Download IIT-JEE 2010 - Career Point Kota

Teachers open the door.
You enter by yourself.
Volume - 5 Issue - 7
January, 2010 (Monthly Magazine)
Editorial / Mailing Office :
112-B, Shakti Nagar, Kota (Raj.) 324009
Editorial
Tel. : 0744-2500492, 2500692, 3040000
e-mail : [email protected]
Editor :
Pramod Maheshwari
[B.Tech. IIT-Delhi]
Cover Design & Layout
Govind Saini, Om Gocher,
Mohammed Rafiq
Circulation & Advertisement
Ankesh Jain, Praveen Chandna
Ph (0744)- 3040007, 9001799502
Subscription
Sudha Jaisingh Ph. 0744-2500492, 2500692
© Strictly reserved with the publishers
• No Portion of the magazine can be
published/ reproduced without the written
permission of the publisher
• All disputes are subject to the exclusive
jurisdiction of the Kota Courts only.
Every effort has been made to avoid errors or
omission in this publication. In spite of this, errors
are possible. Any mistake, error or discrepancy
noted may be brought to our notice which shall be
taken care of in the forthcoming edition, hence any
suggestion is welcome. It is notified that neither the
publisher nor the author or seller will be
responsible for any damage or loss of action to any
one, of any kind, in any manner, there from.
Unit Price Rs. 20/-
Special Subscription Rates
6 issues : Rs. 100 /- [One issue free ]
12 issues : Rs. 200 /- [Two issues free]
24 issues : Rs. 400 /- [Four issues free]
Owned & Published by Pramod Maheshwari,
112, Shakti Nagar, Dadabari, Kota & Printed
by Naval Maheshwari, Published & Printed at
112, Shakti Nagar, Dadabari, Kota.
Editor : Pramod Maheshwari
XtraEdge for IIT-JEE
Dear Students,
Are You an Optimist or a Pessimist ?
I have been giving some thought lately to optimism and pessimism. Basically,
these are attitudes — attitudes that shape and formulate our entire existence. I
mean, have you ever met a happy pessimist?
Of course not.
In short, our optimism or pessimism is this:
The way we interpret the past.
The way we experience and view the present.
The way we imagine the future
Have you given much thought about how your attitude, whether you are an
optimist or a pessimist, affects you business, organization or school? Have you
thought about how it affects you personally? And what about the team you are
a part of?
What is optimism? It is the belief that things in our past were good for us, even
if that means they were hard and taught us lessons. It is also the belief that
things will be better in the future.
Here are some contrasts between optimism and pessimism and how they
affect us:
Optimism breathes life into you each day.
Pessimism drains you.
Optimism helps you to take needed risks.
Pessimism plays it safe and never accomplishes much.
Optimism improves those around you.
Pessimism drags them down.
Optimism inspires people to great heights.
Pessimism deflates people to new lows.
There is only one way that optimism and pessimism are the same, and that is
that they are both self-fulfilling. If you are an optimist, you will generally find
that good things happen to you. And if you are a pessimist, you will find
yourself in the not-so-good situations more often than not.
So can a person just become an optimist? Yes! We can choose to look at the
world any way we want to. We can choose to look at the world and think the
worst, or we can tell ourselves the good things about each situation. As you
find yourself looking at your enterprise, begin to view it through the eyes of an
optimist, and you will reap the rewards listed above, and so will the people
around you.
There are tremendous benefits to being an optimist, as stated above. But there
are some pessimists out there who will say, “But that isn’t realistic.” I say,
“Who cares?” If things go awry, at least I have spent my time beforehand
enjoying life and not worrying about it. And, being an optimist, I would view
the “negative” situation as an opportunity to grow and learn. So I can even
look forward to my failures because they will be steppingstones and learning
tools to be applied to my future success.
Have you ever met a successful pessimist? Become an optimist, and see your
world change before your eyes!
Have a blessed day!
Let's make the uncommon knowledge common
Yours truly
Pramod Maheshwari,
B.Tech., IIT Delhi
1
JANUARY 2010
XtraEdge for IIT-JEE
2
JANUARY 2010
Volume-5 Issue-7
January, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
CONTENTS
INDEX
Regulars ..........
Key Concepts & Problem Solving strategy for IIT-JEE.
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics, Chemistry & Maths
Much more IIT-JEE News.
Xtra Edge Test Series for JEE-2010 & 2011
Mock Test CBSE Pattern Cllass XII
PAGE
NEWS ARTICLE
4
IITian ON THE PATH OF SUCCESS
8
KNOW IIT-JEE
10
IIT boys draw power & water from sewage
Alumni of IIT-Madras to come together on
December 26
Dr. Alok Aggarwal
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS
Success Tips for the Month
•
It is more important to know where you
are going than to get there quickly.
•
The secret of success is constancy to
purpose.
•
8-Challenging Problems [Set# 8]
Students’ Forum
Physics Fundamentals
Refraction at plane & curved surfaces
Properties of Matter
CATALYST CHEMISTRY
of all success.
We cannot discover new oceans unless we
DICEY MATHS
have the courage to lose sight of the
Mathematical Challenges
Students’ Forum
Key Concept
Differential Equations
Trigonomatrical Rations
shore.
•
One person with a belief is equal to 99
who have only interests
•
• Keep your eyes on the stars and your feet
on the ground.
XtraEdge for IIT-JEE
48
Test Time ..........
A thousand mile journey begins with one
step. Start today.
33
Key Concept
Carbohydrates
Salt Analysis
Understanding : Organic Chemistry
Action without planning is the cause of all
failure. Action with planning is the cause
•
17
XTRAEDGE TEST SERIES
59
Class XII – IIT-JEE 2010 Paper
Class XII – IIT-JEE 2011 Paper
Mock Test CBSE Pattern Paper-2 [Class # XII]
Mock Test CBSE Pattern Paper-1 (Solution)[Class # XII]
3
JANUARY 2010
IIT boys draw power &
water from sewage
When the campus placements
Anand (energy engineering) and
a cash prize of Rs 3 lakh. "LOCUS
Mohan Yama (PhD student of
is a green tech development,
biotechnology).
which
Renowned
is
sustainable
both
biotech faculty member Debabrata
economically and environmentally
happen in 2011 at IIT Kharagpur,
Das enthusiastically joined them in
and serves as an ideal integration
these five students will give it a
the pursuit.
to address the key issues of
miss. They would have by then
started their own company. They
have
developed
a
bio
cell'
(battery) that can not only treat
sewage water but also generate
electricity a that could offer a onestop solution to the water and
energy crises.
Their bio-product has won them
The idea is simple. The specially
designed bio cell (LOCUS) will be
set up in the form of a plant,
through which the sewage water
of a housing complex would be
flowed in. The genius of this
invention which is awaiting patent
lies in the design of the cell that
will automatically grow millions of
rave accolades from the ministry
anaerobic bacteria that multiply
of science and technology and a
through respiration. The bacteria
cash award to carry their research
clean up the sewage water and in
forward.
the
process
generate
free
harnessed,
these
It was while working on microbes
electrons.
that can be used as purifying
electrons
agents that Manoj Mandelia, a
electricity.
fourth year student of biotech
"We worked on this concept for
engineering, stumbled upon the
nearly a year before we readied
idea that if a bio cell can be
the cell and applied to the ministry
developed to treat sewage water
If
can
generated
to enter its annual business plan
for use, it would solve one of the
competition that focuses on bio
biggest problems of the present
technology
times. Mandelia, who is pursuing
sustainable development," Mandelia
an integrated M Tech programme
products
for
said.
at the institute, started looking for
Twenty teams, mostly corporate
like minded boys for his project.
houses dealing with bio products,
He
participated
soon
found
Prateek
Jain
(agriculture and food engineering
department),
(electrical
Shobhit
engineering),
XtraEdge for IIT-JEE
Singhal
in
this
premier
competition. The IIT-Kgp team
managed to come second and won
wastewater treatment and energy
gap," reads the award citation.
The cell, at this stage, can clean up
50,000 litres of sewage water,
about the amount generated by
100 flats in a day. The water
produced
this
way
compared
with
can
that
be
supplied
provided by a civic body, the
students say. "The purified water
has been tested and has been
certified to be clean and fit for
household use. It is, however, not
fit
for
drinking,"
Mandelia
explained.
The
IIT-Kgp
team
has
even
produced electricity with the bio
cell. "A township of 100,000
people needs about 2.3 megawatts
of electricity a day. It will be years
before we reach that stage. But
we have already been able to
generate electricity. By next year,
we aim to generate 350 units,
enough to meet 50% of the
demand of a 100-flat complex.
When we say this we are not
taking
airconditioners
into
consideration," said Prateek.
Pulkit
4
JANUARY 2010
Alumni of IIT-Madras to
for the first year, the initiative
visited
come
being executed by the department
College here on Friday. Sources
of management studies is designed
said they discussed the make-shift
to act as a catalyst to provide
requirements and took stock of
initial momentum.
the
together
on
December 26
The "first ever" gathering of the
alumni of IIT-Madras (IITM) will be
held here on December 26.
According to a release here, more
than
32,000
students
have
graduated from the institute in the
past 50 years and most were
expected to attend the reunion,
co-organised by the Office of
Alumni Affairs (OAA) and IIT-M
Alumni Association.
Speaking at the launch of the
initiative here on Friday, L S
Ganesh,
professor-in-charge
of
Cell for Technology, Innovation,
Development
and
Entrepreneurship
(C-TIDES),
Support
said
IIT-Madras
is
perhaps the first institution to
start a formal programme in
entrepreneurship in the 1980s.
The day-long event would see the
Emphasizing the importance of
screening
developing
of
two
films,
one
small
and
medium
highlighting alumni contributions
enterprises, IIT-Madras director M
to the campus and the other their
S Ananth said they generated 10
socially-relevant work, especially
times the employment per unit of
in the rural sector.
investment.
"We
need
to
the
MBM
facilities
Engineering
available
in
the
college. The team will submit the
report to the Director (IIT-R),
who is scheduled to visit the city
in the first week of December to
consolidate
the
ground
for
shifting.
MBM dean Arvind Roy said that
the team was here to oversee the
proposed shifting early next year.
The team had a thorough visit of
the
entire
college
campus
buildings, classrooms, laboratories
and other rooms housing different
departments and collected a map
of the college.
Roy added that their main focus,
decentralise these aspects to get
however,
innovations started with centres
facilities, for which they also
such as this. Also, innovators need
visited the AIIMS site, where the
to be aware of their privileges and
students are initially proposed to
Alumni from different parts of the
what they are entitled to," he said.
be shifted, in case the building
world would make presentations
According to professor Thillai
of the IIT-R is not ready by then.
on various issues, the release said
Rajan,
It is evident that there has been an
adding an "IIT Madras Heritage
department
Quiz" would also be held.
studies, detailed workshops would
IIT-R will be shifted to the said
IIT-M launches seed fund
be
for prospective
college here from next year. A
incubates before issuing a call for
Central team, headed by Union
proposals.
additional secretary (HRD) Ashok
IIT-M director Prof MS Ananth is
scheduled to deliver the inaugural
address, the release said.
of
$
0.8
mn
to
help
budding entrepreneurs
Students, researchers and faculty
assistant
of
conducted
professor,
management
IIT-R shifting to Jodhpur
was on
understanding
Thakur,
the hostel
in-principle
recently
visited
that
the
college and the hostels which are
members from IIT-Madras now
gradually taking shape
have the chance to turn innovative
JODHPUR:
ideas into sustainable businesses
referred to as one step further
arrangements there.
with the Micro, Small and Medium
towards shifting the IIT-R from
The visit of this team of these sites
Enterprises
Kanpur
team
again has further strengthened
administrative
that fact that the plans of shifting
initiative.
(MSME)
Intended
incubation
to provide
In
to
what
Jodhpur,
comprising
can
a
complete
be
and
said
to
have
expressed satisfaction with the
seed-stage funding of up to $ 0.8
in-charge
(Kanpur)
are gradually taking shape. The
million (Rs 3.5 crore) per venture
Niraj Gupta and two others,
team will explore and identify the
XtraEdge for IIT-JEE
of
IIT-R
5
JANUARY 2010
necessary additions and alterations
total number of PD seats in 2009
Signal engineering is the
that will have to be made into the
across all IITs was 251, only 138
existing structure of the college to
candidates could qualify at the JEE,
backbone of Rlys: SCR
accommodate IIT-R and prepare a
despite
report to be submitted to the
candidates taking the exam as
director.
compared to 2008.
IITs worried as reserved
"Despite the 50% relaxation for
seats remain unfilled
PD candidates (from the last
Even as the process for applying to
the Joint Entrance Examination
(JEE) 2010 of the Indian Institutes
of Technology (IITs) has begun,
12%
more
disabled
general candidate), only 44 could
its
productivity to meet the increase
in demand for rail traffic, Sudesh
Kumar,
member
(Electrical)
Railway Board has said.
celerations of the Indian Railways
unfilled," said Kumar.
Institute of Signal Engineering and
scores of PD candidates were
with physical disabilities (PD), got
relaxed by another 50% to enable
wasted. The reason is that not
candidates to qualify for the
enough disabled candidates qualify,
preparatory course, which was
and
started for disabled candidates in
be
enhance
IITs. This left around 207 seats
137 seats meant for candidates
cannot
Railway
Speaking at the 52nd Annual Day
To make up for the shortfall,
seats
Indian
be given admission across all the
figures from JEE 2009 reveal that
these
Technology changes have helped
Telecommunications (IRISET) here
today,
he
said
signalling
has
evolved from an ordinary means
of communication to start a train
and later emerged as a symbol of
trust and safety on railway tracks
over the years.
converted into general seats.
2009. In case of OBCs, 51 seats
"Unless the rules are changed and
across the IITs in 2009 were
Participating in the event, the new
the IITs are allowed to transfer
converted into general seats due
South
the vacant PD seats to general or
to unavailability of OBC students
General Manager M S Jayanth said
non-PD seats within the category,
even
IRISET over a period of time has
we will not be able to stop this
relaxation. Again, over 1,000 seats
loss," said Anil Kumar, JEE 2010
for SCs and STs which remained
chairman, IIT Bombay.
vacant were transferred to the
As per a judgment passed by the
chief commissioner for persons
after
giving
full
10%
preparatory course and were filled
after lowering the bar by 50%.
IIT Guwahati director Gautam
to treat disabled candidates on a
Barua said that a resolution was
par
adopted at the last joint admission
scheduled
caste
tribe
(ST)
(SC)/scheduled
board
meeting
that
the
PD
candidates and give them similar
commissioner
relaxation from 2009, including
approached
admission
conversion of PD seats to general
for
preparatory
courses.
should
for
Railway
(SCR)
evolved as a quality institute.
``Signal engineering happens to be
the backbone of railways, since
this is vital for reliability and safety
of rails,’’ Jayanth said.
with disabilities, IITs were directed
with
Central
be
possible
category in future. "However, the
According to him, advancement
made in telecommunication has
palyed vital role in the information
management of Indian Railways,
especially in passenger reservation
system, fright and other services.
IRISET Director V Balaram said
signalling
is
very
specific
to
each
issue is complex since there are
category [general category, SC, ST
certain PD seats within SC and ST
railways.
and
and these cannot be converted,"
IRISET is the only institute in
said Barua.
railways, which trains both officers
Currently,
other
3%
seats
backward
in
classes
(OBC)] are reserved for disabled
students in each IIT. While the
XtraEdge for IIT-JEE
and supervisory staff. ``During the
6
JANUARY 2010
last 52 years, about 10,036 officers
It helps him determine the amount
site. But with the Australian
and 44,906 signal and telecom
of anaesthesia, and can also be
newspaper for which we made an
supervisiors have been trained,’’
tweaked
app, we not only created an app
Balaram said.
requirements. This app is now
but
V G K Murti, former Dean of IIT
used by around 15,000 doctors
required
to
their
Chennai, said in the earlier days
internationally. In the future, we
operation
so
that
railways used to be the most
have plans to make it possible for
delivered faster to the user. When
preferred place of employment for
doctors
we approach a client we don't
most IIT students. He delivered
profiles," Avinash says.
his
speech
targeting
young
trainees of IRISET, by passing on
knowledge inputs. The function
was attended by many senior
officials of the SCR.
for
to
other
medical
exchange
patient
applications
for
an
Australian newspaper and a top
computer manufacturer, among
others.
Making apps for the enterprise
Bangalore: One of the most
discussed issues when it came to
developing applications for mobile
phones was the lack of any
reasonable returns on apps made
for popular platforms like the
iPhone. A lot of developers said
they hardly made any money from
developing apps for these high-end
instead of selling it to the public
via an appstore is a different
ballgame altogether. The app store
was an instant success as it
allowed a single developer to roll
out an app and be able to sell it
without
worrying
about
marketing. All the developer has
to worry about, is the competition
phones. Turns out they have got
--similar apps available. But while
their strategy wrong. Endeavour, a
making apps for enterprises, it's
city-based company, sees good
the
business
adaptability that counts.
in
making
apps
for
enterprises.
T
functionality
and
the
changes
backend
data
was
developers, but instead, tell them
we can come up with a mobile
strategy for their company. These
enterprise apps will never make it
to
the
app
store,
but
are
a company," he adds.
The company has also developed a
few fun apps like upcoming app
Ambience which allows users mix
and match sound effects like that
of a waterfall, the sea, birds, etc to
create a relaxing atmosphere. "As
of now, most of the enterprises
still use Blackberries.But we're
seeing a slow shift towards touch
screen devices like the iPhone," he
says.
Avinash says they have not made
any apps for Indian companies till
"The cost of an app can range
now, as very few Indians use high-
from
end smartphones or other mobile-
$10,000
to
$3,50,000
he founders, all IIT graduates,
depending on the depth and the
started the company in 2002.
functionality required. We don't
According to Avinash Misra, one
just create an app, we see to it
of the founders, their first app was
that
targeted at doctors. "We made it
operational
in co-ordination with a doctor
environment as well. It's very easy
based in the US using which he
to make a news application by
could make a profile of a patient.
using the RSS feed available on the
XtraEdge for IIT-JEE
the
distributed to all the employees in
Enterprises need mobile
strategy
made
introduce ourselves as mobile app
The company has also developed
mobile
also
their
back-end
for
7
is
such
made
an
enabled devices.
"But in 18 months, I'm sure India
will have one of the largest
markets in terms of mobile usage.
Indians just might skip the desktop
internet revolution and switch
over to
the
mobile
internet
altogether," Avinash adds.
JANUARY 2010
Success Story
This article contains story of a person who get succeed after graduation from different IIT's
Dr. Alok Aggarwal
Electrical Engineering from IIT Delhi in 1980
Ph.D. in Electrical Engineering and Computer Science,
Hopkins University (1984)
Director : IBM Research Division worldwide
Emerging Business Opportunities for IBM Research
Division worldwide.
Dr. Alok Aggarwal received his Bachelor’s Degree in
Electrical Engineering from IIT Delhi in 1980. He
obtained his Ph.D. in Electrical Engineering and Computer
Science from Johns Hopkins University in 1984.
Dr. Alok Aggarwal has published 86 Research papers and
he has also been granted 8 patents from the US Patents
and Trademark Office. Along with his colleagues at
Evalueserve, in 2003, he has pioneered the concept of
“Knowledge Process Outsourcing (KPO)” and wrote the
first article in this regard. Dr. Aggarwal has served as a
Chairperson of the IEEE Computer Society's Technical
Committee on Mathematical Foundations of Computing
and has been on the editorial boards of SIAM Journal of
Computing, Algorithmica, and Journal of Symbolic
Computation. During 1998-2000, Dr. Aggarwal was a
member of Executive Committee on Information
Technology of the Confederation of the Indian Industry
(CII) and also of the Telecom Committee of Federation of
Indian Chamber of Commerce and Industry (FICCI). He is
currently a Chartered Member of The Indus Entrepreneur
(TiE) organization.
Dr. Alok Aggarwal is the Founder and Chairman of
Evalueserve - a company that was started in December
2000 and that provides various kinds of research and
analytics services to clients in North America, Europe and
Asia Pacific from its five research centers in DelhiGurgaon, India; Shanghai, China; Cluj, Romania; SantiagoValparaiso, Chile; and New York, USA.
Dr. Alok Aggarwal joined IBM Research Division in
Yorktown Heights New York in 1984. During the fall of
1987 and 1989, he was on sabbatical from IBM and taught
two courses (in two terms) at the Massachusetts Institute
of Technology (MIT) and also supervised two Ph.D.
students. During 1991 and 1996, along with other
colleagues from IBM, he created and sold a "Supply Chain
Management Solution" for paper mills, steel mills and
other related industries. In July 1997, Dr. Aggarwal
"Founded" the IBM India Research Laboratory that he setup inside the Indian Institute of Technology Delhi. Dr.
Aggarwal started this Laboratory from "ground zero" and
by July 2000, he had built it into a 60-member team (with
30 PhDs and 30 Masters in Electrical Engineering,
Computer Science, and in Business Administration). In
August 2000, Dr. Aggarwal became the Director of
In honouring Dr. Alok Aggarwal, IIT Delhi recognizes the
outstanding contributions made by him as an
Entrepreneur and Researcher. Through his achievements,
Dr. Alok Aggarwal has brought glory to the name of the
Institute.
Anything you can hunt, I can hunt better.
XtraEdge for IIT-JEE
8
JANUARY 2010
XtraEdge for IIT-JEE
9
JANUARY 2010
KNOW IIT-JEE
By Previous Exam Questions
The direction of B is given by Right hand palm rule
no. 1.
PHYSICS
1.
A charge +Q is fixed at the origin of the co-ordinate
system while a small electric dipole of dipole moment
A
P
→
O
P pointing away from the charge along the x-axis is
set free from a point far away from the origin.
(a) Calculate the K.E. of the dipole when it reaches to
a point (d, 0).
(b) Calculate the force on the charge +Q at this
[IIT-2003]
moment.
Sol. (a) Potential energy of the dipole-charge system
U i = 0 (Since the charge is far away)
E=
d
A
→
2pQ
4πε 0 d 3
1 pQ
4πε 0 d 2
f
r
→
→
→
B p = B PA + B PB + B PC
î
where
→
B PA = magnetic field at P due to A
→
B PB = magnetic field at P due to B
Three infinitely long thin wires, each carrying current
i in the same direction, are in the x-y plane of a
gravity free space. The central wire is along the yaxis while the other two are along x = ± d.
(i) Find the locus of the points for which the
[IIT-1997]
magnetic field B is zero.
(ii) If the central wire is displaced along the Zdirection by a small amount and released, show that it
will execute simple harmonic motion. If the linear
mass density of the wires is λ, find the frequency of
oscillation.
Sol. (i) We know that magnetic field due to an infinitely
long current carrying wire at distance r is given by
→
B PC = magnetic field at P due to C.
Bp =
For
µ 0 2I  1
1
1 
(– k̂ )
+ +

4π  d + x x d − x 
→
BP = 0
On solving we get x = ± d 3 .
(ii) The force per unit length between two parallel
current carrying wires is given by
µ 0 2I1I 2
= f(say)
4π r
and is attractive if currents are in the same direction.
So when the wire B is displaced along z-axis by a
small distance z, the restoring force per unit length
µ 0  2I 
 
4π  r 
XtraEdge for IIT-JEE
X
Z
θ θ
Z
Hence in case of three identical wires resultant field
can be zero only if the point P is between the two
wires otherwise field B due to all the wires will be in
the same direction and so resultant B cannot be zero.
Hence, if point P is at a distance x from the central
wire as shown in fig. then,
2.
B=
r
d
B
1 2p
î
4πε 0 d 3
F =QE =
B
f
Now, force on charge Q is given by
→
d
d
(b) Electric field at origin due to dipole
→
X
Z
1 p
Uf = – Q ×
4πε 0 d 2
∴ K.E. = |Uf – Ui | =
C
B
F/l on the wire B due to wires A and C will be
10
JANUARY 2010
The magnetic flux linked with the solenoid
z

as cos θ = r 


µ 2I I
F
z
= 2f cos θ = 2 0 1 2 ×
l
4π r
r
or
F µ0
4I 2
=
z [as I1 = I2 = I and r2 = d2 + z2]
l 4 π (d 2 + z 2 )
or
µ  2I 
F
= 0   z [as d>>z and F is opposite to z]
l
4π  d 
→
2
dφ
= π µ 0 n a2 i0 ω cos ωt
dt
The same rate of change of flux is linked with the
cylindrical shell. By the principle of electromagnetic
induction, the induced emf produced in the cylindrical
shell is
...(1)
Since F ∞ –z the motion is simple harmonic.
Comparing eq. (1) with the standard equation of
S.H.M. which is
F = – mω2z
i.e.,
F
m
= – ω2z = – λω2z, we get
l
l
λω2 =
⇒
3.
µ 0 4I 2
⇒
4π d 2
I
2πn =
d
µ0
⇒
πλ
ω=
××××
µ0 I 2
πd 2 λ
1
n=
2πd
I
TOP VIEWS
µ0
πλ
dφ
= – πµ0 n a2 i0 ω cos ωt
… (i)
dt
The resistance offered by the cylindrical shell to the
flow of induced current I will be
e=–
A long solenoid of radius a and number of turns per
unit length n is enclosed by cylindrical shell of radius
R, thickness d(d <<R) and length L. A variable
current i = i0 sin ωt flows through the coil. If the
resistivity of the material of cylindrical shell is ρ, find
[IIT- 2005]
the induced current in the shell.
R
a
Here,
2πR
Ld
The induced current I will be
∴
R=ρ
I=
L
I=
⇒ I=
Sol. The magnetic field in the solenoid is given by
B = µ0 ni
R
a
… (ii)
[ πµ 0 na 2 i 0 ω cos ωt ] × Ld
|e|
=
R
ρ × 2πR
πµ 0 na 2 Ld i 0 ω cos ωt
2πRρ
µ 0 na 2 Ld i 0 ω cos ωt
2ρR
4.
An object is moving with velocity 0.01 m/s towards a
convex lens of focal length 0.3 m. Find the magnitude
of rate of separation of image from the lens when the
object is at a distance of 0.4 m from the lens. Also
calculate the magnitude of the rate of change of the
[IIT - 2004]
lateral magnification.
Sol. Using lens formula
d
L
1
1
1
=
=
v − 0.4
0.3
B = µ 0 n i0 sin ωt
⇒
[Q i = i0 sin ωt given]
XtraEdge for IIT-JEE
l
A
l = 2 π R,
A= L×d
R=ρ
d
⇒
→
φ= B.A
= B A cos 90º
= (µ 0 n i0 sin ωt) ( πa 2)
∴ The rate of change of magnetic flux through the
solenoid
11
v = 1.2 m
JANUARY 2010
Now we have
A
B
1 1 1
– = , differentiating w.r.t. t
v u f
1 dv 1 du
+
=0
v 2 dt u 2 dt
we have
–
given
du
= 0.01 m/s
dt
⇒
 dv  (120)
× 0.01 = 0.09 m/s
 =
 dt  (0.4) 2
d = 1 cm
(b) Charge on plate B at t = 10 sec
Qb = 33.7 × 10–12 – 5 × 107 × 1.6 × 10–19
= 25.7 × 10 –12 C
also Q a = 8 × 10–12C
2
So, rate of separation of the image (w.r.t. the lens)
= 0.09 m/s
v
m=
u
Now,
udv vdu
−
dm
⇒
= dt 2 dt
dt
u
(0.4)(0.09) − (1.2)(0.01)
(0.4) 2
E=
=
= – 0.35
CHEMISTRY
Two metallic plates A and B, each of area 5 × 10–4 m2,
are placed parallel to each other at a separation of
1 cm. Plate B carries a positive charge of 33.7×10–12C.
A monochromatic beam of light, with photons of
energy 5 eV each, starts falling on plate A at t = 0 so
that 10 16 photons fall on it per square meter per
second. Assume that one photoelectron is emitted for
every 106 incident photons. Also assume that all the
emitted photoelectrons are collected by plate B and
the work function of plate A remain constant at the
[IIT-2002]
value 2eV. Determine
6.
At room temperature, the following reactions proceed
nearly to completion :
2NO + O2 → 2NO2 → N2O 4
The dimer, N2O4, solidified at 262 K. A 250 ml flask
and a 100 ml flask are separated by a stopcock. At
300 K, the nitric oxide in the larger flask exerts a
pressure of 1.053 atm and the smaller one contains
oxygen at 0.789 atm. The gases are mixed by opening
the stopcock and after the end of the reaction the
flasks are cooled to 200 K. Neglecting the vapour
pressure of the dimer, find out the pressure and
composition of the gas remaining at 220 K. (Assume
[IIT-1992]
the gases to behave ideally)
Sol. According to the gas equation,
(a) the number of photoelectrons emitted to t = 10 s,
(b) the magnitude of the electric field between the
plates A and B at t = 10 s, and
(c) the kinetic energy of the most energetic
photoelectron emitted at t = 10 s when it reaches
plate B.
PV = nRT
Neglect the time taken by the photoelectron to reach
plate B. Takes ε0 = × 10–12 C2/N-m2.
or
Sol. (a) Number of electrons falling on the metal plate
16
10
6
XtraEdge for IIT-JEE
PV
RT
For NO, P = 1.053 atm, V = 250 ml = 0.250 L
∴ Number of photoelectrons emitted from metal
plate A upto 10 second is
(5 × 10 −4 ) × 1016
n=
At room temperature,
–4
A = 10 × (5 × 10 )
ne =
17.7 ×10 −12
= 2000 N/C
5 × 10 − 4 × 8.85 ×10 −12
(c) K.E. of most energetic particles
= (hν – φ) + e(Ed) = 23 eV
[(hν – φ) is energy of photo electrons due to light.
e(Ed) is the energy of photoelectrons due to work
done on photoelectrons between the plates].
So magnitude of the rate of change of lateral
magnification = 0.35.
5.
σB
σ
1
– A =
(Q B – QA)
2ε 0
2ε 0
2Aε 0
∴ Number of moles of NO =
1.053 × 0.250
0.0821× 300
= 0.01069 mol
For O2, P = 0.789 atm, V = 100 ml = 0.1L
× 10 = 5 × 10 7
12
JANUARY 2010
∴ Number of moles of O 2 =
0.789 × 0.1
0.0821× 300
= 0.00320 mol
According to the given reaction,
γ=1+
∴ 0.00320 mol of O2 react with = 2 × 0.00320
= 0.0064 mol of NO
Number of moles of NO left = 0.01069 – 0.0064
= 0.00429 mol
Also, 1 mol of O2 yields = 1 mol of N2O4
f=
2
f
2
=5
0.4
P1V1γ = P 2V2γ
P1 = P
V1 = V
V2 = 5.66 V
P
P
P
[using eq.(1)]
=
=
γ
1.4
11
.32
(5.66)
(5.66)
Hence, work done by the gas during adiabatic
expansion
P V −P V
= 1 1 2 2 =
γ –1
An ideal gas having initial pressure P, volume V and
temperature T is allowed to expand adiabatically until
its volume becomes 5.66 V, while its temperature
falls to T/2.
(a) How many degrees of freedom do the gas
molecules have ?
(b) Obtain the work done by the gas during the
expansion as a function of the initial pressure P and
volume V.
[IIT-1990]
Sol. (a) According to adiabatic gas equation,
PV −
P
× 5.66 V
11.32
1.4 – 1
PV
2 = PV = 1.25 PV
0.4
2 × 0.4
PV −
=
8.
(a) A white solid is either Na2O or Na2O2. A piece of
red litmus paper turns white when it is dipped into a
freshly made aqueous solution of the white solid.
(i) Identify the substance and explain with balanced
equation.
(ii) Explain what would happen to the red litmus if
the white solid were the other compound.
(b) A, B and C are three complexes of chromium (III)
with the empirical formula H12O 6Cl3Cr. All the three
complexes have water and chloride ion as ligands.
Complex A does not react with concentrated H2SO4,
whereas complexes B and C lose, 6.75% and 13.5%
of their original mass, respectively, an treatment with
conc. H2SO4. Identity A, B and C.
[IIT-1999]
Sol. (a) The substance is Na2O2
TVγ–1 = constant
T1V1γ–1 = T2V2γ–1
T1 = T ;
T2 = T/2
V1 = V
V2 = 5.66 V
T
× (5.66V)γ–1
2
T
× (5.66)γ–1 × V γ–1
2
or
(5.66)γ–1 = 2
Taking log,
or
or P2 =
7.
=
2
= 1.4 – 1 = 0.4
f
1.4 = 1 +
Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ
nRT
0.00429 × 0.0821× 220
=
= 0.221 atm
V
0.350
Hence, TVγ–1 =
or
or
Here,
∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol
of NO
At T = 220 K,
Pressure of the gas,
and
2
f
(b) According to adiabatic gas equation,
∴ Number of moles of N 2O4 formed = 0.00320 mol
N2O4 condenses on cooling,
or
Here,
log 2
0.3010
=
= 0.4
log 5.66
0.7528
or
γ = 1.4
If f, be the number of degrees of freedom, then
2NO + O2 → 2NO2 → N2O4
Composition of gas after completion of reaction,
Number of moles of O 2 = 0
1 mol of O2 react with = 2 mol of NO
P=
γ–1=
or
...(1)
(i) Na2O2 + 2H2O → 2NaOH
(strong base)
(γ – 1)log 5.66 = log 2
+
H2O2
(Weak acid)
H2O2 + red litmus → White
H2O2 → H2O + [O]
XtraEdge for IIT-JEE
13
JANUARY 2010
Nascent oxygen bleaches the red litumus.
4s
4p
3d
4s
4p
3+
Co ion in
Complex ion
d2sp3 hybridization
H3N
NH3
NH3
3+
18
× 100 = 6.75%
266.5
C = [Cr(H2O)4Cl]Cl2.2H2O
Conc. H2SO 4 removes its 2H2O which are outside of
the coordination sphere.
H3N
H3N
or
Co
% loss =
NH3
NH3
NH3
NH3
Co3+
H3N
NH3
NH3
In [Ni(CN)42– nickel is present as Ni2+ ion and its
coordination numbers is four
Ni28 =1s2, 2s22p 6, 3s23p63d8, 4s2
Ni2+ ion = 1s2, 2s22p 6, 3s23p63d8
3d
4s
4p
2+
Ni ion =
18
× 100 = 13.5 %
266.5
Hence complexes A = [Cr(H2O)6]Cl3
B = [Cr(H2O)5Cl]Cl2.H2O
C = [Cr(H2O)4Cl2]Cl.2H2O
% loss = 2 ×
3d
9.
(a) Write the chemical reaction associated with the
"brown ring test".
(b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2–
and [Ni(CO)4]. Write the hybridization of atomic
orbital of the transition metal in each case.
(c) An aqueous blue coloured solution of a transition
metal sulphate reacts with H2S in acidic medium to
give a black precipitate A, which is insoluble in
warm aqueous solution of KOH. The blue solution on
treatment with KI in weakly acidic medium, turns
yellow and produces a white precipitate B. Identify
the transition metal ion. Write the chemical reaction
involved in the formation of A and B. [IIT-2000]
Sol. (a) NaNO3 + H2SO 4 → NaHSO4 + HNO3
2HNO 3 + 6FeSO4 + 3H2SO4 →
3Fe2(SO4)3 + 2NO + 4H2O
[Fe(H2O)6]SO4.H2O + NO
Ferrous Sulphate
→ [Fe(H2O)5NO] SO4 + 2H2O
(Brown ring)
3+
(b) In [Co(NH3)6] cobalt is present as Co3+ and its
coordination number is six.
Co27 = 1s1, 2s22p6, 3s23p 63d7, 4s2
Co3+ion = 1s2, 2s22p 6, 3s23p63d6
XtraEdge for IIT-JEE
3d
Hence
(ii) Na2O + H2O → 2NaOH
NaOH solution turns colour of red litmus paper into
blue due to stronger alkaline nature.
(b) A = [Cr(H2O)6]Cl3. It has no reaction with conc.
H2SO 4 as its all water molecular are present in
coordination sphere.
B = [Cr(H2O)5Cl]Cl2.H2O
Conc. H2SO4 removes its one mol of H2O as it is
outside the coordination sphere.
Molecular Weight of complex = 266.5
4s
4p
2+
Ni ion in
Complex ion
dsp2 hybridization
Hence structure of [Ni(CN)4]2– is
C≡N
N≡C
Ni2+
C≡N
N≡C
In [Ni(CO)4, nickel is present as Ni atom i.e. its
oxidation number is zero and coordination number is
four.
3d
4s
4p
Ni in
Complex
sp3 hybridization
Its structure is as follows :
CO
Ni
OC
CO
CO
14
JANUARY 2010
(c) The transition metal is Cu2+. The compound is
CuSO4.5H2O
CH3CHCH3
OH
2-propanol
CuSO4 + H2S Acidic
 medium

