tag:blogger.com,1999:blog-44320999750068394202014-10-02T21:17:31.310-07:00IIT JEE 2009KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.comBlogger58125tag:blogger.com,1999:blog-4432099975006839420.post-63806084836557572862009-04-07T23:50:00.000-07:002009-04-08T00:01:55.157-07:00Just Five Days for JEE 2009Relax, Relax and Relax<br /><br />Plan to keep your body physically fit. Mind agile.<br /><br />First believe in your two year effort. You have done your best. Feel happy about it. You are going to demonstrate the competence that you built over the last two years.<br /><br />Think of what works for you. Is it studying intensely for the last five days? Is it more relaxed study? Is it going through core principles of each chapter? Is it solving old difficult problems? Is it solving new problems?<br /><br />You are the best judge. Anything you do now must make you feel more happy, more confident and more engergized. Go to the examination with hope, and answer the question paper with hope. Answer every question with hope. <br /><br />When you are answering a question focus on that question. Don't worry about the rest of the paper. Don't worry the about the result. Think you will get the seat. Till the last second focus on the issue at hand and solve.<br /><br />Remember you have to write more examinations like AIEEE. Don't feel run down at the end of the examination. You should complete the examination with strength and energy and continue your enthusiasm for the next examination. Be cheerful. Learn from the experience to do still better at the next examination. Improvement is the objective. Not dejection.<br /><br />Believe in your effort, believe in your effort.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-54870978233771706082009-03-09T21:07:00.000-07:002009-03-09T21:12:42.536-07:00Last 3 weeks Preparation for JEE 2009My suggestion: <br /><br />Take 6 chapters of a subject each day. <br />Spend half an hour on principles and formulas for each chapter. <br />Revise 30 problems of a chapter in an hour. It will be good if you marked difficult problems in your earlier study. Revise those problems now.<br /><br />That way in 9 hours you can complete 6 chapters in a day. <br /><br />You can complete all the three subjects before the examination. <br /><br />I think this last 3 week preparation will add a lot to your earlier effort and make you do your best in the exam. <br /><br />All the best.<br /><br />Write tests conducted by your coaching institutes. If they tell you to read some chapters in more detail based on those tests, read those chapters. Remember it is time that you put in 12 to 13 hours effort for JEE. So you have lot more time to think your own ideas as well as to follow ideas of your coaching institutes.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-50725280048074749612009-02-16T16:47:00.000-08:002009-02-16T16:48:47.750-08:00Problems-Questions-Chemistry,Maths,Physics, Feb 16Problem-Physics-Friction<br />A rope is kept on the table with one fourth of its lengths hanging over. It begins to slide. What is coefficient friction between the rope and the table.<br /><br />a. 0.14<br />b. 0.21<br />c. 0.33<br />d. 0.67KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1tag:blogger.com,1999:blog-4432099975006839420.post-68511049854155250512009-02-16T08:59:00.000-08:002009-02-16T09:03:52.341-08:00Problems-Questions-Chemistry,Maths, Physics, Feb 16Physics <br />Gravitation chapter<br /><br />The cosmonauts landed on a planet and found that at the pole of the planet the force of gravity was 0.01 of that on earth. The duration of the day on the planet is the same as that on earth. At the equator, the gravity was found to be zero. Then what would be the radius of this planet approximately.<br /><br />a. 18,000 km<br />b. 12,000 km<br />c. 24,000 km<br />d. 30,000 kmKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-857673746129510972009-01-23T22:04:00.000-08:002009-01-23T22:08:03.469-08:00Preparing for Board and IIT JEEToday (24 Jan 2009) Mint paper carried an article titled Tricks from Tutors.<br />You may find it on www.livemint.com.<br /><br />Now you have to think of Boards as well as IIT JEE. Your good performance in Board examination is stepping stone for IIT JEE. There is no conflict between the two. Every concept that you use for board examination is useful for JEE also.