XtraEdge Dec 2010

XtraEdge for IIT-JEE 1 DECEMBER 2010

Dear Students, The difference between success and failure is your attitude towards success and the strategies that you employ to achieve it. The difference between success and failure is only a few minutes or a few hours everyday. You have to keep on striving for success at every conceivable opportunity. Never postpone your happiness and zest for life and work. You should make it a habit to enjoy your profession and your job all the time. Never be a quitter because a quitter can never be a winner. You should always remember that People live not by the reason of any care they have for themselves but by the love for them that is in other people. Have only those people for friends and companions who do their best to bring out the best in you. They will be of unlimited worth to you. Such persons understand what life means to you and your goal. They feel for you as you feel for yourselves. They are the ones who are bound to you in triumph and disaster. They provide a purpose to live and break the spell of loneliness. A true friend is worth befriending as he will always stand by you. But before you expect others to be the right person to be your friend you must also become one. Be always committed to your cause. Be so engrossed in your work that you have hardly any time to think of anything else. The great secret of success is to do whatever you are to do and do it wholeheartedly. Make yourself the star of your workplace. For this you must have clear and precise objectives to be achieved within a definite time-frame. Always respect and value time. Be result-oriented and keep track of the hours. Respect the time of others as well as your own. Be always organized and write down everything you want to accomplish. Always make an assessment of yesterday's "To Do" list to crosscheck how realistic it has turned out to be today. This will help you to avoid or rectify mistakes, if any, in your planning. Keep on visualizing your goals and lists of the task to be done. Forever presenting positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Do not allow the quest for perfection to ruin your life because whatever you do you will always feel that you could have done better

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December, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000) 324009 e-mail : [email protected]

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Volume-6 Issue-6 December, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics,, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

Xtra Edge Test Series for JEE- 2011 & 2012

CBSE Mock Test Paper

S

Success Tips for the Months

• If one asks for success and prepares for failure, he will get the situation he has prepared for.

• Loser's visualize the penalties of failure. Winner's visualize the rewards of success.

• Treat others as if they were what they ought to be and you help them to become what they are capable of being.

• You never achieve real success unless you like what you are doing

• The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself.

• Believe in yourself ! Have faith in your abilities ! without a humble but reasonable confidence in your own powers you can not be successful or happy.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 President inaugurates Pan IIT 2010 conclave Yale University signs MoU with IIT-K, IIM-K

IITian ON THE PATH OF SUCCESS 6 Mr. Sheerang Chhatre

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 60

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper Mock Test CBSE Pattern Paper -1 [Class # XII]

Regulars ..........

DYNAMIC PHYSICS 15

8-Challenging Problems [Set# 7] Students’ Forum Physics Fundamentals Ray Optics Fluid Mechanics & Properties of Matter CATALYSE CHEMISTRY 36

Key Concept Carboxylic Acid Chemical Kinetics Understanding : Physical Chemistry

DICEY MATHS 46

Mathematical Challenges Students’ Forum Key Concept Monotonic Maxima & Minima Function

Study Time........

Test Time ..........


President inaugurates Pan IIT 2010 conclave

New Delhi: President Pratibha Devisingh Patil on Friday inaugurated the PanIIT 2010 conclave, with alumni from all the Indian Institutes of Technology converging for a three-day networking and brainstorming session.

PanIIT is an umbrella organization covering alumni of all Indian Institutes of Technology. Over the years, these conferences have become one of the leading technology summits for business leaders. This year the PanIIT 2010 Global Conference focuses on good governance, knowledge based economy, entrepreneurial, innovative, equivalent and happy society. Speaking to delegates through a video message, President Patil said, "The theme of your conclave, Sustainable Transformation: Our New India, is very relevant to the changes taking place around us.""I am happy to know that you are going to deliver it on the need for better quality of life in the society, environment sustainability and to the imperative of developing low carbon technologies," she added. There are currently 15 IITs in the country.

IT BHU (Banaras Hindu University) is also to be upgraded as an Indian Institute of Technology. This year's conclave will attract coincides with the golden jubilee year of IIT Delhi and IIT Kharagpur.

"Government is undertaking various schemes and initiatives which should lead to the creation of a new India. Your expertise and deliberations at this conclave can help chart out ways in which capacity building and delivery mechanisms in many of these initiatives can be implemented," said the President.

Yale University signs MoU with IIT-K, IIM-K New Delhi: Indian Institute of Management (IIM)-Kozhikode, Indian Institute of Technology (IIT)-Kanpur and Yale University, USA have entered into a partnership to advance higher education in India through academic leadership development programmes for higher education leaders in India and through research on Indian higher education.

A Memorandum of Understanding (MoU) in this regard was signed between Dr. Richard Levin, President of Yale University, Dr. Debashis Chatterjee, Director of IIM - Kozhihode, and Dr. Sanjay Dhande, Director of IIT – Kanpur.

Speaking on the occasion, Kapil Sibal said that this partnership, which will take effect from January 2011, will be sited in two new Centers of Excellence for Academic Leadership (CEAL) to be established at IIM - Kozhikode and IIT -Kanpur."The partnership will begin with a term of five years, and could be renewed thereafter," he added. He also said that a six member committee with equal participation from the three partnering institutes will determine the norms/qualifications for participating in these leadership programmes.

The flagship programme of the partnership will be a new "India - Yale University Leadership Programme," to be developed by Yale University in

consultation with IIM - Kozhikode and IIT - Kanpur, that will expose university and academic leaders in India at the levels of vice-chancellor, director, and deans to the best practices of academic administration and institutional management in the United States.

Yale University, IIM - Kozhikode, and IIT - Kanpur would also engage in joint faculty research on higher education and collaborate to organize workshops and seminars on relevant areas of academic administration and leadership. The first programmes under the agreement would take place in 2011 in New Haven, Connecticut.

Yale President Richard Levin stated, "Yale is pleased to undertake this important and much needed effort on higher education with IIM - Kozhikode and IIT - Kanpur. We look forward to working with them to advance the cause of higher education in India by sharing what we have learned over three centuries as an institution and we similarly look forward to learning from our partners in India in this age of global education."

At IIT-B, juniors give alumni a lesson on altruism... At IIT-B, juniors give alunmi a lesson on altruism...Mihika Basu Inspired by the 2010 batch, more ex-students want to contribute 1% of their salary to the institute

After the graduating batch of 2010 of the Indian Institute of Technology (IIT), Mumbai, pledged 1% of its salary in an uncommon gesture of giving something back to their alma mater, many former students now want to follow suit. Accordingly, the initiative is been scaled up considerably, and a formal


set-up will soon be launched to enable alumni to submit 1% of their salary via various mediums. The initiative, called ‘Give One for IIT Mumbai’, is part of a new fund-raising initiative by the IIT Mumbai Alumni Association (IITBAA). It is a voluntary programme, where each alumnus can contribute 1% of his/her income towards supporting overall improvement at IIT Mumbai. “Several former students are keen to contribute 1% of their incomes to the institute,"said Damayanti Bhattacharya, chief operating officer, IITBAA.

"The IITBAA will collect the funds on behalf of the institute. The set-up will be launched on December 26, on the institute alumni’s day. It will enable former students to choose their own payment mode. They will get periodic updates about how their funds are being used and will be invited to see it in action”.In August this year, over 50% of the 2010 graduating class of IIT-B had committed 1% of their annual salary to the institute. It was the first batch of IIT-B to have committed to start its payback even before getting their first pay cheque. “Their commitment has created a new tradition for all batches,” said Bhattacharya.

The funds will be channelised towards various developmental activities, including aspects like faculty and student development and growth of departments and hostels. “Development of the institute as a whole and alumni benefit are other areas for which the alumni can donate,” said Bhattacharya. Alumni can also choose their own area of contribution. “After fresh graduates made their pledge, even alumni who had graduated nearly 35 years ago were very excited, and said they wanted to be part of the movement. Hence, we are trying to launch an online donation mechanism by December-end for all alumni,” said Bakul Desai, member of the board of directors at IITBAA, in charge of fundraising.

Metallurgist of the Year 2010 The Indian Institute of Metals has elected Dr. N. Eswara Prasad Scientist 'G' & Regional Director Regional

Centre for Military Airworthiness Materials, Hyderabad, who is a graduate of the 1985 batch from IT-BHU, for the Award of Metallurgist of the Year 2010, in the Non-Ferrous Metals Category, in recognition of the developmental work that he conducted for Indian Defence.

SJMSOM, IIT Mumbai hosts the 2nd IPR Researchers' Confluence the 2nd IPR Researchers' Confluence 2nd IPR Researchers' Confluence February 11-12, 2011 SJMSOM, IIT-Mumbai : Welcome back for the 2nd IPR Researchers' Confluence being held at SJMSOM, IIT Mumbai on February 11 - 12, 2011. After a very successful conduct of the 1st confluence in December 2009, where the IPR research and education roadmap evolved, the 2nd confluence is now all set. As the name suggests, this is not a conference! It is an attempt towards creating an environment where people interact, share and unlearn to learn new concepts, appreciate issues and brainstorm to identify suitable approaches to the problems raised. Experts from the technology, legal and management domain across academics, industry are amongst the initial list of speakers and experts who have confirmed to be speakers, panel members and brainstorming session leaders. Refer to the theme document for more details.

IT-An IITian Speaks Out IIT Mumbai's motto is Knowledge is The Supreme Goal. I studied there and in the spirit of its philosophy, am going to present a few essays of mine, to show IIT and IITians from a different perspective than we are used to normally. Hope it will lead to better understanding of IIT and the soci-economic environment it operates in.

Here I have tried to tackle many issues simultaneously. Equal access to education is important to many including women, minorities, and poorer population. How do IITs deal with that? On the other hand, once admitted, do the young minds at IIT

learn skills to develop into responsible citizen or are just trained to become (migrant) workers in economically developed societies? I have also tried to reconcile the vision Nehru had when he conceived of IITs and what they have become today. Are IITs truly what the best what India with population of over one billion capable of?

The IITians: The Story of a Remarkable Indian Institution and How its Alumni Are Reshaping the World IIT (Indian Institute of Technology) is India’s biggest and most powerful brand, and arguably the toughest and most influential engineering school in the world.

Since the first IIT was set up in the 1950s, thousands of initiates have walked out of the campus gates in Kharagpur, Mumbai, Chennai and elsewhere to become leaders in their chosen fields. In India they head many of the biggest and most admired professionally managed companies. Abroad, they lead giant corporations, and their feats figure in the folklore of Silicon Valley. The power that the alumni of this one bunch of undergraduate schools wields in business, academe and research is comparable to that of Cambridge and Oxford in the heyday of the British Empire.

Sandipan Deb, himself an IITian, delves into his own experience and those of scores of alumni to try and explain what makes IITians such outstanding achievers. In part it may be that they cannot be anything else: only one in every hundred applicants gets admitted. Harvard, in comparison, takes one in eight. The unique village-like campuses peopled only by the super-bright and the intensely competitive hone the IITians’ skills further. No wonder then that when they leave the campus, IITians look upon themselves as special people, capable of competing in their field with the best in the world.


IIT gold medallist shares his success story Ever wondered what it would be like to be the gold medallist at an IIT?

For Sheerang Chhatre, this dream recently became reality, as he was named the gold medallist at IIT-Mumbai in this year's graduating class. Now, he's off to MIT in the United States for his PhD, but he plans on returning to India [ Images ] to help the country's growth.

Sheerang chatted with Get Ahead readers on and answered questions regarding academics at the IITs. For those of you who missed the chat, here's the transcript:

Varun asked, Most IIT achievers moves to US or Europe for jobs. Tell me what is the reason that they dont stay here and serve country? Shreerang Chhatre answers, See Varun, young people are more inclined towards moneymaking. So fat pay packages and a comfortable life attracts them to developed countries. But they fail to realise that through their knowledge and expertise they should help their own people to grow. Now, I guess the situation is changing, slowly but surely.

Shreerang Chhatre answers, Thanks for your wishes.

Padmakar asked, Conratulations Shreerang! I would like to know about your school days. Whether you were one of the intllignent/ bright student during those days or you improved yourself afterwardfs....? Shreerang Chhatre answers, I did my schooling from Parle Tilak Vidyalaya, Mumbai [ Images ]. I was bright but lazy; but slowly I realised that you need to work really hard to achieve anything. So I guess that was the only improvement.

Rohan asked, hi Shreerag....congrats mate............I aspire to be an IITian..I am in 10th...when do I start coaching for IIT Shreerang Chhatre answers, After your 10th standard exam, take a break. Relax for a few days. Then start for

JEE. I guess two years are required for a thorough preparation.

Ramesh asked, Hi Shreerang, As of today, which source of energy do you find most promising, and which one will be appropriate for India. Shreerang Chhatre answers, From the Indian point of view, it's solar energy. If a country like germany with a smaller size and much lesser solar radiation can generate so much power from it, then why can't we? Huge amounts of initial investment for solar-cells, invertors and the grid is preventing the commercialisation of solar energy in India.

Hime asked, Congrats, what is your course in Mtech? Shreerang Chhatre answers, I have specialised in Metallurgical process engineering for MTech.

Rutvik asked, Would u suggest going for any branch of IIT or going for a branch of one's choice like Computer Science in other premier institutes like NIT,Bits Pilani, IIIT etc Shreerang Chhatre answers, My personal opinion is that you should go for any branch available in IITs, rather than going for other colleges. The hierarchy or distinction in the branches that people make are not really felt when one is studying in IITs. The academic quality, peer group and facilities that you get in IITs are awesome. Never let the opportunity to get into IIT slip-by.

Kunal asked, hi Shreerang! im kunal from golden gate.. Shreerang, please tell me how u studied 4 chemistry. the portion seems to be too vast..I find it hard to remember each and every reaction Shreerang Chhatre answers, Well, initially it can be a bit difficult, but slowly through revisions and assignments you get used to it. Dont worry, you have a very good organic chem teacher, so just keep pace with the lectures and the portion. You will succeed, dont worry!

Success Story Success Story This article contains storie/interviews of persons who succeed after graduation from different IITs

Mr. Sheerang ChhatreGold Medallist From IIT-Bombay

IIT-Bombay


PHYSICS

1. A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire : Young's modulus = 2.1 × 1011 Pa ; Density = 7860 kg/m3 ; Specific heat = 420 J/kg-K).

[IIT-2001] Sol. When the mass of 100 Kg is attached, the string is

under tension and hence in the deformed state. Therefore it has potential energy (U) which is given by the formula.

U = 21 × stress × strain × volume

= 21 ×

Y)Stress( 2

× πr2l

= 21

Y)r/Mg( 22π × πr2l =

21

YrgM2

22

πl ...(i)

This energy is released in the form of heat, thereby raising the temperature of the wire

Q = mc∆T ...(ii) From (i) and (iii) Since U = Q Therefore

∴ mc∆T = 21

YrgM2

22

πl

∴ ∆T = 21

YcmrgM

2

22

π

l

Here m = mass of string = density × volume of string = ρ × πr2l

∴ ∆T = 21

Ycp)r(gM22

22

π

= 21 ×

7860420101.2)10214.3()10100(

1123–

2

×××××××

= 0.00457º C 2. An unknown resistance X is to be determined using

resistance R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [IIT-2005]

A B C

G

RX R = R1 or R2 or R3

Sol. All Null point, the wheat stone bridge will be balanced

∴ 1r

X = 2r

R

⇒ X = R2

1

rr

where R is a constant r1 and r2 are variable. The maximum fraction error is

A B C

G

R X

r1 r2

R=R1 R=R2 R=R3

N M

XX∆ =

1

1

rr∆ +

2

2

rr∆

Here ∆r1 = ∆r2 = y (say) then

For XX∆ to be minimum r1 × r2 should be max

[Q r1 + r2 = c (Constt.)] Let E = r1 × r2 ⇒ E = r1 × (r1 – c)

∴ 1dr

dE = (r1 – c) + r1 = 0

⇒ r1 = 2c ⇒ r2 =

2c ⇒ r1 = r2

⇒ R2 gives the most accurate value

KNOW IIT-JEE By Previous Exam Questions


3. An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 µF, and the resulting LC circuit is set oscillating at its natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found that the maximum value of Q is 200 µC.

(a) When Q = 100 µC, what is the value of |dI/dt|? (b) When Q = 200 µC, what is the value of I? (c) Find the maximum value of I. (d) When I is equal to one half its maximum value,

what is the value of |Q|? [IIT-1998] Sol. This is a problem of L–C oscillations. Here Q0 = maximum value of Q = 200 µC = 2 × 10–4 C

ω = LC1 =

)F100.5)(H102(

16–3– ××

= 10+4 s–1

Let at t = 0, Q = Q0 then Q(t) = Q0 cos ωt ...(1)

I(t) = dtdQ = – Q0 ω sin ωt ...(2)

L=2.0 mH

C=5.0 µF

dt

)t(dI = – Q0 w2 cos (ωt) ...(3)

(a) For Q = 100 µC

2

Qor 0

For (1) 100 = 200 cos ωt

or cos (ωt) = 21 , From equation (3) :

dtdI = (2.0 × 10–4C) (10+4 s–1)2

21

dtdI = 10+4 A/s

(b) Q = 200 µC when cos(ωt) = 1 i.e. ωt = 2π ... At this time I(t) = – Q0 sin ωt or I (t) = 0 (c) I(t) = – Q0ω sin ωt ∴ Maximum value of I is Q0ω or Imax = Q0ω = (2.0 × 10–4 C) (10+4 s–1) Imax = 2.0 A (d) From energy conservation.

21 2

maxLI = 21 LI2 +

21

CQ2

or Q = )I–I(LC 22max

I = 2

Imax = 1.0 A

∴ Q = )1–2)(100.5)(100.2( 226–3– ××

Q = 3 × 10–4 C or Q = 1.732 × 10–4 C 4. Shown in the figure is a prism of an angle 30º and

refractive index µp = 3 . Face AC of the prism is covered with a thin film of refractive index µf = 2.2. A monochromatic light of wavelength λ = 550 nm fall on the face AB at an angle of incidence of 60º.

[IIT-2003] A

C B

30º60º

3µp =

µf = 2.2 Calculate (a) angle of emergence. (b) min. value of thickness t so that intensity of

emergent ray is maximum. Sol. (a) Using snell's law at surface AB

µair sin 60º = µp sin r ⇒ 23 = 3 sin r ⇒ r = 30º

Now, NN' is the normal to surface AB. ∴ ∠AMN = 90º But ∠QMN = 30º ⇒ ∠AMQ = 60º

A

C B

30º

60º60º30º

M N

N' Q

In ∆AMQ ∠AQM = 180º – (60º + 30º) = 90º The refracted ray inside the prism hits the other face

at 90º ; hence deviation produced by this face is zero and hence angle of emergence is zero.

(b) Multiple reflection occurs between the surfaces of the prism for minimum thickness.

∆x = 2µt = λ, where

t = thickness ⇒ t = µ2λ = 125 nm

5. Highly energetic electrons are bombarded on a target

of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is (14)1/3. Find (a) atomic number of the nucleus. (b) the frequency of Kα line of the X-ray produced. (R = 1.1 × 107 m–1 and c = 3 × 108 m/s) IIT-2005]


Sol. (a) We know that radius of nucleus is given by the formula

r = r0 A1/3 where r0 = constt. and A = mass number. For the nucleus r1 = r0 41/3

For the Nucleus r1 = r0 (4)1/3

∴ 1

2

rr =

3/1

4A

⇒ (14)1/3 = 3/1

4A

⇒ A = 56

∴ No. of proton = A – no. of neutrons = 56 – 30 = 26 ∴ Atomic number = 26

(b) We know that v = Rc (z – b)2

21

21 n

1–n1

Here, R = 1.1 × 107 , c = 3 × 108 , Z = 26 b = 1 (for Kα), n1 = 1, n2 = 2

∴ ν = 1.1 × 107 × 3 × 108 [26 – 1]2

41–

11

= 3.3 × 1015 × 25 × 25 × 43 = 1.546 × 1018 Hz.

CHEMISTRY

6. One litre of a mixture of O2 and O3 at NTP was

allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 ml of M/10 sodium thiosulphate solution for titration. What is the weight percent of ozone in the mixture ? Ultraviolet radiation of wavelength 300 nm can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? [IIT-1997]

Sol. The concerned chemical reaction are : O3 + 2KI + H2O → 2KOH + I2 + O2 I2 + 2Na2S2O3 → Na2S4O6 + 2NaI Millimoles of ozone = Millimoles of I2

mM of O3 = mM of I2 = 21 × mM of Na2S2O3

= 21 × 40 ×

101 = 2 mM = 0.002 mole

Calculation of total number of moles of O2 and O3 PV = nRT 1 × 1 = n × 0.0821 × 273 n = 0.044 mole ∴ Mole of O2 = 0.044 – 0.002 =- 0.042 ∴ Wt. of O2 = No of moles × Mol. wt. = 0.042 × 32

= 1.344 g Similarly, Wt. of O3 = 0.002 × 48 = 0.096 g

∴ % of O3 = 44.1096.0 × 100 = 6.6%

No. of photons or molecules of O3 = 4810023.6096.0 23××

= 1.2 × 1021

7. Give reasons for the following : (i) Methane does not react with chlorine in the dark

[IIT-1983] (ii) Propene react with HBr to give isopropyl

bromide but does not give n-propyl bromide. [IIT-1983]

(iii) Although benzene is highly unsaturated, normally it does not undergo addition reaction.

[IIT-1983] (iv) Toluene reacts with bromine in the presence of

light to give benzyl bromide while in presence of FeBr3 it gives p-bromotoluene. Give explanation for the above observations.

[IIT-1996] (v) Explain very briefly why alkynes are generally

less reactive than alkenes towards electrophilic reagents such as H+. [IIT-1997]

(vi) The central carbon-carbon bond in 1, 3-butadiene is shorter than that in n-butane.

[IIT-1998] (vii) tert-Butylbenzene does not give benzoic acid

on treatment with acidic KMnO4. [IIT-2000] (viii) 7-Bromo-1, 3, 5-cycloheptatriene exists as

ionic compound, while 5-bromo-1, 3-cyclopentadiene does not ionise even in presence of Ag+ ion. Explain. [IIT-2004]

(ix) CH3

CH3

Br → OHHC.aq 52 Acidic solution

CH3

CH3 Br → OHHC.aq 52 Neutral

solution. Explain. [IIT-2005]

(x)

(A)

→ Pd/H3 3

(B)

but not


(C)

[IIT-2005]

Sol. (i) Tips/Formula : Chlorination of methane is a

free radical substitution reaction. Solution : In dark, Chlorine is unable to be

converted into free radicals, hence the reaction does not occur.

(ii) Tips/Formula : Addition of unsymmetrical addendum (HBr in present case) to unsymmetrical olefin (CH3CH=CH2, in present case) takes place according to Markownikoffrule.

Solution : According to this rule "the negative part of reagent (i.e., Br–) adds on the carbon atom having minimum number of hydrgon atoms". Hence isopropyl bromide will be formed in the present case.

CH3CH=CH2 + HBr → CH3.CHBr.CH3 Propene iso-Propyl bromide (iii) Unlike olefins, π-electrons of benzene are

delocalised (resonance) and hence these are unreactive towards addition reactions.

(iv) Tips/Formula : In presence of FeBr3 it gives o and p derivative and in absence of FeBr3 it gives side chain derivative.

Solution : In presence of light, toluene undergoes side chain bromination through a free radical mechanism.

CH3

hv

Br2→

CH2Br

Benzyl bromide

In presence of FeBr3, toluene undergoes electrophilic substitution in the benzene ring.

CH3

3

2

FeBr

Br→

CH3

Br

p-bromotoluene (v) The low reactivity of alkynes towards

electrophilic addition reaction is believed to be due to following two factors.

C C— + E → C ⊕

C—

E ⊕

Highly strained bridged carbocation

(a) The bridged intermediate cation formed by the initial attack of electrophile on the triple bond is less stable because it is a highly strained system. Due to formation of cyclic intermediate carbocation, the olefinic intermediate products would invariably be trans.

(b) In acetylenic carbon atoms, the π-electrons are more tightly held by the carbon nuclei and hence they are less easily available for reaction with electrophiles.

