XtraEdge Dec 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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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...

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 Chhatre

Gold Medallist From 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 = 2 1

× stress × strain × volume

= 2 1 × Y ) Stress ( 2 × πr2 l = 2 1 Y ) r / Mg ( π 2 2 × πr2 l = ₂1 Y r g M 2 2 2 π l _...(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 = 2 1 Y r g M 2 2 2 π l ∴ ∆T = 2 1 Ycm r g M 2 2 2 π l Here

m = mass of string = density × volume of string

= ρ × πr2 l ∴ ∆T = 2 1 Ycp ) r ( g M 2 2 2 2 π = 2 1 _× 7860 420 10 1 . 2 ) 10 2 14 . 3 ( ) 10 100 ( 11 2 3 – 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 R

X R = R1 or R2 or R3

Sol. All Null point, the wheat stone bridge will be balanced ∴ 1 r X = 2 r R ⇒ X = R 2 1 r r

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 X X ∆ = 1 1 r r ∆ + 2 2 r r ∆

Here ∆r1 = ∆r2 = y (say) then

For X

X ∆

to be minimum r1 × r2 should be max

[Q r1 + r2 = c (Constt.)] Let E = r1 × r2 ⇒ E = r1 × (r1 – c) ∴ 1 dr dE _{= (r} 1 – c) + r1 = 0 ⇒ r1 = 2 c _{⇒ r} 2 = 2 c _{⇒ r} 1 = r2

⇒ R2 gives the most accurate value

KNOW IIT-JEE

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 = LC 1 ₌ ) F 10 0 . 5 )( H 10 2 ( 1 6 – 3 – _× × = 10 +4 _s–1 Let at t = 0, Q = Q0 then Q(t) = Q0 cos ωt ...(1) I(t) = dt dQ = – Q0 ω sin ωt ...(2) L=2.0 mH C=5.0 µF dt ) t ( dI _{= – Q} 0 w2 cos (ωt) ...(3) (a) For Q = 100 µC       2 Q or 0 For (1) 100 = 200 cos ωt or cos (ωt) = 2 1_{, From equation (3) :} dt dI _{= (2.0 × 10}–4_{C) (10}+4 _s–1₎2 _      2 1 dt dI = 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. 2 1 2 max LI = 2 1 _LI2 ₊ 2 1 C Q2 or Q = LC(I2 –I2) max I = 2 I_max = 1.0 A ∴ Q = (2.0×10–3)(5.0×10–6)(22–12) 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 30º 60º 3 µ_p = 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 ⇒

2 3

= 3 sin r ⇒ r = 30º Now, NN' is the normal to surface AB.

∴ ∠AMN = 90º But ∠QMN = 30º ⇒ ∠AMQ = 60º A C B 30º 60º 60º 30º 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 = µ 2 λ = 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.

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 r r ₌ 1/3 4 A_      ⇒ (14)1/3 ₌ 3 / 1 4 A       _{⇒ A = 56} ∴ No. of proton = A – no. of neutrons

= 56 – 30 = 26 ∴ Atomic number = 26 (b) We know that v = Rc (z – b)2         2 1 2 1 n 1 – n 1 Here, R = 1.1 × 107 _{, c = 3 × 10}8 _{, Z = 26} b = 1 (for Kα), n1 = 1, n2 = 2 ∴ ν = 1.1 × 107 _{× 3 × 10}8 _{[26 – 1]}2     4 1 – 1 1 = 3.3 × 1015 _{× 25 × 25 ×} 4 3 = 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 = 2 1 _{× mM of Na} 2S2O3 = 2 1 × 40 × 10 1 = 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 . 1 096 . 0 × 100 = 6.6%

No. of photons or molecules of O3 =

48 10 023 . 6 096 . 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 aq.C2H5OH→ _{Acidic solution} CH3 CH3 Br aq.C2H5OH→ _Neutral solution. Explain. [IIT-2005]