<br /><br />Some suggestions<br /><br />If you are academically bright student, the next couple of weeks is the time for you to concentrate on concepts that you have not revised earlier. Don't waste time revisiting waht you alread know.<br /><br />Weaker students have to revise what they know and perfect that.<br /><br />If you have to appear for competitive examinations, devote 20% of time to prepare for these examinations until mid-February.<br /><br />5 R's for success: Reorganize, Revise, Rest, Relax and Reassure<br /><br />Reorganize: Collect and organize you study material for each retrieval.<br /><br /><br />Connect the course content to everyday ideas to have a greater recall. Try acronyms, mnemonics, rhymes that associate with the concepts.<br /><br />I personally advocate somebody in the family asking some questions everyday. When you are not able to answer, revise the topic and natural association between that event and the topic will be there.<br /><br />First understand and then memorize. If you memorize without understanding, examination anxiety may prevent you from recollecting it.<br /><br />Practice writing answers and attempt some mock tests. It will tell you how much time you are taking to write answers.<br /><br />Have a timetable for a day, but be flexible around it. Don't get stressed. You may stretch on a day or slightly relax on another day. Stay refreshed and start everyday with enthusiasm.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-37976987800290600302009-01-01T00:54:00.000-08:002009-01-01T00:57:10.261-08:00Ask questions and answer questions about IIT JEE SubjectsKNOL QUESTION AND ANSWER BOARD<br /><a href="http://knol.google.com/k/narayana-rao-kvss/-/2utb2lsm2k7a/654#">http://knol.google.com/k/narayana-rao-kvss/-/2utb2lsm2k7a/654#</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-72188052569099715382008-12-23T20:57:00.000-08:002008-12-23T21:00:14.136-08:00Nomenclature of Organic Compounds - Practice setNomenclature when two or more substituents are in the compound<br /><br />Visit<br /><br /><a href="http://iit-jee-chemistry-ps.blogspot.com/2008/12/nomenclature-of-organic-compounds.html">http://iit-jee-chemistry-ps.blogspot.com/2008/12/nomenclature-of-organic-compounds.html</a><br /><br />The exercise has compound names and you have to visualize the structural formula and line formula.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-30417919475354506992008-12-23T01:34:00.000-08:002008-12-23T01:45:03.770-08:00Sets - Example Revision - Number of Elements in a SetFormulas used to solve the example problems.<br /><br />n(A U B) = n(A)+n(B) - N(A∩B)<br /><br />n(A-B) = n(A)-n(A∩B)<br /><br />n(A'∩B') = n(U) - n(A∩B) = n((A∩B)')KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-55710639661031650102008-12-08T07:06:00.000-08:002008-12-08T07:10:15.754-08:00Happy Graduate from IIITM GwaliorAnswered with data about my IIT JEE rank, admission offer from IIT, refusal of the offer to take up course of interest at IIITM Gwalior, top grades throughout college, internship selection through a national level contest, pre-placement offer, achievements at workplace, recognition received etc. And oh yeah, the GMAT score too (though it was present in the application.)<br /><br />Received an acceptance call on 20th November from ISB Hyderabad.<br /><br />I work as a software development engineer for a dream software company and located in Hyderabad. I have been working here since May 2007 (first and only job and 18 months experience as of writing this post.) My GMAT score is 760<br /><br />Sid's Scribble <br /><br /><a href="http://sidsscribble.blogspot.com/2008/12/road-to-isb.html">http://sidsscribble.blogspot.com/2008/12/road-to-isb.html</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-47755272337658775542008-11-30T04:20:00.000-08:002008-11-30T04:27:08.973-08:00Strategy from December 2008 - April 2009Many have to learn some new materials/chapters. Some institutes have completed the full syllabus.<br /><br />If you have not yet completed full syllabus, you have to learn daily some new materials by sitting in the class, studying at home and doing problems in these chapters.<br /><br />Revision: Have a plan to complete one full revision of all chapters by December end.<br /><br />Mastering chapters: Have a plan to master at least one chapter for day. I earliest said five point revision per hour. Do five point revision for the chapter you want to master on a day.