Download IIT-JEE 2011 - Career Point

Do not allow the quest for perfection to ruin your
life because whatever you do you will always feel
that you could have done better
Volume - 6 Issue - 6
December, 2010 (Monthly Magazine)
Editorial / Mailing Office :
112-B, Shakti Nagar, Kota (Raj.)
Tel. : 0744-2500492, 2500692, 3040000)
324009
e-mail : [email protected]
Editor :
Pramod Maheshwari
[B.Tech. IIT-Delhi]
Cover Design
Om Gocher, Govind Saini
Layout
Rajaram Gocher
Circulation & Advertisement
Ankesh Jain, Praveen Chandna
Ph 0744-3040000, 9672977502
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 ` 20/Special Subscription Rates
6 issues : ` 100 /- [One issue free ]
12 issues : ` 200 /- [Two issues free]
24 issues : ` 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.
Editorial
Dear Students,
The difference between success and failure is your attitude towards
success and the strategies that you employ to achieve it. The difference
between success and failure is only a few minutes or a few hours
everyday. You have to keep on striving for success at every conceivable
opportunity. Never postpone your happiness and zest for life and work.
You should make it a habit to enjoy your profession and your job all the
time. Never be a quitter because a quitter can never be a winner.
You should always remember that People live not by the reason of any
care they have for themselves but by the love for them that is in other
people. Have only those people for friends and companions who do
their best to bring out the best in you. They will be of unlimited worth
to you. Such persons understand what life means to you and your goal.
They feel for you as you feel for yourselves. They are the ones who are
bound to you in triumph and disaster. They provide a purpose to live
and break the spell of loneliness. A true friend is worth befriending as
he will always stand by you. But before you expect others to be the right
person to be your friend you must also become one.
Be always committed to your cause. Be so engrossed in your work that
you have hardly any time to think of anything else. The great secret of
success is to do whatever you are to do and do it wholeheartedly. Make
yourself the star of your workplace. For this you must have clear and
precise objectives to be achieved within a definite time-frame. Always
respect and value time. Be result-oriented and keep track of the hours.
Respect the time of others as well as your own. Be always organized
and write down everything you want to accomplish.
Always make an assessment of yesterday's "To Do" list to crosscheck
how realistic it has turned out to be today. This will help you to avoid or
rectify mistakes, if any, in your planning. Keep on visualizing your
goals and lists of the task to be done.
Forever presenting positive ideas to your success.
Yours truly
Pramod Maheshwari,
B.Tech., IIT Delhi
Editor : Pramod Maheshwari
XtraEdge for IIT-JEE
1
DECEMBER 2010
XtraEdge for IIT-JEE
2
DECEMBER 2010
Volume-6 Issue-6
December, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
CONTENTS
INDEX
PAGE
Regulars ..........
Much more IIT-JEE News.
Know IIT-JEE With 15 Best Questions of IIT-JEE
NEWS ARTICLE
Challenging Problems in Physics,, Chemistry & Maths
President inaugurates Pan IIT 2010 conclave
Yale University signs MoU with IIT-K, IIM-K
Key Concepts & Problem Solving strategy for IIT-JEE.
Xtra Edge Test Series for JEE- 2011 & 2012
CBSE Mock Test Paper
4
IITian ON THE PATH OF SUCCESS
6
Mr. Sheerang Chhatre
KNOW IIT-JEE
7
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS
S
Success Tips for the Months
• If one asks for success and prepares for
failure, he will get the situation he has
prepared for.
8-Challenging Problems [Set# 7]
Students’ Forum
Physics Fundamentals
Ray Optics
Fluid Mechanics & Properties of Matter
CATALYSE CHEMISTRY
• You never achieve real success unless you
like what you are doing
• The first step toward success is taken when
you refuse to be a captive of the
environment in which you first find
yourself.
• Believe in yourself ! Have faith in your
abilities ! without a humble but reasonable
confidence in your own powers you can not
be successful or happy.
XtraEdge for IIT-JEE
36
Key Concept
Carboxylic Acid
Chemical Kinetics
Understanding : Physical Chemistry
• Loser's visualize the penalties of failure.
Winner's visualize the rewards of success.
• Treat others as if they were what they
ought to be and you help them to become
what they are capable of being.
15
DICEY MATHS
46
Mathematical Challenges
Students’ Forum
Key Concept
Monotonic Maxima & Minima
Function
Test Time ..........
XTRAEDGE TEST SERIES
60
Class XII – IIT-JEE 2011 Paper
Class XI – IIT-JEE 2012 Paper
Mock Test CBSE Pattern Paper -1 [Class # XII]
3
DECEMBER 2010
President inaugurates
IIT 2010 conclave
Pan
"Government is undertaking various
schemes and initiatives which should
lead to the creation of a new India.
Your expertise and deliberations at this
conclave can help chart out ways in
which capacity building and delivery
mechanisms in many of these initiatives
can be implemented," said the
President.
Yale University signs MoU
with IIT-K, IIM-K
New Delhi: President Pratibha
Devisingh Patil on Friday inaugurated
the PanIIT 2010 conclave, with
alumni from all the Indian Institutes of
Technology converging for a threeday networking and brainstorming
session.
PanIIT is an umbrella organization
covering alumni of all Indian
Institutes of Technology. Over the
years, these conferences have become
one of the leading technology summits
for business leaders. This year the
PanIIT 2010 Global Conference
focuses
on
good
governance,
knowledge
based
economy,
entrepreneurial, innovative, equivalent
and happy society. Speaking to
delegates through a video message,
President Patil said, "The theme of
your
conclave,
Sustainable
Transformation: Our New India, is
very relevant to the changes taking
place around us.""I am happy to know
that you are going to deliver it on the
need for better quality of life in the
society, environment sustainability
and to the imperative of developing
low carbon technologies," she added.
There are currently 15 IITs in the
country.
IT BHU (Banaras Hindu University)
is also to be upgraded as an Indian
Institute
of
Technology.
This year's conclave will attract
coincides with the golden jubilee year
of IIT Delhi and IIT Kharagpur.
XtraEdge for IIT-JEE
New Delhi: Indian Institute of
Management (IIM)-Kozhikode, Indian
Institute of Technology (IIT)-Kanpur
and Yale University, USA have entered
into a partnership to advance higher
education in India through academic
leadership development programmes
for higher education leaders in India
and through research on Indian higher
education.
A Memorandum of Understanding
(MoU) in this regard was signed
between Dr. Richard Levin, President
of Yale University, Dr. Debashis
Chatterjee, Director of IIM Kozhihode, and Dr. Sanjay Dhande,
Director of IIT – Kanpur.
Speaking on the occasion, Kapil Sibal
said that this partnership, which will
take effect from January 2011, will be
sited in two new Centers of Excellence
for Academic Leadership (CEAL) to be
established at IIM - Kozhikode and
IIT -Kanpur."The partnership will
begin with a term of five years, and
could be renewed thereafter," he added.
He also said that a six member
committee with equal participation
from the three partnering institutes will
determine the norms/qualifications for
participating in these leadership
programmes.
The flagship programme of the
partnership will be a new "India - Yale
University Leadership Programme," to
be developed by Yale University in
4
consultation with IIM - Kozhikode
and IIT - Kanpur, that will expose
university and academic leaders in
India at the levels of vice-chancellor,
director, and deans to the best
practices of academic administration
and institutional management in the
United States.
Yale University, IIM - Kozhikode,
and IIT - Kanpur would also engage
in joint faculty research on higher
education and collaborate to organize
workshops and seminars on relevant
areas of academic administration and
leadership. The first programmes
under the agreement would take
place in 2011 in New Haven,
Connecticut.
Yale President Richard Levin stated,
"Yale is pleased to undertake this
important and much needed effort on
higher education with IIM Kozhikode and IIT - Kanpur. We
look forward to working with them
to advance the cause of higher
education in India by sharing what
we have learned over three centuries
as an institution and we similarly
look forward to learning from our
partners in India in this age of global
education."
At IIT-B, juniors give alumni
a lesson on altruism...
At IIT-B, juniors give alunmi a
lesson on altruism...Mihika Basu
Inspired by the 2010 batch, more
ex-students want to contribute 1%
of their salary to the institute
After the graduating batch of 2010 of
the Indian Institute of Technology
(IIT), Mumbai, pledged 1% of its
salary in an uncommon gesture of
giving something back to their alma
mater, many former students now
want
to
follow
suit.
Accordingly, the initiative is been
scaled up considerably, and a formal
DECEMBER 2010
set-up will soon be launched to enable
alumni to submit 1% of their salary
via various mediums. The initiative,
called ‘Give One for IIT Mumbai’, is
part of a new fund-raising initiative by
the IIT Mumbai Alumni Association
(IITBAA). It is a voluntary
programme, where each alumnus can
contribute 1% of his/her income
towards supporting overall improvement
at IIT Mumbai. “Several former
students are keen to contribute 1% of
their incomes to the institute,"said
Damayanti
Bhattacharya,
chief
operating officer, IITBAA.
"The IITBAA will collect the funds on
behalf of the institute. The set-up will be
launched on December 26, on the
institute alumni’s day. It will enable
former students to choose their own
payment mode. They will get periodic
updates about how their funds are being
used and will be invited to see it in
action”.In August this year, over 50% of
the 2010 graduating class of IIT-B had
committed 1% of their annual salary to
the institute. It was the first batch of IITB to have committed to start its payback
even before getting their first pay
cheque. “Their commitment has created
a new tradition for all batches,” said
Bhattacharya.
The funds will be channelised towards
various
developmental
activities,
including aspects like faculty and
student development and growth of
departments
and
hostels.
“Development of the institute as a
whole and alumni benefit are other
areas for which the alumni can donate,”
said Bhattacharya. Alumni can also
choose their own area of contribution.
“After fresh graduates made their
pledge, even alumni who had
graduated nearly 35 years ago were
very excited, and said they wanted to
be part of the movement. Hence, we
are trying to launch an online donation
mechanism by December-end for all
alumni,” said Bakul Desai, member of
the board of directors at IITBAA, in
charge of fundraising.
Metallurgist of the Year 2010
The Indian Institute of Metals has
elected Dr. N. Eswara Prasad Scientist
'G' & Regional Director Regional
XtraEdge for IIT-JEE
Centre for Military Airworthiness
Materials, Hyderabad, who is a
graduate of the 1985 batch from ITBHU, for the Award of Metallurgist of
the Year 2010, in the Non-Ferrous
Metals Category, in recognition of the
developmental work that he conducted
for Indian Defence.
SJMSOM, IIT Mumbai hosts
the 2nd IPR Researchers'
Confluence
the 2nd IPR Researchers' Confluence
2nd IPR Researchers' Confluence
February 11-12, 2011
SJMSOM, IIT-Mumbai : Welcome
back for the 2nd IPR Researchers'
Confluence being held at SJMSOM, IIT
Mumbai on February 11 - 12, 2011.
After a very successful conduct of the
1st confluence in December 2009,
where the IPR research and education
roadmap evolved, the 2nd confluence is
now all set.
As the name suggests, this is not a
conference! It is an attempt towards
creating an environment where people
interact, share and unlearn to learn new
concepts, appreciate issues and
brainstorm
to
identify
suitable
approaches to the problems raised.
Experts from the technology, legal and
management domain across academics,
industry are amongst the initial list of
speakers and experts who have
confirmed to be speakers, panel
members and brainstorming session
leaders. Refer to the theme document
for more details.
IT-An IITian Speaks Out
IIT Mumbai's motto is Knowledge is
The Supreme Goal. I studied there and
in the spirit of its philosophy, am going
to present a few essays of mine, to
show IIT and IITians from a different
perspective than we are used to
normally. Hope it will lead to better
understanding of IIT and the socieconomic environment it operates in.
Here I have tried to tackle many issues
simultaneously. Equal access to
education is important to many
including women, minorities, and
poorer population. How do IITs deal
with that? On the other hand, once
admitted, do the young minds at IIT
5
learn skills to develop into
responsible citizen or are just trained
to become (migrant) workers in
economically developed societies? I
have also tried to reconcile the vision
Nehru had when he conceived of
IITs and what they have become
today. Are IITs truly what the best
what India with population of over
one billion capable of?
The IITians: The Story of a
Remarkable Indian Institution
and How its Alumni Are
Reshaping the World
IIT (Indian Institute of Technology)
is India’s biggest and most powerful
brand, and arguably the toughest and
most influential engineering school
in the world.
Since the first IIT was set up in the
1950s, thousands of initiates have
walked out of the campus gates in
Kharagpur, Mumbai, Chennai and
elsewhere to become leaders in their
chosen fields. In India they head
many of the biggest and most
admired professionally managed
companies. Abroad, they lead giant
corporations, and their feats figure in
the folklore of Silicon Valley. The
power that the alumni of this one
bunch of undergraduate schools
wields in business, academe and
research is comparable to that of
Cambridge and Oxford in the heyday
of the British Empire.
Sandipan Deb, himself an IITian,
delves into his own experience and
those of scores of alumni to try and
explain what makes IITians such
outstanding achievers. In part it may
be that they cannot be anything else:
only one in every hundred applicants
gets
admitted.
Harvard,
in
comparison, takes one in eight. The
unique village-like campuses peopled
only by the super-bright and the
intensely competitive hone the
IITians’ skills further. No wonder
then that when they leave the campus,
IITians look upon themselves as
special people, capable of competing
in their field with the best in the
world.
DECEMBER 2010
Success Story
This article contains storie/interviews of persons who succeed after graduation from different IITs
Mr. Sheerang Chhatre
Gold Medallist From IIT-Bombay
IIT-Bombay
JEE. I guess two years are required for a thorough
preparation.
IIT gold medallist shares his success story
Ever wondered what it would be like to be the gold
medallist at an IIT?
Ramesh asked, Hi Shreerang, As of today, which
source of energy do you find most promising, and
which one will be appropriate for India.
For Sheerang Chhatre, this dream recently became
reality, as he was named the gold medallist at IIT-Mumbai
in this year's graduating class. Now, he's off to MIT in the
United States for his PhD, but he plans on returning to
India [ Images ] to help the country's growth.
Shreerang Chhatre answers, From the Indian point of view,
it's solar energy. If a country like germany with a smaller
size and much lesser solar radiation can generate so much
power from it, then why can't we? Huge amounts of initial
investment for solar-cells, invertors and the grid is
preventing the commercialisation of solar energy in India.
Sheerang chatted with Get Ahead readers on and
answered questions regarding academics at the IITs. For
those of you who missed the chat, here's the transcript:
Hime asked, Congrats, what is your course in Mtech?
Varun asked, Most IIT achievers moves to US or
Europe for jobs. Tell me what is the reason that they
dont stay here and serve country?
Shreerang Chhatre answers, I have specialised
Metallurgical process engineering for MTech.
Shreerang Chhatre answers, See Varun, young people are
more inclined towards moneymaking. So fat pay packages
and a comfortable life attracts them to developed countries.
But they fail to realise that through their knowledge and
expertise they should help their own people to grow. Now,
I guess the situation is changing, slowly but surely.
Rutvik asked, Would u suggest going for any branch of
IIT or going for a branch of one's choice like Computer
Science in other premier institutes like NIT,Bits Pilani,
IIIT etc
Shreerang Chhatre answers, My personal opinion is that
you should go for any branch available in IITs, rather than
going for other colleges. The hierarchy or distinction in the
branches that people make are not really felt when one is
studying in IITs. The academic quality, peer group and
facilities that you get in IITs are awesome. Never let the
opportunity to get into IIT slip-by.
Shreerang Chhatre answers, Thanks for your wishes.
Padmakar asked, Conratulations Shreerang! I would
like to know about your school days. Whether you were
one of the intllignent/ bright student during those days
or you improved yourself afterwardfs....?
Kunal asked, hi Shreerang! im kunal from golden gate..
Shreerang, please tell me how u studied 4 chemistry.
the portion seems to be too vast..I find it hard to
remember each and every reaction
Shreerang Chhatre answers, I did my schooling from Parle
Tilak Vidyalaya, Mumbai [ Images ]. I was bright but lazy;
but slowly I realised that you need to work really hard to
achieve anything. So I guess that was the only
improvement.
Shreerang Chhatre answers, Well, initially it can be a bit
difficult, but slowly through revisions and assignments you
get used to it. Dont worry, you have a very good organic
chem teacher, so just keep pace with the lectures and the
portion. You will succeed, dont worry!
Rohan asked, hi Shreerag....congrats mate............I
aspire to be an IITian..I am in 10th...when do I start
coaching for IIT
Shreerang Chhatre answers, After your 10th standard
exam, take a break. Relax for a few days. Then start for
XtraEdge for IIT-JEE
in
6
DECEMBER 2010
KNOW IIT-JEE
By Previous Exam Questions
PHYSICS
A 5 m long cylindrical steel wire with radius 2 × 10–3 m
is suspended vertically from a rigid support and
carries a bob of mass 100 kg at the other end. If the
bob gets snapped, calculate the change in
temperature of the wire ignoring radiation losses.
(For the steel wire : Young's modulus = 2.1 × 1011 Pa ;
Density = 7860 kg/m3 ; Specific heat = 420 J/kg-K).
[IIT-2001]
Sol. When the mass of 100 Kg is attached, the string is
under tension and hence in the deformed state.
Therefore it has potential energy (U) which is given
by the formula.
1
× stress × strain × volume
U=
2
1.
=
(Stress)
1
×
2
Y
2
X
R = R1 or R2 or R3
R
G
B
C
A
Sol. All Null point, the wheat stone bridge will be
balanced
X
R
∴
=
r1
r2
⇒X=R
r1
r2
where R is a constant r1 and r2 are variable. The
maximum fraction error is
× πr2l
1 (Mg / πr 2 ) 2
1 M 2g 2l
× πr2l =
...(i)
2
Y
2 πr 2 Y
This energy is released in the form of heat, thereby
raising the temperature of the wire
Q = mc∆T
...(ii)
From (i) and (iii) Since U = Q Therefore
=
X
r1
M
1 M 2g 2l
∴ mc∆T =
2 πr 2 Y
∴ ∆T =
=
1
M 2g 2
2 (πr 2 ) 2 Ycp
(100 × 10) 2
1
×
–3 2
2
(3.14 × 2 × 10 ) × 2.1× 1011 × 420 × 7860
⇒
An unknown resistance X is to be determined using
resistance R1, R2 or R3. Their corresponding null
points are A, B and C. Find which of the above will
give the most accurate reading and why? [IIT-2005]
XtraEdge for IIT-JEE
r2
N
B
C
A
R=R1 R=R2 R=R3
Here ∆r1 = ∆r2 = y (say) then
∆X
For
to be minimum r1 × r2 should be max
X
[Q r1 + r2 = c (Constt.)]
Let E = r1 × r2
⇒
E = r1 × (r1 – c)
dE
∴
= (r1 – c) + r1 = 0
dr1
= 0.00457º C
2.
G
∆r
∆r
∆X
= 1 + 2
X
r1
r2
1 M 2g 2l
2 πr 2 Ycm
Here
m = mass of string = density × volume of string
= ρ × πr2l
∴ ∆T =
R
⇒
7
c
c
⇒ r2 =
⇒ r 1 = r2
2
2
R2 gives the most accurate value
r1 =
DECEMBER 2010
3.
An inductor of inductance 2.0 mH is connected
across a charged capacitor of capacitance 5.0 µF,
and the resulting LC circuit is set oscillating at its
natural frequency. Let Q denote the instantaneous
charge on the capacitor, and I the current in the
circuit. It is found that the maximum value of Q is
200 µC.
(a) When Q = 100 µC, what is the value of |dI/dt|?
(b) When Q = 200 µC, what is the value of I?
(c) Find the maximum value of I.
(d) When I is equal to one half its maximum value,
what is the value of |Q|?
[IIT-1998]
Sol. This is a problem of L–C oscillations.
Here Q0 = maximum value of
Q = 200 µC = 2 × 10–4 C
1
1
ω=
=
= 10+4 s–1
–
3
–
6
LC
(2 × 10 H)(5.0 × 10 F)
Let at t = 0, Q = Q0 then
Q(t) = Q0 cos ωt
...(1)
dQ
I(t) =
...(2)
= – Q0 ω sin ωt
dt
L=2.0 mH
C=5.0 µF
dI( t )
= – Q0 w2 cos (ωt)
dt
 Q 
(a) For Q = 100 µC  or 0 
2 

∴ Q=
Q=
4.
I=
60º
µp = 3
B
C
µf = 2.2
Calculate
(a) angle of emergence.
(b) min. value of thickness t so that intensity of
emergent ray is maximum.
Sol. (a) Using snell's law at surface AB
3
= 3 sin r ⇒ r = 30º
µair sin 60º = µp sin r ⇒
2
Now, NN' is the normal to surface AB.
∴ ∠AMN = 90º
But ∠QMN = 30º ⇒ ∠AMQ = 60º
A
...(3)
30º
N'
60º
Q
60º
30º
M N
B
C
In ∆AMQ
∠AQM = 180º – (60º + 30º) = 90º
The refracted ray inside the prism hits the other face
at 90º ; hence deviation produced by this face is zero
and hence angle of emergence is zero.
(b) Multiple reflection occurs between the surfaces
of the prism for minimum thickness.
∆x = 2µt = λ, where
λ
= 125 nm
t = thickness ⇒ t =
2µ
5.
2
–I )
I max
= 1.0 A
2
XtraEdge for IIT-JEE
Shown in the figure is a prism of an angle 30º and
30º
dI
= 10+4 A/s
dt
(b) Q = 200 µC when
cos(ωt) = 1 i.e. ωt = 2π ...
At this time I(t) = – Q0 sin ωt or I (t) = 0
(c) I(t) = – Q0ω sin ωt
∴ Maximum value of I is Q0ω
or Imax = Q0ω = (2.0 × 10–4 C) (10+4 s–1)
Imax = 2.0 A
(d) From energy conservation.
1
1
1 Q2
LI 2max = LI2 +
2
2
2 C
or Q =
3 × 10–4 C or Q = 1.732 × 10–4 C
refractive index µp = 3 . Face AC of the prism is
covered with a thin film of refractive index µf = 2.2.
A monochromatic light of wavelength λ = 550 nm
fall on the face AB at an angle of incidence of 60º.
[IIT-2003]
A
For (1) 100 = 200 cos ωt
1
or cos (ωt) = , From equation (3) :
2
dI
1
= (2.0 × 10–4C) (10+4 s–1)2  
dt
2
LC(I 2max
(2.0 × 10 –3 )(5.0 × 10 –6 )(2 2 – 12 )
8
Highly energetic electrons are bombarded on a target
of an element containing 30 neutrons. The ratio of
radii of nucleus to that of helium nucleus is (14)1/3.
Find (a) atomic number of the nucleus. (b) the
frequency of Kα line of the X-ray produced.
(R = 1.1 × 107 m–1 and c = 3 × 108 m/s) IIT-2005]
DECEMBER 2010
Sol. (a) We know that radius of nucleus is given by the
formula
r = r0 A1/3 where r0 = constt. and A = mass number.
For the nucleus r1 = r0 41/3
For the Nucleus r1 = r0 (4)1/3
∴ % of O3 =
0.096
× 100 = 6.6%
1.44
0.096 × 6.023 × 10 23
48
= 1.2 × 1021
No. of photons or molecules of O3 =
1/ 3
∴
r2
A
=  
r1
4
7.
1/ 3
A
(14)1/3 =  
⇒ A = 56
4
∴ No. of proton = A – no. of neutrons
= 56 – 30
= 26
∴
Atomic number = 26
1
1 
(b) We know that v = Rc (z – b)2  2 – 2 
 n1 n1 
7
8
Here, R = 1.1 × 10 , c = 3 × 10 , Z = 26
b = 1 (for Kα), n1 = 1, n2 = 2
1 1 
∴ ν = 1.1 × 107 × 3 × 108 [26 – 1]2  – 
1 4 
⇒
= 3.3 × 1015 × 25 × 25 ×
(iv) Toluene reacts with bromine in the presence of
light to give benzyl bromide while in presence
of FeBr3 it gives p-bromotoluene. Give
explanation for the above observations.
[IIT-1996]
(v) Explain very briefly why alkynes are generally
less reactive than alkenes towards electrophilic
reagents such as H+.
[IIT-1997]
(vi) The central carbon-carbon bond in 1,
3-butadiene is shorter than that in n-butane.
[IIT-1998]
(vii) tert-Butylbenzene does not give benzoic acid
on treatment with acidic KMnO4. [IIT-2000]
(viii) 7-Bromo-1, 3, 5-cycloheptatriene exists as
ionic
compound,
while
5-bromo-1,
3-cyclopentadiene does not ionise even in
presence of Ag+ ion. Explain.
[IIT-2004]
CH3
aq. C 2 H 5OH
(ix)
→ Acidic solution
Br  
CH3
CH3
aq. C 2 H 5OH
Br
 
→
Neutral
CH3
[IIT-2005]
solution. Explain.
3
= 1.546 × 1018 Hz.
4
CHEMISTRY
One litre of a mixture of O2 and O3 at NTP was
allowed to react with an excess of acidified solution
of KI. The iodine liberated required 40 ml of M/10
sodium thiosulphate solution for titration. What is
the weight percent of ozone in the mixture ?
Ultraviolet radiation of wavelength 300 nm can
decompose ozone. Assuming that one photon can
decompose one ozone molecule, how many photons
would have been required for the complete
decomposition of ozone in the original mixture?
[IIT-1997]
Sol. The concerned chemical reaction are :
O3 + 2KI + H2O → 2KOH + I2 + O2
I2 + 2Na2S2O3 → Na2S4O6 + 2NaI
Millimoles of ozone = Millimoles of I2
1
mM of O3 = mM of I2 =
× mM of Na2S2O3
2
1
1
=
× 40 ×
= 2 mM = 0.002 mole
10
2
Calculation of total number of moles of O2 and O3
PV = nRT
1 × 1 = n × 0.0821 × 273
n = 0.044 mole
∴ Mole of O2 = 0.044 – 0.002 =- 0.042
∴ Wt. of O2 = No of moles × Mol. wt. = 0.042 × 32
= 1.344 g
Similarly, Wt. of O3 = 0.002 × 48 = 0.096 g
6.
XtraEdge for IIT-JEE
Give reasons for the following :
(i) Methane does not react with chlorine in the dark
[IIT-1983]
(ii) Propene react with HBr to give isopropyl
bromide but does not give n-propyl bromide.
[IIT-1983]
(iii) Although benzene is highly unsaturated,
normally it does not undergo addition reaction.
[IIT-1983]
3H / Pd
3
→
(x)
(A)
but not
(B)
9
DECEMBER 2010
(a) The bridged intermediate cation formed by
the initial attack of electrophile on the triple
bond is less stable because it is a highly
strained system. Due to formation of cyclic
intermediate carbocation, the olefinic
intermediate products would invariably be
trans.
(b) In acetylenic carbon atoms, the π-electrons
are more tightly held by the carbon nuclei
and hence they are less easily available for
reaction with electrophiles.
Perhaps both the above factors, steric and
electronic, play their in diminishing the
reactivity of alkynes towards electrophiles.
(vi) Tips/Formula : 1, 3-Butadience is a
conjugated diene and is a reasonance hybrid :
Solution :
–
–