(x) (A)    → 3H3/Pd (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 Br→2  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–C C– –C+–C C–C– –C–C C–C+– C – | | | | • • | • • | | | | | | | – –

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

C6H5–C–Br | | CH3 CH3 ) 1 S ( ) aq ( OH H C N 5 2_  →  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 4p Co3+ ⇒ 3d d2_sp3 4p NH3 Co NH3 NH3 NH3 NH3 NH3 3+

Octahedral complex, d2_sp3 _{hybridisation}

[Ni(CN)4]2– 3d 4s 4p Ni2+ 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 4p Ni = Ni (after rearrangement) 3d sp3 4p pairing due to CO Ni CO CO CO 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. ) solution colourless ( ) colourless ( SO H . dil ) white ( A B C 4 2_{ +}   →  K2Cr2O7/H+

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

(coloured)

E↑ + B↑ → D + Colourless liquid anhy.CuSO4→ _Blue C NaOH or NH . aq_3_→  Precipitate reagent of excess_→  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

(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↑ 2H2S(B)→2H2O + 3S↓

(D) (E) (colourless liq) D CuSO4(White)→ _CuSO

4.5H2O

(blue) ZnSO4 + 2NaOH → Zn (OH)2↑

(C) white.ppt ) excess ( NaOH 2_→  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 HI of excess with flux Re_  →

 2RI hydrolysis→ _2ROH (X) (Y) (Z) P + I2 ) O ( KMnO_4  →  –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– _2C

2H5OH KMnO4→ CH3COOH

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

MATHEMATICS

11. Prove that there exists no complex number z such that |z| < 3 1 and

∑

= n 1 r r rz a = 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| > 3 1 + 3 2 |z|n+1 ⇒ |z| > 3 1 , which contradicts (1)

∴ There exists no complex number z such that

|z| < 3 1 and

∑

= n 1 r r rz a = 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 x6 x5 x7 x4 x1 x2 x3 n m x1 are {x2, x3, x4, x5} x5 are {x1, x6, x7} x7 are {x5, x4} ⇒ x1 = 4 x x x x₂+ ₃+ ₄+ ₅ , x5 = 3 x x x₁+ ₆+ ₇ and x7 = 2 x x₄+ ₅

∴ 4x1 = x2 + x3 + x4 + 3 7 6 1 x x x + + ⇒ 12x1 = 3x2 + 3x3 + 3x4 + x1 + x6 + 2 x x4+ 5 ⇒ 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.       + 2 y x f = 2 ) y ( f ) x ( f + ∀ x, y ∈ R (given) Putting y = 0, we get f       2 x ₌ 2 ) 0 ( f ) x ( f + = 2 1 [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 – ) h 0 ( f lim 0 h + → = – 1 ⇒ h 1 – ) h ( f lim 0 h→ = – 1

Now, let x ∈ R then applying formula of differentiability. f '(x) = h ) x ( f – ) h x ( f lim 0 h + → = h ) x ( f – 2 h 2 x 2 f lim 0 h       + → = h ) x ( f – 2 ) h 2 ( f ) x 2 ( f lim 0 h + → = h ) x ( f – 1 – 2 h 2 f 2 1 – 2 x 2 f 2 2 1 lim 0 h             +       → [Using equation (1)] = h ) x ( f – } 1 – ) h ( f 2 1 – ) x ( f 2 { 2 1 lim 0 h + → = h ) x ( f – 1 – ) h ( f ) x ( f lim 0 h + → = _h 1 – ) h ( f lim 0 h→ = – 1 Therefore f ' (x) = – 1 ∀ x ∈ R ⇒

∫

f'(x)dx =

∫

–1dx ⇒ 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

0f(t)dt → ∞ as |x| → ∞, then show that every line y = mx intersects the curve y2 ₊

∫

0xf(t)dt = 2! [IIT-1991] (0, 2) A B O (xp,0) X (0, – 2)