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-38781234066468103582008-11-23T06:22:00.000-08:002008-11-23T06:25:02.607-08:00Study Plan for Examination - JEE 2009Complete revision of all chapters once before Dec 31.<br /><br /><strong>Jan 1 to Feb 15.</strong> Complete revision of all principles at the rate of 3 chapters per day.<br /><br />Daily solve <strong>100</strong> problems including some difficult examples<br /><br /><strong>Second revision</strong> of principles Feb 16 to 31 March<br /><br />Daily solve <strong>100</strong> problems including some difficult examples<br /><br /><strong>Third revision</strong> 1 April to 7th April - 15 Chapters a day. Revise some difficult problems.<br /><br />Relax for one or two days.<br /><br />Do revision as you feel without tiring yourself before the examination.<br /><br />All the best for the examination. Go with all the confidence. You have done your best preparation.<br /><br />Your comments and plan welcomeKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-16299118081585895332008-11-09T02:49:00.000-08:002008-11-09T02:50:17.546-08:00Pursuit of Goals - Motivation EssentialBonnie St. John Deane<br /> <br />Winner in Disabled Olympics, Harvard and Oxford Degree holder, White House Official on National Economic Council<br /> <br />Who or what motivates you?<br /> <br />Her answer: "I motivate me."<br /> <br />Her method: Five step exercise.<br /> <br />Five steps<br /> <br />1.What are your payoffs? List down the benefits or rewards from achieving or completing the task.<br /> <br />2. Find deeper meaning in your goals. Find compelling reasons for completing your goal.<br /> <br />3. Get your dreams at half price. Find more goals which can be reached with the work that you are putting in for reaching this goal. Work with a friend. Develop a plan to achieve the goal efficiently with less cost.<br /> <br />4. Stop underestimating your odds for success: Underestimating your odds to succeed can stop you dead in your tracks. Believe in your success till the end of the task.<br /> <br />Err on the side of optimism. Spend less time with people who discourage you on the task or goal. But don't just believe in positive thinking. Investigate the odds, have a real confidence that they are within your teach and always work on improving the odds.<br /> <br />5. Change your real odds or winning or improving odds: Learn from people who have succeeded in reaching the goal before.<br /><br />http://<a href="http://knol.google.com/k/kvssn-rao/pursuit-of-excellence-and-success/1zb6eis38d7or/8#">knol.google.com/k/kvssn-rao/pursuit-of-excellence-and-success/1zb6eis38d7or/8#</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com2tag:blogger.com,1999:blog-4432099975006839420.post-37645350013051315072008-10-13T03:55:00.000-07:002008-10-13T04:03:31.422-07:00JEE 2009 - Get ready to increase your effortIn the last lap of the preparation you have to put in more intensive and focused effort.<br /><br />13 hours per day from 1st January to 31 March 2009.<br /><br />12 hours per day December<br /><br />11 hours per day in November<br /><br />10 hours per day so far.<br /><br /><strong>Targets:</strong> Complete study of all the chaters one more time before 31st December 2008.<br /><br />From January: Every day <br /><br />Revision of important points 3 chapters 1.5 hours<br />Problems and questions from 3 chapters 5 hours - target time for each question 2 minutes<br />preparation for the board examination 3.5 hours<br /><br />Attending the coaching class 3 hours (if coaching class is not there additional preparation in studying the details of the three chapters/revision of difficult problems in the chapters)<br /><br />By 15th February all chapters revision of important points and formulas to be over.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1tag:blogger.com,1999:blog-4432099975006839420.post-12468669813111256252008-09-20T22:19:00.001-07:002008-09-20T22:21:16.285-07:00IIT JEE 2009 ENTRANCE EXAM NOTIFICATIONIIT Joint Entrance Examination (JEE 2009) Schedule<br /><br />Examination<br /><br /><strong>April 12, 2009 (Sunday)</strong><br /><br />09.00 –12.00 hrs Paper – 1<br />14.00 - 17.00 hrs Paper - 2<br /><br />Paper – 1 and Paper – 2 will each have three separate sections on Physics, Chemistry, Mathematics. Both the papers will be objective types, designed to test comprehension, reasoning and analytical ability of candidates.<br /><br />Eligibility requirements for this examination and syllabus for Physics, Chemistry, Mathematics and Aptitude Test will be available on the websites of all IITs and will also be given in the Information Brochure of JEE – 2009.