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XtraEdge for IIT-JEE 1 SEPTEMBER 2011

Dear Students,

Everyone knows that setting goals will help you achieve more and adds excitement and meaning to life. But setting a goal is only the beginning. We often fail to follow through and our goals turn into unfulfilled daydreams. To eliminate that pitafall, here is a systematic, approach that will help you turn your goals into realities.

• Decide what you want to achieve. Determine exactly what you want. Be specific. Be sure your goal is measurable, so you can tell when you're making progress. Pick a target date for a achieving it. Be sure it is realistically achievable.

• Ask yourself why it is important for your to achieve this goal. How you will benefit from reaching this goal ? Knowing why you want something raises your level of motivation. The higher your motivation level, the more likely you are to act on your goal.

• Consider what obstacles, problems or personal shortcomings might block your progress. List every one you can think of some obstacles will be real, others may be only imaginary. You must conquer both.

• Examine the obstacles one at a time, and think about how you might solve each problem.

• List the people or organizations who could help you achieve your goal. Decide specifically what you will ask them to do.

• Consider what information you need that you don't have now. Where will you get it? What could you read? Who could you talk to? What seminars could you attend?

• Write out a detailed action plan for achieving your goal. What are the priorities involved. ? Which tasks must be done first? When will different actions take place?

Setting a goal is a good step, but it is only the beginning. It takes all seven steps to make sure you actually follow through, and by so doing achieve your goal.

I guarantee that you will succeed and will secure a good rank in your exams if you make a habit of never to postpone your work.

Forever presenting positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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• All disputes are subject to the exclusive jurisdiction of the Kota Courts only.

Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari

To determine your priorities, examine your goals Volume - 7 Issue - 3

September, 2011 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected]

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Volume-7 Issue-3 September, 2011 (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- 2012 & 2013

S

Success Tips for the Months

• If you don’t notice when you win, you will only notice when you lose.

• It’s not bragging if you can do it.

• Feel the power of yet. As in “I don't know how to do this yet.”

• The difficult we do immediately. The impossible takes a bit longer.

• Some look down the rapids and see the rocks. Hunters look down the rapids and see the flow around the rocks.

• To know what you are doing is an advantage. To look like you know what you are doing is essential.

• First law of expertise: Never ask a barber if you need a haircut.

• If you think you can, you are probably right. If you think you can't, you are certainly right.

• Don't do modesty unless you have earned it.

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 • IIT Kharagpur presents Development Plan for Cuttach • IIT Mandi to rid the peasantry off agrarian crisis Shimla

IITian ON THE PATH OF SUCCESS 5 Mr. Prassanna Pathmanathan

KNOW IIT-JEE 6 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 54

Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper

Regulars ..........

DYNAMIC PHYSICS 13

8-Challenging Problems [Set # 5] Students’ Forum Physics Fundamentals • Current Electricity • Circular Motion, Rotational Motion

CATALYSE CHEMISTRY 33

Key Concept • Aliphatic Hydrocarbon • Oxygen & Hydrogen Family Understanding : Inorganic Chemistry

DICEY MATHS 42

Mathematical Challenges Students’ Forum Key Concept

• Probability • Binomial Theorem

Study Time........

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


IIT Kharagpur presents Development Plan for Cuttack Bhubaneswar: The long wait is finally over. The IIT, Kharagpur presented the final draft for the Comprehensive Development Plan (CDP) Vision 2030 for Cuttack City. The plan was presented to Chief Minister Naveen Patnaik. The proposed Cuttack Development Plan Area (CDPA) will include 48 wards of Cuttack Municipal Corporation, 28 wards of Choudwar Municipality and 60 adjacent villages under Tangi and Salepur areas. At least 60 villages will merge into Cuttack. The plan visualises the total population under the CDPA at 15 lakh by 2030. The CDPA will comprise of three distinct regions - Northern Fringe, Central Millennium City and Southern Fringe. The plan sought setting up of industrial centres, logistic centres, residential enclaves, commercial complexes and cultural centres in Charbatia, Nirgundi, Chhatisha, Choudwar, Nimpur, Bidanasi, Sikharpur, Mundali, Barang, Gopalpur and old Cuttack area. The plan also includes supply of drinking water, education, waste management and sewarage system. The report also sought setting up of Bhubaneswar-Cuttack Urban Conglomerate (BCUC) Metro Authority.

IIT Mandi to rid the peasantry off agrarian crisis Shimla : Recently founded elite technology institution IIT Mandi has shown its eagerness to rid the farmers off the agrarian crisis after setting up Joint Working Group(JWG) with government department to sort out their key problems, state official disclosed.

State Secretary Agriculture Ram Subhag Singh said in a statement issued that state Agriculture Department and IIT Mandi would join hands after formulating a scheme as JWG would be formed for sorting out the problems relating to farm technology. Director IIT Mandi T.A. Gonsalvej and Ram Subhag Singh choked out the joint strategy extending cooperation and providing assistance to farmers pertaining to information of agriculture. IIT has also shown eagerness to develop a Mobile Phone ‘Voice Base’ for this purpose. IIT is emulating this scheme as it had been started earlier at Chennai (Madras) on pilot basis, which had yielded good results. Those farmers of panchayats in the state who have not been linked with Broad band Internet connectivity would be connected through ‘Voice Base’ network, by being providing information through Mobile Phones. Farmers in this hill state are facing number of problems like low production, monkey and wild animal menace and sharp decline in size of holdings of arable lands. Monkeys and wild boars were even forcing number of them to leave this profession for ever.

"Smart Cane" research gets major funding from Wellcome Trust

A group of students under the guidance of Prof. M. Balakrishnan (CSE) and Prof. P.V.M. Rao (Mech Engg.) have been working on designing a number of assistive devices for the visually impaired.

These devices target independent and safe mobility as well as education of the visually impaired. One of their first devices namely “Smart Cane” that helps detect knee-above obstacles has received major funding from Wellcome Trust for translational research. The project is being done in collaboration with Phoenix Medical Systems, Chennai (Industrial partner) and Saksham Trust, Delhi (NGO working for the visually impaired).

Tissue engineering research at IIT Delhi

Recent advances of Tissue engineering could develop several tissues in laboratory, but regenerating tissue-to-tissue interface is still a challenging proposition. Intervertebral disc has a gradient of cartilagenous interface which plays important role in biomechanics of Spine. A biomimetic scaffold has been developed by covalent tethering of chondroitin sulphate with Silk fibrous scaffold. Culturing human chondrocytes on such lamellar scaffold could replicate complex tissue interface, as evidenced by staining with Safranin-O dye, showing fibrocartilagenous tissue at outer layers and hyaline cartilage at inner layers of Tissue Engineered construct.

IIT Kanpur Wins First Prize in NASA Competition In order to engage and inspire the next generation of engineers and scientists, NASA offered technical


challenges for both high school and college level students in the area of environmentally responsible aviation. These challenges were open to both US citizens and foreign students. The goal of the competition was to submit ideas and designs for aircraft or engine concepts and technologies that would assist in meeting the project’s goals for more environmentally friendly aviation by the year 2020. Two senior students of Aerospace engineering department of Indian Institute of Technology Kanpur secured first position in foreign division of NASA’s Environmentally Responsible (Green) Aviation College Student Challenge. Their design “Vaayu-An effort towards greener aviation” proposed a novel engine concept which bagged the first prize. This would probably be the first time that teams from India swept all awards in foreign division category for a NASA competition as two seniors from Anna University in Chennai came in second for their blended wing body design. Third place went to four undergraduates from SRM University in Kattankulathur for their airliner concept. Among high school category students from India and Romania took top honors among the foreign entries. First place went to 11th grader Nitish Kulkarni, from the Oakridge International School Hyderabad, India. Twelfth grader teams from Tudor Vianu National High School of Computer Science in Bucharest, Romania came in second and third. This certainly bodes well for the future of science and technology in India

Revisiting protein folding

Protein folding is at least a six decade old problem, since the times of

Pauling and Anfinsen. At IIT Delhi, a completely new direction has been provided towards obtaining a solution to this problem. Rigorous analyses of several thousand crystal structures of folded proteins reveal a surprisingly simple unifying principle of backbone organization. We find that protein folding is a direct consequence of a narrow band of stoichiometric occurrences of amino-acids in primary sequences, regardless of the size and the fold of the protein. We call this narrow band of stoichiometric occurrences as the "margin of life". Our findings present a compelling case for a newer view of protein folding which takes into account solvent mediated and amino acid shape and size assisted optimization of the tertiary structure of the polypeptide chain to make a functional protein that leads to survival of living systems over evolutionary timescales.

Ion Beam Complex At IIT Kanpur for Micro and Nanoscale Science, Engineering & Technology An Ion Beam Complex for Science Engineering and Technology (IBC-SET), equipped with different types of low energy ion beam facilities ranging from few keV to few MeV, is being set-up at IIT Kanpur for carrying out research in the interdisciplinary areas leading to technology development and prototype device fabrication. Under this complex a state of art focused ion beam system was Tandetron accelerator equipped with micro beam facility was inaugurated. Prof. V. S. Ramamurthy, Prof. G. K. Mehta, Prof. R. M. Singru, Dr. S. Kailas and other eminent scientists, faculty members of our institute and other institutions graced the occasion. Prof. S. G. Dhande presided over the function. Following research and technology development areas will be pursued by the researchers within and outside the institute using this facility: i. Microfabrication /Micromachining

(MEMS/NEMS) ii. Ion Beam synthesis of nano phases /

surface engg.

iii. Surface/interface studies by RBS/ERDA/PIXE/Channelling.

iv. Defect and damage studies in materials.

v. Bio Materials; damage studies and 3D mapping.

vi. Process optimization and system automation.

Mr. Narayana Murthy as chief guest inaugurated the Golden Jubilee celebrations. A documentary on IITK premiered during the function. Detials of the events and projects for this period are at the website.

Inauguration of DST Unit on Nanosciences & Technologry Forum Initiative on Fabrionics Dr. T. Ramasami, Secretary, Department of Science & Technology, Government of India, inaugurated DST Unit on Nanosciences & Launching of Indo-US Science & Technologry Forum Initiative of Fabrionics at Nanosciences Laboratory Building, IIT Kanpur. Dr. Samir K. Brahmachari, Director General, Council of Scientific & Industrial Research, Government of India was the Guest of Honor.

IIT Guwahati Celebrates "Techniche 2011" from 1st - 4th september 2011 Indian Institute of Technology (IIT), Guwahati is celebrating the Techno- Management extravaganza- “Techniche 2011” from 1st -4th September 2011. Charmed by the beauty of innovation and the underlying essence of growth, IIT Guwahati yet again brings the Techno Management extravaganza, the quintessential celebration of innovation, for everyone to enjoy. Techniche 2011 offers perfect opportunity to stretch one’s imagination, flaunt skills and prove mettle in the toughest of competitions and be inspired by the leaders of innovation. In the lectures series, some of the biggest names of today enlighten all with their words of wisdom for everyone to appreciate and reflect in awe..


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

This alumnus of Informatics Institute of Technology (IIT) is a PR specialist, lecturer, supervisor, research scholar, trainer, entertainer and journalist. Pathmanathan, a 25-year-old past graduate of IIT’s BSc in Information Systems with Business Management (first class) from the University of Westminster UK and the gold medallist in his year by toping the batch, now lectures other students studying for the same degree and guides them through their final year research/thesis projects as a project supervisor as well. The experience and knowledge gained from this degree, Prassanna was able to move into other disciplines such as business administration, marketing and human resource management as well. He’s a Chartered Marketer and currently studying for a degree in psychology and mastering in organisational psychology as well. His first venture into the business world was in the fashion industry as a writer. In 2009 he was the youngest editor-in-chief of a publication when he held that position at Agora – South Asia’s first online magazine. What started out as a hobby turned him towards fashion journalism and he has had the privilege of interviewing and featuring many prominent personalities in the field of fashion, sports and entertainment and he has now moved into fashion styling as well. He was invited as a member of the jury this year to judge Mr. Earth 2011 this October in Central America. He is currently working as the Head of Public relations at Cameron Pale and Medina (Pvt) Ltd., an all-inclusive Communication Agency. Success story is just one example of many other similar stories related to Informatics Institute of Technology (IIT). In fact Prassanna stated: “As a project supervisor I have guided a few students in their final year projects. The topics and solutions that these students have come up with on their own are amazing in their complexity and advanced thinking capabilities. All these students received an ‘A’ and what’s more industry representatives who had the opportunity of going through these projects are now making arrangements to purchase these projects and implement them in a real-time environment. This is made possible only through the practical and advanced education

system implemented by IIT in BSc (Hons) Information System with Business Management degree.” Is also in the process of writing research papers together with his students to present at international business research conferences. As a result, he is presenting three research papers with his students in the International Conference on Business and Technology in November 2011 at FRI University, Dehradun, New Delhi, India. This kind of exposure is made available to the students so that they receive local as well as international experience in their studies. passionately about his work at IIT, Prassanna said: “It is no secret that every industry, no matter what business it is engaged in, ultimately needs the assistance of information technology in order to take the venture into the future. In varying degrees of importance, information technology has become a part and parcel of our everyday life. This is where the Information Systems with Business Management BSc (Hons) Degree that IIT offers comes into play. The degree is engineered in such a unique manner so as to enable the students to gain a well rounded knowledge of necessary IT components as well as a sound business acumen that will provide a stepping stone into whatever industry they wish to move into.” “popular perception regarding IIT’s Information Systems with Business Management degree is that it is a technical degree for extreme IT savvy people. However, this is not so. Instead this degree gives more knowledge on business management essentials together with a view point of IT harmonisation. The syllabus demands a complete immersion in the studies. However what sets this degree option apart is the practical angle that is inherent in every aspect of the programme. “third year of the four year course is the placement year. During this year, students are assisted in getting into a company of their choice in order that they might develop and acquire a practical understanding of the modules that they are learning about, and its application in the real world. The students get immense experience in working for the multinational giants. “The internship has proven to be a resounding success as the depth of experience a student gains at that crucial moment in their studies is an essential step. It has also enabled the students to better apply themselves to the compulsory final year project that needs to be completed within a timeframe of eight to nine months.” So think big. Think different,” were Prassanna’s final words of encouragement to hopeful students.

Mr. Prassanna PathmanathanIIT success story


PHYSICS

1. A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s and maximum transverse acceleration is 90 m/s2. If the wave velocity is 20 m/s then find the waveform.

[IIT-2005] Sol. The wave form of a transverse harmonic disturbance y = a sin (ωt ± kx ± φ) Given vmax = aω = 3 m/s ...(i) Amax = aω2 = 90 m/s2 ....(ii) Velocity of wave v = 20 m/s ...(iii) Dividing (ii) by (i)

ω

ωa

a 2 =

390 ⇒ ω = 30 rad/s ...(iv)

Substituting the value of ω in (i) we get

a = 303 = 0.1 m ...(v)

Now

k = λπ2 =

v/v2π =

vv2π =

vω =

2030 =

23 ...(vi)

From (iv), (v) and (vi) the wave form is

y = 0.1 sin

φ±± x

23t30

2. A 5m 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 × stain × 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 ...(iii)

From (i) and (iii) Since U = Q Therefore

∴ mc∆T = Yr

gM21

2

22

πl

∴ ∆T = YcmrgM

21

2

22

πl

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

∴ ∆T = ρπ Yc)r(

gM21

22

22

= 21 ×

7860420101.2)10214.3()10100(

1123

2

×××××××

−

= 0.00457ºC

3. The x – y plane is the boundary between two transparent media. Medium –1 with z ≥ 0 has a refractive index 2 and medium –2 with z ≤ 0 has a refractive index 3 . A ray of light in medium –1

given by the vector A = i36 + ^j38 – 10

^k is

incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium –2. [IIT-2003]

Sol.

Y

^^ j38i36 + ^j38

^i36

M'

M

XO

Z M'

XO

–10 ^K

^^^ k10–j38i36A +=→

^^ j38i36 +

Fig(1) Fig(2)

Figure 1 shows vector i36 + ^j38

Figure 2 shows vector →A = ^i36 + ^j38 – ^k10

The perpendicular to line MOM' is Z-Axis which has

a unit vector of ^k . Angle between vector →IO and

→ZO can be found by dot product

→IO .

→ZO = (IO) (ZO) cos i

2222

^^^^

)1(–)10(–)38()36(

)k).(–k10–j38i36(

++

+ = cos i

⇒ i = 60 Unit vector in the direction MOM' from figure (1) is

KNOW IIT-JEE By Previous Exam Questions


2/122

^^^

])38()36[(j38i36n

+

+=

^^^ j54i

83n +=

To find the angle of refraction, we use snell's law

23 =

rsinisin =

rsinº60sin ⇒ r = 45º

From the triangle ORS

^r = (sin r) ^n – ( cos r) ^k

= (sin 45º)

+ ^^ j

54i

53 – (cos 45º)

^k

= ]k5–j4i3[25

1 ^^^+

4. Along horizontal wire AB, which is free to move in a

vertical plane and carries a steady current of 20A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30A, as shown in figure. Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations. [IIT-2003]

A B

DC Sol. When AB is steady, Weight per unit length = Force per unit length

weight per unit length = π

µ4

0 rII2 21 ...(i)

when the rod is depressed by a distance x, then the force acting on the upper wire increases and behave as a restoring force

A

Fmag

I1 = 20A x

B

B'mg r = 0.01 m A'

C I2 = 30A D

Restoring force/length = π

µ4

0 x–rII2 21 –

πµ4

0 rII2 21

= π

µ4

0 2I1I2

r1–

x–r1

⇒ Restoring force/length = π

µ4

0 2I1I2

r)x–r()x–r(–r

= π

µ4

0 )x–r(r

xII2 21

when x is small i.e., x <<r then r – x ≈ r

Restoring force/length F = π

µ4

0 221

rII2 x

Since F ∝ x ∴ The motion is simple harmonic

∴ π

µ4

0 221

rII2 = (mass per unit length) ω2 ...(ii)

From (i) (Mass per unit length) × g = π

µ4

0 rII2 21

Mass per unit length = π

µ4

0

rgII2 21 ...(iii)

From (ii) and (iii)

π

µ4

0 221

rII2 =

πµ4

0

rgII2 21 × ω2

⇒ ω = rg ⇒

T2π =

rg

⇒ T = 2π gr = 2π

8.901.0 = 0.2 sec

5. In the figure both cells A and B are of equal emf.

Find R for which potential difference across battery A will be zero, long time after the switch is closed. Internal resistance of batteries A and B are r1 and r2 respectively (r1 > r2).s [IIT-2005]

C R

L RR

R

RR

S

BA

r1 r2

Sol. After a long time capacitor will be fully charged, hence no current will flow through capacitor and all the current will flow from inductor. Since current is D.C., resistance of L is zero.

∴ Reg =

+ R

2R ×

21 + r1 + r2

= 4R3 + r1 + r2

I = eqRε+ε ⇒ I =

egR2ε =

21 rr4/R33

++ε

Potential drop across A is ε – I r1 = 0

⇒ ε = 21 rr4/R3

2++

ε r1

⇒ r1 = r2 + 3R/4

or R = 34 (r1 – r2)


CHEMISTRY

6. A hydrated metallic salt A, light green in colour, gives a white anhydrous residue B after being heated gradually. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue and a mixture of two gases E and F. The gaseous mixture, when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Identify A, B, C, D, E and F.

[IIT-1988] Sol. The given observations are as follows. (i)

)A(saltmetallicHydrated →heat

)B(

residueanhydrouswhite

(ii) Aqueous solution of B →NO

)C(compoundbrowndark

(iii) Salt B heatingStrong →

)D(residueBrown +

)F()E(gasesTwo+

(iv)

Gaseous mixture (E) + (F)

acidified KMnO4

BaCl2 solution

Pink colour is discharged White precipitate

The observation (ii) shows that B must be ferrous sulphate since with NO, it gives dark brown compound according to the reaction

[Fe(H2O)6]2+ + NO → browndark

252 )]NO()OH(Fe[ + + H2O

Hence, the salt A must be FeSO4 . 7H2O The observation (iii) is

2FeSO4 → Fe2O3 + SO2 + SO3 (D)

brown (E) + (F) The gaseous mixture of SO2 and SO3 explains the

observation (iv), namely,

colourpink

4MnO2 − + 5SO2 + 2H2O → colourno

24

2 SO5Mn2 −+ + + 4H+

2H2O + SO2 + SO3 4H+ + SO32– + SO4

2–

Ba2+ + SO32– →

.pptwhite3BaSO ; Ba2+ + SO4

– → .pptwhite

4BaSO

Hence, the various compounds are A. FeSO4 . 7H2O B. FeSO4 C. [Fe(H2O)5NO]SO4 D. Fe2O3 E and F SO2 and SO3

7. An organic compound (A) C8H6, on treatment with dil. H2SO4 containing HgSO4 gives a compound (B), which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous AlCl3. The compound (B) when treated with iodine in aq. KOH, yields (C) and a yellow compound (D). Identify (A), (B), (C) and (D) with justification. Show, how (B) is formed from (A).

[IIT-1994] Sol. The given reactions may be formulated as follows :

C8H6(A)

Dil H2SO4

∆ (B)

I2 + KOH

(C) + (D)

AlCl3

∆ C6H6 + Acid chlorideHgSO4

The reaction of compound (B) with I2 in KOH is

iodoform reaction. The compound (B) must have a –COCH3 group so as to exhibit iodoform reaction. Since (B) is obtained from benzene by Friedal-Crafts reaction, it is an aromatic ketone (C6H5COCH3). The compound (C) must be potassium salt of an acid.

The compound (A) may be represented as C6H5C2H. Since it gives C6H5COCH3 on treating with dil. H2SO4 and HgSO4, it must contain a triple bond (–C ≡ CH) in the side chain. Here, the given reactions may be formulated as follows :

C≡CH

(A)

dil H2SO4

HgSO4; H2O

C = CH2

OH

COCH3

Acetophenone(B)

Benzene

CH3COCl AlCl3; – HCl

– C – CH3

O

+ 3I2 + 4KOH –3KI;–3H2O

∆

(C) Potassium benzoate

COOK + CHI3

(D)

(B)

Hence. (A) C≡CH

Phenyl acetylene

(B)COCH3

Acetophenone

(C)

COOK

Potassium benzoate

(D) Idoform

3CHI


8. (a) Write the chemical reaction associated with the "brown ring test".

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

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

(c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000]

Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its

coordination number is six. Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6

3d 4s 4p

3d 4s 4p

d2sp3 hybridization

Hence

Co3+ion in Complex ion

3+

Co

NH3

NH3

NH3 H3N

H3N NH3

or

H3N

H3N

NH3 NH3

NH3NH3

Co3+

In [Ni(CN)4

2– nickel is present as Ni2+ ion and its coordination numbers is four

Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8

3d 4s 4p

3d 4s 4p

dsp2 hybridization

Ni2+ ion =

Ni2+ion in Complex ion

Hence structure of [Ni(CN)4]2– is

Ni2+

N ≡ C

N ≡ C

C ≡ N

C ≡ N

In [Ni(CO)4, nickel is present as Ni atom i.e. its

oxidation number is zero and coordination number is four.

