Career Point XtraEdge for IIT-JEE 2010

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Career Point XtraEdge for IIT-JEE 2010 Monthly Magzine.

Transcript of Career Point XtraEdge for IIT-JEE 2010

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XtraEdge for IIT-JEE 1 AUGUST 2010

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XtraEdge for IIT-JEE 2 AUGUST 2010

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XtraEdge for IIT-JEE 3 AUGUST 2010

"People with integrity do what they say they are going to do. Others have excuses." Rudyard Kipling, the celebrated English author and poet, once said, "We have forty million reasons for failure, but not a single excuse." Yet today we are literally inundated with a tidal wave of excuses from every direction. In fact, it seems everyone has a reason, explanation or justification for not doing what they were supposed to do.

Why do so many of us crank out one excuse after another for virtually everything we fail to do? Well, for starters, excuses are easy. In fact, they're way too easy. After all, making excuses doesn't require any effort or commitment on our part. All we have to do is toss out excuse after excuse and we feel we're off the hook, since the best excuses always absolve us of any personal responsibility whatsoever.

While getting in the habit of making of excuses is easy, excuse making doesn't get any of us anywhere close to where we want to go in life. Sooner or later all of our years of excuses eventually catch up with us. Before we realize it, the best of life has slipped away in a lazy, hazy, crazy blur of excuses.

Ninety-nine percent of the failures come from people who have the habit of making excuses. Hold yourself responsible for a higher standard than anybody else expects of you, never excuse yourself. The person who really wants to do something invariably finds a way to get it done. And for those who don't want to do something; well, one excuse is just as good as another I suppose. What it all boils down to is simply this: what kind of person do you really want to be? Do you want to make excuses - or make something happen instead?

It's time to turn all of your excuses loose once and for all. Each and every time you fall short, pick yourself up, learn from your mistakes and immediately get going again. No complaining, no explaining and absolutely no excuses allowed. You will find that the minute you stop making excuses and start finding a way to get the job done, you'll start making your life everything it could be and should be... and so much more.

Get rid of the excuses and you can get anywhere you've ever dreamed of going. Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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XtraEdge for IIT-JEE 4 AUGUST 2010

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Volume-6 Issue-2 August, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

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

Challenging Problems in Physics,, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

Xtra Edge Test Series for JEE- 2011 & 2012

S

Success Tips for the Months

• " Always bear in mind that your own resolution to succeed is more important than any other thing."

• "God gave us two ends. One to sit on and one to think with. Success depends on which one you use; head you win -- tails, you lose."

• "The ladder of success is best climbed by stepping on the rungs of opportunity."

• "Success is getting what you want. Happiness is wanting what you get."

• "The secret of success in life is for a man to be ready for his opportunity when it comes."

• "I don't know the key to success, but the key to failure is trying to please everybody."

• "The secret of success is to be in harmony with existence, to be always calm… to let each wave of life wash us a little farther up the shore."

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 22-year-old becomes youngest IIT teacher Delhi power firm, IIT tie-up to reduce power loss

IITian ON THE PATH OF SUCCESS 6 Prof. Mohit Renderia & Dr. Rajiv Laroia

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 54

Class XII – IIT-JEE 2011 Paper

Class XI – IIT-JEE 2012 Paper

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 4] Students’ Forum Physics Fundamentals Capacitor - 2 Work power energy & Conservation Law

CATALYSE CHEMISTRY 30

Key Concept Reaction Mechanism Solid State Understanding : Physical Chemistry

DICEY MATHS 43

Mathematical Challenges Students’ Forum Key Concept Vector Permutation & Combination

Study Time........

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

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22-year-old becomes youngest IIT teacher

MUMBAI: IITians often liken the generation gap between themselves and their teachers to that between MS-DOS and Windows. This semester, however, the students on the Powai campus can look forward to someone much closer to their age: a physics teacher who has just entered his 20s.

At 22, Tathagat Avatar Tulsi, who has never studied in a classroom, plans to ask his students how they would want to be taught. "I have never taught in a class. But I believe I can come down to the level of a student and help them understand the subject," he said.

Having completed high school when he was nine, his graduation in science at 10, an MSc in Physics at 12, and his PhD in Quantum Computing from the Indian Institute of Science (IISc), Bangalore, at 21, Tulsi says he is going to write to the Limca Book of Records to include him as the youngest faculty member in the country. Having achieved a lot pretty early in life, Tulsi may seem like a young man in hurry, but he has set a huge task for himself— to come up with an important scientific discovery, which will probably lead him to his ultimate dream: to own that shining piece of gold with Alfred Nobel on the obverse.

The "wonder boy", who suffered humiliation in August 2001 when a delegation of scientists taken by the department of science & technology to Lindau in Germany for an interaction with Nobel laureates, suggested that he was not a thinker, but a "fake prodigy" who had "mugged up" theories. Putting that behind, the Patna boy will stay on the Powai campus in the faculty quarters and work towards achieving that dream.

That "not-so-distant" goal is probably why Tulsi chose teaching over a vocation. "I want to pursue my research and at IIT-B, I will have the leisure to continue my research and one day set up a lab focused on quantum computation in our country." Going to foreign shores is currently not on Tulsi’s plans. He chose the Powai college over Waterloo University, Canada, and the Indian Institute of Science Education & Research (IISER), Bhopal, both of which had also offered him teaching jobs.

Delhi power firm, IIT tie-up to reduce power loss New Delhi: The BSES Yamuna Power Limited (BYPL) and the Indian Institute of Technology (IIT)-Delhi have come together to improve the quality and reliability of power supply by reducing transmission and distribution losses.

A memorandum of understanding (MoU) was signed Monday between BYPL CEO Ramesh Narayanan and Anil Wali, managing director, Foundation for Innovation and Technology Transfer (FITT) - a society established by IIT-Delhi, to focus on how to bring the next-level or

"SMART" technology to the power distribution business and to keep pace with technological innovations taking place in the transmission business.

Both BYPL and IIT-Delhi will appoint one principal project investigator each. The cost of the projects will be borne by BYPL. IANS.

Forests ministry teams with IITs for Ganga management plan

New Delhi: The Indian Institutes of Technology (IIT), the premier higher technical educational institutions of the Ministry of Human Resource Development (HRD) have committed themselves to the responsibility of development of a management plan of the Ganga river basin called the National Ganga River Basin Management Plan Project (NGBRM).

Seven IITs have come together for this purpose. The IITs have accepted this societal challenge as part of their response to the present-day challenges of the Indian society.

"It is required to ensure that the flow of the river Ganga must be continuous (Aviral Dhara), the river must have longitudinal and lateral connectivity, the river must have adequate space for its various

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functions and the river must not be seen as a carrier of waste loads (Nirmal Dhara)," stated an official press release. The management plan will outline the strategy and the actions that need to be undertaken for the maintenance and restoration of the Ganga basin. The management plan should take into account the constraints of population, urbanization, industrialization and agriculture activities.

The IITs will form several thematic groups and each group will develop a detailed outline for the improvement of ecological health of the basin system. Besides the thematic groups, the IITs will also integrate in a holistic manner, all the issues into a comprehensive management plan.

In order to develop this plan, discussions will be held with local, state and other agencies who have to deal with the maintenance of the basin system. The management plan will also take into account the experience of earlier attempts of Ganga Action Plans. The HRD ministry and the Ministry of Environment & Forests are coming together to support the initiatives of the IITs. The work is estimated to be carried out in a period of 18 months. The funding for this project is estimated to be about Rs.15 crores. An agreement has been signed between the Directors of seven IITs and the Ministry of Environment & Forests in the presence of Minister of State for HRD, Kapil Sibal and Mister of State for Environments and Forest, Jairam Ramesh.

This initiative will involve not only faculty and students of seven IITs but will also take help from experts from other institutes and universities also.

IIT seeks robotic solution in conflict KOLKATA: A group of students from IIT-Kharagpur is working on a prototype of an Autonomous

Ground Vehicle (AGV) which, if it proves successful, may be developed by the Defence Research and Development Organisation (DRDO) for use by security agencies for Low-Intensity Conflict (LIC).

The six-member IIT team has bagged the first rank in Phase-I in the Student Robot Competition, 2010, organised by DRDO. As many as 240 colleges and institutes from across the country participated in the first phase of the contest, where third- and fourth-year students were asked to design an Autonomous Ground Vehicle for Low Intensity Conflict'. Only 14 teams were shortlisted for Phase-II. Among the entries from the east, the Indian School of Mines team from Dhanbad was ranked fourth. The National Institute of Technology, Rourkela, ranked 14th.

"Participants were told that the AGV will be a combat vehicle of the future and assigned tasks that a conventional manned vehicle cannot perform. These are basically autonomous robots to be used by security agencies engaged in LIC in urban and unstructured environments. These robots would be used in LIC and Explosive Ordnance Disposal (EOD) programmes in undesirable, hazardous and potentially life-threatening environments," said a senior DRDO official.

The AGV would have to be armed with sensors, software and other equipment to help it negotiate harsh terrain, identify and designate targets, engage and neutralise them. The vehicle would also have to detect minefields and neutralise them. In short, the AGV would be an autonomous off-road robotic platform that would navigate rough terrain and avoid natural and man-made obstacles in the shortest possible time.

According to the team of experts who judged the entries, the IIT-Kharagpur team, comprising Nalin

Gupta, Sarbartha Banerjee, Subhagato Dutta, Rahul Das, Anindita Bhattacharya and A Srinivas Reddy submitted an excellent design. Officials are waiting to find out how the prototype performs.

"The idea of the competition is to harness the innovative ideas of our student community to the National Robotics Program of India. The demands made from the participants are enormous. The robot will have to complete a closed loop obstacle course of 500 metres within an anticipated time of 20 minutes, using autonomous navigation. For the first 350 metres, the robot would have to navigate with the help of lane-following' by colour detection. While doing this, it would have to avoid static positive obstacles, cross over slopes, staircases and corrugations. It would have to navigate the remaining 150 metres with the help of GPS waypoints. Maximum width of the robot would have to be 1 metre, maximum speed of 10 km per hour and minimum turning radius of 5 metres. It would have to carry an additional payload of 10 kg. The robot would have to be self-powered in all respects," the official said. In Phase-I, teams submitted designs with system configuration details. In the next phase, the selected teams would have to build a prototype and make it perform before the judges. Initially, 10 teams were supposed to be shortlisted for Phase-II. Given the nature of the papers submitted, it was finally decided to shortlist 14. Each team received a cash award of Rs 1,00,000.

The teams are now busy building their prototypes. The final competition will be held at the Combat Vehicles Research and Development Establishment (CVRDE), Chennai, between September 27-29, where the prototypes would be tried out.

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Prof. Mohit Randeria received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained M.S. Degree from the California Institute of Technology, USA in 1982 and Ph.D. Degree from Cornell University, USA in 1987. Prof. Mohit Randeria is presently Professor of Physics at the Ohio State University in Columbus, Ohio. Prof. Randeria started his academic career as an Assistant Professor of Physics at SUNY Stony Brook in 1989, after a few years of post-doctoral work at Cornell and Illinois. He then joined the Argonne National Lab in Illinois as a Staff Scientist in the Materials Science Division, and after four years there, resumed his academic career at the Tata Institute of Fundamental Research (TIFR), Mumbai in 1995. He spent nine years in the Theoretical Physics Department at TIFR as Reader, Associate Professor, and Professor. Since 2004, he has been a Professor of Physics at Ohio State University. Prof. Randeria is an acknowledged expert in the area of Theoretical Condensed Matter Physics. His research interests are focused, at present, on high temperature superconductivity, strongly correlated systems, and ultra-cold atomic gases. Prof. Randeria is the author of over hundred research papers in Condensed Matter Physics. He is the recipient of a prize in Condensed Matter Physics, awarded in honour of Nobel Laureate Phillip Anderson by the International Center of Theoretical Physics, Trieste, Italy in 2002. He has also been awarded the Swarnajayanti Fellowship in 1998, the B.M. Birla Science prize in 1999, the S.S. Bhatnagar Award in 2002, and the George A. Miller Visiting Professorship at University of Illinois, Urbana-Champaign in 2002-2003. In honouring Prof. Mohit Randeria, IIT Delhi recognizes the outstanding contributions made by him as a Researcher and Scientist. Through his achievements, Prof. Mohit Randeria has brought glory to the name of the Institute.

Dr. Rajiv Laroia received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1985. He obtained his Ph.D. in Electrical Engineering from University of Maryland, College Park in 1992. Dr. Laroia is presently Senior Vice President of Technology at Qualcomm, USA. Dr. Laroia joined Siemens Research Labs in Munich, Germany after graduating from IIT Delhi. In 1992, he joined AT&T Bell Labs after obtaining his Ph.D. In 2000, Dr. Laroia founded Flarion Technologies, a venture backed company to develop and commercialize a novel all-IP mobile wireless broadband technology. He served as the CTO of Flarion and in that role provided the vision and led the development of technology and products for the company. In 2006, Flarion was acquired by the wireless technology giant Qualcomm, where he currently serves as Senior Vice President (Technology). Dr. Laroia is one of the world’s leading researchers and innovators in the field of communication. He holds more than 50 patents and has more than 100 others pending. His early research focused on wire-line communication. He has significant technology contributions to V.34 and V.90 International Standards for sending data over telephone lines. The technology he invented is used in virtually all dial-up modems in the world. Since 1997, Dr. Laroia has been working in the field of mobile wireless communication. The technology he and his team developed at Flarion is now incorporated in all three major next generation international wireless standards UMTS LTE, UMB and Wimax. Dr. Laroia has won numerous industry awards. In 2006, he was inducted to the Innovation Hall of Fame at the University of Maryland. He is a fellow of the IEEE. In honouring Dr. Rajiv Laroia, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Technologist. Through his achievements, Dr. Rajiv Laroia has brought glory to the name of this Institute.

Success Story Success Story This articles contains stories of persons who have succeed after graduation from different IIT's

Dr. Rajiv Laroia B.Tech., IIT – Delhi

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PHYSICS

1. A small body attached to one end of a vertically hanging spring is performing SHM about it's mean position with angular frequency ω and amplitude a. If at a height y* from the mean position, the body gets detached from the spring calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it's subsequent motion after detachment. (aω2 > g)

[IIT-2005]

m

y0

Sol. The total energy of the spring-mass system at any

position of mass above the mean position is the sum of the follows.

(a) Gravitation potential energy of mass (b) Kinetic energy of mass (c) Elastic potential of spring. The mass will reach the highest point when its

mechanical energy [Sum of (a) and (b)] is maximum. This is possible when elastic potential energy of system is zero.

⇒ The mass should detach when the spring is at its natural length.

Let L = Natural length of spring when mass m is hanging at equilibrium the

K

mg

l Kl

L L

Mean Positionof oscillation

mg = kl ; l = k

mg

⇒ y = k

mg

⇒ y = 2g

ω [Q K = mω2]

where 2g

ω < a (given)

2. One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate. [IIT-1998]

B

C A

2V0 V0

P3P0

P0

(a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the

heat absorbed by the gas in the path AB; (c) the net heat absorbed by the gas in the path BC; (d) the maximum temperature attained by the gas

during the cycle. Sol. n = 1 = no. of moles, For monoatomic gas :

Cp = 2R5 , Cv =

2R3

Cyclic process A → B ⇒ Isochoric process C → A ⇒ Isobaric compression (a) Work done = Area of closed curve ABCA during

cyclic process. i.e. ∆ABC

∆W = 21 × base × height =

21 V0 × 2P0 = P0V0

(b) Heat rejected by the gas in the path CA during Isobaric compression process

∆QCA = nCp∆T = 1 × (5R/2)(TA – TC)

TC = RIVP2 00

×, TA =

RIVP 00

×,

∆QCA =

−R

VP2RVP

2R5 0000 = –

25 P0V0

Heat absorbed by the gas on the path AB during Isochoric process

∆QAB = nCv∆T = 1 × (3R/2) (TB – TA)

=

×

−× R1

VPR1VP3

2R3 0000 = 3P0V0

(c) As ∆U = 0 in cyclic process, hence ∆Q = ∆W ∆QAB + ∆QCA + ∆QBC = ∆W,

∆QBC = P0V0 – 2VP 00 =

2VP 00

As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C.

KNOW IIT-JEE By Previous Exam Questions

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(d) Equation for line BC

P = –

0

0

VP2

V + 5P0, As PV = RT hence,

P = V

RT [For one mole] [as y = mx + c]

∴ RT = –0

0

VP2

V2 + 5P0V ...(1)

For maximum; dVdT = 0, –

0

0

VP2

× 2V + 5P0 = 0;

∴ V = 4V5 0 ...(2)

Hence from equation (1) and (2)

RTmax = –0

0

VP2

× 2

0

4V5

+ 5P0

4V5 0

= –2P0V0 × 1625 +

4VP25 00 =

825 P0V0

∴ Tmax = 825

RVP 00

3. Two isolated metallic solid spheres of radii R and 2R

are charged such that both of these have same charge density σ. The spheres ares are located far away from each other, and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

[IIT-1996]

Sol.

R 2R q2

σ σ

q1

V

q'2 σ

q'1

V

Connecting Wire

For sphere of radius R σ = 21

R4qπ

∴ q1 = σ × 4πR2

For sphere of radius 2R σ = 22

)R2(4q

π

⇒ q2 = σ × 16πR2 When the two spheres are connected then the

potential on the two spheres will be same. There will be a rearrangement of charge for this to happen.

Let q1' and q2' be the new charges on the two spheres. Since the total charge remains the same q'1 + q'2 = q1 + q2 = σ × 20 πR2 ...(1) Also Since V1 = V2

04

1πε R

'q1 = 02

1πε R2

'q2

⇒ q1' = 2

'q2 . ..(2)

Substituting the value of q1' from (2) in (1)

2

'q1 + q2' = σ × 20 πR2

⇒ 2

'q3 2 = σ × 20 πR2

⇒ 22

)R2(4'q

π =

3σ ×

25

⇒ New charge density on bigger sphere

= 22

)R2(4'q

π =

65σ

4. In the given circuit E1 = 3E2 = 2E3 = 6 volts ; R1 = 2R4 = 6 ohms R3 = 2R2 = 4 ohms ; C = 5µF Find the current in R3 and the energy stored in the

capacitor. [IIT-1988] R1

R2 R3

E1

E3 R4

E2

C

Sol. R1 = 6Ω

R2 = 2Ω

R3 = 4Ω

E1 = 6V

E3 = 3V R4 = 4Ω

G A

5µF = C

E F 2V = E2

i2

i2 i2+i1 i1

i2

B

D C

Applying Kirchoff's law in ABFGA 6 – (i1 + i2) 4 = 0 …(1) Applying Kichoff's law in BCDEFB I2 × 3 – 3 – 2 + 2i2 + (i2 + i1) 4 = 0 …(2) Putting the value of 4 (i1 + i2) = 6 in (2) 3i2 – 5 + 2i2 + 6 = 0

∴ i2 = – 51 A

Sybstituting this value in (i) we get

i1 = 1.5 –

51– = 1.7 A

Therefore current in R3 = i1 + i2 = 1.7 – 0.2 = 1.5 A To find the p.d. across the capacitor VE – 2 – 0.2 × 2 = VG ∴ VE – VG = 2.4 V

∴ Energy stored in capacitor = 21 CV2

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= 21 × 5 × 10–6 × (2.4)2

= 1.44 × 10–5 J 5. A wire loop carrying a current I is placed in the x-y

plane as shown in figure. [IIT-1991]

x

y

O

vM

+Q P

a

120º

N

I

(a) If a particle with charge +Q and mass m is placed

at the centre P and given a velocity →V along NP (see

figure), find its instantaneous acceleration. (b) If an external uniform magnetic induction field

→B = B i is applied, find the force and the torque acting on the loop due to this field.

Sol. (a) Magnetic field at the centre P due to arc of circle, Subtending an angle of 120º at centre would be :

x

y M

+Q P

a

60º

N

I 60º

a

r

x

y

60º

v

B1 = 31 (field dut to circle) =

31

a2Iµ0

=

a6Iµ0 (outwards) =

aIµ16.0 0 (outwards)

or 1B→

= a

Iµ16.0 0 k

Magnetic field due to straight wire NM at P :

B2 = π4

µ0

rI (sin 60º + sin 60º)

Here, r = a cos 60º

∴ B2 = π4

µ0

º60cosaI (2 sin 60º)

or B2 = aI

2µ0

πtan 60º =

aIµ27.0 0 (inwards)

or 2B→

= –a

Iµ27.0 0 k

∴→

netB = →

1B +→

2B = –a

Iµ11.0 0 k

Now, velocity of particle can be written as,

→v = v cos 60º i + v sin 60º j =

2v i +

2v3 j

Magnetic force

mF = Q(→v ×

→B )

= a2

IQvµ11.0 0 j – ia2

IQvµ311.0 0

∴ Instantaneous acceleration

→a =

mFm→

= am2

IQvµ11.0 0 )i3j( −

(b) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop

will be, →M =(IA) k

Here, A is the area of the loop.

