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Students can Download Chapter 9 String Handling and I/O Functions Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 9 String Handling and I/O Functions

Summary
String handling using arrays:
A string is a combination of characters hence char data type is used to store string. A string should be enclosed in double quotes. In C++ a variable is to be declared before it is used.Eg. “BVM HSS KALPARAMBU”.

Memory allocation for strings:
To store “BVM” an array of char type is used. We have to specify the size. Remember each and every string is end with a null (\0) character. So we can store only size- 1 characters in a variable. Please note that \0 is treated as a single character. \0 is also called as the delimiter.
char school_name[4]; By this we can store a maximum of three characters.

Consider the following declarations

• char my_name[10] = ”Andrea”;
• char my_name2[ ] = ”Andrea”;
• char str[ ] = ”Hello World”

In the first declaration 10 Bytes will be allocated but it will use only 6 + 1 (one for ‘\0’) = 7 Bytes the remaining 3 Bytes will be unused. But in the second declaration the size of the array is not mentioned so only 7 Bytes will be allocated and used hence no wastage of memory.

Similarly in the third declaration the size of the array is also not mentioned so only 12( one Byte for space and one Byte for ‘\0’) Bytes will be allocated and used hence no wastage of memory

Input/output operations on strings:
Consider the following code
#include<iostream>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
cin>>name;
cout<<“Hello “<<name;
}
If you run the program you will get the prompt as follows
Enter your name: Alvis Emerin
The output will be displayed as follows and the “Emerin” will be truncated.
Hello Alvis
This is because of cin statement that will take upto the space. Here space is the delimiter. To resolve this gets() function can be used. To use gets() and puts() function the header file stdio.h must be included. gets() function is used to get a string from the keyboard including spaces.

puts() function is used to print a string on the screen. Consider the following code snippet that will take the input including the space.
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char name[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello “<<name;
}

More console functions:

Stream functions for I / O operations:
Somefunctions that are available in the header file iostream.h to perforrn I/O operations on character and strings(stream of characters). It transfers streams of bytes between memory and objects. Keyboard and monitor are considered as the objects in C++.

Input functions:
The input functions like get( )(to read a character from the keyboard) and getline() (to read a line of characters from the keyboard) is used with cin and dot(.) operator.

eg:
#include<iostream>
using namespace std;
int main()
{
char str[80],ch=’z’;
cout<<“enter a string that end with z:”;
cin.getline(str,80,ch);
cout<<str;
}
If you run the program you will get the prompt as follows
Enter a string that end with z: Hi I am Jobi. I am a teacher. My school is BVM HSS The output will be displayed as follows and the string after ‘z’ will be truncated.
Hi, I am Jobi. I am a teacher

Output function:
The outputt functions like put() (to print a character on the screen) and write() (to print a line of characters on the screen) is used with cout and dot(.) operator.

 

Students can Download Chapter 4 Chemical Bonding and Molecular Structure Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 4 Chemical Bonding and Molecular Structure

Introduction
Matter is made up of different type of elements. The attractive force which holds the constituents together is called a chemical bond.

Kossel-Lewis Approach To Chemical Bonding
The bond between constituents are formed by the sharing of a pair of electrons or their transfer. G.N. Lewis introduced simple notations to represent these outer shell electrons in an atom. These notations are called Lewis symbols.

Significance of Lewis Symbols :
The number of dots around the symbol represents the number of valence electrons. This number of valence electrons helps to calculate the common or group valence of the element. The group valence of the elements is generally either equal to the number of dots in Lewis symbols or8 minus the number of dots or valence electrons.

Kossel, in relation to chemical bonding, drew attention to the following facts:
The bond formed, as a result of the electrostatic attraction between the positiveand negative ions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge (s) on the ion.
In terms of Lewis structures

1. According to electronic theory of chemical bond¬ing, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.

2. Covalent Bond, Langmuir in 1919 refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet, and by introducing the term covalent bond.

By Lewis – Langmuir theory the formation of chlorine molecule is as follows :

In water molecule covalent bond is as follows:

when two atoms share one electron pair they are said to be joined by a single covalent bond. If two atoms share two pairs of electrons, the covalent bond between them is called a double bond. And when combining atoms share three electron pairs as in the case of N₂ molecule a triple bond a triple bond is formed.

• The total number of electrons required for writing the structures are obtained by adding the valence electrons of the combining atoms.
• For anions, each negative charge would mean addition of one electron. For cations, each positive charge would result in subtraction of one electron from the total number of valence electrons.
• The least electronegative atom occupies the central position in the molecule/ion.

Formal charge
Formal charge (F.C.) on an atom in a Lewis structure = total number of valence electrons in the free atom— total number of non bonding (lone pairjelectrons—(1/2) total number of bonding(shared)electrons.

Let us consider the ozone molecule (O₃).
The Lewis structure of O₃ may be drawn as:

The atoms have been marked as 1,2 and 3. The formal charge on:

• The central O atom marked 1 = 6 – 2- \(\frac{1}{2}\)(6) = +1
• The end O atom marked 2 = 6 – 4 – \(\frac{1}{2}\)(4) = o
• The end O atom marked 3 = 6 – 6 – \(\frac{1}{2}\)(2) = -1

Hence, we represent O₃ along with the formal changes as follows:

Limitations of the Octet Rule
There are three types of exceptions to the octet rule. The incomplete octet of the central atom In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons.
Some compounds are BCl₃, AlCl₃ and BF₃.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO₂, the octet rule is not satisfied for all the atoms.

The expanded octet
In a number of compounds of these elements there are more than eight valence electrons around the central atom.Some examples are given below.

Ionic Or Electrovalent Bond
The formation of a positive ion involves ionization, i.e., removal of electrons from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.

A qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.

Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituentions.

Bond Parameters

Bond Length
It may be defined as the equilibrium distance between the centres of the nuclei of the two bonded i atoms in a molecule. Bond length are measured by spectroscopic, X-ray diffraction and electron diffraction techniques. It is usually expressed in Angstrom units (A°) or picometres (pm)
1 A° = 10^-10m and 1 pm = 10^-12m

Bond Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule.

Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

Bond Order:
In the Lewis description of covalent bond, the bond order is given by the number of bonds between the two atoms in a molecule. For example, the bond order in H₂ is one, in O₂ is two and in N₂ is three. Isoelectronic molecules and ions have identical bond orders. For example N₂, CO and NO⁺ have bond order 3. It is found that as the bond order increases, bond enthalpy increases and bond length decreases.

Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. As in the case of O₃.

O₃ is represented by the above 3 structures. These are called canonical structures.

Experimentally determined oxygen-oxygen bond lengths in the O₃ molecule are same (128 pm). Thus the oxygen-oxygen bonds in the O₃ molecule are intermediate between a double and a single bond. According to the concept of resonance, the canonical structures of the hybrid describes the molecule accurately.

Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.

Polarity of Bonds
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character. As a result of polarisation, the molecule possesses the dipole moment. Which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter Mathematically, it is expressed as follows: Dipole moment (µ) = change (Q) X distance of separation (r)

In case of polyatomic molecules, the dipole moment not only depend upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule.lt is due to the shifting of electrons to the side of more eletro negative element. For example,

The shifting of electrons is represented by an arrow. In case of H₂O the resultant dipole moment is given by the following figure:

Fajans Rules:
Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:

• The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
• The greater the charge on the cation, the greater the covalent character of the ionic bond.
• For cations of the same size and charge, the one, with electronic configuration (n-1)d”ns°, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.

The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond.

The Valence Shell Electron Pair Repul-Sion (Vspert) Theory
The main postulates of VSEPR theory are as follows:

• The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.
• Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
• These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them.
• The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.
• A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
• Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.

The repulsive interaction of electron pairs de-crease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)

Valence Bond Theory
Valence bond theory was introduced by Heitlerand London (1927) and developed further by Pauling and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition. First, we consider the formation of H2. When the attractive forces become greater than the repulsive forces, the molecule is formed and the system gets minimum energy. Because energy is released when a bond is formed. The energy so released is called bond enthalpy.

Orbital Overlap Concept
When two atoms approach each other, their atomic orbitals undergo partial interpenetration. This partial interpenetration of atomic orbitals is called overlapping of atomic orbitals. The electrons belonging to these orbitals are said to be shared and this results in the formation of a covalent bond. The main ideas of orbital of overlap concept of formation of covalent bonds are

• Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the atoms taking part in bonding.
• The orbitals undergoing overlapping must have electrons with opposite spins.
  Overlapping of atomic orbitals results in decrease of energy and formation of covalent bond.
• The strength of a covalent bond depends upon the extent of overlapping. The greater the overlapping, the stronger is the bond formed.

The above treatment of formation of covalent bond involving the overlap of half-filled atomic orbitals is called valence bond theory.

Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
(i) Sigma(σ) bond, and
(ii) pi(π) bond

(i) Sigma( σ) bond :
This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.
s-s overlapping:
In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below:

s-p overlapping:
This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.

p-p overlapping :
This type of overlap takes place between half filled p-orbitals of the two approaching atoms.

(ii) pi(π) bond :
In the formation of n bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to side wise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms.

Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double ortriple bonds).

Hybridisation
Hybridisationis defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.

The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.

These hybrid orbitals are stable due to their arrangement which provides minimum repulsion between electron pairs. Therefore, the type of hybridisation indicates the geometry of the molecules.

It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Types of Hybridisation
There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under:

(I) sp hybridisation:
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp- hybridised and linked directly to two other central atoms possesses linear geometry.The two sp hybrids point in the opposite direction which provides more effective overlapping resulting in the formation of stronger bonds.

Example of molecule having sp hybridisation BeCl₂:
The ground state electronic.configuration of Be is 1s²2s². In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its divalency. One 2s and one 2p-orbitalsget hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite direction forming an angle of 180°. Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be-Cl sigma bonds.

II) sp² hybridisation :
In this hybridisation there is involvement of one s and two p-orbitals in orderto form three equivalent sp² hybridised orbitals. For example, in BCl₂ molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s
electrons is promoted to vacant 2p orbital as a result boron has three unpaired electrons.

These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals. The three hybrid orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl₃ the geometry is trigonal planar with ClBCl bond angle of 120°

III) sp³ hybridisation:
This type of hybridisation can be explained by taking the example of CH₄ molecule in which there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp³ hybrid orbital of equivalent energies and shape.

There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron.

The angle between sp³ hybrid orbital is 109.5°

The structure of NH₃ and H₂O molecules can also be explained with the help of sp3hybridisation. In NH₃, the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s²\(p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) having three unpaired electrons in the sp³ hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N-H sigma bonds. Due to the force of repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramidal.

Other Examples of sp³, sp² and sp Hybridisation:
sp³ Hybridisation in C₂H₆ molecule:
In ethane molecule both the carbon atoms assume sp3 hybrid state. One of the four sp³ hybrid orbitals of carbon atom overlaps axially with similar orbitals of other atom to form sp³-sp³ sigma bond while the other three hybrid orbitals of each carbon atom are used in forming sp³-s sigma bonds with hydrogen atoms Therefore in ethane C-C bond length is 154 pm and each C-H bond length is 109 pm.

sp² Hybridisation in C₂H₄:
In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C-C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp²-s sigma bond with two hydrogen atoms. The unhybridised orbital (2p_x or 2p_y) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond, which consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms.