→ CuS ↓ + H2SO4
Black ppt
and hence the structure of (A) should be
2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4
CH3CHCH3
(B) white
NH2
Propan-2-amine
I2 + I– → I3– (yellow solution)
10. A basic volatile, nitrogen compound gave a foul
smelling gas when treated with CHCl3 and alcoholic
KOH. A 0.295 g sample of the substance dissolved in
aqueous HCl and treated with NaNO 2 solution at 0ºC
liberated a colourless; odourless gas whose volume
corresponds to 112 ml at STP. After the evolution of
the gas was complete, the aqueous solution was
distilled to give an organic liquid which did not
contain nitrogen and which on warming with alkali
and iodine gave yellow precipitate. Identify the
original substance. Assume it contains one N-atom
[IIT-1993]
per molecule.
MATHEMATICS
11. Find the values of a and b so that the function
 x + a 2 sin x,
0 ≤ x ≤ π/ 4

f(x) =  2x cot x + b,
π/4 ≤ x ≤ π/2
a cos 2 x − b sin x , π / 2 < x ≤ π

is continuous for 0 ≤ x ≤ π
Sol. As, f(x) is continuous for 0 ≤ x ≤ π
π
π


∴ R.H.L.  at x =  = L.H.L.  at x = 
4
4


Sol. Clue 1. Nitrogen compound gave foul smelling gas
when treated with CHCl3 and alc. KOH (carbylamine
reaction), thus it is a primary amine.
π
π
 π
 π
⇒  2. cot + b  =  + a 2 . sin 
4
4
4
4
 

Clue 2. This compound when treated with HCl +
NaNO2 solution (nitrous acid test) at 0ºC liberates
colourless and odourless gas.
⇒
π
π
+b=
+a
2
4
⇒ a–b=
HCl + NaNO 2
CnH2n+1NH2   → ROH + N 2 ↑
Alcohol
[IIT-1989]
π
4
....(i)
π
π


also, R.H.L  at x =  = L.H.L  at x = 
2
2


Nitrogen
At STP,
112 ml of N2 is evolved from = 0.295 g CnH2n+1NH2
2π
π  π
π


⇒  a cos
− b sin  =  2. . cot + b 
2
2  2
2


∴ 22400 ml of N2 is evolved from
⇒ –a–b =b
⇒ a + 2b = 0
=
0.295 × 22400
= 59 g CnH2n+1NH2
112
∴ CnH2n+1NH2 = 59
or n × C + (2n + 1) × H + N + 2 × H = 59
12. Find
or 12n + 2n + 1 + 14 + 2 × 1 = 59
or n =
42
=3
14
dy
at x = –1, when
dx
(sin y)
Thus the molecular formula of nitrogen compound is
C3H7NH2.
π
sin x
2
+
π
sin x
2
+
Sol. Here,
Clue 3. Alcohol obtained gives iodoform test
positive, thus it is a secondary alcohol and its
structure should be
XtraEdge for IIT-JEE
...(ii)
3π
−3π
and b =
From (i) and (ii), a =
2
4
3
sec–1(2x) + 2x tan ln (x + 2) = 0
2
[IIT-1991]
3
sec–1(2x) + 2x tan (log (x + 2)) = 0
2
Differentiating both sides, we get
(sin y)
15
JANUARY 2010
π
sin x
(sin y) 2
14. If exp {(sin2x + sin4x + sin6x + ...... ∞). ln 2} satisfies
the equation x2 – 9x + 8 = 0, find the value of
π
cos x
,0<x< .
[IIT-1991]
cos x + sin x
2
Sol. exp {(sin2x + sin4x + sin6x + ...... ∞) loge2
π
π
. log(sin y) . cos x .
2
2

π 
 sin x  −1
dy
π 

+  sin x  (sin y)  2  . cos y .
dx
2


2
2 x . sec 2 (log( x + 2))
3
+
+
.
( x + 2)
2 ( 2 | x |) 4 x 2 − 2
x
+ 2 log 2 . tan (log(x + 2)) = 0

3 
, we get
putting,  x = −1, y = −

π 

 dy 
 
 dx   −1, −


3 
π 


− 3 
 π 


=
2
 3

1− 
 π 


2
=
3
sin(2A + B) = sin(C – A) = –sin(B + 2C) =
e log e 2
⇒
⇒
2 tan x satisfy x 2 – 9x + 8 = 0
x = 1, 8
∴
⇒
2 tan x = 1 and 2 tan x = 8
tan2x = 0 and tan2x = 3
⇒
π

x = nπ and tan2x =  tan 
3

and
x = nπ ±
1
2
⇒
2
2
2
x=
2
π
3
π
2
π
 π
∈  0, 
3
 2
1
cos x
1
2
∴
=
×
=
cos x + sin x
1+ 3
1
3
+
2 2
Sol. Given that in ∆ABC, A, B and C are in A.P.
A + C = 2B
A + B + C = 180º
⇒
tan 2 x
Neglecting x = nπ as 0 < x <
If A, B and C are in Arithmetic Progression,
determine the values of A, B and C. [IIT-1990]
=
B = 60º
Also given that,
∴
sin (2A + B) = sin (C – A) = – sin (B + 2C) = 1/2
...(1)
3 −1
3 −1
3 −1
2
cos x
=
cos x + sin x
3 −1
2
15. Find the value of :
1
⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) =
2
cos (2 cos–1 x + sin–1x) at x =
⇒ 2A + 60º = 30º, 150º
and –π/2 ≤ sin–1x ≤ π/2
Sol. cos{2cos–1x + sin–1x}
{neglecting 30º, as not possible}
⇒
⇒
π π2 − 3
13. ABC is a triangle such that
also
⇒
sin 2 x
. log e 2
2
e 1−sin x
2A + 60º = 150º
1
, where 0 ≤ cos–1x ≤ π
5
[IIT-1981]
again from (1), sin (60º + 2c) = –1/2
π
π

= cos cos −1 x +  , as cos–1x + sin–1x =
2
2


–1
= – sin(cos x )
⇒
60º + 2C = 210º, 330º
= – sin(sin–1
⇒
C = 75º or 135º

1 
= – sin  sin −1 1 − 2 

5 

⇒
A = 45º
Also from (1) sin (C – A) = ½
C – A = 30º, 150º, 195º

2 6 
2 6
= – sin  sin −1
=


5
5 

for A = 45º, C = 75º and C = 195º (not possible)
∴
1− x 2 )
C = 75º
Hence, A = 45º, B = 60º, C = 75º
XtraEdge for IIT-JEE
16
JANUARY 2010
Physics Challenging Problems
Set # 9
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in physics that would be very helpful in facing IIT
JEE. Each and every problem is well thought of in order to strengthen the concepts and we
hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma
Director Academics, Jodhpur Branch
So lu tio n s wi ll b e p ub lish ed in n ex t is su e
Q.1 An imaginary closed loop is shown with current
carrying conductors then
4A
c
d
mv
then the particle will hit the upper
qB
plate at P 1 if it is negatively charged
(C) If d ≥
mv
then it will never hit upper plate and
qB
it's motion will be clockwise for positively
charged particle as seen in plane mirror
(D) If d ≥
b
a
3A
Closed Loop
1A
(A) Line integral of magnetic field over the closed
loop abcda is zero
(B) Surface integral of the magnetic field over the
closed loop abcda is non zero
(C) Line integral of magnetic field and surface
integral of magnetic field over the closed loop
abcda both are non zero
(D) Surface integral of magnetic field over the
closed loop is zero but the line integral of
1
magnetic field over the closed loop is 2
C .ε 0
Here C - speed of light in air / free space / vacuum
ε0 - Absolute permittivity of air / free space / vacuum
Q.3 The principal axis of the given concave mirror is
along X-axis. The details about the mirror are also
shown. In the shaded portion there is the coexistence
of uniform magnetic field and electric field the
details are written below
→
E = E0 î ,
Plate-2
d
Plate-1
×
×
×
P
×
×
×
×
×
×
P1
×
×
×
E 0 : B0 = 19.6 : 1
A charged particle is projected from the point
→
(30 cm, 0, 0) with the velocity of v = v 0 ˆj , v0 is a
positive constant. If the particle moves undeviated
then the particle is Y-axis
→
×
×
×
B
×
×
×
Plane Mirror
Concave
mirror
(A)
(B)
(C)
(D)
(A) If the particle is positively charged it's motion
as seen in plane mirror is anticlockwise and on
circular path
mv
(B) If d ≤
then the particle will hit the upper
qB
plate at P 2 if it is negatively charged
XtraEdge for IIT-JEE
C
F
10cm 20cm
Z-axis
Charged
particle
B = B 0 k̂
E0 and B0 are positive constants
Q.2 A charged particle is entering through the tiny hole,
in the given magnetic field between the plates
P2
→
17
30cm
40cm
X-axis
F = Focus
C= centre of curvature
Positively charged
Negatively charged
May be positive of negative
Particle can not pass undeviated through the
pair of transverse magnetic field and electric
field
JANUARY 2010
Q.4 In previous ques.(Q.3) if charged particle is
projected from point (30cm, 0, 0) with the velocity
Particle-A
→
of v = v 0 ˆj and it goes un deviated then (here v0 is
positive constant)
(A) Particle should be negatively charged and
v0 = 19.6 m/s
(B) Particle should be positively charged and
Particle-B
(A) Proton
Electron
(B) Deutron
Electron
(C) Monoionized
helium atom
Electron
(D) Doubly ionized
Electron
v0 = 19.6 m/s
helium atom
(C) Particle is negatively charged and it will reach
up to the maximum height of 10m parallel to
Y-axis if gravity is taken into account
Q.7 A circular current carrying coil is placed in uniform
magnetic field as shown in left column
(D) Particle is negatively charged and when viewed
through the concave mirror it is going parallel
to negative Y-axis maximum 20m below
Column-I
i
(A)
Q.5 An R-L series circuit is shown in figure
Column-II
→
r
o
B
The R-L circuit is in discharging mode and current
i = 18 amp. then
o
r
(B)
B
A
3R
(C)
L↑i
6R
R2
R1
L
×
×
×
i
i
(P) Magnetic force τm = 0
→
B
(Q) Magnetic torque τm = 0
× × × B→
×o r × ×
× × ×
(R) Expansion of coil
i
R1 = R2 = R = 1Ω L = 1/3 Henry
r o
(D)
A, B = Terminals of resistance R 1
(S) Compression of coil
G = Ground terminal
In column II, quantities are in SI units
(A)
(B)
(C)
(D)
Column-I
Total energy dissipated
in resistances
Time constant for
discharging mode
Potential drop across R1
initially
Potential drop across R2
initially
Q.8 Match the followings
Column-I
Column-II
Column-II
(P) 0.25
(A)
(Q) 12
O
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
(x, 0, 0)
×
×
×
×
×
×
s
(P) C2µ0 q net Gauss law
(C = speed of light in
Air/Free space/Vacuum)
(R) 108
(S) 18
Q.6 Two charged particles having same de-Broglie
wavelengths enters in given transverse magnetic
field as shown below
y-axis
→
B
×
×
×
×
×
×
→ →
∫ E. d
→ →
(B)
∫ B. d
(C)
∫ B. d
s
→ →
e
(Q) Zero
(R) Magnetic monopole is
impossible
(D)
×
×
×
×
×
×
→ →
∫ E. d
e
(S) Induced emf Faraday's law of
electromagnetic induction
(T)
x-axis
1
C 2 .ε 0
.i net Ampere's circuital law
z-axis
XtraEdge for IIT-JEE
18
JANUARY 2010
1.
8
Solution
Physics Challenging Problems
Q u es tio n s wer e Pu b lish ed in D ecem b er I ss u e
As information given in:
1st Bright Fringe occurs in front of a slit so,
x
As 2d. n = nλ for nth Bright Fringe
D
x
d
d2
= λb ⇒ λb = 2.
So, 2d. 1 = 1λb ⇒ 2d
D
D
D
For the missing wavelength, destructive interference
should occur in front of slit
λ missing
d
So, 2d. = (2n – 1).
D
2
2d 2
1
⇒ λmissing = 2.
.
D ( 2n − 1)
=
If R eq = 1 =
(a , b )
Slit s1
=
5.
6.
vab = i. R eq = 1 (1) = 1 volt
ε1 ε1
ε
ε
R
ε
–
=0 ⇒ 1 = 2 ⇒ 1 = 1
R1 R1
R1 R 2
R 2 ε2
ε1
R1
R
C
ε 2 R2
Option (A) is correct
D
For n = 1 λmissing = 2λb ≡ = more than λb
2
n = 2 λmissing = . λb ≡ less than λb
3
(λmissing) max. = 2λb , Option (C) is correct
2.
As(lmissing) max. = 2λb
Option (A) is correct
3.
If Slit width are not equal then I1 ≠ I2 and
a1 ≠ a2
So Imin. = Intensity of Dark fringe
= (a1 ~ a2)2
≠ 0 So, dark fringe will be of blue colour
Option (B) is correct
7.
R1R2/R1+R2
R
XtraEdge for IIT-JEE
⇒
C
R2
C
R2
Option (D) is correct
8.
R
To calculate time constant
Replace voltage source by short circuit mean by zero
resistance and then find Req with C and time
constant τ = Req.C
 R R

τ =  1 2 + R  C
 R1 + R 2

R1
The equivalent circuit is
R
(a , b )
(a , b )
If vab= 0, then irrespective to the value of capacitance
C energy stored will be zero.
ε / R + (− ε 2 / R 2 )
=0
vab = 1 1
1 / R1 + 1 / R 2
So,
2
R eq
R eq = 1Ω
So,
x1
R
R 5R
=
= 2R +
2
2
2
Ω,
5
R=
(a , b )
1st B/F
R
2
Ω
5
Option (B) is correct
Slit =s
a
So R =
As,
O
4.
5R
2
Option (A) is correct
2
2
2d 2
=
.
. λb
2n − 1 D
2n − 1
Screen
d
Set # 8
b
19
Maximum current through the resistance
Imax
v
(ε / R ) + ( −ε 2 / R 2 ) / 1 / R 1 + 1 / R 2 ε eq
=
= ab = 1 1
R
R
R
Option (C) is correct.
JANUARY 2010
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
PHYSICS
1.
A sphere of mass 50 g is attached to one end of a
steel wire, 0.315 mm diameter and one metre long. In
order to form a conical pendulum, the other end is
attached to a vertical shaft which is set rotating about
its axis. Calculate the number of revolutions
necessary to extend the wire by 1 mm. Young's
modulus of elasticity of steel = 2 × 1012 dynes/cm 2
and g = 980 cm/sec2.
T sin θ =
or,
1.559 × 106 × 1 =
50 × v 2
100.1
or
1.559 ×10 6 × 100.1
50
v = 1766 cm/sec
∴
Period of revolution =
∴
Sol. Let T be the tension in the wire, when the extension
is 1 mm. According to definition,
mv 2
r
Now,
v2 =
2πr
2π × 100.1
=
v
1766
= 0.3561 s
So, Frequency of revolution =
θ
θ
T cos θ
θ
= 2.808/sec
L = 1m
T
2.
T sin θ
mg
Y=
A wave travels out in all directions from a point
source. Justify the expression y = (a0/r) sin K(r – vt),
at a distance r from the source. Find the speed,
periodicity and intensity of the wave.
Sol. If P be the power of the source then intensity
Tensile stress
T/A
T
L
=
×
=
Tensile strain
l/L
A
l
I=
L
=
×
2
l
πr
T
∴ T=
2 × 1012 × π × (0.01575) 2 × 0.1
Yπr 2 l
=
L
100
= 1.559 × 10 6 dynes
∴ sin θ =
or cos θ =
4πr 2
1
r2
or,
I∝
But
I ∝ a2,
so
a∝
a
1
or a = 0
r
r
a0
sin K(r – vt)
r
comparing this equation with
y = a sin (Kr – ωt)
y=
mg
50 × 980
=
T
1.559 ×10 6
Now,
(1 − cos 2 θ) = 0.9998 = 1.00 (nearly)
We have, ω = Kv or
∴ Radius of the circle described = r = (L + l) sin θ
K=
= 100.1 cm
XtraEdge for IIT-JEE
P
Where a0 is constant.
The equation in standard form is, y = a sin K (r – vt)
Therefore, above equation is written as :
When the sphere is revolving, it is acted upon by two
forces namely the tension T along the wire and its
weight mg acting vertically downwards. Resolving T
into vertical and horizontal components, we get
T cos θ = mg
1
1
=
period
0.3561
20
2π
λ
or
n=
Kv
2π
λ=
2π
K
and
JANUARY 2010
4.
 Kv   2π 
Speed c = nλ = 
 ×  = v
 2π   K 
1 2π
=
n Kv
Thus, intensity is given by
Also,
or
3.
T=
I=
1 2 2
1 a2
ρa ω c = ρ 20 . K2v2 . v
2
2 r
I=
1 ρ a 20 K 2 v 3
2
r2
Two coherent light sources emit light of wavelength
550 nm which produce an interference pattern on a
screen. The sources are 2.2 mm apart and 2.2 m from
the screen. Determine whether the interference at the
point O is constructive or destructive. Calculate the
fringe width.
S1
D
l
A conducting bar of mass m, length l is pushed with a
speed v0 on a smooth horizontal conducting rail
containing an inductance L. If the applied magnetic
field has inward field of induction B, find the
maximum distance covered by the bar before it stops.
S2
2d
O
Sol. The path difference at O is given by
∆ = S2O – S1O
From figure, S2O = [l 2 + (2d)2]1/2
m
L
B
⊗
v0
l
Sol. If the bar slides a distance dx, the flux linkage
– dφ = Bldx
The induced e.m.f. =
or,
dx
dI
=–L
dt
dt
LdI = Bl dx
or,
LI = Bl x
or,
Bl
x
I=
L
∴
– Bl
∴
B2 l 2
v dv = –
v0
ML
or,
∫
∴
v
s = ( mL ) 0
Bl
XtraEdge for IIT-JEE
n=
β=
∆ 1.1× 10 −6
=2
=
λ 5.5 × 10 −7
∫
λD 5.5 ×10 −7 × 2.2
=
2d
2.2 × 10 −3
= 5.5 × 10–4 m = 0.55 mm
dv
B2 l 2
=–
x
dx
L
0
(2.2 ×10 −3 ) 2
= 1.1 × 10–6 m
2 × 2.2
Fringe width,
5.
Light from a discharge tube containing hydrogen
atoms falls on the surface of a piece of sodium. The
kinetic energy of the fastest photoelectrons emitted
from sodium is 0.73 eV. The work function for
sodium is 1.82 eV. Find
(a)
the energy of the photons causing photoelectric
emission
 Bl 
F = IlB = –  x  l B
 L 
mv
 1  2 d  2  (2 d ) 2
∆ = l 1 +   − 1 =
2l
 2  l 

The difference will be constructive if path
difference is an integral multiple of wavelength i.e.,
n = 1, 2, 3, …….
This induced current interacts with the applied
magnetic field of induction and imparts a restoring
(magnetic) force
or,
Now,
=
dI
dt
 1  2d  2 
= l 1 +   
 2  l  
  2d  2 
S2O = l 1 +   
  l  
dφ
dx
= – Bl
dt
dt
Since the induced e.m.f. across the inductor = – L
1/ 2
∴
B2 l 2 s 2
x dx = –
0
2mL
s
(b) the quantum numbers of the two levels involved in
the emission of these photons
21
JANUARY 2010
(c)
the change in the angular momentum of the hydrogen
atom in the above transition, and
p Ph =
(d) the recoil speed of the emitting atoms assuming it to
be at rest before the transition.
hν 2.55 × 1.6 × 10 −19 J
= 1.36 × 10–27 Kg.m/s
=
c
3 ×10 8 m / s
According to the law of conservation of momentum,
the recoil momentum of a hydrogen atom will be
equal and opposite to the momentum of the emitted
photon.
(Ionization potential of hydrogen is 13.6 volt and the
mass of the hydrogen atom is 1.67 × 10 –27 Kg, 1 eV =
1.6 × 10 –19 J)
→
→
→
( p Ph + p A = 0 or
Sol. (a) According to Einstein’s photo-electric equation,
the maximum kinetic energy Ek of the emitted
electrons is given by
→
p A = – p Ph )
Hence, the recoil speed of the atoms is :
→
→
| Momentum | | p A | | p Ph |
=
=
V=
mass
mA
mA
EKmax = hν – W,
Where hν is the energy of photons causing the photoelectric emission and W is the work-function of the
emitting surface.
=
1.36 × 10 −27 kg − m / s
1.67 × 10 − 27 kg
= 0.814 m/s
Given that, EKmax = 0.73 eV and W = 1.82 eV
∴ hν = E Kmax + W = 0.73 eV + 1.82 eV = 2.55 eV
SCIENCE TIPS
(b) These photons (where energy is 2.55 eV) are emitted
by hydrogen atoms.
As
• A porcelain funnel used for filtration by suction is
known as
® Bucher Funnel
• What is diazomethane ?
(I.E.)H = 13.6 eV, hence
E1H
= – (I.E.)H = – 13.6 eV
The energy of higher levels is given by
E aH =
13.6
= – 3.4 eV
4
• Reforming of a gasoline fraction to increase
branching in presence of AlCl3 is known as
® Isomerization
13.6
–
= – 1.5 eV
9
• A condenser consisting of glass tube surrounded by
another glass tube through which cooling water
® Liebig condenser
flows is known as
13.6
= – 0.85 eV
16
The energy of the emitted photon is 2.55 eV
(c)
–
• A drying chamber, containing chemicals such as
concentrated sulphuric acid or silica gel is known as
® Desiccator
n2
Hence, E H
2 =–
E 3H =
+
® [CH 2 = N = N or CH 2 N 2 ]
E1H
and
EH
4 =–
Now
H
EH
4 – E 2 = – 0.85 – (– 3.4) = 2.55 eV
• For wattles current what should be the value of the
® Zero
power factor of the circuit ?
Thus, the quantum numbers of two levels involved in
the emission of photon of energy 2.55 eV are 4 and 2.
• For which colour is the critical angle of light, pasing
® Violet
from glass to air, minimum ?
The electron transition causing the emission of
photon of energy 2.55 eV is from n = 4 level to n = 2
level. Now, according to Bohr’s 2 nd postulate, the
angular momentum of electron in the hydrogen atom
is (n h/2π). Thus, the change in angular momentum in
above transition is
• Give an example of application of mutual induction
in any device.
® Transformer
• What is the correct sequence of the semiconductors
silicon, tellunium and germanium in the increasing
order of their energy gap ?
® Tellurium, germanium, silicon
• Which ammeter is used to measure alternating
current ?
® Hot wire ammeter
4h 2h h
–
=
∆L =
2π 2π π
• What quantity has the ampere-second as its unit ?
(d) The momentum of the photon emitted from the
hydrogen atom
XtraEdge for IIT-JEE
® Quantity of electricity
22
JANUARY 2010
XtraEdge for IIT-JEE
23
JANUARY 2010
P HYSICS F UNDAMENTAL F OR IIT-J EE
Refraction at plane & curved surface
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Laws of Refraction :
(e) For a given time, optical path remains constant.
The incident ray, the refracted ray and normal on
incidence point are coplanar.
i.e., µ1x1 = µ2x2 = ... constant
µ 1 sin θ1 = µ 2 sin θ2 = ... = constant.
θ1
µ1
∴
µ1c1 = µ2c2
∴
µ2
c
= 1
µ1
c2
i.e.,
µ∝
µ1
µ2
θ2
Snell's law in vector form :
(where c1 and c2 are speed
of light in respective mediums)
1
c
(f) The frequency of light does not depend upon
medium.
n̂
ê1
dx1
dx
= µ2 2
dt
dt
∴
µ1
ê 2
∴
c1 = fλ 1,
∴
µ1
c
λ
= 2 = 2
µ2
c1
λ1
∴
µ∝
µ2
c2 = fλ 2
1
λ
When observer is rarer medium and object is in
denser medium :
Let, ê1 = unit vector along incident ray
ê 2 = unit vector along refracted.
Then
n̂ = unit vector along normal on incidence point.
µ=
real depth
apparent depth
Then µ1( ê1 × n̂ ) = µ 2( ê 2 × n̂ )
Air
Observer
Some important points :
(a) The value of absolute refractive index µ is always
greater or equal to one.
(b) The value of refractive index depends upon
material of medium, colour of light and
temperature of medium.
Denser medium
(µ)
P
Object
(c) When temperature increases, refractive index
decreases.
When object is in rarer and observer is in denser
medium :
(d) Optical path is defined as product of geometrical
path and refractive index.
µ=
i.e., optical path = µx
XtraEdge for IIT-JEE
Apparent
depth P
Real
depth
24
apparent position
real position
JANUARY 2010
 1
The shift of object due to slab is x = t 1 – 
 µ
90º
t
µ
P
Q
P´
Mathematically,
Object
shiftness
c
Denser
µ2
sin c =
µ1
µ2
Rarer medium
(µ1)
(a) This formula is only applicable when observer is
in rarer medium.
r
(b) The object shiftiness does not depend upon the
position of object.
(c) Object shiftiness takes place in the direction of
incidence ray.
number of slabs for normal incidence is µ =
µ1
t1
µ2
t2
i
c
i<c
i=c
i i
Denser medium
(µ2)
(i) When angle of incidence is lesser than critical
The equivalent refractive index of a combination of a
angle, refraction takes place. The corresponding
deviation is
Σt i
t
Σ i
µi
µ

δ = sin–1  2 sin i  – i
µ
 1

for i < c
(ii) When angle of incidence is greater than critical
angle, total internal reflection takes place. The
corresponding deviation is
δ = π – 2i
Σti = t1 + t2 + ...
Here,
Rarer
µ1
when i < c
The δ – i graph is :
t
t
t
Σ i = 1 + 2 + ...
µi
µ1
µ2
(i) Critical angle depends upon colour of light,
material of medium, and temperature of medium.
The apparent depth due to a number of media is Σ
ti
µi
(ii) Critical angle does not depend upon angle of
incidence
The lateral shifting due to a slab is d = t sec r sin(i – r).
i
µ
δ
t
r
i
d
c
π/2
Refractive surface formula,
µ2
µ
µ − µ1
– 1 = 2
v
u
r
Critical angle : When a ray passes from denser
medium (µ2) to rarer medium (µ 1), then for 90º angle
of refraction, the corresponding angle of incidence is
critical angle.
Here, v = image distance,
u = object distance,
r = radius of curvature of spherical surface.
XtraEdge for IIT-JEE
25
JANUARY 2010
(a) For plane surface , r = ∞
(a) Thin lens formula is only applicable for paraxial
ray.
(b) Transverse magnification,
m=
(b) This formula is only applicable when medium on
both sides of lens are same.
Im age size
µv
= 1
object size
µ2u
(c) Intensity
aperture.
(c) Refractive surface formula is only applicable for
paraxial ray.
is
proportional
to
square
of
(d) When lens is placed in a medium whose refractive
Lens :
index is greater than that of lens. i.e., µ 1 > µ2.
Then converging lens behaves as diverging lens
and vice versa.
Lens formula :
1
1
1
–
=
v
u
f
(e) When medium on both sides of lens are not same.
Then both focal lengths are not same to each
other.
(a) Lens formula is only applicable for thin lens.
(b) r = 2f formula is not applicable for lens.
(c) m =
image size
v
=
object size
u
(f) If a lens is cut along the diameter, focal length
does not change.
(d) Magnification formula is only applicable when
object is perpendicular to optical axis.
(g) If lens is cut by a vertical, it converts into two
lenses of different focal lengths.
(e) lens formula and the magnification formula is
only applicable when medium on both sides of
lenses are same.
i.e.,
1
1
1
=
+
f1
f2
f
(f)
+
f(+ve)
f(–ve)
(i)
(ii)
f1
f
f1
(h) If a lens is made of a number of layers of different
refractive index (shown in figure)
f(+ve)
f(–ve)
(iii)
µ1
µ2
µ3
(iv)
+++
+++
µ4
µ5
f(–ve)
f(+ve)
µ6
(v)
(vi)
Then number of images of an object by the lens is
equal to number of different media.
(g) Thin lens formula is applicable for converging as
well diverging lens. Thin lens maker's formula :
 µ − µ1 
1

=  2
f
 µ1 
(i) The minimum distance between real object and
real image in is 4f.
1 1
 − 
 r1 r2 
µ1
(j) The
µ1
focal
length
of
co-axial
1
1
1
d
=
+
–
F
f1 f 2
f1f 2
µ2
XtraEdge for IIT-JEE
equivalent
combination of two lenses is given by
26
JANUARY 2010
f1
f2
d<f1
In 1 st case,
d<f2
o2
o1
In 2 nd case,
d
(k) If a number of lenses are in contact, then
1
1
1
=
+
+ ......
F
f1 f 2
(l) (i) Power of thin lens, P =
1
F
(ii) Power of mirror is P = –
1
F
2.
u − a q +1
=
f
q
(Q m = p)
or,
or,
p +1 a q +1
– =
p
f
q
or,
a pq + q − pq − p q − p
=
=
f
pq
pq
∴
f=
u a q +1
– =
f f
q
or, =
a p +1 q +1
–
–
f
p
q
apq
q−p
A convex refracting surface of radius of curvature
30 cm separates two media of refractive indices
n1 = 4/3 and n2 = 3/2 respectively. Find the position
of image formed by refraction of an object placed at a
(m) If a lens silvered at one surface, then the system
behaves as an equivalent mirror, whose power
P = 2P L + Pm
distance of (i) 280 cm and (ii) 80 cm, from the
surface.
Here, PL = Power of lens
 µ − µ1 

=  2
 µ1 
u p +1
=
f
p
1 1
 − 
 r1 r2 
Pm = Power of silvered surface = –
Sol. (i) Given that n1 = 4/3,
n2 = 3/2, |u| = 280 cm, |R| = 30cm
1
Fm
For refraction through a spherical surface :
n 2 n 1 (n 2 − n 1 )
–
=
v
u
R
Here, Fm = r2/2, where r2 = radius of silvered surface.
Here,
u = – 280 cm, R = + 30 cm. Hence
3
4
[(4 / 3) − (3 / 2)]
–
=
2 v − 3 × 280
+3
P = – 1/F
Here, F = focal length of equivalent mirror.
Solved Examples
1.
object magnified p times. The magnification becomes
q when the lens is moved nearer to the object by a
distance a. Calculate the focal length of the lens.
XtraEdge for IIT-JEE
or
∴
3
1
1
1
=
–
=
2 v 180 210 1260
or,
v = (3/2) × 1260 = 1890 cm = 18.9 m
(ii) In this case, u = – 80 cm, R = + 30 cm
Again from the formula for refraction through a
surface,
(3/2v) – [4/–3 × 80)] = [{(4/3) – (3/2)} / + 30]
or
(3/2v) + (1/60) = (1/180)
or
3/2v = [(1/180) – (1/60)]
Sol. The magnification (m) produced by a lens in terms of
u and f i : given by
f
u−f
3
1
1
+
=
2 v 210 180
As v is positive, hence the image is real and is
formed in second medium at a distance of 18.9 m
from the refracting surface.
A thin converging lens forms the image of a certain
m=
or
u m +1
=
f
m
or
27
v = (3/2) × (– 90) = – 135 cm
JANUARY 2010
Screen
As v is negative, hence the image is virtual and is
formed in the first medium of refractive index 4/3 at
a distance of 135 cm from the pole.
3.
O
I1
I
There is a small air bubble in side a glass sphere (n =
10 cm
1.5) of radius 10 cm. The bubble is 4 cm below the
surface and is viewed normally from the outside
1 
 1

