Students' Forum - Career Point

Volume - 5 Issue - 8
February, 2010 (Monthly Magazine)
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[B.Tech. IIT-Delhi]
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Editor : Pramod Maheshwari
Dear Students,
Success is the result of inspiration, aspiration, desperation and
perspiration. Your luck generally changes as a result of the above
inputs. There is always a fierce struggle on the road leading to your
desired goal. No triumph comes without effort and changing obstacles
into tools for success. For achieving your desired goal you must have a
fierce and a strong desire to succeed.
Success begins in the mind. Whatever the mind of man can believe and
conceive it can achieve. You should make a commitment to yourself to
succeed. Do not believe that success is the result of either an accident
or a fluke. Nothing is ever achieved without working for it with
integrity, wisdom, commitment and a success oriented attitude. You
have to accept responsibility for your decisions which will ultimately
determine your destiny. The best way to succeed is to accept and learn
from your mistakes. It is a fact that luck only follows hard work.
Master the fundamentals of whatever you are required to do keep on
developing your character confidence, values, beliefs and personality by
always keeping in view the examples and lives of successful people. You
will notice that all successful people combine in themselves integrity,
unselfishness, patience, understanding, conviction, courage, loyalty and
self-esteem. They have both ability and character. Hence, they are
successful in whatever they undertake.
No success is possible unless you believe that you can succeed.
Positive faith in yourself is both vital and crucial for success. This
attitude will help your competence reach visibly successful levels of
performance and prepare you for hard work.
Look at the task and ask yourself whether your output can be further
improved. There is little room at the top. The top is always rarefied
and limited space. Only one person can be there at a time. Do not be
the one who gets left out for want of persistence, determination and
commitment.
Forever presenting positive ideas to your success.
Yours truly
Pramod Maheshwari,
B.Tech., IIT Delhi
XtraEdge for IIT-JEE 1 FEBRUARY 2010

XtraEdge for IIT-JEE 2 FEBRUARY 2010

Volume-5 Issue-8
February, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Key Concepts & Problem Solving strategy for IIT-JEE.
INDEX
CONTENTS
Regulars ..........
PAGE
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics, Chemistry & Maths
Much more IIT-JEE News.
IIT-JEE Mock Test Paper with Solution
AIEEE & BIT-SAT Mock Test Paper with Solution
Success Tips for the Month
• If you haven nothing else to do, look about
you and see if there is not something close
at hand that you can improve !
• He has achieved success who has worked
well, laughed often, and loved much.
• You always pass failure on the way to
success.
• A journey of a thousand miles begins with
a single step.
• Your success will be largely determined by
your ability to concentrates singlemindedly
on one thing at a time.
• Success is a journey, not a destination.
• Success comes in "Cans". Failure comes in
"Can't".
• Success seems to be largely a matter of
hanging on after others have let go.
NEWS ARTICLE 4
IIT Madras team wins New York competition
Bio-energy centre launched at IIT-Kharagpur
IIT-Kharagpur best technology school in India-survey
IITian ON THE PATH OF SUCCESS 8
Dr. Pradip K. Dutta
KNOW IIT-JEE 10
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS 15
8-Challenging Problems [Set# 10]
Students’ Forum
Physics Fundamentals
Prism & Wave Nature of Light
Waves & Doppler Effect
CATALYST CHEMISTRY 34
Key Concept
Carbonyl Compounds
Co-ordination Compound & Metallurgy
Understanding: Physical Chemistry
DICEY MATHS 42
Mathematical Challenges
Students’ Forum
Key Concept
Integration
Trigonometrical Equation
Test Time ..........
XTRAEDGE TEST SERIES 57
Class XII – IIT-JEE 2010 Paper
Class XII – IIT-JEE 2011 Paper
Mock Test CBSE Pattern Paper-3 [Class # XII]
Mock Test CBSE Pattern Paper-2 & 3 (Solution) [Class # XII]
XtraEdge for IIT-JEE 3 FEBRUARY 2010

IIT Madras team wins New
York competition
New York: An entrepreneur team
from the Indian Institute of
Technology, Madras (IIT-M) has
won the NYC Next Idea 2009-
2010, an inaugural global business
plan competition launched by
New York City last...
Organs may soon be
grown like nails at IIT Delhi
New Delhi: "Science is a cemetery
of dead ideas," said an eminent
scientist but engineers at the
Indian Institute of Technology,
Delhi (IIT-D) believe in re-creating
those dead ideas and harvest
new. ...
India developing e-dog to
sniff out explosives
Thiruvananthapuram:
Indian scientists are developing an
electronic device that will sniff out
explosives like RDX, which remain
undetected by existing security
equipments."A prototype of the
e-device.
India develops affordable
nano sensors to detect
heart attack
Thiruvananthapuram: A team of
Indian scientists and engineers has
developed affordable sensors using
nano materials to detect a heart
attack quickly.
Bio-energy centre launched
at IIT-Kharagpur
West Bengal: The country's first
bio-energy centre was launched at
the Indian Institute of Technology-
Kharagpur (IIT-K) on Tuesday for
undertaking research, teaching and
technological implementati.
When there are no
teachers to teach at IITs
Mumbai: Luring requires no effort
if the perks are high and in the
field of employment what does
one need if he is placed in one of
the prestigious centrally funded
technical institutions (collectivel).
'Share and Share Alike'- IIT
Bombay's motto
Mumbai: 'Share and share alike' is
what the Indian Institute of
Technology, Bombay (IIT-B) is
spreading across the message to
reach out to the smaller
engineering colleges to share their
exper.
DRDO, IIT-D joins hands
for weather forecast system
Chandigarh: In order to develop
an indigenous capability and
methodology for long term
forecast of weather, the Defence
Research and Development
Organization (DRDO) in its first
kind of venture.
IIT–Kharagpur best
technology school in
India – survey
New Delhi: The premier Indian
Institute of Technology (IIT)
Kharagpur is the best technology
school in the country followed by
IIT-Delhi, revealed a new survey
released recently.
Crack the IIT code, it's too
easy
Kolkata: From next year, the IIT
entrance test is likely to get
simpler.
Concerned over the immense
stress that IIT-JEE puts on
thousands of students, the Union
HRD ministry has set up a highlevel
panel to modify the test
pattern.
Of the 1.5 lakh aspirants every
year, only 3,500 make it to the
seven IITs. To tone down the
gruelling test, the ministry has
formed a committee with teachers
from IITs and representatives of
the two +2 level national boards -
CBSE and ICSE.
According to sources, the first
change may be to limit questions
to the +2 syllabus. "The HRD
ministry feels many of the IIT-JEE
questions are based on topics that
are not taught at the +2 stage, and
are, in fact, of a far advanced
standard. This forces candidates to
start preparing at least three years
in advance - from Class IX itself.
They overload themselves and this
leads to depression, which
sometimes leads to suicides,"
IIT-Kharagpur director SK Dubey
told TOI.
XtraEdge for IIT-JEE 4 FEBRUARY 2010

Dubey and IIT-JEE chairman VK
Tewari are on the committee,
which is expected to submit its
recommendations by July. The
changes could be introduced from
IIT-JEE, 2006.
The government believes the
tougher syllabus forces students
to neglect their board
examinations for IIT-JEE. They also
end up spending a lot on coaching
classes that claim to be tailormade
for the entrance test, said
Dubey.
The committee, therefore, is
singling out topics that are... ...not
taught in the +2 stage anywhere in
the country.
These topics will probably be
deleted. And every topic that is
now included in IIT-JEE will be
vetted by the representatives of
CBSE and ICSE.
The two-tier exam system
includes objective-type questions
in the prelims and subjective ones
in the finals. The panel will
examine whether it should be
replaced with a uniform system.
In the finals, candidates have to
answer three papers - physics,
chemistry and mathematics -
through a gruelling six hours on a
single day.
The ministry feels it is too taxing
and has asked the committee to
work out a better "fatigue and
rest cycle."
The committee will also compare
IIT-JEE question papers with those
of the All India Engineering
Entrance Examination (AIEEE) and
other engineering entrance
examinations conducted by
various states to check if there is
too wide a gap in their standards.
IITs move to hike fee,
adopt IIM fee strategy
New Delhi : Taking a cue from the
Indian Institutes of Management,
the IIT bosses are drawing a cautious
plan to gradually equate their fee
structure with that of the IIMs.
According to sources the exercise
is to make the Indian Institutes
Technology self-reliant and to cut
dependence on state subsidy,
which the IIT dons say, would
gradually taper off in the coming
years.
A panel set up by the IIT Council
— the apex decision making body
— headed by atomic energy chief
Anil Kakodkar has been asked to
draft the roadmap for gradual fee
hikes, the sources said.
Drafting the fee hike roadmap for
the IITs is one of the components
of the mandate of the Kakodkar
panel set up at the Council
meeting on October 19. The
Kakodkar panel has been asked to
submit its report in six months.
The IIT Council, which met here
on October 19, discussed the feehike
possibility in view of the
government starting a loan
scheme with subsidised interest
rate to help poor students in
higher studies, sources said. The
Kakodkar panel will also suggest
how the IITs should increase the
number of scholarships,
fellowships and other financial aid
to ensure that deserving but
economically weak students do
not suffer from the hike, sources
said.
The new fee-hike strategy aims at
following the IIM practice of a
gradual but regular fee hike
supported by an increase in
financial assistance for those
students who cannot afford the
new fee structure.
“The strategy of gradual fee hikes
will allow us, for the first time, an
opportunity to hike fees
commensurate with rising costs,”
an IIT director said.
The IITs had a fixed tuition fee of
Rs 25,000 per annum for
undergraduate and post-graduate
science students for 10 years
before the fees were doubled last
year — to Rs 50,000 a year. But
even with the new fee structure,
the IITs earn only Rs 2 lakh for
four years of undergraduate
teaching, or Rs 1 lakh for two
years of the masters in science
programme from each student.
The top IIMs — which typically
raise their fees each year — in
contrast earn around 10 times as
much through tuition fees from
each student over comparable
course lengths.
IIM Ahmedabad, for instance
raised the fees for its two-year
postgraduate diploma in
management to Rs 12.5 lakh this
year, from Rs 11.5 lakh last year.
The IIMs in Bangalore and
Calcutta charge Rs 9.5 lakh and Rs
9 lakh for their two year
postgraduate diploma courses
respectively.
Six IITs figure among top
ten technology institutes
PTI, 14 January 2010, 12:19pm IST
KOLKATA: As many as six among
the top 10 technology institutes in
the country are from the
XtraEdge for IIT-JEE 5 FEBRUARY 2010

prestigious Indian Institute of
Technology, a survey has revealed.
The top slot in the pecking order
is occupied by the IIT- Kharagpur
followed by IIT-Delhi, IIT-Madras,
IIT-Kanpur, IIT- Roorkee and IIT-
Guwahati, according to the 5th
IDC-Dataquest T-School 2009
Survey of 111 engineering colleges
across the country.
IIT-Bombay did not participate in
the survey, global market
intelligence provider IDC said.
The seventh best technology
institute of the country is IIIT,
Hyderabad followed by BITS Pilani,
the survey report said.
IIIT, Hyderabad is the youngest
T-School, set up in 1998, in the
top 10 list.
Close behind BITS Pilani is the
National Institute of Technology
(NIT), Surathkal and the Institute
of Technology (IT) of the Banaras
Hindu University (BHU), Varanasi,
it said.
IIT Bombay exhibits
masterpieces
From collapsible bamboo bar
stools to workstations for people
with cerebral palsy, the Degree
Design Show by IIT-Bombays
Industrial Design Centre has it all.
A number of designs have been
sanctioned and designed by
corporates.
The youngsters behind the
machines have designed a number
of socially relevant products for
the less privileged.
Sarabjit Singh Kalsis workstation
for cerebral palsy patients has a
circular grip at the base of the
rotating chair that keeps it in
place, a C-shaped table that helps
the patient maintain the right
posture and an abductor in the
chair that separates the persons
legs, as crossing the legs is harmful
to those with cerebral palsy.
IIT Kharagpur ties up with
Taiwan school
IIT Kharagpur ties up with Taiwan
school The Indian Institute of
Technology at Kharagpur (IIT-
KGP) has is in the process of
signing a memorandum of
understanding with National Chiao
Tung University (NCTU) of
Taiwan which will facilitate
research collaboration, joint
research, joint mentoring for
students and student and faculty
exchange especially in the field of
chip designing and fabrication.
While the Indian market has
facilities and trained manpower for
chip designing, those for chip
fabrication are not available.
Although IIT-KGP has a VLSI (very
large scale integration) designing
laboratory, we are still trying to
set up a chip laboratory.
Ajai Chowdhry to chair IIT
Hyderabad's BoG
New Delhi, India: Hardware,
services and ICT system
Integration company HCL
Infosystems Ltd today announced
that Ajai Chowdhry, the founder
chairman and CEO of the
company has been nominated by
the Honorable President of India
to be the Chairman of the Board
of Governors, IIT Hyderabad.
Accepting his nomination, Ajai
Chowdhry said it is a matter of
immense pride and privilege to be
associated with this esteemed
knowledge body.
"I look forward to furthering IIT
Hyderabad's objectives of setting
new standards in engineering
practice in India contribute
actively to growth of India in the
decades to come," he added.
Commenting on the appointment
Prof. U. B. Desai, Director, IIT
Hyderabad expressed confidence
that under his Chairmanship IIT
Hyderabad would become a
world-class institute.
"We are very privileged and
enthused that Ajai Chowdhry has
been appointed as chairman BoG
for IIT Hyderabad in its formative
years. His vision will add new and
challenging dimensions to IIT."
An engineer by training,
Chowdhry is one of the six
founder members of HCL,
according to a press release. He is
currently the Chairman of
Confederation of Indian Industry's
(CII) National Committee on
Technology and innovation,
besides being a part of the IT
Hardware Task Force set up by
the Prime Minister of India.
Chowdhry also has chaired CII's
National Committee for IT, ITES
& E-Commerce, where he actively
encouraged the deployment of IT
in Indian SMEs to increase their
productivity and to make them
globally competitive.
Number of women at IITs
triples in 5 years
In the last five years, women’s
presence in certain IITs has gone
as high as 10 percent. In 2005,five
per cent of the total number of
those who cleared the JEE were
women(381 out of 6,433). In 2009,
this number has increased to 10
per cent (1,048 of 10,035).
“The number of applications from
women has also increased and
courses at IIT are no longer
viewed as only-for-men. Even
women are interested in technical
fields,” said Anil Kumar, IIT-
Bombay’s JEE chairperson.
XtraEdge for IIT-JEE 6 FEBRUARY 2010

IBM collaborates with
engineering colleges to
set up COE
IBM India has collaborated with six
engineering colleges of Tamil
Nadu for setting up Centre of
Excellence (COE) to promote high
quality education by providing
state-of the-art technologies in
colleges with the objective of
nurturing highly skilled computer
professionals.The colleges will
provide infrastructure and high
end systems while IBM will extend
its entire range of software suite
free of charge. This will enable
students to learn new skill sets on
IBM software products including
DB2, WebSphere, Lotus, Rational,
and Tivoli.
HRD suggests Engineers
Bill to standardize
engineers
In order to standardize the Indian
engineers globally, the HRD
ministry is planning to introduce
the Engineers Bill, 2009, that
would require obligatory
certification of professional
engineers. The bill aims to
streamline the quality of engineers
in the country and plans to set up
the ICE, which will maintain a
national and international record
of professional engineers and
associate professional engineers
and will standardize the
engineering profession.
To Set up a Campus in
Quatar
India’s premier technological
institutes, Indian Institutes of
Technology (IITs) has reportedly
been given a nod by the central
government to set up their first
offshore campus in Qatar.
The new institute to be set
up in Quatar will be called
the International Institute of
Technology and this will be set up
all the IITs in a concerted effort.
The entire process of setting up a
new institute in Quatar will be coordinated
by IIT Council, the top
decision-making body for all IITs.
Though the HRD ministry has
initially opposed to IITs or Indian
Institute of Managements
venturing abroad, arguing that
“elite” educational institutions
must stay focused on India alone,
the new dispensation in the
ministry is keen that Brand India
makes a mark abroad and for this
reason IITs has got the go-ahead
signal from the Indian Human
Resource Development Ministry.
IIT-Qatar will be set up in Qatar
Foundation’s Education City,
which already has branch
campuses of six noted universities.
Top IT companies continue
to flock IITs
With the economic recession
taking backstage, top IT companies
are quite happy to offer huge
salaries to the IIT graduates .
Training and Placement head at IIT
Kharagpur, Suneel Srivastava, said
that the highest number of offers
(13) and the highest salary package
of Rs 22 lakh was offered by
Barclays Singapore. The institute
also had about 27 pre-placement
offers made to the students, about
10 of which were made by
Reliance Industries (RIL).
IITs to reduce foreign
student fee at PG level
Under the chairmanship of HRD
minister Kapil Sibal, the first IIT
Council is eyeing for more foreign
admissions at the PG level by
framing a policy that includes
introducing scholarships for
studying at IIT and fee reduction
for PG students in the IITs.
Permitting IITs to create extra
seats for foreign PG students to
ensure that youth from other
countries take part in R&D in a big
way is also being considered by
the IIT council.
IIT Kanpur to hold B-Plan
Competition in Techkriti
IIT Kanpur is going to organize
its annual festival Techkriti on
11-14 February 2010. One of the
flagship events of Techkriti will be
‘Ideas 10’, the international
business plan competition. The
event will witness participation
from leading MBA schools,
such as IIMs and XLRI as well as
IITs and many international
institutes like Stanford University,
Cornell University, University of
Purdue and the National
University of Singapore. Cut
throat competition from the
best in the world will give
participants tremendous exposure
and will be an excellent test of
their business acumen. There are
high chances for the participants
to start a company based on the
B-plan.
IDEAS is the platform where the
seeds of future business tycoons
will be laid down. Ideas will have
best prizes for Bio Business plan
Competition, Web Business idea,
Clean Energy Solutions and Social
Business Plan. This year IDEAS has
American Embassy as its cosponsor
and NEN, VC Hunt,
Rajeev Kumar Foundation as its
associates.
XtraEdge for IIT-JEE 7 FEBRUARY 2010

Success Story
This article contains story of a person who get succeed after graduation from different IIT's
Dr. Pradip K. Dutta
B.Tech -Electronics Engineering IIT Kharagpur,
MS & Ph.D.
Corporate Vice President & Managing Director,
Synopsys (I) Pvt. Ltd.
Dr. Pradip K. Dutta is the Corporate VP & MD of
Synopsys (India) Private Limited, a wholly owned
subsidiary of Synopsys Inc., a world leader in
Electronic Design Automation (EDA) software. His
primary focus is to position Synopsys as a leader in
the Indian semiconductor eco-system and help foster
its growth through partnership with government,
academia and industry. Dr. Dutta has been heading
the India operations since 2000, overseeing the
growth from a little over 50 employees operating from
one small office in Bangalore to more than 600 highly
skilled employee base spread across Bangalore,
Hyderabad and Noida.
Prior to joining Synopsys, Dr. Dutta started his career
in the field of automotive electronics with General
Motors in USA and held a variety of positions in
engineering and management both in the US and in
the Asian-Pacific region.
Dr. Dutta has earned his B.Tech in Electronics
Engineering from IIT Kharagpur followed by MS and
Ph.D. in Electrical Engineering from the University of
Maryland, College Park under a US government
fellowship grant from National Institute of Standards
and Technology. He sits on the Advisory Board of the
Govt. of West Bengal (IT Ministry) and is a member
of Executive Committees of several industry
associations
Adventure :
• Adventure is not outside man; it is within.
• There are two kinds of adventures: those who go truly hoping to find adventure and those who go secretly.
hoping they won't.
• Life is either a daring adventure or nothing.
• Some people dream of worthy accomplishments while others stay awake and do them.
• Life is an adventure. The greatest pleasure is doing what people say you cannot do.
XtraEdge for IIT-JEE 8 FEBRUARY 2010

KNOW IIT-JEE
By Previous Exam Questions
PHYSICS
1. A small ball of mass 2 × 10 –3 kg having a charge of
1 µC is suspended by a string of length 0.8 m.
Another identical ball having the same charge is kept
at the point of suspension. Determine the minimum
horizontal velocity which should be imparted to the
lower ball so that it can make complete revolution.
[IIT-2001]
Sol. This is a case of vertical of circular motion. The body
undergoing vertical circular motion is moving under
the action of three forces as shown
(i) mg (Gravitation pull)
(ii) Electrostatic force of repulsion
(iii) Tension of the string
V
F e
P
mgcosθ
lcosθ θ
θ l
T mgsinθ
mg
l
Reference level
V
for P.E.
b
B
For the body to move in circular motion, a centripetal
force is required. Therefore at P
mv 2
(T + mg cos θ) – Fe = … (i)
r
Applying conservation of mechanical energy.
Total mechanical energy at B
1 2 1 2
= mV b + mg (0) = mV b
2 2
Total mechanical energy at P
=
2
1 mV 2 + mg(l + l cos θ)
1 2
1
∴ mV b = mV 2 + mg(l + l cos θ) …(ii)
2 2
On using the eq. (i) and (ii) for the condition of just
completing a circle we get for eq. (i)
T = 0, θ = 0º
2
mv
∴ mg – Fe = l
1 2
… (iii)
1
and mv b = mv 2 + mg(2l)
2 2
2
∴ v b = v 2 + 4gl
…(iv)
Putting the value of v from (iii) in (iv) we get
∴
2 Fe
v b = gl – l + 4gl m
= 5gl –
m
Fe l
−6
9×
10 × 10 × 10
v b = 5 × 10 × 0.8 –
0.8×
0.8
∴
v b = 5.86 m/s.
9
−6
⎡ Kq1q
⎢Q
Fe =
2
⎣ r
0. 8
× 0.
002
2. Shown in the figure is a container whose top and
bottom diameters are D and d respectively. At the
bottom of the container, there is a capillary tube of
outer radius b and inner radius a.
D
P
h
ρ
d
The volume flow rate in the capillary is Q. If the
capillary is removed the liquid comes out with a
velocity of v 0 . The density of the liquid is given as in Fig.
Calculate the coefficient of viscosity η. [IIT-2003]
Sol. When the tube is not there, using Bernoulli's theorem
1 2 1
P + P 0 + ρν 1 + ρgH = ρν
2
0 + P0
2
2
1 2 2 1
⇒ P + ρgH = ρ ( ν0 − ν )
2
But according to equation of continuity
2
⎤
⎥
⎦
XtraEdge for IIT-JEE 9 FEBRUARY 2010

A 2v
2
A 1 v 1 = A 2 v 2 or v 1 =
A1
1
⎡
2
∴ P + ρgH = ρ 2 ⎥ ⎥ ⎤
⎢
2 ⎛ A ⎞
⎢
⎜
2
v 0 − v0
⎟
⎣ ⎝ A1
⎠ ⎦
1
⎡
2
2 ⎛ A ⎞ ⎤
P + ρgH = ρv 0
⎢ ⎥
2 ⎢
⎜
2
1 −
⎟
⎥
⎣ ⎝ A1
⎠ ⎦
Here P + ρgH = ∆P
According to Poisseuille's equation
4
4
π(
∆P)a
π(
∆P)a
Q =
∴ η =
8ηl
8Ql
π(P
+ ρgH)a
∴ η =
8Ql
Where
1
4
2
A 2 b =
A D 2
π
⎡
2
1
= ×
8Ql
⎥ ⎥ ⎤
2 ⎛ A ⎞
⎢
⎢
⎜
2
ρv0
1−
⎟ × a 4
2
⎣ ⎝ A1
⎠ ⎦
3. Two moles of an ideal monatomic gas is taken
through a cycle ABCA as shown in the P-T diagram.
During the process AB, pressure and temperature of
the gas vary such that PT = Constant. If T 1 = 300 K,
calculate
[IIT-2000]
P
2P 1
B C
P 1
T 1 2T 1
A
(a) the work done on the gas in the process AB and
(b) the heat absorbed or released by the gas in each
of the processes.
Give answer in terms of the gas constant R.
Sol. (a) Number of moles, n = 2, T 1 = 300 K
During the process A → B
PT = constant or P 2 V = constant = K (say)
Therefore P =
V
K .
B
Therefore, W A → B =
∫
P.dV
=
∫
V
VA
T
= K[ V V ]
2 B − A
= [ KV KV ]
2 B − A
2
VB
VA
K
dV
V
= 2 [ (PBVB
)VB
− (PA
VA
)VA
]
(K = P 2 V)
2
= 2[P B V B – P A V A ] = 2[nRT B – nRT A ]
= 2nR[T 1 – 2T 1 ] = (2) (2) R [300 – 600]
= – 1200R
Therefore work done on the gas in the process AB is
1200 R.
(b) Heat absorbed/released in different processes.
Since the gas is monatomic, therefore
C V =
2
3 R and CP =
2
5 R and γ =
3
5 .
Process A – B :
⎛ 3 ⎞
∆U = nC V ∆T = (2) ⎜ R ⎟ (T B – T A )
⎝ 2 ⎠
⎛ 3 ⎞
= (2) ⎜ R ⎟ (300 – 600) = – 900 R
⎝ 2 ⎠
Q A → B = W A → B + ∆U = (– 1200R) – (900R)
Q A → B = – 2100 R (Heat released)
4. A metal bar AB can slide on two parallel thick
metallic rails separated by a distance l. A resistance R
and an inductance L are connected to the rails as
shown in the figure. A long straight wire carrying a
constant current I 0 is placed in the plane of the rails
and perpendicular to them as shown. The bar AB is
held at rest at a distance x 0 from the long wire. At
t = 0, it is made to slide on the rails away from the
wire. Answer the following questions. [IIT-2002]
A
I 0
l
x 0
di
(a) Find a relation among i, dt
R
L
B
dφ
and , where i is
dt
the current in the circuit and φ is the flux of the
magnetic field due to the long wire through the
circuit.
(b) It is observed that at time t = T, the metal bar AB
is at a distance of 2x 0 from the long wire and the
resistance R caries a current i 1 . Obtain an
expression for the net charge that has flown
through resistance R from t = 0 to t = T.
XtraEdge for IIT-JEE 10 FEBRUARY 2010

(c) The bar is suddenly stopped at time T. The
current through resistance R is found to be
4
i 1 at
time 2T. Find the value of R
L in terms of the
other given quantities.
Sol. As the metal bar AB moves towards the right, the
magnetic flux in the loop ABCD increases in the
downward direction. By Lenz's law to oppose this,
current will flow in anticlockwise direction as shown
in figure.
D
A
I 0
i
R
L
C
B
x
Applying Kirchoff's loop law is ABCD we get
⎡ di ⎤
E induced – iR – L ⎢ ⎥ = 0 …(i)
⎣dt
⎦
dφ di ⎡
dφ⎤
⇒ – = iR + L dt dt
⎢Q
E induced = − ⎥
⎣
dt ⎦
Let AB be at a distance x from the long straight wire
at any instant of time t during its motion. The
magnetic field at that instant at AB due to long
straight current carrying wire is
µ
B = 0 0
2πI x
The change in flux through ABCD in time dt is
dφ = B (dA) = Bldx
Therefore the total flux change when metal bar moves
from a distance x 0 to 2x 0 is
∆φ =
∫
2x 0
x 0
Bl
dx =
∫
2x 0
x0
V
µ 0I0
µ 0I0l
dx = [loge
x]
2πx
2π
2x 0
x 0
µ
= 0 I 0 l loge 2 … (ii)
2π
The charge flowing through resistance R in time T is
q =
∫
T
0
⎡
⎤
=
∫
T 1
di
idt ⎢Einduced
− L ⎥dt
[from eq. (i)]
0 R ⎣ dt ⎦
1
R
T
=
∫
Einduceddt
−
∫
1
R
0
L
R
= ( ∆ φ)
− i1
L
R
i1
0
di
q =
1 ⎡µ
I0
R
⎢
⎣ 2π
0 l
⎤
loge
2⎥ ⎦
L
R
– i1
from eq. (ii)
(c) When the metal bar AB is stopped, the rate of
change of magnetic flux through ABCD becomes
zero.
di
From (i) iR = – L dt
⇒
2T L
∫
dt =
T R
∫
i1
/ 4
0
di
i
L i1
/ 4
T = – loge
R i1
L T =
R 2log 2
e
5. In hydrogen-like atom (z = 11), nth line of Lyman
series has wavelength λ. The de-Broglie's wavelength
of electron in the level from which it originated is
also λ. Find the value of n
[IIT-2006]
Sol. nth line of lyman series means electron jumping from
(n + 1)th orbit to Ist orbit. for an electron to move in
(n + 1)λ
2π 2π
⇒ λ = × r =
(n + 1) (n + 1)
= [0.529 × 10 –10 (n + 1)
]
z
1 Z
⇒ =
−10
λ 2π[0.529×
10 ](n + 1)
Also we know that when electron jumps from
(n + 1)th orbit to Ist orbit.
1 = RZ
2
⎡ 1 1 ⎤
⎢ − ⎥
λ
2
2
⎣1
(n + 1)
⎦
⎡
= 1.09 × 10 7 Z 2 1
⎥ ⎤
⎢1
−
2
⎣ (n + 1) ⎦
From (i) and (ii)
Z
2π(0.529×
10
−10
)(n + 1)
⎡
= 1.09 × 10 7 Z 2 1
⎥ ⎤
⎢1
−
2
⎣ (n + 1) ⎦
on solving, we get n = 24.
2
…(i)
XtraEdge for IIT-JEE 11 FEBRUARY 2010

8. Hydrogen peroxide acts both as an oxidising and as a
CHEMISTRY
reducing agent in alkaline solution towards certain
first row transition metal ions. IIIustrate both these
properties of H 2 O 2 using chemical equations.
[IIT-1998]
[IIT-2004] Sol. When H 2 O 2 acts as oxidising agent, therefore,
following reaction takes place :
H 2 O 2 + 2e → 2OH –
⎛ ⎞
⎜
a
P + ⎟ (V 2
m – b) = RT
While regarding its action as reducing agent, the
⎝ V m ⎠
following reaction takes place :
a ab
H
PV m – Pb + –
V
2 = RT
2 O 2 + 2OH – → O 2 + 2H 2 O + 2e
m V m
Examples of oxidising Character of H 2 O 2 in alkaline
a ab
medium
PV m = RT + Pb – + ....(i)
V
2
m V
2Cr(OH) 3 + 4NaOH + 3H 2 O 2 → 2Na 2 CrO 4 + 8H 2 O
m
Here Fe 3+ (Fe is a first row transition metal) is
reduced to Fe 2+ .
Example of reducing character of H 2 O 2 in alkaline
medium
2K 3 Fe(CN) 6 + 2KOH + H 2 O 2 → 2K 4 [Fe(CN) 6 ] +
2H 2 O + O 2
Here Cr 3+ (Cr is a first row transition metal) is
oxidised to Cr 6+
9. Write the structures of (CH 3 ) 3 N and (Me 3 Si) 3 N. Are
they isostructural Justify your answer. [IIT-2005]
Sol. (CH 3 ) 3 N and (Me 3 Si) 3 N are not isostructural, the
former is pyramidal while the latter is trigonal planar.
Silicon has vacant d orbitals which can accommodate
[IIT-1991] lone pair of electrons from N(back bonding) leading
to planar shape.
SiMe 3
..
N
N
left
H 3 C CH
0.1162
3
= 0.2324 moles l –1
CH SiMe3 SiMe 3
3
0.5
0.0358
10. How is boron obtained from borax Give chemical
0.5
equations with reaction conditions. Write the
structure of B 2 H 6 and its reaction with HCl.
[IIT-2002]
Sol. When hot concentrated HCl is added to borax
(Na 2 B 4 O 7 .10H 2 O) the sparingly soluble H 3 BO 3 is
formed which on subsequent heating gives B 2 O 3
1.29
× 10 –11
which is reduced to boron on heating with Mg, Na or
0.2324
K
Na 2 B 4 O 7 (anhydrous) + 2HCl(hot, conc.)
−11
1.29×
10
→ 2NaCl + H 2 B 4 O 7
× 0.0716
0.2324
H 2 B 4 O 7 + 5H 2 O → 4H 3 BO 3 ↓
= 3.794 × 10 –12 mol 3 l –3
6. A graph is plotted between PV m along Y-axis and P
along X-axis, where V m is the molar volume of a real
gas. Find the intercept along Y-axis.
Sol. The van der Waal equation (for one mole) of a real
gas is
To calculate the intercept P → 0, hence V m → ∞ due
to which the last two terms on the right side of the
equation (i) can be neglected.
∴ PV m = RT + Pb
When P = 0, intercept = RT
7. The solubility product of Ag 2 C 2 O 4 at 25ºC is
1.29 × 10 –11 mol 3 l –3 . A solution of K 2 C 2 O 4
containing 0.1520 mole in 500 ml water is shaken at
25ºC with excess of Ag 2 CO 3 till the following
equilibrium is reached :
Ag 2 CO 3 + K 2 C 2 O 4 Ag 2 C 2 O 4 + K 2 CO 3
At equilibrium the solution contains 0.0358 mole of
K 2 CO 3 . Assuming the degree of dissociation of
K 2 C 2 O 4 and K 2 CO 3 to be equal, calculate the
solubility product of Ag 2 CO 3 .
Sol. Ag 2 CO 3 + K 2 C 2 O 4 → Ag 2 C 2 O 4 + K 2 CO 3
Moles at start Excess 0.1520 0 0
Moles after reaction
0.1520 – 0.0358 0.0358 0.0358
= 0.1162
2–
Molar concentration of K 2 C 2 O 4 or C 2 O 4
unreacted =
[K 2 CO 3 ] = [CO 3 2– ] at equilibrium =
= 0.07156 moles l –1
Given that K sp for Ag 2 C 2 O 4 = 1.29 × 10 –11 mol 3 l –3 at
25ºC
So, [Ag + ] 2 [C 2 O 4 2– ] = 1.29 × 10 –11
or [Ag + ] 2 × 0.2324 = 1.29 × 10 –11
Hence [Ag+] 2 =
Then K sp for
Ag 2 CO 3 = [Ag + ] 2 [CO 3 2– ] =
strong heating
2H 3 BO 3 ⎯⎯
⎯⎯⎯→
B 2 O 3 + 3H 2 O
B 2 O 3 + 6K → 2B + 3K 2 O or
XtraEdge for IIT-JEE 12 FEBRUARY 2010