Sol. We are given that f is continuous function and

∫

0fx(t)dt → ∞, as |x| → ∞

To show that every line y = mx intersects the curve y2 ₊

∫

x

0f(t)dt = 2

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

m2_x2 ₊

∫

x

0f(t)dt = 2 Consider, F(x) = m2_x2 ₊

∫

0fx(t)dt – 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       π π 2 , 2 – satisfying the equation

(1 – tan θ )(1 + tan θ) sec2_{θ + tan}2θ

2 = 0 [IIT-1996] Sol. (1 – tan θ) (1 + tan θ) sec2_{θ + tan}2θ

2 = 0 ⇒ (1 – tan2_{θ).(1 + tan}2_{θ) + tan}2θ

2 = 0 ⇒ 1 – tan4 _{θ + tan}2θ 2 = 0 put tan2_{θ = x} ⇒ 1 – x2 _{+ 2}x _{= 0 ⇒ x}2 _{– 1 = 2}x –1 –1 1 y O x Curves y = x2 _{– 1and y = 2}x

intersect at one point (negative value will not consider) x = 3, y = 8

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 tanφ=b 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 1 0 c t c ) t (

b + where c0 and c1 are positive constants

then find the slope of dt

df _{v/s t graph where f is the} degrees of freedom for the gas.

5. A particle enters in the given magnetic field ∧

→ =B k

B ₀ whre B0 is a constant with the velocity of

∧ ∧ →

+ =ai bj

v 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 2 2 B . q b a m r= +

(B) Helix and the pitch of the Helix is .a B . q m 2 0 π

(C) Helix and the pitch of the Helix is .b B . q m 2 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 R R ε ε = (B) 1 2 2 1 R R ε ε = (C) 0ε₁+ε₂ = (D) ε₁R₁+ε₂R₂ =0 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

Solutions will be published in next issue

7. Time constant for the circuit (A) RC (B) R1C if ε1 >ε2 (C) R2C if ε1<ε2 (D) _{R R} C R R R 2 1 2 1       + + Where 2 1 2 1 1 1 eq _{1/R 1/R} R / R / + ε − ε = ε 2 1 2 1 eq _{R R} R R R + =

8. Maximum current passing through resistance R (A) eq eq R ε (B) eq eq R R+ ε (C) R eq ε (D) R | |ε₁−ε₂

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       _{+ + +} = ... 4 1 2 1 1 R R 2 2 / 1 1 1 R =      − =

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 λ=3x

dq = 3xdx

Equivalent current due to element of length dx ) xdx 3 ( 2 dq . di π ω = ω =

Total equivalent current

∫ ∫

π ω = = (3xdx) 2 di i b a ) a b ( 2 . 2 3 2 a b 2 3 2 x 2 3 2 2 2 2 b a 2 − π ω =         ₋ π ω =         π ω = ) a b ( 4 3 2₋ 2 π ω = Option A is correct Equivalent current ) a b )( a b ( 4 3 ) a b ( 4 3 2 2 _{− +} π ω = − π ω =

Solution

Physics Challenging Problems

Set # 7

l ). a b .( 4 3 ) a b )( a b ( 4 3 _{ω +} π = − + π ω = As ω=4π/3 So, Equivalent current .(b a).l 3 4 . 4 3 π ₊ π = = (b + a). l = const. l i ∝ l Option B is correct. Charge on rod

∫ ∫

= −         = = = b a 2 2 b a 2 ) a b ( 2 3 2 x . 3 xdx 3 dq q 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 i π ω = q . 2 i π ω = ; .q 2 i π ω =

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

i .q 2π ω = (closed cone) ; .( ) 2 i σ π 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) ) C cone ( ) coneC ( 1 1 q . 2 i − π ω = and ) C cone ( ) coneC ( 2 2 q . 2 i − π ω = i (cone C1) = i (cone C2) Option B is correct

As charge on cone C₃ ≠ 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 mA 5 10 mV 50 i_g = Ω =

For terminals CT and a range is 5V so using Ω = − × = ⇒ − = ₋ 10 990 10 5 5 R G i V R _{1 3} g Ω = 990 R₁

