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Logs without tables
Easiest Way to Find LOG OF ANY NUMBER WITHOUT USING TABLES
This method has been found by a Mr. Satish sharma. This way is applicable only to the
common logarithms (i.e. to base 10) First of all you will have to remember only 4 values.
Log 2 = 0.3010 Log 3 = 0.4771 Log 5 = 0.6989 Log 7 = 0.8450
Let me take an example to illustrate the rest –
Log 945 = Log(9.45 * 10^2)
= Log(9.45) + Log10^2 (since LogMN = LogM + LogN) = Log(9 + 0.45) + 2Log10 = Log(9 + 0.45) + 2 (since Log10 = 1)
Now the formula to be used is this Log(x + y) = Logx + y(with the decimal)/ 2.42x
In our example Log(9 + 0.45) = Log9 + 0.45 / 2.42(9)
= 2Log3 + 0.45 / 2.42(9)
Log945 = 2(0.4771) + 0.45 / 2.42(9) + 2 (since we have remembered the value
of Log3) now if we proceed we get 2.9748612 which is approximately equal to 2.9749
which is equal to the value of Log945.
Now I will take a bigger number – 999374309911034587984641915580 which has 30
digits. Log(9.9937…………………….. * 10^29)
= Log10^29 + Log9 + 0/9937………………/9(2.42)
= 29 + 2(0.4771) + 0.0413
= 29 + 0.9542 + 0.0413
= 29.9955
Hence you have learnt the easiest and quickest way to find the log of any number to the
base 10 without using the log table.
Trigonometry : Truly difficult Practice Questions. . .
:-)
Q) If and are positive integers such that
,
compute the ordered pair .
Q) Find y :
Q) Find the value of
Q) Prove that
Q) Prove that if , then .
Q) Calculate
Q) evaluate :
Q) evaluate :
Q) Solve the equation
Q) Evaluate:
Q) solve for x where .
Q) Find the value of:
Q)
Prove that :
Q) find
Q)
Prove that .
here Q stands for the set of rational numbers.
Important questions on Permutation and combination by Anil Verma,Director GIITJEE Chandigarh ( SCO 382,Sec 37 D,Above HDFC Bank,Chandigarh Tel : 0172-4652111 )
1. Find the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time.
2. If all the letters of the word “GIITJEE” are written in all possible orders and these words are written out as in a
dictionary, then find the rank of the following word “GIITJEE”3. A man has 8 children to taken them to a zoo. He takes three of them at a time to the zoo as often as he can
without taking the same 3 children together more than once. How many times will he have to go to zoo ? How
many times a particular child will go ?4. How many five digits numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, 7 and 8 if each digit is
to be used at most once.5. A shop sells 6 different flavours of ice–cream. In how many ways can a customer choose 4 ice–cream cones
if(i) they are all of different flavours
(ii) they are non necessarily of different flavours
(iii) they contain only 3 different flavours
(iv) they contain only 2 or 3 different flavours ?
6. In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the
front and 4 at the back ? How many seating arrangements are possible if 3 girls sit together in a back row on
adjacent seats ?7. A regular polygon of 10 sides is constructed. In how many ways can 3 vertices be selected so that no two
vertices are consecutive ?8. Find the number of ways in which four S come consecutively in the word MISSISSIPPI.
9. There are ten points in a plane. Of these ten points four points are in a straight line and with the exception of
these four points, no other three points are in the same straight line. Find
(i) the number of triangles formed
(ii) the number of straight lines formed
(iii) the number of quadrilaterals formed, by joining these ten points
10. In a plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B.
Besides, no three lines pass through one point, no line passes through both points A and B, and no two are
parallel. Find the number of points of intersection of the straight lines.
(iv) There are r distinct objects and n distinct boxes such that n > r. Find the number of ways of
(A) putting r objects in n boxes if each box is sufficient to contain r objects.
(B) arranging r objects in n boxes so that each box contains only one object.
(C) arranging r objects in n boxes if, a box may contain any number of objects.
11. In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the
front and 4 at the back ? How many seating arrangements are possible if 3 girls sit together in a back row on
adjacent seats ?
12. A boat is to be manned by 9 crew with 4 on the stroke side, 4 on the row side and one to steer. There are 11
crew of which 2 can stroke only, 1 can row only while 3 can steer only. In how many ways the crew can be
arranged for the boat ?