I1I = t 1 −  = (1.5 cm) 1 −
 = 0.5 cm.
 n
 1.5 
(Fig.). Find the apparent depth of the air bubble.
P
I
A
Thus, the lens forms the image at a distance of 9.5 cm
n2 = 1
from itself. Using
O
C
1 1 1
– = , we get
v u f
n1 = 1.5
or
1 1 1
1
1
= – =
–
u v f 9.5 10
u = – 190 cm.
i.e. the object should be placed at a distance of
Sol. The observer sees the image formed due to refraction
190 cm. from the lens.
at the spherical surface when the light from the
bubble goes from the glass to air.
Here u = – 4.0 m,
We have
5.
R = – 10 cm, n1 = 1.5 and n2 = 1
Where must a convex lens of focal length 8 inches be
[(n2/v) – (n1/u) = (n2 – n1)/R
or
(1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm)
or
(1/v) = (0.5/10 cm) – (1.5/4.0 cm)
or
v = – 3.0 cm
A candle is placed at a distance of 3 ft from the wall.
placed so that a real image is formed on the wall ?
Sol. According to formula for refraction though a lens
36 – v
v
Thus, the bubble will appear 3.0 cm below the
surface.
f = 8"
d = 3 ft = 36"
4.
A convex lens focuses a distance object on a screen
1 1 1
– =
v u f
placed 10 cm away from it. A glass plate (n = 1.5) of
thickness 1.5 is inserted between the lens and the
screen. Where should the object be placed so that its
or
image is again focused on the screen ?
1
1
1
+
=
v 36 − v 8
or
1
1
1
–
=
v − (36 − v) 8
or
36 − v + v 1
=
v(36 − v) 8
or, v2 – 36 v + 8 × 36 = 0
Sol. The situation when the glass plate is inserted between
the lens and the screen, is shown in fig. The lens
or v = 12"
or
24" = 1 ft or 2 ft.
forms the image of object O at point I1 but the glass
∴
or
12" = 2 ft or 1 ft
plate intercepts the rays and forms the final image at I
Hence, lens should be placed at either 1 ft or 2 ft
on the screen. The shift in the position of image after
away from the wall.
insertion of glass plate
XtraEdge for IIT-JEE
u = 24"
28
JANUARY 2010
P HYSICS F UNDAMENTAL F OR IIT-J EE
Properties of Matter
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Key Concepts :
Stress :
The restoring force setup inside the body per unit
area is known as stress.
Restoring forces : If the magnitude of applied
deforming force at equilibrium = F
φ
Shear strain
Stress-strain graph :
From graph, it is obvious that in elastic limit, stress is
proportional to strain. This is known as Hooke's law.
∴ Stress ∝ Strain
F
A
In SI system, unit of stress is N/m2.
Difference between pressure and stress :
(a) Pressure is scalar but stress is tensor quantity.
(b) Pressure always acts normal to the surface, but
stress may be normal or tangential.
(c) Pressure is compressive in nature but stress may
be compressive or tensile.
Strain :
Then,
∴ Stress = E .strain
∴ E=
change in dimension
original dimension
(a) Longitudinal strain =
stress
strain
where E is proportionality dimensional constant
known as coefficient of elasticity.
Plastic
region
Breaking
B C
strength
Stress
Strain =
Stress =
∆L
L
Elastic
limit
A
L
F
F
O
Longitudinal strain is in the direction of
deforming force but lateral strain is in
perpendicular direction of deforming force.
Poisson ratio :
σ=
Types of coefficient of elasticity :
(a) Young's modulus = Y =
lateral strain
∆d/D
=
longitudinal strain
∆L/L
∴
Here ∆d = change in diameter.
(b) Volumetric strain =
Strain
Y=
logitudinal stress
longitudinal strain
F
FL
=
∆L
A∆L
A
L
∆V
V
L
F
∆L
F
F
F
(c) Shear strain = φ
XtraEdge for IIT-JEE
F
(b) Bulk modulus = B =
V
volumetric stress
volumetric strain
Compressibility = 1/B
29
JANUARY 2010
(c) Modulus of rigidity = η =
Surface tension :
F
shear stress
=
Aφ
shear strain
F
L
Here L = length of imaginary line drawn at the
surface of liquid. and F = force acting on one side of
line (shown in figure)
(a) Surface tension does not depend upon surface
area.
(b) When temperature increases, surface tension
decreases.
(c) At critical temperature surface tension is zero.
T=
(d) For isothermal process, B = P.
F
φ
φ
F
(e) For adiabatic process, B = γP
(f)
Adiabatic bulk modulus
=γ
Isothermal bulk modulus
(g) Esolid > E liquid > E gas
F
(h) Young's modulus Y and modulus of rigidity η
exist only for solids.
(i) Bulk modulus B exist for solid, liquid and gas.
(j) When temperature increases, coefficient of
elasticity (Y, B, η) decreases.
(k)
L
F
1
3
9
+
=
B
η
Y
Rise or fall of a liquid in a capillary tube :
h=
(l) Y = 2(1 + σ)η
(m) Poisson's ratio σ is unitless and dimensionless.
Theoretically,
Here
1
–1 < σ <
2
1
1
1
× load × extension = Fx = kx2
2
2
2
= stress × strain × volume
For twisting motion,
U=
(a) For a drop of radius R, W = 4πR2T
(b) For a soap bubble, W = 8πR2T
Excess pressure :
1
U=
× torque × angular twist
2
1
1
τ × θ = cθ2
2
2
Elastic energy density,
(a) For drop, P =
=
2T
R
(b) For soap bubble, P =
1
1
× stress × strain J/m3 = Y × strain2J/m3
2
2
Thermal stress = Yα∆θ and Thermal strain = α∆θ
Work done in stretching a wire :
u=
4T
R
Viscosity :
(a) Newton's law of viscous force :
F = – ηA
1
F∆L
2
where
1
× stress × strain
2
(c) Breaking weight = breaking stress × area
(b) Work done per unit volume =
XtraEdge for IIT-JEE
θ = angle of contact.
r = radius of capillary tube
ρ = density of liquid
For a given liquid and solid at a given place,
hr = constant
Surface energy :
Surface energy density is defined as work done
against surface tension per unit area. It is numerically
equal to surface tension.
W = work = surface tension × area
1
Practically,
0<σ<
2
(n) Thermal stress = Yα∆θ
(o) Elastic energy stored,
(a) W =
2T cos θ
rρg
dv
dy
dv
= velocity gradient
dy
A = area of liquid layer
η = coefficient of viscosity
The unit of coefficient of viscosity in CGS is poise.
30
JANUARY 2010
or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.
(b) SI unit of coefficient of viscosity
= poiseuille = 10 poise.
(c) In the case of liquid, viscosity increases with
density.
(d) In the case of gas, viscosity decreases with
density.
(e) In the case of liquid, when temperature increases,
viscosity decreases.
(f) In the case of gas, when temperature increases,
viscosity increases.
Poiseuille's equation :
V=
Y=
or l =
Pπr 4
8ηL
FL
=
πr 2 Y
Hence,
or d'2 =
5.0 ×1000 × 980 × 200
3.142 × (0.25) 2 × 1.1× 10 2
4F
πd ' 2
= 1.5 × 10 9
4F
2.
F = 6πηrv
where r = radius of spherical body
Determination of η :
2r 2 (ρ − σ)g
9v
where r = radius of spherical body moving with
constant velocity v in a viscous liquid of coefficient
of viscosity η and density ρ
9
A uniform horizontal rigid bar of 100 kg in supported
horizontally by three equal vertical wires A, B and C
each of initial length one meter and cross-section I
mm2. B is a copper wire passing through the centre of
the bar; A and C are steel wires and are arranged
symmetrically one on each side of B YCu = 1.5 × 1012
dynes / cm2, Y s = 2 × 10 12 dynes/cm2. Calculate the
tension in each wire and extension.
kη
ρr
As Y =
where k = Reynold's number for narrow tube, k ≈ 1000.
(a) For stream line motion, flow velocity v < v0.
(b) For turbulant motion, flow velocity v > v0.
Stress
Strain
A
S
Solved Examples
B
Cu
C
S
100 Kg
A mass of 5 kg is suspended from a copper wire of 5
mm diameter and 2 m in length. What is the
extension produced in the wire ? What should be the
minimum diameter of the wire so that its elastic limit
is not exceed ? Elastic limit for copper = 1.5 × 10 9
dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2.
FCu / A
Strain
Hence,
YCu =
and
Ys =
∴
YCu FCu 1.5 3
=
=
=
YS
FS
2 4
… (1)
Fs / A
Strain
… (2)
or 4FCu = 3FS
...(3)
According to figure, we can write
2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g
Sol. Given that Y = 1.1 × 10 12 dynes/cm2,
L = 2m = 200 cm, d = 5 mm = 0.5 cm
XtraEdge for IIT-JEE
4 × 5.0 × 1000 × 980
Sol. The situation is shown in figure. Because the rod is
horizontally supported, hence extensions in all the
wires must be equal i.e., strains in all the wires are
equal as initial lengths are also equal.
and
σ = density of spherical body
Critical velocity (v 0) :
1.
=
π × 1.5 × 10
3.142 × 1.5 × 10 9
–4
= 41.58 × 10
d' = 0.0645 cm.
η = coefficient of viscosity
and P = pressure difference between ends of the tube
Stoke's law :
The viscous force acting on a spherical body moving
with constant velocity v in a viscous liquid is
v0 =
πr 2 l
= 4.99 × 10–3 cm
Also, elastic limit for copper = 1.5 × 10 9 dynes/cm2
If d' is the minimum diameter, then maximum stress
F
4F
=
on the wire =
πd '2 / 4 πd' 2
where V = the volume of liquid flowing per second
through a capillary tube of length L and radius r
η=
FL
or
31
[(8/3) + 1] FCu = 100 g
JANUARY 2010
∴
4.
FCu = (3/11) × 100g
= (3/11) × 100 Kgwt = 27.28 Kgwt
FS = (4/3) FCu = (4/3) × (3/11) × 100g
and
= (400/11)g = 36.36 Kgwt
Extension in each wire,
l=
3.
FCu L 27280 × 980 × 100
= 0.178 cm
=
AYCu
10 − 2 × 1.5 × 1012
A ring is cut from a platinum tube 8.5 cm internal
and 8.7 cm external diameter. It is supported
horizontally from a pan of a balance so that it comes
in contact with the water in a glass vessel. What is
the surface tension of water if an extra 3.97 gm
weight is required to pull it away from water (g = 980
cm/sec2).
Sol. The ring is in contact with water along is inner and
outer circumference. So when pulled out the total
force on it due to surface tension will be
A copper rod of length L and radius r is suspended
from the ceiling by one of its ends. Find: (a) the
elongation of the rod due to its own weight when ρ
and Y are density and Young's modulus of the copper
respectively, (b) the elastic potential energy stored in
the rod due to its own weight.
F
T
Sol. (a) Consider any length x of the rod from the fixed
end. Weight of lower portion of rod will exert
stretching force on the upper portion.
∴
T=
where m is the mass per unit length of the rod
(m = πr 2ρ)
5.
[Q Strain = (Stress/Y)]
Hence, increase in length in elementary length dx at
πr 2 Y
dx =
L
0
( L − x )mg
h=
πr 2 Y
L2 πr 2ρgL2 ρgL2
=
=
2Y
πr 2 Y 2
2πr 2 Y
mg
(b) Energy density at x =
=
∫
1
× Strain × Stress
2
∴ h=
hr =
1 (L − x ) 2 ρ 2 g 2
× πr2dx
2
Y
∴Total potential energy stored in the rod of length L,
dx =
XtraEdge for IIT-JEE
∫
L
0
(L − x ) 2 dx =
… (1)
2 × (7.0 × 10 −2 N / m )
( 2.5 ×10 − 3 ) × (1× 10 3 Kg / m 3 ) × (10 N / Kg )
= 5.6 mm
According to equation (1), for the same liquid, we
have
1 (L − x ) 2 ρ 2 g 2
= .
2
Y
∴ Energy stored in the volume of length element
1 πr 2ρ 2 g 2
2
Y
2T cos θ
rρg
where T is surface tension, ρ is density and θ is angle
of contact of water-glass which can be assumed zero.
For the first tube, r = 2.5 mm = 2.5 × 10 –3 m
1 (L − x ) mg ( L − x )mg
×
×
2
πr 2
πr 2 Y
U=
Two long capillary tubes of diameter 5.0 mm and 4.0
mm are held vertically inside water one by one. How
much high the water will rise in each tube ? (g = 10
m/s 2, surface tension of water = 7.0 × 10–2 N/m).
Sol. Height of water column in a capillary tube of radius r
is given by
dx
∴ Total increase in length =
mg
3.97 × 980
=
2π(r1 + r2 ) 2 × 3.14 × (8.5 + 8.7)
= 72.13 dyne/cm.
∴ Strain at x = (L – x)mg/πr2Y
(L − x ) mg
T
F = T (2πr1 + 2πr2)
∴ Stress over the portion at a distance x from fixed
end = (L – x) mg/πr2
x=
F
2T cos θ
= Constant
ρg
If a liquid rises to a height h1 in a capillary tube of
radius r 1 and to a height of h2 in a capillary tube of
radius r 2, then
1 πr 2ρ 2 L3g 2
6
Y
h1r1 = h2r2
32
or
h2 =
h1r1 5.6 × 2.5
=
= 7.0 mm
r2
2.0
JANUARY 2010
KEY CONCEPT
Organic
Chemistry
Fundamentals
CARBOHYDRATES
By 1895 it had become clear that the picture of D(+)-glucose as a pentahydroxy aldehyde had to be
modified.
Among the facts that had still to be accounted for
were the following:
(a) D-(+)-Glucose fails to undergo certain
reactions typical of aldehydes. Although it is
readily oxidized, it gives a negative Schiff test and
does not form a bisulfite addition product.
(b) D-(+)-Glucose exists in two isomeric forms
which undergo mutarotation. When crystals of
ordinary D-(+)-glucose of m.p. 146ºC are dissolved
in water, the specific rotation gradually drops from an
initial + 112º to + 52.7º. On the other hand, when
crystals of D-(+)-glucose of m.p. 150ºC (obtained by
crystallization at temperatures above 98ºC) are
dissolved in water, the specific rotation gradually
rises from an initial + 19º to + 52.7º. The form with
the higher positive rotation is called α-D-(+)-glucose
and that with lower rotation β-D-(+)-glucose. The
change in rotation of each of these to the equilibrium
value is called mutarotation.
(c) D-(+)-Glucose forms two isomeric methyl Dglucosides. Aldehydes react with alcohols in the
presence of anhydrous HCl to form acetals. If the
alcohol is, say methanol, the acetal contains two
methyl groups :
H
H
H
Definition and Classification :
are
polyhydroxy
aldehydes,
Carbohydrates
polyhydroxy ketones, or compounds that can be
hydrolyzed to them. A carbohydrate that cannot be
hydrolyzed to simpler compounds is called a
monosaccharide. A carbohydrate that can be
hydrolyzed to two monosaccharide molecules is
called a disaccharide. A carbohydrate that can be
hydrolyzed to many monosaccharide molecules is
called a polysaccharide.
A monosaccharide may be further classified. If it
contains an aldehyde group, it is known as an aldose;
if it contains a keto group, it is known as a ketose.
Depending upon the number of carbon atoms. It
contains, a monosaccharide is known as a triose,
tetrose, pentose, hexose, and so on. An aldohexose,
for example, is a six-carbon monosaccharide
containing an aldehyde group; a ketopentose is a
five-carbon monosaccharide containing a keto group.
Most naturally occurring monosaccharides are
pentoses or hexoses.
Carbohydrates that reduce Fehling’s (or Benedict’s)
or Tollens’ reagent are known as reducing sugars.
All monosaccharides, whether aldose or ketose, are
known as reducing sugars. Most disaccharides are
reducing sugars; sucrose (common table sugar) is a
notable exception, for it is a non-reducing sugar.
(+)-Glucose : an aldohexose :
Because it is the unit of which starch, cellulose, and
glycogen are made up, and because of its special role
in biological processes, (+)-glucose is by far the most
abundant monosaccharide- there are probably more
(+)-glucose units in nature than any other organic
group–and
by
far
the
most
important
monosaccharide.
Cyclic structure of D-(+)-glucose. Formation of
glucosides :
D-(+)-glucose is a pentahydroxy aldehyde. D-(+)glucose had been definitely proved to have structure.
CHO
H
OH
H
HO
OH
H
OH
H
–C=O
–C–OCH3
OH
CH3 OH,H+
–C–OCH3
OCH3
Aldehyde
Hemiacetal
Acetal
When D-(+)-glucose is treated with methanol and
HCl, the product, methyl D-glucoside, contains only
one –CH3 group; yet it has properties resembling
those of a full acetal. It does not spontaneously revert
to aldehyde and alcohol on contact with water, but
requires hydrolysis by aqueous acids.
Furthermore, not just one but two of these
monomethyl derivatives of D-(+)-glucose are known,
one with m.p. 165ºC and specific rotation + 158º, and
the other with m.p. 107 ºC and specific rotation –33º.
The isomer of higher positive rotation is called
methyl α-D-glucoside, and the other is called methyl
β-D-glucoside. These glucosides do not undergo
mutarotation, and do not reduce Tollens’ or Fehling’s
reagent.
CH2 OH
D-(+)-Glucose
XtraEdge for IIT-JEE
CH3 OH,H+
33
JANUARY 2010
We shall study four disaccharides : (+)-maltose (malt
sugar), (+)-cellobiose, (+)-lactose (milk sugar), and
(+)-sucrose (cane or beet sugar).
D-(+)-Glucose has the cyclic structure represented
crudely by IIa and IIIa, more accurately by IIb and IIIb.
1
2
3
4
5
H
H
HO
H
H
6
OH
OH O
H
OH
6
CH2OH
4
H
OH
H
3
HO
H
(+)-Maltose :
O H
5
2
H
(+)-Maltose can be obtained, among other products,
by partial hydrolysis of starch in aqueous acid. (+)Maltose is also formed in one stage of the
fermentation of starch to ethyl alcohol; here
hydrolysis is catalyzed by the enzyme diastase, which
is present in malt (sprouted barley).
1
OH
OH
CH2OH
IIa
IIb
α-D-(+)-Glucose (m.p. 146 ºC, [α] = +112º)
1 HO
2 H
3 HO
4 H
5 H
6
H
OH O
H
OH
6
CH2OH
4
H
OH
H
HO
5
3
H
CH2OH
IIIa
Let us look at some of the facts from which the
structure of (+)-maltose has been deduced.
O OH
2
H
(+)-Maltose has the molecular formula C12H22O 11. It
reduces Tollens’ and Fehling’s reagents and hence is
a reducing sugar. It reacts with phenylhydrazine to
yield an osazone, C12H20O9(=NNHC6H5)2. It is
oxidized by bromide water to a monocarboxylic acid,
(C11H21O10)COOH, maltobionic acid. (+)-Maltose
exists in alpha ([α] = + 168º) and beta ([α] = + 112º)
forms which undergo mutarotation in solution
(equilibrium [α] = + 136º).
1
H
OH
IIIb
β-D-(+)-Glucose (m.p. 150 ºC, [α] = +19º)
D-(+)-Glucose is the hemiacetal corresponding to
reaction between the aldehyde group and the C-5
hydroxyl group of the open-chain structure. It has a
cyclic structure simply because aldehyde and alcohol
are part of the same molecule.
(+)-Cellobiose :
When cellulose (cotton fibers) is treated for several
days with sulfuric acid and acetic anhydride, a
combination of acetylation and hydrolysis takes
place; there is obtained the octaacetate of (+)cellobiose. Alkaline hydrolysis of the octaacetate
yields (+)-cellobiose itself.
There are two isomeric forms of D-(+)-glucose
because this cyclic structure has one more chiral
centre than Fisher’s original open-chain structure. αD-(+)-Glucose
and
β-D-(+)-glucose
are
diastereomers, differing in configuration about C-1.
Such a pair of distereomers are called anomers.
Like (+)-maltose, (+)-cellobiose has the molecular
formula C12H22O 11, is a reducing sugar, forms an
osazone, exists in alpha and beta forms that undergo
mutarotation, and can be hydrolyzed to two
molecules of D-(+)-glucose. The sequence of
oxidation, methylation, and hydrolysis (as described
for (+)-maltose) shows that (+)-cellobiose contains
two pyranose rings and glucoside linkage to an –OH
group on C–4.
As hemiacetals, α-and β-D-(+)- glucose are readily
hydrolyzed by water. In aqueous solution either
anomer is converted –via the open-chain form–into
an equilibrium mixture containing both cyclic
isomers. This mutarotation results from the ready
opening and closing of the hemiacetal ring.
The typical aldehyde reactions of D-(+)-glucose –
osazone formation, and perhaps reduction of Tollens’
and Fehling’s reagents– are presumably due to a
small amount of open-chain compound, which is
replenished as fast as it is consumed. The
concentration of this open-chain structure, however,
is too low (less than 0.5%) for certain easily
reversible aldehyde reactions like bisulfite addition
and the Schiff test.
(+)-Cellobiose differs from (+)-maltose in one
respect : it is hydrolyzed by the enzyme emulsin
(from bitter almonds), not by maltase. Since emulsin
is known to hydrolyze only β-glucoside linkages.
(+)-Lactose :
(+)-Lactose makes up about 5% of human milk and
of cow’s milk. It is obtained commercially as a byproduct of cheese manufacture, being found in the
whey, the aqueous solution that remains after the
milk proteins have been coagulated. Milk sours when
lactose is converted into lactic acid (sour, like all
acids) by bacterial action (e.g., by Lactobacillus
bulgaricus).
Disaccharides :
Disaccharides are carbohydrates that are made up of
two monosaccharide units. On hydrolysis a molecule
of disaccharide yields two molecules of
monosaccharide.
XtraEdge for IIT-JEE
34
JANUARY 2010
(+)-Lactose has the molecular formula C12H22O11, is a
reducing sugar, forms an osazone, and exists in alpha
and beta forms which undergo mutarotation. Acidic
hydrolysis or treatment with emulsin (which splits β
linkages only) converts (+)-lactose into equal
amounts of D-(+)-glucose and D-(+)-galactose. (+)Lactose is evidently a β-glycoside formed by the
union of a molecule of D-(+)glucose and a molecule
of D-(+)-galactose.
influence of enzymes, the components of starch are
hydrolyzed progressively to dextrin (a mixture of
low-molecular-weight polysaccharides), (+)-maltose,
and finally D-(+)-glucose. (A mixture of all these is
found in corn sirup, for example.) Both amylose and
amylopectin are made up of D-(+)-glucose units, but
differ in molecular size and shape.
Cellulose :
Cellulose is the chief component of wood and plant
fibers; cotton, for instance, is nearly pure cellulose. It
is insoluble in water and tasteless; it is a nonreducing carbohydrate. These properties, in part at
least, are due to its extremely high molecular weight.
(+)-Sucrose :
(+)-Sucrose is our common table sugar, obtained
from sugar cane and sugar beets. Of organic
chemicals, it is the one produced in the largest
amount in pure form.
Cellulose has the formula (C6H10O5)n. Complete
hydrolysis by acid yields D-(+)-glucose as the only
monosaccharide.
Hydrolysis
of
completely
methylated cellulose gives a high yield of 2, 3, 6-triO-methyl-D-glucose. Like starch, therefore, cellulose
is made up of chains of D-glucose units, each unit
joined by a glycoside linkage to C–4 of the next.
(+)-Sucrose has the molecular formula C12H22O11. It
does not reduce Tollen’s or Fehling’s reagent. It is a
non-reducing sugar, and in this respect it differs from
the other disaccharides we have studied. Moreover,
(+)-sucrose does not form an osazone, does not exist
in anomeric forms, and does not show mutarotation
in solution. All these facts indicate that (+)-sucrose
does not contain a “free”aldehyde or ketone group.
Cellulose differs from starch, however, in the
configuration of the glycoside linkage. Upon
treatment with acetic anhydride and sulfuric acid,
cellulose yields octa-O-acetylcellobiose.
(+)-Sucrose is made up of a D-glucose unit and a Dfructose unit; since there is no “free” carbonyl group,
if must be both a D-glucoside and a D-fructoside.
Reactions of cellulose :
Polysaccharides :
Like any alcohol, cellulose form esters. Treatment
with a mixture of nitric and sulfuric acid converts
cellulose into cellulose nitrate. The properties and
uses of the product depend upon the extent of
nitration.
Polysaccharides are compounds made up of manyhundreds or even thousands-monosaccharide units
per molecule.
Polysaccharides are naturally occurring polymers,
which can be considered as derived from aldoses or
ketoses by polymerization with loss of water. A
polysaccharide derived from hexoses, for example,
has the general formula (C6H10O5)n.
In the presence of acetic anhydride, acetic acid, and a
little sulfuric acid, cellulose is converted into the
triacetate. Partial hydrolysis removes some of the
acetate groups, degrades the chains to smaller
fragments (of 200–300 units each), and yields the
vastly important commercial cellulose acetate
(roughly a diacetate). Cellulose acetate is less
flammable than cellulose nitrate and has replaced the
nitrate in many of its applications, in safety-type
photographic film, for example. When a solution of
cellulose acetate in acetone is forced through the fine
holes of a spinnerette, the solvent evaporates and
leaves solid filaments. Threads from these filaments
make up the material known as acetate rayon.
The most important polysaccharides are cellulose and
starch. Both are produced in plants from carbon
dioxide and water by the process of photosynthesis.
Starch :
Starch occurs as granules whose size and shape are
characteristic of the plant from which the starch is
obtained. When intact, starch granules are insoluble
in cold water; if the outer membrane has been broken
by grinding, the granules swell in cold water and
form a gel.
Industrially, cellulose is alkylated to ethers by action
of alkyl chlorides (cheaper than sulfates) in the
presence of alkali. Considerable degradation of the
long chains is unavoidable in these reactions. Methyl,
ethyl, and benzyl ethers of cellulose are important in
the production of textiles, films, and various plastic
objects.
In general, starch contains about 20% of a watersoluble fraction called amylose, and 80% of a waterinsoluble fraction called amylopectin. These two
fractions appear to correspond to different
carbohydrates of high molecular weight and formula
(C6H10O5)n. Upon treatment with acid or under the
XtraEdge for IIT-JEE
35
JANUARY 2010
KEY CONCEPT
Inorganic
Chemistry
Fundamentals
SALT ANALYSIS
2.
Identification of acidic radicals
For the identification of the acidic radicals, the
following scheme is followed.
Group I : The radicals which are analysed by dilute
H2SO 4 or dilute HCl. These are (i) carbonate (ii)
sulphite, (iii) sulphide, (iv) nitrite, and (v) acetate
Group II : The radicals which are analysed by
concentrated H2SO4 . These are (i) chloride, (ii)
bromide, (iii) iodide (iv) nitrate, and (v) oxalate
Group III : The radicals which are not analysed by
dilute and concentrated H2SO4. These are (i)
sulphate, (ii) Phosphate, (iii) borate, and (iv) fluoride.
Group I :
Add dilute HCl or H2SO4 to a small amount of
substance and warm gently, observe.
1. Carbonate or CO32– :
The carbonates are decomposed with the
effervescence of carbon dioxide gas.
Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 ↑
When this gas is passed through lime water, it
turns milky with the formation of calcium
carbonate.
Ca(OH)2 + CO2 →
CaCO3 + H2O
Lime water
K2Cr2O7 + H2SO4 + 3SO2 →
K2SO4 + Cr2 (SO4)3 + H2O
The sulphite also gives white precipitate with
BaCl2, Soluble in dil. HCl
3.
Na2SO3 + BaCl2 → 2 NaCl + BaSO 3 ↓
Sulphide, S–2:
The sulphide salts form H2S which smells like
rotten eggs.
Na2S + H2SO4 → Na2SO4 + H2S ↑
On exposure to this gas, the lead acetate paper
turns black due to the formation of lead sulphide.
Pb(CH3COO)2 + H2S → PbS ↓ + 2CH3COOH
black ppt.
The sulphides also turn sodium nitroprusside
solution violet (use sodium carbonate extract for
this test).
White ppt.
If the CO2, gas is passed in excess, the milky
solution becomes colourless due to the formation
of soluble calcium bicarbonate.
CaCO3 + H2O + CO 2 → Ca(HCO3)2
White ppt.
Sulphite :
The sulphites give out sulphur dioxide gas having
suffocating smell of burning sulphur.
CaSO3 + H2SO4 → CaSO4 + H2O + SO2 ↑
When acidified potassium dichromate paper is
exposed to the gas, it attains green colour due to
the formation of chromic sulphate.
Na2S + Na2[FeNO(CN)5] →
Na4 [Fe(NOS) (CN)5]
Sulphides of lead, calcium, nickel, cobalt,
antimony and stannic are not decomposed with
dilute H2SO4. Conc. HCl should be used for their
test.However brisk evolution of H2S takes place
even by use of dilute H2SO 4 if a pinch of zinc
dust is added.
Soluble
Note :
Carbonates of bismuth and barium are not easily
decomposed by dilute H2SO 4. Dilute HCl should
be used.
Sulphur dioxide evolved from sulphites also turns
lime water milky.
Ca(OH)2 + SO2 → CaSO3 + H2O
Zn + H2SO4 → ZnSO4 + 2H
HgS + 2H →Hg + H2S ↑
4.
White ppt.
However SO 2 can be identified by its pungent
odour of burning sulphur.
PbCO3 reacts with HCl or H2SO4 to give in the
initial stage some effervescence but the reaction
slows down due to formation of a protective
insoluble layer of PbCl2 or PbSO4 on the surface
of remaining salt or mixture.
–
Nitrite, NO2 :
The nitrites yield a colourless nitric oxide gas
which in contact with oxygen of the air becomes
brown due to the formation of nitrogen dioxide.
2KNO2 + H2SO4 →`K2SO4 + 2HNO2
Nitrous acid
3HNO2 → H2O + 2NO + HNO3
2 NO + O2 →
2NO 2 ↑
brown coloured gas
XtraEdge for IIT-JEE
36
JANUARY 2010
On passing the gas through dilute FeSO4 solution,
brown coloured complex salt is formed.
1.
FeSO4.7H2O + NO →
[Fe(H2O)5NO].SO4 + 2H2O
NaCl + H2SO4 → NaHSO4 + HCl ↑
The gas evolved forms white fumes of ammonium
chloride with NH4OH.
Brown coloured
(panta aquo nitroso ferrous sulphate)
When a mixture of iodide and nitrite is treated
with dilute H2SO4, the iodides are decomposed
giving violet vapours of iodine, which turns
starch iodide paper blue.
NH4OH + HCl
White fumes
2KI + H2SO4 → K2SO 4 + 2HI
+
AgNO3 + HCl → AgCl ↓ + HNO3
Yellowish : green chlorine gas with suffocating
odour is evolved on addition of MnO2 to the
above reaction mixture.
2NO
Violet vapours
I2 + Starch → Blue colour
5.
→ NH4Cl + H 2O
The gas evolved or solution of chloride salt forms
a curdy precipitate of silver chloride with silver
nitrate solution.
2NaNO2 + H2SO4 → Na2SO4 + 2HNO2
2HNO 2+ 2HI → 2H2O + I2
Chloride Cl–:
Colourless pungent fumes of hydrogen chloride are
evolved.
NaCl + H2SO4
Acetate :
Acetates decompose to give acetic acid vapours
having characteristic smell of vinegar.
–→
NaHSO4 + HCl
MnO2 + 4HCl
–→
MnCl2 + 2H2O + Cl2
Note :
The curdy precipitate of AgCl dissolves in
ammonium hydroxide forming a complex salt.
2CH3COONa + H2SO4 →
2CH3COOH + Na2SO4
All acetates are soluble in water and their aqueous
solution on addition to neutral FeCl3 solution
develops a blood red colour due to the formation
of ferric acetate.
Ag(NH3)2Cl + 2H2O
AgCl + 2NH4OH →
The solution having the silver complex on
acidifying with dilute nitric acid gives again a
white precipitate of silver chloride.
Ag(NH3)2Cl + 2HNO3 → AgCl + 2NH4NO3
Chromyl chloride Test : When solid chloride is
heated with conc. H2SO4 in presence of K2Cr2O7,
deep red
vapours of chromyl chloride are
evolved.
FeCl3 + 3CH3COONa →
(CH3COO)3Fe + 3NaCl
Blood Red colour
Acetates are also decomposed with oxalic acid
and give off acetic acid.
NaCl + H2SO4 → NaHSO4 + HCl
2CH3COONa + H2C2O4 →
Na2C2O4 + 2CH3COOH
K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O
CrO3 + 2HCl → CrO2Cl2 + H2O
Note :
Chromyl chloride
The ferric chloride solution supplied in the
laboratory is always acidic due to hydrolysis. It is
made neutral by the addition of dilute solution
of NH4OH drop by drop with constant stirring till
the precipitate formed does not dissolve. The
filtrate is called neutral ferric chloride solution.
Before testing acetate in the aqueous solution by
FeCl3, it must be made sure that the solution does
not contain
(i) CO3–2,
(ii) SO 3–2
(iii) PO4–3,
(iv) I–
Since these also combine with Fe+3. Therefore ,
the test of acetate should be performed by neutral
ferric chloride solution only after the removal of
these ions by AgNO 3 solution.
These vapours on passing through NaOH
solution, form the yellow solution due to the
formation of sodium chromate.
CrO2Cl2 + 4NaOH → Na2CrO4 +2NaCl+ 2H2O
Yellow colour
The yellow solution neutralised with acetic acid
gives a yellow precipitate of lead chromate with
lead acetate.
Na2CrO4 + Pb(CH3COO)2 –→
PbCrO4 + 2CH3COONa
Yellow ppt.
Note :
This test is not given by the chloride of mercuric,
tin, silver, lead and antimony.
The chromyl chloride test is always to be
performed in a dry test tube otherwise the
Group II:
Add concentrated H2SO4 to a small amount of the salt
or mixture and warm gently, observe.
XtraEdge for IIT-JEE
37
JANUARY 2010
chromyl chloride vapours will be hydrolysed in
the test tube.
4.
CrO2Cl2 + 2H2O →H2CrO4 + 2HCl
Bromides and iodides do not give this test.
2.
NaNO3 + H2SO4 → NaHSO4 + HNO3
4 HNO 3 → 2H2O + 4 NO2 + O 2
These fumes intensify when copper turnings are
added.
Bromide, Br– :
Reddish- brown fumes of bromine are formed.
NaBr + H2SO4 → NaHSO4 + HBr
Cu + 4HNO3 → Cu(NO3)2 + 2NO 2 + 2H2O
Ring Test : An aqueous solution of salt is mixed
with freshly prepared FeSO4 solution and conc.
H2SO4 is poured in test tube from sides, a brown
ring is formed on account of the formation of a
complex at the junction of two liquids.
2HBr + H2SO4 →Br2 + 2H2O + SO2
More reddish brown fumes of bromine are
evolved when MnO2 is added.
2NaBr + MnO2 + 3H2SO4 →
2NaHSO 4 + MnSO4 + 2H2O + Br2
The aqueous solution of bromide or sodium
carbonate extract gives pale yellow precipitate of
silver bromide which partly dissolves in excess of
NH4OH forming a soluble complex.
NaBr + AgNO3 → AgBr ↓ +
NaNO3 + H2SO4 → NaHSO4 + HNO3
6 FeSO4 + 2HNO3 + 3H2SO4 →
3Fe2 (SO4)3 + 4H2O + 2NO
[Fe(H2O)6]SO4. H2O + NO →
NaNO 3
Ferrous sulphate
Pale yellow ppt.
3.
[Fe(H2O)5 NO]SO 4 + 2H2O
AgBr +2NH4OH → Ag(NH3)2Br + 2H2O
Iodide, I– :
Violet vapours of iodine are evolved.
Brown ring
The nitrates can also be tested by boiling nitrate
with Zn or Al in presence of concentrated NaOH
solution when ammonia is evolved which can be
detected by the characteristics odour.
2KI + H2SO4 → 2KHSO4 + 2HI
2 HI + H2SO4 → I2 + SO2 + 2H2O
Violet vapours with starch produce blue colour.
Zn + 2NaOH → Na2ZnO 2 + 2H
Al + NaOH + H2O → NaAlO 2 + 3H
I2 + Starch → Blue colour
More violet vapours are evolved when MnO2 is
added.
2KI + MnO2 + 3H2SO 4 –→
2KHSO4 + MnSO 4 + 2H2O + I2
5.
Aqueous solution of the iodide or sodium
carbonate extract gives yellow precipitate of AgI
with silver nitrate solution which does not
dissolve in NH4OH.
NaNO3 + 8H → NaOH + 2H2O + NH3
Note : Ring test is not reliable in presence of
nitrite, bromide and iodide.
Oxalate, C2O4–2 :
A mixture of CO and CO2 is given off. The CO
burns with blue flame.
Na2C2O4 + H2SO4 → Na2SO4 + H2C2O4
H2C2O4 + [H2SO4] → CO + CO2 + H2O + [H2SO4]
NaI + AgNO3 → AgI + NaNO3
2CO + O2 → 2CO 2
Yellow ppt.
A solution of oxalates give the white precipitate
with CaCl2 solution. This precipitate get dissolved
in dil. H2SO4 and decolourises KMnO4 (acidified)
solution.
Note :
Sodium carbonate extract of bromide and iodide
on addition of CHCl3 and chlorine water gives
brown or violet layer to CHCl3 respectively.
Na2C2O4 + CaCl2 → CaC2O4 ↓ + 2NaCl
2NaBr + Cl2 → 2NaCl + Br2 ;
CaC2O4 + H2SO 4 → CaSO4 + H2C2O4
Br2 + CHCl3 → Brown
5H2C2O4 + 2KMnO4 + 3H2SO4 →
2 MnSO4 + K2SO4+ 8 H2O + 10CO2
2NaI + Cl2 → 2NaCl + I2 ;
I2 + CHCl3 → Violet
Excess of chlorine water should be avoided as the
layer may become colour less due to conversion
of Br2 into HBrO and I2 into HIO 3.
Group III :
1. Sulphate ,SO4–2 :
Add conc. HNO3 to a small amount of substance or
take sodium carbonate extract and then add BaCl2
solution. A white precipitate of BaSO 4 insoluble in
conc. acid is obtained.
Br2 + 2H2O + Cl2 → 2HBrO + 2HCl
I2 + 5Cl2 + 6H2O → 2HIO3 + 10 HCl
XtraEdge for IIT-JEE
Nitrate, NO3– :
Light brown fumes of nitrogen dioxide are
evolved.
38
JANUARY 2010
Na2SO 4 + BaCl2 → 2NaCl + BaSO4
3SiF4 + 4H2O → H4SiO4 + 2H2SiF6
White ppt.
2.
Silicic acid (white)
Note : Silver and lead if present, may be precipitated
as silver chloride and lead chloride by the addition of
barium chloride. To avoid it, barium nitrate may be
used in place of barium chloride.
Borate :
To a small quantity of the substance (salt or mixture),
add few ml. of ethyl alcohol and conc. H2SO4. Stir
the contents with a glass rod. Heat the test tube and
bring the mouth of the test tube near the flame. The
formation of green edged flame indicates the
presence of borate.
2Na3BO3 + 3H2SO4 → 3Na2SO 4 + 2H3BO3
Note :
The test should be performed in perfectly dry test
tube otherwise waxy white deposit will not be
formed on the rod.
HgCl2 and NH4Cl also give white deposits under
these conditions, but these are crystalline in
nature.
Sodium carbonate extract :
One part of the given substance is mixed with
about 3 parts of sodium carbonate and nearly 10
to 15 ml. of distilled water. The contents are then
heated for 10-15 minutes and filtered. The filtrate
is known as sodium carbonate extract or soda
extract and this contains soluble sodium salts due
to exchange of partners in between sodium
carbonate and salts.
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
H3BO3 + 3C2H5OH → (C2H5)3BO3 + 3H2O
Ethyl borate
3.
Phosphate :
Add conc. HNO 3 to a small amount of substance or
take sodium carbonate extract, heat and then add
ammonium molybdate. A canary yellow precipitate
of ammonium phospho molybdate is formed.
Ca3(PO4)2 + 6HNO 3 → 3Ca (NO 3)2 + 2H3PO4
Insoluble
(soluble)
PbSO4 + Na2CO3 → PbCO3 + Na2SO4
Insoluble
H3PO 4 + 12 (NH4)2 MoO4 + 21 HNO 3 →
(NH4)3 PO4. 12 MoO 3 + 21 NH4NO3 + 12 H2O
Sodium sulphate
(Soluble)
BaCl2 + Na2CO3 → BaCO3 + 2 NaCl
Canary yellow ppt.
4.
Sodium chloride
Insoluble Sodium chloride
Note :
Arsenic also yields a yellow precipitate of
(NH4)3. AsO4.12 MoO3 (Ammonium arseno
molybdate).Thus in presence of As, phosphate is
tested in the filtrate of second group.
The precipitate of ammonium phosphomolybdate
dissolves in excess of phosphate. Thus, the
reagent (ammonium molybdate) should always be
added in excess.
HCl interferes in this test. Hence, before the test
of phosphate is to be performed, the solution
should be boiled to remove HCl.
Reducing agents such as sulphites, sulphides, etc.,
interfere as they reduce Mo+6 to molybdenum
blue (Mo3O8.xH2O). The solution, therefore, turns
blue. In such cases, the solutions should be boiled
with HNO 3 so as to oxidise them before the
addition of ammonium molybdate.
Fluoride :
Take small amount of the substance in dry test tube
and add an equal amount of sand and conc.
H2SO 4.Heat the contents and place a glass rod
moistened with water over the mouth of the test tube.
A gelatinous waxy white deposit on the rod is
formed.
2NaF + H2SO4 → Na2SO4 + H2F2
(Soluble)
The carbonates of the cations of the mixtures are
mostly insoluble in water and are obtained in the
residue. On the other hand, sodium salts of the
anions (acidic radicals) of the mixture being
soluble in water are obtained in the filtrate.
The sodium carbonate extract is basic in nature
and before it is used for the analysis of a
particular acidic radical, it is first neutralised by
the addition of small quantity of an appropriate
acid. The acid is added to the extract till the
effervescence cease to evolve.
Advantages of preparing sodium carbonate
extractThe preparation of sodium carbonate extract
affords a convenient method for bringing the
anions of the mixture into solution which were
otherwise insoluble with cation of salt.
It removes the basic radicals (usually coloured)
which interferes in the usual tests of some of the
acidic radicals.
The residue can be used for the tests of basic
radicals of I to VI groups. Such a solution does
not involve the problem of removing interfering
radicals like oxalate, fluoride, borate and
phosphate.
SiO2 + 2H2F2 → SiF4 + 2H2O
XtraEdge for IIT-JEE
39
JANUARY 2010
UNDERSTANDING
Organic Chemistry
1.
A hydrocarbon (A) [C = 90.56%, V.D. = 53] was
subjected to vigrous oxidation to give a dibasic acid
(B). 0.10 g of (B) required 24.10 ml of 0.05 N NaOH
for complete neutralization. When (B) was heated
strongly with soda-lime it gave benzene. Identify (A)
and (B) with proper reasoning and also give their
structures.
Sol. Determination of empirical formula of (A) :
Element
%
C
90.56
H
9.44
CH3
Phthalic acid
Isophthalic acid
CH3
COOH
+ 2H2O
CH3
COOH
m-xylene
COOH
CH3
6[O]
+ 2H2O
CH3
COOH
All the above three acids on heating with soda-lime
yields only benzene.
COOH
COOH
,
,
COOH
COOH
COOH
NaOH + CaO
∆
+ 2CO2
COOH
Of the three acids, one which on heating gives an
anhydride, is o-isomer.
COOH
CO
∆
O
CO
COOH
–H2O
One acid which on nitration gives a mono nitro
compounds is p-dicarboxylic acid.
COOH
COOH
NO2
HNO3
∆; H2SO4
COOH
COOH
One acid which on nitration gives three mono nitro
compounds will be the m-isomer.
COOH
COOH
COOH
HNO3
NO2
COOH ,
COOH ,
COOH H2SO4
NO2
COOH
COOH
NO2
Terphthalic acid
2.
All the above three acids are obtained by the
oxidation of respectively xylenes.
XtraEdge for IIT-JEE
6[O]
CH3
The empirical formula of (A) = C4H5
Empirical formula weight = 48 + 5 = 53
Molecular weight = V.D. × 2 = 53 × 2 = 106
Molecular wt. 106
Hence, n =
=
=2
Empirical wt.
53
Molecular formula = 2 × C4H5 = C8H10
The given equation may be outlined as follows :
HOOC
oxidation
C 8 H 10 Vigrous