3d 4s 4p

sp3 hybridization

Ni in Complex

Its structure is as follows :

Ni

CO

CO

CO

OC

(c) The transition metal is Cu2+. The compound is

CuSO4.5H2O

CuSO4 + H2S → mediumAcidic pptBlack

CuS ↓ + H2SO4

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white I2 + I– → I3

– (yellow solution)

9. The solubility product of Ag2C2O4 at 25ºC is 1.29 × 10–11 mol3 l–3. A solution of K2C2O4 containing 0.1520 mole in 500 ml water is shaken at 25ºC with excess of Ag2CO3 till the following equilibrium is reached :

Ag2CO3 + K2C2O4 Ag2C2O4 + K2CO3 At equilibrium the solution contains 0.0358 mole of

K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3. [IIT-1991]

Sol. Ag2CO3 + K2C2O4 → Ag2C2O4 + K2CO3 Moles at start Excess 0.1520 0 0 Moles after reaction 0.1520 – 0.0358 0.0358 0.0358 = 0.1162 Molar concentration of K2C2O4 or C2O4

2– left

unreacted = 5.0

1162.0 = 0.2324 moles l–1

[K2CO3] = [CO32–] at equilibrium =

5.00358.0

= 0.07156 moles l–1


Given that Ksp for Ag2C2O4 = 1.29 × 10–11 mol3 l–3 at 25ºC

So, [Ag+]2[C2O42–] = 1.29 × 10–11

or [Ag+]2 × 0.2324 = 1.29 × 10–11

Hence [Ag+]2 = 2324.029.1 × 10–11

Then Ksp for

Ag2CO3 = [Ag+]2 [CO32–] =

2324.01029.1 11−× × 0.0716

= 3.794 × 10–12 mol3 l–3 10. The molar volume of liquid benzene (density = 0.877 g ml–1) increases by a factor of 2750

as it vaporizes at 20ºC and that of liquid toluene (density = 0.867 g ml–1) increases by a factor of 7720 at 20ºC. A solution of benzene and toluene at 20ºC has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in vapour above the solution. [IIT-1996]

Sol. Given that, Density of benzene = 0.877 g ml–1 Molecular mass of benzene (C6H6) = 6 × 12 + 6 × 1 = 78

∴ Molar volume of benzene in liquid form =877.078 ml

= 877.078 ×

10001 L = 244.58 L

And molar volume of benzene in vapour phse

= 877.078 ×

10002750 L = 244.58 L

Density of toluene = 0.867 g ml–1 Molecular mass of toluene (C6H5CH3) = 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92 ∴ Molar volume of toluene in liquid form

= 867.092 ml =

867.092 ×

10001 L

And molar volume of toluene in vapour phase

= 867.092 ×

10007720 L = 819.19 L

Using the ideal gas equation, PV = nRT At T = 20ºC = 293 K

For benzene, P = 0BP =

VnRT

= 58.244

293082.01 ×× = 0.098 atm

= 74.48 torr (Q 1 atm = 760 torr) Similarly, for toluene,

P = 0TP =

VnRT

= 19.819

293082.01 ×× = 0.029 atm

= 22.04 torr (Q 1 atm = 760 torr) According to Raoult's law, PB = 0

BP xB = 74.48 xB

PT = 0TP xT = 22.04 (1 – xB)

And PM = 0BP xB + 0

TP xT or 46.0 = 74.48 xB + 22.04 (1 – xB) Solving, xB = 0.457 According to Dalton's law, PB = PM '

Bx (in vapour phase) or mole fraction of benzene in vapour form,

'Bx =

M

B

PP =

0.46457.048.74 × = 0.74

MATHEMATICS

11. For any real t, x = 2

–tt ee + , y = 2– –tt ee is a point

on the hyperbola x2 – y2 = 1. Find the area bounded by this hyperbola and the lines joining its centre to the points corresponding to t1 and – t1. [IIT-1982]

Sol. We have to find the area of the region bounded by the curve x2 – y2 = 1 and the lines joining the centre x = 0, y = 0 to the point (t1) and (– t1)

y

CA 1 N

x

P(t1)

–1

Required area

= 2

∆ ∫+2

1

1–1

–PCNofarea

tt ee

dxy

= 2

+∫11111

1

––.–

2–

221

tttttdt

dtdxyeeee

=

∫111

0

2–2–2

2––

8–

t ttttdteeee


= 4– 11 2–2 tt ee – ∫ +

1

0

2–2 )2–(21

ttt ee dt

= 4– 11 2–2 tt ee –

1

0

2–22–

2–

221

ttttee

= 4– 11 2–2 tt ee – 4––

41

12–2 11 tee tt = t1

Hence, the required area bounded by this hyperbola and the lines joining its centre is 't1'.

12. A swimmer S is in the sea at a distance d km from

the closest point A on a straight spere. The house of the swimmer is on the shore at a distance L km from A. He can swim at a speed of u km/hr and walk at a speed of v km/hr (v > u). At what point on the shore should be land so that he reaches his house in the shortest possible time? [IIT-1983]

Sol. Let the house of the swimmer be at B. ∴ d (AB) = L km. Let the swimmer land at C, on the shore and let d (AC) = x km

S

d

A x C (L–x) B

22 dx +

∴ d(SC) = 22 dx + and d (CB) = (L – x)

time = speed

cetandis

Time from S to B = time from S to C + time from C to B.

∴ T = u

dx 22 + + v

xL –

Hence, we take f (x) = 221 dxu

+ + vL –

vx

⇒ f ' (x) = 222

2.1.1

dx

xu +

+ 0 – v1

For either a maximum or minimum f, f '(x) = 0 ⇒ v2x2 = u2(x2 + d2)

i.e., x2 = 22

22

– uvdu

∴ f ' (x) = 0 at x = ± 22 – uv

ud .(v > u)

But x ≠ 22 –

–

uv

ud

∴ we consider x = 22 – uv

ud

f '' (x) = )(

12222

2

dxdx

du ++

> 0 for all x.

∴ f has minimum at x = 22 – uv

ud

13. Let Im = ∫π

0cos–1

cos–1x

mx dx

Use mathematical induction to prove that Im = mπ, m = 0, 1, 2... [IIT-1995]

Sol. Im = ∫π

0cos–1

cos–1x

mx dx (given)

for m = 0,

I0 = ∫π

0cos–1

)0cos(–1x

dx

= ∫π

0cos–1

1–1x

dx

I0 = ∫π

0cos–10

x dx = 0

for m = 1,

I1 = ∫π

0cos–1cos–1

xx dx

= ∫π

0

1dx = π

Therefore, the result is true for m = 0 and m = 1 Assume that the result is true for all m ≤ k, k > 0.

Let now m = k + 1. We have

Ik+1 = ∫π

+

0cos–1

)1cos(–1x

xk dx

1 – cos (k + 1) x = 1 – cos kx cos x – sin kx sinx = 1 + cos kx cos x + sin kx sin x – 2 cos kx cos x

(add and subtract cos kx cos x) = 1 + cos (k – 1) x – 2cos kx cos x = 2 – [1 – cos (k – 1) x] – 2cos kx cos x = 2 – [1 – cos (k –1)x ] – 2 cos kx cos x

+ 2 cos kx – 2 cos x = 2(1 – cos kx) – [1 – cos (k – 1)x]

+ 2 cos kx (1 – cos kx) Hence,

Ik+1 = ∫π

+

0cos–1

)1cos(–1x

xk dx


= ∫π

0cos–1

)cos–1(2xkx dx – ∫

π

0cos–1

)1–cos(–1x

xk dx

+ ∫π

0)cos–1(

))(coscos–1(2x

kxx dx

= 2Ik – Ik – 1 + [ ]π0sin2 kxk

= 2Ik – Ik–1 + 0 = 2 kπ – ( k – 1) π = (k + 1)π This shows that the result is true for m = k + 1. By the principle of mathematical a induction the

result is true for all non-negative integers m.

14. A bird flies in a circle on a horizontal plane. An observer stands at a point on the ground. Suppose 60º and 30º are the maximum and the minimum angles of elevation of the bird and that they occur when the bird is at the points P and Q respectively on its path. Let θ be the angle of elevation of the bird when it is at a point on the arc of the circle exactly midway between P and Q. Find the numerical value of tan2θ. (Assume that the observer is not inside the vertical projection of the path of the bird.) [IIT-1998]

Sol. Let OM = d, PU = MT = UQ = TN = TS = r Also, let PM = RS = QN = h In ∆ POM

P r U r QR

O

60º30º d

M r r

T N

S

We have tan 60º = OMPM =

dh

⇒ h = d3 In ∆QON

tan 30º = rd

h2+

⇒ d + 2r = h3 = 3d ⇒ d = r Also, OS2 = OT2 + TS2 = (d + r)2 + r2 = 5d2 In ∆ ROS, we have

tan θ = OSRS =

dd

53 =

53

⇒ tan2θ = 3/5

15. A =

bdbc

a

11

10, B =

hgfcd

a0

11,

U =

hgf

, V =

00

2a

If there is a vector matrix X, such that AX = U has infinitely many solutions, then

prove that BX = V cannot have a unique solution. If a + d ≠ 0. Then prove that BX = V has no solution.

[IIT-2004] Sol. Since AX = U has infinitely many solutions ⇒ |A| = 0

bdbc

a

11

10 = 0

⇒ a(bc – bd) + 1(d – c) = 0 ⇒ (d – c) (ab – 1) = 0 ∴ ab = 1 or d = c

Again, |A1| = hdgcfa

11

0 = 0 ⇒ g = h

|A2| = bhbg

fa

11

1 = 0 ⇒ g = h

and |A3| = bdhbcg

f 10 = 0 ⇒ g = h

∴ g = h, c = d and ab = 1 ...(i) Now, BX = V

|B| = hgfcd

a0

11 = 0

(Since C2 and C3 are equal) using (i) ∴ BX = V has no solution.

|B1| = hgcd

a

00

112

= 0

(Since c = d and g = h) using (i)

|B2| = hfc

aa

000

12

= a2 cf = a2 df

(Since c = d ) Since, adf ≠ 0 ⇒ |B2| ≠ 0 ∴ |B| = 0 and |B2| ≠ 0; ∴ BX = V has no solution.


1. Two point monochromatic and coherent sources of

light of wavelength λ each are placed as shown. If initial phase difference between the sources is zero and (D >> d) then –

(A) If ,2

7d λ= O will be minima

(B) If λ=d , only one maxima can be observed on screen

(C) If ,8.4d λ= the total 10 minima will be observed on screen

(D) If ,2

5d λ= the intensity at O will be minimum

2. Resonance occurs in a series LCR circuit when the frequency of applied emf is 1000 Hz. Then – (A) When frequency is 900Hz, then current through

the voltage source will be ahead of emf of source (B) The impedance of circuit will be minimum at 100Hz (C) Only at resonance voltage across L and current in

C differ in phase by 180º (D) If the value of C is doubled resonance occurs at

2000Hz 3. For the following

situation which of the following is/are correct –

(A) Potential of

conductor is )Rd(4

q

0 +∈π

(B) Potential of conductor is d4

q

0∈π

(C) Potential of conductor cannot be determined as distribution of induced charged is not known

(D) Potential at B due to induced charges on conductor

is d)Rd(4

qR

0 +∈π−

Passage # (Q. No. 4 to Q. No. 6) A non conducting vessel containing n moles of a diatomic gas ions fitted with a conducting piston. The cross-sectional area, thickness and thermal conductivity of piston are A, l

and k respectively. The right side of piston is open to atmosphere at temperature T0. Heat is supplied to the gas by means of an electric heater at a constant rates q.

4. Temperature of gas as a function of time is (if initial temperature is T0)

(A) ]e1[kAqTT )nR7/KAt2(

0ll −−+=

(B) ]e1[kAqTT )KAt2/nR7(

0ll −−+=

(C) )]e1[kAqTT nR7/KAt2(

0ll −−−=

(D) None of these 5. Maximum temperature of gas –

(A)kAqTT 0maxl

+= (B) ]e1[kAqTT 0max −+=l

(C) lq

kATT 0max += (D) kAqTmaxl

=

6. The ratio of the maximum volume to the minimum volume is –

(A)0kAT

q1 l+ (B)

lqkAT

1 0+ (C) 0kAT

ql (D) lq

kAT0

7. The electric field in a region is radially outward with magnitude rE α= . Calculate the charge contained in a sphere in 10–10 C of radius R centred at origin. Use

2mV100=α and R = 0.30m.

8. Magnetic flux linked with a stationary loop of resistance R varies with time during time period T as follows )tT(at −=φ Then the amount of heat generated in the loop during time T is (assume inductance of coil is negligible)

(A)R3

aT3 (B)

R3Ta 22

(C)RTa 32

(D) R3Ta 32

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 # 5

D

OS1 S2

d screen

ROB+q

d Hollow neutralconductor


1. Conceptual Option [A,B,C,D] is correct 2. Conceptual Option [A,C,D] is correct

3. (A) RC2RKC ==τ

(C,D) R/vtv

II

Vtan1

1

R

C ω==φ

RCtan ω=φ

Option [A,C,D] is correct

4. Option [A,B,C] is correct 5. Option [A] is correct

6. Final current is RV that distributes in inverse ratio to

inductance. Option [C] is correct 7. Option [B] is correct 8. Option [B] is correct Transitions emitting photons of energy more than

work function W = 2.3 eV can result in a photoelectric.

max

minmK2

hmK2h

ph

=λ⇒==λ

Kmax = 13.6 – 0.85

Solution Physics Challenging Problems

Set # 4

8 Questions were Published in August Issue

-: CHEMISTRY JOKE :-

If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for you to learn a little chemistry.

Chemistry Joke : This is no joke but a call to *BAN* dihydrogen monoxide, otherwise know as the invisible, killer substance. Jupiter Scientific's science joke webpage is probably not the place to post this protest, but the JS staff feels very strongly about this issue. For your information, dihydrogen monoxide (DHMO) is colorless, odorless, tasteless, and kills thousands of people every year. Most of these deaths are caused by accidental inhalation of DHMO in its liquid form, but the dangers of dihydrogen monoxide do not end there. Prolonged exposure to its solid form causes tissue damage and contact with its gaseous form causes burns. DHMO use is widespread. For those who have become dependent on it, DHMO withdrawal means death. DHMO can be an environmental hazard: it is a major component of acid rain, contributes to the "greenhouse effect", leads to the erosion of natural landscapes and hastens the corrosion of most metals. Being so prevalent (quantities are found in every stream, lake and reservoir), DHMO contamination is at epidemic proportions. Despite the dangers, DHMO is often used as an industrial solvent, as a fire retardant, in nuclear power plants and (can you believe this) in certain food products. Companies dump waste dihydrogen monoxide into rivers and the ocean, and nothing can be done to stop them because this practice is still legal. STOP THE HORROR NOW! The American government and the United Nations have refused to ban the production, distribution or use of this chemical due to its "economic importance." The navy and certain other military organizations are highly dependent on DHMO for various purposes. Military facilities receive tons of it through a sophisticated underground distribution network. It is also stored in large quantities for military emergencies. BUT IT'S NOT TOO LATE! You can help. Act *NOW* to prevent further contamination. Write your representatives. Start and sign petitions. Send e-mails. Inform your friends about the dangers. What you don't know *CAN* hurt you and every individual throughout the world.


1. Consider a sphere of radius R0 and mass M0. Inside it is a spherical cavity of radius b. The distance between the two centers is R(see figure). Calculate the gravitational force acting on a mass m inside the cavity.]

R

O R0

M0

O´

b

Sol. The main principle which we use in the solution of

the problem is the superposition principle, where we consider the spherical cavity as a sphere of negative mass. We denote the position vector of a point inside the cavity as r

r and the position vector as measured

from the solid sphere's center O as r (see figure). The force acting upon m is

r´r

R F

r = 0F

r – 1F

r ...(1)

where 0Fr

is the force of the solid sphere and 1Fr

is the force of the cavity (which is filled with negative mass).

According to Newton's gravitation law,

0Fr

= )r(r

m)r(GM2 − ...(2)

where M(r) is the mass up to radius r (see figure).

R0 r

M(r)

Note that this is the only mass which affects m positioned at r. The mass density is uniform, thus

ρ = 30

0

R3

4Mπ

. Therefore,

M(r) = 3

4π r3

30

0

R3

4Mπ

= M0

3

0Rr

Substituting M(r) in Eq. (2), we obtain :

0Fr

= – rrR

mGM30

0 = – rR

mGM30

0 r

Similarly,

1Fr

= rR

mGM30

0 r

Now, Fr

= 0Fr

– 1Fr

= )rr(R

mGM30

0 rr−

Since Rr

= rr

– rr

,

Fr

= 30

0

RmGM

Rr

≡ –C Rr

where C is a constant. We have arrived at the interesting result that the gravitational force inside the cavity is constant and is directed towards the sphere's center. Notice that for the special case Rr

= 0 the force vanishes, as expected. 2. Two masses, m1 and m2, are tied to the ends of a

spring whose force constant is k, and whose natural length is a. This system placed horizontally on a perfectly smooth table, as shown in fig. At t = 0, m1 is bumped and receives a linear momentum of

1pr

= p0 x , where p0 is a constant. m2 k m1

a (i) Write the equations of motion for m1 and m2.

What is the velocity of the center of mass?

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

Students' ForumPHYSICS


(ii) Prove that the harmonic oscillation equation of the system is : µ( 2x&& – 1x&& ) = – k(x2 – x1)

where µ = 21

21

mmmm

+

(iii) What is the oscillation amplitude of (x2 – x1) ? Sol. The motion of the system of the two bodies can be

conveniently described by using the center of mass frame of reference. The centre of mass moves in a straight line with constant velocity, due to the conservation of linear momentum. In the centre of mass frame the two bodies perform simple harmonic oscillations. Denoting the position of the masses m1 and m2 by x1 and x2, respectively, we can express the distance between the masses as x2 – x1. The change in the length of the spring is then x = x2 – x1 – a.

(i) The forces applied by the spring on the two systems are :

m1 1x&& = kx ...(1)

m2 2x&& = – kx ...(2)

The signs are used according to the position of the mass relative to the spring. Multiplying Eq.(1) by m2, Eq. (2) by m1, and subtracting Eq. (1) from Eq. (2) we have

m1m2( 2x&& – 1x&& ) = – k(m1 + m2)x ...(3)

Since x = x2 – x1 – a, we have x& = 2x& – 1x& . And

x&& = 2x&& – 1x&& . Therefore,

m1m2 x&& = –k(m1 + m2)x ...(4) and the solution to this equation is : x(t) = A cos(ω0t + φ) ...(5)

where ω0 = kmmmm

21

21 + . Notice that x does not

denote the position of any of the masses. It denotes the difference between the distance between the masses and the initial state, so that x = (x2 – x1) – a. The velocity of the centre of mass is given by :

cmvr

= 21

0

mmxp

+ ...(6)

(ii) The equation specified in the problem is easily derived from Eq. (3), which we found in the first section. The constant µ is called the "reduced mass" of the system, and is defined as

µ1 ≡

1m1 +

2m1 ...(7)

or in a different form,

µ = 21

21

mmmm

+ ...(8)

(iii) Taking energy into consideration, we have

Ek = 21 m1

2

1

0

mp

=

21

1

20

mp

...(9)

where Ek is the initial kinetic energy. The kinetic energy of the center of mass is :

Ek(cm) = 21

20

mmp

21

+ ...(10)

and therefore, the total kinetic energy in the centre of mass frame becomes :

E´k(cm) = Ek – Ek(cm) = 20

211

2 p)mm(m2

m+

The kinetic energy is proportional to the square of

ampitude, E´k = 21 kA2

Therefore, A = )mm(km

pm

211

202

+

3. (i) A mass m with kinetic energy Ek collides with another mass M, initially at rest, and sticks to it at the moment of contact. What is the total kinetic energy immediately after the collision ?

(ii) The mass M is used as the weighting surface of a spring-scale whose spring is ideal (a mass less spring). A body of known mass m is released from a certain height from where it falls to hit M. The two masses M and m stick together at the moment they touch, and move together from then on. The oscillations they perform reach to height a above the original level of the scale, and depth b below it (see figure).

(a) Find the constant of force of the spring. (b) Find the oscillation frequency. (c) What is the height above the initial level from

which the mass m was released ?

m

M

bm

M

m

Ma

Before After

Sol. (i) The initial kinetic energy is given by Ek = m21 p2,

and the final value of the kinetic energy is given by

E´k = )Mm(2

1+

p´2.

Using linear-momentum conservation (p´ = p), we have:

E´k = Mm

m+

Ek


(ii) (a) Since the point of equilibrium changes, we know that a ≠ b (it moves down from the original level due to the extra mass m). Furthermore, using a – (–b) = a + b = 2A, where A is the amplitude of oscillations, and a – y = A, where y is the height of the new point of equilibrium relative to the original one (y < 0), we find that the point of equilibrium is

2ab − below the original level of the scale. Applying

the equilibrium of forces, we can write :

k|y| = k

−

2ab = mg

And, therefore, k = ab

mg2−

.

(b) The oscillation frequency is v = π

ω2

,

where ω is defined as :

ω = Mm

k+

= )ab)(Mm(

mg2−+

(c) The potential energy in harmonic motion is

known to be 21 kx2. Therefore, the law of

conservation of energy yields

E(t = 0) = E´k + 21 ky2 = E´k +

2k 2

2ab

−

= E´´ = 2

2ab

2k

+

where E´´ denotes the total energy when the mass stops; i.e., when the amplitude is maximal and the kinetic energy vanishes. Using the result of the first section (Ek = mgh and the relation between Ek and E´k) along with Eq. we obtain :

E´k = mghMm

m+

= 2

2ab

2k

+ –

2

2ab

2k

−

= 2k ab

Plugging in the value of k, we arrive at :

h = ab

abm

Mm−

+

4. A constant current I flows through a cable consisting

of two thin co-axial metallic cylinders of radii R and 2R. Calculate

(i) energy stored in it per unit length and (ii) inductance per unit length Sol. Due to flow of current through the cable, show in

figure (A), magnetic field is established in the space between cylinders. Energy is stored in this space due to magnetic field.

I

Fig.(A) Since, energy stored per unit volume in magnetic

fields of induction B is equal to B2/2µ0 , therefore, magnetic induction in the space must be known. But magnetic field in the space is not uniform.

Hence, consider a thin cylindrical coaxial shell in the space. Let radius of the shell be x and radial thickness dx as shown in figure(B).

dx

x

R 2R

Fig.(B) According to Ampere's circuital law, magnetic

induction B is given by B 2πx = µ0I

or B = x2I0

πµ

∴ Energy stored per unit length of the shell considered is

dU = 0

2

2Bµ

(2πx dx) = x4I2

0

πµ

dx

∴ Energy stored per unit length of the cable,

U = ∫dU = ∫=

=π

µR2x

Rx

20

x4I dx =

πµ4

I20 loge 2 Ans. (i)

Since, energy stored in magnetic field is U = 21 LI2

where L is self inductance, therefore, inductance per unit length of the cable is

L = 2IU2 =

πµ2

0 loge 2 Ans. (ii)

5. A non-conducting thin spherical shell of radius R has uniform surface charge density σ. The shell rotates about a diameter with constant angular velocity ω. Calculate magnetic induction B at the centre of the shell.

Sol. When the shell rotates, current is induced due to motion of charge. To calculate magnetic induction at centre of the shell, rotating shell can be assumed to be composed of thin circular current carrying rings. Such a ring can be assumed as follows:


ω

θ

Consider a radius of the shell inclined at angle 'θ'

with the axis of rotation. This radius is rotated about the axis keeping θ constant. Thus a circle is traced as shown in figure.

Its radius, r = R sin θ Distance of its centre from centre of the shell,

x = R cos θ, Now consider another radius inclined at angle

(θ + dθ). It is also rotated in the same way and another circle is traced. The portion between two circles forms a circular ring.