A = 31 (πa2) –

21 [2 × a sin 60º] [a cos 60º]

= 3a 2π –

2a 2

sin 120º = 0.61 a2

∴ →M =(0.61 Ia2) k Given,

→B = B i

∴ →τ =

→M ×

→B = (0.61 Ia2B) j

CHEMISTRY

6. A sample of hard water contains 96 ppm of SO42– and

183 ppm of HCO3– with 60 ppm of Ca2+ as the only

cation. How many moles of CaO will be required to remove HCO3

2– from 100 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca2+ ions ? (Assume CaCO3 to be completely insoluble in water). If the Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (one ppm means one part of the substance in one million part of water, mass/mass) [IIT-1997]

Sol. In 106 g(= 1000 kg) of the given hard water, we have amount of SO4

2– ions = 96 g amount of HCO3

– ions = 183 g

So amount of SO42– ions = 1molg96

g96−

= 1 mol

and amount of HCO3– ions = 1molg61

g183−

= 3 mol

These ions are present as CaSO4 and Ca(HCO3)2.

Hence, amount of Ca2+ ions =

+

231 = 2.5 mol

The addition of CaO causes the following reactions: CaO + Ca(HCO3)2 → 2CaCO3 + H2O 1.5 mol of CaO will be required for the removal of

1.5 mol of Ca(HCO3)2 in form of CaCO3. In the treated water, only CaSO4 is present now.

Thus, 1 mol of Ca2+ ions will be present in 106 g of water. Hence, its concentration will be 40 ppm.

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Molarity of Ca2+ ions in the treated water will be 10–3 mol l–1.

If the Ca2+ ions are exchanged by H+ ions then, Molartiy of H+ in the treated water = 2 × 10–3 M Thus, pH = – log(2 × 10–3) = 2.7 7. The vapour pressure of ethanol and methanol are 44.5

mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. [IIT-1986]

Sol. Given that, For ethanol (C2H5OH), 0

eP = 44.5 mm Hg M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g

∴ Moles of ethanol, ne = Mm =

4660 = 1.3

For methanol (CH3OH), 0mP = 88.7 mm Hg

M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g

∴ Moles of methanol, nm = Mm =

3240 = 1.25

∴ xe = me

e

nnn+

= 25.13.1

3.1+

= 55.23.1

xm = me

m

nnn+

= 25.13.1

25.1+

= 55.225.1

According to Raoult's law,

Pe = 0eP xe =

55.23.15.44 × = 22.69 mm Hg

and Pm = 0mP xm =

55.225.17.88 × = 43.48 mm Hg

Hence, total vapour pressure of the solution, PT = Pe + Pm = 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law, Pm = PTx´m (in vapour form) Hence, mole fraction of methanol in vapour form,

x´m = T

m

PP =

17.6648.43 = 0.66

8. An alkyl halide X, of formula C6H13Cl on treatment with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y and Z. [IIT-1996]

Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes.

X136 ClHC

HCl–;

butoxidetK

−− → 126HCZY +

Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.

Y and Z (C6H12) Ni

H2→ CH3 – CH – CH – CH3

CH3 CH3

2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3

CH3 CH32,3-dimethyl

butene-2 (Y)

and CH3 – CH – C = CH2

CH3 CH32,3-dimethyl butene-1

(Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3

CH3 CH3(Y)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

CH3 – CH – C = CH2

CH3 CH3(Z)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

Both, Y and Z can be obtained from following alkyl

halide :

CH3 – C – CH – CH3

CH3 CH32-chloro-2,3-dimethyl butane

(X)

K-t-butoxide

∆; –HCl

CH2 = C — CH – CH3

CH3 CH3

Cl

+ CH3 – C = C – CH3

CH3 CH3

(Z) 20% (Y) 80%

Hence, X,

CH3 – C – CH – CH3

CH3 CH3

Cl

Y, CH3 – C = C – CH3

CH3 CH3

Z, CH3 – CH – C = CH2

CH3 CH3

9. An organic compound (X), C5H8O, does not react appreciably with Lucas reagent at room temperatures but gives a precipitate with ammonical AgNO3 solution. With excess CH3MgBr; 0.42 g of (X) gives 224 ml of CH4 at STP. Treatment of (X) with H2 in the presence of Pt catalyst followed by boiling with excess HI gives n-pentane. Suggest structure of (X) and write the equations involved. [IIT-1992]

Sol. Lucas test sensitive test for the distinction of p, s, and t-alcohol. A t-alcohol gives cloudiness immediately, while s-alcohol within 5 minutes. A p-alcohol does not react with the reagent at room temperature. Thus, the present compound (X) does not react with this reagent, hence it is a p-alcohol.

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XtraEdge for IIT-JEE 13 AUGUST 2010

(X) = C4H6.CH2OH(p-alcohol) Since the compound gives a ppt. with ammonical

AgNO3, hence it is an alkyne containing one –C≡ CH, thus (X) may be written as :

HC≡C –C2H4 – CH2OH (X) It is given that 0.42 g of the compound (which is

0.005 mol) produces 22.4 ml of CH4 at STP (which is 0.01 mol) with excess of CH3MgBr. This shows that the compound (X) contains two active H atoms (H atom attached to O, S, N and –C≡CH is called active). Of these, one is due to the p-alcoholic group (–CH2OH) and the other is due to the –C≡CH bond, since both these groups are present in (X), hence it evolves two moles of CH4 on reaction with CH3MgBr.

H – C≡C.)X(

42HC – CH2OH + 2CH3MgBr →

BrMgC≡C–C2H4 – CH2OMgBr + 2CH4 Moreover, the treatment of (X) with H2/Pt followed

by boiling with excess of HI gives n-pentane (remember that 2HI are required to convert one –CH2OH into CH3). This shows that the compound (X) contains a straight chain of five carbon atoms.

H – C≡C–C2H4 – CH2OH → Pt/H2 2 CH3CH2.C2H4 – CH2OH

∆ → HI2 CH3CH2CH2CH2CH3 + H2O + I2

n-pentane On the basis of abvoe analytical facts (X) has the

structure :

HC≡C.CH2 CH2 – CH2OH (X) 5 4 3 2 1

4-pentyne-1-ol The different equations of (X) are :

)X(222 OHCHCHCHCCH −≡− .tempRoom

HClZnCl2 → + No reaction

AgNO3 Ag – C≡C – CH2CH2CH2OH + NH4NO3

White ppt. NH3

2CH3MgBr Br MgC≡C.CH2CH2CH2OMgBr + 2CH4

2H2/Pt CH3CH2CH2CH2CH2OH

Pentanol-1

CH3CH2CH2CH2CH3

n-pentane

2 HI ∆, –H2O; –I2

The production of 2 moles of CH4 is confirmed as the

reactions give 224 ml of CH4. Q 84 g(X) gives = 2 × 22.4 litre CH4

∴ 0.42 g (X) gives = 84

42.04.222 ××

= 224 ml of CH4

10. 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

MATHEMATICS

11. Find the values of a and b so that the function

f(x) =

π≤<π−π≤≤π+

π≤≤+

x2/,xsinbx2cosa2/x4/,bxcotx2

4/x0,xsin2ax

is continuous for 0 ≤ x ≤ π [IIT-1989] Sol. As, f(x) is continuous for 0 ≤ x ≤ π

∴ R.H.L.

π

=4

xat = L.H.L.

π

=4

xat

+

ππ b4

cot4

.2 =

π

4sin.2a

4

⇒ 2π + b =

4π + a

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XtraEdge for IIT-JEE 14 AUGUST 2010

⇒ a – b = 4π ....(i)

also, R.H.L

π

=2

xat = L.H.L

π

=2

xat

π

−π

2sinb

22cosa =

+

ππ b2

cot.2

.2

⇒ – a – b = b ⇒ a + 2b = 0 ...(ii)

From (i) and (ii), a = 2

3π and b = 43π−

12. Find dxdy at x = –1, when

x

2sin

)y(sinπ

+ 23 sec–1(2x) + 2x tan ln (x + 2) = 0

[IIT-1991] Sol. Here,

x

2sin

)y(sinπ

+ 23 sec–1(2x) + 2x tan (log (x + 2)) = 0

Differentiating both sides, we get

x

2sin

)y(sinπ

. log(sin y) . cos2π x .

+

π x

2sin

1x2

sin)y(sin

π

. cos y . dxdy

+ 2x4|)x|2(

2.23

2 −+

)2x())2x(log(sec.2 2x

++

+ 2x log 2 . tan (log(x + 2)) = 0

putting,

π−=−=

3y,1x , we get

π−

3,1dxdy =

2

2

31

3

π−

π−

= 3

32 −ππ

13. ABC is a triangle such that

sin(2A + B) = sin(C – A) = –sin(B + 2C) = 21

If A, B and C are in Arithmetic Progression, determine the values of A, B and C. [IIT-1990]

Sol. Given that in ∆ABC, A, B and C are in A.P. A + C = 2B also A + B + C = 180º ⇒ B = 60º Also given that, sin (2A + B) = sin (C – A) = – sin (B + 2C) = ½ ...(1)

⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) = 21

⇒ 2A + 60º = 30º, 150º

neglecting 30º, as not possible ⇒ 2A + 60º = 150º ⇒ A = 45º again from (1), sin (60º + 2c) = –1/2 ⇒ 60º + 2C = 210º, 330º ⇒ C = 75º or 135º Also from (1) sin (C – A) = ½ C – A = 30º, 150º, 195º for A = 45º, C = 75º and C = 195º (not possible) ∴ C = 75º Hence, A = 45º, B = 60º, C = 75º

14. If exp (sin2x + sin4x + sin6x + ...... ∞). ln 2 satisfies the equation x2 – 9x + 8 = 0, find the value of

xsinxcosxcos

+, 0 < x <

2π . [IIT-1991]

Sol. exp (sin2x + sin4x + sin6x + ...... ∞) loge2

⇒ 2log.

xsin1xsin

e2

2

e − ⇒ x2tan

e 2loge

⇒ xtan22 satisfy x2 – 9x + 8 = 0 ⇒ x = 1, 8

∴ xtan22 = 1 and xtan2

2 = 8 ⇒ tan2x = 0 and tan2x = 3

⇒ x = nπ and tan2x = 2

3tan

π

and x = nπ ± 3π

Neglecting x = nπ as 0 < x < 2π

⇒ x = 3π ∈

π

2,0

∴xsinxcos

xcos+

=

23

21

21

+=

311

1313

− =2

13 −

∴ xsinxcos

xcos+

= 2

13 −

15. Find the value of : cos (2 cos–1 x + sin–1x) at x = 51 ,

where 0 ≤ cos–1x ≤ π and –π/2 ≤ sin–1x ≤ π/2 [IIT-1981] Sol. cos2cos–1x + sin–1x

= cos

π

+−

2xcos 1 , as cos–1x + sin–1x =

= – sin(cos–1x )

= – sin(sin–1 2x1− )

= – sin

−−

21

511sin

= – sin

562sin 1 =

562

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XtraEdge for IIT-JEE 15 AUGUST 2010

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XtraEdge for IIT-JEE 16 AUGUST 2010

1. In the circuit shown in figure F2CC 21 µ== . Then charge stored in (steady state)

C1 C2

2Ω 1Ω 3Ω

120 V

1Ω 2Ω 3Ω

(A) capacitor C1 is zero (B) capacitor C2 is zero (C) both capacitors is zero (D) capacitor C1 is C40µ

Passage # (Q. No. 2 to Q. No. 4) A charged metal sheet is placed into uniform electric field E, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet will be E1 = 5.6 × 105 V/m and on the right it will be E2 = 3.1 × 105 V/m.

Q. 2 Find the total charge of the sheet if a electric force

of 0.08N is exerted on it (A) C28.0 µ (B) C32.0 µ (C) C24.0 µ (D) C38.0 µ

Q. 3 Find the area of sheet of one side (A) 0.02m2 (B) 0.03m2 (C) 0.04 m2 (D) 0.05 m2

Q. 4 Find the value of E (A) 2.5 × 104 V/m (B) 12.5 × 104 V/m (C) 3.5 × 104 V/m (D) 8.7 × 104 V/m

Q. 5 A capacitor consists of two parallel metal plate of area A separated by a distance d. A dielectric slab of area A, thickness b & dielectric constant K is placed inside the capacitor. If CK is the capacitance of capacitor with dielectric. How much K and b are restricted so that CK = 2C, where C is capacitance without dielectric

(A) db3d&

db2b4K ≤<−

=

(B) db2d&

db2b2K ≤<−

=

(C) d2b2d&

db2b2K ≤≤−

=

(D) db4d&

db2b2K ≤≤−

=

Q. 6 In the circuit shown in figure S1 S1

1Ω 2Ω 3Ω10V

20Vi (A) i = 2.5A when S1 is closed and S2 is open

(B) A3

20i = when S1 is open and S2 is closed

(C) A35i = when S1 and S2 both are open

(D) i = 20A when both S1 and S2 are closed

Q. 7 A charged particle of unit mass and unit charge

moves with velocity s/m)j6i8(v∧∧→

+= in a

magnetic field of Tk2B∧

= . Choose the correct alternative(s)

(A) The path of the particle may be x2 + y2 – 4x – 24 = 0 (B) The path of the particle may be x2 + y2 = 25 (D) The time period of the particle will be 3.14s

Q. 8 In the diagram shown, the wires P1Q1 and P2Q2 are made to slide on the rails with same speed of 5m/s. In this region a magnetic field of 1T exists. The electric current in Ω9 resistor is (A) zero if both wires slide towards left

(B) zero if both wires slide in opposite direction (C) 20mA if both wires move towards left (D) 20mA if both wires move in opposite direction

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 # 4

E1 E2

×P1 P2

Q1 Q2

2Ω2Ω

× × × × × × ×

×××××××××

× × × × × × × ×

×

×

× × × × × × × × ×

4 cm

Page 17: Career Point XtraEdge for IIT-JEE 2010

XtraEdge for IIT-JEE 17 AUGUST 2010

1. [A,C] )e1(REI /t τ−−=

E and R is constant. ,RL

=τ if L increases then τ

will increase hence the curve will shift towards right if E and R are halved then τ will increase hence the curve will shift towards right.

2. [B,C] (Moderate) As the length is double, the cross section area of the wire becomes half, thus the

resistance of the wire ALR ρ= becomes four

times the previous value. Hence after the wire is elongated the current becomes one fourth. Electric field is potential difference per unit length and hence becomes half the initial value.

The power delivered to resistance is R

Vp2

= and

hence becomes one fourth. 3. [A,B,C] Charge is distributed over the surface of

conductor in such a way that net field due to the charge and outside charge q is zero inside. Field due to only q is non-zero.

4. [A] RRR

R.RR

V

VA <

+=

5. [A] RRRR GB >+= 6. [A] (Tough)% error in case A

1001RR

R100

RRR

V

VA ×

+=×

%1100RRR

V−=×

+−

=

%error in case B

%10100R

R100

RRR GB =×=×

Hence percentage error in circuit B is more than that in A.

7. [D] Order of the fringe can be counted on either side of the central maximum for example fringe no. 3 is first order bright fringe.

8. [C]

Solution Physics Challenging Problems

Set # 3

8 Questions were Published in July Issue

• 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.

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XtraEdge for IIT-JEE 18 AUGUST 2010

1. A block B of mass m = 0.5 kg is attached with upper end of a vertical spring of force constant K = 1000 Nm–1 as shown in Figure. Another identical block A falls from a height h = 49.5 cm on the block B and gets stuck with it. The combined body starts to perform vertical oscillations.

Calculate amplitude of these oscillations. (g = 10 ms–2)

Sol. Since block A falls freely under gravity through a

height h before colliding with block B, therefore, its velocity just before collision is v0 = gh2 .

Let velocity of combined body just after collision be v.

Applying law of conservation of momentum for collision,

2mv = mv0 or v = 2

v0 = 21 gh2

Since, before collision, block B was in static equilibrium, therefore, compressive force in spring was exactly equal to its weight.

Suppose, initially spring was compressed through y0, then

Ky0 = mg or y0 = K

mg

Hence, initial strain energy in spring,

U0 = 21 Ky0

2 = K2gm 22

After collision, combined body starts to move vertically downward due to velocity v. Therefore, spring is further compressed. Let maximum contraction of spring be y.

Then according to law of conservation of energy, Maximum strain energy in spring = initial strain

energy U0 + kinetic energy of combined body just after collision + further loss of potential energy of combined body.

∴ 21 Ky2 =

K2gm 22

+ 21 (2m) v2 + 2mg (y – y0)

or 21 Ky2 – 2mgy +

222

0 mv–K2gm–mgy2 = 0

or 500 y2 – 10y – 1.2 = 0 From above equation y = + .6 or – .04

It means during vertical oscillations of combined body, its lowest position corresponds to a contraction of 0.06 m or 6 cm of spring and uppermost position corresponds to an elongation of 0.04 m or 4 cm of the spring.

Distance between these two extreme positions is 2a = (6 + 4) cm

Or amplitude of oscillations, a = 5 cm Ans. 2. Two identical blocks A and B of mass m = 3 kg are

attached with ends of an ideal spring of force constant K = 2000 Nm–1 and rest over a smooth horizontal floor. Another identical block C moving with velocity v0 = 0.6 ms–1 as shown in fig. strikes the block A and gets stuck to it. Calculate for subsequent motion

(i) velocity of centre of mass of the system, (ii) frequency of oscillations of the system, (iii) oscillation energy of the system, and (iv) maximum compression of the spring.

m m mA BC v0

Sol. When block C collides with A and get stuck with it,

combined body moves to the right, due to which spring is compressed. Therefore, the combined body retards and block B accelerates. In fact, deformation of spring varies with time and the system continues to move rightwards. In other words, centre of mass of the system moves rightwards and combined body and block B oscillate about the centre of mass of the system.

Let just after the collision velocity of combined body formed by blocks C and A be v. Then, according to law of conservation of momentum,

(m + m)v = mv0

or v = 2

v0 = 0.3 ms–1

∴ Velocity of centre of mass of the system,

vc = mm2

0mvm2+

×+× = 0.2 ms–1

Now the system is as shown in fig. 2m m

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

Students' ForumPHYSICS

h

A

B

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XtraEdge for IIT-JEE 19 AUGUST 2010

Its reduced mass, m0 = mm2

)m)(m2(+

= 3m2

∴ Frequency of oscillations,

f = 0m

K21π

= π105 Hz. Ans.

Since, just after the collision, combined body has velocity v, therefore, energy of the system at that

instant, E = 21 (2m)v2 = 0.27 joule

Due to velocity vC of centre of mass of the system, translational kinetic energy,

Et = 21 (3m) 2

cv = 0.18 joule

But total energy E of the system = its translational kinetic (Et) + oscillation energy (E0)

∴ E0 = E – Et = 0.09 joule At the instant of maximum compression, oscillation

energy is stored in the spring in the form of its strain energy. Let maximum compression of spring be x0.

then 20Kx

21 = E0

∴ x0 = 90 × 10–3 m or 103 mm Ans. 3. An artificial satellite of mass m of a planet of mass

M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite to depend on velocity as F = a.v2 where a is a constant, calculate how long the satellite will stay in orbit before it falls onto the planet's surface.

Sol. Due to slight resistance offered by cosmic dust, energy of satellite decreases slowly but continuously and setellite follows a spiral path of decreasing radius and ultimately falls onto the planet's surface as shown in Figure

Planet

Since, energy of satellite decreases slowly,

therefore, its radial velocity is negligible in comparison to tangential velocity.

Let at an instant, distance of satellite from centre of planet be x,

Velocity of satellite is v = x

GM …(1)

and energy of satellite is E = – 21

xm.GM ...(2)

Resisting force, F = av2 = x

aGM

During a very small time interval dt,

Distance travelled by the satellite is vdt = x

GM dt

∴ ` Work done by satellite against resistance of cosmic dust is

dW = F . (v.dt) = a 2/3

xGM

dt

But due to the work done by stellite, its energy decreases by the same amount and radius x also decreases simultanceously.

Or Increase in energy of satellite is dE = – dW. But from equation (2)

dE = 21

2xGMm dx.

∴ 21 2x

GMm . dx = – a 2/3

xGM

dt

or dt = – GMa2

mx

dx …(3)

At initial moment (t = 0), x = nR and we have to calculate time t when x becomes equal to R. Integrating equation (3) and substituting above limits.

Gma2m–dt

?t

0t=∫

=

=. ∫

=

=

Rx

nRx xdx

or t = GMa

)1–n(Rm Ans.

4. Each plate of a parallel plate air capacitor has are area

S = 5 × 10–3 m2 and are d = 8.85 mm apart as shown in fig. Plate A has a positive charge q1 = 10–10coulomb and plate B has charge q2 = +2 × 10–10 coulomb. Calculate energy supplied by a battery of emf E = 10 volt when its positive terminal is connected with plate A and negative terminal with plate B.