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

Using 10 10 5 10 R R G i V R _{1 2 3} g − × = + ⇒ − = ₋ 990 + R2 = 2000 – 10 Ω = − =2000 1000 1000 R2 Ω = 1000 R₂

8. Range between CT and c is V so

Using G i V R g − = 10 10 5 V R R R1 2 3 ₃ − × = + + ₋ 10 10 5 V 3000 1000 990 ₃ − × = + + ⇒ ₋ 3 10 5 V 5000 ₋ × = ⇒ volt 25 V= ⇒

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 = 2 1 0 2 1 m m x m 0 . 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 = – 2 0 2 2 1 x R m Gm +

Gravitational energy of the system when particle is at the centre of the ring U2 =

R m Gm1 2 ∴ 2 1 1v m 2 1 + m₂v2₂ 2 1 = 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 1 K F 2 _, 2 K F 2 _, 3 K F 2 _and 4 K F 2 _{respectively.} According to geometry of the arrangement,

downward displacement of the block from its equilibrium position is y = 2 _      + + + 4 3 2 1 K F 2 K F 2 K F 2 K F 2 ...(1)

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

Students' Forum

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 =       + + + 4 3 2 1 K 1 K 1 K 1 K 1 4 y

∴ Restoring acceleration of block = m F ₌       + + + 4 3 2 1 K 1 K 1 K 1 K 1 m 4 y

Since, acceleration of block is restoring and is directly proportional to displacement y, therefore, the block performs SHM. Its period T = on accelerati nt displaceme 2π ∴ T = _      + + + π 4 3 2 1 K 1 K 1 K 1 K 1 m 4 2 = _      + + + π 4 3 2 1 K 1 K 1 K 1 K 1 m 4 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 0 πε = – 2 ₀ q ε ρ x dx or total initial potential energy of particle, U0 = – 0 2 q ε ρ

∫

== R x 0 xx.dx = ₀ 2 4 R q – ε ρ

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

∴ Work done = Increase in potential energy = U – U0 = 0 2 4 R q ε 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 = _l2+y2 ∴ Magnitude of magnetic induction at R due to each wire is B' = r 2 I µ0 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 = 2 0 r Iy µ π = ( y ) Iy µ 2 2 0 + π _l ...(1) For B to be maximum dy dB = 0 or y = l ∴ Bmax = l π 2 I µ₀

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

∴ Fmax = l π 2 I µ0 2 _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 = Y/ρ = 4000 ms–1 _.

Since points A and B are clamped, therefore, nodes are formed at these points or rod oscillates with integer number of loops in the middle part. Let number of these loops be m.

Since, length of each loop is 2 λ , therefore, m. 2 λ = (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 

     2 1 – n 2 where n is an integer. Then,       2 1 – n 2 _. 2 λ = a or (2n – 1) λ = 129 cm ...(2) Dividing equation (1) by (2), ) 1 – n 2 ( m = 3 2 ...(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 v λ = 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 =

λ v

= 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.

P Virtual Object P Real Object P Real

Object Virual _Object P n

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.

v Object Rest Image v v Object Rest Image θ vcosθ vsinθ vcosθ vsinθ v Object Image vm _2v m–v

Ray Optics

P

HYSICS

F

UNDAMENTAL

F

OR

IIT-J

EE

Object In rest Image vm _2v m Object Image vm _2v m+v v

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

θ º 360

where θ = angle between mirrors. If

θ º 360

is even number, the number of images is n – 1.

If θ

º

360 _{is odd number and object is placed on} bisector of angle between mirror, then number of images is n – 1.

If θ

º

360 _{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 eˆ1 = unit vector along incident ray.

2

eˆ = unit vector along reflected ray nˆ = unit vector along normal on point of

Incidence

Then eˆ₂ = eˆ₁−2(eˆ₁.nˆ)nˆ n

nˆ

1

eˆ eˆ₂

(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

f 1 u 1 v 1_{+ =} 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 = size object size image ₌ u v −

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 = y D ₌ f d 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