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Chapter-1 Introduction to Fundamental Concepts,Chapter-2 Three States Of Matter,Chapter-3 Atomic Structure,Chapter-5 Chemical Bond,Chapter-5 Energetics Of Chemical Reaction,Chapter-6 Chemical Equilibrium,Chapter-8 Chemical Kinetics

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  • Chapter-1 Introduction to Fundamental Concepts...
  • Chapter-2 Three States Of Matter
  • Chapter-3 Atomic Structure
  • Chapter-5 Chemical Bond
  • Chapter-5 Energetics Of Chemical Reaction
  • Chapter-6 Chemical Equilibrium
  • Chapter-8 Chemical Kinetics

Chapter-1 Introduction to Fundamental Concepts of Chemistry


Atom

It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.



Molecule

When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.



Formula Weight

It is the sum of the weights of the atoms present in the formula of a substance.



Molecular Weight

It is the sum of the atomic masses of all the atoms present in a molecule.



Chemistry

It is a branch of science which deals with the properties, composition and the structure of matter.



Empirical Formula

Definition
It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.

Examples
The empirical formula of benzene is "CH". It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.
The empirical formula of glucose is "CH2O". This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.



Determination of Empirical Formula

To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.



Illustrated Example of Empirical Formula

Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I - Determination of the Elements
By performing test it is found that the compound contains magnesium and oxygen elements.

Step II - Determination of the Masses
Masses of the elements are experimentally determined which are given below.
Mass of Mg = 2.4 gm
Mass of Oxygen = 1.6 gm

Step III - Estimation of the Percentage
The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% Mg = Mass of Mg / Mass of Compound x 100
= 2.4 / 4.0 x 100
= 60%
% O = Mass of Oxygen / Mass of Compound x 100
= 1.6 / 4.0 x 100
= 40%

Step IV - Determination of Mole Composition
Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
= 60 / 24
= 2.5
Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen
= 40 / 16
= 2.5

Step V - Determination of Simplest Ratio
To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Mg = 2.5 / 2.5 = 1
O = 2.5 / 2.5 = 1
Thus the empirical formula of the compound is MgO

Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.



Molecular Formula

Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.
OR
It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.

Examples
The molecular formula of benzene is "C6H6". It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.
The molecular formula of glucose is "C6H12O6". The formula represents that
1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

Determination of Molecular Formula

The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula)n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
n = Molecular Mass / Empirical Formula Mass

Example
The empirical formula of a compound is CH2O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass
Empirical formula mass of CH2O = 12 + 1 x 2 + 16
= 30
n = Molecular Mass / Empirical Formula Mass
= 180 / 30
= 6
Molecular formula = (Empirical formula)n
= (CH2O)6
= C6H12O6



Molecular Mass

Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.
OR
It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.

Example
The molecular mass of CO2 may be calculated as
Molecular mass of CO2 = Mass of Carbon + 2 (Mass of Oxygen)
= 12 + 2 x 16
= 44 a.m.u
Molecular mass of H2O = (Mass of Hydrogen) x 2 + Mass of Oxygen
= 1 x 2 + 16
= 18 a.m.u
Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
= 1 + 35.5
= 36.5 a.m.u



Gram Molecular Mass

Definition
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.

Examples
1. The molecular mass of H2O is 18. If we take 18 gm H2O then it is called 1 gm molecular mass of H2O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.



Mole

Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.
OR
The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.

Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.
From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO2 is 44. If we take 44 gm of CO2 it is called one mole of CO2 or the molecular mass of H2O is 18. If we take 18 gm of H2O it is called one mole of H2O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contain 6.02 x 10(23) atoms
While
One gram molecule contain 6.02 x 10(23) molecules.



Avagadro's Number

An Italian scientist, Avagadro's calculated that the number of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10(23). This number is known as Avogadro's number and represented as N(A).