B 2 O 3 + 6Na → 2B + 3Na 2 O
B 2 O 3 + 3Mg → 2B + 3MgO
Structure of B 2 H 6
H
H
1.19Å
B
or
Hydrogen bridge
bonding (3C-2e bond)
H
97º
B
H
1.37Å
1.77Å
H
122º
B 2 H 6 + HCl → B 2 H 5 Cl + H 2
Normally this reaction takes place in the presence of
Lewis acid (AlCl 3 ).
MATHEMATICS
11. The curve y = ax 3 + bx 2 + cx + 5, touches the x-axis
at P(–2, 0) and cuts the y axis at a point Q, where its
gradient is 3. Find a, b, c.
[IIT-1994]
Sol. It is given that y = ax 3 + bx 2 + cx + 5 touches x-axis
at P(–2, 0) which implies that x-axis is tangent at
(–2, 0) and the curve is also passes through (–2, 0).
The curve cuts y-axis at (0, 5) and gradient at this
point is given 3 therefore at (0, 5) slope of the tangent
is 3.
dy
Now, = 3ax 2 + 2bx + c
dx
since x-axis is tangent at (–2, 0) therefore
dy
dx
x=−2
= 0
⇒ 0 = 3a(–2) 2 + 2b(–2) + c
⇒ 0 = 12a – 4b + c ...(1)
again slope of tangent at (0, 5) is 3
dy
⇒ = 3
dx
(0,5)
⇒ 3 = 3a(0) 2 + 2b(0) + c
⇒ 3 = c ...(2)
Since, the curve passes through (–2, 0), we get
0 = a(–2) 3 + b(–2) 2 + c(–2) + 5
0 = –8a + 4b – 2c + 5 ...(3)
from (1) and (2), we get
12a – 4b = –3 ...(4)
from (3) and (2), we get
–8a + 4b = 1 ...(5)
adding (4) and (5), we get
4a = –2
H
⇒ a = –1/2
Putting a = –1/2 in (4), we get
12(–1/2) – 4b = –3
⇒ –6 – 4b = –3
⇒
–3 = 4b
⇒ b = –3/4
Hence, a = –1/2, b = –3/4 and c = 3
12. Prove that :
3x(x + 1)
⎡ π⎤
sin x + 2x ≥ , ∀ x ∈
π
⎢0
, ⎥
⎣ 2 ⎦
(justify the inequality, if any used). [IIT-2004]
3x(x + 1)
Sol. Let f(x) = sin x + 2x –
π
(6x + 3)
f´(x) = cos x + 2 –
π
6 ⎡ π⎤
f´´(x) = –sin x – < 0 for all x ∈
π
⎢0
, ⎥
⎣ 2 ⎦
⎡ π⎤
∴ f´(x) is decreasing for all x ∈ ⎢0
, ⎥
⎣ 2 ⎦
⇒ f´(x) > 0 {as, x < π/2
⇒ f´(x) > f´(π/2)}
∴ f(x) is increasing
Thus, when x ≥ 0
f(x) ≥ f(0)
3x(x + 1)
sin x + 2x – ≥ 0
π
3x(x + 1)
or sin x + 2x ≥
π
13. A window of perimeter (including the base of the
arch) is in the form of a rectangle surrounded by a
semi-circle. The semi-circular portion is fitted with
coloured glass while the rectangular part is fitted with
clear glass. The clear glass transmits three times as
much light per square meter as the coloured glass
does.
What is the ratio for the sides of the rectangle so that
the window transmits the maximum light[IIT-1991]
Sol. Let '2b' be the diameter of the circular portion and 'a'
be the lengths of the other sides of the rectangle.
Total perimeter = 2a + 4b + πb = K (say) ...(1)
Now, let the light transmission rate (per square
metre) of the coloured glass be L and Q be the total
amount of transmitted light.
XtraEdge for IIT-JEE 13 FEBRUARY 2010

a
Coloured
glass
Clear glass
Then, Q = 2ab(3L) + 2
1 πb 2 (L)
Q = 2
L {πb 2 + 12ab}
Q =
2
L {πb 2 + 6b (K – 4b – πb)}
Q = 2
L {6Kb – 24b 2 – 5πb 2 }
dQ L = {6K – 48b – 10πb} = 0
db 2
⇒ b =
2
6K
48 +10π
a
...(2)
d Q L
and =
2 {–48 + 10π}La
db 2
Thus, Q is maximum and from (1) and (2),
(48 + 10π) b = 6K and K = 2a + 4b + πb
⇒ (48 + 10π) b = 6{2a + 4b + πb}
Thus, the ratio =
2b
a
=
6
6 + π
14. With usual notation, if in a triangle ABC
Sol. Let
b + c c + =
11 12a
cos A
7
=
=
cos B
19
b + c c + =
11 12a
a + b , then prove that
13
=
=
cosC
25
a + b
13
= λ
[IIT-1984]
⇒ (b + c) = 11λ, c + a = 12λ, a + b = 13λ
⇒ 2(a + b + c) = 36λ
or a + b + c = 18λ
Now, b + c = 11λ and a + b + c = 18λ ⇒ a = 7λ
c + a = 12, and a + b + c = 18λ ⇒ b = 6λ
a + b = 13λ and a + b + c = 18λ ⇒ c = 5λ
∴ cos A =
=
b
2
2
+ c − a
2bc
2
36λ
2
2
+ 25λ
2
2(30) λ
2
− 49λ
= 5
1
cos B =
=
cos C =
=
a
2
2
+ c − b
2ac
2
25λ
a
2
2
2
+ 49λ
2
2
70λ
+ b − c
2ab
2
49λ
2
2
+ 36λ
2
84λ
2
− 36λ
2
− 25λ
19
= 35
= 7
5
1 19 5
∴ cos A : cos B : cos C = : : = 7 : 19 : 25
5 35 7
15. Let ABC be a triangle with incentre I and inradius r.
Let D, E, F be the feet of the perpendiculars from I to
the sides BC, CA and AB respectively. If r 1 , r 2 and r 3
are the radii of circles inscribed in the quadrilaterals
AFIE, BDIF and CEID respectively, prove that :
r
r −
1
r1
+
r2
r − r
2
+
r3
r − r
3
=
r r r
(r − r )(r − r )(r − r )
1
1 2 3
[IIT-2000]
Sol. The quadrilateral HEKJ is a square because all four
angles are right angle and JK = JH.
A
F
Circle
A/2 A/2
r 1
K
J 90º
90º
r 1
E
H
Circle
B
D
C
90º
Therefore, HE = JK = r 1 and IE = r (given)
⇒ IH = r – r 1
Now, in right angled triangle IHJ, ∠JIH = π/2 – A/2
[Q ∠IEA = 90º, ∠IAE = A/2 and ∠JIH = ∠AIE] in
triangle JIH
tan(π/2 – A/2) =
⇒ cot A/2 =
r1
r − r
r1
r − r
1
r2
r3
Similarly, cot B/2 = and cot C/2 =
r − r2
r − r3
adding above results, we obtain
A B C A B C
cot + cot + cot = cot cot cot 2 2 2 2 2 2
⇒
r1
+
r − r
1
r2
r − r
2
+
r
3
1
r − r
3
=
(r − r )(r − r )(r − r )
1
2
r r r
1 2 3
2
3
3
XtraEdge for IIT-JEE 14 FEBRUARY 2010

Physics Challenging Problems
Set #10
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in physics that would be very helpful in facing IIT
JEE. Each and every problem is well thought of in order to strengthen the concepts and we
hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma
Solutions will be published in next issue
Director Academics, Jodhpur Branch
Passage # 1 (Q. No. 1 to 4)
A circular coil is placed in a time varying
magnetic field as shown in figure. The numeral
relation between the radius and the resistance of
the coil is - r = π
R where r = radius of coil
R = resistance of the coil
the magnetic field is perpendicular to the plane of
paper and inwards and given by B = B 0 + B 1 t 2
where B 0 is a positive and B 1 is a negative
constant. The numeral values of B 0 and B 1 are
1 and 2 respectively.
→
B
× × × ×
× × × ×
r
× × × ×
× × × ×
× × × ×
Q.1 The value of induced electric field in the circular
coil
(A) Depends only on radius of circular coil in
linear manner
(B) Independent of the radius of circular coil
(C) Varies nonlinearly with respect to time
(D) Numeral value depends on the radius of
circular coil and for given radius it varies
linearly with respect to time
Q.2 Induced current in the coil at t = 0
(A) 4 amp. (B) zero (C) 1 amp. (D) π amp.
Q.3 RMS value of the induced current for the time
interval 0 ≤ t ≤ 2s
(A) 8 amp.
(B) 4 amp.
(C) 8/ 3 amp. (D) 8/3 amp.
Q.4 Induced charge for time period as stated above
(A) 8x Faraday (B) 2x Faraday
(C) x Faraday (D) 4x Faraday
1
Here x = 96500
Passage # 2 (Q. No. 5 to 8)
A rod ab of mass M, resistance R and length L is
supported by two supports SP-1 and SP-2 against
gravity At t = 0 both the supports get removed
and the rod starts falling under the gravity in a
very long spread uniform magnetic field
perpendicular to the plane of paper and directed
inwards then
×
×
t = 0 ×
×
× SP-1
×
×
×
×
a
B →
× × ×
× × ×
× × b × x = 0
× × ×
× × SP-2
× × ×
× × ×
L
× × ×
× × ×
SP-1 and SP-2 Rigid supports
Q.5 Expression for the speed at time t is
(A) k
g (1 – e –kt )
(C)
g
k (1 – e –kt )
B 2 L
2
where k = MR
(B) k
g .e
–kt
(D)
g
k .e
–kt
Q.6 Time after which the acceleration of rod is 37%
of the maximum acceleration
(A)
MR
B
2 L 2
(C) 0.37
MR
B
2 L 2
(B)
1 MR
2 B
2 L 2
(D) 0.63
MR
B
2 L 2
XtraEdge for IIT-JEE 15 FEBRUARY 2010

Q.7 Time after which the velocity of the rod is 63% of
terminal velocity
(A)
MR
B
2 L 2
(C) 0.37
MR
B
2 L 2
(B)
1 MR
2 B
2 L 2
(D) 0.63
MR
B
2 L 2
Q.8 Expression for a real velocity at any instant t
(A)
K
gL (1 – e –kt ) (B) gL(1 – e –kt )
(C) gK(1 – e –kt ) (D)
K
L (1 – e –kt )
B 2 L
2
Where K =
MR
ELECTRONIC NOSE
NASA researchers are developing an exquisitely sensitive artificial nose for space exploration.
Onboard the space station, astronauts are surrounded by ammonia. It flows through pipes, carrying heat
generated inside the station (by people and electronics) outside to space. Ammonia helps keep the station
habitable.
But it's also a poison. And if it leaks, the astronauts will need to know quickly. Ammonia becomes dangerous at
a concentration of a few parts per million (ppm). Humans, though, can't sense it until it reaches about 50 ppm.
Ammonia is just one of about forty or fifty compounds necessary on the shuttle and space station, which cannot
be allowed to accumulate in a closed environment.
And then there's fire. Before an electrical fire breaks out, increasing heat releases a variety of signature
molecules. Humans can't sense them either until concentrations become high.
Astronauts need better noses!
That's why NASA is developing the Electronic Nose, or ENose for short. It's a device that can learn to recognize
almost any compound or combination of compounds. It can even be trained to distinguish between Pepsi and
Coke. Like a human nose, the ENose is amazingly versatile, yet it's much more sensitive.
"ENose can detect an electronic change of 1 part per million," says Dr. Amy Ryan who heads the project at JPL.
She and her colleagues are teaching the ENose to recognize those compounds - like ammonia - that cannot be
allowed to accumulate in a space habitat.
Here's how it works: ENose uses a collection of 16 different polymer films. These films are specially designed to
conduct electricity. When a substance - such as the stray molecules from a glass of soda - is absorbed into
these films, the films expand slightly, and that changes how much electricity they conduct.
Because each film is made of a different polymer, each one reacts to each substance, or analyte, in a slightly
different way. And, while the changes in conductivity in a single polymer film wouldn't be enough to identify an
analyte, the varied changes in 16 films produce a distinctive, identifiable pattern.
Electronic Noses are already being used on Earth. In the food industry, for example, they can be used to detect
spoilage. There's even an Electronic Tongue, which identifies compounds in liquids. The ENose needs to be
able to detect lower concentrations than these devices.
E-Nose
Right now, Ryan is working on a stand-alone version of ENose. "Everything is in one package," she explains:
polymer films, a pump to pull air (and everything in the air) through the device, computers to analyze data, the
energy source. The noses could simply be posted, like smoke detectors, at various points around the habitat.
XtraEdge for IIT-JEE 16 FEBRUARY 2010

XtraEdge for IIT-JEE 17 FEBRUARY 2010

8 Questions
Solution
Set # 9
Physics Challenging Problems
were Published in January Issue
1. Question is based on Ampere's circuital law
As line integral of magnetic field over the closed
→
→
loop
∫
B .d1 = µ 0.
inet
here i net = 4 – 3 = 1 amp.
→ →
1 ⎡
2 1 ⎤
∫
B .d1
= µ 0(1)
= µ 0 = ⎢As
c =
2
⎥
c . ε0
⎣ µ 0.
ε0
⎦
where c is the speed of light
so line integral of magnetic field over the closed
→ →
1
loop abcda
∫
B.ds
= µ 0 =
2
c ε0
As we know that surface integral of magnetic
field over the closed loop,
∫
B.d s
is always zero
according to Gauss law for magnetostatics.
Option D is correct.
2. Positively charged particle will turn towards left
and moves anticlockwise. In mirror it is seen
clockwise. Just opposite for negatively charged
particle.
mv
Radius of circular path r = qB
mv
To avoid hitting the upper plate d ≥ r, d ≥ qB
A positively charged particle will never hit the
mv
upper plate if d ≥ it rotates anticlockwise but
qB
if viewed in mirror it is clockwise.
Option D is correct.
3. Electric force on particle F e
→
→
→
=
→
± q E
= ± qE0 ( −î)
= ± qE0î
→
F m
Magnetic force on particle = ± q( v×
B)
F m
→
→
= ± [v ĵ B (kˆ )]
q 0 × 0
= ± qv0 B0(ĵ×
kˆ )
→
= ± .v .B (î)
q 0 0
→ →
As F m and F e are in opposite direction so either
the particle is positively charged or negatively it
can go undeviated if
→
=
→
| F m | | Fe
|
Option C is correct
→ →
| m = e
4. As F | | F |
qv 0 .B 0 = qE 0
v 0 =
E 0 = ( 19.6)m / s
B0
Height achieved by the particle
v 2 19.6
h = = = 1m
2g
2(9.8)
As from mirror point of view focal length
f = 20cm (Negative for concave mirror)
Using mirror formula. 1/v + 1/u = 1/f
1 1 1 1 1 1 2 − 3 1
+ = ⇒ = − = = −
v − 30 − 20 v 30 20 60 60
So image is at, v = -60cm
Height achieved by the particle behaves like
object so mirror forms it's image and the
magnification
−60
m = v/u = – = 2
− 30
As height achieved by the charged particle
upward = Height of object = 1m
XtraEdge for IIT-JEE 18 FEBRUARY 2010

Height achieved by the charged particle
downward = Height of image
HOI
as m = So HOI = m (HOO) = 2(1) = 2m
HOO
Option D is correct
5. (1). Initial current is 18 amp.
Total energy stored in the inductor
= 2
1 Li 2 = 2
1 (2/3)(18)
2
= 3
1 (18) 2 = 108 Joule
{L = L 1 + L 2 = 1/3 + 1/3 = 2/3(As two inductor
are in series)}
(1). Total energy dissipated in resistors = 108 J
(2). Time constant
The equivalent circuit is
1/3H 1/3H
L
τ =
R
eq
eq
a
1Ω
1Ω
6Ω
3Ω
between terminals a and b.
2 / 3 2
= = = 0.4
5 / 3 5
[2.Time constant of the circuit = 0.4 sec.]
(3). Potential drop across R 1 initially
v = i.R 1 = 18(1) = 18 volt
(4). Using current division formula.
Current passing through the R 2 initially is
= 12 amp.
So potential drop v = i.R = R(1) = 12 volt.
6. As the particles are having same de-Broglie wave
lengths so
λ A = λ B
h =
P A
h
PB
b
P A = P B Means momentum of both the particles is
same.
As the radius of circular path in magnetic field is
r = p/qB so r A =
P A PB
, r B =
q .B q .B
A
Particle moving left Particle moving right
Positively charged Negatively charged
radius of path less - Particle B electron
- Particle A radius of path is more
as shown in figure
Particle A should have more charge as compared
to electron so it is doubly ionized Helium atom.
Option D is correct.
7. 1. Force on each and every coil is zero
2. Torque on coil of option A and option B is
maximum because angle between M → and → B is
90º.
τ max. = MB τ m ≠ 0
3. Torque on coil of option C and option D is zero
because angle between M → →
and B is zero
So τ min. = MB sin 0 = 0
4. Coil in option C and option D are having the
tendency of compression.
8. Conceptual problem
1. Gauss law for electrostatics
φ
→ →
1
e∫ E .ds
= . qnet
= µc 2 ⎡ 2 1 ⎤
q net ⎢c
= ⎥
ε0
ε0µ
0 ⎦
2. Gauss law for magnetistatics
∫
→
→
φ m E .d = 0 , Magnetic monopole is impossible
s
3. Ampere's circuital law
1
B = µ 0 i net =
2
C ε
→ →
∫
.d1
0
.i net
4. Faraday's law of electromagnetic induction
Induced emf e =
∫
→
→
E .d
e
⎣
B
XtraEdge for IIT-JEE 19 FEBRUARY 2010

PHYSICS
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. AB is a horizontal diameter of a ball of mass m = 0.4 kg
and radius R = 0.10 m. At time t = 0, a sharp impulse
is applied at B at angle of 45º with the horizontal, as
shown in Fig. So that the ball immediately starts to
move with velocity v 0 = 10 ms –1 .
A
B
45º
(i) Calculate the impulse
If coefficient of kinetic friction between the floor
and the ball is µ = 0.1, calculate
(ii) Velocity of ball when it stops sliding,
(iii) time t at that instant,
(iv) horizontal distance traveled by the ball upto that
instant,
(v) Angular displacement of the ball about horizontal
diameter perpendicular to AB, upto that instant,
and
(vi) energy lost due to friction. (g = 10 ms –2 )
Sol. Since, the impulse applied is sharp and its line of
action does not pass through centre of mass of the
sphere, therefore, (just after application of impulse),
sphere starts to move, both translationally and
rotationally,. Translational motion is produced by
horizontal component of the impulse, while rotational
motion is produced by moment of the impulse. Let
the impulse applied be J.
lα
O
mg
ma
µN
N
Then its horizontal component,
J. cos 45º = Initial horizontal momentum (m.v 0 ) of
the ball.
∴ J = 4 2 kg ms –1 Ans. (i)
Moment of inertia of ball about centroidal axis is
I = 5
2 mR 2 = 1.6 × 10 –3 kg m 2
Initial angular momentum of ball (about centre)
= J (R. sin 45)
or Iω 0 = J.R. sin 45º
∴ ω 0 = 250 rad sec –1 (clockwise)
Now sphere slides on floor (to the left). Therefore,
friction on it acts towards right. Considering free
body diagram of sphere Fig. (while it is sliding).
(Note: Since, the sphere is sliding on the floor,
therefore, A is not an instantaneous axis of rotation.
Hence, we can not take moments about A)
For vertical forces, N = mg …(1)
For horizontal forces, µN = ma or a = µg = 1 ms –2
Now taking moments (about O) of forces acting on
sphere,
µN . R = Iα …(2)
From equations (1) and (2) α = 25 rad/sec 2
(anticlockwise)
Let sliding continue for a time 't'.
At that instant, translational velocity, v = v 0 – at
or
v = (10 – t)ms –1 (towards left)
and angular velocity, ω = (–ω 0 )+ αt (anticlockwise)
or ω = (25t – 250) rad s –2
But when sliding stops, v = rω
∴ (10 – t) = 0.1 (25t – 205) or t = 10 sec Ans. (iii)
∴ At that instant v = 10 – t = 0 Ans. (ii)
Considering leftward translational motion of ball (for
first 10 second),
Distance moved by the ball is s = v 0 t – 2
1 at
2
or s = 50 m Ans. (iv)
Now considering clockwise rotational motion of the
ball (about its centroidal axis),
Angular displacement, θ = ω 0 t – 2
1 αt
2
or θ = 1250 radian (clockwise) Ans. (v)
XtraEdge for IIT-JEE 20 FEBRUARY 2010

Since, the ball has stopped, it means whole of its
initial kinetic energy is lost against friction and that is
⎛ 1 2 1 2 ⎞
⎜ mv0 + Iω0
⎟
⎝ 2 2 ⎠
∴ Energy lost against friction = 70 joule Ans. (vi)
2. A rectangular tank having base 15 cm × 20 cm is
filled with water (density ρ = 1000 kg m –3 ) upto
20 cm height. One end of an ideal spring of natural
length h 0 = 20 cm and force constant K = 280 Nm –1 is
fixed to the bottom of a tank so that spring remains
vertical. This system is in an elevator moving
downwards with acceleration a = 2 ms –2 . A cubical
block of side l = 10 cm and mass m = 2 kg is gently
placed over the spring and released gradually, as
shown in Fig.
20cm
(i) Calculate compression of the spring in
equilibrium position.
(ii) If block is slightly pushed down from equilibrium
position and released, calculate frequency of its
vertical oscillations. (g = 10 ms –2 )
Sol. Let, in equilibrium position, compression of spring be
x. Liquid of volume l 2 x is displaced from its original
position and level of liquid in tank rises as shown in
2
l x
Fig. This rise in level, ∆ x =
2
A − l
where A = 15 cm × 20 cm (Base area of tank)
Increased level of
water surface
Original level of
water surface
K
K
a
∆x
x
Upthrust exerted by water = apparent weight of water
displaced,
∴ Upthrust F 1 = 1.5x (g – a) = 120.x newton
Upward force exerted by spring
F 2 = Kx = 280.x.
Considering free body diagram of the block, (Fig.)
mg
m.a
(F 1 + F 2 )
Fig. : 2
Mg – (F 1 + F 2 ) = ma
Substituting values of F 1 and F 2 , x = 0.04 m
= 4 cm Ans. (i)
If the block is slightly pushed downward by dx, both
F 1 and F 2 increase.
Increase in F 1 is dF 1 = 120 dx
Increase in F 2 is dF 2 = 280.dx
restoring force on block = increase in F 1 + increase in
F 2 = dF 1 + dF 2 = (120 dx +280.dx) = 400.dx
400.dx
or Restoring acceleration = = 200.dx
m
Since, restoring acceleration ∝ displacement (dx)
Therefore, block performs SHM.
Hence, frequency,
f =
1
2π
5 2
= π
acceleration
displacement
per second
1
= 2π
200
Ans. (ii)
3. A two way switch S is used in the circuit shown in
Fig. First, the capacitor is charged by putting the
switch in position 1.
Calculate heat generated across each resistor when
switch is in position 2.
60V
+ – 10Ω
1
0.1F
Fig. : 1
or ∆x = 0.5x
∴ Mass of water displaced by the block
= l 2 (x + ∆x)ρ
= 15x kg
2
S
4Ω
6Ω
3Ω
XtraEdge for IIT-JEE 21 FEBRUARY 2010

Sol. Initially the switch was in position 1. Therefore,
initially potential difference across capacitor was
equal to emf of the battery i.e. 60 volt.
∴ Initially energy stored in the capacitor was
U =
2
1 CV 2 =
2
1 × 0.1 × 60 2 J
z
R
O
y
i
i
= 180 J
q
+ –
6Ω
4Ω
3Ω
When switch is shifted to position 2, capacitor begins
to discharge and energy stored in it is dissipated in
the form of heat across resistances. Let at some
instant discharging current through the capacitor be i
as shown in Fig.
According to Kirchhoff's laws,
i 1 + i 2 = i … (1)
6i 1 – 3i 2 = 0 or i 2 = 2i 1 … (2)
From above two equations,
i
2
i 1 = and i 2 = i
3 3
But thermal power generated in a resistance R is
P = i 2 R where i is current flowing through it.
Therefore, heat generated P 1 , P 2 and P 3 across 4Ω,
6Ω and 3Ω resistances is in ration
i 1
i 2
4 i
or P 1 : P 2 : P 3 = 4 :
3
2 :
3
4 = 6 : 1 : 2
2
: 6i
2
1
: 3i
But total heat generated is P 1 + P 2 + P 3 = U
∴ Heat generated across 4Ω is P 1 = 120 J Ans.
Heat generated across 6Ω is P 2 = 20 J Ans.
Heat generated across 3Ω is P 3 = 40 J Ans.
Since, during discharging, no current flows through
10Ω, therefore heat generated across it is equal to
zero.
Ans.
4. Calculate magnetic induction at point O if the
wire carrying a current I has the shape shown in
(i) Fig. (a) and (ii) Fig.(b).
z
2
2
x
45º
Fig. (b)
The radius of the curved part of the wire is equal to R
and linear parts of the wire are very long.
Sol. (i) Current carrying wire shown in figure (a) can be
considered in four parts.
(1) A straight part along y-axis.
Since, point O lies on its axis, therefore magnetic
induction due to it is B → 1 = 0.
(2) A semi-circular part in y – z plane.
Magnetic induction due to it is
→ µ (1/ 2)I
B 2 = 0 µ (– î ) = – 0 I î
2R
4R
(3) One fourth circle in x – y plane.
Magnetic induction due to it is
→ µ (1/ 4)I
B 3 = 0 µ (– kˆ ) = – 0 I kˆ
2R
8R
(4) A straight part in x – y plane carrying current
along negative y-directions.
Magnetic induction due to it is
→ µ
B 4 = 0 I
4π R
(– kˆ µ ) = – 0 I
kˆ
4πR
→
∴ B= B → 1 + B → 2 + B → 3 + B
→ 4
µ
= – 0 I µ î – 0 I
4R
8π R
(π+2) kˆ Ans. (i)
(ii) Circuit segment shown in figure (b) can be
considered in three parts.
z
R
O
R
y
r
45º
O
R
y
x
45º I
Fig. (a)
XtraEdge for IIT-JEE 22 FEBRUARY 2010

(1) A circular loop in y – z plane. Since, this loop is
made of uniform wire, therefore, magnetic
induction at O due to it is B 1 = 0
(2) A straight part, parallel to x-axis. Magnetic
µ
induction due to it is B 2 = 0 I
4π R
(– kˆ µ ) = – 0 I
kˆ .
4πR
(3) A straight part in y – z plane. Perpendicular
distance of O from axis of this straight part is r =
R cos 45º as shown in fig (a). Angles subtended
by lines joining O and ends of this straight part
with perpendicular drawn from O are α = – 45º
and β = 90º.
Magnetic induction at O due to this part is
µ
B 3 = 0 I
(sin α + sin β)
4π r
→ µ
or B 3 = 0 I
4π R
( 2 – 1)î
→
∴ B= B → 1 + B → 2 + B
→ 3
µ
= 0 I
4π R
( 2 – 1) î – µ 0 I
kˆ Ans. (ii)
4πR
5. In a Young's double slit experiment a parallel light
beam containing wavelength λ 1 = 4000 Å and
λ 2 = 5600 Å is incident on a diaphragm having two
narrow slits. Separation between the slits is
d = 2 mm. If distance between diaphragm and screen
is D = 40 cm, calculate
(i) distance of first black line from central bright
fringe and
(ii) distance between two consecutive black lines.
Sol. When a monochromatic light of wavelength λ is used
to obtain interference pattern in Young's double slit
λD
experiment, fringe width is given by ω = where d
D is distance of screen from slits and d is distance
between the slits.
Hence, fringe width for light of wavelength λ 1 ,
λ D
ω 1 = 1
d
∴ ω 1 = 80 µm
and fringe width for light of wavelength, λ 2 ,
λ
ω 2 = 2 D
= 112 µm
d
Since, the incident light beam has both the
wavelength λ 1 and λ 2 , therefore, interference patterns
are formed on the screen for both the wavelengths. A
black line is formed at the position where dark
fringes are formed for both of the wavelengths.
Let first black line be formed at distance y from
central bright fringe. Let at this position there be mth
dark fringe of wavelength λ 1 and nth dark fringe of
wavelength λ 2 .
∴ Distance of first black line, from central bright
line,
⎛ 1 ⎞ ⎛ 1 ⎞
y = ⎜ m − ⎟ ω 1 = ⎜ n − ⎟ ω 2 … (1)
⎝ 2 ⎠ ⎝ 2 ⎠
2m − 1 ω
or
=
2
…(2)
2n −1
ω1
For first black line, y should be minimum possible
which corresponds to least possible integer values of
m and n.
2m − 1 7
Hence,
= or m = 4, n = 3
2n −1
5
∴ Position of first black line
⎛ 1 ⎞
y = ⎜ m − ⎟ ω 1 = 280 µm Ans. (i)
⎝ 2 ⎠
Since, interference pattern is always symmetric about
central bright fringe, therefore, there are two first
black lines one is at height y from central bright
fringe and the other at a depth y from it.
Hence, distance between two consecutive black lines
= 2y = 560 µm Ans. (ii)
SCIENCE TIPS
• By seeing a glowing electric bulb can you say if it is
being fed by A.C. or D.C.
No
• What does the sudden burst of a cycle tyre
represent
Adiabatic process
• What happens to the velocity and wavelength of
light when it enters a denser medium
Both decrease
• The skylab space station did not have a safe landing.
Why
Because its remote control system failed
• What happens when even a small bird hits a flying
aeroplane
It causes heavy damage
• When does the lunar eclipse occur
It occurs when the earth comes
in between the moon and the sun
XtraEdge for IIT-JEE 23 FEBRUARY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
Prism & Wave Nature of Light
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Prism :
(i)
Deviation 'δ' produced by the prism,
Normal
i
A
δ
P r r'
Q
i'
Normal
B
C
δ = i + i' – A
and A = r + r'
(ii) For minimum deviation 'δ m '
i = i' and r = r' and also PQ||BC and the refractive
index for the material of prism is given by
⎛ A + δm
⎞
sin⎜
⎟
µ =
⎝ 2 ⎠
⎛ A ⎞
sin⎜
⎟
⎝ 2 ⎠
(iii) δ – i graph for prism
δ
δ m
i
(iv) For not transmitting the ray from prism,
⎛ A ⎞
µ > cosec ⎜ ⎟
⎝ 2 ⎠
(v) For grazing incidence i = 90º and for grazing
emergence i' = 90º. For maximum deviation i = 90º or
i' = 90º
(vi) The limiting angle of prism = 2C
when i = i' = 90º
If the angle of prism A > 2C, then the rays are totally
reflected.
(vii) Right-angled prism : These prisms are used to turn a
light beam to 90º or 180º. These are usually made of
crown glass for which
⎛
µ g = 1.5 and C = tan –1 ⎟ ⎞
⎜
1
= 42º.
⎝
µ g ⎠
Such prisms are used in binoculars and submarine
periscopes.
(viii) Deviation produced by a thin prism δ = (µ – 1)A
(ix) Angular dispersion D = δ v – δ R = (µ V – µ R )A
Where V and R stand for violet and red colours
respectively.
Mean deviation δ Y = (µ Y – 1)A
where µ Y is the refractive index of mean yellow
colour.
Angular dispersion
(x) Dispersive Power, ω =
Mean deviation
µ V − µ
ω =
µ −1
Y
R
δV
− δ
=
δ
Y
R
µ V + µ
where µ Y = R
2
(xi) Pair of prisms (or crossed prism) : Two thin prisms
of different material when placed crossed, i.e., with
their refracting edges parallel and pointing in
opposite directions as shown in figure, produce a
total deviation δ given by δ = δ 1 ~ δ 2
A
Crown
glass
Flint
glass
where δ 1 and δ 2 are the mean deviations produced by
the first and second prism respectively.
Total angular dispersion
D = D 1 ~ D 2
where D 1 and D 2 are the angular dispersions
produced by respective prisms.
(xii) Dispersion without deviation : If the angle of two
prisms A and A' are so adjusted that the deviation
produced by the mean ray by the first prism is equal
and opposite to that produced by the second prism,
then the total final beam will be parallel to the
A'
XtraEdge for IIT-JEE 24 FEBRUARY 2010

incident beam and there will be dispersion without
deviation.
Here,
δ = δ 1 – δ 2 = 0 or δ 1 = δ 2
i.e., (µ 1 – 1)A = (µ 2 – 1)A'
This combination produces total angular dispersion.
D = D 1 – D 2 = (µ 1V – µ 1R ) A – (µ 2V – µ 2R )A'
(xiii) Deviation without dispersion :
If the combination is such that D = D 1 ~ D 2 = 0
or D 1 = D 2
or (µ 1V – µ 1R ) A = (µ 2V – µ 2R )A'
The combination is said to be achromatic and the
total mean deviation will be
δ = δ 1 ~ δ 2 = (µ 1 – 1)A ~ (µ 2 – 1)A'
Wave nature of light ::
Wave front :
A point source produces a spherical wave front
⎛ 1⎞
⎡ 1 ⎤
⎜ A ∝ ⎟ or
⎝ r
⎢1
∝ ⎥
⎠ ⎣ r 2 ⎦
Where A = Amplitude, I = intensity and
r = distance of point of observation from source.
A line source produces a cylindrical wave front
⎡ 1 ⎤ ⎡ 1⎤
⎢A ∝ ⎥ or ⎢I ∝ ⎥ .
⎣ r ⎦ ⎣ r ⎦
Wave front is locus of points in the same phase.
A distance source produce a plane wave front.
Wave front for a parallel beam of light is plane.
The angle between ray and wave front is 90º
Huygen's principle:
Huygen's principle is a geometrical method to find
secondary wave front produced by a primary wave
front.
Interference of light :
(a) Redistribution of light energy i.e. alternate
maximum and minima).
Conditions for two light waves producing
interference is that
(i) Wave should be of same wavelength/frequency.
(ii) Waves should be travelling in the same direction.
(iii)Wave should have a constant phase difference
For the above conditions the two source must be
coherent and that is possible when we make two
sources out of a single source of light.
For monochromatic light we get alternate maxima
and minima of same colour. For white light we
get white central fringe flanked by coloured
fringes because fringe width of different colour is
different due to different wavelengths.
(b) Resultant intensity at a point is
I = I 1 + I 2 + 2 I1 I2
cos φ
When I 1 = I 2 = I 0 then I = 4I 0 cos 2 φ/2
For constructive interference
φ = ± 2nπ and ∆x = ±nλ
I max = ( I 1 + I 2 ) 2 ∝ (A 1 + A 2 ) 2 [Q I ∝ A 2 ]
For destructive interference
⎛ 1 ⎞
φ = (2n + 1)π and ∆x = ⎜ n − ⎟ λ
⎝ 2 ⎠
I min = ( I 1 – I 2 ) 2 ∝ (A 1 – A 2 ) 2
Imax
( I1
+ I2
) (A1
+ A 2)
⇒ =
=
I
2
min ( I1
− I2
) (A1
− A 2)
The energy remains conserved during the process
of interference.
P
S 1
α
2
S 2
Thin lines shows the rays of light.
Dotted line shows the wavefronts.
Intensity of light at any point P as shown the
figure I = I 0 cos 2 ⎛ πd
tan α ⎞
⎜ ⎟
⎝ λ ⎠
(c) The fringe width β =
λD
d
Angular width θ =
D
β =
d
λ
⇒ θ does not depend on D
XtraEdge for IIT-JEE 25 FEBRUARY 2010

(d) When the source of light is placed asymmetrical
with respect to the slits then the central maxima
also shifts.
S
x
D x
α
S 1
S 2
y x = and θ = – α
D y Dx
(e) If young's double slit experiment is done in a
liquid of refractive index medium µ then the
fringe width β´ = β/µ
(f)
S 1
S 2
t
θ
D y
P
O´
O
If a transparent sheet of thickness t is placed in
front of upper slit then the central maxima shift
upside. The new optical path becomes µt instead
of t and the increase in optical path is (µ – 1)t.
The shift = d
D (µ – 1)t = λ
β (µ – 1)t
(g) Interference in thin films :
y
Diffraction :
Bending of light through an aperture / corner when
the dimension of aperture is comparable to the
wavelength of light is called diffraction.
Fraunhoffer diffraction at a single slit
Condition for minima :
a sin θ n = n λ
Condition of secondary maxima :
⎛ 1 ⎞ λ
a sin θ n = ⎜n
+ ⎟ Where n = n = 1, 2 ...
⎝ 2 ⎠ 2
Width of central maxima = 2λD/a
a
Polarisation :
θ
Angular width of central maxima = 2λ/a
Angular width of secondary maxima = λ/a
2
⎡sin
α ⎤
Intensity at any point P = I 0 ⎢ ⎥
⎣ α ⎦
where α = λ
π (a sin θ)
The ratio of intensities of secondary maxima are
1 1 1 , , , ...
22 61 121
For a path difference of λ, the phase difference is
2π radian.
I 0
I = I 0 /2
P
O
I cos 2 θ
i
1
2
t
r
r
r
r
Unpolarised
light
Polarised
light
Transmitted rays
For reflected rays interference
λ
Maxima 2 µt cos r = (2n – 1) 2
Minima 2 µt cos r = nλ
r i p
Medium 1
Medium 2 ½ µ
i p = Angle of polarization, i p + r = 90º, µ = tan i p
XtraEdge for IIT-JEE 26 FEBRUARY 2010

Solved Examples
1. (i) A ray of light incident normally on one of the
faces of a right-angled isosceles prism is found to be
totally reflected. What is the minimum value of the
refractive index of the material of prism
(ii) When the prism is immersed in water, trace the
path of the emergent ray for the same incident ray
indicating the values of all the angles (µ ω = 4/3)
Sol. (i) According to the problem, ∠A = 90º, ∠B = ∠C = 45º.
At face BC, incident ray PQ is totally reflected
therefore i ≥ C fig.
Here
R
A
C
N
r=i
i
P
Q
45º
N´
i = 45º, ∴ C max = 45º; ∴ µ = 1/sin C
or µ min = (1/sin C max ) = (1/sin 45º) = 2 = 1.414
(ii) When the prism is immersed in water, then for
normal incident ray, the ray passes undeviated up to
PQ and becomes incident at face BC at angle of
incidence 45º(fig.) The ray travels from glass to
water, therefore from Snell's law,
sin i
sin r
=
∴ sin r =
sin 45º µ
=
µ
µ 2 , we have
µ 1
sin r
=
sin 45º
gµ w
=
sin 45º
× µ g
µ w
1 1.414 3 × = = 0.75
2 4 / 3 4
∴ r = sin –1 (0.75) = 48º36´
P
A
90º 45º
R
45º
Q
45º
r
R
C
The path of light ray is shown in fig.
w
g
B
B
= g µ w
2. The cross-section of a glass prism has the form of an
isosceles triangle. One of the equal faces is coated
with silver. A ray is normally incident on another
unsilvered face and being reflected twice emerges
through the base of the prism perpendicular to it.
Find the angles of the prism.
Sol. Suppose refracting angle of prism be α and other two
base angles of the isosceles prism be β. The light ray
PQ, incident normally on the face AB, is refracted
undeviated along QR. The refracted ray QR strikes
the silvered face AC and gets reflected from it. The
reflected ray RS now strikes the face AB from where
it is again reflected along ST and emerges
perpendicular to base BC.
P
N´2
B
β
T
S
A
α
Q
i = α
N´1
90º
N 2 β
C
N 1 i R
θ=β
It follows from fig. that angle of incidence on face
AC = i = α and also angle of incidence of face
AB = θ = β As N 2 N 2´ is parallel to PR, hence θ = 2i
i.e.,
β = 2α
Also α + 2β = 180º or α + 2(2α) = 180º
or α = 36º so β = 2α = 72º
3. Two coherent light sources A and B with separation
2λ are placed on the x-axis symmetrically about the
origin. They emit light of wavelength λ. Obtain the
positions of maximum on a circle of large radius,
lying in the xy-plane and with centre at the origin.
Sol. Distance between two coherent light sources = 2λ.
Consider the interference of waves at some point C of
the circumference of circle.
2
BC = (r + λ − 2rλ
cosθ)
2
AC = (r + λ + 2rλ
cosθ)
2
2
∴ Path difference = AC – BC = λ (For Maxima)
It is clear from figure, that
Path difference = AC – BC = AP + OM = 2λ cos θ
XtraEdge for IIT-JEE 27 FEBRUARY 2010