Perhaps both the above factors, steric and electronic, play their in diminishing the reactivity of alkynes towards electrophiles.

(vi) Tips/Formula : 1, 3-Butadience is a conjugated diene and is a reasonance hybrid :

Solution :

=↔=↔==

++–C–CC–C––C–CC–C––CC–CC–

||||

••

|

••

|||||||

––

The charged structures induce some double

bond character in the central C–C bond leading to the shortening of this bond. Alternatively, all the four C atoms of 1, 3-butadiene are sp2 hybridised and thus their C–C bond length will be lower than that of n-butane in which all the four c-atoms are sp3 hybridised.

(vii) tert-Butylbenzene does not give benzoic acid on treatment with acidic KMnO4 ause it does not contain any hydrogen atom on the key carbon atom.

(viii)Tips/Formula : 7-Bromo-1,3,5-cycloheptatriene is aromatic whereas 5-Bromo-1,3-Cycloheptadiene is non aromatic.

Solution :

7-Bromo-1,3,5-cycloheptatriene

(Triopylium bromide)

Br

Its corresponding cation is

+

7-Bromo-1,3,5-cyclo heptarienyl cation

(Triopy lium cation) It has 6π electrons, hence

aromatic and easily formed

+ Br–

Br

5-Bromo-1,3-cyclopentadiene

+ Its corresponding cation is

Cyclopentadienyl cation(It has 4π electrons,

hence not aromatic, thus not easily formed)

+ Br–

(ix) The halide is a 3º halide, hence it undergoes

SN1 reaction forming HBr, as one of the products which make solution acidic.


C6H5–C–Br |

| CH3

CH3

)1S(

)aq(OHHC

N

52 →

C6H5–C–OC2H5 + HBr|

| CH3

CH3

(acidic)

A 3º bromide

CH(CH3)2 Br

is an aryl halide so it

does not undergo nucleophilic substitution reactions. Hence the solution will remain neutral.

(x) Reduction of cental ring to form A reduces all the three cyclobutadiene rings (which are antiaromatic as they have 4π electrons each), i.e. antiaromatic rings are converted into nonaromatic rings. On the other hand, reduction of the terminal ring to form B reduces only one antiaromatic ring and two antiaromatic cyclobutadiene rings remain intact. Remember that antiaromatic rings impart unstability.

8. Draw the structures of [Co(NH3)6]3+ , [Ni(CN)4]2–

and [Ni(CO)4]. Write the hybridisation of atomic orbitals of the transition metal in each case.

[IIIT-2000] Sol. [Co(NH3)6]3+

3d 4s 4pCo3+

⇒ 3d

d2sp3

4p

NH3

Co

NH3

NH3

NH3

NH3

NH3

3+

Octahedral complex, d2sp3 hybridisation [Ni(CN)4]2–

3d 4s 4pNi2+

Ni2+ (after rearrangement)

3d

dsp2

4p

pairing due to CN–

Ni

CN

CN

CN

CN

2–

Square planar dsp2 hybridisation [Ni(CO)4]

3d 4s 4pNi =

Ni (after rearrangement)

3d

sp3

4p

pairing dueto CO

Ni

CO

COCO

CO

Tetrahedral (sp3 hybridisation) 9. A white substance (A) reacts with dilute H2SO4 to

produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K3Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reaction involved.

[IIT-2001] Sol.

)solutioncolourless()colourless(

SOH.dil

)white(CBA 42 + →

K2Cr2O7/H+

Green solution + D ↓(burns in air to form E) (coloured)

E↑ + B↑ → D + Colourless liquid → 4CuSO.anhy Blue C

NaOHor

NH.aq 3 → Precipitate reagent

ofexcess → Clar solution

The above set leads to following conclusions. (i) Since the gas (B) is colourless and turns

acidified K2Cr2O7 solution given, it should be H2S.


(ii) Since H2S gas is obtained by the reaction of dil. H2SO4 on A, the latter must be sulphide.

(iii) The white colour of the sulphide (A) points out towards ZnS.

Thus the various reactions can be written as below. ZnS + H2SO4 (dil) → ZnSO4 + H2S↑ (A) (C) (B) 3H2S + K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 7H2O + 3S (green) (D) S + O2 → SO2↑ → )B(SH2 2 2H2O + 3S↓ (D) (E) (colourless liq) D → )White(CuSO4 CuSO4.5H2O (blue) ZnSO4 + 2NaOH → Zn (OH)2↑ (C) white.ppt

)excess(

NaOH2 → Na2ZnO2 + 2H2O

(soluble)

10. A compounds (X) containing C, H and O is unreactive towards sodium. It does not add bromine. It also does not react with Schiff's reagent. On refluxing with an excess of hydriodic acid, (X) yields only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60. What are the compounds (X), (Y) and (Z)? Write chemical equations leading to the conversion of (X) to (Y).

[IIT-1981] Sol. Tips/Formula : The unreactivity of the compound

(X) towards sodium indicates that it is neither an acid nor an alcohol, further its unreactivity towards Schiff's base indicates that it is not an aldehyde. The reaction of compound (X) with excess of HI to form only one product indicates that it should be an ether.

Solution : Hence its other reactions are sketched as below.

R–O–R HIofexcess

withfluxRe → 2RI →hydrolysis 2ROH

(X) (Y) (Z)

P + I2

)O(

KMnO4 → –COOH

eq. wt. 60 Since the carboxylic acid has equivalent weight of

60, it must be acetic acid (CH3COOH), hence Z must be ethyl alcohol, (Y) ethyl iodide and (X) diethyl ether.

C2H5–O–C2H5 + 2HI →reflux 2C2H5I Ethyl iodide (Y)

→–OH 2C2H5OH → 4KMnO CH3COOH

Ethyl alcohol (Z) Acetic acid (Eq. wt. = 60)

MATHEMATICS

11. Prove that there exists no complex number z such

that |z| < 31 and ∑

=

n

1r

rrza = 1, where |ar| < 2.

[IIT-2003] Sol. Given : a1z + a2z2 + ... + anzn = 1 and |z| < 1/3 ...(1) |a1z + a2z2 + a3z3 + ... + anzn| = 1 using |z1 + z2| ≤ |z1| + |z2| ⇒ |a1z| + |a2z2| + |a3z3| + ... + | anzn| ≥ 1 ⇒ 2|z| + |z|2 + |z|3 + ... + |z|n > 1 [using |ar| < 2]

⇒ |z|–1

)|z|–1(|z|2 n > 1

using sum of n terms of G.P. ⇒ 2|z| – 2|z|n+1 > 1 – |z| ⇒ 3 |z| > 1 + 2 |z|n+1

⇒ |z| > 31 +

32 |z|n+1

⇒ |z| > 31 , which contradicts (1)

∴ There exists no complex number z such that

|z| < 31 and ∑

=

n

1r

rrza = 1

12. mn squares of equal size are arranged to form a rectangle of dimension m by n where m and n are natural numbers. Two squares will be called 'neighbours' if they have exactly one common side. A natural number is written in each square such that the number in written any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the numbers used are equal. [IIT-1982]

Sol. Let mn squares of equal size are arrange to form a rectangle of dimension m by n. Shown as, from figure, neighbours of

x6x5

x7 x4

x1

x2x3

n

m x1 are x2, x3, x4, x5 x5 are x1, x6, x7 x7 are x5, x4

⇒ x1 = 4

xxxx 5432 +++, x5 =

3xxx 761 ++

and x7 = 2

xx 54 +


∴ 4x1 = x2 + x3 + x4 + 3

761 xxx ++

⇒ 12x1 = 3x2 + 3x3 + 3x4 + x1 + x6 + 2

xx 54 +

⇒ 24x1 = 6x2 + 6x3 + 6x4 + 2x1 + 2x6 + x4 + x5

⇒ 22x1 = 6x2 + 6x3 + 7x4 + x5 + 2x6 where, x1, x2, x3, x4, x5, x6 are all the natural

numbers and x1 is linearly expressed as the sum of x2, x3, x4, x5, x6

where sum of coefficients are equal only if, all observations are same.

⇒ x2 = x3 = x4 = x5 = x6 ⇒ all the numbers used are equal. 13. Let f [(x + y)/2] = f(x) + f(y)/2 for all real x and y.

If f ' (0) exists and equals –1 and f(0) = 1, find f(2). [IIT-1995]

Sol.

+

2yxf =

2)y(f)x(f + ∀ x, y ∈ R (given)

Putting y = 0, we get

f

2x =

2)0(f)x(f + =

21 [1 + f(x)] [Q f(0) = 1]

⇒ 2f(x/2) = f(x) + 1 ⇒ f(x) = 2 f(x/2) – 1 ∀ x, y ∈ R ...(1) Since f ' (0) = – 1, we get

h

)0(f–)h0(flim0h

+→

= – 1 ⇒ h

1–)h(flim0h→

= – 1

Now, let x ∈ R then applying formula of differentiability.

f '(x) = h

)x(f–)hx(flim0h

+→

=h

)x(f–2

h2x2flim

0h

+

→

= h

)x(f–2

)h2(f)x2(f

lim0h

+

→

= h

)x(f–1–2h2f21–

2x2f2

21

lim0h

+

→

[Using equation (1)]

= h

)x(f–1–)h(f21–)x(f221

lim0h

+

→

= h

)x(f–1–)h(f)x(flim0h

+→

= h

1–)h(flim0h→

= – 1

Therefore f ' (x) = – 1 ∀ x ∈ R ⇒ ∫ dx)x('f = ∫ dx1–

⇒ f(x) = – x + k where k is a constant. But f(0) = 1, therefore f(0) = – 0 + k ⇒ 1 = k ⇒ f(x) = 1 – x ∀x ∈ R ⇒ f(2) = – 1

14. If 'f ' is a continuous function with ∫x

0dt)t(f → ∞ as

|x| → ∞, then show that every line y = mx intersects

the curve y2 + ∫x

0dt)t(f = 2! [IIT-1991]

(0, 2 )A

B

O(xp,0)

X

(0, – 2 )

Sol. We are given that f is continuous function and

∫x

0dt)t(f → ∞, as |x| → ∞

To show that every line y = mx intersects the curve

y2 + ∫x

0dt)t(f = 2

If possible, let y = mx intersects the given curve, then substituting y = mx in the curve, we get

m2x2 + ∫x

0dt)t(f = 2

Consider, F(x) = m2x2 + ∫x

0dt)t(f – 2 ...(1)

Then F(x) is a continuous function as F(x) is given to be continuous.

Also F(x) → ∞ as |x| → ∞ But F(0) = – 2 Thus F(0) = – ve and F(b) = +ve, where b is some

value of x and F(x) is continuous. ∴ F(x) = 0 for some value of x ∈ (0, b) or equation

(1) is solvable for x. Hence, y = mx intersects the given curve.

15. Find all values of θ in the interval

ππ

2,

2–

satisfying the equation

(1 – tan θ )(1 + tan θ) sec2θ + θ2tan2 = 0 [IIT-1996]

Sol. (1 – tan θ) (1 + tan θ) sec2θ + θ2tan2 = 0

⇒ (1 – tan2θ).(1 + tan2θ) + θ2tan2 = 0

⇒ 1 – tan4 θ + θ2tan2 = 0 put tan2θ = x ⇒ 1 – x2 + 2x = 0 ⇒ x2 – 1 = 2x

–1 –1 1

y

O x

Curves y = x2 – 1and y = 2x intersect at one point (negative value will not

consider) x = 3, y = 8 Therefore, tan2θ = 3 ⇒ tan θ = ± 3 ⇒ θ = ± π/3


Passage # 1 (Que. 1 to 3)

The internal energy ‘U’ v/s PV graph where P is the pressure and V is the volume of an ideal gas filled up in a piston cylinder system is shown below If btan =φ then

1. What is the atomocity and the shape of the Gaseous molecule if b= 3 and a = 2.

2. Write the relation of adiabatic index of the gas in terms of a or b or interms of both a and b.

3. If a start varying with respect to time as a(t) = 2(3+t) and b remains constant then draw the graph of CV v/s a where CV is the molar specific heat at constant volume.

4. If b start varying with respect to time as 2

10 tcc)t(b + where c0 and c1 are positive constants

then find the slope of dtdf v/s t graph where f is the

degrees of freedom for the gas. 5. A particle enters in the given magnetic field

∧→= kBB 0 whre B0 is a constant with the velocity of

∧∧→+= jbiav where a and b are the positive

constants. The place where the magnetic field exists and the

particle moves is filled with the resistive medium then path followed by the particle is-

(Charge on particle q and mass m)

(A) Circular path with radius of the circular path

is 0

22

B.qbamr +

=

(B) Helix and the pitch of the Helix is a.B.qm2

0

π

(C) Helix and the pitch of the Helix is b.B.qm2

0

π

(D) Same path as followed by circulating electrons which is responsible for the unstable Rutherford atomic model, Means spiral path of decreasing radius.

Passage # II (Que. 6 to 8) Behaviour of capacitor in electric circuits is very

typical because of it's energy storing nature. Capacitor behaves in just opposite manner to inductor, Inductor 'L' which is measured in Henary in SI system stores the energy in magnetic field instead of capacitor which stores in electric field Inductor opposes the change in current and capacitor opposes change in voltage.

Behaviour of inductor:

For the electric circuit shown

6. If capacitor C varies even after that energy stored in capacitor is zero at steady state then

(A) 2

1

2

1

RR

εε

= (B) 1

2

2

1

RR

εε

=

(C) 021 =ε+ε (D) 0RR 2211 =ε+ε

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 8


7. Time constant for the circuit (A) RC (B) R1C if 21 ε>ε

(C) R2C if 21 ε<ε (D) CRR

RRR

21

21

+

+

Where 21

2111eq R/1R/1

R/R/+

ε−ε=ε

21

21eq RR

RRR

+=

8. Maximum current passing through resistance R

(A) eq

eq

Rε

(B) eq

eq

RR +

ε

(C) Reqε

(D) R

|| 21 ε−ε

What do Aliens Look Like?

Aliens are the extraterrestrial beings believed to exist. Some give accounts of having seen them visit our world. Then, what do aliens look like? Want to know? The read on…

Aliens have always aroused the interest for many. With new discoveries in astronomy, man has been able to explore the extraterrestrial world and examine the chances of the existence of aliens.

On one hand, the existence of extraterrestrial life is considered hypothetical while on the other hand, aliens have been sighted on a few occasions. There have been news about the aliens visiting Earth; there have been some people claiming to have seen the aliens. The concept of ‘aliens’ remains alien! The sightings of aliens have brought about descriptions of their appearance. What they look like, has been a question in the minds of one and all and news have many a time answered it by giving accounts of people witnessing aliens. We know of films and television shows, which have depicted aliens as being humanoid in appearance.

What do Aliens Look Like? Aliens are largely described as resembling human beings. Their height is approximately same as the average height of human beings. Like any normal human beings, aliens have a pair of eyes, a nose, a mouth, a pair of arms and a pair of feet. There are citations of aliens having wings or wheels instead of feet and other such abnormalities. It is believed that aliens have a rough lizard-like skin. Their skin colors are believed to vary from gray, white, tan to gold, pink or red. Their skin is believed to glow in the dark. Their eyes are considered to resemble those of humans, lizards or insects. Some have documented aliens as having webbed fingers while others believe that aliens have suction cups for fingertips or claws. Aliens have been documented as being variedly sized and shaped. Some have documented them as 3 inches tall while others say that they are about 15 feet tall. In some places aliens have been documented as being shaped like balls of light, while in other places they have been shown as resembling robots or metal objects. Some believe that aliens look like animals or large insects while some think of aliens as human-like figures clothed in uniforms. Many believe that aliens can float through walls.

Human Quick Facts

1. The hardest bone in the human body is the jawbone.

2. The number of eye blinks varies greatly from about 29 blinks each minute if you are talking to someone to only 4 blinks each minute if you are reading.

3. The average human blinks 25 times per minute.

4. A nail takes around 6 months to grow from base to the tip.

5. Each second 10,000,000 cells die and are replaced in your body.

6. Your liver performs over 500 functions in your body.

7. The average person spends 1/3 of their lifetime sleeping.

8. More germs are transferred when shaking hands than kissing.

9. The average person (from western culture) consumes 10 liters of alcohol per year.

10. Roughly 75% of people who play the radio in their car sing along to it.

11. Human thigh bones are stronger than concrete. 12. Your right lung takes in more air than your left

one does.

13. The human brain is composed of 75% water.

14. 70% of the composition of dust in your home is made up of shed human skin and hair.

15. The tooth is the only part of the human body that can’t repair itself.


1. As the resistances of voltmeters in upper branch are R, R/2, R/4 …………………. The equivalent circuit is as shown below

The resistance of upper branch is = R + R/2 + R/4+……….. up to infinite

+++= ......

41

211R

R22/11

1R =

−=

further the equivalent circuit is

the resistance of voltmeter V should be 2R so that current in upper and lower branch is same.

2. Entire upper branch is having the resistance of 2R and voltmeter V1 is having the resistance of R so we can conclude that equivalent resistance of all the voltmeters in upper branch except V1 is R and the upper branch is as follows:

As reading of voltmeter V1 is X = i.R Sum of the readings of voltmeters is Y = i.R Except V1 in upper branch So, X = Y

3. From current division formula we can conclude that current in upper and lower branch are in the ratio of 1 : 2.

Reading of voltmeter V1 is i.R Reading of voltmeter V is (2i.)R So V = 2V1

4.

l = length of rod = b – a charge on element of length dx is dq dq = λdx as x3=λ dq = 3xdx Equivalent current due to element of length dx

)xdx3(2

dq.diπ

ω=ω=

Total equivalent current ∫ ∫ πω

== )xdx3(2

diib

a

)ab(2

.23

2ab

23

2x

23 22

22b

a

2−

πω

=

−πω

=

πω

=

)ab(43 22 −

πω

=

Option A is correct Equivalent current

)ab)(ab(43)ab(

43 22 +−

πω

=−πω

=

Solution Physics Challenging Problems

Set # 7

8 Questions were Published in November Issue


l).ab.(43)ab)(ab(

43

+ωπ

=−+πω

=

As 3/4π=ω So,

Equivalent current l).ab.(3

4.43

+π

π=

= (b + a). l = const. l i ∝ l Option B is correct.

Charge on rod

∫ ∫ −=

===

b

a

22b

a

2)ab(

23

2x.3xdx3dqq

Option D is correct Ans. A, B, D

5. For part B q (closed cone) > q (open cone) for part A q (close cone) = q (open cone)

Equivalent current q.2

iπ

ω=

q.2

iπ

ω= ; q.

2i

πω

=

cone –C1(closed cone) cone-C3(closed cone)

i q.2πω

= (closed cone) ; ).(2

i σπ

ω= (Surface area

of closed cone) If σ varies then charge on cone C1 differs from C3 So their currents will be different. Option A incorrect q (cone- C1) = q (cone-C2)

)Ccone()coneC(

11

q.2

i−π

ω= and

)Ccone()coneC(22

q.2

i−π

ω=

i (cone C1) = i (cone C2) Option B is correct As charge on cone ≠3C charge on cone C4 Option C correct Part-A and Part –B will have different charges so Option C correct Part –A and Part – B will have different charges so Option D incorrect Ans. B, C

6. The circuit is as follows

Full scale deflection current for galvanometer is

mA510

mV50ig =Ω

=

For terminals CT and a range is 5V so using

Ω=−×

=⇒−=−

990101055RG

iVR

31g

Ω= 990R1

7. Range between CT and b is 10 volt so,

Using 1010510RRG

iVR

321g

−×

=+⇒−=−

990 + R2 = 2000 – 10 Ω=−= 100010002000R 2 Ω= 1000R 2

8. Range between CT and c is V so

Using GiVRg

−=

10105VRRR

3321 −×

=++−

10105V30001000990

3−

×=++⇒

−

3105V5000

−×=⇒

volt25V =⇒ So range between CT and C is 25 volts.

GLOBAL WARMING IS REAL

The arctic ice is receding and global warming is no longer a theory but a reality. Scientists predict that by the year 2100, the average surface temperature will jump up by 6 degrees Fahrenheit. Nighttime temperatures will be higher and there will be hotter days.

Since air temperature is a powerful component of climate, there will be unavoidable climate changes in the future. Some climate changes involve extreme weather disturbances such as more severe hurricanes and longer droughts. There will be an increased precipitation of snow and rain during winter. The faster melting of snow during the spring will result in flooding. All these climate changes are predicted based on the assumption that changes will be relatively gradual.


1. A ring of radius R = 4 m is made of a highly dense material. mass of the ring is m1 = 5.4 × 109 kg. distributed uniformly over its circumference. A highly dense particle of mass m2 = 6 × 108 kg is placed on the axis of the ring at a distance x0 = 3m from the centre. Neglecting all other forces, except mutual gravitational interaction of the two, calculate

(i) displacement of the ring when particle is closest to it, and

(ii) speed of the particle at that instant. Sol. Since, there is no external force on the system of ring

and particle, therefore, centre of mass of the system remains stationary.

R m2

m1 x0 Particle is closest to ring when it is at its centre. At

this instant centre of mass of the system is at centre of the ring. It means displacement of ring is equal to the distance of centre of mass of the system from initial position of the ring.

i.e., x = 21

021

mmxm0.m

++

= 0.3 m Ans. (i) Since, initially ring and particle both are stationary,

therefore, momentum of the system is zero and it always remains zero because there is no external force on the system.

If velocities of ring and particle are v1 (rightward) and v2 (leftward) respectively and particle reaches the centre of ring, then according to law of conservation of momentum,

m1v1 = m2v2 or v1 = 9

v2 ...(1)

But when ring and particle move towards each other, gravitational potential energy of their system decreases and converts into kinetic energy. Therefore, according to law of conservation of energy,

kinetic energy of the system = loss of gravitational potential energy

Initial gravitational energy of the system,

U1 = – 20

221

xR

mGm

+

Gravitational energy of the system when particle is at

the centre of the ring U2 = R

mGm 21

∴ 211vm

21 + 2

22vm21 = U1 – U2

or v2 = 0.18 ms–1 or cm/sec Ans. (ii) 2. In the arrangement shown in Figure pulleys are small

and light and springs are ideal. K1, K2, K3 and K4 are force constant of the springs.

Calculate period of small vertical oscillations of block of mass m.

K3 K1

m

K4 K2

Sol. In static equilibrium of block, tension in the string is exactly equal to its weight. Let a vertically downward force F be applied on the block to pull it downwards. Equilibrium is again restored when tension in string is increased by the same amount F. Hence, total tension string becomes equal to (mg + F).

Strings are further elongated due to extra tension F. Due to this extra tension F in strings, tension in each spring increase by 2F. Hence increases in elongation

of springs is 1KF2 ,

2KF2 ,

3KF2 and

4KF2 respectively.

According to geometry of the arrangement, downward displacement of the block from its

equilibrium position is y = 2

+++

4321 KF2

KF2

KF2

KF2

...(1)

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICS


If the block is released now, it starts to accelerate upwards due to extra tension F in string. It means restoring force on the block is equal to F.

From equation (1), F =

+++

4321 K1

K1

K1

K14

y

∴ Restoring acceleration of block = mF =

+++

4321 K1

K1

K1

K1m4

y

Since, acceleration of block is restoring and is directly proportional to displacement y, therefore, the block performs SHM.

Its period T = onacceleratintdisplaceme2π

∴ T =

+++π

4321 K1

K1

K1

K1m42

=

+++π

4321 K1

K1

K1

K1m4 Ans.

3. A Solid non-conducting hemisphere of radius R has a

uniformly distributed positive charge of density ρ per unit volume. A negatively charged particle having charge q is transferred from centre of its base to infinity. Calculate work performed in the process.

