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					                                                                       WORRY IS A MISUSE OF IMAGINATION.
          Volume - 7 Issue - 6
  December, 2011 (Monthly Magazine)
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                                                                                             Editorial
                                                    Dear Students,
 Pramod Maheshwari
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    Editor : Pramod Maheshwari                      Pramod Maheshwari,
                                                    B.Tech., IIT Delhi

XtraEdge for IIT-JEE                                              1                                           DECEMBER 2011
                 Volume-7 Issue-6
           December, 2011 (Monthly Magazine)                                  CONTENTS
                                                               INDEX                                             PAGE
            NEXT MONTHS ATTRACTIONS

        Much more IIT-JEE News.
                                                                   Regulars ..........
        Know IIT-JEE With 15 Best Questions of IIT-JEE
                                                               NEWS ARTICLE                                           3
        Challenging Problems in Physics,, Chemistry & Maths    IIT-B'S Plan : Job support to Entrepreneurial Spirit
                                                               IIT Students' Research Work to Be Available Online
        Key Concepts & Problem Solving strategy for IIT-JEE.

        Xtra Edge Test Series for JEE- 2011 & 2012             IITian ON THE PATH OF SUCCESS                          6
                                                               Mr. Amitabha Ghosh
        CBSE Mock Test Paper
                                                               KNOW IIT-JEE                                           7
                                                               Previous IIT-JEE Question

                                                                   Study Time........
                                                               DYNAMIC PHYSICS                                        15
                                                               8-Challenging Problems [Set# 8]
    S
                                                               Students’ Forum
        Success Tips for the Months                            Physics Fundamentals
                                                               Ray Optics
                                                               Fluid Mechanics & Properties of Matter
    • "The way to succeed is to double your error
      rate."
                                                               CATALYSE CHEMISTRY                                     34
    • "Success is the ability to go from failure to
      failure without losing your enthusiasm."                                Key Concept
                                                                              Carboxylic Acid
    • "Success is the maximum utilization of the                              Chemical Kinetics
      ability that you have."                                                 Understanding : Inorganic Chemistry

    • We are all motivated by a keen desire for
      praise, and the better a man is, the more he             DICEY MATHS                                            45
      is inspired to glory.
                                                               Mathematical Challenges
    • Along with success comes a reputation for                Students’ Forum
      wisdom.                                                  Key Concept
                                                               Monotonicity,
    • They can, because they think they can.                   Maxima & Minima
                                                               Function
    • Nothing can stop the man with the right
      mental attitude from achieving his goal;                     Test Time ..........
      nothing on earth can help the man with the
      wrong mental attitude.                                   XTRAEDGE TEST SERIES                                   58
    • Keep steadily before you the fact that all               Class XII – IIT-JEE 2012 Paper
      true success depends at last upon yourself.              Class XI – IIT-JEE 2013 Paper
                                                               Mock Test CBSE Pattern Paper -1 [Class # XII]




XtraEdge for IIT-JEE                                           2                                        DECEMBER 2011
IIT-B'S Plan : Job support to                 IIT Students' Research Work to               witnessed tremendous competition
Entrepreneurial Spirit                        Be Available Online                          between the energy of the present
                                                                                           generation and that of the alumni. The
IIT-Bombay has hit upon an idea               Students of Indian Institutes of             event was full of activities comprising
that could boost the spirit of                Technology (IITs) will soon share            of talks, alumni -students matches and
entrepreneurship among its students.          their research details for all institutes    Inter-Sports Arm Wrestling and Tug of
Its placement cell is weighing the            via a common website to be launched          war competition which was followed
option of helping students whose              in November.                                 by dinner and group discussions in
start-ups have not fired to be placed         The two day long 'Inter-IIT Gymkhana         SAC Lawns.
in jobs after two years of                    Summit 2011' being held at IIT-
experimenting with their ideas.                                                            The post match fundae sessions (be it
                                              Gandhinagar saw students of various          related to the life post IITs or the
As part of the scheme, students keen          consortium discuss issues common to          prospects and benefits of sports in
on their own start-ups will be assigned       all.     The      students    included       everyday life) were undoubtedly the
mentors after graduating. These               representatives from 10 IITs including       highlight of the complete event. The
experts-either people who have                Madras, Mandi, Rajasthan, Mandi,             commitment and dedication shown
successfully floated their own                Hyderabad, with IIT Kanpur and IIT           by the Old Machine Guns from IITB
companies or those with enough                Bombay participating in online               left the students sports fraternity
exposure to new businesses-will               sessions.                                    speechless.
evaluate their ideas to see if there's        Among the things discussed, students
any potential. Once the ideas are             particularly discussed methods to share      Alumni showed the perfect blend of
approved, students can float their own        research works of B.Tech, M.Tech and         attitude, experience and sporting
companies. After two years, if a start-up     Phd students. Students are seen to           spirit with some exceptional
fails to take off, the student-entrepreneur   repeat researches, being unaware of          performances in all the sports.
can participate in the regular placement      past researchers. Sharing of past            Results of this year’s meet like last
process and get a job.                        researches will help avoid such              year were pretty much dominated by
                                              repetitions.                                 the alumni, winning in almost all the
Ravi Sinha, professor in charge of                                                         sports.
placements, said the idea, which is at a      Sport's Alumni Student Meet
nascent stage, can give students the                                                       IITians' Nano Satellite Launched
                                              The 2nd official annual Alumni
assurance to float their ideas without                                                     IIT-Kanpur has made the 3kg
                                              Sports Day was full of tremendous
hesitation. "Very few start-up ideas on                                                    satellite 'Jugnu'. It will generate
                                              competition between the energy of
the campus turn out to be successful                                                       real time data on droughts, floods,
                                              the present generation and that of the
ventures. Often, many good ideas are                                                       vegetation and forestation.
                                              alumni. The event was full of
not commercially-viable. So, students
                                              activities comprising of talks, alumni       Jugnu, the Nano satellite designed by
are apprehensive about floating their
                                              -students matches and Inter-Sports           IIT Kanpur students has been
ideas," added Sinha.
                                              Arm Wrestling and Tug of war                 launched via PSLV C-18 rocket from
The office of Society for Innovation          competition which was followed by            Shrihari Kota, Andhra Pradesh 11:00
and Entrepreneurship ( SINE), which           dinner and group discussions in SAC          hrs today, reports TNN. This is the
promotes entrepreneurship on the              Lawns.                                       country's first Nano satellite and
campus by promoting business                                                               weighs just 3 Kgs incorporated with
                                              “Good, better, best. Never let it rest.
incubation, and the Entrepreneurship                                                       PSLV C-18.
                                              Until your good is better and your
cell (E-cell) is working with the
                                              better is best. ” this quote perfectly fit   The mission of developing Jugnu
placement cell to check the feasibility
                                              the enthralling display of experience,       was done under the guidance of Prof.
of the project. "We have been working
                                              wit and never say die attitude of the        N.S vyas. this is extremely a
on the initiative for a couple of
                                              alumni at this year’s Alumni Sports          complex process and this was a hard
months. If it works out, student-
                                              Day.                                         exercise. The Nano satellite has been
entrepreneurs will be reasonably
assured of getting a job through the          Brainchild of the basketball fraternity of   designed ethnically.
institute's placement office," said           IIT Bombay, this was the 2nd official        "We have done our best. Jugnu has
Sinha.                                        annual Alumni Sports Day which               undergone laborious tests like thermo


XtraEdge for IIT-JEE                                              3                                         DECEMBER 2011
vacuum tests, vibration tests and           arbitrary    or     unreasonable   or        may be made at a meeting between
electronic magnetic interference test.      discriminatory,”   Justice Raveendran        senior HRD ministry officials and
Students have worked very hard in           said.                                        the IIT directors on May 5, the
developing Jugnu. They have worked          The apex court passed the judgment           sources said.
round-the-clock. We are hopeful of its      while dismissing an appeal filed by          The Council alone is empowered to
successful launch," said Vyas.
                                            aspirant Sanchit Bansal, son of an IIT       appoint IIT directors and take any
The payload of the satellite has            Kharagpur professor, who appeared in         policy decisions binding across the
uniquely designed camera for infra red      IIT-JEE 2006 as a general category           top engineering schools -- including
remote sensing, a GPS receiver and an       candidate.                                   their fees, reforms, administrative
inertial measurement unit.                                                               structure and any amendments to the
                                            Sanchit had secured 75 marks in
Shatanu Agarwal, the team head of           mathematics, 104 in physics and 52 in        aw governing the Institutes.
Jugnu said "About 25 minutes after          chemistry, aggregating to 231.               “The IIT Council -the highest
the launch, Jugnu would will click the                                                   decision making body of the IITs -at
                                            The board had fixed the cut-off marks
first image of its launch vehicle and its   for admission at 37 for math, 48 for         present depends on the HRD
antennas would be deployed. It will         physics and 55 for chemistry and the         ministry to even invite members for
then stabilize and start transmitting                                                    meetings or to prepare the Council's
                                            aggregate cut-off at 154.
images to the ground station."                                                           agenda. All this will change,“ a
                                            As Sanchit did not secure the minimum        source said.
After 115 minutes of the take off           cut-off in chemistry, he failed to qualify
Jugnu was visible at the ground             even though his aggregate was higher         The IITs have decided to approach D
station.                                                                                 Udaya Kumar, assistant professor at
                                            than required.
"Jugnu's beacon (blinking signal at all                                                  IIT Guwahati and the designer of the
                                            He then challenged the procedure on          new rupee symbol, to design a logo
times all over the earth) will get          the ground that candidates with              for the IIT Council.
switched on after 30 minutes of             aggregates lower than his were
separation from the launch vehicle.                                                      The move is a component of HRD
                                            selected.
Five seconds after its separation,                                                       minister Kapil Sibal's larger plan to
Jugnu will click its image," said           Rejecting his plea, the court said: “For     increase the functional autonomy of
Shantanu.                                   a layman, the above procedure may            the IITs. The IIT Council --
                                            appear to be highly cumbersome and           consisting of all IIT directors and
It took two years to develop this Nano
                                            complicated.                                 chairman of boards, other eminent
satellite and as people working on it
                                            But the object of the aforesaid              academic        administrators      and
used non space grade, commercial off
the shelf (COTS) to make the research       procedure for arriving at the cut-off        scientists -- at present does not even
                                            marks is to select candidates well           have an office of its own. It will now
on Jugnu affordable. Jugnu has
minimum number of redundant                 equipped in all the three subjects, with     have an office - a location has been
systems when compared to other              reference to their merit, weighed            identified     in    South     Delhi's
                                            against the average merit of all the         Chittaranjan Park.
conventional satellites.
                                            candidates who appeared in the               But this may only be the first step
SC refuses to adjudicate on IIT-            examination.”                                towards greater autonomy for the
JEE selections                                                                           IITs, sources indicated. “Once the
                                            IIT Council free from HRD
NEW DELHI: The Supreme Court has                                                         independent IIT Council is capable
                                            ministry clutches
refused to interfere with the ranking                                                    of handling matters, there is a
and selection procedure adopted for         NEW DELH: The apex decision                  possibility that we will empower it
the IIT-JEE saying there was no             making body of the Indian Institutes of      with far greater powers and withdraw
arbitrariness or ulterior motives in        Technology (IITs) has broken free of         from many administrative aspects of
fixing the methodology, says a PTI          the Human Resource Development               the IIT governance system,“ a source
report.                                     (HRD) Ministry in a move that could          said.
                                            be the first step towards allowing the       (Courtesy : Hindustan Times)
A bench of Justices R.V. Raveendran         IITs to govern themselves.
and A.K. Patnaik said courts would                                                       No permanent foreign faculty
interfere with the procedure only if        Empowered by independent staff and           for IITs
there was proven malafide, caprice or       with a identity of its own, the IIT
arbitrariness, which it said was            Council will now no longer need the          NEW DELHI: In a major setback to
lacking in the present system adopted       HRD ministry to take its administrative      the Indian Institutes of Technology
                                            decisions under the move, government         (IITs) plan, the Ministry for External
by the Joint Admission Board, which
conducts the exams.                         sources told HT.                             Affairs (MEA) has rejected a
                                                                                         proposal to liberalise visa norms to
“The fact that the procedure was            The final decision on the plan -- aimed      allow foreign teachers to take up
complicated would not make it               at creating an IIT Council Secretariat --    permanent posts at the IITs.

XtraEdge for IIT-JEE                                            4                                         DECEMBER 2011
The MEA has refused to change the         fold increase in fee for undergraduate      “This will ensure timely selection of
rules under which each foreign faculty    programme of the IITs.                      professors,” the Council noted.
member at the IITs needs to re-obtain                                                 It also decided that the appointment
                                          The Kakodkar committee, set up by the
a work visa every five years, top         government in October 2009 to study         of directors should be through
government and IIT sources have           the roadmap for the autonomy and            advertisements so that a wider base
confirmed to HT.                                                                      was created.
                                          future of the IITs, had recommended
Human resource development minister       that the fee be raised from Rs 50,000 to    “It was decided that in principle
Kapil Sibal had on September 11,          Rs 2 to Rs 2.5 lakh per annum.              approval may be granted for setting
2010 announced the plan to allow the      As the committee report came for            up an institute in Mauritius with the
IITs to fill up to 10% of their           discussion at the 42nd meeting of the       help of the IITs,” an official said.
permanent teaching posts with foreign     IIT Council Sibal rejected the fee hike     At the meeting, a presentation was
faculty.                                  proposal saying “such a hike would          made on adopting cyber security as
The proposal -first reported by HT on     prove a deterrent to a large number of      part of the curriculum for the IITs.
September 2, 2010 -was approved by        IIT aspirants,” a ministry official         So it was decided that a committee
the IIT Council -the highest decision     said. The Council asked the committee       be set up to develop a roadmap for
making body of the IITs -and is aimed     to rework the fee structure taking into     the future and give a report in next
at reducing a massive faculty crunch      account the aspirations of all sections.    three months.
plaguing the IITs.                        During the meeting, Sibal announced         “The committee would involve all
But the MEA's refusal to allow            setting up 50 research parks at a cost of   educational institutions as well as
foreign faculty to join with visas of     Rs 200 crore during the 12th Five Year      government departments,” the HRD
longer duration than five years means     Plan period.                                Ministry official said.
that the IITs will not be able to offer   Under the programme, industry will          The meeting also could not discuss
permanent posts to foreign faculty.       undertake research on various subjects      reform in the Joint Entrance
“We will need to continue to offer        with the support of experts from the        Examination and curriculum as T
contractual appointments something        IITs.                                       Ramaswamy, secretary, Department
we wanted to, and quite frankly, need     The research parks have been proposed       of Science & Technology, was not
to change,“ an IIT Director said.         to be set up on public-private-             present. Ramaswamy had prepared a
                                          partnership (PPP) model. One such           report on the two issues.
Each IIT is facing a faculty crunch
                                          research park has already come up in        "The Indian Statistics Institute has
between 15 and 40% with a total of
                                          Chennai.                                    arrived at a formula for equalising
over 1,000 faculty posts vacant across
the premier engineering schools. The      The meeting took note of the fact that      marks in all boards. If one board
Institutes have over the past year        credit-based practices were being           gives 90 percent as highest marks,
however received a number of              followed by the IITs to promote             and the other gives 75 percent, the
applications from foreign faculty,        students from one semester to the next,     marks will be equalized on the basis
including Persons of Indian Origin        and agreed that academic bodies of the      of a formula," Sibal said.
(PIOs) keen to teach at the IITs. The     IITs should consider acquisition of         Producing      more research       scho-
IITs are arguing that permanent posts     credits as a criteria for students and      lars was one of the key issues taken
would help them lure the best of          granting of degrees to bring uniformity.    up during the meeting, with the
foreign teachers.                         The issue came up following submission      council deciding to enhance the
All foreign teachers are at present       of Dhande committee report on uniform       capacity of IITs to produce 10,000
                                          and homogeneous criteria for promoting      Ph.D. graduates annually            from
required to teach as visiting or ad-hoc
faculty.                                  students in the IITs.                       around 1,000 presently and increase
(Courtesy : Hindustan Times)              Kanpur IIT director Dhande, who             faculty strength from around 4,000
                                          headed two committees, presented            presently to 16,000 by 2020. IANS
Sibal rejects steep fee hike for          reports on a “uniform criteria for
IIT students                              promoting students from one semester                  Science Fact:-
NEW DELHI: There will be no steep         to the next in the IITs and on the
fee hike for the students of the Indian   “requirement of infrastructure for           •    The brain uses over a quarter
Institutes of Technology, according to    research”. Both reports have been                 of the oxygen used by the
a decision taken by the IIT Council on    accepted. Each IIT at present has its             human body.
January 21.                               own criteria for promotion.
                                          The Council decided that a panel for         •    Your heart beats around
Chairing the IIT Council meet here,                                                         100000     times   a    day,
Human      Resource      Development      visitor’s nominee for a particular
                                          department would be created which all             36500000 times a year and
Minister Kapil Sibal rejected the Anil                                                      over a billion times if you
Kakodkar committee proposal for five-     IITs could use for faculty selection.
                                                                                            live beyond 30.


XtraEdge for IIT-JEE                                          5                                         DECEMBER 2011
                                                             Success Story
                         This article contains storie/interviews of persons who succeed after graduation from different IITs




                                Dr. Amitabha Ghosh
                                •    Post graduation in applied geology from IIT Kharagpur,
                                •    Working at NASA


Dr Amitabha Ghosh was the only Asian on NASA's                             How India can we develop science and technology
Mars Pathfinder mission. At present, he is a member of the               sector :
Mars Odyssey Mission and the Mars Exploration Rover                      It should be treated as a business. There should be more
Mission.                                                                 private participation. We must have an external review to
During the Mars Pathfinder Mission, he conducted                         evaluate the system and make changes as science and
chemical analysis of rocks and soil on the landing site. The             technology can take the country forward.
simple and unassuming 34-year-old planetary geologist                    We must check brain drain. About 80,000 students migrate
has won several accolades, which include the NASA Mars                   to the US for further studies, and settle there. They find the
Pathfinder Achievement Award in 1997 and the NASA
                                                                         facilities much better abroad. We need to reverse brain
Mars Exploration Rover Achievement Award in 2004.                        drain by enhancing and upgrading institutes in India.
The journey from India to NASA.                                            The state of space research in India :
It has been an intriguing experience. I was keen on                      I don't want to make controversial statements. All I can say
geologic research data interpretation and solar system
                                                                         is India is not at the frontier of space research. We have
formation. During my geological research days in India, I                made commendable progress but there is a long way to go.
had slept in railway stations while traveling to various                 We can do much better. I would be glad to be of help in
places.
                                                                         any way. Investment in research is investment in
After my post graduation in applied geology from IIT                     imagination. It is a matter of national pride and internal
Kharagpur, I wrote a letter to a professor at NASA                       recognition. We need to allocate more funds to enhance
expressing a desire to work at the space agency.                         research and development work.
I made certain suggestions; in fact, it was a critical letter.           We need good educational institutes like IITs and IIMs,
In India, you can never imagine criticising your professor.              but IITians don't rule the world. You must remember that
My suggestions were approved, while I got an opportunity                 Microsoft co-founder (Bill Gates does not have a college
to work at NASA.                                                         degree.

I think one requires luck and to put in sincere effort to                Youngsters must look around for role models and see what
                                                                         it is that they are doing right. Individuals must make use of
achieve one's goals. Being in the right place at the right
time is also important.                                                  their inherent strengths to succeed.

In Mumbai for the Pravasi Bharatiya Divas, he spoke                        How can India become a leading global player :
about his work at NASA and his vision for India.                         Globalisation will reap huge and long-term benefits and India
                                                                         must make the best use of the opportunities. At the PBD
The Vision for India :
                                                                         seminar, I found people presenting grandiose plans. Instead,
I feel there India has a great future. We have world-class               we should look at the realities and immediate solutions.
companies. Today, companies like Infosys can be
compared with world leaders like Oracle. Like the                        The private sector has to be actively involved in the
                                                                         development of the country and the government has to
Information Technology revolution, we can have a science
or space revolution. We have the potential to bring about                respond to the needs of the people. Fifteen years ago, we
revolutions in other sectors as well.                                    didn't have an Infosys, today we have many global
                                                                         companies.

XtraEdge for IIT-JEE                                                 6                                              DECEMBER 2011
                              KNOW IIT-JEE
                                                                By Previous Exam Questions



                  PHYSICS                                             2.   A cubical box of side 1 meter contains Helium gas
                                                                           (atomic weight 4) at a pressure of 100 N/m2.
1.   A non-viscous liquid of constant density 1000 kg/m3                   During an observation time of 1 second, an atom
     flows in a streamline motion along a tube of variable                 travelling with the root-mean-square speed parallel
     cross section. The tube is kept inclined in the vertical              to one of the edges of the cube, was found to make
     plane as shown in figure. The area of cross section of                500 hits with a particular wall, without any
     the tube two points P and Q at heights of 2 metres                                                           25
     and 5 metres are respectively 4 × 10–3 m2 and 8 × 10–3                collision with other atoms. Take R =        J/mol-K
     m2. The velocity of the liquid at point P is 1 m/s. Find                                                       3
     the work doen per unit volume by the pressure and                     and k = 1.38 × 10–23 J/K                [IIT-2002]
     the gravity forces at the fluid flows from point P to Q.              (a) Evaluate the temperature of the gas.
                                                  [IIT-1997]               (b) Evaluate the average kinetic energy per atom.
                                                                           (c) Evaluate the total mass of helium gas in the
                                           Q                                   box.
                                                                      Sol. The distance travelled by an atom of helium in
                                                                             1
                                                                                 sec (times between two successive collision)
             P                                                              500
                                        5m                                 is 2m. Therefore root mean square speed
                  2m


Sol. Given that
     ρ = 1000 Kg/m3, h1 = 2 m, h2 = 5 m
     A1 = 4 × 10–3 m2 , A2 = 8 × 10–3 m2 . v1 = 1 m/s
                                                                                                   1m
     Equation of continuity
                                                                                     dis tan ce       2
     A1 = 4 × 10–3 m2, A2 = 8 × 10–3 m2 , v1 = 1 m/s                        Crms =              =         = 1000 m/s
                                                                                       time       1 / 500
     Equation of continuity
                                  A1v1                                                          3RT
     A1v1 = A1v2,       ∴ v2 =         = 0.5 m/s                            (a) But Crms =
                                   A2                                                            M
     According to Bernouilli's theorem,                                                      3 × 25 / 3 × T
                                                                            ⇒ 1000 =                          ⇒ T = 160 K
                               1                                                               4 × 10 – 3
     (p1 – p2) = ρg (h2 – h1) – ρ (v22 – v12)
                               2                                            (b) Average kinetic energy of an atom of a
     where (p1 – p2) = work done/vol. [by the pressure]                                      3
                                                                            monoatomic gas = kT
     ρg (h2 – h1) = work done/vol. [by gravity forces.]                                      2
     Now, work done/vol by gravity forces                                               3
                                                                            ∴ Eav =       kT = 3.312 × 10–12 Joules
     = ρg (h2 – h1) = 103 × 9.8 × 3 = 29.4 × 103 J/m3                                   2
     and
                                                                                                              m
      1                 1       1        3                                (c) From gas equation PV =          RT
        ρ(v22 – v12) =    × 103  – 1 = –   × 103                                                            M
      2                 2       4        8
                                                                            ⇒ m = 0.3012 gm
     J/m3
     = – 0.375 × 103 J/m3
     ∴ Work done/vol. by pressure
     = 29.4 × 103 – 0.375 × 103 J/m3 = 29.025 × 103 m3.

XtraEdge for IIT-JEE                                              7                                             DECEMBER 2011
3.     A thin equiconvex lens of glass of refractive index                                   1           2
       µ = 3/2 and of focal length 0.3 m in air is sealed into                      ⇒                      ⇒ R = 0.3 m
                                                                                                = (1.5 – 1)
                                                                                            0.3          R
       an opening at one end of a tank filled with water
       (µ = 4/3). On the opposite side of the lens, a mirror is                     Substituting the value of R in equation (iii) we get
       placed inside the tank on the tank wall perpendicular                                              4          3
                                                                                                        – –1+ 2×
       to the lens axis, as shown in figure. The separation                             1       4/3       3          2
                                                                                    –        +       =
       between the lens and the mirror is 0.8 m. A small                              (0.9)      v           0.3
       object is placed outside the tank in front of the lens at
       a distance of 0.9 m from the lens along its axis. Find                                         –4 – 3 + 9
       the position (relative to the lens) of the image of the                        1     4             3         2
                                                                                         +    =                  =
       object formed by the system.               [IIT-1997]                         0.9   3v            0.3       0.9
                    0.9 m           0.8 m                                           ⇒ v = 0.2 m
                                                                                    As shown in the figure this image will form as I"
                                                                                    behind the mirror. But the ray will get reflected
                                                                                    from the mirror in between and the final image
                                                                                    formed will be I.
                                                                                    Since CI" = 0.2 m and CM = 0.8 m

Sol.                                                                                ∴ MI" = 0.2 m
                               v'                                                   ⇒ MI = 0.4 m                   [Q MI = MI"]
                                                                                    ∴ CI = 0.4 m

                                v                                             4.   A 5 m long cylindrical steel wire with radius 2 × 10–3
                                                                                   m is suspended vertically from a rigid support and
                                                                                   carries a bob of mass 100 kg at the other end. If
             O        C             I M                  I''     I'                the bob gets snapped, calculate the change in
                                                                                   temperature of the wire ignoring radiation losses.
                  u                              v                                 (For the steel wire : Young's modulus = 2.1 × 1011
                                                                                   Pa ; Density = 7860 kg/m3 ; Specific heat = 420
       The first refraction of the ray coming form the object                      J/kg-K).                                 [IIT-2001]
       is suffered on the left side of convex lens. For this we               Sol. When the mass of 100 Kg is attached, the string is
       can apply the equation                                                      under tension and hence in the deformed state.
                  µg     µg – µa                                                   Therefore it has potential energy (U) which is
          µ
       – a +          =                          ...(i)                            given by the formula.
           u      V'         R
                                                                                            1
       The image formed by this can be treated as a                                 U=        × stress × strain × volume
       virtual object and the refracting surface now is the                                 2
       right of the convex lens.                                                            1   (Stress) 2
                                                                                        =     ×            × πr2l
          µg      µ       µw – µg                                                           2       Y
       –       + w =                             ...(ii)
           v'      v         –R                                                           1 (Mg / πr 2 ) 2          1 M 2g 2l
       Adding (i) and (ii)                                                              =                  × πr2l =            ...(i)
                                                                                          2       Y                 2 πr 2 Y
           µa        µg        µg        µw       µg – µa
       –         +         –         +         =              +                     This energy is released in the form of heat, thereby
            u         v'        v'        v            R                            raising the temperature of the wire
        µw – µg                                                                                     Q = mc∆T
           –R                                                                                       ...(ii)
             µa    µ       –µ w – µ a + 2µ g                                        From (i) and (iii) Since U = Q Therefore
       ⇒ –       + w =
             u      v             R                                                                1 M 2g 2l
                                                                                    ∴ mc∆T =
       But according to lens formula                                                               2 πr 2 Y
                          1
                          f
                            =   ( µ – 1)  R
                                    a
                                    g    
                                         
                                           1
                                                             –
                                                                 1
                                                                 R2
                                                                      
                                                                      
                                                                                   ∴ ∆T =
                                                                                                 1 M 2g 2l
                                                        1                                      2 πr 2 Ycm

       ⇒
           1
           f
             =   ( µ – 1)  R1
                  a
                  g       
                                       –
                                            1
                                            R2
                                                 
                                                 
                                                 
                                                                                    Here
                                                                                    m = mass of string = density × volume of string
                               1                
                                                                                                      = ρ × πr2l

XtraEdge for IIT-JEE                                                      8                                           DECEMBER 2011
               1   M 2g 2                                                           c         c
     ∴ ∆T =                                                            ⇒       r1 =   ⇒ r2 =     ⇒ r 1 = r2
               2 (πr 2 ) 2 Ycp                                                      2         2
                                                                       ⇒       R2 gives the most accurate value
         1                   (100 × 10) 2
     =     ×
         2                 –3 2
             (3.14 × 2 × 10 ) × 2.1× 1011 × 420 × 7860
     = 0.00457º C
                                                                                 CHEMISTRY
5.   An unknown resistance X is to be determined
     using resistance R1, R2 or R3. Their corresponding          6.   From the following data, form the reaction between
     null points are A, B and C. Find which of the                    A and B.                                [IIT-1994]
     above will give the most accurate reading and
     why?                                  [IIT-2005]                     [A]               [B]           Initial rate (mol L–1s–1)
                                                                        mol L–1           mol L–1           300 K          320 K
                                                                       2.5 ×10–4         3.0 ×10–5        5.0 ×10   –4
                                                                                                                         2.0 × 10–3
                                                                       5.0 × 10–4        6.0 × 10–5       4.0 × 10–3         –
                                                                       1.0 × 10–3        6.0 × 10–5       1.6 × 10  –2
                                                                                                                             –
          X           R            R = R1 or R2 or R3                 Calculate
                                                                      (a) the order of reaction with respect to A and with
                G
                                                                          respect to B,
         A      B          C                                          (b) the rate constant at 300 K,
                                                                      (c) the energy of activation,
Sol. All Null point, the wheat stone bridge will be                   (d) the pre exponential factor.
     balanced
                                                                 Sol. Rate of reaction = k[A]l [B]m
        X     R
     ∴      =                                                         where l and m are the order of reaction with respect
         r1   r2                                                      to A and B respectively. From the given data, we
                r1                                                    obtain following expressions :
     ⇒X=R
                r2                                                    5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m
                                                                          ..(i)
     where R is a constant r1 and r2 are variable. The
     maximum fraction error is                                        4.0 × 10–3 = k[5.0 × 10–4] l [6.0 × 10–5]m
                                                                          ...(ii)
                                                                      1.6 × 10–2 = k[1.0 × 10–3]l [6.0 × 10–5]m
                                                                          ..(iii)
                                                                      From eq. (ii) and eq. (iii), we get
                                                                                                              l
                      X             R                                      4.0 × 10 −3     5.0 × 10 −4   
                                                                                        = 
                                                                                           1.0 × 10 −3
                                                                                                          
                                                                                                          
                               G                                           1.6 × 10 − 2                  
                      r1           r2
               M                            N                         or 0.25 = (0.5)l
                      A        B        C                             or (0.5)2 = (0.5) l
                     R=R1 R=R2 R=R3                                   or      l=2
      ∆X  ∆r  ∆r                                                      From eq. (i) and eq. (ii), we get
         = 1 + 2
      X    r1  r2                                                                                                     2                       m
                                                                           5.0 × 10 −4      2.5 × 10 −4                  3.0 × 10 −5   
                                                                                          =
                                                                                            5.0 × 10 − 4
                                                                                                                  
                                                                                                                  
                                                                                                                          
                                                                                                                           6.0 × 10 −5
                                                                                                                                          
                                                                                                                                          
     Here ∆r1 = ∆r2 = y (say) then                                         4.0 × 10 −3                                                 
         ∆X
     For      to be minimum r1 × r2 should be max                               1   1  1
                                                                                                      m
          X                                                           or          =   × 
                                                                                8   4  2
                                 [Q r1 + r2 = c (Constt.)]
                                                                                             m
     Let E = r1 × r2                                                           1   1
                                                                      or         = 
     ⇒     E = r1 × (r1 – c)                                                   2   2
            dE                                                        or       m=1
     ∴           = (r1 – c) + r1 = 0
            dr1

