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AUGUST 2010

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AUGUST 2010

WORRY IS A MISUSE OF IMAGINATION.

Volume - 6 Issue - 2 August , 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : xtraedge@gmail.com Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Cover Design Om Gocher, Govind Saini Layout Rajaram Gocher Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040000, 9001799502 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

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"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

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

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AUGUST 2010

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

CONTENTS INDEX

Regulars ..........

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

PAGE

NEWS ARTICLE

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IITian ON THE PATH OF SUCCESS

6

KNOW IIT-JEE

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22-year-old becomes youngest IIT teacher Delhi power firm, IIT tie-up to reduce power loss Prof. Mohit Renderia & Dr. Rajiv Laroia Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS 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."

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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

Test Time .......... XTRAEDGE TEST SERIES

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Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper

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AUGUST 2010

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 himselfâ€” 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.

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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 Tulsiâ€™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 tieup 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 6

"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 AUGUST 2010

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

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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).

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 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.

"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 lanefollowing' 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 selfpowered 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.

"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 lifethreatening 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 IITKharagpur team, comprising Nalin 7

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.

AUGUST 2010

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

Dr. Rajiv Laroia B.Tech., IIT – Delhi

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.

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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. 8

AUGUST 2010

KNOW IIT-JEE By Previous Exam Questions

2.

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]

P0

m 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 L

L Kl

l

mg = kl ; ⇒y= ⇒ y=

where

mg mg l= k

mg k g

ω2 g ω2

Mean Position of oscillation

C

(c) As ∆U = 0 in cyclic process, hence ∆Q = ∆W ∆QAB + ∆QCA + ∆QBC = ∆W, PV PV ∆QBC = P0V0 – 0 0 = 0 0 2 2 As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C.

[Q K = mω2] < a (given)

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A

2V0 V0 (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 : 5R 3R Cp = , Cv = 2 2 Cyclic process A → B ⇒ Isochoric process C → A ⇒ Isobaric compression (a) Work done = Area of closed curve ABCA during cyclic process. i.e. ∆ABC 1 1 ∆W = × base × height = V0 × 2P0 = P0V0 2 2 (b) Heat rejected by the gas in the path CA during Isobaric compression process ∆QCA = nCp∆T = 1 × (5R/2)(TA – TC) 2P0 V0 PV TC = , TA = 0 0 , I×R I×R 5 5R P0 V0 2P0 V0 = – P0V0 ∆QCA = − 2 2 R R Heat absorbed by the gas on the path AB during Isochoric process ∆QAB = nCv∆T = 1 × (3R/2) (TB – TA) 3R 3P0 V0 P0 V0 − = = 3P0V0 2 1× R 1× R

y0

K

One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate. [IIT-1998] P B 3P0

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AUGUST 2010

(d) Equation for line BC 2P P = – 0 V + 5P0, As PV = RT hence, V0 RT P= [For one mole] [as y = mx + c] V 2P ...(1) ∴ RT = – 0 V2 + 5P0V V0

For maximum;

⇒ ⇒

2P0 dT = 0, – × 2V + 5P0 = 0; dV V0

5V0 4 Hence from equation (1) and (2)

∴ V=

...(2)

Substituting the value of q1' from (2) in (1) q1 ' + q2' = σ × 20 πR2 2 3q 2 ' = σ × 20 πR2 2 q2 ' 5 σ = × 2 3 2 4π(2R )

⇒

New charge density on bigger sphere q2 ' 5σ = = 2 6 4π(2R )

4.

In the given circuit R1 = 2R4 = 6 ohms E1 = 3E2 = 2E3 = 6 volts ; ; C = 5µF R3 = 2R2 = 4 ohms Find the current in R3 and the energy stored in the [IIT-1988] capacitor. E1 R1

2

2P0 5V 5V × 0 + 5P0 0 V0 4 4 25P0 V0 25 25 = –2P0V0 × + = P0V0 16 8 4 25 P0 V0 ∴ Tmax = 8 R

RTmax = –

C

Sol.

R1 = 6Ω

σ

q'1

For sphere of radius R

For sphere of radius 2R

R2 = 2Ω i2 i2+i1 F R3 = 4Ω 2V = E

D

σ=

σ=

A i1 B i2

4πR 2 q2 4π(2R ) 2

⇒

q1' =

.

E3 = 3V

R4 = 4Ω

C

Applying Kirchoff's law in ABFGA 6 – (i1 + i2) 4 = 0 Applying Kichoff's law in BCDEFB I2 × 3 – 3 – 2 + 2i2 + (i2 + i1) 4 = 0 Putting the value of 4 (i1 + i2) = 6 in (2) 3i2 – 5 + 2i2 + 6 = 0 1 ∴ i2 = – A 5 Sybstituting this value in (i) we get 1 i1 = 1.5 – – = 1.7 A 5 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 1 CV2 ∴ Energy stored in capacitor = 2

q1

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 ...(1) q'1 + q'2 = q1 + q2 = σ × 20 πR2 Also Since V1 = V2 1 q1 ' 1 q2 ' = 4πε 0 R 2πε 0 2R

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E1 = 6V

i2

q'2

⇒

q2 ' 2

G

2

q1 = σ × 4πR2

∴

R4

5µF = C

V

Connecting Wire

R3

E3

E

V

R2

E2

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] σ σ q2 Sol. q1 2R R 3.

..(2) 10

…(1) …(2)

AUGUST 2010

→

1 × 5 × 10–6 × (2.4)2 2 = 1.44 × 10–5 J

=

5.

I

120º

0.11µ 0 IQv ˆ 0.11 3µ 0 IQv ˆ j– i 2a 2a ∴ Instantaneous acceleration

O

0.11µ 0 IQv ˆ F a= m = ( j − 3 ˆi ) 2am m (b) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop

x

P

→

will be, M =(IA) kˆ Here, A is the area of the loop. 1 1 A = (πa2) – [2 × a sin 60º] [a cos 60º] 3 2

N (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

=

∴ M =(0.61 Ia2) kˆ

B = B ˆi 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 : y M a a

N

→

y

→

B = B ˆi

→

CHEMISTRY 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 HCO32– 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 [IIT-1997] water, mass/mass) Sol. In 106 g(= 1000 kg) of the given hard water, we have amount of SO42– ions = 96 g amount of HCO3– ions = 183 g 96 g So amount of SO42– ions = = 1 mol 96 g mol −1 6.

v 60º

x

1 1 µ I (field dut to circle) = 0 3 3 2a 0.16µ 0 I µ I (outwards) = 0 (outwards) = a 6a

B1 =

→

0.16µ 0 I ˆ k a Magnetic field due to straight wire NM at P : µ I B2 = 0 (sin 60º + sin 60º) 4π r Here, r = a cos 60º µ I ∴ B2 = 0 (2 sin 60º) 4π a cos 60º µ I 0.27µ 0 I (inwards) or B2 = 0 tan 60º = 2π a a → 0.27µ 0 I ˆ or B 2 = – k a → → → 0.11µ 0 I ˆ ∴ B net = B1 + B 2 = – k a Now, velocity of particle can be written as, → 3v ˆ v ˆ j v = v cos 60º ˆi + v sin 60º ˆj = i+ 2 2 Magnetic force

or B1 =

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→

Given,

∴ τ = M × B = (0.61 Ia2B) ˆj

x

+Q P

πa 2 a2 – sin 120º = 0.61 a2 3 2

→

→

60º r 60º

→

→

a

I

→

=

A wire loop carrying a current I is placed in the x-y plane as shown in figure. [IIT-1991] y v M +Q

→

Fm = Q( v × B )

and amount of HCO3– ions =

183 g 61 g mol −1

= 3 mol

These ions are present as CaSO4 and Ca(HCO3)2. 3 Hence, amount of Ca2+ ions = 1 + = 2.5 mol 2 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. 11

AUGUST 2010

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

2 CH3 – CH – CH – CH3 Y and Z (C6H12) H→

Ni

CH3 CH3 2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3 and CH3 – CH – C = CH2

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 [IIT-1986] of methanol in the vapour. Sol. Given that, Pe0 = 44.5 mm Hg For ethanol (C2H5OH), M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g 60 m ∴ Moles of ethanol, ne = = = 1.3 46 M 7.

Ni

CH3 CH3

(Y) H2 Ni

CH3 – CH – C = CH2 CH3 CH3

CH3 – CH – CH – CH3 CH3 CH3

(Z)

Both, Y and Z can be obtained from following alkyl halide : Cl CH3 – C – CH – CH3

K-t-butoxide

CH3 CH3

∆; –HCl

2-chloro-2,3-dimethyl butane (X)

nm 1.25 1.25 = = 1.3 + 1.25 2.55 ne + nm According to Raoult's law, 44.5 × 1.3 Pe = Pe0 xe = = 22.69 mm Hg 2.55 88.7 × 1.25 = 43.48 mm Hg and Pm = Pm0 xm = 2.55 Hence, total vapour pressure of the solution, PT = Pe + Pm = 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law, (in vapour form) Pm = PTx´m Hence, mole fraction of methanol in vapour form, P 43.48 x´m = m = = 0.66 66.17 PT

CH2 = C — CH – CH3 + CH3 – C = C – CH3 CH3 CH3 (Z) 20%

CH3 CH3 (Y) 80%

Cl

Hence,

X, CH3 – C – CH – CH3 CH3 CH3 Y, CH3 – C = C – CH3 CH3 CH3

Z, CH3 – CH – C = CH2 CH3 CH3

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) [IIT-1992] and write the equations involved. 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.

9.

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 [IIT-1996] and Z. Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes. 8.

K − t − butoxide C 6 H13Cl → Y + Z C 6 H12

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.

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(Z)

CH3 CH3

xm =

∆ ; – HCl

2,3-dimethyl

The hydrogenation of Y and Z is shown below : H2 CH3 – C = C – CH3 CH3 – CH – CH – CH3

For methanol (CH3OH), Pm0 = 88.7 mm Hg M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g 40 m ∴ Moles of methanol, nm = = = 1.25 32 M ne 1.3 1.3 ∴ xe = = = 1.3 + 1.25 2.55 ne + nm

X

CH3 CH3 butene-1

CH3 CH3 2,3-dimethyl butene-2 (Y)

12

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. C 2 H 4 – CH2OH + 2CH3MgBr →

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: 4BX 3 + 3LiAlH4 → 2B 2 H 6 + 3LiX + 3AlX3 (X)

(X = Cl or Br) B 2 H 6 + 3O2 → B2O3 + 3H2O + heat (Y)

(X)

Structure of B2H6 is as follows: Hb

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.