••
••
+
+
– C = C– C = C – ↔ – C– C = C– C – ↔ – C– C = C– C–
 |
|
|
|
|
|
|
|
|
|
|
| 


[IIT-2005]
(C)
Sol.
(i) Tips/Formula : Chlorination of methane is a
free radical substitution reaction.
Solution : In dark, Chlorine is unable to be
converted into free radicals, hence the reaction
does not occur.
(ii) Tips/Formula : Addition of unsymmetrical
addendum (HBr in present case) to
unsymmetrical olefin (CH3CH=CH2, in present
case)
takes
place
according
to
Markownikoffrule.
Solution : According to this rule "the negative
part of reagent (i.e., Br–) adds on the carbon
atom having minimum number of hydrgon
atoms". Hence isopropyl bromide will be
formed in the present case.
CH3CH=CH2 + HBr → CH3.CHBr.CH3
Propene
iso-Propyl bromide
(iii) Unlike olefins, π-electrons of benzene are
delocalised (resonance) and hence these are
unreactive towards addition reactions.
(iv) Tips/Formula : In presence of FeBr3 it gives o
and p derivative and in absence of FeBr3 it
gives side chain derivative.
Solution : In presence of light, toluene
undergoes side chain bromination through a
free radical mechanism.
CH3
CH2Br
The charged structures induce some double
bond character in the central C–C bond leading
to the shortening of this bond. Alternatively, all
the four C atoms of 1, 3-butadiene are sp2
hybridised and thus their C–C bond length will
be lower than that of n-butane in which all the
four c-atoms are sp3 hybridised.
(vii) tert-Butylbenzene does not give benzoic acid
on treatment with acidic KMnO4 ause it does
not contain any hydrogen atom on the key
carbon atom.
(viii)Tips/Formula
:
7-Bromo-1,3,5cycloheptatriene is aromatic whereas 5-Bromo1,3-Cycloheptadiene is non aromatic.
Solution :
Br
+
Its corresponding
+ Br–
cation is
Br
2
→

hv
Benzyl bromide
In presence of FeBr3, toluene undergoes
electrophilic substitution in the benzene ring.
CH3
CH3
7-Bromo-1,3,5cycloheptatriene
(Triopylium bromide)
Br
2
→

FeBr3
Br
Br
+
Its corresponding
cation is
p-bromotoluene
(v) The low reactivity of alkynes towards
electrophilic addition reaction is believed to be
due to following two factors.
E
⊕
⊕
C— + E → C
C
C—
5-Bromo-1,3cyclopentadiene
+ Br–
Cyclopentadienyl cation
(It has 4π electrons,
hence not aromatic, thus
not easily formed)
(ix) The halide is a 3º halide, hence it undergoes
SN1 reaction forming HBr, as one of the
products which make solution acidic.
Highly strained bridged
carbocation
XtraEdge for IIT-JEE
7-Bromo-1,3,5-cyclo
heptarienyl cation
(Triopy lium cation)
It has 6π electrons, hence
aromatic and easily formed
10
DECEMBER 2010
2–
CH3
CH3
|
|
C 2 H 5OH ( aq )
 → C6H5–C–OC2H5 + HBr
C6H5–C–Br 
(S N 1)
|
|
(acidic)
CH3
CH3
A 3º bromide
Br
CN
CN
Ni
CN
CN
CH(CH3)2 is an aryl halide so it
does not undergo nucleophilic substitution
reactions. Hence the solution will remain
neutral.
(x) Reduction of cental ring to form A reduces all
the three cyclobutadiene rings (which are
antiaromatic as they have 4π electrons each),
i.e. antiaromatic rings are converted into
nonaromatic rings. On the other hand,
reduction of the terminal ring to form B
reduces only one antiaromatic ring and two
antiaromatic cyclobutadiene rings remain
intact. Remember that antiaromatic rings
impart unstability.
Square planar dsp2 hybridisation
[Ni(CO)4]
4s
3d
Ni (after rearrangement)
3d
sp3
CO
Draw the structures of [Co(NH3)6]3+ , [Ni(CN)4]2–
and [Ni(CO)4]. Write the hybridisation of atomic
orbitals of the transition metal in each case.
[IIIT-2000]
Sol. [Co(NH3)6]3+
4s
3d
4p
3+
Co
Ni
CO
Tetrahedral (sp3 hybridisation)
9.
⇒
d2sp3
3+
NH3
NH3
Co
NH3
NH3
NH3
Octahedral complex, d2sp3 hybridisation
[Ni(CN)4]2–
3d
4s
CO
CO
4p
NH3
4p
pairing due
to CO
8.
3d
4p
Ni =
Sol.
4p
Ni2+
A white substance (A) reacts with dilute H2SO4 to
produce a colourless gas (B) and a colourless
solution (C). The reaction between (B) and acidified
K3Cr2O7 solution produces a green solution and a
slightly coloured precipitate (D). The substance (D)
burns in air to produce a gas (E) which reacts with
(B) to yield (D) and colourless liquid. Anhydrous
copper sulphate is turned blue on addition of this
colourless liquid. Addition of aqueous NH3 or
NaOH to (C) produces first a precipitate, which
dissolves in the excess of the respective reagent to
produce a clear solution in each case. Identify (A),
(B), (C), (D) and (E). Write the equations of the
reaction involved.
[IIT-2001]
dil.H SO
2
4
A  


→
( white )
B
( colourless)
+
C
( colourless solution )
K2Cr2O7/H+
Ni2+ (after rearrangement)
3d
4p
Green solution + D ↓(burns in air to form E)
(coloured)
pairing due
to CN–
. CuSO 4
E↑ + B↑ → D + Colourless liquid anhy

→ Blue
dsp2
aq. NH
of
3

→ Precipitate excess
 
→ Clar solution
C  
or NaOH
reagent
The above set leads to following conclusions.
(i) Since the gas (B) is colourless and turns
acidified K2Cr2O7 solution given, it should be
H2S.
XtraEdge for IIT-JEE
11
DECEMBER 2010
(ii) Since H2S gas is obtained by the reaction of
dil. H2SO4 on A, the latter must be sulphide.
(iii) The white colour of the sulphide (A) points out
towards ZnS.
Thus the various reactions can be written as below.
ZnS + H2SO4 (dil) → ZnSO4 + H2S↑
(A)
(C)
(B)
3H2S + K2Cr2O7 + 4H2SO4
→ K2SO4 + Cr2(SO4)3 + 7H2O + 3S
(green)
(D)
11.
MATHEMATICS
Prove that there exists no complex number z such
1
and
3
that |z| <
n
∑a z
r
r
= 1, where |ar| < 2.
r =1
[IIT-2003]
Sol. Given : a1z + a2z2 + ... + anzn = 1
and
|z| < 1/3
...(1)
|a1z + a2z2 + a3z3 + ... + anzn| = 1
{using |z1 + z2| ≤ |z1| + |z2|}
⇒ |a1z| + |a2z2| + |a3z3| + ... + | anzn| ≥ 1
⇒ 2{|z| + |z|2 + |z|3 + ... + |z|n} > 1
[using |ar| < 2]
n
2 | z | (1– | z | )
⇒
>1
1– | z |
{using sum of n terms of G.P.}
n+1
⇒ 2|z| – 2|z| > 1 – |z| ⇒ 3 |z| > 1 + 2 |z|n+1
1
2 n+1
⇒ |z| > +
|z|
3
3
1
⇒ |z| > , which contradicts (1)
3
∴ There exists no complex number z such that
H 2S( B)
S + O2 → SO2↑ 2
→ 2H2O + 3S↓
(D)
(E)
(colourless liq) D
( White )
CuSO
 4 

→ CuSO4.5H2O
(blue)
ZnSO4 + 2NaOH → Zn (OH)2↑
(C)
white.ppt
NaOH
2

→ Na2ZnO2 + 2H2O
( excess )
(soluble)
10. A compounds (X) containing C, H and O is
unreactive towards sodium. It does not add bromine.
It also does not react with Schiff's reagent. On
refluxing with an excess of hydriodic acid, (X)
yields only one organic product (Y). On hydrolysis,
(Y) yields a new compound (Z) which can be
converted into (Y) by reaction with red phosphorus
and iodine. The compound (Z) on oxidation with
potassium permanganate gives a carboxylic acid.
The equivalent weight of this acid is 60. What are
the compounds (X), (Y) and (Z)? Write chemical
equations leading to the conversion of (X) to (Y).
[IIT-1981]
Sol. Tips/Formula : The unreactivity of the compound
(X) towards sodium indicates that it is neither an
acid nor an alcohol, further its unreactivity towards
Schiff's base indicates that it is not an aldehyde. The
reaction of compound (X) with excess of HI to form
only one product indicates that it should be an ether.
Solution : Hence its other reactions are sketched as
below.
|z| <
1
3
n
and
∑a z
r
r
=1
r =1
mn squares of equal size are arranged to form a
rectangle of dimension m by n where m and n are
natural numbers. Two squares will be called
'neighbours' if they have exactly one common side.
A natural number is written in each square such that
the number in written any square is the arithmetic
mean of the numbers written in its neighbouring
squares. Show that this is possible only if all the
numbers used are equal.
[IIT-1982]
Sol. Let mn squares of equal size are arrange to form a
rectangle of dimension m by n. Shown as, from
figure, neighbours of
12.
R–O–R Re
flux
with

→ 2RI hydrolysis
 → 2ROH
excess of HI
(X)
(Y)
(Z)
n
P + I2
x6 x2
x5 x1 x3
x7 x4
KMnO
4
 

→ –COOH
(O)
eq. wt. 60
Since the carboxylic acid has equivalent weight of
60, it must be acetic acid (CH3COOH), hence Z
must be ethyl alcohol, (Y) ethyl iodide and (X)
diethyl ether.
m
x1 are {x2, x3, x4, x5}
x5 are {x1, x6, x7}
x7 are {x5, x4}
x + x6 + x7
x + x3 + x4 + x5
⇒ x1 = 2
, x5 = 1
3
4
x4 + x5
and x7 =
2
C2H5–O–C2H5 + 2HI reflux
→ 2C2H5I
Ethyl iodide (Y)
–
4
OH
→ 2C2H5OH KMnO


→ CH3COOH
Ethyl alcohol (Z)
Acetic acid
(Eq. wt. = 60)
XtraEdge for IIT-JEE
12
DECEMBER 2010
∴
4x1 = x2 + x3 + x4 +
x1 + x6 + x7
3
14.
x + x5
⇒ 12x1 = 3x2 + 3x3 + 3x4 + x1 + x6 + 4
2
⇒ 24x1 = 6x2 + 6x3 + 6x4 + 2x1 + 2x6 + x4 + x5
⇒ 22x1 = 6x2 + 6x3 + 7x4 + x5 + 2x6
where, x1, x2, x3, x4, x5, x6 are all the natural
numbers and x1 is linearly expressed as the sum of
x2, x3, x4, x5, x6
where sum of coefficients are equal only if, all
observations are same.
⇒ x 2 = x3 = x4 = x 5 = x6
⇒ all the numbers used are equal.
If 'f ' is a continuous function with
x
∫ f (t)dt → ∞ as
0
|x| → ∞, then show that every line y = mx intersects
x
∫ f (t)dt = 2!
the curve y2 +
[IIT-1991]
0
(0, 2 )
A
O
B
(xp,0)
X
(0, – 2 )
Sol. We are given that f is continuous function and
x
∫ f (t) dt → ∞, as |x| → ∞
Let f [(x + y)/2] = {f(x) + f(y)}/2 for all real x and y.
If f ' (0) exists and equals –1 and f(0) = 1, find f(2).
[IIT-1995]
f ( x ) + f ( y)
x+y
∀ x, y ∈ R
(given)
Sol. f 
 =
2
2


Putting y = 0, we get
f ( x ) + f (0)
1
x
f  =
=
[1 + f(x)] [Q f(0) = 1]
2
2
2
13.
0
To show that every line y = mx intersects the curve
y2 +
m2x2 +
∫
x
f ( t ) dt = 2
0
Consider, F(x) = m2x2 +
x
∫ f (t) dt – 2
...(1)
0
Then F(x) is a continuous function as F(x) is given
to be continuous.
Also F(x) → ∞ as |x| → ∞
But F(0) = – 2
Thus F(0) = – ve and F(b) = +ve, where b is some
value of x and F(x) is continuous.
∴ F(x) = 0 for some value of x ∈ (0, b) or equation
(1) is solvable for x.
Hence, y = mx intersects the given curve.
15.
 π π
Find all values of θ in the interval  – , 
 2 2
satisfying the equation
(1 – tan θ )(1 + tan θ) sec2θ + 2 tan
2
Sol. (1 – tan θ) (1 + tan θ) sec θ + 2
⇒
⇒
⇒
∫ f ' (x)dx = ∫ – 1 dx
2
2
4
2
θ
= 0 [IIT-1996]
2
tan θ
(1 – tan θ).(1 + tan θ) + 2
=0
tan 2 θ
=0
tan 2 θ
1 – tan θ + 2
=0
put tan2θ = x
2
x
2
1 – x + 2 = 0 ⇒ x – 1 = 2x
y
–1
O
x
1
–1
⇒ f(x) = – x + k where k is a constant.
But f(0) = 1, therefore f(0) = – 0 + k
⇒ 1=k
⇒ f(x) = 1 – x ∀x ∈ R
⇒ f(2) = – 1
XtraEdge for IIT-JEE
0
If possible, let y = mx intersects the given curve,
then substituting y = mx in the curve, we get
⇒ 2f(x/2) = f(x) + 1
⇒ f(x) = 2 f(x/2) – 1 ∀ x, y ∈ R
...(1)
Since f ' (0) = – 1, we get
f (0 + h ) – f (0)
f (h ) – 1
= – 1 ⇒ lim
=–1
lim
h →0
h→0
h
h
Now, let x ∈ R then applying formula of
differentiability.
 2 x + 2h 
f
 – f (x)
f (x + h) – f (x)
2 

= lim
f '(x) = lim
h →0
h →0
h
h
f ( 2 x ) + f ( 2h )
– f (x)
2
= lim
h →0
h
1   2x 
 2h  
2f   – 1 + 2f   – 1 – f ( x )
2  2 
 2  
= lim
h →0
h
[Using equation (1)]
1
{2f ( x ) – 1 + 2f (h ) – 1} – f ( x )
= lim 2
h →0
h
f ( x ) + f (h ) – 1 – f ( x )
f (h ) – 1
= lim
=–1
= lim
h →0
h→0
h
h
Therefore f ' (x) = – 1 ∀ x ∈ R
⇒
x
∫ f (t) dt = 2
Curves
y = x2 – 1and y = 2x
intersect at one point (negative value will not
consider) x = 3, y = 8
Therefore, tan2θ = 3 ⇒ tan θ = ±
13
3
⇒ θ = ± π/3
DECEMBER 2010
XtraEdge for IIT-JEE
14
DECEMBER 2010
Physics Challenging Problems
Set # 8
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in physics that would be very helpful in facing IIT
JEE. Each and every problem is well thought of in order to strengthen the concepts and we
hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma
Director Academics, Jodhpur Branch
So lutions will b e p ub lished in nex t issue
(A) Circular path with radius of the circular path
Passage # 1 (Que. 1 to 3)
The internal energy ‘U’ v/s PV graph where P is the
pressure and V is the volume of an ideal gas filled up
in a piston cylinder system is shown below
If tan φ = b then
is r =
m a 2 + b2
q.B 0
(B) Helix and the pitch of the Helix is
2πm
.a
q.B 0
2πm
.b
q.B 0
(D) Same path as followed by circulating
electrons which is responsible for the
unstable Rutherford atomic model, Means
spiral path of decreasing radius.
(C) Helix and the pitch of the Helix is
1.
What is the atomocity and the shape of the Gaseous
molecule if b= 3 and a = 2.
2.
Write the relation of adiabatic index of the gas in
terms of a or b or interms of both a and b.
3.
If a start varying with respect to time as
a(t) = 2(3+t) and b remains constant then draw the
graph of CV v/s a where CV is the molar specific heat
at constant volume.
4.
If b start varying with respect to time as
b( t )c 0 + c1 t 2 where c0 and c1 are positive constants
Passage # II (Que. 6 to 8)
Behaviour of capacitor in electric circuits is very
typical because of it's energy storing nature.
Capacitor behaves in just opposite manner to
inductor, Inductor 'L' which is measured in
Henary in SI system stores the energy in magnetic
field instead of capacitor which stores in electric
field Inductor opposes the change in current and
capacitor opposes change in voltage.
Behaviour of inductor:
df
v/s t graph where f is the
dt
degrees of freedom for the gas.
then find the slope of
5.
For the electric circuit shown
A particle enters in the given magnetic field
→
∧
B = B 0 k whre B0 is a constant with the velocity of
→
∧
∧
v = a i + b j where a and b are the positive
constants.
The place where the magnetic field exists and the
particle moves is filled with the resistive medium
then path followed by the particle is(Charge on particle q and mass m)
6.
If capacitor C varies even after that energy stored
in capacitor is zero at steady state then
R
ε
R
ε
(A) 1 = 1
(B) 1 = 2
R 2 ε2
R 2 ε1
(C) ε1 + ε 2 = 0
XtraEdge for IIT-JEE
15
(D) ε1 R 1 + ε 2 R 2 = 0
DECEMBER 2010
7.

R 1R 2
(D)  R +
R1 + R 2

ε / R 1 − ε1 / R 2
= 1
1/ R1 +1/ R 2
(C) R2C if ε1 < ε 2
Where ε eq
R eq =
8.
What do Aliens Look Like?
Time constant for the circuit
(A) RC
(B) R1C if ε1 > ε 2

C

Aliens are the extraterrestrial beings believed to exist.
Some give accounts of having seen them visit our
world. Then, what do aliens look like? Want to know?
The read on…
R 1R 2
R1 + R 2
Maximum current passing through resistance R
ε eq
ε eq
(B)
(A)
R eq
R + R eq
(C)
ε eq
R
(D)
Aliens have always aroused the interest for many. With
new discoveries in astronomy, man has been able to
explore the extraterrestrial world and examine the
chances of the existence of aliens.
| ε1 − ε 2 |
R
On one hand, the existence of extraterrestrial life is
considered hypothetical while on the other hand,
aliens have been sighted on a few occasions. There
have been news about the aliens visiting Earth; there
have been some people claiming to have seen the
aliens. The concept of ‘aliens’ remains alien!
The sightings of aliens have brought about
descriptions of their appearance. What they look like,
has been a question in the minds of one and all and
news have many a time answered it by giving
accounts of people witnessing aliens. We know of
films and television shows, which have depicted aliens
as being humanoid in appearance.
Human Quick Facts
1. The hardest bone in the human body is the
jawbone.
2. The number of eye blinks varies greatly from
about 29 blinks each minute if you are talking to
someone to only 4 blinks each minute if you are
reading.
3. The average human blinks 25 times per minute.
What do Aliens Look Like?
4. A nail takes around 6 months to grow from base
to the tip.
Aliens are largely described as resembling human
beings. Their height is approximately same as the
average height of human beings. Like any normal
human beings, aliens have a pair of eyes, a nose, a
mouth, a pair of arms and a pair of feet. There are
citations of aliens having wings or wheels instead of
feet and other such abnormalities. It is believed that
aliens have a rough lizard-like skin. Their skin colors
are believed to vary from gray, white, tan to gold, pink
or red. Their skin is believed to glow in the dark. Their
eyes are considered to resemble those of humans,
lizards or insects. Some have documented aliens as
having webbed fingers while others believe that aliens
have suction cups for fingertips or claws. Aliens have
been documented as being variedly sized and shaped.
Some have documented them as 3 inches tall while
others say that they are about 15 feet tall. In some
places aliens have been documented as being shaped
like balls of light, while in other places they have been
shown as resembling robots or metal objects. Some
believe that aliens look like animals or large insects
while some think of aliens as human-like figures
clothed in uniforms. Many believe that aliens can float
through walls.
5. Each second 10,000,000 cells die and are replaced
in your body.
6. Your liver performs over 500 functions in your
body.
7. The average person spends 1/3 of their lifetime
sleeping.
8. More germs are transferred when shaking hands
than kissing.
9. The average person (from western culture)
consumes 10 liters of alcohol per year.
10. Roughly 75% of people who play the radio in
their car sing along to it.
11. Human thigh bones are stronger than concrete.
12. Your right lung takes in more air than your left
one does.
13. The human brain is composed of 75% water.
14. 70% of the composition of dust in your home is
made up of shed human skin and hair.
15. The tooth is the only part of the human body that
can’t repair itself.
XtraEdge for IIT-JEE
16
DECEMBER 2010
1.
8
Solution
Set # 7
Physics Challenging Problems
Qu e s tio ns we r e Pub lis he d in Nov emb er I ss ue
As the resistances of voltmeters in upper branch
are R, R/2, R/4 ………………….
The equivalent circuit is as shown below
3.
From current division formula we can conclude that
current in upper and lower branch are in the ratio of
1 : 2.
The resistance of upper branch is
= R + R/2 + R/4+……….. up to infinite
 1 1

= R 1 + + + ...... 
 2 4

Reading of voltmeter V1 is i.R
Reading of voltmeter V is (2i.)R
So V = 2V1
 1 
= R
 = 2R
 1−1 / 2 
further the equivalent circuit is
4.
the resistance of voltmeter V should be 2R so that
current in upper and lower branch is same.
2.
l = length of rod = b – a
charge on element of length dx is dq
dq = λdx
as λ = 3x
dq = 3xdx
Equivalent current due to element of length dx
Entire upper branch is having the resistance of 2R
and voltmeter V1 is having the resistance of R so
we can conclude that equivalent resistance of all
the voltmeters in upper branch except V1 is R and
the upper branch is as follows:
di = ω.dq =
ω
(3xdx )
2π
∫
Total equivalent current i = di =
b
=
3ω  x 2 
3ω  b 2 − a 2
  =
2π  2 
2π 
2
a
b
ω
∫ 2π (3xdx)
a
 3 ω 2
 = . (b − a 2 )
 2 2π

3ω 2
(b − a 2 )
4π
Option A is correct
Equivalent current
=
As reading of voltmeter V1 is X = i.R
Sum of the readings of voltmeters is Y = i.R
Except V1 in upper branch
So, X = Y
XtraEdge for IIT-JEE
=
17
3ω
3ω 2
(b − a 2 ) =
(b − a )(b + a )
4π
4π
DECEMBER 2010
Full scale deflection current for galvanometer is
50mV
ig =
= 5mA
10Ω
For terminals CT and a range is 5V so using
V
5
R = − G ⇒ R1 =
− 10 = 990Ω
ig
5 × 10 −3
3ω
3
(b + a )(b − a ) =
ω.(b + a ).l
4π
4π
As ω = 4π / 3 So,
3 4π
. .(b + a ).l
Equivalent current =
4π 3
= (b + a). l = const. l
i∝l
Option B is correct.
Charge on rod
=
b
R 1 = 990Ω
7.
b
x2 
3
q = dq = 3xdx = 3.  = (b 2 − a 2 )
 2  a 2
a
Option D is correct
Ans. A, B, D
∫
∫
990 + R2 = 2000 – 10
R 2 = 2000 − 1000 = 1000Ω
R 2 = 1000Ω
8.
5. For part B
q (closed cone) > q (open cone)
for part A
q (close cone) = q (open cone)
ω
Equivalent current i =
.q
2π
ω
ω
i=
.q
;
i=
.q
2π
2π
cone –C1(closed cone) cone-C3(closed cone)
ω
ω
i =
.q (closed cone)
; i=
.(σ) (Surface area
2π
2π
of closed cone)
If σ varies then charge on cone C1 differs from C3 So
their currents will be different.
Option A incorrect
=
q (cone-C2)
q (cone- C1)
ω
ω
i =
. q
and
i =
. q
( coneC1 )
(
coneC
)
2π ( cone −C1 )
2π ( cone − C 2 )
2
Range between CT and c is V so
V
Using R = − G
ig
R1 + R 2 + R 3 =
V
5 × 10 −3
⇒ 990 + 1000 + 3000 =
⇒ 5000 =
− 10
V
5 × 10 −3
− 10
V
5 × 10 −3
⇒ V = 25volt
So range between CT and C is 25 volts.
GLOBAL WARMING IS REAL
=
i (cone C2)
i (cone C1)
Option B is correct
As charge on cone C 3 ≠ charge on cone C4
The arctic ice is receding and global warming is no
longer a theory but a reality. Scientists predict that
by the year 2100, the average surface temperature
will jump up by 6 degrees Fahrenheit. Nighttime
temperatures will be higher and there will be hotter
days.
Option C correct
Part-A and Part –B will have different charges so
Option C correct
Part –A and Part – B will have different charges so
Option D incorrect
Ans. B, C
Since air temperature is a powerful component of
climate, there will be unavoidable climate changes
in the future. Some climate changes involve
extreme weather disturbances such as more severe
hurricanes and longer droughts. There will be an
increased precipitation of snow and rain during
winter. The faster melting of snow during the spring
will result in flooding. All these climate changes are
predicted based on the assumption that changes will
be relatively gradual.
6. The circuit is as follows
XtraEdge for IIT-JEE
Range between CT and b is 10 volt so,
V
10
− 10
Using R = − G ⇒ R 1 + R 2 =
ig
5 × 10 −3
18
DECEMBER 2010
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
PHYSICS
Initial
A ring of radius R = 4 m is made of a highly dense
material. mass of the ring is m1 = 5.4 × 109 kg.
distributed uniformly over its circumference. A
highly dense particle of mass m2 = 6 × 108 kg is
placed on the axis of the ring at a distance x0 = 3m
from the centre. Neglecting all other forces, except
mutual gravitational interaction of the two, calculate
(i) displacement of the ring when particle is closest
to it, and
(ii) speed of the particle at that instant.
Sol. Since, there is no external force on the system of ring
and particle, therefore, centre of mass of the system
remains stationary.
1.
U1 = –
2.
m2
of
the
system,
In the arrangement shown in Figure pulleys are small
and light and springs are ideal. K1, K2, K3 and K4 are
force constant of the springs.
Calculate period of small vertical oscillations of
block of mass m.
K2
x0
Particle is closest to ring when it is at its centre. At
this instant centre of mass of the system is at centre
of the ring. It means displacement of ring is equal to
the distance of centre of mass of the system from
initial position of the ring.
m .0 + m 2 x 0
i.e., x = 1
m1 + m 2
= 0.3 m
Ans. (i)
Since, initially ring and particle both are stationary,
therefore, momentum of the system is zero and it
always remains zero because there is no external
force on the system.
If velocities of ring and particle are v1 (rightward)
and v2 (leftward) respectively and particle reaches the
centre of ring, then according to law of conservation
of momentum,
v
...(1)
m1v1 = m2v2 or v1 = 2
9
But when ring and particle move towards each other,
gravitational potential energy of their system
decreases and converts into kinetic energy.
Therefore, according to law of conservation of
energy,
kinetic energy of the system = loss of gravitational
potential energy
XtraEdge for IIT-JEE
energy
R 2 + x 02
Gravitational energy of the system when particle is at
Gm1m 2
the centre of the ring U2 =
R
1
1
∴ m1v12 + m 2 v 22 = U1 – U2
2
2
Ans. (ii)
or v2 = 0.18 ms–1 or cm/sec
R
m1
gravitational
Gm1m 2
K4
m
K1
K3
Sol. In static equilibrium of block, tension in the string is
exactly equal to its weight. Let a vertically downward
force F be applied on the block to pull it downwards.
Equilibrium is again restored when tension in string
is increased by the same amount F. Hence, total
tension string becomes equal to (mg + F).
Strings are further elongated due to extra tension F.
Due to this extra tension F in strings, tension in each
spring increase by 2F. Hence increases in elongation
2F 2F 2F
2F
,
,
and
respectively.
of springs is
K1 K 2 K 3
K4
According to geometry of the arrangement,
downward displacement of the block from its
 2F 2F 2F 2F 

+
+
+
equilibrium position is y = 2 

 K1 K 2 K 3 K 4 
...(1)
19
DECEMBER 2010
When particle reaches infinity, its potential energy U
becomes equal to zero.
∴ Work done = Increase in potential energy
qρR 2
= U – U0 =
Ans.
4ε 0
If the block is released now, it starts to accelerate
upwards due to extra tension F in string. It means
restoring force on the block is equal to F.
y
From equation (1), F =
 1
1
1
1 

+
+
+
4
 K1 K 2 K 3 K 4 
∴ Restoring acceleration of block =
F
m
Each of two long parallel wires carries a constant
current I along the same direction. The wires are
separated by a distance 2l. Calculate maximum
magnitude of magnetic induction in the symmetry
plane of this system located between the wires.
Calculate also, the maximum force experienced by
unit length of a third wire carrying the same current
along the same direction if third wire is parallel to
and in the symmetry plane of other two wires.
Sol. In Figure, points P and Q represent two wires, each
carrying current along inward normal to plane of the
paper. It is given that each of these two wires carries
a current I and separation between the wires is 2l. In
the figure, dotted line PQ represents the plane of
wires and firm line normal to PQ represents the plane
of symmetry.
B'
B' θ θ
B
R
4.
=
y
 1
1
1
1 