Di-electric constant of material of hemisphere is unity

Sol. When negative charge q is displaced from centre of base to infinity, its electrical potential energy increases.Work is to be performed to increase this energy. To calculate initial potential energy of the particle, first a thin hemispherical shell of radius x and radial thickness dx is considered as shown in Figure

Volume of material of the shell = 2πx2.dx ∴ Charge on shell is dQ = ρ(2πx2 dx) Since, every element of this shell is at a constant

distance x from centre of curvature, therefore, potential energy of the particle, due to charge of the shell considered is

dU = x

)dQ)(q(–4

1

0πε = –

02qερ x dx

or total initial potential energy of particle,

U0 = – 02

qερ

∫=

=

Rx

0xdx.x =

0

2

4Rq–ε

ρ

When particle reaches infinity, its potential energy U becomes equal to zero.

∴ Work done = Increase in potential energy

= U – U0 = 0

2

4Rqε

ρ Ans.

4. Each of two long parallel wires carries a constant current I along the same direction. The wires are separated by a distance 2l. Calculate maximum magnitude of magnetic induction in the symmetry plane of this system located between the wires. Calculate also, the maximum force experienced by unit length of a third wire carrying the same current along the same direction if third wire is parallel to and in the symmetry plane of other two wires.

Sol. In Figure, points P and Q represent two wires, each carrying current along inward normal to plane of the paper. It is given that each of these two wires carries a current I and separation between the wires is 2l. In the figure, dotted line PQ represents the plane of wires and firm line normal to PQ represents the plane of symmetry.

y

× ×

B'

B B' θ θ

θ θ

l l

r r

R

P Q

Let magnetic induction in plane of symmetry be maximum at point R, at a distance y from plane of wires P and Q.

Distance of this point from each wire is r = 22 y+l ∴ Magnitude of magnetic induction at R due to

each wire is B' = r2Iµ0

π

Directions of these two magnetic inductions at R are as shown in figure. Their components in the plane of symmetry neutralise each other. Therefore, at R, resultant magnetic induction is normal to the plane of symmetry.

The resultant magnetic induction, B = 2B' sinθ

∴ B = 20

rIyµ

π =

)y(Iyµ

220

+π l ...(1)

For B to be maximum dydB = 0 or y = l

∴ Bmax = lπ2Iµ0

Maximum force experienced by unit length of the third wire, Fmax = Bmax . I Nm–1

∴ Fmax = lπ2

Iµ 20 Nm–1


According to Flemming's left hand rule vector of force F lies in the symmetry plane and towards plane of wires.

5. A metal rod of length l = 100 cm is clamped at two points A and B as shown in Figure. Distance of each clamp from nearer end is α = 30 cm. It density and Young's modulus of elasticity of rod material are ρ = 9000 kg m–3 and Y = 144 GPa res-pectively, calculate minimum and next higher frequency of natural longitudinal oscillations of the rod.

A A

30 cm 30 cm

l = 100 cm Sol. Speed of longitudinal waves in the rod is v = ρ/Y

= 4000 ms–1 . Since points A and B are clamped, therefore, nodes

are formed at these points or rod oscillates with integer number of loops in the middle part. Let number of these loops be m.

Since, length of each loop is 2λ ,

therefore, m. 2λ = (l – 2a)

or mλ = 80 cm ...(1) Since ends of the rod are free, therefore, antinodes

are formed at each end of the rod or at one end of each end part is an antinode and at the other end is a node. It means that number of loops in each end part

will be an odd multiple of half. Let these be

21–n2

where n is an integer.

Then,

21–n2 .

2λ = a or (2n – 1) λ = 129 cm

...(2) Dividing equation (1) by (2),

)1–n2(

m = 32 ...(3)

Minimum possible frequency corresponds to maximum possible wavelength, hence, minimum number of loops.

Hence, from equation (3), for minimum frequency m should be equal to 2 and (2n – 1) should be equal to 3 or n = 2.

Substitution m = 2 in equation (1), Maximum wavelength, λ0 = 40 cm

fo = 0

vλ

= 10 kHz Ans.

Next higher frequency corresponds to next higher integer values of m and n which satisfy equation (3). Hence, for this case m = 6 and (2n – 1 = 9 or n = 5

Substitution m = 6 in equation (1),

It means rod oscillates with odd harmonics

λ = 6

80 cm or 3

40 cm

∴ Next higher frequency, f1 = λv = 30 kHz Ans.

COMPLEMENTARY COLOURS

If you arrange some colours in a circle, you get a "colour wheel". The diagram shows one possible version of this. An internet search will throw up many different versions!

Colours directly opposite each other on the colour wheel are said to be complementary colours. Blue and yellow are complementary colours; red and cyan are complementary; and so are green and magenta.

Mixing together two complementary colours of light will give you white light.

What this all means is that if a particular colour is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary colour. Copper(II) sulphate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum. Cyan is the complementary colour of red.

The origin of colour in complex ions

Transition metal v other metal complex ions


Reflection : Key Concepts : (a) Due to reflection, none of frequency, wavelength

and speed of light change. (b) Law of reflection : Incident ray, reflected ray and normal on incident

point are coplanar. The angle of incidence is equal to angle of

reflection

θ θ

Plane surface

Incident Ray

Reflected Ray

n

θ θ

Convex surface

Tangent at point P

n

P

α α

Convex surface Tangent

at point P

n

A

Some important points : In case of plane mirror For real object, image is virtual. For virtual object, image is real. The converging point of incident beam behaves as a

object. If incident beam on optical instrument (mirror, lens

etc) is converging in nature, object is virtual. If incident beam on optical instrument is diverging in

nature, the object is real. The converging point of reflected or refracted beam

from an optical instrument behaves as image. If reflected beam or refracted beam from an optical

instrument is converging in nature, image is real.

PVirtual Object

P

Real Object

P

Real Object

Virual Object

Pn

n

If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual.

P

Real Object

Virual Object

P' n

n

α α

α α

For solving the problem, the reference frame is

chosen in which optical instrument (mirror, lens, etc.) is in rest.

The formation of image and size of image is independent of size of mirror.

Visual region and intensity of image depend on size of mirror.

α α

θ θ n

P P'

If the plane mirror is rotated through an angle θ, the

reflected ray and image is rotated through an angle 2θ in the same sense.

If mirror is cut into a number of pieces, then the focal length does not change.

The minimum height of mirror required to see the full image of a man of height h is h/2.

vObject

Rest

Image v

v

Object

Rest

Image θ

vcosθ

vsinθ

vcosθ

vsinθ

vObject Image

vm 2vm–v

Ray Optics

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Object In rest

Image vm 2vm

Object

Image vm 2vm+vv

(c) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirror are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle.

Here, n = θ

º360

where θ = angle between mirrors.

If θ

º360 is even number, the number of images is

n – 1.

If θ

º360 is odd number and object is placed on

bisector of angle between mirror, then number of images is n – 1.

If θ

º360 is odd and object is not situated on

bisector of angle between mirrors, then the number of images is equal to n.

(d) Law of reflection in vector form : Let 1e = unit vector along incident ray.

2e = unit vector along reflected ray

n = unit vector along normal on point of Incidence Then 2e = n)n.e(2e 11 −

nn

1e 2e

(e) Spherical mirrors : It easy to solve the problems in geometrical optics

by the help of co-ordinate sign convention.

y

y' x' x

y

y' x' x

y

y'x' x

y

y'x' x

y

y' x' x

The mirror formula is f1

u1

v1

=+

Also, R = 2f These formulae are only applicable for paraxial

rays. All distances are measured from optical centre. It

means optical centre is taken as origin. The sign conventions are only applicable in given

values. The transverse magnification is

β = sizeobjectsizeimage =

uv−

1. If object and image both are real, β is negative. 2. If object and image both are virtual, β is negative. 3. If object is real but image is virtual, β is positive. 4. If object is virtual but image is real, β is positive. 5. Image of star; moon or distant object is formed at

focus of mirror. If y = the distance of sun or moon from earth. D = diameter of moon or sun's disc f = focal length of the mirror d = diameter of the image θ = the angle subtended by sun or moon's disc

Then tan θ = θ = yD =

fd

Here, θ is in radian.

θ θ F D

d

Sun

Problem solving strategy : Image formation by mirrors Step 1: Identify the relevant concepts : There are

two different and complementary ways to solve problems involving image formation by mirrors. One approach uses equations, while the other involves drawing a principle-ray diagram. A successful problem solution uses both approaches.

Step 2: Set up the problem : Determine the target variables. The three key quantities are the focal length, object distance, and image distance; typically


you'll be given two of these and will have to determine the third.

Step 3: Execute the solution as follows : The principal-ray diagram is to geometric optics

what the free-body diagram is to mechanics. In any problem involving image formation by a mirror, always draw a principal-ray diagram first if you have enough information. (The same advice should be followed when dealing with lenses in the following sections.)

It is usually best to orient your diagrams consistently with the incoming rays traveling from left to right. Don't draw a lot of other rays at random ; stick with the principal rays, the ones you know something about. Use a ruler and measure distance carefully ! A freehand sketch will not give good results.

If your principal rays don't converge at a real image point, you may have to extend them straight backward to locate a virtual image point, as figure (b). We recommend drawing the extensions with broken lines. Another useful aid is to color-code the different principal rays, as is done in figure(a) & (b).

1

4 2 Q'

P' C 3 2 4

v

3

(a)

F

I Q

P

Q'

v

(b)

C

Q

P 4

P' F

1 1

4 2 2 3

Check your results using Eq. f1

's1

s1

=+ and the

magnification equation m = y'y =

s's

− . The

results you find using this equation must be consistent with your principal-ray diagram; if not, double-check both your calculation and your diagram.

Pay careful attention to signs on object and image distances, radii or curvature, and object and image heights. A negative sign on any of these quantities always has significance. Use the equations and the sign rules carefully and consistently, and they will tell you the truth !

Note that the same sign rules (given in section) work for all four cases in this chapter : reflection and refraction from plane and spherical surfaces.

Step 4: Evaluate your answer : You've already checked your results by using both diagrams and equations. But it always helps to take a look back and ask yourself. "Do these results make sense ?".

Refraction : Laws of Refraction :

The incident ray, the refracted ray and normal on incidence point are coplanar.

µ1 sin θ1 = µ2 sin θ2 = ... = constant.

µ1

µ2

θ2

θ1

Snell's law in vector form :

µ1

µ2 2e

1e

n

Let, 1e = unit vector along incident ray

2e = unit vector along refracted.

n = unit vector along normal on incidence point.

Then µ1( 1e × n ) = µ2( 2e × n )

Some important points :

(a) The value of absolute refractive index µ is always greater or equal to one.

(b) The value of refractive index depends upon material of medium, colour of light and temperature of medium.

(c) When temperature increases, refractive index decreases.

(d) Optical path is defined as product of geometrical path and refractive index.

i.e., optical path = µx


(e) For a given time, optical path remains constant.

i.e., µ1x1 = µ2x2 = ... constant

∴ µ1 dtdx1 = µ2 dt

dx2

∴ µ1c1 = µ2c2 (where c1 and c2 are speed of light in respective mediums)

∴ 1

2

µµ =

2

1

cc

i.e., µ ∝ c1

(f) The frequency of light does not depend upon medium.

∴ c1 = fλ1, c2 = fλ2

∴ 2

1

µµ =

1

2

cc =

1

2

λλ

∴ µ ∝ λ1

When observer is rarer medium and object is in denser medium :

Then µ = depthapparent

depth real

When object is in rarer and observer is in denser medium :

µ = position real

positionapparent

The shift of object due to slab is x = t

µ1–1

(a) This formula is only applicable when observer is in rarer medium.

(b) The object shiftiness does not depend upon the position of object.

(c) Object shiftiness takes place in the direction of incidence ray.

The equivalent refractive index of a combination of a

number of slabs for normal incidence is µ =

i

i

i

µtt

Σ

Σ

Here, Σti = t1 + t2 + ...

Σi

i

µt =

1

1

µt +

2

2

µt + ...

The apparent depth due to a number of media is Σi

i

µt

The lateral shifting due to a slab is d = t sec r sin(i – r).

Critical angle : When a ray passes from denser medium (µ2) to rarer medium (µ1), then for 90º angle of refraction, the corresponding angle of incidence is critical angle.

Mathematically, sin c = 2

1

µµ

(i) When angle of incidence is lesser than critical angle, refraction takes place. The corresponding deviation is

δ = sin–1

isin

µµ

1

2 – i for i < c

(ii) When angle of incidence is greater than critical angle, total internal reflection takes place. The corresponding deviation is

δ = π – 2i when i < c

The δ – i graph is :

(i) Critical angle depends upon colour of light, material of medium, and temperature of medium.

(ii) Critical angle does not depend upon angle of incidence

π/2 c i

δ

Refractive surface formula,

v

µ2 – uµ1 =

rµµ 12 −

Here, v = image distance,

u = object distance,

r = radius of curvature of spherical surface.

(a) For plane surface , r = ∞

(b) Transverse magnification,


m = sizeobject size ageIm =

uµvµ

2

1

(c) Refractive surface formula is only applicable for paraxial ray.

Lens :

Lens formula :

v1 –

u1 =

f1

(a) Lens formula is only applicable for thin lens.

(b) r = 2f formula is not applicable for lens.

(c) m = sizeobject size image =

uv

(d) Magnification formula is only applicable when object is perpendicular to optical axis.

(e) lens formula and the magnification formula is only applicable when medium on both sides of lenses are same.

(f)

f(+ve)

(i)

f(–ve)

(ii)

f(–ve) f(+ve)

(iii) (iv)

f(–ve)

(v)

f(+ve)

(vi) (g) Thin lens formula is applicable for converging as

well diverging lens. Thin lens maker's formula :

f1 =

−

1

12

µµµ

−

21 r1

r1

µ1 µ1

µ2

(a) Thin lens formula is only applicable for paraxial

ray.

(b) This formula is only applicable when medium on both sides of lens are same.

(c) Intensity is proportional to square of aperture.

(d) When lens is placed in a medium whose refractive index is greater than that of lens. i.e., µ1 > µ2. Then converging lens behaves as diverging lens and vice versa.

(e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other.

(f) If a lens is cut along the diameter, focal length does not change.

(g) If lens is cut by a vertical, it converts into two lenses of different focal lengths.

i.e., f1 =

1f1 +

2f1

(h) If a lens is made of a number of layers of different refractive index number of images of an object by the lens is equal to number of different media.

(i) The minimum distance between real object and real image in is 4f.

(j) The equivalent focal length of co-axial combination of two lenses is given by

F1 =

1f1 +

2f1 –

21ffd

(k) If a number of lenses are in contact, then

F1 =

1f1 +

2f1 + ......

(l) (i) Power of thin lens, P = F1

(ii) Power of mirror is P = –F1

(m) If a lens silvered at one surface, then the system behaves as an equivalent mirror, whose power

P = 2PL + Pm

Here, PL = Power of lens

=

−

1

12

µµµ

−

21 r1

r1


Pm = Power of silvered surface = –mF1

Here, Fm = r2/2, where r2 = radius of silvered surface.

P = – 1/F

Here, F = focal length of equivalent mirror. 1. Rays of light strike a horizontal plane mirror at an

angle of 45º. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror.

Sol. The situation is shown in figure

C GD

N

QB P

A S

45º 45º

θ θ

The ray AB strikes the first plane mirror PQ at an

angle of 45º. Now, we suppose that the second mirror SG is arranged such that the ray BC after reflection from this mirror is horizontal.

From the figure we see that emergent ray CD is parallel to PQ and BC is a line intersecting these parallel lines.

So, ∠DCE = ∠CBQ = 180º ∠DCN + ∠NCB + ∠CBQ = 180º θ + θ + 45º = 180º ∴ θ = 67.5º As ∠NCS = 90º, therefore the second mirror should

be inclined to the horizontal at an angle 22.5º. 2. An object is placed exactly midway between a

concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror.

Sol. The image formation is shown in figure.

P1 25cm

F

r = 40 cm r = 30 cm

P2 C I2

I1

50cm

(i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ?

Now 1f1 =

11 v1

u1

+

or 201 =

1v1

251

+

v1 = 100 cm. As v1 is positive, hence the image is real. In the

absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to the presence of convex mirror, the rays are reflected and appear to come from I2.

(ii) For convex mirror, In this case, I1 acts as virtual object and I2 is the

virtual image. The distance of the virtual object from the convex

mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm. As focal length of convex mirror is negative and

hence f2 = –30/2 = –15 cm. Here we shall calculate the value of v2. Using the mirror formula, we have

151

− = 2v

1501

+−

or v2 = –21.42 cm As v2 is negative, image is virtual. So image is

formed behind the convex mirror at a distance of 21.43 cm.

3. There is a small air bubble in side a glass sphere (n =

1.5) of radius 10 cm. The bubble is 4 cm below the

surface and is viewed normally from the outside

(Fig.). Find the apparent depth of the air bubble.

n2 = 1A

C

P

n1 = 1.5

O

I

Sol. The observer sees the image formed due to refraction at the spherical surface when the light from the bubble goes from the glass to air.

Here u = – 4.0 m, R = – 10 cm, n1 = 1.5 and n2 = 1

We have [(n2/v) – (n1/u) = (n2 – n1)/R

or (1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm)

or (1/v) = (0.5/10 cm) – (1.5/4.0 cm)

or v = – 3.0 cm

Thus, the bubble will appear 3.0 cm below the surface.

Solved Examples


4. A convex lens focuses a distance object on a screen placed 10 cm away from it. A glass plate (n = 1.5) of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen ?

Sol. The situation when the glass plate is inserted between the lens and the screen, is shown in fig. The lens forms the image of object O at point I1 but the glass plate intercepts the rays and forms the final image at I on the screen. The shift in the position of image after insertion of glass plate

II1 O

Scre

en

10 cm

I1I = t

−

n11 = (1.5 cm)

−

5.111 = 0.5 cm.

Thus, the lens forms the image at a distance of 9.5 cm from itself. Using

v1 –

u1 =

f1 , we get

u1 =

v1 –

f1 =

5.91 –

101

or u = – 190 cm. i.e. the object should be placed at a distance of

190 cm. from the lens.

5. A candle is placed at a distance of 3 ft from the wall. Where must a convex lens of focal length 8 inches be placed so that a real image is formed on the wall ?

Sol. According to formula for refraction though a lens

f = 8"

36 – v v

d = 3 ft = 36"

v1 –

u1 =

f1 or

v1 –

)v36(1

−−=

81

or v1 +

v361−

=81 or

)v36(vvv36

−+− =

81

or, v2 – 36 v + 8 × 36 = 0 or v = 12" or 24" = 1 ft or 2 ft. ∴ u = 24" or 12" = 2 ft or 1 ft Hence, lens should be placed at either 1 ft or 2 ft

away from the wall.

IIT -JEE 2011 Few important Instructions to fill the IIT-JEE 2011 Form: 1. Use Black ballpoint pen to fill the entire

form 2. Use pencil to darken the bubbles in the OMR

sheet 3. Use CAPIATL LETTERS to fill the form. 4. Name of the candidate should be written in

Capital Letters as given in your 10th Class Certificate. Leave only one blank box between consecutive words of your name. If by mistake you had made wrong entry of name then use whitener, do not use blade or rubber.

5. Enter Date , Month & Year of your birth as per 10th class marksheet. Use numerals 01 to 31 for DATE, 01 to 12 for MONTH and last two digits of the YEAR of birth (the digit 19 is already there). For example: if you were born on 6th march 1993 then DATE should be entered as: “06 03 93”

6. Only those candidate which are opting IIT KHARAGPUR ZONE or IIT ROORKEE ZONE and they come under the Category of SC,ST,OBC,PD or DS may opt for either of the option mention below from their respective zones : IT-BHU VARANASI or ISM DHANABAD.

7. If you wish to obtain the question paper in HINDI darken the bubble corresponding to “HINDI” And If you wish to obtain the question paper in ENGLISH darken the bubble corresponding to “ENGLISH”. Note: Please choose only one option from the above choices

8. Write Complete name & address to which any communication is to be sent in capital Letters Please do not give address of your school/college, hostel or any other institute here. Please write Only those phone / mobile numbers which can be made available at the time of necessary information & counselling.

9. The Candidate must sign the declaration. The place and date should also be mentioned. The two signatures, the one below this declaration and the other in the box below your photograph (S.No 16) should be identical.

10. Put your signature within the box provided. Your signature must not overflow or touch the border of the box provided. Please do not sign in Capital letters.


Fluid Mechanics : Fluid statics : Pressure at a point inside a Liquid : p = p0 + ρgh where p0 is the atmospheric pressure, ρ is the density

of the liquid and h is the depth of the point below the free surface.

p0

h p

ρ

Pressure is a Scalar : The unit of pressure may be

atmosphere or cm of mercury. These are derived units. The absolute unit of pressure is Nm–2. Normal atmospheric pressure, i.e, 76 cm of mercury, is approximately equal to 105 Nm–2.

Thrust : Thrust = pressure × area. Thrust has the unit of force.

Laws of liquid pressure (a) A liquid at rest exerts pressure equally in all

directions. (b) Pressure at two points on the same horizontal line

in a liquid at rest is the same. (c) Pressure exerted at a point in a confined liquid at

rest is transmitted equally in all directions and acts normally on the wall of the containing vessel. This is called Pascal's law. A hydraulic press works on this principle of transmission of pressure.

The principle of floating bodies (law of flotation) is that W = W´, that is, weight of body = weight of displaced liquid or buoyant force. The weight of the displaced liquid is also called buoyancy or upthrust. Hydrometers work on the principle of floating bodies. This principle may also be applied to gases (e.g., a balloon).

Liquids and gases are together called fluids. The important difference between them is that liquids cannot be compressed, while gases can be compressed. Hence, the density of a liquid is the same everywhere and does not depend on its pressure. In the case of a gas, however, the density is proportional to the pressure.

Fluid dynamics :

Bernoulli's Theorem : 21 v2 + gh +

ρp = a constant

for a streamline flow of a fluid (liquid or gas). Here, v is the velocity of the fluid, h is its height

above some horizontal level, p is the pressure and ρ is the density.

p1

v1

h1

v2

p2

h2

v2 > v1 p2 < p1 According to this principle, the greater the velocity,

the lower is the pressure in a fluid flow. It would be useful to remember that in liquid flow,

the volume of liquid flowing past any point per second is the same for every point. Therefore, when the cross-section of the tube decreases, the velocity increases.

Note : Density = relative density or specific gravity × 1000 kg m–3.

Surface tension and surface energy : Surface Tension : The property due to which a

liquid surface tends to contract and occupy the minimum area is called the surface tension of the liquid. It is caused by forces of attraction between the molecules of the liquid. A molecule on the free surface of a liquid experiences a net resultant force which tends to draw it into the liquid. Surface tension is actually a manifestation of the forces experienced by the surface molecules.

If an imaginary line is drawn on a liquid surface then the force acting per unit length of this line is defined as the surface tension. Its unit is, therefore, newton / metre. This force acts along the liquid surface. For curved surfaces, the force is tangent to the liquid surface at every point.

Surface Energy : A liquid surface possesses potential energy due to surface tension. This energy per unit area of the surface is called the surface energy of the liquid. Its units is joule per square metre. The surface energy of a liquid has the same numerical values as the surface tension. The surface

Fluid Mechanics & Properties of Matter

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


tension of a liquid depends on temperature. It decreases with rise in temperature.

Excess of Pressure : Inside a soap bubble or a gas bubble inside a liquid, there must be pressure in excess of the outside pressure to balance the tendency of the liquid surface to contract due to surface tension.

p(excess of pressure) = T

+

21 r1

r1 in general

where T is surface tension of the liquid, and r1 and r2 are the principal radii of curvature of the bubble in two mutually perpendicular directions.

For a spherical soap bubble, r1 = r2 = r and there are two free surfaces of the liquid.

∴ p = rT4

For a gas bubble inside a liquid, r1 = r2 = r and there is only one surface.

∴ p = rT2

For a cylindrical surface r1 = r and r2 = ∞ and there are two surfaces.

∴ p = rT2

Angle of Contact : The angle made by the surface of a liquid with the solid surface inside of a liquid at the point of contact is called the angle of contact. It is at this angle that the surface tension acts on the wall of the container.

The angle of contact θ depends on the natures of the liquid and solid in contact. If the liquid wets the solid (e.g., water and glass), the angle of contact is zero. In most cases, θ is acute (figure i). In the special case of mercury on glass, θ is obtuse (figure ii).