XtraEdge for IIT-JEE                                         9                                                        DECEMBER 2011
     (b) At T1 = 300 K,                                             Hence, difference in energy between first and
              Rate of reaction                                      second Bohr orbit for a H-atom is given by,
         k1 =                                                          ∆E = E n i – E n f = E2 – E1
                  [A]2 [B]1
                    5.0 × 10 −4                                                   21.76 × 10 −19         21.76 × 10 −19
          =                                                                  =–                     +
            [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ]                                            22            12
           = 2.67 × 108 L2 mol–2 s–1                                                          1    1
                                                                            = – 21.76 × 10–19  2 − 2  = 16.32 ×10–19J
                                                                                              2 1 
                                                                                        –8
     (c) At T2 = 320 K,                                             (b) For λ = 3.0 × 10 m
              Rate of reaction                                                hc    6.626 × 10 −34 × 3 × 108
         k2 =                                                          ∆E =       =
                  [A]2 [B]1                                                   λ            3.0 × 10 −8
                       2.0 × 10 −3                                                = 6.626 × 10–18 J           ....(i)
           =                                                        We know that, for H-like atoms,
               [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ]
                                                                       En for H-like atom = En for H-atom × Z2
           = 1.067 × 109 L2 mol–2 s–1                               ∴ ∆E for H-like atom = Z2 × ∆E for H-atom
                                k2  E T −T                                                        1      1
     We know, 2.303 log            = a  2 1                               = –Z2 × 21.76 × 10–19  2 − 2 
                                k1   R  T1T2 
                                                                                                  2     1 
                      1.067 × 109    E a  320 − 300                       = 16.32 × 10–19 Z2                ...(ii)
     or 2.303 log                 =                               From eq. (i) and (ii),
                       2.67 × 108   8.314  320 × 300 
                                                                       16.32 × 10–19 Z2 = 6.626 × 10–18
                                 Ea       20                      or           Z=2
     or 2.303 × 0.6017 =                         
                                8.314  320 × 300                  Thus, hydrogen atom like species for Z = 2 is He+.
              2.303 × 0.6017 × 8.314 × 320 × 300               8.   An organic compound A, C8H4O3, in dry benzene
     or Ea =
                              20                                    in the presence of anhydrous AlCl3 gives compound
           = 55.3 kJ mol –1                                         B. The compound B on treatment with PCl5
     (d) According to Arrhenius equation,                           followed by reaction with H2/Pd(BaSO4) gives
                                                                    compound C, which on reaction with hydrazine
        k = Ae − E a / RT                                           gives a cyclised compound D(C14H10N2). Identify
                                                 Ea                 A, B, C and D. Explain the formation of D from C.
     or 2.303 log k = 2.303 log A –
                                                 RT                                                        [IIT-2000]
     At 300 K,                                                 Sol. The given reactions are as follows.
                                            55.3 × 10 3                   O                            O
     2.303 log (2.67 × 108) = 2.303 log A –
                                           8.314 × 300                                      AlCl3
                                                                             O +
     or 2.303 × 8.4265 = 2.303 log A – 22.17                                                                     OH
                19.4062 + 22.17    41.5762                               O                                       O
     or logA =                   =           = 18.0531
                     2.303          2.303                                                      C6H5
                                                                                                                             C6H5
     A = Antilog 18.0531 = 1.13 × 1018 s–1
                                                                                               C     O H2NNH2
                                                                         PCl5                                                    N
7.   Estimate the difference in energy between 1st and                                               O
                                                                     H2/Pd (BaSO4)             C
     2nd Bohr orbit for a H atom. At what minimum                                                                                N
     atomic no., a transition from n=2 to n = 1 energy
                                                                                        H
     level would result in the emission of X-rays with
                                                                    The formation of D from C may be explained as
     λ = 3.0×10–8 m. Which hydrogen atom like species
                                                                    follows.
     does this atomic no. corresponds to ? [IIT-1993]
                                                                     C6H5                                O–                  O–
Sol. (a) For H atom,                                                                     C6H5                        C6H5
                                                                                                         +
         Z=1                                                            O    NH2                         NH2                     N–H
                                                                        O                                +
         ni = 2                                                              NH2                         NH2                  N–H
         nf = 1                                                                                     O–                      OH
                 21.76 × 10 −19                                                                                             C6H5
        En = –              2
                                   J
                        n                                                                                                        N
                                                                                                                                 N


XtraEdge for IIT-JEE                                      10                                                   DECEMBER 2011
9. (a) Write the chemical reaction associated with the                                       N≡C                   C≡N
           "brown ring test".                                                                                2+
     (b)       Draw the structures of [Co(NH3)6]3+,                                                     Ni
           [Ni(CN)4]2– and [Ni(CO)4]. Write the                                              N≡C                   C≡N
           hybridization of atomic orbital of the transition
           metal in each case.
     (c) An aqueous blue coloured solution of a                                 In [Ni(CO)4, nickel is present as Ni atom i.e. its
           transition metal sulphate reacts with H2S in                         oxidation number is zero and coordination number
           acidic medium to give a black precipitate A,                         is four.
           which is insoluble in warm aqueous solution of                                              3d           4s           4p
           KOH. The blue solution on treatment with KI                          Ni in
           in weakly acidic medium, turns yellow and                            Complex
           produces a white precipitate B. Identify the
           transition metal ion. Write the chemical                                                                 sp3 hybridization
           reaction involved in the formation of A and B.                       Its structure is as follows :
                                               [IIT-2000]
                                                                                                            CO
Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3
          2HNO3 + 6FeSO4 + 3H2SO4 →
                            3Fe2(SO4)3 + 2NO + 4H2O
                                                                                                            Ni
          [Fe(H2O)6]SO4.H2O + NO
                                                                                              OC                   CO
          Ferrous Sulphate
                       → [Fe(H2O)5NO] SO4 + 2H2O
                            (Brown ring)                                                                    CO
      (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its                     (c) The transition metal is Cu2+. The compound is
      coordination number is six.                                               CuSO4.5H2O
          Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2
                                                                                CuSO4 + H2S Acidic medium→ CuS ↓ + H2SO4
                                                                                                    
          Co3+ion = 1s2, 2s22p6, 3s23p63d6                                                                           Black ppt
                              3d         4s        4p
                                                                                2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4
      Hence
                                                                                                        (B) white
                                                                                         –         –
                              3d              4s          4p                        I2 + I → I3 (yellow solution)
     Co3+ion in
     Complex ion                                                           10. An organic compound A, C6H10O, on reaction with
                                     2    3                                     CH3MgBr followed by acid treatment gives
                                    d sp hybridization
                                                                                compound B. The compound B on ozonolysis gives
               NH3            3+                    NH3                         compound C, which in presence of a base gives
      H3N               NH3          H3N                      NH3
                                                                                1-acetyl cyclopentene D. The compound B on
                                                                                reaction with HBr gives compound E. Write the
              Co               or                                               structures of A, B, C and E. Show how D is formed
                                                   Co3+
                                                                                from C.                              [IIT-2000]
                                                                           Sol. The given reactions are as follows.
     H3N                NH3
               NH3                       H3N                   NH3            O                 OMgBr           CH3        CH3
                                                        NH3
                   2–                              2+                                               CH3                        Br
     In [Ni(CN)4 nickel is present as Ni                ion and its
                                                                                    CH3MgBr                  H+            HBr
     coordination numbers is four
                                                                                                            –H2O
         Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2
                                                                             (A)                                   (B)                (E)
         Ni2+ ion = 1s2, 2s22p6, 3s23p63d8
                             3d         4s                4p
        2+
     Ni ion =                                                                 COCH3              COCH3             CH3
                                                                                                  O
                                                                                       Base                              O O
                              3d              4s          4p
     Ni2+ion in                                                               (D)                                  (C)
     Complex ion
                                                                                The conversion of C into D may involve the
                                    dsp2 hybridization                          following mechanism.
     Hence structure of [Ni(CN)4]2– is

XtraEdge for IIT-JEE                                                  11                                            DECEMBER 2011
  COCH3                    COCH3                    COCH3                       12.  Solve for x the following equation         [IIT-1987]
                                                                                                            2
                                                                                              log(2x +3)(6x + 23x + 21)
  CH2        O             HC          O        HC            O–
              B+                                                    BH+                       = 4 – log(3x + 7) (4x2 + 12x + 9)
             –BH+                                                  –B           Sol. log(2x + 3) (6x2 + 23x + 21)
  (C)                                                                                        = 4 – log(3x + 7) (4x2 + 12x + 9)
                                       COCH3                                         ⇒       log(2x + 3) (2x + 3).(3x + 7)
         COCH3                                                     COCH3
                                        –                                                    = 4 – log(3x + 7) (2x + 3)2
                   OH +B                       OH –OH–
                                                                                     ⇒       1 + log(2x + 3) (3x + 7)
                     –BH+                                                                    = 4 – 2 log(3x + 7) (2x + 3)
                                                                   (D)               Put log(2x + 3) (3x + 7) = y
                                                                                                   2
                                                                                     ⇒
             MATHEMATICS                                                                     y+
                                                                                                   y
                                                                                                      –3=0

                                                                                      ⇒      y2 – 3y + 2 = 0        ⇒ (y – 1) (y – 2) = 0
11.     Let the three digit numbers A28, 3B9, and 62C,
        where A, B and C are integers between 0 and 9, be                             ⇒      y = 1 or               y=2
        divisible by a fixed integer K. Show that the                                 ⇒      log(2x +3) (3x + 7) = 1
        determinant                                                                   or     log(2x + 3) (3x + 7) = 2
        A    3     6                                                                  ⇒      3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2
        8    9     C is divisible by K.                  [IIT-1990]                   ⇒      x = – 4 or 3x + 7 = 4x2 + 12x + 9
         2   B     2                                                                         4x2 + 9x + 2 = 0
                                                                                             4x2 + 8x + x + 2 = 0
Sol. We know,
                                                                                             (4x + 1)(x + 2) = 0
            A28 = A × 100 + 2 × 10 + 8
                                                                                             x = – 2, – 1/4
            3B9 = 3 × 100 + B × 10 + 9
     and 62C = 6 × 100 + 2 × 10 + C                                                   ∴      x = – 2, – 4, – 1/4               ...(i)
                                                                                                                      2
     Since; A28, 3B9 and 62C are divisible by K,                                      But, log exists only when, 6x + 23x + 21 > 0,
     therefore there exist positive integers m1, m2 and                                      4x2 + 12x + 9 > 0,
     m3 such that,                                                                           2x + 3 > 0 and 3x + 7 > 0
            100 × A + 10 × 2 + 8 = m1K                                                ⇒      x > – 3/2                         ...(ii)
            100 × 3 + 10 × B + 9 = m2K                                                ∴      x = – 1/4 is the only solution.
     and, 100 × 6 + 10 × 2 + C = m3K             ...(i)
                       A   3       6                                            13.  Find the equation of the normal to the curve
        ∴        ∆= 8      9       C                                                 y = (1 + x)y + sin–1 (sin2x) at x = 0.     [IIT-1993]
                                                                                Sol. y = (1 + x)y + sin–1 (sin2x)               (given)
                    2      B       2
                                                                                     Let y = u + v, where u = (1 + x)y, v = sin–1 (sin2 x).
        Applying R2 → 100R1 + 10R3 + R2                                              Differentiationg
                               A                3                                            dy     du      dv
                                                                                                 =      +                       ...(i)
        ⇒        ∆ = 100 A + 2 × 10 100 × 3 + 10 × B + 9                                     dx     dx      dx
                           2                B                                        Now, u = (1 + x)y
                                                         6                           take logarithm of both sides
                                                                                            loge u = loge(1 + x)y
                                               100 × 6 + 10 × 2 + C
                                                                                     ⇒      loge u = y loge(1 + x)
                                                         2
                                                                                             1 du       y       dy
           A   3   6                                                                 ⇒             =         +      . {loge(1 + x)}
                                                                                             u dx     1+ x      dx
        = A28 3B9 62C , Using (i)
                                                                                              du    y     dy            
           2   B   2                                                                  ⇒          =u      + log e (1 + x)
                                                                                              dx    1 + x dx            
             A         3           6       A        3     6                                   du             y    dy            
        = m1 K      m2 K       m3 K = K m1          m2   m3                           ⇒          = (1 + x)y      + log e (1 + x)    ...(ii)
                                                                                              dx            1 + x dx            
           2           B           2       2        B     2
                                                                                      Again,          v = sin–1sin2 x
        ∴ ∆ = mK, Hence determinant is divisible by K.                                ⇒     sin v = sin2x

XtraEdge for IIT-JEE                                                       12                                           DECEMBER 2011
                         dv                                                                        π/3
       ⇒                                                                                                                 dx
                cos v
                         dx
                            = 2 sin x cos x                                               I = 2π    ∫ 2 – cos( x + π / 3)
                                                                                                     0
                 dv     1
       ⇒            =       [2sin x cos x]                                                         2π / 3
                                                                                                                  dt                 π
                 dx   cos v
                    2 sin x cos x         2 sin x cos x
                                                                                            = 2π
                                                                                                   π/3
                                                                                                       ∫       2 – cos t
                                                                                                                         , where x +
                                                                                                                                     3
                                                                                                                                       =t
       ⇒        =                     =                             ...(iii)
                      1 – sin v   2                4
                                           1 – sin x                                                           t
                                                                                                   2π / 3      sec 2
                                                                                                                 dt
       Put these values in equation (i)                                                     = 2π       ∫       2
                                                                                                 π / 3 1 + 3 tan
                                                                                                                 2 t
       dy             y     dy             2 sin x cos x                                                         2
          = (1 + x)y       + log e (1 + x) +
       dx             1 + x dx               1 – sin 4 x                                                 3


                 dy y (1 + x) y –1 + 2 sin x cos x / 1 – sin 4 x
                                                                                            = 2π       ∫
                                                                                                                   2du
                                                                                                               1 + 3u    2
                                                                                                                              =
                                                                                                                                  4π
                                                                                                                                   3
                                                                                                                                     .   { 3 tan   –1
                                                                                                                                                        3u   }   3
                                                                                                                                                                 1
       ⇒            =                                                                              1/ 3                                                          3
                 dx           1 – (1 + x) y ln (1 + x)
                                                                                                4π                                        4π        1
       At x = 0,                                                                            =            (tan–1 3 – tan–1 1) =                tan–1  
                                                                                                   3                                        3       2
             y = (1 + 0)y + sin–1 sin (0) = 1
                                                                                                       π/3
                                                                                                                 π + 4 x3           4π        1
       ⇒
                dy 1(1 + 0)1–1 + 2 sin 0. cos 0 / (1 – sin 4 0)
                   =
                                                                                          ∴              ∫                 π
                                                                                                    – π / 3 2 – cos | x | + 
                                                                                                                               dx =
                                                                                                                                     3
                                                                                                                                       tan –1   .
                                                                                                                                              2
                dx           1 – (1 + 0)1 ln (1 + 0)
                                                                                                                           3
                 dy
       ⇒            =1
                 dx                                                                 15.  The position vectors of the vertices A, B and C of a
       Again the slope of the normal is                                                  tetrahedron ABCD are i + j + k, i and 3i, respectively.
                       1                                                                 The altitude from vertex D to the opposite face ABC
             m=–            =–1                                                          meets the median line through A of the triangle ABC
                    dy / dx                                                              at a point E. If the length of the side AD is 4 and the
       Thus, the required equation of the normal is                                                                      2 2
             y – 1 = (–1) (x – 0)                                                        volume of the tetrahedron is          , find the position
                                                                                                                          2
       i.e., y + x – 1 = 0.                                                              vector of the point E for all its possible positions.
                                                                                                                                      [IIT-1996]
                    π/3
                                 π + 4x3                                                                                         i + 3i
14.    Evaluate      ∫                     π
                    – π / 3 2 – cos | x | + 
                                               dx.        [IIT-2004]                Sol. F is mid-point of BC i.e., F =
                                                                                                                                   2
                                                                                                                                         = 2i and

                                            3                                           AE ⊥ DE (given)
                                   
                                                                                                A(i+j+k)
Sol.                                                                                                                               D
          π/3                                π/3
                         πdx                           x 3dx                                               λ
Let, I=    ∫                     π
          – π / 3 2 – cos | x | + 
                                     + 4      ∫                 π
                                         – π / 3 2 – cos | x | + 
                                                                                                               E
                                 3                            3                                                1

           a
                       0,                    f (– x) = – f ( x)
                   a                                           

      –a
           ∫
(Using f ( x)dx = 
                             ∫
                    2 f ( x)dx,                f (– x) = f ( x) 
                                                                
                                                                                                 B(i)                             F(2i)        C(3i)
                   0                                                                    Let E divides AF in λ : 1. Then position vector of
                        π/3                                                               E is given by
                                     πdx
       ∴        I= 2     ∫                   π
                                                 +0                                        2 λi + 1(i + j + k )
                                                                                                                =
                                                                                                                   2λ + 1 
                                                                                                                           i+
                                                                                                                                 1
                                                                                                                                     j+
                                                                                                                                         1
                                                                                                                                             k
                          0   2 – cos | x | +                                                   λ +1                          λ +1    λ +1
                                             3                                                                   λ +1 
                                                                                         Now, volume of the tetrahedron
                  π/3            3                                                                 1
                                x dx                
                 as    ∫                 π
                  – π / 3 2 – cos | x | + 
                                              is odd 
                                                     
                                                                                                  =
                                                                                                     3
                                                                                                        (area of the base) (height)

                 
                                         3        
                                                                                                   2 2  1
                                                                                          ⇒             = (area of the ∆ ABC) (DE)
                                                                                                     3   3


XtraEdge for IIT-JEE                                                           13                                                           DECEMBER 2011
                                       1 → →
     But area of the ∆ ABC =             | BC × BA |                            WHAT ARE EARTHQUAKES?
                                       2
                 1
             =     |2i × (j + k)| = |i × j + i × k|
                 2
             = |k – j| =     2
                 2 2      1
     Therefore,         = ( 2) (DE)
                  3       3
     ⇒     DE = 2
     Since ∆ ADE is a right angle triangle,
           AD2 = AE2 + DE2
     ⇒     (4)2 = AE2 + (2)2
                                                                           Earthquakes like hurricanes are not only super
     ⇒     AE2 = 12
                                                                           destructive forces but continue to remain a mystery
          →      2λ + 1      1        1
     But AE =           i+       j+       k – (i + j + k)                  in terms of how to predict and anticipate them. To
                 λ +1       λ +1     λ +1                                  understand the level of destruction associated with
                      λ       λ       λ                                    earthquakes you really need to look at some
                 =        i–      j–      k
                     λ +1    λ +1    λ +1                                  examples of the past.
                 →            1                             3λ2            If we go back to the 27th July 1976 in Tangshan,
     ⇒       | AE | 2 =                [λ2 + λ2 + λ2] =
                           (λ + 1) 2                      (λ + 1) 2        China, a huge earthquake racked up an official
                             3λ2                                           death toll of 255,000 people. In addition to this an
     Therefore, 12 =                                                       estimated 690,000 were also injured, whole
                           (λ + 1) 2
                                                                           families, industries and areas were wiped out in the
     ⇒     4(λ + 1)2 = λ2
                                                                           blink of a second. The scale of destruction is hard to
     ⇒     4λ2 + 4 + 8λ = λ2
                                                                           imagine but earthquakes of all scales continue to
     ⇒     3λ2 + 8λ + 4 = 0
                                                                           happen all the time.
     ⇒     3λ2 + 6λ + 2λ + 4 = 0
     ⇒     3λ(λ + 2) + 2 (λ + 2) = 0                                       So what exactly are they ? Well the earths outer
     ⇒     (3λ + 2) (λ + 2) = 0                                            layer is made up of a thin crust divided into a
     ⇒     λ = – 2/3, λ = – 2                                              number of plates. The edges of these plates are
     Therefore, when λ = – 2/3, position vector of E is                    referred to as boundaries and it’s at these
     given by                                                              boundaries that the plates collide, slide and rub
      2λ + 1     1       1                                               against each other. Over time when the pressure at
             i+      j+      k
      λ +1      λ +1    λ +1                                             the plate edges gets too much, something has to
                                                                           give which results in the sudden and often violent
         2.(–2 / 3) + 1        1             1
     =                  i+            j+            k                      tremblings we know as earthquakes.
          – 2 / 3 +1       – 2 / 3 +1    – 2 / 3 +1
        –4 / 3 + 1        1            1                                   The strength of an earthquake is measured using a
     =             i+         j+           k
          – 2+3        – 2+3        – 2+3                                  machine called a seismograph. It records the
             3            3            3                                   trembling of the ground and scientists are able to
         –4 + 3                                                            measure the exact power of the quake via a scale
     =      3 i+ 1 j+ 1 k                                                  known as the richter scale. The numbers range from
          1/ 3      1/ 3     1/ 3                                          1-10 with 1 being a minor earthquake (happen
     = – i + 3j + 3k                                                       multiple times per day and in most case we don’t
     and when λ = – 2,                                                     even feel them) and 7-10 being the stronger quakes
     Position vector of E is given by,                                     (happen around once every 10-20 years). There’s a
      2 × (–2) + 1       1          1        –4 + 1                        lot to learn about earthquakes so hopefully we’ll
                   i+        j+          k=         i–j–k
         – 2 +1       – 2 +1      – 2 +1      –1                           release some more cool facts in the coming months.
     = 3i – j – k
     Therefore, – i + 3j + 3k and +3i – j – k are the
     answer.

XtraEdge for IIT-JEE                                                  14                                       DECEMBER 2011
              Physics Challenging Problems
                                                                                                                        Set # 8

 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
 of possible twists and turns of problems in physics that would be very helpful in facing IIT
 JEE. Each and every problem is well thought of in order to strengthen the concepts and we
 hope that this section would prove a rich resource for practicing challenging problems and
 enhancing the preparation level of IIT JEE aspirants.
                                                                                                    By : Dev Sharma
            Solutions will be published in next issue                              Director Academics, Jodhpur Branch

Passage # 1 (Que. 1 to 3)                                          4.    When barium is irradiated by a light of λ = 4000Å all
1.    In the circuit shown C1 = 6µF, C 2 = 3µF and battery               the photoelectrons emitted are bent in a circle of radius
                                                                         50cm by a magnetic field of 5.26 × 10-6 T. Then
      E = 20V. The switch S1 is first closed. It is then
      opened and afterwards S2 is closed them the final                  (A) the kinetic energy of fastest photoelectrons is 0.6eV
      charge on C2 is                                                    (B) work function of the metal is 2.5eV
                          C2                                             (C) the maximum velocity of photoelectrons is
                                                                             0.46 × 106 m/s
                           C1        S2                                  (D) the stopping potential for photoelectric effect is 0.6V

                                                                   5.    Suppose the potential energy between on electron
                                     S1                                                                                     ke 2
                                                                         and a proton at a distance r is given by                .
                        E = 20 V                                                                                            3r 3
      (A) 120µC                     (B) 80µC                             Application of Bohr’s theory to hydrogen atom in
                                                                         this case shows that
      (C) 40µC                      (D) 20µC
                                                                         (A) Energy in nth orbit is proportional to n6
2.    A current in a coil of self-inductance 2H is                       (B) Energy is proportional to m-3 (m = mass of electron)
      increasing as i = 2 sin( t 2 ) . The amount of energy              (C) Energy in nth orbit is proportional to n-2
      spent during the period when current changes from                  (D) Energy is proportional to m3 (m = mass of electron)
      0 to 2A is
      (A) 1J         (B) 2J      (C) 3J       (D) 4J               Passage # (Q. No. 6 to Q. No. 8)
3.    A particle of charge q and mass m is projected with               An ideal gas is taken round a cyclic process ABCA
      a velocity v forwards a circular region having                    as shown. If the internal energy of the gas at point A
      uniform magnetic field B perpendicular and into the               is assumed zero while at B it is 50J. The heat
      plane of paper from point P as shown. R is the                    absorbed by the gas in the process BC is 90J.
      radius and O is the centre of the circular region. If
      the line OP makes an angle θ with the direction of                                 30                      C
      v then the value of v so that particle passes through
      O is                                                                               20
                                                                               P(N/m2)            A
                                                                                         10                 B
                              × × ×
                           ×    ×                                                            D              E
                              × O× ×                                                   O       1     2     3
                          ×
                             R × ×
                    q   v × θ× ×                                                              V(m)3
                    m      P × ×                                   6.    Heat energy absorbed by the gas in process AB is
           qBR                      qBR                                  (A) 50J      (B) 70J (C)30J           (D) None
      (A)                      (B)
          m sin θ                  2m sin θ                        7.    Heat energy rejected by gas in process CA is
          2qBR                      3qBR                                 (A) 180J      (B) 140J (C) 220J       (D) None
      (C)                      (D)
          m sin θ                  2m sin θ                        8.    The net work done by gas in the complete cycle
                                                                         ABCA is
                                                                         (A) 20J     (B) zero (C) 40J     (D) -20J

XtraEdge for IIT-JEE                                          15                                             DECEMBER 2011
           8                                                     Solution
                                                                                                                            Set # 7




                                           Physics Challenging Problems
                                                        Qu e s tio ns we r e Pub lis he d in Nov emb er I ss ue

           3R               R                                              B 0 cos r = A 0 (1 + cos φ + cos 2φ)
1.   I AB =   I and I DC =    I
           4R              4R                                              B 0 sin r = (A 0 (sin φ + sin 2φ))
     Option [C] is correct                                                     2     2
                                                                           ∴ B 0 = A 0 (1 + 4 cos φ + 4 cos 2 φ)
2.   L.∆I = area under v – t curve                                         ∴ I θ = I 0 (1 + 4 cos φ + 4 cos 2 φ)
               1                                                           ∴ (I θ ) max = 9I 0
     2[i − 0] = × 10 x 2
               2                                                                   Iθ     1 1
     Option [A,C] is correct                                               also          = = (1 + 4 cos φ + 4 cos 2 φ)
                                                                               (I θ ) max 2 9
3. By conservation of energy                                               ⇒ cos φ = 0.56
                                              
                    2
            gR   − GMm  1                 
      1                                    GMm 
        × m       +           = mV + 
                                       2
            2   R  2
                                                                      8.   A → P; B → R ; C → Q; D → S
      2                                   R
                                         R+ 
                                            4                                       5          5  GM 
                                                                           v net =      v0 =           
     by conservation of angular momentum                                              4          4  r0 
                                                                                                       
          gR                   R                                                             −GMm 1        5 GM −3GMm
     m.      .R sin θ = mV R +                                           ∴ total energy = −        + m× .        =
           2                   4                                                               r0    2     4 r0       8r0
     Option [A] is correct                                                 when particle is at maximum or minimum distance r,
                                                                           then energy conservation
           2q 2  1 1        1   1                                         1        GMm        3GMm
4.   U=             − +    −   +   + .....                                   mv 2 −        =−           …….(1)
         4π ∈0 d  d 2d 3d 4d
                                                                          2          r         8r0
     Option [C] is correct                                                 Angular momentum conservation mvr
                                                                                  5
Passage Based Question:                                                     = m. v 0 .r0                 …….(2)
                                                                                  4
5. Option [A] is correct
6. Option [D] is correct
7. Option [A] is correct                                                                       Science Jokes
   Let A and w be the amplitude and angular frequency of
                                                                           A chemistry professor couldn't resist interjecting a
   the wave incident on the three slit grating. Let φ be
                                                                           little philosophy into a class lecture. He interrupted
     the phase difference between the diffracted waves                     his discussion on balancing chemical equations,
     from S1 and S2 and φ between those from S2 and S2. If                 saying, "Remember, if you're not part of the
     θ is angle of diffraction then using.                                 solution, you're part of the precipitate!"
          2πd sin θ                                                        1.     1 couldn't resist interjecting a little philosophy
     φ=
             λ                                                                    into a class lecture. He interrupted his
      A 1 = A 0 sin ωt                                                            discussion on balancing chemical equations,
                                                                                  saying, "Remember, if you're not part of the
     A 2 = A 0 sin(ωt + φ)                                                        solution, you're part of the precipitate!".
     A 3 = A 0 sin(ωt + 2φ)
                                                                           2.     Q. What is volume of a person who lost all his
     A = A 1 + A 2 + A 3 = A 0 [sin ωt (1 + cos φ + cos 2φ)                       memory ? A. 1/3 πr2h
     + cos ωt (sin φ + sin 2 φ)]
                                                                                  Because he keeps on saying, “main CONE
     A = B 0 sin(ωt + r )                                                         hu!"
     where

XtraEdge for IIT-JEE                                             16                                                DECEMBER 2011
XtraEdge for IIT-JEE   17   DECEMBER 2011
                                 Students Forum
                                         Expert’s Solution for Question asked by IIT-JEE Aspirants
     PHYSICS
1.    Oxygen is used as working substance in an engine                   Heat   supplied   to     gas during the process,
      working on the cycle shown in Figure                                                       7
                                                                         Q23 = nCP(T3 – T2) =       nRT0
                                                                                                 2
                       2     3
                                                                         Work done by gas during the process,
                                                                         W23 = nR(T3 – T2) = nRT0
                  P                  4                                   Now considering adiabatic process 3 → 4,
                                 1                                              V3 = 2V0 ,         T3 = 2T0
                                                                                V4 = 5V0 ,         T4 = ?
                             V                                                     γ–1
     Processes 1-2, 2-3, 3-4 and 4-1 are isothermal,                     Using T.V = constant
     isobaric, adiabatic and isochoric, respectively. If                        (2T0) (2V0)γ – 1 = T4(5V0)γ – 1
     ratio of maximum to minimum volume of oxygen                        or     T4 = 2 (0.4)0.4 T0
     during the cycle is 5 and that of maximum to                        Work done by the gas during the process,
     minimum absolute temperature is 2, assuming                                       P V – P4 V4        nR (T3 – T4 )
     oxygen to be an ideal gas, calculate efficiency of the                     W24 = 3 3              =
     engine.                                                                                γ –1               γ –1
     Given, (0.4)0.4 = 0.693 and loge 5 1.6094                                       = 5 nRT0 (1 – (0.4)0.4) T0
Sol. Volume of gas is minimum at state 2 during the                      During isochoric process 4 → 1, no work is done by
     cycle. Let it be V0. Then maximum volume of gas                     the gas and heat is rejected from the gas.
     during the cycle will be equal to 5V0 which is at                   Hence,          W41 = 0 and Q41 is negative
     states 4 and 1. Therefore, V4 = V1 = 5V0.                           ∴ Net work done by the gas during the cycle,
     Temperature during the cycle is maximum at the end                     W = W12 + W23 + W34 + W41 =
     of isobaric process 2 → 3 i.e. state 3 and minimum
                                                                                              nRT0 {6 – loge 5 – 5 × (0.4)0.4}
     at the end of isochoric cooling process 4 → 1 i.e.
     state 1. Let minimum absolute temperature be T0.                    Heat supplied to the gas during heating process,
     Then T1 = T0 and T3 = 2T0.                                                      7
                                                                         QS = Q23 =     ( 6 – loge 5 – 5 × 0.40.4) = 0.2642
     Since gas is Oxygen which is di-atomic, therefore,                              2
           5           7            7                                            W     2
     Cv = R, Cp =         R and γ =                                      or η =      =     ( 6 – loge 5 – 5 × 0.40.4) = 0.2642
           2           2            5                                            Qs    7
      Since, process 1 → 2 is isothermal, therefore,                     or η = 26.42 %                        Ans.
      temperature during the process remains constant.
      Hence temperature T2 is also equal to T0.                    2.   Calculate speed of sound in a mixture containing n1
      Considering n mole of the gas,                                    mole of He, n2 mole of H2 and n3 mole of CO2 at
      Work done by the gas during isothermal process                    temperature T
      1 → 2,                                                            (i) when n1 = 1, n2 = 2, n3 = 3 and T = 434 K,
                     V                                                  (ii) when n1 = 3, n2 = 2, n3 = 1 and T = 415 K
      W12 = nRT1.log 2 = – nRT0 loge 5
                     V1                                                      Gas constant, R = 8.3 J (mole)–1 K–1
      But for isothermal process Q = W, therefore,                 Sol. Sound waves in gases are longitudinal waves and
      Q12 = – nRT0 loge 5                                               speed of longitudinal waves in a gas is given by
      Now considering isobaric process 2 → 3                                  γRT
                                                                        v=         .
                      V3    T                                                  M
                          = 3 =2          or V3 = 2V0
                      V2    T2                                          Hence, to calculate speed of sound waves in the
                                                                        mixture, its mean molar mass M and adiabatic
                                                                        exponent γ must be known.