+
+
+
4m

 K1 K 2 K 3 K 4 
Since, acceleration of block is restoring and is
directly proportional to displacement y, therefore, the
block performs SHM.
Its period T = 2π
displacement
acceleration
 1
1
1
1 

∴ T = 2π 4m
+
+
+

 K1 K 2 K 3 K 4 
 1
1
1
1 

+
+
+
= 4π m

 K1 K 2 K 3 K 4 
Ans.
r
A Solid non-conducting hemisphere of radius R has a
uniformly distributed positive charge of density ρ per
unit volume. A negatively charged particle having
charge q is transferred from centre of its base to
infinity. Calculate work performed in the process.
Di-electric constant of material of hemisphere is
unity
Sol. When negative charge q is displaced from centre of
base to infinity, its electrical potential energy
increases.Work is to be performed to increase this
energy. To calculate initial potential energy of the
particle, first a thin hemispherical shell of radius x
and radial thickness dx is considered as shown in
Figure
3.
×
P
θ
y
×
Q
l
l
Let magnetic induction in plane of symmetry be
maximum at point R, at a distance y from plane of
wires P and Q.
Distance of this point from each wire is r =
l2 + y2
∴ Magnitude of magnetic induction at R due to
µ I
each wire is B' = 0
2πr
Directions of these two magnetic inductions at R are
as shown in figure. Their components in the plane of
symmetry neutralise each other. Therefore, at R,
resultant magnetic induction is normal to the plane of
symmetry.
The resultant magnetic induction, B = 2B' sinθ
µ 0 Iy
µ Iy
...(1)
∴ B= 02 =
πr
π( l 2 + y 2 )
Volume of material of the shell = 2πx2.dx
∴ Charge on shell is dQ = ρ(2πx2 dx)
Since, every element of this shell is at a constant
distance x from centre of curvature, therefore,
potential energy of the particle, due to charge of the
shell considered is
qρ
1 (– q)(dQ)
=–
x dx
dU =
4πε 0
x
2ε 0
or total initial potential energy of particle,
qρR 2
qρ x = R
U0 = –
x.dx = –
2ε 0 x = 0
4ε 0
For B to be maximum
dB
= 0 or y = l
dy
µ0I
2πl
Maximum force experienced by unit length of the
third wire, Fmax = Bmax . I Nm–1
µ I2
∴ Fmax = 0 Nm–1
2πl
∴ Bmax =
∫
XtraEdge for IIT-JEE
θ
r
20
DECEMBER 2010
5.
According to Flemming's left hand rule vector of
force F lies in the symmetry plane and towards plane
of wires.
A metal rod of length l = 100 cm is clamped at two
points A and B as shown in Figure. Distance of each
clamp from nearer end is α = 30 cm. It density and
Young's modulus of elasticity of rod material are
ρ = 9000 kg m–3 and Y = 144 GPa res-pectively,
calculate minimum and next higher frequency of
natural longitudinal oscillations of the rod.
A
v
= 10 kHz
Ans.
λ0
Next higher frequency corresponds to next higher
integer values of m and n which satisfy equation (3).
Hence, for this case m = 6 and (2n – 1 = 9 or n = 5
Substitution m = 6 in equation (1),
fo =
A
It means rod oscillates with odd harmonics
80
40
cm or
cm
λ=
6
3
v
∴ Next higher frequency,
f1 =
= 30 kHz Ans.
λ
30 cm
30 cm
l = 100 cm
Sol. Speed of longitudinal waves in the rod is v =
Y/ρ
= 4000 ms–1 .
Since points A and B are clamped, therefore, nodes
are formed at these points or rod oscillates with
integer number of loops in the middle part. Let
number of these loops be m.
λ
Since, length of each loop is ,
2
λ
= (l – 2a)
therefore,
m.
2
or
mλ = 80 cm
...(1)
Since ends of the rod are free, therefore, antinodes
are formed at each end of the rod or at one end of
each end part is an antinode and at the other end is a
node. It means that number of loops in each end part
 2n – 1 
will be an odd multiple of half. Let these be 

 2 
where n is an integer.
 2n – 1  λ
Then, 
= a or (2n – 1) λ = 129 cm
.
 2  2
...(2)
Dividing equation (1) by (2),
2
m
...(3)
=
(2n – 1)
3
Minimum possible frequency corresponds to
maximum possible wavelength, hence, minimum
number of loops.
COMPLEMENTARY COLOURS
If you arrange some colours in a circle, you get a
"colour wheel". The diagram shows one possible
version of this. An internet search will throw up many
different versions!
Colours directly opposite each other on the colour
wheel are said to be complementary colours. Blue
and yellow are complementary colours; red and cyan
are complementary; and so are green and magenta.
Mixing together two complementary colours of light
will give you white light.
What this all means is that if a particular colour is
absorbed from white light, what your eye detects by
mixing up all the other wavelengths of light is its
complementary colour. Copper(II) sulphate solution
is pale blue (cyan) because it absorbs light in the red
region of the spectrum. Cyan is the complementary
colour of red.
Hence, from equation (3), for minimum frequency m
should be equal to 2 and (2n – 1) should be equal to 3
or n = 2.
Substitution m = 2 in equation (1),
Maximum wavelength,
λ0 = 40 cm
XtraEdge for IIT-JEE
The origin of colour in complex ions
Transition metal v other metal complex ions
21
DECEMBER 2010
P HYSICS F UNDAMENTAL F OR IIT-J EE
Ray Optics
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Reflection :
Key Concepts :
(a) Due to reflection, none of frequency, wavelength
and speed of light change.
(b) Law of reflection :
Incident ray, reflected ray and normal on incident
point are coplanar.
The angle of incidence is equal to angle of
reflection
Incident n Reflected
Ray
Ray
θ θ
P
Real n
Object
Tangent
at point P
P
Convex surface
Plane surface
α
α
n
P
Virtual
Object
Real
Object
P
Real
Object
n
n
If the plane mirror is rotated through an angle θ, the
reflected ray and image is rotated through an angle 2θ
in the same sense.
If mirror is cut into a number of pieces, then the focal
length does not change.
The minimum height of mirror required to see the full
image of a man of height h is h/2.
Rest
v
Object v
P
vsinθ
Virual
Object
Object
v
θ
vcosθ
If reflected beam or refracted beam from an optical
instrument is diverging in nature, image is virtual.
Object v
XtraEdge for IIT-JEE
θ
θ
α
α
A
Tangent
at point P
Some important points : In case of plane mirror
For real object, image is virtual.
For virtual object, image is real.
The converging point of incident beam behaves as a
object.
If incident beam on optical instrument (mirror, lens
etc) is converging in nature, object is virtual.
If incident beam on optical instrument is diverging in
nature, the object is real.
The converging point of reflected or refracted beam
from an optical instrument behaves as image.
If reflected beam or refracted beam from an optical
instrument is converging in nature, image is real.
P
Virual
Object
n
αα
Convex
surface
P'
For solving the problem, the reference frame is
chosen in which optical instrument (mirror, lens, etc.)
is in rest.
The formation of image and size of image is
independent of size of mirror.
Visual region and intensity of image depend on size
of mirror.
P
P'
n
θ θ
n αα
22
Image
Rest
vsinθ
vcosθ Image
vm
Image
2vm–v
DECEMBER 2010
y
vm
Image
Object
In rest
2vm
y
x
x'
y'
y'
The mirror formula is
v
Object
vm
Image
360º
θ
β=
where θ = angle between mirrors.
2. If object and image both are virtual, β is negative.
3. If object is real but image is virtual, β is positive.
4. If object is virtual but image is real, β is positive.
5. Image of star; moon or distant object is formed at
focus of mirror.
If y = the distance of sun or moon from earth.
D = diameter of moon or sun's disc
f = focal length of the mirror
d = diameter of the image
360º
is odd number and object is placed on
θ
bisector of angle between mirror, then number of
images is n – 1.
If
360º
is odd and object is not situated on
θ
bisector of angle between mirrors, then the
number of images is equal to n.
(d) Law of reflection in vector form :
If
θ = the angle subtended by sun or moon's disc
Then tan θ = θ =
Let ê1 = unit vector along incident ray.
n̂ = unit vector along normal on point of
Incidence
Sun
D
ê 2 = ê1 − 2(ê1.n̂ ) n̂
ê 2
y'
XtraEdge for IIT-JEE
x
x'
y'
x
x'
θ
θ
Problem solving strategy :
Image formation by mirrors
Step 1: Identify the relevant concepts : There are
two different and complementary ways to solve
problems involving image formation by mirrors. One
approach uses equations, while the other involves
drawing a principle-ray diagram. A successful
problem solution uses both approaches.
Step 2: Set up the problem : Determine the target
variables. The three key quantities are the focal
length, object distance, and image distance; typically
(e) Spherical mirrors :
It easy to solve the problems in geometrical optics
by the help of co-ordinate sign convention.
y
y
y
x
F
d
n
n̂
x'
D
d
=
y
f
Here, θ is in radian.
ê 2 = unit vector along reflected ray
ê1
−v
image size
=
u
object size
1. If object and image both are real, β is negative.
360º
If
is even number, the number of images is
θ
n – 1.
Then
1 1 1
+ =
v u f
Also, R = 2f
These formulae are only applicable for paraxial
rays.
All distances are measured from optical centre. It
means optical centre is taken as origin.
The sign conventions are only applicable in given
values.
The transverse magnification is
2vm+v
(c) Number of images formed by combination of
two plane mirrors : The images formed by
combination of two plane mirror are lying on a
circle whose centre is at the meeting points of
mirrors. Also, object is lying on that circle.
Here, n =
x
x'
y'
23
DECEMBER 2010
Note that the same sign rules (given in section)
work for all four cases in this chapter : reflection
and refraction from plane and spherical surfaces.
Step 4: Evaluate your answer : You've already
checked your results by using both diagrams and
equations. But it always helps to take a look back
and ask yourself. "Do these results make sense ?".
you'll be given two of these and will have to
determine the third.
Step 3: Execute the solution as follows :
The principal-ray diagram is to geometric optics
what the free-body diagram is to mechanics. In
any problem involving image formation by a
mirror, always draw a principal-ray diagram first
if you have enough information. (The same
advice should be followed when dealing with
lenses in the following sections.)
It is usually best to orient your diagrams
consistently with the incoming rays traveling
from left to right. Don't draw a lot of other rays at
random ; stick with the principal rays, the ones
you know something about. Use a ruler and
measure distance carefully ! A freehand sketch
will not give good results.
If your principal rays don't converge at a real
image point, you may have to extend them
straight backward to locate a virtual image point,
as figure (b). We recommend drawing the
extensions with broken lines. Another useful aid
is to color-code the different principal rays, as is
done in figure(a) & (b).
Q
I
4
2
C P' F
3
P
2
4
Q'
Refraction :
Laws of Refraction :
The incident ray, the refracted ray and normal on
incidence point are coplanar.
µ1 sin θ1 = µ2 sin θ2 = ... = constant.
θ1
µ1
µ2
θ2
Snell's law in vector form :
n̂
v
ê1
3
1
ê 2
(a)
Q
2 4
ê 2 = unit vector along refracted.
Q'
v
P' F
n̂ = unit vector along normal on incidence point.
C
Then µ1( ê1 × n̂ ) = µ2( ê 2 × n̂ )
4
(b)
Some important points :
1 1 1
+ =
and the
s s' f
s'
y'
magnification equation m =
= − . The
s
y
results you find using this equation must be
consistent with your principal-ray diagram; if not,
double-check both your calculation and your
diagram.
Pay careful attention to signs on object and image
distances, radii or curvature, and object and image
heights. A negative sign on any of these
quantities always has significance. Use the
equations and the sign rules carefully and
consistently, and they will tell you the truth !
(a) The value of absolute refractive index µ is always
greater or equal to one.
Check your results using Eq.
XtraEdge for IIT-JEE
µ2
Let, ê1 = unit vector along incident ray
1 1
3
2
P
µ1
(b) The value of refractive index depends upon
material of medium, colour of light and
temperature of medium.
(c) When temperature increases, refractive index
decreases.
(d) Optical path is defined as product of geometrical
path and refractive index.
i.e., optical path = µx
24
DECEMBER 2010
Σ
(e) For a given time, optical path remains constant.
i.e., µ1x1 = µ2x2 = ... constant
∴
dx
dx
µ1 1 = µ2 2
dt
dt
∴
µ1c1 = µ2c2
∴
µ2
c
= 1
µ1
c2
i.e.,
µ∝
The apparent depth due to a number of media is Σ
Critical angle : When a ray passes from denser
medium (µ2) to rarer medium (µ1), then for 90º angle
of refraction, the corresponding angle of incidence is
critical angle.
1
c
Mathematically,
∴
c1 = fλ1,
∴
λ
µ1
c
= 2 = 2
µ2
c1
λ1
∴
1
µ∝
λ
c2 = fλ2
µ1
µ2
µ

δ = sin–1  2 sin i  – i
 µ1

for i < c
(ii) When angle of incidence is greater than critical
angle, total internal reflection takes place. The
corresponding deviation is
real depth
apparent depth
δ = π – 2i
when i < c
The δ – i graph is :
When object is in rarer and observer is in denser
medium :
µ=
sin c =
(i) When angle of incidence is lesser than critical
angle, refraction takes place. The corresponding
deviation is
When observer is rarer medium and object is in
denser medium :
µ=
ti
µi
The lateral shifting due to a slab is d = t sec r sin(i – r).
(where c1 and c2 are speed
of light in respective mediums)
(f) The frequency of light does not depend upon
medium.
Then
ti
t
t
= 1 + 2 + ...
µi
µ1
µ2
(i) Critical angle depends upon colour of light,
material of medium, and temperature of medium.
apparent position
real position
(ii) Critical angle does not depend upon angle of
incidence
 1
The shift of object due to slab is x = t 1 – 
 µ
(a) This formula is only applicable when observer is
in rarer medium.
δ
(b) The object shiftiness does not depend upon the
position of object.
i
µ2
µ
µ − µ1
– 1 = 2
v
u
r
The equivalent refractive index of a combination of a
Here,
π/2
Refractive surface formula,
(c) Object shiftiness takes place in the direction of
incidence ray.
number of slabs for normal incidence is µ =
c
Σt i
t
Σ i
µi
Here, v = image distance,
u = object distance,
r = radius of curvature of spherical surface.
Σti = t1 + t2 + ...
(a) For plane surface , r = ∞
(b) Transverse magnification,
XtraEdge for IIT-JEE
25
DECEMBER 2010
m=
(b) This formula is only applicable when medium on
both sides of lens are same.
µv
Im age size
= 1
object size
µ2u
(c) Intensity
aperture.
(c) Refractive surface formula is only applicable for
paraxial ray.
proportional
to
square
of
(d) When lens is placed in a medium whose refractive
index is greater than that of lens. i.e., µ1 > µ2.
Then converging lens behaves as diverging lens
and vice versa.
Lens :
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 =
is
v
image size
=
u
object size
(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.,
(h) If a lens is made of a number of layers of different
refractive index number of images of an object by
the lens is equal to number of different media.
(f)
f(+ve)
f(–ve)
(i)
(ii)
1
1
1
=
+
f
f1
f2
(i) The minimum distance between real object and
real image in is 4f.
(j) The equivalent focal length of
combination of two lenses is given by
f(+ve)
f(–ve)
1
1
1
d
=
+
–
F
f1 f 2
f1f 2
(iv)
(iii)
co-axial
(k) If a number of lenses are in contact, then
f(–ve)
f(+ve)
(v)
(vi)
1
1
1
=
+
+ ......
F
f1 f 2
(l) (i) Power of thin lens, P =
(g) Thin lens formula is applicable for converging as
well diverging lens. Thin lens maker's formula :
 µ −µ 
1
=  2 1 
f
 µ1 
(ii) Power of mirror is P = –
1 1
 − 
 r1 r2 
µ1
1
F
(m) If a lens silvered at one surface, then the system
behaves as an equivalent mirror, whose power
µ1
P = 2PL + Pm
µ2
Here, PL = Power of lens
 µ −µ 
=  2 1 
 µ1 
(a) Thin lens formula is only applicable for paraxial
ray.
XtraEdge for IIT-JEE
1
F
26
1 1
 − 
 r1 r2 
DECEMBER 2010
Pm = Power of silvered surface = –
1
Fm
Now
1
1
1
=
+
20
25 v1
v1 = 100 cm.
As v1 is positive, hence the image is real. In the
absence of convex mirror, the rays after reflection
from concave mirror would have formed a real image
I1 at distance 100 cm from the mirror. Due to the
presence of convex mirror, the rays are reflected and
appear to come from I2.
(ii) For convex mirror,
In this case, I1 acts as virtual object and I2 is the
virtual image.
The distance of the virtual object from the convex
mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm.
As focal length of convex mirror is negative and
hence f2 = –30/2 = –15 cm. Here we shall calculate
the value of v2. Using the mirror formula, we have
1
1
1
−
= − +
15
50 v 2
or
v2 = –21.42 cm
As v2 is negative, image is virtual. So image is
formed behind the convex mirror at a distance of
21.43 cm.
Here, Fm = r2/2, where r2 = radius of silvered surface.
or
P = – 1/F
Here, F = focal length of equivalent mirror.
Solved Examples
Rays of light strike a horizontal plane mirror at an
angle of 45º. At what angle should a second plane
mirror be placed in order that the reflected ray finally
be reflected horizontally from the second mirror.
Sol. The situation is shown in figure
G
C
D
A
S
θ θ
1.
45º
1
1
1
=
+
f1
u1 v1
N
45º
Q
B
The ray AB strikes the first plane mirror PQ at an
angle of 45º. Now, we suppose that the second
mirror SG is arranged such that the ray BC after
reflection from this mirror is horizontal.
From the figure we see that emergent ray CD is
parallel to PQ and BC is a line intersecting these
parallel lines.
So,
∠DCE = ∠CBQ = 180º
∠DCN + ∠NCB + ∠CBQ = 180º
θ + θ + 45º = 180º
∴ θ = 67.5º
As ∠NCS = 90º, therefore the second mirror should
be inclined to the horizontal at an angle 22.5º.
P
3.
There is a small air bubble in side a glass sphere (n =
1.5) of radius 10 cm. The bubble is 4 cm below the
surface and is viewed normally from the outside
(Fig.). Find the apparent depth of the air bubble.
A
n1 = 1.5
Sol. The observer sees the image formed due to refraction
at the spherical surface when the light from the
bubble goes from the glass to air.
Here u = – 4.0 m,
R = – 10 cm, n1 = 1.5 and n2 = 1
50cm
P2 C
F
We have
P1
25cm
I1
r = 30 cm
r = 40 cm
[(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
Thus, the bubble will appear 3.0 cm below the
surface.
(i) For concave mirror,
u1 = 25 cm, f1 = 20 cm and v1 = ?
XtraEdge for IIT-JEE
n2 = 1
O
C
An object is placed exactly midway between a
concave mirror of radius of curvature 40 cm and a
convex mirror of radius of curvature 30 cm. The
mirrors face each other and are 50 cm apart.
Determine the nature and position of the image
formed by the successive reflections, first at the
concave mirror and then at the convex mirror.
Sol. The image formation is shown in figure.
2.
I2
P
I
27
DECEMBER 2010
4.
A convex lens focuses a distance object on a screen
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
image is again focused on the screen ?
Sol. The situation when the glass plate is inserted between
the lens and the screen, is shown in fig. The lens
forms the image of object O at point I1 but the glass
plate intercepts the rays and forms the final image at I
on the screen. The shift in the position of image after
insertion of glass plate
IIT -JEE 2011
Screen
Few important Instructions to fill the IIT-JEE
2011 Form:
1.
Use Black ballpoint pen to fill the entire
form
2.
Use pencil to darken the bubbles in the OMR
sheet
3.
Use CAPIATL LETTERS to fill the form.
4.
Name of the candidate should be written in
Capital Letters as given in your 10th Class
Certificate. Leave only one blank box
between consecutive words of your name. If
by mistake you had made wrong entry of
name then use whitener, do not use blade or
rubber.
5.
Enter Date , Month & Year of your birth as
per 10th class marksheet. Use numerals 01 to
31 for DATE, 01 to 12 for MONTH and last
two digits of the YEAR of birth (the digit 19
is already there). For example: if you were
born on 6th march 1993 then DATE should
be entered as: “06 03 93”
6.
Only those candidate which are opting IIT
KHARAGPUR ZONE or IIT ROORKEE
ZONE and they come under the Category of
SC,ST,OBC,PD or DS may opt for either of
the option mention below from their
respective zones : IT-BHU VARANASI or
ISM DHANABAD.
7.
If you wish to obtain the question paper in
HINDI darken the bubble corresponding to
“HINDI” And If you wish to obtain the
question paper in ENGLISH darken the
bubble corresponding to “ENGLISH”. Note:
Please choose only one option from the
above choices
8.
Write Complete name & address to which
any communication is to be sent in capital
Letters Please do not give address of your
school/college, hostel or any other institute
here. Please write Only those phone / mobile
numbers which can be made available at the
time of necessary information & counselling.
9.
The Candidate must sign the declaration. The
place and date should also be mentioned. The
two signatures, the one below this
declaration and the other in the box below
your photograph (S.No 16) should be
identical.
10. Put your signature within the box provided.
Your signature must not overflow or touch
the border of the box provided. Please do not
sign in Capital letters.
O
I1
I
10 cm
1 
 1

