

× 10–8

Among the following, the number of elements showing only one non-zero oxidation state is O, Cl, F, N. P, Sn, Tl, Na, TI [2] F & Na only show one non zero oxidation state that are – 1 & + 1 respectively. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line paths is wd, then the integer closest to the ratio wd\ws is

Sol.

For solid line path show approxy isothermal process 5.5 ∴ work done |WS| = 2.303 (PV) log .5 = 2.303 × 4 × .5 × log 11 ~ – 4.79 for dashed line path work done wd = 4 × |2 – .5| + 1 × |3 – 2| + .5 × |5.5 –3| = 6 + 1 + 1.25 = 8.25 wd 8.25 ∴ = = 1.71 ~ – 2 4.8 ws

10.

The total number of diprotic acids among the following is H2SO4 H3PO3 H2CO3 H2S2O7 H3PO4 H3BO3 H3PO2 H2CrO4 H2SO3

This section contains a group of 5 questions. The answer to each questions is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 7.

2/3

4. a 4. 3. P 3.0 (atm)2. 2.0 1.5 1.0 0.5 b 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 V(lit) Ans. [2]

hyb = sp3 no. of unpaired electrons = 2 2(4) =

× 10–8

Thus, x = 7

The complex showing a spin-only magnetic moment of 2.82 B.M. is (A) Ni(CO)4 (B) [NiCl4]2– (C) Ni((PPh3)4 (D) [Ni(CN)4]2– [B] [NiCl4]–2 Ni+2 1s2 2s2 2p6 3s2 3d8 4s°

µ=

2/3

 10.5 × 6.02 × 10 23   =    108   = 1.5 × 107

NH2 O || NH–C

Cl–

10.5 × NA 108 2/3

CH3

6.

3

  10.5 In 1cm2, number of atoms of Ag =  × NA    108 –12 2 –8 2 In 10 m or 10 cm , number of atoms of

[C] O

CH3

In 1cm, number of atoms of Ag =

88

MAY 2010

Ans. Sol.

CH3

[6]

O || HO – S – OH || O

OH OH

O O || || H–O – S – O– S – O – H || || O O

OH

Cr O

O

CH

(B) H3C

O || C

O || HO – P – OH

O

OH

(C) H3 C

O || S OH OH

Ans. Sol. Cl

PPh3 Cl

(D) H3 C

and

H3 C

CH2

CH

H

C O

H

C

and

H

C

H

O

O

[B] The compound R is

Rh

O

PPh3

NH3 CO

CO Cl

H3 C

PPh3 Rh

C

(A) H3C

CO

NH3

CH2

SECTION – III Paragraph Type

H3 C

(B) H3C H3 C

C

Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below – H3C OH H3C

(C) H3C

C O

XtraEdge for IIT-JEE

C

H

OH O

CH

CH

C

H

CH2 OH

O

H and H3C

O

CH3

(D) H3C (A) H3C

OH

CH3

The compounds P and Q respectively are CH3 CH

H

CH

Paragraph for questions No. 12 to 14

O

C

O

This section contains 2 paragraphs. Based upon each of the paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

12.

H

O

CH3

Ans. 13.

H

C O

C

CH3

NH3

Rh

CH2

CH

Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is [3]

11.

H and H

C

C

CH

CH

C

H

H H3 C

O

Ans.

89

CH

OH

[A] MAY 2010

14.

The compound S is

CH3 CH3

CH

(A) H3C

H O+

O

CH3 CH

C

(iv) CH3 – C – CH2 3 → ∆ OH H – C – CN

H

CH3 – C — CH2 O

HC OH C

O Final product

OH

CH2 CN

Paragraph for Questions No. 15 to 17

O H3 C

(B)

C

C

H3 C

The hydrogen like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

H

CH2 CN

CH3 CH

(C) H3C

CH

CH

Ans.

The state S1 is – (A) 1s (C) 2p [B]

Sol.

Q One radial node

15.

CN

OH

CH2

∴n–l–1=1

OH

or n – l = 2

CN

(D)

H3 C H3 C

l=0 n=2 Orbital name = 2s

CH

C

OH

CH2

Ans.

Energy of the state S1 in units of the hydrogen atom ground state energy is – (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50 [C]

Sol.

S1 = Energy of e of H in ground state ×

16.

OH

Ans. [D] Sol. (12 to 14) CH3

CH3

(i) CH3 – CH

OH –

C

Θ H → CH3 – C

C

H

O

O

17.

(ii) CH3 CH3 – CΘ C

+ H

O

CH3

H C H O

CH3 CH3 – C – CH2 OH C H O [R]

Sol.