Thus, in ethene molecule, the carbon-carbon bond consists of one sp²-sp² sigma bond and one pi (π) bond between p orbitals which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length is 134 pm. The C-H bond is sp^s sigma with bond length 108 pm. The H-C-H bond angle is 117.6° while the H-C-C angle is 121°.

sp Hybridisation in C₂H₂:
In the formation of ethyne molecule, both the carbon atoms undergo sp- hybridisation having two unhybridised orbital i.e., 2p_y and 2p_x. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming σ bonds. Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds

Hybridisation of Elements involving d-Orbitals
The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the hybridisation involving either 3s, 3p, and 3d or 3d, 4s and 4p is possible. However, since the difference in energies of 3p and 4s orbitals is significant, no hybridisation involving 3p, 3d and 4s orbitals is possible.

1. Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below.

Now the five orbitals (i.eone s, three p, and one d orbitals) are available for hybridisation to yield a set of five sp3d hybrid orbitals which are directed towards the five comers of a trigonal bipyramidal.

Three sigma bond known as equatorial bonds lie in one plane and make an angle of 120° with each other.
The remaining two P-Cl bonds(called axial bonds)-one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane.

As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl₅ molecule more reactive.

2. Formation of SF₆ (sp³d² hybridisation):
In SF₆ the central sulphur atom has the ground state outer electronic configuration 3s²3p⁴. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp³d² hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF₆. These six sp³d² hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus SF₆ molecule has a regular octahedral geometry.

The structure of SF₆ is given below.

Molecular Orbital Theory
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of this theory are :

• The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.
• The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric.
• The number of molecular orbital formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
• The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
• Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
• The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’srule.

Formation of Molecular Orbitals
Linear Combination of Atomic Orbitals (LCAO)
The atomic orbitals of these atoms may be represented by the wave functions ψ_A and ψ_B. The formation of molecular orbitals is the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals as shown below.

Energy Level Diagram for Molecular orbitals

The increasing order of energies of various molecular orbitals for O₂ and F₂ is given below:
σ1s < σ*1s < σ2s < σ*2s < σ2p_x <(π2p_x = π2p_y) <(π*2p_x = π*2p_z) < σ*2p_x

This sequence of energy levels of molecular orbitals is not correct for the remaining molecules Li₂, Be₂, B₂, C₂, N₂. For molecules such as B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is
σ1s < σ*1s < σ2s < σ*2s <(π2p_x = π2p_y) < σ2p_x <(π*2p_x = π*2p_z) < σ*2p_z

The important characteristic feature of this order is that the energy of σ 2p_z molecular orbital is higher than that of π2p_x and π2p_y molecular orbitals. If the bonding influence is stronger a stable molecule results and if the antibonding influence is stronger,the molecule is unstable.

Bonding In Some Homonuclear Diatomic Molecules
Bond Order:
Bond order is defined as half of the difference between the number of electrons in the bonding molecular orbitals and that in the antibonding molecular orbitals.
Nh – N
i.e. Bond Order= \(\frac{N_{b}-N_{a}}{2}\). Where N_b is the number of electrons in the bonding molecular orbitals and Na is the number of electrons in the antibonding mo-lecular orbitals.

Significance of bond order:
Bond order conveys the following important informations about a molecule.
i) If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is negative or zero, the molecule is unstable and is not formed.
ii) Bond dissociation energy of a diatomic molecule is directly proportional to the bond order of the molecule. The greater the bond order, the higher is the bond dissociation energy.
iii) Bond order is inversely proportional to the bond length. The higherthe bond ondervalue, smaller is the bond length. For example, the bond length in N₂ molecule (having bond order 3) is less than that in O₂ molecule (having bond order 2).

Magnetic character:
If all the electrons in the mol-ecules of a substance are paired, the substance will be diamagnetic. On the other hand, if there are un-paired electrons in the molecule, the substance will be paramagnetic.

Hydrogen Bonding
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O orN) of another molecule. When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes, highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ⁺) while ‘X’ attain fractional negative charge (δ^–). This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as: H^δ+ – X^δ-

The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.

Types of Hydrogen Bonds
There are two types of hydrogen bonds

1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond

1. Intermolecular hydrogen bond:
It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

2. Intramolecular hydrogen bond:
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-Nitrophenol the hydrogen is in between the two oxygen atoms as shown below:

Students can Download Chapter 4 Motion in a Plane Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 4 Motion in a Plane

Summary
Introduction
In this chapter, we will study, about vector, its ’ addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.

Scalars And Vectors

a. Scalars:
A quantity which has only magnitude and no direction is called a scalar quantity.
Eg: length; volume, mass, time, work etc.

b. Vectors:
(i) The need for vectors:
In one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.

(ii) Vector:
A physical quantity which has both magnitude and direction is called a vector quantity.
Eg: Displacement, Velocity, Acceleration, Force, momentum.

1. Position and Displacement Vectors:
Position vector:
Consider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.

A vector representing the position of an object P with respect to an origin O is called position vector \(\overrightarrow{\mathrm{OP}}\) of the object. This position vector may be represented
by an arrow with tail at O and head at P.

The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\(\overrightarrow{\mathrm{OP}}\)|.

Displacement vector:

Consider the motion of an object in a plane. Let P be the position of a moving object at a time t and p¹ that at a later time t¹. \(\overrightarrow{\mathrm{OP}}\) and \(\overrightarrow{\mathrm{OP}^{1}}\) are the position vectors at time t and t¹ respectively. So the vector \(\overrightarrow{\mathrm{PP}^{1}}\) is called displacement vector corresponding to the motion in the time interval (t – t¹).

2. Equality of vectors:
Two vectors are said to be equal if they have the same magnitude and direction.

The above figure shows two vectors \(\vec{A}\) and \(\vec{B}\) having the same magnitude and direction.
∴ \(\vec{A}\) = \(\vec{B}\).

Question 1.
Observe the following figures (a) and (b) and find which pair does represents equal vectors?

Answer:
Figure a represent that A and B are equal vectors. Two vectors A¹ and B¹ are unequal, because they were in different directions.

Multiplication Of Vectors By Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number I gives a vector whose magnitude is changed by the factor λ.

The direction λ\(\vec{A}\) is the same as that of \(\vec{A}\).
Examples:

A vector \(\vec{A}\) and the resultant vector after multiplying \(\vec{A}\) by a positive number 2.

A vector A and resultant vector after multiplying it by a negative number-1 and -1.5.

Addition And Subtraction Of Vectors – Graphical Method
Vectors representing physical quantities of the same dimensions can be added or subtracted. The sum of two or more vectors is known as their resultant.

1. When two vectors are acting in the same direction:

2. When two vectors act in opposite direction:
In this case, the angle between the vectors is 180°.

The resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.

3. When two vectors are inclined to each other:
The sum of two vectors inclined at an angle q can be obtained either by

• the law of triangle of vectors
• the parallelogram law of vectors

(i) Triangle method:
This law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

(ii) Parallelogram law of vector addition:
This law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

To find \(\vec{A}\) + \(\vec{B}\), we bring theirtails to a common origin Q as shown below.


The diagonal of parallelogram OQSP, gives the resultantof (\(\vec{R}\) = \(\vec{A}\) + \(\vec{B}\)) of two vectors \(\vec{A}\) and \(\vec{B}\).
Note: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.

4. Substraction of vectors:

To substract \(\vec{B}\) from \(\vec{A}\), reverse the direction of \(\vec{B}\).

Then add –\(\vec{B}\) with \(\vec{A}\) using parallelogram law or tri¬angle law.

The resultant of \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{R}\).
Null vector or zero vector:
A vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \(\vec{O}\). Since the magnitude is zero, we don’t have to specify its direction.
Properties of null vector:

Question 2.
Explain a zero vector using an example.
Answer:
Suppose that an object which is at P at time t, moves to p¹ and then comes back to P. In this case displacement is a null vector.

Resolution Of Vectors Unit Vectors
A vector divided by its magnitude is called unit vector along the direction of that vector. A unit vector in the direction of \(\vec{A}\) is written as \(\hat{A}\).

Orthogonal unit vectors:

In the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) respectively and are known as orthogonal unit vectors.
For unit vectors

Resolution of vector into rectangular components:
The components of a vector in two mutually perpendicular directions are called its rectangular components.
Explanation

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in x-y plane as shown in figure. To resolve \(\overrightarrow{\mathrm{A}}\), draw lines from the head of \(\overrightarrow{\mathrm{A}}\) perpendicularto the coordinate axes as shown below.

The quantities A_x and A_y are called x and y components of the vector \(\overrightarrow{\mathrm{A}}\). Hence the vector \(\overrightarrow{\mathrm{A}}\) can be written in terms of rectangular components as

Magnitude of \(\overrightarrow{\mathrm{A}}\):

From the figure, the magnitude of \(\overrightarrow{\mathrm{A}}\) can be written as,

Question 3.
A vector \(\overrightarrow{\mathrm{A}}\) in xyz plane is given below. A_x, A_y and A_z are the perpendicular components in x,y and z directions respectively.

1. Write \(\overrightarrow{\mathrm{A}}\) in terms of rectangular components.
2. Write the magnitude of \(\overrightarrow{\mathrm{A}}\).

Answer:

The magnitude of vector \(\overrightarrow{\mathrm{A}}\) is

Vector Addition – Analytical Method
The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use analytical method to add vectors.
Explanation

The vectors obey commutative and associative laws. Hence

Question 4.
Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{A}}\) in terms of their magnitudes and angle between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle q. Using the parallelogram method of
vectors, the resultant vector \(\overrightarrow{\mathrm{R}}\) can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosq and SN = Bsinq
Substituting these values in eq.(1), we get
OS² = (OP + Bcosq)² + (Bsinq)²
But OS = R and OP = A
R² = (A + Bcosq)² + B²sin²q
= A² + 2ABcosq + B²cos²q + B²sin²q
R² = A² + 2 ABcosq + B²

The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \overrightarrow{\mathrm{A}}. From the right angled triangle OSN,

But SN = Bsinq PN = Bcosq

Motion In A Plane

1. Position vector and displacement vector Position vector:

Consider a small body located at P with reference to the origin O. The position vector of the point ‘P’

Displacement vector


where Dx = x¹ – x¹, Dy = y¹ – y
Velocity:
If Dt is the time taken to reach from P to P¹
The average velocity, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\Delta r}}{\Delta \mathrm{t}}\) ____(3)
Substitute eq.(2) in eq.(3), we get

The direction of average velocity is the same as that of \(\overrightarrow{\Delta r}\).
The instantaneous velocity can be written as

Acceleration:
If the velocity of an object changes from \(\overrightarrow{\mathrm{v}} \text { to } \overrightarrow{\mathrm{v}^{1}}\) in time Dt, then its average acceleration is given by

Instantaneous acceleration:
The acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Motion In A Plane With Constant Acceleration
Consider an object moving in xy plane with constant acceleration ‘a’. Let \(\vec{u}\) be the initial velocity at t=0 and \(\vec{v}\) be the final velocity at time t.
Then by definition acceleration

In terms of components
v_x = u_x + a_xt
v_y = u_y + a_yt
Displacement in a plane
If \(\overrightarrow{\mathrm{r}_{0}}\) and \(\vec{r}\) be position vectors of particle at t = 0 and time t respectively, then
displacement = \(\vec{r}-\vec{r}_{0}\) _______(1)
For uniformly accelerated motion, displacement,

In terms of components
x = x₀ + u_xt + 1/2 a_xt²
y = y₀ + u_yt + 1/2 a_yt²
The eq.(2) shows that, the above motion in xy plane can be treated as two separate one dimensional motions along two perpendicular directions.

Relative Velocity In Two Dimensions
Consider two bodies A and B moving along a plane with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\). Then velocity of A relative to that of B is,

Similarly velocity of B relative to that of A

Projectile Motion
Projectile:
A body is projected into air and is allowed to move under the influence of gravity is called projectile.

Consider a body which is projected into air with a velocity u at an angle q. The initial velocity ‘u’ can be divided into two components ucosq along horizontal direction and using along vertical direction.

