XtraEdge for IIT-JEE 1 JULY 2010 - Career Point

XtraEdge for IIT-JEE 1 JULY 2010

XtraEdge for IIT-JEE 2 JULY 2010

Teachers open the door. You enter by yourself
Volume - 6 Issue - 1
July, 2010 (Monthly Magazine)
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Editor : Pramod Maheshwari
Dear Students,
Is examination a common cause of stress
In most Asian cultures, the great emphasis on academic achievement and high
expectations of success make it especially stressful for students. The strong
negative stigma attached to failure also adds to the pressure.
Like it or not, we have to accept that examinations are necessary in any
educational system. Even though it is debatable whether they are accurate
measures of actual ability, no better alternatives have been proposed.
Examinations remain necessary to motivate students’ learning, measure their
progress and ultimately, serve as evidence of attainment of certain skills,
standards or qualifications.
Success at examinations provides opportunities to proceed with higher
education and improves employment prospects, underlining their importance.
No matter how well prepared, many factors may influence one’s performance
at the time of the examination and there is seemingly, no definite guarantee of
success. Essentially, it is this vital importance attached to success at
examinations coupled with the element of uncertainty that makes them so
stressful.
As with other sources of stress, the stress of examinations is not all bad. It is a
strong incentive for students to study and poses a challenge for individual
achievement. However, when stress becomes excessive, performance begins
to suffer. There is thus a need to control levels of stress before it becomes
overwhelming and detrimental. Reliase of stress is necessary for optimum
performance, the means of which is relasing.
Learn to relax
The stress responses produces muscle tension, which you would commonly
experience as backache, neck ache or tension headache at the end of the day.
Often this is unconscious. So to relax these muscles, you need to consciously
practice relaxation exercises. These could involve muscle relaxation, deep
breathing exercises, body massage or guided imagery. Like any particular skill,
you need to practice them regularly in order to reap the benefits.
Another way to relax is to maintain a quiet time as part of the daily routine.
Quiet time refers to a time for you with no interruption from external sources
or distractions. This is a time where you may choose to just think of nothing
and relax. Finally, you can always take up a hobby to help you relax. Do
something you enjoy, be it listening to music.
Ideally, the drive to study should be internally driven by a desire to achieve
one’s own personal goals. Instead, many are driven more by the fear of failure,
which is more stress-provoking and leads easily to discouragement.
Attending school should not merely revolve around preparation for
examinations. Interacting with teachers, socializing with friends, participating in
sports or other extra-curricular activities are all valuable aspects of a ‘wellrounded’
education.
Instead of wishing things would get easier, start looking at how you can get
better...
Yours truly
Pramod Maheshwari,
B.Tech., IIT Delhi
Editorial
XtraEdge for IIT-JEE 1 JULY 2010

XtraEdge for IIT-JEE 2 JULY 2010

Volume-6 Issue-1
July, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Much more IIT-JEE News.
INDEX
CONTENTS
Regulars ..........
PAGE
Know IIT-JEE With 15 Best Questions of IIT-JEE
Challenging Problems in Physics,, Chemistry & Maths
Key Concepts & Problem Solving strategy for IIT-JEE.
Xtra Edge Test Series for JEE- 2011 & 2012
S
Success Tips for the Months
• If you don’t notice when you win, you will
only notice when you lose.
• It’s not bragging if you can do it.
• Feel the power of yet. As in “I don't know
how to do this yet.”
• The difficult we do immediately.
impossible takes a bit longer.
The
• Some look down the rapids and see the
rocks. Hunters look down the rapids and
see the flow around the rocks.
• To know what you are doing is an
advantage. To look like you know what
you are doing is essential.
• First law of expertise: Never ask a barber
if you need a haircut.
• If you think you can, you are probably
right. If you think you can't, you are
certainly right.
• Don't do modesty unless you have earned
it.
NEWS ARTICLE 4
Are Nanoparticles health hazard
Nano is the new black
IITian ON THE PATH OF SUCCESS 6
Ms. Padmasree Warrior & Dr. Krishan K. Sabnani
KNOW IIT-JEE 7
Previous IIT-JEE Question
Study Time........
DYNAMIC PHYSICS 14
8-Challenging Problems [Set# 3]
Students’ Forum
Physics Fundamentals
Capacitor - 1
Friction
CATALYSE CHEMISTRY 29
XTRAEDGE TEST SERIES 52
Class XII – IIT-JEE 2011 Paper
Class XI – IIT-JEE 2012 Paper
Key Concept
Reaction Mechanism
Energetics
Understanding : Organic Chemistry
DICEY MATHS 42
Mathematical Challenges
Students’ Forum
Key Concept
3-Dimensional Geometry
Progression & Mathematical Induction
Test Time ..........
XtraEdge for IIT-JEE 3 JULY 2010

Are Nanoparticles a
Health Hazard
Nanoparticles are a mega business
opportunity for multinationals but
they may pose a health hazard to
users
There is a new industrial
revolution taking place all around
us. The only problem is we can’t
see it. The building blocks, being
developed at the cost of billions of
dollars by scientists, governments
and multinational corporations, are
just a few atoms or molecules
thick — nanoparticles. Many are
less than 100 nanometres (nm) —
one-billionth of a metre — thick. A
single human red blood cell in
comparison is around 500 nm in
diametre. It’s a pity though that
our eyesight isn’t good enough at
nanometre level, for if it were, we
would see that nanoparticles of
precious metals like gold, silver and
titanium have already made the
jump from research labs to our
homes. Manufactured nanoparticles
are today present in thousands of
consumer products around the
world — silver in washing
machines and water purifiers to kill
bacteria, zinc in cosmetics to
protect against ultraviolet rays,
carbon nano-tubes in tennis
rackets to make them stronger and
lighter, titanium in household
paints to decompose dust and
grime without human intervention.
Nano is the New Black
“There’s Plenty of Room at the
Bottom” So went the classic
lecture by the Nobel prize-winning
physicist Richard Feynman in 1959
that many nano-ficionados now
consider the conceptual sun of the
nanotechnology universe. “Why
cannot we write the entire 24
volumes of the Encyclopedia
Britannica on the head of a pin”
he asked.
“Because there isn’t much of a
point, or money, in doing so!” is
the answer he would have got
today from nanotechnology
researchers. Instead their time is
mostly spent figuring out newer
properties for nanoparticles which
can then be embedded into
commercial applications.
Nanoparticles are highly reactive
and prone to unusual properties.
Describing gold, a metal that is
normally inert to all other
chemicals, Prof. C.N.R Rao,
Honorary President and Linus
Pauling Professor at the Jawaharlal
Nehru Centre for Advanced
Scientific Research (JNCASR) and
the head of its Nanoscience centre,
says “At 200-300 nm thickness,
gold is not metallic, it does not
shine — in fact it is not gold. And
at 1.5-2 nm, it reacts like mad!”
Gold that is not gold when shrunk
to nanometer size might sound like
an absurdity to many, but it’s
exactly this change in physical
properties that make nanoparticles
popular.
For example zinc oxide (ZnO) and
titanium dioxide (TiO2) have been
used as active ingredients in
sunscreens for decades because of
their ability to absorb ultraviolet
rays and reflect back much of the
other remaining sunlight. But they
are both white — the reason many
sunscreens leave a white residue
on the face. When shrunk to
nanometre size however, they
become transparent without losing
their light reflecting or absorbing
abilities.
Silver, an ornamental metal and a
powerful bactericide, can be
reduced to nanoparticle form to
destroy disease-causing bacteria
from all kinds of places — kitchen
counters, contaminated water,
dirty clothes and stinky underarms.
Samsung claims its Silver Nano
range of washing machines release
hundreds of billions of silver nanoions
with each wash to kill over 99
percent of the bacteria found in
dirty clothes, while the same
technology when lined on the
doors of their refrigerators kill
bacteria that could spoil stored
food. Eureka Forbes’ water
purifiers use nanosilver-coated
filters, developed by Prof. Pradeep,
head of IIT-Madras’ Nanoscience
department, to destroy harmful
bacteria from drinking water.
Swach, the mass-market water
filter introduced by the Tata
Group, also uses nanosilver
(coated on rice husk particles) to
purify drinking water. During the
last flu pandemic threat authorities
in Hong-Kong sprayed subways
with nanosilver to disinfect them.
L’Oreal, the world’s largest
cosmetics company, reportedly
spends over $600 million each year
researching and patenting
nanoparticles. The head of its
nanotechnology unit also sits on
the management board.
XtraEdge for IIT-JEE 4 JULY 2010

Therefore a metal that is a poor
second cousin to gold in a world
where we value yellow over white,
is the undisputed metal of choice in
the nanoparticle world. In fact,
silver is more popular than any
other material, according to the
database of consumer products
using nanoparticles maintained by
the Woodrow Wilson
International Center for Scholars.
Obama admn
nominates IIT alumnus
for post of NSF Director
IIT Madras alumnus Subra Suresh,
popularly known as 'Bakthi Suresh'
during his student days, has been
nominated for the post of director
of the National Science Foundation
(NSF) by the Barrack Obama
administration. An official relese
from IIT Madras Alumni
Association here said ''when
confirmed by the Senate, Mr
Suresh will become one of the
highest ranking Indian-Americans
ever to serve in an administration.''
An Indian-American technocrat,
53-year-old Subra Suresh
completed his B.Tech Mechanical
Engineering in 1977. Currently the
dean of the MIT engineering
school, he received the
distinguished alumnus award in
1997. In a statement, President
Obama said ''I am proud that such
experienced and committed
individuals have agreed to take on
these important roles in my
administration. I look forward to
working with them in the coming
months and years.'' The National
Science Foundation is the funding
source for nearly 20 per cent of all
federally supported basic research
and was an independent federal
agency created by US Congress in
1950. Subra Suresh has been
elected to the US National
Academy of Engineering, the Indian
National Academy of Engineering,
the American Academy of Arts and
Sciences, the Indian Academy of
Sciences in Bangalore, the German
National Academy of Sciences, the
Royal Spanish Academy of Sciences
and the Academy of Sciences of
the Developing World based in
Trieste, Italy.
Major decision regarding
Ganga at IIT-K
KANPUR: A collaboration
between the consortium of seven
IITs (IIT-Kanpur, Madras, Bombay,
Delhi, Kharagpur, Guwahati and
Roorkee) and Ministry of
Environment and Forest,
Government of India, is being
worked out for the purpose of
cleaning the national river Ganga.
The two are expected to sign an
important memorandum of
understanding (MoU) in this regard
during Prime Minister Manmohan
Singh's visit to the Indian Institute
of Technology-Kanpur. Singh is
scheduled to visit IIT-K on July 3 to
take part in its convocation
ceremony.
It will be worth mentioning here
that the Central government aims
at meeting the formidable
challenge of cleaning the Ganga.
With this goal in mind, it had
launched a new initiative and
established the National Ganga
River Basin Authority (NGRBA)
last year. The Prime Minister, who
is the Chairman of the NGRBA,
asked the Ministry of Environment
and Forest to involve IITs in the
mega project. A joint meeting of all
the seven IITs was convened on
March 12, 2010 in which IIT-
Kanpur was represented by Prof
Vinod Tare, also the convener of
the mission. It is for the very first
time that the Central government
has involved the seven IITs
together in one single project of
such a large magnitude
Prof Tare further informed TOI
that the 'zero discharge' of both
treated and untreated sewage
waste into the river had been
proposed to the government
under which the waste would not
be allowed into the Ganga.
"We will be doing this mega
project in phases. The first phase
will come to an end in 18 months
wherein the concept of 'zero
discharge' will be put into
application. We also plan to apply
the 'zero discharge' formula in four
cities initially, viz. Hardwar,
Rishikesh, Kanpur and Allahabad. If
we are able to do so in these four
cities, water of the river will
become clean up to Allahabad and
a major work will come to an end
in the first phase."
Meanwhile, Prof SG Dhande,
director, IIT-Kanpur, and Prof
Vinod Tare from IIT-Kanpur, Prof
Devang Khakhar, director, IIT-
Bombay, took part in a meeting
with the officials of the Ministry of
Environment and Forest in New
Delhi on May 19 and discussed all
important aspects of the mega
project. On the occasion, the team
also handed over a set of proposals
to the officials of the ministry. "A
detailed project report will be
given later as several social, legal
aspects will have to be examined,"
said Prof Tare.
The consortium of the seven IITs
have been sanctioned Rs 16 crore
by the government for formulating
a proper plan of action for the
purpose of cleaning the Ganga. The
authority formed by the Central
government has both regulatory
and developmental functions. The
authority will take measures for
effective abatement of pollution
and conservation of the Ganga in
keeping with sustainable development
needs.
Science Research : Conventional solar cell efficiency could be increased from the current limit of 30 percent to more than 60
percent, suggests new research on semiconductor nanocrystals, or quantum dots, led by chemist Xiaoyang Zhu at The University
of Texas at Austin.
The scientists have discovered a method to capture the higher energy sunlight that is lost as heat in conventional solar cells.
XtraEdge for IIT-JEE 5 JULY 2010

Success Story
This articles contains stories of person who have succeed after graduation from different IIT's
Ms. Padmasree Warrior
B.Tech, IIT Madras
Dr. Krishan K. Sabnani
B.Tech, IIT – Kanpur
Padmasree Warrior is senior vice president and chief
technology officer for Motorola, with responsibility for
Motorola Labs, the global software group and emerging
early-stage businesses. Warrior's operational
responsibilities include leading a global team of 4,600
technologists, prioritizing technology programs, creating
value from intellectual property, guiding creative research
from innovation through early-stage commercialization,
and influencing standards and roadmaps. She also serves
as a technology advisor to the office of the chairman and
to the board's technology and design steering committee.
Before assuming her current position in January 2003,
Warrior was corporate vice president and general
manager of Motorola's energy systems group, where she
was responsible for profit and loss, sales, marketing,
engineering and manufacturing. She also was general
manager of Thought beam, Inc., a wholly owned
subsidiary of Motorola, where she led the
commercialization evaluation team related to compound
semiconductor materials research.
Prior to these assignments, Warrior was corporate vice
president and chief technology officer for Motorola's
Semiconductor Products Sector (SPS).
A Motorola since 1984, she has been instrumental in
driving innovative methods for technology
commercialization realizing early "time to revenue" for
the corporation. She has held many leadership positions
within Motorola, was appointed vice president in 1999
and was elected a corporate officer in 2000.
Warrior received a M.S. degree in chemical engineering
from Cornell University, and a B.S. degree in chemical
engineering from the Indian Institute of Technology (IIT)
in New Delhi, India.
Warrior served on the Texas Governor's Council for
Digital Economy, and is a member of the Texas Higher
Education Board review panel. She was one of six women
nationwide selected to receive the "Women Elevating
Science and Technology" award from Working Woman
magazine in 2001. She also is a director of Ferro
Corporation.
Krishan Sabnani is Senior Vice President of the
Networking Research Laboratory at Bell Labs in New
Jersey. For the past 23 years Krishan has been a member
of Bell Labs Research. Krishan has conceived and launched
several systems projects in the areas of Internetworking
and wireless networking, led successful transfers of
research ideas to products in Lucent and AT&T business
units and conducted extensive personal research in data
and wireless networking. He has built organizations
known for technical excellence by recruiting and coaching
the best people in the industry.
Krishan has received the 2005 IEEE Eric E. Sumner Award
and the 2005 IEEE W. Wallace McDowell Award - the
only person ever to receive both awards. Krishan is a Bell
Labs Fellow. He is also a fellow of the Institute of
Electrical and Electronic Engineers (IEEE) and the
Association of Computing Machinery (ACM). He received
the Leonard G. Abraham Prize Paper Award from the
IEEE Communications Society in 1991. Krishan will
receive the 2005 Distinguished Alumni Award from Indian
Institute of Technology (IIT), New Delhi, India. He has
also won the 2005 Thomas Alva Edison Patent Award
from the R&D Council of New Jersey. He holds 37
patents and has published more than 70 papers.
In his personal research, Krishan has made major
contributions to the communications protocols area. He
has designed several protocols such as SNR, RMTP, and
Airmail. He has also made significant contributions to
conformance test generation, protocol validation,
automated converter generation, and reverse engineering.
Krishan received his Ph.D. in electrical engineering from
Columbia University, New York, in 1981. He joined Bell
Labs in 1981.
Key Awards and Honors
1. 2005 IEEE Eric E. Sumner Award, received for
seminal contributions to networking protocols
2. 2005 IEEE Computer Society W. Wallace McDowell
Award
3. 2005 IIT Delhi Outstanding Alumni Award
4. 2005 Thomas Alva Edison Patent Award
5. 1991 Leonard G. Abraham Prize Paper Award
6. 1997 Bell Labs Fellow.
XtraEdge for IIT-JEE 6 JULY 2010

KNOW IIT-JEE
By Previous Exam Questions
PHYSICS
1. Masses M 1 , M 2 and M 3 are connected by strings of
negligible mass which pass over massless and friction
less pulleys P 1 and P 2 as shown in fig. The masses
move such the portion of the string between P 1 and P 2
in parallel to the inclined plane and the portion of the
string between P 2 and M 3 is horizontal. The masses
M 2 and M 3 are 4.0 kg each and the coefficient of
kinetic friction between the masses and the surfaces
is 0.25. The inclined plane makes an angle of 37º
with the horizontal.
[IIT-1981]
P 1
M 1
M 2
P 2 M 3
37º
If the mass M 1 moves downwards with a uniform
velocity, find
(a) the mass of M 1
(b) The tension in the horizontal portion of the string
(g = 9.8 m/sec 2 , sin 37º ≈ 3/5)
Sol. (a) Applying Fnet = ma on M 1 we get
T – m 1 . g = M 1 × 0 = 0 ⇒ T = M 1 g ...(i)
Applying Fnet = Ma on M 2 we get
T – (T´ + M 2 g sin θ – f) = M 2 × a
T = T´ + M 2 g sin θ + f = T´ + M 2 g sin θ + µN
[Q f = µN = µM 2 g cos θ]
∴ T = T´ + M 2 g sin θ + µM 2 g cos θ ...(ii)
P
V
1
T M 2
M
T´ P 2
1
T
V
θ M 2gsinθ N
M
T´
2gcosθ
M 1g
M
f 2g θ
M 3g
Applying F net = Ma for M 3 we get
T´ – f ´ = M 3 × 0
⇒ T´ = f ´ = µN´ = µM 3 g
...(iii)
Putting the value of T and T´ from (i) and (iii) in (ii)
we get
M 1 g = µM 3 g + M 2 g sin θ – µ M 2 g cos θ
M 1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º
= 4.2 kg
(b) The tension in the horizontal string will be
T ´ = µM 3 g = 0.25 × 4 × 9.8 = 9.8 N
2. A 0.5 kg block slides from the point A (see fig.) on a
horizontal track with an initial speed of 3 m/s towards
a weightless horizontal spring of length 1 m and force
constant 2 Newton/m. The part AB of the track is
frictionless and the part BC has the coefficients of
static and kinetic friction as 0.22 and 0.2 respectively.
If the distances AB and BD are 2m and 2.14 m
respectively, find the total distance through which the
block moves before it comes to rest completely.
(Take g = 10 m/s 2 )
[IIT-1983]
A B D C
Sol. K.E. of block = work against friction + P.E. of spring
1 mv 2 1
= µ k mg (2.14 + x) + kx
2
2
2
1 × 0.5 × 3 2 1
= 0.2 × 0.5 × 9.8(2.14 + x) + 2 × x
2
2
2
2.14+ x + x 2 = 2.25
∴ x 2 + x – 0.11 = 0
11
On solving we get x = – 10
1
or x = = 0.1 (valid answer)
10
Here the body stops momentarily.
Restoring force at y = kx = 2 × 0.1 = 0.2 N
Frictional force at
y = µ s mg × x = 0.22 × 0.5 × 9.8 = 1.078 N
Since friction force > Restoring force the body will
stop here.
∴ The total distance travelled
= AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.
A B D C Rough L
2m
2.14m
x
Y
3. A small sphere rolls down without slipping from the
top of a track in a vertical plane. The track in a
vertical plane. The track has an elevated section and a
horizontal part, The horizontal part is 1.0 meter
above the ground level and the top of the track is 2.4
metres above the ground. Find the distance on the
ground with respect to the point B(which is vertically
XtraEdge for IIT-JEE 7 JULY 2010

elow the end of the track as shown in fig.) where the
sphere lands. During its flight as a projectile, does the
sphere continue to rotate about its centers of mass
Explain.
[IIT-1987]
speed of 0.001 ms –1 . Calculate the current drawn
from the battery during the process. (Dielectric
constant of oil = 11, ε 0 = 8.85 × 10 –12 C 2 N –1 m –1 )
[IIT-1994]
Sol. The adjacent figure is a case of parallel plate
capacitor. The combined capacitance will be
v
2.4 m
A
1.0m
+
B
Sol. Applying law of conservation of energy at point D
and point A
P.E. at D = P.E. at A + (K.E.) T + (K.E.) R
(K.E.) T = Translational K.E.
mg (2.4) = mg (1) + 2
1 mv 2 + 2
1 Iω
2
(K.E.) R = Rotational K.E.
Since the case is of rolling without slipping
D
2.4m
∴ v = rω
A
1m
B C
∴ ω = r
v where r is the radius of the sphere Also
I = 5
2 mr
2
1
∴ mg(2.4) = mg(1) + mv 2 1 2
+ × mr 2 v
×
2 2 5
2
r
⇒ v = 4.43 m/s
After point A, the body takes a parabolic path. The
vertical motion parameters of parabolic motion will
be
u y = 0 S = ut + 2
1 at
2
S y = 1m
2
1 = 4.9 t y
a y = 9.8 m/s 2
∴ t y = t y =
1
4.9
= 0.45 sec
Applying this time in horizontal motion of parabolic
path, BC = 4.43 × 0.45 = 2m
During his flight as projectile, the sphere continues to
rotate because of conservation of angular momentum.
4. Two square metal plates of side 1 m are kept 0.01 m
apart like a parallel plate capacitor in air in such a
way that one of their edges is perpendicular to an oil
surface in a tank filled with an insulating oil. The
plates are connected to a battery of emf 500 V. The
plates are then lowered vertically into the oil at a
2
1–x
x
1m
d
C = C 1 + C 2
kε
(x 1)
= 0 × ε
+
d
ε 0
C = [kx + 1 – x]
d
[(1 − x)
d
0 ×
After time dt, the dielectric rises by dx. The new
equivalent capacitance will be
C + dC = C 1´ + C 2´
kε = 0
ε
[(x + dx) × 1] +
0 [1 − x − dx) × 1]
d
d
dC = Change of capacitance in time dt
ε 0
= [kx + kdx + 1 – x – dx – kx – 1 + x]
d
ε 0
= d
(k – 1)dx
dC ε =
0 dx ε (k – 1) =
0 (k – 1)v ...(i)
dt d dt d
dx
where v = dt
We know that q = CV
dq dC = V ...(ii)
dt dt
ε 0
⇒ I = V (k – 1)v d
From (i) and (ii)
−12
500×
8.85×
10
I =
(11 – 1) × 0.001
0.01
= 4.425 × 10 –9 Amp.
5. Two resistors, 400 ohms, and 800 ohms are
connected in series with a 6-volt battery. It is desired
to measure the current in the circuit. An ammeter of a
10 ohms resistance is used for this purpose. What
will be the reading in the ammeter Similarly, If a
voltmeter of 10,000 ohms resistance is used to
measure the potential difference across the 400-ohms
resistor, What will be the reading in the voltmeter.
[IIT-1982]
1]
XtraEdge for IIT-JEE 8 JULY 2010

Sol. Applying Kirchoff's law moving in clockwise
direction starting from battery we get
400Ω 800Ω
10Ω
A
6 volt
+ 6 – 10I – 400 I – 800 I = 0
∴ 6 = 1210 I
6
∴ I = = 4.96 × 10 –3 A
1210
The voltmeter and 400 Ω resistor are in parallel and
hence p.d. will be same
∴ 10,000 I 1 = 400 I 2
...(i)
Applying Kircoff's law in loop ABCDEA starting
from A in clockwise direction.
– 400 I 2 – 800 I + 6 = 0
∴ 6 = 400 I 2 + 800 (I 1 + I 2 )
∴ 6 = 400 I 2 + 800(0.04 I 2 + I 2 )
From (i) putting the value of I 1
∴ 6 = 1232 I 2
10,000Ω
B
I
A
F
V
400Ω
G
C
I
800Ω
6 volt
∴ I 2 = 4.87 × 10 –3 Amp.
Potential drop across 400 Ω resistor
= I 2 × 400
= 4.87 × 10 –3 × 400
= 1.948 volt ≈ 1.95 volt
∴ The reading measured by voltmeter = 1.95 volt
D
CHEMISTRY
6. At constant temperature and volume, X decomposes as
2X(g) → 3Y(g) + 2Z(g); P x is the partial pressure of X.
Observation Time (in minute) Rx (in mm of Hg)
No.
1 0 800
2 100 400
3 200 200
(i) What is the order of reaction with respect to X
(ii) Find the rate constant.
(iii) Find the time for 75% completion of the reaction.
(iv) Find the total pressure when pressure of X is
700 mm of Hg. [IIT-2005]
E
Sol.
(i) From the given data, it is evident that the t 1/2 (half-life
period)for the decomposition of X (g) is constant
(100 minutes) therefore the order of reaction is one.
0.693
(ii) Rate constant, K =
t 1/ 2
0.693
= = 6.93 × 10 –3 min –1
100
(iii) Time taken for 75% completion of reaction
= 2t 1/2 = 2 × 100 = 200 minutes
(iv) 2x ⎯→ 2Y + 2Z
Initial pressure 800 0 0
Ater time t (800 – 2p) 3P 2p
when the pressure of X is 700 mm of Hg the, 800 –
2P = 700
2P = 100; P = 50 mm of Hg
Total pressure = 800 – 2P + 3P + 2P = 800 + 150
= 950 mm of Hg.
7. A acid solution of Cu 2+ salt containing 0.4 g of Cu 2+
is electrolysed until all the copper is deposited. The
electrolysis is continued for seven more minutes with
the volume of solution kept at 100 ml. and the current
at 1.2 amp. Calculate the volume of gases evolved at
NTP during the entire electrolysis. [IIT-89]
Sol. The chemical reactions taking place at the two
electrodes are
At cathode : Cu 2+ + 2e – → Cu
H 2 O H + + OH –
However, note that only Cu 2+ ions will be discharged
so as these are present in solution and H + ions will be
discharged only when all the cu 2+ ions have been
deposited.
Atcathode : 2OH – → H 2 O + O + 2e –
O + O → O 2
Thus in first case, Cu 2+ ion will be discharged at the
cathode and O 2 gas at the anode. Let us calculate the
volume of gas (O 2 ) discharged during electrolysis.
According to Faraday's cecond law
31.75 g Cu ≡ 8 g of oxygen
≡ 5.6 litres of O 2 at NTP
5.6
0.4 g Cu = × 0.4 litres of O 2 at NTP
31.75
= 0.07055 litres = 70.55 ml
As explained earlier, when all the Cu 2+ ion will be
deposited at cathode, H + ions will start going to
cathode liberating hydrogen (H 2 ) gas i.e.
H + + e – H H + H → H 2
However, the anode reaction remains same as
previous. Thus in the second (latter) case, amount of
H 2 collected at cathode should be calculated.
8 g of O 2 = 1 g of H 2
5.6 litres of O 2 at NTP = 11.2 litres of hydrogen
Quantity of electricity passed after 1 st electrolysis,
i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs
XtraEdge for IIT-JEE 9 JULY 2010