32

22 = 2.25 × energy of e of H in ground state

The orbital angular momentum quantum number of the state S2 is – (A) 0 (B) 1 (C) 2 (D) 3 [B] For S2 = n – l – 1 n–l=2

CH3

n = 3, l = 1 Orbital = 3p ∴ l = 1

C – CH2 (iii) CH3 – C – CH2 + HCN CH3 OH OH C H – C – CN H O OH [S]

XtraEdge for IIT-JEE

(B) 2s (D) 3s

[ S2 = energy of e of H in ground state ×

90

32 n2

, n = 3]

MAY 2010

SECTION – IV Matrix Type This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 18.

Ans.

[A → r,s,t; B → t; C → p,q; D → r]

19.

All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II. Column I (A) (CH3)2SiCl2 (B) XeF4 (C) Cl2 (D) VCl5

Ans.

Match the reactions in Column I with appropriate options in Column II.

SECTION – I

Single Correct Choice Type

NaOH / H O

2 → OH   

N2Cl +

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

0° C

N=N

OH

20.

OH OH

(B)

H3C – C – C – CH3

H SO

2  4 →

CH3 CH3

O C H3C

O

(C)

[A → p,s; B → p,q,r,t; C → p,q,t; D → p]

MATHEMATICS

Column I

(A)

Column II (p) Hydrogen halide formation (q) Redox reaction (r) Reacts with glass (s) Polymerization (t) O2 formation

C CH3

CH3 C

CH3 CH3

.LiAlH4 1  → + 2.H 3O

Ans. Sol.

OH CH CH3

(D) HS

Base → Cl 

S

Column II (p) Racemic mixture (q) Addition reaction (r) Substitution reaction (s) Coupling reaction (t) Carbocation intermediate

XtraEdge for IIT-JEE

91

A signal which can be green or red with 4 1 and respectively, is received by probability 5 5 station A and then transmitted to station B. The probability of each station receiving the signal 3 correctly is . If the signal received at station B 4 is green, then the probability that the original signal was green is 3 6 (B) (A) 5 7 20 9 (D) (C) 23 20 [C] Event (1) : original signal OG : Original signal is green OR : Original signal is red Event (2) : Signal received by A. AG : A received green AR : A received Red Event (3) : Signal received by B BG : B received green BR : B received Red  BG  P(OG ).P   OG   OG  P =  BG  P(OG ).P BG  + P(OR ).P BG       OG   OR  MAY 2010

4 3 3 1 1 . + . 20 5  4 4 4 4  = = 4  3 3 1 1  1  1 3 3 1  23 . + . + . + . 5  4 4 4 4  5  4 4 4 4 

21.

If the distance of the point P (1, –2, 1) from the plane x + 2y –2z = α, where α > 0, is 5, then the foot of the perpendicular from P to the plane is 8 4 7  4 4 1 (B)  ,− ,  (A)  , ,−  3 3 3    3 3 3

 1 2 10  2 1 5 (D)  , − ,  (C)  , ,  3 3 3  3 3 2 Ans. [A] |1− 4 − 2 − α | x −1 y + 2 z −1 Sol. = = =λ =5 3 1 2 −2 foot (1 + λ, –2 + 2λ, 1 –2λ) |α + 5| = 15 (1 + λ) + 2(–2 + 2λ) – 2 (1 –2λ) = 10 α = 10 (correct), –20 (wrong) 1 + λ – 4 + 4λ –2 + 4λ = 10 9λ = 15 , ⇒ λ = 5/3 8 4 7 foot =  , , −  3 3 3

22.

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to (A) 25 (B) 34 (C) 42 (D) 41

Ans.

[D]

Sol.

S = {1, 2, 3, 4} Possible subsets

24.

∫

2

2

=

4! =3 2 ! 2 ! 2!

Ans. Sol.

f(x) = 2ex + ex

x

∫

t 4 + 1 dt

0

; f(0) = 2 x

f ′(x) = 2ex + ex

∫

t 4 + 1 dt + ex 1+ x 4

0

Sol.

f ′(0) = 2 + 1 = 3

C

25.

α  π  − α  2

1 1 = f ′(0) 3

10

2 ˆi + 10 ˆj + 11 kˆ

Then

B

∑ A r (B10 Br − C10 A r ) is equal to r =1

8 40 AB . AD π  cos  − α  = = = 2  | AB | | AD | 3 (15) 9 8 ⇒ cos α = 9

(f–1)′ (2) =

For r = 0, 1, …, 10, let Ar, Br and Cr denote, respectively, the coefficient of xr in the expansions of (1 + x)10, (1 + x)20 and (1+ x)30.

