IIT-JEE 2011 - Career Point

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
Tel. : 0744-2500492, 2500692, 3040000
e-mail : xtraedge@gmail.com
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.
Unit Price Rs. 20/-
Special Subscription Rates
6 issues : Rs. 100 /- [One issue free ]
12 issues : Rs. 200 /- [Two issues free]
24 issues : Rs. 400 /- [Four issues free]
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
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
Editorial
XtraEdge for IIT-JEE 1 MAY 2010

XtraEdge for IIT-JEE 2 MAY 2010

Volume-5 Issue-11
May, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Key Concepts & Problem Solving strategy for IIT-JEE.
INDEX
CONTENTS
Regulars ..........
PAGE
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
NEWS ARTICLE 4
IIT-Develops technology to produce stealth aircraft
Urine-processing technologies yield rich cash flow potential
IITian ON THE PATH OF SUCCESS 8
Mr. Sujal Patel
KNOW IIT-JEE 10
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS 17
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."
• "A strong, positive self-image is the best
possible preparation for success."
• "Failure is success if we learn from it."
• "The first step toward success is taken
when you refuse to be a captive of the
environment in which you first find
yourself."
8-Challenging Problems [Set# 1]
Students’ Forum
Physics Fundamentals
Electrostatics-I
1- D Motion, Projectile Motion
CATALYSE CHEMISTRY 33
XTRAEDGE TEST SERIES 49
Class XII – IIT-JEE 2011 Paper
Class XI – IIT-JEE 2012 Paper
Key Concept
Gaseous State & Real Gases
General organic Chemistry
Understanding : Physical Chemistry
DICEY MATHS 41
Mathematical Challenges
Students’ Forum
Key Concept
Complex Number
Matrices & Determinants
Test Time ..........
IIT - 2010 Examination Paper with Solution 66
XtraEdge for IIT-JEE 3 MAY 2010

IIT develops technology
to produce stealth
aircraft
Materials scientists at the Indian
Institute of Technology in
Roorkee (IIT-R) have developed
microwave absorbing nanocomposite
coatings that could make
aircraft almost invisible to radar.
The technology for building
invisible, or stealth aircraft, is a
closely guarded secret of
developed countries and a handful
of laboratories in India are doing
research in this area.
Radars that emit pulses of
microwave radiation identify flying
aircraft by detecting the radiation
reflected by the aircraft’s metallic
body. The nanocomposite coatings
developed by Rahul Sharma, R.C.
Agarwala and Vijaya Agarwala at
IIT-R absorb most of the incident
radiation and reflect very little.
Sharma, who revealed his team’s
work at an international
nanomaterials conference held
recently at the Indian Institute of
Science in Bangalore, believes
their nano-product is a significant
step in developing a technology to
enable aircraft escape radar
surveillance and protect its
equipment from electronic
“jamming”.Nanoparticles - so
called because of their very small
size - are known to exhibit unique
physical and chemical properties.
The IIT team found that crystals of
“barium hexaferrite” with particle
size of 10-15 nanometres have the
ability to absorb microwaves.
(Human hair, for comparison, is
100,000 nanometres thick). They
developed special processes for
synthesizing the nanopowder and
formulating it as a coating.
Sharma said that the
nanocomposite coating on the
aluminium sheet absorbed 89
percent of incident microwaves at
15 giga hertz - the frequency
normally used by radars —
reflecting only 11 percent. A
stealth aircraft should ideally
absorb all the incident radiation
and reflect nothing.
Urine-processing
technologies yield rich
cash flow potential
The stink is out of urine, literally
and metaphorically, with a growing
number of researchers spotting
commercial and ecological value in
a liquid most people consider
waste.
The Indian Institute of Technology
(IIT) Delhi, for instance, is working
to harvest this human waste and
convert it into fertiliser. The Delhi
government is willing to consider
a revenue-share commercial
venture selling the phosphates and
nitrates in urine.
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.
Director Madhab Nayak says the
foundation is working towards
making farmers aware of its
potential as replacement for
expensive urea.
"There is no such thing as waste,"
says Vijayaraghavan M Chariar,
assistant professor at the Centre
for Rural Development and
Technology at IIT. "Urine consists
of a lot of inorganic salts, which
produce gases only when mixed
with water. It is, in fact, pure
fertiliser," he added.
IIT has come up with a cheap,
odour-free, urinal which it has
successfully tested on campus. The
odour-free urinal combines
technology with simple science to
XtraEdge for IIT-JEE 4 MAY 2010

translate into a significant watersaving
initiative (urine smells only
when mixed with water, which
this technology eliminates).
Urine is collected through a tank
placed 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.
The public urinal at IIT uses a
simple technology, called Zerodor,
developed by Chariar, that fits
into the waste coupler in the pan
and diverts the urine through a
drain where it is collected and
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.
Meanwhile, the Delhi government,
which has already installed 200
such odourless urinals in different
parts of the city, uses a different
and perhaps more expensive
technology. Amiya Chandra,
deputy commissioner of the city’s
municipal corporation, says,
"Other than problems of
vandalism, these urinals are
working perfectly."
In preparation for the
Commonwealth Games, the Delhi
government is planning to install
1,000 such urinals at a nominal
cost of Rs 3 lakh.
Chariar is already working on the
second phase of his project, which
was initiated by Unicef and
Stockholm Environment Institute,
for setting up a small reactor to
extract nitrates and phosphates
from urine. "This could become a
micro-enterprise from the urinal,"
says Chariar.
The Delhi government is also
looking at installing Chariar’s
technology at a few parks in the
city, while harvesting urine in
those places.
Chariar has even designed similar
urinals for women. "We have filed
for trademark registration and we
are in discussion with companies
for marketing it," he says. With a
little more investment, he says, a
hydrophobic coating on pans
could make it water resistant and
completely drain the urine, leaving
no room for any oxidisation,
which can also cause odour.
In the developed 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
Chariar, explaining how it could
be diverted for use as a nutrient
by a simple plumbing.
The urine tank could deliver the
liquid nutrient directly to plants
about two to three metres below
the soil, he says.
The Centre for Banana Research
in Trichy is already using it for
banana plantations and the
University of agriculture Sciences,
Bangalore, too is looking at its
varied uses.
IIT student produces
electricity from waste
water
Kolkata: Waste water
management is a big issue world
wide and specially in India where
there is acute shortage of the
precious resource in many places
but a 23-year-old student of IIT
Kharagpur claims he has found a
solution.
Apart from finding solutions
management of waste water he
has also demonstrated producing
electricity from it, which could go
a long way in protecting the
earth's resources.
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
XtraEdge for IIT-JEE 5 MAY 2010

waste as part of a cycle of
production-consumption-ecovery.
"Waste management still
constituted a linear system of
collection and disposal which
creates health and environmental
hazards," he said.
"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 Water Diviner
The project, named LOCUS which
stands for Localised Operation of
Bio-cells Using Sewage, can
achieve chemical oxygen demand
(COD) reduction levels in waste
water to about 60-80 per cent.
IIT Kharagpur Calls for
Nominations for the
Nina Saxena Excellence
in Technology Award
IIT Kharagpur announces the
fourth edition of Nina Saxena
Excellence in Technology Award
to invite entries of this year in
areas of technical innovations.
Nominations for entries to the
award are open until April 30,
2010. Entries can be submitted
either by post to the Director, IIT
Kharagpur, West Bengal,PIN -
721302, India or by email at
director@iitkgp.ernet.in. The
nomination form is also available
on the official award website
The objective behind the award is
to commemorate the spirit and
drive of Dr. Nina Saxena, who
personified technical excellence.
The fourth edition will continue
the tradition of rewarding
pioneering innovations for
betterment of 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
Deans and selected faculty
members of IIT Kharagpur and
well known alumnus, based in
India and in the US.
IIT’s New On-Campus
Wind Turbine to
Support Green Jobs,
Research, and Education
The consortium members will
research the wind energy
challenges identified in the U.S.
Department of Energy's "20%
Wind Energy by 2030" report,
including wind technology, grid
system integration, and workforce
challenges. The consortium's plan
relies on IIT experts in electrical
and computer engineering;
mechanical, materials, and
aerospace
engineering;
architecture; business; and
members of the Wanger Institute
for Sustainable Energy Research to
tackle these challenges.
Many of the 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 consultation with
industry. To ensure student
involvement in the project,
fellowships will be offered annually
to undergraduate and graduate
students in wind energy
engineering fields of study. Faculty
and students from international
university consortium members
will also be invited to IIT to attend
workshops and to share ideas with
their American counterparts.
The wind energy consortium will
work with small wind turbine
manufacturer Viryd Technologies
to procure and install an 8KW
Viryd wind turbine on IIT's Main
Campus, and to deliver a second
turbine to one of IIT's engineering
laboratories to perform turbine
reliability studies. The consortium
will also work with wind energy
developer Invenergy to install a
1.5MW GE wind turbine adjacent
to a wind farm in Marseilles, Ill.
The close proximity of IIT's
XtraEdge for IIT-JEE 6 MAY 2010

Marseilles turbine to an existing
wind farm provides an ideal
opportunity to study turbine-toturbine
wake interaction, wind
farm interaction, and wind energy
efficiencies in addition to turbine
reliability studies.
Hyderabad boy tops in
GATE 2010
HYDERABAD : Malladi
Harikrishna, a final year computer
science engineering student from
the city, has topped the national
level Graduate Aptitude Test in
Engineering (GATE-2010). He
achieved the feat in his first
attempt, scoring 99.99 percentile
by scoring 83.55 per cent in
GATE.
The results were announced on
March 15 by IIT-Guwahati, which
conducted
GATE this
year. Mallad,
who topped
the nationallevel
Graduate Aptitude Test in
Engineering (GATE-2010), said,
“My aim is to join ME at the Indian
Institute of Science, Bengaluru. I
want to become a scientist,”
Harikrishna, 21, said.
Students who clear GATE are
eligible for admission to masters’
degree courses in engineering,
technology, architecture,
pharmacy, science in premier
institutes like IITs and NITs.
Many students from the state
made it to the top-100.
Srujana (JNTU, Kukatpally) ranked
22, Karthik Nagarjuna 44,
Mufaquam Ali 51 and Pavan
Kishore got the 102nd rank in
ECE stream.
Srinivas Reddy got the 68th rank
in EEE, and V. Suryanarayana 31 in
Mechanical.
GATE 2010 score is valid for two
years from the date of
announcement of the results,
according to the details published
on GATE website
Terminated IIT students
seek intervention of
Prez,PM
Terminated for "bad
performance", 38 students of IIT
Kanpur have taken the issue to
President Pratibha Patil and Prime
Minister Manmohan Singh with a
plea that the institute reconsider
the decision.
The students have written to
Prime Minister and the PMO has
forwarded the matter to the HRD
Ministry for "appropriate action".
The Ministry has again forwarded
it to the institute for action, a
ministry official said.
The 38 students, including 24 from
under-graduate and 14 postgraduate
levels, were denied
admission into fresh semester in
January this year for "bad
academic record"
.Prof V N Pal, who is an alumni of
the institute, has taken the
students' issue to President Patil,
who is the Visitor of the institute.
Prof Pal met the President on
April one at Rashtrapati Bhawan
and discussed the issue of
termination of students of IIT
Kanpur at length.
He explained the socio-economic
condition of these students.
Vidya Balan to address
IIT
She has been invited by one of the
Indian Institutes of
Technology to be a keynote
speaker at an upcoming seminar.
In an interview to a leading daily,
she confirmed her invitation from
one of the IITs. She said, “I have
been approached by one of the
IITs for being a keynote speaker at
a seminar they are holding. The
topic is ‘The changing face of the
Indian heroine’. I am excited about
it, but am figuring out dates and
prior commitments at the
moment. But, I hope this comes
through.”
A MA in sociology, Vidya Balan
believes her education has
groomed her and given her the
confidence to address seminars at
these prestigious universities.
XtraEdge for IIT-JEE 7 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
with. His company Isilon Systems, in the clustered storage
space, has not only earned its position as the fastest
growing technology company in North America, but in
the five years of selling its products, Patel has transformed
the company from zero sales, into a company with a $100
million run rate, $80 million in cash, and no debt.
“Slightly better is not a good term,” says Patel, the CEO
of Isilon Systems and a pioneer in the clustered storage
space. “For a technology to be adopted in an existing
market, you really need to have a technology that is
substantially better — 10 (times) better than what is in
the marketplace today,” adds Patel. “It has to be so much
better that it is overwhelming for people who buy that
product and service.” When Patel says so, he is not being
merely theoretical. The unsatiated desire to bring out the
best led this 35 year old entrepreneur to steer his
company successfully, in a market space, which was
crowded by biggies like EMC and NetApp; not to mention
the 250 odd startups in the storage space.
Patel founded Isilon at the age of 26. Prior to this, Patel
spent his initial career days at Real Networks. As an
engineer, Patel used to solve some of Real Networks’
most complex back-end operational challenges. That
experience gave him the insight for a new type of solution,
a type of virtualized storage optimized for media. His
experience gave him the insight to a real customer need,
and his deep technical knowledge gave him the ability to
spot a solution.
I did not want to wait 10-15 years, treading cautiously at
every step before taking the plunge. So when I got the
chance to found Isilon, I jumped at the opportunity,”
beams Patel.
That’s also a reason which led venture capitalists to take
this 26 year old lad seriously. A few months after Patel
founded Isilon in 2001, the NASDAQ came down
crashing, bringing the ‘Dot Com Burst’. The venture
capital market was in disarray. With the existing
companies dropping their revenues, there was not much
hope for new companies to find potential investment. To
make matters worse, there were about 250 other
startups in the storage space. Patel was undeterred. Sure
about his ideas, he approached close to 40 venture
capitalists, and with perseverance, eventually he managed
to gain their confidence. Five months after the company’s
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
money.
But all was not rosy yet. The road ahead proved to be
more challenging than raising the funds. The next three
years were spent in building the products. Developing the
company’s IP not only took a tremendous amount of
money, but also ate in to the time to get into the market.
By the time the product was ready to debut in
the markets in 2003, the company’s debt was near
$20 million.
Cruising over Obstacles
Not only confident and unshakable, Patel is also a man of
clear vision, with no illusions about his capabilities. At 26,
as the Founder of the company, he served as the CEO for
the initial three years. But he knew that, if he wanted to
make Isilon the next big thing in storage space, he needed
someone who knew the dynamics of running a large
organization. He promptly stepped down and appointed
as a new CEO, who knew what it takes to grow into the
larger league. “I floated the company but knew my
limitations in the business front. We had three products
ready to debut in the market and if we wanted then to
succeed, we needed an expert who knew the right strings
to pull,” says Patel. His focus and timely decisions were
fruitful in the subsequent years when Isilon was on a
dream run, literally growing at 200 percent year-on-year.
By the time the company went public, it was a $60 million
company.
“My goal was to see Isilon become a $100 million
company by 2007 and become a player to be reckoned in
the $4 billion global storage market for its technology.
And we were still $40 million short. I knew that despite
the economic instability, the company I had founded had
great potential if one could maximize it,” says Patel.
To start with, he re-structured the entire organization,
replacing every executive in the management, including
CTO, CFO, Head of engineering, Head of operations and
others. Such an action is quite unheard of, especially
immediately after a company had gone public.
Next, Patel began revamping the business strategy by
reaching out to the broader enterprise segments. It was
not easy. The segment he wanted to target comprised of
Fortune 50 companies who did not have much
expectation from a startup like Isilon. With the
organizational re-structuring, he completely overhauled
the company’s services and products to meet the
expectations of the large enterprise customers.
Eventually, more and more Fortune 50 companies began
to adopt Isilon’s products.
Finally he decided to increase the company’s focus on
R&D. Innovation has always been an essential part of his
life. A very innovative and inquisitive person, Patel is
known to get at least six ideas every day. Even as a
kid he was known for asking around 500 questions
about everything under the Sun. Thus, Patel diverted
about 25 percent of the company’s revenue
towards R&D and fostered innovation within the
organization.
So what drives this confident and zealous man? It is the
self-belief, passion and the creative way of looking at
problems and coming up with solutions, says Patel. Even
during his college days, while working on the Internet, he
looked for opportunities to think about new ways, solve
problems, or to bring innovative techniques/technologies
to the market place.
To run a business successfully, it is also important that
one should be honest with oneself, the team and the
stakeholders. The foremost thing that Patel did after
taking over the company was to communicate with
customers and partners and update them about the
happenings within the organization, reinstating their faith
in him and the company. “I went and talked to each of our
investors and customers, telling what we planned
and how much earnings we expected through the
quarters. I believe that apart from your technology
offerings, one reason companies want to do business
with you is the goodwill you develop in tough times,”
says Patel.
XtraEdge for IIT-JEE 9 MAY 2010

KNOW IIT-JEE
By Previous Exam Questions
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
W L
Y = ×
2
πD
X
4
⎛ ∆Y ⎞
⎜ ⎟⎠
⎝ Y
max.
⎛ ∆D
⎞
= 2 ⎜ ⎟
⎝ D ⎠
∆ x + x
∆L
+ L
⎛ 0.001⎞
⎛ 0.001 ⎞ ⎛ 0.1 ⎞
= 2 ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 0.0489
⎝ 0.05 ⎠ ⎝ 0.125 ⎠ ⎝110
⎠
So maximum percentage error = 4.89%.
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]
P
Q
45º
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/s 2 .
Sol. m p = 20 g m Q = 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).
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
49m/s
135º
45º
A 49 / 2m / s 49 / 2m / s B
Horizontal direction
S x = 122.5
T x = ?
v x = 49 2
∴ velocity =
⇒
49 122.5 =
2 t x
245m
(122.5) 2
∴ t x =
49
Vertical Direction v y = ?
49
u y =
2
(122.5) 2
t y =
49
a y = – 9.8 m/s 2
displacement
time
49 (122.5)
∴ v y = u y + at y = – 9.8 ×
2
49
m p V p m Q V Q – +
Before collision
m p V p m Q V Q
′
After collision
Also, v 2 – u 2 = 2as
⎛ 49 ⎞
∴ –
⎜
3
⎟
⎝ ⎠
2
⇒ s = 61.25
= 2 (– 9.8) × s
2
= 0
XtraEdge for IIT-JEE 10 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
∴ m p m p – m Q v Q = – m P v P + m Q v′ Q
2mPvP
− mQvQ
∴ v′ Q =
m
2×
0.02×
(49 / 2) − 0.040(49 /
=
0.040
49 ⎡ 0.04 ⎤ 49
= ⎢ −1⎥ = × 0 = 0
2 ⎣0.040
⎦ 2
New Path of Q after Collision
Considering vertical Motion of Q
u y = 0
s y = – 61.25
a y = – 9.8
t y = ?
Q
2)
S = ut + 2
1 at 2 = 2
1 × (– 9.8) × t 2 = (–61.25)
∴ t = 3.53 sec
Considering Horizontal motion of Q :
'
Since the V Q = 0, therefore the particle Q falls down
vertically so it falls down on the mid point of AB.
3. 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
period of the circular motion. [IIT-1988]
Sol. The radius of the circle
2
r =
3
a
=
3
a
2 −
2
a
4
m
Let v be the velocity given. The centripetal force is
provided by the resultant gravitational attraction of
the two masses
2 2 2
F R = F + F + 2F cos60º
m× m
= 3 F = 3 G
2
a
2
m mv 2
∴ 3 G
a 2 = r
⇒ v 2 3G ma
=
a 3
2 ×
Gm
⇒ v =
a
Time period of circular motion
2π r 2πa 3
T = = = 2π
v Gm
a
a 3
3Gm
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
P(x,y)
∴
A
B
X′ (–3a,0) O x +Q C
(–3a,0)
(5a,0)
(3a+x) (3a-x)
Y′
K(2Q)
=
K( + Q)
2 2
(3a + x) + y
2 2
(3a − x) + y
y
X
m
F
v
F R
F
a/ 3
a
m
⇒ 2
2
2
( 3a − x) + y =
2
( 3a + x) + y
⇒ 4[(3a – x) 2 + y 2 ] = [(3a + x) 2 + y 2 ]
⇒ 4[6a 2 + x 2 – 12ax + y 2 ] = [6a 2 + x 2 + 12ax + y 2 ]
⇒ 3x 2 + 3y 2 – 30ax + 27a 2 = 0
⇒ x 2 + y 2 – 10ax + 9a 2 = 0
⇒ (x – 5a) 2 + (y – 0) 2 = (4a) 2
2
XtraEdge for IIT-JEE 11 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.
–2Q
+Q
(–3a,0) O (3a,x) M
x
The potential at M will be
(c)
K( −2Q)
K( + Q)
V(x) = +
x + 3a (x − 3a)
where k =
1
4πε
⎡ 1 2 ⎤
∴ V(x) = KQ ⎢ − ⎥ For |x| > 3a
⎣ x − 3a x + 3a
⎦
Similarly, for
⎡ 1 2 ⎤
0 < |x| < 3a V(x) = KQ ⎢ − ⎥
⎣3a
− x 3a + x ⎦
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
x → – 3a.
• V(x) → 0 as x → + ∞
1
• V(x) varies at in general. x
V
–3a
X
a 3a
Applying Energy Conservation
(K.E. + P.E.) centre = (KE. + P.E.) circumference
⎡Qq
2Qq ⎤ 1
0 + K ⎢ − ⎥ = mv 2 ⎡Qq
2Qq ⎤
+ K
⎣ 2a 8a ⎦ 2
⎢ − ⎥
⎣ 6a 12a ⎦
⇒ 2
1 mv -2 =
KQq
4a
⇒ v =
KQq =
2ma
1
4πε
0
0
⎛ Qq ⎞
⎜ ⎟⎠
⎝ 2ma
5. 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]
X 12Ω
A B C 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 AJ 0.6ρ
12Ω
(b) & (c) Q Bridge is balanced = =
R 0.4ρ
X
⇒ X = 8 Ω
where ρ is the resistance per unit length.
A
J
G
X
C
JB
12Ω
CHEMISTRY
6. 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 Na 2 O and K 2 O in
the mixture ?
[IIT-1979]
Sol. Mass of sample of feldspar containing Na 2 O and
K 2 O = 0.5 g.
According to the question,
Na 2 O + 2HCl → 2NaCl + H 2 O ..(1)
2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g
K 2 O + 2HCl → 2KCl + H 2 O ...(2)
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 + AgNO 3 → AgCl + NaNO 3 ...(3)
23 + 35.5 = 58.5g 108 + 35.5 = 143.5g
KCl + AgNO 3 → AgCl + KNO 3 ...(4)
39 + 35.5 = 74.5g 108 + 35.5 = 143.5g
Total mass of AgCl obtained = 0.2451 g
Step 1. From eq. (3)
D
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 Na 2 O
∴ 0.0342 g NaCl is obtained from
62
= × 0.032 = 0.018 g Na2 O
117
From eq. (2),
149 g of KCl is obtained from = 94 g K 2 O
∴ 0.0838 g of KCl is obtained from
94
= × 0.0838 = 0.053 g K2 O
149
0.018
Step 3. % of Na 2 O in feldspar = × 100 = 3.6%
0.5
0.053
% of K 2 O 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
Number of atoms per unit cell = 8
8 + 1 = 2
O
a
60º
N b
3
Area of the base = b × ON = b × a sin 60º = a
2
2
( Q b = a)
Volume of the hexagonal cell
3
= Area of the base × height = a 2 . c
2
But c =
2 2
a
3
c
β α
b
a γ
∴ Volume of the hexagonal cell
3
= a 2 2 2
. a = a 3 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
4 4 ⎛ a ⎞
3
2×
πr
2 × π⎜
⎟
=
3 3 2
=
⎝ ⎠ π
=
3
3
a 2 a 2 3 2
= 0.74 = 74%
∴ The percentage of void space = 100 – 74 = 26%
8. A basic nitrogen compound gave a foul smelling gas
when treated with CHCl 3 and alc. KOH. A 0.295 g
sample of the substance, dissolved in aq. HCl and
treated with NaNO 2 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 I 2 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 CHCl 3 and alc.
KOH gives foul smelling gas, it should be any
primary amine.
RNH 2 + CHCl 3 + 3KOH
3
⎯⎯→
∆ RN C
(Alkyl isocyanide
(foul smelling gas)
+ 3KCl + 3H 2 O
Since the compound on treating with NaNO 2 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
XtraEdge for IIT-JEE 13 MAY 2010

NaNO 2 + HCl → NaCl + HNO 2
RNH 2 + HNO 2 → ROH + N 2 + H 2 O
Thus, the gas liberated is N 2 .
Amount of gas produced =
112 ml
22,400 ml
=
1
200
From the above equation, it is obvious that amount of
1
compound RNH 2 = mol. 200
If M is the molar mass of RNH 2 , then
0.295g
Molar mass (M)
1
= 200
mol
ml
∴ M = 0.295 × 200 g mol –1 = 59 g mol –1
Thus, the molar mass of alkyl group (R—) will be,
59 – 16 = 43 g mol –1 .
Hence, R = C 3 H 7 —, i.e., CH 3 CH 2 CH 2 –
or CH 3
CH–
CH 3
The original compound may be either
CH 3 CH 2 CH 2 NH 2 or CH 3 – CH – CH 3
NH 2
From equation, it is clear that the liquid obtained
after distillation is ROH. Since it gives yellow ppt.
with NaOH and I 2 , it must have CH 3 — C — group.
OH
Hence, it is concluded that ROH is CH 3 – CH – CH 3
.
OH
Thus, the original compound is CH 3 – CH – CH 3
.
NH 2
The different equations are :
CH 3
CH 3
CH– NH 2 + CHCl 3 + 3KOH
∆
Isopropyl amine
CH 3
CH 3
CH – N C + 3KCl +
Isopropyl isocyanide
CH 3
CH 3
CH – NH 2 + HNO 2 ⎯→
CH 3
CH – OH + N 2 + H 2 O
CH 3
Isopropyl alcohol
CH 3 – CH – CH 3 + I 2 + 2NaOH ⎯→
OH
CH 3 – C – CH 3
O
Acetone
CH 3 – C – CH 3 + 3I 2 + 3NaOH ⎯→
O
CI 3 – C – CH 3
O
+ 2NaI + 2H 2 O
+ 3NaI + 3H 2 O
∆
CI 3 – C – CH 3 + NaOH ⎯→ CH 3 COONa + CHI 3
Yellow ppt.
O
9. An organic compound (A) C 8 H 6 , on treatment with
dil. H 2 SO 4 containing HgSO 4 gives a compound (B),
which can also be obtained from a reaction of
benzene with an acid chloride in the presence of
anhydrous AlCl 3 . 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 :
C 8 H 6
(A)
Dil H 2SO 4
HgSO 4
AlCl
(B)
3
∆
∆ I 2 + KOH
C 6 H 6 + Acid chloride
(C) + (D)
The reaction of compound (B) with I 2 in KOH is
iodoform reaction. The compound (B) must have a
–COCH 3 group so as to exhibit iodoform reaction.
Since (B) is obtained from benzene by Friedal-Crafts
reaction, it is an aromatic ketone (C 6 H 5 COCH 3 ). The
compound (C) must be potassium salt of an acid.
The compound (A) may be represented as C 6 H 5 C 2 H.
Since it gives C 6 H 5 COCH 3 on treating with dil.
H 2 SO 4 and HgSO 4 , it must contain a triple bond
(–C ≡ CH) in the side chain. Here, the given reactions
may be formulated as follows :
OH
C≡CH
dil H 2SO 4
HgSO 4; H 2O
(A)
C = CH 2
CH 3COCl
AlCl 3; – HCl
Benzene
COCH 3
Acetophenone
(B)
XtraEdge for IIT-JEE 14 MAY 2010

(B)
Hence. (A)
(C)
O
– C – CH 3
+ 3I 2 + 4KOH
C≡CH
Phenyl acetylene
COOK
Potassium benzoate
∆
–3KI;–3H 2O
COOK
+ CHI3
(D)
(C)
Potassium benzoate
COCH 3
(B)
(D)
Acetophenone
CHI 3
Idoform
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
CuSO 4 blue. The aqueous solution of (Y) is alkaline
to litmus and gives 19.7 g of a white precipitate (Z)
with BaCl 2 solution. (Z) gives CO 2 with an acid.
Identify (A), (B), (X), (Y) and (Z). [IIT-1989]
Sol. (i) Since acidic gas (A) turns lime water milky hence
it is CO 2 or SO 2 , both of which form white insoluble
compound with Ca(OH) 2
(ii) Neutral gas (B) condenses into a liquid which
turns anhydrous CuSO 4 (white) into blue
(CuSO 4 .5H 2 O), hence (B) is H 2 O.
(iii) (Y) gives alkaline solution and its solution forms
white precipitate (Z) with BaCl 2 solution. (Z) on
heating gives the acid gas CO 2 , hence (Z) is BaCO 3
and therefore (Y) is a metal carbonate.
(iv) Since (Y) and (A) are produced from (X), thus
(X) is a metal bicarbonate.
( X) →
20.02 g
( A) +
4.4 g
( B) +
1.80g
( Y)
13.8g
From the above values we may write a general
equation for a bicarbonate.
⎯ ∆ 2
(A)
2MHCO 3 ⎯→
(X)
CO + H 2 O +
(B)
M 2CO 3
(Y)
Q 4.4g CO 2 is obtained from 20.02 g of MHCO 3
∴ 4g CO 2 is obtained from 200.2 g of MHCO 3
200.2
Molecular weight of MHCO 3 = = 100.1
2
Atomic weight of M = 39
Thus, the metal M is potassium and then (X) is
KHCO 3 . The equations are :
⎯ ∆ (Y)
2 KHCO 3 ⎯→
(X)
K 2CO 3 +
K 2CO 3 + BaCl 2 ⎯→ 2KCl +
(Y)
BaCO
(Z)
3
⎯⎯→
∆ BaO +
CO 2 +
(A)
BaCO
(Z)
CO2
↑
(A)
3
H 2O
(B)
Hence, (A) is CO 2 (B) is H 2 O (X) is KHCO 3 (Y) is
K 2 CO 3 and (Z) is BaCO 3 .
MATHEMATICS
11. From a point A common tangents are drawn to the
circle x 2 + y 2 = a 2 /2 and parabola y 2 = 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, y 2 = 4ax is
y = mx + a/m.
This line will touch the circle x 2 + y 2 = a 2 /2
A(–a, 0)
If
⇒
⎛ a ⎞
⎜ ⎟⎠
⎝ m
2
=
πx = – a/2
y
B
C
O
a 2 (m 2 + 1)
2
E
D
x = a
1 1
2 = (m 2 + 1) ⇒ 2 = m 4 + m 2
m 2
⇒ m 4 + m 2 – 2 = 0
⇒ (m 2 – 1)(m 2 + 2) = 0
⇒ m 2 – 1 = 0, m 2 = – 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
x 2 + y 2 = a 2 /2 is
(–a)x + 0.y = a 2 /2 or x = – a/2
and chord of contact of A(–a, 0) for the parabola
y 2 = 4ax is
0.y = 2a(x – a) or x = a
Again length of BC = 2BK
L
x
XtraEdge for IIT-JEE 15 MAY 2010