1. Time of flight:
The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinq, we get
0 = usinqt – 1/2gt²
1/2 gt² = usinqt

2. Vertical height:
Vertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v² = u² + 2as
When we substitute v=0, a = -g, s = H and u = usinq, we get
0 = (usinq)² + 2 – g × H
2gH = u²sin²q

3. Horizontal Range:
If we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

The eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q₀ = 45°.
The maximum horizontal range

Equation for path of projectile

Question 5.
What is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
Answer:
S = ut+ 1/2at²
y = usinqt – 1/2gt²
But we know horizontal displacement, x = ucosq × t

In this equation g, q and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

Uniform Circular Motion
The motion of an object along the circumference of a circle is called circular motion.
Uniform circular motion:
When an object follows a circular path at a constant speed, the motion is called uniform circular motion.
Period:
The time taken by the object to complete one full revolution is called the period.
Frequency:
The number of revolutions completed per second is called the frequency u of the circular motion.
If the period of a circular motion isT, its frequency

Angular Displacement (Dq):
The angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
Angular velocity:
The rate of change of angular displacement is called the angular velocity.

If T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
Therefore in one second it sweeps out an angle \(\frac{2 \pi}{T}\).
∴ Angular velocity of the object

Expression for velocity and acceleration in uniform circular motion:

The direction of velocity is in the direction of tangent at that point. The change in velocity vectors \((\overrightarrow{\Delta v})\) is obtained by triangle law of vector as shown in figure (b).

a. Speed and angular speed in uniform circular motion:
Let the Dq be the angle constructed by the body during the time interval ∆t. The angular velocity can be written as

If the distance travelled by the object during the time Dt is Dr (ie. PP¹ = Ds) then speed

But Ds = RDq
where R = \(|\vec{r}|=|\overrightarrow{r^{\prime}}|\)
Substituting Dr = RDq in eq.(1)
we get

b. Acceleration in uniform circular motion:

The direction of this acceleration should be in the direction of \(\overrightarrow{\Delta V}\). The fig(b) shows that \(\overrightarrow{\Delta V}\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration.

The force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as
F = ma_c

But ω = \(\frac{V}{R}\). Hence we get K

Students can Download Chapter 4 Motion in a Plane Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 4 Motion in a Plane

Plus One Physics Motion in a Plane One Mark Questions and Answers

Question 1.
A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?
(a) 200m
(b) 150m
(c) 100m
(d) 50m
Answer:
(d) 50m
R_max = \(\frac{u^{2}}{g}\)
100 = \(\frac{u^{2}}{g}\) or u² = 100g
Using, v² = u² + 2as
0 = (100g) + 2(-g)h or h = 50m.

Question 2.
If vectors \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) and \(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\mathrm{ak}\) are equal vectors, then the value of a is
(a) 5
(b) 2
(c) -3
(d) -5
Answer:
(d) Comparing vector, we get = \(+5 \hat{k}=-a \hat{k}\)
a = -5.

Question 3.
State which of the following algebraic operations are not meaningful
(a) Addition of a scalar to a vector
(b) Multiplication of any two scalars.
(c) Multiplication of vector by scalar
(d) Division of a vector by scalar Addition of a scalar to a vector
Answer:
(a) Addition of a scalar to a vector

Question 4.
What is the acceleration of train travelling at 40ms^-1 as it goes round a curve of 160m radius?
Answer:
a = \(\frac{v^{2}}{r}=\frac{40 \times 40}{160}\) = ms^-2.

Question 5.
What provides centripetal force in the following cases.

1. Electron revolving around nucleus.
2. Earth revolving around sun

Answer:

1. Electrostatic force
2. Gravitational force

Question 6.
Why a cyclist has to bend inwards while going on a circular track?
Answer:
The cyclist bends inwards to provide required centripetal force.

Question 7.
A body executing uniform circular motion has constant
(i) velocity
(ii) acceleration
(iii) speed
(iv) angular velocity
Answer:
(iii) speed

Question 8.
Name a quantity which remains unchanged during projectile motion.
Answer:
Horizontal component.

Question 9.
What is the effect of air resistance in time of flight and horizontal range?
Answer:
The effect of air resistance is to increase time of flight and decrease horizontal range.

Question 10.
What is the angle between directions of velocity and acceleration at the highest point of trajectory of projectile?
Answer:
At the highest point velocity is horizontal and acceleration is vertical. So angle is 90°.

Question 11.
Can a body have constant velocity and still have a varying speed?
Answer:
No.
If velocity is constant, speed also will be constant.

Question 12.
Can a body have zero velocity, still accelerating?
Answer:
Yes.
When a body is at highest point of motion, its velocity is zero but acceleration is equal to acceleration due to gravity.

Question 13.
A quantity has both magnitude and direction. Is it necessarily a vector? Give an example.
Answer:
No. The given quantity will be a vector only if it obeys laws of vector addition.
Example: Current.

Question 14.
What is the angle between \(\vec{A} \times \vec{B}\) and \(\vec{B} \times \vec{A}\) ?
Answer:
These two vectors will be antiparallel. Hence θ = 180°.

Plus One Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
A particle is projected with a velocity u so that its horizontal range is twice the greatest hieght attained. The horizontal range is
(a) \(\frac{u^{2}}{g}\)
(b) \(\frac{2 u^{2}}{g}\)
(c) \(\frac{4 u^{2}}{5g}\)
(d) None of these
Answer:
(c) Horizontal range, R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\);
Maximum height, H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
As per question, R = 2H

or sin2θ = sin²θ
or 2sinθcosθ = sin²θ or tanθ = 2
Hence, sinθ = \(\left(\frac{2}{\sqrt{5}}\right)\) and cosθ = \(\left(\frac{1}{\sqrt{5}}\right)\)
Horizontal range, R = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)

Question 2.

Answer:

Question 3.
Choose the correct alternative given below. A Particle executing uniform circular motion. Then its
(a) Velocity and acceleration are radial.
(b) Velocity and acceleration are tangential.
(c) Velocity is tangential, acceleration is radial.
(d) Velocity is radial, acceleration is tangential.
(e) In a circus, a rider rides in a circular track of radius Y in a vertical plane. The minimum velocity at the highest point of the track will be

• \(\sqrt{2 g r}\)
• \(\sqrt{ g r}\)
• \(\sqrt{3 g r}\)
• 0

Answer:
(c) Velocity is tangential, acceleration is radial.
(e) \(\sqrt{ g r}\)

Question 4.
Two non-zero vectors \(\bar{A}\) and \(\bar{B}\) are such that \(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\). Find the angle between them.
Answer:
\(|\bar{A}+\bar{B}|=|\bar{A}-\bar{B}|\)
|A² + B² + 2ABcosθ| = |A² + B² – 2ABcosθ|
|4ABcosθ| = 0
Since A and B are non-zero we get, cos θ = 0 or θ = 90°.

Question 5.
Consider a particle moving along the circumference of a circle of radius R with constant speed with a time period T.

1. During T, what is the distance coverd and displacement?
2. What is the direction of the velocity at each point?

Answer:

1. Distance = 2πR. Displacement = 0
2. Tangent to the circle at every point.

Question 6.
Classify into scalars and vectors. Frequency, velocity gradient, instantaneous velocity, Area.
Answer:

Scalars	Vectors
Frequency	Instantaneous velocity
Velocity gradient	Ares

Question 7.
A body is projected so that it has maximum range R. What is the maximum height reached during the fight?
Answer:
At maximum range, θ = 45°

Plus One Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
An electron of mass ‘m’ moves with a uniform speed v around the nucleus along a circular radius Y.

1. Derive an expression for the acceleration of the electron.
2. Explain why the speed of electron does not increase even though it is accelerated by the above acceleration.

Answer:
1.

The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a₀ = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
a_c = Rω²

2. The direction of centripetal force is towards the centre. The angle between force and displacement is 90°. Hence the work done by the centripetal force is zero. So speed does not increase.

Question 2.
A boy pulls his friend in a home made trolley by means of a rope inclined at 30° to the horizontal. If the tension in the rope is 400N.

1. Draw the vertical and horizontal components of tension in the rope.
2. Find the effective force pulling the trolley along the ground.
3. Find the force tending to lift the trolley off the ground.

Answer:
1.

2. Effective horizontal force = T Cos30°
= 400 × Cos 30°

3. Vertical force = T Sin 30°
= 400 × Sin 30°

Question 3.
A stone tied to the end of a string is whirled in a horizontal circle with constant speed.

1. Name the acceleration experienced by the stone.
2. Arrive at an equation for magnitude of acceleration experienced by the stone.

Answer:
1. Centripetal acceleration

2.

The acceleration is directed towards the centre of the circle and is called centripetal acceleration.
a₀ = \(\frac{v^{2}}{R}\)
But V = Rω
Substituting we get
a_c = Rω²

Plus One Physics Motion in a Plane Four Mark Questions and Answers

Question 1.
Two balls are released simultaneously from a certain height, one is allowed to fall freely and other thrown with some horizontal velocity.

1. Will they hit the ground together?
2. At any time during the fall will the velocities of the balls are same?
3. How does the path of the balls appear to a person standing on the ground?

Answer:

1. Both balls will reach at same time.
2. Total velocity of first body and second body is different. First ball has only downward velocity but second ball has both downward and horizontal velocity.
3. The path of first ball appears to be straight line and that of second ball appears to be parabola.

Question 2.
A ball is thrown straight up.

1. Obtain a mathematical expression for the height to which it travels.
2. What js its velocity and acceleration at the top?
3. Draw the velocity-time graph for the ball showing its motion up and down.

Answer:
1. u = u, v = o, a = -g, h = ?
We can find maximum height using the equation
v² = u² + 2as
0 = u² + 2 × – g × H
2gh = u²
H = \(\frac{u^{2}}{2 g}\)

2. Velocity is zero, but it has a acceleration and its value g = 9.8 m/s².

3.

Question 3.

1. Parallelogram law helps to find the magnitude and direction of the resultant of two forces. State the law.
2. For two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are acting at a point with an angle a between them, find the magnitude and direction of the resultant vector.
3. What will be the angle between two vectors of equal magnitude for their resultant to have the same magnitude as one of the vectors?

Answer:
1. Law of parallelogram of vectors:
If two vectors acting simultaneously at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram then the diagonal of the parallelogram passing through that point represents the resultant in magnitude and direction.

2.

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle θ. Using the parallelogram method of vectors, the resultant vector R can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosθ and SN = Bsinθ
Substituting these values in eq.(1), we get
OS² = (OP + Bcosθ)² + (Bsinθ)²
But OS = R and OP = A
= A² + 2ABcosθ + B²cos²θ + B²sin²θ
² = A² +2 ABcosθ + B²
R = \(\sqrt{A^{2}+2 A B \cos \theta+B^{2}}\)
The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \(\overrightarrow{\mathrm{R}}\). From the right angled triangle OSN,

But SN = Bsinθ and PN = Bcosθ

3. 120°

Question 4.
A ball of mass m is projected at an angle with the ground and it is found that its kinetic energy at the highest point is 75% of that at the point of projection.

1. Is it a one dimensional or a two-dimensional motion? Why?
2. Find the angle of projection
3. Determine another angle of projection which produces the same range.

Answer:
1. Two-dimensional motion. Projectile has two dimensions.

2. The K.E at highest point
E_k = E cos²θ,
where E = initial K.E
0.75 E = E cos²θ
cos²θ = 0.75 = 3/4
cosθ = \(\sqrt{3} / 2\)
θ = 60°

3. When an object is projected with velocity ‘u’ making an angle θ to with the horizontal direction, the horizontal range will be
R₁ = \(\frac{u^{2} \sin 2 \theta}{g}\) ____(1)
when an object is projected with velocity u making an angle (90° – θ) with the horizontal direction, then horizontal range will be

From eq (1) eq (2), we get R₁ = R₂, which means that we get same horizontal range for two angles θ and (90 – θ).