Area of this ring = 2πr R dθ = 2πR2 sin θ dθ Charge on this ring , = dθ = σ.2πr2 sin θ dθ Since, angular velocity of the shell is ω, therefore, it

completes π

ω2

revolutions per second.

Hence, current associated with the ring considered,

i = π

ω2

dQ = σωR2 sin θ dθ

Since, centre of the shell is a point lying on the axis of a circular coil of radius r, carrying current i at a distance x from centre of the coil, therefore, magnetic induction at centre of the shell due to this coil is

dB = 2/322

20

)xr(2ir

+µ

= 21 µ0 σω R sin3 θ dθ

Hence. resultant magnetic induction at centre of the shell

B = ∫dB

= 21

µ0σωR ∫π

θ0

3sin dθ

= 32 µ0σωR Ans.

-: Niobium :-

Brief description : the name niobium was adopted officially by IUPAC in 1950, but a few commercial producers still like to refer to it as columbium. Niobium is a shiny, white, soft, and ductile metal, and takes on a bluish tinge when exposed to air at room temperatures for a long time. The metal starts to oxidize in air at high temperatures, and when handled hot must be done so under a protective atmosphere so as to minimize oxide production.

Table : basic information about and classifications of niobium.

• Name : Niobium

• Symbol : Nb

• Atomic number : 41

• Atomic weight : 92.90638 (2)

• Standard state : solid at 298 K

• Group in periodic table : 5

• Group name : (none)

• Period in periodic table : 5

• Block in periodic table : d-block

• Colour : grey metallic

• Classification : Metallic

ISOLATION Isolation : isolation of niobium appears to be complicated. Niobium minerals usually contain both niobium and tantalum. Since they are so similar chemically, it is difficult to separate them. Niobium can be extracted from the ores by first fusing the ore with alkali, and then extracting the resultant mixture into hydrofluoric acid, HF. Current methodology involves the separation of tantalum from these acid solutions using a liquid-liquid extraction technique. In this process tantalum salts are extracted into the ketone MIBK (methyl isobutyl ketone, 4-methyl pentan-2-one). The niobium remains in the HF solution. Acidification of the HF solution followed by further extraction in MIBK gives an organic solution containing niobium.

After conversion to the oxide, metallic niobium can be made by reduction with sodium or carbon. Electrolysis of molten fluorides is also used.


Review of Concepts : Electric current is the rate of transfer of charge

through a certain surface. The direction of electric current is as that of flow

of positive charge. If a charge ∆q cross an area in time ∆t, then the

average current = ∆q/∆t Its unit is C/s or ampere. Electric current has direction as well as

magnitude but it is a scalar quantity. Electric current obeys simple law of algebra. i.e., I = I1 + I2

α I1

I2

I1

Types of Current : Steady state current or constant current : This

type of current is not function of time. Transient or variable current : This type of current

passing through a surface depends upon time.

i.e., I = f(t) or I = tqlim

0t ∆∆

→∆ ⇒

dtdq

Electric charge passing a surface in time

t = q = ∫t

0dtI

Average current I =

∫∫

t

0

t

0

dt

dtI

Convection Current : The electric due to mechanical transfer of charged particle is called convection current. Convection current in different situation.

Case I : If a point charge is rotating with constant angular velocity ω.

I = Tq ; T =

ωπ2 ⇒ I =

πω

2q

Case II : If a non-conducting ring having λ charge per unit length is rotating with constant angular velocity ω about an axis passing through centre of ring and perpendicular to the plane of ring.

I = R λω

θ J

∆S

or J = θ∆

∆cosSI

Its unit A/m2 Electric current can be defined as flux of current

density vector.

i.e., i = ∫→→dS.j

Relation between drift velocity and current

density dv = – enj

→

Here, negative sign indicates that drifting of electron takes place in the opposite direction of current density.

The average thermal velocity of electron is zero. Electric resistance : Electric resistance (R) is

defined as the opposition to the flow of electric charge through the material.

It is a microscopic quantity. Its symbol is Its unit is ohm.

(a)

(b)

R = Alρ

where, R = resistance, ρ = resistivity of the material, l = length of the conductor, A = area of cross section Continuity Equation :

∫→→

cdS.j = –

dtdq

The continuity equation is based on conservation principle of charge.

Drift Velocity (vd) : When a potential difference is applied between ends of metallic conductor, an

Current Electricity

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


electric field is established inside the metallic conductor. Due to this, electron modify their random motion and starts to drift slowly in the opposite direction of electric field. The average velocity of drifting possessed by electron is known as drift velocity.

→

dv = →

τ E

me

where, dv→

= drift velocity, e = electron, τ = relaxation time, m = mass of electron

→E = electric field

Variation of Resistance with Temperature : Let a metallic conductor of length l and cross-sectional area A.

Rt = R0(1 + αt) where, Rt = resistance of conductor at temperature tºC, R0 = resistance of conductor at 0ºC, α = temperature coefficient.

A

i

Some Important Points : (a) 'α' is proportionality constant known as

temperature coefficient of resistance variation. (b) The value of α does not depend upon initial and

final resistance of the conductor. (c) The value of α depends upon the unit which is

chosen. (d) The value of α may by negative. Electric Conductance (G) :

It is reciprocal of resistance, G = R1

Its unit is per ohm.

Electric conductivity σ = ρ1

Ohm's law in vector form :

E = ρ→i

where, ρ = τ2ne

m = receptivity of material

According to ohm's law, electric current passing through a conductor is proportional to the potential difference between end of the conductor

i.e., V = IR In case of ohm´s law, V-I graph is straight line.

I

V

α

Ohm's Law fails in tube, crystal diodes, thyristors

etc. EMF and PD of a Cell : A device which supplies

electric energy is called a seat of emf. The seat of emf is also called a cell.

A battery is a device which manages a potential difference between its two terminals.

e = EMF of the battery is the work done by the force per unit charge.

When the terminals of a cell are connected to an external resistance, the cell is said to be in closed circuit.

E.M.F. has no electrostatic origin. Internal Resistance of a Cell (r) : Internal

resistance of a cell is the resistance of its electrolyte.

The internal resistance of cell : (a) Varies directly as concentration of the solution of

the cell. (b) Varies directly as the separation between

electrodes i.e., length of solution between electrodes.

(c) Varies inversely as the area of immersed electrodes.

(d) is independent of the material of electrodes. Potential difference across the cell : Potential difference across the first cell V1 = E1 + Ir1 (discharging of cell)

R

i

E1r1 E2r2

Potential difference across the second cell V2 = E2 – Ir2 (charging of cells)

Concept of Rise up and Drop up of voltage: (a) Ideal cell

Rise up

+E

Drop up

–E


(b) Real cell

Rise up

E – ir

Drop up

–E – ir

r,E

i r,E i

(c) Electric resistance

Drop up

–IR

Rise up

+IR

R i R i

When a battery being charged, the terminal

voltage is greater than its emf V = E + Ir. Kirchhoff's Law : Kirchhoff's law is able to

solve complicated circuit problems. (i) First Law : Incoming current = Outgoing current I1 + I2 = I3 + I4 + I5

I2 I5

I1 I3

I4

This law is based upon conservation principle of

charge. (ii) Second Law : (Loop rule or voltage law.) This

law is based upon conservation principle of energy.

Grouping of resistors : Case I : Resistors in series RMN = Req = R1 + R2

M R1 R2 N

In general, Req = R1 + R2 + ... + Rn Case II : Resistors in parallel

MNR1 =

eqR1 =

1R1 +

2R1

M

R1

R2

N

In general,

MNR1 =

1R1 +

2R1 + ... +

nR1

Problem solving strategy. : Power and Energy in circuits

Step 1 Identify the relevant concepts : The ideas of electric power input and output can be

applied to any electric circuit. In most cases you’ll know when these concepts are needed, because the

problem will ask you explicitly to consider power or energy.

Step 2 Set up the problem using the following steps : Make a drawing of the circuit. Identify the circuit elements, including sources of

emf and resistors. Determine the target variables. Typically they

will be the power input or output for each circuit element, or the total amount of energy put into or taken out of a circuit element in a given time.

Step 3 Execute the solution as follows : A source of emf ε delivers power εI into a circuit

when the current I runs through the source from – to +. The energy is converted from chemical energy in a battery, from mechanical energy in a generator, or whatever. In this case the source has a positive power output to the circuit or, equivalently, a negative power input to the source.

A source of emf power εI from a circuit – that is, it has a negative power output, or, equivalently, a positive power input–when currents passes through the source in the direction from + to –.This occurs in charging a storage battery, when electrical energy is converted back to chemical energy. In this case the source has a negative power output to the circuit or, equivalently, a positive power input to the source.

No matter what the direction of the current through a resistor, It removes energy from a circuit at a rate given by VI = I2R = V2/R, where V is the potential difference across the resistor.

There is also a positive power input to the internal resistance r of a source, irrespective of the direction of the current. The internal resistance always removes energy from the circuit, converting it into heat at a rate I2r.

You may need to calculated the total energy delivered to or extracted from a circuit element in a given amount of time. If integral is just the product of power and elapsed time.

Step 4 Evaluate your answer : Check your results, including a check that energy is conserved. This conservation can be expressed in either of two forms: “net power input = net power output” or “the algebraic sum of the power inputs to the circuit elements is zero.”

Problem solving strategy : Series and Parallel

Step 1 Identify the relevant concepts : Many resistor networks are made up of resistors in series, in parallel, or a combination of the two. The key concept is such a network can be replaced by a single equivalent resistor.


Step 2 Set up the problem using the following steps: Make a drawing of the resistor network. Determine whether the resistors are connected in

series or parallel. Note that you can often consider networks such as combinations of series and parallel arrangements.

a R1 x R2 y R3 b

I I(a) R1, R2, and R3 in series

R1

I

(b) R1, R2, and R3 in parallel

R2

R3 I

a b

Determine what the target variables are. They could include the equivalent resistance of the network, the potential difference across each resistor, or the current through each resistor.

Step 3 Execute the solution as follows : Use Eq. Req = R1 + R2 + R3 (resistors in series)

or eqR

1 =1R

1 +2R

1 +3R

1 +...(resistors in parallel)

to find the equivalent resistance for a series or a parallel combination, respectively.

If the network is more complex, try reducing in to series and parallel combinations.

When calculating potential differences, remember that when resistors are connected in series, the total potential differences across the combination equals the sum of the individual potential differences. When they are connected in parallel, the potential difference across the parallel combination.

Keep in mind the analogous statements for current. When resistors are connected in series, the current is the same through every resistor and equals the current through the series combination. When resistors are connected in parallel the total current through the combination equals the sum of the currents through the individual resistors.

Step 4 Evaluate your answer : Check whether your results are consistent. If resistors are connected in series, the equivalent resistance should be greater than that of any individual resistor; if they are connected in parallel, the equivalent resistance should be less than that of any individual resistor.

Problem solving st. : Kirchhoff’s Rules : Step 1 Identify the relevant concepts : Kirchhoff’s

rules are important tools for analyzing any circuit more complicated than a single loop.

Step 2 Set up the problem using the following steps : Draw a large circuit diagram so you have plenty

of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. Often you will not know in advance the actual direction of an unknown current or emf, but this does not matter. If the the actual direction of a particular quantity is opposite to your assumption, the result will come out with a negative sign. If you are Kirchhoff’s rules correctly, they will give you the directions as well as the magnitudes of unknown currents and emfs.

When you label currents, it is usually best to use the junction rule immediately to express the currents of as few quantities as possible. For example, fig (a) shows a circuit correctly labeled; fig. (b) shows the same circuit, relabeled by applying the junction rule to point a to eliminate I3.

r1 ε1 r2 ε2

I1

I1

I3 R3

I2

I2

R1 a R2

+ +

(a)

r1 ε1 r2 ε2

I1

I1

I1 + I2 R3

I2

I2

R1 a R2

+ +

(b) Determine which quantities are the target

variables. Step 3 Execute the solution as follows : Choose any closed loop in the network and

designate a direction (clockwise or counterclockwise) to travel around the loop when applying the loop rule. The direction does not have to be the same as any assumed current direction.


Travel around the loop in the designated direction, adding potential differences as you cross them. Remember that a positive potential and a negative potential difference corresponds to a decrease in potential. An emf is counted as positive when you traverse it from (–) to (+), and negative when you go from (+) to (–). An IR term is negative if you travel through the resistor in the same direction as the assumed current and positive if you pass it in the opposite direction. Figure. summarizes these sign conventions. In each part of the figure “travel” is the direction that we imagine going around a loop while using Kirchhoff’s loop law, not necessary the direction of current.

Equate the sum is Step 2 to zero. If necessary, choose another loop to get a

different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in a at least one of the chosen loops.

Solve the equations simultaneously to determine the unknowns. This step involves algebra, not physics, but it can be fairly complex. Be careful with algebraic manipulations; one sign error will prove fatal to the entire solution.

You can use this same bookkeeping system to find the potential Vab of any point a with respect to any other point b. Start at b and add the potential changes you encounter in going from b to a, using the same sign rules as in Step 2. The algebraic sum of the these changes is vab = Va

– Vb.

Step 4 Evaluate your answer : Check all the step in your algebra. A useful strategy is to consider a loop other than the ones you used to solve the problem; if the sum of potential drops around this loop is not zero, you made an error somewhere in your calculations. As always, ask yourself whether is answer make sense.

Travel

+ – + ε

Travel

– + – ε

Travelε ε

Travel

– + –IR

TravelR

I

Travel

+ – +IR

TravelR

I When using Kirchhoff’s rules, follow these sign

conventions as you travel around a circuit loop.

1. In a circuit shown in fig. (i) find the current drawn from the accumulator. (ii) find the current through the 3 ohm resistor, (iii) What happens when 3 ohm resistor is removed

from the circuit ? 2V

1Ω

3Ω2Ω 2Ω

4Ω

A

C B

D

Sol. The equivalent Wheatstone's bridge network of the

given circuit is shown in fig. B

2Ω 4Ω

3Ω

1Ω 2Ω

D

2 Volt

A C

Here the points B and D are at the same potential as

the bridge is balanced. So the 3Ω resistance in BD arm is ineffective and can be omitted from the circuit. The resistance of ABC branch is 2Ω + 4Ω = 6Ω as AB and BC are in series. Similarly the resistance of A D C branch is 1Ω + 2Ω = 3Ω.

The two resistances, i.e., 6 ohm and 3 ohm are in parallel. The equivalent resistance R is given by

R1 =

61 +

31 =

21 ∴ R = 2Ω

(i) The current drawn from 2 volt accumulator is

i = RE =

22 = I amp.

(ii) The current through 3Ω resistor is zero. (iii) When the 3Ω resistor is removed from the

circuit, there will be no change. 2. A battery of e.m.f. 5 volt and internal resistance 20Ω

is connected with a resistance R1 = 50 Ω and a resistance R2 = 40Ω. A voltmeter of resistance 1000 Ω is used to measure the potential difference across R1. What percentage error is made in the reading ?

Solved Examples


Sol. The circuit is shown in fig.

V 1000Ω

20Ω

R2 = 50Ω

5V

R1 = 50Ω

When voltmeter is not connected

current in the circuit i = 21 RRr

E++

∴ i = 405020

5++

= 110

5 = 221 A

Potential difference across R1 = i × R1

= 221 × 50 = 2.27 volt.

When the voltmeter is connected across R1. In this case the galvanometer resistance is in parallel

with R1. Hence

Equivalent resistance = 501000501000

+× = 47.62 ohm

Current in the circuit

= 62.474020

5++

= 62.107

5 A

Potential difference measured by voltmeter

= 62.107

5 × 47.62 = 2.21 volt.

Percentage error = 27.2

21.227.2 − × 100 = 2.6%

3. In the circuit fig. a voltmeter reads 30 V when it is

connected across 400 ohm resistance. Calculate what the same voltmeter will read when it is connected across the 300 Ω resistance ?

V

30 V

300 Ω

400 Ω

60 V Sol. Potential difference across 400 ohm = 30 V Potential difference across 300 ohm = (60 – 30) = 30 V This shows that the potential difference is equally

shared.

Let R be the voltmeter resistance. The resistance 400 and voltmeter resistance R are in parallel. Their equivalent resistance R´ is given by

´R

1 = R1 +

4001 =

R400R400 + or

R400R400

+

But R´ should be equal to 300 ohm. Hence

R400

R400+

= 300 ∴ R = 1200 ohm

Thus, voltmeter resistance is 1200 ohm. When the voltmeter is connected across 300 ohm, the

effective resistance R" is given by

"R

1 = 1200

1 + 300

1 = 1200

41+ = 1200

5

∴ R´´ = 5

1200 = 240 ohm.

Now the potential difference is shared between 240 ohm and 400 ohm.

Potential diff. across 240 ohm : Potential difference across 400 ohm

= 240 : 400 = 3 : 5 As total potential is 60 V, hence potential difference

across 240 ohm, i.e., across resistance 300 ohm will be

83 × 60 = 22.5 V.

4. In the circuit shown in fig. E, F, G and H are cells of

e.m.f. 2, 1, 3 and 1 volt and their internal resistances are 2, 1, 3 and 1 ohm respectively. Calculate

E – +

+–

–

–+

+

A B

D CG

F H2Ω

(i) The potential difference between B and D and (ii) the potential difference across the terminals of

each of the cells G and H. Sol. Fig. shows the current distribution. Applying Kirchhoff's first law at point D, we have i = i1 + i2 ...(1) Applying Kirchhoff's second law to mesh and

ADBA, we have 2i + 1i + 2i1 = 2 – 1 = 1 or 3i + 2i1 = 1 ...(2)

2V A B

D C3V

1V

1V

2Ω

2Ω

3Ω

1Ω

ii2

1Ωi1


Applying Kirchhoff's second law to mesh DCBD, we get

3i2 – 1i2 – 2i1 = 3 – 1 or 4i2 – 2i1 = 2 ...(3) Solving eqs. (1), (2) and (3), we get

i1 = 131 amp., i2 =

136 amp. and i =

135 amp.

(i) Potential difference between B and D

= 2i1 = 2

131 =

132 volt.

(ii) Potential difference across G

= E – i2R = 3 – 13

36× = 1.61 V

Potential difference across H

= 1 –

−

136 (1) = 1.46 V.

5. Twelve equal wires, each of resistance 6 ohm are

joined up to form a skeleton cube. A current enters at one corner and leaves at the diagonally opposite corner. Find the joint resistance between the corners.

Sol. The skeleton ABCDEFGH, is shown in fig.

A

E F i/6

B

GH

i/3 i/3 i/6

i/6i/3

C D i/6

i/3

i/6

i i/6 i/3

i/3

This skeleton consists of twelve wires. Let the

resistance of each wire be r. Here the current i enters at corner A and leaves at corner G. The current i at corner A is divided into three equal parts (i/3) because the resistance of each wire is the same. At B, D and E, the current i/3 is divided into two equal parts each having magnitude i/6. At the corners C, F and H, the currents again combine to give currents, each of magnitude i/3 along CG, FG and HG respectively. At corner G, all these currents combine so that the current leaving at G is i.

Let R be the equivalent resistance between the corners A and G. Taking any one of the paths say ABCG, we have

VAG = VAB + VBC + VCG

iR = 3i r +

6i r +

3i r

or R = 65 r

According to given problem r = 6 ohm

∴ R = 65 × 6 = 5 ohm.

• In late 2001, Associated Press reported, "NASA might allow McDonald's to put its logo on the international space station galley in exchange for McDonald's promoting space exploration to kids". Err...Mine's a Big Mac Please.

• A 10 pound sack of flour on the moon would bake six times as much bread as a sack weighing 10 pounds on earth.

• The Comets that pass close to the Sun originally came from one of two places; either the Oort Cloud or the Kuiper Belt. Approximately a dozen 'new' Comets are discovered every year. Because they are so far from the Sun, the Comets in the Oort Cloud take over 1 million years to make a single revolution around the Sun.

• There are stars as much as 400,000 brighter than the sun and others as much as 400,000 time fainter if they could all be seen at the same distance.

• A pulsar is a small star made up of neutrons so densely packed together that if one the size of a silver dollar landed on Earth, it would weigh approximately 100 million tons.

• An exploding supernova can outshine an entire galaxy of stars.

• There are 17 bodies in the solar system whose radius is greater than 1000 km.

• Over 90 per cent of the Universe consists of invisible 'dark matter'.

• In 1719 Mars was closer to Earth than it would be until the year 2003.

• By 2100, in the absence of emissions control policies, carbon dioxide concentrations are projected to be 30-150% higher than today's levels.


Circular Motion : When a particle moves on a circular path with

uniform speed, its is said to execute a uniform circular motion.

Angular Velocity : It is the rate of change of angular displacements of the body. If the radial line in the adjoining figure rotates through an angle θ(radian) in time t (seconds) then its angular velocity.

O θ

ω = tθ radian / second

If it takes the radial line a time T to complete one revolution, then

ω = T2π

and if n revolutions are made in 1s then

n = T1 and ω = 2πn

The angular acceleration of the particle is given by

α = t∆ω∆

Linear Velocity : Linear velocity = angular velocity × radius

v = ω × r linear acceleration of particle (a) = a × r Centripetal Acceleration : When a particle moves

with uniform speed v in a circle of radius r it is acted upon by an acceleration v2/r in the direction of centre. It is called centripetal acceleration. The acceleration has a fixed magnitude but its direction is continuously changing. It is always directed towards the centre of the circle.

Centripetal Forces : If the particle of mass m moves with uniform velocity v in circle of radius r, then

force acting on it towards the centre is r

mv2. This is

called centripetal force. It has a fixed magnitude and is always directed towards the centre.

Without centripetal force, a body can not move on a circular path. Earth gets this force from the gravitational attraction between earth and sun; electron moves in circular path due to electrostatic attraction between it and nucleus. A cyclic or car while taking turn, gets the centripetal force from the friction between road and type. To create this force, the vehicle tilts itself towards the centre. If it makes angle θ with the vertical in tilted position then than θ = v2/rg. where v is its velocity and r is the radius of the path. In order to avoid skidding (or slipping), the angle of tilt θ with vertical should be less than angle of friction λ. i.e. tan θ < tan λ

or rgv2

< µ (since coefficient of friction µ = tan λ)

In limiting condition rgv2

= µ or v = g.r.µ

This is the maximum safe speed at the turn. Since centripetal force is provided by the friction, it

can never be more than the maximum value µR = (µmg) or frictional force.

Motion in a vertical circle : When a body tied at one end of a string is revolved in a vertical circle, it has different speed at different points of the circular path. Therefore, the centripetal force and tension in the string change continuously. At the highest point A of motion.

mg Ta

r

Tbvb B

mg

A va

C

Ta + mg = r

mv2a or Ta =

rmv2

a – mg

This tension, at highest point will be zero, for a minimum velocity vc given by

0 = r

mv2c – mg or vc = gr

Circular Motion, Rotational Motion PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


This minimum speed is called critical speed (vc). If the speed at A is less than this value, the particle will not reach up to the highest point. To reach with this speed at A, the body should have speed at B given by the conservation laws viz.

Decrease in kinetic energy = increase in potential energy

21 mvb

2 – 21 mva

2 = mg.2r

vb2 = va

2 + 4gr

for critical speed va = vc = gr

∴ vb2gr + 4gr or vb = gr5

Therefore, the body should have speed at B at least gr5 , so that it can just move in vertical circle.

Tension in string at B is given by.

Tb – mg = r

mv2b or Tb = mg +

rvgr5m = 6mg

This means that the string should be able to stand to a tension, equal to six times the weight of the body otherwise the string will break.

At any other point P making angle θ with the vertical, from the figure.

