Nomenclature of Organic Compounds - Practice set

Nomenclature when two or more substituents are in the compound

Visit

http://iit-jee-chemistry-ps.blogspot.com/2008/12/nomenclature-of-organic-compounds.html

The exercise has compound names and you have to visualize the structural formula and line formula.

Sets - Example Revision - Number of Elements in a Set

Formulas used to solve the example problems.

n(A U B) = n(A)+n(B) - N(A∩B)

n(A-B) = n(A)-n(A∩B)

n(A'∩B') = n(U) - n(A∩B) = n((A∩B)')

Happy Graduate from IIITM Gwalior

Answered 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.)

Received an acceptance call on 20th November from ISB Hyderabad.

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

Sid's Scribble

http://sidsscribble.blogspot.com/2008/12/road-to-isb.html

Strategy from December 2008 - April 2009

Many have to learn some new materials/chapters. Some institutes have completed the full syllabus.

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.

Revision: Have a plan to complete one full revision of all chapters by December end.

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.

Study Plan for Examination - JEE 2009

Complete revision of all chapters once before Dec 31.

Jan 1 to Feb 15. Complete revision of all principles at the rate of 3 chapters per day.

Daily solve 100 problems including some difficult examples

Second revision of principles Feb 16 to 31 March

Daily solve 100 problems including some difficult examples

Third revision 1 April to 7th April - 15 Chapters a day. Revise some difficult problems.

Relax for one or two days.

Do revision as you feel without tiring yourself before the examination.

All the best for the examination. Go with all the confidence. You have done your best preparation.

Your comments and plan welcome

Pursuit of Goals - Motivation Essential

Bonnie St. John Deane

Winner in Disabled Olympics, Harvard and Oxford Degree holder, White House Official on National Economic Council

Who or what motivates you?

Her answer: "I motivate me."

Her method: Five step exercise.

Five steps

1.What are your payoffs? List down the benefits or rewards from achieving or completing the task.

2. Find deeper meaning in your goals. Find compelling reasons for completing your goal.

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.

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.

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.

5. Change your real odds or winning or improving odds: Learn from people who have succeeded in reaching the goal before.

http://knol.google.com/k/kvssn-rao/pursuit-of-excellence-and-success/1zb6eis38d7or/8#

JEE 2009 - Get ready to increase your effort

In the last lap of the preparation you have to put in more intensive and focused effort.

13 hours per day from 1st January to 31 March 2009.

12 hours per day December

11 hours per day in November

10 hours per day so far.

Targets: Complete study of all the chaters one more time before 31st December 2008.

From January: Every day

Revision of important points 3 chapters 1.5 hours
Problems and questions from 3 chapters 5 hours - target time for each question 2 minutes
preparation for the board examination 3.5 hours

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)

By 15th February all chapters revision of important points and formulas to be over.

IIT JEE 2009 ENTRANCE EXAM NOTIFICATION

IIT Joint Entrance Examination (JEE 2009) Schedule

Examination

April 12, 2009 (Sunday)

09.00 –12.00 hrs Paper – 1
14.00 - 17.00 hrs Paper - 2

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.

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.

Candidates will have the option of submitting either on – line through internet) or paper application form- offline.

Important dates regarding Application Form and Brochure:

Sale at designated branches of Banks and at all IITs: Nov 19 - Dec 24 2008
Postal Request of Application form: Nov 19 - Dec 16 2008
Last date of receipt of completed application forms at IITs : Dec 24 2008

Inorganic Chemistry Past JEE Fill Blanks Questions

Fill in the blanks

1. The increase in solubility of iodine in an aqueous solution of potassium iodide is due to the formation of ___________ . (1982)

2. Galvanization of iron denotes coating with ___________ .(1983)

3. If metal ions of group III are precipitated by NH4Cl and NH4OH without prior oxidation by conc. HNO3 ___________ in not completely precipitated. (1984)

4. Sodium dissolved in liquid ammonia conducts electricity because _________. (1985)

5. Hydrogen gas is liberated by the action of aluminum with concentrated solution of ___________. (1987)

6. Silver chloride is sparingly soluble in water because its lattice energy is greater than __________ energy. (1987)

7. _______ phosphorus is reactive because of its highly strained tetrahedral structure.