+10–10C +2 × 10–10C

A B d

Sol. Charges q1 and q2 get distributed such that charges appearing on inner surfaces of two plates become numerically equal but opposite in nature. Since charge q1 on plate A is less than charge q2 on plate B, therefore inner surface of plate. A becomes negatively charged and that of B become positively charged.

Let magnitude of this charge be q. Then distribution of charge on various surfaces will be as shown in fig.

But the plates are metallic, therefore electric field inside the plates will be zero.

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XtraEdge for IIT-JEE 20 AUGUST 2010

+(10

–10 +

q)

+(2

× 10

–10 –

q)

–q +qp

Considering a point P inside the plate B, Electric field on it is

E = S2

)q10(

0

10

ε+−

– S2

q

0ε –

S2q

– S2

)q102(

0

10

ε−× −

= 0

or q = 5 × 10–11 coulomb or 50 pC Hence, the charges are as shown in fig.

150 pC

– – –

+ + +

+ + +

+ + +

50 pC150 pC

When battery is connected with the plates, a charge flows through the circuit. Due to flow of this charge, charges on inner surfaces are changed while charges on outer surfaces remain unchanged.

Let charge flowing through the battery be q´. Then charges on various surfaces become as shown in fig.

150 pC

– – –

+ + +

+ + +

+ + +

(50×10–12–q´) 150 pC

+ – q E

Capacitance of the capacitor is

C = dS0ε

= 5 × 10–12 F

Applying Kirchhoff's voltage law,

– C

´)q1050( 12 −× −

– E = 0

∴ q´ = 1 × 10–10 coulomb ∴ Energy supplied by battery = q´E = 10–9 joule Ans.

5. Nine identical capacitors, each of capacitance

C = 15 µF are connected as shown in fig. Calculate equivalent capacitance between terminals 1 and 4.

1 6 5

43 2

Sol. Given arrangement of capacitors is symmetric about mid-point of arm 3–6. If the arrangement is rotated through 180º about this point, given arrangement is obtained again. Let a battery of emf V be connected across terminals 1 and 4 of the arrangement. Then, in steady state, charges on various capacitors will be as shown in fig.

1 6 5

43 2

q2 –+

+ ++–

–+

+ – + –

+ –+ –

q1 (q2 – q3)(q1 + q2)(q1 + q2)

+ –

(q1 – q2 + 2q3)

q2 – q3 q1

q3

q3

q2

Applying Kirchhoff´s voltage law on mesh

1 – 2 – 6 –1,

Cq2 +

Cq3 –

Cq1 = 0

or q1 = (q2 + q3) ...(i) For mesh 2 – 3 – 6 – 2,

C

qq 32 − –

Cq2qq 321 +−

– Cq3 = 0

or q1 = (2q2 – 4q3) ...(2) From equation (1) and (2), q2 = 5q3 and q1 = 6q3 Now applying Kirchhoff's voltage law on mesh 1 – 6 – 5 – 4 – V – 1,

Cq1 +

Cqq 32 −

+ Cq2 – V = 0

Substituting q1 = 6q3 and q2 = 5q3, q3 = 151 CV.

But charge drawn by the arrangement from battery is

q = (q1 + q2) = 11q3 = 1511 CV

∴ Equivalent capacitance = Vq =

15C11 = 11µF Ans.

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Capacitors in Series :

V1 V2 V3

V

A

+Q –Q +Q –Q +Q –Q

C1 C2 C3B

In this arrangement of capacitor the charge has no

alternative path(s) to flow. (a) The charges on each capacitor are equal i.e. Q = C1V1 = C2V2 = C3V3 ...(1) (b) The total potential difference across AB is shared

by the capacitors in the inverse ratio of the capacitances.

V = V1 + V2 + V3 ...(2) If Cs is the net capacitance of the series combination,

then

sC

Q = 321 C

QCQ

CQ

++

⇒ sC

1 = 321 C

1C1

C1

++

Further V1 = 1C

Q and V = sC

Q

Capacitors in Parallel :

V

A

+Q1 –Q1

+Q2 –Q2

C1

C2

+Q3 –Q3

C3

B

In such an arrangement of capacitors the charge has

an alternative path(s) to flow (a) The potential difference across each capacitor is

same and equals the total potential applied. i.e. V = V1 = V2 = V3 ...(1)

⇒ V = 1

1

CQ =

2

2

CQ =

3

3

CQ

...(2)

(b) The total charge Q is shared by each capacitor in the direct ratio of the capacitances.

⇒ Q = Q1 + Q2 + Q3

If CpV is the net capacitance for the parallel combination of capacitors then

CpV = C1V + C2V + C3V ⇒ Cp = C1 + C2 + C3 Important terms : (a) If C1, C2, C3 .... are capacitors connected in series

and if total potential across all is V, then potential across each capacitor is

V1 =

s

1

C1

C1

V; V2 =

s

2

C1

C1

V; V3 =

s

3

C1

C1

V

and so on, where sC

1 = n321 C

1....C1

C1

C1

++++

(b) If C1, C2, C3 ... are capacitors connected in parallel and if Q is total charge on the combination, then charge on each capacitor is

Q1 =

p

1

CC Q; Q2 =

p

2

CC Q; Q3 =

p

3

CC

Q

and so on, where Cp = C1 + C2 + C3 + ... + Cn Energy Density : For a parallel plate capacitor

U = 21 CV2

where C = dA0ε

and V = Ed

εσ

=0

Ewhere

⇒ U = dA

21 0ε

E2d2

⇒ U =

ε 2

0E21 (Ad) ⇒ U =

21

ε0E2τ

where τ is volume of the capacitor

⇒ τU = Ue =

VolumeEnergy ticElectrosta

= Electrostatic Pressure

= 21

ε0E2 = 0

2

2εσ

εσ

=0

EQ

Capacitor-2 PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY

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Energy for series and parallel combinations : Series Combination : For a series combination of

capacitor Q = constant and

sC

1 = ...C1

C1

C1

321+++

⇒ s

2

C2Q =

1

2

C2Q +

2

2

C2Q +

3

2

C2Q + ....

⇒ Us = U1 + U2 + U3 + ...... Parallel Combination : For a parallel combination

of capacitors V = constant and Cp = C1 + C2 + C3 + ....

⇒ 21 CPV2 =

21 C1V2 +

21 C2V2 +

21 C3V2 + ...

⇒ Up = U1 + U2 + U3 + .... Electrostatic force between the plates of a parallel plate capacitor : The plates of the capacitor each carry equal and

opposite charges, hence they must attract each other with a force, say F.

+ + + + + +

– – – – – –

+Q –Q

At any instant let the plate separation be x, then

C = xA0ε

Also U = C2

Q2 ⇒ U =

ε A2Q

0

2x

Let the plates be moved towards each other through dx, such that the new separation between the plates is (x – dx). If Uf is the final potential energy, then

Uf = ´C2

Q2 =

A2Q

0

2

ε(x – dx)

If dU is the change in potential energy, then dU = Uf – Ui

⇒ dU = A2

Q

0

2

ε(x – dx) –

A2Q

0

2

εx

⇒ dU = – A2

Q

0

2

εdx

Further since

F = – dxdU

⇒ F = A2

Q

0

2

ε =

εσ

0

2

2A =

ε 2

0E21 A

εσ

=σ=0

E,AQQ

Kirochhoff's laws for capacitor circuits : Kirchhoff's first law or junction law : Charge can

never accumulate at a junction i.e. at the junction

∑q = 0

Important terms : This law is helpful in determining the nature of charge on an unknown capacitor plate. Charge on capacitor C can be determined by using this rule. As no charge must accumulate at the junction O, so if x is charge on plate 1 of C, then

–q1 + q2 + x = 0 ⇒ x = q1 – q2

+ – – +

+q1

–q1 +q2 –q2

B12

A

C

i.e. plate 1 has a charge (q1 – q2) and plate 2 has a charge –(q1 – q2).

Kirchhoffs second law or loop law : In a closed loop (a closed loop is the one which starts

and ends at the same point), the algebraic sum of potential differences across each element of a closed circuit is zero.

⇒ ∑V = 0

Conventions followed to apply loop law : (a) In a loop, across a battery, if we travel from

negative terminal of battery to the positive terminal then there is a potential rise and a +ve sign is applied with voltage of the battery.

(b) In a loop, across a battery, if we travel from positive terminal of the battery to the negative terminal then there is a potential fall and a –ve sign is applied with voltage of the battery.

(c) In a loop, across a capacitor, if we go from negative plate to the positive plate of the capacitor then there is a potential rise and a +ve sign is to be taken with potential difference across the

capacitor i.e. ∆V = + Cq .

(d) In a loop, across a capacitor, if we go from positive plate to the negative plate of the capacitor then there is a potential fall and a –ve sign is to be taken with the potential difference across the

capacitor i.e. ∆V = –Cq .

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Finding net capacitance of circuits : A. Simple Circuits : Analyse the circuit carefully to conclude which pair

of capacitors are in series and which are in parallel (This all should be done keeping in mind the points across which net capacitance has to be calculated). Find their net capacitance and again draw an equivalent diagram to apply the above specified technique repeatedly so as to get the total capacitance between the specified points.

B. Concept of line of symmetry : Line of symmetry (L.O.S.) is an imagination of our

mind to divide a highly symmetric circuit into two equal halves such that the points of the circuit through which LOS passes are at equal potential.

Solved Examples

1. Find the net capacitance of the circuit shown between the points A and B.

C

C

CC

C C

C

A B

Sol. This circuit is highly symmetric and so we can

consider the line of symmetry to pass through the circuit to divide it into two equal (identical) halves. If line of symmetry passes through a branch possessing a capacitor, then on each side of line of Symmetry the capacitance will become 2C (2C and 2C in series will gives C), as shown.

2C1

C

C

C

3

4 A P P C B

2C

CC

LOS Now, the concept of line of Symmetry makes our job

easy to calculate capacitance across AP. (1) and (2) are in parallel further in series with (3), whose resultant capacitance is in parallel with (4).

Resultant of (1) and (2) is 3C

Resultant of 3C and (3) is 4C3

Resultant of 4C3 and (4) is

4C7

So total capacitance across AB is

CAB = 2

CAP ⇒ CAB = 8C7

2. Find the equivalent capacitance between the point A and B in figure.

A

C2 C1

B C3

C2C1

Sol. Let us connect a battery between the points A and B.

The charge distribution is shown in figure. Suppose the positive terminal of the battery supplies a charge +Q and the negative terminal a charge –Q. The charge Q is divided between plates a and e.

A

C2 C1

BC3

C2C1Q1 –Q1 Q–Q1 –(Q–Q1)

Q1 –Q1Q–Q1 –(Q–Q1)

(2Q1–Q)–(2Q1–Q)

i

j

e f E g h

a b D

Let a charge Q1 goes to the plate a and the rest Q – Q1

goes to the plate e. The charge –Q supplied by the negative terminal is divided between plates d and h. Using the symmetry of the figure, charge –Q1 goes to the plate h (as it has a capacitance C1) and –(Q – Q1) to the plate d (as it has a capacitance C2). This is because if we look into the circuit from A or from B, the circuit looks identical. The division of charge at A and at B should, therefore, be similar. The charges on the other plates may be written easily. The charge on the plate i is 2Q1 – Q which ensures that the total charge on plates b, c and i remains zero as these three plates form an isolated system.

We have VA – VB = (VA – VD) + (VD – VB)

or VA – VB = 1

1

CQ +

2

1

CQQ − ...(1)

Also, VA – VB = (VA – VD) + (VD – VE) + (VE – VB)

or VA – VB = 1

1

CQ +

3

1

CQQ2 − +

1

1

CQ ...(2)

We have to eliminate Q1 from these equation to get

the equivalent capacitance )VV(

Q

BA −.

The first equation may be written as

VA – VB = Q1

21 C1

C1 +

2CQ

or 12

21

CCCC−

(VA – VB) = Q1 + 12

1

CCC−

Q ...(3)

The second equation may be written as

VA – VB = 2Q1

+

31 C1

C1 –

3CQ

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or )CC(2

CC

31

21

+(VA – VB) = Q1 –

)CC(2C

31

1

+Q ...(4)

Subtracting (4) from (3)

(VA – VB)

+

−− )CC(2

CCCC

CC

31

31

12

21

=

+

+− )CC(2

CCC

C

31

1

12

1 Q

or (VA – VB)[2C1C2(C1 + C3) – C1C3(C2 – C1)] = C1[2(C1 + C3) + (C2 – C1)]Q

or C = BA VV

Q−

= 321

133221

C2CCCCCCCC2

++++

3. Five identical conducting plates 1, 2, 3, 4 and 5 are

fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 the plate 4 are connected to a source of constant e.m.f. V0. Find

(i) The effective capacity of the system between the terminals of the source

(ii) the charge on plates 3 and 5. Given d = distance between any two successive

plates and A = are of either face of each plate. Sol. (i) The equivalent circuits is shown in fig. The

system consists of four capacitors. 5

4 32 1

(–)

(+)

(a)

(b)

1 2

3 2

3 4

5 4

(Q2/2)

(Q2/2)

Q2

Q1

Q

(–)(+)

i.e., C12, C32, C34 and C54. The capacity of each

capacitor is

εd

AK 0 = C0. The effective capacity

across the source can be calculated as follows : The capacitors C12 and C32 are in parallel and hence

their capacity is C0 + C0 = 2C0. The capacitor C54 is in series with effective capacitor of capacity 2C0. Hence the resultant capacity will be

00

00

C2CC2C

Further C34 is again in parallel. Hence the effective capacity

= C0 + 00

00

C2CC2C

= 35 C0 =

35 Kε0 d

A .

(ii) Charge on the plate 5 = charge on the uper half of parallel combination

∴Q5 = V0

0C32 =

dAVK

32 00ε

Charge on plate 3 on the surface facing 4

∴ V0C0 = dAVk 00ε

Charge on plate 3 on the surface facing 2 = [potential difference across (3 – 2)]C0

= V000

0

C2CC+

C0 = Kε0 d3AV0

∴ Q3 = dAVK 00ε

+ Kε0 d3AV0

= dAVK 00ε

+

311 =

34 Kε0 d

A V0

4. In diagram find the potential difference between the

points A and B and between the points B and C in the steady state.

B3µF

3µF 1µF

1µF

1µF20Ω

10Ω100 V

AC

Sol. The circuit is redrawn in fig (a, b, c)

1µF20Ω 10Ω

AC

3µF 1µF

3µF 1µF

B

100 VFig.(a)

1µF

20Ω 10Ω

A C100 V

Fig.(b)

6µF 2µF

P Q

B

1µF

20Ω 10Ω

A C100 V

Fig.(c)

R

P Q

3/2 µFS

From fig. (c). potential difference between P and Q = Potential difference between R and S = 100 volt

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∴ Q = capacity × volt = 23 × 10–6 × 100

= 150 × 10–6 coulomb Now according to fig.(b), the charge flowing through

capacitors of capacity 6 µF and 2 µF is 150 × 10–6 coulomb because they are connected in series.

Potential difference between A and B = Potential difference across the two ends of condenser of capacity 6 µF.

∴ V1 = capacity

Q = 6

6

10610150

×

× = 25 volt.

Again potential difference between C and D = potential difference across the two ends of condenser of capacity 2µF

V2 = 6

6

10210150

×× = 75 volt

5. Fig. shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

V A C B C

S

Sol. Initially the charge on either capacitor, i.e. qA or qB is

CV coulomb. When dielectric is introduced, the new capacitance of

either capacitor

C1 = KK1 C = 3C.

After the opening of switch S, the potential across capacitor A is volt.

Let the potential across capacitor B is V1 ∴ qB = CV = C1V1 or CV = 3CV1

∴ V1 = 3V volt

Initial energy of capacitor A = 21 CV2

energy of capacitor B = 21 CV2

∴ Total energy Ei = 21 CV2 +

21 CV2 = CV2

Final energy of capacitor A

= 21 × (3C)V2 =

23 CV2

Final energy of capacitor B

= 21 × (3C)

2

3V

=

6CV2

∴ Total final energy

Ef = 23 CV2 +

6CV2

= 35 CV2

∴ f

i

EE = 2

2

CV)3/5(CV =

53

Believe

• Four things for success: work and pray, think and believe.

• The future belongs to those who believe in the beauty of their dreams.

• I believe in being an innovator.

• You can do it if you believe you can!

• Believe deep down in your heart that you're destined to do great things.

• He who believes is strong; he who doubts is weak. Strong convictions precede great actions.

• I can't believe that God put us on this earth to be ordinary.

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

• When you believe in a thing, believe in it all the way, implicitly and unquestionable.

• To love means loving the unlovable. To forgive means pardoning the unpardonable. Faith means believing the unbelievable. Hope means hoping when everything seems hopeless.

• If you want to be confident, but don’t normally act that way, today, just this once, act in the physical world the way you believe a confident person would.

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Work, Energy and Power : Work is done when a force (F) is displaced.

F

dr

θ

The work done is dW = F dr cos θ Using vector notation

dW = rd.Frr

When the force and the displacement are in the same

direction, θ = 0, cos θ = +1, work done is positive. When the force and the displacement are in opposite

directions, θ = 180º, cos θ = –1, work done is negative.

When the displacement is perpendicular to the direction of the force, θ = 90º, cos θ = 0, no work is done.

For a system of particles the quantity ∫ cmxd.Frr

is

called pseudo work. At times actual work may be zero but not pseudo work.

Work is a scalar quantity. Its unit is joule. Power is the rate of doing work. Thus

Power = takentime

donework

The unit of power is the watt (= joules/second).

The power of an agent is given by P = v.Frr

where F is the force applied by the agent and v is the

velocity of the body on which the agent applies the force.

The energy of a system is its capacity of doing work. Mechanical energy may be of two types : (i) kinetic energy and (ii) potential energy.

The kinetic energy of a particle is T = 21 mv2.

The kinetic energy of a system is T = 2cmMv

21

The potential energy of particle in the gravitational field is given by

U = U0 + mgh where U0 = potential energy of the body at the

ground level. This is true only for objects near the surface of the

earth because g is uniform only near the surface of the earth.

The strain potential energy of a spring is given by U

= 21 kx2, where k is the force constant of the spring

and x is the charge in length of the spring. This change in length may be either a compression or on extension.

Potential Energy and force

Fx = –xU

∂∂

Principle of Conservation of energy : Conservative and Non-conservative Force : If the

work done by a force in moving a body from one point to another depends only on the positions of the body and not on the process or the path taken, the force is said to be conservative. Gravitational force, spring force, elastic forces, electric and magnetic forces are examples of conservative forces. If the work done depends on the paths taken, the force is said to be non-conservative. Frictional force is a non-conservative force.

Work-energy Theorem : The work by external forces on a body is equal to the change of kinetic energy of the body. This is true for both constant forces and variable forces (variable in both magnitude and direction).

For a particle W = ∆K. For a system of particles Wnet = Wreal + Wpseudo = ∆Kcm

Principle of Conservation of Energy : Energy can neither be created nor destroyed by any process.

For a particle K + U = a constant. For a system of particles Kcm + Uext + Eint = a constant.

However, energy can be transformed from one form into another.

Work, Power, Energy & Conservation Law

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY

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Collision of Bodies : Elastic Collision : When two bodies meet a with

certain relative speed they are said to collide with each other. In a collision, kinetic energy is transferred, from one body to another. When the transfer of kinetic energy takes place in such a way that the total kinetic energy is conserved, the collision is said to be perfectly elastic, or simply elastic. When kinetic energy is not conserved the collision is said to be inelastic. Further, in a collision, if one body gets embedded in the other and kinetic energy is not conserved, it is a completely inelastic collision.

In inelastic and completely inelastic collisions there is always a loss of kinetic energy and this energy is converted into other forms of energy, mostly heat.

A collision is said to be direct or head-on if the relative motion before and after the collision is in the same direction; if not it an oblique collision.

Remember the following points while solving problems on the collision of bodies.

(i) Apply the principle of conservation of momentum. In one-dimensional direct collisions, one equation is obtained by equating momenta before and after collision in the direction of motion. In two-dimensional collisions, select the line of impact as the X-axis and the line perpendicular to it as the Y-axis and obtain two equations by equation by equating momenta before and after the collision along the X- and Y-axes.

Remember that momentum is a vector quantity. It may be positive or negative depending on the direction. Choose any one direction as positive; the opposite will be negative.

(ii) If it is an elastic collision, apply the principle of conservation of kinetic energy. For inelastic collisions, apply the principle of conservation of energy to obtain an additional equation.

(iii) Remember there is no change in momentum along the common tangent to the colliding bodies.

Coefficient of restitution : According to Newton, the relative velocity of a body after collision is proportional to its relative velocity in the same direction before collision, with a reversal of sign.

Here, relative velocity means the velocity of any one of the colliding bodies (say A) with respect to the other colliding body (say B).

The constant of proportionality is called the coefficient of restitution (e).