A

C

B mg

mg cos θ Q

P

vp T θ

T – mg cos θ = r

mv2p or T = m

θ+ cosg

rv2

r

At point A, θ = 180º; Ta = m

− g

rv2

a

At point B, θ = 0º; Tb = m

+ g

rv2

b

Conical pendulum : A conical pendulum consists of a string AB (fig.)

whose upper end is fixed at A and other and B is tied with a bob. When the bob is drawn aside and is given a horizontal push. Let it describe a horizontal circle with constant angular speed ω in such a way that AB makes a constant angle θ with the vertical. As the string traces the surface of a cone, it is known as conic pendulum.

Let l be the length of string AB. The forces acting on the bob are (i) weight mg acting downwards, (ii) tension T along the sting (horizontal) component is T sin θ and vertical component is T cos θ).

T cos θ = mg The horizontal component is equal to the centripetal

force i.e.,

A

h

r T sin θ T

T cosθ

B

mg

O

Rotational Motion : Centre of mass of a system of particles : The point at which the whole mass of the body may

be supposed to be concentrated is called the centre of mass.

Consider the case of a body of an arbitrary shape of n XY plane as shown in fig. Let the body consist of number of

P1(x1, y1)(x2, y2)

P2(x, y )

P3(x3, y3)

X O

Y

particles P1, P2, P3, .... of masses m1, m2, m3, ..... and

coordinates (x1, y1), (x2, y2), (x3, y3), ..... If )y,x( be the coordinates of centre of mass, then

x = .....mmm

....xmxmxm

321

332211

++++++

= n

nn

mxm

ΣΣ

and y = ....mmm

...ymymym

321

332211

++++++

= n

nn

mym

ΣΣ

When there is a continuous distribution of mass instead of being discrete, we treat an infinitesimal element of the body of mass dm whose position is (x, y, z). In such a case, we replace summation by integration in above equations. Now we have,

x = ∫∫

dm

dmx =

M

dmx∫

y = ∫∫

dm

dmy =

M

dmy∫


z = ∫∫

dm

dmz =

M

dmz∫

where M is the total mass. Motion of centre of mass : Consider two particles of masses m1 and m2 located

at position vectors r1 and r2 respectively with respect to origin. Now the position vector r of the centre of mass is given by

(m1 + m2)r = m1r1 + m2r2 ...(1) Thus, the product of the total mass of the system and

position vector of the centre of mass is equal to the sum of the products of the individual masses and their respective position vectors. Hence

r = 21

2211

mmrmrm

++ ...(2)

Now the velocity of centre of mass of the system is

given by v = dtdr

The acceleration of the centre of mass is given by

a = dtdv =

dtd

dtd = 2

2

dtxd

The equation describing the motion of the centre of mass may be written as

f(total) = Mdtdv

When no external force acts on the system, then

0 = Mdtdv or

dtdv = 0

∴ v = constant Therefore, when no external force acts on the system,

the centre of mass of an isolated system move with uniform velocity.

Moment of inertia and radius of gyration : Moment of Inertia : The moment of inertia of a

body about an axis is defined as the sum of the products of the masses of the particles constituting the body and the square of their respective distance from the axis.

Radius of Gyration : If we consider that the whole mass of the body is concentrated at a distance K from the axis of rotation, then moment of inertia I can be expressed as

I = MK2 where M is the total mass of the body and K is the

radius of gyration. Thus the quantity whose square when multiplied by the total mass of the body gives the moment of inertia of the body about that axis is known as radius of gyration.

Theorems on moment of inertia : Theorem of parallel axes : According to this

theorem, the moment of inertia I of a body about any axis is equal to its moment of inertia about a parallel axis through centre of mass IG plus Ma2 where M is the mass of the body and a is the perpendicular distance between the axes, i.e., I = IG + Ma2

Theorem of perpendicular axes : According to this theorem, the moment of inertia I of the body about a perpendicular axis is equal to the sum of moment of inertia of the body about two axes right angles to each other in the plane of the body and intersecting at a point where the perpendicular axis passes, i.e.,

I = Ix + Iy Table of moment of inertia :

Body Axis Moment of inertia

1. Thin uniform rod of length l

Through its centre and perpendicular to its length

12M 2l

2. Thin rectangular sheet of sides a and b.

Through its centre and perpendicular to its plane

M

+

12b

12a 22

3. Thick rectangular bar of length l, breadth b and thickness t.

Through its midpoint and perpendicular to its length

M

+

12b

12

22l

4. Uniform solid sphere of radius R

About a diameter 52 MR2

5. Circular ring of radius R.

Through its centre and perpendicular to its plane

MR2

6. Disc of radius R. Through its centre and perpendicular to its plane

21 MR2

7. Solid cylinder of length l and radius R.

(i) Through its centre and parallel to its length (ii) Through its centre and perpendicular to its length.

21 MR2

M

+

124R 22 l

Angular momentum of a rotating body : In case of rotating body about an axis, the sum of the

momentum of the linear momentum of all the


particles about the axis of rotation is called angular momentum about the axis.

Q Also the angular momentum of rigid body about an axis is the product of moment of inertia and the angular velocity of the body about that axis.

L = r × p = Iω Translational and rotational quantities :

Translational Motion Rotational Motion Displacement = s Angular displacement = θ

Velocity = v Angular velocity = ω

Acceleration = a Angular acceleration = α

Inertia = m Moment of inertia = I

Force = F Torque = τ

Momentum = mv Angular momentum = Iω

Power = Fv Rotational power = τω

Kinetic energy = 21 mv2 Rotational K.E. =

21 Iω2

Kinematics equation of a rotating rigid body : The angular velocity of a rotating rigid body is

defined as the rate of change of angular displacement,

i.e., →ω = )dt/d(

→θ

Similarly, the angular acceleration is defined as the rate of change of angular velocity, i.e.,

→α =

dtd

→ω = 2

2

dtd

→θ

Let a body be rotating with constant angular

acceleration →α with initial angular velocity

→ω0 . If θ

is the initial angular displacement, then its angular

velocity →ω and angular displacement θ at any time is

given by the following equations ω = ω0 + αt

0 = ω0t + 21

αt2

and ω2 = ω02 + 2 αθ

These equations are similar to usual kinematics equation of translatory motion.

v = u + at, s = ut + 21 at2

and v2 = u2 + 2as Problem Solving Strategy : Rotational Dynamics for Rigid Bodies : Our strategy for solving problems in rotational

dynamics is very similar to the strategy for solving problems that in involve Newton’s second law.

Step-1 : Identify the relevant concepts : The equation Στ = Iαz is useful whenever torques act on a rigid body–that is, whenever forces act on a rigid body in such a way as to change the state of the body’s rotation.

In some cases you may be able to use an energy approach instead. However, if the target variable is a force, a torque, an acceleration, an angular acceleration, or an elapsed time, the approach using Στ = Iα2 is almost always the most efficient one.

Step-2 : Setup the problem using the following steps: Draw a sketch of the situation and select the body

or bodies to be analyzed. For each body, draw a free-body diagram

isolating the body and including all the forces (and only those forces) that act on the body, including its weight. Label unknown quantities with algebraic symbols. A new consideration is that you must show the shape of the body accurately, including all dimensions and angles you will need for torque calculations.

Choose coordinate axes for each body and indicate a positive sense of rotation for each rotating body. If there is a linear acceleration, it’s usually simplest to pick a positive axis in its direction. If you know the sense of αz in advance, picking it as the positive sense of rotation simplifies the calculations. When you represent a force in terms of its components, cross out the original force to avoid including it twice.

Step-3 : Execute the solution as follows : For each body in the problem, decide whether it

under goes translational motion, rotational motion, or both. Depending on the behavior of the body in question, apply ΣF = m a

r, Στz = Iαz, or

both to the body. Be careful to write separate equations of motion for each body.

There may be geometrical relations between the motions of two or more bodies, as with a string that unwinds from a pulley while turning it or a wheel that rolls without slipping. Express these relations in algebraic form, usually as relations between two linear accelerations or between a linear acceleration and an angular acceleration.

Check that the number of equations matches the number of unknown quantities. Then solve the equations to find the target variable(s).

Step-4 : Evaluate your answer : Check that the algebraic signs of your results make sense. As an example, suppose the problem is about a spool of thread. If you are pulling thread off the spool, your answers should not tell you that the spool is turning in the direction the results for special cases or intuitive expectations. Ask yourself : Does this result make sense ?”


Problem Solving Strategy: Equilibrium of a Rigid Body Step-1 : Identify the relevant concepts : The first and

second conditions for equilibrium are useful whenever there is a rigid body that is not rotating and not accelerating in space.

Step-2 : Set up the problem using the following steps: Draw a sketch of the physical situation, including

dimensions, and select the body in equilibrium to be analyzed.

Draw a free-body diagram showing the forces acting on the selected body and no others. Do not include forces exerted by this body on other bodies. Be careful to show correctly the point at which each force acts; this is crucial for correct torque calculations. You can't represent a rigid body as a point.

Choose coordinate axes and specify a positive sense of rotation for torques. Represent forces in terms of their components with respect to the axes you have chosen; when you do this, cross out the original force so that you don't included it twice.

In choosing a point to compute torques, note that if a force has a line of action that goes through a particular point, the torque of the force with respect to that point is zero. You can often eliminate unknown forces or components from the torque equation by a clever choice of point for your calculation. The body doesn't actually have to be pivoted about an axis through the chosen point.

Step-3 : Execute the solution as follows : Write equations expressing the equilibrium

conditions. Remember that ΣFx = 0, ΣFy = 0, and Στz = 0 are always separate equations; never add x-and y-components in a single equation. Also remember that when a force is represented in term of its components, you can compute the torque of that force by finding the torque of each component separately, each with its appropriate lever arm and sign, and adding the results. This is often easier than determining the lever arm of the original force.

You always need as many equations as you have unknowns. Depending on the number of unknowns, you may need to compute torques with respect to two or more axes to obtain enough equations. Often, there are several equally good sets of force and torque equations for a particular problem; there is usually no single "right" combination of equations. When you have as many independent equations as unknowns, you can solve the equations simultaneously.

Step-4 : Evaluate your answer : A useful way to check your results is to rewrite the second condition for equilibrium, Στz = 0, using a different choice of origin. If you've done everything correctly, you'll get the same answers using this new choice of origin as you did with your original choice

1. A particle a moves along a circle of radius R = 50 cm

so that its radius vector r relative to point O (fig.) rotates with the

A

R

CO

r

A

R

C O

r

Y

Xθ

θ

(a) (b) constant angular velocity ω = 0.40 rad/sec. Find the

modulus of the velocity of the particle and modulus and direction of its total acceleration.

Sol. Consider X and Y axes as shown in fig. Using sine law in triangle CAO, we get

)2sin(

rθ−π

= θsin

R or θθcossin2

r = θsin

R

∴ r = 2 R cos θ Now r = r cos θ i + r sin θ j = 2 R cos2θ i + 2R cos θ sin θ j

Now, vdtdr = – 4R cos θ sin θ

dtdθ i + 2R cos 2θ

dtdθ j

= – 2 R sin 2 θ ω i + 2 R cos 2θ ω j ∴ |v| = 2 ω R Further

a = dtdv = 4 R cos 2 θ

dtdθ i – 4 R ω sin 2 θ

dtdθ j

= –4 R ω2 cos 2 θ i – 4R ω2 sin 2θ j |a| = 4 Rω2

2. A particle describes a horizontal circle on the smooth inner surface of a conical funnel as shown in fig. If the height of the plane of the circle above the vertex 9.8 mark cm, find the speed of the particle.

Sol. The forces acting on the particle are shown in fig. They are

R sin α

Rα

r h=9.8 cmα

mg

mv2/r

R c

os α

(i) weight m g acting vertically downwards. (ii) normal reaction R of smooth surface of the cone. (iii) reaction of the centripetal force (mv2/r) acting

radially outward. Hence, R sin α = m g ...(1) and R cos α = (mv2/r) ...(2)

Solved Examples


Dividing eq. (1) by eq. (2), we get

tan α =

2vrg ...(3)

From figure, tan α = (r/h) ...(4) From eqs. (3) and (4), we get

hr = 2v

rg or v = )hg(

∴ v = [9.8 × (9.8 × 10–2)]1/2 = 9.98 m/s

3. A particle of 10 kg mass is moving in a circle of 4m radius with a constant speed of 5m/sec. What is its angular momentum about (i) the centre of circle (ii) a point on the axis of the circle and 3 m distant from its centre ? Which of these will always be in same direction ?

Sol. The situation is shown in fig. L1

L2

3 0 4

5

(a) We know that L = r × mv

L = m v r sin θ Here m = 10 kg, r = 4 m, v = 5 m/sec and θ = 90º ∴ L = 10 × 5 × 4 × 1 = 200 kg-m2/sec.

(b) In this case r = )34( 22 + = 5m. ∴ L = 10 × 5 × 5 = 250 kg-m2/sec. From figure it is obvious that angular momentum in

first case always has same direction but in second case the direction changes.

4. A symmetrical body is rotating about its axis of symmetry, its moment of inertia about the axis of rotation being 1 kg-m2 and its rate of rotation 2 rev./sec. (a) what is its angular momentum ? (b) what additional work will have to be done to double its rate of rotation ?

Sol. (a) As the body is rotating about its axis of symmetry, the angular momentum vector coincides with the axis of rotation.

∴ Angular momentum L = Iω ...(1)

Kinetic energy of rotation E = 21 Iω2

or 2E = Iω2 ∴ I2ω2 = 2IE or Iω = )IE2( ...(2)

From eqs. (1) and (2), L = )IE2( ...(3) ω = 2 rev/sec = 2 × 2π or 4π radian/sec.

∴ E = 21 × 1 × (4π)2 = 8π2 joule

Now L = )IE2( = )812( 2π××

= )16( 2π = 4π = 12.57 kg.m2/sec. (b) When the rate of rotation is doubled, i.e., 4

rev/sec or 8π radians/sec, the kinetic energy of rotation is given by

E = 21 × 1 × (8π)2 = 32π2 joule

Additional work required = Final K.E. of rotation – Initial K.E. of rotation = 32π2 – 8π2 = 24 π2 = 236.8 joule

5. A thin horizontal uniform rod AB of mass m and length l can rotate freely about a vertical axis passing through its end A. At a certain moment the end B starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed in the horizontal plane. Find the angular velocity of the rod as a function of its rotation angle φ counted relative to the initial position.

Sol. The situation of the rod at an angle φ is shown in fig. Here

r = i l cos φ + + j l sin φ and F = j F (Force is always perpendicular to rod)

A B X

F

F Y

θ l

→τ = r × F = (i l cos φ + j l sin φ) × (j F)

= l F cos φ k

| →τ | = l F cos φ

We know that τ = 1 α

Here I = 31 m l2 (for rod) and α = ω (dω/dφ)

∴ l F cos φ = 31 m l2 . ω (dω/dφ)

or l F cos φ dφ = 31 m l2 . ω dω

Integrating within proper limits, we have

l F ∫φ

φφ0

dcos = 31 m l2 ∫

ωωω

0d

l F sin φ = 31 m l2(ω2/2) ∴ ω =

φ

lmsinF6


Organic Chemistry

Fundamentals

Addition of hydrogen halides to Alkenes : Markovnikov’s Rule Hydrogen halides (HI, HBr, HCl, and HF) add to the

double bond of alkenes :

C = C + HX → – C – C –

H H These additions are sometimes carried out by

dissolving the hydrogen halide in a solvent, such as acetic acid or CH2Cl2, or by bubbling the gaseous hydrogen halide directly into the alkene and using the alkene itself as the solvent. HF is prepared as polyhydrogen fluoride in pyridine. The order of reactivity of the hydrogen halides is HI > HBr > HCl > HF, and unless the alkene is highly substituted, HCl reacts so slowly that the reaction is not one that is useful as a preparative method. HBr adds readily, the reaction may follow an alternate course. However, adding silica gel or alumina to the mixture of the alkene and HCl or HBr in CH2Cl2 increases the rate of addition dramatically and makes the reaction an easy one to carry out.

The addition of HX to an unsymmetrical alkene could conceivably occur in two ways. In practice, however, one product usually predominates. The addition of HBr to propene, for example, could conceivably lead to either 1-bromopropane or 2-bromopropane. The main product, however is 2-bromopropane :

CH2 = CHCH3 + HBr → CH3CHCH3

Br2-Bromopropane

When 2-methylpropene reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide :

C = CH2 + HBr → CH3 – C – CH3

Brtert-Butyl bromide

CH3 H3C

H3C

2-Methylpropene (isobutylene)

Consideration of many examples like this led the Russian chemist Vladimir Markovnikov in 1870 to formulate what is now known as Markovnikov’s rule. One way to state this rule is to say that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that

already has the greater number of hydrogen atoms. The addition of HBr to propene is an illustration :

CH2 = CHCH3 → CH2 – CHCH3

H BrMarkovnikov addition

product H Br

Carbon atom with the greater number of hydrogen atoms

Reactions that illustrate Markovnikov’s rule are said

to be Markovnikov additions. A mechanism for addition of a hydrogen halide to an

alkene involves the following two steps : Step 1 :

C = C + H – X → +C – C – + X

H

– slow

The π electrons of the alkene form a bond with a proton from HX to form a carbocation and a halide ion

Step 2 :

X + +C – C – → – C – C –

H

fast

The halide ion reacts with the carbocation bydonating an electron pair; the result is an alkyl halide

–H

X

Modern Statement of Markovniov’s Rule : According to Modern statement of Markovnikov’s

rule, In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the adding reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as an intermediate. Because this is the step that occurs first (before the addition of the nucleophilic portion of the adding reagent), it is the step that determines the overall orientation of the reaction.

Notice that this formulation of Markovnikov’s rule allows us to predict the outcome of the addition of a

such as ICl. Because of the greater electro negativity of chlorine, the positive portion of this molecule is iodine. The addition of ICl to 2-methylpropene takes place in the following way and produces 2-chloro-1-iodo-2-methylpropane :

ALIPHATIC

HYDROCARBON


C = CH2 + I – Cl →

H3C

H3C

2-Methylpropene

C – CH2 – I H3C

H3C +

Cl–

δ+ δ–

→ CH3 – C – CH2 – I

CH3

2-Chloro-1-iodo-2-methylpropane

Cl

An Exception to Markovnikov’s Rule : This rule exception concerns the addition of HBr to

alkenes when the addition is carried out in the presence of peroxides (i.e., compounds with the general formula ROOR). When alkenes are treated with HBr in the presence of peroxides, an anti-Markovnikov addition occurs in the sense that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms. With propene, for example, the addition takes place as follows :

CH3CH = CH2 + HBr →ROOR CH3CH2CH2Br This addition occurs by a radical mechanism, and not

by the ionic mechanism. This anti-Markovnikov addition occurs only when HBr is used in the presence of peroxides and does not occur significantly with HF, HCl, and HI even when peroxides are present.

Alcohols from Alkenes through Oxymercuration-Demercuration Markovnikov Addition : A useful laboratory procedure for synthesizing

alcohols from alkenes that avoids rearrangement is a two-step method called oxymercuration -demercuration.

Alkenes react with mercuric acetate in a mixture of tetrahydrofurane (THF) and water to produce (hydroxyalkyl) mercury compounds. These (hydroxyalkyl) mercury compounds can be reduced to alcohols with sodium borohydride :

Step 1 : Oxymercuration

C = C THF + H2O + Hg 23OCCH

||O

– C – C –

HO Hg – OCCH3

+ CH3COH

O

O

Step 2 : Demercuration

– C – C –

HO Hg – OCCH3

+ OH– + NaBH4 O

– C – C – + Hg + CH3CO–

HO H

O

In the first step, oxymercuration, water and mercuric

acetate add to the double bond; in the second step, demercuration, sodium borohydride reduces the acetoxymercury group and replaces it with hydrogen. (The acetate group is often abbreviated – OAc.)

Both steps can be carried out in the same vessel, and both reactions take place very rapidly at room temperature or below. The first step–oxymercuration–usually goes to completion within a period of 20s – 10 min. The second step – demercuration – normally requires less than an hour. The overall reaction gives alcohols in very high yields, usually greater than 90%.

Oxymercuration–demercuration is also highly regioselective. The net orientation of the addition of the elements of water, H – and –OH, is in accordance with Markovnikov’s rule. The H– becomes attached to the carbon atom of the double bond with the greater number of hydrogen atoms :

R– C – C – H

HO H

(1) Hg(OAc)2/THF–H2O C = C

(2) NaBH4, OH–

H H H H

H R

+ HO – H

The following are specific examples :

Pentene1

CHCH)CH(CH 2223−

=)s15(

OHTHF

)OAc(Hg

2

2

− →

CH3(CH2)2

HgOAcOH||CHCH 2−

)h1(OH

NaBH4− →

CH3(CH2)2

OH|CHCH3 + Hg

2-Pentanol (93%)

CH3

Hg(OAc)2

THF-H2O (20 s)

1-Methylcyclopentanol

OH H3C HgOAc

H

NaBH4

OH–

(6 min)

OHH3C

+ Hg

1-Methylcyclopentene

Rearrangements of the carbon skeleton seldom occur

in oxymercuration - demercuration. The


oxymercuration - demercuration of 3, 3-dimethyl-1-butene is a striking example illustrating this feature. It is in direct contrast to the hydration of 3, 3-dimethyl-1-butene.

3

23

3

CH|

CHCHCCH|

CH

=− OH,NaBH)2(

OHTHF/)OAc(Hg)1(

4

22 →−

3, 3-Dimethyl-1-butene

OHCH||CHCH—CCH

|CH

3

33

3

3,3-Dimethyl-2-butanol (94%) Analysis of the mixture of products by gas

chromatography failed to reveal the presence of any 2, 3-dimethyl-2-butanol. The acid-catalyzed hydration of 3, 3-dimethyl-1-butene, by contrast, gives 2, 3-dimethyl-2-butanol as the major product.

A mechanism that accounts for the orientation of addition in the oxymercuration stage, and one that also explains the general lack of accompanying rearrangements. Central to this mechanism is an

electrophilic attack by the mercury species, gOAcH+

, at the less substituted carbon of the double bond (i.e., at the carbon atom that bears the greater number of hydrogen atoms), and the formation of a bridged intermediate.

Hydroboration : Synthesis of Alkylboranes Hydroboration of an alkene is the starting point for a

number of useful synthetic procedures, including the anti-Markovnikov syn hydration procedure. Hydroboration was discovered by Herbert C. Brown, and it can be represented in its simplest terms as follows :

C = C + H — B hydroboration — C — C —

H BAlkene Boron hydride

Alkylborane Hydroboration can be accomplished with diborane

(B2H6), which is a gaseous dimer of borane (BH3), or more conveniently with a reagent prepared by dissolving diborane in THF. When diborane is introduced to THF, it reacts to form a Lewis acid–base complex of borane (the Lewis acid) and THF. The complex is represented as BH3 : THF.

B2H6 + 2 O 2H – B – O

H

H

– +

Diborane THF (tetrahydrofuran)

BH3 : THF

Solutions containing the BH3 : THF complex can be obtained commercially. Hydroboration reactions are usually carried out in ethers : either in diethyl ether (CH3CH2)2O, or in some higher molecular weight ether such as “diglyme” [(CH3OCH2CH2)2O, diethylene glycol dimethyl ether]. Great care must be used in handling diborane and alkylboranes because they ignite spontaneously in air (with a green flame). The solution of BH3: THF must be used in an inert atmosphere (e.g., argon or nitrogen) and with care.