(A) B10 – C10

XtraEdge for IIT-JEE

1

t 4 + 1 dt , for all x ∈ (–1, 1) , and let f–1 be the

x=0

D

sin α =

3

= 4C4 = 1 4! = =4 3! 1!

0

D′

A

0

inverse function of f. Then (f–1)′ (2) is equal to 1 (A) 1 (B) 3 1 1 (C) (D) 2 e [B]

4 5 (D) 9

– ˆi + 2 ˆj + 2 kˆ

4

Let f be a real-valued function defined on the interval (–1, 1) such that e–x f(x) = 2 + x

Two adjacent sides of a parallelogram ABCD are given by AB = 2 ˆi + 10 ˆj + 11 kˆ and AD = – ˆi +

1 (C) 9 [B]

No. of elements in Ways Set A Set B 0 0 =1 1 0 = 4C1 = 4 2 0 = 4C2 = 6 1 1 = 4C2 = 6 3 0 = 4C3 = 4 4 2 2 1 = C2. C1 = 12

Total ⇒ 1 + 4 + 6 + 6 + 4 + 12 + 1 + 4 + 3 = 41

2 ˆj + 2 kˆ The side AD is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD′. If AD′ makes a right angle with the side AB, then the cosine of the angle α is given by 17 8 (B) (A) 9 9

Ans.

23.

Ans. Sol.

(

2 (B) A10 B10 − C10 A10

)

(C) 0 (D) C10– B10 [D] Ar = 10Cr, Br = 20Cr , Cr = 30Cr 10

∑ 10 C r (20C10 20Cr – 30C10 10Cr)

17 9

r =1

92

MAY 2010

= 20C10

10

10

∑ 10 C r 20C20–r – 30C10 ∑ 10 C r .10 C r r =1

20

r =1

30

10

20

Ans. Sol.

[1] g(x) = ef(x) g′(x) = ef(x) f ′ (x) g′(x) = 0 ⇒ f ′ (x) = 0 ⇒ x = 2009, 2010, 2011, 2012 Points of local maxima = 2009, ⇒ only one point

28.

Let k be a positive real number and let

30

= C10 [ C20 – C0 C20] – C10 [20C10 – (10C0)2] = 20C1030C20 – 20C10 – 30C10 20C10 + 30C10 = 30C10 – 20C10 = C10 – B10

SECTION – II Integer type

 2k − 1  A=  2 k − 2 k 

This section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled. 26.

Let a1, a2, a3, …….., a11 be real numbers satisfying a1 = 15, 27 –2a2 > 0 and ak =2ak–1 – ak–2 for k = 3, 4, …., 11. 2 a12 + a 22 + ..... + a 11 = 90, then the value of 11 a 1 + a 2 + ..... + a 11 is equal to 11 [0] a + a k −2 Q ak–1 = k 2

If

Ans. Sol.

Ans. Sol.

2 k

2 k

1

− 2k

2k

−1

+ 2 k [4k k + 2 k ] det (A) = (2k –1) (4k –1) + 4 k (2k + 1) + 4k (2k +1) = (2k –1) (4k2 –1) + 8k (2k + 1) det (B) = 0 det (adj A) = (det A)2 = 106 det A = 103 3 2 2 3 8k + 1 –2k –4k + 16k + 8k = 10 8k3 + 12k2 + 6k – 999 = 0 k = 2 → 64 + 48 + 12 – 999 < 0 k = 3 → 8(27) + 109 + 18 – 999 < 0 k = 4 → 8(64) + 12 (16) + 24 – 999 512 + 192 + 24 – 999 < 0 k = 5 → 8(125) + 12 (25) + 6(5) – 999 > 0 so [k] = 4

10 × 11 10 × 11× 21 2 + d = 11 × 90 6 2 so on solving d = –3 a 1 + a 2 + ...... + a 11 11

11 1 . . (2 × a1 + (11 –1) (–3)) 2 11 1 = (30 – 30) = 0 2

=

29.

Let f be a function defined on R (the set of all real numbers) such that f ′(x) = 2010 (x –2009) (x –2010)2 (x –2011)3 (x –2012)4, for all x ∈ R. If g is a function defined on R with values in the interval (0, ∞) such that

Two parallel chords of a circle of radius 2 are at a distance

3 + 1 apart. If the chords subtend at the

center, angles of

f(x) = ln {g(x)}, for all x ∈ R, then the number of points in R at which g has a local maximum is

XtraEdge for IIT-JEE

2k − 1 det (A) = 2 k −2 k

2

11a2 + 2ad

27.