θ

fig. (i)

θ

fig. (ii) Rise of Liquid in a Capillary Tube : In a thin

(capacity) tube, the free surface of the liquid becomes curved. The forces of surface tension at the edges of the liquid surface then acquire a vertical component.

TT θ θ

θ θ

h

r

meniscus

The upward force by which a liquid surface is pulled up in a capillary tube is 2πrTcos θ, and the downward force due to the gravitational pull on the mass of liquid in the tube is (πr2h + v)ρg, where v is the volume above the liquid meniscus. If θ = 0º, the meniscus is hemispherical in shape. Then v = difference between the volume of the cylinder of radius r and height r and the volume of the hemisphere of radius r

= πr3 – 32

πr3 = 31

πr3

When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For equilibrium

(πr2h + v) ρg = 2πrT cos θ When a glass capillary tube is dipper in mercury, the

meniscus is convex, since the angle of contact is obtuse. The surface tension forces now acquire a downward component, and the level of mercury inside the tube the falls below the level outside it. the relation 2T cos θ = hρgr may be used to obtain the fall in the mercury level.

Problem Solving Strategy Bernoulli's Equations : Bernoulli's equation is derived from the work-energy

theorem, so it is not surprising that much of the problem-solving strategy suggested in W.E.P. also applicable here.

Step 1: Identify the relevant concepts : First ensure that the fluid flow is steady and that fluid is incompressible and has no internal friction. This case is an idealization, but it hold up surprisingly well for fluids flowing through sufficiently large pipes and for flows within bulk fluids (e.g., air flowing around an airplane or water flowing around a fish).

Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1

and 2 referred to in Bernoulli's equation. Define your coordinate system, particular the

level at which y = 0. Make lists of the unknown and known quantities

in Eq. p1 + ρgy1 + 21

ρv12 = p2 + ρgy2 +

21

ρv22

(Bernoulli's equation) The variables are p1, p2, v1, v2, y1 and y2, and the

constants are ρ and g. Decide which unknowns are your target variables.

Step 3: Execute the solutions as follows : Write Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v2 (continuity equation, incompressible fluid), to get a relation between the two speeds in terms of cross-sectional areas of pipes


or containers. Or perhaps you will know both speeds and need to determine one of the areas. You may also

need to use Eq. dtdV = Av (volume flow rate) to find

the volume flow rate. Step 4: Evaluate your answer : As always, verify that

the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is in pascals, density in kilograms per cubic meter, and speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures.

Properties of matter : Key Concepts : Stress : The restoring force setup inside the body per unit

area is known as stress. Restoring forces : If the magnitude of applied

deforming force at equilibrium = F

Then, Stress = AF

In SI system, unit of stress is N/m2. Difference between pressure and stress : (a) Pressure is scalar but stress is tensor quantity. (b) Pressure always acts normal to the surface, but

stress may be normal or tangential. (c) Pressure is compressive in nature but stress may

be compressive or tensile. Strain :

Strain = dimension originaldimensionin change

(a) Longitudinal strain = LL∆

L F F

Longitudinal strain is in the direction of deforming force but lateral strain is in perpendicular direction of deforming force.

Poisson ratio :

σ = strain allongitudin

strain lateral = L/Ld/D

∆∆

Here ∆d = change in diameter.

(b) Volumetric strain = VV∆

F

F F

F V (c) Shear strain = φ

Shear strain

φ

Stress-strain graph : From graph, it is obvious that in elastic limit, stress is

proportional to strain. This is known as Hooke's law. ∴ Stress ∝ Strain ∴ Stress = E .strain

∴ E = strainstress

where E is proportionality dimensional constant known as coefficient of elasticity.

O

A

B C

Plasticregion Breaking

strength Elasticlimit

Strain

Stre

ss

Types of coefficient of elasticity :

(a) Young's modulus = Y = strain allongitudinstress ogitudinall

∴ Y =

LLA

F∆

= LA

FL∆

L

∆L

F

(b) Bulk modulus = B = strain volumetricstress volumetric

Compressibility = 1/B


(c) Modulus of rigidity = η = φA

F = strainshear stress shear

(d) For isothermal process, B = P. F

φφ

F (e) For adiabatic process, B = γP

(f) modulusbulk Isothermalmodulusbulk Adiabatic = γ

(g) Esolid > Eliquid > Egas (h) Young's modulus Y and modulus of rigidity η

exist only for solids. (i) Bulk modulus B exist for solid, liquid and gas. (j) When temperature increases, coefficient of

elasticity (Y, B, η) decreases.

(k) B1 +

η3 =

Y9

(l) Y = 2(1 + σ)η (m) Poisson's ratio σ is unitless and dimensionless.

Theoretically, –1 < σ < 21

Practically, 0 < σ < 21

(n) Thermal stress = Yα∆θ (o) Elastic energy stored,

U = 21 × load × extension =

21 Fx =

21 kx2

= stress × strain × volume For twisting motion,

U = 21 × torque × angular twist

= 21

τ × θ = 21 cθ2

Elastic energy density,

u = 21 × stress × strain J/m3 =

21 Y × strain2J/m3

Thermal stress = Yα∆θ and Thermal strain = α∆θ Work done in stretching a wire :

(a) W = 21 F∆L

(b) Work done per unit volume = 21 × stress × strain

(c) Breaking weight = breaking stress × area

Surface tension :

T = LF

Here L = length of imaginary line drawn at the surface of liquid. and F = force acting on one side of line (shown in figure)

(a) Surface tension does not depend upon surface area.

(b) When temperature increases, surface tension decreases.

(c) At critical temperature surface tension is zero.

F

F L

Rise or fall of a liquid in a capillary tube :

h = gr

cosT2ρ

θ

Here θ = angle of contact. r = radius of capillary tube ρ = density of liquid For a given liquid and solid at a given place, hr = constant Surface energy : Surface energy density is defined as work done

against surface tension per unit area. It is numerically equal to surface tension.

W = work = surface tension × area (a) For a drop of radius R, W = 4πR2T (b) For a soap bubble, W = 8πR2T Excess pressure :

(a) For drop, P = RT2

(b) For soap bubble, P = RT4

Viscosity : (a) Newton's law of viscous force :

F = – ηAdydv

where dydv = velocity gradient

A = area of liquid layer η = coefficient of viscosity The unit of coefficient of viscosity in CGS is poise.


(b) SI unit of coefficient of viscosity = poiseuille = 10 poise. (c) In the case of liquid, viscosity increases with

density. (d) In the case of gas, viscosity decreases with

density. (e) In the case of liquid, when temperature increases,

viscosity decreases. (f) In the case of gas, when temperature increases,

viscosity increases. Poiseuille's equation :

V = L8rP 4

ηπ

where V = the volume of liquid flowing per second through a capillary tube of length L and radius r

η = coefficient of viscosity and P = pressure difference between ends of the tube Stoke's law : The viscous force acting on a spherical body moving

with constant velocity v in a viscous liquid is F = 6πηrv where r = radius of spherical body Determination of η :

η = v9

g)(r2 2 σ−ρ

where r = radius of spherical body moving with constant velocity v in a viscous liquid of coefficient of viscosity η and density ρ

and σ = density of spherical body Critical velocity (v0) :

v0 = r

kρη

where k = Reynold's number for narrow tube, k ≈ 1000. (a) For stream line motion, flow velocity v < v0. (b) For turbulant motion, flow velocity v > v0. 1. A vertical U-tube of uniform cross-section contains

mercury in both arms. A glycerine (relative density 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6 × 103 kg m–3.

Sol. Draw a horizontal line through the mercury-glycerine surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the points A and B must be the same.

h

10 c

m

(1 –

h)

A B

Pressure at B = p0 + 0.1 × (1.3 × 1000) × g Pressure at A = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g ∴ p0 + 0.1 × 1300 × g = p0 + 800gh + 1360g – 13600 × g × h ⇒ 130 = 800h + 1360 – 13600h

⇒ h = 128001230 = 0.096 m = 9.6 cm

2. A liquid flows out of a broad vessel through a narrow

vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the length of the pipe is h ?

Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure.

Applying Bernoulli's theorem to points 1, 2 and 3

hHx

•1

•2

•3

ρ0p + 2

1v21 + gH =

ρ

2p + 22v

21 + g (h – x)

=ρ0p + 2

3v21 + 0

By continuity equation v 1A1 = A2v 2 = A2v 3 Since A1 >> A2,v1 is negligible and v2 = v3 = n (say).

∴ ρ0p + gH =

ρ2p +

21 v2 + g (h – x)

=ρ0p +

21 v2

∴ v = gH2 (i)

Solved Examples


and ρ0p + gH =

ρ2p + gH + g (h – x)

⇒ p0 + p2 + ρg (h – x) ⇒ p2 = p0 – ρg (h – x) (ii) Thus pressure varies with distance from the upper

end of the pipe according to equation (ii) and velocity is a constant and is given by (i).

3. Calculate the difference in water levels in two

communicating tubes of diameter d = 1 mm and d = 1.5 mm. Surface tension of water = 0.07 Nm–1 and angle of contact between glass and water = 0º.

Sol. Pressure at A = p0 – 2r

cosT2 θ

(Q pressure inside a curved surface is greater than that outside)

Pressure at B = p0 – 1r

cosT2 θ

∴ pressure difference = 2T cos θ

−

21 r1

r1

B A

Let this pressure difference correspond to h units of

the liquid.

Then 2T cos θ

−

21 r1

r1 = ρgh

⇒ h =

−

ρθ

21 r1

r1

gcosT2

∴ h =

×−

×××

−− 33 105.11

1011

8.9100007.02 = 4.76 mm

4. A mass of 5 kg is suspended from a copper wire of 5

mm diameter and 2 m in length. What is the extension produced in the wire ? What should be the minimum diameter of the wire so that its elastic limit is not exceed ? Elastic limit for copper = 1.5 × 109 dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2.

Sol. Given that Y = 1.1 × 1012 dynes/cm2, L = 2m = 200 cm, d = 5 mm = 0.5 cm or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.

Y =l2r

FLπ

or l =Yr

FL2π

= 22 101.1)25.0(142.320098010000.5×××

×××

= 4.99 × 10–3 cm Also, elastic limit for copper = 1.5 × 109 dynes/cm2 If d' is the minimum diameter, then maximum stress

on the wire =4/'d

F2π

= 2'dF4

π

Hence, 2'dF4

π= 1.5 × 109

or d'2 = 9105.1F4××π

= 9105.1142.398010000.54

×××××

= 41.58 × 10–4 d' = 0.0645 cm.

5. A uniform horizontal rigid bar of 100 kg in supported horizontally by three equal vertical wires A, B and C each of initial length one meter and cross-section I mm2. B is a copper wire passing through the centre of the bar; A and C are steel wires and are arranged symmetrically one on each side of B YCu = 1.5 × 1012 dynes / cm2, Ys = 2 × 1012 dynes/cm2. Calculate the tension in each wire and extension.

Sol. The situation is shown in figure. Because the rod is horizontally supported, hence extensions in all the wires must be equal i.e., strains in all the wires are equal as initial lengths are also equal.

As Y =StrainStress

AS

B Cu

C S

100 Kg

Hence, YCu = StrainA/FCu … (1)

and Ys = StrainA/Fs … (2)

∴ S

Cu

YY =

S

Cu

FF =

25.1 =

43 or 4FCu = 3FS ...(3)

According to figure, we can write 2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g or [(8/3) + 1] FCu = 100 g ∴ FCu = (3/11) × 100g = (3/11) × 100 Kgwt = 27.28 Kgwt and FS = (4/3) FCu = (4/3) × (3/11) × 100g = (400/11)g = 36.36 Kgwt Extension in each wire,

l =Cu

Cu

AYLF = 122 105.110

10098027280××

××−

= 0.178 cm


Acidity of carboxylic acids. Fatty acids are weak acids as compared to inorganic

acids. The acidic character of fatty acids decreases with increase in molecular weight. Formic acid is the strongest of all fatty acids.

The acidic character of carboxylic acids is due to resonance in the acidic group which imparts electron deficiency (positive charge) on the oxygen atom of the hydroxyl group (structure II).

R C O H

O

INon-equivalent structures

R C O H

O

II

– +

(Resonance less important)

R C O + H+

O–

The positive charge (electron deficiency) on oxygen

atom causes a displacement of electron pair of the O—H bond towards the oxygen atom with the result the hydrogen atom of the O—H group is eliminated as proton and a carboxylate ion is formed.

Once the carboxylate ion is formed, it is stabilised by means of resonance.

R C O

Resonating forms of carboxylate ion (Equivalent structures)(Resonance more important)

O– R C

O

O–

R C O

Resonance hybrid of carboxylate ion O

–

Due to equivalent resonating structures, resonance in

carboxylate anion is more important than in the parent carboxylic acid. Hence carboxylate anion is more stabilised than the acid itself and hence the equilibrium of the ionisation of carboxylic acids to the right hand side.

RCOOH RCOO– + H+ The existence of resonance in carboxylate ion is

supported by bond lengths. For example, in formic acid, there is one C=O double bond (1.23 Å) and one C—O single bond (1.36Å), while in sodium formate both of the carbon-oxygen bond lengths are identical

(1.27Å) which is nearly intermediate between C O and C—O bond length values. This proves resonance in carboxylate anion.

H CO

OHH C

O

O

Na+

Formic acid Sodium formate

–

It is important to note that although carboxylic acids

and alcohols both contain –OH group, the latter are not acidic in nature. It is due to the absence of resonance (factor responsible for acidic character of –COOH) in both the alcohols as well as in their corresponding ions (alkoxide ions).

R—O—H R—O– + H+ Alcohol Alkoxide ion (No resonance) (No resonance)

Relative acidic character of carboxylic acids with common species not having —COOH group.

RCOOH > Ar—OH > HOH > ROH >

HC CH > NH3 > RH Effect of Substituents on acidity. The carboxylic acids are acidic in nature because of

stabilisation (i.e., dispersal of negative charge) of carboxylate ion. So any factor which can enhance the dispersal of negative charge of the carboxylate ion will increase the acidity of the carboxylic acid and vice versa. Thus electron-withdrawing substitutents (like halogens, —NO2, —C6H5, etc.) would disperse the negative charge and hence stabilise the carboxylate ion and thus increase acidity of the parent acid. On the other hand, electron-releasing substituents would increase the negative charge, destabilise the carboxylate ion and thus decrease acidity of the parent acid.

X C O

O

–

The substituent X withdraws electrons, disperses negative charge, stabilises the ion and hence increases acidity

Y C O

O

–

The substituent Y releases electrons, intensifies negative charge, destabilises the ion and hence decreases acidity

Organic Chemistry

Fundamentals

CARBOXYLIC ACID KEY CONCEPT


Now, since alkyl groups are electron-releasing, their presence in the molecule will decrease the acidity. In general, greater the length of the alkyl chain, lower shall be the acidity of the acid. Thus, formic acid (HCOOH), having no alkyl group, is about 10 times stronger than acetic acid (CH3COOH) which in turn is stronger than propanoic acid (CH3CH2COOH) and so on. Similarly, following order is observed in chloro acids.

Cl C

Cl

Cl

CO2H > Cl C

Cl

H

CO2H

pKa 0.70 1.48

Cl C

H

H

CO2H > H C

H

H

CO2H

pKa 2.86 4.76

>

Decreasing order of aliphatic acids (i) O2NCH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH > (CH3)3CCOOH (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH > ClCH2CHClCH2COOH (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH Benzoic acid is somewhat stronger than simple

aliphatic acids. Here the carboxylate group is attached to a more electronegative carbon (sp2 hybridised) than in aliphatic acids (sp3 hybridised).

HCOOH > C6H5COOH > CH3COOH. Nucleophilic substitution at acyl carbon : It is important to note that nucleophilic substitution

(e.g., hydrolysis, reaction with NH3, C2H5OH, etc.) in acid derivatives (acid chlorides, anhydrides, esters and amides) takes place at acyl carbon atom (difference from nucleophilic substitution in alkyl halides where substitution takes place at alkyl carbon atom). Nucleophilic substitution in acyl halides is faster than in alkyl halides. This is due to the presence of > CO group in acid chlorides which facilitate the release of halogen as halide ion.

R C

O δ–

δ+ Cl δ–

Acid chloride

R δ+

Cl δ–

Alkyl chloride

Comparison of nucleophilic substitution (e.g., hydrolysis) in acid derivatives. Let us first study the mechanism of such reaction.

R C Z + Nu (i) Addition step R C

O

Nu

Z

O

(ii) Elimination step

R C

O

Nu + Z (where Z= —Cl, —OCOR, —OR, —NH2 and Nu =

A nucleophile) Nucleophilic substitution in acid derivatives

R C R' R C

O

Nu

R'

ONu R C

OH

Nu

R'

H+

(where R' = H or alkyl group)

Nucleophilic addition on aldehydes and ketones The (i) step is similar to that of nucleophilic addition

in aldehydes and ketones and favoured by the presence of electron withdrawing group (which would stabilise the intermediate by developing negative charge) and hindered by electron-releasing group. The (ii) step (elimination of the leaving group Z) depends upon the ability of Z to accommodate electron pair, i.e., on the basicity of the leaving group. Weaker bases are good leaving groups, hence weaker a base, the more easily it is removed. Among the four leaving groups (Cl–, –OCOR, –OR, and –NH2) of the four acid derivatives, Cl– being the weakest base is eliminated most readily. The relative order of the basic nature of the four groups is

–NH2 > –OR > –O.COR > Cl– Hence acid chlorides are most reactive and acid

amides are the least reactive towards nucleophilic acyl substitution. Thus, the relative reactivity of acid derivatives (acyl compounds) towards nucleophilic substitution reactions is

ROCl > RCO.O.COR > RCOOR > RCONH2 Acid Acid Esters Acid chlorides anhydrides amides

OH– being stronger base than Cl–, carboxylic acids (RCOOH) undergo nucleophilic substitution (esterfication) less readily than acid chlorides.


Physical Chemistry

Fundamentals

The temperature dependence of reaction rates : The rate constants of most reactions increase as the

temperature is raised. Many reactions in solution fall somewhere in the range spanned by the hydrolysis of methyl ethanoate (where the rate constant at 35ºC is 1.82 times that at 25ºC) and the hydrolysis of sucrose (where the factor is 4.13).

(a) The Arrhenius parameters : It is found experimentally for many reactions that a

plot of ln k against 1/T gives a straight line. This behaviour is normally expressed mathematically by introducing two parameters, one representing the intercept and the other the slope of the straight line, and writing the Arrhenius equaion.

ln k = ln A – RTEa ......(i)

The parameter A, which corresponds to the intercept of the line at 1/T = 0(at infinite temperature, shown in figure), is called the pre-exponential factor or the 'frequency factor'. The parameter Ea, which is obtained from the slope of the line (–Ea/R), is called the activation energy. Collectively the two quantities are called the Arrhenius parameters.

Slope = –Ea/R

ln A

ln k

1/T A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the Arrhenius equation. The slope gives –Ea/R and the intercept at 1/T = 0 gives ln A.

The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of the rate constant (that is, the steeper the slope). A high activation energy signifies that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see that such

behaviour is a signal that the reaction has a complex mechanism.

The temperature dependence of some reactions is non-Arrhenius, in the sense that a straight line is not obtained when ln k is plotted against 1/T. However, it is still possible to define an activation energy at any temperature as

Ea = RT2

dTknd l .......(ii)

This definition reduces to the earlier one (as the slope of a straight line) for a temperature-independent activation energy. However, the definition in eqn.(ii) is more general than eqn.(i), because it allows Ea to be obtained from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a straight line. Non-Arrhenius behaviour is sometimes a sign that quantum mechanical tunnelling is playing a significant role in the reaction.

(b) The interpretation of the parameters : We shall regard the Arrhenius parameters as purely

empirical quantities that enable us to discuss the variation of rate constants with temperature; however, it is useful to have an interpretation in mind and write eqn.(i) as

k = RT/EaAe− .......(iii) To interpret Ea we consider how the molecular

potential energy changes in the course of a chemical reaction that begins with a collision between molecules of A and molecules of B(shown in figure).

Ea

Reactants

Products

Progress of reaction

Pote

ntia

l ene

rgy

A potential energy profile for an exothermicreaction. The height of the barrier betweenthe reactants and products is the activationenergy of the reaction

CHEMICAL KINETICS

KEY CONCEPT


As the reaction event proceeds, A and B come into contact, distort, and begin to exchange or discard atoms. The reaction coordinate is the collection of motions, such as changes in interatomic distances and bond angles, that are directly involved in the formation of products from reactants. (The reaction coordinate is essentially a geometrical concept and quite distinct from the extent of reaction.) The potential energy rises to a maximum and the cluster of atoms that corresponds to the region close to the maximum is called the activated complex. After the maximum, the potential energy falls as the atoms rearrange in the cluster and reaches a value characteristic of the products. The climax of the reaction is at the peak of the potential energy, which corresponds to the activation energy Ea. Here two reactant molecules have come to such a degree of closeness and distortion that a small further distortion will send them in the direction of products. This crucial configuration is called the transition state of the reaction. Although some molecules entering the transition state might revert to reactants, if they pass through this configuration then it is inevitable that products will emerge from the encounter.

We also conclude from the preceding discussion that, for a reaction involving the collision of two molecules, the activation energy is the minimum kinetic energy that reactants must have in order to form products. For example, in a gas-phase reaction there are numerous collisions each second, but only a tiny proportion are sufficiently energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess of an energy Ea is given by the Boltzmann distribution as RT/Eae− . Hence, we can interpret the exponential factor in eqn(iii) as the fraction of collision that have enough kinetic energy to lead to reaction.

The pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Hence, the product of A and the exponential factor,

RT/Eae− , gives the rate of successful collisions. Kinetic and thermodynamic control of reactions : In some cases reactants can give rise to a variety of

products, as in nitrations of mono-substituted benzene, when various proportions of the ortho-, meta-, and para- substituted products are obtained, depending on the directing power of the original substituent. Suppose two products, P1 and P2, are produced by the following competing reactions :

A + B → P1 Rate of formation of P1 = k1[A][B] A + B → P2 Rate of formation of P2 = k2[A][B] The relative proportion in which the two products

have been produced at a given state of the reaction (before it has reached equilibrium) is given by the

ratio of the two rates, and therefore of the two rate constants :

]P[]P[

1

2 = 1

2

kk

This ratio represents the kinetic control over the proportions of products, and is a common feature of the reactions encountered in organic chemistry where reactants are chosen that facilitate pathways favouring the formation of a desired product. If a reaction is allowed to reach equilibrium, then the proportion of products is determined by thermodynamic rather than kinetic considerations, and the ratio of concentration is controlled by considerations of the standard Gibbs energies of all the reactants and products.

The kinetic isotope effect The postulation of a plausible mechanism requires

careful analysis of many experiments designed to determine the fate of atoms during the formation of products. Observation of the kinetic isotope effect, a decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope, facilitates the identification of bond-breaking events in the rate-determining step. A primary kinetic isotope effect is observed when the rate-determining step requires the scission of a bond involving the isotope. A secondary isotope effect is the reduction in reaction rate even though the bond involving the isotope is not broken to form product. In both cases, the effect arises from the change in activation energy that accompanies the replacement of an atom by a heavier isotope on account of changes in the zero-point vibrational energies.

First, we consider the origin of the primary kinetic isotope effect in a reaction in which the rate-determining step is the scission of a C–H bond. The reaction coordinate corresponds to the stretching of the C–H bond and the potential energy profile is shown in figure. On deuteration, the dominant change is the reduction of the zero-point energy of the bond (because the deuterium atom is heavier). The whole reaction profile is not lowered, however, because the relevant vibration in the activated complex has a very low force constant, so there is little zero-point energy associated with the reaction coordinate in either isotopomeric form of the activated complex.