5 .6×
504
504 coulombs will liberate = 96500
= 29.24 ml
of O 2 . Similarly, H 2 liberated by 504 coulombs
504
= 11.2 × 96500
= 58.45 ml
Total volume of O 2 liberated = 70.55 + 29.24 = 99.79
ml vol. of H 2 liberated = 58.48 ml.
8. Cyclobutyl bromide on treatment with magnesium in
dry ether forms an organometallic (A). The
organometallic reacts with ethanal to give an alcohol
(B) after mild acidification Prolonged treatment of
alcohol (B) with an equivalent amount of HBr gives
1-bromo 1-methylcyclopentane (C). Write the
structures of (A), (B) and explain how (C) is obtained
from (B).
[IIT-2001]
Br
MgBr
Sol.
+
CH CHO,H3O
3
⎯⎯⎯⎯⎯⎯
→
H
⎯ ⎯→
+
dry
ether
⎯ ⎯→
Cyclobutymagnesium
Bromide
(A)
CH–OH
CH3
1-Cyclobutylethanol
(B)
CH–CH 3
OH 2
⊕
Oxonium ion
ringexpansion
⎯⎯⎯⎯→
through1,2−alkylshift
(B)
CH–CH 3
OH
CH–CH 3
– H O
⎯ 2
⎯ →
⎯ +
⎯
H
⊕ H
CH 3
H
H
⊕
CH 3
(3º carbocation)
(ring expansion)
(2º carbocation in
5 membered ring
H
Br
⎯ ⎯→
–
+
(2º carbocation
4-membered ring)
,2–
⎯
1 ⎯→
Hydrideshift
H
Br
⊕
CH 3
1-bromo-1-methyl
Cyclopentane
(C)
9. A basic, volatile nitrogen compound gave a foul
smelling gas when treated with chloroform and
alcoholic potash. A 0.295 g sample of the substance.
Dissolved in aq. HCl and treated with NaNO 2
solution at 0ºC, liberated a colorless, odourless gas
whose volume corresponded to 112 ml at STP, After
the evolution of the gas was complete, the aqueous
solution was distilled to give an organic liquid which
did not contain nitrogen and which on warming with
alkali and iodine gave a yellow precipitate. Identify
the original substance. Assume that it contains one N
atom per molecule.
[IIT-93]
Sol. Let us summarise the given facts.
112 ml of
colourless,
odourless
gas at S.T.P
+ Residue
Distilled aq. Sol.
(i) aq. HCl
(ii) NaNO 2 0ºC
Basic
Nitrogen
Compound
(0.295 g)
CHCl 3
KOH
Foul
smelling
gas
Organic liquid OH – /I 2
Yellow ppt.
(no N)
Reaction of the original compound with alcoholic
potash and chloroform to give foul smelling gas
indicates that it contains a primary –NH 2 group.
R–NH 2 + CHCl 3 + KOH ⎯→ R–NC↑
(Basic compound)
Carbylamine
(foul smelling)
Determination of mol. Weight of the amine.
112 ml. of gas is evolved at S.T.P. by 0.295 g of
amine
0.295
22400 ml. of gas is evolved by = × 22400 = 59
112
Hence the mol. wt. of the amine = 59
∴ Mol. wt. of the alkyl group = 59 – 16 = 43
–
Nature of alkyl gp. of mol. wt.= 43 = C 3 H 7
Thus the amine may be either
CH 3
CH 3 CH 2 CH 2 NH 2 or CHNH 2
CH 3
The reaction of amine with NaNO 2 at 0ºC and all
other reactions may thus be written as below.
(i) HCl
CH 3 CH 2 CH 2 NH 2 ⎯⎯⎯→
CH 3 CH 2 CH 2 OH +
(ii) NaNO2
/ 0ºC
N 2 ↑
n-Propylamine
⎯
aq ⎯
.sol. ⎯
distill ⎯⎯ → CH 3 CH 2 CH 2 OH
⎯ OH
–
⎯ ⎯⎯ , I 2 → No yellow ppt.
(CH 3 )CHNH 2 ⎯→ (CH 3 ) 2 CHOH + N 2 ↑
Isopropylamine
OH ,I
⎯→ (CH 3 ) 2 CHOH ⎯ ⎯
– ⎯⎯ 2→
CHI3 ↓
(Haloformreaction)
(yellow)
Since the given reactions correspond to
isopropylamine, the original compound is
isopropylamine, (CH 3 ) 2 CHNH 2
10. Interpret the non-linear shape of H 2 S molecule and
non-planar shape of PCl 3 using valence shell electron
pair repulsion (VSEPR) theory. (Atomic numbers :
H = 1, P = 15, S = 16, Cl = 17.) [IIT-98]
Sol. In H 2 S, no. of hybrid orbitals = 2
1 (6 + 2 – 0 + 0) = 4
Hence here sulphur is sp 3 hybridised, so
16S = 1s 2 , 2s 2 2p 6 2 2 1 1
, 3 s 3p
x
3p
y
3p
z
144
243
4
3
sp hybridisation
XtraEdge for IIT-JEE 10 JULY 2010

x + y = 1
a b
S
or
H H
H H
Due to repulsion between lp - lp; the geometry of
H 2 S is distorted from tetrahedral to V-shape.
In PCl 3 , no. of hybrid orbitals = 2
1 [5 + 3 – 0 + 0] = 4
Hence, here P shows sp 3 -hybridisation
15P = 1s 2 , 2s 2 2p 6 2 1 1 1
, 3 s 3p
x
3p
y
3p
z
144
243
4
Cl
P
Cl
3
sp hybridisation
Cl
or
Thus due to repulsion between lp – bp, geometry is
distorted from tetrahedral to pyramidal.
Cl
S
P
Cl
MATHEMATICS
11. The circle x 2 + y 2 – 4x – 4y + 4 = 0 is inscribed in a
triangle which has two of its sides along the
coordinate axes. If the locus of the circumcentre of
the triangle is
2 2
x + y – xy + k x + y = 0,
find the value of k.
[IIT-1987]
Sol. Let OAB be the triangle in which the circle
x 2 + y 2 – 4x – 4y + 4 = 0 is inscribed. Let the
x y
equation of AB be + = 1
a b
y
B(0,b)
x´ O
y´
C
2
x y
+ = 1
a b
(a, 0)A
Since AB touches the circle x 2 + y 2 – 4x – 4y + 4 = 0.
There fore,
2 2
+ −1
a b
1 1
+
2 2
a b
⎛ 2 2 ⎞
⎜ + −1⎟
a b
= 2 ⇒ –
⎝ ⎠
= 2
1 1
+
2 2
a b
x
Cl
[Q O(0, 0) and C(2, 2) lie on the same side of AB
Therefore, a
2 + b
2 – 1 < 0]
(2b + 2a – ab)
⇒ –
= 2
2 2
a + b
⇒ 2a + 2b – ab + 2
2 2
a + b = 0 ...(i)
Let P(h, k) be the circumcentre of ∆OAB. Since
∆ OAB is a right angled triangle. So its circumcentre
is the mid-point of AB.
a b
∴ h = and k = 2 2
⇒ a = 2h and b = 2k
...(ii)
From (i) and (ii), we get
2 2
4h + 4k – 4hk + 2 4 h + 4k = 0
2
⇒ h + k – hk + h + k = 0
So, the locus of P(h, k) is
2 2
x + y – xy + x + y = 0
But, the locus of the circumcentre is given to be
2 2
x + y – xy + k x + y = 0
Thus, the value of k is 1
12. If A, B, C are the angles of a triangle ABC and the
system of linear equations
x sin A + y sin B + z sin C = 0
x sin B + y sin C + z sin A = 0
x sin C + y sin A + z sin B = 0
has a non trivial solution, prove that
sin 2 A + sin 2 B + sin 2 C – (cos A + cos B + cos C
+ cos A cos B + cos B cos C + cos C cos A) = 0
[IIT-2002]
Sol. The given system of linear equations has a non-trivial
solution. Therefore,
sin A sin B sin C
⇒
sin B
sin C
sin C
sin A
sin A
sin B
sin A + sin B + sin C
sin B + sin C + sin A
sin C + sin A + sin B
2
= 0
sin B
sin C
sin A
sin C
sin A
sin B
Applying C 1 → C 1 + C 2 + C 3
⇒ (sin A + sin B + sin C)
⇒
1
1
1
sin B
sin C
sin A
1
1
1
sin B
sin C
sin A
= 0
sin C
sin A
sin B
= 0
sin C
⎡Qsin A + sin B + sin C ⎤
sin A = 0 ⎢ A B C ⎥
⎢=
4cos cos cos ≠ 0⎥
sin B ⎣ 2 2 2 ⎦
XtraEdge for IIT-JEE 11 JULY 2010

⇒
1
0
0
sin B
sin C − sin B sin A − sin C = 0
sin A − sin B
sin C
sin B − sin C
Applying R 2 → R 2 – R 1 , R 3 → R 3 – R 1
⇒ –(sin B – sin C) 2 – (sin A – sin C)
(sin A – sin B) = 0
⇒ sin 2 B + sin 2 C – 2 sin B sin C + sin 2 A
– sin A sin B – sin C sin A + sin B sin C = 0
⇒ sin 2 A + sin 2 B + sin 2 C – sin A sin B – sin B sin C
– sin C sin A = 0
⇒ sin 2 A + sin 2 B + sin 2 C – cos A cos B
– cos B cos C – cos C cos A + cos (A + B)
+ cos (B + C) + cos (C + A) = 0
⇒ sin 2 A + sin 2 B + sin 2 C – cos A cos B
– cos B cos C – cos C cos A – cos A
– cos B – cos C = 0
13. Determine the name of the name of the curve
described parametrically by the equations
x = t 2 + t + 1, y = t 2 – t + 1 [IIT-1998]
Sol. We have,
x = t 2 + t + 1 and, y = t 2 – t + 1
⇒ x + y = 2(t 2 + 1) and, x – y = 2t
⎪⎧
2
⎛ x − y ⎞ ⎪⎫
⇒ x + y = 2 ⎨⎜
⎟ + 1⎬
⎪⎩ ⎝ 2 ⎠ ⎪⎭
⇒ 2(x + y) = (x – y) 2 + 4
⇒ x 2 + y 2 – 2xy – 2x – 2y + 4
Comparing this equation with the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, we get
a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1
∴ abc + 2fgh – af 2 – bg 2 – ch 2 = 4 – 2 – 1 – 1 – 4 ≠ 0
and , h 2 – ab = 1 – 1 = 0
Thus, we have
∆ ≠ 0 and h 2 = ab
So, the given equations represent a parabola.
14. Let C be any circle with centre (0, 2 ). Prove that at
most two rational points can be there on C.
(A rational points is a point both of whose
coordinates are rational numbers) [IIT-1997]
Sol. The equation of any circle C with centre (0, 2 ) is
given by
(x – 0) 2 + (y – 2 ) 2 = r 2 , where r is any positive real
number.
or, x 2 + y 2 – 2 2 y = r 2 – 2
If possible, let P(x 1 , y 1 ), Q(x 2 , y 2 ) and R(x 3 , y 3 ) be
three distinct rational points on circle C. Then,
2 2
x1 + y1
− 2 2 y 1 = r 2 – 2 ...(ii)
2 2
x 2 + y2
− 2 2 y 2 = r 2 – 2 ...(iii)
2 2
x + y 2 2 y 3 = r 2 – 2 ...(iv)
3 3 −
We claim that at least two y 1 , y 2 , and y 3 are distinct.
For if y 1 = y 2 = y 3 , then P, Q and R lie on a line
parallel to x-axis and a line parallel to x-axis does not
cross the circle in more than two points. Thus, we
have either y 1 ≠ y 2 or, y 1 ≠ y 3 or, y 2 ≠ y 3 .
Subtracting (ii) from (iii) and (iv), we get
2 2 2 2
(x 2 + y2)
– (x1 + y1
) – 2 2 (y 2 – y 1 ) = 0
2 2 2 2
and, (x3 + y3
) – (x1 + y1
) – 2 2 (y 3 – y 1 ) = 0
⇒ a 1 – 2b 1 = 0 and a 2 – 2b 2 = 0 ...(v)
where,
2 2 2 2
a 1 = (x 2 + y2)
– (x1 + y1
) , b 1 = 2(y 2 – y 1 )
2 2 2 2
a 2 = (x3 + y3
) – (x1 + y1
) , b 2 = 2(y 3 – y 1 )
Clearly, a 1 , a 2 , b 1 , b 2 are rational numbers as x 1 , x 2 ,
x 3 , y 1 , y 2 , y 3 are rational numbers.
Since either y 1 ≠ y 2 or, y 1 ≠ y 3
∴ Either b 1 ≠ 0 or, b 2 ≠ 0
If b 1 ≠ 0, then
a 1 – 2 b 1 = 0 [From (v)]
⇒
a
b
1
1
= 2,
a1
which is not possible because is a rational
b1
number and 2 is an irrational number.
If b 2 ≠ 0, then
a 2
a 2 – 2b 2 = 0 ⇒ = 2,
b
2
a 2
which is not possible because is a rational
b2
number and 2 is an irrational number.
Thus, in both the cases we arrive at a contradiction.
This means that our supposition is wrong. Hence,
there can be at most two rational points on circle C.
15. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices P, Q and S lie on the lines y = a,
x = b and x = –b, respectively. Find the locus of the
vertex R.
[IIT-1996]
Sol. Let the coordinates of R be (h, k). It is given that P
lies on y = a. So, let the coordinates of P be (x 1 , a).
Since PQ is parallel to the line y = mx. Therefore,
Slope of PQ = (Slope of y = mx) = m
1
And, Slope of PS = –
(Slope of y = mx)
= – m
1
[∴ PS ⊥ PQ]
Now, equation of PQ is
y – a = m(x – x 1 )
...(i)
XtraEdge for IIT-JEE 12 JULY 2010

x = –b
x´ S
(0, – b)
(0, a)
R
O
y
y = 0
P
y´
x = b
Q
(0, b) x
It is given that Q lies on x = b. So, Q is the point of
intersection if (i) and x = b.
Putting x = b in (i), we get
y = a + m(b – x 1 )
So, coordinates of Q are (b, a + m(b – x 1 )).
Since PS passes through P(x 1 , a) and has slope – m
1 .
So, Equation of PS is y – a = – m
1 (x – x1 ) ...(ii)
It is given that S lies on x = – b. So, S is the point of
intersection of (ii) and x = –b.
Solving (ii) and x = – b, we get y = a + m
1 (b + x1 )
⎛ 1 ⎞
So, coordinates of S are ⎜− b,a
+ (b + x1
) ⎟
⎝ m ⎠
1
k − a − (b + x1)
Now, Slope of RS =
m
= m
h + b
But RS is parallel to PQ.
1
k − a − (b + x1)
∴
m
= m
h + b
⇒ b + x 1 = m(k – a) – m 2 (h + b) ...(iii)
Similarly,
k − a − m(b − x1)
Slope of RQ =
h − b
But, RQ is perpendicular to PQ whose slope is m.
k − a − m(b − x1)
1
∴
= –
h − b m
1 1
⇒ b – x 1 = (k – a) + m
2 (h – a) ...(iv)
m
We have only one variable x 1 . To eliminate x 1 , add
(iii) and (iv) to obtain
⎛ 1 ⎞
2b = (k – a) ⎜m + ⎟ – m 2 1
(h + b) +
⎝ m ⎠
2 (h – b)
m
⎛ ⎞
⇒ 2b = (k – a) ⎜
m 2 + 1 ⎛
4
⎞
⎟ – h ⎜
m + 1 ⎛
4
⎞
⎟ – b ⎜
m + 1
⎟
⎝ m
2
⎠ ⎝ m
2
⎠ ⎝ m ⎠
⎛ ⎞
⇒ (k–a) ⎜
m 2 2 2
2
+ 1
⎟
h (m −1)(m
+ 1) b (m + 1)
–
–
2
2
⎝ m ⎠ m
m
2
=0
h(m 2 −1)
b(m 2 + 1)
⇒ (k – a) – – = 0
m m
⇒ m(k – a) – h(m 2 – 1) – b(m 2 + 1) = 0
Hence, the locus of R(h, k) is
m(y – a) – x(m 2 – 1) – b(m 2 + 1) = 0
BEWARE OF THE BATTERIES
THAT YOU USE!
Have you ever noticed that the batteries
are becoming smaller and smaller day after day
Many scientists and researchers have been finding
the effective way to shrink the batteries into the
smallest size as possible!
In this case, Jae Kwon, an assistant professor of
Electrical and computer engineering has recently
developed a nuclear energy source, which is smaller,
lighter and more efficient than the common
batteries.
Kwon’s described that the new discovered
radioisotope battery can provide power density as
much as six orders of magnitude higher than
chemical batteries.
Kwon and his research team members have been
cooperated and working on building a small nuclear
battery. According to the information, the
radioisotope batteries are having the size and
thickness of a penny, but it’s powerful enough to
power various micro or nanoelectromechanical
systems.
Even though the nuclear power sources have always
been a safety concern, they’ve claimed to be safe, as
the nuclear power sources have been used for
powering many types of devices, including the pacemakers,
space satellites and underwater systems.
Kwon’s battery is in a liquid semiconductor rather
than a solid semiconductor, as he believed that the
liquid semiconductor can overcome the problem,
where the lattice structure of the semiconductor
being damaged, if it’s in the solid semiconductor
form!
XtraEdge for IIT-JEE 13 JULY 2010

Physics Challenging Problems
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.
Solutions will be published in next issue
1. A circuit consisting of a constant e.m.f. ‘E’, a self
induction ‘L’ and a resistance ‘R’ is closed at
t = 0. The relation between the current I in the
circuit and time t is as shown by curve ‘a’ in the
figure. When one or more of parameters E, R and
L are changed, the curve ‘b’ is obtained. The
steady state current is same in both the cases.
Then it is possible that
I
(a)
(b)
t
(A) E and R are kept constant and L is increased
(B) E and R are kept constant and L is decreased
(C) E and R are both halved and L is kept constant
(D) E and L are kept constant and R is decreased
2. Consider a resistor of uniform cross section area
connected to a battery of internal resistance zero.
If the length of the resistor is doubled by
stretching it then
(A) current will become four times
(B) the electric field in the wire will become half
(C) the thermal power produced by the resistor
will become one fourth
(D) the product of the current density and
conductance will become half
3. In front of an earthed conductor a point charge +q
is placed as shown in figure
+q
Set # 3
By : Dev Sharma
Director Academics, Jodhpur Branch
Passage # (Q. No. 4 to Q. No. 6)
Resistance value of an unknown resistor is
calculated using the formula R= V/I where V and
I be the readings of the voltmeter and the
ammeter respectively. Consider the circuits
below. The internal resistances of the voltmeter
and the ammeter (R V and R G respectively) are
finite and non zero.
V
V
A
A
R
R
E r E r
Fig. (A)
Fig. (B)
Let R A and R B the calculated values in the two
cases A and B respectively.
4. The relation between R A and the actual value R is
(A) R > R A (B) R < R A
(C) R = R A (D) dependent upon E and r
5. The relation between R B and the actual value R is
(A) R< R B (B) R > R B
(C) R = R B (D) dependent upon E and R
6. If the resistance of voltmeter is R V = 1 KΩ and
that of ammeter is R G = 1Ω, the magnitude of the
percentage error in the measurement of R (the
value of R is nearly 10 Ω) is
(A) zero in both cases
(B) non-zero but equal in both cases
(C) more in circuit A
(D) more in circuit B
Passage # (Q. No. 7 to Q. No. 8)
The figure shows the interference pattern obtained in
a double-slit experiment using light of wavelength
600 nm. 1,2,3,4 and 5 are marked on five fringes.
(A) On the surface of conductor the net charge is
always negative
(B) On the surface of conductor at the same points
charges are negative and at some points charges
may be positive distributed non uniformly
(C) Inside the conductor electric field due to point
charge is non-zero
(D) None of these
7. The third order bright fringe is
(A) 2 (B) 3 (C) 4 (D) 5
8. Which fringe results from a phase difference of 4 π
between the light waves incidenting from two slits
(A) 2 (B) 3 (C) 4 (D) 5
XtraEdge for IIT-JEE 14 JULY 2010

1.[C]
8
Questions were Published in June Issue
Initially the potential at centre of sphere is
⎛ x − e ⎞ ⎛ x − 0 ⎞
4 ⎜ ⎟ + ⎜ ⎟ = 0
1 Q 1 2Q 1 3Q
⎝ r ⎠ ⎝ r ⎠
VC
= + =
4πε0
x 4πε0
x 4πε0
x
5x = 4e x = 4e/5
2
After the sphere grounded, potential at centre
x 2Bωa
and
⎧ e 4e ⎫
i = =
⎨i
= = ⎬
becomes zero. Let the net charge on sphere finally
r 5 ⎩ r + r / 4 5r ⎭
be q.
1 q 1 3Q 3Qr
6. (A) → R , (B) → Q,S, (C) → P, (D) → Q,R
∴ + = 0 or q =
→
4πε0
r 4πε0
x
x
(A) F = constant and → u × →
F = 0
3Qr
∴The charge flowing out of sphere is
x
2. [D] The velocity is maximum at mean position.
Hence the magnetic force on block is maximum,
at its mean position.
The magnetic force on the block while it crosses
the mean position towards right and left is as
shown
N 1
v max v max
qv max B
Mg + qv max B mg
Case-1 Case-2
Hence normal reaction is maximum in case-1 and
minimum in case-2. Hence correct option is D.
3. (A) → Q,R, (B) → P,S, (C) → P,R, (D) → Q,S
4. [A,B,C]
Total charge =
∫ Idt = area under curve = 10C
Average current =
∫
∫
Idt
= 5A
dt
Total heat produced =
∫ I 2 Rdt
2
2 200
=
∫
( −5t
+ 10) .1.dt = J
3
0
Maximum power = I 2 R, when I is maximum current
= 100 × 1 =100W.
5. [B,D] Equivalent circuit
r/2 ε r/2
x
x
ε r
x
2 2
Bωr
Bωa
Induced emf e = = (∴ Radius = a)
2 2
By nodal equation, nodal
Solution
Set # 2
Physics Challenging Problems
N 2
Therefore initial velocity is either in direction of
constant force or opposite to it. Hence the particle
will move in straight line and speed may increase
or decrease.
XtraEdge for IIT-JEE 15 JULY 2010
7. [B]
8. [B,C]
→ →
→
and F = constant
(B) u . F = 0
Initial velocity is perpendicular to constant force,
hence the path will be parabolic with speed of
particle increasing.
→ →
(C) v . F = 0 means instantaneous velocity is
always perpendicular to force. Hence the speed
→
|
will remain constant. And also | F = constant.
Since the particle moves in one plane, the
resulting motion has to be circular.
→
∧
∧
→
(D) u = 2 i−
3 j and a = 6 i − 9 j . Hence initial
velocity is in same direction of constant
acceleration, therefore particle moves in straight
line with increasing speed.
C r
r θ
R
q
φ
total
O
v 0
v 0
r
mv0
r = ; R = 2r sin θ
qB
=
qBR
2m sin
×
× × × ×
× h
dr
⎛ µ 0N
× h R + b ⎞
= ⎜ log ⎟ × I
⎝ 2π
R ⎠
dφ e = total
dt
θ
∧
∧
µ
dφ
= B × hdr =
max
0
NIhdr
2πr
× sin ωt