Stereochemistry of Hydroboration :

H B

HH

H — B

+

We can see the results of a syn addition in our examples

involving the hydroboration of 1-methylcyclopentene ring :

Space Quick Facts

1. There has only been one satellite destroyed by a meteor, it was the European Space Agency’s Olympus in 1993.

2. The International Space Station orbits at 248 miles above the Earth.

3. The Earth orbits the Sun at 66,700mph.

4. Venus spins in the opposite direction compared to the Earth and most other planets. This means that the Sun rises in the West and sets in the East.

5. The Moon is moving away from the Earth at about 34cm per year.

6. The Sun, composed mostly of helium and hydrogen, has a surface temperature of 6000 degrees Celsius.

7. A manned rocket reaches the moon in less time than it took a stagecoach to travel the length of England.

8. The nearest known black hole is 1,600 light years (10 quadrillion miles/16 quadrillion kilometers) away.


Inorganic Chemistry

Fundamentals

Oxygen : Oxygen occurs as two non-metallic forms, dioxygen

O2 and ozone O3. Dioxygen O2 is stable as a diatomic molecule, which accounts for it being a gas. The bonding in the O2 molecules is not as simple as it might at first appear. If the molecule had two covalent bonds, then all electron would be paired and the molecule should be diamagnetic.

O + O → O O or O = O Dioxygen is paramaginetic and therefore contains

unpaired electrons. The explanation of this phenomenon was one of the early successes of the molecular orbital theory.

Liquid dioxygen is pale blue in colour, and the solid is also blue. The colour arises from electronic transitions which excite the ground state (a triplet state) to a singlet state. This transition is 'forbidden' in gaseous dioxygen. In liquid or solid dioxygen a single photon may collide with two molecules simultaneously and promote both to excited states, absorbing red – yellow – green light, so O2 appears blue. The origin of the excited singlet states in O2 lies in the arrangement of electrons in the antibonding π*2py and π*2pz molecular orbitals, and is shown below.

Second excited state (electrons have opposite

spins

π*pyπ*pz

singlet

State 1Σg

+ Energy /kJ

157

First excited state (electrons

paired )

singlet 1∆g 92

Ground state (electrons have parallel spins)

triplet 3Σg– 0

Singlet O2 is excited, and is much more reactive than normal ground state triplet dioxygen. Singlet dioxygen can be generated photochemically by irradiating normal dioxygen in the presence of a sensitizer such as fluorescein, methylence blue or some polycyclic hydrocarbons. Singlet dioxygen can also be made chemically :

H2O2 + OCl– →EtOH O2(1∆g) + Η2Ο + Cl–

Singlet dioxygen can add to a diene molecule in the 1, 4 positions, rather like a Diels–Alder reaction. It may add 1, 2 to an alkene which can be cleaved into two carbonyl compounds.

CH – CH

CH2 CH2 + singlet O2 O – O

Singlet dioxygen may be involved in biological

oxidations. Ozone O3 is the triatomic allotrope of oxygen. It is

unstable, and decomposes to O2. The structure of O3 is angular, with an O – O – O bond angle of 116º48´. Both O – O bond lengths are 1.28 Å, which is intermediate between a single bond (1.48 Å in H2O2) and a double bond (1.21 Å in O2). The older valence bond representation as resonance hybrid now seldom used. The structure is described as the central O atom using sp2 hybrid orbitals to bond to the terminal O atoms. The central atom has one lone pair, and the terminal O atoms have two lone pairs. This leaves four electrons for π bonding. The pz atomic orbitals from the three atoms form three delocalized molecular orbitals covering all three atoms. One MO is bonding, one non-bonding, and one antibonding. The four π electron fill the bonding and non-bonding MOs and thus contribute one delocalized π bond to the molecule in addition to the two σ bonds. Thus the bond order is 1.5, and the π system is described as a four-electron three-centre bond.

Trioxides of Sulphur : The only important trioxide in this group, SO3, is

obtained by reaction of sulfur dioxide with molecular oxygen, a reaction that is thermodynamically very favorable but extremely slow in the absence of a catalyst. Platinum sponge, V2O5, and NO serve as catalysts under various conditions. Sulfur trioxide reacts vigorously with water to form sulfuric acid. Commercially, for practical reasons, SO3 is absorbed in concentrated sulfuric acid, to give oleum, which is then diluted. Sulfur trioxide is used as such for preparing sulfonated oils and alkyl arenesulfonate detergents. It is also a powerful but generally indiscriminate oxidizing agent; however, it will selectively oxidize pentachlorotoluene and similar compounds to the alcohol.

OXYGEN & HYDROGEN FAMILY

KEY CONCEPT


The free molecule, in the gas phase, has a planar, triangular structure that may be considered to be a resonance hybrid involving pπ – pπ S – O bonding, with additional π bonding via overlap of filled oxygen pπ orbitals with empty sulfur dπ orbitals, to account for the very short S – O distance of 1.41 Å

S

:O:

O O S

:O:

O S

:O:

O O O In view of this affinity of S in SO3 for electrons, it is

not surprising that SO3 functions as a fairly strong Lewis acid toward the bases that it does not preferentially oxidize. Thus the trioxide gives crystalline complexes with pyridine, trimethylamine, or dioxane, which can be used, like SO3 itself, as sulfonating agents for organic compounds.

The structure of solid SO3 is complex. At least three well-defined phase are known. γ-Sulfur trioxide, formed by condensation of vapors at – 80ºC or below, is an icelike solid containing cyclic trimers with structure.

S S

S O

O

O

O

O

O

O

O

O

A more stable, asbestos-like phase (β-SO3) has

infinite helical chains of linked SO4 tetrahedra and the most stable form, α-SO3, which also has an asbestos-like appearance, presumably has similar chains crosslinked into layers.

– S – O – S – O – S – O –

O

O

O

O

O

O Liquid γ-SO3, which is a monomer-trimer mixture,

can be stabilized by the addition of boric acid. In the pure state it is readily polymerized by traces of water.

Compounds of Sulphur and Nitrogen : A number of ring and chain compounds containing S

and N exist. The elements N and S are diagonally related in the periodic table, and have similar charge densities. Their electronegativities are close (N 3.0, S 2.5) so covalent bonding is expected. The compounds formed have unusual structures which cannot be explained by the usual bonding theories.

Attempting to work out oxidation states is unhelpful or misleading.

The best known is tetrasulphur tetranitride S4N4, and this is starting point for many other S – N compounds. S4N4 may be made as follows :

6SCl2 + 16NH3 → S4N4 + 2S + 14NH4Cl

6S2Cl2 + 16NH3 → 4CCl S4N4 + 8S + 12NH4Cl 6S2Cl2 + 4NH4Cl → S4N4 + 8S + 16HCl

S S

NN NN

S S S4N4 is a solid, m.p. 178ºC, It is thermochromic, that

is it changes colour with temperature. At liquid nitorgen temperatures it is almost colourless, but at room temperature it is orange-yellow, and at 100ºC it is red. It is stable in air, but may detonate with shock, grinding or sudden heating. The structure is a heterocylic ring. This is cradle shaped and differs structurally from the S8 ring, which is crown shaped. The X-ray structure shows that the average S – N bond length is 1.62 Å. Since the sum of the covalent radii for S and N is 1.78 Å, the S – N bonds seem to have some double bond character. The fact that the bonds are of equal length suggest that this is delocalized. The S.....S distances at the top and bottom of the cradle are 2.58 Å. The van der waals (non-bonded) distance S ... S is 3.30 Å, and the single bond distance S – S is 2.08 Å.

This indicates weak S – S bonding, and S4N4 is thus a cage structure.

Many different sizes of rings exist, for example cyclo-S2N2, cyclo-S4N2, cyclo-S4N3Cl, cyclo-S3N3Cl3. In addition bicyclo compounds S11N2, S15N2, S16N2,S17N2 and S19N2 are known. The last four may be regarded as two heterocyclic S7N ring, with the N atoms joined through a chain of 1 – 5S atoms.

S4N4 is very slowly hydrolysed by water, but reacts rapidly with warm NaOH with the break-up of the ring:

S4N4 + 6NaOH + 3H2O → Na2S2O3 + 2Na2SO3 + 4NH3 If S4N4 is treated with Ag2F in CCl4 solution then

S4N4F4 is formed. This has an eight-membered S – N ring, with the F atoms bonded to S. This results from breaking the S – S bonds across the ring. Similarly the formation of adducts such as S4N4.BF3 or S4N4.SbF5 (in which the extra group is bonded to N) breaks the S – S bonds and increases the mean S – N distance from 1.62 Å to 1.68 Å. This is presumably because the electron attracting power of BF3 or SbF5 withdraws some of the π electron density.

Reduction of S4N4 with SNCl2 in MeOH gives tetrasulphur tetraimide S4(NH4). Several imides can be made by reacting S4N4 with S, or S2Cl2 with NH3. These imides are related to an S8 ring in which one or more S atoms have been replaced by imide NH groups, for example in S7NH, S6(NH)2, S5(NH)3 and S4(NH)4.


If S4N4 is vaporized under reduced pressure and passed through silver wool, then disulphur dinitrogen S2N2 is formed.

S4N4 + 4Ag → S2N2 + 2Ag2S + N2 S2N2 is a crystalline solid, which is insoluble in water

but soluble in many organic solvents. It explodes with shock or heat. The structure is cyclic and the four atoms are very nearly square planar.

The most important reaction of S2N2 is the slow polymerization of the solid or vapour to form polythiazyl (SN)x. This is a bronze coloured shiny solid that looks like a metal. It conducts electricity and conductivity increases as the temperature decreases, which is typical of a metal. It becomes a superconductor at 0.26 K. The crystal structure shows that the four-membered rings in S2N2 have opened and polymerized into a long chain polymer. The atoms have a zig-zag arrangement, and the chain is almost flat. Conductivity is much greater along the chains than in other directions, and so the polymer behaves as a one-dimensional metal. The resistivity is quite high at room temperature.

Ortho and Para Hydrogen : The hydrogen molecule H2 exists in two different

forms known as ortho and para hydrogen. The nucleus of an atom has nuclear spin, in a similar way to electrons having a spin. In the H2 molecule, the two nuclei may be spinning in either the same direction, or in opposite directions. This gives rise to spin isomerism, that is two different forms of H2 may exist. These are called ortho and para hydrogen. Spin isomerism is also found in other symmetrical molecules whose nuclei have spin momenta, e.g. D2, N2, F2, Cl2. There are considerable differences between the physical properties (e.g. boiling points, specific heats and thermal conductivities) of the ortho and para forms, because of differences in their internal energy. There are also difference in the band spectra of the ortho and para forms of H2.

The para form has the lower energy, and at absolute zero the gas contains 100% of the para form. As the temperature is raised, some of the para form changes into the ortho form. At high temperatures the gas contains about 75% ortho hydrogen.

Para hydrogen is usually prepared by passing a mixture of the two forms of hydrogen through a tube packed with charcoal cooled to liquid air temperature. Para hydrogen prepared in this way can be kept for weeks at room temperature in a glass vessel, because the ortho-para conversion is slow in the absence of catalysts. Suitable catalysts include activated charcoal, atomic hydrogen, metals such as Fe, Ni, Pt and W and paramagnetic substances or ions (which contain unpaired electrons) such as O2, NO, NO2, Co2+ and Cr2O3.

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.


1. For SO2(g) at 273 K and 1 atm pressure, the dielectric

constant (or relative permittivity) is 1.00993. This molecule has a permanent dipole moment of 1.63 D. Assuming that SO2 behaves as an ideal gas, calculate per mol of (a) total, (b) orientation, (c) induced polarizations, and (d) distortion polarizability.

Sol. We have

εr = 0εε = 1.00993

p = 1.63 D = 1.63(3.3356 × 10–30 Cm) Vm = 22414 cm3 mol–1 at 1 atm and 273 K (a) Total polarization,

Ptotal = 21

r

r

+ε−ε

ρM

= 200993.1100993.1

+− × 22.414 cm3 mol–1

= 73.95 cm3 mol–1 (b) Orientation polarization,

P0 =

ε kT3p

3N 2

0

A

=

××

−−−

−

21212

123

mNC10854.8(3mol10023.6

×××

−−

−

)K273)(KJ1038.1(3)Cm103356.363.1(

123

230

= 59.31 × 10–6 m3 mol–1 = 59.31 cm3 mol–1 (c) Induced polarization, Pind = Ptotal – P0 = 73.95 cm3 mol–1 – 59.31 cm3 mol–1 = 14.64 cm3 mol–1 (d) Distortion polarizability,

αd = A0

ind

N)3/1(Pε

= )mol10023.6)(mNC10854.83/(1

molm1064.1412321212

136–

−−−−

−

××××

= 6.46 × 10–40 C2 N–1 m

2. A green coloured compound (A) gave the following reactions :

(i) (A) dissolves in water to give a green solution. The solution on reaction with AgNO3 gives a white ppt. (B) which dissolves in NH4OH solution and reappears on addition of dil. HNO3. It on heating with

K2Cr2O7 and conc. H2SO4 produced a red gas which dissolves in NaOH to give yellow solution (C). Addition of lead acetate solution to (C) gives a yellow ppt. which is used as a paint.

(ii) The hydroxide of cation of (A) in borax bead test gives brown colour in oxidising flame and grey colour in reducing flame.

(iii) Aqueous solution of (A) gives a black ppt. on passing H2S gas. The black ppt. dissolves in aquaregia and gives back (A).

(iv) (A) on boiling with NaHCO3 and Br2 water gives a black ppt. (D)

(v) (A) on treatment with KCN gives a light green ppt. (E) which dissolves in excess of KCN to give (F). (F) on heating with alkaline bromine water gives the same black ppt. as (D).

Identify compounds (A) to (F) and give balanced equations of the reactions.

Sol. Reaction (i) indicates that (A) contains Cl– ions because, it gives white ppt. soluble in NH4OH. It is again confirmed because it gives chromyl chloride test. The colour of oxidising and reducing flames indicate that (A) also contains Ni2+ ions. Hence, (A) is NiCl2. The different reactions are :

(i) NiCl2 + 2AgNO3 → 2AgCl + Ni(NO3)2 AgCl + 2NH3 →

lelubSo23 Cl])NH(Ag[

Ag(NH3)2Cl + 2HNO3 → )B(.pptwhite

AgCl ↓ + 2NH4NO3

The equations of chromyl chloride tests are : NiCl2 + Na2CO3 → 2NaCl + NiCO3 4NaCl + K2Cr2O7 + 6H2SO4 → 4NaHSO4 + 2KHSO4 + 3H2O +

gasdRe22ClCrO2

CrO2Cl2 + 4NaOH →)C(solutionYellow

42CrONa + 2NaCl + 2H2O

Na2CrO4 + (CH3COO)2Pb → .pptYellow

4PbCrO + 2CH3COONa

(ii) Na2B4O7 . 10H2O ∆ Na2B4O7 + 10H2O

Na2B4O7 ∆ 444 3444 21

beadtTransparen322 OBNaBO2 +

NiO + B2O3 ∆

)Brown(boratemetaNickel

22 )BO(Ni [Oxidising flame]

Ni(BO2)2 + C ∆ GreyNi + B2O3 + CO

[Reducing flame]

UNDERSTANDINGInorganic Chemistry


(iii) NiCl2 + H2S → 2HCl + .pptBlack

NiS ↓

NiS + 2HCl + [O] → )A(

2NiCl + H2S ↑

(iv) )A(

2NiCl + 2NaHCO3 → NiCO3 + 2NaCl

+ CO2 + H2O

2NiCO3 + 4NaOH + [O] ∆

)D(.pptBlack

32ONi ↓

+ 2Na2CO3 + H2O (v)

)A(2NiCl + 2KCN →

)E(.pptGreen2)CN(Ni + 2KCl

Ni(CN)2 + 2KCN → )F(

42 ])CN(Ni[K

NaOH + Br2 → NaOBr + HBr

2K2[Ni(CN)4] + 4NaOH + 9NaOBr ∆

)D(32ONi ↓ + 4KCNO + 9NaBr + 4NaCNO

3. A metal (A) gives the following observations : (i) It gives golden yellow flame. (ii) It is highly reactive and used in photoelectric cells

as well as used in the preparation of Lassaigane solution.

(iii) (A) on fusion with NaN3 and NaNO3 separately, yields an alkaline oxide (B) and an inert gas (C). The gas (C) when mixed with H2 in Haber's process gives another gas (D). (D) turns red litmus blue and gives white dense fumes with HCl.

(iv) Compound (B) react with water forming on alkaline solution (E). (E) is used for the saponification of oils and fats to give glycerol and a hard soap.

(v) (B) on heating at 670 K give (F) and (A). The compound (F) liberates H2O2 on reaction with dil. mineral acids. It is an oxidising agent and oxidises Cr(OH)3 to chromate, manganous salt to manganate, sulphides to sulphates.

(vi) (B) reacts with liquid ammonia to give (G) and (E). (G) is used for the conversion of 1, 2 dihaloalkanes into alkynes.

What are (A) to (G)? Explain the reactions involved. Sol. (i) (A) appears to be Na as it gives the golden yellow

flame. It is also used in the preparation of Lassaigane solution which is sodium extract of organic compounds.

Na + C + N → NaCN Na + Cl → NaCl 2Na + S → Na2S (ii) Compound (B) is Na2O and (C) is N2 while (D) is

NH3, as (D) is alkaline and turns red litmus blue and gives white fumes with HCl

(C) + H2 → NH3 N2 + 3H2

)D(3NH2

NH3 + HCl → NH4Cl White fumes

(iii) is prepared from Na as follows. 2NaNO3 + 10 Na →

)B(2ONa6 +

)C(2N

3NaN3 + NaNO2 → )B(2ONa2 +

)C(2N

(iv) Compound (E) is NaOH as it is used in the preparation of soaps.

)B(2ONa + H2O →

)E(NaOH2

CH2OOCC17H35

CHOOCC17H35 + 3NaOH

CH2OOCC17H35

∆

CH2OH

CH2OH + 3C17H35COONa

CH2OH (soap)

(v) (F) is sodium peroxide as only peroxides gives H2O2 on reaction with dil. acids.

)B(2ONa2

∆ → K670

)F(22ONa +

)A(Na2

)F(

22ONa + .dil

42SOH → H2O2 + Na2SO4

(F) gives the following oxidations : Cr(OH)3 + 5OH– → CrO4

2– + 4H2O + 3e–

Mn2+ + 8OH– → MnO4– + 4H2O + 5e–

S2– + 8OH– → SO42– + 4H2O + 8e–

The reduction equation of (F) is O2

2– + 2H2O + 2e– → 4OH– (vi) (G) is sodamide because it is used in the

dehydrohalogenation reactions.

)B(2ONa + NH3(l) →

)G(2NaNH +

)E(NaOH

CH3 – CH – CH2 + 2NaNH2

Br Br

∆ CH3 – C ≡ CHPropyne

+ 2NaBr + 2NH3 4. A white amorphous powder (A) when heated, gives a

colourless gas (B), which turns lime water milky which dissolves on passing excess of gas (B) and the residue (C) which is yellow while hot but white when cold. The residue (C) dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dil. HCl with the evolution of a gas which is identical in all respects to gas (B). The solution of (A) in dil. HCl gives a white ppt. (D) on addition of excess of NH4OH and on passing H2S gas. Another portion of this solution gives initially a white ppt. (E) on addition of NaOH solution which dissolves in excess of NaOH, the solution on passing again H2S gas gives back the white ppt. of (E), the white ppt. on heating with dil. H2SO4 give a gas used in II and IV group analysis. What are (A) to (F) ? Give balanced chemical equations of the reaction.


Sol. All the above information can be formulated as follows :

(a)

coldin whiteandhot in yellow residue A

milkywater lime turning

gas ColurlesspowderWhite

CBA +→∆

(b) C → HCl.Dil a solution → 64 )CN(FeK a white ppt. (c)

Solution + B

NH4OH + H2S NaOH

(D) White ppt.

(E) White ppt.

NaOH

dissolves → (F) H2S

A dil.HCl

From observations of section (a) one may conclude

that the colourless gas is CO2 because it turns lime water milky, due to formation of insoluble CaCO3.

)B(2

waterLime2 CO)OH(Ca + →

milky3CaCO ↓ + H2O

CaCO3 is soluble in excess of CO2 due to formation of soluble calcium bicarbonate.

CaCO3 + CO2 + H2O → lelubSo

23 )HCO(Ca

The compound (C) is zinc oxide (ZnO) because it is yellow in hot and white in cold, hence the initial compound (A) is zinc carbonate (ZnCO3).

From section (b), it is inferred that (C) is a salt of Zn (II) which dissolves in dil. HCl and white ppt. obtained after addition of K4Fe(CN)6 is due to zinc ferrocyanide, a test of Zn++ cation.

The data of collection (C) proved that (A) is ZnCO3 because on treatment with dil. HCl it gives gas (B), i.e., CO2, while Zn (II) goes in solution, i.e., ZnCl2, on passing H2S gas in presence of NH4OH, it gives a white ppt. of ZnS(D). ZnS on heating with dil. H2SO4 evolves H2S, which is used for the precipitation of sulphides of group II in acidic medium and of IV group in alkaline medium. ZnCl2, reacts with NaOH to give a ppt. of Zn(OH)2 which dissolves in NaOH, as Zn(OH)2 is amphoteric in nature. The solution Na2ZnO2 again gives ZnS on passing H2S gas into it. Different chemical equations concerned to the above numerical are given below :

)B(2

)C()A(3 COZnOZnCO +→∆

)C(

ZnO + .dil

HCl2 → Solution

2ZnCl + H2O

2ZnCl2 + K4Fe(CN)6 → .pptWhite

62 ])CN(Fe[Zn ↓ + 4KCl

)A(

3ZnCO + .dil

HCl2 →∆ ZnCl2 + )B(2CO + H2O

ZnCl2 + H2S → OHNH4 )D(

.pptWhiteZnS + 2HCl

ZnS + .dil

42SOH →∆ ZnSO4 + H2S ↑

ZnCl2 + 2NaOH → )E(

2)OH(Zn ↓ + 2NaCl

)E(Amphoteric

2)OH(Zn + 2NaOH → lelubSo

22ZnONa + 2H2O

Na2ZnO2 + Gas

2SH → )F(

ZnS ↓ + 2NaOH

5. 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:

)X(

3BX4 + 3LiAlH4 → )Y(

62HB2 + 3LiX + 3AlX3

(X = Cl or Br)

)Y(

62HB + 3O2 → B2O3 + 3H2O + heat

Structure of B2H6 is as follows:

B

Ht

BHt

Ht

Ht

Hb

Hb or

B

Ht

Ht Hb

Hb

B

Ht

Ht

121.5º1.33Å

97º

1.19Å1.77Å

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– bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively.


1. Show that the conics through the intersection of two rectangular hyperbolas are also rectangular hyperbolas. If A, B, C & D be the four points of intersection of these two rectangular hyperbolas, then find the orthocentre of the triangle ABC.

2. Find the area of a right angle triangle if it is known that the radius of circle inscribed in the triangle is r and that of the circumscribed circle is R.

3. Q is any point on the line x = a. If A is the point (a, 0) and QR, the bisector of the angle OQA, meets OX in R, then prove that the locus of the foot of the perpendicular from R to OQ has the equation

(x – 2a) (x2 + y2) + a2x = 0

4. Show that the equation z4 + 2z3 + 3z2 + 4z + 5 = 0 with (z ∈ C) have no

purely real as well as purely imaginary root.