Pote

ntia

l ene

rgy Ea(C–H)

Ea(C–D)

C–H C–D

Reaction coordinate


1. Arrange the following in :

(i) Decreasing ionic size : Mg2+, O2–, Na+, F– [IIT-1985]

(ii) Increasing acidic property : ZnO, Na2O2, P2O5, MgO [IIT-1985] (iii) Increasing first ionization potential : Mg, Al, Si, Na [IIT-1985] (iv) Increasing size : Cl–, S2–, Ca2+, Ar [IIT-1986] (v) Increasing order of ionic size : N3–, Na+, F–, O2–, Mg2+ [IIT-1991] (vi) Increasing order of basic character : MgO, SrO, K2O, NiO, Cs2O [IIT-1991] (vii) Arrange the following ions in order of their

increasing radii : Li+, Mg2+, K+, Al3+. Sol. (i) O2– > F– > Na+ > Mg2+ Note that all the above ions are isoelectronic

having 10 electron each. In such a case the greater the nuclear charge, the

greater is the attraction for electrons and smaller is the ionic radius. Hence O2– has the highest and Mg2+ has the least ionic size.

(ii) 221

ONa+

< OMg2+

< OZn2+

< 5

5

2 OP+

Among oxides the acidic strength increases with

oxidation state. So Na2O2 is least acidic and P2O5 is most acidic. Further Na2O2 and MgO are basic, ZnO is amphoteric and P2O5 is acidic.

(iii) The first ionization potential of the 3rd period elemens follows the order

Na < Al < Mg < Si ; Ionisation energy increases across a period but not regularly. Mg (1s2, 2s2p6, 3s2) is more stable because the electron is to be removed from 3s which is difficult as compared to Al (1s2, 2s2p6, 3s2p1) where electron is to be removed from 3p.

(iv) Ca2+ < Ar > Cl– < S2– ; All of these are isoelectronic. In such cases the greater the nuclear charge, the greater is the attraction for electrons and smaller is ionic size.

ionic radius ∝ eargchnucleareffective

1

(v) Increasing order of ionic size Mg2+ < Na+ < F– < O2– < N3– Note that all the above ions are isoelectronic

having 10 electron each. In such a case the greater the nuclear charge, the

greater is the attraction for electrons and smaller is the ionic radius. Hence N3– has the highest and Mg2+ has least ionic size.

(vi) Increasing order of basic character : NiO < MgO < SrO < K2O < Cs2O The basic character of oxides increases when

we move down the group. So, K2O < Cr2O and MgO < SrO.

Further higher the group number lesser is the basic character. Hence NiO is the least basic.

(vii) Al3+ < Mg2+ < Li+ < K+ In these Al3+ & Mg2+ are isoelectronic species,

so in these size decreases with increase in atomic number because increase in atomic number decreases Zeff.

Size ∝ .effZ

1

In Li+ & K+, K+ is in size than Li+ because on moving from top to bottom in a group, the group size increases.

2. A white amorphous powder (A) on heating yields a

colourless, non-combustible gas (B) and solid (C). The latter compound assumes a yellow colour on heating and changes to white on cooling. 'C' dissolves in dilute acid and the resulting solution gives a white precipitate on adding K4Fe(CN)6 solution.

'A' dissolves in dilute HCl with the evolution of gas, which is identical in all respects with 'B'. The gas 'B' turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of 'A', as obtained above, gives a white precipitate (D) on the addition of excess of NH4OH and and passing H2S. Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on futher addition of the base. Identify the compounds A, B, D and E. [IIT-1979]

Sol. (i) ZnCO3 →∆ ZnO + CO2 (A) (B) (ii) ZnO + 2HCl → H2O + ZnCl2 (C) (soluble) (iii) 2ZnCl2 + K4[Fe(CN)6] → 4KCl + Zn2[Fe(CN)6]↓ (white ppt) (iv) ZnCO3 + HCl → CO2 + ZnCl2 (A) (soluble) (v) CO2 + Ca(OH)2 → CaCO3 + H2O (B) (Milky) (vi) CaCO3 + CO2 + H2O → Ca(HCO3)2 (soluble)

UNDERSTANDINGPhysical Chemistry


(vii) ZnCl2 + H2S → OHNH4 2HCl + ZnS↓ (white) (viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓ (white) (ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2 sod. ziniate (soluble) 3. Write the lewis dot structural formula for each of the

following. Give, also, the formula of a neutral molecule, which has the same geometry and the same arrangement of the bonding electrons as in each of the following. An example is given below in the case of H3O+ :

H

•• • O • • H •

H

Lewis dot structure

+

H

•• • N • • H •

H

Neutral molecule

(i) O22– ; (ii) CO3

2– ; (iii) CN– ; (iv) NCS– [IIT-1983] Sol. Lewis dot structure Neutral molecule

O2

2– ; O O××

×× ××

•• ••

• × • • 2–

F2 ; F F

••

•• • •

••••

• • ••

Co32– ; O C O

×× × ×

• ×

2–

×× ×

× • O × • ×

• • × × ××

× × •

SO3 ; O S O

× ×

• ×

O

•• ××××× •

××••

×× ××

CN– ; C N• •

× × • • –

• ×× ×

N2 ; N N• • × ×• • • ×× ×

NCS– ; N C S• • •

–

• •• •• × × × ×× ×

••

CO2 ; O C O• • ××

•• ••

×× ••••••

4. Compound (X) on reduction with LiAlH4 gives a

hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air. Draw the structure of (Y).

Sol. Since B2O3 is formed by reaction of (Y) with air, (Y) therefore should be B2H6 in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH4 gives B2H6. Thus it is boron trihalide. The reactions are shown as :

4BX3 + 3LiAlH4 → 2B2H6 + 3LiX + 3AlX3 (X) (Y) (X = Cl or Br) B2H6 + 3O2 → B2O3 + 3H2O + heat (Y)

Structure of B2H6 is as follows :

BB

••

••

Ht

Ht Ht

Ht

Hb

Hb

or B B

Hb

Hb Ht

Ht Ht

Ht

121.5º97º

1.33Å 1.19Å

1.77Å

Thus the diborane molecule has four two-centre-two-electron bonds (2c – 2e bonds) also called usual bonds and two three-centre-two-electron bonds (3c – 2e) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively.

5. Draw the structure of XeF4 and OSF4 according to

VSEPR theory, clearly indicating the state of hybridisation of the central atom and lone pair of electrons (if any) on the central atom. [IIT-2004]

Sol. First determine the total number of electron pairs around the central atom.

XeF4 = 2N =

248 + = 6

Thus in XeF4, Xe is sp3d2 hybridised. The structure of the molecule is octahedral and shape is square planer with two lone pair of electrons.

F

FF

F

Xe

••

••

OF

F

S F

F

For OSF4 : 2N =

246 + = 5

Thus the central atom (S) is sp3d hybridised leading to trigonal bipyramidal structure with no lone pair of electrons.


1. Show that the six planes through the middle point of

each edge of a tetrahedron perpendicular to the opposite edge meet in a point.

2. Prove that if the graph of the function y = f (x), defined throughout the number scale, is symmetrical about two lines x = a and x = b, (a < b), then this function is a periodic one.

3. Show that an equilateral triangle is a triangle of maximum area for a given perimeter and a triangle of minimum perimeter for a given area.

4. Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region

| z | ≤ ab +

bc

5. An isosceles triangle with its base parallel to the

major axis of the ellipse 2

2

ax + 2

2

by = 1 is

circumscribed with all the three sides touching the ellipse. Find the least possible area of the triangle.

6. If one of the straight lines given by the equation ax2 + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′xy + b′y2 = 0 and the other lines represented by them be perpendicular, show that

´a´b´b´ha

− =

ab´ab´h

−

7. Prove that

0n

nm

+

1n

+n

1m +

2n

+n

2m + .........

.... to (n + 1) terms

=

0n

0m

+

1n

1m

2 +

2n

2m

22 + ........

..... to (n + 1) terms

8. If n ≥ 2 and In = ∫−

−1

1

n2 )x1( cos mx dx, then show that

m2In = 2n(2n – 1) In–1 – 4n(n – 1) In–2.

9. Find the sum to infinite terms of the series

43 +

365 +

1447 +

4009 +

90011 + ........ ∞

10. ABC is a triangle inscribed in a circle. Two of its

sides are parallel to two given straight lines. Show that the locus of foot of the perpendicular from the centre of the circle on to the third side is also a circle, concentric to the given circle.

`tàxÅtà|vtÄ VtÄÄxÇzxá 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 Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

8Set

MEMORABLE POINTS

• The vector relation between linear velocity and

angular velocity is →v =

→ω ×

→r

• In the case of uniform circular motion the angle between →ω and

→r is always 90º(hence |

→v | = ωr

• The relation between Faraday constant F, Avogadro number N and the electronic charge e is F = Ne

• Depolariser used in Lechlanche cell is

Manganese dioxide

• The absorption or evolution of heat at a junction of two dissimilar metals when a current is passed is known as Peltier effect

• The part of the human ear where sound is transduced is the Cochlea

• Similar trait resulting from similar selection pressure acting on similar gene pool is termed

Parallel evolution

• Group of related species with the potential, directly or indirectly, of forming fertile hybrids with one another is called Coenospecies


1. Let the line be y = 2x + c

Point A

+−

3c29,

6c9

Point B

−−+

−−

36c,

33c2

Point C

++

312c5,

36c

mid point of B & C is 21 .

+

+−

−3

6c3

3c2 ,

+

++

+−3

12c53

6c21 =

+−

39c2,

6c9

which is point A, so AB and AC are equal.

2.

A

b

C D B

a

ba + = AB1 .

ABAB +

AC1 .

ACAC

= 2AB1 AB + AC

AC1

2

= )DBAD(AB

12 + + )DCAD(

AC1

2 +

=

+ 22 AC

1AB

1 AD + DC.BD

DB + CB.CD

DC

=

+ 22 AC

1AB

1 AD +

+

CDDC

BDDB

BC1

= AD .

+

CB.CD1

DC.BD1

= CDAD

.

+

CD1

BD1

= CD.BDBDDC.

BCAD +

= CD.BD

AD = 2AD

AD =

ADAD

.AD1

so it is vector along BA with magnitude AD1 .

| ba + | = AD1

3. The line PQ always passes through (α, β) so it is y –β = m(x – α)

Let the circle be x2 + y2 – 2hx – 2ky = 0

Joint equation of OP and OQ.

x2 + y2 – 2 (hx + ky) α−β

−m

)mxy( = 0

P

(h,k)

Q

O

α−β

−mk21 y2 – 2

α−β

−mmkh xy +

α−β

+m

hn21 x2 = 0

It must represent y2 – x2 = 0

so α−β

−mmkh = 0 ⇒ m = h/k ...(1)

and 1 – α−β m

k2 = –1 –α−β m

hm2

⇒ β – mα – 2k = –β + mα – 2hm

⇒ –β + mα + k – hm = 0

⇒ –β + k + h/k(α – h) = 0 (using (1) in it)

⇒ k2 – βy + αh – h2 = 0 so required locus is

x2 – y2 – αx + βy = 0

MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 7)


4. As |f(x)| ≤ |tan x| for ∀ x ∈

ππ−

2,

2

so f(0) = 0

so |f(x) – f(0)| ≤ |tan x|

divides both sides by |x|

⇒ x

)0(f)x(f −≤

xxtan

⇒ 0x

lim→ x

)0(f)x(f −≤

0xlim

→ xxtan

⇒ |f´(0)| ≤ 1

⇒ n321 an1.....a

31a

21a ++++ ≤ 1

⇒ ∑=

n

1i

i

ia ≤ 1

5. Let the number is xyz, here x < y and z < y.

Let y = n, then x can be filled in (n – 1) ways. (i.e. from 1 to (n – 1)) and z can be filled in n ways (i.e. from 0 to (n – 1))

here 2 ≤ n ≤ 9

so total no. of 3 digit numbers with largest middle digit

= ∑=

−9

2n

)1n(n = ∑=

9

2n

2n – ∑=

9

2n

n

= 6

19.10.9 – 210.9

= 285 – 45 = 240

required probability = 10109

240××

= 308

= 154

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

(1,0)

(3/2,–1)(0,–1)

(–1,0)

(–1,log23)

= ∫−

−1

12 dx)x2(log + ∫

−

−0

1

y dy)22( + 41

π

= 3log22n

23log 22 +−l

+ 2 – 2n2

1l

+ 4π

= – log2 27ee2

+ 2 + 4π sq units

7. A

E F

B D

M

C

∠BMC = 2∠BAC = 2∠BMD

so tan A = MDBD =

MD2BC =

1r4BC =

1r4a

so 21

2

ra = tan2A

so 21

2

ra + 2

2

2

rb + 2

3

2

rc

= 16 (tan2A + tan2B + tan2C) ...(1) Now as tan A + tan B + tan C ≥ 3 (tan A . tan B . tan C)1/3 and for a triangle tan A + tan B + tan C = tan A . tan B . tan C so (tan A . tan B . tan C)2/3 ≥ 3

⇒ tan A . tan B . tan C ≥ 3 3

⇒ tan2A + tan2B + tan2C ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3

so from (1), 21

2

ra + 2

2

2

rb + 2

3

2

rc ≥ 144.


8. Z1, Z2, Z3 are centroids of equilateral triangles ACX, ABY and BCZ respectively.

Z1 – ZA = (ZC – ZA) AC

A1

ZZZZ

−− eiπ/6

x

y A

ZA

ZB Z3

ZC

z

C B

Z2

Z1

Z1 – ZA = (ZC – ZA)

+

2i

23

31 ...(1)

similarly,

Z2 – ZA = (ZB – ZA)

−

2i

23

31 ...(2)

So, Z1 – Z2 = 21 (ZC – ZB) +

32i (ZC + ZB – 2ZA)

...(3)

similarly Z2 – Z3 = 21 (ZA – ZC)

+ 32

i (ZA + ZC – 2ZB) ..(4)

To prove ∆xyz as equilateral triangle, we prove that (Z3 – Z2)eiπ/3 = Z1 – Z2

So, (Z3 – Z2)eiπ/3 = (21 (ZC – ZA)

– 32

i (ZA + ZC – 2ZB))

+ i

23

21

= 21 (ZC – ZB) +

32i (ZC + ZB – 2ZA)

= Z1 – Z2

9. Tr = 2 ∫ +−−

a

02rucosr21

ucosr1 du. ...(1)

= ∫ +−+−+−

a

02

22

rucosr211rrucosr21 du

= ∫

+−−

+a

02

2

rucosr21r11 du

= a + (1 – r2)

∫ −−++

a

0222

2

)2/utan1(r2)2/utan1)(r1(2/usec

= a + (1 – r2) ∫ −++

a

0222

2

)r1(2/utan)r1(2/usec

= a + 2

2

)r1(r1

+−

∫+−

+

a

02

22

2

)r1()r1(2/utan

du2/usec

Let tan u/2 = t

so, Tr = a + 2

2

)r1(r1

+−

∫+−

+

2/atan

02

2

r1r1t

dt2

= a + 2

2

)r1()r1(2

+−

r1r1

−+

2/atan

0

1

r1r1ttan

−+−

Now +→1r

lim Tr = a – )1r)(r1()1r)(r1(2

−+−+

2π = a – π

and +→1r

lim Tr = a + )1r)(r1()1r)(r1(2

−++−

2π = a + π

and (from (1)) T1 = ∫a

0

du = a

Hence +→1r

lim Tr, T1, −→1r

lim Tr form an A.P. with common

difference π.

10. Let α, β, γ be the three real roots of the equation

without loss of generality, it can be assumed that α ≤ β ≤ γ.

so

x2 + ax2 + bx + c = (x – γ) (x2 + (a + γ) x + (γ2 + aγ + b)) where – γ (γ2 + aγ + b) = c, as γ is the root of given equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must have two roots i.e. α and β. So its discriminant is non negative, thus

(γ + a)2 – 4(γ2 + aγ + b) ≥ 0

3γ2 + 2aγ – a2 + 4b ≤ 0

so γ ≤ 3

b3a2a 2 −+−

so greatest root is also less than or equal to

3b3a2a 2 −+− .


1. Find the range of the function

f(x) =

++

1xexlogsin 2

2

e +

++

1xexlogcos 2

2

e

Sol. we have,

1xex

2

2

++ = 1 +

1x1–e

2 +

⇒ 1xex

2

2

++ > 1

>

+xallfor0

1x1–e

2Q

Again,

1xex

2

2

++ = 1 +

1x1–e

2 +

⇒ 1xex

2

2

++ ≤ 1 + e – 1 = e

=+

0xatvalueimummax

itsassumes1x

1–e2Q

Thus,

1 < 1xex

2

2

++ ≤ e for all x ∈ R.

⇒ 0 < loge

++

1xex

2

2 ≤ 1 for all x ∈ R

⇒ 0 < θ ≤ 1, where θ = loge

++

1xex

2

2.

∴ f(x) = θsin + θcos

Let g (θ) = θsin + θcos , where 0 ≤ θ ≤ 1. Clearly, the range of f(x) will be same as that of g(θ). Now, g(θ) = θsin + θcos

⇒ g'(θ) = θ

θ

sin2cos –

θ

θ

cos2sin

⇒ g'(θ) = θθ

θθ

cossin2)(sin–)(cos 2/32/3

⇒ g'(θ) ≥ 0 for 0 < θ ≤ π/4 and g'(θ) < 0 for π/4 < θ ≤ 1 ⇒ g(θ) is increasing on (0, π/4] and decreasing on

(π/4,1] ⇒ g(θ) ∈ ( ])4/(g),0(g π for 0 ∈ (0, π/4] and, g(0) ∈ [ ))4/(g),1(g π for θ ∈ (π/4, 1]

⇒ g(θ) ∈ (1, 23/4] for θ ∈ (0, π/4] and, g(θ) ∈ )2,1cos1sin[ 4/3+ for θ ∈ (π/4, 1] ⇒ g(θ) ∈ (1, 23/4], for θ ∈ (0, 1] ]11cos1sin[ >+Q Hence, range of f(x) is (1, 23/4] 2. A ladder 15 m long, leans against a wall 7 m high,

and a portion of the ladder protrudes over the wall such that its projection along the vertical is 3 m. How fast does the bottom start to slip away from the wall if the ladder slides down along the top edge of the wall at 2 m/sec.

Sol. Let OC be the wall and AB be the ladder of which the portion BC is protruded over the wall OC. The projection of BC on the wall is CD = 3 m (given).

Let the foot A of the ladder be x metres away from the wall and the protruded portion of the ladder be y m in length. Further, let ∠OCA = θ.

Now, AB = 15 m and BC = y m ∴ AC = (15 – y) m.

B Dy m

Cθ

7m

O x m

(15–y) m

A

It is given that dtdy = – 2 m/sec.

We have to find dtdx .

In ∆OCA, we have

tan θ = 7x and cos θ =

y–157

⇒ x = 7 tan θ and, 15 – y = 7 sec θ. In ∆BCD, we have

cos θ = y3

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumMATHS


⇒ y = 3 sec θ. ∴ 15 – 3 sec θ = 7 sec θ

⇒ sec θ = 1015

⇒ sec θ = 23

⇒ cos θ = 32

Now, 15 – y = 7 sec θ ⇒ y = 15 – 7 sec θ

⇒ dtdy = – 7 sec θ tan θ

dtdθ

= )given(2–

dtdy

Q

⇒ – 2 = – 7 sec θ tan θ dtdθ

⇒ dtdθ =

θθ tansec72 ...(1)

Now, x = 7 tan θ

⇒ dtdx = 7 sec2θ

dtdθ

⇒ dtdx = 7 sec2θ ×

θθ tansec72 [Using (1)]

⇒ dtdx =

θsin2

⇒ dtdx =

94–1

2

=θ

32cosQ

⇒ dtdx =

56

Hence , the bottom starts to slip away from the

wall at the rate of 5

6 m/sec.

3. If In = ∫1

0

1–n dxxtanx , prove that

(n + 1) In + (n – 1) In–2 = 2π –

n1

Sol. We have,

In = dxxtanx1

0I

1–

II

n∫

⇒ In = 1

0

1–1n

xtan1n

x

+

+

– 1n

1+

∫ +

+1

02

1n

x1x dx

⇒ (n + 1) In = [ ]10

1–1n xtanx + – ∫ +

+1

02

1n

x1x dx

⇒ (n + 1) In = 4π – ∫ +

+1

02

1n

x1x dx ...(i)

⇒ (n – 1) In–2 = 4π – ∫ +

1

02

1–n

x1x dx ...(ii)

)i(in)2–n(

bynplacingRe

Adding (i) and (ii), we get

(n + 1) In + (n – 1) In –2 = 2π – ∫ +

++1

02

1–n1ndx

x1xx

⇒ (n + 1) In + (n – 1) In–2 = 2π – ∫ +

+1

02

21–ndx

x1)1x(x

⇒ (n + 1) In + (n – 1) In–2 = 2π – dxx

1

0

1–n∫

⇒ (n + 1) In + (n – 1) In–2 = 2π –

n1

4. Prove that the internal bisector of an angle of a

triangle and the external bisector of the other two are concurrent.

Sol. Referred to the vertex C as the origin, let →a ,

→b be

the position vectors of the vertices A and B respectively. Let a, b, c be the lengths of the sides BC, CA and AB respectively.

C(origin)

→)a(A

→)b(B

D Vector equation of the internal bisector of ∠C is

→r = λ1

+

→→

CBCB

CACA

or →r = λ1

+

→→

ab

ba

or →r = λ1

+→→

abbbaa

or →r =

ab1λ (

→→+ bbaa ) ...(i)


The external bisector of ∠A is the internal bisector of

the angle between vectors →

CA and →

AB . Therefore, vector equation of the external bisector of ∠A is

→r =

→a + λ2

+

→→

CACA

ABAB

or →r =

→a + λ2

+

→→→

ba

ca–b ...(ii)

Similarly the external bisector of ∠B is the internal

bisector of the angle between the vector →

CB and →

BA . Therefore, the vector equation of the external bisector of ∠B is

→r =

→b + λ3

+

→→

BABA

CBCB

or, →r =

→b + λ3

+

→→→

cb–a

ab ...(iii)

Suppose (ii) and (iii) intersect at D. Then, for the point D, we have

→a + λ2

+→

→→→

b

ac

a–b = →b + λ3

+

→→→

cb–a

ab

⇒ 1 – c2λ +

b2λ =

c3λ

and, c2λ = 1 +

a3λ

– c3λ

[On equating the coefficient of →a and

→b ]

⇒ c3λ

= 1 + bc

2λ (c – b) and c2λ = 1 +

ac3λ

(c – a)

⇒ c2λ = 1 +

a1 (c – a)

λ

+ )b–c(bc

1 2

[Substituting the value of in second equation]

⇒ c2λ = 1 +

aa–c +

abc2λ (c – a) (c – b)

⇒ c2λ =

ac +

abc2λ (c2 – ac – bc – ab)

⇒ abc

2λ = (ab – c2 + ac + bc – ab) = ac

⇒ ab

2λ (a + b – c) = ac

⇒ λ2 = c–ba

bc+

Putting this value of λ2 in c3λ

= 1 + bc

2λ (c – b), we

get

c3λ

= 1 + c–ba

b–c+

⇒ λ3 = c–ba

ac+

Now, putting the value of λ2 in (ii) or that of λ3 in

(iii), we get the position vector →

1r of D as

→

1r = →a +

cbabc

++

+

→→→

ba

ca–b

= c–ba

ac)a–b(b)c–ba(a+

+++→→→→

= c–babbaa

++

→→

Clearly, →

1r = c–babbaa

++

→→

satisfies equation of CD i.e.

(i) for λ1 = c–ab

ab+

Hence the internal bisector of ∠C and external bisector of ∠A and ∠B are concurrent.

5. Find the solutions of the equation x2 – 3

π

6–xsin = 3,

where [.] denotes the greatest integer function. Sol. The given equation can be written as

x2 – 3 = 3

π

6–xsin ...(i)

Clearly, right hand side can take only three values –3, 0, 3.

CASE I When

π

6–xsin3 = 3 :

In this case, equation (i) reduces to x2 – 3 = 3 ⇒ x = ± 6

But, for x = ± 6 , we have

π

6–xsin3 =

π±

6–6sin3 ≠ 3.

So, x = ± 6 is not a solution of the given equation

CASE II When

π

6–xsin3 = 0 :

In this case, the given equation reduces to x2 – 3 = 0 ⇒ x = ± 3

When x = 3 , we have,


x – 6π = 3 –

6π <

2π ⇒

π

6–xsin = 0

∴ x = 3 is a solution.