I1I = t 1 −  = (1.5 cm) 1 −
 = 0.5 cm.
 n
 1.5 
Thus, the lens forms the image at a distance of 9.5 cm
from itself. Using
1 1 1
– = , we get
v u f
1 1 1
1
1
= – =
–
u v f 9.5 10
or
u = – 190 cm.
i.e. the object should be placed at a distance of
190 cm. from the lens.
5.
A candle is placed at a distance of 3 ft from the wall.
Where must a convex lens of focal length 8 inches be
placed so that a real image is formed on the wall ?
Sol. According to formula for refraction though a lens
36 – v
v
f = 8"
d = 3 ft = 36"
1 1 1
– =
v u f
or
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
or v = 12"
or
24" = 1 ft or 2 ft.
∴
u = 24"
or
12" = 2 ft or 1 ft
Hence, lens should be placed at either 1 ft or 2 ft
away from the wall.
XtraEdge for IIT-JEE
28
DECEMBER 2010
XtraEdge for IIT-JEE
29
DECEMBER 2010
P HYSICS F UNDAMENTAL F OR IIT-J EE
Fluid Mechanics & Properties of Matter
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Fluid dynamics :
Fluid Mechanics :
p
1 2
v + gh +
= a constant
ρ
2
for a streamline flow of a fluid (liquid or gas).
Here, v is the velocity of the fluid, h is its height
above some horizontal level, p is the pressure and ρ
is the density.
p1
Fluid statics :
Bernoulli's Theorem :
Pressure at a point inside a Liquid : p = p0 + ρgh
where p0 is the atmospheric pressure, ρ is the density
of the liquid and h is the depth of the point below the
free surface.
p0
h
v1
p
h1
ρ
v2 h2
v2 > v1
Pressure is a Scalar : The unit of pressure may be
atmosphere or cm of mercury. These are derived
units. The absolute unit of pressure is Nm–2. Normal
atmospheric pressure, i.e, 76 cm of mercury, is
approximately equal to 105 Nm–2.
Thrust : Thrust = pressure × area. Thrust has the unit
of force.
Laws of liquid pressure
(a) A liquid at rest exerts pressure equally in all
directions.
(b) Pressure at two points on the same horizontal line
in a liquid at rest is the same.
(c) Pressure exerted at a point in a confined liquid at
rest is transmitted equally in all directions and
acts normally on the wall of the containing vessel.
This is called Pascal's law. A hydraulic press
works on this principle of transmission of
pressure.
The principle of floating bodies (law of flotation) is
that W = W´, that is, weight of body = weight of
displaced liquid or buoyant force. The weight of the
displaced liquid is also called buoyancy or upthrust.
Hydrometers work on the principle of floating
bodies. This principle may also be applied to gases
(e.g., a balloon).
Liquids and gases are together called fluids. The
important difference between them is that liquids
cannot be compressed, while gases can be
compressed. Hence, the density of a liquid is the
same everywhere and does not depend on its
pressure. In the case of a gas, however, the density is
proportional to the pressure.
XtraEdge for IIT-JEE
p2
p2 < p1
According to this principle, the greater the velocity,
the lower is the pressure in a fluid flow.
It would be useful to remember that in liquid flow,
the volume of liquid flowing past any point per
second is the same for every point. Therefore, when
the cross-section of the tube decreases, the velocity
increases.
Note : Density = relative density
or
specific gravity × 1000 kg m–3.
Surface tension and surface energy :
Surface Tension : The property due to which a
liquid surface tends to contract and occupy the
minimum area is called the surface tension of the
liquid. It is caused by forces of attraction between the
molecules of the liquid. A molecule on the free
surface of a liquid experiences a net resultant force
which tends to draw it into the liquid. Surface tension
is actually a manifestation of the forces experienced
by the surface molecules.
If an imaginary line is drawn on a liquid surface then
the force acting per unit length of this line is defined
as the surface tension. Its unit is, therefore, newton /
metre. This force acts along the liquid surface. For
curved surfaces, the force is tangent to the liquid
surface at every point.
Surface Energy : A liquid surface possesses
potential energy due to surface tension. This energy
per unit area of the surface is called the surface
energy of the liquid. Its units is joule per square
metre. The surface energy of a liquid has the same
numerical values as the surface tension. The surface
30
DECEMBER 2010
tension of a liquid depends on temperature. It
decreases with rise in temperature.
Excess of Pressure : Inside a soap bubble or a gas
bubble inside a liquid, there must be pressure in
excess of the outside pressure to balance the tendency
of the liquid surface to contract due to surface
tension.
The upward force by which a liquid surface is pulled
up in a capillary tube is 2πrTcos θ, and the downward
force due to the gravitational pull on the mass of
liquid in the tube is (πr2h + v)ρg, where v is the
volume above the liquid meniscus. If θ = 0º, the
meniscus is hemispherical in shape. Then v =
difference between the volume of the cylinder of
radius r and height r and the volume of the
hemisphere of radius r
1 1
p(excess of pressure) = T  +  in general
 r1 r2 
= πr3 –
where T is surface tension of the liquid, and r1 and r2
are the principal radii of curvature of the bubble in
two mutually perpendicular directions.
For a spherical soap bubble, r1 = r2 = r and there are
two free surfaces of the liquid.
When θ ≠ 0, we cannot calculate v which is generally
very small and so it may be neglected. For
equilibrium
(πr2h + v) ρg = 2πrT cos θ
When a glass capillary tube is dipper in mercury, the
meniscus is convex, since the angle of contact is
obtuse. The surface tension forces now acquire a
downward component, and the level of mercury
inside the tube the falls below the level outside it. the
relation 2T cos θ = hρgr may be used to obtain the
fall in the mercury level.
Problem Solving Strategy
Bernoulli's Equations :
Bernoulli's equation is derived from the work-energy
theorem, so it is not surprising that much of the
problem-solving strategy suggested in W.E.P. also
applicable here.
Step 1: Identify the relevant concepts : First ensure
that the fluid flow is steady and that fluid is
incompressible and has no internal friction. This case
is an idealization, but it hold up surprisingly well for
fluids flowing through sufficiently large pipes and for
flows within bulk fluids (e.g., air flowing around an
airplane or water flowing around a fish).
Step 2: Set up the problem using the following steps
Always begin by identifying clearly the points 1
and 2 referred to in Bernoulli's equation.
Define your coordinate system, particular the
level at which y = 0.
Make lists of the unknown and known quantities
1
1
in Eq. p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22
2
2
(Bernoulli's equation)
The variables are p1, p2, v1, v2, y1 and y2, and the
constants are ρ and g. Decide which unknowns
are your target variables.
Step 3: Execute the solutions as follows : Write
Bernoulli's equation and solve for the unknowns. In
some problems you will need to use the continuity
equation, Eq. A1v1 = A2v2 (continuity equation,
incompressible fluid), to get a relation between the
two speeds in terms of cross-sectional areas of pipes
4T
∴
p=
r
For a gas bubble inside a liquid, r1 = r2 = r and there
is only one surface.
∴
p=
2T
r
For a cylindrical surface r1 = r and r2 = ∞ and there
are two surfaces.
2T
r
Angle of Contact : The angle made by the surface of
a liquid with the solid surface inside of a liquid at the
point of contact is called the angle of contact. It is at
this angle that the surface tension acts on the wall of
the container.
∴
p=
The angle of contact θ depends on the natures of the
liquid and solid in contact. If the liquid wets the solid
(e.g., water and glass), the angle of contact is zero. In
most cases, θ is acute (figure i). In the special case of
mercury on glass, θ is obtuse (figure ii).
θ
θ
fig. (i)
fig. (ii)
Rise of Liquid in a Capillary Tube : In a thin
(capacity) tube, the free surface of the liquid becomes
curved. The forces of surface tension at the edges of
the liquid surface then acquire a vertical component.
T θ meniscus θ T
θ
θ
h
r
XtraEdge for IIT-JEE
2 3 1 3
πr = πr
3
3
31
DECEMBER 2010
or containers. Or perhaps you will know both speeds
and need to determine one of the areas. You may also
dV
= Av (volume flow rate) to find
need to use Eq.
dt
the volume flow rate.
Step 4: Evaluate your answer : As always, verify that
the results make physical sense. Double-check that
you have used consistent units. In SI units, pressure is
in pascals, density in kilograms per cubic meter, and
speed in meters per second. Also note that the
pressures must be either all absolute pressure or all
gauge pressures.
F
F
F
V
F
(c) Shear strain = φ
φ
Properties of matter :
Shear strain
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
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
∴ Stress = E .strain
F
Then,
Stress =
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 :
Strain =
∴ E=
Stress
where E is proportionality dimensional constant
known as coefficient of elasticity.
Plastic
region
Breaking
B C
strength
change in dimension
original dimension
∆L
(a) Longitudinal strain =
L
L
F
Elastic
limit
A
Strain
O
Types of coefficient of elasticity :
(a) Young's modulus = Y =
F
∴
Longitudinal strain is in the direction of
deforming force but lateral strain is in
perpendicular direction of deforming force.
Poisson ratio :
σ=
stress
strain
Y=
logitudinal stress
longitudinal strain
F
FL
=
∆L
A∆L
A
L
∆d/D
lateral strain
=
∆L/L
longitudinal strain
L
Here ∆d = change in diameter.
(b) Volumetric strain =
∆L
∆V
V
F
(b) Bulk modulus = B =
volumetric stress
volumetric strain
Compressibility = 1/B
XtraEdge for IIT-JEE
32
DECEMBER 2010
(c) Modulus of rigidity = η =
Surface tension :
shear stress
F
=
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 > Eliquid > Egas
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
9
3
+
=
η
B
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,
1
–1 < σ <
2
Practically,
1
0<σ<
2
Here
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
× torque × angular twist
2
1
1
τ × θ = cθ2
2
2
Elastic energy density,
(a) For drop, P =
=
u=
1
1
× stress × strain J/m3 = Y × strain2J/m3
2
2
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
2T
R
(b) For soap bubble, P =
Thermal stress = Yα∆θ and Thermal strain = α∆θ
Work done in stretching a wire :
(a) W =
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
(n) Thermal stress = Yα∆θ
(o) Elastic energy stored,
U=
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.
33
DECEMBER 2010
(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=
Sol. Draw a horizontal line through the mercury-glycerine
surface. This is a horizontal line in the same liquid at
rest namely, mercury. Therefore, pressure at the
points A and B must be the same.
(1 – h)
A
Pπr 4
8ηL
B
Pressure at B
= p0 + 0.1 × (1.3 × 1000) × g
Pressure at A
= p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g
∴ p0 + 0.1 × 1300 × g
= p0 + 800gh + 1360g – 13600 × g × h
⇒ 130 = 800h + 1360 – 13600h
1230
⇒ h=
= 0.096 m = 9.6 cm
12800
where V = the volume of liquid flowing per second
through a capillary tube of length L and radius r
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
A liquid flows out of a broad vessel through a narrow
vertical pipe. How are the pressure and the velocity
of the liquid in the pipe distributed when the height
of the liquid level in the vessel is H from the lower
end of the length of the pipe is h ?
Sol. Let us consider three points 1, 2, 3 in the flow of
water. The positions of the points are as shown in the
figure.
Applying Bernoulli's theorem to points 1, 2 and 3
2.
F = 6πηrv
where r = radius of spherical body
Determination of η :
η=
10 cm
h
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
•1
and
σ = density of spherical body
Critical velocity (v0) :
v0 =
kη
ρr
H
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.
h
•3
p0 1 2
1
p
+ v 1 + gH =
+ v 22 + g (h – x)
ρ 2
ρ 2
p
1
= 0 + v 32 + 0
ρ 2
By continuity equation
v 1A1 = A2v 2 = A2v 3
Since A1 >> A2,v1 is negligible and v2 = v3 = n (say).
p0
p
1
∴
+ gH = 2 + v2 + g (h – x)
ρ
ρ 2
p0 1 2
+ v
=
ρ 2
A vertical U-tube of uniform cross-section contains
mercury in both arms. A glycerine (relative density
1.3) column of length 10 cm is introduced into one of
the arms. Oil of density 800 kg m–3 is poured into the
other arm until the upper surface of the oil and
glycerine are at the same horizontal level. Find the
length of the oil column. Density of mercury is 13.6
× 103 kg m–3.
XtraEdge for IIT-JEE
•2
2
Solved Examples
1.
x
∴
34
v = 2gH
(i)
DECEMBER 2010
= 4.99 × 10–3 cm
Also, elastic limit for copper = 1.5 × 109 dynes/cm2
If d' is the minimum diameter, then maximum stress
F
4F
=
on the wire =
πd '2 / 4 πd' 2
p0
p
+ gH = 2 + gH + g (h – x)
ρ
ρ
⇒
p0 + p2 + ρg (h – x)
(ii)
⇒
p2 = p0 – ρg (h – x)
Thus pressure varies with distance from the upper
end of the pipe according to equation (ii) and velocity
is a constant and is given by (i).
and
Calculate the difference in water levels in two
communicating tubes of diameter d = 1 mm and
d = 1.5 mm. Surface tension of water = 0.07 Nm–1
and angle of contact between glass and water = 0º.
2T cos θ
Sol. Pressure at A = p0 –
r2
3.
πd' 2
or d'2 =
= 1.5 × 109
4F
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, Ys = 2 × 1012 dynes/cm2. Calculate the
tension in each wire and extension.
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.
A
As Y =
Let this pressure difference correspond to h units of
the liquid.
1 1
Then 2T cos θ  −  = ρgh
 r1 r2 
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 × 109
dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2.
Sol. Given that Y = 1.1 × 1012 dynes/cm2,
L = 2m = 200 cm, d = 5 mm = 0.5 cm
or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.
or l =
πr 2 Y
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)
FCu = (3/11) × 100g
= (3/11) × 100 Kgwt = 27.28 Kgwt
and
FS = (4/3) FCu = (4/3) × (3/11) × 100g
= (400/11)g = 36.36 Kgwt
Extension in each wire,
5.0 × 1000 × 980 × 200
l=
3.142 × (0.25) 2 × 1.1× 10 2
XtraEdge for IIT-JEE
S
∴
πr 2l
=
C
According to figure, we can write
2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g
or
[(8/3) + 1] FCu = 100 g
FL
FL
B
Cu
100 Kg



4.
Y=
Stress
Strain
A
S
2 × 0.07  1
1

−

 = 4.76 mm
−3
−3
1000 × 9.8  1× 10
1.5 × 10 
∴ h=
4 × 5.0 × 1000 × 980
5.
1 1
∴ pressure difference = 2T cos θ  − 
 r1 r2 
2T cos θ  1 1
 −
⇒ h=
ρg  r1 r2
=
π × 1.5 × 10
3.142 × 1.5 × 109
–4
= 41.58 × 10
d' = 0.0645 cm.
(Q pressure inside a curved surface is greater than
that outside)
2T cos θ
Pressure at B = p0 –
r1
B
4F
Hence,
35
FCu L 27280 × 980 × 100
=
= 0.178 cm
AYCu
10 − 2 × 1.5 × 1012
DECEMBER 2010
KEY CONCEPT
Organic
Chemistry
Fundamentals
CARBOXYLIC ACID
(1.27Å) which is nearly intermediate between C O
and C—O bond length values. This proves resonance
in carboxylate anion.
Acidity of carboxylic acids.
Fatty acids are weak acids as compared to inorganic
acids. The acidic character of fatty acids decreases
with increase in molecular weight. Formic acid is the
strongest of all fatty acids.
The acidic character of carboxylic acids is due to
resonance in the acidic group which imparts electron
deficiency (positive charge) on the oxygen atom of
the hydroxyl group (structure II).
O–
O
R
C
O
R
H
I
Non-equivalent structures
C
+
O
O
H
C
R
O–
C
O+H
R
O
–
C
O
Resonance hybrid of carboxylate ion
Due to equivalent resonating structures, resonance in
carboxylate anion is more important than in the
parent carboxylic acid. Hence carboxylate anion is
more stabilised than the acid itself and hence the
equilibrium of the ionisation of carboxylic acids to
the right hand side.
RCOO– + H+
RCOOH
The existence of resonance in carboxylate ion is
supported by bond lengths. For example, in formic
acid, there is one C=O double bond (1.23 Å) and one
C—O single bond (1.36Å), while in sodium formate
both of the carbon-oxygen bond lengths are identical
XtraEdge for IIT-JEE
O
Alkoxide ion
(No resonance)
Relative acidic character of carboxylic acids with
common species not having —COOH group.
RCOOH > Ar—OH > HOH > ROH >
HC CH > NH3 > RH
Effect of Substituents on acidity.
The carboxylic acids are acidic in nature because of
stabilisation (i.e., dispersal of negative charge) of
carboxylate ion. So any factor which can enhance the
dispersal of negative charge of the carboxylate ion
will increase the acidity of the carboxylic acid and
vice versa. Thus electron-withdrawing substitutents
(like halogens, —NO2, —C6H5, etc.) would disperse
the negative charge and hence stabilise the
carboxylate ion and thus increase acidity of the
parent acid. On the other hand, electron-releasing
substituents would increase the negative charge,
destabilise the carboxylate ion and thus decrease
acidity of the parent acid.
Resonating forms of carboxylate ion (Equivalent structures)
(Resonance more important)
O
Na+
Sodium formate
Alcohol
(No resonance)
+
The positive charge (electron deficiency) on oxygen
atom causes a displacement of electron pair of the
O—H bond towards the oxygen atom with the result
the hydrogen atom of the O—H group is eliminated
as proton and a carboxylate ion is formed.
Once the carboxylate ion is formed, it is stabilised by
means of resonance.
O
O–
R
OH
C
–
It is important to note that although carboxylic acids
and alcohols both contain –OH group, the latter are
not acidic in nature. It is due to the absence of
resonance (factor responsible for acidic character of
–COOH) in both the alcohols as well as in their
corresponding ions (alkoxide ions).
R—O– + H+
R—O—H
H
C
H
Formic acid
II
(Resonance less important)
O–
R
C
O
O
X
C
–
O
The substituent X withdraws electrons, disperses negative
charge, stabilises the ion and hence increases acidity
O
Y
C
–
O
The substituent Y releases electrons, intensifies negative
charge, destabilises the ion and hence decreases acidity
36
DECEMBER 2010
Now, since alkyl groups are electron-releasing, their
presence in the molecule will decrease the acidity. In
general, greater the length of the alkyl chain, lower
shall be the acidity of the acid. Thus, formic acid
(HCOOH), having no alkyl group, is about 10 times
stronger than acetic acid (CH3COOH) which in turn
is stronger than propanoic acid (CH3CH2COOH) and
so on. Similarly, following order is observed in
chloro acids.
Cl
Cl
Cl
pKa
CO2H > Cl
C
C
Comparison of nucleophilic substitution (e.g.,
hydrolysis) in acid derivatives. Let us first study the
mechanism of such reaction.
O
R
R C Nu + Z
(where Z= —Cl, —OCOR, —OR, —NH2 and Nu=
A nucleophile)
Nucleophilic substitution in acid derivatives
O
O
OH
C
δ–
Cl
δ+
Acid chloride
XtraEdge for IIT-JEE
δ+
R
Nu
R
C
Nu
H+
R
C
Nu
R'
Nucleophilic addition on aldehydes and ketones
The (i) step is similar to that of nucleophilic addition
in aldehydes and ketones and favoured by the
presence of electron withdrawing group (which
would stabilise the intermediate by developing
negative charge) and hindered by electron-releasing
group. The (ii) step (elimination of the leaving group
Z) depends upon the ability of Z to accommodate
electron pair, i.e., on the basicity of the leaving
group. Weaker bases are good leaving groups,
hence weaker a base, the more easily it is removed.
Among the four leaving groups (Cl–, –OCOR, –OR,
and –NH2) of the four acid derivatives, Cl– being the
weakest base is eliminated most readily. The relative
order of the basic nature of the four groups is
–
NH2 > –OR > –O.COR > Cl–
Hence acid chlorides are most reactive and acid
amides are the least reactive towards nucleophilic
acyl substitution. Thus, the relative reactivity of acid
derivatives (acyl compounds) towards nucleophilic
substitution reactions is
ROCl > RCO.O.COR > RCOOR > RCONH2
Acid
Acid
Esters
Acid
chlorides
anhydrides
amides
OH– being stronger base than Cl–, carboxylic acids
(RCOOH)
undergo
nucleophilic
substitution
(esterfication) less readily than acid chlorides.
δ–
R C
R'
(where R' = H or alkyl group)
Decreasing order of aliphatic acids
(i) O2NCH2COOH > FCH2COOH > ClCH2COOH
> BrCH2COOH
(ii) HCOOH > CH3COOH > (CH3)2 CHCOOH
> (CH3)3CCOOH
(iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH
> ClCH2CHClCH2COOH
(iv) F3CCOOH > Cl3CCOOH > Br3CCOOH
Benzoic acid is somewhat stronger than simple
aliphatic acids. Here the carboxylate group is
attached to a more electronegative carbon (sp2
hybridised) than in aliphatic acids (sp3 hybridised).
HCOOH > C6H5COOH > CH3COOH.
Nucleophilic substitution at acyl carbon :
It is important to note that nucleophilic substitution
(e.g., hydrolysis, reaction with NH3, C2H5OH, etc.) in
acid derivatives (acid chlorides, anhydrides, esters
and amides) takes place at acyl carbon atom
(difference from nucleophilic substitution in alkyl
halides where substitution takes place at alkyl carbon
atom). Nucleophilic substitution in acyl halides is
faster than in alkyl halides. This is due to the
presence of > CO group in acid chlorides which
facilitate the release of halogen as halide ion.
O
C
R'
CO2H
H
4.76
H
2.86
pKa
R
H
CO2H > H
C
Nu
C
O
1.48
> Cl
R
(ii) Elimination step
CO2H
H
Z + Nu
Z
H
Cl
0.70
C
O
(i) Addition step
δ–
Cl
Alkyl chloride
37
DECEMBER 2010
KEY CONCEPT
Physical
Chemistry
Fundamentals
CHEMICAL KINETICS
behaviour is a signal that the reaction has a complex
mechanism.
The temperature dependence of some reactions is
non-Arrhenius, in the sense that a straight line is not
obtained when ln k is plotted against 1/T. However,
it is still possible to define an activation energy at any
temperature as
 d ln k 
.......(ii)
Ea = RT2 

 dT 
The temperature dependence of reaction rates :
The rate constants of most reactions increase as the
temperature is raised. Many reactions in solution fall
somewhere in the range spanned by the hydrolysis of
methyl ethanoate (where the rate constant at 35ºC is
1.82 times that at 25ºC) and the hydrolysis of sucrose
(where the factor is 4.13).
(a) The Arrhenius parameters :
It is found experimentally for many reactions that a
plot of ln k against 1/T gives a straight line. This
behaviour is normally expressed mathematically by
introducing two parameters, one representing the
intercept and the other the slope of the straight line,
and writing the Arrhenius equaion.
This definition reduces to the earlier one (as the slope
of a straight line) for a temperature-independent
activation energy. However, the definition in eqn.(ii)
is more general than eqn.(i), because it allows Ea to
be obtained from the slope (at the temperature of
interest) of a plot of ln k against 1/T even if the
Arrhenius plot is not a straight line. Non-Arrhenius
behaviour is sometimes a sign that quantum
mechanical tunnelling is playing a significant role in
the reaction.
(b) The interpretation of the parameters :
We shall regard the Arrhenius parameters as purely
empirical quantities that enable us to discuss the
variation of rate constants with temperature;
however, it is useful to have an interpretation in mind
and write eqn.(i) as
Ea
......(i)
RT
The parameter A, which corresponds to the intercept
of the line at 1/T = 0(at infinite temperature, shown in
figure), is called the pre-exponential factor or the
'frequency factor'. The parameter Ea, which is
obtained from the slope of the line (–Ea/R), is called
the activation energy. Collectively the two quantities
are called the Arrhenius parameters.
ln k = ln A –
ln A
.......(iii)
k = Ae − E a / RT
To interpret Ea we consider how the molecular
potential energy changes in the course of a chemical
reaction that begins with a collision between
molecules of A and molecules of B(shown in figure).
ln k
Slope = –Ea/R
Potential energy
1/T
A plot of ln k against 1/T is a straight line when
the reaction follows the behaviour described by
the Arrhenius equation. The slope gives –Ea/R
and the intercept at 1/T = 0 gives ln A.
The fact that Ea is given by the slope of the plot of
ln k against 1/T means that, the higher the activation
energy, the stronger the temperature dependence of
the rate constant (that is, the steeper the slope). A
high activation energy signifies that the rate constant
depends strongly on temperature. If a reaction has
zero activation energy, its rate is independent of
temperature. In some cases the activation energy is
negative, which indicates that the rate decreases as
the temperature is raised. We shall see that such
XtraEdge for IIT-JEE
Ea
Reactants
Products
Progress of reaction
A potential energy profile for an exothermic
reaction. The height of the barrier between
the reactants and products is the activation
energy of the reaction
38
DECEMBER 2010
As the reaction event proceeds, A and B come into
contact, distort, and begin to exchange or discard
atoms. The reaction coordinate is the collection of
motions, such as changes in interatomic distances and
bond angles, that are directly involved in the
formation of products from reactants. (The reaction
coordinate is essentially a geometrical concept and
quite distinct from the extent of reaction.) The
potential energy rises to a maximum and the cluster
of atoms that corresponds to the region close to the
maximum is called the activated complex. After the
maximum, the potential energy falls as the atoms
rearrange in the cluster and reaches a value
characteristic of the products. The climax of the
reaction is at the peak of the potential energy, which
corresponds to the activation energy Ea. Here two
reactant molecules have come to such a degree of
closeness and distortion that a small further
distortion will send them in the direction of products.
This crucial configuration is called the transition
state of the reaction. Although some molecules
entering the transition state might revert to reactants,
if they pass through this configuration then it is
inevitable that products will emerge from the
encounter.
We also conclude from the preceding discussion that,
for a reaction involving the collision of two
molecules, the activation energy is the minimum
kinetic energy that reactants must have in order
to form products. For example, in a gas-phase
reaction there are numerous collisions each second,
but only a tiny proportion are sufficiently energetic to
lead to reaction. The fraction of collisions with a
kinetic energy in excess of an energy Ea is given by
ratio of the two rates, and therefore of the two rate
constants :
[P2 ]
k
= 2
[P1 ]
k1
This ratio represents the kinetic control over the
proportions of products, and is a common feature of
the reactions encountered in organic chemistry where
reactants are chosen that facilitate pathways
favouring the formation of a desired product. If a
reaction is allowed to reach equilibrium, then the
proportion of products is determined by
thermodynamic rather than kinetic considerations,
and the ratio of concentration is controlled by
considerations of the standard Gibbs energies of all
the reactants and products.
The kinetic isotope effect
The postulation of a plausible mechanism requires
careful analysis of many experiments designed to
determine the fate of atoms during the formation of
products. Observation of the kinetic isotope effect, a
decrease in the rate of a chemical reaction upon
replacement of one atom in a reactant by a heavier
isotope, facilitates the identification of bond-breaking
events in the rate-determining step. A primary
kinetic isotope effect is observed when the ratedetermining step requires the scission of a bond
involving the isotope. A secondary isotope effect is
the reduction in reaction rate even though the bond
involving the isotope is not broken to form product.
In both cases, the effect arises from the change in
activation energy that accompanies the replacement
of an atom by a heavier isotope on account of
changes in the zero-point vibrational energies.
First, we consider the origin of the primary kinetic
isotope effect in a reaction in which the ratedetermining step is the scission of a C–H bond. The
reaction coordinate corresponds to the stretching of
the C–H bond and the potential energy profile is
shown in figure. On deuteration, the dominant
change is the reduction of the zero-point energy of
the bond (because the deuterium atom is heavier).
The whole reaction profile is not lowered, however,
because the relevant vibration in the activated
complex has a very low force constant, so there is
little zero-point energy associated with the reaction
coordinate in either isotopomeric form of the
activated complex.
the Boltzmann distribution as e − E a / RT . Hence, we
can interpret the exponential factor in eqn(iii) as the
fraction of collision that have enough kinetic energy
to lead to reaction.
The pre-exponential factor is a measure of the rate at
which collisions occur irrespective of their energy.
Hence, the product of A and the exponential factor,
e − E a / RT , gives the rate of successful collisions.
Potential energy
Kinetic and thermodynamic control of reactions :
In some cases reactants can give rise to a variety of
products, as in nitrations of mono-substituted
benzene, when various proportions of the ortho-,
meta-, and para- substituted products are obtained,
depending on the directing power of the original
substituent. Suppose two products, P1 and P2, are
produced by the following competing reactions :
A + B → P1 Rate of formation of P1 = k1[A][B]
A + B → P2 Rate of formation of P2 = k2[A][B]
The relative proportion in which the two products
have been produced at a given state of the reaction
(before it has reached equilibrium) is given by the
XtraEdge for IIT-JEE
C–H
C–D
Ea(C–H)
Ea(C–D)
Reaction coordinate
39
DECEMBER 2010
UNDERSTANDING
Physical Chemistry
Arrange the following in :
(i) Decreasing ionic size : Mg2+, O2–, Na+, F–
[IIT-1985]
(ii) Increasing acidic property :
[IIT-1985]
ZnO, Na2O2, P2O5, MgO
(iii) Increasing first ionization potential :
[IIT-1985]
Mg, Al, Si, Na
(iv) Increasing size : Cl–, S2–, Ca2+, Ar [IIT-1986]
(v) Increasing order of ionic size :
[IIT-1991]
N3–, Na+, F–, O2–, Mg2+
(vi) Increasing order of basic character :
[IIT-1991]
MgO, SrO, K2O, NiO, Cs2O
(vii) Arrange the following ions in order of their
increasing radii : Li+, Mg2+, K+, Al3+.
Sol. (i) O2– > F– > Na+ > Mg2+
Note that all the above ions are isoelectronic
having 10 electron each.
In such a case the greater the nuclear charge, the
greater is the attraction for electrons and smaller
is the ionic radius. Hence O2– has the highest
and Mg2+ has the least ionic size.
(vi) Increasing order of basic character :
NiO < MgO < SrO < K2O < Cs2O
The basic character of oxides increases when
we move down the group. So, K2O < Cr2O and
MgO < SrO.
Further higher the group number lesser is the
basic character. Hence NiO is the least basic.
(vii) Al3+ < Mg2+ < Li+ < K+
In these Al3+ & Mg2+ are isoelectronic species,
so in these size decreases with increase in
atomic number because increase in atomic
number decreases Zeff.
1
Size ∝
Z eff .
In Li+ & K+, K+ is in size than Li+ because on
moving from top to bottom in a group, the
group size increases.
1.
+1
+2
+2
2.
+5
Na 2 O 2 < Mg O < Zn O < P2 O 5
Among oxides the acidic strength increases with
oxidation state. So Na2O2 is least acidic and
P2O5 is most acidic. Further Na2O2 and MgO
are basic, ZnO is amphoteric and P2O5 is acidic.
(iii) The first ionization potential of the 3rd period
elemens follows the order
Na < Al < Mg < Si ; Ionisation energy increases
across a period but not regularly. Mg (1s2, 2s2p6,
3s2) is more stable because the electron is to be
removed from 3s which is difficult as compared
to Al (1s2, 2s2p6, 3s2p1) where electron is to be
removed from 3p.
(iv) Ca2+ < Ar > Cl– < S2– ; All of these are
isoelectronic. In such cases the greater the
nuclear charge, the greater is the attraction for
electrons and smaller is ionic size.
1
ionic radius ∝
effective nuclear ch arg e
(v) Increasing order of ionic size
Mg2+ < Na+ < F– < O2– < N3–
Note that all the above ions are isoelectronic
having 10 electron each.
In such a case the greater the nuclear charge, the
greater is the attraction for electrons and smaller
is the ionic radius. Hence N3– has the highest
and Mg2+ has least ionic size.
(ii)
XtraEdge for IIT-JEE
A white amorphous powder (A) on heating yields a
colourless, non-combustible gas (B) and solid (C).
The latter compound assumes a yellow colour on
heating and changes to white on cooling. 'C' dissolves
in dilute acid and the resulting solution gives a white
precipitate on adding K4Fe(CN)6 solution.
'A' dissolves in dilute HCl with the evolution of gas,
which is identical in all respects with 'B'. The gas 'B'
turns lime water milky, but the milkiness disappears
with the continuous passage of gas. The solution of
'A', as obtained above, gives a white precipitate (D)
on the addition of excess of NH4OH and and passing
H2S. Another portion of the solution gives initially a
white precipitate (E) on the addition of sodium
hydroxide solution, which dissolves on futher
addition of the base. Identify the compounds A, B, D
[IIT-1979]
and E.
∆
ZnCO3 →
ZnO + CO2
(A)
(B)
(ii) ZnO + 2HCl → H2O + ZnCl2
(C)
(soluble)
(iii) 2ZnCl2 + K4[Fe(CN)6]
→ 4KCl + Zn2[Fe(CN)6]↓
(white ppt)
(iv) ZnCO3 + HCl → CO2 + ZnCl2
(A)
(soluble)
(v) CO2 + Ca(OH)2 → CaCO3 + H2O
(B)
(Milky)
(vi) CaCO3 + CO2 + H2O → Ca(HCO3)2
(soluble)
Sol. (i)
40
DECEMBER 2010
XtraEdge for IIT-JEE
41
DECEMBER 2010
XtraEdge for IIT-JEE
42
DECEMBER 2010
XtraEdge for IIT-JEE
43
DECEMBER 2010
XtraEdge for IIT-JEE
44
DECEMBER 2010
Structure of B2H6 is as follows :
Hb
••
Ht
Ht
B
B
Ht
Ht
••
Hb
Hb
Ht
Ht
4 OH
(vii) ZnCl2 + H2S NH