B

2

1

ZnCl + HCl

(X)

Ag – C≡C – CH2CH2CH2OH + NH4NO3

Br MgC≡C.CH2CH2CH2OMgBr + 2CH4

Pentanol-1 CH3CH2CH2CH2CH3 n-pentane

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 2 × 22.4 × 0.42 ∴ 0.42 g (X) gives = 84 = 224 ml of CH4

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121.5º

11. Find the values of a and b so that the function x + a 2 sin x, 0 ≤ x ≤ π/ 4 f(x) = 2 x cot x + b, π/ 4 ≤ x ≤ π/2 a cos 2 x − b sin x , π / 2 < x ≤ π [IIT-1989] is continuous for 0 ≤ x ≤ π Sol. As, f(x) is continuous for 0 ≤ x ≤ π π π ∴ R.H.L. at x = = L.H.L. at x = 4 4

CH3CH2CH2CH2CH2OH

∆, –H2O; –I2

B

MATHEMATICS

White ppt.

2 HI

1.33Å

Hb Ht 1.77Å Thus, the diborane molecule has four two-centre-two electron bonds (2c-2e– bonds) also called usual bonds and two three-centre-two-electron bonds (3c-2e– bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively

2 Room → No reaction H − C ≡ C − CH 2 CH 2 CH 2 OH temp.

2H2/Pt

97º

Ht

Ht 1.19Å

4-pentyne-1-ol

2CH3MgBr

Hb B

The different equations of (X) are :

NH3

Hb Ht

HC≡C.CH2 CH2 – CH2OH (X)

AgNO3

Ht

or

CH3CH2CH2CH2CH3 + H2O + I2

4 3

B

Ht

n-pentane On the basis of abvoe analytical facts (X) has the structure : 5

Ht

Ht

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

(Y)

π π π π ⇒ 2. cot + b = + a 2 . sin 4 4 4 4 π π ⇒ +b= +a 2 4

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AUGUST 2010

{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

π ....(i) 4 π π also, R.H.L at x = = L.H.L at x = 2 2

⇒ a–b=

π π 2π π − b sin = 2. . cot + b ⇒ a cos 2 2 2 2 ⇒ –a–b=b ⇒ a + 2b = 0 ...(ii) 3π −3π From (i) and (ii), a = and b = 2 4 dy at x = –1, when 12. Find dx (sin y)

π sin x 2

+

14. If exp {(sin2x + sin4x + sin6x + ...... ∞). ln 2} satisfies the equation x2 – 9x + 8 = 0, find the value of cos x π ,0<x< . [IIT-1991] cos x + sin x 2 Sol. exp {(sin2x + sin4x + sin6x + ...... ∞) loge2

3 sec–1(2x) + 2x tan ln (x + 2) = 0 2 [IIT-1991]

Sol. Here, π sin x 2

3 sec–1(2x) + 2x tan (log (x + 2)) = 0 2 Differentiating both sides, we get (sin y)

π sin x (sin y) 2

+

. log(sin y) . cos

π π x. 2 2

=

π

2

3 1− π

2

=

3

2 tan x = 1 and 2 tan x = 8 tan2x = 0 and tan2x = 3

∴

π π2 − 3

x

satisfy x2 – 9x + 8 = 0 ⇒ x = 1, 8

2

cos x = cos x + sin x

15. Find the value of :

13. ABC is a triangle such that

2

3 −1 2

cos (2 cos–1 x + sin–1x) at x =

1 , 5

where 0 ≤ cos–1x ≤ π and –π/2 ≤ sin–1x ≤ π/2 [IIT-1981] Sol. cos{2cos–1x + sin–1x} π π = cos cos −1 x + , as cos–1x + sin–1x = 2 2 –1 = – sin(cos x )

1 2 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) 1 ⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) = 2 ⇒ 2A + 60º = 30º, 150º

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

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∴ ⇒

2

2

2 x . sec 2 (log( x + 2)) 2 3 + . ( x + 2) 2 (2 | x |) 4x 2 − 2 x + 2 log 2 . tan (log(x + 2)) = 0 3 putting, x = −1, y = − , we get π

3 π

2 tan

tan 2 x

π x = nπ and tan2x = tan 3 π and x = nπ ± 3 π Neglecting x = nπ as 0 < x < 2 π π ⇒ x= ∈ 0, 3 2 1 1 3 −1 3 −1 cos x 2 ∴ = = × = 2 cos x + sin x 1 3 1+ 3 3 −1 + 2 2

+

dy dx −1, −

⇒

⇒ e log e 2

⇒

sin x −1 dy π + sin x (sin y) 2 . cos y . dx 2

− 3 π

⇒

sin 2 x . log e 2 2 e 1−sin x

= – sin(sin–1 1 − x 2 ) 1 = – sin sin −1 1 − 2 5 2 6 2 6 = – sin sin −1 = 5 5

14

AUGUST 2010

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15

AUGUST 2010

Physics Challenging Problems

Set # 4

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. So lutions will b e p ub lished in n ex t issue 1.

d 4b & 2b − d 3 d 2b (B) K = & 2b − d 2 d 2b (C) K = & 2b − d 2 d 2b (D) K = & 2b − d 4

In the circuit shown in figure C1 = C 2 = 2µF . Then charge stored in (steady state)

(A) K =

1Ω 2Ω 3Ω C1 C 2 2Ω 1Ω 3Ω

120 V

(A) capacitor C1 is zero (B) capacitor C2 is zero (C) both capacitors is zero (D) capacitor C1 is 40µC

Q. 6

<b≤d <b≤d ≤ b ≤ 2d ≤b≤d

In the circuit shown in figure S1

Q. 2

Find the total charge of the sheet if a electric force of 0.08N is exerted on it (B) 0.32µC (A) 0.28µC (D) 0.38µC (C) 0.24µC

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 (C) 3.5 × 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

S1

2Ω 10V 3Ω

1Ω

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. E1 E2

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By : Dev Sharma Director Academics, Jodhpur Branch

i

20V

(A) i = 2.5A when S1 is closed and S2 is open 20 (B) i = A when S1 is open and S2 is closed 3 5 (C) i = A when S1 and S2 both are open 3 (D) i = 20A when both S1 and S2 are closed Q. 7

A charged particle of unit mass and unit charge moves with velocity

→

∧

∧

v = (8 i + 6 j)m / s

in a

∧

magnetic field of B = 2 k T . 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

(B) 12.5 × 104 V/m (D) 8.7 × 104 V/m

16

In the diagram × × × × × × × × × P1 P2 shown, the wires P1Q1 and 4 cm × × × × × × × × × 2Ω 2Ω 9Ω P2Q2 are made × × × × × × × × × to slide on the Q × × × 1× ×Q2 × × × × rails with same speed of 5m/s. In this region a magnetic field of 1T exists. The electric current in 9Ω 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 AUGUST 2010

8

1. [A,C] I =

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Ju ly Is su e

E (1 − e − t / τ ) R

L , if L increases then τ R 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.

E and R is constant. τ =

2. [B,C] (Moderate) As the length is double, the cross section area of the wire becomes half, thus the L becomes four resistance of the wire R = ρ A 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. V2 The power delivered to resistance is p = and R 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. R.R V <R R + RV

4. [A]

RA =

5. [A]

RB = R + RG > R

6. [A]

(Tough)% error in case A RV RA − R × 100 = − 1 × 100 R R + RV −R = × 100 = −1% R + RV %error in case B R RB − R × 100 = G × 100 = 10% R R Hence percentage error in circuit B is more than that in A.

7. [D]

Set # 3

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

•

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.

8. [C]

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17

AUGUST 2010

Students' Forum PHYSICS 1.

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

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)

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 Ans. Or amplitude of oscillations, a = 5 cm

A h B

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 =

∴ or or

2gh .

Let velocity of combined body just after collision be v. Applying law of conservation of momentum for collision, v 1 2mv = mv0 or v= 0 = 2gh 2 2 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 mg Ky0 = mg or y0 = K Hence, initial strain energy in spring, 1 m 2g 2 U0 = Ky02 = 2 2K 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. m 2g 2 1 1 Ky2 = + (2m) v2 + 2mg (y – y0) 2K 2 2 1 m 2g 2 Ky2 – 2mgy + 2mgy 0 – – mv 2 = 0 2K 2 2 500 y – 10y – 1.2 = 0 From above equation y = + .6 or – .04

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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 m C

v0

A

B

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 v or v = 0 = 0.3 ms–1 2 ∴ Velocity of centre of mass of the system, 2m × v + m × 0 vc = = 0.2 ms–1 2m + m Now the system is as shown in fig. 2m m

18

AUGUST 2010

2m (2m)(m) = 3 2m + m ∴ Frequency of oscillations, 1 K 5 10 f= = Hz. 2π m 0 π

Its reduced mass,

During a very small time interval dt,

m0 =

GM dt x Work done by satellite against resistance of cosmic dust is

Distance travelled by the satellite is vdt = ∴`

Ans.

3/ 2

GM dW = F . (v.dt) = a dt x 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) 1 GMm dx. dE = 2 x2

Since, just after the collision, combined body has velocity v, therefore, energy of the system at that 1 instant, E = (2m)v2 = 0.27 joule 2 Due to velocity vC of centre of mass of the system, translational kinetic energy, 1 Et = (3m) v c2 = 0.18 joule 2 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. 1 then Kx 02 = E0 2 ∴ x0 = 90 × 10–3 m or 3 10 mm Ans.

3/ 2

∴ or

1 GMm GM . dx = – a dt 2 2 x x m dx dt = – …(3) 2a GM x 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. t =? x = R dx m dt = – . t =0 2a Gm x = nR x

∫

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 3.

∫

or 4.

t=

m R ( n – 1) a GM

Ans.

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

Planet

A

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, GM …(1) Velocity of satellite is v = x 1 GM.m ...(2) and energy of satellite is E = – 2 x aGM Resisting force, F = av2 = x

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d

B

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.