→
C6H4+ 2H2O
6[O ]
HOOC
(A)
(B)
Meq. of dicarboxylic acid = Meq. of NaOH
0.1× 1000
= 24.1 × 0.05
E
Equivalent of acid = 83
Molecular wt. = Basicity × Equivalent weight
= 2 × 83 = 166
Since (B) on heating with soda-lime gives benzene,
the C6H4 represents to benzene nucleus having two
side chains, thus (B) is a benzene dicarboxylic acid.
There are three benzene dicarboxylic acids.
COOH
COOH
COOH
COOH
COOH
+ 2H2O
COOH
o-xylene
Atomic Relative no. Simplest ratio
wt. of atoms
12
90.56
7.55
= 7.55
= 1 or 4
12
7.55
1
9.44
9.44
= 9.44
= 1.25
1
7.55
or 5
COOH
6[O]
40
COOH
An organic compound (A) contains 69.42% C, 5.78%
H and 11.57% N. Its vapour density is 60.5. It
evolves NH3 when boiled with KOH. On heating
with P2O5, it gives a compound (B) having C =
81.55%, H = 4.85% and N= 13.59%. On reduction
JANUARY 2010
with Na + C2H5OH (B) gives a base, which reacts
with HNO 2 giving off N2 and yielding an alcohol (C).
The alcohol can be oxidised to benzoic acid. Explain
the above reactions and assign structural formulae to
(A), (B) and (C)
Sol. (i) Calculation of empirical formula of (A) :
Element
C
At.
wt.
12
%
69.42
H
1
5.78
N
14
11.57
O
16
13.23
Relative no. of
atoms
69.42
= 5.785
12
5.78
= 5.78
1
11.57
= 0.826
14
13.23
= 1.827
16
The formula of benzoic acid indicates that the
compound (A) is an aromatic amide. Hence, the
reactions are :
COOK
CONH2
KOH
Boil
Benzamide
(A)
Simplest
ratio
5.785
=7
0.826
5.78
=7
0.826
0.826
=1
0.826
0.827
=1
0.826
CONH2
C
At.
wt.
12
%
81.55
H
1
4.85
N
14
13.59
Relative no. of
atoms
81.55
= 6.80
12
4.85
= 4.85
1
13.59
= 0.97
14
+ 4[H]
XtraEdge for IIT-JEE
N2 + (C)
CH2OH
HNO2
+ N2 + H2O
Benzyl alcohol
(C)
COOH
CH2OH
2[O]
– H2O
(C)
3.
Simplest
ratio
6.80
=7
0.97
4.85
=5
0.97
0.97
=1
0.97
A (C 6 H12 ) HCl

→
Benzoic acid
B+C
(C 6 H13Cl)
KOH
B alc
. 
→ D isomer of A
D Ozonolysis
 
→ E (it gives negative test with Fehling
solution but responds to
iodoform test)
A Ozonolysis
 
→ F + G Both gives positive Tollen's
test but do not give iodoform test.
. NaOH
F + G Conc
→ HCOONa + primary alcohol
Identify to A to G
Sol. A (C 6 H12 ) HCl

→
B+C
(C 6 H13Cl)
B Ozonolysis
 
→ E (it gives negative test with Fehling
solution but responds to
iodoform test)
A Ozonolysis
 
→ F + G Both gives positive Tollen's
test but do not give iodoform test.
( B)
(B)
Benzyl amine
(Base)
CH2NH2
. NaOH
F + G Conc

→ HCOONa + primary alcohol
Both F and G are aldehydes because they give
positive Tollen's test and do not give iodoform test.
These aldehydes give Cross Cannizzaro's reaction. So
they do not have α-hydrogen atoms. In cross
Cannizzaro's reaction HCOONa is formed along with
p-alcohols. So in these an aldehyde is HCHO and
The above reaction confirms that (A) is an amide,
and the remaining reaction are :
COOH
HNO2
Na +C2H5OH
(B)
C 7 H 7 ON → C 7 H 5 N + H 2 O
[H]
CH2NH2
C≡N
P2 O5
C7H5N
+ H2O
Benzonitrile
(B)
(A)
Hence, empirical formula of (B) = C7H5N
(v) Determination of structural formulae :
(a) Since compound (A) on heating with KOH gives
NH3, a characteristic test of amide, hence the
compound (A) is an amide (–CONH2).
(b) Since compound (B) is obtained by heating (A)
with P2O5, a dehydrating agent.
(A)
C≡N
P2O5
∆
Hence, empirical formula of (A) = C7H7NO
Empirical formula wt. = 84 + 7 + 14 + 16 = 121
(ii) Calculation of molecular weight of (A) :
Molecular weight = 2 × V.D. = 2 × 60.5 = 121
(iii) Determination of molecular formula of (A):
Molecular wt. 121
=
=1
n=
Empirical wt.
121
Hence, molecular formula = empirical formula
i.e.,
C7H7NO
(iv) Calculate of empirical formula of (B) :
Element
+ NH3 ↑
[O]
Alcohol
41
JANUARY 2010
Hence,
another is (CH3) 3C.CHO. F and G are obtained by
ozonolysis of A. Therefore compound 'A' is
CH2 = CH – C(CH3)3.
Structure of compound 'A' is
CH3
CH3 – C – CH = CH2
CH3
Compound 'A' on reaction with HCl gives comp. B
and C which have molecular formula C 6H13Cl. Thus,
CH3
CH3
CH3
Compound 'A' = CH3 – C – CH = CH2 (C6H12)
CH3
CH3
Compound 'B' = CH3 – C – CH – CH3
CH3 Cl
CH3
HCl
CH3 – C – CH = CH2 → CH3 – C — CH – CH3
CH3
Compound 'C' = CH3 – C – CH2 – CH2Cl
CH3 Cl
Comp. 'B'
+
CH3
CH3
Compound 'D' =
CH3 – C – CH2 – CH2Cl
CH3
–
Boil (–Cl )
Compound 'F' =
+
CH3 – C — CH – CH3
CH3 Cl
Compound 'G' = CH3 – C – CH2OH
CH3
CH3 – C – CH
+
CH3 H+ CH3
CH3
CH3
C=C
CH3
4.
CH3
CH3
Compound 'D'
Compound 'D' on ozonolysis to give compound 'E'
CH3 Ozonolysis
CH3
C=C
2CH3 – C – CH3
CH3
CH3
O
Sol.
Compound 'E'
'F'
(ii ) H 2 O
( B)
2H
H – C ≡ C – H −
→ C3H5 – C ≡ C – C3H5
+ C 6 H10
the C3H5 – corresponds to cyclopropyl (∆ ) radical,
hence compund (A) is
CH2
CH2
CH – C≡C – CH
CH2
CH2
Ozonolysis
CH3 – C — CHO + CH2O + conc. NaOH →
'F'
Comp. 'G'
i ) O3
A(C8H10) (
→ C 4 H 6 O 2
C=C
or a – C ≡ C – bond.
should have either
If it was alkene its formula should be C8H16 (CnH2n),
and if it was alkyne it should have the formula C8H14;
it means it is neither a simple alkenen or simple
alkyne. However it is definite that the compound has
an unsaturated group, it appears that it is a
cyclosubstituted ethyne.
Compound G and F gives crossed Cannizzaro's
reaction with conc. NaOH solution.
CH3
CH3
A hydrocarbon (A) of the formula C8H10, on
ozonolysis gives compound (B), C4H6O2, only. The
compound (B) can also be obtained from the alkyl
bromide, (C) (C3H4Br) upon treatment with Mg in
dry ether, followed by treatment with CO2 and
acidification. Identify (A), (B) and (C) and also
equations for the reactions.
Since compound (A) adds one mol of O3, hence it
Compound 'E' has methyl ketonic groups (–COCH3)
so it gives positive iodoform test and does not give
the test with Fehling solution due to absence of
–CHO group.
Compound 'A' on ozonolysis to give compounds F
and G as follows :
Ozonolysis
(CH3)3CCH = CH2 → (CH3)3C – CHO + CH2O
Comp. 'G'
CH3
H–C–H
CH3
Sec. carbonium ion
CH3
CH3
Compound 'E' = CH3 – C – CH3
O
Compound 'B' gives 'D' on dehydrohalogenation with
alc. KOH.
CH3
CH3
alc. KOH
CH3
C=C
O
Comp. 'C'
CH3 – C — CH – CH3
CH3
1,2-dicyclopropyl ethyne
CH3
The ozonolysis of above compound would give two
moles of cyclopropane carboxylic acid (C4H 6O2).
HCOONa + CH3 – C – CH2OH
CH3
XtraEdge for IIT-JEE
42
JANUARY 2010
CH2
CH2
CH2
CH – C≡C – CH
CH2
(A)
O
CH2
CH2
CH2
CH2
CH2
CH – C — C – CH
(A)
O
CH – C – C – CH
O
2
H2O
CH2 Warm
O
O
compound (E) gives Cannizaro's reaction with
NaOH. So, (E) is an aldehyde which does not contain
α–H atom. Hence it is HCHO. Compound (D) can
also be prepared by the hydration of propyne in the
presence of acidic solution and Hg++.
++
CH3 – C ≡ CH + H2O Hg

→ CH 3 − C = CH 2
+
H
|
OH
→ CH 3 − C − CH 3
||
O
(i) O3
CH2
CH2
CH2
CH2
+ H2O2
(D)
Hence (D) is acetone and (E) is formaldehyde.
Therefore, alkene (C) is 2-methyl propene.
(CH3)2–C=CH2
(D) reacts with hydroxyl amine (NH2OH) to form
oxime (F).
CH3
–H2O CH3
C = O + H2 NOH
C = NOH
CH3
CH3
CH – COOH
(B)
Compound (B) is prepared from cyclopropyl bromide
as follows :
CH2
CH2
CH – Br
Mg
ether
O
CH2
CH . MgBr
CH2
C=O
∆
(D)
Cyclopropyl
magnesium bromide
CH2
CH2
CH .COOMgBr
HOH
dil. HCl;
–MgBrOH
CH2
CH2
CH–COOH
Addition compound
5.
Sol.
An organic compound (A), C4H9Cl, on reacting with
aqueous KOH gives (B) and on reaction with
alcoholic KOH gives (C) which is also formed on
passing vapours of (B) over heated copper. The
compound (C) readily decolourises bromine water.
Ozonolysis of (C) gives two compounds (D) and (E).
Compound (D) reacts with NH2OH to gives (F) and
the compound (E) reacts with NaOH to give an
alcohol (G) and sodium salt (H) of an acid. (D) can
also be prepared from propyne on treatment with
water in presence of Hg++ and H 2SO4. Identify (A) to
(H) with proper reasoning.
C4H9Cl
Alc. KOH
∆; –KCl
(A)
(Alkyl halide)
Aq.KOH
∆; –KCl
(A )
( B)
Cu / 300 º C
 → CH 3 − C = CH 2 + H 2 O
− H 2O
|
CH 3
( C)
Alc.KOH / ∆
→ CH 3 − C = CH 2
− KCl; − H 2O
|
CH 3
( C)
C4H8
O
(C)
(Alkene)
Cu
C4H9OH
∆; –H2O
(B)
(Alcohol)
CH3 – C = CH2
CH3
(I) O3
CH3
C=O+H–C–H
(II) H2O/Zn CH3
(E)
(D)
(C)
We know that p-alcohol on heating with Cu gives
aldehyde while s-alcohol under similar conditions
gives ketone. Thus, (B) is a t-alcohol because it, on
heating with Cu gives an alkene (C). Since a talcohol is obtained by the hydrolysis of a t-alkyl
halide, hence (A) is t-butyl chloride.
Cl
OH
|
|
(A) = CH 3 − C − CH 3 and (B) = CH 3 − C − CH 3
|
|
CH 3
CH 3
The alkene (C) on ozonolysis gives (D) and (E),
hence (C) is not symmetrical alkene. In these
XtraEdge for IIT-JEE
(F)
OH
Cl
|
|
(B) = CH 3 − C − CH 3 and (A) = CH 3 − C − CH 3
|
|
CH 3
CH 3
Reactions :
OH
Cl
|
|
CH 3 − C − CH 3
CH 3 − C − CH 3
Aq.KOH
 →
|
|
∆;– KCl
CH 3
CH 3
CH3
CH3
C = O + H2NOH
(D)
CH3
∆
–H2O
CH3
C = NOH
(F)
2HCHO + NaOH → CH 3OH + HCOONa
(E)
(G)
( H)
O
Hg + +
CH3 – C ≡ CH + H2O → CH3 – C – CH3
+
H
43
(D)
JANUARY 2010
Set
9
`tà{xÅtà|vtÄ V{tÄÄxÇzxá
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in mathematics that would be very helpful in facing
IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and
we hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Shailendra Maheshwari
So lu tio n s w ill b e p ub lis h ed in n ex t iss u e
Joint Director Academics, Career Point, Kota
1.
( x − b ) ( x − c)
( x − b ) (a − c )
Given that φ (x) =
Show that the dip θ (angle with horizontal) of the oil
f (a) +
bed which is assumed to be a plane is given by tan θ .
a < c < b and f ′′(x) exists at all points in (a, b) .
lengths of the sides CA and AB respectively and A is
a<
that
there
exists
a
number
c
2
+
b
2
−
2 yz
cos A where b and c are the
bc
the angle between CA and AB.
µ,
µ < b , such that
f (a )
f ( b)
+
(a − b ) (a − c )
( b − c) ( b − a )
1
f ( c)
+
f ′′(µ).
=
(c − a ) (c − b )
2
∫
cos 8x − cos 7 x
1 + 2 cos 5x
6.
Evaluate :
7.
Let f (x) be an even function such that f ′ (x) is
continuous, find y for which
d2y
dx 2
x
=
∫ f (t) dt
−x
An unbiased die is tossed until it lands the same way
8.
up twice running. Find the probability that it requires
Prove the inequality (aα + bα)1/α < (aβ + bβ)1/β,
for a > 0, b > 0 & α > β > 0.
r tosses.
3.
z2
sin A =
Prove
2.
y2
( x − c) ( x − a )
( x − a ) ( x − b)
f (b)+
f (c) - f (x) where
( b − c) ( b − a )
(c − a ) (c − b )
9.
Given the base of a triangle and the sum of its sides
A circle of radius 1 rolls (without sliding) along
prove that the locus of the centre of its incircle is an
the x-axis so that its centre is of the form (t, 1)
ellipse.
with t increasing. A certain point P touches the
x-axis at the origin as the circle rolls. As the circle
4.
Let f (x) = ax2 + bx + c & g (x) = cx2 + bx + a, such
rolls further, the point P passes through the point
that | f (0) | ≤ 1, | f (1) | ≤ 1 and |f (-1) | ≤ 1 , prove that |
(x, 1/2). Find x, when it passes through (x, 1/2)
f (x) | ≤ 5/4 and | g (x) | ≤ 2.
5.
first time.
In order to find the dip of an oil bed below the
surface of the ground, vertical borings are made from
10. Find all positive integers n for which
n −1 +
the angular points, A, B, C of a triangle ABC which
n + 1 is rational.
is in horizontal plane. The depth of the bed at these
points are found to be x, x + y and x + z respectively.
XtraEdge for IIT-JEE
44
JANUARY 2010
MATHEMATICAL CHALLENGES
SOLUTION FOR NOVEMBER ISSUE (SET # 8)
1.
3.
If A is the area of the triangle with sides a, b and c,
then A2 = s (s − a) (s − b) (s − c) ;
A a
where 2s = a + b + c.
using AM - GM inequality for s − a, s − b, s − c, we
have
dd
Bb
3
 (s − a ) + (s − b) + (s − c) 
A2 ≤ s 

3


C c
Plane through mid pt of AB, ⊥ to CD is
r
r
r
r
r
( r – 1/2 ( a + b )).( c – d ) = 0; Let centroid
r r r r
a +b+c+d
= 0 at origin
4
r
r
r
r
r
( r + 1/2 ( c + d )). ( c – d ) = 0
r
r
r
r
| r + c |2 = | r + d |2
Let 2s = p , then A ≤
r
r
it is the locus of pt. equidistance from – c & – d
s − a = s − b = s − c which happen if a = b = c.
3
s4
 3s − 2s 
A2 ≤ s 
 = 3
3
 3 
similarly.
r
r
r
r
r r
r
r
| r + c |2 = | r + d |2 = | r + a |2 = | r + b |2
r
r
r
r
so the pt. is equidistant from – a , – b , – c , – d
r
r
r
r
(i.e. circumcentre of tetrahedron a , – b , – c , – d )
2.
................(1)
f (b + x ) = f (b − x)
................(2)
12 3
so Amax =
p min =
12 3
s2
3 3
p2
12 3
, As condition of equality holds iff
; for a = b = c
12 3 A
12 3 A , and
again equality holds if a = b = c.
b 2 − 4ac ≤ | b |
4.
&

2ac
≤ | b | 1 + 2
b

As it is defined for x ∈ R.
Let x = b − a − t in (1)
use (2) in it
so that −
f (b + t) = f (2a − 2b + b + t)
so the function is periodic & its possible period
=
may be |2a − 2b| = 2b - 2a (as b > a).
45
1−
4ac
b
2
≤|b|
1+
4ac
b2



b 2 − 4ac ≤ | b | +
so
f (b − t) = f (2a − b + t)
XtraEdge for IIT-JEE
p2
Now again p ≥
As the function is symmetrical about x = a & x = b
lines
so f (a + x) = f (a − x)
p2
Amax =
⇒ A≤
2ac
b
b
b 2 − 4ac
b
b
c
≤
+
+
±
2a
2a
b
2a
2a
b
c
+
a
b
JANUARY 2010
Hence the solutions of az2 + bz + c = 0 satisfy
condition | z | ≤ +
5.
=
b
c
+
.
a
b
sin 2 θ cos 2 θ
[– cos2θ + sin2θ + sin3θ + cos 2θ sinθ]
for max / min .
P (a cos θ , b sin θ)
Equation of AC ⇒
ab (1 + sin θ)
x
y
cos θ +
sin θ = 1
a
b
dA
=0
dθ
sinθ (cos2θ + sin2θ) + sin2θ − cos2θ = 0
sinθ + sin2θ − (1 − sin2θ) = 0
A
⇒ 2sin2θ + sinθ – 1 = 0
(2 sinθ − 1) (sin θ + 1) = 0
as sin θ ≠ −1
sin θ = 1/2 ; θ = π/6
P(θ)
C
B
D
Point A : (0 , b cosec θ)
⇒
Equation of BC
Point C =
x=
when θ > π/6 ;
dA
>0
dθ
when θ < π/6 ;
dA
<0
dθ
y = −b
x
cos θ − sin θ = 1
a
min. area.
(1 + sin θ) a
cos θ
Amin =
 a (1 + sin θ)

,−b 
Point C 
 cos θ

Area A =
=
=3
6.
ab (1 + sin θ) 2
sin θ cos θ
A = ab
(1 + sin θ) 2
sin θ cos θ
1 3
4. .
2 2
3 ab sq. units.
ax2 + 2hxy + by2 = 0; (y – M1x) (y – M2x) = 0
2h
a
& M1M2 =
b
b
M2y2 + (1 – M1M2)xy – M1x2 = 0
Compare it with a´x 2 + 2h´xy + b´y2 = 0
−M1
M2
1 − M1M 2
=
=
b´
2h´
a´
2
sin θ cos θ
sin 2 θ cos 2 θ
ab × 9
(y – M1x)(M2y + x) = 0
2
ab (1 + sin θ)
=
Now as given the second pair must be given by
2 (1 + sin θ) cos 2 θ sin θ − (1 + sin θ) 2 (cos 2 θ − sin 2 θ)
=
1/ 2 . 3 / 2
where M1 + M2 = –
dA
=
dθ
ab .
ab (1 + sin θ) 2
sin θ cos θ
ab . (1 + 1 / 2) 2
1
AD . BC = AD . DC
2
b  a (1 + sin θ)

= b+
.
sin θ 
cos θ

=
π
; is the pt of min.
6
so θ =
so
[2cos2θ sinθ − (1 + sinθ) cos2θ
+ (1 + sinθ) sin3θ)]
XtraEdge for IIT-JEE
=
46
−M1
M2
M + M2
−2h
= 1
=–
=
b´
a´
b´−a´
b( b´−a´)
1 − M1M 2
1− a / b
=
2h´
2h´
JANUARY 2010
M2 = –
2hb´
(b − a )a´
& M1 = –
b( b´−a´)
2h´b
9.
a
2hb´
(b − a )a´
a
so
.
=
b
b( b´−a´)
2h´b
b
Since M1M2 =
=
ha´b´
h´ab
=
b´−a´
b−a
Thus
LHS = coeff. of xn in [ nC0(1 + x)m + nC1(1 + x)m+1 +
7.
n
m+n
.... + Cn(1 + x)
tn =
]
2n + 1
n . (n + 1) 2
2
1
n
2
Sn =
1−
Sn =
1−
−
1
( n + 1) 2
1
1
1
1
1
+ 2 – 2 + 2 − 2 .................
22
2
3
3
4
1
( n + 1) 2
Required sum = Lim S n = 1.
n
m n
n
n →∞
n
= coeff. of x in (1 + x) [ C0 + C1(1 + x) + ..... + C0
(1 + x)n]
10. Let the given circle be x2 + y2 = r2 & parametric
= coeff. of xn in (1 + x)m(2 + x)n
= coeff of xn in (1 + x)m
n
∑
n
angles of A, B, C are respectively θ1 , θ2 & θ3. Let
the slopes of the given two lines are m1 & m2. Sides
AB & BC are parallel to these lines.
C r x n − r .2 r
r =0
n
m
n
A(θ1)
m
= C0 . C0 + C1 . C1 . 2 + nC 2 . mC 2 . 22 + .... + nCnmC n . 2n
1
8.
∫ (1 − x
In =
2 n
cos mx dx
)
−1
∫
Equation of AB;
2n
=0+
m

 − x (1 − x 2

(
)
n −1
x cos
1
1
cos mx 
1
 +
n  −1 m
∫ (2 (n − 1)x
−1
C(θ3)
B(θ2)
1
1
sin mx 
2n

+
= (1 − x 2 ) n
x (1 − x 2 ) n −1 sin mx dx
m  −1 m

−1
2

(1 − x 2 ) n −2 + (1 − x 2 ) n −1 cos mx dx 

θ1 + θ 2
θ + θ2
θ − θ2
+ y sin 1
= r cos 1
2
2
2
so m1 = – cot
θ1 + θ 2
= θ1 + θ2 = α
2
θ2 + θ3
=θ2 + θ3 = β
2
Here α, β are constants as m1 & m2 are constants.
similarly : m2 = − cot
=
=
=

(2 n − 1)


2n
m2
2n
m
2
1
∫ (1− x
2 n −2
)
[(−2n + 2) x
2
]
+ 1 − x 2 cosmx dx
−1
Now equation of AC ;
1
∫ (1− x
2 n −2
)
[(−2n + 1) x + 1] cos mx dx
θ +θ
θ − θ3
 θ + θ3 
x cos  1
 + y sin 1 3 = r cos 1
2
2
 2 
2
−1
θ +θ
 θ + θ3 
x cos  1
 + y sin 1 3 = rk
2
2


2n
m2
1
∫ (1 − x
1
2 n −1
)
cos mx dx − (2 n − 2)
−1
∫ (1 − x
2
) n −2
−1
α−β
(i . e. constant)
2
so foot of the perpendicular from centre of given
θ + θ3
θ +θ 

circle on AC  r k cos 1
, r k 1 3  is
2
2 

which lies on x2 + y2 = (rk)2.
where k = cos

cos mx dx 


m2In = (2n (2n − 1) In–1 − 4n (n − 1) In–2.
Hence proved.
XtraEdge for IIT-JEE
47
JANUARY 2010
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
MATHS
1.
 x + x 5 + x 6 y 4 + y5 + y 6 
G is  4
,

3
3


There are two die A and B both having six faces. Die
A has 3 faces marked with 1, 2 faces marked with 2
and 1 face marked with 3. Die B has 1 face marked
with 1, 2 faces marked with 2 and 3 faces marked
with 3. Both dice are thrown randomly once. If E is
the event of getting sum of the numbers appearing on
top faces equal to x and let P(E) be the probability of
event E, then
 α − α1 β − β1 
G is 
,