Example
1 gm mole of Na contain 6.02 x 10(23) atoms of Na.
1 gm mole of Sulphur = 6.02 x 10(23) atoms of Sulphur.
1 gm mole of H2SO4 = 6.02 x 10(23) molecules H2SO4
1 gm mole of H2O = 6.02 x 10(23) molecules of H2O
On the basis of Avogadro's Number "mole" is also defined as
Mass of 6.02 x 10(23) molecules, atoms or ions in gram is called mole.

Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance

Example 1
Calculate the number of atoms in 9.2 gm of Na.

Solution
Atomic mass of Na = 23 a.m.u
If we take 23 gm of Na, it is equal to 1 mole.
23 gm of Na contain 6.02 x 10(23) atoms
1 gm of Na contain 6.02 x 10(23) / 23 atoms
9.2 gm of Na contain 9.2 x 6.02 x 10(23) /23
= 2.408 x 10(23) atoms of Na

Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance

Example 2
Calculate the mass in grams of 3.01 x 10(23) molecules of glucose.

Solution
Molecular mass of glucose = 180 a.m.u
So when we take 180 gm of glucose it is equal to one mole So,
6.02 x 10(23) molecules of glucose = 180 gm
1 molecule of glucose = 180 / 6.02 x 10(23) gm
3.01 x 10(23) molecules of glucose = 3.01 x 10(23) x 180 / 6.02 x 10(23)
= 90 gm



Stoichiometry

(Calculation Based On Chemical Equations)

Definition
The study of relationship between the amount of reactant and the products in chemical reactions as given by chemical equations is called stoichiometry.
In this study we always use a balanced chemical equation because a balanced chemical equation tells us the exact mass ratio of the reactants and products in the chemical reaction.
There are three relationships involved for the stoichiometric calculations from the balanced chemical equations which are
1. Mass - Mass Relationship
2. Mass - Volume Relationship
3. Volume - Volume Relationship

Mass - Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of teh substance involved in the chemical reaction by using a balanced chemical equation.

Example
Calculate the mass of CO2 that can be obtained by heating 50 gm of limestone.

Solution
Step I - Write a Balanced Equation
CaCO3 ----> CaO + CO2

Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CaCO3 ----> CaO + CO2

Method I - MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO3 gives 1 mole of CO2
0.5 mole of CaCO3 will give 0.5 mole of CO2
Mass of CO2 = Moles x Molecular Mass
= 0.5 x 44
= 22 gm

Method II - FACTOR METHOD
From equation we may write as
100 gm of CaCO3 gives 44 gm of CO2
1 gm of CaCO3 will give 44/100 gm of CO2
50 gm of CaCO3 will give 50 x 44 / 100 gm of CO2
= 22 gm of CO2

Mass - Volume Relationship
The major quantities of gases can be expressed in terms of volume as well as masses. According to Avogardro One gm mole of any gas always occupies 22.4 dm3 volume at S.T.P. So this law is applied in mass-volume relationship.
This relationship is useful in determining the unknown mass or volume of reactant or product by using a given mass or volume of some substance in a chemical reaction.

Example
Calculate the volume of CO2 gas produced at S.T.P by combustion of 20 gm of CH4.

Solution
Step I - Write a Balanced Equation
CH4 + 2 O2 ----> CO2 + 2 H2O

Step II - Write Down The Molecular Masses And Moles Of Reactant & Product
CH4 + 2 O2 ----> CO2 + 2 H2O

Method I - MOLE METHOD
Convert the given mass of CH4 in moles
Number of moles of CH4 = Given Mass of CH4 / Molar Mass of CH4
From Equation
1 mole of CH4 gives 1 moles of CO2
1.25 mole of CH4 will give 1.25 mole of CO2
No. of moles of CO2 obtained = 1.25
But 1 mole of CO2 at S.T.P occupies 22.4 dm3
1.25 mole of CO2 at S.T.P occupies 22.4 x 1.25
= 28 dm3