2
2
= 2 OB − OK
= 2
2 2
a a a 2
− = 2
2 4 4
= a
and we know that DE is the latus rectum of the
parabola so its length is 4a.
Thus area of the trapezium
BCDE = 2
1 (BC + DE) (KL)
1 ⎛ 3a ⎞
= (a + 4a) ⎜ ⎟⎠ =
2 ⎝ 2
15a
2
4
12. Let V be the volume of the parallelopiped formed by
the vectors
→
a = a1 î + a 2 ĵ + a 3 kˆ ;
→
b = b1 î + b 2 ĵ + b 3 kˆ
→
c = c1 î + c 2 ĵ + c 3 kˆ
If a r , b r , c r , where r = 1, 2, 3 are non-negative real
3
numbers and ∑ ( a r + br
+ cr
) = 3L. Show that
r=
1
V ≤ L 3 .
[IIT-2002]
Sol. V = |
→ →
a .(b × →
c ) | ≤ 2 2
a + a +
2
Now, L =
1 2 a 3
2 2 2 2 2 2
1 + b2
b3
c 1 c2
+ c3
b + + ...(1)
( a1 + a 2 + a 3)
+ (b1
+ b2
+ b3)
+ (c1
+ c2
+ c3)
3
[(a 1 + a 2 + a 3 ) (b 1 + b 2 + b 3 ) (c 1 + c 2 + c 3 )] 1/3
∴ L 3 ≥ [(a 1 + a 2 + a 3 )(b 1 + b 2 + b 3 )(c 1 + c 2 + c 3 )] ..(2)
Now, (a 1 + a 2 + a 3 ) 2
=
2
a 1 + a 2 2 + a 3
2 + 2a 1 a 2 + 2a 1 a 3 + 2a 2 a 3 ≥
⇒ (a 1 + a 2 + a 3 ) ≥
Similarly,
and
2 2 2
1 + a 2 a 3
a +
(b 1 + b 2 + b 3 ) ≥
(c 1 + c 2 + c 3 ) ≥
2 2 2
1 + b2
b3
b +
2 2 2
1 + c2
c3
c +
2
a 1 + a 2 2 + a 3
2
∴ from (1) and (2)
L 3 2
≥ [( a 1 + a 2 2 + a 3
2 2 2 2 2 2 2
)( b 1 + b2
+ b3
)( c 1 + c2
+ c3
)] 1/3 ≥ V
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.
⇒
| aα + bβ + cγ + d |
= 90%. h
2 2 2
a + b + c
⇒ aα + bβ + cγ + d = ± 0.9h
2
2
a + b + c
⇒ locus is, ax + by + cz + d = ±0.9h
2
2
2
a + b + c
⇒ 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 : [r 1 , r 2 ... r n ]
to B : [1, 2, 3] where r 1 r 2 .... r n 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 = 3 n , n(1) = 3.2 n , n(2) = 3, n(3) = 0
⇒ M = 3 n – 3.2 n + 3
Hence the total number of favourable cases
= (3 n – 3.2 n + 3). 6 C 3
⇒ required probability =
n
n
( 3 − 3.2 + 3) ×
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 L 1 and L 2 are
drawn, parallel to 2x – y = 5 and 3x + y = 5
respectively. Lines L 1 and L 2 intersect at R, shown
that the locus of R as L varies, is a straight line.
[IIT-2002]
Sol. Let the equation of straight line L be y = mx
⎛
P ≡ ⎜
⎝
1
m
m ⎞ ⎛ 3 3m ⎞
, ⎟ ; Q ≡ ⎜ , ⎟
+ 1 m + 1 ⎠ ⎝ m + 1 m + 1 ⎠
m − 2
Now equation of L 1 : y – 2x = ...(1)
m + 1
3m + 9
equation of L 2 : y + 3x =
...(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.
6
n
6
C
3
2
XtraEdge for IIT-JEE 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
Solutions will be published in next issue
Director Academics, Jodhpur Branch
1. Two capacitors C 1 and C 2 , can be charged to a
potential V/2 each by having
C 1 C 2
O
V R R
S 1 S 2
(A) S 1 closed and S 2 open
(B) S 1 open and S 2 closed
(C) S 1 and S 2 both closed
(D) cannot be charged at V/2
2. Energy liberated in the de-excitation of hydrogen
atom from 3 rd level to 1 st level falls on a photocathode.
Later when the same photo-cathode is
exposed to a spectrum of some unknown
hydrogen like gas, excited to 2 nd energy level, it is
found that the de-Broglie wavelength of the
fastest photoelectrons, now ejected has decreased
by a factor of 3. For this new gas, difference of
energies of 2 nd Lyman line and 1 st Balmer line if
found to be 3 times the ionization potential of the
hydrogen atom. Select the correct statement(s)
(A) The gas is lithium
(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
3. In the figure shown there exists a uniform time
varying magnetic field B = [(4T/s) t + 0.3T] in a
cylindrical region of radius 4m. An equilateral
triangular conducting loop is placed in the
magnetic field with its centroide on the axis of the
field and its plane perpendicular to the field.
+
+
+
+
B
+
+
+
+
+
+
+
A
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+
+
+
C
(A) e.m.f. induced in any one rod is 16V
(B) e.m.f. induced in the complete ∆ ABC is
48
3V
(C) e.m.f. induced in the complete
(D) e.m.f. induced in any one rod is 16
∆ ABC is 48V
3V
4. 6 parallel plates are arranged as shown. Each
plate has an area A and Distance between them is
as shown. Plate 1-4 and plates 3-6 are connected
equivalent capacitance across 2 and 5 can be
nA ∈
writted as 0 . Find min value of n. (n, d are
d
natural numbers)
1
2 d
3 d
2d
4
d 5
d 6
XtraEdge for IIT-JEE 17 MAY 2010

5. Match the following
Column – I
Column – II
(A) A light conducting (P) Magnetic field B
circular flexible is doubled.
loop of wire of
radius r carrying
current I is placed
in uniform magnetic
field B, the tension
in the loop is doubled
if
(B) Magnetic field at a (Q) Inductance is
point due to a long increased by four
straight current times.
carrying wire at a
point near the wire
is doubled if
(C) The energy stored (R) Current I is
in the inductor will doubled
become four times
(D) The force acting on a (S) Radius r is
moving charge, doubled
moving in a constant
magnetic field will be
doubled if
(T) Velocity v is
Doubled
Passage # (Q. No. 6 to Q. No. 8 )
A solid, insulating ball of radius ‘a’ is surrounded
by a conducting spherical shell of inner radius ‘b’
and outer radius ‘c’ as shown in the figure. The
inner ball has a charge Q which is uniformly
distribute throughout is volume. The conducting
spherical shell has a charge –Q.
Answer the following questions.
–Q
b c
Q
a
6. Assuming the potential at infinity to be zero, the
potential at a point located at a distance a/2 from
the centre of the sphere will be
Q ⎡2
1 ⎤
Q ⎡11
1 ⎤
(A) ⎢ −
4πε
⎥ (B)
⎣a
b
⎢ − ⎥ ⎦ 4πε
⎣8a
b ⎦
(C)
Q
4πε
0
0
⎡1
1 ⎤
⎢ − ⎥
⎣a
b ⎦
0
(D) None of these
7. Work done by external agent in taking a charge q
slowly from inner surface of the shell to surface
of the sphericalball will be
⎡1
1⎤
⎡1
1⎤
(A) kQq ⎢ − ⎥ (B) kQq
⎣a
c
⎢ − ⎥ ⎦ ⎣b
a ⎦
⎡1
1 ⎤
⎡1
1⎤
(C) kQq ⎢ − ⎥ (D) kQq
⎣a
b
⎢ − ⎥ ⎦ ⎣c
a ⎦
8. Now the outer shell is grounded, i.e., the outer
surface is fixed to be zero. Now the charge on the
inner ball will be
(A) zero
(B) Q
(C)
Q ⎛ 1
⎜ +
C ⎝ a
1
c
1 ⎞
− ⎟
b ⎠
(D)
Q ⎛ 1
⎜ +
b ⎝ a
1
c
Cartoon Law of Physics
1 ⎞
− ⎟
b ⎠
Any body passing through solid matter will leave a
perforation conforming to its perimeter.
Also called the silhouette of passage, this
phenomenon is the specialty of victims of directedpressure
explosions and of reckless cowards who
are so eager to escape that they exit directly
through the wall of a house, leaving a cookiecutout-perfect
hole. The threat of skunks or
matrimony often catalyzes this reaction.
XtraEdge for IIT-JEE 18 MAY 2010

PHYSICS
Students'Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
⎡ m ⎤
1. 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
⎦
⎡ 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 v s the velocity of the plane from which the
shots are fired, by v t 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
t 1 =
…(1)
v + v s − 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:
vs vt
L' = L –
−
…(2)
r
Thus, the time it takes the second projectile to arrive
at the target plane is:
v s − v t
L −
t 2 =
r
…(3)
v + v s − v t
which is
∆t = t 2 + r
1 – t1
1 v
= –
s − vt
v
=
…(4)
r r(v + vs
− vt
) r(v + vs − vt
)
after the first shot. Naturally, the time increment does
not depend on the initial distance; thus the rate of
hitting is:
1
r′ =
∆ = r ( v + vs − vt
)
t v
1300 ⎡ hits ⎤
= 40 × = 43.33 1200
⎢ ⎥ … (5)
⎣ second ⎦
(b) Using the same reasoning for this case, we obtain:
v − vs + vt
r′ = r
v
1100 ⎡ hits ⎤
= 40 × = 36.66 1200
⎢ ⎥ … (6)
⎣ second ⎦
(c) In this case:
v + vs + vt
r′ = r
v
1700 ⎡ hits ⎤
= 40 × = 56.67 1200
⎢ ⎥ … (7)
⎣ second ⎦
(d) Here,
v v v
r′ = r
− s − t
v
700 ⎡ hits ⎤
= 40 × = 23.33 1200
⎢ ⎥ … (8)
⎣ second ⎦
2. Consider the system described in figure.
m 1
m 2
(a) Use the equations of energy conservation to find
the velocities of masses m 1 and m 2 after they are
released from rest and pass a distance y
(assume m 2 > m 1 ).
(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 :
m 1 : = ∆E p = m 1 gy ...(1)
m 2 : ∆E p = – m 2 gy ...(2)
The change in the kinetic energy is
∆E k = 2
1
m1 v 1 2 + 2
1
m2 v 2
2
...(3)
XtraEdge for IIT-JEE 19 MAY 2010

Because | v r 1 | = | v r 2 | ≡ v, we obtain
∆E k = 2
1 (m1 + m 2 )v 2 ...(4)
Since there are no external forces except gravity,
which is a conservative force, we know, using energy
conservation that :
∆E p + ∆E k = 0 ...(5)
By substituting
(m 1 – m 2 )gy + 2
1 (m1 + m 2 )v 2 = 0 ...(6)
1/ 2
⎛ 2g(m2
− m1)
⎞
Hence, v(y) =
⎜
m1
m
⎟
⎝ + 2 ⎠
y ...(7)
(b) By definition,
1/ 2
dv ⎛ 2g(m
a = =
2 − m1)
⎞ d y
dt
⎜
m1
m
⎟ .
⎝ + 2 ⎠ dt
⎛ 2g(m2
− m1)
⎞
=
⎜
m1
m
⎟
⎝ + 2 ⎠
1/ 2
2
1
y
dy
. dt
...(8)
dy
and because = v we substitute Eq. (7) into Eq.
dt
(8) and obtain.
m2
− m1
a = g
m + m
2
1
3. 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
a
mg
m
(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,
m a r = mg xˆ ´ sin α – mg ŷ ´ cos α + N ŷ ´ – m a r
a r is expressed by the unit vectors of the accelerated
system as :
a r = – a xˆ ´ cos α – a ŷ ´ sin α
and in component form :
⎧a
⎪
x´ = g sin α + a cos α
⎨ N
a y´ = − g cos α + a sin α (N ≥ 0)
⎪⎩ m
y´
α
x´
(b) The maximal value is obtained when N = 0 and
a y´ = 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 :
x(t) = x 0 + v 0 t + 2
1 at
2
1
In our case, we have L = a x ´ t 2 . Therefore,
2
2L
t =
gsin α + a cosα
(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
4. 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,
+ –
Its area = 2π x × 1
Charge enclosed within it = λ
+
+
+
x
∴ Flux passing through the cylindrical surface =
Electric field,
–
–
–
E = Flux passing per unit area.
( λ / ε0)
λ
= =
2π
x 2πε0x
Magnitude of charge on unit length of second wire = λ
∴ Force of attraction per unit length is F = λ E
λ
ε 0
XtraEdge for IIT-JEE 20 MAY 2010

2
λ
or F =
2πε0x
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
or Total work done =
∫ = x=
x
2
a
λ
=
2πε
2a
0
x
F dx =
∫ = 2a λ
x= a 2πε
log e 2
2
0
dx
x
5. Two short electric dipoles having dipole moment p 1
and p 2 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 p 2 consist of
charges (+ q 2 ) and (– q 2 ) which are separated by an
elemental distance 2dr as shown in Fig.
Then p 2 = q 2 2dr …(1)
→
p 1
r
2.dr
– +
q 2 q 2
Since, dipoles are separated by a distance r, it means
distance between their centres is r. Distance of
charges (– q 2 ) & (+ q 2 ) 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 p 1 is E, then electric field strength at
position of two charges will be (E – dE) & (E + dE),
respectively.
1 2p1
Where E =
(rightward)
3
4πε
r
0
∴
1 2p1
dE = – 3 . . dr
4
4πε0
r
Force on charge (– q 2 ) is F 1 = (E – dE) q 2 (leftward)
and that on charge (+ q 2 ) is F 2 = (E + dE) q 2
(rightwards)
Hence, net force on second dipole is
F = F 2 – F 1 (rightward)
or
1 2p1
F = dE 2q 2 = – 3 . dr 2q
4 2
4πε0
r
But 2q 2 dr = p 2
∴
1 6p1p2
Net force = –
4
4πε0
r
(–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
1
F =
4πε 6p1p2
4
r
0
1. 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 moon, is completely
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.
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.
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.
XtraEdge for IIT-JEE 21 MAY 2010

XtraEdge for IIT-JEE 22 MAY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
Electrostatics-I
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
• Coulomb's Law :
1 q1q
2
F 0 =
(in vacuum)
2
4πε0 r
Vectorially → 1 q1q
2
F= rˆ
2
4πε
0 r
1 q1q
2
In any material medium F =
2
4πε0εr
r
where ε r is a constant of the material medium called
its relative permittivity, and ε 0 is a universal constant,
called the permittivity of free space.
ε 0 = 8.85 × 10 –12 1
or
4πε = 9 × 109
0
The unit of ε 0 is C 2 N m –2 or farad per metre.
1 q1q
2
Also F =
2
4πε0εr
r
Where ε is called the absolute permittivity of the
medium.
Obviously, ε r = F 0 /F. Remember ε r = ∞ 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
∆ F = ∆q
F or E =
∆q
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
E =
1
4πε
q
2
r
Vectorially → 1 q
E = 4πε
2 rˆ
r
With reference to any origin
→
E =
q
4πε
→
R−
r
→
→
3
→
R−
r
Where R → is the position vector of the field point and
→
r , the position vector of q.
Due to a number of discrete charges
i=
N
→
→
→
q E =
1 R−
ri
∑
3
4πε
→ →
i=
1
R−
r
i
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∞→
(by an external agent)
V P = P
volt(V) or JC –1
∆q
The potential difference between two points P and Q
is given by
∆W Q → P (by agent)
V P – V Q =
volt (V)
∆q
The potential at a distance r from a point charge q in
a medium of permittivity ε is
ϕ or V =
1
4πε
q 1 =
r 4πε
q
→ →
R− r
with reference to any arbitrary origin.
Due to a number of charges
i=
N
1 q1
ϕ or V = ∑
4πε
→ →
i= 1 R−
r
i
XtraEdge for IIT-JEE 23 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 q 1 , q 2
are separated by distance r, the mutual potential
q
energy of the system is 1 q 2
4πε r
.
• 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
E r = – along r
∂ r
∂V and E ⊥ = perpendicular to r
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 × 10 9 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,
dϕ = (E cos θ) ds = E n ds = → E . → ds
ds
O
The total electric flux over a surface is obtained by
summing :
→ →
→ →
ϕ E = ∑ 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/ε 0
where q is the total charge enclosed by the surface.
E
N
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/ε 0 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/2ε 0 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
r
0
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 =
2
4πε0 r
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
XtraEdge for IIT-JEE 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 = 1/4πε 0 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 –6 C and 1nC = 10 –9 C.
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
electric field E r ). Sometimes the field point will
be at some arbitrary position along a line. For
example, you may be asked to find E r at point on
the x-axis.
Step 3 : The solution as follows :
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.
In working out the directions of E r
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
easiest to do this for each component of E r
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.
Step 4 : your answer : Check that the direction of E r
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
we determine the direction of E r from the symmetry
of the charge distribution. If we are given the charge
distribution. we can use Gauss's law to find the the
magnitude of E r . Alternatively, if we are given the
field, we can use Gauss's law to determine the details
XtraEdge for IIT-JEE 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.
ΦE =
∫
E cosφ dA =
∫
E dA =
∫
E
r .dA
r Q
= encl
ε0
(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
∫ 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.
If E r 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
instead E r is perpendicular and inward, then E ⊥ =
– E and
∫ E ⊥dA = – EA.
If E r is tangent to a surface at every point, then E ⊥
= 0 and the integral over that surface is zero.
If E r = 0 at every point on a surface, the integral
is zero.
In the integral
∫ E ⊥dA , 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
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.
1. Supposing that the earth has a charge surface density
of 1 electron/metre 2 , 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 6.4×10 6 metre (ε 0 = 8.9 × 10 –12 coul 2 /nt–m 2 ).
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 R 2 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
1 4πR
2 σ R. σ
= . =
4πε0
R ε0
Substituting the given values
−6
−19
2
(6.4×
10 metre) × ( −1.6×
10 coul / metre )
V =
−12
2 2
(8.9×
10 coul / nt − m )
(ii) E =
Solved Examples
nt − m
= – 0.115 coul
1 q 1
2 =
4πε0
R 4πε . 4πR
σ
2
0 R
−19
joule
= – 0.115 = – 0.115 volt.
coul
−1.6×
10 coul / metre
=
= – 1.8 × 10 –8 nt/coul.
−12
2 2
8.9×
10 coul / nt − m
The negative sign shows that E is radially inward.
2. Determine the electric field strength vector if the
potential of this field depends on x, y co-ordinates as
(a) V = a(x 2 – y 2 ) and (b) V = axy.
Sol. (a) V = a(x 2 – y 2 )
∂V
∂V
Hence, E x = – = – 2ax, E y = – = + 2ay
∂ x ∂ y
∴ E = – 2axi + 2ayj
or E = – 2a(xi – yj)
(b) V = a x y
2
2
=
σ
ε 0
XtraEdge for IIT-JEE 26 MAY 2010

Hence, E x = –
∴
∂V
∂ x
E = – ayi – axj
= – a[yi + xj]
∂V
= –ay, E y = –
∂ y
= – ax
3. 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
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 2
4πr
4πR
(∴ Surface density = charge/area)
∴ q R 2 = q′ r 2
…(2)
From eq. (1) q′ = (Q – q), hence
q R 2 = (Q – q)r 2
q(R 2 + r 2 ) = Q r 2
2
2
Q r
Q R
∴ q = and q′ = Q – q =
2 2
2 2
R + r
R + r
Now potential at O is given by
1 q 1 q'
V = +
4πε0
r 4πε0
r
=
=
1
4πε
0
0
2
Q r 1
+
2 2
(R + r ) r 4πε
Q (r + R)
2
4πε (R r )
2 +
r
0
(R
Q r
2
2
2
+ r ) r
4. S 1 and S 2 are two parallel concentric spheres enclosing
charges q and 2q respectively as shown in fig.
(a) What is the ratio of electric flux through S 1 and S 2 ?
(b) How will the electric flux through the sphere S 1
change, if a medium of dielectric constant 5 is
introduced in the space inside S 1 in place of air ?
S 2
Sol. (a) Let Φ 1 and Φ 2 be the electric flux through spheres
S 1 and S 2 respectively.
q
2q
S 1
∴
q
Φ 1 =
ε
0
q / ε
Φ 1 =
0
Φ 2 3q / ε0
and Φ 2 =
= 3
1
q + 2q 3q
=
ε ε
(b) Let E be the electric field intensity on the surface of
sphere S 1 due to charge q placed inside the sphere.
When dielectric medium of dielectric constant K is
introduced inside sphere S 1 , then electric field
intensity E′ is given by
E′ = E/K
Now the flux Φ′ through S 1 becomes
' 1 q
Φ′ =
∫
E .dS =
∫
E.dS =
K Kε
∴ Φ′ =
q
5ε
0
5. 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
5cm
Φ =
2cm
P
Fig. (a) Fig. (b)
∫
∫
E.dS = E dS = 4π(2×
10 ) E
0
0
0
−2
2
According to Gauss's law, Φ = q/ε 0
∴ 4π × (2 × 10 –2 ) 2 E = q/ε 0
9
−8
q (9×
10 ) × (4×
10 )
or E =
=
−2
2
−4
4πε0 × (2×
10 ) 4×
10
= 9 × 10 5 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.
XtraEdge for IIT-JEE 27 MAY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
1-D Motion, Projectile Motion
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Kinematics :
Velocity (in a particular direction)
=
Displacement(in that direction)
Time taken
( V r AB ) x = V r Ax – V r Bx and ( V r AB ) x =
dx
t
Where dx is the displacement in the x direction in
time t.
Swimmer crossing a river
d
v s
θ
v s cosθ
v s sinθ
d
Time taken to cross the river =
Vs
cosθ
For minimum time, θ should be zero.
x
v s
v r
in this case resultant velocity
2 2 d
V R = V s + vr
and t = .
v s
Also x = v r × t
For reaching a point just opposite the horizontal
component of velocity should be zero.
v sinθ = V r
| Displacement |
|Average Velocity| =
time
a r v
r – u
r r r
=
t
= v + (–u)
t
2
2
v + u – 2uvcosθ
⇒ |a| =
t
Where θ is the angle between v and u.
The direction of acceleration is along the resultant of
v r and (– u r ).
v r
v R
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)]
Y
m = tanθ
θ is acute and
m is positive
Y
m = tanθ
θ is abtuse and
m is positive
θ
θ
X
X
(i) fig.(a) (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
Y
c is negative
m is '+' ve
m is '+' ve
c
(i)
Y
X
fig(b)
c
(ii)
c is positive
m is negative
(ii) X
fig(c)
For y = kx 2 , where k is a constant, we get parabola
[see fig (d)]
Y
Parabola
Fig.(d)
x 2 + y 2 = r 2 is equation of a circle with centre at
origin and radius r.
X
X
XtraEdge for IIT-JEE 28 MAY 2010

For (x – a) 2 + (y – b) 2 = r 2 , the motion is in a circular
path with centre at (a, b) and radius r
2
2
x y
2 +
2 = 1 is equation of an ellipse
a b
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)
s = f(t)
Differentiate w.r.t
time
v = f(t)
integrate w.r.t.
time
Differentiate w.r.t
time
a = f(t)
integrate w.r.t.
time
S stand for displacement
2
dv dv d s
a = v = = ds dt 2
dt
dx dy
Also v x = ⇒ vy = dt dt
2
dv
a x = x d x dv 2 y d y
= and ay = =
dt 2
dt
dt 2
dt
The same concept can be applied for z-co-ordintae.
Projectile motion :
P
ucosθ
u ucosθ
usinθ ucosθ
θ
ucosθ
ucosθ
Q θ
u
usinθ
g
g F
F
g
F
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
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
v r = u r + at ; s r 1 r
= ut + a t
2
; v r . v r – u r . u = 2 a r s r
2
s = 2
1 ( u
r + v
r )t
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
dυ
a x = x d ⎛ dx ⎞ d x
= ⎜ ⎟ =
dt dt
2
⎝ dt ⎠ 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 x 0 = 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 v x and a x 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 v x has the specified value)? "where is the
motorcyclist when his velocity is 25m/s?"
2
XtraEdge for IIT-JEE 29 MAY 2010

Translated into symbols, this becomes "What is
the value of x when v x = 25 m/s?"
Make a list of quantities such as x, x 0 ,v x ,v 0x ,a x 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 v 0x = 0.
Step 3 : Execute the solution :
Choose an equation from Equation v x = v 0x + a x t
x = x 0 + v 0x t + 2
1
ax t 2
2
v x =
(constant acceleration only)
2
v 0x
+ 2a x (x – x 0 ) (constant accelerations only)
⎛ v0x
+ vx
⎞
x – x 0 = ⎜ ⎟ 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?
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.
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 x 0 = 0 and y 0 = 0, and the
components of the (constant) acceleration are a x = 0,
a y = – g.
List the unknown and known quantities, and
decide which unknowns are your target variables.
In some problems you'll be given the initial
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
x = (v 0 cosα 0 )t (projectile motion) through ...(1)
v y = v 0 sin α 0 – gt (projectile motion) ...(2)
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 v x or
v y has the specified value ?) At the highest point
in a trajectory, v y = 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 = y 0 ?"
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)
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 : 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 v C/B .
Step 3 : Execute the solution : Solve for the target
variable using equation
v P/A = v P/B + v B/A (relative velocity along a line) ...(1)
XtraEdge for IIT-JEE 30 MAY 2010

(If the velocities are not along the same direction, (As the ball returns to its initial position, the change
→
(d) horizontal range of the projectile
→
Since s 1 = s 2 = d and s net = s + s | = 0
you'll need to use the vector from of this equations,
derived later in this section.) It's important to note the
in position, the change in position vector of the ball,
that is the net displacement will be zero).
order of the double subscripts in equation (1) v A/B
→
always means "velocity of A relative to B." These ∴ | VaV
| = 0.
subscripts obey an interesting kind of algebra, as
equation (1) shown. If regard each one as a fraction,
2. A long belt is moving horizontally with a speed of 4
then the fraction on the left side is the product of the
Km/hour. A child runs on this belt to and fro with a
fractions on the right sides : P/A = (P/B) (B/A). This
speed of 9 Km/hour (with respect to the belt) between
is a handy rule you can use when applying Equation
his father and mother located 50 m apart on the
(1) to any number of frames of reference. For
moving belt. For an observer on a stationary platform
example, if there are three different frames of
outside, what is the
reference A, B, and C, we can write immediately. (a) speed of the child running in the direction of motion
v P/A = v P/C + v C/B + V B/A
of the belt,
Step 4 : Evaluate your answer : Be on the lookout for
stray minus signs in your answer. If the target
(b) speed of the child running opposite to the direction of
motion of the belt and
variable is the velocity of a car relative to a bus
(v V/B ), make sure that you haven't accidentally
calculated the velocity of the bus relative of the car
(c) time taken by the child in case (a) and (b) ?
Which of the answers change, if motion is viewed by
one of the parents ?
(v B/C ). If you have made this mistake, you can
recover using equation.
v A/B = – v B/A
Sol. Let us consider positive direction of x-axis from left
to right
(a) Here, v B = + 4 Km/hour
Speed of child w.r.t. belt, v C = = 9 Km/hour
Solved Examples
∴ Speed of child w.r.t. stationary observer,
v C ′ = v C + v B or v C ′ = 9 + 4 = 13 Km/hour
1. 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.
(b) Here, v B = + 4 Km/hour, v C = – 9 Km/hour
∴ Speed of child w.r.t. stationary observer,
v C ′ = v C + v B or v C ′ = – 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.
A V 0.5V B
(c) 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
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
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),
from B to A with a constant speed V/2.
0.50 km
∴ V 1 = V and V 2 = V/2
t = = 20 sec.
9 km / hour
d1
+ d 2
Using the formula, V aV =
If the motion is observed by one of parents, answer to
(d1
/ V 1)
+ (d 2 / V2
)
case (a) case (b) gets altred. It is because the speed
Putting d 1 = d 2 = d; V 1 = V and V 2 = V/2
of the child w.r.t. either of mother or father is
d1
+ d 2 2V
9 Km/hour.
We obtain, V aV =
=
(dV) + (d / 0.5V) 3
3. A particle is projected with velocity v
From the formula,
0 = 100 m/s at
an angle θ = 30º with the horizontal. Find :
→ → →
→
| s1+
s2
| | snet
| (a) velocity of the particle after 2 sec.
average velocity = VaV
= =
s1
s2
t
(b) angle between initial velocity and the velocity after 2 sec.
+
net
V1
V2
(c) the maximum height reached by the projectile
| 1 2
XtraEdge for IIT-JEE 31 MAY 2010

Sol. (a)
(b)
(c)
→
v
t
→
= v
xt
→
î + v
yt
ĵ
where î and ĵ are the unit vectors along +ve x and
+ve y-axis respectively
→
t
v =(u x + a x t) î + (u y + a y t) ĵ
Here, u x = v 0 cos θ = 50 3 m/s, a x = 0
u y = v 0 sin θ = 50 m/s, a y = – g
(Q g acts downwards)
→
t
v = 50
3 î + (50 – 10 × 2) ĵ
=[50 3 î + 30 ĵ ] m/s
∴ | → 2 2
v 2 | = (vx + v y ) = 2 2
( 50 3) + (30)
∴
→
0
→
2
→
0
v = 50
v = 50
3 î + 50 ĵ
3 î + 30 ĵ
v . v → 2 = 7500 + 1500 = 9000
If α is the angle between v → 0 and v
→ 2
→
v0
.v2
9000
Then, cos α = =
→ →
100×
91.65
| v0
| × | v2
|
α = cos –1 (0.98) = 10.8º
v 2 y – u 2 y = 2a y y
At y = y max , v y = 0
∴ 0 – v 2 0 sin 2 θ = 2 (–g)y max
2 2
v0 sin θ
∴ y max = = 125 m
2g
→
u 2 sin 2θ (d) R = = 1732 m
g
4. 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
Sol. Just before impact magnitude of velocity of the ball,
v = ( 2gh)
α
α
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)t 2
and y = (v cos α) t – ½ (g cos α)t 2
When the ball hits the plane for a second time,
y = 0, (v cos α)t – ½(g cos α)t 2 or t = (2v/g)
Putting this value of t in x,
4v
2 sin α
x = = 8h sin α
g
5. 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/s 2
v0
2 sin 2θ
Sol. R(range) =
g
or,
1.22m
2 Rg
v 0 = = Rg as θ = 45º
sin 2θ
A
45º
v 0
106.68m
or, v 0 = ( Rg)
…(1)
Equation of trajectory
or, y = x –
y = x tan 45º –
2v
gx 2
2Rg.½
2
0
gx
2
cos
2
gx 2
= x – Rg
B
45º
Putting x = 97.53, we get
2
10×
(97.53)
y = 97.53 –
= 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.
XtraEdge for IIT-JEE 32 MAY 2010