A
Y
P
θ
O
λ
M
r
θ
B
λ
C
X
Thus there is brightness at O of nth order. Since the
path difference decreases, the other fringes will be of
lower order. The next bright fringe will be of
(n – 1)th order. Hence for the next bright fringe
D 2 – D 2 = (n – 1)λ
x 2
d – d = (n – 1)λ
2D(D + d)
∴ 2λ cos θ = λ or cos θ = 1/2
∴ Possible values of angle θ = 60º, 120º, 180º, 240º,
300º, 360º.
4. Two point coherent sources are on a straight line
d = nλ apart. The distance of a screen perpendicular
to the line of the sources is D >> d from the nearest
source. Calculate the distance of the point on the
screen where the first bright fringe is formed.
Sol. Consider any point P on the screen at a distance x
from O. Then
S 1
d
S 2
D
1/ 2
2
D 1 = D 2 + x 2 ⎛
2
x ⎞ ⎛
2
or D 1 = D ⎜1
+ ⎟
2 = D
D
⎟ ⎞
⎜
x
1 + ;
2
⎝ ⎠ ⎝ 2D ⎠
∴ D 1 = D +
x 2
2D
x 2
Similarly, D 2 = (D + d) +
2(D + d)
x 2
∴ D 2 – D 1 = (D + d) +
2(D + d)
x 2
= d + 2
⎛ 1 1 ⎞
⎜ − ⎟ = d – d
⎝ D + d D ⎠
D 2
O
– D – D 2
x 2
x 2
2D(D +
D 1
x
D O
S
d
1 S 2
For the point O, D 2 – D 1 = d = nλ (given).
P
d)
∴ x =
x 2
nλ – nλ
= (n – 1)λ
2D(D + nλ)
2D(D
+ nλ)
n
5. One slit of a Young's experiment is covered by a
glass plate (n = 1.4) and the other by another glass
plate (n' = 1.7) of the same thickness t. The point of
central maximum on the screen, before the plates
were introduced, is now occupied by the previous
fifth bright fringe. Find the thickness of the plates
(λ = 4800 Å)
Sol. Path of the wave from slit S 1 = D 1 + n't – t
Path of the wave from slit S 2 = D 2 + nt – t
∴ Path difference = D 2 + nt – t – D 1 – n't + t
= (D 2 – D 1 ) + (n – n')t
But
∴
S 1
D 2
O′
D 1
O
d
S 2 D
D 2 – D 1 =
D
xd
Path difference = D
xd + (n – n')t
Let O' be the point where paths difference is zero.
xd
∴ = (n' – n)t
D
D (n' −n)
tβ ⎡
or, x = (n' – n) t =
d λ
⎢Q
β =
⎣
(n' −n)
tβ
Given that x = 5β ∴ 5β =
λ
5λ
or, t =
n' −n
5×
4800×
10
or, t =
1.7×
1.4
−10
x
λD
⎤
d
⎥
⎦
= 8 µm.
XtraEdge for IIT-JEE 28 FEBRUARY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
Waves & Doppler Effect
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Key Concepts :
1. Equation of a harmonic wave is y = a sin(kx ± ωt ± φ).
Here y is measure of disturbance from zero level.
y may represent as electric field, magnetic field,
pressure etc. Also K = 2π/λ = wave number.
Note : The positive sign between kx and ωt shows
that the wave propagates is the +x direction. If the
wave travels in the –x direction then negative sign is
used between kx and ωt.
2. Particle Velocity :
dy
v = = aω cos (kx ± ωt ± φ)
dt
∴ Maximum particle velocity = aω = velocity amplitude
Particle velocity is different from wave velocity.
The wave velocity v = vλ.
3. Particle acceleration :
2
dv d y
A = = = – aω 2 cos(kx + ωt ± φ) = – ω 2 y
dt
2
dt
Max acceleration = acceleration amplitude = –ω 2 a
T
4. Velocity of transverse wave on a string =
m
πD 2
Where m = mass per unit length = ρ ×
4
Where ρ = density of the wire material and
D = diameter of wire
More the tension, more is the velocity
5. A wave, after reflection from a free end, suffers
change of π.
A wave, after reflection from a free end, suffers no
change in the phase.
6. Velocity of sound in a fluid =
For air B = γP ∴ v =
γP =
ρ
B
ρ
γRT
M
Velocity of sound in general follows the order
V solid > V liquid > V gas
⇒ Velocity of sound ∝ T
Also velocity of sound ∝
γ / M
and velocity of sound ∝ 1 / ρ .
But velocity of sound does not depend on pressure
because P/ρ becomes constant.
Velocity of sound depend on the frame of reference.
7. (a) According to principle of superposition
→ →
y = y 1 + y
→ 2 + ….
(b) Interference of waves y 1 = A 1 sin(kx – ωt)
y 2 = A 2 sin(kx – ωt + φ)
For constructive interference φ = 2nπ
n = 0, 1, 2, ……
(i) I max = ( I 1 + I 2 ) 2
(waves should be in same phase)
(ii) A max = A 1 + A 2
For destructive interference φ = (2n + 1)π
n = 0, 1, 2, ……
(i) I min = ( I 1 – I 2 ) 2
(waves should be in opposite phase)
(ii) A min = A 1 – A 2
(c) I = I 1 + I 2 + 2 I1I2
cos φ
Where φ is the phase difference between the two
waves.
8. Beats : When two waves of same amplitude with
slight difference in frequency (

9. Standing waves (stationary)
When two waves of same amplitude and frequency
moving in opposite direction superimpose, standing
waves are produced.
Nodes are the point where the displacement is
always zero.
The amplitudes of different particles different and
is maximum at antinodes.
The equation of standing waves is
y = [2A sin kx]cos ωt where amplitude = 2A sin kx
The above expression shows that the amplitude is
different for different values of x and varies
sinusoidally.
For a node to occur at position x, y = 0 ⇒ kx = 0
For an antinode two occur at position x, y should
be max ⇒ kx = π/2 , ….
In terms of pressure ∆P = ∆P 0 cos kx cos ωt.
10. For standing waves on strings (and both end open
organ pipe)
v v
Fundamental frequency v 0 = =
λ0
2l
v
First mode of vibration v 1 =
λ = 2 ⎛ v
⎟ ⎞
⎜
1 ⎝ 2l ⎠
= 2v 0 = 2 nd harmonic
⎛ v ⎞
nth mode of vibration v n = n ⎜ ⎟ = nv 0
⎝ 2l ⎠
T
= nth harmonic where v = for string.
m
Also more the tension in the same string, higher is
the value of v 0
11. For closed organ pipe :
v v
Fundamental frequency v 0 = =
λ0
4l
v
First mode of vibration v 1 =
λ = 3 ⎛ v
⎟ ⎞
⎜ = 3v 0
1 ⎝ 4l ⎠
= Third harmonic
n th v
mode of vibration v n = (2n + 1)
4l
where n = 1, 2, …..
In case where end correction is taken replace l by
(l + e)
12. (a) Intensity of sound at a distance r from a point
P
source is I = where P = power of source.
2
4πr
P
(b) For a line source I =
πrl
where l is the length of source
(c) I =
2
1 ρv(4π 2 v 2 )A 2 =
(Pr essure amplitude)
2ρν
2
13. Doppler's effect :
⎡ v ± vL
⎤
v = v 0 ⎢ ⎥ v L = velocity of listener
⎣ v ± vs
⎦
The above formula is valid when v s < v
Replace v by (v ± v m ) if it is given that the
medium also moving.
When listener and source are not moving along
the line joining the two, then the component of
velocity along the line joining the two are taken
as velocity of listener or source.
14. If the source and listener are on the same vehicle and
the sound is reflected from a stationary object
towards which the vehicle is approaching then the
frequency of sound as heard by the observer is
⎡ v + vL
⎤
v´ = v 0 ⎢ ⎥
⎣ v + vs
⎦
15. For a path difference of λ, the phase difference is 2π
for harmonic waves.
16. For a transverse wave the energy per unit length
possessed by a string is given as
dE = m(4π 2 f 2 )A 2 cos 2 (kx – ωt)
dl
17. Equation for a wave pulse is y = f(x + vt)
18. When a wave on reaching on interface is partly
reflected and partly transmitted then for no power
loss.
P i = P t + P r where P i = Power of incident wave
P t = Power of transmitted wave
P r = Power of reflected wave.
⎛ v
Also in this case A r =
⎜
⎝ v
2
2
− v
+ v
1
1
⎞ ⎛
⎟ A i ; A t =
⎜
⎠ ⎝
2v
v
2
1 + v 2
Where A i , A r and A t are amplitudes of incident
reflected and transmitted waves v 1 is the velocity in
the medium of incidence and v 2 is the velocity in the
medium where transmitted wave is present.
Problem Solving Strategy : Mechanical Waves
Identify the relevant concepts : Wave problems fall
into two broad categories. Kinematics problems are
concerned with describing wave motion; they involve
wave speed v, wave length λ(or wave number k),
frequency f (or angular frequency ω), and amplitude
A. They may also involve the position, velocity, and
acceleration of individual particles in the medium.
Dynamics problems also use concepts from Newton's
laws such as force and mass. In this chapter we'll
encounter problems that involve the relation of wave
speed to the mechanical properties of the wave
medium. We'll get into these relations.
As always, make sure that you identify the target
variable(s) for the problem. In some cases it will be
⎞
⎟
⎠
A i
XtraEdge for IIT-JEE 30 FEBRUARY 2010

A
A
y
the wavelength, frequency, or wave speed; in other
cases you'll be asked to find an expression for the
wave function.
Set up the problem using the following steps :
Make a list of the quantities whose value are
given. To help you visualize the situation, you'll
find it useful to sketch graphs of y versus x (fig.
a) and of y versus (fig. b). Label your graphs with
the values of the known quantities.
Wave displacement
versus coordinate x
at time t = 0
Wavelength λ
(a)
t
A
A
y
Wave displacement
versus time t at
coordinate x = 0
Period T
(b)
Decide which equations you'll need to use. If any
two of v, f, and λ are given, you'll need to use eq.
v = λf (periodic wave) to find the third quantity.
If the problem involves the angular frequency ω
and / or the wave number k, you'll need to use the
definitions of those quantities and eq. (ω = vk).
You may also need the various forms of the wave
function given in Eqs.
⎡ ⎛ x ⎞⎤
⎛ x ⎞
y(x, t) = A cos ⎢ω⎜
− t ⎟⎥ = A cos 2πf ⎜ − t ⎟ ,
⎣ ⎝ v ⎠ ⎦ ⎝ v ⎠
⎛ x
y(x, t) = A cos 2π ⎜ −
⎝ λ
and y(x, t) = A cos (kx – ωt).
If the wave speed is not given, and you don't have
enough information to determine it using v = λf,
you may be able to find v using the relationship
between v and the mechanical properties of the
system.
Execute the solution as follows : Solve for the
unknown quantities using the equations you've
selected. In some problems all you need to do is find
the value of one of the wave variables.
If you're asked to determine the wave function, you
need to know A and any two of v, λ and f(or v, k and
ω). Once you have this information, you can use it in
eq. (ω = vk). You may also need the various forms of
the wave function given in Eqs.
⎡ ⎛ x ⎞⎤
⎛ x ⎞
y(x, t) = A cos ⎢ω⎜
− t ⎟⎥ = A cos 2πf ⎜ − t ⎟ ,
⎣ ⎝ v ⎠ ⎦ ⎝ v ⎠
⎛ x
y(x, t) = A cos 2π ⎜ −
⎝ λ
t
T
t
T
⎟
⎠
⎞
⎞
⎟ and y(x, t) = A cos (kx – ωt)
⎠
to get the specific wave function for the problem at
hand. Once you have that, you can find the value of y
t
at any point (value of x) and at any time by
substituting into the wave function.
Evaluate your answer : Look at your results with a
critical eye. Check to see whether the values of v, f,
and λ (or v, ω, and k) agree with the relationships
given in eq. . v = λf or w = vk. If you've calculated
the wave function, check one or more special cases
for which you can guess what the results ought to be.
Problem Solving Strategy : Standing waves
Identify the relevant concepts : As with traveling
waves, it's useful to distinguish between the purely
kinematic quantities, such as wave speed v,
wavelength λ, and frequency f, and the dynamic
quantities involving the properties of the medium,
such as F and µ for transverse waves on a string.
Once you decide what the target variable is, try to
determine whether the problem is only kinematic in
nature or whether the properties of the medium are
also involved.
Set up the problem using the following steps :
In visualizing nodes and antinodes in standing
waves, it is always helpful to draw diagrams. For
a string you can draw the shape at one instant and
label the nodes N and antinodes A. The distance
between two adjacent nodes or two adjacent
antinodes is always λ/2, and the distance between
a node and the adjacent antinode is always λ/4.
Decide which equation you'll need to use. The
wave function for the standing wave is almost
always useful ex. y(x, t) = (A SW sin kx) sin ωt.
You can compute the wave speed if you know
either λ and f (or, equivalently, k = 2π/λ and
ω = 2πf) or the properties of the medium (for a
string. F and µ.)
Execute the solution as follows: Solve for the
unknown quantities using the equations you've
selected. Once you have the wave function, you can
find the value of the displacement y at any point in
the wave medium (value of x) and at any time. You
can find the velocity of a particle in the wave
medium by taking the partial derivative of y with
respect to time. To find the acceleration of such a
particle, take the second partial derivative of y with
respect to time.
Evaluate your answer : Compare your numerical
answers with your diagram. Check that the wave
function is compatible with the boundary conditions
(for example, the displacement should be zero at a
fixed end).
Problem Solving Strategy : Sound Intensity
Identify the relevant concepts : The relationships
between intensity and amplitude of a sound wave are
rather straightforward. Quite a few other quantities
are involved in these relationships, however, so it's
particularly important to decide which is your target
variable.
XtraEdge for IIT-JEE 31 FEBRUARY 2010

Set up the problem using the following steps :
Sort the various physical quantities into
categories. The amplitude is described by A or
p max , and the frequency f can be determined from
ω, k, or λ. These quantities are related through the
wave speed v, which in turn is determined by the
properties of the medium: B and ρ for a liquid; γ,
T, and M for a gas.
Determine which quantities are given and which
are the unknown target variables. Then start
looking for relationships that take you where you
want to go.
Execute the solution as follows: Use the equations
you've selected to solve for the target variables. Be
certain that all of the quantities are expressed in the
correct units. In particular, if temperature is used to
calculate the speed of sound in a gas, make sure that
it is expressed in Kelvins (Celsius temperature plus
273.15).
Evaluate your answer: There are multiple
relationships among the quantities that describe a
wave. Try using an alternative one to check your
results.
Problem Solving Strategy : Doppler Effect
Identify the relevant concepts : The Doppler effect is
relevant whenever the source of waves, the wave
detector (listener), or both are in motion.
Set up the problem using the following steps :
Establish a coordinate system. Define the positive
direction to be the direction from the listener to
the source, and make sure you know the signs of
all relevant velocities. A velocity in the direction
from the listener toward the source is positive; a
velocity in the opposite direction is negative.
Also, the velocities must all be measured relative
to the air in which the sound is traveling.
Use consistent notation to identify the various
quantities: subscript S for source, L for listener.
Determine which unknown quantities are your
target variables.
Execute the solutions :
v + v L
Use eq. f L = f S to relate the frequencies at
v + vS
the source and the listener, the sound speed, and
the velocities of the source and the listener. If the
source is moving, you can find the wavelength
measured by the listener using Eq.
v v
λ = –
S v − vS
v + vS
= or λ = .
f S fS
fS
fS
When a wave is reflected from a surface, either
stationary or moving, the analysis can be carried
out in two steps. In the first, the surface plays the
role of listener; the frequency with which the
wave crests arrive at the surface is f L . Then think
of the surface as a new source, emitting waves
with this same frequency f L . Finally, determine
what frequency is heard by a listener detecting
this new wave.
Evaluate your answer: Ask whether your final result
makes sense. If the source and the listener are
moving towards each other, f L > F S ; if they are
moving apart, f L < f S . If the source and the listener
have no relative motion, f L = f S .
Solved Examples
1. A stationary wave is given by
πx
y = 5 sin cos 40 πt
3
where x and y are in cm and t is in seconds.
(a) What are the amplitude and velocity of the
component waves whose superposition can give rise
to this vibration
(b) What is the distance between the nodes
(c) What is the velocity of a particle of the string at
the position x = 1.5 cm when t = 9/8 s
Sol. Using the relation 2 sin C cos D = sin (C + D) +
sin(C – D)
πx
5 πx
y = 5 sin cos 40 πt = × 2 sin cos 40πt
3 2 3
5 ⎡ ⎛ πx
⎞ ⎛ πx
⎞⎤
⇒ y = ⎢sin⎜
+ 40πt
⎟ + sin⎜
− 40πt⎟⎥ 2 ⎣ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎦
5 ⎛ πx
⎞ 5 ⎛ πx
⎞
= .sin ⎜ 40 πt
+ ⎟⎠ – sin ⎜ 40 πt
− ⎟⎠ 2 ⎝ 3 2 ⎝ 3
5 ⎛ πx
⎞ 5 ⎛ πx
⎞
= .sin ⎜ 40 πt
+ ⎟⎠ + sin ⎜40
πt
− + π ⎟⎠ 2 ⎝ 3 2 ⎝ 3
Thus, the given stationary wave is formed by the
superposition of the progressive waves
5 ⎛ πx
⎞ 5 ⎛ πx
⎞
y 1 = sin ⎜ 40 πt
+ ⎟⎠ and y 2 = sin ⎜40
πt
− + π ⎟⎠ 2 ⎝ 3
2 ⎝ 3
Comparing each wave with the standard form of the
progressive wave
⎛ π ⎞
y = a sin ⎜ω −
2
5
t + α⎟ ; a = = 2.5 cm
⎝ λ ⎠ 2
ω = 40π or n = 20
2π 2π
and = or λ = 6 cm = 0.06 m
λ 3
∴ c = nλ = 20 × 0.06 = 1.2 ms –1
λ 0.06
Distance between the nodes = = = 0.03 m
2 2
πx
Q y = 5 sin cos 40 πt
3
XtraEdge for IIT-JEE 32 FEBRUARY 2010

dy πx
v = = – 5 × 40π sin sin 40 πt
dt
3
πx
⇒ v = – 200 π sin sin 40πt 3
∴ At x = 1.5 cm and t = 8
9 s
v = – 200π sin 45π = 0
2. An engine blowing a whistle of frequency 133 Hz
moves with a velocity of 60 m s –1 towards a hill from
which an echo is heard. Calculate the frequency of
the echo heard by the driver. (Velocity of sound in air
= 340 ms –1 .)
Sol. The 'image' of the source approaches the driver at the
same speed. Here, the image or echo is the source.
∴ v s = + 60 ms –1 , v 0 = – 60 ms –1
c − v0
n´ = × n
c − vs
∴
340 − ( −60)
n´ =
× 133 = 190 Hz
340 − 60
3. A source of sound of frequency 1000 Hz moves to
the right with a speed of 32 ms –1 relative to the
ground. To its right is reflecting surface moving to
the left with a speed of 64 ms –1 relative to the ground.
Take the speed of sound in air to be 332 m s –1 and
find
(a) the wavelength of the sound emitted in air by the
source
(b) the number of waves per second arriving at the
reflecting surface
(c) the speed of the reflected waves, and
(d) the wavelength of the reflected waves
Sol. (a) Due to the motion of the source, the wavelength
(and hence, the frequency) is actually changed from λ
to λ´ such that if n = actual frequency
c − vS
332 − 32
λ´ = = = 0.3 m
n 1000
(b) The number of waves arriving at the reflecting
surface is the same as the number of waves received
by an observer moving towards the source. This is
given by the apparent frequency.
c − v0
332 − ( −64)
n´ = × n =
× 1000 = 1320 Hz
c − vS
332 − 32
(c) Same as that of the incident wave because the
speed of a wave depends only on the characteristics
of the medium.
∴ speed of the reflected wave = 332 ms –1
(d) To calculate the wavelength of the reflected wave,
we may consider the source to be stationary and
emitting waves of wavelength 0.3 m. If the reflector
were stationary, waves in a tube of length c would
reach the reflector and the same number of reflected
waves would be contained in a tube of the same
length, so the wavelength of the reflected wave
would also be the same as that of the incident wave.
But when the reflector moves towards the source
with speed v ref´ it would reflect additional waves
contained in v ref and the total number of waves
reflected would be contained in a tube of length
c – v ref . If λ´is the changed wavelength of the wave
due to the motion of the source
⎛ c vref
⎞
λ ´´ = ( c − vref
) ⎜ + ⎟
⎝ λ´
λ´
⎠
c − vref
332 − 64
or λ´´ = × λ´ = × 0.3 = 0.2 m
c + v 332 + 64
ref
4. Find the ratio of the fundamental frequencies of two
identical strings after one of them is stretched by 2%
and the other by 4%.
Sol. n =
1
2l
T
m
. If l 0 be the initial length and f be
fractional increase in length, l = l 0 + fl 0 . Since
tension is proportional to the increase in length,
T = k × fl 0 where k is a constant.
m =
∴ n =
M
l 0 + fl
0
where M is the mass of the string
1 kfl
0 / M
2l (1 f ) l (1 + f )
0 +
0
=
2l
Since l 0 , k and M are constants n ∝
∴
n
n
1
2
=
f1(1
+ f 2)
=
f (1 + f )
2
1
kl
f
1 0
0 M(1 +
f
1+
f
0.02(1 + 0.04)
0.04(1 + 0.02)
f )
= 0.71
5. An open organ pipe has a fundamental frequency of
300 Hz. The first overtone of a closed organ pipe has
the same frequency as the first overtone of the open
pipe. How long is each pipe The velocity of sound
in air = 350 ms –1 .
c
Sol. For a closed pipe n = and 3n, 5n, 7n, ... are the
4l
c
overtones. For an open pipe n = and 2n, 3n, 4n,
2l
... are the overtones.
c 350
⇒ l = = = 0.58 m
2n 2×
300
The frequency of the first overtone
= 2n = 2 × 300 = 600 Hz
∴ the frequency of the first overtone of the closed
pipe = 600 = 3n
∴ n = 200 Hz
350 350
∴ 200 = or l = = 0.44 m
4l
4×
200
XtraEdge for IIT-JEE 33 FEBRUARY 2010

KEY CONCEPT
Organic
Chemistry
Fundamentals
CARBONYL
COMPOUNDS
The Diels-Alder reaction :
α, β-Unsaturated carbonyl compounds undergo an
exceedingly useful reaction with conjugated dienes,
known as the Diels-Alder reaction. This is an
addition reaction in which C-1 and C-4 of the
conjugated diene system become attached to the
doubly bonded carbons of the unsaturated carbonyl
compound to form a six membered ring.
C
C
C
C
+
C
C
O
C
Diene Dienophile
(Greek: diene-loving)
C
C
C
C
C
C
O
C
Adduct
Six-membered ring
A concerted, single-step mechanism is almost
certainly involved; both new carbon-carbon bonds
are partly formed in the same transition state,
although not necessarily to the same exent. The
Diels-Alder reaction is the most important example
of cycloaddition, Since reaction involves a system of
four π electrons (the diene) and a system of two π
electrons (the dienophile), it is known as a [4 + 2]
cycloaddition.
The Diels-Alder reaction is useful not only because a
ring is generated, but also because it takes place so
readily for a wide variety of reactants. Reaction is
favored by electron-withdrawing substituents in the
dienophile, but even simple alkenes can react.
Reaction often takes place with the evolution of heat
when the reactants are simply mixed together. A few
examples of the Diels-Alder reaction are:
O
O
CH 2
H
HC
C C
C
benzene, 20 ºC
+ O
O
quantitative
HC
C
C
C
CH 2 H
O
O
1,3-Butadiene Maleic anhydride cis-1,2,3,6-Tetrahydrophthalic
anhydride
H
CH 2 C O
HC HC 100 ºC
CHO
+
quantitative
HC HC
CH 2 H 1,2,3,6-
1,3-Butadiene Acrolein Tetrahydrobenzaldehyde
O
O
HC
CH 2
benzene,35ºC
+
HC
quantitative
CH 2
O
O
1,3- p-Benzoqpuinone 5,8,9,10-Tetrahydro-
Butadiene
1,4-naphthoquinone
1,3-butadiene,
100 ºC
O
O
1,4,5,8,11,12,13,14
-Octahydro-9,10-anthraquinone
Cannizzaro reaction :
In the presence of concentrated alkali, aldehydes
containing no α-hydrogens undergo self-oxidationand
-reduction to yield a mixture of an alcohol and a
salt of a carboxylic acid. This reaction, known as the
Cannizzaro reaction, is generally brought about by
allowing the aldehyde to stand at room temperature
with concentrated aqueous or alcoholic hydroxide.
(Under these conditions an aldehyde containing
α-hydrogens would undergo aldol condensation faster)
2HCHO ⎯
50 ⎯
% ⎯
NaOH ⎯→
CH 3 OH + HCOO – Na +
Formaldehyde Methanol Sodium formate
O 2 N CHO 35%NaOH
p-Nitrobenzaldehyde
O 2 N
CH 2 OH
p-Nitrobenzl alcohol
+
O 2 N COO – Na +
Sodium p-nitrobenzoate
XtraEdge for IIT-JEE 34 FEBRUARY 2010

In general, a mixture of two aldehydes undergoes a
Cannizzaro reaction to yield all possible products. If
one of the aldehydes is formaldehyde, however,
reaction yields almost exclusively sodium formate
and the alcohol corresponding to the other aldehyde:
ArCHO + HCHO ⎯
conc. ⎯⎯
NaOH ⎯ →
ArCH 2 OH + HCOO – Na +
Such a reaction is called a crossed Cannizzaro
reaction. For example :
CHO
CH 2 OH
conc. NaOH
+ HCHO
+ HCOO – Na +
OCH 3
Anisaldehyde
(p-Methoxy
benzaldehyde)
OCH 3
p-Methoxybenzyl alcohol
Evidence, chiefly from kinetics and experiments with
isotopically labeled compounds, indicates that even
this seemingly different reaction follows the familiar
pattern for carbonyl compounds: nucleophilic
addition. Two successive additions are involved:
addition of hydroxide ion (step 1) to give
intermediate I;
(1)
(2)
H
Ar–C=O + OH – Ar–C–O –
H
H
Ar–C = O + Ar–C–O –
OH
I
H
Ar–C–O –
H
OH
I
+ Ar–C=O
H OH
+H + –H +
ArCH 2 OH ArCOO –
and addition of a hydride ion from I (step 2) to a
second molecule of aldehyde. The presence of the
negative charge on I aids in the loss of hydride ion.
Reduction :
Aldehydes can be reduced to primary alcohols, and
ketones to secondary alcohols, either by catalytic
hydrogenation or by use of chemical reducing agents
like lithium aluminum hydride, LiAlH 4 . Such
reduction is useful for the preparation of certain
alcohols that are less available than the corresponding
carbonyl compounds, in particular carbonyl
compounds that can be obtained by the aldol
condensation. For example :
O
Cyclopentanone
2−Butenal
Crotonaldehyde
Fromaldolcondensation
of acetaldehyde
LiAIH 4 H + H OH
H ,Ni
Cyclopentanol
2
CH3CH
= CHCHO ⎯⎯⎯→
CH CH CH CH OH
CH=CHCHO
3-Phenylpropenal
Cinnamaldehyde
From aldol condensation
of benzaldehyde and acetaldehyde
HOCH 2CH 2NH 2
9-BBN
3 2 2 2
n−Butylalcohol
CH=CHCH 2 OH
3-Phenyl-2-propen-1-ol
Cinnamyl alcohol
To reduce a carbonyl group that is conjugated with a
carbon-carbon double bond without reducing the
carbon-carbon double bond, too, requires a
regioselective reducing agent.
Aldehydes and ketones can be reduced to
hydrocarbons by the action (a) of amalgamated zinc
and concentrated hydrochloric acid, the Clemmensen
reduction; or (b) of hydrazine, NH 2 NH 2 , and a strong
base like KOH or potassium tertbutoxide, the Wolff-
Kishner reduction. These are particularly important
when applied to the alkyl aryl ketones obtained from
Friedel – Crafts acylation, since this reaction
sequence permits, indirectly, the attachment of
straight alkyl chains to the benzene ring. For example
OH
OH
CH 3(CH 2) 4COOH, ZnCl 2
Zn(Hg),HCl
OH
OH
OH
CO(CH 2 ) 4 CH 3
OH
CH 2 (CH 2 ) 4 CH 3
4-n-Hexylresorcinol
Used as an antiseptic
Alcohols are formed from carbonyl compounds,
smoothly and in high yield, by the action of such
compounds as lithium aluminum hydride, LiAlH 4 .
4R 2 C=O + LiAlH 4 ⎯→ (R 2 CHO) 4 AlLi
⎯
H 2
⎯⎯ O →4R 2 CHOH + LiOH + Al(OH) 3 .
XtraEdge for IIT-JEE 35 FEBRUARY 2010

KEY CONCEPT
Inorganic
Chemistry
Fundamentals
CO-ORDINATION
COMPOUND & METALLURGY
Tetragonal distortion of octahedral complexes (Jahn-
Teller distortion) :
The shape of transition metal complexes are affected
by whether the d orbitals are symmetrically or
asymmetrically filled.
Repulsion by six ligands in an octahedral complex
splits the d orbitals on the central metal into t 2g and e g
levels. It follows that there is a corresponding
repulsion between the d electrons and the ligands. If
the d electrons are symmetrically arranged, they will
repel all six ligands equally. Thus the structure will
be a completely regular octahedron. The symmetrical
arrangements of d electrons are shown in Table.
Symmetrical electronic arrangements :
Electronic
configuration
d 5
d 6
d 8
d 10
t 2g
All other arrangements have an asymmetrical
arrangement of d electrons. If the d electrons are
asymmetrically arranged, they will repel some
ligands in the complex more than others. Thus the
structure is distorted because some ligands are
prevented from approaching the metal.
as closely as others. The e g orbitals point directly at
the ligands. Thus asymmetric filling of the e g orbitals
in some ligands being repelled more than others. This
causes a significant distortion of the octahedral
shape. In contrast the t 2g orbitals do not point directly
at the ligands, but point in between the ligand
directions. Thus asymmetric filling of the t 2g orbitals
has only a very small effect on the stereochemistry.
Distortion caused by asymmetric filling of the t 2g
orbitals is usually too small to measure. The
electronic arrangements which will produce a large
distortion are shown in Table.
The two e g orbitals d 2 2
x − y
and d are normally
z
2
degenerate. However, if they are asymmetrically
filled then this degeneracy is destroyed, and the two
e g
orbitals are no longer equal in energy. If the d 2
z
orbital contains one.
Asymmetrical electronic arrangements :
Electronic
configuration
d 4
d 7
d 9
t 2g
more electron than the d 2 2
x − y
orbital then the ligands
approaching along +z and –z will encounter greater
repulsion than the other four ligands. The repulsion
and distortion result in elongation of the octahedron
along the z axis. This is called tetragonal distortion.
Strictly it should be called tetragonal elongation. This
form of distortion is commonly obsered.
If the d 2 2
x − y
orbital contains the extra electron, then
elongation will occur along the x and y axes. This
means that the ligands approach more closely along
the z-axis. Thus there will be four long bonds and
two short bonds. This is equivalent to compressing
the octahedron along the z axis, and is called
tetragonal compression, and it is not possible to
predict which will occur.
For example, the crystal structure of CrF 2 is a
distorted rutile (TiO 2 ) structure. Cr 2+ is octahedrally
surrounded by six F – , and there are four Cr–F bonds
of length 1.98 – 2.01 Å, and two longer bonds of
length 2.43 Å. The octahedron is said to be
tetragonally distorted. The electronic arrangement in
Cr 2+ is d 4 . F – is a weak field ligand, and so the t 2g
level contains three electrons and the e g level contains
one electron. The d 2 2
x − y
orbital has four lobes whilst
the d 2 orbital has only two lobes pointing at the
z
ligands. To minimize repulsion with the ligands, the
single e g electron will occupy the d 2 orbital. This is
z
equivalent to splitting the degeneracy of the e g level
so that d 2 is of lower energy, i.e. more stable, and
z
d is of higher energy, i.e. less stable. Thus the
−
2 2
x y
e g
XtraEdge for IIT-JEE 36 FEBRUARY 2010

two ligands approaching along the +z and –z
directions are subjected to greater repulsion than the
four ligands along +x, –x, +y and –y. This causes
tetragonal distortion with four short bonds and two
long bonds. In the same way MnF 3 contains Mn 3+
with a d 4 configuration, and forms a tetragonally
distorted octahedral structure.
Many Cu(+II) salts and complexes also show
tetragonally distorted octahedral structures. Cu 2+ has
a d 9 configuration :
t 2g
To minimize repulsion with the ligands, two
electrons occupy the d 2 orbital and one electron
z
occupies the d 2 2
x − y
orbital. Thus the two ligands
along –z and –z are repelled more strongly than are
the other four ligands.
The examples above show that whenever the d 2 and
z
d 2 2
x − y
orbitals are unequally occupied, distortion
occurs. This is know as Jahn–Teller distortion.
Leaching :
It involves the treatment of the ore with a suitable
reagents as to make it soluble while impurities
remain insoluble. The ore is recovered from the
solution by suitable chemical method. For example,
bauxite ore contains ferric oxide, titanium oxide and
silica as impurities. When the powdered ore is
digested with an aqueous solution of sodium
hydroxide at about 150ºC under pressure, the alumina
(Al 2 O 3 ) dissolves forming soluble sodium metaaluminate
while ferric oxide (Fe 2 O 3 ), TiO 2 and silica
remain as insoluble part.
Al 2 O 3 + 2NaOH → 2NaAlO 2 + H 2 O
Pure alumina is recovered from the filtrate
NaAlO 2 + 2H 2 O ⎯→ Al(OH) 3 + NaOH
2Al(OH) 3
Ignited ⎯⎯
(autoclave)
e g
⎯ → Al 2 O 3 + 3H 2 O
Gold and silver are also extracted from their native
ores by Leaching (Mac-Arthur Forrest cyanide
process). Both silver and gold particles dissolve in
dilute solution of sodium cyanide in presence of
oxygen of the air forming complex cyanides.
4Ag + 8NaCN + 2H 2 O + O 2
⎯→ 4NaAg(CN) 2 + 4NaOH
Sod. argentocyanide
4Au + 8NaCN + 2H 2 O + O 2
⎯→ 4NaAu(CN) 2 + 4NaOH
Sod. aurocyanide
Ag or Au is recovered from the solution by the
addition of electropositive metal like zinc.
2NaAg(CN) 2 + Zn ⎯→ Na 2 Zn(CN) 4 + 2Ag ↓
2NaAu(CN) 2 + Zn ⎯→ Na 2 Zn(CN) 4 + 2Au ↓
Soluble complex
Special Methods :
Mond's process : Nickel is purified by this method.
Impure nickel is treated with carbon monoxide at 60–
80º C when volatile compound, nickel carbonyl, is
formed. Nickel carbonyl decomposes at 180ºC to
form pure nickel and carbon monoxide which can
again be used.
Impure nickel + CO 60–80ºC NI(CO) 4
Gaseous compound
180ºC
Ni + 4CO
Zone refining or Fractional crystallisation :
Elements such as Si, Ge, Ga, etc., which are used as
semiconductors are refined by this method. Highly
pure metals are obtained. The method is based on the
difference in solubility of impurities in molten and
solid state of the metal. A movable heater is fitted
around a rod of the impure metal. The heater is
slowly moved across the rod. The metal melts at the
point of heating and as the heater moves on from one
end of the rod to the other end, the pure metal
crystallises while the impurities pass on the adjacent
melted zone.
Pure metal
Moving circular
heater
Molten zone
containing
impurity
Impure
zone
Different metallurgical processes can be broadly
divided into three main types.
Pyrometallurgy : Extraction is done using heat
energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr,
Hg, etc., which are found in nature in the form of
oxides, carbonates, sulphides are extracted by this
process.
Hydrometallurgy : Extraction of metals involving
aqueous solution is known as hydrometallurgy.
Silver, gold, etc., are extracted by this process.
Electrometallurgy : Extraction of highly reactive
metals such as Na, K, Ca, Mg, Al, etc., by
carrying electrolysis of one of the suitable
compound in fused or molten state.
XtraEdge for IIT-JEE 37 FEBRUARY 2010

UNDERSTANDING
Physical Chemistry
1. Calculate ∆ r U, ∆ r H and ∆ r S for the process
T2
∆ r S = C p,m (g) ln
1 mole H 2 O (1,293 K, 101.325 kPa) →
T1
= (35.982 J K –1 mol –1 ) (150 K) = 5397.3 J mol –1 ∆ r H 1 = Cp ,m (1)
dT
1 mol H 2 O (g, 523 K, 101.325 kPa)
Given the following data :
= (35.982 J K –1 mol –1 ⎛ 523K ⎞
) × 2.303 × log
⎜
⎟
C p,m (1) = 75.312 J K –1 mol –1 ⎝ 373K ⎠
;
C p,m (g) = 35.982 J K –1 mol –1
∆ vap H at 373 K, 101.325 kPa = 40.668 kJ mol –1
Sol. The changes in ∆ r U, ∆ r H and ∆ r S can be calculated
following the reversible paths given below.
Step I: 1 mole H 2 O(1,293 K, 101.325 kPa) →
1 mole H 2 O(1,373 K, 101.325 kPa)
q p = ∆ r H = C p,m (1) ∆T
= (75.312 J K –1 mol –1 ) (80 K)
= 6024.96 J mol –1
= (35982 J K –1 mol –1 ) × 2.303 × 0.1468
= 12.164 J K –1 mol –1
∆ r U = ∆ r H – R(∆T)
= 5397.3 J mol –1 – (8.314 J K –1 mol –1 ) (150 K)
= 5397.3 J mol –1 – 1247.1 J mol –1
= 4 150.2 J mol –1
Thus ∆U total = (6024.96 + 37565 + 4150.2) J mol –1
= 47740.16 J mol –1
∆H total = (6024.96 + 40668 + 5397.3) J mol –1
= 52090.26 J mol –1
T2
∆ r S = C p,m ln
T
∆S total = (18.184 + 109.03 + 12.164) J K –1 mol –1
1
= 139.378 J K –1 mol –1
= (75.312 J K –1 mol –1 ⎛ 373K ⎞
) × 2.303 × log
⎜
⎟
⎝ 293K ⎠ 2. It is possible to supercool water without freezing. 18
= 18.184 J K –1 mol –1
g of water are supercooled to 263.15 K(–10ºC) in a
thermostat held at this temperature, and then
∆ r U = ∆ r H – p∆ r V – ~ ∆ r H
Step II: 1 mol H 2 O(1,373 K, 101.325 kPa)
→ 1 mol H 2 O (g, 373K, 101.325 kPa)
q p = ∆ vap H = 40.668 kJ mol –1
crystallization takes place.
Calculate ∆ r G for this process. Given:
C p (H 2 O,1) = 75.312 J K –1 mol –1
C p (H 2 O,s) = 36.400 J K –1 mol –1
–1
40668J mol
∆ r S =
= 109.03 J K –1 mol –1
∆ fus H (at 0ºC) = 6.008 kJ mol –1
373K
Sol. The process of crystallization at 0ºC and at 101.325
kPa pressure is an equilibrium process, for which
∆ r U = ∆ r H – p∆ r V
= 40668 J mol –1 ∆G = 0. The crystallization of supercooled water is a
– (101.325 kPa)
spontaneous phase transformation, for which ∆G
⎛
3 –1 373K ⎞
⎜(22.414dm
mol ) ×
⎟ must be less than zero. Its value for this process can
⎝
273K ⎠ be calculated as shown below.
= 40668 J mol –1 – 3 103 J mol –1
= 37565 J mol –1
Step III: 1 mol H 2 O(g, 373 K, 101.325 kPa)
→ 1 mol H 2 O(g, 523 K, 101.325 kPa)
∆ r H = C p,m (g) ∆T
The given process
H 2 O(1, – 10ºC) → H 2 O(s, –10ºC)
is replaced by the following reversible steps.
(a) H 2 O(1, – 10ºC) → H 2 O(1, 0ºC)
273.15K
...(1)
∫
263.15K
XtraEdge for IIT-JEE 38 FEBRUARY 2010