PHYSICS
Students'Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
1. A particle of mass m moves along a horizontal circle
of radius R such that normal acceleration of particle
varies with time as a n = Kt 2 , where K is a constant.
Calculate
(i) tangential force on particle at time t,
(ii) total force on particle at time t,
(iii) Power developed by total force at time t, and
(iv) average power developed by total force over
first t second.
Sol. Since, the particle is moving along a circle,
therefore, its normal acceleration is centripetal
acceleration i.e. v 2 /R, where v is velocity of particle
at time t.
v 2
∴ = Kt 2 or v = t. KR
R
…(1)
Due to centripetal acceleration, particle follows a
circular path but due to it velocity its magnitude
does not change. Velocity magnitude increases due
to tangential acceleration alone.
d
∴ Tangential acceleration, a t = . v = KR
dt
∴ Tangential force, F t = ma t = m KR
Ans. (i)
Normal force, F n = ma n = mKt 2
∴ Resultant force on particle,
2 2
F = F + F
t
= m K(R + K t )
Ans. (ii)
Since, power developed by force → F is given by
→ →
v
P = F , therefore, power developed by normal
force F n is always zero because its direction is
always perpendicular to the instantaneous direction
of motion of the particle. Hence, power is developed
by tangential force alone. Figure
F t
F n
i.e.
P = F → → t v = mkRt Ans.(iii)
Since, resultant force equals (mass × acceleration),
therefore, resultant force is used to accelerate the
v
n
4
body. It means that velocity of the body increases
due to resultant force. Hence power developed by
resultant force is used to increase kinetic energy of
the body.
∴ Average power developed by resultant
force = Average rate of increase of KE
Initial kinetic energy (at t = 0), E 0 = 0
Kinetic energy at time t, E = 2
1 mv
2
= 2
1 mKRt
2
…(2)
∴
E – E
Average power =
t
1
= mKRt` 2 Ans. (iv)
2. Figure Shows a particle of mass m = 100 gm,
attached with four identical springs, each of length
l = 10 cm. Initial tension in each spring is
F 0 = 25 newton. Neglecting gravity, calculate period
of small oscillations of the particle along a line
perpendicular to the plane of the figure
B
A
P
m
D
Sol. Let the particle be displaced slightly through x along
a line normal to plane of the figure. Then each
spring is further elongated. Since, springs are
identical, therefore, increase in tension of each
spring will be the same. Let this increase be dE 0 .
l
l
A
θ
θ
C
C
(F 0 + dF 0 ) (F 0 + dF 0 )
P
First considering forces exerted by spring AP and
CP only as shown in Figure.
Restoring force produced by these two springs = (F 0
+ dF 0 ) 2 sin θ
x
Since x is very small, therefore, sin θ ≈
l
Neglecting product of very small quantities,
restoring force produced by these two springs
C
XtraEdge for IIT-JEE 16 JULY 2010

= 2F 0
l
x
Similarly, restoring force produced by two
remaining springs BP and DP will also be equal to
⎛ 2F
⎞
⎜ 0 x
⎟
⎝ l ⎠
⎛ 2F
⎞
∴ Resultant restoring force, F = 2 × ⎜ 0 x 4F
⎟ = 0
.x
⎝ l ⎠ l
∴ Restoring acceleration is directly proportional to
displacement x, therefore, the particle executes
SHM,
∴ Its period T = 2π
displacement
acceleration
ml
or T = 2π =
4F 0
ml
π = 0.02π sec
F 0
Ans.
3. Find the electric potential, at any point on the axis of
a uniformly charged circular disc, whose surface
charge density is σ, radius, a.
Sol. Let us consider a small elemental thin ring of width
dy.
Area of the ring = 2πy dy
Charge on this elemental ring = (2πy dy)σ
Again, we can consider that this ring is divided into
a large number of small elements. Each such
element e is at the same distance from P. Hence,
potential produced by this ring of width dy at the
point P is given by dV, where
1 2πyσdy
dV =
4πε
2 2
0 r + y
e
y
2 2
r + y
O
a
r
P
Again, each ring as we go from centre to rim,
produces different contributions. Since the distance
of each ring from P changes as y increases from 0 to
a, hence, total potential produced by the whole ring,
a
V =
∫
dV
y = 0
a 2πσ
ydy
=
∫
.
0 4πε
2 2
0 r + y
σ a ydy
=
2ε
∫ 2
r +
0 2
0 y
Put (r
2 2
+ y ) = P
∴ r 2 + y 2 = p 2
or 2ydy = 2pdp
∴
∫
ydy pdp
=
∫
r 2 + y
2 p
σ a ydy
∴
2ε
∫ 2
r +
=
∫ dp = p = 2 2
r + y
0 2
0 y
=
σ
2ε
0
⎡
⎢⎣
=
r
2
σ
ε
0
⎡
⎢⎣
+ a
σ
∴ V =
⎡ 2 2
r + a − r
⎤
2ε0
⎢⎣ ⎥⎦
As a special case, if r >> a
2
r
1/ 2
2
− r
⎤
⎥⎦
+ y
⎡
2
2 2 ⎛ a ⎞ ⎤ ⎡
2
1 a ⎤
r + a = r ⎢1
+ ⎜ ⎟ ⎥ ≈ r ⎢1
+
2
⎥
⎢⎣
⎝ r ⎠ ⎥⎦
⎢⎣
2 r ⎥⎦
σ a 2 π q
or, V = . × =
2ε 0 2r
π 4πε 0 r
[Q πa 2 = A and Aσ = q]
i.e., the result is the same as if all the charge is
concentrated at the centre of the ring.
4. Three concentric, conducting spherical shells A,B
and C have radii a = 10 cm, b = 20 cm and c = 30
cm respectively. The innermost shell A is earthed
and charges q 2 = 4µC and q 3 = 3 µC are given to
shells B and C respectively. Calculate charge q 1
induced on shell A and energy stored in the system.
Sol. System of three concentric shells is as shown in
Figure(A) Since, Shell A is earthed, therefore its
potential is zero. But its potential is
∴
q 3
q 2
q 1
V =
a
4πε
C
B
A
(A)
b
⎡q1
⎢
⎣ a
c
q 2
+
b
C B
q ⎤
+
c
⎥ ⎦
1 3
0
(–3µC)
A
a
2
⎤
⎥⎦
(B)
b
a
0
(+4µC)
(–1µC) (+1µC)
(+3µC)
⎛ q1 q 2 q3
⎞
⎜ + + ⎟ = 0
⎝ a b c ⎠
or q 1 = – 3µC
Ans.
Now, charges on different surface will be as shown
in Figure(B) to calculate energy stored in the
system, it can be as considered in three parts :
(i) a spherical capacitor having radii a and b and
having charge |q 1 | 3µC.
c
XtraEdge for IIT-JEE 17 JULY 2010

∴
4πε0 Its capacitance, C 1 =
ab
(b – a)
(ii) a spherical capacitor having radii b and c and
having charge (q 2 + q 1 ) = 1 µC.
4πε0 Its capacitance. C 2 =
bc
(c – b)
(iii) an isolated sphere of radius c and having charge
(q 3 + q 2 + q 1 ) = 4µC
Energy stored in the system
2
1
1
2
q ( q1
+ q 2)
( q1
+ q 2 + q3)
= +
+
2C 2C2
2C3
= 0.45 joule Ans.
5. In the circuit shown in Figure emf of each battery is
E = 20 volts and capacitance is C 1 = 5 µF, C 2 = 3 µF
and C 3 = 6 µF. Calculate charge on capacitor C 3
when switch S is closed and steady state is reached.
Calculate also, heat generated in the circuit.
2
E
C 1 C 3 C 1
+ – + – + –
+ + +
– C 2 – (q 1 –q 2 ) C 2
q
– 2
(B)
S
+
– E
Applying Kirchhoff's voltage law on left mesh, of
Figure (B)
q1
q1
– q 2
+ – E = 0 …(1)
C1
C2
For middle mesh,
q 2 q 2 q1
– q 2
+ – = 0 …(2)
C3
C2
C2
From equation (1) and (2), q 1 = 50 µC and
q 2 = 20 µC
Now consider the circuit when switch S is closed
and steady state is reached. The circuit will be as
shown in Figure (C)
Sol.
E + –
C 1
C 2
C 3 C 1
C 2
S
+
–
E
When Switch S is closed and steady state is
reached, the circuit becomes symmetric about the
dotted line shown in Figure(A).
E + –
C 1
C 2
C 3 C 1
C 2
(A)
+
–
E
Right part of the circuit is exactly mirror image of
the left part. Hence, charges on both plates of
capacitor C 3 should be identical. But charges on
plates of a capacitor are always opposite to each
other. It means one of the plates is always positively
charged and the other is negatively charged. Both
these conditions can be satisfied only if charge on
capacitor C 3 is zero.
To calculate heat generated in the circuit initial and
final charges on all the capacitors must be known.
First analyse the circuit when switch S was open
Charges on capacitors will be as shown in
Figure (B)
E
C C 3 1 C 1
+ – – + q
– C + +
2 C + q
–
q 2 – q
(C)
+
–
E
Applying Kirchhoff's voltage law on left mesh of
Figure(C)
q q + – E = 0 or q = 37.5 µC.
C 1 C2
After shorting of switch S, increase in charge on
capacitor C 1 of
Left mesh, ∆q 1 = (q – q 1 ) = – 12.5 µC
That for capacitor C 2 of left mesh,
∆q 2 = q – (q 1 –q 2 ) = 7.5 µC
That for capacitor C 3 , ∆q 3 = 0 – q 2 = – 20µC
That for capacitor C 2 of right mesh,
∆q 4 = (q –q 2 ) = 17.5 µC
That for capacitor C 1 or right mesh,
∆q 5 = q – 0 = 37.5 µC
Since, heat generated in the circuit is given by
2
( ∆q)
H = ∑
2C
H =
(∆q1)
2C
1
2
+
(∆q 2)
2C
2
2
+
(∆q3)
2C
(∆q 4)
(∆q5)
+ +
2C2
2C1
= 250 × 10 –6 joule Ans.
3
2
2
2
XtraEdge for IIT-JEE 18 JULY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
Capacitor-1
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Capacitance :
Whenever charge is given to a conductor of any
shape its potential increases. The more the charge (Q)
given to the conductor the more is its potential (V)
i.e. Q ∝ V
⇒ Q = CV
where C is constant of proportionality called
capacitance of the conductor C = Q/V, C = Q
SI unit of capacitance is farad (F) and 1 F = 1
coulomb/volt (1CV –1 )
Energy stored in a charged capacitor :
1 2 Q 2 1
W = CV 0 = = QV0
2 2C 2
Capacitance of an isolated sphere :
Let a conducting sphere of radius a acquire a
potential V when a charge Q is given to it. The
potential acquired by the sphere is
Q Q
V = ⇒ C = = 4πε0 a
4πε 0 a V
Charge sharing Between two charged conductors :
C 1
V 1
C 2
V 2
q 1 = C 1 V 1 q 2 = C 2 V 2
(Initially)
C 1
V
C 2
V
q´1 = C 1 V q´2 = C 2 V
(Finally)
C1V1
+ C2V2
V =
C1
+ C2
There is always a loss in energy during the sharing
process as some energy gets converted to heat.
Loss = – ∆U =
1 ⎛ C ⎜
1C2
2 ⎝ C 1 C
+ 2
⎞
⎟ (V 1 – V 2 ) 2
⎠
Capacitor or Condenser :
An arrangement which has capability of collecting
(and storing) charge and whose capacitance can be
varied is called a capacitor (or condenser)
The capacitance of a capacitor depends.
(a) directly on the size of the conductors of the
capacitor.
(b) directly on the dielectric constant K of the
medium between the conductors.
(c) inversely on the distance of separation between
the conductor.
Principle of a condenser :
Consider a conducting plate A which is given a
charge Q such that its potential rises to V. Then
C = Q/V
Let us place another identical conducting plate B
parallel to it such that charge is induced on plate B
(as shown in figure).
A
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Q
If V – is the potential at A due to induced negative
charge on B and V + is the potential at A due to
induced positive charge on B, then
A B
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
Q Q
C´ = =
V´ V + V+ − V−
Since V´ < V (as the induced negative charge lies
closer to the plate A in comparison to induced
positive charge).
⇒ C´ > C
Further, if B is earthed from the outer side
(see figure) then V n = V – V – as the entire positive
charge flows to the earth. So
C" =
Q Q =
V − V−
n
V
+
+
+
+
+
+
+
+
⇒ C n >> C
So, if an identical earthed conductor is placed in the
viscinty of a charged conductor then the capacitance
of the charged conductor increases appreciably. This
is the principle of a parallel plate capacitor.
XtraEdge for IIT-JEE 19 JULY 2010

Parallel Plate Capacitor :
A
–σ
d
B
+ σ
A
+
+
A = Area of plate +
d = Separation +
between the +
plates +
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
B
σ
E = =
σ
ε kε0
It consists of two metallic plates A and B each of area
A at separation d. Plate A is positively charged and
plate B is earthed. If K is the dielectric constant of
the material medium and E is the field that exists
between the two plates, then
⇒
σ σ
E = = ε Kε0
V =
d
q
K
ε 0
A
⎧ V
⎨Q
E = and σ =
⎩ d
q Kε
⇒ C = = 0 A
V d
If medium between the plates is air or vacuum, then
K = 1
⇒ C 0 =
ε 0 A
d
Special Case I :
When the space between the parallel plate capacitor
is partly filled with a dielectric of thickness t(

1
⇒ E =
ε (σ – σ Case II : When battery is connected to A, then C 1
p) ...(1)
and C 2 are in series.
0
1 1 1
σ
⇒ = +
Also E =
...(2)
C C1
C2
Kε 0
Compare (1) and (2), we get
1 b − a 1 1
⇒ = +
C ab 4πε 1
ε (σ – σ σ
0 K 4πε 0 Kb
p) =
0 Kε 0
1 1 ⎛ b − a ⎞
⇒ = ⎜ + 1 ⎟⎠
⎛ 1 ⎞
C 4πε0Kb
⎝ a
⇒ σ p = σ ⎜1
− ⎟
⎝ K ⎠ 1 1 ⎛ b ⎞
⇒ = ⎜ ⎟⎠
q p q ⎛ 1 ⎞
C 4πε0Kb
⎝ a
⇒ = ⎜1
− ⎟⎠
A A ⎝ K ⇒ C = 4πε 0 Ka
⎛ 1 ⎞
Case III : When battery connected to A and B is
⇒ q p = q ⎜1
− ⎟ earthed. Then C
⎝ K 2 can be omitted as it will not receive
⎠ any charge.
Spherical capacitor :
So, C = C 1
B
b
⎛ ab ⎞
⇒ C = 4πε 0 K ⎜ ⎟
⎝ b − a ⎠
a C
C 2
1
Case IV : When battery connected to B and A is
A
open circuited (or made non conducted) then C 1 can
be omitted (as it is open circuited). So,
let C 1 be the capacitance in between the two
C = C 2 ⇒ C = 4πε 0 Kb
conductors and C 2 be capacitance out side both.
Cylindrical capacitor :
To find C 1 :
Let inner cylinder be given a charge per unit length
Imagine the outer surface of B to be earthed. Then –q
is the charge induced on the inner surface of B.
⎛ q ⎞
of λ ⎜ = ⎟ . A charge – q is induced on length l at
If V is the potential difference between the two
⎝ l ⎠
surfaces, then
inner surface of outer cylinder
q −q
V = +
4πε 0 Ka 4πε 0 Kb
b
q
a
–q
q ⎛ 1 1 ⎞
⇒ V = ⎜ − ⎟
4πε 0 K ⎝ a b ⎠
l
q ⎛ ab ⎞
⇒ C = = 4πε0 K ⎜ ⎟ ...(1)
V ⎝ b − a ⎠
To find C,
λ
E = for a < r < b
Imagine A to be made open circuited (i.e. made non
2πε0r
conducting), then
dV λ
C 2 = 4πε 0 Kb ...(2)
⇒ – = dr 2πε0Kr
Case I : When battery is connected to B and A is
outer
earthed. Then C 1 and C 2 are in parallel
surface
r=
b
λ dr
⇒ C = C 1 + C 2
⇒
∫
dV = –
2πε
∫
⎛ ab ⎞
inner
0 K r r=
a
⇒ C = 4πε 0 K ⎜ ⎟ + 4πε 0 Kb
surface
⎝ b − a ⎠
λ ⎛ b ⎞
⎛ ⎞
⇒ C = 4πε 0 K ⎜
b 2 ⇒ V inner surface – V outer surface = log e ⎜ ⎟
⎟
2πε0K
⎝ a ⎠ ⎝ b − a ⎠
Since, inner surface is at higher potential and outer at
lower potential, so
XtraEdge for IIT-JEE 21 JULY 2010

1
Energy U' = C'V'
2
–q
2
b
(b) If a dielectric slab (dielectric constant k) is
a
introduced between two plates then
+q
E
V' = V, C' = kC, q' = kq, E' = ; U' = KU k
(p.d.) will not change)
Gaussian surface
λ ⎛ b ⎞
⇒ V outer surface – V inner surface = log e ⎜ ⎟ Solved Examples
2πε0K
⎝ a ⎠
q ⎛ b ⎞
1. A capacitor of 20 µF and charged to 500 volt is
⇒ V inner surface – V outer surface = log e ⎜ ⎟ connected in parallel with another capacitor of 10 µF
2πε 0 lK
⎝ a ⎠
charged to 200 volt. Find the common potential.
q
2πε0 K
⇒ C =
=
l
Sol. Charge on one capacitor q 1 = C 1 V 1
V inner surface − Vouter
surface ⎛ b ⎞
∴ q 1 = 20 × 10 –6 × 500 = 0.01 coulomb
loge⎜
⎟
Charge on second capacitor
⎝ a ⎠
q 2 = 10 × 10 –6 × 200 = 0.002 coulomb
2πε
K
⇒ C =
0l
The charge on the two capacitors
⎛ b ⎞
q = q 1 + q 2 = 0.01 + 0.002 = 0.003 coulomb
loge⎜
⎟
Total capacity C = C
⎝ a ⎠
1 + C 2
= 20 × 10 –6 + 10 × 10 –6
1
(A) Energy stored in a capacitor E = CV
2 1 = 30 × 10 –6 Farad.
= QV
2 2
Common potential = q/C
2
1 Q
1
0.012
= and energy stored per unit volume = ε0 E 2
= = 400 Volt.
−6
2 C
2 30×
10
Note: The energy is stored in a capacitor is in the
2. A battery of 10V is connected to a capacitor of
form of electric field between the plates.
capacity of 0.1 F. The battery is now removed and
(B) A parallel plate capacitor is charged by a battery and
this capacitor is connected to a second uncharged
then the capacitor disconnected from the battery
capacitor. If the charge distributes equally on these
(a) If the distance between plates of the capacitor is
two capacitors, find the total energy stored in the two
increased then the new parameter of the capacitors as
capacitors. Further, compare this energy with the
compared to the previous parameters is
initial energy stored in the first capacitor.
Sol. The initial energy stored in the first capacitor.
ε
q' = q; C' = 0 A q '
, V' = ,
1
d' C'
U 0 = CV
2
2
V '
1
E' = (charge will not change) Energy = C'V'
2
1
= × 0.1 × (10) 2 = 5.0 J
d'
2 2
If a dielectric slab (dielectric constant k) is
When this capacitor is connected to the second
introduced between the plates then
uncharged capacitor, the charge distributes equally.
This shows that the capacitance of the second
V E
U
q' = q, C' = kC, V' = , E' = U'(Energy) = capacitor is also C. The voltage across each capacitor
k k k will be V/2. If U be the energy stored in the two
(charge will not change)
capacitors, then
2
2
1 ⎛ V ⎞ 1 ⎛ V ⎞
(C) A parallel plate capacitor is charged by a battery.
U = C ⎜ ⎟⎠ + C ⎜ ⎟⎠ 2 ⎝ 2 2 ⎝ 2
(a) If the distance between plates of the capacitor is
Increased (with the battery connected) then the new
1
= CV 2 = 2.5 J
parameters of the capacitors as compared to the
4
previous parameters is
U 2.5 1 = =
ε
V' = V, C' = 0 A
V '
U
, q' = C'V', E' = , 0 5. 0 2
d'
d
XtraEdge for IIT-JEE 22 JULY 2010

3. Two isolated metallic solid spheres of radii R and 2R
are charged such that both of these have same charge
density σ. The spheres are located far away from
each other, and connected by a thin conducting wire.
Find the new charge density on the bigger sphere.
Sol. Charge on smaller sphere
Q 1 = 4πR 2 . σ
Charge on bigger sphere
Q 2 = 4π(2R) 2 σ = 16πR 2 σ
∴ Total charge Q = Q 1 + Q 2 = 20πR 2 σ ...(1)
Capacitances of two spherical conductors are
C 1 = 4πε 0 R and C 2 = 4πε 0 (2R)
∴ Total capacitance
C = C 1 + C 2 = 12πε 0 R ...(2)
After connection, the common potential V is given by
Q 20πR
σ 5Rσ
V = = =
C 12πε0R
3ε
0
New charge on bigger sphere
Q 2´
= C 2 V
Surface density
2
= 4πε 0 R(2R) × (5Rσ/3ε 0 ) =
40πR
3
2 σ
⎛
2
40 R ⎞
⎜
π σ
⎟
Q ´
σ 2´ = 2
3
=
⎝ ⎠ 5
=
2 σ.
surface area 4π(2R)
6
4. A 8 µF capacitor C 1 is charged to V 0 = 120 volt. The
charging battery is then removed and the capacitor is
connected in parallel to an uncharged 4 µF capacitor
C 2 (a) what is the potential difference V across the
combination (b) What is the stored energy before
and after the switch S is thrown
S
V 0 C1 C 2
Sol. (a) Let q 0 be the charge on C 1 initially. Then
q 0 = C 1 V 0
when C 1 is connected to C 2 in parallel, the charge q 0
is distributed between C 1 and C 2 . Let q 1 and q 2 be the
charges on C 1 and C 2 respectively. Now let V be the
potential difference across each condenser.
Now q 0 = q 1 + q 2
or C 1 V 0 = C 1 V + C 2 V
C1
8µF
∴ V = V 0 = (120 V)
C + C 8µF + 4µF
1
= 80 volt.
2
(b) Initial energy stored
U 0 = 2
1
C1 V 0
2
= 2
1 (8 × 10 –6 ) (120) 2
= 5.76 × 10 –2 Joule
Final energy stored
U = 2
1
C1 V 2 + 2
1
C2 V 2
= 2
1 (8 × 10 –6 )(80) 2 + 2
1 (4 × 10 –6 )(80) 2
= 3.84 × 10 –2 joule
Final energy is less than the initial energy. The loss
of energy appears as heat in connecting wires.
5. Calculate the capacitance of a parallel plate
condenser, with plate area A and distance between
plates d, when filled with a dielectric whose dielectric
constant varies as
d
ε(x) = ε 0 + βx
0 < x < 2
d
ε(x) = ε 0 + β(d – x) < x < d
2
For what value of β would the capacity of the
condenser be twice that when it is without any
dielectric.
Sol. The capacitance in series is given by
1 1 1 = +
C´ C1
C2
=
∴
1 1 ⎡
⎤
= × ⎢
⎥
C´ A ∫
d / 2 dx d dx
+
⎣
0 ε + β ∫
0 x d / 2 ε0
+ β(d
− x)
⎦
=
1 [{log(ε0 + βx)} d / 2
0 – {log(ε 0 + β(d – x) d / 2
Aβ
d ]
1 ⎡⎧
⎛ d ⎞ ⎫ ⎧ ⎛ d ⎞⎫⎤
⎢⎨log⎜ε
0 + β ⎟ − log ε 0 ⎬ − ⎨log
ε 0 − log⎜ε
0 + β ⎟⎬⎥
Aβ ⎣⎩
⎝ 2 ⎠ ⎭ ⎩ ⎝ 2 ⎠⎭
⎦
2 ⎡ ⎛ d ⎞ ⎤
= ⎢log⎜ε0 + β ⎟ − log ε0
⎥
Aβ
⎣ ⎝ 2 ⎠ ⎦
2 ⎛ ε
= log
Aβ
⎟ ⎞
⎜
0 + βd / 2
⎝ ε0
⎠
The capacitance C of a condenser without dielectric
is given by
Aε
C = 0
d
According to the question, C´ = 2C
2 ⎛ ε ⎞
∴ log ⎜
0 + βd / 2 d
⎟
Aε
=
0 ⎝ ε0
⎠ 2ε 0 A
β =
4ε 0
⎛ ε log
d
⎟ ⎞
⎜
0 + βd / 2
⎝ ε0
⎠
XtraEdge for IIT-JEE 23 JULY 2010

XtraEdge for IIT-JEE 24 JULY 2010

PHYSICS FUNDAMENTAL FOR IIT-JEE
Friction
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
Friction :
Whenever there is a relative motion between two
surfaces in contact with each other, an opposing force
comes into play which forbids the relative motion of
two bodies. This opposing force is called the force of
friction.
Ex. : If a book on a table slides from left to right
along the surface of a table, a frictional force directed
from right to left acts on the book.
Frictional force may also exist between the surfaces
when there is no relative motion. Frictional forces
arise due to molecular interactions.
Static and Kinetic Friction :
The frictional force between two surface before the
relative motion actually starts is called static
frictional force or static friction, While the
frictional force between two surfaces in contact and
in relative motion is called kinetic frictional force or
kinetic friction.
Static friction is a self adjusting force and it adjusts
both in magnitude and direction automatically. Its
magnitude is always equal to external effective
applied force, tending to cause the relative motion
and its direction is always opposite to that of external
applied force.
So, when a body is not in motion or equilibrium, then
Force of static Friction = Applied External Force
Limiting friction, coefficients of friction and angle of
friction :
Consider a block resting on a rough horizontal
surface. The forces acting on the block are its weight
mg downwards and normal reaction N acting upward.
Such that N = mg.
f
R
θ
N
P(