When x = – 3 , we have

x – 6π = – 3 –

6π

⇒ – π < x – 6π < 0

⇒

π

6–xsin = – 1

∴ x = – 3 in not a solution.

CASE III When

π

6–xsin3 = – 3

In this case, the given equation reduces to x2 – 3 = – 3 ⇒ x = 0 When x = 0, we have

π

6–xsin =

π

6–sin = – 1.

⇒

π

6–xsin3 = – 3.

So, x = 0 is a solution of the given equation. Hence, the given equation has only two solution

x = 0, 3 . 6. A condolence meeting is being held in a big hall

which has 7 doors by which the mourners enter the hall. One can use any one of the 7 doors to enter and can come at any time during meeting. At each door a register is kept in which a mourner has to put down his signature while entering the hall. If 200 people attend the meeting, how many different sets of 7 lists of signatures can arise?

Sol. Suppose xi persons enter into the hall through ith door, where i = 1, 2, ..., 6, 7. Then,

x1 + x2 + x3 + x4 + x5 + x6 + x7 = 200, ...(i) where xi ≥ 0; i = 1, 2, ... , 7 The total number of non-negative integral solutions

of equation (i) is 200+7–1 C7–1 = 206C6

Thus, there are 206C6 ways in which 200 persons can enter into the hall through 7 doors. But corresponding to each way of entering into the hall the signatures of 200 persons can be arranged in 200! ways.

Hence, the required number of lists

= 206C6 × 200! = !6

!206

Know about Pie

3.14

= Perimeter / Diagonal, of any circle. Pi expanded to 45 decimal places: 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 Pi expanded to 52 binary places: 11.0010010000111 1110110101010 0010001000010 1101000111001 You cannot square a disc using just a ruler and compasses because is a transcendental number. = 4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... ) = 2(2/1 x 2/3 x 4/3 x 4/5 x 6/5 x 6/7 x 8/7 x 8/9 x ... ) ≈ 355/113 (a real good rational approximation of ) ≈ (6 2)/5 In the late 18th century, James Stirling, a Scottish mathematician, developed an approximation for factorials using the transcendental numbers 'Pi' and 'e': n! ≈ (2 n)1/2 (n/e)n The most famous formula for calculating Pi is Machin's formula:

/4 = 4 arctan(1/5) – arctan(1/239) This formula, and similar ones, were used to push the accuracy of approximations to Pi to over 500 decimal places by the early 18th century (this was all hand calculation!). Interestingly, there are no occurrences of the sequence 123456 in the first million digits of Pi. - posted by George Pantazis Bamboozlement with Pi Does Pi equal 3? No? Then have a look on the algebraic equation below: x = ( + 3)/2 2x = + 3 2x( - 3) = ( + 3)( - 3) 2 x - 6x = 2 - 9 9 - 6x = 2 - 2 x 9 - 6x + x2 = 2 - 2 x + x2 (3 - x)2 = ( - x)2 3 - x = - x 3 = We use Pi to: Describe the DNA double helix, Determining the distribution of primes - the probability that two randomly selected integers are relatively prime (i.e. have no common factors) is 6 / p2, Analyzing the ripples on water Checking for accuracy - as there are now millions upon millions of known decimal places of Pi, by asking a super computer to compute this many figures its accuracy can be tested.


Monotonic Functions : A function f(x) defined in a domain D is said to be (i) Monotonic increasing :

⇔

≥⇒>≤⇒<

)x(f)x(fxx)x(f)x(fxx

2121

2121 ∀ x1, x2 ∈ D

y

O x

y

O x

i.e., ⇔

<⇒>>⇒<

)x(f/)x(fxx)x(f/)x(fxx

2121

2121 ∀ x1, x2 ∈ D

(ii) Monotonic decreasing :

⇔

≤⇒>≥⇒<

)x(f)x(fxx)x(f)x(fxx

2121

2121 ∀ x1, x2 ∈ D

y

O x

y

O x

i.e., ⇔

>⇒><⇒<

)x(f/)x(fxx)x(f/)x(fxx

2121

2121 ∀ x1, x2 ∈ D

A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain.

Note : If x1 < x2 ⇒ f(x1) < f(x2) ∀ x1, x2 ∈ D, then f(x) is called strictly increasing in domain D and similarly decreasing in D.

Method of testing monotonicity : (i) At a point : A function f(x) is said to be

monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing (decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may observer that if f(x) is monotonic increasing at x = a then at this point tangent to its graph will make an acute angle with x-axis where as if the function is monotonic decreasing there then tangent will make an obtuse angle with x-axis. Consequently f´(a) will be positive or negative

according as f(x) is monotonic increasing or decreasing at x = a.

So at x = a, function f(x) is monotonic increasing ⇔ f´(a) > 0 monotonic decreasing ⇔ f´(a) < 0 (ii) In an interval : In [a, b], f(x) is

=⇔≤⇔≥⇔

0 f´(x)constant 0 f´(x) decreasing monotonic0 f´(x) increasing monotonic

∀ x ∈ (a, b)

Note : (i) In above results f´(x) should not be zero for all

values of x, otherwise f(x) will be a constant function.

(ii) If in [a, b], f´(x) < 0 at least for one value of x and f´(x) > 0 for at least one value of x, then f(x) will not be monotonic in [a, b].

Examples of monotonic function : If a functions is monotonic increasing (decreasing ) at

every point of its domain, then it is said to be monotonic increasing (decreasing) function.

In the following table we have example of some monotonic/not monotonic functions Monotonic increasing

Monotonic decreasing

Not monotonic

x3 1/x, x > 0 x2 x|x| 1 – 2x |x| ex e–x ex + e–x log x log2x sin x sin h x cosec h x, x > 0 cos h x [x] cot hx, x > 0 sec h x

Properties of monotonic functions : If f(x) is strictly increasing in some interval, then in

that interval, f–1 exists and that is also strictly increasing function.

If f(x) is continuous in [a, b] and differentiable in (a, b), then

f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic increasing in [a, b]

f´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f(x) is monotonic decreasing in [a, b]

MONOTONICITY, MAXIMA & MINIMA

Mathematics Fundamentals MATHS


If both f(x) and g(x) are increasing (or decreasing) in [a, b] and gof is defined in [a, b], then gof is increasing.

If f(x) and g(x) are two monotonic functions in [a, b] such that one is increasing and other is decreasing then gof, it is defined, is decreasing function.

Maximum and Minimum Points : The value of a function f(x) is said to be maximum at

x = a if there exists a small positive number δ such that f(a) > f(x)

a b c

y

x O

( ) ( ) ( )

Also then the point x = a is called a maximum point for the function f(x).

Similarly the value of f(x) is said to be minimum at x = b if there exists a small positive number δ such that

f(b) < f(x) ∀ x ∈ (b – δ, b + δ) Also then the point x = b is called a minimum point

for f(x) Hence we find that : (i) x = a is a maximum point of f(x)

⇔

>>+

0 h)– f(a– f(a)0 h)f(a– )a(f

(ii) x = b is a minimum point of f(x)

⇔

><+

0 h)– f(b– f(b)0 h)f(b– )b(f

(iii) x = c is neither a maximum point nor a minimum point

⇔

−−

+

)hc(f)c(fand

)hc(f–)c(f have opposite signs.

Where h is a very small positive number. Note : The maximum and minimum points are also

known as extreme points. A function may have more than one maximum

and minimum points. A maximum value of a function f(x) in an interval

[a, b] is not necessarily its greatest value in that interval. Similarly a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function.

The greatest and least values of a function f(x) in an interval [a, b] may be determined as follows :

Greatest value = max. f(a), f(b), f(c) Least value = min. f(a), f(b), f(c)

where x = c is a point such that f´(c) = 0. If a continuous function has only one maximum

(minimum) point, then at this point function has its greatest (least) value.

Monotonic functions do not have extreme points. Conditions for maxima and minima of a function Necessary condition : A point x = a is an extreme

point of a function f(x) if f´(a) = 0, provided f´(a) exists. Thus if f´(a) exists, then

x = a is an extreme point ⇒ f´(a) = 0 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

not an extreme point. Sufficient condition : For a given function f(x), a

point x = a is a maximum point if f´(a) = 0 and f´´(a) < 0 a minimum point if f´(a) = 0 and f´´(a) > 0 not an extreme point if f´(a) = 0 = f´´(a) and

f´´´(a) ≠ 0. Note : If f´(a) = 0, f´´(a) = 0, f´´´(a) = 0 then the sign

of f(4)(a) will determine the maximum or minimum point as above.

Working Method : Find f´(x) and f´´(x). Solve f´(x) = 0. Let its roots be a, b, c, ... Determine the sign of f´´(x) at x = a, b, c, .... and

decide the nature of the point as mentioned above. Properties of maxima and minima : If f(x) is continuous function, then

Between two equal values of f(x), there lie atleast one maxima or minima.

Maxima and minima occur alternately. For example if x = –1, 2, 5 are extreme points of a continuous function and if x = –1 is a maximum point then x = 2 will be a minimum point and x = 5 will be a maximum point.

When x passes a maximum point, the sign of dy/dx changes from + ve to – ve, where as when x passes through a minimum point, the sign of f´(x) changes from –ve to + ve.

If there is no change in the sign of dy/dx on two sides of a point, then such a point is not an extreme point.

If f(x) is maximum (minimum) at a point x = a, then 1/f(x), [f(x) ≠ 0] will be minimum (maximum) at that point.

If f(x) is maximum (minimum) at a point x = a, then for any λ ∈ R, λ + f(x), log f(x) and for any k > 0, k f(x), [f(x)]k are also maxmimum (minimum) at that point.


Definition of a Function :

Let A and B be two sets and f be a rule under which every element of A is associated to a unique element of B. Then such a rule f is called a function from A to B and symbolically it is expressed as

f : A → B

or A →f B

Function as a Set of Ordered Pairs

Every function f : A → B can be considered as a set of ordered pairs in which first element is an element of A and second is the image of the first element. Thus

f = a, f(a) /a ∈ A, f(a) ∈ B.

Domain, Codomain and Range of a Function :

If f : A → B is a function, then A is called domain of f and B is called codomain of f. Also the set of all images of elements of A is called the range of f and it is expressed by f(A). Thus

f(A) = f(a) |a ∈ A

obviously f(A) ⊂ B.

Note : Generally we denote domain of a function f by Df and its range by Rf.

Equal Functions :

Two functions f and g are said to be equal functions if

domain of f = domain of g

codomain of f = codomain of g

f(x) = g(x) ∀ x.

Algebra of Functions :

If f and g are two functions then their sum, difference, product, quotient and composite are denoted by

f + g, f – g, fg, f/g, fog

and they are defined as follows :

(f + g) (x) = f(x) + g(x)

(f – g) (x) = f(x) – g(x)

(fg) (x) = f(x) f(g)

(f/g) (x) = f(x)/g(x) (g(x) ≠ 0)

(fog) (x) = f[g(x)]

Formulae for domain of functions :

Df ± g = Df ∩ Dg

Dfg = Df ∩ Dg

Df/g = Df ∩ Dg ∩ x |g(x) ≠ 0

Dgof = x ∈ Df | f(x) ∈ Dg

fD = Df ∩ x |f(x) ≥ 0

Classification of Functions

1. Algebraic and Transcendental Functions :

Algebraic functions : If the rule of the function consists of sum, difference, product, power or roots of a variable, then it is called an algebraic function.

Transcendental Functions : Those functions which are not algebraic are named as transcendental or non algebraic functions.

2. Even and Odd Functions : Even functions : If by replacing x by –x in f(x)

there in no change in the rule then f(x) is called an even function. Thus

f(x) is even ⇔ f(–x) = f(x)

FUNCTION Mathematics Fundamentals M

ATHS


Odd function : If by replacing x by –x in f(x) there is only change of sign of f(x) then f(x) is called an odd function. Thus

f(x) is odd ⇔ f(–x) = – f(x)

3. Explicit and Implicit Functions :

Explicit function : A function is said to be explicit if its rule is directly expressed (or can be expressed( in terms of the independent variable. Such a function is generally written as

y = f(x), x = g(y) etc.

Implicit function : A function is said to be implicit if its rule cannot be expressed directly in terms of the independent variable. Symbolically we write such a function as

f(x, y) = 0, φ(x, y) = 0 etc.

4. Continuous and Discontinuous Functions :

Continuous functions : A functions is said to be continuous if its graph is continuous i.e. there is no gap or break or jump in the graph.

Discontinuous Functions : A function is said to be discontinuous if it has a gap or break in its graph atleast at one point. Thus a function which is not continuous is named as discontinuous.

5. Increasing and Decreasing Functions :

Increasing Functions : A function f(x) is said to be increasing function if for any x1, x2 of its domain

x1 < x2 ⇒ f(x1) ≤ f(x2)

or x1 > x2 ⇒ f(x1) ≥ f(x2)

Decreasing Functions : A function f(x) is said to be decreasing function if for any x1, x2 of its domain

x1 < x2 ⇒ f(x1) ≥ f(x2)

or x1 > x2 ⇒ f(x1) ≤ f(x2)

Periodic Functions :

A functions f(x) is called a periodic function if there exists a positive real number T such that

f(x + T) = f(x) ∀ x

Also then the least value of T is called the period of the function f(x).

Period of f(x) = T

⇒ Period of f(nx + a) = T/n

Periods of some functions :

Function Period

sin x, cos x, sec x, cosec x, 2π

tan x, cot x π

sinnx, cosn x, secn x, cosecn x 2π if n is odd

π if n is even

tann x, cotnx π ∀ n ∈ N

|sin x|, |cos x|, |sec x|, |cosec x| π

|tan x|, |cot x|, π

|sin x| + |cos x|, sin4x + cos4x

|sec x| + |cosec x| 2π

|tan x| + |cot x| 2π

x – [x] 1

Period of f(x) = T ⇒ period of f(ax + b) = T/|a|

Period of f1(x) = T1, period of f2(x) = T2

⇒ period of a f1(x) + bf2(x) ≤ LCM T1, T2

Kinds of Functions :

One-one/ Many one Functions :

A function f : A → B is said to be one-one if different elements of A have their different images in B.

Thus

f is one-one ⇔

=⇒=

≠⇒≠

ba)b(f)a(for

)b(f)a(fba

A function which is not one-one is called many one. Thus if f is many one then atleast two different elements have same f-image.

Onto/Into Functions : A function f : A → B is said to be onto if range of f = codomain of f

Thus f is onto ⇔ f(A) = B


Hence f : A → B is onto if every element of B (co-domain) has its f–preimage in A (domain).

A function which is not onto is named as into function. Thus f : A → B is into if f(A) ≠ B. i.e., if there exists atleast one element in codomain of f which has no preimage in domain.

Note :

Total number of functions : If A and B are finite sets containing m and n elements respectively, then

total number of functions which can be defined from A to B = nm.

total number of one-one functions from A to B

=

>≤

nmif0nmifPm

n

total number of onto functions from A to B (if m ≥ n) = total number of different n groups of m elements.

Composite of Functions :

Let f : A → B and g : B → C be two functions, then the composite of the functions f and g denoted by gof, is a function from A to C given by gof : A → C, (gof) (x) = g[f(x)].

Properties of Composite Function :

The following properties of composite functions can easily be established.

Composite of functions is not commutative i.e.,

fog ≠ gof

Composite of functions is associative i.e.

(fog)oh = fo(goh)

Composite of two bijections is also a bijection.

Inverse Function :

If f : A → B is one-one onto, then the inverse of f i.e., f–1 is a function from B to A under which every b ∈ B is associated to that a ∈ A for which f(a) = b.

Thus f–1 : B → A,

f–1(b) = a ⇔ f(a) = b.

Domain and Range of some standard functions :

Function Domain Range

Polynomial function

R R

Identity function x

R R

Constant function c

R c

Reciprocal function 1/x

R0 R0

x2, |x| R R+ ∪ 0

x3, x |x| R R

Signum function

R –1, 0, 1

x + |x| R R+ ∪ 0

x – |x| R R– ∪ 0

[x] R Z

x – [x] R [0, 1)

x [0, ∞) [0, ∞)

ax R R+

log x R+ R

sin x R [–1, 1]

cos x R [–1, 7]

tan x R – ± π/2, ± 3π/2, ... R

cot x R – 0, ± π. ± 2π, ..... R

sec x R – (± π/2, ± 3π/2, ..... R – (–1, 1)

cosec x R – 0, ±π, ± 2π, ...... R –(–1, 1)

sinh x R R

cosh x R [1, ∞)

tanh x R (–1, 1)

coth x R0 R –[1, –1]

sech x R (0, 1]

cosech x R0 R0

sin–1 x [–1, 1] [–π/2, π/2]

cos–1x [–1, 1] [0, π]

tan–1x R (–π/2, π/2

cot–1 x R (0, π)

sec–1 x R –(–1, 1) [0, π] – π/2

cosec–1x R – (–1, 1) (– π/2, π/2] – 0


a

PHYSICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates, as shown. Potential difference between the plates after the dielectric slab has been introduced is –

(A) 18.5 V (B) 13.5 V (C) 11.5 V (D) 6.5 V

2. Reading of ammeter is 1 A. If each of the 4 Ω resistor is replaced by 2 Ω resistor, the reading of ammeter will become nearly –

4Ω

4Ω 2Ω

15Ω

15Ω

15Ω 10V

A(ammeter)

(A) 1 .11 A (B) 1.25 A (C) 1.34 A (D) 1.68 A 3. In an insulating medium (K = 1) volumetric charge

density varies with y-coordinates according to the law ρ = a.y. A particle of mass m having positive charge q is at point A(0, y0) and projected with velocity

ivv 0=r

as shown in figure. At y = 0 electric field is zero. Neglect the gravity and frictional resistance, the slope of trajectory of the particle as a function of y (E is only along y-axis) is –

(A) )y–y(vm

qa 30

3200ε

(B) )y–y(vm3

qa 30

3200ε

(C) 200

30

3

vm5)y–y(qa

ε (D) 2

00

30

3

vm2)y–y(qa

ε

4. Average torque on a projectile of mass m, initial

speed u and angle of projection θ between initial and final positions P and Q as shown in figure about the point of projection is –

θ

u

Q x P

y

IIT-JEE 2011

XtraEdge Test Series # 8

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 7 to 10 are multiple choice questions with multiple correct answer. +4 marks and No negative mark for

wrong answer. Section - II • Question 11 to 14 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for

wrong answer. • Question 15 to 23 are passage based single type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.

3mm5mm

(0, y0)

y

xA


(A) 2

2sinmu2 θ (B) mu2 cosθ

(C) mu2 sinθ (D) 2cosmu2 θ

5. Portion AB of the wedge shown in figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC, then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is –

A

B

C D

(A) 3/5 (B) 5 (C) 7/5 (D) 8/3 6. Two bodies of masses m1 and m2 are initially at rest

placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity when they are r distance apart is –

(A) r

)mm(G2 21 + (B) r)mm(

mmG2

21

21

+

(C) r

)mm(G 21 + (D) ( )rmmmmG

21

21+

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer.

7. Point 1 is at middle of solenoid, point (2) at an end face and point (3) is outside the solenoid at a distance a. Plane of coil and plane of cross-section of solenoid are parallel –

Solenoid

i Coil

2a a

3 1 2

(A) Force between coil and solenoid is attractive at

all three points (i.e. 1, 2, 3) (B) Force between coil and solenoid at the point 1 is

zero (C) Among these three point force between coil and

solenoid is maximum at point 2 (D) Among these three point force between coil and

solenoid is maximum at point 1

8. Four identical bulbs A, B, C, D are connected in a circuit as shown in figure. Now whenever any bulb fails, then it cannot conduct current through it. Then –

DC

A B

Ideal Battery

(A) Brightness of bulb C is highest (B) If C fails, brightness of bulb D increases (C) If C fails, brightness of all bulbs remain same (D) If A fails, B will not glow 9. A gas is kept in a closed container at temperature T.

If temperature of gas is increased then according to Maxwell theory of molecular speed distribution, choose the correct alternatives :

(A) Number of molecules moving with vrms may increase

(B) Number of molecules moving w3ith vmp must decrease

(C) Number of molecular moving with vav may decrease

(D) Number of molecules moving with 2vrms may increase

10. The speed v of a particle moving along a straight line,

when it is at a distance x from a fixed point on the line is v2 = 144 – 9x2. Select the correct alternative(s):

(A) The motion of the particle is SHM with time

period T = 3

2π unit

(B) The maximum displacement of the particle from the fixed point is 4 unit

(C) The magnitude of acceleration at a distance 3 units from the fixed point is 27 unit

(D) The motion of the particle is periodic but not simple harmonic

This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of

an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.


(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

11. Assertion (A) : When a current flow in the coil of a

transformer then its core becomes hot. Reason (R) : The core of transformer is made of iron. 12. Assertion (A) : The stream of water flowing at high

speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

Reason (R) : In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

13. Assertion (A) : Doppler effect for sound wave is

symmetric w.r.t. speed of source and speed of observer.

Reason (R) : Change in frequency due to motion of source w.r.t. stationary observer is not same with that due to motion of observer w.r.t. stationary source.

14. Assertion (A) : In simple harmonic motion A is the

amplitude of oscillation. If t1 be the time to reach the

particle from mean position to 2

A and t2 the time to

reach from 2

A to A. Then t1 = 2

t2 .

Reason (R) : Equation of motion for the particle starting from mean position is given by x = ± A sin ωt and of the particle starting from extreme position is given by x = ± A cos ωt.

This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 15 to 17)

R 6Ω

24Ω R b R d R R1 R1

R R R1

12Ω

12Ω R2 R1

4Ω x y

V

15Ω

6Ω R

R

R1

R1 A B

a c 24Ω

It is observed that change of R2 has no effect on the

equivalent resistance between x & y. Also if we connect a wire between a & b, there is no flow of charge between a & b. However if a wire is

connected between c & d although we shall actually connect none in the circuit, electrons will flow in a definite direction between c & d. Potential difference between A & a is 6V.

15. Resistance R is – (A) 4 Ω (B) 8 Ω (C) 6 Ω (D) 9 Ω

16. Equivalent resistance of the circuit is nearly – (A) 5.95 Ω (B) 5.25 Ω (C) 6.45 Ω (D) 7.85 Ω 17. Assuming internal resistance of cell to be zero,

potential difference applied between x & y is nearly– (A) 82.6 V (B) 94.2 V (C) 106.5 V (D) 112.4 V Passage # 2 (Ques. 18 to 20) A narrow beam of electrons, of radius r and all

moving at the same velocity v much less than speed of light (c), produce a charge current I. Assume that the beam has cylindrical symmetry. me = mass of electron & e = charge of electron.

18. If electron moving along the edge of beam experience magnetic force (FM) as well as electric force (Fe), then –

(A) Fe & FM are in radial and in the same direction (B) Fe & FM are in radial direction and in opposite

direction (C) Fe & FM are perpendicular to each other (D) Fe & FM are tangential to surface of beam 19. What is the change in radial velocity of electrons at

the border of the beam after, the beam has travelled a longitudinal distance 100 times of r i.e. 100 r. [Here

assume that v = 2c ]

(A) 2e0 cm

Ie50πε

(B) 2e0 cmIe100

πε

(C) 2e0 cm

Ieπε

(D) 2e0 cmIe150

πε

20. If it is assumed that beam start to move with zero

radial velocity of electron. If 'θ' is the angle between velocity of electron at the edge of beam and edge of beam. Then find tan θ after the beam has travelled

distance 100 r – [assume v = 2c ]

(A) 3e0 cmIe100

πε (B) 3

e0 cmIe200

πε

(C) 3e0 cmIe300

πε (D) 3

e0 cmIe400

πε


Passage # 3 (Ques. 21 to 23) Two identical blocks P and Q have mass m each.

They are attached to two identical springs initially unstretched. Now the left spring (along with P) is

compressed by 2A and the right spring (along with

Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the block was T.