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 ____________ .
(smelting, calcining, roasting, sintering) (1988)

9. The salts _________ and _______________ are isostructural.
(FeSO4.7H2O; CuSO4.5H2O; MnSO4.4H2O; ZnSO4.7H2O)

10. ______________ acid gives hypo ____________ ion.
(hydrobromic, hypobromous, perbromic, bromide, bromite, perbromate) (1988)

11. Sulphur acts as ___________ agent in vulcanization of rubber. (1989)

12. The basicity of phosphorus acid (H3PO3) is ____________ . (1990)

13. In P4O10, the number of oxygen atoms bonded to each phosphorus atom is _________ . (1992)

14. The lead chamber process involves oxidation of SO2 by atmospheric oxygen under the influence of __________ as catalyst. (1992)

15. Ca²⁺ has a smaller ionic radius than K⁺ because it has __________- . (1993)

16. The formula of the deep red liquid formed on warming dichromate with KCl in concentrated sulphuric acid is ______________ . (1993)

17. The two types of bonds present in B2H6 are covalent and __________ . (194).

18. The type of magnetism exhibited by [Mn(H2O)6] ²⁺ is ________________ . (1994)

19. One recently discovered allotrope of carbon (e.g., C60) is commonly known as _________ . (1994)

20. Among PCl3, CH3⁺, NH2^-, and NF3, ___________ is least reactive towards water. (1997)

21. A solution of sodium in liquid ammonia at -33˚C conducts electricity. On cooling, the conductivity of this solution ______________ .

22. When an aqueous solution of sodium fluoride is electrolysed, the gas liberated at the anode is ___________ .

23. Silver jewellery items tarnish slowly in the air due to their reaction with ___________. (1997).

24. Compounds that formally contain Pb⁴⁺ are easily reduced to Pb²⁺. The stability of the lower oxidation state is due to _____ . (1997)



Answers some time later

INorganic Chemistry - Past JEE - True or False questions

State whether the following statements are true or false

1. Red phosphorus is less volatile than white phosphorus because the former has a tetrahedral structure. (1982)

2. MgCl2.6H2O on heating gives anhydrous MgCl2. (1982)

3. When PbO2 reacts with a dilute acid, it gives hydrogen peroxide. (1982)

4. Copper metal reduces Fe²⁺ in an acid medium. (1982)

5. Silver fluoride is fairly soluble in water. (1982)

6. Dilute HCl oxidizes metallic Fe to Fe²⁺. (1983)

7. In an aqueous solution chlorine is a stronger oxidizing agent than fluorine. (1984)

8. Silver chloride is more soluble in very concentrated sodium chloride solution than in pure water. (1984)

9. Solubility of sodium hydroxide increases with increase in temperature. (1985)

10. Sodium when burnt in excess of oxygen gives sodium oxide. (1987).

11. Both potassium ferrocyanide and potassium ferricyanide are diamagnetic. (1989)

12. Cu⁺ disproportionates to Cu²⁺. (1991)

13. Nitrous oxide, though an odd electron molecule, is diamagnetic in liquid state. (1991)

14. Diamond is harder than graphite. (1993)

15. The basic nature of the hydroxides of Group 13 (Gr. III B) decreases progressively down the group. (1993)

16. The tendency for catenation is much higher for C than for Si. (1993)

17. HBr is stronger acid than HI because of hydrogen bonding. (1993)

Answers some time later

Laws and Theories - Revision Part 2

MaxWell's Speed Distribution Law

It is an equation giving the distribution of molecules in different speeds in a gas at a temperature.

If dN represents the number of molecules with speeds between v and v+dv then

dN = 4πN[m/2πkT]^3/2v²e^-mv²/2kTdv

where
dN represents the number of molecules with speeds between v and v+dv
N = total number of molecules in the gas
m = mass of a molecule
T = absolute temperature of the gas
v = velocity of the molecules

The speed v_p at which dN/dv is maximum is called the most probable speed.
Its value is given by

v_p = √(2kT/m)

(Topic: Kinetic theory of gases)


Avogadro's Law

Under similar conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules.

(Basic concepts of chemistry)


Vectors

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.

Explanation

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.

(Topic: Vectors)

Revision - Laws and Theories - 1

Laws of Thermodynamics

Zeroth law: 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.



First law of thermodynamics

∆U = ∆Q -∆W
Where
∆U = change in internal energy of a thermodynamic system
∆Q = Heat given to the system
∆W = work done by the system

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.

Second law of thermodynamics

Kelvin-Planck statement

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.