∆ r S 1 =
= (75.312 J K –1 mol –1 ) (10 K)
= 753.12 J mol –1
273.15K
∫
263.15K
C
p,m (1)
R
dT
= (75.312 J K –1 mol –1 ⎛ 273.15K ⎞
) × ln
⎜
⎟
⎝ 263.15K ⎠
= 2.809 J K –1 mol –1
(b) H 2 O(1, 0ºC) → H 2 O(s, 0ºC) ...(2)
∆ r H 2 = – 6.008 kJ mol –1
–1
(6008 J mol )
∆ r S 2 = –
= – 21.995 J K –1 mol –1
(273.15K)
(c) H 2 O(s, 0ºC) → H 2 O(s, –10ºC) ...(3)
∆ r H 3 =
∆ r S 3 =
263.15K
Cp ,m (s)
∫
273.15K
dT
= (36.400 J K –1 mol –1 )(–10 K)
= – 364.0 J mol –1
263.15K
∫
273.15K
Cp,m
(s)
dT
T
= (36.400 J K –1 mol –1 ⎛ 263.15K ⎞
) ×ln
⎜
⎟
⎝ 273.15K ⎠
= – 1.358 J K –1 mol –1
The overall process is obtained by adding Eqs. (1),
(2) and (3), i.e.
H 2 O(1, –10ºC) → H 2 O(s, –10ºC)
The total changes in ∆ r H and ∆ r S are given by
∆ r H = ∆ r H 1 + ∆ r H 2 + ∆ r H 3
=(753.12 – 6008 – 364.0) J mol –1
= – 5618.88 J mol –1
∆ r S = ∆ r S 1 + ∆ r S 2 + ∆ r S 3
= (2.809 – 21.995 – 1.358) J K –1 mol –1
= – 20.544 J K –1 mol –1
Now ∆ r G of this process is given by
∆ r G = ∆ r H – T∆ r S
= – 5618.88 J mol –1 – (263.15 K)( –20.544 J K –1 mol –1 )
= – 212.726 J mol –1
3. Given a solution that is 0.5 M CH 3 COOH. To what
volume at 25ºC must one dm 3 of this solution be
diluted to (a) double the pH; (b) double the
hydroxide-ion concentration
Given that K a = 1.8 × 10 –5 M.
Sol. If α is the degree of dissociation of acetic acid of
concentration c then the concentrations of various
species in the solution are
CH 3 COOH + H 2 O CH 3 COO – + H 2 O +
c(1 – α) cα cα
With these concentrations, the equilibrium constant
becomes
[CH3COO
][H3O
]
K a =
=
[CH COOH]
3
K
or = α = a
c
Substituting the values, we have
–5
–
+
(cα)(cα)
= cα 2
c(1– α)
(1.8 × 10 M)
α =
= 6 × 10 –3
(0.5 M)
The concentration of hydrogen ions is given as
[H 3 O + ] = cα = (0.5 M) (6 × 10 –3 ) = 3 × 10 –3 M
Hence, pH = – log {[H 3 O + ]/M}
= – log (3 × 10 –3 ) = 2.52
(a) To double the pH
Thus pH = 5.04
Since pH = – log {[H 3 O + ]/M}
therefore [H 3 O + ]/M= 10 –pH . Substituting the value of
pH , we have
[H 3 O + ]/M = 10 –5.04 = 0.912 × 10 –5 = 9.12 × 10 –6
Thus, c 1 α = 9.12 × 10 –6 M
In dilution, α will increase, and its value will not be
negligible in comparison to one. Thus, we shall have
to use the expression
2 2
–6
(c1α)
c1 α ( c
K a = = = 1 α)
α (9.12×
10 M) α
=
c1(1–
α)
1– α 1– α 1– α
or (1.8 × 10 –5 M) (1 – α) = ( 9.12 × 10 –6 M) α
which gives
(9.12 × 10 –6 M + 1.8 × 10 –5 M) α = 1.8 × 10 –5 M
–5
1.8×
10 M
or a =
= 0.6637
–6
27.12×
10 M
Since c 1 α = 9.12 × 10 –6 M and α = 0.6637
Therefore,
–6
9.12×
10 M
c 1 =
= 1.374 × 10 –5 M
0.6637
Volume to which the solution should be diluted
cV (0.5M) (1dm )
= = = 3.369 × 10 4 dm 3
–5
(1.374×
10 M)
c 1
(b) To double the hydroxyl-ion concentration
Since [H 3 O + ] in 0.5 M acetic acid is 3 × 10 –3 M,
therefore
[OH – ] =
(1.0 × 10
(3×
10
–14
–3
M
M)
2
)
3
XtraEdge for IIT-JEE 39 FEBRUARY 2010

In the present case, the concentration of hydroxyl
becomes
–14 2
[OH – 2(1.0 × 10 M )
] =
–3
(3×
10 M)
which gives
–3
[H 3 O + (3×
10 M)
] =
= 1.5 × 10 –3 M
2
For the concentration, we can use
K a = c 2 α 2 = (c 2 α) (α)
–5
Ka
or α =
(c α )
= (1.8 × 10 M)
= 1.2 × 10 –2
–3
(1.5 × 10 M)
2
–3
(1.5 × 10 M)
Thus, c 2 =
= 1.25 × 10 –1 M = 0.125 M
–2
1.2 × 10
Volume to which the solution should be diluted
cV (0.5M)(1dm )
= = = 4 dm 3
(0.125M)
c 2
3
4. The freezing point of an aqueous solution of KCN
containing 0.189 mol kg –1 was – 0.704 ºC. On adding
0.095 mol of Hg(CN) 2 , the freezing point of the
solution became –0.530ºC. Assuming that the
complex is formed according to the equation
Hg(CN) 2 + x CN – x–
→ Hg (CN) x + 2
Find the formula of the complex.
Sol. Molality of the solution containing only KCN is
(–∆Tf
) (0.704 K)
m = =
= 0.379 mol kg –1
K
–1
f (1.86 K kg mol )
This is just double of the given molality
( = 0.189 mol kg –1 ) of KCN, indicating complete
dissociation of KCN. Molality of the solution after
the formation of the complex
(–∆T ) (0.530 K)
m = =
K (1.86 K kg mol
f
f
–1
)
= 0.285 mol kg –1
If it be assumed that the whole of Hg(CN) 2 is
converted into complex, the amounts of various
species in 1 kg of solvent after the formation of the
complex will be
n(K + ) = 0.189 mol,
n(CN – ) = (0.189 – x) mol
x –
n(Hg(CN) x+ 2)
= 0.095 mol
Total amount of species in 1 kg solvent becomes
n total = [0.189 + (0.189 – x) + 0.095] mol
= (0.473 – x) mol Equating this to 0.285 mol,
we get
(0.473 – x) mol = 0.285 mol
i.e. x = (0.473 – 0.285) = 0.188
Number of CN – units combined =
Thus, the formula of the complex is
0.188 mol
0.095 mol
= 2
2–
Hg (CN) 4 .
5. From the standard potentials shown in the following
º º
diagram, calculate the potentials E 1 and E 2 .
E 1 º
BrO 3
– 0.54 V BrO
–
0.45 V 1 1.07 V Br2 Br
–
2
0.17 V
E 2 º
Sol. The reaction corresponding to the potential Eº 1 is
BrO 3 – + 3H 2 O + 5e – = 2
1 Br2 + 6OH – ...(1)
This reaction can be obtained by adding the
following two reduction reactions:
BrO 3 – + 2H 2 O + 4e – = BrO – + 4OH – ...(2)
BrO – + H 2 O + e – = 2
1 Br2 + 2OH – ...(3)
Hence the free energy change of reaction (1) will be
º
∆ G reaction(1) =
º
∆ G reaction(2) +
º
∆ G reaction(3)
Replacing ∆Gºs in terms of potentials, we get
– 5FE 1 º = – 4F(0.54 V) – 1F (0.45 V)
= (–2.61 V) F
2.61V
Hence E 1 º = = 0.52 V
5
Now the reaction corresponding to the potential E 2 º is
BrO – 3 + 2H 2 O + 6e – = Br – + 6OH – ...(4)
This reaction can be obtained by adding the
following three reactions.
BrO – 3 + 2H 2 O + 4e – = BrO – + 4OH – (Eq.2)
BrO – + H 2 O + e – =
2
1 Br2 + 2OH –
(Eq.3)
1 Br2 + e – = Br – ...(5)
2
Hence
º
∆ G reaction(4) =
º
∆ G reaction(2) +
º
∆ G reaction(3)
º
+ ∆ G reaction(5)
or – 6F(E 2 º) = – 4F(0.54 V) – 1F(0.45 V)
– 1F (1.07 V)
= (– 3.68 V) F
3.68
or E 2 º = = 0.61 V.
6
XtraEdge for IIT-JEE 40 FEBRUARY 2010

XtraEdge for IIT-JEE 41 FEBRUARY 2010

`tà{xÅtà|vtÄ V{tÄÄxÇzxá
10
Set
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in mathematics that would be very helpful in facing
IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and
we hope that this section would prove a rich resource for practicing challenging problems and
enhancing the preparation level of IIT JEE aspirants.
By : Shailendra Maheshwari
Solutions will be published in next issue
Joint Director Academics, Career Point, Kota
1. Let f(x) = sinx and
⎧{max f (t); 0 ≤ t ≤ x ; for 0 ≤ x ≤ π
g(x) = ⎨ 2
Discuss
⎩ sin x / 2 ; x > π
the continuity and differentiability of g(x) in (0, ∞)
2. Is the inequality sin 2 x < x sin(sinx) true for
0 < x < π/2 Justify your answer.
3. A shop sells 6 different flavours of ice-cream. In how
many ways can a customer choose 4 ice-cream cones
if
(i) they are all of different flavours;
(ii) they are not necessarily of different flavours;
(iii) they contain only 3 different flavoures;
(iv) they contain only 2 or 3 different flavoures
4. Using vector method, show that the internal
(external) bisector of any angle of a triangle divides
the opposite side internally (externally) in the ratio of
the other two sides containing the triangle.
5. Prove that
(a) cos x + n C 1 cos 2x + n C 2 cos 3x + ............. + n C n
cos(n + 1)x = 2 n . cos n ⎛ n + 2 ⎞
x/2. cos ⎜ x ⎟
⎝ 2 ⎠
(b) sin x + n C 1 sin 2x + n C 2 sin 3x + .............. + n C n
sin(n + 1)x = 2 n . cos n x/2 . sin ⎛ n + 2 ⎞
⎜ x ⎟
⎝ 2 ⎠
6. In a town with a population of n, a person sands two
letters to two sperate people, each of whom is asked
to repeat the procedure. Thus, for each letter
received, two letters are sent to separate persons
chosen at random (irrespective of what happened in
the past). What is the probability that in the first k
stages, the person who started the chain will not
receive a letter
7. Prove the identity :
x 2
2 x 2
zx−z
x 4 −z
4
e dz =
0 ∫
e
0
∫
function f(x) =
∫
and solving it.
e dz, deriving for the
0
x
e
2
zx−z
8. Prove that
∫
sin nθsecθ
dθ
dz a differential equation
2cos(n −1)
θ
= –
–
n −1
∫sin(
n – 2) θ secθdθ
dθ.
Hence or otherwise evaluate
∫ π / 2 cos5θ
sin 3θ
dθ.
0 cos θ
9. Find the latus rectum of parabola
9x 2 – 24 xy + 16y 2 – 18x – 101y + 19 = 0.
10. A circle of radius 1 unit touches positive x-axis and
positive y-axis at A and B respectively. A variable
line passing through origin intersects the circle in two
points in two points D and E. Find the equation of the
lines for which area of ∆ DEB is maximum.
XtraEdge for IIT-JEE 42 FEBRUARY 2010

MATHEMATICAL CHALLENGES
SOLUTION FOR JANUARY ISSUE (SET # 9)
−2 s + a + 2b b − c
1. as φ (a) = φ (b) = φ (c)
=
= & r = k.
2 2
so by Rolle’s theorem there must exist at least a point
∆ s(s
− a)(s − b)(s − c)
x = α & x = β each of intervals (a, c) & (c, b) such so r = k = =
that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem,
s s
a ⇒ 2 | g (x) | < − 2x 2 + 4 ≤ 4.
so h = ON = − (s − b) 2 ⇒ |g (x) | ≤ 2.
there must exist at least a point x = µ such that
s(s
− a)(s − b)(s − c)
r = k =
α < µ < β where φ′(µ) = 0
s
2f (a)
2f (b)
2sk = s(s
− a)(a − b + c)(a + b − c)
so
+
(a − b) (a − c) (b − c)(b − a)
= s (s − a)(a − 2x)(a + 2x)
2f (c)
+
- f ′′ (µ) = 0
2
(c − a) (c − b)
2sk = s(s − a)(a
− 4h )
f (a)
f (b)
required locus is
so
+
4s
(a − b) (a − c) (b − c) (b − a)
y 2 = A(a 2 – 4x 2 )
f (c) 1 ⇒ s 2 y 2 + Ax 2 Aa 2
=
+
= f ′′ (µ)
4
(c − a) (c − b) 2 where A is = s (s – a)
where a < µ < b.
here h 2 < as so it is an ellipse
2. Required probability
4. f (0) = c
r 2
5 5 5 5 1 ⎛ 5 ⎞
−
1
f (1) = a + b + c & f (−1) = a − b + c
1 . . . ........ . = ⎜ ⎟ . (r – 2) times solving these,
6 6 6 6 6 ⎝ 6 ⎠ 6
1
Note : any number in 1st loss
a = [f (1) + f (−1) − 2 f (0)] ,
2
same no. does not in 2nd (any other comes).
Now 3rd is also diff. (and in same r − 2 times)
1
b = [f (1) − f (−1)] & c = f (0)
2
Now (r − 1) th & r th must be same.
x (x + 1)
so f (x) = f (1) + (1− x 2 x(x
−1)
) f (0) +
2
2
3. 2s = a + b + c
f (−1) 2 | f (x) | < | x | | x + 1 | + 2 | 1 − x 2 | + |x|
ON = − BN + BO
| x − 1| ; as | f (1) | , | f (0) |, | f (−1) | ≤ 1.
Let BN = x
2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x 2 ) + | x | (1 − x) as
2BN + 2CN + 2AR = 2s
x ∈ [−1, 1]
x + (a − x) + (b − a + x) = s
x = s − b
A
so 2 | f (x) | ≤ 2 (|x| + 1 − x 2 5
) ≤ 2 . 4
5
so | f (x) | ≤ 4
M I (h,k)
Now as g (x) = x 2 1
f (1/x) = (1 + x) f (1)
R
2
+ (x 2 1
− 1) f (0) + (1 − x) f (−1)
2
r
so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x 2 | + | 1 − x|
B
C
⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x 2 ) | + 1 − x ;
N O
as x ∈ [−1, 1]
XtraEdge for IIT-JEE 43 FEBRUARY 2010

5. Oil bed is being shown by the plane A′ PQ. θ be the
angle between the planes A′ PQ & A′ B′ C′ Let A′ B′
C′ be the x − y plane with x axis along A′ C′ and
origin at A′. The P.V.s of the various points are
defined as follows
B
A
C
2sin 5x sin
=
∫ − 5x
2cos
2
x
2
dx
⎛ 5x x ⎞
= − 2
∫
⎜sin
sin ⎟ dx
⎝ 2 2 ⎠
⎛ 6x 4x ⎞
=
∫
⎜cos − cos ⎟ dx
⎝ 2 2 ⎠
=
∫
(cos 3x − cos 2x) dx
A´
B´
P
point C′ : b î , point B′ : cos A î + c sin A ĵ , point Q :
bî – z kˆ , point P : i cos A î + c sin A ĵ – ykˆ
normal vector to the plane A′ B′ C′
= n r
1 = bc sin A kˆ
normal vector to the plane A’PQ = n r
2
= cz sin A î + (by - cz cos A) ĵ + bc sin A kˆ
r r
n .n 2
so cos θ = r
1 r
| n || n |
=
[c
cos θ =
2
z
[b
2
2
sin
c
2
so tan θ =
2
1
1
bcsin A
A + (by − cz cos A)
sin
[c
2
2
A + (c
z
2
so tan θ . sin A =
cos8x − cos7x
6.
∫
.
1+
2cos5x
2
2
z
2
+ b
2
y
2
b csin A
2
2
C´
Q
+ b
2
c
2
sin
2
A]
1/ 2
− 2byczcos A)]
+ b y − 2byczcosA]
bc sin A
z
b
2
2
y
+
c
2
2
2sin 5x
2sin 5x
1/ 2
2yz
− cos A
bc
dx
sin13x −sin 3x −sin12x
+ sin 2x
=
∫
dx
2(sin 5x + sin10x)
sin13x + sin 2x −sin 3x – sin12x
=
∫
dx
2(sin 5x + sin10x)
=
∫
=
∫
15x 11x 15x
2sin cos − 2sin
2 2 2
15x 5x
2.2.sin cos
2 2
11x 9x
cos −cos
2 2
5x
2cos
2
dx
9x
cos
2
dx
1/ 2
7.
=
sin 3x
3
−
sin 2x
2
+ C
2 x
d y
= 2
2
dx ∫
f (t) dt
0
integrate using by parts method
dy ⎡
x
x
= 2
dx
⎥ ⎥ ⎤
⎢x
−
⎢ ∫
f (t) dt
∫
x . f (x) dx
⎣ 0
0 ⎦
⎡
⎤
= 2 ⎢
⎥
⎢∫ x ( x − t) f (t) dt
⎥
⎣ 0
⎦
again integrating,
⎡ x
x ⎛ x ⎞ ⎤
y = 2 ⎢ x
⎜ ⎟ ⎥
⎢ ∫(x
− t) f (t) dt −∫
x −
⎜∫
f (t) dt 0 dx
⎟ ⎥
⎣ 0
0 ⎝ 0 ⎠ ⎦
⎡
= 2 ⎢x
⎢
⎣
x
x
∫
0
x
(x − t) f (t) dt −
2
2 x
∫
0
f (t) dt +
2
2
=
∫
2 (x − xt) f (t) dt −
∫
x f (t) dt +
∫
t
0
x
y =
∫
( x − 2xt + t ) f (t) dt =
∫
( x − t)
8. To prove that
0
2
Let
b
a = c > 0
2
⎛
⎜⎛
a ⎞
⎜ ⎟
⎜
⎝⎝
b ⎠
α
x
0
⎞
+ 1⎟
⎟
⎠
1/ α
x
0
⎛
< ⎜⎛
a ⎞
⎜ ⎟
⎜
⎝⎝
b ⎠
so (c α + 1) 1/α < (c β + 1) 1/β .
Let f (x) = (c x + 1) 1/x ; x > 0
f ′(x) = (c x + 1) 1/x ln (c x ⎛ 1 ⎞
+ 1) ⎜ − ⎟
⎝ x 2 ⎠
β
x
∫
0
2
2
x ⎤
f (x) dx⎥
2 ⎥
⎦
x
0
2
f (t) dt
f (t) dt
⎞
+ 1⎟
⎟
⎠
1/ β
1 1
+ (c x + 1) –1
x . c x ln c
x
XtraEdge for IIT-JEE 44 FEBRUARY 2010

=
x
1
−1
x
(c + 1)
x
x
2
x
so f (x) is decreasing function
so f (α) < f (β). Hence proved.
[ − (c + 1) l n (c + 1) + c ln
c ] < 0
9. Point P (x, 1/2) under the given condition are length
PB = OB
O
P
C
(t – 1)
A
θ
B (t, 1)
rθ = t ; so θ = t
PB θ
from ∆PAB : = PA sin
2 2
t
⇒ PB = 2 sin
2
Now ∠ PBC =
2
θ =
2
t ;
so from ∠ PCB ;
2
θ =
2
t
1/ 2
so from ∆ PCB ;
PB
= sin
2
t
x
x
........(1)
........(2)
from (1) & (2) PB = 1 ; so θ = t = π/3
thus | PB | 2 = (t − x) 2 +
4
1 = 1.
3
| t − x | = ; t − x = ; as t > x
23
2
π 3
so x = − 3 2
10. Let x n = n − 1 + n + 1 be rational, then
1 1 = is also rational
n −1
+ n + 1
x n
1 n + 1 − n −1
= is also rational
x n 2
n + 1 − n − 1 is also rational
as n + 1 + n − 1 & n + 1 − n − 1 are rational
so n + 1 + n − 1 must be rational
i.e. (n + 1) & (n – 1) are perfect squares.
This is not possible as any two perfect squares differe
at least by 3. Hence there is not positive integer n for
which n − 1 + n + 1 is a rational.
TRUE OR FALSE
1. The thrust exerted by a liquid on the base of a
vessel does not depend on the mass of the liquid
but depends on the area of the base and height of
the liquid.
2. The path of one projectile as seen from another
projectile is a straight line.
3. The arithmetic logic shift unit (ALSU) is
combinational circuit that performs a number of
arithmetic, logic and shift micro-operations.
4. A thin circular disc of mass M and radius R is
rotating in a horizontal plane about an axis
passing through its centre and perpendicular to its
plane with an angular velocity ω. Another disc of
the same dimensions but of mass
4
M is gently
placed on the first disc coaxially. The angular
velocity of the system now is 2ω/ 2 .
5. The rms speed of oxygen molecules (O 2 ) at a
certain temperature T (absolute) is v. If the
temperature is doubled and oxygen gas dissociates
into atomic oxygen, the rms speed remains
unchanged.
Sol.
1. [True]
2. [True] Let u 1 and θ 1 be the initial speed and angle
of projection of the projectile and u 2 and θ 2 be the
corresponding quantities, respectively, for the
other projectile.
Then the coordinates of one as seen from the other
projectile are
x = (u 1 cos θ 1 – u 2 cos θ 2 ) t,
y = (u 1 sin θ 1 – u 2 cos θ 2 ) t
x u
∴ =
1 cosθ1
− u 2 cosθ2
= m (say)
y u1
sin θ1
− u 2 sin θ2
or x = my,
which is the equation of a straight line.
3. [True]
4. [False]
1 ⎛ M ⎞
⎜M + ⎟ R 2 1
ω′ = MR 2 ω
2 ⎝ 4 ⎠ 2
⇒
5. [False] v rms = const.
v
'
rms
= const.
T
M
2T
M / 2
ω′ = 5
4 ω
= const. × 2 M
T = 2 vrms
XtraEdge for IIT-JEE 45 FEBRUARY 2010

MATHS
Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. An urn containing '14' green and '6' pink ball. K
(< 14, 6) balls are drawn and laid a side, their colour
being ignored. Then one more ball is drawn. Let P(E)
be the probability that it is a green ball, then
20 P(E) = ..............
Sol. Let E i denote the event that out of the first k balls
drawn, i balls are green. Let A denote the event that
(k + 1)th ball drawn is also green.
14 6
C
P(E i ) = i × Ck−i
0 ≤ i ≤ k
20
Ck
14 − i
and P(A/E i ) =
20 − k
k 14 6
C j × Ck−i
14 − j
Now P(A) = ∑
+
20
C 20 − k
j=
0 k
Also (1 + x) 14 – 1 (1 + x) 6
= ( 14–1 C 0 + 14–1 C 1 x +.......+ 14 – 1 C 14 – 1 x 14–1 )
( 6 C 0 + 6 C 1 x + .......+ 6 C 6 x 6 )
k
14−1
6
⇒ ∑( Cj
+ Ck−
j)
= co-efficient of x k
j = 0
14 14
∴ P(E) = =
6 + 14 20
∴ 20P(Ε) = 14
2. If f(x + y + z) = f(x) + f(y) + f(z) with f(1) = 1 and
f(2) = 2 and x, y, z ∈ R, then evaluate
n
∑(4r)f (3r)
lim r=
1 is equal to__________
n→∞
3
n
Sol. f(3) = 3f(1) = 3, f(4) = f(2 + 1 + 1) = 2 + 1 + 1 = 4
and so on. In general, we get f(r) = r for r ∈ N
n
∑ (4r)f (3r)
⇒ r=
1
12n(n + 1) (2n + 1)
lim
= lim
n→∞
3 n
3
n
→∞ 6n
= 4
3. Six points (x i , y i ), i = 1, 2,..., 6 are taken on the circle
6
6
x 2 + y 2 = 4 such that ∑x i = 8 and ∑ y i = 4 . The
i=
1
i=
1
line segment joining orthocentre of a triangle made
by any three points and the centroid of the triangle
made by other three points passes through a fixed
points (h, k), then h + k is_________
6
6
Sol. Let ∑ x i = α and ∑ y i = β .
i=
1
i=
1
Let O be the orthocentre of the triangle made by
(x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 )
⇒ O is (x 1 + x 2 + x 3 , y 1 + y 2 + y 3 ) ≡ (α 1 , β 1 )
Similarly let G be the centroid of the triangle made
by other three points
⎛ x 4 + x 5 + x 6 y 4 + y 5 + y 6 ⎞
⇒ G is ⎜
,
⎟
⎝ 3
3 ⎠
⎛ α − α ⎞
⇒ G is 1 β − β1
⎜ , ⎟ .
⎝ 3 3 ⎠
The point dividing OG in the ratio 3 : 1 is
⎛ α β ⎞
⎜ , ⎟ ≡ (2, 1) ⇒ h + k = 3
⎝ 4 4 ⎠
4. Let P(x) = x 4 + ax 3 + bx 2 + cx + d where a, b, c, d are
constants. If P(1) = 10, P(2) = 20 and P(3) = 30,
P(2)
+ P( −8)
compute
10
Sol. Let Q(x) = P(x) – 10 x
Q(1) = P(1) – 10 = 0
Q(2) = P(2) – 20 = 0
Q(3) = P(3) – 30 = 0
∴ Q(x) is divisible by (x – 1) (x – 2) (x – 3)
But Q (x) is a 4th degree polynomial
∴ Q(x) = (x – 1) (x – 2) (x – 3) (x – K)
∴ P(x) = (x – 1) (x – 2) (x – 3) (x – K) + 10x
P(12) = (11) (10) (9) (12 – K) + 120
P (–8) = (–9) (–10) (–11) (–8 – K) – 80
XtraEdge for IIT-JEE 46 FEBRUARY 2010

⎞
⎟
⎠
P(12)
+ P( −8)
∴
6. If sin –1 ⎛ π
x ∈ ⎜0 ,
10
⎝ 2
, then the value of
⎡ 25 7 3 1⎤
37 (9.1 meters) long but had a brain the size of a
= 2
⎢ + × − ⎥
= = A, then 96A = 37 walnut.
⎣128
8 16 6⎦
96
990 (12 − K) + 120 + 990 (8 + K) − 80
⎡ −1
−1
−1
−1
=
cos (sin(cos x)) + sin (cos(sin x)) ⎤
tan
10
⎢
⎥ is___
⎢⎣
2
⎥⎦
11880 + 120 − 990K + 7920 + 990K − 80
=
10
Sol. As sin –1 ⎛ π ⎞
x ∈ ⎜0 , ⎟ and cos –1 π
x = – sin –1 x
⎝ 2 ⎠ 2
12000 + 7840 19840
=
= = 1984
10 10
⇒ cos –1 ⎛ π ⎞
x ∈ ⎜0
, ⎟
⎝ 2 ⎠
5. If A be the area bounded by y = f(x), y = f –1 (x) and
⇒ sin(cos –1 x) = cos (sin –1 1
x) =
line 4x + 4y – 5 = 0 where f(x) is a polynomial of 2 nd
2
1−
x
degree passing through the origin and maximum
value of 1/4 at x = 1, then 96A is equal to______
Sol. Let f(x) = ax 2 + bx
1 = a + b ...(1)
Thus, cos –1 (sin(cos –1 x)) + sin –1 (cos(sin –1 π
x)) = . 2
π
⇒ required value = tan = 1
4
4
f '(x) = 2ax + b
⇒ 2a + b = 0 ...(2)
Do you know
Y
y = x
• The smallest bone in the human body is the stapes
or stirrup bone located in the middle ear. It is
approximately .11 inches (.28 cm) long.
B
• The longest cells in the human body are the
C
motor neurons. They can be up to 4.5 feet (1.37
A
meters) long and run from the lower spinal cord
to the big toe.
• There are no poisonous snakes in Maine.
O P Q
X • The blue whale can produce sounds up to 188
decibels. This is the loudest sound produced by a
living animal and has been detected as far away as
From (1) and (2),
530 miles.
1 1
• The largest man-made lake in the U.S. is Lake
a = – , b = Mead, created by Hoover Dam.
4 2
2
• The poison arrow frogs of South and Central
2x − x
f(x) =
America are the most poisonous animals in the
4
world.
Since 4x + 4y – 5 = 0 passes through
• A new born blue whale measures 20-26 feet
⎛ 1 ⎞ ⎛ 1 ⎞
(6.0 - 7.9 meters) long and weighs up to 6,614
A ⎜1 , ⎟ and B⎜
, 1⎟ so area bounded is
⎝ 4 ⎠ ⎝ 4 ⎠
pounds (3003 kg).
OAB = 2 × OAC
• The first coast-to-coast telephone line was
= 2 [area (OCP) + area(CPQA) – OAQ]
established in 1914.
⎡ ⎛ ⎞ ⎛ ⎞ 1 2
• The Virginia opossum has a gestation period of
1 5 5 5 1 1 5 2x − x ⎤
= 2 ⎢ × × + ⎜ + ⎟ × × ⎜1
− ⎟ − ∫ dx⎥
only 12-13 days.
⎢ 2 8 8
⎣ ⎝ 8 4 ⎠ 2 ⎝ 8 ⎠ 0 4 ⎥⎦
• The Stegosaurus dinosaur measured up to 30 feet
XtraEdge for IIT-JEE 47 FEBRUARY 2010

MATH
INTEGRATION
Mathematics Fundamentals
Integration :
If d f(x) = F(x), then ∫ F(x
) dx = f(x) + c, where
dx
c is an arbitrary constant called constant of
integration.
1.
∫
x n dx =
1
2.
∫
dx
x
x n + 1
(n ≠ –1)
n + 1
= log x
3.
∫
e x dx = e x
4.
∫
a x dx =
5.
∫sin
x dx
x
a
log
e
a
= – cos x
6.
∫
cos x dx = sin x
7.
∫
sec 2 x dx
2
= tan x
8.
∫
cosec x dx = – cot x
9. sec x tan x dx = sec x
∫
10. cosec x cot x dx = – cosec x
∫
11.
∫ sec x dx = log(sec x + tan x) = log tan ⎛ x π
⎟ ⎞
⎜ +
⎝ 2 4 ⎠
12.
∫ cosec x dx = – log (cosec x + cot x) = log tan ⎛ x
⎟ ⎞
⎜
⎝ 2 ⎠
13. tan x dx = – log cos x
∫
14. cot x dx = log sin x
∫
dx
15.
∫
a 2 − x
2
dx
16.
∫ a 2 + x
2
dx
17.
∫
x x 2 − a
2
= sin –1 x = – cos
–1 x
a a
1
= tan
–1 x 1 = – cot
–1 ⎛ x ⎞
⎜ ⎟
a a a ⎝ a ⎠
1
= sec
–1 x 1 = – cosec
–1 ⎛ x ⎞
⎜ ⎟
a a a ⎝ a ⎠
1
18.
∫ x 2 − a
2
1
19.
∫ a 2 − x
2
dx
20.
∫
x 2 − a
2
dx
21.
∫
x 2 + a
2
=
dx =
1 x − a log , when x > a
2a x + a
= log
1 a + x log , when x < a
2a a − x
⎧ 2 2 ⎫
⎨x + x − a ⎬ = cos h –1 ⎛ x ⎞
⎜ ⎟
⎩
⎭ ⎝ a ⎠
= log
⎧ 2 2 ⎫
⎨x + x + a ⎬ = sin h –1 ⎛ x ⎞
⎜ ⎟
⎩
⎭ ⎝ a ⎠
22.
∫
a 2 − x
2 1
dx = x
2 2 1
a − x + a 2 sin –1 ⎛ x ⎞
⎜ ⎟ 2 2 ⎝ a ⎠
23.
∫
x 2 − a
2 1
dx = x
2
x 2 − a 2
1
– a 2 log
⎧
⎨x
+ 2 ⎩
24.
∫
x 2 + a
2 1
dx = x
2
x 2 + a 2
f´(x)
25.
∫
dx = log f(x)
f (x)
f´(x)
26.
∫
f (x)
dx = 2 f (x)
1
+ a
2
log
⎧
⎨x
+
2 ⎩
x
x
2
2
− a
2
+ a
Integration by Decomposition into Sum :
1. Trigonometrical transformations : For the
integrations of the trigonometrical products such as
sin 2 x, cos 2 x, sin 3 x, cos 3 x, sin ax cos bx, etc., they
are expressed as the sum or difference of the sines
and cosines of multiples of angles.
2. Partial fractions : If the given function is in the
form of fractions of two polynomials, then for its
integration, decompose it into partial fractions (if
possible).
Integration of some special integrals :
dx
(i)
∫ 2
ax + bx + c
This may be reduced to one of the forms of the
above formulae (16), (18) or (19).
⎫
⎬
⎭
2
⎫
⎬
⎭
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dx
dx
1
(ii) 4.
∫ 2
∫
, at first x = and then a + ct 2 = z 2
2 2
ax + bx + c
(px + r) ax + c
t
This can be reduced to one of the forms of the above Some Important Integrals :
formulae (15), (20) or (21).
(iii)
∫
ax 2
dx
⎛ x − α ⎞
+ bx + c dx
1. To evaluate , ⎜ ⎟
∫
(x − α)(x
− β)
∫
dx,
⎝ β − x ⎠
This can be reduced to one of the forms of the above
formulae (22), (23) or (24).
∫
( x − α)(
β − x)
dx. Put x = α cos 2 θ + β sin 2 θ
(px + q)dx (px + q)dx
(iv)
∫
, dx dx
2
ax + bx + c ∫ 2
2. To evaluate , ,
ax + bx + c
∫ a + b cos x ∫ a + bsin x
For the evaluation of any of these integrals, put
dx
px + q = A {differentiation of (ax 2 + bx + c)} + B ∫ a + bcos x + csin x
Find A and B by comparing the coefficients of like
powers of x on the two sides.
⎛ x ⎞
⎛ ⎞
⎜2 tan ⎟ ⎜ −
2 x
1 tan ⎟
1. If k is a constant, then
Replace sin x =
⎝ 2 ⎠
and cos x =
⎝ 2 ⎠
∫ k dx = kx and ∫
k f (x) dx = k
∫
f (x) dx
⎛ ⎞
⎜ +
2 x
⎛ ⎞
1 tan ⎟
⎜ +
2 x
1 tan ⎟
⎝ 2 ⎠
⎝ 2 ⎠
2.
∫
{ f1 (x) ± f2(x)}
dx =
∫
f 1 (x) dx ±
∫
f 2 (x) dx
x
Then put tan = t.
Some Proper Substitutions :
2
1.
∫
f(ax + b) dx, ax + b = t
pcos x + qsin x
3. To evaluate dx
∫ a + bcos x + csin x
2.
∫ f(axn + b)x n–1 dx, ax n + b = t
Put p cos x + q sin x = A(a + b cos x + c sin x)
+ B. diff. of (a + b cos x + c sin x) + C
3. f{φ(x)} φ´(x) dx, φ(x) = t
∫ A, B and C can be calculated by equating the
coefficients of cos x, sin x and the constant terms.
f´(x)
4.
∫
dx , f(x) = t
f (x)
dx
4. To evaluate
∫
,
2
2
a cos x + 2bsin x cos x + csin x
5.
∫
a
2 − x
2 dx, x = a sin θ or a cos θ
dx
dx
∫
,
2
6.
∫
a
2 + x
2
a cos x + b ∫ a + bsin
2 x
dx, x = a tan θ
In the above type of questions divide N r and D r by
2 2
cos 2 x. The numerator will become sec 2 x and in the
a − x
7. dx, x
∫ 2 = a 2 cos 2θ
denominator we will have a quadratic equation in tan
2 2
a + x
x (change sec 2 x into 1 + tan 2 x).
8.
∫
a ± x dx, a ± x = t 2
Putting tan x = t the question will reduce to the form
dt
a − x
∫ 2
9.
∫
dx, x = a cos 2θ
at + bt + c
a + x
5. Integration of rational function of the given form
10.
∫
2ax
− x
2
2 2
2 2
dx, x = a(1 – cos θ)
x + a
x − a
(i) dx, (ii) dx, where
∫ 4 2 4
x + kx + a ∫ 4 2 4
x + kx + a
11.
∫
x
2 − a
2 dx, x = a sec θ
k is a constant, positive, negative or zero.
These integrals can be obtained by dividing
Substitution for Some irrational Functions :
numerator and denominator by x 2 , then putting
dx
1.
∫
, ax + b = t 2
a 2 a 2
(px + q) ax + b
x – = t and x + = t respectively.
x
x
dx
1
2.
∫
, px + q = Integration of Product of Two Functions :
2
(px + q) ax + bx + c t
1.
dx
∫ f 1(x) f 2 (x) dx = f 1 (x)
∫ f '
2(x) dx –
∫[ ( f1(x)
∫f2(x)
dx]
dx
3.
∫
, ax + b = t 2 2
(px + qx + r) ax + b
XtraEdge for IIT-JEE 53 FEBRUARY 2010

