March 2012 - Career Point

XtraEdge for IIT-JEE 2 MARCH 2012

Volume - 7 Issue - 9
March, 2012 (Monthly Magazine)
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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
"Faliure is Success if we learn from it"
Editorial
XtraEdge for IIT-JEE 1 MARCH 2012

S
Volume-7 Issue-9
March, 2012 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Much more IIT-JEE News.
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics,, Chemistry & Maths
Key Concepts & Problem Solving strategy for IIT-JEE.
IIT-JEE Mock Test Paper with Solution
AIEEE & BIT-SAT Mock Test Paper with Solution
INDEX
CONTENTS
Regulars ..........
PAGE
NEWS ARTICLE 3
• Wavelet & Fractal
• IITian bags job with Rs 73 lc salary
IITian ON THE PATH OF SUCCESS 5
Mr. Pramod Maheshwari
ALL ABOUT ISEET 7
Frequently Asked Questions
KNOW IIT-JEE 9
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS 18
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 # 11]
Students’ Forum
Physics Fundamentals
Calorimetry, K.T.G., Heat transfer
Atomic Structure, X-Ray & Radio Activity
CATALYSE CHEMISTRY 32
Key Concept
Purification of Organic Compounds
Boron Family & Carbon Family
Metallurgy
Understanding : Inorganic Chemistry
DICEY MATHS 39
Mathematical Challenges
Students’ Forum
Key Concept
Definite Integrals & Area under curves
Probability
Test Time ..........
XTRAEDGE TEST SERIES 50
Mock Test IIT-JEE Paper-1 & Paper-2
Mock Test AIEEE
Mock Test BIT SAT
SOLUTIONS 85
XtraEdge for IIT-JEE 2 MARCH 2012

Wavelet & Fractal
The subject of wavelet and fractal
analyses is fast developing and has
drawn a great deal of attention of
scientists in varied disciplines of
science and engineering. The wavelet
transformation is a localized
transformation of signals in spacetime
and time-frequency domains.
This property can be effectively
utilized to extract information from
signals that is not possible with the
conventional signal processing tools.
Over the past decade, wavelets,
multiresolution and multifractal
analyses have been formalized into a
thorough mathematical framework
and have found a variety of
applications with significant impact in
the analyses of several geophysical
processes such as geomagnetism,
atmospheric turbulence, space-time
rainfall, ocean wind waves, fluid
dynamics, seafloor bathymetry, welllogging
and climate change studies
among others. It is likely that there
will be a variety of applications of
wavelets and fractals in geophysics in
the years to come.
This workshop aims to create a
platform to discuss the developments
in wavelet-based and fractal-based
data analysis techniques and their
applications in various processes of
Earth, Ocean and Atmospheric
sciences. Papers related to but not
limited to the following themes are
welcome.
• Construction of wavelets
• Discrete and Continuous wavelet
transforms in geophysics
• Multiresolution and Multifractal
Analysis in geophysics
• Wavelet-based optimization
• Wavelet-based data processing
techniques
• Wavelets and Fractals in Earth,
Ocean and Atmospheric Sciences
IIT-ian bags job with Rs 73 lc
salary
Bagging a job in US company, Pocket
Gems with a magnificient salary (per
annum) of $137,000 (over Rs 73 lakh), an
IIT student from Kanpur, Karan Narain
made a record.
With these huge amount of salary, Narain
became one of the first IIT-ian and one of
the Indians who have offered such a huge
huge salary during their campus selection.
Speaking about his selection in Pocket
Gems, Narain said, "Pocket Gems is into
mobile application development for
Android and iOS. The interviews held in
Kanpur were completely technical."
The IIT student gave all credit to his
schooling in different cities (Delhi,
Chennai, Hyderabad and Agra) in India and
claimed that the schools, where he has done
his schooling, explored a new world of
knowledge to Narain who is pursuing M
Tech in the IIT-Kanpur.
Ex-IITian NRI's 5 mn dollar
gift to cancer esearch project
Living up to his promise made nearly
four years ago, a US-based Indian and
IT entrepreneur came to his almamater
Indian Institute of
Technology (IIT) here to launch
a research centre for biosciences and
bio-engineering with the focus
on cancer.
An alumnus of the 1969 batch of IIT,
Romesh Wadhwani, founder of
Symphony Technology Group and
chairman of Wadhwani Foundation,
along with his wife Kathy and IIT
Bombay director Devang Khakhar,
inaugurated the Wadhwani
Research Centre in Biosciences &
Bioengineering (WRCBB) at the IIT-
Bombay campus
The WRCBB follows a generous gift of
US$ 5 million announced by
Wadhwani in New York, towards the
project, designed to focus on
understanding cell and cancer invasion.
The IIT-B's Faculty of Department of
Biosciences & Bioengineering will be
affiliated to the WRCBB for
the research projects.
The Department of Biosciences &
Bioengineering is relatively a young
department but has already achieved
recognition for its research
excellence in multiple areas.
These include -- cancer cell biology,
signalling mechanisms in immune
cells, computational biology,
computational neurobiology, bionanotechnology,
biosensors and drug
delivery systems, among others.
In addition, WRCBB will focus on
understanding cell motility
and cancer invasion as its research
area with the ultimate goal to build a
better knowledge base in this field.
IIT to design early earthquake
warning system
Indian Institute of Technology,
Gandhinagar, Ahmedabad has taken
up a project to design a 'slightly early'
earthquake warning system. The IIT-
GN researchers have partnered with
California Institute of Technology
(CIT) , Pasadena , USA, to develop
the system which will have a strong
network of low cost motion sensor
and the ability to detect earthquake's
early seismic activity.
"The goal of our project is to put in
place a network of small devices
called accelerometers near an active
fault line which can pick up earth
vibrations. They can be standalone
devices transmitting data to a central
server," said Prof Girish Singhal,
project in-charge at IIT-GN.
Singhal clamied that "A mesh of
very low-cost sensors in that area
XtraEdge for IIT-JEE 3 MARCH 2012

shall be able to pick up velocity of
shock waves, issuing slight early
alerts of an earthquake".
According to experts, the alert from the
system will help to close down big
machines, which are major sources of
secondary losses during an earthquake.
Meanwhile, the project is expected to
be of significant help to first
responders during a natural calamity.
IIT KGP to develop technology
for Coal India Ltd.
Coal India Limited (CIL) has
approached IIT Kharagpur for
developing technology to improve
safety standards at mines. CIL had
searched for appropriate technology at
Indian and overseas institutes and
universities without much success.
Currently CIL has adopted the same
practices used in Australian mines,
where safety standards are considered
to be the best in the world.The other
idigenous partner in this project along
with IIT KGP is Jadavpur University.
The outgoing Chairman of CIL
commented that there has been
progress in this regard. The
organiszation is ready to spend
considerable amount for developing
the new technology.
IIT honors Digvijay Singh for
inventing call records eraser
technology
Kanpur : Mighty impressed with his
ability to erase call records from
telecom companies’ database with
help of some secret technology, IIT
here has decided to confer an honorary
degree on Digvijay Singh, senior
Congress leader and former CM of
Madhya Pradesh.
The secret technology came to public
knowledge when it was found that call
records didn’t confirm the claims of
Digvijay Singh about him talking to
Hemant Karkare hours before he (Mr.
Karkare, unfortunately) was killed
fighting the terrorists during the Mumbai
terror attacks in November 2008.
Digvijay Singh apparently has a
special phone with his secretly
developed technology that seamlessly
erases the call records as soon as the
conversation is over
“Of course he erased the call records;
he can’t be lying,” Professor P K Dhutt
of IIT Kanpur said.
“It’s amazing,” Professor Dhutt added,
“it surpasses the hacking abilities of
both Mark Zuckerberg and Julian
Assange, and since he (Digvijay Singh)
is from a party that idolizes Mahatma
Gandhi, we have no option but to
conclude that it was a case of ethical
hacking.”
If Diggy Raja, as Digvijay Singh is
respectfully called, accepts this offer from
IIT Kanpur, he’d be awarded PhD in
Electronics and Communications
Engineering. In fact, IIT administration has
suggested that Diggy Raja should file an
application for patent over the technology
that he had developed.
“Need for such a technology is being
felt strongly all over the world,”
Professor Dhutt explained, “imagine if
all those who have been caught on
tapes talking to Niira Radia could erase
all the call records and claim that the
released tapes were work of some
mimicry artists.”
Professor Dhutt further informed that
the institute was willing to spend on
research and development and “enrich”
the technology developed by Digvijay
Singh so that in could be applied in
other areas too.
“It could help us in curbing student
suicides as well; badly performing
students could simply erase their bad
grades from our database with this
technology and choose to start a new
life,” Prof Dhutt wondered about the
possibilities.
Meanwhile sources at IIT Kanpur reveal
that Diggy Raja had almost lost this
honorary degree to some unidentified
“capable people” in BJP, whom he
thought could tamper with WikiLeaks,
rendering the whistleblower website to
some kind of wiki site. But the latest
revelations of missing called records
clinched the deal for Diggy Raja.
Faking News tried to call up Digvijay
Singh many times to get his reactions
on this development, but all our calls
were unanswered. Strangely, none of
our mobile phones showed Digvijay
Singh’s number in its outgoing call
records list afterwards
IITs will set papers for ISEET:
Kapil Sibal
Kapil Sibal in a conference with state
ministers said that the prestigious
IITs (Indian Institutes of
Technology) of the country will
prepare questions papers for
the ISEET (Indian Science
Engineering Eligibility Test) for next
year. This is the first time when the
government has decided to hold a
common entrance test for all IITs,
NITs of the country rather than
conducting individual tests for them.
Central Board of Secondary
Education (CBSE) will take care of
management and also conduct the
entrance test.
HRD Ministry want to replace the
existing exam AIEEE and IIT-
JEE by ISEET. Both the exams are
conducted by CBSE for engineering
college admission.JEE exam is
conducted by IITs for UG and
Integrated PG engineering courses
for admission in IITs only. This
aspiring exam is going to replace all
state level entrance exams too for
engineering courses.
The ISEET 2013 will be held in two
phases: Main exam and advance
examination. IITs and other Central
Educational Institutions were also
proposing to accept a weightage of
40% for state Board marks.
Some states are not happy with this
change and not willing to accept this
exam. They said that states conduct
the exam in regional language, but
this ISEET exam will be held only in
Hindi or in English. This will create
problem for those students who do
not know both the languages. The
States of Tamil Nadu, Himachal
Pradesh, Odisha, Puducherry and
West Bengal asked for some more
time to study the proposal in detail.
However, the ministry want to reach
up to some conclusion over this
proposal within two months.
XtraEdge for IIT-JEE 4 MARCH 2012

Success Story
This article contains story/interview of persons who succeed after graduation from different IITs
“Pursue Excellence and all else shall follow….”
Mr. Pramod Maheshwari
B-Tech, IIT-Delhi
CMD & CEO, Career Point Infosystems Ltd., Kota
He holds a B.Tech Degree from IIT Delhi. He is a first
generation entrepreneur and the key founder member of
the company. Recently he has been awarded as the
'Star CEO' of the country. Pramod Maheshwari has over
15 years of experience in developing and implementing
training methodologies. He plays a major role in providing
thought, leadership and strategic guidance, in addition to
supervising the functional heads. He is responsible for the
overall operation and growth of the company.
Career Point was established in 1993, to impart quality
education to students preparing for various competitive
examinations. With the sky-high ideals and commitment to
excellence, now Career Point has taken a shape of vibrant,
dynamic and responsible institute of the country. Today,
Career Point stands apart and well above the rest on a
distinguished platform, as an epitome of success. Since
beginning Career Point’s objective is to enable each
aspirant to achieve success in different competitive
examinations. In the pursuit of which, Career Point has to
its credit a team of outstanding faculty members including
IITians, NITians and Doctors, added with the complete &
finest study material, excellent coaching methodology and
a stimulating academic environment. Career Point believes
that effective guidance is the primary need of every
student, which would create motivation and instill courage
and confidence to face all challenges. And that is exactly
what Career Point imparts in all its coaching programmes.
For every course, Career Point has a strategic & a well
charted programme, which aims at skills in a well
organized manner so that it leads them like a self guided
missile to unfailingly hit the target in the bull’s eye.
Sir, can you tell us what the major competitive strength
of Career Point are?
We believe the following competitive strengths contribute
to our success and differentiate us from our competitors:
Commitment to offering quality courses and student success
We offer quality tutorial courses, and intend to improve
the learning experience for our students. We believe
offering quality academic courses is contingent upon
recruiting and retaining experienced faculty members,
providing updated educational content and effective
academic administration and control on content delivery.
We retain faculty and instructors with relevant industry
experience and appropriate academic credentials. Our
Research and Development Cell helps in reassessing and
updating our tutorial courses on a regular basis which also
helps us in designing new academic courses. Regular
feedbacks from our students are also an inbuilt standard
procedure for our content delivery.
Strong brands and geographic presence
We believe that our training centres have established a
competitive position and brand recognition among
students. We currently have presence across 13 states and
providing our expert services to them. In our Kota centre
students from across the country and even from Singapore
and the Middle East are also there, which in turn reinforces
the brand equity and our geographical reach.
Qualified faculty team
We believe that our qualified and experienced faculty
members contribute to our success. Our faculty members
are graduates in engineering and science from Indian
Institute of Technology, National Institute of Technology
and other colleges in India. Our faculty members are well
equipped with subject knowledge guiding and tutoring
students. We also have an ongoing in-house faculty
training facility which ensures that all our faculty members
undergo training on our teaching methodologies and skills
and subject matter of relevant courses and to keep them
abreast of the changes in competitive entrance examination
trends and changing student needs.
Experience management team
Our senior management team, comprising of senior vice
presidents and above, has collective experience of over 65
years and over 13 years of average experience in the
education industry. We believe our management led by our
Promoters, some of whom have extensive tutorial
XtraEdge for IIT-JEE 5 MARCH 2012

experience, have deep understanding of the education
industry, which enables us to successfully manage our
operations and facilitate our growth.
Quality teaching methodology
We have over a period of time developed a scientific
teaching methodology and system of teaching, which we
believe is essential for success in any competitive entrance
examination. We understand that in order to achieve
success, one needs knowledge which should be acquired
through a comprehensive systematic approach, rigorous
practice, time management and confidence.
Our focus is to train our students by developing necessary
conceptual knowledge base, enhance speed and accuracy
levels, infuse confidence and build the right temperament
to face the competitive entrance examination. In such
competitive examinations, we believe our teaching
methodology plays a key role in enhancing students’
overall performance.
Sir, recently Career Point has penetrated in capital
market through IPO? Can you tell us what your future
plans are?
The IPO proceeds would be utilised for constructing and
developing an integrated campus facility; secondly for the
expansion of classroom infrastructure and office facility;
thirdly, for acquisitions and other strategic initiatives; and
to meet expenses towards general corporate purposes.
The entire requirement of funds set forth will be met from
the proceeds of the issue. We intend to set-up an integrated
campus facility at Kota, for 3,000 students which in
addition to providing tutoring facility will also provide
facilities such as accommodation for students, library,
guest house for visiting parents, primary health centre,
auditorium, canteen, cafeteria, indoor and outdoor
recreation, staff quarters, provision for utilities like
departmental store, bank facilities etc.
Sir, brief us about your recent foray into Education
Consultancy and Management Services and formal
education?
We plan to participate in formal education through
ECAMS, we believe there is a huge potential for ECAMS
in the K-12 and Higher Education segment in India. We
will explore opportunities to provide ECAMS to a number
of privately and/or Government-run schools, colleges and
universities. We also intend to enter into partnerships with
the Governments under the PPP model to manage schools,
colleges and universities in rural and/or urban areas.
Sir, can you suggest some study pattern to the students
at home after taking class in coaching?
Ans. This study plan is meant for the students to save their
time and decide the strategy to study the current day topic
and go through the next day topic. Regular revision is very
important to understand the topic and the subject in depth.
DAY 1 DAY 2 DAY 3
Class of ABC
Revision
of ABC
Questions
of ABC
Question and
Pending work of
ABC
Quick Reviev
for the class of
PQR
Class of PQR
Revision of PQR
Questions of
PQR
Question and
Pending work of
PQR
Quick Review
for the class of
ABC
Class of ABC
Revision of ABC
Questions of
ABC
Next
Class
DAY1
On day 1, suppose topic ABC is being taught in
the class, now after the class student should
revise the ABC at home (shown as ABC-
Revision) in the diagram.
After revision of ABC student should do the
questions of ABC (shown as Questions of
ABC) to make the ABC topic perfect.
DAY2
Now on day 2, student should complete the
questions and pending work of ABC
(Questions and Pending work of ABC).
Before the class of PQR on this day student
should study the PQR which is going to be
taught in the class (shown as Quick Review
for the class of PQR).
After the class of PQR student will do the
revision and problems of PQR (shown as
Revision of PQR and Questions of PQR).
What is your take on the new education bill?
The initiatives taken by the government are applaudable.
In my opinion, the government should introduce more
education reforms. We would be happy to take part in
anything which is for the benefit of Indian education
system.
XtraEdge for IIT-JEE 6 MARCH 2012

ALL ABOUT ISEET
Frequently Asked Questions
What is ISEET - Indian Science-Engineering Eligibility
Test and when it will be in operation?
• ISEET (Indian Science-Engineering eligibility Test) is
the proposed Single National level entrance test.
• It is proposed to be conducted from year 2013 for all
students who seek admissions in central engineering
institutions i.e. IITs, NITs, IIITs and IISERs
• As of now ISEET 2013 will replace IIT-JEE & AIEEE but
it is proposed that ISEET will gradually replace all State
level exams like RPET, MH-CET, Karnataka CET, etc.
Why has the pattern changed?
• ISEET 2013 has been proposed by the Ministry of
Human Resource Development, Government of India
to reduce the burden on aspirants due to multiple
entrance examinations conducted across India each
year.
• The burden in terms of time, money (examination &
Transit fee) and the stress caused in scheduling and
preparing for each examination syllabus.
Is this change beneficial for students in general?
• Yes, it is definitely beneficial for students. Earlier they
had to prepare for many engineering entrance tests.
• ISEET implementation will reduce mental and
financial burden on the student/parent and save time as
well.
• Students can concentrate on the single test. It will
improve the chances of getting admission in a good
college.
• As ISEET will be conducted twice or more during a
year and the scores will be valid for two years, students
will have multiple chances to improve their scores
What major changes come with the new pattern?
• With emphasis on School Board marks a minimum
40% weightage will be mandatorily given to School
Board marks in the admission process across India
• ISEET Main (Objective Aptitude Test -
comprehension, critical thinking, logical reasoning)
made a mandatory part of admission process with a
minimum of 30% weightage
• ISEET Advance (Advanced to School Board
curriculum & in-between AIEEE and IIT-JEE pattern
basic science subject knowledge objective test) with a
maximum of 30% weightage in admission process
made a mandatory part of for taking admissions into
IITs and NITs while it is optional for other central and
state level institutions
What is the pattern of the ISEET 2013?
• ISEET 2013 will be conducted twice in 2013, first in
the month of April or May and second in November or
December and the ultimate aim is to conduct it thrice
or four times in a year
• Results will be valid for two years while multiple
attempts are allowed for students to improve scores by
getting multiple attempts
• ISEET 2013 will be a single day exam in two sessions
o Morning Session – 10:00 AM to 01:00 PM (3
Hours)
o Afternoon Session – 02:00 PM to 05:00 PM (3
Hours)
• ISEET 2013 will have two papers
o ISEET Main – Objective type aptitude test to be
held in the morning session to access the abilities
of comprehension, critical thinking, logical
reasoning of students
o ISEET Advance– Objective type test to be held in
the afternoon session to access the problem
solving abilities of a student in basic science
subjects i.e. Physics, Chemistry & Mathematics
What will be the admission process from 2013 for IITs?
• Board Percentage of Class 12 th will be given a
minimum of 40% weightage
• Different State Boards and Central Board results
normalized on the basis of percentile formula.
• ISEET will replace IITJEE and will come into place
from academic session 2012-13
• ISEET Main and Advance will be compulsory.
• Main will have 30% to 60% weightage and Advance
exam will have 0% to 30% weightage. The actual
weightage within this limit will be decided by the
board of directors if IITs.
• Merit list for the admissions will be prepared by the
admission committee for IITs
• All other institutes who took admissions based on JEE
will also follow the same pattern
What will be that admission process from 2013 for
institutes currently using AIEEE rank?
• Board Percentage of class 12 th will be given a
minimum 40% weightage.
• Different State Boards and Central Board results
normalized on the basis of percentile formula.
• ISEET will replace AIEEE and will come into place
from academic session 2012-13
• ISEET Main and Advance will be compulsory.
XtraEdge for IIT-JEE 7 MARCH 2012

• Main will have 30% to 60% weightage and Advance
exam will have 0% to 30% weightage. The actual
weightage within this limit will be decided by the
admission committee for NITs..
• Merit list for the admissions will be prepared by the
common admission committee for NITs.
• All other institutes who took admissions based on
AIEEE will also follow the same pattern
What is the selection process for admissions to other
institutes accepting ISEET 2013 scores?
• The final cut-off list for admissions with ISEET score
will be generated in three steps. Each state government
or institute will be able to decide the specific
weightage for Board, Main and Advance exam scores.
• Following are guidelines mentioned below:
o Class XII Board Score: weightage not less than
40% and can go up to 100% of the total score
o ISEET Main - weightage not less than 30% and can
go up to 60% of the total score
o ISEET Advance - weightage not more than 30% of
the total score.
o Combined Score- of ISEET Main & Advance does
not exceed weightage 60% of the total score
• It would be up to each institution/ groups of
institutions/State agencies to carry out the task of
counseling and finally the admission in a coordinated
manner
What is ISEET Main?
• ISEET Main is an Objective Aptitude Test like CAT
for IIMs, SAT in the USA, partly BITSA.
• Main will test the inherent intelligence of the student
What is ISEET Advance?
• ISEET Advance is expected to be Advance to School
Board curriculum & its difficulty level will be some
what in between AIEEE and IIT-JEE.
• This is a mandatory part of for taking admissions into
IITs and NITs while it is optional for other central and
state level institutions
• Since HRD Ministry of India has indicated to pay more
emphasis on board education, ISEET Advance exam
level is likely to be closer to AIEEE which has the
difficulty level advanced to board curriculum
Who will conduct ISEET?
• CBSE in collaboration with State Boards will
physically conduct and manage the examinations
across India
• CBSE is the body which currently conducts AIEEE
• Question paper will be set by IITs.
Will ISEET replace all State level engineering and
science entrance exams?
• All states have accepted the new pattern of common
entrance exam except Tamil Nadu, Himachal Pradesh,
Odisha, Puducherry and West Bengal which will take
the decision by the end of April 2012
• Further, It is proposed that ISEET will gradually
replace all entrance exams within a couple years and
aspirants will have to prepare for only one national
entrance exams to take admissions into all science and
engineering colleges in India vis a vis IITs, NITs,
IIITs, IISc, IISERs, other Technical and Deemed
Universities as well as all Private colleges
• States which will base its admissions on ISEET as of
now are Delhi, Haryana, Chandigarh and Uttarakhand
• The decision to implement ISEET by the states of
Gujarat, Madhya Pradesh and Chhattisgarh is final
stages
• The state of Tamil Nadu has rejected ISEET and
admissions to state government colleges will be done with
100% weightage for Tamil Nadu State Board results
Will I have to take coaching for the same pattern?
Yes, you will require coaching to score good marks in
boards, ISEET main (aptitude test) and advance. Now
onwards ISEET will be the only exam. To take admission
in a good engineering colleges like IITs, NITs good
preparation is mandatory. “Early Start” to the preparation
will help the student.
I am an intelligent student, will this change be beneficial
for me?
Yes, this change will increase your chances to get IITs
instead of NITs because selection criteria for both the
institutes will be same in general.
I am an average/ below average student, how it will be
beneficial for me?
It will be even beneficial for you since you will be free
from tension of preparing different test syllabus for
different colleges and can concentrate for only one test.
Your chances of getting good college will increase.
Will Career Point be able to prepare me according to
that examination pattern?
• Yes, Career point has already completed its preparation
for the change.
• We have already included 12 th Board (CBSE/ State
Board) in our academic curriculum in 2007 and we are
providing the complete study material for board
examinations. In fact we have adopted these patterns
far before than other institutes in Kota.
• Our students have been performing tremendously good
in board examination nation wide along with NTSE,
Olympiads & other reputed national level scholarship
tests.
• We have been conducting National Science Proficiency
Test (NSPT) every year at national level where we
have tested more that 4.5 lacs students for their of
comprehension, critical thinking, logical reasoning of
students
• Secondly, the pattern of ISEET will be more close to
AIEEE in which Career Point has proved its leadership
in the very first year of AIEEE Examination in year
2001-02 and maintaining the leading position since then.
XtraEdge for IIT-JEE 8 MARCH 2012

KNOW IIT-JEE
By Previous Exam Questions
PHYSICS
1. A magnetic field B = B 0 (y/a) ^K is into the paper in
the +z direction. B 0 and a are positive constants. A
square loop EFGH of side a, mass m and
resistance R, in x – y plane, starts falling under the
influence of gravity see figure) Note the direction
of x and y axes in figure
[IIT-1999]
O
x
⊗ ⊗ ⊗
E F
⊗ ⊗ ⊗
H G g
⊗ ⊗ ⊗
y
Find
(a) the induced current in the loop and indicate its
direction.
(b) the total Lorentz force acting on the loop and
indicate its direction, and
(c) an expression for the speed of the loop, v(t) and
its terminal value.
Sol. Suppose at t = 0, y = 0 and t = t, y = y
(A) Total magnetic flux = → AB
Where → A =
2 ^
a k and → B =
B0y
^
k
a
B y
∴ φ = 0
.a 2 = B 0 ya
a
dφ dy
Net emf., e = – = – B0 a = – B0 av(t)
dt dt
As total resistance = R
| e | B
∴ | i | = = 0 av(t)
R R
→
X
F 1
y
Y
→
F 4
y+a
→
F 3
→
F 2
Now as loop goes down, magnetic flux linked with
it increases, hence induced current flows in such a
direction so a to reduce the magnetic flux linked
with it. Hence induced current flows in
anticlockwise direction.
(B) Each side of the cube will experience a force
as shown (since a current carrying segment in a
magnetic field experience a force).
→ → →
⎛ ^ B ^ ⎞
F 1 = i( l × B)
=
0y
i ⎜ – a i×
k ⎟ = B 0 yi ^
j ;
⎝ a ⎠
→
⎛ ^ B0
(y + a) ^ ⎞
F 3 = i ⎜+
a i×
k⎟
= – iB 0 (y + a) ^
j
⎝ a ⎠
Please note that F → 2 = – F → 4 and hence will cancel
out each other.
→ → → → →
F = F + F + F + F
∴
1
2
= – iB 0 a ^
j = –
dv
m = mg –
dt
3
2
0
2
4
B a v(t) ^
j ;
R
B 2 2
0
v
Integrating it, we get,
∫
0
or
or 1 –
⎡
2 2
B a v(t) ⎤
0
log⎢g – ⎥
⎢⎣
mR ⎥⎦
2 2
– B0a
mR
a v(t)
R
g –
(v)t
⎡
2 2
B ⎤
0a
v(t)
⎢g – ⎥
log⎢
mR ⎥ = –
⎢ g ⎥
⎢
⎣
⎥
⎦
B 2 2
0
0
dv
2 2
B0a
v(t)
mR
= t
B0
2 2
a v(t) 2 2
–B a t
= e
0
mR
mgR
B0
2 2
2 2
–B a t a
or 1 – e
0
= v(t)
;
mgR
∴ v(t) =
2
⎡ –B
mgR e
0a
⎢1–
2 2
B0a
⎢ mR
⎣
2
t
⎤
⎥
⎥
⎦
a t
mR
t
=
∫
dt
0
XtraEdge for IIT-JEE 9 MARCH 2012

When terminal velocity is attained, V(t) does not
depend on t
∴ V(t) =
2 2
0 a
B mgr
2. In Young's experiment, the upper slit is covered by a
thin glass plate of refractive index 1.4 while the lower
slit it covered by another glass plate, having the same
thickness as the first one but having refractive index
1.7. Interference pattern is observed using light of
wavelength 5400 Å. It is found that the point P on the
screen where the central maximum (n = 0) fell before
the glass plates were inserted now has 3/4 the original
intensity. It is further observed that what used to be
the fifth maximum earlier, lies below the point P
while the sixth minimum lies above P. Calculate the
thickness of the glass plate. (Absorption of light by
glass plate may be neglected.) [IIT-1997]
Sol. The time taken by the ray to reach P' from S 1
d d
air plate
= +
V V
=
=
air
S 1 P'–t
+
c
P'–t + t
c
plate
S1 µ 1
t
c / µ
1
Effective path travelled = S 1 P' – t + tµ 1
where c is the speed of light in air.
P'
d
S 1
S 2
t
µ 1
µ 2
D
Similarly the time taken by the ray to reach P' from
S 2
S2 P'–t + tµ
2
=
c
Effective path travelled = S 2 P' – t + tµ 2
∴ Path difference
= S 2 P' – t + tµ 2 – S 1 P' + t – tµ 1
= (S 2 P' – S 1 P') + t(µ 2 – µ 1 )
Also when there ware no plates infront of the slits.
xd
= S 2 P' – S 1 P' = D
S 2 P' – S 1 P' = D
xd
∴ Path difference = D
xd + t(µ2 – µ 1 )
P
x
For the point P, x = 0
∴ Path difference
= t (µ 2 – µ 1 ) = t(1.7 – 1.4) = 0.3 t ...(i)
But the point P lies between the 5 th maximum and
6 th minimum (given). Therefore the path difference
= 5λ + ∆ ...(ii)
Equating equations (i) and (ii), we get
0.3t = 5λ + ∆ ...(iii)
The path difference ∆ can be determined from the
I 3
given intensity at P, which is = . 4
The expression of I/I 0 in terms of ∆ is
I
I 0
2⎛
π∆ ⎞
= cos ⎜ ⎟
⎝ λ ⎠
I 0
⎛ π∆ ⎞
For I/I 0 = 3/4 , we get cos ⎜ ⎟ =
⎝ λ ⎠
π∆ π λ
or = or ∆ =
λ 6 6
Hence, the thickness of the glass plates (Eq. 3) is
λ
0.3t = 5λ + λ/6 or ∆ = 6
Hence, the thickness of the glass plates (Eq. 2) is
⎛ 1 ⎞⎛
31 ⎞
0.3t = 5λ + λ/6 or t = ⎜ ⎟⎜
λ⎟ ⎝ 0.3 ⎠⎝
6 ⎠
⎛ 1 ⎞⎛
31
= ⎜ ⎟⎜ × 5400 Å = 9.3 × 10 4 Å = 9.3×10 –6 m
⎝ 0.3 ⎠ ⎝ 6
3. A 3.6 m long vertical pipe resonates with a source of
frequency 212.5 Hz when water level is at certain
height in the pipe. Find the height of water level
(from the bottom of the pipe) at which resonance
occurs. Neglect end correction. Now, the pipe is
filled to a height H(≈ 3.6m). A small hole is drilled
very close to its bottom and water is allowed to leak.
Obtain an expression for the rate of fall of water level
in the pipe as a function of H. If the radii of the pipe
and the hole are 2 × 10 –2 m and 1 × 10 –3 m
respectively, calculate the time interval between the
occurance of first two resonances. Speed of sound in
air is 340 m/s and g = 10 m/s 2 . [IIT-2000]
Sol. Speed of sound, V = 340 m/s.
Let l 0 be the length of air column corresponding to
the fundamental frequency. Then
V
= 212.5
4l
0
or l 0 =
V
4(212.5)
=
340
4(212.5)
3
2
= 0.4 m.
XtraEdge for IIT-JEE 10 MARCH 2012

In close pipe only odd harmonics are obtained.
Now, let l 1 , l 2 , l 3 , l 4 etc. be the lengths
corresponding to the 3 rd harmonic, 5 th harmonic,
7 th harmonic etc. Then
⎛ V ⎞
3
⎜
⎟ = 212.5 ⇒ l 1 = 1.2 m ;
⎝ 4l1
⎠
⎛ V
5
⎜
⎝ 4l
2
⎛ V
7⎜
⎝ 4l
3
⎛ V
9
⎜
⎝ 4l
4
⎞
⎟ = 212.5 ⇒ l 2 = 2.0 m
⎠
⎞
⎟
= 212.5 ⇒ l 3 = 2.8 m;
⎠
⎞
⎟ = 212.5 ⇒ l 4 = 3.6 m
⎠
–6
–dH 3.14×
10
= 2×
10×
H
–3
dt 1.26×
10
–dH
⇒ = (1.11 × 10 –2 H
H
Between first two resonances, the water level falls
from 3.2 m to 2.4 m.
dH
∴ = – 1.11 × 10 –2 dt
H
⇒
2.4
∫
3.2
dH
H
⇒ [ 2.4 – 3.2]
1
= – (1.11 × 10 –2 )
∫
dt
0
2 = – (1.11 × 10 –2 ) . t
⇒ t ≈ 43 second
0.4m
3.2m
2.0m
1.2m
3.4m
2.8m
4. Three particles A, B and C, each of mass m, are
connected to each other by three massless rigid rods
to form a rigid, equilateral triangular body of side l.
This body is placed on a horizontal frictioness table
(x-y plane) and is hinged to it at the point A so that
it can move without friction about the vertical axis
through A (see figure). The body is set into
rotational motion on the table about A with a
constant angular velocity ω. [IIT-2002]
y
A
ω
x
1.6m
0.8m
or heights of water level are (3.6 – 0.4) m,
(3.6 – 1.2) m (3.6 – 2.0) m and (3.6 – 2.8) m.
Therefore heights of water level are 3.2 m, 2.4 m,
2.4 m, 1.6 m and 0.8 m.
Let A and a be the are of cross-sections of the pipe
and hole respectively. Then
A = π (2 × 10 –2 ) 2 = 1.26 × 10 –3 m –2 .
and a = π(10 –3 ) 2 = 3.14 × 10 –6 m 2
Velocity of efflux, V = 2 gH
Continuity equation at 1 and 2 gives,
⎛ – dH ⎞
a 2gH = A ⎜ ⎟
⎝ dt ⎠
Therefore rate of fall of water level in the pipe,
⎛ – dH ⎞ a
⎜ ⎟ = 2gH
⎝ dt ⎠ A
Substituting the values, we get
→
F
B l C
(a) Find the magnitude of the horizontal force
exerted by the hinge on the body.
(b) At time T, when the side BC is parallel to the
x-axis, a force F is applied on B along BC (as
shown). Obtain the x-component and the
y-component of the force exerted by the hinge
on the body, immediately after time T.
Sol. The mass B is moving in a circular path centred at
A. The centripetal force (mlω 2 ) required for this
circular motion is provided by F′. Therefore a force
F′ acts on A (the hinge) which is equal to mlω 2 . The
same is the case for mass C. Therefore the net force
on the hinge is
2
2
F net = F' + F' + 2F' F'cos60º
2 2 1
F net = 2F' + 2F' × = 3 F′ = 3 mlω 2
2
XtraEdge for IIT-JEE 11 MARCH 2012

Y
Sol. Let the piston be displaced by a distance x.
l
A
F′ 60º
F′
F net
l
X
⎛
Then ⎜
⎝
Mg ⎞ ⎛ Mg ⎞
+ − Ax)
A ⎠ ⎝ A ⎠
γ
p0
⎟V0
= ⎜p0
+ + p⎟(v0
Q Initial pressure on the gas P 1 = p 0 + A
Mg
γ
F′
F′
Final pressure on the gas
P 2 = p 0 + A
Mg + p
B l C
(b) The force F acting on B will provide a torque
to the system. This torque is
P 0
A
V 0
x
l 3
F × 2
= Iα
F × 2
3l = (2ml 2 )α ⇒ α =
The total force acting on the system along
x-direction is
F + (F net ) x
This force is responsible for giving an acceleration
a x to the system.
c.m
F
Therefore
F + (F net ) x = 3m(a x ) c.m.
F
= 3m Q a x = αr =
4m
3F
= 4
∴
(F net ) x = 4
F
l
3
2
3
4
F
ml
3
4
F
ml
l
×
3
= 4
F
(F net ) y remains the same as before = 3 mlω 2 .
5. An ideal gas is enclosed in a vertical cylindrical
container and supports a freely moving piston of
mass M. The piston and the cylinder have equal
cross-sectional area A. Atmospheric pressure is P 0 ,
and when the piston is in equilibrium, the volume of
the gas is V 0 . The piston is now displaced slightly
from its equilibrium position. Assuming that the
system is completely isolated from its surroundings,
show that the piston executes simple harmonic
motion and find the frequency of oscillation.
[IIT-1981]
⎛
⎜p
=
⎝
where p is the extra pressure due to which the
compression x takes place.
Final volume of the gas V 2 = V 0 – Ax
The above equation can be rearranged as
0
Mg ⎞
+ + p⎟(V0
− Ax)
A ⎠
⎛ Mg ⎞ γ
⎜p0
+ ⎟V0
⎝ A ⎠
⇒ 1 = 1 +
p
∴
0 +
γ
⎡
⎢1
+
= ⎢ p
⎢
⎣
0
p ⎤ ⎡
⎥ ⎢1
−
Mg
+ ⎥ ⎣
A ⎥
⎦
γ
Ax ⎤
⎥
V0
⎦
⎛ ⎞
p γAx
⎜ p ⎟⎛ γAx
– + ⎟ ⎞
⎜
⎜ Mg ⎟
Mg V ⎝ ⎠
0 ⎜ p +
V
⎟
0
0
A
A ⎝ ⎠
Negligible as p and x are small
p
p γAx
=
Mg V 0
A
0 +
⎛
∴ p = ⎜p
⎝
⇒
F
A
0
⎛
= ⎜p
⎝
⎛
⇒ F = ⎜p
⎝
0
Mg ⎞ γAx
+ ⎟
A ⎠ V
0
0
Mg ⎞ γAx
+ ⎟
A ⎠ V
0
0
2
Mg ⎞ γA
x
+ ⎟
A ⎠ V
⎛ Mg ⎞ γA
x
⇒ Ma = ⎜p
0 + ⎟
⎝ A ⎠ V0
2
⎛ Mg ⎞ γA
x
⇒ a = ⎜p
0 + ⎟
⎝ A ⎠ V0M
Comparing it with a = ω 2 x we get
2
ω 2 ⎛ Mg ⎞ γA
x
= ⎜p
0 + ⎟
⎝ A ⎠ V M
0
2
XtraEdge for IIT-JEE 12 MARCH 2012

∴ ω =
If
ω =
⎛
⎜p
⎝
0
Mg ⎞ γA
x
+ ⎟
A ⎠ V M
Mg is small as compared to p0 then
A
p0γA
V M
0
A
∴ f = 2π
2
= 2πf
p0γ
V M
0
CHEMISTRY
6. (a) Write the chemical reaction associated with the
"brown ring test".
(b) Draw the structures of [Co(NH 3 ) 6 ] 3+ , [Ni(CN) 4 ] 2–
and [Ni(CO) 4 ]. Write the hybridization of atomic
orbital of the transition metal in each case.
(c) An aqueous blue coloured solution of a
transition metal sulphate reacts with H 2 S in acidic
medium to give a black precipitate A, which is
insoluble in warm aqueous solution of KOH. The
blue solution on treatment with KI in weakly acidic
medium, turns yellow and produces a white
precipitate B. Identify the transition metal ion.
Write the chemical reaction involved in the
formation of A and B. [IIT-2000]
Sol. (a) NaNO 3 + H 2 SO 4 → NaHSO 4 + HNO 3
2HNO 3 + 6FeSO 4 + 3H 2 SO 4 →
3Fe 2 (SO 4 ) 3 + 2NO + 4H 2 O
[Fe(H 2 O) 6 ]SO 4 .H 2 O + NO
Ferrous Sulphate
⎯→ [Fe(H 2 O) 5 NO] SO 4 + 2H 2 O
(Brown ring)
(b) In [Co(NH 3 ) 6 ] 3+ cobalt is present as Co 3+ and its
coordination number is six.
Co 27 = 1s 1 , 2s 2 2p 6 , 3s 2 3p 6 3d 7 , 4s 2
Co 3+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 6
Hence
Co 3+ ion in
Complex ion
0
2
3d 4s 4p
3d 4s 4p
d 2 sp 3 hybridization
H 3 N
NH 3
3+
Co
or
H 3 N NH 3
NH 3
NH 3
H 3 N
H 3 N
NH 3
Co 3+
NH3
NH3
NH 3
In [Ni(CN) 4 2– nickel is present as Ni 2+ ion and its
coordination numbers is four
Ni 28 =1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8 , 4s 2
Ni 2+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8
Ni 2+ ion =
Ni 2+ ion in
Complex ion
3d 4s 4p
3d 4s 4p
dsp 2 hybridization
Hence structure of [Ni(CN) 4 ] 2– is
N ≡ C
N ≡ C
Ni 2+
C ≡ N
C ≡ N
In [Ni(CO) 4 , nickel is present as Ni atom i.e. its
oxidation number is zero and coordination number
is four.
Ni in
3d 4s 4p
Complex
Its structure is as follows :
OC
CO
Ni
sp 3 hybridization
CO
CO
(c) The transition metal is Cu 2+ . The compound is
CuSO 4 .5H 2 O
CuSO 4 + H 2 S ⎯
Acidic ⎯⎯⎯
medium ⎯ → CuS ↓ + H 2 SO 4
Black ppt
2CuSO 4 + 4KI ⎯→ Cu 2 I 2 + I 2 + 2K 2 SO 4
(B) white
I 2 + I – ⎯→ I – 3 (yellow solution)
XtraEdge for IIT-JEE 13 MARCH 2012

7. (a) Write the chemical reaction associated with the
"brown ring test".
(b) Draw the structures of [Co(NH 3 ) 6 ] 3+ ,
[Ni(CN) 4 ] 2– and [Ni(CO) 4 ]. Write the hybridization
of atomic orbital of the transition metal in each
case.
(c) An aqueous blue coloured solution of a
transition metal sulphate reacts with H 2 S in acidic
medium to give a black precipitate A, which is
insoluble in warm aqueous solution of KOH. The
blue solution on treatment with KI in weakly acidic
medium, turns yellow and produces a white
precipitate B. Identify the transition metal ion.
Write the chemical reaction involved in the
formation of A and B.
[IIT-2000]
Sol. (a) NaNO 3 + H 2 SO 4 → NaHSO 4 + HNO 3
2HNO 3 + 6FeSO 4 + 3H 2 SO 4 →
3Fe 2 (SO 4 ) 3 + 2NO + 4H 2 O
[Fe(H 2 O) 6 ]SO 4 .H 2 O + NO
Ferrous Sulphate
⎯→ [Fe(H 2 O) 5 NO] SO 4 + 2H 2 O
(Brown ring)
(b) In [Co(NH 3 ) 6 ] 3+ cobalt is present as Co 3+ and its
coordination number is six.
Co 27 = 1s 1 , 2s 2 2p 6 , 3s 2 3p 6 3d 7 , 4s 2
Co 3+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 6
Hence
Co 3+ ion in
Complex ion
H 3 N
NH 3
3d 4s 4p
3d 4s 4p
3+
Co
or
H 3 N NH 3
NH 3
NH 3
d 2 sp 3 hybridization
H 3 N
H 3 N
NH 3
Co 3+
NH3
NH3
NH 3
In [Ni(CN) 4 2– nickel is present as Ni 2+ ion and its
coordination numbers is four
Ni 28 =1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8 , 4s 2
Ni 2+ ion = 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 8
Ni 2+ ion =
Ni 2+ ion in
Complex ion
3d 4s 4p
3d 4s 4p
dsp 2 hybridization
Hence structure of [Ni(CN) 4 ] 2– is
N ≡ C
N ≡ C
Ni 2+
C ≡ N
C ≡ N
In [Ni(CO) 4 , nickel is present as Ni atom i.e. its
oxidation number is zero and coordination number
is four.
Ni in
3d 4s 4p
Complex
Its structure is as follows :
CO
OC
Ni
sp 3 hybridization
CO
CO
(c) The transition metal is Cu 2+ . The compound is
CuSO 4 .5H 2 O
CuSO 4 + H 2 S ⎯
Acidic ⎯⎯⎯
medium ⎯ → CuS ↓ + H 2 SO 4
Black ppt
2CuSO 4 + 4KI ⎯→ Cu 2 I 2 + I 2 + 2K 2 SO 4
(B) white
I 2 + I – ⎯→ I – 3 (yellow solution)
8. (a) A white solid is either Na 2 O or Na 2 O 2 . A piece
of red litmus paper turns white when it is dipped
into a freshly made aqueous solution of the white
solid.
(i) Identify the substance and explain with balanced
equation.
(ii) Explain what would happen to the red litmus if
the white solid were the other compound.
(b) A, B and C are three complexes of chromium
(III) with the empirical formula H 12 O 6 Cl 3 Cr. All the
three complexes have water and chloride ion as
ligands. Complex A does not react with
concentrated H 2 SO 4 , whereas complexes B and C
lose, 6.75% and 13.5% of their original mass,
respectively, an treatment with conc. H 2 SO 4 .
Identity A, B and C.
[IIT-1999]
Sol. (a) The substance is Na 2 O 2
(i) Na 2 O 2 + 2H 2 O ⎯→ 2NaOH + H 2 O 2
(strong base) (Weak acid)
H 2 O 2 + red litmus ⎯→ White
H 2 O 2 ⎯→ H 2 O + [O]
Nascent oxygen bleaches the red litumus.
(ii) Na 2 O + H 2 O ⎯→ 2NaOH
XtraEdge for IIT-JEE 14 MARCH 2012