→ 2HCl + ZnS↓
(white)
(viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓
(white)
(ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2
sod. ziniate
(soluble)
3.
Write the lewis dot structural formula for each of the
following. Give, also, the formula of a neutral
molecule, which has the same geometry and the same
arrangement of the bonding electrons as in each of
the following. An example is given below in the case
of H3O+ :
+
H
•• •
•
H •O •H
Ht
× N×
CN– ; •• C •• ×
•× ×
×× –
••
•× S
NCS– ; •• N ••
C
•
× ××
••
N2 ;
CO2 ;
Ht
Hb
Draw the structure of XeF4 and OSF4 according to
VSEPR theory, clearly indicating the state of
hybridisation of the central atom and lone pair of
electrons (if any) on the central atom.
[IIT-2004]
Sol. First determine the total number of electron pairs
around the central atom.
N
8+ 4
XeF4 =
=
=6
2
2
Thus in XeF4, Xe is sp3d2 hybridised. The structure of
the molecule is octahedral and shape is square planer
with two lone pair of electrons.
F
××
ו S• O×
O×
• •• • ×××
×O×
× ×
××
• N • × N×
× ×
• •• ×
F
••
••
•O
O •• ×
C×
•
×
×
••
••
••
F
F
Xe
F
Compound (X) on reduction with LiAlH4 gives a
hydride (Y) containing 21.72% hydrogen along with
other products. The compound (Y) reacts with air
explosively resulting in boron trioxide. Identify (X)
and (Y). Give balanced reactions involved in the
formation of (Y) and its reaction with air. Draw the
structure of (Y).
Sol. Since B2O3 is formed by reaction of (Y) with air, (Y)
therefore should be B2H6 in which % of hydrogen is
21.72. The compound (X) on reduction with LiAlH4
gives B2H6. Thus it is boron trihalide. The reactions
are shown as :
4BX3 + 3LiAlH4 → 2B2H6 + 3LiX + 3AlX3
(X)
(Y)
(X = Cl or Br)
B2H6 + 3O2 → B2O3 + 3H2O + heat
(Y)
4.
XtraEdge for IIT-JEE
121.5º
5.
[IIT-1983]
Neutral molecule
•• ••
F2 ; •• F •• F ••
•• ••
–
B
Thus the diborane molecule has four two-centre-twoelectron bonds (2c – 2e bonds) also called usual
bonds and two three-centre-two-electron bonds (3c –
2e) also called banana bonds. Hydrogen attached to
usual and banana bonds are called Ht (terminal H)
and Hb (bridged H) respectively.
Neutral
Lewis dot
molecule
structure
(i) O22– ; (ii) CO32– ; (iii) CN– ; (iv) NCS–
2–
•×
×O
×××
SO3 ;
•
Co32– ; × ××
O×
• O × • C •• ×
×
×
××
××
1.19Å
97º
1.33Å
1.77Å
H
••
H •• N •• H
Sol. Lewis dot structure
×× •• 2–
O ו O ••
O22– ; ×
× ××
••
B
or
••
S
F
F
O
F
N
6+4
=
=5
For OSF4 :
2
2
Thus the central atom (S) is sp3d hybridised
leading to trigonal bipyramidal structure with no
lone pair of electrons.
45
DECEMBER 2010
Set
8
`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 t ion s wi l l b e p ub lished in nex t issue
Joint Director Academics, Career Point, Kota
1.
2.
Show that the six planes through the middle point of
each edge of a tetrahedron perpendicular to the
opposite edge meet in a point.
Prove that if the graph of the function y = f (x),
defined throughout the number scale, is symmetrical
about two lines x = a and x = b, (a < b), then this
function is a periodic one.
3.
Show that an equilateral triangle is a triangle of
maximum area for a given perimeter and a triangle of
minimum perimeter for a given area.
4.
Let az2 + bz + c be a polynomial with complex
coefficients such that a and b are non zero. Prove
that the zeros of this polynomial lie in the region
b
c
+
|z|≤
a
b
5.
6.
7.
9.
10. ABC is a triangle inscribed in a circle. Two of its
sides are parallel to two given straight lines. Show
that the locus of foot of the perpendicular from the
centre of the circle on to the third side is also a circle,
concentric to the given circle.
MEMORABLE POINTS
• The vector relation between linear velocity and
→
An isosceles triangle with its base parallel to the
y2
x2
+
= 1 is
major axis of the ellipse
a2
b2
circumscribed with all the three sides touching the
ellipse. Find the least possible area of the triangle.
→
→
® v =ω× r
angular velocity is
• In the case of uniform circular motion the angle between
→
→
ω and r is always
→
® 90º(hence | v | = ωr
• The relation between Faraday constant F, Avogadro
number N and the electronic charge e is ® F = Ne
If one of the straight lines given by the equation
ax2 + 2hxy + by2 = 0 coincides with one of those
given by a′x2 + 2h′xy + b′y2 = 0 and the other lines
represented by them be perpendicular, show that
ha´b´
h´ab´
=
b´−a´
b−a
• Depolariser used in Lechlanche cell is
® Manganese dioxide
• The absorption or evolution of heat at a junction of
two dissimilar metals when a current is passed is
® Peltier effect
known as
Prove that
n m
 n   m + 1
 n   m + 2
 +   
 + .........
    +   
0  n 
1   n 
 2  n 
.... to (n + 1) terms
n  m
n m
n  m
=     +     2 +     22 + ........
0  0 
1  1 
 2  2 
..... to (n + 1) terms
• The part of the human ear where sound is transduced
® Cochlea
is the
• Similar trait resulting from similar selection pressure
acting on similar gene pool is termed
® Parallel evolution
• Group of related species with the potential, directly
or indirectly, of forming fertile hybrids with one
® Coenospecies
another is called
1
8.
Find the sum to infinite terms of the series
5
9
11
3
7
+
+
+ ........ ∞
+
+
36
400
900
4
144
∫
If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that
−1
m2In = 2n(2n – 1) In–1 – 4n(n – 1) In–2.
XtraEdge for IIT-JEE
46
DECEMBER 2010
MATHEMATICAL CHALLENGES
SOLUTION FOR NOVEMBER ISSUE (SET # 7)
1.
Let the line be y = 2x + c
 9 − c 9 + 2c 
Point A 
,

3 
 6
=
AD DC + BD
.
BC BD.CD
 2c − 3 + c − 6 
,
Point B 

−3 
 −3
=
AD
AD
AD 1
=
=
.
2
BD.CD
AD AD
AD
so it is vector along AB with magnitude
 c + 6 5c + 12 
Point C 
,

3 
 3
mid point of B & C is
1
2
|a+b | =
 2c − 3 c + 6 
.
,
+
3 
 −3
3.
1
AD
The line PQ always passes through (α, β) so it is
y –β = m(x – α)
1  − c + 6 5c + 12   9 − c 2c + 9 
,
=
+
2  + 3
3   6
3 
Let the circle be x2 + y2 – 2hx – 2ky = 0
which is point A, so AB and AC are equal.
Joint equation of OP and OQ.
x2 + y2 – 2 (hx + ky)
2.
A
( y − mx )
=0
β − mα
P
b
a
O
B
D
=
1
AB
2
1
AB
2
Q
AB +
1
AC 2

 h − mk 
2k  2
 y – 2 
 xy +
1 −
 β − mα 
 β − mα 
AC
(AD + DB) +
1
AC

2hn  2
 x = 0
1 +
 β − mα 
It must represent y2 – x2 = 0
2
h − mk
= 0 ⇒ m = h/k
β − mα
so
(AD + DC)
DB
DC
1 
 1
+
= 
+
 AD +
2
2
BD.DC
CD.CB
AC 
 AB
and
1–
...(1)
2k
2hm
= –1 –
β − mα
β − mα
⇒ β – mα – 2k = –β + mα – 2hm
 DB DC  1
1 
 1


+
= 
AD
+
+

 BD CD  BC
 AB2 AC 2 


⇒ –β + mα + k – hm = 0
⇒ –β + k + h/k(α – h) = 0
1 
 1
= AD . 
+

 BD.DC CD.CB 
=
(h,k)
C
1 AC
1 AB
a+b =
.
+
.
AC AC
AB AB
=
1
.
AD
(using (1) in it)
⇒ k2 – βy + αh – h2 = 0 so required locus is
x2 – y2 – αx + βy = 0
AD  1
1 
.
+

CD  BD CD 
XtraEdge for IIT-JEE
47
DECEMBER 2010
4.
 π π
As |f(x)| ≤ |tan x| for ∀ x ∈  − , 
 2 2
(–1,log23)
so f(0) = 0
so |f(x) – f(0)| ≤ |tan x|
divides both sides by |x|
⇒ lim
x →0
f ( x ) − f (0)
tan x
≤ lim
x →0
x
x
1
=
ai
∑i
∫
0
log 2 (2 − x ) dx +
−1
y
) dy +
1
π
4
1
2
π
+ 2 log 2 3 + 2 –
+
ln 2
2ln 2
4
e2 e
π
+2+
sq units
27
4
= – log2
≤1
∫ (2 − 2
−1
= log 2 3 −
1
1
1
⇒ a1 + a 2 + a 3 + ..... + a n ≤ 1
2
3
n
⇒
(3/2,–1)
(0,–1)
⇒ |f´(0)| ≤ 1
n
(1,0)
(–1,0)
f ( x ) − f (0)
tan x
≤
⇒
x
x
i =1
7.
5.
A
Let the number is xyz, here x < y and z < y.
Let y = n, then x can be filled in (n – 1) ways.
(i.e. from 1 to (n – 1)) and z can be filled in n ways
(i.e. from 0 to (n – 1))
F
here 2 ≤ n ≤ 9
B
so total no. of 3 digit numbers with largest middle
digit
=
=
9
9
n =2
n =2
∑ n(n − 1) = ∑ n – ∑ n
so tan A =
n =2
9.10
9.10.19
–
6
2
so
= 285 – 45 = 240
240
required probability =
9 × 10 × 10
so
a2
r12
a2
r12
BD
BC
BC
a
=
=
=
MD
2MD
4r1
4r1
= tan2A
+
b2
r22
+
c2
r32
= 16 (tan2A + tan2B + tan2C)
8
=
30
=
C
D
∠BMC = 2∠BAC = 2∠BMD
9
2
E
M
...(1)
Now as tan A + tan B + tan C ≥
3 (tan A . tan B . tan C)1/3
and for a triangle tan A + tan B + tan C
= tan A . tan B . tan C
4
15
so (tan A . tan B . tan C)2/3 ≥ 3
6.
The region bounded by the curve y = log2(2 – x) and
2
⇒ tan A . tan B . tan C ≥ 3 3
2
the inequality (x – |x|) + (y – |y|) ≤ 4 is required area
⇒ tan2A + tan2B + tan2C
≥ 3(tan A. tan B tan C)2/3 ≥ 3.3
is
so from (1),
XtraEdge for IIT-JEE
48
a2
r12
+
b2
r22
+
c2
r32
≥ 144.
DECEMBER 2010
8.
= a + (1 – r2)
Z1, Z2, Z3 are centroids of equilateral triangles ACX,
ABY and BCZ respectively.
∫ (1 + r
Z1 − Z A iπ/6
e
ZC − ZA
Z1 – ZA = (ZC – ZA)
a
= a + (1 – r )
=a+
C
B
Z3
1− r2
(1 + r ) 2
a
∫
0
ZC
so, Tr = a +
...(1)
similarly,
1  3 i 
−
3  2 2 
tan 2 u / 2 + (1 − r ) 2
sec 2 u / 2 du
tan 2 u / 2 +
1− r2
tan a / 2
(1 − r ) 2
(1 + r ) 2
...(2)
Now
and
...(3)
1
(ZA – ZC)
2
2 dt
∫
(1 + r ) 2
t2 +
0
2(1 − r 2 ) 1 + r
=a+
(1 + r ) 2 1 − r
1
i
(ZC – ZB) +
(ZC + ZB – 2ZA)
2
2 3
+
2
Let tan u/2 = t
1  3 i 
Z1 – ZA = (ZC – ZA)
+
3  2 2 
similarly Z2 – Z3 =
∫ (1 + r)
sec 2 u / 2
0
z
So, Z1 – Z2 =
)(1 + tan 2 u / 2) − 2r (1 − tan 2 u / 2)
Z1
A
Z2 – ZA = (ZB – ZA)
2
2
Z2
ZB
sec 2 u / 2
0
x
ZA
y
a
1− r
1+ r
2
 −1  1 + r
 tan  t

 1− r
tan a / 2


 0
lim Tr = a –
2(1 + r )(r − 1) π
= a–π
(1 + r )(r − 1) 2
lim+ Tr = a +
2(1 − r )(r + 1) π
= a+π
(1 + r )(r − 1) 2
r →1+
r →1
a
and (from (1)) T1 =
∫ du = a
0
i
2 3
(ZA + ZC – 2ZB) ..(4)
Hence lim+ Tr, T1, lim− Tr form an A.P. with common
r →1
To prove ∆xyz as equilateral triangle, we prove that
r →1
difference π.
(Z3 – Z2)eiπ/3 = Z1 – Z2
So, (Z3 – Z2)eiπ/3 = (
–
1
(ZC – ZA)
2
10. Let α, β, γ be the three real roots of the equation
without loss of generality, it can be assumed that
α ≤ β ≤ γ.
1
3 
(ZA + ZC – 2ZB))  +
i
2 2 
2 3


i
so
x2 + ax2 + bx + c = (x – γ) (x2 + (a + γ) x + (γ2 + aγ + b))
where – γ (γ2 + aγ + b) = c, as γ is the root of given
equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must
have two roots i.e. α and β. So its discriminant is non
negative, thus
1
i
= (ZC – ZB) +
(ZC + ZB – 2ZA)
2
2 3
= Z1 – Z2
a
9.
Tr = 2
(γ + a)2 – 4(γ2 + aγ + b) ≥ 0
1 − r cos u
∫ 1 − 2r cos u + r
2
du.
...(1)
3γ2 + 2aγ – a2 + 4b ≤ 0
0
a
=
∫
0
1 − 2r cos u + r 2 − r 2 + 1
1 − 2r cos u + r 2
a
=

1− r
1 +
 1 − 2r cos u + r 2
0
∫
XtraEdge for IIT-JEE
2
so γ ≤
du
− a + 2 a 2 − 3b
3
so greatest root is also less than or equal to
− a + 2 a 2 − 3b
.
3

 du


49
DECEMBER 2010
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
MATHS
1.
⇒ g(θ) ∈ (1, 23/4] for θ ∈ (0, π/4]
and,
Find the range of the function



 2
 x 2 + e 
 + coslog e  x + e 
sin log e  2

 x 2 + 1 


 x + 1 


Sol. we have,
x2 + e
e –1
=1+ 2
2
x +1
x +1
x2 + e
 e –1

⇒ 2
>1
> 0 for all x 
Q 2
x +1
 x +1

Again,
x2 + e
e –1
=1+ 2
2
x +1
x +1

 e –1
Q 2
assumes its
x2 + e


⇒ 2
≤1+e–1=e

 x +1
x +1
maximum value at x = 0
Thus,
x2 + e
1< 2
≤ e for all x ∈ R.
x +1
 x2 + e 
 ≤ 1 for all x ∈ R
⇒ 0 < loge  2

 x +1 
f(x) =
g(θ) ∈ [ sin 1 + cos 1,2 3 / 4 ) for θ ∈ (π/4, 1]
⇒ g(θ) ∈ (1, 23/4], for θ ∈ (0, 1]
Hence, range of f(x) is (1, 23/4]
A ladder 15 m long, leans against a wall 7 m high,
and a portion of the ladder protrudes over the wall
such that its projection along the vertical is 3 m. How
fast does the bottom start to slip away from the wall
if the ladder slides down along the top edge of the
wall at 2 m/sec.
Sol. Let OC be the wall and AB be the ladder of which
the portion BC is protruded over the wall OC. The
projection of BC on the wall is CD = 3 m (given).
Let the foot A of the ladder be x metres away from
the wall and the protruded portion of the ladder be
y m in length. Further, let ∠OCA = θ.
Now, AB = 15 m and BC = y m
∴ AC = (15 – y) m.
B
D
2.
ym
C
x +e
.
⇒ 0 < θ ≤ 1, where θ = loge  2
 x +1 


2
∴ f(x) =
sin θ +
θ
(15–y) m
7m
cos θ
Let g (θ) = sin θ + cos θ , where 0 ≤ θ ≤ 1.
Clearly, the range of f(x) will be same as that of g(θ).
Now,
O
A
It is given that
(cos θ) 3 / 2 – (sin θ) 3 / 2
2 sin θ cos θ
⇒ g'(θ) ≥ 0 for 0 < θ ≤ π/4 and g'(θ) < 0 for π/4 < θ ≤ 1
⇒ g(θ) is increasing on (0, π/4] and decreasing on
(π/4,1]
⇒ g(θ) ∈ (g (0), g (π / 4)] for 0 ∈ (0, π/4]
and,
g(0) ∈ [g (1), g (π / 4) ) for θ ∈ (π/4, 1]
XtraEdge for IIT-JEE
xm
dy
= – 2 m/sec.
dt
dx
We have to find
.
dt
In ∆OCA, we have
x
7
tan θ =
and cos θ =
7
15 – y
sin θ + cos θ
cos θ
sin θ
⇒ g'(θ) =
–
2 sin θ
2 cos θ
g(θ) =
⇒ g'(θ) =
[Q sin 1 + cos1 > 1]
⇒ x = 7 tan θ
and, 15 – y = 7 sec θ.
In ∆BCD, we have
3
cos θ =
y
50
DECEMBER 2010
⇒ y = 3 sec θ.
∴ 15 – 3 sec θ = 7 sec θ
15
⇒ sec θ =
10
3
⇒ sec θ =
2
2
⇒ cos θ =
3
Now,
15 – y = 7 sec θ
⇒ y = 15 – 7 sec θ
dy
dθ  dy

= – 7 sec θ tan θ
⇒
Q
= –2(given)
dt
dt  dt

dθ
⇒ – 2 = – 7 sec θ tan θ
dt
dθ
2
⇒
=
...(1)
dt
7 sec θ tan θ
Now,
x = 7 tan θ
dx
dθ
= 7 sec2θ
⇒
dt
dt
dx
2
⇒
= 7 sec2θ ×
[Using (1)]
dt
7 sec θ tan θ
dx
2
⇒
=
dt
sin θ
dx
2
2

⇒
=
Q cos θ = 

dt
3
4

1–
9
dx
6
=
⇒
dt
5
Hence , the bottom starts to slip away from the
6
m/sec.
wall at the rate of
5
⇒ (n + 1) In =
⇒ (n – 1) In–2 =
If In =
∫x
n
0
n
...(i)
0
π
–
4
1
x n –1
∫ 1+ x
2
dx
...(ii)
0
⇒ (n + 1) In + (n – 1) In–2 =
⇒ (n + 1) In + (n – 1) In–2 =
4.
1
∫
x n +1 + x n –1
π
–
2
∫
x n –1 ( x 2 + 1)
π
–
2
∫x
π
–
2
0
1
0
1
1+ x2
1+ x2
n –1
dx
dx
dx
0
1
π
–
2
n
Prove that the internal bisector of an angle of a
triangle and the external bisector of the other two are
concurrent.
→
→
Sol. Referred to the vertex C as the origin, let a , b be
the position vectors of the vertices A and B
respectively. Let a, b, c be the lengths of the sides
BC, CA and AB respectively.
C(origin)
→
A (a )
→
B ( b)
D
Vector equation of the internal bisector of ∠C is
→ 
 →
→
 CA CB 
r = λ1 
+

 CA CB 


1
π
–
2
n
→ → 
a b
or r = λ1  + 
b a 


→
1
II
dx
2
⇒ (n + 1) In + (n – 1) In–2 =
Sol. We have,
∫x
∫ 1+ x
(n + 1) In + (n – 1) In –2 =
0
In =
x n +1
Adding (i) and (ii), we get
tan –1 x dx , prove that
(n + 1) In + (n – 1) In–2 =
1
Re placing n by 
(n – 2)in (i)



1
3.
π
–
4
tan –1x dx
I
1
 x n +1

1
⇒ In = 
tan –1 x  –
n +1
 n + 1
 0
[
⇒ (n + 1) In = x n +1 tan –1 x
1
x n +1
∫ 1+ x
0
1
] – ∫ 1x+ x
1
0
2
 → →
a a + b b 
or r = λ1 

 ab 


→
λ1 → →
or r =
(a a+b b )
ab
→
dx
n +1
2
dx
...(i)
0
XtraEdge for IIT-JEE
51
DECEMBER 2010
The external bisector of ∠A is the internal bisector of
→
Putting this value of λ2 in
→
the angle between vectors CA and AB . Therefore,
vector equation of the external bisector of ∠A is
→ 
 →
→
→
 AB CA 
r = a + λ2 
+

 AB CA 


get
λ3
c–b
=1+
c
a+b–c
ac
⇒ λ3 =
a+b–c
Now, putting the value of λ2 in (ii) or that of λ3 in
→ → → 
b– a a 
...(ii)
or r = a + λ2 
+ 
b
 c


Similarly the external bisector of ∠B is the internal
→
→
→
(iii), we get the position vector r1 of D as
→ → → 
bc
b– a a 
r1 = a +
+ 

b
a+b+c  c


→
→
bisector of the angle between the vector CB and
→
BA . Therefore, the vector equation of the external
bisector of ∠B is
→ 
 →
→
→
 CB BA 
+
r = b + λ3 

 CB BA 


⇒
⇒
⇒
⇒
⇒
⇒
⇒
→
→
→
→
a a+bb
=
a+b–c
→
→
→
a a+bb
satisfies equation of CD i.e.
Clearly, r1 =
a+b–c
ab
(i) for λ1 =
b+a –c
Hence the internal bisector of ∠C and external
bisector of ∠A and ∠B are concurrent.
 
π 
Find the solutions of the equation x2 – 3 sin  x –  = 3,
6 
 
where [.] denotes the greatest integer function.
Sol. The given equation can be written as
 
π 
x2 – 3 = 3 sin  x – 
...(i)
6 
 
5.
→
[On equating the coefficient of a and b ]
λ3
λ
λ
λ
= 1 + 2 (c – b) and 2 = 1 + 3 (c – a)
c
bc
c
ac
λ2
1
 λ

=1+
(c – a) 1 + 2 (c – b)
c
a
bc


[Substituting the value
of in second equation]
λ2
λ2
c–a
=1+
+
(c – a) (c – b)
c
a
abc
λ2
λ
c
+ 2 (c2 – ac – bc – ab)
=
c
a
abc
λ2
c
= (ab – c2 + ac + bc – ab) =
abc
a
λ2
c
(a + b – c) =
ab
a
bc
λ2 =
a+b–c
XtraEdge for IIT-JEE
→
a ( a + b – c) + b ( b – a ) + c a
=
a+b–c
→
→
→
→
→ → → 
b a– b
...(iii)
or, r = b + λ3  +

c 
a


Suppose (ii) and (iii) intersect at D. Then, for the
point D, we have
→ → →  →
→ → → 
→
b– a a 
b a– b
a + λ2 
+ →  = b + λ3  +

c 
 c

a
b




λ
λ
λ
⇒ 1– 2 + 2 = 3
c
b
c
λ
λ
λ
and, 2 = 1 + 3 – 3
c
a
c
→
λ3
λ
= 1 + 2 (c – b), we
c
bc
Clearly, right hand side can take only three values –3,
0, 3.
 
π 
CASE I
When 3sin  x –  = 3 :
6 
 
In this case, equation (i) reduces to
x2 – 3 = 3 ⇒ x = ± 6
But, for x = ±
6 , we have
 
 
π 
π 
3sin  x –  = 3sin  ± 6 –  ≠ 3.
6 
6 
 
 
So, x = ±
6 is not a solution of the given equation
 
π 
CASE II When 3sin  x –  = 0 :
6 
 
In this case, the given equation reduces to
x2 – 3 = 0 ⇒ x = ±
When x =
52
3
3 , we have,
DECEMBER 2010
x–
π
=
6
∴ x=
3 –
 
π 
π
π
<
⇒ sin  x –  = 0
6
6 
2
 
Know about Pie
3 is a solution.
3.14
When x = – 3 , we have
π
π
=– 3 –
x–
6
6
⇒ –π<x–
= Perimeter / Diagonal, of any circle.
Pi expanded to 45 decimal places:
3.14159 26535 89793 23846 26433 83279 50288 41971
69399
Pi expanded to 52 binary places:
11.0010010000111 1110110101010 0010001000010
1101000111001
You cannot square a disc using just a ruler and compasses
because is a transcendental number.
= 4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... )
= 2(2/1 x 2/3 x 4/3 x 4/5 x 6/5 x 6/7 x 8/7 x 8/9 x ... )
≈ 355/113 (a real good rational approximation of )
≈ (6 2)/5
In the late 18th century, James Stirling, a Scottish
mathematician, developed an approximation for factorials
using the transcendental numbers 'Pi' and 'e':
n! ≈ (2 n)1/2 (n/e)n
The most famous formula for calculating Pi is Machin's
formula:
/4 = 4 arctan(1/5) – arctan(1/239)
This formula, and similar ones, were used to push the
accuracy of approximations to Pi to over 500 decimal
places by the early 18th century (this was all hand
calculation!).
Interestingly, there are no occurrences of the sequence
123456 in the first million digits of Pi. - posted by
George Pantazis
Bamboozlement with Pi
Does Pi equal 3? No? Then have a look on the algebraic
equation below:
x = ( + 3)/2
2x = + 3
2x( - 3) = ( + 3)( - 3)
2 x - 6x = 2 - 9
9 - 6x = 2 - 2 x
9 - 6x + x2 = 2 - 2 x + x2
(3 - x)2 = ( - x)2
3-x= -x
3=
We use Pi to:
Describe the DNA double helix,
Determining the distribution of primes - the probability
that two randomly selected integers are relatively prime
(i.e. have no common factors) is 6 / p2,
Analyzing the ripples on water
Checking for accuracy - as there are now millions upon
millions of known decimal places of Pi, by asking a super
computer to compute this many figures its accuracy can
be tested.
π
<0
6
 
π 
⇒ sin  x –  = – 1
6 
 
∴ x=–
3 in not a solution.
 