XtraEdge for IIT-JEE                                          18                                         DECEMBER 2011
     Molar mass of He is M1 = 4 gm/mole and its molar                Sol. First, incident waves are received by the wall and
                                                3                         then reflected back. These reflected waves are
     specific heat at constant volume is C v1 =     R.                    received by the source.
                                                2
                                                  5                       First, consider incident waves received by the wall.
     These for H2 are M2 = 2 gm/mole and C v 2 = R                        Since, source and wall (observer) both are
                                                  2
     respectively and for CO2, M3 = 44 gm/mole and                        approaching towards each other, therefore,
      C v3 = 3R respectively.                                             frequency received by the wall is given by,
                                                                                        v + v0       v+u
     Mean molar mass of the mixture,                                          n 1 = n0         
                                                                                        v – v  = n0  v – u  = 272 Hz
               M m + m 2 n 2 + M 3n 3                                                        s             
          M= 1 1                                       ...(i)
                   n1 + n 2 + n 3                                             Now waves are reflected by the wall. For reflected
                                                                              waves wall works as a sonic source of frequency
     and   Mean molar specific heat at constant volume,
                                                                              n1 = 272 Hz, which is approaching towards the
                n1C v1 + n 2 C v 2 + n 3C v3                                  receiver R with velocity u. Receiver is also
           Cv =                                     ...(ii)
                      n1 + n 2 + n 3                                          approaching the wall (source) with the same speed.
     (i) substituting n1 = 1, n2 = 2, n3 = 3 in equation (i)                  Therefore, frequency n2 of reflected waves received
        and (ii)                                                              by the receiver is given by
                  140                  140                                              v + v0        v+u
            M=         gm/mole or           × 10–3 kg/mole                    n 2 = n1         
                                                                                        v – v  = n1  v – u  = 289 Hz     Ans.
                    6                   6                                                    s              
                   31                                                         Since, receiver is approaching the wall with velocity
            Cv =      R
                  12                                                          u to receive reflected waves, therefore, velocity of
                                         43                                   wave propagation relative to the receiver is
     But Cp = Cv + R, therefore, Cp =       R                                                   v' = (v + u) = 340 ms–1.
                                         12
                                        C      43                             Let wavelength of reflected waves by λ'.
     ∴      Adiabatic exponent, γ = P =                                                                           20
                                        Cv     31                             Using,            v' = n2λ' or λ' =    m       Ans.
                                                                                                                  17
                                              γRT
     ∴     Speed of longitudinal waves =                             4.   Both ends of a solid cylindrical glass rod of diameter
                                               M
                                                                          d = 10 cm are made hemispherical. When a luminous
            = 214140 ms–1                                                 object is placed on axis of the rod at a distance
     or               462.75 ms–1                      Ans.               a = 20 cm from one end, its real image is obtained at a
     (ii) Substituting n1 = 3, n2 = 2, n3 = 1 in equation (i)             distance b = 40 cm from the other end. If the refractive
          and (ii)                                                        index of glass µ = 1.5. Calculate length l of the rod.
            M = 10 gm/mole or 10 × 10–3 kg/mole                      Sol. Since image is real, therefore, it is formed by
                   25                            37                       convergence of transmitted rays. If object is placed
            Cv =       R and Cp = Cv + R =          R
                   12                           12                        on left of left end of the rod, image will be on right
                 Cp     37                                                of the right end of rod.
            γ=        =
                 Cv     25                                                Since diameter of the rod is d and ends are made
                                                                          hemispherical, radius of curvature of each end is
                                      γRT                                 equal to d/2.
     Speed of longitudinal waves
                                       M                                  First considering refraction at left end,
     = 83 74 ms–1 or 714 ms–1                          Ans.                                                      d
                                                                          u = – a, µ1 = 1, µ2 = µ = 1.5, R = + , v = v1 (let)
3.   A sonic source of natural frequency n0 = 256 Hz and                                                         2
     a receiver are moving along the same line with speed                                         l
     u = 10 ms–1 towards a reflecting surface. Their line of                                        (µ)
     motion is normal to the surface and the surface is also
     approaching towards them with the same speed as                      O                               I1                    I
     shown in figure. If speed of sound in air is v = 330
     ms–1, calculate frequency and wavelength of reflected                                     v1
     waves received by the receiver.                                                          µ2     µ     µ – µ1
             u          u                                                     Using formula,      – 1 = 2
                                                                                               v      u       R
             S                                                                               v1 = + 30 cm
                       R
                                                                              It means that image I1 formed after refraction at left
                                         u                                    end is on right of left end at a distance of 30 cm.

XtraEdge for IIT-JEE                                            19                                             DECEMBER 2011
      This image works as an object for refraction at the                               dφ 1
      other end.                                                                 e=–       = π ωa2 B sin (ωt)
                                                                                        dt    2
      Then, for refraction at right end,                                     since, resistance of the circuit is negligible,
                       µ1 = 1.5, µ2 = 1, v = + b = + 40 cm,                  therefore, Potential difference across capacitor is
               d                                                             equal to induced emf in the circuit.
      R=–        = – 5 cm, u = ?
               2                                                             ∴ Charge on the capacitor at time t is q = C.e.
                        µ     µ      µ – µ1                                                     1
      Substituting in 2 – 1 = 2                                                              = πωa2 CB sin (ωt)
                         v     u         R                                                      2
                       u = – 20 cm                                                             dq    1
                                                                             But current I =       = π ω2 a2 CB cos (ωt) Ans.
      Negative sign indicates that object for right end is on                                   dt   2
      its left or image I1 is on left of right end of the rod                Due to flow of current, semicircle experience a
      and at a distance of 20 cm.                                            moment. Therefore, power is required to keep the
      ∴ Length of the rod = v1 + 20 cm = 50 cm Ans.                          semi circle rotating with constant angular velocity.
                                                                             In fact, power required to rotate the semicircle is
5.    A copper rod is bent into a semi-circle of radius a and                equal to electrical power generated in the circuit.
      at ends straight parts are bent along diameter of the
      semi-circle and are passed through fixed, smooth and                                                 1
                                                                             ∴ Power required, P = e.I = π2ω3a4 CB2 sin (2ωt)
      conducting rings O and O' as shown in figure. A                                                      8
      capacitor having capacitance C is connected to the                                                                      Ans.
      rings. The system is located in a uniform magnetic field
      of induction B such that axis of rotation OO' is
      perpendicular to the field direction. At initial moment
      of time (t = 0), plane of semi-circle was normal to the
      field direction and the semi-circle is set in rotation with
                                                                           Dimensional Formulae of Some
      constant angular velocity ω. Neglecting resistance and                     Physical Quantities
      inductance of the circuit, calculate current I flowing
      through the circuit at time t and instantaneous power                                                       Dimensional
                                                                         Physical Quantity
      required to rotate the semi-circle.                                                                          Formulae
                                                                         Work (W)                                 [ML2T–2]
                O                      O'                                Stress                                   [ML–1T–2]
                               a
                                                                         Torque (τ)                               [ML2T–2]
                                                                         Moment of Inertia (I)                    [ML2]
                           ×B
                                                                         Coefficient of viscosity (η)             [ML–1T–1]
                           C                                             Gravitational constant (G)               [M–1L3T–2]
Sol. When the copper rod is rotated, flux linked with the                Specific heat (S)                        [L2T–2θ–1]
     circuit varies with time. Therefore, an emf is                      Coeficient of thermal conductivity (K)   [MLT–3θ–1]
     induced in the circuit.                                             Universal gas constant (R)               [ML2T–2θ–1]
     At time t, plane of semi-circle makes angle ωt with
                                                                         Potential (V)                            [ML2T–3A–1]
     the plane of rectangular part of the circuit. Hence,
     component of the magnetic induction normal to                       Intensity of electric field (E)          [MLT–3A–1]
     plane of semi circle is equal to B.cos ωt.                          Permittivity of free space (ε0)          [M–1L–3T4A2]
     ∴ Flux linked with semicircular part is                             Specific resistance (ρ)                  [ML3T–3A2]
               1                                                         Magnetic Induction (B)                   [MT–2A–1]
         φ1 = πa2.B cos ωt
               2
                                                                         Planck's constant (h)                    [ML2T–1]
     Let area of rectangular part of the circuit be A.
                                                                         Boltzmann's constant (k)                 [ML2T–2θ–1]
     Then flux linked with this part is φ2 = BA
     ∴ Total flux linked with the circuit is φ = φ1 + φ2                 Entropy (S)                              [ML2T–2θ–1]
               1                                                         Decay constant (λ)                       [T–1]
     or φ = πa2 B.cos(ωt) + B.A
               2                                                         Bohr magnetic (µB)                       [L2A]
     ∴ Induced emf in the circuit,                                       Thermmionic current density (J)          [AL–2]



XtraEdge for IIT-JEE                                                20                                       DECEMBER 2011
                           P HYSICS F UNDAMENTAL F OR IIT-J EE

                                                    Ray Optics
                                       KEY CONCEPTS & PROBLEM SOLVING STRATEGY

Reflection :
Key Concepts :                                                                                 n αα
    (a) Due to reflection, none of frequency, wavelength                                   P                      P'
        and speed of light change.                                                                α
                                                                                      Real n      α           Virual
    (b) Law of reflection :                                                           Object                  Object
       Incident ray, reflected ray and normal on incident
        point are coplanar.                                              For solving the problem, the reference frame is
        The angle of incidence is equal to angle of                      chosen in which optical instrument (mirror, lens, etc.)
        reflection                                                       is in rest.
                                                                         The formation of image and size of image is
         Incident n Reflected             n                              independent of size of mirror.
           Ray        Ray
                θ θ                      θ θ
                                                    Tangent              Visual region and intensity of image depend on size
                                                   at point P            of mirror.
                                           P
           Plane surface             Convex surface                                           P                     P'
                                n
                                                                                                      n      θ
                                                                                                              θ
                                αα
                                                                                                             α
                      Convex     A                                                                           α
                      surface      Tangent
                                  at point P

Some important points : In case of plane mirror
    For real object, image is virtual.
                                                                         If the plane mirror is rotated through an angle θ, the
    For virtual object, image is real.
                                                                         reflected ray and image is rotated through an angle 2θ
    The converging point of incident beam behaves as a                   in the same sense.
    object.
                                                                         If mirror is cut into a number of pieces, then the focal
    If incident beam on optical instrument (mirror, lens                 length does not change.
    etc) is converging in nature, object is virtual.
                                                                         The minimum height of mirror required to see the full
    If incident beam on optical instrument is diverging in               image of a man of height h is h/2.
    nature, the object is real.
                                                                                           Rest
    The converging point of reflected or refracted beam
    from an optical instrument behaves as image.
                                                                                                  v
    If reflected beam or refracted beam from an optical                     Object v                   Image
    instrument is converging in nature, image is real.


                                               n                                                             vsinθ
                      P                   P                     P                          v          Rest
            P                                                               vsinθ
                                     Real      n                                      θ
         Virtual       Real                               Virual
         Object        Object        Object               Object             Object       vcosθ       vcosθ Image

     If reflected beam or refracted beam from an optical
     instrument is diverging in nature, image is virtual.                                             vm
                                                                                                                  2vm–v
                                                                            Object v                   Image


XtraEdge for IIT-JEE                                                21                                            DECEMBER 2011
                                                                                   y                  y
                        vm
                                        2vm                                   x'           x     x'            x
       Object             Image
       In rest                                                                     y'                 y'
                                                                                                      1 1 1
                                                                             The mirror formula is     + =
                                                                                                      v u f
           Object        vm
       v                                 2vm+v                               Also, R = 2f
                           Image                                             These formulae are only applicable for paraxial
                                                                             rays.
    (c) Number of images formed by combination of                            All distances are measured from optical centre. It
        two plane mirrors : The images formed by                             means optical centre is taken as origin.
        combination of two plane mirror are lying on a                       The sign conventions are only applicable in given
        circle whose centre is at the meeting points of                      values.
        mirrors. Also, object is lying on that circle.                       The transverse magnification is
                    360º                                                                 image size    −v
       Here, n =                                                                   β=                =
                     θ                                                                   object size   u
       where θ = angle between mirrors.
                                                                          1. If object and image both are real, β is negative.
          360º                                                            2. If object and image both are virtual, β is negative.
       If      is even number, the number of images is
            θ
                                                                          3. If object is real but image is virtual, β is positive.
       n – 1.
                                                                          4. If object is virtual but image is real, β is positive.
          360º
       If        is odd number and object is placed on                    5. Image of star; moon or distant object is formed at
            θ                                                                focus of mirror.
       bisector of angle between mirror, then number of
       images is n – 1.                                                      If y = the distance of sun or moon from earth.
                                                                                  D = diameter of moon or sun's disc
            360º
       If        is odd and object is not situated on                             f = focal length of the mirror
             θ
        bisector of angle between mirrors, then the                               d = diameter of the image
        number of images is equal to n.                                            θ = the angle subtended by sun or moon's disc
    (d) Law of reflection in vector form :                                                      D   d
                                                                             Then tan θ = θ =     =
             ˆ
        Let e1 = unit vector along incident ray.                                                y   f
             ˆ
             e 2 = unit vector along reflected ray                           Here, θ is in radian.
            ˆ
           n = unit vector along normal on point of
               Incidence                                                                Sun
                                                                                           D     F
       Then    e 2 = e1 − 2(e1.n ) n
               ˆ     ˆ      ˆ ˆ ˆ                                                                     θ
                                                                                                           θ
                              n                                                                 d
                              nˆ

                        ˆ
                        e1          ˆ
                                    e2                               Problem solving strategy :
                                                                     Image formation by mirrors
    (e) Spherical mirrors :                                              Step 1: Identify the relevant concepts : There are
                                                                         two different and complementary ways to solve
        It easy to solve the problems in geometrical optics              problems involving image formation by mirrors. One
        by the help of co-ordinate sign convention.                      approach uses equations, while the other involves
             y                  y             y                          drawing a principle-ray diagram. A successful
                                                                         problem solution uses both approaches.
        x'          x        x'               x   x'        x            Step 2: Set up the problem : Determine the target
                                                                         variables. The three key quantities are the focal
             y'                    y'                  y'                length, object distance, and image distance; typically


XtraEdge for IIT-JEE                                            22                                             DECEMBER 2011
    you'll be given two of these and will have to                          Note that the same sign rules (given in section)
    determine the third.                                                   work for all four cases in this chapter : reflection
    Step 3: Execute the solution as follows :                              and refraction from plane and spherical surfaces.
       The principal-ray diagram is to geometric optics                 Step 4: Evaluate your answer : You've already
       what the free-body diagram is to mechanics. In                   checked your results by using both diagrams and
       any problem involving image formation by a                       equations. But it always helps to take a look back
       mirror, always draw a principal-ray diagram first                and ask yourself. "Do these results make sense ?".
       if you have enough information. (The same
       advice should be followed when dealing with                  Refraction :
       lenses in the following sections.)
                                                                    Laws of Refraction :
       It is usually best to orient your diagrams
       consistently with the incoming rays traveling                    The incident ray, the refracted ray and normal on
       from left to right. Don't draw a lot of other rays at            incidence point are coplanar.
       random ; stick with the principal rays, the ones
       you know something about. Use a ruler and                        µ1 sin θ1 = µ2 sin θ2 = ... = constant.
       measure distance carefully ! A freehand sketch
       will not give good results.
                                                                                                θ1
       If your principal rays don't converge at a real                                                     µ1
       image point, you may have to extend them
       straight backward to locate a virtual image point,                                                  µ2
       as figure (b). We recommend drawing the
       extensions with broken lines. Another useful aid                                               θ2
       is to color-code the different principal rays, as is
       done in figure(a) & (b).
                                                                        Snell's law in vector form :
           Q             I
                                                                                                     ˆ
                                                                                                     n
                                 4
                              2
                      3         C P' F                  v
            P
                          2                                                                ˆ
                                                                                           e1
                          4               Q'                                                               µ1
                                                    3
                                                                                                                µ2
                               1                                                                           ˆ
                                                                                                           e2
                                    (a)

                                                                             ˆ
                                                                        Let, e1 = unit vector along incident ray
        Q                 1 1
                            3                                                 ˆ
                                                                              e 2 = unit vector along refracted.
                                               Q'
                2 4       2
                                     v                                        ˆ
                                                                              n = unit vector along normal on incidence point.
        P                                  P' F         C
                  4                                                              ˆ    ˆ         ˆ     ˆ
                                                                        Then µ1( e1 × n ) = µ2( e 2 × n )
                              (b)                                       Some important points :
                                        1 1 1                           (a) The value of absolute refractive index µ is always
       Check your results using Eq.       + =        and the
                                        s s' f                              greater or equal to one.
                                         y'       s'
       magnification equation m =           = − . The                   (b) The value of refractive index depends upon
                                         y        s
                                                                            material of medium, colour of light and
       results you find using this equation must be
       consistent with your principal-ray diagram; if not,                  temperature of medium.
       double-check both your calculation and your                      (c) When temperature increases, refractive index
       diagram.
                                                                            decreases.
       Pay careful attention to signs on object and image
       distances, radii or curvature, and object and image              (d) Optical path is defined as product of geometrical
       heights. A negative sign on any of these                             path and refractive index.
       quantities always has significance. Use the
       equations and the sign rules carefully and                           i.e., optical path = µx
       consistently, and they will tell you the truth !

XtraEdge for IIT-JEE                                           23                                                DECEMBER 2011
                                                                                      ti   t   t
                                                                                  Σ      = 1 + 2 + ...
    (e) For a given time, optical path remains constant.                              µi  µ1  µ2

        i.e., µ1x1 = µ2x2 = ... constant                                                                                          ti
                                                                         The apparent depth due to a number of media is Σ
              dx     dx                                                                                                           µi
    ∴       µ1 1 = µ2 2
               dt     dt
                                                                         The lateral shifting due to a slab is d = t sec r sin(i – r).
    ∴       µ1c1 = µ2c2       (where c1 and c2 are speed
                                                                         Critical angle : When a ray passes from denser
                             of light in respective mediums)
                                                                         medium (µ2) to rarer medium (µ1), then for 90º angle
             µ2   c
    ∴           = 1                                                      of refraction, the corresponding angle of incidence is
             µ1   c2
                                                                         critical angle.
                  1
    i.e.,   µ∝                                                                                             µ1
                  c                                                      Mathematically,         sin c =
                                                                                                           µ2
    (f) The frequency of light does not depend upon
        medium.                                                          (i) When angle of incidence is lesser than critical
    ∴       c1 = fλ1,       c2 = fλ2                                         angle, refraction takes place. The corresponding
             µ1   c  λ                                                       deviation is
    ∴           = 2 = 2
             µ2   c1  λ1                                                               µ        
                                                                             δ = sin–1  2 sin i  – i
                                                                                       µ                 for i < c
               1                                                                        1       
    ∴       µ∝
               λ
                                                                         (ii) When angle of incidence is greater than critical
    When observer is rarer medium and object is in
                                                                              angle, total internal reflection takes place. The
    denser medium :
                                                                              corresponding deviation is
                          real depth
    Then         µ=                                                          δ = π – 2i          when i < c
                        apparent depth
                                                                    The δ – i graph is :
    When object is in rarer and observer is in denser
                                                                         (i) Critical angle depends upon colour of light,
    medium :
                                                                             material of medium, and temperature of medium.
                  apparent position
            µ=                                                           (ii) Critical angle does not depend upon angle of
                    real position
                                                                              incidence
                                              1
    The shift of object due to slab is x = t 1 – 
                                              µ
                                                 
    (a) This formula is only applicable when observer is
        in rarer medium.                                                               δ
    (b) The object shiftiness does not depend upon the                                               c
                                                                                                 i              π/2
        position of object.
    (c) Object shiftiness takes place in the direction of           Refractive surface formula,
        incidence ray.                                                       µ2  µ   µ − µ1
                                                                                – 1 = 2
    The equivalent refractive index of a combination of a                     v   u     r
                                                      Σt i               Here, v = image distance,
    number of slabs for normal incidence is µ =
                                                       t
                                                      Σ i                u = object distance,
                                                       µi
                                                                         r = radius of curvature of spherical surface.
    Here,   Σti = t1 + t2 + ...
                                                                         (a) For plane surface , r = ∞
                                                                         (b) Transverse magnification,


XtraEdge for IIT-JEE                                           24                                               DECEMBER 2011
                   Im age size  µv                               (b) This formula is only applicable when medium on
              m=               = 1
                   object size  µ2u                                  both sides of lens are same.
    (c) Refractive surface formula is only applicable for        (c) Intensity     is   proportional     to   square   of
        paraxial ray.                                                aperture.
Lens :                                                           (d) When lens is placed in a medium whose refractive
    Lens formula :                                                   index is greater than that of lens. i.e., µ1 > µ2.
                   1   1   1                                         Then converging lens behaves as diverging lens
                     –   =                                           and vice versa.
                   v   u   f
    (a) Lens formula is only applicable for thin lens.           (e) When medium on both sides of lens are not same.
    (b) r = 2f formula is not applicable for lens.                   Then both focal lengths are not same to each
                                                                     other.
              image size    v
    (c) m =               =                                      (f) If a lens is cut along the diameter, focal length
              object size   u
                                                                     does not change.
    (d) Magnification formula is only applicable when
        object is perpendicular to optical axis.                 (g) If lens is cut by a vertical, it converts into two
                                                                     lenses of different focal lengths.
    (e) lens formula and the magnification formula is
        only applicable when medium on both sides of                        1   1    1
                                                                    i.e.,     =    +
        lenses are same.                                                    f   f1   f2

                                                                 (h) If a lens is made of a number of layers of different
    (f)
                                                                     refractive index number of images of an object by
                                                                     the lens is equal to number of different media.
                f(+ve)                  f(–ve)
                                                                 (i) The minimum distance between real object and
                   (i)                   (ii)
                                                                     real image in is 4f.
                                                                 (j) The equivalent focal length of               co-axial
                                                                     combination of two lenses is given by
                     f(–ve)                f(+ve)                    1   1   1      d
                                                                       =   +    –
                                                                     F   f1 f 2   f1f 2
                (iii)                   (iv)
                                                                 (k) If a number of lenses are in contact, then
                                                                     1   1   1
                                                                       =   +    + ......
                                                                     F   f1 f 2
                f(–ve)                     f(+ve)
                   (v)                          (vi)                                               1
                                                                 (l) (i) Power of thin lens, P =
                                                                                                   F
    (g) Thin lens formula is applicable for converging as
        well diverging lens. Thin lens maker's formula :                                            1
                                                                    (ii) Power of mirror is P = –
                                                                                                    F
              1   µ − µ1      1 1
                = 2
                  µ      
                                − 
                                r r                            (m) If a lens silvered at one surface, then the system
              f     1          1 2
                                                                    behaves as an equivalent mirror, whose power
                         µ1        µ1                                        P = 2PL + Pm
                              µ2                                    Here, PL = Power of lens

                                                                                   µ − µ1    1 1
                                                                                 = 2
                                                                                   µ      
                                                                                               − 
                                                                                               r r 
    (a) Thin lens formula is only applicable for paraxial                            1        1 2
        ray.

XtraEdge for IIT-JEE                                        25                                         DECEMBER 2011
                                                         1                              1    1   1
                   Pm = Power of silvered surface = –                             Now      =   +
                                                        Fm                              f1   u1 v1
                                                                                        1      1    1
         Here, Fm = r2/2, where r2 = radius of silvered surface.                  or        =    +
                                                                                        20     25 v1
         P = – 1/F
                                                                                  v1 = 100 cm.
         Here, F = focal length of equivalent mirror.                        As v1 is positive, hence the image is real. In the
                                                                             absence of convex mirror, the rays after reflection
                                                                             from concave mirror would have formed a real image
                                                                             I1 at distance 100 cm from the mirror. Due to the
                  Solved Examples                                            presence of convex mirror, the rays are reflected and
                                                                             appear to come from I2.
1.   Rays of light strike a horizontal plane mirror at an                    (ii) For convex mirror,
     angle of 45º. At what angle should a second plane                       In this case, I1 acts as virtual object and I2 is the
     mirror be placed in order that the reflected ray finally                virtual image.
     be reflected horizontally from the second mirror.                       The distance of the virtual object from the convex
Sol. The situation is shown in figure                                        mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm.
                                    C     G                                  As focal length of convex mirror is negative and
             A                                  D                            hence f2 = –30/2 = –15 cm. Here we shall calculate
                             S     θ θ                                       the value of v2. Using the mirror formula, we have
                       45º                 N                                          1        1   1
                                                                                   −     = −     +
                            45º                                                      15       50 v 2
         P                              Q
                       B                                                     or        v2 = –21.42 cm
     The ray AB strikes the first plane mirror PQ at an                      As v2 is negative, image is virtual. So image is
     angle of 45º. Now, we suppose that the second                           formed behind the convex mirror at a distance of
     mirror SG is arranged such that the ray BC after                        21.43 cm.
     reflection from this mirror is horizontal.
                                                                        3.   There is a small air bubble in side a glass sphere (n =
     From the figure we see that emergent ray CD is
     parallel to PQ and BC is a line intersecting these                      1.5) of radius 10 cm. The bubble is 4 cm below the
     parallel lines.
                                                                             surface and is viewed normally from the outside
     So,      ∠DCE = ∠CBQ = 180º
         ∠DCN + ∠NCB + ∠CBQ = 180º                                           (Fig.). Find the apparent depth of the air bubble.
              θ + θ + 45º = 180º     ∴ θ = 67.5º                                                         P
                                                                                                A             n2 = 1
     As ∠NCS = 90º, therefore the second mirror should
                                                                                                         I
     be inclined to the horizontal at an angle 22.5º.
                                                                                                         O
2.   An object is placed exactly midway between a                                                        C
     concave mirror of radius of curvature 40 cm and a                                                    n1 = 1.5
     convex mirror of radius of curvature 30 cm. The
     mirrors face each other and are 50 cm apart.
     Determine the nature and position of the image
     formed by the successive reflections, first at the                 Sol. The observer sees the image formed due to refraction
     concave mirror and then at the convex mirror.                           at the spherical surface when the light from the
Sol. The image formation is shown in figure.                                 bubble goes from the glass to air.
                                                                             Here u = – 4.0 m,    R = – 10 cm, n1 = 1.5 and n2 = 1
                                    50cm
                                                                             We have       [(n2/v) – (n1/u) = (n2 – n1)/R
                  I2
                             P2 C      F                                     or        (1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm)
                                               P1
                                       25cm                                  or        (1/v) = (0.5/10 cm) – (1.5/4.0 cm)

                                      r = 40 cm                              or        v = – 3.0 cm
                        r = 30 cm
             I1
                                                                             Thus, the bubble will appear 3.0 cm below the
                                                                             surface.
     (i) For concave mirror,
         u1 = 25 cm, f1 = 20 cm and v1 = ?

XtraEdge for IIT-JEE                                               26                                            DECEMBER 2011
4.   A convex lens focuses a distance object on a screen
     placed 10 cm away from it. A glass plate (n = 1.5) of           DEEPEST LAKE IN THE WORLD ?
     thickness 1.5 is inserted between the lens and the
     screen. Where should the object be placed so that its
     image is again focused on the screen ?
Sol. The situation when the glass plate is inserted between
     the lens and the screen, is shown in fig. The lens
     forms the image of object O at point I1 but the glass
     plate intercepts the rays and forms the final image at I
     on the screen. The shift in the position of image after
     insertion of glass plate




                                                    Screen
                                                                     Lake Baikal (Baikal) in Siberia, Russia is the deepest
          O                                   I1     I               lake in the world measuring 1620m deep at its
                                                                     deepest point. This makes it not only deep but also
                                                                     the oldest lake in the world estimated to be around 25
                                  10 cm                              million years old. At over 636 kilometers long and 80
                                                                     kilometers wide this fresh water lake holds over 20
              1                  1                               percent of all the fresh water in the world and is
     I1I = t 1 −  = (1.5 cm) 1 −    = 0.5 cm.
              n               1.5                                second in size only to the Caspian Sea (the caspian is
                                                                     called a sea but is technically a lake).
     Thus, the lens forms the image at a distance of 9.5 cm
     from itself. Using                                              To put things into perspective the lake is so big that if
                                                                     all the rivers in the world flowed into its basin it
      1 1 1                  1 1 1  1   1                            would take almost 1 year to fill. We all know Siberia
       – = , we get           = – =   –
      v u f                  u v f 9.5 10                            isn’t the warmest of places so you can imagine what
     or      u = – 190 cm.                                           a phenomenal site it is when in the winter months the
                                                                     lake freezes over holding ice up to 115 meters thick.
     i.e. the object should be placed at a distance of               Now that’s a lot of ice!
     190 cm. from the lens.

5.   A candle is placed at a distance of 3 ft from the wall.         WHICH  IS    THE   HIGHEST
     Where must a convex lens of focal length 8 inches be            WATERFALL IN THE WORLD ?
     placed so that a real image is formed on the wall ?
Sol. According to formula for refraction though a lens
                   36 – v                v




                                  f = 8"

                        d = 3 ft = 36"

      1 1 1                                1     1       1           The highest waterfall in the world is the Angel Falls
       – =                  or               –         =
      v u f                                v − (36 − v) 8            in Venezuela. At a towering height of 979m did you
                                                                     know that each drop of water takes 14 seconds to fall
          1   1     1                      36 − v + v 1
     or     +     =         or                       =               from the top to the bottom. The water flows from the
          v 36 − v 8                       v(36 − v) 8               top of a “Tepui” which is a flat topped mountain with
     or, v2 – 36 v + 8 × 36 = 0                                      vertical sides.
     or v = 12"            or            24" = 1 ft or 2 ft.         The waterfall which despite being known to the local
                                                                     indians for thousands of years was originally called
     ∴      u = 24"      or      12" = 2 ft or 1 ft
                                                                     the “Churun Meru” but for some reason they were
     Hence, lens should be placed at either 1 ft or 2 ft             renamed by an American bush pilot called Jimmy
     away from the wall.                                             Angel, who noticed them in 1935 whilst flying over
                                                                     the area looking for gold.