 0 2k − 1 k    B = 1 − 2k 0 2 k − k − 2 k 0    If det (adj A) + det(adj B) = 106, then [k] is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k] [4]

= (2k –1) [–1 + 4k2] –2 k [–2 k –4k k ]

2 a 2 + a 22 + ..... + a11 = 90 so 1 11 ⇒ Σ(a + (r –1) d)2 = 11 × 90 ⇒ Σ(a2 + 2ad (r –1) + (r –1)2d2) = 11 × 90

so

2 k  − 2k  and −1  

2 k 1 2k

Ans.

93

π 2π and , where k > 0, then k k

the value of [k] is [Note : [k] denotes the largest integer less than or equal to k] [3]

MAY 2010

Paragraph for Questions No. 31 to 33 Sol.

Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum of all distinct real roots of f(x) and let t = |s|.

π/k π/2k

31.

 1  (A)  − ,0   4 

π π + 2 cos k 2k 3 +1 1 3 +1 3π π π 3π = cos cos ⇒ cos cos = . 4 4k 4k 4k 4k 2 2 2 π π = 4k 12 4k = 12 k=3

d = 2 cos

30.

Ans.

Ans. Sol.

Q

32. 10 C

∆=

the equation have only one real root so

The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

3  (A)  , 3  4 

 21 11  (B)  ,   64 16 

(C) (9, 10)

 21  (D)  0,   64 

Ans.

[A]

Sol.

A(t) =

t

3 /2

1− t4 f ( x ) d ( x ) = t4 + t3 + t2 + t = t   1− t  0

∫

   

 175  A(1/2) = 15/16 & A (3/4) = 3    256 

100 + 36 − c 2 ⇒ C2 = 136 + 120° (1/2) 2.10.6 ⇒ C2 = 196 ⇒ C =14 s = 15 ∆ r= = 3 s 2 r =3

cos C =

3  So A(t) ∈  , 3  4  33.

The function f ′(x) is (A) increasing in

SECTION – III

1   − t , −  and decreasing 4 

 1  in  − , t   4 

Paragraph Type This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

XtraEdge for IIT-JEE

1 2

 − 3 −1  1 3 s∈  ,  and t ∈  ,   4 2  2 4

Sol.

1 ab sin C 2 1 15 3 = 6(10) sin C ⇒ sin C = 2 ⇒ C = 120°

 3 1  1 (D)  0,  (C)  − ,−  4 2    4 [C] f(x) = 4x3 + 3x2 + 2x + 1 f ′(x) = 12x2 + 6x + 2 is always positive

 − 3 −1  so root ∈  ,   4 2 

area of the triangle is 15 3 . If ∠ACB is obtuse and if r denotes the radius of the in-circle of the triangle, then r2 is equal to [3] A

6

3  (B)  − 11,−  4 

f(0) = 1, f(–1/2) = 1/4, f(–3/4) = –

Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the

B

The real number s lies in the interval

(B) decreasing in

1   − t , −  and increasing 4 

 1  in  − , t   4 

94

MAY 2010

Ans. Sol.

(C) increasing in (–t, t) (D) decreasing in (–t, t) [B] f ′′(x) = 6 (4x + 1)

36.

The equation of the locus of the point whose distances from the point P and the line AB are equal, is (A) 9x2 + y2 – 6xy –54 x – 62 y + 241 = 0 (B) x2 + 9y2 + 6xy –54x + 62 y –241 = 0 (C) 9x2 + 9y2 –6xy –54 x –62 y – 241 = 0 (D) x2 + y2 –2xy + 27x + 31y – 120 = 0

Ans. Sol.

[A] Equation of AB is x + 3y – 3 = 0 so required locus will be ( x + 3y – 3) 2 (x – 3)2 + (y –4)2 = 10 2 2 ⇒ 9x + y – 6xy –54x –62y + 241 = 0

Paragraph for questions No. 34 to 36

Tangents are drawn from the point P(3, 4) to the

34.

x 2 y2 + = 1, touching the ellipse at points ellipse 9 4 A and B. The coordinates of A and B are (A) (3, 0) and (0, 2)  8 2 161   and  − 9 , 8  (B)  − ,  5 15   5 5  

SECTION – IV Matrix Type

 8 2 161   and (0, 2) (C)  − ,  5 15   

Ans. Sol.

This section contains 2 questions. Each question has four statements (A, B, C and D)given in Column I and five statements (p, q, r, s and t) in column-II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column-II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

 9 8 (D) (3, 0) and  − ,   5 5 [D] Equation of tangent y = mx ± 9m 2 + 4 as it passes through (3, 4)

37.

9m 2 + 4

so 4 = 3m ±

1 and undefined. 2 So equation of the tangents will be x –2y + 5 = 0 and x = 3

m=

(A) The set of points

 −9 8 so point of contacts are (3, 0) and  ,   5 5

35.