KEY CONCEPT
Physical
Chemistry
Fundamentals
GASEOUS STATE
& REAL GASES
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
p
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 V m to
the ideal molar volume V m,ideal (= RT/p) as a function
of pressure at constant temperature. This ratio is
called the compressibility factor Z and can be
expressed as
Z =
V
V
m
m,ideal
p
= RT
Vm
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 H 2 .
(2) For N 2 , 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
H 2
increases continuously with pressure in the higher
pressure region.
(3) For CO 2 , 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.
1.0
t = 0ºC
H 2
N 2
CH 4
ideal gas
CO 2
Z
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 CO 2 , but when far away, the curves are
like those for H 2 (below fig.)
Z
1.0
T 1 >T 2 >T 3 >T 4
ideal gas
0 200 400 600
p/101.325 kPa
T 4
T 3
T 2
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:
T 1
XtraEdge for IIT-JEE 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
p i V i = nRT. His corrections are given below.
Correction for volume :
V i 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 V i 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 V i is
given by
V i = V – nb ...(1)
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.
excluded
volume
2r
Excluded volume per pair of molecules
Thus, excluded volume per pair of molecules
4
= π(2r) 3 ⎛ 4 ⎞
= 8 ⎜ πr
3 ⎟ 3 ⎝ 3 ⎠
Excluded volume per molecule
1 ⎡ ⎛ 4 ⎞⎤
= ⎢ ⎜ π 3 ⎛ 4 ⎞
8 r ⎟⎥ = 4 ⎜ πr
3 ⎟
2 ⎣ ⎝ 3 ⎠ ⎦ ⎝ 3 ⎠
= 4 (volume occupied by a molecule)
Since b represents excluded volume per mole of the
gas, it is obvious that
⎡ ⎛ 4 ⎞⎤
b = N A ⎢4⎜
πr
3 ⎟⎥ ⎣ ⎝ 3 ⎠ ⎦
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
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.
p i = p + correction term ...(2)
This correction term depends upon two factors:
(i) 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
nN
N' = A n
or N' ∝
V
V
Thus, the correction term is given as :
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:
Correction term ∝ n / V
...(2b)
Taking both these factors together, we have
B
XtraEdge for IIT-JEE 34 MAY 2010

⎛ n ⎞ ⎛ n ⎞
Correction term ∝ ⎜ ⎟ ⎜ ⎟
⎝ V ⎠ ⎝ V ⎠
2
n
or Correction term = a
...( 3)
2
V
Where a is the proportionality constant and is a
measure of the forces of attraction between the
molecules. Thus
2
n
p i = p + a
...(4)
2
V
The unit of the term an 2 /V 2 will be the same as that of
the pressure. Thus, the SI unit of a will be Pa m 6 mol –2 .
It may be conveniently expressed in kPa dm 6 mol –2 .
When the expressions as given by Eqs (1) and (4) are
substituted in the ideal gas equation p i V i = nRT, we
get
⎛
2
⎞
⎜
n a
p + ⎟ (V – nb) = nRT ...(5)
2
⎝ V ⎠
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
a
Gas 6 2
kPa dm mol –
H 2
He
N 2
O 2
Cl 2
NO
NO 2
H 2 O
CH 4
C 2 H 6
C 3 H 8
C 4 H 10 (n)
C 4 H 10 (iso)
C 5 H 12 (n)
CO
CO 2
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
b
dm
3 mol –1
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
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
⎛ ⎞
⎜
a
p + ⎟ (V 2
m – b) = RT ..(i)
⎝ V m ⎠
At low pressure When pressure is low, the volume is
sufficiently large and b can be ignored in comparison
to V m in Eq. (i). Thus, we have
⎛ ⎞
⎜
a
⎟
a
p + V 2
m = RT or pV m + =RT
⎝ V m ⎠
V m
a
or Z = 1 –
...(ii)
VmRT
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/V m RT) increase as V is inversely proportional to p.
Consequently, Z decreases with increase of p.
At high pressure When p is large , V m will be small
and one cannot ignore b in comparison to V m .
2
However, the term a / V m may be considered
negligible in comparison to p in Eq. (i) Thus,
pb
p(V m – b) = RT or Z = 1 + ...(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, V m will also be sufficiently large
2
and thus the term a / V m will be negligibly small. At
this stage, b may also be negligible in comparison to
V m . Under these conditions, Eq. (i) reduces to an
ideal gas equation of state:
pV m = 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(V m – b) = RT,
obtained from the van der Waals equation by
2
ignoring the term a / V m . 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.
XtraEdge for IIT-JEE 35 MAY 2010

XtraEdge for IIT-JEE 36 MAY 2010

KEY CONCEPT
Organic
Chemistry
Fundamentals
GENERAL ORGANIC
CHEMISTRY
Stability of different types of carbocations in
decreasing order :
⊕
C
⊕
> CH
⊕
⊕ ⊕
(Ph) 3 C > (Ph)2 CH > Ph – C H2 ≥
⊕ ⊕ ⊕
CH 2 = CH – CH 2 ≥ R – C – R > R – CH – R
>
R
⊕ ⊕
> R – CH 2 > CH 2 = CH
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
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 :
H 2 O < NH 3
CH 3 OH ≈ H 2 O < CH 3 CO Θ 2 < CH 3 O Θ ≈ OH Θ
⇒
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
R O <
Θ
R S
R 3 N < R 3 P
C
H
H
⊕
>
Θ
F < Θ Cl < Θ Br < Θ I
increasing nucleophilic strength
decreasing base strength
⇒ Steric bulk decreases nucleophilicity. For
example :
CH 3
Θ
H 3 C – C – O < HO Θ
CH 3
weaker nucleophile
Stronger base
stronger nucleophile
weaker base
Leaving Groups :
A good leaving groups is the one which becomes a
stable ion after its departure. As most leaving groups
leave as a negative ion, the good leaving groups are
those ions which stabilize this negative charge most
effectively. The weak bases do this best, thus the best
groups are weak bases. If a group is a weak base i.e.,
the conjugate base of a strong acid, it will generally
be a good leaving group. In an S N 2 reaction the
leaving group begins to gain negative charge as the
transition state is reached. The more the negative
charge is stabilized, the lower is the energy of the
transition state; this lowers the energy of activation
and thereby increases the rate of reaction.
The acids HCl, HBr, HI and H 2 SO 4 are all strong
acids since the anions Cl – , Br – , I – and HSO – 4 are
stable anions these anions (weak bases) are also good
leaving groups in S N 2 reactions. Of the halogens, an
iodide ion is the best leaving group and the fluoride
ion is the poorest :
I – > Br – > Cl – > F –
The order of basicity is opposite : F – > Cl – > Br – > I – ,
the reason that alkyl fluorides are ineffective
substrates in S N 2 reactions is related, to the relatively
low acidity of HF (pK a = 3). Sulfonic acids, R
SO 2 OH are similar to sulfuric acid in acidity and the
sulfonate ion RSO – 3 is a very good leaving group.
Alky benzenesulfonates, alkyl p-toluenesulfonates
are therefore, very good substrates in S N 2 reactions.
The triflate ion (CF 3 SO – 3 ) is one of the best leaving
groups known, it is the anion of CF 3 SO 3 H which is a
strong acid much stronger than sulfuric acid.
XtraEdge for IIT-JEE 37 MAY 2010

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 k H /k D , 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 k H /k D was found to be around 7.
CH 3 COCH 3 + Br 2 ⎯→ CH 3 COCH 2 Br
rate-determining step
Bromoacetone
OH
CH 3 COCH 3 CH 3 C = CH 2
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 k H /k D are
generally between 0.6 and 2.0.
(CZ 3 ) 2 CHBr + H 2 O → (CZ 3 ) 2 CHOH + HBr
the solvolysis of isopropyl bromide where Z = H or D, k H /k D is
1.34 Secondary isotope effect.
The substitution of tritium for hydrogen gives isotope
effects which are numerically larger (k H /k T = 16).
E2 elimination like S N 2 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
Base
– C – CH 2 Br – CH = CH 2
H
1-Bromo-2-phenylethane (I)
D
Faster reaction
Base
– C – CH 2 Br – CD = CH 2
D
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
S N 2
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.
S N 2 and E2
Secondary or primary substrate
Strong unhindered
base/nucleophile leads to S N 2
Strong hindered
base/nucleophile leads to E2
Low temperatrue (S N 2)/high
temperature (E2)
S N 1 and E 1
Tertiary substrate
Carbocation intermediate
Weak nucleophile/base (e.g.,
solvent)
Unimolecular in ratedetermining
step
Racemization if S N 1
Removal of β-hydrogen if E1
Protic solvent assists ionization
of LG
Low temperature (S N 1)/high
temperature (E2)
E2
Tertiary or secondary substrate
Concerted anti-coplanar TS
Bimolecular in ratedetermining
step
Strong hindered base
High temperature
XtraEdge for IIT-JEE 38 MAY 2010

UNDERSTANDING
Physical Chemistry
1. What is the solubility of AgCl in 0.20 M NH 3 ?
Given : K sp (AgCl) = 1.7 × 10 –10 M 2
K 1 = [Ag(NH 3 ) + ] / [Ag + ] [NH 3 ] = 2.33 × 10 3 M –1 and
K 2 = [Ag(NH 3 ) + 2 ]/[Ag(NH 3 ) + ][NH 3 ] = 7.14 × 10 3 M –1
Sol. If x be the concentration of AgCl in the solution, then
[Cl – ] = x
From the K sp for AgCl, we derive
K −10 2
[Ag + sp 1.7×
10 M
] = =
−
[Cl ] x
If we assume that the majority of the dissolved Ag +
goes into solution as Ag(NH 3 ) + 2 then [Ag(NH 3 ) + 2 ] = x
Since two molecules of NH 3 are required for every
Ag(NH 3 ) + 2 ion formed, we have [NH 3 ] = 0.20 M – 2x
Therefore,
⎛
⎜
.7× 10
−
+ 2
[Ag ][NH ]
x
3
K inst =
=
⎝
+
[Ag(NH3)
2 ]
= 6.0 × 10 –8 M 2
From which we derive
2
2
M ⎞
⎟
(0.20M − 2x)
⎠
x
10
1 2
−8
( 0.20M − 2x) 6.0×
10 M
=
= 3.5 × 10 2
2
−10
2
x 1.7×
10 M
which gives x = [Ag(NH 3 ) + 2 ] = 9.6 × 10 –3 M, which
is the solubility of AgCl in 0.20 M NH 3
2. The values of Λ ∞ for HCl, NaCl and NaAc (sodium
acetate) are 420, 126 and 91 Ω –1 cm 2 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 cm 2 mol –1 ,
∞
Λ m (NaCl) = 126 Ω –1 cm 2 mol –1 ,
and Λ m (NaAc) = 91 Ω –1 cm 2 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
G = – = 0.00627 Ω –1
122 Ω 520 Ω
Conductivity of 0.1 M NaCl solution
k = Λ m c = (126 Ω –1 cm 2 mol –1 )(0.1 mol dm –3 )
= 12.6 Ω –1 cm 2 dm –3 = 12.6 Ω –1 cm 2 (10 cm) –3
= 0.0126 Ω –1 cm –1
Cell constant,
2
−1
−1
k (0.0126 Ω cm )
K = = = 2.01 cm –1
G
−1
(0.00627Ω
)
Conductivity of 0.1 M HAc solution
−1
K 2.01 cm
k = = R 520 Ω
Molar conductivity of 0.1 M HAc solution
−1
−1
k (2.01/ 520) Ω cm
Λ m (HAc) = = c
−3
(0.1 mol dm )
= 0.038 65 Ω –1 cm –1 dm 3 mol –1
= 38.65 Ω –1 cm 2 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 cm 2 mol –1
= 385 Ω –1 cm 2 mol –1
Therefore, the degree of dissociation of acetic acid is
given as
α =
Λ
Λ
m =
∞
m
(38.65 Ω
(385Ω
−1
−1
cm
cm
2
2
mol
mol
−1
−1
)
)
≈ 0.1
and the hydrogen-ion concentration of 0.1 M HAc
solution is
[H + ] = cα = (0.1 M)(0.1) = 0.01 M
Thus, its pH is pH = – log{[H + ]/M} = – log(0.01) = 2
3. Potassium alum is KA1(SO 4 ) 2 .12H 2 O. As a strong
electrolyte, it is considered to be 100% dissociated
into K + , Al 3+ , and SO 2– 4 . The solution is acidic
because of the hydrolysis of Al 3+ , but not so acidic as
might be expected, because the SO 2– 4 can sponge up
some of the H 3 O + by forming HSO – 4 . Given a
solution made by dissolving 11.4 g of
KA1(SO 4 ) 2 .12H 2 O in enough water to make 0.10 dm 3
of solution, calculate its [H 3 O + ] :
(a) Considering the hydrolysis
Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +
with K h = 1.4 × 10 –5 M
(b) Allowing also for the equilibrium
HSO – 4 + H 2 O H 3 O + 2–
+ SO 4
with K 2 = 1.26 × 10 –2 M
11.4 g
Sol. (a) Amount of alum =
= 0.024 mol
−1
474.38 g mol
0.024 mol
Molarity of the prepared solution =
3
0.1dm
= 0.24 M
Hydrolysis of Al 3+ is
XtraEdge for IIT-JEE 39 MAY 2010

Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +
2+
+
[Al(OH) ][H3O
]
K h =
3+
[Al ]
If x is the concentration of Al 3+ that has hydrolyzed,
we have
(x)(x)
K h =
0.24M − x
= 1.4 × 10 –5 M
Solving for x, we get
[H 3 O + ] = x = 1.82 × 10 –3 M
(b) We will have to consider the following equilibria.
Al 3+ + 2H 2 O Al(OH) 2+ + H 3 O +
H 3 O + 2–
+ SO 4 HSO – 4 + H 2 O
Let z be the concentration of SO 2– 4 that combines
with H 3 O + and y be the net concentration of H 3 O +
that is present in the solution. Since the concentration
z of SO 2–
4 combines with the concentration z of
H 3 O + , it is obvious that the net concentration of H 3 O +
produced in the hydrolysis reaction of Al 3+ is (y + z).
Thus, the concentration (y + z) of Al 3+ out of 0.24 M
hydrolyzes in the solution. With these, the
concentrations of various species in the solution are
3+
2
Al + 2H 2 O Al(OH)
+ 3 O 0.24 M−y−z
y+
z
y
H 3 O 2−
+ SO 4 HSO −
4 + H 2 O
y 0.48 M−z
z
(y + z)(y)
Thus, K h =
= 1.4 × 10 –5 M ...(i)
(0.24M − y − z)
z
1
K 2 =
=
...(ii)
−2
y(0.48M − z) 1.26×
10 M
From Eq. (ii), we get
(0.48M)y
z =
2
(1.26× 10
− 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 −
−2
⎝ (1.26×
10 M) + y ⎠
Making an assumption that y

`tà{xÅtà|vtÄ V{tÄÄxÇzxá
1 Set
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in mathematics that would be very helpful in facing
IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and
we hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Shailendra Maheshwari
Solutions will be published in next issue
Joint Director Academics, Career Point, Kota
Passage :
A bag contains ‘n’ cards marked 1, 2, 3, ......, n. ‘X’
draws a card from the bag and the card is put back
into the bag. Then ‘Y’ draws a card. The probability
that ‘X’ draws.
1. The same card as ‘Y’ is –
1
1
(A) (B) n 2n
(C)
1
2
n
(D) n
2
2. a higher card than ‘Y’ is –
(A)
(C)
n −1
n
n −1
2
n
(B)
(D)
3. a lower card than ‘Y’ is –
n −1
(A)
(B)
n
(C)
n −1
2
n
1
0
1
0
∫ x
4. Evaluate : 228
∫ x
25
24
(d)
(1 − x)
(1 − x)
5. Find the minimum value of
⎛
2
(x 1 – x 2 ) 2 + ⎜
x
− (17 − x
⎝ 20
n −1
2n
n −1
2
2n
n −1
2n
n − 1
2
2n
50
49
dx
dx
= ?
1
2)(x
2 −13
)
where x 1 ∈ R + and x 2 ∈ (13, 17).
⎞
⎟
⎠
2
6. Let f(x) = a 1 tan x + a 2 tan 2
x + a3 tan 3
x + ...... + an
tan n
x , where a1 , a 2 , a 3 , ... a n ∈ R and n ∈ N. If | f(x) |
n
≤ | tan x | for ∀ x ∈ ⎛ π π ⎞
a
⎜−
, ⎟ , Prove that ∑
⎝ 2 2 ⎠ i=
1
i
i
≤ 1
7. Let az 2 + bz + c be a polynomial with complex
coefficients such that a and b are non zero. Prove that
the zeros of this polynomial lie in the region.
b c
| z | ≤ + + a b
8. Find the fifth degree polynomial which leaves
remainder 1 when divided by (x – 1) 3 and remainder
–1 when divided by (x + 1) 3 .
9. A quadrilateral ABCD is inscribed in a circle of
radius R such that AB 2 + CD 2 = 4R 2 . Using vector
method prove that its diagonals are at right angle.
10. Through a focus of an ellipse two chords are drawn
and a conic is described to pass through their
extremities, and also through the centre of the ellipse.
Prove that it cuts the major axis in another fixed
point.
High Speed Avalanches
Although an avalanche can mean the
fall of any material e.g. snow, soil and
even rocks, in common usage it
generally refers to a falling mass of
ice and snow which breaks away
from the side of a mountain or cliff
and surges down at great speed.
XtraEdge for IIT-JEE 41 MAY 2010

θ 2
θ1
MATHS
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. An ellipse of eccentricity 2/3 is inscribed in an ellipse
of equal eccentricity and area equals to 9 square units
in such a way that both the ellipses touch each other
at one end of their common major axis. If length of
major axis of smaller ellipse is equal to length of
minor axis of bigger ellipse, find the area of the
bigger ellipse outside the smaller ellipse.
Sol. The required figure will be drawn as follows
y
2. Given a point P on the circumference of the circle
|z| = 1, and vertices A 1 , A 2 , ......, A n of an inscribed
regular polygon of n sides. Prove using complex
numbers that
(PA 1 ) 2 + (PA 2 ) 2 + ......... + (PA n ) 2 is a constant.
Sol. Without loss of generality we can take P as
1 + 0i.
i.e., P ≡ C is 0
A 3
A 2
A 1
x
θ n
P
A n
and we can redraw the figure for our purpose (i.e.
keeping the area out side the smaller ellipse and
inside the bigger ellipse same) as
(b, 0)
(c, 0)
(–a, 0) (–b, 0) (–c, 0)
(–b, 0)
Therefore, we can let the ellipses be
2
2
2
y
(b, 0) (a, 0)
x
2 2
x
2 +
2
x y
and +
2 = 1
b c
Required area = π ab – π bc
= π b (a – c)
Now b 2 = a 2 (1 – e 2 ) and c 2 = b 2 (1 – e 2 ) (1 – e 2 ) 2
⇒ c = a (1 – e 2 )
⇒ a – c = ae 2
Thus required area = πb (ae 2 )
= πabe 2
2
⎛ 2 ⎞
= 9 × ⎜ ⎟⎠ = 4 sq. units.
⎝ 3
a
y = 1
b
Let A r ≡ C is θ r , r = 1, 2, ......, n.
PA r = |Cis θ r – Cis 0| = |(cosθ r – 1) + i(sinθ r )|
PA 2 r = (cos θ r – 1) 2 + (sinθ r ) 2
= 2 – 2cos θ r
n
⇒ ∑
2
( PA r ) = 2n – 2
r=
1
∑ cos θr
r=
1
n
⎡
n
⎤
Now, ∑ cos θr
= Re ⎢∑Cis
θr
⎥
r=
1
⎢⎣
r=
1 ⎥⎦
iθ1 iθ2
iθn
= Re [ e + e + ....... + e ]
⎡ ⎛
n
⎞⎤
⎢ ⎜ ⎛
2π
i ⎞
iθ
⎟⎥
⎢
e ⎜ ⎟
1
⎜1−
e
n
⎟⎥
⎢
⎜ ⎟
⎥
= Re
⎝ ⎝ ⎠ ⎠
⎢
2π
⎥
⎢
i ⎥
⎢ 1−
e
n
⎥
⎢
⎥
⎣
⎦
Q θ 2 – θ 1 = θ 3 – θ 2 = ..... = θ n – θ n–1 =
⎡
iθ
⎤
= Re
⎢e 1 (1 −1)
⎥
= 0
⎢ 2π
⎥
i
⎢⎣
1−
e
n ⎥⎦
n
Hence, ∑
=
r 1
2
( PA r ) = 2n = constant.
n
2π
n
XtraEdge for IIT-JEE 42 MAY 2010

3. In a class of 20 students, the probability that exactly x
students pass the examination is directly proportional
to x 2 (0 ≤ x ≤ 20). Find out the probability that a
student selected at random has passed the
examination. If a selected students has been found to
pass the examination find out the probability that
he/she is only student to have passed the
examination.
Sol. Let E x : event that exactly x out of 20 students
pass the examination
and A : event that a particular student passes
the examination
⇒ P(E x ) = kx 2 (k is the proportionality constant)
Now, E 0 , E 2 , ....., E 20 are mutually exclusive and
exhaustive events.
⇒ P(E 0 ) + P(E 1 ) + P(E 2 ) + ... + P(E 20 ) = 1
⇒ 0 + k(1) 2 + k(2) 2 + .... + k(20) 2 = 1
⎡ (20)(20 + 1)(40 + 1) ⎤
⇒ k ⎢
⎥ = 1
⎣ 6 ⎦
1
⇒ k = 2870
20
Now, P(A) = ∑
=
20
= ∑
=
x
0
x
0
P (E
2 x
kx . = 20
x ).P(A / E x )
k
20
20
∑
x=
0
1 ⎡ 20(20 + 1) ⎤ 63
=
20 2870
⎢
2
⎥ =
× ⎣ ⎦ 82
P(E1
).P(A / E1)
and P(E 1 /A) =
P(A)
1 1
(1)
2 .
=
2870 20 1
=
63 44100
82
4. Find the set of values of ‘a’ for which minimum
value of x 3 – 6ax 2 + 9a 2 x + 7, x ∈ [–1, 2] is 3.
Sol. Let f(x) = x 3 – 6ax 2 + 9a 2 x + 7
a ≠ 0, otherwise f(x) = x 3 + 7, which is always
increasing and hence min f = f(–1) = 6 ≠ 3.
Now f´(x) = 3x 2 – 12ax + 9a 2 = 0 for stationary points
⇒ x = a, 3a
CaseI : a > 0
⇒ –1 is always in the left of a.
Case I. (a) : 2 ≤ a, then
3 = min f = f(–1) = –1 – 6a – 9a 2 + 7
⇒ 3a 2 + 2a – 1 = 0, no admissible value of a is
obtained.
(a,f(a))
x
2
3
(3a,f(3a))
Case I. (b) : –1 < a < 2 < 3a
i.e., 3
2 < a < 2, then
3 = min f = min{f(–1), f(2)}
= min {–1 –6a – 9a 2 + 7, 8 – 24 a + 18a 2 + 7}
= –1 – 6a – 9a 2 + 7
as – 1 – 6a – 9a 2 + 7 < 8 – 24a + 18a 2 + 7
i.e., 3a 2 – 2a + 1 > 0, which is true
Hence 3 = –1 – 6a – 9a 2 + 7
⇒ a = –1 or 3
1 , none of which is possible.
Case I(c) : 3a ≤ 2
⇒ 3 = min f = min{f(–1), f(3a)}
= {–1 – 6a –9a 2 + 7, 18a 3 + 7}
= –1 – 6a – 9a 2 + 7,
as 18a 3 + 77 – 1 – 6a – 9a 2 + 7
i.e., 18a 3 + 9a 2 + 6a + 170
which is true as a > 0. Hence a = – 1 or 3
1 ,
in which a = 3
1 is permissible.
Case II : a < 0
⇒ 2 is always in the right of a
Case II (a) a ≤ –1
⇒ 3 = min f = f(–1)
⇒ a = –1, as a = 3
1
Hence a = –1 is one possibility
(3a,f(3a))
(a,f(a))
Case II (b) 3a ≤ –1 < a
1
⇒ –1, as a = 3
⇒ 3 = min f = f(a) = a 3 – 6a 3 + 9a 3 + 7
⇒ 4a 3 = – 4 ⇒ a = – 1, not possible
Case II(c) –1 < 3a ⇒ 3 = min f = min {f(–1), f(a) }
= min {–1 –6a – 9a 2 + 7, 4a 3 + 7}
= 4a 3 + 7,
as 4a 3 + 7 < –1 – 6a – 9a 2 + 7
as (a + 1) 2 (4a + 1) < 0. Hence a = –1, not possible
Hence a = –1 or a = 1/3
5. A man standing at a distance 5m in front of the base
of a building 10m high on which a flagstaff is
mounted observes that the top of the building and the
top of a mountain behind the building are along the
same straight line. When he recedes by a distance of
48 m he observes that now the top of the flagstaff and
the top of the mountain are along the same straight
line. If at both the locations, the flagstaff subtends the
same angle at the man’s eye, find the height of
mountain.
XtraEdge for IIT-JEE 43 MAY 2010

Sol. CD : Flagstaf
DE : Building
KF : Mountain (height = h say)
The figure illustrates the situation.
Since, ∠CBD = ∠CAD = α say, points A, B, C and
D are concyclic.
⇒ ∠ABD = ∠ACD = 90º – (α + β)
⇒ ∠ABC = 90º – (α + β) + α = 90º – β = ∠KCH
K
90º – β
α 10
α
90º – β
β
B 48
Now, h = KH + HF
A 5 E
F
= (CH) tan (90º – β) + (BE) tan(90º – β)
(Q HF = CE)
= [DG + (BA + AE) cot β
= [KG cot β + (48 + 5)] cot β
⇒ h = [(h – 10)cotβ + 53] cot β
(Q KG = KF – GF)
5
Putting cot β = 10
1 = , we get 2
⎛ h −10
⎞ 1
h = ⎜ + 53⎟ ⎝ 2 ⎠ 2
⇒ 4h = h – 10 + 106
⇒ 3h = 96
⇒ h = 32 m
6. If a, b, c and n are positive integers such that
a + b + c = n, show that
(a a b b c c ) 1/n + (a b b c c a ) 1/n + (a c b a c b ) 1/n ≤ n.
Sol. Since a, b, c are integers, from A.M. – G.M.
inequality we can write
(a + a + ....a times) + (b + b + ...b times) + (c + c + ...c times)
a + b + c
1/a +b+c
≥ [(a.a....a times)(b.b....b times)(c.c...c times)]
1
a.a + b.b + c.c a b c
⇒
≥ ( a b c ) a+
b+
c
a + b + c
1
c.a + a.b + b.c c a c
Similarly,
≥ ( a b c ) a+
b+
c
c + a + b
1
b.a + c.b + a.c b c a
and
≥ ( a b c ) b+
c+
a
b + c + a
Adding these three inequalities, we get
2 2 2
a + b + c + 2ab + 2bc + 2ca
≥ (a a b b c c ) 1/n
a + b + c
+ (a c b a c b ) 1/n + (a b b c c a ) 1/n
2
(a + b + c)
where LHS =
= a + b + c Hence proved
a + b + c
C
D
β
H
G
How do Satellites
Stay Up?
Satellites orbit the earth because of the force of
gravity. To understand why this happens and why the
satellite does not get pulled in and fall, we have to
understand what forces do. A force will change the motion
of an object; it might speed it up, slow it down or change
its direction. For example, if you are running and someone
pushes you from behind, you speed up (the force is in the
direction of your motion). But if someone pushes you in
the chest when you are running, you slow down (the force
is in the opposite direction to your motion). If you are
running and someone pushes you from the side, you move
away from them, changing your direction. (the force is at
right angles to the motion). This idea is called Newton’s
First Law.
To make something move in a circle it must be
moving and have a force that is always at right angles to the
motion so that it constantly changes direction. This force is
called the centripetal force.
Imagine swinging a rock on a string around your
head. The tension in the string pulls the rock round in a
circle (this is the force at right angles to the motion). So
the tension is the centripetal force. If we cut the string, the
rock will continue in a straight line because there is no
longer a force to change its direction. For a satellite, the
centripetal force is the gravitational force, the pull of the
earth. If we could switch gravity off, we would lose all our
satellites as they move off in straight lines!
Going back to the rock example, we need to put
energy into keeping the rock moving because the rock is
moving through air and is losing energy constantly because
of air resistance. We don’t need to do this with satellites
because they are moving through space where there is no
air, so no air resistance acts on the satellites and they don’t
slow down.
Many people think there is a centrifugal force
acting which pulls the satellite (or rock) outwards. This is
not the case; there is no such thing as a centrifugal force.
Imagine riding in the back seat of a car as it turns the
corner. Let us assume the back seat is very slippery and
you don’t have a seatbelt on. As the car turns the corner,
you slide from the inside to the outside. You could argue
that a force is pushing you outwards. In fact the reason you
move outwards is that there is no force keeping you
moving in the same circle as the car. What you really do is
continue in a straight line. Eventually you will hit the far
door and the door then pushes you providing the
centripetal force to keep you moving in the same circle as
the car. If the car door was open you would not have a
centripetal force acting and you would continue in a
straight line out of the car! (Don’t try this at home!)
XtraEdge for IIT-JEE 44 MAY 2010