5. Prove that

x

xnax

xaf

l

+∫

∞

0

dx = lnax

dxax

xaf

+∫

∞

0

6. A straight line moves so that the product of the perpendiculars on it form two fixed points is a constant. Prove that the locus of the foot of the perpendiculars from each of these points upon the straight line is a circle, the same for each.

7. Prove the identity :

∫ −x

zzxe0

2dz = ∫ −

xz

x

ee0

4/42

2

dz, deriving for the

function f (x) = ∫ −x

zzxe0

2dz a differential equation and

solving it. 8. Let α, β be the roots of a quadratic equation, such

that αβ = 4 and 1−α

α + 1−β

β = 47

2

2

−

−

aa Find the set

of values of a for which α, β ∈ (1, 4)

9. Investigate the function f (x) = x5/3 – 5x2/3 for points of extremum and find the values of k such that the equation x5/3 – 5x2/3 = k has exactly one positive root.

10. Let A = 1, 2, 3, ....., 100. If X is a subset of A containing exactly 50 elements then show that

∑∈xp

pmin = 101C51.

`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

5Set

Do you know

• The largest meteorite crater in the world is in Winslow, Arizona. It is 4,150 feet across and 150 feet deep.

• The human eye blinks an average of 4,200,000 times a year.

• Skylab, the first American space station, fell to the earth in thousands of pieces in 1979. Thankfully most over the ocean.

• It takes approximately 12 hours for food to entirely digest.

• Human jaw muscles can generate a force of 200 pounds (90.8 kilograms) on the molars.

• The Skylab astronauts grew 1.5 - 2.25 inches (3.8 - 5.7 centimeters) due to spinal lengthening and straightening as a result of zero gravity.

• An inch (2.5 centimeters) of rain water is equivalent to 15 inches (38.1 centimeters) of dry, powdery snow.

• Tremendous erosion at the base of Niagara Falls (USA) undermines the shale cliffs and as a result the falls have receded approximately 7 miles over the last 10,000 years.

• 40 to 50 percent of body heat can be lost through the head (no hat) as a result of its extensive circulatory network.


1. Let x2 + y2 = a2 ....(1) and x2 + y2 + 2gx + c = 0 ....(2) are two circles they cut orthogonally. Hence c – a2 = 0 so from (2) x2 + y2 + 2gx + a2 = 0 ...(3) Let P (a cos α, a sin α) be any point on 1st circle. It’s

polar w.r.t. 2nd circle is ax cos α + ay sin α + g(x + a cos α) + a2 = 0 ....(4) other end of diameter of 1st circle through P is Q(– a cos α, – a sin α) This satisfies eqn.(4) Hence proved. 2. Let A be at origin & position vectors of B, C, D are ,

br

, cr

& dr

respectively. Perpendicular from B and C to the faces ACD and ABD meet at H with position vector h

r,

A

B

C

D

dc b

so BH ⊥ AC ⇒ cbh

rrr).( − = 0 ⇒ ch

rr. = cb

rr.

BH ⊥ AD ⇒ dbhrrr

).( − = 0 ⇒ dhrr

. = dbrr

.

CH ⊥ AB ⇒ bchrrr

).( − = 0 ⇒ bhrr

. = bcrr

.

and CH ⊥ AD ⇒ d).ch(rrr

− = 0 ⇒ d.hrr

= bcrr

.

so dbrr

. = dcrr

. ⇒ )( cbrr

− . dr

= 0 ...(1) so BC ⊥ AD , proved. Now, let any point M (with position vector m ) be

on BCD such that AM ⊥ plane BCD then . m . )( bc

rr− = 0 = m . )( cd

rr−

so m . br

= m . cr

= m . dr

...(2) Now, let P be any point on AM with position vector

t m such that DP is perpendicular to ABC, then (t m – dr

). br

= 0 = (t m – dr

). cr

so t m . br

= br

. dr

& t m . cr

= d

r. cr

which are same equations using (1) and (2) in

them. So such a scalar t can be obtained as t = bm

dbr

rr

.

.

3. I (u) = ∫π

+−0

2 )cos21( dxuxunl ;

I (– u) = ∫π

++0

2 )cos21( dxuxunl

Use ∫a

dxxf0

)( = ∫ −a

dxxaf0

)(

I (u) = I (– u) I (u) + I (– u)

= ∫π

+++−0

22 )cos21()cos21( dxuxuuxunl

= ∫π

−−0

2222 ]cos4)1[( dxxuunl

= ]cos421[ 2224

0

xuuun −++∫π

l dx

= ]2cos21[ 42

0

uxun +−∫π

l dx

Now let 2x = y

⇒ Ι (u) + I (– u) = ]cos21[21 42

2

0

uyun +−∫π

l dy

=21 I (u2) + ]cos21[

21 42

2

0

uyun +−∫π

l dy

Now let y = 2π – t

I (u) + I (– u) =21 I (u2) + ]cos21[

21 42

0

utun +−∫π

l (– dt)

=21 I (u2) +

21 I (u2)

2I (u) = I (u2) as I(u) = I (– u) (or using f (2a – x) = f (x) Prop).

so I (u) =21 I(u2)

similarly find ]cos21[3 22

0

uxun +−∫π

l dx

& show I (u) = I (– u) = 21 I (u2) = )(

2

1 2 nuI

n

MATHEMATICAL CHALLENGES SOLUTION FOR AUGUST ISSUE (SET # 4)


4. Let f (x) = (x – α) (x – β) so f (n) f (n + 1) = (n – α) (n –β) (n + 1 – α)(n + 1– β) = (n – α) (n + 1 – β) (n – β) (n + 1 – α) = [n (n + 1) – n(α + β) – α + αβ] [n(n + 1) – n (α + β) – β + αβ] = [n (n + 1) + na + b – α] [n (n + 1) + an + b – β] = (m – α) (m – β) ; let m = n(n + 1) + an + b = f (m) 5. Let xy = c2 be rectangular hyperbola. Let A (ct1, c/t1) and b (ct2, c/t2) be two fixed points on

it. and P (ct, c/t) be any variable point. Line AP : x + y t1t = c (t1 + t) Line BP : x + y t2 t = c (t2 + t) These lines intersect with x- axis at M (c(t1 + t), 0)

and N(c(t2 + t), 0). Length MN = |c (t1 – t2) | which is a constant.

Similarly intercept on y-axis can be obtained as

−

21

11tt

c . Hence proved.

6. xy

dyxdxy − + 2

22

)( yxdxydyx

−

− = 0

⇒ 2)(

yxdyxdxyy −

+

2

22

)( yxxydyxydydxydxxydxxydyx

−

+−−−+

⇒ xy . d(x/y) +

2)()()()(

yxdydxxydyxydxydxydyxx

−

−−+−+= 0

⇒ y/x

)y/x(d + 2)()()()(.

yxyxdxyxydyxydx

−

−−− = 0

⇒ yxyxd

/)/(

+ 2)()()()(

yxyxdxyxydyx

−

−−−= 0

⇒ yxyxd

/)/(

+ d

− yxyx = 0

⇒ ln (x/y) + yx

yx−

= c

7. Let 2

2

ax + 2

2

by = 1 be the ellipse and

y = m1 (x – ae) and y = m2 (x – ae) are two chords through its focus (ae, 0). Any conic through the extremities of these chords can be defined as y – m1 (x – ae) y – m2 (x – ae) +

λ

−+ 12

2

2

2

by

ax = 0 ...(1)

It it passes through origin, then m1m2 a2 e2 – λ = 0 ...(2) Solving (1) with x- axis

m1m2 (x – a e)2 + λ

−12

2

ax = 0

using (2) in it (x – ae)2 + e2 (x2 – a2 ) = 0 (1 + e2) x2 – 2ae x = 0

x = 0 & x = 212

eae

+

so other point on x-axis through which this conic

passes is

+

0,1

22e

ae which is a fixed point.

Hence proved.

8. Let x = c ∈ R

f ′ (c ±) = 0

lim→h h

cfhcf±

−± )()(

= 0

lim→h h

cfchcf

±

−

± )(1

; c ≠ 0

= 0

lim→h h

cfchfcf

±

−+ )()/1(2

)2(

= 0

lim→h h

cfchfcf

2

)(21)2(

±

−

±

= 0

lim→h h

fcfchfcf

2

)1()2(1)2(

±

−

±

;

(using x = 2c & y = 1 in

2xyf =

2)().( yfxf

)

= f (2c) 0

lim→h c

ch

fchf

.2

)1(1

±

−

±

= c

fcf2

)1(')2(

= c

fcf2

)1(')2(; as given f ′(1) = f (1)

f ′ (c ±) = ccfx

2)(

So f (x) is differentiable for ∀ x ∈ R except x = 0

Now f ′(x) = xxf )(

⇒ )()('

xfxf =

x1

so ln f(x) = ln x + ln c ⇒ f (x) = cx

Now as f

2xy =

2)()( yfxf


let y = 1 in it 2. f (x/2) = f (x) (f (1)

22xc =cx f(1)

so f (1) = 1 as c ≠ 0 so f (x) = x 9. Let i lines are there, no two of which are parallel and

no three of which are coincident. Introduction of (i + 1)th line will introduce (i + 1) new parts. Let Pi denotes the number of parts in which plane is being divided by i lines, then

Pi + 1 = Pi + (i + 1)

Pi + 1 – Pi = i + 1 using i = 1, 2, 3, ......, n – 1 P2 – P1 = 2, P1 – P2 = 3 M M Pn – Pn–1 = n Add these equation Pn – Pi = 2 + 3 +.....+ n Pn = Pi + 2 + 3 + ....+ n

= 2 + 2 + 3+ .....+ n = 1 + 2

)1( +nn

Pn = 21 (n2 + n + 2)

10. 21 tan x +

xx3cos

sin + xx

9cos3sin +

xx

27cos9sin =

21 tan 27x

L.H.S. : Consider on

21 tan x +

xx3cos

sin = 21

xx

cossin +

xx3cos

sin

= xx

xxxx3coscos2

cossin23cossin +

= xx

xxx3coscos4

2sin23cossin2 +

= xx

xxx3coscos4

2sin22sin4sin +−

= xxxx

3coscos42sin4sin +

= xxxx

3coscos4cos3sin2 =

21 tan 3x

similarly 21 tan 3x +

xx

9cos3sin =

21 tan 9x

and 21 tan 9x +

xx

27cos9sin =

21 tan 27

on adding all these we get

xx

3cos3sin +

xx

9cos3sin +

xx

27cos9sin =

21 (tan 27x – tan x)

Proved.

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1. The sum of the digits of a seven-digit number is 59.

Find the probability that this number is divisible by 11.

Sol. As 7 × 8 = 56 and the sum of seven digits is 59, clearly at least three of the digits must be 9.

Obviously, the seven digits of the number will be as follows;

(a) 9,9,9,8,8,8,8 (b) 9,9,9,9,8,8,7 (c) 9,9,9,9,9,7,7 (d) 9,9,9,9,9,8,6 (e) 9,9,9,9,9,9,5 ∴ the total number of ways to form a seven-digit

number whose sum of digits is 59

= !4!3

!7 + !2!4

!7 + !2!5

!7 + !5!7 +

!6!7

= 2.35.6.7 +

25.6.7 +

26.7 + 7.6 + 7

= 210 ...(i) A number is divisible by 11 if the difference of the

sum of the digits in odd places and that of the digits in even places is divisible by 11.

As the number is of seven digits we must have (for favourable cases), sum of four digits in odd places –sum of three digits in even places

= 0, 11, 22, 33, 44, 55. If the two sums are denoted by x and y respectively

then x + y = 59 x + y = 59 x + y = 59 x – y = 0 or x – y = 11 or x – y = 22 (i) (ii) (iii) x + y = 59 x + y = 59 x + y = 59 or x – y = 33 or x – y = 44 or x – y = 5 5 (iv) (v) (vi) Clearly, (i), (iii) and (iv) do not give integral values

of x and y. (ii) ⇒ x = 35, y = 24; (iv) ⇒ x = 46, y = 13; (vi) ⇒ x = 57, y = 2. Obviously, from (a), (b), (c), (d) and (e) we get, sum

of three digits y cannot be 2 or 13. Hence, only favourable case takes place when the

sum of four digits in the odd places = 35 and the sum of the three digits in even places = 24.

∴ in the favourable numbers we will get, 9,9,9,8 in odd places and 8,8,8 in even places or 9,9,9,8 in odd places and 9,8,7 in even places or 9,9,9,8 in odd places and 9,9,6 in even places ∴ the number of numbers divisible by 11

= !3!4 +

!3!4 × 3 +

!3!4 ×

!2!3

= 4 + 24 + 12 = 40 ...(ii) ∴ from (i) and (ii)

the required probability = 21040 =

214 .

2. The medians of a triangle ABC make angles α,β,γ

with each other. Prove that cot α + cot β + cot γ + cot A + cot B + cot C = 0. Sol. Here, G is the centroid of the ∆ABC and ∠BGC =

α, etc. We know from geometry, AB2 + AC2 = 2(AD2 + BD2), etc.,

and AG = 23 AD, BG =

32 BE, CG =

32 CF.

A

B C

E F

D

G

α

Now, BC2 + BA2 = 2 (BE2 + CE2)

or a2 + c2 = 2

+

22

223 bBG

= 21 (9BG2 + b2)

∴ 9BG2 = 2(a2 + c2) – b2

or BG2 = 91 (2a2 + 2c2 – b2).

Similarly, CG2 = 91 (2a2 + 2b2 – c2)

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

Students' ForumMATHS


Now, cos α = .CG2BG

BC–CGBG 222 +

= CG.2BG

–)–22(91)–22(

91 2222222 acbabca +++

= 91 .

α

+

sin.BG.CG21.2

5– 222 acb . 21 sin α

= 91 .

41 .

BGC)ar(sin)5–( 222

∆α+ acb

= 361 .

ABC)ar(31

sin)5–( 222

∆

α+ acb

Q ar(∆ABC) = 3 . ar(∆BGC)

∴ αα

sincos =

∆+

125– 222 acb

∴ cot α = ∆

+12

5– 222 acb

Similarly, cot β = ∆

+12

5– 222 bac ,

cot γ = ∆

+12

5– 222 cba

∴ cot α + cot β + cot γ = – ∆

++4

222 cba .

Again cot A + cot B + cot C = AA

sincos +

BB

sincos +

CC

sincos

= bc

acb2

– 222 + .aR2 +

cabac

2– 222 + .

bR2

+ ab

cba2

– 222 + . cR2

= abcR

22 (b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2)

= abcR (a2 + b2 + c2)

= ∆41 (a2 + b2 + c2)

∆=

4abcRQ

∴ cot α + cot β + cot γ + cot A + cot B + cot C

= – ∆

++4

222 cba + ∆

++4

222 cba = 0.

3. A ray of light is coming along the line y = b, (b > 0) from the positive direction of the x-axis and strikes a concave mirror whose intersection with the x-y plane is the parabola y2 = 4ax, (a > 0). Find the equation of the reflected ray and show that it passes through the focus of the parabola.

Sol. Let P be the point of incidence Then P is the intersection of the line y = b and the

parabola y2 = 4ax.

Y

Q y2 = 4ax

P

θ

θ y = b

X

T

∴ P =

b

ab ,4

2

∴ the equation of the tangent PT at P is

y . b =

+

abxa4

22

or by = 2ax + 2

2b ...(i)

'm' of (i) is ba2 . So, tan θ =

ba2

Let the slope of the reflected ray PQ be m.

∴ tan θ =

bam

bam

2.1

2–

+

or ba2 =

bam

bam

2.1

2–

+

∴

bam

bam

2.1

2–

+ = ±

ba2

or m – ba2 = ±

+ 2

24.2bam

ba

∴ m – ba2 =

ba2 + m . 2

24ba

and m – ba2 = –

ba2 – m . 2

24ba

∴

2

24–1bam =

ba4 and

+ 2

241bam = 0.


But m ≠ 0;

∴

2

24–1bam =

ba4

∴ m = 22 4–4

abab

∴ the equation of the reflected ray PQ is

y – b =

abx

abab

4–

4–4 2

22

or (b2 – 4a2)y – b(b2 – 4a2) = 4abx – b3 or (b2 – 4a2)y – 4abx + 4a2b = 0. This will pass through the focus (a, 0) if (b2 – 4a2)0 – 4ab . a + 4a2b = 0, which is true ∴ the reflected ray passes through the focus.

4. If a, b ∈ R be fixed positive numbers such that f (a + x)

= b + [b3 + 1 – 3b2. f (x) + 3b f (x)2 – f (x)3]1/3 for all x ∈ R then prove that f (x) is a periodic

function. Sol. Here, f (a + x) – b3 = b3 + 1 – 3b2f (x) + 3bf (x)2 – f (x)3 = 1 – [f (x)3 – 3b . f (x)2 + 3b2 . f (x) – b3] = 1 – f (x) – b3 ∴ f (a + x) – b3 + f (x) – b3 = 1 ...(i) This is true for all x. Putting a + x for x in (i), we get, f (2a + x) – b3 + f (a + x) – b]3 = 1 ...(ii) Subtracting (i) from (ii), f (2a + x) – b3 – f (x) – b3 = 0 or f (2a + x) – b3 = f (x) – b3 or f (2a + x) – b = f (x) – b or f (2a + x) = f (x) ∴ f (x) is a periodic function

5. Let PM be the perpendicular from the point P(1,2,3) to the x-y plane. If OP makes an angle θ with the positive direction of the z-axis and OM makes an angle φ with the positive direction of the x-axis, where O is the origin, then find θ and φ.

Z

X M

P(1,2,3)

Y O φ

θ r

Sol. We know that if P = (x, y, z) then x = r sin θ . cos φ, y = r sin θ. sin φ, z = r cos θ.

∴ 1 = r sin θ . cos φ, 2 = r sin θ . sin φ, 3 = r cos θ. ⇒ 12 + 22 = r2sin2θ . cos2φ + r2sin2θ . sin2φ = r2sin2θ(cos2φ + sin2φ) = r2sin2θ.

∴ 5 = r2sin2θ. ∴ r sinθ = 5 (clearly, θ and φ are acute).

∴ θθ

cossin

rr =

35 or tan θ =

35 .

∴ θ =

35tan 1– .

Also, 21 =

φθφθ

sin.sincos.sin

rr

∴ cot φ = 21 . ∴ tan φ = 2 ∴ φ = tan–1 2.

6. Take two sets of m and n parallel lines, lines of one

set cutting those of the other. Thus m × n small rhombuses are formed. Two rhombuses are called neighbouring rhombuses if they have a common side. In each of these rhombuses a natural number is put such that the number written in a rhombus is the AM of the numbers written in all its neighbouring members. Show that all the numbers put in the rhombuses are equal.

Sol. There are two possibilities : (i) all the numbers are equal (ii) at least two numbers are different If all the numbers are equal, there is nothing to

prove. If at least two numbers are different, there will be

the smallest number. Select the rhombus with the smallest number p in it. As p is the AM of the numbers in the neighbouring rhombuses, p must be greater than at least one of the numbers in the neighbouring rhombuses.

(Q AM of two numbers a, b is 2

ba + which is

bigger than one but smaller than the other; AM of

three numbers a, b, c is 3

cba ++ which cannot be

smaller than a, b as well as c, etc.). In that case p is not the least among the numbers put in all the rhombuses.

Thus the possibility (ii) is absurd. Hence, all the numbers are equal.


Some Definitions : Experiment : A operation which can produce some

well defined outcomes is known as an experiment. Random experiment : If in each trail of an

experiment conducted under identical conditions, the outcome is not unique, then such an experiment is called a random experiment.

Sample space : The set of all possible outcomes in an experiment is called a sample space. For example, in a throw of dice, the sample space is 1, 2, 3, 4, 5, 6. Each element of a sample space is called a sample point.

Event : An event is a subset of a sample space. Simple event : An event containing only a single

sample point is called an elementary or simple event. Events other than elementary are called composite or compound or mixed events.

For example, in a single toss of coin, the event of getting a head is a simple event.

Here S = H, T and E = H In a simultaneous toss of two coins, the event of

getting at least one head is a compound event. Here S = HH, HT, TH, TT and E = HH, HT, TH Equally likely events : The given events are said to

be equally likely, if none of them is expected to occur in preference to the other.

Mutually exclusive events : If two or more events have no point in common, the events are said to be mutually exclusive. Thus E1 and E2 are mutually exclusive in E1 ∩ E2 = φ.

The events which are not mutually exclusive are known as compatible events.

Exhaustive events : A set of events is said to be totally exhaustive (simply exhaustive), if no event out side this set occurs and at least one of these event must happen as a result of an experiment.

Independent and dependent events : If there are events in which the occurrence of one does not depend upon the occurrence of the other, such events are known as independent events. On the other hand, if occurrence of one depend upon other, such events are known as dependent events.

Probability : In a random experiment, let S be the sample space

and E ⊆ S, then E is an event. The probability of occurrence of event E is defined as

P(E) = Sin element distinct ofnumber Ein elementsdistinct ofnumber =

n(S)n(E)

= outcomes possible all ofnumber

E of occurrence tofavourable outocomes ofnumber

Notations : Let A and B be two events, then A ∪ B or A + B stands for the occurrence of at

least one of A and B. A ∩ B or AB stands for the simultaneous

occurrence of A and B. A´ ∩ B´ stands for the non-occurrence of both A

and B. A ⊆ B stands for "the occurrence of A implies

occurrence of B". Random variable : A random variable is a real valued function whose

domain is the sample space of a random experiment. Bay’s rule : Let (Hj) be mutually exclusive events such that

P(Hj) > 0 for j = 1, 2, ..... n and S = Un

1jjH

=. Let A be

an events with P(A) > 0, then for j = 1, 2, .... , n

P

AH j =

∑=

n

1kkk

jj

)H/A(P)H(P

)H/A(P)H(P

Binomial Distribution : If the probability of happening of an event in a single

trial of an experiment be p, then the probability of happening of that event r times in n trials will be nCr pr (1 – p)n – r.

Some important results :

(A) P(A) = cases ofnumber Total

Aevent tofavourable cases ofNumber

= )S()A(

nn

PROBABILITY

Mathematics Fundamentals MA

TH

S


)AP( = cases ofnumber Total

Aevent tofavourablenot cases ofNumber

= )S()A(

nn

(B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E.

Thus odds in favour of an event E

= cases leunfavourab ofNumber

cases favourable ofNumber = ba

Similarly, odds against an event E

= cases favorable ofNumber

cases leunfavourab ofNumber = ab

Note : If odds in favour of an event are a : b, then the

probability of the occurrence of that event is

baa+

and the probability of non-occurrence of

that event is ba

a+

.

If odds against an event are a : b, then the probability of the occurrence of that event is

bab+

and the probability of non-occurrence of

that event is ba

a+

.

(C) P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1 P(φ) = 0 P(S) = 1 If S = A1, A2, ..... An, then P(A1) + P(A2) + .... + P(An) = 1 If the probability of happening of an event in one

trial be p, then the probability of successive happening of that event in r trials is pr.

(D) If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or

P(A + B) = P(A) + P(B) If A and B are any two events, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B)

If the probabilities of happening of n independent events be p1, p2, ...... , pn respectively, then

(i) Probability of happening none of them = (1 – p1) (1 – p2) ........ (1 – pn) (ii) Probability of happening at least one of them = 1 – (1 – p1) (1 – p2) ....... (1 – pn) (iii) Probability of happening of first event and not

happening of the remaining = p1(1 – p2) (1 – p3) ....... (1 – pn) If A and B are any two events, then

P(A ∩ B) = P(A) . P

AB or

P(AB) = P(A) . P

AB

Where P

AB is known as conditional probability

means probability of B when A has occured. Difference between mutually exclusiveness and

independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments.