Ans.

z satisfying

The orthocentre of the triangle PAB is  8 (A)  5,   7

 7 25  (B)  ,  5 8 

 11 8  (C)  ,   5 5 [C]

 8 7 (D)  ,   25 5 

|z – i|z|| = |z + i|z|| is contained in or e qual to

θ

A(3,0)

H

 −9 8 B ,   5 5

Equation of two altitudes PH and AQ are 3x – y – 5 = 0 and 2x + y – 6 = 0 respectively

(q) the set of points z satisfying Im z = 0 (r) the set of points z satisfying |Im z| ≤ 1

1 is contained w in or equal to z=w+

 11 8  so orthocentre will be  ,   5 5

XtraEdge for IIT-JEE

(p) an ellipse with 4 eccentricity 5

(B) The set of points z Satisfying |z + 4| + |z –4| = 10 is contained in or equal to (C) If |w| = 2, then the set (s) the set of points z of points satisfying |Re z| ≤ 2 1 z=w– is contained w in or equal to (D) If |w| =1, then the set (t) the set of points z of points satisfying |z| ≤ 3

P(3,4)

Sol.

Match the statements in Column-I with those in Column-II. [Note : Here z takes values in the complex plane and lm z and Re z denote, respectively, the imaginary part and the real part of z] COLUMN-I COLUMN-II

95

MAY 2010

Ans. Sol.

38.

A → q, r ; B → p ; C → p, s, t ; D → q, r, s, t

Sol. (A) Let P ≡ (λ + 2, 1 –2 λ, λ – 1)

(A) Let |Z| = r ∀ r ∈ R Z − ir = 1Which is the equation of line of Z + ir perpendicular bisector of y = r & y = – r that is y = 0 (B) |Z + 4| + |Z – 4| = 10 it will represent on ellipse having foci (–4, 0), (4, 0) x 2 y2 so its equation will be + =1 25 9 whose eccentricity is 4/5 (C) Let w = 2eiθ. 3 5 z= cos + i sin θ 2 2 (D) Let w = eiθ Z = eiθ + e–iθ = 2 cos θ. Match the statements in Column-I with the values in Column-II. COLUMN-I COLUMN-II (A) A line from the origin (p) –4 meets the lines x − 2 y −1 z +1 = = and 1 −2 1 8 x− 3 = y + 3 = z −1 2 −1 1 at P and Q respectively. If length PQ = d, then d2 is (B) The values of x satisfying (q) 0 tan–1 (x + 3) – tan–1 (x –3) 3 = sin–1   are, 5 r r (C) Non-zero vectors a , b (r) 4 r r r and c satis1fy a . b = 0, r r r r ( b – a ). ( b + c ) = 0 and r r r r 2 | b + c |= | b – a |. r r r If a = µ b + 4 c , then the possible values of µ are

2 ; – µ –3, µ + 1) 3 equation line PQ

Q ≡ (2µ +

→

r = (λ + 2) ˆi + (1 – 2λ) ˆj + (λ – 1)

+ α ((2µ – λ +

2 π

∴ This line passing through origin so. λ + 2 + α (2µ − λ +

on solving above three µ = So P ≡ (5, – 5, 2) & Q ≡ ( So PQ =

1 &λ=3 3

10 −10 4 , , ) 3 3 3

6 ⇒ (PQ)2 = 6

(B) tan–1(x + 3) – tan–1 (x –3) = sin–1

tan–1

6

= tan–1

2

3 5

3 4

x −8 ⇒ x2 – 8 = 8 ⇒ x2 = 16 ⇒ x = ± 4 r r r r r (C) | b |2 + b . c = a . c …. (1) r r r r r r r µ r put a = µ b + 4 c ∀ a . b = 0 ⇒ b . c = – | b |2 4 …(2) from (1) and (2) b2 16 = … (3) 2 c 4 − µ + µ2 r r r r r r r ∴ 2| b + c | = | b – a | and a = µ b + 4 c b2 c

2

=

12 3 − 2µ + µ 2

… (4)

from (3) and (4) m = 0,5 9x sin 2 = sin 5x + sin 4 x (D) f(x) = x sin x sin x sin 2

π

∫ f (x) dx is

−π

I=

A → t ; B → p, r ; C → q, s ; D → r

XtraEdge for IIT-JEE

2 )=0 3

1 – 2λ + α(2λ – µ – 4) = 0 λ – 1 + α(µ – λ + 2) = 0

(t) 6 Ans.