XtraEdge for IIT-JEE                                            27                                         DECEMBER 2011
                           P HYSICS F UNDAMENTAL F OR IIT-J EE

    Fluid Mechanics & Properties of Matter
                                  KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Fluid Mechanics :                                                   Fluid dynamics :
Fluid statics :                                                                                     1 2           p
                                                                         Bernoulli's Theorem :        v + gh +       = a constant
     Pressure at a point inside a Liquid : p = p0 + ρgh                                             2             ρ
     where p0 is the atmospheric pressure, ρ is the density              for a streamline flow of a fluid (liquid or gas).
     of the liquid and h is the depth of the point below the             Here, v is the velocity of the fluid, h is its height
     free surface.                                                       above some horizontal level, p is the pressure and ρ
                                p0                                       is the density.
                                                                                      p1
                       h                                                               v1                       p2
                             p
                                                                             h1
                                  ρ
                                                                                                                v2 h2

                                                                                     v2 > v1      p2 < p1
     Pressure is a Scalar : The unit of pressure may be
     atmosphere or cm of mercury. These are derived                      According to this principle, the greater the velocity,
     units. The absolute unit of pressure is Nm–2. Normal                the lower is the pressure in a fluid flow.
     atmospheric pressure, i.e, 76 cm of mercury, is                     It would be useful to remember that in liquid flow,
     approximately equal to 105 Nm–2.                                    the volume of liquid flowing past any point per
     Thrust : Thrust = pressure × area. Thrust has the unit              second is the same for every point. Therefore, when
     of force.                                                           the cross-section of the tube decreases, the velocity
     Laws of liquid pressure                                             increases.
     (a) A liquid at rest exerts pressure equally in all                 Note : Density = relative density
         directions.                                                     or       specific gravity × 1000 kg m–3.
     (b) Pressure at two points on the same horizontal line         Surface tension and surface energy :
         in a liquid at rest is the same.                                Surface Tension : The property due to which a
     (c) Pressure exerted at a point in a confined liquid at             liquid surface tends to contract and occupy the
         rest is transmitted equally in all directions and               minimum area is called the surface tension of the
         acts normally on the wall of the containing vessel.             liquid. It is caused by forces of attraction between the
         This is called Pascal's law. A hydraulic press                  molecules of the liquid. A molecule on the free
         works on this principle of transmission of                      surface of a liquid experiences a net resultant force
         pressure.                                                       which tends to draw it into the liquid. Surface tension
     The principle of floating bodies (law of flotation) is              is actually a manifestation of the forces experienced
     that W = W´, that is, weight of body = weight of                    by the surface molecules.
     displaced liquid or buoyant force. The weight of the                If an imaginary line is drawn on a liquid surface then
     displaced liquid is also called buoyancy or upthrust.               the force acting per unit length of this line is defined
     Hydrometers work on the principle of floating                       as the surface tension. Its unit is, therefore, newton /
     bodies. This principle may also be applied to gases                 metre. This force acts along the liquid surface. For
     (e.g., a balloon).                                                  curved surfaces, the force is tangent to the liquid
     Liquids and gases are together called fluids. The                   surface at every point.
     important difference between them is that liquids                   Surface Energy : A liquid surface possesses
     cannot be compressed, while gases can be                            potential energy due to surface tension. This energy
     compressed. Hence, the density of a liquid is the                   per unit area of the surface is called the surface
     same everywhere and does not depend on its                          energy of the liquid. Its units is joule per square
     pressure. In the case of a gas, however, the density is             metre. The surface energy of a liquid has the same
     proportional to the pressure.                                       numerical values as the surface tension. The surface

XtraEdge for IIT-JEE                                           28                                           DECEMBER 2011
    tension of a liquid depends on temperature. It                        The upward force by which a liquid surface is pulled
    decreases with rise in temperature.                                   up in a capillary tube is 2πrTcos θ, and the downward
    Excess of Pressure : Inside a soap bubble or a gas                    force due to the gravitational pull on the mass of
    bubble inside a liquid, there must be pressure in                     liquid in the tube is (πr2h + v)ρg, where v is the
    excess of the outside pressure to balance the tendency                volume above the liquid meniscus. If θ = 0º, the
    of the liquid surface to contract due to surface                      meniscus is hemispherical in shape. Then v =
    tension.                                                              difference between the volume of the cylinder of
                                                                          radius r and height r and the volume of the
                              1 1
    p(excess of pressure) = T  +  in general                            hemisphere of radius r
                              r r 
                               1 2                                                             2 3 1 3
                                                                                       = πr3 –     πr = πr
    where T is surface tension of the liquid, and r1 and r2                                      3     3
    are the principal radii of curvature of the bubble in
    two mutually perpendicular directions.                                When θ ≠ 0, we cannot calculate v which is generally
                                                                          very small and so it may be neglected. For
    For a spherical soap bubble, r1 = r2 = r and there are                equilibrium
    two free surfaces of the liquid.
                                                                             (πr2h + v) ρg = 2πrT cos θ
                 4T
    ∴        p=                                                          When a glass capillary tube is dipper in mercury, the
                  r                                                      meniscus is convex, since the angle of contact is
    For a gas bubble inside a liquid, r1 = r2 = r and there              obtuse. The surface tension forces now acquire a
    is only one surface.                                                 downward component, and the level of mercury
                   2T                                                    inside the tube the falls below the level outside it. the
    ∴       p=                                                           relation 2T cos θ = hρgr may be used to obtain the
                    r
                                                                         fall in the mercury level.
    For a cylindrical surface r1 = r and r2 = ∞ and there            Problem Solving Strategy
    are two surfaces.
                                                                         Bernoulli's Equations :
                 2T
    ∴       p=                                                           Bernoulli's equation is derived from the work-energy
                   r                                                     theorem, so it is not surprising that much of the
    Angle of Contact : The angle made by the surface of                  problem-solving strategy suggested in W.E.P. also
    a liquid with the solid surface inside of a liquid at the            applicable here.
    point of contact is called the angle of contact. It is at            Step 1: Identify the relevant concepts : First ensure
    this angle that the surface tension acts on the wall of              that the fluid flow is steady and that fluid is
    the container.                                                       incompressible and has no internal friction. This case
    The angle of contact θ depends on the natures of the                 is an idealization, but it hold up surprisingly well for
    liquid and solid in contact. If the liquid wets the solid            fluids flowing through sufficiently large pipes and for
    (e.g., water and glass), the angle of contact is zero. In            flows within bulk fluids (e.g., air flowing around an
    most cases, θ is acute (figure i). In the special case of            airplane or water flowing around a fish).
    mercury on glass, θ is obtuse (figure ii).                           Step 2: Set up the problem using the following steps
                                                                             Always begin by identifying clearly the points 1
               θ                        θ                                    and 2 referred to in Bernoulli's equation.
                                                                             Define your coordinate system, particular the
             fig. (i)                   fig. (ii)                            level at which y = 0.
    Rise of Liquid in a Capillary Tube : In a thin                           Make lists of the unknown and known quantities
    (capacity) tube, the free surface of the liquid becomes                                       1                     1
                                                                             in Eq. p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22
    curved. The forces of surface tension at the edges of                                         2                     2
    the liquid surface then acquire a vertical component.                          (Bernoulli's equation)
                   T θ meniscus θ T                                          The variables are p1, p2, v1, v2, y1 and y2, and the
                                                                             constants are ρ and g. Decide which unknowns
                                                                             are your target variables.
                        θ           θ
                                                                         Step 3: Execute the solutions as follows : Write
                   h                                                     Bernoulli's equation and solve for the unknowns. In
                                                                         some problems you will need to use the continuity
                                                                         equation, Eq. A1v1 = A2v2 (continuity equation,
                                                                         incompressible fluid), to get a relation between the
                                r                                        two speeds in terms of cross-sectional areas of pipes


XtraEdge for IIT-JEE                                            29                                          DECEMBER 2011
     or containers. Or perhaps you will know both speeds                                            F
     and need to determine one of the areas. You may also
                      dV
     need to use Eq.       = Av (volume flow rate) to find
                       dt
                                                                                       F                          F
     the volume flow rate.
     Step 4: Evaluate your answer : As always, verify that
     the results make physical sense. Double-check that
     you have used consistent units. In SI units, pressure is                              F                 V
     in pascals, density in kilograms per cubic meter, and                       (c) Shear strain = φ
     speed in meters per second. Also note that the
     pressures must be either all absolute pressure or all
     gauge pressures.
                                                                                              φ
Properties of matter :
                                                                                                   Shear strain
Key Concepts :
Stress :                                                             Stress-strain graph :
     The restoring force setup inside the body per unit                   From graph, it is obvious that in elastic limit, stress is
     area is known as stress.                                             proportional to strain. This is known as Hooke's law.
     Restoring forces : If the magnitude of applied                         ∴ Stress ∝ Strain
     deforming force at equilibrium = F                                     ∴ Stress = E .strain
                             F                                                       stress
     Then,        Stress =                                                  ∴ E=
                            A                                                        strain
     In SI system, unit of stress is N/m2.                                  where E is proportionality dimensional constant
Difference between pressure and stress :                                    known as coefficient of elasticity.
     (a) Pressure is scalar but stress is tensor quantity.                                                   Plastic
     (b) Pressure always acts normal to the surface, but                                                    region
                                                                              Breaking
         stress may be normal or tangential.                                  strength                               B C
     (c) Pressure is compressive in nature but stress may                        Elastic                     A
                                                                        Stress




         be compressive or tensile.                                              limit
Strain :
                change in dimension
     Strain =
                 original dimension
                                                                                       O                               Strain
                               ∆L                                    Types of coefficient of elasticity :
     (a) Longitudinal strain =
                                L
                                                                                                              logitudinal stress
                        L                                                   (a) Young's modulus = Y =
                                                                                                             longitudinal strain
         F                              F
                                                                                             F     FL
        Longitudinal strain is in the direction of                          ∴        Y=         =
        deforming force but lateral strain is in                                             ∆L   A∆L
                                                                                           A
        perpendicular direction of deforming force.                                           L
        Poisson ratio :
                   lateral strain     ∆d/D
        σ=                          =                                                                        L
                longitudinal strain   ∆L/L
     Here ∆d = change in diameter.
                                   ∆V                                                                        ∆L
        (b) Volumetric strain =
                                    V                                                                    F
                                                                                                        volumetric stress
                                                                            (b) Bulk modulus = B =
                                                                                                        volumetric strain
                                                                                 Compressibility = 1/B

XtraEdge for IIT-JEE                                            30                                                DECEMBER 2011
                                      F    shear stress          Surface tension :
    (c) Modulus of rigidity = η =        =
                                      Aφ   shear strain                     F
                                                                      T=
    (d) For isothermal process, B = P.                                      L
                                                F                     Here L = length of imaginary line drawn at the
                                                                      surface of liquid. and F = force acting on one side of
                                        φ                             line (shown in figure)
                       φ
                                                                      (a) Surface tension does not depend upon surface
               F                                                          area.
    (e) For adiabatic process, B = γP                                 (b) When temperature increases, surface tension
                                                                          decreases.
           Adiabatic bulk modulus
    (f)                           =γ                                  (c) At critical temperature surface tension is zero.
          Isothermal bulk modulus
    (g) Esolid > Eliquid > Egas
                                                                                               F
    (h) Young's modulus Y and modulus of rigidity η
        exist only for solids.
    (i) Bulk modulus B exist for solid, liquid and gas.                                                L
    (j) When temperature increases, coefficient of                                                 F
        elasticity (Y, B, η) decreases.
          1   3   9                                              Rise or fall of a liquid in a capillary tube :
    (k)     +   =
          B   η   Y
                                                                           2T cos θ
    (l) Y = 2(1 + σ)η                                                 h=
                                                                             rρg
    (m) Poisson's ratio σ is unitless and dimensionless.
                                                                      Here    θ = angle of contact.
                                      1                                       r = radius of capillary tube
    Theoretically,           –1 < σ <
                                      2
                                                                              ρ = density of liquid
                                  1                                      For a given liquid and solid at a given place,
    Practically,             0<σ<
                                  2                                      hr = constant
    (n) Thermal stress = Yα∆θ                                    Surface energy :
    (o) Elastic energy stored,                                        Surface energy density is defined as work done
                 1                        1  1                        against surface tension per unit area. It is numerically
           U=       × load × extension = Fx = kx2                     equal to surface tension.
                  2                       2  2
               = stress × strain × volume                                W = work = surface tension × area
          For twisting motion,                                        (a) For a drop of radius R, W = 4πR2T
                   1                                                 (b) For a soap bubble, W = 8πR2T
           U=        × torque × angular twist
                   2                                             Excess pressure :
                  1        1                                                              2T
               =    τ × θ = cθ2                                       (a) For drop, P =
                  2        2                                                              R
          Elastic energy density,                                                                  4T
                                                                      (b) For soap bubble, P =
               1                         1                                                         R
          u=     × stress × strain J/m3 = Y × strain2J/m3
               2                         2                       Viscosity :
       Thermal stress = Yα∆θ and Thermal strain = α∆θ                 (a) Newton's law of viscous force :
Work done in stretching a wire :                                                          dv
                                                                              F = – ηA
                1                                                                         dy
    (a) W =       F∆L
                2                                                             dv
                                                                      where      = velocity gradient
                                      1                                       dy
    (b) Work done per unit volume =     × stress × strain
                                      2                                       A = area of liquid layer
    (c) Breaking weight = breaking stress × area                              η = coefficient of viscosity
                                                                      The unit of coefficient of viscosity in CGS is poise.

XtraEdge for IIT-JEE                                        31                                             DECEMBER 2011
     (b) SI unit of coefficient of viscosity                       Sol. Draw a horizontal line through the mercury-glycerine
                            = poiseuille = 10 poise.                    surface. This is a horizontal line in the same liquid at
                                                                        rest namely, mercury. Therefore, pressure at the
     (c) In the case of liquid, viscosity increases with
                                                                        points A and B must be the same.
         density.
     (d) In the case of gas, viscosity decreases with
         density.                                                                                  h
     (e) In the case of liquid, when temperature increases,




                                                                                                                10 cm
         viscosity decreases.




                                                                                     (1 – h)
     (f) In the case of gas, when temperature increases,
         viscosity increases.
                                                                                               A            B
Poiseuille's equation :
            Pπr 4                                                       Pressure at B
     V=
            8ηL                                                            = p0 + 0.1 × (1.3 × 1000) × g
     where V = the volume of liquid flowing per second                  Pressure at A
     through a capillary tube of length L and radius r                     = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g
                                                                        ∴ p0 + 0.1 × 1300 × g
          η = coefficient of viscosity
                                                                           = p0 + 800gh + 1360g – 13600 × g × h
    and P = pressure difference between ends of the tube                ⇒ 130 = 800h + 1360 – 13600h
Stoke's law :                                                                   1230
                                                                        ⇒ h=           = 0.096 m = 9.6 cm
    The viscous force acting on a spherical body moving                        12800
    with constant velocity v in a viscous liquid is
     F = 6πηrv                                                     2.   A liquid flows out of a broad vessel through a narrow
     where r = radius of spherical body                                 vertical pipe. How are the pressure and the velocity
                                                                        of the liquid in the pipe distributed when the height
Determination of η :                                                    of the liquid level in the vessel is H from the lower
            2r 2 (ρ − σ)g                                               end of the length of the pipe is h ?
     η=                                                            Sol. Let us consider three points 1, 2, 3 in the flow of
                  9v
                                                                        water. The positions of the points are as shown in the
     where r = radius of spherical body moving with                     figure.
     constant velocity v in a viscous liquid of coefficient             Applying Bernoulli's theorem to points 1, 2 and 3
     of viscosity η and density ρ
     and     σ = density of spherical body                                                             •1
Critical velocity (v0) :
            kη                                                                                     x
     v0 =                                                                           H
            ρr                                                                                         •2   h
     where k = Reynold's number for narrow tube, k ≈ 1000.
     (a) For stream line motion, flow velocity v < v0.                                                 •3
                                                                                                   2
     (b) For turbulant motion, flow velocity v > v0.                     p0 1 2           p     1
                                                                            + v 1 + gH =      + v 2 + g (h – x)
                                                                                                   2
                                                                         ρ 2               ρ 2
                                                                                  p    1 2
                                                                               = 0 + v3 + 0
                                                                                   ρ 2
               Solved Examples                                          By continuity equation
                                                                           v 1A1 = A2v 2 = A2v 3
1.   A vertical U-tube of uniform cross-section contains                Since A1 >> A2,v1 is negligible and v2 = v3 = n (say).
     mercury in both arms. A glycerine (relative density                        p0         p     1
     1.3) column of length 10 cm is introduced into one of              ∴           + gH = 2 + v2 + g (h – x)
                                                                                 ρ          ρ 2
     the arms. Oil of density 800 kg m–3 is poured into the
     other arm until the upper surface of the oil and                             p0 1 2
                                                                               =     + v
     glycerine are at the same horizontal level. Find the                          ρ 2
     length of the oil column. Density of mercury is
                                                                        ∴       v = 2gH                                   (i)
     13.6 × 103 kg m–3.


XtraEdge for IIT-JEE                                          32                                                 DECEMBER 2011
              p0          p                                                     = 4.99 × 10–3 cm
     and          + gH = 2 + gH + g (h – x)
               ρ           ρ                                              Also, elastic limit for copper = 1.5 × 109 dynes/cm2
     ⇒       p0 + p2 + ρg (h – x)                                         If d' is the minimum diameter, then maximum stress
     ⇒       p2 = p0 – ρg (h – x)                     (ii)                                  F       4F
                                                                          on the wire =           =
     Thus pressure varies with distance from the upper                                   πd '2 / 4 πd' 2
     end of the pipe according to equation (ii) and velocity
     is a constant and is given by (i).                                               4F
                                                                          Hence,             = 1.5 × 109
                                                                                     πd' 2
3.   Calculate the difference in water levels in two                                         4F               4 × 5.0 × 1000 × 980
     communicating tubes of diameter d = 1 mm and                         or d'2 =                    9
                                                                                                          =
     d = 1.5 mm. Surface tension of water = 0.07 Nm–1                               π × 1.5 × 10    3.142 × 1.5 × 109
                                                                                                 –4
     and angle of contact between glass and water = 0º.                            = 41.58 × 10
                          2T cos θ                                            d' = 0.0645 cm.
Sol. Pressure at A = p0 –
                             r2
                                                                     5.   A uniform horizontal rigid bar of 100 kg in supported
     (Q pressure inside a curved surface is greater than
                                                                          horizontally by three equal vertical wires A, B and C
     that outside)
                                                                          each of initial length one meter and cross-section
                          2T cos θ                                        1 mm2. B is a copper wire passing through the centre
     Pressure at B = p0 –
                             r1                                           of the bar; A and C are steel wires and are arranged
                                      1 1                               symmetrically one on each side of B YCu = 1.5 × 1012
     ∴ pressure difference = 2T cos θ  − 
                                      r r                               dynes / cm2, Ys = 2 × 1012 dynes/cm2. Calculate the
                                       1 2                              tension in each wire and extension.
                                                                     Sol. The situation is shown in figure. Because the rod is
                             B             A                              horizontally supported, hence extensions in all the
                                                                          wires must be equal i.e., strains in all the wires are
                                                                          equal as initial lengths are also equal.
                                                                                                  Stress
                                                                                   As Y =
                                                                                                  Strain

     Let this pressure difference correspond to h units of                                                         C
                                                                                                  A        B
     the liquid.                                                                                  S       Cu           S
                      1 1
     Then 2T cos θ  −  = ρgh
                      r r 
                       1 2                                                                              100 Kg
          2T cos θ  1 1                                                                             FCu / A
     ⇒ h=           −                
            ρg  r1 r2
                   
                                      
                                      
                                                                          Hence,           YCu =
                                                                                                      Strain
                                                                                                                                     … (1)

                2 × 0.07  1                1                                                    Fs / A
     ∴ h=                        −3
                                     −         −3
                                                   = 4.76 mm             and              Ys =                                      … (2)
               1000 × 9.8  1× 10      1.5 × 10                                                  Strain
                                                                                YCu FCu 1.5 3
4.   A mass of 5 kg is suspended from a copper wire of 5                  ∴        =    =  =                      or 4FCu = 3FS         ...(3)
     mm diameter and 2 m in length. What is the                                 YS   FS   2 4
     extension produced in the wire ? What should be the                  According to figure, we can write
     minimum diameter of the wire so that its elastic limit
                                                                          2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g
     is not exceed ? Elastic limit for copper = 1.5 × 109
     dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2.                      or     [(8/3) + 1] FCu = 100 g
Sol. Given that Y = 1.1 × 1012 dynes/cm2,                                 ∴      FCu = (3/11) × 100g
         L = 2m = 200 cm, d = 5 mm = 0.5 cm                                           = (3/11) × 100 Kgwt = 27.28 Kgwt
     or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.                    and    FS = (4/3) FCu = (4/3) × (3/11) × 100g
                FL                                                                   = (400/11)g = 36.36 Kgwt
        Y=                                                                Extension in each wire,
               πr 2l
               FL            5.0 × 1000 × 980 × 200                                     FCu L 27280 × 980 × 100
     or l =            =                                                           l=        =                     = 0.178 cm
              πr 2 Y       3.142 × (0.25) 2 × 1.1× 10 2                                 AYCu   10 − 2 × 1.5 × 1012


XtraEdge for IIT-JEE                                            33                                                         DECEMBER 2011
   KEY CONCEPT

     Organic
    Chemistry                          CARBOXYLIC ACID
  Fundamentals

Acidity of carboxylic acids.                                                        (1.27Å) which is nearly intermediate between C O
     Fatty acids are weak acids as compared to inorganic                            and C—O bond length values. This proves resonance
     acids. The acidic character of fatty acids decreases                           in carboxylate anion.
     with increase in molecular weight. Formic acid is the
                                                                                                      O                            –
     strongest of all fatty acids.                                                                                            O
     The acidic character of carboxylic acids is due to                                    H     C                  H    C          Na+
     resonance in the acidic group which imparts electron                                             OH                       O
     deficiency (positive charge) on the oxygen atom of
                                                                                            Formic acid              Sodium formate
     the hydroxyl group (structure II).
                                                                                    It is important to note that although carboxylic acids
         O                            O–                                            and alcohols both contain –OH group, the latter are
                                                  +
                                       R     C    O    H                            not acidic in nature. It is due to the absence of
   R    C    O     H
                                                                                    resonance (factor responsible for acidic character of
           I                                 II                                     –COOH) in both the alcohols as well as in their
 Non-equivalent structures         (Resonance less important)                       corresponding ions (alkoxide ions).
                                                                                         R—O—H                R—O– + H+
                                                  O–
                                                                                         Alcohol                      Alkoxide ion
                                                 R    C     O+H       +                 (No resonance)               (No resonance)
                                                                                    Relative acidic character of carboxylic acids with
     The positive charge (electron deficiency) on oxygen
                                                                                    common species not having —COOH group.
     atom causes a displacement of electron pair of the
     O—H bond towards the oxygen atom with the result                               RCOOH > Ar—OH > HOH > ROH >
     the hydrogen atom of the O—H group is eliminated                                                               HC CH > NH3 > RH
     as proton and a carboxylate ion is formed.                                Effect of Substituents on acidity.
     Once the carboxylate ion is formed, it is stabilised by                        The carboxylic acids are acidic in nature because of
     means of resonance.                                                            stabilisation (i.e., dispersal of negative charge) of
                      O                  O–                                         carboxylate ion. So any factor which can enhance the
              R     C                R       C                                      dispersal of negative charge of the carboxylate ion
                         O–                      O                                  will increase the acidity of the carboxylic acid and
        Resonating forms of carboxylate ion (Equivalent structures)
                                                                                    vice versa. Thus electron-withdrawing substitutents
                      (Resonance more important)                                    (like halogens, —NO2, —C6H5, etc.) would disperse
                                                                                    the negative charge and hence stabilise the
                                         –
                                   O                                                carboxylate ion and thus increase acidity of the
                        R     C                                                     parent acid. On the other hand, electron-releasing
                                   O                                                substituents would increase the negative charge,
                                                                                    destabilise the carboxylate ion and thus decrease
              Resonance hybrid of carboxylate ion
                                                                                    acidity of the parent acid.
     Due to equivalent resonating structures, resonance in                                                               –
     carboxylate anion is more important than in the                                                                 O
     parent carboxylic acid. Hence carboxylate anion is                                               X        C
     more stabilised than the acid itself and hence the                                                              O
     equilibrium of the ionisation of carboxylic acids to                            The substituent X withdraws electrons, disperses negative
     the right hand side.                                                              charge, stabilises the ion and hence increases acidity
              RCOOH       RCOO– + H+                                                                                      –
                                                                                                                     O
     The existence of resonance in carboxylate ion is                                                  Y        C
     supported by bond lengths. For example, in formic                                                               O
     acid, there is one C=O double bond (1.23 Å) and one
     C—O single bond (1.36Å), while in sodium formate                                The substituent Y releases electrons, intensifies negative
     both of the carbon-oxygen bond lengths are identical                            charge, destabilises the ion and hence decreases acidity


XtraEdge for IIT-JEE                                                      34                                                  DECEMBER 2011
     Now, since alkyl groups are electron-releasing, their                 Comparison of nucleophilic substitution (e.g.,
     presence in the molecule will decrease the acidity. In                hydrolysis) in acid derivatives. Let us first study the
     general, greater the length of the alkyl chain, lower                 mechanism of such reaction.
     shall be the acidity of the acid. Thus, formic acid                               O                                                  O
     (HCOOH), having no alkyl group, is about 10 times
                                                                                                          (i) Addition step
     stronger than acetic acid (CH3COOH) which in turn                         R       C    Z + Nu                                R       C        Nu
     is stronger than propanoic acid (CH3CH2COOH) and
     so on. Similarly, following order is observed in                                                                                     Z
     chloro acids.                                                                                                                    O
                Cl                 Cl                                                                (ii) Elimination step
                                                                                                           R C Nu + Z
           Cl     C        CO2H > Cl          C     CO2H                   (where Z= —Cl, —OCOR, —OR, —NH2 and Nu=
                                                                           A nucleophile)
                 Cl                           H
                                                                             Nucleophilic substitution in acid derivatives
     pKa        0.70                         1.48                              O                 O                  OH
                                  H                        H                                    Nu                           H+
                                                                           R       C       R'         R      C      Nu                R       C     Nu
                          > Cl    C      CO2H > H          C   CO2H
                                                                                                              R'                              R'
                                  H                     H                             (where R' = H or alkyl group)
                    pKa          2.86                  4.76                Nucleophilic addition on aldehydes and ketones
Decreasing order of aliphatic acids                                        The (i) step is similar to that of nucleophilic addition
    (i) O2NCH2COOH > FCH2COOH > ClCH2COOH                                  in aldehydes and ketones and favoured by the
                                     > BrCH2COOH                           presence of electron withdrawing group (which
                                                                           would stabilise the intermediate by developing
    (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH
                                                                           negative charge) and hindered by electron-releasing
                                     > (CH3)3CCOOH                         group. The (ii) step (elimination of the leaving group
    (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH                                 Z) depends upon the ability of Z to accommodate
                                 > ClCH2CHClCH2COOH                        electron pair, i.e., on the basicity of the leaving
                                                                           group. Weaker bases are good leaving groups,
    (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH                                     hence weaker a base, the more easily it is removed.
    Benzoic acid is somewhat stronger than simple                          Among the four leaving groups (Cl–, –OCOR, –OR,
    aliphatic acids. Here the carboxylate group is                         and –NH2) of the four acid derivatives, Cl– being the
    attached to a more electronegative carbon (sp2                         weakest base is eliminated most readily. The relative
    hybridised) than in aliphatic acids (sp3 hybridised).                  order of the basic nature of the four groups is
    HCOOH > C6H5COOH > CH3COOH.                                                     –
                                                                                      NH2 > –OR > –O.COR > Cl–
Nucleophilic substitution at acyl carbon :                                 Hence acid chlorides are most reactive and acid
    It is important to note that nucleophilic substitution                 amides are the least reactive towards nucleophilic
    (e.g., hydrolysis, reaction with NH3, C2H5OH, etc.) in                 acyl substitution. Thus, the relative reactivity of acid
    acid derivatives (acid chlorides, anhydrides, esters                   derivatives (acyl compounds) towards nucleophilic
    and amides) takes place at acyl carbon atom                            substitution reactions is
    (difference from nucleophilic substitution in alkyl                        ROCl > RCO.O.COR > RCOOR > RCONH2
    halides where substitution takes place at alkyl carbon                     Acid             Acid          Esters        Acid
    atom). Nucleophilic substitution in acyl halides is                       chlorides      anhydrides                   amides
    faster than in alkyl halides. This is due to the
                                                                           OH– being stronger base than Cl–, carboxylic acids
    presence of > CO group in acid chlorides which
                                                                           (RCOOH)        undergo      nucleophilic     substitution
    facilitate the release of halogen as halide ion.
                                                                           (esterfication) less readily than acid chlorides.
                      δ–
                      O
                            δ–           δ+    δ–
                R C        Cl            R     Cl
                     δ+
                                      Alkyl chloride
                Acid chloride




XtraEdge for IIT-JEE                                                  35                                                          DECEMBER 2011
   KEY CONCEPT

     Physical
    Chemistry                             CHEMICAL KINETICS
  Fundamentals

The temperature dependence of reaction rates :                            behaviour is a signal that the reaction has a complex
     The rate constants of most reactions increase as the                 mechanism.
     temperature is raised. Many reactions in solution fall               The temperature dependence of some reactions is
     somewhere in the range spanned by the hydrolysis of                  non-Arrhenius, in the sense that a straight line is not
     methyl ethanoate (where the rate constant at 35ºC is                 obtained when ln k is plotted against 1/T. However,
     1.82 times that at 25ºC) and the hydrolysis of sucrose               it is still possible to define an activation energy at any
     (where the factor is 4.13).                                          temperature as
(a) The Arrhenius parameters :                                                                       d ln k 
                                                                                         Ea = RT2                   .......(ii)
     It is found experimentally for many reactions that a                                            dT 
     plot of ln k against 1/T gives a straight line. This                This definition reduces to the earlier one (as the slope
     behaviour is normally expressed mathematically by                   of a straight line) for a temperature-independent
     introducing two parameters, one representing the                    activation energy. However, the definition in eqn.(ii)
     intercept and the other the slope of the straight line,             is more general than eqn.(i), because it allows Ea to
     and writing the Arrhenius equaion.                                  be obtained from the slope (at the temperature of
                             Ea                                          interest) of a plot of ln k against 1/T even if the
               ln k = ln A –                   ......(i)                 Arrhenius plot is not a straight line. Non-Arrhenius
                             RT
                                                                         behaviour is sometimes a sign that quantum
     The parameter A, which corresponds to the intercept                 mechanical tunnelling is playing a significant role in
     of the line at 1/T = 0(at infinite temperature, shown in            the reaction.
     figure), is called the pre-exponential factor or the            (b) The interpretation of the parameters :
     'frequency factor'. The parameter Ea, which is
                                                                         We shall regard the Arrhenius parameters as purely
     obtained from the slope of the line (–Ea/R), is called
                                                                         empirical quantities that enable us to discuss the
     the activation energy. Collectively the two quantities
                                                                         variation of rate constants with temperature;
     are called the Arrhenius parameters.
                                                                         however, it is useful to have an interpretation in mind
                                      ln A                               and write eqn.(i) as
                                                                                  k = Ae − E a / RT             .......(iii)
                               Slope = –Ea/R                              To interpret Ea we consider how the molecular
                                                                          potential energy changes in the course of a chemical
                ln k




                                                                          reaction that begins with a collision between
                                                                          molecules of A and molecules of B(shown in figure).