A great trick for trigonometry
a handy way that will help you to find the value of a lot of compound angles in trigonometry which involve the multiples of 90. This works always.See it is very difficult to remember all this sin (90 + x), cot (270 - x), etc.To make it easy just follow the following steps. First of all see the quadrant in which the given angle lies.For example if we have to find sin (270 + x) we know that the terminal side of the angle will be in the 4thquadrant. Hence we put a negative sign in front of its value (sine is negative in the 4th quadrant). Now we see the line of reference. If the line of reference is vertical we use the complementary of the trigonometric ratio given, if horizontal it remains the same. Here it is 270 degrees so this line is vertical, and hence the value will be the cosine of x with a negative sign. The complementary of sine is cosine, of tangent is cotangent, of sec is cosec and vice -versa for each case (notice here the prefix co). Let me illustrate this for you. Suppose we have to find the value of cosec (180 + x). We can see that the terminal side lies in the 3rd quadrant where cosec is negative. So we the value will be negative. Also the line of reference is horizontal so we use the angle as it is. The value comes out to –cosec x. Just remember thisHorizontal - same angle.Vertical – the complementary of the given angle. This is actually a good method. It can be done mentally too. It saves time, since we don’t have to remember all the values.
I find this similar to this:sin/cos/tan/cosec/sec/cot (n90 +@) =same ratio if n is even, complimentary if n is odd and the sign is decided based on the quadrant
INTEGRALS CONTAINING ln(ax)
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INTEGRALS CONTAINING eax
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Integrals with Inverse Trigonometric Functions
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where the constant is the eulers constant.
18.
where the constant is the EULERs CONSTANT.
19.
where the constant is the EULERs CONSTANT.
20.
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Predicting the products::SUBSTITUTION VERSUS ELIMINATION.Solve Chemistry Short Answer Type Questions
Solve Chemistry Short Answer Type Questions
H
1.) Largest atomic size - cesium
2.) lowest atomic size - H
3.) Largest Anion - At -
4.) smallest anion - F -
5.) Lowest electronegativity - Cs
6.) Lowest electron affinity - noble gases
7.) Metals having highest b.pt and m.pt. = mercury and tungsten
8.) Non metal having highest m.pt and b.pt = Helium
9.) Most reactive solid ekement - Li
10.) Most reactive liquid element - Cs
11.) Most reactive gaseous element - Flourine
12.) Most stable element - Tellurium (half life = 2 * 1021 )
13.) Electro-ive element next to flourine - Oxygen
14.) Grp containing max. no. of gases - 18th
15.) total no. of gaseous element in periodic table - 11
(H , He , N , O , F , Ne , Cl , Ar , Kr , Xe , Ra)
16.) TOTAL NO. OF LIQUID elements - 6
(Ga , Br , Cs , Hg , Fr , EKa)
17.) total no. of solid elements = 89 (can't name them here heheee)
18.) Liquid radioactive element - Francium
19.) N.metal wid highest M.Pt - Carbon
20.) Metal wid highest valency = Plutonium
21.) highest tensile strength - Boron
22.) Most ionic compound = CsF
23.) Strongest base = Cs (OH)
24.) Strongest basic oxide = Cs2 O
25.) Most conducting metal = Ag
26.) insulating = Pb
27.) Elements wich are amphoteric in nature = Al , Zn , An , Pb
28.) N.METALS having ,etallic lusture = Iodine and Graphite
29.) Naturally occuring heaviest element = uranium
30.) most conducting N.metal - C
A new trick for Bond Angle
This method is applicable only for element N O P (mean Nitrogen Oxygen and Phosphorus )
Trick is :-
If central Atom is same ....then Higher the Electronegativity of Surrounding atom...lower the
bond Angle ...and vice-versa
If Surrounding Atom is same...then Lower the Electronegativity of Central atom Lower the
bond angle and vice-versa
Example ;
NF3 NCl3 NBr3 NI3
Central atom is same in all the four case....i.e N...so higher the E.N of Surronding atom..lower the bond
angle...
So EN : F > Cl > Br > I
Hence Bond Angle trend is NF3 < NCl3 < NBr3 < NI3
PH3 PF3 PCl3 PBr3 PI3
Central Atom is same.....again trend of EN of surrounding atom...
F > Cl > Br > I > H ...
So bond angle will be PH3 > PI3 > PBr3 > PCl3 > PF3
NH3 PH3
Here surrounding atom is same....but central is different ...
So EN of N and P ... N > P
So Bond angle of NH3 > PH3 .....( because higher the EN of central atom higher the Bond Angle )
H2O H2F
Again EN of F > O so H2F > H2O ( Bond angle ) ...this question is asked in JEE ( I don't
remember the yr)
Solve for this using the above methods ....
PF3 NF3
OF2 OCl2 OBr2 OH2 NCl3 PCl3
Calculate total no of isomers of a compound
DOUBLE BOND EQUIVALENTS or INDEX NUMBER:
It is a numerical value on the basis of which we can have an idea about the different type of equivalent
structures we can have for a given formula.