Proper choice of the first and second functions : universal substitution. Sometimes it is more
Integration with the help of the above rule is called
x
integration by parts, In the above rule, there are two convenient to make the substitution cot = t for
2
terms on R.H.S. and in both the terms integral of the
0 < x < 2π.
second function is involve. Therefore in the product
of two functions if one of the two functions is not The above substitution enables us to integrate any
directly integrable (e.g. log x, sin –1 x, cos –1 x, tan –1 x function of the form R (sin x, cos x). However, in
etc.) we take it as the first function and the remaining practice, it sometimes leads to extremely complex
function is taken as the second function. If there is no rational functions. In some cases, the integral can be
other function, then unity is taken as the second simplified by –
function. If in the integral both the functions are (i) Substituting sin x = t, if the integral is of the form
easily integrable, then the first function is chosen in
such a way that the derivative of the function is a ∫
R (sin x) cos x dx.
simple functions and the function thus obtained under (ii) Substituting cos x = t, if the integral is of the form
the integral sign is easily integrable than the original
function.
∫
R (cos x) sin x dx.
2.
∫
e ax sin(bx + c)
dx
dt
(iii) Substituting tan x = t, i.e. dx = , if the
2
1+
t
ax
e
integral is dependent only on tan x.
= [a sin (bx + c) – b cos (bx + c)]
2 2
a + b
Some Useful Integrals :
ax
e ⎡
−1
b⎤
dx
= sin
2 2
⎢bx
+ c − tan
1. (When a > b)
⎥
a + b ⎣
a ⎦
∫ a + bcos x
3.
∫
e ax
2 ⎡
cos(bx + c)
dx
= tan –1 a − b x
2 2
a − b
⎥ ⎥ ⎤
⎢ tan + c
⎢⎣
a + b 2 ⎦
ax
e
= [a cos (bx + c) + b sin(bx + c)]
dx
2 2
a + b
2. (When a < b)
∫ a + bcos x
ax
e ⎡
−1
b⎤
= cos
2 2
⎢bx
+ c − tan
x
⎥
a + b ⎣
a
b − a tan − a + b
⎦ 1
= – log
a
4.
∫ ekx {kf(x) + f´(x)} dx = e kx 2 2
b − a
x
f(x)
b − a tan + a + b
a
⎛ x ⎞
dx 1 x
5.
∫
log e x = x(log e x – 1) = x log e ⎜ ⎟ 3. (when a = b)
⎝ e ⎠
∫
= tan + c
a + bcos x a 2
Integration of Trigonometric Functions :
dx
4. (When a > b)
∫
1. To evaluate the integrals of the form
a + bsin x
I =
∫ sinm x cos n x dx, where m and n are rational
⎧ ⎛ x ⎞ ⎫
2
⎪a tan⎜
⎟ + b ⎪
numbers.
= tan –1
⎝ 2 ⎠
⎨
2 2
a − b
(i) Substitute sin x = t, if n is odd;
⎪ ⎪ ⎬ + c
2 2
⎪ a − b
⎪⎩
⎭
(ii) Substitute cos x = t, if m is odd;
(iii) Substitute tan x = t, if m + n is a negative even
dx
5. (When a < b)
∫
integer; and
a + bsin x
1 1 ⎛ x ⎞
(iv) Substitute cot x = t, if (m + 1) + (n – 1) is
2 2
a tan⎜
⎟ + b − b − a
2 2 1
= log
⎝ 2 ⎠
+ c
an integer.
2 2
b − a
⎛ x ⎞
2 2
a tan⎜
⎟ + b + b − a
2. Integrals of the form
∫
R (sin x, cos x) dx, where R is
⎝ 2 ⎠
a rational function of sin x and cos x, are transformed
dx 1
6. (When a = b)
into integrals of a rational function by the substitution
∫
= [tan x – sec x] + c
a + bsin x a
x
tan = t, where –π < x < π. This is the so called
2
XtraEdge for IIT-JEE 54 FEBRUARY 2010

MATH
TRIGONOMETRICAL
EQUATION
Mathematics Fundamentals
Functions with their Periods :
Function
sin (ax + b), cos (ax + b), sec (ax + b),
cosec (ax + b)
tan(ax + b), cot (ax + b)
|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|,
|cosec (ax + b)|
|tan (ax + b)|, |cot (ax + b)|
Trigonometrical Equations with their General
Solution:
Trgonometrical equation
sin θ = 0
General Solution
θ = nπ
cos θ = 0 θ = nπ + π/2
tan θ = 0
θ = nπ
sin θ = 1 θ = 2nπ + π/2
cos θ = 1
sin θ = sin α
cos θ = cos α
tan θ = tan α
sin 2 θ = sin 2 α
tan 2 θ = tan 2 α
cos 2 θ = cos 2 α
sin θ = sin α
*
cosθ = cos α
sin θ = sin α
*
tan θ = tan α
tan θ = tan α
*
cosθ = cos α
θ = 2nπ
θ = nπ + (–1) n α
θ = 2nπ ± α
θ = nπ + α
θ = nπ ± α
θ = nπ ± α
θ = nπ ± α
θ = 2nπ + α
θ = 2nπ + α
θ = 2nπ + α
Period
2π/a
π/a
π/a
π/2a
* If α be the least positive value of θ which satisfy
two given trigonometrical equations, then the
general value of θ will be 2nπ + α.
Note :
1. If while solving an equation we have to square it,
then the roots found after squaring must be
checked whether they satisfy the original
equation or not. e.g. Let x = 3. Squaring, we get
x 2 = 9, ∴ x = 3 and – 3 but x = – 3 does not
satisfy the original equation x = 3.
2. Any value of x which makes both R.H.S. and
L.H.S. equal will be a root but the value of x for
which ∞ = ∞ will not be a solution as it is an
indeterminate form.
3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or
y = z or both. But x
y = x
z ⇒ y = z only and
not x = 0, as it will make ∞ = ∞. Similarly, if ay
= az, then it will also imply y = z only as a ≠ 0
being a constant.
Similarly, x + y = x + z ⇒ y = z and x – y = x – z
⇒ y = z. Here we do not take x = 0 as in the
above because x is an additive factor and not
multiplicative factor.
4. When cos θ = 0, then sin θ = 1 or –1. We have to
verify which value of sin θ is to be chosen which
⎛ 1 ⎞
satisfies the equation cos θ = 0 ⇒ θ = ⎜ n + ⎟ π
⎝ 2 ⎠
If sin θ = 1, then obviously n = even. But if
sin θ = –1, then n = odd.
Similarly, when sin θ = 0, then θ = nπ and cos θ = 1
or –1.
If cos θ = 1, then n is even and if cos θ = –1,
then n is odd.
5. The equations a cos θ ± b sin θ = c are solved as
follows :
Put a = r cos α, b = r sin α so that r =
and α = tan –1 b/a.
The given equation becomes
r[cos θ cos α ± sin θ sin α] = c ;
c c
cos (θ ± α) = provided ≤ 1.
r r
2
a + b
2
XtraEdge for IIT-JEE 55 FEBRUARY 2010

Relation between the sides and the angle of a triangle:
1. Sine formula :
sin A
a
=
sin B
b
=
sin C
c
=
1
2R
Where R is the radius of circumcircle of triangle
ABC.
2. Cosine formulae :
cos A =
b
2
2
2
+ c − a
2bc
2
2
2
, cos B =
a
2
2
+ c − b
2ac
a + b − c
cos C =
2ab
It should be remembered that, in a triangle ABC
If ∠A = 60º, then b 2 + c 2 – a 2 = bc
If ∠B = 60º, then a 2 + c 2 – b 2 = ac
If ∠C = 60º, then a 2 + b 2 – c 2 = ab
3. Projection formulae :
a = b cos C + c cos B, b = c cos A + a cos C
c = a cos B + b cos A
Trigonometrical Ratios of the Half Angles of a Triangle:
a + b + c
If s = in triangle ABC, where a, b and c
2
are the lengths of sides of ∆ABC, then
(a) cos 2
A =
cos 2
C =
(b) sin
2
A =
sin 2
C =
s(s
− a) B s(s
− b)
, cos = ,
bc 2 ac
s(s
− c)
ab
( s − b)(s − c) B
' sin =
bc 2
( s − a)(s − b)
ab
A (s − b)(s − c)
(c) tan = ,
2 s(s − a)
B (s − a)(s − c) C
tan = , tan
2 s(s − b)
2
2
( s − a)(s − c)
,
ac
(s − a)(s − b)
s(s − c)
Napier's Analogy :
B − C b − c A C − A c − a B
tan = cot , tan = cot
2 b + c 2 2 c + a 2
A − B a − b C
tan = cot
2 a + b 2
Area of Triangle :
∆ = 2
1 bc sin A= 2
1 ca sin B = 2
1 ab sin C
∆ =
1 a
2 sin Bsin C
=
1 b
2 sin Csin A
=
1 c
2 sin A sin B
2 sin(B + C) 2 sin(C + A) 2 sin(A + B)
,
2
2∆
sin A = s(s
− a)(s − b)(s − c) =
bc
bc
2∆ 2∆
Similarly sin B = & sin C = ca
ab
Some Important Results :
A B s − c A B
1. tan tan = ∴ cot cot =
2 2 s 2 2
2. tan
2
A + tan
2
B = s
c cot
2
C =
∆
c (s – c)
A B a − b
3. tan – tan = (s – c)
2 2 ∆
A B
tan + tan
A B
4. cot + cot = 2 2 =
2 2 A B
tan tan
2 2
5. Also note the following identities :
c
s
s
s − c
C
cot
− c 2
Σ(p – q) = (p – q) + (q – r) + (r – p) = 0
Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0
Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0
Solution of Triangles :
1. Introduction : In a triangle, there are six
elements viz. three sides and three angles. In
plane geometry we have done that if three of the
elements are given, at least one of which must
be a side, then the other three elements can be
uniquely determined. The procedure of
determining unknown elements from the known
elements is called solving a triangle.
2. Solution of a right angled triangle :
Case I. When two sides are given : Let the
triangle be right angled at C. Then we can
determine the remaining elements as given in
the following table.
Given
(i) a, b
(ii) a, c
Required
tanA = b
a , B = 90º – A, c =
a
sin A
sinA = c
a , b = c cos A, B = 90º – A
Case II. When a side and an acute angle are given
– In this case, we can determine
Given
(i) a, A
(ii) c, A
Required
B = 90º – A, b = a cot A, c =
a
sin A
B = 90º – A, a = c sin A, b = c cos A
XtraEdge for IIT-JEE 56
FEBRUARY 2010

Based on New Pattern
IIT-JEE 2010
XtraEdge Test Series # 10
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 (one or more than one) correct answer. +4 marks and
-1 mark for wrong answer.
• Question 11 to 16 are passage based questions with multiple (one or more than one) correct answer. +5 marks will be
awarded for correct answer and -1 mark for wrong answer.
Section - II
• Question 17 to 19 are Numerical type questions. +6 marks will be awarded for correct answer and No Negative
marks 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.
1. A cylinder of radius R is floating in a liquid as
shown. The work done in submerging the
cylinder completely in the liquid of density ρ is –
R
L
(A) 100 V
(C) 150 V
30°
45°
(B) 50 V
(D) 200 V
3. A plane mirror is inclined at an angle θ with the
horizontal surface. A particle is projected with
velocity v at angle α. Image of the particle is
observed from the frame of the particle projected
path of the image as seen by the particle is –
L/3
θ
(A) parabolic path
(C) circular path
v
α
(B) straight line
(D) helical path
2
(A) ρπR 2 L 2 8
g (B) ρπR 2 L 2 g
9
18
1
(C) ρπR 2 L 2 2
g (D) ρR 2 L 2 g
3 9
2. An electron with a kinetic energy of 100 eV
enters the space between the plates of plane
capacitor made of two dense metal grids at an
angle of 30° with the plates of capacitor and
leaves this space at an angle of 45° with the
plates. What is the potential difference of the
capacitor –
4. The amplitude of wave disturbance propagating in
1
the positive x-axis is given by y =
2
x – 2x + 1
at
1
t = 2 sec and y =
at t = 6 sec, where
2
x + 2x + 5
x and y are in meters. Velocity of the pulse is -
(A) 1 m/s in positive x-direction
(B) + 2 m/s in negative x-direction
(C) 0.5 m/s in negative x-direction
(D) 1 m/s in negative x-direction
XtraEdge for IIT-JEE 57
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5. A straight conductor of mass m and carrying a
current i is hinged at one end and placed in a
plane perpendicular to the magnetic field B as
shown in figure. At any moment if the conductor
is let free, then the angular acceleration of the
conductor will be (neglect gravity) –
× × × ×
Hinged
× × × B ×
end
× × × ×
i
× × × ×
L
(A)
(C)
3iB
2m
iB
2m
(B)
(D)
2 iB
3 m
3i
2mB
6. The velocity of a body moving on a straight line
− t
τ
in v = v 0 e , then the total distance moved by it
before it stops -
(A) v 0 τ
(B) 2 v 0 τ
(C) 3v 0 τ
(D) None of these
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.
7. A solid is heated up and ∆H vs ∆θ (∆H : Heat
given, ∆θ : change in temperature) is plotted as
shown in figure. Material exist in only one phase
in –
∆H
C
B
E
D
F
A
∆θ
(A) AB (B) BC (C) CD (D) EF
8. A container having dimension 5 m × 4 m × 3 m is
accelerated along its breadth in horizontal.
Container is filled with water upto the height of
1.5 m. Container is accelerated with 7.5 m/s 2 .
If g = 10 m/s 2 and density of water is 10 3 kg/m 3 -
A
B
3m
1.5m
D
4 m
C
(A) Gauge pressure at point C is 10 4 Pascal
(B) Gauge pressure at point D is 3 × 10 4 Pascal
(C) Gauge pressure at the middle of the base is
1.5 × 10 4 Pascal
(D) Remaining value of liquid inside the
container is 20 m 3
9. All capacitors were initially uncharged –
10 µF
15Ω
10 Ω
12Ω
50 V
15Ω
5 µF
(A) Battery current just after closing of switch S is
3.42 A
(B) Battery current just after closing of switch S is
0.962 A
(C) Battery current after long time of closing of
switch S is 3.42 A
(D) Battery current after long time of closing of
switch S is 0.962 A
10. R = 10 Ω and ε = 13 V and voltmeter and
ammeter are ideal then –
a
8V R
V
c
3Ω b 6V
ε
A
(A) Reading of ammeter is 2.4 A
(B) Reading of ammeter is 8.4 A
(C) Reading of voltmeter is 8.4 V
(D) Reading of voltmeter is 27 V
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 11 to 13)
A thin super conducting (zero resistance) ring is
held above a vertical, cylindrical metallic rod as
shown in figure. The axis of symmetry of the ring
is the same as that of the rod. The cylindrically
symmetrical magnetic field around the ring can be
described approximately in terms of the vertical
XtraEdge for IIT-JEE 58
FEBRUARY 2010

and radial components of the magnetic field
vector as B z = B 0 (1 – αz) and B r = B 0 βr where B 0 ,
α and β are constants and z and r are the vertical
and radial position co-ordinates respectively.
Initially the ring has no current flowing in it.
When released, it starts to move downward with
its axis still vertical. Consider the effect of self
induction also.
11. As ring will move downward magnetic flux
through the ring -
(A) will increase
(B) will decrease
(C) will remain constant
(D) will increase first and then decrease
12. As ring will move downward after it release
current induced in ring -
(A) will increase (B) will decrease
(C) remain constant (D) will oscillate
13. Lorentz force acting on the ring due to induced
current is -
(A) vertical and constant
(B) horizontal and constant
(C) vertical and depend on vertical displacement
of ring
(D) horizontal and depend on vertical displacement
of ring
Passage : II (No. 14 to 16)
Medical researchers and technicians can track the
characteristic radiation patterns emitted by certain
inherently unstable isotopes as they
spontaneously decay into other elements. The
half-life of a radioactive isotope is the amount of
time necessary for one-half of the initial amount
of its nuclei to decay. The decay curves of
B r
isotopes 39 Y 90 and 39 Y 91 are graphed below as
functions of the ratio of N, the number nuclei
remaining after a given period, to N, the initial
number of nuclei.
1.0
0.9
0.8
0.7
0.6
N/N 0
0.5
0.4
0.3
0.2
0.1
39Y 90
1 2 3 4 5 6
Time (days)
1.0
0.9
39Y 91
0.8
0.7
N/N
0.6
0
0.5
0.4
0.3
0.2
0.1
30 60 90 120 150 180
Time (days)
14. The half-life of 39 Y 90 is approximately –
(A) 2.7 days (B) 5.4 days
(C) 27 days (D) 58 days
15. What will the approximate ratio of 39 Y 90 to 39 Y 91
be after 2.7 days if the initial samples of the two
isotopes contain equal numbers of nuclei
(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 10 : 1
16. Approximately how many 39 Y 91 nuclei will exist
after three half-lives have passed, if there are
1,000 nuclei to begin with
(A) 50 (B) 125 (C) 250 (D) 500
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
17. A thermometer of mass 50 gm and specific heat
0.4 cal/gm/ºC reads 10ºC. It is then inserted into
1 kg of water and reads 40ºC in thermal
equilibrium. The temperature of water before
insertion of thermometer in 10 ºC is (Neglect
other heat losses).
18. A uniform ball of radius R = 10 cm rolls without
slipping between two rails such that the horizontal
distance is d = 16 cm between two contact points
of the rail to the ball. If the angular velocity is
5 rad/s, then find the velocity of centre of mass of
the ball in cm/s.
19. A wedge of mass M = 2 m 0 rests on a smooth
horizontal plane. A small block of mass m 0 rests
over it at left end A as shown in figure. A sharp
impulse is applied on the block, due to which it
starts moving to the right with velocity v 0 = 6 m/s.
At highest point of its trajectory, the block
XtraEdge for IIT-JEE 59
FEBRUARY 2010

collides with a particle of same mass m 0 moving
vertically downwards with velocity v = 2 m/s and
gets stuck with it. If the combined mass lands at
the end point A of the body of mass M, calculate
length l in cm. Neglect friction, take g = 10 m/s 2 .
A
m 0
l
B
20 cm
(B) CH 3 –CH–CH 2
|
CH 3
(C)CH 3 –CH–CH 2
|
CH 3
(D) CH 3 –CH–CH 2
|
CH 3
OH
|
C–COOH
|
CH 3
CH–COOH
|
CH 3
Cl
|
C–COOH
|
CH 3
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.
1. At 25º C, for the reaction
Br 2 (l) + Cl 2 (g) 2BrCl (g)
K p = 2.032 . At the same temperature the vapour
pressure of Br 2 (l) is 0.281 atm. Pure BrCl(g) was
introduced into a closed container of adjustable
volume. The total pressure was kept 1 atm. and
the temperature at 25ºC. What is the fraction of
BrCl originally present that has been converted
into Br 2 and Cl 2 at equilibrium, assuming that the
gaseous species behave ideally
(A) 0.357 (B) 0.667
(C) 0.2
(D) None of these
2. The activation energy of a non-catalysed reaction
at 37ºC is 83.68 kJ mol –1 and the activation
energy of the same reaction catalysed by an
enzyme is 25.10 kJ mol –1 . What is approximate
ratio of the rate constants of the enzyme catalysed
and the non-catalysed reactions
(A) 10 22 (B) 10 10
(C) 10 20 (D) 10 6
3. Consider the following reaction sequence
O
||
CH3CCl
AlCl3
O
||
(CH3)
2 CHCCl
⎯⎯⎯⎯
AlCl3
–
NH2NH
/ OH
⎯ → A
2
⎯⎯⎯⎯⎯⎯
→ B
+
H 2 O/ H
E
⎯ ⎯⎯⎯
→ C ⎯ HCN ⎯ → D ⎯⎯⎯
→
end product (E) is -
(A) CH 3 –CH–CH 2
|
CH 3
CH 2 –CH 3
4. (A) light blue coloured compound on heating will
convert into black (B) which reacts with glucose
gives red compound (C) and (A) reacts with
ammonium hydroxide in excess in presence of
ammonium sulphate give blue compound (D).
What is (A)
(A) CuO (B) CuSO 4
(C) Cu(OH) 2 (D) [Cu(NH 3 ) 4 ] SO 4
5. 0.80 g of impure (NH 4 ) 2 SO 4 was boiled with 100
ml of a 0.2 N NaOH solution till all the NH 3 (g)
evolved. The remaining solution was diluted to
250 ml. 25 ml of this solution was neutralized
using 5 ml of a 0.2 N H 2 SO 4 solution. The
percentage purity of the (NH 4 ) 2 SO 4 sample is.
(A) 82.5 (B) 72.5 (C) 62.5 (D) 17.5
6. Which of the following is incorrect
(A) The kinetic energy of the gas molecules is
higher above T C , is considered as super
critical fluid
(B) At this temperature (T C ) the gas and the liquid
phases have different critical densities
(C) At the Boyle temperature the effects of the
repulsive and attractive inter molecular forces
just offset each other
(D) In the Maxwell's distribution curve of
velocities the fraction of molecules have
different velocities are different at a given
temperature
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.
7. Consider the following compound
CH 3 –CH(OH)–CH=CH–CH 3
Which of the following is/are correct
(A) Cis form is optically active
(B) Trans form is optically active
(C) Total number of stereo isomers are six
(D) Trans form is optically inactive
XtraEdge for IIT-JEE 60
FEBRUARY 2010

8. Which of the following is/are correct
(A) Trialkyl phosphine oxides are more stable
than the corresponding amine oxides due to
pπ-dπ back bonding
(B) (SiH 3 ) 3 N is less basic than (CH 3 ) 3 N
(C) PBr 5 exist in ionic form as [PBr 4 ] + [Br] – in
solid state
(D) CO, CNR, PR 3 and NO all are the π acid
ligands
9. Which of the following statements is/are correct
(A) The conductance of one cm 3 of a solution is
called specific conductance
(B) Specific conductance increases while molar
conductivity decreases on progressive
dilution
(C) The limiting equivalent conductivity of weak
electrolyte cannot be determine exactly by
extraplotation of the plot of Λ eq against c
(D) The conductance of metals is due to the
movement of free electrons
10. Select the correct statement (s)
(A) Radial function [R(r)] a part of wave function
is dependent on quantum number n only
(B) Angular function depends only on the
direction and is independent to the distance
from the nucleus
(C) ψ 2 (r, θ, φ) is the probability density of
finding the electron at a particular point in
space
(D) Radial distribution function (4πr 2 R 2 ) gives
the probability of the electron being present
at a distance r from the nucleus
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONE OR MORE THAN ONE is correct.
Passage : I (Que. No. 11 to 13)
HO
Ascorbic acid, C 6 H 8 O 6 , also known as vitamin C is a
dibasic acid undergoes dissociation as
C 6 H 8 O 6 C 6 H 7 O – 6 + H + ; K 1 = 8 × 10 –5
–
C 6 H 7 O 6 C 6 H 6 O 2– 6 + H + ; K 2 = 1 × 10 –12
The ascorbic acid is readily oxidised to
dehydroascorbic acid as
HO
OH
O
OH
→ HO
O
O
OH
O
O
O
+ 2H + + 2e
The estimation of ascorbic acid in a sample is made
by titrating its solution with KIO 3 solution which acts
as an intermediate and in presence of 1M HCl
solution, the first excess of iodate gives blue colour
with starch due to the redox change given below
3C 6 H 8 O 6 + IO – 3 → 3C 6 H 6 O 6 + I – + 3H 2 O
IO 3
–
+ 5I – + 6H + → 3I 2 + 3H 2 O
excess (generated
in reaction)
However, if 5M HCl is used, the redox change occurs
as follows :
C 6 H 8 O 6 + IO 3 – + H + +Cl –
C 6 H 6 O 6 +ICl + 2H 2 O
11. The 250 mL sample of fruit juice collected by
crushing a fruit is supposed to have only one
ingradient which can react with KIO 3 is taken in
500 mL measuring flask and 250 mL of 2M HCl
is added. A 50 mL solution is now pipette out and
titrated against intermediate KIO 3 of
concentration 4 × 10 –3 M. It was found that 1 mL
of KIO 3 were used. The molarity and strength of
ascorbic acid are.
(A) 4.8 × 10 –4 M, 84.5 mg/litre
(B) 9.6 × 10 –4 M, 169 mg/litre
(C) 4.8 × 10 –4 M, 84.5 × 10 –3 g/litre
(D) 9.6 × 10 –4 M, 169 g/litre
12. The deactivation of ascorbic acid follows first
order kinetics. The 25 mL sample of juice is kept
for 2 months and after 2 months one titration with
same KIO 3 in presence of 1 M HCl solution
requires 0.5 mL of KIO 3 solution. The average life
of fruit juice is -
(A) 60 day
(B) 50 day
(C) 86.5 day (D) 120 day
13. The degree of dissociation of ascorbic acid
solution is -
(A) 0.40 (B) 0.33
(C) 0.20 (D) 0.15
Passage : II (Que. No. 14 to 16)
Black ppt.(U)
Boil with
dil HNO 3
KI solution
Yellow solution
(S)
Red substance(T)
NH 3
H 2S water
BiCl 3 White turbidity(P)
Alkali Conc.H 2SO 4
+Na 2SnO 2
(Q)
Black ppt.(R)
14. Black ppt. (R) is -
(A) Bi 2 O 3 (B) Na 2 SnO 3
(C) Bi(OH) 3
(D) Bi
15. (Q) is -
(A) Bi 2 (SO 4 ) 3 (B) Bi 2 O 3
(C) Bi 2 O 5
(D) Both (A) & (B)
XtraEdge for IIT-JEE 61
FEBRUARY 2010

16. Yellow solution(s) is because of the formation of -
(A) ppt of BiI 3 (B) I 2 in aqueous solution
(C) KI 3
(D) All of these
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
17. A saturated solution of iodine in water contains
0.33g I 2 per dm 3 . More than this can dissolve
in a KI solution as a result of the reaction
I 2 + I – I – 3 . A 0.10 M KI solution actually
dissolves 12.5g I 2 per dm 3 , most of which is
converted into I – 3 . Assuming the concentration of
I 2 in all saturated solution is same, calculate the
equilibrium constant for the above reaction.
18. A 0.138-g sample of solid magnesium (molar
mass = 24.30g mol –1 ) is burned in a constant
volume bomb calorimeter that has a heat capacity
of 1.77 kJ ºC –1 . The calorimeter contains 300 mL
of water (density 1g mL –1 ) and its temperature is
raised by 1.126ºC. The numerical value for the
enthalpy of combustion of the solid magnesium at
298 K in kJ mol –1 is.
19. Two liquids A and B form an ideal solution at
temperature T. When the total vapour pressure
above the solution is 450 torr, the amount fraction
of A in the vapour phase is 0.35 and in the liquid
phase is 0.70. The sum of the vapour pressures of
pure A and pure B at temperature T is.
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.
1. If the last term in the binomial expansion of
n
log 8
3
⎛ 1/ 3 1
2 ⎟
2
⎞ ⎛ 1 ⎞
⎜ − is ⎜
5/3 ⎟⎠
⎝ ⎠ ⎝ 3
, then the 5 th term
from the beginning is -
(A) 210 (B) 420
(C) 105
(D) none of these
⎡ 1 ⎤
2. The matrix product
⎢ ⎥
⎢
2
⎥
[1 2 – 1]
⎢⎣
−1⎥⎦
(A) is not defined (B) equals [–1]
⎡1⎤
(C) equals
⎢ ⎥
⎢
4
⎥
⎢⎣
1⎥⎦
(D) is not invertible
3. The domain of definition of
f(x) =
⎛ x −1
⎞ 1
log 0 . 4 ⎜ ⎟ × is
⎝ x + 5 ⎠ x
− 36
(A) (– ∞, 0) ~ {– 6} (B) (0, ∞) ~ {1, 6}
(C) (1, ∞) ~ {6} (D) [1, ∞) ~ {6}
4. The coordinates of the point on the parabola
y 2 = 8x which is at minimum distance from the
circle x 2 + (y + 6) 2 = 1 are
(A) (2, – 4) (B) (18, –12)
(C) (2, 4)
(D) none of these
5. If
⎧ ⎛ ⎞ ⎛ ⎞⎫
I=
∫ π |cos x| 1
1
e ⎨2sin⎜
cos x⎟ + 3cos⎜
cos x⎟⎬sinxdx,
0
⎩ ⎝ 2 ⎠ ⎝ 2 ⎠⎭
then I equals -
(A) 7 e cos (1/2)
(B) 7 e [cos(1/2) – sin(1/2)]
(C) 0
(D) none of these
6. The solution of the differential equation
2
d y
= sin 3x + e x 2
dx
+ x 2 when y 1 (0) = 1 and
y(0) = 0 is -
−sin 3x
(A) + e x x 4 1
+ + x – 1
9 12 3
−sin 3x
(B) + e x x 4 1
+ + x
9 12 3
− cos3x
(C) + e x x 4 1
+ + x + 1
3 12 3
(D) none of these
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.
7. The determinant
a b
∆ =
b
c
aα + b
bα + c is equal to zero if -
aα + b bα + c 0
(A) a, b, c are in A.P.
(B) a, b, c are in G.P.
(C) a, b, c are in H.P.
(D) α is a root of ax 2 + 2bx + c = 0
XtraEdge for IIT-JEE 62
FEBRUARY 2010

8. For two events A and B, if
P(A) = P(A \ B) = 1/4 and P(B \ A) = 1/2, then
(A) A and B are independent
(B) A and B are mutually exclusive
(C) P (A′ \ B) = 3/4
(D) P (B′ \ A′) = 1/2
8 ⎡ 1 ⎤
9. The lim x
x→0
⎢ 3 ⎥ (where [x] is greatest integer
⎣ x ⎦
function) is -
(A) a nonzero real number
(B) a rational number
(C) an integer
(D) zero
dy x + y + 1
10. The solution of = satisfying
dx 2xy
y(1) = 1 is given by -
(A) a system of hyperbola
(B) a system of circles
(C) y 2 = x(1 + x) – 1
(D) (x – 2) 2 + (y – 3) 2 = 5
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 11 to 13)
C: x 2 + y 2 x
= 9, E:
9
2
2
2
y
+ = 1, L: y = 2x
4
11. P is a point on the circle C, the perpendicular PL
to the major axis of the ellipse E meets the ellipse
ML
at M, then is equal to -
PL
(A) 1/3 (B) 2/3
(C) 1/2
(D) none of these
12. If L represents the line joining the point P on C to
its centre O, then equation of the tangent at M to
the ellipse E is -
(A) x + 3y = 3 5 (B) 4x + 3y = 3 5
(C) x + 3y + 3 5 = 0 (D) 4x + 3 + 5 = 0
13. Equation of the diameter of the ellipse E
conjugate to the diameter represented by L is -
(A) 9x + 2y = 0 (B) 2x + 9y = 0
(C) 4x + 9y = 0 (D) 4x – 9y = 0
Passage : II (No. 14 to 16)
Integrals of class of functions following a definite
pattern can be found by the method of reduction
2
and recursion. Reduction formulas make it
possible to reduce an integral dependent on the
index n > 0, called the order of the integral, to an
integral of the same type with a smaller index.
Integration by parts helps us to derive reduction
formulas.
dx
14. If I n =
∫ 2 2
(x + a )
equal to -
x
(A)
2 2 n
(x + a )
(C)
1
2n a
2
.
(x
2
x
+ a
n
n
2
)
1− 2n 1
then I n + 1 + . I
2n
2 n is
a
n
(B)
(D)
1
2n a
1
2n a
2
2
(x
(x
2
2
1
+ a
+ a
2 n−1
x
)
2 n+
1
sin x
n −1
15. If I n, –m =
∫
dx then I
m
n,–m + In–2, 2–m , is
cos x
m −1
equal to -
n−1
n−1
sin x
1 sin x
(A)
(B)
m−1
cos x
(m −1)
m−1
cos x
(C)
1 sin
(n −1)
cos
n−1
m−1
n
x
x
(D)
n −1
sin
m −1
cos
n−1
m−1
x
16. If u n = dx, then
∫ 2
ax + 2bx + c
(n + 1)au n+1 + (2n + 1)bu n + nc u n–1 is equal to -
(A) x n–1
(C)
ax
2
ax 2 + bx + c (B)
x
n
+ bx + c
(D) x
ax
n
x
2
n−2
x
x
+ bx + c
ax
2
)
+ bx + c
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
17. If sec A tan B + tan A sec B = 91, then the value
of (sec A sec B + tan A tan B) 2 is equal to …
18. If
∫ π / 2 x cos x
0 (1 + sin x)
2
2
A
dx = π – π 2 then A is …
498
19. Two circles are inscribed and circumscribed about
a square ABCD, length of each side of the square
is 32. P and Q are two points respectively on
these circles, then Σ(PA) 2 + Σ(QA) 2 is equal to …
XtraEdge for IIT-JEE 63
FEBRUARY 2010

Based on New Pattern
IIT-JEE 2011
XtraEdge Test Series # 10
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 (one or more than one) correct answer. +4 marks and
-1 mark for wrong answer.
• Question 11 to 16 are passage based questions with multiple (one or more than one) correct answer. +5 marks will be
awarded for correct answer and -1 mark for wrong answer.
Section - II
• Question 17 to 19 are Numerical type questions. +6 marks will be awarded for correct answer and No Negative
marks for wrong answer.
PHYSICS
A
u = 0
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.
1. Ice point and steam point on a particular scale
reads 10º and 80º respectively. The temperature
on ºF scale when temperature on new scale is 45º
is -
(A) 50º F (B) 112ºF
(C) 122ºF (D) 138ºF
2. In a system of four unequal particles located in an
arrangement of non-linear, coplanar system -
(A) The centre of mass must lie within the closed
figure formed by joining the extreme
particles by straight line
(B) The centre of mass may lie within or outside
the closed figure formed by joining the
extreme particles by straight lines
(C) The centre of mass must lie within or at the
edge of at least one of the triangles formed by
any three particles.
(D) None of these
3. A body starts slipping on a smooth track from
point A and leaves the track from point B as
shown. The part OB of track is straight at angle
37º with horizontal. The maximum height of body
from ground when it is in air is : (g = 10 m/s 2 )
B
H 1 = 15m
H 2 = 10m
37º
ground O
(A) 16.8 m
(B) 13.6 m
(C) 11.8 m
(D) None of these
4. A particle is moving along x-axis and graph
between velocity of the particle and position is
given in figure :
v (m/s)
6
2
4 x (m)
Acceleration of particle at x = 2 m is –
(A) 2 m/s 2 (B) 1 m/s 2
(C) 4 m/s 2 (D) 3 m/s 2
5. Select the incorrect statement –
(A) It is possible to transfer heat to a gas without
raising its temperature
(B) It is possible to raise temperature of gas
without transfer heat to gas
XtraEdge for IIT-JEE 64
FEBRUARY 2010

(C) It is possible to raise temperature by
expanding the volume at keeping pressure
constant
(D) It is not possible to calculate the work done
by gas if we do not know the initial and final
value of pressure and volume when process is
isobaric
6. Select the incorrect statement –
(A) The pressure on the bottom of a vessel filled
with liquid does not depend upon the area of
liquid surface
(B) Buoyancy occurs because, as the depth in a
fluid increase, the pressure increases
(C) The output piston of a hydraulic press cannot
exceed the input piston's work
(D) The pressure of atmosphere at sea level
corresponds to 101.3 millibar
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.
7. For two different gases x and y, having degrees
of freedom f 1 and f 2 and molar heat capacities at
constant volume CV 1
and CV 2
respectively, the
lnP versus lnV graph is plotted for adiabatic
process as shown, then –
lnP y
x
lnV
(A) f 1 > f 2 (B) f 2 > f 1
(C) C < C (D) C > C
V 2
V 1
V1 V 2
8. Which of the following is valid wave equation
traveling on string -
–b(x + vt)
(A) Ae (B) A sec (kx – ωt)
1
(C)
2
1+
{x(1 + t / x)}
(D) A sin (x 2 – vt 2 )
9. Two sphere of same radius and material, one solid
and one hollow are heated to same temperature
and kept in a chamber maintained at lower
temperature at t = 0 -
(A) Rate of heat loss of the two sphere will be
same at t = 0
(B) Rate of temperature loss of the two sphere
will be same at t = 0
(C) Rate of heat loss of solid sphere will be more
than hollow sphere at t > 0
(D) Rate of temperature loss of the two sphere
may be same at t > 0
10. A stick is tied to the floor of the water tank with a
string as shown. The length of stick is 2 m and its
area of cross-section is 10 –3 m 2 . If specific gravity
of stick is 0.25 and g = 10 m/s 2 , then –
(A) tension in the string is 5 N
(B) buoyancy force acting on stick is 10 N
(C) length of stick immersed in water is 1 m
(D) tension in the string is zero
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 11 to 13)
A completely inelastic collision takes place
between two body A and B of masses m and 2 m
moving respectively with speed v each as shown.
The collision is oblique and before collision A is
moving along positive x-axis while B is moving at
angle θ with x-axis as shown. Then -
y
A v r
m θ
v r
B
2m
11. Speed of composite body after collision is -
v
v
(Α) 5 + 4cosθ
(B) 5 + 4 sin θ
3
3
v
(C) 4 + 5cosθ
(D) None of these
3
12. The angle α that the velocity vector of composite
body makes with x-axis after collision is -
− ⎡ 2sin θ ⎤
(A) tan 1 − ⎡ 2cosθ
⎤
⎢ ⎥ (B) tan 1
⎣1+
2cosθ
⎢ ⎥ ⎦ ⎣1+
2sin θ ⎦
− ⎡ 2sin θ ⎤
(C) tan 1
⎢ ⎥
⎣1+
2sin θ ⎦
x
(D) None of these
13. The loss in kinetic energy in collision is -
(A)
(C)
2
2mv
(1 – cos θ) (B)
6
2
4mv
(1 – cos θ) (D)
6
2
2mv
(1 – sin θ)
6
2
4mv
(1 – sin θ)
6
XtraEdge for IIT-JEE 65
FEBRUARY 2010