Read the problem Carefully
Find the value of applied
force F app and limiting force
of Static Friction (f s )
If F app < f s then body does not
move and the force of friction
f = F app
if F app > f s
then body moves
If F app = f s the body is on the
verge of motion (still in
equilibrium)
EITHER
with Constant Velocity
OR
with an Acceleration a
F app – f k = 0
or F app = f k
F app – f k = ma
On a Level
Track
On an inclined
Plane
On a Level
Track
On an inclined
Plane
Applied Pull = f k
f k = mg sinθ
F app – µ k mg = ma
or F app = m(a + µ k g)
F app – µ k mg cosθ = ma
or F app = m(a + µ k g cosθ)
Laws of static and kinetic friction :
(a) The force of limiting friction is directly
proportional to normal reaction for the same two
surfaces in contact and acts opposite to direction
of pull.
The kinetic friction is also proportional to
normal reaction and acts opposite to direction of
instantaneous relative motion. The kinetic
friction is less than the static friction.
(b) The force of limiting (or static) friction is
independent of area of contact of bodies as long
as normal reaction remains the same.
The kinetic friction (to a good approximation) is
independent of velocity, provided the velocity is
neither too large nor too small.
Angle of repose (α)
This is concerned with an inclined plane on which a
block rests, exerting its weight on the plane.
The angle of repose α is the angle which an inclined
plane makes with the horizontal such that a body
placed on it is on the verge of motion (is just about to
loose the state of rest).
Under this condition the forces acting on the block are:
(a) its weight mg, downward,
(b) normal reaction N, normal to plane,
(c) a force of friction f s , parallel and tangential to
plane upward.
Taking α as angle of inclination of the plane with the
horizontal and resolving mg, parallel and
perpendicular to inclined plane, then for equilibrium,
we get
N = mg cos α and f s = mg sin α
⇒ tan α = N
f s
N
mg sinα
Tendency
mg mg cos α
to slide
α
Frictional force on a bicycle in motion :
(a) When a wheel is rotated about its axle without
sliding, the frictional force acting on it is the
rolling friction and it acts opposite to the direction
of tendency of motion of a points of its contacts
with the ground. In case the wheel rotates
clockwise and frictional force (f) on wheel is
forward. In case the wheel rotates anticlockwise,
the frictional force (f) on wheel is backward.
(b) When the bicycle is pedalled, the force exerted on
the rear wheel through the pedal-chain-axle
system is in backward direction, therefore force of
friction on rear wheel is forward. The front wheel
of cycle moves by itself in forward direction,
hence the force of friction of front-wheel is in
backward direction.
f s
XtraEdge for IIT-JEE 26 JULY 2010

(c) When the bicycle is not pedalled, no external
force is being exerted, both wheels move forward
by itself due to inertia and so the net frictional
force on both wheels is in backward direction.
Solved Examples
1. A block of mass 5 kg is placed on a slope which
makes an angle of 20º with the horizontal and is
given a velocity of 10 m/sec up the slope. Assuming
that the coefficient of sliding friction between the
block and the slope is 0.20, find how far the block
travels up the slope Take g = 10 m/sec 2 .
Sol. This situation is shown in fig.
R u = 10 m/s
mg sin 20º
x
mg
mg cos 20º
20º
The component of the weight perpendicular to plane
= mg cos 20º = 5 × 10 × 0.9397 = 46.98 N
The component of the weight parallel to the plane
= mg sin 20º = 5 × 10 × 0.3420 = 17.10 N
From figure R = mg cos 20º = 46.98 N
Here the coefficient of kinetic friction = 0.2
Thus the frictional force X = 0.2 × 46.98 = 9.39 N
The frictional force will be downward because the
motion is in the upward direction.
The resultant force parallel to the plane is given by
= X + mg sin 20º = 9.39 + 17.10 = 26.49 N
From Newton's law F = ma, i.e., 26.49 = 5 × a
26.49
∴ a = = 5.29 m/s 2 downward
5
When the block is given a velocity 10 m/s in the
upward direction we have
u = 10 m/s, v = 0, a = – 5.9 m/s 2 .
(Taking the direction up the plane as positive)
Let s be the distance traveled by the block.
Using the formula v 2 = u 2 + 2a s, we have
0 = (10) 2 – 2 × 5.29 × s
100
or s = = 9.45 m.
2×
5.29
2. A block is projected up with 10 m/s along a fixed
inclined plane of inclination 37º with the horizontal.
If the time of ascend from the point of projection is
half the time of descend to the same point, find the
distance travelled by the block during the up and
down journey.
Sol. Let µ, t 1 and t 2 be the coefficient of friction between
the plane and the block, time of ascend and time of
descend respectively.
The retardation while going up
⎛ 3 4 ⎞
a 1 = g (sin θ + µ cos θ) = 10 ⎜ + µ ⎟
⎝ 5 5 ⎠
The acceleration while descending
⎛ 3 4 ⎞
a 2 = g(sin θ – µ cos θ) = 10 ⎜ −µ
⎟
⎝ 5 5 ⎠
Now, s = distance of ascend = distance of descend.
As final velocity is zero, we have
0 = u – a 1 t 1 or u = a 1 t 1
Now s = a 1 t 1 2 – 2
1
a1 t 1 2 = 2
1
a1 t 1
2
s = a 1 t 1 2 = 2
1
a2 t 2 2 and t 2 = 2t 1
⎛ 3 4 ⎞
⎛
2
2 ⎜ −µ
⎟
⎞
∴ ⎜
a ⎛
⎟
t1
⎞
=
⎝ a
⎜
1 ⎠ t
⎟ or
⎝ 5 5 ⎠ ⎛ 1 ⎞
= ⎜ ⎟⎠
⎝ 2 ⎠
⎛ 3 4 ⎞ 2
⎜ + µ
⎝
⎟
⎝ 5 5 ⎠
Solving we get µ = (9/20)
⎛ 3 9 4 ⎞
Again a 1 = 10 ⎜ + × ⎟ = 9.6 m/sec 2
⎝ 5 20 5 ⎠
2
2
u (10)
∴ s = = = 5.21 meter
2a1
2×
9.6
So total distance = 2s = 10.42 metre
3. A block weighing 20 nt is at rest on a horizontal
table. The coefficient of static friction between block
and table is 0.50. (a) What is the magnitude of the
horizontal force that will just start the block moving
(b) What is the magnitude of a force acting upward
60º from the horizontal that will just start the block
moving (c) If the force acts down at 60º from the
horizontal how large can it be without causing the
block to move
Sol. (a) As shown in fig. the horizontal force F that will
just start the block moving is equal to the maximum
force of static friction. Thus,
R
µR
W
F = µR = µW = 0.50 × 20 nt. = 10.0 nt.
(b) The forces acting on the block are shown in fig.
R
F sin θ
F
µR
W
The applied force is inclined at an angle θ in the
upward direction. Its horizontal and vertical
F
θ
Fcos θ
2
XtraEdge for IIT-JEE 27 JULY 2010

components are F cos θ and F sin θ respectively. In
equilibrium.
F cos θ = µR and F sin θ + R = W
or R = (W – F sin θ)
∴ F cos θ = µ(W – F sin θ) = µ W – µF sin θ
F (cos θ + µ sin θ) = µW
µW
or F =
cosθ
+ µ sin θ
Here µ = 0.50, W = 20 nt. and θ = 60º
0.50×
20
10
F =
=
cos60º + 0.5sin 60º
= 10.72 nt.
(c) In this case,
µR
R
Fcos θ
θ
F
0.50 + 0.5×
0.866
F sin θ
W
F cos θ = µR and R = W + F sin θ
Solving we get,
µW
0.50×
20
F =
=
cosθ
−µ
sin θ 0.50 − 0.5×
0.866
= 149.2 nt.
4. Two blocks, m 1 = 2kg and m 2 = 4kg, are connected
with a light string that runs over a frictionless peg to
a hanging block with a mass M as shown in fig. (a).
The coefficient of sliding friction between block m 2
and the horizontal surface at the speeds involved is
µ k = 0.2. The coefficient of static friction between the
two blocks is µ S = 0.4. What is the maximum mass M
for the hanging block if the block m 1 is not to slip on
block m 2 while m 2 is sliding over the surface
Sol. The relevant free body diagrams are shown in fig.(b)
Using two body system, we have
N T
m s = 0.4 N 1
+
m 1
m 2
T
f 1 f
µ k = 0.2
m 1 g (m 1 +m 2 )g
a
Mg
(a) M
(b)
N – (m 1 + m 2 )g = 0 ...(1)
T – F = (m 1 + m 2 )a ...(2)
For hanging block
Mg – T = Ma ...(3)
From eqs. (2) and (3),
Mg – f = (M + m 1 + m 2 )a
But f = µ k N = µ k (m 1 + m 2 )g [Q using eq. (1)]
∴ Mg – µ k (m 1 + m 2 )g = (m + m 1 + m 2 )a
{M −µ
k (m1
+ m2
)}g
or a =
(M + m + m )
1
2
...(4)
From free body diagram of mass m 1 , we have
N 1 – m 1 g = 0 and f 1 = m 1 a
It should be noticed that the force f 1 accelerates m 1 to
the right. Just before slipping occurs, we find
f1
m1 a a
= µ S or µ S = =
N
m g g
1
{M −µ
k (m1
+ m2
)}
∴ µ S =
(M + m1
+ m2
)
Solving eq. (5) for M, we have
(µ S + µ k (m1
+ m2
)
M =
1−µ
(0.4 + 0.2)(2kg + 4kg)
or M =
= 6 kg.
(1 − 0.4)
S
1
....(5)
5. In fig.(a) the blocks A, B and C weight are 3kg, 4kg
and 8kg respectively. The coefficient of sliding
friction between any two surfaces is 0.25. A is held at
rest by a massless rigid rod fixed to the wall, while B
and C are connected by a light flexible cord passing
around a fixed frictionless pulley. Find the force P
necessary to drag C along the horizontal surface to
the left at a constant speed. Assume that the
arrangement shown in the diagram, B on C and A on
B is maintained all the throughout.
P
A
B
C
Sol. When block C moves towards left, B moves towards
right, while A is fixed. There would be a tension T in
the string. Under this condition, let us consider the
frictional forces between different surfaces.
Frictional force between A and B
= µR = 0.25 × 3
Frictional force between C and B
= µR = 0.25 (3 + 4) = 0.25 × 7
Frictional force between C and surface
= 0.25(3 + 4 + 8) = 0.25 × 15
Considering fig. (b)
0.25 × 3
0.25(3 + 4)
B
0.25(3 + 4)
T P
C
T
0.25 × 15
Fig (b)
Tension in the string = Frictional forces at upper and
lower surfaces of block B
or T = 0.25 × 3 + 0.25 × 7 = 2.5 kg wt.
For block C,
P = T + Frictional force between C and B + Frictional
force between C and surface
= 2.5 + 0.25 × (3 + 4) + 0.25 × (15) = 8 kg wt.
= 8 × 9.8 = 78.4 newton
XtraEdge for IIT-JEE 28 JULY 2010

KEY CONCEPT
Organic
Chemistry
Fundamentals
REACTION
MECHANISM
Elimination reactions :
The elimination reactions are reverse of addition
reactions. In these reactions two atoms or group
attached to the adjacent carbon atoms of the substrate
molecule are eliminated to form a multiple bond. In
these reactions a atom or group from α-carbon atom
and a proton from the β-carbon are eliminated.
X
β
α
– C – C – –HX – C = C –
H
In eliminations reactions, the presence of one
hydrogen on the β-carbon atom is necessary. In
general the elimination reactions are divided into two
types, i.e., bimolecular elimination reactions (E 2 ) and
unimolecular elimination reactions (E 1 ).
Bimolecular elimination reactions (E 2 ) :
In these elimination reactions, the rate of elimination
depends on the concentration of the substrate and the
nucleophile and the reaction is of second order. It is
represented as E2. Like S N 2 reaction, the E2 reaction
is also one step process. In these reactions abstraction
of proton from the β-carbon atom and the expulsion
of an atom or group from the α-carbon atom occur
simultaneously. The mechanism of this reaction is
represented as follows:
δ+
B: H
B H
R – CH – CH 2 R — CH — CH 2
β α
X
X
δ –
Transition state
⊕ Θ
RCH = CH 2 + BH + X
The above reaction is a one step process and passes
through a transition state. This reaction is also known
as 1, 2-elimination or simply β-elimination. In these
reactions, the two groups to be eliminated (i.e., H and
X) are trans to each other and hence E2 reactions are
generally trans elimination.
The second-order elimination reaction may also
proceed in two steps (as in E1 elimination which will
be discussed subsequently). In this mechanism, the
base removes the hydrogen in the first step to form an
intermediate carbanion. In the second step, the
intermediate carbanion looses the leaving group. The
second step is slow and is rate determining step.
H
–
– C – C – + OEt
Fast
Θ
– C – C –
(First step)
Br
Br
Θ
– C – C – Slow – C = C – + Br – (Second step)
Br
The rate of this reaction is dependent on the
carbanion (conjugate base of the substrate). So this
mechanism is called ElcB mechanism (Elimination,
Unimolecular from conjugate base).
E1cB mechanism is not common for the E2 reactions.
The carbanion mechanism occurs only where the
carbanion from the substrate is stabilized and where
the leaving group is a poor leaving group. A typical
example, which follows E1cB mechanism is the
formation of 1,1-dichloro-2,2-difluoroethene from
1,1-dichloro-2,2,2-trifluoroethane in presence of
sodium ethoxide.
C 2 H 5 ONa –
CHCl 2 – CF 3
Cl 2 C – CF –F– 3 Cl 2 C = CF 2
1,1-Dichloro-2,2,2-
trifluoroethane
Carbanion
1,1-Dichloro-2,2-
difluoroethane
In the above case the carbanion is strongly stabilized
due to –I effect of halogens. Also F – is a poor leaving
group.
A distinction between the E2 and E1cB mechanism
can be made by tracer experiments. Thus, the
reaction of 1-bromo-2-phenylethane (this substrate
was selected as Ph group is expected to increase the
acidity of β-hydrogen and also to stabilize the
carbanion) with C 2 H 5 OD gives back the starting 1-
bromo-2-phenylethane. If the carbanion mechanism
had operated, the deuterium would have been found
in the recovered 1-bromo-2-phenylethane, which is
not the case.
C 6 H 5 CH 2 CH 2 Br + C 2 H 5 O –
1-Bromo-2-phenylethane
C 6 H 5 CHCH 2 Br + C 2 H 5 OH
C 6 H 5 CHCH 2 Br + C 2 H 5 OD
C 6 H 5 CHCH 2 Br + OC 2 H 5
In case the above reaction is allowed to go to
completion, the product obtained will be
D
XtraEdge for IIT-JEE 29 JULY 2010

EtO –
H
PhCH – CH 2 Br
Fast
C 2H 5OD
PhCD = CH 2 + Br
–
Ph – CH – CH 2 Br
D
Ph – CH – CH 2 Br
OEt
PhCD – CH 2 Br
Styrene
The styrene obtained does not contain any deuterium
(contrary to what has been shown in the above E1cB
mechanism). So in the above reaction E2 mechanism
operates.
The E2 mechanism is supported by the following
evidences.
(i) During elimination, there is no rearranged product
obtained. This is due to the fact that E2 is a single
step process and does not involve the formation of
intermediate carbocation (the carbocations are known
to undergo rearrangement).
(ii) The E2 mechanism finds support from isotope
labeling experiments. Dehydrohalogenation of
unlabelled 1-bromopropane is seven times faster than
the dehydrohalogenation of CH 3 CD 2 CH 2 Br.
CH 3 CH 2 CH 2 Br ⎯⎯→
E2
CH 3 CH = CH 2
Br
CH 3 CD – CH 2
E2 CH 3 CD = CH 2
D
In E2 mechanism a hydrogen (from CH 3 CH 2 CH 2 Br)
or a deuterium (from CH 3 CD 2 CH 2 Br) has to be
abstracted. It is known that the C – D bond is
stronger than the C – H bond and requires more
energy to be broken. Therefore, rate of elimination in
CH 3 CD 2 CH 2 Br should be slower. In fact, it has been
found that in the unlabelled alkyl halides the
elimination rate is seven times more than in labelled
alkyl halides.
Unsymmetrical substrate which has hydrogen
attached to two different β-carbons can affored two
alkenes. For example, 2-bromobutane on
dehydrohalogenation may give 1-butene or 2-butene.
Br
CH 3 – CHCH 2 CH 3
2-Bromobutane
–HBr
CH 2 = CHCH 2 CH 3 + CH 3 CH=CHCH 3
1-Butene 2-Butene
In a similar way, decomposition of sec-butyltrimethylammonium
hydroxide may give a mixture
of two alkenes. The question arises as to which
alkene will be obtained in major amount in the above
dehydrohalogenation. The orientation of the reaction
is determined by Hafmann and Saytzeff Rule.
Hofmann Rule : This rule is applicable for those
substrates in which α-carbon atom is attached to a
positively charged atom. According to this rule, in
the elimination reaction of positively charged species,
the major product will be the alkene which is least
substituted.
CH 3 OH
–
+ Heating
CH 3 CH 2 NCH 2 CH 2 CH 3
CH 3
CH 2 = CH 2 + CH 3 CH 2 CH 2 N(CH 3 ) 2
+
CH 3 CH 2 S(CH 3 ) 2
–
C 2H 5O
CH 2 = CH 2 + S(CH 3 ) 2
Saytzeff Rule : In case of unsymmetrical alkyl
halides, for example in 2-bromobutane, the course of
elimination is determined by Saytzeff Rule.
According to this rule, hydrogen is eliminated
preferentially from the carbon atom which has less
number of hydrogen atoms and so the highly
substituted alkene is the major product.
Br
alk.KOH
CH 3 CH 2 – CH – CH 3
2-Bromobutane
CH 3
CH 3 CH 2 – C – CH 3
Br
2-Bromo-2-methylbutane
CH 3 CH = CHCH 3 + CH 3 CH 2 CH = CH 2
2-Butene(80%) 1-Butene (20%)
C 2H 5O –
CH 3 CH 3
CH 3 CH = C – CH 3 + CH 3 CH 2 C = CH 2
2-methyl-2-butene 2-methyl-1-butene
(71%)
(29%)
The formation of highly substituted alkene can be
explained as follows.
The transition states of less substituted and more
substituted alkenes from an alkyl halide are
represented as shown below:
H
CH 3 CH 2 CH — CH 2
δ–
OR
Br
δ–
T.S. of less substituted alkene
δ–
RO
H
CH 3 CH — CHCH 3
Br
δ–
T.S. of more substituted alkene
Both the transition states have partial double bond
character. However, the transition state leading to
more stable alkene is more stabilized and is of lower
energy. Thus, the more stable alkene is formed as the
major product.
XtraEdge for IIT-JEE 30 JULY 2010

E
RX + base
less substitued alkene
more substitued alkene
predominant product
Reaction progress
Energy diagram for a typical E2 reaction, showing
why the more substituted alkene predominates
Hofmann rule can be understood by considering the
mechanism of elimination reaction of quaternary
ammonium hydroxide.
H CH 3
β´ β´´
+
H – C – CH 2 – N – CH 2 CH 2 CH 3
H CH 3
B – Route a
CH 2 = CH 2 + (CH 3 ) 2 N(CH 2 ) 2 CH 3
Another possibility is :
CH 3 H
β´
+
β´´
CH 3 CH 2 – N – CH 2 – C – CH 3
CH 3 H B –
Route b
(CH 3 ) 2 NCH 2 CH 3 + CH 3 CH = CH 2
In the above reaction the strong electron-withdrawing
group makes the hydrogens of the β-carbons more
acidic for facile abstraction by the base. In this
compound, with alternate β-hydrogens (marked β´
and β´´), the β" hydrogen are less acidic due to +I
effect of the adjacent methyl group. Hence β´hydrogen
is relatively more acidic and is removed to
give the alkene (ethene) by route a.
In elimination reactions steric effect also plays an
important role. Thus, dehydrohalogenation of alkyl
halide using the bulky base leads to the formation of
terminal alkene as the major product.
Br
–
t-BuO
CH 3 CH 2 CHCH 3
2-Bromobutane CH 3 CH 2 CH=CH 2 + CH 3 CH=CHCH 3
(73%) (27%)
Unimolecular elimination reactions (E1) :
In these reactions the rate of elimination is dependent
only on the concentration of the substrate and is
independent of the concentration of the nucleophile
and the reaction is of first order, (E1). Like S N 1
reaction the E1 reaction is also a two step process.
The first step is the slow ionization of alkyl halide to
give the carbocation. The second step involves the
fast abstraction of a proton from the adjacent
β-carbon atom giving rise to the formation of an
alkene.
CH 3
CH 3 — C — X
CH 3
Slow
CH 3 CH 2 – H
Fast
C :B –
CH 3
Carbocation
CH 3 +
CH 3 — C + X –
CH 3
Carbocation
CH 3 C = CH 2 + BH
CH 3
2-Methylpropene
In case the substrate is such that more than one
alkenes can be formed, that alkene will predominate
which has larger number of alkyl groups on the
double bonded carbon (this is as per Saytzeffs rule.
This can be visualised since the substituted alkyl
groups will stabilise the alkene by hyperconjugation.
CH 3 Br
CH 3
–HBr
CH 3 — C — C — CH 3 CH 3 — C == C — CH 3
H
H
2-Bromo-3-methylbutane
CH 3
H
2-Methyl-2-butene (major)
+ CH 3 — CH — CH == CH 2
3-Methyl-1-butene (minor)
The acid catalysed dehydration of alcohols also
follows E1 mechanism.
⊕
(CH 3 ) 3 COH H2SO4 –H
(CH 3 ) 3 C—OH 2O
2 (CH3 ) 3 C ⊕
t-Butyl alcohol
CH 3
CH 3
⊕
H 3 C — C
CH 3 — C == CH 2
2-Methylpropene
CH 2 – H
In the E1 mechanism the rate of reaction is
determined by the rate of formation of carbocation,
which in turn depends on the stability of carbocation.
Due to the formation of carbocation, these may
undergo rearrangements. This has been
experimentally confirmed.
Br
(CH 3 ) 3 CCHCH 3
C 2H 5OH
CH 3
CH 3
(CH 3 ) 3 CCH = CH 2 + CH .3 C = CCH 3
XtraEdge for IIT-JEE 31 JULY 2010

KEY CONCEPT
Physical
Chemistry
Fundamentals
ENERGETICS
Thermodynamics deals with the transfer of heat
between a chemical system and its surroundings
when a reaction or phase change takes place within
the system. The entire formulation of
thermodynamics is based on two fundamental laws
which have been established on the basis of
experimental on the basis of experimental behaviour
of macroscopic aggregates of matter, collected over a
long period of time.
Since
First Law of thermodynamics
The internal energy of a system can be changed by
transferring heat to/from the system from/to the
surroundings. It can also be changed by doing the
mechanical work on/by the system by/on the
surroundings. These facts are represented in the form
of the first law of thermodynamics as
dU = dq + dw or ∆U = q + w
Since heat given to the system and work done on the
system raise the internal energy of the system, these
two operations are assigned positive values. The
converse of the two operations, viz., heat given out
and work done by the system are assigned negative
values.
The expression of work done by/on a gaseous system
is given by
dw = – p ext dV
Where p ext is the external pressure against which the
volume gaseous system is changed by an amount dV.
For a constant external pressure, we have
w = – nRT ln (V 2 /V 1 )
where V 1 and V 2 are the initial and final volumes of
the gaseous system.
If p ext differs from the pressure of the gas by
infinitesimal amount, the work is said to be carried
out under reversible condition. In this case, the
expression of work under constant temperature
condition is given by
w = – nRT ln (V 2 /V 1 ) '
Note that for V 2 > V 1 , there occurs an expansion of
the gas. The work is done by the system on the
surroundings and it carries a negative sign.
For V 1 > V 2 , there occurs compression of the gaseous
system. The work is done by the surroundings on the
system and it carries a positive sign.
Internal energy and enthalpy
From the first law of thermodynamics, it can be
shown that the heat transferred at constant volume
changes the internal energy of the system, whereas
that at constant pressure changes the enthalpy of the
system .
∆U = nC v , m (T 2 – T 1 ) ; and ∆H = nC p, m (T 2 – T 1 )
where C v,m and C p,m are the molar heat capacities at
constant volume and constant pressure, respectively.
In the laboratory, the majority of chemical reactions
are carried out under the condition of constant
pressure, and thus the heat transferred in such a
system is equal to the enthalpy change in a chemical
reaction. Since the enthalpy of a system can also
change due to the variation in temperature and
pressure, it is, therefore essential that the reactants
and products in a chemical reaction must have the
same temperature and pressure.
Enthalpy change of a chemical equation
The enthalpy change of thermochemical equation is
∆H = ∑ v jH m, j – ∑ v iH m,i
(products)
(reactan ts)
where H m,i refers to the molar enthalpy of species i in
the balanced chemical equation and v i the
corresponding stoichiometric coefficient. The unit of
∆H are kJ mol –1 .
Two types of reactions may be distinguished.
(a) Exothermic reactions For these ∆H is negative,
which implies negative q p and hence release of heat
when reactants are converted into products. In this
case
ΣH(products) < ΣH(reactants)
(b) Endothermic reactions For these ∆H is is positive,
which implies positive q p and hence absorption of
heat when reactants are converted into products. In
this case
ΣH(products) > ΣH(reactants)
Molar enthalpies of formations
It is not possible to determine the absolute value of
enthalpy of a substance. However, based on the
following convention, the relative values of standard
molar enthalpies of formation (the term standard
indicates of pressure of 1 bar) other substances can be
determined.
The enthalpy of formation of every element in its
stable states of aggregation at 1 bar and 25ºC is
assigned a zero value.
For example, ∆ f Hº (graphite) = 0 ∆ f Hº (Br 2 , 1) = 0
∆ f Hº (S, rhombic) = 0 ∆ f Hº (H 2 , g) = 0 and so on.]
XtraEdge for IIT-JEE 32 JULY 2010