That is VAB (after collision) = –e × V´AB (before collision)

This is Newton's law of collision. For elastic collisions, e = 1. For inelastic collisions,

e < 1

(A) Problem solving strategy :

Work and Kinetic Energy : Step 1 : Identify the relevant concepts : The work-

energy theorem is extremely useful in situations where you want to relate a body’s speed v1 at one point in its motion to its speed v2 at a different point. This approach is less useful for problems that involve time, such as finding the time it takes a body to go from point 1 to point 2. The reason is that the work-energy theorem. Wtot = K2 – K1, doesn’t involve time at all. For problems that involve time, it’s usually best to use the relationships among time, position, velocity, and acceleration

Step 2 : Set up the problem using the following steps :

Choose the initial and final positions of the body, and draw a free-body diagram showing all the forces that act on the body.

Choose a coordinate system. (If the motion is along a straight line, it’s usually easiest to have both the initial and final positions lie along the x-axis.)

List the unknown and known quantities, and decide which unknowns are your target variables. In some cases the target variable will be the body’s initial or final speed; in other cases it will be the magnitude of one of the forces acting on the body.

Step 3 : Execute the solution : Calculate the work done by each force. If the force is constant and the displacement is a straight line, you can use Eq. W = Fs cos φ or W = S.F

rr. (Latter in this chapter

we’ll see how to handle varying forces and curved trajectories.) Be sure to check the sign of the work for each force; it must be positive if the force has a component in the direction of the displacement, negative if it has a component opposite the displacement, and zero if the force and displacement are perpendicular.

Add the amounts of work done by each force to find the total work Wtot. Be careful with signs! Sometimes it may be easier to calculate the vector sum of the forces (the net force), then find the work done by the net force; this value is also equal to Wtot.

Write expressions for the initial and final kinetic energies, K1 and K2. Note that kinetic energy involves mass, not weight; if you are given the body’s weight, you’ll need to use the relationship W = mg to find the mass.

Finally, use the relationship Wtot = K2 – K1 to solve for the target variable. Remember that the right-hand side of this equation is the final kinetic energy minus the initial kinetic energy, never the other way around.

Step 4 : Evaluate your answer : Check whether your answer makes physical sense. A key item to

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remember is that kinetic energy K = 21 mv2 can never

be negative. If you come up with a negative value of K, you’ve made a mistake. Perhaps you interchanged the initial and final kinetic energies in Wtot = K2 – K1 or made a sign error in one of the work calculations.

(B) Problems using Mechanical Energy : Step 1 : Identify the relevant concepts : First decide

whether the problem should be solved by energy methods, by using amF

rr=∑ directly, or by a

combination of these strategies. The energy approach is particularly useful when the problem involves motion with varying forces, along a curved path (discussed later in this section), or both. But if the problem involves elapsed time, the energy approach is usually not the best choice, because this approach doesn’t involve time directly.

Step 2 : Set up the problem using the following steps :

When using the energy approach, first decide what the initial and final states (the positions and velocities) of the system are Use the subscript 1 for the initial state and the subscript 2 for the final state. It helps to draw sketches showing the initial and final states.

Define your coordinate system, particularly the level at which y = 0. You will use it to compute gravitational potential energies. Equation U = mgy (gravitational potential energy) assumes that the positive direction for y is upward; we suggest that you use this choice consistently.

Identify all no gravitational forces that do work. A free body diagram is always helpful. It some of the quantities you need are unknown, represent them by algebraic symbols.

List the unknown and known quantities, including the coordinates and velocities at each point. Decide which unknowns are your target variables.

Step 3 : Execute the solution : Write expressions for the initial and final kinetic and potential energies – that is K1, K2, U1, and U2. In general some of these quantities will be known and some will be unknown. Then relate the kinetic and potential energies and the no gravitation work Wother using eq.

K1 + U1 + Wother = K2 + U2

(you will have to calculate Wother in terms of the nongravitational forces.) If there is no nongravitational work, this expression becomes eq.

21 mv1

2 + mgy1 = 21 mv2

2 + mgy2

It’s helpful to draw bar graphs showing the initial and final values of K, U, and E = K + U. Then solve to find whatever unknown quantity is required.

Step 4 : Evaluate your answer : Check whether your answer makes physical sense. Keep in mind, here and in later sections, that the work done by each force must be represented either in U1 – U2 = –∆U, so make sure you did not include in ∆U, so make sure you did not include it again in Wother.

(C) Problem solving strategy : Conservation of Momentum : Step 1 : Identify the relevant concepts : Before

applying conservation of momentum to a problem, you must first decide whether momentum is conserved. This will be true only if the vector sum of the external forces acting on the system of particles is zero. If this is not the case, you can’t use conservation of momentum.

Step 2 : Set up the problem using the following steps :

Define a coordinate system. Make a sketch showing the coordinate axes, including the positive direction for each. Often it is easiest to choose the x-axis to have the direction of one of the initial velocities. Make sure you are using an inertial frame of reference. Most of the problems in this chapter deal with two-dimensional situations, in which the vectors have only x- and y-components; all of the following statements can be generalized to includes-components when necessary.

Treat each body as a particle. Draw “before” and “after” sketches, and include vectors on each to represent all known velocities. Label the vectors with magnitudes, angles, components, or whatever information is given, and give each unknown magnitude, angle, or component an algebraic symbol. You may find it helpful to use the subscripts 1 and 2 for velocities before and after the interaction, respectively; if you use these subscripts, use letters (not numbers) to label each paritcle.

As always, identify the target variable(s) from among the unknowns.

Step 3 : Execute the solution as follows : Write an equation in terms of symbols equating

the total initial x-component of momentum (that is, before the interaction) to the total final x-component of momentum (that is, after the interaction), using px = mvx for each particle. Write another equation for the y-components, using py = mvy for each particle. Remember that the x- and y-components of velocity or momentum are never added together in the same equation ! Even when all the velocities lie along a line (such as the x-axis), the components of velocity along this line can be positive or negative; be careful with signs !

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Solve these equations to determine whatever results are required. In some problems you will have to convert from the x-and y-components of a velocity to its magnitude and direction, or the reverse.

In some problems, energy considerations give additional relationships among the various velocities.

Step 4 : Evaluate your answer: Does your answer make physical sense ? If your target variable is a certain body’s momentum, check that the direction of the momentum is reasonable.

Solved Examples

1. A bus of mass 1000 kg has an engine which produces a constant power of 50 kW. If the resistance to motion, assumed constant is 1000 N, find the maximum speed at which the bus can travel on level road and the acceleration when it is traveling at 25 m/s.

Sol. At maximum speed all the power is used to overcome the resistance to motion. Hence if the maximum speed is v, then

50,000 = 1000 × v or v = 50 m/s The maximum speed = 50 m/s At 25 m/s, let the pull of the engine be P. Then the

power 50,000 = P × 25

or P = 25

50000 = 2000 N

1000 N 2000 N

a Now resultant force = 2000 – 1000 = 1000 N Applying Newton's law, F = ma, we have 1000 = 1000 a or a = 1.0 m/s2

2. A man is drawing water from a well with a bucket which leaks uniformly. The bucket when full weighs 20 kg and when it arrives the top only half the water remains. The depth of the water is 20 metres. If g = 10 m/sec2, what is the work done ?

Sol. When the bucket arrives at the top, the mass is 10 kg. Hence loss in mass = 20 – 10 = 10 kg. The depth of the well is 20 metres.

∴ mass lost per unit distance = 2010 =

21 kg

Consider a point at a height x from the bottom of the well.

At height x from the bottom, the bucket weighs =

2x20 kg. The work done against the force during

elementary displacement dx =

2x20 dx.g

∴ Total work done

W = ∫

20

0 2x20 g dx = g ∫

20

0xdx

21dx20

= g

−20

0

2200 4

xx20

= g[400 – 100] = 300g = 300 × 10 = 3000J 3. A body of mass m is thrown at an angle α to the

horizontal with an initial velocity v0. Find the mean power developed by gravity over the whole time of motion of the body and the instantaneous power of gravity as a function of time.

Sol. We know that Pinstan = F.v The velocity of the particle after time t is given by v = v0 cos α i + (v0 sin α – g t)j) and F = – mg j ∴ Pinstan. = (–mg j) . v0 cos α i + (v0 sin α – g t)j = – mg(v0 sin α – g t) The average power is given by

(P) =

∫∫

T

0

T

0

dt

dt)t(P=

T

dt)gtsinv(mgT

00∫ −α−

= T

]2/gTTsinv[mg 20 −α−

Here T is total time of flight T = (2 v0 sin α)/g Substituting this value, we have

(P) = – mg

αα−αα

g/)sinv2()g2/sinv4(gg/)sinv2(sinv

0

222000

Solving we get (P) = 0 4. Two blocks of masses m1 = 2 kg and m2 = 5 kg are

moving in the same direction along a frictionless surface with speeds 10 m/s and 3m/s respectively, m2 being ahead of m1. An ideal spring with k = 1120 nt/m is attached to the back side of m2. Find the maximum compression of the spring when the blocks collide.

Sol. The situation is shown in fig.

u1

m1 m2

u2

Let v be the speed of the system after collision. Applying the law of conservation of energy, we have m1u1 + m2u2 = (m1 + m2)v Substituting the given values (2 × 10) + (5 × 3) = (2 + 5)v

v = 7

1520 + = 5 m/s

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Now applying the law of conservation of kinetic energy, we get

211um

21 + 2

22um21 =

21 (m1 + m2)v2 +

21 kx2

or m1u12 + m2u2

2 = (m1 + m2)v2 + kx2 (2 × 100) + (5 × 9) = (7 × 25) + (1120 × x2) 200 + 45 = 175 + 1120 x2

x2 = 1120

17545200 −+ = 161

x = 41 = 0.25 m

5. A wooden block of mass 10g is dropped from the top

of a cliff 100 metres high. simultaneously, a bullet of mass 10 g is fired from the foot of the cliff vertically upwards with a velocity of 100 m/sec.

(i) Where and after what time will they meet ? (ii) If the bullet, after striking the block, gets

embedded in it, how high will it rise above the cliff before it starts falling ?

Sol. (i) Let the wooden block and bullet meet after a time t seconds. The distance s1 moved by the block is given by

s1 = 21 gt2 ...(1)

The distance s2 moved by the bullet in time t second is given by

s2 = ut – 21 gt2 = 100t –

21 gt2 ...(2)

Adding eqs. (1) and (2) s1 + s2 = 100 t or 100 = 100t (Q s1 + s2 = 100 m) ∴ t = 1 sec.

Now s1 = 21 gt2 =

21 × 9.8 × 1 = 4.9 m

Thus the two meet after 1 sec. at distance of 4.9 m from the top of the cliff.

(ii) The velocity of the block before impact u1 = 0 + gt = 9.8 m/s The velocity of the bullet before impact u2 = u – gt ∴ u2 = 100 – (9.8 × 1) = 90.2 m/s Let after the impact. v be the velocity of combined

mass. Applying the law of conservation of linear momentum, we have

m1u1 + m2u2 = (m1 + m2)v 10 × 9.8 + 10 × (– 90.2) = (10 + 10)v Here we have taken the velocity positive in

downward direction.

v = 2

2.908.9 − = – 40.2 m/s

The negative sign shows that the velocity of combined mass is in the upward direction.

The height to which the combined mass rises after impact

= g2

v2 =

8.92)2.40( 2

×= 82.45 m

The height to which it rises above the cliff is = 82.45 – 4.90 = 77.5 m

Regents Physics You Should Know

Modern Physics : • The particle behavior of light is proven by the

photoelectric effect. • A photon is a particle of light wave packet. • Large objects have very short wavelengths

when moving and thus can not be observed behaving as a wave. (DeBroglie Waves)

• All electromagnetic waves originate from accelerating charged particles.

• The frequency of a light wave determines its energy (E = hf).

• The lowest energy state of a atom is called the ground state.

• Increasing light frequency increases the kinetic energy of the emitted photo-electrons.

• As the threshold frequency increase for a photo-cell (photo emissive material) the work function also increases.

• Increasing light intensity increases the number of emitted photo-electrons but not their KE.

Mechanics : • Centripetal force and centripetal acceleration

vectors are toward the center of the circle- while the velocity vector is tangent to the circle.

• An unbalanced force (object not in equilibrium) must produce acceleration.

• The slope of the distance-tine graph is velocity. • The equilibrant force is equal in magnitude but

opposite in direction to the resultant vector. • Momentum is conserved in all collision

systems. • Magnitude is a term use to state how large a

vector quantity is.

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SUBSTITUTION REACTION

In a substitution reaction one atom or group of atoms in a molecule is replaced by another. They are known to proceed by a free radical or ionic mechanism.

Free Radical Substitution : CH4 + Cl2 → νh CH3Cl + HCl

+ Br2 → νh Br

+ HBr

CH2 = CHCH3 + NBS → CH2 = CHCH2Br C—H bond is replaced by a C—X bond in free

radical halogenation. Reaction takes place through free radical intermediates and is thus a homolytic substitution reaction.

CH3CH3 + Cl2 → CH3CH2Cl + HCl

Step I (initiation) Cl—Cl → Cl + Cl

• •

Step II (propagation)

CH3CH2—H + Cl → CH3CH2 + HCl• •

CH3CH2+Cl—Cl → CH3CH2Cl + Cl• •• •

Step III (termination) Cl + Cl → Cl2 • •

CH3CH2 + Cl → CH3CH2Cl• •

CH3CH2 + CH2CH3

• •

→coupling CH2CH2CH2CH3

CH3CH2 + CH3CH2 • •

→ ionationdisproport CH2 = CH2 + CH3 —CH3 Reactivity of the halogens for free radical substitution

is in order F2 > Cl2 > Br2 > I2

For a given halogen, abstraction of H is in the following order :

allylic 3º > allylic 2º > 2º > 1º (CH3) > vinylic

For

H H H H H CH2—H 1º

H H H H

Allylic 2º

vinylic

Allylic 3º

Abstraction of H from allylic carbon or benzylic carbon takes place using NBS (N-bromosuccinimide) in which weak nitrogen-bromine bond can be cleaved homolytically into radical upon warming or exposure to visible light.

∆ →NBS

Br

ν →h

NBS

Br

IONIC SUBSTITUTION Substitution reaction may be brought about by (a) replacement by electrophiles-called Electrophilic

Substitution reaction (SE) (b) replacement by nucleophiles-called Nucleophilic

Substitution reaction (SN) Electrophilic Substitution Reaction (SE) This SE reaction takes place in benzene nucleus

(aromatic compounds) in which π elelctrons are highly delocalised and an electrophile can attack this region of high elelctron density.

+ E+ → E

E

E

+ CH3Cl → 3AlCl

CH3

+ HCl

Step I. Formation of an electrophile

CH3Cl + AlCl3 → 3–4 HCAlCl

⊕+

Step II. Attack of electerophile on benzene when resonance-stabilised σ complex is formed.

+ CH3

H ⊕

CH3

H

⊕ CH3

(carbocation-an arenium ion)

HCH3⊕

≡ H

CH3 +

σ-complex Step III. Loss of H+ from σ complex to form end

product H⊕

CH3 →

CH3

+ H+

Step IV. AlCl4– + H+ → AlCl3 + HCl

Organic Chemistry

Fundamentals

REACTION MECHANISM

KEY CONCEPT

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An additional reagent (called Lewis acid) is always required that can help in the formation of an electrophilic

+ NHO3 → 42SOH

NO2

+ H2O

+ Br2 →Fe

Br

+ HBr

Nitration would not take place in the absence of H2SO4 . It helps in the formations NO2

+ (nitronium ion). Similarly Br+ electrophile is formed when Fe or FeCl3 is present.

H ⊕

CH3 + ΘuN →

H CH3

H Nu

If there is attack of nucelophile on positive site,

aromaticity is lost, hence this is not desirable. Nucleophilic Substitution Reactions When a substitution reaction is brought about by a

nucleophile, the reaction is called SN (nucleophilic (N) substitution (S) reaction.

R—X + OHΘ → R—OH + XΘ Substrate nucleophile leaving group

1NS Reaction If rate of subsititution depends on the concentration

of the substrate, then it is said to be the Unimolecular (1) Nucleophilic (N) Substitution (S) reaction, written as 1N

S

R — X Slow R⊕ + XΘ

Carbocation

dtdx = k[R—X]

In this R — X bond breaking slow, is rate determining step.

R⊕ + OHΘ →fast R — OH Carbocation formed can undergo rearrangement to

give more stable carbocation before attack of the nucleophile

CH3CCH2Cl |

| CH3

CH3

slow

S 1N →

CH3CCH2 |

| CH3

CH3 ⊕

→ − shiftmethyl2,1

1º (less stable)

CH3CCH2CH3

|CH3

⊕ 3º (more stable)

CH3CO + CH3CCl || O

Θ |

| CH3

CH3

→ COHCH||O

3

CH3CO—CCH3 + Cl||

CH3 Θ |

CH3

O

|

Rate = k [t-BuCl]

The fact that the rate law depends only on the concentration of tert-butyl chloride means that only tert-butyl chloride is present in the transition state that determines tha rate of the reaction. There must be more than one step in the mechanism because the acetate ion must bnot be involoved until after the step with this transition state. Because only one molecule (tert-butyl chloride) is present in the step involving the transition state that determines the state of reaction, this step is said to be Unimolecular. The reaction is therefore, described as a 1N

S reaction.

CH3C—Cl |

|CH3

CH3

Istep

slow →

CH3C ⊕ + Cl|

| CH3

CH3 Θ

CH3C ⊕ + O—CCH3 |

|CH3

CH3||O

Θ

IIstep

fast→

CH3C—O—CCH3 |

|CH3

CH3||O

As a result of SN1 reaction, there can be

recemisation and inversion.. When (–) 2-bromo-butane having chiral centre is treated with low [OH–] such that SN1 reaction is followed, (+) 2-butanol is obtained. There is also loss in activity. This loss in optical activity is due to formation of d-and l-isomers by SN1 reaction.

CH3 H

CH2CH3 BrΘ

H2O• •••

CH3

H

CH2CH3

HO

CH3

HBr

CH2CH3(–)

CH3

H

CH3CH2

OH

(+) Inversion(–) Retention

II I

I II

Nucleophilic reagent attacks both (I) backside and

(II) front side of the carbocation. In back side attack (I), configuration is retained but in front attack (II), inversion takes place. There can be recemisation if d-and l-are formed in equal amounts. Attack is preferred on the side opposite to where leaving group Br– exists since it shields the passage of nucleophile for attack. (In this case front attack is preferred, hence mixture is not purely racemic and some optical actvity exists)

In SN1 reaction the order of reactivity of RX is Allyl or benzyl > 3º > 2º > 1º > CH3X SN2 Reaction If substrate nucleophile bothe are involved in the

rate-determining step then this is called as

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Bimolecular (2) Nucleophilic (N) Substitution (S) reaction indicated as SN2.

R—X + OH– → ROH + X–

dtdx = k[R—X] [OH–]

Step I. OH– attacks the side opposite to that where halide exists to give an intermediate (transition state)

H—C—Br + OH– |

| H

H

→slow

HO C Brδ–δ– HH

H

Intermediate (A) Step II. HBr is stronger acid than H2O therefore, Br–

is a better leaving group than OH–, hence Br–, is lost from the intermediate

Intermediate (A) →fast HO—C

HHH

How do we confirm this type of attack? Again we take (+)-2-bromo-butane (with one chiral carbon). It is treated with conc. aq. KOH (OH– being strong nucleophile). We get (–)-2-butanaol and no (+) isomer. There is thus complete stereochemical inversion.

H—C—Br + OH– |

| CH2CH3

CH3

→slow

HO C Br

δ–δ– CH3H

CH2CH3

Θ→Br–

fast

HO—C—H |

| CH2CH3

CH3

(–) For SN2 reaction the order of reactivity is

CH3X > 1º > 2º > 3º (alkyl halide)

CH2Br > CHBr

CH3

> CBr

CH3

CH3

Thus, in SN1 reaction, recemisation as well as

inversion is observed, while in case of SN2 reaction, completee inversion takes place (where chiral carbon exists).

Rearrangement of the carbocation (formed in SN1 reaction) leading to more stable carbocation is also observed in SN1 reaction (solved example)

SN2 reaction RX = CH3X 1º 2º 3º

SN1 reaction CH3X > 1º > 2º > 3º SN2 SN2 mixed SN1

High concentration of the nucleophile favours SN2 reaction while low concentration favours SN1 reaction.

CH3CCH2Br|

|CH3

CH3

shows SN2 reaction with C2H5O–, but

SN1 reaction with C2H5OH. The higher the polarity of the solvent, the greater is

the tendency for SN1 reaction Elimination VS Substitution

CH3CCH2CH3

|

CH3

OH– SN CH3CCH2CH3

| CH3

| OH

CH3C=CHCH3

|CH3

E2 OH–

In the above example we find that a carbocation can show SN2 as well a E2 reaction. Where substitution and elimination are competing reaction, the proportion of elimination increases as the structure of alkyl halide is changed from primary to secondary to tertiary.

