IIT-JEE 2010 - Career Point

Volume - 5 Issue - 6
December, 2009 (Monthly Magazine)
Editorial / Mailing Office :
112-B, Shakti Nagar, Kota (Raj.) 324009
Tel. : 0744-2500492, 2500692, 3040000
e-mail : xtraedge@gmail.com
Editor :
Pramod Maheshwari
[B.Tech. IIT-Delhi]
Analyst & Correspondent
Mr. Ajay Jain
Cover Design & Layout
Mohammed Rafiq
Om Gocher, Govind Saini
Circulation & Advertisement
Ankesh Jain, Praveen Chandna
Ph (0744)- 3040007, 9001799502
Subscription
Sudha Jaisingh Ph. 0744-2500492, 2500692
© Strictly reserved with the publishers
• No Portion of the magazine can be
published/ reproduced without the written
permission of the publisher
• All disputes are subject to the exclusive
jurisdiction of the Kota Courts only.
Every effort has been made to avoid errors or
omission in this publication. In spite of this, errors
are possible. Any mistake, error or discrepancy
noted may be brought to our notice which shall be
taken care of in the forthcoming edition, hence any
suggestion is welcome. It is notified that neither the
publisher nor the author or seller will be
responsible for any damage or loss of action to any
one, of any kind, in any manner, there from.
Unit Price Rs. 20/-
Special Subscription Rates
6 issues : Rs. 100 /- [One issue free ]
12 issues : Rs. 200 /- [Two issues free]
24 issues : Rs. 400 /- [Four issues free]
Owned & Published by Pramod Maheshwari,
112, Shakti Nagar, Dadabari, Kota & Printed
by Naval Maheshwari, Published & Printed at
112, Shakti Nagar, Dadabari, Kota.
Editor : Pramod Maheshwari
Dear Students,
All of us want to get organized. The first thing in getting organized is to
find our obstacles and conquer them. Make a beginning by conquering
obstacles for starting in right earnest.
Sometimes the quest for perfectionism holds us back. We occasionally
feel that we should start when we have enough time to do a job
thoroughly. One way to tackle this kind of mindset is to choose
smaller projects or parts of projects that can be completed within 15
minutes to one hour or less. It is important to keep yourself
motivated. Approach your projects as something which are going to
give you pleasure and fun. Reward yourself for all that you accomplish
no matter how small they may be. Never hesitate to ask for help from
a friend. Make all efforts to keep your motivational level high. You
might feel overwhelmed because you are focusing on every trivial thing
that needs to be got done.
How do you act or react to your life? When you are merely reacting
to events in your life, you are putting yourself in a weak position. You
are only waiting for things to happen in order to take the next step in
your life. On the other hand when you are enthusiastic about your
happiness you facilitate great things to happen. It is always better to act
from a position of power. Never be a passive victim of life. Be
someone who steers his life in exactly the direction he wants it to go.
It is all upto you now.
If you do what you have always done and in the way you have done it
you shall get only such results which you have always got. Getting
organized requires that not doing things that cause clutter, waste of
time and hurt your chances adversely of realizing your goals. You
should concentrate only on doing things that eliminate clutter, waste of
times and hurt your chances adversely of realizing your goals. You
should concentrate only on doing things that eliminate clutter, increase
your productivity and provide the best chance for achieving your goals.
The first step should be to stop leaving papers and other things on
tables, desks, counter tops and in all kind of odd place. The more
things you leave around in places other than rightful places the quicker
the clutter will accumulate. Keep things in their assigned places after
you have finished using them. It does not take along to put something
away. If you leave things lying around they will build into a mountain of
clutter. It could take hours if not weeks or months to trace them and
declutter the atmosphere.
Presenting forever positive ideas to your success.
Yours truly
Pramod Maheshwari,
B.Tech., IIT Delhi
If you can't make a mistake,
you can't make anything.
Editorial
XtraEdge for IIT-JEE 1 DECEMBER 2009

XtraEdge for IIT-JEE 2 DECEMBER 2009

Volume-5 Issue-6
December, 2009 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Key Concepts & Problem Solving strategy for IIT-JEE.
INDEX
CONTENTS
Regulars ..........
PAGE
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics, Chemistry & Maths
Much more IIT-JEE News.
Xtra Edge Test Series for JEE-2010 & 2011
NEWS ARTICLE 4
IIT-Jodhpur to begin functioning from
February - March 2010
IIT-K students develop autonomous vehicle
IITian ON THE PATH OF SUCCESS 8
Dr. Rajeewa Arya
KNOW IIT-JEE 10
Previous IIT-JEE Question
Study Time........
S
Success Tips for the Month
• The greatest adventure is what lies ahead.
• Fixers believe they can fix. Complainers
believe they can complain. They are both
right.
• The tire model for motivation: People
work best at the right pressure.
• Trust the force, Luke.
• Use your feelings or your feelings will use
you.
• People who expect to fail are usually right.
• The path to success is paved with
mistakes.
• You've got to cross that lonesome valley.
You've got to cross it by yourself.
• Appreciate what your brain does. In case
nobody else does.
• Learn to mock the woe-mongers.
• Be confident. Even if you are not, pretend
to be. No one can tell the difference.
DYNAMIC PHYSICS 18
8-Challenging Problems [Set# 8]
Students’ Forum
Physics Fundamentals
Reflection at plane & curved surfaces
Fluid Mechanics
CATALYST CHEMISTRY 35
Key Concept
Carboxylic Acid
Chemical Kinetics
Understanding : Inorganic Chemistry
DICEY MATHS 46
Mathematical Challenges
Students’ Forum
Key Concept
Monotonicity, Maxima & Minima
Function
Test Time ..........
XTRAEDGE TEST SERIES 58
Class XII – IIT-JEE 2010 Paper
Class XII – IIT-JEE 2011 Paper
Mock Test CBSE Pattern Paper -1 [Class # XII]
XtraEdge for IIT-JEE 3 DECEMBER 2009

IIT-Jodhpur to begin
functioning from February
- March 2010
Indian Institute of Technology in
Rajasthan is all set to begin
functioning from the MBM
Engineering College, here from
next academic session
commencing in February-March
2010.
The decision to this effect was
made after the visit of a central
team headed by additional
secretary, ministry of Human
Resource Development, Ashok
Thakur during recent visit to
different sites, which included the
sites proposed for the
construction of the IIT-Jodhpur
and the existing engineering
college of the city.
So, if everything goes well, the
next session of IIT-J, which is
presently being run from the IIT-
Kanpur campus, will start
functioning from the campus of
this college here.
Thakur, who himself approved the
MBM college during a visit on
Saturday, said, "The final decision
is to be taken by the ministry, to
whom we will submit the report
in a 2-3 days."
Agarwal, who was quite ecstatic
following the visit of the team,
expressed hope that the next
session of the IIT will start here
from February–March 2010.
He added that the new building of
the IIT will take atleast 2-3 years,
but owing to repeated reminders
by IIT-Kanpur citing its inability to
continue running the session of
IIT-J from its own campus, there is
a growing pressure to shift it to
Jodhpur.
IIT-K students develop
autonomous vehicle
True to its reputation of being
ahead in technological
developments, the Indian Institute
of Technology, Kanpur, in
collaboration with the Boeing
Company is all set to unveil a new
autonomous vehicle `Abhyast' as
part of its Golden Jubilee
celebrations.
Prof Shantanu Bhattacharya,
faculty in mechanical engineering
department of IIT-K and coordinator
of the project, while
talking to TOI said, "The vehicle
will set new landmarks in terms of
applied robotics research. Besides
being unique, it will probably be
one of its kind in India. Such
vehicles play important role in
defence applications and disaster
management plans." He added that
quite a few modules of such
vehicles have already been
developed by the United States
and some other countries. But,
India has so far not produced
vehicles of this nature.
Further, Prof Bhattacharya
informed, `Abhyast' will serve as a
first running prototype for such
vehicles in the country. "It has a
very small footprint -- 30x30x15
cm -- and thus it can be easily
carried by soldiers and relief
workers for disaster management
operations. This technology was
widely used by the US in
investigating the World Trade
Centre attack as well as in the war
zones of Afghanistan and Iraq," he
added.
Earlier this year, as a part of its
University relations programme,
Boeing decided to actively
collaborate with IIT-K to foster
research and innovation among
undergraduate students. As part of
this endeavour, eight students of
IIT-K -- Abhilash Jindal, Ankur Jain,
Faiz Ahmed, Gaurav Dhama,
Palash Soni, Shishir Pandya and
Sriram Ganesan -- were selected
by Prof Bhattacharya to work on
the project. "My role was mainly
confined to that of a mentor.
`Abhyast' in fact the result of
students' labour and skills and
should serve as an inspiration for
other students," said a beaming
Prof Bhattacharya.
`Abhyast' is a fully autonomous
vehicle capable of navigating in
unstructured and unknown
environments. The user needs to
specify only the end coordinates
where he wants the vehicle to
reach, and the task of reaching
there would be taken by the
vehicle itself, requiring no
intervention by the user. The
XtraEdge for IIT-JEE 4 DECEMBER 2009

vehicle is equipped with high end
sensors like GPS, IMU and
SONARS to navigate and avoid
obstacles in its vicinity. The vehicle
has a tank like chassis design that
allows it to move even on uneven
and slippery terrain, thus making it
a robust vehicle for warlike and
other disastrous situations.
IIT forecast system for
Oman, Maldives
Ocean State Forecasting System
(OSFS), a technology developed by
Indian Institute of Technology,
Kharagpur, will now be adopted
by Oman and Maldives. The
World Meteorological Organisation
(WMO) feels that OSFS should be
immediately adopted by other
nations that have sea coasts.
The system, that will continuously
measure height, direction and
period of waves, will help shipping
and navigation.
IIT-Kgp was given the prestigious
project jointly by the union
ministry of ocean development
and the department of science and
technology.
It was headed by the institute's
former director, S K Dubey.
The model was developed by the
department of ocean engineering
and naval architecture. It was
completed about a year ago and
handed over to the India
Meteorological Department by the
institute for adoption. Dubey is
presently in IIT Delhi to
coordinate the project.
"WMO was so impressed that it
immediately referred the
technology to all the coastal
nations. But it cannot be
transferred without proper
training to end users, so we will
be training meteorologists from
other countries in batches. This
will start with Oman and Maldives.
Their scientists will be on the
campus for an intensive training,"
Dubey said.
IITs come up with their RTI
'shield'
Stung by the exposure of
admission anomalies in recent
years, the IIT system has come up
with an innovative method of
blocking transparency even as it
agreed to give data under RTI on
the marks obtained by the four
lakh candidates in this year’s joint
entrance examination (JEE). It
insisted on giving the data only in
the hard copy running into
hundreds of thousands of pages
rather than in the more
convenient form of a CD.
The information seeker, Rajeev
Kumar, a computer science
professor in IIT Kharagpur, is
crying foul. For, the hard copy
would not only result in a steep
increase in the cost of information
(running into six figures) but also
make it almost impossible for him
to detect irregularities in the latest
JEE as he did in the three previous
ones by analyzing the electronic
data that had then by given to him
under RTI.
As a result of this change in the
strategy of the IIT system, central
information commissioner Shailesh
Gandhi fixed a hearing for
November 6 specially to resolve
this soft vs hard debate. The
hearing follows the unusual
reasons given by Gautam Barua,
director of IIT Guwahati and
overall in-charge of JEE 2009, for
his failure to comply with the
CIC’s disclosure direction passed
on July 30.
In his first mail to CIC on October
2, Barua said that as there were a
number of RTI applications
seeking the CD, “we are
apprehensive that this request for
electronic data is to profit from it
by using it for IIT JEE coaching
purposes (planning, targeting
particular cities, population
segments, etc).”
The reference to the coaching
institutes is reminiscent of the
recent controversy over the move
to raise the bar on 12th class
marks to be eligible for IIT
selection.
Asserting that IITs had “nothing to
hide regarding the results”, Barua
said, “We are ready to show the
running of the software with the
original data to the CIC, if it so
desires.”
As a corollary, Barua made an
issue of the fact that Kumar “has
not asked to see the data, but he
wants an electronic version
delivered to him. Why is this so?”
Kumar responded to that by
pointing out that the irregularities
he had uncovered in the JEE of the
previous three years was on the
basis of “compute intensive
scientific calculations and analysis,
which could not have been done
just by looking at the data.”
Barua’s explanation in his
subsequent mail on October 3 is:
“By seeing, I meant that the
appellant could come to IIT
Guwahati and view the data, see
the software being run, etc.” He
added that if this option was
unacceptable to CIC, “we will
wish to provide the data in hard
copy form, the costs of printing
having to be borne by the
appellant.”
XtraEdge for IIT-JEE 5 DECEMBER 2009

If Kumar is pressing that the data
be given to him “in the form in
which it is originally available”, it is
because the access to the
electronic data of the previous
three years helped him unearth,
for instance, the shocking fact that
general category candidates got
into IITs after scoring in JEE as
little as little as 5% in Physics and
6% in Mathematics.
Kerala seeks second IIT
report on record
Kerala has filed an application in
the apex court urging it to take on
record the second part of the
report titled Seismic Stability of
Mullaiperiyar Composite Dam,
submitted by D K Paul, Professor
of earthquake engineering
department, Indian Institute of
Technology, Roorkee.
The first part of the report
Structural Stability of Mullaiperiyar
Dam Considering Seismic Effects
— Part I — Seismic Hazard
Assessment was submitted by IIT-
Roorkee in May 2008. The said
report is already before the apex
court, it averred.
Kerala would like to file the
second part of the report in
support of its argument that
Mullaiperiyar dam is not safe for
storage, the application stated.
The report stated that the
earthquake safety of old concrete
or masonry gravity dams under
moderate to strong ground
motions is of great concern.
Although there is no evidence of
catastrophic failure of gravity
dams, yet the possibility of tensile
cracking is never ruled out.
The finite element analysis of dam
subjected to static and seismic
loading shows tensile stresses at
the heel of dam-foundation
interface, the report indicated.
“The Mullaiperiyar dam is a
composite gravity dam built during
1887-1895. The front and rear
faces of the dam were built with
un-coursed rubble masonry in
lime surkhi mortar. The hearting is
constructed of lime surkhi
concrete. It lies in seismic zone III
as per seismic zoning map of India.
The 176 feet-high composite
gravity dam is now over 114 years
old,” the report went on to say.
IITs want to be accredited
by statutory body
As the government wants to make
accreditation mandatory for all
institutions, the IITs have said they
would like to be accredited by a
statutory body and not by the
National Board of Accreditation
(NBA).
The IIT directors have told the
government that they have no
objection to accreditation of the
institutes, but insisted that the
accreditation agency should be a
statutory and autonomous
organisation.
They expressed these views at the
meeting of the IIT Council, the
highest decision making body for
the IITs, held here last month.
HRD Minister Kapil Sibal, who is
the chairman of the council, told
them that the government would
set up an accreditation agency by
introducing a bill in the Parliament
soon.
"The directors said the
accreditation should be conducted
by a statutory body
IITs told to reveal
candidate details
The Central Information
Commission has ordered the IITs
to disclose most details of
candidates who sat the 2009
entrance examination, rejecting
the institutes’ argument that
revealing candidates’ names would
be a breach of their privacy.
India’s apex watchdog for the
Right to Information Act has
ordered the IITs to reveal the
names, addresses, pin codes and
marks of all students who
appeared in the Joint Entrance
Examination this year.
In its November 6 order, the
commission asked IIT Guwahati,
the chief organiser among the IITs
of the 2009 examination, to
disclose by November 25 the
information sought by the
appellant.
The order follows efforts by the
IIT to withhold information on
candidates who appeared in the
2009 JEE despite earlier orders
mandating the release of similar
data on IIT candidates over the
past three years.
The order is significant because a
similar disclosure in 2006 revealed
discrepancies between cutoff
marks used by the IITs that year
and the cutoffs arrived at by using
the formula the institutes claimed
to have used.
At least 994 students, who cleared
the cutoffs arrived at by using the
formula the IITs claimed to have
used, were denied admission
because the institutes used
different cutoffs.
XtraEdge for IIT-JEE 6 DECEMBER 2009

The IITs have till now been unable
to explain how they arrived at the
cutoffs — by using the formula
they claim to have used.
The appellant in the 2009 case is
the parent of a student who
appeared in the controversial
2006 examination and is trying to
use the RTI Act to ensure that the
IITs do not repeat their errors.
The IITs, in the 2009 case, argued
that the release of candidate
details sought by the appellant
could compromise the privacy of
these candidates.
The appellant had sought the
registration numbers, names,
gender, parents’ names, pin codes
and marks in physics, chemistry
and mathematics of all students
who appeared in the 2009
examination.
The appellant has expressed
concern that the IITs may be
admitting the children of institute
administrators or certain faculty
members despite poor marks, and
has argued that details he has
sought would help clarify his
doubts.
The commission, in its order, has
argued that while providing email
addresses and mobile phone
numbers of candidates would
constitute a violation of privacy,
merely disclosing their names and
addresses would not.
Open source software
needs marketing
PUNE: There is a need for greater
promotion of the use of open
source software for information
and communication technology
(ICT)-based teaching and learning.
Professor Kannan M Moudgalya of
the Indian Institute of Technology,
Bombay (IIT-B), highlighted this on
Monday. Moudgalya, who heads
the Centre for Distance
Engineering Education Programme
(CDEEP) at the IIT, was delivering
the keynote address at the launch
of kPoint, a software solution for
interactive learning and training.
kPoint, developed by city-based
Great Software Laboratory (GSL),
was launched by noted computer
expert Vijay Bhatkar, creator of
India's Param series of
supercomputers. Heads and
professionals from leading IT
companies as well as principals of
engineering institutions were
present at the occasion.
Open source software refers to
computer software provided
under a license that is in the public
domain. "Open source software
has a distinct cost advantage over
the expensive commercial
software packages. However, a
considerable marketing effort is
required to secure a greater and
wider audience of students for
courses transmitted live using ICT
tools based on open source
software," Moudgalya said.
"Open source software is often
sufficient in most distance
education programmes, except for
some niche academic segments.
However, academic institutions
don't train students in using good
open source software," he further
stated.
Moudgalya gave an overview of
the CDEEP's involvement in the
Talk to teacher' programme,
which is funded by the Union
human resource development
ministry and aims to train students
as well as teachers in higher
education. IIT-B started
disseminating its courses live on
the internet nearly a decade ago.
For the last two years, he stated,
the CDEEP has been using the
education satellite Edusat,
provided by the Indian Space
Research Organisation, and has
raised a network of 75 centres for
transmission of live courses.
In his brief address, Bhatkar made
out a strong case for Indian ICT
professionals acknowledging and
adopting technologies and
innovations
developed
indigenously.
Sunil Gaitonde, chief executive of
GSL, said, "Technology must
bridge the gap between the
growing number of learners and
lesser number of teachers. The
prevailing knowledge economy
needs highly skilled workers and
the existing faculty crunch can be
tackled only through apt use of
technology."
Gaitonde said: "Factors like
grassroots videos, collaborations,
mobile broadband, data mash-ups,
collective intelligence and social
operating systems are bound to
make a sea change in the way
education is delivered."
College of Engineering, Pune
(CoEP) principal Anil
Sahasrabudhe, Vishwakarma
Institute of Technology
principal Hemant Abhyankar,
Persistent Systems chief
Anand Deshpande and founder-
CEO of music education
web portal ShadjaMadhyam,
Nandu Kulkarni, were
among those present at the
event.
XtraEdge for IIT-JEE 7 DECEMBER 2009

Success Story
This article contains story of a person who get succeed after graduation from different IIT's
Dr. Rajeewa Arya
B.E., M.E.(IIT – Kanpur)
Chief Executive Officer, Moser Bear (I), Ltd.
Dr. Rajeewa (Rajiv) Arya, M. Tech., Ph. D is presently
the Chief Executive Officer l at Moserbaer Photovoltaic
(MBPV) in New Delhi, India. He was previously the COO
& CTO for the Thin Film Vertical. He joined MBPV as a
Senior Vice-President & CTO (Thin Film) in September,
2007. and Electrical Engineers (IEEE) and the Materials
Research Society (MRS). Prior to that Dr. Arya was a
founder, Vice-President and CTO at Optisolar (previously
called Gen3Solar) in Hayward, California. Before founding
Gen3solar, he was the Director of Oregon Renewable
Energy Center (OREC), an academic/research center at
the Oregon Institute of Technology (OIT).
Dr. Arya launched Arya International, Inc., a Solar
Technology and Business Consulting firm, in 2003. Prior
to that Dr. Arya worked at Solarex/ BPSolar for 19 years
in various capacities, from Scientist to Executive Director,
thin-film technology.
He has over 25 years experience in thin-film solar cells
and modules. His R&D activities have centered on
material and device aspects of three types of thin-film
solar cells and modules – amorphous silicon, copperindium-diselenide,
and cadmium telluride. His work
includes product design, process scale-up, process
transfer, piloting and start-up of a thin-film solar module
plant. He has maintained a professional interest in many
aspects of renewable energy components and systems.
Dr. Arya holds a Masters of Science degree in Solid-State
Physics from Jadavpur University, Calcutta, India and a
Master of Technology degree in Material Science from the
Indian Institute of Technology, Kanpur, India. He obtained
his Ph.D. in Engineering from Brown University, Rhode
Island, in 1983. He has extensive management training in
Total Quality Management, Finance, Project management,
and Technology Innovation Management.
Dr. Arya has authored and co-authored over 100
technical papers and holds 6 U.S. Patents. He is the
recipient of the “Outstanding Paper Award” at the 7th
PVSEC, the “Team of the Year” award from Solarex
Quality Process, and his group received an R&D 100
award for the Power view Product. He chaired the
Program Committee for the 29th IEEE Photovoltaic
Specialists Conference in 2002. He is a member of the
Institute of Electronics.
XtraEdge for IIT-JEE 8 DECEMBER 2009

XtraEdge for IIT-JEE 9 DECEMBER 2009

KNOW IIT-JEE
By Previous Exam Questions
PHYSICS
1. A circular disc with a groove along its diameter is
placed horizontally on a rough surface. A block of
mass 1 kg is placed as shown. The co-efficient of
friction in contact is µ = 2/5. The disc has an
acceleration of 25 m/s 2 towards left. Find the
acceleration of the block with respect to disc. Given
cos θ = 4/5, sin θ = 3/5.
[IIT-2006]
25 m/s 2
Sol. Applying pseudo force ma and resolving it. Applying
F net = ma x for x-direction
ma cos θ – (f 1 + f 2 + = ma x
ma cos θ – µN 1 – µN 2 = ma x
ma cos θ – µma sin θ – µmg = ma x
⇒ a x = a cos θ – µa sin θ – µg
= 25 × 5
4 – 5
2 × 25 × 5
3 – 5
2 × 10 = 10 m/s
2
2. A wooden stick of length L, radius R and density ρ
has a small metal piece of mass m (of negligible
volume) attached to its one end. Find the minimum
value for the mass m (in terms of given parameters)
that would make the stick float vertically in
equilibrium in liquid of density σ(>p). [IIT-1999]
Sol. For the wooden stick-mass system to be in stable
equilibrium the centre of gravity of stick-mass system
should be lower than the centre of buoyancy. Also in
equilibrium the centre of gravity (G) and the centre of
buoyancy (B) lie in the same vertical axis.
The above condition 1 will be satisfied if the mass is
towards the lower side of the stick as shown in the
figure.
The two forces will create a torque which will bring
the stick-mass system in the vertical position of the
stable equilibrium
Let l be the length of the stick immersed in the
liquid.
θ
Then
F B =πR 2 hσg
θ
mg
h/2
L/2
θ
OB = 2
l . Let OG = y
For vertical equilibrium
F G = F B
⇒ (M + m)g = F B
⇒ πR 2 Lρg + mg = πR 2 l σ g
2
C
(πR 2 Lρ)g
πR
Lρ + m
l =
...(1)
2
πR
σ
Now using the concept of centre of mass to find y.
Then
My1
+ my2
y =
M + n
Since mass m is at O the origin, therefore y 2 = 0
M(L / 2) + m×
O ML
∴ y =
=
M + m 2(M + m)
2
( πR
Lρ)L
=
2
2( πR
Lρ + m)
Therefore for stable equilibrium
l > y
2
2
2
πR
Lρ + m ( πR
Lρ)L
∴
>
2
2
2( πR
σ)
2( πR
Lρ + m)
⇒ m ≥ π R 2 L ( ρσ – ρ)
...(2)
∴ minimum value of m is πr 2 L ( ρσ – ρ)
3. A gaseous mixture enclosed in a vessel of volume V
consists of one mole of a gas A with λ (=C p /C v ) = 5/3
and another gas B with λ = 7/5 at a certain
temperature T. The relative molar masses of the gases
A and B are 4 and 32, respectively. The gases A and
B do not react with each other and are assumed to be
ideal. The gaseous mixture follows the equation
PV 19/13 = constant, in adiabatic processes. [IIT-1995]
XtraEdge for IIT-JEE 10 DECEMBER 2009

(a) Find the number of moles of the gas B in the
gaseous mixture.
(b) Compute the speed of sound in the gaseous
mixture at T = 300 K.
(c) If T is raised by 1 K from 300 K, find the %
change in the speed of sound in the gaseous mixture.
(d) The mixture is compressed adiabatically to 1/5 of
its initial volume V. Find the change in its adiabatic
compressibility in terms of the given quantities.
Sol. (a) The ratio of specific heat of mixture of gases
(C p ) m
γ m =
* m stands for mixture.
(Cv
) m
n ACpA
+ n BCpB
Also (C p ) m =
n A + n B
n ACVA
+ n BCvB
and (C v ) m =
n A + n B
Also according to Mayer's relationship
C p – C v = R
C p R R R
⇒ – 1 = ⇒ γ – 1 = ⇒ Cv =
Cv
Cv
Cv
γ −1
R
For Gas A (C v ) A =
5 /3 − 1
= 3R
2
3R 5R
∴ (C p ) A = R + (C v ) A = R + =
2 2
R
For Gas B (C v ) B =
7 /5 − 1
= 5R
2
5R 7R
∴ (C p ) B = R + =
2 2
(C p ) m n ACpA
+ n BCpB
γ m = =
(Cv
) m n ACvA
+ n BCvB
1×
(5R / 2) + n B(7R / 2) 5 + 7n B
=
=
1×
(3R / 2) + n B (5R / 2) 3+
5n B
According to the relationship
PV 19/13 = constant we get γ m = 19/13
5 + 7n B 19
∴ = ⇒ nB = 2 mol.
3+
5n B 13
Alternatively we may use the following formula
n1
n1
n 2
= +
γ m −1
γ1
−1
γ 2 −1
where γ m = Ratio of specific heats of mixture
(b) We know that velocity of sound in air is given by
the relationship
v =
γP
d
m
where d = density = v
Also, PV = (n A + n B )RT
⇒
( n A n B)
PV =
RT
V
∴ v =
γ( n A + n B )RT
m
V ×
V
=
γ ( n A + n B )RT
m
Mass of the gas,
m = n A M A + n B M B = 1 × 4 + 2 × 32
= 68 g/mol mol = 0.068 kg/mol
19(1 + 2) × 8.314×
300
∴ v =
= 400.03 ms –1
13×
0.068
(c) We know that the velocity of sound
⇒
v =
γP
d
v + ∆v
v
=
=
γRT
M
T + ∆T
T
∆ v 1
⇒ 1 + = 1 + v 2
and v + ∆v =
⎛1+
∆T
⎞
= ⎜ ⎟
⎝ T ⎠
∆T
T
∆ T