MATHS
COMPLEX NUMBER
Mathematics Fundamentals
− 1 is denoted by ‘i’ and is pronounced as ‘iota’.
i = − 1 ⇒ i 2 = –1, i 3 = –i, i 4 = 1.
If a, b ∈ R and i = − 1 then a + ib is called a
complex number. The complex number a + ib is also
denoted by the ordered pair (a, b)
If z = a + ib is a complex number, then :
(i) a is called the real part of z and we write
Re (z) = a.
(ii) b is called the imaginary part of z and we write
Im (z) = b
Two complex numbers z 1 and z 2 are said to be equal
complex numbers if Re (z 1 ) = Re (z 2 ) and Im (z 1 ) =
Im (z 2 ).
If z = x + iy is a non zero complex number, then 1/z
is called the multiplicative inverse of z.
If x + iy is a complex number, then the complex
number x – iy is called the conjugate of the complex
number x + iy and we write x + iy = x – iy.
Algebra of Complex Numbers
(i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d)
(ii) Subtraction :
(a + ib) – (c + id) = (a – c) + i(b – d)
(iii) Multiplication :
(a + ib) + (c + id) = (ac – bd) + i(ab + bc)
(iv) Division by a non-zero complex number :
a + ib ac + bd bc − ad
= + i , (c + id) ≠ 0
2 2 2 2
c + id c + d c + d
Properties : If z 1 , z 2 are complex numbers, then
(i) ( z 1 ) = z 1
(ii) z + z = 2 Re (z)
(iii) z – z = 2i Im (z)
(iv) z = z iff z is purely real
(v) z = z iff z is purely imaginary
(vi) z 1 + z2
= z1
+ z2
(vii) z 1 – z2
= z1
– z2
(viii) z 1 .z2
= z1
. z2
⎛ z
(ix)
⎜
⎝ z
1
2
⎞
⎟ =
⎠
z1
z2
provided z 2 ≠ 0
If x + iy is a complex number, then the non-negative
2 2
ral number x + y is called the modulus of the
complex number x + iy and write
| x + iy| =
2 2
x + y
Properties : If z 1 , z 2 are complex numbers, then
(i) | z 1 | = 0 iff z 1 = 0
(ii) | z 1 | = | z 1 | = | – z 1 |
(iii) – | z 1 | ≤ Re (z 1 ) ≤ | z 1 |
(iv) – | z 1 | ≤ Im (z 1 ) ≤ | z 1 |
(v) | z 1 z 1 | = | z 1 | 2
(vi) | z 1 + z 2 | ≤ | z 1 | + | z 2 |
(vii) | z 1 – z 2 | ≥ | z 1 | – | z 2 |
(viii) | z 1 z 2 | = | z 1 | | z 2 |
z1
| z
(ix) =
1 |
, provided z 2 ≠ 0
z2
| z2
|
(x) | z 1 + z 2 | 2 = | z 1 | 2 + | z 2 | 2 + 2 Re (z 1 z 2 )
(xi) | z 1 – z 2 | 2 = | z 1 | 2 + | z 2 | 2 – 2 Re (z 1 z 2 )
(xi) | z 1 + z 2 | 2 + | z 1 – z 2 | 2 = 2 [| z 1 | 2 + | z 2 | 2 ].
De Moivre’s Theorem
(i) If n is any integer (positive or negative), then
(cos θ + i sin θ) n = cos nθ + i sin nθ
(ii) If n is a rational number, then the value or one of
the values of (cos θ + i sin θ) n is cos nθ + i sin nθ
Euler’s Formula
e iθ = cos θ + i sin θ and e –iθ = cos θ – i sin θ
Square root of complex number
Square root of z = a + ib are given by
⎡
± ⎢
⎢⎣
⎡
± ⎢
⎢⎣
⎛ | z | + a ⎞
⎜ ⎟ + i
⎝ 2 ⎠
⎛ | z | +
⎜
⎝ 2
a
⎞
⎟
⎠
⎛ | z | −a
⎞⎤
⎜ ⎟⎥
for b > 0 and
⎝ 2 ⎠⎥⎦
⎛ | z | −a
⎞⎤
– i ⎜ ⎟⎥
for b < 0.
⎝ 2 ⎠⎥⎦
XtraEdge for IIT-JEE 45 MAY 2010

−1+ i 3
If ω = , then the cube roots of unity are 1, ω
2
and ω 2 . We have:
(i) 1 + ω + ω 2 = 0 (ii) ω 3 = 1
Let z = x + iy be any complex number.
Let z = r (cos θ + i sin θ) where r > 0.
∴ x = r cos θ and y = r sin θ
∴ x 2 + y 2 = r 2
⇒ r =
∴ cos θ =
x
2
2
y
x + (Q r > 0)
2
x
+ y
2
and sin θ =
x
2
y
+ y
The value of θ is found by solving these equations. θ
is called the argument (or amplitude) of z.
If – p < θ ≤ π, then θ is called the principal argument
of z.
Identification of θ –
x y arg(z) Interval of θ
⎛ π ⎞
+ + θ ⎜0
< θ < ⎟
⎝ 2 ⎠
⎛ – π ⎞
+ – –θ ⎜ < θ < 0⎟ ⎝ 2 ⎠
⎛ π ⎞
– + (π – θ) ⎜ < θ < π⎟ ⎝ 2 ⎠
⎛ – π ⎞
– – –(π – θ) ⎜ – π < θ < ⎟
⎝ 2 ⎠
If z 1 and z 2 are two complex numbers then
(i) | z 1 – z 2 | is the distance between the points with
affixes z 1 and z 2 .
mz
(ii)
2 + nz1
is the affix of the point dividing the
m + n
line joining the points with affixes z 1 and z 2 in the
ratio m : n internally.
mz
(iii)
2 – nz1
is the affix of the point dividing the
m – n
line joining the points with affixes z 1 and z 2 in the
ratio m : n externally where m ≠ n.
(iv) If z 1 , z 2 , z 3 are the affixes of the vertices of a
z1 + z2
+ z3
triangle then the affix of its centroid is
.
3
(v) z = tz 1 + (1 – t)z 2 is the equation of the line
joining points with affixes z 1 and z 2 . Here ‘t’ is a
parameter.
2
z − z
(vi)
1 z − z
=
1
is the equation of the line
z2
− z1
z2
− z1
joining points with affixes z 1 and z 2 .
Three points with affixes z 1 , z 2 , z 3 are collinear if
z1
z1
1
z2
z2
1 = 0.
z3
z3
1
The general equation of a straight line is
a z + az + b = 0 , where b is any real number.
(i) | z – z 1 | < r represents the circle with centre z 1
and radius r.
(ii) | z – z 1 | < r represents the interior of the circle
with centre z 1 and radius r.
z − z1
= k represents a circle line which is the
z − z1
perpendicular bisector of the line segment joining
points with affixes z 1 and z 2 .
(z – z 1 ) ( z − z2)
+ ( z − z1)
+ (z – z 2 ) = 0 represents
the circle with line joining points with affixes z 1 and
z 2 as a diameter.
| z – z 1 | + | z – z 2 | = 2k, k ∈ R + represents the ellipse
with foci at points with affixes z 1 and z 2 .
If z 1 , z 2 , z 3 be the affixes of the points A, B, C
respectively, then the angle between AB and AC is
⎛ z ⎞
given by arg
⎜
3 − z1
⎟ .
⎝ z2
− z1
⎠
If z 1 , z 2 , z 3 , z 4 are the affixes of the points A, B, C, D
respectively, then the angle between AB and CD is
⎛ z ⎞
given by arg ⎜ 2 − z1
⎟
.
⎝ z4
− z3
⎠
nth roots of a complex number
Let z = r (cos θ + i sin θ), r > 0 be any complex
number. nth root o z = z 1/n
= r 1/n ⎛ 2kπ + θ 2kπ + θ ⎞
⎜cos + isin ⎟ ,
⎝ n
n ⎠
where k = 0, 1, 2, ………, n – 1.
There are n distinct values and sum of all these
values is 0.
Logarithm of a complex number
Let z = re iθ be any complex number.
Then log z = log re iθ = log r + log e iθ
= log r + iθ log e = log r + iθ.
∴ log z = log | z | + i amp (z).
XtraEdge for IIT-JEE 46 MAY 2010

MATHS
MATRICES &
DETERMINANTS
Mathematics Fundamentals
Matrices :
An m × n matrix is a rectangular array of mn
numbers (real or complex) arranged in an ordered set
of m horizontal lines called rows and n vertical lines
called columns enclosed in parentheses. An m × n
matrix A is usually written as :
⎡ a
⎢
⎢
a
⎢ M
A = ⎢
⎢ a
⎢ M
⎢
⎢⎣
a
11
21
i1
m1
a12
a 22
M
ai2
M
a m2
...
...
...
...
a1j
a 2 j
aij
a mj
...
...
...
...
a1n
a 2n
ain
a mn
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
Where 1 ≤ i ≤ m and 1 ≤ j ≤ n
and is written in compact form as A = [a ij ] m× n
A matrix A = [a ij ] m × n is called
(i) a rectangular matrix if m ≠ n
(ii) a square matrix if m = n
(iii) a row matrix or row vector if m = 1
(iv) a column matrix or column vector if n = 1
(v) a null matrix if a ij = 0 for all i, j and is denoted by
O m× n
(vi) a diagonal matrix if a ij = 0 for i ≠ j
(vii) a scalar matrix if a ij = 0 for i ≠ j and all diagonal
elements a ii are equal
Two matrices can be added only when thye are of same
order. If A = [a ij ] m × n and B = [b ij ] m × n , then sum of A
and B is denoted by A + B and is a matrix [a ij + b ij ] m × n
The product of two matrices A and B, written as AB,
is defined in this very order of matrices if number of
columns of A (pre factor) is equal to the number of
rows of B (post factor). If AB is defined , we say that
A and B are conformable for multiplication in the
order AB.
If A = [a ij ] m × n and B = [b ij ] n × p , then their product AB
is a matrix C = [c ij ] m × p where
C ij = sum of the products of elements of ith row of A
with the corresponding elements of jth column of B.
Types of matrices :
(i) Idempotent if A 2 = A
(ii) Periodic if A k+1 = A for some positive integer k.
The least value of k is called the period of A.
(iii) Nilpotent if A k = O when k is a positive integer.
Least value of k is called the index of the
nilpotent matrix.
(iv) Involutary if A 2 = I.
The matrix obtained from a matrix A = [a ij ] m × n by
changing its rows into columns and columns of A into
rows is called the transpose of A and is denoted by A′.
A square matrix a = [a ij ] n × n is said to be
(i) Symmetric if a ij = a ji for all i and j i.e. if A′ = A.
(ii) Skew-symmetric if
a ij = – a ji for all i and j i.e., if A′ = –A.
Every square matrix A can be uniquely written as sum
of a symmetric and a skew-symmetric matrix.
A = 2
1 (A + A′) + 2
1 (A – A′) where 2
1 (A + A′) is
symmetric and 2
1 (A – A′) is skew-symmetric.
Let A = [a ij ] m × n be a given matrix. Then the matrix
obtained from A by replacing all the elements by their
conjugate complex is called the conjugate of the matrix
A and is denoted by A = [a ] .
Properties :
(i) ( A ) = A
(ii) ( A + B)
= A + B
(iii) (λ A)
= λ A , where λ is a scalar
(iv) ( A B) = A B .
Determinant :
Consider the set of linear equations a 1 x + b 1 y = 0 and
a 2 x + b 2 y = 0, where on eliminating x and y we get
the eliminant a 1 b 2 – a 2 b 1 = 0; or symbolically, we
write in the determinant notation
ij
a1
b1
≡ a 1 b 2 – a 2 b 1 = 0
a 2 b2
Here the scalar a 1 b 2 – a 2 b 1 is said to be the expansion
a1
b1
of the 2 × 2 order determinant having 2
a 2 b2
rows and 2 columns.
Similarly, a determinant of 3 × 3 order can be
expanded as :
XtraEdge for IIT-JEE 47 MAY 2010

a
a
a
1
2
3
b
b
b
1
2
3
c
c
c
1
2
3
b2
c2
= a 1
b3
c3
a 2 c2
a 2 b2
– b 1 + c 1
a 3 c3
a 3 b3
= a 1 (b 2 c 3 – b 3 c 2 ) – b 1 (a 2 c 3 – a 3 c 2 ) + c 1 (a 2 b 3 – a 3 b 2 )
= a 1 (b 2 c 3 – b 3 c 2 ) – a 2 (b 1 c 3 – b 3 c 1 ) + a 3 (b 1 c 2 – b 2 c 1 )
= Σ(± a i b j c k )
To every square matrix A = [a ij ] m × n is associated a
number of function called the determinant of A and is
denoted by | A | or det A.
Thus, | A | =
a11
a 21
M
a n1
a12
a 22
M
a n2
...
...
...
a1n
a 2n
M
a nn
If A = [a ij ] n × n , then the matrix obtained from A after
deleting ith row and jth column is called a submatrix
of A. The determinant of this submatrix is called a
minor or a ij .
Sum of products of elements of a row (or column) in
a det with their corresponding cofactors is equal to
the value of the determinant.
n
i.e., ∑
=
i 1
aij
C ij = | A | and ∑
j=
1
n
a C ij = | A |.
(i) If all the elements of any two rows or two columns
of a determinant ate either identical or
proportional, then the determinant is zero.
(ii) If A is a square matrix of order n, then
| kA | = k n | A |.
(iii) If ∆ is determinant of order n and ∆′ is the
determinant obtained from ∆ by replacing the
elements by the corresponding cofactors, then
∆′ = ∆ n–1
(iv) Determinant of a skew-symmetric matrix of odd
order is always zero.
The determinant of a square matrix can be evaluated
by expanding from any row or column.
If A = [a ij ] n × n is a square matrix and C ij is the
cofactor of a ij in A, then the transpose of the matrix
obtained from A after replacing each element by the
corresponding cofactor is called the adjoint of A and
is denoted by adj. A.
Thus, adj. A = [C ij ]′.
Properties of adjoint of a square matrix
(i) If A is a square matrix of order n, then
A . (adj. A) = (adj . A) A = | A | I n .
(ii) If | A | = 0, then A (adj. A) = (adj. A) A = O.
(iii) | adj . A | = | A | n –1 if | A | ≠ 0
(iv) adj. (AB) = (adj. B) (adj. A).
(v) adj. (adj. A) = | A | n – 2 A.
ij
Let A be a square matrix of order n. Then the inverse of
A is given by A –1 1
= adj. A.
| A |
Reversal law : If A, B, C are invertible matrices of same
order, then
(i) (AB) –1 = B –1 A –1
(ii) (ABC) –1 = C –1 B –1 A –1
Criterion of consistency of a system of linear equations
(i) The non-homogeneous system AX = B, B ≠ 0 has
unique solution if | A | ≠ 0 and the unique solution is
given by X = A –1 B.
(ii) Cramer’s Rule : If | A | ≠ 0 and X = (x 1 , x 2 ,..., x n )′
| A
then for each i =1, 2, 3, …, n ; x i =
i |
where
| A |
Ai is the matrix obtained from A by replacing the
ith column with B.
(iii) If | A | = 0 and (adj. A) B = O, then the system
AX = B is consistent and has infinitely many
solutions.
(iv) If | A | = 0 and (adj. A) B ≠ O, then the system
AX = B is inconsistent.
(v) If | A | ≠ 0 then the homogeneous system AX = O
has only null solution or trivial solution
(i.e., x 1 = 0, x 2 = 0, …. x n = 0)
(vi) If | A | = 0, then the system AX = O has non-null
solution.
(i) Area of a triangle having vertices at (x 1 , y 1 ), (x 2 , y 2 )
and (x 3 , y 3 ) is given by
(ii) Three points A(x 1 , y 1 ), B(x 2 , y 2 ) and C(x 3 , y 3 ) are
collinear iff area of ∆ABC = 0.
A square matrix A is called an orthogonal matrix if
AA′ = AA′ = I.
A square matrix A is called unitary if AA θ = A θ A = I
(i) The determinant of a unitary matrix is of modulus
unity.
(ii) If A is a unitary matrix then A′, A , A θ , A –1 are
unitary.
(iii) Product of two unitary matrices is unitary.
Differentiation of Determinants :
Let A = | C 1 C 2 C 3 | is a determinant then
dA = | C′1 C 2 C 3 | + | C 1 C′ 2 C 3 | + | C 1 C 2 C′ 3 |
dx
Same process we have for row.
Thus, to differentiate a determinant, we differentiate one
column (or row) at a time, keeping others unchanged.
1
2
x
x
x
1
2
3
y
y
y
1
2
3
1
1
1
XtraEdge for IIT-JEE 48 MAY 2010

a
Based on New Pattern
IIT-JEE 2011
XtraEdge Test Series # 1
Time : 3 Hours
Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile
motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current,
Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical
Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics:
Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of
Circle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal,
Monotonicity, Maxima & Minima)
Instructions :
Section - I
• Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for
correct answer and -1 mark for wrong answer.
• Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for
correct answer and -1 mark for wrong answer.
• Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer
and -1 mark for wrong answer.
Section - II
• Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly
matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.
PHYSICS
Question 1 to 8 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. A target is made of two plates, one of wood and the
other of iron. The thickness of the wooden plate is 4
cm and that of iron plate is 2 cm. A bullet fired
goes through the wood first and then penetrates 1
cm into iron. A similar bullet fired with the same
velocity from opposite direction goes through iron
first and then penetrates 2 cm into wood. If a 1 and
a 2 be the retardations offered to the bullet by wood
and iron plates respectively then -
(A) a 1 = 2a 2 (B) a 2 = 2a 1
(C) a 1 = a 2
(D) Data Insufficient
2. The cone falling with a speed v 0 strikes and
penetrates the block of packing material. The
acceleration of the cone after impact is a = g – cx 2 ,
where c is a positive constant and x is the
penetration distance. If the maximum penetration
depth is x m . Then c equals -
v 0
x
(A)
(C)
2 gx + v
m
2
x m
m
3
2x m
2
0
6gx
− 3v
2
0
(B)
(D)
2gx
− v
m
2
x m
m
3
2x m
2
0
6 gx + 3v
3. A dipole consists of two particles, one of charge Q,
mass m and the other of charge –Q and mass 2m
separated by a distance L. For small oscillations
about the equilibrium position the time after which
the dipole will align itself in the direction of the
uniform field E is -
(A) 2π
(C) 2
π
2m
2mL
3QE
2mL
3QE
–Q
m
(B) π
Q
(D) 4
π
2mL
3QE
2
0
→
E
2mL
3QE
4. In the circuit shown, a potential difference of 60 V
is applied across AB. The potential difference
between the points P and Q is -
XtraEdge for IIT-JEE 49
MAY 2010

A
90V
S
2C
B
Q
R
(A) 15 V (B) 30 V (C) 45 V (D) 60 V
C
2C
5. A long straight wire along the z-axis carries a
current I in the negative z direction. The magnetic
vector field → B at a point having coordinates (x, y)
in the z = 0 plane is -
µ ⎛ ⎞
0I
(A) ⎜
yî − xĵ
⎟
µ ⎛ ⎞
0I
(B) ⎜
xî + yĵ
⎟
π
2 2
2
⎝ x + y
2 2
⎠ 2π
⎝ x + y ⎠
µ ⎛ ⎞
0I
(C) ⎜
xĵ−
yî
⎟
µ ⎛ ⎞
0I
(D) ⎜
xî − yĵ
⎟
π
2 2
2
⎝ x + y
2 2
⎠ 2π
⎝ x + y ⎠
6. Two coils, X and Y, are linked such that emf E is
induced in Y when the current in X is changing at
⎛ dI ⎞
the rate I • ⎜ = ⎟ . If a current I 0 is now made to
⎝ dt ⎠
flow through Y, the flux linked with X will be -
•
⎛
(A) EI 0 I (B)
⎟ ⎟ ⎞
⎜ E
I0
⎜ •
⎝ I ⎠
•
(C) (E I)
I 0
(D)
C
I
E
P
•
0 I
7. In the circuit shown, A is joined to B for a long
time, and then A is joined to C. The total heat
produced in R is -
R
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. Two objects have life times given by t 1 and t 2 . if t is
the life time of an object lying midway between
these two times on the logarithmic scale then -
(A) log 10 (t) = 2
1 [log10 (t 1 ) + log 10 (t 2 )]
t t 2
(B) t =
2
(C) t = t 1 t 2
1 1 ⎛ 1
(D) = t 2 ⎜ +
⎝ t
1 +
1 t 2
1 ⎞
⎟
⎠
10. The position of a particle traveling along x-axis is
given by x t = t 3 – 9t 2 + 6t where x t is in cm and t is
in second Then -
(A) the body comes to rest firstly at (3 – 7 ) s and
then at (3 + 7 ) s
(B) the total displacement of the particle in
traveling from the first zero of velocity to the
second zero of velocity is zero
(C) the total displacement of the particle in
traveling from the first zero of velocity to the
second zero of velocity is – 74 cm.
(D) the particle reverses its velocity at (3 – 7 ) s
and then at (3 + 7 ) s and has a negative
velocity for (3 – 7 ) < t < (3 + 7 )
11. A projectile is fired upward with velocity v 0 at an
angle θ and strikes a point P(x, y) on the roof of the
building (as shown). Then,
y
Roof
(A)
(C)
LE
R
LE
4R
2
2
2
2
+
2L
E
–
2R
(B)
A
LE
2R
2
2
2
LE
(D)
8R 2
8. In a transformer, N P and N S are 1000 and 3000
respectively. If the primary is connected across 80
V A.C., the potential difference across each turn of
the secondary will be -
(A) 240 V
(B) 0.24 V
(C) 0.9 V
(D) 0.08 V
C
B
h
θ
v 0
P(x,y)
(A) the projectile hits the roof in minimum time if
π
θ + α = . 2
(B) the projectile hits the roof in minimum time if
π
θ + α = . 4
(C) the minimum time taken by the projectile to hit
2
2
v0
− v0
− 2gh cos α
the roof is t min =
.
g cosα
(D) the projectile never reaches the roof for
v 0 < 2 gh cosα
α
x
XtraEdge for IIT-JEE 50
MAY 2010

12. In the given circuit the point A is 9 V higher than
point B -
A B C D
6V
1Ω
15V
2Ω
24V
1Ω
R
(A) R = 1 Ω
(B) R = 7 Ω
(C) Potential difference between B and D is 30 V
(D) Potential difference between B and C is 15 V
This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
Passage : I (No. 13 to 15)
In the diagram (given below), the broken lines
represent the paths followed by particles W, X, Y
and Z respectively through the constant field E.
The numbers below the field represent meters.
E
X
Z
Y
W
0 1 2 3 4
13. If all particles started from rest, and all are
positively charged, which particles must have been
acted upon by a force other than that produced by
the electric field –
(A) W and Y (B) X and Z
(C) X, Y and Z (D) W, X, Y and Z
14. If the particles are positively charged, which
particles increased their electric potential energy –
(A) X and Z
(B) Y and Z
(C) W, X, Y and Z
(D) Since the electric field is constant, none of the
particles increased their electric potential
energy
15. Suppose that the field strength E is 10 N/C and
particle Y has a charge of –10 C. When particle
Y is released from rest, it follows the path as
shown and accelerates to a velocity of 10 m/s.
What is the mass of particle Y –
(A) 1 kg
(B) 2 kg
(C) 3 kg
(D) 4 kg
Passage : II (No. 16 to 18)
A set of experiments in the physics lab is designed
to develop understanding of simple electrical circuit
principles for direct current circuits . The student is
given a variety of batteries, resistors, and DC
meters ; and is directed to wire series and parallel
combinations of resistors and batteries making
measurements of the currents and voltage drops
using the ammeters and voltmeters. The student
calculates expected current and voltage values
using Ohm’s law and Kirchoff’s circuit rules and
then checks the results with the meters.
16. Resistors of 4 ohms and 8 ohms are connected in
series. A battery of 6 volts is connected across the
series combination . How much power, in watts, is
consumed in the 8-ohm resistor ?
(A) 0.67 W
(B) 2 W
(C) 12 W
(D) 24 W
17. Two 4-ohm resistors are connected in series and
this pair is connected in parallel with an 8-ohm
resistor. A 12 volt battery is connected across the
ends of this parallel set. What power, in watts, is
consumed in the 8-ohm resistor in this case ?
(A) 0.9 W
(B) 2.0 W
(C) 4.4 W
(D) 18 W
18. A 6-volt battery is connected across a 2- ohm
resistor. What is the heat energy dissipated in the
resistor in 5 minutes ?
(A) 430 joules
(B) 560 joules
(C) 4300 joules
(D) 5400 joules
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
XtraEdge for IIT-JEE 51
MAY 2010

19. A circular current carrying loop is placed in x-y 2. The potential of the Daniell cell,
(C) 2 × 10 –2 Ms –1 (D) 7 × 10 –3 Ms –1 OA = 2L mol –1
Hence, rate at the start of the reaction is –
plane as shown. A uniform magnetic field B = B 0 kˆ
ZnSO
Zn 4 CuSO 4
Cu was reported by Buckbee,
is present in the region. Match the following :
(1M) (1M )
y
Surdzial, and Metz as
Eº = 1.1028 – 0.641 × 10 –3 T + 0.72 × 10 –5 T 2 ,
x
where T is the celcius temperature. Calculate ∆Sº
for the cell reaction at 25º C -
(A) – 45.32 (B) – 34.52
(C) – 25.43 (D) – 54.23
Column-I
Column-II
(A) Magnetic moment of loop (P) Zero
3. In a hypothetical solid C atoms form CCP lattice
(B) Torque on the loop (Q) Maximum
with A atoms occupying all the Tetrahedral voids
and B atoms occupying all the octahedral voids. A
(C) Potential energy of loop (R) Along +ve
and B atoms are of the appropriate size such that
z-axis
there is no distortion in the CCP lattice. Now
(D) Equilibrium of loop (S) Stable
if a plane is cut (as shown) then the cross
(T) Minimum
section would like –
20. An L-C circuit consists of an inductor with L = 0.09
H and a capacitor of C = 4 × 10 –4 F. The initial
Plane
charge on the capacitor is 5 µC, and the initial
current in the inductor is zero. Match the following:
Column-I
Column-II
(A) Maximum voltage across (P) 8.33 × 10 –4
CCP unit cell
capacitor
(S.I. unit)
(B) Maximum current in the (Q) 3.125 × 10 –8
C B
inductor
(S.I. unit)
A
C C C C C
(C) Maximum energy stored (R) 4.33 × 10 –6
(A) B B B (B) B B B
A
In the inductor
(S.I. unit)
(D) Charge on the capacitor (S) 1.25 × 10 –2
C B C C C C
when current in the inductor (S.I. unit)
has half its maximum value
C C C C C C C
(T) None
A A
(C) B B B (D) B B B
A A
CHEMISTRY
C C C C C C
Question 1 to 8 are multiple choice questions. Each 4. The melting point of RbBr is 682ºC while that of
questions has four choices (A), (B), (C) and (D), out of
NaF is 988ºC. The principal reason that the melting
which ONLY ONE is correct.
point of NaF is much higher than that of RbBr is
that :
1. A reaction follows the given concentration (C) vs
time graph. The rate for this reaction at 20 seconds
(A) the molar mass of NaF is smaller than that of
will be –
RbBr
0.5
(B) the bond in RbBr has more covalent character
0.4
than the bond in NaF
(C) the difference in electronegativity between Rb
0.3
and Br is smaller than the difference between
0.2
Na and F
0.1
(D) the internuclear distance, r c + r a is greater for
RbBr than for NaF
0 20 40 60 80 100
Time/second
5. Following is the graph between (a – x) –1 and time t
for a second order reaction
(A) 4 × 10 –3 Ms –1 (B) 8 × 10 –2 Ms –1
θ = tan –1 (0.5)
XtraEdge for IIT-JEE 52
MAY 2010

(a–x) –1
A
O time t
(A) 1.25 mol –1 min –1
(B) 0.5 mol L –1 min –1
(C) 0.125 mol L –1 min –1
(D) 1.25 mol L –1 min –1
θ
6. A solution contains Na 2 CO 3 and NaHCO 3 . 10 ml of
the solution requires 2.5 ml of 0.1 M H 2 SO 4 for
neutralization using phenolphthalein as the
indicator. Methyl orange is then added when a
further 2.5 ml of 0.2 M H 2 SO 4 was required. The
amount of Na 2 CO 3 in the g/litre is
(A) 5.3 (B) 4.2 (C) 10.6 (D) 8.4
7. 1 mole mixture of CH 4 & air (containing 80% N 2 ,
20% O 2 by volume) of a composition such that
when underwent combustion gave maximum heat
(assume combustion of only CH 4 ). Then which of
the statements are correct, regarding composition of
initial mixture:
1 2 8
(A) XCH = ,X ,X
4 O =
2
N =
2
11 11 11
3 1 1
(B) XCH = ,X ,X
4 O =
2 N =
2
8 8 2
1 1 2
(C) XCH = ,X , X
4 O =
2 N =
2
6 6 3
(D) Data insufficient
8. Consider the following I st order competing
reactions :
k
X ⎯⎯→
1
k
A + B and Y ⎯⎯→
2 C + D
if 50% of the reaction of X was completed when
96% of the reaction of Y was completed, the ratio
of their rate constants (k 2 / k 1 ) is :
(A) 4.06 (B) 0.215 (C) 1.1 (D) 4.65
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. Select the correct statements from the following
regarding sols –
(A) Viscosity of lyophilic sols (emulsoid) is much
higher than that of solvent
(B) Surface tension of lyophobic sols (suspensoid)
is usually low
(C) The particles of lyophilic sols always carry a
characteristics charge either positive or
negative
(D) Hydrophobic sols can easily be coagulated by
addition of electrolytes
10. Which of the following is/are correct ?
(A) α-rays are more penetrating than β-rays
(B) α-rays have greater ionizing power than β-rays
(C) β-particles are not present in the nucleus, yet
they are emitted from the nucleus
(D) γ-rays are not emitted simultaneously with α
and β-rays.
11. Choose the correct statement(s) -
(A) At the anode, the species having minimum
reduction potential is formed from the
oxidation of corresponding oxidizable species
(B) In highly alkaline medium, the anodic process
during the electrolytic process is
4OH – → O 2 + 2H 2 O + 4e –
(C) The standard potential of Cl – | AgCl | Ag half–
cell is related to that of Ag + | Ag through the
expression
º
Ag
E + =
/ Ag
º
E
Cl – +
|AgCl|Ag
RT ln Ksp (AgCl)
F
(D) Compounds of active metals (Zn, Na, Mg) are
reducible by H 2 whereas those of noble metals
(Cu, Ag, Au) are not reducible.
12. The co-ordination number of FCC structure for
metals is 12, since -
(A) each atom touches 4 others in same layer, 3 in
layer above and 3 in layer below
(B) each atom touches 4 others in same layer, 4 in
layer above and 4 in layer below
(C) each atom touches 6 others in same layer, 3 in
layer above and 3 in layer below
(D) each atom touches 3 others in same layer, 6 in
layer above and 6 in layer below
This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
Passage : I (No. 13 to 15)
The cell potential for the unbalanced chemical reaction :
Hg 2
2+
+ NO 3
–
+ H 3 O + ⎯→ Hg 2+ +HNO 2 + H 2 O
is measured under standard state conditions in the
electrochemical cell shown in the accompanying
diagram. The cell voltage is positive: Eº Cell = 0.02V
Eº = 0.02 V
Dish A anode Dish B cathode
Given
NO 3
–
+ 3H 3 O + 2e – ⎯→ HNO 2 + 4H 2 O
Eº = 0.94 V
XtraEdge for IIT-JEE 53
MAY 2010

13. Which of the following statements must be true of
the solutions in order for the cell to operate with the
voltage indicated ?
(A) The solution in Dish A must be acidic
(B) The solution in Dish B must be acidic
(C) The solutions in both Dish A and Dish B must
be acidic
(D) No acid may be in either Dish A or Dish B
14. At what pH will the cell potential be zero if the
activity of other components are equal to one ?
(A)
(C)
0.02
2×
0.059
0.04
0.059
(B) –
(D)
0.02
0.059
0.02
0.059
× 3
2
15. How many moles of electrons pass through the
circuit when 0.6 mole of Hg 2+ and 0.30 mole of
HNO 2 are produced in the cell that contains 0.5
2+
mole of Hg 2 and 0.40 mole of NO – 3 at the
beginning of the reaction ?
(A) 0.6 mole (B) 0.8 mole
(C) 0.3 mole (D) 1 mole
Passage : II (No. 16 to 18)
At any fixed temperature, the vapour phase is
always richer in more volatile component as
compared to the solution phase. In other words,
mole fraction of more volatile component is always
greater in vapour phase than in solution phase. We
can also say that vapour phase is relatively richer in
the component whose addition to liquid mixture
results in an increase in total vapour pressure.
16. If 2 moles of A and 3 moles of B are mixed to
form an ideal solution, vapour pressures of A and B
are 120 and 180 mm of Hg respectively, the total
vapour pressure of solution will be
(A) 48 mm Hg (B) 108 mm Hg
(C) 156 mm Hg (D) 15.6 mm Hg
17. From the statement in question 75, the composition
of A and B in the vapour phase when the first traces
of vapours are formed is :
(A) A = 0.407, B = 0.592
(B) A = 0.8, B = 0.1
(C) A = 0.109, B = 0.791
(D) A = 0.307, B = 0.692
18. From the statement in question 75, at what pressure
will the last traces of liquid disappear ?
(A) 100 mm Hg (B) 130 mm Hg
(C) 140 mm Hg (D) 150 mm Hg
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
19. Match the following :
Column-I
Column-II
(A) Intermolecular (P) Ne
H-bonding
(B) Intramolecular (Q) NaCl
H-bonding
(C) Vander Waal's forces (R) H 2 O
OH
(D) Strongest bonding
(S)
CHO
(T) Chloroal
hydrate
20. Match the following :
Column-I
Column-II
(A) PH 3 (P) ≈ 90º or = 90º
(B) H 2 O
(Q) 100º < B.A.