π 
CASE III When 3sin  x –  = – 3
6 
 
In this case, the given equation reduces to
⇒x=0
x2 – 3 = – 3
When x = 0, we have
 
  π 
π 
sin  x –  = sin  –  = – 1.
6

 
  6 
 
π 
⇒ 3sin  x –  = – 3.
6 
 
So, x = 0 is a solution of the given equation.
Hence, the given equation has only two solution
x = 0,
3.
A condolence meeting is being held in a big hall
which has 7 doors by which the mourners enter the
hall. One can use any one of the 7 doors to enter and
can come at any time during meeting. At each door a
register is kept in which a mourner has to put down
his signature while entering the hall. If 200 people
attend the meeting, how many different sets of 7 lists
of signatures can arise?
Sol. Suppose xi persons enter into the hall through ith door,
where i = 1, 2, ..., 6, 7. Then,
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 200, ...(i)
where xi ≥ 0; i = 1, 2, ... , 7
The total number of non-negative integral solutions
of equation (i) is
200+7–1
C7–1 = 206C6
206
Thus, there are C6 ways in which 200 persons can
enter into the hall through 7 doors. But corresponding
to each way of entering into the hall the signatures of
200 persons can be arranged in 200! ways.
Hence, the required number of lists
206!
= 206C6 × 200! =
6!
6.
XtraEdge for IIT-JEE
53
DECEMBER 2010
MATHS
MONOTONICITY,
MAXIMA & MINIMA
Mathematics Fundamentals
Monotonic Functions :
A function f(x) defined in a domain D is said to be
(i) Monotonic increasing :
according as f(x) is monotonic increasing or
decreasing at x = a.
So at x = a, function f(x) is
monotonic increasing ⇔ f´(a) > 0
x < x 2 ⇒ f ( x1 ) ≤ f ( x 2 )
⇔  1
∀ x1, x2 ∈ D
x1 > x 2 ⇒ f ( x1 ) ≥ f ( x 2 )
y
monotonic decreasing ⇔ f´(a) < 0
(ii) In an interval : In [a, b], f(x) is
y
O
x
monotonic increasing ⇔ f´(x) ≥ 0

monotonic decreasing ⇔ f´(x) ≤ 0 ∀ x ∈ (a, b)
constant ⇔ f´(x) = 0
O
x
Note :
(i) In above results f´(x) should not be zero for all
values of x, otherwise f(x) will be a constant
function.
(ii) If in [a, b], f´(x) < 0 at least for one value of x and
f´(x) > 0 for at least one value of x, then f(x) will
not be monotonic in [a, b].
Examples of monotonic function :
If a functions is monotonic increasing (decreasing ) at
every point of its domain, then it is said to be
monotonic increasing (decreasing) function.
In the following table we have example of some
monotonic/not monotonic functions
Monotonic
Monotonic
Not
increasing
decreasing
monotonic
x3
1/x, x > 0
x2
x|x|
1 – 2x
|x|
x
–x
e
e
ex + e–x
log x
log2x
sin x
sin h x
cosec h x, x > 0
cos h x
[x]
cot hx, x > 0
sec h x
x < x 2 ⇒ f ( x1 ) >/ f ( x 2 )
i.e., ⇔  1
∀ x1, x2 ∈ D
x1 > x 2 ⇒ f ( x1 ) </ f ( x 2 )
(ii) Monotonic decreasing :
x < x 2 ⇒ f ( x 1 ) ≥ f ( x 2 )
⇔  1
∀ x1, x2 ∈ D
x 1 > x 2 ⇒ f ( x 1 ) ≤ f ( x 2 )
y
y
O
x
O
x
 x < x 2 ⇒ f ( x1 ) </ f ( x 2 )
i.e., ⇔  1
∀ x1, x2 ∈ D
/ f (x 2 )
x1 > x 2 ⇒ f ( x1 ) >
A function is said to be monotonic function in a
domain if it is either monotonic increasing or
monotonic decreasing in that domain.
Note : If x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ D, then
f(x) is called strictly increasing in domain D and
similarly decreasing in D.
Method of testing monotonicity :
(i) At a point : A function f(x) is said to be
monotonic increasing (decreasing) at a point x = a of
its domain if it is monotonic increasing (decreasing)
in the interval (a – h, a + h) where h is a small
positive number. Hence we may observer that if f(x)
is monotonic increasing at x = a then at this point
tangent to its graph will make an acute angle with xaxis where as if the function is monotonic decreasing
there then tangent will make an obtuse angle with xaxis. Consequently f´(a) will be positive or negative
XtraEdge for IIT-JEE
Properties of monotonic functions :
If f(x) is strictly increasing in some interval, then in
that interval, f–1 exists and that is also strictly
increasing function.
If f(x) is continuous in [a, b] and differentiable in
(a, b), then
f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic increasing
in [a, b]
f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic decreasing
in [a, b]
54
DECEMBER 2010
If both f(x) and g(x) are increasing (or decreasing) in
[a, b] and gof is defined in [a, b], then gof is
increasing.
If f(x) and g(x) are two monotonic functions in [a, b]
such that one is increasing and other is decreasing
then gof, it is defined, is decreasing function.
Maximum and Minimum Points :
The value of a function f(x) is said to be maximum at
x = a if there exists a small positive number δ such
that
f(a) > f(x)
y
O
( )
a
( )
b
( )
c
where x = c is a point such that f´(c) = 0.
If a continuous function has only one maximum
(minimum) point, then at this point function has
its greatest (least) value.
Monotonic functions do not have extreme points.
Conditions for maxima and minima of a function
Necessary condition : A point x = a is an extreme
point of a function f(x) if f´(a) = 0, provided f´(a)
exists. Thus if f´(a) exists, then
x = a is an extreme point ⇒ f´(a) = 0
f´(a) ≠ 0 ⇒ x = a is not an extreme point
But its converse is not true i.e.
f´(a) = 0 ⇒
/ x = a is an extreme point.
For example if f(x) = x3, then f´(0) = 0 but x = 0 is
not an extreme point.
Sufficient condition : For a given function f(x), a
point x = a is
a maximum point if f´(a) = 0 and f´´(a) < 0
a minimum point if f´(a) = 0 and f´´(a) > 0
not an extreme point if f´(a) = 0 = f´´(a) and
f´´´(a) ≠ 0.
Note : If f´(a) = 0, f´´(a) = 0, f´´´(a) = 0 then the sign
of f(4)(a) will determine the maximum or minimum
point as above.
Working Method :
Find f´(x) and f´´(x).
Solve f´(x) = 0. Let its roots be a, b, c, ...
Determine the sign of f´´(x) at x = a, b, c, .... and
decide the nature of the point as mentioned above.
Properties of maxima and minima :
If f(x) is continuous function, then
Between two equal values of f(x), there lie atleast one
maxima or minima.
Maxima and minima occur alternately. For example
if x = –1, 2, 5 are extreme points of a continuous
function and if x = –1 is a maximum point then x = 2
will be a minimum point and x = 5 will be a
maximum point.
When x passes a maximum point, the sign of dy/dx
changes from + ve to – ve, where as when x passes
through a minimum point, the sign of f´(x) changes
from –ve to + ve.
If there is no change in the sign of dy/dx on two sides
of a point, then such a point is not an extreme point.
If f(x) is maximum (minimum) at a point x = a, then
1/f(x), [f(x) ≠ 0] will be minimum (maximum) at that
point.
If f(x) is maximum (minimum) at a point x = a, then
for any λ ∈ R, λ + f(x), log f(x) and for any k > 0, k
f(x), [f(x)]k are also maxmimum (minimum) at that
point.
x
Also then the point x = a is called a maximum point
for the function f(x).
Similarly the value of f(x) is said to be minimum at x
= b if there exists a small positive number δ such that
f(b) < f(x) ∀ x ∈ (b – δ, b + δ)
Also then the point x = b is called a minimum point
for f(x)
Hence we find that :
(i) x = a is a maximum point of f(x)
f (a ) – f(a + h) > 0
⇔ 
 f(a) – f(a – h) > 0
(ii) x = b is a minimum point of f(x)
f (b) – f(b + h) < 0
⇔ 
 f(b) – f(b – h) > 0
(iii) x = c is neither a maximum point nor a minimum
point
f (c) – f (c + h )
⇔ 
and
 have opposite signs.
f (c) − f (c − h ) 
Where h is a very small positive number.
Note :
The maximum and minimum points are also
known as extreme points.
A function may have more than one maximum
and minimum points.
A maximum value of a function f(x) in an interval
[a, b] is not necessarily its greatest value in that
interval. Similarly a minimum value may not be
the least value of the function. A minimum value
may be greater than some maximum value for a
function.
The greatest and least values of a function f(x) in
an interval [a, b] may be determined as follows :
Greatest value = max. {f(a), f(b), f(c)}
Least value = min. {f(a), f(b), f(c)}
XtraEdge for IIT-JEE
or
55
DECEMBER 2010
MATHS
FUNCTION
Mathematics Fundamentals
Definition of a Function :
If f and g are two functions then their sum,
difference, product, quotient and composite are
denoted by
Let A and B be two sets and f be a rule under which
every element of A is associated to a unique element
of B. Then such a rule f is called a function from A to
B and symbolically it is expressed as
f + g, f – g, fg, f/g, fog
and they are defined as follows :
f:A→B
or
(f + g) (x) = f(x) + g(x)
f
→ B
A 
(f – g) (x) = f(x) – g(x)
Function as a Set of Ordered Pairs
(fg) (x) = f(x) f(g)
Every function f : A → B can be considered as a set
of ordered pairs in which first element is an element
of A and second is the image of the first element.
Thus
(f/g) (x) = f(x)/g(x)
(fog) (x) = f[g(x)]
Formulae for domain of functions :
f = {a, f(a) /a ∈ A, f(a) ∈ B}.
Df ± g = Df ∩ Dg
Domain, Codomain and Range of a Function :
Dfg = Df ∩ Dg
If f : A → B is a function, then A is called domain of
f and B is called codomain of f. Also the set of all
images of elements of A is called the range of f and it
is expressed by f(A). Thus
Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0}
Dgof = {x ∈ Df | f(x) ∈ Dg}
f(A) = {f(a) |a ∈ A}
obviously
D
f(A) ⊂ B.
f
= Df ∩ {x |f(x) ≥ 0}
Classification of Functions
Note : Generally we denote domain of a function f by
Df and its range by Rf.
1.
Algebraic and Transcendental Functions :
Algebraic functions : If the rule of the function
consists of sum, difference, product, power or
roots of a variable, then it is called an algebraic
function.
Transcendental Functions : Those functions
which are not algebraic are named as
transcendental or non algebraic functions.
Equal Functions :
Two functions f and g are said to be equal functions
if
domain of f = domain of g
codomain of f = codomain of g
2.
f(x) = g(x) ∀ x.
Algebra of Functions :
XtraEdge for IIT-JEE
(g(x) ≠ 0)
56
Even and Odd Functions :
Even functions : If by replacing x by –x in f(x)
there in no change in the rule then f(x) is called an
even function. Thus
f(x) is even ⇔ f(–x) = f(x)
DECEMBER 2010
Odd function : If by replacing x by –x in f(x)
there is only change of sign of f(x) then f(x) is
called an odd function. Thus
Period of f(x) = T
⇒ Period of f(nx + a) = T/n
Periods of some functions :
f(x) is odd ⇔ f(–x) = – f(x)
3.
Function
Explicit and Implicit Functions :
Explicit function : A function is said to be
explicit if its rule is directly expressed (or can be
expressed( in terms of the independent variable.
Such a function is generally written as
sin x, cos x, sec x, cosec x,
2π
tan x, cot x
π
sinnx, cosn x, secn x, cosecn x
2π if n is odd
π if n is even
y = f(x), x = g(y) etc.
4.
Implicit function : A function is said to be
implicit if its rule cannot be expressed directly in
terms of the independent variable. Symbolically
we write such a function as
tann x, cotnx
π∀n∈N
|sin x|, |cos x|, |sec x|, |cosec x|
π
|tan x|, |cot x|,
π
f(x, y) = 0, φ(x, y) = 0 etc.
|sin x| + |cos x|, sin4x + cos4x
π
2
|sec x| + |cosec x|
Continuous and Discontinuous Functions :
Continuous functions : A functions is said to be
continuous if its graph is continuous i.e. there is
no gap or break or jump in the graph.
Discontinuous Functions : A function is said to
be discontinuous if it has a gap or break in its
graph atleast at one point. Thus a function which
is not continuous is named as discontinuous.
5.
Period
π
2
x – [x]
1
Period of f(x) = T ⇒ period of f(ax + b) = T/|a|
Period of f1(x) = T1, period of f2(x) = T2
⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T2}
Increasing and Decreasing Functions :
Kinds of Functions :
Increasing Functions : A function f(x) is said to
be increasing function if for any x1, x2 of its
domain
One-one/ Many one Functions :
A function f : A → B is said to be one-one if
different elements of A have their different
images in B.
x1 < x2 ⇒ f(x1) ≤ f(x2)
or x1 > x2 ⇒ f(x1) ≥ f(x2)
Thus
Decreasing Functions : A function f(x) is said to
be decreasing function if for any x1, x2 of its
domain
⇒ f (a ) ≠ f ( b)
 a≠b

f is one-one ⇔ 
or
f (a ) = f (b) ⇒
a=b

x1 < x2 ⇒ f(x1) ≥ f(x2)
or x1 > x2 ⇒ f(x1) ≤ f(x2)
A function which is not one-one is called many
one. Thus if f is many one then atleast two
different elements have same f-image.
Periodic Functions :
A functions f(x) is called a periodic function if there
exists a positive real number T such that
Onto/Into Functions : A function f : A → B is
said to be onto if range of f = codomain of f
f(x + T) = f(x) ∀ x
Thus f is onto ⇔ f(A) = B
Also then the least value of T is called the period of
the function f(x).
XtraEdge for IIT-JEE
|tan x| + |cot x|
57
DECEMBER 2010
Hence f : A → B is onto if every element of B
(co-domain) has its f–preimage in A (domain).
Domain and Range of some standard functions :
Function
A function which is not onto is named as into
function. Thus f : A → B is into if f(A) ≠ B. i.e.,
if there exists atleast one element in codomain of
f which has no preimage in domain.
Note :
Total number of functions : If A and B are finite
sets containing m and n elements respectively,
then
total number of functions which can be defined
from A to B = nm.
total number of one-one functions from A to B
 n P
=  m
 0
Domain
Range
Polynomial
function
R
R
Identity
function x
R
R
Constant
function c
R
{c}
Reciprocal
function 1/x
R0
R0
x2, |x|
R
R+ ∪ {0}
x3, x |x|
R
R
Signum
function
R
{–1, 0, 1}
if
m≤n
x + |x|
R
R+ ∪ {0}
if
m>n
x – |x|
R
R– ∪ {0}
[x]
R
Z
x – [x]
R
[0, 1)
x
[0, ∞)
[0, ∞)
total number of onto functions from A to B
(if m ≥ n) = total number of different n groups of
m elements.
Composite of Functions :
Let f : A → B and g : B → C be two functions, then
the composite of the functions f and g denoted by
gof, is a function from A to C given by gof : A → C,
(gof) (x) = g[f(x)].
ax
R
R+
log x
R+
R
sin x
R
[–1, 1]
cos x
R
[–1, 7]
tan x
R – {± π/2, ± 3π/2, ...} R
cot x
R – {0, ± π. ± 2π, .....
R
The following properties of composite functions can
easily be established.
sec x
R – (± π/2, ± 3π/2, .....
R – (–1, 1)
cosec x
R – {0, ±π, ± 2π, ......} R –(–1, 1)
Composite of functions is not commutative i.e.,
sinh x
R
R
cosh x
R
[1, ∞)
tanh x
R
(–1, 1)
coth x
R0
R –[1, –1]
sech x
R
(0, 1]
cosech x
R0
R0
Properties of Composite Function :
fog ≠ gof
Composite of functions is associative i.e.
(fog)oh = fo(goh)
Composite of two bijections is also a bijection.
Inverse Function :
–1
If f : A → B is one-one onto, then the inverse of f i.e.,
f–1 is a function from B to A under which every b ∈ B
is associated to that a ∈ A for which f(a) = b.
Thus
[–1, 1]
[–π/2, π/2]
cos–1x
[–1, 1]
[0, π]
–1
R
(–π/2, π/2}
–1
R
(0, π)
tan x
cot x
f–1 : B → A,
–1
f–1(b) = a ⇔ f(a) = b.
XtraEdge for IIT-JEE
sin x
sec x
R –(–1, 1)
[0, π] – {π/2}
–1
R – (–1, 1)
(– π/2, π/2] – {0}
cosec x
58
DECEMBER 2010
XtraEdge for IIT-JEE
59
DECEMBER 2010
Based on New Pattern
a
IIT-JEE 2011
XtraEdge Test Series # 8
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
•
Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
•
Question 7 to 10 are multiple choice questions with multiple correct answer. +4 marks and No negative mark for
wrong answer.
Section - II
•
Question 11 to 14 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for
wrong answer.
•
Question 15 to 23 are passage based single type questions. +4 marks will be awarded for correct answer and
-1 mark for wrong answer.
PHYSICS
(A) 1 .11 A
3.
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1.
Separation between the plates of a
parallel plate capacitor is 5 mm. This
capacitor, having air as the dielectric
medium between the plates, is
charged to a potential difference 25
3mm
V using a battery. The battery is then
5mm
disconnected and a dielectric slab of
thickness 3 mm and dielectric constant K = 10 is
placed between the plates, as shown. Potential
difference between the plates after the dielectric slab
has been introduced is –
(A) 18.5 V
2.
(B) 13.5 V
(C) 11.5 V
(B) 1.25 A
In an insulating medium (K = 1) volumetric charge
density varies with y-coordinates according to the
law ρ = a.y. A particle of mass m having positive
charge q is at point A(0, y0)
y
and projected with velocity
r
v = v 0 î as shown in figure. At
y = 0 electric field is zero.
Neglect the gravity and A
(0, y0)
frictional resistance, the slope
x
of trajectory of the particle as a
function of y (E is only along y-axis) is –
(A)
(C)
qa
mε 0 v 02
( y 3 – y 30 )
qa ( y 3 – y 30 )
(B)
(D)
5mε 0 v 02
qa
3mε 0 v 02
( y 3 – y 30 )
qa ( y 3 – y 30 )
2mε 0 v 02
(D) 6.5 V
4.
Reading of ammeter is 1 A. If each of the 4 Ω resistor
is replaced by 2 Ω resistor, the reading of ammeter
will become nearly –
15Ω
4Ω
2Ω
15Ω
4Ω
15Ω
10V
Average torque on a projectile of mass m, initial
speed u and angle of projection θ between initial and
final positions P and Q as shown in figure about the
point of projection is –
y
u
P
A(ammeter)
XtraEdge for IIT-JEE
(C) 1.34 A (D) 1.68 A
60
θ
Q
x
DECEMBER 2010
(A)
mu 2 sin 2θ
2
(C) mu2 sinθ
5.
8.
(B) mu2 cosθ
(D)
mu 2 cos θ
2
Portion AB of the wedge shown in figure is rough
and BC is smooth. A solid cylinder rolls without
slipping from A to B. If AB = BC, then ratio of
translational kinetic energy to rotational kinetic
energy, when the cylinder reaches point C is –
A
C
(A) Brightness of bulb C is highest
(B) If C fails, brightness of bulb D increases
(C) If C fails, brightness of all bulbs remain same
(D) If A fails, B will not glow
C
D
6.
9.
A gas is kept in a closed container at temperature T.
If temperature of gas is increased then according to
Maxwell theory of molecular speed distribution,
choose the correct alternatives :
(A) Number of molecules moving with vrms may
increase
(B) Number of molecules moving w3ith vmp
must decrease
(C) Number of molecular moving with vav may
decrease
(D) Number of molecules moving with 2vrms may
increase
10.
The speed v of a particle moving along a straight line,
when it is at a distance x from a fixed point on the
line is v2 = 144 – 9x2. Select the correct
alternative(s):
(A) The motion of the particle is SHM with time
2π
period T =
unit
3
(B) The maximum displacement of the particle from
the fixed point is 4 unit
(C) The magnitude of acceleration at a distance 3
units from the fixed point is 27 unit
(D) The motion of the particle is periodic but not
simple harmonic
(B) 5
(D) 8/3
Two bodies of masses m1 and m2 are initially at rest
placed infinite distance apart. They are then allowed
to move towards each other under mutual
gravitational attraction. Their relative velocity when
they are r distance apart is –
2 G m1 m 2
2G (m1 + m 2 )
(B)
(A)
r
(m1 + m 2 )r
(C)
G (m1 + m 2 )
r
(D)
G m1 m 2
(m1 + m 2 )r
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
7.
Point 1 is at middle of solenoid, point (2) at an end
face and point (3) is outside the solenoid at a distance
a. Plane of coil and plane of cross-section of solenoid
are parallel –
Solenoid
2
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
Coil
i
1
3
a
2a
(A) Force between coil and solenoid is attractive at
all three points (i.e. 1, 2, 3)
(B) Force between coil and solenoid at the point 1 is
zero
(C) Among these three point force between coil and
solenoid is maximum at point 2
(D) Among these three point force between coil and
solenoid is maximum at point 1
XtraEdge for IIT-JEE
D
Ideal
Battery
B
(A) 3/5
(C) 7/5
Four identical bulbs A, B, C, D are connected in a
circuit as shown in figure. Now whenever any bulb
fails, then it cannot conduct current through it.
Then –
B
A
61
DECEMBER 2010
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
15.
connected between c & d although we shall actually
connect none in the circuit, electrons will flow in a
definite direction between c & d. Potential difference
between A & a is 6V.
Resistance R is –
(B) 8 Ω
(A) 4 Ω
(C) 6 Ω
(D) 9 Ω
11.
Assertion (A) : When a current flow in the coil of a
transformer then its core becomes hot.
Reason (R) : The core of transformer is made of iron.
16.
Equivalent resistance of the circuit is nearly –
(B) 5.25 Ω
(A) 5.95 Ω
(C) 6.45 Ω
(D) 7.85 Ω
12.
Assertion (A) : The stream of water flowing at high
speed from a garden hose pipe tends to spread like a
fountain when held vertically up, but tends to narrow
down when held vertically down.
Reason (R) : In any steady flow of an incompressible
fluid, the volume flow rate of the fluid remains
constant.
17.
Assuming internal resistance of cell to be zero,
potential difference applied between x & y is nearly–
(A) 82.6 V
(B) 94.2 V
(C) 106.5 V
(D) 112.4 V
13.
14.
Passage # 2 (Ques. 18 to 20)
A narrow beam of electrons, of radius r and all
moving at the same velocity v much less than speed
of light (c), produce a charge current I. Assume that
the beam has cylindrical symmetry. me = mass of
electron & e = charge of electron.
Assertion (A) : Doppler effect for sound wave is
symmetric w.r.t. speed of source and speed of
observer.
Reason (R) : Change in frequency due to motion of
source w.r.t. stationary observer is not same with that
due to motion of observer w.r.t. stationary source.
Assertion (A) : In simple harmonic motion A is the
amplitude of oscillation. If t1 be the time to reach the
A
particle from mean position to
and t2 the time to
2
A
t
to A. Then t1 = 2 .
reach from
2
2
Reason (R) : Equation of motion for the particle
starting from mean position is given by
x = ± A sin ωt and of the particle starting from
extreme position is given by x = ± A cos ωt.
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
Passage # 1 (Ques. 15 to 17)
R
x
4Ω
R1
6Ω
R2
12Ω
12Ω
R
24Ω
R c R1
R
A
B R1
R
6Ω
R1
R1 d R
R1 b R
V
R a 24Ω
15Ω
y
It is observed that change of R2 has no effect on the
equivalent resistance between x & y. Also if we
connect a wire between a & b, there is no flow of
charge between a & b. However if a wire is
XtraEdge for IIT-JEE
62
18.
If electron moving along the edge of beam
experience magnetic force (FM) as well as electric
force (Fe), then –
(A) Fe & FM are in radial and in the same direction
(B) Fe & FM are in radial direction and in opposite
direction
(C) Fe & FM are perpendicular to each other
(D) Fe & FM are tangential to surface of beam
19.
What is the change in radial velocity of electrons at
the border of the beam after, the beam has travelled a
longitudinal distance 100 times of r i.e. 100 r. [Here
c
assume that v = ]
2
50 Ie
100 Ie
(B)
(A)
2
πε0 m e c
πε0 m e c 2
Ie
150 Ie
(C)
(D)
2
πε0 m e c
πε0 m e c 2
20.
If it is assumed that beam start to move with zero
radial velocity of electron. If 'θ' is the angle between
velocity of electron at the edge of beam and edge of
beam. Then find tan θ after the beam has travelled
c
[assume v = ]
distance 100 r –
2
100 Ie
200 Ie
(B)
(A)
πε0 m e c3
πε0 m e c3
300 Ie
400 Ie
(C)
(D)
3
πε0 m e c
πε0 m e c3
DECEMBER 2010
Passage # 3 (Ques. 21 to 23)
Two identical blocks P and Q have mass m each.
They are attached to two identical springs initially
unstretched. Now the left spring (along with P) is
A
compressed by
and the right spring (along with
2
Q) is compressed by A. Both the blocks are released
simultaneously. They collide perfectly inelastically.
Initially time period of both the block was T.
A
A
2
P Q
21.
22.
23.
4.
The structure
COOH
(A) geometrical isomersism
(B) optical isomerism
(C) geometrical & optical isomerism
(D) tautomerism
5.
CH MgBr
Ethyl ester 3 → P. The product P will be
excess
H3C
CH3
(A)
What is energy of oscillation of the combined mass ?
1
1
(A) kA2
(B) kA2
2
4
1 2
1
(C) kA
(D)
kA2
8
16
C2H5
H 3C
OH
H5C2
OH
H5C2
C2H5
H5C2
C2H5
H7C3
OH
(D)
H5C2
6.
H3C
(B)
(C)
OH
KF combines with HF to form KHF2. The compound
contains the species (B) K+, F– and HF
(A) K+, F– and H+
+
–
(C) K and [HF2]
(D) [KHF]+ and F–
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
7.
Refer to the figure given :
Which of the following statements is/are correct ?
gas C
gas A
Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02
mole of [Co(NH3)5Br]SO4 was prepared in 2 litre of
solution.
1 litre of mixture X + excess AgNO3 → Y.
1 litre of mixture X + excess BaCl2 → Z
No. of moles of Y and Z are
(A) 0.01, 0.01
(B) 0.02, 0.01
(C) 0.01, 0.02
(D) 0.02, 0.02
XtraEdge for IIT-JEE
shows :
H
C
CH3
The amplitude of combined mass is –
A
A
(A)
(B)
4
2
2A
3A
(C)
(D)
3
4
Among the following species, identify
isostructural pairs. NF3, NO3–, BF3, H3O+, HN3
(A) [NF3, NO3–] and [BF3, H3O+]
(B) [NF3, HN3] and [NO3–, BF3]
(C) [NF3, H3O+] and [NO3–, BF3]
(D) [NF3, H3O+] and [HN3, BF3]
C=C
H 3C
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
2.
The following equilibrium is established when
hydrogen chloride is dissolved in acetic acid.
HCl + CH3COOH
Cl– + CH3COOH2+
The set that characterises the conjugate acid-base
pairs is (A) (HCl, CH3COOH) and (CH3COOH2+,Cl–)
(B) (HCl, CH3COOH2+) and (CH3COOH,Cl–)
(C) (CH3COOH2+, HCl) and (Cl–, CH3COOH)
(D) (HCl, Cl–) and (CH3COOH2+, CH3COOH)
H 3C
H
The time period of oscillation of combined mass is –
T
(A)
(B) 2 T
2
T
(C) T
(D)
2
CHEMISTRY
1.
3.
Z
the
1
ideal gas
gas B
P
(A) For gas A, a = 0 and Z will linearly depend on
pressure
(B) For gas B, b = 0 and Z will linearly depend on
pressure
63
DECEMBER 2010
(C) Gas C is a real gas and we can find 'a' and 'b' if
intersection data is given
(D) All van der Waal gases will behave like gas C
and give positive slope at high pressure
8.
The following is (are) endothermic reaction (s) :
(A) Combustion of methane
(B) Decomposition of water
(C) Dehydrogenation of ethane to ethylene
(D) Conversion of graphite to diamond
9.
When zeolite, which is hydrated sodium aluminium
silicate, is treated with hard water the sodium ions
are exchanged with (B) Ca++ ions
(A) H+ ions
––
(C) SO4 ions
(D) Mg++ ions
14.
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
Passage # 1 (Ques. 15 to 17)
The coordination number of Ni2+ is 4.
NiCl2 + KCN (excess) → A (cyano complex)
NiCl2 + Conc. HCl (excess) → B (chloro complex)
15. The IUPAC name of A and B are
(A) Potassium tetracyanonickelate (II),
potassium tetrachloronickelate (II)
(B) Tetracyanopotassiumnickelate (II),
tetrachloropotassiumnickelate (II)
(C) Tetracyanonickel (II), tetrachloronickel (II)
(D) Potassium tetracyanonickel (II),
potassium tetrachloronickel (II)
16. Predict the magnetic nature of A and B
(A) Both are diamagnetic
(B) A is diamagnetic and B is paramagnetic with one
unpaired electron
(C) A is diamagnetic and B is paramagnetic with two
unpaired electrons
(D) Both are paramagnetic
CH3
10.
H3C
, hv
Cl
2 
→ N (isomeric products) ;
CH3
on
C5H11Cl fractional
  distillati
 