XtraEdge for IIT-JEE 33 JULY 2010

XtraEdge for IIT-JEE 34 JULY 2010

XtraEdge for IIT-JEE 35 JULY 2010

XtraEdge for IIT-JEE 36 JULY 2010

The enthalpy change of a chemical equation can be
computed by using the expression
0
0
∆ r Hº = ∆ f Hi
– ∑ ∆f
Hi
∑
( products) (reactors)
Hess's law of constant heat summation
Since the molar enthalpies of formation of reactants
and products involved in a chemical equation have
definite values, the enthalpy change of (or heat
involved in) the chemical equation will have a
definite value, irrespective of the fact whether the
reaction is carried out in one step or more than one
step. This fact is known as Hess's law of constant
heat summation. For example,
(i) C(graphite) + O 2 ((g) → CO 2 (g)
∆H 1 = – 393.51 kJ mol –1
(ii) C(graphite) + 2
1
O2 (g) → CO(g)
CO(g) + 2
1
O2 (g) → CO 2 (g)
∆H 2 = – 110.52 kJ mol –1
∆H 3 = – 282.99 kJ mol –1
Obviously, ∆H 1 = ∆H 2 + ∆H 3
Types of reactions and corresponding enthalpy changes
The enthalpy change in a reaction is suitable named
according to the type of reaction in question. Two
types of reaction are specifically defined as follows.
Enthalpy of formation: the enthalpy of combustion
of a given substance is defined as the enthalpy
change when 1 mole of a given substance is formed,
starting from the elements in their stable states of
aggregation. A few examples are
H 2 (g) + 2
1
O2 (g) → H 2 O(1)
∆ f Hº(H 2 O, 1) = – 285.77 kJ mol –1
11
12C(graphite) + 11H 2 (g) + O2 (g) → C 12 H 22 O 11 (s)
2
∆ f Hº(C 12 H 22 O 11 ,s) = –2218 kJ mol –1
Enthalpy of Combustion: The enthalpy of
combustion of a given substance is defined as the
enthalpy change when one mole of this substance
combines with requisite amount of oxygen to form
products in their stable states of aggregation. A few
examples are
CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l)
∆ c H(CH 4 , g) = –74.85 kJ mol –1
C 12 H 22 O 11 (s) + 12O 2 (g) → 12CO 2 (g) + 11H 2 O(l)
∆ c H(C 12 H 22 O 11 , s) = – 5644 kJ mol –1
Similarly, one can mane the enthalpy change based
on the type of reaction. A few examples are
Enthalpy of fusion: H 2 O(s) ⎯ 273
⎯ .15
⎯
K
→ H 2 O(l)
Enthalpy of vaporization :
∆ fus H = 6 kJ mol –1
H 2 O(1 ) ⎯ 373
⎯ .15
⎯
K
→ H 2 O(g)
∆ vap H = 40.6 kJ mol –1
Enthalpy of sublimation : I 2 (s) → I 2 (g)
∆ sub H = 63.4 kJ mol –1
Enthalpy of transition : C(graphite) → C(diamond)
∆ trs H = 1.90 kJ mol –1
Enthalpy of neutralization : H + (aq)+OH – (aq)→H 2 O(l)
∆ neut H = – 57.3 kJ mol –1
Enthalpy of ionization : HCN(aq) → H + (aq)+CN – (aq)
∆ ioniz H = 45.17 kJ mol –1
Relation between ∆H and ∆U of a chemical equation
Since, H = U + pV, we have
∆H = ∆U + ∆(pV) = ∆U + (∆v g ) RT
where ∆v g is the change in the stoichiometric number
of gaseous molecules in converting reactants to
products and is given as
∆v g = v – ∑ v g,i
∑ g,i
(products)
(reac tan ts)
rFor a reaction involving condensed phases
∆H – ~ ∆U
Bond Enthalpies: Bond enthalpy of a given bond is
defined as the average enthalpies required to
dissociate the said bond present in different gaseous
compounds into free atoms in the gaseous phase. The
bond enthalpy may be distinguished from bond
dissociation enthalpy which is enthalpy required to
dissociate a given bond of some specific molecule. It
is possible to construct a table listing the average
bond enthalpies of different types of bonds and with
the help of this, one can estimate the enthalpy change
of a chemical equation involving gaseous species.
For example, for a reaction
H 2 (g) + Cl 2 (g) → 2HCl(g)
we can write ∆H = + ε(H–H) + ε(Cl–Cl) –2ε(H–Cl)
Second Law of thermodynamics
The second law of thermodynamics identifies a state
function, called the entropy, which provides a
criterion for identifying reversible or irreversible
nature of the given process undergone by a system.
The entropy of the universe (system + surroundings)
increases for irreversible processes whereas it
remains constant for reversible processes.
The entropy function has been identified with the
disorderliness of the system–larger the disorderliness,
larger the entropy of the system. Foe example, for a
substance in three states of matter we have
S(gaseous state) > > S(liquid state) > (solid state)
Expression of Entropy Function
For a system which involves transferring
infinitesimal heat at constant temperature, the entropy
change of the system is given by
dS =
dq rev
T
XtraEdge for IIT-JEE 37 JULY 2010

For finite heat transferred at constant temperature, we
have
q
∆S = rev
T
For example, for a pure substance we have
∆ vapH
∆fusH
∆ vap S = and ∆ fus S =
Tb
Tm
where the subscripts vap and fus represent
vaporization and fusion, respectively.
Gibbs Function
Gibbs function (or energy) or simply free energy is
defined as
G = H – TS
For a process occurring at constant T and P, the
change in Gibbs function is given by
∆G = ∆H – T ∆S
For a process to be spontaneous, the value of ∆G is
negative. For a nonspontaneous reaction, ∆G is
positive. For a reaction at equilibrium, ∆G = 0 and
temperature at which the system occurs at
equilibrium is given by
T eq = ∆H/∆S
Pressure-Volume Work
An ideal gas can undergo expansion of compression
under isothermal or adiabatic conditions. The
expansion ant compression may be carried out under
reversible or irreversible conditions. We give below
the expressions of p-V work under different
conditions.
Isothermal p–V Work
In this case, temperature of the system remains
constant, ie. ∆T = 0
For irreversible condition: w = – P ext (V 2 – V 1 )
For reversible condition: w = – nRT In (V 2 /V 1 )
Adiabatic p–V Work
In this case, heat can neither enter to or leave from
the system, i.e. q = 0. From first law of
thermodynamics, it follows that
∆U = w
where ∆U is given by
∆U = C v (T 2 – T 1 )
For a gas undergoing adiabatic irreversible volume
change, the expression of work is given by
w = – P ext (V 2 –V 1 )
For an ideal gas undergoing adiabatic reversible
expansion/compression, we also have
pV γ = constant
pT γ(1–γ) = constant
and TV γ–1 = constant
here γ = C p,m /C v,m The symbols C p,m and C v,m represent
molar heat capacities at constant pressure and volume
conditions, respectively.
For a monatomic ideal gas:
C v,m = (3/2)R; C p,m = (5/2)R; and γ = 5/3
For a diatomic ideal gas:
C v,m = (5/2)R; C p,m = (7/2)R; and γ = 7/5
CHEMISTRY JOKES
If you didn't get the joke, you probably didn't
understand the science behind it. If this is the case,
it's a chance for you to learn a little chemistry.
Chemistry Joke 1:
Outside his buckyball home, one molecule overheard
another molecule saying, "I'm positive that a free
electron once stripped me of an electron after he
lepton me. You gotta keep your ion them."
Chemistry Joke 2:
A chemistry professor couldn't resist interjecting a
little philosophy into a class lecture. He interrupted
his discussion on balancing chemical equations,
saying, "Remember, if you're not part of the solution,
you're part of the precipitate!"
Chemistry Joke 3:
One day on the Tonight Show, Jay Leno showed a
classified add that read: "Do you have mole
problems If so, call Avogadro at 602-1023."
Chemistry Joke 4:
A student comes into his lab class right at the end of
the hour. Fearing he'll get an "F", he asks a fellow
student what she's been doing. "We've been
observing water under the microscope. We're
suppose to write up what we see." The page of her
notebook is filled with little figures resembling circles
and ellipses with hair on them. The panic-stricken
student hears the bell go off, opens his notebook and
writes, "During this laboratory, I examined water
under the microscope and I saw twice as many H's as
O's."
XtraEdge for IIT-JEE 38 JULY 2010

U n d e r s t a n d i n g
UNDERSTANDING
Organic Chemistry
1. An organic compound (A), C 10 H 15 N, undergoes
carbylamine reaction but no diazotization. It reacts
with HNO 2 giving off N 2 and a compound (B),
C 10 H 14 O. (B) reacts with Lucas reagent immediately,
but no colour in Victor meyer's test. (B) on heating
with conc. H 2 SO 4 eliminates water to give (C),
C 10 H 12 , which decolourises Br 2 /CCl 4 and cold dilute
neutral KMnO 4 solution. (C) on ozonolysis gives (D),
C 7 H 6 O and (E), C 3 H 6 O. Compound (E) on heating
with I 2 and NaOH produced yellow precipitate and
sodium acetate. Compound (D) reacts with conc.
NaOH to give (F) and (G). Compound (G) on heating
with sodalime gives benzene. Compound (F) gives a
red colour with ceric ammonium nitrate, and on
oxidation and heating the product with sodalime
produced benzene. What are (A) to (G)
Sol. The given data are :
HNO 2
C 10 H 15 N
–N
(A)
2;–H 2O
C 7 H 6 O
(D)
3H6O
(E)
Conc. NaOH
Conc. H2SO4
C 10 H 14 O
∆;–H
(B)
2O
(I) O 3
(II) H 2/Pd
C 10 H 12
(C)
C 7 H 6 O + C 3 H 6 O
(D) (E)
(F) + (G) Sodalime C 6 H 6
[O]
Sodalime
Product C
∆ 6 H 6
C ⎯
2 ⎯
I 2 +NaOH ⎯⎯
→ CHI 3 ↓ + CH 3 COONa
+ 3NaI + 3H 2 O
Since (C) decolourise Br 2 /CCl 4 and KMnO 4 colour,
hence it has C=C bond. Its ozonolysis gives (D)
and (E). Among these (D) undergoes Cannizaro's
reaction, while (E) gives iodoform test, hence (D) is
benzaldehyde and (E) acetone. Now joining (D) and
(E), the structure of (C) can be determined.
H
C=O + O =C
CH 3
–2[O]
(D) (E) CH 3
H
C=C
CH 3
(C)
CH 3
Since (C) is produced from (B), which is a t-alcohol,
as it gives Lucas test immediately, hence (B) is.
H
CH 3
C–CH
CH 3
OH
(B)
As (B) is obtained by the action of HNO 2 on (A),
hence (A) would be
H
CH 3
C–CH
CH 3
NH 2
(A)
2. Two isomeric compounds (A) and (B) have the
molecular formula C 7 H 9 N. (A) being soluble in
water, the solution being alkaline to litmus It does not
undergoes diazotization, but show carbylamine
reaction and mustard oil reaction, it reacts with acetyl
chloride and acetic anhydride. Its product with
benzene sulphonyl chloride dissolves in KOH. (B) on
the other hand, does not dissolve in water, but
undergoes diazotization. Its product with C 6 H 5 SO 2 Cl
dissolves in KOH. Its salt undergo hydrolysis in
aqueous solution showing an acidic test. What are
(A) and (B)
Sol. As both (A) and (B) give products with C 6 H 5 SO 2 Cl,
which are soluble in KOH, they contain –NH 2 group.
(B) can be diazotized so contains – NH 2 in the
nucleus. (A) cannot be diazotized, hence contains
–NH 2 in the side chain. The number of carbon and
hydrogen atoms also indicates aromatic character.
On the basis of above considerations we may show
that (A) is benzylamine and (B) o – , m – or p-toludine.
CH 2 NH 2 NaNO 2 + HCl CH 2 OH
Benzylamine
(A)
CHCl 3 + 3KOH
CS 2 + HgCl 2
C 6H 5SO 2Cl
– HCl
CH 2 NC
CH 2 NCS
CH 2 NHSO 2 C 6 H 5
KOH CH 2 N–SO 2 C 6 H 5
–H 2O
K
Soluble
C 6 H 5 CH 2 NH 2 + HOH ⎯→ C 6 H 5 CH 2 N + H 3 OH –
XtraEdge for IIT-JEE 39 JULY 2010

C 6 H 4
(B)
CH 3
NH 2
NaNO 2 + HCl
C 6 H 4
CH 3
N 2 Cl
(A)
– CH.CH 3
HOH/H +
Cl –HCl
(B)
– CH–CH 3
OH
C 6H 5SO 2Cl
–HCl
C 6 H 4
CH 3 KOH
NHSO 2 C 6 H 5
C 6 H 4
CH 3
NKSO 2 C 6 H 5
Soluble
[O]
–H 2O
(C)
– COCH 3
3. An organic compound (A) of molecular weight
140.5, has 68.32% C, 6.4% H and 25.26% Cl.
Hydrolysis of (A) with dilute acid gives compound
(B), C 8 H 10 O. (B) can be oxidised under milder
conditions to (C), C 8 H 8 O. (C) forms a phenyl
hydrazone (D) with C 6 H 5 NHNH 2 and gives positive
iodoform test. What are (A) to (D)
Sol. (i) Calculation of empirical formula of (A)
Element %
C 68.32
H 6.4
Cl 25.26
Relative no.
of atoms
68.32
= 5.59
12
6.4
= 6.40
1
25.26
= 0.71
35.5
Simplest
ratio
5.59
= 8
0.71
6.40
= 9
0.71
0.71
= 1
0.71
Empirical formula = C 8 H 9 Cl
Empirical formula wt. = 140.5
Molecular weight = Emp. formula weight
Hence, Molecular formula = Empirical formula
= C 8 H 9 Cl
(ii) Given that
C8H
Cl
9
(A)
⎯ HOH ⎯ → C8H9OH
(B)
[O]
⎯⎯→ C H
−H2O
8 8O
(C)
(iii) (C) reacts with C 6 H 5 NHNH 2 to give
C 8 H 8 =N.NHC 6 H 5 , hence (C) contains a C = O
group.
(iv) (C) on heating with I 2 + NaOH gives CHI 3 , hence
(C) contains –COCH 3 group. Thus (C) is
O
– C – CH 3
(v) Oxidation of (B) gives (C), hence (B) is a
secondary alcohol, i.e.,
– CH – CH 3
OH
(vi) (B) is obtained by the hydrolysis of (A), hence it
is :
– CH.CH 3
Cl
1-Chloro-1-phenyl ethane
Now, different reactions are as follows :
– C = O + H 2 N . NH . C 6 H 5
CH 3
(C) – C=N.NH.C 6 H 5
(C)
–H 2O CH 3
∆
– C – CH 3 + 3I 2 + 4NaOH ⎯→ CHI 3 + 3NaI
Yellow ppt
O
– COONa
+ 3H 2 O +
4. An organic compound (A), C 4 H 9 Cl, on reacting with
aqueous KOH gives (B) and on reaction with
alcoholic KOH gives (C) which is also formed on
passing vapours of (B) over heated copper. The
compound (C) readily decolourises bromine water.
Ozonolysis of (C) gives two compounds (D) and (E).
Compound (D) reacts with NH 2 OH to gives (F) and
the compound (E) reacts with NaOH to give an
alcohol (G) and sodium salt (H) of an acid. (D) can
also be prepared from propyne on treatment with
water in presence of Hg ++ and H 2 SO 4 . Identify (A) to
(H) with proper reasoning.
Sol.
Alc. KOH
C 4 H 9 Cl
(A)
∆; –KCl
(Alkyl halide)
Aq.KOH
C 4 H 8
(C)
(Alkene)
Cu
C
∆; –KCl 4 H 9 OH
∆; –H
(B)
2O
(Alcohol)
We know that p-alcohol on heating with Cu gives
aldehyde while s-alcohol under similar conditions
gives ketone. Thus, (B) is a t-alcohol because it, on
heating with Cu gives an alkene (C). Since a t-
alcohol is obtained by the hydrolysis of a t-alkyl
halide, hence (A) is t-butyl chloride.
(A) = CH
3
Cl
|
− C − CH3
and (B) = CH
|
CH
3
3
OH
|
− C − CH
|
CH
The alkene (C) on ozonolysis gives (D) and (E),
hence (C) is not symmetrical alkene. In these
compound (E) gives Cannizaro's reaction with
NaOH. So, (E) is an aldehyde which does not contain
α–H atom. Hence it is HCHO. Compound (D) can
also be prepared by the hydration of propyne in the
presence of acidic solution and Hg ++ .
3
3
XtraEdge for IIT-JEE 40 JULY 2010

CH 3 – C ≡ CH + H 2 O
Hg
++
⎯⎯⎯
→
H
+
CH3 − C = CH
OH
| 2
⎯→ CH3 − C − CH3
||
O
(D)
Hence (D) is acetone and (E) is formaldehyde.
Therefore, alkene (C) is 2-methyl propene.
(CH 3 ) 2 –C=CH 2
(D) reacts with hydroxyl amine (NH 2 OH) to form
oxime (F).
CH 3
CH 3
(B) = CH
C = O + H 2 NOH –H 2O CH 3
CH 3
C = NOH
(D) (F)
3
Reactions :
Cl
|
CH 3 − C − CH 3
|
CH 3
(A)
Cu /300ºC
⎯ ⎯⎯⎯→
−H2O
Alc.KOH / ∆
⎯ ⎯⎯⎯⎯
→
−KCl;
−H2O
(I) O
CH 3 – C = CH 3
2
(II) H 2 O/Zn
CH 3
CH 3
CH 3
(C)
C = O + H 2 NOH
(D)
2 HCHO + NaOH →
(E)
CH 3 – C ≡ CH + H 2 O
OH
|
− C − CH3
and (A) = CH
|
CH
3
3
Cl
|
− C − CH
|
CH
OH
|
Aq.KOH CH
⎯⎯ ⎯⎯ →
3 − C − CH 3
∆;–KCl
|
CH 3
(B)
CH 3 − C = CH 2 + H 2O
|
CH 3
(C)
CH 3 − C = CH 2
|
CH 3
(C)
∆
–H 2 O
CH 3
CH 3
(D)
CH 3
CH 3
CH 3OH +
(G)
++
Hg
C = O + H – C – H
3
C = NOH
(F)
HCOONa
(H)
O
O
(E)
⎯⎯⎯
→CH + 3 – C – CH 3
H
(D)
3
5. A hydrocarbon (A) of the formula C 8 H 10 , on
ozonolysis gives compound (B), C 4 H 6 O 2 , only. The
compound (B) can also be obtained from the alkyl
bromide, (C) (C 3 H 4 Br) upon treatment with Mg in
dry ether, followed by treatment with CO 2 and
acidification. Identify (A), (B) and (C) and also
equations for the reactions.
(i) O
Sol. A(C 8 H 10 ) ⎯⎯⎯ 3 → C H
(ii) H2O
4 6O2
(B)
Since compound (A) adds one mol of O 3 , hence it
should have either C = C or a – C ≡ C – bond.
If it was alkene its formula should be C 8 H 16 (C n H 2n ),
and if it was alkyne it should have the formula C 8 H 14 ;
it means it is neither a simple alkenen or simple
alkyne. However it is definite that the compound has
an unsaturated group, it appears that it is a
cyclosubstituted ethyne.
2H
H – C ≡ C – H ⎯ − ⎯⎯→C 3 H 5 – C ≡ C – C 3 H 5
+ C 6 H 10
the C 3 H 5 – corresponds to cyclopropyl (∆) radical,
hence compund (A) is
CH 2
CH 2
CH – C≡C – CH
CH 2 CH 2
1,2-dicyclopropyl ethyne
The ozonolysis of above compound would give two
moles of cyclopropane carboxylic acid (C 4 H 6 O 2 ).
CH 2
CH 2
(i) O
CH – C≡C – CH
3
CH 2 CH 2
(A)
O
CH 2
CH 2
CH – C — C – CH H 2O
CH 2 CH Warm 2
(A) O O
CH 2
CH 2
CH – C – C – CH + H 2 O 2
CH 2 CH 2
O O
CH 2
2 CH – COOH
CH 2
(B)
Compound (B) is prepared from cyclopropyl bromide
as follows :
O
CH 2
CH
Mg 2
C=O
CH – Br
CH . MgBr
CH ether
2 CH ∆
2
Cyclopropyl
magnesium bromide
CH 2
CH
HOH 2
CH .COOMgBr
CH–COOH
CH dil. HCl;
2 CH 2
–MgBrOH
Addition compound
XtraEdge for IIT-JEE 41 JULY 2010

`tà{xÅtà|vtÄ V{tÄÄxÇzxá
3 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 y = f(x) be a curve satisfying
dy – y ln 2 = 2
sin x (cos x – 1). ln2, then
dx
(A) y is bounded when x → ∞
(B) f(x) = 2 sin x + c . 2 x , where c is an arbitrary
constant
(C) y = 2 sinx , y is bounded when x → ∞
(D) f(x) = 2 sinx does not have any solution if y is not
bounded.
2. In a right angled triangle the length of its hypotenuse
is four times the length of the perpendicular drawn
from its orthocentre on the hypotenuse. The acute
angles of the triangle can be
π π
π 3π
(A) , (B) , 6 3 8 8
(C) 6
π , 4
π
π 5π
(D) , 12 12
7. Find all the real values of a, for which the roots of the
equation x 2 – 2x – a 2 + 1 = 0 lie between the roots of
equation
x 2 – 2(a + 1) x + a(a – 1) = 0
8. Given the base of a triangle and the sum of its sides
prove that the locus of the centre of its incircle is an
ellipse.
9. A bag contains 7 tickets marked with the number 0,
1, 2, 3, 4, 5, 6 respectively. A ticket is drawn and
replaced. Then the chance that after 4 drawings the
sum of the numbers drawn is 8, is –
10. A polynomial in x of degree greater than 3 leaves
remainders 2, 1 and – 1 when divided by (x – 1), (x + 2)
and (x + 1) respectively. What would be the remainder
if the polynomial is divided by (x 2 – 1) (x + 2)
3. Let a, b ∈ R such that 0 < a < 1 and 0 < b < 1. The
values of a and b such that the complex number
z 1 = –a + i, z 2 = –1 + bi and z 3 = 0 form an equilateral
triangle are
(A) 2 − 3, 3 (B) 2 − 3, 2 − 3
(C) 3,
2 − 3 (D) None of these
4. If c 1 is a fixed circle and c 2 is a variable circle with
fixed radius. The common transverse tangents to c 1
and c 2 are perpendicular to each other. The locus of
the centre of variable circle is :
(A) circle
(B) ellipse
(C) hyperbola (D) parabola
5. The length of the latus rectum of the parabola
169 {(x – 1) 2 + (y – 3) 2 } = (5x – 12y + 17) 2 is –
14 56 28
(A) (B) (C) (D) None
13 13 13
cos5x + cos 4x
6. Evaluate :
∫
dx
1−
2cos3x
• William Bottke at Cornell University in the US
has calculated that at least 900 asteroids of a
kilometre or more across regularly sweep
across Earth's path.
• The Dutch astronomer Christiaan Huygens
(1629 - 1695) drew Mars using an advanced
telescope of his own design. He recorded a
large, dark spot on Mars, probably Syrtis Major.
He noticed that the spot returned to the same
position at the same time the next day, and
calculated that Mars has a 24 hour period. (It is
actually 24 hours and 37 minutes)
• Space debris travels through space at over
18,000 mph.
• The nucleus of Comet Halley is approximately
16x8x8 kilometers. Contrary to prior
expectations, Halley's nucleus is very dark: its
albedo is only about 0.03 making it darker than
coal and one of the darkest objects in the solar
system.
XtraEdge for IIT-JEE 42 JULY 2010