4. A square loop of side 'a' with a capacitor of capacitance
C is located between two current carrying
long parallel wires as shown. The value of I in the
wires is given as I = I 0 sin ωt. [IIT-2003]
a
a
I
I
a
(a) Calculate maximum current in the square loop.
(b) Draw a graph between charges on the upper plate
of the capacitor vs time.
Sol. (a) Let us consider a small strip of thickness dx as
shown in the figure.
The magnetic field at this strip
B = B A + B B (Perpendicular to the plane of
paper directed upwards)
µ 0 I µ
= +
0 I
2 π x 2 π (3a − x)
B A = Magnetic field due to current in wire A
µ 0
= I ⎡1
1 ⎤
2 π
⎢ + ⎥
⎣x
3a − x ⎦
B B = Magnetic field due to current in wire B
dx
I
x
Small amount of magnetic flux passing through
the strip of thickness dx is
dφ = B × adx
µ 0 Ia×
3a dx
=
2πx (3a − x)
Total flux through the square loop
2a
2
µ ×
φ =
0 I 3a dx µ 0 Ia
∫
= ln 2
a 2π
x(3a − x) π
µ 0 a ln(2)
= (I 0 sin ωt)
π
The emf produced
dφ µ 0
e = − =
aI 0 ω ln(2) cos ωt
dt π
Charged stored in the capacitor
µ 0 0
q = C × e = C ×
aI ω ln(2) cos ωt ... (i)
π
∴ Current in the loop
dq
i = C
2
×µ ω
=
0 aI 0
ln(2) sin ωt
dt π
I
2
µ 0 aI0ω
C ln(2)
∴ i ma x =
π
(b) From (1), the graph between charge and time is
Q
Q 0
–Q 0
π/2ω
Here q 0 =
π/ω
2π/ω
3π/2ω
C 0
× µ 0 aI ωln(2)
π
5. Highly energetic electrons are bombarded on a
target of an element containing 30 neutrons. The
ratio of the radii of nucleus to that of Helium
nucleus is (14) 1/3 . Find (a) atomic number of the
nucleus. (b) the frequency of K a line of the X-ray
produced. (R = 1.1 × 10 7 m –1 and c = 3 × 10 8 m/s)
[IIT-2005]
Sol. (a) We know that radius of nucleus is given by the
formula
r = r 0 A 1/3 where r 0 = constant and A = mass
number.
For the Nucleus r 1 = r 0 4 1/3
∴
r
r
2
1
⎛ A ⎞
= ⎜ ⎟⎠
⎝ 4
1/3
⇒ (14) 1/3 ⎛ A ⎞
= ⎜ ⎟⎠
⎝ 4
1/3
⇒ A = 56
∴ No. of proton = A – no. of neutrons
= 56 – 30
= 26
∴ Atomic number = 26
⎡
(b) We know that ν = Rc (z – b) 2 1 1
⎥ ⎥ ⎤
⎢ −
2 21
⎢⎣
n1 n ⎦
Here, R = 1.1 × 10 7 , C = 3 × 10 8 , Z = 26
b = 1 (for K a ), n 1 = 1, n 2 = 2
∴ ν = 1.1 × 10 7 × 3 × 10 8 [26 – 1] 2 ⎡1
1⎤
⎢ − ⎥
⎣4
4⎦ = 3.3 × 10 15 3
× 25 × 25 × = 1.546 × 10 18 Hz.
4
CHEMISTRY
6. The standard reduction potential of Ag + /Ag electrode
at 298 K is 0.799V. Given that for AgI,
K sp = 8.7 × 10 –17 , evaluate the potential of Ag + /Ag
electrode in a saturated solution of AgI. Also
calculate the standard reduction potential of
I – electrode.
[IIT-1994]
t
XtraEdge for IIT-JEE 12 DECEMBER 2009

Sol. In the saturated solution of AgI, the half cell
reactions are
At anode : Ag ⎯→ Ag + + e –
At cathode : AgI + e – ⎯→ Ag + I –
Cell reaction AgI ⎯→ Ag + + I –
On applying Nernst equation
0.0591
E cell = Eº cell – log [Ag + ] [I – ]
n
For electrode
Ag + + e – → Ag
∴
E
Ag + = º
/ Ag Ag / Ag
K sp of AgI = [Ag + ] [I – ]
Q [Ag + ] = [I – ]
∴ K sp of AgI = [Ag + ] 2
∴ [Ag + ] of AgI =
E – 0.0591 1
log
+
n [Ag
K sp of AgI
−17
[Ag + ] = 8.7 × 10
= 9.3 × 10 –9 M
So E Ag + / Ag
= 0.799 – 0.0591 1
log
1
−9
9.3×
10
= + 0.799 + 0.0591 log 9.3 – 0.0591 × 9 log 10
= + 0.799 + 0.0591 × 0.9785 – 0.0591 × 9
= 0.325 V
For above cell reaction
0.0591
E cell = Eº cell – log [Ag + ] [I – ]
n
0.0591
= Eº cell – log (K sp of AgI)
n
At equilibrium E cell = 0
0.0591
∴ Eº cell = log(8.7 × 10 –17 ) = –0.95 volt
1
Eº cell = Eº cathode + Eº anode
–
–0.95 = –0.799 + Eº Ag/AgI/I
(In form of cell reaction)
Eº – Ag/AgI/I = – 0.95 + 0.799 = –0.151 V
–
or Eº I /AgI/Ag = + 0.151 V
7. A sample of hard water contains 96 ppm of SO 2– 4 and
183 ppm of HCO – 3 with 60 ppm of Ca 2+ as the only
cation. How many moles of CaO will be required to
remove HCO 2– 3 from 100 kg of this water ? If 1000
kg of this water is treated with the amount of CaO
calculated above, what will be the concentration (in
ppm) of residual Ca 2+ ions ? (Assume CaCO 3 to be
completely insoluble in water). If the Ca 2+ ions in one
litre of the treated water are completely exchanged
with hydrogen ions, what will be its pH ? (one ppm
means one part of the substance in one million part of
water, mass/mass)
[IIT-1997]
+
]
Sol. In 10 6 g(= 1000 kg) of the given hard water, we have
amount of SO 4 2– ions = 96 g
amount of HCO 3 – ions = 183 g
So amount of SO 4 2– ions =
96 g
96 g mol
−1
= 1 mol
and amount of HCO – 183 g
3 ions =
− 1
61 g mol
= 3 mol
These ions are present as CaSO 4 and Ca(HCO 3 ) 2 .
Hence, amount of Ca 2+ ⎛ 3 ⎞
ions = ⎜1 + ⎟ = 2.5 mol
⎝ 2 ⎠
The addition of CaO causes the following reactions:
CaO + Ca(HCO 3 ) 2 → 2CaCO 3 + H 2 O
1.5 mol of CaO will be required for the removal of
1.5 mol of Ca(HCO 3 ) 2 in form of CaCO 3 .
In the treated water, only CaSO 4 is present now.
Thus, 1 mol of Ca 2+ ions will be present in 10 6 g of
water. Hence, its concentration will be 40 ppm.
Molarity of Ca 2+ ions in the treated water will be 10 –3
mol l –1 .
If the Ca 2+ ions are exchanged by H + ions then,
Molartiy of H + in the treated water = 2 × 10 –3 M
Thus, pH = – log(2 × 10 –3 ) = 2.7
8. A white precipitate was formed slowly when AgNO 3
was added to compound (A) with molecular formula
C 6 H 13 Cl. Compound (A) on treatment with hot
alcoholic KOH gave a mixture of two isomeric
alkenes (B) and (C), having formula C 6 H 12 . The
mixture of (B) and (C) on ozonolysis, furnished four
compounds (i) CH 3 CHO, (ii) C 2 H 5 CHO,
(iii) CH 3 COCH 3 and
(iv) CH 3 – CH(CH 3 )–CHO. What are the structures
of (A) and (C) ?
[IIT-1986]
Sol. It is given that,
Alcoholic
C 6 H 13 Cl
(A) KOH; –HCl
Alkyl chloride
(i) O 3
C 6 H 12
((ii) H2O/Zn
(B) and (C)
Two alkenes (B) + (C) with
Formula C 6 H 12
CH 3 CHO + C 2 H 5 CHO
+ CH 3 COCH 3 + CH 3 – CH – CHO
CH 3
It is observed that during ozonolysis, no loss of
carbon takes place, it may be concluded that
CH 3 CHO and CH 3 – CH(CH 3 ) – CHO are the
products of one alkene (B) and C 2 H 5 CHO and
CH 3 COCH 3 are the products of other alkene (say)
(C). Thus, from the above we have :
XtraEdge for IIT-JEE 13 DECEMBER 2009

H
CH 3 – C = O + O = HC – CH – CH 3
CH 3
–2[O]
CH 3 – CH = CH – CH – CH 3
(B)
CH 3
(i) O 3
CH 3 CHO + OHC – CH – CH 3
(ii) H 2 O/Zn
CH 3
Similarly alkene (C) will be derived as :
CH 3
C = O + O = CH.CH 2 CH 3
CH 3
–2[O]
CH 3 – C = CH.CH 2 CH 3
CH 3
(i) O 3
CH 3 – C = O + OHC.CH 2 CH 3
(ii) H 2 O/Zn
CH 3
Since the compounds (B) and (C) are obtained when
(A), C 6 H 13 Cl, is dehydrohalogenated by heating it
with alcoholic KOH, as follows :
CH 3 CH 2 – CH . CH – CH 3
Cl CH 3
(C)
Alc.KOH
CH 3 – CH = CH – CH – CH 3 + CH 3 CH 2 CH = C – CH 3
CH 3
(B) (20%) (minor)
∆; –HCl
CH 3
(C) (80%) (major)
Since the Cl atom in (A) is an aliphatic chlorine, and
it is attached to a secondary carbon atom which is
adjacent to a tertiary cabon atom and one secondary
carbon atom – CH2 – CH . CH – , it will react
Cl CH 3
slowly with AgNO 3 to give a white precipitate.
Thus,
A, CH 3 – CH 2 – CH – CH – CH 3
B,
C,
Cl
CH 3
3-chloro-2-methyl pentane
CH 3 CH = CH – CH – CH 3
CH 3
4-methyl pentene -2
CH 3 CH 2 CH = C – CH 3
CH 3
2-methyl pentene-2
9. A hydrated metallic salt A, light green in colour,
gives a white anhydrous residue B after being heated
gradually. B is soluble in water and its aqueous
solution reacts with NO to give a dark brown
compound C. B on strong heating gives a brown
residue and a mixture of two gases E and F. The
gaseous mixture, when passed through acidified
permanganate, discharges the pink colour and when
passed through acidified BaCl 2 solution, gives a
white precipitate. Identify A, B, C, D, E and F.
[IIT-1988]
Sol. The given observations are as follows.
(i) Hydrated metallic salt ⎯ →
(A)
⎯ heat
(B)
white anhydrous residue
(ii) Aqueous solution of B ⎯→
dark brown compound
Strong
heating
(iii) Salt B ⎯ ⎯⎯→
(iv)
Gaseous mixture
(E) + (F)
⎯ NO (C)
Brown residue +
(D)
acidified KMnO 4
BaCl 2 solution
Two gases
(E) + (F)
Pink colour is
discharged
White precipitate
The observation (ii) shows that B must be ferrous
sulphate since with NO, it gives dark brown
compound according to the reaction
[Fe(H 2 O) 6 ] 2+ 2
+ NO → [Fe(H2 O) 5(NO)]
+ + H 2 O
dark brown
Hence, the salt A must be FeSO 4 . 7H 2 O
The observation (iii) is
2FeSO 4 → Fe 2 O 3 + SO 2 + SO 3
(D)
brown
(E) + (F)
The gaseous mixture of SO 2 and SO 3 explains the
observation (iv), namely,
2 2
+ 5SO 2 + 2H 2 O → 2Mn
+ −
+ 5SO + 4H +
2MnO −
4
pink colour
no colour
2H 2 O + SO 2 + SO 3 4H + + SO 2– 2–
3 + SO 4
Ba 2+ + SO 2– 3 → BaSO ; Ba 2+ + SO – 4 → BaSO
3
white ppt.
Hence, the various compounds are
A. FeSO 4 . 7H 2 O B. FeSO 4
C. [Fe(H 2 O) 5 NO]SO 4 D. Fe 2 O 3
E and F SO 2 and SO 3
4
4
white ppt.
10. A white amorphous powder A when heated gives a
colourless gas B, which turns lime water milky and
the residue C which is yellow when hot but white
when cold. The residue C dissolves in dilute HCl and
the resulting solution gives a white precipitate on
addition of potassium ferrocyanide solution. A
dissolves in dilute HCl with the evolution of a gas
XtraEdge for IIT-JEE 14 DECEMBER 2009

a+
which is identical in all respects with B. The solution
of A as obtained above gives a white precipitate D on
addition of excess of NH 4 OH and on passing H 2 S.
Another portion of this solution gives initially a white
precipitate E on addition of NaOH solution, which
dissolves on further addition of the base. Identify the
compound A to E.
[IIT-1979]
Sol. The given information is as follows.
(a)
A
white powder
diluteHCl
⎯ heat ⎯→
(b)C ⎯⎯⎯⎯
→ solution
(c) A
dilute HCl
(i) NH4OH
(ii) H 2 S
D
white precipitate
B +
turns
colourless
milky lime water
gas
K
6
⎯ 4 Fe(CN )
C
hot yellow
residue
white when when
cold
⎯⎯ ⎯⎯ → white precipitate
Solution + B
(i) NaOH
E
white precipitate
dissolves
NaOH
From part (a), we conclude that B is CO 2 as it turns
lime water milky :
Ca(OH 2 ) + CO 2 → CaCO + H 2 O
milky
to this
due
3
and C is ZnO as it becomes yellow on heating and is
white in cold. Hence, the salt A must be ZnCO 3 .
From part (b), it is confirmed that C is a salt of zinc
(II) which dissolves in dilute HCl and white
precipitate obtained after adding K 4 [Fe(CN) 6 is due
to Zn 2 [Fe(CN) 6 ].
From part (c), it is again confirmed that A is ZnCO 3
as on adding dilute HCl, we get CO 2 and zinc (II)
goes into solution. White precipitate is of ZnS which
is precipitated in ammonical medium as its solubility
product is not very low. White precipitate E is of
Zn(OH) 2 which dissolves as zincate, in excess of
NaOH. Hence the given information is explained as
follows.
(a) ZnCO
(b) ZnO
⎯ heat ⎯→ CO
(B) 2 +
3
(A)
⎯ dil ⎯
HCl ⎯→
ZnCl
(C)
solution 2
(c) ZnCO 3
ZnO
(C)
K
6
⎯ 4 Fe(CN )
⎯ dil ⎯
HCl ⎯→
Solution 2
ZnCl 2 + S 2– → ZnS ↓ + 2Cl –
(D)
Zn 2+ + 2OH – → Zn(OH)
⎯⎯
⎯⎯ → Zn [Fe(CN)
ZnCl + CO 2 + H 2 O
2
(E)
2-
2
dissolves
Zn(OH) 2 + 2OH – → ZnO + 2H 2 O
2 6 ]
White precipitate
MATHEMATICS
11. Determine the name of the name of the curve
described parametrically by the equations
x = t 2 + t + 1, y = t 2 – t + 1 [IIT-1998]
Sol. We have,
x = t 2 + t + 1 and, y = t 2 – t + 1
⇒ x + y = 2(t 2 + 1) and, x – y = 2t
⎪⎧
2
⎛ x − y ⎞ ⎪⎫
⇒ x + y = 2 ⎨⎜
⎟ + 1⎬
⎪⎩ ⎝ 2 ⎠ ⎪⎭
⇒ 2(x + y) = (x – y) 2 + 4
⇒ x 2 + y 2 – 2xy – 2x – 2y + 4
Comparing this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, we get
a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1
∴ abc + 2fgh – af 2 – bg 2 – ch 2 = 4 – 2 – 1 – 1 – 4 ≠ 0
and , h 2 – ab = 1 – 1 = 0
Thus, we have
∆ ≠ 0 and h 2 = ab
So, the given equations represent a parabola.
12. The circle x 2 + y 2 – 4x – 4y + 4 = 0 is inscribed in a
triangle which has two of its sides along the
coordinate axes. If the locus of the circumcentre of
the triangle is
2
x + y – xy + k x + y = 0,
find the value of k.
[IIT-1987]
Sol. Let OAB be the triangle in which the circle
x 2 + y 2 – 4x – 4y + 4 = 0 is inscribed. Let the
x y
equation of AB be + = 1
a b
y
B(0,b)
C
2
x y
+ = 1
a b
2
x´ O (a, 0)A x
y´
Since AB touches the circle x 2 + y 2 – 4x – 4y + 4 = 0.
There fore,
2 2
+ −1
a b
1 1
+
2 2
a b
⎛ 2 2 ⎞
⎜ + −1⎟
a b
= 2 ⇒ –
⎝ ⎠
= 2
1 1
+
2 2
a b
XtraEdge for IIT-JEE 15 DECEMBER 2009

[Q O(0, 0) and C(2, 2) lie on the same side of AB
Therefore, a
2 + b
2 – 1 < 0]
(2b + 2a – ab)
⇒ –
= 2
2 2
a + b
2
⇒ 2a + 2b – ab + 2 a + b = 0 ...(i)
Let P(h, k) be the circumcentre of ∆OAB. Since
∆ OAB is a right angled triangle. So its circumcentre
is the mid-point of AB.
a b
∴ h = and k = 2 2
⇒ a = 2h and b = 2k
From (i) and (ii), we get
4h + 4k – 4hk + 2
2
2
2
4 h + 4k = 0
⇒ h + k – hk + h + k = 0
So, the locus of P(h, k) is
2
2
...(ii)
x + y – xy + x + y = 0
But, the locus of the circumcentre is given to be
x + y – xy + k x + y = 0
Thus, the value of k is 1
2
13. Let C be any circle with centre (0, 2 ). Prove that at
most two rational points can be there on C.
(A rational points is a point both of whose
coordinates are rational numbers) [IIT-1997]
Sol. The equation of any circle C with centre (0, 2) is
given by
2
2
(x – 0) 2 + (y – 2) 2 = r 2 , where r is any positive real
number.
or, x 2 + y 2 – 2 2 y = r 2 – 2
If possible, let P(x 1 , y 1 ), Q(x 2 , y 2 ) and R(x 3 , y 3 ) be
three distinct rational points on circle C. Then,
2 2
x + y 2 2 y 1 = r 2 – 2 ...(ii)
x
1 1 −
2 2
2 2 −
2
+ y 2 2 y 2 = r 2 – 2 ...(iii)
2
x 3 + y3
− 2 2 y 3 = r 2 – 2 ...(iv)
We claim that at least two y 1 , y 2 , and y 3 are distinct.
For if y 1 = y 2 = y 3 , then P, Q and R lie on a line
parallel to x-axis and a line parallel to x-axis does not
cross the circle in more than two points. Thus, we
have either y 1 ≠ y 2 or, y 1 ≠ y 3 or, y 2 ≠ y 3 .
Subtracting (ii) from (iii) and (iv), we get
2 2 2 2
(x 2 + y 2)
– (x1 + y1
) – 2 2 (y 2 – y 1 ) = 0
and,
2 2 2 2
(x 3 + y3
) – (x1 + y1
) – 2 2 (y 3 – y 1 ) = 0
⇒ a 1 – 2b 1 = 0 and a 2 – 2b 2 = 0 ...(v)
2
where,
2 2 2 2
a 1 = (x 2 + y 2)
– (x1 + y1
) , b 1 = 2(y 2 – y 1 )
2 2 2 2
a 2 = (x 3 + y3
) – (x1 + y1
) , b 2 = 2(y 3 – y 1 )
Clearly, a 1 , a 2 , b 1 , b 2 are rational numbers as x 1 , x 2 ,
x 3 , y 1 , y 2 , y 3 are rational numbers.
Since either y 1 ≠ y 2 or, y 1 ≠ y 3
∴ Either b 1 ≠ 0 or, b 2 ≠ 0
If b 1 ≠ 0, then
a 1 – 2b 1 = 0 [From (v)]
a1
⇒ = 2,
b
1
a1
which is not possible because is a rational
b1
number and 2 is an irrational number.
If b 2 ≠ 0, then
a 2
a 2 – 2b 2 = 0 ⇒ = 2,
b
2
a 2
which is not possible because is a rational
b2
number and 2 is an irrational number.
Thus, in both the cases we arrive at a contradiction.
This means that our supposition is wrong. Hence,
there can be at most two rational points on circle C.
14. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices P, Q and S lie on the lines y = a,
x = b and x = –b, respectively. Find the locus of the
vertex R.
[IIT-1996]
Sol. Let the coordinates of R be (h, k). It is given that P
lies on y = a. So, let the coordinates of P be (x 1 , a).
Since PQ is parallel to the line y = mx. Therefore,
Slope of PQ = (Slope of y = mx) = m
1
And, Slope of PS = –
(Slope of y = mx)
1
= – [∴ PS ⊥ PQ]
m
Now, equation of PQ is
y – a = m(x – x 1 )
x = –b
x´ S
(0, – b)
(0, a)
R
O
y
y = 0
P
x = b
Q
(0, b)
x
...(i)
y´
It is given that Q lies on x = b. So, Q is the point of
intersection if (i) and x = b.
XtraEdge for IIT-JEE 16 DECEMBER 2009

Putting x = b in (i), we get
y = a + m(b – x 1 )
So, coordinates of Q are (b, a + m(b – x 1 )).
Since PS passes through P(x 1 , a) and has slope – m
1 .
So, Equation of PS is
y – a = – m
1 (x – x1 )
...(ii)
It is given that S lies on x = – b. So, S is the point of
intersection of (ii) and x = –b.
Solving (ii) and x = – b, we get
y = a + m
1 (b + x1 )
⎛ 1 ⎞
So, coordinates of S are ⎜− b,a
+ (b + x1
) ⎟
⎝ m ⎠
1
k − a − (b + x1)
Now, Slope of RS = m = m
h + b
But RS is parallel to PQ.
1
k − a − (b + x1)
∴
m = m
h + b
⇒ b + x 1 = m(k – a) – m 2 (h + b) ...(iii)
Similarly,
k − a − m(b − x1)
Slope of RQ =
h − b
But, RQ is perpendicular to PQ whose slope is m.
k − a − m(b − x1)
1
∴
= –
h − b m
1 1
⇒ b – x 1 = (k – a) + m
2 (h – a) ...(iv)
m
We have only one variable x 1 . To eliminate x 1 , add
(iii) and (iv) to obtain
⎛ 1 ⎞
2b = (k – a) ⎜m + ⎟ – m 2 1
(h + b) +
⎝ m ⎠
2 (h – b)
m
⎛ m ⎞
⇒ 2b = (k – a) ⎜
2 + 1 ⎛
4
m ⎞
⎟
– h ⎜
+ 1 ⎛
4
m ⎞
⎟
⎝ m
2
– b ⎜
+ 1
⎟
⎠ ⎝ m
2
⎠ ⎝ m ⎠
⎛ m ⎞
⇒ (k – a) ⎜
2 + 1
⎟
–
⎝ m ⎠
2
h (m −1)(m
+ 1)
m
–
2
2
2
b (m + 1)
h(m 2 −1)
b(m 2 + 1)
⇒ (k – a) – – = 0
m m
⇒ m(k – a) – h(m 2 – 1) – b(m 2 + 1) = 0
Hence, the locus of R(h, k) is
m(y – a) – x(m 2 – 1) – b(m 2 + 1) = 0
m
2
2
= 0
15. If A, B, C are the angles of a triangle ABC and the
system of linear equations
x sin A + y sin B + z sin C = 0
x sin B + y sin C + z sin A = 0
x sin C + y sin A + z sin B = 0
has a non trivial solution, prove that
sin 2 A + sin 2 B + sin 2 C – (cos A + cos B + cos C
+ cos A cos B + cos B cos C + cos C cos A) = 0
[IIT-2002]
Sol. The given system of linear equations has a non-trivial
solution. Therefore,
⇒
sin A
sin B
sin C
sin B
sin C
sin A
sin C
sin A
sin B
sin A + sin B + sin C
sin B + sin C + sin A
sin C + sin A + sin B
= 0
sin B
sin C
sin A
sin C
sin A
sin B
Applying C 1 → C 1 + C 2 + C 3
⇒ (sin A + sin B + sin C)
⇒
⇒
1
1
1
1
0
0
sin B
sin C
sin A
sin B
sin C
sin A
sin B
sin A − sin B
= 0
1
1
1
sin B
sin C
sin A
= 0
sin C
sin A
sin B
= 0
⎡Qsin A + sin B + sin C ⎤
⎢ A B C ⎥
⎢=
4cos cos cos ≠0
⎣ 2 2 2 ⎥
⎦
sin C
sin C − sin B sin A − sin C = 0
sin B − sin C
Applying R 2 → R 2 – R 1 , R 3 → R 3 – R 1
⇒ –(sin B – sin C) 2 – (sin A – sin C)
(sin A – sin B) = 0
⇒ sin 2 B + sin 2 C – 2 sin B sin C + sin 2 A
– sin A sin B – sin C sin A + sin B sin C = 0
⇒ sin 2 A + sin 2 B + sin 2 C – sin A sin B – sin B sin C
– sin C sin A = 0
⇒ sin 2 A + sin 2 B + sin 2 C – cos A cos B
– cos B cos C – cos C cos A + cos (A + B)
+ cos (B + C) + cos (C + A) = 0
⇒ sin 2 A + sin 2 B + sin 2 C – cos A cos B
– cos B cos C – cos C cos A – cos A
– cos B – cos C = 0
XtraEdge for IIT-JEE 17 DECEMBER 2009

Physics Challenging Problems
Set # 8
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
Passage # 1 (Ques. 1 to 3)
Young's double slit experiment is conducted with the
following conditions
1. Slit S 1 and slit S 2 are of same width
2. Slits are illuminated by monochromatic light source
of wave length 'λ b ' which is of blue color.
3. Distance between slits and screen D. Distance
between slits is 2d and D > > 2d
It is observed that 1st bright fringe is observed in
front of one of the slit.
1. If monochromatic light source of blue color is
replaced by the white colored light source then
maximum wavelength which is missing in front of
one of the slit is -
(A) Never of indigo and violet colors
(B) It is always of less than blue color
(C) Missing wave lengths can have wave length
more or less than blue color
(D)
λ
2 = λ
max . .
( missing ) 3
b
2. Relation of the maximum wavelength missing with
wave length of Blue light is -
(A) 2λ b
(B) 3λ b
(C) 4λ b
(D) 5λ b
3. With blue light if the slit widths are made unequal
then -
(A) Position of 1st bright fringe will not be in front
of one of the slit
(B) Dark fringe which was of black colour earlier
with same slit widths now is of blue colour
(C) YDSE can not be conducted with unequal slit
widths
(D) Dark fringe is always black either the slit widths
are equal or not
Passage # 2 (Ques. 4 to 5)
For the given circuit
a
R
8R
R
R
4R
3R
12R
4. What should be the value of R so that equivalent
resistance between terminals a and b is 1Ω -
2 5
(A) Ω (B) Ω
5 2
(C) 2Ω
2R
R
(D) 15Ω
5. If current passing through the circuit is 1 amp then-
(A) Potential difference across 4R and 8R is in the
ratio of 1 : 2
(B) Potential difference across ab is 1 volt when
measured by an ideal voltmeter if R = 5
2 Ω
(C) Maximum potential difference will appear
across 12R resistance
(D) Maximum potential difference will appear
across 3R resistance
R
b
XtraEdge for IIT-JEE 18 DECEMBER 2009