Method II - FACTOR METHOD
Molecular mass of CH4 = 16
Molecular mass of CO2 = 44
According to the equation
16 gm of CH4 gives 44 gm of CO2
1 gm of CH4 will give 44/16 gm of CO2
20 gm of CH4 will give 20 x 44/16 gm of CO2
= 55 gm of CO2
44 gm of CO2 at S.T.P occupy a volume 22.4 dm3
1 gm of CO2 at S.T.P occupy a volume 22.4/44 dm3
55 gm of CO2 at S.T.P occupy a volume 55 x 22.4/44
= 28 dm3

Volume - Volume Relationship
This relationship determine the unknown volumes of reactants or products from a known volume of other gas.
This relationship is based on Gay-Lussac's law of combining volume which states that gases react in the ratio of small whole number by volume under similar conditions of temperature & pressure.
Consider this equation
CH4 + 2 O2 ----> CO2 + 2 H2O
In this reaction one volume of CH4 gas reacts with two volumes of oxygen gas to give one volume of CO2 and two volumes of H2O

Examples
What volume of O2 at S.T.P is required to burn 500 litres (dm3) of C2H4 (ethylene)?

Solution
Step I - Write a Balanced Equation
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

Step II - Write Down The Moles And Volume Of Reactant & Product
C2H4 + 3 O2 ----> 2 CO2 + 2 H2O

According to Equation
1 dm3 of C2H4 requires 3 dm3 of O2
500 dm3 of C2H4 requires 3 x 500 dm3 of O2
= 1500 dm3 of O2

Limiting Reactant

In stoichiometry when more than one reactant is involved in a chemical reaction, it is not so simple to get actual result of the stoichiometric problem by making relationship between any one of the reactant and product, which are involved in the chemical reaction. As we know that when any one of the reactant is completely used or consumed the reaction is stopped no matter the other reactants are present in very large quantity. This reactant which is totally consumed during the chemical reaction due to which the reaction is stopped is called limiting reactant.
Limiting reactant help us in calculating the actual amount of product formed during the chemical reaction. To understand the concept the limiting reactant consider the following calculation.

Problem
We are provided 50 gm of H2 and 50 gm of N2. Calculate how many gm of NH3 will be formed when the reaction is irreversible.
The equation for the reaction is as follows.
N2 + 3 H2 ----> 2 NH3

Solution
In this problem moles of N2 and H2 are as follows
Moles of N2 = Mass of N2 / Mol. Mass of N2
= 50 / 28
= 1.79
Moles of H2 = Mass of H2 / Mol. Mass of H2
= 50 / 2
= 25
So, the provided moles for the reaction are
nitrogen = 1.79 moles and hydrogen = 25 moles
But in the equation of the process 1 mole of nitrogen require 3 mole of hydrogen. Therefore the provided moles of nitrogen i.e. 1.79 require 1.79 x 3 moles of hydrogen i.e. 5.37 moles although 25 moles of H2 are provided but when nitrogen is consumed the reaction will be stopped and the remaining hydrogen is useless for the reaction so in this problem N2 is a limiting reactant by which we can calculate the actual amount of product formed during the reaction.
N2 + 3 H2 ----> 2 NH3

1 mole of N2 gives 2 moles of NH3
1.79 mole of N2 gives 2 x 1.79 moles of NH3
= 3.58 moles of NH3

Mass of NH3 = Moles of NH3 x Mol. Mass
= 3.58 x 17
= 60.86 gm of NH3

Chapter-2 Three States Of Matter


Matter

It is defined as any thing which has mass and occupies space is called matter.
Matter is composed of small and tiny particles called Atoms or molecules. It exist in three different states which are gaseous, liquid & solid.



Properties of Gas

1. It has no definite shape.
2. It has no definite volume, so it can be compressed or expanded.
3. A gas may diffuse with the other gas.
4. The molecules of a gas are in continuous motion.