MATHEMATICAL CHALLENGES
SOLUTION FOR JUNE ISSUE (SET # 2)
Hence plane given by (3) is bisecting the acute angle
1. 1 st box can be filled in 4 ways.
⇒ tan θ =
1037 < 1 ⇒ 47 + 1 + 1 = 49
61
Next each box can be filled in 3 ways (except the ball
of colour in previous box).
between given two planes also. Hence the conclusion
holds true.
Hence the required no. of ways = 4 × 3 5 = 972
f (b)
−1
2 2
5. Let I
2. Given |A| ≠ 0; AA –1 = I ⇒ (AA –1 ) T = I T
2 =
∫
(( f (y)) − a ) dy
f (a)
(A –1 ) T A T = I (as A is symmetric)
Let f –1 (y) = x
(A –1 ) T A = I
⇒ f(x) = y
so by the definition of inverse A –1 = (A –1 ) T
b
Hence A –1 2 2
is also symmetric.
I 2 =
∫
( x − a ) f´(x) dx
a
Using by parts
3. The normal to hyperbola at the point
2 2
P(a sec θ, b tan θ) is
I 2 = ( x − a )f (x)) b a
–
∫ b
2x
f(x) dx
a
ax cos θ + by cot θ = a 2 + b 2
If it passes through (h, k) then
= (b 2 – a 2 ) f(b) –
∫ b
2x
f(x) dx
a h cos θ + b k cot θ = a 2 + b 2 a
...(1)
Let z = e iθ z 2 +1
=
= cos θ + i sin θ then put cos θ =
∫ b
2x
f(b) dx –
a ∫ b
2x
f(x) dx
a
2z
⎛
2
⎞
and cot θ = i ⎜
z + 1
=
⎟ in the equation (1).
∫ b
2 x (f(b) – f(x)) dx
a
2
⎝ z −1⎠
I
ahz 4 + 2(i bk – (a 2 + b 2 )) z 3
1 1
Hence =
I
+ 2(i bk + (a 2 + b 2 2 2
))z – ah = 0
z 1 , z 2 , z 3 , z 4 are its four solutions so
i(
θ1
+θ2
)
Σ z 1 z 2 = 0 = Σ e = 0
6.
1
y + = 2 y
Σ (cos (θ 1 + θ 2 ) + i sin (θ 1 + θ 2 )) = 0
⇒ y = 1
Hence Σ cos(θ 1 + θ 2 ) = Σ sin(θ 1 + θ 2 ) = 0
1
x + = 5 + 2
4. Planes are – x – 2y – 2z + 9 = 0 ....(1)
x
and 4x – 3y + 12z + 13 = 0 ...(2)
⇒ x 2 1
+
2
The plane bisecting the angle b/w these planes
x
containing origin is
= ( 5 + 2) – 2 = 5
−x − 2y − 2z + 9 4 x − 3y + 12z + 13
= +
3
13
x 4 1
+
4 = 5 – 2
x
i.e. 25x + 17y + 62z – 78 = 0 ...(3)
If θ be the angle between (1) & (3) then
⇒ x 8 1
+
8 = 9 – 2
x
61
cos θ =
4758
⇒ x 16 1
+ = 49 – 2
16
x
XtraEdge for IIT-JEE 43 JULY 2010

7. Let the radius of S 2 is r
6
r
r
6
6
2 r + r + 6 = 2 6
⎛ ⎞
r = 6 ⎜
2 −1
⎟
⎝ 2 + 1⎠
= 6(3 – 2 2 )
= 18 – 12 2
8. S 1 = 2 + 4 + 6 + .... + 120
60
= (2 + 120)
2
= 30 × 122 = 3660
S 2 = 7 + 14 + 21 + ..... + 119
17
= (7 + 119) 2
= 17 × 63 = 1071
S 3 = 14 + 28 + ..... + 102
= 2
8 (14 + 112)
= 4 × 126 = 504
120×121
Ans. = – 3660 – 1071 + 504
2
= 7260 – 4731 + 504
= 2529 + 504
= 3033
9. Here F(x) is even function
so f(x) = F(–x) = F(x)
⇒ f(–x) = f(x)
g(x) = –F(x) = – f(x) = –f(–x)
h(x) = –F(–x) = – F(x) = –f(x)
Ans. (C)
10. f(x) + h(x) = f(x) – f(x) = 0
g(x) – h(x) = – f(x) + f(x) = 0
F(x) + f(x) ≠ 0
f(x) – g(x) = f(x) + f(x) ≠ 0
Ans. (B)
6
Brief Description : zirconium is a grayish-white lustrous
metal. The finely divided metal can ignite spontaneously
in air, especially at elevated temperatures. The solid
metal is much more difficult to ignite. The inherent
toxicity of zirconium compounds is low. Hafnium is
invariably found in zirconium ores, and the separation is
difficult. Commercial grade zirconium contains from 1 to
3% hafnium. The hafnium is removed from the zirconium
used in the nuclear power industry.
Zirconium is found in S-type stars, and has been
identified in the sun and meteorites. Analyses of lunar
rock samples show a surprisingly high zirconium oxide
content as compared with terrestrial rocks. Some forms
of zircon (ZrSiO4) have excellent gemstone qualities.
Table: basic information about and classifications of
zirconium.
• Name : Zirconium
• Symbol : Zr
• Atomic number : 40
• Atomic weight : 91.224 (2)
• Standard state : solid at 298 K
• CAS Registry ID : 7440-67-7
• Group in periodic table : 4
• Group name : (none)
• Period in periodic table : 5
• Block in periodic table : d-block
• Colour : silvery white
• Classification : Metallic
ISOLATION
Isolation : zirconium is available from commercial
sources so preparation in the laboratory is not normally
required. In industry, reduction of ores with carbon is
not a useful option as intractable carbides are produced.
As for titanium, the Kroll method is used for zirconium
and involves the action of chlorine and carbon upon
baddeleyite (ZrO2). The resultant zirconium
tetrachloride, ZrCl4, is separated from the iron
trichloride, FeCl3, by fractional distillation. Finally ZrCl4
is reduced to metallic zirconium by reduction with
magnesium, Mg. Air is excluded so as to prevent
contamination of the product with oxygen or nitrogen.
ZrO2 + 2Cl2 + 2C (900°C) → ZrCl4 + 2CO
ZrCl4 + 2Mg (1100°C) → 2MgCl2 + Zr
Excess magnesium and magnesium dichloride is removed
from the product by treatment with water and
hydrochloric acid to leave a zirconium "sponge". This
can be melted under helium by electrical heating.
XtraEdge for IIT-JEE 44 JULY 2010

Students' Forum
Expert’s Solution for Question asked by IIT-JEE Aspirants
MATHS
1. Let f(x) be a function which satisfy the equation
f(xy) = f(x) + f(y) for all x > 0, y > 0 such that
f ´(1) = 2. Find the area of region bounded by the
curves y = f(x), y = |x 3 – 6x 2 + 11x – 6| and x = 0
Sol. Take x = y = 1 ⇒ f(1) = 0
⎛ 1 ⎞ ⎛ 1 ⎞
Now f ⎜ x . ⎟ = f(x) + f ⎜ ⎟
⎝ x ⎠ ⎝ x ⎠
⎛ 1 ⎞
⇒ f ⎜ ⎟ = – f(x)
⎝ x ⎠
⎛ x
f ⎟ ⎞ ⎛ 1
⎜ = f(x) + f
⎝ y
⎟ ⎞
⎜ = f(x) – f(y)
⎠ ⎝ y ⎠
f (x + h) − f (x)
f ´(x) = lim
h→0
h
⎛ h ⎞
f ⎜1
+ ⎟ − f (1)
1 ⎛ x + h ⎞ ⎝ x ⎠
= lim f ⎜ ⎟ = lim
h→0
h ⎝ h ⎠ h→0
h
. x
x
f´(1) 2
= =
x x
⇒ f(x) = 2 log |x| + c ⇒ c = 0
{when x = 1; as f(1) = 0}
⇒ f(x) = 2logx
∴ Required area
1
=
∫
( x − 6x + 11x − 6)dx + e dy
0
3
= 4
7 sq units
2
0
y / 2
∫–
∞
⎛
= cos x . ⎜
⎜
⎝
n
∑
k=
1
cosa
2
k
k−1
⎞ ⎛
⎟ – sin x ⎜
⎟ ⎜
⎠ ⎝
n
∑
k=
1
= A cos x – B sin x, where A = ∑
k=
n
sin a
B = ∑
k
k−1
k=
1
2
since f(x 1 ) = f(x 2 ) = 0
⇒ A cos x 1 – B sin x 1 = 0
and A cos x 2 – B sin x 2 = 0
A
⇒ tan x 1 = B
⇒ tan x 2 = B
A
⇒ tan x 1 = tan x 2
⇒ (x 2 – x 1 ) = mπ
n
1
sin a
2
k
k−1
cosa
2
⎞
⎟
⎟
⎠
k
k−1
and
3. Let a variable chord from (–1, 0) point to the circle
(x – 2) 2 + y 2 = 1, makes a intercept of length 'l' on the
circle and length of perpendicular from centre of the
circle to chord is 'p'. find the range of 'λ' such that
l 2 + 3λ p 2 + 5 = 0.
Sol. We have OB 2 = OD 2 + BD 2
A
(–1, 0) 0
D
p
B
2. Let a 1 , a 2 , ......, a n be real constant, x be a real variable
and f(x) = cos(a 1 + x) + 2
1 cos(a2 + x) + 4
1 cos(a3 + x)
1
+...... + cos(a
n−1
n + x). Given that f(x 1 ) = f(x 2 ) = 0,
2
prove that (x 2 – x 1 ) = mπ for integer m.
Sol. f(x) may be written as,
n
f(x) = ∑
k=
n
= ∑
k=
1
1
k
2
1
k
2
1 cos(ak + x)
−1
−1
{cosa k . cos x – sin a k . sin x}
2
2
1 = p 2 l
+ ⇒ p 2 4 − l
= 4 4
we have been given, l 2 + 3λp 2 + 5 = 0
l 2 3λ ⎛
2
⎞
+ ⎜
4 − l
⎟ + 5 = 0
4
⎝ 4 ⎠
l 2 12λ + 20
=
3λ − 4
clearly 0 ≤ l 2 < 4
4( λ + 5/ 3)
⇒ 0 ≤
< 4
( λ − 4 / 3)
⇒ λ ∈ (– ∞, –5/3)
[as λ can not be + ve]
XtraEdge for IIT-JEE 45 JULY 2010

4. Find all possible negative real values of 'a' such that
∫
a
0
−2t
Sol. Here,
∫
−t
( 9 − 2.9 ) dt ≥ 0
a
0
−2t
( 9 − 2.9 ) dt ≥ 0
⎛
−2t
−t
9 2.9 ⎞
⇒ ⎜
⎟
− ≥ 0
2log9 log9
⎝ − − ⎠a
−2t
−t
⇒ ( − 9 + (49) ) 0 a
≥ 0
⇒ 9 –2a – 4.9 –a + 3 ≥ 0
t 2 – 4t + 3 ≥ 0
where t = 9 –a and t ∈(1, ∞)
(t – 1) (t – 3) ≥ 0
⇒ t ≤ 1 or t ≥ 3
⇒ t ≥ 3 is possible as t > 1
9 –a 1
≥ 3 ⇒ a ≤ – 2
t
0
⇒
2
⎛ a − a ⎞
⎜ ⎟
⎝ 2 ⎠
2
⎛ a + a ⎞
+ ⎜ ⎟ = 0
⎝ 2 ⎠
2
⎛ a − a ⎞ ⎛ a + a ⎞
⇒ – ⎜ ⎟ + ⎜ ⎟ = 0
⎝ 2i ⎠ ⎝ 2 ⎠
⇒ – sin 2 θ + cos θ = 0
⇒ cos θ = sin 2 θ
...(iii)
Now, f(x) = x 3 – 3x 2 + 3(1 + cos θ)x + 5
f ´(x) = 3x 2 – 6x + 3 (1 + cos θ)
∴ Discriminate (D)
= 36 – 36(1 + cos θ) = – 36 cos θ
= – 36 sin 2 θ < 0
⇒ f(x) is increasing ∀ x ∈ R
2
5. Let a 0 , a 1 , .... a n – 1 be real numbers where n ≥ 1 and
het f(x) = x n + a n – 1 x n –1 + ..... + a 0 be such that :
|f(0)| = f(1) and each root of f(x) = 0 is real and lies
between 0 and 1. Prove that the product of the roots
1
does not exceed
n .
2
Sol. Let, f(x) = (x – α 1 ) (x – α 2 ) ..... (x – α n )
where α 1 , α 2 ........, α n are the roots of f(x) = 0
since |f(0)| = f(1)
∴ α 1 . α 2 ...... α n = (1 – α 1 ) (1 – α 2 ) ...... (1 – α n )
⇒ (α 1 . α 2 ....... α n ) 2
⇒ α 1 (1 – α 1 ) α 2 (1 – α 2 ) ...... α n (1 – α n )
⇒ (Π α i ) 2 = II α i (1 – α i ) . {i = 1, 2, ..... n}
Now, ( Π α i ) 2 = Π α i (1 – α i ) ≤ Π
1
=
2n
2
Since GM ≤ AM
1
⇒ ( Π α i ) ≤
n
2
⎧α + (1 − αi
) ⎫
⎨ ⎬
⎩ 2 ⎭
6. If the equation az 2 + z + 1 = 0 has a purely imaginary
root where a = cos θ + i sin θ, i = − 1 . Then find
the interval in which the function, f(x) = x 3 – 3x 2 +
3(1 + cos θ)x + 5 is increasing.
Sol. We have, az 2 + z + 1 = 0
...(i)
⇒ az 2 + z + 1 = 0 {Taking conjugate on both sides}
⇒ a z 2 + z + 1 = 0
...(ii)
Eliminating z from eq. (i) and (ii) by cross
multiplication rule,
(a – a) 2 + 2(a + a ) = 0
2
Regents Physics
You Should Know Electricity
1. A coulomb is charge, an amp is current
[coulomb/sec] and a volt is potential difference
[joule/coulomb].
2. Short fat cold wires make the best conductors.
3. Electrons and protons have equal amounts of
charge (1.6 x 10 -19 coulombs each).
4. Adding a resistor in parallel decreases the total
resistance of a circuit.
5. Adding a resistor in series increases the total
resistance of a circuit.
6. All resistors in series have equal current (I).
7. All resistors in parallel have equal voltage (V).
8. If two charged spheres touch each other add
the charges and divide by two to find the final
charge on each sphere.
9. Insulators contain no free electrons.
10. Ionized gases conduct electric current using
positive ions, negative ions and electrons.
11. Electric fields all point in the direction of the
force on a positive test charge.
12. Electric fields between two parallel plates are
uniform in strength except at the edges.
13. Millikan determined the charge on a single
electron using his famous oil-drop experiment.
14. All charge changes result from the movement
of electrons not protons (an object becomes
positive by losing electrons)
XtraEdge for IIT-JEE 46 JULY 2010

MATHS
3-DIMENSIONAL
GEOMETRY
Mathematics Fundamentals
Coordinates of a point :
Z
L
O
x
y
P (x,y,z)
Y
x-coordinate = perpendicular distance of P from
yz-plane
y-coordinate = perpendicular distance of P from
zx-plane
z-coordinate = perpendicular distance of P from
xy-plane
Coordinates of a point on the coordinate planes and axes:
yz-plane : x = 0
zx-plane : y = 0
xy-plane : z = 0
x-axis : y = 0, z = 0
y-axis : y = 0, x = 0
z-axis : x = 0, y = 0
Distance between two points :
If P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) are two points, then
distance between them
z
N
M
2
2
2
1 − x2)
+ (y1
− y2)
+ (z1
z2)
PQ = ( x
−
Coordinates of division point :
Coordinates of the point dividing the line joining two
points P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) in the ratio
m 1 : m 2 are
(i) in case of internal division
⎛ m ⎞
⎜ 1x2
+ m2x1
m1y2
+ m2y1
m1z
2 + m2z
,
,
1
⎟
⎝ m1
+ m2
m1
+ m2
m1
+ m2
⎠
(ii) in case of external division
⎛ m1x
2 − m 2x1
m1y
2 − m 2 y1
m1z
2 − m 2z1
⎞
⎜
,
,
⎟
⎝ m1
− m 2 m1
− m 2 m1
− m 2 ⎠
X
Note: When m 1 , m 2 are in opposite sign, then
division will be external.
Coordinates of the midpoint:
When division point is the mid-point of PQ, then
ration will be 1 : 1; hence coordinates of the midpoint
of PQ are
⎛ 1 + x2
y + y z + z
⎜ ,
1 2
,
1
⎝ 2 2 2
x 2
Coordinates of the general point :
The coordinates of any point lying on the line joining
points P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) may be taken as
⎛
⎜
⎝
⎞
⎟
⎠
kx2
+ x1
ky + y kz + z
,
2 1
,
2 1
k + 1 k + 1 k + 1
which divides PQ in the ratio k : 1. This is called
general point on the line PQ.
Division by coordinate planes :
The ratios in which the line segment PQ joining
P(x 1 , y 1 , z 1 ) and Q(x 2 , y 2 , z 2 ) is divided by coordinate
planes are as follows :
(i) by yz-plane : –x 1 /x 2 ratio
(ii) by zx-plane : – y 1 /y 2 ratio
(iii) by xy-plane : –z 1 /z 2 ratio
Coordinates of the centroid :
(i) If (x 1 , y 1 , z 1 ); (x 2 , y 2 , z 2 ) and (x 3 , y 3 , z 3 ) are
vertices of a triangle then coordinates of its centroid
are
⎞
⎟
⎠
⎛ 1 + x 2 + x 3 y1
+ y 2 + y3
z1
+ z 2 + z
⎜
,
,
⎝ 3
3
3
x 3
(ii) If (x r , y r , z r ); r = 1, 2, 3, 4 are vertices of a
tetrahedron, then coordinates of its centroid are
⎛ 1 + x 2 + x 3 + x 4 y1
+ y 2 + y3
+ y 4 z1
+ z 2 + z 3 + z
⎜
,
,
⎝ 4
4
4
x 4
Direction cosines of a line [Dc's] :
The cosines of the angles made by a line with
positive direction of coordinate axes are called the
direction cosines of that line.
Let α, β, γ be the angles made by a line AB with
positive direction of coordinate axes then cos α, cos
β, cos γ are the direction cosines of AB which are
generally denoted by l, m, n. Hence
l = cos α, m = cos β, n = cos γ
⎞
⎟
⎠
⎞
⎟
⎠
XtraEdge for IIT-JEE 47 JULY 2010

x-axis makes 0º, 90º and 90º angles with three
coordinate axes, so its direction cosines are cos 0º,
cos 90º, cos 90º i.e. 1, 0, 0. Similarly direction
cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1
respectively. Hence
dc's of x-axis = 1, 0, 0
dc's of y-axis = 0, 1, 0
dc's of z-axis = 0, 0, 1
Relation between dc's
∴ l 2 + m 2 + n 2 = 1
Direction ratios of a line [DR's] :
Three numbers which are proportional to the
direction cosines of a line are called the direction
ratios of that line. If a, b, c are such numbers which
are proportional to the direction cosines l, m, n of a
line then a, b, c are direction ratios of the line. Hence
a
⇒ l = ±
,
2 2 2
a + b + c
b
c
m = ±
, n = ±
2 2 2
2 2 2
a + b + c a + b + c
Direction cosines of a line joining two points :
Let ≡ (x 1 , y 1 , z 1 ) and Q ≡ (x 2 , y 2 , z 2 ); then
(i) dr's of PQ : (x 2 – x 1 ), (y 2 – y 1 ), (z 2 – z 1 )
x z z
(ii)dc's of PQ :
2 − x1
y y
,
2 − 1
,
2 − 1
PQ PQ PQ
i.e.,
x
2
Σ(x
− x
−
2
1
2
x1)
,
y
2
Σ(x
− y
2
−
1
2
x1)
,
z
2
Σ(x
− z
−
2
1
2
x1)
Angle between two lines :
Case I. When dc's of the lines are given
If l 1 , m 1 , and l 2 ,m 2 n 2 are dc's of given two lines, then
the angle θ between them is given by
cos θ = l 1 l 2 + m 1 m 2 + n 1 n 2
sin θ = 2
2
2
( l1 m2
− l 2m1)
+ (m1n
2 − m2n1)
+ (n1l
2 − n2l1)
The value of sin θ can easily be obtained by the
following form :
sin θ =
l1
l 2
m
m
2
1
2
m
+
m
1
2
n
n
2
1
2
n
+
n
1
2
2
l1
l 2
Case II. When dr's of the lines are given
If a 1 , b 1 , c 1 and a 2 , b 2 , c 2 are dr's of given two lines,
then the angle θ between them is given by
cos θ =
sin θ =
a
a
2
1
2
1
a a
+
1 2
2
b1
+ b
2
1
+ b b
+ c
Σ(a
b
1
1
2
1
2
2
1
+ c
+ c c
2
2
a2
− a
a
2
2
1 2
2
b2
+
2
2b1)
+ b
2
2
+ c
+ c
2
2
2
2
Conditions of parallelism and perpendicularity of two
lines :
Case I. When dc's of two lines AB and CD, say l 1 ,
m 1 ,n 1 and l 2, m 2 , n 2 are known
AB || CD ⇔ l 1 = l 2, m 1 = m 2 , n 1 = n 2
AB ⊥ CD ⇔ l 1 l 2 + m 1 m 2 + n 1 n 2 = 0.
Case II. When dr's of two lines AB and CD, say : a 1 ,
b 1 , c 1 and a 2 , b 2 , c 2 are known
a
AB || CD ⇔
1 b1
c
= =
1
a2
b2
c2
AB ⊥ CD ⇔ a 1 a 2 + b 1 b 2 + c 1 c 2 = 0.
Area of a triangle :
Let A(x 1 , y 1 , z 1 ); B(x 2 , y 2 , z 2 ) and C(x 3 , y 3 , z 3 ) are
vertices of a triangle. Then
dr's of AB = x 2 – x 1 , y 2 – y 1 , z 2 – z 1
= a 1 , b 1 , c 1 (say)
2 2 2
and AB = a 1 + b1
+ c1
dr's of BC = x 3 – x 2 , y 3 – y 2 , z 3 – z 2
= a 2 , b 2 , c 2 (say)
2 2 2
and BC = a 2 + b2
+ c2
Now sin B =
=
2
Σ(b1c
2 − b2c1)
2
Σa1
2
Σa2
2
Σ(b1c
2 − b2c1)
AB.BC
∴ Area of ∆ABC = 2
1 AB. BC sin B
=
1
2
2
Σ (b1 c2
− b2c1)
Projection of a line segment joining two points on a line :
Let PQ be a line segment where P ≡ (x 1 , y 1 , z 1 ) and
Q ≡ (x 2 , y 2 , z 2 ); and AB be a given line with dc's as l,
m, n. If P'Q' be the projection of PQ on AB, then
P'Q' = PQ cos θ
where θ is the angle between PQ and AB. On
replacing the value of cos θ in this, we shall get the
following value of P'Q'.
P'Q' = l (x 2 – x 1 ) + m(y 2 – y 1 ) + n (z 2 – z 1 )
Projection of PQ on x-axis : a = |x 2 – x 1 |
Projection of PQ on y-axis : b = |y 2 – y 1 |
Projection of PQ on z-axis : c = |z 2 – z 1 |
Length of line segment PQ = a + b + c
*
x − α y − β z − γ
If the given lines are = =
l m n
and
x − α´
y − β´ z − γ´
= = ,
l´
m´ n´
then condition for
intersection is
2
2
2
XtraEdge for IIT-JEE 48 JULY 2010

If the given lines are
x − α´
y − β´
= =
l´
m´
intersections is
α – α´
l
l
β − β´
m
m´
γ – γ´
n
n´
x − α y − β z − γ
= = and
l m n
z − γ´
, then condition for
n´
= 0
Plane containing the above two lines is
x – α
l
l´
y − β
m
m´
z –
n
n´
γ
= 0
Condition of coplanarity if both the lines are in general
form:
Let the lines be
ax + by + cz + d = 0 = a´x + b´y + c´z + d´
and αx + βy + γz + δ = 0 = α´x + β´y + γ´z + δ´
These are coplanar if
a
a´
α
α´
b
b´
β
β´
c
c´
γ
γ´
d
d´
δ
δ´
= 0
Reduction of non-symmetrical form to symmetrical form:
Let equation of the line in non-symmetrical form be'
a 1 x + b 1 y + c 1 z + d 1 = 0; a 2 x + b 2 y + c 2 z + d 2 = 0.
To find the equation of the line in symmetrical form,
we must know (i) its direction ratios (ii) coordinates
of any point on it.
Direction ratios : Let l, m, n be the direction ratios
of the line. Since the line lies in both the planes, it
must be perpendicular to normals of both planes. So
a 1 l + b 1 m + c 1 n = 0; a 2 l + b 2 m + c 2 n = 0
From these equations, proportional values of l, m, n
can be found by cross-multiplication as
or
b
l
−
1c2
b2c1
=
c
m
−
1a
2 c2a1
=
a
n
−
1b2
a 2b1
Point on the line : Note that as l, m, n cannot be
zero simultaneously, so at least one must be nonzero.
Let a 1 b 2 – a 2 b 1 ≠ 0, then the line cannot be
parallel to xy-plane, so it intersect it. Let it intersect
xy-plane in (x 1 ,y 1 , 0). Then
a 1 x 1 + b 1 y 1 + d 1 = 0 and a 2 x 1 + b 2 y 1 + d 2 = 0
Solving these, we get a point on the line. Then its
equation becomes
x − x1
b c − b c
1
1
2
2
2
2
1
1
b1d
2 − b2d
x −
a1b2
− a 2b
b c − b c
1
1
=
y − y
c a − c
1
2
1
2
a
1
=
d1a
2 − d 2a
y −
a1b2
− a 2b
=
c a − c a
1
2
2
1
1
1
a
z − 0
1b2
− a 2b1
z − 0
=
a b − a
1 2 2b1
Note : If l ≠ 0, take a point on yz –plane as (0, t 1 , z 1 )
and if m ≠ 0, take a point on xz-plane as (x 1 , 0, z 1 )
Skew lines : The straight lines which are not parallel
and non-coplanar i.e. non-intersecting are called
skew lines.
If ∆ =
x – α
l
l´
y − β
m
m´
z –
n
n´
γ
≠ 0, the lines are skew.
Shortest distance : Suppose the equation of the lines
x − α y − β z − γ
are = =
l m n
and
S.D. =
=
x − α´
=
l´
α – α´
l
l´
y − β´
=
m´
z − γ´
. Then
n´
( α − α´)(mn´
−m´n)
+ ( β − β´)(nl´
−n´
l)(
lm´
−l´m)
β − β´
m
m´
γ – γ´
n
n´
Σ(mn´
−m´n)
Some results for plane and straight line:
(i) General equation of a plane :
ax + by + cz + d = 0
where a, b, c are dr's of a normal to this plane.
(ii) Equation of a straight line :
a1x
+ b1y
+ c1z
+ d1
= 0 ⎫
General form :
⎬
a2x
+ b2y
+ c2z
+ d2
= 0⎭
(In fact it is the straight line which is the intersection
of two given planes)
x − x y y z z
Symmetric form :
1 − 1 −
= =
1
a b c
where (x 1 , y 1 , z 1 ) is a point on this line and a, b, c are
its dr's
(iii) Angle between two planes :
If θ be the angle between planes a 1 x + b 1 y c 1 z + d 1 = 0
and a 2 x + b 2 y + c 2 z + d 2 = 0, then
cos θ =
a
2
1
a a
+
1 2
2
b1
+ b b
+ c
1
2
1
+ c c
2
2
a2
1 2
2
b2
+
2
+ c
(In fact angle between two planes is the angle
between their normals.)
Further above two planes are
a1
b1
c1
parallel ⇔ = =
a2
b2
c2
perpendicular ⇔ a 1 a 2 + b 1 b 2 + c 1 c 2 = 0
2
2
XtraEdge for IIT-JEE 49 JULY 2010