Knowing the DBE the problems on structural isomerism can be more easily solved.
Mathematically,
DBE= {( n(v-2))/2}+1
where n is the number of atoms
v is the valency of the atom.
Eg 1):
Consider C4H6
DBE={4(4-2)/2+6(1-2)/2}+1
={(8-6)/2}+1
=2
This implies that the molecule may contain,
1)two double bonds or one triple bond.
2)one ring and one double bond.
3)two rings.
Thus by knowing the DBE total number of structural isomers can be formulated.
Eg 2):
Consider C3H6O
DBE={3(4-2)/2+6(1-2)/2+1(2-2)/2}+1
=1
So the isomers may contain either a double bond or a ring.
No. of Geometrical Isomers
1. If a compound contains only 1 d.b. then no. of geometical isomers are 2
2. if a compound contains 2 or more than 2 d.b. and end groups are different , then no. of geometrical
isomers will be where n is no. of d.b.
3. If a compound contains 2 or more than 2 d.b. and end groups are same , then no. of geometrical
isomers will be where
p =n/2 if d.b. are even and p=(n+1)/2 if d.b. are odd
these formulas have limitations ...as in if there are more than 4 chiral centres these will not work..even for 2 chiral centres it will not work becuase you will have to face the case of psuedo chiral
Identifying that the compound are identical or enantimers to each other
1. If a compound is rotated in a odd multiple of 90° or any odd interchanges between groups then the
compounds so formed are enantiomer of the original compound
2. If a compound is rotated in a even multiple of 90° or any even interchanges between groups then the
compounds so formed are identical to each other
Formulae to find stereomers of the compound
1. If a compound is non symmetrical
no. of d/l pair - , no of mesomers - 0 , no. of total stereomers = no. of d/l pair + no. of mesomers
2. If a compound is symmetrical and containing even no. of chirale carbons then
no. of d/l pair - , no. of mesomers = , no. of total stereomers = no. of d/l pair + no. of
mesomers
3. If a compound is symmetrical and containing odd no. of chirale carbons then
no. of d/l pair - , no. of mesomers = , no. of total stereomers = no. of d/l pair +
no. of mesomers
Formula to find enantionmers of non symmentrical compound
If a compound is a symmetrical or non symmetrical and doesn't contain plane of symmetry , centre of
symmetry or alternating axis of symmetry and contain n chirale center then the total enantiomers of the
compound will be
Types of Inorganic Chemical ReactionsFour General Categories
Elements and compounds react with each other in numerous ways. Memorizing every type of reaction would be challenging and also unnecessary, since nearly every inorganic chemical reaction falls into one or more of four broad categories.
1. Combination ReactionsTwo or more reactants form one product in a combination reaction. An example of a combination reaction is the formation of sulfur dioxide when sulfur is burned in air:
S (s) + O2 (g) → SO2 (g)
2. Decomposition Reactions
In a decomposition reaction, a compound breaks down into two or more substances. Decomposition usually results from electrolysis or heating. An example of a decomposition reaction is the breakdown of mercury (II) oxide into its component elements.
2HgO (s) + heat → 2Hg (l) + O2 (g)
3. Single Displacement Reactions
A single displacement reaction is characterized by an atom or ion of a single compound replacing an atom of another element. An example of a single displacement reaction is the displacement of copper ions in a copper sulfate solution by zinc metal, forming zinc sulfate:
Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)
Single displacement reactions are often subdivided into more specific categories (e.g.,redox reactions).
4. Double Displacement Reactions
Double displacement reactions also may be called metathesis reactions. In this type of reaction, elements from two compounds displace each other to form new compounds. Double displacement reactions may occur when one product is removed from the solution as a gas or precipitate or when two species combine to form a weak electrolyte that remains undissociated in solution. An example of a double displacement reaction occurs when solutions of calcium chloride and silver nitrate are reacted to form insoluble silver chloride in a solution of calcium nitrate.
CaCl2 (aq) + 2 AgNO3 (aq) → Ca(NO3)2 (aq) + 2 AgCl (s)
A neutralization reaction is a specific type of double displacement reaction that occurs when an acid reacts with a base, producing a solution of salt and water. An example of a neutralization reaction is the reaction of hydrochloric acid and sodium hydroxide to form sodium chloride and water:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
Remember that reactions can be belong to more than one category. Also, it would be possible to present more specific categories, such as combustion reactions or precipitation reactions. Learning the main categories will help you balance equations and predict the types of compounds formed from a chemical reaction.