NaOH solution turns colour of red litmus paper into
2
blue due to stronger alkaline nature.
or f = = 5
0.4
(b) A = [Cr(H 2 O) 6 ]Cl 3 . It has no reaction with conc.
(b) According to adiabatic gas equation,
H 2 SO 4 as its all water molecular are present in
γ γ
coordination sphere.
P 1 V 1 = P 2 V 2
B = [Cr(H 2 O) 5 Cl]Cl 2 .H 2 O
Here, P 1 = P
or γ = 1.4
molecular formula of X = C 2 H 4 Cl 2
If f, be the number of degrees of freedom, then
The two isomers of X are :
H
2
2
H
γ = 1 + or 1.4 = 1 + | |
f f Y = ClCH 2 CH 2 Cl → Cl − C − C − Cl
| |
2
or = 1.4 – 1 = 0.4 H H
f
Conc. H 2 SO 4 removes its one mol of H 2 O as it is
outside the coordination sphere.
Molecular Weight of complex = 266.5
V 1 = V
V 2 = 5.66 V
Hence, PV γ = P 2 × (5.66V) γ = P 2 ×(5.66) γ × V γ
18
P P P
% loss = × 100 = 6.75%
or P 2 = = =
γ
1.4
266.5
(5.66) (5.66) 11.32
[using eq.(1)]
C = [Cr(H 2 O) 4 Cl]Cl 2 .2H 2 O
Conc. H 2 SO 4 removes its 2H 2 O which are outside
Hence, work done by the gas during adiabatic
expansion
of the coordination sphere.
P
PV − × 5.66V
18
P1
V1
− P2V2
% loss = 2 × × 100 = 13.5 %
=
=
11.32
266.5
γ –1
1.4 –1
Hence complexes A = [Cr(H 2 O) 6 ]Cl 3
PV
PV −
B = [Cr(H 2 O) 5 Cl]Cl 2 .H 2 O
= 2 PV
= = 1.25 PV
C = [Cr(H 2 O) 4 Cl 2 ]Cl.2H 2 O
0.4 2×
0.4
9. An ideal gas having initial pressure P, volume V
10. An organic compound X, on analysis gives 24.24
and temperature T is allowed to expand
percent carbon and 4.04 percent hydrogen. Further,
adiabatically until its volume becomes 5.66 V,
sodium extract of 1.0 g of X gives 2.90 g of silver
while its temperature falls to T/2.
chloride with acidified silver nitrate solution. The
(a) How many degrees of freedom do the gas
compound X may be represented by two isomeric
molecules have ?
structures, Y and Z. Y on treatment with aqueous
(b) Obtain the work done by the gas during the
expansion as a function of the initial pressure P and
potassium hydroxide solution gives a dihydroxy
compound, while Z on similar treatement gives
volume V.
[IIT-1990]
ethanal. Find out the molecular formula of X and
Sol. (a) According to adiabatic gas equation,
give the structures of Y and Z. [IIT-1989]
TV γ–1 = constant
Sol. C = 24.24%, H = 4.04%
or T 1 V γ–1 γ–1
35.5 2.90g
1 = T 2 V 2 Percentage of Cl = × × 100 = 71.74%
143.5 1g
Here, T 1 = T ; T 2 = T/2
V 1 = V
24.24
Relative number of C atoms =
and V 2 = 5.66 V
12
= 2.02
Hence, TV γ–1 T
4.04
= × (5.66V)
γ–1
Relative number of H atoms =
2
1
= 4.04
71.74
T
= × (5.66) γ–1 × V γ–1
Relative number of Cl atoms = = 2.02
35.5
2
Atomic ratio = C : H : Cl = 2.02 : 4.04 : 2.02
or (5.66) γ–1 = 2 ...(1)
= 1 : 2: 1
Taking log,
(γ – 1)log 5.66 = log 2
Empirical formula of X = CH 2 Cl
It is given that isomer of Y of the compound X
or
log 2 0.3010
gives a dihydroxy compound on treatment with
γ – 1 = = = 0.4
aqueous potassium hydroxide. Therefore, the given
log5.66 0.7528
compound should contain two Cl atoms. Thus
XtraEdge for IIT-JEE 15 MARCH 2012

H H | |
H − C − C − Cl
Z = CH 3 CHCl 2 → | |
H Cl
The relevant reactions are :
(i) ClCH 2 –CH 2 Cl ⎯⎯ KOH(aq) ⎯⎯
→
HO–CH 2 –CH 2 –OH
(ii) CH 3 CHCl 2 ⎯⎯ KOH(aq) ⎯⎯
→ CH 3 CHO
MATHEMATICS
11. Let ABC be an equilateral triangle inscribed in the
circle x 2 + y 2 = a 2 . Suppose perpendiculars from A,
x y
B, C to the major axis of the ellipse + = 1,
2 2
a b
(a > b) meets the ellipse respectively at P, Q, R so
that P, Q, R lie on the same side of the major axis
as A, B, C respectively. Prove that the normals to
the ellipse drawn at the points P, Q and R are
concurrent.
[IIT-2000]
Sol. Let A ≡ (a cosθ, b sinθ) so the coordinates of
B ≡ {a cos(θ + 2π/3), a sin (θ + 2π/3)}
and C ≡ {a cos(θ + 4π/3), a sin (θ + 4π/3)}.
y
B
Q
C
O
R
P
2
A(a cos θ, b sinθ)
Ellipse
Circle
According to the given condition, coordinates of P
are (a cosθ, b sinθ), and that of Q are
(a cos(θ + 2π/3), b sin(θ + 2π/3)) and that of R are
(a cos(θ + 4π/3), b sin (θ + 4π/3))
(It is given that P, Q, R are on the same side of
x-axis as A, B and C) Equation of the normal to the
ellipse at P is
ax by
– = a 2 – b 2
cosθ sin θ
or ax sin θ – by cos θ = 2
1 (a 2 – b 2 ) sin 2θ ...(1)
Equation of normal to the ellipse at Q is
⎛ 2π
⎞ ⎛ 2π
⎞
ax sin ⎜θ + ⎟ – by cos ⎜θ + ⎟
⎝ 3 ⎠ ⎝ 3 ⎠
2
x
∆ 1 =
1
= (a 2 – b 2 ⎛ 4π
⎞
) sin ⎜2 θ + ⎟ ...(2)
2 ⎝ 3 ⎠
Equation of normal to the ellipse at R is
ax sin(θ + 4π/3) – by cos(θ + 4π/3)
= 2
1 (a 2 – b 2 ) sin (2θ + 8π/3) ...(3)
But sin (θ + 4π/3) = sin(2π + θ – 2π/3)
= sin(θ – 2π/3)
and cos (θ + 4π/3) = cos (2π + θ – 2π/3)
and sin (2θ + 8π/3) = sin (4π + 2θ – 4π/3)
= sin (2θ – 4π/3)
Now, (3) can be written as
ax sin (θ – 2π/3) – by cos (θ – 2π/3)
= 2
1 (a 2 – b 2 ) sin (2θ – 4π/3) ...(4)
For the lines (1), (2) and (4) to be concurrent, we
must have the determinant
1 2 2
( a − b
asin
θ − bcosθ
2
⎛ 2π
⎞ ⎛ 2π
⎞ 1 2 2 ⎛
asin⎜θ + ⎟ − bcos⎜θ + ⎟ ( a − b )sin⎜2θ +
⎝ 3 ⎠ ⎝ 3 ⎠ 2
⎝
⎛ 2π
⎞ ⎛ 2π
⎞ 1 2 2 ⎛
asin⎜θ − ⎟ − b⎜θ − ⎟ ( a − b )sin⎜2θ −
⎝ 3 ⎠ ⎝ 3 ⎠ 2
⎝
Thus line (1), (2) and (4) are concurrent.
)sin 2θ
4π
⎞
⎟
3 ⎠
4π
⎞
⎟
3 ⎠
= 0
12. A right circular cone with radius R and height H
contains a liquid which evaporates at a rate
proportional to its surface area in contact with air
(proportionality constant = k > 0). Find the time
after which the cone is empty. [IIT-2003]
Sol. Given : liquid evaporates at a rate proportional to
its surface area
dv
⇒ ∝ – S ...(1)
dt
We know, volume of cone = 3
1 πr 2 h and
surface area = πr 3 (of liquid in contact with air)
or V = 3
1 πr 2 h and S = πr 2 ...(2)
where tan θ = H
R and h
r = tan θ ...(3)
from (2) and (3)
V = 3
1 πr 3 cotθ and S = πr 2 ...(4)
h
R
Substituting (4) in (1), we get
1 cotθ . 3r 2 dr
. = – kπr
2
3
dt
θ
r
H
XtraEdge for IIT-JEE 16 MARCH 2012

⇒ cot θ
∫ 0 R
dr = – k
∫ T dt
0
⇒ cot θ(0 – R) = – k(T – 0)
⇒ R cot θ = kT
⇒ H = kT (using (3))
⇒ T = k
H
∴ required time after which the cone is empty
H
= T = k
13. Sketch the curves and identify the region bounded
by x = 1/2, x = 2, y = ln x and y = 2 x . Find the area
of this region.
[IIT-1991]
Sol. The required area is the shaded portion in following
figure.
y
y = 2 x
O
1/2
1
2
y = log e x
In the region 2
1 ≤ x ≤ 2 the curve y = 2 x lies above
as compared to y = log e x
Hence, the required area
2
=
∫
( 2 − log x)
dx
1/ 2
x
⎛
x
2
⎞
= ⎜ − ( x log x − x)
⎟
log 2
⎝
⎠
=
2
1/ 2
4 − 2 5 3
– log 2 +
log 2 2 2
14. ABC is a triangle such that
1
sin(2A + B) = sin(C – A) = –sin(B + 2C) = 2
If A, B and C are in Arithmetic Progression,
determine the values of A, B and C. [IIT-1990]
Sol. Given that in ∆ABC, A, B and C are in A.P.
A + C = 2B
also A + B + C = 180º
⇒ B = 60º
Also given that,
sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1/2
...(1)
⇒ sin (2A + 60º) = sin (C – A) = – sin(60º + 2C) = 2
1
⇒ 2A + 60º = 30º, 150º
{neglecting 30º, as not possible}
⇒ 2A + 60º = 150º
⇒ A = 45º
again from (1), sin (60º + 2c) = –1/2
⇒ 60º + 2C = 210º, 330º
⇒ C = 75º or 135º
Also from (1) sin (C – A) = 1/2
C – A = 30º, 150º, 195º
for A = 45º, C = 75º and C = 195º (not possible)
∴ C = 75º
Hence, A = 45º, B = 60º, C = 75º
15. Find the centre and radius of the circle formed by
all the points represented by z = x + iy satisfying the
z − α
relation =k(k ≠ 1), where α and β are
z − β
constant complex numbers given by α = α 1 + iα 2 ,
β = β 1 + iβ 2 .
[IIT-2004]
Sol. As we know; |z| 2 = z. z
2
| z − α |
⇒
= k 2
2
| z − β |
⇒ (z – α)( z – α ) = k 2 (z – β)( z – β )
|z| 2 – α z – α z + |α| 2 = k 2 (|z| 2 – β z – β z+ |β| 2 )
or |z| 2 (1 – k 2 ) – (α – k 2 β) z – ( α – β k 2 ) z
+ (|α| 2 – k 2 |β| 2 ) = 0
2
2
⇒ |z| 2 ( α − k β)
( α − βk
)
– z – z
2
2
(1 − k ) (1 − k )
| α | −k
| β |
+
= 0 ...(i)
2
(1 − k )
On comparing with equation of circle,
|z| 2 + a z + α z + b = 0
whose centre is (– a) and radius =
∴ centre for (i)
2
α − k β
= and radius
2
1−
k
=
2
2
2
| a |
2
−b
⎛
2
2
2
k ⎞⎛
k ⎞
⎜
α − β
⎟⎜
α − β
⎟
αα − k ββ
−
2
2
2
1 k
1 k
⎝ − ⎠⎝
− ⎠ 1−
k
k(
α − β)
radius =
2
1−
k
XtraEdge for IIT-JEE 17 MARCH 2012

Physics Challenging Problems
Set # 11
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. A metallic conductor of irregular cross section is as
shown in figure A constant potential difference is
applied across the ends (A) and (B). Then
A
P
Q
(A) Electric current at cross section P is equal to
that of cross section Q
(B) Electric field intensity at P is less than that at Q
(C) The number of electrons crossing per unit area
per unit time at cross section P is less than that
at Q
(D) The rate of heat generating per unit time at Q is
greater than that of P
2. A circular ring of radius R with uniform positive
charge density λ per unit length is located in the y-z
plane with its centre at the origin O. A particle of
charge –q 0 is released from x = 3R
on x-axis at t
= 0 then kinetic energy of particle when it passes
through origin, is
(A)
(C)
λq
0
0
2∈
q 0 λ
∈
0
(B)
(D)
q
0
λ
3∈
q
0
0
λ
4∈
3. Missile is fired for maximum range at your town
from a place in the enemy country at a distance ‘x’
from your town. The missile is first detected at its
half-way point. Then
(A) the velocity with which the missile was
projected is gx
(B) you have a warning time of
x
2g
0
B
(C) the speed of the missile when it was detected
gx
is
2
x
(D) the maximum height attained by the missile is 4
4. Figure shows a square loop being pulled out with a
constant speed out of region of uniform magnetic
field. The induced emf in the loop
B
× × × ×
× l × l
× × × × v
× × × ×
× × × l
× × × ×
(A) first increases, then decreases
(B) first decreases, then increases
(C) has a maximum value Bvl 2
(D) has a maximum value 2 Bvl
5. The variation of pressure versus volume is shown in
the figure. The gas is diatomic and the molar
specific heat capacity for the process is found to be
xR. Find the value of x.
P
6. Figure shows a parabolic reflector in x-y plane
given by y 2 = 8x. A ray of light traveling along the
line y = a is incident on the reflector. Find where the
ray intersects the x-axis after reflection.
y-axis y 2 =8x
line y = a
P(0,a)
incident ray
x-axis
V
XtraEdge for IIT-JEE 18 MARCH 2012

7. A hydrogen like atom (atomic number Z) is in a
higher excited state of quantum number n. This
excited atom can make a transition to the first
excited state by successively emitting two photons
of energies 10.20 eV and 17.00 eV respectively.
Alternatively, the atom from the same excited state
can make a transition to the second excited state by
successively emitting two photons of energies 4.25
eV and 5.95 eV respectively. Determine the value of
Z. (Ionisation energy of hydrogen atom is 13.6 eV)
8. Consider the circuit showing in figure. There are
three switches S 1 , S 2 , S 3 . Match the columns.
Column-I
Column-II
(A) If S 2 and S 3 are opened and S 1 is (P) CV/4
closed then in steady state, charge
on the capacitor is
(B) If switch S 2 only is closed then (Q) 2CV/5
maximum charge on the capacitor is
(C) If switch S 3 only is closed then (R) CV/3
maximum charge on the capacitor is
(D) If all the switches are closed then (S) CV
maximum charge on the capacitor is
(T) zero
Cartoon Law of Physics
As speed increases, objects can be in several places
at once.
This is particularly true of tooth-and-claw fights, in
which a character's head may be glimpsed emerging
from the cloud of altercation at several places
simultaneously. This effect is common as well
among bodies that are spinning or being throttled.
A `wacky' character has the option of self- replication
only at manic high speeds and may ricochet off walls
to achieve the velocity required.
Puzzle : Marble Mix Up
• Years ago, to puzzle his friends, a scientist gave
one of four containers containing blue and/or
yellow marbles to each of the friends; Tom,
Dick, Harry, and Sally.
• There were 3 marbles in each container, and the
number of blue marbles was different in each
one. There was a piece of paper in each container
telling which color marbles were in that
container, but the papers had been mixed up and
were ALL in the wrong containers.
• He then told all of his friends to take 2 marbles
out of their container, read the label, and then tell
him the color of the third marble.
• So Tom took two blue marbles out of his
container and looked at the label. He was able to
tell the color of the third marble immediately.
• Dick took 1 blue marble and 1 yellow marble
from his container. After looking at his label he
was able to tell the color of his remaining
marble.
• Harry took 2 yellow marbles from his container.
He looked at the label in his container, but could
not tell what color the remaining marble was.
• Sally, without even looking at her marbles or her
label, was able to tell the scientist what color her
marbles were. Can you tell what color marbles
Sally had? Can you also tell what color marbles
the others had, and what label was in each of
their containers?
XtraEdge for IIT-JEE 19 MARCH 2012

8 Questions
1. Ans. Remain Same
Hint: KE = QU
Magnetic moment = i × Area
Q
= × π
2
R
T
2πm
2mKE
Q T = R = =
qB qB
2MU
qB
Q × B 2m×
U
Magnetic moment = × π×
2πm
2 2
Q B
Magnetic moment
B v
2. Ans. 0
λ
Hint:
U
=
B
3B
l
= 0 v
i 3λ l
B0
v
i =
λ
3. Ans. Zero if both wires slide in opposite direction,
0.2mA if both wires move towards left
Hint:
2
Solution
Set # 10
Physics Challenging Problems
were Published in February Issue
4. Ans. None
Hint:
Both dF get cancel out net force on the loop is zero.
5. Ans. 2q Ea
Hint: Work done by field = – (U B – U A )
= −q[V
V ]
2B qR
6. Ans. 0π
3
Hint:
2 2π
φ = R × × B0
(p + qt)
3
dφ
2 2π
e = = R × × B0q
dt 3
7. Ans. 4
Hint:
2
B −
= +q [E × d]
= qE × 2a
= 2qEa
A
When both are moving in same direction
BLv
i = = 0. 2mA
(9 + 1)
When both are moving in opposite direction equation
emf of battery = 0 ∴i = 0
2
2 2
0 1 − B0t(
π 2 − π1
2
2
0t
× π1
− B0t
× r2
φ = 3B t × π
φ = 2B
π
dφ
2 2
= 2B0π1
− B0πr2
dt
dφ
E×
2πr
=
dt
As E = 0
d =
2
0 Q 4 r 1 = r
dt
2
⎛ r1
⎞ 1
⎜ =
r
⎟
⎝ 2 ⎠ 4
φ 2
2
8. Ans. Α→P,Q,R; B→P,Q,S; C→Q,T; D→P,Q,S
)
XtraEdge for IIT-JEE 20 MARCH 2012

XtraEdge for IIT-JEE 21 MARCH 2012

Students Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
PHYSICS
1. A triangular prism of mass M = 1.12 kg having base
angle 37º is placed on a smooth horizontal floor. A
solid cylinder of radius R = 20 cm and mass m = 4 kg
is placed over the inclined surface of the prism. If
sufficient friction exists between the cylinder surface
and the prism, so that cylinder does not slip, calculate
acceleration of prism when the system is released.
Calculate also, force of friction existing between the
cylinder and the prism. (g = 10 ms –2 )
m
M
37º
Sol. Let angular acceleration of cylinder be α clockwise
and acceleration of prism be a leftwards.
Acceleration of cylinder axis (relative to prism) is
Rα = 0.2 α down the plane.
Its horizontal and vertical components are
0.2α .cos37º (rightwards) and 0.2αsin37º
(downwards) respectively. But resultant acceleration
of cylinder axis is vector sum of Rα and a, therefore
horizontal and vertical components of resultant
acceleration of cylinder axis become
(0.2α cos37º – α) rightwards and (0.2α sin 37º)
downwards respectively.
Considering free body di agrams (fig. (a) and Fig.
(b))
F
M.g
N
M.a
v
Fig. (a)
37º
Iα
O
Mg
m (0.2 α cos 37º – a)
N F m (0.2 a sin 37º)
Fig. (b)
For horizontal forces acting on prism,
N sin 37º – F cos 37º = Ma
For horizontal forces acting on cylinder,
N sin 37º – F cos 37º = m (0.2α cos 37 – a)
...(i)
...(ii)
For vertical forces on cylinder,
mg – N cos37º – F sin37º = m (0.2α . sin37º) ...(iii)
Taking moments (about O) of forces acting on the
cylinder,
FR = Iα
...(iv)
mR 2
where I = = 0.08 kg m 2
2
From above equation N = 23 newton
α = 30 rad/sec 2
a = 3.75 ms –2
F = 12 newton
Ans.
Ans.
2. A cylindrical tank of base area A has a small orifice
of area a at the bottom. At time t = 0, a tap starts to
supply water into the tank at a constant rate Q m 3 s –1 .
Calculate relation between height h of water in the
tank and time t.
Sol. When water supply is started, water starts to
accumulate in the tank. But leakage of water through
orifice at bottom also start simultaneously.
Let at instant t, height of water in the tank be y as
shown in figure
Then flow velocity through orifice,
v =
2 gy
∴ Volume flowing out per second through orifice,
q = a
2gy
But rate of supply to the tank is Q. Therefore net of
increase of volume in tank = (Q – q) m 3 s –1 . Since,
y
XtraEdge for IIT-JEE 22 MARCH 2012

area of tank base is A, therefore, net rate of increase
of height of water in tank,
dy =
dt
(Q – q)
A
=
Q –
A
2gy
integrating above equation with limits, at t = 0, y = 0
and at t, y = h,
∴
h
∫
0
t =
Q –
dy
2gy
A ⎡
⎢–
g ⎢
⎣
t
= A
∫.dt
2h +
0
Q ⎪⎧
Q – 2gh ⎪⎫
⎤
loge
⎨ ⎬⎥
g ⎪⎩
Q ⎪⎭ ⎥
⎦
3. Distance between centres of two stars is 10 α. Mass of
these stars in M and 16 M and their radii are a and 2a
respectively. A body of mass m is fired straight from
the surface of larger star directly towards the smaller
star. Calculate minimum initial speed of the body so
that it can reach the surface of smaller star. Obtain the
expression in terms of G, M, and a.
Sol. Since the body is projected from surface of large star
towards smaller star, therefore, the body follows a
straight line path AB, as shown in figure (a). Near
point A, magnitude of gravitational force exerted by
larger star on the body is greater than that exerted by
smaller star. Therefore, near point A, the body
experiences a resultant force directed towards larger
star. Hence, the body retards till this resultant force
becomes zero. It means velocity of body is
minimum at that point where magnitudes of
gravitational force exerted by two stars are equal. If
initial velocity of star is such that it crosses this
point, then it will reach the smaller star.
2a
16M
2a
A
10.a
Fig.(a)
x
P
B
a
M
a
(10a–x)
Fig.(b)
Let distance of this point P from centre of larger star
be x.
Then,
or
G(16M)m
2
x
x = 8a
=
GMm
2
(10a – x)
Gravitational potential energy of body at A = that
due to larger star + that due to smaller star.
∴ U 1 = –
G(16M)m
2a
–
G(M)m
8a
= –
Similarly, gravitational potential energy at P,
U 2 = –
G(16M)m
8a
–
GMm
a
65 GMm
8 a
5 GMm
= – 2 a
Minimum kinetic energy required at A = Increase in
potential energy from A to P
∴
1 2
mv 0 = U 2 – U 1
2
∴ v 0 =
45GM
4a
Ans.
4. A non-conducting piston of mass m and area S divides
a non-conducting, closed cylinder into two parts as
shown in figure. Piston is connected with left wall of
cylinder by a spring of force constant K. Left part is
evacuated and right part contains an ideal gas at
pressure P. Adiabatic constant of the gas is γ and in
equilibrium length of each part is l.
Calculate angular frequency of small oscillations of the
piston.
K
Pressure P
Sol. If the piston is slightly displaced leftwards from its
equilibrium position, spring is further compressed
and gas expands. Due to expansion of gas, its
pressure decreases. Piston is restored due to both the
reasons, i.e., increase in compression and decrease
in pressure.
Let the piston be displaced through dx.
Then increase in compression in spring = Kdx
Increase in volume of gas is dV = Sdx
Since piston and cylinder both are non-conducting,
therefore, gas undergoes an adiabatic expansion.
Hence, it obeys the law PV γ = constant.
Taking log, log P + γ. log V = constant.
Differentiating the above equation.
dP +
P
or dP = –
dV
γP
γ = 0 or dP = – dV
V
V
γP
(S.dx) = –
(Sl)
γP
dx
l
XtraEdge for IIT-JEE 23 MARCH 2012

Restoring force,
γPS
F = K.dx + S. |dP | = K.dx + dx
l
Kl
+ γPS
or F = dx
l
F ⎛ Kl
+ γPS
⎞
∴ Restoring acceleration = = ⎜ ⎟⎠ dx
m ⎝ ml
...(i)
Since acceleration of piston is restoring and is
directly proportional to displacement dx, therefore, it
performs SHM.
Comparing equation (i) with Restoring acceleration
= ω 2 . (displacement)
Kl
+ γPS
Angular frequency, ω =
Ans.
ml
5. A steady beam of α-particles travelling with kinetic
energy E = 83.5 ke V carries a current of I = 0.2 µA.
(i) If this beam strikes a plane surface at an angle
θ = 30º with normal to the surface, how many
α-particles strike the surface in t = 4 second ?
(ii) How many α-particles are there in length l = 20 cm
of the beam?
(iii) Calculate power of the source used to accelerate
these α-particles from rest.
(Mass of α-particle = 6.68 × 10 –27 kg)
Sol. Since, current is rate of flow of charge through a
section, therefore, a current I = 0.2µA means that a
charge 0.2 µC is flowing per second.
Charge of an α-particle is q = 2e = 3.2 × 10 –19 C
∴ Rate of flow of α-particles, n
= q
I = 6.25 × 10
11
per second
∴ Number of α-particles striking against a surface
in t = 4 second
= n × t = 6.25 × 10 11 × 4 = 2.5 × 10 12 Ans.(i)
(Not : these is no significance of angle θ for
calculation of number of α-particles striking the
surface.)
Kinetic energy of each α-particle is E = 83.5 Ke V
or E = (83.5 × 10 3 ) (1.6 × 10 –19 ) J
But E = 2
1 mv
2
where m = 6.68 × 10 –27 kg
∴ Velocity of α-particles is v = 2 × 10 6 ms –1 .
It means a beam of length v = 2 × 10 6 m crosses a
section in one second. But number of α-particles
passing through a section in one second in
n = 6.25 × 10 11
∴ Number of α-particles in unit length of the
beam = v
n = 3.125 × 10
5
per m.
∴ Number of α-particle in length l of the beam
n
= l
v
= 6.25 × 10 4 Ans.(ii)
Let potential difference of the source be V volt.
Kinetic energy of α-particle accelerated by this
source,
E = qV or V = q
E = 41.75 kV
Power supplied by the source to accelerate
α-particles,
P = VI = 8.35 × 10 –3 watt Ans.(iii)
GLOBAL WARMING IS REAL
The arctic ice is receding and global warming is no
longer a theory but a reality. Scientists predict that
by the year 2100, the average surface temperature
will jump up by 6 degrees Fahrenheit. Nighttime
temperatures will be higher and there will be hotter
days.
Since air temperature is a powerful component of
climate, there will be unavoidable climate changes
in the future. Some climate changes involve
extreme weather disturbances such as more severe
hurricanes and longer droughts. There will be an
increased precipitation of snow and rain during
winter. The faster melting of snow during the spring
will result in flooding. All these climate changes are
predicted based on the assumption that changes will
be relatively gradual.
XtraEdge for IIT-JEE 24 MARCH 2012

PHYSICS FUNDAMENTAL FOR IIT-JEE
Calorimetry, K.T.G., Heat transfer
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Calorimetry :
The specific heat capacity of a material is the amount
of heat required to raise the temperature of 1 kg of it
by 1 K. This leads to the relation
Q = ms θ
where Q = heat supplied, m = mass, θ = rise in
temperature.
The relative specific heat capacity of a material is the
ratio of its specific heat capacity to the specific heat
capacity of water (4200 J kg –1 K –1 ).
Heat capacity or thermal capacity of a body is the
amount of heat required to raise its temperature by 1
K. [Unit : J K –1 ]
Thus heat capacity = Q/θ = ms
dθ 1 dQ
Also = ×
dt ms dt
i.e., the rate of heating (or cooling) of a body depends
inversely on its heat capacity.
The water equivalent of a body is that mass of water
which has the same heat capacity as the body itself.
[Unit : g or kg] This is given by
m×s
W =
s w
where m = mass of body, s = specific heat capacity of
the body, s w = specific heat capacity of water.
Principle of Calorimetry : The heat lost by one
system = the heat gained by another system. Or, the
net heat lost or gainsed by an isolated system is zero.
It system with masses m 1 , m 2 , ...., specific heat
capacities s 1 , s 2 , ...., and initial temperatures θ 1 , θ 2 , ....
are mixed and attain an equilibrium temperature θ
then
θ =
Σmsθ
, for equal masses θ =
Σms´
Σsθ
Σs
Newton's law of cooling :
The rate of loss of heat from a body in an
environment of constant temperature is proportional
to the difference between its temperature and that of
the surroundings.
If θ = temperature of the surroundings then
dθ
– ms = C´(θ – θ0 )
dt
where C´ is a constant that depends on the nature and
extent of the surface exposed. Simplifying
dθ C
= –C(θ – θ0 ) where C = = constant
dt
mś
Kinetic theory of gases :
The pressure of an ideal gas is given by p = 3
1 µnC
2
where µ = mass of each molecule, n = number of
molecules per unit volume and C is the root square
speed of molecules.
p = 3
1 ρC
2
or pV = 3
1 mC
2
where ρ is the density of the gas and m = mass of the
gas.
Root Mean Square Speed of Molecules :
This is defined as
2
2
2
C1 + C2
+ C3
+ ... + CN
C =
N
where N = total number of molecules. It can be
obtained through these relations
C =
3p
ρ
=
3RT
M
Total Energy of an ideal gas (E) :
This is equal to the sum of the kinetic energies of all
the molecules. It is assumed that the molecules do not
have any potential energy. This follows from the
assumption that these molecules do not exert any
force on each other.
1
E = mC 2 3 m 3
= RT = pV
2 2 M 2
Thus, the energy per unit mass of gas = 2
1 C
2
The energy per unit volume = 2
3 p
The energy per mole = 2
3 pV = 2
3 RT
2
XtraEdge for IIT-JEE 25 MARCH 2012

Perfect gas equation :
From the kinetic theory of gases the equation of an
ideal gas is
pV = RT for a mole
and
pV = M
m RT for any mass m
Avogadro number (N) and Boltzmann constant (k) :
The number of entities in a mole of a substance is
called the Avogadro number. Its value is
6.023 × 10 23 mol –1 .
The value of the universal gas constant per molecular
is called Boltzmann constant (k). Its value is
1.38 × 10 –23 J K –1 .
Degrees of Freedom : Principle of equipartition of
energy :
The number of ways in which energy may be stored
by a system is called its degrees of freedom.
Principle of Equipartition of Energy : This
principle states that the total energy of a gas in
thermal equilibrium is divided equally among its
degrees of freedom and that the energy per degree of
freedom is kT/2 where T is the temperature of the
gas. For a monoatomic atom the number of degrees
of freedom is 3, for a diatomic atom it is 5, for a
polyatomic atom it is 6.
Hence the energy of a mole of a monoatomic gas is
⎛ 1 ⎞ 3
µ = N ⎜3 × kT ⎟ = RT
⎝ 2 ⎠ 2
Which is the same as that given by the kinetic theory.
For a mole of diatomic gas µ
⎛ 1 ⎞ 5
= N⎜5 × kT ⎟ = RT
⎝ 2 ⎠ 2
For a mole of polyatomic gas µ
⎛ 1 ⎞
= N⎜6 × kT ⎟ = 3RT
⎝ 2 ⎠
When the irrational degrees of freedom are also taken
into account, the number of degrees of freedom
= 6n – 6 for non-linear molecules
= 6n – 5 for linear molecules
where n = number of atoms in a molecule.
Kinetic Temperature :
The kinetic temperature of a moving particle is the
temperature of an ideal gas in thermal equilibrium
whose rms velocity equals the velocity of the given
particle.
Maxwellian distribution of velocities :
In a perfect gas all the molecules do not have the
same velocity, rather velocities are distributed among
them. Maxwell enunciated a law of distribution of
velocities among the molecules of a perfect gas.
According to this law, the number of molecules with
velocities between c and c + dc per unit volume is
dn = 4πna 3 2
bc
e − c 2 dc where
m m
b = and a =
2kT 2πkT
and the number of molecules with the velocity c per
unit volume is
n c = 4πna 3 2
bc
e − c 2
The plot of n c and c is shown in the figure. The
velocity possessed by the maximum number of
molecules is called the most probable velocity
α = 2 kT / m
The mean velocity
c =
α c C rms
8 kT / mπ
and v rms = 3 kT / mπ
Conduction :
The transfer of heat through solids occurs mainly by
conduction, in which each particle passes on thermal
energy to the neighboring particle but does not move
from its position. Very little conduction occurs in
liquids and gases.
θ 1 θ 2
Q
d A
Consider a slab of area A and thickness d, whose
opposite faces are at temperature θ 1 and θ 2 (θ 1 > θ 2 ).
Let Q heat be conducted through the slab in time t.
⎛ θ1 − θ2
⎞
Then Q = λA ⎜ ⎟ t
⎝ d ⎠
where λ = thermal conductivity of the material.
This has a fixed value for a particular material, being
large for good conductors (e.g., Cu, Ag) and low for
insulators (e.g., glass, wood).
Heat Current : The quantity Q/t gives the heat flow
per unit time, and is called the heat current.
In the steady state, the heat current must be the same
across every cross-section. This is a very useful
principle, and can be applied also to layers or slabs in
contact.
Q dθ dθ θ
= – λA where the quantity =
1 − θ 2
t dx
dx d
called the temperature gradient.
Q
is
XtraEdge for IIT-JEE 26 MARCH 2012

Unit of λ : Different units are used,
e.g., cal cm s ºC –1 , cal m –1 s –1 ºC –1 , jm´1 s –1 ºC –1 .
Convection :
It is a process by which heat is conveyed by the
actual movement of particles. Particles closest to the
source receive heat by conduction through the wall of
the vessel. They rise up-wards and are replaced by
colder particles from the sides. Thus, a circulation of
particles is set up – hot particles constitute the
upward current and cold particles, the side and
downward current.
The transfer of heat by convection occurs only in
fluids, and is the main mode of heat transfer in them.
Most fluids are very poor conductors.
Radiation :
Thermal Radiation : Thermal radiations are
electromagnetic waves of long wavelengths.
Black Body : Bodies which absorb the whole of the
incident radiation and emit radiations of all
wavelengths are called black bodies.
It is difficult to realize a perfect black body in
practice. However, a cavity whose interior walls are
dull black does behave like a black body.
Absorption : Every surface absorbs a part or all of
the radiation falling on it. The degree of absorption
depends on the nature and colour of the surface. Dull,
black surfaces are the best absorbers. Polished, white
surfaces absorb the least. The coefficient of
absorption for a surface is
radiation absorbed
a λ =
radiation incident
The suffix λ denotes the wavelength of the radiation
being considered, Clearly, a λ = 1 for a black body, for
all values of λ.
Emission : Each surface emits radiation (radiates)
continuously. The emissive power (e λ ) is defined as
the radiation emitted normally per second per unit
solid angle per unit area, in the wave-length range λ
and λ + dλ. Clearly, the emissive power of a black
body (denoted by E λ ) is the maximum.
Kirchhoff's Law : According to this law, for the
same conditions of temperature and wavelength, the
ratio e λ /a λ is the same for all surfaces and is equal to
E λ . This simply means that good absorbers are good
emitters. Hence, a black body is the best emitter, and
a polished white body, the poorest emitter.
Prevost's Theory of Exchanges : All bodies emit
radiations irrespective of their temperatures. They
emit radiations to their environments and receive
radiations from their environments simultaneously. In
the equilibrium state the exchange between a body
and the environment of energy continues in equal
amounts.
Stefan-Boltzmann Law : If a black body at an
absolute temperature T be surrounded by another
black body at an absolute temperature T 0 , the rate of
loss of radiant energy per unit area is
E = σ(T 4 – T 4 0 )
where σ is a constant called Stefan constant and its
value is 5.6697 × 10 –8 W m –2 K –4
The total energy radiated by a black body at an
absolute temperature T is given by
E = σT 4 × surface area × time
Note : Remember that rate of generation of heat by
electricity is given by H = I 2 V 2
R or or VI Js –1 or W.
R
Solved Examples
1. An earthenware vessel loses 1 g of water per second
due to evaporation. The water equivalent of the
vessel is 0.5 kg and the vessel contains 9.5 kg of
water. Find the time required for the water in the
vessel to cool to 28ºC from 30ºC. Neglect radiation
losses. Latent heat of vaporization of water in this
range of temperature is 540 cal g –1 .
Sol. Here water at the surface is evaporated at the cost of
the water in the vessel losing heat.
Heat lost by the water in the vessel
= (9.5 + 0.5) × 1000 × (30 – 20) = 10 5 cal
Let t be the required time in seconds.
Heat gained by the water at the surface
= (t × 10 –3 ) × 540 × 10 3
(Q L = 540 cal g –1 = 540 × 10 3 cal kg –1 )
∴ 10 5 = 540t or t = 185 s = 3 min 5s
2. 15 gm of nitrogen is enclosed in a vessel at
temperature T = 300 K. Find the amount of heat
required to double the root mean square velocity of
these molecules.
Sol. The kinetic energy of each molecule with mass m is
given by
1 m
2 3
v rms = kT ...(1)
2 2
If we want to increase the r.m.s. speed to η times,
then the temperature has to be raised to T´. Then,
1 2 3 1
mv rms = kT´ or mη
2 2 3
v rms = kT´
2 2 2 2 ...(2)
From eqs. (1) and (2), T´ = η 2 T ...(3)
XtraEdge for IIT-JEE 27 MARCH 2012

The internal energy of n molecules at temperature T
is given by
Similarly,
U = 2
5 nRT
U´ = 2
5 nRT´
∴ Change in internal energy ∆U = 2
5 nR[T´ – T]
or ∆U = 2
5 nRT[η 2 – 1]
= 2
5
= 2
5
⎛ m ⎞
⎜ ⎟ RT[η 2 – 1]
⎝ M ⎠
⎛ 15 ⎞
⎜ ⎟ (8.31) (300) [4 – 1] = 10 4 J
⎝ 28 ⎠
3. 10 gm of oxygen at a pressure 3 × 10 5 N/m 2 and
temperature 10ºC is heated at constant pressure and
after heating it occupies a volume of 10 litres (a) find
the amount of heat received by the gas and (b) the
energy of thermal motion of gas molecules before
heating.
Sol. (a) The states of the gas before and after heating are
PV 1 = µ
M RT1 and PV 2 = µ
M RT2
Solving these equations for T 2 , we have
−3
5
µV P
T 2 = 2 32×
(10×
10 )(3×
10 )
= = 1156 K
−3
3
MR (10×
10 )(8.31×
10 )
Now T 2 – T 1 = 1156 – 283 = 873 K
The amount of heat received by the gas is given by
∆Q = µ
M
CP (T 2 – T 1 )
3
× −
(10 10 )29.08×
10 × 873
=
32
= 7.9 × 10 3 J
(b) The energy of the gas before heating
M i
E 1 = × × RT 1
µ 2
where i = number of degrees of freedom
= 5 (for oxygen)
−3
−3
(10×
10 )5×
(8.31×
10 )(283)
=
2×
32
= 1.8 × 10 3 J
4. A slab of stone of area 3600 sq cm and thickness 10
cm is exposed on the lower surface of steam 100ºC.
A block of ice at 0ºC rests on upper surface of the
slab. In one hour 4800 gm of ice is melted. Calculate
the thermal conductivity of the stone.
3
Sol. The quantity of heat Q passing across the stone is
given by
KA(T1 − T2
)t
Q =
d
Here A = 3600 sq. cm = 0.36 m 2
d = 10 cm = 0.10 m, (T 1 – T 2 ) = 100 – 0 = 100ºC and
t = 1 hour = 3600 sec.
K × 0.36×
100×
3600
∴ Q =
kilo-calories ...(1)
0.10
Now heat gained by the ice in one hour
= mass of the ice × latent heat of ice
= 4.8 × 80 kilo calories ...(2)
From eqs. (1) and (2)
K × 0.36×
100×
3600
4.8 × 80 =
0.10
4.8×
80×
0.10
or K =
0.36×
100×
3600
= 3 × 10 –4 kilo cal m –1 (ºC) –1 s –1
5. A flat bottomed metal tank of water is dragged along
a horizontal floor at the rate of 20m/sec. The tank is
of mass 20 kg and contains 1000 kg of water and all
the heat produced in the dragging is conducted to the
water through the bottom plate of the tank. If the
bottom plate has an effective area of conduction 1 m 2
and the thickness 5 cm and the temperature of water
in the tank remains constant at 50ºC, calculate the
temperature of the bottom surface of the tank, given
the coefficient of friction between the tank and the
floor is 0.343 and K for the material of the tank is 25
cal m –1 s –1 K –1 .
Sol. Frictional force = µ m g
= 0.343 × (1000 + 20) × 9.81 = 3432 N
The rate of dragging, i.e., the distance travelled in
one second = 20 m.
∴ Work done per second
= (3432 × 20) Nm/sec.
This work done appears as heat at the bottom plate of
the tank. Hence
3432× 20
H =
cal/sec
4.18
KA(T1 − T2
)
But H =
(Q t = 1 sec)
d
3432× 20 25×
1×
(T1 − T2
)
Now
=
4.18 0.05
3432×
20×
0.05
∴ T 1 – T 2 =
= 32.84
4.18×
25×
1
Temp. of bottom surface T 1 = 50 + 32.84
= 82.84ºC
XtraEdge for IIT-JEE 28 MARCH 2012

PHYSICS FUNDAMENTAL FOR IIT-JEE
Atomic Structure, X-Ray & Radio Activity
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Atomic Structure :
According to Neil Bohr's hypothesis is the angular
momentum of an electron is quantised.
⎛ h ⎞ h
mvr = n ⎜ ⎟ or L = n
⎝ 2π ⎠ 2π
2πr = nλ
h ⎛ c ⎞ z
v n = Zn = ⎜ ⎟ × ms
–1
2πmr
⎝137
⎠ n
n = ∞
n = 7
n = 6
n = 5
n = 4
⎛
2
r n = ⎜
h
2
⎝ 4π
mke
2
⎞
⎟
⎠
n 2
Z
⎛ ⎞
f n = ⎜
ke 2 ⎟
1 6.58×
× = ⎝ hr ⎠ n n10
n 2
= 0.529 Å where k =
Z
15
Hz
1
4πε
1 ke 2 Z − ke 2
ke 2
K.E. = ; P.E. = × Z; T.E. = – × Z
2 r
r
2r
n = 3
n = 2
2
−13.6Z
T.E. = ev/atom where –13.6
2
n
= Ionisation energy
+P.E.
⇒ +T.E. = = – K.E.
2
Note : If dielectric medium is present then ε r has to
be taken into consideration.
v 1 = = v =
c λ
⎡
= RZ 2 1 1
⎥ ⎥ ⎤
⎢ −
2 22
⎢⎣
n1 n ⎦
K β
K γ
K δ
L γ
L β
4 2
me z ⎡ 1 1
⎥ ⎥ ⎤
⎢ −
2 3 2 2
8ε0
h c ⎢⎣
n1
n 2 ⎦
L α
Balmer
(Visible)
p mv
= = h h
Paschen
(I.R.)
Pfund
Brackett (I.R.)
(I.R.)
Limiting line of Lyman series
n = 1
Lyman Series
(U.V. rays)
–0.85 eV
–1.5 eV
–3.4 eV
–13.6 eV
0
The maximum number of electrons that can be
accommodated in an orbit is 2n 2 .
X-rays :
When fast moving electron strikes a hard metal,
X-rays are produced. When the number of electrons
striking the target metal increases, the intensity of X-
rays increases. When the accelerating voltage/kinetic
energy of electron increases λ min decreases. X-rays
have the following properties :
(a) Radiations of short wavelength (0.01 Å – 10Å);
high pentrating power; having a speed of 3 × 10 8 m/s
in vacuum.
Intensity
Continuous spectrum
(Varies & depends on
accelerating voltage)
λ min
hc
(b) λ min = = eV
K β
K α
L γ
hc
K. E
Characteristic spectrum
(fixed for a target material)
L β
L α
=
λ
12400 Å
V
1
(c) = R(Z – b)
2 ⎡ 1 ⎤
λ
⎢1
− ⎥
⎣ n 2
⎦
b = 1 for k-line transfer of electron
(d) Moseley law ν = a(z – b)
R = R 0 A 1/3 where R 0 = 1.2 × 10 –15 m
R = radius of nucleus of mass number A.
* Nucleus density is of the order of 10 17 kg/m 3
Isomers are nuclides which have identical atomic
number and mass number but differ in their energy
states.
Nuclear binding energy ∆mc 2
=
Nucleon
A
where ∆m = mass defect
2
[Zmp
+ (A − Z)mn
− M]c
=
A
XtraEdge for IIT-JEE 29 MARCH 2012

* The binding energy per nucleon is small for small
nuclei.
* For 2 < A < 20, there are well defined maxima
which indicate that these nuclei are more stable.
* For 30 < A < 120 the average B.E./A is 8.5 MeV /
nucleon with a peak value of 8.8 MeV for Iron.
* For A > 120, there is a gradual decreases in
B.E./nucleon.
* More the B.E./A, more is the stability.
Radioactivity :
β particles are electrons emitted from the nucleus.
(n → p + β)
(a) N = N 0 e –λt
(b)
−dN
dt
⎛ 1 ⎞
(c) N = N 0 ⎜ ⎟⎠
⎝ 2
⎛ 1 ⎞
⇒ A = A 0 ⎜ ⎟⎠
⎝ 2
(d) T 1/2 =
(e) τ = λ
1
dN
= λN where = activity level
dt
n
n
0.693
λ
(f) τ = 1.4 T 1/2
(g) t =
⎛ 1 ⎞ T 1/ 2
= N 0 ⎜ ⎟⎠
⎝ 2
2.303 N
log 0
10 =
λ N
t
where A = activity level
2.303 A
log 0 10
λ A
2.303 m
= log 0
λ m
(h) If a radioactive element decays by simultaneous
−dN
emission of two particle then = λ 1 N + λ 2 N
dt
The following parameters remain conserved during a
nuclear reaction
(a) linear momentum
(b) Angular momentum
(c) Number of nucleons
(d) Charge
(e) The energy released in a nuclear reaction
X + P → Y + Z + Q
Q = [m x + m p ) – (m y + m z )]c 2 = ∆m × c 2
Q = ∆m × 931 MeV
(f) In a nuclear fusion reaction small nuclei fuse to
give big nuclei whereas in a nuclear fusion reaction a
big nuclei breaks down.
Thermal neutrons produce fission in fissile nuclei.
Fast moving neutrons, when collide with atoms of
comparable masses, transfer their kinetic energy to
colliding particle and slow down.
According to Doppler's effect of light
Power, P = t
E =
η =
out put
In put
nhν =
t
nhc
λt
∆ λ
λ
Solved Examples
=
c
v
1. The energy of an excited hydrogen atom is –3.4 eV.
Calculate the angular momentum of the electron
according to Bohr theory.
Sol. The energy of the electron in the n th orbit is
2
n
E n = –
13.6
Here, – = –3.4
13.6
eV
2
n
or n = 2
nh 2×
6.63×
10
Angular momentum = =
2π 2×
3. 14
= 2.11 × 10 –34 Js.
−34
2. The wavelength of the first member of the Balmer
series in the hydrogen spectrum is 6563 Å. Calculate
the wavelength of the first member of the Lyman
series.
Sol. For the first member of the Balmer series
1 ⎡ 1 1 ⎤ 5R
= R
λ
⎢ − ⎥
⎣ 2 2
3 2
=
⎦ 36
For the first member of the Lyman series
1
λ ´
= R ⎡ 1 1 ⎤
⎢ − ⎥
⎣1
2
2 2 ⎦
=
3R
4
Dividing Eq. (1) by Eq. (2)
λ´ 5×
4 5
= =
λ 36×
3 27
or λ´ = 27
5 λ = 27
5 × 6563 = 1215 Å
...(1)
...(2)
XtraEdge for IIT-JEE 30 MARCH 2012