P 2A

Q A

21. The time period of oscillation of combined mass is –

(A) 2

T (B) T2

(C) T (D) 2T

22. The amplitude of combined mass is –

(A) 4A (B)

2A

(C) 3A2 (D)

4A3

23. What is energy of oscillation of the combined mass ?

(A) 21 kA2 (B)

41 kA2

(C) 81 kA2 (D)

161 kA2

CHEMISTRY


1. Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02

mole of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution.

1 litre of mixture X + excess AgNO3 → Y. 1 litre of mixture X + excess BaCl2 → Z No. of moles of Y and Z are (A) 0.01, 0.01 (B) 0.02, 0.01 (C) 0.01, 0.02 (D) 0.02, 0.02

2. Among the following species, identify the isostructural pairs. NF3, NO3

–, BF3, H3O+, HN3 (A) [NF3, NO3

–] and [BF3, H3O+] (B) [NF3, HN3] and [NO3

–, BF3] (C) [NF3, H3O+] and [NO3

–, BF3] (D) [NF3, H3O+] and [HN3, BF3]

3. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid.

HCl + CH3COOH Cl– + CH3COOH2+

The set that characterises the conjugate acid-base pairs is -

(A) (HCl, CH3COOH) and (CH3COOH2+,Cl–)

(B) (HCl, CH3COOH2+) and (CH3COOH,Cl–)

(C) (CH3COOH2+, HCl) and (Cl–, CH3COOH)

(D) (HCl, Cl–) and (CH3COOH2+, CH3COOH)

4. The structure C = C C

H H3C

H3CH

COOHCH3

shows :

(A) geometrical isomersism (B) optical isomerism (C) geometrical & optical isomerism (D) tautomerism

5. Ethyl ester excess

MgBrCH3 → P. The product P will be

(A)

H3C

H3C

CH3

OH

(B)

H3C

H5C2

C2H5

OH

(C)

H5C2

H5C2

C2H5

OH

(D)

H5C2

H7C3

C2H5

OH

6. KF combines with HF to form KHF2. The compound

contains the species - (A) K+, F– and H+ (B) K+, F– and HF (C) K+ and [HF2]– (D) [KHF]+ and F– Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer.

7. Refer to the figure given : Which of the following statements is/are correct ?

Z1

gas Cgas A

ideal gas

gas B

P (A) For gas A, a = 0 and Z will linearly depend on

pressure (B) For gas B, b = 0 and Z will linearly depend on

pressure


(C) Gas C is a real gas and we can find 'a' and 'b' if intersection data is given

(D) All van der Waal gases will behave like gas C and give positive slope at high pressure

8. The following is (are) endothermic reaction (s) : (A) Combustion of methane (B) Decomposition of water (C) Dehydrogenation of ethane to ethylene (D) Conversion of graphite to diamond

9. When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water the sodium ions are exchanged with -

(A) H+ ions (B) Ca++ ions (C) SO4

– – ions (D) Mg++ ions

10.

CH3

CH3 H3C

→ hv,Cl2 N (isomeric products) ;

C5H11Cl → ondistillatifractional M (isomeric products) Identify N and M

(A) 6, 4 (B) 6, 6 (C) 4, 4 (D) 3, 3 This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of





11. Assertion (A) : Micelles are formed by surfactant

molecules above the critical micellar concentration (CMC).

Reason (R) : The conductivity of a solution having surfactant molecules decreases sharply at the CMC.

12. Assertion (A) : Boron always forms covalent bond

because. Reason (R) : The small size of B3+ favours formation

of covalent bond.

13. Assertion (A) : Band gap in germanium is small. Reason (R) : The energy spread of each germanium

atomic energy level is infinitesimally small.

14. Assertion (A) : p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.

Reason (R) : o-Hydroxybenzoic acid has intramolecular hydrogen bonding.

This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 15 to 17) The coordination number of Ni2+ is 4. NiCl2 + KCN (excess) → A (cyano complex) NiCl2 + Conc. HCl (excess) → B (chloro complex) 15. The IUPAC name of A and B are (A) Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II) (B) Tetracyanopotassiumnickelate (II), tetrachloropotassiumnickelate (II) (C) Tetracyanonickel (II), tetrachloronickel (II) (D) Potassium tetracyanonickel (II), potassium tetrachloronickel (II)

16. Predict the magnetic nature of A and B (A) Both are diamagnetic (B) A is diamagnetic and B is paramagnetic with one

unpaired electron (C) A is diamagnetic and B is paramagnetic with two

unpaired electrons (D) Both are paramagnetic 17. The hybridization of A and B are (A) dsp2, sp3 (B) sp3, sp3 (C) dsp2, dsp2 (D) sp3d2, d2sp3 Passage # 2 (Ques. 18 to 20) The conversion of an amide to an amine with one

carbon atom less by the action of alkaline hydrohalite is known as Hofmann bromamide degradation.

O

NH2

(i)

→ O

NH–Br(ii)

→ O

N–Br(iii) ••

–

→

C

N

(iv)

O

→

HO

N

(v)

O

H

→

NH2

(vi)

In this reaction, RCONHBr is formed from which the reaction has derived its name. Hofmann reaction is accelerated if the migration group is more electron-releasing. Hofmann degradation reaction is an intramolecular reaction.


18. How can the conversion of (i) to (ii) be brought about?

(A) KBr (B) KBr + CH3ONa (C) KBr + KOH (D) Br2 + KOH

19. Which is the rate determining step in Hofmann bromamide degradation?

(A) Formation of (i) (B) Formation of (ii) (C) Formation of (iii) (D) Formation of (iv) 20. What are the constituent amines formed when the

mixture of (i) and (ii) undergoes Hofmann bromamide degradation?

CONH2

D (i)

&

CONH2

(ii)

15

(A)

NH2

D ,

NH2

,

NH2 15

&

NH2

D

15

(B)

NH2

D

and

NH2 15

(C)

NH2 15

and

NH2

(D)

NHD 15

and

Passage # 3 (Ques. 21 to 23) Chemical reactions involve interaction of atoms and

molecules A large number of atoms/molecules (approximately 6.023 × 1023 ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200 ; ] Faraday = 96500 coulombs).

21. The total number of moles of chlorine gas evolved is- (A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0

22. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is -

(A) 200 (B) 225 (C) 400 (D) 446 23. The total charge (coulombs) required for complete

electrolysis is - (A) 24125 (B) 48250 (C) 96500 (D) 193000

MATHEMATICS


1. If S is the set of values of θ for which (0, 0) and (sin θ, cos θ) lie on opposite sides of y = |x – 1| then S contains -

(A)

ππ

3,

3– (B)

ππ

2,

3

(C)

ππ

23,

2 (D)

ππ

2,

2–

2. 0x

lim→

xsin

xtan–x–1cos3

1–21– is -

(A) 2 (B) 1

(C) 0 (D) 21

3. If [y] = [sin x] ; ([.] denotes the greatest integer

function) and y = cos x are two given equations, then the number of ordered pairs (x, y) is -

(A) 2 (B) 4 (C) 0 (D) infinitely many 4. Let 20 distinct balls have been randomly distributed

into 4 distinct boxes, 5 into each. Let 'A' be the event that two specific balls have been put into a particular box. The probability of occurrence of event 'A' is :

(A) 191 (B)

194

(C) 198 (D)

192


5. Let x2 ≠ nπ – 1, n ∈ N, then

∫ +++++

)1x(2sin)1xsin(2)1x(2sin–)1xsin(2x 22

22 dx is equal to :

(A) )1xsec(21n 2 +l + C

(B)

+2

1xsecn2

l + C

(C) 21 ln |sec (x2 + 1)| + C

(D) )1xsec(

2n21

2 +l + C

6. If z = 1mzmz 21

++ , then distance of point z from the

line joining z1 + 1 and z2 + 1 is (A) 0 (B) 1

(C) 1m

m2+

(D) 1m

m+


7. The pairs of straight lines ax2 + 2hxy – ay2 = 0 and hx2 – 2axy – hy2 = 0 are such that -

(A) one pair bisects the angle between the other pair (B) the lines of one pair are equally inclined to the

lines of the other pair (C) the lines of one pair are perpendicular to the lines

of the other pair (D) these four lines are forming a rectangle

8. The equation x4 – (m – 3) x2 + m = 0 has - (A) No real root iff m ∈ (0, 1) (B) Four real and distinct roots iff m > 9 (C) No real root iff m ∈ (0, 9) (D) Zero as its roots if m = 0

9. The differential equation of the curve for which intercept cut by any tangent on y-axis is equal to the length of the sub normal :

(A) is linear (B) is homogeneous of first degree (C) has separable variables (D) is of first order

10. If in a triangle ABC, a, b, c are in A.P. and p1, p2, p3 are the altitudes from the vertices A, B, C respectively then

(A) p1, p2, p3 are in A.P. (B) p1, p2, p3 are in H.P.

(C) p1 + p2 + p3 ≤ ∆R3 (D)

1p1 +

2p1 +

3p1

≤ ∆R3





(C) If (A) is true but (R) is false.

(D) If (A) is false but (R) is true.

11. Consider the curve f(x) = x ln x Assertion (A) : f(x) = k will have no solution if

k ∈ (– ∞, –1/e) Reason (R) : f(x) is decreasing in (0, 1/e) and

increasing in (1/e, ∞)

12. Assertion (A) : ABCDA1B1C1D1 is a cube of edge 1 unit. P and Q are the mid points of the edges B1A1 and B1C1 respectively. Then the distance of the vertex D from the plane PBQ is 8/3

Reason (R) : Perpendicular distance of point (x1, y1, z1) from the plane ax + by + cz + d = 0 is

given by 222

111

cba

dczbyax

++

+++

13. Consider the function f(x) = cos(sin–1x),

differentiable in (–1, 1) Assertion (A) : f(x) is bounded in [–1, 1] Reason (R) : If a function is differentiable in (a, b)

then it is bounded in [a, b]

14. Assertion (A) : F(x) = ∫ ++

x

1 2tt1tlog dt then

F(x) = – F(1/x)

Reason (R) : If F(x) = ∫ +

x

1 1ttlog dt then

F(x) + F(1/x) = (1/2) (log x)2 This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.


Passage # 1 (Ques. 15 to 17) A spherical ball of diameter 210 cm is floating so

that the top of the ball is 2 cm above the smooth surface of the pond. Two point light sources are placed on it along a vertical line such that one is at 1 cm above the water surface and another is 9 cm below the water surface. The light sources emittes the light rays which move tangentially.

15. The angle between the line of directions of light rays is -

(A) 45º (B) 90º

(C) 2221 º (D) 60º

16. What is the circumference in centimeters of the circle formed by the contact of the water surface with the ball -

(A) π23 (B) π26 (C) π29 (D) π212

17. If the ball is rotated in the vertical plane in which two light sources lie, then the intersection point of light rays will describe -

(A) linear path (B) circular path (C) elliptical path (D) parabolic path Passage # 2 (Ques. 18 to 20)

Consider N = 61234429411823301878143

= pa qb rc sd where a, b,

c, d ∈ I+ and p, q, r, s are prime numbers such that p < q < r < s.

Let L1 and L2 are two lines defined as

L1 : [p q]

yx

= [a]

L2 : [r s]

yx

= [– 20b]

18. p + q + r + s is equal to : (A) 23 (B) 31 (C) 25 (D) 33 19. Product of all factors of N is (A) N4 (B) N6 (C) N8 (D) N16

20. Set of values of β for which point (0, β) lies in the triangle formed by the lines L1 = 0, L2 = 0 and L3 : x – y – 1 = 0 is :

(A)

71,1– (B)

71,

1320–

(C)

21,

1120– (D)

1,

1120–

Passage # 3 (Ques. 21 to 23) ax + by + c = 0 is a line touching the parabola

y2 = 4dx + e at (1/2, – 2). a & 3b are two A.M. between c & 2d, where a, b, c are positive integers. Consider a function y = f(x), f : R → R, is a polynomial not more than 2 degree and f(x) satisfies the following conditions

f(e) = 0, f(b) = c, f(a) = d 21. Area bounded between y2 = 4dx + e and y = f(x) will

be (in sq. units)

(A) 38 (B)

316

(C) 3

34 (D) 3

32

22. The line ax + by + c = 0 will touch : (A) x2 + (y + 1)2 = 1 (B) (x – 1)2 + (y + 3)2 = 2 (C) (x – 1)2 + (y – 2)2 = 5 (D) x2 + y2 = 8

23. y = f(x), cuts y2 = 4dx + e at two points A & B. Circum centre of the triangle formed by chord AB and tangents at A & B on the parabola is :

(A) (2, 2) (B) (6, 2) (C) (8, 8) (D) (0, 0)

Maths Facts • 40 when written "forty" is the only number

with letters in alphabetical order, while "one" is the only one with letters in reverse order.

• 1 googol = 10100;

1 googolplex = 10googol = 1010100 .

• 111 111 111 x 111 111 111

= 12345678 9 87654321

• Pi (3.14159...) is a number that cannot be written as a fraction.

• If you add up the numbers 1-100 consecutively (1+2+3+4+5...) the total is 5050.

• The billionth digit of Pi is 9.

• 1 and 2 are the only numbers where they are the values of the numbers of factors they have.

• 2 and 5 are the only primes that end in 2 or 5.

• The largest prime number is 9,808,358 digits

long; more than the number of atoms in the universe.


PHYSICS


1. A ball of mass 1 kg is released from position P inside a wedge with a hemispherical cut of radius 0.5 m as shown in the figure. The force exerted by the vertical wall OA on wedge, when the ball is in position Q is (neglect friction everywhere) (Take g = 10 m/s2) –

A

O B

C

Q

P

60º

(A) 2

315 N (B) 15 N

(C) 2

15 N (D) None

2. Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as figure. (initial velocity of particle is zero) Time after which total acceleration of particle makes an angle of 30º with radial acceleration is –

60º

at (m/s2)

time (s)

(A) 4s (B) 34 s

(C) 22/3 s (D) 2 s 3. A man is standing on a rough (µ = 0.5) horizontal

disc rotating with constant angular velocity of 5 rad/s. At what distance from centre should he stand so that he does not slip on the disc ?

(A) R ≤ 0.2 m (B) R > 0.2 m (C) R > 0.5 m (D) R > 0.3 m 4. Heat is being supplied at a constant rate to a sphere of

ice which is melting at the rate of 0.1 g/s. It melts completely in 100 s. The rate of rise of temperature thereafter will be (Assume no loss of heat) -

(A) 0.8 ºC/s (B) 5.4 ºC/s (C) 3.6 ºC/s (D) None of these

5. A 2100 W continuous flow geyser (instant geyer) has water inlet temperature = 10ºC while the water flows out at the rate of 20 g/s. The outlet temperature of water must be about -

(A) 20ºC (B) 30ºC (C) 35ºC (D) 40ºC

IIT-JEE 2012

XtraEdge Test Series # 8

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 7 to 10 are multiple choice questions with multiple correct answer. +4 marks and No negative mark for

wrong answer. Section - II • Question 11 to 14 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for

wrong answer. • Question 15 to 23 are passage based single type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.


6. Two boys are separated by a distance 50 m and are standing away from a vertical wall. When one of the bopys claps, the other boy hears the echo after 1 s. If velocity of sound in air is 330 ms–1 then distance of the boy from the wall is –

Q BA

P

(A) 330 m (B) 818 m (C) 188 m (D) 881 m Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer.

7. A bus is moving with a velocity of 30 m/s towards a huge wall. The driver sounds a horn of frequency 120 Hz. If the speed of sound in air = 330 m/s. Select the correct option -

(A) Frequency received by wall is 120 Hz (B) Frequency received by wall is 132 Hz (C) Frequency at reflected wave observed by the

driver is 240 Hz (D) Frequency of reflected wave observed by the

driver is 144 Hz

8. A cyclist rides along the circumference of a circular horizontal plane of radius R, the friction coefficient being dependent only on distance r from the centre O

of the plane as µ = µ0

Rr–1 , where µ0 is a

constant. Which of the following is/are correct ? (A) The radius 'r' of the circle with the centre at the

point along which the cyclist can ride with the

maximum velocity, r = 2R

(B) The radius 'r' of the circle with the centre at the point along which the cyclist can ride with the

maximum velocity vmax = 21 gRµ

(C) The maximum velocity vmax= gRµ (D) none 9. A bob of mass m is projected with an upward

velocity v0 so that it moves in a vertical circle of radius R in a vertical smooth circular tube. The normal reaction on it is zero, the velocity of bob will be –

R

v0

m

(A) 3

v0 (B)

31 v0

(C) 2

v0 (D) 3

v0

10. A particle of mass 'm' is moving with an acceleration

3a towards left relative to the trolley car. James bond

accelerates towards left with an acceleration of

magnitude 3a . If the car moves with a rightward

acceleration 'a', the magnitude of pseudo-force acting on 'm' viewed by James bond will be –

car

a/3 m a/3

James bond

a

(A) 3

mg (B) 3ma2

(C) ma (D) 3ma4






(D) If (A) is false but (R) is true.


11. Assertion (A) : While drawing a line on paper, friction force acts on paper in the same direction along which line is drawn on the paper.

Reason (R) : Friction always opposes motion. 12. Assertion (A) : Work done by a conservative force is

always zero in round trip of the point of application of force.

Reason (R) : Sometimes, a conservative force does positive work, negative work and zero work; as a whole, the net work done must be zero for a round trip.

13. Assertion (A) : A particle of mass 'm' is moving with

constant speed v along circular path of constant radius. The net acceleration of the particle is constant in this case.

Reason (R) : The tangential acceleration of the particle moving along circular path is defined as the rate of change of speed of the particle with time.

14. Assertion (A) : The potential energy is only defined

for the conservative forces. Reason (R) : In case of uniform circular motion, the

change in kinetic energy of the moving object is zero. This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) A uniform rod of mass 'm' length 'l' is sliding along

its length on a horizontal table whose top is partly smooth and rest rough with friction coefficient µ. If the rod after moving through smooth part, enters the rough with velocity v0.

l v0

Smooth A B µ (rough)

15. The magnitude of the friction force when its 'x' length (<L) lies in the rough part during sliding will be -

(A) x2

gm 2lµ (B) x

mg 2lµ

(C) l2

mgx2µ (D) l

mgxµ

16. The minimum velocity v0 with which it must enter so that it lies completely in rough region before coming to rest, is -

(A) lgµ (B) lg2µ

(C) 2 lgµ (D) 2glµ

17. If the velocity is double the minimum velocity (v0) as calculated in above question then what distance does its front end A would have travelled in rough region before rod comes to rest ?

(A) 23l (B)

25l

(C) 2l (D) 3l Passage # 2 (Ques. 18 to 20) With the help of Archimede's principle, one can

understand the floating nature and defects in metal formation. One can easily find out the amount of space left hollow in a sphere. For a body to float, there should be a balance between the weight of the body and the upthrust. The apparent weight felt differs based on the volume immersed in the liquid. More than one liquid may also balance the mass while floating. In a frame accelerated down with 'a' any mass will experience a normal force of m(g – a).

18. For a cubical block (ρ) to float in a pair of liquids of density ρ1 and ρ2 as show, the relation between ρ, ρ1 and ρ2 is –

x2x

ρ1

ρ2

(A) ρ = 2

21 ρ+ρ (B) ρ = 3

2 12 ρ+ρ

(C) ρ = 2

2 21 ρ+ρ (D) ρ = 3

21 ρ+ρ

19. If the container in which a body floats in a liquid falls under gravity, the upthrust felt by the body will be (symbols carry usual meaning) -

(A) zero (B) Vlρl g

(C) 2

Vl ρl g (D) Vbρb g

20. An ice cube holding a steel ball floats in a cup of

water. After all the ice melts, the level in the cup will-

(A) rise (B) fall (C) remains unchanged

(D) cannot be confirmed without know8ing density of steel ball

Passage # 3 (Ques. 21 to 23) In espresso coffee machines steam is passed into milk

at room temperature for a brief time interval. Some of the steam condenses and the temperature or milk rises. Since the time for which the steam is passed is brief, one can ignore the heat lost to the environment and assume that the usual assumption of calorimetry :

Heat lost = Heat gain is valid.


21. Steam at 100ºC is passed into milk to heat it. The amount of heat required to heat 150 g of milk from room temperature (20ºC) to 80ºC is (specific heat of capacity of milk = 4.0 kJ kg–1 K–1 specific latent heat of steam = 2.2 MJ kg–1 , specific heat capacity of water = 4.2 × 103 J kgK–1)

(A) 3.6 × 104 J (B) 3.6 × 103 J (C) 3.6 × 102 J (D) None of these 22. How many grams of steam condensed into water in

above question - (A) 1.57 g (B) 15.7 g (C) 157 g (D) None of these 23. If some of heat is allowed to escape to surrounding

(temperature of surrounding is 20ºC) then this amount of steam (mentioned in question 22) is increase the temperature to -

(A) greater than 80ºC (B) less than 80ºC (C) equal to 80ºC (D) can't say anything

CHEMISTRY


1. The equivalent weight of MnSO4 is half its molecular

weight when it is converted to : (A) Mn2O3 (B) MnO2 (C) MnO4

– (D) MnO42–

2. The increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (α) is :

(A) e, p, n, α (B) n, p, e, α (C) n, p, α, e (D) n, α, p, e 3. The molecule having one unpaired electron is : (A) NO (B) CO (C) CN– (D) O2 4. For the chemical reaction 3X(g) + Y(g) X3 Y(g)

the amount of X3 Y at equilibrium is affected by (A) temperature and pressure (B) temperature only (C) pressure only (D) temperature, pressure and catalyst

5. The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is

(A) C > N > O > F (B) O > N > F > C (C) O > F > N > C (D) F > O > N > C

6. Which of the following has the maximum number of unpaired electrons?

(A) Mg2+ (B) Ti3+ (C) V3+ (D) Fe2+ Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer. 7. The statements that are true for the long form of the

periodic table are : (A) it reflects the sequence of filling the electrons in

the order of sub-energy level s, p, d and f. (B) it helps to predict the stable valency states of the

elements (C) it reflects trends in physical and chemical

properties of the elements (D) it helps to predict the relative ionicity of the bond

between any two elements.

8. Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with?

(A) SO2Cl2 (B) SOCl2 (C) Cl2 (D) NaOCl 9. Which of the following are examples of aldol

condensation? (A) 2CH3CHO → NaOH.dil CH3CHOHCH2CHO

(B) 2CH3COCH3 → NaOH.dil CH3COH(CH3)CH2COCH3 (C) 2HCHO → NaOH.dil CH3OH (D) C6H5CHO + HCHO → NaOH.dil C6H5CH2OH 10. An aromatic molecule will - (A) have 4n π electrons (B) have (4n + 2) π electrons (C) be planar (D) be cyclic This section contains 4 questions numbered 11 to 14, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of






(D) If (A) is false but (R) is true. 11. Assertion (A) : Addition of bromine to trans-2-butene

yields meso-2,3-dibromobutane. Reason (R) : Bromine addition to an alkene is an

electrophilic addition. 12. Assertion (A) : In water, orthoboric acid behaves as

a weak monobasic acid. Reason (R) : In water, orthoboric acid acts as a

proton donor 13. Assertion (A) : Nuclide 30

13 Al is less stable than 4020

Ca Reason (R) : Nuclides having odd number of protons

and neutrons are generally unstable. 14. Assertion (A) : The electronic structure of O3

is

O O • •

• • ••

O • • • • • • Θ

⊕

Reason (R) :

O O • •

• • ••

O • • • •

structure is not allowed

because octet around O cannot be expanded. This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 15 to 17) The noble gases have closed-shell electronic

configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

15. Argon is used in arc welding because of its - (A) low reactivity with metal (B) ability to lower the melting point of metal (C) flammability (D) high calorific value

16. The structure of XeO3 is (A) linear (B) planar (C) pyramidal (D) T-shaped 17. XeF4 and XeF6 are expected to be (A) oxidizing (B) reducing (C) unreactive (D) strongly basic Passage # 2 (Ques. 18 to 20) Riemer-Tiemann reaction introduces an aldehyde

group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehyde as depicted below.