Passage # 3 (Ques. 6 to 8)
Behaviour of capacitor in electric circuits is very
typical because of it's energy storing nature.
Capacitor behaves in just opposite manner to
inductor, Inductor 'L' which is measured in Henary
in SI system stores the energy in magnetic field
instead of capacitor which stores in electric field
Inductor opposes the change in current and capacitor
opposes change in voltage.
Behaviour of inductor:
t = 0
t → ∞
Steady
State
open switch
Closed switch
For the electric circuit shown
ε 1
R
L/R is known as time constant
of R-L series circuit which is
measured in ohm
R 1
C
ε 2
R 2
6. If capacitance C varies even after that energy stored
in capacitor is zero at steady state then -
R1
ε1
R1
ε 2
(A) = (B) =
R 2 ε 2
R 2 ε1
(C) ε 1 + ε 2 = 0 (D) ε 1 R 1 + ε 2 R 2 = 0
7. Time constant for the circuit -
(A) RC (B) R 1 C if ε 1 > ε 2
⎛ R
(C) R 2 C if ε 1 < ε 2 (D) ⎟ ⎞
⎜
1R
2
R+
C
⎝ R1
+ R2
⎠
ε1
/ R1
− ε1
/ R 2
where ε eq =
1/ R1
+ 1/ R 2
R1R
2
R eq =
R + R
1
2
8. Maximum current passing through resistance R -
εeq
εeq
(A)
(B)
R
R + R
(C)
ε eq
R
eq
(D)
−
R
eq
| ε1 ε2
|
SCIENCE TIPS
• What is the expression for growing current, in LR
⎛ R
− t ⎞
circuit ? I = I 0
⎜1
− e
L ⎟
⎜ ⎟
⎝ ⎠
• What is the range of infrared spectrum ?
This covers wavelengths
from 10 –3 m down to 7.8 × 10 –7 m
• What is the nature of graph between electric field
and potential energy (U) ?
The nature of the graph
will be parabola having
symmetry about U-axis
• Why no beats can be heard if the frequencies of
the two interfering waves differ by more than ten ?
this is due to persistence
of hearing
• Why heating systems based on steam are more
efficient than those based on circulation of hot
water ?
This is because steam
has more heat than water
a the same temperature
• Can the specific heat of a gas be infinity ? Yes
• What is the liquid ascent formula for a capillary ?
2T cosθ r
h = –
γpg
3
where h is the height through
which a liquid of density ρ and
surface tension T rises in a
capillary tube of radius r
• What is the expression for total time of flight (T)
2u sin θ
for oblique projection ? T =
g
• The space charge limited current i P in the diode
3/2
value is given by i P = k V p
• What is an ideal gas ?
An ideal gas is one in
which intermolecular
forces are absent
• Can a rough sea be calmed by pouring oil on its
surface ?
Yes
• What is the expression for fringe width (β) in
Young's double slit experiment? β=Dλ/d where
D is the distance between the
source and screen and d is
distance between two slits
XtraEdge for IIT-JEE 19 DECEMBER 2009

8
Solution
Set # 7
Physics Challenging Problems
Questions were Published in November Issue
1. As the resistances of voltmeters in upper branch are
R, R/2, R/4 ......................
the equivalent circuit is as shown below
3. From current division formula we can conclude that
current in upper and lower branch are in the ratio of
1 : 2.
a
R
R/2 R/4
V
Lower Branch
upper Branch
...................
the resistance of upper branch is
= R + R/2 + R/4 + ............. up to infinite
⎛ 1 1 ⎞
= R ⎜1
+ + + ..... ⎟
⎝ 2 4 ⎠
⎛ 1 ⎞
= R ⎜ ⎟ = 2R
⎝1−1/
2 ⎠
further the equivalent circuit is
a
R
V
upper branch
Lower Branch
the resistance of voltmeter V should be 2R so that
current in upper and lower branch is same.
2. Entire upper branch is having the resistance of 2R
and voltmeter V 1 is having the resistance of R so we
can conclude that equivalent resistance of all the
voltmeters in upper branch except V 1 is R and the
upper branch is as follows:
a V 1 V 2 V 3
.....up to infinite b
a
i
R
C
V 1 =X V 2 =Y
As reading of voltmeter V 1 is X = i.R
sum of the readings of voltmeters is Y = i.R
Except V 1 in upper branch
So,
X = Y
R
b
b
b
4.
a
i
2i
R
C
R′ = R
voltmeter V
Reading of voltmeter V 1 is i.R
Reading of voltmeter V is (2i.)R
So V = 2V 1
a.
x.
A
b.
d x.
l = length of rod = b – a
l.
B
charge on element of length d x is d q
d q = λd x as λ = 3x
d q = 3xd x
Equivalent current due to element of length d x
ω
d i = ω.d q =
2 (3xd x) π
ω
Total equivalent current i =
∫di
=
∫
(3xd x )
2 π
3ω
=
2 π
⎡
2
x ⎤
⎢ ⎥
⎣ 2
⎦
b
a
3ω
=
2 π
⎛ b
⎜
⎝
3 ω
= (b 2 – a 2 )
4π
Option A is correct
(B) Equivalent current
2
R
− a
2
2
⎞
⎟
⎠
b
a
b
3 ω
= . 2 2 π
(b2 – a 2 )
XtraEdge for IIT-JEE 20 DECEMBER 2009

=
3 ω (b 2 – a 2 ) =
4π
3 ω (b – a)(b + a)
4π
3 ω
3
= (b + a)(b – a) = ω. (b + a).l
4 π 4π
As ω = 4π/3 So,
3 4π
Equivalent current = . . (b + a).l
4π
3
= (b + a).l = const.l
i ∝ l
Option B is correct.
(D) Charge on rod
=
b
=
⎡x
q =
∫d
q ∫3xd
x 3. ⎢
⎣ 2 ⎥ ⎦
a
= 2
3 (b 2 – a 2 )
Option D is correct
Ans. A, B, D
5. For part B
q > q
closed cone
open cone
for part A
q = q
closed cone
open cone
ω
Equivalent current i =
2 .q π
2
⎤
ω
i =
2 .q , i = ω
π 2 π
.q
cone - C 1 (closed cone) cone-C 3 (closed cone)
ω
=
2 .q i = ω
π 2 π
.(σ)
(closed cone) (Surface area of
closed cone)
If σ varies then charge on cone C 1 differs from C 3 So
their currents will be different.
Option A incorrect
q = q
(cone C 1 ) (cone - C 2 )
ω
i =
( coneC 1)
2 . ω
q and i = π ( ConeC1
) ( coneC2
) 2 . q π ( ConeC2
)
i = i
(cone C 1 ) (cone C 2 )
b
a
Option B is correct
As charge on cone C 3 ≠ charge on cone C 4
Option C correct
Part-A and part-B will have different charges so
option D incorrect
Ans. B, C
6. The circuit is as follows
CT
10Ω
R 1
Full scale deflection current for galvanometer is
50m
i g = = 5mA
10Ω
For terminals CT and a range is 5V so
V 5
using R = – G ⇒ R1 = – 10 = 990Ω
i
−3
g
5 × 10
R 1 = 990Ω
7. Range between CT and b is 10 volt so,
V 10
Using R = – G ⇒ R1 + R 2 = – 10
i
−3
g
5×
10
990 + R 2 = 2000 – 10
R 2 = 2000 – 1000 = 1000Ω
R 2 = 1000Ω
8. Range between CT and c is V so
V
Using R = – G
i g
R 1 + R 2 + R 3 =
V
5 × 10
−3
a
– 10
R 2
V
⇒ 990 + 1000 + 3000 =
5 × 10
b
−3
R 3
c
– 10
V
⇒ 5000 =
−3
5 × 10
⇒ V = 25 volt
So range between CT and C is 25 volts.
XtraEdge for IIT-JEE 21 DECEMBER 2009

PHYSICS
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. A homogeneous sphere of radius r rolls without
slipping with constant angular speed ω' over bigger
sphere, of radius R, which is in pure rotation with
constant angular speed ω about its centre O. (Fig.)
Find the time taken.
ω
R
O
ω′
r
C
(i) for the centre C of the rolling sphere to return to its
initial position (with respect to O), and
(ii) for the point of contact of the rolling sphere to make
one full revolution over the bigger rotating sphere.
(b)(i) Determine the acceleration of the contact point of the
rolling sphere, and
(ii) the point of greatest acceleration of the rolling sphere
(both w.r.t. the centre O)
Sol. Suppose that at t = 0, the contact points of lie along
the fixed line (reference line) OX. Let at time t the
line OC makes the angle θ with OX (Fig.)
ω
O
^
t
C
P
θ P′
It is better to express the velocity and acceleration of
any point including contact point P (say) at an
arbitrary instant of time t as,
v po = v co and a po = a co = a pc + a pc + a co (1)
We know that in the case of pure rolling the velocity
of contact point of the rolling body has zero velocity
and zero tangential acceleration relative to the contact
point of the surface on which it rolls. So if P′ be the
contact point of rotating sphere at time t, then
v p′o = v po
So, v p'o = v pc + v co (2)
ω′
X
(a) If t is the direction of common tangent, then from
eqn. (2)
v p'o (t) = v pc (t) + v co (t)
⎛ dθ
⎞
ωR = – ω′R + ⎜ ⎟ (R + r)
⎝ dt ⎠
dθ ωR
+ ω'r
or,
=
(3)
dt R + r
R + r
so, dt =
dθ (4)
( ωR
+ ω'r)
(ii) It is simple to observer if rotating sphere were at rest,
the CM of rolling sphere C will turn by the angle 2 π,
⎛ dθ
⎞
with angular speed ⎜ ⎟ – ω to satisfy the condition
⎝ dt ⎠
of the problem.
dθ
dt
ω
Hence the sought time t′ (say) is given by
2π
2π(R
+ r)
t′ = = [using Eq. (4)]
⎛ dθ
⎞ r( ω−ω ' )
⎜ ⎟ − ω
⎝ dt ⎠
(b) From a po = a pc + a co = a pc (tangential) + a pc (normal) + a co
In our case a pc (tangential) = 0, because ω′ = constant
Hence, a po = a pc (normal) + a co (5)
⎡
2
⎤
2 ⎛ dθ
⎞
a po = ⎢− ω'
r + ⎜ ⎟ (R + r)
⎥
⎢⎣
⎝ dt ⎠ ⎥⎦
dθ ωR
+ ω'r
(using = from equation (3) or part (a))
dt R + r
(ii) From eqn. (5) it is obvious that the maximum value
|apo | ma x = |a pc (normal) | + | a co |
2
= – ω′ 2 ⎛ dθ
⎞
r + ⎜ ⎟⎠ (R + r)
⎝ dt
dθ
where will be substituted from Eqn. (3).
dt
XtraEdge for IIT-JEE 22 DECEMBER 2009

2. Show that the temperature of a planet varies inversely
as the square root of its distance from the Sun.
Sol. Let R s be the radius of the Sun. Consider the Sun as a
black body.
Energy emitted per sec by it equals 4 π R σ T
4
This energy falls uniformly on the inner surface of
spheres centred on the Sun. If d is the distance of the
planet from the Sun, then energy falling on unit area
of the sphere of radius d is :
2 4
sσTs
2
4πR
σR sTs
=
2
4πd
d
This energy received by the planet is given by
2
4
2
Q = πr 2 σR sTs
= where r is the radius of planet.
2
d
If T is the temperature of the planet, then energy lost
by it per sec is 4 π r 2 σ T 4
In the steady state the rate of reception of energy is
equal to the loss of energy
Hence, 4 π r 2 σ T 4 πr σR
=
2
d
1
Thus, T ∝
d
4
2
2 4
s Ts
3. A sphere of specific gravity s just fits into a vertical
cylinder with lower end closed. The sphere is allowed
to drop slowly until it is held in equilibrium by the
thrust of the compressed air. There is no leakage of
air. If the diameter of the sphere is d, the length of
the cylinder is L and the height of the water
barometer is h, then what will be the position of
sphere ?
Sol. Initially, the cylinder contained air at atmospheric
pressure. When the sphere comes down into the
cylinder by the action of its own weight, it presses the
air downwards. Suppose the sphere comes to position
C which is at height x above the closed end. Let the
sphere remain in equilibrium in this position.
Volume of sphere = (4/3)πr 3
Weight of the sphere = (4/3)πr 3 sg
Volume of cylinder = πr 2 L
Volume of air inside the cylinder when the sphere is
in position A = πr 2 L – (2/3)πr 3
2
s
s
Volume of compressed air when the sphere is in
position C = [πr 2 x – (2/3)πr 3 ]
Atmospheric pressure = h cm of water
Let the pressure of compressed air be p cm of water.
L
d
According to Boyle's Law (assuming no change in
the temperature of compressed air)
p[πr 2 x – (2/3)πr 3 ] = h[πr 2 L – (2/3)πr 3 ]
or p[x – (2r/3) = h[L – (2r/3)] ...(1)
In the equilibrium position C, the weight of the
sphere is balanced by the difference of vertical thrust
on either side due to atmospheric and compressed air.
Hence, πr 2 (p – h)g = mg = (4/3)πr 3 sg
or p – h = (4/3)rs ...(2)
From equation (1) and (2), we get
h(3L
− 2r)
x =
+
3p
h(3L − 2r)
or x =
+
3h + 4rs
or x =
2
9hL + 2d s
9h + 6ds
2r
3
A
C
x
2 r 9hL + 8r s
=
3 3(3h + 4rs)
4. Given the position of the object O and the image I as
shown in the figure. Find (a) the position of the
convex lens (b) its focal length and the magnification
of the image. Verify graphically.
2
1
0
–1
–2
Y
O
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Sol. A ray of light from the object passes undeviated
through the optic centre of the lens (C) and also the
image I. So join OI.
So it cuts the principal axis XY and C. So AB is the
position of the lens. A ray parallel to the principal
axis from the object after refraction meets the
principal axis at F. F is the focus.
2
I
X
XtraEdge for IIT-JEE 23 DECEMBER 2009

X
O
4
C
A
8 F
B
10.4
14
F is at a distance 2.4 cm from C.
∴ Focal length of lens = 2.4 × 10 = 24 cm
By calculation
1 1 1 = –
f v u
1 1
= – 60 − 40
∴ f = 24 cm
100 1
= = 2400 24
v 60 3
Magnification = = = u 40 2
=
=
length of image
length of object
3 mm
2 mm
= 2
3
5. Calculate the magnetic field B at the point P shown
in the figure. Assume that i = 10 A and a = 8.0 cm.
B a/4
a
4
i
A
P
i
Sol. First of all, we determine the expression of B at a
distance R from a straight conductor of length l.
x
x 2
x 1
φ
dx
i
R
Consider a typical element dx. The magnitude of the
contribution dB of this element to the magnetic field
at P as found from Biot-Savart's law is
r
C
i
D
I
X
µ 0idxsin
θ
dB =
...(i)
2
4πr
Since, the direction of the contribution dB at the
point P for all elements are identical viz, at right
angle into the plane of the figure, the resultant field is
obtained by simply integration equation (i), which
gives
x
B =
∫ 2 µ 0i
dB =
4π
x1
x 2
∫
x1
sin θdx
where, sin θ = r
R and r = (x 2 + R 2 ) 1/2
µ 0iR
∴ B =
4π
x2
r
∫
+
x1
2
dx
2 2
( x R )
3/ 2
Let x = R tan φ, such that dx = R sec 2 φ dφ
At x = x 1 φ 1 = tan –1 ⎛ x ⎞
⎜ 1
⎟ and x = x 2 ,
⎝ R ⎠
x 2
φ 2 = tan –1 ⎛ ⎞
⎜ ⎟
⎝ R ⎠
Also, x 2 + R 2 = R 2 tan 2 φ + R 2 = R 2 sec 2 φ
Hence, equation (ii) becomes
B =
µ 0iR
4π
∫ − 1 x
tan
2
R
−1
x
tan
1
R
µ tan
=
∫ −
0iR
−
4π
tan
1 x2
R
1 x1
R
R sec
R
3
sec
2
2
φ
φ
cosφ
dφ
...(ii)
µ 0 iR ⎡ ⎛ −1
x 2 ⎞ ⎛ −1
x1
⎞⎤
= ⎢sin⎜
tan ⎟ − sin⎜
tan ⎟⎥ ...(iii)
4π
⎣ ⎝ R ⎠ ⎝ R ⎠ ⎦
Let, z = tan –1 ⎛ x ⎞
⎜ ⎟ ,
⎝ R ⎠
x
Thus, tan z = R
⇒
sin z
1−
sin
2
z
= R
x
⇒ R 2 sin 2 z = x 2 (1 – sin 2 z)
⇒ sin 2 z(R 2 + x 2 ) = x 2
⇒ sin z =
x
2
x
+ R
⎛
⇒ z = sin –1 ⎟ ⎟ ⎞
⎜ x
⎜
⎝ x 2 + R
2 ⎠
2
XtraEdge for IIT-JEE 24 DECEMBER 2009

Hence, equation (iii) becomes
µ
⎡
⎤
0i
B = ⎢
x 2 x1
− ⎥ ...(iv)
4πR
⎢ 2 2 2 2 ⎥
⎣ x 2 + R x1
+ R ⎦
Applying the equation (iv) to the given problem
For AB, we get
⎛ 3 ⎞ a a
x 1 = – ⎜ ⎟ a, x2 = and R =
⎝ 4 ⎠ 9 4
Hence, B 1 =
For BC :
⎡
⎢
µ 0i
⎢
⎛ a ⎞ ⎢
4π⎜
⎟ ⎢
⎝ 4 ⎠ ⎢⎣
a
4
2 2
a a
+
16 16
µ
= 0 i ⎛ 1 3
⎟ ⎞
⎜ +
πa
⎝ 2 10 ⎠
− a 3a
x 1 = , x2 =
4 4
This also gives
For CD :
∴ Β 3 =
For DA :
µ
B 2 = 0 i ⎛ 1 3
⎜ +
πa
⎝ 2 10
x 1 = 4
a , x2 =
⎡
⎢
µ 0i
⎢
⎛ − 3a ⎞ ⎢
4π⎜
⎟ ⎢
⎝ 4 ⎠ ⎢⎣
−3a
4
= µ i ⎡ 1 1 ⎤
0
⎢ + ⎥
3πa
⎣ 10 2 ⎦
x 1 =
−
and R = 4
a
⎟ ⎞
⎠
and R =
− 3a
4
2 2
9a 9a
+
16 16
−
3 a −a
3a
, x2 = and R =
4 4 4
⎛ − 3 ⎞ ⎤
⎜ ⎟a
⎥
⎝ 4 ⎠ ⎥
2 2 ⎥
9a a
+ ⎥
16 16 ⎥⎦
−3a
4
a
⎤
⎥
4 ⎥
2 2 ⎥
a 9a
+ ⎥
16 16 ⎥⎦
This also gives B 4 = µ i ⎡ 1 1 ⎤
0
⎢ + ⎥
3πa
⎣ 10 2 ⎦
Total magnitude at magnetic field,
B = B 1 + B 2 + B 3 + B 4
= µ i ⎡ 1 3 1 3 1 1 1 1 ⎤
0
⎢ + + + + + + + ⎥
πa
⎣ 2 10 2 10 3 10 3 2 3 10 3 2 ⎦
2µ
= 0 i
( 10 + 2 2)
3πa
= 2.0 × 10 4 T
MEMORABLE POINTS
• Can a current be measured by a voltameter ?
Yes, direct current can be
measured by a voltameter
• The equivalent resistance of n resistances each equal to r
and connected in parallel is given by
r/n
• Repeated use of which digital gate or gates can
produce all the three basic gates (OR, AND and
NOT)
NAND gate and NOR gate
• What is Turnbull's blue ? Fe 4 [Fe(CN) 6 ] 3
• An hypothesis tested by experiments is known as
Theory
• What is magnesia alba ? Mg(OH) 2 .MgCO 3 .3 H 2 O
• The humidity of air is measured by
Hygrometer
• What is oleum ? H 2 S 2 O 7
Also known as fuming sulphuric acid
• Vulcanised rubber was invented by
Charies Goodyear (1839)
• An amino acid which does not contain a chiral
centre.
Glycine [NH 2 –CH 2 –COOH]
• Scientist who perfected the technique for converting
pig iron into steel. Hynry Besemer (1856)
• When the pH of the blood is lower than the normal
value, this condition is known as Acidosis
• The electrolytic method of obtaining aluminium
from bauxite was first developed by
Charles Hall (1886)
• Which compound possesses characteristic smell like
that of mustard oil ?
Ethyl isothiocyanate [C 2 H 5 N = C = S]
• First solar battery was developed in the
Bell Telephone Laboratory (1954)
• What is Wilkinson's catalyst ?
tris (triphenylphosphine) chlororhodlum (I)
• In 1836 the galvanised iron was introduced first in
France
• What is caro's acid ?
Permonosulphuric acid [H 2 SO 5 ]
XtraEdge for IIT-JEE 25 DECEMBER 2009

PHYSICS FUNDAMENTAL FOR IIT-JEE
Reflection at plane & curved surfaces
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Key Concepts :
(a) Due to reflection, none of frequency, wavelength
and speed of light change.
(b) Law of reflection :
Incident ray, reflected ray and normal on incident
point are coplanar.
The angle of incidence is equal to angle of
reflection
Incident n Reflected
Ray Ray
θ θ
Plane surface
Convex
surface
n
α α
n
Tangent
θ θ
at point P
P
Convex surface
A
Tangent
at point P
Some important points : In case of plane mirror
For real object, image is virtual.
For virtual object, image is real.
The converging point of incident beam behaves as a
object.
If incident beam on optical instrument (mirror, lens
etc) is converging in nature, object is virtual.
If incident beam on optical instrument is diverging in
nature, the object is real.
The converging point of reflected or refracted beam
from an optical instrument behaves as image.
If reflected beam or refracted beam from an optical
instrument is converging in nature, image is real.
P
Virtual
Object
P
Real
Object
P
Real
Object
n
n
P
Virual
Object
If reflected beam or refracted beam from an optical
instrument is diverging in nature, image is virtual.
For solving the problem, the reference frame is
chosen in which optical instrument (mirror, lens, etc.)
is in rest.
The formation of image and size of image is
independent of size of mirror.
Visual region and intensity of image depend on size
of mirror.
P
n
If the plane mirror is rotated through an angle θ, the
reflected ray and image is rotated through an angle 2θ
in the same sense.
If mirror is cut into a number of pieces, then the focal
length does not change.
The minimum height of mirror required to see the full
image of a man of height h is h/2.
Object
vsinθ
Object
Object
v
Rest
v
θ
vcosθ
v
v
θ
θ
α
α
Image
Rest
vsinθ
vcosθ Image
v m
2v m –v
Image
P'
P
n
Real
n
Object
α
α
α
α
P'
Virual
Object
Object
In rest
v m
Image
2v m
XtraEdge for IIT-JEE 26 DECEMBER 2009

Object
v
v m
2v m +v
Image
(c) Number of images formed by combination of
two plane mirrors : The images formed by
combination of two plane mirror are lying on a
circle whose centre is at the meeting points of
mirrors. Also, object is lying on that circle.
360º
Here, n =
θ
where θ = angle between mirrors.
360º
If is even number, the number of images is
θ
n – 1.
360º
If θ
is odd number and object is placed on
bisector of angle between mirror, then number of
images is n – 1.
360º
If is odd and object is not situated on
θ
bisector of angle between mirrors, then the
number of images is equal to n.
(d) Law of reflection in vector form :
Let ê 1 = unit vector along incident ray.
ê 2 = unit vector along reflected ray
nˆ = unit vector along normal on point of
Incidence
Then ê 2 =
ê 2(ê1
1 −
n
nˆ
.nˆ ) nˆ
ê1
ê 2
(e) Spherical mirrors :
It easy to solve the problems in geometrical optics
by the help of co-ordinate sign convention.
x'
y
y'
y
x
x'
y
y'
x' x x'
y'
y'
1 1
The mirror formula is +
v u
Also, R = 2f
y
x
x
=
1
f
x'
y
y'
x
These formulae are only applicable for paraxial
rays.
All distances are measured from optical centre. It
means optical centre is taken as origin.
The sign conventions are only applicable in given
values.
The transverse magnification is
image size −v
β =
=
object size u
1. If object and image both are real, β is negative.
2. If object and image both are virtual, β is negative.
3. If object is real but image is virtual, β is positive.
4. If object is virtual but image is real, β is positive.
5. Image of star; moon or distant object is formed at
focus of mirror.
If y = the distance of sun or moon from earth.
D = diameter of moon or sun's disc
f = focal length of the mirror
d = diameter of the image
θ = the angle subtended by sun or moon's disc
D d
Then tan θ = θ = = y f
Here, θ is in radian.
Sun
D
d
F
Problem solving strategy :
Image formation by mirrors
Step 1: Identify the relevant concepts : There are
two different and complementary ways to solve
problems involving image formation by mirrors. One
approach uses equations, while the other involves
drawing a principle-ray diagram. A successful
problem solution uses both approaches.
Step 2: Set up the problem : Determine the target
variables. The three key quantities are the focal
length, object distance, and image distance; typically
you'll be given two of these and will have to
determine the third.
Step 3: Execute the solution as follows :
The principal-ray diagram is to geometric optics
what the free-body diagram is to mechanics. In
any problem involving image formation by a
mirror, always draw a principal-ray diagram first
if you have enough information. (The same
θ
θ
XtraEdge for IIT-JEE 27 DECEMBER 2009

advice should be followed when dealing with
lenses in the following sections.)
It is usually best to orient your diagrams
consistently with the incoming rays traveling
from left to right. Don't draw a lot of other rays at
random ; stick with the principal rays, the ones
you know something about. Use a ruler and
measure distance carefully ! A freehand sketch
will not give good results.
If your principal rays don't converge at a real
image point, you may have to extend them
straight backward to locate a virtual image point,
as figure (b). We recommend drawing the
extensions with broken lines. Another useful aid
is to color-code the different principal rays, as is
done in figure(a) & (b).
Q
P
Q
P
4
3
I
2
4
1
4
2 2
2
1
(b)
4
C
1
3
(a)
v
P'
Q'
F
Q'
3
P' F
1 1 1
Check your results using Eq. + = and the
s s' f
y'
s'
magnification equation m = = − . The
y s
results you find using this equation must be
consistent with your principal-ray diagram; if not,
double-check both your calculation and your
diagram.
Pay careful attention to signs on object and image
distances, radii or curvature, and object and image
heights. A negative sign on any of these
quantities always has significance. Use the
equations and the sign rules carefully and
consistently, and they will tell you the truth !
Note that the same sign rules (given in section)
work for all four cases in this chapter : reflection
and refraction from plane and spherical surfaces.
Step 4: Evaluate your answer : You've already
checked your results by using both diagrams and
equations. But it always helps to take a look back
and ask yourself. "Do these results make sense ?".
v
C
1. How will you arrange the two mirrors so that
whatever may be the angle of incidence, the incident
ray and the reflected ray from the two mirrors will be
parallel to each other.
Sol.
B
A
Solved Examples
θ
P
Q
i 1 i 2
i 1 i 2
The total deviation of the ray is given by
δ =180 – 2i 1 + 180 – 2i 2
= 360 – 2(i 1 + i 2 )
For the resultant ray to be parallel, δ should be 180º
∴ 360 – 2(i 1 + i 2 ) = 180
i.e., i 1 + i 2 = 90º
From the geometry of the figure i 1 + i 2 = θ
θ Angle between the mirrors should be 90º.
2. Rays of light strike a horizontal plane mirror at an
angle of 45º. At what angle should a second plane
mirror be placed in order that the reflected ray finally
be reflected horizontally from the second mirror.
Sol. The situation is shown in figure
C G
A
S θ θ
D
P
45º
45º
B
N
Q
The ray AB strikes the first plane mirror PQ at an
angle of 45º. Now, we suppose that the second
mirror SG is arranged such that the ray BC after
reflection from this mirror is horizontal.
From the figure we see that emergent ray CD is
parallel to PQ and BC is a line intersecting these
parallel lines.
So, ∠DCE = ∠CBQ = 180º
∠DCN + ∠NCB + ∠CBQ = 180º
θ + θ + 45º = 180º ∴ θ = 67.5º
As ∠NCS = 90º, therefore the second mirror should
be inclined to the horizontal at an angle 22.5º.
3. An object is placed exactly midway between a
concave mirror of radius of curvature 40 cm and a
convex mirror of radius of curvature 30 cm. The
mirrors face each other and are 50 cm apart.
Determine the nature and position of the image
formed by the successive reflections, first at the
concave mirror and then at the convex mirror.
C
XtraEdge for IIT-JEE 28 DECEMBER 2009