3. Hydrogen atom in its ground state is excited by
means of a monochromatic radiation of wavelength
970.6 Å. How many different wavelengths are
possible in the resulting emission spectrum ? Find the
longest wavelength amongst these.
Sol. Energy the radiation quantum
−34
hc 6.6×
10 × 3×
10
E = hv = = λ
−10
−19
970.6×
10 × 1.6×
10
= 12.75 eV
Energy of the excited sate
E n = – 13.6 + 12.75 = – 0.85 eV
13.6
Now, we know that E n = –
2
n
or n 2 13.6 −13.6
= – = = 16
E n − 0.85
or n = 4
The number of possible transition in going to the
ground state and hence the number of different
wavelengths in the spectrum will be six as shown in
the figure.
n
4
3
2
1
The longest wavelength corresponds to minimum
energy difference, i.e., for the transition 4 → 3.
Now E 3 = –
hc
λ
max
13.6
2
3
= E 4 – E 3
= – 1.51 eV
−34
6.6×
10 × 3×
10
or λ max =
−19
(1.51−
0.85) × 1.6×
10
= 18.75 × 10 –7 m = 18750 Å
4. X-rays are produced in an X-ray tube by electrons
accelerated through a potential difference of 50.0 kV.
An electron makes three collisions in the target
before coming to rest and loses half its kinetic energy
in each of the first two collisions. Determine the
wavelengths of the resulting photons. Neglect the
recoil of the heavy target atoms.
8
8
Sol. Initial kinetic energy of the electron = 50.0 keV
Energy of the photon produced in the first collision,
E 1 = 50.0 – 25.0 = 25.0 keV
Wavelength of this photon
−34
hc 6.6×
10 × 3×
10
λ 1 = =
−19
3
E 1 1.6×
10 × 12.5×
10
= 0.99 × 10 –10 m
= 0.99 Å
Kinetic energy of the electron after third collision = 0
Energy of the photon produced in the third collision ,
E 3 = 12.5 – 0 = 12.5 keV
This is same as E 2 . Therefore, wavelength of this
photon, λ 3 = λ 2 = 0.99 Å
5. In an experiment on two radioactive isotopes of an
elements (which do not decay into each other), their
mass ratio at a given instant was found to be 3. The
rapidly decaying isotopes has larger mass and an
activity of 1.0 µCi initially. The half lives of the two
isotopes are known to be 12 hours and 16 hours.
What would be the activity of each isotope and their
mass ratio after two days ?
Sol. We have, after two days, i.e., 48 hours,
N 1 =
4
0⎛
1 ⎞
N1
⎜ ⎟
⎝ 2 ⎠
3
=
0⎛
1 ⎞
N 2 = N 2⎜
⎟ =
⎝ 2 ⎠
Mass ratio =
0
N 1 /16
0
N 2 /8
0
N 1 =
1
0
N 2 N 2
0 0
Now, A 1 = λ 1 N 1 = 1.0 µCi
After two days,
But
or
N 8 3× 8 3
. = = 16 162 2
0
A 1 = λ 1 N 1 = λ 1 N 1 /16 =
0
A 2 = λ 2 N 2 = λ 2 N 2 /8
λ
λ
2
1
=
T 1 = 16
T
2
λ 2 = 4
3
λ1
⎛ 3
A 2 = ⎜
⎝ 4
λ 1
12 = 4
3
⎞ ⎛ 1 0 ⎞ 1
⎟ × ⎜ N 1 ⎟ ×
⎠ ⎝ 3 ⎠ 8
1 0 1 0
= λ1 N 1 = A 1
32 32
= (1/32) µCi
8
0
A 1 /16 = (1/16)µCi
XtraEdge for IIT-JEE 31 MARCH 2012

KEY CONCEPT
Organic
Chemistry
Fundamentals
PURIFICATION OF
ORGANIC CHEMISTRY
Qualitative Analysis :
Qualitative analysis of an organic compound involves
the detection of various elements present in it. The
elements commonly present in organic compounds
are carbon, hydrogen, oxygen, nitrogen, halogens,
sulphur and sometimes phosphorus.
Detection of carbon and Hydrogen :
Principle. Carbon and hydrogen are detected by
strongly heating the organic compound with cupric
oxide, (CuO). The carbon present in the organic
compound is oxidised to carbon dioxide and
hydrogen is oxidised to water. Carbon dioxide is
tested by lime water test, whereas water is tested by
anhydrous copper sulphate test.
Mixture of orgainc compoud
and dry copper oxide (CuO)
Anhydrous
copper sulphate
(white)
Guard tube
containing sodalime
Cotton plug
Lime water
Reactions :
C + 2CuO ⎯→ CO 2 + 2Cu
in the compound
2H + CuO ⎯→ H 2 O + Cu
in the compound
CO 2 + Ca(OH) 2 ⎯→ CaCO 3 + H 2 O
limewater milky
5H 2 O + CuSO 4 (anhyd) ⎯→ CuSO 4 .5H 2 O
white
blue
Process : The given organic compound is mixed with
dry cupric oxide (CuO) and heated in a hard glass
tube. The products of the reaction are passed over
(white) anhydrous copper sulphate and then bubbled
through limewater. The copper sulphate turns blue
(due to the formation of CuSO 4 .5H 2 O) by water
vapour, showing that the compound contains
hydrogen. The limewater is turned milky by CO 2 ,
showing that the compound contains carbon.
Detection of Nitrogen, Sulphur and Halogens :
Nitrogen, sulphur and halogens in any organic
compound are detected by Lassaigne's test.
Preparation of Lassaigne's Extract (or sodium extract):
A small piece of sodium is gently heated in an
ignition tube till it melts. The ignition tube is
removed from the flame, about 50–60 mg of the
organic compound added and the tube heated
strongly for 2–3 minutes to fuse the material inside it.
After cooling , the tube is carefully broken in a china
dish containing about 20–30 mL of distilled water.
The fused material along with the pieces of ignition
tube are crushed with the help of a glass rod and the
contents of the china dish are boiled for a few
minutes. The sodium salts formed in the above
reactions (i.e., NaCN, Na 2 S, NaX or NaSCN)
dissolve in water. Excess of sodium, if any, reacts
with water to give sodium hydroxide. This alkaline
solution is called Lassaigne's extract or sodium
extract. The solution is then filtered to remove the
insoluble materials and the filtrate is used for making
the tests for nitrogen, sulphur and halogens.
Reactions : An organic compound containing C, H,
N, S, halogens when fused with sodium metal gives
the following reactions.
C + N + Na ⎯⎯→
NaCN
in organic compound
sodium cyanide
X(Cl, Br, I) + Na ⎯ fusion ⎯⎯→NaX(X=Cl,Br, I)
from organic compound
sodium halide
⎯ fusion
S + 2Na ⎯⎯→
Na 2 S
from organic compound
sodium sulphide
If nitrogen and sulphur both are present in any
organic compound, sodium thiocyanate (NaSCN) is
formed during fusion which in the presence of excess
sodium, forms sodium cyanide and sodium sulphide.
⎯ fusion
Na + C + N + S ⎯⎯
→ NaCNS
in organic compound sodium thiocyanate
Detection of Nitrogen :
Take a small quantity of the sodium extract in a test
tube. If not alkaline, make it alkaline by adding 2–3
drops of sodium hydroxide (NaOH) solution. To this
solution, add 1 mL of freshly prepared solution of
ferrous sulphate. Heat the mixture of the two
solutions to boiling and then acidify it with dilute
sulphuric acid. The appearance of prussion blue or
green colouration or precipitate confirms the
presence of nitrogen in the given organic compound.
⎯ fusion
XtraEdge for IIT-JEE 32 MARCH 2012

Chemistry of the test : The following reactions
describe the chemistry of the tests of nitrogen. The
carbon and nitrogen present in the organic compound
on fusion with sodium metal give sodium cyanide
(NaCN). NaCN being ionic salt dissolves in water.
So, the sodium extract contains sodium cyanide.
Sodium cyanide on reaction with ferrous sulphate
gives sodium ferrocyanide. On heating, some of the
ferrous salt is oxidised to the ferric salt and this reacts
with sodium ferrocyanide to form ferric-ferrocyanide.
6 NaCN + FeSO 4 ⎯→ Na 4 [Fe(CN) 6 ] + Na 2 SO 4
sodium ferrocyanide
3Na 4 [Fe(CN) 6 ] + 2Fe 2 (SO 4 ) 3
formed during boiling of the solution
⎯→ Fe 4 [Fe(CN) 6 ] 3 + 6Na 2 SO 4
prussian blue
When nitrogen and sulphur both are present in any
organic compound, sodium thiocyanate is formed
during fusion. When extracted with water sodium
thiocynate goes into the sodium extract and gives
blood red colouration with ferric ions due to the
formation of ferric thiocyanate
Na + C + N + S NaCNS
from organic Sod. thiocyanate
3NaCNS + Fe 3+ ⎯→ Fe(CNS) 3 + 3Na +
ferric thiocyanate
(blood red)
Detection of Sulphur :
The presence of sulphur in any organic compound is
detected by using sodium extract as follows :
(a) Lead acetate test : Acidify a small portion of
sodium extract with acetic acid and add lead acetate
solution to it. A black precipitate of lead sulphide
indicates the presence of sulphur.
H
(CH 3 COO) 2 Pb + Na 2 S ⎯ ⎯→
+
PbS + 2CH 3 COONa
lead acetate in sodium black ppt
extract
(b) Sodium nitroprusside test : To a small quantity
of sodium extract taken in a test tube, add 2-3 drops
of sodium nitroprusside solution. A violet colour
indicates the presence of sulphur. This colour fades
away slowly on standing.
Na 2 S + Na 2 [Fe(CN) 5 NO] ⎯→ Na 4 [Fe(CN) 5 NOS]
sodium nitroprusside violet or purple colour
Detection of Halogens :
The presence of halogens in any organic compound is
detected by using sodium extract (Lassaigne's
extract) by silver nitrate test.
(a) Silver nitrate test : Sodium extract
(or Lassaigne's extract) is boiled with dilute nitric
acid to decompose sodium cyanide or sodium
sulphide (if present) to hydrogen cyanide and
hydrogen sulphide gases, respectively. This solution
is cooled and silver nitrate solution added. A white
precipitate soluble in ammonia shows chlorine, a
yellowish precipitate sparingly soluble in ammonia
indicates bromine, and a yellow precipitate insoluble
in ammonia shows the presence of iodine in the
given organic compound.
NaCl(aq) + AgNO 3 (aq) ⎯→ AgCl(s) + NaNO 3 (aq)
white precipitate
(soluble in ammonia)
NaBr(aq) + AgNO 3 (aq) ⎯→ AgBr(s) + NaNO 3 (aq)
light yellow ppt.
(sparingly soluble in ammonia)
NaI(aq) + AgNO 3 (aq) ⎯→ AgI(s) + NaNO 3 (aq)
yellow precipitate
(insoluble in ammonia)
(b) CS 2 layer test for detecting bromine and iodine :
Boil a small quantity of sodium extract with dilute
HNO 3 for 1–2 min and cool the solution. To this
solution, add a few drops of carbon disulphide (CS 2 )
and 1–2 mL fresh chlorine water, and shake.
Appearance of orange colour in the CS 2 layer
confirms the presence of bromine, whereas that of a
violet/purple colouration confirms the presence of
iodine in the compound.
2NaBr(aq) + Cl 2
in sodium extract
CS
⎯⎯→
2
2NaI(aq) + Cl 2
in sodium extract
CS
2NaCl(aq) + Br 2
dissolves in CS 2 to
give orange colour.
⎯⎯→
2 2NaCl(aq) + I 2
dissolves in CS 2
to give purple/violet colour
Detection of Phosphorus :
The organic compound is fused with sodium
peroxide. The fused mass is then extracted with
water. The aqueous solution so obtained is boiled
with concentrated nitric acid, and ammonium
molybedate solution is added to it.
A yellow solution or precipitate indicates the
presence of phosphorus in the organic compound.
The yellow precipitate is of ammonium
phosphomolybedate (NH 4 ) 3 [PMo 12 O 40 ] or
(NH 4 ) 3 PO 4 .12MoO 3 .
XtraEdge for IIT-JEE 33 MARCH 2012

KEY CONCEPT
Inorganic
Chemistry
Fundamentals
BORON FAMILY &
CARBON FAMILY
Boron Trihalides :
The trihalides of boron are electron deficient
compounds having a planar structure as shown. They
act as Lewis acids because of incomplete octet.
BF +
3
Lewis acid
BF +
3
Lewis acid
X
Lewis base
X
B
120º
Planar structure of
boron trihalides
: NH 3 →
: F
−
Lewis base
→
3B
Addition
product
X
F ← NH 3
BF −
Fluorobora 4 te ion
The acid strength of trihalides decreases as :
BF 3 < BCl 3 < BBr 3 < BI 3
Explanation :
This order of acid strength is reverse of what may
normally be expected on the basis of
electronegativity of halogens. Since F is most
electronegative, hence BF 3 should be most electron
deficient and thus should be strongest acid. The
anomalous behaviour is explained on the basis of
tendency of halogen atom to back-donate its electrons
to boron atom. For example, in BF 3 one of the
2p-orbital of F atom having lone pair overlaps
sidewise with the empty 2p-orbital of boron atom to
form pπ-pπ back bonding. This is also known as back
donation. Further, due to back-π donation of three
surrounding fluorine atoms. BF 3 can be represented
as a resonance hybrid of following three structures.
+
F
F
B – = F + B – F
F
– F B – — F
F
F
F + ≡ B – — F
F
Resonating forms of BF 3 Probable hybrid
structure
As a result of this back donation, the electron
deficiency of boron gets compensated and its Lewis
acid character decreases.
Now, the tendency for back donation is maximum in
the case of fluorine due to its small size and more
interelectronic repulsions, therefore, it is the least
acidic. The tendency of back bonding falls as we
move from BF 3 to BCl 3 and BCl 3 to BBr 3 due to
increase in the size of halogen atoms consequently,
the acidic character increase accordingly.
F
F
B
π
π
Empty
2p-orbital
2p-orbital with lone pair
pπ-pπ back bonding
Acidic nature of H 3 BO 3 or B(OH) 3 :
Since B(OH) 3 only partially reacts with water to form
H 3 O + and [B(OH) 4 ] – , it behaves as a weak acid. Thus
H 3 BO 3 or (B(OH) 3 ) cannot be titrated satisfactorily
with NaOH, as a sharp end point is not obtained. If
certain organic polyhydroxy compounds such as
glycerol, mannitol or sugars are added to the titration
mixture, then B(OH) 3 behaves as a strong monobasic
acid. It can now be titrated with NaOH, and the end
point is detected using phenolphthalein as indicator
(indicator changes at pH 8.3 – 10.0).
2B(OH) 3 + 2NaOH
F
Na[B(OH) 4 ] +
NaBO +
2 2H 2O
sodium metaborate
The added compound must be a cis-diol, to enhance
the acidic properties in this way. (This means that it
has OH groups on adjacent carbon atoms in the cis
configuration.) The cis-diol forms very stable
complexes with the [B(OH) 4 ] – formed by the forward
reaction above, thus effectively removing it from
solution. The reaction is reversible. Thus removal of
one of the products at the right hand side of the
equation upsets the balance, and the reaction
proceeds completely to the right. Thus all the B(OH) 3
reacts with NaOH : in effect it acts as a strong acid in
the presence of the cis-diol.
XtraEdge for IIT-JEE 34 MARCH 2012

– C – OH
+
– C – OH
HO
HO
B
OH
OH
–
–2H 2 O
–2H 2 O
– C – O
– C – O
B
– C – O
– C – O
OH
OH
B
–
HO – C –
+
HO – C –
O – C –
O – C –
Borax :
The most common metaborate is borax
Na 2 [B 4 O 5 (OH) 4 ] . 8H 2 O. It is a useful primary
standard for titrating against acids.
(Na 2 [B 4 O 5 (OH) 4 ] . 8H 2 O) + 2HCl →
2NaCl + 4H 3 BO 3 + 5H 2 O
One of the products H 3 BO 3 is itself a weak acid. Thus
the indicator used to detect the end point of this
reaction must be one that is unaffected by H 3 BO 3 .
Methyl orange is normally used, which changes in
the pH range 3.1 – 4.4.
One mole of borax reacts with two moles of acid.
This is because when borax is dissolved in water both
B(OH) 3 and [B(OH) 4 ] – are formed, but only the
[B(OH) 4 ] – reacts with HCl.
[B 4 O 5 (OH) 4 ] 2– + 5H 2 O 2B(OH) 3 + 2[B(OH) 4 ] –
2[B(OH) 4 ] – + 2H 3 O + → 2B(OH) 3 + 4H 2 O
The last reaction will titrate at pH 9.2, so the
indicator must have pK a < 8. Borax is also used as a
buffer since its aqueous solution contains equal
amounts of weak acid and its salt.
Structures of the Boranes :
The bonding and structures of the boranes are of
great interest. They are different from all other
hydrides. There are not enough valency electrons to
form conventional two-electron bonds between all of
the adjacent pair of atoms, and so these compounds
are termed as electron dificient.
In diborane there are 12 valency electrons, three from
each B atom and six from the H atoms. Electron
diffraction results indicate the structure shown in fig.
H H 1.33Å H
1.19Å
B B 1.19Å
H H 1.33Å H
The two bridging H atoms are in a plane
perpendicular to the rest of the molecules and prevent
rotation between the two B atoms. Specific heat
measurements confirm that rotation is hindered. Four
of the H atoms are in a different environment from
the other two. This is confirmed by Raman spectra
and by the fact that diborane cannot be methylated
–
beyond Me 4 B 2 H 2 without breaking the molecule into
BMe 3 .
The terminal B – H distance are the same as the bond
lengths measured in non-electron-deficient
compounds. These are assumed to be normal
covalent bonds, with two electrons shared between
two atoms. We can describe these bonds as twocentre
two-electron bonds (2c-2e).
Thus the electron deficiency must be associated with
the bridge groups. The nature of the bonds in the
hydrogen bridges is now will established. Obviously
they are abnormal bonds as the two bridges involve
only one electron from each boron atom and one
from each hydrogen atom, making a total of four
electrons. An sp 3 hybrid orbital from each boron
atom overlaps with the 1s orbital of the hydrogen.
This gives a delocalized molecular orbital covering
all three nuclei, containing one pair of electrons and
making up one of the bridges. This is a three centre
two-electron bond (3c-2e). A second three-centre
bond is also formed.
H
H
H
H
B
B
H
H
H
H
B
B
Overlap of approximately sp 2 hybrid orbitals from B with
an s orbital from H to give a banana-shaped three-centre
two-electron bond.
The higher boranes have an open cage structure. Both
normal and multi-centre bonds are required to explain
these structures.
Terminal B–H bonds. These are normal covalent
bonds, that is two centre two-electron (2c-2e)
bonds.
B – B bonds. These are also normal 2c-2e bonds.
Three-centre bridge bonds including B ... H ... B
as in diborane. These are 3c-2e bonds.
H
H
Three-centre bridge bonds including B....B.....B,
similar to the hydrogen bridge. These are called
H
H
XtraEdge for IIT-JEE 35 MARCH 2012

'open boron bridge bonds' and are of the type
3c-2e.
Closed 3c-2e bonds between three B atoms.
B
B
B
Silicones :
These are organosilicon polymers containing Si – O – Si
linkages. These are formed by the hydrolysis of alkyl
or aryl substituded chlorosilanes and their subsequent
polymerisation. The alkyl or aryl substitued
chlorosilanes are prepared by the reaction of
Grignard reagent and silicon tetrachloride.
RMgCl + SiCl 4 ⎯→ R – SiCl 3 + MgCl 2
Grignard reagent
2RMgCl + SiCl 4 ⎯→ R 2 SiCl 2 + 2MgCl 2
3RMgCl + SiCl 4 ⎯→ R 3 SiCl + 3MgCl 2
R stands for – CH 3 , –C 2 H 5 or –C 6 H 5 groups
Hydrolysis of substituted chlorosilanes yield
corresponding silanols which udergo polymerisation.
R
R
Si
Cl H OH
Cl H + –2HCl
R
Si OH
OH R OH
Dialky silandiol
Polymerisation of dialkyl silandiol yields linear
thermoplastic polymer.
R
HO – Si – OH + H O – Si – OH
R
R
R
R
HO – Si – O – Si – OH
R R
Polymerisation continues on both the ends and thus
chain increases in length.
RSiCl 3 on hydrolysis gives a cross linked silicone.
The formation can be explained in three steps :
Cl
OH
3H
(i) R – Si – Cl 2 O
R – Si – OH
–3HCl
Cl
OH
R
R
R
(ii) HO – Si – OH + H O – Si – OH + H O – Si – OH
OH
OH
R
R
HO – Si – O – Si – O – Si – OH
OH
R
OH
OH
R
OH
R R R
(iii) HO – Si – O – Si – O – Si – OH
–3H 2 O
OH OH OH
H O H O H O
HO – Si – O – Si – O – Si – OH
R
R
R
– O – Si – O – Si – O – Si – O –
O
R
R
O
R
O
– O – Si – O – Si – O – Si – O –
R R R
Cross linked silicone
Cyclic (ring) silicones are formed when water is
eliminated from the terminal –OH group of linear
silicones.
R R
Si
O O
R
R
Si Si
R
R
O
R 3 SiCl on hydrolysis forms only a dimer
R 3 Si OH + OH Si R 3
R 3 Si – O – Si R 3
SCIENCE TIPS
• A porcelain funnel used for filtration by suction is
known as
Bucher Funnel
• What is diazomethane ?
–
[ 2 = +
=
2 2
CH N N or CH N ]
• A drying chamber, containing chemicals such as
concentrated sulphuric acid or silica gel is known as
Desiccator
• Reforming of a gasoline fraction to increase
branching in presence of AlCl 3 is known as
Isomerization
• A condenser consisting of glass tube surrounded by
another glass tube through which cooling water
flows is known as
Liebig condenser
• What is
• Hot wire ammeter
• What quantity has the
XtraEdge for IIT-JEE 36 MARCH 2012

UNDERSTANDING
Inorganic Chemistry
1. (a) When Mn(OH) 2 is made by adding an alkali to a
solution containing Mn 2+ ions, the precipitate quickly
darkens, and eventually goes black. What might be
the chemical giving the black colour, and how is it
made ?
(b) Dimercury (I) iodide, Hg 2 I 2 is a greenish colour
and is precipitated if iodide ions are added to a
solution of dimercury (I) sulphate, Hg 2 SO 4 . Likewise
the red mercury (II) iodide, HgI 2 , is precipitated from
a solution of mercury (II) sulphate, HgSO 4 . However,
both precipitates dissolve in excess iodide solution.
What might be the reason for this ?
Sol. (a) The black colour is due to the manganese (IV)
oxide, MnO 2 . It is made by the Mn(OH) 2 being
oxidised by oxygen in the air :
Mn(OH) 2 ⎯→ MnO + H 2 O
MnO + ½O 2 ⎯→ MnO 2
air black
2–
(b) It is due to formation of HgI 4 (a soluble
complex) in both the cases with HgI 2 :
HgI 2 + 2I – 2–
→ HgI 4
But in Hg 2 I 2 , first there is oxidation of Hg(I) to
Hg(II) and then complex formation takes place; it is
by following disproportionation reaction :
2
Hg + 2 + 4I – 2
⎯→ HgI + 4 + Hg
0
+ 1
+ 2
2. Calculate mol of Ca(OH) 2 required to carry out
following conversion taking one mol in each case :
COOH
(a) into COO
Ca
COOH COO
(b) H 3 PO 4 into CaHPO 4
(c) NH 4 Cl into NH 3
(d) NaHCO 3 into CaCO 3
COOH
Sol. (a) is a dibasic acid
COOH
COOH
COOH + Ca(OH) COO
2
Ca
COO
1 mo l 1 mo l
Ca(OH) 2 required = 1 mol
(b) H 3 PO 4 + Ca(OH) 2 ⎯→ CaHPO 4 + 2H 2 O
1 mol of H 3 PO 4 ≡ 2H + neutralised by 1 mol of
Ca(OH) 2
Ca(OH) 2 required = 1 mol
(c) 2NH 4 Cl + Ca(OH) 2 ⎯→ CaCl 2 + 2NH 3 + 2H 2 O
2 mol NH 4 Cl ≡ 1 mol Ca(OH) 2
1 mol NH 4 Cl ≡ 0.5 mol Ca(OH) 2
(d) 2NaHCO 3 + Ca(OH) 2 ⎯→ Na 2 CO 3 + CaCO 3 +
2H 2 O
2 mol NaHCO 3 ≡ 1 mol Ca(OH) 2
1 mol NaHCO 3 ≡ 0.5 mol Ca(OH) 2
3. A colourless salt (A), soluble in water, gives a
mixture of three gases (B), (C) and (D) along with
water vapours. Gas (B) is blue towards litmus paper,
gas (C) red and gas (D) is neutral. Gas (B) is also
obtained when (A) is heated with NaOH and gives
brown ppt. with K 2 HgI 4 . Solution thus obtained gives
white ppt. (E) with CaCl 2 solution in presence of
−
CH 3 COOH. Precipatete (E) decolorises MnO 4 /H + .
Gas (C) turns lime water milky while gas (D) burns
with blue flame and is fatal when inhaled. Identify
(A) to (D) and explain chemical reactions.
Sol. Gas (B) gives brown ppt. with K 2 HgI 4
+
⇒ gas (B) is NH 3 ⇒ gas (A) has NH 4
(C) turns lime water milky
⇒ gas (C) can be SO 2 or CO 2
Gas (D) is also obtained along with (C). Gas (D)
burns with blue flame and is fatal when inhaled
⇒ gas (D) is CO ⇒ gas (C) is CO 2
2–
⇒ (A) has C 2 O 4
It is confirmed by the fact that CaCl 2 gives white ppt.
CaC 2 O 4 (E) which decolourises MnO – 4 /H +
⇒ (A) is (NH 4 ) 2 C 2 O 4
Explanation :
(NH 4 ) 2 C 2 O 4 ⎯⎯→
∆ 2NH 3 + CO 2 + CO + H 2 O
(A) (B) (C) (D)
(B) is blue towards litmus (basic)
(C) is red toward litmus (acidic)
(D) is neutral
(NH 4 ) 2 C 2 O 4 +2NaOH ⎯⎯→
∆ Na 2 C 2 O 4 + 2NH3 +2H 2 O
(B)
Na 2 C 2 O 4 + CaCl 2 ⎯→ CaC 2 O 4 ↓ + 2NaCl
White ppt. (E)
XtraEdge for IIT-JEE 37 MARCH 2012

Hg
NH 3 + K 2 HgI 4 ⎯→ O NH 2 I
Hg
brown ppt
(Iodide of Million’s base)
2MnO – 4 +16H + +5C 2 O 2– 4 → 10CO 2 +2Mn 2+ + H 2 O
violet
colourless
4. A solution of a salt (A) when treated with calculated
quantity of sodium hydroxide gave a green coloured
ppt (B), which dissolve in excess of NaOH. (B) acts
as a weak base and loses water on heating to give a
green powder (C). The green powder is used as
refractory material. When (C) is fused with an alkali
in presence of air or oxidising agent, a yellow
coloured solution (D) is obtained. Identify the
compounds from (A) to (D) -
Sol. The compound (A) is chromic salt. The chemical
reactions are as under -
(i) With calculated quantity of sodium hydroxide -
CrCl 3 + 3NaOH ⎯→ Cr(OH) 3 + 3NaCl
green ppt (B)
(ii) In excess of sodium hydroxide, soluble NaCrO 2 is
formed
Cr(OH) 3 + NaOH ⎯→ NaCrO 2 + 2H 2 O
(sod. chromite)
(iii) Since Cr(OH) 3 contains -OH group, so it will act
as a base. On heating it will lose water to give Cr 2 O 3
powder (C)
2Cr(OH) 3 ⎯→ Cr 2 O 3 + 3H 2 O
(C)
(iv) On fusing Cr 2 O 3 with an alkali in presence of
oxygen or oxidising agent, a yellow soluble chromate
will be formed -
2Cr 2 O 3 + 8NaOH + 3O 2 ⎯→ 4Na 2 CrO 4 + 4H 2 O
yellow soln. (D)
5. Two moles of an anhydrous ester (A) are condensed
in presence of sodium ethoxide to give a β-keto ester
(B) and ethanol. On heating in an acidic solution
compound (B) gives ethanol and a β-keto acid (C).
(C) on decarboxylation gives (D) of molecular
formula C 3 H 6 O. Compound (D) reacts with sodamide
to give a sodium salt (E), which on heating with CH 3 I
gives (F), C 4 H 8 O, which reacts with phenyl hydrazine
but not with Fehling reagent. (F) on heating with I 2
and NaOH gives yellow precipitate of CHI 3 and
sodium propionate. Compound (D) also gives
iodoform, but sodium salt of acetic acid. The sodium
salt of acetic acid on acidification gives acetic acid
which on heating with C 2 H 5 OH in presence of conc.
H 2 SO 4 gives the original ester (A). What are (A) to
(F) ?
Sol. (i) Acetic acid on heating with C 2 H 5 OH gives
original compound (A).
CH 3 COOH + C 2 H 5 OH
⎯ H ⎯ 2 SO ⎯
4
∆
→
CH
3COOC2H5
(A)
+ H 2 O
(ii) CH 3 COOC 2 H 5 (A) on heating with C 2 H 5 ONa
undergoes Claisen condensation to give (B), which is
aceto acetic ester.
CH 3 CO OC 2 H 5 + H CH 2 COOC 2 H 5
(A)
C 2H 5ONa
Reflux
+ C 2 H 5 OH + CH 3 COCH 2 COOC 2 H 5
(B)
(iii) (B) on heating in acidic solution gives (C) and
ethyl alcohol.
CH + HOH ⎯→
3COCH
2COOC2H5
(B)
2
(C)
H
⎯ +
CH 3COCH
COOH + C 2 H 5 OH
(iv) (C) on decarboxylation gives acetone (D).
CH 3COCH
COOH
2
(C)
∆
⎯ ⎯→
−CO 2
CH 3COCH 3
(D)
(v) (D) reacts with NaNH 2 to form sodium salt (E),
which on heating with CH 3 I gives butanone (F).
CH 3COCH 3 + NaNH 2
(D)
∆
⎯ ⎯→
−NH 3
CH
3COCH
2Na
(E)
CH 3⎯ I ⎯ ⎯
–NaI → CH 3COCH
2CH3
(F)
(vi) CH 3COCH
2CH3
+ 3I 2 + 4NaOH ⎯→
(vii)
(F)
⎯ ∆
CHI 3 + CH 3 CH 2 COONa + 3NaI + 3H 2 O
CH 3COCH 3 + 3I 2 + 4NaOH ⎯⎯→
∆
(D)
CHI 3 + CH 3 COONa + 3NaI + 3H 2 O
CH 3 COONa
⎯ HCl ⎯→
Thus, (A) CH 3 COOC 2 H 5
(B) CH 3 COCH 2 COOC 2 H 5
(C) CH 3 COCH 2 COOH
(D) CH 3 COCH 3
(E) CH 3 COCH 2 Na
(F) CH 3 COCH 2 CH 3
CH 3 COOH + NaCl
XtraEdge for IIT-JEE 38 MARCH 2012

XtraEdge for IIT-JEE 39 MARCH 2012

`tà{xÅtà|vtÄ V{tÄÄxÇzxá
11
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
1. For complex numbers z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2
we write z 1 ∩ z 2 , if x 1 ≤ x 2 and y 1 ≤ y 2 . The for all
1−
z
complex numbers z with 1 ∩ z, we have ∩ 0,
1+
z
Justify the result.
2. AP and BQ are fixed parallel tangents to a circle, and
a tangent at any point C cuts them at P and Q
respectively. Show that CP.CQ is independent of the
position of C on the circle.
3. Let f(x) = ax 2 + bx + c & g(x) = cx 2 + bx + a, such
that |f(0)| ≤ 1, |f(1)| ≤ 1 and |f(–1)| ≤ 1. Prove that
|f(x)| ≤ 5/4 and |g(x)| ≤ 2
4. A straight line is drawn throguh the origin
and parallel to the tangent to the curve
x +
a
a
2 −
y
2
⎛
= ln
⎜ a +
⎜
⎝
2
2
a − y
y
⎞
⎟
⎟
at an arbitrary
⎠
point M. Show that the locus of the points P of
intersection of this straight line and the straight line
parallel to the x-axis and passing through the point M
is a circle.
5. Show that ∑
r=
n
0
⎧ 1
n
r Cr
⎪ ,
( −2)
=
r+
2
⎨
n + 1
C 1
r ⎪ ,
⎩n
+ 2
If n is even
If n is odd
⎛ x + y ⎞ 2 + f ( x)
+ f ( y)
7. If f ⎜ ⎟ =
⎝ 3 ⎠ 3
for all real x and y. If f ´(2) = 2, then f(2) is -
Passage :
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.
8. g(f(x)) is -
(A) many one into
(B) many one onto
(C) one one onto
(D) one one into
9. f(g(x)) = p has -
(A) no real root
(B) at least one real root
(C) infinity many roots
(D) exactly one real root
10. g(f(x)) is –
(A) non periodic function
(B) odd function
(C) even function
(D) None of these
6. Let I n =
∫
of I n–2 .
1
x n
0
tan
−1
x dx , then expression I n in terms
XtraEdge for IIT-JEE 40 MARCH 2012

MATHEMATICAL CHALLENGES
SOLUTION FOR FEBRUARY ISSUE (SET # 10)
1. g(x) = sin x ; 0 ≤ x < π/2
1 ; π/2 ≤ x ≤ π
sin 2 x/2 ; π < x
lim g(x) = lim g(x) = g(π/2) = 1
−
x→π
2
lim g(x) =
−
x→π
+
x→π
2
lim g(x) = g(π) = 1
+
x→π
g´(π–) = g´(π +) = 0
and g´(π/2–) = g´(π/2+) = 0
Hence g(x) is continuous and differentiable in (0,∞)
2.
sin x sin(sin x)
<
x sin x
sin θ
Let f(θ) = ;
θ
0 < θ < π/2
f ´(θ) =
θcosθ − sin θ
2
θ
cosθ.(
θ − tan θ)
=
2
θ
< 0 as tan θ > θ
so f(θ) ↓
so f(x) < f(sinx) as sin x < x
3. (i) 6 6.5
C 4 =
2
= 15
(ii) coeff. of x 4 in (1 – x) –6
= 4 + 6 – 1 C 6 – 1 = 9 9.8.7.6
C 5 = = 126
4.3.2
(iii) select 3 different flavours : 6 C 3 ways
choose (at least one from each) 4 cones :
4 – 1 C 3 – 1 = 3 C 2 = 3 ways
so required ways = 6 6.5.4
C 3 × 3 = × 3 = 60
3.2
(iv) Select 2 different flavours : 6 C 2 ways
choose (at least one from each) 4 cones ;
4 – 1 C 2 – 1 = 3 C 1 = 3
so required ways (either 2 or 3 different flavours)
= 60 + 6 6.5
C 2 3 = 60 +
2
× 3 = 105
4. Let A at origin & P.V. of B & C are
b & c .
⎛ b c ⎞
So line AD ⇒ r = t
⎜ +
⎟
⎝ | b | | c | ⎠
& line BC ⇒ r = b + ∆ ( b – c )
solve them together to find pt. D
⎛ b c ⎞
t
⎜ +
⎟ = b + s ( b – c )
⎝ | b | | c | ⎠
E B D C
t
= 1 + s
| b |
...(1)
t
= –s
| c |
...(2)
so
t t | b || c |
= 1 – ⇒ t =
| b | | c | | b | + | c |
use it in line AD .
pt D :
| b || c | ⎛ b c ⎞ b | c | + c | b |
. ⎜ ⎟
| b | + | c |
+
=
⎝ | b | | c | ⎠ | b | + | c |
which divides BC in ratio of |c| : |b|
similary use eq. of external angle bisector line AE
⎛ b c ⎞
⇒ r = p ⎜ ⎟
−
⎝ | b | | c | ⎠
solve it with BC to find pt. E.
5. Consider
e ix (1 + e ix ) n = e ix [1 + n C 1 e ix + n C 2 e i2x + .... + n C n e inx ]
⎛ n+ 2 ⎞
i⎜
x ⎟
⎝ 2 ⎠
e . 2cos n x
2
....+ n i(n+ 1)x
C n e
Compare real parts & get (a)
Compare imaginary. parts & get (b)
A
= e ix + n C 1 e i2x + n C 2 e i3x +...
6. Let E i = the event that originator will not receive a
letter in the ith stage.
Originator sands letters to two persons so in 1st stage
he will not get letter.
Prob. that letter sent by 1st received is not received
n−2
C2
( n − 2)( n − 3) n − 3
by originator is =
=
n−1
C1
( n −1)(
n − 2) n −1
similarly prob. that letter sent by 2nd receipiant is not
n − 3
received by originator is
n −1
so P(E 2 ) = prob. that originator not received letter in
2
2 nd ⎛ n − 3 ⎞
stage is = ⎜ ⎟⎠ .
⎝ n −1
XtraEdge for IIT-JEE 41 MARCH 2012

similarly P(E 3 ) = prob. that originator not receive
letter sent by the four person getting letters from two
recipients is
2
4
2
⎛ n − 3 ⎞ ⎛ n − 3 ⎞ ⎛ n − 3 ⎞ ⎛ n − 3 ⎞
⎜ ⎟ . ⎜ ⎟ . ⎜ ⎟ . ⎜ ⎟ =
⎛ n − 3 ⎞ ⎛ n − 3 ⎞
⎜ ⎟⎠ = ⎜ ⎟⎠
⎝ n −1
⎠ ⎝ n −1
⎠ ⎝ n −1
⎠ ⎝ n −1
⎠ ⎝ n −1
⎝ n −1
3
8
2
⎛ n − 3 ⎞ ⎛ n − 3 ⎞
Similarly, P(E 4 ) = ⎜ ⎟⎠ = ⎜ ⎟⎠
⎝ n −1
⎝ n −1
–1
2
⎛ n − 3 ⎞
k
Similarly, P(E k ) = ⎜ ⎟
⎝ n −1
⎠
So the required prob. is
P(E) = prob. the originator not receive letter in 1st k
stages
= P(E 1 ) . P(E 2 ) . ........ P(E k )
2 3 k−1
2+
2 + 2 + ....2
⎛ n − 3 ⎞
= ⎜ ⎟
⎝ n −1
⎠
⎛ n − ⎞
= ⎜ ⎟
⎝ n −1
⎠
7. y = f(x) =
∫
x
k−1
2 −1
3 2.
2 − 1
x
zx−z
0
e
⎛ n − 3 ⎞
= ⎜ ⎟
⎝ n −1
⎠
2
dz =
∫
0
x
k
(2 −2)
2
−z
e zx . e dz
2
−z
. dz + 1 = 1 x
–
zx −z
∫
e ( −2ze
2
0
y´ =
∫
e zx e
) dz + 1
0
2
= – 1 ⎡ 2
x
z zx x
⎤
⎢ −
−z
zx
( e . e ) 0 ⎥
⎣
∫
xe
2
1
. e dz + 1 = xy + 1
2
0 ⎦ 2
dy 1 – xy = 1
dx 2
− x dx
I.F. = e ∫ / 2
=
e −x
2
/ 4
2
x / 4
Sol is y . e − =
∫ − 2
x
e
/ 4 dx =
∫
e
y =
e
2
x / 4
∫
x 2
−z
/ 4
0
e
dz.
x 2
−z
/ 4
8. ∫ sin n θ sec θ dθ = ∫ sin (n –1 + 1) θ sec θ dθ
= ∫ sin (n – 1)θ + cos (n – 1)θ sin θ sec θ ) dθ
= ∫ sin (n – 1)θ + [ sin (n – 1)θ cos θ
– sin (n – 2)θ sec θ ] dθ
= ∫ (2 sin (n – 1)θ – sin (n – 2)θ sec θ ) dθ
2cos( n −1)
θ
= –
– ∫ sin (n – 2)θ secθ dθ
n −1
1
=
∫ π 2 sin 8θ − sin 2θ
dθ
2 0 cos θ
⎡
π / 2
1 ⎛ 2 ⎞
= ⎢⎜−
θ⎟
− θ θ θ
⎢⎣
⎝ ⎠
∫ π 2
cos7
sin 6 sec d
2 7
0
0
⎤
−
∫ π 2
sin 2θsec
θd
θ⎥
0
⎦
1 ⎡ 2 2 2 π
⎢ − − θ − θ θ θ
⎣
∫ π / 2
/ 2
(cos3 ) 0 sin 2 sec d
2 7 5 3
0
⎤
−
∫ π 2
sin 2θsecθdθ
0
⎥
⎦
0
dz
1 1 1 – + – ∫ π / 2
sin 2θsecθd
7
θ
5 3 0
= 29 + π /
) 2
181
2(cosθ 0
– 105 ∫
0 dθ
= – 105
9. 9x 2 – 24xy + 16y 2 – 18x – 101y + 19 = 0
(3x – 4y) 2 = 18x + 101y – 19.
Let the vertex of the parabola be A(α, β). Shift origin
to A and y-axis along the tangent at vertex (3x – 4y + l).
So the axis of parabola be 4x + 3y + m = 0 (along x
axis) If L.R. of parabola be a then it’s equation is
2
⎛ 3x
− 4y
+ l ⎞ ⎛ 4x
+ 3y
+ m ⎞
⎜ ⎟ = a ⎜
⎟
⎝ 5 ⎠ ⎝ 5 ⎠
(3x – 4y + l) 2 – 5a(4x + 3y + m) = 0
9x 2 – 24xy + 16y 2 + (6l – 20a)x + (–8l – 15a)y
+ (l 2 – 5am) = 0
comp. it with given equation.
6l – 20a = –18
⇒ 24l – 80a = –72 ...(1)
–8l –15a = –101
⇒ –24l – 45a = –303 ...(2)
From (1) & (2)
⇒ 125a = –375 ⇒ a = 3
10. circle : (x – 1) 2 + (y – 1) 2 = 1
⇒ x 2 + y 2 – 2x – 2y + 1 = 0
(0,1)B
D
A(1,0)
Let the line be y = mx
1
Altitude of ∆ =
2
1 + m
For DE length : solve line with circle.
x 2 + m 2 x 2 – 2x – 2mx + 1 = 0
(1 + m 2 )x 2 – 2(1 + m)x + 1 = 0
2
|x 1 – x 2 | = ( x1 + x2
) − 4x1x2
=
4(1 + m)
(1 + m
2
)
2
2
E
1
− 4
1 + m
|DE| = x 2
1 + 1 |x 1 – x 2 | = 2 2m
2
1 + m
2
2
= 2m
2
1 + m
XtraEdge for IIT-JEE 42 MARCH 2012

Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
MATHS
1. 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
θ n
A 2
A 1
θ 2
θ1
A n
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 ⎢∑Cisθr
⎥
r=
1
⎢⎣
r=
1 ⎥⎦
iθ
= Re [
1 iθ2
iθ
e + e + ....... + e
n
]
⎡ ⎛
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)
⎥
⎢ π ⎥
= 0
2
i
⎢⎣
1−
e
n ⎥⎦
n
Hence, ∑
=
r 1
2
n
P
( PA r ) = 2n = constant.
2π
n
2. Find the point inside a triangle from which the sum
of the squares of distance to the three side is
minimum. Find also the minimum value of the sum
of squares of distance.
Sol. If a, b, c are the lengths of the sides of the ∆ and x, y,
z are length of perpendicular from the points on the
sides BC, CA and AB respectively, we have to
minimise : ∆ = x 2 + y 2 + z 2
we have, 2
1 ax + 2
1 by + 2
1 cz = ∆
⇒ ax + by + cz = 2∆
A
z
y
x
B
C
where ∆ is the area of ∆ABC.
We have the identity :
⇒ (x 2 + y 2 + z 2 ) (a 2 + b 2 + c 2 ) – (ax + by + cz) 2
= (ax – by) 2 + (by – cz) 2 + (cz – ax) 2
⇒ (x 2 + y 2 + z 2 )(a 2 + b 2 + c 2 ) ≥ (ax + by + cz) 2
⇒ (x 2 + y 2 + z 2 ) (a 2 + b 2 + c 2 ≥ 4∆ 2
2
⇒ x 2 + y 2 + z 2 4∆
≥
2 2 2
a + b + c
Equality holds only when
x y z ax + by + cz 2∆
= = = =
a b c
2 2 2 2 2 2
a + b + c a + b + c
∴ The minimum value of ∆ is ;
2
4∆
4( s − a)(
s − b)(
s − c)
s
=
2 2 2
2 2 2
a + b + c a + b + c
3. Let a 1 , a 2 , ......, a n be real constant, x be a real variable
and f (x) = cos(a 1 + x) + 2
1 cos(a2 + x) + 4
1 cos(a3 + x)
1
+...... + cos(a
n−1
n + x). Given that f (x 1 ) = f (x 2 ) = 0,
2
prove that (x 2 – x 1 ) = mπ for integer m.
Sol. f (x) may be written as,
n
1
f (x) = ∑ cos(ak + x)
k −1
k = 1
2
XtraEdge for IIT-JEE 43 MARCH 2012

n
1
= ∑
k = 1
2
k −1
⎛
= cos x . ⎜
⎜
⎝
n
∑
k = 1
{cosa k . cos x – sin a k . sin x}
cos
2
a k
k −1
⎞ ⎛
⎟ – sin x ⎜
⎟ ⎜
⎠ ⎝
n
∑
k = 1
= A cos x – B sin x, where A = ∑
k =
n
sin a
B = ∑
k
k −1
k = 1
2
since f (x 1 ) = f (x 2 ) = 0
⇒ A cos x 1 – B sin x 1 = 0
and A cos x 2 – B sin x 2 = 0
A
⇒ tan x 1 = B
⇒ tan x 2 = B
A
⇒ tan x 1 = tan x 2
⇒ (x 2 – x 1 ) = mπ
n
sin
1
2
a k
k −1
cos
2
⎞
⎟
⎟
⎠
a k
k −1
and
4. If (a, b, c) is a point on the plane 3x + 2y + z = 7, then
find the least value of a 2 + b 2 + c 2 , using vector
methods.
Sol. Let → A = a î + b ĵ + c kˆ
⇒ → B = 3 î + 2 ĵ + kˆ
⇒
→ →
(A.B)
2
3a + 2b + c ≤
≤ | → A| 2 | → B| 2
(7) 2 ≤ (a 2 + b 2 + c 2 ) (14)
2
2
a + b + c 14
{Q 3a + 2b + c = 7, point lies on the plane}
a 2 + b 2 + c 2 49 7
≥ = 14 2
5. If parameters p, r, q are in H.P. and d be the length of
perpendicular from origin to any member of family
of lines xr(p + q – 2pq) – 2pq(y – 5r) – 3pqr = 0 then
7
show that |d| ≤ .
2
Sol. Given family of line is
xr(p + q – 2pq) – 2pq(y – 5r) – 3pqr = 0
dividing by pqr, we get
⎡ 1 1 ⎤ ⎡ y ⎤
x ⎢ + − 2⎥ – 2
⎣ q p
⎢ − 5⎥ – 3 = 0
⎦ ⎣ r ⎦
⎛ 2 ⎞ ⎛ y ⎞
x ⎜ − 2⎟ – 2 ⎜ − 5⎟ – 3 = 0
⎝ r ⎠ ⎝ r ⎠
2
(7 – 2x) + r
2 (x – y) = 0
⇒ Given family of line passes through fixed point
(7/2, 7/2)..
equation of line in normal form x cos α + y sin α = p
⎛ 7 7 ⎞
passes through ⎜ , ⎟
⎝ 2 2 ⎠
2d
⇒ cos α + sin α =
7
but – 2 ≤ cos α + sin α ≤ 2
2d 7
≤ 2 ⇒ |d| ≤
7
2
6. There are n straight lines in a plane such that n 1 of
them are parallel in one direction, n 2 are parallel in
different direction and so on, n k are parallel in
another direction such that n 1 + n 2 + ... + n k = n. Also
no three of the given lines meet at a point. Prove that
the total number of points of intersection is
⎡
k
1
⎤
⎢n
2 2
– ∑n r ⎥
2 ⎢⎣
r=
1 ⎥⎦
Sol. If no two of n given lines are parallel and no three of
them meet at a point, then the total number of points
of intersection is n C 2 . But it is given that there are k
sets of n 1 , n 2 , n 3 , ... , n k parallel lines such that no line
in one set is parallel to a line in any other set. Also,
lines of one set do not intersect with each other.
Therefore, lines of one set do not provide any point
of intersection. Hence,
Total number of points of intersection
= n n
C 2 – (
2
)
1 n
n
C + C + +
k
C
2 2 ...
n(n –1) ⎧
⎫
= – ⎨
n 1( n 1 –1)
+
n 2(
n 2 –1) ( –1)
+ ... +
n k n k
⎬
2 ⎩ 2 2
2 ⎭
n(n –1) 1
= – 2 {(n1 + n 2 2 + ... + n 2 k ) – (n 1 + n 2 + ... + n k )}
2 2
n(n –1) 1
= –
2 {(n1 + n 2 2 + ... + n 2 k ) – n}
2 2
2
n
1
= – 2 (n1 + n 2 2 + ... + n 2 k )
2 2
1 ⎡
k
⎤
= ⎢n
2 2
– ⎥
2
∑ n r
⎢⎣
r=
1 ⎥⎦
2
XtraEdge for IIT-JEE 44 MARCH 2012