OH

CH3

[I]

O Na

CH3

Θ ⊕

CHO aq.HCl

OH

CH3

CHO

(I) (II) (III)

18. Which one of the following reagents is used in the above reaction?

(A) aq.NaOH + CH3Cl (B) aq.NaOH + CH2Cl2 (C) aq.NaOH + CHCl3 (D) aq. NaOH + CCl4

19. The electrophile in the reaction is - (A) :CHCl (B) +CHCl2 (C) :CCl2 (D) CCl3 20. The structure of the intermediate I is

(A)

O Na

CH3

Θ ⊕

CH2Cl (B)

O Na

CH3

Θ ⊕

CHCl2

(C)

O Na

CH3

Θ ⊕

CCl3

(D)

O Na

CH3

Θ ⊕

CH2OH

Passage # 3 (Ques. 21 to 23) Several short-lived radioactive species have been

used to determine the age of wood or animal fossils. One of the most interesting substances is 6C14 (half-life 5760 years) which is used in determining the age of carbon-bearing materials (e.g. wood, animal fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays).

7N14 + 0n1 → 6C14 + 1H1


Thus carbon-14 is oxidised to CO2 and eventually ingested by plants and animals. The death of plants or animals put an end to the intake of C14 from the atmosphere. After this the amount of C14 in the dead tissues starts decreasing due to its disintegration as per the following reaction :

6C14 → 7N14 + –1β0 The C14 isotope enters the biosphere when carbon

dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale C14 as CO2. Eventually, C14 participates in many aspects of the carbon cycle. The C14 lost by radioactive decay is constantly replenished by the production of new isotopes in the atmosphere. In this decay-replenishment process, a dynamic equilibrium is established whereby the ratio of C14 to C12 remains constant in living matter. But when an individual plant or an animal dies, the C14 isotope in it is no longer replenished, so the ratio decreases as C14

decays. So, the number of C14 nuclei after time t (after the death of living matter) would be less than in a living matter. The decay constant can be calculated using the following formula,

t1/2 = λ693.0

The intensity of the consmic rays have remain the same for 30,000 years. But since some years the changes in this are observed due to excessive burning of fossil fuel and nuclear tests.

21. Why do we use the carbon dating to calculate the age

of the fossil? (A) Rate of exchange of carbon between atmosphere

and living is slower than decay of C14 (B) It is not appropriate to use C14 dating to

determine age (C) Rate of exchange of C14 dating to determine age

organism is so fast that an equilibrium is set up between the intake of C14 by organism and its exponential decay

(D) none of the above

22. What should be the age of the fossil for meaningful determination of its age?

(A) 6 years (B) 6000 years (C) 60,000 years (D) can be used to calculate any age 23. A nuclear explosion has taken place leading to

increase in concentration of C14 in nearby areas. C14 concentration is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T2 at the respective places then

(A) The age of the fossil will increase at the place where explosion has taken place and

T1 – T2 = λ1 ln

2

1

CC

(B) The age of the fossil will decrease at the place where explosion has taken place and

T1 – T2 = λ1 ln

2

1

CC

(C) The age of fossil will be determined to be same

(D) 2

1

TT =

2

1

CC

MATHEMATICS


1. z is a complex number satisfying |z – 3| ≤ 4 and |ωz – 1 – ω2| = a (where ω is complex cube root of unity) then :

(A) 0 ≤ a ≤ 2 (B) 0 ≤ a ≤ 8 (C) 2 ≤ a ≤ 8 (D) 2 ≤ a ≤ 4

2. If α and β are roots of the equation ax2 + bx + c = 0 then roots of the equation a(2x + 1)2 – b(2x + 1) (3 – x) + c(3 – x)2 = 0 are;

(A) 3–12

α+α ,

3–12

β+β (B)

2–13

α+α ,

2–13

β+β

(C) 2–12

α+α ,

2–12

β+β (D) None of these

3. The set x : |1 – log1/5 x| + 2 = |3 – log1/5 x| is

equal to (A) (0, ∞) (B) [1/5, ∞) (C) [1/5, 5] (D) (0, 1/5]

4. If tan–1

xa + tan–1

xb + tan–1

xc + tan–1

xd =

2π then

x4 – x2 Σab + abcd is equal to (A) –1 (B) 0 (C) 1 (D) 2

5. If E = 41 .

62 .

83 .

104 ...

6230 .

6431 = 8x, then value of x is

(A) – 7 (B) – 9 (C) – 10 (D) – 12 6. The position vector of a point P is r = xi + yj + zk,

where x, y, z ∈ N and u = i + j + k. If r.u = 10 then the possible positions of P are

(A) 72 (B) 36 (C) 60 (D) 108



7. Tangents are drawn from the point (α, 3) to the circle 2x2 + 2y2 = 25 will be perpendicular to each other if α equals :

(A) 5 (B) – 4 (C) 4 (D) – 5

8. In the expansion of 30

3/2

x1–x

, a term

containing the power x13 : (A) does not exist (B) exists and the coefficient is divisible by 29 (C) exists and the coefficient is divisible by 63 (D) exists and the coefficient is divisible by 65 9. Suppose a1, a2, ... real numbers, with a1 ≠ 0. If a1, a2,

a3, ... are in A.P. then

(A) A =

765

654

321

aaaaaaaaa

is singular

(B) the system of equation a1x + a2y + a3z = 0, a4x + a5y + a6z = 0, a7x + a8y + a9z = 0 has infinite number of solutions

(C) B =

12

21

aiaiaa

is non singular

(D) none of these 10. An equation of a circle through the origin, making an

intercept of 10 on the line y = 2x + 5/ 2 , which subtends an angle of 45º at the origin is

(A) x2 + y2 – 4x – 2y = 0 (B) x2 + y2 – 2x – 4y = 0 (C) x2 + y2 + 4x + 2y = 0 (D) x2 + y2 + 2x + 4y = 0






11. Lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect the line

L3 : y + 2 = 0 at P and Q respectively. The bisector of the acute angle between L1 and L2 intersect at R.

Assertion (A) : The ratio PR : RQ equals 22 : 5 Reason (R) : In any triangle bisector of an angle of

divides the triangle into two similar triangles. 12. Assertion (A) : Let p < 0 and α1, α2, ... α9 be the

nine roots of x9 = p, then

∆ =

987

654

321

ααααααααα

= 0

Reason (R) : If two rows of a determinant are identical then determinant equals zero.

13. Assertion (A) : If 2 sin2 ((π/2) cos2x ) = 1 – cos (π sin 2x), x ≠ (2n + 1) π/2,

n is a integer, then sin 2x + cos 2x is equal to 1/5.

Reason (R) : sin 2x + cos 2x = xtan1

xtan–xtan212

2

++

14. Consider the planes 3x – 6y – 2z = 15 & 2x + y – 2z = 5 Assertion (A) : The parametric equations of the line

of intersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t.

Reason (R) : The vector ^i14 +

^j2 + 15

^k is

parallel to the line of intersection of the given planes. This section contains 3 paragraphs, each has 3 multiple choice questions. (Questions 15 to 23) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 15 to 17) A and B are two points on the boundary of a circular

field of radius R and centre O. ∠AOB = θ. A circle with centre A and radius AB meets the circular field again at C and the line AO produced at E. L, M are points on the boundary of the field lying between C and A, A and B, respectively.

15. AB is equal to (A) R sin θ (B) 2R sin (θ/2) (C) R cos θ (D) 2R cos (θ/2)


16. Area of the segment AMB is equal to (A) (1/2) R2 θ (B) (1/2) R2 sin θ (C) (1/2) R2 (θ – sin θ) (D) none of these 17. If the area AMBECL is 1/nth of the field, then

sin θ + (π – θ) cos θ is equal to

(A) nπ (B) n

1–nπ

(C) (n – 1)π (D) (n + 1)π Passage # 2 (Ques. 18 to 20) Let ABCD be a square with each side of length 2

units. C2 is the circle through vertices A, B, C, D and C1 is the circle touching all sides of the square ABCD. L is a line through A.

18. If P is a point on C1 and Q is a point on C2, then

2222

2222

QDQCQBQAPDPCPBPA

++++++ is equal to

(A) 0.75 (B) 1.25 (C) 1 (D) 0.5

19 A circle touches the line L and the circle C1 externally such that both the circles are on the same side of the line, then the locus of the centre of the circle is-

(A) ellipse (B) hyperbola (C) parabola (D) parts of straight line

20. A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts M at T2 and T3 and AC at T1, then area of ∆T1T2T3 is

(A) 1/2 sq. units (B) 2/3 sq. units (C) 1 sq. unit (D) 2 sq. units Passage # 3 (Ques. 21 to 23) The sum of three terms of a strictly increasing G.P. is

αS and sum of the squares of these terms is S2. 21. α2 lies (A) (1/3, 2) (B) (1, 2) (C) (1/3, 3) (D) none of these

22. If α = 1/2, S = 20, then the greatest value of the first term is

(A) 10/3 (B) 7/3 (C) 1/3 (D) 3 23. If we drop the condition that the G.P. is strictly

increasing and take α2 = 3, then common ratio is given by

(A) 2± (B) ± 1

(C) 0 (D) 3±

At a Glance

Some Important Practical Units 1. Par sec : It is the largest practical unit of

distance.

1 par sec = 3.26 light year

2. X-ray unit : It is the unit of length.

1 X-ray unit = 10–13 m

3. Slug : It is the unit of mass.

1 slug = 14.59 kg

4. Chandra Shekhar limit : It is the largest practical unit of mass.

1 Chandra Shekhar limit = 1.4 × Solar mass

5. Shake : It is the unit of time.

1 Shake = 10–6 second

6. Barn : It is the unit of area.

1 barn = 10–28 m2

7. Cusec : It is the unit of water flow.

1 cusec = 1 cubic foot per second flow

8. Match No. : This unit is used to express velocity of supersonic jets.

1 match no. = velocity of sound

= 332 m/sec.

9. Knot : This unit is used to express velocity of ships in water.

1 knot = 1.852 km/hour

10. Rutherford : It is the unit of radioactivity.

1 rutherford (rd) = 1 × 106 disintegrations/sec

11. Dalton : It is the unit of mass.

1 dalton = 121 mass of C12 = 931 MeV

= 1 a.m.u.

12. Curie : It is the unit of radioactivity.

1 curie = 3.7 × 1010 disintegration / sec


PHYSICS 1. Define principal axis of a spherical mirror.

2. Which of the following quantity – speed, wavelength, frequency does not change when light goes from one medium to another ?

3. Write the S.I. unit of (i) electric field intensity and (ii) electric dipole moment .

4. State and explain Lenz's law.

5. In an electromagnetic wave, at any point, the amplitude of electric field is 3 × 106 N/C. Find amplitude of magnetic field at that point.

6. How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers ?

7. Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its,

(i) Second permitted energy level to the first level, and (ii) Highest permitted energy level to the second

permitted level

8. Why are infrared radiations referred to as heat waves

also? Name the radiations which are next to these radiations in electromagnetic spectrum having

(i) Shorter wavelength. (ii) Longer wavelength.

9. What would be the change in focal length of a concave mirror when it is held in water ?

10. How width of central maxima changes in single slit diffraction pattern when

(a) slit width is decreased (b) distance between slit & screen is increased (c) light of smaller wavelength is used.

11. Calculate the potential at the centre of square of side 2 m, which carries at its four corners charges of +2 nC, + 1 nC, – 2 nC and –3 nC respectively.

12. A voltage of 30 V is applied across a carbon resistor

with first, second and third rings of blue, black and yellow colours respectively. Calculate the value of current, in mA, through the resistor.

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.

MOCK TEST PAPER-1

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solut ions wil l be published in next issue


13. Write the order of frequency range and one use of each of the following electromagnetic radiations:

(i) Microwaves (ii) Ultra-violet rays (iii) Gamma rays.

14. Prove that the instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of its half life.

15. The following table gives the values of work function for a few photo sensitive metals

17.415.292.1

MoKNa

.3

.2.1

)eV(FunctionWorkMetal.No.S

If each of these metals is exposed to radiations of wavelength 300 nm, which of them will not emit photo electrons and why?

16. By how much would the stopping potential for a given photo sensitive surface go up if the frequency of the incident radiations were to be increased from 4 x 1015 Hz to 8 x 1015 Hz ?

Given h = 6.4 x 10–34 J-s, e = 1.6 x 10–19 C and c = 3 x 108 ms–1

17. How is the band gap, Eg, of a photo diode related to the maximum wavelength λm, that can be detected by it ?

18. What does the term LOS communication mean ? Name the types of waves that are used for this communication. Which of the two-height of transmitting antenna and height of receiving antenna-can affect the range over which this mode of communication remains effective ?

19. In a double slit interference experiment, two coherent beams has intensity I & I + δI where δI << I. Then show that intensity of maxima and minima are

4I & I4)I( 2δ respectively.

20. Using Gauss's theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the center. Why do you except the electric field inside the shell to be zero according to this theorem ?

21. Calculate the current shown by the ammeter A in the circuit diagram given below:

A 5 Ω

10 Ω

5 Ω

10 Ω

10 Ω 10 Ω

10 Ω

22. A bar magnet of magnetic moment M is aligned with the direction of a uniform magnetic field B. What is the work done, to turn the magnet, so as to align its magnetic moment :

(i) opposite to the field direction and (ii) normal to the field direction ?

23. A circular coil of N turns and radius R, is kept normal to a magnetic field, given by B = B0 cos ωt. Deduce an expression for e.m.f. induced in this coil. State the rule which helps to detect the direction of induced current.

24. When an inductor L and a resistor R in series are connected across a 12 V, 50 Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by π/3 radian. Calculate the value of R.

Or A 0.5 m long metal rod PQ completes the circuit as

shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the total circuit is 3Ω, calculate the force needed to move the rod in the

direction as indicated with a constant speed of 2 ms–1

× × × × × × × × × × × × ×

× × ×

× × × × ××××

××××× × × × ×

× × Q

P

25. The nucleus of an atom of Y23592 , initially at rest,

decays by emitting an α-particle as per the equation EnergyHeXY 4

223190

23592 ++→

It is given that the binding energies per nucleon of the parent and the daughter nuclei are 7.8 MeV and 7.835 MeV respectively and that of α-particle is 7.07MeV/nucleon. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, calculate the speed of the emitted α-particle.Take mass of α-particle to be 6.68 x 10–27 kg.

26. Define the term ‘modulation index’ for an AM wave. What would be the modulation index for an AM wave for which the maximum amplitude is ‘a’ while the minimum amplitude is ‘b’ ?

27. What are the main assumptions of Bohr's Atomic Model ? How it explain the linear spectrum of H-atom ? Write the drawbacks of Bohr's model.

28. (a) What are coherent sources of light ? Why is no interference pattern observed when two slits of YDSE are illuminated by two different sodium lamps ?

(b) Obtain conditions of getting bright & dark fringes YDSE and write the expression of fringe width.


Or

Derive the lens formula, v1 –

u1 =

f1 , for concave

lens, using the necessary ray diagram. Two lenses of power + 10D & – 5D are placed in contact,

(a) calculate power of the combination. (b) find object position to obtain image position with

magnification 2. 29. Explain with the help of a labelled diagram the

underlying principle and working of a step-up transformer. Why cannot such a device be used to step-up d.c. voltage ? [5]

or Draw a labelled diagram of an a.c. generator. Explain

briefly its principle and working. 30. A cube has side a and charge q at its each corner.

Find potential energy of charge –q placed at its centre. or

Derive Gauss's theorem in Electrostatics.

CHEMISTRY 1. What is holme signal ? 2. Give two ores of copper ? 3. Name the simplest amino acid ? 4. Draw a graph between t1/2 & [A0] for Ist order

reaction 5. Calculate the standard cell potential of the following H2 / O2 fuel cell. Given that; O2 + 4H+ + 4e– → 2H2O Eº = 1.229 V 2 H2 → 4H+ + 4e¯ Eº = 0.0 V 6. (i) What is meant by the term critical micellization

concentration (CMC) ? (ii) What is the function of emulsifying agent ? 7. Name the reagent used for producing fluorobenzene

from benzene diazonium chloride. What is the name of this reaction ?

8. Give the major products that are formed by heating

following ether with HI.

CH3 – CH2 – CH – CH2 – O – CH2 – CH3

CH3

9. Give the structure of N2O5 ? 10. Why d-block elements are used as catalysts ?

11. Explain Van arkel method ? 12. What is specific rate or rate constant on which factors

it depends ? 13. For a Ist order chemical reaction rate constant is 2.303

sec–1. Determine the time required for 90% completion of a chemical reaction

14. Name the reagents used to convert – (i) Propene to allyl bromide (ii) Methyl iodide to propyne 15. The two isomeric aromatic compounds A and B have

formula C7H7OH. A gives purple colour with ferric chloride solution while B does not. Suggest the structures of A and B.

16. Name the following compounds according to IUPAC

system : (i) CH3 – CH = C – CH – CH2 – CH3

OH CH3

(ii) C6H5 – CH – CH2 – CH2 –CH2OH

Cl

17. Identify A and B – R2CO → 3NH A → 2H/Ni B 18. Complete the following equations supplying A & B

(i)

OH

Zn dust A BH2SO4 SO3

(ii) C6H5

–O

+Na + CH3Cl →

19. What is lanthanide contraction and explain its causes ? 20. Why [NiCl4]2– is paramagnetic but [Ni(CN)4]2– is

diamagnetic ? Explain 21. Give the structures of three synthetic rubbers ? 22. Classify amino acids with examples on the basis of

their production in body ? 23. What are tranquilizers. Give the types with examples. 24. Calculate the equilibrium constant for a reaction Ni(s) + Cu+2(aq) → Cu(s) + Ni+2(aq) Given the values of 0

Ni/Ni 2E + and 0Cu/Cu 2E +

as – 0.25 and 0.34 V respectively


25. Describe the following terms while stating the properties of colloids :

(i) Brownian movement (ii) Tyndall Effect; (iii) Electrophoresis 26. An alkene [A] on reductive ozonolysis yields acetone

and an aldehyde. The aldehyde is easily oxidised to B. [B] on treatment with Br2 in the presence of phosphorus yields a compound [C] which on further hydrolysis gives a hydroxy acid [D]. The acid [D] can also be produced from acetone by treatment with HCN followed by subsequent acidic hydrolysis. Identify A, B, C, D.

27.

A N3H / H2SO4

Heat (An acid) B

CHCl3 / KOH C

H3O+ B + HCOOH

HNO2 D + E

Reduction of C produces (CH3)3CNHCH3. Identify A,B,C,D & E

28. Explain the structure of diborane (by orbital diagram)? 29. (i) Differentiate between reaction rate and reaction

rate constant. (ii) An antifreeze solution is prepared from 222.6 g of

ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1 then what shall be the molarity of the solution ?

(iii) What is azeotrope and what are the types of azeotrope. Explain

30. (a) What is the relationship between lattice parameter

a and r of the atom in. (i) B.C.C. (ii) F.C.C. (iii) H.C.P. arrangements (b) Explain the non-stoichiometric point defect

responsible for colour in alkali metal. (c) Explain the following (i) Ferromagnetic (ii) Ferrimagnetic substance

MATHEMATICS 1. Construct a matrix of order 2 × 2 whose element

aij = 2

j3i2 −

2. Find x, y, a, b so that

++

−−b4a3yx1

3bay3x2 =

−2961321

3. Evaluate : alog1

1blog

b

a

4. Let a relation R1 on the set R of real numbers be

defined as (a, b) ∈ R1 ⇔ 1 + ab > 0 for all a, b ∈ R. Show that R1 is reflexive.

5. If y = ax + ex + xa + xx then find dxdy .

6. Evaluate : dx222 x22 xx2

∫

7. Form the differential equation representing the family

of curves y = A cos (x + B), where A and B are parameters.

8. Find the value of λ so that the vectors

ar

= 2 i + λ j + k and br

= i – 2 j + 3 k are perpendicular to each other.

9. Find the angle between two vectors a

r and b

r having

the same length 2 and their scalar product is –1.

10. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines

1x =

2y =

3z and

32x

−+ =

21y − =

51z + .

Section B 11. Using properties of determinant prove that

yxxzzyqpprrqbaaccb

+++++++++

= 2 zyxrqpcba

12. From a well shuffled pack of 52 cards, a card is drawn at random, find the probability that it is either a heart or a queen.

Or A football match may be either won, drawn or lost by

the host country team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.

13. f(x) =

=

≠+−

+−−

0x,k

0x,xcos12

14936 xxx

f(x) is continuous at x = 0 then k = ?


14. If y = (x + 22 ax + )n then prove that

dxdy

= 22 xa

ny

+

15. If y = tan–1

xaxa

+− , – a < x < a then find

dxdy

.

Or If x = a (θ – sin θ) and y = a (1 – cos θ),

find θatdx

yd2

2 =

2π .

16. If the tangent to curve y = x3 + ax + b at (1, –6) is parallel to line x – y + 5 = 0 then find a, b.

17. Show that f : R – –1 → R – 1 given by

f(x) = 1x

x+

is invertible. Also find f–1.

18. Evaluate : dxxsin4∫

19. Evaluate : dxx4cosx2cos∫

20. Obtain the differential equation of all circles of radius r.

Or

Solve the differential equation : 0x2ydxdyx 3 =−−

21. If a

r, b

r, c

r are vectors that a

r. br

= ar

. cr

, ar

× br

= ar

× cr

, ar

≠ 0r

, then show that br

= cr

. Or

If →a and

→b are unit vectors and θ is the angle

between them, then prove that cos →→

+=θ ba

21

2

22. Show that the lines 3

1x − = 2

1y + = 5

1z − and 4

2x + =

31y −

= 21z

−+ do not intersect.

Section C

23. Evaluate : ∫ +−−1

0

21 dx)xx1(cot

24. A man speaks truth in 80% of the cases and another in 90% of the cases. While stating the same fact, what is the probability that they

(A) contradict (B) Agree

25. Using integration, find the area of the region bounded by the line 2y = – x + 8, x-axis and the lines x = 2 and x = 4.

26. Find the image of the point having position vector i + 3 j + 4 k in the plane r

r. (2 i – j + k ) + 3 = 0.

27. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs.5760.00 to invest and has space for at most 20 items. A fan costs him Rs.360.00 and a sewing machine Rs.240.00. His expectation is that he can sell a fan at a profit of Rs.22 and a sewing machine at a profit of Rs.18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Translate this problem mathematically and then solve it.

28. If A =

−− 120102112

, then find A–1.Using A–1, solve

the following system of linear equations 2x + y + z = 3 2x + z = 5 –2y –z = 1

Or

Find A–1 where A =

−−

−

433232321

. Using

A–1 solve the following system of linear equations x + 2y –3z = –4 2x + 3y + 2z = 2 3x –3y –4z = 11

29. Find the local maxima or local minima for f(x) = sin x + cos x, 0 < x < π/2.

Or Find two positive numbers whose sum is 14 and the

sum of whose squares is minimum.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 Ans C A B A B A A,B,C A,C,D Ques 9 10 11 12 13 14 15 16 Ans A,B,C,D A,B,C B A D D C D Ques 17 18 19 20 21 22 23 Ans B B D C C A D

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 Ans A C D B A C A,C,D B,C,D Ques 9 10 11 12 13 14 15 16 Ans B,D A B A B D A C Ques 17 18 19 20 21 22 23 Ans A D D B B D D

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 Ans B D D A B B A,B B,C,D Ques 9 10 11 12 13 14 15 16 Ans A,B,D B,D A D C D B B Ques 17 18 19 20 21 22 23 Ans B D C A D C B

PHYSICS

Ques 1 2 3 4 5 6 7 8 Ans A C A A C C B,D A,B Ques 9 10 11 12 13 14 15 16 Ans B B C A D B D A Ques 17 18 19 20 21 22 23 Ans B B A B A A B

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 Ans B D A A C D A,C,D C Ques 9 10 11 12 13 14 15 16 Ans A,B B,C,D B C C A A C Ques 17 18 19 20 21 22 23 Ans A C C B C B A

MATHEMATICS Ques 1 2 3 4 5 6 7 8 Ans B B B B D B B,C B,C,D Ques 9 10 11 12 13 14 15 16 Ans A,B,C B,D C D D D B C Ques 17 18 19 20 21 22 23 Ans B A C C D A B

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