Many tertiary alkyl halides yield exclusively alkenes

under these conditions Elimination increases

Substitution increases RX = 1º 2º 3º

CH3CBr + C2H5ONa|

|CH3

CH3

→ 2E

CH3C ||

| CH3

CH2

CH3C—ONa + C2H5Br |

|CH3

CH3

→ 2SN

CH3C—OC2H5

|

|CH3

CH3

The general pattern of reactivity expected from

various structural classes of alky halides in reactions with a representative range of nucleophiles (which may behave as bases).

Effect of Solvent SN1 reaction are faster in the more polar solvents. SN2 reaction involving a negative nucleophile is

slower in more polar solvent and that involving a neutral nucleophile is faster in more polar solvent.

In additions to these polarity effects, the ability of certain solvents to form hydrogen bonds to the nucleophile also affects the rate of the SN2 reaction. Such solvents are termed protic solvents and have a hydrogen bonded to nitrogen or oxygen (H2O, ROH, RCOOH are some examples of protic solvents).

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Density of Cubic Crystals : The density based on the structure can be calculated

from the mass contained in a unit cell and its volume. If N is the number of molecules per unit cubic cell of edge length a, then the mass and volume per unit cell are

Mass =

ANM N Volume = a3

Therefore, Density = volumemass =

A3NaNM

The value of N for the three cubic cells can be calculated as follows :

Primitive cubic cell : In a primitive cubic cell, atoms are present at the corners of the cube. There are eight corners of a cube and thus eight atoms are present at these corners. Now, any particular corner of the cube is actually shared amongst eight such cubic unit cells placed adjacent to one another. Thus, the contribution of the atom placed at one of the corners to the single cubic unit cell is 1/8. Since there are eight corners of a cube, the number of atoms associated with a single primitive unit cell is 8/8 = 1.

Body-centred cubic cell : In a body-centred cubic unit cell, besides atoms being present at the corners, there is one atom in the centre of the cube which belongs exclusively to this cubic unit cell. Therefore, number of atoms per unit cell are two.

Face-centred cubic cell : Here, atoms, besides being at the corners, are also present at the centre of the six faces. Each of these atoms is shared between two such unit cells. Thus, their contribution to the unit cell is 6/2 = 3 atoms, making a total of 4 atoms per cubic unit cell

Packing in a simple Cubic Lattice : In a lattice of this type, the spheres are packed in the

form of a square array by laying down a base of spheres and then piling upon the base other layers in such a way that each sphere is immediately above the other sphere, as shown in fig.

Packing in a simple cubic lattice

In this structure, each sphere is in contact with six nearest neighbours (four in the same base, one above and one below). The percentage of occupied volume in this structure can be calculate as follows:

The edge length a of the cube will be twice the radius of the sphere, i.e. a = 2r. Since in the primitive cubic lattice, there is only one sphere present in the unit lattice, the volume occupied by the sphere is

V = 34

πr3 or V = 34

π3

2a

The fraction of the total volume occupied by the sphere is

φ = 3

3

a2a

34

π

= 6π = 0.5236

or 52.36 percent Thus, the structure is relatively open since only

52.36% (π/6) of the total volume is occupied by the spheres. The remainder, i.e. 0.4764 of the total volume is empty space or void volume.

No crystalline element has been found to have this structure.

Closest Packing : In closest packing arrangements, each sphere is in

contact with the maximum possible number of nearest neighbours. Fig. shows a closest packed layer of spheres. Each sphere is surrounded by six nearest neighbours lying in the plane, three spheres Just above it and three below it, thus making the total number of nearest neighbours equal to twelve.

If the spheres are packed in the same plane, then just above these spheres

A

B

A A A

B B B

C C C CA A A

A A AA

A

A

A

B B

Fig. (a) Closest packed layers of spheres

B

C C C C

Physical Chemistry

Fundamentals

SOLID STATE

KEY CONCEPT

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A BA

B

A B A

B

A

CB

A

CBA

Fig. (b) Two types of packing there exist two different types of voids, pointing in

different directions as shown in fig. (a). Thus, we can have three different types of locations as shown by A, B and C in fig. (a). Location A is occupied by the spheres while B and C are the two different types of voids. But because of the size of the spheres, both types of voids cannot be occupied simultaneously.

The third layer of closest-packed sphere can be formed in two different ways. If, for example, we choose to place the spheres of the second layer in B sites, one of the available sets of voids for the third layer will be directly above the spheres in the original layer. These are A sites. The other set of voids will be directly above the voids designated by C in the original layer.

Types of Packing : Thus, two types of packing (fig. b) are possible ABABA.... or ABCABC .... We can have many other varieties of patterns such as

ABCACB....., ABAC .... etc. But for many of the common substances that form closest-packed structures, one of the above two symmetrical arrangements is observed.

Hexagonal Closest Packed Structure : The packing ABAB.... is known as a hexagonal

closest-packed structure (HCP). The unit cell of shown in figure.

A

B

A

Exploded view Hexagonal closest -packed

Unit cell formed by ABA packing

The fraction of the volume occupied in HCP can be calculated as described in the following.

The distance C/2 (in figure) is the distance between the layers A and B. This distance will be from the centre of a sphere to the plane of the three spheres that are in contact with it. This distance can be

determined by reference to a face centred cubic lattice with unit cell length a. In such a lattice, the distance between closest-packed layers (Miller indices 111)is one –third of the body diagonal, i.e.

3 a./3.

Thus, 2C =

3a3

Layer A

Layer B

Layer A

C/2

a2r

Hexagonal closest-packed structure

Now, in the face-centred lattice spheres touch one another along the face diagonal. Thus, we have

4r = 2 a or a = 2

4 r

With this, the distance C becomes

C = 2

a

33 = 2

2r4

33 =

68 r

The hexagonal base consists of six equilateral triangles, each with side 2r and with an altitude of 2r sin 60º, i.e. 3 r. Therefore,

Area of the base = 6 ( )

)r2(r3

21 = 36 r2

Volume of the prism = ( )2r36

r

68 = 24 2 r3

Number of spheres belonging to this prism 3 spheres in B layers exclusively belong to this prism. 1 from the centre of the base. There are two spheres

of this type and each is shared by two prisms. 2 from the corners. There are twelve such spheres

and each is shared amongst six prisms of this type. Thus, the total number of spheres is 6. The fraction of volume of the prism actually

occupied by the spheres is

3

3

r224

r346

π

= 62π = 0.7405

or 70.05 percent Example of HCP are Ca, Cd, Cr, Mg and Zn.

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Cubical Closest-Packed Structure The packing ABCABC, .... is a cubical closest-

packing (CCP) or face-centred cubic packing. The fraction of volume occupied in CCP can be calculated as follows :

The radius of the sphere in terms of the unit length of the face-centred cube is given by

r = 4

a2

since the sphere will be touching each other along the diagonal of the face of the cube.

In the face-centred cubic lattice, there are four spheres per unit cell. Therefore, fraction of volume occupied by the spheres is

3

3

a

4a2

344

π

= 62π = 0.7405

or 74.05 percent

A

C

B

Exploded view Cubical closest-packed stricture

Face-cented cubic unit cell formed by ABCA packing

A

out of all these packings, HCP and CCP are more

common for uniform spheres. In general, the packing fraction, i.e. fraction of

volume occupied, is independent of the radius of the sphere and depends only on the nature of packing. From the values of packing fractions, it follows that the density of a substance in HCP and CCP structures will be more than in the other two packings.

Packing in a Body Centred Cubic Lattice : Here the packing consists of a base of spheres,

followed by a second layer where each sphere rests in the hollow at the junction of four spheres below it, as shown in figure.

Packing in a body-centredcubic lattice

The third layer then rests on these in arrangement which corresponds exactly to that in the first layer. In

this arrangement, spheres are touching one another along the cross diagonal of the cube, making its distance equal to 4r. This must be equal to 3 a.

Thus, 4r = 3 a,

i.e., r = 43 a

Volume of the cube = a3

Volume of one sphere = 34

πr3 = 34

π3

a43

Since there are two spheres in each unit cell, the total volume occupied will be

2

π

3

a43

34

The fraction of the volume occupied by the spheres

φ = 3

3

a

a43

342

π

= 83π = 0.6802

or 68.02 percent In this arrangement each sphere has eight nearest

neighbours

PERSONAL GROWTH • The searching-out and thorough investigation of

truth ought to be the primary study of man. • The only journey is the journey within. • Know thyself means this, that you get

acquainted with what you know, and what you can do.

• Yes, know thyself: in great concerns or small, Be this thy care, for this, my friend, is all.

• Collect as precious pearls the words of the wise and virtuous.

• If we do not plant knowledge when young, it will give us no shade when we are old.

• If you have an hour, will you not improve that hour, instead of idling it away?

• Follow your honest convictions, and stay strong.

• He that will not reflect is a ruined man. • Every day do something that will inch you

closer to a better tomorrow. • God ever works with those who work with

will. • Insist on yourself. Never imitate. • Heaven never helps the man who will not act.

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1. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ......... .

Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ? [IIT-1996]

Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in figure (i) & (ii). Three such cells form one hcp unit.

For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside,

Number of atoms per unit cell = 88 + 1 = 2

Area of the base = b × ON [From fig.(ii)] = b × a sin 60º

= 23 a2 (Q b = a)

a γ b α

Figure (i)

Volume of the hexagonal cell

= Area of the base × height = 23 a2.c

But c = 322 a

∴ Volume of the hexagonal cell

= 23 a2 .

322 a = a3 2

and radius of the atom, r = 2a

Hence, fraction of total volume or atomic packing factor

= cell hexagonal theof Volume

atoms 2 of Volume

60º

N b

O

a

figure (ii)

= 2a

r342

3

3π× =

2a2a

342

3

3

π×

= 23

π = 0.74 = 74%

∴ The percentage of void space = 100 – 74 = 26% 2. The vapour pressure of ethanol and methanol are 44.5

mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. [IIT-1986]

Sol. Given that, For ethanol (C2H5OH), 0

eP = 44.5 mm Hg M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g

∴ Moles of ethanol, ne = Mm =

4660 = 1.3

For methanol (CH3OH), 0mP = 88.7 mm Hg

M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g

∴ Moles of methanol, nm = Mm =

3240 = 1.25

∴ xe = me

e

nnn+

= 25.13.1

3.1+

= 55.23.1

xm = me

m

nnn+

= 25.13.1

25.1+

= 55.225.1

According to Raoult's law,

Pe = 0eP xe =

55.23.15.44 × = 22.69 mm Hg

and Pm = 0mP xm =

55.225.17.88 × = 43.48 mm Hg

Hence, total vapour pressure of the solution, PT = Pe + Pm = 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law,

UNDERSTANDINGPhysical Chemistry

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XtraEdge for IIT-JEE 43 AUGUST 2010

Pm = PTx´m (in vapour form) Hence, mole fraction of methanol in vapour form,

x´m = T

m

PP =

17.6648.43 = 0.66

3. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 g ml–1. Sulphuric acid density 1.294 g ml–1 is 39% H2SO4 by weight and that of density 1.139 g ml–

1 is 20% H2SO4 by weight. The battery holds 3.5 L of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere hours for which the battery must have been used. The charging and discharging reactions are :

Pb + −24SO → PbSO4 + 2e– (Charging)

PbO2 + 4H+ + −24SO + 2e– → PbSO4 + 2H2O

(Discharging) [IIT-1986]

Sol. We know, Volume = Density

Mass

Volume of 100 g of 39% H2SO4 = 294.1

100 = 77.28ml

77.28 ml of H2SO4 acid solution contains = 39 g H2SO4 ∴ 3.5 L of sulphuric acid solution contains

= 28.77

5.3100039 ×× = 1766.3 g H2SO4

Volume of 100 g of 20% H2SO4 = 139.1

100 = 87.79ml

87.79 ml of H2SO4 acid solution contains = 20 g H2SO4 ∴ 3.5 L of sulphuric acid solution contains

= 79.87

5.3100020 ×× = 797.3 g H2SO4

∴ Amount of H2SO4 used during discharge = 1766.3 – 797.3 = 969 g The overall reaction is Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O 98g H2SO4 (2 × 1 + 32 + 4 × 16) requires = 1 F electricity

∴ 969 g H2SO4 requires = 981 × 969 F

= 981 × 969 × 96500 amp-sec

= 981 ×

360096500969× amp-hr

= 265 amp-hour 4. It is possible to supercool water without freezing. 18

g of water are supercooled to 263.15 K(–10ºC) in a thermostat held at this temperature, and then crystallization takes place.

Calculate ∆rG for this process. Given: Cp(H2O,1) = 75.312 J K–1 mol–1

Cp (H2O,s) = 36.400 J K–1 mol–1 ∆fusH (at 0ºC) = 6.008 kJ mol–1

Sol. The process of crystallization at 0ºC and at 101.325 kPa pressure is an equilibrium process, for which ∆G = 0. The crystallization of supercooled water is a spontaneous phase transformation, for which ∆G must be less than zero. Its value for this process can be calculated as shown below.

The given process H2O(1, – 10ºC) → H2O(s, –10ºC) is replaced by the following reversible steps. (a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1)

∆rH1 = ∫K15.273

K15.263m,p )1(C dT

= (75.312 J K–1 mol–1 ) (10 K) = 753.12 J mol–1

∆rS1 = ∫K15.273

K15.263

m,p

R)1(C

dT

= (75.312 J K–1mol–1) × ln

K15.263K15.273

= 2.809 J K–1 mol–1 (b) H2O(1, 0ºC) → H2O(s, 0ºC) ...(2) ∆rH2 = – 6.008 kJ mol–1

∆rS2 = – )K15.273(

)molJ6008( 1– = – 21.995 J K–1 mol–1

(c) H2O(s, 0ºC) → H2O(s, –10ºC) ...(3)

∆rH3 = ∫K15.263

K15.273m,p )s(C dT

= (36.400 J K–1 mol–1)(–10 K) = – 364.0 J mol–1

∆rS3 = ∫K15.263

K15.273

m,p

T)s(C

dT

= (36.400 J K–1 mol–1) ×ln

K15.273K15.263

= – 1.358 J K–1 mol–1 The overall process is obtained by adding Eqs. (1),

(2) and (3), i.e. H2O(1, –10ºC) → H2O(s, –10ºC) The total changes in ∆rH and ∆rS are given by ∆rH = ∆rH1 + ∆rH2 + ∆rH3 =(753.12 – 6008 – 364.0) J mol–1

= – 5618.88 J mol–1 ∆rS = ∆rS1 + ∆rS2 + ∆rS3 = (2.809 – 21.995 – 1.358) J K–1 mol–1 = – 20.544 J K–1 mol–1 Now ∆rG of this process is given by ∆rG = ∆rH – T∆rS = – 5618.88 J mol–1 – (263.15 K)( –20.544 J K–1 mol–1 ) = – 212.726 J mol–1

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XtraEdge for IIT-JEE 44 AUGUST 2010

5. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2.

(a) How many degrees of freedom do the gas molecules have ?

(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990]

Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1

γ–1 = T2V2γ–1

Here, T1 = T ; T2 = T/2 V1 = V and V2 = 5.66 V

Hence, TVγ–1 = 2T × (5.66V)γ–1

= 2T × (5.66)γ–1 × Vγ–1

or (5.66)γ–1 = 2 ...(1) Taking log, (γ – 1)log 5.66 = log 2

or γ – 1 = 66.5log2log =

7528.03010.0 = 0.4

or γ = 1.4 If f, be the number of degrees of freedom, then

γ = 1 + f2 or 1.4 = 1 +

f2

or f2 = 1.4 – 1 = 0.4

or f = 4.0

2 = 5

(b) According to adiabatic gas equation, P1V1

γ = P2V2γ

Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ

or P2 = γ)66.5(

P = 4.1)66.5(P =

32.11P [using eq.(1)]

Hence, work done by the gas during adiabatic expansion

= 1–

VPVP 2211

γ− =

1–4.1

V66.532.11

PPV ×−

= 4.0

2PVPV −

= 4.02

PV×

= 1.25 PV

Galena

Galena is the natural mineral form of lead sulfide. It is the most important lead ore mineral.

Galena is one of the most abundant and widely distributed sulfide minerals. It crystallizes in the cubic crystal system often showing octahedral forms. It is often associated with the minerals sphalerite, calcite and fluorite. Galena deposits often contain significant amounts of silver as included silver sulfide mineral phases or as limited solid solution within the galena structure. These argentiferous galenas have long been the most important ore of silver in mining. In addition zinc, cadmium, antimony, arsenic and bismuth also occur in variable amounts in lead ores. Selenium substitutes for sulfur in the structure constituting a solid solution series. The lead telluride mineral altaite has the same crystal structure as galena. Within the weathering or oxidation zone galena alters to anglesite (lead sulfate) or cerussite (lead carbonate). Galena exposed to acid mine drainage can be oxidized to anglesite by naturally occurring bacteria and archaea, in a process similar to bioleaching [3] Galena uses : One of the earliest uses of galena was as kohl, which in Ancient Egypt, was applied around the eyes to reduce the glare of the desert sun and to repel flies, which were a potential source of disease.[4] Galena is a semiconductor with a small bandgap of about 0.4 eV which found use in early wireless communication systems. For example, it was used as the crystal in crystal radio sets, in which it was used as a point-contact diode to detect the radio signals. The galena crystal was used with a safety pin or similar sharp wire, which was known as a "cat's whisker". Making such wireless sets was a popular home hobby in the North of England during the 1930s. Derbyshire was one of the main areas where Galena was mined. Scientists that were linked to this application are Karl Ferdinand Braun and Sir Jagdish Bose. In modern wireless communication systems, galena detectors have been replaced by more reliable semiconductor devices, though silicon point-contact microwave detectors still exist in the market.

Page 45: Career Point XtraEdge for IIT-JEE 2010

XtraEdge for IIT-JEE 45 AUGUST 2010

1. If two circles cut orthogonally, prove that the polar of

any point P on the first circle with respect to the second passes through the other end of diameter of the first circle which goes through P.

2. Let ABCD be a tetrahedron. If perpendiculars from B

and C to the opposite faces intersect, then show that BC is perpendicular to AD and the perpendiculars from A and D to the opposite faces will also intersect.

3. For a real number u ;

I(u) = ∫π

+−0

2 )uxcosu21(nl dx;

prove that I (u) = I (– u) = 21

I(u2).

Generalize the result as )u(I21 n2n

.

4. Let f(x) = x2 + ax + b be a quadratic polynomial

where a and b are integers. Let n be an integer, show that f(n). f(n + 1) = f(m) for some integer m.

5. Show that the straight lines joining any two fixed

points on a rectangular hyperbola to any variable point on it intercept a constant length on either asymptote.

6. Solve :

−−

2

2

)yx(y

x1 dx +

− y1

)yx(x

2

2dy = 0

7. Through a focus of an ellipse two chords are drawn

and a conic is described to pass through their extremities, and also through the center of the ellipse. Prove that it cuts the major axis in another fixed point.

8. Let f

2xy =

2)y(f).x(f for all real x & y. If

f ′(1) = f(1) ≠ 0 then show that f(x) is

differentiable ∀ x ∈ R except zero. Also find f(x) for all x ≠ 0.

9. Let there be n straight lines in a plane, no two of

which being parallel or coincident and no three of them meet at a point, then show that they divide the plane in

21 (n2 + n + 2) parts.

10. Prove that

x3cos

xsin + x9cosx3sin +

x27cosx9sin

= 21 [tan 27 x – tan x]

`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

4Set

NaN (Not a Number) ..... is not really a number but a symbol that represents a numerical quantity whose magnitude cannot be determined by the operating system.