<br /><br />Candidates will have the option of submitting either on – line through internet) or paper application form- offline.<br /><br />Important dates regarding Application Form and Brochure:<br /><br />Sale at designated branches of Banks and at all IITs: Nov 19 - Dec 24 2008<br />Postal Request of Application form: Nov 19 - Dec 16 2008<br />Last date of receipt of completed application forms at IITs : <strong>Dec 24 2008</strong>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com1tag:blogger.com,1999:blog-4432099975006839420.post-85871713488036804042008-08-17T02:36:00.000-07:002008-08-17T02:37:11.707-07:00Inorganic Chemistry Past JEE Fill Blanks QuestionsFill in the blanks<br /><br />1. The increase in solubility of iodine in an aqueous solution of potassium iodide is due to the formation of ___________ . (1982)<br /><br />2. Galvanization of iron denotes coating with ___________ .(1983)<br /><br />3. If metal ions of group III are precipitated by NH4Cl and NH4OH without prior oxidation by conc. HNO3 ___________ in not completely precipitated. (1984)<br /><br />4. Sodium dissolved in liquid ammonia conducts electricity because _________. (1985)<br /><br />5. Hydrogen gas is liberated by the action of aluminum with concentrated solution of ___________. (1987)<br /><br />6. Silver chloride is sparingly soluble in water because its lattice energy is greater than __________ energy. (1987)<br /><br />7. _______ phosphorus is reactive because of its highly strained tetrahedral structure.<br /><br />8. In extractive metallurgy of zinc, partial fusion of ZnO with coke is called _______ and reduction of the ore to the molten metal is called ____________ .<br />(smelting, calcining, roasting, sintering) (1988)<br /><br />9. The salts _________ and _______________ are isostructural.<br />(FeSO4.7H2O; CuSO4.5H2O; MnSO4.4H2O; ZnSO4.7H2O)<br /><br />10. ______________ acid gives hypo ____________ ion.<br />(hydrobromic, hypobromous, perbromic, bromide, bromite, perbromate) (1988) <br /><br />11. Sulphur acts as ___________ agent in vulcanization of rubber. (1989)<br /><br />12. The basicity of phosphorus acid (H3PO3) is ____________ . (1990)<br /><br />13. In P4O10, the number of oxygen atoms bonded to each phosphorus atom is _________ . (1992)<br /><br />14. The lead chamber process involves oxidation of SO2 by atmospheric oxygen under the influence of __________ as catalyst. (1992)<br /><br />15. Ca<sup>2+</sup> has a smaller ionic radius than K<sup>+</sup> because it has __________- . (1993)<br /><br />16. The formula of the deep red liquid formed on warming dichromate with KCl in concentrated sulphuric acid is ______________ . (1993)<br /><br />17. The two types of bonds present in B2H6 are covalent and __________ . (194).<br /><br />18. The type of magnetism exhibited by [Mn(H2O)6] <sup>2+</sup> is ________________ . (1994)<br /><br />19. One recently discovered allotrope of carbon (e.g., C60) is commonly known as _________ . (1994)<br /><br />20. Among PCl3, CH3<sup>+</sup>, NH2<sup>-</sup>, and NF3, ___________ is least reactive towards water. (1997)<br /><br />21. A solution of sodium in liquid ammonia at -33˚C conducts electricity. On cooling, the conductivity of this solution ______________ .<br /><br />22. When an aqueous solution of sodium fluoride is electrolysed, the gas liberated at the anode is ___________ .<br /><br />23. Silver jewellery items tarnish slowly in the air due to their reaction with ___________. (1997).<br /><br />24. Compounds that formally contain Pb<sup>4+</sup> are easily reduced to Pb<sup>2+</sup>. The stability of the lower oxidation state is due to _____ . (1997)<br /><br /><br /><br />Answers some time laterKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-39710832715147646562008-08-17T00:46:00.001-07:002008-08-17T00:47:24.267-07:00INorganic Chemistry - Past JEE - True or False questionsState whether the following statements are true or false<br /><br />1. Red phosphorus is less volatile than white phosphorus because the former has a tetrahedral structure. (1982)<br /><br />2. MgCl2.6H2O on heating gives anhydrous MgCl2. (1982)<br /><br />3. When PbO2 reacts with a dilute acid, it gives hydrogen peroxide. (1982)<br /><br />4. Copper metal reduces Fe<sup>2+</sup> in an acid medium. (1982)<br /><br />5. Silver fluoride is fairly soluble in water. (1982)<br /><br />6. Dilute HCl oxidizes metallic Fe to Fe<sup>2+</sup>. (1983)<br /><br />7. In an aqueous solution chlorine is a stronger oxidizing agent than fluorine. (1984)<br /><br />8. Silver chloride is more soluble in very concentrated sodium chloride solution than in pure water. (1984)<br /><br />9. Solubility of sodium hydroxide increases with increase in temperature. (1985)<br /><br />10. Sodium when burnt in excess of oxygen gives sodium oxide. (1987).