Question 5.
“The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But, sometimes, it is tedious and has limited accuracy”.

1. Name the alternative method of vector addition.
2. Write a mathematical expression to find resultant of two vectors.
3. A particle is moving eastward with a velocity of 5m/s. |f in 10s, the velocity changes by 5 m/s northwards, what is the average acceleration in this time.

Answer:
1. Analytical method of vector addition.

2. R = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

3. Change in velocity

Average acceleration

Question 6.

Answer:

Question 7.
A projectile is an object projected into air with a velocity V so that it is moving under the influence of gravity.

1. What is the shape of the path of projectile?
2. As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are perpendicular to each other
3. If E is energy with a projectile is projected.

(i) What is the Kinetic energy at the highest point.
(ii) What is P.E at highest point?

Answer:
1. Parabola

2. Yes, highest point

3. Energy with a projectile is projected:
(i) Kinetic energy E_k = 1/2mv²
Velocity at the highest point = Vcosθ
∴ K.E. at highest point = 1/2m(Vcosθ)²
= 1/2 mv²cos²θ
K.E. = E_k cos²θ

(ii) Potential energy at highest point,
P.E. = mgh
= mg\(\frac{v^{2} \sin ^{2} \theta}{2 g}\) = 1/2mv²sin²θ
P.E. = E_k sin²θ.

Plus One Physics Motion in a Plane Five Mark Questions and Answers

Question 1.
An object is projected with velocity U at an angle θ to the horizontal.

1. Obtain a mathematical expression for the range in the horizontal plane.
2. What are the conditions to obtain maximum horizontal range?
3. Find the maximum height of the object when its path makes an angle of 30° with the horizontal (velocity of projection = 8 ms^-1)

Answer:
1. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

2. when θ = 45° we get maximum horizontal range

3. Height H

Question 2.
The path of projectile from A is shown in the figure. M is the mass of the particle.

When the particle moves from A to M.

1. a) What is the change in vertical velocity?
2. b) What is the change in speed?
3. What is the change in linear momentum?
4. The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms^-1 can go without hitting the ceiling of the hall?

Answer:
1.

When the particle reaches at M, the vertical component of velocity becomes zero.
change in vertical velocity = u sinθ – 0
= u sinθ

2. change in speed = u cosθ – u

3. change in momentum = mu sinθ – 0
= mu sinθ

4.

θ = 33°48¹
The horizontal range of the ball is

Question 3.
A particle moving uniformly along a circle, experiences a force directed towards the centre and an equal and opposite force directed away from the centre.

1. Name the two forces directed towards and away from the centre.
2. Obtain an expression for the force directed towards the centre.
3. An aircraft executes a horizontal loop at a speed of 720 km hr^-1 with its wings banked at 15°. What is the radius of the loop?

Answer:
1. centripetal force and centrifugal force.

2.


force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as F = ma.
F = m\(\frac{V^{2}}{R}\)

3. Speed of aircraft, 720 × \(\frac{5}{18}\) = 200 m/s
The velocity of aircraft, ν = \(\sqrt{r g \tan \theta}\)

Question 4.
A body is projected with a velocity ‘u’ in a direction making an angle θ with the horizontal.

1. Derive the mathematical equation of the path followed.
2. Draw the velocity-time graphs for the horizontal and vertical components of velocity of the projectile.
3. Obtain an expression for the time of flight of the projectile.

Answer:
1. The vertical displacement of projectile at any time t, can be found using the formula.
S = ut + 1/2at²
y = usinθt – 1/2gt²
But we know horizontal displacement,
x = ucosθxt

In this equation g, θ and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

2.

3. The time taken by the projectile to cover the horizontal range is called the time of flight.
Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt²
1/2 gt² = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

Question 5.
An object projected into air with a velocity is called a projectile.

1. What will be the range when the angle of projections are zero degrees and ninety degrees?
2. Show that fora projectile, the upward time of flight is equal to the downward time of flight.
3. At what angles will a projectile have the same range fora velocity?

Answer:
1. When θ = 0°
R = 0
When θ = 90°

2. Upward motion
u = u sinθ, a = -g, v = 0
V = u + at
0 = u sinθ + ^–gt
t_a = \(\frac{u \sin \theta}{g}\) ____(1)
Downward motion
u = 0, V= u sinθ, a = +g
V = u + at
u sinθ = 0 + gt
t_a = \(\frac{u \sin \theta}{g}\) ____(2)
eq (1) and (2), shows up ward time of flight is equal to downward time of flight.

3. θ, 90 – θ.

Question 6.
“An object that is in flight after being thrown or projected is called a projectile”.
1. Which of the following remains constant throughout the motion of the projectile?
(i) Vertical component of velocity
(ii) Horizontal component of velocity
2. Derive an expression for maximum range of a projectile.
3. Show that range of projection of a projectile for two angles of a projection a and (3 is same where α + β = 90°.

Answer:
1. (ii)

2. If we neglect the air resistance, the horizontal velocity (ucosθ) of projectile will be a constant. Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

3. Range of projectile R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\)
Case – 1
at angle α,
R_α = \(\frac{\mathrm{u}^{2} \sin 2 \alpha}{\mathrm{g}}\) ____(1)
Case – 2
at an angle β,

From (1) and (2), we get
R_α = R_β

Question 7.
A bullet is fired with a velocity ‘u’ at an angle ‘θ’ with the horizontal such that it moves under the effect of gravity.

1. What is the nature of its trajectory.
2. Arrive at an expression for time of flight of the bullet.
3. What is the relation between time of ascent and time of decent, when air resistance is neglected.
4. How the relation is affected when air resistance . is considered.

Answer:
1. Parabola

2. The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinθ. The time of flight can be found using the formula s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinθ, we get
0 = usinθt – 1/2gt²
1/2 gt² = usinθt
t = \(\frac{2 u \sin \theta}{g}\).

3. Time of ascent = time of descent

4. Time of descent > time of ascent

Plus One Physics Motion in a Plane NCERT Questions and Answers

Question 1.
State for each of the following physical quantities, if it is a scalar or a vector:
Volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:

• Scalars: Volume, mass, speed, density, number of moles and angular frequency.
• Vectors: acceleration, velocity, displacement, and angular velocity.

Question 2.
Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Work, current.

Question 3.
State, with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

1. Adding any two scalars.
2. Adding a scalar to a vector of the same dimensions
3. Multiplying any vector by any scalar
4. Multiplying any two scalars
5. Adding any two vectors
6. Adding a component of a vector to the same vector.

Answer:

1. No. Scalars must represent same physical quantity.
2. No. Vector can be added only to another vector.
3. Yes. When a vector is multiplied by any scalar, we get a vector.
4. Yes. In multiplication, scalars may not represent the same physical quantity.
5. No. The two vectors must represent the same physical quantity.
6. Yes. The vector and its component must have the same dimensions.

Question 4.
Pick out the only vector quantity in the following list: temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Impulse.

Question 5.
Read each statement below carefully and state with reasons, if it is true or false:

1. The magnitude of a vector is always a scalar.
2. Each component of a vector is always a scalar.
3. The total path length is always equal to the magnitude of the displacement vector of a particle.
4. The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time.
5. Three vectors not lying in a plane can never add up to give a null vector.

Answer:

1. True
2. False
3. False
4. True
5. True.

In order to give a null vector, the third vector should have the same magnitude and opposite direction to the resultant of two vectors.

Question 6.
A cyclist starts from the centre O of a circular park, then cycles along the circumference, and returns to the centre along QO as shown. If the round trip takes 10 minute, what is the

1. Net displacement.
2. Average velocity and
3. Average speed of the cyclist?

Answer:
1. Since both the initial and final positions are the same therefore the net displacement is zero.

2. Average velocity is the ratio of net displacement and total time taken. Since the net displacement is zero therefore the average velocity is also zero.

3. Average Speed

Question 7.
A passenger arriving in a new town wishes to go from the station to hotel located 10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 minute, what is

1. The average speed of the taxi
2. The magnitude of average velocity? Are the two equal?

Answer:
Magnitude of displacement = 10km
Total path length = 23km
Time taken = 28 min

Average Speed = \(\frac{23 \mathrm{km}}{\frac{7}{15} \mathrm{h}}\)
= 49.3 kmh^-1
Magnitude of average velocity

Students can Download Chapter 1 Nature and Purpose of Business Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 1 Nature and Purpose of Business

Contents

• Economic and non-economic activities
• Business and its characteristics
• Business, Profession and Employment
• Classification of business activities
• Industry and commerce
• Objectives of business
• Business risk
• Factors to be considered before starting a business.

Activities which human beings undertake are known as human activities. We can divide these activities into two categories.

1. Economic activities
2. Non-economic activities

1. Economic Activities:
The human activities that are undertaken with an objective to earn money or livelihood is known as economic activities. eg: A worker working in a factory, a doctor operating in his hospital, a manager working in the office, a teacher teaching in a school etc.

2. Non-economic activities:
Activities are undertaken to derive psychological satisfaction are known as non economic activities. eg: Mother preparing food for her children, Praying, listening to radio or watching television, playing football with friends, etc.

Types of economic activities:

• Business
• Employment
• Profession

Business:
Business may be defined as an economic activity involving the production or purchase and sale of goods and services with the main object of earning profit by satisfying human needs in the society.
Characteristics of business

1. Business is an economic activity with the object of earning profit.
2. Business includes all the activities concerned with the production or procurement of goods and services.
3. There should be sale or exchange of goods and services for the satisfaction of human needs.
4. Business involves dealings in goods or services on a regular basis. Normally, one single transaction of sale or purchase is not treated as business.
5. One of the main objectives of business is to earn maximum profit.
6. Business involves risk and uncertainty of income. Risk means the possibility of loss due to change in consumer taste and fashion, strike, lockout competition, fire, theft etc.

Employment:
Employment refers to that type of economic activity in which people engage in some work for others regularly and get salary or wages in return of their services.
Characteristics of Employment

1. There must exist employer-employee relationship.
2. There must be a service contract between the employer and employee.
3. Employees get salary or wages for their services
4. Regularity in service.

Profession:
Profession refers to an occupation which requires specialized knowledge, skip and training. Its objective is to provide service to the society.
Characteristics of Profession

1. A profession requires specialized knowledge, training and skill
2. The membership of a professional body is a must
3. Professionals have a code of conduct
4. They charge fee in return of their service.

Comparison of Business, Profession and Employment:

Classification of Business Activities: Business activities may be classified into two categories

• Industry
• Commerce.

Chart showing business activities

Industry:
Industry refers to economic activities, which are connected with conversion of resources into useful goods. Industries may be divided into 3 categories.
They are
1. Primary industries:
Primary industries are connected with the extraction and production of natural resources and reproduction and development of living organisms, plants, etc. Such industries are further divided into two.
(i) Extractive industries:
These industries extract products from natural resources. eg: mining, farrqing, hunting, fishing etc.

(ii) Genetic industries:
These industries are engaged in activities like rearing and breeding of animals, birds and plants. eg: diary faming, paultry farming, floriculture, pisciculture etc.

2. Secondary industries:
Secondary industries deal with materials extracted at the primary stage. Such goods may be used for consumption or for further production. Secondary industries are classified into two.
They are:
(i) Manufacturing industries:
Manufacturing industries engage in converting raw materials into finished goods. eg: Conversion of rubber into cotton, timber into furniture rubber into tyres etc. Manufacturing industries may be further divided into four categories. They are,

• Analytical industry which analyses and separates different elements from the same materials. eg: Oil refinery
• Synthetical industry which combines various ingredients into a new product. eg: cement
• Processing industry which involves successive stages for manufacturing finished products. eg: Sugar and paper industry.
• Assembling industry which assembles different component parts to make a new product. eg: television, car, computer, etc.