• -1 • log (-n) = ln (-n)

• 0 / 0

• 00

• 1∞ • ∞0 • ∞ / ∞ = ∞ / -∞ = -∞ / ∞ = -∞ / -∞ • 0 x ∞ = 0 x -∞ • (-∞) + ∞ = ∞ + (-∞) • ln |0| / ln |±∞| • e±∞ x ln |0|

• (m / ±∞) x (n / 0) if m ±∞ and n 0

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XtraEdge for IIT-JEE 46 AUGUST 2010

1. dxdy – y ln 2 = 2sin x(cos x – 1) ln 2

I.F. = dx2ln

e ∫− = 2–x

so y2–x = ∫ −xxsin2 . (cos x – 1). ln 2 dx

y2–x = 2sin x – x + c y = 2sin x + c.2x Ans (B) y . 2–x = 2sin x – x + c Now if x → ∞ and y is bounded (finite) then c = 0 Ans (C) For option (D) : f(x) = 2sin x c.2x = 0, so c = 0 but in this case y is bounded so this is not

correct option. Hence correct answers are (B, C) 2. AC = 4p

A

B C

D

α α

p

tan α = DCp =

pAD = 4

cot α + tan α = p

DC + p

AD = 4

tan2α – 4 tan α + 1 = 0

tan α = 2

4164 −± = 2 ± 3

α = 15º & 75º Hence correct answer is (D)

3. z2 = z1iπ/3

z3

z1(–a + i)

z2(–1+bi)

O

–1 + bi = (–a + i)

+ i

23

21

=

−−

23a

21 + i

+−

21a

23

–1 = –21 a –

23

⇒ a = 2 – 3

b = – 23 a +

21

⇒ b = – 3 +23 +

21 = 12 – 3

Hence correct answer is (B) 4. c1 ≡ x2 + y2 = a2 c2 ≡ (x – h)2 + (y – k)2 = r2

c1

c2P

Director circle x2 + y2 = 2a2 ...(1)

Pt. P

++

++

ra0.rak;

ra0.rah

it lies on eqn. (1) a2h2+ a2k2 = (a + r)2 2a2 h2 + k2 = 2(a + r)2 x3 + y3 = 2(a + r)2 Hence correct answer is (A)

5. 4a = 2(2a) = 2 . 13

17365 +−

= 2 . 1314 =

1328

Hence correct answer is (C)

6. ∫ − x6sinx3sin

x3sin.2xcos

2x9cos2

= ∫−

2x3sin

2x9cos2

x3sin.2xcos

2x9cos2

MATHEMATICAL CHALLENGES SOLUTION FOR JULY ISSUE (SET # 3)

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XtraEdge for IIT-JEE 47 AUGUST 2010

= – ∫ 2x3cos

2xcos2 dx

= – ∫ + )xcosx2(cos dx

= – 2

x2sin – sin x + c

7. (x – 1)2 = a2

a = 1 ± |a| f(x) = x2 – 2(a + 1)x + a(a – 1) = 0

1 2

D = 4(a + 1)2 – 4a (a – 1) > 0 a2 + 2a + 1 – a2 + a > 0

a > –31 ...(1) not essential

f(1 ± |a|) < 0 ⇒ (1 ± |a|)2 – 2(a + 1) (1 ± |a|) + a(a – 1) < 0 ⇒ 1 + a2 ± 2|a| – 2a m 2a|a| – 2 m 2|a| + a2 – a < 0 ⇒ 2a2 m 2a|a| – 3a – 1 < 0 if a ≥ 0 2a2 – 2a2 – 3a – 1 < 0

⇒ a > – 31 ⇒ a ≥ 0

2a2 + 2a2 – 3a – 1 < 0 ⇒ 4a2 – 3a – 1 < 0 (4a + 1) (a – 1) < 0

⇒ –41 < a < 1

⇒ 0 ≤ a < 1 So, similarly if a < 0 then f(1 ± |a|) < 0 ⇒ f(1 m a) < 0

so f(1 ± |a|) < 0 ⇒ –41 < a < 1 ...(2)

eqn. (1) ∩ (2) –41 < a < 1

8. 2r = a + b + c ON = – BN + BO Let BN = x 2BN + 2CN + 2AR = 2s x + (a – x) + (b – a + x) = s x = s – b

A

B N O C

R

r

M I(h,k)

so h = ON = 2a – (s – b)

= 2

b2as2 ++− = 2

cb −

& r = k

so r = k = s∆ =

s)cs)(bs)(as(s −−−

r = k = s

)cs)(bs)(as(s −−−

2sk = )cba)(cba)(as(s −++−−

= )x2a)(x2a)(as(s +−−

2sk = )h4a)(as(s 22 −− required lows is

4s2y2 = A(a2 – 4x2) ⇒ s2y2 + A x2 = 4

Aa2

where A is = s(s – a) here h2 < ab so it is an ellipse. 9. Total no of ways of drawing = 74 = 2401 favorable no of drawings = coeffi. of x8 in the

expansion of (x0 + x1 + x2 + ...... x6)4 = coeffi. of x8 is (1 – x7)4 (1 – x)–4

= coeffi of x8 in (1 – 4x7 + ....) (1 + 4x + ..... + 165x8 + ..........) = 165 – 16 = 149

so required probability = 2401149

10. f(x) = (x – α)Q(x) + R f(+1) = 2; f(–2) = 1, f(–1) = –1. If f(x) is divided by (x2 – 1) (x + 2) which is a cubic

expression then remainder should be a quadratic expression.

f(x) = (x2 – 1) (x + 2) Q(x) + (ax2 + bx + c) f(1) = a + b + c = 2 f(–2) = 4a – 2b + c = 1 f(–1) = a – b + c = –1

Solving these eqn, a = 67 , b =

23 , c = –

32

so the remainder is 67 x2 +

23 x –

32

Page 48: Career Point XtraEdge for IIT-JEE 2010

XtraEdge for IIT-JEE 48 AUGUST 2010

1. Suppose f(x) = x3 + ax2 + bx + c, where a, b, c are

chosen respectively by throwing a die three times. Find the probability that f(x) is an increasing function.

Sol. f´(x) = 3x2 + 2ax + b

y = f(x) is strictly increasing

⇒ f´(x) > 0 ∀ x

⇒ (2a)2 – 4.3.b < 0

This is true for exactly 15 ordered pairs (a, b); 1 ≤ a,

b ≤ 6, so probability = 3615 =

125

2. If (a, b, c) is a point on the plane 3x + 2y + z = 7, then find the least value of a2 + b2 + c2, using vector methods.

Sol. Let →A = a i + b j + c k

⇒ →B = 3 i + 2 j + k

⇒ 2)B.A(→→

≤ |→A |2 |

→B |2

3a + 2b + c ≤ 222 cba ++ 14

(7)2 ≤ (a2 + b2 + c2) (14)

Q 3a + 2b + c = 7, point lies on the plane

a2 + b2 + c2 ≥ 1449 =

27

3. Let g be a real valued function satisfying g(x) + g(x +

4) = g(x + 2) + g(x + 6), then prove that ∫+8x

x)t(g dt

is a constant function.

Sol. given that g(x) + g(x + 4) = g(x + 2) + g(x + 6) ...(1)

putting x = x + 2 in (1) ........

g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8) ...(2)

from (1) & (2)

g(x) = g(x + B)

Now, f(x) = ∫+8x

x)t(g dt

f´(x) = g(x + 8) – g(x) = 0

⇒ g is constant function

4. If exactly three distinct chords from (h, 0) point to the circle x2 + y2 = a2 are bisected by the parabola y2 = 4ax, a > 0, then find the range of 'h' parameter.

Sol. Let M(at2, 2at) is mid-point of chord AB, then chord AB = T = S1

B

MA

AB : x.at2 + y.2at = a2t4 + 4a2t2

since AB chord passes through (h, 0)

so, h.at2 = a2t4 + 4a2t2

at2 [at2 + (4a – h)] = 0

If a > 0 ⇒ 4a – h < 0

⇒ h > 4a ...(i)

Now point (at2,2at) must lie inside the circle, on solving

a2t4 + 4a2t2 – a2 < 0

we get, h < a ( 5 + 2) ...(ii)

from (i) & (ii)

4a < h < a ( 5 + 2)

5. Find the sum of the terms of G.P. a + ar + ar2 + ..... + ∞

where a is the value of x for which the function 7 + 2x loge25 – 5x – 1 – 52–x has the greatest value and

r is the ∫ +π→

x

0 2

2

0x )xtan(xdttLt

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

Students' ForumMATHS

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XtraEdge for IIT-JEE 49 AUGUST 2010

Sol. S = r1

a−

,

To get the greatest value f´(x) = 2log e25 – 5x – 1 log 5 + 52–x log 5

f´(x) = 4 loge5 – 5x–1 loge5 + 5.51 – xloge5

⇒ f´(x) = 0 put 5x – 1 t(> 0)

t2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5x –1 = 5 ⇒ x = 2

to evaluate r :

r = ∫ +π→

x

0 2

2

0x )xtan(xdttLt =

π1

since a = 2, r = π1 ⇒ sum of G.P. =

12−ππ

6. The tangent and normal at a point P on the ellipse

2

2

ax + 2

2

by = 1 meets the y-axis at A and B

respectively. Find the angle subtended by AB at the points of intersection of the circle (through A,S,B) and the ellipse. S being one of the foci.

Sol. Let P be (x1, y1),

B

M

S

P AQ

N

Points of intersection of tangent and normal at P

points with y-axis and A

1

2

yb,0 , B

− 2

12

1 byay,0 ,

S : (ae, 0) slope (SA). slope (SB) = –1 ⇒ ∠ASB = 90º (PA and PB are tangent and normal) P must lie on the circle with AB as diameter.

Hence the point of intersection of the ellipse and the circle is P. Due to symmetry the angles made by AB at P,Q,M, N are all 90º.

Confidence Tips

• Can you enjoy success if you don’t know what you wanted?

• Can't is the worst of the 4-letter words.

• Can you tell whether someone else lacks self-confidence?

• Can you act confident even when you are not?

• Meal check: How many successes have you had since your last meal?

• Meal check: What will you accomplish in the next hour?

• Meal check: Take the next thing you will do. How will you see that it is well done?

• Meal check: How many times have you thought “I can’t” since you last ate?

• You already have all to brain tools you need. You just need to find the tools that fit the job.

• You don't have to know everything. You just have to know how to find out anything.

• You will know more tomorrow. How will you use that insight today?

• If you want to be smart, find friends who are smarter than you are.

• List 5 ways to undermine your own self-confidence.

• Will what you seek be worth the work?

• Pay attention to what you say about yourself. Would you say that about someone else?

• Pay attention to what you say about what you can do. Why do you believe it?

• Those who say it can't be done should not interrupt those who are doing it.

• Don't should on yourself .

• Some focus on what they can. Others focus on what they can't. What do you do you?

• Tell yourself what you can't do. Hear a stop sign. Tell yourself what you can do. What do you hear?

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XtraEdge for IIT-JEE 50 AUGUST 2010

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XtraEdge for IIT-JEE 51 AUGUST 2010

Representation of vectors : Geometrically a vector is represent by a line segment.

For example, a = AB. Here A is called the initial point and B, the terminal point or tip.

Magnitude or modulus of a is expressed as

|a| = |AB| = AB.

B

a

A Types of Vector: Zero or null vector : A vector whose magnitude

is zero is called zero or null vector and it is represented by O .

Unit vector : A vector whose modulus is unity, is called a unit vector. The unit vector in the direction of a vector a is denoted by a , read as “a cap”. Thus, | a | = 1.

a = || a

a = aofMagnitude

aVector

Like and unlike vectors : Vectors are said to be like when they have the same sense of direction and unlike when they have opposite directions.

Collinear or parallel vectors : Vectors having the same or parallel supports are called collinear or parallel vectors.

Co-initial vectors : Vectors having the same initial point are called co-initial vectors.

Coplanar vectors : A system of vectors is said to be coplanar, if their supports are parallel to the same plane.

Two vectors having the same initial point are always coplanar but such three or more vectors may or may not be coplanar.

Negative of a vector : The vector which has the same magnitude as the vector a but opposite direction, is called the negative of a and is denoted by –a. Thus, if PQ = a, then QP = –a.

Properties of vectors : (i) Addition of vectors Triangle law of addition : If in a ∆ABC,

AB = a, BC = b and AC = c, then

AB+ BC = AC i.e., a + b = c

A B

C

b

a

c = a + b

Parallelogram law of addition : If in a

parallelogram OACB, OA = a, OB = b and

OC = c

C

AaO

b

B

c = a + b

Then OA + OB = OC i.e., a + b = c, where OC is a diagonal of the parallelogram OABC.

Addition in component form : If the vectors are defined in terms of i, j, and k, i.e.,

if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then their sum is defined as

a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k. Properties of vector addition : Vector addition has the following properties. Binary operation : The sum of two vectors is

always a vector. Commutativity : For any two vectors a and b,

a + b = b + a. Associativity : For any three vectors a, b and c,

a + (b + c) = (a + b) + c Identity : Zero vector is the identity for addition.

For any vector a, 0 + a = a = a + 0 Additive inverse : For every vector a its negative

vector –a exits such that a + (–a) = (–a) + a = 0 i.e., (–a) is the additive inverse of the vector a.

VECTOR Mathematics Fundamentals M

ATHS

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Subtraction of vectors : If a and b are two vectors, then their subtraction a – b

is defined as a – b = a + (–b) where –b is the negative of b having magnitude equal to that of b and direction opposite to b. If

a = a1i + a2j + a3k, b = b1i + b2j + b3k Then a – b = (a1 – b1)i + (a2 – b2)j + (a3 – b3)k.

B

b

A

–b

a

a + (–b) = a – b

O

a + b

Properties of vector subtraction : (i) a – b ≠ b – a (ii) (a – b) – c ≠ a – (b – c) (iii) Since any one side of a triangle is less than the

sum and greater than the difference of the other two sides, so for any two vectors a and b, we have

(a) |a + b| ≤ |a| + |b| (b) |a + b| ≥ |a| – |b| (c) |a – b| ≤ |a| + |b| (d) |a – b| ≥ |a| – |b| Multiplication of a vector by a scalar : If a is a

vector and m is a scalar (i.e., a real number) then ma is a vector whose magnitude is m times that of a and whose direction is the same as that of a, if m is positive and opposite to that of a, if m is negative.

Properties of Multiplication of vector by a scalar : The following are properties of multiplication of vectors by scalars, for vector a, b and scalars m, n.

(i) m(–a) = (–m)a = –(ma) (ii) (–m) (–a) = ma (iii) m(na) = (mn)a = n(ma) (iv) (m + n)a = ma + na (v) m(a + b) = ma + mb Position vector :

AB in terms of the position vectors of points A and B : If a and b are position vectors of points A and B respectively. Then, OA = a, OB = b

∴ AB = (Position vector of B)– (Position vector of A)

= OB – OA = b – a Position vector of a dividing point : The

position vectors of the points dividing the line AB in the ratio m : n internally or externally are

nmnm

++ ab or

nmnm

−− ab .

Scalar or Dot product Scalar or Dot product of two vectors : If a and b

are two non-zero vectors and θ be the angle between them, then their scalar product (or dot product) is denoted by a . b and is defined as the scalar |a| |b| cosθ, where |a| and |b| are modulii of a and b respectively and 0 ≤ θ ≤ π. Dot product of two vectors is a scalar quantity.

A

B

b

aOθ

Angle between two vectors : If a, b be two vectors

inclined at an angle θ, then a . b = |a| |b| cos θ

⇒ cos θ = ||||

.ba

ba

⇒ θ = cos–1

||||

.ba

ba

If a = a1i + a2j + a3k and b = b1i + b2j + b3k; then

θ = cos–1

++++

++23

22

21

23

22

21

332211

bbbaaa

bababa

Properties of scalar product Commutativity : The scalar product of two

vector is commutative i.e., a . b = b . a Distributivity of scalar product over vector

addition : The scalar product of vectors is distributive over vector addition i.e.,

(a) a . (b + c) = a . b + a . c, (Left distributivity) (b) (b + c) . a = b . a + c . a, (Right distributivity) Let a and b be two non-zero vectors a . b = 0 ⇔ a ⊥ b. As i, j, k are mutually perpendicular unit vectors

along the coordinate axes, therefore, i . j = j . i = 0; j . k = k . j = 0; k . i = i . k = 0. For any vector a . a . a = |a|2. As i. j. k are unit vectors along the co-ordinate

axes, therefore i . i = |i|2 = 1, j . j = |j|2 = 1 and k . k = |k|2 = 1

If m, n are scalars and a . b be two vectors, then ma . nb = mn(a . b) = (mna) . b = a .(mnb)

For any vectors a and b, we have (a) a . (–b) = –(a . b) = (–a). b (b) (–a) . (–b) = a . b For any two vectors a and b, we have (a) |a + b|2 = |a|2 + |b|2 + 2a.b (b) |a – b|2 = |a|2 + |b|2 – 2a.b

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(c) (a + b) . (a – b) = |a|2 – |b|2 (d) |a + b| = |a| + |b| ⇒ a | | b (e) |a + b|2 = |a|2 + |b|2 ⇒ a ⊥ b (f) |a + b| = |a – b| ⇒ a ⊥ b Vector or Cross product Vector product of two vectors : Let a, b be two

non-zero, non-parallel vectors.

O a

b

θ

Then a × b = |a| |b| sin θ η , and |a × b| = |a| |b| sin θ,

where θ is the angle between a and b, η is a unit vector perpendicular to the plane of a and b such that a, b, η form a right-handed system.

Properties of vector product : Vector product is not commutative i.e., if a and b

are any two vectors, then a × b ≠ b × a, however, a × b = –(b × a)

If a, b are two vectors and m, n are scalars, then ma × nb = mn(a × b) = m(a × nb) = n(ma × b).

Distributivity of vector product over vector addition. Let a, b, c be any three vectors. Then

(a) a × (b + c) = a × b + a × c (left distributivity) (b) (b + c) × a = b × a + c × a (Right disributivity) For any three vectors a, b, c we have

a × (b – c) = a × b – a × c. The vector product of two non-zero vectors is

zero vector iff they are parallel (Collinear) i.e., a × b = 0 ⇔ a| | b, a, b are non-zero vectors. It follows from the above property that a × a = 0

for every non-zero vector a, which in turn implies that i × i = j × j = k × k = 0.

Vector product of orthonormal triad of unit vectors i, j, k using the definition of the vector product, we obtain i × j = k, j × k = i, k × i = j, j × i = –k, k × j = –i, i × k = –j.

Vector product in terms of components : If a = a1i + a2j + a3k and b = b1i + b2j + b3k.

Then, a × b = 321

321bbbaaakji

Angle between two vectors :

If θ is the angle between a and b then sin θ = ||||||

baba×

Right handed system of vectors : Three mutually perpendicular vectors a, b, c form a right handed system of vector iff a × b = c, b × c = a, c × a = b

Left handed system of vectors : The vectors a, b, c mutually perpendicular to one another form a left handed system of vector iff c × b = a, a × c = b, b × a = c.

Area of parallelogram and triangle : The area of a parallelogram with adjacent sides a

and b is |a × b|. The area of a plane quadrilateral ABCD is

|BDAC|21

× , where AC and BC are its

diagonals.

The area of a triangle ABC is |ACAB|21

× or

|BABC|21

× or |CACB|21

×

Scalar triple product Scalar triple product of three vectors : If a, b, c are

three vectors, then scalar triple product is defined as the dot product of two vectors a and b × c. It is generally denoted by a . (b × c) or [a b c].

Properties of scalar triple product : If a, b, c are cyclically permuted, the value of scalar triple product remains the same. i.e.,

(a × b) . c = (b × c) . a = (c × a). b or [a b c] = [b a c] = [c a b] Vector triple product Let a, b, c be any three vectors, then the vectors

a × (b × c) and (a × b) × c are called vector triple product of a, b, c.

Thus, a × (b × c) = (a . c) b – (a . b)c Properties of vector triple product : The vector triple product a × (b × c) is a linear

combination of those two vectors which are within brackets.

The vector r = a × (b × c) is perpendicular to a and lies in the plane of b and c.

The formula a × (b × c) = (a . c)b – (a . b)c is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula.

Thus, (b × c) × a = –a × (b × c) = (a . c)b – (a . b)c = (a . b)c – (a . c)b Vector triple product is a vector quantity. a × (b × c) ≠ (a × b) × c

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Permutation : Definition : The ways of arranging or selecting a

smaller or an equal number of persons or objects at a time from a given group of person or objects with due regard being paid to the order of arrangement or selection are called the (different) permutations.

Number of permutations without repetition : Arranging n objects, taken r at a time equivalent

to filling r places from n things.

r-places :

Number of choice

1 2 3 4 rn (n–1)(n–2) (n–3) n–(r – 1)

The number of ways of arranging = The number

of ways of filling r places. = n(n – 1) (n – 2) ....... (n – r + 1)

= !)rn(

)!)rn)((1rn)....(2n)(1n(n−

−+−−− = !)rn(

!n−

= nPr The number of arrangements of n different

objects taken all at a time = nPn = n !

(i) nP0 = !n!n = 1; nPr = n. n–1Pr – 1

(ii) 0 ! = 1; !)r(–

1 = 0 or (–r) ! = ∞ (r ∈ N)

Number of permutations with repetition : The number of permutations (arrangements) of n

different objects, taken r at a time, when each object may occur once, twice, thrice, ....... upto r times in any arrangement = The number of ways of filling r places where each place can be filled by any one of n objects.

r-places :

Number of choices :

1 2 3 4 rn (n) (n) (n) n

The number of permutations = The number of

ways of filling r places = (n)r. The number of arrangements that can be formed

using n objects out of which p are identical (and of one kind) q are identical (and of another kind), r are identical (and of another kind) and the rest

are distinct is !r!q!p

!n .

Condition permutations : Number of permutations of n dissimilar things

taken r at a time when p particular things always occur = n – pCr – p r !.

Number of permutations of n dissimilar things taken r at a time when p particular things never occur = n – pCr r !.

The total number of permutations of n different things taken not more than r at a time, when each thing may be repeated any number of times, is

1n)1n(n r

−− .