                                1/T                                                                              Ea
                                                                                  Potential energy




         A plot of ln k against 1/T is a straight line when
         the reaction follows the behaviour described by                                             Reactants
         the Arrhenius equation. The slope gives –Ea/R
         and the intercept at 1/T = 0 gives ln A.
     The fact that Ea is given by the slope of the plot of
     ln k against 1/T means that, the higher the activation
     energy, the stronger the temperature dependence of                                                               Products
     the rate constant (that is, the steeper the slope). A
     high activation energy signifies that the rate constant                                           Progress of reaction
     depends strongly on temperature. If a reaction has
                                                                            A potential energy profile for an exothermic
     zero activation energy, its rate is independent of
                                                                            reaction. The height of the barrier between
     temperature. In some cases the activation energy is
                                                                            the reactants and products is the activation
     negative, which indicates that the rate decreases as
                                                                            energy of the reaction
     the temperature is raised. We shall see that such



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XtraEdge for IIT-JEE                                                                                           38                                                                              DECEMBER 2011
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XtraEdge for IIT-JEE   40   DECEMBER 2011
     As the reaction event proceeds, A and B come into                    ratio of the two rates, and therefore of the two rate
     contact, distort, and begin to exchange or discard                   constants :
     atoms. The reaction coordinate is the collection of                           [P2 ]  k
     motions, such as changes in interatomic distances and                               = 2
     bond angles, that are directly involved in the                                [P1 ]   k1
     formation of products from reactants. (The reaction                 This ratio represents the kinetic control over the
     coordinate is essentially a geometrical concept and                 proportions of products, and is a common feature of
     quite distinct from the extent of reaction.) The                    the reactions encountered in organic chemistry where
     potential energy rises to a maximum and the cluster                 reactants are chosen that facilitate pathways
     of atoms that corresponds to the region close to the                favouring the formation of a desired product. If a
     maximum is called the activated complex. After the                  reaction is allowed to reach equilibrium, then the
     maximum, the potential energy falls as the atoms                    proportion of products is determined by
     rearrange in the cluster and reaches a value                        thermodynamic rather than kinetic considerations,
     characteristic of the products. The climax of the                   and the ratio of concentration is controlled by
     reaction is at the peak of the potential energy, which              considerations of the standard Gibbs energies of all
     corresponds to the activation energy Ea. Here two                   the reactants and products.
     reactant molecules have come to such a degree of                The kinetic isotope effect
     closeness and distortion that a small further
     distortion will send them in the direction of products.             The postulation of a plausible mechanism requires
     This crucial configuration is called the transition                 careful analysis of many experiments designed to
     state of the reaction. Although some molecules                      determine the fate of atoms during the formation of
     entering the transition state might revert to reactants,            products. Observation of the kinetic isotope effect, a
     if they pass through this configuration then it is                  decrease in the rate of a chemical reaction upon
     inevitable that products will emerge from the                       replacement of one atom in a reactant by a heavier
     encounter.                                                          isotope, facilitates the identification of bond-breaking
                                                                         events in the rate-determining step. A primary
     We also conclude from the preceding discussion that,                kinetic isotope effect is observed when the rate-
     for a reaction involving the collision of two                       determining step requires the scission of a bond
     molecules, the activation energy is the minimum                     involving the isotope. A secondary isotope effect is
     kinetic energy that reactants must have in order                    the reduction in reaction rate even though the bond
     to form products. For example, in a gas-phase                       involving the isotope is not broken to form product.
     reaction there are numerous collisions each second,                 In both cases, the effect arises from the change in
     but only a tiny proportion are sufficiently energetic to            activation energy that accompanies the replacement
     lead to reaction. The fraction of collisions with a                 of an atom by a heavier isotope on account of
     kinetic energy in excess of an energy Ea is given by                changes in the zero-point vibrational energies.
     the Boltzmann distribution as e − E a / RT . Hence, we              First, we consider the origin of the primary kinetic
     can interpret the exponential factor in eqn(iii) as the             isotope effect in a reaction in which the rate-
     fraction of collision that have enough kinetic energy               determining step is the scission of a C–H bond. The
     to lead to reaction.                                                reaction coordinate corresponds to the stretching of
     The pre-exponential factor is a measure of the rate at              the C–H bond and the potential energy profile is
     which collisions occur irrespective of their energy.                shown in figure. On deuteration, the dominant
     Hence, the product of A and the exponential factor,                 change is the reduction of the zero-point energy of
     e − E a / RT , gives the rate of successful collisions.             the bond (because the deuterium atom is heavier).
                                                                         The whole reaction profile is not lowered, however,
Kinetic and thermodynamic control of reactions :                         because the relevant vibration in the activated
    In some cases reactants can give rise to a variety of                complex has a very low force constant, so there is
    products, as in nitrations of mono-substituted                       little zero-point energy associated with the reaction
    benzene, when various proportions of the ortho-,                     coordinate in either isotopomeric form of the
    meta-, and para- substituted products are obtained,                  activated complex.
    depending on the directing power of the original
    substituent. Suppose two products, P1 and P2, are
    produced by the following competing reactions :                                                             Ea(C–H)
                                                                                     Potential energy




                                                                                                        C–H
     A + B → P1 Rate of formation of P1 = k1[A][B]                                                     C–D
                                                                                                              Ea(C–D)
     A + B → P2 Rate of formation of P2 = k2[A][B]
     The relative proportion in which the two products
     have been produced at a given state of the reaction
     (before it has reached equilibrium) is given by the                                                 Reaction coordinate


XtraEdge for IIT-JEE                                            41                                                             DECEMBER 2011
                                  UNDERSTANDING
                                                                                  Inorganic Chemistry
1.   An inorganic halide (A) gives the following                                              Bi + NaSnO3 + H2O + 3NaCl
     reactions.                                                                         (G) Black ppt.
     (i) The cation of (A) on raction with H2S in HCl                        2Bi + 6HCl ∆   → 2BiCl3 + 3H2
          medium, gives a black ppt. of (B). (A) neither
          gives ppt. with HCl nor blue colour with                           (G)                    (A)
          K4Fe(CN)6.                                                     Hence,
     (ii) (B) on heating with dil.HCl gives back                         (A) is BiCl3,
          compound(A) and a gas (C) which gives a black                  (B) is Bi2S3,
          ppt. with lead acetate solution.                               (C) is H2S,
     (iii) The anion of (A) gives chromyl chloride test.                 (D) is Bi(NO3)2,
     (iv) (B) dissolves in hot dil. HNO3 to give a solution,             (E) is Bi(OH)3, (F) is BiOCl and (G) is Bi
          (D). (D) gives ring test.
     (v) When NH4OH solution is added to (D), a white               2.   A colourless solid (A) on heating gives a white solid
          precipitate (E) is formed. (E) dissolves in                    (B) and a colourless gas (C). (B) gives off reddish-
          minimum amount of dil. HCl to give a solution of               brown fumes on treating with H2SO4. On treating
          (A). Aqueous solution of (A) on addition of water              with NH4Cl, (B) gives a colourless gas (D) and a
          gives a whitish turbidity (F).                                 residue (E). The compound (A) on heating with
     (vi) Aqueous solution of (A) on warming with                        (NH4)2SO4 gives a colourless gas (F) and white
          alkaline sodium stannite gives a black precipitate             residue (G). Both (E) and (G) impart bright yellow
          of a metal (G) and sodium stannate. The metal                  colour to Bunsen flame. The gas (C) forms white
          (G) dissolves in hydrochloride acid to give                    powder with strongly heated Mg metal which on
          solution of (A).                                               hydrolysis produces Mg(OH)2. The gas (D) on
     Identify (A) to (G) and give balanced chemical                      heating with Ca gives a compound which on
     equations of reactions.                                             hydrolysis produces NH3. Identify compounds (A) to
Sol. Observation of (i) indicates that cation (A) is Bi3+                (G) giving chemical equations involved.
     because it does not give ppt. with HCl nor blue                Sol. The given information is as follows :
     colour with K4Fe(CN)6, hence it is neither Pb2+ nor                 (i) A       Heat → B
                                                                                                + C
     Cu2+. Since anion of (A) gives chromyl chloride test,                 Colourless       Solid Colourless
     hence it contains Cl– ions. Thus, (A) is BiCl3. Its                    Solid                   gas
     different reactions are given below :
     (i) 2BiCl3 + 3H2S → Bi2S3 + 6HCl                                   (ii) B + H2SO4 ∆ Reddish brown gas
                                                                                         →
            (A)                   (B)                                    (iii) B + NH4Cl ∆
                                                                                          →     D     + E
     (ii) Bi2S3 + 6HCl → 3H2S + 2 BiCl3                                                    Colourless gas
           (B)                   (C)     (A)                            (iv) A + (NH4)2SO4 ∆→ F            + G
                              ∆
     (iii) Bi2S3 + 8HNO3  → 2Bi(NO3)3 + 2NO                                                 olourless gas      White
           (B)                   (D)     + 3S + 4H2O                                                            Residue
     (iv) Bi(N O3)3 + 3NH4OH →                                     (v) E and G imparts yellow colour to the flame.
              (D)              Bi(OH)3 ↓ + 3NH4NO3                  (vi) C + Mg Heat → White powder
                                                                                 
                             (E) White ppt.
                              ∆
                                                                                                O
                                                                                               H→ Mg(OH)2
                                                                                                 2

         Bi(OH)3 + 3HCl      → BiCl3 + 3H2O                                                             H O
                                                                    (vii) D + Ca Heat → Compound  → NH3
                                                                                                           
                                                                                                             2
                       Dil.          (A)
                                                                          Information of (v) indicates that (E) and (G) and also
         BiCl3 + H2O → BiOCl + 2HCl
                                                                          (A) are the salts of sodium because Na+ ions give
          (A)                  (F)
                                                                          yellow coloured flame. Observations of (ii) indicate
                            Bismuth oxychloride                           that the anion associated with Na+ in (A) may be
                            (White turbidity)                             NO3–. Thus, the compound (A) is NaNO3.
         (v) BiCl3 + 2NaOH +Na2[SnO2] →                                  The reactions involved are as follows :
             (A)

XtraEdge for IIT-JEE                                           42                                          DECEMBER 2011
     (i) 2NaNO3 ∆ → 2NaNO2 + O2 ↑                                        H2S + 2HNO3 → 2NO2 + 2H2O + S (D)
            (A)          (B) (C)                                                     (C)                         White turbidity
     (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2                                 CuSO4 + H2S → CuS ↓ + H2SO4
             (B)    Dil.                                                     (E)       (B)         (F)
                 3HNO2 → HNO3 + H2O + 2NO↑                                                      Black ppt.
                 2NO + O2 →     2NO2 ↑                                    3CuS + 8HNO3 →
                               Reddish brown                                           Dil. 3Cu(NO3)2 + 2NO + 3S + 4H2O
                                  Fumes                                    Cu++ + 4NH3 → [Cu(NH3)4]2+
     (iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O                                                   (G) Blue colour
           (B)               (E)    (D)                                    [Cu(NH3)4]2+ + 4CH3COOH →
                                                                                                 Cu2+ + 4CH3COONH4
     (iv) 2NaNO3 + (NH4)2SO4 ∆
                              → Na2SO4 + 2NH3
                                                                           Cu2+ + [Fe(CN)6]4– → Cu2[Fe(CN)6]
          (A)                     (G)    (F)                                                              (H)
                                         2HNO3                                                       Chocolate colour
     (v) O2 + 2Mg ∆
                                       H O
                            
                   → 2MgO  → Mg(OH)2
                             2
                                                                           CuSO4(aq) + BaCl2(aq) → BaSO4 ↓ + CuCl2
         (C)                                                                  (E)                         White ppt.
     (vi) N2 + 3Ca ∆→ Ca3N2                                                                           Insuluble in HNO3
          (D)                                                              Hence,
         Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑                                 (A) is FeS, (B) is H2S, (C) is HNO3, (D) is S,
     Hence,                                                                (E) is CuSO4, (F) is CuS, (G) is [Cu(NH3)4]SO4 and
                                                                           (H) is Cu2[Fe(CN)6]
     (A) is NaNO3,
     (B) is NaNO2,
                                                                      4.   (i) An inorganic compound (A) is formed on passing
     (C) is O2,
                                                                                a gas (B) through a concentrated liquor containing
     (D) is N2,                                                                 Na2S and Na2SO3.
     (E) is NaCl,
                                                                           (ii) On adding (A) into a dilute solution of AgNO3, a
     (F) is NH3 and (G) is Na2SO4.                                              white ppt. appears which quickly changes into
3.   A black coloured compound (A) on reaction with dil.                        black coloured compound (C).
     H2SO4 gives a gas (B) which on passing in a solution                  (iii) On adding two or three drops of FeCl3 into
     of an acid (C) gives a white turbidity (D). Gas (B)                        excess of solution of (A), a violet coloured
     when passed through an acidified solution of a                             compound (D) is formed. This colour disappears
     compound (E), gives ppt.(F) which is soluble in                            quickly.
     dilute nitric acid. After boiling this solution an excess             (iv) On adding a solution of (A) into the solution of
     of NH4OH is added, a blue coloured compound (G) is                         CuCl2, a white ppt. is first formed which dissolves
     produced. To this solution, on addition of CH3COOH                         on adding excess of (A) forming a compound (E).
     and aqueous K4[Fe(CN)6], a chocolate ppt. (H) is                      Identify (A) to (E) and give chemical equations for
     produced. On addition of an aqueous solution of                       the reactions at steps (i) to (iv)
     BaCl2 to aqueous solution of (E), a white ppt.                   Sol. (i) The compound (A) appears to be Na2S2O3 from its
     insoluble in HNO3 is obtained. Identify compounds                     method of preparation given in the problem.
     (A) to (H).
                                                                                Na2S + Na2SO3 + I2 → 2NaI + Na2S2O3
Sol. From the data on compounds (G) and (H), it may be
                                                                                                    (B)               (A)
     inferred that the compound (E) contains cupric ions
     (Cu2+), i.e., (E) is a salt of copper. Since the addition             or Na2SO3 + 3Na2S + 3SO2 → 3Na2S2O3
     of BaCl2 to (E) gives a white ppt. insoluble in HNO3,                                           (B)              (A)
     it may be said that the anion in the salt is sulphate ion             (ii) White ppt. of Ag2S2O3 is formed which is
     (SO42–). Hence, (E) is CuSO4.                                         hydrolysed to black Ag2S
     The gas (B) which is obtained by adding dil. H2SO4                    Na2S2O3 + 2AgNO3 → 2NaNO3 + Ag2S2O3 ↓
     to a black coloured compound (A), may be H2S since                                                               White ppt
     it can cause precipitation of Cu2+ ions in acidic                     Ag2S2O3 + H2O → Ag2S + H2SO4
     medium. The black coloured compound (A) may be                                                  (C)
     ferrous sulphide (iron pyrite).                                       (iii) A violet ferric salt is formed.
     Hence, the given observation may be explained from                    3Na2S2O3 + 2FeCl3 → Fe2(S2O3)3 + 6NaCl
     the following equations.
                                                                                                         (D)(violet)
     Fe S + H2 SO4 → FeSO4 + H2S
                                                                           (iv) 2CuCl2 + 2Na2S2O3 → 2CuCl + Na2S4O6 + 2NaCl
        (A) Dil.                            (B)
                                                                                                              White ppt.

XtraEdge for IIT-JEE                                             43                                          DECEMBER 2011
     2CuCl + Na2S2O3 → Cu2S2O3 + 2NaCl
     3Cu2S2O3 + 2Na2S2O3 → Na4[Cu6(S2O3)5]
                                 (E)                                                 SCIENCE TIPS
     or 6CuCl + 5Na2S2O3 → Na4[Cu6(S2O3)5] + 6NaCl
                              (E)
                                                                     • What is the expression for growing current, in LR
     Hence,
     (A) is Na2S2O3,                                                                                                − t
                                                                                                                      R
                                                                       circuit ?                       I = I0  1 − e L 
     (B) is I2 or SO2,                                                                                                 
                                                                                                                       
     (C) is Ag2S,
     (D) is Fe2(S2O3)3 and                                           • What is the range of infrared spectrum ?
     (E) is Na4[Cu6(S2O3)5].                                                                     This covers wavelengths
                                                                                       from 10–3 m down to 7.8 × 10–7 m
5.   A white amorphous powder (A) on heating yields a                • What is the nature of graph between electric field
     colourless, non-combustible gas (B) and solid (C).                and potential energy (U) ?
     The latter compound assumes a yellow colour on                                             The nature of the graph
     heating and changes to white on cooling. 'C' dissolves                                      will be parabola having
     in dilute acid and the resulting solution gives a white                                     symmetry about U-axis
     precipitate on adding K4Fe(CN)6 solution.
     'A' dissolves in dilute HCl with the evolution of gas,          • Why no beats can be heard if the frequencies of
     which is identical in all respects with 'B'. The gas 'B'          the two interfering waves differ by more than ten ?
     turns lime water milky, but the milkiness disappears                                      this is due to persistence
     with the continuous passage of gas. The solution of                                                       of hearing
     'A', as obtained above, gives a white precipitate (D)           • Why heating systems based on steam are more
     on the addition of excess of NH4OH and and passing                efficient than those based on circulation of hot
     H2S. Another portion of the solution gives initially a            water ?                    This is because steam
     white precipitate (E) on the addition of sodium                                          has more heat than water
     hydroxide solution, which dissolves on futher                                              a the same temperature
     addition of the base. Identify the compounds A, B, D
     and E.                                       [IIT-1979]         • Can the specific heat of a gas be infinity ?    Yes
                      ∆
Sol. (i)   ZnCO3   → ZnO + CO2                                     • What is the liquid ascent formula for a capillary ?
              (A)        (B)                                                                               2T cos θ     r
                                                                                                       h=            –
     (ii) ZnO + 2HCl → H2O + ZnCl2                                                                           γpg       3
            (C)                  (soluble)                                                where h is the height through
     (iii) 2ZnCl2 + K4[Fe(CN)6]                                                          which a liquid of density ρ and
                           → 4KCl + Zn2[Fe(CN)6]↓                                           surface tension T rises in a
                                           (white ppt)                                         capillary tube of radius r
     (iv) ZnCO3 + HCl → CO2 + ZnCl2
                                                                     • What is the expression for total time of flight (T)
             (A)                   (soluble)
                                                                                                                 2u sin θ
     (v) CO2 + Ca(OH)2 → CaCO3 + H2O                                  for oblique projection ?             T=
                                                                                                                    g
            (B)                (Milky)
     (vi) CaCO3 + CO2 + H2O → Ca(HCO3)2                             • The space charge limited current iP in the diode
                                                                       value is given by                      iP = k Vp3/2
                                     (soluble)
     (vii) ZnCl2 + H2S   → 2HCl + ZnS↓
                           4 NH OH
                                                                    • What is an ideal gas ?        An ideal gas is one in
                                       (white)                                                       which intermolecular
                                                                                                         forces are absent
     (viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓
                                       (white)                       • Can a rough sea be calmed by pouring oil on its
     (ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2                              surface ?                                 Yes
                                sod. ziniate                         • What is the expression for fringe width (β) in
                                 (soluble)                             Young's double slit experiment?     β=Dλ/d where
                                                                                           D is the distance between the
                                                                                              source and screen and d is
                                                                                               distance between two slits



XtraEdge for IIT-JEE                                            44                                        DECEMBER 2011
                                                                                                                              Set

     `tà{xÅtà|vtÄ V{tÄÄxÇzxá                                                                                                  8
 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
 of possible twists and turns of problems in mathematics that would be very helpful in facing
 IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and
 we hope that this section would prove a rich resource for practicing challenging problems and
 enhancing the preparation level of IIT JEE aspirants.
                                                                                        By : Shailendra Maheshwari
       Solutions will be p ublished in next issue                       Joint Director Academics, Career Point, Kota

1.   Show that the six planes through the middle point of                                      1
     each edge of a tetrahedron perpendicular to the
     opposite edge meet in a point.
                                                                    8.                         ∫
                                                                             If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that
                                                                                               −1
                                                                            m2In = 2n(2n – 1) In–1 – 4n(n – 1) In–2.
2.   Prove that if the graph of the function y = f (x),
     defined throughout the number scale, is symmetrical            9.      Find the sum to infinite terms of the series
     about two lines x = a and x = b, (a < b), then this                     3     5     7        9       11
     function is a periodic one.                                               +      +       +        +       + ........ ∞
                                                                             4    36    144      400     900
3.   Show that an equilateral triangle is a triangle of             10. ABC is a triangle inscribed in a circle. Two of its
     maximum area for a given perimeter and a triangle of               sides are parallel to two given straight lines. Show
     minimum perimeter for a given area.                                that the locus of foot of the perpendicular from the
                                                                        centre of the circle on to the third side is also a circle,
4.   Let az2 + bz + c be a polynomial with complex
                                                                        concentric to the given circle.
     coefficients such that a and b are non zero. Prove
     that the zeros of this polynomial lie in the region
     |z|≤
            b
               +
                  c                                                                   MEMORABLE POINTS
            a     b
                                                                         • The vector relation between linear velocity and
5.   An isosceles triangle with its base parallel to the                                                                 →    →   →
                                        x2         y2                      angular velocity is                           v =ω× r
     major axis of the ellipse                +       = 1 is
                                        a2        b2                     • In the case of uniform circular motion the angle between
     circumscribed with all the three sides touching the                    →     →                                           →
                                                                            ω and r is always                   90º(hence | v | = ωr
     ellipse. Find the least possible area of the triangle.
                                                                         • The relation between Faraday constant F, Avogadro
6.   If one of the straight lines given by the equation                    number N and the electronic charge e is    F = Ne
     ax2 + 2hxy + by2 = 0 coincides with one of those
     given by a′x2 + 2h′xy + b′y2 = 0 and the other lines                • Depolariser used in Lechlanche cell is
     represented by them be perpendicular, show that                                                            Manganese dioxide
      ha´b´    h´ab´
             =                                                           • The absorption or evolution of heat at a junction of
      b´− a´   b−a
                                                                           two dissimilar metals when a current is passed is
7.   Prove that                                                            known as                                Peltier effect
       n  m         n   m + 1  n  m + 2                      • The part of the human ear where sound is transduced
         +  
      0 n          1   n  +  2   n  + .........
                                               
                                                               is the                                      Cochlea
     .... to (n + 1) terms                                               • Similar trait resulting from similar selection pressure
            n  m       n  m      n  m                          acting on similar gene pool is termed
     =     +     2 +     22 + ........
            0  0      1  1       2  2 
                                                                                                      Parallel evolution
     ..... to (n + 1) terms                                              • Group of related species with the potential, directly
                                                                           or indirectly, of forming fertile hybrids with one
                                                                           another is called                     Coenospecies

XtraEdge for IIT-JEE                                           45                                                DECEMBER 2011
     MATHEMATICAL CHALLENGES
                                SOLUTION FOR NOVEMBER ISSUE (SET # 7)


                                                             3.   The line PQ always passes through (α, β) so it is
1.   Let the line be y = 2x + c
                                                                  y –β = m(x – α)
               9 − c 9 + 2c                                     Let the circle be x2 + y2 – 2hx – 2ky = 0
     Point A         ,      
               6         3                                      Joint equation of OP and OQ.
               2c − 3 + c − 6                                                          ( y − mx)
     Point B           ,                                        x2 + y2 – 2 (hx + ky)            =0
               −3        −3                                                             β − mα
               c + 6 5c + 12                                                                  P
     Point C        ,        
               3         3 
                             1  2c − 3 c + 6                                          O
     mid point of B & C is .            +         ,
                             2  −3          3  
                                                                                                       (h,k)
     1  − c + 6 5c + 12        9 − c 2c + 9 
        +3 +
      2                3  =  6 , 3                                                            Q
                                              
     which is point A, so AB and AC are equal.                           2k  2         h − mk             2hn  2
                                                                   β − mα  y – 2  β − mα  xy + 1 + β − mα  x = 0
                                                                  1 −                                            
         A                                                                                                       
                                                                                      2     2
                                                                  It must represent y – x = 0
2.               b                                                     h − mk
     a                                                            so            = 0 ⇒ m = h/k                ...(1)
                                                                       β − mα
                            C                                                     2k               2hm
     B   D                                                        and      1–             = –1 –
                                                                                β − mα           β − mα
                1 AB         1 AC                                 ⇒ β – mα – 2k = –β + mα – 2hm
     a +b =        .     +     .
               AB AB        AC AC                                 ⇒ –β + mα + k – hm = 0
             1            1                                       ⇒ –β + k + h/k(α – h) = 0         (using (1) in it)
         =        AB +        AC
           AB2           AC 2                                     ⇒ k2 – βy + αh – h2 = 0 so required locus is
             1                   1                                    x2 – y2 – αx + βy = 0
         =       (AD + DB) +        (AD + DC)
           AB 2                AC 2                                                                π π
                                                             4.   As |f (x)| ≤ |tan x| for ∀ x ∈  − , 
            1        1            DB       DC                                                    2 2
         =       +       AD +          +
            AB 2      2
                     AC           BD.DC   CD.CB                  so f (0) = 0
                                                                  so |f (x) – f (0)| ≤ |tan x|
            1       1         DB DC  1                        divides both sides by |x|
         =     2
                  +   2
                         AD +    +   
                                BD CD  BC
            AB     AC                                                f ( x ) − f ( 0)   tan x
                                                                  ⇒                       ≤
                 1       1                                                     x            x
         = AD .       +      
                 BD.DC CD.CB                                                 f ( x ) − f ( 0)        tan x
                                                                  ⇒ lim                         ≤ lim
             AD  1   1                                              x →0             x          x →0   x
         =     .   +   
             CD  BD CD                                          ⇒ |f (0)| ≤ 1
                                                                           1    1           1
             AD DC + BD                                           ⇒ a1 + a 2 + a 3 + ..... + a n ≤ 1
         =     .                                                           2    3           n
             BC BD.CD
                                                                        n
                                                                              ai
         =
            AD
           BD.CD
                 =
                   AD
                      2
                        =
                          AD 1
                            .
                          AD AD
                                                                  ⇒   ∑i           ≤1
                   AD                                                  i =1
                                                  1
     so it is vector along AB with magnitude        .        5.   Let the number is xyz, here x < y and z < y.
                                                 AD               Let y = n, then x can be filled in (n – 1) ways.
                  1                                               (i.e. from 1 to (n – 1)) and z can be filled in n ways
     |a +b | =
                 AD                                               (i.e. from 0 to (n – 1))

XtraEdge for IIT-JEE                                    46                                                 DECEMBER 2011
     here 2 ≤ n ≤ 9                                                                                          3 (tan A . tan B . tan C)1/3
     so total no. of 3 digit numbers with largest middle                             and for a triangle tan A + tan B + tan C
     digit                                                                               = tan A . tan B . tan C
           9                        9           9                                    so (tan A . tan B . tan C)2/3 ≥ 3
     =    ∑
          n=2
                 n(n − 1) =      ∑n=2
                                        n2 –   ∑n
                                               n=2
                                                                                     ⇒ tan A . tan B . tan C ≥ 3 3
                                                                                     ⇒ tan2A + tan2B + tan2C
        9.10.19    9.10
     =          –                                                                            ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3
           6         2
                                                                                                  a2     b2     c2
     = 285 – 45 = 240                                                                so from (1), 2 + 2 + 2 ≥ 144.
                                               240                                                r1      r2    r3
     required probability =
                                           9 × 10 × 10
                                                                                8.   Z1, Z2, Z3 are centroids of equilateral triangles ACX,
                                            8                                        ABY and BCZ respectively.
                                         =
                                           30                                                              Z − Z A iπ/6
                                            4                                        Z1 – ZA = (ZC – ZA) 1          e
                                         =                                                                 ZC − ZA
                                           15
                                                                                                               ZA                  x
6.   The region bounded by the curve y = log2(2 – x) and the
     inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is required area is                                                A         Z1
                                                                                                  y
                                                                                                      Z2
          (–1,log23)
                                                                                                           B                  C
                                                                                                      ZB                           ZC
                                                                                                                    Z3

                 (–1,0)                                   (1,0)                                                      z
                                                              (3/2,–1)
                                (0,–1)                                                                         1  3 i
                                                                                                                 
                                                                                     Z1 – ZA = (ZC – ZA)             +                 ...(1)
          1                                0                                                                    3 2
                                                                                                                     2
                                                                                                                       
                                                                  1
          ∫ log                           ∫ (2 − 2
                                                     y
     =             2 (2   − x) dx +                      ) dy +     π                similarly,
                                                                  4
          −1                              −1
                                                                                                             1  3 i
                                                                                                                 
                  2                       1    π                                     Z2 – ZA = (ZB – ZA)               −        ...(2)
     = log 2 3 −      + 2 log 2 3 + 2 –      +                                                                3 2
                                                                                                                         2
                                                                                                                           
                 ln 2                   2ln2   4
                                                                                                     1                  i
                      e2 e      π                                                    So, Z1 – Z2 = (ZC – ZB) +              (ZC + ZB – 2ZA)
     = – log2               +2+   sq units                                                           2                2 3
                       27       4                                                                                                ...(3)
                          A                                                                                1
                                                                                     similarly Z2 – Z3 = (ZA – ZC)
                                                                                                           2
                   F                E                                                                              i
7.                                                                                                           +        (ZA + ZC – 2ZB) ..(4)
                                M                                                                                2 3
                                                                                     To prove ∆xyz as equilateral triangle, we prove that
          B                                C                                         (Z3 – Z2)eiπ/3 = Z1 – Z2
                            D
                                                                                                            1
                                                                                     So, (Z3 – Z2)eiπ/3 = ( (ZC – ZA)
                                                                                                            2
     ∠BMC = 2∠BAC = 2∠BMD                                                                                                  1
                                                                                                   i                               3 
                BD    BC   BC     a                                                          –         (ZA + ZC – 2ZB))  +          i
     so tan A =    =     =     =                                                                2 3                        2 2 
                MD   2MD   4r1   4r1                                                                                                 
                                                                                           1                   i
          a2                                                                             = (ZC – ZB) +              (ZC + ZB – 2ZA)
     so           = tan2A                                                                  2                 2 3
           r12
                                                                                         = Z1 – Z2
          a2           b2        c2
     so           +         +
       r12    r22    r32
                  2
       = 16 (tan A + tan2B + tan2C)                               ...(1)
     Now as tan A + tan B + tan C ≥

XtraEdge for IIT-JEE                                                       47                                                     DECEMBER 2011
              a
                         1 − r cos u
9.   Tr = 2   ∫ 1 − 2r cos u + r                         2
                                                              du.                     ...(1)                 FRACTIONAL DISTILLATION
              a
               0
                                                                                                                     OF AIR
                   1 − 2r cos u + r 2 − r 2 + 1
         =    ∫
              0
                          1 − 2r cos u + r 2
                                                                           du

              a
                       1− r2                                       
         =    ∫
             1 +
              1 − 2r cos u + r 2
            0
                                                                     du
                                                                    
                                                                    
         = a + (1 – r2) ×
                    a
                                                              sec 2 u / 2
                   ∫ (1 + r
                    0
                                  2
                                      )(1 + tan 2 u / 2) − 2r (1 − tan 2 u / 2)
                                          a
                                                                    sec 2 u / 2
         = a + (1 – r2)               ∫ (1 + r )
                                          0
                                                              2
                                                                  tan 2 u / 2 + (1 − r ) 2
                                              a
                        1− r2                                sec 2 u / 2 du
         =a+
                        (1 + r ) 2            ∫                          (1 − r ) 2
                                                                                                            Did you know that the air we breathe isn’t just
                                              0   tan 2 u / 2 +                                             oxygen, infact it’s made up of a number of different
                                                                         (1 + r ) 2                         gases such as nitrogen, oxygen, carbon dioxide,
     Let tan u/2 = t
                                                                                                            argon, neon and many others. Each of these
                                                      tan a / 2
                              1− r2                                    2 dt                                 gases carry useful properties so separating them
     so, Tr = a +
                              (1 + r )        2          ∫               1− r
                                                                                  2
                                                                                                            from the air around us is extremely beneficial.
                                                         0        t2 +
                                                                         1+ r                               The process is called fractional distillation and
                                                                                           tan a / 2
                        2(1 − r ) 1 + r  −1  1 + r
                                      2
                                                                                                          consists of two steps, the first relies on cooling the
         =a+                              tan  t
                                                1− r                                   
                                                                                                           air to a very low temperature (i.e. converting it into
                        (1 + r ) 2 1 − r 
                                                                                       0
                                                                                          
                                                                                                            a liquid), the second involves heating it up thus
                                                       2(1 + r )(r − 1) π
     Now            lim Tr = a –                                          = a–π                             allowing each gas within the mixture to evaporate at
                   r →1   +                            (1 + r )(r − 1) 2
                                                                                                            its own boiling point. The key to success here is that
                                                       2(1 − r )(r + 1) π                                   every element within air has its own unique boiling
     and            lim Tr = a +                                          = a+π
                   r →1+                               (1 + r )(r − 1) 2                                    temperature. As long as we know these boiling
                                                  a
                                                                                                            temperatures we know when to collect each gas.
     and (from (1)) T1 =                          ∫ du = a
                                                  0                                                         So what are the real world benefits of separating
     Hence          lim Tr, T1,                          lim Tr form an A.P. with                           and extracting these gases? Well liquid oxygen is
                    r →1+                               r →1−
                                                                                                            used to power rockets, oxygen gas is used in
     common difference π.
                                                                                                            breathing apparatus, nitrogen is used to make
10. Let α, β, γ be the three real roots of the equation                                                     fertilizers, the nitric acid component of nitrogen is
    without loss of generality, it can be assumed that
                                                                                                            used in explosives.
    α ≤ β ≤ γ. so
    x2 + ax2 + bx + c = (x – γ) (x2 + (a + γ) x + (γ2 + aγ + b))                                            The other gases all have their own uses too, for
    where – γ (γ2 + aγ + b) = c, as γ is the root of given                                                  example argon is used to fill up the empty space in
    equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must                                                    most light bulbs (thanks to its unreactive nature).
    have two roots i.e. α and β. So its discriminant is non                                                 Carbon dioxide is used in fire extinguishers and is
    negative, thus
                                                                                                            great for putting out fires in burning liquids and
    (γ + a)2 – 4(γ2 + aγ + b) ≥ 0 ⇒ 3γ2 + 2aγ – a2 + 4b ≤ 0
                                                                                                            electrical fires. There really are too many uses to list
            − a + 2 a 2 − 3b
     so γ ≤                                                                                                 but suffice it to say that fractional distillation is an
                   3
                                                                                                            extremely useful process for humans the world
     so greatest root is also less than or equal to
                                                                                                            over.
      − a + 2 a 2 − 3b
                       .
             3


XtraEdge for IIT-JEE                                                                                   48                                         DECEMBER 2011
                                        Students' Forum
                                           Expert’s Solution for Question asked by IIT-JEE Aspirants
      MATHS
1.   A man parks his car among n cars standing in a row,                 The line of intersection OC of the planes OAC and
     his car not being parked at an end. On his return he                OBC, i.e., y + z = 0 & x + y = 0, is given by x = – y = z,
     finds that exactly m of the n cars are still there. What                  x–0      y–0      z–0
     is the probability that both the cars parked on two                 i.e.,        =       =        .                     ...(ii)
                                                                                1        –1        1
     sides of his car, have left?
                                                                         The line of intersection OB of the planes OBC and
Sol. Clearly, his car is at one of the crosses (×).
                                                                         OAB, i.e., x + y = 0 and z + x = 0, is given by
                       |× × × . . . × × ×|                               – x = y = z,
     The number of the ways in which the remaining
                                                                               x–0      y–0      z–0
     m–1 cars can take their places (excluding the car of                i.e.,        =       =        .                     ...(iii)
     the man) = n–1Cm–1 {Q there are n – 1 places for the                       –1        1        1
     m – 1 cars}. The number of ways in which the                        The line BC is the intersection of the planes
     remaining m – 1 cars can take places keeping the two                                              O (0,0,0)
     places on two sides of his car vacant = n–3Cm–1                                      1,1,–1 P
     ∴ the required probability                                                      A
                        n –3
           n( E )              C m –1                                                                         C
         =        =     n –1
           n( S )              C m –1
                                                                                                       (0,0,a)
                   (n – 3) !         (m – 1) ! (n – m) !
         =                         ×
             (m – 1)!(n – m – 2) !       (n – 1) !                                                Q
                                                                                                 1,–1,0
             (n – m)(n – m – 1)                                                            B
         =                      .
               (n – 1)(n – 2)                                               x + y = 0 and x + y + z = a
                                                                         ⇒ y = – x, z = a
2.   Show that the shortest distance between any two                         x       y     z–a
     opposite edges of the tetrahedron formed by the                     ⇒       =      =         .                 ...(iv)
                                                                             1      –1       0
     planes y + z = 0, z + x = 0, x + y = 0 and x + y + z = a
        2a                                                               Similarly, CA has the equations
     is    .                                                                y + z = 0, x + y + z = a
         6
                                                                             x–a        y      z
Sol. Clearly, the planes y + z = 0, z + x = 0 and x + y = 0              ⇒           =     =                        ...(v)
     pas through the origin O(0, 0, 0,).                                       0       1      –1
                        O                                                AB has the equation z + x = 0, x + y + z = a
                                                                              x       y–a       z
                                                                         ⇒        =         =
                                                                             –1        0       1
                                          C
                                                                         Let PQ be the shortest distance between OA and BC,
                    A                                                    and its line has the direction cosines l, m, n.
                                                                         Then using perpendicularity, l .1 + m. 1 + n . (–1) = 0
                                    B
                                                                                             and l . 1 + m (–1) + n . 0 = 0
     The line of intersection OA of the planes OAC and
     OAB,                                                                ⇒ l + m – n = 0 and l – m = 0
     i.e., y + z = 0 and z + x = 0,                                          l   m   n           l2 + m2 + n2         1
                                                                         ⇒     =   =   =                          =        .
     is given by x = y = – z, i.e.,                                          1   1   2             2      2   2        6
                                                                                                  1 +1 + 2
      x–0   y–0   z–0
          =     =     .                                    ...(i)
       1     1     –1