(ii) Construction industries:
These industries are involved in the construction of buildings, dams, bridges, roads etc.

3. Tertiary industries:
These are concerned with providing support services to primary and secondary industries. eg: Transport, banking, insurance, warehousing, communication, advertising etc.

Commerce:
Commerce is defined as all activities involving the removal of hindrances in the process of exchange of goods. It includes all those activities, which are necessary for the free flow of goods and services from the producer to the consumer. Commerce includes trade and auxiliaries to trade.
Commerce = Trade + auxiliaries to trade
Functions of commerce:

Various Hindrances	Remedies
Hindrance of person	Trade
Hindrance of place	Transportation
Hindrance of time	Warehousing
Hindrance of risk	Insurance
Hindrance of knowledge	Advertising
Hindrance of finance	Banking

Trade:
Trade refers to sale, transfer or exchange of goods. Trade may be classified into two broad categories
They are:

1. Internal trade
2. External trade

1. Internal, domestic ot dome trade:
is concerned with the buying and selling of goods and services within the geographical boundaries of a country. This may further be divided into two. They are:-
(a) Wholesale trade:
Under wholesale trade, the trader purchases goods in large quantities from the producers, and sells them in smaller quantities to the retailers.

(b) Retail trade:
Under the retail trade, the trader buys in comparatively smaller quantities from the wholesalers or producers and sells them to ultimate consumers.

2. External or Foreign trade:
Foreign trade consists of exchange of goods and services between two or more countries. Foreign trade may be divided in to three.

• Import trade: If goods are purchased from a foreign country, it is called import trade.
• Export trade: When goods are sold to a foreign country, it is known as export trade.
• Entrepot trade: When goods are imported for export to other countries, it is known as entrepot trade.

Auxiliaries to Trade (Aids to trade):
Activities which assist trade are called aids to trade or auxiliaries to trade.

1. Transport & Communication:
Transport facilitates the movement of raw material to the place of production and the finished products from factories to the place of consumption. Communication helps the producers, traders and consumers to exchange information with one another.

2. Banking & Finance:
Banking helps business activities to overcome the problem of finance. Commercial banks lend money in the form of overdraft, cash credit, loans and advances etc… and they also provide many services required for the business activity.

3. Insurance:
The goods may be destroyed while in production process or in transit due to accidents, or in storage due to fire or theft, etc. Insurance provides protection in all such cases.

4. Warehousing:
The goods should be stored carefully from the time they are produced till the time they are sold. This function is performed by warehouses.

5. Advertising:
Advertising helps in providing information about available goods and services and create in them a strong desire to buy the product.

Multiple Objectives of Business:
The main objectives of a business are:
1. Market standing:
A business firm can succeed only when it has a good market standing. Market standing refers to the position of an enterprise in relation to its competitors.

2. Innovation:
Innovation means developing new product or services orfinding new ideas and new methods of production and distribution. Innovation accelerates the growth of an enterprise.

3. Productivity:
Productivity is ascertained by comparing the value of output with the value of input. Every enterprise must aim at greater productivity through the best use of available resources.

4. Physical and financial resources:
The business must aim at maximum utilization of available physical and financial resources, i.e. men, material, money and machine in the best possible manner.

5. Earning Profit:
Earning maximum profit is the primary objective of every business. Profit is required for survival and growth of a business.

6. Manager performance and development:
Efficient managers are needed to conduct and co-ordinate business activities. So it is the objective of an enterprise to implement various programs for motivating the managers.

7. Worker performance and attitude:
Every enterprise must aim at improving its workers performance by providing fair salary, incentives, good working conditions, medical and housing facilities.

8. Social responsibility:
It refers to the obligation of business firm to contribute resources for solving social problems and work in a socially desirable manner.

Business Risks:
The term ‘business risks’ refers to the possibility of inadequate profits or even losses due to uncertainties or unexpected events. Business enterprises may face two types of risk, i.e. speculative risk and pure risk.

Speculative risks involve both the possibility of gain as well as the possibility of loss. It arise due to change in demand, change in price etc. Pure risks involve only the possibility of loss or no loss. The chance of fire, theft or strike is examples of pure risks.
Nature of Business Risks:

1. Business risks arise due to uncertainties.
2. Risk can be minimized, but cannot be eliminated. It is an essential part of business.
3. Degree of risk depends mainly upon the nature and size of business.
4. Profit is the reward for risk taking.

Causes of Business Risks: Business risks arise due to a variety of causes.
They are:

1. Natural Causes: it includes natural calamities like flood, earthquakes, lightning, heavy rains, famine, etc.
2. Human Causes: Human causes include dishonesty, carelessness or negligence of employees, strikes, riots, management inefficiency, etc.
3. Economic causes: These include change in demand, change in price, competition, technological changes etc.
4. Political Causes: Change in Govt, policies, taxation, licensing policy etc.

Starting a Business – Basic Factors:
Factors to be considered for starting a business:
1. Selection of line of business:
The first thing to be decided by any entrepreneur of a new business is the nature and type of business to be undertaken.

2. Size of the firm:
If the market conditions are favorable, the entrepreneur can start the business at a large scale. If the market conditions are uncertain and risks are high, a small size business would be better choice.

3. Choice of form of ownership:
The selection of a suitable form of business enterprise i.e. Sole proprietorship, Partnership or a Joint stock company is an important management decision. It depends on factors like nature of business, capital requirements, liability of owners, legal formalities, continuity of business etc.

4. Location of business enterprise:
Availability of raw materials and labour, power supply and services like banking, transportation, communication, warehousing, etc., are important factors while making a choice of location.

5. Financing:
Proper financial planning must be done to determine (a) the requirement of capital, (b) source from which capital will be raised and (c) the best ways of utilizing the capital in the firm.

6. Physical facilities:
Availability of physical facilities including machines and equipment, building and supportive services is a very important factor to be considered at the start of the business.

7. Plant layout:
Layout means the physical arrangement of machines and equipment needed to manufacture a product.

8. Competent worked force:
Every enterprise needs competent and committed employees to perform various activities so that physical and financial resources are converted into desired outputs.

9. Tax planning:
The promoter must consider in advance the tax liability under various tax laws and its impact on business decision.

Students can Download Chapter 8 Arrays Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 8 Arrays

Summary
An array is a collection of elements with same data type Or with the same name we can store many elements, the first or second or third, etc can be distinguished by using the index(subscript). The first element’s index is 0, the second elements index is 1, and so on.

Declaring arrays:
Suppose we want to find the sum of 100 numbers then we have to declare 100 variables to store the values. It is a laborious work. Hence the need for array arises.
Syntax: data_type array_name[size];
To store 100 numbers the array declaration is as follows
int n[100]; By this we store 100 numbers. The index of the first element is 0 and the index of last element is 99.

Memory allocation for arrays:
The amount of memory requirement is directly related to its type and size,

• int n[100]; It requires 2Bytes(for each integer) × 100 = 200 Bytes.
• float d[100]; It requires 4Bytes(for each float) × 100=400 Bytes.

Array initialization:
Array can be initialized in the time of declaration. eg: int age[4] = {16, 17, 15, 18};

Accessing elements of arrays:
Normally loops are used to store and access elements in an array.
eg:
int mark[50], i;
for(i=0;i<50;i++)
{
cout<<“Enter value for mark”<<i+1;
cin>>mark[i];
}
cout<<“The marks are given below:”;
for(i=0;i<50;i++)
cout<<mark[i];

Array operations:
Traversal:
Accessing all the elements of an array is called traversal.

Sorting:
Arranging elements of an array in an order(ascending or descending)
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. first smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Searching:
It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Two dimensional (2D) arrays:
Some occasions we have to store 6 different marks of 50 students. For this we use 2D arrays. An array with two subscripts is used.
eg: int mark[r][c]; Here r is the row and c is the column.

Declaring 2D arrays:
Syntax: datatype array_name[rows][columns];
The elements of this array is referred as mark[0][0], mark[0][1], mark[r – 1][c – 1].
eg: int m[5][5]; This array can store 5 × 5 = 25 elements.

Matrices as 2D arrays:
Matrix is a concept in mathematics that can be represented by 2D array with rows and columns. A nested loop(a loop contains another loop) is used to store and access elements in an array.

Multi-dimensional arrays:
3 Dimensional(3D) array is an example for this.
Syntax: data_type array_name[size1 ][size2][size3];
eg: int m[5][5][5]; This array can store 5 × 5 × 5 = 125 elements.

Students can Download Chapter 12 Internet and Mobile Computing Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 12 Internet and Mobile Computing

Summary
History of Internet:
Internet means international network of networks. The first form of Internet is ARPANET(Advanced Research Project Agency Network) started by US Department of Defence for their military during 1970’s. In 1989 a team lead by Tim Berners Lee introduced WWW(World Wide Web) by using the protocol HTTP. In 1998, Internet Corporation for Assigned Names and Numbers (ICANN) was established.

Internet:
It is a network of networks. It means that international network. We can transfer information between computers within nations very cheaply and speedily.

Intranet:
A private network inside a company or organisation is called intranet.

Extranet:
It allows vendors and business partners to access the company resources.

The hardware and software requirement for internet:

• A computer with a modem (internal/external)
• A telephone connection
• An account with an ISP
• A browser S/W eg: Internet Explorer or Mozilla…

Types of connectivity:
There are two ways to connect to the internet. First one dialing to an ISP’s computer or with a direct connection to an ISP.

Dial-up Connection:
Here the internet connection is established by dialing into an ISP’s computer. If ISP is not busy they verify the user name and password if it is valid they will connect our computer to the internet.lt uses Serial Line Internet Protocol (SLIP) or Point to Point Protocol (PPP). It is slower and has a higher error rate.

Direct connection:
In direct connection there is a fixed cable or dedicated phone line to the ISP. Here it uses ISDN (Integrated Services Digital Network) a high speed version of a standard phone line. Another method is leased lines that uses fibre optic cables.

Digital Subscribers Line (DSL) is another direct connection, this uses copper wires instead of fibre optic for data transfer. Direct connection provides high speed internet connection and error rate is less. Fibre To The Home(FTTH) uses optical fibers for data transmission.

Wireless broadband connectivity:
1. Mobile broadband:
Accessing Internet using wireless devices like mobile phones, tablet, USB dongles.

2. Wi MAX(Wireless Microwave Access):
It uses micro waves to transmit information across a network in a range 2 GHz to 11 GHz over very long distance.

3. Satellite broadband:
Accessing internet through satellite. A Very Small Aperture Terminal(VSAT) dish antenna and transceiver and modem are required at the user’s location. Expensive and high speed.

Internet access sharing methods:
One Internet connection can be shared among several computers using a LAN, Wi Fi or Li Fi.
1. Using LAN:
The Internet connection in a LAN can be shared among other computers in the network

2. Using Wi Fi(Wireless Fidelity):
It uses radio waves to transmit information across a network in a range 2.4 GHz to 5 GHz in short distance. Nowadays this technology is used to access internet in campuses, hyper markets, hotels by using Laptops, Desktops, tablet, mobile phones etc

3. Using Li Fi(Light Fidelity) network:
It is a fast optical(uses visible light for data transmission) version of Wi Fi. Its main component is a LED lamp that can transmit data and a photodiode that acts as a receiver.

Services on Internet:
1. www(World Wide Web):
This means this website address is unique and can be accessed each nook and corner of the world.