(Topic: Laws of Thermodynamics)



1. Law of conservation of mass

During any physical or chemical change, the total mass of the products is equal to the total mass of reactants.

2. Law of constant proportions

A pure chemical compound always contains same elements combined together in the same definite proportion by weight.

3. Law of multiple proportions

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.

4. Law of reciprocal proportions

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.

5. Gay Lussac’s law of combing volumes

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.

Lagrange's Identity

(a × b)² = a²b² - (a.b)

a and b are vectors
a and b are magnitudes of a and b respectively.

Lagrange's identity is a relation between the cross product and the dot product.

(Topic: Vectors)

Electricty questions from physicsgoeasy blog

Electricty questions from physicsgoeasy blog


http://physicsgoeasy.blogspot.com/2008/05/iit-jee-test-serieselectricity-1.html
dated 18 May 2008


IIT JEE Test series (ELectricty) solutions
dated 18 June 2008
http://physicsgoeasy.blogspot.com/2008/06/iit-jee-test-series-electricty.html

Answers to 8 June 2008 Practice paper

Mathematics

1. if |z-1| /|z-4| = 2, the locus of z = x+iy in Argand plane will be:

a. parabola
b. hyperbola
c. circle
d. straight line

Answer (c)

z = x+iy
| (x+iy-1) |/| (x+iy-4) | = 2

(x-1) ² + y² = 4[(x-4) ² +y²]

x² -2x+1+y² = 4{x-8x+16+y²]
x² -2x+1+y² = 4x²-32x+64+4y²
3x²-30x+63+3y² = 0
x²+y²-10x+21 = 0
Equation represents a circle with (5,0) as centre and SQRT(5²-21) as radius.
Radius is 2






2.

Let a,b,c be three non-coplanar vectors and p,q,r are vectors defined by the relations

p = b×c/[abc]
q = c×a/[abc]
r = a×b/[abc]

then the value of the expression
[(a+b).p +(b+c).q +(c+a).r] is equal to

a. 3
b. 1
c. 0
d. 2



Answer: (a)

[(a+b).p +(b+c).q +(c+a).r] = [1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)]

The property of distributivity of scalar product over vector addition is
a.(b+c) = a.b+a.c and
(b+c).a = b.a+b.c

[1/(abc)][( [(a+b).(b×c)+ (b+c) .( c×a ) +(c+a).( a×b)] =

[1/(abc)][( a .(b×c)+ b.(b×c)+b .( c×a ) + b.(b×c)+c.( a×b) +a.( a×b)]

If a┴b, a.b = 0

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.

Hence the expression reduces to [1/(abc)][( a .(b×c)+ b .( c×a ) + c.( a×b)]

As a .(b×c) = b .( c×a ) = c.( a×b) = (abc)
The expression reduces to 3/(abc) /(abc) = 3



Physics
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

a. zero
b. 1/SQRT(2) m/s² towards north-west
c. ½ m/s² toward north-west
d. ½ m/s² towards north



Answer (b)

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

Hence it will act in the direction of northwest.

Westward component of average acceleration = (0-5)/10 = -½ m/s²
Northward component of average of acceleration = (5-0)/10 = ½ m/s²

Hence magnitude of average acceleration = Sqrt( (-1/2) ²+(1/2) ²) = 1/SQRT(2)



Chemistry

4. Which of the following has maximum number of atoms?

a. 24 g of C (M=12 g mol^-1
b. 23 g of Na (M = 23 g mol^-1)
c. 48 g of S (M = 32 b mol^-1)
d. 108 g of Ag (M = 108 mol ^-1)



Answer: (a)

Number of moles (N) = m/M
Where m = given mass
M = molar mass


Hence Number of moles of carbon = 2
Number of moles of Na = 1
Number of moles of S = 1.5
Number of moles of Ag = 1

So 24 g of C having 2 moles has the maximum number of atoms.



5. The wave number of the shortest wavelength transition in the Balmer series of atomic hydrogen is.

a. 27420 cm^-1
b. 28420 cm^-1
c. 29420 cm^-1
d. 12186 cm^-1


Answer: (a)

Wave number = 1/ λ = R(1/nf² - 1/ni²)
Wave length is shortest, when energy emitted is highest. This will happen when the electron jumps from n = ∞ to n=2.

So wave number = 109677 cm^-1(1/2² - 1/∞²)
= 109677(1/4)
= 27419.25 cm^-1

IIT JEE June 8 2008 Practice paper

Mathematics

1. if |z-1| /|z-4| = 2, the locus of z = x+iy in Argand plane will be:

a. parabola
b. hyperbola
c. circle
d. straight line





2.