MATHS
PROGRESSION & MATHEMATICAL
INDUCTION
Mathematics Fundamentals
Arithmetic Progression (AP)
AP is a progression in which the difference between
any two consecutive terms is constant. This constant
difference is called common difference (c.d.) and
generally it is denoted by d.
Standard form: Its standard form is
a + (a + d) + (a + 2d) +..........
General term :
T n = a + (n – 1) d
If T n = l then it should be noted that
l − a
l − a
(i) d =
(ii) n = 1+
n −1
d
Note:
a ,b,c are in AP ⇔ 2b = a + c
Sum of n terms of an AP :
S n
=
n
(a
2
+ l)
where l is last term (nth term). Replacing the value of
l, it takes the form
S n
=
n
[2a
2
+ (n −1)d]
Arithmetic Mean :
(i) If A be the AM between two numbers a and b,
1
then A = (a + b)
2
(ii) The AM of n numbers a 1 , a 2 ,..............,a n
= n
1 (a1 + a 2 +........+ a n )
(iii) n AM's between two numbers
If A 1 , A 2 ,....., A n be n AM's between a and b then
a A 1 , A 2 ,....., A n , b is an AP of (n + 2) terms. Its common
difference d is given by
b − a
T n+2 = b = a + (n + 1)d ⇒ d =
n + 1
so A 1 = a + d, A 2 = a + 2d,....., A n = a + nd.
Sum of n AM's between a and b
∴ ΣA n = n(A)
Assuming numbers in AP :
(i) When number of terms be odd
Three terms : a –d, a, a + d
Five terms : a – 2d, a–d, a, a + d, a + 2d
................ ....... ....... ....... .......
(ii) When number of terms be even
Four terms: a – 3d, a – d, a + d, a + 3d
Six terms : a –5d, a – 3d, a –d, a + d, a+3d,
a + 5d
............... ........ ...... ...... ...... ......
Geometrical Progression (GP) :
A progression is called a GP if the ratio of its each
term to its previous term is always constant. This
constant ratio is called its common ratio and it is
generally denoted by r.
Standard form : Its standard form is
a + ar + ar 2 +.........
General term : T n = ar n–1
a, b, c are in GP ⇔
b c = ⇔ b 2 = ac
a b
Sum of n terms of a GP :
The sum of n terms of a GP a + ar + ar 2 +....... is
given by
⎧ n
a(1 − r ) a − lr
⎪ = ,when r < 1
S n = ⎨
1−
r 1−
r
n
⎪a(r
−1)
lr
− a
= ,when r > 1
⎪⎩
r −1
r −1
when l = T n .
Sum of an infinite GP :
(i) When r > 1, then r n → ∞, so S n → ∞ Thus when
r > 1, the sum S of infinite GP = ∞
(ii) When | r | < 1, then r n → 0, so
a
S =
1−
r
(iii) When r = 1, then each term is a so S = ∞.
Geometric Mean :
(i) If G be the GM between a and b then
G = ab
(ii) G.M. of n numbers a 1 , a 2 ......, a n = (a 1 a 2 a 3 .....a n ) 1/n
(iii) n GM’s between two numbers
⇒ r = (b/a) 1/n+1
XtraEdge for IIT-JEE 50 JULY 2010

Product of n GM's between a and b
Product of GM's = (ab) n/2 = G n
Assuming numbers in GP :
(i) When number of terms be odd
Three terms : a/r, a, ar
Five terms : a/r 2 , a/r, a, ar, ar 2
............... .. ..... ..... ..... ..... .....
(ii) When number of terms be even
Four terms : a/r 3 , a/r, ar, ar 3
Six terms : a/r 5 , a/r 3 , a/r, ar, ar 3 , ar 5
Arithmetic-Geometric Progression :
If each term of a progression is the product of the
corresponding terms of an AP and a GP, then it is
called arithmetic-geometric progression (AGP). For
example:
a, (a + d)r, (a + 2d)r 2 .......
T n = [a + (n – 1)d] r n–1
S n =
a
1
n−1
2
dr(1 − r
+
− r (1 − r)
a dr
S ∞ = +
1−
r (1 − r)
2
)
–
[a + (n −1)d]r
1−
r
| r | < 1
Harmonic Progression :
A progression is called a harmonic progression (HP)
if the reciprocals of its terms are in AP.
1 1 1
Standard form : + + +.............
a a + d a + 2d
General term :
T n
1
=
a + (n −1)d
2 1 1 2ac
∴ a, b, c are in HP ⇔ = + ⇔ b =
b a c a + c
Harmonic Mean :
(i) If H be a HM between two numbers a and b, then
2ab
H = or
a + b
2
H
=
1
a
(ii) To find n HM's between a and b we first find n
AM's between 1/a and 1/b, then their reciprocals will
be the required HM's.
Relations between AM, GM and HM :
G 2 = AH
A > G > H, when a, b > 0.
If A and AM and GM respectively between two
positive numbers, then those numbers are
A +
A
2
− G
2
+
1
b
,A −
A
2
− G
Some Important Results :
If number of terms in an AP/GP/HP is odd then
its mid term is the AM/GM/HM between the first
and last term.
2
n
If number of terms in an AP/GP/HP is even the
AM/GM/HM of its two middle terms is equal to
the AM/GM/HM between the first and last term.
a, b, c are in AP, GP and HP ⇔ a = b = c
a, b, c are in AP and HP ⇒ a, b,c are in GP.
a, b, c are in AP
⇔
1 1
,
bc ca
1
,
ab
are in AP. ⇔ bc, ca, ab are in HP.
a, b, c are in GP ⇔ a 2 , b 2 , c 2 are in GP.
a, b, c are in GP ⇔ loga, logb, logc are in AP.
a, b, c are in GP ⇔ log a m log b m, log c m are in HP.
a, b, c d are in GP ⇔ a + b, b + c, c + d are in GP.
a, b, c are in AP ⇔ α a , α b , α c are in GP (α ∈ R 0 )
Principle of Mathematical Induction :
It states that any statement P(n) is true for all positive
integral values of n if
(i) P(1) is true i.e., it is true for n = 1.
(ii) P(m) is true ⇒ P(m + 1) is also true
i.e., if the statement is true for n = m then it must also
be true for n = m + 1.
Some Formula based on the Principle of Induction :
n (n + 1)
Σn = 1 + 2 + 3 +....... + n =
2
(Sum of first n natural numbers)
Σ(2n – 1) = 1 + 3 + 5 + ... + (2n – 1) = n 2
(Sum of first n odd numbers)
Σ2n = 2 + 4 + 6 + ...... + 2n = n(n + 1)
(Sum of first n even numbers)
Σn 2 = 1 2 + 2 2 + 3 2 +.......+ n 2 n (n + 1) (2n + 1)
=
6
(Sum of the squares of first n natural numbers)
Σn 3 = 1 3 + 2 3 + 3 3 +.......+ n 3 n (n + 1)
=
4
(Sum of the cubes of first n natural numbers)
Application in Solving Objective Question :
For solving objective question related to natural
numbers we find out the correct alternative by
negative examination of this principle. If the given
statement is P(n), then by putting n = 1, 2, 3, ..... in
P(n), we decide the correct answer.
We also use the above formulae established by this
principle to find the sum of n terms of a given series.
For this we first express T n as a polynomial in n and
then for finding S n , we put Σ before each term of this
polynomial and then use above results of Σn, Σn 2 , Σn 3
etc.
2
2
XtraEdge for IIT-JEE 51 JUNE 2010

Based on New Pattern
IIT-JEE 2011
XtraEdge Test Series # 3
Time : 3 Hours
Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound
wave, Doppler's effect, Atomic Structure, Radioactivity, X-ray, Nuclear Physics, Matter Waves, Photoelectric Effect,
Practical Physics. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature,
Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements, Nitrogen Family, Oxygen Family,
Halogen Family & Noble Gas, Salt Analysis, Metallurgy, Co-ordination Compounds, Transitional Elements. Mathematics:
Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D, Probability, Determinants, Matrices.
Instructions :
Section - I
• Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer and
-2 mark for wrong answer.
Section - II
• Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and
-1 mark for wrong answer.
Section - III
• Question 13 to 14 are Column Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong answer.
Section - IV
• Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer
and 0 mark for wrong answer.
PHYSICS
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1. The figure shows two isotherms at temperatures T 1
and T 2 . A gas is taken from one isotherm to another
isotherm through different processes. Then change in
internal energy ∆U has relation –
P b
3.
(A)
R
1– γ
R
(B)
γ – 1
(C) R (D) R/2
T
A B C
(a)
x
T
T
A B C
(b)
x
a
c
d
e
T 1
(A) ∆U ab > ∆U ac > ∆U ad > ∆U ac
(B) ∆U ab = ∆U ac > ∆U ad > ∆U ac
(C) ∆U ab = ∆U ac = ∆U ad = ∆U ac
(D) ∆U ab < ∆U ac < ∆U ad < ∆U ac
2. An ideal gas whose adiabatic exponent is γ is
expanded so that the amount of heat transferred to the
gas is equal to the decrease of its internal energy.
Molar heat capacity of the gas for this process is -
T 2
V
A B C
(c)
The above graphs shows conduction of heat through
materials A,B,C connected in series. Graph shows
variations of temperature with distnace x-axis. Which
of the above graph are not possible -
(A) a, b, c
(B) a, b
(C) a, c (D) b, c
x
XtraEdge for IIT-JEE 52
JULY 2010

4. Figure shows a rectangular pulse and a triangular
pulse approaching each other along x-axis. The pulse
speed is 0.5 cm/s. What is the resultant displacement
of medium particles due to superposition of waves x
= 0.5 cm and t = 2 sec.
y (cm)
0.5 cm/s 0.5 cm/s
2
–2 –1 0 1 2
(A) 3.5 cm
(B) 2.5 cm
(C) 4 cm (D) 3 cm
1
3
x (cm)
5. Choose the correct statement (s) related to the
photocurrent and the potential difference between the
plate and the collector -
(A) Photocurrent always increase with the increase in
potential difference
(B) when the potential difference is zero, the
photocurrent is also zero
(C) Photocurrent attain a saturation value of some
positive value of the potential difference
(D) None of these
6. Binding energy per nucleon of 1 H 2 and 2He 4 are
1.1 MeV and 7.0 MeV respectively. Energy released
in the process 1 H 2 + 1 H 2 = 2He 4 is -
(A) 20.8 MeV (B) 16.6 MeV
(C) 25.2 MeV (D) 23.6 MeV
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
Passage : I (Ques. 7 to 9)
Two parallel plates in vacuum, separated by a small
distance which is small compared with their linear
dimensions, are at temperatures T 1 & T 2 respectively
(T 1 > T 2 ). The plates are black bodies. Another plate
(black body) at temperature T 0 is a kept in between
the two plates.
7. Temperature of the plate kept i.e. T 0 is -
4 4
T1 T2
(A)
+ 4 4
4 T1 – T2
4
= T 0 (B) = T 0
2
2
(C) T 1 T 2 = T 0 (D) 2T 4 1 – T 4 4
2 = T 0
8. Energy absorbed by the plate having temperature T 0 ,
per sec per unit area is -
(A) σ(T 1 4 –T 2 4 )
4
1
(T
(C) σ
4
– T2
)
2
4
1
(T
(B) σ
4
0
(T
(D) σ
4
– T0
)
2
4
– T2
)
2
9. If 'n' black Body plates are placed in between the two
plates having temperature T 1 & T 2 , then the net rate
of emission of radiant energy by first plate is -
σ
(A) 4 (T1 – T 4 σ
2 ) (B)
4 (T1 – T 4 2 )
n n +1
(C) σ
n
n
(T 4 1 – T 4 2 ) (D)
+ 1
Passage: II (Ques. 10 to 12)
σ 4 (T1 – T 4 2 )
n – 1
Two hydrogen like atoms A and
equal number of protons and neutrons. The energy
difference between the radiation corresponding to
first Balmer lines emitted A and B is 5.667 eV. When
the atoms A and B moving with the dame velocity,
strikes a heavy target they rebound back with the
same velocity. In this process the atom B imparts
twice the momentum to the target than the A imparts.
10. Ionization energy of Atom B is -
(A) 27.2 eV (B) 13.6 eV
(C) 10.2 eV (D) 54.4 eV
11. Atomic number of atom A is -
(A) 1 (B) 2
(C) 3 (D) 4
12. Mass number of atom B & atom A
(A) 2, 4 (B) 4, 2
(C) 2, 1 (D) 4, 1
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
XtraEdge for IIT-JEE 53
JULY 2010

13. Match the Column-I with Column-II :
Column-I
Column-II
(A) An electron moves (P) Total Energy
Potential Energy
in an orbit in a =
2
Bohr atom
(B) As a satellite moves (Q) Kinetic Energy =
in a circular orbit Magnitude of Energy
around total earth
(C) In Rutherford's (R) Motion is under a
α-scattering experiment central force
as an α-particle moves
in the electric field of
a nucleus
(D) As an object, released (S) Mechanical energy
from some height above is conserved
ground, falls towards
earth, assuming negligible
air resistance
(T) None of these
14. Match the following :
Column-I`
Column-II
(A) Steady state (P) A blackened platinum
wire, when gradually
heated appear first red
and then blue.
(B) Wein's displacement (Q) Radiated power is
law
proportional to fourth
power of absolute
temperature of body
(C) Stefan's law (R) Energy absorbed
is equal to energy
emitted
(D) Black body (S) Absorptive power of
body is unity
(T) None of these
This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
15. Four cylindrical rods of same material with length
and radius (l,r), (2l,r), (2l,2r) and (l,2r) are
connected between two reservoirs at 0ºC and 100ºC.
Find the ratio of the maximum to minimum rate of
conduction in them.
16. Using solar constant S = 20 kilo cal/mm –m 2 and
Joule's constant J = 4.2 J/cal, find the pressure
exerted by sunlight (Ans. in …… × 10 –6 N/m 2 )
17. The minimum intensity of audible sound is 10 –12
W/m 2 sec and density of air is 1.3 kg/m 3 . If the
frequency of sound is 1000 Hz, find the amplitude
(Ans. in …… × 10 –11 m) of vibration [Speed of
sound = 332 ms –1 ]
18. The size of a nucleus is of the order of –14 m.
Calculate the velocity with which protons move
inside the nucleus. The mass of a proton
= 1.675 × 10 –27 kg. [Ans. in …… × 10 7 ms –1 ]
19. Find the change in frequency of red light whose
original frequency is 7.3 × 10 14 Hz when it falls
through 22.5 m losing gravitational potential energy.
[Ans. in …… ×10 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. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1. The optically active tartaric acid is named as D–(+)–
tartaric acid because it has a positive -
(A) optical rotation and is derived from D-glucose
(B) pH in organic solvent
(C) optical rotation and is derived from D–(+)–
glyceraldehyde
(D) optical rotation only when substituted by
deuterium
2. Which of the following compounds is not coloured
(A) Na 2 [CuCl 4 ] (B) Na 2 [CdCl 2 ]
(C) K 4 [Fe(CN) 6 ] (D) K 3 [Fe(CN) 6 ]
3. The brown ring complex compound is formulated as
[Fe(H 2 O) 5 (NO)]SO 4 . The oxidation state of iron is -
(A) 1 (B) 0 (C) 2 (D) 3
4. The following equilibrium is established when HCl is
dissolved the acetic acid,
HCl + CH 3 COOH Cl – + CH 3 COOH 2
+
the set that characterises the conjugate acid-base pairs is-
(A) (HCl, CH 3 COOH) and (Cl – , CH 3 COOH 2 + )
(B) (HCl, CH 3 COOH 2 + ) and (CH 3 COOH, Cl – )
(C) (CH 3 COOH 2 + ) HCl) and (Cl – , CH 3 COOH)
(D) (HCl, Cl – ) and (CH 3 COOH 2 + , CH 3 COOH)
XtraEdge for IIT-JEE 54
JULY 2010

5. Pure ammonia is placed in a vessel at a temperature
where its dissociation constant(α) is appreciable. At
equilibrium -
(A) k p does not change significantly with pressure
(B) does not change with pressure
(C) concentration of NH 3 does not change with
pressure
(D) concentration of hydrogen is less than that of
nitrogen
6. At constant temperature, the equilibrium constant (k p )
for the decomposition reaction N 2 O 4 2NO 2 is
expressed by k p = (4x 2 P)/(1–x 2 ), where P = pressure,
x = extent of decomposition. Which one of the
following statement is true
(A) k p increases with increase of P
(B) k p increases with increase of x
(C) k p increases with decrease of x
(D) k p remains constant with change in p and x
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
Passage : I (Ques. 7 to 9)
The IUPAC has set guidelines for logical and
methodical naming of organic compounds. The
complex substituents are written in small brackets
and their numbering is done separately. The bivalent
radicals are named by adding 'idene' to the name of
alkyl group. In polyfunctional compounds all lower
priority groups are written in prefix. Now name the
following compounds.
7. CHCHCH 2 OH is -
(CH 3 ) 2 CHOOC
Br
(A) 3-(3'-isopropoxycarbonyl cyclopentylidene
propane-1-ol
(B)3-(2'-bromo-3'-hydroxypropylidene) cyclopentane
carboxylate
(C) Iso-propyl-3-(2'-bromo-3'-hydroxy propylidenyl)
cyclopentane carboxylate
(D) Iso-propyl-3-(2'-bromo-3' hydroxypropylidene)
cyclopentane carboxylate
C 2 H 5
8. CH 3 CH 2 O CH 3 CH 2
is -
(A) 2-(3'-Ethylphenyl)-1-(4'-ethoxyphenyl) ethane
(B) 1-Ethyl-3-(2'-(4''-ethoxyphenyl) ethyl) benzene
(C) 1-(3'-Ethylphenyl)-2-(4'-ethoxylphenyl) ethane
(D) None of these
9.
Cl
O
OH
N
is -
O
(A) 3-chlorocarbonyl-6-(N, N-diethylamino) hex-4-
ene-1-oic acid
(B)4-chlorocarbonyl-3-(N, N-diethylamino) butanoic
acid
(C) 3-chlorocarbonyl-3-(3-N, N-diethylamino prop-
1'-enyl) butane-1-oic acid
(D) 3-chlorocarbonylmethyl-6- (N, N-diethylamino)
hex-4-en-1-oic acid
Passage: II (Ques. 10 to 12)
The property of hydrides of p-block elements mostly
depend on
(i) Electronegatively difference between central atom
and hydrogen
(ii) Size of central atom
(iii) Number of valence electrons in central atom. some
undergo hydride in which central atom is less
electronegative, react with OH – to given hydrogen
while acidic property of hydride in a period depends
on electronegativity of central atom i.e. more
electronegative is the atom, more acidic is hydride. In
a group, acidic property is proportional to size of
central atom.
Some electron deficient hydride behaves as lewis
acid while only one hydride of an element in p-plock
behaves as lewis base with lone pair of electrons.
Hydrides in which central atom's electronegativity to
close to hydrogen has no reaction with water.
10. The hydride which do not react with water is -
(A) NH 3 (B) PH 3 (C) B 2 H 6 (D) AsH 3
11. Which one undergoes spontaneous combustion with
exposure to air
(A) PH 3 (B) P 2 H 4 (C) N 2 H 4 (D) NH 3
12. Which one is strongest base
(A) OH – (B) HS – (C) HSe – (D) HTe –
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
A
B
C
D
P Q R S T
P Q R
P Q R
P Q R
P Q R
S T
S T
S T
S T
XtraEdge for IIT-JEE 55
JULY 2010

Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
13. Match the extraction processes listed in Column-I
with metals listed in Column-II :
Column-I Column-II
(A) Self reduction (P) Lead
(B) Carbon reduction (Q) Silver
(C) Complex formation (R) Copper
and displacement by
metal
(D) Decomposition of (S) Boron
iodide
(T) Au
14. Match statements in Column-I with appropriate
solution in Column-II
Column-I
Column-II
(A) A solution having (P) 5M HCl
pH less then 7
(B) A solution having (Q) 1M NaCl
pH more than 7
(C) A solution having (R) 0.1 M Na 2 CO 3
pH almost equal to 7
(D) solution having (S) 0.1 M CaCl 2
negative value of pH
(T) 1M H 2 SO 4
This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
15. A solution is prepared by mixing 50 mL 0.1 M HCl
with 50 mL 2.9 M CH 3 CH 2 COOH and 100 mL 0.2
M CH 3 CH 2 COONa. Find pH of the resulting
solution. K a for CH 3 CH 2 COOH is 1 × 10 –5 .
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
16. Calculate the change in pressure (in atm) when 2
mole of NO and 16 gram O 2 in a 6.25 litre originally
at 27ºC react to produce the maximum quantity of
NO 2 possible according to the equation –
2NO(g) + O 2 (g) → 2NO 2 (g)
(Take R = 12
1 ltr. Atm/mol-K)
17. the number of isomers for the compound with
molecular formula C 2 BrCIFI is.
18. In P 4 O 10 each P atom is linked with ...........O atoms.
19. 0.15 mole of Pyridinium chloride has been added
into 500 cm 3 of 0.2 M pyridine solution. Calculate pH
(K b for pyridine = 1.5 × 10 –9 M)
MATHEMATICS
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1.
⎡ 0 α α ⎤
If A =
⎢
⎥
⎢
2β
β – β
⎥
⎢⎣
γ – γ γ ⎥⎦
is an orthogonal matrix, then
the number of possible triplets (α,β,γ) is -
(A) 8 (B) 6
(C) 4 (D) 2
n
2. If ∑ αn = an 2 + + bn, where a, b are constants and
n=
1
α 1 , α 2 , α 3 ∈ {1, 2, 3........9} and 25α 1 , 37α 2 , 49α 3 be
⎡ α1
α2
α3
⎤
three digit number then
⎢
⎥
⎢
5 7 9
⎥
is
⎢⎣
25α1
37α2
49α3⎥⎦
equal to -
(A) α 1 + α 2 + α 3 (B) α 1 – α 2 + α 3
(C) 7 (D) 0
3. Consider the matrix A, B, C, D with order 2 × 3,
3 × 4, 4 × 4, 4 × 2 respectively. Let x = (αABγC 2 D) 3
where α & γ are scalars. Let |x| = k|ABC 2 D| 3 , then k
is -
(A) αγ (B)α 2 γ 2
(C) α 4 γ 4 (D) α 6 γ 6
4. Three numbers are selected at random from the set
{1, 2, 3...........N}, one by one without replacement. If
the first number is known to be smaller than second,
then the probability that third selected number lies
between the first two numbers is -
XtraEdge for IIT-JEE 56
JULY 2010

(A) 2
1
(B) 3
1
(C) 6
1
(D) 8
1
Passage: II (Ques. 10 to 12)
consider the equation of two straight lines
5. A bag 'A' contains 2 white and 3 red balls, another
bag 'B' contains 4 white and 5 red balls. If one ball is
drawn at random from one of the bag and it is found
to be red, the probability that it was drawn from the
bag B is -
89
(A) 245
25
(B) 52
93
(C) 256
24
(D) 663
6. Vectors → a , → b and → c with magnitude 2, 3 & 4
→
respectively are coplanar. A unit vector d is
perpendicular to all of them. If ( → a × → b ) × ( → c × → d ) =
^
i j^
k^
– + and the angle between → a and → b is 30º,
6 3 3
then
(A) 3
5
→ → →
c .^i + c .
^j + c .
^k is equal to -
(B) 9
5
(C) 12
5
(D) 18
5
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
Passage : I (Ques. 7 to 9)
There are four boxes A 1 , A 2 , A 3 and A 4 . Box A i has i
cards and on each card a number is printed, the
numbers are from 1 to i, A box is selected randomly,
1
the probability of selection of box A i is and then 10
a card is drawn. Let E i represents the event that a card
with number 'i' is drawn.
7. P(E 1 ) is equal to -
1 1
(A) (B) 5 10
8. P(A 3 /E 2 ) is equal to -
1 1
(A) (B) 4 3
(C) 5
2
(C) 2
1
(D) 4
1
(D) 3
2
9. Expectation of the number on the card is -
(A) 2 (B) 2.5 (C) 3 (D) 3.5
→
r = 3 ^i + 5 ^j + 7 ^k + λ( ^i – 2 ^j + ^k ) and
x +1 y + 1 z + 1
= = .
7 – 6 1
10. The equation of the line of shortest distance between
the given lines is -
(A) → r = ( 3
^i + 5
^j + 7
^k ) – δ( ^i + ^j + ^k )
(B) → r = ( ^i +
^j + ^k ) + δ(3 ^i + 5^j + 7 ^k )
(C) → r = ( 3
^i + 5
^j + 7
^k ) – δ(4 ^i + 6 ^j + 8 ^k )
(D) → r = ( 4
^i + 6
^j + 8
^k ) – δ( ^i + ^j + ^k )
11. The length of shortest distance between the lines is -
(A) 29 (B) 2 29
(C) 3 29 (D) 4 29
12. The point of intersection of the lines is -
⎛ –10 53 – 4 ⎞ ⎛10
– 53 4 ⎞
(A) ⎜ , , ⎟ (B) ⎜ , , ⎟
⎝ 3 3 3 ⎠ ⎝ 3 3 3 ⎠
(C) (–10, 53, –4) (D) None of these
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
13. Match the following :
Column-I
Column-II
(A) Through (γ, γ + 1) there (P) 2 < γ < 8
can not be more than one
normal to parabola y 2 = 4x if
(B) If two cicles (x–1) 2 + (y–1) 2 = γ 2 (Q) – 2 < γ < 2
and x 2 + y 2 – 8x + 2y + 8 = 0
intersect at two points, then
XtraEdge for IIT-JEE 57
JULY 2010