Sol. The image formation is shown in figure.
I 1
I 2
P 2 C
r = 30 cm
50cm
F
25cm
P 1
r = 40 cm
(i) For concave mirror,
u 1 = 25 cm, f 1 = 20 cm and v 1 = ?
1 1 1
Now = +
f 1 u1 v1
1 1 1
or = +
20 25 v1
v 1 = 100 cm.
As v 1 is positive, hence the image is real. In the
absence of convex mirror, the rays after reflection
from concave mirror would have formed a real image
I 1 at distance 100 cm from the mirror. Due to the
presence of convex mirror, the rays are reflected and
appear to come from I 2 .
(ii) For convex mirror,
In this case, I 1 acts as virtual object and I 2 is the
virtual image.
The distance of the virtual object from the convex
mirror is 100 – 50 = 50 cm. Hence u 2 = –50 cm.
As focal length of convex mirror is negative and
hence f 2 = –30/2 = –15 cm. Here we shall calculate
the value of v 2 . Using the mirror formula, we have
1 1 1
− = − +
15 50 v 2
or v 2 = –21.42 cm
As v 2 is negative, image is virtual. So image is
formed behind the convex mirror at a distance of
21.43 cm.
4. A convex and a concave mirror each 30 cm in radius
are placed opposite to each other 60 cm apart on the
same axis. An object 5 cm in height is placed
midway between them. Find the position and size of
the image formed by reflection, first at convex and
then at the concave mirror.
Sol. The image formation is shown in figure.
P 2
I 2
O
P 1 I 1
r = 30 cm r = 30 cm
(i) For convex mirror,
u 1 = +30 m, f 1 = –15 cm and v 1 = ?
1 1 1
∴ − = +
15 50 v 1
or v 1 = –10 cm
The image will be virtual. This is formed at I 1 behind
the convex mirror at a distance of 10 cm. The image
I 1 acts as an object for convex mirror.
(ii) For concave mirror,
u 2 = P 2 I 1 = 60 + 10 = 70 cm,
f 2 = +15 cm and v 2 = ?
1 1 1
∴ = +
15 70 v2
Solving we get, v 2 = (210/11) cm.
As v 2 is positive, the image I 2 is formed in front of
concave mirror at a distance of (210/11) cm.
Magnification m 1 for first reflection
v 1 10 1
= = =
u1
30 3
Magnification m 2 for second reflection
v 2 (210/11) 3
= = =
u 2 70 11
1 3 1
Final magnification = m 1 × m 2 = × = 3 11 11
∴ Size of the image = 5 × 11
1 = 11
5
5. An object is placed infront of a convex mirror at a
distance of 50 cm. A plane mirror is introduced
covering the lower half of the convex mirror. If the
distance between the object and the plane mirror is 30
cm it is found that there is no parallex between the
images formed by the two mirrors. What is the
radius of curvature of the convex mirror ?
Sol. Let O be the object placed infront of a convex mirror
MM' at a distance of 50 cm as shown in figure. The
distance of the plane mirror NN' from the object is
30. We know that in a plane mirror the image is
formed behind the mirror at the same distance as the
object infront of it. It is also given that there is no
parallax between the images formed by the two
mirrors, i.e., the image is formed at a distance of 30
cm behind the plane mirror.
For convex mirror,
u = 50 cm, v = 10 cm [Q QP = QN – PN]
as v is negative in convex mirror.
M
Q
M'
P
20cm
50cm
N
N'
30cm
O
1 1 1
Using the mirror formula = + , we have
f u v
1 1 1 4
50
= − = − , ∴ f = −
f 50 10 50
4
50×2
Now v = 2f = − = –25 m
4
The radius of curvature of convex mirror is 25 cm.
XtraEdge for IIT-JEE 29 DECEMBER 2009

PHYSICS FUNDAMENTAL FOR IIT-JEE
Fluid Mechanics
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Hydrostatics :
Pressure at a point inside a Liquid : p = p 0 + ρgh
where p 0 is the atmospheric pressure, ρ is the density
of the liquid and h is the depth of the point below the
free surface.
h
p
p 0
ρ
Hydrodynamics :
Bernoulli's Theorem : 2
1 v 2 + gh + ρ
p = a constant
for a streamline flow of a fluid (liquid or gas).
Here, v is the velocity of the fluid, h is its height
above some horizontal level, p is the pressure and ρ
is the density.
p 1
v 1
p 2
h 2
h 1
v 2
Pressure is a Scalar : The unit of pressure may be
atmosphere or cm of mercury. These are derived
units. The absolute unit of pressure is Nm –2 . Normal
atmospheric pressure, i.e, 76 cm of mercury, is
approximately equal to 10 5 Nm –2 .
Thrust : Thrust = pressure × area. Thrust has the unit
of force.
Laws of liquid pressure
(a) A liquid at rest exerts pressure equally in all
directions.
(b) Pressure at two points on the same horizontal line
in a liquid at rest is the same.
(c) Pressure exerted at a point in a confined liquid at
rest is transmitted equally in all directions and
acts normally on the wall of the containing vessel.
This is called Pascal's law. A hydraulic press
works on this principle of transmission of
pressure.
The principle of floating bodies (law of flotation) is
that W = W´, that is, weight of body = weight of
displaced liquid or buoyant force. The weight of the
displaced liquid is also called buoyancy or upthrust.
Hydrometers work on the principle of floating
bodies. This principle may also be applied to gases
(e.g., a balloon).
Liquids and gases are together called fluids. The
important difference between them is that liquids
cannot be compressed, while gases can be
compressed. Hence, the density of a liquid is the
same everywhere and does not depend on its
pressure. In the case of a gas, however, the density is
proportional to the pressure.
v 2 > v 1 p 2 < p 1
According to this principle, the greater the velocity,
the lower is the pressure in a fluid flow.
It would be useful to remember that in liquid flow,
the volume of liquid flowing past any point per
second is the same for every point. Therefore, when
the cross-section of the tube decreases, the velocity
increases.
Note : Density = relative density
or specific gravity × 1000 kg m –3 .
Surface tension and surface energy :
Surface Tension : The property due to which a
liquid surface tends to contract and occupy the
minimum area is called the surface tension of the
liquid. It is caused by forces of attraction between the
molecules of the liquid. A molecule on the free
surface of a liquid experiences a net resultant force
which tends to draw it into the liquid. Surface tension
is actually a manifestation of the forces experienced
by the surface molecules.
If an imaginary line is drawn on a liquid surface then
the force acting per unit length of this line is defined
as the surface tension. Its unit is, therefore, newton /
metre. This force acts along the liquid surface. For
curved surfaces, the force is tangent to the liquid
surface at every point.
Surface Energy : A liquid surface possesses
potential energy due to surface tension. This energy
per unit area of the surface is called the surface
energy of the liquid. Its units is joule per square
metre. The surface energy of a liquid has the same
numerical values as the surface tension. The surface
XtraEdge for IIT-JEE 30 DECEMBER 2009

tension of a liquid depends on temperature. It
decreases with rise in temperature.
Excess of Pressure : Inside a soap bubble or a gas
bubble inside a liquid, there must be pressure in
excess of the outside pressure to balance the tendency
of the liquid surface to contract due to surface
tension.
⎛
p(excess of pressure) = T
⎜
⎝
1
r
1
+
1 r 2
⎟ ⎞
⎠
in general
where T is surface tension of the liquid, and r 1 and r 2
are the principal radii of curvature of the bubble in
two mutually perpendicular directions.
For a spherical soap bubble, r 1 = r 2 = r and there are
two free surfaces of the liquid.
4T
∴ p = r
For a gas bubble inside a liquid, r 1 = r 2 = r and there
is only one surface.
2T
∴ p = r
For a cylindrical surface r 1 = r and r 2 = ∞ and there
are two surfaces.
2T
∴ p = r
Angle of Contact : The angle made by the surface of
a liquid with the solid surface inside of a liquid at the
point of contact is called the angle of contact. It is at
this angle that the surface tension acts on the wall of
the container.
The angle of contact θ depends on the natures of the
liquid and solid in contact. If the liquid wets the solid
(e.g., water and glass), the angle of contact is zero. In
most cases, θ is acute (figure i). In the special case of
mercury on glass, θ is obtuse (figure ii).
θ
fig. (i)
θ
fig. (ii)
Rise of Liquid in a Capillary Tube : In a thin
(capacity) tube, the free surface of the liquid becomes
curved. The forces of surface tension at the edges of
the liquid surface then acquire a vertical component.
h
T θ
θ
meniscus
r
θ
θ
T
The upward force by which a liquid surface is pulled
up in a capillary tube is 2πrTcos θ, and the downward
force due to the gravitational pull on the mass of
liquid in the tube is (πr2h + v)ρg, where v is the
volume above the liquid meniscus. If θ = 0º, the
meniscus is hemispherical in shape. Then v =
difference between the volume of the cylinder of
radius r and height r and the volume of the
hemisphere of radius r
= πr 3 – 3
2 πr 3 = 3
1 πr
3
When θ ≠ 0, we cannot calculate v which is generally
very small and so it may be neglected. For
equilibrium
(πr 2 h + v) ρg = 2πrT cos θ
When a glass capillary tube is dipper in mercury, the
meniscus is convex, since the angle of contact is
obtuse. The surface tension forces now acquire a
downward component, and the level of mercury
inside the tube the falls below the level outside it. the
relation 2T cos θ = hρgr may be used to obtain the
fall in the mercury level.
Problem Solving Strategy
Bernoulli's Equations :
Bernoulli's equation is derived from the work-energy
theorem, so it is not surprising that much of the
problem-solving strategy suggested in W.E.P. also
applicable here.
Step 1: Identify the relevant concepts : First ensure
that the fluid flow is steady and that fluid is
incompressible and has no internal friction. This case
is an idealization, but it hold up surprisingly well for
fluids flowing through sufficiently large pipes and for
flows within bulk fluids (e.g., air flowing around an
airplane or water flowing around a fish).
Step 2: Set up the problem using the following steps
Always begin by identifying clearly the points 1
and 2 referred to in Bernoulli's equation.
Define your coordinate system, particular the
level at which y = 0.
Make lists of the unknown and known quantities
in Eq. p 1 + ρgy 1 + 2
1
ρv1 2 = p 2 + ρgy 2 + 2
1
ρv2
2
(Bernoulli's equation)
The variables are p 1 , p 2 , v 1 , v 2 , y 1 and y 2 , and the
constants are ρ and g. Decide which unknowns
are your target variables.
Step 3: Execute the solutions as follows : Write
Bernoulli's equation and solve for the unknowns. In
some problems you will need to use the continuity
equation, Eq. A 1 v 1 = A 2 v 2 (continuity equation,
incompressible fluid), to get a relation between the
two speeds in terms of cross-sectional areas of pipes
XtraEdge for IIT-JEE 31 DECEMBER 2009

or containers. Or perhaps you will know both speeds
and need to determine one of the areas. You may also
dV
need to use Eq. = Av (volume flow rate) to find
dt
the volume flow rate.
Step 4: Evaluate your answer : As always, verify that
the results make physical sense. Double-check that
you have used consistent units. In SI units, pressure is
in pascals, density in kilograms per cubic meter, and
speed in meters per second. Also note that the
pressures must be either all absolute pressure or all
gauge pressures.
Solved Examples
1. A vertical U-tube of uniform cross-section contains
mercury in both arms. A glycerine (relative density
1.3) column of length 10 cm is introduced into one of
the arms. Oil of density 800 kg m –3 is poured into the
other arm until the upper surface of the oil and
glycerine are at the same horizontal level. Find the
length of the oil column. Density of mercury is 13.6
× 10 3 kg m –3 .
Sol. Draw a horizontal line through the mercury-glycerine
surface. This is a horizontal line in the same liquid at
rest namely, mercury. Therefore, pressure at the
points A and B must be the same.
(1 – h)
A
h
B
10 cm
Pressure at B
= p 0 + 0.1 × (1.3 × 1000) × g
Pressure at A
= p 0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g
∴ p 0 + 0.1 × 1300 × g
= p 0 + 800gh + 1360g – 13600 × g × h
⇒ 130 = 800h + 1360 – 13600h
1230
⇒ h = = 0.096 m = 9.6 cm
12800
H
x
•1
•2
•3
2
p 0 1 +
2 p 1
v 1 + gH = +
2
v 2 + g (h – x)
ρ 2 ρ 2
p 0 1
= +
2
v 3 + 0
ρ 2
By continuity equation
v 1A 1 = A 2 v 2 = A 2 v 3
Since A 1 >> A 2 ,v 1 is negligible and v 2 = v 3 = n (say).
∴
p p + gH =
2 1 + v 2 + g (h – x)
ρ 2
0
ρ
p 0 1
= + v
2
ρ 2
∴ v = 2 gH
(i)
p 0 p
and + gH =
2 + gH + g (h – x)
ρ ρ
⇒ p 0 + p 2 + ρg (h – x)
⇒ p 2 = p 0 – ρg (h – x) (ii)
Thus pressure varies with distance from the upper
end of the pipe according to equation (ii) and velocity
is a constant and is given by (i).
3. A rod of length 6m has a mass of 12 kg. It is hinged
at one end at a distance of 3 m below the water
surface.
(i) What weight must be attached to the other end so
that 5 m of the rod is submerged ?
(ii) Find the magnitude and direction of the force
exerted by the hinge on the rod. The specific gravity
of the material of the rod is 0.5
Sol. Mass per unit length of the rod is 2 kg. Therefore,
mass of the submerged portion of the rod is 10 kg and
10
its volume = m 3 (using the simple formula,
500
mass
volume = and density = specific gravity ×
density
h
2. A liquid flows out of a broad vessel through a narrow
vertical pipe. How are the pressure and the velocity
of the liquid in the pipe distributed when the height
of the liquid level in the vessel is H from the lower
end of the length of the pipe is h ?
Sol. Let us consider three points 1, 2, 3 in the flow of
water. The positions of the points are as shown in the
figure.
Applying Bernoulli's theorem to points 1, 2 and 3
1000 kg m –3 ).
3m
H
A
N
F b
θ
O
O´
12 kgf
C
B
W
XtraEdge for IIT-JEE 32 DECEMBER 2009

Therefore, buoyant force
10
F b = × 1000 = 20 kgf
500
Let N and H be the vertical downward and horizontal
reactions of the hinge on the rod. Considering
horizontal and vertical translational equilibrium of
the rod, N + 12 + W = 20 (where W is the weight to
be attached and H = 0).
N + W = 8 and H = 0 ..(i)
Considering the rotational equilibrium of the rod
about A
–20 × g × 2.5 cos θ + 12 × g × 3 cos θ + W × 6 cos θ = 0
⇒ 6W = 50g – 36g = 14g
⇒ W = 3
7 g = 3
7 kgf
7 17
Ν = 8 – = kgf 3 3
4. The end of a capillary tube with a radius r is
immersed in water. Is mechanical energy conserved
when the water rises in the tube ? The tube is
sufficiently long. If not, calculate the energy change.
Sol. In the equilibrium position (θ = 0º for pure water and
glass)
2πrT cos 0º = πr 2 hρg
2T
or h =
ρgr
2
4πT
Work done by surface tension = (2πrT) × h =
ρg
U, potential energy of water in the tube
= (πr 2 hρ)gh/2; it is multiple by h/2 because the cg of
the water in the capillary tube is at a height h/2.
2
2πT
⇒ U =
ρg
Thus it is seen that the mechanical energy is not
conserved.
2 2
4πT
2πT
∴ mechanical energy loss = –
ρg
ρg
2πT
=
ρg
This energy is converted into heat.
5. Calculate the difference in water levels in two
communicating tubes of diameter d = 1 mm and
d = 1.5 mm. Surface tension of water = 0.07 Nm –1
and angle of contact between glass and water = 0º.
2T cosθ
Sol. Pressure at A = p 0 –
r 2
(Q pressure inside a curved surface is greater than
that outside)
2T cosθ
Pressure at B = p 0 –
r 1
2
⎛
∴ pressure difference = 2T cos θ
⎜
⎝
B
A
1
r
1
−
1 r 2
Let this pressure difference correspond to h units of
the liquid.
⎛ 1 1
Then 2T cos θ ⎟ ⎞
⎜ − = ρgh
⎝ r1 r 2 ⎠
2T cosθ
⎛ 1 1
⇒ h =
⎟ ⎞
⎜ −
ρg
⎝ r1 r 2 ⎠
2×
0.07 ⎛ 1 1 ⎞
∴ h = ⎜ − ⎟ = 4.76 mm
1000×
9.8
−3
−3
⎝1×
10 1.5×
10 ⎠
Interesting Science Facts
• The dinosaurs became extinct before the
Rockies or the Alps were formed.
• Female black widow spiders eat their males after
mating.
• When a flea jumps, the rate of acceleration is 20
times that of the space shuttle during launch.
• The earliest wine makers lived in Egypt around
2300 BC.
• If our Sun were just inch in diameter, the nearest
star would be 445 miles away.
• The Australian billy goat plum contains 100
times more vitamin C than an orange.
• Astronauts cannot belch - there is no gravity to
separate liquid from gas in their stomachs.
• The air at the summit of Mount Everest, 29,029
feet is only a third as thick as the air at sea level.
• One million, million, million, million, millionth of
a second after the Big Bang the Universe was the
size of a …pea.
• DNA was first discovered in 1869 by Swiss
Friedrich Mieschler.
• The molecular structure of DNA was first
determined by Watson and Crick in 1953.
• The thermometer was invented in 1607 by
Galileo.
• Englishman Roger Bacon invented the magnifying
glass in 1250.
⎟ ⎞
⎠
XtraEdge for IIT-JEE 33 DECEMBER 2009

KEY CONCEPT
Organic
Chemistry
Fundamentals
CARBOXYLIC ACID
Acidity of carboxylic acids.
Fatty acids are weak acids as compared to inorganic
acids. The acidic character of fatty acids decreases
with increase in molecular weight. Formic acid is the
strongest of all fatty acids.
The acidic character of carboxylic acids is due to
resonance in the acidic group which imparts electron
deficiency (positive charge) on the oxygen atom of
the hydroxyl group (structure II).
O
R C O H
I
Non-equivalent structures
O –
R C
+
O H
II
(Resonance less important)
O –
R C O + H +
The positive charge (electron deficiency) on oxygen
atom causes a displacement of electron pair of the
O—H bond towards the oxygen atom with the result
the hydrogen atom of the O—H group is eliminated
as proton and a carboxylate ion is formed.
Once the carboxylate ion is formed, it is stabilised by
means of resonance.
O
O –
R C
R C
O – O
Resonating forms of carboxylate ion (Equivalent structures)
(Resonance more important)
R
C
O
O
Resonance hybrid of carboxylate ion
Due to equivalent resonating structures, resonance in
carboxylate anion is more important than in the
parent carboxylic acid. Hence carboxylate anion is
more stabilised than the acid itself and hence the
equilibrium of the ionisation of carboxylic acids to
the right hand side.
RCOOH RCOO – + H +
The existence of resonance in carboxylate ion is
supported by bond lengths. For example, in formic
acid, there is one C=O double bond (1.23 Å) and one
C—O single bond (1.36Å), while in sodium formate
both of the carbon-oxygen bond lengths are identical
–
(1.27Å) which is nearly intermediate between C O
and C—O bond length values. This proves resonance
in carboxylate anion.
H
C
O
Formic acid
OH
H
C
O
O
–
Sodium formate
Na +
It is important to note that although carboxylic acids
and alcohols both contain –OH group, the latter are
not acidic in nature. It is due to the absence of
resonance (factor responsible for acidic character of
–COOH) in both the alcohols as well as in their
corresponding ions (alkoxide ions).
R—O—H R—O – + H +
Alcohol
Alkoxide ion
(No resonance)
(No resonance)
Relative acidic character of carboxylic acids with
common species not having —COOH group.
RCOOH > Ar—OH > HOH > ROH >
HC CH > NH 3 > RH
Effect of Substituents on acidity.
The carboxylic acids are acidic in nature because of
stabilisation (i.e., dispersal of negative charge) of
carboxylate ion. So any factor which can enhance the
dispersal of negative charge of the carboxylate ion
will increase the acidity of the carboxylic acid and
vice versa. Thus electron-withdrawing substitutents
(like halogens, —NO 2 , —C 6 H 5 , etc.) would disperse
the negative charge and hence stabilise the
carboxylate ion and thus increase acidity of the
parent acid. On the other hand, electron-releasing
substituents would increase the negative charge,
destabilise the carboxylate ion and thus decrease
acidity of the parent acid.
X
C
The substituent X withdraws electrons, disperses negative
charge, stabilises the ion and hence increases acidity
–
O
Y C
O
The substituent Y releases electrons, intensifies negative
charge, destabilises the ion and hence decreases acidity
O
O
–
XtraEdge for IIT-JEE 34 DECEMBER 2009

Now, since alkyl groups are electron-releasing, their
presence in the molecule will decrease the acidity. In
general, greater the length of the alkyl chain, lower
shall be the acidity of the acid. Thus, formic acid
(HCOOH), having no alkyl group, is about 10 times
stronger than acetic acid (CH 3 COOH) which in turn
is stronger than propanoic acid (CH 3 CH 2 COOH) and
so on. Similarly, following order is observed in
chloro acids.
Cl
Cl
Cl
C
CO 2 H > Cl
Cl
H
pKa 0.70 1.48
H
> Cl
C
C
CO 2 H
CO 2 H > H
H
H
pKa 2.86 4.76
Decreasing order of aliphatic acids
(i) O 2 NCH 2 COOH > FCH 2 COOH > ClCH 2 COOH
> BrCH 2 COOH
(ii) HCOOH > CH 3 COOH > (CH 3 ) 2 CHCOOH
> (CH 3 ) 3 CCOOH
(iii) CH 3 CH 2 CCl 2 COOH > CH 3 CHCl.CHCl.COOH
> ClCH 2 CHClCH 2 COOH
(iv) F 3 CCOOH > Cl 3 CCOOH > Br 3 CCOOH
Benzoic acid is somewhat stronger than simple
aliphatic acids. Here the carboxylate group is
attached to a more electronegative carbon (sp 2
hybridised) than in aliphatic acids (sp 3 hybridised).
HCOOH > C 6 H 5 COOH > CH 3 COOH.
Nucleophilic substitution at acyl carbon :
It is important to note that nucleophilic substitution
(e.g., hydrolysis, reaction with NH 3 , C 2 H 5 OH, etc.) in
acid derivatives (acid chlorides, anhydrides, esters
and amides) takes place at acyl carbon atom
(difference from nucleophilic substitution in alkyl
halides where substitution takes place at alkyl carbon
atom). Nucleophilic substitution in acyl halides is
faster than in alkyl halides. This is due to the
presence of > CO group in acid chlorides which
facilitate the release of halogen as halide ion.
O δ–
R C Cl
δ–
δ+
Acid chloride
R δ+ δ–
Cl
Alkyl chloride
H
C
CO 2 H
Comparison of nucleophilic substitution (e.g.,
hydrolysis) in acid derivatives. Let us first study the
mechanism of such reaction.
R
O
C Z + Nu (i) Addition step R
O
C Nu
Z
(ii) Elimination step
R
O
C Nu + Z
(where Z= —Cl, —OCOR, —OR, —NH 2 and Nu =
A nucleophile)
Nucleophilic substitution in acid derivatives
O
O
OH
R C R'
Nu
R C
H
Nu
R C Nu
R'
R'
(where R' = H or alkyl group)
Nucleophilic addition on aldehydes and ketones
The (i) step is similar to that of nucleophilic addition
in aldehydes and ketones and favoured by the
presence of electron withdrawing group (which
would stabilise the intermediate by developing
negative charge) and hindered by electron-releasing
group. The (ii) step (elimination of the leaving group
Z) depends upon the ability of Z to accommodate
electron pair, i.e., on the basicity of the leaving
group. Weaker bases are good leaving groups,
hence weaker a base, the more easily it is removed.
Among the four leaving groups (Cl – , – OCOR, – OR,
and – NH 2 ) of the four acid derivatives, Cl – being the
weakest base is eliminated most readily. The relative
order of the basic nature of the four groups is
– NH 2 > – OR > – O.COR > Cl –
Hence acid chlorides are most reactive and acid
amides are the least reactive towards nucleophilic
acyl substitution. Thus, the relative reactivity of acid
derivatives (acyl compounds) towards nucleophilic
substitution reactions is
ROCl > RCO.O.COR > RCOOR > RCONH 2
Acid Acid Esters Acid
chlorides anhydrides amides
OH – being stronger base than Cl – , carboxylic acids
(RCOOH) undergo nucleophilic substitution
(esterfication) less readily than acid chlorides.
XtraEdge for IIT-JEE 35 DECEMBER 2009

XtraEdge for IIT-JEE 36 DECEMBER 2009

CAREER POINT’ s
Correspondence & Test Series Courses Information
Courses of IIT-JEE
COURSE
INFORMATION
Eligibiltiy for
Admission
Study Material
Package
IIT-JEE 2010
Class 12th or
12th pass
students
All India Test
Series IIT JEE 2010
(AT CENTER)
Class 12th or
12th pass
students
Postal All India
Test Series
IIT-JEE 2010
Class 12th or
12th pass
students
Major Test Series
IIT JEE 2010
(AT CENTER)
Class 12th or
12th pass
students
Postal Major Test
Series IIT JEE
2010
(By PO ST)
Class 12th or
12th pass
students
CP Ranker's
Package
IIT-JEE 2010
Class 12th or
12th pass
students
Study Material
Package
IIT-JEE 2011
Class 11th
students
All India Foundation
Test Series IIT JEE
2011 (AT CENTER)
Class 11th
students
Postal All India
Test Series
IIT-JEE 2011
Class 11th
students
Medium English OR Hindi English English English English English English OR Hindi English English
Test Center Postal Visit our website Postal
Issue of
Application Kit
Date of Dispatch/
First Test
Course Fee
COURSE
INFORMATION
Eligibiltiy for
Admission
Immediate
Dispatch
Immediate
Dispatch
Rs.8,500/-
including S.Tax
Study Material
Package
AIEEE 2010
Class 12th or 12th
pass students
Immediate
Dispatch
25-Oct-09
Rs. 3,500/-
including S.Tax
Immediate
Dispatch
Immediate
Dispatch
Rs. 1,500/-
including S.Tax
All India
Test Series
AIEEE 2010
(AT CENTER)
Class 12th or 12th
pass students
For Class 12th / 12th Pass Students
Visit our
website
Immediate
Dispatch
17-Jan-10
Rs. 1,550/-
including S.Tax
Postal Postal Postal Visit our website Postal
Immediate
Dispatch
Dispatch
17-Jan-10
Onwards
Rs. 700/-
including S.Tax
Courses of AIEEE
Postal All India Test
Series
AIEEE 2010
Class 12th or 12th
pass students
Major Test Series
AIEEE 2010
(AT CENTER)
Class 12th or 12th
pass students
Immediate
Dispatch
Immediate
Dispatch
Rs. 850/-
including S.Tax
Immediate
Dispatch
Immediate
Dispatch
Rs. 9,500/-
including S.Tax
Postal Major Test
Series
AIEEE 2010
(BY POST)
Class 12th or 12th
pass students
For Class 11th Students
Immediate
Dispatch
25-Oct-09
Rs. 4,500/-
including S.Tax
CP Ranker's Package
IIT-JEE 2010
Class 12th or 12th
pass students
Immediate
Dispatch
Immediate
Dispatch
Rs. 2,500/-
including S.Tax
For Class 11th
Students
Study Material
Package
AIEEE 2011
Class 11th
students
Medium English OR Hindi English/Hindi English/Hindi English/Hindi English/Hindi English English OR Hindi
Test Center Postal Visit our website Postal Visit our website Postal Postal Postal
Issue of
Application Kit
Date of Dispatch/
First Test
Course Fee
COURSE
INFORMATION
Immediate
Dispatch
Immediate
Dispatch
Rs. 7,500/-
including S. Tax
Study Material
Package
Pre-Medical 2010
Immediate
Dispatch
25-Oct-09
Rs. 4,000/-
including S. Tax
All India Test Series
Pre-Medical 2010
(AT CENTER)
For Class 12th / 12th Pass Students
Immediate
Dispatch
Immediate
Dispatch
Rs. 1450/-
including S. Tax
Immediate
Dispatch
17-Jan-10
Rs. 1000/-
including S. Tax
Courses of Pre-Medical
For Class 12th / 12th Pass Students
Postal Test
Series
Pre-Medical 2010
Major Test Series
Pre-Medical 2010
(AT CENTER )
Immediate
Dispatch
Dispatch
17-Jan-10
Onwards
Rs. 550/-
including S. Tax
Postal Major Test
Series Pre-Medical
2010 (BY POST)
Immediate
Dispatch
Immediate
Dispatch
Rs. 850/-
including S.Tax
CP Ranker's Package
IIT-JEE 2010
Immediate
Dispatch
Immediate
Dispatch
Rs. 8,500/-
including S.Tax
For Class 11th
Students
Study Material
Package
Pre-Medical 2011
Eligibiltiy for
Admission
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Class 11th
students
Medium English OR Hindi English/ Hindi English/ Hindi English/ Hindi English English English OR Hindi
Test Center
Postal
Visit our wesite for
list of test centers
Postal
Visit our wesite for
list of test centers
Postal Postal Postal
Issue of
Application Kit
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Date of Dispatch
Commencement /
First Test
Immediate
Dispatch
25-Oct-09
Immediate
Dispatch
17-Jan-10
Dispatch
17-Jan-10
Onwards
Immediate
Dispatch
Immediate
Dispatch
Course Fee
Rs.7,200/-
including S. Tax
Rs. 3,600/-
including S. Tax
Rs. 1400/-
including S. Tax
Rs. 1100/- including
S. Tax
Rs. 600/- including
S. Tax
Rs. 900/-
including S.Tax
Rs. 8,000/-
including S.Tax
◆For details about all Correspondence & Test Series Course Information, please visit our website.: www.careerpointgroup.com
XtraEdge for IIT-JEE 37 DECEMBER 2009