Let a,b,c be three non-coplanar vectors and p,q,r are vectors defined by the relations

p = b×c/[abc]
q = c×a/[abc]
r = a×b/[abc]

then the value of the expression
[(a+b).p +(b+c).q +(c+a).r] is equal to

a. 3
b. 1
c. 0
d. 2


Physics
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

a. zero
b. 1/SQRT(2) m/s² towards north-west
c. ½ m/s² toward north-west
d. ½ m/s² towards north

Chemistry

4. Which of the following has maximum number of atoms?

a. 24 g of C (M=12 g mol^-1
b. 23 g of Na (M = 23 g mol^-1)
c. 48 g of S (M = 32 b mol^-1)
d. 108 g of Ag (M = 108 mol ^-1)


5. The wave number of the shortest wavelength transition in the Balmer series of atomic hydrogen is.

a. 27420 cm^-1
b. 28420 cm^-1
c. 29420 cm^-1
d. 12186 cm^-1

Solutions to June 1 2008 practice set

1. State whether the statement is true or false

A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.


Answer: true

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.



2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is

a. three
b. two
c. one
d. infinite
e. none of these

Answer (b)

If vector (x,y,z) is a unit vector perpendicular to (1,1,0) their scalar product is zero
=> x +y = 0
=> y = -x


If vector (x,y,z) is a unit vector perpendicular to (0,1,1) their scalar product is zero
=> y+z = 0
=> y = -z

As it is a unit vector x²+y²+z² = 1
=> 3y² = 1
=> y = ±1/SQRT(3)

This gives two value of y and hence there are two vectors







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

a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these


Answer: (b) and (d)

Assume z1 = a+ib, a is positive means a>0,
z2 = c+id, d<0

|z1| = |z2|
=> a²+b² = c²+d²

(z1+z2)/(z1-z2) = [(a+c) + i(b+d)]/ [(a-c) + i(b-d)]

Taking the multiplicative inverse of denominator and multiplying the numerator

= [(a+c) + i(b+d)]* [(a-c) - i(b-d)]/[(a-c) ² + (b-d) ²]


= [(a²-c²)+(b²-d²)]+i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]

=[{(a²+b²)-(c²+d²)}+i{(b+d)(a-c)-(a+c)(b-d)}]/ [(a-c) ² + (b-d) ²]

As a²+b² = c²+d² the first term of the numerator is zero.

Hence
(z1+z2)/(z1-z2)
= i[(b+d)(a-c)-(a+c)(b-d)]/ [(a-c) ² + (b-d) ²]

So this is a pure imaginary number.
If b+d as well as a+c are equal zero, the expression can be zero also.





4. If two compounds have the same empirical formula but different molecular formulae they must have ------

a. different percentage composition
b. different molecular weights
c. same viscosity
d. same vapour density

answer (b)


Different molecular formulae but same empirical formula imply same empirical mass but different molecular masses

IIT JEE June 1 st Weekly Cumulative Practice Test

1. State whether the statement is true or false

A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.


2. The number of vectors of unit length perpendicular to vectors a = (1,1,0) and b = (0,1,1) is

a. three
b. two
c. one
d. infinite
e. none of these






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

a. real and positive
b. zero
c. real and negative
d. purely imaginary
e. none of these



4. If two compounds have the same empirical formula but different molecular formulae they must have ------

a. different percentage composition
b. different molecular weight
c. same viscosity
d. same vapour density

Solutions for May 25 Practice set

covers topics
Complex Numbers
Vectors (common to physics and mathematics)
Mole concept

1. The inequality |z-4|<|z-2| represents the region given by

a. Re(z)>0
b. Re(z)<2
c. Re(z)<0
d. none of these

Answer (d)

Answer

Let z = x+iy then

|z-4|<|z-2|

=> |z-4²|<|z-2|²
=> (x-4) ²+y²<(x-2) ²+y²
=> -8x+16<-4x+4
=>12<4x
=> x>3
Means Re(z)>3





2. The points with the position vectors 60i +3j, 40i-8j, ai-52j are collinear if

a. a = -40
b. a = 40
c. a = 20
d. none of these



If the vectors x and y are collinear,
x = py
If we take given position vectors m.n,o, it implies
o-m = p(n-m)

[(ai-52j)-( 60i +3j)] = k[(40i-8j)-( 60i +3j)]

=> (a-60)i – 55j = k(-20i-11j)
=> a-60 = -20k and -55 = -11k
=> k = 55/11
=> a = -20*5 + 60
=> a = -40


3. One molal solution is one that contains one mole of a solute in :

a. 1000 g of the solvent
b. one litre of the solvent
c. one litre of the solution
d. 22.4 litres of the solution

Answer (a)

As per the definition, molality is amount of solute per kg of solvent.