(E) P(A A ) = 0

P(AB) + P( AB ) = 1

P( A B) = P(B) – P(AB)

P(A B ) = P(A) – P(AB)

P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards : Coins : A coin has a head side and a tail side. If

an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated.

Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated.

Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards.

In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack.


Binomial Theorem (For a positive Integral Index) : If n is a positive integer and x, a are two real or

complex quantities, then (x + a)n = nC0 xn + nC1xn – 1 a + nC2 xn – 2 a2 + ... + nCrxn–r ar + .... + nCn – 1x an–1 + nCnan ..(1) The coefficient nC0, nC1, ......, nCn are called binomial

coefficients. Properties of Binomial Expansion : There are (n + 1) terms in the expansion of

(x + a)n, n being a positive integer. In any term of expansion (1), the sum of the

exponents of x and a is always constant = n. The binomial coefficients of term equidistant

from the beginning and the end are equal, i.e. nCr = nCn – r (0 ≤ r ≤ n).

The general term of the expansion is (r + 1)th term usually denoted by Tr + 1 = nCr xn – r ar (0 ≤ r ≤ n).

The middle term in the expansion of (x + a)n (a) If n is even then there is just one middle term, i.e.

thn

+1

2 term.

(b) if n is odd, then there are two middle terms, i.e. thn

+1

2 term and

thn

+

23 term.

The greatest term in the expansion of (x + a)n, x, a ∈ R and x, a > 0 can be obtained as below :

∴ r

r

TT 1+ =

xa

rrn 1+−

or r

r

TT 1+ – 1 =

rxxaran )()1( +−+

=

−

+++ r

xaan

rxxa )1()( =

rxxa + |k – r|,

where k = xaan

++ )1(

Now, suppose that

(i) k = xaan

++ )1( is an integer. We have

Tr + 1 > Tr ⇔ r

r

TT 1+ > 1 ⇔ r < k (i.e. 1 ≤ r < k)

Along , Tr + 1 = Tr ⇔ r

r

TT 1+ = 1 ⇔ r = k,

i.e. Tk + 1 = Tk > Tk–1 > .... > T3 > T2 > T1 In this case there are two greatest terms Tk and

Tk+1.

(ii) k = xaan

++ )1( is not an integer. Let [k] be the

greatest integer in k. We have

Tr+1 > Tr ⇔ r

r

TT 1+ ⇔ r < k = [k] + (fraction)

⇔ r ≤ [k] i.e. T1 < T2 < T3 < ..... < T[k] – 1 < T[k] < T[k] + 1 In this case there is exactly one greatest term viz.

([k] + 1)th term. Term independent of x in the expansion of

(x + a)n – Let Tr + 1 be the term independent of x. Equate to zero the index of x and you will find the value of r.

The number of term in the expansion of

(x + y + z)n is 2

)2)(1( ++ nn , where n is a positive

integer. Pascal Triangle In(x + a)n when expanded the various coefficients

which occur are nC0, nC1, nC2, .... The Pascal triangle gives the values of these coefficients for n = 0, 1, 2, 3, 4, 5, ....

n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 n = 6 1 6 15 20 15 6 1 n = 7 1 7 21 35 35 21 7 1

n = 8 1 8 28 56 70 56 28 8 1

BINOMIAL THEOREM Mathematics Fundamentals M

AT

HS


Rule : It is to be noted that the first and least terms in each row is 1. The terms equidistant from the beginning and end are equal. Any number in any row is obtained by adding the two numbers in the preceding row which are just at the left and just at the right of the given number, e.g. the number 21 in the row for n = 7 is the sum of 6 (left) and 15 (right) which occur in the preceding row for n = 6.

Important Cases of Binomial Expansion : If we put x = 1 in (1), we get (1 + a)n = nC0 + nC1a + nC2a2 + .................. + nCrar + ........... nCnan ...(2) If we put x = 1 and replace a by – a, we get (1 – a)n = nC0 – nC1a + nC2a2 – .................. + (–1)r nCrar + .... + (–1)n nCnan ...(3) Adding and subtracting (2) and (3), and then

dividing by 2, we get

21 (1 + a)n + (1 – a)n = nC0 + nC2a2

+ nC4a4 + .... ...(4)

21 (1 + a)n – (1 – a)n = nC1a + nC3a3

+ nC5a5 + ...... ...(5) Properties of Binomial Coefficients : If we put a = 1 in (2) and (3), we get 2n = nC0 + nC2 + ...... + nCr + ....+ nCr + ..... nCn–1 + nCn and 0 = nC0 – nC1 + nC2 – ...... + ...... + (–1)n nCn

∴ nC0 + nC2 + .... = nC1 + nC3 + .... = 21 [2n ± 0]

= 2n – 1 ...(6) Due to convenience usually written as C0 + C2 + C4 + ... = C1 + C3 + C5 + .... = 2n – 1 and C0 + C1 + C2 + C3 + ...... + Cn = 2n

Where nCr ≡ Cr = !)(!

!rnr

n−

= !

)1)...(2)(1(r

rnnnn +−−−

Some other properties to remember : C1 + 2C2 + 3C3 + ... + nCn = n . 2n – 1 C1 – 2C2 + 3C3 – .... = 0 C0 + 2C1 + 3C2 + ... + (n + 1)Cn = (n + 2) 2n –1

C0Cr + C1Cr+1 + ... + Cn – rCn = !)(.!)(

!)2(rnrn

n+−

C02 + C1

2 + C22 + .... + Cn

2 = 2)!(!)2(

nn

C02 – C1

2 + C22 – C3

2 + ...

=

evenisnif,C.(–1)odd isn if,0

2/nnn/2

Binomial Theorem for Any Index : The binomial theorem for any index states that

(1 + x)n = 1 + !1

nx + !2

)1( −nn x2

+ !3

)2)(1( −− nnn x3 +..... ...(7)

Where n is any index (positive or negative) The general term in expansion (7) is

Tr + 1 = !

)1)......(1(r

rnnn +−− xr

In this expansion there are infinitely many terms. This expansion is valid for |x| < 1 and first term

unity. When x is small compared with 1, we see that the

terms finally get smaller and smaller. If x is very small compared with 1, we take 1 as a first approximation to the value of (1 + x)n or 1 + nx as a second approximation.

Replacing n by – n in the above expansion, we get

(1 + x)–n = 1 – nx + !2

)1( +nn x2 – !3

)2)(1( ++ nnn x3

+ ... + (–1)r

!)1)...(2)(1(

rrnnnn −+++ xr +

... Replacing x by – x in this expansion, we get

(1 – x)–n = 1 + nx + !2

)1( +nn x2 + !3

)2)(1( ++ nnn x3 +

....... + !

)1)...(2)(1(r

rnnnn −+++ xr + ....

Important expansions for n = –1, –2 are : (1 + x)–1 = 1 – x + x2 – x3 + ...+ (–1)rxr + .... to ∞ (1 – x)–1 = 1 + x + x2 + x3 + ..... + xr + .... to ∞ (1 + x)–2 = 1 – 2x + 3x2 – .... + (–1)r(r + 1)xr +…. (1 – x)–2 = 1 + 2x + 3x2 + ..... + (r + 1)xr + .... to ∞ (1 + x)–3 = 1 – 3x + 6x2 – 10x3 + ...

+ (–1)r

!2)2)(1( ++ rr xr + .....

(1 – x)–3 = 1 + 3x + 6x2 + 10x3 + ...

...... + !2

)2)(1( ++ rr xr + ....


PHYSICS

Questions 1 to 8 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. A block of mass M is kept in gravity free space and

touches the two springs as shown in the figure. Initially springs are in its natural length. Now, the block is shifted by small displacement 'x' from the given position in such a way it compresses a spring and released. The time-period of oscillation of mass will be –

l0 2l0

4 K K M

(A) KM

2π (B)

K5M2π

(C) KM

23π (D)

K2M

π

2. With what minimum speed should m be projected from point C in presence of two fixed masses M each at A and B as shown in the figure such that mass m should escape the gravitational attraction of A and B ?

30º

R

Cm

Vmin

M

R RAM

B

(A) R

GM2 (B) RGM22

(C) R

GM2 (D) R

GM22

3. If the optic axis of convex and concave lenses are

separated by a distance 5 mm as shown in the figure. Find, the co-ordinate of the final image formed by the combination if parallel beam of light is incident on lens, origin is at the optical center of convex lens.

IIT-JEE 2012

XtraEdge Test Series # 5

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity, Refl. At Plane surface, Ref. at Curved surface, Refraction at Plane surface, Prism (Deviation & Dispersion), Refraction at Curved surface, Wave Nature of Light: Interference. Chemistry : Gaseous state, Chemical Energetics, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis, Reaction Mechanism, Alkane, Alkene, Alkyne, Alcohol, Ether & Phenol, Practical Organic Chemistry, Aromatic Hydrocarbons, Halogen Derivatives, Carboxylic Acid & Its Derivatives, Nitrogen Compounds, Amines, Carbohydrates, Amino Acid, Protein & Polymers. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number, Indefinite Integration, Definite Integration, Area Under the Curve, Differential Equations.

Instructions : [Each subject contain] Section – I : Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be

awarded for correct answer and -1 mark for wrong answer. Section – II : Question 9 to 12 are multiple choice question with multiple correct answer. +4 marks will be awarded for

correct answer and -1 mark for wrong answer.

Section – III : Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer

Section – III : Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.


5mm

f = 20 cm f = –10

Principle axis of concave lensPrinciple axis of convex lens

30 cm

(0, 0)

(A) (25 cm, 0.5 cm) (B) (25 cm, 0.25 cm) (C) (25 cm, 0.5 cm) (D) (25 cm, –0.25 cm)

4. A non viscous and incompressible liquid is flowing through a section of tube as shown. Area of cross-section at A is 3 cm2 whereas it is 1 cm2 at B. Velocity of liquid at B is 6 m/s. Find the force exerted by liquid on curved surface of conical section. Given the density of liquid is 103 kg/m2, pressure at A is PA = 18 × 103 M/m2 and pressure at B is PB.

3 cm2

PA

A

B

PB

6 m/s1 cm2

(A) 2.8 N (B) 2.4 N (C) 6.4 N (D) none of the above

5. Let S1 and S2 be the two slits in Young's double slit experiment. If central maxima is observed at P and angle S1 P S2 = θ, (θ is small) find the y-coordinates of the 3rd minima assuming the origin at the central maxima. (λ = wavelength of monochromatic light used)

(A) ± θλ2 (B)

θλ

±25 (C) λθ±

25 (D) ± 2 λθ

6. The mirror of length 2l makes 10 revolutions per minute about the axis crossing its mid point O and perpendicular to the plane of the figure There is a light source in point A and an observer at point B of the circle of radius R drawn around centre O

(∠AOB = 90º). What is the proportion l

R if the

observer B first sees the light source when the angle of mirror ψ = 15º ?

A

R Bl

l

O

ψ

(A) 2 (B) 2

1 (C) 22 (D) 22

1

7. A capillary tube is made of glass with the index of refraction 3, outer radius of the tube is 30 cm. The tube is filled with a liquid with the index of refraction 2. What should be the minimum internal radius of the tube so that any ray that hits the tube would enter the liquid -

(A) 15 cm (B) 10 cm (C) 20 cm (D) 45 cm

8. A ray of light when incident upon a prism suffers a minimum deviation of 39°. If the shaded half portion of the prism is removed, then the same ray will -

(A) suffer a deviation of 19.5° (B) suffer a deviation of 39°

(C) not suffer any deviation (D) will be totally internally reflected

Questions 9 to 12 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 – 1 mark for each wrong answer.

9. A smooth inclined plane has an angle of inclination θ from the horizontal. From point A on the inclined plane, body 1 is projected horizontally with speed u and from point B on the inclined plane, body 2 is released from rest. Distance AB is equal to l. If the magnitude of relative velocity of body 2 w.r.t. body 1 and magnitude of relative acceleration of body 2 w.r.t body 1 at time t are v and ‘a’, respectively, then which of the following is/are true ?

A

θ

Bl 2

u y

x

1

(A) v = 2222 tcosgtcossinug2u θ+θθ−

(B) a = zero

(C) a = g cos θ

(D) v = 222 )gttcossing()utsing( −θθ+−θ


10. A particle is performing linear SHM. When its speed is v1, then its acceleration is a1. At another instant, its speed is v2 and acceleration is a2. Find the amplitude and angular frequency of SHM.

(A) ω = 22

21

22

21

vvaa

−

− (B) ω = 22

21

21

22

vvaa

−

−

(C) A = )vava()aa()vv( 2

221

21

2222

122

22

21 −−

−

(D) A = )vava()aa()vv( 2

221

21

2222

122

21

22 −−

−

11. If a force field →E is defined as

→E )1r(r

rk

1n >−=→

+,

then - (A) the change in potential energy corresponding to

this force field can be written as,

U(r2) – U(r1) =

−

−−

−− 1n1

1n2 r

1r

1)1n(

k

(B) the potential energy corresponding to this force field can be written as,

U(r) = 1nr)1n(k

−−− , with U(∞) = 0

(C) the force attracts the particle towards the

origin 0r =→

(D) the force repels the particle away from the

origin 0r =→

12. Coefficient of friction between the two block is 0.3 whereas the surface AB is smooth. (g = 10m/s2) :

10kg

T2

T1

3kg

2kg

B A

(A) Acceleration of 10 kg mass is 5.86 m/s2 (B) Acceleration of 10 kg mass is 7.55 m/s2 (C) Tension T1 in the string is 17.7 N (D) Tension T2 in the string is about 41.4 N This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15) The following figure shows a horizontal pipe of

varying cross-sectional area. The water flows in the pipe and flow remains laminar for whole the length. The pipe is completely smooth, i.e. no pressure energy losses occur in the pipe.

Here small tubes are connected at each cross-section to measure the pressure at each cross- section. Neglect the capillary rise. Also, the difference in heights of different cross-section are small, so we may assume that they are at the same height.

(The flow in pipe is assumed incompressible, non-viscous and irrotational.)

1 2 3A1 = 0.4 m2 v2 = 5 m/s ρ = 1000 kg/m3

A2 = 0.1 m2 A3 =1 m2

13. What is the mass flow rate in the pipe ? (A) 2 kg/s (B) 20 kg/s (C) 200 kg/s (D) 2000 kg/s 14. What is the difference in kinetic energy per unit

volume between sections 2 and 3, i.e. K2 – K3 ? (A) 98 kJ (B) 198 kJ (C) 396 kJ (D) 1.98 ×105 kJ 15. What is the difference in water levels of small tubes

connected at sections 2 and 3 ? (A) 9.8 m (B) 19.8 m (C) 20 m (D) 39.6 m Passage # 2 (Ques. 16 to 18)

The diagram shows an equilateral prism. The medium on one side of the prism is µ1. The

refractive index of the prism is µ = 3

4 . The

diagram shows variation of magnitude of angle of deviation with respect to µ1. Consider the light ray to be normally incident on the first face.

β1

β2

0 1 k1 k2

β

µ1 (Ang

le o

f dev

iatio

n)


16. Value of k2 is –

(A) 3

8 (B) 23 (C)

34 (D)

36

17. Value of k1 is –

(A) 2 (B) 38 (C)

35 (D)

34

18. Value of β1 – β2 is – (A) 60° (B) 30° (C) 20° (D) 90° This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

19. A particle moving on the smooth horizontal table strikes the rod AB kept freely at an end A perpendicularly. Match the following –

Column-I Column-II

(A) Angular momentum of system (particle + rod) about any point is

(P) Conserved for elastic collision

(B) Linear momentum of system (particle + rod) is

(Q) Not conserved for elastic collision

(C) Angular momentum of rod about its COM is

(R) Conserved for perfect inelastic collision

(D) Kinetic energy of system (particle + rod) except during collision is

(S) Not conserved for perfect inelastic collision

(T) Conserve in any type of collision

20. Match the column : Column-I Column-II

(P) Converging (A)

µ RR

(Q) Concave-convex(B) µR

(R) Convex-concave (C) µR 2R

(S) Diverging (D) µ2R R

(T) None

CHEMISTRY

Questions 1 to 8 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. Select the correct statement - (A) When attractive forces are dominant in a real

gas U internal energy increases by increasing the average molecular separation

(B) Isothermal contraction decreases the internal energy when repulsive forces between molecules exist

(C) For an ideal gas at constant T internal energy is function of volume only.

(D) In case of attractive forces between particle, if the volume of the sample expands isothermally its dU< 0

2. A piece of plumber’s solder (containing Pb & Sn only)

weighing 3 gm was dissolved in dil HNO3 and then treated with dilute H2SO4 — the ppt. of PbSO4 is produced , which after washing and drying weighed 2.93 gm. The remaining solution was neutralized – a white ppt of stannic acid is produced , which on heating yields 1.27 gm of SnO2. Therefore mass % of Pb and Sn in the plumber’s solder is

At wt. Pb = 207 ; Sn = 118.7


(A) % Pb = 66.67 % , % Sn = 33.33 % (B) % Pb = 33.33 % , Sn = 66.67 % (C) % Pb = 50% , % Sn = 50 % (D) % Pb = 75 % , % Sn = 25 % 3. The major product of following reaction is:

O

CF3CO3H

(A)

O

O

(B)

O

O

(C)

O

(D) None of above

4. In the reaction MnO4

– + SO32– + H+ → SO4

2– + Mn2+ + H2O

(A) MnO4– and H+ both are reduced

(B) MnO4– is reduced and H+ is oxidised

(C) MnO4– is reduced and SO3

2– is oxidised

(D) MnO4– is oxidised and SO3

2– is reduced

5.

D D

→ /NiH2 A. A is -

(A) CH3 – (CH2)4 – CH3 (B) D D

H H

(C) D H

D H

(D)

H H

6. Which of the following alcohols would you expect

to form a corbocation most readily in H2SO4 -

(A)

OH

(B)

OH

(C)

OH

O

(D)

OH

7. 15 ml of gaseous hydrocarbon required for

complete combustion 357 ml of air containing O2, 21 % by volume and the gaseous products occupied 327 ml. If all volumes are measured at STP, the molecular formula of the hydrocarbon is -

(A) C3H6 (B) C4H10 (C) C4H8 (D) C3H8 8. Which of the following is nylon 6, 6 -

(A)

N

O 4

H

N

H

(B)

O O

N 4H

N

H

(C)

O O

N

H

N

H

(D)

O

4N H

C


9. The graph represent Boyle's law is/are -

(A)

P1

V1

(B)

PV

P

(C) logP

logV

(D)

V

1/P


10. 0.5 M KI solution reacts with excess of H2SO4 and

KIO3 solution according to the equation - 6H+ + −I5 + 3IO− –→ 3I2 + 3H2O which of the following statements is /are true - (A) 200 ml of KI solution reacts with 0.10 mole

KIO3 (B) 0.5 litre of KI solution produces 0.15mole of I2 (C) Volume of 0.5M KIO3 used to react completely

with 1 litre KI solution is 200 ml (D) 100 ml of the KI solution reacts with 0.06 M

H2SO4 solution 11. When nitrobenzene is treated with Br2 in presence

of FeBr3 the major product formed is m-bromonitrobenzene. Statements which are related obtain the m-isomer are -

(A) The electron density on meta carbon is more than on ortho and para position.

(B) The intermediate carbonium ion formed after initial attack of Br⊕ attack the meta position is least destabilized

(C) Loss of aromaticity when Br+ attcks at the ortho and para positions and not at meta position.

(D) Easier loss of H+ to regain aromaticity form the meta position than from ortho and para positions. [A, B]

12. Reaction (1)

Cl

Cl

Cl

Cl

Cl

Cl

∆ → KOH.alcmol3 [A]

Reaction (2)

Cl

Cl

Cl

Cl

Cl

Cl → KOH.alcmol3 [B]

Reaction (3)

Cl

Cl

Cl

Cl

Cl

Cl → KOH.alcmol3 [C]

The correct statement is –

(A) A and B are same (B) B and C are different

(C) B and C are same (D) A and B are different This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15) In the following reaction sequence, products I, J

and L are formed. K represents a reagent.

Hex-3-ynal 3

4

PBr.2

NaBH.1 → I + →

OH.3

CO.2ether/Mg.1

3

2 J

→K Me

O quinolineBaSO/Pd

H

4

2→ L

13. The structure of the product I is-

(A) Me Br (B) Me

Br

(C) Me Br (D) Me Br

14. The structures of compounds J and K, respectively, are-

(A) Me COOH and SOCl2

(B) MeOH

O and SOCl2

(C) MeCOOH

and SOCl2

(D) Me COOH and CH3SO2Cl

15. The structure of product L is- (A) Me CHO

(B) Me CHO

(C) MeCHO

(D) Me CHO Passage # 2 (Ques. 16 to 18) When (C–H) σ electrons are in conjugation to pi

bond, this conjugation is known as hyperconjugation.For any compound to show hyper conjugation

(i) Compound should have one sp2-hybridised carbon

(ii) α-carbon with respect to sp2 should be sp3


(iii) α-carbon should contain at least one hydrogen atom No. of α-carbon ∝ stability of cation and alkene

16. Which of the following cations is hyperconjugation destabilized ?

(A) CH3– 2HCΘ

(B) CH3–CH–CH3 Θ

(C) CH3–C⊕

CH3

CH3 (D) ⊕

17. Which of the following alkyl benzene has

maximum electron density ?

(A) CH3

(B) CH2CH3

(C) H3C–CH–CH3

(D)

H3C–C–CH3

CH3

18. Which of the following orders is correct for magnitude

of +M power among these compounds ?

O 'X'

O

'Y'

N

'Z' CH3

(A) Z > Y > X (B) Y > X > Z (C) X > Y > Z (D) Z >X > Y

This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer

(i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

19. Match the column : Column-I Column-II (A) Hydrogen gas (p) Compressibility factor ¹ 1 (P = 200 atm And T = 273 K) (B) Hydrogen gas (Q) attractive forces are (P ~ O and T = 273 K) dominant (C) CO2 (P = 1atm (R) PV = nRT and T = 273 K) (D) Real gas with (S) P ( V – nb) = nRT very large Molar volume (T) P (V + nb) = nRT 20. Given below are the reactions of oxidising agents

with excess KI and the liberated iodine is titrated against the standard reducing agent, sodiumthio sulphate-

Column-I Column-II (A) K2Cr2O7 + KI + H2SO4 (P) 20 milli moles of (20ml, 0.2 M) thiosulphate → Cr2(SO4)3 + K2SO4 + H2O + I2 (B) CuSO4 + KI (Q) 24 millimoles of (20 ml, 1.2M) thiosulphate → Cu2I2 + K2SO4 + I2 (C) KMnO4 + KI + H2SO4→ (R) 20 milli equivalents (20 ml, 0.2 M) K2SO4 + MnSO4 + H2O + I2 (D) H2O2 + KI + H2SO4 → (S) 24 milli

equivalents (20 ml, 0.2 M) K2SO4 + H2O + I2 (T) 36 milli

MATHEMATICS

Questions 1 to 8 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. The number of odd consecutive integers whose sum is 452 – 212 is :

(A) 24 (B) 23 (C) 25 (D) None


2. The remainder when 599 is divided by 13 is : (A) 6 (B) 8 (C) 9 (D) 10

3.