19

AUGUST 2010

+q

p

+(2 × 10–10 – q)

+(10–10 + q)

–q

2

1

Considering a point P inside the plate B, Electric field on it is (10 −10 + q ) q q E= – – 2ε 0 S 2ε 0 S 2ε 0 S –

–

(2 × 10 −10 − q) =0 2ε 0S

150 pC +

–

+

–

+

q2 – +

+ 150 pC

+ +

q2 – q3

2

+

50 pC

5

6

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.

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

4

3

1

+

+

–

+

q1

–

+

–

+

6

–

(q2 – q3)

– +

+ 150 pC + +

–

+

+

+

+ 150 pC

+

+

(50×10–12–q´)

– –

(q1 + q2) +

q

+ – E

Capacitance of the capacitor is ε S C = 0 = 5 × 10–12 F d Applying Kirchhoff's voltage law, (50 × 10 −12 − q´) –E=0 – C ∴ q´ = 1 × 10–10 coulomb ∴ Energy supplied by battery = q´E = 10–9 joule 5.

–

Applying Kirchhoff´s voltage 1 – 2 – 6 –1, q q2 q + 3 – 1 =0 C C C or q1 = (q2 + q3)

q´

∴

Ans.

q2

5

(q1 + q2)

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.

4

–

– + q3 + – (q1 – q2 + 2q3)

+ q3

q1

3

law

on

mesh

...(i)

For mesh 2 – 3 – 6 – 2, q 2 − q3 q − q 2 + 2q 3 q – 1 – 3 =0 C C C ...(2) or q1 = (2q2 – 4q3) From equation (1) and (2), q2 = 5q3 and q1 = 6q3 Now applying Kirchhoff's voltage law on mesh 1 – 6 – 5 – 4 – V – 1, q − q3 q1 q + 2 + 2 –V=0 C C C 1 Substituting q1 = 6q3 and q2 = 5q3, q3 = CV. 15 But charge drawn by the arrangement from battery is 11 CV q = (q1 + q2) = 11q3 = 15 q 11C Equivalent capacitance = = = 11µF Ans. V 15

Nine identical capacitors, each of capacitance C = 15 µF are connected as shown in fig. Calculate equivalent capacitance between terminals 1 and 4.

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20

AUGUST 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Capacitor-2 KEY CONCEPTS & PROBLEM SOLVING STRATEGY Capacitors in Series : +Q –Q

+Q –Q

+Q –Q

A C1

C2

V1

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

C3

V2

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

B

V3

V

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

V1 =

C Q1 = 1 Cp

U=

Capacitors in Parallel :

A

–Q1 C1

+Q2

–Q2 C2

+Q3

–Q3 C3

B

⇒ U=

Q Q1 Q = 2 = 3 C1 C2 C3

V3 =

1 C3 1 Cs

V

1 1 1 1 1 = + + + .... + C1 C 2 C 3 Cn Cs

Q;

C Q2 = 2 Cp

Q;

C Q3 = 3 Cp

Q

ε0A σ and V = Ed where E = ε0 d

1 ε0A 2 2 Ed 2 d

1 ⇒ U = ε 0 E 2 (Ad) 2

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=

V;

1 CV2 2

where C =

V

⇒

1 C2 1 Cs

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

Q Q and V = C1 Cs

+Q1

V2 =

(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

1 1 1 1 = + + Cs C1 C 2 C 3

Further V1 =

V;

and so on, where

Q Q Q Q = + + Cs C1 C 2 C 3

⇒

1 C1 1 Cs

⇒

U=

1 ε0E2τ 2

where τ is volume of the capacitor ⇒

U Electrostatic Energy = Ue = τ Volume

= Electrostatic Pressure

...(2)

=

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

σ2 1 ε0E2 = 2 2ε 0

σ Q E = ε0

⇒ Q = Q1 + Q2 + Q3

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21

AUGUST 2010

Energy for series and parallel combinations : Series Combination : For a series combination of capacitor Q = constant and

⇒

F=

σ2 Q2 = 2ε 2ε 0 A 0

σ Q Q = σA, E = ε0

1 1 1 1 = + + + ... C1 C 2 C 3 Cs

⇒

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

Q2 Q2 Q2 Q2 = + + + .... 2C s 2C1 2C 2 2C 3

⇒ Us = U1 + U2 + U3 + ...... Parallel Combination : For a parallel combination of capacitors V = constant and Cp = C1 + C2 + C3 + .... ⇒

∑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

1 1 1 1 CPV2 = C1V2 + C2V2 + C3V2 + ... 2 2 2 2

⇒ 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 + – + – + – + – + – – +

⇒ x = q1 – q2 +

Also

U=

Q x 2ε A 0

Q2 Q2 = (x – dx) 2C´ 2ε 0 A

If dU is the change in potential energy, then dU = Uf – Ui Q2 Q2 (x – dx) – x 2ε 0 A 2ε 0 A

⇒

dU =

⇒

dU = –

Q2 dx 2ε 0 A

dU dx

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+

C

∑V = 0

(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 q capacitor i.e. ∆V = – . C

Further since F=–

–

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 q capacitor i.e. ∆V = + . C

2

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 =

–

B

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. ⇒

Q ⇒U= 2C

+q1 –q1 +q2 –q2

1 2

ε0A x 2

–

A

At any instant let the plate separation be x, then C=

A = 1 ε 0 E 2 A 2

22

AUGUST 2010

2.

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.

Find the equivalent capacitance between the point A and B in figure. C1

C2

A

C C B C

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. 1 C 3

A

2C

2C

C C

C 4

C

P

P

C

B

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 3C Resultant of 3C and (3) is 4 3C 7C Resultant of and (4) is 4 4 So total capacitance across AB is C 7C CAB = AP ⇒ CAB = 8 2

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C2 C1 Q1 –Q1 Q–Q1 –(Q–Q1) a b D (2Q1–Q) i –(2Q1–Q) C3 j

B

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) Q Q − Q1 or VA – VB = 1 + ...(1) C1 C2 Also, VA – VB = (VA – VD) + (VD – VE) + (VE – VB) Q 2Q1 − Q Q or VA – VB = 1 + + 1 ...(2) C1 C3 C1 We have to eliminate Q1 from these equation to get Q the equivalent capacitance . (VA − VB ) The first equation may be written as 1 1 Q + − VA – VB = Q1 C2 C1 C 2

C

C

C1

e f E g h Q–Q1 –(Q–Q1) Q1 –Q1 C2 C1

C

A

B

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.

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

C

C3

A

Solved Examples 1.

C2

C1C 2 C1 (VA – VB) = Q1 + Q C 2 − C1 C 2 − C1 The second equation may be written as 1 1 Q – VA – VB = 2Q1 + C1 C 3 C 3

or

23

...(3)

AUGUST 2010

Further C34 is again in parallel. Hence the effective capacity C × 2C 0 5 5 A = C0 + 0 = C0 = Kε0 . 3 3 d C 0 + 2C 0 (ii) Charge on the plate 5 = charge on the uper half of parallel combination 2 Kε 0 AV0 2 ∴Q5 = V0 C 0 = 3 d 3 Charge on plate 3 on the surface facing 4 kε 0 AV0 ∴ V0C0 = d Charge on plate 3 on the surface facing 2 = [potential difference across (3 – 2)]C0 AV0 C0 = V0 C0 = Kε0 3d C 0 + 2C 0

C1C 2 C1 (VA – VB) = Q1 – Q ...(4) 2(C1 + C 3 ) 2(C1 + C 3 ) Subtracting (4) from (3) CC C1C 3 (VA – VB) 1 2 − C 2 − C1 2(C1 + C 3 )

or

C1 C1 + = Q C 2 − C1 2(C1 + C 3 ) or (VA – VB)[2C1C2(C1 + C3) – C1C3(C2 – C1)] = C1[2(C1 + C3) + (C2 – C1)]Q 2C1C 2 + C 2 C 3 + C 3C1 Q or C = = VA − VB C1 + C 2 + 2C 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 3.

Kε 0 AV0 AV0 + Kε0 d 3d Kε 0 AV0 1 4 A = 1 + 3 = 3 Kε0 d V0 d

∴ Q3 =

4.

3µF

3µF

20Ω

C

A

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

2

3µF

1µF B

3

2

3

4

5

4

3µF

Q2 (–)

(+)

A

(b)

10Ω C

100 V Fig.(a) B 2µF

6µF

i.e., C12, C32, C34 and C54. The capacity of each Kε A capacitor is 0 = C0. The effective capacity d 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 C 0 × 2C 0 C 0 + 2C 0

1µF 1µF

20Ω

Q1

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10Ω

100 V

(Q2/2)

Q

1µF

1µF

(+)

(a)

(Q2/2)

1µF

B

(–)

4 3 2 1 1

In diagram find the potential difference between the points A and B and between the points B and C in the steady state.

3/2 µF

R

1µF

1µF

P

Q 20Ω

A

P

10Ω C 100 V

Fig.(b)

S

Q 20Ω

A

10Ω C 100 V

Fig.(c)

From fig. (c). potential difference between P and Q = Potential difference between R and S = 100 volt 24

AUGUST 2010

2

3 × 10–6 × 100 2 = 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. Q 150 × 10 −6 = ∴ V1 = = 25 volt. capacity 6 × 10 −6 Again potential difference between C and D = potential difference across the two ends of condenser of capacity 2µF 150 × 10 −6 = 75 volt V2 = 2 × 10 −6

∴ Q = capacity × volt =

5.

=

∴ Total final energy CV 2 5 3 = CV2 Ef = CV2 + 6 3 2 ∴

S

A

C

B

C

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 K C1 = 1 C = 3C. K 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 V ∴ V1 = volt 3 1 Initial energy of capacitor A = CV2 2 1 energy of capacitor B = CV2 2 1 1 ∴ Total energy Ei = CV2 + CV2 = CV2 2 2 Final energy of capacitor A 1 3 × (3C)V2 = CV2 = 2 2 Final energy of capacitor B

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Ei 3 CV 2 = = 2 5 Ef (5 / 3)CV

Believe

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

CV 2 1 V × (3C) = 6 2 3

25

•

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.