May 25 2008 Question set 1

covers topics
Complex Numbers
Vectors (common to physics and mathematics)
Mole concept

1. The inequality |z-4|<|z-2| represents the region given by

a. Re(z)>0
b. Re(z)<2
c. Re(z)<0
d. none of these


2. The points with the position vectors 60i +3j, 40i-8j, ai-52j are collinear if

a. a = -40
b. a = 40
c. a = 20
d. none of these

3. One molal solution is one that contains one mole of a solute in :

a. 1000 g of the solvent
b. one litre of the solvent
c. one litre of the solution
d. 22.4 litres of the solution


For solution see
http://iit-jee-nrao.blogspot.com/2008/05/solutions-for-may-25-practice-set.html

IIT JEE Model Paper Eenadu April 2008

Solutions to model paper

http://www.eenadu.net/iitjeepaper1.pdf

http://www.eenadu.net/iitjeepaper2.pdf

New Book for Physical Chemistry IIT JEE

TMH published a new book

Physical Chemistry : For IIT JEE & other Engineering Entrance Examinations

--------------------------------------------------------------------------------

PREM DHAWAN, Delhi Public School, R K Puram, Delhi



ISBN: 0070655461
Copyright year: 2008
Table of Contents




1. Stoichiometry - I
2. Gaseous State
3. Atomic Structure
4. Oeriodic Properties
5. Chemical Bonding
6. Chemical Thermodynamics
7. Chemical Kinetics
8. Chemical Equilibrium
9. Ionic Equilibrium
10. Stoichiometry - II: Redox Reaction and Oxidation Numbers
11. Solutions
12. Electrochemistry
13. The Solid State
14. Surface Chemistry


Book Preface


--------------------------------------------------------------------------------
The objective of this book is to make the study of physical chemistry systematic and enjoyable. Apart from a thorough grounding in the fundamental concepts, emphasis is also on enhancing your ability to solve problems and to use relationships between concepts and information that would strengthen your mastery of the subject.

The book is tailored to meet the requirements of those appearing in IIT-JEE entrance and similar engineering entrance examinations and also classes XI & XII students.

As you begin reading this book, you will soon see that there are many examples given and each topic ends with a practice test along with their solutions. A large number of exercises are given at the end of each chapter. Written in a simple and systematic manner, the books' highlights are as follows:

A large number of examples to supplement the text
Practice test at the end of each topic
Large number of chapter-end exercises divided into
(i) Subjective Problems (ii) Multiple choice questions with one answer correct (iii) Multiple choice questions with multiple correct answers (iv) Fill in the blanks, True/False questions, Match the following (v) A special section on comprehension-based questions.
With such an exhaustive coverage of the topic, I am sure that students would find the book an ideal companion to understand and master the subject.

You can download solution manual of the book from

http://highered.mcgraw-hill.com/sites/0070655461/student_view0/solution_manual.html

For study guides on various chapters of JEE Chemistry visit

www.iit-jee-chemistry.blogspot.com

My IIT JEE Blogs

www.iit-jee-chemistry.blogspot.com

www.iit-jee-physics.blogspot.com

www.iit-jee-maths.blogspot.com

TIPS for JEE

The IIT/JEE : there is a difference between your preparations for CBSE exams and IIT/JEE. Here what matter most is not the quantum of books one has grasped or gone through. But the no. of correct answers, he gets. Students must change their mind sets while appearing for IIT/JEE.Here you have to clear the section cut off of the subject.Trends in the last 2 years show that the section cut off for Physic, Chemistry and Maths stands at 30, 35 and 25 percent. If a student gets this percentage of the questions correct, they have all chances to get a seat.

Never give priority to any section during preparation. While Maths, physic and Chemistry stnd in the hierarchy of difficulty.Students must not ignore chemistry. It may seem boring nd but it is a rewarding section.

Source:

http://fromprincipaloffice.blogspot.com/2008/01/tracking-cracking-engineering-entexams.html