→ M (isomeric products)
Identify N and M
(A) 6, 4
(B) 6, 6
(C) 4, 4
(D) 3, 3
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
11.
12.
13.
17.
The hybridization of A and B are
(A) dsp2, sp3
(B) sp3, sp3
2
2
(C) dsp , dsp
(D) sp3d2, d2sp3
Passage # 2 (Ques. 18 to 20)
The conversion of an amide to an amine with one
carbon atom less by the action of alkaline hydrohalite
is known as Hofmann bromamide degradation.
O
O
→
NH2
NH–Br
(i)
(ii)
O
C
O
→
→
N
–
N–Br
••
(iv)
(iii)
Assertion (A) : Micelles are formed by surfactant
molecules above the critical micellar concentration
(CMC).
Reason (R) : The conductivity of a solution having
surfactant molecules decreases sharply at the CMC.
HO
→
Assertion (A) : Boron always forms covalent bond
because.
Reason (R) : The small size of B3+ favours formation
of covalent bond.
N
O
H
→
NH2
(v)
(vi)
In this reaction, RCONHBr is formed from which the
reaction has derived its name. Hofmann reaction is
accelerated if the migration group is more electronreleasing. Hofmann degradation reaction is an
intramolecular reaction.
Assertion (A) : Band gap in germanium is small.
Reason (R) : The energy spread of each germanium
atomic energy level is infinitesimally small.
XtraEdge for IIT-JEE
Assertion (A) : p-Hydroxybenzoic acid has a lower
boiling point than o-hydroxybenzoic acid.
Reason (R) : o-Hydroxybenzoic acid has
intramolecular hydrogen bonding.
64
DECEMBER 2010
18.
How can the conversion of (i) to (ii) be brought
about?
(A) KBr
(B) KBr + CH3ONa
(C) KBr + KOH
(D) Br2 + KOH
19.
Which is the rate determining step in Hofmann
bromamide degradation?
(A) Formation of (i)
(B) Formation of (ii)
(C) Formation of (iii)
(D) Formation of (iv)
20.
What are the constituent amines formed when the
mixture of (i) and (ii) undergoes Hofmann
bromamide degradation?
21.
The total number of moles of chlorine gas evolved is(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
22.
If the cathode is a Hg electrode, the maximum weight
(g) of amalgam formed from this solution is (A) 200
(B) 225
(C) 400
(D) 446
23.
The total charge (coulombs) required for complete
electrolysis is (A) 24125
(B) 48250
(C) 96500
(D) 193000
15
CONH2
CONH2
MATHEMATICS
&
D
NH2
15
,
15
NH2
NH2
(A)
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
(ii)
(i)
,
D
NH2
&
D
15
NH2
1.
NH2
(B)
and
D
15
NH2
(C)
 π 3π 
(C)  , 
2 2 
NH2
and
2.
15
NHD
(D)
If S is the set of values of θ for which (0, 0) and
(sin θ, cos θ) lie on opposite sides of y = |x – 1| then
S contains  π π
π π
(A)  – , 
(B)  , 
 3 3
3 2
lim
x →0
(A) 2
(C) 0
and
Passage # 3 (Ques. 21 to 23)
Chemical reactions involve interaction of atoms and
molecules A large number of atoms/molecules
(approximately 6.023 × 1023 ) are present in a few
grams of any chemical compound varying with their
atomic/molecular masses. To handle such large
numbers conveniently, the mole concept was
introduced. This concept has implications in diverse
areas such as analytical chemistry, biochemistry,
electrochemistry and radiochemistry. The following
example illustrates a typical case, involving
chemical/electrochemical reaction, which requires a
clear understanding of the mole concept. A 4.0 molar
aqueous solution of NaCl is prepared and 500 mL of
this solution is electrolysed. This leads to the
evolution of chlorine gas at one of the electrodes
(atomic mass : Na = 23, Hg = 200 ; ] Faraday =
96500 coulombs).
XtraEdge for IIT-JEE
65
 π π
(D)  – , 
 2 2
cos –1 1 – x 2 – tan –1 x
sin 3 x
is -
(B) 1
1
(D)
2
3.
If [y] = [sin x] ; ([.] denotes the greatest integer
function) and y = cos x are two given equations, then
the number of ordered pairs (x, y) is (A) 2
(B) 4
(C) 0
(D) infinitely many
4.
Let 20 distinct balls have been randomly distributed
into 4 distinct boxes, 5 into each. Let 'A' be the event
that two specific balls have been put into a particular
box. The probability of occurrence of event 'A' is :
1
4
(B)
(A)
19
19
8
2
(D)
(C)
19
19
DECEMBER 2010
5.
Let x2 ≠ nπ – 1, n ∈ N, then
∫x
2
2 sin( x + 1) – sin 2( x + 1)
2 sin( x 2 + 1) + sin 2( x 2 + 1)
(A) ln
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
2
dx is equal to :
1
sec( x 2 + 1) + C
2
 x2 +1
 +C
(B) ln sec
 2 


6.
(C)
1
ln |sec (x2 + 1)| + C
2
(D)
1
2
+C
ln
2 sec( x 2 + 1)
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
mz1 + z 2
, then distance of point z from the
m +1
line joining z1 + 1 and z2 + 1 is
(A) 0
(B) 1
2m
m
(C)
(D)
m +1
m +1
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
If z =
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
7.
The pairs of straight lines ax2 + 2hxy – ay2 = 0 and
hx2 – 2axy – hy2 = 0 are such that (A) one pair bisects the angle between the other pair
(B) the lines of one pair are equally inclined to the
lines of the other pair
(C) the lines of one pair are perpendicular to the lines
of the other pair
(D) these four lines are forming a rectangle
11.
Consider the curve f(x) = x ln x
Assertion (A) : f(x) = k will have no solution if
k ∈ (– ∞, –1/e)
Reason (R) : f(x) is decreasing in (0, 1/e) and
increasing in (1/e, ∞)
12.
Assertion (A) : ABCDA1B1C1D1 is a cube of edge 1
unit. P and Q are the mid points of the edges B1A1
and B1C1 respectively. Then the distance of the
vertex D from the plane PBQ is 8/3
Reason (R) : Perpendicular distance of point
(x1, y1, z1) from the plane ax + by + cz + d = 0 is
given by
8.
The equation x4 – (m – 3) x2 + m = 0 has (A) No real root iff m ∈ (0, 1)
(B) Four real and distinct roots iff m > 9
(C) No real root iff m ∈ (0, 9)
(D) Zero as its roots if m = 0
13.
9.
The differential equation of the curve for which
intercept cut by any tangent on y-axis is equal to the
length of the sub normal :
(A) is linear
(B) is homogeneous of first degree
(C) has separable variables
(D) is of first order
14.
10.
If in a triangle ABC, a, b,
are the altitudes from
respectively then
(A) p1, p2, p3 are in A.P.
3R
(C) p1 + p2 + p3 ≤
∆
XtraEdge for IIT-JEE
ax1 + by1 + cz1 + d
a 2 + b2 + c2
Consider the function f(x) = cos(sin–1x),
differentiable in (–1, 1)
Assertion (A) : f(x) is bounded in [–1, 1]
Reason (R) : If a function is differentiable in (a, b)
then it is bounded in [a, b]
x log t
Assertion (A) : F(x) =
dt then
1 1+ t + t2
F(x) = – F(1/x)
x log t
Reason (R) : If F(x) =
dt then
1 t +1
F(x) + F(1/x) = (1/2) (log x)2
∫
∫
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
c are in A.P. and p1, p2, p3
the vertices A, B, C
(B) p1, p2, p3 are in H.P.
3R
1
1
1
(D)
+
+
≤
p1
p 2 p3
∆
66
DECEMBER 2010
Passage # 1 (Ques. 15 to 17)
Passage # 3 (Ques. 21 to 23)
ax + by + c = 0 is a line touching the parabola
y2 = 4dx + e at (1/2, – 2). a & 3b are two A.M.
between c & 2d, where a, b, c are positive integers.
Consider a function y = f(x), f : R → R, is a
polynomial not more than 2 degree and f(x) satisfies
the following conditions
f(e) = 0, f(b) = c, f(a) = d
21. Area bounded between y2 = 4dx + e and y = f(x) will
be (in sq. units)
8
16
34
32
(A)
(B)
(C)
(D)
3
3
3
3
A spherical ball of diameter 10 2 cm is floating so
15.
16.
17.
that the top of the ball is 2 cm above the smooth
surface of the pond. Two point light sources are
placed on it along a vertical line such that one is at
1 cm above the water surface and another is 9 cm
below the water surface. The light sources emittes
the light rays which move tangentially.
The angle between the line of directions of light rays
is (A) 45º
(B) 90º
1
(C) 22 º
(D) 60º
2
What is the circumference in centimeters of the circle
formed by the contact of the water surface with the
ball (A) 3 2 π
(B) 6 2 π
(C) 9 2 π
(D) 12 2 π
The line ax + by + c = 0 will touch :
(B) (x – 1)2 + (y + 3)2 = 2
(A) x2 + (y + 1)2 = 1
2
2
(C) (x – 1) + (y – 2) = 5 (D) x2 + y2 = 8
23.
y = f(x), cuts y2 = 4dx + e at two points A & B.
Circum centre of the triangle formed by chord AB
and tangents at A & B on the parabola is :
(C) (8, 8) (D) (0, 0)
(A) (2, 2) (B) (6, 2)
If the ball is rotated in the vertical plane in which two
light sources lie, then the intersection point of light
rays will describe (A) linear path
(B) circular path
(C) elliptical path
(D) parabolic path
Passage # 2 (Ques. 18 to 20)
143 78 18
Consider N = 330 182 41 = pa qb rc sd where a, b,
429 234 61
Maths Facts
c, d ∈ I+ and p, q, r, s are prime numbers such that
p < q < r < s.
Let L1 and L2 are two lines defined as
x 
L1 : [p q]   = [a]
 y
x 
L2 : [r s]   = [– 20b]
 y
18.
22.
p + q + r + s is equal to :
(A) 23
(C) 25
(B) 31
(D) 33
Product of all factors of N is
(B) N6
(A) N4
8
(C) N
(D) N16
20.
Set of values of β for which point (0, β) lies in the
triangle formed by the lines L1 = 0, L2 = 0 and
L3 : x – y – 1 = 0 is :
1

 20 1 
(B)  – , 
(A)  – 1, 
7

 13 7 
XtraEdge for IIT-JEE
40 when written "forty" is the only number
with letters in alphabetical order, while "one"
is the only one with letters in reverse order.
•
1 googol = 10100;
1 googolplex = 10googol = 1010100 .
•
111 111 111 x 111 111 111
= 12345678 9 87654321
19.
 20 1 
(C)  – , 
 11 2 
•
 20 
(D)  – ,1
 11 
67
•
Pi (3.14159...) is a number that cannot be written
as a fraction.
•
If you add up the numbers 1-100
consecutively (1+2+3+4+5...) the total is
5050.
•
The billionth digit of Pi is 9.
•
1 and 2 are the only numbers where they are
the values of the numbers of factors they have.
•
2 and 5 are the only primes that end in 2 or 5.
•
The largest prime number is 9,808,358 digits
long; more than the number of atoms in the
universe.
DECEMBER 2010
XtraEdge for IIT-JEE
68
DECEMBER 2010
Based on New Pattern
IIT-JEE 2012
XtraEdge Test Series # 8
Time : 3 Hours
Syllabus :
Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
Instructions :
Section - I
•
Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct
answer and -1 mark for wrong answer.
•
Question 7 to 10 are multiple choice questions with multiple correct answer. +4 marks and No negative mark for
wrong answer.
Section - II
•
Question 11 to 14 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for
wrong answer.
•
Question 15 to 23 are passage based single type questions. +4 marks will be awarded for correct answer and
-1 mark for wrong answer.
PHYSICS
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1.
at
(m/s2)
60º
time (s)
A ball of mass 1 kg is released from position P inside
a wedge with a hemispherical cut of radius 0.5 m as
shown in the figure. The force exerted by the vertical
wall OA on wedge, when the ball is in position Q is
(neglect friction everywhere) (Take g = 10 m/s2) –
A
P
2.
XtraEdge for IIT-JEE
5.
A 2100 W continuous flow geyser (instant geyer) has
water inlet temperature = 10ºC while the water flows
out at the rate of 20 g/s. The outlet temperature of
water must be about (A) 20ºC
(B) 30ºC
(C) 35ºC
(D) 40ºC
(B) 15 N
(D) None
69
2 s
Heat is being supplied at a constant rate to a sphere of
ice which is melting at the rate of 0.1 g/s. It melts
completely in 100 s. The rate of rise of temperature
thereafter will be (Assume no loss of heat) (A) 0.8 ºC/s
(B) 5.4 ºC/s
(C) 3.6 ºC/s
(D) None of these
B
Tangential acceleration of a particle moving in a
circle of radius 1 m varies with time t as figure.
(initial velocity of particle is zero) Time after which
total acceleration of particle makes an angle of 30º
with radial acceleration is –
(D)
4.
Q
15 3
N
2
15
(C)
N
2
(C) 22/3 s
4
s
3
A man is standing on a rough (µ = 0.5) horizontal
disc rotating with constant angular velocity of
5 rad/s. At what distance from centre should he stand
so that he does not slip on the disc ?
(B) R > 0.2 m
(A) R ≤ 0.2 m
(C) R > 0.5 m
(D) R > 0.3 m
60º
(A)
(B)
3.
C
O
(A) 4s
DECEMBER 2010
6.
Two boys are separated by a distance 50 m and are
standing away from a vertical wall. When one of the
bopys claps, the other boy hears the echo after 1 s. If
velocity of sound in air is 330 ms–1 then distance of
the boy from the wall is –
P
v0
(A)
A
(A) 330 m
(C) 188 m
(C)
B
Q
(B) 818 m
(D) 881 m
10.
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
7.
A bus is moving with a velocity of 30 m/s towards a
huge wall. The driver sounds a horn of frequency
120 Hz. If the speed of sound in air = 330 m/s. Select
the correct option (A) Frequency received by wall is 120 Hz
(B) Frequency received by wall is 132 Hz
(C) Frequency at reflected wave observed by the
driver is 240 Hz
(D) Frequency of reflected wave observed by the
driver is 144 Hz
8.
2
A particle of mass 'm' is moving with an acceleration
a
towards left relative to the trolley car. James bond
3
accelerates towards left with an acceleration of
a
. If the car moves with a rightward
magnitude
3
acceleration 'a', the magnitude of pseudo-force acting
on 'm' viewed by James bond will be –
a/3
m
car
(A)
mg
3
(C) ma
a
2ma
3
4ma
(D)
3
(B)
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
µgR
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
A bob of mass m is projected with an upward
velocity v0 so that it moves in a vertical circle of
radius R in a vertical smooth circular tube. The
normal reaction on it is zero, the velocity of bob will
be –
XtraEdge for IIT-JEE
v0
a/3
(D) none
9.
 1
 v
(B) 
 3 0


v
(D) 0
3
v0
3
James bond
A cyclist rides along the circumference of a circular
horizontal plane of radius R, the friction coefficient
being dependent only on distance r from the centre O
r 

of the plane as µ = µ0 1 –  , where µ0 is a
 R
constant. Which of the following is/are correct ?
(A) The radius 'r' of the circle with the centre at the
point along which the cyclist can ride with the
R
maximum velocity, r =
2
(B) The radius 'r' of the circle with the centre at the
point along which the cyclist can ride with the
1
maximum velocity vmax =
µgR
2
(C) The maximum velocity vmax=
m
R
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
70
DECEMBER 2010
11.
Assertion (A) : While drawing a line on paper,
friction force acts on paper in the same direction
along which line is drawn on the paper.
Reason (R) : Friction always opposes motion.
12.
Assertion (A) : Work done by a conservative force is
always zero in round trip of the point of application
of force.
Reason (R) : Sometimes, a conservative force does
positive work, negative work and zero work; as a
whole, the net work done must be zero for a round
trip.
13.
Assertion (A) : A particle of mass 'm' is moving with
constant speed v along circular path of constant radius.
The net acceleration of the particle is constant in this
case.
Reason (R) : The tangential acceleration of the
particle moving along circular path is defined as the
rate of change of speed of the particle with time.
14.
Assertion (A) : The potential energy is only defined
for the conservative forces.
Reason (R) : In case of uniform circular motion, the
change in kinetic energy of the moving object is zero.
17.
Passage # 2 (Ques. 18 to 20)
With the help of Archimede's principle, one can
understand the floating nature and defects in metal
formation. One can easily find out the amount of
space left hollow in a sphere. For a body to float,
there should be a balance between the weight of the
body and the upthrust. The apparent weight felt
differs based on the volume immersed in the liquid.
More than one liquid may also balance the mass
while floating. In a frame accelerated down with 'a'
any mass will experience a normal force of m(g – a).
18. For a cubical block (ρ) to float in a pair of liquids of
density ρ1 and ρ2 as show, the relation between ρ, ρ1
and ρ2 is –
ρ1
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
ρ2
ρ1 + ρ 2
2
2ρ1 + ρ 2
(C) ρ =
2
(A) ρ =
Passage # 1 (Ques. 15 to 17)
A uniform rod of mass 'm' length 'l' is sliding along
its length on a horizontal table whose top is partly
smooth and rest rough with friction coefficient µ. If
the rod after moving through smooth part, enters the
rough with velocity v0.
l
v0
B Smooth A
µ (rough)
15. The magnitude of the friction force when its 'x' length
(<L) lies in the rough part during sliding will be µml 2 g
µmgl 2
(B)
(A)
2x
x
2
µmgx
µmgx
(C)
(D)
2l
l
16.
(C) 2
µgl
µgl
XtraEdge for IIT-JEE
(B)
2µgl
(D)
µgl
x
2x
2ρ 2 + ρ1
3
ρ + ρ2
(D) ρ = 1
3
(B) ρ =
19.
If the container in which a body floats in a liquid falls
under gravity, the upthrust felt by the body will be
(symbols carry usual meaning) (A) zero
(B) Vlρl g
Vl
(C)
ρl g
(D) Vbρb g
2
20.
An ice cube holding a steel ball floats in a cup of
water. After all the ice melts, the level in the cup
will(A) rise
(B) fall
(C) remains unchanged
(D) cannot be confirmed without know8ing density
of steel ball
Passage # 3 (Ques. 21 to 23)
In espresso coffee machines steam is passed into milk
at room temperature for a brief time interval. Some
of the steam condenses and the temperature or milk
rises. Since the time for which the steam is passed is
brief, one can ignore the heat lost to the environment
and assume that the usual assumption of calorimetry :
Heat lost = Heat gain is valid.
The minimum velocity v0 with which it must enter so
that it lies completely in rough region before coming
to rest, is (A)
If the velocity is double the minimum velocity (v0) as
calculated in above question then what distance does
its front end A would have travelled in rough region
before rod comes to rest ?
3l
5l
(B)
(A)
2
2
(C) 2l
(D) 3l
2
71
DECEMBER 2010
21.
Steam at 100ºC is passed into milk to heat it. The
amount of heat required to heat 150 g of milk from
room temperature (20ºC) to 80ºC is (specific heat of
capacity of milk = 4.0 kJ kg–1 K–1 specific latent heat
of steam = 2.2 MJ kg–1 , specific heat capacity of
water = 4.2 × 103 J kgK–1)
(B) 3.6 × 103 J
(A) 3.6 × 104 J
2
(D) None of these
(C) 3.6 × 10 J
22.
How many grams of steam condensed into water in
above question (A) 1.57 g
(B) 15.7 g
(C) 157 g
(D) None of these
23.
If some of heat is allowed to escape to surrounding
(temperature of surrounding is 20ºC) then this
amount of steam (mentioned in question 22) is
increase the temperature to (A) greater than 80ºC
(B) less than 80ºC
(C) equal to 80ºC
(D) can't say anything
6.
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
CHEMISTRY
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
1.
2.
Which of the following has the maximum number of
unpaired electrons?
(B) Ti3+
(A) Mg2+
3+
(D) Fe2+
(C) V
7.
The statements that are true for the long form of the
periodic table are :
(A) it reflects the sequence of filling the electrons in
the order of sub-energy level s, p, d and f.
(B) it helps to predict the stable valency states of the
elements
(C) it reflects trends in physical and chemical
properties of the elements
(D) it helps to predict the relative ionicity of the bond
between any two elements.
8.
Benzyl chloride (C6H5CH2Cl) can be prepared from
toluene by chlorination with?
(B) SOCl2
(A) SO2Cl2
(C) Cl2
(D) NaOCl
9.
Which of the following are examples of aldol
condensation?
dil
. NaOH
→ CH3CHOHCH2CHO
The equivalent weight of MnSO4 is half its molecular
weight when it is converted to :
(B) MnO2
(A) Mn2O3
(D) MnO42–
(C) MnO4–
(A) 2CH3CHO
The increasing order (lowest first) for the values of
e/m (charge/mass) for electron (e), proton (p),
neutron (n) and alpha particle (α) is :
(A) e, p, n, α
(B) n, p, e, α
(C) n, p, α, e
(D) n, α, p, e
(C) 2HCHO dil
. NaOH
→ CH3OH
3.
The molecule having one unpaired electron is :
(A) NO
(B) CO
(D) O2
(C) CN–
4.
For the chemical reaction 3X(g) + Y(g)
X3 Y(g)
the amount of X3 Y at equilibrium is affected by
(A) temperature and pressure
(B) temperature only
(C) pressure only
(D) temperature, pressure and catalyst
5.
The correct order of second ionisation potential of
carbon, nitrogen, oxygen and fluorine is
(A) C > N > O > F
(B) O > N > F > C
(C) O > F > N > C
(D) F > O > N > C
XtraEdge for IIT-JEE
(B) 2CH3COCH3 dil
. NaOH
→
CH3COH(CH3)CH2COCH3
. NaOH
→ C6H5CH2OH
(D) C6H5CHO + HCHO dil
10.
An aromatic molecule will (A) have 4n π electrons
(B) have (4n + 2) π electrons
(C) be planar
(D) be cyclic
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
72
DECEMBER 2010
16.
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
17.
(C) If (A) is true but (R) is false.
The structure of XeO3 is
(A) linear
(C) pyramidal
(B) planar
(D) T-shaped
XeF4 and XeF6 are expected to be
(A) oxidizing
(B) reducing
(C) unreactive
(D) strongly basic
(D) If (A) is false but (R) is true.
11.
12.
13.
14.
Passage # 2 (Ques. 18 to 20)
Riemer-Tiemann reaction introduces an aldehyde
group, on to the aromatic ring of phenol, ortho to the
hydroxyl group. This reaction involves electrophilic
aromatic substitution. This is a general method for
the synthesis of substituted salicylaldehyde as
depicted below.
Assertion (A) : Addition of bromine to trans-2-butene
yields meso-2,3-dibromobutane.
Reason (R) : Bromine addition to an alkene is an
electrophilic addition.
Assertion (A) : In water, orthoboric acid behaves as
a weak monobasic acid.
Reason (R) : In water, orthoboric acid acts as a
proton donor
30
Assertion (A) : Nuclide 13
Al is less stable than 40
20
Ca
Reason (R) : Nuclides having odd number of protons
and neutrons are generally unstable.
Assertion (A) : The electronic structure of O3
•• ⊕
O
is •
•
•O
••O•• Θ
•
••
••
Reason (R) : • O • structure is not allowed
•O
••O•
••
because octet around O cannot be expanded.
OH
CHO
[I]
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
CHO
aq.HCl
CH3
CH3
CH3
(I)
(II)
(III)
18.
Which one of the following reagents is used in the
above reaction?
(B) aq.NaOH + CH2Cl2
(A) aq.NaOH + CH3Cl
(D) aq. NaOH + CCl4
(C) aq.NaOH + CHCl3
19.
The electrophile in the reaction is (A) :CHCl
(B) +CHCl2
(D) CCl3
(C) :CCl2
20.
The structure of the intermediate I is
Θ ⊕
Θ ⊕
O Na
O Na
CH2Cl
(A)
Passage # 1 (Ques. 15 to 17)
The noble gases have closed-shell electronic
configuration and are monoatomic gases under
normal conditions. The low boiling points of the
lighter noble gases are due to weak dispersion forces
between the atoms and the absence of other
interatomic interactions.
The direct reaction of xenon with fluorine leads to a
series of compounds with oxidation numbers +2, +4
and +6. XeF4 reacts violently with water to given
XeO3. The compounds of xenon exhibit rich
stereochemistry and their geometries can be deduced
considering the total number of electron pairs in the
valence shell.
15. Argon is used in arc welding because of its (A) low reactivity with metal
(B) ability to lower the melting point of metal
(C) flammability
(D) high calorific value
XtraEdge for IIT-JEE
Θ ⊕
O Na
OH
CHCl2
(B)
CH3
CH3
Θ ⊕
Θ ⊕
O Na
O Na
CCl3
(C)
CH2OH
(D)
CH3
CH3
Passage # 3 (Ques. 21 to 23)
Several short-lived radioactive species have been
used to determine the age of wood or animal fossils.
One of the most interesting substances is 6C14 (halflife 5760 years) which is used in determining the age
of carbon-bearing materials (e.g. wood, animal
fossils, etc.). Carbon-14 is produced by the
bombardment of nitrogen atoms present in the upper
atmosphere with neutrons (from cosmic rays).
14
1
14
1
7N + 0n → 6C + 1H
73
DECEMBER 2010
Thus carbon-14 is oxidised to CO2 and eventually
ingested by plants and animals. The death of plants
or animals put an end to the intake of C14 from the
atmosphere. After this the amount of C14 in the dead
tissues starts decreasing due to its disintegration as
per the following reaction :
14
14
0
6C → 7N + –1β
14
The C isotope enters the biosphere when carbon
dioxide is taken up in plant photosynthesis. Plants are
eaten by animals, which exhale C14 as CO2.
Eventually, C14 participates in many aspects of the
carbon cycle. The C14 lost by radioactive decay is
constantly replenished by the production of new
isotopes in the atmosphere. In this decayreplenishment process, a dynamic equilibrium is
established whereby the ratio of C14 to C12 remains
constant in living matter. But when an individual
plant or an animal dies, the C14 isotope in it is no
longer replenished, so the ratio decreases as C14
decays. So, the number of C14 nuclei after time t
(after the death of living matter) would be less than
in a living matter. The decay constant can be
calculated using the following formula,
0.693
t1/2 =
λ
The intensity of the consmic rays have remain the
same for 30,000 years. But since some years the
changes in this are observed due to excessive burning
of fossil fuel and nuclear tests.
21.
22.
23.
(A) The age of the fossil will increase at the place
where explosion has taken place and
1 C
T1 – T2 = ln 1
λ C2
(B) The age of the fossil will decrease at the place
where explosion has taken place and
1 C
T1 – T2 = ln 1
λ C2
(C) The age of fossil will be determined to be same
T
C
(D) 1 = 1
T2
C2
MATHEMATICS
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct
answer and – 1 mark for each wrong answer.
Why do we use the carbon dating to calculate the age
of the fossil?
(A) Rate of exchange of carbon between atmosphere
and living is slower than decay of C14
(B) It is not appropriate to use C14 dating to
determine age
(C) Rate of exchange of C14 dating to determine age
organism is so fast that an equilibrium is set up
between the intake of C14 by organism and its
exponential decay
(D) none of the above
What should be the age of the fossil for meaningful
determination of its age?
(A) 6 years
(B) 6000 years
(C) 60,000 years
(D) can be used to calculate any age
A nuclear explosion has taken place leading to
increase in concentration of C14 in nearby areas. C14
concentration is C1 in nearby areas and C2 in areas
far away. If the age of the fossil is determined to be
T1 and T2 at the respective places then
XtraEdge for IIT-JEE
74
1.
z is a complex number satisfying |z – 3| ≤ 4 and
|ωz – 1 – ω2| = a (where ω is complex cube root of
unity) then :
(A) 0 ≤ a ≤ 2
(B) 0 ≤ a ≤ 8
(C) 2 ≤ a ≤ 8
(D) 2 ≤ a ≤ 4
2.
If α and β are roots of the equation
ax2 + bx + c = 0 then roots of the equation
a(2x + 1)2 – b(2x + 1) (3 – x) + c(3 – x)2 = 0 are;
2α + 1 2β + 1
3α + 1 3β + 1
(A)
,
,
(B)
α–3 β–3
α–2 β–2
2α + 1 2β + 1
,
(D) None of these
(C)
α–2 β–2
3.
The set {x : |1 – log1/5 x| + 2 = |3 – log1/5 x|} is
equal to
(A) (0, ∞)
(B) [1/5, ∞)
(C) [1/5, 5]
(D) (0, 1/5]
4.
If tan–1
5.
If E =
6.
The position vector of a point P is r = xi + yj + zk,
where x, y, z ∈ N and u = i + j + k. If r.u = 10 then
the possible positions of P are
(A) 72
(B) 36
(C) 60
(D) 108
a
b
c
d
π
+ tan–1
+ tan–1
+ tan–1
=
then
x
x
x
x
2
x4 – x2 Σab + abcd is equal to
(A) –1
(B) 0
(C) 1
(D) 2
30 31
1 2 3 4
. . . ...
.
= 8x, then value of x is
4 6 8 10 62 64
(A) – 7
(B) – 9
(C) – 10
(D) – 12
DECEMBER 2010
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) is correct. Mark
your response in OMR sheet against the question
number of that question. + 4 marks will be given for
each correct answer and NO NEGATIVE marks for
wrong answer.
7.
Tangents are drawn from the point (α, 3) to the circle
2x2 + 2y2 = 25 will be perpendicular to each other if
α equals :
(A) 5
(B) – 4
(C) 4
(D) – 5
11.
Assertion (A) : The ratio PR : RQ equals 2 2 : 5
Reason (R) : In any triangle bisector of an angle of
divides the triangle into two similar triangles.
30
8.
In the expansion of
 2/3
1 
 x
 , a term
–
x