AUGUST 2010

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26

AUGUST 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Work, Power, Energy & Conservation Law KEY CONCEPTS & PROBLEM SOLVING STRATEGY Work, Energy and Power :

The potential energy of particle in the gravitational field is given by U = U0 + mgh

Work is done when a force (F) is displaced. dr

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 1 = kx2, where k is the force constant of the spring 2 and x is the charge in length of the spring. This change in length may be either a compression or on extension.

θ F

The work done is dW = F dr cos θ Using vector notation rr dW = F.dr 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. r r For a system of particles the quantity F.dx cm is

Potential Energy and force

Fx = –

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 nonconservative force.

∫

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 =

work done time taken

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).

The unit of power is the watt (= joules/second). rr The power of an agent is given by P = F.v where F is the force applied by the agent and v is the velocity of the body on which the agent applies the force.

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.

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 =

1 mv2. 2

The kinetic energy of a system is T =

1 2 Mv cm 2

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∂U ∂x

27

AUGUST 2010

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 Yaxes. 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

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(A) Problem solving strategy : Work and Kinetic Energy : Step 1 : Identify the relevant concepts : The workenergy 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 workenergy 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 xaxis.) 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. rr W = Fs cos φ or W = F.S . (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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AUGUST 2010

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 xcomponent 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 !

1 mv2 can never 2 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 r r methods, by using F = ma directly, or by a

remember is that kinetic energy K =

∑

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. 1 1 mv12 + mgy1 = mv22 + mgy2 2 2 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.

XtraEdge for IIT-JEE

29

AUGUST 2010

∴ Total work done 20 x W= 20 − g dx = g 0 2

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.

∫

1 20dx − 2 xdx

0

20 x 2 = g {20 x}020 − 4 0 = g[400 – 100] = 300g = 300 × 10 = 3000J

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 3.

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 50000 = 2000 N or P = 25 1000 N 2000 N

(P) =

∫

T

0

− mg

P( t ) dt

∫

T

0

= dt

∫

T

0

( v 0 sin α − gt )dt T

− mg[ v 0 sin αT − gT 2 / 2] T Here T is total time of flight T = (2 v0 sin α)/g Substituting this value, we have v sin α(2v 0 sin α) / g − g (4 v 02 sin 2 α / 2g 2 ) (P) = – mg 0 (2 v 0 sin α) / g Solving we get (P) = 0

=

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

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 u2 4.

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. 10 1 ∴ mass lost per unit distance = = kg 20 2 Consider a point at a height x from the bottom of the well. At height x from the bottom, the bucket weighs = x 20 − kg. The work done against the force during 2 2.

m1

m2

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 20 + 15 = 5 m/s v= 7

x elementary displacement dx = 20 − dx.g 2

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∫

20

30

AUGUST 2010

Now applying the law of conservation of kinetic energy, we get 1 1 1 1 m1u12 + m 2 u 22 = (m1 + m2)v2 + kx2 2 2 2 2 or m1u12 + m2u22 = (m1 + m2)v2 + kx2 (2 × 100) + (5 × 9) = (7 × 25) + (1120 × x2) 200 + 45 = 175 + 1120 x2 200 + 45 − 175 1 x2 = = 1120 16 1 x= = 0.25 m 4

(40.2) 2 v2 = = 82.45 m 2 × 9.8 2g The height to which it rises above the cliff is = 82.45 – 4.90 = 77.5 m

=

Regents Physics

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 1 s1 = gt2 ...(1) 2 The distance s2 moved by the bullet in time t second is given by 1 1 s2 = ut – gt2 = 100t – gt2 ...(2) 2 2 Adding eqs. (1) and (2) s1 + s2 = 100 t or 100 = 100t (Q s1 + s2 = 100 m) ∴ t = 1 sec. 1 2 1 Now s1 = gt = × 9.8 × 1 = 4.9 m 2 2 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. 9.8 − 90.2 v= = – 40.2 m/s 2 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

5.

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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 :

31

•

Centripetal force and centripetal acceleration vectors are toward the center of the circlewhile 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.

AUGUST 2010

KEY CONCEPT

REACTION MECHANISM

Organic Chemistry Fundamentals

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 :

∆

Br

hν CH4 + Cl2 → CH3Cl + HCl Br hν + Br2 → + HBr

NBS → hν

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 → ↔ ≡

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

CH3

• • CH3CH2 + Cl → CH3CH2Cl

AlCl

+ CH3Cl 3 →

• • CH3CH2 + CH2CH3 coupling → CH2CH2CH2CH3 • • CH3CH2 + CH3CH2 disproport ionation → 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

+ HCl

Step I. Formation of an electrophile ⊕

CH3Cl + AlCl3 → AlCl4– + C H 3 Step II. Attack of electerophile on benzene when resonance-stabilised σ complex is formed. H ⊕ H ⊕ CH3 + CH3 ↔ ↔ CH3 ↔ ⊕ (carbocation-an arenium ion)

H vinylic For CH2—H 1º H Allylic H HH H 3º 2º 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. H

H CH3 ≡

Allylic H 2º H

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Br

NBS →

⊕

+

H CH3

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

Step IV.

32

H CH3 →

CH3

+ H+

AlCl4– + H+ → AlCl3 + HCl

AUGUST 2010

An additional reagent (called Lewis acid) is always required that can help in the formation of an electrophilic NO2 4 + NHO3 H2SO + H2O → Fe + Br2 →

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 S N1 reaction.

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 Nu ⊕ H H Θ CH3 CH3 + N u →

CH3 | CH3C—Cl | CH3

If there is attack of nucelophile on positive site, aromaticity is lost, hence this is not desirable.

CH3 CH3 O O | | || || Θ fast CH3C—O—CCH3 CH3C ⊕ + O—CCH3 → step II | | CH3 CH3 As a result of SN1 reaction, there can be recemisation and inversion.. When (–) 2-bromobutane 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. •• H2O •• CH3 I CH3 ⊕ II H Br H Θ Br CH2CH3 CH2CH3 (–) I II CH3 CH3 HO H H OH CH2CH3 CH3CH2 (–) Retention (+) Inversion 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

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 S N1 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 S N1 R—X

Slow

R⊕ + XΘ Carbocation

dx = k[R—X] dt In this R — X bond breaking slow, is rate determining step. → R — OH R⊕ + OHΘ fast Carbocation formed can undergo rearrangement to give more stable carbocation before attack of the nucleophile CH3 CH3 | |⊕ S 1 , 2 − methylshift N → CH3CCH2 1 → CH3CCH2Cl slow | | CH3 CH3 1º (less stable) CH3 | CH3CCH2CH3 ⊕

3º (more stable) O O CH3 CH3 O || || Θ | || | Θ CH COH CH3CO + CH3CCl 3→ CH3CO—CCH3 + Cl | | CH3 CH3 Rate = k [t-BuCl]

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CH3 | Θ slow → CH3C ⊕ + Cl step I | CH3

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Bimolecular (2) Nucleophilic (N) Substitution (S) reaction indicated as SN2. R—X + OH– → ROH + X– dx – = k[R—X] [OH ] dt Step I. OH– attacks the side opposite to that where halide exists to give an intermediate (transition state)

High concentration of the nucleophile favours SN2 reaction while low concentration favours SN1 reaction. CH3 | CH3CCH2Br shows SN2 reaction with C2H5O–, but | CH3 SN1 reaction with C2H5OH. The higher the polarity of the solvent, the greater is the tendency for SN1 reaction Elimination VS Substitution

H H δ– δ– H HO C Br | H—C—Br + OH– slow → | H 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 H Intermediate (A) fast → HO—C H H 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.

CH3 | OH– CH3CCH2CH3 SN ⊕ E2 OH–

CH3 | CH3C=CHCH3 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 RX = 1º 2º 3º Substitution increases CH3 CH2 | || E2 CH3CBr + C2H5ONa → CH3C | | CH3 CH3

CH3 CH3 H δ– | δ– slow – C Br H—C—Br + OH → HO | CH2CH3 CH2CH3 CH3 | fast → HO—C—H | – Br Θ CH2CH3 (–) For SN2 reaction the order of reactivity is CH3X > 1º > 2º > 3º (alkyl halide) CH3

CH2Br >

CHBr >

CH3 CH3 | | SN 2 CH3C—ONa + C2H5Br → CH3C—OC2H5 | | 1º 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).

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 SN2

> 1º > 2º > 3º SN2 mixed SN1

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CH3 | CH3CCH2CH3 | OH

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AUGUST 2010

KEY CONCEPT

Physical Chemistry Fundamentals

SOLID STATE 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

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 M Mass = NA

N

Therefore, Density =

Volume = a3 NM mass = 3 volume a NA

V=

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.

4 3 πr 3

or

V=

4 a π 3 2

3

The fraction of the total volume occupied by the sphere is

φ=

4 a π 3 2 3

3

=

π = 0.5236 6

a 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 C A

A B C

A B A

B C

B

B

B C

A

C A

A

C

A

A

C A

A C

A

A

Packing in a simple cubic lattice

Fig. (a) Closest packed layers of spheres

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AUGUST 2010

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.

B

A B

A

A

C B

B

3 a./3.

A

A B A

Thus,

C B A

C = 2

3a 3 Layer A

Fig. (b) Two types of packing

Layer B

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.