XtraEdge for IIT-JEE 38 DECEMBER 2009

XtraEdge for IIT-JEE 39 DECEMBER 2009

KEY CONCEPT
Physical
Chemistry
Fundamentals
CHEMICAL KINETICS
The temperature dependence of reaction rates :
The rate constants of most reactions increase as the
temperature is raised. Many reactions in solution fall
somewhere in the range spanned by the hydrolysis of
methyl ethanoate (where the rate constant at 35ºC is
1.82 times that at 25ºC) and the hydrolysis of sucrose
(where the factor is 4.13).
(a) The Arrhenius parameters :
It is found experimentally for many reactions that a
plot of ln k against 1/T gives a straight line. This
behaviour is normally expressed mathematically by
introducing two parameters, one representing the
intercept and the other the slope of the straight line,
and writing the Arrhenius equaion.
E
ln k = ln A – a
......(i)
RT
The parameter A, which corresponds to the intercept
of the line at 1/T = 0(at infinite temperature, shown in
figure), is called the pre-exponential factor or the
'frequency factor'. The parameter E a , which is
obtained from the slope of the line (–E a /R), is called
the activation energy. Collectively the two quantities
are called the Arrhenius parameters.
ln k
ln A
Slope = –E a /R
behaviour is a signal that the reaction has a complex
mechanism.
The temperature dependence of some reactions is
non-Arrhenius, in the sense that a straight line is not
obtained when ln k is plotted against 1/T. However,
it is still possible to define an activation energy at any
temperature as
E a = RT 2 ⎛ dln k ⎞
⎜ ⎟ .......(ii)
⎝ dT ⎠
This definition reduces to the earlier one (as the slope
of a straight line) for a temperature-independent
activation energy. However, the definition in eqn.(ii)
is more general than eqn.(i), because it allows E a to
be obtained from the slope (at the temperature of
interest) of a plot of ln k against 1/T even if the
Arrhenius plot is not a straight line. Non-Arrhenius
behaviour is sometimes a sign that quantum
mechanical tunnelling is playing a significant role in
the reaction.
(b) The interpretation of the parameters :
We shall regard the Arrhenius parameters as purely
empirical quantities that enable us to discuss the
variation of rate constants with temperature;
however, it is useful to have an interpretation in mind
and write eqn.(i) as
E / RT
k = Ae − a
.......(iii)
To interpret E a we consider how the molecular
potential energy changes in the course of a chemical
reaction that begins with a collision between
molecules of A and molecules of B(shown in figure).
1/T
A plot of ln k against 1/T is a straight line when
the reaction follows the behaviour described by
the Arrhenius equation. The slope gives –E a /R
and the intercept at 1/T = 0 gives ln A.
The fact that E a is given by the slope of the plot of
ln k against 1/T means that, the higher the activation
energy, the stronger the temperature dependence of
the rate constant (that is, the steeper the slope). A
high activation energy signifies that the rate constant
depends strongly on temperature. If a reaction has
zero activation energy, its rate is independent of
temperature. In some cases the activation energy is
negative, which indicates that the rate decreases as
the temperature is raised. We shall see that such
Potential energy
Reactants
E a
Products
Progress of reaction
A potential energy profile for an exothermic
reaction. The height of the barrier between
the reactants and products is the activation
energy of the reaction
XtraEdge for IIT-JEE 40 DECEMBER 2009

As the reaction event proceeds, A and B come into
contact, distort, and begin to exchange or discard
atoms. The reaction coordinate is the collection of
motions, such as changes in interatomic distances and
bond angles, that are directly involved in the
formation of products from reactants. (The reaction
coordinate is essentially a geometrical concept and
quite distinct from the extent of reaction.) The
potential energy rises to a maximum and the cluster
of atoms that corresponds to the region close to the
maximum is called the activated complex. After the
maximum, the potential energy falls as the atoms
rearrange in the cluster and reaches a value
characteristic of the products. The climax of the
reaction is at the peak of the potential energy, which
corresponds to the activation energy E a . Here two
reactant molecules have come to such a degree of
closeness and distortion that a small further
distortion will send them in the direction of products.
This crucial configuration is called the transition
state of the reaction. Although some molecules
entering the transition state might revert to reactants,
if they pass through this configuration then it is
inevitable that products will emerge from the
encounter.
We also conclude from the preceding discussion that,
for a reaction involving the collision of two
molecules, the activation energy is the minimum
kinetic energy that reactants must have in order
to form products. For example, in a gas-phase
reaction there are numerous collisions each second,
but only a tiny proportion are sufficiently energetic to
lead to reaction. The fraction of collisions with a
kinetic energy in excess of an energy E a is given by
e −
E a / RT
the Boltzmann distribution as . Hence, we
can interpret the exponential factor in eqn(iii) as the
fraction of collision that have enough kinetic energy
to lead to reaction.
The pre-exponential factor is a measure of the rate at
which collisions occur irrespective of their energy.
Hence, the product of A and the exponential factor,
E a / RT
e − , gives the rate of successful collisions.
Kinetic and thermodynamic control of reactions :
In some cases reactants can give rise to a variety of
products, as in nitrations of mono-substituted
benzene, when various proportions of the ortho-,
meta-, and para- substituted products are obtained,
depending on the directing power of the original
substituent. Suppose two products, P 1 and P 2 , are
produced by the following competing reactions :
A + B ⎯→ P 1 Rate of formation of P 1 = k 1 [A][B]
A + B ⎯→ P 2 Rate of formation of P 2 = k 2 [A][B]
The relative proportion in which the two products
have been produced at a given state of the reaction
(before it has reached equilibrium) is given by the
ratio of the two rates, and therefore of the two rate
constants :
[P2]
k 2
=
[P1
] k1
This ratio represents the kinetic control over the
proportions of products, and is a common feature of
the reactions encountered in organic chemistry where
reactants are chosen that facilitate pathways
favouring the formation of a desired product. If a
reaction is allowed to reach equilibrium, then the
proportion of products is determined by
thermodynamic rather than kinetic considerations,
and the ratio of concentration is controlled by
considerations of the standard Gibbs energies of all
the reactants and products.
The kinetic isotope effect
The postulation of a plausible mechanism requires
careful analysis of many experiments designed to
determine the fate of atoms during the formation of
products. Observation of the kinetic isotope effect, a
decrease in the rate of a chemical reaction upon
replacement of one atom in a reactant by a heavier
isotope, facilitates the identification of bond-breaking
events in the rate-determining step. A primary
kinetic isotope effect is observed when the ratedetermining
step requires the scission of a bond
involving the isotope. A secondary isotope effect is
the reduction in reaction rate even though the bond
involving the isotope is not broken to form product.
In both cases, the effect arises from the change in
activation energy that accompanies the replacement
of an atom by a heavier isotope on account of
changes in the zero-point vibrational energies.
First, we consider the origin of the primary kinetic
isotope effect in a reaction in which the ratedetermining
step is the scission of a C–H bond. The
reaction coordinate corresponds to the stretching of
the C–H bond and the potential energy profile is
shown in figure. On deuteration, the dominant
change is the reduction of the zero-point energy of
the bond (because the deuterium atom is heavier).
The whole reaction profile is not lowered, however,
because the relevant vibration in the activated
complex has a very low force constant, so there is
little zero-point energy associated with the reaction
coordinate in either isotopomeric form of the
activated complex.
Potential energy
C–H
C–D
E a (C–H)
E a (C–D)
Reaction coordinate
XtraEdge for IIT-JEE 41 DECEMBER 2009

UNDERSTANDING
Inorganic Chemistry
1. A black coloured compound (A) on reaction with dil.
H 2 SO 4 gives a gas (B) which on passing in a solution
of an acid (C) gives a white turbidity (D). Gas (B)
when passed through an acidified solution of a
compound (E), gives ppt.(F) which is soluble in
dilute nitric acid. After boiling this solution an excess
of NH 4 OH is added, a blue coloured compound (G) is
produced. To this solution, on addition of CH 3 COOH
and aqueous K 4 [Fe(CN) 6 ], a chocolate ppt. (H) is
produced. On addition of an aqueous solution of
BaCl 2 to aqueous solution of (E), a white ppt.
insoluble in HNO 3 is obtained. Identify compounds
(A) to (H).
Sol. From the data on compounds (G) and (H), it may be
inferred that the compound (E) contains cupric ions
(Cu 2+ ), i.e., (E) is a salt of copper. Since the addition
of BaCl 2 to (E) gives a white ppt. insoluble in HNO 3 ,
it may be said that the anion in the salt is sulphate ion
(SO 2– 4 ). Hence, (E) is CuSO 4 .
The gas (B) which is obtained by adding dil. H 2 SO 4
to a black coloured compound (A), may be H 2 S since
it can cause precipitation of Cu 2+ ions in acidic
medium. The black coloured compound (A) may be
ferrous sulphide (iron pyrite).
Hence, the given observation may be explained from
the following equations.
Fe S + H 2 SO 4 ⎯→ FeSO 4 + H 2 S
(A) Dil. (B)
H 2 S + 2HNO 3 ⎯→ 2NO 2 + 2H 2 O + S (D)
(C)
White turbidity
CuSO 4 + H 2 S ⎯→ CuS ↓ + H 2 SO 4
(E) (B) (F)
Black ppt.
3CuS + 8HNO 3 ⎯→
Dil. 3Cu(NO 3 ) 2 + 2NO + 3S + 4H 2 O
Cu ++ + 4NH 3 ⎯→ [Cu(NH 3 ) 4 ] 2+
(G) Blue colour
[Cu(NH 3 ) 4 ] 2+ + 4CH 3 COOH ⎯→
Cu 2+ + 4CH 3 COONH 4
Cu 2+ + [Fe(CN) 6 ] 4– ⎯→ Cu 2 [Fe(CN) 6 ]
(H)
Chocolate colour
CuSO 4 (aq) + BaCl 2 (aq) ⎯→ BaSO 4 ↓ + CuCl 2
(E)
White ppt.
Insuluble in HNO 3
Hence,
(A) is FeS, (B) is H 2 S, (C) is HNO 3 , (D) is S,
(E) is CuSO 4 , (F) is CuS, (G) is [Cu(NH 3 ) 4 ]SO 4 and
(H) is Cu 2 [Fe(CN) 6 ]
2. An unknown inorganic compound (X) gives the
following observations :
(i) When added to CuSO 4 solution it liberates iodine
and a white ppt. (Y) is formed. The liberated I 2
reacts with Na 2 S 2 O 3 solution to give NaI and a
colourless compound (Z).
(ii) When CHCl 3 and Cl 2 water is added to aqueous
solution of (X), a violet layer of chloroform is
formed.
(iii)(X) gives a violet colour flame when heated in
Bunsen burner flame.
(iv)When aqueous solution of (X) is added to
aqueous lead nitrate, a yellow ppt. (M) is formed.
(v) Addition of (X) to HgCl 2 gives a red ppt. which
dissolves in excess of (X) to give Nessler's
reagent.
(vi) On heating (X) with dil. HCl and KNO 2 , violet
vapours of a compound (N) are formed which
condenses on the wall of test tube.
What are (X), (Y), (Z), (M) and (N) ? Explain the
reactions.
Sol. Observation (iii) indicates that the compound (X)
contains K + , it is because it gives a violet coloured
flame. On the other hand set (ii) confirmed that (X)
contains I – ions, thus (X) is KI. Now the different
reactions may be formulated as follows.
CHCl
2KI + Cl 2 ⎯⎯⎯
3 → I 2 + 2KCl
Violet layer
(i) When CuSO 4 reacts with KI, I 2 is liberated and a
ppt. of cuprous iodide (Y) is formed.
[CuSO 4 + 2KI ⎯→ K 2 SO 4 + CuI 2 ] × 2
2CuI 2 ⎯→ Cu 2 I 2 + I 2
Unstable
On adding,
2CuSO 4 + 4KI ⎯→ 2K 2 SO 4 + Cu 2 I 2 + I 2
(Y) white ppt.
The librated iodine is titrated against standard hypo
solution when a colourless sodium tetrathionate (Z) is
formed.
2Na 2 S 2 O 3 + I 2 ⎯→ Na 2 S 4 O 6 + 2NaI
(Z) (colourless)
(iv) Pb(NO 3 ) 2 + 2KI ⎯→ PbI 2 + 2KNO 3
(M) yellow ppt
(v) HgCl 2 + 2KI ⎯→ HgI 2 ↓ + 2KCl
Red ppt.
HgI 2 + 2KI ⎯→ K 2 HgI 4
(Soluble)
K 2 HgI 4 + NaOH ⎯→ Nessler's reagent
(vi) HCl + KNO 2 ⎯→ HNO 2 + KCl
HCl + KI ⎯→ KCI + HI
XtraEdge for IIT-JEE 42 DECEMBER 2009

2HNO 2 + 2HI ⎯→ I 2 + 2H 2 O + 2NO
(N)
Hence,
(X) is KI,
(Y) is Cu 2 I 2 ,
(Z) is Na 2 S 4 O 6 ,
(M) is PbI 2 and (N) is I 2 .
3. An inorganic halide (A) gives the following
reactions.
(i) The cation of (A) on raction with H 2 S in HCl
medium, gives a black ppt. of (B). (A) neither
gives ppt. with HCl nor blue colour with
K 4 Fe(CN) 6 .
(ii) (B) on heating with dil.HCl gives back
compound(A) and a gas (C) which gives a black
ppt. with lead acetate solution.
(iii) The anion of (A) gives chromyl chloride test.
(iv) (B) dissolves in hot dil. HNO 3 to give a solution,
(D). (D) gives ring test.
(v) When NH 4 OH solution is added to (D), a white
precipitate (E) is formed. (E) dissolves in
minimum amount of dil. HCl to give a solution of
(A). Aqueous solution of (A) on addition of water
gives a whitish turbidity (F).
(vi) Aqueous solution of (A) on warming with
alkaline sodium stannite gives a black precipitate
of a metal (G) and sodium stannate. The metal
(G) dissolves in hydrochloride acid to give
solution of (A).
Identify (A) to (G) and give balanced chemical
equations of reactions.
Sol. Observation of (i) indicates that cation (A) is Bi 3+
because it does not give ppt. with HCl nor blue
colour with K 4 Fe(CN) 6 , hence it is neither Pb 2+ nor
Cu 2+ . Since anion of (A) gives chromyl chloride test,
hence it contains Cl – ions. Thus, (A) is BiCl 3 . Its
different reactions are given below :
(i) 2BiCl 3 + 3H 2 S ⎯→ Bi 2 S 3 + 6HCl
(A)
(B)
(ii) Bi 2 S 3 + 6HCl ⎯→ 3H 2 S + 2 BiCl 3
(B) (C) (A)
(iii) Bi 2 S 3 + 8HNO 3 ⎯⎯→
∆ 2Bi(NO 3 ) 3 + 2NO
(B) (D) + 3S + 4H 2 O
(iv) Bi(N O 3 ) 3 + 3NH 4 OH ⎯→
(D) Bi(OH) 3 ↓ + 3NH 4 NO 3
(E) White ppt.
Bi(OH) 3 + 3HCl ⎯⎯→
∆ BiCl 3 + 3H 2 O
Dil. (A)
BiCl 3 + H 2 O ⎯→ BiOCl + 2HCl
(A)
(F)
Bismuth oxychloride
(White turbidity)
(v) BiCl 3 + 2NaOH +Na 2 [SnO 2 ] ⎯→
(A)
Bi + NaSnO 3 + H 2 O + 3NaCl
(G) Black ppt.
2Bi + 6HCl ⎯⎯→
∆ 2BiCl 3 + 3H 2
(G)
(A)
Hence,
(A) is BiCl 3 ,
(B) is Bi 2 S 3 ,
(C) is H 2 S,
(D) is Bi(NO 3 ) 2 ,
(E) is Bi(OH) 3 , (F) is BiOCl and (G) is Bi
4. (i) An inorganic compound (A) is formed on passing
a gas (B) through a concentrated liquor containing
Na 2 S and Na 2 SO 3 .
(ii) On adding (A) into a dilute solution of AgNO 3 , a
white ppt. appears which quickly changes into
black coloured compound (C).
(iii) On adding two or three drops of FeCl 3 into
excess of solution of (A), a violet coloured
compound (D) is formed. This colour disappears
quickly.
(iv) On adding a solution of (A) into the solution of
CuCl 2 , a white ppt. is first formed which dissolves
on adding excess of (A) forming a compound (E).
Identify (A) to (E) and give chemical equations for
the reactions at steps (i) to (iv)
Sol. (i) The compound (A) appears to be Na 2 S 2 O 3 from its
method of preparation given in the problem.
Na 2 S + Na 2 SO 3 + I 2 ⎯→ 2NaI + Na 2 S 2 O 3
(B)
(A)
or Na 2 SO 3 + 3Na 2 S + 3SO 2 ⎯→ 3Na 2 S 2 O 3
(B)
(A)
(ii) White ppt. of Ag 2 S 2 O 3 is formed which is
hydrolysed to black Ag 2 S
Na 2 S 2 O 3 + 2AgNO 3 ⎯→ 2NaNO 3 + Ag 2 S 2 O 3 ↓
White ppt
Ag 2 S 2 O 3 + H 2 O ⎯→ Ag 2 S + H 2 SO 4
(C)
(iii) A violet ferric salt is formed.
3Na 2 S 2 O 3 + 2FeCl 3 ⎯→ Fe 2 (S 2 O 3 ) 3 + 6NaCl
(D)(violet)
(iv) 2CuCl 2 + 2Na 2 S 2 O 3 → 2CuCl + Na 2 S 4 O 6 + 2NaCl
White ppt.
2CuCl + Na 2 S 2 O 3 ⎯→ Cu 2 S 2 O 3 + 2NaCl
3Cu 2 S 2 O 3 + 2Na 2 S 2 O 3 ⎯→ Na 4 [Cu 6 (S 2 O 3 ) 5 ]
(E)
or 6CuCl + 5Na 2 S 2 O 3 ⎯→ Na 4 [Cu 6 (S 2 O 3 ) 5 ] + 6NaCl
(E)
Hence,
(A) is Na 2 S 2 O 3 ,
(B) is I 2 or SO 2 ,
(C) is Ag 2 S,
(D) is Fe 2 (S 2 O 3 ) 3 and
(E) is Na 4 [Cu 6 (S 2 O 3 ) 5 ]
5. A colourless solid (A) on heating gives a white solid
(B) and a colourless gas (C). (B) gives off reddishbrown
fumes on treating with H 2 SO 4 . On treating
with NH 4 Cl, (B) gives a colourless gas (D) and a
XtraEdge for IIT-JEE 43 DECEMBER 2009

esidue (E). The compound (A) on heating with
(NH 4 ) 2 SO 4 gives a colourless gas (F) and white
residue (G). Both (E) and (G) impart bright yellow
colour to Bunsen flame. The gas (C) forms white
powder with strongly heated Mg metal which on
hydrolysis produces Mg(OH) 2 . The gas (D) on
heating with Ca gives a compound which on
hydrolysis produces NH 3 . Identify compounds (A) to
(G) giving chemical equations involved.
Sol. The given information is as follows :
(i) A ⎯ Heat ⎯⎯ → B + C
Colourless Solid Colourless
Solid
gas
(ii) B + H 2 SO 4 ⎯⎯→
∆ Reddish brown gas
(iii) B + NH 4 Cl ⎯⎯→
∆ D + E
Colourless gas
(iv) A + (NH 4 ) 2 SO 4 ⎯⎯→
∆ F + G
olourless gas White
Residue
(v) E and G imparts yellow colour to the flame.
(vi) C + Mg ⎯ Heat ⎯⎯ →White powder
⎯ H 2
⎯⎯ O →Mg(OH) 2
(vii) D + Ca ⎯ Heat ⎯⎯ →Compound ⎯ H 2
⎯⎯ O →NH 3
Information of (v) indicates that (E) and (G) and also
(A) are the salts of sodium because Na + ions give
yellow coloured flame. Observations of (ii) indicate
that the anion associated with Na + in (A) may be
NO – 3 . Thus, the compound (A) is NaNO 3 .
The reactions involved are as follows :
(i) 2NaNO 3 ⎯⎯→
∆ 2NaNO 2 + O 2 ↑
(A) (B) (C)
(ii) 2NaNO 2 + H 2 SO 4 ⎯→ Na 2 SO 4 + 2HNO 2
(B) Dil.
3HNO 2 ⎯→ HNO 3 + H 2 O + 2NO↑
2NO + O 2 ⎯→ 2NO 2 ↑
Reddish brown
Fumes
(iii) NaNO 2 + NH 4 Cl ⎯→ NaCl + N 2 ↑ + 2H 2 O
(B) (E) (D)
(iv) 2NaNO 3 + (NH 4 ) 2 SO 4 ⎯⎯→
∆ Na 2 SO 4 + 2NH 3
(A) (G) (F)
2HNO 3
(v) O 2 + 2Mg ⎯⎯→
∆ 2MgO ⎯ H 2
⎯⎯ O →Mg(OH) 2
(C)
(vi) N 2 + 3Ca ⎯⎯→
∆ Ca 3 N 2
(D)
Ca 3 N 2 + 6H 2 O ⎯→ 3Ca(OH) 2 + 2NH 3 ↑
Hence,
(A) is NaNO 3 ,
(B) is NaNO 2 ,
(C) is O 2 ,
(D) is N 2 ,
(E) is NaCl,
(F) is NH 3 and (G) is Na 2 SO 4 .
TRUE OR FALSE
1. The magnitude of charge on one gram of
electrons is 1.60 × 10 –19 coulomb.
2. Chromyl chloride test of Cl – radical is not given
by HgCl 2 .
3. The energy levels in a hydrogen atom can be
compared with the steps of a ladder placed at
equal distance.
4. In S N 1 mechanism, the leaving group in the
molecule, leaves the molecule, well before
joining of an attacking group.
5. Metamerism is special type of isomerism where
isomers exist simultaneously in dynamic
equilibrium.
6. Addition of HCN with formaldehyde is an
example of electrophilic addition reaction.
7. Ligroin is essentially petroleum ether containing
aliphatic hydrocarbons and is generally used in
dry cleaning clothes.
Sol.
1. [False] Thomson through his experiment
determined the charge to mass ratio of an
electron and the value of e/m is equal to 1.76 ×
10 8 coulomb/gm. Hence one gm of electrons
have charge 1.76 × 10 8 C.
2. [True]
3. [False]
4. [True] S N 1 reaction mechanism takes place in
two steps as :
R—X
R + + OH –
⎯ Slow ⎯→ ⎯ R + + X –
⎯ Fast ⎯→
ROH
5. [False] In metamerism isomers differ in structure
due to difference in distribution of carbon atoms
about the functional group.
For example :
CH 3 CH 2 –O–CH 2 CH 3 and CH 3 –OCH 2 CH 2 CH 3
Conditions mentioned in the statement are
associated with phenomenon of trautomerism.
6. [False]
H
H – C = O + H + CN –
7. [True]
H
H – C = OH
CN
XtraEdge for IIT-JEE 44 DECEMBER 2009

`tà{xÅtà|vtÄ V{tÄÄxÇzxá
Set
8
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. Show that the six planes through the middle point of
each edge of a tetrahedron perpendicular to the
opposite edge meet in a point.
2. Prove that if the graph of the function
y = f (x), defined throughout the number scale, is
symmetrical about two lines x = a and x = b, (a < b),
then this function is a periodic one.
3. Show that an equilateral triangle is a triangle of
maximum area for a given perimeter and a triangle of
minimum perimeter for a given area.
4. Let az 2 + bz + c be a polynomial with complex
coefficients such that a and b are non zero. Prove
that the zeros of this polynomial lie in the region
| z | ≤ a
b + b
c
5. An isosceles triangle with its base parallel to the
x y
major axis of the ellipse +
2 2 = 1 is
a b
circumscribed with all the three sides touching the
ellipse. Find the least possible area of the triangle.
6. If one of the straight lines given by the equation ax 2 +
2hxy + by 2 = 0 coincides with one of those given by
a′x 2 + 2h′xy + b′y 2 = 0 and the other lines represented
ha´b´ h´ab´
by them be perpendicular, show that =
b´ −a´
b − a
7. Prove that
⎛n
⎞ ⎛m ⎞ ⎛n
⎜
⎟
⎝0
⎜
⎟ +
⎠ ⎝n
⎟ ⎞ ⎛ m + 1⎞
⎛n
⎜
⎠ ⎝1
⎜
⎟ +
⎠ ⎝ n ⎟ ⎞ ⎛ m + 2⎞
⎜
⎠ ⎝ 2
⎜
⎟ + ....... to
⎠ ⎝ n ⎠
(n + 1) terms
⎛n
= ⎟ ⎞ ⎛m ⎞ ⎛n
⎜
⎝0
⎜
⎟ +
⎠ ⎝ 0 ⎟ ⎞ ⎛m⎞
⎛n
⎜
⎠ ⎝1
⎜
⎟ 2 +
⎠ ⎝ 1 ⎟ ⎞ ⎛m⎞
⎜
⎠ ⎝ 2
⎜
⎟ 2 2 + ..... to (n +
⎠ ⎝ 2 ⎠
1) terms
1
8. If n ≥ 2 and I n =
∫
( 1−
x
2 ) n cos mx dx, then show that
−1
m 2 I n = 2n(2n – 1) I n–1 – 4n(n – 1) I n–2 .
2
2
9. Find the sum to infinite terms of the series
3 5 7 9 11 + + + + + ........ ∞
4 36 144 400 900
10. ABC is a triangle inscribed in a circle. Two of its
sides are parallel to two given straight lines. Show
that the locus of foot of the perpendicular from the
centre of the circle on to the third side is also a circle,
concentric to the given circle.
Dimensional Formulae of Some
Physical Quantities
Physical Quantity
Dime nsional
Formulae
Work (W) [ML 2 T –2 ]
Stress [ML –1 T –2 ]
Torque (τ) [ML 2 T –2 ]
Moment of Inertia (I) [ML 2 ]
Coefficient of viscosity (η) [ML –1 T –1 ]
Gravitational constant (G) [M –1 L 3 T –2 ]
Specific heat (S) [L 2 T –2 θ –1 ]
Coeficient of thermal conductivity (K) [MLT –3 θ –1 ]
Universal gas constant (R) [ML 2 T –2 θ –1 ]
Potential (V) [ML 2 T –3 A –1 ]
Intensity of electric field (E) [MLT –3 A –1 ]
Permittivity of free space (ε 0 ) [M –1 L –3 T 4 A 2 ]
Specific resistance (ρ) [ML 3 T –3 A 2 ]
Magnetic Induction (B) [MT –2 A –1 ]
Planck's constant (h) [ML 2 T –1 ]
Boltzmann's constant (k) [ML 2 T –2 θ –1 ]
Entropy (S) [ML 2 T –2 θ –1 ]
Decay constant (λ) [T –1 ]
Bohr magnetic (µ B ) [L 2 A]
Thermmionic current density (J) [AL –2 ]
XtraEdge for IIT-JEE 45 DECEMBER 2009