MATHS
DEFINITE INTEGRALS &
AREA UNDER CURVES
Mathematics Fundamentals
Properties 1 :
If
∫
f (x)
dx = F(x), then
∫ b a
f ( x)
dx = F(b) – F(a), b ≥ a
Where F(x) is one of the antiderivatives of the
function f(x), i.e. F´(x) = f(x) (a ≤ x ≤ b).
Remark : When evaluating integrals with the help of
the above formula, the students should keep in mind
the condition for its legitimate use. This formula is
used to compute the definite integral of a function
continuous on the interval [a, b] only when the
equality F´(x) = f(x) is fulfilled in the whole interval
[a, b], where F(x) is antiderivative of the function
f(x). In particular, the antiderivative must be a
function continuous on the whole interval [a, b]. A
discontinuous function used as an antiderivative will
lead to wrong result.
If F(x) =
∫ x
f ( t)
dt, t ≥ a, then F´(x) = f(x)
a
Properties of Definite Integrals :
or
If f(x) ≥ 0 on the interval [a, b], then
∫ b a
∫ b a
∫ a
b
∫ b a
∫ a
0
∫ b
a
∫ −
aa
2
∫ a
0
f ( x)
dx =
∫ b f ( t)
dt
a
f ( x)
dx = –
∫ b a
f ( x)
dx =
∫ c a
f ( x)
dx
f ( x)
dx +
∫ b
c
a
f ( x)
dx =
∫
f ( a − x)
dx
0
b
f (x) dx =
∫
f ( a + b − x)
dx
a
⎧
⎪2 f ( x)
dx = ⎨ ∫
b f (x)dx
a
⎪⎩ 0
⎧
⎪2 f ( x)
dx = ⎨ ∫
b f (x)dx
a
⎪⎩ 0
f ( x)
dx ≥ 0
f ( x)
dx, a < c < b
if f(–x) = f(x)
if f(–x) = –f (x)
if f(2a – x)
if f(2a – x)
= f(x)
= –f (x)
Every continuous function defined on [a, b] is
integrable over [a, b].
Every monotonic function defined on [a, b] is
integrable over [a, b]
If f(x) is a continuous function defined on [a, b], then
there exists c ∈ (a, b)such that
∫ b a
f ( x)
dx = f(c) . (b – a)
1
The number f(c) =
( b − a)
∫ b f ( x)
dx is called the
a
mean value of the function f(x) on the interval [a, b].
If f is continous on [a, b], then the integral function g
defined by g(x) =
∫ x
f ( t)
dt for x ∈ [a, b] is derivable
a
on [a, b] and g´(x) = f(x) for all x ∈ [a, b].
If m and M are the smallest and greatest values of a
function f(x) on an interval [a, b], then
m(b – a) ≤
∫ b a
f ( x)
dx ≤ M(b – a)
If the function φ(x) and ψ(x) are defined on [a, b] and
differentiable at a point x ∈ (a, b) and f(t) is
continuous for φ(a) ≤ t ≤ ψ(b), then
d ⎛
⎞
⎜
⎟
dx ⎝∫ ψ ( x)
f ( t)
dt = f(ψ(x)) ψ´(x) – f(φ(x)) φ´(x)
φ(
x)
⎠
∫ b a
f ( x)
dx ≤
∫ b | f ( x)
| dx
a
If f 2 (x) and g 2 (x) are integrable on [a, b], then
∫ b a
⎛
f ( x)
g(
x)
dx ≤ ⎜
⎝
∫ b a
f
2
⎞
( x)
dx⎟ ⎠
1/ 2
⎛
⎜
⎝
∫ b a
2 ⎞
g ( x)
dx⎟ ⎠
1/ 2
Change of variables : If the function f(x) is
continuous on [a, b] and the function x = φ(t) is
continuously differentiable on the interval [t 1 , t 2 ] and
a = φ(t 1 ), b = φ(t 2 ), then
∫ b a
t2
t1
f ( x)
dx =
∫
f ( φ(
t
)) φ´(t) dt
Let a function f(x, α) be continuous for a ≤ x ≤ b and
c ≤ α ≤ d. Then for any α ∈ [c, d], if
b
b
I(α) =
∫
f ( x,
α)
dx, then I´(α) =
∫
f ´( x,
α)
dx,
a
a
XtraEdge for IIT-JEE 45 MARCH 2012

Where I´(α) is the derivative of I(α) w.r.t. α and
f ´(x, α) is the derivative of f(x, α) w.r.t. α, kepping x
constant.
Integrals with Infinite Limits :
If a function f(x) is continuous for a ≤ x < ∞, then by
definition
∫ ∞ b
f ( x)
dx = lim
a
b→∞∫
f ( x)
dx ....(i)
a
If there exists a finite limit on the right hand side of
(i), then the improper integrals is said to be
convergent; otherwise it is divergent.
Geometrically, the improper integral (i) for f(x) > 0,
is the area of the figure bounded by the graph of the
function y = f(x), the straight line x = a and the x-axis.
Similarly,
∫ ∞ −
b
∫ ∞ −∞
( dx = lim
a→−∞∫
f x)
properties :
∫ a
0
and
∫
∫ π / 2
0
a
f (x) dx =
∫ −∞
b
a
f ( x)
dx and
f ( x)
dx +
∫ ∞ a
f ( x)
dx
1
x f ( x)
dx = a 2 ∫ a x f ( x)
if f(a – x) = f(x)
a
f ( x)
f ( x)
+ f ( a − x
0 )
0
dx = 2
a
logsin
x dx =
∫ π / 2
logcos
x dx
0
= – 2
π log 2 = 2
π log 2
1
⎛ 1 ⎞
Γ(n + 1) = n Γ (n), Γ(1) = 1, Γ ⎜ ⎟ = π
⎝ 2 ⎠
If m and n are non-negative integers, then
∫ π / 2
0
sin
m
x cos
n
⎛ m + 1⎞
⎛ n + 1⎞
Γ⎜
⎟Γ⎜
⎟
x dx =
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ m + n + 2 ⎞
2Γ⎜
⎟
⎝ 2 ⎠
Reduction Formulae of some Define Integrals :
∫ ∞
0
∫ ∞
0
∫ ∞
0
−ax
e cos bx dx =
−ax
e sin bx dx =
−ax
e x n dx =
If I n =
∫ π / 2
0
sin
n
a
n!
n+ 1
x dx
a
a
2
2
a
+ b
b
+ b
, then
2
2
⎧ n −1
n − 3 n − 5 2
⎪ . . .....
I n = n n − 2 n − 4 3
⎨
n −1
n − 3 n − 5 1 π
⎪ . . ..... .
⎩ n n − 2 n − 4 2 2
If I n =
∫ π / 2
0
cos
n
x dx , then
⎧ n −1
n − 3 n − 5 2
⎪ . . .....
I m = n n − 2 n − 4 3
⎨
n −1
n − 3 n − 5 1 π
⎪ . . ..... .
⎩ n n − 2 n − 4 2 2
( when n is odd)
( when n is even)
( when n is odd)
( when n is even)
Leibnitz's Rule :
If f(x) is continuous and u(x), v(x) are differentiable
functions in the interval [a, b], then
d
dx
∫
v(
x)
u(
x)
f ( t)
dt
d d
= f{v(x)} v(x) – f{u(x) u(x)
dx
dx
Summation of Series by Integration :
n
∑ − 1
lim
n→∞
r=
0
⎛ r ⎞ 1
f ⎜ ⎟ . =
⎝ n ⎠ n ∫ 1 f ( x)
dx
0
Some Important Results :
n
∑ − 1
r=
0
n
∑ − 1
r=
0
2
1
sin( α + r β)
=
cos( α + r β)
=
2
1 1 1 π
–
2 +
2 – .... =
2 3 12
1 1 +
2 +
2 + .... =
2 31
6
2
1
⎧ 1 ⎫ ⎛ 1 ⎞
sin⎨α + ( n −1)
β⎬sin⎜
nβ⎟
⎩ 2 ⎭ ⎝ 2 ⎠
⎛ 1 ⎞
sin⎜
β⎟
⎝ 2 ⎠
⎧ 1 ⎫ ⎛ 1 ⎞
cos⎨α + ( n −1)
β⎬sin⎜
nβ⎟
⎩ 2 ⎭ ⎝ 2 ⎠
⎛ 1 ⎞
sin⎜
β⎟
⎝ 2 ⎠
Area under Curves :
Area bounded by the curve y = f(x), the x-axis and the
ordinates x = a, x = b
=
∫ b y dx =
a ∫ b a
Y
f ( x)
dx
2
π
y = f (x)
y
x = b
O δx X
Area bounded by the curve x = f(y), the y-axis and the
abscissae y = a, y = b
XtraEdge for IIT-JEE 46 MARCH 2012

=
∫ b x dy =
a ∫ b f ( y)
dy
a
Y
δy
y = b
x
y = a
x = f (y)
O
X
The area of the region bounded by y 1 = f 1 (x), y 2 = f 2 (x)
and the ordinates x = a and x = b is given by
=
∫ b f 2 ( x)
dx –
a ∫ b f 1(
x)
dx
a
Y
A
x = a
x = b
O
X
where f 2 (x) is y 2 of the upper curve and f 1 (x) is y 1 of
the lower curve, i.e. the required area
b
=
∫
[ f 2 ( x)
− f1(
x)]
dx =
∫
( y 2 − y1)
dx
a
b
a
f(x) ≤ 0 for all x in a ≤ x ≤ b, then area bounded by
x-axis, the curve y = f(x) and the ordinates x = a, x = b
is given by
B
Black Holes-The Most Efficient
Engines in the Universe
The scientists have just found the most energyefficient
engines in the universe. Black holes,
whirling super dense centres of galaxies that suck in
nearly everything. Jets of energy spurting out of older
ultra-efficient black holes also seem to be playing a
crucial role as zoning police in large galaxies
preventing to many stars from sprouting. This
explains why there are fewer burgeoning galaxies
chock full of stars than previously expected.
For the first time, the scientists have measured both
the mas of hot gas that is being sucked into nine older
black holes and the unseen super speedy jets of high
energy particles split out, which essentially form a
cosmic engine. Then they determined a rate of how
efficient these older black hole engines are and were
awe-struck. These black holes are 25 times more
efficient than anything man has built, with nuclear
power being the most efficient of man-made efforts,
said the research's lead author, Professor Steve Allen
of Stanford University.
The galaxies in which these black holes live are
bigger than the Milky way, which is the Earth's
galaxy and are 50 million to 400 million light-years
away.
= –
∫ b a
f ( x)
dx
Y
O
D
C
X
A
If f(x) ≥ 0 for a ≤ x ≤ c and f(x) ≤ 0 for c ≤ x ≤ b, then
area bounded by y = f(x), x-axis and the ordinates
x = a, x = b is given by
=
∫ c f ( x)
dx +
a ∫ b
− f ( x)
dx =
c ∫ c f ( x)
dx –
a ∫ b
f (x)dx
O
x = a
B
A
f (x)≥0
C
M
f (x)≤0
N
x = b
B
c
Black holes are the most fuel-efficient engines in
the universe.
The results were surprising because the types of black
holes studied were older, less powerful and generally
considered boring, scientists said. But they ended up
being more efficient than originally thought, possibly
as efficient as their younger, brighter and more potent
black hole siblings called quasars. One way the
scientists measured the efficiency of black holes was
by looking at the jets of high energy spewed out.
XtraEdge for IIT-JEE 47 MARCH 2012

MATHS
PROBABILITY
Mathematics Fundamentals
Some Definitions :
Experiment : A operation which can produce some
well defined outcomes is known as an experiment.
Random experiment : If in each trail of an
experiment conducted under identical conditions, the
outcome is not unique, then such an experiment is
called a random experiment.
Sample space : The set of all possible outcomes in an
experiment is called a sample space. For example, in a
throw of dice, the sample space is {1, 2, 3, 4, 5, 6}.
Each element of a sample space is called a sample
point.
Event :
An event is a subset of a sample space.
Simple event : An event containing only a single
sample point is called an elementary or simple event.
Events other than elementary are called composite or
compound or mixed events.
For example, in a single toss of coin, the event of
getting a head is a simple event.
Here S = {H, T} and E = {H}
In a simultaneous toss of two coins, the event of
getting at least one head is a compound event.
Here S = {HH, HT, TH, TT} and E = {HH, HT, TH}
Equally likely events : The given events are said to
be equally likely, if none of them is expected to occur
in preference to the other.
Mutually exclusive events : If two or more events
have no point in common, the events are said to be
mutually exclusive. Thus E 1 and E 2 are mutually
exclusive in E 1 ∩ E 2 = φ.
The events which are not mutually exclusive are
known as compatible events.
Exhaustive events : A set of events is said to be
totally exhaustive (simply exhaustive), if no event out
side this set occurs and at least one of these event
must happen as a result of an experiment.
Independent and dependent events : If there are
events in which the occurrence of one does not
depend upon the occurrence of the other, such events
are known as independent events. On the other hand,
if occurrence of one depend upon other, such events
are known as dependent events.
Probability :
In a random experiment, let S be the sample space
and E ⊆ S, then E is an event.
The probability of occurrence of event E is defined as
=
P(E) =
number of distinct elements in E
number of distinct element in S
n(E)
= n(S)
number of outocomes favourable to occurrence of
number of all possible outcomes
Notations :
Let A and B be two events, then
A ∪ B or A + B stands for the occurrence of at
least one of A and B.
A ∩ B or AB stands for the simultaneous
occurrence of A and B.
A´ ∩ B´ stands for the non-occurrence of both A
and B.
A ⊆ B stands for "the occurrence of A implies
occurrence of B".
Random variable :
A random variable is a real valued function whose
domain is the sample space of a random experiment.
Bay’s rule :
Let (H j ) be mutually exclusive events such that
n
P(H j ) > 0 for j = 1, 2, ..... n and S = U . Let A
j=1
H j
be an events with P(A) > 0, then for j = 1, 2, .... , n
⎛ H ⎞
P⎜
j P(H j)P(A / H j)
⎟
=
n
⎝ A ⎠ ∑ P(H ) P(A / H )
k=
1
k
Binomial Distribution :
If the probability of happening of an event in a single
trial of an experiment be p, then the probability of
happening of that event r times in n trials will be
n C r p r (1 – p) n – r .
Some important results :
Number of cases favourable to event A
(A) P(A) =
Total number of cases
n(A)
= n(S)
k
E
XtraEdge for IIT-JEE 48 MARCH 2012

P( A) =
Number of cases not favourable to event A
Total number of cases
n(A)
= n(S)
(B) Odd in favour and odds against an event : As a
result of an experiment if “a” of the outcomes are
favourable to an event E and b of the outcomes are
against it, then we say that odds are a to b in favour
of E or odds are b to a against E.
Thus odds in favour of an event E
Number of favourablecases a
=
=
Number of unfavourable cases b
Similarly, odds against an event E
Number of unfavourable cases b
=
=
Number of favorablecases a
Note :
If odds in favour of an event are a : b, then the
probability of the occurrence of that event is
a
and the probability of non-occurrence of
a + b
b
that event is
a + b
. a
a + b
If odds against an event are a : b, then the
probability of the occurrence of that event is
b
and the probability of non-occurrence of
a + b
a
that event is
a + b
.
(C) P(A) + P( A ) = 1
0 ≤ P(A) ≤ 1
P(φ) = 0
P(S) = 1
If S = {A 1 , A 2 , ..... A n }, then
P(A 1 ) + P(A 2 ) + .... + P(A n ) = 1
If the probability of happening of an event in one
trial be p, then the probability of successive
happening of that event in r trials is p r .
(D) If A and B are mutually exclusive events, then
P(A ∪ B) = P(A) + P(B) or
P(A + B) = P(A) + P(B)
If A and B are any two events, then
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or
P(A + B) = P(A) + P(B) – P(AB)
If A and B are two independent events, then
P(A ∩ B) = P(A) . P(B) or
P(AB) = P(A) . P(B)
If the probabilities of happening of n independent
events be p 1 , p 2 , ...... , p n respectively, then
(i) Probability of happening none of them
= (1 – p 1 ) (1 – p 2 ) ........ (1 – p n )
(ii) Probability of happening at least one of them
= 1 – (1 – p 1 ) (1 – p 2 ) ....... (1 – p n )
(iii) Probability of happening of first event and not
happening of the remaining
= p 1 (1 – p 2 ) (1 – p 3 ) ....... (1 – p n )
If A and B are any two events, then
⎛ B ⎞
P(A ∩ B) = P(A) . P ⎜ ⎟ or
⎝ A ⎠
⎛ B ⎞
P(AB) = P(A) . P ⎜ ⎟
⎝ A ⎠
⎛ B ⎞
Where P ⎜ ⎟ is known as conditional probability
⎝ A ⎠
means probability of B when A has occurred.
Difference between mutually exclusiveness and
independence : Mutually exclusiveness is used
when the events are taken from the same
experiment and independence is used when the
events are taken from the same experiments.
(E) P(A A ) = 0
P(AB) + P( AB ) = 1
P( A B) = P(B) – P(AB)
P(A B ) = P(A) – P(AB)
P(A + B) = P(A B ) + P( A B) + P(AB)
Some important remark about coins, dice and playing
cards :
Coins : A coin has a head side and a tail side. If
an experiment consists of more than a coin, then
coins are considered to be distinct if not otherwise
stated.
Dice : A die (cubical) has six faces marked 1, 2,
3, 4, 5, 6. We may have tetrahedral (having four
faces 1, 2, 3, 4,) or pentagonal (having five faces
1, 2, 3, 4, 5) die. As in the case of coins, If we
have more than one die, then all dice are
considered to be distinct if not otherwise stated.
Playing cards : A pack of playing cards usually
has 52 cards. There are 4 suits (Spade, Heart,
Diamond and Club) each having 13 cards. There
are two colours red (Heart and Diamond) and
black (Spade and Club) each having 26 cards.
In thirteen cards of each suit, there are 3 face cards or
coart card namely king, queen and jack. So there are
in all 12 face cards (4 kings, 4 queens and 4 jacks).
Also there are 16 honour cards, 4 of each suit namely
ace, king, queen and jack.
XtraEdge for IIT-JEE 49 MARCH 2012

MOCK TEST FOR IIT-JEE
PAPER - I
Time : 3 Hours Total Marks : 240
Instructions :
• This question paper contains 69 questions in Chemistry (23), Mathematics (23) & Physics (23).
• In section -I (7 Ques. SCQ Type) of each paper +3 marks will be given for correct answer & –1 mark for wrong
answer.
• In section -II (4 Ques. MCQ Type) of each paper +4 marks will be given for correct answer no negative marking
for wrong answer.
• In section -III contains 2 groups of questions [Pass. 1 (2 Ques.) + Pass. 2 (3 Ques.) = 5 Ques.] of each paper +3
marks will be given for each correct answer & –1 mark for wrong answer.
• In section -IV contain (7 Ques. of Numerical Response with single-digit Ans.) of each paper +4 marks will be
given for correct answer & No Negative marking for wrong answer.
CHEMISTRY
SECTION – I
Straight Objective Type
Questions 1 to 7 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. Arrange the following compounds according to
decreasing order of combustion -
CH 3
OH
OH
(C)
(D)
3. Assign double bond configurations to the following–
COOH
C
NC
CH
C = C
2 OH
NH 2 –CH 2
CN
(A) E (B) Z (C) E.E. (D) Z, Z
4. Product (P) in the following reaction is -
O
⊕
H / H 2 O
(P)
CH 3
OH
CH 2 –OH
2.
I
III
(A) II > IV > I > III
(C) I > II > III > IV
O
⊕
H
form) among the following is -
(A)
OH
II
IV
CH 3
CH 3
CH 3
(B) I > IV > III > II
(D) IV > I > III > II
(B)
major tautomer (enol
OH
(A)
OH
CH 3 HO CH 3
OH
OH
(C)
(D)
O
OH CH 3
CH 3 CH 3
5. V 1 mL of NaOH of normality X and V 2 mL of
Ba(OH) 2 of normality Y are mixed together. the
mixture is completely neutralised by 100 mL of 0.1
N HCl. If V 1 / V 2 = 4
1 and Y
X = 4, what fraction of
the acid is neutralized by Ba(OH) 2 :
(A) 0.5 (B) 0.25 (C) 0.33 (D) 0.67
XtraEdge for IIT-JEE 50
MARCH 2012

6. Equal moles of CO, B 2 H 6 , H 2 and CH 4 are placed in
a container. If a hole was made in container, after 5
minute partial pressure of gases in container would
be :
(at wt. of C, O, B and H are 12, 16, 11 and 1
respectively)
(A) P CO > P > P > P
(A) Cr 3+ (B) Mn 2+ (C) Fe 3+ (D) Cu 2+ I II III
IV
11. Which of the following metals become passive when
dropped into conc. HNO 3 ?
(A) Cu (B) Fe (C) Cr (D) Al
SECTION – III
Comprehension Type
B 2 H 6 H 2 CH 4
(B) P CO = P B 2 H 6
> P CH 4
> P
This section contains 2 paragraphs; passage- I has 2
H 2
multiple choice questions (No. 12 & 13) and passage- II
(C) P CO > P B 2 H 6
= P H 2
> P CH 4
has 3 multiple (No. 14 to 16). Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
(D) P B 2 H 6
> P H 2
> P CH 4
> P CO
is correct. Mark your response in OMR sheet against
the question number of that question. +3 marks will be
given for each correct answer and –1 mark for each
wrong answer.
Paragraph # 1 (Ques. 12 to 13)
De-Broglie proposed dual nature for electron by
h
putting his famous equation λ = . Later on
SECTION – II
mu
Multiple Correct Answers Type
Heisenburg proposed uncertainty principle as ∆p.∆x ³
h ⎛ h ⎞
⎜h
= ⎟ . On the contrary particle nature of
2 ⎝ 2 π ⎠
electron was established on the basis of photoelectric
effect. When a photon strikes the metal surface,, it
gives up its energy to the electron. Part of this energy
(say W) is used by the electrons of escape from the
metal and the remaining imparts the kinetic energy
(½ mu 2 ) to the photoelectron. The potential applied
may be named as -
on the surface to reduce the velocity of photoelectron
to zero is known as stopping potential.
(A) 3– ethyl, 2-methyl oxirane
12. With what velocity must an electron travel so that its
(B) 1, 2 – epoxy – 2 – methyl butane
momentum is equal to that of photon of wavelength
(C) 1, 2 – oxa pentane
of λ = 5200 Å :
(D) 2 – methyl –1, 2– butoxide
(A) 800 m s –1 (B) 1400 m s –1
(C) 400 m s –1 (D) 200 m s –1
⊕
13. The wavelength of helium atom whose speed is equal
(A)
H
MgBr + HCOCl ⎯⎯→
to its rms speed at 27ºC :
(A) 7.29 × 10 –11 m (B) 4.28 × 10 –10 m
(excess)
(C) 5.31 × 10 –11 m (D) 6.28 × 10 –11 m
O
⊕
H
(B) Ph–Mg–Br + CH 3 —C—Cl → ⎯⎯→
Paragraph # 2 (Ques. 14 to 16)
(excess)
From following sets of compounds give answer of
⊕
H
(C) CH 3 –Mg–Br + (CH 3 CO 2 )O → ⎯⎯→
following question
O O
O
O
(excess)
O
O
⊕
Set – A
H
(D) CH 3 –Mg – Br + Cl–C–OC 2 H 5 → ⎯⎯→
O O
(excess)
I
II III
IV
O
Set – B
7. The dipole moment of HBr is 2.6 × 10 –30 c–m and
the interatomic spacing is 1.41Å. The percentage of
ionic character in HBr is :
(A) 10.5 (B) 11.5 (C) 12.5 (D) 13.5
Questions 8 to 11 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and no negative marks.
8. The compound C2 H 5
CH 3
O
9. In which of the following reactions 3º alcohol will
be obtained as a product -
10. Which of the following species has same number of
unpaired electron ?
XtraEdge for IIT-JEE 51
MARCH 2012

Set – C
O
O
O
I
O
O
O
III
IV
14. Correct statement regarding set A -
(A) I is stronger acid than III
(B) II is stronger acid then I
(C) III & IV are equal acidic strength
(D) IV is weaker acidic then I
O
O
15. Correct statement regarding set B -
(A) I & II compound show both resonance and
hyperconjugation
(B) III compound show five hyperconjugation
structure
(C) IV is more stable than II
(D) II is more stable than I
16. Correct statement regarding set C -
(A) II is stronger acid than I
(B) I is stronger acid than IV
(C) II and III are equal acidic
(D) In IV acid anion is not stablises by resonance
II
O
O
SECTION – IV
Numerical Response Type
This section contains 7 questions (Q.17 to 23). +4 marks
will be awarded for each correct answer and no
negative marking for wrong answer. The answer to
each question is a single-digit integer, ranging from 0 to
9. The bubble corresponding to the correct answer is to
be darkened in the OMR.
17. How many of the following compound contain chiral
atom.
CH 3
CH 3
(i)
N (ii) ⊕ N
H T
Ph C 2 H 5
D
CH 2 =CH–CH 2
CH 3 O H
O
(iii)
(iv) CH 3 S CH 3
(v)
H
O
P
T
(vi)
D
CH 3
H
Si
D
H
CH 3
CH 3 Ge
(vii) D C CH 3
(viii)
Ph
H
C 2 H
CH 3
5
14
18. In the following reaction double bond equivalent of
D is –
OH
OH
dil.HNO 3 (1) Sn / HCl
A + B
(2)NaNO 2 / HCl
More Less
volatile volatile
C
(mild-basic)
condition
19. How many groups are o/p director in electrophilic
aromatic substitution reaction –
(i) — NH 2
(ii) — CHO
(iii) — COOH (iv) —OMe
(v) —O—C—Me (iv) — Et
O
(vii) —C—NH—Me
O
(ix) —N = O (x) —N = NH
(viii) – SO 3 H
20. How many compound which is given below is
isomer of D-glucose.
D-Mannose, D – Fructose, D-Idose,
D-Galactose, D-Arabinose, D-Ribose
21. What is oxidation state of sulphur in Caro's acid?
22. How many π-bonds are present in Marhall's acid?
23. How many P —O—P bonds are present in P 4 O 8 ?
MATHEMATICS
SECTION – I
Straight Objective Type
Questions 1 to 7 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. Let f (x) =
sin π x
x
x
3
1
2x
3x
2
4
D
1
1 , f (x) be an odd function
1
and its odd value is equal to g(x), then
f (1) g(1) is -
(A) –1 (B) – 4 (C) – 5 (D) 1
XtraEdge for IIT-JEE 52
MARCH 2012

2.
sin3α
< 0 if α lies in
cos2α
(A) (13π/48, 14π/48)
(B) (14π/48, 18π/48)
(C) (18π/48, 23π/48)
(D) any of these intervals
(C) f (x) is an increasing function in the interval
– ∞ < x ≤ 0 and decreasing in the interval
0 ≤ x < ∞
(D) f (x) is a decreasing function in the interval
– ∞ < x ≤ 0 and increasing in the interval
0 ≤ x < ∞
3. If x 1 , x 2 , x 3 , x 4 are roots of the equation
x 4 – x 3 sin 2β + x 2 cos 2β – x cos β – sin β = 0, then
4.
4
∑
i=
1
tan
−1
(A) π – β
(C) π/2 – β
−
π
x→ 2
x i is equal to
(B) π – 2β
(D) π/2 – 2β
Lim [1 + (cos x) cos x ] 2 is equal to
(A) Does not exist (B) 1
(C) e (D) 4
5. If f (x) = log x (ln (x)), then f '(x) at x = e is
(A) 0 (B) 1
(C) e (D) 1/e
6. If the function f (x) = |x 2 + a |x| + b| has exactly three
points of non-differentiability, then which of the
following may hold
(A) b = 0, a < 0 (B) b < 0, a ∈ R
(C) b > 0, a ∈ R (D) b < 0, a ∈ R –
7. The inequality log 2 (x) < sin –1 (sin(5)) holds if x ∈
(A) (0, 2 5–2π ) (B) (2 5–2π , ∞)
(C) (2 2π–5 , ∞) (D) (0, 2 2π–5 )
SECTION – II
Multiple Correct Answers Type
Questions 8 to 11 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and no negative marks.
8. 2 tan –1 (–3) is equal to
(A) – cos –1 (–4/5) (B) – π + cos –1 (4/5)
(C) – π/2 + tan –1 (–4/3) (D) cot –1 (4/3)
9. Let [x] denotes the greatest integer less than or equal
to x. If f (x) = [x sin πx], then f (x) is
(A) continuous at x = 0
(B) continuous in (–1, 0)
(C) differentiable at x = 1
(D) differentiable at (–1, 1)
⎛
2
x ⎞
10. Let f be the function f (x) = cos x – ⎜ ⎟
1 − , then
⎝ 2 ⎠
(A) f (x) is an increasing function in (0, π/2)
(B) f (x) is a decreasing function in (–∞, ∞)
2
11. If u = a cos θ + b sin θ +
(A) max. u 2 =
2
2
2
a
a + b
(B) max. u 2 = 2 a + b
(C) min. u 2 = 2(a + b) 2
(D) min u 2 = (a + b) 2
2
2
2
2
2
sin
2
θ + b
2
cos
2
θ , then
SECTION – III
Comprehension Type
This section contains 2 paragraphs; passage- I has 3
multiple choice questions (No. 12 to 14) and passage- II
has 2 multiple (No. 15 & 16). Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. +3 marks will be
given for each correct answer and –1 mark for each
wrong answer.
Passage # 1 (Ques. 12 to 14)
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)
12. The value of x for which f (x) = 0 is -
(A) – 1/2 (B) 0
(C) 1/2 (D) 1
13. If f (x) = 0, then a 2 is equal to -
(A) 1/2 (B) 2/3
(C) 5/9 (D) 9/5
14. If a 2 = 26/51, then b 2 is equal to -
(A) 1/25 (B) 24/25
(C) 25/26 (D) 50/51
Passage # 2 (Ques. 15 & 16)
v(
x)
dy
If y =
∫
f ( t)
dt, let us define in different
u(
x)
dx
dy
manner = v'(x) f (v(x)) – u'(x) f (u(x)) and the
dx
equation of tangent at (a, b) is.
(y – b) =
⎛ dy ⎞
⎜ ⎟⎠
⎝ dx
( a,
b)
(x – a)
XtraEdge for IIT-JEE 53
MARCH 2012

x
15. If y =
∫ 2 2
t dt , then equation of the tangent at
x
x = 1 is -
(A) y = x + 1 (B) x + y = 1
(C) y = x – 1
(D) y = x
16 If f (x) =
∫ x t
e
1
2
/ 2
(1 – t 2 )dt, then
at x = 1 is -
(A) 0 (B) 1
(C) 2 (D) – 1
d f (x),
dx
SECTION – IV
Numerical Response Type
This section contains 7 questions (Q.17 to 23). +4 marks
will be awarded for each correct answer and no
negative marking for wrong answer. The answer to
each question is a single-digit integer, ranging from 0
to 9. The bubble corresponding to the correct answer is
to be darkened in the OMR.
17. The number of elements in the range of
⎡2
⎤
f (x) = [x] + [2x] + ⎢ x⎥ + [3x] + [4x] + [5x]
⎣3
⎦
for 0 ≤ x < 3 is m + 21. Then the value of m is -
18.
⎡ ⎛ 1 ⎞⎛
1 ⎞⎤
⎢(
h + 1) ⎜h
+ ⎟⎜h
+ ⎟⎥
2
h
Lim h −
2
⎢ ⎝ 2 ⎠⎝
2 ⎠⎥
h→∞ ⎢
⎛ 1 ⎞
h
⎥
⎢
..................... ⎜ + ⎟
h−1
⎥
⎣
⎝ 2 ⎠⎦
find the value of S.
h
= e S , then
19. The number of polynomials of the form
x 3 + ax 2 + bx + c which are divisible by x 2 + 1, where
a, b, c ∈ {1, 2, 3, ….. 10} must be 2k. Value of k will
be.
20. If e y d y
+ xy = e, then at x = 0 is e –λ , then
2
dx
numerical quantity λ should be equal to……….
21. If a circle S (x, y) = 0 touches at the point (2, 3) of the
line x + y = 5 and S (1, 2) = 0, then
( 2 × Radius) of such circle is.
2 2
x y
22. A line through P(λ, 3) meets the ellipse + 16 9
2
= 1
at A and D meets the x-axis and y-axis at B and C
respectively, so that PA.PD = PB.PC, then | λ | is
greater than or equal to ……
23. If | → a | = 3, | → b | = 4, | → c | = 5 and → a ⊥ ( → b + → c ), → b ⊥
( → c + → a ), → c ⊥ ( → a + → b ), then | → a + → b + → c | is……
PHYSICS
SECTION – I
Straight Objective Type
Questions 1 to 7 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. A block of mass 'm 1 ' attached to the free end of a spring
of force constant 'k' is mounted on a smooth horizontal
surface as shown in figure. The block execute S.H.M.
with amplitude A and frequency 'f '. If an object of
mass 'm 2 ' is put on it, when the block was passing
through mean position and both move together, then the
new amplitude of oscillation is -
smooth
m 2
m 1
m1
(A) A (B) A
m
(C) A
2
m1
(m + m
1
2
)
(D) A
m2
(m + m
2. Two smooth, identical billiard balls A and B collide
with B at rest and A in motion as shown. If speed of
A is 'v' before collision and coefficient of restitution
for the collision is 'e', the speed of B after collision
will be-
B
(A) v cos θ
( 1−
e)vcosθ
(C)
2
v
θ
A
(B) ev cosθ
( 1+
e)vcosθ
(D)
2
3. The moment of inertia of a flat annular ring is having
mass M, inner radius 'a' and outer radius 'b' about the
perpendicular axis through the centre is-
(A) M (b 2 – a 2 ) (B) M/2 (b 2 – a 2 )
(C) M/2 (b 2 + a 2 ) (D) M/2 (b – a)
1
2
)
XtraEdge for IIT-JEE 54
MARCH 2012

4. A solid sphere of mass 2 kg is pulled by a constant
force acting at its centre on a rough surface having
co-efficient of friction 0.5. The maximum value of F
so that the sphere rolls without slipping is-
9. A metallic conductor of irregular cross section is as
shown in figure A constant potential difference is
applied across the ends (A) and (B). Then-
F
A
P
Q
B
(A) 70 N
(C) 40 N
(B) 25 N
(D) 35 N
5. A tunnel is dug across the diameter of Earth. A ball
is released from the surface of Earth into the tunnel.
The velocity of ball when it is at a distance R/2 from
centre of Earth is (where R = radius of earth and M =
mass of Earth)
(A)
(C)
3GM
4R
GM
2R
(B)
(D)
2GM
3R
2GM
R
6. The work done to break a spherical drop of radius R in
n drops of equal size is proportional to-
1
1
(A) − 1
(B) − 1
2/3
1/3
n
n
(C) n 1/3 –1 (D) n 4/3 – 1
7. A rod of length L kept on a smooth horizontal
surface is pulled along its length by a force F. The
area of cross-section is A and Young's modulus is Y.
The extension in the rod is-
F
FL
(A) AY
2FL
(B) AY
(C)
FL
2AY
(D) zero
SECTION – II
Multiple Correct Answers Type
Questions 8 to 11 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and no negative marks.
8. Under the action of a force, 2 kg body moves such
that its position x as a function of time is given by x
= 3
t 3 , x is in meter and t in second-
(A) body is acted by constant force
(B) body is acted by a force which is proportional to
time
(C) body is acted by a force which is proportional to
square of time
(D) work done by the force in first 2 second is 16 J
(A) Electric current at cross section P is equal to that
of cross section Q
(B) Electric field intensity at P is less than that at Q
(C) The number of electrons crossing per unit area
per unit time at cross section P is less than that at
Q
(D) The rate of heat generating per unit time at Q is
greater than that of P
10. The horizontal distance x and the vertical height y of
a projectile at time t are given by
x = at and y = bt 2 + ct, where a, b and c are constant.
Then-
(A) the speed of the projectile 1 second after it is
fired is (a 2 + b 2 + c 2 ) 1/2
(B) the angle with the horizontal at which the
projectile is fired is tan –1 ⎡ c ⎤
⎢ ⎥
⎣a
⎦
(C) the acceleration due to gravity is –2b
(D) the initial speed of the projectile is
(a 2 + c 2 ) 1/2
11. A block having mass m and charge q is connected by
spring of force constant k. The block lies on a
frictionless horizontal track and a uniform electric
field E acts on system as shown. The block is
released from rest when spring is unstretched (at x =
0). Then-
E
q,m
(A) maximum elongation in the spring is
2qE
k
(B) at equilibrium position, elongation in the spring is
(C) amplitude of oscillation of block is
(D) amplitude of oscillation of block is
qE
k
2qE
k
qE
k
XtraEdge for IIT-JEE 55
MARCH 2012

SECTION – III
Comprehension Type
This section contains 2 paragraphs; passage- I has 3
multiple choice questions (No. 12 & 14) and passage- II
has 2 multiple (No. 15 to 16). Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. +3 marks will be
given for each correct answer and –1 mark for each
wrong answer.
Passage # 1 (Ques. 12 & 14)
Consider a circuit shown in figure, then
23µF 7µF
A
12µF 5µF
1µF
10µF 1µF
12V
12. Equivalent capacitance of the circuit between A and
B is-
(A) 10 µF (B) 7.5 µF
(C) 5 µF (D) 14 µF
13. Charge stored by 5 µF capacitor is-
(A) 20 µC (B) 70 µC
(C) 10 µC
(D) zero
14. Energy stored by 10 µF capacitor is-
(A) 20 µJ (B) 40 µC
(C) 400 µJ (D) 245 µF
Passage # 2 (Ques. 15 & 16)
Potential energy of particle is given by
U = 5 + (x – 1) 2 . Particle have kinetic energy at x =
2m is 10 J. Then
15. Total mechanical energy is-
(A) 16 J (B) 26 J (C) 6 J (D) 5 J
16. Minimum potential energy is-
(A) 5 J (B) 10 J (C) 6 J (D) 9 J
SECTION – IV
Numerical Response Type
This section contains 7 questions (Q.17 to 23). +4 marks
will be awarded for each correct answer and no
negative marking for wrong answer. The answer to
each question is a single-digit integer, ranging from 0
to 9. The bubble corresponding to the correct answer is
to be darkened in the OMR.
B
17. Two simple pendulum have lengths l and 25l. At
t = 0 they are in same phase, after how many
oscillations of smaller pendulum will they be again
in phase for first time ?
18. In the resonance tube experiment, the first and
second states of resonance are observed at
20 cm and 66 cm. Find the value of end correction
(in cm).
19. The variation of pressure versus volume is shown in
the figure. The gas is diatomic and the molar
specific heat capacity for the process is found to be
xR. Find the value of x.
P
20. Three identical rods are joined at point O as shown in
the figure. In the steady state, find the ratio of
thermal current through rod AO and OC.
C
20ºC
100ºC
A
O
V
0ºC
21. A rod of length 20 cm is placed along the optical axis
of a concave mirror of focal length 30 cm. One end
of the rod is at the centre of curvature and other end
lies between F and C. Calculate the linear
magnification of the rod.
22. Figure shows a parabolic reflector in x-y plane given
by y 2 = 8x. A ray of light traveling along the line y =
a is incident on the reflector. Find where the ray
intersects the x-axis after reflection.
y-axis
y 2 =8x
line y = a
P(0,a)
incident ray
x-axis
23. A hydrogen like atom (atomic number Z) is in a
higher excited state of quantum number n. This
excited atom can make a transition to the first excited
state by successively emitting two photons of
energies 10.20 eV and 17.00 eV respectively.
Alternatively, the atom from the same excited state
can make a transition to the second excited state by
successively emitting two photons of energies 4.25
eV and 5.95 eV respectively. Determine the value of Z.
(Ionisation energy of hydrogen atom is 13.6 eV).
XtraEdge for IIT-JEE 56
MARCH 2012

XtraEdge for IIT-JEE 57
MARCH 2012

MOCK TEST FOR IIT-JEE
PAPER - II
Time : 3 Hours Total Marks : 240
Instructions :
• This question paper contains 60 questions in Chemistry (20,) Mathematics (20) & Physics (20).
• In section -I (8 Ques. SCQ Type) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer.
• In section -II (4 Ques. MCQ Type) of each paper +4 marks will be given for correct answer no negative marking
for wrong answer.
• In section -III (2 Ques. Column Matching Type) of each paper +8(2×4) marks will be given for correct answer.
No Negative marking for wrong answer.
• In section -IV contain (6 Ques. of Numerical Response with single-digit Ans.) of each paper +4 marks will be
given for correct answer & No Negative marking for wrong answer.
CHEMISTRY
SECTION – I
Straight Objective Type
Questions 1 to 8 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. In the enolisation of the given molecule the,
H-atom involved is :
O
Ha
Ha
Hc c
Hb
(A) Ha (B) Hb (C) Hc (D) Hd
2. The no. of position isomers for
CH 3 —CH—CH 2 —C≡CH are (give answer
CH 3
including this structure) :
(A) 1 (B) 2 (C) 0 (D) 3
3. Rank the following alkyl halides in order of
increasing SN 2 reactivity:
Br
Br
Br
I
II
(A) I < III < II (B) III < I < II
(C) II < I < III (D) I < II < III
III
4. The two compounds given below are :
D
Cl
H
Br
H Cl
I
(A) Identical
(B) Diastereomers
(C) Optically inactive
(D) enantiomers
5. The metals present in insulin, haemoglobin and
vitamin B 12 are respectively :
(A) Zn, Hg, Cr (B) Co, Fe, Zn
(C) Mg, Fe, Co (D) Zn, Fe, Co
6.
I
D
Br
(A) Fused with Na2CO3 (B) (H 2SO 4 + H 2O)
Green solid
Yellow
Evaporisation
solution
(D) Yellow
H
H
(CH 3 COO) 2 Pb
(C)
Orange
Here, A, B, C and D are respectively :
A B C D
(A) FeSO 4 FeCO 3 Fe(OH) 3 PbCO 3
(B) Cr 2 O 3 Na 2 CrO 4 Na 2 Cr 2 O 7 PbCrO 4
(C) FeCl 2 FeSO 4 PbSO 4 Fe(OH) 3
(D) FeSO 4 FeCl 3 Fe(OH) 3 PbCl 2
7. The negative charge on As 2 S 3 sol is due to absorption
of :
(A) H¯
(B) OH¯
(C) O 2– (D) S 2–
XtraEdge for IIT-JEE 58
MARCH 2012

8. pH of a mixture of 1M benzoic acid (pK a = 4.20) and
1M sodium benzoate is 4.5 In 300 mL buffer, benzoic
acid is (log 2 = 0.3) :
(A) 200 mL
(B) 150 mL
(C) 100 mL
(D) 50 mL
SECTION – II
Multiple Correct Answers Type
Questions 9 to 12 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) may be correct.
Mark your response in OMR sheet against the question
number of that question. +4 marks will be given for each
correct answer and no negative marks for wrong answer.
9. Which of the following pair have same IUPAC
naming :
Br
COOCH 3 COOH
(A)
&
Br
COOH
OCOCH 3
I
Br
Br F Cl
(B)
&
I Cl I I
F
(C) `
&
(D)
F
H 2 N
OH
Cl
H 2 N
&
Cl
10. Benzene can be nitration by :
(A) NO 2 BF 4 (B) NO 2 ClO 4
(C) conc. HNO 3 + H 2 SO 4 (D) Cl 2 / AlCl 3
11. In which of the following salt bridge is not needed :
(A) Pb(s) |PbSO 4(s) | H 2 SO 4 |PbO 2(s) |Pb (s)
(B) Zn(s)|ZnSO 4 |CuSO 4 |Cu(s)
(C) Cd(s)|CdO (s) |KOH (aq) |NiO 2(s) |Ni(s)
(D) Fe (s) | FeO (s) |KOH (aq) |Ni 2 O 3(s) |Ni(s)
F
OH
12. Which of the following is (are) true :
(A) H 2 S + H 2 O H 3 O + + HS¯ ;
K c acidity constant of H 2 S
(B) AgCl + 2NH 3 Ag(NH 3 ) 2 Cl;
K c is stability constant for Ag(NH 3 ) 2 Cl
(C) H 2 O H + + OH¯ ;
K c is equilibrium constant for dissociation of water
+
(D) RNH 2 + H 2 O RNH 3 + OH¯;
SECTION – III
Matrix Match 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 the particular question, against
statement B, darken the bubbles corresponding to Q
and R in the OMR. +8 marks will be given for each
correct answer (i.e. +2 marks for each correct row) and
no negative marking for each wrong answer.
13. Match the column:
Column -I
Column-II
(A) Electro chamical cell (P) ∆G = +ve
(B) I st law of faraday (Q) ∆G = –ve
(C) Electrolytic cell (R) w = z . i . t.
(D) lead acid storage cell (S) salt bridge
(T) rechargable cell
14. Match the column:
Column -I
(A) lithium
(B) Barium
(C) calcium
(D) magnesium
Column-II
(P) violet
(Q) Brick red
(R) Apple green
(S) Crimson red
(T) No flame colour
SECTION – IV
Numerical Response Type
This section contains 6 questions (Q.15 to 20). The
answer to each of the questions is a Single-digit integer,
ranging from 0 to 9. The bubble corresponding to the
correct answer is to be darkened in the OMR. +4
marks will be given for each correct answer and no
negative marking for each wrong answer.
15. How many acidic group is present in given
compound :
⊕ NH 3 —CH—CH 2 —CH 2 —COOH
COO
Θ
16. How many isomers of ‘X’ C 8 H 10 when reacts with
hot KMnO 4 give only aromatic dicarboxylic acid ?
How many isomers of 'Y' C 4 H 8 when reacts with hot
alkaline KMnO 4 give carbondioxide sum of X + Y = ?
K C is basicity constant for RNH 2
XtraEdge for IIT-JEE 59
MARCH 2012