C/2 Layer A a 2r Hexagonal closest-packed structure

Now, in the face-centred lattice spheres touch one another along the face diagonal. Thus, we have 4r = 2 a

4 2

r

3 3 4r = 8 r a = 2 C = 2 3 3 2 6

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,

1 Area of the base = 6 2

( 3r )(2r) (

= 6 3 r2

)

8 r = 24 2 r3 Volume of the prism = 6 3r 2 6

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

B A 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

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a=

With this, the distance C becomes

A

Exploded view

or

4 6 πr 3 3 = 24 2 r 3

2π = 0.7405 6

or 70.05 percent Example of HCP are Ca, Cd, Cr, Mg and Zn. 36

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this arrangement, spheres are touching one another along the cross diagonal of the cube, making its

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

distance equal to 4r. This must be equal to Thus, 4r =

3 a,

3 a 4 Volume of the cube = a3

i.e.,

2a 4 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

r=

3 4 2a 4 π 3 4 = a3

3 a.

r=

4 4 3 Volume of one sphere = πr3 = π a 3 3 4

3

Since there are two spheres in each unit cell, the total volume occupied will be 3 4 3 2 π a 3 4

2π = 0.7405 6

The fraction of the volume occupied by the spheres 3 4 3 2 π a 3 4 φ = = a3

or 74.05 percent A C

3π = 0.6802 8

or 68.02 percent In this arrangement each sphere has eight nearest neighbours

B A Exploded view

PERSONAL GROWTH

Cubical closest-packed stricture

•

Face-cented cubic unit cell formed by ABCA packing

• •

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-centred cubic lattice

• •

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

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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. AUGUST 2010

UNDERSTANDING

Physical Chemistry

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 [IIT-1996] empty space ? 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, 8 Number of atoms per unit cell = + 1 = 2 8 Area of the base = b × ON [From fig.(ii)] = b × a sin 60º 3 2 a (Q b = a) = 2

O

1.

a 60º N b figure (ii) 3

4 a 4 2 × π 2 × πr 3 3 2 3 = = 3 a 2 a3 2 π = 0.74 = 74% = 3 2 ∴ The percentage of void space = 100 – 74 = 26%

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 [IIT-1986] of methanol in the vapour. Sol. Given that, Pe0 = 44.5 mm Hg For ethanol (C2H5OH), M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g 60 m = 1.3 ∴ Moles of ethanol, ne = = M 46 For methanol (CH3OH), Pm0 = 88.7 mm Hg M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g 40 m ∴ Moles of methanol, nm = = = 1.25 32 M ne 1.3 1.3 ∴ xe = = = 1.3 + 1.25 2.55 ne + nm 2.

c β αb a γ Figure (i)

Volume of the hexagonal cell = Area of the base × height =

3 2 a .c 2

2 2

a 3 ∴ Volume of the hexagonal cell 3 2 2 2 = a . a = a3 2 2 3 But c =

nm 1.25 1.25 = = 1.3 + 1.25 2.55 ne + nm According to Raoult's law, 44.5 × 1.3 = 22.69 mm Hg Pe = Pe0 xe = 2.55 88.7 × 1.25 = 43.48 mm Hg and Pm = Pm0 xm = 2.55 Hence, total vapour pressure of the solution, PT = Pe + Pm = 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law,

xm =

a 2 Hence, fraction of total volume or atomic packing factor Volume of 2 atoms = Volume of the hexagonal cell

and radius of the atom, r =

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Pm = PTx´m (in vapour form) Hence, mole fraction of methanol in vapour form, P 43.48 = 0.66 x´m = m = 66.17 PT 3.

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)

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 + SO 24 − → PbSO4 + 2e– (Charging)

273.15 K

= (75.312 J K–1 mol–1 ) (10 K) = 753.12 J mol–1 273.15K

C p, m (1)

∫

∆rS1 =

R

263.15K

dT

273.15 K = (75.312 J K–1mol–1) × ln 263.15 K = 2.809 J K–1 mol–1 (b) H2O(1, 0ºC) → H2O(s, 0ºC) ∆rH2 = – 6.008 kJ mol–1

100 = 77.28ml 1.294 77.28 ml of H2SO4 acid solution contains = 39 g H2SO4 ∴ 3.5 L of sulphuric acid solution contains 39 × 1000 × 3.5 = = 1766.3 g H2SO4 77.28 100 Volume of 100 g of 20% H2SO4 = = 87.79ml 1.139 87.79 ml of H2SO4 acid solution contains = 20 g H2SO4 ∴ 3.5 L of sulphuric acid solution contains 20 × 1000 × 3.5 = = 797.3 g H2SO4 87.79 ∴ 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 1 ∴ 969 g H2SO4 requires = × 969 F 98 1 × 969 × 96500 amp-sec = 98 1 969 × 96500 × amp-hr = 98 3600 = 265 amp-hour Volume of 100 g of 39% H2SO4 =

∆rS2 = –

...(2)

(6008 J mol –1 ) = – 21.995 J K–1 mol–1 (273.15 K )

(c) H2O(s, 0ºC) → H2O(s, –10ºC)

...(3)

263.15 K

∆rH3 =

∫C

p , m (s)

dT

273.15 K

= (36.400 J K–1 mol–1)(–10 K) = – 364.0 J mol–1 263.15 K

∆rS3 =

∫

273.15 K

C p, m (s) T

dT

263.15 K = (36.400 J K–1 mol–1) ×ln 273.15 K = – 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

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

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p , m (1) dT

263.15 K

PbO2 + 4H+ + SO 24 − + 2e– → PbSO4 + 2H2O (Discharging) [IIT-1986] Mass Sol. We know, Volume = Density

4.

∫C

∆rH1 =

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AUGUST 2010

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 [IIT-1990] volume V. Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1γ–1 = T2V2γ–1 T2 = T/2 Here, T1 = T ; V1 = V and V2 = 5.66 V T Hence, TVγ–1 = × (5.66V)γ–1 2 T × (5.66)γ–1 × Vγ–1 = 2 or (5.66)γ–1 = 2 ...(1) Taking log, (γ – 1)log 5.66 = log 2 0.3010 log 2 or γ–1= = = 0.4 0.7528 log 5.66 5.

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.

or γ = 1.4 If f, be the number of degrees of freedom, then 2 2 γ=1+ or 1.4 = 1 + f f 2 = 1.4 – 1 = 0.4 or f 2 =5 or f= 0.4 (b) According to adiabatic gas equation, P1V1γ = P2V2γ Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ P P P or P2 = = = [using eq.(1)] γ 11.32 (5.66) (5.66)1.4 Hence, work done by the gas during adiabatic expansion P PV − × 5.66V P1V1 − P2 V2 11 .32 = = γ –1 1.4 – 1 PV 2 = PV = 1.25 PV 0.4 2 × 0.4

PV −

=

XtraEdge for IIT-JEE

44

AUGUST 2010

Set

4

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

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 So lu t ion s wi l l b e p ub lished in nex t issue Joint Director Academics, Career Point, Kota 1.

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

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 ;

10. Prove that sin x sin 3x sin 9 x + + cos 3x cos 9x cos 27 x 1 = [tan 27 x – tan x] 2

π

∫

I(u) = ln (1 − 2 u cos x + u 2 ) dx; 0

1 2 I(u ). 2

prove that I (u) = I (– u) = Generalize the result as 4.

5.

6.

7.

8.

1 2

n

n

I( u 2 ) .

NaN (Not a Number)

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.

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

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.

•

Solve : 1 x2 y2 1 dx + − − dy = 0 2 2 y x ( x − y) ( x − y) 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. xy f ( x ).f ( y) for all real x & y. If Let f = 2 2 f ′(1) = f(1) ≠ 0 then show that f(x) is

XtraEdge for IIT-JEE

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 1 2 (n + n + 2) parts. 2

9.

45

-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

AUGUST 2010

MATHEMATICAL CHALLENGES SOLUTION FOR JULY ISSUE (SET # 3)

1.

dy – y ln 2 = 2sin x(cos x – 1) ln 2 dx

1 3 3 1 = − a − + i − a+ 2 2 2 2

− ln 2 dx I.F. = e ∫ = 2–x –x

so y2 =

∫2

sin x − x

–1 = –

. (cos x – 1). ln 2 dx

⇒ a=2– 3

y2–x = 2sin x – x + c Ans (B) y = 2sin x + c.2x y . 2–x = 2sin x – x + c Now if x → ∞ and y is bounded (finite) then c = 0 Ans (C) sin x For option (D) : f(x) = 2 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.

1 3 a– 2 2

1 3 a+ 2 2 3 1 ⇒ b = – 3 + + = 12 – 3 2 2 Hence correct answer is (B)

b=–

4.

c1 ≡ x2 + y2 = a2 c2 ≡ (x – h)2 + (y – k)2 = r2

AC = 4p A

P

α

p

B

tan α =

c1

D α

Director circle x2 + y2 = 2a2 ah + r.0 ak + r.0 Pt. P ; a+r a+r 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)

C

p AD =4 = p DC

cot α + tan α =

c2

DC AD + =4 p p

tan2α – 4 tan α + 1 = 0 4 ± 16 − 4 =2± 3 2 α = 15º & 75º Hence correct answer is (D)

...(1)

tan α =

3.

5.

4a = 2(2a) = 2 .

5 − 36 + 17 13

14 28 = 13 13 Hence correct answer is (C)

=2.

z2 = z1iπ/3 z1(–a + i) z2 (–1+bi)

6. O z3

1 3 –1 + bi = (–a + i) + i 2 2

XtraEdge for IIT-JEE

∫ =

46

9x x cos . sin 3x 2 2 sin 3x − sin 6 x 9x x 2 cos cos . sin 3x 2 2 9x 3x − 2 cos sin 2 2

2 cos

∫

AUGUST 2010

x

A

3x

∫ 2 cos 2 dx = – ∫ (cos 2x + cos x ) dx = – 2 cos

=– 7.

R I(h,k)

M

sin 2 x – sin x + c 2

r B

(x – 1)2 = a2 a = 1 ± |a| f(x) = x2 – 2(a + 1)x + a(a – 1) = 0

N

O

C

a – (s – b) 2 b−c −2s + a + 2b = = 2 2

so

h = ON =

& r=k 2

1

so r = k = r=k= 2sk =

D = 4(a + 1)2 – 4a (a – 1) > 0 a2 + 2a + 1 – a2 + a > 0 1 a>– ...(1) not essential 3 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 1 ⇒ a>– ⇒a≥0 3 2a2 + 2a2 – 3a – 1 < 0 ⇒ 4a2 – 3a – 1 < 0 (4a + 1) (a – 1) < 0 1 ⇒ – <a<1 4 ⇒ 0≤a<1 So, similarly if a < 0 then ⇒ f(1 m a) < 0 f(1 ± |a|) < 0 1 so f(1 ± |a|) < 0 ⇒ – < a < 1 ...(2) 4 1 eqn. (1) ∩ (2) – < a < 1 4 8.

= 2sk =

s(s − a )(s − b)(s − c) s

s(s − a )(s − b)(s − c) s

s(s − a )(a − b + c)(a + b − c) s(s − a )(a − 2x )(a + 2x ) s(s − a )(a 2 − 4h 2 )

required lows is 4s2y2 = A(a2 – 4x2) ⇒ s2y2 + A x2 =

Aa 2 4

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 = 149 149 so required probability = 2401

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 3 7 2 Solving these eqn, a = , b = , c = – 6 3 2 7 2 3 2 so the remainder is x + x – 6 3 2

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

XtraEdge for IIT-JEE

∆ = s

47

AUGUST 2010

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

MATHS 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.