MATHEMATICAL CHALLENGES
SOLUTION FOR NOVEMBER ISSUE (SET # 7)
1. Let the line be y = 2x + c
⎛ 9 − c 9 + 2c ⎞
Point A ⎜ , ⎟
⎝ 6 3 ⎠
2.
⎛ 2c − 3 + c − 6 ⎞
Point B ⎜ , ⎟
⎝ − 3 − 3 ⎠
⎛ c + 6 5c + 12 ⎞
Point C ⎜ , ⎟
⎝ 3 3 ⎠
1 ⎡ 2c − 3 c + 6 ⎤
mid point of B & C is . 2
⎢ + ⎥ ,
⎣ − 3 3 ⎦
1 ⎡ − c + 6 5c + 12⎤
⎡ 9 − c 2c + 9⎤
⎢ +
2
⎥ =
⎣ + 3 3
⎢ , ⎥
⎦ ⎣ 6 3 ⎦
which is point A, so AB and AC are equal.
a
B
A
D
b
C
1 AB 1 AC
a + b = . + . AB AB AC AC
=
1
2
AB
AB +
1
AC
2
AC
1
1
= (AD + DB)
+ (AD + DC)
2
2
AB
AC
⎛ 1 1 ⎞ DB DC
= ⎜ + ⎟ AD +
2 2
+
⎝ AB AC ⎠ BD.DC CD. CB
⎛ 1 1 ⎞ ⎛ DB DC ⎞
= ⎜ + ⎟ AD + ⎜ ⎟
2 2
⎝ AB AC ⎠ + ⎝ BD CD ⎠
⎛ 1 1 ⎞
= AD . ⎜ + ⎟
⎝ BD.DC CD.CB ⎠
AD ⎛
= . ⎜ CD ⎝
1
BD
+ CD
1
⎟⎠
⎞
1
BC
AD DC + BD
= .
BC BD.CD
=
AD AD =
BD.CD
2
AD
so it is vector along
| a + b
1
| = AD
AD 1 = . AD AD
1
A B with magnitude . AD
3. The line PQ always passes through (α, β) so it is
y –β = m(x – α)
Let the circle be x 2 + y 2 – 2hx – 2ky = 0
Joint equation of OP and OQ.
x 2 + y 2 (y − mx)
– 2 (hx + ky) = 0
β − mα
O
Q
P
(h,k)
⎛ 2k ⎞
⎜1−
⎟ y 2 ⎛ h − mk ⎞ ⎛ 2hn ⎞
– 2 ⎜ ⎟ xy + ⎜1+
⎟ x 2 = 0
⎝ β − mα
⎠ ⎝ β − mα
⎠ ⎝ β − mα
⎠
It must represent y 2 – x 2 = 0
h − mk
so = 0 ⇒ m = h/k ...(1)
β − mα
2k 2hm
and 1 – = –1 –
β − mα
β − mα
⇒ β – mα – 2k = –β + mα – 2hm
⇒ –β + mα + k – hm = 0
⇒ –β + k + h/k(α – h) = 0 (using (1) in it)
⇒ k 2 – βy + αh – h 2 = 0 so required locus is
x 2 – y 2 – αx + βy = 0
XtraEdge for IIT-JEE 46 DECEMBER 2009

⎛ π π ⎞
4. As |f(x)| ≤ |tan x| for ∀ x ∈ ⎜−
, ⎟
⎝ 2 2 ⎠
(–1,log 2 3 )
so f(0) = 0
so |f(x) – f(0)| ≤ |tan x|
divides both sides by |x|
f (x) − f (0) tan x
⇒
≤
x x
(–1,0)
(0,–1)
(1,0)
(3/2,–1)
⇒ lim f (x) − f (0)
≤
x→0
x
⇒ |f´(0)| ≤ 1
⇒
lim x→0
tan x
x
1 1 1
a 1 + a 2 + a 3 + ..... + a n ≤ 1
2 3 n
n
a
⇒ ∑
i=
1
i
i
≤ 1
5. Let the number is xyz, here x < y and z < y.
Let y = n, then x can be filled in (n – 1) ways.
(i.e. from 1 to (n – 1)) and z can be filled in n ways
(i.e. from 0 to (n – 1))
here 2 ≤ n ≤ 9
so total no. of 3 digit numbers with largest middle
digit
n
9
= ∑ n (n −1)
=
=
∑ n –
n = 2
∑ n
n = 2
=
2
9.10.19
6
–
9
9.10
2
= 285 – 45 = 240
required probability =
2
9
240
9×
10×
10
= 30
8
= 15
4
6. The region bounded by the curve y = log 2 (2 – x) and
the inequality (x – |x|) 2 + (y – |y|) 2 ≤ 4 is required area
is
7.
1
=
∫
log 2 (2 − x) dx +
∫
−
y 1
( 2 2 ) dy + π 4
−1
0
−1
2
= log 2 3 − + 2log2
3 + 2 –
l n2
e 2 e
= – log 2
27
B
+ 2 + 4
π sq units
F
A
D
E
M
∠BMC = 2∠BAC = 2∠BMD
BD
so tan A = = MD
so
a
2
2
1
r
2
= tan 2 A
2
2
1 π
+
2ln2
4
C
BC a = =
2BC
MD 4r
1 4r
1
a b c
so + +
2 2 2
r1
r2
r3
= 16 (tan 2 A + tan 2 B + tan 2 C) ...(1)
Now as tan A + tan B + tan C ≥
3 (tan A . tan B . tan C) 1/3
and for a triangle tan A + tan B + tan C
= tan A . tan B . tan C
so (tan A . tan B . tan C) 2/3 ≥ 3
⇒ tan A . tan B . tan C ≥ 3 3
⇒ tan 2 A + tan 2 B + tan 2 C
≥ 3(tan A. tan B tan C) 2/3 ≥ 3.3
so from (1),
a
2
2
1
r
+
b
2
2
2
r
+
c
2
2
3
r
≥ 144.
XtraEdge for IIT-JEE 47 DECEMBER 2009

8. Z 1 , Z 2 , Z 3 are centroids of equilateral triangles ACX,
ABY and BCZ respectively.
Z 1 – Z A = (Z C – Z A )
y
Z 1 – Z A = (Z C – Z A )
similarly,
Z 2 – Z A = (Z B – Z A )
Z
Z
Z A
1
C
− Z
− Z
A
A
A Z 1
Z 2
e iπ/6 x
B C
Z B Z C
Z 3
1 ⎛
⎜
3
⎝
1 ⎛
⎜
3
⎝
So, Z 1 – Z 2 = 2
1 (ZC – Z B ) +
z
3 i ⎞
+ ⎟
2 2 ⎠
3 i ⎞
− ⎟
2 2 ⎠
similarly Z 2 – Z 3 = 2
1 (ZA – Z C )
...(1)
...(2)
i
(Z C + Z B – 2Z A )
2 3
...(3)
i
+ (Z A + Z C – 2Z B ) ..(4)
2 3
To prove ∆xyz as equilateral triangle, we prove that
(Z 3 – Z 2 )e iπ/3 = Z 1 – Z 2
So, (Z 3 – Z 2 )e iπ/3 = ( 2
1 (ZC – Z A )
i
⎛
– (Z A + Z C – 2Z B )) ⎜
1
2 3
⎝ 2
= 2
1 (ZC – Z B ) +
= Z 1 – Z 2
a
1−
r cosu
9. T r = 2
∫ 1−
2r cosu + r
0
a
3
+ i
2
i
(Z C + Z B – 2Z A )
2 3
2
⎞
⎟
⎠
du. ...(1)
1−
2r cosu + r − r + 1
=
∫
du
2
1−
2r cosu + r
0
a
⎛
2
1
=
∫ ⎟ ⎞
⎜
− r
1+
du
2
⎝ 1−
2r cosu + r ⎠
0
2
2
= a + (1 – r 2 )
a
∫
0
(1 + r
2
)(1 + tan
a
2
sec
2
u / 2
u / 2) − 2r(1 − tan
= a + (1 – r 2 sec u / 2
)
∫ 2 2
(1 + r) tan u / 2 + (1 − r)
= a +
1−
r
2
(1 + r)
Let tan u/2 = t
so, T r = a +
Now
and
= a +
2
1−
r
0
2
(1 + r)
2(1 − r
2
(1 + r)
2
)
a
∫
0
2
lim T
+ r = a –
r→1
lim T
+ r = a +
r→1
tan
2
2
2
sec u / 2 du
tana/ 2
∫
0
1+
r
1−
r
(1 − r)
u / 2 +
(1 + r)
t
2
⎡
⎢tan
⎣
2dt
1−
r
+
1+
r
−1
2(1 + r)(r −1)
(1 + r)(r −1)
2(1 − r)(r + 1)
(1 + r)(r −1)
and (from (1)) T 1 =
∫ a 0
du = a
Hence
difference π.
lim T
+ r , T 1 ,
r→1
2
2
2
⎛ 1+
r ⎞⎤
⎜t
⎥
1 r
⎟
⎝ − ⎠
⎦
2
u / 2)
2
tana/ 2
0
π = a – π
2
π = a + π
2
lim T
− r form an A.P. with common
r→1
10. Let α, β, γ be the three real roots of the equation
without loss of generality, it can be assumed that
α ≤ β ≤ γ.
so
x 2 + ax 2 + bx + c = (x – γ) (x 2 + (a + γ) x + (γ 2 + aγ + b))
where – γ (γ 2 + aγ + b) = c, as γ is the root of given
equation, so x 2 + (a + γ) x + (γ 2 + aγ + b) = 0 must
have two roots i.e. α and β. So its discriminant is non
negative, thus
(γ + a) 2 – 4(γ 2 + aγ + b) ≥ 0
3γ 2 + 2aγ – a 2 + 4b ≤ 0
2 −
− a + 2 a 3b
so γ ≤
3
so greatest root is also less than or equal to
− a + 2 a
2 − 3b
.
3
XtraEdge for IIT-JEE 48 DECEMBER 2009

MATHS
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. Suppose f(x) = x 3 + ax 2 + bx + c, where a, b, c are
chosen respectively by throwing a die three times.
Find the probability that f(x) is an increasing
function.
Sol. f´(x) = 3x 2 + 2ax + b
y = f(x) is strictly increasing
⇒ f´(x) > 0 ∀ x
⇒ (2a) 2 – 4.3.b < 0
This is true for exactly 15 ordered pairs (a, b); 1 ≤ a,
15 5
b ≤ 6, so probability = = 36 12
3. Let g be a real valued function satisfying g(x) + g(x +
4) = g(x + 2) + g(x + 6), then prove that
∫ x +8
x
is a constant function.
g(t)
dt
Sol. given that g(x) + g(x + 4) = g(x + 2) + g(x + 6) ...(1)
putting x = x + 2 in (1) ........
g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8) ...(2)
from (1) & (2)
g(x) = g(x + B)
Now, f(x) =
∫ x +8
x
g(t)
dt
2. 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 = 3î + 2 ĵ + kˆ
f´(x) = g(x + 8) – g(x) = 0
⇒ g is constant function
4. If exactly three distinct chords from (h, 0) point to the
circle x 2 + y 2 = a 2 are bisected by the parabola
y 2 = 4ax, a > 0, then find the range of 'h' parameter.
Sol. Let M(at 2 , 2at) is mid-point of chord AB, then chord
⇒
→ →
(A.B)
2
≤ | → A | 2 | → B| 2
AB = T = S 1
3a + 2b + c ≤
(7) 2 ≤ (a 2 + b 2 + c 2 ) (14)
2
2
a + b + c 14
2
B
M
A
{Q 3a + 2b + c = 7, point lies on the plane}
a 2 + b 2 + c 2 49 7
≥ = 14 2
AB : x.at 2 + y.2at = a 2 t 4 + 4a 2 t 2
since AB chord passes through (h, 0)
XtraEdge for IIT-JEE 49 DECEMBER 2009

so, h.at 2 = a 2 t 4 + 4a 2 t 2
Sol. Let P be (x 1 , y 1 ),
at 2 [at 2 + (4a – h)] = 0
If a > 0 ⇒ 4a – h < 0
Q
A
P
⇒ h > 4a
...(i)
Now point (at 2 ,2at) must lie inside the circle, on
solving
a 2 t 4 + 4a 2 t 2 – a 2 < 0
we get, h < a ( 5 + 2) ...(ii)
from (i) & (ii)
4a < h < a ( 5 + 2)
5. Find the sum of the terms of G.P. a + ar + ar 2 + ..... + ∞
where a is the value of x for which the function
7 + 2x log e 25 – 5 x – 1 – 5 2– x has the greatest value and
x t dt
r is the Lt
→0∫ 0
2
x tan( π + x
a
Sol. S =
1− r
,
x )
2
To get the greatest value f´(x) = 2log e 25 – 5 x – 1 log 5
+ 5 2– x log 5
f´(x) = 4 log e 5 – 5 x– 1 log e 5 + 5.5 1 – x log e 5
⇒ f´(x) = 0 put 5 x – 1 t(> 0)
t 2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5 x –1 = 5 ⇒ x = 2
to evaluate r :
x t dt 1
r = Lt
→0∫ x 0
2
x tan( π + x )
= π
2
1 2π
since a = 2, r = ⇒ sum of G.P. = π π −1
6. The tangent and normal at a point P on the ellipse
2
x y
= 1 meets the y-axis at A and B
2
a b
respectively. Find the angle subtended by AB at the
points of intersection of the circle (through A,S,B)
and the ellipse. S being one of the foci.
+ 2
2
N
B
Points of intersection of tangent and normal at P
⎛
2
b ⎞ ⎛
2
a y ⎞
points with y-axis and A ⎜ ⎟
0 ,
, B⎜
1 ⎟
⎝ y
0 , y1
−
2
,
1 ⎠ ⎝ b ⎠
S : (ae, 0)
slope (SA). slope (SB) = –1
⇒ ∠ASB = 90º (PA and PB are tangent and normal)
P must lie on the circle with AB as diameter.
Hence the point of intersection of the ellipse and the
circle is P. Due to symmetry the angles made by AB
at P,Q,M, N are all 90º.
M
Do you know
• 100 years ago: The first virus was found in both
plants and animals.
• 90 years ago: The Grand Canyon became a
national monument & Cellophane is invented.
• 80 years ago: The food mixer and the domestic
refrigerator were invented.
• 70 years ago: The teletype and PVC (polyvinylchloride)
were invented.
• 60 years ago: Otto Hahn discovered nuclear
fission by splitting uranium, Teflon was invented.
• 50 years ago: Velcro was invented.
• 40 years ago: An all-female population of lizards
was discovered in Armenia.
• 30 years ago: The computer mouse was
invented.
• 20 years ago: First test-tube baby born in
England, Pluto’s moon, Charon, discovered.
• 10 years ago: First patent for a geneticallyengineered
mouse was issued to Harvard
Medical School.
S
XtraEdge for IIT-JEE 50 DECEMBER 2009

MATH
MONOTONICITY,
MAXIMA & MINIMA
Mathematics Fundamentals
Monotonic Functions :
A function f(x) defined in a domain D is said to be
(i) Monotonic increasing :
⎧x1
< x 2 ⇒ f (x1)
≤ f (x 2 )
⇔ ⎨
∀ x 1 , x 2 ∈ D
⎩x1
> x 2 ⇒ f (x1)
≥ f (x 2)
y
x
x
O
O
⎧x1
< x 2 ⇒ f (x1)
> / f (x 2)
i.e., ⇔ ⎨
∀ x 1 , x 2 ∈ D
⎩x1
> x 2 ⇒ f (x1)
< / f (x 2)
(ii) Monotonic decreasing :
⎧x1
< x 2 ⇒ f (x1
) ≥ f(x 2 )
⇔ ⎨
∀ x 1 , x 2 ∈ D
⎩x1
> x 2 ⇒ f (x 1)
≤ f(x 2 )
y
x
x
O
O
⎧x1
< x 2 ⇒ f (x1)
< / f (x 2)
i.e., ⇔ ⎨
∀ x 1 , x 2 ∈ D
⎩x1
> x 2 ⇒ f (x1)
/ > f(x 2)
A function is said to be monotonic function in a
domain if it is either monotonic increasing or
monotonic decreasing in that domain.
Note : If x 1 < x 2 ⇒ f(x 1 ) < f(x 2 ) ∀ x 1 , x 2 ∈ D, then
f(x) is called strictly increasing in domain D and
similarly decreasing in D.
Method of testing monotonicity :
(i) At a point : A function f(x) is said to be
monotonic increasing (decreasing) at a point x = a of
its domain if it is monotonic increasing (decreasing)
in the interval (a – h, a + h) where h is a small
positive number. Hence we may observer that if f(x)
is monotonic increasing at x = a then at this point
tangent to its graph will make an acute angle with x-
axis where as if the function is monotonic decreasing
there then tangent will make an obtuse angle with x-
axis. Consequently f´(a) will be positive or negative
y
y
according as f(x) is monotonic increasing or
decreasing at x = a.
So at x = a, function f(x) is
monotonic increasing ⇔ f´(a) > 0
monotonic decreasing ⇔ f´(a) < 0
(ii) In an interval : In [a, b], f(x) is
monotonic increasing ⇔ f´(x) ≥ 0⎫
⎪
monotonic decreasing ⇔ f´(x) ≤ 0⎬
∀ x ∈ (a, b)
constant ⇔ f´(x) = 0⎪
⎭
Note :
(i) In above results f´(x) should not be zero for all
values of x, otherwise f(x) will be a constant
function.
(ii) If in [a, b], f´(x) < 0 at least for one value of x and
f´(x) > 0 for at least one value of x, then f(x) will
not be monotonic in [a, b].
Examples of monotonic function :
If a functions is monotonic increasing (decreasing ) at
every point of its domain, then it is said to be
monotonic increasing (decreasing) function.
In the following table we have example of some
monotonic/not monotonic functions
Monotonic
increasing
Monotonic
decreasing
Not
monotonic
x 3 1/x, x > 0 x 2
x|x| 1 – 2x |x|
e x e –x e x + e –x
log x log 2 x sin x
sin h x cosec h x, x > 0 cos h x
[x] cot hx, x > 0 sec h x
Properties of monotonic functions :
If f(x) is strictly increasing in some interval, then in
that interval, f –1 exists and that is also strictly
increasing function.
If f(x) is continuous in [a, b] and differentiable in
(a, b), then
f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic increasing
in [a, b]
f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic decreasing
in [a, b]
XtraEdge for IIT-JEE 51 DECEMBER 2009

If both f(x) and g(x) are increasing (or decreasing) in
[a, b] and gof is defined in [a, b], then gof is
increasing.
If f(x) and g(x) are two monotonic functions in [a, b]
such that one is increasing and other is decreasing
then gof, it is defined, is decreasing function.
Maximum and Minimum Points :
The value of a function f(x) is said to be maximum at
x = a if there exists a small positive number δ such
that f(a) > f(x)
y
( ) ( ) ( ) x
O a b c
Also then the point x = a is called a maximum point
for the function f(x).
Similarly the value of f(x) is said to be minimum at x
= b if there exists a small positive number δ such that
f(b) < f(x) ∀ x ∈ (b – δ, b + δ)
Also then the point x = b is called a minimum point
for f(x)
Hence we find that :
(i) x = a is a maximum point of f(x)
⇔ ⎨ ⎧ f (a) – f(a + h) > 0
⎩ f(a) – f(a – h) > 0
(ii) x = b is a minimum point of f(x)
⇔ ⎨ ⎧ f (b) – f(b + h) < 0
⎩ f(b) – f(b – h) > 0
(iii) x = c is neither a maximum point nor a minimum
point
⎪⎧
f (c) – f(c + h) ⎪⎫
⇔ ⎨ and ⎬ have opposite signs.
⎪⎩ f(c) − f(c − h) ⎪⎭
Where h is a very small positive number.
Note :
The maximum and minimum points are also
known as extreme points.
A function may have more than one maximum
and minimum points.
A maximum value of a function f(x) in an interval
[a, b] is not necessarily its greatest value in that
interval. Similarly a minimum value may not be
the least value of the function. A minimum value
may be greater than some maximum value for a
function.
The greatest and least values of a function f(x) in
an interval [a, b] may be determined as follows :
Greatest value = max. {f(a), f(b), f(c)}
Least value = min. {f(a), f(b), f(c)}
where x = c is a point such that f´(c) = 0.
If a continuous function has only one maximum
(minimum) point, then at this point function has
its greatest (least) value.
Monotonic functions do not have extreme points.
Conditions for maxima and minima of a function
Necessary condition : A point x = a is an extreme
point of a function f(x) if f´(a) = 0, provided f´(a)
exists. Thus if f´(a) exists, then
x = a is an extreme point ⇒ f´(a) = 0 or
f´(a) ≠ 0 ⇒ x = a is not an extreme point
But its converse is not true i.e.
f´(a) = 0 /⇒ x = a is an extreme point.
For example if f(x) = x 3 , then f´(0) = 0 but x = 0 is
not an extreme point.
Sufficient condition : For a given function f(x), a
point x = a is
a maximum point if f´(a) = 0 and f´´(a) < 0
a minimum point if f´(a) = 0 and f´´(a) > 0
not an extreme point if f´(a) = 0 = f´´(a) and
f´´´(a) ≠ 0.
Note : If f´(a) = 0, f´´(a) = 0, f´´´(a) = 0 then the sign
of f (4) (a) will determine the maximum or minimum
point as above.
Working Method :
Find f´(x) and f´´(x).
Solve f´(x) = 0. Let its roots be a, b, c, ...
Determine the sign of f´´(x) at x = a, b, c, .... and
decide the nature of the point as mentioned above.
Properties of maxima and minima :
If f(x) is continuous function, then
Between two equal values of f(x), there lie atleast one
maxima or minima.
Maxima and minima occur alternately. For example
if x = –1, 2, 5 are extreme points of a continuous
function and if x = –1 is a maximum point then x = 2
will be a minimum point and x = 5 will be a
maximum point.
When x passes a maximum point, the sign of dy/dx
changes from + ve to – ve, where as when x passes
through a minimum point, the sign of f´(x) changes
from –ve to + ve.
If there is no change in the sign of dy/dx on two sides
of a point, then such a point is not an extreme point.
If f(x) is maximum (minimum) at a point x = a, then
1/f(x), [f(x) ≠ 0] will be minimum (maximum) at that
point.
If f(x) is maximum (minimum) at a point x = a, then
for any λ ∈ R, λ + f(x), log f(x) and for any k > 0, k
f(x), [f(x)] k are also maxmimum (minimum) at that
point.
XtraEdge for IIT-JEE 52 DECEMBER 2009

MATH
FUNCTION
Mathematics Fundamentals
Definition of a Function :
Let A and B be two sets and f be a rule under which
every element of A is associated to a unique element
of B. Then such a rule f is called a function from A to
B and symbolically it is expressed as
f : A → B
If f and g are two functions then their sum,
difference, product, quotient and composite are
denoted by
f + g, f – g, fg, f/g, fog
and they are defined as follows :
(f + g) (x) = f(x) + g(x)
or
A ⎯→
f
B
(f – g) (x) = f(x) – g(x)
Function as a Set of Ordered Pairs
Every function f : A → B can be considered as a set
of ordered pairs in which first element is an element
of A and second is the image of the first element.
Thus
f = {a, f(a) /a ∈ A, f(a) ∈ B}.
Domain, Codomain and Range of a Function :
If f : A → B is a function, then A is called domain of
f and B is called codomain of f. Also the set of all
images of elements of A is called the range of f and it
is expressed by f(A). Thus
f(A) = {f(a) |a ∈ A}
obviously f(A) ⊂ B.
Note : Generally we denote domain of a function f by
D f and its range by R f.
Equal Functions :
Two functions f and g are said to be equal functions
if
domain of f = domain of g
codomain of f = codomain of g
f(x) = g(x) ∀ x.
Algebra of Functions :
(fg) (x) = f(x) f(g)
(f/g) (x) = f(x)/g(x) (g(x) ≠ 0)
(fog) (x) = f[g(x)]
Formulae for domain of functions :
D f ± g = D f ∩ D g
D fg = D f ∩ D g
D f/g = D f ∩ D g ∩ {x |g(x) ≠ 0}
D gof = {x ∈ D f | f(x) ∈ D g }
D = D
f f ∩ {x |f(x) ≥ 0}
Classification of Functions
1. Algebraic and Transcendental Functions :
Algebraic functions : If the rule of the function
consists of sum, difference, product, power or
roots of a variable, then it is called an algebraic
function.
Transcendental Functions : Those functions
which are not algebraic are named as
transcendental or non algebraic functions.
2. Even and Odd Functions :
Even functions : If by replacing x by –x in f(x)
there in no change in the rule then f(x) is called an
even function. Thus
f(x) is even ⇔ f(–x) = f(x)
XtraEdge for IIT-JEE 53 DECEMBER 2009