(C) Point (γ, γ + 1) lies inside the (R) γ < – 2
circle x 2 + y 2 = 1 for
(D) Both equation x 2 + y 2 + 2γx + 4 = 0 (S) – 1 < γ < 0
and x 2 + y 2 – 4γy + 8 = 0 represent
real circles if
(T) γ > 8
⎡3
– 4⎤
⎡1/
2 3⎤
14. Consider the matrix A = ⎢ ⎥ ; B =
⎣1
–1
⎢ ⎥ .
⎦ ⎣ 0 1 ⎦
Let P be an orthogonal matrix and Q = PAP T ,
R k = P T Q k .P, S = PBP T & T k = P T S k P where k ∈ N
Column-I
Column-II
5
(A) ∑ a k , where a k represents the element (P) –9
k=
1
of first row & first column in matrix R k
3
(B) ∑ b k , where b k represents the element (Q) 10
k=
1
of second row & second column in
matrix R k
∞
(C) ∑ X , where X k represents the element (R) 35
k=1
k
of first row & first column in matrix T k
10
(D) ∑ y k , where y k represents the element (S) 1
k=
1
of second row and second column in
Matrix T K
(T) 15
15. There are N + 1 identical boxes each containing N
wall clocks. The first box contains zero defective
clocks. The second box contains one defective and
(N – 1) effective clocks, in general r th box contains
(r – 1) defective and (N – r + 1) effective clocks
(1 ≤ r ≤ N + 1). Thus, the (N + 1) th box contains all
defective clocks. A wall clock is selected and found
an effective one. The probability that it is from k th
γN – γK
+ γ
box is
find γ.
N
2 + N
16. If a determinant of order 3 × 3 is formed by using the
numbers 1 or – 1then it minimum value of
determinant is – γ find the value of γ.
17. If equation of the plane through the straight line
x –1 y + 2 z
= = and perpendicular to the plane
2 – 3 5
x – y + z + 2 = 0 ia ax – by + cz + 4 = 0, then find the
3 2
10 a + 10 b + 10c
value of
.
342
⎡3
– 2
18. If A =
⎢
⎢
2 1
⎢⎣
4 – 3
⎡3
0 3⎤
⎡x⎤
⎢ ⎥
⎢
2 1 0
⎢ ⎥
⎥ ⎢
y
⎥
⎢⎣
4 0 2⎥⎦
⎢⎣
z⎥⎦
y z
of x + + 2 3
3 ⎤
–1
⎥
⎥
. Solve the system of equations
2 ⎥⎦
⎡8⎤
⎡2y⎤
=
⎢ ⎥
⎢
1
⎥
+
⎢ ⎥
⎢
z
⎥
, then find the value
⎢⎣
4⎥⎦
⎢⎣
3y⎥⎦
This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
19. Find the coefficient of x in the determinant
(1 + x)
(1 + x)
(1 + x)
a1b1
a2b1
a3b1
(1 + x)
(1 + x)
(1 + x)
a1b2
a2b2
a3b2
(1 + x)
(1 + x)
(1 + x)
a1b3
a2b3
a3b3
where a i , b j ∈N
How to Handle Difficult People
A bully at your work is difficult for you to face. He is
demanding you do part of his job without pay or
credit. How do you handle it
Your neighbors are constantly fighting. They wake
you up in the middle of the night with their screams
and curses. What do you say to them
Your father is unhappy about your career choice. He
constantly criticizes your work and points out what
he thinks you should do. How do you deal with him
XtraEdge for IIT-JEE 58
JULY 2010

Based on New Pattern
IIT-JEE 2012
XtraEdge Test Series # 3
Time : 3 Hours
Syllabus : Physics : Calorimetry, K.T.G.,Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave,
Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification &
Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics:
Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D
Instructions :
Section - I
• Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer
and -2 mark for wrong answer.
Section - II
• Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and
-1 mark for wrong answer.
Section - III
• Question 13 to 14 are Column Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong
answer.
Section - IV
• Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer
and 0 mark for wrong answer.
PHYSICS
(A) 52 W
(C) 512 W
(B) 256 W
(D) 215 W
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1. The molar heat capacity for a process is :
R α
C = + , then process equation is -
1– γ T
(A) Ve –(α/RT)T = constant
(B) Ve (α/R)T = constant
(C) VT = constant
(D) Ve α/RT = constant
2. Three black bodies 1,2,3 have radius r 1 < r 2 < r 3 .
Emissive powers of black bodies at Temperature T
are E 1 , E 2 , E 3 , Then correct relation between them is -
(A) E 1 < E 2 < E 3 (B) E 1 > E 2 > E 3
(C) E 1 = E 2 = E 3 (D) E 1 > E 2 < E 3
3. A taut string for which µ = 5 × 10 –2 kg/m is under
tension of 80 N. How much is the average rate of
transport of potential energy if the frequency is 60 Hz
and amplitude 6 cm - (Given 4π 2 = 39.5)
4. In case of standing waves -
(A) At nodes particles displacement is time
dependent
(B) At antinodes displacement of particle may or
may not be zero
(C) Wave does not travel but energy is transmitted
(D) Components waves traveling in same direction
having same amplitude and same frequency are
superimposed
5. For an ideal gas graph is shown for three processes.
Processes 1, 2 and 3 are respectively –
Work done (magnitude)
1
Temperature change
(A) Isochoric, isobaric, adiabatic
(B) Isochoric, adiabatic, isobaric
(C) isobaric, adiabatic, isochoric
(D) Adiabatic, isobaric, isochoric
3
2
XtraEdge for IIT-JEE 59
JULY 2010

6. Figure shows cyclic process. From c to b, 40 J is
transferred as heat from b to a, 130 J is transferred as
heat, and work done is 80 J from a to c, 400 J is
transferred as heat then –
P
c
a
V
(A) Work done in process a to c is 310 J
(B) Net work done is cycle is 230
(C) Net change in internal energy in cycle is 130 J
(D) None of these
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
Passage : I (Ques. 7 to 9)
Many waveforms are described in terms of
combinations of travelling waves. Superposition
principle is used to analyse such wave combinations.
Two pulses travelling on same string are described
by-
5
y 1 =
(3x – 4t)
2 + 2
, y –5
2 =
(3x + 4t – 6)
2 + 2
7. The direction in which each pulse is travelling is -
(A) y 1 is in positive x-axis, y 2 is in positive x-axis
(B) y 1 is in negative x-axis, y 2 is in negative x-axis
(C) y 1 is in positive x-axis, y 2 is in negative x-axis
(D) y 1 is in negative x-axis, y 2 is in positive x-axis
8. The time when the two waves cancel everywhere -
(A) 1 sec (B) 0.5 sec (C) 0.25 sec (D) 0.75 sec
9. The point where two waves always cancel-
(A) 0.25 m (B) 0.5 m (C) 0.75 m (D) 1 m
Passage: II (Ques. 10 to 12)
One mole of monoatomic gas is taken through above
cyclic process. T A = 300 K
Process AB is defined as PT = constant
P
B C
3P 0
P 0
T
10. Work done in process AB is -
(A) 400 R
(B) – 400 R
(C) 200 R
(D) – 300 R
b
11. Change in internal energy in process CA
(A) 900 R
(B) 300 R
(C) 1200 R
(D) zero
12. Heat transferred in the process BC is -
(A) 1000 R (B) 500 R
(C) 2000 R
(D) 1500 R
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
13. Match the standing waves formed in column-II due to
plane progressive waves in Column-I and also with
conditions in Column-I
Column-I
Column-II
(A) Incident wave is (P) y = 2A cos kx sin ωt
y = A sin (kx – ωt)
(B) Incident wave is (Q) y = 2A sin kx cos ωt
y = A cos (kx – ωt)
(C) x = 0 is rigid support (R) y = 2A sin kx cos ωt
(D) x = 0 is flexible support (S) y = 2A cos kx cos ωt
(T) None of these
14. Column-I Column-II
(A) Specific heat capacity S (P) l 1 – l 2 = constant
for l 1 α 1 = l 2 α 2
(B) Two metals (l 1 , α 1 ) and (Q) Y is same
(l 2 , α 2 ) are heated uniformly
(C) Thermal stress (R) S = ∞ for ∆T = 0
(D) Four wires of same (S) Y α ∆t
material
(T) None of these
XtraEdge for IIT-JEE 60
JULY 2010

This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
15. If the volume of a block of metal changes by 0.12 %
when heat is changed from 40ºC to 60ºC, find the
linear expansion coefficient of the metal
[Ans. in …… × 10 –5 /ºK]
16. Calculate the pressure exerted by a mixture of 8 g of
oxygen, 14 g of nitrogen and 22 g of carbon di-oxide
in a container of 30 litres at a temperature of 27ºC.
[Ans. in …… × 10 5 N/m 2 ]
17. A sphere and a cube of same material and total
surface area placed in an evacuated chamber turn by
turn and heated to the same temperature. Calculate
the ratio of the rate of cooling of spherical to cubical
surface. [Ans. in …… × 10 –1 ]
18. Two oscillating waves have a phase difference of 2
π
0
1
2
3
4
5
6
7
8
9
is 25 oscillations. What is the percentage difference
in their frequency
19. For a certain organ pipe, three successive resonance
observed are 425, 595 and 765 Hz. Taking the speed
of sound to be 340 ms –1 , find the length of the pipe,
in metre.
0
1
2
3
4
5
6
7
8
9
CHEMISTRY
Questions 1 to 6 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1. The IUPAC name of the given compound
OHC CONH 2
| |
CH 3 –CH–CH–CH–CH–COOH is -
|
|
Br COCl
(A) 2-Bromo-4-carbamoyl-5-chloroformyl-3-formyl
hexanoic acid
(B) 5-Bromo-3-carbamoyl-2-chloroformyl-4-formyl
hexanoic
(C) 4-Formyl-2-chloroformyl-3-carbamoyl-5-bromo
hexanoic acid
(D) 2-Chloroformyl-3-carbamoyl-4-formyl-5-bromo
hexanoic acid
2. Geometry in the given compound is
CH 3
H
H
CH 3
(A) cis
(B) trans
(C) cis as well as trans (D) no geometrical isomerism
3. The structure of spiro [3,3] heptane is -
(A)
(B)
(C)
(D)
4. The pH of a 10 –8 molar solution of HCl in water is -
(A) 8 (B) – 8
(C) between 7 and 8 (D) between 6 and 7
5. Consider the following equilibrium in a closed
container N 2 O 4 (g) 2NO 2 (g). At a fixed
temperature, the volume of the reaction container is
halved. For this change, which of the following
statements holds true regarding the equilibrium
constant (k p ) and degree of dissociation (α)
(A) neither k p nor α changes
(B) both k p and α change
(C) k p changes, but α does not change
(D) k p does not change, but α changes
6. For H 3 PO 3 and H 3 PO 4 the correct choice is -
(A) H 3 PO 3 is dibasic and reducing
(B) H 3 PO 3 is dibasic and non-reducing
(C) H 3 PO 4 is tribasic and reducing
(D) H 3 PO 3 is tribasic and non-reducing
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
XtraEdge for IIT-JEE 61
JULY 2010

Passage : I (Ques. 7 to 9)
In a reversible chemical reaction, the rate of forward
reaction decreases and that of backward reaction
increases with the passage of time; at equilibrium the
rate of forward and backward reaction become same.
Let us consider the formation of SO 3 (g) in the
following reversible reaction :
2SO 2 (g) + O 2 (g) 2SO 3 (g)
Following graphs are plotted for this reactions
Conc.
t 1 t 2 t 3
time
t 4
A.
B.
C.
7. In the above graph, A,B & C respectively are -
(A) SO 3 , SO 2 and O 2 (B) SO 3 , O 2 and SO 2
(C) SO 2 , O 2 and SO 3 (D) O 2 , SO 2 and SO 3
8. In the above graph, the equilibrium state is attained at
time -
(A) t 1 (B) t 2
(C) t 3 (D) t 4
9. Which of the following represent the rates of forward
reaction (r f ) and rates of backward reaction (r b ) at
equilibrium
(A)
(C)
rate of
reaction
rate of
reaction
r b
r f
time
r f
r b
time
Passage: II (Ques. 10 to 12)
(B)
(D)
rate of
reaction
rate of
reaction
time
time
Different spatial arrangements of the atoms that result
from restricted rotation about a single bond are
conformers. n-Butane has four conformers eclipsed,
fully eclipsed, gauche and anti. The stability order of
these conformers are as follows:
anti > gauche > partial eclipsed > fully eclipsed
Although anti is more stable than gauche but in some
cases gauche is more stable than anti.
r b
r b
r f
r f
10. Which one of the following is most stable
conformer
Cl
Cl
CH 3 H Cl CH 3
(A)
(B)
H CH
H
3
CH 3
Cl
H
Cl
Cl
H Cl
Cl
(C)
H
CH 3
CH 3
(D)
H
H
CH 3
CH 3
11. Which one of the following is the most stable
conformer
CH 3
CH 3
HO H H CH 3
(A)
(B)
H
OH
H
OH
CH 3 OH
CH 3
OH
H OH H CH 3
(C)
H OH
(D)
H CH 3
CH 3
OH
12. Number of possible conformers of n-butane is -
(A) 2 (B) 4
(C) 6 (D) infinite
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
XtraEdge for IIT-JEE 62
JULY 2010

13. Match the following :
Column-I
(A) N 2(g) + 3H 2(g) 2NH 3(g) ; ∆H = –ve
(B) N 2(g) + O 2(g) 2NO (g) ; ∆H = +ve
(C) A (g) + B (g) 2C (g) + D (g) ; ∆H = +ve
(D) PCl 5(g) PCl 3(g) + Cl 2(g) ; ∆H = +ve
Column-II
(P) K increases with increase in temperature
(Q) K decreases with increase in temperature
(R) Pressure has no effect
(S) Product concentration, increases due to addition
of inert gas at constant pressure
(T) Product concentration, increases due to addition
of inert gas at constant volume
14. Match the following :
Column-I
Column-II
(A) Bi 3+ → (BiO) + (P) Heat
(B) [AlO 2 ] – → Al(OH) 3 (Q) Hydrolysis
(C) [SiO 4 ] 4– → [Si 2 O 7 ] 6– (R) Acidification
(D) [B 4 O 7 ] 2– → [B(OH) 3 ] (S) Dilution by water
(T) Basification
This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
15. Calculate the pH at which the following conversion
(reaction) will be at equilibrium in basic medium
I 2 (s) I – (aq.) + IO – 3 (aq.)
When the equilibrium concentrations at 300 K are [I – ]
= 0.10 M and [IO – 3 ] = 0.10 M.
{Given → ∆G f º (I – , aq.) = – 50 kJ./mol,
∆G f º (IO – 3 , aq) = – 123.5 KJ/mol,
∆G f º (H 2 O, l) = –233 KJ/mol
∆G f º (OH – , aq.) = – 150 KJ/mol
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
25
R(Gas constant) = J/mol–K
3
Log e = 2.3}
16. Number of configurational isomers of 2,3-
dibromocinnamic acid is .
17. Consider the reaction AB 2(g) AB g + B (g) . It the
initial pressure of AB 2 is 100 torr and equilibrium
pressure is 120 torr. The equilibrium constant Kp in
terms of torr is.
18. Dissociation of H 3 PO 3 occurs in ......... stages.
19. The number of hydroxyl groups in pyrophosphoric
acid 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. Mark your response in
OMR sheet against the question number of that
question. + 5 marks will be given for each correct
answer and – 2 mark for each wrong answer.
1. A circle C 1 of radius b touches the circle x 2 + y 2 = a 2
externally and has its centre on the positive x-axis;
another circle C 2 of radius c touches the circle C 1
externally and has its centre on the positive x-axis.
Given a < b < c, then the three circles have a
common tangent if a, b, c are in -
(A) A.P.
(B) G.P.
(C) H.P. (D) None of these
2. P is a point on the axis of the parabola y 2 = 4ax;
Q and R are the extremities of its latus rectum, A is
its vertex. If PQR is an equilateral triangle lying
within the parabola and ∠AQP = θ, then cos θ =
2 – 3
9
(A)
(B)
2 5
8 5
(C)
5 – 2
2
3
(D) None of these
x 2 y 2
3. The length of the diameter of the ellipse + = 1,
25 9
perpendicular to the asymptote of the hyperbola
x 2 y 2
– = 1 passing through the first and third
16 9
quadrants is :
100
150
(A)
(B)
431
481
(C)
25
3
(D) 11 2
XtraEdge for IIT-JEE 63
JULY 2010

→ → → → → →
4. If a , b , c are such that [ a , b , c ] = 1,
→ → → → →
2π
→ →
c = γ( a × b ), a ^ b < , and | a | = 2 , | b |
3
= 3 , | → c | =
is -
π
(A) 6
1
→ →
, then the angle between a and b
3
(B) 4
π
(C) 3
π
(D) 2
π
5. Equation of a plane which passes through the point of
x –1 y – 2 z – 3
intersection of lines = = and
3 1 2
x – 3 y –1 z – 2
= = and at greatest distance
1 2 3
from the point (0, 0, 0) is -
(A) 4x + 3y + 5z = 25
(B) 4x + 3y + 5z = 50
(C) 3x + 4y + 5z = 49
(D) x + 7y – 5z = 2
6. p 1 , p 2 are lengths of perpendicular from foci on
tangent to ellipse and p 3 , p 4 are perpendiculars from
extremities of major axis and p from centre of ellipse
p1p2
– p
on same tangent, then equals -
2
p3p4
– p
(A) e (B) e
(C) e 2
(D) None of these
This section contains 2 paragraphs, each has 3 multiple
choice questions. (Questions 7 to 12) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against
the question number of that question. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
Passage : I (Ques. 7 to 9)
Let C : x 2 + y 2 x 2
= 9, E : +
9 4
= 1, L : y = 2x
7. P is a point on the circle C, the perpendicular PQ to
the major axis of the ellipse E meets the ellipse at M,
then
MQ is equal to -
PQ
(A) 1/3 (B) 2/3
(C) 1/2 (D) none of these
8. 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 = 5
(C) x + 3y + 5 = 0 (D) 4x +3y + 5 = 0
2
y 2
9. 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 (Ques. 10 to 12)
In a parallelogram OABC vector → a , → b , → c are
respectively the position vectors of vertices A, B, C
with reference to O as origin. A point E is taken on
the side BC which divides it in the ratio of 2 : 1. Also
the line segment AE intersects the line besecting the
angle O internally in point P. If CP when extended
meets AB in point F, then
10. The position vector of point P, is -
→ →
⎛
→ →
⎞
3 | a || c | ⎜ a c ⎟
(A)
⎜ +
→ → → → ⎟
3 | c | + 2 | a |
⎝ | a | | c | ⎠
(B)
→ →
⎛
→ →
⎞
| a || c | ⎜ a c ⎟
⎜ +
→ → → → ⎟
3 | c | + 2 | a |
⎝ | a | | c | ⎠
→ →
⎛
→ →
⎞
2 | a || c | ⎜ a c ⎟
(C)
⎜ +
→ → → → ⎟
3 | c | + 2 | a |
⎝ | a | | c | ⎠
(D) none of these
11. The position vector of point F is -
→
→
1 | a |
→
(A) a +
→ c (B)
3
| c |
→
→
2 | a |
→
(C) a +
→ c (D)
| c |
12. The vector → AF , is given by -
(A)
→
1 | a |
→
c
→
3
| c |
→
(B)
| a |
→
(C) 2 c (D)
→
| c |
→
a +
→
| a |
→
→ c
| c |
→
→
| a |
→
a – c
→
| c |
→
| a |
→
c
→
| c |
→
| a |
→
– c
→
| c |
This section contains 2 questions (Questions 13, 14).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-
S, D-T then the correctly bubbled 4 × 5 matrix should
be as follows :
XtraEdge for IIT-JEE 64
JULY 2010

A
B
C
D
P Q R S T
P Q R S T
P Q R S T
P Q R S T
P Q R S T
Mark your response in OMR sheet against the question
number of that question in section-II. + 8 marks will be
given for complete correct answer (i.e. +2 marks for
each correct row) and NO NEGATIVE MARKING for
wrong answer.
13.
Column-I
Column-II
(A) If lines x + 2y – 1 = 0, ax + y + 3 = 0 (P) 4
and bx – y + 2 = 0 are concurrent,
the least distance from origin to
(a, b) is S. The value of 58 . S is
(B) A,B are two fixed points on a line (Q) 5
L. Let locus of point P such that
PA = 2PB be a curve cutting line
L at R and S. If slope of PR is – 2
1 ,
then slope of PS is
(C) Let tangents at P and Q to curve (R) 2
y 2 – 4x – 2y + 5 = 0 intersect at T.
If S(2, 1) is a point such that
SP.QS = 16 then the length ST is
(D) Let the double ordinate PNP' of the (S) π – 1
2
x y
hyperbola – = 1 is
2
( π –1) ( π –1)
preduced both side to meet
asymptotes in Q and Q'. The
product PQ. PQ' is equal to
2
(T) (π–1) 2
14.
Column-I
Column-II
(A) The distance of the point (1, – 2, 3) (P) 1
from the plane x – y + z = 5
measured parallel to the line
1 1 ⎛ 1 ⎞
x = y = ⎜ – ⎟⎠ z is
2 3 ⎝ 6
(B) The shortest distance between
the lines
and
x – 2
3
x –1
2
=
=
y – 4
4
y – 2
3
=
=
x – 5
5
z – 3
4
is
(Q)
(C) The points (0, –1, –1), (4, 5, 1), (R) 4
(3, 9, 4) and (–4, 4, k) are coplanar
then k =
1
6
(D) The radius of the circular section (S) 3
of the sphere | → r | = 5 by the plane
→
r . ( ^i + ^j + ^k ) = 3 3 is
(T) 2
This section contains 5 questions (Q.15 to 19).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions
is a SINGLE-DIGIT INTEGER, ranging from 0 to 9.
The appropriate bubbles below the respective question
numbers in the OMR has to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
15. Two circle of radii 'a' and 'b' touching externally are
2
inscribed in area bounded by y = 1 – x and
x-axis. If b = 2
1 and a = k
1 , then k is ...............
16. If a circle S (x, y) = 0 touches at the point (2, 3) of
the line x + y = 5 and S (1, 2) = 0, then 2 × radius
of such circle is.
17. Consider two concentric circles C 1 : x 2 + y 2 = 1,
C 2 : x 2 + y 2 = 4. Tangents are drawn to C 1 from any
point P on C 2 . These tangents again meet circle C 2 at
A & B. It can be proved that locus of point of
intersection of tangents drawn to C 2 at A and B is a
circle, what is the radius of that circle.
18. In a regular tetrahedron let θ be the angle between
any edge and a face not containing the edge.
If cos 2 a
θ = where a, b ∈ I + also a and b are
b
coprime, then find the value of 13
5 (10a + b)
19. Let A (1, 2), B (3, 4) be two point and C (x,y) be a
point such that (x – 1) (x – 3) + (y – 2) (y – 4) = 0. If
area of ∆ABC is 1 sq, unit. Then maximum number
of positions of C in xy plane is.
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
XtraEdge for IIT-JEE 65
JULY 2010

XtraEdge Test Series
ANSWER KEY
IIT- JEE 2011 (July issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans C A B D C D A C B D A B
Column 13 A → P, Q, R, S B → P, Q, R, S C → R, S D → R, S
Match 14 A → R B → P C → Q D → P, S
Numerical Ques 15 16 17 18 19
Response Ans 8 5 1 4 2
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans C B C A A D D B D D B A
Column 13 A → P, R B → P, R C → Q, T D → S
Match 14 A → P, S, T B → R, S C → Q D → P
Numerical Ques 15 16 17 18 19
Response Ans 4 2 6 4 5
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans A D D C B D C B A C B D
Column 13 A → Q, R, S B → P C → S D → P, R, T
Match 14 A → R B → P C → S D → Q
Numerical Ques 15 16 17 18 19
Response Ans 2 4 5 3 0
IIT- JEE 2012 (July issue)
PHYSICS
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans D C B B A A C D D B A C
Column 13 A → P, R B → Q, S C → Q, R D → P, S
Match 14 A → R B → P C → S D → Q
Numerical Ques 15 16 17 18 19
Response Ans 2 3 7 5 7
CHEMISTRY
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans B B D D D A A C A A C D
Column 13 A → Q B → P, R C → P, S D → P, S
Match 14 A → Q B → S C → P D → R
Numerical Ques 15 16 17 18 19
Response Ans 8 4 5 2 4
MATHEMATICS
Ques 1 2 3 4 5 6 7 8 9 10 11 12
Ans B A B B B C B A B A A A
Column 13 A → Q B → R C → P D → S
Match 14 A → P B → Q C → R D → R
Numerical Ques 15 16 17 18 19
Response Ans 4 1 4 5 4
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