3 
 3
The point dividing OG. in the ratio 3 : 1 is
α β
 ,  ≡ (2,1) ⇒ h + k = 3
 4 4
3.
(i) find x, when P(E) is maxm.
(ii) find x, when P(E) is minm
conditions f(x + y) =
Sol. X can be 2, 3, 4, 5, 6.
the set of values of x where f(x) is differentiable and
can occur are the coefficients of x2, x3, x4, x5, x6 is
(3x + 2x2 + x3) (x + 2x2 + 3x3)
3
4
5
also find the value of lim [f(x)]x.
x →∞
6
Sol. First put x = 0, y = 0 ⇒ f(0) = 0
= 3x + 8x + 14x + 8x + 3x
This shows that sum that occurs most often is 4, and
sum that occurs minimum times is 2 or 6.
2.
Now,
6
∑x
i =1
6
i
= 8 and
∑y
i
 f ( h ) − f (0)  1 − {f ( x )}2 
= lim 


x →0
h − 0  − 1 + f ( x ).f (h ) 
= 4.
i =1
The line segment joining orthocentre of a ∆ made by
= 1 – {f2(x)}
any three points and the contrioid of the ∆ made by
other three points passes through a fixed points (h, k),
then find h + k.
6
Sol. Let
∑
integrating we get
6
x i = α and
i =1
∑
⇒ f(x) =
yi = β
i =1
1 1 + f ( x ) 
ln 
 =x+c
2 1 − f ( x) 
ex − e−x
e x + e−x
clearly f(x) is differentiable for all x ∈R.
let 0 be orthocentre of ∆ made by (x1, y1), (x2, y2) and
(x3, y3)
 e x − e−x
lim [f ( x )] x = lim  x
x →∞
x →∞ e + e − x

⇒ 0 is (x1 + x2 + x3, y1 + y2 + y3) = (αi , β 1)
similarly let G be the centroid of the ∆ made by other
three points.
XtraEdge for IIT-JEE
f (x + h) − f (x)
x →0
h
f´(x) = lim
f (x) + f (h)
− f ( x)
1 + f ( x ).f (h )
= lim
x →0
h
Six points (xi, yi), i = 1, 2, 3, .... 6 are taken on the
circle x2 + y2 = 4 such that
f ( x ) + f ( y)
∀ x, y and
1 + f ( x ).f ( y)
f´(0) = 1. Also – 1 < f(x) < 1 for all x ∈R, then find
The no of ways in which sum 2, 3, 4, 5, 6.
2
Suppose a function f(x) satisfies the following
= e
48
 ex −e− x
lim 

x →∞  e x + e − x








x
x=1
JANUARY 2010
XtraEdge for IIT-JEE
49
JANUARY 2010
CAREER POINT’s
Correspondence & Test Series Courses Information
Courses of IIT-JEE
For Class 12th / 12th Pass Students
CP Ranker's
Package
IIT-JEE 2010
Study Material
Package
IIT-JEE 2011
All India Foundation
Test Series IIT JEE
2011 (AT CENTER)
Postal All India
Test Series
IIT-JEE 2011
Class 12th or
12th pass
students
Class 12th or
12th pass
students
Class 11th
students
Class 11th
students
Class 11th
students
English
English
English
English OR Hindi
English
English
Postal
Postal
Postal
Visit our website
Postal
Immediate
Dispatch
Dispatch
17-Jan-10
Onwards
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
25-Oct-09
Immediate
Dispatch
Rs. 9,500/including S.Tax
Rs. 4,500/including S.Tax
Rs. 2,500/including S.Tax
COURSE
INFORMATION
All India Test
Series IIT JEE 2010
(AT CENTER)
Postal All India
Test Series
IIT-JEE 2010
Major Test Series
IIT JEE 2010
(AT CENTER)
Eligibiltiy for
Admission
Class 12th or
12th pass
students
Class 12th or
12th pass
students
Class 12th or
12th pass
students
Medium
English OR Hindi
English
English
Test Center
Postal
Visit our website
Postal
Issue of
Application Kit
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Date of Dispatch/
First Test
Immediate
Dispatch
25-Oct-09
Immediate
Dispatch
Course Fee
Rs.8,500/including S.Tax
Rs. 3,500/including S.Tax
For Class 11th Students
Postal Major Test
Series IIT JEE
2010
(By POST)
Class 12th or
12th pass
students
Study Material
Package
IIT-JEE 2010
Visit our
website
Immediate
Dispatch
17-Jan-10
Rs. 1,500/Rs. 1,550/Rs. 700/Rs. 850/including S.Tax including S.Tax including S.Tax including S.Tax
Courses of AIEEE
For Class 11th
Students
For Class 12th / 12th Pass Students
COURSE
INFORMATION
Study Material
Package
AIEEE 2010
Eligibiltiy for
Admission
Class 12th or 12th
pass students
All India
Test Series
AIEEE 2010
(AT CENTER)
Class 12th or 12th
pass students
Postal All India Test
Series
AIEEE 2010
Major Test Series
AIEEE 2010
(AT CENTER)
Class 12th or 12th
pass students
Class 12th or 12th
pass students
Postal Major Test
Series
AIEEE 2010
(BY POST)
Class 12th or 12th
pass students
CP Ranker's Package
IIT-JEE 2010
Study Material
Package
AIEEE 2011
Class 12th or 12th
pass students
Class 11th
students
English OR Hindi
Medium
English OR Hindi
English/Hindi
English/Hindi
English/Hindi
English/Hindi
English
Test Center
Postal
Visit our website
Postal
Visit our website
Postal
Postal
Postal
Issue of
Application Kit
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Immediate
Dispatch
Date of Dispatch/
First Test
Immediate
Dispatch
25-Oct-09
Immediate
Dispatch
17-Jan-10
Immediate
Dispatch
Immediate
Dispatch
Course Fee
Rs. 7,500/including S. Tax
Rs. 4,000/including S. Tax
Rs. 1450/including S. Tax
Rs. 1000/including S. Tax
Immediate
Dispatch
Dispatch
17-Jan-10
Onwards
Rs. 550/including S. Tax
Rs. 850/including S.Tax
Rs. 8,500/including S.Tax
Courses of Pre-Medical
For Class 11th
Students
For Class 12th / 12th Pass Students
COURSE
INFORMATION
Study Material
Package
Pre-Medical 2010
All India Test Series
Pre-Medical 2010
(AT CENTER)
Postal Test
Series
Pre-Medical 2010
Major Test Series
Pre-Medical 2010
(AT CENTER)
Postal Major Test
CP Ranker's Package
Series Pre-Medical
IIT-JEE 2010
2010 (BY POST)
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 Class 12th or 12th
pass students
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. 900/including S.Tax
Rs. 8,000/including S.Tax
Rs. 1100/- including Rs. 600/- including
S. Tax
S. Tax
◆For details about all Correspondence & Test Series Course Information, please visit our website.: www.careerpointgroup.com
XtraEdge for IIT-JEE
50
JANUARY 2010
XtraEdge for IIT-JEE
51
JANUARY 2010
XtraEdge for IIT-JEE
52
JANUARY 2010
4.
Let z1, z2, z3 be the complex nos. represent the
given that
vertices of the ∆ABC which is circumscribed by the
circle |z| = 1. Altitude from A meets the side BC at D
a=–
(ii) the complex no. of point E.
1
1
,b=
4
2
2x − x 2
4
f(x) =
A(z1)
 1
since 4x + 4y – 5 = 0 passes through A 1,  and
 4
O
D
1 
B  ,1 so area bounded is OAB = 2OAC
4 
C(z3)
= 2[ar(OCP) + ar(CAQP) – ar(OAQ)]
E(z4)
1 5 5 1  5 1  3
=2 × × +  +  −
 2 8 8 2  8 4  8
Sol. (i) we know that the image of orthocentre about any
side of the ∆ lies on the circum circle of ∆.
Point
...(2)
from (1) and (2)
(i) the complex no. of point P.
B
(z2)
...(1)
0 = 2a + b
and circum circle at E. Let P be the image of E about
BC, then find
P
1
=a+b
4
P = z1 + z2 + z3
=
(ii) Let O (origin) be the circum centre of ∆.
∫
1 2x − x 2
0
4

dx 

37
(unit)2
96
∠BOE = π – 2B and ∠AOC = 2B
6.
z1
= ei 2B
z3
...(1)
z4
= ei(π–2B)
z2
...(ii)
i
for i = 0, 1, 2 ..... n. If n is odd than find
i +1
the value of P(n + 1).
P(i) =
Sol. Let Q(x) = (x + 1) P(x) – x
z z
z1 z 4
.
= –1 ⇒ z4 = – 3 2
z3 z 2
z1
5.
clearly Q(x) is polynomial of degree n + 1. Also
Q(i) = (i + 1)
If A be the area bounded by y = f(x), y = f–1(x) and
the line 4x + 4y – 5 = 0 where f(x) is a polynomial of
2 nd degree passing through the origin and having
maximum value of
y = f (x)
Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a
constant.
1
at x = 1, then find A.
4
Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n)
1 = (–1)n + 1 k(n + 1) !
y=x
⇒ k=
1
( n + 1) !
Thus
P(x) =
A
y=f(x)
O
i
– i = 0 for i = 1, 2, 3 .....n
i +1
Thus we can assume
–1
BC
Let P(x) be a polynomial of degree n such that
P Q x=1

1  x ( x − 1)(x − 2)....( x − n )
+ x ,

x +1 
( x + 1) !

where n is odd
∴ P(n + 1) = 1
Sol. Let f(x) = ax 2 + bx
XtraEdge for IIT-JEE
(Q n is odd)
53
JANUARY 2010
MATH
DIFFERENTIAL
EQUATIONS
Mathematics Fundamentals
Differential Equation :
Formation of Differential Equation :
An equation involving independent variable x,
dependent variable y and the differential coefficients
(1) Write down the given equation.
(2) Differentiate it successively with respect to x that
number of times equal to the arbitrary constants.
dy d 2 y
,
, .... is called differential equation.
dx dx 2
(3) And hence on eliminating arbitrary constants
results a differential equation which involves x, y,
Examples :
(1)
dy
=1+x+y
dx
(2)
dy
+ xy = cot x
dx
dy d 2 y
,
.....
dx dx 2
Solution of Differential Equation :
A solution of a differential equation is any function
which when put into the equation changes it into an
identity.
3
 d4 y 
dy
+ 4y = 5 cos 3x
(3)  4  – 4
 dx 
dx


d2y
(4) x2 2 +
dx
 dy 
1+  
 dx 
General and particular solution :
The solution which contains a number of arbitrary
constant equal to the order of the equation is called
general solution by giving particular values to the
constants are called particular solutions.
2
=0
Order of a Differential Equation :
Several Types of Differential Equations and their
Solution :
The order of a differential equation is the order of the
highest derivative occurring in the differential
equation. For example, the order of above differential
equations are 1, 1, 4 and 2 respectively.
(1) Solution of differential equation
dy
= f(x) is y =
dx
Degree of a Differential Equation :
(2) Solution of differential equation
The degree of the differential equation is the degree
of the highest derivative when differential
coefficients are free from radical and fraction. For
example, the degree of above differential equations
are 1, 1, 3 and 2 respectively.
dy
= f(x) g(y) is
dx
dy
= f(ax + by + c) by
dx
dy
1  dv

putting ax + by + c = v and
= 
−a
dx
b  dx

A differential equation in which the dependent
variable and its differential coefficients occurs only
in the first degree and are not multiplied together is
called a linear differential equation. The general and
nth order differential equation is given below :
dn y
dx
n
+ a1(x)
d n −1y
dx
n −1
+ .... + an – 1
dy
∫ g( y) = ∫ f ( x) dx + c
(3) Solution of diff. equation
Linear and Non-linear Differential Equation :
a0(x)
∫ f (x)dx + c
dv
= dx
a + bf ( v)
dy
dx
Thus solution is by integrating
dv
∫ a + bf (v) = ∫ dx
+ an(x)y + φ(x) = 0
Those equations which are not linear are called nonlinear differential equations.
XtraEdge for IIT-JEE
54
JANUARY 2010
(4) To solve the homogeneous differential equation
f ( x, y)
dy
, substitute y = vx and so
=
g ( x, y)
dx
Solve
a b


obviously = = 2
a´ b´


dy
dv
=v+x
.
dx
dx
Put x – 3y = v
dv
Thus v + x
= f(v)
dx
⇒
⇒1–3
dx
dv
=
x
f ( v) − v
Therefore solution is
dy
2x − 6 y + 7
2( x − 3y) + 7
=
=
dx
x − 3y + 4
x − 3y + 4
∫
dx
=
x
∫
Solution of the linear differential equation :
dv
+c
f ( v) − v
dy
+ Py = Q, where P and Q are either constants or
dx
functions of x, is
Equation reducible to homogeneous form :
A differential equation of the form
ye ∫
dy
a x + b1y + c1
= 1
,
dx
a 2x + b2 y + c2
x = X + h, y = Y + k,
so that
dY
dy
=
dX
dx
P dx

∫  Qe ∫
=
Where e ∫
a
b
where 1 ≠ 1 , can be reduced to homogeneous
a2
b2
form by adopting the following procedure :
Put
dy
dv
=
(Now proceed yourself)
dx
dx
P dx
P dx

 dx + c

is called the integrating factor.
Equations reducible to linear form :
Bernoulli's equation : A differential equation of
dy
the form
+ Py = Qyn, where P and Q are
dx
functions of x alone is called Bernoulli's equation.
Dividing by yn, we get y–n
dy
+ y–(n – 1). P = Q
dx
a X + b1Y + (a 1h + b1k + c1 )
dY
= 1
dX
a 2 X + b 2 Y + (a 2 h + b 2 k + c 2 )
Putting y–(n – 1) = Y, so that
(1 − n ) dy
dY
=
,
dx
y n dx
Now choose h and k such that a1h + b1k + c1 = 0 and
a2h + b2k + c2 = 0. Then for these values of h and k,
the equation becomes
we get
The equation then transformed to
dY
+ (1 – n)P. Y = (1 – n)Q
dx
dY
a X + b1Y
= 1
dX
a 2X + b2Y
which is a linear differential equation.
This is a homogeneous equation which can be solved
by putting Y = vX and then Y and X should be
replaced by y – k and x – h.
dy
+ P. f(y) = Q . g(y), where P and Q are
dx
functions of x alone, we divide the equation by
g(y) and get
If the given equations is of the form
Special case :
1 dy
f ( y)
+ P.
=Q
g( y) dx
g ( y)
ax + by + c
dy
a
b
=
=
= m (say), i.e.
and
dx
a´x + b´y + c´
a´
b´
when coefficient of x and y in numerator and
denominator are proportional, then the above
equation cannot be solved by the discussed before
because the values of h and k given by the equations
will be indeterminate.
If
f ( y)
= v and solve.
g ( y)
Solution of the differential equation :
Now substitute
d2y
= f(x) is obtained by integrating it with respect
dx 2
to x twice.
In order to solve such equations, we proceed as
explained in the following example.
XtraEdge for IIT-JEE
55
JANUARY 2010
MATH
TRIGONOMETRICAL
RATIOS
Mathematics Fundamentals
Some Important Definitions and Formulae :
Trigo.
Function
sin x
Measurement of angles : The angles are measured
in degrees, grades or in radius which are defined as
follows:
Degree : A right angle is divided into 90 equal
parts and each part is called a degree. Thus a right
angle is equal to 90 degrees. One degree is denoted
by 1º.
A degree is divided into sixty equal parts is called a
minute. One minute is denoted by 1´.
A minute is divided into sixty equal parts and each
parts is called a second. One second is denoted by 1´´.
Thus,
1 right angle = 90º (Read as 90 degrees)
1º = 60´ (Read as 60 minutes)
1´ = 60´´ (Read as 60 seconds).
Grades : A right angle is divided into 100 equal
parts and each part is called a grade. Thus a right
angle is equal to 100 grades. One grade is denoted
by 1g.
A grade is divided into 100 equal parts and each
part is called a minute and is denoted by 1´.
A minute is divided into 100 equal parts and each
part is called a second and is denoted by 1"
Thus,
1 right angled = 100g (Read as 100 grades)
1 g = 100´ (Read as 100 minutes)
1´ = 100´´ (Read as 100 seconds)
Radians : A radian is the angle subtended at the
centre of a circle by an arc equal in length to the
radius of the circle.
Domain and Range of a Trigono. Function :
If f : X → Y is a function, defined on the set X,
then the domain of the function f, written as Domf
is the set of all independent variables x, for which
the image f(x) is well defined element of Y, called
the co-domain of f.
Range
R, the set of all the real
number
–1 ≤ sin x ≤ 1
cos x
R
– 1 ≤ cos x ≤ 1
tan x
π


R – ( 2n + 1) , n ∈ I
2


R
cosec x
R – {n π, n ∈ I}
R – {x : –1 < x < 1}
sec x
π


R – ( 2n + 1) , n ∈ I
2


R – {x : –1 < x < 1}
cot x
R – {n π, n ∈ I}
R
Relation between Trigonometrically Ratios and
identities:
sin θ
cos θ
; cot θ =
cos θ
sin θ
sin A cosec A = tan A cot A = cos A sec A = 1
tan θ =
sin2θ + cos2θ = 1
or sin2θ = 1 – cos2θ or cos2θ = 1 – sin2θ
1 + tan2θ = sec2θ
or sec2θ – tan2θ = 1 or sec2θ – 1 = tan2θ
1 + cot2θ = cosec2θ
or cosec2θ – cot2θ = 1 or cosec2θ – 1 = cot2θ
Since sin 2A + cos2A = 1, hence each of sin A
and cos A is numerically less than or equal to
unity. i.e.
| sin A| ≤ 1 and | cos A | ≤ 1
or –1 ≤ sin A ≤ 1 and – 1 ≤ cos A ≤ 1
Note : The modulus of real number x is defined as
|x| = x if x ≥ 0 and |x| = – x if x < 0.
Since sec A and cosec A are respectively
reciprocals of cos A and sin A, therefore the values
of sec A and cosec A are always numerically
greater than or equal to unity i.e.
Range of f : X → Y is the set of all images f(x)
which belongs to Y, i.e.,
sec A ≥ 1 or sec A ≤ – 1
and cosec A ≥ 1 or cosec A ≤ – 1
In other words, we never have
–1 < cosec A < 1 and –1 < sec A < 1.
Range f = {f(x) ∈ Y : x ∈ X} ⊆ Y
The domain and range of trigonmetrical functions
are tabulated as follows :
XtraEdge for IIT-JEE
Domain
56
JANUARY 2010
Trigonometrical Ratios for Various Angles :
Formulae Involving Double, Triple and Half Angles :
θ
0
π
6
π
4
π
3
π
2
π
3π
2
2π
sin θ
0
1
2
1
3
2
1
0
–1
0
cos θ
1
3
2
1
1
2
0
–1
0
1
tan θ
0
3
∞
0
∞
0
2
2
1
1
3
2 tan θ
1 + tan 2 θ
cos 2θ = cos2 θ – sin 2 θ = 2 cos2θ – 1
sin 2θ = 2 sin θ cos θ =
= 1 – 2 sin2θ =
Trigonometrical Ratios for Related Angles :
θ
–θ
π±θ
π
±θ
2
1 − tan 2 θ
1 + tan 2 θ
sin
θ
=±
2
θ
1 − cos θ
; cos
=±
2
2
tan
θ
=±
2
1 − cos θ
1 + cos θ
tan 2θ =
2 tan θ
1 − tan 2 θ
sin 3θ = 3 sin θ – 4 sin3θ
2π ± θ
3π
±θ
2
sin
– sin θ
cos θ
m sin θ
– cos θ
± sin θ
cos
cos θ
m sin θ
– cos θ
± sin θ
cos θ
tan
– tan θ
m cot θ
± tan θ
m cot θ
± tan θ
1
(3 sin θ – sin 3θ)
4
cos 3θ = 4 cos3θ – 3 cos θ
cot
– cot θ
m tan θ
± cot θ
m tan θ
± cot θ
or cos3θ =
or sin3θ =
Addition and Subtraction Formulae :
sin (A ± B) = sin A cos B ± cos A sin B
tan 3θ =
cos (A ± B) = cos A cos B m sin A sin B
tan (A ± B) =
1
(3 cos θ + cos 3θ)
4
3 tan θ − tan 3 θ
1 − 3 tan 2 θ
π

θ ≠ nπ + 6 


Trigonometrical Ratios for Some Special Angles :
tan A ± tan B
1 m tan A tan B
θ
cot A cot B m 1
cot B ± cot A
sin (A + B) sin (A – B) = sin2A – sin2B
= cos2B – cos2A
cos (A + B) cos (A – B) = cos2A – sin 2B
= cos2B – sin2A
Formulae for Changing the Sum or Difference into
Product :
cot (A ± B) =
sin θ
cos θ
tan θ
7
1º
2
4− 2 − 6
22
3 −1
2− 2
2
2 2
2 2
3 +1
4+ 2 + 6
2 2
( 3– 2)
( 2 –1)
2 2
2+ 2
2
2– 3
2 –1
C+D
C−D
cos
2
2
θ
sin C – sin D = 2 cos
C+D
C−D
sin
2
2
sin θ
5 −1
4
10 − 2 5
4
cos θ
10 + 2 5
4
5 +1
4
tan θ
25 − 10 5
5
5−2 5
C+D
C−D
cos
2
2
C+D
D−C
sin
2
2
Formulae for Changing the Product into Sum or
Difference :
2 sin A cos B = sin (A + B) + sin (A – B)
2 cos A sin B = sin (A + B) – sin (A – B)
2 cos A cos B = cos (A + B) + cos (A – B)
2 sin A sin B = cos (A – B) – cos(A + B)
cos C – cos D = 2 sin
XtraEdge for IIT-JEE
18º
1º
2
15º
sin C + sin D = 2 sin
cos C + cos D = 2 cos
1 + cos θ
2
36º
Important Points to Remember :
Maximum and minimum values of
a sin x + b cos x are +
respectively.
57
a 2 + b2 , –
a 2 + b2
JANUARY 2010
sin2x + cosec2x ≥ 2 for every real x.
2
cos x + sec x ≥ 2 for every real x.
tan2x + cot2x ≥ 2 for every real x
If x = sec θ + tan θ, then
A
B
C
A
B
C
+ cos2 – cos2 = 2cos cos sin
2
2
2
2
2
2
5. If x + y + z = π/2, then
sin2x + sin2y + sin2z = 1 – 2 sin x sin y sin z
cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z
sin 2x + sin 2y + sin 2z = 4 cos x cos y cos z
cos2
1
= sec θ – tan θ
x
1
= cosec θ – cot θ
x
cos θ . cos 2θ . cos 4θ . cos 8θ
If x = cosec θ + cot θ, then
6. If A + B + C = π, then
tan A + tan B + tan C = tan A tan B tan C
cot B cot C + cot C cot A + cot A cot B = 1
sin 2 n θ
2 n sin θ
.... cos 2n–1θ =
1
sin 3θ
4
sin θ sin (60º – θ) sin (60º + θ) =
tan
1
cos 3θ
4
tan θ tan (60º – θ) tan (60º + θ) = tan 3θ
cos θ cos (60º – θ) cos (60º + θ) =
A
B
C
cos cos
2
2
2
sin A + sin B – sin C = 4 sin
A
B
C
sin sin
2
2
2
cos A+ cos B+ cos C = 1 + 4 sin
A
B
C
sin sin
2
2
2
cos A+ cos B – cos C = –1 + 4cos
A
B
C
cos sin
2
2
2
A
B
C
A
B
C
+ cot + cot
= cot cot cot
2
2
2
2
2
2
7. (a) For any angles A, B, C we have
sin (A + B + C)
= sin A cos B cos C + cos A sin B cos C
+ cos A cos B sin C – sin A sin B sin C
cos (A + B + C)
= cos A cos B cos C – cos A sin B sin C
– sin A cos B sin C – sin A sin B cos C
tan(A + B + C)
=
sin(A + B + C) = sin π = 0 and
cos (A + B + C) = cos π = –1
then (a) gives
sin A sin B sin C
= sin A cos B cos C + cos A sin B cos C
+ cos A cos B sin C
and (a) gives
1 + cos A cos B cos C
= cos A sin B sin C + sin A cos B sin C
+ sin A sin B cos C
Method of Componendo and Dividendo :
3. If A + B + C = π, then
sin2A + sin 2B – sin2C = 2 sin A sin B cos C
cos2A + cos2B + cos2C = 1– 2 cos A cos B cos C
sin2A + sin 2B + sin2C = 2 + 2 cos A cos B cos C
cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C
p a
= , then by componendo and dividendo we
q b
can write
If
p−q a −b
q−p
b−a
=
=
or
p+q a +b
q+p
b+a
4. If A + B + C = π, then
A
B
C
A
B
C
+sin2 + sin2 = 1– 2 sin sin sin
2
2
2
2
2
2
cos2
A
B
C
A
B
C
+cos2 +cos2 =2 + 2 sin sin sin
2
2
2
2
2
2
XtraEdge for IIT-JEE
tan A + tan B + tan C – tan A tan B tan C
1 − tan A tan B − tan B tan C − tan C tan A
(b) If A,B, C are the angles of a triangle, then
cos A
cos B
cos C
+
+
=2
sin B sin C
sin C sin A
sin A sin B
sin2
B
C
C
A
A
B
tan + tan tan
+ tan tan = 1
2
2
2
2
2
2
cot
Conditional Identities :
1. If A + B + C = 180º, then
sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C
sin (B + C – A) + sin (C + A – B) + sin (A+B – C)
= 4 sin A sin B sin C
cos 2A + cos 2B + cos 2C
= –1 – 4 cos A cos B cos C
cos 2A+ cos 2B – cos 2C = 1 – 4 sin A sinB cosC
2. If A + B + C = 180º, then
sin A + sin B + sin C = 4 cos
A
B
C
A
B
C
+sin2 –sin2 = 1 – 2cos cos cos
2
2
2
2
2
2
sin2
2
or
58
p+q
a+b
q+p
b+a
=
or
=
p−q
a−b
q−p
b−a
JANUARY 2010
Based on New Pattern
IIT-JEE 2010
XtraEdge Test Series # 9
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
•
Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
•
Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and
-1 mark for wrong answer.
•
Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and
-1 mark for wrong answer.
Section - II
•
Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly
matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly
marked answer in any row.
(A)
PHYSICS
(C)
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
a sin θ + b cos θ
If x =
, then –
1.
a+b
(A) The dimension of a and x are same
(B) The dimension of b and x are same
(C) Both (A) and (B)
(D) x is dimensionless
2.
ABCD is a smooth horizontal fixed plane on
which mass m1 = 0.1 kg is moving in a circular
path of radius 1 m. It is connected by an ideal
string which is passing through a smooth hole and
1
kg at the other end as
connects mass m2 =
2
shown. m2 also moves in a horizontal circle of
same radius of 1 m with a speed of 10 m/s. If
g = 10 m/s2, then the speed of m1 is –
A
B
m1
D
C
m2
XtraEdge for IIT-JEE
59
10 m/s
1
m/s
10
(B) 10 m/s
(D) None of these
3.
If x grams of steam at 100ºC becomes water at
100ºC which converts y grams of ice at 0ºC into
water at 100ºC, then the ratio x/y will be –
1
27
(B)
(A)
3
4
4
(C) 3
(D)
27
4.
A gas is at pressure P and temperature T.
Coefficient of volume expansion of one mole of
gas at constant pressure is –
1
1
(D) T2
(B) T
(C) 2
(A)
T
T
5.
A source of light is placed at double focal length
from a convergent lens. The focal length of the
lens is f = 30 cm. At what distance from the lens
should a flat mirror be placed so that ray reflected
from the mirror are parallel after passing through
the lens for the second time ?
(A) Beyond 2 F
(B) Between lens and F
(C) Between F and 2F near 2F
(D) Between F and 2F equidistant from F and 2F
JANUARY 2010
6.
Consider a toroid of circular cross-section of
radius b, major radius R much greater than minor
radius b, (see diagram) find the total energy
stored in magnetic field of toroid –
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
10.
Path of three projectiles are shown. If T1, T2 and
T3 are time of flights and ignoring air resistances y
1
2
3
x
7.
(A)
B2 π 2 b 2 R
2µ 0
(B)
B2 π 2 b 2 R
4µ 0
(C)
B2 π 2 b 2 R
8µ 0
(D)
B2 π 2 b 2 R
µ0
(A) T1 > T3
T +T
(C) T2 = 1 3
2
11.
A small charged ball is hovering in the state of
equilibrium at a height h over a large horizontal
uniformly charged dielectric plate. What would
be the instantaneous acceleration of the ball if a
disc of radius r = 0.001 h is removed from the
plate directly underneath the ball –
g r
(A)
 
2 h
2
g r
 
4 h
2
(C)
8.
R
1.5R
2R
g h
(B)
 
2 r
2
g h
 
4 r
2
(D)
(A) K = 8J
(C) e = 0.5
a =
13.
A solenoid is connected to a source of constant
emf for a long time. A soft iron piece is inserted
into it. Then –
(A) self inductance of the solenoid gets increased
(B) flux linked with the solenoid increases hence
steady state current gets decrease
(C) energy stored in the solenoid gets increased
(D) magnetic moment of the solenoid increased
→
→
3 î + ĵ and | b | = 10 units while θ = 23º,
→
→
then the value of R = | a + b | is nearly –
y
→
b
→
a
θ
O
(A) 12
x
(B) 13
XtraEdge for IIT-JEE
(B) K = 16J
(D) e = 0
A constant voltage is applied between two ends of
a uniform conducting wire. If both the length and
radius of the wire are doubled –
(A) the heat produced in the wire will be doubled
(B) the electric field across the wire will be
doubled
(C) the heat produced will remain unchanged
(D) the electric field across the wire will become
half
For the vectors a and b shown in figure,
→
t
12.
(D) 10 30/sec
→
Two blocks of masses 2 kg each are moving in
opposite direction with equal speed collides at
t = 5 sec. The magnitude of relative velocity (v) is
plotted against time 't'. The loss in kinetic energy
is K and coefficient of restitution in e, then –
v
t = 5 sec
electron and positron incident photon must have
minimum frequency of the order of –
(B) 10 21/sec
(A) 10 18/sec
9.
(D) T1 = T2 = T3
4 m/s
For pair production i.e. for the production of
(C) 10 25/sec
(B) T1 < T3
(C) 14
(D) 15
60
JANUARY 2010
14.
In radioactivity decay according to law N = N0e–πt
which of the following is/are true ?
(A) Probability that a nucleus will decay is 1 – e–λt
(B) Probability that a nucleus will decay four half
lives is 15/16
(C) Fraction nuclei that will remain after two half
lives is zero
(D) Fraction of nuclei that will remain after two
half-lives is 1/4
Scientists can observe the spectral lines of atoms
that are dominant in far-away galaxies. Due to
the speed at which these galaxies are travelling,
these lines are shifted, but their pattern remains
the same. This allows researchers to use the
spectral pattern to determine which atoms they
are seeing. Table - 1 below shows spectroscopic
measurements made by researchers trying to
determine the atomic makeup of a particular faraway galaxy. Light energy is not measured
directly, but rather is determined from measuring
the frequency of light, which is proportional to
the energy.
Table – 1
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 15 to 17)
ε
10 Ω
Frequencies Measured
868440
880570
879910
856390
V
6
10×106Ω
Al foil
Voltmeter is ideal and two aluminum foil is given at
certain separation. In this set up upper aluminum foil
jumps to the lower Al foil at the potential difference
between the plates (or foils) of 500 V.
15.
When emf ε of the battery is 400 volt and foil has
not jumped, approximate reading of the voltmeter
is (A) 500 V
(B) 400 V
(C) 0 V
(D) 250 V
16.
What is the reading of voltmeter just after the foil
has jumped and connected the two plates (A) 500 V
(B) less than 500 V
(C) more than 500 V
(D) 600 V
17.
What is the emf of the battery just after the foil
has jumped and connected the two plates (A) 500 V
(B) 600 V
(C) 700 V
(D) 800 V
18.
For each of three hypothetical atoms (Atom 1,
Atom 2 and Atom 3), Figure 1 depicts the (A) number of electrons and the amount of
energy the atom contains
(B) distance an electron travels from one part of
the atom to another
(C) energy released by the atom as an electron as
it moves from one energy state to another
(D) frequency with which the atom’s electrons
move from one energy state to another
19.
Based on the spectroscopic measurements shown
in Table - 1, which of the atoms in Figure (i)
(Atom 1, Atom 2, or Atom 3) is most similar to
the one the scientists were observing, and why ?
(A) Atom 2, because it contains four different
energy levels
(B) Atom 3, because it contains four different
energy levels
(C) Atom 1, because the frequencies listed in
Table 1 indicate a high level of atomic
activity
(D) Atom 3, because there is a comparatively
small difference between exactly two of the
four frequencies listed in Table 1
20.
The laws of atomic physics prohibit electron
movements between certain energy states. In
atomic physics, these prohibitions are called
“forbidden transitions.” Based on Figure (i),
which of the following is most accurate ?
(A) Atom 2 has the same number of forbidden
transitions as Atom 1
Passage : II (No. 18 to 20)
Figure (i), below depicts three hypothetical
atoms.
Energy levels are represented as
horizontal segments. The distance between the
segments is representative of the energy
difference between the various levels.
All
possible transitions between energy levels are
indicated by arrows.
Atom # 1
Atom # 2
Atom # 3
Fig (i)
XtraEdge for IIT-JEE
61
JANUARY 2010
(B) Atom 2 has more forbidden transitions than
Atom 3
(C) Atom 3 has the same number of forbidden
transitions as Atom 1
(D) Atom 1 has fewer forbidden transitions than
Atom 2
CHEMISTRY
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
1.
Brown ppt. (A) dissolve in HNO3 gives (B) which
gives white ppt. (C) with NH4OH. (C) on reaction
with HCl gives solution (D) which gives white
turbidity on addition of water. What is (D) ?
(A) BiCl3
(B) Bi(OH)3
(C) BiOCl
(D) Bi(NO3)3
2.
Histidine (A) has pKa values as indicated
O
CH2
6.04
N
C P Q R S
D P Q R S
21.
CH C OH
+
NH
NH3
9.17
1.82
(A)
What will be its form at pH = 4 ?
Column I
Column II
(A) Pair production
(P) Few MeV
(B) Inverse photoelectric
(Q) 20 KeV
effect
(C) De-excitation of Be+3 (R) 54 eV
atom from second
excited state
(S) 0.1 eV
(D) Kα – X-ray photons
of molybdenum
Z = 42
O
CH2–CHCOH
(A) HN
+
⊕
NH3
NH
O
–
CH2CHC O
(B) HN
+
+
NH
NH3
O
CH2CHC
(C) N
22.
+
NH
–
O
NH3
O
µ1
µ2
60°
30°
(A)
µ1
µ1
µ2
µ2
θc
(B)
Column I
(Α) θc
 1 