Odd function : If by replacing x by –x in f(x)
there is only change of sign of f(x) then f(x) is
called an odd function. Thus
f(x) is odd ⇔ f(–x) = – f(x)
3. Explicit and Implicit Functions :
Explicit function : A function is said to be
explicit if its rule is directly expressed (or can be
expressed( in terms of the independent variable.
Such a function is generally written as
y = f(x), x = g(y) etc.
Implicit function : A function is said to be
implicit if its rule cannot be expressed directly in
terms of the independent variable. Symbolically
we write such a function as
f(x, y) = 0, φ(x, y) = 0 etc.
4. Continuous and Discontinuous Functions :
Continuous functions : A functions is said to be
continuous if its graph is continuous i.e. there is
no gap or break or jump in the graph.
Discontinuous Functions : A function is said to
be discontinuous if it has a gap or break in its
graph atleast at one point. Thus a function which
is not continuous is named as discontinuous.
5. Increasing and Decreasing Functions :
Increasing Functions : A function f(x) is said to
be increasing function if for any x 1 , x 2 of its
domain
x 1 < x 2 ⇒ f(x 1 ) ≤ f(x 2 )
or x 1 > x 2 ⇒ f(x 1 ) ≥ f(x 2 )
Decreasing Functions : A function f(x) is said to
be decreasing function if for any x 1 , x 2 of its
domain
x 1 < x 2 ⇒ f(x 1 ) ≥ f(x 2 )
or x 1 > x 2 ⇒ f(x 1 ) ≤ f(x 2 )
Periodic Functions :
A functions f(x) is called a periodic function if there
exists a positive real number T such that
f(x + T) = f(x).
∀ x
Also then the least value of T is called the period of
the function f(x).
Period of f(x) = T
⇒ Period of f(nx + a) = T/n
Periods of some functions :
Function
sin x, cos x, sec x, cosec x,
tan x, cot x
sin n x, cos n x, sec n x, cosec n x
tan n x, cot n x
|sin x|, |cos x|, |sec x|, |cosec x|
|tan x|, |cot x|,
2π
|sin x| + |cos x|, sin 4 x + cos 4 x π
|sec x| + |cosec x| 2
|tan x| + |cot x|
x – [x] 1
π
Period
2π if n is odd
π if n is even
π ∀ n ∈ N
π
π
π
2
Period of f(x) = T ⇒ period of f(ax + b) = T/|a|
Period of f 1 (x) = T 1 , period fo f 2 (x) = T 2
⇒ period of a f 1 (x) + bf 2 (x) ≤ LCM {T 1 , T 2 }
Kinds of Functions :
One-one/ May one Functions :
A function f : A → B is said to be one-one if
different elements of A have their different
images in B.
Thus
⎧ a ≠ b
⎪
f is one-one ⇔ ⎨
⎪
⎩f
(a) = f (b)
⇒
or
⇒
f (a)
≠ f (b)
a = b
A function which is not one-one is called many
one. Thus if f is many one then atleast two
different elements have same f-image.
Onto/Into Functions : A function f : A → B is
said to be onto if range of f = codomain of f
Thus f is onto ⇔ f(A) = B
XtraEdge for IIT-JEE 54 DECEMBER 2009

Hence f : A → B is onto if every element of B
(co-domain) has its f–preimage in A (domain).
A function which is not onto is named as into
function. Thus f : A → B is into if f(A) ≠ B. i.e.,
if there exists atleast one element in codomain of
f which has no preimage in domain.
Note :
Total number of functions : If A and B are finite
sets containing m and n elements respectively,
then
total number of functions which can be defined
from A to B = n m .
total number of one-one functions from A to B
=
⎪⎩
⎪ ⎨
⎧
n
p m
0
if
if
m ≤ n
m > n
total number of onto functions from A to B (if m
≥ n) = total number of different n groups of m
elements.
Composite of Functions :
Let f : A → B and g : B → C be two functions, then
the composite of the functions f and g denoted by
gof, is a function from A to C given by gof : A → C,
(gof) (x) = g[f(x)].
Properties of Composite Function :
The following properties of composite functions can
easily be established.
Composite of functions is not commutative i.e.,
fog ≠ gof
Composite of functions is associative i.e.
(fog)oh = fo(goh)
Composite of two bijections is also a bijection.
Inverse Function :
If f : A → B is one-one onto, then the inverse of f i.e.,
f –1 is a function from B to A under which every b ∈ B
is associated to that a ∈ A for which f(a) = b.
Thus f –1 : B → A,
f –1 (b) = a ⇔ f(a) = b.
Domain and Range of some standard functions :
Function Domain Range
Polynomial
function
Identity
function x
Constant
function c
Reciprocal
function 1/x
R
R
R
R
R
{c}
R 0 R 0
x 2 , |x| R R + ∪ {0}
x 3 , x |x| R R
Signum
function
R {–1, 0, 1}
x + |x| R R + ∪ {0}
x – |x| R R – ∪ {0}
[x] R Z
x – [x] R [0, 1)
x [0, ∞) [0, ∞)
a x R R +
log x R + R
sin x R [–1, 1]
cos x R [–1, 7]
tan x
R – {± π/2, ± 3π/2, ...} R
cot x R – {0, ± π. ± 2π, ..... R
sec x R – (± π/2, ± 3π/2, ..... R – (–1, 1)
cosec x R – {0, ±π, ± 2π, ......} R –(–1, 1)
sinh x R R
cosh x R [1, ∞)
tanh x R (–1, 1)
coth x R 0 R –[1, –1]
sech x R (0, 1]
cosech x R 0 R 0
sin –1 x [–1, 1] [–π/2, π/2]
cos –1 x [–1, 1] [0, π]
tan –1 x R (–π/2, π/2}
cot –1 x R (0, π)
sec –1 x R –(–1, 1) [0, π] – {π/2}
cosec –1 x R – (–1, 1) (– π/2, π/2] – {0}
XtraEdge for IIT-JEE 55 DECEMBER 2009

Based on New Pattern
IIT-JEE 2010
XtraEdge Test Series # 8
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
• Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
• Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer.
Section - II
• Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly
matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly
marked answer in any row.
Section - III
• Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark
for wrong answer.
PHYSICS
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. A deuterium plasma is a neutral mixture of
negatively charged electrons and positively charged
deuterium charged electrons and positively charged
deuterium nuclei. Temperature required to produced
fusion in deuterium plasma is : (Take the range of
nuclear forces 2 fermi)
(A) 835 K
(B) 835 × 10 7 K
(C) 8.35 × 10 7 K (D) 273 K
2. As observed in the laboratory system, a 6 MeV
proton is incident on a stationary 12 C target velocity
of center of mass of the system is : (Take mass of
proton to be 1 amu)
(A) 2.6 × 10 6 m/s (B) 6.2 × 10 6 m/s
(C) 10 × 10 6 m/s (D) 10 m/s
3. Find the de Broglie wavelength of Earth. Mass of
Earth is 6 × 10 24 kg. Mean orbital radius of Earth
around Sun is 150 × 10 6 km -
(A) 3.7 m
(B) 3.7 × 10 –63 m
(C) 3.7 × 10 63 m (D) 3.7 × 10 –63 cm
4. A particle starts from rest and travels a distance x
with uniform acceleration, then moves uniformly a
distance 2x and finally comes at rest after moving
further 5x distance with uniform retardation. The
ratio of maximum speed to average speed is -
5
5
(A) (B) 2 3
(C) 4
7
(D) 5
7
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
5. A body moves in a circular path of radius R with
deceleration so that at any moment of time its
tangential and normal accelerations are equal in
magnitude. At the initial moment t = 0, the velocity
of body is v 0 then the velocity of body will be-
(A)
(B) v =
v0
v = at time t
⎛ v0t
⎞
1 + ⎜ ⎟
⎝ R ⎠
v 0 e
−S/
R
after it has moved S meter
(C) v = v 0 e –SR after it has moved S meter
(D) None of these
XtraEdge for IIT-JEE 56 DECEMBER 2009

6. Three identical rods of same material are joined to
form a triangular shape ABC as shown. Angles at
edge A and C are respectively θ 1 and θ 2 as shown.
When this triangular shape is heated then -
A
B
θ 1
(A) θ 1 decreases and θ 2 increases
(B) θ 1 increases and θ 2 decreases
(C) θ 1 increases
(D) θ 2 increases
7. Direction of current in coil (2) is opposite to
direction of current in coil (1) and coil (3). All three
coils are coaxial and equidistant coil (1) and coil (3)
are fixed while coil (2) is suspended thus able to
move freely. Then -
R
θ 2
1
2
3
(A) coil (2) is in equilibrium
(B) equilibrium state of coil (2) is stable
equilibrium along axial direction
(C) equilibrium state of coil (2) is unstable
equilibrium along axial direction
(D) if direction of current in coil (2) is same as that
of coil (1) and coil (3) then state of equilibrium
of coil (2) along axial direction is unstable
8. Which of the following statements are correct about
the circuits shown in the figure where 1 Ω and 0.5
Ω are internal resistances of the 6 V and 12 V
batteries respectively –
P
R
4Ω
6V, 1Ω
R
C
12V, 0.5Ω
Q
0.5Ω
(A) The potential at point P is 6 V
(B) The potential at point Q is – 0.5 V
(C) If a voltmeter is connected across the 6 V
battery, it will read 7 V
(D) If a voltmeter is connected across the 6 V
battery, it will read 5 V
S
R
9. In passing through a boundary refraction will not
take place if -
(A) light is incident normally on the boundary
(B) the indices of refraction of the two media are same
(C) the boundary is not visible
(D) angle of incidence is lesser than angle of
refraction but greater then sin –1 ⎛ µ
⎟ ⎞
⎜
R
⎝ µ D ⎠
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
A
B
C
D
P Q R S T
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. Match column I with column II in the light of
possibility of occurrence of phenomena listed in
column I using the systems in column II -
Column-I Column-II
(A) Interference
(B) Diffraction
(C) Polarisation
(D) Reflection
(P) Non-mechanical
waves
(Q) Electromagnetic
waves
(R) Visible light
(S) Sound waves
(T) None
11. Column-I Column-II
(A) 5000 Å (P) De-Broglie wavelength of
electron in x-ray tube
(B) 1 Å
(Q) Photoelectric threshold
wavelength
(C) 0.1 Å (R) x-ray wavelength
(D) 10 Å (S) De-Broglie wavelength of
most energetic photoelectron
emitted from metal surface
(T) None
XtraEdge for IIT-JEE 57 DECEMBER 2009

This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbe rs in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. The minimum speed in m/s with which a projectile
must be thrown from origin at ground so that it is able to
pass through a point P (30 m, 40 m) is : (g = 10 m/s 2 )
(Ans. in .............. × 10)
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
between the particles in process of their motion in
meters is (g = 10 m/s 2 ).
u = 2 m/s
A 45°
9m v = 14 m/s
B
22m
45°
17. Consider the circuit shown in figure. What is the
current through the battery just after the switch is
closed.
18V
S
2mH
9Ω
9Ω
9Ω
2mH
18. A rectangular plate of mass 20 kg is suspended from
points A and B as shown. If the pin B is suddenly
removed then the angular acceleration in rad/sec 2 of
the plate is : (g = 10 m/s 2 ).
A
B
13. In U 238 ore containing Uranium the ratio of U 234 to
Pb 206 nuclei is 3. Assuming that all the lead present in
the ore is final stable product of U 238 . Half life of
U 238 to be 4.5 × 10 9 years and find the age of ore.
(in 10 9 years)
l =0.2m
b =0.15m
14. Under standard conditions the gas density is
1.3 mg/cm 3 and the velocity of sound propagation in
it is 330 m/s, then the number of degrees of freedom
of gas is.
15. A single conservative force acts on a body of mass
1 kg that moves along the x-axis. The potential energy
U(x) is given by U (x) = 20 + (x – 2) 2 , where x is in
meters. At x = 5.0 m the particle has a kinetic energy of
20 J, then the maximum kinetic energy of body in J is.
(Ans. in .............. × 10)
16. Two particles are simultaneously thrown from top of
two towers with making angle 45º with horizontal.
Their velocities are 2 m/s and 14 m/s. Horizontal and
vertical separation between these particles are 22 m
and 9 m respectively. Then the minimum separation
19. If the temperature of a gas is raised by 1 K from
27ºC. Find the percentage change in speed of sound.
(Speed = 300 ms –1 )
(Ans. in .............. × 10 2 )
CHEMISTRY
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. In the precipitation of sulphides of second group
basic radicals. H 2 S is passed into acidified solution
with dilute HCl. If the solution is not acidified, then
which of the following is correct ?
XtraEdge for IIT-JEE 58 DECEMBER 2009

(A) Only the sulphides of second group get
precipitated
(B) Only the sulphides of fourth group get
precipitated
(C) Neither of the sulphides of second and fourth
groups get pricipitated
(D) Sulphides of both the groups second and fourth
get precipitated
2. The geometrical shapes of XeF + 2–
5 , XeF 6 and XeF 8
respectively are -
(A) trigonal bipyramidal, octahedral and square
planar
(B) square based pyramidal, distorted octahedral and
octahedral
(C) planar pentagonal, octahedral and square anti
prismatic
(D) square based pyramidal, distorted octahedral and
square anti prismatic
3. Regarding graphite the following informations are
available :
3.35Å
Top view
5. The difference between ortho and para hydrogen
is/are -
(A) they are electron spin isomer, where the spins of
electrons are opposite
(B) ortho hydrogen is more stable at lower
temperature
(C) they are nuclear spin isomer, where the spins of
protons are same in para and different in ortho
isomer
(D) they are nuclear spin isomers, where the spins of
protons are opposite in para but same in ortho
isomer
6. NGP assistance support for S N 2 reaction will be seen
in -
(A)
(C)
OH
Cl
OH
Θ
COO
(B) CH 3 –S–CH 2 –CH 2 –Cl
(D)
OH
7. When the compound called isoborneol is heated with
50% sulfuric acid the product of the reaction is/are ?
HO
Cl
Isoborneol
The density of graphite = 2.25 gm/cm 3 . What is C–C
bond distance in graphite ?
(A) 1.68Å (B) 1.545Å
(C) 2.852 Å (D) 1.426Å
(A)
(C)
(B)
(D)
4. The molecular formula of a non-stoichiometric tin
oxide containing Sn(II) and Sn (IV) ions is Sn 4.44 O 8 .
Therefore, the molar ratio of Sn(II) to Sn(IV) is
approximately -
(A) 1 : 8 (B) 1 : 6
(C) 1 : 4 (D) 1 : 1
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
8. Regarding the radial probability distribution
(4nr 2 R 2 nl) vs r plot which of the following is/are
correct ?
(A) The number of maxima is (n-l)
(B) The number of nodal points is (n-l-1)
(C) The radius at which the radial probability density
reaches to maxima is 3s < 3p < 3d
(D) The number of angular nodes is l
XtraEdge for IIT-JEE 59 DECEMBER 2009

9. H 2 C 2 O 4 and NaHC 2 O 4 behave as acids as well as
reducing agents which is/are correct statement (s) ?
(A) Equivalent wt. of H 2 C 2 O 4 and NaHC 2 O 4 are
equal to their molecular weights when behaving
as reducing agent
(B) 100 mL of 1N solution of each is neutralised by
equal volumes of 1M Ca(OH) 2
(C) 100 mL of 1N solution of each is neutralised by
equal volumes of 1N Ca(OH) 2
(D) 100 mL of 1M solution of each is oxidised by
equal volumes of 1M KMnO 4
(B)
(C)
Cl
Me
Cl
Me
Me
Me
Me
Me
Me
Me
Me
Br
Me
Me
Me
Me
Br
Cl
Me
Me
Me
Cl
Me
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. Column-I Column-II
⎛ a ⎞
(A) If force of attraction (P) ⎜P
+ ⎟
⎝ V 2
(V–b)= RT
⎠
among the gas molecules
be negligible
(B) If the volume of the (Q) PV = RT –a/V
gas molecules be
negligible
(C) At STP (for real gas) (R) PV = RT + Pb
(D) At low pressure and (S) PV = RT
at high temperature
(T) PV/RT = 1–a/VRT
11. Column-I
Me
Me
Me
Me
Cl
(D)
Me
Cl
Me
Me
Br
Me
Me
Br
Me
Me
Column-II
(P) Optically active
(Q) Cis compound
(R) Trans compound
(S) Optically inactive
(T) Chiral axis (element of chirality)
Cl
Me
This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
(A)
Me
Me
Me
Me
Me
Me
12. What is the number of lone pair on the molecule
XeO 2 F 2 ?
XtraEdge for IIT-JEE 60 DECEMBER 2009

13. 15g of a solute in 100g water makes a solution freeze
at –1ºC. What will be the depression in freezing point
if 30g of a solute is dissolved in 100g of water ?
14. The half lives of decomposition of gaseous CH 3 CHO
at constant temperature but at initial pressure of
364mm and 170mm Hg were 410 second and 880
second respectively. Hence what is the order of
reaction.
15. The number of S–S bonds in sulphur trioxide trimer
(S 3 O 9 ) is ....... .
16. The number of peroxo linkages present in the
[H 4 B 2 O 8 ] 2– is ..... .
17. The maximum number of structural isomers (acyclic
and cyclic) possible for C 4 H 8 are ....... .
18. What is the no. of lone pair of electrons present on N
in Trisilylamine ?
19. A small amount of solution containing 24 Na with
activity 2 × 10 3 dps was administered into the blood
of patient in a hospital. After 5 hour a sample of the
blood drawn out from the patient shared an activity
of 16 dps per cm 3 . (t 1/2 of 24 Na = 15 hrs.) Find the
volume (in L) of blood in patient.
[Given : log 1.2598 = 0.1003]
MATHEMATICS
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. The largest term in the expansion of (3 + 2x) 50 , where
x = 1/5, is -
(A) 5th
(B) 6th
(C) 8th
(D) 9th
1
2. If sin
–1 ⎡ 3sin 2θ
⎤
2
⎢ ⎥ = tan –1 x, then x =
⎣5+
4cos 2θ⎦
(A) tan 3θ
(B) 3 tan θ
(C) (1/3) tan θ
(D) 3 cot θ
3.
(cos x −1)(cos x − e
The integer n for which lim
x→0
n
x
a finite nonzero number is -
(A) 1 (B) 2
(C) 3 (D) 4
x
)
is
x
4. If f(x) = and g(x) =
sin x
then in this interval -
x
tan x
where 0 < x ≤ 1,
(A) both f(x) and g(x) are increasing functions
(B) both f(x) and g(x) are decreasing functions
(C) f(x) is an increasing function
(D) g(x) is an increasing function
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
5. If a, b, c are in A.P., and a 2 , b 2 , c 2 are in H.P., then
(A) a = b = c
(C) a, b, c are in H.P.
(B) a, b, – 2
1 c are in G.P.
(D) – 2
1 a, b, c are in G.P.
6. The system of equations
–2x + y + z = a
x – 2y + z = b
x = y – 2z = c
has
(A) no solution if a + b + c ≠ 0
(B) unique solution if a + b + c = 0
(C) infinite number of solutions if a + b + c = 0
(D) none of these
7.
x
The function f(x) =
1+
| x |
is differentiable on -
(A) (0, ∞) (B) [0, ∞)
(C) (–∞, 0)
(D) (– ∞,∞)
4 2
x sinx + cos x⎛ x cos x x sin x cos x ⎞
8. If l = e ⎜
− +
⎟
∫
dx
2 2
x cos x
⎝
⎠
then l equals -
(A) e x s in x ⎛ sec x ⎞
+ cos x ⎜x
− ⎟ + C
⎝ x ⎠
(B) e x s in x ⎛ cosx
⎞
+ cos x ⎜x
sin x − ⎟
⎝ x ⎠
(C) e x s in x ⎛
+ cos x ⎜
⎝
(D) xe x sin x+cos x –
∫
x
tan x
e
secx
⎞
− ⎟ + C x ⎠
x sinx + cos x
⎛ cos x − x sin x ⎞
⎜1−
⎟ dx
2 2
⎝ x cos x ⎠
XtraEdge for IIT-JEE 61 DECEMBER 2009

9.
y ⎛ x
If for the differential equation y′ = + φ x ⎟ ⎞
⎜
⎝ y ⎠
the
x
general solution is y =
log | Cx |
then f (x / y) is
given by -
(A) – x 2 / y 2 (B) y 2 / x 2
(C) x 2 / y 2 (D) – y 2 / x 2
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. The domain of the functions
Column-I
Column-II
(A) sin –1 (x/2 – 1) (P) (3 – 2π, 3 – π) ∪ (3,4]
+ log (x – [x])
2
(B) e x + 5sin π / 16 − x (Q)(0, 4) – {1, 2, 3}
(C) log 10 sin (x – 3) (R)[– π/4, π/4]
2
+ 16 − x
(D) cos –1 1− 2x
4
2
(S) [– 3/2, 5/2]
(T) None
11. Column-I Column-II
I denotes an integral
(A)
∫ π x log sin x dx (P) I = (π/8) log 2
0
(B)
∫ ∞ 2
log (x+x –1 dx − π
) (Q) I = log 2
0
2
1+
x
2
(C)
∫ π / 4
log (1+ tan x)dx (R) I = π log 2
0
(D)
∫ π log (1 – cos x)dx (S) I = π log 2
0
(T) None
This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. Suppose X follows a binomial distribution with
parameters n = 6 and p. If 9P(X = 4) = P(X = 2), find
4p.
13. If Q is the foot of the perpendicular from the point
x − 5 y + z − 6
P(4, –5,3) on the line = = then
3 − 42
5
100 (PQ) 2 is equal to
457
14. If a = (0, 1, –1) and c = (1, 1, 1) are given vectors,
then |b| 2 where b satisfies a × b + c = 0 and a . b = 3
is equal to
15. ABC is an isosceles triangle inscribed in a circle of
radius r. If AB = AC and h is the altitude from A to
BC. If the triangle ABC has perimeter P and area ∆
then lim 512r ∆
h→0
3 is equal to
p
16. If f(x) = sin x, x ≠ n π, n = 0, ± 1, ± 2, .....
= 0 otherwise
and g(x) = x 2 + 1, x ≠ 0, 2
= 4 x = 0
= 5 x = 2
then lim g(f(x)) is .....
x→0
17. If y = (1 + 1/x) x 2 y 2(2)
+ 1/8
then
is equal to
(log 3/ 2 −1/3)
18. If the greatest value of y = x/log x on [e, e 3 ] is u then
e 3 /u is equal to
19. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the
value of 2(z + z ) – |z| 2 .
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
XtraEdge for IIT-JEE 62 DECEMBER 2009

Based on New Pattern
IIT-JEE 2011
XtraEdge Test Series # 8
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
• Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
• Question 5 to 9 are multiple choice questions with multiple correct answer. +4 marks and -1 mark for wrong answer.
Section - II
• Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly
matched answer and No Negative marks for wrong answer. However, +2 marks will be given for a correctly
marked answer in any row.
Section - III
• Question 12 to 19 are numerical response type questions. +4 marks will be awarded for correct answer and -1 mark
for wrong answer.
PHYSICS
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. Velocity of block 2 shown in figure is -
u
(A)
3u
2
(B)
3u
2
u
1
(C)
60º
3
3u
2
2
(D) 2u
2. Which of the following restoring force can give rise
to S.H.M -
(A) F = 2x
(B) F = 2 – 4x
(C) F = – 2x 2
(D) None of these
3. A disk of mass M and radius R is rotating about its
axis with angular velocity ω. Axis of disk is rotated
by 90º in time ∆t. Average torque acting on disk is -
2
2Mω
(A)
∆t
MR
(C)
2∆ω
2
t
(B)
MR 2 ω
2∆t
(D) Zero
4. Select the incorrect statement -
(A) The velocity of the centre of mass of an isolated
system must stay constant
(B) Only a net external force can change the
velocity of the center of mass of a system
(C) A system have non-zero kinetic energy but zero
linear momentum
(D) F
→ ext
d → v
= m + → v dt
dm is true for all situation
dt
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
5. Velocity of a particle moving on straight line varies
as 2
1 th
power of displacement. Then -
(A) K.E. ∝ S (B) P ∝ S 1/2
(C) a = constant (D) S ∝ t 2
XtraEdge for IIT-JEE 63 DECEMBER 2009

6. A block of density ρ is floating in a liquid X kept in
container. A liquid Y of density ρ′ (< ρ) is slowly
poured into container –
(A) The block will move up if liquid X and Y are
immiscible
(B) The block will sink more if liquid X and Y are
immiscible
(C) The block will sink more if liquid X and Y are
miscible
(D) The block will not move if liquid X and Y are
miscible
7. Six identical rod are connected as shown in figure
and temperature difference of 100ºC is maintained
across P and Q –
A
B
C
D
P Q R S T
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. There are four identical rod having thermal
resistance 10 Ω, each column I contains various
arrangement of rod. Column II contains current
flowing across point C when a temperature
difference of 100ºC is maintained across A & B.
Match them.
Column-I
Column-II
P A B Q
(A) A
C
B
(P) 2 J/sec
C
(A) Temperature of point 'A' is 50ºC
(B) A
C
B
(Q) 4 J/sec
200
(B) Temperature of point A is ºC
3
(C) Thermal current passing through B is zero
(D) Thermal current passing through A is twice of
that through C
8. A solid iron cylinder A rolls down a ramp and an
identical iron cylinder B slides down the same ramp
without friction –
(A) B reaches the bottom first
(B) A and B have the same kinetic energy
(C) B has greater translational kinetic energy than
that of A
(D) Linear speed of centre of mass of B is greater
than that of A
9. A mass and spring system oscillates with amplitude
A and angular frequency ω –
(A) The average speed during one complete cycle
2Aω
of oscillation is
π
(B) Maximum speed is ωA
(C) Average velocity of particle during one
complete cycle of oscillation is zero
(D) Average acceleration of particle during one
complete cycle of oscillation is zero
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
(C) A
(D)
A
C
C
B
B
(R) 6 J/sec
(S) 7
20 J/sec
(T) None
11. A car of mass 500 kg is moving in a circular road
of radius 35 / 3 . Angle of banking of road is
30º. Coefficient of friction between road and tyres
is µ =
1 . Match the following:
2 3
Column-I
Column-II
(A) Maximum speed (in m/s) of (P) 5 2
car for safe turning
(B) Minimum speed (in m/s) of (Q) 12.50
car for safe turning
(C) Speed (in m/s) at which friction (R) 210
force between tyres and road
is zero
(D) Friction force (in 10 2 Newton) (S)
350
3
between tyres and road if
speed is
350 m/s
6
(T) None
XtraEdge for IIT-JEE 64 DECEMBER 2009

This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbe rs in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. Pulley 'P' shown in figure is pulled upward with
F = 2t N, where t is time in sec. Velocity of block
of mass 1 at the time block 2 is about to lift is
(in cm/sec). (Ans. in ................. × 10 1 )
F = 2t
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
14. A longitudinal wave of frequency 220 Hz travels
down a copper rod of radius 8.00 mm. The average
power in the wave is 6.50 µW. The amplitude of the
wave is n × 10 –8 m. Find n.
15. A piston-cylinder device with air at an initial
temperature of 30ºC undergoes an expansion
process for which pressure and volume are related
as given below
P (kPa) 100 25 6.25
V (m 3 ) 0.1 0.2 0.4
The work done by the system is n × 10 3 J. Find n.
16. The block connected with spring is pushed to
compress the spring by 10 cm and then released. All
surfaces are frictionless and collision are elastic.
Time period of the motion in sec (mass of block = 9 kg
and spring constant 4π 2 N/m).
5 cm
17. RMS velocity of gas at 27ºC is 300× 381 m/s.
RMS velocity (in m/s) when temperature is
increased four times is. (Ans. in ................. × 10 2 )
2 1
m 1 = 0.5 kg
m 2 = 1 kg
13. The upper edge of a gate in a dam runs along the
water surface. The gate is 2.00 m high and 4.00 m
wide and is hinged along a horizontal line through
its center. The torque about the hinge arising from
the force due to the water is (n × 10 4 Nm). Find
value of n.
18. A block of mass 2 kg is placed on a wedge of mass
10 kg kept on a horizontal surface. Coefficient of
friction between all surfaces is µ = 0.2. If block is
slipping down the wedge with constant speed then
friction force on wedge due to horizontal surface is
(in Newton) :
19. A particular quantity 'y' varies as 'x' as shown in
figure. RMS value of y with respect to x for large
values of 'x' is.
2 m
y
60º 60º
1 2 3 4
x
XtraEdge for IIT-JEE 65 DECEMBER 2009