Passage : II (No. 14 to 16)
A disc of radius 20 cm is rolling with slipping on
a flat horizontal surface. At a certain instant, the
velocity of its centre is 4 m/s and its angular
velocity is 10 rad/s. The lowest contact point is O.
10 rad/s
P
C
4 m/s
O
14. Velocity of point O is -
(A) 0
(B) 2 m/s
(C) 4 m/s
(D) 8 m/s
15. Velocity of point P is -
(A) 2 5 m/s (B) 5 2 m/s
(C) 2 2 m/s (D) 8 m/s
16. The distance of instantaneous center of rotation
from the point O is -
(A) 0.2 m below (B) 0.2 m above
(C) 0.4 m below (D) 0.4 m above
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
17. The upper edge of a gate in a dam runs along the
water surface. The gate is 2.00 m high and 4.00 m
wide and is hinged along a horizontal line through
its center. The torque about the hinge arising from
the force due to the water is (n × 10 4 Nm). Find
value of n.
2 m
18. A longitudinal wave of frequency 220 Hz travels
down a copper rod of radius 8.00 mm. The
average power in the wave is 6.50 µW. The
amplitude of the wave is n × 10 –8 m. Find n.
(Density of copper is 8.9 × 10 3 kg/m 3 , young's
modulus of copper Y cu = 1.1 × 10 11 Pa).
19. A piston-cylinder device with air at an initial
temperature of 30ºC undergoes an expansion
process for which pressure and volume are related
as given below
P (kPa) 100 25 6.25
V (m 3 ) 0.1 0.2 0.4
The work done by the system is n × 10 3 J. Find n.
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.
1. The equilibrium constant for the reaction in
aqueous solution
H 3 BO 3 + glycerin (H 3 BO 3 – glycerin) is
0.90. How many moles of glycerin should be
added per litre of 0.10 M H 3 BO 3 so that 80% of
the H 3 BO 3 is converted to the boric acid glycerin
complex
(A) 4.44 (B) 4.52
(C) 3.62 (D) 0.08
2. Liquid NH 3 ionises to a slight extent. At a certain
temperature it's self ionization constant(K SIC ) is
10 –30 +
. The number of NH4
ions are present per
100 cm 3 of pure liquid are -
(A) 1 × 10 –15 (B) 6.022 × 10 8
(C) 6.022 × 10 7 (D) 6.022 × 10 6
3. The preparation of SO 3 (g) by reaction
SO 2 (g) + 2
1
O2 (g)
SO 3 (g) is an exothermic
reaction. If the preparation follows the following
temperature-pressure relationship for its % yield,
then for temperature T 1 , T 2 and T 3 which of the
following is correct -
50
40
T 3
30
T 2
20
T 1
10
% yield
1 2 3 4
P(atm)
(A) T 1 > T 2 > T 3
(B) T 3 > T 2 > T 1
(C) T 1 = T 2 = T 3
(D) None is correct
XtraEdge for IIT-JEE 66
FEBRUARY 2010

4. Which of the following would be optically
inactive
H
CH 3
Cl
CH 3
CH
C C
3
OH
CH 3
Cl
H H
H
OH
(I)
(II)
OH
H
CH 3
(A) Only I
(C) Only II & III
CH 3
H
OH
(III)
(B) Only II
(D) I, II & III
5. A molecule may be represented by three
structures having energies Q 1 , Q 2 and Q 3
respectively. The energies of these structures
follow the order Q 1 > Q 2 > Q 3 respectively. If the
experimental bond energy of the molecule is Q E ,
the resonance energy is -
(A) (Q 1 + Q 2 + Q 3 ) – Q E (B) Q E – Q 3
(C) Q E – Q 1 (D) Q E – Q 2
6. In a compound
NC
M(CO) 3
C C
NC C 4 H 3
The number of sigma and pi bonds respectively
are -
(A) 19, 11 (B) 19, 10
(C) 13, 11 (D) 19, 14
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.
7. Select the correct statement(s) about the
compound NO[BF 4 ].
(A) It has 5σ and 2π bond
(B) Nitrogen-oxygen bond length is higher than
nitric oxide
(C) It is a diamagnetic species
(D) B-F bond length in this compound is lower
than in BF 3
8. Which of the following process is/are associated
with change of hybridization of the underlined
compound
(A) Solidification PCl 5 vapour
(B) SiF 4 vapour is passed through liquid HF
(C) B 2 H 6 is dissolved in THF
(D) Al(OH) 3 ppt. dissolved in NaOH
9. S, T and U are the aqueous chlorides of the
elements X, Y and Z respectively. X, Y and Z are
in the same period of the periodic table. U gives a
white precipitate with NaOH but this white
precipitate dissolves as more NaOH is added.
When NaOH is added to T, a white precipitate
forms which does not dissolve when more base is
added. S does not give precipitate with NaOH.
Which of the following statements are correct
(A) The three elements are metals
(B) The electronegativity decreases from X to Y
to Z
(C) X, Y and Z could be sodium, magnesium and
aluminium respectively
(D) The first ionization increases from X to Y to Z
10. Which of the following enol form dominate over
keto form
O
(A)
(C)
O
O
N O (B)
H
O
(D)
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONE OR MORE THAN ONE is correct.
Passage : I (Que. No. 11 to 13)
Alkenes undergo electrophilic addition reaction
with Hg(OAc) 2 , BH 3 and H 2 O. In all these cases
reaction is regioselective reaction. BH 3 gives
addition reaction via formation of four centred
cyclic transition state. Hg(OAc) 2 gives addition
reaction via formation of bridge carbocation as
reaction intermediate whereas water gives
addition reaction via formation of classical
carbocation.
11. Alkene can be converted into alcohol by which of
the following reagents -
(A) Hg(OAc) 2 /HOH followed by NaBH 4
(B) BH 3 /THF followed by H 2 O 2 /NaOH
(C) H 2 O/H 2 SO 4
(D) None of these
12. In the gives reaction :
CH 3 − C = CH
|
CH3
[X] is/are -
2
O
⎯ [ ⎯→
X]
CH3
− CH
|
CH3
O
− CH
2
OH
XtraEdge for IIT-JEE 67
FEBRUARY 2010

(A) Hg(OAc) 2 /HOH followed by NaBH 4
(B) BH 3 followed by H 2 O 2 /NaOH
(C) H 2 O/H 2 SO 4
(D) None of these
13. In the given reaction
CH3
|
CH3
−C
− CH = CH2
⎯ ⎯→
X]
CH
|
CH
3
O
| H
[
3 − C — CH −
| |
CH3
CH3
[X] is/are -
(A) H 2 O/H 2 SO 4
(B) Hg(OAc) 2 /HOH followed by NaBH 4
(C) BH 3 followed by H 2 O 2 /NaOH
(D) None of these
Passage : II (Que. No. 14 to 16)
CH
Heat of neutralization is heat evolved when 1g
equivalent of acid and 1g equivalent of base react
together to form salt and water. Heat of
neutralization is –57.1 kJ mol –1 for strong acid
and strong base. In case of weak acid or weak
base, it is less than 57.1 kJ mol –1 .
14. 400 ml of 0.1 M NaOH is mixed with 300 ml of
0.1 M H 2 SO 4 . The heat evolved will be -
(A) 2.284 kJ (B) 1.713 kJ
(C) 9.59 kcal (D) 7.1946 kcal
15. A solution of 200 ml of 1 M KOH is added to 200
ml of 1M HCl and the mixture is well shaken.
This rise in temperature T 1 is noted. The
experiment is repeated by using 100 ml of each
solution and increase in temperature T 2 is again
noted. Which of the following is/are incorrect -
(A) T 1 = T 2
(B) T 2 is twice as large as T 1
(C) T 1 is twice as large as T 2
(D) T 1 is four times as large as T 2
16. Which of the following will not produce
maximum energy except one
(A) Ba(OH) 2 + H 2 SO 4 (B) NH 4 OH + HCl
(C) (COOH) 2 + NaOH (D) H 3 PO 4 + NaOH
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
17. Calculate the % of free SO 3 in an oleum that is
labelled '109% H 2 SO 4 '.
18. A mixture of NH 3 (g) and N 2 H 4 (g) is placed in a
sealed container at 300 K. The total pressure is
0.5 atm. The container is heated to 1200 K at
which time both substances decompose
completely according to the equations
3
2NH 3 (g) → N 2 (g) + 3H 2 (g) and
N 2 H 4 (g) → N 2 (g) + 2H 2 (g)
After decomposition is complete, the total
pressure at 1200 K is found to be 4.5 atm. Find
the mole % of N 2 H 4 in the original mixture.
19. A 200 g sample of hard water is passed through
the column of cation exchange resin, in which H +
is exchanged by Ca 2+ . The outlet water of column
required 50 ml of 0.1 M NaOH for complete
neutralization. What is the hardness of Ca 2+ ion in
ppm
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.
1. If w is an imaginary cube root of unity, then value
of the expression
2(1 + w) (1 + w 2 ) + 3(2 + w) (2 + w 2 )
+ ….. + (n + 1) (n + w) (n + w 2 ) is
(A) 4
1 n 2 (n + 1) 2 + n
(C) 4
1 n(n + 1) 2 – n
(B) 4
1 n 2 (n + 1) 2 – n
(D) none of these
2. In a triangle PQR, ∠ R = π/4. If tan (P/3)
and tan(Q/3) are the roots of the equation
ax 2 + bx + c = 0, then -
(A) a + b = c (B) b + c = 0
(C) a + c = b (D) b = c
3. Sum of all three digit numbers (no digit being
zero) having the property that all digits are perfect
squares, is -
(A) 3108 (B) 6210
(C) 13986
(D) none of these
4. In a triangle ABC,
r 1 r + 2 r + 3 is equal to
bc ca ab
1 1
(A) –
2R r
(B) 2R – r
(C) r – 2R
1 1
(D) –
r 2R
5. If an ellipse slides between two perpendicular
straight lines, then the locus of its centre is -
(A) a parabola (B) an ellipse
(C) a hyperbola (D) a circle
6. If the lines whose vector equations are
r = a + tb, r = c + t'd are coplanar then -
(A) (a – b). c × d = 0 (B) (a – c). b × d = 0
(C) (b – c). a × d = 0 (D) (b – d). a × c = 0
XtraEdge for IIT-JEE 68
FEBRUARY 2010

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.
7. For a positive integer n, let
1 1 1 1
a(n) = 1 + + + + … + 2 3 4 (2
n ) − 1
. Then -
(A) a(n) < n
(C) a(2n) > n
(B) a(n) >
2
n
(D) a(2n) < 2n
8. If log 2 (3 2x – 2 + 7) = 2 + log 2 (3 x – 1 + 1) then x is -
(A) 0 (B) 1
(C) 2
(D) none of these
9. Let P (a sec θ, b tan θ) and Q (a sec φ, b tan θ)
where θ + φ = π/2, be two points on the hyperbola
x 2 /a 2 – y 2 /b 2 = 1. If (h, k) is the point of
intersection of normals at P and Q, then k is equal
to -
a
2 2
+ b
⎡
2 2
a + b ⎤
(A)
(B) – ⎢ ⎥
a
⎢⎣
a ⎥⎦
(C)
a
2 +
b
b
2
⎡
2 2
a + b ⎤
(D) – ⎢ ⎥
⎢⎣
b ⎥⎦
10. If a, b, c are three unit vectors such that
1
a × (b × c) = b and c being non parallel then -
2
(A) angle between a and b is π/2
(B) angle between a and c is π/4
(C) angle between a and c is π/3
(D) angle between a and b is π/3
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 11 to 16) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.
Passage : I (No. 11 to 13)
α, β, γ, δ are angles in I, II, III and IV quadrant
respectively and no one of them is an integral
multiple of π/2. They form an increasing arithmetic
progression.
11. Which statement are true -
(A) cos (α + δ) > 0 (B) cos (α + δ) = 0
(C) cos (α + δ) < 0 (D) none of these
12. Which statement are true -
(A) sin (β + γ) = sin (α + δ)
(B) sin (β – γ) = sin (α – δ)
(C) tan 2(α – β) = tan (β – δ)
(D) cos (α + γ) = cos 2β
13. If α + β + γ + δ = θ and α = 70º
(A) 400º < θ < 580º
(B) 470º < θ < 650º
(C) 680º < θ < 860º
(D) 540º < θ < 900º
Passage : II (No. 14 to 16)
P is a point on the circle C 1 : q 2 (x 2 + y 2 ) = a 2 p 2
Q is a point on the circle C 2 : x 2 + y 2 = a 2
14. If the coordinates of P are (h, k) then the locus of
the point which divides the join of PQ in the ratio
p : q is a circle C 3 , whose centre is at the point -
⎛
(A)
⎜
⎝
⎛
(C)
⎜
⎝
hp
p
kq
,
+ q p + q
hq
p
kq
,
+ q p + q
⎟ ⎞
⎠
⎟ ⎞
⎠
⎛
(B)
⎜
⎝
⎛
(D)
⎜
⎝
h
p
k
,
+ q p + q
hp
p
kp
,
+ q p + q
15. Locus of the centre of C 3 as P moves on the circle
C 1 is a circle C 4
(A) concentric with C 1
(B) concentric with C 2
(C) having radius equal to the radius of C 3
(D) having area equal to the area of C 1
16. If the point (p, q) lies on the line y = 2x, then the
radius of C1
radius of C4
is equal to -
(A) 2/3 (B) 3/2
(C) 3 (D) 1/3
Numerical response questions (Q. 17 to 19). Answers to
this Section are to be given in the form of nearest
integer-in four digits. Please follow as per example :
(i.e. for answer : 1492.2 write 1492; 491.8 write 0492;
92.5 write 0092; 2.1 write 0002)
n
17. If z n = ( 1+ i 3) , find the value of 3 lm (z 5 z 4 ).
⎡ −1
1 −1
4 ⎤
18. If x = tan ⎢cos
− sin ⎥
⎣ 5 2 17 ⎦
equal to.
⎟ ⎞
⎠
⎟ ⎞
⎠
90
then
x 2 is
19. If Q is the foot of the perpendicular from the
x − 5 y + z − 6
point P(4, –5, 3) on the line = =
3 − 42
5
then 100(PQ) 2 is equal to.
XtraEdge for IIT-JEE 69
FEBRUARY 2010

MOCK TEST PAPER-3
CBSE BOARD PATTERN
CLASS # XII
SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS
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. Write the formula for the force 'F' experienced by a
particle carrying a charge 'q' moving with velocity 'v'
in a uniform magnetic field 'B'. Under what condition
is this force zero
2. Two metals A and B have a work function 4eV and
10eV respectively. Which metal has a higher
threshold wavelength
3. Why is the transmission of signals using ground
waves restricted to frequencies less than about 1500
kHz
4. Name the phenomenon responsible for the reddish
appearance of the sun at sunrise and sunset.
5. Why is the penetrating power of gamma rays very
large
6. What are the two main considerations that have to be
kept in mind while designing the 'objective' of an
astronomical telescope
7. Is Young's experiment interference or diffraction
experiment
8. Light passes from air into glass. Which of the
following quantities namely, velocity, frequency and
wavelength change during the process
9. Draw the graphs showing variation of resistivity with
temperature for (i) nichrome and (ii) silicon.
10. The circuit shown in the diagram contains a battery
'B', a rheostat 'Rh' and identical lamps P and Q. What
will happen to the brightness of the lamps, if the
resistance through the rheostat is increased Give
reasons.
B
P
Q
Rh
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11. A circular coil of 30 turns and radius 8.0 cm carrying
a current of 6.0 A is suspended vertically in a
uniform horizontal magnetic field of magnitude 1.0
T. The field lines make an angle of 60º with the
normal to the coil. Calculate the magnitude of
counter torque that must be applied to prevent the
coil from turning.
12. A uniform magnetic field exists normal to the plane
of the paper over a small region of space. A
rectangular loop of wire is slowly moved with a
uniform velocity across the field as shown. Draw the
graph showing the variation of
(i) magnetic flux linked with the loop and (ii) the
induced e.m.f. in the loop with time.
× × × × × × × × × ×
× × × × × × × × ×
× × × × × × × × ×
× × × × × × × × ×
Stage-1 Stage-2 Stage-3
OR
A bar magnet is dropped so that it falls vertically
through the coil C. The graph obtained for the
voltage produced across the coil versus time is as
shown in figure (b) (i) Explain the shape of the graph
and (ii) why is the negative peak longer than the
positive peak
Magnet
v R Coil
C
(a)
p.d/mV
(b)
Time/ms
13. Violet light is incident on a thin convex lens. If this
light is replaced by red light, explain with reason,
how the power of the lens would change.
14. (a) Draw a graph showing the variation of potential
energy of a pair of nucleons as a function of their
separation. Indicate the regions in which nuclear
force is
(i) attractive, and (ii) repulsive.
(b) Write two characteristic features of nuclear force
which distinguish it from the Coulomb force.
15. Distinguish between 'point to point' and 'broadcast'
communication modes. Give one example of each.
16. 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
17 For a CE-transistor amplifier, the audio signal
voltage across the collector resistance of 2kΩ is 2V.
Suppose the current amplification factor of the
transistor is 100, find the input signal voltage and
base current, if the base resistance is 1kΩ.
18. Give the logic symbol for an OR gate. Write its truth
table. Draw the output wave form for input wave
forms shown for this gate.
A
(Inputs)
B
19. Define mutual inductance of a pair of coils. Deduce
an expression for the mutual inductance between a
pair of coils having number of turns N 1
and N 2
wound over an air core.
20. In the given circuit, the potential difference across the
inductor L and resistor R are 120V and 90V respectively
and the rms value of current is 3A. Calculate (i) the
impedance of the circuit and (ii) the phase angle
between the voltage and the current.
L
~
21. With the help of a labelled circuit diagram, explain
how an n-p-n transistor is used to produce selfsustained
oscillations in an oscillator.
OR
Draw a labelled circuit diagram to show how an n-pn
transistor can be used as an amplifier in common
emitter configuration. For the given input waveform,
, draw the corresponding output waveform.
22. Two point charges 5 × 10 -8 C and –2 × 10 -8 C are
separated by a distance of 20cm in air as shown in the
figure.
5×10 –6 C –2×10 –8 C
A
20 cm
B
(i) Find at what distance from point A the electric
potential would be zero.
(ii) Also calculate the electrostatic potential energy of
the system.
R
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23. Which constituent radiation of the electromagnetic
spectrum is used
(i) in radar,
(ii) to photograph internal parts of a human body, and
(iii) for taking photographs of the sky during light
and foggy conditions
Give one reason for your answer in each case.
24. In the potentiometer circuit shown, the balance (null)
point is at X. State with reason, where the balance
point will be shifted when
(i) Resistance R is increased, keeping all parameters
unchanged.
(ii) Resistance S is increased, keeping R constant.
(iii) Cell P is replaced by another cell whose e.m.f. is
lower than that of cell Q.
A
Q
P
S
X
25. The work function of caesium is 2.14eV. Find (i) the
threshold frequency for caesium, and (ii) the
wavelength of incident light if the photocurrent is
brought to zero by stopping potential of 0.60V.
26. Complete the following decay process for β-decay of
Phosphorus 32:
32
15 P → S +
.......
The graph shows how the activity of a radioactive
nucleus changes with time. Using the graph,
determine (i) half-life of the nucleus and (ii) its decay
constant.
80
Activity/Bq
60
40
20
G
R
B
0 50 100 150 200 250
Time/s
27. In Young's double-slit experiment, explain with
reason what happens to the interference fringes, when
(i)widths of the slits are increased,
(ii) mono-chromatic light source is replaced by a
white light source, and (iii) one of the slits is closed.
28. Draw electric field lines between the plates of a
parallel plate capacitor with (i) air and (ii) dielectric
as the medium. A parallel plate capacitor with air as
dielectric is connected to a power supply and charged
to a potential difference V 0
. After disconnecting from
power supply, a sheet of insulating material is
inserted between the plates completely filling the
space between them. How will its (i) capacity, (ii)
electric field and (iii) energy change Given that the
capacity of capacitor with air as medium is C 0 and
permittivity for air and medium are ε and ε 0
respectively.
OR
Derive an expression for the electric potential at a
point along the axial line of an electric dipole. At a
point due to a point charge, the values of electric field
and electric potential are 32 NC -1 and 16JC -1
respectively. Calculate (i) magnitude of the charge
and (ii) distance of the charge from the point of
observation.
29. (i) With the help of a schematic sketch of a cyclotron
explain its working principle.
Mention its two applications. What is the important
limitation encountered in accelerating a light
elementary particle such as electron to high energies.
(ii) A particle of mass m and charge q moves at right
angles to a uniform magnetic field. Plot a graph
showing the variation of the radius of the circular
path described by it with the increase in its (a)
charge, (b) kinetic energy, where, in each case other
factors remain constant. Justify your answer.
OR
(i) Using Biot-Savart's law derive an expression for
the magnetic field due to a current carrying loop at a
point along the axis of the loop.
(ii) A long straight conductor carries a steady current
I. The current is distributed uniformly across its
cross-section of radius 'a'. Plot a graph showing the
variation of magnetic field 'B' produced by the
conductor with the distance 'r' from the axis of the
conductor in the region
(i) r < a and (ii) r > a.
30. How would you estimate rough focal length of a
converging lens Draw a ray diagram to show
image formation by a diverging lens. Using this
diagram, derive the relation between object distance
'u' image distance 'v' and focal length 'f' of the lens,
Sketch the graph between 1/u and 1/v for this lens.
OR
Define magnifying power of an optical telescope.
Draw a ray diagram for an astronomical refracting
telescope in normal adjustment showing the paths
through the instrument of three rays from a distant
object. Derive an expression for its magnifying
power. Write the significance of diameter of the
objective lens on the optical performance of a
telescope.
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1. What are F- centres
CHEMISTRY
2. 2.46 g of NaOH (molar mass = 40) are dissolved in
water and solution is made to 100 cm 3 . Calculate
molarity of solution.
3. How are gold and Pt sol prepared.
4. Why is PbO 2 and PbCl 2 are good oxidising agent
5. Arrange the following in decreasing order
(a) F 2 , Cl 2 , Br 2 , I 2 [Bond energy.]
(b) MF, MCl, MBr, MI [Ionic character]
6. Account for the following
(a) NH 3 has higher boiling point than PH 3 .
(b) H 3 PO 3 is diprotic acid.
7. Why Ce +3 can be easily oxidised to Ce +4
8. What is oxoprocess For what purpose is it used.
9. A decimolar solution of K 4 [Fe(CN) 6 ] is 50%
dissociate at 300 K. Calculate the osmotic pressure of
solution (R = 0.0821 L atm K –1 mol –1 ).
10. E º = – 0.76 V. Write the reactions occuring at
Zn + 2 / Zn
the electrodes when coupled with SHE (Standard
Hydrogen Electrode).
11. Convert :
(a) 2-propanol to chloroform
(b) Acetone to iodoform
12. Complete the following reactions
CH2OH
(a) | + HNO
CH OH Conc 3 →
2
CH 2OH
|
(b) CHOH
|
CH OH
2
⎯
KHSO
⎯ 4⎯
Heat
→
13. Convert :
(a) Aniline to benzonitrile
(b) Aniline to phenyl isocyanide.
14. Name two principal ways by which cell obtain
energy for the synthesis of ATP.
15. Name the vitamins, the deficiency of which causes
the following disease.
(a) Beri-beri
(b) Night blindness
(c) Poor coagulation of blood
(d) Pernicous anaemia
16. KF has NaCl structure. What is the distance between
K + and F – in KF, if the density is 2.48 g cm –3
17. Derive the relationship between activation energy
and rate constant.
18. Define heterogeneous catalysis. Give four example.
19. Complete the following :
(a) XeF 4 + SbF 5 →
(b) Cl 2 + NaOH →
(Cold & dil.)
(c) F 2 +
(d) F 2 +
H →
2O
( Hot)
NaOH →
(Hot&conc)
(e) XeF 6 + KF →
(f) BrO 3 – + F 2 + 2OH – →
20. Discuss :
(a) Catenation
(b) Thermal stability of hydride
(c) Reducing power of hydrides with respect to group
15, 16 and 17.
21. Work out the following chemical equations :
(i) In moist air copper corrodes to produce green
layer on its surface.
(ii) Chlorination of Ca(OH) 2 produces bleaching
powder.
(iii) Copper sulphate from metallic copper.
22. Using VBT predict the shape and magnetic behaviour
of
(i) [Ni(CO) 4 ] (ii) [NiCl 4 ] 2–
23. Mention one method of preparation of following
organometallics
(a) Zeise's salt (b) Dibenzene chromium
(c) n-butyl lithium
24. What is Lucas reagent For what purpose is it used
and how
25. (a) What is Buna-S Name the monomers used in its
preparation. Mention its use.
(b) Differentiate between elastomer and fibres on the
basis of intermolecular forces.
(c) Give an example of step growth polymer.
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26. Give an example of
(a) Triphenyl methane dye
(b) Azo dye
(c) Anthraquinone dye
27. Using IUPAC names write the formula for the
following :
(a) Tetrahydrozincate (II)
(b) Hexammine cobalt (III) sulphate
(c) Potassium tetracyanonickelate (II)
(d) Potassium tetrachloro palladate (II)
(e) Potassium tri (oxalato) chromate (III)
(f) Diammine dichloro platinum (II)
28. Calculate the EMF of the following cell at 298 K
Fe/Fe +2 (0.1 M) || Ag + (0.1 M) | Ag (s).
[Given E º = – 0.44 V, E º = + 0.80V]
+ +
Fe 2 / Fe
R = 8.31 J/K/mol, F = 96500 C.
Ag / Ag
29. In a reaction between A & B, the initial rate of
reaction was measured for different initial
concentration of A & B as given below :
A/M 0.20 0.20 0.40
B/M 0.20 0.10 0.05
r 0 /Ms –1 5.07 × 10 –5 5.07 × 10 –5 7.6 × 10 –5
What is the order of reaction with respect to A & B.
30. Draw the structure of all isomeric form of alcohol of
molecular formula C 5 H 12 O and give their IUPAC
names. Classify them as primary, secondary and
tertiary alcohols.
MATHEMATICS
Section A
⎧ 3⎫
1. Let f : R – ⎨ − ⎬ → R be a function defined as
⎩ 5 ⎭
f(x) =
2x
5x + 3
, find f–1 : Range of f → R –
⎧ 3⎫
⎨ − ⎬
⎩ 5 ⎭
2. Write the range of one branch of sin –1 x, other than
the Principal Branch.
⎛ cos x sin x
3. If A = ⎟ ⎞
π ⎜
, find x, 0 < x < when
⎝ − sin x cos x⎠
2
A + A´ = I
4. If B is a skew symmetric matrix, write whether the
matrix (ABA´) is symmetric or skew symmetric.
5. On expanding by first row, the value of a third order
determinant is a 11 A 11 + a 12 A 12 + a 13 A 13 . Write the
expression for its value on expanding by 2 nd column.
Where A ij is the cofactor of element a ij .
1+
cot x
6. Write a value of
∫
dx.
x + logsin x
π/
2
⎡3+
5cos x ⎤
7. Write the value of
∫
log⎢
⎥ dx
⎣ 3+
5sin x ⎦
0
8. Let → a and → b be two vectors such that | → a | = 3 and
| → 2 → →
b | = and a × b is a unit vector. Then what is
3
the angle between → a and → b
9. Write the value of
î .( ĵ × kˆ ) + ĵ.( kˆ × î ) + kˆ .( ĵ × î )
10. For two non zero vectors → a and → b write when
| → a + → b | = | → a | + | → b | holds.
Section B
11. Show that the relation R in the set
A = {x|x ∈W, 0 ≤ x ≤ 12|} given by
R = {(a, b) : (a – b) is a multiple of 4} is an
equivalence relation. Also find the set of all elements
related to 2.
OR
Let * be a binary operation defined on N × N, by
(a, b) * (c, d) = (a + c, b + d). Show that * is
commutative and associative. Also find the identity
element for * on N × N, if any.
12. Solve for x :
tan –1 ⎛ x −1
⎞
⎜ ⎟ + tan –1 ⎛ x + 1 ⎞
⎜ ⎟ =
⎝ x − 2 ⎠ ⎝ x + 2 ⎠
13. If a, b and c are real numbers and
b + c c + a a + b
c + a a + b b + c = 0
a + b b + c c + a
π , |x| < 1
4
Show that either a + b + c = 0 or a = b = c.
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⎡ x − 5
⎢ + a, if x < 5
| x − 5 |
⎢
14. If f(x) = ⎢ a + b, if x = 5
⎢ x − 5
⎢
+ b, if x < 5
⎣|
x − 5 |
is a continuous function. Find a, b.
15. If x y + y x dy
= log a, find . dx
16. Use lagrange's Mean Value theorem to determine a
point P on the curve y = x − 2 where the tangent is
parallel to the chord joining (2, 0) and (3, 1).
1
17. Evaluate :
∫
dx
cos(x − a) cos(x − b)
OR
2 + sin x x / 2
Evaluate :
∫
.e .dx
1 + cos x
18. If
→
a and
→
b are unit vectors and θ is the angle
between them, then prove that cos 2
θ = 2
1 |
→
a +
→
b |.
OR
If are the diagonals of a parallelogram with sides,
→ →
a and b in d the area of parallelogram in terms of
and hence find the area with d → 1 = i + 2 ĵ + 3 kˆ and
→
d = 3i – 2 ĵ + k.
2
19. Find the shortest distance between the lines, whose
equations are
x − 8 y 9 10 − z
= = &
3 −+
16 − 7
x −15
3
58 − 2y
= =
−16
z − 5
− 5
20. A bag contains 50 tickets numbered 1, 2, 3, ..... , 50
of which five are drawn at random and arranged in
ascending order of the number appearing on the
tickets (x 1 < x 2 < x 3 < x 4 < x 5 ). Find the probability
that x 3 = 30
21. Show that the differential equation
2ye x/y dx + (y – 2xe x/y )dy = 0 is
homogeneous and find its particular solution given
that x = 0 when y = 1.
OR
Find the particular solution of the differential
dx
equation + y cot x = 2x + x 2 cot x, x ≠ 0 given
dy
π
that y = 0, when x =
2
22. From the differential equation representing the family
of ellipses having foci on x-axis and centre at origin.
Section C
23. A letter is known to have come from either
TATANAGAR or CALCUTTA. On the envelope
just two consecutive letters TA are visible. What is
the probability that the letter has come from
(i) Tata Nagar (ii) Calcutta
OR
Find the probability distribution of the number of
white balls drawn in a random draw of 3 balls
without replacement from a bag containing 4 white
and 6 red balls. Also find the mean and variance of
the distribution.
24. Find the distance of the point (3, 4, 5) from the plane
x + y + z = 2 measured parallel to the line 2x = y = z.
25. Using integration, compute the area bounded by the
lines x + 2y = 2, y – x = 1 and 2x + y = 7
OR
Find the ratio of the areas into which curve y 2 = 6x
divides the region bounded by x 2 + y 2 = 16
−
tan
e
26. Evaluate :
∫
1 + x
1
x
2
( )
2
dx
27. A point the hypotenuse of a right triangle is at a
distance 'a' and 'b' from the sides of the triangle.
Show that the minimum length of the hypotenuse is
2 /3 2/3
[ b ] 3/ 2
a + .
28. Using elementary tranformations, find the inverse of
⎛ 1 3 − 2⎞
⎜ ⎟
the matrix ⎜−
3 0 − 5⎟
⎜ ⎟
⎝ 2 5 0 ⎠
29. A furniture firm manufactures chairs and tables, each
requiring the use of three machines A, B and C.
Production of one chair requires 2 hours on machine
A, 1 hour on machine B and 1 hour on machine C.
Each table requires 1 hour each on machine A and B
and 3 hours on machine C. The profit obtained by
selling one chair is Rs. 30 while by selling one table
the profit is Rs. 60. The total time available per week
on machine A is 70 hours, on machine B is 40 hours
and on machine C is 90 hours. How many chairs and
tables should be made per week so as to maximize
profit Formulate the problem as L.P.P. and solve it
graphically.
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MOCK TEST PAPER SOLUTION
FOR PAPER – 2 PUBLISHED IN JANUARY ISSUE
PHYSICS
2. For diamagnetic materials like Bi
3. Zero
12.27 12.27
4. For an electron, λ = Å = = 3.9 Å
V 10
5. If a thin foil is introduced parallel to plates than
capacity remains reflected.
6. Core is laminated to present eddy current losses
7. Photodiode
11.
A
B
B
A
Y
Y
A B Y
0 0 1
0 1 1
1 0 0
1 1 1
A B Y
0 0 0
0 1 0
1 0 0
1 1 0
13. No. of field lines emitted by a charge =
for a proton =
−19
1.6×
10
−12
8.854×
10
q
∈0
8. Maxwell’s fourth equation is Ampere's law and
→ →
dθ
E
according to it
∫
B.d l = µ 0i+ ∈0
µ 0
dt
14. The magnetic force on sides AB and BC due to
magnetic field of current carrying wires is equal and
opposite thus they balance each other while on AC it
is towards the wire. Thus the loop will move
towards the conductor
A
10. Let the charge on β particle be -e then charge on
deutron is +e and on α -particle is +2e
2e
a
–e
a e
2 2 2
− ke 2ke 2ke
P.E. = + −
a a a
P.E
P.E
= −
F
2
ke
a
− ke
=
2a
=
2a
2
− ke 2
2ke
+
2a
2
ke
W = P.E F − P.E i =
2a
a
2
2ke
−
2a
2
16.
i 1 i 2
C
15.
C
C
A
C 5V
B
2C 2C
+Q Q +Q Q
B = A B
5V
Charge remains same in series combination
Q 1
thus P.D. on 2 C = = (5V) = 2.5 volt
2C 2
P.D. across AB = 5 + 2.5 = 7.5 V
R
R
A
X
R
R
R
R
B
R
R
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2R 2R
=
A
X
B
2R 2R
=
R R
A X B
= V A
– V X
= IR ...(1)
V X
– V B
= IR = 10 V ...(2)
Equation (1) + (2)
V A
– V B
= 2IR = 20 V
17. V = IR
V 100
I = = = 10 Amp
R 10
V V 100 2
I = ⇒ Z = = = 10 2 Ω
Z I 10
R 10 1
cos θ = = =
Z 10 2 2
18. Einstein’s equation
hv = hv 0 + KE max
or hv = hv 0 + eV
v = frequency of incident light
v 0 = Threshold frequency,
V = Stopping potential
hv = 5 eV, hv 0 = 2 eV
5 = 2 + eV
St. p. = 3 eV
Stopping potential = – 3 volt
19. (i) t = 1 sec.
I ∝ w (width of slit)
t = 1 sec
y = x
I 1
= I 2
a 1 = a 2 = a [I ∝ a 2 ]
a = 2a,
max
I max = 4a 2
a min = 0, I min = a 2
(ii) t = 4 sec
2
I ⎛
max I1
I ⎞
⎜
+ 2
=
⎟
Imin
⎜ I1
I ⎟
⎝ − 2 ⎠
I 1 4
= = 9 : 1
I2
1
20. According to shell’s law
As light ray pass from rare to denser medium it bands
towards the normal and when passes from denser to
rarer bends away from normal
Thus µ 3 > µ 1
µ 2 > µ 3
Medium-2 is most dense
µ 2 > µ 3 > µ 1
21. (i) tan i p = a µ w = 4/3
i p = tan –1 (4/3)
(ii) θ = 90º
(iii) Reflected ray is plane polarized
1 1
(iv) sin θ c = =
a
µ tan i
w p
22.
23.
2Ω
7Ω
03.V 1Ω
As the diode is in forward bias total resistance
= 2 + 1 = 3Ω
ν 0.3
i = = = 0.1 amp
R 3
V 0 = i(7) = 0.7 volt
Input
.24 (i) Current Flowing in the circuit.
10V 3Ω
=
R L =2Ω
Output
10
i = = 2amp
5
For battery -1
V = E + Ir = 10 + 2 × 1 = 12 volt
For battery -2
V = E – Ir = 20 – 2 × 2 = 16 volt
(ii) Because battery - 1 is in charging state while
battery -2 is in discharging state
25. When current flows through metallic spring, current
is in same direction thus due to magnetic force
difference coils are attracted towards each other and
spring gets strinked
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26. Inconsistency in Ampere’s law
According to ampere’s law
→ →
∫
B.d
l = µ 0i
and the relation is valid only when the electric field at
the surface does not changes with time and this law
tells is that an electric current produces magnetic
field. If there exists an electric current as well as
changing electric field. The resultant magnetic field
is given by
→ → ∧
⎛ dθE
⎞
∫
B.d
l = µ 0 i+ µ 0 ∈0
⎜ ⎟
⎝ dt ⎠
Suppose that for a parallel plate capacitor.
Q
E = (electric field between plates)
∈ 0 A
Flux of the field through given area
Q Q
θ E = × A =
∈0
A ∈0
dθE
id
=∈ 0
dt
dQ dQ
= ∈ 0 =
dt dt
i d = i c
i d
= displacement current,
i c
= conduction current
27.
q
φ = net
∈0
for a charge placed at corner of cube,
q
φ =
8∈0
∴ thus for given system
(1 − 2 + 3 − 4 − 6 + 7 + 5 −8)
φ =
2 ∈0
φ =
−1
2∈0
CHEMISTRY
1. Co-ordination no. of Ca +2 = 8
F – = 4
2. [Cr(H 2 O) 5 SO 4 ]Br
Pentaaqua sulphato chromium (III) bromide.
3. (a) 4-nitro –1–methoxy benzene.
(b) 4-bromo-3, 3, 4-trimethyl hex-1-ene-2, 5-diol.
4. Aromatic ketones are less reactive, so they do not
react with NaHSO 3 .
5. Sulphanilic exist as zwitter ion, so they are
amphoteric in nature.
6. Ethylene glycol & Phthalic acid.
7. Substance which remove the excess acid and raise the
pH to appropriate level in stomach are Antacids. Eg
Lansoprazole.
8. Carbohydrate having chiral carbon, so they optically
active.
9. Given r + = 95 pm
r – = 181 pm.
r+
95
= = 0.524
r−
181
Since, r + /r – lies between 0.414 to 0.732,
∴ A X has FCC (NaCl type structure) structure
so, the co-ordination number of each ion = 6.
10. In MgO, co-ordination number of Mg +2 is 6 and that
of O 2– is also 6 due to NaCl structure.
In TlCl, co-ordination number Tl + is 8 and that of Cl –
is also 8 due to CsCl type structure.
28.
L'
B
E
C
V BB
1
ƒ =
2π
LC
K
V ∞
11. Mole fraction of solute is defined as ratio of number
of moles of solute to the total number of moles of
solute and solvent
n B WB
/ M B
x B = =
n A + n W
B A WB
+
M M
x A =
n A
n + n
A
B
=
W
W
M
A
A
A
A
/ M A
W
+
M
or x A + x B = 1
Where x A = mole fraction of solvent
x B = mole fraction of solute
B
B
B
XtraEdge for IIT-JEE 78
FEBRUARY 2010