2. A browser is a piece of software that acts as an interface between the user and the internal working of the internet. With the help of a browser the user can search information on the internet and it allows user to navigate through the web pages. The different browsers are

• Microsoft internet explorer
• Mozilla Firefox
• Netscape Navigator
• Google Chrome
• Opera.

3. Web Browsing:

1. The browser determines the URL entered.
2. The browser asks the DNS for URLS corresponding IP address (Numeric address)
3. The DNS returns the address to the browser.
4. The browser makes a TCP connection using the IP address.
5. Then it sends a GET request for the required file to the server.
6. The server collects the file and send it back to the browser.
7. The TCP connection is released.
8. The text and the images in the web pages are displayed in the browser.

Search engines:
By using search engines we will get a variety of information. It is a newly developed tool that helped to search the information on the internet more effectively and easily. Search engines are programs that help people to locate information from crores of website on internet using a database that consists of references.

Users can interact with the search engine through the home page of the search engine. To get the information about artificial intelligence just type this in the box provided for it and click the search button. Search engines searches by using a particular search algorithm then displays the matching documents or web addresses.

Search engine use soft wares called spiders or bots to search documents and their web addresses. Spiders search the internet using the directions given by the search engines and prepare an index and stores it in a database. The searching algorithm searched this database when the users submits a request and create a web page displaying the matching results as hyperlinks.
eg: Google, Yahoo, Rediff etc.

E mail(Electronic mail):
It is used to send text, multi media messages between computers over internet. An example of an email id is jobi_cg@rediffmail.com. Here jobi_cg is the user name, rediffmail is the website address and .com is the top level domain which identifies the types of the organisation. To send an email we require an email address. Some websites provide free email facility.

To send an email first type the recipients address and type the message then click the send button. The website’s server first check the email address is valid, if it is valid it will be sent otherwise the message will not be sent and the sender will get an email that it could not deliver the message.

This message will be received by the recipient’s server and will be delivered to recipient’s mail box. He can read it and it will remain in his mail box as long as he will be deleted. Simple Mail Transfer Protocol(SMTP) is used.
The email message contains the following fields:

1. To: Recipient’s address will be enter here. Multiple recipients are also allowed by using coma.
2. CC: Enterthe address of other recipients to get a carbon copy of the message.
3. bcc: The address to whom blind carbon copies are to be sent. This feature allows people to send copies to third recipient without the knowledge of primary and secondary recipients
4. From: Address of the sender
5. Reply to: The emait address to which replies are to be sent.
6. Subject: Short summary of the message.
7. Body: Here the actual message is to be typed.

The advantages of email are given below:

1. Speed is high
2. It is cheap
3. We can send email to multiple recipients
4. Incoming messages can be saved locally
5. It reduces the usage of paper
6. We can access mail box anytime and from anywhere.

The disadvantages are:

1. It requires a computer, a modem, software and internet connection to check mail.
2. Some mails may contain viruses
3. Mail boxes are filled with junk mail. So very difficult to find the relevant mail.

Social media:
Various social medias are Internet forums, social blogs, microblogs etc.

1. Internet forums: It is an online discussion site where people can exchange information about various issues like social, political, educational etc in the text form.
2. Social blogs: Conducting discussions about . particular subjects by entries or posts. eg: Blogger.com
3. Microblogs: It allows users to exchange short messages, multi media files etc. eg: www.twitter.com
4. Wikis: In this we can give our contributions regarding various topics. eg: www.wikipedia.org
5. Social networks: By using these web sites we can post our data and’ view others data. eg: www.facebook.com
6. Content communities: By using these websites we can share multi media files. eg: www.youtube.com

Advantages of social media:

1. Bring people together: It allows people to maintain the friendship
2. Plan and organize events: It allows users to plan and organize events.
3. Business promotion: It helps the firms to promote their sales.
4. Social skills: There is a key role of the formation of society.

Disadvantages:

1. Intrusion to privacy: Some people may misuse the personal information.
2. Addiction: sometimes it may waste time and money.
3. Spread rumours: The news will spread very quickly and negatively.

Cyber Security:
It is used to provide protection of valuable information such as credit card information from unauthorized access, intentional access, deletion, etc. while shopping on the internet.

Computer virus:
A virus is a bad program or harmful program to damage routine working of a computer system. It reduces the speed of a computer. It may delete the useful system files and make the computer useless.

Worm:
It is a stand alone malware program that replicates itself in order to spread to other computers. It slows down the traffic by consuming the bandwidth. In 2000 a worm called “ILOVEYOU” is affected many computers.

Trojan horse:
It appears as a useful software but it is a harmful software and it will delete useful software or files.

Spams:
Sending an email without recipient’s consent to promote a product or service is called spamming. Such an email is called a spam.

Hacking:
It is a process of trespassing computer networks. Two types white hats and black hats. White hats hack the computer networks to test the security but black hats intentionally stealing valuable data or destroying data.

Phishing (Fishing):
It is an attempt to get others information such as usenames, passwords, bank a/c details etc by acting as the authorized website. Phishing websites have URLs and home pages similar to their original ones and mislead others , it is called spoofing.

Denial of Service(DoS) attack:
Its main target is a Web server. Due to this attack the Web server/computer forced to restart and this results refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server.

Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Man in the Middle attacks:
It is an attack in which an attacker secretly intercepts electronic messages send by the sender to the receiver and then modifies the message and retransmit it to the receiver.

To prevent this type of attack encrypted connections such as HTTPS(HTTP Secure), SFTP(Secure FTP) etc, must be used, that will be displayed in the URL.

Preventing network attacks
Firewall:
It is a system that controls the incoming and outgoing network traffic by analyzing the data and then provides security to the computer network in an organization from other network (internet).

Antivirus scanners:
It is a tool used to scan computer files for viruses, worms and Trojan horses and cure the infected system. If any fault found it stops the file from running and stores the file in a special area called Quarantine(isolated area) and can be deleted later.

Cookies:
Cookies are small text files that are created when we visit a website that keep track of our details. This information will help the hacker to use it for malicious purposes. It acts as a spyware.

Guidelines for using computers over internet:

• Emails may contain Viruses so do not open any unwanted emails
• Download files from reputed sources(sites)
• Avoid clicking on pop up Advt.
• Most of the Viruses spread due to the use of USB drives so use cautiously.
• Use firewall in your computer
• Use anti virus and update regularly
• Take backups in a regular time intervals

Mobile Computing:
The advancements in computing technology have led to the developments of more computing power in hand held devices like laptops, tablets, smart phones, etc. Nowadays people are able to connect to others through internet even when they are in move.

Mobile communication:
The term ‘mobile’ help the people to change their life styles and become the backbone of the society. Mobile communication networks do not require any physical connection.

Generations in mobile communication:
The mobile phone was introduced in the year 1946. Early stage it was expensive and limited services hence its growth was very slow. To solve this problem, cellular communication concept was developed in 1960’s at Bell Lab. 1990’s onwards cellular technology became a common standard in our country.
The various generations in mobile communication are
1.First Generation networks(1 G):
It was developed around 1980, based on analog system and only voice transmission was allowed.

2. Second Generation networks (2G):
This is the next generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) were introduced. GSM and CDMA standards were introduced by 2G.
(i) Global System for Mobile(GSM):
It is the most successful standard. It uses narrow band TDMA(Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).
(a) GPRS(General Packet Radio Services):lt is a packet oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically charged based on volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

(b) EDGE(Enhanced Data rates for GSM Evolution):
It is three times fasterthan GPRS. It is used for voice communication as well as an internet connection.

(ii) Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users, and provide better security.

3. Third Generation networks(3G):
It allows high data transfer rate for mobile devices and offers high speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and hand sets required.

4. Fourth Generation networks(4G):
lt is also called Long Term Evolution(LTE) and also offers ultra broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

Mobile communication services:
1. Short Message Service(SMS):
It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems. It uses the protocol SS7(Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/12/1992 from a PC to a mobile phone on the Vodafone GSM network in UK.

2. Multimedia Messaging Service (MMS):
It allows sending Multi Media(text, picture, audio and video file) content using mobile phones. It is an extension of SMS.

3. Global Positioning System(GPS):
It is a space- based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites. The system provides critical capabilities to military, civil and commercial users around the world.

It is maintained by the United States government and is freely accessible to anyone with a GPS receiver. GPS was created and realized by the U.S. Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

Smart Cards:
A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data. The advantages are it is secure, intelligent and convenient. The smart card technology is used in SIM for GSM phones. A SIM card is used as an identification proof.

Mobile operating system:
It is an OS used in hand held devices such as smart phone, tablet, etc. It manages the hardware, multimedia functions, Internet connectivity,etc. Popular OSs are Android from Google,iOS from Apple, BlackBerry OS from Black Berry and Windows Phone from Microsoft.

Android OS:
It is a Linux based OS forTouch screen devices such as smart phones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends. In 2005, Google acquired this. A team led by Rubin developed a mobile device platform powered by the Linux Kernel.

The interface of Android OS is based on touch inputs like swiping, tapping, pinching in and out to manipulate on screen objects. In 2007 onwards this OS is used in many mobile phones and tablets. Android SDK(Software Development Kit) is available to create applications(apps) like Google Maps, FB, What’s App, etc.

It is of open source nature and many Apps are available for free download from the Android Play Store hence increase the popularity.
Different Android Versions are shown below

Students can Download Chapter 11 Computer Networks Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 11 Computer Networks

Summary
Computer network:
Two or more computers connected through a communication media that allows exchange of information between computers is called a Computer Network. Eg: Internet

Need for network:
The advantages of Networks are given below.
1. Resource sharing:
All the computers in a network can share software (programs, data ) and hardware (printer, scanner, CD drive, etc.).

2. Reliability:
If one computer fails, the other computer can perform the work without any delay. This is very important for banking, air traffic control and other application.

3. Price Vs Performance:
A main frame computer can be 10 times faster than a PC but it costs thousand times a PC. Therefore instead of a main frame 10 personal computers are used with less cost and same performance.

4. Communication Medium:
It is a powerful communication medium. We can exchange information between computers in a network.

5. Scalable:
This means, System performance can be increased by adding computers to a network.

Terminologies:

1. Bandwidth: The maximum amount of data that can be transmitted by the medium measured in Hertz.
2. Noise: It is the unwanted electrical or electromagnetic interferences that adversely affect the transmitted data signals.
3. Node: A computer or an I/O device connected to a network is called Node.

Data communication system:
Communication is the exchange of information between two human beings. But data communication is the exchange of information between two computers(devices).

1. Message: It is the data/information to be transmitted from one computer to another
2. Sender: It is a computer or a device that sends data. It is also called.source or transmitter
3. Receiver: It is a computer ora device that receives data
4. Medium: It is the path through which message transmitted from the sender to the receiver. There are two types Guided and Un Guided media.
5. Protocol: The rules and conventions for transmitting data.

Communication Medium:
There are two types guided and unguided.
Guided Media:
1. Twisted Pair cable:
2 types unshielded twisted pair and shielded twisted pair. Two copper wires individually insulated and twisted around each other and put in a plastic cover.

2. Coaxial cable:
A sturdy copperwire is insulated by plastic, it is covered just like a mesh by a conductor which is enclosed in an protective plastic coating. It is expenssive, less flexible and more difficult to install. But it is more reliable and carry for higher data rates.

3. Optical fibre:
These are made of glass fibres that are enclosed in a plastic jacket. It uses light instead of electrical signals. The light sources are LED or ILD.

Unguided Media:

1. Radio waves: It transmits data at different frequencies ranging from 3 kHz. to 300 GHz.
2. Microwaves: Microwave signals can travel in straight line if there is any obstacle in its path, it can’t bend. So it uses tall towers instead of short one.
3. Infrared waves: These waves are used for transmitting data in short distance and its frequency range is 300 GHz to 400 GHz.