CHEMISTRY
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. An ideal gaseous mixture of ethane (C 2 H 6 ) and
ethene (C 2 H 4 ) occupies 28 litre at STP. The mixture
reacts completely with 128 gm O 2 to produce CO 2
and H 2 O. Mole fraction of C 2 H 4 in the mixture is -
(A) 0.6 (B) 0.4 (C) 0.5 (D) 0.8
2. A bulb of constant volume is attached to a
manometer tube open at other end as shown in figure.
The manometer is filled with a liquid of density
(1/3 rd ) that of mercury. Initially h was 228 cm.
(C) The first ionization energies of elements in a
period do not increase with the increase in
atomic numbers
(D) For transition elements the d-subshells are filled
with electrons monotonically with the increase
in atomic number
7. Which of the following statements regarding
hydrogen peroxide is/are correct ?
(A) Hydrogen peroxide is a pale blue viscous liquid
(B) Hydrogen peroxide can act as oxidising as well
as reducing agent
(C) The two hydroxyl groups in hydrogen peroxide
lie in the same plane
(D) In the crystalline phase, H 2 O 2 is paramagnetic
8. Which of the following is/are state function ?
(A) q (B) q – w (C) q + w (D) q / w
Gas
h
9. The IUPAC name of the following compound is -
OH
Through a small hole in the bulb gas leaked assuming
dp
pressure decreases as = – kP. dt
If value of h is 114 cm after 14 minutes. What is the
value of k (in hour –1 ) ?
[Use : ln(4/3) = 0.28 and density of Hg = 13.6 g/mL]
(A) 0.6 (B) 1.2
(C) 2.4
(D) None of these
3. When 300 mL of 0.2 M HCl is added to 200 mL of
0.1 M NaOH. Resultant solution require how many
equivalent of Ba(OH) 2 ?
(A) 0.06 (B) 0.12 (C) 0.3 (D) 0.04
4. The dipole moment of HCl is 1.03D, if H–Cl bond
distance is 1.26Å, what is the percentage of ionic
character in the H–Cl bond ?
(A) 60% (B) 29% (C) 17% (D) 39%
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
5. Consider the following carbides CaC 2 , BeC 2 , MgC 2
and SrC 2 which of the given carbides on hydrolysis
yield same product -
(A) CaC 2 (B) Be 2 C (C) MgC 2 (D) SrC 2
6. Which of the following is/are correct regarding the
periodic classification of elements ?
(A) The properties of elements are the periodic
function of their atomic number
(B) Non metals are lesser in number than metals
Br
CN
(A) 3-Bromo-3-cyano phenol
(B) 3-Bromo-5-hydroxy benzonitrile
(C) 3-Cyano-3-hydroxybromo benzene
(D) 5-Bromo-3-hydroxy benzonitrile
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. Column-I Column-II
(A) 5.4 g of Al
(P) 0.5 N A electrons
(B) 1.2 g of Mg 2+ (Q) 15.9994 amu
(C) Exact atomic weight (R) 0.2 mole atoms
of mixture of oxygen
isotopes
(D) 0.9 mL of H 2 O (S) 0.05 moles
(T) 3.1 × 10 23 electrons
XtraEdge for IIT-JEE 66 DECEMBER 2009

11. Column-I Column-II
(Ionic species)
(Shapes)
+
(A) XeF 5 (P) Tetrahedral
–
(B) SiF 5 (Q) Square planar
+
(C) AsF 4 (R) Trigonal bipyramidal
–
(D) ICl 4 (S) Square pyramidal
(T) Octahedral
This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbe rs in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. How much volume (in mL) 0.001 M HCl should we
add to 10 cm 3 of 0.001 M NaOH to change its pH by
one unit ?
13. The stopcock, connecting the two bulbs of volumes
5 litres and 10 litres containing an ideal gas at 9 atm
and 6 atm respectively, is opened. What is the final
presure (in atm) in the two bulbs if the temperature
remained the same ?
14. An acid type indicator, HIn differs in colour from its
conjugate base (In – ). The human eye is sensitive to
colour differences only when the ratio [In – ] / [HIn] is
greater than 10 or smaller than 0.1. What should be
the minimum change in the pH of the solution to
osberve a complete colour change ? (K a = 1.5 × 10 –5 )
15. What is the sum of total electron pairs (b.p. + l.p.)
present in XeF 6 molecule ?
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
16. The number of geometrical isomers of
CH 3 CH=CH–CH=CH–CH=CHCl is.
17. At 200ºC, the velocity of hydrogen molecule is
2.0 × 10 5 cm/sec. In this case the de-Broglie
wavelength (in Å) is about.
18. The equivalent weight of a metal is 4.5 and the
molecular weight of its chloride is 80. The atomic
weight of the metal is.
19. No. of π bond in the compound H 2 CSF 4 is.
MATHEMATICS
Questions 1 to 4 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1. If 0 < r < s ≤ n and n P r = n P s , then value of r + s is -
(A) 2n – 2 (B) 2n – 1
(C) 2 (D) 1
2. If sin x + sin 2 x + sin 3 x = 1, then cos 6 x – 4 cos 4 x +
8 cos 2 x is equal to -
(A) 0 (B) 2 (C) 4 (D) 8
3. If x is real, and
2
x − x + 1
k = then
2
x + x + 1
(A) 1/3 ≤ k ≤ 3 (B) k ≥ 5
(C) k ≤ 0
(D) none of these
4. A flagstaff stands in the centre of a rectangular field
whose diagonal is 1200 m, and subtends angles 15º
and 45º at the mid points of the sides of the field. The
height of the flagstaff is -
(A) 200 m (B) 300 2 + 3 m
(C) 300 2 − 3 m (D) 400 m
Questions 5 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
5. If a, b, c are the sides of the ∆ABC and a 2 , b 2 , c 2 are
the roots of x 3 – px 2 + qx – k = 0, then
cos A cos B cosC
p
(A) + + =
a b c 2 k
4q − p
(B) a cos A + b cos B + c cos C =
2 k
2p∆
(C) a sin A + b sin B + c sin C =
k
(D) sin A sin B sin C =
∆
k
8 3
6. The coordinates of the feet of the perpendiculars
from the vertices of a triangle on the opposite sides
are (20, 25), (8, 16) and (8, 9). The coordinates of a
vertex of the triangle are -
(A) (5, 10) (B) (50, –5)
(C) (15, 30) (D) (10, 15)
2
XtraEdge for IIT-JEE 67 DECEMBER 2009

⎡1
1 ⎤ ⎡ ⎤
7. Let E = ⎥ + 1 2
⎢ + ⎢ + ⎥ + ... upto 50 terms, then -
⎣3
50 ⎦ ⎣3
50 ⎦
(A) E is divisible by exactly 2 primes
(B) E is prime
(C) E ≥ 30
(D) E ≤ 35
8. If m is a positive integer, then [( 3 + 1) ] + 1,
where [x] denotes greatest integer ≤ n, is divisible
by-
(A) 2 m (B) 2 m+1 (C) 2 m+2 (D) 2 2m
9. If A and B are acute angles such that sin A = sin 2 B,
2 cos 2 A = 3 cos 2 B; then -
(A) A = π/6 (B) A = π/2
(C) B = π/4 (D) B = π/3
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. For the circle x 2 + y 2 + 4x + 6y – 19 = 0
Column-I
Column-II
72 226
(A) Length of the tangent (P)
113
from (6, 4) to the circle
(B) Length of the chord (Q) 113
of contact from (6, 4)
to the circle
(C) Distance of (6, 4) (R) 113 – 32
from the centre of the
circle
(D) Shortest distance of (S) 9
(6, 4) from the circle
(T) None
11. Value of x when
Column-I
Column-II
(A) 5 2 5 4 5 6 ... 5 2x = (0.04) –28 (P) 3 log 3 5
⎛ 1 1 1 ⎞
(B) x 2 log 5 ⎜ + + + ... ⎟
=
⎝ 4 8 16
( 0.2)
⎠
(Q) 4
⎛ 1 1 1
log
⎞
2.5 ⎜ + + + ... ⎟
⎝ 3 2 3
3 3 ⎠
(C) x = ( 0.16)
(R) 2
2m
(D) 3 x–1 + 3 x–2 + 3 x–3 + ... (S) 7
⎛ 2 1 1 ⎞
= 2 ⎜5
+ 5 + 1+
+ + ... ⎟
⎝ 5
2
5 ⎠
(T) None
This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. Fifteen persons, among whom are A and B, sit down at
random at a round table. if p is The probability that
there are exactly 4 persons between A and B find 14 p.
13. If l is the length of the intercept made by a common
tangent to the circle x 2 + y 2 = 16 and the ellipse
x 2 /25 + y 2 /4 = 1, on the coordinate axes, then
81l
2 + 3
is equal to
1059
14. If x + y = k is a normal to the parabola y 2 = 12x, p is
the length of the perpendicular from the focus of the
3k
3 2
+ 2p
parabola on this normal; then
is equal to
741
15. The volume of the tetrahedron whose vertices are
(0, 1, 2) (3, 0, 1) (4, 3, 6) (2, 3, 2) is equal to
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
16. Let a 1 = 2
1 , ak+1 = a k 2 + a k ∀ k ≥ 1 and
1 1 1
x n = + + ... +
a1 + 1 a 2 + 1 a n + 1
Find [x 100 ] where [x] denotes the greatest integer ≤ x.
17. Find the value of x which satisfy the equation
log 2 (x 2 – 3) – log 2 (6x – 10) + 1 = 0
18. Find the coefficient of x 2009 in the expansion of
(1 – x) 2008 (1 + x + x 2 ) 2007
1/ log x
2
19. Find the value of x satisfying 4 = 2.
XtraEdge for IIT-JEE 68 DECEMBER 2009

MOCK TEST PAPER-1
CBSE BOARD PATTERN
CLASS # XII
SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS
Solutions will be published in next issue
General Instructions : Physics & Chemistry
• Time given for each subject paper is 3 hrs and Max. marks 70 for each.
• All questions are compulsory.
• Marks for each question are indicated against it.
• Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each.
• Question numbers 9 to 18 are short-answer questions, and carry 2 marks each.
• Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each.
• Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
• Use of calculators is not permitted.
General Instructions : Mathematics
• Time given to solve this subject paper is 3 hrs and Max. marks 100.
• All questions are compulsory.
• The question paper consists of 29 questions divided into three sections A, B and C.
Section A comprises of 10 questions of one mark each.
Section B comprises of 12 questions of four marks each.
Section C comprises of 7 questions of six marks each.
• All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
• There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and
2 question of six marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.
PHYSICS
1. q =
q 0
is valid or not, q is the charge on
2
v
1−
2
c
particle when it is moving with velocity v, q 0 is the
rest charge and c is the velocity of light .
2. A concave mirror is dipped inside the liquid of
absolute refractive index 1.25. What will be the
percentage change in its focal length.
7. If nuclear density d ∝ A n , where A is the atomic
number then write the value of n.
8. Why standard resistors are made of alloys.
9. Name the quantities whose SI units are given below :
1. V – m 2. C-m
out of the two also name the vector quantity.
10. A transistor is shown in figure.
3. Write the name of a compound semiconductor.
4. If one of the slit get closed in Young's Double slit
experiment then fringe pattern will be observed or
not on the screen.
5. Write one of the use of Zener Diode.
6. Name the experiment which proves the Dual Nature
of electron.
+2V
+1V +3V
(i) Name the type of transistor
. (ii) Is the transistor is properly biased.
11. A time variant current is given -
i(t) = 1 + 3 2 sin (314 t + 30º)
Find its root mean square value.
XtraEdge for IIT-JEE 69 DECEMBER 2009

12. A metallic conductor of non-uniform cross-sectional
area is shown in figure.
i
a
p 1 p 2 p 3
(i) Out of p 1 , p 2 , and p 3 at which point the drift speed
of the electron is maximum.
(ii) Out of p 1 , p 2 and p 3 at which point the current
density is minimum.
13. Stopping potential versus freq. of the incident light
graph is shown in figure for metal-1 and metal-2
metal-1 metal-2
V 0
θ 1
(i) Which metal will have the higher value of
threshold wavelength.
(ii) Is θ 1 = θ 2 if yes then why ?
14. Draw the circuit diagram for finding the internal
resistance of the cell using potentiometer.
15. (i) Define angle of Dip.
(ii) What is the value of angle of Dip at a place on
earth where Horizontal component and vertical
component of earth magnetic field are equal.
16. (i) State Kirchhoff's current law.
(ii) Find the potential difference across the 2 ohm
resistance for the given circuit diagram.
4A 1Ω 2Ω
θ 2
b
υ
18. (i) Write the order of colors in secondary rainbow.
(ii) Why sun appears reddish at the time of sunrise
and sunset.
19. The charges on the capacitors are Q a , Q b and Q c then
calculate -
1µF
Q a
12µF
30V
Q b
Q c
key-K
3µF
(i) Ratio of energy stored in 1 µF and 3 µF capacitor.
(ii) Potential difference across 12 µF capacitor
(iii) Energy supplied by the battery.
20. (i) A uniform magnetic field of 0.5 T exist in the
given solenoid. If an electron is projected along the
axis of solenoid from a towards b with the speed of
3 × 10 2 m/s then find the Lorentz force working on
electron.
i
a
(ii) When the current flows through the metallic
spring why it get shrinked.
21. (i) For the given circuit diagram find the position of
the null point.
5Ω 10 Ω
b
3Ω
A
B
1A
G
100 cm
17. (i) State Lenz's law.
(ii) If the current i increases then what will be the
direction of induced current in the circular coil for the
given figure.
V bb
R h
Key-k
Rheostat
(ii) For the given figure draw truth table.
A
i
b
B
Y
a
a >> b
XtraEdge for IIT-JEE 70 DECEMBER 2009

22. (i) State Brewster's law for polarization .
(ii) If the Brewster's angle for the given pair of
medium is i p then express the critical angle for the
same pair of media in terms of i p .
(iii) If the lens is made of medium-2 and placed in
medium-1. If the absolute refractive Index of
medium-2 is µ 2 and of medium-1 is µ 1 then µ 2 > µ 1 .
Is it correct or not.
(iii) Where to put the proton that it will have
maximum force on it.
27. For the given configuration find the -
–2q
2a
2a
Incident rays
Medium-1
Refracted rays
Medium -2
Medium-1
+q
+q
2a
(i) Dipole moment of the system.
(ii) Electric potential energy of the system.
23. What is the need of Modulation in communication
system. Draw the shapes of signal, carrier wave and
amplitude Modulated wave.
24. How a Galvanometer can be converted into Ammeter
explain it by drawing the circuit diagram.
⎛ i ⎞
Prove that S = ⎜
g
⎟ G
⎝ i − i g ⎠
G = Resistance of Galvanometer coil
S = value of shunt
i g = Full scale deflection current for Galvanometer
i = Range of Ammeter.
25. (i) Find the equivalent resistance between A and B
for given fig.
R
A
R
R R
R
R R B
(ii) α is the symbol for the temperature coefficient of
resistivity for the given material. If α → 0 then the
material will be copper or constantan ?
26. (i) What is the angle between electric field line and
equi-potential surface.
(ii) If E a , E b and E c are the electric field intensities at
points a, b and c respectively, where to put a proton
that it will have the maximum electric potential
energy.
a
b
c
28. Draw the circuit diagram for common emitter
transistor amplifier. Explain its working. What is the
phase difference between input and output voltage in
case of common emitter transistor amplifier.
29. Explain construction and working of cyclotron. Why
cyclotron can not be used to accelerate the electrons.
30. State Ampere's circuital law. Using Ampere's
circuital law find the magnetic field at the axis of the
long solenoid.
CHEMISTRY
1. Give the IUPAC name of the organic compound
(CH 3 ) 2 C = CH – C – CH 3 .
||
O
2. Complete the following reaction :
CH 3 – CH 2 – CH = CH 2 + HCl → …..
3. State a use for the enzyme streptokinase in medicine.
4. What is copolymerization ?
5. Which type of a metal can be used in cathodic
protection of iron against rusting ?
6. Why is the bond dissociation energy of fluorine
molecule less than that of chlorine molecule ?
7. What is meant by inversion of sugar ?
8. What are the types of lattice imperfections found in
crystals ?
XtraEdge for IIT-JEE 71 DECEMBER 2009

9. Describe the mechanism of the formation of diethyl
ether from ethanol in the presence of concentrated
sulphuric acid.
10. Predict, giving reasons, the order of basicity of the
following compounds in (i) gaseous phase and (ii) in
aqueous solutions (CH 3 ) 3 N, (CH 3 ) 2 NH, CH 3 NH 2 ,
NH 3 .
11. Write names of monomer/s of the following polymers
and classify them as addition or condensation
polymers.
(i) Teflon
(ii) Bakelite
(iii) Natural Rubber
12. Give an example for each of the following reactions :
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
13. Write one distinction test each for :
(i) Ethyl alcohol and 2–propanol
(ii) Acetaldeyde and acetone
14. Distinguish between multimolecular and
macromolecular colloids. Give one example of each
type.
15. Draw the structure of pyrophosphoric acid and how it
is prepared.
16. If E° for copper electrode is +0.34 V how will you
calculate its emf value when the solution in contact
with it is 0.1 M in copper ions ? How does emf for
copper electrode change when concentration of Cu 2+
ions in the solution is decreased ?
17. Physical and chemical adsorptions respond
differently to a rise in temperature. What is this
difference and why is it so ?
18. Using the valence bond approach, predict the shape
and magnetic character of [Ni (CO) 4 ].
(At No. of Ni = 28).
19. Describe the following giving a chemical equation
for each :
(i) Markownikoff’s rule
(ii) Hofmann Bromide Reaction
20 (a) Write chemical equations and reaction conditions
for the conversion of :
(i) Ethene to ethanol
(ii) Phenol to phenyl ethanoate
(iii) Ethanal to 2-propanol
21. Identify the substances A and B in each of the
following sequences of reactions :
(i) C 2 H 5 Br
⎯
alc ⎯
. KOH
Br
⎯⎯
→ A ⎯⎯→
2 B
NaNO2 HCl
(ii) NH + Cu2 (CN)
2
⎯ 2
⎯⎯⎯⎯→ A⎯ ⎯⎯ ⎯⎯ → B
0ºC
(iii) NH H 2 SO 4
2
⎯ ⎯⎯
→ A
Heat
⎯ ⎯→ ⎯ B
22. Give the electronic configuration of the
(a) d-orbitals of Ti in [Ti (H 2 O) 6 ] 3– ion in an
octahedral crystal field.
(b) Why is this complex coloured ? Explain on the
basis of distribution of electrons in the d-orbitals.
23. State reasons for the following :
(a) Rusting of iron is said to be an electrochemical
phenomenon.
(b) For a weak electrolyte, its molar conductance in
dilute solutions increases sharply as its concentration
in solution is decreased.
24. Using the valence bond approach predict the shape
and magnetic character of [Co (NH 3 ) 6 ] 3+ . (Atomic
number of Co is 27).
25. Explain the following :
(a) F-centre
(b) Schottky & Frenkel defect
26. Account for the following :
(i) Ferric hydroxide sol is positively charged.
(ii) The extent of physical adsorption decreases with
rise in temperature.
(iii) A delta is formed at the point where the river
enters the sea.
27. Taking two examples of heterogeneous catalytic
reactions, explain how a heterogeneous catalyst helps
in the reaction.
28. (a) An organic compound ‘A’ with molecular
formula C 5 H 8 O 2 is reduced to n-pentane on treatment
with Zn-Hg/HCl. ‘A’ forms a dioxime with
hydroxylamine and gives a positive lodoform test and
Tollen’s test. Identify the compound A and deduce
its structure.
(b) Write the chemical equations for the following
conversions : (not more than 2 steps)
(i) Ethyl benzene to benzene
(ii) Acetaldehyde to butane – 1, 3–diol
(iii) Acetone to propene
XtraEdge for IIT-JEE 72 DECEMBER 2009

29. Describe how potassium dichromate is made from
chromite ore and give the equations for the chemical
reactions involved.
Write balanced ionic equations for reacting ions to
represent the action of acidified potassium
dichromate solution on :
(i) Potassium iodide solution
(ii) Acidified ferrous sulphate solution
Write two uses of potassium dichromate.
30. Give appropriate reasons for each of the following
observations :
(i) Sulphur vapour exhibits some paramagnetic
behaviour.
(ii) Silicon has no allotropic form analogous to
graphite.
(iii) Of the noble gases only xenon is known to form
real chemical compounds.
(iv) Nitrogen shows only a little tendency for
catenation, whereas phosphorus shows a clear
tendency for catenation.
MATHEMATICS
Section A
1. Show that the relation R in the set {1, 2, 3} is given
by R = {(1, 2), (2, 1)} is symmetric.
−1
2. Evaluate : x.tan x dx.
∫
3. Find the differential equation of the family of curves
given by- x 2 + y 2 = 2ax.
→
→
8. If a = î + ĵ ; b = ĵ+
kˆ ; c = kˆ + î find a unit vector
in the direction of
→
→
→
→
a + b+
c .
9. What is the angle between vector → aand
magnitude 3 and 2 respectively.
→
bwith
10. Find the direction cosines of a line which make equal
angles with the co ordinate axes.
Section B
11. Consider f : N → N, g : N → N and h : N → R
defined as f (x) = 2x, g (y) = 3y + 4 and h (z) = sin
z ∀ x, y and z in N. Show the ho(gof) = (hog)of.
12. Differentiate cot –1 ⎛1−
x ⎞
⎜ ⎟ w.r.t. x.
⎝1+
x ⎠
13. Solve the differential equations :
(1 + e 2x ) dy + e x (1 + y 2 ) dx = 0. Give that y = 1,
when x = 0.
or
dy
Solve the differential equation : x − y − 2x
3 = 0
dx
π
14. Evaluate :
∫
/ 4
0
or
sin
π / 4
+
Evaluate :
∫
log( 1
0
2x sin 3x
dx.
tan x)dx
4. Find the principle value of tan –1 (–1).
5. Find a matrix C such that 2A – B + C = 0
⎡3
1⎤
⎡− 2 1⎤
Where A = ⎢ ⎥ and B =
⎣0
2
⎢ ⎥ ⎦ ⎣ 0 3 ⎦
6. If A is a square matrix of order 3 such that
| adj A | = 64, find | A |.
7. Find the value of x if the matrix A =
singular.
⎡ 4
⎢
⎢
3
⎢⎣
10
3
− 2
−1
5⎤
7
⎥
⎥
x⎥⎦
is
3x + 1
15. Evaluate :
∫
dx .
2
2x − 2x + 3
16. If x = a (θ – sin θ) and y = a (1 – cos θ),
find
d
2
dx
y
2
at θ
= 2
π .
17. Find the value of K so that the function,
⎧Kx
+ 1, if x ≤ π
f (x) = ⎨
⎩cos x , if x > π
or
⎪⎧
3
x + 3, if x ≠ 0
Show that the function f (x) = ⎨
⎪⎩ 1 , if x = 0
XtraEdge for IIT-JEE 73 DECEMBER 2009

18. Solve for x :
x < 1.
19. If A =
⎡4
⎢
⎢
1
⎢⎣
2
tan
− 5
− 3
3
⎛ x −1
⎞
⎜ ⎟ + tan
⎝ x − 2 ⎠
⎛ x + 1 ⎞
⎜ ⎟ = ;
⎝ x + 2 ⎠ 4
−1 −1
π
−11⎤
1
⎥
⎥
find A –1
− 7 ⎥⎦
or
Using the properties of determinates prove that -
a + x y z
x
x
a + y
y
2
z = a (a + x + y + z)
a + z
20. Find the distance of the point (2, 3, 4) from the plane
3x + 2y + 2z + 5 = 0, measured parallel to the line
x + 3 y − 2 z
= =
3 6 2
21. If
→
a and
→
b are unit vectors and θ is the angle
between them, then prove that cos
θ 1 =
2 2
→
+
→
22. A football match may be either won, drawn or lost by
the host country team. So there are three ways of
forecasting the result of any one match, one correct
and two incorrect. Find the probability of forecasting
at least three correct results for four matches.
or
A candidate has to reach the examination centre in
time. Probabilities of him going by bus or scooter or
3 1 3
by other means of transport are , ,
10 10 5
respectively. The probability that he will be late is
1 1 and respectively, if he travels by bus or scooter.
4 3
But he reaches in time if he uses any other mode of
transport. He reached late at the center. Find the
probability that he traveled by bus.
π/
2
Section C
cos x
23. Evaluate :
∫
dx
(1 + sin x)(2 + sin x)
0
a
b
24. Using integration, find the area of the circle
x 2 +y 2 = 16, which is exterior to the parabola y 2 = 6x.
or
Find the area of the smaller region bounded by the
2 2
x y
x y
ellipse + = 1 and the line + = 1.
2 2
a b
a b
25. Show that a right circular cylinder which is open at
the top, and has a given surface area, will have the
greatest volume if its height is equal to the radius of
its base.
⎡1
−1
1 ⎤
26. For A =
⎢ ⎥
⎢
2 1 − 3
⎥
, find A –1 and hence solve the
⎢⎣
1 1 1 ⎥⎦
system of equations.
x + 2y + z = 4
– x + y + z = 0
x – 3y + z = 2
27. A letter is known to have come from either
TATANAGAR or CALCUTTA. On the envelope
just two consecutive letters TA are visible. What is
the probability that the letter has come from-
(i) TataNagar
(ii) Calcutta
or
Find the probability distribution of the number of
white balls drawn in a random draw of 3 balls
without replacement from a bag containing 4 white
and 6 red balls. Also find mean and variance of the
distribution.
28. Find the distance of the point (3, 4, 5) from the plane
x + y + z = 2 measured parallel to the line 2x = y = z.
29. Every gram of wheat provides 0.1 gm of proteins and
0.25 gm of carbohydrates. The corresponding values
for rice are 0.05 gm and 0.5 gm respectively. Wheat
costs Rs. 4 per kg and rice Rs. 6 per kg. The
minimum daily requirements of protein and
carbohydrates for an average child are 50 gms and
200 gms respectively. In what quantities should
wheat and rice be mixed in the daily diet to provide
minimum daily requirements of proteins and
carbohydrates at minimum cost. Frame an L.P.P. and
solve it graphically.
XtraEdge for IIT-JEE 74 DECEMBER 2009

XtraEdge Test Series
ANSWER KEY
IIT- JEE 2010 (December issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9
Ans B A B C A,B C,D A,B,D B,C A,B,C,D
Ques 10 A → P,Q,R,S B → Q,P,Q,R C → P,Q,R,S D → P,Q,R,S
Ques 11 A → Q B → R C → P D → S
Ques 12 13 14 15 16 17 18 19
Ans 3 2 5 2 6 2 1 1
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9
Ans D D D C D A,B,C B A,B,D B,C,D
Ques 10 A → R B → Q,T C → P D → S
Ques 11 A → Q,S B → R,S C → P,Q,T D → R,S
Ques 12 13 14 15 16 17 18 19
Ans 1 2 2 0 2 9 0 6
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9
Ans B C C C A,B,D A,C A,B,C,D A,D D
Ques 10 A → Q B → R C → P D → S
Ques 11 A → Q B → S C → P D → R
Ques 12 13 14 15 16 17 18 19
Ans 1 4 6 4 1 3 3 3
IIT- JEE 2011 (December issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9
Ans D B C D A,B,C,D A,C A,C,D A,B,C,D A,B,C,D
Ques 10 A → Q B → S C → R D → P
Ques 11 A → R B → P C → S D → Q
Ques 12 13 14 15 16 17 18 19
Ans 2 3 3 5 2 3 0 1
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9
Ans A B D C A,C,D A,B,D A,B,D C D
Ques 10 A → R B → P,S,T C → Q D → P,T
Ques 11 A → S B → R C → P D → Q
Ques 12 13 14 15 16 17 18 19
Ans 1 7 2 7 8 1 9 1
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9
Ans B C A C A,B,C,D A,B,C B,D A,B A,C
Ques 10 A → S B → P C → Q D → R
Ques 11 A → S B → R C → Q D → P
Ques 12 13 14 15 16 17 18 19
Ans 2 5 3 6 1 2 0 4
XtraEdge for IIT-JEE 75 DECEMBER 2009

XtraEdge for IIT-JEE 76 DECEMBER 2009