Anurag Pandey

Ever since I have cracked (IES: AIR 14, GATE: Rank 64) these two sacred exams for Engineers, I feel contented as far as clearing exams are concerned. But I have learnt a lot from the competition inherent in these exams. Lacs of candidates, millions of coaching institutes and zillions of aspiring souls in the form of parents of these candidates. These parents spend lacs on the coaching of their budding engineering sons/daughters just to satisfy themselves that they have done their duty.  According to me, if a candidate has completed his engineering or if candidate is a final year student of Engineering - Coaching is not required atleast for cracking GATE or IES !! GATE: Its an aptitude exam for checking that graduates have acquired requisite skills expected from an qualified Engineer.  IES: Its a comprehensive exam which checks the ability of graduates to take up Group A Services of Govt of India. It is more than aptitude exam and to me it appears to me as attitude exam. E

DBASEIII COMMANDS -Notes Database Organization: SORT, INDEX SORT Command ·    The SORT command gives us the ability to copy the record and structure of a database to another database. ·    Files can be arranged in either ascending order or descending order by using SORT command. ·    Sort command can be used in multiple fields for data arrangement. ·    SORT command creates a new database file. But the new file contains records of the original database file only the order is different. INDEX command ·    Index command can also be used to organize a data file. When you index a database, you ask dBASE to create a separate and smaller file based upon designated fields in the database. ·    The index command does not create a new database file. It creates an index file for the database with a default extension ( . NDX). ·    Unlike SORT command however INDEX does not permit the user to arrange the data in descending order DIFFERENCE BETWEEN SORT AND INDEX COMM

O ne of the toughest day- if I can recall. IES exam is spread on three days -Friday, Saturday & Sunday to test your abilities through 5 papers- GAT, Paper-I (Obj), Paper-II (Obj), Conventional Paper-I, Conventional Paper-II. Torture continues till Sunday ! When I concluded my preparations, I had no idea whether I would be able to make it to the elite list of IES qualifiers. But to some extent I was satisfied with my performance, this I thought made me comfortable. Never ever think of end results- just have faith in you and your capabilities which you have build through the preparations in the last 6-7 months or so.  During the preparations I had prepared two small notebooks in which I used to write important concepts and formula to revise them frequently. But before exam day when you have to solve Conventional Papers that too in Mechanical Engineering that too for Engineering Services- everything seems inadequate and so was my note book. I read it, re-read it and kept it as

I present you all with the very best analysis of IES 2009 toppers. Their marks, no of correct questions and their portfolio of scores in various subjects. This will make the complex and hazy picture- ultra clear. This is what I did in very first month of my preparations after cramming the syllabus and before making strategies for cracking the exam.       IES-2009 SCORES OF TOP 20           RANK GAT P-I P-II OBJ CONV-I CONV-II CONV TOTAL INT TOTAL 1 119 105 114 338 106 122 228 566 106 672 2 116 145 104 365 71 113 184 549 120 669 3 100 106 99 305 90 117 207 512 140 652 4 159 134 127 420 56 85 141 561 80 641 5 124 102 85 311 94 117 211 522 116 638 6 132 118 89 339 69 138 207 546 90 636 7 101

Nano composites consists of nano sized particles embedded in some type of matrix -are group of promising new materials being infused in some of our modern technologies.  One type of nano composites is currently being used in high performance tennis balls. These balls retain their original pressure and bounce twice as long as conventional ones. Air permeation through walls is inhibited by a factor of two due to presence of a flexible and very thin (10 to 50 microns) nanocomposite barrier coating that covers the inner core. This nanocomposite coating consists of matrix of butyl rubber, within which is embedded thin platelets of vermiculite, a natural clay mineral. The vermiculite platelets exists as single molecule thin sheets- on the order of nano meter thick-that have a very high aspect ratio.  Furthermore vermiculite platelets are exfoliated that is they remain separated from each other. Also within the butyl rubber, the vermiculite platelets are aligned such that all th

More than a year back I had an extraordinary opportunity to attend the so-called sacred and tough UPSC Interview board after clearing the gruelling written exam (in Gen Category- its always gruesome) . Since I was pretty excited about it and hence planned it well also. I booked a good hotel in the vicinity of UPSC and checked in to it a day before to settle in there. Since I was in job and had nobody to guide me, my plannings and preparations were all mine and since it made me successful I now think yes they were right.  I got up early that day had morning ablutions and got ready in ultra formal attire ! Light blue shirt and dark color trousers and a matching tie. I was looking as if I am heading towards my meeting as I usually do when in BHEL but this time it was something different- that was my rendezvous with UPSC !  I reached the venue quite early to see many veterans ( multiple timers) discussing what was asked last year and what mistakes they committed and how they were

2008 Which one of the following statements about FORTRAN is not correct? a. GO TO 999 b. CONTINUE 999 c. PRINT 999 d. DIMENSION A (999) 119. If A is greater than B then it is expressed in FORTRAN as: a. A > B b. AGTB c. A.GT.B d. AGT.B 2007 87. The arithmetic expression 2.4**3.0 will be computed by the FORTRAN compiler as: a. 2.4*2.4*2.4 b. 103.0 log 102.4 c. e3.0 log e2.4 d. 2.4*2.4*3.0 2006 58. Given J = 2, K = –4, A = –5, B = 0.4, C = 10.5 What are the values of the FORTRAN variables JIP = J*K + J**K, IVAL = A + B + C ? a. 15.3 and 5.9, respectively b. –7.83 and 5.9, respectively c. –7.83 and 5, respectively d. –8 and 5, respectively 2005 60. The initial contents of an array A are: A= [2 7 8 6 15 14 0 50 3] What would be the contents of the array A after executing the following segment of the FORTRAN code? DO 35 J = 1, T,1 TEMP = A(J) A(J) = A(J+2) A(J+1) = TEMP 35 CONTINUE a. A= [7 2 8 6 14 15 50 0 3] b. A= [8 6 2 7 0 50 15 14 3] c. A

FORTRAN @ Glance Declarations Syntax : ·    The type declaration statement indicates the type and name of the variable or constant (note the two colons between the type and the name of the variable or constant). Named constants must be declared with the parameter attribute: real, parameter:: pi = 3.1415927 ·    Like named constants, variables must be declared at the beginning of a program (or subprogram) in a type declaration statement: integer :: total real :: average1, average2 ! this declares 2 real values complex :: cx logical :: done character(len=80) :: line ! a string consisting of 80 characters ·       Constants can be assigned trivially to the complex number cx: cx = (1.0, 2.0) ! cx = 1.0 + 2.0i If you need to assign variables to cx, you need to use cmplx : cx = cmplx (1.0/2.0, -3.0) ! cx = 0.5 – 3.0i cx = cmplx (x, y) ! cx = x + yi The function cmplx is one of the intrinsic functions ·       Arrays real, dimension(5) :: vector ! 1-dim.real