Wireless communication technologies using:
radio waves
1. Bluetooth:
This technology uses radio waves in the frequency range of 2.402 GHz to 2.480 GHz. And transmit data in short distance. Mobile phones, Laptops, tablets etc use Bluetooth technology to transmit data.

2. Wi Fi(Wireless Fidelity):
It uses radio waves to transmit information across a network in a range 2.4 GHz to 5 GHz in short distance. Nowadays this technology is used to access internet in Laptops, Desktops, tablets, Mobile phones etc.

3. Wi MAX(Wireless Microwave Access):
It uses micro waves to transmit information across a network in a range 2 GHz to 11 GHz over very long distance.

4. Satellites:
By using satellite we can communicate from eny part of the world to any other. The ground stations are connected via the satellite. The data signals transmitted from earth to satellite (uplink) and from the satellite to the earth (downlink).

Data communication devices:
It acts as an interface between computer and the communication channel

Network Interface Card (NIC):
This device enables a computer to connect to a network and transmit information.

Hub:
It is a small, simple and inexpensive device used to connect computers(devices) to a network. If a computer wants to transmit data to another computer. First it sends to the hub, the hub retransmits this data to all other computers.

Each and every computer gets the data and check whether it is for them or not. It increases the network traffic and hence the transmission speed is low.

Switch:
It is an expensive device used to connect computers(devices) to a network. Unlike hub, switch transmit data not to all computers, it retransmits data only to the intended computer. So the traffic is less and speed is high

Repeater:
It is a device used to strengthen weak signals on the network.

Bridge:
It is a device used to link same type of networks.

Router:
It is similar to a bridge, but it can connect two networks with different protocols.

Gateway:
It is used to connect two different networks with different protocols.

Data terminal equipments:
These devices are used to control data flow to and from a computer

Modem:
It is a device used to connect the computer to the internet. It converts digital signal into analog signal (modulation) and vice versa (Demodulation)

Multiplexer:
It combines the inputs from different channels of a medium and produces one output.

Network topologies:
Physical or logical arrangement of computers on a network is called structure or topology. It is the geometrical arrangement of computers in a network. The major topologies developed are star, bus, ring, tree and mesh.
1. Star Topology:
A star topology has a server all other computers are connected to it. If computer A wants to transmit a message to computer B. Then computer A first transmit the message to the server then the server retransmits the message to the computer B.

That means all the messages are transmitted through the server. Advantages are added or remove workstations to a star network is easy and the failure of a workstation will not effect the other. The disadvantage is that if the server fails the entire network will fail.

2. Bus Topology:
Here all the computers are attached to a single cable called bus. Here one computer transmits all other computers listen. Therefore it is called broadcast bus. The transmission from any station will travel in both the direction.

The connected computers can hear the message and check whether it is for them or not. Advantages are add or remove computer is very easy. It requires less cable length and the installation cost is less. Disadvantage is fault detection is very difficult because of no central computer.

3. Ring Topology:
Here all the computers are connected in the shape of a ring and it is a closed loop. Here also there is no central computer. Here a computer transmits a message, which is tagged along with its destination computer’s address.

The message travels in one direction and each node check whether the message is for them. If not, it passes to the next node. It requires only short cable length. If a single node fails, at least a portion of the network will fail. To add a node is very difficult.

4. Hybrid Topology:
It is a combination of any two or more network topologies. Tree topology and mesh topology can be considered as hybrid topology.
(a) Tree Topology:
The structure of a tree topology is the shape of an inverted tree with a central node and branches as nodes. It is a variation of bus topology. The data transmission takes place in the way as in bus topology. The disadvantage is that if one node fails, the entire portion will fail.

(b) Mesh Topology:
In this topology each node is connected to more than one node. It is just like a mesh (net). There are multiple paths between computers. If one path fails, we can transmit data through another path.

Types of networks:
The networks are classified into the following based upon the amount of geographical area that covers.
(i) Personal Area Network(PAN):
It is used to connect devices situated in a small radius by using guided media or unguided media

(ii) Local Area Network (LAN):
This is used to connect computers in a single room, rooms within a building or buildings of one location by using twisted pair wire or coaxial cable. Here the computers can share Hardware and software. Data transferrate is high and error rate is less, eg: The computers connected in a school lab.

(iii) Metropolitan Area Network (MAN):
A Metropolitan Area Network is a network spread over a city. For example a Cable TV network. MAN have lesser speed than LAN and the error rate is less. Here optical fiber cable is used.

(iv) Wide Area Network (WAN):
This is used to connect computers over a large geographical area. It is a network of networks. Here the computers are connected using telephone lines or Micro Wave station or Satellites. Internet is an example for this.

LAN and MAN are owned by a single organization but WAN is owned by multiple organization. The error rate in data transmission is high.

Logical classification of networks:
Peer to peer:
In this configuration all the computers have equal priority. That means each computer can function as both a client and a server. There is no dedicated server.

Client-Server:
In this configuration a computer is powerful which acts as a dedicated server and all others are clients (work stations). A Server fulfils the needs of the clients.

1. File Server: A computer that stores and manages files for other devices on a network
2. Web Server: A computer that handles the requests for web pages.
3. Print Server: A computer that handles the print jobs from other computers on a network.
4. Database Server: A computer that manages the database.

Network protocols:
A protocol is a collection of rules and regulations to transfer data from one location to another. Transmission Control Protocol (TCP), which uses a set of rules to exchange messages with other Internet points at the information packet level. Internet Protocol (IP), which uses a set of rules to send and receive messages at the Internet address level
1. FTP:
File Transfer Protocol which is used for transferring files between computers connected to local network or internet.

2. HTTP:
is a protocol used for WWW for enabling the web browse to access web server and request HTML documents.

3. DNS (Domain Name System):
When we type web sites address in the address bar, the browser determines the URL and asks the DNS for URLS corresponding IP address (Numeric address). The DNS returns the address to the browser.

Identification of computers over a network:
A computer gets a data packet on a network, it can identify the sender’s address easily. It is similar to our snails mail, each letter is stamped in sender’s post office as well as receiver’s post office.

Media Access Control(MAC) address:
It is a unique 12 digit hexadecimal number(IMEI for mobile phones, it is a 15 digit decimal number) assigned to each NIC by its manufacturer. This address is known as MAC address and its permanent. It is of the form. MM:MM:MM:SS:SS:SS.

The first MM:MM:MM contains the ID number of the adapter company and the second SS:SS:SS represents the serial number assigned to the adapter by the company.

Internet Protocol (IP) address:
An IP address has 4 parts numeric address. Each parts contains 8 bits. By using 8 bits we can represent a decimal number between 0 to 255(28 = 256 numbers). Each part is separated by dot. A total of 4 × 8 = 32 bits used. But nowadays 128 bits are used for IP address.

Uniform Resource Locator(URL):
Every resource on the internet has a unique URL. Mainly it has three parts
eg: http://www.hscap.kerala.gov.in /index.html.

• http: http means hypertext transfer protocol. It is a protocol used to transfer hypertext.
• www: World Wide Web. With an email address we can open our mail box from anywhere in the world.
• hscap.kerala: It is a unique name. It is the official website name of Single Window System
• gov: It is the top level domain. It means that it is a government organization’s website.
• in: It is the geographical top level domain. It represents the country, in is used for India.
• index.html: It represents the file name.

Top Level Domain Names:

• .com The site register for commercial purpose
• .edu The site register for educational purpose
• .gov The site register by Government agencies
• .mil The site register for military services
• .net The site register for network purpose
• .org The site register by organizations

Country Specific Domain Names:

• .in India
• .au Australia
• .ca Canada
• .ch China
• .jp Japan
• .us United States of America

Students can Download Chapter 11 The p Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Plus One Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The aqueous solution of borax is
a) Acidic
b) Alkaline
c) Neutral
d) Amphoteric
Answer:
b) Alkaline

Question 2.
Say TRUE or FALSE.
Boron in aqueous solution forms B³⁺ ion.
Answer:
False

Question 3.
Which of the halide of group 14 does not exist?
a) CF₄
b) Cl₄
c) SiF₄
d) Pbl₄
Answer:
d) Pbl₄

Question 4.
Orthoboric acid, H₃BO₃ is a
a) Protonic acid
b) Arrhenius acid
c) Lewis acid
d) Bronsted-Lowery acid
Answer:
c) Lewis acid

Question 5.
The zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is _________ .
Answer:
ZSM-5

Question 6.
Dry ice is __________ .
Answer:
Solid CO₂

Question 7.
Thermodynamically most stable allotrope of carbon is __________ .
Answer:
Graphite

Question 8.
The alkalimetal used in solar cells is __________ .
Answer:
Cs (Caesium)

Question 9.
Acidity is in the order
Answer:
BBr₃ > BCl₃ > BF₃

Question 10.
AlCl₃ fumes in most air because
Answer:
HCl is formed due to hydrolysis in moist air

Plus One Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
Silicon belongs to the carbon family. Graphite is an important allotrope of carbon. But Si does not form an analogue of graphite. What are the possible reasons?
Answer:
Silicon atom is much bigger in size than carbon. Si-Si bond energy is less than C-C bond energy and also Si does not form compounds in sp² hybrid state.

Question 2.
Boron is an element with atomic number 5.
a) Write down the electronic configuration of boron.
b) Mention any two uses of boron.
c) Write down some compounds of boron.
Answer:
a) 1s²2s²2p¹
b) To increase the hardness of the steel.
Used as the semiconductor in electronic devices.
c) Diborane, borax, orthoboric acid, boron trifluoride

Question 3.
Answer the questions, with the help of the following:

• Hard solid
• Melting point above 450 K.
• Four allotropic forms are known.
• Low electrical conductivity.
• Mass number! 3

1. Which is the element?
2. Write any two uses of this element.
3.
4. Write any three compounds of this element.
Answer:
1. Boron
2. To increase hardness of steel
As semiconductor in electric devices
3. 2B
4. Diborane, Borax, Boric acid

Question 4.
1. Write any two allotropic forms of carbon.
2. C + ½ O₂ → …………….
3. Write any two uses of carbon monoxide.
4. How is carbon dioxide produced?
5. What is the hardest element/form of an element in the world?
Answer:
1. Diamond, Graphite
2. CO (Carbon monoxide)
3. Reducing agent in metallurgy
Used in the manufacture of methanol
4. Carbon dioxide is formed by the complete combustion of carbon.
C + O₂ → CO₂
5. Diamond

Question 5.
From the compounds of group 14 elements write an appropriate example for each of the following:

No.	Type of compound	Name/ Formulae of example
1.	A strong reducing oxide
2.	A giant covalent oxide
3.	A strongly reducing chloride
4.	A covalent chloride not hydrolysed by water

Answer:
1-CO
2-SiO₂
3-SnCl₂
4-CCl₄

Question 6.
Diborane has an unusual structure. Justify the statement with figure.
Answer:
In diborane, each Batom uses sp³ hybrid orbitals for bonding. Out of the four sp³ hybrid orbitals on each B atom, one is without an electron. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 7.
Match the following:

A	B
CO	i. A semi conductor in electronic devices
H2	ii. Reducing agent in metallurgy
O3	iii. Thermal decomposition of Ammonia
Boron	iv. Used as a chemical reagent in organic chemistry

Answer:
CO – ii,
N₂ – iii,
O₃ – iv,
Boron – i

Question 8.
BCl₃ fumes in moist air.
1. Give reason.
2. Write down a balanced equation that can reveal the answer.
Answer:
1. When water of moist air reacts with boron halide hydrogen chloride is formed which causes fumes,

2. BCl₃ + 3H₂O → H₃BO₃ + 3HCl

Question 9.
‘Generally, non-metal oxides are basic.’
1. Do you agree?
2. What do you meant by oxides?
3. Which are the different types of oxides?
4. Give examples for each type of oxides.
Answer:
1. No. Generally, non metallic oxides are acids.
2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides.
3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide.
4. Acidic oxide → SO₂, NO₂
Basic oxide → MgO
Amphoteric oxide → Al₂O₃, ZnO
Neutral oxide → CO, N₂O

Question 10.
Match the following:
1. Borane – H₃BO₃
2. Boric acid – Na₂B₄O₇.10H₂O
3. Borax – Amphoteric oxide
4. Al₂O₃ – Boron hydride
Answer:
1. Borane – Boron hydride
2. Boric acid – H₃BO₃
3. Borax – Na₂B₄O₇.10H₂0
4. Al₂O₃ – Amphoteric oxide

Question 11.
How diborane reacts with
1. Oxygen?
2. Water?
Answer:
1. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.
B₂H₆ + 3O₂ → B₂O₂ + 3H₂O; ∆_rH^Θ -1976 kJ mol^-1

2. Diborane is readily hydrolysed by water to give boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃(aq) + 6H₂(g)

Question 12.
1. CCl₄ cannot be hydrolysed. Give reason.
2. Draw the structure of the dimer of AlCl₃.
Answer:
1. Carbon has no d orbitals to accommodate the lone pair of electrons from oxygen atom of H₂0.
2.