Now,

∫

x +8

x

g( t ) dt

f´(x) = g(x + 8) – g(x) = 0

2

⇒ g is constant function

Sol. f´(x) = 3x + 2ax + b

y = f(x) is strictly increasing

4.

⇒ f´(x) > 0 ∀ x ⇒ (2a)2 – 4.3.b < 0

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

This is true for exactly 15 ordered pairs (a, b); 1 ≤ a, 15 5 = b ≤ 6, so probability = 36 12 2.

f(x) =

B M

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.

A

→

Sol. Let A = a ˆi + b ˆj + c kˆ →

⇒ B = 3 ˆi + 2 ˆj + kˆ → →

→

AB :

since AB chord passes through (h, 0)

→

⇒ (A . B) 2 ≤ | A |2 | B |2 a 2 + b2 + c2

3a + 2b + c ≤

so, h.at2 = a2t4 + 4a2t2 at2 [at2 + (4a – h)] = 0

14

If a > 0

(7)2 ≤ (a2 + b2 + c2) (14)

3.

Now point (at ,2at) must lie inside the circle, on solving a2t4 + 4a2t2 – a2 < 0

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

∫

x +8

x

we get, h < a ( 5 + 2)

g( t ) dt

4a < h < a ( 5 + 2)

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

5.

putting x = x + 2 in (1) ........ g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8)

...(ii)

from (i) & (ii)

is a constant function.

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 x t 2 dt r is the Lt x →0 0 x 2 tan( π + x )

...(2)

from (1) & (2)

∫

g(x) = g(x + B)

XtraEdge for IIT-JEE

...(i) 2

49 7 = 14 2

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

⇒ 4a – h < 0

⇒ h > 4a

{Q 3a + 2b + c = 7, point lies on the plane} a2 + b2 + c2 ≥

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

48

AUGUST 2010

Sol. S =

a , 1− r

Confidence Tips

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

•

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

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

•

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

to evaluate r :

•

Can you tell whether someone else lacks selfconfidence?

•

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 selfconfidence.

•

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?

⇒ f´(x) = 0 put 5x – 1 t(> 0)

r = Lt

∫

t 2 dt

x

x →0 0

2

x tan(π + x )

=

1 π

2π 1 since a = 2, r = ⇒ sum of G.P. = π π −1 6.

The tangent and normal at a point P on the ellipse x2 2

+

y2 2

= 1 meets the y-axis at A and B

a 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), Q

A P

S M

N B

Points of intersection of tangent and normal at P b points with y-axis and A 0, y 1 2

, B 0, y1 − a y1 , 2 b 2

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.

XtraEdge for IIT-JEE

49

AUGUST 2010

XtraEdge for IIT-JEE

50

AUGUST 2010

MATHS

VECTOR Mathematics Fundamentals Properties of vectors : (i) Addition of vectors

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

Triangle law of addition : If in a ∆ABC,

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

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

|a| = | AB | = AB.

C B

c=a+b

a

B A a Parallelogram law of addition : If in a

A

Types of Vector: Zero or null vector : A vector whose magnitude is zero is called zero or null vector and it is

parallelogram OACB, OA = a, OB = b and OC = c B

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 aˆ , read as “a cap”. Thus, | aˆ | = 1. aˆ =

C c=a+b

b

Vector a a = |a| Magnitude of a

O

a

A

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.

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.

XtraEdge for IIT-JEE

b

51

AUGUST 2010

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 a+b O

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. B

b a

b

A

θ A O a Angle between two vectors : If a, b be two vectors inclined at an angle θ, then a . b = |a| |b| cos θ

–b

a + (–b) = a – b

B´ Properties of vector subtraction :

(i) a – b ≠ b – a

⇒ cos θ =

(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.b θ = cos–1 | a || b | If a = a1i + a2j + a3k and b = b1i + b2j + b3k; then ⇒

(a) |a + b| ≤ |a| + |b| (b) |a + b| ≥ |a| – |b|

a 1 b1 + a 2 b 2 + a 3 b 3 θ = cos–1 2 2 2 2 2 2 a 1 + a 2 + a 3 b1 + b 2 + b 3

(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 :

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

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 mb + na mb − na or . m−n m+n

XtraEdge for IIT-JEE

a.b | a || b |

52

AUGUST 2010

(c) (a + b) . (a – b) = |a|2 – |b|2

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 1 | AC × BD | , where AC and BC are its 2 diagonals.

(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.

b θ O

a

The area of a triangle ABC is

Then a × b = |a| |b| sin θ ηˆ , and |a × b| = |a| |b| sin θ, where θ is the angle between a and b, ηˆ is a unit vector perpendicular to the plane of a and b such that a, b, ηˆ form a right-handed system.

1 1 | BC × BA | or | CB × CA | 2 2 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.

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. i Then, a × b = a 1 b1

j a2 b2

k a3 b3

Angle between two vectors :

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

XtraEdge for IIT-JEE

1 | AB × AC | or 2

| a×b | | a || b |

a × (b × c) ≠ (a × b) × c

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AUGUST 2010

MATHS

PERMUTATION & COMBINATION Mathematics Fundamentals

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 :

1

2

3

4

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 n (n r − 1) . n −1 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 n! . ( m !) × ( n − m ) !

r

Number of choice n (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) =

n! n (n − 1)(n − 2)....(n − r + 1)((n − r )!) = (n − r )! (n − r ) !

= nPr The number of arrangements of n different objects taken all at a time = nPn = n ! (i)

n

P0 =

n! = 1; nPr = n. n–1Pr – 1 n!

(ii) 0 ! = 1;

1 = 0 or (–r) ! = ∞ (r ∈ N) (– r ) !

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 n! . objects is (p1 !) × (p 2 !) × .... × (p r !)

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 :

1

2

3

4

Number of choices : n (n) (n) (n)

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 (n − 1) ! . of n distinct items is 2 Number of circular permutations of n different things, taken r at a time, when clockwise and n pr anticlockwise orders are taken as different is . r

r 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 n! . are distinct is p !q ! r !

XtraEdge for IIT-JEE

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AUGUST 2010

Number or circular permutations of n different things, taken r at a time, when clockwise and n pr . anticlockwise orders are not different is 2r 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 1 necklace, is (n − 1) ! . 2 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 n . nC is always a natural number. r r 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 n! . Also r (0 ≤ r ≤ n) at a time is nCr = r !( n − r ) ! n Cr = 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. n! ⇒ x(r !) = npr ⇒ x = = nCr r !( n − r ) !

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

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.

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 1 1 1 1 n ! 1 − + − + ... + (−1) n . 1 ! 2 ! 3 ! n !

XtraEdge for IIT-JEE

n–p

Cr + n – pCr – 1 +...........+ n – pC0, if r ≤ p and n–p Cr + 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)! =

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

If order of group is not important: The number of ways in which mn different things can be divided (mn ) ! . equally into m groups is ( m !) m 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 (mn)! (mn)! ×m!= m (n !) m ! (n !) m

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AUGUST 2010

Based on New Pattern

IIT-JEE 2011 XtraEdge Test Series # 4

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..

PHYSICS

3.

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.

2.

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

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 – A F X Y B

E

C

π 4 (C) π

(A)

→

For a body moving one dimensionally, ∆S and | ∆ S | are distance travelled and magnitude of displacement respectively, then -

4.

→

(A) ∆S = | ∆ S | ; always true

D π (B) 2 (D) 2π

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 :

→

(B) ∆S > | ∆ S | ; always true →

|∆S| (C) ≤ 1; always true ∆S →

O

(D) ∆S < | ∆ S | ; sometimes true

XtraEdge for IIT-JEE

54

AUGUST 2010

(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) 0.523 A (C) 2.369 A 8.

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

original surface

9. 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

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

+ –

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. I Amp

E

unit length are E, V and H respectively, which of the following graphs are correct ? E H (B)

(A)

Fe

Cu

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 14 2 (D) None of these (C) 7 × 10 m/s

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.

F*

6.

(B) 0.319 A (D) 1.235 A

x

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) (D) ∈0I (ρFe + 2ρCu) (C) ∈0I (ρFe + 3ρCu) 7.

H

0.05m

x

10. The conductor ABCDE has the slope shown in figure. It lies in the y-z plane, with A and E on the yaxis. When it moves with a velocity v in a magnetic field B, an emf e is induced between A and E, then : z

Loop

R

B 0.5m

O x

XtraEdge for IIT-JEE

(D) E

z ω

V

(C)

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. 0.03m

x

55

a

E

C

A

y

a λ

D

AUGUST 2010

(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

Passage: II (Ques. 16 to 18)

Initially circuit is in steady state. 6Ω 18 V a 3 µF

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

18. How much does the charge on capacitor 6µF change when S is closed (B) 18 µC (C) 36 µC (D) 72 µC (A) 16 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 : P Q R S T P Q R S T P Q R S T P Q R S 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 A B C D

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

13. Magnetic field between the plates is µ σV (A) zero (B) 0 4 µ 0 σV (D) µ0σV (C) 2

→

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

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

(A) µ0σ2V2 upward

→

µ 0σ v 2

upward

(P) for uniform motion in any direction

→

µ0σ2 V 2 downward 2 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

(C) 2µ0σ2V2 downward (D)

XtraEdge for IIT-JEE

→

(A) Vinst = Vav

2 2

(B)

3Ω

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

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)

–σ

6 µF b

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

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

+σ

S

(B) | Vinst | = V

(Q) for uniform motion in

(C) Vinst = Vav

given direction (R) Always true

→

(D) | Vinst | < V

(S) Never true (T) None of these

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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

Plane

CCP unit cell C

C

B

C

C

CC

B

B

B

C

C

C

C

C

CC

B

B

B

C

C

C

A

(A)

CHEMISTRY

2.

3.

C

B

C

(C)

7.

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 (D) None of these (C) Na2SO4.10H2O

C A

B

B

C

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

C

C A

B

(D)

A

A

C

C

Regarding graphite the following informations are available : Top view

3.35Å

The correct order of increasing C–O bond length in CO, CO 32– , CO2 is (A) CO 32– < CO2 < CO (B) CO2 < CO 32– < CO

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

(C) CO < CO32– < CO2 (D) CO < CO2 < CO 32– 4.