2 ˆ ) i + (2λ – µ – 4) ˆj 3

+ (µ + 2 – λ) kˆ

(D) Let f be the function on (s) 5 [–π, π] given by f(0) = 9 and f(x)  9x  x = sin   sin   for x ≠ 0  2  2 The value of

kˆ

96

2 π

π

∫ f (x ) dx

−π

MAY 2010

4 π

=

=

4 π

8 = π =

8 π

π/ 2

π

∫

f ( x ) dx

Total force =

π

sin 5x

∫ sin x

40.

0

π/ 2

∫ 0

π/ 2

∫ 0

sin 5x dx sin x 8 sin (3x + 2x ) dx = π sin x

π/ 2

∫ (1 + 2 cos 4x)dx

A block of mass 2 kg is free to move along the xaxis. It is at rest and from t = 0 onwards it is subjected to a time dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is F(t)

0

4N

=4

4.5s

PHYSICS Single Correct Choice Type

Ans. Sol.

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1 2 2 σ R ε0

(B)

1 2 σ R ε0

(C)

1 σ2 ε0 R

(D)

1 σ2 ε0 R 2

4.5

t (sec)

–2N

m= −

4 3

At t = 4.5 sec

→

F = – 2N 1  1  Total Impulse I =  × 3 × 4 –  × 2 × 1.5 2 2     ⇒ = 6 – 1.5 = 4.5 SI unit Impulse = change in momentum 4.5 = 2[v – 0] 4.5 v= = 2.25 m/sec 2 1 × 2 × (2.25)2 = 5.06 J K.E. = 2

(Electrostatic pressure)

41.

A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform 81π electric field of strength × 10 5 Vm–1. When 7 the field is switched off, the drop is observed to fall with terminal velocity 2 × 10–3 ms–1. Given g = 9.8 m s–2, viscosity of the air = 1.8 × 10–5 Nsm–2 and the density of oil = 900 kg m–3, the magnitude of q is (B) 3.2 × 10–19 C (A) 1.6 × 10–19 C –19 (C) 4.8 × 10 C (D) 8.0 × 10–19 C

Ans.

[D]

θ dθ

dA = 2πR sin θ × Rdθ σ2 dF = × dA 2ε 0 Component of dF along vertical axis = dF cos θ

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3

O

[A] σ2 2ε 0

(B) 7.50 J (D) 14.06 J

4N

F

(A)

(A) 4.50 J (C) 5.06 J [C]

t

F

A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to -

F

Ans. Sol.

3s

O

SECTION – I

39.

∫ 0

0

σ2 σ2 πR 2 sin 2θdθ = × πR 2 2ε 0 2ε 0

97

MAY 2010

qE = mg

Sol.

Ans.

4  81π  ⇒ q × 10 5  = 900 × πr3 × 9.8 3  7  q=

900 × 4 × r 3 × 9.8 × 7

3 × 81× 10 5 vT = 2 × 10–3 m/sec

2 × 10–3 =

[B]

Sol.

...... (1)

r 2 × 900 × 9.8 2 × 9 1.8 × 10 −5

6 cm 10 cm

18 × 1.8 × 10 −5 × 10 −3 = 0.1836 × 10– 2 × 900 × 9.8 = 18.36 × 10–12 r = 4.284 × 10–6 m 3600 × 9.8 × 7 q= × 78.62 × 10–18 243 × 10 5 q = 0.799 × 10–18 ≈ 8 × 10–19 C

30cm

r2 =

10

42.

Ans. Sol.

100 =

44.

Ans. Sol.

v 1 T = l l µ

1 50 l m/l

100 50 ⇒ 100 = m × 0.5 m 100 10000 = ⇒ m = 10–2 kg = 10 gm m

A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is (A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm [D] Least Count = M.S. Reading – V.S. Reading .... (1) 20 V.S. = 16 M.S. or 16 mm 16 16 1 V.S. = M.S. or mm 20 20 In equation (1)  16  Least Count = 1 −  mm  20  = 0.2 mm

⇒ 100 =

43.

SECTION – II Integer Type This section contains Five questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is (A) virtual and at a distance of 16 cm from the mirror (B) real and at a distance of 16 cm from the mirror (C) virtual and at a distance of 20 cm from the mirror (D) real and at a distance of 20 cm from the mirror

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10cm

Refraction of reflected light by lens f = + 15 cm u = + 10 cm 1 1 1 1 1 1 – = ⇒ − = v u f v 10 15 v = 6 cm as incident rays are converging so refracted rays will converge more and final image is real.

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string is (A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams [B] v Fundamental frequency of closed pipe = 4l 320 320 = 100 Hz ⇒ = 4 × 0.8 3.2 Frequency of 2nd Harmonic of string =

20 cm

45.

A large glass slab (µ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R ?

Sol.