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                  1                     1                   2                                                                                                ( y '– x' ) 2   1   y '–x'
      ∴ l=             ,m=                  ,n=                 .                                                                                        +                 +   +        =1
                   6                    6                   6                                                                                                      2         4      2
      ∴ the shortest distance PQ                                                                                        1                               1     1
      = projection of OC on the line PQ                                                                            or      {(x' + y')2 + (y' – x')2} +     +     (y' – x' – x' – y')
                                                                                                                        2                               2      2
      = |(x2 – x1) l + (y2 – y1)m + (z2 – z1)n|                                                                                          3       3
                                                                                                                        + 3(y'2 – x'2) –     +      . 2x' = 1
                      1                        1                        2              2a                                                2        2
      = ( 0 – 0)              + (0 – 0)             + ( a – 0)                   =
                       6                        6                        6               6
                                                                                                                   or   x'2 + y'2 –   2 x' + 3y'2 – 3x'2 + 3 2 x' = 2
      Similarly, other shortest distances between opposite
      edges are                                                                                                    or   4y'2 – 2x'2 + 2 2 x' = 2

                          1                         2                        2                   2a                or   x'2 – 2y'2 – 2 x' + 1 = 0
           (0 – 0)                 + ( a – 0)           + (0 – 0)                    ' i.e.,          '
                           6                        6                        6                    6                     ∴ the changed equation of the locus is
                                                                                                                                                (obtained by dropping dashes)
                           2                        1                            1                2a
      and (a – 0)                   + (0 – 0)               + (0 – 0)                  ' i.e.,            .                 x2 – 2y2 –     2 x + 1 = 0.
                               6                        6                        6                    6
      Hence, the problem.                                                                                          4.   A wheel with diameter AB touches the horizontal
                                                                                                                        ground at A. The rod BC is fixed at B, ABC being
3.   Find the changed equation of the locus x2 + 6xy + y2 = 1                                                           vertical. A man from a point P on the ground at a
     when the lines x + y = 0 and x – y + 1 = 0 are taken as                                                            distance d from A, finds that the angle of elevation of
     the new x and y axes respectively.                                                                                 C is α. The wheel turns about the fixed centre O of
Sol. Here the lines x + y = 0 and x – y + 1 = 0 are                                                                     the wheel such that C turns away from the man and
     perpendicular to each other.                                                                                       its angle of elevation is β when it is about to
                                                                                                                        disappear. Find PC when C is about to disappear.
                                   x – y +1
      So, take x' =                                                                              ...(i)            Sol. When C is about to disappear, let it be at C' and then
                               12 + (–1) 2                                                                              PC' touches the wheel at Q.
                                                                                                                          C'               C
                       x– y
      and y' =                                                                                   ...(ii)
                      12 + 12
      From (i), x – y + 1 =                  2 x'                                                ...(iii)
                                                                                                                                               B    Q
      From (ii), x + y =                2 y'                                                     ...(iv)                              B'

      (iii) + (iv) ⇒                  2x + 1 =              2 ( x'+ y ' )                                                                  Ο

                                                                                                                                                    A'
                 x'+ y '            1                                                                                                                           β/2 β α
      ∴ x=                     –
                      2             2                                                                                                      Α                     d           P
                                                                                                                        From the question, we have to find PC'.
      (iv) – (iii)        ⇒                 2y – 1 =                2 (y' – x')
                                                                                                                        Clearly, PQ = PA = d and OC' = OC = AC – r,
                 y '–x'             1                                                                                   r being the radius of the wheel.
      ∴ y=                     +
                      2             2                                                                                                           AC
                                                                                                                        In the ∆CAP, tan α =       ; ∴ AC = d tan α
      ∴ Putting these in the equation of the locus, we get                                                                                       d
       x'+ y ' 1 
                           2
                       x'+ y ' 1 
                                                            2
                                                                 y '–x' 1                                             ∴ OC + r = d tan α;            ∴ OC' = d tan α – r
              –  + 6        –                               
                                                                       + 
      
       2       2   
                       2       2                              2      2                                            ∴ C'Q =        OC' 2 –OQ 2 =           (d tan α – r ) 2 – r 2
                                                                                                 2
                                                                        y '– x' 1                                     ∴ PC' = PQ + C'Q = d +                (d tan α – r ) 2 – r 2 ...(i)
                                                                      +
                                                                               +  =1
                                                                        2       2                                    Also, from the ∆OAP,
           ( x'+ y ' ) 2   1   x'+ y '                                                                                      β     r                   β
      or                 +   –                                                                                          tan    = ;         ∴ r = d tan .
                2          4      2                                                                                         2    d                    2
                                                                                                                                                                                 β
    y ' 2 – x' 2 1
                    1                      
                                                                                                                       ∴ PC' = d +            d 2 tan 2 α – 2d tan α . d tan
+ 6             – +    ( x'+ y '– y '+ x' )                                                                                                                                    2
    2
                 4 2 2                     
                                            

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                                                                                                      {from (i)}                                   1
                                                                                                                          Putting 1 +                     = z,
               
                                         β
                                                                                                                                                  x4
           = d 1 + tan 2 α – 2 tan α. tan 
               
                                         2
                                                                                                                             –4                                 dx           1
                                                                                                                                  5
                                                                                                                                      dx = dz, i.e.,                 5
                                                                                                                                                                         =–     dz
                                                                                                                              x                                  x            4
               
                                           β
                                             
           = d 1 + tan α. 1 – 2 cot α . tan  .                                                                                               1        –1        1 –3 / 4
               
                                           2
                                                                                                                         ∴ I=        ∫z    3/ 4
                                                                                                                                                    .
                                                                                                                                                        4
                                                                                                                                                           dz = –
                                                                                                                                                                  4
                                                                                                                                                                    z     ∫dz

                                                                                                                2                               1
                                                                                             n
                                                       c 
5.     If n ∈ N and ck = Ck find the value ofn

                                              k =1
                                                   k . k  .
                                                      c 
                                                       k –1 
                                                                                             ∑       3
                                                                                                                             =–
                                                                                                                                      1
                                                                                                                                        .
                                                                                                                                      4 1
                                                                                                                                            +c
                                                                                                                                               z4

                            n                                                                                                             4
        ck                      Ck                         n!         (k – 1) !(n – k + 1) !
Sol.         =            n
                                             =                      .                                                                      1
                                                                                                                                                                              1
       ck –1                  C k –1                 k !( n – k ) !             n!                                                                             1 4
                                                                                                                             = –          z4       + c = – 1 + 4  + c
             n – k +1                                                                                                                                       x 
           =
                 k                                                                                                                                      1

               n                             2              n                                2                                  (1 + x 4 ) 4
                               ck                                         n – k +1                                       =–              + c.
       ∴   ∑ k . c
           k =1
                
                
                        3
                                      =
                                     
                                k –1 
                                                          ∑ k .
                                                           k =1
                                                                       3
                                                                                 k
                                                                                     
                                                                                     
                                                                                                                                     x
                                                                                                                          (b) Here we have second and third roots of x.
                    n
           =       ∑ k (n – k + 1)
                   k =1
                                                      2
                                                                                                                          The LCM of 2 and 3 = 6
                                                                                                                          So, Put x = z6; then dx = 6z5 dz
                    n
           =       ∑ k{(n + 1)
                   k =1
                                                 2
                                                     – 2(n + 1)k + k 2 }                                                  ∴ I=        ∫z   3
                                                                                                                                               z3
                                                                                                                                                   + z2
                                                                                                                                                            . 6z5dz

                                         n                                      n             n
                                                                                                                                           z6
           = (n + 1)2                ∑
                                     k =1
                                             k – 2(n + 1)                      ∑
                                                                               k =1
                                                                                      k2 +   ∑
                                                                                             k =1
                                                                                                     k3                      =6       ∫   z +1
                                                                                                                                               dz

                                              n(n + 1)(2n + 1)                                                                            ( z 6 – 1) + 1
            = (n + 1) .
                        n(n + 1)
                           2
                                     2
                                 – 2(n + 1) .
                                                     6
                                                                                                                             =6       ∫        z +1
                                                                                                                                                         dz

                                                                                                     n 2 (n + 1) 2                        ( z 3 + 1)( z 3 – 1)                       dz
                                                                                                 +
                                                                                                           4
                                                                                                                             = 6      ∫           z +1
                                                                                                                                                               dz + 6             ∫ z +1
           =
                   n(n + 1) 2 
                       2
                                         2
                              .(n + 1) – (2n + 1) + 
                                         3
                                                    n
                                                    2
                                                                                                                                 ∫
                                                                                                                             = 6 ( z 2 – z + 1)( z 3 – 1) dz +6log(1 + z)
                               
                                                                                                                             = 6 ∫ (z          5
                                                                                                                                                   – z 4 + z 3 – z 2 + z – 1) dz + 6log (1 + z)
             n(n + 1) 2 6(n + 1) – 4(2n + 1) + 3n
           =           .
                 2                  6                                                                                           z6 z5 z 4 z3 z 2     
                    1                                                                                                        =6 –    +   –   +   – z  + 6 log (1+z)+ c
           =          n(n + 1) 2 . (n + 2)                                                                                     6
                                                                                                                                   5   4   3   2     
                                                                                                                                                      
                   12
                                                                                                                                          6 5/ 6  3
                                                                dx                                                           =x–            x    + x 2 / 3 – 2x1/2 + 3x1/3 – 6x1/6
6.     Evaluate                  (a)         ∫x      2
                                                         (1 + x 4 ) 3 / 4
                                                                                                                                          5       2
                                                                                                                                                                                    + 6 log (1 + x1/6) + c
                                                            x
                                 (b)         ∫        x +3 x
                                                                           dx


                                             dx
Sol. (a) I =        ∫               1    
                                                                3/ 4
                          x 2 .x 3  4 + 1
                                   x     
                                             1                      dx
                     =        ∫     1 
                                                         3/ 4
                                                                .
                                                                    x5
                                                                           .
                                1 + 4 
                                 x 

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  MATHS                     MONOTONICITY,
                           MAXIMA & MINIMA
                                                                                           Mathematics Fundamentals

Monotonic Functions :                                                      or negative according as f (x) is monotonic increasing
   A function f (x) defined in a domain D is said to be                    or decreasing at x = a.
   (i) Monotonic increasing :                                              So at x = a, function f (x) is
             x < x 2 ⇒ f ( x1 ) ≤ f ( x 2 )                                   monotonic increasing ⇔ f ´(a) > 0
          ⇔  1                               ∀ x1, x2 ∈ D
             x1 > x 2 ⇒ f ( x1 ) ≥ f ( x 2 )                                   monotonic decreasing ⇔ f ´(a) < 0
          y                                                                (ii) In an interval : In [a, b], f (x) is
                                      y
                                                                               monotonic increasing ⇔ f '(x) ≥ 0
                                                                               monotonic decreasing ⇔ f '(x) ≤ 0       ∀ x ∈ (a, b)
                                                                                              constant ⇔ f '(x) = 0
                             x                               x            Note :
              O                           O
                                                                          (i) In above results f ´(x) should not be zero for all
              x < x 2 ⇒ f ( x1 ) > f ( x 2 )
                                   /                                           values of x, otherwise f (x) will be a constant
     i.e., ⇔  1                               ∀ x1, x2 ∈ D
              x1 > x 2 ⇒ f ( x1 ) < f ( x 2 )
                                    /                                          function.
     (ii) Monotonic decreasing :                                          (ii) If in [a, b], f ´(x) < 0 at least for one value of x
                                                                               and f ´(x) > 0 for at least one value of x, then f (x)
             x < x 2 ⇒ f ( x1 ) ≥ f ( x 2 )                                   will not be monotonic in [a, b].
          ⇔  1                               ∀ x1, x2 ∈ D
             x1 > x 2 ⇒ f ( x1 ) ≤ f ( x 2 )                         Examples of monotonic function :
          y                           y                                   If a functions is monotonic increasing (decreasing ) at
                                                                          every point of its domain, then it is said to be
                                                                          monotonic increasing (decreasing) function.
                                                                          In the following table we have example of some
                                                                          monotonic/not monotonic functions
                             x                               x
              O                           O                               Monotonic           Monotonic            Not
                                                                          increasing          decreasing           monotonic
              x < x 2 ⇒ f ( x1 ) < f ( x 2 )
                                    /
     i.e., ⇔  1                               ∀ x1, x2 ∈ D               x3                  1/x, x > 0           x2
              x1 > x 2 ⇒ f ( x1 ) > f ( x 2 )
                                   /
                                                                          x|x|                1 – 2x               |x|
     A function is said to be monotonic function in a                     ex
                                                                                              e –x
                                                                                                                   ex + e–x
     domain if it is either monotonic increasing or
     monotonic decreasing in that domain.                                 log x               log2x                sin x
     Note : If x1 < x2 ⇒ f (x1) < f (x2) ∀ x1, x2 ∈ D, then               sin h x             cosec h x, x > 0     cos h x
     f (x) is called strictly increasing in domain D and                  [x]                 cot h x, x > 0       sec h x
     similarly decreasing in D.
     Method of testing monotonicity :                                 Properties of monotonic functions :
     (i) At a point : A function f (x) is said to be                      If f (x) is strictly increasing in some interval, then in
     monotonic increasing (decreasing) at a point x = a of                that interval, f – 1 exists and that is also strictly
     its domain if it is monotonic increasing (decreasing)                increasing function.
     in the interval (a – h, a + h) where h is a small                    If f (x) is continuous in [a, b] and differentiable in
     positive number. Hence we may observer that if f (x)                 (a, b), then
     is monotonic increasing at x = a then at this point                   f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic increasing
     tangent to its graph will make an acute angle with                    in [a, b]
     x-axis where as if the function is monotonic
     decreasing there then tangent will make an obtuse                     f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic
     angle with x-axis. Consequently f ´(a) will be positive               decreasing in [a, b]


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    If both f (x) and g (x) are increasing (or decreasing) in                          Least value = min. {f (a), f (b), f (c)}
    [a, b] and gof is defined in [a, b], then gof is                             where x = c is a point such that f´(c) = 0.
    increasing.
                                                                                 If a continuous function has only one maximum
    If f (x) and g (x) are two monotonic functions in [a, b]                     (minimum) point, then at this point function has
    such that one is increasing and other is decreasing                          its greatest (least) value.
    then gof, it is defined, is decreasing function.
                                                                                 Monotonic functions do not have extreme points.
Maximum and Minimum Points :
                                                                        Conditions for maxima and minima of a function
    The value of a function f (x) is said to be maximum at
                                                                            Necessary condition : A point I = a is an extreme
    x = a if there exists a small positive number δ such
                                                                            point of a function f (x) if f ´(a) = 0, provided f ´(a)
    that      f (a) > f (x)
                                                                            exists. Thus if f ´(a) exists, then
                 y
                                                                                 x = a is an extreme point ⇒ f ´(a) = 0             or
                                                                                 f ´(a) ≠ 0 ⇒ x = a is not an extreme point
                                                                            But its converse is not true i.e.
                                                                            f ´(a) = 0 ⇒ x = a is an extreme point.
                                                                                         /
                                                                            For example if f (x) = x3, then f ´(0) = 0 but x = 0 is
                                              x
                O      (a)    (b)    (c)                                    not an extreme point.
     Also then the point x = a is called a maximum point                    Sufficient condition : For a given function f (x), a
     for the function f (x).                                                point x = a is
     Similarly the value of f (x) is said to be minimum at x = b                 a maximum point if f ´(a) = 0 and f´´(a) < 0
     if there exists a small positive number δ such that                         a minimum point if f´(I) = 0 and f ´´(a) > 0
          f (b) < f (x) ∀ x ∈ (b – δ, b + δ)                                     not an extreme point if f ´(a) = 0 = f ´´(a) and
     Also then the point x = b is called a minimum point                         f ´´´(a) ≠ 0.
     for f (x)                                                              Note : If f ´(a) = 0, f ´´(a) = 0, f ´´´(a) = 0 then the
     Hence we find that :                                                   sign of f(4)(a) will determine the maximum or
                                                                            minimum point as above.
     (i) x = a is a maximum point of f (x)
                                                                        Working Method :
                f ( a ) – f (a + h) > 0
           ⇔                                                                    Find f ´(x) and f ´´(x).
                f ( a ) – f (a – h) > 0                                         Solve f ´(x) = 0. Let its roots be a, b, c, ...
     (ii) x = b is a minimum point of f(x)                                       Determine the sign of f ´´(x) at x = a, b, c, .... and
               f (b) – f (b + h) > 0                                            decide the nature of the point as mentioned above.
          ⇔                                                            Properties of maxima and minima :
               f (b) – f (b – h) > 0
                                                                            If f (x) is continuous function, then
     (iii) x = c is neither a maximum point nor a minimum
          point                                                             Between two equal values of f (x), there lie atleast
                                                                            one maxima or minima.
             f (c ) – f (c + h )                                          Maxima and minima occur alternately. For example
                                 
        ⇔           and           have opposite signs.                    if x = –1, 2, 5 are extreme points of a continuous
             f (c ) – f (c – h )                                          function and if x = –1 is a maximum point then x = 2
                                 
                                                                            will be a minimum point and x = 5 will be a
        Where h is a very small positive number.                            maximum point.
     Note :                                                                 When x passes a maximum point, the sign of dy/dx
        The maximum and minimum points are also                             changes from + ve to – ve, where as when x passes
        known as extreme points.                                            through a minimum point, the sign of f ´(x) changes
        A function may have more than one maximum                           from –ve to + ve.
        and minimum points.                                                 If there is no change in the sign of dy/dx on two sides
        A maximum value of a function f (x) in an                           of a point, then such a point is not an extreme point.
        interval [a, b] is not necessarily its greatest value               If f (x) is maximum (minimum) at a point x = a, then
        in that interval. Similarly a minimum value may                     1/f (x), [f (x) ≠ 0] will be minimum (maximum) at that
        not be the least value of the function. A minimum                   point.
        value may be greater than some maximum value                        If f (x) is maximum (minimum) at a point x = a, then for
        for a function.
                                                                            any λ ∈ R, λ + f (x), log f (x) and for any k > 0, k f (x),
        The greatest and least values of a function f (x) in
        an interval [a, b] may be determined as follows :                   [f (x)]k are also maximum (minimum) at that point.
             Greatest value = max. {f (a), f (b), f (c)}

XtraEdge for IIT-JEE                                               53                                            DECEMBER 2011
 MATHS
                                              FUNCTION
                                                                                            Mathematics Fundamentals

Definition of a Function :                                                  f + g, f – g, fg, f /g, fog
     Let A and B be two sets and f be a rule under which                 and they are defined as follows :
     every element of A is associated to a unique element
                                                                                     (f + g) (x) = f (x) + g(x)
     of B. Then such a rule f is called a function from A to
     B and symbolically it is expressed as                                           (f – g) (x) = f (x) – g(x)

              f :A→B                                                                 (f g) (x) = f (x) f (g)

     or       A f
                 → B                                                                (f /g) (x) = f (x)/g(x)   (g(x) ≠ 0)

Function as a Set of Ordered Pairs                                                   (fog) (x) = f [g(x)]

     Every function f : A → B can be considered as a set            Formulae for domain of functions :
     of ordered pairs in which first element is an element                  Df ± g = Df ∩ Dg
     of A and second is the image of the first element.
     Thus                                                                   Dfg = Df ∩ Dg

          f = {a, f (a) /a ∈ A, f (a) ∈ B}.                                 Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0}

Domain, Codomain and Range of a Function :                                  Dgof = {x ∈ Df | f(x) ∈ Dg}
     If f : A → B is a function, then A is called domain of                  D        = Df ∩ {x |f (x) ≥ 0}
                                                                                 f
     f and B is called codomain of f. Also the set of all
     images of elements of A is called the range of f and it        Classification of Functions
     is expressed by f (A). Thus
                                                                    1.   Algebraic and Transcendental Functions :
          f (A) = {f (a) |a ∈ A}
                                                                            Algebraic functions : If the rule of the function
     obviously     f (A) ⊂ B.                                               consists of sum, difference, product, power or
                                                                            roots of a variable, then it is called an algebraic
     Note : Generally we denote domain of a function f by
                                                                            function.
     Df and its range by Rf.
                                                                            Transcendental Functions : Those functions
Equal Functions :                                                           which are not algebraic are named as
     Two functions f and g are said to be equal functions                   transcendental or non algebraic functions.
     if                                                             2.   Even and Odd Functions :
          domain of f = domain of g                                         Even functions : If by replacing x by – x in f (x)
                                                                            there in no change in the rule then f (x) is called
          codomain of f = codomain of g
                                                                            an even function. Thus
          f (x) = g(x) ∀ x.                                                 f (x) is even ⇔ f (– x) = f (x)
Algebra of Functions :                                                      Odd function : If by replacing x by – x in f (x)
     If f and g are two functions then their sum,                           there is only change of sign of f (x) then f (x) is
     difference, product, quotient and composite are                        called an odd function. Thus
     denoted by
                                                                            f (x) is odd ⇔ f (– x) = – f (x)


XtraEdge for IIT-JEE                                           54                                                 DECEMBER 2011
3.   Explicit and Implicit Functions :                                                       Function                 Period
        Explicit function : A function is said to be                     sin x, cos x, sec x, cosec x,           2π
        explicit if its rule is directly expressed (or can be
        expressed( in terms of the independent variable.
                                                                         tan x, cot x                            π
        Such a function is generally written as                          sinnx, cosn x, secn x, cosecn x         2π if n is odd
        y = f (x), x = g(y) etc.                                                                                 π if n is even
        Implicit function : A function is said to be                     tann x, cotnx                           π∀n∈N
        implicit if its rule cannot be expressed directly in
        terms of the independent variable. Symbolically                  |sin x|, |cos x|, |sec x|, |cosec x|    π
        we write such a function as
                                                                         |tan x|, |cot x|,                       π
        f (x, y) = 0, φ(x, y) = 0 etc.
                                                                         |sin x| + |cos x|, sin4x + cos4x        π
4.   Continuous and Discontinuous Functions :                                                                    2
                                                                         |sec x| + |cosec x|
        Continuous functions : A functions is said to be
                                                                         |tan x| + |cot x|                       π
        continuous if its graph is continuous i.e. there is
                                                                                                                 2
        no gap or break or jump in the graph.
        Discontinuous Functions : A function is said to
                                                                         x – [x]                                 1
        be discontinuous if it has a gap or break in its                    Period of f (x) = T ⇒ period of f (ax + b)= T/|a|
        graph atleast at one point. Thus a function which
        is not continuous is named as discontinuous.                        Period of f1(x) = T1, period of f2(x) = T2

5.   Increasing and Decreasing Functions :                               ⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T2}

        Increasing Functions : A function f (x) is said to
                                                                     Kinds of Functions :
        be increasing function if for any x1, x2 of its
        domain                                                               One-one/ Many one Functions :

             x1 < x2 ⇒ f (x1) ≤ f (x2)                                       A function f : A → B is said to be one-one if
                                                                             different elements of A have their different
        or x1 > x2 ⇒ f (x1) ≥ f (x2)
                                                                             images in B.
       Decreasing Functions : A function f (x) is said to
                                                                             Thus
        be decreasing function if for any x1, x2 of its
        domain                                                                               a≠b           ⇒          f (a) ≠ f (b)
                                                                                            
                                                                             f is one-one ⇔                or
             x1 < x2 ⇒ f (x1) ≥ f(x2)
                                                                                             f (a) = f (b) ⇒             a=b
                                                                                            
        or x1 > x2 ⇒ f (x1) ≤ f (x2)
                                                                             A function which is not one-one is called many
Periodic Functions :
                                                                             one. Thus if f is many one then atleast two
     A functions f (x) is called a periodic function if there                different elements have same f -image.
     exists a positive real number T such that
                                                                             Onto/Into Functions : A function f : A → B is
             f (x + T) = f (x) ∀ x                                           said to be onto if range of f = codomain of f
     Also then the least value of T is called the period of                  Thus f is onto ⇔ f (A) = B
     the function f (x).
                                                                             Hence f : A → B is onto if every element of B
        Period of f (x) = T                                                  (co-domain) has its f –preimage in A (domain).
     ⇒ Period of f (nx + a) = T/n                                            A function which is not onto is named as into
Periods of some functions :                                                  function. Thus f : A → B is into if f (A) ≠ B. i.e.,


XtraEdge for IIT-JEE                                            55                                              DECEMBER 2011
        if there exists atleast one element in codomain of f         Domain and Range of some standard functions :
        which has no preimage in domain.
                                                                     Function               Domain                Range
    Note :
                                                                     Polynomial                 R                 R
        Total number of functions : If A and B are finite            function
        sets containing m and n elements respectively,               Identity                   R                 R
        then                                                         function x
        total number of functions which can be defined               Constant                   R                 {c}
        from A to B = nm.                                            function c
                                                                     Reciprocal                 R0                R0
        total number of one-one functions from A to B                function 1/x
          n P
                       if   m≤n                                     x2, |x|                    R                 R+ ∪ {0}
        =  m
           0
                       if   m>n                                     x3, x |x|                  R                 R
                                                                     Signum                     R                 {–1, 0, 1}
        total number of onto functions from A to B
                                                                     function
        (if m ≥ n) = total number of different n groups of
                                                                     x + |x|                    R                 R+ ∪ {0}
        m elements.
                                                                     x – |x|                    R                 R– ∪ {0}
Composite of Functions :
                                                                     [x]                        R                 Z
    Let f : A → B and g : B → C be two functions, then
                                                                     x – [x]                    R                 [0, 1)
    the composite of the functions f and g denoted by gof,
                                                                          x                   [0, ∞)              [0, ∞)
    is a function from A to C given by gof : A → C,
    (gof) (x) = g[f (x)].                                            ax                         R                 R+
Properties of Composite Function :                                   log x                      R+                R

    The following properties of composite functions can              sin x                      R                 [–1, 1]
    easily be established.                                           cos x                      R                 [–1, 7]

        Composite of functions is not commutative i.e.,              tan x           R – {± π/2, ± 3π/2, ...} R
                                                                     cot x           R – {0, ± π. ± 2π, .....     R
                       fog ≠ gof
                                                                     sec x           R – (± π/2, ± 3π/2, .....    R – (–1, 1)
        Composite of functions is associative i.e.
                                                                     cosec x         R – {0, ±π, ± 2π, ......} R –(–1, 1)
                       (fog)oh = fo(goh)
                                                                     sinh x                     R                 R
        Composite of two bijections is also a bijection.             cosh x                     R                 [1, ∞)
Inverse Function :                                                   tanh x                     R                 (–1, 1)
    If f : A → B is one-one onto, then the inverse of f i.e.,        coth x                     R0                R –[1, –1]
    f –1 is a function from B to A under which every b ∈ B           sech x                     R                 (0, 1]
    is associated to that a ∈ A for which f (a) = b.                 cosech x                   R0                R0
                 –1                                                        –1
    Thus     f        : B → A,                                       sin x                   [–1, 1]              [–π/2, π/2]

             f –1(b) = a ⇔ f (a) = b.                                cos–1x                  [–1, 1]              [0, π]
                                                                           –1
                                                                     tan x                      R                 (–π/2, π/2}
                                                                           –1
                                                                     cot x                      R                 (0, π)
                                                                           –1
                                                                     sec x                 R –(–1, 1)             [0, π] – {π/2}
                                                                                –1
                                                                     cosec x               R – (–1, 1)            (– π/2, π/2] – {0}



XtraEdge for IIT-JEE                                            56                                               DECEMBER 2011
XtraEdge for IIT-JEE   57   DECEMBER 2011
                                                        Based on New Pattern
a

                                     IIT-JEE 2012
                                       XtraEdge Test Series # 8
     Time : 3 Hours
     Syllabus :
     Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
     Instructions :
     Section - I
     •     Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong
           answer.
     •     Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for
           wrong answer.
     •      Question 10 to 15 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for
            correct answer and –1 mark for wrong answer.
     •      Question 16 to 18 are Numerical Response Question (four digit Ans. type) +6 marks will be awarded for
            correct answer and –1 mark for wrong answer.