17. CH 3 —CH–CH—CH 3
18.
Br
CH 3
products
Find total no. of products (including stereoisomer).
‘X ’ is total number of hofmann
N
exhaustive methylation to remove nitrogen from
given compound
Alc.KOH
‘Y’ is total number of possible
Br ∆
product (including stereoisomers) sum of X+ Y = ?
19. If the density of Fe 2 O 3 and Al are 5.2 g/mL and 2.7 g
/ mL respectively. Calculate the fuel value in kcal
mL –1 of mixture according to thermite reaction -
(If ∆H f Al 2 O 3 = – 399 Kcal / mol
& ∆H f Fe 2 O 3 = – 199 Kcal / mol)
dx
20. For a reaction = K[H + ] n . If the pH of reaction
dt
medium changes from two to one, the rate becomes
100 times of the value of at pH = 2. What is the order
of reaction -
MATHEMATICS
SECTION – I
Straight Objective Type
Questions 1 to 8 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. In a triangle, if the sum of two sides is x and their
product is y such that (x + z) (x – z) = y, where z is
the third side of the triangle, then the triangle is
(A) equilateral (B) right angled
(C) obtuse angled (D) none of these
r1 r2
r3
2. In a triangle ABC, + + is equal to -
bc ca ab
1 1
(A) − (B) 2R – r
2R
r
(C) r – 2R
−
1
(D)
r
1
( a + b)sin
θ A + B
2R
(B)
= cos
2 ab
2
3. If f (θ) = cos 2 θ + sin 4 θ. Then minimum value of f (θ) is -
C 2H 5OH ‘X’ (SN1 + E 1 )
Consider all products
(A) 1/4 (B) 3/4 (C) 1 (D) 1/2
4.
3
2 2 − (cos x + sin x)
Let f (x) =
, then Lim f (x) is
1−
sin(2x)
π
x→
4
equal to
(A)
1
3
(B) 2 (C) 1 (D)
2
2
5.
⎧
2 2( x−1)
x e
; x ≤ 1
Let f (x) = ⎨
2
⎩a
cos(2x
− 2) + bx ; x > 1
f (x) will be differentiable at x = 1, if
(A) a = – 1, b = 2 (B) a = 1, b = – 2
(C) a = 1, b = 2 (D) None of these
6 The angle between the tangents at any point P and
the line joining P to origin O, where P is a point on
the curve ln (x 2 + y 2 ) = c tan –1 y/x, c is a constant, is
(A) constant
(B) varies as tan –1 (x)
(C) varies as tan –1 (y) (D) None of these
7. Let f (x) = 2x 3 + ax 2 + bx – 3cos 2 x is an increasing
function for all x ∈ R, then
(A) a 2 – 6b – 18 > 0 (B) a 2 – 6b + 18 < 0
(C) a 2 – 3b – 6 < 0 (D) a > 0, b > 0
8. The function y = f(x) is represented parametrically by
x = t 5 – 5t 3 – 20t + 7 and y = 4t 3 – 3t 2 – 18t + 3,
(–2 < t < 2). The minimum of y = f(x) occurs at
(A) t = – 1 (B) t = 0
(C) t = 1/2 (D) t = 3/2
SECTION – II
Multiple Correct Answers Type
Questions 9 to 12 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) may be correct.
Mark your response in OMR sheet against the question
number of that question. +4 marks will be given for
each correct answer and no negative marks for wrong
answer.
9. 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
XtraEdge for IIT-JEE 60
MARCH 2012

(C)
(D)
c θ A − B
= cos
2sin
ab 2
c θ A + B
= cos
2sin
ab 2
10. Which of the following function are defined for all x
(A) sin[x] + cos [x] ([x] denotes greatest integer ≤ x)
(B) sec –1 (1 + sin 2 x)
(C) tan (log x)
(D)
9
+ cos x + cos2x
8
11. If x + |y| = 2y, then y as a function of x is
(A) defined for all real x
(B) continuous at x = 0
(C) differentiable for all x
(D) such that dy/dx = 1/3 for x < 0
12. Two villages A and B are on the same side of a
straight river. A pump set is to be installed by the
river side at a point P. Then if the villages are
situated at a distance c, then
(where a =distance of village A from river side,
b = distance of village B from river side,
c = distance between villages.)
(A) minimum value of PA + PB is
(B) minimum value of PA + PB is
c 2 + 2ab
c 2 + 4ab
2
(C) minimum value of PA + PB is c + ab
(D) The required location must lie on the imaginary
line joining village A and image of village B in
the river.
SECTION – III
Matrix Match 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 the particular question, against
statement B, darken the bubbles corresponding to Q
and R in the OMR. +8 marks will be given for each
correct answer (i.e. +2 marks for each correct row) and
no negative marking for each wrong answer.
13. Match the column :
Column –I
⎧
⎪
sin 2 x
(A) f (x) =
,
⎨ x
⎪
⎩ 0,
x ≠ 0
x = 0
⎧
3
x − 2x,
(B) f (x) = ⎨ 2
⎩x
− 2sin( x),
⎧
2
4 − x ,
(C) f (x) = ⎨ 2
⎩ 5 − x ,
⎧3
− 2x
,
(D) f (x) = ⎨
⎩7
− 6x,
x < 0
x ≥ 0
x < 1
x ≥ 1
x < 0
x ≥ 0
14. Match the column
Column –I
(A) If x 2 + y 2 = 1, then minimum
value of x + y is
(B) If maximum value of
y = acos(x) – 3
1 cos (3x)
occurs, when x = 6
π , then
value of 'a' is
(C) If f (x) = x – 2sin(x), 0 ≤ x ≤
2π is increasing in interval
(aπ, bπ) then, (a + b) is
(D) If equation of tangent to the
curve y = – e –x/2 , where it
crosses the y-axis is
x/p + y/q = 1, then p – q is
Column-II
(P) continuous
(Q) Discontinuous
(R) Differentiable
(S)Non- differentiable
(T) continuous &
differentiable
Column-II
(P) – 3
(Q) − 2
(R) 3
(S) 2
(T) 1
SECTION – IV
Numerical Response Type
This section contains 6 questions (Q.15 to 20). The
answer to each of the questions is a Single-digit integer,
ranging from 0 to 9. The bubble corresponding to the
correct answer is to be darkened in the OMR. +4
marks will be given for each correct answer and no
negative marking for each wrong answer.
15. An urn containing '14' green and '6' pink ball.
K (< 14, 6) balls are drawn and laid a side, their
colour being ignored. Then one more ball is drawn.
Let P(E) be the probability that it is a green ball, then
10 P(E) = ..............
16. Two lines zi – z i + 2 = 0 and z(1+i) + z (1–i) + 2 = 0
intersect at a point P. Find the sum of minimum and
maximum modulus of complex number of a point on
second line which is at a distance of 2 units from
point P.
XtraEdge for IIT-JEE 61
MARCH 2012

⎡3
−2
3 ⎤
⎢ ⎥
17. If A =
⎢
2 1 −1
⎥
. Solve the system of equations
⎢⎣
4 −3
2 ⎥⎦
⎡3
0 3⎤
⎡x⎤
⎡8⎤
⎡2y⎤
⎢ ⎥ ⎢ ⎥
=
⎢ ⎥
+
⎢ ⎥
⎢
2 1 0
⎥ ⎢
y
⎥ ⎢
1
⎥ ⎢
z
⎥
, then find the value of
⎢⎣
4 0 2⎥⎦
⎢⎣
z⎥⎦
⎢⎣
4⎥⎦
⎢⎣
3y⎥⎦
y z
x + +
2 3
18. If the planes x – cy – bz = 0, cx – y + az = 0 and
bx + ay – z = 0 pass through a straight line, then find
the value of a 2 + b 2 + c 2 + 2abc.
19. If the solution of differential equation
2
x 2 d y dy
+ 2x
2 = 12y is y = Ax m + Bx –n then find the
dx dx
value of m + n, if m & n ∈ N.
20. If f(x) = x +
∫ 1 (xy 2 + x 2 y) f (y) dy, then
0
2
Ax + Bx
A+ B
f(x) = ⇒ then the value of
119
260
PHYSICS
is -
SECTION – I
Straight Objective Type
Questions 1 to 8 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1. A wooden plank of length 0.8 m and uniform crosssection
is hinged at one end to the bottom of a tank as
shown in figure. The tank is filled with water upto a
height of 0.4 m. The specific gravity of the plank is
0.5. Find the angle θ that the plank makes with the
horizontal in the equilibrium position-
(A) 3
π
h
(B) 6
π
θ
(C) 4
π
(D) 5
π
2. 5 gm of water of 30ºC and 5 gm of ice at –20ºC are
mixed together in a calorimeter. Then temperature of
the mixture will be (water equivalent of calorimeter is
negligible, specific heat of ice is 0.5 cal/gmºC and
latent heat of ice is 80 cal/gm)
(A) 10ºC (B) 5ºC (C) zero (D) 25ºC
3. Six resistors each of 5Ω are connected as shown in
figure. Reading of ideal ammeter is-
20V
(A) 4A (B) 2A (C) 8A (D) 6A
4. A particle is moving in parabolic path y = x 2 with
constant speed 'u'. Then acceleration of the particle
when it crosses origin, is-
(A) u 2 (B) 2u 2
(C) zero
(D)
5. The arc AB with centre C and infinitely long wire
have linear charge density λ, are kept in the same
plane as shown. The minimum amount of work to be
done to move a point charge q 0 from point A to B
through a circular path AB radius r is equal to-
(A)
(C)
2
λ
+
+
+
+
+
+
+
q0
⎛ 3 ⎞
loge⎜
⎟
2π∈
⎝ 2 ⎠
0
q0λ
2π∈
0
⎛ 2 ⎞
loge⎜
⎟
⎝ 3 ⎠
B
u 2
2
2r C A
r
(B)
(D)
q0λ
2π∈
q
0
A
0
λ
2π∈
⎛ 3 ⎞
loge⎜
⎟
⎝ 2 ⎠
6. A circular ring of radius R with uniform positive
charge density λ per unit length is located in the y-z
plane with its centre at the origin O. A particle of
charge –q 0 is released from x = 3 R on x-axis at t =
0 then kinetic energy of particle when it passes
through origin, is-
(A)
(C)
λq
2∈
0
0
0
q 0 λ
∈
(B)
(D)
q0λ
3∈
0
q0λ
4∈
0
0
XtraEdge for IIT-JEE 62
MARCH 2012

7. A long current carrying wire is bent as shown. The
magnetic field at O is-
i
(A) zero
(B)
90º
O
L
2 0
2µ i
πL
L
(C)
µ 0 i
πL
i
(D)
2µ
0 i
πL
8. In a given series RLC circuit, average power
dissipated in the circuit is-
50Ω 500mH 100µC
11. Figure shows a square loop being pulled out with a
constant speed out of region of uniform magnetic
field. The induced emf in the loop-
B
× × × ×
× × × l ×
×
l
× × ×
v
× × × ×
l
× × × l ×
(A) first increases, then decreases
(B) first decreases, then increases
(C) has a maximum value Bv l 2
(D) has a maximum value 2Bvl
(A) 200 W
(C) 400 W
~
⎛ π ⎞
V = 200sin⎜100t
+ ⎟
⎝ 4 ⎠
(B) 800 W
(D) 100 W
12. The current a certain circuit varies with time as
shown. The peak value of current is i 0 . If i v and i m
represent the virtual (rms) and mean value of current
for a complete cycle respectively. Theni
i 0
SECTION – II
Multiple Correct Answers Type
Questions 9 to 12 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) may be correct.
Mark your response in OMR sheet against the question
number of that question. +4 marks will be given for
each correct answer and no negative marks for wrong
answer.
9. Missile is fired for maximum range at your town
from a place in the enemy country at a distance 'x'
from your town. The missile is first detected at its
half-way point. Then-
(A) the velocity with which the missile was projected
is gx
(B) you have a warning time of
x
2g
(C) the speed of the missile when it was detected is
gx
2
(D) the maximum height attained by the missile is 4
x
10. A particle A of mass 'm' and charge Q moves directly
towards a fixed particle B, which has charge Q. The
speed of A is 'v' when it is far away from B. The
minimum separation between the particles is
proportional to-
(A) Q 2 (B) 1/v 2
(C) 1/v (D) 1/m
O T/2 T
t
2i
(A) i m = 0
i
(B) i m = 0
π
2
i
(C) i rms = 0 i
(D) i rms = 0
3
2
SECTION – III
Matrix Match 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 the particular question, against
statement B, darken the bubbles corresponding to Q
and R in the OMR. +8 marks will be given for each
correct answer (i.e. +2 marks for each correct row) and
no negative marking for each wrong answer.
13. Consider the circuit showing in figure. There are
three switches S 1 , S 2 , S 3 . Match the columns.
3r
S 1
S 2
C
r
V
S 3
2r
XtraEdge for IIT-JEE 63
MARCH 2012

Column -I
Column-II
(A) If S 2 and S 3 are opened and S 1 is (P) CV/4
closed then in steady state,
charge on the capacitor is
(B) If switch S 2 only is closed then (Q) 2CV/5
maximum charge on the
capacitor is
(C) If switch S 3 only is closed then (R) CV/3
maximum charge on the
capacitor is
(D) If all the switches are closed then (S) CV
maximum charge on the
capacitor is-
(T) zero
14. A particle is projected from ground with velocity 40
m/s at an angle θ = 60º with vertical. Match the
quantities given in Column I to the results in Column II
(g = 10 m/s 2 )
Column -I
Column-II
(A) Time to flight
(P) 40 m
(B) Range
(Q) 80 meter
when θ = 0º
(C) Maximum height (R) 80 3 meter
(D) Maximum possible height
(S) 4 second
(T) 17 second
SECTION – IV
Numerical Response Type
This section contains 6 questions (Q.15 to 20). The
answer to each of the questions is a Single-digit integer,
ranging from 0 to 9. The bubble corresponding to the
correct answer is to be darkened in the OMR. +4
marks will be given for each correct answer and no
negative marking for each wrong answer.
15. In a space, equipotential surfaces are shown below.
80V 60V 40V 20V
30º 30º 30º 30º
10cm 10cm 10cm
then electric field in the space is n × 10 2 N/C. Find
value of n.
16. Two electric bulbs of power rating (200 V, 40 W) and
(200 V, 50 W) are connected in series with the main
supply as shown in figure. The voltage (V 0 ) of the
main supply, so that 40 W bulb glows with full
intensity is 60 n volt. Find value of n.
40 W 50 W
~
V 0
17. Two blocks A and B are connected by a massless
string as shown in figure. Friction coefficient of the
inclined plane is 0.5. The mass of block A is 5 kg. If
minimum and maximum values of mass of the block
B for which the block A remains in equilibrium are
m 1 and m 2 then find the value of (m 2 – m 1 ) in kg.
M=5kg
A
37º
18. A split lens has two parts separated by 4 cm and focal
length is f. An object is placed at a distance 3f/2
from C as shown in figure. The distance the images
formed by the two halves (in cm) is.
O
3f/2
C
f
B
4 cm
19. A solid cylinder is rolling without slipping on a plank
which also moving with speed 10 m/s on a horizontal
surface as shown. The speed of centre of cylinder is
20 m/s. The mass of cylinder is 2 kg. The kinetic
energy of the cylinder when observed from ground is
90 nJ. Then find value of n.
20 m/s
10 m/s
20. The ratio of respective acceleration due to gravity at
the surface of two planets having masses in the ratio
'x' and density in the ratio 'y' is (xy 2 ) 1/n then find value
of n.
Cartoon Law of Physics
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 cookie-cutout-perfect hole.
The threat of skunks or matrimony often catalyzes
this reaction.
XtraEdge for IIT-JEE 64
MARCH 2012

MOCK TEST - AIEEE
Time : 3 Hours Total Marks : 360
Instructions :
• There are three parts in question paper A, B, C consisting of chemistry, Physics & Mathematics having
30 questions in each part of equal weightage. Each question is allotted four marks for each correct response.
• 1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the
total score will be made if no response is indicated for an item in the answer sheet.
CHEMISTRY (Part-A)
1. Consider the following statements :
I. Atomic hydrogen is obtained by passing hydrogen
through an electric arc.
II. Hydrogen gas will not reduce heated aluminium
oxide
III. Finely divided palladium adsorbs large volume
of hydrogen gas
IV. Pure nascent hydrogen is best obtained by
reacting Na with C 2 H 5 OH
Which of the above statements is/are correct ?
(1) I alone (2) II alone
(3) I, II and III (4) II, III and IV
2. When ethanal reacts with CH 3 MgBr and C 2 H 5 OH/dry
HCl, the product formed are -
(1) ethyl alcohol and 2-Propanol
(2) ethane and hemi-acetal
(3) 2-propanol and acetal
(4) propane and ethyl acetate
3. How many moles of CH 3 I will react with one mole of
the ethylamine to form a quarternary salt ?
(1) 2 (2) 3 (3) 4 (4) 5
4. 26.8 g of Na 2 SO 4 ⋅nH 2 O contains 12.6 g of water .
The value of ‘n’ s
(1) 1 (2) 10 (3) 6 (4) 7
5. The concentration of oxalic acid is‘x’ mol litre –1 40
mL of this solution reacts with 16 mL of 0.05 M
acidified KMnO 4 . What is the pH of 'x' M oxalic acid
solution (Assume that oxalic acid dissociates
completely)
(1) 1.3 (2) 1.699 (3) 0.05 (4) 2
6. Which pair of species given below produce bakelite?
(1) phenol, methanol
(2) phenol, NaOH
(3) phenol, urea
(4) phenol, formaldehyde
7. A drug that is antipyretic as well as analgesic is-
(1) Chloroquin
(2) Penicillin
(3) Paracetamol
(4) Chloropromazine hydrochloride
8. Which of the following can possibly be used as
analgesic without causing addiction and
modifications ?
(1) Morphine
(2) N-Acetyl-para-aminophenol
(3) Diazepam
(4) Tetrahydrocatenol
9. Na 2 HPO 4 is used to test -
(1) Ca +2 (2) Ba +2
(3) Ni +2 (4) Mg +2
10. 1 mole of N 2 & 3 mole of H 2 filled in one litre bulb
were allowed to reaction when the reaction attained
equilibrium, two third of N 2 converted to NH 3 . If a
hole is then made in the bulb, the mole ratio of the
gases N 2 , H 2 & NH 3 effusing out initially would be
respectively -
(1) 1 : 3 : 4 (2) 28 : 2 : 17
1 1 1 1 3 4
(3) : : (4) : :
28 2 17 28 2 17
11. Which of the following is not a state function -
(1) q + w (2) Q/T
(3) E + PV (4) Q/W
12. The relative rate of acid catalysed dehydration of
following alcohols would be -
CH 3
{P} Ph – CH – CH – CH 3
OH
CH 3
{Q} Ph – CH – CH 2 – CH 2 – OH
XtraEdge for IIT-JEE 65
MARCH 2012

CH 3
{R} Ph – C – CH 2 – CH 3
OH
CH 3
{S} Ph – C – CH 2 – OH
CH 3
(1) R > P > Q > S (2) P > R > S > Q
(3) R > S > P > Q (4) R > S > Q > P
13. The ether O CH 2 when treated with
HI produces –
[I] CH 2 I [II] CH 2 – OH
[III] I [IV] OH
(1) I and III (2) Only II
(3) I and IV (4) Only III
14. 40% of a mixture of 0.2 mole of N 2 and 0.6 mole of
H 2 react to give NH 3 according to the equation, N 2 (g)
+ 3H 2 (g) 2NH 3 (g) at constant temperature and
pressure. Then the ratio of the final volume to the
initial volume of gases are :
(1) 4 : 5 (2) 5 : 4 (3) 7 : 10 (4) 8 : 5
15. ∆Gº for the reaction, X + Y Z is – 4.606 kcal. The
equilibrium constant for the reaction at 227ºC is
(1) 100 (2) 10 (3) 2 (4) 0.01
16. According to Bronsted Lowry concept, the correct
order of strength of bases follows the order :
(1) CH 3 COO – > OH – > Cl –
(2) OH – > CH 3 COO – > Cl –
(3) CH 3 COO – > Cl – > OH –
(4) OH – > Cl – > CH 3 COO –
17. The element having the highest ionization energy has
the outer shell configuration as -
(1) ns 2 np 3 (2) ns 2 np 6
(3) ns 2 (4) ns 2 np 5
18. Which of the following processes is exothermic ?
(1) EA of N (2) IE of O –
(3) EA of Cl (4) IE of Cl
19. Which of the following is not correct ?
(1) XeO 3 has pyramidal shape
(2) The hybrid state of Xe in XeF 4 is sp 3 d 2
(3) In calcium carbide, between carbon atoms one
sigma and two π-bonds are present
(4) In silica(SiO 2 ), one Si atom is attached with two
oxygen atoms
20. Which of the following contains minimum number of
lone pairs around Xe atom ?
(1) XeF 4 (2) XeF 6
(3) XeOF 2 (4) XeF 2
21. The two isomers X and Y with the formula
Cr(H 2 O) 5 CIBr 2 were taken for experiment on
depression in freezing point. It was found that one
mole of X gave depression corresponding to 2 moles
of particles and one mole of Y gave depression due to
3 moles of particles. The structural formulae of X and
Y respectively are -
(1) [Cr(H 2 O) 5 Cl]Br 2 ; [Cr(H 2 O) 4 Br 2 ]Cl⋅H 2 O
(2) [Cr(H 2 O) 5 Cl]Br 2 ; [Cr(H 2 O) 3 CIBr 2 ] ⋅2 H 2 O
(3) [Cr(H 2 O) 5 Br]BrCl; [Cr(H 2 O) 4 CIBr] Br ⋅H 2 O
(4) [Cr(H 2 O) 4 Br 2 ]Cl⋅H 2 O; [Cr(H 2 O) 5 Cl]Br 2
22 Rates of addition of Cl 2 /H 2 O of the following alkenes
are
O
23.
H 2 C=CH 2
(P)
CH 3 –CH 2 –HC=CH 2
(R)
H 2 C=CH–C–H
(Q)
CH 3
CH 3 –C=CH 2
(S)
(1) S > R > P > Q (2) S > P > Q > R
(3) P > Q > R > S (4) P > Q > S > R
(i) Na/ NH 3(l) (i) O 3
(A) Product
(ii) C 2H 5OH (ii) H 2O-Zn
Product will be -
(1) CHO–CHO
(2) CHO–CH 2 –CHO
O O
(3) CH 3 –C–C–CH 3
(4) CHO–CHO & CHO–CH 2 –CHO
24. In the reaction
Br
CH 3 –CH–CH 2 –Br
(i) X mole NaNH2
(ii) C 2H 5Br
CH 3 –C≡C–C 2 H 5
The value of [X] is
(1) One (2) Two (3) Three (4) Four
25. Relate the following compounds -
F
F
F
Br
C
C
C
Cl
R Br
Cl R
R
Cl
Br
C
F
Cl
Br
S
XtraEdge for IIT-JEE 66
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(1) Identical (2) Enantiomer
∞
n
1
(3) Diastereomers (4) Meso
33. If S λ = ∑
= λ
r ∑( − 1) S λ =
r 0
λ= 1
(1) 8a 2 2
2
a
a
(2) 108 (3) 208 (4) a 2 (1) 1 (2) 2
3 3 (3) 4 (4) 8
26. The correct order of acidic strength of following acid
n( n −1)
n( n +1)
(1)
(2)
is -
2
2
COOH COOH COOH COOH
n( n + 2)
(3)
OH HO OH CH 3
2
(4) none of these
k
⎡ ⎛ 2π
⎞ ⎛ 2π
⎞⎤
(I)
(II) (III) OH
⎢cos⎜
⎟ − sin⎜
⎟⎥
(IV) 34. If ⎢ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎡1
0⎤
⎥ =
⎢ ⎛ 2π
⎞ ⎛ 2π
⎞
⎢ ⎥ , Then the least
⎥
(1) II > I > III > IV (2) I > II > III > IV
⎢
sin⎜
⎟ cos⎜
⎟ ⎣0
1 ⎦
⎥
(3) II > I > IV > III (4) II > IV > I > III
⎣ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎦
positive integral value of k is -
27. In the phenomenon of osmosis, the semipermeable (1) 6 (2) 7 (3) 3 (4) 4
membrane allows the passage of
(1) solute particles
35. The number of 2 digit numbers , which are of the
(2) solvent molecules only
form xy with y < x are given by
(3) both solute and solvent
(1) 45 (2) 55 (3) 17 (4) None
(4) none
28. What is the contribution of the atom present at the
edge centre to the cubic unit cell ?
36. Let A, B, C be three independent events such that
1 1 1
P(A) = , P(B) = , P(C) = . Then probability of
3 2 4
(1) 1/2 (2) 1/4
exactly two events occuring out of three events is -
(3) 1/8 (4) 1
(1) 1/2 (2) 1/3
(3) 1/4 (4) none of these
29. In the cell, Zn | Zn 2+ | | Cu 2+ | Cu, the negative
terminal is -
(1) Cu (2) Cu 2+
37. If a variable x takes values x i such that a ≤ x i ≤ b, for
i = 1, 2, ---- n, then -
(3) Zn (4) Zn 2+
(1) a ≤ Var (x) ≤ b (2) a 2 ≤ Var (x) ≤ b 2
2
a
30. Assign double bond configuration of the following -
(3)
4
COOH
≤ Var (x) (4) (b – a) 2 ≥ Var (x)
CH 2 OH
38. A variable line has it's intercepts on the coordinate
NC
e e'
axes e, e' where , are eccentricies of
H 2 N–H 2 C CN
2 2
(1) E, Z (2) Z, Z
hyperbola and it's conjugate hyperbola then the line
always touches the circle x 2 + y 2 = r 2 , where r = ?
(3) E, E (4) Z, E
(1) 1 (2) 2
(3) 3 (4) cannot be decided
MATHEMATICS (Part-B) 39. If it is possible to draw a line which belongs to all the
given family of lines y – 2x + 1 + λ 1 (2y – x – 1) = 0, 3y
– x– 6 + λ 2 (y – 3x + 6) = 0 and
31. f(x) = log 2 25 and g(x) = log
x
x 5 then f(x) = g(x)
ax + y – 2 + λ 3 (6x + ay – a) = 0 , then -
holds for x belonging to -
(1) a = 4 (2) a = 3
(1) R (2) (0, 1)∪ (1, ∞)
(3) a = – 2 (4) a = 2
(3) φ (4) none
40. If (α, 0) is an interior point of ∆ABC formed by the
32. The area enclosed by the parabola y 2 = 4ax between lines x – y = 0, 4x + 3y – 12 = 0 and y + 2 = 0 then
the ordinates x = a and x = 9a is -
integral values of α are
XtraEdge for IIT-JEE 67
MARCH 2012

41. ABCD is a square of unit area. A circle is tangent to
two sides of ABCD and passes through exactly one of
its vertices. The radius of the circle is -
(1) 2 – 2 (2) 2 – 1
1
1
(3) (4)
2 2
42. If A, B, C, D are four points in space satisfying
AB . CD = k[| AD | 2 + | BC | 2 –| AC | 2 –| BD | 2 ] then
the value of k is
(1) 2 (2) 1/3 (3) 1/2 (4) 1
43. Unit vectors a r , b r and c r are coplanar. A unit vector
d r is perpendicular to them . If
r r r r 1
( a × b)
× ( c × d)
= iˆ
1 1
− ˆj
+ kˆ
6 3 3
and the angle between a r and b r is 30º, then c r is -
( i ˆ − 2 ˆj
+ 2kˆ)
( 2ˆ i + ˆj
− kˆ)
(1)
(2)
3
3
(3)
( −2ˆ
i − 2 ˆj
+ kˆ)
3
(4)
( −iˆ
+ 2 ˆj
− 2kˆ)
3
44. If the shortest distance between the lines
x −1
L 1 : =
1
unity then λ is equal to -
y z = and L2 :
−1 2
x +1
2
y z − 3
= = 2 λ
(1) – 20 ± 222 (2) – 20 ± 221
(3) – 20 ± 224 (4) none of these
45. The direction ratios l, m, n of two lines are connected
by the relations l + m + n = 0 and lm = 0 then angle
between them is -
π
π
(1) (2) 3 4
(3) 2
π
46. Let f(x) =
⎪⎩
⎪ ⎨
⎧
5 1/
x
x
λ[
x]
(4) 0
; x < 0
; λ ∈ R then at x = 0
; ≥ 0
(1) f is discontinuous
(2) f is continuous only if λ = 0
(3) f is continuous whatever λ may be
(4) None
47. If R = {(x, y) : x, y ∈ Z} , x 2 + y 2 ≤ 4 is relation in Z,
then D R is -
(1) {–2, –1, 0, 1, 2} (2) {–2, – 1, 0}
(3) {0, 1, 2} (4) None of these
is
48. If A = {1, 2, 3} and B = { 4, 5, 6} then which of the
following sets are relation from A to B ?
(i) R 1 = {(4, 2) (2, 6) (5, 1) (2, 4)}
(ii) R 2 = {(1, 4) (1, 5) (3, 6) (2, 6) (3, 4)}
(iii) R 3 = {(1, 5) (2, 4) (3, 6)}
(iv) R 4 = {(1, 4) (1, 5) (1, 6)}
(1) R 1 , R 2 , R 3 (2) R 1 , R 3 , R 4
(3) R 2 , R 3 , R 4 (4) R 1 , R 2 , R 3 , R 4
49. If the area of the triangle on the complex plane
formed by the complex numbers z, ωz, z + ωz is
3 square units then |z + ωz| equals
100
(1) 5 (2) 1/5 (3) |z| (4) |ωz|
50. f (x) = x
2
4ax − x , (a > 0) then f (x) is -
(1) Increasing in (0, 3a), decreasing in
(– ∞, 0) ∪ (3a, ∞)
(2) Increasing in (a, 4a) decreasing in (5a, ∞)
(3) Increasing in (0, 4a), decreasing in (– ∞, 0)
(4) None of these
51. Let f(x) be a twice differentiable function for all real
values of x & satisfies f (1) = 1, f (2) = 4, f (3) = 9.
then which of the following is true -
(1) f "(x) = 2 for ∀ x ∈ (1, 3)
(2) f "(x) = f ' (x) = 5 for some x ∈ (2, 3)
(3) f "(x) = 3 ∀ x ∈ (2, 3)
(4) f "(x) = 2 for some x ∈ (1, 3)
2
sin x
52.
∫
dx is a -
6
cos x
(1) polynomial of degree 5 in sin x
(2) polynomial of degree 4 in tan x
(3) polynomial of degree 5 in tan x
(4) polynomial of degree 5 in cos x
53. The real value of m for which the substitution,
y = u m will transform the differential equation,
2x 4 dy
y + y 2 = 4x 6 into a homogeneous equation is -
dx
(1) m = 0 (2) m = 1
(3) m = 3/2 (4) no value of m
54. The mirror image of the parabola y 2 = 4x in the
tangent to the parabola at the point (1, 2) is -
(1) (x – 1) 2 = 4(y + 1) (2) (x + 1) 2 = 4(y + 1)
(3) (x + 1) 2 = 4(y – 1) (4) (x – 1) 2 = 4(y – 1)
XtraEdge for IIT-JEE 68
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55. Let f(x) = minimum ({x + 1}, {x – 1}) ∀ x ∈ R,
4
where{.} denotes the fractional part, then
∫
f (x) d(x) =
(1) 9/2 (2) – 2
1
−5
(3) 9 (4) None
Statement based questions : (Q. No. 56 to 60)
Each of these questions contains two statements.
Statement-I and Statement-II. Each of these has four
alternatives choices. You have to select the correct
choice.
(1) If both statement-I and statement-II are true but
statement-II is not the correct explanation of
statement-I.
(2) If both statement-I and statement- II are true,
and statement-II is correct explanation of
Statement-I.
(3) If statement-I is true but statement-II is false.
(4) If statement-I is false but statement-II is true
56. Statement I : Let h(x) =x m/n for x ∈ R, where m & n
are odd No. & 0 < m < n then y = h (x) has no
Extreme.
Statement II : If h' (x) does not change sign in
neighborhood of x = a then x = a is not an Extreme pt.
57. Statement I : If a 2 x 4 + b 2 y 4 = c 6 then maximum
3
c
value of xy is .
2ab
Statement II : For any + ve f(x), AM ≥ GM
58. Let A and B are two independent events in a sample
space.
Statement I : If P(A) = 0.3, P(B) = 0.4, Then
P (A∩ B ) = 0.18.
Statement II : P(A∩ B ) = P(A) – P(A) P(B)
59. Statement I : If p and q two statements then contra
positive of conditional statement ~ ( p ∧ q)
→ q is
( q →~ p∧
~ q)
∧ .
Statement II : If ~ (p ∧ q) = ~ p∨ ~ q.
60. Let C 1 be the circle with centre O 1 (0, 0) and radius 1
and C 2 be the circle with centre O 2 (t, t 2 + 1),
t ∈ R and radius 2
Statement I : Circle C 1 and C 2 always have at least
one common tangent for any value of t.
Statement II : For the two circles O 1 O 2 ≥ |r 1 – r 2 |
where r 1 and r 2 are their radii for any value of t.
PHYSICS (Part-C)
These questions of two statements each, printed as
statement-1 & statement-2. While answering these
Questions you are required to choose any one of the
following four responses.
(1) If both statement-1 and statement-2 are true but
statement-2 is not a correct explanation of the
statement-1.
(2) If both statement-1 & statement-2 are true & the
statement-2 is a correct explanation of the
statement-1.
(3) If statement-1 is true and statement-2 is false.
(4) If statement-1 is false and statement-2 is true.
61. Statement–1 : Average energy in the interference
pattern is the same as it would be if there were no
interference.
Statement–2 : Interference is the only rare
phenomenon in which law of conservation of energy
does not hold good.
62. A light ray is incident upon a prism in minimum
deviation position and suffers a deviation of 34º. If the
shaded half of the prism is knocked of, the ray will –
(1) suffer a deviation of 34º
(2) suffer a deviation of 68º
(3) suffer a deviation of 17º
(4) not come out of the prism
63. A father (60 kg) and his daughter (20 kg) are both at
rest on a frictionless ice pond. The father lifts a 1 kg
ball and throws it to his daughter with horizontal
speed 5 ms –1 ; the daughter catches it. The speeds of
father and daughter are (in ms –1 ) is
(1) 1/12, 5/21 (2) 5/59, 1/4
(3) 5/61, 1/4 (4) 5/59, 5/21
64. A large sheet carries uniform surface charge density
σ. A rod of length 2l has a linear charge density λ on
one half and −λ on the other half. The rod is hinged
at mid-point O and makes angle θ with the normal to
the sheet. The torque experienced by the rod is -
+λ
O
θ
+σ
–λ
XtraEdge for IIT-JEE 69
MARCH 2012

(1)
(3)
σλl
2ε
0
σλl
2ε
0
2
2
cos θ (2)
sin θ (4)
σλl
ε 0
cos 2 θ
σλl sin 2 θ
ε 0
65. A body cools from 60ºC to 50ºC in 10 minutes. If the
room temperature is 25ºC and assuming Newton's
law of cooling to hold good the temperature of the
body at the end of the next 10 minutes will be -
(1) 38.5ºC (2) 40ºC (3) 42.85ºC (4) 45ºC
66. A passenger is at a distance of x from a bus when the
bus begins to move with constant acceleration a.
What is the minimum velocity with which the
passenger should run towards the bus so as to reach it –
(1) 2 ax (2) 2ax (3) ax (4) ax
67. A bus is moving with a speed of 10 ms –1 on a straight
road. A scooterist wishes to overtake the bus in 100 s.
If the bus is at a distance of 1 km from the scooterist,
with what speed should the scooterist chase the bus ?
(1) 40 ms –1 (2) 25 ms –1
(3) 10 ms –1 (4) 20 ms –1
6'8. For a equilateral glass prism the angle of minimum
deviation is 30°, then refractive index of prism is -
1
1 3
(1)
(2) 2 (3) (4)
2
2 2
69. A zener diode is to be used as voltage regulator.
Identify the correct set up -
(1)
(2)
⊕
⊕
Θ
R S
R S
R L
R L
Passage : (Q. No. 70 to 72)
+ –
12 volt
Bulb 2
Bulb 1
A 12 volt battery is connected in two light bulbs, as
shown in figure. Light bulb 1 has resistance 3Ω while
light bulb 2 has resistance 6Ω. The battery has
essentially no internal resistance and all the wires are
essentially resistanceless too. When a light bulb is
unscrewed, no current flows through that branch of
the circuit. For instance, if light bulb 2 is unscrewed,
current flows only around the lower loop of the
circuit, which consists of the battery and light bulb 1.
When two resistance are joined in series, their
equivalent resistances R eq. = R 1 + R 2 but when two
resistances are wired in parallel. Their net resistance
is given by:
1 1 1
= +
R eq R R
.
70. When bulb 1 is screwed in, but bulb 2 is unscrewed, the
power generated in bulb 1 is -
(1) 4 watt (2) 12 watt
(3) 36 watt (4) 48 watt
71. Bulb 2 is now screwed in, as a result, bulb 1 -
(1) turns off
(2) becomes dimmer
(3) stays about the same brightness
(4) becomes brighter
1
2
Θ
⊕
(3)
Θ
⊕
(4)
Θ
R S
R S
R L
R L
72. When both light bulbs are screwed in, the current
through the battery is -
(1) 1.2 ampere (2) 2 ampere
(3) 4 ampere (4) 6 ampere
73. String 1 has twice the length, twice the radius, twice
the tension and twice the density of another string 2.
The relation between the fundamental frequencies of
1 and 2 is :
(1) f 1 = 2f 2 (2) f 1 = 4f 2
(3) f 2 = 4f 1 (4) f 1 = f 2
XtraEdge for IIT-JEE 70
MARCH 2012

74. When a source of sound of frequency f crosses a
stationary observer with a speed v s (

85. If the space between the lenses in the lens
combination shown were filled with water, what
would happen to the focal length and power of the
lens combination ?
MOTIVATION
Focal length Power
(1) Decreased increased
(2) Decreased unchanged
(3) Increased unchanged
(4) Increased decreased
86. If two coils have self-inductances L 1 and L 2 , the
coefficient of mutual induction will be –
(1) M ∝ L 1 L 2
(2) M ∝ L1
L 2
(3) M ∝
L
L
2
1
(4) None of these
87. Vapour pressure at any temperature is equal to saturated
vapour pressure :
(1) At that temperature (2) At dew point
(3) At boiling point (4) At freezing point
88. The work done in turning a magnet of magnetic moment
M by an angle of 90º from the meridian is n times the
corresponding work done to turn it through an angle of
60º. The value of n is given by –
(1) 1 (2) 1/4
(3) 4 (4) 2
89. As compared to ordinary diode an zener diode is -
(1) also connected in F.B
(2) also connected in F.B as well as RB
(3) always connected in RB
(4) exactly same
90. There are two identical concentric coils X and Y with
their planes at right angles to each other. The coil X lies
in the horizontal plane while coil Y lies in the vertical
plane. If the coil X carries a current of
1 A then what value of current in coil Y be passed so
that the resultant field at the centre of the coils just
balances the earth's magnetic field of 10 –5 tesla inclined
at 30º with the vertical ?
(1) 1 A (2) 3 A
(3) 2 A (4) (1/ 3 )A
• Pull the string, and it will follow wherever
you wish. Push it, and it will go nowhere at
all.
• Be the change that you want to see in the
world.
• Efficiency is doing things right; effectiveness
is doing the right things.
• Formula for success: under promise and over
deliver.
• A life spent making mistakes is not only more
honorable, but more useful than a life spent
doing nothing.
• Discovery consists of seeing what everybody
has seen and thinking what nobody else has
thought.
• The best way to teach people is by telling a
story.
• If you'll not settle for anything less than your
best, you will be amazed at what you can
accomplish in your lives.
• I had to pick myself up and get on with it, do
it all over again, only even better this time.
• Improvement begins with I.
• Success depends upon previous preparation,
and without such preparation there is sure to
be failure.
• The man of virtue makes the difficulty to be
overcome his first business, and success only
a subsequent consideration.
• As a general rule the most successful man in
life is the man who has the best information.
• The secret of success is constancy to purpose.
• One secret of success in life is for a man to be
ready for his opportunity when it comes.
XtraEdge for IIT-JEE 72
MARCH 2012

XtraEdge for IIT-JEE 73
MARCH 2012

MOCK TEST – BIT-SAT
Time : 3 Hours Total Marks : 450
Instructions :
• This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical
Reasoning (10) & English (15). There is Negative Marking
• Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.
• +3 Marks for each correct & – 1 Mark for the incorrect answer.
PHYSICS
1. A particle moving along x-axis has acceleration f, at
⎛ t ⎞
time t, given by f = f 0 ⎜1 – ⎟ , Where f 0 and T are
⎝ T ⎠
constants. The particle at t = 0 has zero velocity. In
the time interval between t = 0 and the instant when
f = 0, the particle's velocity (v x ) is -
(A) 2
1
f0 T 2 (B) f 0 T 2
(C) 2
1
f0 T
(D) f 0 T
2. Two forces P and Q acting at a point are such that if
P is reversed, the direction of the resultant is turned
through 90°. Then -
(A) P = Q
(B) P = 2Q
Q
(C) P = 2
(D) No relation between P and Q
3. Three particles A, B and C of equal mass, move with
equal speed v along the medians of an equilateral
triangle as shown in Fig. They collide at the centroid
G of the triangle. After collision, A comes to rest and
B retraces its path with speed v. What is the speed of
C after collision ?
A
(A) 0
(C) v
B
G
(B) 2
v
(D) 2v
C
4. A man throws bricks to a height of 12 m where they
reach with a speed of 12 ms –1 . If he throws the bricks
such that they just reach that height, what percentage
of energy will be saved ?
(A) 9 % (B) 19 %
(C) 38 % (D) 46 %
5. If the Earth shrinks to half of the present radius,
without any change in mass, then the duration of day
and night becomes -
(A) 24 hours (B) 12 hours
(C) 6 hours (D) 3 hours
6. Two white dots are 1 mm apart on a black paper.
They are viewed by eye of pupil diameter 3 mm.
What is the maximum distance at which these dots
can be resolved by the eye ? (λ =500 nm)
(A) 1 m (B) 5 m (C) 3 m (D) 6 m
7. An object is placed at a distance of 10 cm from a
concave mirror of radius of curvature 0.6 m. Which
of the following statements is incorrect ?
(A) The image is formed at a distance of 15 cm from
the mirror
(B) The image formed is real
(C) The image is 1.5 times the size of the object
(D) The image formed is virtual and erect
8. Two bodies of masses 1 kg and 3 kg have position
vectors î + 2 ĵ +kˆ and – 3 î – 2 ĵ + kˆ , respectively.
The centre of mass of this system has a position
vector -
(A) – î + ĵ + kˆ (B) – 2 î + 2 kˆ
(C) – 2 î – ĵ + kˆ (D) 2 î – ĵ – 2 kˆ
9. A perfect gas at 27°C is heated at constant pressure
so as to triple its volume. The temperature of the gas
will be -
(A) 81°C (B) 900°C
(C) 627°C (D) 450°C
XtraEdge for IIT-JEE 74
MARCH 2012

10. An ideal gas of mass m in a state A goes to another
state B via three different processes as shown in
figure. If Q 1 , Q 2 and Q 3 denote the heat absorbed by
the gas along the three paths, then -
P A
1 2 3
B
V
(A) Q 1 < Q 2 < Q 3 (B) Q 1 < Q 2 = Q 3
(C) Q 1 = Q 2 > Q 3 (D) Q 1 > Q 2 > Q 3
11. Two sound waves (expressed in CGS units) given by
2 π
2 π
y 1 = 0.3 sin (vt – x) and y2 = 0.4 sin (vt – x +
λ λ
θ) interfere. The resultant amplitude at a place where
phase difference is π/2 will be -
(A) 0.7 cm
(B) 0.1 cm
1
(C) 0.5 cm (D) 7 cm 10
12. A tuning fork of frequency 100 when sounded
together with another tuning fork of unknown
frequency produces 2 beats per second. On loading
the tuning fork whose frequency is not known and
sounded together with a tuning fork of frequency 100
produces one beat, then the frequency of the other
tuning fork is -
(A) 102 (B) 98 (C) 99 (D) 101
13. A current of 2 A flows in an electric circuit as shown
in figure. The potential difference
(V R – V S ), in volts (V R and V S are potentials at R and S
respectively) is
R
2A
P
3Ω
7Ω
7Ω
3Ω
Q
2A
S
(A) – 4 (B) + 2 (C) + 4 (D) – 2
14. When a battery connected across a resistor of 16 Ω,
the voltage across the resister is 12 V. When the same
battery is connected across a resistor of 10 Ω, voltage
across it is 11V. The internal resistance of the battery
(in ohm) is
10
(A) 7
(B)
20
7
(C) 7
25
30
(D) 7
15. In a galvanometer 5% of the total current in the
circuit passes through it. If the resistance of the
galvanometer is G, the shunt resistance S connected
to the galvanometer is -
G
G
(A) 19 G (B) (C) 20 G (D) 19 20
16 Two concentric coils of 10 turns each are placed in
the same plane. Their radii are 20 cm and
40 cm and carry 0.2 A and 0.3 A current respectively
in opposite directions. The magnetic induction (in
tesla) at the centre is -
(A) 4
3
µ0 (B) 4
5
µ0
(C) 4
7
µ0 (D) 4
9
µ0
17. The number of turns in primary and secondary coils
of a transformer is 50 and 200 respectively. If the
current in the primary coil is 4 A, then the current in
the secondary coil is
(A) 1 A
(B) 2 A
(C) 4 A
(D) 5 A
18. Which of the following statements is not correct
when a junction diode is in forward bias ?
(A) The width of depletion region decreases
(B) Free electrons on n-side will move towards the
junction.
(C) Holes on p-side move towards the junction.
(D) Electron on n-side and holes on p-side will move
away from junction.
19. The displacement of a charge Q in the electric field
E r = e iˆ
1 + e ˆ
2 j + e3 k ˆ is r = aˆ i + bˆ j . The work done
is -
(A) Q (ae 1 + be 2 )
2 2
1 2)
(B) Q ( ae ) + ( be
(C) Q ( e + e a + b
1 2)
(D) Q ( e + e )( a + )
2
2
2 2
1 2 b
20. An electric line of force in the xy plane is given by
equation x 2 + y 2 = 1. A particle with unit positive
charge, initially at rest at the point x = 1, y = 0 in the
xy plane
(A) not move at all
(B) will move along straight line
(C) will move along the circular line of force
(D) Information is insufficient to draw any conclusion
XtraEdge for IIT-JEE 75
MARCH 2012