(Β) sin–1 
 3
(C) Refractive index of 1
with respect to 2
(D) Total internal reflection
XtraEdge for IIT-JEE
–
CH2CHC O
(D) N
α β
(C) α > θc
The correct order of M–C π bond and strength in
given metal carbonyl is(A) [Fe(CO)4]2– > [Co(CO)4]– > [Ni(CO)4]
(B) [Ni(CO)4] > [Co(CO)4]– > [Fe(CO)4]2–
(C) [Fe(CO)4]2– > [Ni(CO)4] > [Co(CO)4]–
(D) [Ni(CO)4] > [Co(CO)4]– = [Fe(CO)4]2–
4.
Compound 'A' (molecular formula C 3H 8O) is treated
with acidified potassium dichromate to form a
product B (molecular formula C3H6O). 'B' form a
shining silver mirror on warming with ammonical
silver nitrate 'B' when treated with an aqueous
solution of H2NCONHNH2.HCl and sodium acetate
gives a product 'C'. Identify the structure of C.
(Q) Critical angle
(S) α = β
62
NH2
3.
Column II
(P) 45°
 1 

(R) 
 3
NH
JANUARY 2010
(A) CH3C
NCONHNH2
+
CH3
(B) CH3CH2CH
N NHCONH2
(C) CH3 C NNHCONH2
9.
The product 'A' is
(A) Me
CH3
(D) CH3CH2CH = NCONHNH2
5.
(C) Ph


 P + n a  (V – nb) = nRT
2 

V 

2
Q
Q
(D)
(C)
T
6.
T
A solution containing NaOH and Na2CO3 was titrated
against HCl using phenolphthalein as an indicator.
The tire value of HCl solution was found to be x ml.
At the end point, methyl orange was added and the
titration continued. A further y ml of HCl solution
was required to get the end point with methyl orange.
The volume of HCl solution used with Na2CO 3
during the whole process is
(A) 2x
(B) 2y
(C) x
(D) y – x
7.
For crystallisation of a solid from the aqueous
solution, if the values of ∆H and ∆S are –x J mol–1
and – y J K–1 mol–1 respectively, which of the
following relationships is correct
(A) x = T × y
(B) x > T × y
(C) x < T × y
(D) None of these
8.
Reduction of but-2-yne with Na and liquid NH3 gives
an alkene which upon catalytic hydrogenation with
D2/Pt gives an alkane. The alkene and alkane formed
respectively are
(A) cis but-2-ene and racemic-2, 3-dideuterobutane
(B) trans but-2-ene and meso 2, 3-dideuterobutane
(C) trans but-2-ene and racemic 2, 3-dideuterobutane
(D) cis but-2-ene and meso 2, 3-dideuterobutane
XtraEdge for IIT-JEE
Ph
OH
H
Me
(D)
H
Ph
10. The following complexes are given
(1) trans – [Co(NH3)4Cl2]+
(2) cis – [Co(NH3)2 (en)2] 3+
(3) trans – [Co(NH3)2(en)2]3+
(4) NiCl42–
(5) TiF62–
(6) CoF63–
Choose the correct code :
(A) (1), (2) are optically active, (3) is optically inactive
(B) (2) is optically active, (1), (3) are optically inactive
(C) (4), (6) are coloured and (5) is colourless
(D) (4) is coloured and (5), (6) are colourless
(B)
Q
(B)
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
T
T
OH
OH
ab
with temperature is
Plot of quantity Q =
a+b
(A)
Ph
O
Vander Waal's equation for a real gas is
Q
1. AlCl3
A
2. H+/H2O
O
11. Cellulose is made of glucose units joined together by
β-1, 4-glycosidic linkages. These molecules are held
by(A) ionic bonds
(B) hydrogen bonds and van der Waal's forces
(C) weak van der Waal's forces only
(D) All of these
12. Which of the following statements is/are correct. ?
(A) The monoatomic gas He has lower entropy than
the triatomic gas CO2, which has lower entropy
than gaseous benzene
(B) For single atoms, the absolute entropy increases
as the number of electrons and the protons
increases
(C) Among the solid elements the absolute entropy
generally increases as the atomic number
increases
(D) All metallic solids have entropies below 85 J.
mole–1K–1
13. In the Libermman test for phenols, the blue or green
colour produced is due to the formation of -
63
(A) OH
OH
(B) O =
= NOH
JANUARY 2010
(C) O =
=N–
–+
– ONa
(D) O =
=N–
– OH
15. Aldehydes with H at α-carbon do not undergo this
type of reaction
O
R
16. Identify the products in the following reactions ?
O
D
C
(i)
O
(A) D
C
(B) H
H
O
(C) H
(ii) R
C
R
C
(D) D
C
R
C
O
CH3OH H
O
CH2DOH D
C
O– CH2DOH
D
C
O–
CD3OH
O
CCl3
H
C
C
H
(ii)
O
H
C
H
Cl
(iv)
(iii)
The reaction is not given by(A) (i), (ii)
(B) (ii), (iii)
(C) (iii), (i)
(D) only (iii)
–
O
OH + R
C
H
Passage : II (No. 18 to 20)
Equilibrium constant are given (in atm) for the
following reaction at 0ºC :
SrCl2. 6H2O(s)
SrCl2.2H2O(s) + 4H2O(g) :
Kp = 5 × 10–12
Na2HPO4.12H2O(s)
NaSO4.7H2O(s) +5H2O(g) :
CH2O
O
C
O– CH2DOH
C
CH2
H
R
O– CH2DOH
C
O
–
O
O
OH + R
C
O– CH2DOH
C
O
–
(i)
H
O
C
C
H
H
OH
(iii) R
C
(iv)
17. Consider the following aldehydes :
O
O
O
H + R
O– CH2D–OH D
O
OH
O
(iii)
O
O
OH
H + OH
–
C
(ii)
O
R C O + R CH2OH
All non-enolisable aldehydes undergo such type of
ractions in strongly basic medium. These type of
reactions are disproportionation reaction in which
one molecule gets oxidised and other reduced.
–
O
O
(i) R
O + OH → acid salt(iii) + alcohol(iv)
C
2D
Passage : I (No. 15 to 17)
O
O
H+R
D + OH → acid salt(i) + alcohol(ii)
C
2H
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
C
H OH (no reaction)
C
because (A) aldehyde is enolised in basic condition
(B) bond energy of C–H increase
(C) steric hinderance increase
(D) all the above
14. Which of the following statements are correct for SN2
reaction.
(A) Increasing the polarity of solvent causes a large
increase in the rate of SN2 attack by NH3 on alkyl
halide
(B) Increasing the polarity of solvent causes a large
decrease in the rate of SN2 attack by OH– ion on
trimethyl sulphonium ions
(C) Increasing the polarity of solvent causes a small
decrease in the rate of S N2 attack by trimethyl
amine on trimethyl sulphonium ion
(D) Increasing the polarity of solvent causes a large
increase in the rate of S N2 attack by OH– ion on
trimethyl sulphonium ion
R
CH2
O + R CH2OH
Kp = 2.43 × 10–13
XtraEdge for IIT-JEE
64
JANUARY 2010
Na2SO 4.10H2O(s)
Na2SO4 (s) + 10H2O(g) :
MATHEMATICS
Kp = 1.024 × 10 –27
The vapour pressure of water at 0ºC is 4.56 torr.
18.
19.
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
Which is the most effective drying agent at 0ºC ?
(A) SrCl2.2H2O
(B) Na2HPO4.7H2O
(C) Na2SO4
(D) all equal
1.
At what relative humidities will Na2SO4.10H2O
lose water of hydration when exposed to air at
0ºC?
20.
(A) above 33.33%
(B) below 33.33%
(C) above 66.66%
(D) below 66.66%
z1 and z2 lie on a circle with centre at the origin. The
point of intersection z3 of the tangents at z1 and z2 is
given by (A)
1
( z1 + z 2 )
2
(C)
1
2
At what humidities will Na2SO4 absorb moisture
when exposed to air at 0ºC ?
(A) above 33.33%
(C) above 66.66%
2.
(B) below 33.33%
(D) below 66.66%
3.
2z1z 2
z1 + z 2
(D)
z1 + z 2
z1z 2
The set of all x satisfying the equation
x log 3 x
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
 1
1 
 + 
 z1 z 2 
(B)
2
+ (log 3 x ) 2 −10
= 1/x2 is
(A) {1, 9}
(B) {1, 9, 1/81}
(C) {1, 4, 1/81}
(D) {9, 1/81}
Value of
S=1×2×3×4+2×3×4×5
+ … + n (n + 1) (n + 2) (n + 3) is -
C P Q R S
D P Q R S
(A)
1
n (n + 1) (n + 2) (n + 3) (n + 4)
5
(B)
1 n+3
( C5)
5!
(C)
1 n+4
( C4)
5
(D) none of these
−1
21. Column –I
(A) NH3 → NO3–
(B) Fe2S3 → 2FeSO4 + SO2
(C) KMnO4 in acidic medium
(D) CuS → CuSO4
Column-II
(P) M/20
(Q) M/5
(R) M/8
(S) M
22. Column-I
(A) CuCl2.2H2O
Column-II
(P) Colourless and
diamagnetic
(Q) Green coloured
and
paramagnetic
(R) Calamine
(S) Black in colour,
basic in nature
(B) Cu2Cl2
(C) CuO
(D) ZnCO 3
XtraEdge for IIT-JEE
4.
tan θ 
− tan θ   1
 1
 a − b
 
 = 
If 
,
θ
−
θ
tan
1
tan
1



b a 
then (A) a = b = 1
(B) a = cos 2θ, b = sin 2θ
(C) a = sin 2θ, b = cos 2θ
(D) a = 1, b = sin 2θ
5.
A natural number x is chosen at random from the
first one hundred natural numbers. The probability
( x − 20)( x − 40)
< 0 is
that
x − 30
(A) 1/50
(B) 3/50
(C) 3/25
65
(D) 7/25
JANUARY 2010
6.
Two flagstaffs stand on a horizontal plane. A and B
are two points on the line joining their feet and
between them. The angles of elevation of the tops of
the flagstaffs as seen from A are 30º and 60º and as
seen from B are 60º and 45º. If AB is 30 m, then
distance between the flagstaffs in metres is -
12.
(A) 30 + 15
13.
(A) K = –1/4
(C) M = – 2
(D) 60 + 15 3
(C) 60 – 15 3
7.
M cot x + C then -
(B) 45 + 15 3
3
If g(x) is a polynomial satisfying g(x) g(y) = g(x) +
g(y) + g(xy) – 2 for all real x and y and g(2) = 5
then lim g(x) is -
14.
8.
9.
1+ x 2 −1
, then x
(A) y′ (1) = 1
(B) y′ (1) = 1/4
(D) y′ (1) does not exist
x, y, z ∈ {1, 2, …, n, n + 1} is -
(C) 12 + 22 + … + n2
1
n (n + 1) (2n + 1)
6
(D) 2(n+2C3) – n+1C2
1
1
If f(x) = cos 2 x cos 2 x cosec 2 x then 1
π/4
(A)
cos 2 x
cot 2 x
17. Let m, n ∈ N and
p(x) = 1 + x + … + xm.
p(x) will divide p(xn) if
(A) hcf (m, n) = 1
(B) hcf (m + 1, n) = 1
(C) hcf (m, n + 1) = 1
(D) hcf (m+1, n+1) = 1
1
∫ f ( x)dx = 16 (3π + 8)
−π / 4
(B) f ′ (π/2) = 0
(C) Maximum value of f(x) is 1
(D) Minimum value of f(x) is 0
XtraEdge for IIT-JEE
(D) none of these
16. Sum of the rational roots of
133x − 78
x5 =
133 − 78x
is (A) 2/9
(B) 9/2
(C) 13/6
(D) 6/13
sec 2 x
11.
x3/2
15. If p(x) is a reciprocal polynomial of odd degree, then
one of the roots of p(x) = 0 is (A) – 1
(B) 1
(C) 0
(D) (n + 1)/2
The number of ways of choosing triplets (x, y, z)
such that z ≥ max {x, y} and
(B)
x3/2
3 a
A polynomial
p(x) = a0 xn + a1 xn–1 + … + an–1 x + an, a0 ≠ 0 is said
to be a reciprocal equation if ai = an – i for 0 ≤ i ≤ [n/2]
where [x] denote the greatest integer ≤ x.
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
C3
3 a
−2
Passage : I (No. 15 to 17)
(B) 2
(D) none of these
n+2
2
(B) y + C =
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
The degree of the differential equation satisfying
(A) n+1C3 +
(B) a multiple of π/2
(D) a multiple of π
The orthogonal trajectories of the system of curves
(C) y + C =
1 − x 2 + 1 + y 2 = a(x – y) is -
10.
0
(A) 9a(y + C)2 = 4x3
If y = tan –1
(A) 1
(C) 3
π/2
dx
is 1 + tan x
(A) a multiple of π/4
(C) equal to π/4
∫
2
(B) 25
(D) none of these
(C) y′ (1) = 0
(B) L = 2
(D) none of these
 dy 
  = a/x are  dx 
x →3
(A) 9
(C) 10
∫
If I = sec 2 x cosec4 x dx = K cot3 x + L tan x +
66
JANUARY 2010
Passage : II (No. 18 to 20)
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
A straight line is called an asymptote to the curve
y = f(x)( if the distance from the variable point M of
the curve to the straight line approaches zero as the
point M recedes to infinity along some branch of the
curve. We have three kinds of asymptotes; vertical,
horizontal and inclined.
There are three types of asymptotes vertical,
horizontal and inclined.
Vertical asymptotes If at least one of lim f(x) or
x →a +
C P Q R S
D P Q R S
lim f(x) is equal to infinity then x = A is a vertical
x →a −
asymptote. If
lim f(x) = A then y = A is a
x →± a
21.
f ( x)
=k2, lim
x →± a x
x →± a
[f(x) – kx] = b 2 then y = k 2x + b 2 is an inclined
asymptote.
horizontal asymptote. If limits lim
18. Let y =
Column-II
(A) locus of point of intersection (P)
of the lines x = at2, y = 2at
(B) locus of the point of
(Q)
intersection of the
perpendicular tangents to
the circle x 2 + y2 = a2
(C) locus of the point of
(R)
intersection of the lines
x cos θ = y cot θ = a
(D) The locus of the mid
(S)
points of the chords of the
circle x2 + y2 – 2ax = 0
passing through the origin
3x
+ 3x be curve 1 and y = xe1/x be curve 2
x −1
then (A) curve 1 has no horizontal asymptote and curve 2
has no vertical asymptote
(B) y = 3x + 3 and y = x + 2 inclined asymptotes to
curve 1 curve 2
(C) y = 3x + 3 and y = x + 1 are inclined asymptotes
to curve 1 and curve 2
22.
(D) y = x + 1 and y = 3x + 3 are inclined asymptotes
to curve 1 and curve 2
19. Let y =
Column-I
Column-I
(A) f(x) =
5x
then
x −3
x2 + y2 = 2a2
y2 = 4ax
x2 + y2 = ax
x2 – y2 = a2
Column-II
( x + 3) 2
x 2 +1
,
(P) 0 ≤ f(x) ≤ 3
–∞<x<∞
(B) R = {(x, y): x, y ∈ R,
(A) There are no vertical asymptotes
2
(Q) 3 ≤ f(x) ≤ 9
2
x + y ≤ 25}
(B) There are no horizontal asymptotes
R′ = {(x, y): x, y ∈ R,
(C) There are no inclined asymptotes
y ≥ 4x2/9}
(D) x = 3 and y = 5 are the only asymptotes
and let (x, f(x)) = R ∩ R′
2
20. Let y = x + 1 sin 1/x
(C) f(x) =
(A) There are no horizontal asymptotes
9
2 − cos 3x
(S) 0 ≤ f(x) ≤ 5
(D) f(x) =
(B) There is only one horizontal asymptote
(C) There are only two horizontal asymptotes
3 2 sin
(R) 0 ≤ f(x) ≤ 10
( π 2 / 16) − x 2
(D) There is one vertical asymptote
XtraEdge for IIT-JEE
67
JANUARY 2010
Based on New Pattern
IIT-JEE 2011
XtraEdge Test Series # 9
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
• Question 10 to 14 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and
-1 mark for wrong answer.
• Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and
-1 mark for wrong answer.
Section - II
• Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly
matched answer and No Negative marks for wrong answer. However, +1 marks will be given for a correctly
marked answer in any row.
y
PHYSICS
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
L
Two blocks A and B, have equal masses. They are
x
placed next to each other on a horizontal
frictionless fixed table. Compare the velocities of
L
the blocks as each of them reaches the opposite
end of the table –
A
4F
B
F
(A) vA = 2 vB
(C) vA = 8 vB
2.
A
B
XtraEdge for IIT-JEE
68
L L
(C)  , 
3 3
L L
(D)  , 
3 6
A body of mass 1 kg starts moving from rest at
t = 0, in a circular path of radius 8 m. Its kinetic
energy varies as a function of time as :
KE = 2t2 Joules, where t is in seconds. Then (A) Tangential acceleration = 4 m/s2
(B) Power of all forces at t = 2 sec is 8 watt
(C) First round is completed in 2 sec
(D) Tangential force at t = 2 sec is 4 newton
4.
With what minimum velocity should block be
projected from left end A towards end B such that
it reaches the other end B of conveyer belt
moving with constant velocity v. Friction
coefficient between block and belt is µ .
Centre of mass of two thin uniform rods of same
of co-ordinates –
 2L L 
, 
(B) 
 3 2
3.
(B) vA = 4 vB
(D) vA = 16 vB
length but made up of different materials and kept
as shown, can be, if the meeting point is the origin
L L
(A)  , 
2 2
JANUARY 2010
A
v0
m
touching (event 1) or make an elastic head-on
collision (event 2) (A) we can never make out which event has
occurred
(B) we can not make out which event has
occurred only if v 1 = v 2
(C) we can always make out which event has
occurred
(D) we can make out which event has occurred
only if v1 = v2
B
µ
v
L
5.
(A)
µgL
(B)
2µgL
(C)
3µgL
(D) 2 µgL
9.
A block of mass m is attached to a pulley disc of
equal mass m, radius r by means of a slack string
as shown. The pulley is hinged about its centre on
a horizontal table and the block is projected with
an initial velocity of 15 m/s. Its velocity when the
string becomes taut will be –
The escape velocity from the earth is about
11 km/s. The escape velocity from a planet
having twice the radius and the same mean
density as the earth is (A) 22 km/s
(B) 11 km/s
(C) 5.5 km/s
(D) 15.5 km/s
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
(A) 5 m/s
(C) 7.5 m/s
(B) 6 m/s
(D) 10 m/s
6.
A beaker containing water is placed on the
platform of a spring balance. The balance reads
1.5 kg. A stone of mass 0.5 kg and density
500 kg/m3 is completely immersed in water
without touching the walls of beaker. Now the
balance reading will be (A) 2 kg
(B) 1 kg
(C) 2.5 kg
(D) 3 kg
7.
A uniform rod of mass M1 is hinged at its upper
end. A particle of mass M2 moving horizontally
strikes the rod at its mid point elastically. If the
particle comes to rest after collision, the value of
M1
is –
M2
10.
The potential energy of a particle of mass 0.1 kg,
moving along x-axis is given by U = 5x (x – 4) J,
where x is in meters. It can be concluded that (A) The particle is acted upon by a constant force
(B) The speed of the particle is maximum at
x=2m
(C) The speed of the particle is maximum at
x=0m
(D) The period of oscillation of the particle is
π
sec
5
11.
A particle of mass m is at rest in a train moving
with constant velocity with respect to ground.
Now the particle is accelerated by a constant force
F0 acting in the direction of motion of train for
time t0. A girl in the train and a boy on the ground
measure the work done by this force. Which of
the following are incorrect ?
(A) Both will measure the same work
(B) Boy will measure higher value than the girl
(C) Girl will measure higher value than the boy
(D) Data are insufficient for the measurement of
work done by the force F0
12.
In figure, two blocks M1 and m2 are tied together
with an inextensible and light string. The mass M1
is placed on a rough horizontal surface with
coefficient of friction µ and the mass m2 is
hanging vertically against a smooth vertical wall.
The pulley is frictionless –
v
M2
(A)
8.
3
4
M1
4
(B)
3
(C)
2
3
(D)
3
2
Two identical spheres move in opposite directions
with speeds v1 and v2 and pass behind an opaque
screen, where they may either cross without
XtraEdge for IIT-JEE
69
JANUARY 2010
B
M1
1.5 m
Rough (µ)
A
Smooth
C
m2
1m
4m
D
Choose the correct statement (s) related to the
tension T in the string (A) When m2 < µM1, T = m2g
15.
(B) When m2 < µM1, T = M1g
(C) When m2 > µM1, µM1g < T < m2g
(D) When m2 > µM1 , m2g < T < µM1g
13.
14.
Overall changes in volume and radii of a uniform
cylindrical steel wire are 0.2% and 0.002 %
respectively when subjected to some suitable
force. If Young’s modulus of elasticity of steel is
Y = 2.0 × 1011 N/m2, then (A) Longitudinal tensile stress acting on the wire
is 4.08 × 108 N/m2
(B) Longitudinal tensile stress acting on the wire
is 3.92 × 108 N/m2
(C) Longitudinal strain is 0.204 %
(D) Longitudinal strain is 0.196 %
(A) 4
5 m/s
(B)
(C) 5
2 m/s
(D) 10 m/s
130 m/s
16.
Gauge pressure at the highest point B is (A) – 52 kPa
(B) – 44 kPa
(C) – 20 kPa
(D) – 12 kPa
17.
Gauge pressure at point C is (A) – 32 kPa
(B) 8 kPa
(C) 20 kPa
(D) 0
Passage : II (No. 18 to 20)
Two pulse are traveling in opposite direction with
speed 1 m/s. Figure shows the shape of pulse at
t = 0.
y
A particle falls freely near the surface of the earth.
Consider a fixed point O (not vertically below the
particle) on the ground (A) Angular momentum of the particle about O is
increasing
(B) Torque of the gravitational force on the
particle about O is decreasing
(C) The moment of inertia of the particle about O
is decreasing
(D) The angular velocity of the particle about O
is increasing
1 m/s
10 mm
5mm
x
– 5mm
2
4
6
8
10
12 14
16
Distance (in cm)
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 15 to 17)
A siphon tube is discharging a liquid of specific
gravity 0.8 from a reservoir as shown in figure.
(Take g = 10 m/s2).
XtraEdge for IIT-JEE
Velocity of the liquid coming out of siphon at D
is -
70
18.
Speed of particle at x = 2 cm and t = 0 is (A) 1 m/s
(B) 0.75 m/s
(C) 0.5 m/s
(D) 0.25 m/s
19.
Displacement of particle at x = 8 cm and t = 6 sec
is (A) 10 mm
(B) 5 mm
(C) – 5 mm
(D) zero
20.
Speed of particle at x = 8 cm and t = 6 sec (A) zero
(B) 0.125 m/s
(C) 0.25 m/s
(D) None of these
JANUARY 2010
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
Column – I
(A)
(B)
→
v1
(P)
= î + ĵ ,
Acceleration of block
→
Normal force by
22.
Net force on the wedge
= – î – ˆj
log P
A(vA)
log V
(Q)
>θ
(A) number of collision increases
(R)
θ
2.
A compound containing only sodium, nitrogen and
oxygen has 33.33% by weight of sodium. What is the
possible oxidation number of nitrogen in the
compound?
(A) –3
(B) + 3
(C) –2
(D) + 5
3.
How many moles of nitrogen is produced by the
oxidation of one mole of hydrazine by 2/3 mole
bromate ion
1
2
(A)
(B) 1
(C) 1.5
(D)
3
3
4.
For the reaction
(S)
2
y
BaSO4 (s)
3
BaSO4 (aq)
The equilibrium moles of BaSO4(aq) were
0.2 moles. The equilibrium constant of the above
x
4
XtraEdge for IIT-JEE
VA
times
VB
(B) number of moles in this process is constant
(C) it is isothermal process
(D) it is possible for ideal gas
A rigid cylinder is kept on a smooth horizontal
surface as shown. If column I indicates velocities
of various points (3-centre of cylinder, 2-top
point, 4-bottom point, 1-on the level of 3 at the
rim) on it shown. Choose correct state of motion
from column – II.
1
(S) Not possible
A compression of an ideal gas is represented by curve
AB, which of the following is wrong
B(vB)
θ
block on wedge
(D)
(R) Rolling without
slipping to right
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
A relative to wedge
(C)
slipping to left
= – î
A relative to ground
(B)
(Q) Rolling without
CHEMISTRY
1.
Acceleration of the block
centre
→
θ
(A)
(P) Pure rotation about
= 2 î
(D) v 4 = 0,
A
Column-II
= î + ĵ ,
→
In the situation shown, all surfaces are frictionless
and triangular wedge is free to move. In column II
the direction of certain vectors are shown. Match
the direction of quantities in column I with
possible vector in column II.
Column-I
Column – II
(C) v 2 = î , v 3 = 0
C P Q R S
D P Q R S
21.
→
v1
→
v2
→
v1
→
v3
reaction is 71
JANUARY 2010
(A) 0.2
(C) 2 × 10 –4 mol L–1
5.
(B) 0.2 mol L–1
(D) Data insufficient
12. The order of Keq values for the following keto-enol
equilibrium constants is
k
CH3 CHO 1 CH2 CH OH
O
O
k2
CH3 C CH2 C CH3
OH
O
Which of the following correctly explain the nature
of boric acid in aqueous medium 2O
(A) H3BO3 H
→ H3O + + H2BO3–
2 H 2O
(B) H3BO3 
→ 2H3O + + HBO 32–
CH3
O
3–
H 2O
(C) H3BO3 3
→ 3H3O+ + BO 3
CH3
–
2O
(D) H3BO3 H
→ B(OH)4 + H+
6.
7.
(B) Square planar
(D) See-Saw
CH3
C
(B) K2 > K3 > K1
(D) K1 > K3 > K2
the pure Zr is deposited on W
passed over
(B) 2B + 3I2 → 2BI3(g) the
white


→
hot W
the pure B is deposited on W
mixed with W
(C) Zr + 2I2 → ZrI4 (s) &
then

→
heated
ZrI4 is reduced to ZrI2
The correct order of increasing boiling point is (A) NH3 > HF > H2O
(B) H2O > HF > NH3
(D) none of these
(D) HF > H2O > NH3
14. Which of the following statements is correct ?
(A) At 273ºC, the volume of a given mass of a gas at
0ºC and 1 atm. pressure will be twice its volume
(B) At –136.5ºC, the volume of a given mass of a gas
at 0ºC and 1 atm. pressure will be half of its
volume
(C) The mass ratio of equal volumes of NH3 and H2S
under similar conditions of temperature and
pressure is 1 : 2
(D) The molar ratio of equal masses of CH4 and SO2
is 4 : 1
Oxidation states of carbon and nitrogen in KCN are,
respectively (A) – 3, + 2
(B) + 2, – 3
(C) + 1, – 2
(D) zero each
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
10. Which of the following samples of reducing agents
is/are chemically equivalent to 25 mL of 0.2 N
KMnO4, to be reduced to Mn2+ + H2O ?
(A) 25 mL of 0.2 M FeSO4 to be oxidized to Fe3+
(B) 50 mL of 0.1 MH3AsO3 to be oxidized to H3AsO 4
(C) 25 mL of 0.2 M H2O2 to be oxidized to H+ and O2
(D) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 15 to 17)
Entropy is measure of degree of randomness. Entropy
is directly proportional to temperature. Every system
tries to acquire maximum state of randomness or
disorder. Entropy is measure of unavailable energy.
Unavailable energy = Entropy × Temperature
The ratio of entropy of vapourisation and boiling
point of substance remains almost constant.
11. A sample of water has a hardness expressed as 77.5
ppm Ca2+. This sample is passed through an ion
exchange column and the Ca2+ is replaced by H+.
Select correct statement(s)
(A) pH of the water after it has been so treated is 2.4
(B) Every Ca2+ ion is replaced by one H+ ion
(C) Every Ca2+ ion is replaced by two H+ ions
(D) pH of the solution remains unchanged
XtraEdge for IIT-JEE
CH3
passed over
(C) NH4+ < HF < H3O+ < H2O < OH–
(D) H3O+ < NH4+ < HF < OH– < H2O
9.
CH2
C
(A) Zr + 2I2 → ZrI4(g) the
white


→
hot W
Arrange NH4+, H2O, H3O +, HF & OH– in increasing
(C) NH3 > H2O > HF
k3
CH
13. In the purification Zr and B, which of the following
is/are true ?
order of acidic nature (A) OH– < H2O < NH4+ < HF < H3O+
(B) H3O + > HF > H2O > NH4+ > OH–
8.
CH3
(A) K1 > K2 > K3
(C) K2 > K1 > K3
The shape of TeCl4 is (A) Linear
(C) Tetrahedral
C
C
OH
72
JANUARY 2010
15. Which of the following process have ∆S = – ve ?
(A) Adsorption
(B) Dissolution of NH4Cl in water
(C) H2 → 2H
(D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O
16. Observe the
statement(s)
graph
and
identify
the
⇒ log
18.
correct
Entropy
Β ∆Svap
19.
Α ∆Sfusion
 1
1 
 –

 T1 T2 
If standard heat of dissociation of PCl5 is 230 Cal.
1
then the slope of the graph of log K vs
is T
(A) + 50
(B) – 50
(C) 10
T1
T2
Temperature
(A) T 1 is melting point, T2 is boiling point
(B) T 1 is boiling point, T2 is melting point
(C) ∆ Sfusion is more than ∆ S vap
(D) T 2 is lower than T1
(D) None of these
For exothermic reaction of ∆S o < 0 then the
1
sketch of log K vs
may be T
log K
log K
(A)
(B)
1/T
1/T
log K
log K
(C)
(D)
1/T
17. The law of Thermodynamics invented by Nernst,
which helps to determine absolute entropy is
(A) Zeroth law
(B) 1 st law
nd
(D) 3 rd law
(C) 2 law
20.
Passage : II (No. 18 to 20)
Effect of temperature on the equilibrium process
is analysed by using the thermodynamics.
From the thermodynamics relation
∆G° = – 2.3 RT logK........(1) ∆ G° = Standard free
energy change
∆G° = ∆H° – T ∆S°….(2) ∆H° = Standard heat of
the reaction
From (1) & (2)
– 2.3 RT log K = ∆H° – T ∆S° ; ∆S° : Standard
Entropy change,
∆H °
∆S°
+
........(3)
log K =
2.3RT
2.3R
Clearly if a plot of log K vs 1/T is made then it is
– ∆H°
&
a straight line having slope =
2.3R
∆S°
.
y–intercept =
2.3R
If at a temperature T1 equilibrium constant be K1
and at temperature T2 equilibrium constant be K2
then, the above equation reduces to :
– ∆H °
∆S°
⇒ log K1 =
+
........(4)
2.3RT1
2.3R
1/T
If for a particular reversible reaction if Kc = 57 at
355°C and Kc = 69 at 450°C then (A) ∆H < 0
(B) ∆H > 0
(C) ∆H = 0
(D) ∆H sign can't be determined
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
C P Q R S
D P Q R S
21. Match the temperature (in column I) with its value (in
column II)
Column –I
Column II
(A) Critical temperature
(P) a/Rb
(B) Boyle temperature
(Q) 2a/Rb
(C) Inversion temperature
(R) T/Tc
(D) Reduced temperature
(S) 8a/27Rb
∆S°
– ∆H °
+
........ (5)
2.3RT2
2.3R
Subtracting (4) from (5) we get.
⇒ log K2 =
XtraEdge for IIT-JEE
∆H °
K2
=
K1
2.3R
73
JANUARY 2010
22. Match the half-reaction (in column I) with equivalent
mass (molar mass = M) (in column II)
Column –I
Column II
(A) Cr2O72– → Cr3+
(P) M
(B) C2O42– → CO2
(Q) M/2
(C) MnO4– → MnO2
(R) M/6
(D) HC2O 4– → C2O42–
(S) M/3
(A) 3(x – 21) = 3y + 92 = 3z – 32
MATHEMATICS
8.
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
2.
3.
4.
5.
6.
(B) 3
(C) 2
(D) 1
If log3 x + log3 y = 2 + log3 2 and log3 (x + y) = 2
then -
9.
(C)
x − 21 y − (92 / 3) z + (32 / 3)
=
=
1/ 3
1/ 3
1/ 3
(D)
x − 2 y + 3 z −1
=
=
1/ 3
1/ 3
1/ 3
A vector c, directed along the internal bisector of
the angle between the vectors a = 7i – 4j – 4k and
(A)
5
(i – 7j + 2k)
3
(B)
5
(5i + 5j + 2k)
3
(C)
5
(i + 7j + 2k)
3
(D)
5
(–5i + 5j + 2k)
3
If a and b are two unit vectors such that a + 2b and
(A) x = 1, y = 8
(B) x = 8, y = 1
5a – 4b are perpendicular to each other then the
(C) x = 3, y = 6
(D) x = 9, y = 3
angle between a and b is (A) 45º
The exponent of 7 in 100C50 is (A) 0
(B) 2
(C) 4
(D) none of these
–1
(C) cos (1/3)
The equation cos 2x + a sin x = 2a – 7 possesses a
solution if (A) a < 2
(B) 2 ≤ a ≤ 6
(C) a > 6
(D) a is any integer
10.
In a triangle ABC, 2a2 + 4b2 + c2 = 4ab + 2ac, then
numerical value of cos B is equal to (A) 0
(B) 1/8
(C) 3/8
(D) 7/8
11.
(D) cos–1 (2/7)
Suppose a, b, c are positive integers and
f(x) = ax2 – bx + c = 0 has two distinct roots in
(0, 1), then (A) a ≥ 5
(B) b ≥ 5
(C) abc ≥ 25
(D) abc ≥ 250
The coefficient of xk (0 ≤ k ≤ n) in the expansion of
E = 1 + (1 + x) + (1 + x)2 + … + (1 + x)n is -
The tangent at the point P(x1, y1) to the parabola
y2 = 4ax meets the parabola y2 = 4a (x + b) at Q and
R, the coordinates of the mid-point of QR are -
(A) n+1Ck+1
(C)
(A) (x1 – a, y1 + b)
12.
(C) (x1 + b, y1 + a)
(D) (x1 – b, y1 – b)
n+1
Cn–k
(B) nCk
(D) nCn–k–1
The equation of a tangent to the hyperbola
3x2 – y2 = 3, parallel to the line y = 2x + 4 is (A) y = 2x + 3
(B) y = 2x + 1
Equation of the line of shortest distance between the
(C) y = 2x – 1
x
y
z
x − 2 y −1 z + 2
lines =
is =
=
= and
−5
2 −3 1
3
2
(D) y = 2x + 2
XtraEdge for IIT-JEE
(B) 60º
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct.
(B) (x1, y1)
7.
x − (62 / 3) y − 31 z + (31 / 3)
=
=
1/ 3
1/ 3
1/ 3
b = –2i – j + 2k, with |c| = 5 6 , is -
If iz3 + z2 – z + i = 0, then |z| equals (A) 4
(B)
74
JANUARY 2010
13.
The plane passing through the point (–2, –2, 2) and
containing the line joining the points (1, 1, 1) and
(1, –1, 2) makes intercepts of lengths a, b, c
respectively on the axes of x, y and z respectively,
then -
Passage : II (No. 18 to 20)
(A) a = 3b
(B) b = 2c
(C) a + b + c = 12
(D) a + 2b + 2c = 0
18. The value of x for which f(x) = 0 is
(A) –1/2
(B) 0
(C) 1/2
f(x) = sin {cot–1 (x + 1)} – cos (tan–1 x)
a = cos tan–1 sin cot–1 x
b = cos (2 cos–1 x + sin–1 x)
(D) 1
2
14.
19. If f(x) = 0 then a is equal to (A) 1/2
(B) 2/3
(C) 5/9
(D) 9/5
Let a = 4i + 3j and b be two vectors perpendicular
to each other in xy-plane. The vectors c in the same
plane having projections 1 and 2 along a and c are (A) –
2
11
i+ j
3
2
(B) 2 i – j
(C) –
2
11
i+ j
5
5
(D)
20. If a2 = 26/51, then b 2 is equal to (A) 1/25
(B) 24/25
(C) 25/26
(D) 50/51
2
11
i+ j
3
2
This section contains 2 questions (Questions 21 to 22).
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) 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, B-Q, B-R, C-P, C-Q and
D-S, then the correctly bubbled 4 × 4 matrix should be
as follows :
P Q R S
A P Q R S
B P Q R S
C P Q R S
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 15 to 17)
For k, n ∈ N, we define
B(k, n) = 1.2.3 … k + 2.3.4 …(k + 1)
+ … + n(n + 1) … (n + k – 1),
So (n) = n
and S k(n) = 1k + 2k + … + nk
To obtain value of B(k, n), we rewrite B(k, n) as
follows:
B(k, n) = k! kk + kk+1 + kk+ 2 + ... n +kk −1
[( ) ( ) ( ) (
= k! ( )
)]
D P
n+k
k +1
21.
n (n + 1).....(n + k )
=
k +1
(A) B(2, n)
1
(B) B(2, n)
2
1
(C) B(2, n)
6
(D) none of these
(B)
17.
(A) B(3, n)
(B) B(3, n) – 2B(2, n)
(C) B(3, n) – 2B(1, n)
(D) B(3, n) + 2B(1, n)
 z + 3 + 4i 
(D) arg 