XtraEdge for IIT-JEE 79
FEBRUARY 2010

12. With increasing voltage, the sequence of deposition
of metal is
Ag
+ 2
> Hg + 2
2 > Cu
+ + 2
> Mg
( + 0.80)
Mg +2
( + 0.79)
( + 0.34)
( −2.37)
will not be reduced because its reduction
potential value is much lower than water (–0.834).
13. In Cr 2 O –2 7 , all the six normal Cr–O bonds are
equivalent and two bridged Cr– O bonds are
equivalent.
O O
– O
Cr
O
O
Cr
O O–
14. CuSO 4 + 2KCN → Cu(CN) 2 + K 2 SO 4
2Cu(CN) 2 → Cu 2 (CN) 2 + C 2 N 2 (cyanogen)
Cu 2 (CN) 2 + 6 KCN → 2K 3 [Cu(CN) 4 ]
15. Two pairs.
CH 3
*
CH 3 – CH – CH – CH 3
CH 3 – CH – CH 2 – CH 3
Br 2
CH 2 Br – CH – CH 2 .CH 3
CH 3
*
CH 3
Br
16. Fe(s) → Fe +2 (aq) + 2e – (oxidation)
O 2 (g) + 4H + + 4e – → 2H 2 O (Reduction)
Atmospheric oxidation occurs as following
17. (a)
2Fe +2 (aq) + 2H 2 O(l)+ 2
1
O2 (g)
OH
Br 2 /CS 2
0ºC
⎯→ Fe 2 O 3 (s) + 4H + (aq)
OH
Br
+
OH
Br
Mono halogen derivative form.
(b) C 2 H 5 OC 2 H 5 + 2HI ⎯ Heat ⎯⎯ →2C 2 H 5 I + H 2 O
18. Fehling, Tollen's, Schiff's reagent react only with
aldehyde.
Grignard reagent react with both aldehyde & ketones.
O
R – C – H + R´MgBr
OMgBr
R – C – H
R´
OH
R – C – H
R´
O
R – C – R + R´MgBr
OMgBr
R – C – R
R´
OH
R – C – R
19. Iodoform test is given by only those compounds
which having – C – CH 3
or – CH – CH 3
group.
O
OH
Therefore, pentanone-2 give iodoform test
CH 3 – CH 2 – CH 2 – C – CH 3
20. From the Question,
I → Is most basic due to lone pair present in sp 3
hybridised orbital to available for donation.
II → lone pair present in sp 2 orbital but disperse to
small extent.
III → lone pair present is sp 3 orbital but possess
electronegative atom.
IV → lone pair e – present on 'N´ is in p-orbital to
form a part of aromatic sextet. (least basicity) so,
order of basicity = I > III > II > IV
21. HX H + + X –
n = 2
1− α + nα
from the formula i =
1
1 − 0.2 + 2×
0.2
=
= 1.2
1
from the formula
∆T f = k f . m. i
= 1.2 × 1.8 × 0.2 = 0.432
so, the freezing point of solution
= 0 – 0.432 = – 0.432 ºC
22. We know that
Surface volume = area × thick ness
0.005
= 80 × cm 3
10
= 0.04 cm 3
Mass of silver deposited = volume × density
= 0.04 × 10.5 = 0.042 g
The cell reaction is
Ag + + e – ⎯→ Ag
w Q i.t
since = =
E F F
0.042 3× t
∴ =
108 96500
or t = 125.1 sec
O
R´
XtraEdge for IIT-JEE 80
FEBRUARY 2010

23. From the below graph we can say that with increase
of temperature, then occurs a decrease in rate of
physisorption.
x/m
t
Where x/m = Mass of gas adsorbed per unit mass of
adsorbent.
t = Temperature
24. In H 3 BO 3 , 'B´ atom having 6e – (e-deficient) it is a
Lewis acid with one vacant p-orbital and no d-orbital
thus, it can accomodate only one e – pair in its outer
most shell.
OH
OH
H 2 O: + B – OH H 2 O ⎯→ B – OH
OH
OH
[B(OH) 4 ] – + H +
25. (a) The colour of transition metal compound is
obtained due to unpaired e – which gives
d-d-transition.
VOCl 2 & CuCl 2 → V +4 & Cu +2
both having one unpaired e – , so gives same colour in
aqueous medium.
(b) [CuSO 4 + 2KCN → Cu(CN) 2 + K 2 SO 4 ] × 2
2Cu(CN) 2 → Cu 2 (CN) 2 + (CN) 2
Cu 2 (CN) 2 + 6KCN → 2K 3 Cu(CN) 4
2CuSO 4 + 10KCN → 2K 3 Cu(CN) 4
+ 2K 2 SO 4 + (CN) 2
26.
CHO
CHO
COO –
CHO
CHO
CH 2 OH
CH 2 OH COO
–
COOH CH 2OH
CH 2 OH COOH
OH – /100ºC
Intra molecular
Cannizaro reaction
H + /H 2O
Product
27. (i) In aspartame following four functional groups are
present.
(a) (–NH 2 ) Amine (b) –COOH (Carboxylic acid)
O
O
(c) (– C – NH –)
(Amide)
(ii) Its zwitter ion is
(iii)
(d) (– C – O –)
(Ester)
CH 2 – C 6 H 5
+
H 3 N – CH – CONH – CH – COOCH 3
CH 2 – COO –
H 2 N – CH – C – NH – CH – COOCH 3
CH 2 COOH
H 2 N – CH – COOH
CH 2 COOH
(a)
O
CH 2 C 6 H 5
+
Hydrolysis
CH 2 – C 6 H 5
H 2 N – CH – COOH
28. (i) From the equation
k = A.e –Ea/RT
E
or log k = log A – a
2.303RT
Comparing this equation with the given equation
E a = 1.25 × 10
4
2.303R
or E a = 1.25 × 10 4 × 2.303 × 8.314
= 2.39 × 10 5 J/mol
= 239 kJ/mol
(ii) In the question, the unit of rate constant is s –1 ,
therefore, the reaction is first order.
0.693
∴ t 1/2 =
k
0.693 0.693
or k = =
t 1/ 2 256×
60
= 4.51 × 10 –5 s –1
substituting this value in the given expression, we get
4
log(4.51 × 10 –5 1.25×
10
) = 14.34 –
T
1.25×
10
or –4.346 = 14.34 –
T
4
(b)
XtraEdge for IIT-JEE 81
FEBRUARY 2010

⇒
1.25×
10
T
4
1.25×
10
T =
18.686
29. CaO + H 2 O →
= 14.34 + 4.346 = 18.686
4
= 669 K
Ca (OH) 2
(A)
NH 3 + CO 2 + H 2 O →
NH
4 (HCO3)
(B)
NH 4 (HCO 3 ) + NaCl → Na(HCO 3 ) + NH 4Cl
(D)
2NaHCO 3 ⎯⎯→
∆ Na 2 CO 3 + H 2O + CO 2
(C)
Ca(OH) 2 + 2NH 4 Cl → CaCl + 2NH 3 + 2H 2 O
2
(E)
30. Ozonolyis of 'A' to acetone and aldehyde indicated
the presence of the following structure in the
molecule 'A' (alkene).
H 3 C
H 3 C
C = CHR O 3
H 3 C
H 3 C
C = O + RCHO
(aldehyde)
(ketone)
RCHO
⎯ [ O]
(B)
⎯→ RCOOH
Br 2 / P
⎯⎯⎯→
Bromo compound ⎯
2
⎯⎯ →Hydroxy acid
(C)
Hydroxy acid can be determined by following
reaction
H 3 C
H 3 C
C = O HCN
H 3 C
H 3 C
C
OH
CN
H 2 O/H +
H 3 C
H 3 C
(D)
from the above, bromo compound 'C' is –
H 3 C Br
C
H 3 C COOH
C
OH
COOH
(C)
'C' is formed by bromination of (B) so 'B' is
H 3 C
C
H
H 3 C COOH
compound 'B' is formed by oxidation of an aldehyde,
so the structure of the aldehyde is
H 3 C
C
H
H 3 C CHO
The aldehyde and acetone are formed by the
ozonolysis of alkene 'A'. So, the structure of alkene
H 3 C
H 3 C
H 3 C
H 3 C
C = C
H
(A)
C
H 3 C
H 3 C
H
CH 3
CH 3
ozonolysis
H 3 C
H 3 C
C
H
CHO
+ O = C
(Aldehyde) (Ketone)
H H
[O] 3 C
C
H
Br 2 /P
CHO H 3 C COOH
(B)
C
(D)
OH
Hydrolysis
H 3 C
COOH
H 3 C
MATHEMATICS
Section A
1. (a)
∴ for every value of x there is unique y
2. π
3. 135
4. 3
5. (1, 2)
6. π/6
7. (1, –7, 2) or their any multiple
8.
x 8
8
+ c
11
9. 3 î + ĵ+
5kˆ
3
10. order of AB is 2 × 2
order of BA is 3 × 3
2x
− 1
11. f(x) = , x ∈R
3
To show f is one-one
Section B
C
(C)
Br
CH 3
CH 3
COOH
XtraEdge for IIT-JEE 82
FEBRUARY 2010

Let x 1 , x 2 ∈ R s.t. x 1 ≠ x 2
⇒ 2x 1 ≠ 2x 2
⇒ 2x 1 – 1 ≠ 2x 2 – 1
⇒
2x
1 −1
2x
1 ≠ 2 −
3 3
⇒ f(x 1 ) ≠ f(x 2 )
⇒ f is one-one
To show f is onto
Let
2x
− 1
y = , y ∈R(codomain of f)
3
or 3y = 2x – 1
or
3 y + 1
x = ∈ R
2
∴ for all y ∈ R (codomain of f), there exist
3 y + 1
x = ∈ R (codomain of f), such that
2
⎛ 3y + 1⎞
2⎜
⎟ −1
⎛ 3y + 1⎞
2
f(x) = f ⎜ ⎟ =
⎝ ⎠
= y
⎝ 2 ⎠ 3
⇒ every element in codomain of f has its pre-image
in the domain of f,
⇒ f is onto.
To find f –1
Let f(x) = y,
3 y + 1
x =
2
⇒ f –1 (y) = x ⇒ f –1 3 y + 1
(y) =
2
∴ f –1 : R → R given by f –1 3 y + 1
(y) =
2
OR
a + b
(i) a*b = , a, b ∈ N
2
a + b
∀ a, b ∈ N
2
may or may not belong to N.
∴ a*b is not always natural no.
∴ '*' is not a binary operation on N
a + b
(ii) a*b = , a, b ∈ Q
2
∀ a, b ∈ Q;
a + b
2
∈ Q
⇒ a*b ∈ Q
⇒ '*' is a binary operation on Q
a + b
(iii) For a*b = , a, b ∈ Q
2
a + b b + a
a*b = =
2 2
= b*a
⇒ * is commutative
⎛ a + b ⎞
(iv) (a*b)*c = ⎜ ⎟ *C
⎝ 2 ⎠
∀ a, b, c, ∈ Q
a + b
+ c
= 2 a + b + 2c
=
2 4
⎛ b + c ⎞
a*(b*c) = a * ⎜ ⎟
⎝ 2 ⎠
⎛ b + c ⎞
a + ⎜ ⎟
2
=
⎝ ⎠ 2 a + b + c
=
2
4
(a * b) * c ≠ a * (b * c) ∀ a, b, c, ∈ Q
∴ '*' is not associative,
12. Let sin –1 ⎛ 5 ⎞
⎜ ⎟ = x & cos –1 ⎛ 3 ⎞
⎜ ⎟ = y
⎝13
⎠ ⎝ 5 ⎠
⇒ sin x = 13
5
12
& cos x =
13
⇒ tan x = 13
5
tan(x + y) =
63
tan(x + y) =
16
⇒ x + y = tan –1 ⎛ 63 ⎞
⎜ ⎟
⎝ 16 ⎠
& cos y = 5
3
& sin y =
5
4
& tan y = 3
4
tan x + tan y
1−
tan x tan y
⇒ sin –1 ⎛12
⎞
⎜ ⎟ + cos –1 ⎛ 3 ⎞
⎜ ⎟ = tan
–1 ⎛ 63 ⎞
⎜ ⎟
⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 16 ⎠
⎡2
− 6⎤
13. Let = A = ⎢ ⎥
⎣1
− 2 ⎦
A = IA
⎡2
− 6⎤
⎡1
0⎤
⎢ ⎥ =
⎣1
− 2
⎢ ⎥ A
⎦ ⎣0
1 ⎦
R 1 ⇔ R 2
⎡1
− 2⎤
⎡0
1⎤
⇒ ⎢ ⎥ =
⎣2
− 6
⎢ ⎥ A
⎦ ⎣1
0 ⎦
R 2 → R 2 – 2R 1
⎡1
− 2⎤
⎡0
1 ⎤
⇒ ⎢ ⎥ =
⎣0
− 2
⎢ ⎥ A
⎦ ⎣1
− 2 ⎦
R 1 → R 1 – R 2
⎡1
0 ⎤ ⎡−1
3 ⎤
⇒ ⎢ ⎥ =
⎣0
− 2
⎢ ⎥ A
⎦ ⎣ 1 − 2 ⎦
R 2 → – 2
1
R2
XtraEdge for IIT-JEE 83
FEBRUARY 2010

⇒
⎡1
⎢
⎣0
0⎤
⎡ −1
1
⎥ = ⎢ 1
⎦ ⎢−
⎣ 2
3⎤
⎥ A
1⎥
⎦
⎡ −1
3⎤
∴ A –1 = ⎢ 1 ⎥
⎢−
1
⎣ 2 ⎥
⎦
OR
Operate R 1 → aR 1 , R 2 → bR 2 , R 3 → cR 3
=
a
a
1
abc
− bc
2
2
+ ac
+ ab
b
b
− abc
2
2
+ bc
− ac
2
a b + abc
2
a c + abc
+ ab
ab
c
c
2
2
− abc
2
2
+ bc
+ ac
− ab
+ abc
b c + abc
ac
bc
2
2
+ abc
+ bac
− abc
Take, a, b, c common from C 1 , C 2 , C 3 respectively
=
− bc
ab + bc
ac + bc
R 1 → R 1 + R 2 + R 3
=
ab + bc + ac
ab + bc
ac + bc
ab + ac
− ac
bc + ac
ac + ab
bc + ba
− ac
− ab
ab + bc + ac
bc + ac
1
= (ab + bc + ca) ab + bc
ac + bc
C 1 → C 1 – C 3 , C 2 → C 2 – C 3
= (ab + bc + ca)
0
0
ac + bc + ab
On expanding by R 1 we get
= (ab + bc + ca) 3
1
− ac
bc + ac
ab + bc + ac
0
bc + ba
− ab
− (ab + bc + ca)
bc + ac + ab
1
bc + ba
− ab
1
bc + ba
− ab
14. Being a polynomial function f(x) is continuous at all
point for x < 1, 1 < x < 2 and x ≥ 2. Thus the possible
points of discontinuity are x = 1 and x = 2
To check continuity at x = 1
lim f (x) = lim x + 2 = 3⎫
−
x→1
x→1
⎪
lim f (x) = lim x − 2 = −1⎪
x→ 1+
x→1
f (1) = 3⎬
since, lim f (x) = f (1) ≠ lim f (x)
⎪
x→1
−
x→1
+ ⎪
∴ f (x) is not continuous at x = 1⎪
⎭
To check continuity at x = 2
lim f (x) = lim x − 2 = 0
−
x→2
−
x→2
lim f (x) =
x→2
+
since lim f (x)
x→2
−
lim x – 2 = 0
x→2
f(2) = 0
lim f (x) = f(2) = 0
→ 2
x +
∴ f(x) is continuous at x = 2
∴ The only point of discontinuity is x = 1
15. x p y p = (x + y) p + q
Take log on both sides
p log x + q log y = (p + q) log (x + y)
p q dy p + q ⎞
+ . = ⎜
⎛ dy
1 + ⎟⎠
x y dx x + y ⎝ dx
p p + q dy ⎛ p + q q ⎞
or – =
x x + y dx
⎜ −
⎟
⎝ x + y y ⎠
px + py − px − qx dy ⎛ py + qy − qx − qy
or
=
x(x + y)
⎟ ⎞
⎜
dx ⎝ y(x + y) ⎠
py − qx dy ⎛ py − qx
or =
x
⎟ ⎞
⎜
dx ⎝ y ⎠
y dy
or =
x dx
OR
⎡ 2
2
y = tan –1 1+
x + 1−
x
⎥ ⎥ ⎤
⎢
⎢ 2
2
⎣ 1+
x + 1−
x ⎦
Put x 2 = cos θ
⎡
y = tan –1 1+
cosθ + 1−
cosθ
⎥ ⎥ ⎤
⎢
⎢⎣
1+
cosθ
− 1−
cosθ
⎦
⎡ θ θ ⎤ ⎡ θ ⎤
⎢cos
+ sin
= tan –1 2 2 ⎢1+
tan
⎢
= tan –1 2
⎢
θ θ
cos − sin
⎣ 2 2 ⎥ ⎥⎥⎥ ⎢
⎢
θ
1−
tan
⎦
⎣ 2 ⎥ ⎥⎥⎥ ⎦
= tan –1 ⎡ ⎛ π θ ⎞⎤
⎢tan
⎜ + ⎟⎥ ⎣ ⎝ 4 2 ⎠ ⎦
y = 4
π + 2
θ
dy = –
dx
2
1 ⎛
⎜
2 ⎜
⎝
2x
1− x
4
(x + 1)(x + 4)
16.
∫
dx
2 2
(x + 3)(x − 5)
Consider
(x
(x
2
2
+ 1)(x
2
+ 3) (x
2
2
+ 4)
=
− 5)
or y =
⎟ ⎟ ⎞
=
⎠
π 1 + cos –1 (x 2 )
4 2
−x
1−
x
(t + 1)(t + 4)
where t = x 2
(t + 3)(t − 5)
4
XtraEdge for IIT-JEE 84
FEBRUARY 2010

17.
−
7t + 19
2
= 1 +
2
(t + 3)(t − 5)
18. For
∫
( 3x 1)dx
π
∴ 5 = h
2 dh
⎛ 2a
a
⎞
⎜
4 dt
Q
⎟
⎜ ∫f (x)dx = 2∫f (x)dx if f (2a − x) = f (x)
⎟
dh 20
⎝ 0
0
⎠
or = when h = 10 m
2
π/
2
dt π(10)
=
∫
log sin t dt
dh 1
or = m/minute
0
dt 5π
1
Consider
1
7t + 19 A B
a = 1, b = 2, h =
= +
n
(t + 3)(t − 5) t + 3 t − 5
as n → ∞, h → 0
1 27
A = , B =
4 4
b
f(x) = 3x 2 – 1
2 2
(x + 1)(x + 4)
∴
∫
dx
∫
f (x) dx = h[f(a) + f(a + h) + ... + f(a + (n – 1)h)]
h→0
2 2
a
(x + 3)(x − 5)
2
=
∫ dx + 1 dx 27 dx
2
∫
+
4
2
x + 3 4 ∫ 2
∫
( 3x −1) dx = h[3n + 3h 2 (1 2 + 2 2 + .. + (n – 1) 2 )
h→0
x − 5
1
1
= x + tan
–1⎛
x ⎞
x − 5
+ 6h(1 + 2 + .. + (n – 1) – n)]
⎜ ⎟
+ log + c
4 3 ⎝ 3 ⎠ 827
5 x + 5
⎡ 2 (n)(n −1)(2n
−1)
⎤
= lim h ⎢2n
+ 3h
+ 3h(n)(n −1)
h→0
⎥
⎣
6
⎦
⎡
2
2 3h (1 − h)(2 − h) ⎛ 1 ⎞⎛1−
h ⎞⎤
= lim h ⎢ +
+ 3h⎜
⎟⎜
⎟⎥
h→0
⎢⎣
h
3
6h
⎝ h ⎠⎝
h ⎠⎥⎦
r
1
= 2 + (2) + 3 = 6
2
h
α
π/
2
Let r = radius of cone formed by water at any time
19. Given I =
∫
log sin x dx
0
h = height of cone formed by water at any time
π/
2
⎛
a
a
⎞
Given α = tan –1 ⎛ 1 ⎞
I =
⎜ ⎟ ∫
log cos x dx ⎜Q
⎟
⎜ ∫
f (a − x) dx =
∫
f (x)dx
⎟
⎝ 2 ⎠ 0
⎝ 0
0 ⎠
π/
2
1
∴ tan α = ∴ 2I = 2 ∫
(log sin 2x dx − log 2)dx
0
r
Also tan α = ⇒ h = 2r
π/
2
h 1
π
I =
Volume of this cone
∫
log sin 2x dx − log 2
2
4
0
....(1)
2
1
v = πr 2 π ⎛ h ⎞
Consider
h = ⎜ ⎟ h
3 3 ⎝ 2 ⎠
π/
2
π
1
I
π 1 =
v = h
3
∫
log sin 2x dx =
∫
logsin t dt
2
0
0
12
dt π
dv π
⇒ = (3h 2 dh π
) = h
2 dh
[Put 2x = t, dx = ; x = 0 ⇒ t = 0; x = ⇒ t = π]
2
2
dt 12 dt 4 dt
π/
2
dv 1
But = 5 m 3 /minute
= .2
∫
logsin t dt
dt
2
0
XtraEdge for IIT-JEE 85
FEBRUARY 2010

π/
2
21.
I 1 =
∫
log sin x dx
...(2)
E D
∴ Distance between two parallel lines.
∴ By Baye's Theorem
P(A
→ → →
1 /E) =
P(A1)P(E / A1)
⎛ ⎞
P(A1)P(E / A1)
+ P(A 2 )P(E / A 2 ) + P(A 3)P(E / A3
)
b×
⎜a
2 − a1
⎟
⎝ ⎠ 974 3 1
=
= units ×
→
45
b
= 10 4 9
=
3 1 13
+ + 0
40 30
0
From (1) and (2)
π / 2
⎫
1
π ⎪
F
C
I =
∫
log sin x dx − log 2
2
4
⎪
0
⎪
→
1 π
b
I − I = – log 2 ⎬
2 4
⎪
A →
π
⎪
a
B
I = − log 2
⎪
2
⎪⎭
→
From fig. DE = – → a ; EF → = – → b
20. Given line
→ → →
AC = AB + BC = → a + → b
→ → →
x − 1 3 − y z +1
= =
AD = 2 BC = 2 b
5 2 4
→ → →
AD = AC + CD
x − 1 y − z − ( −1)
or, = =
....(i)
→ → →
5 − 23
4
⇒ CD = AD – AC = → b – → a
→ →
is passing through (1, 3, –1) and has D.R. 5, –2, 4.
FA = – CD = → a – → b
Equations of line passing through (3, 0, –4) and
→ →
parallel to given line is
CE = CD + DE → = → b – 2 → a
→ → →
x − 3 y − 0 z + 4
= =
...(ii)
AE = AD + DE = 2 → b – → a
5 − 2 4
Vector equations of line (i) & (ii)
→
r= î + 3 ĵ – kˆ + λ(5 î – 2 ĵ + 4 kˆ )
→
r = 3 î – 4 ĵ + µ (5 î – 2 ĵ + 4 kˆ )
1
22. P(Correct forecast) =
3
2
P(Incorrect forecast) = 3
∴
→
a 2 – a → P (At least three correct forecast for four matches)
1 = 2 î – 3 ĵ – 3 kˆ
= P(3 correct) + P(4 correct)
→
b =
2 2 2
( 5) + ( −2)
+ (4)
3 1
4
= 4 ⎛ 1 ⎞ ⎛ 2 ⎞
C 3 ⎜ ⎟⎠ ⎜ ⎟⎠ + 4 ⎛ 1 ⎞ 8
C 4 ⎜ ⎟⎠ =
⎝ 3 ⎝ 3 ⎝ 3 81
1 1 + = 81 9
= 45 = 3 5
OR
Let E : Candidate Reaches late
Also
î ĵ kˆ
A 1 = Candidate travels by bus
→ ⎛ ⎞
b × ⎜a − →
A 2 : Candidate travels by scooter
2 a1
⎟ = 5 − 2 4
⎝ ⎠
A 3 : Candidate travels by other modes of
2 − 3 − 3
transport
= 18 î + 23 ĵ – 11 kˆ
3 1 3
P(A 1 ) = , P(A2 ) = , P(A3 ) = 10 10 5
∴
→
⎛
→ →
⎞
2 2 2
b × ⎜a 2 − a 1 ⎟ = ( 18) + (23) + (11) =
⎝ ⎠
974
1 1
P(E/A 1 ) = , P(E/A2 ) = , P(E/A3 ) = 0
4 3
XtraEdge for IIT-JEE 86
FEBRUARY 2010

Section C
23. Given
⎡2
1⎤
⎡−
3 2 ⎤ ⎡1
2 ⎤
⎢ ⎥ P
⎣3
2
⎢ ⎥ =
⎦ ⎣ 5 − 3
⎢ ⎥ ⎦ ⎣2
−1 ⎦
Let
⎡2
1⎤
R = ⎢ ⎥ then |R| = 1
⎣3
2 ⎦
⎡−
3 2 ⎤
S = ⎢ ⎥ then |S| = –1
⎣ 5 − 3 ⎦
⎡1
2 ⎤
Q = ⎢ ⎥
⎣2
−1 ⎦
Since R and S are non-singular matrices
∴ R –1 and S –1 exist
R –1 AdjR ⎡2
−1⎤
= = ⎢ ⎥
| R | ⎣3
2 ⎦
S –1 AdjS
= =
| S |
Now given
RPS = Q
∴
⎡3
⎢
⎣5
2⎤
3
⎥ ⎦
⎫
−1
−1
R (RPS) = R Q
⎪
⎪
−1
−1
(R R)PS = R Q⎪
−1
⎬ (Q R –1 R = I I.P = P)
PS = R Q ⎪
−1
−1
−1
PSS = R QS ⎪
−1
−1
⎪
P = R QS
⎭
⎡ 2 −1⎤⎡1
2 ⎤⎡3
2⎤⎫
P = ⎢ ⎥⎢
⎥⎢
⎥⎪
⎣−
3 2 ⎦⎣2
−1⎦⎣5
3⎦
⎬
⎡ 25 15 ⎤
=
⎪
⎢ ⎥
⎣−
37 − 22⎦
⎪
⎭
24. f(x) = sin 2x – x – 2
π < x < 2
π
f´(x) = 2 cos 2x – 1
1
f´(x) = 0 ⇒ cos 2x =
2
or 2x = 3
π , 3
π
or x = –
6
π , 6
π
f´´(x) = –4 sin 2x
π
f´´(x) = 2 3 > 0 at x = –
6
⇒ x = –
6
π is point of local minima
π
f´´(x) = 2 3 < 0 at x = 6
⇒ x = 6
π is point of local maxima
⎛ π ⎞ − 3 π
f⎜
− ⎟ = +
⎝ 6 ⎠ 2 6
Local maximum value is
⎛ π ⎞ 3 π
f⎜
⎟ = –
⎝ 6 ⎠ 2 6
OR
Let h = length of cylinder
r = radius of semi-circular ends of cylinder
v = 2
1 πr 2 h
S = Total surface area of half circular cylinder
= 2(Area of semi circular ends)
+ Curved surface area of half
circular cylinder + Area of rectangular base.
⎛ 1 ⎞
= ⎜ π 2 1
2 r ⎟ + (2πrh) + 2rh
⎝ 2 ⎠ 2
= πr 2 + (π + 2)rh
= πr 2 2v
+ (π + 2)r.
2
πr
ds 2v( π 2)
⎛ 1
= 2πr – ⎜
dr
π+
2
⎝ r
ds = 0 ⇒ r 3 =
dr
2
( π + 2)v
2
π
⎞
⎟
⎠
d S 2v( π + 2)
2
= 2π + .
2
dr
π
3 > 0
r
∴ S is minimum when
r 3 =
( π + 2)v ( π + 2) ⎛ 1 2 ⎞
= ⎜ πr
h⎟ 2
2
π π ⎝ 2 ⎠
π + 2 h π
⇒ r = .h ∴ =
2π
2r
π + 2
Which is required result.
⎧−
25. f(x) ⎨
⎩
(x − 2) + 2
x
2
− 2,
x ≤ 2
x > 2
⎧ 4 − x x ≤ 2
or f(x) = ⎨ 2
⎩x
− 2, x > 2
To sketch the graph of above function following
tables are required.
For f(x) = 4 – x, x ≤ 2 & for f(x) = x 2 – 2, x ≥ 2
x –1 0 1 2
y 5 4 3 2
Also f(x) = x 2 – 2 represent parabolic curve.
x 2 3 4 5 6
y 2 7 14 23 34
∴ Local minimum value is
XtraEdge for IIT-JEE 87
FEBRUARY 2010

y
C
y = 4 – x y = x 2 – 2
A
B
4
Area =
∫
0
O
D
x = 0 2 x = 4
= 4x –
= 6 +
2
f (x)dx =
∫
( 4 − x)dx +
∫
( x − 2)dx
x
2
2
2
0
0
3
x
+ − 2x
3
44 62 = sq. units
3 3
4
2
4
On the graph
∫
f (x) dx represents the area bounded
0
by x-axis the lines x = 0; x = 4 and the curve y = f(x).
i.e. area of shaded region shown in fig.
26. (1 – x 2 dy
) – xy = x
2
dx
or
dy x – .y =
dx
2
1−
x
x
P = –
1−
x
2
x
2
1−
x
2
2
x
, Q =
1−
x
−x
∫
dx
−
I.F. = e ∫ Pdx
= 1 x
2
e = e
2
∴ Solution of diff. equation is
2
y 1− x =
∫
. 1−
x
2
1−
x
2
2
2
x 2
4
2
2
1 log(1−
x
2 )
dx
⎛
=
∫ ⎟ ⎟ ⎞
⎜ 1
2
− 1−
x dx
⎜ 2
⎝ 1−
x ⎠
=
2
1− x
= sin –1 ⎛ 2 1 ⎞
x – ⎜ − + sin − 1
x 1 x x⎟ + c
⎝ 2 ⎠
1
y 1− x = sin –1 x – x 2
When x = 0,
∴ Solution is
2
y = 2 ⇒ 2 = c
1
y 1− x = sin –1 x – x 2
27. The given line is
x − 6 y − 7
= =
3 2
2
1− x + c
2
1− x + 2
z − 7
= λ (say) ...(i)
− 2
Let N be the foot of the perpendicular from P(1, 2, 3)
to the given line
P(1, 2, 3)
A N B
Coordinates of N = (3λ + 6, 2λ + 7, –2λ + 7)
D.R. of NP 3λ + 5, 2λ + 5, – 2λ + 4
D.R. of AB 3, 2, –2
Since NP ⊥ AB
∴ 3(3λ + 5) + 2(2λ + 5) – 2(–2λ + 4) = 0
or λ = –1
∴ Coordinates of foot of perpendicular N are (3, 5, 9)
Equation of plane containing line (i) and point (1, 2, 3)
is
Equation of plane containing point (6, 7, 7) & (1, 2, 3)
and parallel to line with D.R. 3, 2, –2 is
x − 6 y − 7 z − 7
− 5 − 5 − 4 = 0
3 2 − 2
or, 18x – 22y + 5z + 11 = 0
28. Given
x
P(x)
0 0
1 k
2 4k
3 2k
4 k
Σp i = 8k
1
But Σp i = 1 ⇒ k = 8
∴ Probability distribution is
2
x
⎫
i pi
pix
i pix
i
⎪
0 0 0 0 ⎪
1 1 1
1
⎪
8 8 8 ⎪
1
⎪
2 1 2 ⎬
2
⎪
1 3 9
3
⎪
4 4 4 ⎪
1 1 ⎪
4
2 ⎪
8 2 ⎭
Probability of getting admission in two colleges =
2
1
Mean = µ = Σp i x i =
19
8
XtraEdge for IIT-JEE 88
FEBRUARY 2010

29.
2
Variance = σ 2 = Σp i x 2 i – µ 2 51 ⎛19 ⎞ 47
= – ⎜ ⎟⎠ =
8 ⎝ 8 64
OR
W
4 W
W
2 W
3 B
B B 2 B
W
1W 1B
A
R
Three cases arise, when 2 balls from bag A are
shifted to bag B.
Case 1 : If 2 white balls are transferred from bag A.
4 2 2
P(W A W A) = . =
7 6 7
Case 2 : If 2 black balls are transferred from bag A
3 2 1
P(B A B A ) = . = 7 6 7
Case 3 : If 1 white and 1 black ball is transferred
from bag A
⎛ 4 3 ⎞ 4
P(W A B A ) = 2 ⎜ . ⎟ =
⎝ 7 6 ⎠ 7
(a) Probability of drawing 2 white balls from bag B
= P(W A W A ).P(W B W B ) + P(B A B A ).P(W B W B )
+ P(W A B A ).P(W B .W B )
2 ⎛ 2 3 ⎞ 1 ⎛ 2 1 ⎞ 4 ⎛ 3 2 ⎞ 5
= ⎜ . ⎟ + ⎜ . ⎟ + ⎜ . ⎟ =
7 ⎝ 6 5 ⎠ 7 ⎝ 6 5 ⎠ 7 ⎝ 6 5 ⎠ 21
(b) Probability of drawing 2 black balls from B
2 ⎛ 2 1 ⎞ 1 ⎛ 4 3 ⎞ 4 ⎛ 3 2 ⎞ 4
= ⎜ . ⎟ + ⎜ . ⎟ + ⎜ . ⎟ =
7 ⎝ 6 5 ⎠ 7 ⎝ 6 4 ⎠ 7 ⎝ 6 5 ⎠ 21
(c) Probability of drawing 1 white and 1 black ball
from bag B
2 ⎛ 4 2.2 ⎞ 1 ⎛ 2.2 4 ⎞ 4 ⎛ 2.3 3 ⎞ 4
= ⎜ . ⎟ + ⎜ . ⎟ + ⎜ . ⎟ =
7 ⎝ 6 5 ⎠ 7 ⎝ 6 5 ⎠ 7 ⎝ 6 5 ⎠ 7
P
x
40 – x
A
Q
B
y
40 – y
60 –x – y
R
50–(60–x – y)
Let x no. of packets from kitchen A are transported to
P and y of packets from kitchen A to Q. Then only
60 – x – y packets can be transported to R from A.
Similarly from B, 40 – x packets can be transported
to P and 40 – y to Q. Remaining requirement of R i.e.
50 –(60 – x – y) can be transported from B to Q.
x´
∴ Constraints are
40 − x ≥ 0 ⎫
40 − y ≥ 0
⎪
⎪
60 − x − y ≥ 0 ⎬
−10
+ x + y ≥ 0⎪
⎪
x ≥ 0, y ≥ 0 ⎪⎭
Objective function is :
Minimise. z = 5x + 4y + 3(60 – x – y)
+ 4(40 – x) + 2(40 – y) + 5(x + y – 10)
∴ L.P.P. is
To Minimise. z = 3x + 4y + 370
subject to constraints
x+y = 10
x ≤ 40 ⎫
y ≤ 40
⎪
⎪
x + y ≤ 60 ⎬
x + y ≥10
⎪
⎪
x ≥ 0, y ≥ 0⎪⎭
y
B
O
(0, 40)
A(0, 10)
y´
C
x + y = 60
(10, 0)
D(40, 20)
F E(40, 0)
Feasible Region is ABCDEFA with corner points
A(0, 10) z = 3(0) + 4(10) + 370 = 410
B(0, 40) z = 3(10) + 4(40) + 370 = 530
C(20, 40) z = 3(20) + 4(40) + 370 = 590
D(40, 20) z = 3(40) + 4(20) + 370 = 570
E(40, 0) z = 3(40) + 4(0) + 370 = 490
F (10, 0) z = 3(10) + 4(0) + 370 = 400
∴ x = 10, y = 0 gives minimum cost of transportion.
Thus No. of packets can be transported as follows
A
B
P 10 30
Q 0 40
R 50 0
Minimum cost of transportation is Rs. 400
x
XtraEdge for IIT-JEE 89
FEBRUARY 2010