2. If a, b, c, d are the sides of a quadrilateral, then
a + b + c
the minimum value of
2
d
(A) 1 (B) 1/2
(C)1/3 (D) 1/4
3. If the function f(x) = cos |x| – 2ax + b increases
along the entire number scale, the range of
values of a is given by
(A) a ≤ b (B) a = b/2
(C) a ≤ – 1/2 (D) a ≥ – 3/2
⎛ ⎞
4. If φ(x) = 3f ⎜
x 2 ⎟
+ f(3– x 2 ) ∀ x ∈(–3, 4)
⎝ 3 ⎠
where f"(x) > 0 ∀ x ∈ (–3, 4), then φ(x) is
⎛ 3 ⎞
(A) increasing in ⎜ ,4⎟ ⎝ 2 ⎠
⎛ 3 ⎞
(B) decreasing in ⎜−
3 , − ⎟
⎝ 2 ⎠
⎛ 3 ⎞
(C) increasing in ⎜ − , 0 ⎟
⎝ 2 ⎠
⎛ 3 ⎞
(D) decreasing in ⎜0
, ⎟
⎝ 2 ⎠
5. The point in the interval [0, 2π] where f(x) = e x sin x
has maximum slope is
π
π
(A) (B) 4 2
(C) π
2
2
2
is
(D) None of these
a cosA + bcosB + c cos C a + b + c
Q.6 If in ∆ABC,
=
asin B + bsinC + csin A 9R
then the triangle ABC is
(A) isosceles (B) equilateral
(C) right angled (D) None of these
Q.7 The slope of the normal at the point with
abscissa x = – 2 of the graph of the function
f(x) = |x 2 – x| is of form p/q (where p & q are
coprime) then -
(A) p = 4 (B) q = 2
(C) p + q = 6 (D) p – q = – 2
Q.8 In a ∆ABC, b 2 + c 2 = 1999a 2 , then
cot B + cot C
=
cot A
1
1
(A) (B) 999
1999
(C) 999 (D) 1999
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. For what triplets of real numbers (a, b, c)
with a ≠ 0, the function : f(x) =
⎪⎧
x ; x ≤1
⎨ 2
⎪⎩ ax + bx + c ; otherwise
is continous for all real x :
(A) {(a, 1 – 2a ,a)/a ∈ R, a ≠ 0}
(B) {(2a, 1 – 2a ,0)/a ∈ R, a ≠ 0}
(C) {(a, b, c)/a, b, c ∈ R, a + b + c = 1, a ≠ 0}
(D) {(a, 1 – 2a,c)/ a, c ∈ R, a ≠ 0}
⎧– 2 x < 0
10. Let f(x) = ⎨
⎩x – 2 x ≥ 0
and g(x) = |f(x)| then :
(A) g(x) is continuous for all values of x.
(B) g(x) is differentiable for all value of x
(C) g(x) is not differentibale at x = 0
(D) g(x) is not differentiable at x = 2
⎛ 1 ⎞
11. If 5f(x) + 3f ⎜ ⎟ = x + 2 and y = xf(x) then
⎝ x ⎠
⎛ dy ⎞
⎜ ⎟ is equal to :
⎝ dx ⎠
x=
1
(A) 14 (B) 7/8
(C) 1
(D) None of these
12. If f(x) and g(x) are differentiable function in
[1, 5] and φ(x) = max {f(x), g(x)}, f(x) – g(x) = 0
has exactly one root α in [1, 5] then :
(A) φ(x) continuous and differentiable at all
points in [1, 5]
(B) φ(x) differentiable in [1, 5]
(C) φ(x) necessarily differentiable in [1, 5]– {α}
(D) φ(x) is not differentiable at x = α
This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
Passage : I (No. 13 to 15)
⎛ x + y ⎞ f (x) + f (y) + b(n – 2)
Let f ⎜ ⎟ =
and
⎝ n ⎠ n
f '(0) = a ; n ∈ N but n ≠ 2.
Let g(x) = f(x) ; x ≥ 0
= 3x + sin 2x ; x < 0
If f(x) & g(x) both are differentiable function, then -
XtraEdge for IIT-JEE 55
MAY 2010

13. Range of sgn g(|x|) includes :
(A) 2 (B) 1
(C) –1
(D) all of the above
14. Which one of the following statements doesn't
hold good :
(A) g(x) is many one function
(B) g(x) is invertible
(C) f(x) is invertible
(D) g(|x|) is non differentiable at x = 0
15. Let h(x) = |g(x) –5x|, then wrong statement is :
(A) h(|x|) is a single valued function
(B) |h(|x|)| is always positive
(C) |h(|x|)| is differentiable every where
(D) h(x) is differentiable every where
Passage : II (No. 16 to 18)
Let z denotes the set of integers. Let p be a
prime number and let z 1 ≡ {0, 1}. Let f : z → z
and
g : z → z 1 are two functions defined as follows:
f(n) = p n ; if n ∈ z and
g(n) = 1 ; if n is a perfect square
= 0 ; otherwise
16. g(f(x)) is :
(A) Manyone into
(C) one-one onto
17. f(g(x)) = p has :
(A) no real root
(B) at most one real root
(C) infinitely many roots
(D) exactly two real root
(B) Manyone onto
(D) one-one into
18. Wrong statement about g(f((x)) is :
(A) it is non periodic function
(B) it is neither even nor odd function
(C) it is even function
(D) it is many one function
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A P Q R S T
B P Q R S T
C P Q R S T
D P Q R S T
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
19. Match the following :
Column –I
Column –II
(A) f : R → R is defined as (P) 2
⎪⎧
x 2 + kx + 3 for x ≥ 0
f(x) = ⎨ ⎪⎩ 2kx + 3 for x < 0
if f(x) is injective then 'k'
can be equal to
f (x) − 9
(B) If lim = 3 then
x→2
x − 2
(Q) 5
lim f (x) is
x→2
kx
7 + 8
(C) If lim
x →∞ 7 5x
+ 6
does not (R) 9
exists then 'k' can be
(D) Let f(x) & g(x) satisfy the (S) 12
following properties f(3) = 2,
g(3) = 4, g(0) = 3, f ´ (3) = –1,
g´(3) = 0, g´(0) = 2, If
T(x) = f(g(x)) and U(x) = ln(f(x))
then |T´(0) + 6U´3)| can be equal
(T) –1
20. Match the items of column-I with column-II
Column –I
Column –II
⎡ π⎤
(A) Function f : ⎢0,
⎥ → [0, 1]
⎣ 3 ⎦
(P) one-one
defined by f(x) = sin x is function
(B) Function f :(1, ∞) → (1, ∞) (Q) many-one
x + 3
defined by f(x) = is
x −1
function
⎡ π 4π⎤
(C) Function f: ⎢−
, ⎥ →
⎣ 2 3 ⎦
(R) into
[–1, 1] defined by function
f(x) = sin x is
(D) Function f : (2, ∞)→[8, ∞) (S) onto
x 2
defined by f(x) = is
x − 2
function
(T) odd
Function
XtraEdge for IIT-JEE 56
MAY 2010

XtraEdge for IIT-JEE 57
MAY 2010

Based on New Pattern
IIT-JEE 2012
XtraEdge Test Series # 1
Time : 3 Hours
Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion,
Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics:
Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of
Circle
Instructions :
Section - I
• Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
• Question 9 to 12 are multiple choice questions with multiple correct answer. +4 marks will be awarded for correct
answer and -1 mark for wrong answer.
• Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer
and -1 mark for wrong answer.
Section - II
• Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly
matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.
PHYSICS
Question 1 to 8 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. If E, m, L and G denote energy, mass, angular
momentum and universal gravitation constant
2
EL
respectively then dimensions of
5 2 will be that
m G
of -
(A) angle (B) length (C) mass (D) time
2. Three vectors → P , → Q and → R are such that | → P | = | → Q |,
| → R | = 2 | → P | and → P + → Q + → R = → 0 . The angles
between → P & → Q , → Q & → R and → P & → R are respectively
(in degrees) -
(A) 90, 135, 135 (B) 90, 45, 45
(C) 45, 90, 90 (D) 45, 135, 135
3. The velocity of a boat in still water in η times less
than the velocity of flow of the river (η > 1). The
angle with the stream direction at which the boat
must move to minimize drifting is -
(A) sin –1 ⎛ 1
⎟ ⎞
⎜
(B) cot –1 ⎛ 1
⎝ η
⎟ ⎞
⎜
⎠
⎝ η ⎠
π
(C) + sin
–1 ⎛ 1 ⎞ π
2 ⎜
⎟ (D) + cot
–1 ⎛ 1 ⎞
⎝ η ⎠ 2 ⎜
⎟
⎝ η ⎠
4. The position of a particle is given by → r = a cos(ωt)
î + a sin (ωt) ĵ + bt kˆ where ω = T
2π and T is time
period for one revolution of the particle following a
helical path. The distance moved by the particle in
one full turn of the helix is -
4π
2 2 2 2π
2 2 2
(A) a + b ω (B) a ω + b
ω ω
(C) ω
π
2 2 2 2
a
π
+ b ω (D) ω
4 2 2 2
a ω + b
5. If the velocity v of a particle moving along a
straight line decreases linearly with its
displacement s from 20 ms –1 to a value approaching
zero at s = 30 m, then acceleration of the particle at
s = 15 m is -
v(in ms –1 )
20
O
(A) 3
2 ms
–2
(C)
20 ms
–2
3
s (in m)
30
(B) – 3
2 ms
–2
20
(D) – ms
–2
3
XtraEdge for IIT-JEE 58
MAY 2010

6. A self-propelled vehicle of mass M whose engine
delivers constant power P has an acceleration
P
a = . To increase the velocity of the vehicle
Mv
from v 1 to v 2 , the distance traveled by it (assuming
no friction) is -
3 P
(A) s = (
2
v 2 – 2 M
v 1 ) (B) s = (
2
v 2 – v 2 1 )
M 3P
M
(C) s = (
3
v 2 – 3 3 P
v 1 ) (D) s = (
3
v 2 – v 1
3 )
3P
M
7. A particle starts from rest and traverses a distance l
with uniform acceleration, then moves uniformly
over a further distance 2l and finally comes to rest
after moving a further distance 3l under uniform
retardation. Assuming entire motion to be
rectilinear motion the ratio of average speed over
the journey to the maximum speed on its way is -
1 2 3 4
(A) (B) (C) (D) 5 5 5 5
8. An express elevator can accelerate or decelerate
with values whose magnitudes are limited to 0.4g.
The elevator attains a maximum vertical speed of
400 metre per minute. The minimum time required
by the elevator to start from rest from the 10 th floor
and to stop at the 30 th floor, a distance 100 m apart
is -
(A) 1.67 s
(B) 16.7 s
(C) 167 s
(D) 1670 s
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. A physical quantity is measured by using an
instrument having a least count l. Then -
(A) error in the measurement of the physical
quantity can exceed the least count l
(B) error in the measurement of the physical
quantity equals the least count l
(C) error in the measurement of the physical
quantity can be less than the least count l
(D) all (A), (B) and (C) are incorrect
10. A bead is free to slide down a smooth wire tightly
stretched between the points P 1 and P 2 on a vertical
circle of radius R. If the bead starts from rest from
P 1 , the highest point on the circle and P 2 lies
anywhere on the circumference of the circle. Then,
P 1
g
θ
P 2
R
(A) time taken by bead to go from P 1 to P 2 is
dependent on position of P 2 and equals
R
2 cos θ
g
(B) time taken by bead to go from P 1 to P 2 is
R
independent of position of P 2 and equals 2
g
(C) acceleration of bead along the wire is g cos θ
(D) velocity of bead when it arrives at P 2 is
2 gR cos θ
11. A body is moving along a straight line. Its distance
x t from a point on its path at a time t after passing
that point is given by x t = 8t 2 – 3t 3 , where x t is in
metre and t in second.
(A) Average speed during the interval t = 0 s to
t = 4 s is 20.21 ms –1
(B) Average velocity during the interval t = 0 s to
t = 4 s is – 16 ms –1
16
(C) The body starts from rest and at t = s it 9
reverses its direction of motion at x t = 8.43 m
from the start
(D) It has an acceleration of –56 ms –2 at t = 4s
12. A particle is acted upon by a force of constant
magnitude which is always perpendicular to the
velocity of the particle. The motion of the particle
takes place in a plane. It follows that -
(A) its velocity is constant
(B) its acceleration is constant
(C) its kinetic energy is constant
(D) it moves in a circular path
This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
Passage : I (No. 13 to 15)
A student performs an experiment to determine
how the range of a ball depends on the velocity
with which it is released. The “range” is the
distance between where the ball lands and where it
was released, assuming it lands at the same height
from which it was released.
In each trial, the student uses the same baseball, and
launches it at the same angle. Table I shows the
experimental results
Table 1
Trial Launch speed (m/s) Range (m)
1
2
3
4
10
20
30
40
8.0
31.8
70.7
122.5
XtraEdge for IIT-JEE 59
MAY 2010

Based on this data, the student then hypothesizes
that the range, R, depends on the initial speed, v 0 ,
according to the following equation : R = Cv 0 n ,
where C is a constant, and n is another constant.
13. Based on this data, the best guess for the value of n
is –
(A) 1/2 (B) 1 (C) 2 (D) 3
14. The student speculates that the constant C depends
on :
I. The angle at which the ball was launched
II. The ball’s mass
III. The ball’s diameter
If we neglect air resistance, then C actually depends
on –
(A) I only
(B) I and II
(C) I and III (D) I, II and III
15. The student performs another trial in which the ball
is launched at speed 5.0 m/s. Its range is
approximately –
(A) 1.0 meters (B) 2.0 meters
(C) 3.0 meters (D) 4.0 meters
Passage : II (No. 16 to 18)
When an airplane flies, its total velocity with
respect to the ground is :
v total = v plane + v wind
Where v plane denotes the plane's velocity through
motionless air, and v wind denotes the wind's
velocity. Crucially, all the quantities in this
equations are vectors. The magnitude of a velocity
vector is often called the "speed".
Consider an airplane whose speed through
motionless air is 100m/s. To reach its destination,
the plane must fly east.
The "heading" of a plane of the direction in which
the nose of the plane points. So, it is the direction in
which the engines propel the plane.
16. If the plane has an eastward heading, and a 20 m/s
wind blows towards the southwest, then the plane's
speed is :
(A) 80 m/s
(B) more than 80 m/s but less than 100 m/s
(C) 100 m/s
(D) more than 100 m/s
17. The pilot maintains an eastward heading while a
20 m/s wind blows northward. The plane's velocity
is deflected from due east by what angle ?
(A) sin –1 (1/5) (B) cos –1 (1/5)
(C) tan –1 (1/5) (D) None of these
18. Because the 20 m/s northward wind persists, the
pilot adjusts the heading so that the plane's total
velocity is eastward. By what angle does the new
heading differ form due east ?
(A) sin –1 1/5 (B) cos –1 1/5
(C) tan –1 1/5 (D) None of these
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
19. Column-I Column-II
→ →
. b
(A) a (P) Area of the
Parallelogram
formed
(B) L
(Q) Area below y-x
graph within the
limits given
(C)
→ →
a × b
(R) Projection of
→ →
b on a
x
(D)
∫ f ydx
x t
(S) Aerial velocity
(T) Zero for → a ⊥ → b
20. Column-I Column-II
(A) Horizontal range is (P) π/2
maximum, when the
projection is vertical
(B) Vertical height is (Q) π
maximum, when the
angle of projection is
(C) x and y coordinates (R) π/4
changes in projectile
motion with
(D) The magnitude of
velocity with the
time in projection
(S) Time
(T) First decrease then
increase
XtraEdge for IIT-JEE 60
MAY 2010

CHEMISTRY
Question 1 to 8 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. A 0.518 g sample of lime stone is dissolved in HCl
and then the calcium is precipitated as CaC 2 O 4 .
After filtering and washing the precipitate, it
requires 40 ml of 0.250 N KMnO 4 , solution
acidified with H 2 SO 4 to titrate it as,
MnO 4 – + H + + C 2 O 4 –2 → Mn +2 + CO 2 + 2H 2 O
The percentage of CaO in the sample is
(A) 54.0% (B) 27.1% (C) 42% (D) 84%
2. CO 2 is a gas but SiO 2 is a solid, because –
(A) CO 2 has smaller molecular size but SiO 2 has
greater molecular size
(B) In CO 2 pπ – pπ bond exists between C and O
(C) In SiO 2 , pπ – pπ bond exists between Si and O
(D) CO 2 is a discrete molecule where weak
molecular force of attraction exist, while SiO 2 is
a large polymeric molecule
3. In BrF 3 molecule, the lone pairs occupy equatorial
position to minimize –
(A) lone pair-bond pair repulsion only
(B) bond pair-bond pair repulsion only
(C) lone pair-lone pair repulsion and lone pair-bond
pair repulsion
(D) lone pair-lone pair repulsion only
4. Which is/are correct statements ?
(A) A solute will dissolve in water if hydration
energy is greater than lattice energy
(B) If the anion is large compared to the cation, the
lattice energy will remain almost constant
(C) solubility of II A hydroxide is in order
Be(OH) 2 < Mg(OH) 2 < Ca(OH) 2 < Sr(OH) 2
(D) none is correct
5. Sulphur trioxide is prepared by the following two
reactions.
S 8 (s) + 8O 2 (g) → 8SO 2 (g)
2SO 2 (g) + O 2 (g) → 2SO 3 (g)
How many grams of SO 3 are produced from 1 mol
of S 8 ?
(A) 1280.0 (B) 640.0
(C) 960.0 (D) 320.0
6. Which of the following statements is correct ?
(A) The magnitude of the second electron affinity
of Sulphur is greater than that of Oxygen
(B) The magnitude of the second electron affinity
of Sulphur is less than that of Oxygen
(C) The first electron affinities of Bromine and
Iodine are approximately the same
(D) The first electron affinity of Fluorine is greater
than that of Chlorine
7. Iron forms two oxides, in first oxide 56 gram. Iron
is found to be combined with 16 gram oxygen and
in second oxide 112 gram. Iron is found to be
combined with 48 gram oxygen. This data satisfy
the law of -
(A) Conservation of mass
(B) Reciprocal proportion
(C) Multiple proportion
(D) Combining volume
8. Lattice energy of BeCO 3
(I) , MgCO 3
(II) and
CaCO 3
(III) are in the order -
(A) I > II > III (B) I < II < III
(C) I < III < II (D) II < I < III
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. Which of the following statements is/are correct ?
(A) Group 12(IIB) elements do not show
characteristic properties of transition metals
(B) Among transition elements, tungsten has the
highest melting point
(C) Among transition elements, group 3 (IIIB)
elements have lowest densities
(D) Transition metals are more electropositive than
alkaline earth metals.
10. 11.2 g of mixture of MCl (volatile) and NaCl gave
28.7 g of white ppt with excess of AgNO 3 solution.
11.2 g of same mixture on heating gave a gas that
on passing into AgNO 3 solution gave 14.35 g of
white ppt. Hence ?
(A) Ionic mass of M + is 18
(B) Mixture has equal mol fraction of MCl and
NaCl
(C) MCl and NaCl are in 1 : 2 molar ratio
(D) Ionic mass of M + is 10
11. Specify the coordination geometry around the
hybridization of N and B atoms in a 1 : 1 complex
of BF 3 and NH 3 -
(A) N : tetrahedral, sp 3 ;B : tetrahedral, sp 3
(B) N : pyramidal, sp 3 ; B : pyramidal, sp 3
(C) N : pyramidal, sp 3 ; B : planar, sp 2
(D) N : pyramidal, sp 3 ; B : tetrahedral, sp 3
12. The radii of two of the first four Bohr orbits of the
hydrogen atom are in the ratio 1 : 4. The energy
difference between them may be :
(A) Either 12.09 eV or 3.4 eV
(B) Either 2.55 eV or 10.2 eV
(C) Either 13.6 eV or 3.4 eV
(D) Either 3.4 eV or 0.85 eV
This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
XtraEdge for IIT-JEE 61
MAY 2010

Passage : I (No. 13 to 15)
The radius of the nucleus of an atom can be
approximately determined as,
r nu = (1.4×10 –13 )A 1/3
where A is mass number of the atoms and M is the
charge of the electron = 4.8 × 10 –10 esu.
The mass of α-particle = 4 × mass of H-atom
10
mass of hydrogen atom = ×10 –24 gm.
6
Consider during collision kinetic energy of
α-particle just equal to coulombic force of
repulsion.
The mass number of Au = 197
The mass number of He = 4
The atomic number of Au = 79
Given : (4) 1/3 = 1.59 and (197) 1/3 = 5.82
2×
79×
(4.8)
= 351
10.374
3 .51× 3 = 3.245
Plank’s constant, h = 6.625 × 10 –34 JS
13. What is the distance between the α-particle and Au
nucleus during the collision –
(A) 10.374 × 10 –13 cm (B) 10.374 Å
(C) 10.374 × 10 –10 cm (D) 10.374 nm
14. What should be the minimum velocity of the
α-particle to strike the nucleus of 79 Au 197 ?
(A) 3.245 × 10 8 m/s (B) 3.245 × 10 9 m/s
(C) 3.245 × 10 5 m/s (D) 3.245 × 10 7 m/s
15. What is the de-broglie’s wave length associated
with a α-particle while it is moving to colloide with
the Au nucleus ?
6.625× 6
(A) ×10 25 6 .625× 6
m (B) ×10 –15 m
4×
3.245
3.245
6.625× 6
(C) ×10 –15 6.625× 6
m (D) ×10 –15 m
4×
3.245
5×
3.245
Passage : II (No. 16 to 18)
When we use the concept that one mole of one
substance contains the same number of elementary
entities as one mole of any other substance we don't
actually need to know what that number is.
Sometimes however we will need to work with the
actual number of elementary entities in a mole of
substance. This number is called Avogadro's
number.
N A = 6.022137 × 10 23 mol –1
The unit mol –1 'which we read as' per mole signifies
that a collection of N A molecular level entities is
equivalent to one mole at the macroscopic level.
For example a mole of carbon contains 6.02 × 10 23
atoms of C. A mole of oxygen contains 6.02 × 10 23
molecules of O 2 .
2
16. The number of atoms present in 8 g of ozone is
(A) N A
(B) 3N A
(C)
N A
6
(D)
N A
17. Which of the following is a reasonable value for the
number of atoms in 1.00 g of Helium ?
(A) 0.25 (B) 4.0
(C) 4.1 × 10 –23 (D) 1.5 × 10 23
18. How many years it would take to spend Avogadro's
number of rupees at the rate of 10 Lac rupees per
second ?
(A) 1.9 × 10 10 years (B) 1.6 × 10 10 years
(C) 5 × 10 6 years (D) 6.4 × 10 5 years
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
2
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
19. Match the following :
Column-I
Column-II
–
(A) ICl 2 (P) Linear
+
(B) BrF 2 (Q) Pyramidal
–
(C) ClF 4 (R) Tetrahedral
–
(D) AlCl 4 (S) Square planar
(T) Angular
20. Match the following :
Column-I
Column-II
(A) 0.5 mole of SO 2 (g) (P) Occupy 11.2L
at STP
(B) 1g of H 2 (g) (Q) Weighs 24g
(C) 0.5 mole of O 3 (g) (R) Total no. of
atoms = 1.5N A
(D) 1g molecule of O 2 (g) (S) Weighs 32g
(T) Total no. of atoms
≈ 9 × 10 23
XtraEdge for IIT-JEE 62
MAY 2010

MATHEMATICS
Question 1 to 8 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. 3 cosec 20° – sec 20° is equal to :
(A) 1 (B) 2
(C) 4
(D) none of these
2. If cos 20° – sin 20° = p then cos 40° is equal to-
(A)
(C)
2
− p 2 − p (B) p 2 − p
2
p + 2 − p (D) none of these
3. The expression
⎡ 4⎛
3π
⎞ 4 ⎤
3⎢sin
⎜ – α⎟ + sin (3π
– α)
⎥
⎣ ⎝ 2 ⎠
⎦
⎡ 6⎛
π ⎞
– 2⎢sin
⎜ + α⎟ + sin
⎣ ⎝ 2 ⎠
is equal to -
(A) 0 (B) 1
(C) 2 (D) 5
2
6
⎤
(5π
– α)
⎥
⎦
4.
π tan α
If α + β = and β + γ = α, then 2 tanβ + 2 tan γ
(A) 0 (B) 1
(C) 2 (D) 3
5. If 1< x < 2 , then number of solutions of the equation
tan –1 (x – 1) + tan –1 x + tan –1 (x + 1) = tan –1 3x, is/are
(A) 0 (B) 1
(C) 2 (D) 3
6. In a triangle ABC if BC =1 and AC = 2. Then the
maximum possible value of angle A is-
(A) π/6 (B) π/4
(C) π/3 (D) π/2
7. If α ∈
⎛ 3π
⎞
⎜− , − π⎟ ,
⎝ 2 ⎠
then the value of
tan –1 (cot α) – cot –1 (tan α) + sin –1 (sin α) + cos –1 (cos α) is
equal to -
(A) 2π + α
(B) π + α
(C) 0
(D) π – α
=
8. Solution of the equation
3 sin –1 ⎛ 2x ⎞ ⎛ 2
⎜ ⎟
⎝1+ x
2 – 4 cos –1
⎠
⎟ ⎟ ⎞
⎜1−
x
+
⎜ 2
⎝1+
x ⎠
2 tan –1 ⎛ 2x ⎞ π
⎜ ⎟
⎝1− x
2 = is -
⎠ 3
(A) x = 3 (B) x =
3
(C) x = 1 (D) x = 0
Questions 9 to 12 are multiple choice questions. Each
questions has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
9. If sin θ + sin φ = a and cos θ + cos φ = b , then -
⎛ θ – φ ⎞ 1
2
(A) cos⎜
⎟ = ± (a
2 + b )
⎝ 2 ⎠ 2
⎛ θ – φ ⎞ 1 2 2
(B) cos⎜
⎟ = ± (a – b )
⎝ 2 ⎠ 2
⎛ θ – φ ⎞
(C) tan ⎜ ⎟ = ±
⎝ 2 ⎠
(D) cos (θ – φ) =
a
⎛
2
⎜
4 – a – b
2 2
⎝ a + b
2 +
b
2
2
– 2
10. Let y = sin x . sin (60 º + x) . sin(60 º - x). Then for
all real x -
(A) the minimum value of y is 2
(B) the maximum value of y is 1
(C) y ≤ 1/4
(D) y ≥ –1/4
11. If cosec –1 x = sin –1 (1/x), then x may be-
(A) 1 (B) –1/2 (C) 3/2 (D) – 3/2
12. If in a triangle ABC, θ is the angle determined by
cos θ = (a – b)/c, then
( a + b)sin θ A − B
(A)
= cos
2 ab
2
( a + b)sin θ A + B
(B)
= cos
2 ab
2
csin
θ A − B
(C) = cos
2 ab 2
csin
θ A + B
(D) = cos
2 ab 2
2
1
⎞
⎟
⎠
XtraEdge for IIT-JEE 63
MAY 2010

This section contains 2 paragraphs, each has 3
multiple choice questions. (Question 13 to 18) Each
questions has 4 choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
Passage : I (No. 13 to 15)
f(x) = sin {cot –1 (x + 1)} – cos (tan –1 x)
a = cos tan –1 sin cot –1 x
b = cos (2 cos –1 x + sin –1 x)
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
13. The value of x for which f(x) = 0 is
(A) – 1/2 (B) 0
(C) 1/2 (D) 1
14. If f(x) = 0 then a 2 is equal to
(A) 1/2 (B) 2/3
(C) 5/9 (D) 9/5
15. If a 2 = 26/51, then b 2 is equal to
(A) 1/25 (B) 24/25
(C) 25/26 (D) 50/51
Passage : II (No. 16 to 18)
In a ∆ABC, if cos A. cos B. cos C =
3 + 3
sin A. sin B. sin C = , then
8
16. Value of tan A + tan B + tan C is
3 + 3
3 + 4
(A)
(B)
3 −1
3 −1
(C)
6 −
3
3 −1
(D) None of these
17. Value of Σ tan A .tan B =
(A) 5 – 4 3 (B) 5 + 4 3
(C) 6 + 4 3 (D) 6 – 4 3
18. Value of tan A, tan B and tan C are
(A) 1, 3 , 2 (B) 1, 3 , 2
(C) 1, 2, 3 (D) None of these
3 −1
8
Mark your response in OMR sheet against the
question number of that question in section-II. + 8
marks will be given for complete correct answer (i.e.
+2 marks for each correct row) and No Negative
marks for wrong answer.
19. Match the items of Column I with the items of
Column II. The principal value of
Column-I
Column-II
− ⎛ 5π
⎞
(A) sin 1 17π
⎜sin
⎟ is (P)
⎝ 6 ⎠ 20
− ⎛ 7π
⎞
(B) cos 1 1
⎜−
sin ⎟ is (Q)
⎝ 6 ⎠ 2
⎧
(C)
− ⎛ 15π
⎞⎫
π
1
cos⎨tan
⎜ tan ⎟⎬
is (R)
⎩ ⎝ 4 ⎠⎭
3
− ⎛
(D) cos 1
⎜
⎝
1 ⎛ 9π
9π
⎞⎞
π
⎜cos
− sin ⎟
⎟ is (S)
2 ⎝ 10 10 ⎠⎠
6
(T) 2
π
20. Column-I Column-II
(A) If in a ∆ABC the angles are
5π
(P) 12
in AP and b : c = 3 : 2
then ∠A =
(B) If the length of the side of a
2π
(Q)
3
triangle are 3, 5, 7 then largest
angle of the triangle is
(C) If the sides of a triangle are
π
(R) 6
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with
statements (P, Q, R, S, T) in Column II. The answers
to these questions have to be appropriately bubbled
as illustrated in the following example. If the correct
in ratio 2 : 6 : ( 3 + 1) then
largest angle of the triangle is
(D) If in a triangle ABC, b = 3 , (S) π
c = 1 and B – C = 90º then
∠A =
(T) 3π/5
XtraEdge for IIT-JEE 64
MAY 2010