[6]

sin θcr =

98

3 5

⇒

tan θcr =

3 4

MAY 2010

R

47.

element, a student plots a graph of ln

θcr

8 cm

dN( t ) is the rate of dt decay at time t. If the number of nuclei of this element decreases by a after 4.16 years, the value of p is 6 5 dN( t ) 4 ln dt 3 2 1 2 3 4 5 6 7

versus t. Here

R = 8 tan θcr 3 =8× = 6 cm 4 46.

Ans. Sol.

To determine the half life of a radioactive

Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis 25 50 m to m in 30 is observed to move from 3 7 seconds. What is the speed of the object in km per hour ? [3] For position of object initially when image 25 m was at 3 1 1 3 – = − + 10 25 u 1 1 3 – = 25 10 u 1 12 – 10 = 100 u u1 = 50

radioactive radioactive factor of p

8

Years

Ans.

8

Sol.

From graph

slope =

1 = 0.5 year–1 2

dN = Ne–λt dt

 dN  ln   = ln (N) – λt  dt  so comparing we get λ = 0.5 year–1 0.693 t1/2 = year 0.5 t = 4.16 years 4.16 so No. of half lives = × 0.5 = 3 0.693 N N N N0 → 0 → 0 → 0 ⇒ p = 8 2 4 8

O u1

For position of object when image is at

dN ( t ) dt

50 m 7

48.

1 1 7 = − + 10 50 u 1 1 7 – = 10 50 u u2 = 25

–

Ans. Sol.

A diatomic ideal gas is compressed adiabatically 1 to of its initial volume. In the initial 32 temperature of the gas is Ti (in Kelvin) and the final temperature is aTi, the value of a is [4] for adiabatic process TVγ–1 = const.

O

⇒ Ti

u2

7 −1 V5

7

 V 5 = aTi    32 

−1

⇒a=4 50 − 25 25 = m/sec 30 30 25 3600 = × = 3 km/hr 30 1000

49.

Speed of object =

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99

At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4V? (Take : ln 5 = 1.6, ln 3 = 1.1] MAY 2010

2MΩ

Sol.

2µF

A

r

B 2MΩ

Ans.

[2]

Sol.

A

2×10–6

4×10

FT = 2πrT Net vertically upward force 2  r  2πr T ⇒ 2πrT   = R R

B

2×106 6

R

2µF

2×106

10

–6

2×10–6

10V

q = CV0 (1 – e–t/RC) V = V0(1 – e–t/RC) 4 = 10(1 – e–t/4) 3 = 5e–t/4 t log3 = log 5 – 4 t ⇒ t = 2 sec 1.1 – 1.6 = – 4

51.

If r = 5× 10–4 m, ρ = 103 kgm–3, g = 10ms–2, T = 0.11 Nm–1, the radius of the drop when it detaches from the dropper is approximately (A) 1.4 × 10–3 m (B) 3.3 × 10–3 m –3 (D) 4.1 × 10–3 m (C) 2.0 × 10 m

Ans.

[A] 2πr 2 T 4 = πR 3 × ρ × g 3 R

Sol.

⇒

SECTION – III Paragraph Type

4 × 10 4 ⇒ R = 4.125 × 10–12 ⇒ R = 1.4 × 10–3 m

52. Ans. Sol.

When liquid medicine of density ρ is to be put in the eye, it is done with the help of a dropper. As the bulb on the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

Ans.

After the drop detaches, its surface energy is (B) 2.7 × 10–6 J (A) 1.4 × 10–6 J –6 (D) 8.1 × 10–6 J (C) 5.4 × 10 J [B] Surface energy = T(A) = T × 4πR2 ⇒ 0.11 × 4 × 3.14 × 1.96 × 10–6 ⇒ 2.7 × 10–6 J

Paragraph for Questions No. 53 to 55 The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. 53. A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n = 0 is not allowed) is

If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R (assuming r µ2

(q) µ3

µ2

µ1

(C) µ2 = µ3

(r)

µ3

µ2

µ1

(D) µ2 > µ3

(s) µ3

µ2

µ1

(t)

µ2

µ1

3

µ3

µ2

µ1

[A → p,r; B → q,s,t; C → p,r,t D → q,s] For (p) µ2 > µ1 as light rays bend towards normal at first refraction µ2 = µ3 as no refraction occurs at second refraction Option : (A), (C) For (q) µ2 < µ1 as bend away from normal at first refraction µ3 < µ2 as bends away from normal at second refraction Option (B), (D) For (r) µ2 > µ1 as bend towards the normal at first refraction µ2 = µ3 as no refraction occurs at second refraction Option (A), (C) For (s) µ2 < µ1 as bend away from normal at first refraction µ3 < µ2 as bend away from normal at second refraction Option (B), (D) For (t) µ2 < µ1 as bend away from normal at first refraction µ2 = µ3 as no refraction occurs at second refraction Option (B), (C) MAY 2010

57.