21. The relation between voltage sensitivity (σ v ) and
current sensitivity (σ i ) of a moving coil galvanometer
is (resistance of galvanometer is G).
σ i
σ
(A) =
v
σv (B) = σi
G G
(C)
G
v
σ = σ i (D)
G
i
σ = σ v
22. An inductor of 2H and a resistance of 10Ω are
connected in series with a battery of 5V. The initial
rate of change of current is
(A) 0.5 A/s
(B) 2.0 A/s
(C) 2.5 A/s
(D) 0.25 A/s
23. A and B are two radioactive substances whose halflives
are 1 and 2 years respectively. Initially 10 g of
A and 1 g of B is taken. The time (approximate) after
which they will have same quantity remaining is -
(A) 6.62 year (B) 5 year
(C) 3.2 year (D) 7 year
24. In the circuit, the potential difference across PQ will
be nearest to -
100Ω
48V
100Ω
(A) 9.6 V
(C) 4.8 V
80Ω
20Ω
(B) 6.6 V
(D) 3.2 V
25. Radiations of intensity 0.5 W/m 2 are striking a metal
plate. The pressure on the plate is
(A) 0.166 × 10 –8 N/m 2 (B) 0.332 × 10 –8 N/m 2
(C) 0.111 × 10 –8 N/m 2 (D) 0.083 × 10 –8 N/m 2
26. A cell of constant emf first connected to a resistance
R 1 and then connected to a resistance R 2 . If power
delivered in both cases is same then the internal
resistance of the cell is :
(A) R 1R2
(B)
(C)
R1 − R 2
2
(D)
R1
R2
R 1 + R 2
2
27. In a magnetic field of 0.05 T area of coil changes
from 101 cm 2 to 100 cm 2 without changing the
resistance which is 2Ω. The amount of charge that
flow during this period is
(A) 2.5 × 10 –6 C (B) 2 × 10 –6 C
(C) 10 –6 C
(D) 8 × 10 –6 C
P
Q
28. A dielectric is introduced in a charged and isolated
parallel plate capacitor, which of the following
remains unchanged ?
(A) Energy
(B) Charge
(C) Electric field (D) Potential difference
29. Positively charged particles are projected into a
magnetic field. If the direction of the magnetic field
is along the direction of motion of the charge
particles, the particles get :
(A) Accelerated
(B) Decelerated
(C) Deflected
(D) no changed in velocity
30. Fusion reaction takes place at high temperature
because :
(A) KE is high enough to overcome repulsion
between nuclei
(B) Nuclei are most stable at this temperature
(C) Nuclei are unstable at this temperature
(D) None of the above
31. Among the following properties describing
diamagnetism identify the property that is wrongly
stated :
(A) Diamagnetic material do not have permanent
magnetic moment
(B) Diamagnetism is explained in terms of
electromagnetism induction
(C) Diamagnetic materials have a small positive
susceptibility
(D) The magnetic moment of individual electrons
neutralize each other.
32. Electron of mass m and change q is travelling with a
speed v along a circular path of radius r at right
angles to a uniform magnetic field of intensity B. If
the speed of the electron is doubled and the magnetic
field is halved the resulting path would have a radius.
(A) 2r (B) 4r (C) r/4 (D) r/2
33. The density of water at 0°C is 0.998 g/cc. While at
4°C it is 1 g/cc. The average coefficient of volume
expansion of water in the temperature range 0°C to
4°C is -
(A) 5 × 10 – 4 /°C (B) – 5 × 10 – 4 /°C
(C) 6 × 10 – 4 /°C (D) – 6 × 10 – 4 /°C
34. A transverse sinusoidal wave moves along a string in
the positive x-direction at a speed of 10 cm/ s. The
wavelength of the wave is 0.5 m and its amplitude is
10 cm. At a particular time t, the snap-shot of the
wave is shown in the figure. The velocity of point P
when its displacement is 5 cm, is -
XtraEdge for IIT-JEE 76
MARCH 2012

y
P
x
(A)
(C)
ML 2
K
2
KL
2M
(B) zero
(D)
MK L
(A)
(C)
3π
50
ĵ
3π
m/s (B) – ĵ m/s
50
3π 3π
î m/s (D) – î m/s
50
50
40. Given : → A . → B = 0 and → A × → C = 0. The angle
between → B and C → is -
(A) 0° (B) 90°
(C) 180° (D) 270°
35. Which of the following diagrams is a correct
presentation of deviation and dispersion of light by
prism ?
(A)
(C)
R
V
R
V
(B)
(D)
36. A diver is 10 m below the surface of water. The
approximate pressure experienced by the diver is -
(A) 10 5 Pa
(B) 2 × 10 5 Pa
(C) 3 × 10 5 Pa (D) 4 × 10 5 Pa
37. If Earth describes an orbit round the Sun of double its
present radius, the year on Earth will be of -
(A) 365 days (B) 365 × 2 × 2 days
365
(C) days (D) 365 × 4 days
2
38. The moment of inertia of a thin uniform rod of length
L and mass M about an axis passing through a point
at a distance of 3
L from one of its ends and
perpendicular to the rod is -
2
2
ML
ML
(A)
(B)
3
6
2
ML
(C)
9
2
ML
(D)
12
39. The block of mass M moving on the frictionless
horizontal surface collides with the spring of spring
constant K and compresses it by length L. The
maximum momentum of the block after collision is -
M
R
V
R
V
CHEMISTRY
1. The mass of CaCO 3 formed by passing CO 2 gas
through 50 mL of 0.5 M Ca(OH) 2 solution is -
(A) 10 g (B) 2 g (C) 2.5 (D) 5 g
2. 2g of hydrogen diffuse out from a container in 10
min. How many gram of chlorine will diffuse out
from the same container under similar conditions ?
(A) 2× 71 (B)
(C)
71
2
2
71
(D) 71
3. Total volume of atoms present in face-centred cubic
unit cell of a metal is (r = atomic radius)
20
(A) πr
3
3
12
(C) πr
3
3
(B)
24 πr
3
3
16
(D) πr
3
3
4. A ball of 100 g mass is thrown with a velocity of 100
ms -1 . The wavelength of the de Broglie wave
associated with the ball is about -
(A) 6.63 × 10 –35 m (B) 6.63 × 10 –30 m
(C) 6.63 × 10 –35 cm (D) 6.63 × 10 –33 m
5. The unicertainty in the position of an electron
moving with a velocity of 1.0 × 10 4 cm s –1 (accurate
up to 0.011%) will be -
(A) 1.92 cm (B) 7.68 cm
(C) 0.528 cm (D) 3.8 cm
6. ∆Hº for a reaction,
F 2 + 2HCl —→ 2HF + Cl 2
is given to be –352.18 kJ. If
0
∆ H f for HF is –
268.3 kJ mol –1 the ∆H 0 f of HCl would be -
(A) – 22 kJ mol –1 (B) 88.0 kJ mol –1
(C) – 91.9 kJ mol –1 (D) – 183.8 kJ mol –1
XtraEdge for IIT-JEE 77
MARCH 2012

7. If CH 3 COOH(aq) + OH¯ (aq) —→
CH 3 COO¯ + H 2 O + q 1 ,
H + + OH¯ —→ H 2 O (l) + q 2
Then enthalpy change for the reaction,
CH 3 COOH —→ CH 3 COO¯ + H + is
(A) q 1 – q 2 (B) q 1 + q 2
(C) q 2 – q 1 (D) q 1 / q 2
8. In which of the following reaction K p > K c
(A) N 2 + 3H 2 2NH 3
(B) H 2 + I 2 2HI
(C) PCl 3 + Cl 2 PCl 5
(D) 2SO 3 O 2 + 2SO 2
9. If the equilibrium constant for the reaction –
2AB A 2 + B 2 is 49, what is the value of
equilibrium constant for
AB
1 1
A2 + B2
2 2
(A) 49 (B) 2401 (C) 7 (D) 0.02
10. An equimolar solution of CH 3 COOH and
CH 3 COONa has a pH of 6. The value of K α for acetic
acid is -
(A) 10 6 (B) 1 × 10 –6
(C) 2 × 10 –6 (D) cannot be predicted
11. What is the solubility of Al(OH) 3 , K sp =1 × 10 –33 , in a
solution having pH = 4 ?
(A) 10 –3 M
(B) 10 –6 M
(C) 10 –4 M
(D) 10 –10 M
12. Which of the following graph is for a second order
reaction ?
(A)
(C)
Rate
Rate
(A) 2
(A) 2
(B)
(D)
Rate
Rate
(A) 2
(A) 2
13. For an elementary process
2X + Y —® Z + W
(A) 2 (B) 1
(C) 3
(D) unpredictable.
14. The rise in boiling point of a solution containing 1.8
g of glucose in 100 g of solvent is 0.1°C. The molal
elevation constant of the liquid is -
(A) 0.01 K/m (B) 0.1 K/m
(C) 1 K/m
(D) 10 K/m.
15. The value of E + = 0.76 V and that of
° 2
Zn / Zn
E° 2+ = – 0.41 V. The E°
Fe / Fe
cell of the cell with net
cell reaction
Zn + Fe 2+ ⎯→ Zn 2+ + Fe is
(A) – 0.35 V (B) – 1.17 V
(C) + 1.17 V (D) + 0.35 V.
16. The hydrogen electrode can exhibit electrode
potential > 0 if
(A) H 2 is bubbled through the solution at 2 atm.
pressure
(B) concentration of H + ion in solution is increased
(C) concentration of H + ions in solution is decreased
(D) concentration of H + ions is decreased and
simultaneously pressure of H 2 gas is increased.
17. + Br 2 ⎯⎯→
⎯
4 A, A will have
18.
configuration :
(A)
Br
Br
(C) both (a) and (b)
(A) meso diol
(C) both (a) and (b)
OsO 4
⎯
H 2O2
(B)
CCl
Br
Br
(D) none of these
⎯ ⎯ → A, A is -
(B) racemic diol
(D) none of the above
19. Following compound is treated with NBS
Compound formed A is -
CH 2 CH=CH 2 + NBS → A
(A) CHCH=CH 2
(B)
Br
CH = CHCH 2 Br
XtraEdge for IIT-JEE 78
MARCH 2012

(C) CH 2 CH=CH 2
(D)
Br
Br
Br
CH 2 CH = CH 2
20. Bicyclo (1, 1, 0) butane is -
(A)
(B)
27. Pinacol is :
(A) 3-methylbutane-2-ol
(B) 2,3-dimethyl-2,3-butanediol
(C) 2,3-dimethyl-2-propanone
(D) none of the above
28. Aldol condensation will not take place in :
(A) HCHO
(B) CH 3 CH 2 CHO
(C) CH 3 CHO (D) CH 3 COCH 3
29. IUPAC name of the following compound is :
(C)
(D)
21. Grignard reagent reacts with HCHO to produce:
(A) secondary alcohol
(B) anhydride
(C) and acid
(D) primary alcohol
22. Cyanohydrin of which of the following forms lactic
acid :
(A) HCHO
(B) CH 3 CHO
(C) CH 3 CH 2 CHO (D) CH 3 COCH 3
23. Isopropyl bromide on Wurtz reaction gives :
(A) hexane
(B) propane
(C) 2, 3-dimethylbutane
(D) neo-hexane
24. On heating with oxalic acid at 110°C, glycerine gives:
(A) glyceryl trioxalate (B) formic acid
(C) glyceryl dioxalate (D) none of the above
25. The compound, whose stereo-chemical formula is
written below, exhibits x geometrical isomers and y
optical isomers.
CH 3 H OH
C=C
H CH 2 –CH 2 –C–CH 3
H
The values of x and y are :
(A) 4 and 4 (B) 2 and 2
(C) 2 and 4 (D) 4 and 2
26. Which one of the following product is formed when
calcium salt of adipic acid is heated ?
CH 2 –CH 2
(A) O
CH2 –CH 2
CH 2 –CH 2
(B)
CH2 –CH 2
CH 2 CH 2 CO
(C)
CH2 CH 2 CO
C = O
CH 2 CH 2 COOH
(D)
CH2 CH 2 COOH
C = O
H 3 C
CH 3
(A) 3,5-dimethylcyclohexene
(B) 3,5-dimethyl -1-cyclohexene
(C) 1,5-dimethyl-5-cyclohexene
(D) 1,3-dimethyl-5-cyclohexene
30. The ions O 2– , F – , Na + , Mg 2+ and Al 3+ are
isoelectronic. Their ionic radii show.
(A) A decrease from O 2– to F – and then increase from
Na + to Al 3+
(B) A significant increase from O 2– to Al 3+
(C) A significant decrease from O 2– to Al 3+
(D) An increase from O 2– to F – and then decrease
from Na + to Al 3+
31. Which of the following values in electron volt per
atom represent the first ionisation energies of oxygen
and nitrogen atom, respectively ?
(A) 14.6, 13.6 (B) 13.6, 14.6
(C) 13.6, 13.6 (D) 14.6, 14.6
32. Which of the following is not involved in any
diagonal relationship ?
(A) C (B) B (C) Al (D) Si
33. A metal M readily forms water soluble sulphate
MSO 4 , water insoluble hydroxide M(OH) 2 and oxide
MO which becomes insert on heating. The hydroxide
is soluble in NaOH. The M is -
(A) Be
(B) Mg
(C) Ca
(D) Sr
34. Plaster of paris is -
(A) CaSO 4
(C) 2CaSO 4 .H 2 O
(B) CaSO 4 .H 2 O
(D) CaSO 4 .2H 2 O
35. In diborane -
(A) 4-bridged hydrogens and two terminal hydrogens
are present
(B) 2-bridged hydrogens and four terminal
hydrogens are present
(C) 3-bridged hydrogens and three terminal
hydrogens are present
(D) None of these
XtraEdge for IIT-JEE 79
MARCH 2012

36 Which of the following statements is correct ?
(A) BCl 3 and AlCl 3 are both Lewis acids and BCl 3 is
stronger than AlCl 3
(B) BCl 3 and AlCl 3 are both Lewis acids and AlCl 3 is
stronger than BCl 3
(C) BCl 3 and AlCl 3 are both equally strong Lewis
acids
(D) Both BCl 3 and AlCl 3 are not Lewis acids
37. When sodium thiosulphate solution is made to react
with copper sulphate solution, a complex compound
is formed. What is that ?
(A) Na 4 [Cu 3 (S 2 O 3 ) 5 ] (B) Na 2 [Cu 6 (S 2 O 3 ) 4 ]
(C) Na 4 [Cu 6 (S 2 O 3 ) 5 ] (D) Na 4 [Cu 3 (S 2 O 3 ) 4 ]
38. Which one of the following does not contain zinc ?
(A) Brass
(B) German silver
(C) Gun metal (D) Bell metal
39. Stability constants for some copper complexes are
given below :
Cu +2 , + 4NH 3 [Cu(NH 3 ) 4 ] +2 K = 4.5 × 10 11
Cu +2 , + 4CN – [Cu(CN) 4 ] –2 K = 2.0 × 10 27
Cu +2 , + 2en [Cu(en) 2 ] +2 K = 9.5 × 10 15
Cu +2 , + 4H 2 O [Cu(H 2 O) 4 ] +2 K = 9.5 × 10 8
Which is the strongest ligand ?
(A) NH 3 (B) CN –
(C) en
(D) H 2 O
40. The pair [Co(NH 3 ) 5 NO 3 ]SO 4
and [Co(NH 3 ) 5 SO 4 ] NO 3 will exhibit
(A) Hydrate isomerism
(B) Linkage isomerism
(C) Ionisation isomerism
(D) Coordinate isomerism
MATHEMATICS
1. The point on the line 3x + 4y = 5 which is equidistant
from (1, 2) and (3, 4) is
(A) (7, – 4) (B) (15, – 10)
(C) (1/7, 8/7) (D) (0, 5/4)
2. The transformed equation of 3x 2 + 3y 2 + 2xy = 2
when the coordinate axes are rotated through an
angle of 45º, is
(A) x 2 + 2y 2 = 1 (B) 2x 2 + y 2 = 1
(C) x 2 + y 2 = 1 (D) x 2 + 3y 2 = 1
3. The length of the common chord of the ellipse
2
2
( x –1) ( y – 2)
+ = 1 and the circle (x – 1) 2 + (y–2 ) 2 = 1
9 4
is
(A) 0 (B) 3
(C) 4 (D) 5
4. The centres of a set of circles, each of radius 3, lie on
the circle x 2 + y 2 = 25. The locus of any point in the
set is -
(A) 4 ≤ x 2 + y 2 ≤ 64 (B) x 2 + y 2 ≤ 25
(C) x 2 + y 2 ≥ 25 (D) 3 ≤ x 2 + y 2 ≤ 9
5. The pairs of straight lines x 2 – 3xy + 2y 2 = 0 and
x 2 – 3xy + 2y 2 + x – 2 = 0 form a
(A) square but not rhombus
(B) rhombus
(C) parallelogram
(D) rectangle but not a square
6. The distance between the foci of the hyperbola
x 2 – 3y 2 – 4x – 6y – 11 = 0 is
(A) 4 (B) 6
(C) 8 (D) 10
7. If θ is the acute angle of intersection at a real point of
intersection of circle x 2 + y 2 = 5 and the parabola
y 2 = 4x, then tan θ is equal to -
(A) 1 (B) 3
(C) 3 (D) 1 / 3
8. The tangents from a point (2 2 , 1) to the hyperbola
16x 2 – 25y 2 = 400 include an angle equal to
(A) π/2 (B) π/4 (C) π (D) π/3
9. The slope of a common tangent to the ellipse
2 2
x y
2 +
2
= 1 and a concentric circle of radius ‘r’
a b
is -
2 2
2 2
(A) tan –1 r – b
r – b
(B)
2 2
2 2
a – r
a – r
2
2
r – b
(C)
2 2
a – r
(D)
a
r
2
2
– r
– b
10. If f : R → R and g : R → R are defined by f(x) = | x |
and g (x) = [x – 3] for x ∈ R, then
⎧ 8 8⎫
⎨g ( f ( x)) : – < x < ⎬ is equal to -
⎩ 5 5⎭
(A) {0, 1} (B) {1, 2}
(C) {–3, –2} (D) {2, 3}
11. If f : R → R is defined by
⎧cos3x
– cos x
⎪
, for x ≠ 0
f (x) = ⎨
2
x
⎪⎩ λ , for x = 0
and if f is continuous at x = 0, then λ is equal to -
(A) – 2 (B) – 4 (C) – 6 (D) – 8
2
2
XtraEdge for IIT-JEE 80
MARCH 2012

12. The solution of the differential equation
xy 2 dy – (x 3 + y 3 ) dx = 0 is
(A) y 3 = 3x 3 + c (B) y 3 = 3x 3 log (cx)
(C) y 3 = 3x 3 + log (cx) (D) y 3 + 3x 3 = log (cx)
13. If
∫ x e (1 + x).sec 2 (xe x ) dx = f (x) + constant, then
f (x) is equal to -
(A) cos (xe x ) (B) sin (xe x )
(C) 2 tan –1 (x) (D) tan (x e x )
14. If f : R → R is defined by f (x) = [x – 3] + | x – 4 | for
x ∈ R, then lim f (x) is equal to where [.] is G.I.F. -
15.
–
x→3
(A) – 2 (B) – 1 (C) 0 (D) 1
⎧ 2x
–1 ⎫
⎨x
∈ R :
∈ R⎬
equals
3 2
⎩ x + 4x
+ 3x
⎭
(A) R – {0} (B) R – {0, 1, 3}
(C) R – {0, –1, – 3}
⎧ 1 ⎫
(D) R – ⎨0
, –1, – 3,+ ⎬
⎩ 2 ⎭
d ⎡
⎛ –1⎞⎤
16. ⎢ tan – 1 x
a x + blog⎜
⎟⎥ ⎣
⎝ x =
dx
+ 1⎠
⎦
4
x
1
–1
⇒ a – 2b is equal to -
(A) 1 (B) – 1 (C) 0 (D) 2
17. The function f (x) = x 3 + ax 2 + bx + c, a 2 ≤ 3b has
(A) one maximum value
(B) one minimum value
(C) no extreme value
(D) one maximum and one minimum value
18. Area of the region satisfying x ≤ 2, y ≤ | x | and x ≥ 0 is
(A) 4 sq. unit (B) 1 sq. unit
(C) 2 sq. unit (D) None of these
∫ 1 0
2
x
∫ 1 0
19. If I 1 = 2 dx , I 2 = 2 dx , I 3 = 2 dx and
I 4 =
∫ 2
1
3
x
2 dx , then
3
x
(A) I 3 > I 4 (B) I 3 = I 4
(C) I 1 > I 2 (D) I 2 > I 1
∫ 2
1
20. Let A = [–1, 1] and f : A → A be defined as
f (x) = x | x | for all x ∈ A, then f (x) is -
(A) many-one into function
(B) one-one into function
(C) many-one onto function
(D) one-one onto function
2
x
21. The length of the subtangent at (2, 2) to the curve
x 5 = 2y 4 is
5 8 2
5
(A) (B) (C) (D) 2 5 5 8
⎧
⎛ θ ⎞⎫
22. If x = a ⎨cosθ + log tan⎜
⎟⎬
and y = a sin θ, then
⎩
⎝ 2 ⎠⎭
dy is equal to -
dx
(A) cot θ
(B) tan θ
(C) sin θ
(D) cos θ
23. The value of the expression
⎛ 1 ⎞⎛
1 ⎞ ⎛ 1 ⎞⎛
1 ⎞
2 ⎜1+
⎟⎜1+
⎟ + 3 ⎜2
+ ⎟⎜2
+ ⎟ +
2
2
⎝ ω ⎠⎝
ω ⎠ ⎝ ω ⎠⎝
ω ⎠
⎛ 1 ⎞⎛
1 ⎞
⎛ 1 ⎞⎛
1 ⎞
4 ⎜3
+ ⎟⎜3
+ ⎟ + ..... + ( n + 1) ⎜n
+ ⎟⎜n
+ ⎟ ,
2
2
⎝ ω ⎠⎝
ω ⎠
⎝ ω ⎠⎝
ω ⎠
where ω is an imaginary cube root of unity, is
n( n
2 + 2)
n( n
2 – 2)
(A)
(B)
3
3
(C)
2
a2a3
24. If =
a1a4
are in
(A) A.P.
(C) H.P.
2
n ( n + 1) + 4n
4
a2
+ a3
⎛ a
=
a1
+ a
⎜
2 – a
3
4 ⎝ a1
– a
2
(D) None of these
3
4
⎞
⎟ , then a 1 , a 2 , a 3 , a 4
⎠
(B) G.P.
(D) None of these
x –1
25. If f (x) = for every real number x then the
2
x + 1
minimum value of f
(A) does not exist because f is unbounded
(B) is not attained even though f is bounded
(C) is equal to 1
(D) is equal to – 1
26. If x + y and y + 3x are two factors of the expression
λx 3 – µx 2 y + xy 2 + y 3 , then the third factor is
(A) y + 3x
(B) y – 3x
(C) y – x
(D) None of these
27. The number of ways in which 30 marks can be
allotted to 8 questions if each question carries at least
2 marks, is
(A) 115280 (B) 117280
(C) 116280 (D) None of these
28. The numerically largest term in the binomial
expansion of (4 – 3x) 7 , when x = 3
2 is -
(A) 46016 (B) 66016
(C) 86016
(D) None of these
XtraEdge for IIT-JEE 81
MARCH 2012

2 4n
29. The fractional part of is - 15
1
2
(A) (B) 15 15
(C) 15
4
(D) none of these
36. Let a r and b r be two non-collinear unit vectors if
u r = a r – ( a r . b r ) b r and v r = a r × b r , then | v r | is equal
to -
(A) | u r | (B) | u r | + | v r . a r |
(C) 2| v r | (D) | u r | + u r . ( a r + b r )
⎡ 1 2 ⎤
30. If A = ⎢ ⎥
⎣ 2 1 ⎦
⎡1
1⎤
(A) ⎢ ⎥
⎣1
1 ⎦
⎡ 2 2 ⎤
(C) ⎢ ⎥
⎣ 2 2 ⎦
1+ x
and f (x) = , then f (A) is -
1– x
⎡ –1 –1⎤
(B) ⎢ ⎥
⎣ –1 –1 ⎦
(D) None of these
37. The vectors a r . b r , and c r are equal in length and
taken pairwise, they make equal angles. If
a r = iˆ + ˆj
, b r = ˆ j + kˆ
and c r makes an obtuse angle
with x-axis then c r is equal to -
(A) – iˆ
+ 4 ˆj
– kˆ
(B) iˆ + kˆ
1
(C) (–ˆ i + 4 ˆj
– kˆ
) (D)
3
i ˆ – 4 ˆj
+ kˆ
3
31. If f(x) =
5 + sin
sin
sin
2
2
2
x
x
x
cos
5 + cos
cos
(A) domain of function f (x) ∈ (0, ∞)
(B) range of function f (x) ∈ (0, ∞)
(C) period of function f (x) is 2π
(D)
lim
→0 x
f ( x)
– 150
= 200 x
2
2
x
2
x
x
4sin 2x
4sin 2x
5 + 4sin 2x
, then
32. If R be a relation < from A = {1, 2, 3, 4} to B = {1, 3, 5}
i.e. (a, b) ∈ R iff a < b, then RoR –1 is
(A) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(B) {(3, 1), (5, 1), (3, 2), (5, 2), (5,3), (5, 4)}
(C) {(3, 3), (3, 5), (5, 3), (5, 5)}
(D) {(3, 3), (3, 4), (4, 5)}
33. The probability of getting a sum of 12 in four throws
of an ordinary dice, is -
(A)
(C)
3
1 ⎛ 5 ⎞
⎜ ⎟
6 ⎝ 6 ⎠
1
36
⎛ 5 ⎞
⎜ ⎟
⎝ 6 ⎠
2
⎛ 5 ⎞
(B) ⎜ ⎟⎠
⎝ 6
4
(D) None of these
34. Three persons A, B and C are to speak at a function
along with five others. If they all speak in random
order, the probability that A speak before B and B
speaks before C, is -
3
1
(A) (B) 8 6
(C) 5
3
(D) None of these
35. The group of 10 items has arithmetic mean 6. If the
arithmetic mean of 4 of these items is 7.5, then the
mean of the remaining items is -
(A) 6.5 (B) 5.5 (C) 4.5 (D) 5.0
38. In a trapezoid the vector BC = λ AD we will then
find that P = AC + BD is collinear with AD . If
P r = µ AD , then -
(A) µ = λ + 1 (B) λ = µ + 1
(C) λ + µ = 1 (D) µ = 2 + λ
39. The image of the point P (1, 3, 4) in the plane
2x – y + z + 3 = 0 is -
(A) (3, 5, – 2) (B) (–3, 5, 2)
(C) (3, –5, 2) (D) (3, 5, 2)
40. {x ∈ R : cos 2x + 2 cos 2 x = 2} is equal to -
⎧ π ⎫ ⎧ π ⎫
(A) ⎨2 n π + : n∈
Z ⎬ (B) ⎨n π ± : n∈
Z ⎬
⎩ 3 ⎭ ⎩ 6 ⎭
⎧ π ⎫
(C) ⎨n π + : n∈
Z ⎬
⎩ 3 ⎭
⎧ π ⎫
(D) ⎨2
n π – : n∈
Z ⎬
⎩ 3 ⎭
41. If sin –1 ⎛ 3 ⎞
⎜ ⎟ + sin –1 ⎛ 4 ⎞ π
⎜ ⎟ = , then x is equal to -
⎝ x ⎠ ⎝ x ⎠ 2
(A) 3 (B) 5 (C) 7 (D) 11
1 1 3
42. In ∆ABC, if + = , then C is
b + c c + a a + b + c
equal to -
(A) 90º (B) 60º (C) 45º (D) 30º
43. From the top of a hill h metres high the angles of
depressions of the top and the bottom of a pillar are α
and β respectively. The height (in metres) of the
pillar is -
h(tanβ
– tan α)
h(tan α – tanβ)
(A)
(B)
tanβ
tan α
h(tanβ + tan α)
h(tanβ
+ tan α)
(C)
(D)
tanβ
tan α
XtraEdge for IIT-JEE 82
MARCH 2012

44. In a ∆ABC if the sides are a = 3, b = 5 and c = 4, then
sin 2
B + cos 2
B is equal to -
7. Directions : In following question, find out which of
the answer figures (A), (B), (C) and (D) completes
the figure – matrix ?
(A) 2 (B)
3 +1
2
(C)
3 – 1
2
(D) 1
45. If sin θ + cosec θ = 2 then sin 11 θ + cosec 21 θ =
(A) 2 (B) 2 21 (C) 2 32 (D) 1
LOGICAL REASONING
1. Fill in the blank spaces
6, 13, 28, . ?. . .
(A) 56 (B) 57 (C) 58 (D) 59
2. Choose the best alternative
Car : Petrol : : T.V. : ?
(A) Electricity (B) Transmission
(C) Entertainment (D) Antenna
3. Pick the odd one out –
(A) Titan
(B) Mercury
(C) Earth
(D) Jupiter
4. Direction : In questions, find out which of the
figures (A), (B), (C) and (D) can be formed from the
pieces given in (x).
(x)
(A) (B) (C) (D)
5. Directions : In question, choose the set of figures
which follows the given rule.
Rule : Closed figures become more and more open
and open figures more and more closed.
(A)
(C)
(B)
(D)
6. Directions : In question below, you are given a
figure (X) followed by four figures (A), (B), (C) and
(D) such that (X) is embedded in one of them. Trace
out the correct alternative.
(A) (B) (C) (D)
8. Directions : The questions that follow contain a set
of three figure X, Y and Z showing a sequence of
folding of piece of paper. Fig. (Z) shows the manner
in which the folded paper has been cut. These three
figure are followed by four answer figure from which
you have to choose a figure which would most
closely resemble the unfolded form of figure. (Z)
(A)
A
(B)
X Y Z
B
?
(C)
C
(D)
9. Direction : In following questions, complete the
missing portion of the given pattern by selecting from
the given alternatives (A), (B), (C) and (D).
?
(X)
(A) (B) (C) (D)
10. Directions : In question below, you are given a
figure (x) followed by four figures (A), (B), (C) and
(D) such that (X) is embedded in one of them. Trace
out the correct alternative.
(X)
D
(x)
(A) B) (C) (D)
(A) (B) (C) (D)
XtraEdge for IIT-JEE 83
MARCH 2012

ENGLISH
1. Find the correctly spelt word –
(A) Geraff
(B) Giraffe
(C) Giraf
(D) Gerraffe
2. Find out that word where the spelling is wrong –
(A) Puncture (B) Puntuation
(C) Pudding (D) Pungent
3. Pick up the correct synonym for the following words
Plush :
(A) Luxurious (B) Delicious
(C) Comforting (D) Tasty
4. Choose the alternative which can replace the word
printed in underline without changing the meaning of
the sentence.
When he returned, he was accompanied by 'sprightly'
young girl.
(A) Lively
(B) Beautiful
(C) Sportive (D) Intelligent
5. Choose one alternative which is opposite in meaning
to the given word :
Astute :
(A) Wicked (B) Impolite
(C) Cowardly (D) Foolish
6. Choose the word which is closest to the 'opposite' in
meaning of the underlined word
Many snakes are 'innocuous' :
(A) Deadly
(B) Ferocious
(C) Poisonous (D) Harmful
7. Choose the one which can be substituted for the
given words/sentences :
Giving undue favours to one's kith and kin'
(A) Corruption (B) Worldliness
(C) Favouritism (D) Nepotism
8. Find out which one of the words given below the
sentence can most appropriately replace the group of
words underlined in the sentence :
The bus has to "go back and forth" every six hours.
(A) Cross
(B) Shuttle
(C) Travel
(D) Run
9. Read both the sentences carefully and decide on their
correctness on the basis of the underlined words :
1. I am out of practise these days
2. I practice law
(A) Only 1 is correct
(B) Only 2 is correct
(C) Both the sentences 1 & 2 are correct
(D) Both the sentences 1 & 2 are incorrect
10. Which one of the two sentences given below is
wrong on the basis of the underlined words :
1. He is a very "ingenuous" businessman.
2. I like him for his "Ingenious" nature.
(A) Sentence 1 is correct
(B) Sentence 2 is correct
(C) Both the sentences can be made correct by
interchanging the underlined words.
(D) Both the sentences can not be interchanged hence,
both are wrong
11. Choose from the given words below the two
sentences, that word which has the same meaning and
can be used in the same context as the part given
underlined in both the sentences :
1. His "aloof" behaviour is an indication of his
arrogance.
2. During our field visits we visited "remote" parts of
Rajasthan.
(A) Far-off
(B) Introvert
(C) Distant
(D) Depressed
12. Find out which part of the sentence has an error. If
there is no mistake, the answer is 'No error'.
" Meatlessdays" / have been made / int o a film /
(a)
(b)
(c)
(A) Meatless days (B) have been made
(C) into a film (D) No Error
No
Error
(d)
13. Which part of the following sentence has an error ? If
the sentence is correct, the answer will be 'No Error".
Looking forward / to / meet youhere /
(a)
(b) (c)
(A) looking forward (B) to
(C) meet you here (D) No error
No
Error
(d)
14. Choose the one which best expresses the meaning of
the given Idiom/Proverb :
The 'pros and cons'
(A) Good and Evil
(B) Former and Latter
(C) For and Against a thing
(D) Foul and Fair
15. Replace the underlined word with one of the given
options :
The Second World War started in 1939.
(A) Broke out (B) Set out
(C) Took out (D) Went out
XtraEdge for IIT-JEE 84
MARCH 2012

CHEMISTRY
1. [A]
1
HOC ∝ stability
So II > IV > I > III
2. [B]
In acidic médium enol form depend on stability of
alkene
3. [C] Both double are E, E isomer
4. [B]
SOLUTION FOR MOCK TEST
O
⊕
H / H 2 O
IIT-JEE (PAPER - I)
CH 2 –OH
CH 3 –Mg–Br
(excess)
⊕
H
⎯
OH
+ (CH 3 .CO) 2 O ⎯→ CH 3 —C—CH 3
CH 3
CH 3 –Mg – Br + Cl–C–OC 2 H 5
(excess)
O
OH
CH 3 —C—CH 3
CH 3
⊕
H
⎯⎯→
10. [B,C]
Cr 3+ → 3, Mn 2+ → 5, Fe 3+ → 5, Cu 2+ → 1
11. [B,C,D]
Fe, Cr, Al have protected film
HO CH 3
5. [A]
M eq of NaOH + Meq of Ba(OH) 2 = M eq of HCl
XV 1 + YV 2 = 0.1 × 100
V
Also 4Y 2
+ YV 2 = 10 ⇒ YV 2 = 5
4
∴ V 1 X = 5
∴ Fraction of acid by Ba(OH) 2 = 10
5 = 0.5
6. [B] r ∝ M.
CO = 28, B 2 H 6 = 28, H 2 = 2, CH 4 = 16
−30
7.
2.6 × 10
[B]%ionic =
−10
−19
1.41×
10 × 1.6 × 10
× 100
= 11.5%
8. [B] C 2 H 5
CH 3
O
9. [B,C,D]
Ph–Mg–Br
(excess)
O
+ CH 3 —C—Cl
1, 2-epoxy-2-methyl butane
⊕
H
⎯⎯→
OH
Ph—C—Ph
CH 3
12. [B] mu = λ
h ⇒ 9.1 × 10 –31 u =
⇒ u = 1400 m/sec
6.626×
10
5200×
10
−34
−10
13. [A] u =
3RT 3×
8.314×
300
=
−3
M 4×
10
= 1367.8 M/sec.
−34
6.626 × 10
λ −3
λ = =
4×
10 × 1367.8
mu
23
6.023×
10
= 7.29 × 10 –11 m
14. [A,B,D] 15. [A,B,D]
16. [A,B]
17. [6] 18. [9] 19. [6] 20. [4] 21. [6]
22. [4]
O
HO – S – O – O – H
O
O
+6
HO – S – O – O – S – OH
O
O
O
XtraEdge for IIT-JEE 85
MARCH 2012

23. [6]
O
O ← P – O
O
P
O
P
O
O
P → O
MATHEMATICS
1. [B] f(x) is odd function
g(x) = f(–x) = – f(x)
x 1
g(x).f(x) = – sin π x 2x
= –
2
x + 2
xsin(
πx)
+ 2x
4
4x
+ 1
g(1) f (1) = –
2
x
+ 1
3
3
5
3
3x
2
4
1
1
1
xsin(
πx)
+ 2x
sin ( πx)
+ 4x
3
2
2
4
x
sin π x
x
+ 1
+ 1
6
x sin( πx)
+ 6x
3
5
7
5
7
11
= – 4
+ 1
3
3
1
2x
3x
2
4
4x
6
4
1
1
1
+ 1
4
x + 9x
6
x sin( πx)
+ 6x
2.
sin3α
[A]
cos2α
< 0 if sin 3α > 0 and cos 2α < 0
or sin 3α < 0 and cos 2α > 0
i.e. if 3α∈ (0, π) and 2α∈ (π/2, 3π/2)
or 3α∈ (π, 2π) and 2α∈ (–π/2, π/2)
i.e. if α∈ (0, π/3) and α∈ (π/4, 3π/4)
or α∈ (π/3, 2π/3) and α∈ (–π/4, π/4)
i.e. if α∈ (π/4, π/3)
since (13π/48, 14π/48) ⊂ (π/4, π/3)
(A) is correct
3. [C] We have
S 1 = Σ x 1 = sin 2β
S 2 = Σ x 1 x 2 = cos 2β
S 3 = Σ x 1 x 2 x 3 = cos β
S 4 = x 1 x 2 x 3 x 4 = – sin β
4
so that ∑
−1
tan x i = tan –1 S1
− S3
1−
S
i=
1
2 + S4
= tan –1 sin 2β − cosβ
1−
cos 2β − sin β
= tan –1 cosβ(2sinβ −1)
sinβ(2sinβ −1)
= tan cot β = tan –1 (tan (π/2 – β))
= π/2 – β
+ 1
+ 1
4. [D]
y =
Lim [1 + (cos x) cosx ] 2
−
π
x→ 2
Lim (cos x) cos x
−
π
x→ 2
log(y) =
log(y) =
log(y) =
=
−
π
x→ 2
Lim (cos x) log cos x
−
π
x→ 2
log(cos x)
Lim
(∞/∞) L'hospital
sec( x)
−
π
x→ 2
Lim
−
π
x→ 2
Lim – cos x = 0
1 sin x × –
cos x sec x tan x
y = e 0 = 1
Now limit is (1 + 1) 2 = 2 2 = 4
ln(
ln(
x))
5. [D] f(x) =
ln(
x)
1 1
× × ln(
x)
− ln(
ln(
x))
× 1/ x
f '(x) =
lnx
x
2
[ ln(
x)]
f '(x) =
1 − 0
e = 1/e
1
6. [A] b = 0 , a < 0
f(x) = x 2 + ax + b is quadratic polynomial
cut x axis at x = 0 and x = – a
f(x) = x 2 + ax + b
f (| x |) = x 2 + a |x| + b
| f |x|| = | x 2 + a |x| + b|
Exactly at three points function is not differentiable.
XtraEdge for IIT-JEE 86
MARCH 2012

7. [A] Graph of sin –1 sin(x) = f(x)
–π 0 π
5
2π 3π
f(x) = sin –1 (sinx) = x – 2π 3π/2 ≤ x ≤ 5π/2
f(5) = sin –1 (sin 5) = 5 – 2π
log 2 (x) < 5 – 2π
x > 0
x < 2 5–2π
So, (0, 2 5–2π )
8. [A,B,C]
Let tan –1 (–3) = α ⇒ tan α = – 3
and – 2
π < α < 0
⇒ – π < 2α < 0
⇒ cos (– 2α) = cos 2α
2
1 − tan α 1 − 9 4
= = = −
2
1 + tan α 1 + 9 5
⇒ – 2α = cos –1 ⎛ 4 ⎞
⎜ − ⎟
⎝ 5 ⎠
⇒ 2 tan –1 (–3) = – cos –1 (– 4/5)
Again –π < 2α < 0
⇒ 0 < 2α + π < π
4
So cos (π + 2α) = – cos 2α = 5
⇒ π + 2α = cos –1 4
5
⇒ 2 tan –1 (–3) = – π + cos –1 4
5
Finally – π < 2 α < 0
⇒ –π/2 < 2α + π/2 < π/2
So tan (π/2 + 2α) = – cot 2α =
2
−1
tan 2α
− 1 + tan α − 4
=
=
2 tan α 3
∴ π/2 + 2α = tan –1 (–4/3)
⇒ 2 tan –1 (–3) = – π/2 + tan –1 (–4/3)
9. [A,B,D]
f (x) = [x sin πx] = 0 – 1 ≤ x ≤ 1.
So, A, B, D are true
10. [A,D]
⎛
2
x ⎞
f (x) = cos x – ⎜ ⎟
1 − ⎝ 2 ⎠
f '(x) = – sin x + x
sin x < x if x > 0 and sin x > x if x < 0
(A) follow from (D) since (0, π/2) is a subset of (0, ∞)
11. [B,D]
u 2 = a 2 + b 2
2
⎛
2 2
⎞ ⎛
2 2
a + b ( a − b ) ⎞
+2 ⎜ ⎟ − ⎜ ⎟
2
cos 2θ
⎝ 2 ⎠ ⎝ 2 ⎠
⎛
2
max. u 2 = a 2 + b 2 a + b
+ 2 ⎜
⎝ 2
min. u 2 = a 2 + b 2
+ 2
2
⎛
2 2
⎞ ⎛
2 2
a + b a − b
⎜
⎝
2 2 ⎟ ⎞
⎟ ⎜
−
⎠ ⎝ ⎠
Passage # 1 (Q.12 to Q.14)
2
2
⎞
⎟
= 2(a 2 + b 2 )
⎠
2
= (a + b) 2
12. [A] f (x) = 0 ⇒ sin {cot –1 (x + 1)} = cos (tan –1 x)
⇒ sin sin –1 1
= cos cos –1 1
2
2
1+ ( x + 1)
1+ x
⇒
1 1
= ⇒ 1 + x 2 = 2 + x 2 + 2x
2
2
1+ ( x + 1) 1+ x
⇒ x = – 1/2, so f (x) = 0 for x = – 1/2
13. [C] a = cos tan –1 sin cot –1 x = cos tan –1 sin α, where x
= cot α
= cos tan –1 1
= cos β where
2
1+ x
tan β =
1
1+ x
2
2
=
1
1
1+
1+ x
⇒ a 2 x + 1 5
= = for x = – 1/2
2
x + 2 9
14. [B] Now a 2 = 26/51 ⇒ x = ± 1/5
and b = cos (2 cos –1 x + sin –1 x)
= cos (cos –1 x + π/2)
⇒ b = – sin (cos –1 2
x) = – 1−
x
⇒ b 2 = 1 – x 2 = 24/25 for x = ± 1/5
Passage # 2 (Q.15 to Q.16)
dy
15. [C] = 2x (x 4 ) – 1 (x 2 )
dx
⎛ dy ⎞
⎜ ⎟
⎝ dx ⎠
at x = 1
x=1
= 2 – 1 = 1
(y – 0) = 1 (x – 1)
⇒ y = x – 1
1
y =
∫
1
2
2
=
t dt = 0
x
x
2
2
+ 1
+ 2
XtraEdge for IIT-JEE 87
MARCH 2012

16. [A] f '(x) =
2
x /
e (1 – x 2 ) = 0
[f '(x)] x = 1 = e(1 – 1) = 0
2
⎡2
⎤
17. [9] f (x) = [x] + [2x] + ⎢ x⎥ + [3x] +[4x] + [5x] [kx]
⎣3
⎦
changes its value of every integral multiple of 1/k
[x] will change at every integral multiple of 1
[2x] will change at every integral multiple of 1/2
[3x] will change at every integral multiple of 1/3
[4x] will change at every integral multiple of 1/4
[5x] will change at every integral multiple of 1/5
⎡2
⎤
and ⎢ x⎥ will change at every integral multiple of
⎣3
⎦
3/2
They would change all together at every multiple of
LCM of {1, 1/2, 1/3, 1/4, 1/5, 3/2} = 3
No. of total points at which f (x) will changes its value
in the interval [0, 3] will depend on the total number of
different terms in the following cases –
[x] = 0, 1, 2
[2x] = 0, 1/2, 2/2, 3/2, 4/2, 5/2
[3x] = 0, 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 7/3, 8/3
[4x] = 0, 1/4, 2/4, 3/4, 4/4, 5/4, 6/4, 7/4, 8/4, 9/4,
10/4
[5x] = 0, 1/5, 2/5, 3/5, 4/5, 5/5, 6/5, 7/5, 8/5, 9/5,
10/5, 11/5, 12/5, 13/5, 14/5
⎡2
⎤
⎢ x⎥
= 0, 3/2, 6/2
⎣3
⎦
∴ f (x) will change its values in the intervals
14
0 ≤ x < 1/5, 1/5 ≤ x < 1/4, 1/4 ≤ x < 1/3,........... ≤ 5
x < 3
Total no. of different terms in above equation
= 30.
So, no. of terms in the range of
f (x) for 0 ≤ x < 3 is 30 ⇒ m +21 = 30 ⇒ m =9
18. [2]
=
=
Lim
h→∞
Lim
h→∞
Lim
h→∞
2
h
h −
⎡ ⎛ 1 ⎞ ⎛ 1
⎢(
h + 1) ⎜h
+ ⎟.....
⎜h
+
⎣ ⎝ 2 ⎠ ⎝ 2
h−1
⎡ ⎛ 1 ⎞ ⎛ 1
⎢(
h + 1) ⎜h
+ ⎟.......
⎜h
+
⎢ ⎝ 2 ⎠ ⎝ 2
h
⎢
h
⎢
⎣
h
⎛ h +1⎞
⎜ ⎟
⎝ h ⎠
h
h−1
⎛ 1 ⎞ ⎛ 1
⎜ h + ⎟ ⎜ h +
⎜ 2
h
⎟ ......... ⎜ 2
⎜ h ⎟ ⎜ h
⎜
⎝ ⎠ ⎝
⎞⎤
⎟⎥ ⎠ ⎦
⎞⎤
⎟⎥
⎠⎥
⎥
⎥
⎦
−1
h
h
h
⎞
⎟
⎟
⎟
⎟
⎠
=
Lim
h→∞
h
⎛ 1
h ⎟ ⎞ ⎛
⎜1 + . ⎜1
+
⎝ ⎠ ⎝
= e 1 . e 1/2 . e 1/4 ........ e
Lim
h→∞
e
1
2
h ⎟ ⎞
⎠
2h
2
⎞
......... ⎜
⎛ 1
1 + ⎟
n
⎝
− 1
2 h ⎠
ah
1
n−1
2
⎛ 1
ah ⎟ ⎞
⎜1 + = e a
⎝ ⎠
1 1
1+ + + ....... ∞
2 4
1
=
1−1/
2
.......∞
2
h−1
e = e 2 = e S ⇒ S = 2 ⇒ 0002
2
h
h−1
19. [5] x 2 + 1 = (x + i) (x – i)
The cubic polynomial must vanish
for x = i , x = – i
– i – a + bi + c = 0, i – a – bi + c = 0
– a + c = 0, – 1 + b = 0
– a + c = 0, 1 – b = 0
b = 1, a = c ⇒ b is fixed
Now a can be chosen in 10 ways and c = a,
c can be chosen in 1 way only
⇒ Number of ways of choosing
a, b, c = 10 ⇒ 10 = 2k ⇒ k = 5
20. [ 2] e y + xy = e
on putting x = 0, we get e y = e
y = 1 when x = 0
on differentiating the relation (i) we get
dy dy + 1.y + x. = 0
dx dx
on putting x = 0, y = 1 we get
e y ⎛ dy ⎞
dy −
⎜ ⎟ + 1= 0 ⇒ =
⎝ dx ⎠ dx e1
on differentiating solution (ii) we get
2
e y ⎛ dy ⎞
2
⎜ ⎟ + e y d y dy dy d y
+
2 + + x = 0
⎝ dx ⎠ dx dx dx
2
dx
dy −
on putting x = 0, y = 1, = we get
dx e1
2
d y 1
= = e –2 ⇒ λ = 2
2 2
dx e
21. [1] Circle (x – 2) 2 + (y – 3) 2 + λ (x + y – 5) = 0
it passes through (1, 2)
1 + 1 + λ (1 + 2 – 5) = 0 ⇒ λ = 1
x 2 – 4x + 4 + y 2 – 6y + 9 + x + y – 5 = 0
(x – 3/2) 2 + (y – 5/2) 2 = 1/2
So, number of circle is 1.
2
XtraEdge for IIT-JEE 88
MARCH 2012