XtraEdge for IIT-JEE 65
MAY 2010

IIT-JEE 2010
PAPER-I (PAPER & SOLUTION)
Time : 3 Hours Total Marks : 243
Instructions :
• The question paper consists of 3 parts (chemistry, Mathematics and Physics). Each part consists of four sections.
• For each question in Section I, you will be awarded 3 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 one (–1) mark will be
awarded.
• For each question in Section II, you will be awarded 3 marks if you darkem only the bubble corresponding to the
correct answer and zero mark if no bubbles are darkened. Partial marks will be awarded for partially correct
answers. No negative marks will be awarded 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 3 marks if you darken the bubble corresponding to the
correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in this section.
CHEMISTRY
SECTION – I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each
question has 4 choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. The bond energy (in kcal mol –1 ) of a C–C single
bond is approximately
(A) 1 (B) 10
(C) 100 (D) 1000
Ans. [C]
Sol. Value is 82.6 kcal/mol.
2. The species which by definition has ZERO
standard molar enthalpy of formation at 298 K is
(A) Br 2 (g)
(B) Cl 2 (g)
(C) H 2 O (g)
(D) CH 4 (g)
Ans. [B]
Sol. Because standard state of Cl 2 is gas.
3. The ionization isomer of [Cr(H 2 O) 4 Cl(NO 2 )] Cl
is
(A) [Cr(H 2 O) 4 (O 2 N)] Cl 2
(B) [Cr(H 2 O) 4 Cl 2 ](NO 2 )
(C) [Cr(H 2 O) 4 Cl(ONO)]Cl
(D)
[Cr(H 2 O) 4 Cl 2 (NO 2 )]. H 2 O
Ans. [B]
Sol.
Ionization isomer have different ions in solution
4. The correct structure of ethylenediaminetetraacetic
acid (EDTA) is -
HOOC–CH 2
CH 2 –COOH
Ans.
Sol.
(A)
HOOC–CH 2
(B)
HOOC
HOOC
HOOC–CH 2
(C)
HOOC–CH 2
(D)
HOOC–CH 2
N–CH=CH–N
N–CH 2 –CH 2 –N
H
N–CH 2 –CH 2 –N
COOH
CH 2
N–CH–CH–N
HOOC
[C]
Structure is
HOOC–CH 2
HOOC–CH 2
CH 2
CH 2 –COOH
COOH
COOH
CH 2 –COOH
H
N–CH 2 –CH 2 –N
CH 2 –COOH
CH 2 –COOH
CH 2 –COOH
CH 2 –COOH
5. The synthesis of 3-octyne is achieved by adding a
bromoalkane into a mixture of sodium amide and
an alkyne. The bromoalkane and alkyne
respectively are
(A) BrCH 2 CH 2 CH 2 CH 2 CH 3 and CH 3 CH 2 C≡CH
(B) BrCH 2 CH 2 CH 3 and CH 3 CH 2 CH 2 C≡CH
(C) BrCH 2 CH 2 CH 2 CH 2 CH 3 and CH 3 C≡CH
(D) BrCH 2 CH 2 CH 2 CH 3 and CH 3 CH 2 C ≡CH
XtraEdge for IIT-JEE 66
MAY 2010

Ans.
[D]
Sol. (i) CH 3 –CH 2 –C≡CH
NaNH 2
CH 3 CH 2 C≡
C –
C –
+
Na
(ii) CH 3 –CH 2 –CH 2 –CH 2 –Br + CH 3 CH 2 –C≡
+
Na
––→ CH 3 –CH 2 –CH 2 –CH 2 –C ≡C–CH 2 –CH 3
6. The correct statement about the following
disaccharide is
CH 2 OH
H O
CH 2 OH O H
H
H (a)
OH H
(b) OH
OCH 2 CH 2 O
H
OH
CH 2 OH
H OH
OH H
(A) Ring (a) is pyranose with α-glycosidic link
(B) Ring (a) is furanose with α-glycosidic link
(C) Ring (b) is furanose with α-glycosidic link
(D) Ring (b) is Pyranose with β-glycosidic link
Ans. [A]
Sol. Ring (a) is pyranose with α-glycosidic link
7. Plots showing the variation of the rate constant
(k) with temperature (T) are given below. The
plot that follows Arrhenius equation is -
Ans.
Sol.
(A)
k
(C)
k
T
T
[A]
k = Ae –Ea/RT ; k ∝ e
k
T
– E a / RT
(B)
k
(D)
k
8. In the reaction OCH 3 HBr the
products are
(A) Br
OCH 3 and H 2
(B)
Br and CH 3 Br
T
T
Ans.
Sol.
(C)
(D)
[D]
Br and CH 3 OH
OH and CH 3 Br
OCH 3
HBr
SECTION – II
Multiple Correct Choice Type
OH + CH 3 Br
This section contains 5 multiple choice questions. Each
questions has 4 choices (A), (B), (C) and (D), out of
which ONE OR MORE is/are correct.
9. In the reaction
intermediate(s) is (are)
(A)
O
Br
O
(C)
Br
Ans. [A, B, C]
OH
Br
Sol. NaOH(aq)/Br 2
Br
OH
(B)
(D)
NaOH(aq)/Br 2
OH
Br
Br
O
Br Br
O
Br
the
Bromination take place at ortho & para position
due to activation of benzene ring by –OH group.
10. The reagent(s) used for softening the temporary
hardness of water is (are) -
(A) Ca 3 (PO 4 ) 2 (B) Ca(OH) 2
(C) Na 2 CO 3
(D) NaOCl
Ans. [B, C, D]
11. Aqueous solutions of HNO 3 , KOH, CH 3 COOH
and CH 3 COONa of identical concentrations are
provided. The pair(s) of solutions which from a
buffer upon mixing is(are)
XtraEdge for IIT-JEE 67
MAY 2010

(A) HNO 3 and CH 3 COOH
(B) KOH and CH 3 COONa
(C) HNO 3 and CH 3 COONa
(D) CH 3 COOH and CH 3 COONa
Ans. [C, D]
Sol. Acidic buffer is made up of weak acid & it's
conjugate ion
12. In the Newman projection for 2,2-dimethylbutane
X
Ans. [B, D]
H 3 C
CH 3
H H
Y
X and Y can respectively be
(A) H and H (B) H and C 2 H 5
(C) C 2 H 5 and H (D) CH 3 and CH 3
CH 3
Sol. CH 3 –C–CH 2 –CH 3
1 2 3 4
CH 3
along C 2 –C 3 Bond
X = CH 3
Y = CH 3
CH 3
CH 3 –C–CH 2 –CH 3
1 2 3 4
CH 3
along C 1 –C 2
X = H
Y = C 2 H 5
≡
≡
H
C H 3
H
C H 3
CH 3
CH 3
H
C 2 H 5
CH 3
H
Option [D]
CH 3
H
Option [B]
13. Among the following, the intensive property is
(properties are)
(A) molar conductivity
(B) electromotive force
(C) resistance
(D) heat capacity
Ans. [A, B]
SECTION – III
Paragraph Type
This section contains 2 paragraphs. Based upon the
first paragraph 2 multiple choice questions and based
upon the second 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.
Paragraph for Questions No. 14 to 15
The concentration of potassium ions inside a
biological cell is at least twenty times higher than
the outside. The resulting potential difference
across the cell is important in several processes
such as transmission of nerve impulses and
maintaining the ion balance. A simple model for
such a concentration cell involving a metal M is :
M(s) | M + (aq; 0.05 mloar) || M + (aq; 1 molar) |
M(s)
For the above electrolytic cell the magnitude of
the potential |E cell | = 70 mV.
14. For the above cell
(A) E cell < 0; ∆G > 0 (B) E cell > 0; ∆G < 0
(C) E cell < 0; ∆Gº > 0 (D) E cell > 0; ∆Gº < 0
Ans. [B]
Sol. For concentration cell
k 0.05
E cell = – log n 1
[where k is +ve constant]
= + ve
∴ E cell > 0 ; ∆G < 0
15. If the 0.05 molar solution of M + is replaced by a
0.0025 molar M + solution, then the magnitude of
the cell potential would be
(A) 35 mV
(B) 70 mV
(C) 140 mV
(D) 700 mV
Ans. [C]
Sol.
k 0.0025
E cell = – log n 1
⎡ k 0.05⎤
= 2 ⎢– log ⎥ = 2 × 70 = 140
⎣ n 1 ⎦
Paragraph for Question No. 16 to 18
Copper is the most noble of the first row
transition metals and occurs in small deposits in
several countries. Ores of copper include
chalcanthite (CuSO 4 . 5H 2 O), atacamite (Cu 2 Cl
(OH) 3), cuprite (Cu 2 O), copper glance (Cu 2 S) and
malachite (Cu 2 (OH) 2 CO 3 ). However, 18% of the
world copper production come from the ore
chalcopyrite (CuFeS 2 ). The extraction of copper
from chalcopyrite involves partial roasting,
removal of iron and self-reduction.
XtraEdge for IIT-JEE 68
MAY 2010

16. Partial roasting of chalcopyrite produces
(A) Cu 2 S and FeO (B) Cu 2 O and FeO
(C) CuS and Fe 2 O 3 (D) Cu 2 O and Fe 2 O 3
Ans. [A]
17. Iron is removed from chalcopyrite as -
(A) FeO
(B) FeS
(C) Fe 2 O 3 (D) FeSiO 3
Ans. [D]
18. In self-reduction, the reducing species is -
(A) S (B) O 2–
(C) S 2– (D) SO 2
Ans. [C]
Sol.
1. O 3
2. Zn, H 2O
O
O
O
(Only one product)
OH –
Heating
23. Amongst the following, the total number of
compound soluble in aqueous NaOH is
O
O
O
OH
SECTION – IV
Integer Type
This section contains. TEN 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.
235
19. The number of neutrons emitted when 92 U
undergoes controlled nuclear fission to
142
90
54 Xeand 38Sr
is
Ans. [4]
Sol.
235
1
0
142
92 U + n → 54 Xe + 38Sr
+ x n +
235 + 1 = 142 + 90 + x
x = 4
90
1
0
1
1
y p
H 3 C
N CH 3
NO 2
Ans. [4]
Sol.
COOH OCH 2 CH 3
CH 2 OH
OH
N
H 3 C CH 3
COOH
CH 2 CH 3
CH 2 CH 3
OH
OH
COOH
20. The total number of basic groups in the following
form of lysine is
⊕
H 3 N–CH 2 –CH 2 –CH 2 –CH 2 O
Ans. [2]
H 2 N
CH–C
21. The total number of cyclic isomers possible for a
hydrocarbon with the molecular formula C 4 H 6 is
Ans. [5]
Sol.
, , , ,
22. In the scheme given below, the total number of
intramolecular aldol condensation products form
'Y' is
Ans. [1]
1. O 3 1. NaOH(aq)
Y
2. Zn, H 2O 2. heat
O
OH
N
H 3 C CH 3
,
,
COOH
24. Amongst the following, the total number of
compounds whose aqueous solution turns red
litmus paper blue is
KCN K 2 SO 4 (NH 4 ) 2 C 2 O 4 NaCl Zn(NO 3 ) 2
FeCl 3 K 2 CO 3 NH 4 NO 3 LiCN
Ans. [3]
Sol. Basic salt are KCN, LiCN, K 2 CO 3
25. Based on VSEPR theory, the number of 90 degree
F–Br–F angles in BrF 5 is
Ans. [0]
26. The value of n in the molecular formula
Be n Al 2 Si 6 O 18 is
XtraEdge for IIT-JEE 69
MAY 2010

Ans. [3]
Sol. Be 3 Al 2 Si 6 O 18
Si 6 O 18 –12 ion
27. A student performs a titration with different
burettes and finds titre values of 25.2 mL, 25.25
mL and 25.0 mL. The number of significant
figures in the average titre value is
Ans. [3]
Sol.
25 .2 + 25.25 + 25.0
= 25.15
3
Significant figure = 3
Significant figure in the answer can not be more
than least significant figure any given value.
28. The concentration of R in the reaction R → P was
measured as a function of time and the following
data is obtained :
[R] 1.0 0.75 0.40 0.10
(molar)
t (min.) 0.0 0.05 0.12 0.18
The order of the reaction is
Ans. [0]
Sol.
R → P
Assume zero order
R = [R 0 ] – kt
[ R 0 ] −[R]
k =
t
1− 0.75
k 1 = = 5
0.5
1− 0.4
k 2 = = 5
0.12
1− 0.1
k 3 = = 5
0.18
∴ order of reaction should be zero.
MATHEMATICS
SECTION – I
Single Correct Choice Type
This section contains 8 multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
29. Let f , g and h be real-valued functions defined on
2
x x
the interval [0, 1] by f(x) =
2 −
e + e ,
2
x x
g(x) =
2
2
−
2 x x
xe + e and h(x) =
2 −
x e + e .
If a , b and c denote, respectively, the absolute
maximum of f, g and h on [0, 1], then
Ans.[D]
(A) a = b and c ≠ b
(C) a ≠ b and c ≠ b
Sol. f '(x) = 2x( ⎜
⎛ x 2 ⎟
⎞
⎝
− − x
e e
2
⎠
x 2
g'(x) = e ( 2x 2x 1)
2 − +
h'(x) =
3 2
2 x e x
(B) a = c and a ≠ b
(D) a = b = c
Q all f'(x) , g'(x), h'(x) are positive so all attains
absolute maxima at x = 1
So Q f (1) = g(1) = h(1) = e + e –1 = a = b = c
30. Let p and q be real numbers such that p ≠ 0, p 3 ≠
q and p 3 ≠ – q. If α and β are non zero complex
numbers satisfying α + β = – p and α 3 + β 3 = q,
Ans.[B]
then a quadratic equation having β
α and α
β as its
roots is -
(A) (p 3 + q) x 2 – (p 3 + 2q) x + (p 3 + q) =0
(B) (p 3 + q) x 2 – (p 3 – 2q) x + (p 3 + q) =0
(C) (p 3 – q) x 2 – (5p 3 – 2q) x + (p 3 – q) =0
(D) (p 3 – q) x 2 – (5p 3 + 2q) x + (p 3 – q) =0
Sol. α + β = – p ……….(1)
α 3 + β 3 = q
⇒ (α+ β) ( α 2 + β 2 – αβ) = q
⇒ (α+ β) (( α + β) 2 – 3αβ) = q
⇒ (–p) (p 2 – 3αβ) = q
Now S =
αβ =
α β
+
β α
(Sum of root) S =
q + p
3p
=
p
3
p
…..(2)
2
( α + β)
− 2αβ
αβ
3
3
− 2q
+ q
using (1) and (2)
XtraEdge for IIT-JEE 70
MAY 2010

(Product of root) P =
α β . = 1
β α
31. Equation of the plane containing the straight line
x y z
= = and perpendicular to the plane
2 3 4
x y z
containing the straight lines = = and
3 4 2
x y z
= = is
4 2 3
(A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0
Ans.[C]
Sol.
(C) x – 2y + z = 0 (D) 5x + 2y – 4z = 0
Plane passing through origin (0, 0, 0) and normal
vector to plane is perpendicular to 3 î + 4ĵ + 2kˆ
,
4 î + 2ĵ + 3kˆ and 2 î + 3ĵ + 4kˆ
i.e. normal vector
to plane is î − 2ĵ + kˆ so equation to plane is x –
2y + z = 0.
32. If the angle A, B and C of a triangle are in an
arithmetic progression and if a, b and c denote the
lengths of the sides opposite to A, B and C
respectively, then the value of the expression
Ans.[D]
Sol.
a
sin 2C
c
(A) 2
1
c
+ sin 2A is -
a
(B)
3
2
(C) 1 (D) 3
a c
sin 2C + sin 2A
c a
a
c
=
2sin C cosC +
c
a
a cosC + ccos A
R
2sin AcosA
b 2R sin B
= = R R
= 3
33. Let ω be a complex cube root of unity with ω
≠ 1. A fair die is thrown three times If r 1 , r 2 and r 3
are the numbers obtained on the die, then the
r1 r2
r3
probability that ω + ω + ω = 0 is
Sol. n(s) = 6 3
Similarly total number of elements in events set is
48
48 12 2
= = = 216 54 9
34. Let P, Q, R and S be the points on the plane with
position vectors – 2î
− ĵ, 4 î , 3 î + 3ĵ
and
Ans.[A]
Sol.
r 1 r 2 r 3
1 1
2
3
4
5
6
x
3,6
2,5
x
3,6
2,5
− 3 î + 2ĵ respectively. The quadrilateral PQRS
must be a -
(A) Parallelogram, which is neither a rhombus
nor a rectangle
(B) Square
(C) Rectangle, but not a square
(D) Rhombus, but not a square
PQ = 6i +
RS = 6i +
2 1
2
3
4
5
6
3 1
2
3
4
5
6
j
j
RQ = i − 3j
3,6
x
1,4
3,6
x
1,4
2,5
1,4
x
2,5
1,4
x
(A) 18
1
(B) 9
1
(C) 9
2
(D) 36
1
SP = i − 3j
Ans.[C]
PQ ≠ RQ (∴ not a rhombus or a rectangle)
PQ || RS
XtraEdge for IIT-JEE 71
MAY 2010

RQ || SP
Also PQ . RQ ≠ 0
∴ PQRS is not a square
⇒ PQRS is a parallelogram
35. The value of
Ans.[B]
Sol.
x lim →0
t ln(1
+ t)
dt
3 ∫
is
4
t + 4
x1 x 0
(A) 0 (B) 12
1
1
(C) 24
Use L'hospital rule in
1 t ln(1
x +
Lim
x → 0 3 ∫ 0 4
x t + 4
x ln(1
+ x)
= Lim
x → 0 4 2
+
= Lim
=
(x
4)3x
ln(1
+ x)
x
x → 0 .3(x
2 + 4 )
1 1 =
3.4 12
t)
dt
1
(D) 64
36. The number of 3 × 3 matrices A whose entries are
either 0 or 1 and for which the system
⎡x⎤
⎡1⎤
A
⎢ ⎥
=
⎢ ⎥
⎢
y
⎥ ⎢
0
⎥
has exactly two distinct solution, is -
⎢⎣
z⎥⎦
⎢⎣
0⎥⎦
(A) 0 (B) 2 9 – 1
(C) 168 (D) 2
Ans.[A]
SECTION – II
Multiple Correct Choice Type
This section contains 5 multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE may be correct.
37. Let ABC be a triangle such that ∠ ACB = 6
π and
let a, b and c denote the lengths of the sides
opposite to A, B and C respectively. The value(s)
of x for which a = x 2 + x +1, b = x 2 – 1 and c = 2x
+ 1 is (are)
(A) − ( 2 + 3)
(B) 1+ 3
(C) 2 + 3
(D) 4 3
Ans.[B]
Sol.
As sum of two sides is always greater than third
side, So x > 1
Now
2 +
π a b – c
Cos = 6 2ab
(a – b) 2ab – c
=
23 2 +
2ab
2 2
(a – b) – c
3 – 2 =
ab
3 – 2 =
(x
2
(x + 2)
2
2
2
+ x + 1)
2
– (2x + 1)
(x
2
2
–1)
–3
3 – 2 =
2
x + x + 1
x 2 + x + 1 =
3
2 – 3
x 2 + x + 1 = 3(2 + 3 )
x 2 + x – 5 – 3 3 = 0
(x – (1 + 3 ) (x + (2 + 3 )) = 0
x = 1 + 3 , – (2 + 3 )
So x = 3 + 1
38. Let A and B be two distinct points on the
parabola y 2 = 4x. If the axis of the parabola
touches a circle of radius r having AB as its
diameter, then the slope of the line joining A and
B can be -
1
1
(A) − (B)
r
r
2
2
(C) (D) −
r r
Ans.[C, D]
2
2
Sol. Let A ( t 1 ,2t1)
B ( t 2,2t
2 )
2(t
Slope =
2 − t1)
2
=
2 2
t 2 − t1
t 1 + t 2
Equation of circle will be
2 2
(x − t1
) (x − t 2)
+ ( y − 2t1
) ( y − 2t
2)
= 0
2 2 2 2
x + y – x(t1 + t 2)
– 2y(t1 + t 2)
+ 2 2
t 1 t 2 + 4t 1 t 2
= 0
As it touches x axis so
2 2 2
2 2 (t t )
t 1 t 2 + 4t 1 t 2 =
1 + 2
4
2 2
4 4 2 2
4t 1 t 2 + 16 t 1 t 2 = t 1 + t 2 + 2 t 1 t 2
2 2
( ) 2
t1 − t 2 = 16 t 1 t 2 . . . (1)
AB is diameter so
2 2 2
1 t 2)
( t − + 4 (t 1 – t 2 ) 2 = 4r 2 . . . (2)
XtraEdge for IIT-JEE 72
MAY 2010

From (1) and (2)
4 t 1 t 2 + (t 1 – t 2 ) 2 = r 2
(t 1 + t 2 ) 2 = r 2
t 1 + t 2 = ± r
2
∴ Slope = ± r
x (1 − x)
39. The value(s) of
∫ 2
1+
x
(A)
(C) 0
22
− π
7
Ans.[A]
1 4
x (1 − x)
Sol. I =
∫ 2
1+
x
0
4
1
0
4
4
dx is (are) -
2
(B) 105
71 3π
(D) −
15 2
1 8 7 6 5 4
x − 4x + 6x − 4x + x
=
∫
dx
2
0 1+
x
1
6 5 4 2 4
=
∫
( x − 4x + 5x − 4x + 4 − )dx
2
0
1+
x
22
= − π
7
40. Let z 1 and z 2 be two distinct complex numbers let
z = (1–t) z 1 + tz 2 for some real number t with 0 < t
< 1.
If Arg (w) denotes the principal argument of a
nonzero complex number w, then
(A) | z – z 1 | + | z – z 2 | = | z 1 – z 2 |
(B) Arg (z – z 1 ) = Arg (z – z 2 )
(C)
z − z1
z2
− z1
z − z1
z2
− z1
= 0
(D) Arg (z – z 1 ) = Arg (z 2 – z 1 )
Ans.[A,C,D]
Sol. t =
z − z
z −
2
1
z1
z − z1
So, = t e io V t ∈(0, 1)
z2
− z1
So Geometrically
z 1 z z 2
So option A, C, D are true.
41. Let f be a real valued function defined on the
x
interval (0, ∞ ) by f(x) = ln
x +
∫
1 + sin t dt .
0
Then which of the following statement(s) is (are)
true ?
(A) f"(x) exists for all x ∈ (0, ∞ )
(B) f '(x) exists for all x ∈ (0, ∞ ) and f ' is
continuous on (0, ∞ ), but not differentiable
on (0, ∞ )
(C) there exists α > 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]
x
f(x) = ln x +
∫
1+
sin t d t
0
1
1 x x
f ′(x) = + 1+
sin x = + |cos + sin |
x
x 2 2
(A is not correct)
1 x
+ 1+ sin x < ln x + 1 sin t dt
x
∫
+
0
Q ln x > x
1 for some x = α ∀ α > 1
and
x
1+ sin x <
∫
1 + sin t dt
for some x = α ∀ α > 1
so option C is correct
0
SECTION – III
Paragraph Type
This section contains 2 paragraphs. Based upon the
first paragraph 2 multiple choice questions and based
upon the second 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.
Paragraph for Question No. 42 to 43
The circle x 2 + y 2 x 2
– 8x = 0 and hyperbola – 9
y 2
= 1 intersect at the points A and B
4
42. Equation of a common tangent with positive slope
to the circle as well as to the hyperbola is -
(A) 2x – 5 y – 20 = 0
(B) 2x – 5 y + 4 = 0
(C) 3x – 4y + 8 = 0
(D) 4x – 3y + 4 = 0
Ans.[B]
XtraEdge for IIT-JEE 73
MAY 2010

Sol. y = m(x – 4) ± 4
y = mx ± 9m
2 − 4
– 4m ± 4
2
m
2
1+ m
1+ = ± 9m
2 − 4
16m 2 + 16 + 16m 2 m 32 m
2
2
1+ m = 9m 2 – 4
m 32m 1+ m = – 23m 2 – 20
1024m 2 + 1024 m 4 = 529m 4 + 400 + 920 m 2
495 m 4 + 104 m 2 – 400 = 0
(5m 2 – 4) (99m 2 + 100) = 0
∴ m 2 4
2
= ∴ m = ± 5 5
So tangent with positive slope
y =
2 4 x ±
5 5
2x – 5 y ± 4 = 0
43. Equation of the circle with AB as its diameter is
(A) x 2 + y 2 – 12 x + 24 = 0
(B) x 2 + y 2 + 12 x + 24 = 0
(C) x 2 + y 2 + 24 x – 12 = 0
(D) x 2 + y 2 – 24x – 12 = 0
Ans.[A]
Sol. x 2 + y 2 – 8x = 0
4x 2 – 9y 2 = 36
⎛ ⎞
x 2 + ⎜
4x
2 − 36
⎟ – 8x = 0
⎝ 9 ⎠
13x 2 – 72x – 36 = 0
(x – 6) (13x + 6) = 0
−6
x = 6, 13
x = 6, y = ± 12
∴ Equation of required circle
(x – 6) 2 + (y – 12 ) (y + 12 ) = 0
x 2 + y 2 – 12x + 24 = 0
Paragraph for Question No. 44 to 46
Let p be an odd prime number and T p be the
following set of 2 × 2 matrices :
⎪⎧
⎡a
b⎤
⎪⎫
= ⎨A
= ⎢ ⎥ : a, b,c ∈ { 0,1,2,......, p −1} ⎬
⎪⎩ ⎣c
a⎦
⎪⎭
T p
44. The number of A in T p such that A is either
symmetric or skew-symmetric or both, and det
(A) divisible by p is -
(A) (p – 1) 2 (B) 2(p – 1)
(C) (p –1) 2 + 1 (D) 2p –1
Ans.[D]
Sol.
|A | = a 2 – 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 T p 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.]
(A) (p – 1) (p 2 – p + 1) (B) p 3 – (p – 1) 2
(C) (p –1) 2 (D) (p –1) (p 2 – 2)
Ans.[C]
46. The number of A in T p such that det (A) is not
divisible by p is -
(A) 2p 2
(B) p 3 – 5p
(C) p 3 –3p (D) p 3 – p 2
Ans. [D]
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.
47. 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]
dy
Sol. Y – y = ( X – x) dx
dy
y – x = x
3
dx
dy
x – y = – x
3
dx
T y – = – x
2
T x
1
2
I.F. = x
1
y =
x ∫ − xdx + C
y − x 2
= + C
x 2
1
3
1 = − + C ∴ C =
2
2
∴ y =
x( −x
2
2 +
3)
XtraEdge for IIT-JEE 74
MAY 2010

2 +
x( −x
3)
∴ f(x) =
2
−3 ( −9
+ 3)
f(–3) =
= 9
2
48. The number of values of θ in the interval
⎛ π π ⎞
nπ
⎜−
, ⎟ such that θ ≠ for n = 0 , ± 1, ± 2
⎝ 2 2 ⎠
5
and tan θ = cot 5 θ as well as sin 2 θ = cos 4 θ is –
Ans.[3]
π
Sol. tan θ = cot 5θ ⇒ θ = (2n + 1) 12
π π 5π
So θ = ± , ± , ± 12 4 12
1
∴ Sin 2θ = cos 4θ ⇒ sin 2θ = 2
or – 1
π 5π –π
only θ = , , satisfies the given
12 12 4
conditions
So total number of solution = 3
49. The maximum value of the expression
1
is
2
2
sin θ + 3sin θcosθ + 5cos θ
Sol.[2] f (θ) =
sin
2
1
θ + 3sin θcos
θ + 5cos
1
=
2
1+ 4cos θ + 3sin θcosθ
1
=
3
1+ 2(1 + cos 2θ)
+ sin 2θ
2
1
=
3
3 + 2cos 2θ + sin 2θ
2
1
So f(θ) max = = 2
9
3 – 4 +
4
50. If a r and b r are vector is space given by
r î − 2ĵ 2î + ĵ+
3kˆ
a = and bˆ = , then the value of
5
14
r r r r r r
2a + b . a × b × a − 2b is
( ) [( ) ( )]
Ans.[5]
Sol. | a | = | b | = 1& a ⋅ b = 0
( 2a + b) ⋅[(a
× b) × (a – 2b)]
2 2
= ( 2 a + b) ⋅[b
+ 2a] = | b | + 4 | a | = 5
2
θ
51. The line 2x + y = 1 is tangent to the hyperbola
2
2
x y − =1. If this line passes through the point
2 2
a b
of intersection of the nearest directrix and the x-
axis, then the eccentricity of the hyperbola is
Ans.[2]
Sol. 1 = 4a 2 – b 2 ... (1)
2a
= 1
e
e
a = ... (2)
2
also b 2 = a 2 (e 2 – 1) ... (3)
(1) & (3)
1 = 4a 2 – a 2 e 2 + a 2 ⇒ 1 = 5a 2 – a 2 e 2
2
5 e e
⇒ 1 = –
4 4
⇒ e 4 – 5e 2 + 4 = 0
⇒ (e 2 – 4) (e 2 – 1) = 0
∴ e = 2
4
52. 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]
Sol.
î
2
3
6 , then | d | is –
ĵ
3
4
kˆ
4
5
= î(
− 1) − ĵ( −2)
+ kˆ( −1)
Plane is normal to vector î − 2ĵ+
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
Ans.[4]
f (x)
⎧ x −[x]
= ⎨
⎩1
+ [x] − x
Then the value of
if [x]
is odd
if [x]is even
2
π 10
∫ f (x)
10 − 10
⎧ {x} V[x]is odd
Sol. f(x) = ⎨
⎩1
= {x} V[x]is even
graph of y = f(x) is
cos πx dx is
XtraEdge for IIT-JEE 75
MAY 2010