                                                                                 (C) If the speed of water is 4 m/sec then direction is which
                        PHYSICS                                                                                      3
                                                                                     he should row his boat is cos–1   w.r.t. bank
                                                                                                                     4
This section contains 2 paragraphs; each has 3 multiple                          (D) If the speed of water is 4 m/sec then direction in
choice questions. (Questions 1 to 6) Each question has 4
                                                                                                                             3
choices (A), (B), (C) and (D) out of which ONE OR                                    which he should row his boat is sin–1   w.r.t bank
MORE THAN ONE may be correct. Mark your                                                                                      4
response in OMR sheet against the question number of
                                                                            2.   The velocity of boatman is still water is 3m/sec and
that question. + 4 marks will be given for each correct
                                                                                 river is flowing at 2 m/sec. To cross the river in
answer and – 1 mark for each wrong answer.
                                                                                 minimum time.
Passage # 1 (Ques. 1 to 3)                                                       (A) The boatman should move at an angle
                                                           →                                    2
         There is a river which is flowing at the rate v r . If a                     θ = sin–1   w.r.t bank
                                                                                                3
                                                           →
         boatman starts to row his boat at the speed v br with                   (B) The boatman should move at an angle
         respect to river then the velocity of boatman with                                     2
                                                                                      θ = cos–1   w.r.t bank
                                                    →     →    →
         respect to ground can be given by, v b = v br + v r .                                  3
         Now a boat can travel at a speed of 3m/s in still                       (C) the batman should move at an angle 90º w.r.t bank
         water. A boatman has to across the river to reach the                   (D) None of these
         other side. Now give the answer of following                       3.   Two boats A & B move away from a buoy anchored
         question                                                                at the middle of river along the mutually
                                                                                 perpendicular 1 straight lines : the boat A along the
1.       The boatman wants to cross the river in such a way                      river & boat B across the river. Having moved off an
         that he should cover the shortest possible distance.                    equal distance from the buoy the boats returned.
         (A) If the speed of water is 2m/sec then direction in which             Then the ratio of times of boats tA/tB. If the velocity
                                              2                                of each boat w.r.t flow of water ‘x’ times greater than
              he should row his boat is cos–1   w.r.t. bank                    the stream velocity tA/tB is -
                                              3
                                                                                         x                          x
         (B) If the speed of water is 2m/sec then direction in which             (A)                         (B)
                                                                                         2
                                                                                        x −1                      x2 +1
                                             2
             he should row his boat is sin–1   w.r.t bank                              1                          1
                                             3                                 (C)                         (D)
                                                                                         2
                                                                                        x −1                      x2 +1

XtraEdge for IIT-JEE                                                   58                                             DECEMBER 2011
Passage # 2 (Ques. 4 to 6)                                        9.    Assertion (A) : Work done by the static friction is
     If two masses A & B are drawn in their attached                    always zero.
                      0.2t        2                                   Reason (R) : when the body is stationary, there is no
     cables with a =       m / sec where t is in second.                                          → →
                      3                                               displacement. Hence w = F ⋅ S = zero
                                                                  This section contains 6 questions (Q.10 to 15).
                                                                  +4 marks will be given for each correct answer and –1
                                                                  mark for each wrong answer. The answer to each of the
                                  D                               questions is a SINGLE-DIGIT INTEGER, ranging
                                                                  from 0 to 9. The appropriate bubbles below the
                          C                                       respective question numbers in the OMR have to be
                                                                  darkened. For example, if the correct answers to
                                  E                               question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,
                     A        h       B                           respectively, then the correct darkening of bubbles will
4.   The speed of block ‘E’ when it reaches a height of           look like the following :
     h = 4 m starting from rest is                                                      X Y Z W
     (A) 4 m/sec               (B) 2 m/sec                                               0 0 0 0
     (C) 1 m/sec               (D) None of these                                         1 1 1 1
                                                                                         2 2 2 2
5.   The approx speed of pulley D is                                                     3 3 3 3
     (A) – 2.5 m/sec          (B) – 3.5 m/sec                                            4 4 4 4
     (C) – 4.5 m/sec          (D) – 5.5 m/sec                                            5 5 5 5
                                                                                         6 6 6 6
6.   If vA = KvD then K, is
                                                                                         7 7 7 7
     (A) 2      (B) 3             (C) 4    (D) 5
                                                                                         8 8 8 8
                                                                                         9 9 9 9
This section contains 3 questions numbered 7 to 9,
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4                10.   A particle starts oscillating simple harmonically from
choices (A), (B), (C) and (D) out of which ONLY ONE                     its equilibrium position then the KE of the particle is
is correct. Mark your response in OMR sheet against                     ‘n’ times the P.E. of particle at the time T/12, find the
the question number of that question. + 4 marks will be                 value of n (T : time period)
given for each correct answer and – 1 mark for each               11.   A bird flies for 4 sec with a velocity of
wrong answer.                                                           |t – 2| m/sec in a straight line where t is in second the
The following questions given below consist of an                       distance covered by the bird is ….. (in m)
"Assertion" (A) and "Reason" (R) Type questions. Use
the following Key to choose the appropriate answer.               12.   An object of mass 0.2 kg executes SHM along the x-
      (A) If both (A) and (R) are true, and (R) is the                  axis with a frequency 25/π Hz. At the position
          correct explanation of (A).                                   x = 0.04 m, the object has K.E., 0.5 J and P.E.,
      (B) If both (A) and (R) are true but (R) is not the               0.4 J. The amplitude of oscillation in cm will be …...
          correct explanation of (A).                                   (PE is zero at mean position)
      (C) If (A) is true but (R) is false.                        13.      Difference between nth & (n + 1)th Bohr's radius of
      (D) If (A) is false but (R) is true.                                 H atom is equal to its (n – 1)th Bohr's radius. The
                                                                           value of n is ……..
7.   Assertion (A) : A lighter and a heavier bodies
     moving with same momentum and experiencing                   14.   In the arrangement shown in the figure m1 = 1kg m2
     same retarding force have equal stopping times.                    = 2 kg. Pulley are mass less & strings are light for
     Reason (R) : For a given force and momentum,                       what value of M, the mass m1 moves with constant
     stopping time is independent of mass.                              velocity (in kg)
8.   Assertion (A) : The shortest wavelength of x-rays                               M
     emitted from x-ray tube is independent of voltage
     applied to tube.
     Reason (R) : wavelength of characteristic spectrum
     depends upon the atomic number of target.                                                             m1
                                                                                                                m2


XtraEdge for IIT-JEE                                         59                                            DECEMBER 2011
15.   A block of mass 1 kg is attached to one end of a                   electron easily. The oxidation state of any element
      spring of force constant k = 20 N/m. The other end of              can never be in fraction. If oxidation number of any
      the spring is attached to a fixed rigid support. This              element comes out be in fraction, it is average
      spring block system is made to oscillate on a rough                oxidation number of that element which is present in
      horizontal surface with µ = 0.04. The initial                      different oxidation states.
      displacement of block from the (Eq.) mean position            1.   The oxidation state of Fe in Fe3O4 is -
      is a = 30 cm. How many times the block will pass                   (A) 2 and 3              (B) 8/3
      from the mean position before coming to rest ?                     (C) 2                    (D) 3
      (g = 10 m/sec2)
                                                                         1
This section contains Numerical response type questions                  N     3
(Q. 16 to 18). +6 marks will be given for each correct              2.         N–H, In this compound HN3 (hydrazoic acid),
answer and –1 mark for each wrong answer. Answers to                     N
                                                                         2
this Section are to be given in the form of nearest integer-
                                                                         oxidation state of N1, N2 and N3 are -
in four digits. Please follow as per example : (i.e. for
answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write                 (A) 0, 0, 3             (B) 0, 0, –1
0092; 2.1 write 0002)                                                    (C) 1, 1, –3            (D) –3, –3, –3
         235
16.   If 92 U reactor takes 30 days to consume 4 kg of fuel         3.   Equivalent weight of chlorine molecule in the
      and each fission gives 185 MeV of unstable energy                  equation
      then find the output power [× 107 W]                               3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
17.   A bullet in fired at a plank of wood with a speed of               (A) 42.6            (B) 35.5
      200 m/sec. After passing through the plank, its speed              (C) 59.1            (D) 71
      is reduced to 150 m/sec. Another bullet of same mass
      & size but moving with a speed of                             Passage # 2 (Ques. 4 to 6)
      100 m/sec is fired at the same plank. What would be                Secondary and tertiary alcohols always give E1
      the speed of bullet after passing through the plank ?              reaction in dehydration. Primary alcohols whose β-
      Assume that the resistance offered by plank is same                carbon is 3º or 4º also give E1 reaction. However, the
      for both the bullets.                                              primary alcohols whose β-carbon is 1º or 2º give E2
18.   If the end of the cord A is pulled down with 2m/sec                reaction. Dehydrating agents like conc. H2SO4, Al2O3
      then the velocity of block will be :                               anhydrous ZnCl2 are used.
      ( × 10–1 m/sec)                                                    The reactivity of alcohols for elimination reaction lies
                                                                         in following sequence :
                                                                         Tertiary alcohol > secondary alcohol > primary
                                                                         alcohol
                                                                         Electron attracting groups present in alcohols
                                                                         increase the reactivity for dehydration. Greater is the
                      A                                                  –I effect of the group present in alcohol, more will be
                                                                         its reactivity. Both E1 and E2 mechanism give the
                                                                         product according to Saytzeff's rule, i.e., major
                                                                         product is the most substituted alkene.
                    2 m/sec
                                                                                       OH
                               B
                                                                         CH3 – CH – CH – CH3        conc. H2SO4
                                                                                                    Above 413K
               CHEMISTRY                                                         CH3
                                                                             CH3 – C = CH – CH3 + CH3 – CH – CH = CH2
This section contains 2 paragraphs; each has 3 multiple                      CH
                                                                                   CH3                  CH3
choice questions. (Questions 1 to 6) Each question has 4
choices (A), (B), (C) and (D) out of which ONE OR                             Major product        Minor product
MORE THAN ONE may be correct. Mark your
response in OMR sheet against the question number of                4.   Arrange the reactivity of given four alcohols in
that question. + 4 marks will be given for each correct                  decreasing order for dehydration.
answer and – 1 mark for each wrong answer.                                 OH                        OH NO2
Passage # 1 (Ques. 1 to 3)
     Redox reactions are those in which oxidation and                                       NO2
     reduction take place simultaneously. Oxidising agent                          (a)                       (b)
     can gain electron whereas reducing agent can lose

XtraEdge for IIT-JEE                                           60                                          DECEMBER 2011
       OH                          OH                                 7.    Assertion (A) : The value of van der Waals constant
                                                                            'a' is larger for ammonia than for nitrogen.
                                                                            Reason (R) : Hydrogen bonding is present in
                                                                            ammonia.
           NO2                              NO2
           (c)                       (d)                              8.    Assertion (A) : 3-hydroxy - butan-2-one on
     (A) a > b > c > d         (B) d > c > b > a                            treatment with [Ag(NH3)2]⊕ cause precipitation of
     (C) c > b > d > a         (D) b > c > a > d                            silver.
                              CH3                                           Reason (R) : [Ag(NH3)2] ⊕ oxidises 3-hydroxy
                                                                            butan-2-one to butan-2-3-dione
5.   In the reaction, CH3 – C – CH2OH        conc. H2SO4
                                                                      9.    Assertion (A) : HBr adds to 1,4-pentadiene at a
                           CH3                                              faster rate than to 1,3-pentadiene
     the product obtained will be :                                         Reason (R) : 1,4-pentadiene is less stable than
     (A) CH2 = C – CH2 – CH3                                                1,3-pentadiene.
               CH3
                                                                      This section contains 6 questions (Q.10 to 15).
     (B) CH3 – C = CH – CH3                                           +4 marks will be given for each correct answer and –1
                                                                      mark for each wrong answer. The answer to each of the
                 CH3
                                                                      questions is a SINGLE-DIGIT INTEGER, ranging
                 CH3                                                  from 0 to 9. The appropriate bubbles below the
     (C) CH3 – CH – CH = CH2                                          respective question numbers in the OMR have to be
                                                                      darkened. For example, if the correct answers to
     (D) all of these                                                 question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,
6.   Which of the following dehydration product is/are                respectively, then the correct darkening of bubbles will
     correct ?                                                        look like the following :
               CH2OH                                                                        X Y Z W
                          conc.H2SO4                                                         0 0 0 0
     (A)                       ∆                                                             1 1 1 1
              CH3 CH3                              CH3                                       2 2 2 2
     (B) CH3 – C — CH – CH3   conc. H2SO4   CH3 – C – CH = CH2                               3 3 3 3
                                  ∆
              CH3                                  CH3                                       4 4 4 4
                          conc. H2SO4                                                        5 5 5 5
     (C) CH3CH2CH2CH2OH
                               ∆                                                             6 6 6 6
                           CH3 – CH = CH – CH3                                               7 7 7 7
              CH3                      CH3                                                   8 8 8 8
              CH3   conc.H2SO4
     (D)                                                                                     9 9 9 9
              OH         ∆             CH3
                                                                      10. Equal volumes of 0.02 M AgNO3 & 0.02 M HCN
This section contains 3 questions numbered 7 to 9,                        were mixed. If the [Ag+] at equilibrium was 10–n.
(Reason and Assertion type question). Each question                       Find n. Given Ka(HCN) = 4 × 10–10, Ksp(AgCN)
contains Assertion and Reason. Each question has 4                        = 4 × 10–16.
choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against                   11.   Haemoglobin contains 0.25% iron by weight.
the question number of that question. + 4 marks will be                     The molecular weight of haemoglobin is 89600.
given for each correct answer and – 1 mark for each                         Calculate the number of iron atoms per molecule of
wrong answer.                                                               haemoglobin.
The following questions given below consist of an
                                                                      12. Two liquids A and B form an ideal solution at
"Assertion" (A) and "Reason" (R) Type questions. Use
                                                                          temperature T. When the total vapour pressure above
the following Key to choose the appropriate answer.
                                                                          the solution is 600 torr, the amount of A in the
      (A) If both (A) and (R) are true, and (R) is the
                                                                          vapour phase is 0.35 and in the liquid phase is 0.70.
          correct explanation of (A).
                                                                          What is the vapour pressure of pure A ? Express your
      (B) If both (A) and (R) are true but (R) is not the                 answer after divide actual answer by 100.
          correct explanation of (A).
      (C) If (A) is true but (R) is false.                            13. The value of x in the complex Hx[Co(CO)4] is
      (D) If (A) is false but (R) is true.

XtraEdge for IIT-JEE                                             61                                        DECEMBER 2011
14. Calculate the emf of the cell                                        B, one or more than one choice is (are) correct out of
    Cd|Cd2+ (0.10M)1|| H+(0.20M)| Pt, H2(0.5 atm)                        4 choices and the student is awarded 3 marks if he
    [Given: EºCd2+/Cd = – 0.403 V,                                       (she) ticks only the correct choice and all the correct
     2.303RT                                                             choices. There is no negative marking.
               = 0.0591]                                            1.   In how many ways can a student answer to any
        F
    Round off your answer after multiplying actual                       question of section B
    answer by 10.                                                        (A) 11      (B) 5C2 + 5C4 (C) 24 – 1 (D) 15

15. Calculate enthalpy change (in calories) adiabatic               2.   If a student attempts 3 particular questions-one from
    compression of one mole of an ideal monoatomic gas                   section A and two from section B, the probability
    against constant external pressure of 2 atm starting                 that he will get marks in only two questions is
    from initial pressure of 1 atm and initial temperature               (assuming all ways to answer a question to be
    of 300 K. (R = 2 cal/mol degree) Give your answer                    equally likely)
    after divide actual answer by 100.                                         31          23            23
                                                                         (A)         (B)           (C)          (D) None
                                                                              900         900           484
This section contains Numerical response type questions
(Q. 16 to 18). +6 marks will be given for each correct
                                                                    3.   The probability that a student gets 10 marks if he
answer and –1 mark for each wrong answer. Answers to
                                                                         attempts only 4 questions is
this Section are to be given in the form of nearest integer-
                                                                                             3                                    3
in four digits. Please follow as per example : (i.e. for                          1 1                                4 1
answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write
                                                                         (A)                                   (B)      
                                                                                  5  15                              3  11 
0092; 2.1 write 0002)
                                                                                             3                                    3
16.   A current of 4 A flows in a coil when connected to a                      1                                    1 1 
                                                                         (C) 16                                (D)      
      12V dc source. If the same coil is connected to a                         60                                   4  15 
      12V, 50 rad/s ac source a current of 2.4 A flows in
      the circuit. Also find the power developed in the             Passage # 2 (Ques. 4 to 6)
      circuit if a 2500 µF capacitor is connected in series               a                      a
      with the coil.
                                                                          ∫                   ∫
                                                                               f ( x) dx = ( f ( x) + f (− x)) dx
17. A capacitor of capacity 2 µF is charged to a potential               −a                      0
    difference of 12V. It is then connected across an                                                        a

    inductor of inductance 0.6 mH. What is the current in
    the circuit at a time when the potential difference
                                                                         (i)         If f(x) is odd, then   ∫ f ( x) dx = 0
                                                                                                            −a
    across the capacitor is 6.0 V ?                                                                          a                        a

18. A ball of mass 100 g is projected vertically upwards
                                                                         (ii) If f(x) is even, then          ∫
                                                                                                             −a
                                                                                                                                      ∫
                                                                                                                  f ( x) dx = 2 f ( x) dx
                                                                                                                                      0
    from the ground with a velocity of 49 m/s. At the
    same lime another identical ball is dropped from a                   This is one of the important property for the
    height of 98 metre to fall freely along the same path                integrable function f(x)
                                                                          1
    as that followed by the first ball. After some time the
                                                                         ∫ (x         + x | cos x |)(sin −1 x 2 ) dx is equal to :
                                                                                  3
    two balls collide and stick together and finally fall to        4.
    ground. Find the time of flight of the masses.                       −1
    (g = 9.8 m/s2)                                                       (A) 1              (B) – 1              (C) 0                (D) 2

                                                                               1/ 4
                                                                                      
            MATHEMATICS                                             5.   If      ∫     16 x 2 + 16 x + 41 + 40 x 2 + x + 1 +
                                                                                      
                                                                               −1 / 4 
This section contains 2 paragraphs; each has 3 multiple                                                                        1
choice questions. (Questions 1 to 6) Each question has 4                   16 x 2 + 16 x + 41 − 40 x 2 + x + 1  cosx dx = k sin ,
                                                                                                               
choices (A), (B), (C) and (D) out of which ONE OR                                                                              4
MORE THAN ONE may be correct. Mark your                                  then the value of k is :
response in OMR sheet against the question number of                     (A) 10        (B) 0          (C) 20         (D) 30
that question. + 4 marks will be given for each correct
                                                                          π/ 4

                                                                              ∫ ( 1 + sin 2 x +                   )
answer and – 1 mark for each wrong answer.
                                                                    6.                               1 − sin 2 x dx is equal to
Passage # 1 (Ques. 1 to 3)
                                                                         −π / 4
     An objective test contains two sections : A and B
     each consisting of 10 questions. In section A, only                                                                                      4
                                                                         (A)          2     (B) 2 2              (C) 4 2              (D)
     one choice out of 4 choices is correct and student is                                                                                    2
     awarded 1 mark for every correct answer. In section

XtraEdge for IIT-JEE                                           62                                                         DECEMBER 2011
This section contains 3 questions numbered 7 to 9, (Reason             10.   The position vectors of two points A and C are
and Assertion type question). Each question contains                           ˆ j     ˆ        ˆ         ˆ
                                                                             9i − ˆ + 7k and 7i − 2 ˆ + 7 k respectively. The point
                                                                                                    j
Assertion and Reason. Each question has 4 choices (A), (B),
                                                                             of intersection of the lines containing vectors
(C) and (D) out of which ONLY ONE is correct. Mark your
response in OMR sheet against the question number of that                            ˆ j     ˆ              ˆ j     ˆ
                                                                              AB = 4i − ˆ + 3k and CD = 2i − ˆ + 2k is P. If vector
question. + 4 marks will be given for each correct answer
and – 1 mark for each wrong answer.                                          PQ is perpendicular to AB and CD and PQ = 15
The following questions given below consist of an                            units, then possible position vectors of Q are
"Assertion" (A) and "Reason" (R) Type questions. Use                           ˆ             ˆ       ˆ              ˆ
                                                                             x1i + x2 ˆ + x3 k and y1i + y 2 ˆ + y3 k . Find the value
                                                                                      j                      j
the following Key to choose the appropriate answer.                                3
      (A) If both (A) and (R) are true, and (R) is the
           correct explanation of (A).
                                                                             of   ∑ (x + y ) .
                                                                                  i =1
                                                                                          i   i

      (B) If both (A) and (R) are true but (R) is not the
                                                                                                      x −1      y −3    z−4
           correct explanation of (A).                                 11. Let image of the line            =        =         in the
                                                                                                        3         5      2
      (C) If (A) is true but (R) is false.
                                                                             plane 2x – y + z + 3 = 0 be L. A plane
      (D) If (A) is false but (R) is true.
                                                                             7x + By + Cz + D = 0 is such that it contains the line L
7.   Assertion (A) : Let f (x) be an even function which                     and perpendicular to the plane 2x – y + z + 3 = 0 then
                                 x                                           find the value of (B + C + D)/10.
     is periodic, then g (x) =   ∫ f (t ) dt is also periodic.         12.   A circle touches the hypotenuse of a right angled
                                 a                                           triangle at its middle point and passes through the
     Reason (R) : If α(x) is a differentiable and periodic                   middle point of shorter side. If 3 unit and 4 unit be
     function, then α′(x) is also periodic.                                  the length of the sides and 'r' be the radius of the
                                                                             circle, then find the value of '3r'.
8.   Assertion (A) : The locus represented by xy + yz = 0
     is a pair of perpendicular planes.                                13. The remainder when 2740 is divided by 12
     Reason (R) : If a1x + b1y + c1z + d1 = 0 and                      14. Let f (x) = x3 – x2 – 3x – 1 and h(x) = f (x)/g(x)
     a2x + b2y + c2z + d2 = 0 are perpendicular then                       where h is a function such that
     a1a2 + b1b2 + c1c2 = 0                                                (a) it is continuous every where except when x = – 1
9.   Assertion (A) : Locus of center of a variable circle                  (b) lim h(x) = ∞ and (c) lim h(x) = 1/2
                                                                                   x →∞                    x → –1
     touching two circles (x – 1)2 + (y – 2)2 = 25 and
     (x – 2)2 + (y – 1)2 =16 is an ellipse.                                  FInd lim (4h(x) + f (x) + 2g(x)).
                                                                                       x →0
     Reason (R) : If a circle S2= 0 lies completely inside
     the circle S1= 0 then locus of center of a variable circle        15. In a triangle ABC, if sin A cos B = 1/4 and
     S = 0 which touches both the circles is an ellipse.                   3 tan A = tan B, then cot2A =

This section contains 6 questions (Q.10 to 15).                        This section contains Numerical response type questions
+4 marks will be given for each correct answer and –1                  (Q. 16 to 18). +6 marks will be given for each correct
mark for each wrong answer. The answer to each of the                  answer and –1 mark for each wrong answer. Answers to
questions is a SINGLE-DIGIT INTEGER, ranging                           this Section are to be given in the form of nearest integer-
                                                                       in four digits. Please follow as per example : (i.e. for
from 0 to 9. The appropriate bubbles below the
                                                                       answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write
respective question numbers in the OMR have to be
                                                                       0092; 2.1 write 0002)
darkened. For example, if the correct answers to
question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,                16.   The vertices B and C of a triangle ABC lie on the
respectively, then the correct darkening of bubbles will                     lines 3y = 4x and y = 0 respectively and the side BC
look like the following :                                                    passes through the point (2/3, 2/3). If ABOC is a
                      X Y Z W                                                rhombus, O being the origin. If co-ordinates of vertex
                       0 0 0 0                                               A is (α, β), then find the value of 5(α + β).
                       1 1 1 1                                         17.   Number of different words that can be formed using
                       2 2 2 2                                               all the letters of the word "DEEPMALA", I two
                       3 3 3 3                                               vowels are together and the other two are also
                                                                             together but separated from the first two.
                       4 4 4 4
                       5 5 5 5                                         18.   If a complex number z satisfies the conditions
                       6 6 6 6                                                                            4i     1+ i    2 – 3i 
                                                                               z – 4i                                           
                       7 7 7 7                                                        = 1 and z =  x + – 1 + i    – 2i   3 + 4i  ,
                                                                               z + 4i             
                       8 8 8 8
                                                                                                       – 2 – 3i – 3 + 4i   0   
                       9 9 9 9                                               then x =

XtraEdge for IIT-JEE                                              63                                             DECEMBER 2011
XtraEdge for IIT-JEE   64   DECEMBER 2011
                                                   Based on New Pattern


                                    IIT-JEE 2013
                                     XtraEdge Test Series # 8
     Time : 3 Hours
     Syllabus :
     Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus
     Instructions :
     Section - I
     •     Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and
           -1 mark for wrong answer.
     •     Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for
           wrong answer.
     •     Question 10 to 15 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for
           correct answer and –1 mark for wrong answer.
     •     Question 16 to 18 are Numerical Response Question (four digit Ans. type) +6 marks will be awarded for
           correct answer and –1 mark for wrong answer.


                                                                        3.   If some of heat is allowed to escape to surrounding
                       PHYSICS                                               (temperature of surrounding is 20ºC) then this
                                                                             amount of steam (mentioned in question 22) is
This section contains 2 paragraphs; each has 3 multiple                      increase the temperature to -
choice questions. (Questions 1 to 6) Each question has 4                     (A) greater than 80ºC      (B) less than 80ºC
choices (A), (B), (C) and (D) out of which ONE OR
                                                                             (C) equal to 80ºC           (D) can't say anything
MORE THAN ONE may be correct. Mark your
response in OMR sheet against the question number of                    Passage # 2 (Ques. 4 to 6)
that question. + 4 marks will be given for each correct
answer and – 1 mark for each wrong answer.                                   A body of mass 1 kg moving along x-axis has
                                                                             velocity 4 m/s at x = 0. The acceleration and
Passage # 1 (Ques. 1 to 3)                                                   potential energy of body varies as shown in
     In espresso coffee machines steam is passed into milk                   diagrams.
     at room temperature for a brief time interval. Some
     of the steam condenses and the temperature or milk                      a(m/s2)                   U(J)
     rises. Since the time for which the steam is passed is
                                                                                                        120
     brief, one can ignore the heat lost to the environment                        2
     and assume that the usual assumption of calorimetry :
     Heat lost = Heat gain is valid.                                                    4   8                   4      8    x(m)
                                                                                                x(m)
                                                                                                       –120
1.       Steam at 100ºC is passed into milk to heat it. The
         amount of heat required to heat 150 g of milk from
                                                                        4.   Work done by conservative forces when body moves
         room temperature (20ºC) to 80ºC is (specific heat of
                                                                             from x = 0m to x = 8m is –
         capacity of milk = 4.0 kJ kg–1 K–1 specific latent heat
                                                                             (A) 0 J   (B) 120 J      (C) 240 J (D) – 240 J
         of steam = 2.2 MJ kg–1 , specific heat capacity of
         water = 4.2 × 103 J kgK–1)                                     5.   Work done by external forces when body moves
         (A) 3.6 × 104 J            (B) 3.6 × 103 J                          from x = 0m to x = 8m is –
                     2
         (C) 3.6 × 10 J             (D) None of these                        (A) 120 J                (B) – 120 J
                                                                             (C) – 112 J              (D) None of these
2.       How many grams of steam condensed into water in                6.   The change in kinetic energy when body moves from
         above question -                                                    x = 0m to x = 8m is –
         (A) 1.57 g             (B) 15.7 g                                   (A) 256 J                (B) 240 J
         (C) 157 g              (D) None of these                            (C) 128 J                (D) 120 J


XtraEdge for IIT-JEE                                               65                                         DECEMBER 2011
This section contains 3 questions numbered 7 to 9,                10.   If y = 4x2 – 4x + 7. Find the minimum value of 'y'.
(Reason and Assertion type question). Each question
contains Assertion and Reason. Each question has 4                11. A block of mass 1 kg starts slipping on a circular
choices (A), (B), (C) and (D) out of which ONLY ONE                   track of radius 2m and it is observed that when θ =
is correct. Mark your response in OMR sheet against                   60º its speed is 4 m/s as shown in figure. Assuming
the question number of that question. + 4 marks will be               size of block to be                      O r = 2m
given for each correct answer and – 1 mark for each                   negligible and coefficient of
                                                                      friction between block and             θ
wrong answer.
The following questions given below consist of an                     track is 0.5 frictional force
"Assertion" (A) and "Reason" (R) Type questions. Use                  (in N) on block when                 v
the following Key to choose the appropriate answer.                   θ = 60º is (g = 10 m/s2) –
      (A) If both (A) and (R) are true, and (R) is the
          correct explanation of (A).                             12.   A projectile is fired with speed 'u' at angle 60º with
      (B) If both (A) and (R) are true but (R) is not the               horizontal. Velocity of projectile when it makes an
          correct explanation of (A).                                   angle 120º with initial direction of velocity is u' then
      (C) If (A) is true but (R) is false.                              ratio b/w u : u' is .
      (D) If (A) is false but (R) is true.                        13.   A block of mass 'm' is hanged vertically by a wire.
                                                                        Potential energy stored in wire u1. Potential energy
7.   Assertion (A) : In the flow-tube as the cross-section
                                                                        stored in wire is u2. When mass hanged is doubled.
     area decreases the flow velocity increases.
                                                                        Ratio u2 : u1 is.
     Reason (R) : In ideal fluid flow the total energy per
     unit mass remains constant.                                  14.   An electrically heating coil is placed in calorimeter
                                                                        containing 360 gm of H2O at 10ºC. The coil
8.   Assertion (A) : When a spring is elongated work                    consumes energy at the rate of 90 W. The water
     done by spring is negative but when it compressed                  equivalent of calorimeter and the coil is 40 g. The
     work done by spring is positive.                                   temperature of water after 10 minutes will be n/8.5
     Reason (R) : Work done by spring is path                           then find the value of 'n' -
     independent.                                                 15.   Variation of pressure at certain point is space is given
9.   Assertion (A) : If in some case work done by a force               by :
     is path independent then it must be conservative.                  P = P0 is 2πt cos 212 πt sin (220πt – π/2)
     Reason (R) : Work done by conservative forces in a                 The Beat Frequency is -
     round trip must be zero.                                     This section contains Numerical response type questions
                                                                  (Q. 16 to 18). +6 marks will be given for each correct
This section contains 6 questions (Q.10 to 15).                   answer and –1 mark for each wrong answer. Answers to
+4 marks will be given for each correct answer and –1             this Section are to be given in the form of nearest integer-
mark for each wrong answer. The answer to each of the             in four digits. Please follow as per example : (i.e. for
questions is a SINGLE-DIGIT INTEGER, ranging                      answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write
from 0 to 9. The appropriate bubbles below the                    0092; 2.1 write 0002)
respective question numbers in the OMR have to be                 16.   A conical container of radius                 R
darkened. For example, if the correct answers to                        R = 1m and height H = 5m is
question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,                 filled completely with liquid.
respectively, then the correct darkening of bubbles will                There is a hole at the bottom of
look like the following :                                                                                                     H
                                                                        container of area π × 10–3 m2 (see
                     X    Y   Z   W                                     figure). Time taken to empty the
                                                                        conical container (in sec) is……
                      0   0   0   0                                     ………….. Take g = 10 m/s2 –
                      1   1   1   1
                      2   2   2   2                               17.   A pendulum of length l is
                      3   3   3   3                                     given a horizontal velocity
                      4   4   4   4                                       kgl at the lowest point of
                      5   5   5   5                                     vertical circular path as                O
                      6   6   6   6                                     shown. In the subsequent
                          7   7   7                                     motion the string gets slag at           l
                      7
                      8   8   8   8                                     a certain point and the
                                                                        pendulum bob strikes the                       v = kgl
                      9   9   9   9                                     point of suspensión then the
                                                                        value of k is –

XtraEdge for IIT-JEE                                         66                                           DECEMBER 2011
18.   A small body is released from point A of the smooth           2.   Instead of developing a catalyst, why didn’t Haber
      parabolic path y = x2. Where y is vertical axis and x              increase the rate of the reaction by raising the
      is horizontal axis at ground as shown. The body                    temperature ?
      leaves the surface from point B. If g = 10 m/s2 then               (A) The Haber process is exothermic, so raising the
      the total horizontal distance travelled by body before                 temperature would have lowered the rate of the
      it hits ground is –                                                    reaction
                           y                                             (B) The Haber process is exothermic, so raising the
                                                                             temperature would have reduced the yield of the
                  A                                                          reaction
                                                                         (C) Higher temperatures would have caused an
                                B
                                                                             increased in pressure lowering the yield of the
                                                                             reaction
                 –2m       O   +1m               x                       (D) Higher temperatures might have decomposed the
                                                                             hydrogen

                                                                    3.   Which of the following graphs could also accurately
                          CHEMISTRY                                      reflect the establishment of equilibrium between
                                                                         nitrogen, hydrogen, and ammonia ?
This section contains 2 paragraphs; each has 3 multiple
choice questions. (Questions 1 to 6) Each question has 4
choices (A), (B), (C) and (D) out of which ONE OR                                                H2




                                                                               Conc
MORE THAN ONE may be correct. Mark your                                  (A)                     NH3
response in OMR sheet against the question number of                                             N2
that question. + 4 marks will be given for each correct
answer and – 1 mark for each wrong answer.                                            Time
Passage # 1 (Ques. 1 to 3)
     The Haber Process shown below :                                                             N2
                                                                               Conc
                 N2(g) + 3H2(g)         2 NH3(g)                         (B)                     NH3
     In 1912 Fritz Haber developed the Haber process for                                         H2
     making ammonia from nitrogen and hydrogen. His
     development was crucial for the German war effort                                Time
     of world War Ι, providing the Germans with ample
                                                                                                 H2
                                                                               Conc




     fixed nitrogen for the manufacture of explosives.
     The Haber process takes place at 500° C and 200                     (C)                     NH3
     atm. It is an exothermic reaction .The graph below                                          N2
     shows the change in concentrations of reactants and
     products as the reaction progresses.                                             Time
                                                                                                 N2
                                                                               Conc
          Concentration




                                                                         (D)                     NH3

                                             A                                                   H2
                                             B                                        Time
                                             C                      Passage # 2 (Ques. 4 to 6)

                             Time                                        Entropy is measure of degree of randomness. Entropy
                                                                         is directly proportional to temperature. Every system
1.    Even at the high temperatures, the conversion of                   tries to acquire maximum state of randomness or
      nitrogen and hydrogen to ammonia was slow. In                      disorder. Entropy is measure of unavailable energy.
      order to make the process industrially efficient Fritz             Unavailable energy = Entropy × Temperature
      Haber used a metal oxide catalyst. Which of the
                                                                         The ratio of entropy of vapourisation and boiling
      following did not accomplished by use of the metal
                                                                         point of substance remains almost constant.
      oxide catalyst ?
      (A) The rate of production of ammonia increased               4.   Which of the following process have ∆S = – ve ?
      (B) The energy of activation was raised                            (A) Adsorption
      (C) The equilibrium shifted to the right                           (B) Dissolution of NH4Cl in water
      (D) activation energy equals to zero                               (C) H2 → 2H
                                                                         (D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O