22. [8] Let the equation of line through P(λ, 3) be
x − λ y − 3
= = r ⇒ x = λ + r cos θ
cosθ
sin θ
and y = 3 + r sin θ
Line meets the ellipse x 2 2
+ y = 1
16 9
Such that 9x 2 + 16y 2 = 144 at A and D
⇒ 9(λ + r cos θ) 2 + 16(3 + r sin θ) 2 = 144
⇒ 9(λ 2 + r 2 cos 2 θ + 2λr cosθ)
+ 16 (9 + r 2 sin 2 θ + 6r sin θ) = 144
⇒ (9cos 2 θ + 16 sin 2 θ) r 2 +
(18λ cos θ + 96 sin θ)r + 9λ 2 = 0
2
9λ
∴ PA.PD =
… (i)
2
2
9cos θ + 16sin θ
Since line meet the axes at B and C
3λ
So, PB.PC =
…. (ii)
sin θcosθ
from (i) & (ii)
λ ≥ 8
23. [7]
→
a ⊥ (
→
b +
→
c ) ⇒
→
a . (
→
b +
→
c ) = 0
⇒ → a . → b + → a . → c = 0 and two similar results
adding, 2 ( → a . → b + → b . → c + → c . → a ) = 0
Now | → a + → b + → c | 2 = ( → a + → b + → c ). ( → a + → b + → c )
= | → a | 2 + | → b | 2 + | → c | 2 + 2( → a . → b + → b . → c + → c . → a )
= 9 + 16 + 25 + 0 = 50
∴ | → a + → b + → c | = 5 2
PHYSICS
1. [C]
When block m 1 passes through mean position, its
speed is maximum. Let v 1 and v 2 be the speed of
blocks m 1 and (m 1 + m 2 ) respectively at equilibrium
position.
Then m 1 v 1 = (m 1 + m 2 ) v 2
at mean position
v = Aω = A2πf
hence
v 1 =
1 f1
v2
A2
f2
A 2 = A
A × and f1 = 2π
m1
m + m
1
2
f 2 = 2π
m 1
k
m m2
k
1 +
2. [D]
3. [C]
B
ucosθ A
usinθ
θ
v 1
v 2
usinθ
before collision after collision
Apply conservation of momentum
mv 1 + mv 2 = mvcosθ
or v 1 + v 2 = v cosθ ….(i)
and v 1 – v 2 = evcosθ …(ii)
from (i) + (ii)
(1 + e)
v 1 = v cos θ
2
b
b
dI =
∫
a
4. [D]
r
a
2
r
dm and dm =
M 2 +
2
I = (b a )
2
α
f
F – f = ma CM
fR = Iα
Ia CM
or f =
2
R
from (i) & (ii)
F
a CM =
f =
⎛ I
⎜m
+
2
⎝ R
I F
2
R ⎛ I
⎜m
+
⎝ R
….(i)
….(ii)
⎞
⎟
⎠
2
⎞
⎟
⎠
⎛ I
µmg⎜m
+
F max =
⎝ R
⎛ I ⎞
⎜
2
⎟
⎝ R ⎠
F
a CM
and
≤ µmg
2
⎞
⎟
⎠
M(2πrdr)
π(b
2 −
a
2
)
XtraEdge for IIT-JEE 89
MARCH 2012

5. [A]
We have gravitational potential at internal point.
GM 2 2
V = − (3R − r )
3
2R
at surface
3 GM
V 1 = −
2 R
R
x at r = 2
GM ⎛
V 2 = − ⎜
3R
3
2R ⎝
2
2
R
−
4
11GM
V 2 = −
8R
we have
1 mv 2 = m | V 2 – V 1 |
2
6. [C]
W = surface tension (T) × change in surface area
(∆A)
7. [C] Acceleration of the rod (a)
F
a r
F
a = m
⎞
⎟
⎠
Consider an element of length dx at a distance 'x'
from one end of the rod.
x
F
dx
stress developed at a distance x
m( l − x) F
stress =
Al
m
F ⎛ x ⎞
stress = ⎜1
− ⎟
m ⎝ l ⎠
elongation in the dx length
⎛ stress ⎞
dl = ⎜ ⎟ dx
⎝ Y ⎠
or
∫ d l = F ⎞
∫⎜
⎛ x
1 − ⎟ dx
YA ⎝ l ⎠
∆l =
Fl
2YA
l
0
8. [A,B,D]
t 3
x = 3
dx
v = = t
2
dt
2
d x
a = = 2t
2
dt
F = ma
F = 4t or F ∝ t
Apply work – Energy theorem
W = ∆k = 2
1 mv
2
at t = 0, u = 0
at t = 2, v = 4 m/s
during 0 – 2 sec
W = 2
1 × 2 (4)
2
W = 16 J
9. [A,B,C,D]
Here i P = i Q
i
Current density, J P < J Q (Q J = A
I P < E Q (Q J = σE)
ρdx
Resistance, R P < R q (Q R = )
A
or (i 2 R) P < (i 2 R) Q
10. [B,C,D]
Here x = at and y = bt 2 + ct
dx dy
v x = = a and vy = = 2bt + c
dt
dt
| v | =
at t = 1
2
x
v + v
2
2
y
| v | = a + (2b + c)
at t = 0
v x = a, v y = c
c
tanθ = and | u | =
2 2
a + c
a
d 2 x d y
= 0 and = 2b
2
dt dt
i.e. a y = 2b or g = –2b
2
2
)
XtraEdge for IIT-JEE 90
MARCH 2012

11. [A,B,C]
At equilibrium
qE
kx = qE or x =
k
for maximum elongation,
apply Work – Energy theorem
(qE)x 0 – 2
1 kx0 2 = 0
2qE
x 0 =
k
the block perform oscillation about mean position
qE
(i.e. x = )
k
amplitude (A) =
A =
Passage : I
qE
k
2qE
k
–
qE
k
Sol. for Q.No. 12.[B], 13.[D] & 14.[A]
The given circuit can be simplified as
23µF
7µF
A
B
12µF 5µF 1µF
A
10µF
10µF
12V
OR
35µF 7µF
12µF
1µF
12V
⎛ 35×
7 ⎞ 20
C eq = ⎜ ⎟ + = 7.5 µF
⎝ 35 + 7 ⎠ 12
Charge on 5 µF = 0
20
Charge on 10 µF = 12 × = 20 µC 12
20×
20
U 12µF = = 20 µJ
2×
10
B
Passage : II
Sol. for Q.No. 15.[A] & 16.[A]
U = 5 + (x – 1) 2
at x = 2, U = 6 J
KE = 10 J
M.E. = 10 + 6 = 16 J
dU
for minima of U, = 0
dx
2(x – 1) = 0
x = 1
U min = 5 J
17. [5] T 1 = 2π g
l = T and
T 2 = 2π
25l
g
T 2 = 5T
at t = 5T, both pendulum will again be in phase for
1 st time during that time, 5 oscillations are made by
smaller pendulum.
λ
18. [3] Here l 1 e = …(i) 4
3λ
l 2 + e =
4
from (i) & (ii)
l
e = 2 − l
2
e = 3
3 1
…(ii)
19. [3] For a polytropic process
PV n = constant
R
C = C V –
n −1
5 R
= R −
2 −1−1
(Q P ∝ V or PV –1 = constant)
C = 3R
20. [3] Temperature of junction
100 + 20 + 0
T C =
= 40ºC
3
i1
100 − 40
= = 3
i 40 − 20
2
21. [3]
20cm
A B
C f
O
f = 30cm
60cm
XtraEdge for IIT-JEE 91
MARCH 2012

22. [2]
image one end A is formed at same point
(i.e. at C)
for image of other end
1 1 1
+ =
v − 40 − 30
v = –120 cm
length of image l = 120 – 60 = 60 cm
60
linear magnification m l = = 3 20
a
O
N = a 2 /8
P
θ
θ
2θ θ
α
N Q
R
x 0
In ∆PNQ
a
tan 2θ =
⎛
2
a ⎞
⎜ ⎟
x 0 − ⎝ 8 ⎠
y 2 = 8x
dy 4 =
dx y
at P y = a
dy 4 = = tan(90º – θ)
dx a
or tan θ = 4
a
eq.(i) and (ii)
x 0 = 2
y 2 = 8x
Normal
…(i)
…(ii)
Chemistry Facts
• At 0 degress Celsius and 1 atmospheric pressure,
one mole of any gas occupies approximately
22.4 liters.
• Atomic weight is the mass of an atom relative to
the mass of an atom of carbon-12 which has an
atomic weight of exactly 12.00000 amu.
• If the atom were the size of a pixel (or the size of
a period), humans would be a thousand miles
tall.
• It would require about 100 million
(100,000,000) atoms to form a straight line one
centimeter long.
• The weight (or mass) of a proton is
1,836.1526675 times heavier than the weight (or
mass) of an electron.
• The electron was first discovered before the
proton and neutron, in 1897 from English
physicist John Joseph Thomson.
• The neutron was discovered after the proton in
1932 from British physicist James Chadwick,
which proved an important discovery in the
development of nuclear reactors.
23. [3]
We have
⎛
E = 13.6 Z 2 ⎜
1
2
⎝ n
1
− 2
1 n 2
⎟ ⎞
⎠
(in eV)
10.2 + 17 = 13.6 Z 2 ⎛ 1 1 ⎞
⎜ 2
− 2
⎟
⎝ 2 n ⎠
and
4.25 + 5.95 = 13.6 Z 2 ⎛ 1 1 ⎞
⎜ 2
− 2
⎟
⎝ 3 n ⎠
from (i) & (ii)
n = 6 and Z = 3
…(i)
…(ii)
• Carbon dioxide was discovered by Scottish
chemist Joseph Black.
• When silver nitrate is exposed to light, it results
in a blackening effect. (Discovered by Scheele,
which became an important discovery for the
development of photography).
XtraEdge for IIT-JEE 92
MARCH 2012

SOLUTION FOR MOCK TEST
PAPER - II
IIT-JEE (PAPER - II)
1. [C]
CHEMISTRY
D
HC
O
OH
H – OH
⎯⎯⎯
→
2. [B]
CH 3 –CH–CH 2 –C≡CH
CH 3
,
3. [B] III < I < II
4. [D]
D
H Br I
H Cl D
I
Both are enantiomers.
Cl
Br
O
Θ
OH
Θ
+ OH
–
O
CH 3 –CH–C≡C–CH 3
CH 3
H
H
5. [D] Insulin → Zinc (Zn); Haemoglobin → Fe
B 12 → Co
6. [B]
Cr 2 O 3
Green
∆
Na 2CO 3
Na 2 CrO 4
H 2O + H 2SO 4
Na2 CrO 7
(CH 3COO) 2 Pb
Orange
PbCrO 4
Yellow
7. [D] As 2 S 3 sol particles absorb S 2– as common ion
⎡(300 – V ) × 1⎤
8. [C] pH = pK a + log ⎢
⎥
⎣ 1×
V ⎦
⎡(300 – V ) ⎤
4.5 = 4.2 + log ⎢ ⎥ ; V = 100 ml
⎣ V ⎦
9. [C,D] Both C and D pair are same IUPAC name.
10. [A,B,C]
Na 2 BF 3 , Na 2 ClO 4 , conc HNO 3 + H 2 SO 4 used for
nitration of benzene.
11. [A,C,D]
Salt bridge is used to eliminate liquid junction
potential arised due to different speed of ions persent
in cathodic & anodic compartment
12. [A,B,C,D]
(A)
[ H 3O
][ HS¯]
K a =
[ H 2S]
(B) Higher the K c high is stability
(C) Fact
(D)
[ RNH +
3 ][ OH ¯]
K b =
[ RNH ]
13. [A → Q,S; B → R; C → P; D → T; ]
Electrochenical cell → ∆G < 0, Salt bridge
I st law of faraday → W = Zit
Electrolytic cell → ∆G > 0
Lead acid cell → rechargeable
14. [A → S; B → R; C → Q; D → T;]
15. [2]
16. [5]
17. [6]
18. [6]
⊕
NH 3 –CH–CH 2 –CH 2 –COOH
COO Θ
Two group present in above compound.
19. [4] 2Al + Fe 2 O 3 —→ 2Fe + Al 2 O 3
∆H = – 399 – (– 199) = – 200 Kcal / ml
54 160
vol. = + = 50.7 ml
2.7 5. 2
∆H 200
∴ = = 4 Kcal / ml
ml 50.7
20. [2] r λ = 100 r 1 ; C 2 = 10C 1 ⇒ n = 2
2
XtraEdge for IIT-JEE 93
MARCH 2012

MATHEMATICS
1. [C] Let ABC be the triangle with b + c = x and
bc = y, then a = z, and from the given relations we
have (b + c + a) (b + c – a) = bc
⇒ b 2 + c 2 – a 2 = – bc
2 2 2
b + c − a 1
⇒
= −
2bc
2
1
⇒ cos A = – = cos 120º 2
2. [D]
⇒ A = 120º and the triangle is obtuse angled.
⇒ A is an obtuse angle.
r 1 4Rsin(
A/
2)cos( B / 2)cos( C / 2)
=
bc 2R
sin B.2R
sin C
sin( A/
2)
=
4Rsin(
B / 2)sin( C / 2)
=
sin 2 ( A/
2)
4R
sin( A / 2)sin( B / 2)sin( C / 2)
sin 2 ( A / 2)
=
r
r1
r2
r3
So that + +
bc ca ab
1
= [(sin 2 (A/2) + sin 2 (B/2)+ sin 2 (C/2)]
r
1
= (1 – cos A + 1 – cos B + 1 – cos C)
2r
1
= [3 – (cos A + cos B + cos C)]
2r
1
= [3 – (1 + 4 sin (A/2) sin (B/2) sin (C/2))]
2r
1 ⎡ r ⎤ 1 1
= 2
2r ⎢ − = −
R
⎥
⎣ ⎦ r 2R
3. [B]
f (θ) = cos 2 θ + sin 4 θ
f (θ) = cos 2 θ + (1 – cos 2 θ) 2
f (θ) = (cos 2 θ – 1/2) 2 + 3/4
(f (θ)) min = 3/4.
2 2 − (cos x + sin x)
4. [D] f (x) =
1−
sin 2x
L'Hospital Rule
lim f (x)
π
x→
4
=
lim
π
x→
4
–
2
3
(0/0)
3(cos x + sin x)
( −sin
x + cos x)
− 2cos 2x
=
=
lim
π
3(cos x + sin x)
(cos x − sin x)
x → 2(cos x − sin x)
(cos x + sin x)
4
lim
π
x→
2
4
2
3 3 2 3
× (cos x + sin x) = × =
2 2 2
5. [A]
Function is diff. at x = 1 it means function is
continuous at x = 1, diff. at x = 1
R.H.L. at x = 1 = L.H.L. at x = 1, R.H.D. = L.H.D.
a cos (0) + b = 1
a + b = 1 ....(1)
(–2 a sin (2x – 2) + 2bx) x = 1
= (2x 2 e 2(x – 1) + 2xe 2(x – 1) ) x = 1
0 + 2b = 2 + 2
2b = 4
b = 2
so a = 1 – b = 1 – 2 = – 1
6. [A]
P(x, y) be a point on the curve
ln (x 2 + y 2 ) = c tan –1 y/x
differentiating both side with respect to x
2x
+ 2yy'
c(
xy'
−y)
=
2 2 2 2
( x + y ) x + y
2x
+ cy
⇒ y ' = = m 1
cx − 2y
slope of OP = y/x = m 2
so tan θ =
m1
− m
1+
m m
1
2
2
=
2x
+ cy
−
cx − 2y
y
x
2
2xy
+ cy
1+
2
cx − 2xy
= 2/c
θ = tan –1 (2/c) which is independent of x and y
7. [B]
f (x) = 2x 3 + ax 2 + bx – 3cos 2 x
f '(x) = 6x 2 + 2ax + b + 3 sin 2x > 0
6x 2 + 2ax + b – 3 > 0 as sin 2x ≥ – 1
∴ 4a 2 – 4b(b – 3) < 0 ⇒ a 2 – 6b + 18 < 0
8. [D]
x = φ(t) = t 5 – 5t 3 – 20t + 7
dx = φ '(t) = 5t 4 – 15 t 2 – 20 = 5(t 2 – 4) (t 2 + 1) ≠ 0
dt
If – 2 < t < 2
y = ψ(t) = 4t 3 – 3t 2 – 18t + 3
dy = ψ'(t) = 12 t 2 – 6t – 18
dt
XtraEdge for IIT-JEE 94
MARCH 2012

dy = 0 ⇒ t = – 1 or 3/2
dt
2
d y
= ψ"(t) = 24t – 6 ⇒ ψ"(–1) = – 30
2
dt
and ψ"(3/2) = 30
y = f (x) is minimum at t = 3/2
9. [A,D]
a − b sin A − sin B
cos θ = =
c sin C
[by the law of sine]
2 cos[(1/ 2) ( A + B)]sin[(1/
2) ( A − B)]
=
2 sin[(1/ 2) C]cos[(1/
2) C]
=
=
⇒ sin θ =
2 sin[(1/ 2) C]sin[(1/
2) ( A − B)]
2 sin[(1/ 2) C]sin[(1/
2)( A + B)]
sin[(1/ 2) ( A − B)]
sin[(1/ 2)( A + B)]
2
sin [(1/ 2) ( A + B)]
− sin [(1/ 2) ( A − B)]
sin[(1/ 2)
sin Asin
B
=
sin[(1/ 2) ( A + B)]
⇒
=
( a + b) sin θ
2 ab
2
( A + B)]
sin A + sin B sin Asin
B
2 sin Asin
B sin[(1/ 2) ( A + B)]
2 sin(1/ 2)( A + B) cos(1/ 2)( A − B)
=
⎛ 1 ⎞
2 sin⎜
⎟(
A + B)
⎝ 2 ⎠
= cos [(1/2) (A – B)]
c sin θ sin C
and
=
2 ab 2 sin Asin
B
sin Asin
B
sin[(1/ 2) ( A + B)]
=
2 sin( C / 2) cos( C / 2)
= cos ((A + B)/2)
2 sin[( A + B)
/ 2]
10. [A,B,D]
(A) sin[x] + cos[x] is defined for all x, since [x],
sinθ, cosθ are always defined.
(B) sin x is always defined and 1 + sin 2 x ≥ 1
⇒ sec –1 (1+ sin 2 x) is defined for all x
(C) tan(log x) is not defined if logx = (2k + 1) 2
π
(D) Range of function cosx + cos 2x is
put cos x = t, t∈[–1, 1] therefore
9 + cos x + cos 2x > 0 for all x
8
11. [A,B,D]
x + |y| = 2y
3y = x if y < 0
y = x is y ≥ 0
y = x/3
y = x, y ≥ 0
⎡ 9 ⎤
⎢ − , 2 ⎥
⎣ 8 ⎦
y = 1/3 x y < 0
y = x y > 0
(A) domain and range of function is set of real
numbers so (A) is true
(B) f (0) = L.H.L = R.H.L
so (B) is true
(D) L.H.D. = 1/3 and R.H.D. = 1 so (D) is true
12. [B,D]
Take river as x axis, line joining origin and
village A as x axis
A'
P
B(b, k)
(–a, 0) A(a, 0)
k 2 = c 2 – (b – a) 2
image of A in the river, this is A' and BA' must be
minimum value of PA + PB
BA' =
=
2
( b + a)
+ k
2
2
2
( b + a)
+ c − ( b − a)
= c 4ab
2
c
2 +
13. A → P,R,T; B → P,R,T; C → Q,S; D → P,S
(A)
Lim f(x) =
x→0
⇒ continuous
f '(0) =
Lim
x→0
sin 2 (x)
=
x
sin 2 x
− 0
x
x
= 1
Lim sin(x) = 0 = f(0)
x→0
XtraEdge for IIT-JEE 95
MARCH 2012

⇒ differentiable
(B) f(0 – ) = Lim (x 3 – 2x) = 0
x→0
f(0) = f(0 + ) = Lim (x 2 – 2sin (x)) = 0
x→0
⇒ continuous
f '(0) = Lim (3x 2 – 2) = – 2
x→0
f '(0 + ) = Lim (2π – 2cos(x)) = – 2
+
x→0
⇒ differentiable
(C) f (0 – ) = 4 & f (0 + ) = 5 ⇒ discontinuous ⇒ non diff.
(D) At x = 1, f (x) = 3 – 2x, which is polynomial
⇒ continuous & nondifferentiable
14. A → Q, B → S, C → S, D → R
(A) x = sin θ & y = cos θ
x + y = sin θ + cos θ
= 2 sin (θ + π/4) ⇒ min. value = – 2
(B) y = a cos x – 1/3 cos (3x)
y'(π/6) = 0 ⇒ – a sin π/6 + sin π/2 = – 2
9 + 1 = 0
⇒ a = 2
(C) f '(x) = 1 – 2 cosx
f '(x) > 0 ⇒ cos(x) < 1/2 ⇒ x ∈ (π/3, 5π/3)
⇒ a = 1/3, b = 5/3 ⇒ a + b = 2
(D) at x = 0, y = – e 0 = – 1
y' = 1/2 e –x/2 = y' (0) = 1/2
Equation of tangent y + 1 = 1/2 (x – 0)
⇒
x + y ⇒ p = 2, q = – 1
2 −1
So, p – q = 3
15. [7] Let E i denote the event that out of the first k
balls drawn, i balls are green. Let A denote the event
that (k + 1)th ball drawn is also green.
P(E i ) =
and P(A/E i ) =
14 6
Ci
×
20
Ck
14 − i
20 − k
14
C
k−i
; 0 ≤ i ≤ k
k
C j × Ck−i
14 − j
Now P(A) = ∑
+
20
C
k
j
k
20 −
= 0
Also (1 + x) 14 – 1 (1 + x) 6
= ( 14–1 C 0 + 14–1 C 1 x +.......+ 14 – 1 C 14 – 1 x 14–1 )
( 6 C 0 + 6 C 1 x + .......+ 6 C 6 x 6 )
k
⇒ ∑
=
j
14−1
− j
6
( C j + C k ) = co-efficient of x k
0
∴ P(E) =
6
14
=
6 + 14
14
20
⇒ 10P(Ε) = 7
6. [4] zi – z i + 2 = 0
⇒ (z – z ) = 2i ⇒ Im (z) = 1
z (1 + i) + z (1 –i) + 2 = 0
⇒ (z + z ) + i (z – z ) + 2 = 0
⇒ z = – z ⇒ z = i
⇓
Re(z) = 0
Let the point on the line be z so |z –i| = 2
3i
i
–i
|z| max = |3i| = 3
|z| min = | –i| = 1
sum = 4
⎡3
− 2 3 ⎤
⎢ ⎥
17. [3] A =
⎢
2 1 −1
⎥
⎢⎣
4 − 3 2 ⎥⎦
⎡ −1
− 5 −1⎤
⎢
⎥
adj A =
⎢
− 8 − 6 9
⎥
& |A| = – 17
⎢⎣
−10
1 7 ⎥⎦
⎡3
0 3⎤
⎡x⎤
⎡8⎤
⎡2y⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⇒
⎢
2 1 0
⎥ ⎢
y
⎥
=
⎢
1
⎥
+
⎢
z
⎥
is equivalent to
⎢⎣
4 0 2⎥⎦
⎢⎣
z⎥⎦
⎢⎣
4⎥⎦
⎢⎣
3y⎥⎦
⎡3
− 2 3 ⎤ ⎡x⎤
⎡8⎤
⎢ ⎥ ⎢ ⎥
=
⎢ ⎥
⎢
2 1 −1
⎥ ⎢
y
⎥ ⎢
1
⎥
⎢⎣
4 − 3 2 ⎥⎦
⎢⎣
z⎥⎦
⎢⎣
4⎥⎦
∴ solution is given by
⎡x⎤
⎡ −1
− 5 −1⎤
⎡8⎤
⎡−17⎤
⎢ ⎥ 1 ⎢
⎥ ⎢ ⎥ 1
= −
⎢ ⎥
⎢
y
⎥
=
⎢
− 8 − 6 9
⎥ ⎢
1
⎥ ⎢
− 34
−17
17 ⎥
⎢⎣
z⎥⎦
⎢⎣
−10
1 7 ⎥⎦
⎢⎣
4⎥⎦
⎢⎣
− 51⎥⎦
∴x = 1 , y = 2, z = 3
18. [1] Given planes are
x – cy – bz = 0
... (i)
cx – y + az = 0
... (ii)
bx + ay – z = 0
... (iii)
equation of plane passing through the line of
intersection of plane (i) and (ii) may be taken as
(x – cy –bz) + λ (cx – y + az) = 0
⇒ (1 + λc) x – y (c + λ) + z (aλ – b) = 0 ....(iv)
If plane (iii) and (iv) are same then
XtraEdge for IIT-JEE 96
MARCH 2012

⇒ λ = –
1+ λc −( c + λ)
=
b a
a + bc ab + c
= –
2
ac + b 1 − a
=
aλ − b
−1
⇒ a – a 3 + bc – a 2 bc = a 2 bc + ac 2 + ab 2 + bc
⇒ a 2 + b 2 + c 2 + 2abc = 1
19. [7] y = Ax m + Bx –n
dy
⇒ = Amx m–1 – nBx –n–1
dx
from (2) and (3)
180 80
a = and b = 119 119
180
f (x) = x 2 80 Ax + Bx
+ x = 119 119 119
A = 180 B = 80
A + B
A + B = 260 ⇒ = 1
260
2
⇒
d
2
dx
y
2
= Am (m –1) x m–2 –n –2
+ n (n +1) Bx
Putting these values in x 2 d y
2
dx
2
dy
+ 2x = 12y dx
1. [C]
PHYSICS
We have = m (m + 1) Ax m + n (n –1) Bx –n = 12 (Ax m + Bx –n )
⇒ m (m +1) = 12 or n (n –1) = 12
⇒ m = 3, –4 or n = 4, –3
1
20. [1] f (x) = x + x
∫
0
1
y 2 f (y) dy + x 2 ∫
0
⎡
1
⎤ ⎡
1
⎤
= x ⎢ +
2
2
1 ⎥ + ⎢ ⎥
⎢
∫
y f ( y)
dy x
⎣
⎥⎦
⎢∫
yf ( y)
dy
⎣ ⎥
0
0 ⎦
f (x) is quadratic expression
f (x) = ax + bx 2 or f (y) = ay + by 2
a = 1 +
∫ 1 0
= 1 +
∫ 1 0
⎡
= 1 + ⎢
ay
⎣ 4
y 2 f (y) dy
y 2 (ay + by 2 ) dy
4
1
5
⎤
+ by
⎥
5 ⎦
⎛ a b ⎞
a = 1 + ⎜ + ⎟
⎝ 4 5 ⎠
20a = 20 + 5a + 4b
15a – 4b = 20
1
b =
∫
0
1
y f (y) dy =
∫
0
⎛
3 4
=
3 4 ⎟ ⎞
⎜
ay + by ⇒ b =
⎝ ⎠0
12 b = 4a + 3b
9b – 4a = 0
1
0
y (ay + by) 2 dy
a b +
3 4
…(3)
y f (y) dy
…(2)
…(1)
F B
A
0.4 m
O
θ mg
Let x length of plank is inside the water (i.e. AO =
x)
x sinθ = 0.4 …(i)
buoyant force, F B = xA 0 ρ W g
for rotational equilibrium about O
F B
2
x cos θ = mg 2
l cos θ
(xA 0 ρ W g) 2
x = lA0 ρ p g 2
l
x 2 = l 2
x =
0.8
2
ρ
ρ
p
w
= (0.8) 2 × 0.5
from (i) and (ii)
1 π
sin θ = or θ =
2 4
….(ii)
2. [C]
When water cools down to 0ºC then heat released
∆Q = 5 × 1 × 30 = 150 cal. while heat required to
convert ice into water at 0ºC
(ice) –20ºC → (ice) 0ºC → (water) 0ºC
Q 1 = 5 × 0.5 × 20 = 50 cal
Q 2 = 5 × 80 = 400
∆Q′ = 450 cal
here 50 < ∆Q < 450
i.e final temperature is 0ºC with some ice melt.
XtraEdge for IIT-JEE 97
MARCH 2012

3. [B]
The given circuit can be simplified as
i 1
A
20V
Electric potential at A
1 Q
V A =
4π∈
2
R + 3
at origin
1 Q
V 0 =
4π∈ 0 R
We have
W = q(V 0 – V A ) = ∆K
2
0 R
=
1
4π∈ 0
Q
(2R)
4. [B]
20
reading of ammeter, i 1 = = 2A 10
y
a y
x
O u
at origin,
v x = u
dv
a x = x
= 2ux
dt
we have
y = x 2
dy dx = 2x = 0
dt dt
2
d y dx
= 2u
2
dt dt
a y = 2u 2
a =
a = 2u 2
2
x
a + a
2
y
along y-axis
5. [B]
+
+
+
+ r
λ
+ E =
2π∈ 0 r
We have
2r
r
V B – V A = dV = − E.dr
r
W = q dV
λq0
W =
2π∈
6. [D]
+
+ +
+ R +
+ +
+ + +
Q=λ2πR
0
x =
⎛ 3 ⎞
ln⎜
⎟
⎝ 2 ⎠
A
3 R
∫
3r
x
7. [C]
2
90º 3
r
1 i
O
i
L L 4
L
r =
2
Here B 1 = B 4 = 0
µ
B 2 = B 3 = 0 i × (2sin 45 ° )
2π
(L / 2)
µ 0 i
B 2 = B 3 =
2πL
r r r r
B = B + B + B
net
B r net =
1
µ 0
π
i
L
2
3
r
+ B
8. [A] X L = ωL = 50 Ω
1
X C =
ω C
= 100 Ω
2
2
Z = R + (XC
− XL
) = 50 2 Ω
V
i rms = rms
= 2A
Z
P av = i
2 rms R = 200 W
9. [A,B,C,D]
For maximum range, θ = 45º
u 2
i.e. x = ⇒ u = gx
g
warning time t = 2
T
t =
at max. height, v = u cos 45º =
max. height H =
g
u
2
4
=
x
2g
u 2 (sin 45º) 2 =
2g
gx
2
u 2
x
=
4g 4
XtraEdge for IIT-JEE 98
MARCH 2012

10. [A,B,D]
Q
fixed
Q
⇓
r 0
v
fixed
v′=0
Apply conservation of mechanical energy
2
1 mv 2 1 Q
=
2 4π∈
r 0 =
Q
2π∈
0
2
mv
0 r0
r 0 ∝ Q 2 1 1
, r 0 ∝ , r ∝ m
2
v
11. [A,C]
We have
dA
| ε | = B
dt
2
dA
here increases first then decreases.
dt
ε max = Bvl 2
12. [B,C]
We have
i av = T
1
i
i m = 0
2
T
∫
0
T
1
i rms =
∫i 2 dt
T
0
we have
1
i dt = (Area of i – t curve)
T
⎛ 2i
i = 0 ⎞
T
⎜ ⎟ t for 0 ≤ t ≤
⎝ T ⎠ 2
We get
i
i rms = 0
3
T / 2
1
= × 2
∫i 2 dt
T
13. A → P ; B → Q ; C → S ; D → T
When only S 1 → closed then
C
3r
B
A
r
V
0
V AB = 4
V at steady state.
Hence charge on capacitor, q 1 =
When only S 2 → closed then
C
3r
B
A
2r
V
2
V AB = V at steady state
5
2
hence max. charge, q 2 = CV
5
When only S 3 → closed then
3r C
V
q 3 = CV
When all switches are closed then
C
3r
V
no charge appear on capacitor.
CV
4
14. A → S ; B → R ; C → P; D → Q
α = 90º – 60º = 30º
2u sin α 2 × 40 10
T = = × = 4 sec
g 10 2
u 2 sin 2α 40×
R = =
g 1040
R = 80 3 m
H =
u
2
sin
2g
2
×
3
2
α 40×
40 = = 40 m
2 × 10×
4
15. [4]
Electric field lines are perpendicular to equipotential
surfaces and electric potential decreases along
electric field.
80V 60V 40V 20V
90º
30º 30º
B
120º
E r
XtraEdge for IIT-JEE 99
MARCH 2012

From A → B
| dV | = –E dr cos 120º
20 = E × (10 × 10 –2 1
) × 2
E = 400 N/C
V 2
16. [6] We have P = or R =
R
40 W 50 W
V 0
V 2
200 × 200
R 40W =
40
= 1000 Ω
200× 200
R 50W =
50
= 800 Ω
Let max. voltage of main supply is V 0 then
P
18. [6]
O
C
2 cm
2 cm
f
3f/2
For upper part of lens
3f
u = − , h 0 = 2 cm
2
hi
f
m = =
= 0.5
h 0 ⎛ − 3f ⎞
f + ⎜ ⎟
⎝ 2 ⎠
h i = 2 × 0.5 = 1 cm
i.e. image is formed at a height of 3 cm (i.e. 2 +
1) from main principle axis. (above principle axis)
Similarly for lower part, image is formed 3
cm below main principle axis. Hence distance
between image = 3 + 3 = 6 cm.
17. [4]
V 40W = 9
4 × V0 = 200
V 0 = 450 V
and
V 50W = 9
5
V0 = 200 = 360 V
hence for safety of both bulb, V 0 should be
360 V and for this main supply voltage V 40W
= 200V hence it glows with full intensity.
5kg
A
37º
B
f lim = µmgcosθ = 0.5 × 50 × 5
4 = 20 N
max. mass of block B is =
30 + 20 50
m 2 = = = 5 kg
10 10
min. mass of block B
mgsin 30º −flim
m 1 =
g
m 1 = 1 kg
m 2 – m 1 = 4
mgsin 37º + f
g
lim
19. [5]
Velocity of centre of mass of cylinder w.r.t. plank,
v cp = 20 – 10 = 10 m/s
We have v cp = Rω ⇒ ω = R
10
100
2
R
Kinetic energy KE =
= 450 J
i.e. n = 5
20. [3] We have
3
GM 4 πR
ρ
g =
2 = G ×
2
R 3 R
4
g ∝ Rρ and m = πR
3 ρ
3
g1
⎛ R ⎞⎛
ρ ⎞
=
g
⎜
1
⎟
⎜
1
⎟
2 ⎝ R 2 ⎠⎝
ρ2
⎠
⎛ m
and
⎜
⎝ m
1
2
⎞ ⎛ ρ
⎟ ×
⎜
⎠ ⎝ ρ
1
2
2
mv cm
+
1
I ω
2
1 1 R 2
= × 2 × 400 + × 2 × ×
2 2 2
2
1
⎞ ⎛ R
⎟ =
⎜
⎠ ⎝ R
1
2
⎞
⎟
⎠
3
2
XtraEdge for IIT-JEE 100
MARCH 2012

SOLUTION FOR MOCK TEST
PAPER - II AIEEE
CHEMISTRY
1. [3]
2. [3]
O
OH
CH 3 –Mg–Br
CH 3 –C–H
H 3 O ⊕ CH 3 –CH–CH 3
(Propan-2-ol)
C 2 H 5 OH
H ⊕ CH 3 –CH(OC 2 H 5 ) 2
(Acetal)
3. [2]
. .
3CH 3 –I + C 2 H 5 NH 2
4. [4] 5. [3]
6. [4] Laderer monassey reaction
7. [3] Paracetamol
8. [2] Fact
CH 3
⊕
C 2 H 5 –N–CH 3
CH 3
9. [4] Na 2 HPO 4 is group reagent of VIth group
mol
10. [4] r ∝
M.
M
11. [4]
12. [1]
Rate of dehydration ∝ stability of carbocation
of alcohol
13. [3] O CH 2
HI
OH + I – CH 2
14. [1] N 2 (g) + 3H 2 (g)
2NH 3 (g)
Initially at eq. 0.2 0.6 0
(0.2 – a) (0.6 – 3a) 2a
Total mixture is 0.8; 40% of it reacts, i.e.,
0 .8×
40
0 .8×
40 1
reacts to give × mole of NH3
100
100 2
I
or NH 3 formed is 0.16 mole or 2a = 0.16
∴ a = 0.08
∴ Initial mole = 0.8
Final mole = (0.2 – 0.08) + (0.6 – 0.24) + 0.16
= 0.12 + 0.36 + 0.16 = 0.64
0.64 4
∴Ratio of final to initial mole = = 0.8 =
0.8 5
15. [1]
∆Gº = – 2.303 RT logK
– 4.606 × 10 3 = – 2.303 × 2 × 500 log K
K = 100
16. [2]Stronger is acid, weaker is its conjugate base.
17. [3] He has highest ionization energy.
18. [3] Cl(g) + e – → Cl – (g) + EA ; ∆H = – EA
19. [4] In silica, one Si atom is attached with four
oxygen atoms.
20. [2] XeF 4 contains two, XeF 6 one, XeOF 2 two and
XeF 2 three lone pairs of electrons.
21. [4]
22. [1] Addition of Cl 2 /H 2 O is an electrophilic addition
reaction and Rate µ stability of carbocation
formed.
23. [2]
Na-liq.NH 3
C 2H 5OH
24. [3]
Br
CH 3 –CH–CH 2 –Br 2 NaNH2
–2HBr
CHO
(i) O 3
2 (ii) H 2O-Zn CH 2
CHO
H 3 C–C≡CH
–NH 3 NaNH 2
⊕
CH 3 –C≡C Na
C 2H 5Br
CH 3 –C≡C–C 2 H 5
XtraEdge for IIT-JEE 101
MARCH 2012

25. [2] Cl C 1 –––––C 2
AT C 1 : →
(3) F
(4)
Cl C–– CF(Br)Cl
(2)
Br (1)
At C 2 : →
Br(C 1 )FC–––C
(4)
Br
F
Cl
At C 1 : →
(3) F
Br
(1)
Cl
(2)
Cl
(2)
F
Br
Exchange
Method
F (3)
Br
(1)
C–––––C
(4)
C 1 ––CF(Cl)Br
At C 2 : →
(4)
Br(Cl)FC––C
F (3)
Cl (2)
Br(1)
Excharge
Method
(4)
Br
F
Cl
Br
(1)
Cw
C––––(2) R-form
(3)––C
F
Cl
(3)
(4)
Exchange C 1 ––(3)
(2)
Exchange
(1)
(2)––C
1
4
2
1
ACw
S-form
3
4
Cw
R-form
ACw
S-form
26. [1] Due to formation of Intramolecular H.B
Conjugated base if I and II is more stable.
Due to ortho effect III is more acidic than IV
II > I > III > IV
27. [2] SPM allows only solvent molecules to pass
through
28. [2] Edge-centre atom is shared in 4 cubic unit cells.
29. [3] In electrochemical cells, anode = –ve
31. [2] f(x) =
MATHEMATICS
Domain x 2 ≠ 0
x ≠ 0
and
x ≠ 1
log 25
2
log x , g(x) = log5
log x
f(x) =
2log5
2log x
Domain
x > 0, x ≠1
equality hold only for
x > 0, x ≠1
or
(0, 1) ∪ (1, ∞)
= log x 5
32. [3] The given curve is symmetric about the X-axis
as shown below
y
x'
O
y'
y 2 = 4ax
x
9a
∴ The required area = 2
∫
⎡
= ⎢4
⎢⎣
=
33. [2] S λ = 1 +
=
λ
λ −1
∴ ∑
λ=
34. [2]
x x ⎤
a ⎥
3/ 2 ⎥⎦
208a
2
3
1
+
λ λ
1
2
9a
a
⇒ λ = (λ – 1 )S λ
n
n
( λ −1)
Sλ
=
1
∑
λ=
1
=
3
+ ............ ∞
λ
n( n +1)
=
2
⎡ ⎛ 2π
⎞ ⎛ 2π
⎞⎤
⎢cos⎜
⎟ − sin⎜
⎟⎥
⎢ ⎝ 7 ⎠ ⎝ 7 ⎠⎥
⎢ ⎛ 2π
⎞ ⎛ 2π
⎞
sin cos
⎥
⎢ ⎜ ⎟ ⎜ ⎟
7 7 ⎥
⎣ ⎝ ⎠ ⎝ ⎠ ⎦
here k = 7
a
4axdx
8 a [ 27 a a − a a ]
7
⎡1
0⎤
= ⎢ ⎥
⎣0
1 ⎦
XtraEdge for IIT-JEE 102
MARCH 2012

35. [1] We have 10 digits {0, 1, 2, ---, 9}
Select any 2 and write in descending order
n = 10 10.9
C 2 .1 = = 45
2
36. [3] P (Exactly two of A, B and C occur)
= P(B ∩ C) + P(C ∩ A) + P(A ∩ B) –3 P(A ∩ B ∩ C)
= P(B). P(C) + P(C) . P(A) + P(A). P(B)
–3 P(A). P(B). P(C)
= 2
1 × 4
1 + 4
1 × 3
1 + 2
1 × 3
1 – 3× 3
1 × 2
1 × 4
1 = 4
1
37. [4] We have
a ≤ x i ≤ b ; i = 1, 2, 3.........n ...(1)
n
∑
i=
1
n
a ≤ ∑
=
n
i 1
x i
n
≤ ∑
=
i 1
b
na ≤ ∑ x i ≤ nb ⇒ a ≤ 1
i=
1
n
∑ x i ≤ b
i=
1
a ≤ x ≤ b ⇒ – b ≤ – x ≤ – a ...(2)
By (1) and (2), we get
(a – b) ≤ (x i – x ) ≤ (b –a) , i = 1, 2 , ........ n
|x i – x | ≤ (b –a) , i = 1, 2 , ........ n
(x i – x ) 2 ≤ (b – a) 2 ; i = 1, 2, ......n
n
∑
i=
1
38. [2]
2
( x i − x)
≤ n(b – a) 2
var (x) ≤ (b – a) 2
e'
e
x
+ y
e e' = 1
Hyperbola & it's conjugate hyperbola
1 1
+ 1
2 2
e e
1
2
4 4
+ = 1
2 2
e e'
1 1
2 2
e + e'
= 1
4
x 2 + y 2 = r 2
p = r
| 0 + 0 −1|
= r ⇒ r =2
1 1
2 2
e
+ e'
n
....(1)
39. [1] First two family of lines passes through (1, 1)
and (3, 3) respectively.
⇒ Point of intersection of lines belonging to
third family will lie on y = x
⇒ ax + y – 2 = 0, & 6x + ay – a = 0
2
a a −12
solving x = , y =
a 2 2
− 6 a − 6
2
a a −12
⇒
=
a 2 2
− 6 a − 6
⇒ a 2 – a – 12 = 0
⇒ (a – 4) (a + 3) = 0
⇒ a = 4, a = – 3
40. [2] A
41. [1]
(–2,–2)
B
B
(0, 1)
y
⎛12
12 ⎞
⎜ , ⎟
⎝ 7 7 ⎠
(0,0)
(1,0)(2,0)(3,0)
A
2
(r,r)
2
y = x
C
(1, 1)
D (1, 0)
( r −1)
+ ( r −1)
= r
⇒ 2(r – 1 2 ) =r 2
⇒ 2(r 2 – 2r + 1) = r 2
⇒ r 2 – 4r + 2 = 0
4 ± 16 − 8
r =
2
4 ± 2 2
r =
2
r = 2 ± 2
x
4x + 3y = 12
C
⎛ 9 ⎞
⎜ ,−2⎟
⎝ 2 ⎠
y +2 = 0
42. [3]
r r r r
LHS → AB . CD = ( b − a)
. ( d − c)
= b r . d r – b r . c r – a r . d r + a r . c r
RHS → k{| AD | 2 + | BC | 2 – | AC | 2 – | BD | 2 }
r r 2 r r
= k { | d − a | + | c − b |
2 r r r r
– | c − a |
2 – | d − b |
2 }
r r r r r r r r
= k {– 2 ( a.
d + c.
b)
+ 2[ ( a.
c)
+ ( d.
b)
]}
r r r r r r r r
= 2k { a.
c + b.
d − b.
c − a.
d}
...(2)
1
By (1) and (2) ⇒ k = 2
XtraEdge for IIT-JEE 103
MARCH 2012