Properties of Liquids

1. A liquid has no defini...

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Chapter-3 Atomic Structure

April 13, 2010
Introduction
About the structure of atom a theory was put on by John Dalton in 1808. According to this theory matter was made from small indivisible particles called atoms.
But after several experiments many particles have been discovered with in the atom which are electrons, protons, neutrons, positrons etc. For the discovery of these fundamental particles the experiments are as follows.
1. Faraday's experiment indicates the existence of electron.
2. Crook's tube experiment explains the...

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Chapter-5 Chemical Bond


Introduction
Atoms of all the elements except noble gases have incomplete outermost orbits and tends to complete them by chemical combination with the other atoms.
In 1916, W Kossel described the ionic bond which is formed by the transfer of electron from one atom to another and also in 1916 G.N Lewis described about the formation of covalent bond which is formed by the mutual sharing of electrons between two atoms.
Both these scientists based their ideas on the fact that atoms greatest...

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Chapter-5 Energetics Of Chemical Reaction


Thermodynamics

Definition
It is branch of chemistry which deals with the heat energy change during a chemical reaction.

Types of Thermochemical Reactions
Thermo-chemical reactions are of two types.
1. Exothermic Reactions
2. Endothermic Reactions

1. Exothermic Reaction
A chemical reaction in which heat energy is evolved with the formation of product is known as Exothermic Reaction.
An exothermic process is generally represented as
Reactants ----> Products + Heat

2. Endothermic Reaction ...

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Chapter-6 Chemical Equilibrium


Reversible Reactions

Those chemical reactions which take place in both the directions and never proceed to completion are called Reversible reaction.
For these type of reaction both the forward and reverse reaction occur at the same time so these reaction are generally represented as
Reactant □ Product
The double arrow □ indicates that the reaction is reversible and that both the forward and reverse reaction can occur simultaneously.
Some examples of reversible reactions are given bel...

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Chapter-8 Chemical Kinetics


Introduction
The branch of physical chemistry which deals with the speed or rate at which a reaction occurs is called chemical kinetics.
The study of chemical kinetics, therefore includes the rate of a chemical reaction and also the rate of chemical reaction and also the factors which influence its rate.



Slow and Fast Reaction

Those reactions for which short time is required to convert a reactant into product are called fast reaction but if more time is required for the formation of ...

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Scalars & Vectors,Motion,Motion in two Dimension,Torque, Angular, Momentum & Equilibrium,Gravitation,Work, Power & Energy

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  • Scalars and Vectors
  • Motion
  • Motion in two Dimension
  • Torque, Angular, Momentum and Equilibrium
  • Gravitation
  • Work, Power and Energy

Scalars and Vectors


Scalars

Physical quantities which can be completely specified by
1. A number which represents the magnitude of the quantity.
2. An appropriate unit
are called Scalars.
Scalars quantities can be added, subtracted multiplied and divided by usual algebraic laws.

Examples
Mass, distance, volume, density, time, speed, temperature, energy, work, potential, entropy, charge etc.



Vectors

Physical quantities which can be completely specified by
1. A number which represents the magnitude of the quantity.
2. An specific direction
are called Vectors.
Special laws are employed for their mutual operation.

Examples
Displacement, force, velocity, acceleration, momentum.

Representation of a Vector
A straight line parallel to the direction of the given vector used to represent it. Length of the line on a certain scale specifies the magnitude of the vector. An arrow head is put at one end of the line to indicate the direction of the given vector.
The tail end O is regarded as initial point of vector R and the head P is regarded as the terminal point of the vector R.

Diagram Coming Soon

Unit Vector
A vector whose magnitude is unity (1) and directed along the direction of a given vector, is called the unit vector of the given vector.
A unit vector is usually denoted by a letter with a cap over it. For example if r is the given vector, then r will be the unit vector in the direction of r such that
r = r .r
Or
r = r / r
unit vector = vector / magnitude of the vector

Equal Vectors
Two vectors having same directions, magnitude and unit are called equal vectors.