(B)

B

B A

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.

B

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 ? 9x 4x (A) Å (B) Å 4 7 5x 16 x Å (D) Å (C) 4 9

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 –

XtraEdge for IIT-JEE

8.

The geometrical shapes of XeF5+, XeF6 and XeF82– 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.

57

AUGUST 2010

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

1 mv2 , 2 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 :

hν = hν0 +

Kinetic Energy of photoelectrons

9.

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

Frequency ...... (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.

(B) r 1/n

P.E.

(C)

1/n z

n

(D) K.E. n

11. In which of the following molecules there is no S—S bond(s)? (A) S2O42– (B) S2O52– 2– (D) S2O72– (C) S2O3 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

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 (C) ν0 = 2ν1 – ν2

(B) ν0 = 2ν2 – ν1 (D) ν0 = ν1 + ν2

15. The equation of photoelectric effect represents an equation of (A) straight line (B) parabola (C) ellipse (D) None

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage : I (Ques. 13 to 15)

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 Actual number of particles in solution after dissociation i= Number of formula units initially dissolved in solution

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:

XtraEdge for IIT-JEE

v0

i = 1 (for non-electrolyte); i > 1 (for electrolytes, undergoing dissociation) i < 1 (for solutes, undergoing association). 58

AUGUST 2010

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

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.

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

1.

The range of k for which ||x–1|–5| = k have four distinct solutions (A) [0, 5] (B) (–∞, 5) (D) (0, 5) (C) [0, ∞)

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 : P Q R S T

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) 3 3 changes concavity about the point R , f , then 4 4

A B C D

P P P P

Q Q Q Q

R R R R

S S S S

T T T T

which of the following conic section can have eccentricity 'a' (A) circle (B) parabola (C) ellipse (D) hyperbola

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 (Ionic species) (A) XeF5+ (B) SiF5– (C) AsF4+ (D) ICl4–

4.

x 2 + 2mx – 1, x ≤ 0 f(x) = mx – 1 x>0 If f(x) is one-one then set of values of 'm' will be (A) (– ∞, 0) (B) (– ∞, 0] (D) [0, ∞) (C) (0, ∞)

Column-II (Shapes) (P) Tetrahedral (Q) Square planar (R) Trigonal bipyramidal (S) Square pyramidal (T) Octahedral

5.

20. Column-I

(A)

12 1 6 C + 1H

(B)

27 1 13 Al + 1H

(C)

235 92 U

(D)

75 2 33 As + 1 H

→ 13 7 N

x x x If f(n) = Lim 1 + sin 1 + sin 2 ...1 + sin n x →0 2 2 2 then Lim f(n) =

1

x

n →∞

(A) 1

24 → 12 Mg + 42 He

6.

93 1 + 10 n → 140 56 Ba + 36 Kr + 3 0 n 1 1 → 56 25 Mn + 9 1H + 12 0 n

Column-II (P) Projectile capture (Q) Spallation (R) Fusion (S) Projectile capture and particle emission (T) Fission

XtraEdge for IIT-JEE

Let a function is defined as f : R → R

The derivative of 1– x cos 2 tan –1 1+ x 1 (A) 1 – 1– x2 (C) 2 –

59

(B) e

1 1– x2

(D) ∞

(C) 0

– 2 cos –1 1 – x w.r.to x is 2 1 (B) 1 – 1+ x2

(D) 2 –

1 1+ x2

AUGUST 2010

7.

If P = lim (ea 2 .e 3 a 4 .....e n –1a n )1 /( n

2

+1)

n →∞

(A) ea2 (C) ea

11. Which of the following is/are true 2πn (A) Lim sin n =0 n →∞ 3n + 1

, then p4 equals

(B) e2a2 (D) e2a

(B) Lim 8.

x →∞

Maximum value of the expression 10 x12 x 24 + 2x12 + 3x16 + 3x 8 + 1

(A) 1 (C) 10

ln (1 + x 4 + e 2 x )

=1

(C) Lim (tan x)tan 2x = e–1 x →π / 4

is equal to

(D) Lim

cos x – 3 cos x 2

sin x

x →0

(B) 2 (D) not defined

=

–1 12

2

12. If g(x) = 7x2 e – x ∀ 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

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.

ln (1 + x 2 + e x )

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)

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

K(x) is a where 0 a = – 1 2

10. Function whose jump (non-negative difference of LHL & RHL) of discontinuity is greater than or equal to one is/are -

function such that K(f(x)) = a + b + c + d, if

f ( x ) is even

if

f ( x ) is odd

if

f ( x ) is neither even nor odd

(e1/ x + 1) ; x<0 1/ x (A) f ( x ) = (e – 1) (1 – cos x ) ; x > 0 x

if f ( x ) is periodic 3 b= 4 if f ( x ) is non periodic 5 if f ( x ) is one − one c= 6 if f ( x ) is many one

( x1/ 3 – 1) ; 1/ 2 (B) g ( x ) = ( x – 1) lnx ; x – 1

7 if f ( x ) is onto d= 8 if f ( x ) is int o A = {x2, ex, sin x, |x|} all functions in set A are defined from R to R B = {18, 19, 16, 17}

x <1 1 < x <1 2

e 2x + e x + 1 and H : R → R; h(x) = 2 x x + e – e 1

sin –1 2x 1 –1 ; x ∈ 0, 2 (C) u ( x ) = tan 3x | sin x | ; x<2 x

π π φ : – , → R ; φ (x) = tan x 2 2 13. k(φ (x)) is equal to (A) 15 (B) 16 (C) 17 (D) 18

log 3 ( x + 2); x > 2 (D) v( x ) = 2 log1/ 2 ( x + 5); x < 2

14. k(h(x)) is equal to (A) 15 (C) 17

XtraEdge for IIT-JEE

60

(B) 16 (D) 18 AUGUST 2010

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

19. The graph of f and g are given. Use them to evaluate each limit. Y

;

Column-I (A) lim f(g(x)) =

x≤0

; 0 < x <1 ;

x =1

;

x >1

x →1

n ∈ N, k ∈ R

(D) lim

+

x →1

3f ( x ) – g( x ) = f ( x ) + g( x )

(S) 2

20. Column-I Column-II (A) Domain of function (P) 5 |x| 1 f(x) = ln 1 – – 1+ | x | 2 is (p, q), then p + q is equal to

(B) Maximum value of (Q) 1 sin 2x in the f(x) = π sin x + 4 π interval 0, is 2

(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

P Q R S T S S S S

(Q) does not exist

(T) 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 : R R R R

X

f (x) (C) lim + f ( x )g( x ) = (R) 0 x → 0 g ( x )

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)

(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

T T T 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.

XtraEdge for IIT-JEE

–1 O 1 2

Column-II (P) 1

x→ 2

17. Number of point(s) where continuous function f(x) is non differentiable, is (A) 0 (B) 1 (C) 2 (D) 3

Q Q Q Q

1

–1

(B) lim 3f ( x ) – 2 =

16. The value of k + b + λ so that f(x) is continuous in R is (A) 3 (B) 2 (C) 4 (D) 1

P P P P

–2

X

–1

Consider f(x) =

A B C D

y = g(x)

1 –1 O 1 2

–2

Passage: II (Ques. 16 to 18) | – x 2 – x | +λ n –n lim x – x + k n – n →∞ x + x n b x –x sgn(ln (e + e + 1))

Y 2

y = f(x) 2

61

AUGUST 2010

XtraEdge for IIT-JEE

62

AUGUST 2010

Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 4

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..

4.

PHYSICS

given by v = 2x + 9 . 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

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 + (A) x + (C) x –

2.

3.

1 x

1 x2 1

x2

2

+ –

+

1 x

1 x3 1

x3

3

5.

→

→

→

→

→

dy is dx –2 3 (B) 3 – 4 x x –2 3 (D) – 2 x x

(A) R < 0 (C) 0 ≤ R ≤ 2

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 63

(B) 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) π π (A) ωav = rad / s (B) ωav = rad / s 6 3 π rad / s (D) None of these (C) ωav = 12

a 3n 3

. Find c d 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%

XtraEdge for IIT-JEE

→

If a and b are two unit vectors and R = a + b and if | R | = R, then -

then

The relation given the value of x as, x =

The displacement x of body when its velocity v is

AUGUST 2010

8.

(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

d sin 2x dx (A) (sin 2x)–1/2 (B) cos 2x (sin 2x)–1/2 –1/2 (C) 2 cos 2x (sin 2x) (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.

q2 The units of electrical permittivity ε 0 = 4πFr 2 are(A) N–1m–2C2 (B) Nm–2C2 2 2 (C) C /Nm (D) N/Cm2 →

→

→

R →

B

→

→

→

→

→

→

→

→

|B| = b

|A| =a

→

The second law of vector addition is triangle law, (B) 90º

(C) 60º

→

(D) 180º

→

which says that if we take A and B as two vectors acting at point O as shown in figure, then the

11. Two bodies P and Q are moving along positive x-axis

→

→

→

resultant of vector is get by taking A and B as adjacent sides of a triangle and the 3rd side of the

their position-time graph is shown below if VPQ is →

→

velocity of P.w.r.t Q and VQP is velocity of Q w.r.t

triangle as the resultant, then if θ is angle between A

P then –

→

and B then.

x

→

P

13. if α is the angle made by resultant vector with A ; then tan α = A sin θ B sin θ (B) (A) B + A cos θ A + B cos θ A cos θ B cos θ (D) (C) A + B cos θ B + A sin θ

Q t →

→

(A) | VPQ | = | VQP | constant →

→

(B) VPQ is towards origin →

→

→

(D) VPQ and VQP both can be towards origin at same time

→

m u=0 r

XtraEdge for IIT-JEE

O

→

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

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,

h

→

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 (D) 3A cos θ/3 (C) 3A cos θ/2

(C) VQP is towards origin

12.

O

→

A

→

| a + b | + | a – b | = 2 | a | , then angle between a

and b (A) 0º

θ

P

10. If a and b are two vectors with | a | = | b | and

θ

64

AUGUST 2010

Passage: II (Ques. 16 to 18)

Column-I

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.