<br /><br />11. Both potassium ferrocyanide and potassium ferricyanide are diamagnetic. (1989)<br /><br />12. Cu<sup>+</sup> disproportionates to Cu<sup>2+</sup>. (1991)<br /><br />13. Nitrous oxide, though an odd electron molecule, is diamagnetic in liquid state. (1991)<br /><br />14. Diamond is harder than graphite. (1993)<br /><br />15. The basic nature of the hydroxides of Group 13 (Gr. III B) decreases progressively down the group. (1993)<br /><br />16. The tendency for catenation is much higher for C than for Si. (1993)<br /><br />17. HBr is stronger acid than HI because of hydrogen bonding. (1993)<br /><br />Answers some time laterKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-73323895169111336322008-07-08T02:51:00.000-07:002008-07-08T02:52:51.810-07:00Laws and Theories - Revision Part 2<strong>MaxWell's Speed Distribution Law</strong><br /><br />It is an equation giving the distribution of molecules in different speeds in a gas at a temperature.<br /><br />If dN represents the number of molecules with speeds between v and v+dv then<br /><br />dN = 4πN[m/2πkT]<sup>3/2</sup>v²e<sup>-mv²/2kT</sup>dv<br /><br />where<br />dN represents the number of molecules with speeds between v and v+dv<br />N = total number of molecules in the gas<br />m = mass of a molecule<br />T = absolute temperature of the gas<br />v = velocity of the molecules<br /><br />The speed v<sub>p</sub> at which dN/dv is maximum is called the most probable speed.<br />Its value is given by <br /><br />v<sub>p</sub> = √(2kT/m) <br /><br />(Topic: Kinetic theory of gases)<br /><br /><br /><strong>Avogadro's Law</strong><br /><br />Under similar conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules.<br /><br />(Basic concepts of chemistry)<br /><br /><br />Vectors<br /><br />The sum of the resolved parts of two forces acting through a point along any direction is equal to the resolved part of the resultant of the two forces along that direction.<br /><br />Explanation<br /><br />If you find the resolved parts in some direction initially and then find their resultant, it will be equal to the resolved part of the resultant of the two forces in the same direction.<br /><br />(Topic: Vectors)KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-77717659211805328712008-07-08T02:46:00.000-07:002008-07-08T02:51:20.707-07:00Revision - Laws and Theories - 1Laws of Thermodynamics<br /><br /><strong>Zeroth law:</strong> If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium then B and C are also in thermal equilibrium.<br /><br /><br /><br /><strong>First law of thermodynamics</strong><br /><br />∆U = ∆Q -∆W<br />Where<br />∆U = change in internal energy of a thermodynamic system<br />∆Q = Heat given to the system<br />∆W = work done by the system<br /><br />Change in internal energy of a thermodynamic system is equal to the heat given to the system minus the work done by the system on surroundings or environment.<br /><br /><strong>Second law of thermodynamics</strong><br /><br />Kelvin-Planck statement<br /><br />It is not possible to design a heat engine which works in cyclic process and whose only result is to take heat from a body at a single temperature and convert it completely into mechanical work. <br /><br /><br />(Topic: Laws of Thermodynamics)<br /><br /><br /><br /><strong>1. Law of conservation of mass</strong><br /><br />During any physical or chemical change, the total mass of the products is equal to the total mass of reactants.<br /><br /><strong>2. Law of constant proportions</strong><br /><br />A pure chemical compound always contains same elements combined together in the same definite proportion by weight.<br /><br /><strong>3. Law of multiple proportions</strong><br /><br />When two elements combine to form two or more than two compounds, the weights of one of the elements which combine with a fixed weight of the other, bear a simple whole number ratio.