Zero or Null Vector
A vector having zero magnitude and whose initial and terminal points are same is called a null vector. It is usually denoted by O. The difference of two equal vectors (same vector) is represented by a null vector.
R - R - O

Free Vector
A vector which can be displaced parallel to itself and applied at any point, is known as free vector. It can be specified by giving its magnitude and any two of the angles between the vector and the coordinate axes. In 3-D, it is determined by its three projections on x, y, z-axes.

Position Vector
A vector drawn from the origin to a distinct point in space is called position vector, since it determines the position of a point P relative to a fixed point O (origin). It is usually denoted by r. If xi, yi, zk be the x, y, z components of the position vector r, then
r = xi + yj + zk

Diagram Coming Soon

Negative of a Vector
The vector A. is called the negative of the vector A, if it has same magnitude but opposite direction as that of A. The angle between a vector and its negative vector is always of 180º.

Multiplication of a Vector by a Number
When a vector is multiplied by a positive number the magnitude of the vector is multiplied by that number. However, direction of the vector remain same. When a vector is multiplied by a negative number, the magnitude of the vector is multiplied by that number. However, direction of a vector becomes opposite. If a vector is multiplied by zero, the result will be a null vector.
The multiplication of a vector A by two number (m, n) is governed by the following rules.
1. m A = A m
2. m (n A) = (mn) A
3. (m + n) A = mA + nA
4. m(A + B) = mA + mB

Division of a Vector by a Number (Non-Zero)
If a vector A is divided by a number n, then it means it is multiplied by the reciprocal of that number i.e. 1/n. The new vector which is obtained by this division has a magnitude 1/n times of A. The direction will be same if n is positive and the direction will be opposite if n is negative.



Resolution of a Vector Into Rectangular Components

Definition
Splitting up a single vector into its rectangular components is called the Resolution of a vector.

Rectangular Components
Components of a vector making an angle of 90º with each other are called rectangular components.

Procedure
Let us consider a vector F represented by OA, making an angle O with the horizontal direction.
Draw perpendicular AB and AC from point on X and Y axes respectively. Vectors OB and OC represented by Fx and Fy are known as the rectangular components of F. From head to tail rule of vector addition.
OA = OB + BA
F = Fx + Fy

Diagram Coming Soon
To find the magnitude of Fx and Fy, consider the right angled triangle OBA.
Fx / F = Cos θ => Fx = F cos θ
Fy / F = sin θ => Fy = F sin θ


Addition of Vectors by Rectangular Components
Consider two vectors A1 and A2 making angles θ1 and θ2 with x-axis respectively as shown in figure. A1 and A2 are added by using head to tail rule to give the resultant vector A.

Diagram Coming Soon
The addition of two vectors A1 and A2 mentioned in the above figure, consists of following four steps.

Step 1
For the x-components of A, we add the x-components of A1 and A2 which are A1x and A2x. If the x-components of A is denoted by Ax then
Ax = A1x + A2x
Taking magnitudes only
Ax = A1x + A2x
Or
Ax = A1 cos θ1 + A2 cos θ2 ................. (1)

Step 2
For the y-components of A, we add the y-components of A1 and A2 which are A1y and A2y. If the y-components of A is denoted by Ay then
Ay = A1y + A2y
Taking magnitudes only
Ay = A1y + A2y
Or
Ay = A1 sin θ1 + A2 sin θ2 ................. (2)

Step 3
Substituting the value of Ax and Ay from equations (1) and (2) respectively in equation (3) below, we get the magnitude of the resultant A
A = |A| = √ (Ax)2 + (Ay)2 .................. (3)

Step 4
By applying the trigonometric ratio of tangent θ on triangle OAB, we can find the direction of the resultant vector A i.e. angle θ which A makes with the positive x-axis.
tan θ = Ay / Ax
θ = tan-1 [Ay / Ax]
Here four cases arise
(a) If Ax and Ay are both positive, then
θ = tan-1 |Ay / Ax|

(b) If Ax is negative and Ay is positive, then
θ = 180º - tan-1 |Ay / Ax|

(c) If Ax is positive and Ay is negative, then
θ = 360º - tan-1 |Ay / Ax|

(d) If Ax and Ay are both negative, then
θ = 180º + tan-1 |Ay / Ax|



Addition of Vectors by Law of Parallelogram

According to the law of parallelogram of addition of vectors, if we are given two vectors. A1 and A2 starting at a common point O, represented by OA and OB respectively in figure, then their resultant is represented by OC, where OC is the diagonal of the parallelogram having OA and OB as its adjacent sides.