43. [1]
r
Given ( a b
r
× )
r r 1 1 1 × ( c × d ) = î – ĵ + kˆ ...(1)
6 3 3
Q a r . b r = | a r || b r 3
| cos30º = 2
⎡
r r r
Q a,
b,
c are coplaner vectors
⎢ r r r
⎢⎣
∴[
a b c]
= 0
r
( a b)
r r r
× × ( c × d)
r r r r r r r r
= { ( a × b).
d}
c –{ ( a × b).
c}
d
= [ a r b r d r ] c r 1 1 1
= î – ĵ + kˆ
6 3 3
1
= ( i ˆ − 2 ˆj
+ 2kˆ)
6
= c r ⎛ ⎞
= ⎜
iˆ
− 2 ˆj
+ 2kˆ
⎟
⎝ 3 ⎠
44. [3] L 1 :
L 2 :
x +1
2
x −1
1
=
y z =
−1 2
y z − 3
= = 2 λ
{By (1)}
...(1)
...(2)
Shortest distance between the lines (1) & (2) is
r r r r
( a1
− a2).(
b1
× b2
)
⇒ r r = 1
| b1
× b2
|
⇒ λ = ?
45. [1] l + m + n = 0 ...(1)
and lm = 0 ...(2)
⇒ l = 0 or m = 0
By (1) l = – m – n (3)
Case (I)
If m = 0
By (3) ⇒ l = – n
∴ DR's of line (1) is
⇒
⎧ a1,
b1,
c1
⎪
⎨−
n,
0, n
⎪
⎩ −1,
0, 1
Case (II)
If l = 0
By (3) ⇒ m = – n
DR's of line (2) is
⇒
⎧a2,
b2,
c2
⎪
l,
m,
n
⎨
⎪ 0, − n,
n
⎪
⎩ 0, −1,
1
...(2)
⇒ cosθ =
⇒ θ = 3
π
2
1
a
a a
1
2
2
1
+ b
46. [3] f(0) = λ [0] = 0
L.H.L.
R.H.L.
lim
→0 h
+ b b + c c
1
2
1
+ c
1
0− h
2
a
5 = 0
2
2
1
2
+ b
2
2
+ c
2
2
lim λ [0 + h] = 0 ∀ λ ∈ R
h→0
= 2
1
47. [1] x 2 + y 2 ≤ 4
R = {(–2, 0) , (–1, 0), (–1, 1) , (–1, –1), (0, 0), (0, 1),
(0, 2), (0, –1), (0, –2), (0,0), (1, 1), (1, –1), (2, 0)}
⇒ D R = {–2, –1, 0 , 1, 2}
48. [3]
(i) R 1 is not a relation Q 4 ∉ A
(ii) R 2 is subset of A × B, ∴ it is a relation
(iii) R 3 is subset of A × B, ∴ it is relation
(iv) R 4 is subset of A × B, ∴ it is relation
49. [2] Area of ∆ formed by z, ωz, z + ωz
= 2
1 |z| 2 . sin 120º
But it is given 2
1 (z) 2 sin (120º) = 100
3
|z| 2 1
= 25
1
|z | = 5
| z + ωz| = |z| | 1 +ω| = 5
1 | –ω 2 | = 5
1 × 1 = 5
1
(Q 1 + ω = –ω 2 )
50. [1] do your self
51. [4]
Let g(x) = f(x) – x 2 ⇒ [g(1) = 0 , g(2) = 0,
g(3) = 0 as f (1) = 1, f (2) = 4, f (3) = 9]
From RT on g(x), g' (c 1 ) = 0 for at least x ∈ (1, 2)
⇒ c 1 ∈ (1, 2)
RT on g (x), g' (c 2 ) = 0 for at least x ∈ (2, 3)
⇒ c 2 ∈ (2, 3)
∴ Now g' (c 1 ) = g' (c 2 ) = 0
⇒ so between x ∈ [c 1 , c 2 ], g" (x) = 0
⇒ f" (x) –2 = 0 ⇒ f" (x) = 2
XtraEdge for IIT-JEE 104
MARCH 2012

2
sin x
2
52. [3]
∫
dx = x
6
cos x
∫
tan . sec 4 x dx
∫
2
= tan x (1 + tan 2 x). sec 2 xdx
tan x = t
53. [3] y = u m dy
⇒ = mu m–1 du
. dx dx
Hence, 2x 4 .u m .m u m–1 du
. + u 4m = 4x 6
dx
6
4m
du 4x
− u
3 = ⇒ 4m = 6 ⇒ m =
4 2m−1
and
dx 2mx
u
2
3
2m – 1 = 2 ⇒ m = 2
(1,2)
57. [2] Statement I and Statement II both correct and
statement II is correct explanation of Statement I.
58. [2] P(A∩ B ) = P(A) P( B )
= P(A) ⋅ (1–P(B))
59. [4] Contra positive of compound statement
~ (p ∧ q) → q ≡ ~ q → ~ (~( p ∧ q))
~ (p ∧ q) → q ≡ ~ q → (p ∧ q)
so, statement I is wrong.
but statement II is correct.
60. [2] Both statements are correct and statement 2 is
correct explanation of statement 1
PHYSICS
54. [3]
(t 2 , 2t)
61. [3] Statement 1 is true, statement 2 is false.
62. [3] A halved, δ halved.
63.
5
[1] V father = 60
Equation of tangent at (1, 2)
y . 2 = 2 (x + 1)
x – y + 1 = 0
image pt of (t 2 , 2t) about line x – y + 1 = 0
2
x − t y − 2t
− 2(
t
2 − 2t
+ 1)
= =
1 −1
2
x = t 2 – t 2 + 2t – 1, y = 2t + t 2 – 2t + 1
x = 2t – 1 y = t 2 + 1
x +1
= t y – 1 = t 2
2
Eliminating
2
( x + 1)
y – 1 = ⇒ (x + 1) 2 = 4 (y – 1)
4
55. [1] f (x) = min ({x + 1}, {x –1}) = min ({x}, {x}) = {x}
4
so
∫
f ( x)
dx = 9.
∫
x − [ x]
dx
−5
1
0
56. [2] Statement I and Statement II both correct and
statement II is correct explanation of Statement I.
as h' (x) =
m n
n
m − x
n
=
m
.(x)
n
⎛ Even ⎞
−⎜
⎟
⎝ odd ⎠
as h' (x) is undefined at x = 0 so h' (x) does not
change.
sign in neighbour hood ⇒ No extreme
5
For the daughter V = , after catches M = 21
21
64. [3] dτ = (dq)E(2x sin θ)
+
+λ
–λ
⎛ σ ⎞
= λ(dx) ⎜ ⎟
(2x sin θ)
⎝ 2ε0
⎠
σλxsin
θ
= dx
ε
τ =
2
0
σλl sin θ
2ε
65. [3]
According to Newton law of cooling rate of loss of
heat (T – T 0 ), where T is the average temperature
in the given time interval hence.
m C
0
⎛ 60 − 50 ⎞ ⎛ 60 + 50 ⎞
⎜ ⎟ ∝ ⎜ – 25⎟ ⎝ 10 ⎠ ⎝ 2 ⎠
( 50 – T ) ⎛ 50 + T ⎞
and m C ∝ ⎜ 625 ⎟
10 ⎝ 2 ⎠
Solving we get : T = 42.85ºC
XtraEdge for IIT-JEE 105
MARCH 2012

66. [1] x rel = (v 0 ) rel t + 2
1
arel .t 2
x = v 0 t – 2
1 at 2 [a rel = a man – a bus = 0 – a]
f
∴ 1
f2
=
T1
T2
or f 2 = 4f 1
ρ2
ρ1
r2l2
r1 l1
= ( ⎛ 1
2)
⎟ ⎞
⎜
⎝ 2 ⎠
⎛ 1 ⎞
⎜ ⎟
⎝ 4 ⎠
at 2 – 2v 0 t + 2x = 0
t =
t =
2v
0 ±
4v
2a
2
0
a
2
0
– 8ax
v v – 4ax
0 ±
v0 v0
– 2ax
or t =
+
a
For time to be minimum
2
v 0 – 4ax = 0
v 0 = 2 ax
2
74. [1]
s
v s
O
∆f = f 1 – f 2
⎛ v ⎞ ⎛ v ⎞
= f ⎜ ⎟
– f ⎜ ⎟
⎝ v − v
s ⎠ ⎝ v + v s ⎠
⎡ ⎞ ⎛ ⎞
= ⎢⎜
⎛ 1
v − s v
f 1 ⎟ − ⎜1+
s
⎟
⎢⎣
⎝ v ⎠ ⎝ ν ⎠
s
− −1
⎡⎛ v ⎞ ⎛ ⎞⎤
= ⎢⎜
+
s vs
f 1 ⎟ − ⎜1−
⎟⎥ =
⎣⎝
v ⎠ ⎝ ν ⎠ ⎦
⎤
⎥
⎥⎦
2 f vs
v
v s
67. [4] Scooter ←⎯→
1km
Bus
Relative distance = (time) (relative velocity)
(In uniform motion)
1000 = (100) v rel.
∴ 10 = (v s – v b )
v s = 20 m/s
68. [3] Use, δ m = (µ − 1) A
69. [1]
zener diode is used in parallel to load resistance is
connected in R.B.
70. [4] P =
71. [3]
72. [4]
2Ω
V 2 =
R
⇐
12 2 = 48 W
3
6Ω
3Ω
dN
75. [3] From – = λN dt
n
n = λN or λ = N
76. [2] Binding energy
BE = (M nucleus – M nucleuon )c 2
= (M O – 8M p – 9M n )c 2
77. [2] E k =
=
6.6×
10
hc
e
–34
1.6×
10
⎛ 1 1 ⎞
⎜ ⎟
–
(in eV)
⎝ λ λ0
⎠
–19
8
× 3×
10
⎛
10 10
⎞
⎜
10 10
– ⎟
⎝1800
2300 ⎠
78. [2] Voltage gain = β(Resistance gain)
= 1.5 eV
i
12V
12
i = = 6A 2
12V
79. [3] E is always negative
2π
80. [3] ω = 100 π = T
v T / m
73. [3] f ∝ ∝ l l
∝
∝
T / ρs
l
T / ρr
l
f ∝
2
(m = mass per unit length = ρs)
T / ρ
rl
81. [4] F = 2πrT
22
F =2 × × 0.1 × 10 –3 × 0.07
7
82. [3] (K.E.) max = – 2
1 (K.E.) of boy
1 (2m) u 2 1 1
= × mu'
2
2 2 2
(K.E.) man = (K.E.) of boy
1 (2m) (u + 1) 2 1
= mu'
2
2
2
XtraEdge for IIT-JEE 106
MARCH 2012

83. [1] V cm =
m v + m v
1 1 2 2 5
= – m/s
m1
+ m2
3 CHEMISTRY JOKES
⎛ 1 1 ⎞
⎜ –
⎟
⎝ R1
R2
⎠
y
i 2
B R
x
⇐
i 1
Chemistry Joke 7:
B 2
B2
B1
µ 0 i2
3
2R
1 Amp.
84. [4] As no external force is applied
∴ (v cm = constant) = 0
85. [4] P = (µ – 1)
µ decreases, P decreases, f increases.
86. [1] M ∝ k L 1 L 2
87. [2]
88. [4]
W 1 = MB (cos 0º– cos 90º)
W 2 = MB (cos 0º – cos 60º)
W 1 = nW 2
n = 2
89. [3]
Zener diode is always connected in RB and it act
as voltage regulator.
90. [4]
B e
B 1
30º
tan 30º =
B 1 = 3 B 2
µ 0 i1
=
2R
i 2 = i 1
=
3
3
If you didn't get the joke, you probably didn't
understand the science behind it. If this is the case,
it's a chance for you to learn a little chemistry.
Chemistry Joke 1:
Q: Why do chemists call helium, curium and
barium the medical elements?
A: Because if you can't helium or curium, you
barium!
Chemistry Joke 2:
Q: What is the name of the molecule CH 2 O?
A: Seawater
Chemistry Joke 3:
Q: What do you call a joke that is based on cobalt,
radon, and yttrium?
A: CoRnY.
Chemistry Joke 4:
Q: If a mole of moles were digging a mole of holes,
what would you see?
A: A mole of molasses.
Chemistry Joke 5:
Q: What does a teary-eyed, joyful Santa say about
chemistry?
A: HOH, HOH, HOH!
Chemistry Joke 6:
Susan was in chemistry. Susan is no more, for what
she thought was H 2 O was H 2 SO 4 .
Q: Why is potassium a racist element?
A: Because, when you put three of them together,
you get KKK.
Chemistry Joke 8:
An electron sitting in a prison asked a second
electron cellmate, "What are you in for?" To which
the latter replied, "For attempting a forbidden
transition."
Chemistry Joke 9:
Q: What is the dullest element?
A: Bohrium
Chemistry Joke 10:
At the end of the semester, a 10th-grade chemistry
teacher asked her students what was the most
important thing that they learned in lab. A student
promptly raised his hand and said, "Never lick the
spoon."
XtraEdge for IIT-JEE 107
MARCH 2012

PHYSICS
⎛ t ⎞
1. [C] f = f 0 ⎜1
– ⎟
⎝ T ⎠
At t = 0, v = 0
t
When f = 0, then = 1 or t = T
T
dv ⎛ t ⎞ = f0 ⎜1
– ⎟
dt ⎝ T ⎠
or dv = f 0 dt – f 0
T
t dt
v x
∫
dv = f
0
T
0 ∫
0
f
v x = f 0 T – 0 T
T 2
2. [A] tan β =
T
1
dt – f0
T ∫
t dt
2
0
= f 0 T – 2
1
f0 T = 2
1
f0 T
Q sin θ
P + Q cos θ
……. (i)
R
Q
R' Q
Q
θ
β' β
P
Q sin(180º– θ)
tan β' =
……. (ii)
P + Q cos(180º– θ)
90°
180–θ θ
β' β
Qsin
θ
or tan β' =
P – Q cosθ
But β' + 90º + β = 180° or β' = 90° – β
tan β' = tan(90° – β) = cot β
Qsin
θ
∴ cot β =
……… (iii)
P – Q cosθ
Multiplying (i) and (iii),
P
SOLUTION FOR MOCK TEST
PAPER - II BIT-SAT
2
Q
2
– Q
sin
2
2
θ
cos
2
θ
= 1
or Q 2 sin 2 θ = P 2 – Q 2 cos 2 θ
or Q 2 (sin 2 θ + cos 2 θ) = P 2
But sin 2 θ + cos 2 θ = 1
∴ Q 2 = P 2 or Q = P
No need for negative sign.
3. [C] In order to conserve momentum, C should
move with speed v in a direction opposite to that
of B.
4. [C] Percentage energy saved
1 2
mv
2
= 2
v
× 100 = × 100
1
2
2
mv + mgh
v + 2gh
2
12×
12
=
× 100 ≈ 38
12×
12 + 2×
9.8×
12
5. [C] Iω = constant,
2 MR 2 2π
× = constant or
5 T
or
2
R
T '
/ 4
=
2
R
T
2
R
T 4T 24
or
R
2' =
2
R
6. [B] Limit of resolution of eye = θ =
–7
= constant
or T ' = 6 hours
1.22λ
D
1.22×
5×
10
=
= 2.03 × 10 –4 rad
–3
3×
10
If the maximum distance at which dots are
resolved is x, then
θ =
or x =
1 mm
x
=
10
–3
10
–3
x
2.03×
10
0. 6
7. [B] (a) f = – 2
1 1 + =
v – 10
1 1 1 = –
v 10 30
–4
1
– 30
= 2.03 × 10 –4
m ≈ 5m
m = – 0.3 m = – 30 cm
XtraEdge for IIT-JEE 108
MARCH 2012

1 3 – =
v 301
30
or v = cm = 15 cm
2
v 15
(c) m = – = – = 1.5
u – 10
(d) Object lies between principal focus and pole.
So, the image is virtual and erect.
8. [C] R CM =
→
1 r 1
m + m
m + m
1
→
2 r 2
1(ˆ i + 2 ˆj
+ kˆ)
+ 3(–3ˆ i – 2 ˆj
+ kˆ)
=
1+
3
= – 2 î – ĵ + kˆ
V 1 T
9. [C] V∝ T ⇒ =
1 V (273 + 27)
⇒ =
V2
T2
3V
T2
⇒ T 2 = 900 K ⎯→ 627°C [Q T(K) = 273 + t°C]
10. [A] Initial and final states are same in all the
process.
Hence ∆U = 0; in each case
By FLOT; ∆Q = ∆W = Area enclosed by curve
with volume axis.
Q (Area) 1 < (Area) 2 < (Area) 3 ⇒ Q 1 < Q 2 < Q 3
11. [C] Resultant amplitude = a + a + a cosφ
=
2 2
π
2
2 2
1 2 2 1a2
0.3 + 0.4 + 2×
0.3×
0.4×
cos = 0.5 cm
2
12. [A]
Suppose n A = known frequency = 100 Hz. n B = ?
x = 2 = Beat frequency, which is decreasing after
loading (i.e.x ↓)
Unknown tuning fork is loaded so n B ↓
Hence n A – n B ↓ = x↓ …… (i) ⎯→ Wrong
n B ↓ – n A = x ↓ …….(ii) ⎯→ Correct
⇒ n B = n A + x = 100 + 2 = 102 Hz.
13. [C] Current through each arm
PRQ & PSQ = 1 A
V P – V R = 3V
V P – V S = 7V
V R – V S = 4V
14. [B] V < E
12
E = 12 + r _____(i)
6
11
& E = 11 + r _____(ii)
10
20
On solving r =
7
IgG
15. [B] S =
I − Ig
16. [B] B net =
17. [A]
N
N
P =
S
S I P
µ 0N
⎛ i ⎞
⎜ 1 i2
−
⎟
2 ⎝ r1
r2
⎠
I
18. [D] In forward biasing both electrons and protons
move towards the junction and hence the width of
depletion region decreases.
r
19. [A] W = Q ( E
r r r r r
∆ ) = F.
∆
F = Q E
W = Q[e 1 î + e 2 ĵ + e 3 kˆ ] . (a î + b ĵ )
W = Q(e 1 a + e 2 b)
20. [C] Change will move along the circular line of
force because x 2 + y 2 = 1 is the each of circle.
21. [A] σ i = i
θ = iG
θ .G = σv G
σ i
⇒ G
= σv
⎛
Rt
− ⎞
22. [C] i = i ⎜ − L ⎟
0 1 e
⎜ ⎟
⎝ ⎠
23. [A]
di i = 0
e
dt LR −
initially t = 0
di i = 0
dt LR
N A =
N =
N 0
2
10
t /1
N A = N B
10 = 2 t/2
n
L
Rt
1
t /T
=
H
21
2
/ 2
1 t
⎛ ⎞
+ N B = 1 ⎜ ⎟
⎝ 2 ⎠
⇒ log 10 10 = 2
t log10
2
t = 6.62 yr
24. [D] Potential difference across the resistance 20Ω.
Which is V = i × 20
48
i =
100 + 100 + 80 + 20
25. [A] Intensity = Power per limit area.
P = pv
P 0.5
p = = v 8
3×
10
XtraEdge for IIT-JEE 109
MARCH 2012

E
26. [A] I =
R + r
P 1 = I 2 R 1
P 2 = I 2 R 2
Power delivered is same in both cases.
2
2
⎛ E ⎞ ⎛ E ⎞
⎜ R1
R1
r
⎟ = R2
⎝ +
⎜
⎠ R2
r
⎟
⎝ + ⎠
R1
R1
+
( r)
2
=
⇒ r = R 1R2
27. [A] q =
B∆A
R
R2
R2
+
28. [B] F = qvB sinθ
( r)
29. [D] θ = 0 F = 0
Hence no change in the velocity.
2
30. [A]
31. [C] Diamagnetic materials have negative
susceptibility
32.
mV
[D] r = qB
r 1 V =
1 B .
2
r2
V2
B1
ρ0
33. [B] ρ =
1+ γ∆T
or 1(1 + γ × 4) = 0.998
∴ γ = – 5 × 10 –4 /°C
Negative sign tells that for 0 – 4°C, water contracts
on heating.
∂y
34. [A] From V P = – V ×
∂x
∂y
At location of P, is – and V is along +ve
∂x
x-axis.
So, V P is along +ve x-axis
35. [C] Deviation should take place at each face.
Dispersion takes place at first face only.
36. [B] 1 atmosphere ≈ 10 5 Pa
Also, p = hρg
= 10 × 1000 × 10 Pa = 10 5 Pa
So, total pressure is nearly 2 × 10 5 Pa
37. [B] T ∝ (r) 3/2
Since r is doubled therefore T is increased by a
factor of [2] 3/2 or 8 or 2 2
So, the new time period is 365 × 2 2 days.
L/3 L/6
38. [C]
L/2
Using theorem of parallel axes,
2
2
ML ⎛ L
I = + M
12 6 ⎟ ⎞
⎜
⎝ ⎠
I =
39. [D] 2
1 KL 2 =
2
2
ML ML 4ML ML
+ = =
12 36 36 9
2
P
2M
2
or p = L
40. [B] → A . → B = 0 ⇒ → A ⊥ → B
→ → → →
A × C = 0 ⇒ A || C
∴ → B ⊥ → C ∴ θ = 90°
CHEMISTRY
MK
1. [C]
Ca(OH) 2 (aq.) + CO 2 (g) —→ CaCO 3 (s)
1 mol 100 g
given: 0.05 × 0.5 mol ?
0 .05×
0.5×
100
=
1
= 2.5 g
n 1
2. [A] r = ∝ t m
⇒ w ∝
m
⇒ M
w ∝
8 16
3. [D] No. of atoms per unit cell = + = 4 8 2
4
Vol. of 4 atoms = 4 × πr 3 16
= πr
3
3 3
h
4. [C] λ = mv
5. [A] ∆x. ∆v =
Here ∆v = 10 4 ×
h
4πm
0.011
100
2
1
M
XtraEdge for IIT-JEE 110
MARCH 2012

6. [D] ∆ r Hº = ∑ AgH º ( p)
− ∑ AgH º ( R)
7. [C]
8. [C] ∆Hg > 0
⎛ 1 ⎞
9. [C] K 2 =
⎜
⎟
⎝ K1
⎠
1/ 2
[ CH 3COO
]
10. [B] pH = pK a + log
[ CH 3COOH
]
6 = pK a + log 1 ⇒ Ka = 10 –6
11. [A] Al(OH) 3 (s) Al 3+ (aq) + 3OH¯(aq)
S 3S
Given pH = 4 ∴ pm = 10 ∴ [m – ] = 10 –10
k sp = [Al 3+ ][OH¯] 3
⇒ 10 –33 = S × (10 –10 ) 3 ⇒ S = 10 –3 M
12. [B] r = k[A] 2
13. [C]
14. [C] ∆T b = K b × m
1. 8 1000
0.1 = K b × × 180 100
K b = 1K/m
15. [D] E° cell = E° red (c) + E° oxi (A)
= – 0.41 + 0.76 = 0.35 V
16. [B] 2H + + 2e – ⎯→ H 2
2
0.059 [ H + ]
E = E° + log
2 [ H 2]
E can be greater than E° if [H + ]
is increased.
CCl
17. [B] + Br 2 ⎯⎯→
⎯
4
18. [A]
(Anti-add n reaction)
[Syn-add n ]
cis-alkene
O S O4
⎯
H 2O2
−
Br
Br
⎯ ⎯ → meso product
H
C 2 H 5
OH
OH
20. [C] Bicylo[1,1,0]
21. [D]
22. [B]
23. [C]
24. [D]
25. [B]
δ–
O
Θ ⊕
Θ ⊕
R/Mgx + H–C–H ⎯→ R–CH 2 –O/Mg x
δ+
HOH hydrolysis
O
..
⊕ Θ
(i) H/CN
CH 3 – C–H
(NAR)
(ii) H 3 O ⊕
R–CH 2 –OH+ Mg(OH)x
OH
CH 3 – C–COOH
H
Lactic acid
CH 3
CH 3
Ether
CH– Br + 2Na + Br – CH
CH
CH 3
3
CH 3
2,3-dimethylbutane CH–CH
H 3 C
CH 2 –OH
CH–OH
CH 2 –OH
Glycerol
+ COOH ⎯ 110 ⎯⎯
°C → HCOOH
COOH
CH 3
CH 3
formic acid
CH 3
H
C=C
H
OH
|
CH 2 –CH 2 –*C–CH 3
|
H
Chiral centre
(optical activity is shown)
two geometrical isomers will be formed due to
double bond.
two optical isomers with one chiral centre
19. [B]
C 2 H 5
CH 2 –CH=CH 2 ⎯ NBS ⎯⎯ → CH–CH=CH 2
|
Br
[NBS. substitutes bromine at allylic position]
XtraEdge for IIT-JEE 111
MARCH 2012

26. [B] COOH
CH 2
CH 2
CH 2
CH 2
COOH
Adipic acid
Ca(OH) 2
O
||
C–O
C–O
||
O
Ca
–CaCO 3 dry distillation
34. [C] POP ⇒ CaSO 4 . 2
1
H2 O
35. [B]
H
H
B
H
H
B
H
H
Bridge H ⇒ 2
Terminal H ⇒ 4
36. [B] Order of Lewis Acid BCl 3 < AlCl 3
37. [C] Fact
38. [D] Brass ⇒ Cu + Zn
non metal ⇒ Cu + Sn + Zn
German silver ⇒ Cu + Zn + Ni
O
Cyclopentanone
39. [B] CN – is a strongest ligand
40. [C] Ionisation Isomerism
27. [B] Pinacol → (vic-diols)
OH OH
H 3 C—C—C—CH 3
CH 3 CH 3
2,3-dimethyl–2,3-butanediol
28. [A]
O
H–C–H , do not show aldol condensation
αH = 0
1
6 2
29. [A]
3
H CH3
3 C 5
4
(A) 3, 5–dimethyl cyclohexene
30. [C]
L
O –2 F – Na + Mg +2 Al +3 R
Size↓
ENC ↑
∴ Correct order O –2 > F – > Na + > Mg +2 > Al +3
31. [B]
O < N
13.6,14.6
MATHEMATICS
1. [B] Let the point is (x 1 , y 1 ) then
3x 1 + 4y 1 = 5 … (1)
Also, (x 1 – 1) 2 + (y 1 – 2) 2 = (x 1 – 3) 2 + (y 1 –4) 2
⇒ 4x 1 + 4y 1 = 20 … (2)
Solving these we get,
x 1 = 15, y 1 = – 10
X – Y
2. [B] x = X cos 45º – Y sin 45º =
2
X + Y
y = X cos 45º + Y cos 45º =
2
Hence equation be
2
2
3 ⎛ X – Y
2
⎟ ⎞
⎜ + 3 ⎛ X + Y
⎝ ⎠ 2
⎟ ⎞ ⎛
2
⎜ + ⎜
X – Y
2
⎝ ⎠
⎝ 2
3(x – y) 2 + 3(x + y) 2 + 2(x 2 – y 2 ) = 4
⇒ 8x 2 + 4y 2 = 4
⇒ 2x 2 + y 2 = 1
3. [A]
2
⎞
⎟
= 2
⎠
32. [A] (Li – Mg), (Be – Al), (B – Si)
⇒ show Diagonal Relationship
33. [A] Order of solubility of sulphate
BeSO 4 > MgSO 4 > CaSO 4 > SrSO 4
XtraEdge for IIT-JEE 112
MARCH 2012

4. [A]
3
10. [C] f (x) = | x |, g (x) = [x – 3]
– 8 8 < x <
5 5
⇒ 0 ≤ f (x)
Now for 0 ≤ f (x) < 1
= – 3 [Q – 3 ≤ f (x) – 3 < – 2]
again for 1 ≤ f(x) < 1.6
g (f (x)) = – 2 [Q – 2 ≤ f (x) – 3 < 1.4]
required get = {–3, – 2}
5. [C] x 2 – 3xy + 2y 2 = 0
⇒ (x – 2y) (x – y) = 0
⇒ x – 2y = 0 and x – y = 0
Also x 2 – 3xy + 2y 2 + x – 2 = 0
x – 2y + 2 = 0 and x – y – 1 = 0
Clearly it is parallelogram
( x – 2)
6. [C]
12
a 2 e 2 = a 2 + b 2
⇒ ae = 4
⇒ 2ae = 8
2
–
( y + 1)
4
2
= 1
7. [C] x 2 + 4x – 5 = 0
⇒ x 2 + 5x – x – 5 = 0
⇒ x(x + 5) – (x + 5) = 0 ⇒ x = –5, 1
Hence points are (1, 2) and (1, –2)
At (1, 2) tangent on circle be
x + 2y = 5
its slope is (–1/2)
Also at (1, 2) tangent on parabola be
2y = 2(x + 1)
Whose slope = 1
than tan θ =
1
1 +
2
1
1 –
2
= 3
8. [A] As the point (2 2 , 1) satisfies the eq n of
director circle
x 2 + y 2 = 25 – 16
⇒ x 2 + y 2 = 9
9. [B] Eq n of tangent having slope 'm' to ellipse
y = mx +
2
2
a m + b
As it touch circle x 2 + y 2 = r 2 then
a 2 m 2 + b 2 = r 2 + r 2 m 2
⇒ m =
r
a
2
2
– b
– r
2
2
2
11. [B] f (0) = f (0 +)
cos3h
– cos h
λ = lim
h→0
2
h
2sin 2 h.sin
h
= lim –
= λ = – 4
h →0 (2h)
( h)
dy 1+
( y / x)
12. [B] =
2
dx ( y / x)
put y = vx
3
dy 1+ v
v + x = dx
2
v
After solving
y 3 = 3x 3 log cx
13. [D] Put x e x = t
(x + 1) e x dx = dt
2
∴
∫sec t dt = tan t + c =
3
tan ( xe x )
+ c
f ( x)
14. [C] Put x = 3 – h
lim [– h] + [– 1 – h] = –1 + 1 = 0
h →0
15. [C] x 3 + 4x 2 + 3x = x (x 2 + 4x + 3)
= x (x + 1) (x + 3)
x ≠ 0, – 1, – 3
A = R – {0, – 1, – 3}
16. [B] Integrate both sides
a tan –1 x + b [log (x – 1) – ln (x + 1)]
dx
=
∫ 2 2
( x –1)( x + 1)
a tan –1 x + b [log (n – 1) – ln (x + 1)
1 ⎡ dx dx ⎤
= ⎢
⎥
⎣∫ –
2
2
–1
∫ 2
x x + 1 ⎦
1 ⎡1
⎛ x –1⎞
–1 ⎤
= ⎢ l n⎜
⎟ – tan x⎥ 2 ⎣2
⎝ x + 1⎠
⎦
Compare
a = –1/2, b = + 1/4
a – 2b = – 2
1 – 2 (+ 1/4) = – 1
XtraEdge for IIT-JEE 113
MARCH 2012

17. [C] f ′(x) = 3x 2 + 2ax + b
18. [C]
A = 3 > 0
D = 4 (a 2 – 3b) Q a 2 – 3b < 0
∴ D < 0
∴ f ′(x) > 0 ⇒ f is strictly increasing
⇒ no maxi./min. lie.
y = –x
O 2
y = x
A =
∫ 2
y dx =
∫ 2
x dx = 2
19. [C] 0 < x < 1
⇒ x 2 > x 3
0
2
2 x > 2 x3
∫ 1 0
20. [D]
x
2
2 dx >
∫ 1 2 dx
0
–1
x
0
3
1
⇒ I 1 > I 2
One-one function
Range = [–1, 1] = co domain
⇓
onto function
21. [B] x 5 – 2y 4 = 0
dy ⎡
4
5x
=
dx ⎥ ⎥ ⎤
⎢ 3
⎢⎣
8y
⎦
⎛ dy ⎞
⎜ ⎟
⎝ dx ⎠
2,2
length =
22. [B]
dy =
dx
⎡ 80 ⎤ 5
= – ⎢ ⎥ = –
⎣8×8
⎦ 4
y 2 8
= =
5/ 4 5
y 1
dy / dθ
=
dx / dθ
a cosθ
⎡
1 ⎤
a ⎢– sin θ +
⎥
⎣ 2sin θ/
2cosθ/
2⎦
dy =
dθ
(cosθ).sin
θ
cosh
dy
⇒ = tan θ dx
⎛ ⎞
23. [C] t n = (n + 1) ⎜ + 1 ⎛ 1 ⎞
n ⎟ ⎜n
+ ⎟
⎝ ω ⎠ ⎝ ω 2 ⎠
= n 3 + n 2 ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎜ + + 1
2
⎟ + n ⎜1+
+ ⎟ + 1
2
⎝ ω ω ⎠ ⎝ ω ω ⎠
= n 3 + n 2 (ω + ω 2 + 1) + n (ω + ω 2 + 1) + 1
= n 3 + 1
n
∴ S n = ∑
=
r 1
t r
n
= ∑
=
r 1
3 n
( r + 1) =
a 1 + a4
a 2 + a3
24. [C] = ;
a1a4
a2a3
1
so,
a + 1
4 a = 1
1 a + 1
or
3 a2
Also,
25. [D]
3( a
a
a
2 – 3)
2a3
a1 =
– a
a a
1
4
4
1
2
( n + 1)
4
a – 4 3
;
2
+ n
1
a = 1
a – 1
2 a1
… (i)
⎛ 1 1 ⎞ 1
So, 3 ⎜ ⎟
–
=
⎝ a3
a2
⎠ a – 1
… (ii)
4 a1
Clearly, (i) and (ii)
1
⇒
a – 1
2 a = 1
1 a – 1
3 a = 1
2 a – 1
4 a ;
3
1
so,
a , 1
1 a , 1
2 a , 1
3 a
are in A.P. or a 1, a 2 , a 3 , a 4 are
4
in H.P.
2
x + 1– 2 2
f (x) = = 1 – ;
2
x + 1 x 2 + 1
2
f (x) is minimum when is maximum or (x 2
x 2 + 1
+ 1) is minimum
i.e. x 2 + 1 ≥ 1 for all x
⎛ 2 ⎞
Hence ⎜
2
⎟ = 2
⎝ x + 1⎠max.
2
or f (x) = 1– is minimum, when
x 2 + 1
maximum
or f (x) = 1 – 2 = – 1
2
is
x 2 + 1
26. [B] As it is a third-degree homogeneous
expression in x, y, we have
y 3 + y 2 x – µ yx 2 + λ + x 3 =(y + x) (y + 3x)(y + mx)
= y 3 + (m + 3 + 1) y 2 x + (3 + m + 3m) yx 2 + 3mx 3
⇒ 1 = m + 4, – µ = 3 + 4m, λ = 3m,
⇒ m = – 3
∴ y – 3x is third factor
XtraEdge for IIT-JEE 114
MARCH 2012

27. [C] Required number
= coeff. of x 30 in (x 2 + x 3 + …..+ x 16 ) 8
= coeff. of x 30 in x 16 (1 + x + …..+ x 14 ) 8
8
⎛
15
= coeff. of x 14 1– ⎞
in ⎜
x
⎟
1–
⎝ x ⎠
= coeff. of x 14 in (1 – x 2 ) –8 = 21 C 14 = 116280
28. [C] Here |t r + 1 | = 7 C r .4 7–r . (3x) r
and | t r | = 7 C r–1 4 8–r (3x) r–1
| t r 1 |
∴ + 7 ! ( r –1)!(8 – r)!
1
= ×
. .3x
| t | r!(7 – r)
! 7 ! 4
=
r
8 – r
r
∴ |t r + 1 | ≥ | t r |
r ≥
2
2
3
. 3 2 8 – r . = 4 3 2 r
if 8 – r ≥ 2r or 3
8 ≥ r
∴ T [r] + 1 is greatest term
∴ Numerically largest term = | t 3 | = 7 C 2 .4 5 .(3x) 2
=
29. [A]
7!
.4 5 ⎛ 2 ⎞
. ⎜3.
⎟ = 86016.
2! 5! ⎝ 3 ⎠
1+
=
n
1+15k
= 15
= 15
1 + k
2
2 4n 16 n
= =
15 15
n
(1 + 15)
15
C 115
+ C215
+ .... +
15
2
, where k ∈ N,
⎪⎧
2 4n ⎪⎫
⎧ 1 ⎫ 1
∴ ⎨ ⎬ = ⎨ + k⎬
=
⎪⎩ 15 ⎪⎭ ⎩15
⎭ 15
n
n
n
C 15
1+ x
30. [B] Here f (x) = ⇒ f (A) = (I + A) (I – A)
–1
1– x
⎡ 2 2 ⎤ ⎡ 0
= ⎢ ⎥
⎣ 2 2
⎢ ⎦ ⎣ – 2
⎡ –1 –1⎤
= ⎢ ⎥
⎣ –1 –1 ⎦
– 2 ⎤
0
⎥
⎦
–1
n
⎡ 2 2 ⎤ ⎡ 0
= ⎢ ⎥
⎣ 2 2
⎢ ⎦ ⎣ –1/ 2
–1/ 2 ⎤
0
⎥
⎦
31. [D] Applying R 1 → R 1 – R 2 and R 2 → R 2 – R 3
5 – 5 0
f (x) = 0 5 – 5
2
sin x
2
cos x 5 + 4sin 2x
= 25
sin
1
0
2
–1
1
2
0
–1
x cos x 5 + 4sin 2x
⇒ f (x) = 150 + 100 sin 2x
Clearly
(a) domain (– ∞, ∞) (b) range [50,250]
(c) period π
(d)
lim
→0 x
f ( x)
– 150
= 200 x
32. [C] We have
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
∴ R = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
hence R oR –1 = {(3, 3), (3, 5), (5, 3), (5, 5)}
33. [A] n (S) = 6 × 6 × 6 × 6
n(E) = the number of integral solutions of
x 1 + x 2 + x 3 + x 4 = 12,
where 1≤ x 1 ≤ 6, ….,1≤ x 4 ≤ 6
= coefficient of x 12 in (x + x 2 + …. x 6 ) 4
4
⎛
6
= coefficient of x 8 1– ⎞
in ⎜
x
⎟
1–
⎝ x ⎠
= coefficient of x 8 in
(1 – x 6 ) 4 . ( 3 C 0 + 4 C 1 x + 5 C 2 x 2 + ….)
= 11 C 8 – 4. 5 C 2 = 125
.125
∴ P (E) =
6×
6×
6×
6
34. [B] The total number of ways in which 8 persons
can speak is 8 P 8 = 8!. The number of ways in
which A, B and C can be arranged in the specified
speaking order is 8 C 3 . There are 5! ways in which
the other five can speak. So, favourable number of
ways is 8 C 3 × 5!
Hence, required probability =
35. [D] Since,
10
⇒ ∑
=
i 1
4
⇒ ∑
=
i 1
x i = 60
10
∑
i= 1
10
x i
8
C3
× 5! 1
=
8! 6
= 6 (Q A.M. = 6)
and
10
x i = 30, ∴ ∑
i=
5
10
∑
i= 1
4
x i
= 7.5
x i = 60 – 30 = 30
30
⇒ Mean of remaining items = = 5 6
XtraEdge for IIT-JEE 115
MARCH 2012

36. [A] v r = sin θ nˆ
| v r | = sin θ
| u r | 2 = | a r | 2 + cos 2 θ | b r | 2 – 2 cos θ a r . b r
= 1 + cos 2 θ – 2 cos 2 θ
| u r | 2 = sin 2 θ
| u r | = sin θ |v r | = | u r |
37. [C] Let c r = x î + yˆ j + zk
ˆ
|a r | 2 = | b r | 2 = | c r | 2 = x 2 + y 2 + z 2 = 2 … (1)
c r . î < 0
x < 0s
. b
r a r r r r
a.
c . c
r = r r = r b r r
r
| a | .| b | | a | | c | | b | | c |
1 x + y y + z
= =
2 2 2
x = z, y = 1 – z
Put x = z, y = 1 – z in eq n (i) we get
z = 1, – 1/3
1
x = z = –
3
c r = 3
1 (– î + 4 ĵ – kˆ )
38. [A] P r = AC + BD = AC + BC + CD
P r = AC + λ AD + CD
P r = λ AD + AD
P r = (λ +1) AD
µ = λ +1
41. [B] sin –1 x
⇒ sin –1
⇒ sin –1 3 = sin
–1
x
2
3 x – 16
⇒ = x x
⇒ x 2 – 16 = 9
⇒ x = 5
42. [B]
b + c
c + a + b + c
⇒
( b + c)(
c + a)
2 –2 2
a + b – c
⇒
ab
1
⇒ cos c = 2
D α
43. [A]
x
β
A
1 )
h
tan β = AB
In ∆CDE
h – x
tan α =
AB
AB = (h – x) cot θ
h(cot α
⇒ x =
cot α
π , n ∈ z
39. [B]
x – x1
y – y1
z – z1
For image of point = =
a b c
–2( ax + by1
+ cz1
+ d
=
2 2 2
a + b + c
x –1 y – 3 z – 4 –2(2 – 3 + 4 + 3)
= = =
2 –1 1 4 + 1+
1
x = – 3, y =5, z = 2
Image of point (–3, 5, 2)
40. [B] cos 2x + 2 cos 2 x = 2
⇒ 2 cos 2 x –1 + 2 cos 2 x = 2
⇒ 4 cos 2 x = 3
⇒ cos 2 x = 4
3
⇒ cos 2 x = cos 2 π
6
⇒ x = nπ ±
6
3 + sin
–1
3 π = – sin
–1
x 2
4
⇒ sin –1 3 = cos
1
x x
4 π =
x 2
4
XtraEdge for IIT-JEE 116
MARCH 2012
⎛
⎜
⎜
⎝
x
x
2
x – 16
⎞
⎟
⎟
⎠
1 1 3
+ =
c + a a + b + c
3
=
( a + b + c)
⇒ (2c + a + b) (a + b + c) = 3 (b +c) (c + a)
⇒ 2ac + 2bc + 2c 2 + a 2 + ab + ac + ab + b 2 + bc
= 3bc + 3ab + 3c 2 + 3ac
⇒ a 2 + b 2 = ab + c 2
= 1 ⇒
C
a
2 +
E h
B
Let height of piller be x m.
In ∆ABC
b
2 – c
2 1
=
2ab
2
∠c = 60º
AB = h cot β … (1)
from (1) & (2)
h cot β = (h – x) cot α
⇒ h cot β = h cot α – x cot α
… (ii)
– cotβ)
h(tanβ
– tan α)
⇒ x =
tanβ

44. [A] a = 3, b = 5, a = 4
c
2 + a
2 – b
2
cos B =
2ac
16 + 9 – 25
⇒ cos B = = 0
2(3) (4)
⇒ ∠B = 90º
∴ sin 2
β + cos 2
β = sin 45º + cos 45º
=
1 1 + = 2
2 2
45. [A] sin θ + cosec θ = 2
1
⇒ sin θ + = 2
sin θ
⇒ sin 2 θ – 2 sin θ + 1 = 0
⇒ (sin θ – 1) 2 = 0
∴ sin 11 θ + cosec 21 θ = (1) 11 + (1) 21 = 2
LOGICAL REASONING
1. [D] The pattern is x 2 +1, x 2 + 2, . . . .
Missing number = 28 × 2 + 3 = 59
2. [A]A car runs on petrol and a television works by
electricity.
3. [A] All except Titans are planets of the solar
system.
4. [C]; 5. [B]
6. [D]
7. [B] The third figure in each row comprises of
parts which are not common to the first two
figure.
8. [A]
9. [C]
10. [A]
ENGLISH
1. [B] Geraff :
Incorrect spelling.
• 'e' should be replaced with 'i'
• The word should end with 'e' after 'ff'
Giraffe :
Correct spelling.
Giraf :
'fe' is to be added in the end.
Gerraffe :
• 'Ge' is to be replaced with 'Gi' to make the
correct spelling.
2. [B] Puncture :
No error.
It makes the tyre flat.
Puntuation :
Error of spelling
Correct spelling is 'Punctuation'
Hence 'c' is missing.
Pudding :
No error
It is used as 'Dessert'
Pungent :
No Error
It is some what 'sharp' and 'shrill'.
3. [A] Luxurious : (Plush)
Something full of all 'amenities' making life 'cozy'
and 'snug'.
Delicious : Irrelevant as it means 'something very
tasty.'
Comforting : 'Irrelevant' as it means 'giving
necessary comforts', whereas 'Plush' means more
than comforts.
Tasty : (Irrelevant)
It means 'delicious'
4. [A] Lively : Correct synonym to 'sprightly' as both
means, 'someone dashing/energetic/enthusiastic'.
Beautiful : (Irrelevant)
Sportive : (Irrelevant)
Intelligent : (Irrelevant)
5. [D] Wicked : It is almost a synonym to 'Astute'
Impolite : Irrelevant because it is the antonym of
'polite'.
Cowardly : Irrelevant as it is the opposite of
'bravely'.
Foolish : (It's the correct antonym of 'Astute'
which itself means 'clever, shrewd'.
6. [D] Deadly : It means 'Fatal'.
Hence, this is not a proper antonym to 'innocuous'.
Ferocious : It means 'horrible'
Hence, irrelevant to the opposite of 'innocuous'.
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Poisonous : It means 'venomous'.
Hence, an irrelevant 'antonym'.
Harmful : It is a perfect antonym of innocuous
which itself means 'harmless'.
7. [D] Corruption :
Irrelevant
Worldliness :
Irrelevant
Favouritism :
Irrelevant
Nepotism : (Correct Answer) because
It's a kind of corruption in which the authority in
power takes the advantage of giving opportunity
to their relatives in their self interest.
8. [B] Cross : (to pass by, to intersect)
It means different
Hence, irrelevant.
Shuttle : (Proper answer)
It's a kind of "regular beats" of an air flight or bus
service between the two stations.
Travel : It means to journey.
Hence, irrelavent.
Run : (to move regularly)
Hence, irrelevant.
9. [D] Only 1 is correct :
Inappropriate answer because sentence 1 can't be
correct using 'practise' as it is a verb, whereas the
required word should be a noun.
Only 2 is correct :
Sentence 2 is also wrong because the word
'practice' is wrongly used as a verb. It should be a
verb like 'practise'. Hence, incorrect answer.
Both the sentences 1 and 2 are correct.
This is not relevant.
Both the sentences 1 and 2 are not correct.
Correct option, if both the words, i.e. 'practice' and
'practise' are interchanged respectively, it really
makes a meaningful sentence.
10. [C] Sentence 1 is correct :
This option is wrong because the word 'ingenuous'
means 'frank and simple' which is inappropriate.
Sentence 2 is correct :
This option is also wrong because the word
'ingenious' means 'clever or prudent' and this is
inappropriate.
Both the words, i.e. 'ingenuous' and 'ingenious' if
interchanged together respectively, it really makes
both the sentences meaningful.
Hence, appropriate option.
Both the sentences can't be interchanged. This
is an incorrect option because words have been
misinterpreted together.
Incorrect option.
11. [C] Far off :
It can't be used in place of 'aloof' as far off' means
long-long ago.
Hence, incorrect alternative .
Introvert : It means 'self-centred',
Hence, It is an incorrect alternative.
distance : This is an appropriate word because
one of the meaning of 'aloof' is distant also while
keeping distance between two nouns.
Depressed : (it means 'hopeless')
Hence, quite irrelevant.
12. [A] "Meatless days" This is the name of a
novel. Hence, no error is there.
Have been made : (Erroneous)
Because 'have' should be replaced with 'has'
because 'meatless days' is a singular noun.
Into a film :
No error in this part of the sentence.
No error : Incorrect option because there is an
error in the sentence.
13. [C] Looking forward : (No error)
This is a phrase.
'to' (no error)
This is a preposition.
'Meet you here' (erroneous)
Because 'meet will be replaced with 'meeting'
Phrase 'looking forward to' is followed by present
participle (V. I + ing) form of the Verb.
No error : (incorrect option)
Part 'C' is erroneous.
14. [C] Good and Evil
This is a wrong interpretation.
Former and Latter :
Wrong interpretation.
For and against a thing.
Appropriate option as it really suits the Idiom ins
and outs.
Foul and Fair : (by hook or by crook)
This is an inappropriate option.
15. [A] Broke out : (to start suddenly)
'Correct and relevant' option because it is used for
'wars' and 'diseases' e.g. cholera broke out in Surat
in 1985.
Set out : (to start)
it is different because it is used when one leaves
for somewhere
e.g. He set out on his long voyage to Achilese.
took out : (incorrect use)
Because it means differently.
e.g. He took out a one rupee coin to give to the
beggar.
Went out : (Incorrect use) Because meaning is
different
e.g. : The light went out when I was preparing for
my Board Exams.
Hence, inappropriate option.
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