<br /><br /><strong>4. Law of reciprocal proportions</strong><br /><br />When two different elements combine separately with the same weight of a third element, the ratio in which they do so will be the same or some simple multiple of the ratio in which they combine with each other.<br /><br /><strong>5. Gay Lussac’s law of combing volumes</strong><br /><br />Under similar conditions of temperature and pressure, whenever gases react together, the volumes of the reacting gases as well as products (if gases) bear a simple whole number ratio.<br /><br /><strong>Lagrange's Identity</strong><br /><br />(<strong>a</strong> × <strong>b</strong>)² = a²b² - (<strong>a</strong>.<strong>b</strong>)<br /><br /><strong>a</strong> and <strong>b</strong> are vectors<br />a and b are magnitudes of <strong>a</strong> and <strong>b</strong> respectively.<br /><br />Lagrange's identity is a relation between the cross product and the dot product.<br /><br />(Topic: Vectors)KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-3564018237045724952008-06-19T01:22:00.000-07:002008-06-19T01:41:53.811-07:00Electricty questions from physicsgoeasy blogElectricty questions from physicsgoeasy blog<br /><br /><br />http://<a href="http://physicsgoeasy.blogspot.com/2008/05/iit-jee-test-serieselectricity-1.html">physicsgoeasy.blogspot.com/2008/05/iit-jee-test-serieselectricity-1.html</a><br />dated 18 May 2008<br /><br /><br />IIT JEE Test series (ELectricty) solutions <br />dated 18 June 2008<br />http://<a href="http://physicsgoeasy.blogspot.com/2008/06/iit-jee-test-series-electricty.html">physicsgoeasy.blogspot.com/2008/06/iit-jee-test-series-electricty.html</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-7421052469386615972008-06-18T00:05:00.000-07:002008-06-18T00:09:37.114-07:00Answers to 8 June 2008 Practice paperMathematics<br /><br />1. if |z-1| /|z-4| = 2, the locus of z = x+iy in Argand plane will be: <br /><br />a. parabola<br />b. hyperbola<br />c. circle<br />d. straight line<br /><br />Answer (c)<br /><br />z = x+iy<br />| (x+iy-1) |/| (x+iy-4) | = 2<br /><br />(x-1) ² + y² = 4[(x-4) ² +y²]<br /><br />x² -2x+1+y² = 4{x-8x+16+y²]<br />x² -2x+1+y² = 4x²-32x+64+4y²<br />3x²-30x+63+3y² = 0<br />x²+y²-10x+21 = 0<br />Equation represents a circle with (5,0) as centre and SQRT(5²-21) as radius.<br />Radius is 2<br /><br /><br /><br /><br /><br /><br />2. <br /><br />Let a,b,c be three non-coplanar vectors and p,q,r are vectors defined by the relations<br /><br />p = b×c/[abc]<br />q = c×a/[abc]<br />r = a×b/[abc]<br /><br />then the value of the expression<br />[(a+b).p +(b+c).q +(c+a).r] is equal to<br /><br />a. 3<br />b. 1<br />c. 0<br />d. 2<br /><br /><br /><br />Answer: (a)<br /><br />[(a+b).p +(b+c).q +(c+a).r] = [1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)]<br /><br />The property of distributivity of scalar product over vector addition is<br />a.(b+c) = a.b+a.c and<br />(b+c).a = b.a+b.c<br /><br />[1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)] =<br /><br />[1/(abc)][( a .(b×c)+ b.(b×c)+b .( c×a ) + b.(b×c)+c.( a×b) +a.( a×b)] <br /><br />If a┴b, a.b = 0 <br /><br />Hence b.(b×c), and b.(b×c) and a.( a×b) are zero as .(b×c) will be a vector perpendicular to vectors b and c.<br /><br />Hence the expression reduces to [1/(abc)][( a .(b×c)+ b .( c×a ) + c.( a×b)] <br /><br />As a .(b×c) = b .( c×a ) = c.( a×b) = (abc)<br />The expression reduces to 3/(abc) /(abc) = 3<br /><br /><br /><br />Physics<br />3. A particle moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is<br /><br />a. zero<br />b. 1/SQRT(2) m/s² towards north-west<br />c. ½ m/s² toward north-west<br />d. ½ m/s² towards north<br /><br /><br /><br />Answer (b)<br /><br />The average acceleration must have an westward component and northward component to make the velocity towards east as zero and to make velocity towards north as 5 m/s<br /><br />Hence it will act in the direction of northwest.<br /><br />Westward component of average acceleration = (0-5)/10 = -½ m/s²<br />Northward component of average of acceleration = (5-0)/10 = ½ m/s²<br /><br />Hence magnitude of average acceleration = Sqrt( (-1/2) ²+(1/2) ²) = 1/SQRT(2)<br /><br /><br /><br />Chemistry<br /><br />4. Which of the following has maximum number of atoms?<br /><br />a. 24 g of C (M=12 g mol^-1<br />b. 23 g of Na (M = 23 g mol^-1)<br />c. 48 g of S (M = 32 b mol^-1)<br />d. 108 g of Ag (M = 108 mol ^-1)<br /><br /><br /><br />Answer: (a)<br /><br />Number of moles (N) = m/M<br />Where m = given mass<br />M = molar mass<br /><br /><br />Hence Number of moles of carbon = 2<br />Number of moles of Na = 1<br />Number of moles of S = 1.5<br />Number of moles of Ag = 1<br /><br />So 24 g of C having 2 moles has the maximum number of atoms.