Column-II

(A) Velocity – time graph

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

(P) T

2T

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

(B) Acceleration-time (Q) graph

18. The time after which balls will meet in air is (A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec

T

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 : P Q R S T A B C D

P P P P

Q Q Q Q

R R R R

S S S S

20.

sin 2 2θ 2 (T) None of these

(S) 1 +

Parabola

XtraEdge for IIT-JEE

T

2T

(S)

T

2T

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.

The displacement - time graph of a body moving on a straight line is given by x

0

(D) Speed – time graph

CHEMISTRY

Column-II (P) 1 – sin 2θ (Q) 1 + sin 2θ (R) cos 2θ

(D) cos4θ + sin4θ

(R)

(T) None of these

T T T 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 (A) (sin θ + cos θ)2 (B) (sin θ – cos θ)2 (C) cos4θ – sin 4θ)

(C) Distance – time graph

2T

t

65

1.

According to Bohr’s theory, angular momentum of an electron in fourth orbit is h 2h 4h h (B) (C) (D) (A) 2π 4π π π

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 – λ λ (A) λ (B) (C) 2λ (D) 2π 2 AUGUST 2010

4.

5.

6.

7.

8.

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%

r U

(A)

T (C)

v

(D)

n2

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

1/n

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

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

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)

l=n

∑ l =0

(C)

2(2l + 1)

l = n +1

∑ 2(2l + 1) l =0

(B)

l = n –1

∑ 2(2l + 1) l =1

(D)

l = n –1

∑ 2(2l + 1) l =0

12. Which of the following is/are the correct order of mobility ? (A) Li+ < Na+ < K+ (B) Na+ < Mg2+ < Al3+ 3+ 2+ + (C) Al < Mg < Na (D) Li+ > Na+ > K+

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

Specify the coordination geometry around the hybridization of N and B atoms in a 1 : 1 complex of [IIT- 2002] BF3 and NH3 -

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]

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.

13. 'A' is (A) CuS (C) PbS

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

XtraEdge for IIT-JEE

1/n2

n2

(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

9.

(B)

(B) FeS (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

66

AUGUST 2010

15. 'E' is (A) Cu2[Fe(CN)6] (C) Cu3[Fe(CN)6]2

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.

(B) [Cu4[Fe(CN)6] (D) None of these

Passage: II (Ques. 16 to 18)

19. Column-I (Ionic species) (A) XeF5+ (B) SiF5– (C) AsF4+ (D) ICl4–

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. s(s + 1)h

Spin angular momentum =

l(l + 1)h

Orbital angular momentum =

1 . Spin 2 multiplicity is a factor to confirm the electronic configuration of an atom or ion. Spin multiplicity = (2Σs + 1). Answer the following questions :

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

Total spin of an atom or ion is a multiple of

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)

l(l + 1)

MATHEMATICS

(B) l + 2

(C) 2l + 1

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.

(D) 4l + 2

18. The orbital angular momentum for a 2p-electron is :

(A)

3h

(B)

6h

(C) zero

(D)

6

h 2π

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 : P Q R S T A B C D

P P P P

XtraEdge for IIT-JEE

Q Q Q Q

R R R R

S S S S

Column-II (Shapes) (P) Tetrahedral (Q) Square planar (R) Trigonal bipyramidal (S) Square pyramidal (T) Octahedral

2.

T T T T

67

3π If x ∈ ,2π then value of the expression 2 sin–1 (cos(cos–1(cosx) + sin–1(sinx))) equals π π (A) – (B) 2 2 (C) 0 (D) None of these

If median AD of a ∆ABC divides the angle BAC in sin B ratio 1 : 2 then is equal to sin C 1 A 1 A (A) sec (B) cos 2 3 2 3 1 A (D) None of these (C) cos ec 2 3

AUGUST 2010

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)

4.

3 unit 2

1 unit 2

(C) nπ ±

(B) nπ + π 2

2∆ C (D) 2 s Where ∆ is area & s is semi-perimeter of the ∆.

has value equal to (A) 15º (C) 22.5º

π which of the following holds true 7 (A) tanα. tan2α. tan3α = tan3α – tan2α – tan α (B) cosec α = cosec 2α + cosec 4α 1 (C) cos α –cos 2α + cos 3α has value equal to 2 (D) 8 cos α. cos 2α. cos 4α has value equal to 1

10. If α =

π 12

(D) 2nπ

In a ∆ABC if ∠A = 60º,

b = c

3 +1 then ∠B – ∠C 2

11. If tan2θ = 2 tan2φ + 1, then value of cos 2θ + sin2φ is(A) 1 (B) 2 (C) 0 (D) independent of φ

(B) 30º (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 120 (A) 9 (B) 13 (C) 13 (D) 12

7.

If

tan α + 2 tan 2α + 4 tan 4α...... + 2 n –1 tan 2 n –1 α

cot α + 2 n n ∈ N then general solution of α is, π π (A) α = 2 – n nπ – (B) α = 2 n nπ + 4 4

1 2 12. If cos 2 x + (1 + tan 2y) (3 + sin3z) = 4 2 cos x 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

= 1,

π (C) α = 2 – n nπ + (D) None of these 4 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 3 (A) sin 2θ (B) sin 2θ 4 3 (D) sin 4θ sin 4θ (C) 4

In a ∆ABC, if cos A. cos B. cos C =

3 –1 and 8

3+ 3 , then 8 13. Value of tan A + tan B + tan C is

sin A. sin B. sin C =

(A)

XtraEdge for IIT-JEE

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 abc ∆ (A) (B) x 4R 2 (sin A + sin B) (C) x sin

The solution of π 3 sin θ – sin 3θ 3 cos θ + cos 3θ + = 4 2 cos θ + is 1 + cos θ 1 – sin θ 4 (A) nπ

5.

(D)

9.

(C)

3+ 3 3 –1 6– 3 3 –1

(B)

3+4 3 –1

(D) None of these

14. Value of Σ tan A. tan B =

68

(A) 5 – 4 3

(B) 5 + 4 3

(C) 6 + 4 3

(D) 6 – 4 3 AUGUST 2010

15. Value of tan A, tan B and tan C are -

(A) 1,

3,

(C) 1, 2,

2

(B) 1,

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.

3,2

(D) None of these

3

Passage: II (Ques. 16 to 18)

19. Column-I

It is given that A = (tan–1x)3 + (cot–1x)3 , where x > 0 1 & B = (cos–1t)2 + (sin–1t)2, where t ∈ 0, & 2 π sin–1x + cos–1x = for –1 ≤ x ≤ 1 and 2 π tan–1x + cot–1x = ∀x∈R 2

(A)

π3 π3 (B) , 32 8

π π (C) , 10 5

(D) None of these

3

3

π2 8

(B)

π2 (C) 4

π2 16

XtraEdge for IIT-JEE

R R R R

S S S S

(R) 2

(S) 0

(T) 3 20. Column-I

(A) If cos

Column-II

3π 5π π + cos + cos 7 7 7

(P) 3

π 2π 4π cos cos 7 7 7 then value of k is

= k cos

(B) value of expression sin 20º (4 cos 20º +1) is cos 20º. cos 30º

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 : P Q R S T Q Q Q Q

π 0, 2

(D) Number of values of t satisfying the equation cos (sin (cos t) = 1 for t ∈ [0, 2π] is

–π 8 –7 π (D) 8

P P P P

3 , where φ∈ 5

π x = 1 – 4 sin2 – then 4 2 values of sin x is/are

(B)

A B C D

4 1 3π , where θ∈ ,2π (Q) 5 2 2

(C) If sin x cos4x + 2sin22x

18. If least value of A is λ & max. value of B is µ then λ – µπ = cot–1 cot µ π 8 7π (C) 8

(P) 1

then cos (θ – φ) has value

(D) None of these

(A)

cos 1º– cos 2º = 2 sin 3º sin 1º

& cos φ =

17. Maximum value of B is -

(A)

2

(B) If cosθ =

16. The interval in which A lie is π3 π3 (A) , 8 2

Column-II

2

(C) value of

b–c c–a a–b + + r1 r2 r3

(Q) – 4

(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

2

sin A + sin A + 1 is sin A

(T) 1

T T T T

69

AUGUST 2010

XtraEdge Test Series ANSWER KEY IIT- JEE 2011 (August issue) PHYSICS Ques Ans Ques Ans 19 20

1 D 11 B,C A→Q A→R

2 C 12 B,C,D

3 B 13 D B→R B→P

4 A 14 B

Ques Ans Ques Ans 19 20

1 A 11 D A→S A→P

2 C 12 B

3 D 13 D B→R B→S

4 D 14 A

Ques Ans Ques Ans 19 20

1 D 11 A,C,D A→Q A→R

2 A 12 B,C,D

3 D 13 A B→S B→Q

4 A 14 D

5 B 15 B C → P, Q C→Q

6 B 16 D

7 A 17 B D→S D→P

8 D 18 C

9 B,C

10 A,C,D

7 B 17 B D→Q D→Q

8 A 18 C

9 B,C,D

10 A,B

7 C 17 C D→P D→P

8 A 18 B

9 B,C

10 A,C,D

7 A 17 C D→P D→Q

8 B 18 C

9 A,C

10 A,D

7 A 17 C D→Q D→P

8 A 18 D

9 A,B,C,D

7 A 17 C D→R D→P

8 C 18 A

9 A,C

C HE M ISTR Y 5 A 15 D C→P C→T

6 C 16 B

MATHEMATICS 5 B 15 C C→R C→Q

6 A 16 C

IIT- JEE 2012 (August issue) PHYSICS Ques Ans Ques Ans 19 20

1 B 11 A,C A→Q A→S

2 A 12 B, D

3 C 13 B B→P B→R

4 A 14 B

Ques Ans Ques Ans 19 20

1 C 11 D A→S A→R

2 B 12 A,C

3 D 13 A B→R B→S

4 C 14 C

5 C 15 D C→R C→P

6 B 16 C

C HE M ISTR Y 5 A 15 A C→P C→Q

6 D 16 A

10 B,C

MATHEMATICS Ques Ans Ques Ans 19 20

1 B 11 C,D A→Q A→Q

XtraEdge for IIT-JEE

2 A 12 A,C

3 B 13 A B→S B→R

4 D 14 B

5 B 15 D C→P C→S

70

6 C 16 B

10 A,B,C

AUGUST 2010

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