Diagram Coming Soon
If R is the resultant of A1 and A2, then
R = A1 + A2
Or
OC = OA + OB
But OB = AC
Therefore,
OC = OA + AC
β is the angle opposites to the resultant.
Magnitude of the resultant can be determined by using the law of cosines.
R = |R| = √A1(2) + A2(2) - 2 A1 A2 cos β
Direction of R can be determined by using the Law of sines.
A1 / sin γ = A2 / sin α = R / sin β
This completely determines the resultant vector R.

Properties of Vector Addition

1. Commutative Law of Vector Addition (A+B = B+A)
Consider two vectors A and B as shown in figure. From figure
OA + AC = OC
Or
A + B = R .................... (1)
And
OB + BC = OC
Or
B + A = R ..................... (2)
Since A + B and B + A, both equal to R, therefore
A + B = B + A
Therefore, vector addition is commutative.

Diagram Coming Soon

2. Associative Law of Vector Addition (A + B) + C = A + (B + C)
Consider three vectors A, B and C as shown in figure. From figure using head - to - tail rule.
OQ + QS = OS
Or
(A + B) + C = R
And
OP + PS = OS
Or
A + (B + C) = R
Hence
(A + B) + C = A + (B + C)
Therefore, vector addition is associative.

Diagram Coming Soon

Product of Two Vectors
1. Scalar Product (Dot Product)
2. Vector Product (Cross Product)

1. Scalar Product OR Dot Product
If the product of two vectors is a scalar quantity, then the product itself is known as Scalar Product or Dot Product.
The dot product of two vectors A and B having angle θ between them may be defined as the product of magnitudes of A and B and the cosine of the angle θ.
A . B = |A| |B| cos θ
A . B = A B cos θ

Diagram Coming Soon
Because a dot (.) is used between the vectors to write their scalar product, therefore, it is also called dot product.
The scalar product of vector A and vector B is equal to the magnitude, A, of vector A times the projection of vector B onto the direction of A.
If B(A) is the projection of vector B onto the direction of A, then according to the definition of dot product.

Diagram Coming Soon
A . B = A B(A)
A . B = A B cos θ {since B(A) = B cos θ}
Examples of dot product are
W = F . d
P = F . V

Commutative Law for Dot Product (A.B = B.A)
If the order of two vectors are changed then it will not affect the dot product. This law is known as commutative law for dot product.
A . B = B . A
if A and B are two vectors having an angle θ between then, then their dot product A.B is the product of magnitude of A, A, and the projection of vector B onto the direction of vector i.e., B(A).
And B.A is the product of magnitude of B, B, and the projection of vector A onto the direction vector B i.e. A(B).

Diagram Coming Soon
To obtain the projection of a vector on the other, a perpendicular is dropped from the first vector on the second such that a right angled triangle is obtained
In Δ PQR,
cos θ = A(B) / A => A(B) = A cos θ
In Δ ABC,
cos θ = B(A) / B => B(A) = B cos θ
Therefore,
A . B = A B(A) = A B cos θ
B . A = B A (B) = B A cos θ
A B cos θ = B A cos θ
A . B = B . A
Thus scalar product is commutative.

Distributive Law for Dot Product
A . (B + C) = A . B + A . C
Consider three vectors A, B and C.
B(A) = Projection of B on A
C(A) = Projection of C on A
(B + C)A = Projection of (B + C) on A
Therefore
A . (B + C) = A [(B + C}A] {since A . B = A B(A)}
= A [B(A) + C(A)] {since (B + C)A = B(A) + C(A)}
= A B(A) + A C(A)
= A . B + A . C
Therefore,
B(A) = B cos θ => A B(A) = A B cos θ1 = A . B
And C(A) = C cos θ => A C(A) = A C cos θ2 = A . C
Thus dot product obeys distributive law.