∫

∫

π

π

π→

π

2

2 4/

2/

cos–

2/

)2/–(

)1–2(

limx

xt

x

dtt

dt

=

(A) π

2loge (B) π 22 ln

(C) π

2 2ln (D) None

4. A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varieties of perfumes in the showcase is :

(A) 6 (B) 50 (C) 150 (D) None

5. Let f (a) denotes the area of the region bounded by y = a2x2 + ax + 1, co-ordinate axes and the line x = 1. Then f (a) will be least at a =

(A) – 3/4 (B) – 1/4 (C) – 1 (D) None

6. ∫+

)–1()cos1(

sin22 xexxxx dx =

(A) log |xesinx | – 21 log |1 – x2e2sinx| + c

(B) log |xesinx | + 21 log |1 – x2e2sinx| + c

(C) log |xesinx | – log |1 – x2e2sinx| + c (D) None of these

7. The solution of the equation

dxdy + x(x + y) = x(x + y)3 – 1 is ln|f (x)| = x2 + c,

where f (x) =

(A) 3)(

)1–)(1(yx

yxyx+

+++ (B) 2)()1–)(1(

yxyxyx

++++

(C) 2)()1––)(1(

yxyxyx

+++ (D) None of these

8. For a > 0, ≠ 1 the roots of the equation logax a + logx a2 +

xa2log a3 = 0 are given by :

(A) a–3/4 (B) a–4/3, a1/2 (C) a–1/2, a–4/3 (D) None of these


9. Let I = ∫ π+

1

02/1

1x

dx , then

(A) I < π/4 (B) I > π/4 (C) I < loge 2 (D) I < loge 2

10. Let f (x) = 2–

3x

+ 3–

4x

+ 4–

5x

. Then f (x) = 0

has (A) two roots in (2, 4) (B) only one root in (2, 3) (C) only one root in (3, 4) (D) no root in (2, 4)

11. Let z1, z2, z3 be three complex numbers such that |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 = 0. If z1, z2, z3 denote the vertices of a ∆ABC, then :

(A) origin is orthocentre

(B) arg 13

23

––

zzzz = arg

1

2

zz

(C) 23

22

21 zzz ++ = z1z2 + z2z3 + z3z1

(D) area of ∆ABC = 43

12. A point P(x, y) moves in such a way that [|x|] + [|y|] = 1, [.] = G.I.F. Area of the region representing all possible positions of the point P is equal to :

(A) 8 (B) 4 (C) 16 (D) None This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15)

If the length of an arc of a curve measured from a fixed point A(x = a) to another point P(x, y) on the curve be s, then we have the relation

dxds =

2

1

+

dxdy


∴ arc AP = s = ∫

+

x

adxdy 2

1 dx.

P(x,y)

A(x = a)

s

Using the above information, answer the following

questions : 13. Length of the arc of the parabola y2 = 4x cut off by

its latus-rectum is :

(A) 2[ 2 + loge( 2 – 1)]

(B) 2[ 2 + ln( 2 + 1)]

(C) [ 2 + ln( 2 +1)]

(D) [ 2 + ln( 2 –1)] 14. Whole length of the astroid x2/3 + y2/3 = 4 is : (A) 12 (B) 48 (C) 6 (D) none of these 15. Length of the curve y = loge (sec x) between x = 0

to x = π/3 is :

(A) ln(2 + 3 ) (B) ln(1 + 3 )

(C) 21 ln 3 (D) ln(2 – 3 )

Passage # 2 (Ques. 16 to 18) Multinomial theorem states that for n ∈ N and

a1, a2,...........,ap ∈ C, then (a1 + a2 + ...+ ap)n

= ∑ !!.......!!

21 pnnnn pn

pnn aaa .....21

21

where n1, n2,...,np are non-negative integers such that n1 + n2 + .... + np = n

Answer the following questions : 16. If x1, x2,.....xn are independent, then number of

terms in the expansion of (x1 + x2 + .... + xn)5 is : (A) n+1C5 (B) n+4C5 (C) n+5C4 (D) n+4C4 17. Coefficient of x3y3 z in the expansion of (1 + x – y + z)9

is : (A) 2520 (B) 2800 (C) – 2800 (D) – 1400 18. Coefficient of a3b4c5 in the expansion of (ab + bc +

ca)6 is : (A) 60 (B) 40 (C) 0 (D) None


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the column : Column-I Column-II (A) The largest value of n such (P) 2 that 2010n divides 2010!

(B) N = ∑=

6

0

12C)1(–r

rr (Q) 3

(C) The number of positive (R) 5 integer triplets (a,b,c) such that a + b + c ≤ 0 (D) A set contains 6 elements. (S) 7 The number of ordered pairs (A,B) such that A ∩ B = φ (T) 11

20. Match the column : Column-I Column-II (A) Area of the region defined (P) 0 by the inequality

|y| + 21 ≤ ||–1 x is p/q,

then p – q = (B) Sum of the values of α for (Q) 5/6 which the system of inequalities x2 + 2x + α ≤ 0; x2 – 4x – 6α ≤ 0 has unique solution is

(C) If In = ∫ xnx

sincos dx, then (R) 1

I7 – I5 = p

pxcos2 , where p =

(D) The number of 5 digit (S) 4/5 numbers made up of the digits 3, 4, 5, 6,7 without repetitions which are multiple of 11 is ab. Then b – a = (T) 1/6


PHYSICS

Questions 1 to 8 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. A child is standing at one end of a long trolley

moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the centre of mass of the system (trolley + child) will move with a speed -

(A) zero (B) (v + u) (C) (v – u) (D) v

2. A particle strikes a horizontal frictionless floor with a speed u, at an angle θ with the vertical, and rebounds with a speed v, at an angle φ with the vertical. The coefficient of restitution between the particle and the floor is e. The angle φ is equal to –

u θ φ v

(A) θ (B) tan–1[e tan θ]

(C)

θ− tan1tan 1

e (D) (1 + e)θ

3. Two blocks A and B of mass m and 2m are

connected by a massless spring of force constant k. They are placed on a smooth horizontal plane. Spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is –

A B

(A) xmk

23 (B) x

m3k2

(C) mkx2 (D)

xkm2

3

4. Three identical blocks A, B and C are placed on

horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m/s. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is –

A B C

(A) 5.6 m/s (B) 6 m/s (C) 8 m/s (D) 10 m/s

Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity. Chemistry : Gaseous state, Chemical Energetic, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis. Mathematics : Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number. Instructions : [Each subject contain] Section – I : Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded

for correct answer and -1 mark for wrong answer. Section – II : Question 9 to 12 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct

answer and -1 mark for wrong answer.

Section – III : Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer

Section – III : Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

IIT-JEE 2013

XtraEdge Test Series # 5

Based on New Pattern


5. Two bodies of mass m1 = 3 kg and m2 = 2 kg move along mutually perpendicular directions with velocities 4 m/s and 3 m/s respectively as shown in the figure. As a result of collision the bodies stick together. The amount of heat liberated is

m1 4 m/s

3 m/s

m2 (A) 10 J (B) 15 J (C) 20 J (D) 25 J

6. A thin rod of mass m and length l is hinged at one end point which is at a distance h (h < l) above the horizontal surface. The rod is released from rest from the horizontal position. If e is the co-efficient of restitution, the angular velocity of rod just after collision will be (h = 1m, l = 2m, e = 1) –

h

l hinged

(A) 8

g33 (B)

8g36

(C) 8

g35 (D) none of these

7. A body of cross-sectional area A, height H and density ρS, is immersed to depth h in a liquid of density (ρL) and tied to bottom with string. The tension in string is –

String

h

Area = A

H

(A) ρgha (B) (ρS – ρL) ghA (C) (ρL – ρS) ghA (D) (ρLh – ρSH) gA

8. If two wires of same length l and area of cross section A with Young modulus Y and 2Y connect in series and one end is fixed on roof and other end with mass m. Make simple harmonic motion, then the time period is -

(A) YAm2 l

π (B) YA3m2 l

π

(C) YA2m32 l

π (D) YA2m2 l

π


9. A pendulum suspended from the roof of an elevator at rest has a time period T1; when the elevator moves up with an acceleration a its time period becomes T2; when the elevator moves down with an acceleration a; its time period becomes T3 then –

(A) T3 > T2 and T1 (B) T2 > T3 > T1

(C) T1 = 23

22

32

TT

2TT

+ (D) T1 = 2

322 TT +

10. Two concentric spherical shells have masses m1 and m2 and radii r1 and r2. Then–

m1 m3 P

O r

r2

r1

m2

(A) Outer shell will have no contribution in

gravitational field at point (B) Force on P is directed towards O

(C) Force on P is 221

rmGm

(D) Force on P is 231

rmGm

11. A body of mass m was hauled up the hill with

constant speed v by a force F which at each point was directed along tangent to the path. If length of base is l and height of hill is h, then which of the following are correct ?

(A) Work done by gravity is – mgh (B) Work done by friction is – µmgl (C) Work done by gravity is path independent (D) Work done by friction is path independent 12. The potential energy U (in joule) of a particle of

mass 1 kg moving in x-y plane obeys the law U = 3x + 4y, where (x, y) are the co-ordinates of the particle in metre. If the particle is at rest at (6, 4) at time t = 0, then -

(A) the particle has constant acceleration (B) the particle has zero acceleration

(C) the speed of particle when it crosses the y-axis is 10 m/s

(D) coordinates of the particle at t = 1sec are (4.5 , 2)


This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15) ABCD is a vertical frame of mass M. Two strings

between block of mass m and frame are ideal and lies in vertical plane. The frame ABCD is placed on horizontal smooth surface and a horizontal force F is applied as shown. Points P and Q are in same vertical line and two strings are of equal length

m θ

F

P

Q 13. Minimum force F required to make both string tight

is - (A) Mg tanθ (B) Mg cot θ (C) (m + M)g tanθ (D) (m + M)g cot θ

14. If the lower string is just tight then tension in the upper string is -

(A) θsin

mg (B) θcos

mg

(C) mg sinθ (D) mg cosθ

15. For what value of force F tension in upper string is twice than tension in lower string.

(A) 3(m + M)g tanθ (B) 2(m + M)g tanθ (C) 4(m + M)g tanθ (D) None of these

Passage # 2 (Ques. 16 to 18) Two rods 1 and 2 are released from rest as shown in

figure. Given : l1 = 4l, m1 = 2m, l2 = 2l and m2 = m. There is no friction between the two rods. If α be the angular acceleration of rod 1 just after the rods are released. Then :

B

2

1

l

A

16. What is the normal reaction between the two rods at

this instant ?

(A) 16 3 mlα (B) 3

m4 αl

(C) 33

m32 αl (D) 12 3 mlα

17. What is the horizontal force on rod 1 by hinge A at this instant ?

(A)

−

3331232 mlα (B)

−

33216 mlα

(C) (14 + 2 3 ) mlα (D) 3 mlα

18. What is initial angular acceleration of rod 2 in terms of the given parameters in the question ?

(A)

α+ 32

2g32

l (B)

α− 3g33

l

(C)

α+ 35

8g36

l (D)

α−

38

8g33

l


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. A solid sphere is rotating about an axis as shown in figure. An insect follows the dotted path on the circumference of sphere as shown. Match the following:

Column-I Column-II (A) Moment of inertia (P) will remain cons. (B) Angular velocity (Q) will first increase then decrease (C) Angular momentum (R) will first decrease then increase (D) Rotational kinetic (S) will continuously energy decrease (T) will continuously increase

Insect


20. The cubical container filled with water is given

acceleration →a = a0 i + a0 j + a0 k , then : (neglect

the effect of gravity)

B

C

E F

D

G

A z

x

y

H

Column-I Column-II (A) Pressure at point A is (P) less than pressure at point G (B) Pressure at point D is (Q) less than pressure at point F (C) Pressure at point E is (R) Greater than pressure at point C (D) Pressure at point H is (S) Greater than pressure at point B (T) None

CHEMISTRY

Questions 1 to 8 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. For which of the following change ∆H ≠ ∆E ? (A) H2(g) + I2(g) → 2HI(g) (B) HCl(aq) + NaOH(aq) → NaCl(s) + H2O(l) (C) C(s) + O2(g) → CO2(g) (D) N2(g) + 3H2(g) → 2NH3(g) 2. The value of ∆H° for the hypothetical reaction,

21)(2)(2 +→ sAsBA )(2 gB is, + 7.3 kJ

The value of ∆U° for the reaction is - (A) > 7.3 kJ (B) < 7.3 kJ (C) = 7.3 kJ (D) unpredictable 3. 2.8 L of N2 gas at 300 K and 20 atm was allowed to

expand isothermally against the external pressure of 1 atm. The value of ∆U of the process is -

(a) 10 J (B) 0 (C) –10 J (D) 4.18 J

4. What is the volume of O2(g) required at STP for the oxidation of 1L of SO2(g) at 298K and 1 atm ? 2 SO2(g) + O2(g) → 2 SO3(g)

(A) equal to 0.5 L (B) greater than 0.5 L (C) lesser than 0.5 L (D) equal to 1L

5. For a given one mole of ideal gas kept at 6.5 atm in a container of capacity 2.463 L. The Avogadro proportionality constant for the hypothesis is (see figure)

V

n

(A) 0.406 (B) 2.46 (C) 22.4 (D) none of these 6. At low pressure if RT = p.a2 , (a is vander Waal's

constant) then the volume occupied by a real gas is-

(A) PRT2 (B)

RTP2

(C) P2

RT (D) P

RT2

7. According to classical concept, oxidation involves - (A) Addition of oxygen (B) Addition of electronegative radical (C) Removal of either hydrogen or someelectropositive radical (D) All of these

8. According to modern concept, oxidation is - (A) Electronation (B) Deelectronation (C) Addition of oxygen (D) Addition of electronegative element


9. An adiabatic expansion is one in which -

(A) all energy is transferred as heat

(B) no energy is transferred as heat

(C) the temperature of a gas decreases in a reversible adiabatic expansion

(D) du ≠ dw


10. Select the correct statement -

(A) Heat capacity of diatomic gas is higher than that of monoatomic gas

(B) Standard molar enthalpy of formation of CO2 is equal to standard molar enthalpy of combustion of carbon (graphite)

(C) Thermodynamics, a process is called reversible when surrounding and system change into each other

(D) Change in state is completely defined when the initial and final states are specified

11. Which of the following is/are correct regarding

indicators ? (A) pH range of phenolphthalein is 8.3 – 10 (B) pH range of methyl orange is 3.2 – 4.4 (C) By the phenolphthalein equivalence point of

Na2CO3 is indicated (D) For the suitable indicator neutral point of

indicator must coinside with the equivalence point of the titration

12. In which of the following species the oxidation

state of Cr is equal to +6? (A) CrO5 (B) K2CrO4 (C) CrO3 (D) Cr2O3 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15)

For the substance X(g), molar specific heat at constant temperature, Cp, increases linearly with the absolute temperature, T.

Informations regarding X(s) Molar specific heat = 6.4 JK–1 mol–1

The latent heat of fusion of X (s) = 40 KJ/mol. at the melting point. The melting point of X (s) = 100 K Information regarding X(l) Molar specific heat = 4.5 JK–1 mol–1

Boiling point of X (l) = 250 K The latent heat of vapourisation of X(l) = 60 KJ/mol at the boiling point. Information regarding X(g) For X(g) at 1 atm, when T = 200 K, Cp = 0.04 JK–1 mol–1

when T = 300 K, Cp = 0.05 JK–1mol–1

13. What is the molar enthalpy of sublimation of X(s) - (A) 100 kJ/mol (B) 20 kJ/mol (C) 50 kJ/mol (D) – 20 kJ/mol

14. What is the molar enthalpy of vapourisation of X(l) at 400 K ?

(A) Slightly lesser than 60 kJ (B) Double the enthalpy of vapourisation at 250 K (C) Slightly higher than 60 kJ (D) Equal to 60 kJ because enthalpy is a state

function

15. In which of the following phase Transition change of entropy is maximum ?

(A) Heating of solid (B) Fusion of X (s) (C) Vapourisation of X(l) (D) Heating of X (g)

Passage # 2 (Ques. 16 to 18) For a fixed mass of gas under isobaric (constant

pressure) condition, the volume varies with centigrade temperature as follows

V = V0 (1 + α t) V0 represent the volume at 0ºC at constant pressure 16. For every 1ºC rise in temperature volume of the gas

as per charles law -

(A) increases 2731 times of its volume at final

temperature

(B) increases 2731 times of its volume at 0ºC

(C) increases 273 times of its volume at 0ºC

(D) decreases 2731 times of its volume at 0ºC

17. How does pressure vary with the centrigrade temperature (t) for the definite mass of gas under isochoric (constant volume) condition ?

P0 is the pressure of same mass of gas at 0ºC

(A) P ∝ t (B) P = P0

+

t2731

(C) P = P0

+

273t1 (D) P = P0 ×

273t

18. Under the isochoric condition for the definite mass

of gas pressure varies with absolute temperature (T) as follows :

P

V1 V2

V3

T Which of the following is/are correct ? (A) V1 = V2 = V3 (B) V1 > V2 > V3 (C) V1 < V2 < V3 (D) V2 < V1 < V3



A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the column : Column-I Column-II (A) Tc/Pc (P) Z (B) Tc/Vc (Q) a/Rb (C) TB (R) 8b/R (D) Ti/TB (S) 8a/81 Rb2 (T) a/5Rb

20. Match the column : Column- I Column-II (A) Permanent hardness (P) Na+ (aq) (B) Temporary hardness (Q) Ca2+ (aq)

(C) React on heating (R) Mg2+ (aq)

(D) Not responsible for hardness

(S) −3HCO (aq)

(T) Al3+

MATHEMATICS

Questions 1 to 8 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. Solution set of the equation 2|x| – |2x–1 –1| = 2x–1 + 1

is : (A) [1, ∞) (B) (– ∞, – 1] ∪ [1, ∞) (C) [1, ∞) ∪–1 (D) None

2. Let f (x) = ax2 + bx + c, a, b, c ∈ R, a ≠ 0. If f (1) + f (2) = 0, the quadratic equation f (x) = 0 has:

(A) no real root (B) 1 and 2 as real roots (C) two equal roots (D) two distinct real roots

3. If (2n + r)r, n ∈ N, r ∈ N is expressed as the sum of k consecutive odd natural numbers, then k is equal to :

(A) r (B) n (C) r + 1 (D) n + 1

4. Sum of common roots of the equations z3 + 2z2 + 2z + 1 = 0 and z100 + z32 + 1 = 0 is equal to :

(A) 0 (B) – 1 (C) 1 (D) None

5. If x + y = 1, then ∑=

n

r

rnrr

n yxCr0

– equals :

(A) nx (B) ny (C) n (D) None

6. The tens digit of 1! + 2! + 3! + .... + 29! is : (A) 1 (B) 2 (C) 3 (D) 4

7. Number of solutions of the equation

)1–2( 225 +xx + )1–2( 2

9 +xx = 2–215.34 xx

(A) 2 (B) 3 (C) 4 (D) None

8. The equation |x – x2 – 1| = |2x – 3 – x2| has : (A) infinite solution (B) one solution (C) two solutions (D) no solution


9. Solution set of inequation log2log1/2 (2x – 15/16) ≤ 2 is :

(A) (0, log(31/16)) (B) [1, 31/16) (C) [0, log(31/16)) (D) [0, log(31/16)]

10. Let Sn = C0 – 31C +

52C – ......+ (– 1)n

12C

+nn ,

Cr = nCr. Then :

(A) Sn = !)12(

)!(2 22

+nnn

(B) Sn = !)1–2(

)!(2 22

nnn

(C) Sn = 12

2+nn .Sn–1 (D) Sn =

1–22n

n Sn–1

11. In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A.P., then the length of the third side can be :

(A) 5 – 6 (B) 33 (C) 5 (D) 5 + 6

12. If α,β,γ are roots of x3 + ax + 3x – 1 = 0 (α ≤ β ≤ γ) which are in H.P. Then :

(A) one root must be 1 (B) all roots must be equal (C) a must be a negative integer (D) α,β,γ also in A.P.


This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) 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. 13 to 15) Two consecutive numbers from 1,2,3,...n are

removed. A.M. of the remaining numbers is 105/4. Answer the following questions :

13. n = (A) 7 (B) 8 (C) 50 (D) 51 14. G.M. of removed numbers is :

(A) 42 (B) 56 (C) 72 (D) none 15. If roots of the equation x2 + kx + 1 = 0 lies between

removed numbers, then set of values of k is :

(A)

∞

750–,– (B)

865,–

750–

(C)

2,–

757– (D) no such k exists

Passage # 2 (Ques. 16 to 18) Binomial expansion is a powerful tool to solve

divisibility problems, For example, (1 + x)n – 1 – nx = (1 + nx + nC2x2 + ...) – 1 – nx = (nC2 + nC3x+...)x2 ⇒ (1 + x)n – 1 – nx is divisible by x2 for all n ≥ 2.

Answer the following questions : 16. The last three digits of 31000 must be : (A) 249 (B) 001 (C) 003 (D) 279 17. Let f (n) = 52n+2 – 24n – 25, then f (n) is divisible for

all n by (select the best option) (A) 12 (B) 24 (C) 576 (D) 243 18. For n ≥ 5, f (n) = 22n+1 – 9n2 + 3n – 2 must be

divisible by (A) 36 (B) 48 (C) 24 (D) none This section contains 2 questions (Questions 19, 20). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


19. Match the column : Column-I Column-II

(A) If |z1 – 6i| = 3, 0 ≤ arg z2 ≤ π/6 (P) 33 – 3 then minimum value of |z1 – z2| is

(B) If |z – (1 + i)| = r and (Q) 3 π/6 ≤ arg z ≤ π/3, then maximum value of r is

(C) The radius of the circle (R) 2

1–3

zz – (2 – 3i)z – (2 + 3i) z + 4 is

(D) If z satisfies the equations (S) 33

arg z = π/6 and arg

22–z = 2π/3,

then |z| is (T) 3 20. Consider all possible permutations of the letters of

the word ENDEANOEL. Match the Column-I with the Column-II Column-I Column-II (A) The number of permutations (P) 5! containing the word ENDEA is (B) The number of permutations (Q) 2 × 5! in which the letter E occurs in the first and the last postions is (C) The number of permutations in (R) 7 × 5! which none of the letters D, L, N occurs in the last five positions is (D) The number of permutations in (S) 21 × 5! which the letters A, E, O occur only in odd positions is (T) 5 × 4!


XtraEdge Test Series ANSWER KEY

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

Ques 19 20 Column Match

Ans (A) → P,R,T, (B) → P,R,T (C) → Q,S (D) → P,S (A) → P, (B) → P, (C) → P,Q, (D) → R,S

CHEMISTRY

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

Ques 19 20 Column Match Ans (A) → P,S (B) → R (C) →P,Q (D) → P,S (A) →Q,S (B) →Q,S (C) →P,R, (D) →P,R

MATHEMATICS

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

Ques 19 20 Column Match

Ans (A) →P,Q,R, (B) →P,Q,S,T (C) →P,Q,R,S,T (D) →Q (A) → R, (B) → R, (C) → T, (D) → R

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

Ques 19 20 Column Match Ans (A) → Q (B) → R (C) → P (D) → R (A) → P,Q (B) → R,S (C) → Q, (D) → S

CHEMISTRY

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

Ques 19 20 Column Match Ans (A) → R (B) → S (C) → Q (D) → P (A) →Q,R, (B) →Q,R,S (C) →S (D) →P,S

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

Ques 19 20 Column Match Ans (A) → P (B) → R (C) → T (D) → Q (A) → P,T (B) → S, (C) → Q, (D) → Q

IIT- JEE 2012 (September issue)

IIT- JEE 2013 (September issue)


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