<br /><br /><br /><br />5. The wave number of the shortest wavelength transition in the Balmer series of atomic hydrogen is.<br /><br />a. 27420 cm^-1<br />b. 28420 cm^-1<br />c. 29420 cm^-1<br />d. 12186 cm^-1<br /><br /><br />Answer: (a)<br /><br />Wave number = 1/ λ = R(1/nf² - 1/ni²)<br />Wave length is shortest, when energy emitted is highest. This will happen when the electron jumps from n = ∞ to n=2. <br /><br />So wave number = 109677 cm^-1(1/2² - 1/∞²)<br />= 109677(1/4)<br />= 27419.25 cm^-1KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-55931545560721467832008-06-07T21:48:00.000-07:002008-06-07T21:49:26.605-07:00IIT JEE June 8 2008 Practice paperMathematics<br /><br />1. if |z-1| /|z-4| = 2, the locus of z = x+iy in Argand plane will be: <br /><br />a. parabola<br />b. hyperbola<br />c. circle<br />d. straight line<br /><br /><br /><br /><br /><br />2. <br /><br />Let a,b,c be three non-coplanar vectors and p,q,r are vectors defined by the relations<br /><br />p = b×c/[abc]<br />q = c×a/[abc]<br />r = a×b/[abc]<br /><br />then the value of the expression<br />[(a+b).p +(b+c).q +(c+a).r] is equal to<br /><br />a. 3<br />b. 1<br />c. 0<br />d. 2<br /><br /><br />Physics<br />3. A particle moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is<br /><br />a. zero<br />b. 1/SQRT(2) m/s² towards north-west<br />c. ½ m/s² toward north-west<br />d. ½ m/s² towards north<br /><br />Chemistry<br /><br />4. Which of the following has maximum number of atoms?<br /><br />a. 24 g of C (M=12 g mol^-1<br />b. 23 g of Na (M = 23 g mol^-1)<br />c. 48 g of S (M = 32 b mol^-1)<br />d. 108 g of Ag (M = 108 mol ^-1)<br /><br /><br />5. The wave number of the shortest wavelength transition in the Balmer series of atomic hydrogen is.<br /><br />a. 27420 cm^-1<br />b. 28420 cm^-1<br />c. 29420 cm^-1<br />d. 12186 cm^-1KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-309760796365824372008-06-07T21:44:00.000-07:002008-06-07T21:48:00.800-07:00Solutions to June 1 2008 practice set1. State whether the statement is true or false<br /><br />A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.<br /><br /><br />Answer: true<br /><br />At the top of the path, the vertical component of velocity is zero and the particle has only horizontal component. Hence the speed (the magnitude of the velocity) is minimum at the top of the path.<br /><br /><br /><br />2. The number of vectors of unit length perpendicular to vectors <strong>a</strong> = (1,1,0) and <strong>b</strong> = (0,1,1) is<br /><br />a. three <br />b. two<br />c. one<br />d. infinite<br />e. none of these<br /><br />Answer (b)<br /><br />If vector (x,y,z) is a unit vector perpendicular to (1,1,0) their scalar product is zero<br />=> x +y = 0<br />=> y = -x<br /><br /><br />If vector (x,y,z) is a unit vector perpendicular to (0,1,1) their scalar product is zero<br />=> y+z = 0<br />=> y = -z<br /><br />As it is a unit vector x²+y²+z² = 1<br />=> 3y² = 1<br />=> y = ±1/SQRT(3)<br /><br />This gives two value of y and hence there are two vectors<br /><br /><br /><br /><br /><br /><br /><br />3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be<br /><br />a. real and positive<br />b. zero<br />c. real and negative<br />d. purely imaginary<br />e. none of these<br /><br /><br />Answer: (b) and (d) <br /><br />Assume z1 = a+ib, a is positive means a>0, <br />z2 = c+id, d<0<br /><br />|z1| = |z2|<br />=> a²+b² = c²+d²<br /><br />(z1+z2)/(z1-z2) = [(a+c) + i(b+d)]/ [(a-c) + i(b-d)]<br /><br />Taking the multiplicative inverse of denominator and multiplying the numerator<br /><br />= [(a+c) + i(b+d)]* [(a-c) - i(b-d)]/[(a-c) ² + (b-d) ²]<br /><br /><br />= [(a²-c²)+(b²-d²)]+i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]<br /><br />=[{(a²+b²)-(c²+d²)}+i{(b+d)(a-c)-(a+c)(b-d)}]/ [(a-c) ² + (b-d) ²]<br /><br />As a²+b² = c²+d² the first term of the numerator is zero.<br /><br />Hence <br />(z1+z2)/(z1-z2) <br />= i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]<br /><br />So this is a pure imaginary number.<br />If b+d as well as a+c are equal zero, the expression can be zero also.