V1 = I XL; V2 = IR

You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2. (indicated in circuits) are related as shown in Column I. Match the two Column I

So V2 > V1 V2 ∝ I also V1 ∝ I For (s) V1 = I XL V2 = I XC where XC = again

(B) I ≠ 0, V2 > V1

For (t)

V2

(r)

V2

6mH

2Ω

V1

V2

6mH

2Ω

Option (A), (B), (D)

WHICH IS THE HIGHEST WATERFALL IN THE WORLD ?

V2 3µF

6mH

~

V

V1

(t)

V2 3µF

1kΩ

~ Ans.

[A → r,s,t;

B → q,r,s,t;

The highest waterfall in the world is the Angel Falls in Venezuela. At a towering height of

V

C → p,q;

D→

979m did you know that each drop of water takes 14

q,r,s,t] Sol.

1061 Ω

V

V1

(s)

R = 1000 Ω

V1, V2 ∝ I and I ≠ 0

V1

~ (D) I ≠ 0, V2 is proportional to I

V1 = IR when V 2 > V1

V

(C) V1 = 0, V2 = V

1061 Ω

V1 ∝ I; V2 ∝ I, I ≠ 0

V2 = I XC when XC

3µF

6mH V

(q)

1 ωC

Option (A), (B) (D)

V1

(p)

(A),

(B), (D)

Column II

(A) I ≠ 0, V1 is proportional to I

Option

seconds to fall from the top to the bottom. The

For (p) Insteady state when I = constant

water flows from the top of a “Tepui” which is a flat

V L = 0 = V1 So

topped mountain with vertical sides.

V2 = V

The waterfall which despite being known to the local

Option (C)

indians for thousands of years was originally called

For (q) V1 = 0 again as I = constant V2 = V

the “Churun Meru” but for some reason they were

Also V2 = IR ⇒ Propotional to I.

renamed by an American bush pilot called Jimmy Angel, who noticed them in 1935 whilst flying over

Option (B), (C), (D) For (r) XL = ωL = (100 π) 6 × 10–3

1.88 Ω

the area looking for gold.

R = 2Ω

XtraEdge for IIT-JEE

102

MAY 2010

XtraEdge Test Series ANSWER KEY IIT- JEE 2011 (May issue)

Ques Ans Ques Ans 19 20

1 B 11 A,C,D A→R A→S

2 D 12 B,C

3 C 13 C B→P B→P

Ques Ans Ques Ans 19 20

1 D 11 A,B A → P,Q,S A→P

2 D 12 B,C

3 C 13 C B → P,Q,R,S B→Q

PHYSICS

4 C 14 B

5 A 15 D C→T C→Q

6 B 16 B

CHEMISTRY

4 D 14 D

5 C 15 A C→P C → P,R

6 A 16 C

7 C 17 D D→S D→R

8 D 18 D

7 8 A D 17 18 D D D → Q,R,S,T D → S,P

9 A,C

10 A,C,D

9 A,D

10 B,C,D

9 A,B,C

10 A,C,D

9 B

10 B,C,D

MATHEMATICS Ques Ans Ques Ans 19 20

1 C 11 B A → P,Q,R,S A → P,R

2 C 12 C

3 C 13 B B→R B → P,S

4 A 14 A

5 B 15 B C → R,S C → Q,S

6 B 16 B

7 C 17 C D→Q D → Q,S

8 A 18 C

IIT- JEE 2012 (May issue) Ques Ans Ques Ans 19 20

Ques Ans Ques Ans 19 20

1 A 11 A,B,C,D A → R,T A→R

1 A 11 A A→P A → P,R,S,T

2 A 12 C,D

2 D 12 B

3 C 13 C B→S B→P

3 C 13 A B→T B→P

PHYSICS

4 B 14 A

5 D 15 B C→P C→S

6 C 16 B

CHEMISTRY

4 D 14 D

5 6 B B 15 16 C D C→S C → P,Q,R,T

7 C 17 C D→Q D→T

8 B 18 A

7 C 17 D D→R D→S

8 A 18 A

9 A,B,C

10 A,B

7 C 17 B D→P D→R

8 B 18 D

9 A,C,D

10 A,C

MATHEMATICS Ques Ans Ques Ans 19 20

1 C 11 A,C,D A→S A→P

XtraEdge for IIT-JEE

2 B 12 A,D

3 B 13 A B→R B→Q

4 B 14 C

5 A 15 B C→Q C→P

103

6 A 16 A

MAY 2010

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104

MAY 2010