 z + 5 − 2i 
22.
( ) S (n) + ( ) S (n) + … +
( ) S (n) + ( ) S (n) equals k
k–1
k +1
k
(A) (n + 1)k
(C) nk – (n – 1)k
XtraEdge for IIT-JEE
1
k +1
k +1
z −1
=2
z +1
(C) z z – (1 + i)z
– (1 – i) z + 7 = 0
16. S 3(n) + 3S 2(n) equals -
k +1
2
Centre of circle
Column-I
(A) |z – 2|2 + |z – 4i|2 = 20
15. S 3(n) + S1(n) equals -
k +1
1
Q R S
0
(B) (n + 1)k – 1
(D) (n + 1)k – (n – 1)k
(Q) 5/3 + 0i
(R) – 4 – i
(S) 1 + 2i
cos α + cos β = a, sin α + sin β = b
Column-I
Column-II
(A) cos (α + β)
(P) 2ab/(a2 + b2)
(B) sin (α + β)
(C) cos (α – β)
(Q) b/a
(R) (a2 – b2)/(a2 + b 2)
(D) tan
75
Column-II
(P) 1 – i
α+β
2
(S) (a2 – b2 – 2)/2
JANUARY 2010
MOCK TEST PAPER-2
CBSE BOARD PATTERN
CLASS # XII
SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS
S OLU TIONS WI LL BE PU BLI SH ED IN N EXT I S SU E
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.
2.
3.
Write the name of the quantity whose SI Unit is Amp
m–1.
d
→ →
∫ B . ds
→
Here B = magnetic field strength.
→
ds = small area vector over a closed surface.
4.
5.
If an electron is having the energy of 10 eV then find
its De-Broglie wavelength.
6.
Why the 'CORE' of the transformer is laminated.
7.
Draw the symbol of photodiode.
8.
Write the name of maxwell's fourth equation for
electro-magnetic waves.
9.
G/D is known as ground current detector. Find the
reading of it for the given circuit diagram
+3v
+3v
+3v
If a thin foil of metal, parallel to capacitor plates get
introduced between the two plates of air capacitor
then write the effect on capacitance C of air
capacitor.
XtraEdge for IIT-JEE
C = ε0A/d
b
a
Name the material for which magnetic susceptibility
is negative.
Write the value of
A
A
1Ω
G/D
76
JANUARY 2010
10. α particle, β particle and Deutron are placed on the
vertices of an equilateral triangle of side 'a' find the
electric potential energy of the given system. Also
find the work done to place these particles from
equilateral triangle of side 'a' to the equilateral
triangle of side '2a'.
α particle
β particle
15. Find the value of potential difference across tertninals
A and B.
C
C
5V
11. Draw the truth table of the following gates
A
Y
(i)
B
A
B
X
R
R
R
R
10V
A
(ii)
B
16. If potential difference across terminals X and B is
10V then find the potential difference across A and
B.
R
R
R
R
Deutron
a
C
A
17. If the potential difference across the resistance
R = 10 Ω is 100 volt then
(i) find the current passing through R–L–C series
circuit.
(ii) Write the power factor of circuit.
Y
B
R
12. In which case the bulb will glow and why ?
(i) If V A = 0 volt
(ii) If V A = 3 volt
Here V A is the potential of A and B is the bulb
+5V
L
~
V= 100√2 sin 314t volt
B
18. Write Einstein's eqn for photoelectric effect.
A photon of energy 5eV falls on the surface a metal
whose work function is 2eV. Calculate the value of
stopping potential for metal.
RC
A
19. Appratus of YDSE is shown in figure, the
interference fringes are observed on the screen. The
relation between x and y is given below
y → f(t) : y = xt for
0 < t ≤ 4 sec.
13. Calculate the no. of electric field lines emitted by the
stationary proton.
14. An equilateral triangle current carrying coil of side 'b'
is placed near the intinite current carrying conductor
which is stationary then in which direction the
triangular loop will move (towards the conductor or
away from the conductor) and why ?
x
y
S2
Find the intensity ratio of maxima to minima on the
screen at
(i) t = 1s
(ii) t = 4s
a
XtraEdge for IIT-JEE
S1
Screen
i2
i1
C
77
JANUARY 2010
20. A convex lens is made of two different materials
which are having the absolute refractive indices as µ 1
and µ 2, this is placed in a media of ARI µ3 as shown
below, then
Incident rays
Medium - 1
Lens
µ1
Medium - 2
µ3
µ3
µ2
Surrounding
24. (i) Find the terminal voltages of Batteries BT-1 and
BT-2.
(ii) Why terminal voltage of battery BT-1 is more
than it's emf and in case of battery BT-2 the terminal
voltage is less than it's emf.
BT-1
Medium - 3
10V
Incident rays
(i) Which is the most dense media
(ii) Arrange µ1, µ 2 and µ3 in increasing order
i
Medium -1 air
26. State inconsistency in Ampere's circuital law. What is
meant by displacement current? Prove that
displacement current is equals to conduction current.
Refracted ray
ip, polarizing angle for air water
27. Find the electric flux passing through the cube for the
given arrangement of charges.
(i) Calculate the value of ip if wµ a = 3/4
(ii) Find the value of φ, φ is the angle between
reflected ray and refracted ray.
(iii)Out of reflected and refracted ray which of the
following is plane polarized.
(iv)Write the relation between polarizing angle and
critical angle for the two media interface.
+1C
–4C
+3C
+5C
a
V0
23. Draw the wave shape of the output signal Y. The
wave shapes of inputs A and B are shown.
A
1
B
t
0
XtraEdge for IIT-JEE
–8C
29. Explain the principle, construction and working of
Vande Graff Generator.
A
B
1
+7C
28. What type of feedback is used in transistor Amplifier.
Draw the circuit diagram of transistor oscillator and
explain it's working. What type of feedback is used in
transistor oscillator. Write the expression for
frequency of signals generated by transistor
oscillator.
10.3V 1Ω
0
–2C
-6C
22. Calculate the output voltage V 0
2Ω
7Ω
2Ω
25. Explain the followings
(i) Why a metallic spring get shrinked when current
is flown through it
(ii) Why N-P-N transistor is preferred over P-N-P
transistor in electronic industry
Medium -2 water
Ge. Diode
20V
RL = 2Ω
Reflected ray
ip
1Ω
Load Resistance
21. A monochromatic light Ray is incident as shown in
figure then
Normal
incident ray
BT-2
30. Explain construction and working of Cyclotron.
Why cyclotron can not be used to accelerate the
electrons.
Y
t
78
JANUARY 2010
16. Write the chemical equations for all the steps
involved in the rusting of iron.
CHEMISTRY
1.
What are the co-ordination no. of Ca+2 & F– ions in
CaF2 structure ?
2.
Name of ionisation isomer of [Cr(H2O)5Br]SO4.
3.
Write IUPAC name of following
(a) O2N – C6H4 – OCH3(p)
17. Give the reactions :
(a) When phenol is treated with excess of Br2 water.
(b) Diethyl ether heated with conc. HI.
18. From Fehling's solution, Schiff's reagent, Tollen's
reagent & Grignard reagent, which reagent react with
both aldehyde & ketone.
CH3 CH3 H
(b) CH2 = C — C — C — C — OH
OH CH3 Br
19. Why 2-pentanone give iodoform test but 3-pentanone
not ?
CH3
20. Give the order of basicity in the following compound.
O
4.
Why aromatic ketones are not react with NaHSO3 ?
5.
Why sulphanilic are amphoteric in nature ?
6.
Give monomer of Glyptal.
N
7.
Explain terms Antacids.
H
(I)
8.
Why carbohydrates are optically active ?
9.
N
(II)
H
(III)
N
H
(IV)
21. 0.2 molal acid HX is 20% ionised in solution,
Kf = 1.86 K molality–1. Calculate the freezing point
of the solution.
+
In a compound AX, the radius of A ion is 95 pm and
that of X– is 181 pm. Predict the crystal structure of
AX and write the co-ordination number of each of the
ions.
22. How long a current of 3-ampere has to be passed
through a solution of AgNO3 to coat a metal surface
of 80 cm2 with a 0.005 mm thick layer. Density of
silver is 10.5 g/cm3.
10. MgO has structure of NaCl and TlCl has the structure
of CsCl. What are the co-ordination number of ions
in MgO & TlCl ?
23. Explain that the rate of physisorption increases with
decrease in temperature.
11. What is meant by mole fraction of solute & solvent?
12. A solution containing one mole per litre of each
Cu(NO3)2, AgNO3, Hg2(NO3)2 is being electrolyzed
by using inert electrodes, the value of standard
electrode potential in volts (reduction potential) are
2Hg/Hg2++ = + 0.79 V
Ag/Ag+ = 0.80 V,
+2
Cu/Cu = 0.34 V,
Mg/Mg++ = – 2.37 V
with increasing voltage, give the sequence of
deposition of metal on cathode ?
24. Explain that boric acid is monobasic & weak lewis
acid.
25. Give the reason :
(a) VOCl2 & CuCl2 give same colour in aqueous
solution.
(b) CuSO4 decolourise on addition of KCN.
13. Give the structure of dichromate dianion.
26. Complete the reaction.
CHO CHO
14. Which compound is form when excess of KCN is
added to aqueous solution of CuSO4.
(i) NaOH/100ºC
(ii) H+/H2O
15. Give the number & structure of possible enantiomeric
pairs that can be produced during monochlorination
of 2-methyl butane.
XtraEdge for IIT-JEE
N
CHO
79
CHO
JANUARY 2010
27. Aspartame, an artificial sweetner, is a peptide having
following structure.
CH2 – C6H5
y
y
x
x
H2N – CH – CONH – CH – COOCH3
(a)
CH2 – COOH
(b)
(i) Identify the four functional group.
(ii) Write the zwitter ionic structure.
(iii) Write the structure of the amino acids obtained
from the hydrochloride of aspartame.
2.
28. The rate constant for the first order decomposition of
a certain reaction is described by the equation
3.
A matrix A of order 3 × 3 has determinant 5. What is
the value of |3A| ?
4.
For what value of x, the following matrix is singular?
2π 
2π 


cos–1  cos  + sin–1  sin
 ?
3 
3 


1.25 × 10 4
.
T
(i) What is the energy of activation for this reaction.
(ii) At what temperature will its half-life period be
256 minutes.
log (k) = 14.34 –
29.
5 − x x + 1
 2
4 

The Haber process can be represented by following
NH3 + H2O
B
CaCO3 → CaO + CO2
A
5.
Find the point on the curve y = x2 – 2x + 3, where the
tangent is parallel to x-axis.
6.
What is the angle between vectors a & b with
NaCl
H2O
What is the principal value of
→
magnitude
→ →
3 and 2 respectively ? Given a . b = 3.
NaHCO3 + D
C + H2O + CO2
7.
Cartesian equations of a line AB are.
2x − 1 4 − y
z +1
=
=
2
7
2
NH3 + H2O + E
Identify A, B, C, D & E.
Write the direction ratios of a line parallel to AB.
30. An alkene 'A' on ozonolysis yields acetone and an
aldehyde, the aldehyde is easily oxidised to an acid
(B). When (B) is treated with Br2 in presence of
phosphorus it yields a compound (C) which on
hydrolysis give a hydroxy acid (D). This acid can
also be obtained from acetone by the reaction with
HCN followed by hydrolysis. Identify compound A,
B, C, & D.
3 log x
(x ) dx
4
Write a value of
9.
Write the position vector of a point dividing the line
→
→
b externally in the ratio 1 : 4,
→
→
where a = 2î + 3 ĵ + 4k̂ and b = − î + ĵ + k̂
 2 1 4
10. If A = 
 and B =
 4 1 5
Section A
 3 − 1
2 2  .


1 3 
Write the order of AB and BA.
Which one of the following graphs represent the
function of x ? Why ?
XtraEdge for IIT-JEE
∫e
8.
segment joining A and B with position vectors a &
MATHEMATICS
1.
→
80
JANUARY 2010
1
most. Its semi vertical angles is tan–1   . Water is
 2
poured into it at a constant rate of 5 cubic meter per
minute. Find the rate at which the level of the water
is rising at the instant when the depth of water in the
tank is 10m.
Section B
11. Show that the function f : R → R defined by
2x − 1
, x ∈ R is one-one function. Also find
f(x) =
3
the inverse of the function f.
OR
Examine which of the following is a binary operation
a+b
, a, b ∈ N
(i) a * b =
2
a+b
, a, b ∈ Q
(ii) a * b =
2
for binary operation check the commutative and
associative property.
18. Evaluate
sum
∫
2
1
the
following
integral
as
limit
of
2
(3x − 1) dx
19. Evaluate
∫
π/ 2
0
log sin x dx
20. Find the vector equation of the line parallel to the line
x −1 3 − y
z +1
=
=
and passing through (3, 0, –4).
5
2
4
Also find the distance between these two lines.
12. Prove that
 63 
5
 3
tan–1   = sin–1   + cos–1  
 16 
 13 
 5
→
→
21. In a regular hexagon ABCDEF, if AB =
→
→
→
→
→
a and
→
→
BC = b , then express CD , DE , EF , FA , AC ,
→
13. Using elementary transformations, find the inverse of
 2 − 6
1 − 2 


a 2 + ac
b 2 + bc c 2 + bc
− ac
a 2 + ab b 2 + ab
c 2 + ac = (ab + bc + ca)3
− ab
14. Find all the points of discontinuity of the function f
defined by
x + 2,
x ≤1
f(x) = x − 2, 1 < x < 2
0,
x≥2
15. If x pyq = (x + y)p+q, prove that
→
→
→
22. A football match may be either won, drawn or lost by
the host country's 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. Probability of him going by bus or scooter or by
3 1 3
,
, respectively.
other means of transport are
10 10 5
1
1
and
The probability that he will be late is
4
3
respectively, if he travels by bus or scooter. But he
reaches in time if the uses any mode of transport. He
reached late at the centre. Find the probability that he
travelled by bus.
OR
Using properties of determinants, prove that
− bc
→
AD , AE and CE in terms of a and b .
dy
y
=
dx
x
OR
Find
 1 + x 2 + 1− x 2
dy
, if y = tan–1 
dx
 1 + x 2 − 1− x 2

16. Evaluate
( x 2 + 1)( x 2 + 4)
∫ (x
2
+ 3)( x 2 − 5)

 , 0 < |x| < 1


Section C
23. Find the matrix P satisfying the matrix equation
2 1   − 3 2  1 2 
3 2 P  5 − 3 =  2 −1

 
 

dx
24. Find all the local maximum values and local
minimum values of the function
π
π
f(x) = sin 2x – x, – < x <
2
2
17. A water tank has the shape of an inverted right
circular cone with its axis vertical and vertex lower
XtraEdge for IIT-JEE
81
JANUARY 2010
29. A catering agency has two kitchens to prepare food at
two places A and B. From these places 'Mid-day
Meal' is to be supplied to three different schools
situated at P, Q, R. The monthly requirements of the
schools are respectively 40, 40 and 50 food packets.
A packet contains lunch for 1000 students. Preparing
capacity of kitchens A and B are 60 and 70 packets
per month respectively. The transportation cost per
packet from the kitchens to schools is given below :
OR
A given quantity of metal is to be cast into a solid
half circular cylinder (i.e., with rectangular base and
semicircular ends). Show that in order that the total
surface area may be minimum, the ratio of the length
of the cylinder to the diameter of its circular ends is
π : (π + 2).
25. Sketch the graph of
| x − 2 | +2, x ≤ 2
f(x) =  2
x>2
 x − 2,
Evaluate
∫
4
0
Transportation cost per packet (in rupees)
To
f ( x ) dx. What does the value of this
integral represent on the graph ?
26. Solve the following differential equation
dy
– xy = x2,
dx
given y = 2 when x = 0
(1 – x2)
A
B
P
5
4
Q
4
2
R
3
5
How many packets from each kitchen should be
transported to school so that the cost of transportation
is minimum ? Also find the minimum cost.
27. Find the foot of the perpendicular from P(1, 2, 3) on
the line
x −6
y−7
z−7
=
=
3
2
−2
MEMORABLE POINTS
Also obtain the equation of the plane containing the
line and the point (1, 2, 3)
1.
2.
3.
28. Let X denote the number of colleges where you will
apply after your result and P(X = x) denotes your
probability of getting admission in x number of
colleges. It is given that
4.
5.
if x = 0 or 1
kx

P(X = x) = 2kx
, k is +ve constant
if x = 2
k(5 – x) if x = 3 or 4

6.
7.
8.
(a) Find the value of k.
(b) What is the probability that you will get
admission in exactly two colleges ?
(c) Find the mean and variance of the probability
distribution.
9.
OR
The bags A and B contain 4 white 3 black balls and 2
white and 2 black balls respectively. From bag A two
balls are transferred to bag B. Find the probability of
drawing
10.
11.
12.
(a) 2 white balls from bag B ?
13.
14.
(b) 2 black balls from bag B ?
(c) 1 white & 1 black ball from bag B ?
XtraEdge for IIT-JEE
From
82
MECHANICS
Weight (force of gravity) decreases as you move
away from the earth by distance squared.
Mass and inertia are the same thing.
Constant velocity and zero velocity means the
net force is zero and acceleration is zero.
Weight (in newtons) is mass x acceleration
(w=mg). Mass is not weight!
Velocity, displacement [s], momentum, force
and acceleration are vectors.
Speed, distance [d], time, and energy (joules) are
scalar quantities.
The slope of the velocity-time graph is acceleration.
At zero (0) degrees two vectors have a
resultant equal to their sum. At 180 degrees
two vectors have a resultant equal to their
difference. From the difference to the sum is the
total range of possible resultants.
Centripetal force and centripetal acceleration
vectors are toward the center of the circlewhile the velocity vector is tangent to the circle.
An unbalanced force (object not in equilibrium)
must produce acceleration.
The slope of the distance-tine graph is velocity.
The equilibrant force is equal in magnitude but
opposite in direction to the resultant vector.
Momentum is conserved in all collision systems.
Magnitude is a term use to state how large a
vector quantity is.
JANUARY 2010
XtraEdge for IIT-JEE
83
JANUARY 2010
MOCK TEST PAPER SOLUTION
FOR PAPER – 1 PUBLISHED IN DECEMBER ISSUE
13. From Einsten's equation
Kmax = hv – hv0
eV0 = hv – hv0
hν 0
φ
hν
hν
V0 =
–
=
–
e
e
e
e
φ = work function.
From the graph. metal-2 has high value of φ
∴ threshold wavelength of metal-1 is high.
(ii) θ1 = θ2 because slope of the graph is constant
for all metals.
PHYSICS
1.
Relation is not valid because charge is independent
of motion of particle.
2.
No change because focal length of mirror is
independent of refractive index.
3.
Galium Aresnide
4.
Yes, it get observed but the fringes are of
diffraction type. These are not the interference
fringes.
5.
Zener diode is used as a voltage regulator to obtain
constant voltage output.
6.
Davison – Germer experiment.
7.
n=0
8.
Because for alloys temperature coefficient of
resistivity is nearly constant for wide range of
temperature.
9.
14.
Ep
(
R.B
T
∫
0
i = 1 + 3 2 sin (314 t + 30º)
G = 1, i0 = 3 2
from the above formulae
17. (i) Lenz's law : The direction of magnetic induction
in a circuit is such that so as to oppose the cause of
change in magnetic flux.
(ii) On increasing current i inwards magnetic field
increases. Direction of induced current is such that
to produce outwards magnetic field i.e.
anticlockwise direction.
1/ 2
= (1 + 9)1/2 =
10
12. At P1 drift speed of the electron is maximum for a
1
conductor vd ∝
(I = neAVd)
A
At P3 current density is minimum.
XtraEdge for IIT-JEE
)
K2
16. (i) Algebric sum of currents meeting at a point is
equal to zero.
(ii) Current in 2 Ω resistor is 3A
∴ Potential difference is = 2 × 3 = 6V
1 2
i (f ) dt
T




(
15. (i) Angle of Dip: It is the angle which the direction
of resultant intensity of earth's magnetic field
subtends with horizontal line in magnetic meridian
at the given place.
B
(ii) BH = B cos φ ⇒ tan φ = v
BH
Bv = B sin φ
Bv = BH ⇒ tan φ = 1
If φ = 45º
10. (i) NPN transistor
(ii) Yes, the transistor is properly biased because
emitter-base junction is in forward bias and the
collector base junction is in reverse bais

i2
irms  G 2 + 0

2

Rh
G
(i) V-m = electric flux
(ii) C-m = Dipole moment
11. irms =
K1
)
18. (i) The order of colours in secondary rainbow is
opposite to that of primary rainbow
(ii) This is due to total internal reflection.
84
JANUARY 2010
19. (i) Potential difference on 1µF and 3µF is same
1
∴ from energy = CV2
2
E∝C
E2
1
=
E1
3
(ii) Parallel combination of 1 µF and 3 µF is 4 µF
12
3µF
4
30V
Q = CV = 90µC
The refractive index is related to the angle ip called
the polarising angle by a relation known as
Brewster's law
µ = tan ip.
Plane polarised
light
ip
µ
30V
(ii) sin θc =
d
90
30
=
=
= 7.5 Volt
c
12
4
(iii) Energy supplied by battery
E = QV = 9 ×10–6 × 30
= 2.7 × 10–3 J
P.d. on 12µF ⇒ V =
θc = sin–1(cot ip)
(iii) It is not correct.
23. Need for modulation
(i) Frequency of signal : The audio frequency signal
(20 Hz to 20 KHz) cannot the transmitted without
distortion over long distance due to less energy
carried by low frequency audio waves.
(ii) Number of channels: Audio frequencies are
concentrated in the range 20 Hz to 20 kHz. This
range is so narrow that there will be overlapping of
signals. In order to separated the various signals it
is necessary to convert all of them to different
portions of the electromagnetic spectrum
20. (i) Magnetic field in the solenoid is along the axis
→
→
∴ angle between v and B is 0º
→
→
1
1
=
µ
tan i p
∴ F = q( v × B ) = 0
(ii) When current flows through the spring current
in different coils of spring flows in same direction.
Therefore due to magnetic force spring gets
compressed.
21. (i)
10Ω
5Ω
EC
EC
x
Carrier
Mo dulating signal
(100 – x)
1
x
=
2
100 – x
2x = 100 – x, x = 33.33 cm
A
let P be the null point
(ii)
A.M. wave
Y
B
A
0
0
1
1
B
0
1
0
1
24. To convert a galvanometer into an ammeter of
range I a small resistance S is connected in parallel
with the galvanometer so that the current passing
through the galvanometer G becomes equal to its
null scale deflection value Ig.
S
Y
1
0
1
1
(I – Ig)
22. (i) When unpolarised light is incident on a
transparent surface of refractive index µ at a certain
angle ip such that the refracted light ray and the
reflected light ray are perpendicular to each other,
the reflected light is plane polarised as shown here.
XtraEdge for IIT-JEE
I
85
G
Ig
I
JANUARY 2010
28. Working : During the positive half cycle of the
input signal, the forward bias across the emitterbase junction will be increased while during the
negative half cycle of the signal, the forward bias
across emitter-base junction is decreased. Hence
more electrons flow from the emitter to the
collector via the base during positive half cycle.
The increased collector current will produce a large
voltage drop across the load resistance RL.
However during the negative half cycle of the
collector, current decreases resulting in the
decreased output voltage. Hence an amplified
output is obtained across the load.
IgG
I − Ig
25. (i) Equivalent ckt is
R/2 R/2
R
R
R
R
R
4R/3
4R/3
R
Ans. 8R/7
(ii) Material will be constantan because for alloys
value of α varies very slightly with temperature as
compared to metals.
27. (i)
BE
O
VCE
high frequency
AC source
D2
2a
D1
path of
accelerated
proton
+q
P net = P 2 + P 2 + 2P 2 cos 60º =
(ii) Electric potential energy
–2q
3P=
+q
+q
2a
kq (−2q ) kq (−2q ) kq 2
+
+
2a
2a
2a
=
− 2kq 2 − 2kq 2 kq 2
–
+
2a
2a
2a
=
− 3kq 2
2a
B
positive charged
beam
Principle : Cyclotron device is based on the fact
that heavy positive ions can be accelerated to the
high energies with a comparatively smaller
alternating potential difference by making them to
cross the field again and again using strong
magnetic field. Here the magnetic field used in
cyclotron maintain the charged particles in circular
paths while the electric field imparts them energy
periodically.
Construction : Cyclotron consists of two D shaped
hollow metallic enclosures D1 and D 2 called dees.
These dees have their diometric edge parallel to one
other and are separated by a small gap. These dees
are connected to the terminals of a high frequency
alternating potential difference. This potential
difference creates an electric field of high
frequency in the gap between the dees. The whole
apparatus is placed between N-S poles of a
powerful electro magnet which produces strong
magnetic field.
3 q(2a)
2a
2a
XtraEdge for IIT-JEE
IC
29. Cyclotron
2a
–2q can be assumed as two –q charges placed at the
point p = q(2a)
U =
IC
IE
t
Pnet
+q
RL
~
–q –q
Pt
IB
Signal
26. (i) Angle between electric field line and
equipotential surface is 90º.
(ii) Electric field directed in the direction of
decreasing potential. So electric potential is
maximum at point a and proton will have
maximum potential energy at point a .
(iii) Electric field is maximum at point c. Thus
proton will have maximum force at c.
2a
IC
Vi
Output
IgG = (I – Ig) × S ⇒ S =
86
JANUARY 2010
t
Cyclotron can not be used to accelerate electrons
because it is used to accelerate heavy ions.
4.
When two different molecules participate in the
polymerization process it is called copolymerization.
5.
A metal which is more electropositive than iron
such as Al, Zn, Mg can be used in cathodic
protection of iron against rusting.
6.
Lower value of bond dissociation energy of F2 is
due to the strong repulsion between the non
bonding electrons of F atoms in the small sized F 2
molecule. Also there is no multiple bonding due to
absence of
d-orbitals.
7.
Hydrolysis of sucrose produces change in optical
nature form dextro rotatory to laevorotatory, the
process is called inversion of sugar.
→
30. Ampere's circuital law : The circulation of B along
a closed loop of any arbitary shap is µ0 times the
algebric sum of current embraced by the loop.
Determination of magnetic field for a solenoid :
l
b
c
P
a
d
b
c
B
C6H12O6 + C6H12O6
Glucose
Fructose
Turns of
Solenoid
Axis
a
H+
C12H22O11 ((cane sugar) + H2O ) →

Invert sugar
d
dl
Let p be the point where B is to be determined.
From Ampere's law
→ →
∫ B . dl = µ i ∫
0 net
[
∫
b→ →
B . dl =
a
and
∫
∫
∫
d→ →
c
c
→
B . dl
a→ →
d
9.
Mechanism of the formation of diethyl ether from
ethanol : The formation of ether is a nucleophilic
bimolecular reaction (SN2) involving the attack of
alcohol molecule on a protonated alcohol, as
indicated below
B . dl
→
B . dl = 0; B ⊥ dl
(i)
→
H
CH3 – CH2 – O – H + H+ → CH3 – CH2 – +O – H
Ethanol
B . dl = 0, Q B = 0]
a
∫
B . dl = B dl cos 0 = BI ⇒ µ0(nil)
H
+
→
(ii) CH3 CH2 – O + CH3 – CH2 – O
|
H
H
d
⇒ B = µ0ni
CHEMISTRY
1.
Three types of lattice imperfections are possible
(a) Schottky defect (b) Frenkel defect
(c) Interstitial defects
c→ →
b
∫
B . dl +
d→ →
a→ →
d
d→ →
c
∫
B . dl +
a
∫
+
b→ →
8.
5
4
3
2
1
CH3 – C = CH – C – CH3
||
|
O
CH3
4-methyl-pent-3-en-2-one
2.
CH3 – CH2 – CH = CH2 + HCl
→ CH3 – CH2 – CH – CH3
|
Cl
3.
Enzyme streptokinase can dissolve blood clots. So
it is useful in medicines for checking heart attacks
caused by blood clotting.
XtraEdge for IIT-JEE
+
CH3 CH2 – O – CH2 CH3 + H2O
|
H
(iii)
CH3 CH2 – +O – CH2 CH3 →
|
H
CH3 CH2 – O – CH2 CH3 + H+
Diethyl ether
(Ethoxyethane)
87
JANUARY 2010
10. The order of basicity in gaseous phase is
(i) (CH3)3 N > (CH3)2 NH > CH3NH2 > NH3 due to
+I effect of alkyl group, there is more density at
N at tertiary amine.
(ii) The order of basicity in aqueous state is
(CH3)2 NH > CH3 NH2 > (CH3) 3 N > NH3
The inductive effect, solvation effect, Hbonding and steric hinderance of the alkyl group
decides the basic strength of alkyl amines in the
aqueous state.
13. (i) Treat the compound with Lucas reagent (conc.
HCl + anhy. ZnCl2) 2-propanol gives turbidity
in 5 min whereas ethanol gives no turbidity at
room temperature
ZnCl
2 → No reaction
CH3CH2OH + HCl  

ZnCl
2→
 

CH3CHCH3 + HCl
|
OH
CH3 – CH – CH3 + H2O
|
Cl
Turbidity appears in 5 min
(ii) Acetaldehyde reduces Tollen’s reagent to silver
mirror but acetone does not.
11. (i) Monomer of Teflon is tetrafluro ethylene.
Teflon is a addition polymer.
(ii) Monomer of Bakelite is formaldehyde and
phenol. Bakelite is a condensation polymer.
(iii)Monomer of natural rubber is isoprene (2methyl-1, 3-butadiene). Natural rubber is a
addition polymer.
CH3CHO + 2 [Ag (NH3)2]+ + OH– –→
CH3COO + 2H2O + 2Ag ↓ + 4NH3
Silver mirror
ONa
Tollen’s reagent
CH3COCH3     → No action
4 − 7 atm
+ CO2   →
12. (i)
400 K
14.
Sod. phenoxide
OH
Multimolecular
Colloids
The particles of this type
of
colloids
are
aggregates of atoms or
molecules with diameter
less than 1 nm.
Examples : Sol of
sulphur
conists
of
colloidal particles which
are aggregate of 58
molecules.
The atoms of molecules
are held together with
Van der Waal’s forces
+
COONa
H

→
H O
2
OH
COOH
sod. salicylate
salicylic acid
OH
CHCl , aq NaOH
 3   →
340 K
(ii)
Phenol
ONa
CHCl2
Examples
:
proteins etc.
Starch,
Covalent
bonds
are
present in one chain and
different chains have the
force
like
H-bonds,
dipole-dipole interaction
and salt bridge etc.
15. Pyrophosphoric acid is prepared by the removal of
H2O from two molecules of orthophosphoric acid
(having tetrahedral shape). Hence two tetrahedra
are attached through an oxygen.
OH
OH
ONa
CHO H3O +
 
→
NaOH
 →
Macromolecular
Colloids
The particles of this type
of colloids are themselves
large
molecules
of
colloidal dimension.
OH
O
P
O
P
O
CHO
O
Salicylaldehyde
XtraEdge for IIT-JEE
88
O
JANUARY 2010
OR
Formation of Ni (CO)4
OH
OH
|
|
O=P–O–P=O
|
|
OH
OH
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
0.059
[Cu 2+ ]
log
2
[Cu ]
19. (i) Markownikoff’s rule : According to this rule,
when addition across an unsymmetrical double
bond takes place, the positive part of the
addendum goes to the carbon atom with the
larger number of hydrogen atoms.
0.059
0.1
0.059
1
= 0.34 +
log
log
2
1
2
10
0.059
[Q log 10 = 1]
= 0.34 +
× (−1)
2
= 0.34 – 0.0295 = 0.3105 V
When the concentration of Cu2+ ions is decreased,
the emf for copper electrode decreases.
= 0.34 +
Absence of