<br /><br /><br /><br /><br /><br />4. If two compounds have the same empirical formula but different molecular formulae they must have ------<br /><br />a. different percentage composition<br />b. different molecular weights<br />c. same viscosity<br />d. same vapour density<br /><br />answer (b)<br /><br /><br />Different molecular formulae but same empirical formula imply same empirical mass but different molecular massesKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-46718289558603592532008-05-31T05:26:00.000-07:002008-05-31T05:29:27.920-07:00IIT JEE June 1 st Weekly Cumulative Practice Test1. State whether the statement is true or false<br /><br />A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.<br /><br /><br />2. The number of vectors of unit length perpendicular to vectors <strong>a</strong> = (1,1,0) and <strong>b</strong> = (0,1,1) is<br /><br />a. three <br />b. two<br />c. one<br />d. infinite<br />e. none of these<br /><br /><br /><br /><br /><br /><br />3. Let z1 and z2 be complex numbers such that z1≠z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative imaginary part, then (z1+z2)/(z1-z2) may be<br /><br />a. real and positive<br />b. zero<br />c. real and negative<br />d. purely imaginary<br />e. none of these<br /><br /><br /><br />4. If two compounds have the same empirical formula but different molecular formulae they must have ------<br /><br />a. different percentage composition<br />b. different molecular weight<br />c. same viscosity<br />d. same vapour densityKVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-13320040751191260282008-05-27T10:35:00.000-07:002008-05-27T10:41:08.263-07:00Solutions for May 25 Practice setcovers topics<br />Complex Numbers<br />Vectors (common to physics and mathematics)<br />Mole concept<br /><br />1. The inequality |z-4|<|z-2| represents the region given by<br /><br />a. Re(z)>0<br />b. Re(z)<2<br />c. Re(z)<0<br />d. none of these<br /><br />Answer (d)<br /><br />Answer<br /><br />Let z = x+iy then<br /><br />|z-4|<|z-2|<br /> <br />=> |z-4²|<|z-2|²<br />=> (x-4) ²+y²<(x-2) ²+y²<br /> => -8x+16<-4x+4<br />=>12<4x<br />=> x>3<br />Means Re(z)>3<br /><br /><br /><br /><br /><br />2. The points with the position vectors 60<strong>i</strong> +3<strong>j</strong>, 40<strong>i</strong>-8<strong>j</strong>, a<strong>i</strong>-52<strong>j</strong> are collinear if<br /><br />a. a = -40<br />b. a = 40<br />c. a = 20<br />d. none of these<br /><br /><br /><br />If the vectors <strong>x</strong> and <strong>y</strong> are collinear, <br />x = py<br />If we take given position vectors m.n,o, it implies<br /><strong>o</strong>-<strong>m</strong> = p(<strong>n</strong>-<strong>m</strong>)<br /><br />[(a<strong>i</strong>-52<strong>j</strong>)-( 60<strong>i</strong> +3<strong>j</strong>)] = k[(40<strong>i</strong>-8<strong>j</strong>)-( 60<strong>i</strong> +3<strong>j</strong>)]<br /><br />=> (a-60)<strong>i</strong> – 55<strong>j</strong> = k(-20<strong>i</strong>-11<strong>j</strong>)<br />=> a-60 = -20k and -55 = -11k<br />=> k = 55/11 <br />=> a = -20*5 + 60<br />=> a = -40<br /><br /><br />3. One molal solution is one that contains one mole of a solute in :<br /><br />a. 1000 g of the solvent<br />b. one litre of the solvent<br />c. one litre of the solution<br />d. 22.4 litres of the solution<br /><br />Answer (a)<br /><br />As per the definition, molality is amount of solute per kg of solvent.KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0tag:blogger.com,1999:blog-4432099975006839420.post-31038466233468515282008-05-24T22:29:00.000-07:002008-05-27T22:01:09.586-07:00May 25 2008 Question set 1covers topics<br />Complex Numbers<br />Vectors (common to physics and mathematics)<br />Mole concept<br /><br />1. The inequality |z-4|<|z-2| represents the region given by<br /><br />a. Re(z)>0<br />b. Re(z)<2<br />c. Re(z)<0<br />d. none of these<br /><br /><br />2. The points with the position vectors 60<strong>i</strong> +3<strong>j</strong>, 40<strong>i</strong>-8<strong>j</strong>, a<strong>i</strong>-52<strong>j</strong> are collinear if<br /><br />a. a = -40<br />b. a = 40<br />c. a = 20<br />d. none of these<br /><br />3. One molal solution is one that contains one mole of a solute in :<br /><br />a. 1000 g of the solvent<br />b. one litre of the solvent<br />c. one litre of the solution<br />d. 22.4 litres of the solution<br /><br /><br />For solution see<br />http://<a href="http://iit-jee-nrao.blogspot.com/2008/05/solutions-for-may-25-practice-set.html">iit-jee-nrao.blogspot.com/2008/05/solutions-for-may-25-practice-set.html</a>KVSSNraohttp://www.blogger.com/profile/05748254811752425330noreply@blogger.com0