12.
containing the power x13 :
(A) does not exist
(B) exists and the coefficient is divisible by 29
(C) exists and the coefficient is divisible by 63
(D) exists and the coefficient is divisible by 65
9.
Lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line
L3 : y + 2 = 0 at P and Q respectively. The bisector of
the acute angle between L1 and L2 intersect at R.
Assertion (A) : Let p < 0 and α1, α2, ... α9 be the
nine roots of x9 = p, then
α1 α 2 α 3
∆ = α4 α5 α6 = 0
α 7 α8 α9
Reason (R) : If two rows of a determinant are
identical then determinant equals zero.
Suppose a1, a2, ... real numbers, with a1 ≠ 0. If a1, a2,
a3, ... are in A.P. then
a1 a 2 a 3 
(A) A = a 4 a 5 a 6  is singular
a 5 a 6 a 7 
13.
(B) the system of equation a1x + a2y + a3z = 0,
a4x + a5y + a6z = 0, a7x + a8y + a9z = 0 has
infinite number of solutions
ia 2 
a
(C) B =  1
 is non singular
ia
 2 a1 
Assertion (A) : If
2 sin2 ((π/2) cos2x ) = 1 – cos (π sin 2x), x ≠ (2n + 1) π/2,
n is a integer, then sin 2x + cos 2x is equal to 1/5.
1 + 2 tan x – tan 2 x
Reason (R) : sin 2x + cos 2x =
1 + tan 2 x
14.
Consider the planes 3x – 6y – 2z = 15 & 2x + y – 2z = 5
Assertion (A) : The parametric equations of the line
of intersection of the given planes are x = 3 + 14t,
y = 1 + 2t, z = 15t.
(D) none of these
^
10.
intercept of 10 on the line y = 2x + 5/ 2 , which
subtends an angle of 45º at the origin is
(A) x2 + y2 – 4x – 2y = 0
(B) x2 + y2 – 2x – 4y = 0
(C) x2 + y2 + 4x + 2y = 0
(D) x2 + y2 + 2x + 4y = 0
^
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) Each question
has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct. Mark your response in OMR sheet
against the question number of that question. + 4
marks will be given for each correct answer and – 1
mark for each wrong answer.
Passage # 1 (Ques. 15 to 17)
A and B are two points on the boundary of a circular
field of radius R and centre O. ∠AOB = θ. A circle
with centre A and radius AB meets the circular field
again at C and the line AO produced at E. L, M are
points on the boundary of the field lying between C
and A, A and B, respectively.
15. AB is equal to
(A) R sin θ
(B) 2R sin (θ/2)
(C) R cos θ
(D) 2R cos (θ/2)
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
XtraEdge for IIT-JEE
^
Reason (R) : The vector 14 i + 2 j + 15 k is
parallel to the line of intersection of the given planes.
An equation of a circle through the origin, making an
75
DECEMBER 2010
16.
Area of the segment AMB is equal to
(B) (1/2) R2 sin θ
(A) (1/2) R2 θ
(D) none of these
(C) (1/2) R2 (θ – sin θ)
17.
If the area AMBECL is 1/nth of the field, then
sin θ + (π – θ) cos θ is equal to
n –1
(A) nπ
(B)
π
n
(C) (n – 1)π
(D) (n + 1)π
At a Glance
Some Important Practical Units
1.
Passage # 2 (Ques. 18 to 20)
Let ABCD be a square with each side of length 2
units. C2 is the circle through vertices A, B, C, D and
C1 is the circle touching all sides of the square
ABCD. L is a line through A.
18. If P is a point on C1 and Q is a point on C2, then
1 par sec = 3.26 light year
2.
3.
is equal to
QA 2 + QB 2 + QC 2 + QD 2
(A) 0.75
(B) 1.25
(C) 1
(D) 0.5
20.
23.
4.
5.
6.
7.
XtraEdge for IIT-JEE
Cusec : It is the unit of water flow.
1 cusec = 1 cubic foot per second flow
8.
Match No. : This unit is used to express velocity
of supersonic jets.
1 match no. = velocity of sound
= 332 m/sec.
9.
Knot : This unit is used to express velocity of
ships in water.
1 knot = 1.852 km/hour
10. Rutherford : It is the unit of radioactivity.
1 rutherford (rd) = 1 × 106 disintegrations/sec
11. Dalton : It is the unit of mass.
1 dalton =
If we drop the condition that the G.P. is strictly
increasing and take α2 = 3, then common ratio is
given by
(D) ± 3
Barn : It is the unit of area.
1 barn = 10–28 m2
If α = 1/2, S = 20, then the greatest value of the first
term is
(A) 10/3
(B) 7/3
(C) 1/3
(D) 3
(C) 0
Shake : It is the unit of time.
1 Shake = 10–6 second
A line M through A is drawn parallel to BD. Point S
moves such that its distances from the line BD and
the vertex A are equal. If locus of S cuts M at T2 and
T3 and AC at T1, then area of ∆T1T2T3 is
(A) 1/2 sq. units
(B) 2/3 sq. units
(C) 1 sq. unit
(D) 2 sq. units
(B) ± 1
Chandra Shekhar limit : It is the largest
practical unit of mass.
1 Chandra Shekhar limit = 1.4 × Solar mass
A circle touches the line L and the circle C1
externally such that both the circles are on the same
side of the line, then the locus of the centre of the
circle is(A) ellipse
(B) hyperbola
(C) parabola
(D) parts of straight line
(A) ± 2
Slug : It is the unit of mass.
1 slug = 14.59 kg
Passage # 3 (Ques. 21 to 23)
The sum of three terms of a strictly increasing G.P. is
αS and sum of the squares of these terms is S2.
21. α2 lies
(A) (1/3, 2)
(B) (1, 2)
(C) (1/3, 3)
(D) none of these
22.
X-ray unit : It is the unit of length.
1 X-ray unit = 10–13 m
PA 2 + PB 2 + PC 2 + PD 2
19
Par sec : It is the largest practical unit of
distance.
1
mass of C12 = 931 MeV
12
= 1 a.m.u.
12. Curie : It is the unit of radioactivity.
1 curie = 3.7 × 1010 disintegration / sec
76
DECEMBER 2010
MOCK TEST PAPER-1
CBSE BOARD PATTERN
CLASS # XII
SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS
So lu t ion s wi l l b e p ub lished in nex t issue
General Instructions : Physics & Chemistry
•
Time given for each subject paper is 3 hrs and Max. marks 70 for each.
•
All questions are compulsory.
Marks for each question are indicated against it.
•
Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each.
•
Question numbers 9 to 18 are short-answer questions, and carry 2 marks each.
•
•
Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each.
•
Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
Use of calculators is not permitted.
•
General Instructions : Mathematics
•
Time given to solve this subject paper is 3 hrs and Max. marks 100.
•
All questions are compulsory.
•
The question paper consists of 29 questions divided into three sections A, B and C.
Section A comprises of 10 questions of one mark each.
Section B comprises of 12 questions of four marks each.
Section C comprises of 7 questions of six marks each.
All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
•
•
There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and
2 question of six marks each. You have to attempt only one of the alternatives in all such questions.
Use of calculators is not permitted.
•
PHYSICS
1.
Define principal axis of a spherical mirror.
2.
Which of the following quantity – speed, wavelength,
frequency does not change when light goes from one
medium to another ?
3.
8. Why are infrared radiations referred to as heat waves
also? Name the radiations which are next to these
radiations in electromagnetic spectrum having
(i) Shorter wavelength.
(ii) Longer wavelength.
9.
Write the S.I. unit of
(i) electric field intensity and
(ii) electric dipole moment .
4.
State and explain Lenz's law.
5.
In an electromagnetic wave, at any point, the
amplitude of electric field is 3 × 106 N/C. Find
amplitude of magnetic field at that point.
6.
How is a sample of an n-type semiconductor
electrically neutral though it has an excess of negative
charge carriers ?
7.
Calculate the ratio of energies of photons produced due
to transition of electron of hydrogen atom from its,
(i) Second permitted energy level to the first level, and
(ii) Highest permitted energy level to the second
permitted level
XtraEdge for IIT-JEE
What would be the change in focal length of a
concave mirror when it is held in water ?
10. How width of central maxima changes in single slit
diffraction pattern when
(a) slit width is decreased
(b) distance between slit & screen is increased
(c) light of smaller wavelength is used.
11. Calculate the potential at the centre of square of
side 2 m, which carries at its four corners charges of
+2 nC, + 1 nC, – 2 nC and –3 nC respectively.
12. A voltage of 30 V is applied across a carbon resistor
with first, second and third rings of blue, black and
yellow colours respectively. Calculate the value of
current, in mA, through the resistor.
77
DECEMBER 2010
13. Write the order of frequency range and one use of
each of the following electromagnetic radiations:
(i) Microwaves
(ii) Ultra-violet rays
(iii) Gamma rays.
22. A bar magnet of magnetic moment M is aligned with
the direction of a uniform magnetic field B. What is
the work done, to turn the magnet, so as to align its
magnetic moment :
(i) opposite to the field direction and
(ii) normal to the field direction ?
14. Prove that the instantaneous rate of change of the
activity of a radioactive substance is inversely
proportional to the square of its half life.
23. A circular coil of N turns and radius R, is kept
normal to a magnetic field, given by B = B0 cos ωt.
Deduce an expression for e.m.f. induced in this coil.
State the rule which helps to detect the direction of
induced current.
15. The following table gives the values of work function
for a few photo sensitive metals
S. No. Metal Work Function (eV)
1.92
Na
1.
2.
K
2.15
3.
Mo
4.17
24.
If each of these metals is exposed to radiations of
wavelength 300 nm, which of them will not emit
photo electrons and why?
16. By how much would the stopping potential for a
given photo sensitive surface go up if the frequency
of the incident radiations were to be increased from 4
x 1015 Hz to 8 x 1015 Hz ?
Given h = 6.4 x 10–34 J-s, e = 1.6 x 10–19 C and
c = 3 x 108 ms–1
× ×
× ×
× ×
× ×
× ×
17. How is the band gap, Eg, of a photo diode related to
the maximum wavelength λm, that can be detected
by it ?
18. What does the term LOS communication mean ?
Name the types of waves that are used for this
communication. Which of the two-height of
transmitting antenna and height of receiving antennacan affect the range over which this mode of
communication remains effective ?
5Ω
×
×
×
×
× ×
× ×
× ×
× P×
28. (a) What are coherent sources of light ? Why is no
interference pattern observed when two slits of
YDSE are illuminated by two different sodium
lamps ?
(b) Obtain conditions of getting bright & dark fringes
YDSE and write the expression of fringe width.
10 Ω
A
10 Ω
XtraEdge for IIT-JEE
×
×
×
×
× ×Q×
27. What are the main assumptions of Bohr's Atomic
Model ? How it explain the linear spectrum of H-atom
? Write the drawbacks of Bohr's model.
10 Ω
5Ω
×
×
×
×
×
26. Define the term ‘modulation index’ for an AM wave.
What would be the modulation index for an AM wave
for which the maximum amplitude is ‘a’ while the
minimum amplitude is ‘b’ ?
21. Calculate the current shown by the ammeter A in the
circuit diagram given below:
10 Ω
×
235
Y , initially at rest,
25. The nucleus of an atom of 92
decays by emitting an α-particle as per the equation
235
231
4
92 Y → 90 X + 2 He + Energy
It is given that the binding energies per nucleon of
the parent and the daughter nuclei are 7.8 MeV and
7.835 MeV respectively and that of α-particle is
7.07MeV/nucleon. Assuming the daughter nucleus to
be formed in the unexcited state and neglecting its
share in the energy of the reaction, calculate the speed
of the emitted α-particle.Take mass of α-particle to be
6.68 x 10–27 kg.
19. In a double slit interference experiment, two coherent
beams has intensity I & I + δI where
δI << I.
Then show that intensity of maxima and minima are
(δI) 2
respectively.
4I &
4I
20. Using Gauss's theorem, show mathematically that for
any point outside the shell, the field due to a
uniformly charged thin spherical shell is the same as
if the entire charge of the shell is concentrated at the
center. Why do you except the electric field inside the
shell to be zero according to this theorem ?
10 Ω
When an inductor L and a resistor R in series are
connected across a 12 V, 50 Hz supply, a current of
0.5 A flows in the circuit. The current differs in
phase from applied voltage by π/3 radian. Calculate
the value of R.
Or
A 0.5 m long metal rod PQ completes the circuit as
shown in the figure. The area of the circuit is
perpendicular to the magnetic field of flux density
0.15 T. If the resistance of the total circuit is
3Ω,
calculate the force needed to move the rod in the
direction as indicated with a constant speed of 2 ms–1
78
DECEMBER 2010
11. Explain Van arkel method ?
Or
1
1
1
–
= , for concave
Derive the lens formula,
v
u
f
lens, using the necessary ray diagram. Two lenses of
power + 10D & – 5D are placed in contact,
(a) calculate power of the combination.
(b) find object position to obtain image position with
magnification 2.
12. What is specific rate or rate constant on which factors
it depends ?
13. For a Ist order chemical reaction rate constant is 2.303
sec–1. Determine the time required for 90%
completion of a chemical reaction
14. Name the reagents used to convert –
(i)
Propene to allyl bromide
(ii)
Methyl iodide to propyne
29. Explain with the help of a labelled diagram the
underlying principle and working of a step-up
transformer. Why cannot such a device be used to
[5]
step-up d.c. voltage ?
or
Draw a labelled diagram of an a.c. generator. Explain
briefly its principle and working.
15. The two isomeric aromatic compounds A and B have
formula C7H7OH. A gives purple colour with ferric
chloride solution while B does not. Suggest the
structures of A and B.
30. A cube has side a and charge q at its each corner.
Find potential energy of charge –q placed at its
centre.
or
Derive Gauss's theorem in Electrostatics.
16. Name the following compounds according to IUPAC
system :
(i) CH3 – CH = C – CH – CH2 – CH3
OH CH3
CHEMISTRY
1.
What is holme signal ?
2.
Give two ores of copper ?
3.
Name the simplest amino acid ?
4.
Draw a graph between t1/2 & [A0] for Ist order
reaction
5.
6.
7.
8.
Cl
(ii) C6H5 – CH – CH2 – CH2 –CH2OH
17. Identify A and B –
NH
H2
3 → A Ni
/ 
→ B
R2CO 
18. Complete the following equations supplying A & B
OH
(i)
Calculate the standard cell potential of the following
H2 / O2 fuel cell. Given that;
Eº = 1.229 V
O2 + 4H+ + 4e– → 2H2O
2 H2 → 4H+ + 4e¯
Eº = 0.0 V
Zn dust
–
A
H2SO4
SO3
B
+
(ii) C6H5 O Na + CH3Cl →
(i) What is meant by the term critical micellization
concentration (CMC) ?
(ii) What is the function of emulsifying agent ?
19. What is lanthanide contraction and explain its causes ?
20. Why [NiCl4]2– is paramagnetic but [Ni(CN)4]2– is
diamagnetic ? Explain
Name the reagent used for producing fluorobenzene
from benzene diazonium chloride. What is the name
of this reaction ?
21. Give the structures of three synthetic rubbers ?
Give the major products that are formed by heating
following ether with HI.
CH3
22. Classify amino acids with examples on the basis of
their production in body ?
23. What are tranquilizers. Give the types with examples.
CH3 – CH2 – CH – CH2 – O – CH2 – CH3
9.
24. Calculate the equilibrium constant for a
reaction Ni(s) + Cu+2(aq) → Cu(s) + Ni+2(aq)
Given the values of E 0Ni +2 / Ni and E 0Cu + 2 / Cu
Give the structure of N2O5 ?
10. Why d-block elements are used as catalysts ?
XtraEdge for IIT-JEE
as – 0.25 and 0.34 V respectively
79
DECEMBER 2010
25. Describe the following terms while stating the
properties of colloids :
(i) Brownian movement (ii) Tyndall Effect;
(iii) Electrophoresis
26. An alkene [A] on reductive ozonolysis yields acetone
and an aldehyde. The aldehyde is easily oxidised to B.
[B] on treatment with Br2 in the presence of
phosphorus yields a compound [C] which on further
hydrolysis gives a hydroxy acid [D]. The acid [D] can
also be produced from acetone by treatment with
HCN followed by subsequent acidic hydrolysis.
Identify A, B, C, D.
27.
CHCl3 / KOH
N3H / H2SO4
A
(An acid) Heat
B
HNO2
C
H3O
+
B + HCOOH
1
1
log b a
Evaluate :
4.
Let a relation R1 on the set R of real numbers be
defined as (a, b) ∈ R1 ⇔ 1 + ab > 0 for all
a, b ∈ R. Show that R1 is reflexive.
5.
If y = ax + ex + xa + xx then find
6.
Evaluate : ∫ 2 2 2 2 2 x dx
7.
Form the differential equation representing the family
of curves y = A cos (x + B), where A and B are
parameters.
8.
Find the value of λ so that the vectors
r
r
a = 2 î + λ ĵ + k̂ and b = î – 2 ˆj + 3 k̂ are
perpendicular to each other.
D+E
Reduction of C produces (CH3)3CNHCH3. Identify
A,B,C,D & E
28. Explain the structure of diborane (by orbital diagram)?
log a b
3.
9.
2x
dy
.
dx
x
r
r
Find the angle between two vectors a and b having
29. (i) Differentiate between reaction rate and reaction
rate constant.
(ii) An antifreeze solution is prepared from 222.6 g of
ethylene glycol (C2H6O2) and 200 g of water.
Calculate the molality of the solution. If the
density of the solution is 1.072 g mL–1 then what
shall be the molarity of the solution ?
(iii) What is azeotrope and what are the types of
azeotrope. Explain
10. Find the equation of the line passing through the
point (–1, 3, –2) and perpendicular to the lines
x + 2 y −1 z +1
x y z
= = and
=
=
.
−3
5
1 2 3
2
30. (a) What is the relationship between lattice parameter
a and r of the atom in.
(i) B.C.C.
(ii) F.C.C. (iii) H.C.P.
arrangements
(b) Explain the non-stoichiometric point defect
responsible for colour in alkali metal.
(c) Explain the following
(i) Ferromagnetic (ii) Ferrimagnetic substance
11. Using properties of determinant prove that
b+c c+a a+b
a b c
q+r r+p p+q =2 p q r
y+z z+x x+y
x y z
the same length
Section B
12. From a well shuffled pack of 52 cards, a card is
drawn at random, find the probability that it is either
a heart or a queen.
Or
A football match may be either won, drawn or lost by
the host country team. So there are three ways of
forecasting the result of any one match, one correct
and two incorrect. Find the probability of forecasting
at least three correct results for four matches.
MATHEMATICS
1.
Construct a matrix of order 2 × 2 whose element
2i − 3 j
aij =
2
2.
Find x, y, a, b so that
3  1 − 2 3 
 2 x − 3y a − b
=
 1
x + y 3a + 4b 1 6 29

XtraEdge for IIT-JEE
2 and their scalar product is –1.
 36 x − 9 x − 4 x + 1

, x≠0
13. f(x) =  2 − 1 + cos x

k,
x=0

f(x) is continuous at x = 0 then k = ?
80
DECEMBER 2010
dy
=
dx
25. Using integration, find the area of the region bounded
by the line 2y = – x + 8, x-axis and the lines x = 2
and x = 4.
x 2 + a 2 )n then prove that
14. If y = (x +
ny
2
a + x2
26. Find the image of the point having position vector
r
î + 3 ĵ + 4 k̂ in the plane r . (2 î – ĵ + k̂ ) + 3 = 0.
dy
a−x
, – a < x < a then find
.
a+x
dx
Or
If x = a (θ – sin θ) and y = a (1 – cos θ),
15. If y = tan–1
find
d2y
dx
2
at θ =
27. A dealer wishes to purchase a number of fans and
sewing machines. He has only Rs.5760.00 to invest
and has space for at most 20 items. A fan costs him
Rs.360.00 and a sewing machine Rs.240.00. His
expectation is that he can sell a fan at a profit of
Rs.22 and a sewing machine at a profit of Rs.18.
Assuming that he can sell all the items that he can
buy, how should he invest his money in order to
maximize his profit? Translate this problem
mathematically and then solve it.
π
.
2
16. If the tangent to curve y = x3 + ax + b at (1, –6) is
parallel to line x – y + 5 = 0 then find a, b.
17. Show that f : R – {–1} → R – {1} given by
x
f(x) =
is invertible. Also find f–1.
x +1
18. Evaluate :
∫ sin
19. Evaluate :
∫ cos 2x cos 4x dx
4
1
2 1

28. If A = 2 0
1  , then find A–1.Using A–1, solve
0 − 2 − 1
x dx
the following system of linear equations
2x + y + z = 3
2x + z = 5
–2y –z = 1
Or
20. Obtain the differential equation of all circles of radius r.
Or
dy
Solve the differential equation : x
− y − 2x 3 = 0
dx
1 2 − 3 
Find A
where A = 2 3
2  . Using
3 − 3 − 4
–1
A solve the following system of linear equations
x + 2y –3z = –4
2x + 3y + 2z = 2
3x –3y –4z = 11
–1
r r r
r r r r r r r r
21. If a , b , c are vectors that a . b = a . c , a × b = a × c ,
r r
r r
a ≠ 0 , then show that b = c .
Or
→
→
If a and b are unit vectors and θ is the angle
θ 1→ →
between them, then prove that cos = a + b
2 2
22. Show that the lines
29. Find the local maxima or local minima for
f(x) = sin x + cos x, 0 < x < π/2.
Or
Find two positive numbers whose sum is 14 and the
sum of whose squares is minimum.
x − 1 y + 1 z −1
x+2
=
=
and
=
3
2
5
4
y −1 z + 1
=
do not intersect.
3
−2
Section C
1
23. Evaluate :
∫ cot
−1
(1 − x + x 2 ) dx
0
24. A man speaks truth in 80% of the cases and another
in 90% of the cases. While stating the same fact,
what is the probability that they
(A) contradict
(B) Agree
XtraEdge for IIT-JEE
81
DECEMBER 2010
XtraEdge Test Series
ANSWER KEY
IIT- JEE 2011 (December issue)
PHYSICS
Ques
Ans
Ques
Ans
Ques
Ans
1
C
9
A,B,C,D
17
2
A
10
A,B,C
18
3
B
11
B
19
4
A
12
A
20
5
B
13
D
21
6
A
14
D
22
7
A,B,C
15
C
23
B
B
D
C
C
A
D
Ques
Ans
Ques
Ans
Ques
Ans
1
A
9
B,D
17
2
C
10
A
18
3
D
11
B
19
4
B
12
A
20
5
A
13
B
21
6
C
14
D
22
7
A,C,D
15
A
23
A
D
D
B
B
D
D
8
A,C,D
16
D
C HE M ISTR Y
8
B,C,D
16
C
MATHEMATICS
Ques
Ans
Ques
Ans
Ques
Ans
1
B
9
A,B,D
17
2
D
10
B,D
18
3
D
11
A
19
4
A
12
D
20
5
B
13
C
21
6
B
14
D
22
7
A,B
15
B
23
B
D
C
A
D
C
B
8
B,C,D
16
B
IIT- JEE 2012 (December issue)
Ques
Ans
Ques
Ans
Ques
Ans
1
A
9
B
17
2
C
10
B
18
3
A
11
C
19
B
B
A
Ques
Ans
Ques
Ans
Ques
Ans
1
B
9
A,B
17
2
D
10
B,C,D
18
3
A
11
B
19
A
C
C
Ques
Ans
Ques
Ans
Ques
Ans
1
B
9
A,B,C
17
2
B
10
B,D
18
3
B
11
C
19
B
A
C
XtraEdge for IIT-JEE
PHYSICS
4
A
12
A
20
5
C
13
D
21
6
C
14
B
22
7
B,D
15
D
23
B
A
A
B
C HE M ISTR Y
4
A
12
C
20
5
C
13
C
21
6
D
14
A
22
7
A,C,D
15
A
23
B
C
B
A
MATHEMATICS
4
B
12
D
20
5
D
13
D
21
6
B
14
D
22
7
B,C
15
B
23
C
D
A
B
82
8
A,B
16
A
8
C
16
C
8
B,C,D
16
C
DECEMBER 2010
Subscription Offer for Students
'XtraEdge for IIT-JEE
C
IIT JEE becoming more competitive examination day by day.
Regular change in pattern making it more challenging.
"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront.
Every month get the XtraEdge Advantage at your door step.
✓
✓
✓
✓
✓
✓
Magazine content is prepared by highly experienced faculty members on the latest trend of IIT JEE.
Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.
Take advantage of experts' articles on concepts development and problem solving skills
Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.
Confidence building exercises with Self Tests and success stories of IITians
Elevate you to the international arena with international Olympiad/Contests problems and Challenging Questions.
SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE
The Manager-Subscription,
“XtraEdge for IIT-JEE”
Career Point Infosystems Ltd,
4th Floor, CP-Tower,
IPIA, Kota (Raj)-324005
Special
Offer
I wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”
Half Yearly Subscription (Rs. 100/-)
One Year subscription (Rs. 200/-)
Two year Subscription (Rs. 400/-)
Money Order (M.O)
Bank Demand Draft of No………………………..Bank………………………………………………………..Dated
(Note: Demand Draft should be in favour of "Career Point Infosystems Ltd" payable at Kota.)
Name:
_______________________________________________________________________________________
Father's Name:
_______________________________________________________________________________________
Address: _______________________________________________________________________________________
________________________City_____________________________State__________________________
PIN_________________________________________Ph with STD Code __________________________
Class Studying in ________________E-Mail: ________________________________________________
From months:
XtraEdge for IIT-JEE
____________________to ________________________________________________
83
DECEMBER 2010
C
I am paying R. …………………….through
Subscription Offer for Schools
XtraEdge for IIT-JEE
C
IIT JEE becoming more competitive examination day by day.
Regular change in pattern making it more challenging.
"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront.
Every month get the XtraEdge Advantage at your door step.
✓
✓
✓
✓
✓
✓
Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE.
Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.
Take advantage of experts' articles on concepts development and problem solving skills
Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.
Confidence building exercises with Self Tests and success stories of IITians
Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions.
FREE SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE
C
The Manager-Subscription,
“XtraEdge for IIT-JEE”
Career Point Infosystems Ltd,
4th Floor, CP-Tower,
IPIA, Kota (Raj)-324005
We wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”
Half Yearly Subscription
One Year subscription
Two year Subscription
Senior Secondary School
Higher Secondary School
Institution Detail:
Graduate Collage
Name of the Institute:
_____________________________________________________________________________
Name of the Principal:
_____________________________________________________________________________
Mailing Address:
_____________________________________________________________________________
__________________City_________________________State__________________________
PIN_____________________Ph with STD Code_____________________________________
Fax_______________________________ E-Mail_____________________________________
Board/ University: _____________________________________________________________________________________
School Seal with Signature
XtraEdge for IIT-JEE
84
DECEMBER 2010