 
→
CH3 – CH = CH2 + HBr
Peroxide
propene
Br
|
CH3 – CH – CH3
2-Bromopropane
(ii) Hofmann Bromide Reaction : When amide is
treated with bromide in on alkaline solution, an
amide yields an amine containing one carbon
less than the starting amide.
O
||
R – C – NH2 + Br2 + 4 KOH
Amide
RNH2 + K2 CO3 + 2KBr + 2H2O
Amine
For example :
O
||
CH3 CH – C – NH2 + Br2 + 4 KOH
Propanamide
CH3 CH2 NH2 + K2 CO3 + 2KBr + 2H2O
Ethylamine
17. Adsorption isobar for physical adsorption shows
that the extent of adsorption decreases with the
increase in temperature. The adsorption isobar of
chemical adsorption shows that the extent of
adsorption first increases and then decreases with
the increase in temperature. The initial unexpected
increase in the extent of adsorption with
temperature is due to the fact that the heat supplied
acts as activation energy required for chemical
adsorption which is much more than that of
physical adsorption.
Extent of
adsorption
Physical adsorption
isobar
Extent of
adsorption
Temperature
Chemical adsorption
isobar
20. (a) (i) CH2 = CH2 + H2SO4 –→ CH3CH2HSO4
conc.
Ethene
Ethyl
Sulphuric
hydrogen
acid
Temperature
H O
18. In the complex [Ni (CO)4], the oxidation state of Ni
is ‘0’. Its electronic configurations is [Ar] 3d8 4s2.
A.O. of Ni (28)
3d
4s
4p
↑↓
↑↓ ↑↓ ↑↓ ↑
↑
2→ CH3CH2OH + H2SO4
Boil
Ethanol
(ii)
OCOCH3
OH
+ CH3 COCl
sp hybridised obritals of Ni
Phenol
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
sp
(iii)
3
hybridisation
89
CH3
H
Pyridine
Acetyl
chloride
3
XtraEdge for IIT-JEE
× × ×
Four electrons pairs from
four CO molecules
The resulting complex has tetrahedral shape and is
diamagnetic due to absence of unpaired electrons.
16. Cu2+ (aq) + 2e– → Cu (s)
E Cu 2+ / Cu = E°Cu 2+ / Cu +
×
Phenyl ethanoate
+ CH MgI
CH3
Dry ether
H
C = O  3→
Ethanal
+ HCl
C
OMgI
CH3
JANUARY 2010
H + , H 2O
CH3
   →
− Mg(OH) I
H
C
(b) On dilution, the degree of ionization of the
weak electrolyte increases. This increases the
molar conductance of the solution sharply.
OH
CH3
2-Propanol
24. In [Co (NH3)6]3+, the oxidation state of Co is + 3.
Co (Z = 27) atom is ground state
21. (i) A is CH2 = CH2
B is CH2 – CH2
|
|
Br
Br
3d
↑↓ ↑↓
↑↓
(Benzene diazonium chloride)
↑↓
↑↓
NH3+ HSO4–
(Anilium hydrogen
↑
↑
sp3 d2 hybridisation
↑
↑
↑
↑
↑
↑
↑
↑
×
× × ×
×
Due to sp 3 d2 hybridisation, the complex ion has
octahedral shape. Due to the presence of four
unpaired electrons, the complex ion is
paramagnetic.
NH2
(Sulphanilic acid)
26. (i) A deep red sol of ferric hydroxide is obtained
by the hydrolysis of ferric chloride. The sol
particles are propositively charged because of
preferential adsorption of Fe3+ ions.
(ii) Adsorption is an endothermic process. So the
rate of physical adsorption decreases with the
rise in temperature in accordance with Le
Chatelier’s principle.
(iii)River water contains charged colloidal particles
of sand, clay, etc. As river water comes in
contact with saline sea water, the electrolytes of
sea water coagulate the suspended colloidal
particles which settle down at the point of
contact resulting in the rise of river bed. So
water adopts a different course and a delta is
formed in due course of time.
SO3H
22. (a) Electronic configuration of Ti in [Ti (H2O)6]3+
is Ti3+ (d1)
Two vacant d orbitals are available for
octahedral hybridization with 4s and 4p orbitals.
(b) The colour of the complex is purple.
The colour of complex is due to the jumping of
electron from lower level to higher level. When
an electron from a lower energy of orbital is
excited to higher energy of level, the energy of
excitation corresponds to the frequency of light
absorbed. This frequency lies in the visible
region. The colour observed corresponds to the
complementary colour of the light absorbed.
27. Solid catalysts are used in a number of gaseous
reactions. Such catalytic reactions called
heterogeneous
reactions.
Examples
of
heterogeneous catalysis are
(i) Manufacture of ammonia from N2 and H2 by
Haber’s process in the presence of catalyst.
23. (a) According to electrochemical theory during the
formation of just the impure iron surface
behaves like a small electrochemical cell in the
presence of water containing dissolved oxygen
or carbon dioxide. In such alls, pure iron acts as
anode and impure surface acts as cathode.
Moisture containing dissolved oxygen or CO 2 in
the electrolytic solution. Hence rusting is an
electrochemical phenomenon.
XtraEdge for IIT-JEE
↑
Six pairs of electrons from
six NH3 molecules
sulphate)
B is
↑
4d
↑↓
Formation of [Co(NH3)6]3+ ion
(Benzonitrile)
(iii) A is
↑
Hybridisation
CN
B is
↑
Co3+ ion
+
N2Cl–
(ii) A is
4p
4s
↑
Fe(s)
N2(g) + 3H2(g) → 2NH3(g)
(ii) V2O5 catalyst is used in the manufacture of
H2SO4 by contact process
90
JANUARY 2010
2Na2 CrO4 + H2 SO4 –→ Na2 Cr2 O7 + Na2 SO4 +
H2O
Na2 SO4 is separated by fractional crystallisation.
Sod. dichromate is converted into potassium
dichromate by heating with KCl.
Na2 Cr2 O7 + 2KCl –→ K2 Cr2 O7 + 2NaCl
Potassium dichromate being less soluble is obtained
by fractional crystallisation.
(i) Cr2 O72– + 14 H+ + 6I– –→ 2Cr3+ + 7H2 O + 3I2
(ii) Cr 2 O72– + 4Fe2+ + 14 H+ – →
2Cr3+ + 6Fe3+ + 7H2 O
Uses the potassium dichromate :
(i) In volumetric analysis for the estimation of Fe2+
and I– ions.
(ii) In chrome tanning in leather industry.
V O (s)
2SO 2(g) + O2 2 5
→ 2SO3(g)
Solid catalyst helps in the following ways :
(a) Simultaneous adsorption of reactants increases
the concentration at the surface of the catalyst
which increases the reaction rate.
(b) Adsorption of reactant molecules makes the
attack of other molecules on it easier.
(c) Some adsorbed molecules dissociate into atoms
which are very reactive.
(d) Heat of adsorption released provides activation
energy for the reaction.
28. (a) A is aldehyde or Ketone. A gives Tollen’s test
hence it is an aldehyde
CH3 – CH = CH – CH2 – CHO
CH2 CH3
(b)
30. (i) At elevated temperatures, sulphur vapours
exists, as S2 molecules which are paramagnetic
like O2.
(ii) This is due to reluctance of silicon to form
pπ – pπ multiple bonds because of large size of
silicon atom. Hence, silicon exists only in
diamond structure.
(iii) Xe has relatively lower ionization energy
among inert gases and thus the outermost shell
electrons of Xe are excited to d-subshell and
thereby showing unpaired electronic structure.
(iv) Nitrogen shows a little tendency for catenation,
due to weakness of N – N single bond whereas
phosphorus shows a clear tendency for
catenation due to its unexpectedly high bond
energy.
KMn O / OH −
  4 →
(i)
Ethylbenzene
COOH
Benzoic acid
CaO + NaOH
    →
Benzene
Dil NaOH
(ii) 2CH3CHO
  →
Acetaldehyde (Aldol condensati on )
4
3
2
1
NaBH 4
→
CH 3 − CHOH − CH 2 − CHO  
(Reduction )
4
3
2
1
CH3 – CHOH – CH2 CH2OH
Butane-1, 3-diol
CH3
C = O + H2 –→ CH3 CHOH CH3
(iii)
CH3
Acetone
MATHEMATICS
Section – A
1.
Conc. H SO
4
  2 

→ CH3 CH = CH2
Dehydration
Propene
2.
29. Preparation of K2 Cr2 O7 : Chromite ore is fused
with molten NaOH in the presence of air to get
sodium dichromate.
4 Fe Cr2 O4 + 16 NaOH + 7O2 – →
Chromite ore
8 Na2 Cr O4 + 2Fe2 O3 + 8H2 O
Sod. chromate
The fused mass is dissolved in water. The filtrate is
treated with dil H2 SO4
XtraEdge for IIT-JEE
Let (a, b) ∈ R
∴ (b, a) ∈ R
Hence R is symmetric.
[Q (1, 2) ∈ R ]
[Q (2, 1) ∈ R ]
x2
1
x2
.
tan −1 x –
dx
2 x2 +1
2
(Integrating by parts)
∫
x. tan −1 x dx =
=
1 2
1
x tan −1 x −
2
2
∫
∫
x 2 + 1−1
dx
x 2 +1
1
1 
1 
1−
= x 2 tan −1 x −
dx
2
2  x 2 + 1 
∫
=
91
1 2
1
x tan −1 x − ( x − tan −1 x ) + C.
2
2
JANUARY 2010
3.
The given curves are
x2 + y2 = 2ax
Here, a is arbitrary constant
Diff. (1) w.r.t. x
dy
2x + 2y
= 2a
dx
Substituting for 2a in (1), we get
 4 3 5
∴ | A | = O ⇒  3 − 2 7 
10 − 1 x 
⇒ 4 (–2x + 7) –3 (3x – 70) + 5 (–3 + 20) = 0
⇒ x = 19
….(1)
dy 

x2 + y2 = x  2x + 2 y 
dx 

8.
[Q tan( π – θ) = –
9.
7.
→
→
→
→
→
→
→
a + b+ c
→
3π
4
θ=
3π
4
→ →
a .b
→
3
=
→
3.2
=
3
2
π
3
10. Let the equal angle = x
∴ l = cos α ; m = cos α and n = cos α
∴ l2 + m 2 + n 2 = 1
cos2 α + cos2 α + cos2 α = 1
⇒ 3cos2α = 1
1
⇒ cos α = ±
Hence direction cosines
3
1
3
1
3
±
− 2  − 2 1
+
− 4  0 3
− 1
− 1
1
3
, ±
1
3
, ±
1
3
Section – B
11. We have f(x) = 2x, g (y) = 3y + 4 and h (z) = sin z
L.H.S. = ho (gof) (x) = h ( g ( f ( x )) = h ( g (2x))
= h (3 (2x) +4) = h (6x + 4) = sin (6x + 4)
R.H.S. = (hog) of (x) = (hog) (f (x)) = (hog) 2x
= h (g (2x) = h (3 (2x) +4) = h (6x + 4)
= sin (6x + 4)
∴ L.H.S. = R.H.S.
{Order of matrix n=3}
 1− x 
12. Let y = cot–1 

1+ x 
1− x
Put :
=u
1+ x
 4 3 5
Let A =  3 − 2 7 
10 − 1 x 
Q the matrix A is singular
XtraEdge for IIT-JEE
2 3
3 and | b | = 2 ; a . b = 3
| a || b |
Q | adj A | = | A | n – 1
64 = | A | n – 1
64 = | A |3 – 1
| A |2 = 64
|A|=8
2(î + ĵ + k̂ )
3
→
Here | a | =
cos θ =
3 1 
− 2
and B = 
A= 

0 2 
0
2A – B + C = 0
⇒ C = –2A + B
3 1  − 2
⇒ C = –2 
+ 
0 2   0
=
î + ĵ + k̂
→ →
3π
4
− 6
⇒ C= 
0
− 8
C= 
0
6.
→
=
∴ the principle value of tan–1 (–1) is
5.
→
| a + b+ c |
tanθ]
x=
→
∴ unit vector =
Let x = tan–1 (–1)
tan x = –1
π
tan x = – tan
4
π

tan x = tan  π − 
4

tan x = tan
→
a + b + c = î + ĵ + ĵ + k̂ + k̂ + î = 2 ( î + ĵ + k̂ )
a + b + c = 2 1+ 1 + 1 = 2 3
dy
+ x2 – y2 = 0
⇒ 2xy
dx
which is the reqd. diff. eq.
4.
→
92
...(1)
JANUARY 2010
∴ y = cot–1 u
Diff. w.r.t. u
−1
dy
=
=
du 1 + u 2
=
∴ The reqd. sol. of eq. (1) is
∫ 2x .x dx + c
= ∫ 2 x dx + c = x + c
1− x 
1+ 

1+ x 
− (1 + x )
2
2
(1 + x ) + (1 − x )
2
2
=
2
(1 + x )
2(1 + x 2 )
...(2)
π/4
∫ sin 2 x sin 3 x dx
14.
…(3)
0
From (2) and (3),
dy dy du
=
⋅
dx du dx
13. The given diff. eq. is
(1 + e2x) dy + ex (1 + y2) dx = 0
dy
1+ y
2
+
e
1 + e 2x
∫
0
sin 5x  π / 4

sin x − 5 

0
=
1
2
5π
 π 1

sin 4 − 5 sin 4 − 0


1 π 1
π 1  1  1
3
sin + sin  = 1 + 
=

2 4 5
4 2  5 2 5 2
OR
(where t = ex)
π/4
Sol. Let I =
∫ log(1 + tan x )dx
…..(1)
0
a
By the formula,
∫
a
∫
f ( x )dx = f (a − x )dx
0
π/4
=
0

π

1 − tan x 

2

∫ log 1 + tan 4 − x dx
0
π/4
=
∫ log 1 + 1 + tan x dx
0
π/ 4
=

∫ log 1 + tan x dx
0
π/ 4
...(1)
=
∫ [log 2 − log(1 + tan x)]dx
0
π/ 4
=
log 2
∫
π/4
dx −
0
= log2 [x ]
1
−1
∫ (cos x − cos 5x)
1
2
=
∴
Pdx
∫ − dx
I.F. = e ∫
= e x = e − log x
= e log x
0
π/4
=
dx = 0
Integrating, we get
dt
tan–1 y +
=c
1+ t2
⇒ tan–1 y + tan –1 t = c
⇒ tan–1 y + tan –1 ex = c
when x = 0, y =1
⇒ tan–1 + tan–1 e° =c
π π π
⇒ c= + =
4 4 2
π
tan–1 y + tan–1 ex = .
2
This is the reqd. sol of (1)
OR
Sol. The given diff. eq. is
dy
x
− y − 2x 3 = 0.
dx
dy 1
− y = 2x 2
⇒
dx x
This is a linear diff. eq.
dy
On comparing by,
+ Py = Q
dx
1
Here, P = – ,
Q = 2x2
x
∫ (2 sin 3x sin 2x)dx
1
2
=
...(1)
π/4
1
2
=
−2
(1 + x ) 2
1
⋅
=
=–
.
2
2(1 + x ) (1 + x )
1+ x2
⇒
y = x3 + cx.
⇒
2
Diff. (1) w.r.t. x
du (1 + x ).(−1) − (1 − x ).1
2
=
=
2
dx
(1 + x )
(1 + x ) 2
x
−1
2
y . x–1 =
−1
2I =
∫ log(1 + tan x )dx
0
π/4
0 −
I
[By (1)]
π
π

log2  − 0  ⇒ I = log 2
8
4

= x −1
XtraEdge for IIT-JEE
93
JANUARY 2010
15
Let I =
∫ 2x
3x + 1
2
17
dx
− 2x + 3
3 d
5
(2x2 – 2x + 3) +
Here : 3x + 1 = .
4 dx
2
3
5
( 4x − 2) +
2 dx
∴ I= 4 2
2x − 2x + 3
3
4x − 2
5
dx
dx +
=
4 2x 2 − 2 x + 3
2 2x 2 − 2 x + 3
3
5
= I1 + I 2 + c
…(1)
4
2
4x − 2
dx
where I1 =
2
2 x − 2x + 3
Put : 2x2 – 2x + 3 = t ⇒ (4x – 2)dx = dt
dt
∴ I2 =
= log | t |
t
....(3)
= log | 2x2 – 2x + 3 |
dx
1
dx
=
and I 2 =
2
3
2
2
2x 2 − x +
1   5 

−
+
x


2
2   2 

x →π −
x →π −
lim f ( x ) = lim x 3 + 3 = 3
∫
x →0
x →0
Hence f(x) continuous at x = 0.
π
 x −1 
−1  x + 1 
18. tan–1 
=
 + tan 
 x−2
 x+2 4
∫
 x −1 x + 1 
+
 π

tan −1  x − 2 x + 2  =
x
1
x
1
−
+
 4
1 −
.
 x − 2 x + 2 
1

 x− 
1
1
−1 
2
= ⋅
tan
2 ( 5 / 2)
 5/2




∫x
dx
2
+a
2
=
 2x − 1 

tan −1 
5
 5 
From (1), (2) and (3), we get
x2 + x − 2 + x2 − x − 2
…(3)
2
x − 4 − x +1
d2 y
dx
At θ =
2
=
d 
θ  dθ
 cot  .
dθ 
2  dx
=–
1
θ
1
cosec2 .
2 a (1 − cos θ)
2
= −
1
θ
cos ec 4 .
4a
2
19.
 4 − 5 − 11
A = 1 − 3 1 
 2 3 − 7 
| A | = 4 (21 – 3) + 5 (–7 – 2) –11 ( 3 + 6)
= 72 – 45 – 99 = – 72
9
9 
 18

Adj. A =  − 68 − 6 − 22
 − 38 − 15 − 7 
2
1
π d y
1
π
;
= − cos ec 4   = – .
4a
2 dx 2
a
4
XtraEdge for IIT-JEE
 π 
Q 4 = 1


=1
2x − 4
=1
−3
2
2x – 4 = –3
2x2 = –3 + 4 = 1
2x2 = 1
1
x2 =
2
1
⇒x=±
2
Given : x = a (θ – sin θ);
y = a (1 – cosθ)
dx
dy
= a (1– cos θ);
= a sin θ
⇒
dθ
dθ
θ
dy dy / dθ
a sin θ
⇒
= cot .
=
=
dx dx / dθ a (1 − cos θ)
2
Now
2
2
 2x − 1 
3
5
 + c
log | 2 x 2 − 2x + 3 | +
tan −1
4
2
 5 
I=
16
 ( x − 1)( x + 2) + ( x + 1)( x − 2) 


π
( x − 2)( x + 2)

 = tan
4
 ( x − 2)( x + 2) − ( x − 1)(x + 1) 


( x − 2)( x + 2)
1
 x 
tan −1  
a
 a 
1
=
x →0
Also f(0) = 1
∴ lim f ( x ) = f (0)
∫

 By the formula,

x → π+
K · π + 1= cos π
πK+1 =–1
π K = –2
−2
K =
π
OR
Sol. Since the function is defined at x = 0
∫
∫
x →π +
lim Kx + 1 = lim cos x
∫
∫
Q f(x) is continuous at x = π
∴ lim f ( x ) = lim f ( x ) = f (π)
94
T
JANUARY 2010
θ
2
→ →
θ
| a + b | = 2 cos
2
→
→
θ
1
cos = | a + b |
2
2
→
18 − 68 − 38
∴ adj A =  9 − 6 − 15 
 9 − 22 − 7 
19 − 68 − 38
adjA − 1 
=
9 − 6 − 15 
A
72 
 9 − 22 − 7 
OR
y
z 
a + x
Sol. ∆ =  x
a+y
z 
 x
y
a + z
∴ A –1 =
 2
= 4C3  
 3
operate C2 → C2 + C 1
o
o 
a

∆ = x a + y + x
z 
 x
y+ x
a + z
∆ = a [(a + y + x) (a + z) – z (y + x)]
= a [a2 + az + ( y + x) a + (y + x ) z – z (y + x)]
= a2 ( a + x + y + z)
Proved.
→
→
→
3
4
[Q P (r) = nCr qn– r Pr]
4
3
3 1
·
40
10 4
=
=
3
1
3 1 1 1 3
+
· + · + ·0
40 30
10 4 10 3 5
3
3 120 9
= 40 =
×
= .
9 + 4 40 13 13
120
→→
Section – C
→
= 1 + 1 + 2 | a | | b | cos θ
π/2
23. Let
= 2 + 2 (1) (1) cos θ
= 2 (1 + cosθ)
θ
= 2 . 2 cos2
2
XtraEdge for IIT-JEE
1
 
 3
2 1
1
·
+1 · 1 ·
3 27
81
8 1
9 1
=
+ =
=
81 81 81 9
OR
Sol. Let A, B and C be the events of candidate going by
bus, scooter and other means of transport.
Let E be the event of getting late.
3
1
3
P (A) =
, P (B) =
, P (C) =
10
10
5
1
1
P (E/A) =
, P(E/B) = , P(E/C) = 0
4
3
P (that he traveled by bus) = P (A/E)
P( A ) P( E / A )
=
P( A) P( E / A) + P( B) P( E / B) + P (C)P (E / C)
| a + b |2 = | a |2 + | b | 2 +2 a b
→
0
4−4
=4·
21. Q | a | = | b | = 1 (Given)
→
3
1 4  2
  + C4 
 3
 3
1
( 2 + 1) 2 + (3 + 3) 2 + ( 4 − 2) 2 = 49 = 7
→
4 −3
 2 1
 2 1
= 4C1     + 4 C 0    
 3  3
 3  3
20. The given plane is
3x + 2y + 2z + 5 = 0
...(i)
line through P (2, 3, 4) and parallel to the line :
x+3 y−2 z
=
= is
3
6
2
x −2 y−3 z −4
=
=
= k (say)
….(ii)
3
6
2
Any point on it is Q (3k +2, 6k + 3, 2k + 4)
Let it lie on (i)
∴ 3(3k + 2) +2 (6k +3) +2 (2k + 4) +5 = 0
⇒ 25 k + 25 = 0
⇒ k=–1
∴ Q (–1, –3, 2)
∴ The required distance = PQ
→
Proved.
22. p = P (correct forcasting) = 1/3
q = P (two incorrect forecasting) = 2/3
n=4
Let r be the number of correct forecast
P (at least three correct results)
= P ( r = 3) + P (r = 4)
operate R1 → R1 – R2
o 
a − a

∆ = x a + y
z 
y
a + z 
 x
=
→
| a + b |2 = 4 cos2
I=
∫
0
cos x
dx
(1 + sin x )(2 + sin x )
⇒ cos x dx = dt
π
Also x = 0 ⇒ t = 0 and x =
⇒t=1
2
Put : sin x = t
95
JANUARY 2010
1
∴ I=
∫
0
dt
=
(1 + t )(2 + t )
1
 4 3 16 
∴ Reqd. Area = Area of circle – 
+ π
 3
3 

1 
 1
1 + t − 2 + t dt


0
∫
(Resolving into partial fractions)
= [log | 1 + t | – log | 2 + t
= 16 π –
|] 10
= (log 2 – log 3) – (log 1 – log 2)
= 2 log 2 – log 3 = log 2 2 – log 3
4
= log .
3
2
32
4 3 4
π−
= (8π − 3 ) sq. units.
3
3
3
OR
=
Sol.
2
24. x +y = 16
y2 = 6x.
(1) and (2) intersect, where
…(1)
...(2)
x2
4 3 16
− π
3
3
y2
+
=1
a2 b2
x y
+ =1
a b
….(1)
…..(2)
Y
2
Y
B (0, b)
2
X'
1
A (2, 2 3 )
1
X'
x=4
x=2
x=0
O
D
C
Y'
We shall find the shaded area
(Area of the smaller region)
a
∫
a
(Q x ≠ –8)
2
0
b
=
a
a

∫
0
1−
 x 
− b1 −  dx
a
 a 
x2
2
a
 x
a − x dx − b 1 − dx
 a
0
2
∫
2
a

16 − x 2 dx 


=

b a2
a2 
−1 
 sin 1 − b a − 
a  2
2a 
 
=
ab π ab 1
. −
= ab(π − 2)sq.units
2 2 2 4
25. Let S, V, r and h be the surface area, volume, radius
of the base and height of the given cylinder
respectively. Then
S = 2 πrh + πr2
(Given)
[Q Cylinder is open at the top]
2 6
 2.2 3
1 
= 2
.2 2 + {0 + 8 sin −1 1} − 
+ 8 sin −1 
2 
 2
 3
16 3
π
π
+ 16. − (4 3 + 16. )
3
2
6
⇒
 4 3 16 
= 
+ π  sq. units.
 3
3 

XtraEdge for IIT-JEE
− y 2 )dx
a

b  x a2 − x2 a2
x
x2 

=
+ sin −1  − b  x − 
a
2
2
a
2a  0


0
4

2
 2 3/ 2   x 16 − x 2 16 −1 x  

= 2 6. x  + 
+ sin

 3
2
2
4 
0 
2 

=

∫ b
=
∫
∫
1
0
4
2

= 2  y 2 dx + y1dx 
0

2


4
∫ (y
=
Q A (2, 2 3 ) and B (2, –2 3 )
Also C (4, 0).
Area OBCAO = 2 (Area ODA + Area DCA)
2
= 2 6 x dx +
0

A
(a, 0)
x=0
Y'
∫
X
x=a
X
B (2, –2 3 )
x2 + 6x – 16 = 0
⇒ (x + 8) (x – 2) = 0
⇒ x=2
O
h=
S − πr 2
2πr
V = πr2h = πr 2
96
S − πr 2
1
= [Sr − πr 3 ]
2
2πr
JANUARY 2010
Diff. w.r.t r
x 
 4 − 5 1   4
 y  = 1  2 0 − 2  0 
  10 
 
 z 
 2 5
3   2
2
dV
1
d V
= [S − 3πr 2 ] and 2 = −3πr
2
dr
dr
For max. or min.,
dV
= 0
dr
⇒ S – 3πr2 = 0
S
3π
⇒ r=
∴ x=
For this value of r,
d 2V
dr 2
∴ V is maximum and
h=
16 − 0 + 2
18
1 
1  

=
8 + 0 − 4 =  4 
10 
10
 8 + 0 + 6
14 
<0
3πr 2 − πr 2
=r
2πr
27. Let A and B the events of getting letter from Tata
Nagar and Calcutta
1
1
∴ P(A) = , P (B) =
2
2
Let E be the event of visibility of letter TA
2
7
2.2
1
=
=
P (E | A) =
7.6.5.4.3.2.1
1260
2
[Using (1) and (2) ]
26. Sol. Part-I
1 − 1 1 
Given matrix A =  2 1 − 3
1 1
1 
1 −1
[Q Total no. of events in Tata Nagar =
1
|A|= 2
1
−3
1
1
1
TA
= 1 (1 + 3) + 1 ( 2 + 3) +1 ( 2 – 1 )
=4+5+1
= 10
∴ A–1 exists
C11 = (–1)2 (1 + 3) = 4
C12 = (–1)3 (2 + 3) = –5
C13 = (–1)4 (2 – 1) = 1
C21 = (–1)3 (–1 –1) = 2
C22 = (–1)4 (1 – 1) = 0
C23 = (–1)5 (1 + 1) = –2
C31 = (–1)4 (3 – 1) = 2
C32 = (–1)5 (–3 –2) = 5
C33 = (–1)6 (1 + 2) = 3
2 2
4
1
1 
–1
A =
adj A =
− 5 0 5
A
10 
 1 − 2 3
P (E | B) =
4 − 5 1 
1 
∴ A =  2 0 − 2
10
 2 5
3 
XtraEdge for IIT-JEE
TA
1
7
2
=
NAGAR
7
2
as
has only 2 A's ]
2
1
=
7.6.5.4.3.2.1
2520
[Q Total no. of events in Calcutta =
7
2
as
CALCUT
TA has only 2 C's ]
(i) P (that letter has come from Tata Nagar)
= (A | E) =
P (A ).P( E | A)
P ( A ) P ( E | A ) + P ( B) P ( E | B)
1
1 1
·
1260
2
1260
=
=
2 +1
1 1
1 1
·
+ ·
2520
2 1260 2 2520
1
2520 2
×
=
1260
3
3
(ii) P (that letters has come from Calcutta) = P (B |
E)
= 1 – P (A | E)
2
1
=1–
=
3
3
OR
Sol. White balls = 4
Red balls = 6
Total balls = 4 + 6 = 10
Let X be the number of drawing 3 white balls
=
Part II
The given equation can be written as
AX = B ⇒ X = A–1 B
2 1
1
x 
4 




Where A =  − 1 1 1 x =  y  B = 0 
 1 − 3 1
z 
2
–1
18 9
4 2
14 7
= , y=
= , z=
= .
10 5
10 5
10 5
[Using part I]
97
JANUARY 2010
∴ X = 0, 1, 2, 3
4
P
(X
=
0)
=
5λ = 2 – 12
5λ = –10
⇒ λ = –2
Putting the value of λ, we get.
The coordinate of P (–2 + 3, –4 + 4, –4 + 5)
= P (1, 0 1)
The required distance = |AP|
6
C 0 · C3
10
=
C3
1
.
5
6.5.4 3.2.1 15
.
=
=
3.2.1 10.9.8 90 30
4
P (x = 1) =
4
P (x = 2)
C1·6 C2
10
C3
C 2 · 6 C1
10
C3
=4·
=
6.5 3.2.1 45 15
.
=
=
2.1 10.9.8 90 30
4.3 6.3.2.1 27 9
=
=
2.1 10.9.8 90 30
4
P
(X
=
3)
=
(3 − 1) 2 + ( 4 − 0) 2 + (5 − 1) 2
=
4 + 16 + 16
=
36 = 6 unit.
6
C3 · C 0
10
C3
=
29. Given problem can be tabulated as
3.2.1
3
1
4.3.2
.1.
=
=
3.2.1 10.9.8 90 30
Therefore, required probability distribution is
X
0 1 2 3
5 15 9 1
P (X)
30 30 30 30
Wheat
Rice
28. Equation of the given line is
2x = y = z
x y z
= =
1 2 2
Equation of line passes through
A (3, 4, 5) and parallel to line ( 1 ) is
x −3 y− 4 z −5
=
=
= λ(say)
1
2
2
∴ Any point on the line (2) is
P (λ + 3, 2λ + 4, 2λ + 5)
Since the point P lies on the plane
x+ y+ z= 2
∴ (λ + 3) + (2λ + 4) + (2λ + 5) = 2
λ + 3 + 2λ + 4 + 2λ + 5 = 2
Proteins
0.01
0.05
Min50g
Carbohydrates
0.025
0.5
Min200 g
Cost
Rs. 4kg = 0.4 P/g
Rs. 6 kg = 0.6 P/g
Let the quantity of wheat = x gms
and the quantity of rice = y gms
Min cost z = 0.4x + 0.6y
subject to constraints
0.1x + 0.05y ≥ 50
0.25x + 0.5y ≥ 200
x, y ≥ 0
Table for 0.1x + 0.05y = 50
Calculation for mean and variance
X
P(x)
XP(x) X2P(x)
5
0
0
0
30
15
15
15
1
30
30
30
9
18
36
2
30
30
30
1
3
9
3
30
30
30
36 6 60
Total 1
=
=2
30 5 30
6
Mean µ = Σ X P(x) =
= 1.2
5
Variance = Σ X2 P (x) – [ Σ X P (X)]2
= 2 – (1.2)2
= 2 – 1.44 = 0.56
XtraEdge for IIT-JEE
=
x
y
0
500
1000
0
Table for 0.25x + 0.5y = 200
x
y
0
400
800
0
X
1000
800
(0, 500)
600
400
(1000, 0)
200
…..(1)
200 400 600 800 1000
the point
The corner points of feasible region is (0, 500) and
(1000, 0) Now evaluate z at the corner points
…..(2)
Corner point
(0, 500)
(1000, 0)
….(3)
Z = 0.4x + 0.6y
Z = 0 + 300.0 = 300 ← Min
Z = 400.0 + 0 = 400
Minimum cost = 300 paise = Rs. 3.
when x = 0 gms, y = 500 gms.
98
JANUARY 2010
XtraEdge Test Series
ANSWER KEY
IIT- JEE 2010 (January issue)
PHYSICS
Ques
An s
Ques
An s
21
22
1
D
11
A,D
A→P
A→Q
2
B
12
A,D
3
A
13
A ,C , D
B→Q
B→Q
4
A
14
A ,B , D
5
D
15
C
C→R
C→R
6
D
16
B
7
A
17
A
D→Q
D → Q,S
8
B
18
C
9
A
19
D
10
C,D
20
B
C HEM I ST RY
Ques
An s
Ques
An s
21
22
1
A
11
B
A→R
A→Q
2
B
12
A ,B ,C ,D
3
A
13
C
B→P
B→P
4
B
14
A,B,C
5
A
15
A
C→Q
C→S
6
B
16
D
7
B
17
D
D→R
D→R
8
C
18
A
9
A
19
B
10
B ,D
20
A
MATHEMATICS
Ques
An s
Ques
An s
21
22
1
B
11
A,B,C,D
A→Q
A→R
2
B
12
A ,C
3
A
13
A ,C
B→P
B→S
4
B
14
A,B,C
5
D
15
A
C→S
C→Q
6
D
16
C
7
C
17
B
D→R
D→P
8
B
18
C
9
A
19
D
7
A
17
A
D→S
D → P,Q,R
8
A
18
D
9
A
19
D
10
A ,B ,C ,D
20
C
IIT- JEE 2011 (January issue)
PHYSICS
Ques
An s
Ques
An s
21
22
1
A
11
A,C
A→Q
A→S
2
D
12
A,C
3
B
13
A,C
B→P
B→S
4
B
14
A ,C , D
5
D
15
D
C→R
C→P
6
C
16
A
10
B,D
20
B
C HEM I ST RY
Ques
An s
Ques
An s
21
22
1
A
11
A,C
A→S
A→R
2
B
12
B
3
B
13
A,B
B→P
B → P,S
4
D
14
A , B ,C ,D
5
D
15
A
C→Q
C→Q
6
D
16
A
7
A
17
D
D→R
D→Q
8
B
18
B
9
B
19
B
10
A ,D
20
B
MATHEMATICS
Ques
An s
Ques
An s
21
22
1
D
11
A,C
A→S
A→R
XtraEdge for IIT-JEE
2
C
12
B,C
3
A
13
A ,B , C
B→Q
B→P
4
B
14
B,C
5
D
15
A
C→P
C→S
99
6
B
16
C
7
A
17
B
D→R
D→Q
8
A
18
A
9
B
19
C
10
A ,B ,C
20
B
JANUARY 2010
XtraEdge for IIT-JEE
100
JANUARY 2010