56. The number of all possible values of θ, where
0 < θ < π, for which the system of equations
–5 –4 –3 –2 –1 0 1 2 3
(y + z) cos 3 θ = (xyz) sin 3θ
Q f(x) & cos πx both are even functions
2cos3θ
2sin 3θ
x sin 3θ = +
2 10
y z
π
So, I = 10 ∫
f (x) cos πx dx
(xyz) sin3θ = (y + 2z) cos 3θ + y sin 3θ)
−10
have a solution (x
2 10
0 , y 0 , z 0 ) with y 0 z 0 ≠ 0, is
π
=
5 ∫
f (x)cos( π x)dx
Ans.[3]
0
Sol. (xyz) sin 3θ + y (– cos 3θ) + z (– cos 3θ) = 0
∴ f(x) & cos πx both are periodic then
(xyz) sin 3θ + y (–2 sin 3θ) + z (–2 cos 3θ) = 0
2
(xyz) sin 3θ + y (– cos 3θ − sin 3θ ) + z (– 2cos
I = π 2 ∫
f (x) cos(πx) dx
3θ) = 0
0
For y
⎡
1
2
0 z 0 ≠ 0 ⇒ Nontrivial solution
⎤
= π
2
⎢ − π + − π ⎥
⎢∫(
1 x)cos x( x)dx
∫(x
1)cos( x)dx
sin 3θ
− cos3θ
− cos3θ
⎥
⎣ 0
1
⎦
sin 3θ
− 2sin 3θ
− 2cos3θ
= 0
2 ⎡2
+ 2⎤
sin 3θ
− cos3θ − sin 3θ
− 2cos3θ
= π ⎢ 2 ⎥ = 4
⎣ π ⎦
1 cos3θ
1
2 π 2
+ isin
sin 3θ cos3θ 1 2sin 3θ
2 = 0
.
3 3
1 cos3θ + sin 3θ
2
Then the number of distinct complex number z
sin3θ cos3θ [(4sin 3θ – 2 cos 3θ – 2sin 3θ) –
2
z + 1 ω ω
(2cos 3θ – cos 3θ – sin 3θ) + 2 cos 3θ – 2 sin 3θ]
2
satisfying ω z + ω 1 = 0 is equal to =
= 0
2
ω 1 z + ω
⇒ (sin3θ cos3θ) [2 sin3θ – 2cos 3θ – cos 3θ + sin
3θ + 2cos 3θ – 2sin 3θ] = 0
On solving the determinant
It become z 3 = 0
⇒ (sin3θ cos3θ) (sin3θ – cos 3θ) = 0
So no. of solutions = 1
π 2π
⇒ sin 3θ = 0 ⇒ θ = , 3 3
These two donot
satisfy system
infinite geometric series whose first term is k
π π 5π 7π of equations
k −1
1 ⇒ cos 3θ = 0 ⇒ θ = , , , and the common ratio is . Then the value
6 2 6 6
k!
k
100 2 100
2
of + ∑ ( k − 3k + 1S ) k
100!
k=
1
is –
π 5π 9π
⇒ sin 3θ = cos 3θ ⇒ 3θ = , , 4 4 4
K
π 5π 3π
⇒ θ = , , = 3
12 12 4
K
No. of solutions = 3
100
∑ |(k 2 – 3k + 1)S k |
K=
1
PHYSICS
100 ( k
– 3k + 1)
= 1 + 1 + ∑
k –1
K=
3
SECTION – I
= 2 + ∑
k – 1 k
Single Correct Choice Type
–
k – 2 k – 1
54. Let ω be the complex number cos
Ans.[1]
Sol.
55. Let S k , k = 1, 2, ……, 100, denote the sum of the
Ans.[4]
Sol. S k =
= 2 + 2 –
99
This section contains 8 multiple choice questions. Each
question has 4 choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
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 is-
3R
P
2GM
2GM
(A) (4 2 − 5)
(B) − (4 2 − 5)
7R
7R
GM
2GM
(C)
(D) ( 2 − 1)
4R
5R
Ans. [A]
Sol. ∆W ext = U 2 – U 1
for unit +ve mass
U 1 = V 1 and U 2 = V 2 = 0
V 1 =
∫ dV = ∫ − Gdm
2 2 1/
(r + 16R )
2
=
∫ − GM2πrdr
2 2 2 1/
7πR
(r + 16R )
2
Put r 2 + 16R 2 = t 2
4 2R
2GM 2GM
V 1 = −
2 ∫
dt = − (4 2 − 5)
7R
7R
5R
2GM
∆W ext = U 2 – U 1 = (4 2 − 5)
7R
58. 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
P 1 = mg(sinθ – µ cosθ) to P 2 = mg(sinθ + µ cosθ),
the frictional force f versus P graph will look
like –
f
P
θ
(A) P 2
P 1
(C)
f
P 1
P 2
P
P
4R
4R
(B)
f
f
P 1
P 2
(D) P 1 P 2
P
P
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
Ans.
Sol.
[D]
Reason : PV = nRT holds true in case of low
pressure and high temperature conditions.
60. 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-
Ans.
Sol.
t
(A) directly proportional to L
(B) directly proportional to t
(C) independent of L
(D) independent of t
[C]
ρL
R = A
R = t
ρ
ρL
= Lt
L
R is independent of L
61. 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 R 100 , R 60 and R 40 ,
respectively, the relation between these
resistances is-
1 1 1
(A) = +
R R
(B) R 100 = R 40 + R 60
Ans.
100 40
R
60
(C) R 100 > R 60 > R 40
[D]
V 2
Sol. Rated power =
R
1
R ∝
Rated power
∴
1
R
1
R
>
1 2
R
3
>
1
R
1
R
>
>
1
1
100 60
R
40
(D)
1
R
>
P 1 > P 2 > P 3
1
R
+
1
100 60
R
40
XtraEdge for IIT-JEE 77
MAY 2010

62. To verify Ohm's law, student is provided with a
test resistor R T , a high resistance R 1 , a small
resistance R 2 , two identical galvanometers G 1 and
G 2 , and a variable voltage source V. The correct
to carry out the experiment is-
Ans.
(A)
(B)
(C)
(D)
[C]
G 1
G 1
G 1
R 2
R T
G 2
R 1
V
G 1
R 1
R T
G 2
R 2
V
V
V
R 1
R T
G 2
R 2
Sol. Converted ammeter =
Converted ammeter =
R 2
R T
G 2
Voltmeter should be connected in parallel to R T
and Ammeter should be connected in series with
R T .
G
R 1
R 2
G
R 1
63. 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
Ans.
Sol.
(A) IBL
(C)
[C]
T
IBL
2π
dθ
T cos
2
i
dθ
dθ
2T sin
2
T = tension
dθ
iB × R dθ = 2Tsin
2
T = iBR
2πR = L
L
R =
∴ T =
2π
x x x x x x x
(B)
(D)
dθ
T cos
2
IBL
π
IBL
4π
F magnetic = iB×Rdθ
iBL
2π
64. An AC voltage source of variable angular
frequency ω and fixed amplitude V 0 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
Ans. [B]
R C
Sol.
V 0 , ω
~
T
XtraEdge for IIT-JEE 78
MAY 2010

2 ⎛ 1 ⎞
Z = R + ⎜ ⎟
⎝ ωC
⎠
ω increased z decreased
∴ current in circuit increase
∴ Bulb glow brighter.
2
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. 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 %
Ans. [A,C]
Total time(t)
Sol. Time period (T) =
no. of oscillations
So
∆ T ∆ 1sec
= =
T tt
40sec
1
∆T = × 2 = 0.05 sec
40
l
T = 2π ; T 2 =
g
4π
2 l
g
∆ g ∆ = + 2
g ll ∆ ;
TT
∆T = 0.05 sec ;
∆ l 2× 0.05
Putting we get =
l 2
2
4π
l
; g =
2
T
T = 2 sec.
∆ g × 100 = 5 %
g
∆ l = 0
l
= 0.05
66. A few electric field lines for a system of two
charges Q 1 and Q 2 fixed at two different points on
the x-axis are shown in the figure. These lines
suggest that
;
Ans.
Sol.
Q
(A) |Q 1 | > |Q 2 |
(B) |Q 1 | < |Q 2 |
(C) at a finite distance to the left of Q 1 the electric
field is zero
(D) at a finite distance to the right of Q 2 the
electric field is zero
[A,D]
Number of field lines emitting from Q 1 is more
than number of field lines reaching at Q 2
So | Q 1 | > | Q 2 |
and if so E r at a point which is right to Q 2 will be
zero.
67. One mole of an ideal gas in initial state A
undergoes a cyclic process ABCA, As shown in
figure. Its pressure at A is P 0 . Choose the correct
option(s) from the following
V
4V
B
Ans.
Sol.
V
C A
T
T
(A) Internal energies at A and B are the same
(B) Work done by the gas in process AB is P 0 V 0
ln 4
(C) Pressure at C is P 0 /4
T
(D) Temperature at C is 0
4
[A,B]
From figure
AB → isothermal process
So T A = T B ⇒ Internal energies will be same.
⎛ V ⎞
2
W AB = nRT 0 ln
⎜
⎟ = P0 V 0 ln 4
⎝ V1
⎠
It is not given that line BC passes through origin.
So we can't find pressure or temperature at point C.
68. 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
Q
XtraEdge for IIT-JEE 79
MAY 2010

Ans.
Sol.
[A,C]
1kg
v 1
5kg
Before collision
1kg 5kg
2m/sec
v2
After collision
Collision is elastic so
v 2 + 2 = v 1
......(i)
Conservation of momentum,
1 × v 1 + 0 = – 2 × 1 + 5 × v 2
....(ii)
Solving
v 1 = 3 m/sec
v 2 = 1 m/sec
Total momentum p r
system
= 1 × v 1 = 3 kg-m/sec
Momentum of 5 kg = 5 × v 2 = 5 kg-m/sec
1 × 3 + 5×
0
v CM =
= 0.5 m/sec
6
1
2
k CM = × (M1 + M
2
) vCM
= 0.75 joule
2
1
k system = × 1 × 9 = 4.5 joule.
2
69. 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
index of the material of the prism is 3 , which
of the following is (are) correct ?
B
Ans.
60º
60º
135º
90º 75º
A
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 :
C
sin 60º
sin r
= 1
3
O
60º
B
A
60º
90º
C
135º
30º = r
60º
r 1
E
r 2
45º
it hits at E
By geometry angle r 1 = 45º, r 2 = 45º
1 1
We know sinθ C = and sin 45º =
3
2
1 1 >
2 3
so 45º > θ C
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.
Final Ray
60º
60º30º
Incident Ray
SECTION – III
Paragraph Type
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.
Paragraph for Question No. 70 to 71
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 T C (0). An
interesting property of superconductors is that
their critical temperature becomes smaller than T C
(0) if they are placed in a magnetic field, i.e. the
critical temperature T C (B) is a function of the
magnetic field strength B. The dependence of T C
(B) on B is shown in the figure.
T C (B)
T C (0)
D
r = 30º
O
B
XtraEdge for IIT-JEE 80
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 B 1
(solid line) and B 2 (dashed line). If B 2 is larger
than B 1 , which of the following graphs shows the
correct variation of R with T in these fields ?
(A)
R
R
O
B 2
B 2
B 1
T
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) = kx 2 , it performs
simple harmonic motion. The corresponding time
m
period is proportional to , as can be seen
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 kx 2 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) = αx 4
(α > 0) for | x | near the origin and becomes a
constant equal to V 0 for | x | ≥ X 0 (see figure)
V(x)
(B)
(C)
(D)
R
R
O
O
O
B 1
B 1
Ans. [A]
Sol. B 2 > B 1
so T C (B 2 ) < T C (B 1 )
Dashed Solid
line line
Resistance ∝ Temperature above critical .
B 1
B 2
B 2
71. A superconductor has T C (0) = 100 K. When a
magnetic field of 7.5. Tesla is applied, its T C
decreases to 75 K. For this material one can
definitely say that when
(A) B = 5 Tesla, T C (B) = 80 K
(B) B = 5 Tesla, 75 K < T C (B) < 100 K
(C) B = 10 Tesla, 75 K < T C (B) < 100 K
(D) B = Tesla, T C (B) = 70 K
Ans. [B]
Sol. T C (0) = 100 K, B = 0
B = 7.5 tesla
T C (B) = 75
If B = 5 Tesla, T C (B) should be greater than 75 K
T
T
T
72. If the total energy of the particle is E, it will
perform periodic motion only if-
(A) E < 0 (B) E > 0
(C) V 0 > E > 0 (D) E > V 0
Ans.
Sol.
[C]
Energy Total should be less than maximum
potential energy
so E < V 0 and E > 0.
73. For periodic motion of small amplitude A, the
time period T of this particle is proportional to-
(A)
(C)
Ans [B]
Sol. V(x) = α x 4
[V(x)]
[α] =
4
[x]
X 0
V 0
m
A (B)
α
A α (D)
m
=
[ML T
4
[L ]
2 −2
]
1
A
x
m
α
1 α
A m
= [ML –2 T –2 ]
Time period ∝ (Amplitude) x (α) y (Mass) z
[T] = [L] x [ML –2 T –2 ] y [M] 2
1 1
Solving x = –1, y = − , z =
2 2
T = A –1 α –1/2 M 1/2 =
1
A
M
α
74. The acceleration of this particle for | x | > X 0 is -
(A) proportional to V 0
V0
(B) proportional to
mX
(C) proportional to
0
V0
mX
0
(D) zero
XtraEdge for IIT-JEE 81
MAY 2010

Ans. [D]
Sol. for |x| > x 0
U = constant
dU
F = – = 0
dx
acceleration = zero.
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.
75. A stationary source is emitting sound at a fixed
frequency f 0 , which is reflected by two cars
approaching the source. The difference between
the frequencies of sound reflected from the cars is
1.2% of f 0 . 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.
Ans. [7]
v 1 v 2
Sol.
v
2
+ v(v
f 0
*
S
2
⎛ v + v1
⎞
f 1 = f 0 ⎜ ⎟
⎝ v ⎠
⎛ v ⎞ f
f 2 = f 1
⎜
⎟
0(v
+ v1)
=
⎝ v − v 1 ⎠ (v − v1)
f0(v
+ v2)
f 2'
=
(v − v )
f '
2
−f
f
0
2
2
(v + v2
) (v + v1)
= – = 0.012
(v − v ) (v − v )
(v + v2)(v
− v1)
− (v + v1)(v
− v2)
= 0.012
(v − v )(v − v )
− v1)
− v1v2
− v + v1v
(v − v )(v − v )
2
2
2
1
2
2
1
− v(v1
− v2)
= 0.012
2v(v2
− v1)
(v − v )(v − v
1
2
1
)
= 0.012
2
(v2 v1)
v
− = 0.012
18
(v 2 – v 1 ) = 0.006 × 330 × 5
= 7.128
76. 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
25
image changes from m 25 to m 50 . The ratio
m50
is ?
Ans. [6]
Sol. m =
f
f + u
20
m 25 = = –4 ; m 50 =
20 − 25
m 25 = 6
m
50
20 2
= –
20 − 50 3
77. An α-particle and a proton are accelerated from
rest by a potential difference of 100V. After this,
their de-Broglie wavelengths are λ α and λ p
respectively. The ratio
λ p
λ α
, to the nearest integer
is ?
Ans. [3]
Sol. After accelerating through V 0 KE of a particle
becomes = qV 0 evolts
so
KE α = 200 eV
KE p = 100 eV
h
λ debroglied =
2MKE
λ P
=
λ α
MαKE
M
α
p KE p
=
= 2 × 1.414 ~ = 3
4×
200
1×
100
= 2 2
78. 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 J 1 . When the
same batteries are connected in parallel across R,
the rate is J 2 . If J 1 = 2.25 J 2 then the value of R is
Ω is ?
Ans. [4]
Sol.
R
i 1
1Ω
R
1Ω
E E
2E
i 1 =
R + 2
2
⎛ 2E ⎞
J 1 = ⎜ ⎟ . R
⎝ R + 2 ⎠
i 2
E
E eq = E
E
r eq = 0.5Ω
1Ω
1Ω
XtraEdge for IIT-JEE 82
MAY 2010

i 2 =
From J 1 = 2.25 J 2
2
2
⎛ 2E ⎞ ⎛ E ⎞
⎜ R
R 2
⎟ = 2.25 R
⎝ +
⎜
⎠
R 0.5
⎟
⎝ + ⎠
2
R + 2
= 1.5
R + 0. 5
2R + 1 = 1.5 R + 3
0.5 R = 2
R =
2 = 4 Ω
0.5
E
R + 0.5
⎛ E ⎞
J 2 =
⎜ R
R 0.5
⎟
⎝ + ⎠
79. Two spherical bodies A(radius 6 cm) and B
(radius 18 cm) are at temperature T 1 and T 2 ,
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 ?
Ans. [9]
Sol. λ A = 500 nm
λ B = 1500 nm
λ A T A = λ B T B
T =
T T
1
2
T A = 3 … (i)
B
rA
1
=
rB
3
From Stefan's law
4
E σ(4πr
2 )(T )
=
A 1 ⎛ 1 ⎞
= ⎜
4 ⎟⎠ × (3) 4 = 9
E σ(4gr
2 )(T ) ⎝ 3
A
B
B
2
80. When two progressive waves y 1 = 4sin(2x – 6t)
⎛ π ⎞
and y 2 = 3 sin ⎜2 x − 6t − ⎟ are superimposed,
⎝ 2 ⎠
the amplitude of the resultant wave is ?
Ans. [5]
Sol. y 1 = 4 sin (2x –6t)
y 2 = 3 sin (2x –6t –π/2)
π
φ = 2
2
2
A res =
2 2
A1
2 1 2
2
+ A + 2A A cosφ
A res = 3 + 4 + 0 = 5
2
81. 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 m 2 . 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 × 10 9 Nm –2 , the value of n is ?
Ans. [4]
Sol.
x
strain = l
x
stress = Y
strain
stress = Yx (l = 1 m)
F = Yx
A
F = Ayx
Ayx = ma
a =
140 =
AYx
m
140 = 70 n
n = 4
; ω =
–7
4.9×
10 × n × 10
0.1
AY
m
82. A binary star consists of two stars (mass 2.2 M S )
and B(mass 11 M S ), where M S 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. [6]
Sol.
r A
d
r B
9
XtraEdge for IIT-JEE 83
MAY 2010

2
L A = m A ω r A
2
L B = m B ωr B
Ratio (K) =
r
r
m
=
m
A
L A + L
L
11
=
2.2
B
A B
=
B
5
B
=
Ratio (K) = 5
1 × 5 2 + 1 = 6
2
L A m + 1 =
ArA
2
LB
mBrB
+ 1
Ans. [8]
Sol. The amount of heat required to raise the temp
from –5°C to 0°C.
Q 1 = m × 2100 × 10 –3 × 5 = 10.5 m Joule
The amount of heat required to melt 1 gm
ice = 10 –3 × 3.36 × 10 5 = 336 J
420 = 336 + 10.5 m
10.5 = 84
m = 8 gm.00
83. Gravitational acceleration on the surface of a
planet is 11
6 g, where g is the gravitational
acceleration on the surface of the earth. The
2
average mass density of the planet is 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 ?
Ans. [3]
Sol.
υ p 2g pR
p =
υc
2geR
e
(1)
⇒
M
4
πR
3
p
3
p
= 3
2
=
g p R p × …
g
e
Me
4
πR
3
3
e
R
e
M P 2 R
⇒ =
M 3 R
GM p 6 GM M
2 =
e p 6 R
R p 11
2 ⇒ =
R M
e
c 11 R
from (2) and (3)
3
p
2 R 6
3 =
3 R e 11 R
from (1) and (4)
2
p
2
e
R R p 3 ⇒ =
R e 226
υ p 6 3 18 3
= × = =
υe
11 226
242 11
v p = 3 km /sec.
e
2
p
2
e
3
p
3
e
… (4)
… (2)
…(3)
84. A piece of ice (heat capacity = 2100 Jkg –1 ºC –1 and
latent heat = 3.36 × 10 8 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 ?
ATTITUDE
• 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 aptitude, will
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!
XtraEdge for IIT-JEE 84
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
SECTION – I
Single Correct Choice Type
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 B 2 is
(A) 1 and diamagnetic
(B) 0 and diamagnetic
(C) 1 and paramagnetic
(D) 0 and paramagnetic
Ans. [A]
Sol. 2
1s
*
1s
σ σ
2
σ
2
2 σ
2s
*
2s
2
p
π =
2 x
B. Ο = 2
1 (6 – 4) = 1
Diamagnetic
2. The compounds P, Q and S
COOH
π 2p y
OCH 3
were separately subjected to nitration using
HNO 3 /H 2 SO 4 mixture. The major product formed
in each case respectively, is
(A)
(B)
HO
HO
NO 2
COOH
O 2 N
COOH
NO 2
H 3 C
H 3 C
NO 2
O
||
C
O
NO 2
OCH 3
OCH 3
HO
P
H 3 C
Q
O
||
C
O
O
||
C
NO 2
O
S
XtraEdge for IIT-JEE 86
MAY 2010

(C)
HO
NO 2
COOH
O
||
C
H 3 C
O
OCH 3
NO 2
NO 2
(A) 39.27 % (B) 68.02%
(C) 74.05% (D) 78.54%
Ans. [D]
Sol. Area of square = L 2
(D)
HO
NO 2
COOH
H 3 C
O
||
C
O
OCH 3
NO 2
NO 2
& 4R =
R =
2 L
L
2 2
area of circle
% packing efficiency (η) = × 100
area of square
Ans.
Sol.
[C]
COOH
HNO 3 /H 2 SO 4
COOH
NO 2
2× π R
= × 100
2
L
L
2× π
= 4×
2 × 100
2
L
2
2
= 78. 5%
OH
OH
OCH 3 OCH 3
NO
HNO 3 /H 2 SO 2
4
CH 3 CH 3
4. The species having pyramidal shape is
(A) SO 3 (B) BrF 3
Ans.
(C)
[D]
2–
SiO 3
(D) OSF 2
O
||
C HNO 3 /H 2 SO 4
O
3. The packing efficiency of the two-dimensional
square unit cell shown below is
O
||
C
O
NO 2
Sol.
F
S
F
O
5. In the reaction
O (1) NaOH/Br 2
H 3 C C O
NH 2 (2) C
Cl
the structure of the product T is
O
(A) H 3 C C O
O– C
T
L
(B)
NH
C
O
CH 3
XtraEdge for IIT-JEE 87
MAY 2010

(C) H 3 C
NH
10 .5
3 × N
C
108
A
Number of atoms of Ag in 1cc ⇒
.5
H 3 PO 4 H 2 SO 4 H 3 PO 3 H 2 CO 3 H 2 S 2 O 7
× N A
H
108
3 BO 3 H 3 PO 2 H 2 CrO 4 H 2 SO 3
(D) H 3 C
2/3
O
In 1cm
O
⎛10.5
⎞
, number of atoms of Ag = ⎜ × N A ⎟
⎝ 108 ⎠
C O
In 10 –12 m 2 or 10 –8 cm 2 , number of atoms of
NH–C
2/3
⎛10.5
⎞
Ag = ⎜ N A ⎟ × 10 –8
⎝ 108 ⎠
Ans. [C]
2/3
⎛
23
10.5 6.02 10 ⎞
Sol.
= ⎜
× ×
⎟ × 10 O
108
⎝
⎠
NaOH/Br 2
CH 3 C CH NH
Hoffmann's 3
2
= 1.5 × 10 7
NH 2 degradation
O
||
Thus, x = 7
Cl O
C
NH–C
8. Among the following, the number of elements
CH 3 NH 2 +
showing only one non-zero oxidation state is
–HCl
O, Cl, F, N. P, Sn, Tl, Na, TI
CH 3
Ans. [2]
Sol. F & Na only show one non zero oxidation state
that are – 1 & + 1 respectively.
6. The complex showing a spin-only magnetic
moment of 2.82 B.M. is
9. One mole of an ideal gas is taken from a to b
along two paths denoted by the solid and the
(A) Ni(CO) 4 (B) [NiCl 4 ] 2–
dashed lines as shown in the graph below. If the
(C) Ni((PPh 3 ) 4 (D) [Ni(CN) 4 ] 2–
work done along the solid line path is w s and that
Ans. [B]
along the dotted line paths is w d , then the integer
Sol. [NiCl 4 ] –2 closest to the ratio w d \w s is
Ni +2 1s 2 2s 2 2p 6 3s 2 3d 8 4s°
4.
4. a
3.
Cl – Cl – Cl – Cl –
P 3.0
(atm) 2.
hyb = sp 3
2.0
1.5
no. of unpaired electrons = 2
1.0
0.5
b
µ = 2 (4)
= 8 = 2.82 BM
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]
SECTION – II
Sol. For solid line path show approxy isothermal
Integer Type
process
This section contains a group of 5 questions. The
5.5
answer to each questions is a single digit integer
∴ work done |W S | = 2.303 (PV) log
ranging from 0 to 9. The correct digit below the
.5
question number in the ORS is to be bubbled.
= 2.303 × 4 × .5 × log 11
~– 4.79
7. Silver (atomic weight = 108 g mol –1 ) has a
for dashed line path work done
density of 10.5 g cm –3 . The number of silver
w
atoms on a surface of area 10 –12 m 2 d = 4 × |2 – .5| + 1 × |3 – 2| + .5 × |5.5 –3|
can be
= 6 + 1 + 1.25 = 8.25
expressed in scientific notation as y × 10 x . The
w d 8.25
value of x is
∴ = = 1.71 ~– 2
Ans. [7]
w s 4. 8
Sol.
m 10. The total number of diprotic acids among the
d = ⇒ 10.5 g/cc
V following is
XtraEdge for IIT-JEE 88
MAY 2010

Ans. [6]
Sol.
O
||
HO – S – OH
||
O
O O
|| ||
H–O – S – O– S – O – H
||
O
||
O
O
||
HO – P – OH
O
OH
Cr
OH
O
OH
O
||
S
OH
O
||
C
OH
11. Total number of geometrical isomers for the
complex [RhCl(CO)(PPh 3 )(NH 3 )] is
Ans. [3]
Sol.
Cl
PPh 3
Cl
NH 3
Rh
Rh
CO
Cl
NH 3
NH 3
Rh
CO
PPh 3
CO
PPh 3
OH
(B) H 3 C
(C)
H 3 C
(D)
H 3 C
CH
CH 3
CH
C
O
CH 3
CH 2
O
CH
C
CH 3
CH 2
O
Ans. [B]
13. The compound R is
(A) H 3 C
H 3 C
C
CH 2
O
C
OH
C
H
H and H
C
O
H H 3 C
and
H H
and
H
C
O
C
O
H
H
SECTION – III
Paragraph Type
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.
Paragraph for questions No. 12 to 14
(B) H 3 C
H 3 C
H 3 C
C
CH
O
C
OH
H
Two aliphatic aldehydes P and Q react in the
presence of aqueous K 2 CO 3 to give compound R,
which upon treatment with HCN provides
compound S. On acidification and heating, S
gives the product shown below –
H 3 C OH
H 3 C
12. The compounds P and Q respectively are -
CH 3
(A) H 3 C
CH
O
C
O
O
H and H 3 C
C
O
H
Ans.
(C) H 3 C
(D) H 3 C
[A]
CH 3
CH
CH
CH 2
CH 3
CH
H 3 C
CH
CH
O
C
OH
O
C
OH
H
H
XtraEdge for IIT-JEE 89
MAY 2010

14. The compound S is
CH 3
(A) H 3 C
CH
CH
O
C
H
CH 3
(iv) CH 3 – C – CH 2
OH
H – C – CN
OH
⎯ ⎯ H +
3 →
O ⎯ ∆
CH 3 – C — CH 2
HC O
OH C
O
Final product
H 3 C
(B)
H3 C
(C) H 3 C
H 3 C
(D)
H3 C
CH 3
CH 2
CN
O
C
C
H
CH 2
CN
CH 3 CN
CH
CH
CH
CH 2
OH
CN
C
CH
OH
OH
Paragraph for Questions No. 15 to 17
The hydrogen like species Li 2+ is in a spherically
symmetric state S 1 with one radial node. Upon
absorbing light the ion undergoes transition to a
state S 2 . The state S 2 has one radial node and its
energy is equal to the ground state energy of the
hydrogen atom.
15. The state S 1 is –
(A) 1s
(C) 2p
Ans. [B]
Sol. Q One radial node
∴ n – l – 1 = 1
or n – l = 2
l = 0
n = 2
Orbital name = 2s
(B) 2s
(D) 3s
CH 2
Ans. [D]
Sol. (12 to 14)
CH 3
(i) CH 3 – CH
C
O
H
OH
OH
⎯ –
⎯→ ⎯
CH 3
CH 3 – C Θ C
O
H
16. Energy of the state S 1 in units of the hydrogen
atom ground state energy is –
(A) 0.75 (B) 1.50
(C) 2.25 (D) 4.50
Ans. [C]
3
Sol. S 1 = Energy of e of H in ground state ×
2
= 2.25 × energy of e of H in ground state
2
2
(ii)
CH 3
CH 3
H
CH 3 – C Θ + C H CH 3 – C – CH 2
C H
O
C OH
H
O
O
[R]
CH 3
CH 3
(iii) CH 3 – C – CH 2 + HCN
C OH
H
O
C – CH 2
CH 3 OH
H – C – CN
OH
[S]
17. The orbital angular momentum quantum number
of the state S 2 is –
(A) 0 (B) 1
(C) 2 (D) 3
Sol. [B]
For S 2 = n – l – 1
n – l = 2
n = 3, l = 1
Orbital = 3p ∴ l = 1
2
3
[ S 2 = energy of e of H in ground state ×
2 , n = 3]
n
XtraEdge for IIT-JEE 90
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. Match the reactions in Column I with appropriate
options in Column II.
Column I
NaOH/
H 2 O
0°
C
(A) N 2 Cl + OH ⎯ ⎯⎯⎯⎯
→
N = N
OH
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) (CH 3 ) 2 SiCl 2
(B) XeF 4
(C) Cl 2
(D) VCl 5
Column II
(p) Hydrogen halide formation
(q) Redox reaction
(r) Reacts with glass
(s) Polymerization
(t) O 2 formation
Ans. [A → p,s; B → p,q,r,t; C → p,q,t; D → p]
MATHEMATICS
SECTION – I
Single Correct Choice Type
This section contains 6 multiple choice questions. Each
question has 4 choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
20. A signal which can be green or red with
OH OH
(B) H 3C – C – C – CH 3
CH 3 CH 3
H 2 SO
⎯ 4
⎯⎯→
O
probability 5
4 and 5
1 respectively, is received by
station A and then transmitted to station B. The
probability of each station receiving the signal
correctly is 4
3 . If the signal received at station B
H 3 C
C CH 3
C
CH3
CH 3
is green, then the probability that the original
signal was green is
3
6
(A) (B) 5 7
(C)
C
O
CH 3
1. LiAlH
2.H
O
4
⎯⎯⎯ +
→
(D) HS Cl ⎯ Base ⎯⎯ →
Column II
(p) Racemic mixture
(q) Addition reaction
(r) Substitution reaction
(s) Coupling reaction
(t) Carbocation intermediate
3
S
OH
CH
CH 3
Ans.
Sol.
20
(C) 23
9
(D) 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 ⎞
P⎜
⎟ =
⎝ OG ⎠
⎝ BG ⎠ ⎛ BG ⎞ ⎛ BG ⎞
P(OG).P⎜
⎟ + P(OR).P⎜
⎟
⎝ OG ⎠ ⎝ OR ⎠
XtraEdge for IIT-JEE 91
MAY 2010