XtraEdge for IIT-JEE                                           67                                        DECEMBER 2011
5.   Observe the graph and identify the incorrect                  This section contains 6 questions (Q.10 to 15).
     statement(s)                                                  +4 marks will be given for each correct answer and –1
                                                                   mark for each wrong answer. The answer to each of the
                                   Β ∆Svap                         questions is a SINGLE-DIGIT INTEGER, ranging
                                                                   from 0 to 9. The appropriate bubbles below the
              Entropy                                              respective question numbers in the OMR have to be
                         Α ∆Sfusion                                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 :
                            T1         T2                                                X Y Z W
                        Temperature                                                       0 0 0 0
     (A) T1 is melting point, T2 is boiling point                                         1 1 1 1
     (B) T1 is boiling point, T2 is melting point                                         2 2 2 2
     (C) ∆Sfusion is more than ∆Svap                                                      3 3 3 3
     (D) T2 is lower than T1                                                              4 4 4 4
                                                                                          5 5 5 5
6.   The law of Thermodynamics invented by Nernst,                                        6 6 6 6
     which helps to determine absolute entropy is                                         7 7 7 7
     (A) Zeroth law          (B) 1st law
          nd
                                                                                          8 8 8 8
     (C) 2 law               (D) 3rd law
                                                                                          9 9 9 9
This section contains 3 questions numbered 7 to 9, (Reason         10. For the Ca atom calculate total No. of e– which have
and Assertion type question). Each question contains                   m = –1
Assertion and Reason. Each question has 4 choices (A), (B),
                                                                   11. The molecular formula of a non-stoichiometric tin
(C) and (D) out of which ONLY ONE is correct. Mark your
response in OMR sheet against the question number of that              oxide containing Sn (II) and Sn (IV) ions is Sn4.44O8.
question. + 4 marks will be given for each correct answer              Therefore, the molar ratio of Sn (II) to Sn (IV) is
and – 1 mark for each wrong answer.                                    approximately
The following questions given below consist of an                  12.   How much volume (in mL) 0.001 M HCl should we
"Assertion" (A) and "Reason" (R) Type questions. Use                     add to 10 cm3 of 0.001 M NaOH to change its pH by
the following Key to choose the appropriate answer.                      one unit ?
      (A) If both (A) and (R) are true, and (R) is the
           correct explanation of (A).                             13.   What is the sum of total electron pairs (b.p. + l.p.)
                                                                         present in XeF6 molecule ?
      (B) If both (A) and (R) are true but (R) is not the
           correct explanation of (A).                             14. The number of geometrical isomers of
      (C) If (A) is true but (R) is false.                             CH3CH=CH–CH=CH–CH=CHCl is.
      (D) If (A) is false but (R) is true.                         15. No. of π bond in the compound H2CSF4 is.
7.   Assertion (A) : At zero degree Kelvin the volume
                                                                   This section contains Numerical response type questions
     occupied by a gas is negligible.
                                                                   (Q. 16 to 18). +6 marks will be given for each correct
     Reason (R) : All molecular motion ceases at 0 K.              answer and –1 mark for each wrong answer. Answers to
                                                                   this Section are to be given in the form of nearest integer-
8.   Assertion (A) : Compressibility factor for hydrogen           in four digits. Please follow as per example : (i.e. for
     varies with pressure with positive slope at all               answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write
     pressures.                                                    0092; 2.1 write 0002)
     Reason (R) : Even at low pressures, repulsive forces
                                                                   16.   The ion An+ is oxidised to AO3– by MnO4–, changing
     dominate hydrogen gas.
                                                                         to Mn2+ in acidic solution. Given that 2.68 × 10–3 mol
                                                                         of An+ requires 1.61 × 10–3 mol of MnO4–. What is
9.   Assertion (A) : Enthalpy of graphite is lower than
                                                                         the value of n ?
     that of diamond.
     Reason (R) : Entropy of graphite is greater than that         17.   Standard heat of formation of HgO(s) at 298 K and at
     of diamond.                                                         constant pressure is –90.8 kJ mol–1. Excess of
                                                                         HgO(s) absorbs 41.84 kJ of heat. Calculate the mass
                                                                         of Hg (in g) that can be obtained at constant volume
                                                                         and at 298 K. (Hg = 200.6g mol–1)

XtraEdge for IIT-JEE                                          68                                          DECEMBER 2011
18.   A near ultraviolet photon of 300 nm is absorbed by a            6.   If BL2 + CM2 + AN2 = x and CL2 + AM2 + BN2 = y
      gas and then re-emitted as two photons. One photon                   then
      is red with wavelength 760 nm. What would be the                     (A) x + y = 0
      wavelength (in nm) of the second photon ?                            (B) x – y = 0
                                                                           (C) x + y = 24R2 – 2(a2 + b2 + c2)
                                                                           (D) none of these
            MATHEMATICS
                                                                      This section contains 3 questions numbered 7 to 9,
This section contains 2 paragraphs; each has 3 multiple               (Reason and Assertion type question). Each question
choice questions. (Questions 1 to 6) Each question has 4              contains Assertion and Reason. Each question has 4
choices (A), (B), (C) and (D) out of which ONE OR                     choices (A), (B), (C) and (D) out of which ONLY ONE
MORE THAN ONE may be correct. Mark your                               is correct. Mark your response in OMR sheet against
response in OMR sheet against the question number of                  the question number of that question. + 4 marks will be
that question. + 4 marks will be given for each correct               given for each correct answer and – 1 mark for each
answer and – 1 mark for each wrong answer.                            wrong answer.
                                                                      The following questions given below consist of an
Passage # 1 (Ques. 1 to 3)                                            "Assertion" (A) and "Reason" (R) Type questions. Use
     A circle C1 of radius 2 units rolls on the outerside of          the following Key to choose the appropriate answer.
     the circle C2 : x2 + y2 + 4x = 0, touching it externally.
                                                                            (A) If both (A) and (R) are true, and (R) is the
1.   If the line joining the centres of C1 and C2 makes an                      correct explanation of (A).
     angle of 60º with the x-axis, equation of a common
                                                                            (B) If both (A) and (R) are true but (R) is not the
     tangent to them is -
                                                                                correct explanation of (A).
      (A) x +   3y–2=0                                                      (C) If (A) is true but (R) is false.
      (B)   3x–y+4+2 3 =0                                                   (D) If (A) is false but (R) is true.

      (C)   3x–y–4+2 3 =0                                             7.   Assertion (A) : If a, b > 0 and a3 + b3 = a – b, then
                                                                           a2 + b2 < 1
      (D)   3x–y–4–2 3 =0                                                                                        1
                                                                           Reason (R) : If a, b > 0, then ab <     (a + b)
                                                                                                                 2
2.    Area of the quadrilateral formed by a pair of tangents
      from a point on C3 to the circle C2 with a pair of radii
      at the points of contact of the tangents is -                   8.   Assertion (A) : The line bx – ay = 0 will not meet the
                                                                                      x2     y2
      (A) 2 3 sq. units          (B) 4 3 sq. units                         hyperbola 2 – 2 = 1 (a > b > 0)
                                                                                      a      b
      (C)   3 sq. units          (D) 3 3 sq. units                         Reason (R) : The line y = mx + c does not meet the
                                                                                     x2 y2
3.    If the line joining the centres of C1 and C2 is                      hyperbola 2 – 2 = 1 if c2 = a2 m2 – b2.
      perpendicular to the x-axis; equation of the chord of                          a     b
      contact of the tangents drawn from the centre of C2 to
      the circle C1 is -                                              9.   Assertion (A) : The solution set of the inequality
      (A) y – 2 = 0             (B) y + 2 = 0                                              x2 + x 
      (C) y – 3 = 0              (D) y + 3 = 0                              log 0.7  log 6         < 0 is (– 4, – 3) ∪ (8, ∞).
                                                                                            x+4 
                                                                                                  
Passage # 2 (Ques. 4 to 6)                                                 Reason (R) : For x > 0, loga x is an increasing function
     AL, BM and CN are diameter of the circumcircle of a                   if a > 1 and a decreasing function if 0 < a < 1.
     triangle ABC. ∆1, ∆2, ∆3, and ∆ are the areas of the
     triangles BLC, CMA, ANB and ABC respectively,                    This section contains 6 questions (Q.10 to 15).
                                                                      +4 marks will be given for each correct answer and –1
4.   ∆1 is equal to
                                                                      mark for each wrong answer. The answer to each of the
     (A) 2R2 sin A cos B cos C                                        questions is a SINGLE-DIGIT INTEGER, ranging
     (B) 2R2 sin A sin B cos C                                        from 0 to 9. The appropriate bubbles below the
     (C) 2R2 cos A cos B sin C                                        respective question numbers in the OMR have to be
     (D) 2R2 sin A sin B sin C                                        darkened. For example, if the correct answers to
                                                                      question numbers X, Y, Z and W (say) are 6, 0, 9 and 2,
5.    ∆1 + ∆2 + ∆3 is equal to                                        respectively, then the correct darkening of bubbles will
      (A) 2∆                     (B) 3∆                               look like the following :
      (C) ∆                      (D) none of these


XtraEdge for IIT-JEE                                             69                                          DECEMBER 2011
                         X    Y   Z    W
                          0
                          1
                              0
                              1
                                  0
                                  1
                                       0
                                       1
                                                                            Elements Named for Places
                          2   2   2    2
                                                                            This is an alphabetical list of element toponyms or
                          3   3   3    3
                                                                            elements named for places or regions. Ytterby in
                          4   4   4    4                                    Sweden has given its name to four elements: Erbium,
                          5   5   5    5                                    Terbium, Ytterbium and Yttrium.
                          6   6   6    6
                          7   7   7    7                                    •   Americium : America, the Americas
                          8   8   8    8                                    •   Berkelium : University of California at Berkeley
                          9   9   9    9
                                                                            •   Californium : State of California and University
10.   If two of the lines represented by                                        of California at Berkeley
      x4 + x3 y + cx2 y2 – xy3 + y4 = 0
      bisect the angle between the other two, then the value                •   Copper : probably named for Cyprus
      of |c| is                                                             •   Darmstadtium : Darmstadt, Germany
11.   For n > 3, a, b ∈ R, let
                                                                            •   Dubnium : Dubna, Russia
                    n
                              n(n – 1)...(n – r + 1)
      S(n, a, b)= ∑
                  r =0
                       (–1) r
                                        r!
                                                     (a – r) (b – r)        •   Erbium : Ytterby, a town in Sweden
                                                                            •   Europium : Europe
              1 1
      Find S  5, , 
              2 7                                                         •   Francium : France

12.   Find the degree of the remainder when x2007 – 1 is                    •   Gallium : Gallia, Latin for France. Also named
      divided by (x2 + 1) (x2 + x + 1).                                         for Lecoq de Boisbaudran, the element's
                                                                                discoverer (Lecoq in Latin is gallus)
13.   Let l be the length of the interval satisfying the
      inequality                                                            •   Germanium : Germany
                                             1
      log6(x + 2) (x + 4) + log1/6 (x + 2) <   log 6 (7).                   •   Hafnium : Hafnia, Latin for Copenhagen
                                             2
      Find the value of l.                                                  •   Hassium : Hesse, Germany
14.   The number of pairs (x, y) satisfying the equation                    •   Holmium : Holmia, Latin for Stockholm
      sin x + sin y = sin (x + y) and |x| + |y| = 1 is.
                                                                            •   Lutetium : Lutecia, ancient name for Paris
15.   If a = ( 0, 1, – 1) and c = (1, 1, 1) are given vectors,
      the |b|2 where b satisfies a × b + c = 0 and a . b = 3 is.            •   Magnesium : Magnesia prefecture in Thessaly,
                                                                                Greece
This section contains Numerical response type questions
(Q. 16 to 18). +6 marks will be given for each correct                      •   Polonium : Poland
answer and –1 mark for each wrong answer. Answers to
this Section are to be given in the form of nearest integer-                •   Rhenium : Rhenus, Latin for Rhine, a German
in four digits. Please follow as per example : (i.e. for                        province
answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write
0092; 2.1 write 0002)                                                       •   Ruthenium : Ruthenia, Latin for Russia
16.   The number of natural numbers which are smaller                       •   Scandium : Scandia, Latin for Scandinavia
      than 2.108 and which can be written by means of the
      digits 1 and 2 is ............. .                                     •   Strontium : Strontian, a town in Scotland
             1                                                              •   Terbium : Ytterby, Sweden
17.   If z =   ( 3 – i), find the smallest value of positive
             2                                                              •   Thulium : Thule, a mythical island in the far
      integer n for which (z89 + i97)94 = zn.
                                                                                north (Scandinavia?)
18.   If α + β = γ and tan γ = 22, a is the arithmetic and b is
                                                                            •   Ytterbium : Ytterby, Sweden
      the geometric mean respectively between tan α and
                                    a3                                      •   Yttrium : Ytterby, Sweden
      tan β, then the value of               is equal to -
                                (1 – b 2 ) 3

XtraEdge for IIT-JEE                                                   70                                      DECEMBER 2011
                                 MOCK TEST PAPER-1
                                       CBSE BOARD PATTERN
                                                  CLASS # XII
           SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS
                                    Solutions will be p ublished in next issue
 General Instructions : Physics & Chemistry
 •   Time given for each subject paper is 3 hrs and Max. marks 70 for each.
 •   All questions are compulsory.
 •   Marks for each question are indicated against it.
 •   Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each.
 •   Question numbers 9 to 18 are short-answer questions, and carry 2 marks each.
 •   Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each.
 •   Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
 •   Use of calculators is not permitted.
 General Instructions : Mathematics
 •   Time given to solve this subject paper is 3 hrs and Max. marks 100.
 •   All questions are compulsory.
 •   The question paper consists of 29 questions divided into three sections A, B and C.
     Section A comprises of 10 questions of one mark each.
     Section B comprises of 12 questions of four marks each.
     Section C comprises of 7 questions of six marks each.
 •   All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
 •   There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and
     2 question of six marks each. You have to attempt only one of the alternatives in all such questions.
 •   Use of calculators is not permitted.


                   PHYSICS                                         10. Explain how the interference fringe width in young's
                                                                       double slits experiment will change if
1.   What is the phase difference between two points on a              (i) separation between the two slits is decreased
     wave front                                                        (ii) wavelength of light increased.
2.   Why stars seems to be twinkling ?                             11. In an atom, two electrons moves around the nucleus
3.   Name the characteristics of electromagnetic wave that             in a circular orbit of radius R and 4R. Calculate the
     (i) increases                                                     ratio of speeds of two electrons.
     (ii) remains constant
                                                                   12. Which type of semiconductor is better out of
     in the electromagnetic spectrum as one moves from
                                                                       p-type and n-type ?
     infrared to ultraviolet region.
4.   What is total energy of an electron revolving in first        13. Relate input frequency and output frequency of a half
     orbit of Hydrogen atom .                                          wave rectifier and a full wave rectifier
5.   What are digital and analog signals ?
                                                                   14. Define the term modulation index for an AM wave.
6.   Name any two basic properties of electric charge.                 What would be the modulation index for an AM
                                                                       wave for which the maximum amplitude is a while
7.   Out of the two bulbs marked 25 W and 100 W which
                                                                       the minimum amplitude is 'b' ?
     one has higher resistance ?
8.   What is displacement current ?.                               15. Distinguish between point to point and broadcast
                                                                       communication modes. Give one example of each
9.   A thin glass prism has a minimum deviation δm in air.
     State with reason, how the angle of minimum                   16. Why does the conductivity of a semi conductor
     deviation will change if the prism is immersed in a               change with the rise in temperature ?
     liquid of refractive index greater than 1

XtraEdge for IIT-JEE                                          71                                        DECEMBER 2011
17. An electric dipole is held in uniform electric field.               Why objective lens of an astronomical telescope is of
    (i) Show that no translatory force acts on it.                      large size ? what defects can be remoned by using
    (ii) Derive an expression for the torque acting on it.              reflecting telescope in place of refracting telescope ?
                                                                        IF length of an astronomical telescope is 100 cm and
18. Define angle of dip at a given place. What is the                   its magnifying power is 19 then calculate focal
    value of angle of dip on the equator ?                              lengths of objective and eye-piece lens.
19. When a Uranium nucleus (U238) originally at rest
    decays by emitting alpha particle having speed u.              29. Explain with the help of a labelled diagram, the
    Find the recoil speed of residual nucleus.                         principle, construction and working of a transformer.
                                                                                               OR
20. If focal lengths of objective and eye-piece lens of a              An a.c. generator consists of a coil of 50 turns and
    compound microscope is 2cm and 3cm respectively                    area 2.5 m2 rotating at an angular speed of 60 rad s–1
    and distance between both the lenses is 15cm then                  in a uniform magnetic field B = 0.30 T between two
    calculate distance of object from objective lens if                fixed pole pieces. The resistance of the circuit
    final image forms at infinity.                                     including that of the coil is 500 Ω. (i) Find the
                                                                       maximum current drawn from the generator. (ii) What
21. If a gas is at temp T, then derive an expression for               will be the orientation of the coil w.r.t. the magnetic
    debroglie wavelength of its molecule of mass m.                    field to have (a) maximum (b) zero magnetic flux. (iii)
                                                                       Would the generator work if the coil were stationary
22. For a common emitter transistor amplifier current                  and instead the poles were rotated with same speed as
    gain is 72. Calculate the base current for which                   above.
    emitter current is 8.9mA.
                                                                   30. Calculate the electric field intensity for following points
23. Give one use of each of the following                              due to a uniformly charged non-conducting sphere.
    (i) microwaves                                                     Represent the results by graph –
    (ii) infra-red waves                                               (A) at any point outside the sphere
    (iii) ultraviolet radiation                                        (B) at any point on the surface of sphere
    (iv) gamma rays.                                                   (C) at any point inside the sphere.
                                                                                                OR
24. State the principle of potentiometer with the help of
    circuit diagram, describe a method to find the                     What is a Capacitor ? Explain its principles. Derive
    internal resistance of a primary cell.                             the relations for equivalent capacity of series and
                                                  r                    parallel combinations of capacitors.
25. A proton is shot into the magnetic field B = 0.8ˆ T j
    with a velocity (2 × 106 ˆ + 3 × 106 ˆ) ms–1. Calculate
                              i          j                                         CHEMISTRY
      the radius and pitch of the helix path followed by
      proton.                                                      1.   Write the IUPAC names of the following :
                                                                                                Br
26. Derive an expression for the torque on a rectangular
    coil of area A, carrying a current I placed in a
    magnetic field B. The angle between the direction of                                             OH
    B and vector perpendicular to plane of coil is θ.                                         COOC2H5
27. Using Kirchhoff's laws in the given network, calculate
    the values of I1, I2 and I3.                                   2.   How would         you    convert    methylamine      into
                                                                        ethylamine ?
         A                     B                C
                                                                   3.   What are essential amino acids ? Give two example.
       5Ω       I1                             3Ω       I2
                         2Ω       I3
                                                                   4.   What is meant by the term peptization ?
      12V                                               6V
                                                                   5.   Identify the reaction order if the unit of rate constant
            F                 E                     D                   is sec–1.

28. With the help of ray diagram, explain the                      6.   In an alloy of gold and cadmium if gold crystallizes
    phenomenon of total interval reflection. Obtain the                 in cubic structure occupying the corners only and
    relation between critical angle and refractive index of             cadmium fits into edge centre voids, what is the
    the medium.                                                         formula of the alloy ?
                           OR


XtraEdge for IIT-JEE                                          72                                            DECEMBER 2011
7.   Which oxide of nitrogen has oxidation number of N                 19. Name the reagents which are used in the following
     same as that in nitric acid.                                          conversions :
                                                                           (i) A primary alcohol to an aldehyde
8.   What is froth floatation for which ores it is used ?                  (ii) Butan-2-one to butan-2-ol
                                                                           (iii) Phenol to 2,4,6-tribromopheno
9.   Explain as to why haloarenes are much less reactive
     than haloalkanes towards nucleophilic substitution                20. Give the structures of following polymers.
     reactions.                                                            (A) Perlon      (B) Orlon       (C) Neoprene

10. Assign a reason for each of the following statements :             21. What are tranquilizers ? Explain with example
    (i) Alkylamines are stronger bases than arylamines                     tranquilizers are neurologically active drugs which are
    (ii) Acetonitrile is preferred as solvent for carrying                 used to reduce strain or anxiety.
        out several organic reactions.
                                                                       22. Name any three fat soluble vitamins & their
11. Write one chemical equation each to exemplify the                      deficiency disease.
    following reactions :
    (i) Carbylamine reaction                                           23. (a) State the products of electrolysis obtained on the
    (ii) Hofmann bromamide reaction                                             cathode and the anode in the following
                                                                                cases :                                [2 + 1]
12. Identify A and B in the following :                                    (i) A dilute solution of H2SO4 with platinum
          CH2Br                                                                 electrodes.
                                                                           (ii) An aqueous solution of AgNO3 with silver
                   CN–           LiAlH4
     (i)                  A                B                                    electrodes
                                                                           (b) Write the cell formulation and calculate the
     (ii) R2CO      NH3
                           A
                                 Ni/H2
                                          B                                     standard cell potential of the galvanic cell in
                                                                                which the following reaction takes place :
13. Write any two feature which                    distinguish                  Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s)
    physisorption from Chemisorption.                                      Calculate ∆rGº for the above reaction.
                                                                           [Given : E o + / Ag = + 0.80 V ;
                                                                                        Ag
14. The decomposition of a compound is found to follow
     a first order rate law. If it takes 15 minutes for 20% of              E o 3+ / Fe 2 + = + 0.77 V ; 1 F = 96500 C mol–1]
                                                                              Fe
     the        original         material       to       react,
     calculate-                                                        24. (a) Explain each of the following with a suitable
     (i) Specific rate constant.                                           example :
    (ii) the time at which 10% of the original material                    (i) Paramagnetism (ii) Frenkel defect in crystals
          remains unreacted.                                               (b) An element occurs in bcc structure with cell edge
                                                                           300 pm. The density of the element is
15. Prove that the time required for the completion of                     5.2 g cm–3. How many atoms of the element does 200
    3/4th of the reaction of first order is twice the                      g of the element contain ?
    time required for the completion of half of the
    reaction.                                                          25. (a) Describe the preparation of KMnO4 from
                                                                                pyrolusite ore.
16. When a certain conductivity cell was filled with 0.1 M                 (b) Among ionic species, Sc+3, Ce+4 and Eu+2, which
    KCl, it has a resistance of 85 Ω at 25ºC. When the                          one is good oxidizing agent ?
    same cell was filled with an aqueous solution of 0.052                  (Atomic numbers : Sc = 21, Ce = 58, Eu = 63 )
    M unknown electrolyte the resistance was 96 Ω.
                                                                       26. (a) Assign a reason for each of the following
    Calculate the molar conductivity of the unknown
                                                                                statements :
    electrolyte   at    this    concentration.   (Specific
                                                                           (i) Di-methyl amine is stronger base than
    conductivity of 0.1 M KCl = 129 × 10–2 ohm–1cm–1).
                                                                                tri-methyl amine in aq. solution.
                                                                           (ii) Explain why ? Benzamide is more basic in
17. Give methods of preparation of XeO3 and XeOF4.
                                                                                comparision to acetamide.
18. Explain giving reason :                                                 (b) How would you convert aryl amine into
    (i) Copper (I) is diamagnetic whereas copper (II) is                        cynobenzene.
         paramagnetic
    (ii) K2PtCl6 is a well known compound whereas the
         corresponding Ni compound does not exist.


XtraEdge for IIT-JEE                                              73                                            DECEMBER 2011
27. (a) Why chelated complexes are more stable than
          unchelated complexes ?                                                 MATHEMATICS
     (b) Write IUPAC names of :
     (i) K3 [Al(C2O4)3]
     (ii) [Mn (H2O)6] S                                                                              Section A

28. You are provided with four reagents :
     LiAlH4, I2/NaOH, NaHSO3 and Schiff's reagent                   1.   Find order and degree of diff. equation
                                                                                 2
     (a) Write which two reagents can be used to                          dy  1
          distinguish between the compounds in each of the                 +     =2
                                                                          dx  dy
          following pairs :
                                                                                dx
          [3+2=5]
     (i) CH3CHO and CH3COCH3
                                                                                      ∫ tan
                                                                                                −1
     (ii) CH3CHO and C6H5CHO                                        2.   Evaluate :                  x dx .
     (iii)     C6H5COCH3 and C6H5COC6H5
     (b) Account for the following :
     (i) The order of reactivity of halogen acids with ether        3.   Find unit vector in the direction of vector
                                                                         →
               is HI > HBr > HCl.                                                      ˆ
                                                                         a = 2i + 3 ˆ +k .
                                                                              ˆ     j
     (ii) The pKa value of chloroacetic acid is lower than
               the pKa value of acetic acid.
                                                                             →                            →                        →    →
                              Or                                    4.          ˆ j       ˆ               ˆ
                                                                         If a = i + ˆ – 3 k and b = ˆ + 2 k . Find |2 b × a |
                                                                                                    j
     (a) An organic compound contains 69.77% carbon,
           11.63% hydrogen and the rest is oxygen. The
           molecular mass of the compound is 86. It does            5.   Find the angle between the                          two       planes
           not reduce Tollens' reagent but forms an addition             3x – 6y – 2z = 7 and 2x + 2y – 2z = 5
           compound with sodium hydrogen sulphite and
           gives a positive iodoform test. On vigorous              6.   Construct a matrix of order 3 × 3, whose element aij
           oxidation it gives ethanoic and propanoic acids.              is given by rule aij = (i + j)2.
           Deduce the possible structure of the organic
           compound.                                                7.   Find a matrix X such that A + 2B + X = 0, where
    (b) State reasons for the following :                                    2 − 1       − 1 1
                                                                         A=          ;B=         .
          (i) Monochloroethanoic acid has a higher pKa
                                                                              3 5        0 2
               value than dichloroethanoic acid.
          (ii) Ethanoic acid is a weaker acid than benzoic
               acid.
                                                                                               sin 10º − cos 10º
                                                                    8.   Find value of                           .
29. (a) Density of 0.8 M aqueous solution of H2SO4 is                                          sin 80º cos 80º
        1.06 g mL–1. Calculate the concentration of
        solution in (i) mol fraction (ii) molality (molar           9.   Let f = {(1, 3), (2, 1), (2, 1), (3, 2)} and
        mass of H2SO4 = 98 g mol–1).                                     g = {(1, 2), (2, 3), (3, 1)} then find gof (1).
    (b) Heptane and octane form ideal solution. At 373
        K, the vapour pressures of the two liquid                   10. Find derivative of sin–1 (2 x                   1 − x 2 ) w.r.to x.
        components are 105.2 kPa and 46.8 kPa,
        respectively. What will be the vapour pressure,
        in bar, of a mixture of 25.0 g of heptane and 35.0
        g of octane ?                                                                                Section B

30. (a) Explain the following giving reasons :                      11. A and B are mutually exclusive events of an
        (i) H3PO3 is diprotic.                                          experiment. If P (Not A) = 0.65, P (A ∪ B) = 0.65
        (ii) Nitrogen does not form pentahalides                        and P (B) = p. Find the value of p.
        (iii) SF6 is well known but SH6 is not known
    (b) Complete the equations :                                                          x 1 + sin    x cos x 
        (i) XeF6 + PF5 →
                                                                    12. Evaluate     ∫e    
                                                                                                    cos 2 x
                                                                                                                 dx.
                                                                                                                
        (ii) AlN + H2O →
                                                                                                     cos θ
                                                                    13. Evaluate     ∫ (2 + sin θ)(3 + sin θ) dθ
                                                                                                         OR

XtraEdge for IIT-JEE                                           74                                                   DECEMBER 2011
                         3x + 1
     Evaluate :   ∫ 2x   2
                             − 2x + 3
                                         dx .                                                             Section C
                                                                             23. The bag A contains 5 red and 3 green balls and bag B
14. Find the diff. equation of the family of curves                              contains 3 red and 5 green balls. One ball is drawn
    y = a sin (bx + c), a and c being parameter.                                 from bag A and two from bag B. Find the Probability
                            OR                                                   that of the three balls drawn two are red and one is
                                                                                 green.
                                        dy
    Solve the differential equation : x    − y − 2x 3 = 0 .
                                        dx                                   24. Find area between x2 + y2 = 4 and line x = 3 y in
          →                      →                      →
           ˆ j       ˆ        ˆ j       ˆ       ˆ    j ˆ
15. If a = i – ˆ + 2 k , b =2 i + ˆ – 3 k , c = i +2 ˆ – k ,                     first quadrant
                                                                                                                 OR
                  →          →   →         → →      →   → →   →
     verify that a × ( b × c ) = ( a . c ) b –( a . b ) c .                      Find the area of the region
                                                                                 [(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9]
16. Find the vector equation of the plane passing through
                                                →                            25. Find the image of the point (1, 6, 3) in the line
                                          ˆ          ˆ
     the intersection of the planes r .(2 i –7 ˆ + 4 k ) =3
                                               j
                                                                                  x y −1 z − 2
           →                                                                        =     =     .
                ˆ          ˆ
     and r .(3 i – 5 ˆ + 4 k ) + 11=0 and passing through
                       j                                                         1     2     3
     the point (–2, 1, 3).                                                   26. A dealer wishes to purchase a number of fans and
                            OR                                                   radios. He has only Rs 5,760 to invest and has a
     Find the distance of the point (2, 3, 4) from the plane                     space for at most 20 items. A fan costs him Rs. 360
     3x + 2y + 2z + 5 = 0, measured parallel to the line                         and a radio Rs. 240. His expectation is that he can
      x+3 y−2 z                                                                  sell a fan at a profit of Rs. 22 and a radio at a profit of
           =        =                                                            Rs. 18. Assuming that he can sell all the items he
        3       6       2
                                                                                 buys, how should he invest his money for maximum
                                                                                 profit ? Translate the problem as LPP and solve it
17. Using properties of determinats, prove that                                  graphically.
     1+ a    1      1
                                1 1 1                                      27. Evaluate the following integrals as limit of sums
      1    1+ b     1 = abc 1 + + + 
                                a b c                                            2
             1 1+ c
                                                                                   ∫ (x
      1                                                                                   2
                                                                                              + 3) dx .
                                 OR                                                0
            4 − 5 − 11                                                                            1 − 1 0 
     If A = 1 − 3 1  find A–1
                                                                           28.   Given that A = 2 3 4 and
                                                                                                            
            2 3 − 7 
                                                                                                  0 1 2
                                                                                                            
                                                                                         2     2 − 4
18. If f (x) = cos x, g(x) = x2, then show that
                                                                                   B = − 4 2 − 4 find A.B. Use this to solve the
                                                                                                      
    fog (x) ≠ gof (x)
                                                                                         2 −1 5 
                                                                                                      
19. Show that curve xy = a2 and x2 + y2 = 2a2 touch each                           following system of equations
    other.                                                                         x–y=3
                                                                                   2x + 3y + 4z = 17
20. Find the derivative of xex from first principle.                               y + 2z = 7
                                                                                                               OR
21. If cos y = x cos (a + y), then prove that                                      Using matrix method solve the following system of
                                                                                   linear equations
     dy cos 2 (a + y )
        =               .                                                          2x – 3y + 5z = 11
     dx        sin a                                                               3x + 2y – 4z = –5
22. Determine f (0) so that          function f (x) is defined by                  x + y – 2z = –3
                  (4 x − 1) 3                                                29. If the sum of the lengths of the hypotenuse and a side
    f (x) =                              becomes continuous at x = 0.
                 x         x2                                                of right angled triangle is given, show that the area of
            sin   log1 +          
                4                 
                              3                                                  triangle is maximum when the angle between them is
                                                                               π/3.




XtraEdge for IIT-JEE                                                    75                                            DECEMBER 2011
                            XtraEdge Test Series
                                ANSWER KEY
                            IIT- JEE 2012 (December issue)
                                      PHYSICS
  Ques        1        2    3        4           5       6       7      8       9
  Ans        A,C       C    A        D           A       A       A      D       D
  Ques       10        11   12       13          14      15      16     17      18
  Ans         3        4    6        4           8       7      0012   0049    0005

                                    C HE M ISTR Y
  Ques       1         2    3        4           5        6      7      8       9
  Ans        A         B    A        C          A,B     A,B,C    A      A       D
  Ques       10        11   12       13         14       15      16     17      18
  Ans        4         4    3        1           4        6     0108   0001    0041

                                   MATHEMATICS
  Ques        1        2     3       4           5       6       7      8       9
  Ans       B,C,D      A    C,D      C           C      B,D      D      C       D
  Ques       10        11   12       13          14     15       16     17      18
  Ans         6        3     5       9           6       3      0012   1440    0000



                            IIT- JEE 2013 (December issue)
                                      PHYSICS
  Ques       1         2    3        4           5       6       7      8       9
  Ans        A         A    B        C           B       D       B      D       D
  Ques       10        11   12       13          14      15      16     17      18
  Ans        6         7    1        4           5       4      0200   0004    0008

                                    C HE M ISTR Y
  Ques        1        2    3        4            5      6       7      8       9
  Ans       B,C,D      B    A        A          B,C,D    D       C      A       B
  Ques       10        11   12       13          14      15      16     17      18
  Ans         4        1    1        6            1      4      0002   0094    0496

                                   MATHEMATICS
  Ques        1        2     3       4           5       6       7      8       9
  Ans       A,B,C      B    C,D      A           C      B,C      B      C       A
  Ques       10        11   12       13          14     15       16     17      18
  Ans         6        0     3       5           6       6      0766   0010    1331




XtraEdge for IIT-JEE                       76                          DECEMBER 2011

				
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