Diagram Coming Soon

2. Vector Product OR Cross Product
When the product of two vectors is another vector perpendicular to the plane formed by the multiplying vectors, the product is then called vector or cross product.
The cross product of two vector A and B having angle θ between them may be defined as "the product of magnitude of A and B and the sine of the angle θ, such that the product vector has a direction perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B".
A x B = |A| |B| sin θ u
Where u is the unit vector perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B.
Examples of vector products are
(a) The moment M of a force about a point O is defined as
M = R x F
Where R is a vector joining the point O to the initial point of F.

(b) Force experienced F by an electric charge q which is moving with velocity V in a magnetic field B
F = q (V x B)

Physical Interpretation of Vector OR Cross Product
Area of Parallelogram = |A x B|
Area of Triangle = 1/2 |A x B|

Motion

Definition
If an object continuously changes its position with respect to its surrounding, then it is said to be in state of motion.

Rectilinear Motion
The motion along a straight line is called rectilinear motion.



Velocity

Velocity may be defined as the change of displacement of a body with respect time.
Velocity = change of displacement / time
Velocity is a vector quantity and its unit in S.I system is meter per second (m/sec).

Average Velocity
Average velocity of a body i...

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Motion in two Dimension


Projectile Motion

A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its trajectory.
Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.

Analysis of Projectile Motion
Let us consider a body of mass m, projected an angle θ with the horizonta...

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Torque, Angular, Momentum and Equilibrium


Torque or Moment of Force

Definition
If a body is capable of rotating about an axis, then force applied properly on this body will rotate it about the axis (axis of rotating). This turning effect of the force about the axis of rotation is called torque.
Torque is the physical quantity which produces angular acceleration in the body.

Explanation
Consider a body which can rotate about O (axis of rotation). A force F acts on point P whose position vector w.r.t O is r.



Diagram Coming So...

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Gravitation

The property of all objects in the universe which carry mass, by virtue of which they attract one another, is called Gravitation.



Centripetal Acceleration of the Moon

Newton, after determining the centripetal acceleration of the moon, formulated the law of universal gravitation.
Suppose that the moon is orbiting around the earth in a circular orbit.
If V = velocity of the moon in its orbit,
Rm = distance between the centres of earth and moon,
T = time taken by the moon to c...

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Work, Power and Energy


Work

Work is said to be done when a force causes a displacement to a body on which it acts.
Work is a scalar quantity. It is the dot product of applied force F and displacement d.



Diagram Coming Soon

W = F . d
W = F d cos θ .............................. (1)
Where θ is the angle between F and d.
Equation (1) can be written as
W = (F cos θ) d
i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of displacement d.
equ...

Physics Notes for Metric (x) Class (Karachi Board)

Physics Notes for Metric Class (Karachi Board)

Dear Students, you are welcome to Study Guide Centre. We are presenting the Notes for Metric class, Karachi Board.
Here are the notes for Physics.
Metric Class:
Chapter # 1: Introduction
Chapter # 2: Measurments
Chapter #3 : Scalars and Vectors
Chapter #4 : Kinematics
Chapter #5 : Force and Motion
Chapter # 6: Statics
Chapter # 7: Circular Motion and Gravitation
Chapter # 8: Work Energy and Power
Chapter # 9: Machines
Chapter # 10: Matter
Chapter # 11: Heat
Chapter # 12: Waves and Sounds
Chapter # 13: Propagation and Reflection of Light
Chapter # 14: Reflection of Light and Optical Instruments
Chapter # 15: Nature of Light and Electromagnetics spectrum
Chapter # 16: Electricity
Chapter # 17: Magnetic and Electromagnetism
Chapter # 18: Electronics
Chapter # 19: Nuclear Physics