=
4 ⎡3
⎢ .
5 ⎣4
3
4
4 ⎡3
3 1 1 ⎤
⎢ . + .
5
⎥
⎣4
4 4 4⎦
1 1 ⎤ 1 ⎡1
3
+ . ⎥ + ⎢ . +
4 4⎦
5 ⎣ 4 4
3
.
4
1 ⎤
4⎥
⎦
20
= 23
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 ⎞
(A) ⎜ , ,− ⎟ (B) ⎜ , − , ⎟
⎝ 3 3 3 ⎠ ⎝ 3 3 3 ⎠
⎛ 1 2 10 ⎞
⎛ 2 1 5 ⎞
(C) ⎜ , , ⎟ (D) ⎜ , − , ⎟
⎝ 3 3 3 ⎠ ⎝ 3 3 2 ⎠
Ans. [A]
x − 1 y + 2 z − |1−
4 − 2 − α |
Sol. = = = λ = 5
1 2 − 21
3
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. Two adjacent sides of a parallelogram ABCD are
given by AB= 2 î + 10 ĵ + 11 kˆ and AD = – î +
2 ĵ + 2 kˆ The side AD is rotated by an acute angle
Ans.
Sol.
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
(A) 9
8
(C) 9
1
[B]
D′
– î + 2 ĵ + 2 kˆ
A
α
D
⎛ π ⎞
⎜ − α⎟
⎝ 2 ⎠
(B)
(D)
2 î + 10 ĵ + 11 kˆ
17
9
4 5
9
B
C
23. 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 No. of elements in Ways
Set A Set B
0 0 = 1
1 0 = 4 C 1 = 4
2 0 = 4 C 2 = 6
1 1 = 4 C 2 = 6
3 0 = 4 C 3 = 4
2 1 = 4 C 2 . 2 C 1 = 12
4 0 = 4 C 4 = 1
3 1
4!
= = 4
3! 1!
2 2
4!
= = 3
2! 2! 2!
Total ⇒ 1 + 4 + 6 + 6 + 4 + 12 + 1 + 4 + 3 = 41
24. Let f be a real-valued function defined on the
interval (–1, 1) such that e –x f(x) = 2 +
Ans.
x
4
∫
t + 1 dt , for all x ∈ (–1, 1) , and let f –1 be the
0
inverse function of f. Then (f –1 )′ (2) is equal to
(A) 1 (B) 3
1
(C) 2
1
[B]
Sol. f(x) = 2e x + e x ∫
t + 1 dt
x = 0 ; f(0) = 2
x
0
x
4
(D) e
1
f ′(x) = 2e x + e x 4
∫
t + 1 dt + e x 1+ x
4
0
f ′(0) = 2 + 1 = 3 (f –1 1 1
)′ (2) = =
f ′(0)
3
25. For r = 0, 1, …, 10, let A r , B r and C r denote,
respectively, the coefficient of x r in the
expansions of (1 + x) 10 , (1 + x) 20 and (1+ x) 30 .
10
Then ∑ Ar (B10Br
− C10Ar
) is equal to
r=
1
2
(A) B 10 – C 10 (B) A 10 ( B − C )
10 10A10
⎛ π ⎞
cos ⎜ − α⎟ =
⎝ 2 ⎠
AB.AD
| AB || AD |
8 17
sin α = ⇒ cos α = 9 9
=
40 8 =
3(15) 9
Ans.
Sol.
(C) 0 (D) C 10 – B 10
[D]
A r = 10 C r , B r = 20 C r , C r = 30 C r
10
∑
r=
1
10 C ( 20 C 10 20 C r – 30 C 10 10 C r )
r
XtraEdge for IIT-JEE 92
MAY 2010

10
= 20 C 10 ∑
10 Cr
20 C 20–r – 30 10 10
C 10 ∑ Cr.
Cr
r=
1
10
r=
1
= 20 C 10 [ 30 C 20 – 10 C 0 20 C 20 ] – 30 C 10 [ 20 C 10 – ( 10 C 0 ) 2 ]
= 20 C 10 30 C 20 – 20 C 10 – 30 C 10 20 C 10 + 30 C 10
= 30 C 10 – 20 C 10
= C 10 – B 10
SECTION – II
Integer type
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 a 1 , a 2 , a 3 , …….., a 11 be real numbers
satisfying a 1 = 15, 27 –2a 2 > 0 and a k =2a k–1 – a k–2
for k = 3, 4, …., 11.
2 2
2
a1 + a 2 + ..... + a11
If
= 90, then the value of
11
a1 + a 2 + ..... + a11
is equal to
11
Ans. [0]
a k + a k−2
Sol. Q a k–1 =
2
2 2
2
a1 + a 2 + ..... + a11
so
= 90
11
⇒ Σ(a + (r –1) d) 2 = 11 × 90
⇒ Σ(a 2 + 2ad (r –1) + (r –1) 2 d 2 ) = 11 × 90
11a 2 10×11 10 × 11×
21
+ 2ad +
d 2 = 11 × 90
2 6
so on solving d = –3
a1 + a 2 + ...... + a11
so
11
11 1
= . . (2 × a1 + (11 –1) (–3))
2 11
= 2
1 (30 – 30) = 0
27. 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
f(x) = ln {g(x)}, for all x ∈ R, then the number of
points in R at which g has a local maximum is
Ans. [1]
Sol. g(x) = e f(x)
g′(x) = e f(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
⎡ 2k −1
⎢
A = ⎢ 2 k
⎢−
2 k
⎣
2
1
2k
k
2 k ⎤
⎥
− 2k⎥
and
−1
⎥
⎦
⎡ 0 2k −1
k ⎤
⎢
⎥
B = ⎢1
− 2k 0 2 k ⎥
⎢ − k − 2 k 0 ⎥
⎣
⎦
If det (adj A) + det(adj B) = 10 6 , 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]
Ans. [4]
Sol. det (A) =
2k −1
2
− 2
k
k
2
1
2k
k
2 k
− 2k
−1
= (2k –1) [–1 + 4k 2 ] –2 k[–2 k –4k k]
+ 2 k [4k k + 2 k ]
det (A) = (2k –1) (4k 2 –1) + 4 k (2k + 1) + 4k (2k +1)
= (2k –1) (4k 2 –1) + 8k (2k + 1)
det (B) = 0
det (adj A) = (det A) 2 = 10 6 det A = 10 3
8k 3 + 1 –2k –4k 2 + 16k 2 + 8k = 10 3
8k 3 + 12k 2 + 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
29. Two parallel chords of a circle of radius 2 are at a
distance 3 + 1 apart. If the chords subtend at the
π 2π
center, angles of and , where k > 0, then
k k
the value of [k] is
[Note : [k] denotes the largest integer less than or
equal to k]
Ans. [3]
XtraEdge for IIT-JEE 93
MAY 2010

Sol.
π/k
π/2k
d = 2 cos k
π + 2 cos
π
2k
3 + 1 3π 3π 3 + 1 1
= cos cos ⇒ cos cos = .
4 4k
4π
k 4π
k 4k
2 2 2
π π =
4k 12
4k = 12
k = 3
30. 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
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 r 2 is equal to
Ans. [3]
Sol.
B 6 C
∆ = 2
1 ab sin C
10
15 3 = 2
1 6(10) sin C ⇒ sin C = 3 /2
⇒ C = 120°
2
100 + 36 − c
cos C =
⇒ C 2 = 136 + 120° (1/2)
2.10.6
⇒ C 2 = 196 ⇒ C =14
s = 15
r = s
∆ = 3
r 2 = 3
SECTION – III
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.
A
Paragraph for Questions No. 31 to 33
Consider the polynomial f(x) = 1 + 2x + 3x 2 +
4x 3 . Let s be the sum of all distinct real roots of
f(x) and let t = |s|.
31. The real number s lies in the interval
⎛ 1 ⎞
⎛ 3 ⎞
(A) ⎜ − , 0 ⎟ (B) ⎜−11
, − ⎟
⎝ 4 ⎠ ⎝ 4 ⎠
⎛ 3 1 ⎞
⎛ 1 ⎞
(C) ⎜−
, − ⎟ (D) ⎜0
, ⎟
⎝ 4 2 ⎠ ⎝ 4 ⎠
Ans. [C]
Sol. f(x) = 4x 3 + 3x 2 + 2x + 1
f ′(x) = 12x 2 + 6x + 2 is always positive
1
f(0) = 1, f(–1/2) = 1/4, f(–3/4) = – 2
Q
⎛ − 3 −1⎞
so root ∈ ⎜ , ⎟
⎝ 4 2 ⎠
the equation have only one real root so
⎛ − 3 −1⎞
⎛ 1 3 ⎞
s ∈ ⎜ , ⎟ and t ∈ ⎜ , ⎟
⎝ 4 2 ⎠ ⎝ 2 4 ⎠
32. The area bounded by the curve y = f(x) and the
lines x = 0, y = 0 and x = t, lies in the interval
⎛ 3 ⎞
⎛ 21 11 ⎞
(A) ⎜ , 3⎟ (B) ⎜ , ⎟
⎝ 4 ⎠ ⎝ 64 16 ⎠
Ans.
Sol.
⎛ 21⎞
(C) (9, 10) (D) ⎜0
, ⎟
⎝ 64 ⎠
[A]
t
⎛
4
⎞
A(t) =
∫
f (x)d(x) = t 4 + t 3 + t 2 + t = t ⎜
1−
t
⎟
0
⎝ 1−
t ⎠
⎛ 175 ⎞
A(1/2) = 15/16 & A (3/4) = 3 ⎜ ⎟
⎝ 256 ⎠
⎛ 3 ⎞
So A(t) ∈ ⎜ , 3⎟ ⎝ 4 ⎠
33. The function f ′(x) is
(A) increasing in
in
⎛
⎜ −
⎝
1 ,
4
⎞ t ⎟
⎠
(B) decreasing in
in
⎛
⎜ −
⎝
1 ,
4
⎞ t ⎟
⎠
⎛ 1 ⎞
⎜−
t,
− ⎟ and decreasing
⎝ 4 ⎠
⎛ 1 ⎞
⎜−
t,
− ⎟ and increasing
⎝ 4 ⎠
XtraEdge for IIT-JEE 94
MAY 2010

(C) increasing in (–t, t)
(D) decreasing in (–t, t)
Ans. [B]
Sol. f ′′(x) = 6 (4x + 1)
Paragraph for questions No. 34 to 36
Tangents are drawn from the point P(3, 4) to the
2 2
x y
ellipse + = 1, touching the ellipse at points
9 4
A and B.
34. The coordinates of A and B are
(A) (3, 0) and (0, 2)
⎛ ⎞
(B) ⎜
8 2 161
⎟
⎞
− , and
⎜
⎛ 9 8 − , ⎟
⎝ 5 15 ⎠ ⎝ 5 5 ⎠
Ans.
Sol.
⎛ ⎞
(C) ⎜
8 2 161 − , ⎟ and (0, 2)
⎝ 5 15 ⎠
⎞
(D) (3, 0) and ⎜
⎛ 9 8 − , ⎟
⎝ 5 5 ⎠
[D]
Equation of tangent
y = mx ± 9m
2 + 4
as it passes through (3, 4)
so 4 = 3m ± 9m
2 + 4
m = 2
1 and undefined.
So equation of the tangents will be
x –2y + 5 = 0 and x = 3
⎛ − 9 8 ⎞
so point of contacts are (3, 0) and ⎜ , ⎟
⎝ 5 5 ⎠
35. The orthocentre of the triangle PAB is
⎛ 8 ⎞
⎛ 7 25 ⎞
(A) ⎜5 , ⎟ (B) ⎜ , ⎟
⎝ 7 ⎠ ⎝ 5 8 ⎠
Ans.
Sol.
⎛11
8 ⎞
(C) ⎜ , ⎟
⎝ 5 5 ⎠
[C]
A(3,0)
H
P(3,4)
θ
⎛
(D) ⎜
⎝
⎛ − 9 8 ⎞
B⎜
, ⎟
⎝ 5 5 ⎠
8
25
7 ⎞
, ⎟
5 ⎠
Equation of two altitudes PH and AQ are
3x – y – 5 = 0 and 2x + y – 6 = 0 respectively
⎛11
8 ⎞
so orthocentre will be ⎜ , ⎟
⎝ 5 5 ⎠
36. The equation of the locus of the point whose
distances from the point P and the line AB are
equal, is
(A) 9x 2 + y 2 – 6xy –54 x – 62 y + 241 = 0
(B) x 2 + 9y 2 + 6xy –54x + 62 y –241 = 0
(C) 9x 2 + 9y 2 –6xy –54 x –62 y – 241 = 0
(D) x 2 + y 2 –2xy + 27x + 31y – 120 = 0
Ans. [A]
Sol. Equation of AB is x + 3y – 3 = 0
so required locus will be
2
(x – 3) 2 + (y –4) 2 (x + 3y – 3)
=
10
⇒ 9x 2 + y 2 – 6xy –54x –62y + 241 = 0
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.
37. 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
(A) The set of points (p) an ellipse with
z satisfying
4
eccentricity 5
|z – i|z|| = |z + i|z||
is contained in or e
qual to
(q) the set of points z
satisfying Im z = 0
(B) The set of points z (r) the set of points z
Satisfying satisfying |Im z| ≤ 1
|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
z = w – w
1 is contained
in or equal to
(D) If |w| =1, then the set (t) the set of points z
of points satisfying |z| ≤ 3
z = w + w
1 is contained
in or equal to
XtraEdge for IIT-JEE 95
MAY 2010

Ans.
Sol.
A → q, r ; B → p ; C → p, s, t ; D → q, r, s, t
(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)
2 2
x y
so its equation will be + = 1
25 9
whose eccentricity is 4/5
(C) Let w = 2e iθ .
z = 2
3 cos + 2
5 i sin θ
(D) Let w = e iθ
Z = e iθ + e –iθ
= 2 cos θ.
38. 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 z + 1
= and
21
Ans.
1
8
x −
3
2
=
−− =
1
y
3 =
−+
1 1
z −1
at P and Q respectively.
If length PQ = d, then d 2 is
(B) The values of x satisfying (q) 0
tan –1 (x + 3) – tan –1 (x –3)
= sin –1 ⎛ 3 ⎞
⎜ ⎟ are,
⎝ 5 ⎠
(C) Non-zero vectors a r , b r (r) 4
and c r satis1fy a r . b r = 0,
(b r – a r ). ( b r + c r ) = 0 and
2 | b r + c r |= | b r – a r |.
If a r = µ b r + 4 c r , then the
possible values of µ are
(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 ⎠
π
2
The value of π ∫
f (x)dx is
−π
(t) 6
A → t ; B → p, r ; C → q, s ; D → r
Sol.
(A) Let P ≡ (λ + 2, 1 –2 λ, λ – 1)
Q ≡ (2µ + 3
2 ; – µ –3, µ + 1)
equation line PQ
→
r = (λ + 2) î + (1 – 2λ) ĵ + (λ – 1) kˆ
+ α ((2µ – λ + 3
2 ) î + (2λ – µ – 4) ĵ
+ (µ + 2 – λ) kˆ
∴ This line passing through origin so.
λ + 2 + α (2µ − λ + 3
2 ) = 0
1 – 2λ + α(2λ – µ – 4) = 0
λ – 1 + α(µ – λ + 2) = 0
on solving above three µ = 3
1 & λ = 3
10 −10 4
So P ≡ (5, – 5, 2) & Q ≡ ( , , )
3 3 3
So PQ = 6 ⇒ (PQ) 2 = 6
(B) tan –1 (x + 3) – tan –1 (x –3) = sin –1 3
5
tan –1 6
= tan –1 3
x
2 − 8 4
⇒ x 2 – 8 = 8
⇒ x 2 = 16 ⇒ x = ± 4
(C) | b r | 2 + b r . c r = a r . c r …. (1)
put a r = µ b r + 4 c r ∀ a r . b r = 0 ⇒ b r . c r = – 4
µ | b
r
|
2
from (1) and (2)
2
b 16
2 =
2
…(2)
…(3)
c 4 − µ + µ
∴ 2| b r + c r | = | b r –a r | and a r = µ b r + 4 c r
2
b 12
2 =
2
c
3 − 2µ + µ
from (3) and (4)
m = 0,5
9x
sin
(D) f(x) =
2
=
x
sin
2
π
2
I = π ∫
f (x)dx
−π
sin 5x
+
sin x
sin 4x
sin x
… (4)
XtraEdge for IIT-JEE 96
MAY 2010

π
4
= π ∫
f (x)dx
0
π
4 sin 5x
= π ∫ sin x
0
π/
2
= π
8
∫
0
π/
2
= π
8
∫
= 4
0
sin 5x
dx
sin x
sin (3x + 2x)
dx
sin x
π/
2
8
= π ∫
( 1+
2cos 4x)dx
0
Total force =
π/
2
∫
2
0 0
σ
πR
2ε
σ
sin 2θdθ
=
2ε
2
2
0
× πR
40. A block of mass 2 kg is free to move along the x-
axis. 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)
4N
2
PHYSICS
O
3s
4.5s
t
SECTION – I
Single Correct Choice Type
This section contains 6 multiple choice questions. Each
question has 4 choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
39. 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 -
Ans.
Sol.
F
(A)
1 σ
ε
0
2
2
R
1 σ
(C)
ε0
R
[A]
θ
dθ
2
2
σ
2ε
0
(B)
(D)
dA = 2πR sin θ × Rdθ
2
ε
1 2
0
σ
1 σ
ε
2
R
2
0 R
F
(Electrostatic pressure)
σ
dF = × dA
2ε0
Component of dF along vertical axis = dF cos θ
Ans.
Sol.
(A) 4.50 J
(C) 5.06 J
[C]
F
4N
–2N
O 3
(B) 7.50 J
(D) 14.06 J
4.5
t (sec)
4
m = −
3
→
At t = 4.5 sec F = – 2N
⎡1
⎤ ⎡1
⎤
Total Impulse I = ⎢ × 3×
4⎥ –
⎣ 2
⎢ × 2×
1. 5⎥ ⎦ ⎣ 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
K.E. = × 2 × (2.25) 2 = 5.06 J
2
41. A tiny spherical oil drop carrying a net charge q is
balanced in still air with a vertical uniform
81 π 5
electric field of strength × 10 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 -
(A) 1.6 × 10 –19 C (B) 3.2 × 10 –19 C
(C) 4.8 × 10 –19 C (D) 8.0 × 10 –19 C
Ans.
[D]
XtraEdge for IIT-JEE 97
MAY 2010

Sol.
qE = mg
⎡81π 5 ⎤ 4
⇒ q ⎢ × 10 ⎥ = 900 × πr 3 × 9.8
⎣ 7 ⎦ 3
3
900×
4×
r × 9.8×
7
q =
...... (1)
5
3×
81×
10
v T = 2 × 10 –3 m/sec
2
2 × 10 –3 2 r × 900×
9.8
= × 9
−5
1.8×
10
−5
r 2 18×
1.8×
10 × 10
=
2×
900×
9.8
10 = 18.36 × 10 –12
−3
= 0.1836 × 10 –
r = 4.284 × 10 –6 m
3600×
9.8×
7
q =
5
243×
10
× 78.62 × 10 –18
q = 0.799 × 10 –18 ≈ 8 × 10 –19 C
42. 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
Ans. [B]
v
Sol. Fundamental frequency of closed pipe =
4l
320
⇒ =
4×
0.8
320 = 100 Hz
3.2
Frequency of 2 nd v 1 T
Harmonic of string = = l l µ
100 =
⇒ 100 =
1 50
l m / l
50
m×
0.5
⇒ 100 =
100
m
100
10000 = ⇒ m = 10
–2
kg = 10 gm
m
43. 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
Ans.
Sol.
[B]
30cm
6 cm
10 cm
10cm
20 cm
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.
44. 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
Ans.
Sol.
[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
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.
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]
3
sin θ cr = 5
⇒ tan θ cr = 4
3
XtraEdge for IIT-JEE 98
MAY 2010

8 cm
R = 8 tan θ cr
R
47. To determine the half life of a radioactive
dN(t)
element, a student plots a graph of l n
θ
dt
cr
versus t. Here (t)
is the rate of radioactive
dt
decay at time t. If the number of radioactive
nuclei of this element decreases by a factor of p
after 4.16 years, the value of p is -
6
5
50 m in 30
dN(t) 4
ln
dt 3
2
1
2 3 4 5 6 7 8
Years
Ans. 8
Sol. From graph
1
slope = = 0.5 year
–1
2
dN = Ne
–λt
dt
⎛ dN ⎞
ln ⎜ ⎟
⎝ dt ⎠
= ln (N) – λt
so comparing we get λ = 0.5 year –1
= 0.693
t 1/2
0.5
year
t = 4.16 years
4.16
so No. of half lives =
0.693
u 1
× 0.5 = 3
N
N 0 → 0 N
→ 0 N
→ 0
2 4 8
⇒ p = 8
48. A diatomic ideal gas is compressed adiabatically
to
1 of its initial volume. In the initial
32
temperature of the gas is T i (in Kelvin) and the
final temperature is aT i , the value of a is -
Ans. [4]
Sol. for adiabatic process
TV γ–1 = const.
7
7
1
⇒ T i V
5
1
⎛ V ⎞ 5
= aT i ⎜ ⎟
u 2
⎝ 32 ⎠
⇒ a = 4
50 − 25 25
= m/sec
30 30
= 8 × 4
3 = 6 cm
46. Image of an object approaching a convex mirror
of radius of curvature 20 m along its optical axis
Ans. [3]
Sol.
25
is observed to move from m to 3 7
seconds. What is the speed of the object in km
per hour ?
For position of object initially when image
25
was at m
3
1 3 1
– = − + 10 25 u
3 1 1 – =
25 10 u
12 –10 1
= u
100
u 1 = 50
O
50
For position of object when image is at m 7
1 7 1
– = − + 10 50 u
7 1 1 – =
50 10 u
u 2 = 25
O
Speed of object =
25
= × 30 1000
49. 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
4 V ?
(Take : ln 5 = 1.6, ln 3 = 1.1]
XtraEdge for IIT-JEE 99
MAY 2010

2MΩ
2µF
Sol.
F T
A
Ans. [2]
2MΩ
2×10 6
2×10 –6
2µF
B
r
R
Sol.
A
2×10 6 B
2×10 –6
10 6 4×10 –6
10V
F T = 2πrT
Net vertically upward force
⎛ r ⎞ 2πr
2 T
⇒ 2πrT ⎜ ⎟ =
⎝ R ⎠ R
q = CV 0 (1 – e –t/RC )
V = V 0 (1 – e –t/RC )
4 = 10(1 – e –t/4 )
3 = 5e –t/4
t
log3 = log 5 – 4
1.1 – 1.6 = – 4
t
⇒ t = 2 sec
SECTION – III
Paragraph Type
This section contains 2 paragraphs.. Based upon each
of 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.
Paragraph for Questions No. 50 to 52
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.
50. 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

Ans.
Sol.
[D]
Bohr quantization principle
nh nh
L = = Iω ⇒ ω =
2π
2πI
2
1 2 1 ⎛ nh ⎞
Rotational KE = I ω = I⎜
⎟ =
2 2 ⎝ 2πI
⎠
2
2
n h
2
8π
I
54. It is found that the excitation frequency from
ground to the first excited state of rotation for the
CO moelcule is close to π
4 × 10 11 Hz. Then the
moment of inertia of CO molecule about its
center of mass is close to (Take h = 2π × 10 –34 J s)
(A) 2.76 × 10 –46 kg m 2 (B) 1.87 × 10 –46 kg m 2
(C) 4.67 × 10 –47 kg m 2 (D) 1.17 × 10 –47 kg m 2
Ans. [B]
Sol. ∆E = E 2 – E 1
2 2 2 2 2
2 h 1 h 3h
= – = = hν
2 2 2
8π
I 8π
I 8π
I
56. Two transparent media of refractive indices µ 1
and µ 3 have a solid lens shaped transparent
material of refractive index µ 2 between them as
shown in figures in Column II. A ray traversing
these media is also shown in the figures. In
Column I different relationships between µ 1 , µ 2
and µ 3 are given. Match them to the ray diagrams
shown in Column II.
Column I
(A) µ 1 < µ 2
Column II
(p)
µ 3
µ 2 µ 1
(B) µ 1 > µ 2 (q) µ 3 µ 2 µ 1
(C) µ 2 = µ 3 (r) µ 3 µ 2 µ 1
When ν = π
4 × 10 11 Hz
Solving I = 1.87 ×10 –46 kg – m 2
55. In a CO molecule, the distance between C (mass
= 12 a.m.u.) and O (mass = 16 a.m.u.) where 1
a.m.u. = 3
5 ×10 –27 kg, is close to
(A) 2.4 × 10 –10 m (B) 1.9 × 10 –10 m
(C) 1.3 × 10 –10 m (D) 4.4 × 10 –11 m
Ans. [C]
2 2
Sol. I = m 1r 1 + m 2r 2
Where m 1 = 12 amu
m 2 = 16 amu
m 1 r 1 = m 2 r 2
r 1 + r 2 = r where r → distance between C & O.
Putting and solving
r = 1.279 × 10 –10 m
~_ 1.3 × 10 –10 m
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.
(D) µ 2 > µ 3 (s) µ 3 µ 2 µ 1
(t) µ 3 µ 2 µ 1
Ans. [A → p,r; B → q,s,t; C → p,r,t D → q,s]
Sol. For (p) µ 2 > µ 1
as light rays bend towards normal at first refraction
µ 2 = µ 3 as no refraction occurs at second refraction
Option : (A), (C)
For (q)
µ 2 < µ 1 as bend away from normal at first refraction
µ 3 < µ 2 as bends away from normal at second refraction
Option (B), (D)
For (r)
µ 2 > µ 1 as bend towards the normal at first refraction
µ 2 = µ 3 as no refraction occurs at second refraction
Option (A), (C)
For (s)
µ 2 < µ 1 as bend away from normal at first refraction
µ 3 < µ 2 as bend away from normal at second refraction
Option (B), (D)
For (t)
µ 2 < µ 1 as bend away from normal at first refraction
µ 2 = µ 3 as no refraction occurs at second refraction
Option (B), (C)
XtraEdge for IIT-JEE 101
MAY 2010

57. 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 V 1 and V 2 . (indicated in
circuits) are related as shown in Column I. Match
the two
Column I
(A)
I ≠ 0, V 1 is
proportional to I
Column II
(p)
V 1
6mH
V 1
V
V 2
3µF
V 2
V 1 = I X L ; V 2 = IR
So V 2 > V 1
V 2 ∝ I
also V 1 ∝ I Option (A),
(B), (D)
For (s) V 1 = I X L
1
V 2 = I X C where X C =
ωC
1061 Ω
again V 1 ∝ I; V 2 ∝ I, I ≠ 0
Option (A), (B) (D)
For (t) V 1 = IR when R = 1000 Ω
V 2 = I X C when X C 1061 Ω
V 2 > V 1
V 1 , V 2 ∝ I and I ≠ 0
Option (A), (B), (D)
(B) I ≠ 0, V 2 > V 1
(q)
6mH
V 1
2Ω
V
V 2
WHICH IS THE HIGHEST
WATERFALL IN THE WORLD ?
(C) V 1 = 0, V 2 = V
(r)
6mH
2Ω
~ V
V 1
V 2
(D)
I ≠ 0, V 2 is
proportional to I
(s)
6mH
~ V
3µF
V 1
V 2
(t)
1kΩ 3µF
~ V
Ans. [A → r,s,t; B → q,r,s,t; C → p,q; D →
q,r,s,t]
Sol. For (p) Insteady state when I = constant
V L = 0 = V 1
So V 2 = V
Option (C)
For (q) V 1 = 0 again as I = constant
V 2 = V
Also V 2 = IR ⇒ Propotional to I.
Option (B), (C), (D)
For (r) X L = ωL = (100 π) 6 × 10 –3 1.88 Ω
R = 2Ω
The highest waterfall in the world is the
Angel Falls in Venezuela. At a towering height of
979m did you know that each drop of water takes 14
seconds to fall from the top to the bottom. The
water flows from the top of a “Tepui” which is a flat
topped mountain with vertical sides.
The waterfall which despite being known to the local
indians for thousands of years was originally called
the “Churun Meru” but for some reason they were
renamed by an American bush pilot called Jimmy
Angel, who noticed them in 1935 whilst flying over
the area looking for gold.
XtraEdge for IIT-JEE 102
MAY 2010

XtraEdge Test Series
ANSWER KEY
IIT- JEE 2011 (May issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9 10
Ans B D C C A B C D A,C A,C,D
Ques 11 12 13 14 15 16 17 18
Ans A,C,D B,C C B D B D D
19 A → R B → P C → T D → S
20 A → S B → P C → Q D → R
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9 10
Ans D D C D C A A D A,D B,C,D
Ques 11 12 13 14 15 16 17 18
Ans A,B B,C C D A C D D
19 A → P,Q,S B → P,Q,R,S C → P D → Q,R,S,T
20 A → P B → Q C → P,R D → S,P
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9 10
Ans C C C A B B C A A,B,C A,C,D
Ques 11 12 13 14 15 16 17 18
Ans B C B A B B C C
19 A → P,Q,R,S B → R C → R,S D → Q
20 A → P,R B → P,S C → Q,S D → Q,S
IIT- JEE 2012 (May issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9 10
Ans A A C B D C C B B B,C,D
Ques 11 12 13 14 15 16 17 18
Ans A,B,C,D C,D C A B B C A
19 A → R,T B → S C → P D → Q
20 A → R B → P C → S D → T
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9 10
Ans A D C D B B C A A,B,C A,B
Ques 11 12 13 14 15 16 17 18
Ans A B A D C D D A
19 A → P B → T C → S D → R
20 A → P,R,S,T B → P C → P,Q,R,T D → S
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9 10
Ans C B B B A A C B A,C,D A,C
Ques 11 12 13 14 15 16 17 18
Ans A,C,D A,D A C B A B D
19 A → S B → R C → Q D → P
20 A → P B → Q C → P D → R
XtraEdge for IIT-JEE 103
MAY 2010

XtraEdge for IIT-JEE 104
MAY 2010