MOCK TEST PAPER SOLUTION
FOR PAPER – 3 PUBLISHED IN THIS ISSUE
PHYSICS
22.
x
(r–x)
q 1
r
q 2
2. hv = hv 0 + K.E max
v = frequency of incident light
v 0 = threshold frequency.
kq1
kq2
=
x (r − x)
5. γ-rays have maximum penetrating power and
minimum ionising power.
q1
q 2
= ;
x r − x
5×
10
x
−8
=
−8
− 2×
10
(20 − x)
8. Wavelength of the light increases because light ray
travels from denser to rarer medium as it bends
away from normal.
Velocity and wavelength of light changes as it
passes from air to glass.
11. τ = MB sinθ
M = NIA
= 30 × 6 × 3.14 × (0.08) 2
τ = 3. 62 × 1 × sin 60º
τ = 3.135 N-M
14. (b) (i) Nuclear force is a short range force
(ii) It does not varies as inverse of the square of the
distance
19. Then magnetic flux produced in a coil is associated
with another coil and change in magnetic flux of a
coil induces emf in other coil, this phenomenan is
called mutual inductance.
Derivation
r 2
S 2
suppose a current i is passed through s 1 then
B = µ 0 n 1 i(n 1 = no. of turns/length in s 1 )
Then flux through each turn of s 2
Bπr 1 2 = µ 0 n 1 iπr 1
2
φ total = n 2 l(µ 0 n 1 iπr 1 2 )
φ total = µ 0 n 1 n 2 liπr 1
2
M = µ 0 n 1 n 2 πr 1 lI
Erms
20. Z = =
irms
= i = 90º
2
l
v + v
i
rms
2
R
S 1
r 1
100 – 5x = –2x ; x =
kq q
(ii) U = 2
r
= –4.5 × 10 5 J
100 cm
3
1
=
−12
9
− 9×
10 × 5×
2×
10
20×
10
23. (i) Microwaves are used in radars because of their
short wavelength.
(ii) x-rays are used because of their penetration
power
24. (i) Null point is shifted towards B
(ii) Null point is shifted towards A
(iii) Null point is not obtained
25. (i) φ = hv 0
v 0 = h
φ
−19
10
−34
2.14×
1.6×
=
6.6×
10
= 5.18 × 10 14 Hz
hc
(ii) = hv0 + eV 0 = 2.14 + 0.60
λ
hc = 2.74 eV
λ
λ =
6.6 × 10
−34
× 3×
10
2.74×
1.6×
10
−19
32 32
26. (i) 15 P → 16 S + –1 eº + ν
(antineutrino)
(ii) half life T 1/2 = 50 sec. λ =
8
−5
= 5 × 10 –7
0.693
T 1/ 2
27. (i) width of interference fringes will reduce.
(ii) if white light is used then due to overlapping of
pattern central fringe will be white with red edges.
(iii) No interference pattern is obtained.
XtraEdge for IIT-JEE 90
FEBRUARY 2010

28.
–q
V p =
2a
(r + a)
kq kq
–
(r − a) (r + a)
r
q
(r – a)
⎡ 1 1 ⎤ ⎡ 2a ⎤
= kq ⎢ + ⎥ = kq
⎣ r − a r + a
⎢ ⎥ ⎦ ⎣ r 2 − a
2 ⎦
kp
V p =
2 2
r − a
xp
a Br2
> F2
> I 2
⎯⎯⎯⎯⎯
⎯ →
decreasing order of bond energy
MF>
MCl>
MBr>
MI
(b) ⎯⎯⎯⎯⎯⎯⎯→
decreasing order of ionic character
p
6. (a) NH 3 molecules are associated with
intermolecular H-bonding where as PH 3 is not.
(b) H 3 PO 3 is diprotic acid because it has two
replacable hydrogen i.e.
O
HO
P
OH
H
7. It is because Ce +4 has stable electronic
configuration.
8. CH 2 = CH 2 + CO + H 2 [Co(CO) 4] 2
CH 3 CH 2 CHO
H 2 /Ni
CH 3 CH 2 CH 2 OH
It is used to convert alkene to higher aldehydes and
alcohols.
9. K 4 (Fe(CN) 6 ] → 4K + + [Fe(CN) 6 ] 4–
n = 5
πV = i nRT
π = i V
n RT
π = iCRT where 'C' is molarity.
i −1
50 i −1
α = ⇒ = ⇒ i = 3
n −1
100 5 −1
π = 3 × 10
1 × 0.0821 × 300
= 7.386 atm
10. At anode Zn ⎯→ Zn +2 + 2e –
At cathode 2H + + 2e – ⎯→ H 2 (g)
11. (a) CH 3 – CH – CH 3
OH
Cl 2
Ca(OH) 2 + CCl 3 – C – CH 3
CHCl 3 + (CH 3 COO) 2 Ca
Chloroform
O
O
CH 3 – C – CH 3
O
3Cl 2
(b) CH 3 – C – CH 3 + 3I 2 + 4NaOH →
(Acetone)
CHI 3 + 3NaI + CH 3 COONa + 2H 2 O
CH 2 OH
12. (a)
CH 2 OH + 2HNO CH 2 ONO 2
3
+ 2H 2 O
CH 2 ONO 2
Glycol dinitrate
CH 2 OH
CH 2
KHSO 4
(b)
CHOH
CH
Heat
+ 2H 2 O
CH 2 OH
CHO
Acrolein (prop-2-en-1-al)
XtraEdge for IIT-JEE 91
FEBRUARY 2010

13. (a) C 6 H 5 NH 2
NaNO HCl
⎯⎯⎯
2 + ⎯
0−5ºC
⎯ → C 6 H 5 N + 2 Cl –
CuCN ⎯⎯
KCN
⎯ → C 6 H 5 C≡N + N 2
(b) C 6 H 5 NH 2 + CHCl 3 + 3KOH
⎯→ C 6 H 5 N — → C + 3KCl + 3H 2 O
Phenyl isocyanide
14. The principal ways by which cells obtain energy for
synthesis of ATP are –
(a) Photo synthesis (b) Catabolism of nutrients
such as carbohydrate, proteins & lipid.
15. (a) Beri-beri is caused by deficiency of Vit.-B.
(b) Night blindness is caused by deficiency of Vit.-
A.
(c) Vit.-K
(d) Vit. – B 12
16. a = ; d = 2.48 g cm –3
N 0 = 6.023 × 10 23 , z = 4
z×
M
d =
3
a × N 0
or a 3 z × M 4×
58
= =
N0 × d
23
2.48×
6.023×
10
232
232
=
= × 10 –23
23
2.48×
6.023×
10 14.937
= 15.53 × 10 –23
= 155.3 × 10 –24
a = (155.3) 1/3 × 10 –8
= 5.375 × 10 –8
= 5.375 × 10 –8 × 10 10 pm
= 537.5 pm.
a = 2(r + + r – )
Distance between K + & F – a
= 2
=
537.5
2
= 268.75 pm
17. Arrhenius equation is k = Ae –Ea/RT
Where A = Frequency factor
E a = Activation Energy
R = 8.314 J/K/mol
T = Temperature in Kelvin
E
lnk = lnA – a
RT
Ea
lnk 1 = lnA –
RT1
lnk 2 = lnA – E a /RT 2
k 2 E
ln = a
⎛ 1 1
k1
R ⎟ ⎞
⎜ −
⎝ T1 T 2 ⎠
k 2 E
= log =
a
⎛ T −
k
⎟ ⎞
⎜
2 T1
1 2.303 ⎝ T1
T2
⎠
18. When catalyst and reactions are in different
physical states. It is called heterogenous catalysis.
(a) 2SO 2 (g) + O 2 (g)
V 2 O 5 2SO3 (g)
[Contact process]
(b) N 2 (g) + 3H 2 (g)
Fe
2NH 3 (g)
[Haber's process]
(c) 4NH 3 (g) + 5O 2 (g) ⎯⎯→
Pt 4NO + 6H 2 O
[Ostwald process ]
(d)
n
CH = CH 2
CH 3
TiCl 4 + Al(C 2H 5) 3
Ziglar Natta catalyst
CH – CH 2
CH 3
Polymer
19. (a) XeF 4 + SbF 5 → [XeF 3 ] + [SbF 6 ] –
(b) Cl 2 + NaOH → NaCl + NaClO + H 2 O
(c) 2F 2 + 2H 2 O → 4HF + O 2
(d) 2F 2 + 4NaOH → 4NaF + O 2 + 2H 2 O
(e) XeF 6 + KF → K + [XeF 7 ] –
(f) BrO 3 – + F 2 + 2OH – → BrO 4 – + 2F – + H 2 O
20. (a) Tendency to show catenation process decreases
down the group due to increase in atomic size. In
group 15, (P), in 16 (S), show catenation to
maximum extent because 'N' & 'O' form multiple
bond.
21. (i) 2Cu + O 2 → 2CuO
CuO + H 2 O → Cu(OH) 2
Cu(OH) 2 + CO 2 → CuCO 3 + H 2 O
Green layer is due to formation of basic copper
carbonate.
(ii) Ca(OH) 2 + Cl 2 → CaOCl 2 + H 2 O
(Bleaching powder)
(iii) Cu +
2H 2 SO 4 ⎯Heat ⎯⎯ → CuSO 4 + 2H 2 O + SO 2
(conc.)
22. (i) Ni(28) ⇒ [Ar] 4s 2 3d 8
Ni (0) ⇒ [Ar] 4s 0 3d 10
[Ni(CO) 4 ] =
sp 3 hybridisation
⇒ Tetrahedral
⇒ Diamagnetic
(ii) Ni +2 = [Ar] 4s 0 3d 8 sp 3
'Cl' does not cause pairing of electron because it is a
weak field ligand.
[NiCl 4 ] 2– is Tetrahedral & Paramagnetic.
23. (a) CH 2 = CH 2 + K 2 [PtCl 4 ] → K[PtCl 3 (C 2 H 4 ) + KCl
Zeise's salt
XtraEdge for IIT-JEE 92
FEBRUARY 2010

(b) 2C 6 H 6 + Cr → [(C 6 H 6 ) 2 Cr] dibenzene
chromium
(vapours)
(c) FeCl 2 + 2C 5 H 5 MgBr → [(C 5 H 5 ) 2 FCl] + 2Mg(Br)Cl
Ferrocene
24. Lucas reagent is a mixture of conc. HCl and Anhyd.
ZnCl 2 . It is used to distinguish between 1º, 2º and 3º
alcohols.
Primary alcohols do not react with Lucas reagent at
room temperature. Secondary alcohols react with
Lucas reagent and turbidity (milkyness) appears
after 5-minutes. 3º Alcohols react immediately
forming milkyness.
CH 3 – CH – CH 3 + HCl
OH
(conc)
ZnCl 2(Anhy.)
CH 3 – CH – CH 3 + H 2 O
Cl
2-chloropropane
CH 3
ZnCl
CH 3 – C – CH 3 + HCl 2(Anhy.)
OH (conc)
CH 3
CH 3 – C – CH 3 + H 2 O
Cl
2-chloro-2-methylpropane
25. (a) Buna-S is synthetic rubber. Its monomers are
butadiene and styrene, Na is polymerising agent. It
is used for making automobile tyres.
(b) Elastomers have less force of attraction as
compared to fibres. Elastomers regain their shape
after the stress removed.
Eg. Buna-S, Vulcanised rubber are elastomers
whereas nylon, terylene are example of fibers.
(c) Nylon - 66
26. (a) Malachite green (b) Methyl Orange
(c) Alizarine
27. (a) [Zn(OH) 4 ] 2–
(b) [Co(NH 3 ) 6 ] 2 (SO 4 ) 3
(c) K 2 [Ni(CN) 4 ]
(d) K 2 [PdCl 4 ]
(e) K 3 [Cr(C 2 O 4 ) 3 ]
(f) [Pt (NH 3 ) 2 Cl 2 ]
28. Fe ⎯→ Fe +2 + 2e –
2Ag + + 2e – ⎯→ 2Ag(s)
+
2
Fe(s) + 2Ag (aq) ⎯→ Fe +
(aq) + 2 Ag(s)
E cell =
E º cell –
0.0591
n
+ 2
[Fe ]
log
+ 2
[Ag ]
29.
= [ E º Eº
+ 2 ]
+ − –
Ag / Ag
Fe
/ Fe
= + 0.80 – (– 0.44) –
= + 1.24 – 0.0295
E cell = + 1.205 V
0.0591 0.1
log
2
2
(0.1)
0.0591
log 10
2
dx = k[A] x [B] y
dt
5.07 × 10 –5 = k[0.2] x [0.2] y ...(1)
5.07 × 10 –5 = k[0.2] x [0.1] y ...(2)
on dividing (1) & (2), we get
1 = 2y ⇒ 2 0 = 2 y i.e. y = 0
5.07 × 10 –5 = k[0.2] x [0.1] y
7.60 × 10 –5 = k[0.4] x [0.05] y
1 1 1
=
1.5
y ⇒ =
1/ 2
2 2
1
x = = 0.5 2
1
x
2
The order of reaction is 0.5 with respect to (A) and
zero with respect to (B).
30. (a) CH 3 CH 2 CH 2 CH 2 CH 2 OH
Pentan-1-ol (1º alcohol)
(b) CH 3 – CH 2 – CH – CH 2 – OH
CH 3
2-Methyl butan-1-ol (1º alcohol)
(c) CH 3 – CH – CH 2 – CH 2 – OH
CH 3
3-Methyl butan-1-ol (1º alcohol)
CH 3
(d) CH 3 – C – CH 2 – OH
CH 3
2, 2-Dimethyl propan-1-ol (1º alcohol)
(e) CH 3 – CH 2 – CH 2 – CH – CH 3
OH
Pentan-2-ol (2º alcohol)
(f) CH 3 – CH 2 – CH – CH 2 – CH 3
OH
Pentan-3-ol (2º alcohol)
(g) CH 3 – CH – CH – CH 3
CH 3 OH
3-Methyl butan-2-ol (2º alcohol)
CH 3
(h) CH 3 – C – CH 2 – CH 3
OH
2-Methyl butan-2-ol (3º alcohol)
XtraEdge for IIT-JEE 93
FEBRUARY 2010

1. f –1 (x) =
MATHEMATICS
3x
2 − 5x
Section A
⎡ π 3π
⎤
2. ⎢ , ⎥ (or any other equivalent)
⎣ 2 2 ⎦
3. x = 3
π
4. Skew symmetric
5. a 12 A 12 + a 22 A 22 + A 32 A 32
6. log |x + log sin x| + c
7. Zero
8.
9. 1
10.
π
4
→
a and
→
b are like parallel vectors.
Section B
11. (i) since (a – a) = 0 is a multiple of 4, ∀ a ∈A
∴ R is reflexive
(ii) (a, b) ∈ R ⇒ (a – b) is a multiple of 4
⇒ (b – a) is also a multiple of 4
⇒ (b, a) ∈ R ∀ a, b ∈ A ⇒ R is Symmetric
(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a – b) = 4k, k ∈ z
(b – c) = 4m, m ∈ z ∀ a, b, c ∈ A
⇒ (a – c) = 4(k + m), (k + m) ∈ Z
∴ (a, c) ∈ R
⇒ R is transitive
Set of all elements related to 2 are {2, 6, 10}
OR
(i) ∀ a, b, c, d ∈ N, (a, b)* (c, d) = (a + c, b + d)
= (c + a, d + b)
= (c, d) * (a, b)
⇒ * is commutative
(ii) [(a, b)*(c, d)]*(e, f) = (a + c, b + d) * (e, f)
= ((a + c) + e, (b + d) + f)
= (a + c + e, b + d + f)
= (a + (c + e), b + (d + f))
∀ a, b, c, d, e, f, ∈ N
⇒ = (a, b) * [(c, d) * (e, f)]* is associative
(iii) Let (e, f) be the identity element, then
(a, b) * (e, f) = (a, b)
⇒ (a + e, b + f) = (a, b) ⇒ e = 0, f = 0
but (0, 0) ∉ N × N
So, identity element does not exist
⎧ x −1
x + 1 ⎫
⎪ + ⎪
12. We have tan –1 x − 2 x + 2 π
⎨
⎪ ⎬ =
⎪
x −1
x + 1
1−
. 4
⎩ x − 2 x + 2 ⎭
13.
⇒ tan –1 ⎪⎧
x
⎨
⎪⎩
2
+ x − 2 + x
x
2
− 4 − x
2
2
− x − 2⎪⎫
π
⎬ =
+ 1 ⎪ ⎭ 4
2x
2 − 4
⇒ = 1 ⇒ 2x 2 = 1
− 3
⇒ x 2 1 1
= , x = ± 2 2
b + c
c + a
c + a
a + b
a + b
b + c = 0
a + b b + c c + a
C 1 → C 1 + C 2 + C 3
1 c + a
⇒ 2(a + b + c) 1
1
a + b
b + c
a + b
b + c = 0
c + a
R 2 → R 2 – R 1 , R 3 → R 3 – R 1
1 c + a a + b
⇒ 2(a + b + c)
0
0
b − c
b − a
c − a
c − b
= 0
⇒ 2(a + b + c) (–a 2 – b 2 – c 2 + ab + bc + ca) = 0
⇒ –(a + b + c) [(a – b) 2 + (b – c) 2 + (c – a) 2 ] = 0
⇒ a + b + c = 0 or a = b = c
14. lim f (x)
= –1 + a
−
x→5
f(5) = a + b
lim f (x) = 7 + b
+
x→5
⇒ – 1 + a = a + b = 7 + b ⇒ b = –1, a = 7
15. Let u = x y and v = y x ⇒ u + v = log a
⇒
du dv + = 0
dx dx
...(i)
log u = y log x
⇒
1 du y dy
= + log x
u dx x dx
⇒
⇒
⇒
du = x
y ⎡ y dy ⎤
dx
⎢ + log x ⎥
⎣ x dx ⎦
log v = x log y
1 dv x dy
= + log y
v dx y dx
dv = y
x ⎡ x
⎢
dx ⎣ y
dy
dx
⎤
+ log y⎥
⎦
(i) ⇒ yx y–1 + x y dy
log x + xy
x–1 dy + y x log y = 0
dx dx
⇒
dy ⎡ y.x = – ⎢
dx ⎢⎣
x.y
y−1
x−1
+ y
+ x
x
y
.log y ⎤
.log x ⎥ ⎥ ⎦
XtraEdge for IIT-JEE 94
FEBRUARY 2010

16. (i) Since (x – 2) ≥ 0 in [2, 3]
so f(x) = x − 2 is continuous
1
(ii) f ´(x) = exists for all x ∈(2, 3)
2 x − 2
∴ f(x) is differentiable in (2, 3)
Thus lagrang's mean value theorem is applicable;
∴ There exists at least one real number in (2, 3)
such that
f (3) − f (2)
f´(c) =
3 − 2
1 ( 1) − 0
or = ⇒ 2 c − 2 = 1
2 c − 2 1
1
c = 2 + = 2.25 ∈(2, 3)
4
⇒ LMV is verified and the
req. point is (2.25, 0.5)
1
17. I =
∫
dx
cos(x − a)cos(x − b)
1 sin((x − a) − (x − b))
=
− ∫
dx
sin(b a) cos(x − a)cos(x − b)
1
=
∫[tan(x
− a) − tan(x − b)] dx
sin(b − a)
1
= [log | sec(x – a)| – log |sec (x – b)|]
sin(b − a)
1 ⎡ sec(x − a)
=
sin(b − a)
⎥ ⎤
⎢log
+ c
⎣ sec(x − b) ⎦
OR
2 + sin x
I =
∫
e x/2 dx
1 + cos x
⎡ 2 sin x ⎤
=
∫ ⎢ + ⎥ e x/2 dx
⎣1+ cos x 1+
cos x ⎦
18.
⎡
⎢ 2
=
∫ ⎢
⎢ 2 x
cos
⎣ 2
+
x
2sin
2
2cos
x
cos
2
2
x
2
⎛ x x ⎞
=
∫
⎜sec 2 + tan ⎟ e x/2 dx
⎝ 2 2 ⎠
x
2 tan .e x/2 + c 2
→ →
2
2
⎛ ⎞
a + b = ⎜
→ a + → →
2
b ⎟ =
⇒
⎝
⎠
⎤
⎥
⎥ e x/2 dx
⎥
⎦
→
2
a + b + 2 → a . → b
= | → a | 2 + | → b | 2 + 2| → a | | → b | cos θ
= 1 + 1 + 2.1.1. cos θ
= 2(1 + cos θ) = 2.2cos 2 θ
2
1 → 2
→
a + b = cos
2
4
θ
2
⇒ cos 2
θ =
1
→ →
a + b
2
OR
Let ABCD be the parallelogram with sides → a and → b
D
C
∴
Now
→
d 1 =
A
b
→
AC = → a +
a
d 2
d 1
B
→ →
b and d 2 = → a – → b
→ → ⎛ → →⎞
⎛ → →⎞
→ →
d 1× d 2 = ⎜ a + b ⎟× ⎜ a − b ⎟ = 2 a × b
⎝ ⎠ ⎝ ⎠
→ → 1 → →
⇒ area | | gm = a × b = d 1× d 2
2
When d → 1 = î + 2 ĵ + 3 kˆ and d → 2 = 3 î – 2 ĵ + kˆ
⇒
→
d 1 × → 2
d =
î
1
3
ĵ
2
− 2
1
∴ area of || gm =
⎡
8
2 ⎢⎣
kˆ
3
1
2
= 8 î + 8 ĵ – 8 kˆ
+ 8
2
+ 8
2
⎤
⎥⎦
= 4 3 sq.u
19. The given equations can be written as
x −8 y + 9 z −10
= = and
3 −16
7
x −15 y − 29 z − 5
= =
3 8 − 5
The shortest distance between two lines
→ →
r= a 1 + λ b → 1 and → r= a → 2 + µ b → 2 is given by
S.D. =
⎛
⎜
⎝
→
a 2
−
→
a1
→
b1
⎞ ⎛
⎟.
⎜
⎠ ⎝
×
→
b1
→
b2
×
→
b2
⎞
⎟
⎠
Here a → 1 = (8, –9, 10), a → 2 = (15, 29, 5)
→ →
2 − a 1
⇒ a = (7, 38, –5)
and b → 1 = (3, –16, 7) and b → 2 = (3, 8, –5)
⇒
→
b 1 × b → 2 =
i
3
3
j
− 16
8
= 24 î + 36 ĵ + 72 kˆ
∴ S.D. =
k
7
− 5
168 + 1368 − 360
=
576 + 1296 + 5184
1176 = 14 units.
84
XtraEdge for IIT-JEE 95
FEBRUARY 2010

20. Since x 3 = 30
∴ x 1 , x 2 < 30 and x 4 , x 5 > 30
∴ Required Probability is
= 29 C 2 . 1 C 1 . 20 C 2 / 50 C 5
=
dx
21. Here = dy
29.28 1 20.19
. .
2.1 1 2.1
50.49.48.47.46
5.4.3.2.1
Let F(x, y) =
x / y
2xe
− y
2ye
x / y
x / y
2xe
− y
2ye
x / y
551
=
15134
x / y
λ(2xe
− y)
then F(λx, λy) =
x / y
λ(2ye
)
F(x, y) is a homogeneous function of degree zero,
thus the given differential equation is a
homogeneous differential equation
dx dv
Put x = vy to get = v + y dy dy
dv
∴ v + y = dy
dv
or y = dy
v
2ve
−1
v
2e
2e
v
2ve
−1−
2ve
v
v
= –
1
v
2e
∴ 2e v dy
dv = – ⇒ 2e v + log|y| = c
y
x = 0, y = 1 or, 2e x/y + log|y| = c
⇒ c = 2 ∴ 2e x/y + log |y| = 2
OR
Here integrating factor = ∫
cot xdx
e = e log sinx = sin x
∴ the solution of differential equation is given by
2
y.sin x =
∫
(2x + x cot x)
sin x dx
=
∫
2 x sin x dx +
∫
cos x dx
=
∫
2 x sin x dx + x 2 sin x –
∫
2 x sin x dx + c
= x 2 sin x + c ...(1)
Substituting y = 0 and x = π/2, we get
0 =
2
π
4
x 2
2
π
+ c or c = – 4
∴ (i) ⇒ y sin x = x 2 sin x –
or y = x 2 –
2
π
4
cosec x
22. Equation of the said family is
2
2
2
π
x y
+ = 1
2 2
a b
Differentiating w.r.t. x, we get
4
2
2x 2y dy y dy b
+
2 2 = 0 or = –
a b dx x dx a 2
dy
⎫
2 x − y
⎛ y ⎞ d y ⎛ ⎞ ⎪
⎜ ⎟ +
dx dy
⎜ ⎟ = 0 ⎪
⎝ x
2 2
⎠ dx x ⎝ dx ⎠ ⎬
2
2
d y ⎛ dy ⎞ dy ⎪
or xy + x⎜
⎟ − y = 0⎪
2
dx ⎝ dx ⎠ dx ⎭
Section C
23. Let E 1 : Letter has come from tatanagar ∴ P(E 1 )=1/2
E 2 : Letter has come from calcutta P(E 2 ) = 1/2
A : Obtaining two consecutive letters "TA"
∴ P(A|E 1 ) =
8
2 =
4
1
{ Total possible TA, AT, TA, AN, NA, AG, GA,
AR = 8, favourable = 2}
P(A|E 2 ) = 7
1 {Total possibilities CA, AL, LC, CU,
UT, TT, TA = 7 favourable = 1}
P(E1)P(A | E1)
∴ P(E 1 |A) =
P(E1)P(A | E1)
+ P(E2)P(A | E2)
1 1
.
= 2 4 7
=
1 1 1 1 11
. + .
2 4 2 7
7 4
∴ P(E 2 |A) = 1 – =
11 11
OR
Let x = Number of while balls.
6
C
⎫
3 6.5.4 1
P(x = 0) = = =
10
⎪
C3
10.9.8 6 ⎪
4 6
C
⎪
1.
C2
4.3.6.3 1
P(x = 1) = = = ⎪
10
C3
10.9.8 2 ⎪
4 6
⎬
C2.
C1
4.3.6.3 3
P(x = 2) = = = ⎪
10
C 10.9.8 10 ⎪
3
4
⎪
C3
4.3.2 1
P(x = 3) = = =
⎪
10
C 10.9.8 30 ⎪
3
⎭
Thus we have
x P(x) xP(x) x 2 P(x)
0 1/6 0 0
1 1/2 1/2 1/2
2 3/10 6/10 12/10
3 1/30 3/30 9/30
1 36/30 60/30 = 2
36 18 6
Mean = ΣxP(x) = = = = 1.2
30 15 5
Variance = Σx 2 P(x) – [ΣxP(x) 2 ] 2
36 14
= 2 – = or 0.56
25 25
XtraEdge for IIT-JEE 96
FEBRUARY 2010

24. AB is parallel to the line 2x = y = z
A(3, 4, 5)
2x= y = z
B
x + y + z = 2
x y z
or = =
1/ 2 1 1
x y z
or = =
1 2 2
x − 3
⇒ Equation of AB is
1
=
y − 4
2
=
z − 5
2
For some value of λ, B is (λ + 3, 2λ + 4, 2λ + 5)
B lies on the plane
∴ λ + 3 + 2λ + 4 + 2λ + 5 – 2 = 0
⇒ 5λ = –10 ⇒ λ = 2
∴ B is (1, 0, 1)
ΑΒ =
2
( 3−
1) + (4 − 0) + (5 −1)
2
2
= 6 units.
25. Solving the equations in pairs to get the vertices of
∆ as (0, 1), (2, 3) and (4, – 1)
For correct figure
y
B (2, 3)
3
–x + y = 1
A
(0, 1)
2x + y = 7
x
0
x + 2y = 2
–1 C(4, –1)
Required area
3
1
3
1
=
∫
(7 − y)dy –
2 ∫
( 2 − 2y)dy –
∫
( y −1)dy
⎡
Area A 1 = 2 ⎢
⎢
⎣
−1
3
−1
2
1 ⎡ y ⎤
= ⎢7y
⎥
2 ⎢⎣
2 ⎥⎦
− 1
= 12 – 4 – 2 = 6 sq. U
OR
2
∫
0
2
2y
y
−
− – [ ] 1 1
6 x
A 2
⎡
2
y ⎤
− – ⎢ − y⎥
⎢⎣
2 ⎥⎦
A1
(0, 0) (2, 0) (4, 0)
dx +
4
∫
2
⎤
16 − x 2 dx⎥
⎥
⎦
0
3
1
⎡⎛
= 2 ⎢⎜
⎢
⎣⎝
6 .
2x
3
3/ 2
⎞
⎟
⎠
2
0
⎛ x
+ ⎜
⎝ 2
16 − x
4 3 + 16π
= sq. U.
3
A 2 = Area of circle – shaded area
4 3 + 16π
16π –
=
3
∴
26. I =
∫
e
27.
A 1 16π + 4 3
=
A2
32π + 4 3
tan
−1
x
1
(1 + x
32π
− 4 3
3
2 ) 2
=
dx
4π +
8π −
2
3
3
+ 8sin
−1
θ 2
Put x = tan θ to get I =
∫
e .cos θ dθ
1 θ
=
∫e
(1 + cos 2θ)
dθ
2
1
= e θ 1 θ
+ 2 ∫e
.cos 2θ
dθ
2
1
= e θ + I 1 ...(1)
2
4 ⎤
x ⎞
⎟ ⎥
4 ⎠2
⎥
⎦
1 θ
I 1 =
∫
e .cos2θ
dθ
2
1 θ
θ
= [ e .cos 2θ −∫
− 2sin 2θ.e
dθ]
2
1 θ
θ
θ
= [ e .cos 2θ + 2{ sin 2θ.e
−∫
2cos2θ.e
dθ}
]
2
1
= [e θ cos 2θ + 2 sin 2θ e θ ] –
2
1
4. cos 2θ.e
θ dθ
2
∫
I 1 = 2
1 e θ cos 2θ + sin 2θ e θ – 4I 1
⇒ I 1 = 10
1 e θ cos 2θ + 5
1 sin 2θ e
θ
Putting in (i) we get
I =
2
1 e θ +
10
1 e θ cos 2θ +
5
1 sin 2θ e θ + c
A
L
B
=
10
1 e θ [5 + cos 2θ + 2sin 2θ] + c
=
1 −
tan 1 x
e
10
a
⎡
2
1−
x 4x ⎤
. ⎢5 + + ⎥ + c
2 2
⎢⎣
1+
x 1+
x ⎥⎦
P
b
M
θ
C
XtraEdge for IIT-JEE 97
FEBRUARY 2010

Let ∠C = θ.
∴ AC = AP + PC = S (say)
∴ S = a sec θ + b cosec θ
ds
∴ = a sec θ tan θ – b cosec θ cot θ
dθ
ds a sin θ b cosθ
= 0 ⇒ =
dθ
2
2
cos θ sin θ
3
or tan 3 b ⎛ b ⎞
1/
θ = or tan θ = ⎜ ⎟⎠ a ⎝ a
2
d s
2
dθ = a[sec3 θ + secθ tan 2 θ]
+ b[cosec 3 θ + cosec θ cot 2 θ]
Which is +ve as a, b > 0 and θ is acute
1/ 3
⎛ b ⎞
∴ S is minimum when tan θ = ⎜ ⎟⎠
⎝ a
2
∴ Minimum S = AC = a 1+
tan θ +
2
b 1+
cot
θ
2/3
2/ 3
⎛ b ⎞
⎛ a ⎞
= a 1+ ⎜ ⎟ + b 1+
⎜ ⎟
⎝ a ⎠ ⎝ b ⎠
= a 2/3 a 2/
3 + b
2/ 3 + b 2/3 b 2 /3 + a
2/ 3
= (a 2/3 + b 2/3 ) 3/2
⎛ 1 3 − 2⎞
⎜ ⎟
28. Let A = ⎜−
3 0 − 5⎟
⎜ ⎟
⎝ 2 5 0 ⎠
Writing A = IA
⎛ 1 3 − 2⎞
⎛1
0 0⎞
⎜ ⎟ ⎜ ⎟
or ⎜−
3 0 − 5⎟
= ⎜0
1 0⎟
A.
⎜ ⎟
⎝ 2 5 0
⎜ ⎟
⎠ ⎝0
0 1⎠
R 2 → R 2 + 3R 1 , R 3 → R 3 – 2R 1
⎛1
3 − 2 ⎞ ⎛ 1 0 0⎞
⎜
⎟ ⎜ ⎟
⎜0
9 −11⎟
= ⎜ 3 1 0⎟
A
⎜
⎟
⎝0
−1
4
⎜ ⎟
⎠ ⎝ − 2 0 1⎠
R 1 → R 1 + 3R 3
⎛1
0 10 ⎞ ⎛ − 5 0 3⎞
⎜
⎟ ⎜ ⎟
⎜0
9 −11⎟
= ⎜ 3 1 0⎟
A
⎜
⎟
⎝0
−1
4
⎜ ⎟
⎠ ⎝ − 2 0 1⎠
R 2 → R 2 + 8R 3
⎛1
0 10⎞
⎛ − 5 0 3⎞
⎜ ⎟ ⎜ ⎟
⎜0
1 21⎟
= ⎜ −13
1 8⎟
A
⎜ ⎟
⎝0
−1
4
⎜ ⎟
⎠ ⎝ − 2 0 1⎠
R 3 → R 3 + R 2
⎛1
0 10 ⎞ ⎛ − 5 0 3⎞
⎜ ⎟ ⎜ ⎟
⎜0
1 21⎟
= ⎜−13
1 8⎟
A
⎜ ⎟
⎝0
0 25
⎜ ⎟
⎠ ⎝−15
1 9⎠
1
R 3 → R3
25
⎛1
⎜
⎜0
⎜
⎝0
0
1
0
10⎞
⎟
0 ⎟
1
⎟
⎠
⎛
⎜ − 5
=
⎜
⎜
−13
⎜
15
−
⎝ 25
0
1
1
25
⎞
3 ⎟
8
⎟
⎟
A
9
⎟
25 ⎠
R 1 → R 1 – 10R 3
⎛ −10
−15
⎞
⎜ 1
⎟
⎛1
0 0⎞
⎜ 25 25 ⎟
⎜ ⎟
⎜0
1 0⎟
= ⎜ −10
4 11 ⎟ A
⎜
⎟
⎜ ⎟ 25 25 25
⎝0
0 1⎠
⎜ −15
1 9 ⎟
⎜
⎟
⎝ 25 25 25 ⎠
⎛ −10
−15
⎞
⎜ 1
⎟
⎜ 25 25 ⎟
∴ A –1 = ⎜ −10
4 11 ⎟
⎜ 25 25 25 ⎟
⎜ −15
1 9 ⎟
⎜
⎟
⎝ 25 25 25 ⎠
⎛ 25 −10
−15⎞
1 ⎜
⎟
or ⎜−10
4 11 ⎟
25 ⎜
⎟
⎝ −15
1 9 ⎠
29. Let number of chairs = x
number of tables = y
∴ LPP is Maximise P = 30x + 60y
⎧2x
+ y ≤ 70⎫
⎪ ⎪
Subject to ⎨ x + y ≤ 40 ⎬
⎪ ⎪
⎩x
+ 3y ≤ 90⎭
x ≥ 0, y ≥ 0
For correct graph
80
60
70
40
30
20
n
(15, 25)
C
(30, 10)
20 35 40 60 80 100
y–40
P = 30(x + 2y)
P (A) = 30(60)
P B = 30(65)
P C = 30(50)
P D = 30(35)
∴ For Max Profit (30 × 65)
No. of chairs = 15
No. of tables = 25
XtraEdge for IIT-JEE 98
FEBRUARY 2010

XtraEdge Test Series
ANSWER KEY
IIT- JEE 2010 (February issue)
PHYSICS
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. D B C C A A A,C,D B,C A,D B,D
Ques. 11 12 13 14 15 16 17 18 19
Ans. C D C A B B 0041 0030 0040
CHEMISTRY
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. A B B C A B A,B,C A,B,C,D A,C,D B,C,D
Ques. 11 12 13 14 15 16 17 18 19
Ans. A,C C B D A C 0707 0602 1200
MATHEMATICS
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. A D C A D A B,D A,C,D B,C,D A,C
Ques. 11 12 13 14 15 16 17 18 19
Ans. B A,C B C B D 8282 0996 7168
IIT- JEE 2011 (February issue)
PHYSICS
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. C D C C D D B,C A,C A,C A,B,C
Ques. 11 12 13 14 15 16 17 18 19
Ans. A A C B A A 0003 0003 0005
CHEMISTRY
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. B C A D B D A,C A,B,D A,C A,B,C,D
Ques. 11 12 13 14 15 16 17 18 19
Ans. A,B,C B A A,C B,C,D B,C,D 0040 0025 0500
MATHEMATICS
Ques. 1 2 3 4 5 6 7 8 9 10
Ans. A A C D D B A,B,C,D B,C D A,C
Ques. 11 12 13 14 15 16 17 18 19
Ans. A A,C,D C C A,B,C B 1536 8410 1828
XtraEdge for IIT-JEE 99
FEBRUARY 2010

XtraEdge for IIT-JEE 100
FEBRUARY 2010