Number of permutations of n different things, taken all at a time, when m specified things always come together is m ! × (n – m + 1) !.

Number of permutation of n different things, taken all at a time, when m specified things never come together is n ! – m! × (n – m + 1) !.

Let there be n objects, of which m objects are alike of one kind, and the remaining (n – m) objects are alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from these objects is

!)mn()!m(!n

−×.

The above theorem can be extended further i.e., if there are n objects, of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of 3rd kind; .....; pr are alike of rth kind such that p1 + p2 + .... + pr = n; then the number of permutations of these n

objects is )!p(....)!p()!p(

!n

r21 ×××.

Circular permutations : Difference between clockwise and anti-clockwise

arrangement : If anti-clockwise and clockwise order of arrangement are not distinct e.g., arrangement of beads in a necklace, arrangement of flowers in garland etc. then the number of circular permutations

of n distinct items is 2

!)1n( − .

Number of circular permutations of n different things, taken r at a time, when clockwise and

anticlockwise orders are taken as different is rpr

n

.

PERMUTATION & COMBINATION

Mathematics Fundamentals MATHS

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Number or circular permutations of n different things, taken r at a time, when clockwise and

anticlockwise orders are not different is r2pr

n

.

Theorems on circular permutations : Theorem (i) : The number of circular

permutations on n different objects is (n – 1) !. Theorem (ii) : Then number of ways in which n

persons can be seated round a table is (n – 1) !. Theorem (iii) : The number of ways in which n

different beads can be arranged to form a

necklace, is !)1n(21

− .

Combinations : Definition : Each of the different groups or selection

which can be formed by taking some or all of a number of objects, irrespective of their arrangements, called a combination.

Notation : The number of all combinations of n things, taken r at a time is denoted by C(n, r) or nCr or

rn . nCr is always a natural number.

Difference between a permutation and combination :

In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered.

Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, CBA and CAB correspond to the same combination ABC.

Number of combinations without repetition The number of combinations (selections or groups)

that can be formed from n different objects taken

r (0 ≤ r ≤ n) at a time is nCr = !)rn(!r

!n−

. Also

nCr = nCn –r. Let the total number of selections (or groups) = x.

Each group contains r objects, which can be arranged in r ! ways. Hence the number of arrangements of r objects = x × (r!). But the number of arrangements = npr.

⇒ x(r !) = npr ⇒ x = !)rn(!r

!n−

= nCr

Number of combinations with repetition and all possible selections : The number of combinations of n distinct objects

taken r at a time when any object may be repeated any number of times.

= Coefficient of xr in (1 + x + x2 + ...... + xr)n = Coefficient of xr in (1 – x)–n = n + r – 1Cr The total number of ways in which it is possible

to form groups by taking some or all of n things at a time is nC1 + nC2 + .... + nCn = 2n – 1.

The total number of ways in which it is possible to make groups by taking some or all out of n = (n1 + n2 + .....) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is (n1 + 1) (n2 + 1) ..... – 1.

The number of selections taking at least one out of a1 + a2 + a3 + .... + an + k objects, where a1 are alike (of one kind), a2 are alike (of second kind) and so on ............... an are alike (of nth kind) and k are distinct

= [(a1 + 1) (a2 + 1) (a3 + 1) ......... (an + 1)]2k – 1 Conditional combinations : The number of ways in which r objects can be

selected from n different objects if k particular objects are

(i) Always included = n – kCr–k

(ii) Never included = n – kCr The number of combinations of n objects, of

which p are identical, taken r at a time is n – pCr + n – pCr – 1 +...........+ n – pC0, if r ≤ p and n – pCr + n – pCr – 1 + ........... + n – pCr – p, if r > p.

Division into groups The number of ways in which n different things

can be arranged into r different groups is n + r – 1Pn or n ! n –1Cr – 1 according as blank group are or are not admissible.

Number of ways in which m × n different objects can be distributed equally among n persons (or numbered groups) = (number of ways of dividing into groups) × (number of groups)!

= !n)!m(!n!)mn(

n = n)!m(!)mn(

If order of group is not important: The number of ways in which mn different things can be divided

equally into m groups is !m)!m(

!)mn(m .

If order of groups is important : The number of ways in which mn different things can be divided equally into m distinct groups is

!m)!n(

!)mn(m × m ! = m)!n(

!)mn(

Derangement : Any change in the given order of the things is called

a derangement. If n things form an arrangement in a row, the number

of ways in which they can be deranged so that no one of them occupies its original place is

n !

−++−+−

!n1.)1(...

!31

!21

!111 n

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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. Two men B and C are watching man A. B watches A to be stationary and C watches A moving. Then -

(A) Man A may be at absolute rest (B) Man B may be at absolute rest (C) Man C may be at absolute rest (D) None of these

2. For a body moving one dimensionally, ∆S and |S|→

∆ are distance travelled and magnitude of displacement respectively, then -

(A) ∆S = |S|→

∆ ; always true

(B) ∆S > |S|→

∆ ; always true

(C) S

|S|∆∆

≤ 1; always true

(D) ∆S < |S|→

∆ ; sometimes true

3. Two particles X and Y are respectively moving on the circular path and regular hexagon as shown. O is centre of circle and hexagon both. When both X and Y have moved from point A to point D, the ratio of distance moved by X to magnitude of displacement of Y is –

AX Y

F

E

D C

B

(A) 4π (B)

(C) π

(D) 2π

4. If the magnetic lines of force are shaped like arcs of concentric circles with their centre at point O in a certain section of magnetic field :

O

IIT-JEE 2011

XtraEdge Test Series # 4

Based on New Pattern

Time : 3 Hours Syllabus (Revision – 1): Physics: Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current, Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal, Monotonicity, Maxima & Minima)

Instructions : 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 passage based single correct type questions. +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 - IV • 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..

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(A) The intensity of the field in this section should at each point be inversely proportional to its distance from point O

(B) The intensity of the field in this section should at each point be inversely proportional to square of its distance from point O

(C) The intensity of the field in this section should at each point be inversely proportional to cube of its distance from point O

(D) Nothing can be said

5. A closed body, whose surface F is made of metal foil, has an electrical capacitance C (if potential at infinity is assumed zero) The foil is now dented in such a way that the new surface F* is entirely inside of an the original surface as shown in the figure. Then –

F*

F

original surface

new surface

(A) Capacitance of F* > capacitance of F (B) Capacitance of F* < capacitance of F (C) Capacitance of F* = capacitance of F (D) Nothing can be concluded from given information

6. A current of I Amp flow through a wire made of a piece of copper and a piece of iron of identical cross sections welded end to end as shown in the figure.

Cu

I Amp

Fe

How much electric charge accumulates at the

boundary between the two metals? ρFe & ρCu are resistivity of Fe & Cu respectively -

(A) ∈0I (ρFe + ρCu) (B) ∈0I (ρFe – ρCu) (C) ∈0I (ρFe + 3ρCu) (D) ∈0I (ρFe + 2ρCu)

7. Rectangular loop is rotating about z-axis, at 500 rev/min. Then find the current in loop. Where R = 0.20 Ω and B is in radial outward direction and its value is 0.2 T.

0.05m

0.03m

z

ω

R

Loop

0.5m

(A) 0.523 A (B) 0.319 A (C) 2.369 A (D) 1.235 A

8. An electron flies into a homogeneous magnetic field perpendicular to the force lines. The velocity of the electron is v = 4 × 107 m/s. The induction of the field is 10–3T. What is tangential acceleration of electron in the magnetic field –

(A) 7 × 1015 m/s2 (B) 7 × 1013 m/s2 (C) 7 × 1014 m/s2 (D) None of these

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 conductor is made of an isotropic material and has

shape of a truncated cone. A battery of constant emf E is connected across it and its left end is earthed as show in figure. If at a section distance x from left end, electric field intensity, potential and the rate of generation of heat per

E + –

unit length are E, V and H respectively, which of the

following graphs are correct ?

(A)

E

x

(B)

H

x

(C)

H

E

(D)

V

x

10. The conductor ABCDE has the slope shown in

figure. It lies in the y-z plane, with A and E on the y-axis. When it moves with a velocity v in a magnetic field B, an emf e is induced between A and E, then :

B

z

O

x

aA

D

a

E y

λ

C

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XtraEdge for IIT-JEE AUGUST 2010 56

(A) e = 0, if v is in the y-direction and B is in the x-direction (B) e = 2Bav, if v is in the y-direction and B is in the x-direction (C) e = Bλv, if v is in the z-direction and B is in the x-direction (D) e = Bλv, if v is in the x-direction and B is in the z-direction 11. Instantaneous velocity of a particle - (A) depends on instantaneous position (B) depends on instantaneous speed (C) independent of instantaneous position (D) independent of instantaneous speed 12. Two bodies A and B are moving with speeds V and

2v respectively, then - (A) distance moved by A must be greater than that of B (B) distance moved by A must be smaller than that of B (C) displacement of A may be greater than that of B (D) displacement of A may be smaller than that of B 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 : I (Ques. 13 to 15)

A large parallel plate capacitor with uniform surface charge σ on the upper plate & – σ on the lower is moving with a constant speed V as shown in figure.

V

V

+ σ – σ

13. Magnetic field between the plates is

(A) zero (B) 4

V0σµ

(C) 2

V0σµ

(D) µ0σV

14. Magnetic force per unit area on the upper plate, including its direction -

(A) µ0σ2V2 upward (B) 2

v220σµ upward

(C) 2µ0σ2V2 downward (D) 2

V220σµ downward

15. At what speed V would the magnetic force balance the electrical force, where C is the speed of light -

(A) 0. 5 C (B) 0.2 C (C) C (D) 0.9 C

Passage: II (Ques. 16 to 18)

Initially circuit is in steady state.

S 6 µF6 Ω 18 V

3 µF a

b

3 Ω

16. When switch S is open, then Vb– Va is - (A) 6V (B) – 6 V (C) 18 V (D) – 18 V

17. When switch S is closed then Vb – Va is - (A) 6V (B) 0 V (C) 18 V (D) – 6 V

18. How much does the charge on capacitor 6µF change when S is closed -

(A) 16 µC (B) 18 µC (C) 36 µC (D) 72 µC 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

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. For one dimensional motion if

Vav = average speed →

avV = average velocity Vinst = instantaneous speed Vinst = instantaneous velocity; v = speed Then match the following Column-I Column-II

(A) →

instV = →

avV (P) for uniform motion in any direction

(B) |→

instV | = V (Q) for uniform motion in given direction (C) Vinst = Vav (R) Always true

(D) |→

instV | < V (S) Never true (T) None of these

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XtraEdge for IIT-JEE AUGUST 2010 57

20. Match the following Column-I Column-II (A) Motion of dropped (P) Two dimensional ball motion (B) Motion of a snake (Q) Three dimensional motion (C) Motion of a bird (R) One-Dimensional motion (D) Earth (S) Absolute rest (T) None of these

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. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is -

(A) 0 (B) 2.84 (C) 4.90 (D) 5.92 2. The crystalline salt, Na2SO4xH2O on heating loses

55.9% of its weight. The formula of the crystalline salt is -

(A) Na2SO4.5H2O (B) Na2SO4.7H2O (C) Na2SO4.10H2O (D) None of these 3. The correct order of increasing C–O bond length in

CO, –23CO , CO2 is -

(A) –23CO < CO2 < CO (B) CO2 < –2

3CO < CO

(C) CO < –23CO < CO2 (D) CO < CO2 < –2

3CO

4. If the wavelength of series limit of Lyman series for He+ ion is x Å then what will be the wavelength of series limit of Balmer series for Li+2 ion ?

(A) 4x9 Å (B)

7x4 Å

(C) 4x5 Å (D)

9x16 Å

5. The correct order of acidic strength is – (A) Cl2O7

>SO2>P4O10

(B) CO2 >N2O5 <SO3

(C) Na2O >MgO >Al2O3

(D) K2O >CaO >MgO

6. In a hypothetical solid C atoms form CCP lattice with A atoms occupying all the Tetrahedral voids and B atoms occupying all the octahedral voids. A and B atoms are of the appropriate size such that there is no distortion in the CCP lattice. Now if a plane is cut (as shown) then the cross section would like –

Plane

CCP unit cell

(A)

C C

C C

B

BBBA

A

B

(B)

C C

C

BB B

C C

CC

(C)

C

C

BBB

C C

CC

A

A A

A

(D)

C C

C

BB B

C C

CC

7. Regarding graphite the following informations are available :

3.35Å

Top view

The density of graphite = 2.25 gm/cm3. What is C–C bond distance in graphite ?

(A) 1.68Å (B) 1.545Å (C) 2.852 Å (D) 1.426Å

8. The geometrical shapes of XeF5+, XeF6 and XeF8

2– respectively are -

(A) trigonal bipyramidal, octahedral and square planar

(B) square based pyramidal, distorted octahedral and octahedral

(C) planar pentagonal, octahedral and square anti prismatic

(D) square based pyramidal, distorted octahedral and square anti prismatic

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.

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9. Which of the following is/are correct ? (A) α-rays are more penetrating than β-rays (B) α-rays have greater ionizing power than β-rays (C) β-particles are not present in the nucleus, yet they

are emitted from the nucleus (D) γ-rays are not emitted simultaneously with α and

β-rays 10. Select the correct curve (s) If v = Velocity of electron in Bohr's orbit r = Radius of electron in Bohr's orbit P.E. = Potential energy of electron in Bohr's orbit K.E. = Kinetic energy of electron in Bohr's orbit

(A)

v 1/n

(B)

r n

(C)

1/n z

P.E.

(D)

K.E. n

11. In which of the following molecules there is no S—S bond(s)?

(A) S2O42– (B) S2O5

2– (C) S2O3

2–

(D) S2O72–

12. During the electrolysis of AgNO3 (using Pt electrodes) concentration around cathode as well as anode falls from 4 M to 3 M. What will happen if this happened with Ag electrodes ?

(A) Result will remain same (B) Concentration around cathode will fall from 4 M

to 3 M but around anode will increase from 4 M to 5 M

(C) Reverse of statement 'b' (D) Concentration increases from 4 M to 5 M on both

the electrodes

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 : I (Ques. 13 to 15)

According to Einstein, when light of some particular frequency falls on surface of metal, the photon give its entire energy to the electron of the metal atom. The electron will be dislodged or detached form the metal atom only if the energy of the photon is sufficient to over come the force of attraction of the electron by the nucleus. That is why photoelectrons are ejected only when the incident light has a certain minimum frequency. According to Einstein photoelectric explanation:

hν = hν0 + 21 mv2 ,

Where ν is the frequency of its incident light and ν0 is the minimum energy and v is the velocity of electron. A graph can be plotted between the kinetic energy and frequency of absorbed photons :

Frequency

Kin

etic

Ene

rgy

of p

hoto

elec

trons

v0

...... (K.E.) = hν – hν0

13. As per the experiments carried out for PEE; we can say -

(A) As the intensity of light increases, velocity of photoelectron increases.

(B) As the intensity of light increases, energy of photoelectron increases.

(C) As the intensity of light increases, no. of photoelectrons decreases.

(D) Kinetic energy of photoelectron is proportional to frequency of incident light.

14. When a light of frequency ν1 is incident on a metal surface the photoelectron emitted had twice the kinetic energy as did the photoelectrons emitted when the same metal had irradiated with light of frequency ν2. What will be threshold frequency.

(A) ν0 = ν1 – ν2 (B) ν0 = 2ν2 – ν1 (C) ν0 = 2ν1 – ν2 (D) ν0 = ν1 + ν2

15. The equation of photoelectric effect represents an equation of -

(A) straight line (B) parabola (C) ellipse (D) None Passage: II (Ques. 16 to 18)

The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of non-electrolytes. The electrolytes dissociate into ions in solution. It is the number of solute particles that determines the colligative properties of a solution. The electrolyte solution therefore, show abnormal colligative properties. To account for this effect we define a quantity called the van't Hoff factor, given by

i = solutionindissolvedinitiallyunitsformulaofNumber

ondissociatiaftersolutioninparticlesofnumberActual

i = 1 (for non-electrolyte); i > 1 (for electrolytes, undergoing dissociation) i < 1 (for solutes, undergoing association).

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16. Certain substance trimerises when dissolved in a solvent A. The van't Hoff factor 'i' for the solution is :

(A) 1 (B) 1/3 (C) 3

(D) unpredictable

17. For a solution of a non-electrolyte in water, the van't Hoff factor is :

(A) always equal to 0 (B) ≤ 1 (C) always equal to 2 (D) > 1 but < 2

18. 0.1 M K4[Fe(CN)6] is 60% ionized. What will be its van't Hoff factor ?

(A) 1.4 (B) 2.4 (C) 3.4

(D) 4.4

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. Column-I Column-II (Ionic species) (Shapes) (A) XeF5

+ (P) Tetrahedral (B) SiF5

– (Q) Square planar (C) AsF4

+ (R) Trigonal bipyramidal (D) ICl4

– (S) Square pyramidal (T) Octahedral

20. Column-I (A) NHC 13

711

126 →+

(B) HeMgHAl 42

2412

11

2713 +→+

(C) n3KrBanU 10

9336

14056

10

23592 ++→+

(D) n12H9MnHAs 10

11

5625

21

7533 ++→+

Column-II (P) Projectile capture (Q) Spallation (R) Fusion (S) Projectile capture and particle emission (T) Fission

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 range of k for which ||x–1|–5| = k have four distinct solutions -

(A) [0, 5] (B) (–∞, 5) (C) [0, ∞)

(D) (0, 5)

2. Let f(x) be a second degree polynomial function such that lnf(x) > 0 ∀ x ∈ R & the equation f '(x) + 786 f(x), has no real roots. If g(x) = e786x f(x), then -

(A) g(x) is an increasing function (B) f(x) is a decreasing function (C) g(x) is an even function (D) the graph of f(x) cuts x-axis at least once.

3. The function f(x) = ax2 – βx –4x3 + γ, where α, β, γ ∈ R, has local maxima at P(log2a, f(log2a)) & Local minima at Q (log2a2, f(log2a2)). If the graph of f(x)

changes concavity about the point R

43f,

43 , then

which of the following conic section can have eccentricity 'a' -

(A) circle (B) parabola (C) ellipse

(D) hyperbola

4. Let a function is defined as f : R → R

f(x) =

>≤+

0x1–mx0x,1–mx2x2

If f(x) is one-one then set of values of 'm' will be (A) (– ∞, 0) (B) (– ∞, 0] (C) (0, ∞)

(D) [0, ∞)

5. If f(n) = x1

n20x 2xsin1...

2xsin1

2xsin1Lim

+

+

+

then ∞→n

Lim f(n) =

(A) 1 (B) e (C) 0

(D) ∞

6. The derivative of

cos

+ 2x–1cos2–

x1x–1tan2 1–1– w.r.to x is

(A) 1 – 2x–1

1 (B) 1 – 2x1

1

+

(C) 2 – 2x–1

1 (D) 2 – 2x1

1

+

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7. If P = )1n/(1n1–n432n

2)ae.....ae.ea(lim +

∞→, then p4 equals

(A) ea2 (B) e2a2 (C) ea (D) e2a 8. Maximum value of the expression

1x3x3x2x

x108161224

12

++++ is equal to

(A) 1 (B) 2 (C) 10

(D) not defined 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. If f(x) = (h1(x) – h1(–x))(h2(x) – h2(–x))…… (h2n+1(x) – h2n+1(–x)) Where h1(x), h2(x), ……. hn(x) are defined

everywhere & f(200) = 0, then f(x) is (A) one-one (B)many one (C) odd

(D) even 10. Function whose jump (non-negative difference of

LHL & RHL) of discontinuity is greater than or equal to one is/are -

(A)

>

<+

=0x;

x)xcos–1(

0x;)1–e()1e(

)x(fx/1

x/1

(B)

<<

<=

1x21;

1–xnx

1x;)1–x()1–x(

)x(g2/1

3/1

l

(C)

<

=2x;

x|xsin|

21,0x;

x3tanx2sin

)x(u1–

1–

(D)

<+>+

=2x);5x(log2x);2x(log

)x(v 22/1

3

11. Which of the following is/are true

(A)

∞→ 1n3n2sinLim n

n = 0

(B) )ex1(n)ex1(nLim x24

x2

x ++++

∞→ l

l = 1

(C) 4/x

Limπ→

(tan x)tan 2x = e–1

(D) xsin

xcos–xcosLim 2

3

0x→ =

121–

12. If g(x) = 7x2 2x–e ∀ x ∈ R, then g(x) has

(A) local maxima at x = 0 (B) local minima at x = 0 (C) local maxima at x = –1 (D) two local maxima and one local minima

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 : I (Ques. 13 to 15)

K(x) is a function such that K(f(x)) = a + b + c + d, where

a =

oddnorevenneitheris)x(fif2oddis)x(fif1–evenis)x(fif0

b =

periodicnonis)x(fif4periodicis)x(fif3

c = −

onemanyis)x(fif6oneoneis)x(fif5

d =

ointis)x(fif8ontois)x(fif7

A = x2, ex, sin x, |x| all functions in set A are defined from R to R

B = 18, 19, 16, 17

H : R → R; h(x) =

+++

1e–e1ee

xx2

xx2 and

φ :

ππ

2,

2– → R ; φ (x) = tan x

13. k(φ (x)) is equal to (A) 15 (B) 16 (C) 17

(D) 18

14. k(h(x)) is equal to (A) 15 (B) 16 (C) 17

(D) 18

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15. If k(x) is a function such that k : A → B , y = k(x) where x ∈ A, y ∈ B, then k(x) is -

(A) one-one-onto (B) one-one into (C) many one into (D) many one onto Passage: II (Ques. 16 to 18)

Consider f(x) =

>++=

<<++

≤λ+

∞→

1x;))1ee(nsgn(1x;b

1x0;kxxx–xlim

0x;|x–x–|

x–x

n–n

n–n

n

2

l

n ∈ N, k ∈ R

16. The value of k + b + λ so that f(x) is continuous in R is -

(A) 3 (B) 2 (C) 4

(D) 1 17. Number of point(s) where continuous function f(x) is

non differentiable, is - (A) 0 (B) 1 (C) 2

(D) 3 18. If f(x) is continuous then set of values of x for which

f '(x) is decreasing, is - (A) (– ∞, – 1) (B) (– 1, 0) (C) (0, 1)

(D) (– 1, 1)

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. The graph of f and g are given. Use them to evaluate each limit.

y = f(x)

Y

21

–1–2 O 1 2 X–1

y = g(x)

Y

1

–1 –2 O 1 2 X–1

2

Column-I Column-II (A)

1xlim

→f(g(x)) = (P) 1

(B) 2–)x(f3lim2x→

= (Q) does not exist

(C)

+

→)x(g)x(f

)x(g)x(flim

0x = (R) 0

(D) )x(g)x(f)x(g–)x(f3lim

1x ++→ = (S) 2

(T) 4 20. Column-I Column-II (A) Domain of function (P) 5

f(x) = ln

+ 21–

|x|1|x|–1

is (p, q), then p + q is equal to

(B) Maximum value of (Q) 1

f(x) =

π

+4

xsin

x2sin in the

interval

π

2,0 is

(C) Let f(x) = x3 + ax + b (R) 0 with a ≠ b and suppose the tangent lines to the graph of f at x = a and x = b have the same gradient. Then the value of f(1) is equal to

(D) If f is a differentiable (S) – 1 function for all real x and f '(x) ≤ 5, ∀ x ∈ R. If f(2) = 0 and f(5) = 15 then f(3) is equal to (T) 2

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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. If y = 1 + 2x1 + 3x

1 then dxdy is -

(A) x + 2x1 + 3x

1 (B) 3x2– – 4x

3

(C) x – 2x1 – 3x

1

(D) x2– – 2x

3

2. The relation given the value of x as, x = dc

na 33. Find

the maximum percentage error in x, if the percentage error a, b, c, d are 2%, 1%, 2% & 4% respectively -

(A) 13% (B) 5% (C) 9%

(D) 8% 3. There are two different quantities A and B having

different dimensions. Then which of the following operation is dimensionally correct ?

(A) A + B (B) A – B (C) A/B

(D) eA/B

4. The displacement x of body when its velocity v is given by v = 9x2 + . Its acceleration and initial velocity are -

(A) 1 units and 3 units (B) 2 units and 9 units (C) 4 units and 81 units (D) 9 units and 2 units

5. If →a and

→b are two unit vectors and

→R =

→a +

→b

and if |R|→

= R, then - (A) R < 0 (B) R > 2 (C) 0 ≤ R ≤ 2

(D) R must be 2

6. A boat travels upstream in a river and at t = 0 a wooden cork is thrown over the side with zero initial velocity. After 7.5 minutes the boat turns and starts moving downstream catches the cork when it has drifted 1 km downstream. Then the velocity of water current is -

(A) 2 Km/hr (B) 4 Km/hr (C) 6 Km/hr (D) 8 Km/hr

7. A projectile is thrown with speed 20 m/s at an angle 30º with horizontal from ground. Then the average angular velocity of projectile in its time of light is (g = 10 m/s2) -

(A) ωav = s/rad6π (B) ωav = s/rad

(C) ωav = s/rad12π (D) None of these

IIT-JEE 2012

XtraEdge Test Series # 4

Based on New Pattern

Time : 3 Hours Syllabus (Revision – 1) : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle

Instructions : 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 passage based single correct type questions. +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 - IV • 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..

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8. x2sindxd

(A) (sin 2x)–1/2 (B) cos 2x (sin 2x)–1/2 (C) 2 cos 2x (sin 2x)–1/2 (D) cos 2x (sin 2x)1/2

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. The units of electrical permittivity

π=ε 2

2

0 Fr4q

are- (A) N–1m–2C2 (B) Nm–2C2 (C) C2/Nm2

(D) N/Cm2

10. If →a and

→b are two vectors with |a|

→= |b|

→ and

|ba|→→

+ + |b–a|→→

= 2 |a|→

, then angle between →a

and →b -

(A) 0º (B) 90º (C) 60º

(D) 180º

11. Two bodies P and Q are moving along positive x-axis

their position-time graph is shown below if →

PQV is

velocity of P.w.r.t Q and →

QPV is velocity of Q w.r.t P then –

x

t

PQ

(A) |V| PQ

→ = |V| QP

→ constant

(B) →

PQV is towards origin

(C) →

QPV is towards origin

(D) →

PQV and →

QPV both can be towards origin at same time

12. A particle of mass m is released from height h on a smooth curved surface which ends into a vertical loop of radius r as shown. If h = 2r then,

O r

θ h

m u = 0

(A)The particle reaches the top of the loop with zero velocity

(B) The particle cannot reach the top of the loop (C) The particle breaks off at a height h = r from base (D) The particle breaks off at a height r < h < 2r 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 : I (Ques. 13 to 15)

P O

R

θ

→B

→A

|A|→

= a |B|→

= b The second law of vector addition is triangle law,

which says that if we take →A and

→B as two vectors

acting at point O as shown in figure, then the

resultant of vector is get by taking →A and

→B as

adjacent sides of a triangle and the 3rd side of the

triangle as the resultant, then if θ is angle between →A

and →B then.

13. if α is the angle made by resultant vector with →A ;

then tan α =

(A) θ+

θcosAB

sinA (B) θ+

θcosBA

sinB

(C) θ+

θcosBA

cosA

(D) θ+

θsinAB

cosB

14. If the magnitude of both the vector |A|→

& |B|→

is A, then the resultant will have magnitude -

(A) A cos θ/2 (B) 2A cos θ/2 (C) 3A cos θ/2

(D) 3A cos θ/3

15. If |A|→

= |B|→

= a and θ = 120º, then the two vectors and the resultant will form a -

(A) Acute angle triangle (B) Obtuse angle triangle (C) Right angle triangle (D) Equilateral triangle

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XtraEdge for IIT-JEE AUGUST 2010 65

Passage: II (Ques. 16 to 18)

Height of a tower is 80 m. Two balls A and B are thrown simultaneously. Ball A is thrown upwards with speed u from top of tower while ball B is thrown upwards with speed of 50 m/s from the foot of tower.

16. Ball A will reaches ground in 8 sec if ball do not collide then u is equal to -

(A) 20 m/s (B) 25 m/s (C) 30 ms/s (D) 35 m/s

17. If the balls meets in air then the height from foots of tower is -

(A) 40 m (B) 80 m (C) 120 m

(D) 160 m 18. The time after which balls will meet in air is - (A) 3 sec (B) 4 sec (C) 5 sec

(D) 6 sec 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) (sin θ + cos θ)2 (P) 1 – sin 2θ (B) (sin θ – cos θ)2 (Q) 1 + sin 2θ (C) cos4θ – sin 4θ) (R) cos 2θ

(D) cos4θ + sin4θ (S) 1 + 2

2sin 2 θ

(T) None of these

20. The displacement - time graph of a body moving on a straight line is given by

x

t 0 T 2T

Parabola

Column-I Column-II

(A) (P)

T 2T

(B) (Q)

T 2T

(C) (R)

(D) (S)

T 2T

(T) None of these

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. According to Bohr’s theory, angular momentum of an electron in fourth orbit is -

(A) π2

h (B) π4

h (C) πh2 (D)

πh4

2. Calculate the maximum no. of possible e– for which 4 < n + l ≤ 6 -

(A) 18 (B) 36 (C) 72 (D) 4 3. If the De –Broglie wavelength of an electron in first

Bohr's orbit be λ then the minimum radial distance between the electrons in the first and second Bohr's orbit is –

(A) λ (B) 2λ (C) 2λ (D)

πλ2

Velocity – timegraph

Acceleration-timegraph

Distance – time graph

Speed – time graph

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4. Hexamethylenediamine, C6H16N2, is one of the starting materials for the production of nylon. It can be prepared from adipic acid C6H10O4, by the following reaction

C6H10O4(l) + 2NH3(g) + 4H2(g) → C6H16N2(l) + 4H2O(l) If 385 g of hexamethylenediamine is made from

5.00 × 102 g of adipic acid, the percent yield is (A) 24.2 % (B) 75.0% (C) 96.9 % (D) 99.9% 5. One mole of a mixutre of CO and CO2 requires

exactly 20 g of NaOH to convert all the CO2 into Na2CO3. How many more grams of NaOH would it require for conversion into Na2CO3, if the mixture (one mole) is completely oxidised to CO2 ?

(A) 60 g (B) 80 g (C) 40 g (D) 20 g 6. Rutherford’s experiment, which estabilished the

nuclear model of the atom, used a beam of - (A) β–particles, which impinged on a metal foil and

got absorbed (B) γ–rays, which impinged on a metal foil and

ejected electrons (C) helium atoms, which impinged on a metal foil

and got scattered (D) helium nuclei, which impinged on a metal foil

and got scattered 7. Lattice energy of BeCO3 (I) , MgCO3 (II) and CaCO3

(III) are in the order - (A) I > II > III (B) I < II < III (C) I < III < II (D) II < I < III 8. Specify the coordination geometry around the

hybridization of N and B atoms in a 1 : 1 complex of BF3 and NH3 - [IIT- 2002]

(A) N : tetrahedral, sp3 ;B : tetrahedral, sp3

(B) N : pyramidal, sp3 ; B : pyramidal, sp3 (C) N : pyramidal, sp3 ; B : planar, sp2 (D) N : pyramidal, sp3 ; B : tetrahedral, sp3

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. Select the correctly presented graph if v = velocity of e in Bohr's orbit r = radius of Bohr's orbit U = potential energy of e– in Bohr's orbit T = kinetic energy of e– in Bohr's orbit

(A)

n2

r

(B)

1/n2

U

(C)

n2

T

(D)

1/n

v

10. Which of the following have the same mass ? (A) 0.1 mole of O2 gas (B) 0.1 mole of SO2 gas (C) 6.023 × 1022 molecules of SO2 gas (D) 1.204 × 1023 molecules O2 gas

11. If n and l are principal and azimuthal quantam no. respectively, then the expression of calculating the total no. of electrons in any energy level is :

(A) ∑=

=

+n

0

)12(2l

l

l (B) ∑=

=

+1–n

1

)12(2l

l

l

(C) ∑+=

=

+1n

0

)12(2l

l

l (D) ∑=

=

+1–n

0

)12(2l

l

l

12. Which of the following is/are the correct order of mobility ?

(A) Li+ < Na+ < K+ (B) Na+ < Mg2+ < Al3+ (C) Al3+ < Mg2+ < Na+ (D) Li+ > Na+ > K+

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 : I (Ques. 13 to 15)

A black coloured compound (A) on reaction with dil H2SO4 form a gas 'B' and a solution of compound (C). When gas B is passed through solution of compound (C), a black coloured compound 'A' is obtained which is soluble in 50% HNO3 and forms blue coloured complex 'D' with excess of NH4OH and chocolate brown ppt. 'E' with K4[Fe(CN)6]

13. 'A' is (A) CuS (B) FeS (C) PbS (D) HgS

14. 'D' is (A) Cu(OH)2 (B) [Cu(NH3)2]SO4 (C) [Cu(NH3)4](NO.3)2 (D) [Cu(NH3)6]SO4

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XtraEdge for IIT-JEE AUGUST 2010 67

15. 'E' is (A) Cu2[Fe(CN)6] (B) [Cu4[Fe(CN)6] (C) Cu3[Fe(CN)6]2 (D) None of these Passage: II (Ques. 16 to 18)

Spin angular momentum of an electron has no analog in classical mechanics. However, it turns out that the treatment of spin angular momentum is closely analogous to the treatment of orbital angular momentum.

Spin angular momentum = h)1s(s +

Orbital angular momentum = hll )1( +

Total spin of an atom or ion is a multiple of 21 . Spin

multiplicity is a factor to confirm the electronic configuration of an atom or ion.

Spin multiplicity = (2Σs + 1). Answer the following questions : 16. Which of the following electronic configurations

have four spin multiplicity ? (A) (B)

(C)

(D)

17. In any subshell, the maximum number of electrons

having same value of spin quantum number is : (A) )1( +ll (B) l + 2

(C) 2l + 1

(D) 4l + 2 18. The orbital angular momentum for a 2p-electron is : (A) h3 (B) h6

(C) zero

(D) π2

h6

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. Column-I Column-II (Ionic species) (Shapes) (A) XeF5

+ (P) Tetrahedral (B) SiF5

– (Q) Square planar (C) AsF4

+ (R) Trigonal bipyramidal (D) ICl4

– (S) Square pyramidal (T) Octahedral 20. Column –I Column II (A) If P.E. = –13.6 eV (P) 21 (B) Ionization energy of (Q) 10 electron from 2nd shell of Na10+ (C) Number of spectral line (R)T. energy when electron jumps form = – 6.8 eV 7th to 3rd shell (D) Number of spectral lines (S) 411.4 eV when electron comes form 7th shell to 1st shell (T) zero

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. If x ∈

π

π 2,2

3 then value of the expression

sin–1 (cos(cos–1(cosx) + sin–1(sinx))) equals

(A) – 2π (B)

(C) 0 (D) None of these 2. If median AD of a ∆ABC divides the angle BAC in

ratio 1 : 2 then CsinBsin is equal to

(A) 3Asec

21 (B)

3Acos

21

(C) 3Aeccos

21 (D) None of these

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XtraEdge for IIT-JEE AUGUST 2010 68

3. If distance between incentre & one of the excentre of equilateral triangle is 4 unit. Then inradius of triangle is -

(A) 2 unit (B) 1 unit

(C) 23 unit (D)

21 unit

4. The solution of

θ+

θθcos1

3sin–sin3 +θ

θ+θsin–1

3coscos3 =

π

+θ4

cos24 is

(A) nπ (B) nπ + 12π

(C) nπ ± 2π

(D) 2nπ

5. In a ∆ABC if ∠A = 60º, cb =

213 + then ∠B – ∠C

has value equal to - (A) 15º (B) 30º (C) 22.5º

(D) 45º

6. A circle is inscribed in a triangle ABC touching the side AB at D such that AD = 5, BD = 3. If ∠A = 60º then length BC equals -

(A) 9 (B) 13

120

(C) 13

(D) 12

7. If n

1–n1–n

2cot2tan2......4tan42tan2tan

+αα+α+α+α = 1,

n ∈ N then general solution of α is,

(A) α =

π

π4

–n2 n– (B) α =

π

+π4

n2n

(C) α =

π

+π4

n2 n– (D) None of these

8. If (p, q) is at a distance θ from (1, 0) along circumference in anticlockwise direction & (r, s) is at a distance of 2θ from (p, q) along circumference in anticlockwise direction, then expression sp3 + q3r equals -

(A) 43 sin 2θ (B) sin 2θ

(C) 43 sin 4θ

(D) sin 4θ

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. In a ∆ABC a semi-circle is inscribed, whose diameter lies on side c. If x is length of angle bisector through angle C, then radius of semi-circle is -

(A) )BsinA(sinR4

abc2 +

(B) x∆

(C) x sin 2C (D)

s2∆

Where ∆ is area & s is semi-perimeter of the ∆.

10. If α = 7π which of the following holds true

(A) tanα. tan2α. tan3α = tan3α – tan2α – tan α (B) cosec α = cosec 2α + cosec 4α

(C) cos α –cos 2α + cos 3α has value equal to21

(D) 8 cos α. cos 2α. cos 4α has value equal to 1 11. If tan2θ = 2 tan2φ + 1, then value of cos 2θ + sin2φ is- (A) 1 (B) 2 (C) 0

(D) independent of φ

12. If

+

xcos1xcos 2

2 (1 + tan22y) (3 + sin3z) = 4

then - (A) x is a multiple of π (B) x is a multiple of 2π (C) y is a multiple of π/2

(D) None of these

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 : I (Ques. 13 to 15)

In a ∆ABC, if cos A. cos B. cos C = 8

1–3 and

sin A. sin B. sin C = 8

33 + , then

13. Value of tan A + tan B + tan C is

(A) 1–333 + (B)

1–343 +

(C) 1–33–6 (D) None of these

14. Value of Σ tan A. tan B = (A) 5 – 4 3 (B) 5 + 4 3

(C) 6 + 4 3 (D) 6 – 4 3

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XtraEdge for IIT-JEE AUGUST 2010 69

15. Value of tan A, tan B and tan C are - (A) 1, 3 , 2 (B) 1, 3 , 2

(C) 1, 2, 3 (D) None of these Passage: II (Ques. 16 to 18)

It is given that A = (tan–1x)3 + (cot–1x)3 , where x > 0

& B = (cos–1t)2 + (sin–1t)2, where t ∈

21,0 &

sin–1x + cos–1x = 2π for –1 ≤ x ≤ 1 and

tan–1x + cot–1x = 2π ∀ x ∈ R

16. The interval in which A lie is -

(A)

ππ2

,8

33 (B)

ππ8

,32

33

(C)

ππ5

,10

33 (D) None of these

17. Maximum value of B is -

(A) 8

2π (B) 16

(C) 4

(D) None of these

18. If least value of A is λ & max. value of B is µ then

cot–1

πλµµ–cot =

(A) 8π (B)

8–π

(C) 8

(D) 87– π

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. Column-I Column-II

(A) º1sinº3sin2

º2cosº–1cos 22 = (P) 1

(B) If cosθ = 54 , where θ∈

π

π 2,2

3 (Q) 21

& cos φ = 53 , where φ∈

π

2,0

then cos (θ – φ) has value

(C) If sin x cos4x + 2sin22x (R) 2

= 1 – 4 sin2

π

2x–

4 then

values of sin x is/are

(D) Number of values of t satisfying (S) 0 the equation cos (sin (cos t) = 1 for t ∈ [0, 2π] is (T) 3 20. Column-I Column-II

(A) If cos 7π + cos

73π + cos

75π (P) 3

= k cos 7π cos

72π cos

74π

then value of k is

(B) value of expression (Q) – 4

º30cosº.20cos

)1º20cos4(º20sin + is

(C) value of 1r

c–b + 2ra–c +

3rb–a (R) 2

is where r1, r2, r3 are exradii of ∆. corresponding to ∠A, ∠B & ∠C respectively

(D) In any ∆ABC, minimum value of (S) 0

Asin

1AsinAsin 2 ++ is

(T) 1

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XtraEdge for IIT-JEE AUGUST 2010 70

XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans D C B A B B A D B,C A,C,D Ques 11 12 13 14 15 16 17 18 Ans B,C B,C,D D B B D B C 19 A → Q B → R C → P, Q D → S 20 A → R B → P C → Q D → P

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans A C D D A C B A B,C,D A,B Ques 11 12 13 14 15 16 17 18 Ans D B D A D B B C 19 A → S B → R C → P D → Q 20 A → P B → S C → T D → Q

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans D A D A B A C A B,C A,C,D Ques 11 12 13 14 15 16 17 18 Ans A,C,D B,C,D A D C C C B 19 A → Q B → S C → R D → P 20 A → R B → Q C → Q D → P

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B A C A C B A B A,C A,D Ques 11 12 13 14 15 16 17 18 Ans A,C B, D B B D C C C 19 A → Q B → P C → R D → P 20 A → S B → R C → P D → Q

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans C B D C A D A A A,B,C,D B,C Ques 11 12 13 14 15 16 17 18 Ans D A,C A C A A C D 19 A → S B → R C → P D → Q 20 A → R B → S C → Q D → P

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans B A B D B C A C A,C A,B,C Ques 11 12 13 14 15 16 17 18 Ans C,D A,C A B D B C A 19 A → Q B → S C → P D → R 20 A → Q B → R C → S D → P

IIT- JEE 2011 (August issue)

IIT- JEE 2012 (August issue)

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"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. Every month get the XtraEdge Advantage at your door step.

Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice.

Take advantage of experts' articles on concepts development and problem solving skills

Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda.

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