Question 13.
1. Draw the structure of boric acid.
2. Starting from borax how will you prepare boric acid? (Write the chemical equation).
Answer:
1.

2. Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃.

Question 14.
CO₂ is a gas but SiO₂ is a solid. Give reason.
Answer:
CO₂ exists as discrete molecules due to the formation of pπ -pπ double bond between carbon and oxygen. But SiO₂ has a three dimensional network structure in which each Si atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

Question 15.
1. What are zeolites?
2. What is ZSM-5?
Answer:
1. Zeolites are aluminosilicates with three-dimensional network structure in which Al atoms replace few Si atoms. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge of aluminosilicate anion.

2. ZSM-5 is a zeolite catalyst used in petrochemical industry to convert alcohols directly into gasoline.

Plus One Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Explain the following:
1. Allotropy
2. Coke and Charcoal
Answer:
1. The phenomenon of existence of an element in two or more forms, which have different physical properties but almost similar chemical properties, is called allotropy and the different forms are called allotropes.

2. Coke and Charcoal are amorphous forms of carbon. Coke is formed-by the destructive distillation of coal. It is used as a fuel and also as a reducing agent in metallurgy.

Question 2.
Give justifications.
1. The first ionisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.
2. Graphite is a better lubricant on moon than that on earth.
Answer:
1. This is because carbon has greater nuclear charge. For second ionisation enthalpy, an electron is to be removed from 2p of carbon while in boron the second electron is in 2s orbital. Removal of a 2s electron is more difficult due to the high penetrating power of 2s orbital.

2. Graphite contains hexagonal sheets of carbon held together by weak van der Waals’ forces and low density. In the moon, the different layers experience less weight due to less gravitational force and the density is still reduced. The lighter layers can easily slide over another to make graphite more lubricating on the moon than on earth.

Question 3.
1. Identify the compound with the following structure:

2. Mention the angles a & b.
3. Write down the corresponding elements in the figure.
1……………..
2……………..
3……………..
4……………..
5……………..
6……………..
Answer:
a) Diborane(B₂H₆).
b) a-97°
b-120°
c) 1-H, 2-H, 3-B, 4-B, 5-H, 6-H

Question 4.
The simplest boron hydride is diborane.
1. Write down the molecular formula of diborane.
2. Distinguish between terminal hydrogen and bridging hydrogen atoms of diboran.
Answer:
1. B₂H₆

2. In diborane, four hydrogen atoms and two boron atoms are present in a single plane. These hydrogen atoms are called terminal atoms. The terminal B-H bonds are regular two centre-two electron bonds. The remaining two hydrogen atoms above and below this plane are called bridging hydrogen atoms. The two bridge bonds (B-H-B) are three centre-two electron bonds.

Question 5.
1. How is orthoboric acid prepared?
2. Account for the acidic nature of orthoboric acid.
Answer:
1. Orthoboric acid is prepared by acidifying an
aqueous solution of borax.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃

2. Orthoboric acid is a weak nonobasic acid. It acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)₃ + 2HOH → [B(OH)₃]^– + H₃O⁺

Question 6.
1. How is diborane prepared in the laboratory?
2. BCl₃ is a good Lewis acid. Why?
Answer:
1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine.
2NaBH₄ + l₂ → B₂H₆ + 2Nal + H₂

2. In BCl₃, the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

Question 7.
1. Name the allotropes of carbon.
2. Carbon monoxide is highly poisonous. Do you agree? Justify.
Answer:
1. Graphite, diamond and fullerene.

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxyhaemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

Question 8.
1. Diamond is hard and non conducting while graphite is soft and conducting. Why?
2. Explain the action of heat on boric acid.
3. What is inorganic benzene? How is it formed?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. Due to this closely packed arrangement, it is hard. Since, there are no free electrons it is an insulator. But in graphite, the carbon atoms are in sp² hybridised state. It is a good conductor due to the presence of delocalised electrons between the layers. Graphite cleaves easily between the layers. Therefore, It is very soft and slippery.

2. On heating, orthoboric acid above 370 K forms metaboric acid, HBO₂ which on further heating yields boric oxide, B₂O₃.

3. B₃N₃H₆ is called Inorganic benzene. It is obtained by heating diborane with ammonia.

Question 9.
1. Explain the difference in properties of diamond and graphite on the basis of their structures.
2. How do you explain the lower atomic radius of gallium as compared to aluminium?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. As a result of this hybridisation, a closely packed arrangement is present in diamond and it explains the hardness of diamond. Due to the absence of electrons, diamond is an insulator.

In graphite, the carbon atoms are in sp² hybridised state. As a result of this hybridisation, graphite has a layered structure with hexagonal rings. Each layer can slide over the other which explains the lubricating property of graphite.

2. This can be explained on the basis of the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).

Question 10.
1. Boron resembles silicon in many of its properties. What is this resemblance generally known as?
2. What is dry ice? What is it used for?
3. What are silicones? How do they differ from silicates?
Answer:
1. Diagonal relationship

2. Carbon dioxide can be obtained as a solid in the form of dry ice, by allowing the liquified CO₂ to expand rapidly.
Dry ice is used as a refrigerant for ice-cream and frozen food.

3. Silicones are a group of organosilicon polymers containing R₂SiO repeating units.
Silicates are minerals with SiO₄^4- as the basic structural unit. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units.

Plus One Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
Match the following:

Answer:
1 – c – f
2 – a – h
3 – b – g
4 – d – e

Question 2.
a) Elements of group 13 are
1) Al, Cr, Cd, Ga, Ti
2) B, C, Si, Ga, Ti
3) B, Al, Ga, Ti, In
4) B, Al, Cr, Ca, Ti
b) Which of the following is not a mineral of boron? (borax, bauxite, colemanite, tincal)
c) In the reaction between NH₃ and BF₃ ammonia acts as
(Lewis base, Lewis acid, brownsted base)
d) Diamond and Graphite are carbon’s
(isotropic forms, allotropic forms, isotopic forms, amorphous form)
Answer:
a) 3) B, Al, Ga, TI, In
b) Bauxite
c) Lewis base
d) allotropic forms

Question 3.
The hydrides of boron are called boranes.
1. How diborane reacts with ammonia?
2. Account for the exceptional hardness of diamond.
Answer:
1. Diborane reacts with ammonia to give B₂H₆.2NH₃
initially, which can be formulated as [BH₂(NH₃)₂]⁺[BH₄]^–. Further heating gives borazine, B₃N₃H₆ known as ‘inorganic benzene’ in view of its ring structure with alternate B-H and N-H groups,

2. In diamond, all the carbon atoms are in sp3 hybridised state. As a result of this hybridisation, a rigid three dimensional network of carbon atom is generated with directional covalent bonds throughout the lattice. It is very difficult to break this network structure.

Plus One Chemistry The p Block Elements NCERT Questions and Answers

Question 1.
Why does boron trifluoride behave as a Lewis acid? (2)
Answer:
The B atom in BF₃ has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F^–, NH₃, (C₂H₅)₂O, RCH₂OH etc. and thus behaves as a Lewis acid.

Question 2.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? (2)
Answer:
Due to increase in atomic size and screening effect the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result, there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.

Question 3.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? (2)
Answer:
The B atom in BCl₃ has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.
BCl₃ + 3H₂O → H₃BO₃ + 3HCl

In contrast, C atom in CCl₄ has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H₂O molecule and hence CCl₄ does not undergo hydrolysis in water.

Question 4.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is buddled through. Give reasons. (3)
Answer:
Anhydrous HF is a covalent compound and is strongly H-bonded. Therefore, it does not give free F^– ions and hence AlF₃ does not dissolve in HF. In contrast, NaF is an ionic compound and hence F^– ions are easily available. As a result, it combines with AlF₃ to form the soluble complex.
3NaF + AlF₃ → Na₃[AlF₃]

On bubbling gaseous BF₃, AlF₃ is precipitated. It is because BF₃ is a stronger Lewis acid than AlF₃. This is attributed to smaller size and higher electronegativity of Boron. As a result of this B has much higher tending to form complexes than Al. Therefore, when BF₃ is added to the above solution, AlF₃ gets precipitated.
Na₃[AlF₆] + 3BF₃ → 3Na[BF₄] + AlF₃(s)

Question 5.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences. (3)
Answer:
Aluminium shows a uniform oxidation state of+3 in its compounds. Like aluminium, thallium also shows +3 oxidation state in some of its compounds like TlCl₃, Tl₂O₃, etc. Al is known to form octahedral complexes like [AlF₆]^3-. Similarly, Tl also forms octahedral complexes as [TlF₆]^3-.

Thallium also resembles group 1 metals. Like group 1 metals which show a stable oxidation state of +1 in their compounds Tl, due to inert pair effect, also shows +1 oxidation state in some of its compounds such as Tl₂O, TlCl, TlClO₄, etc. Similarly, like group 1 oxides, Tl₂O is strongly basic.

Students can Download Chapter 3 Components of the Computer System Notes, Plus One Computer Science Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Notes Chapter 3 Components of the Computer System

Summary
Hardware:
The tangible parts of a computer that we can touch and see are called hardware. Eg: Monitor, Keyboard, Mouse, CPU, Etc.

Processors:
It is the brain of the computer and consists of three components

• Arithmetic Logic Unit(ALU) – As the name implies it performs all calculations and comparison operations.
• Control Unit(CU) – It controls overall functions of a computer
• Registers – It stores the intermediate results temporarily.

A CPU is an Integrated Circuit(IC) package contains millions of transistors and other components. Popular Processors are Intel core i3, core i5, core i7, AMD Quadcore etc.

Important registers inside a CPU are:

1. Accumulator: After performing an operation (arithmetic or logical) the result is stored in the accumulator
2. Memory Address Register(MAR): It stores the address of memory location to which data is either read or written by the processor.
3. Memory Buffer Register (MBR): It stores the data, either to be written to or read from the memory by the processor.
4. Instruction Register(IR): It stores the instructions to be executed by the processor.
5. Program Counter(PC): It stores the address of the next instruction to be executed by the processor.

Motherboard:
It is a Printed Circuit Board(PCB). All the major components like Processor (Remember the processor must be compatible with the Motherboard), RAM, ROM, HDD, Graphics card, Sound card, etc are connected to the Motherboard.

Peripherals and Ports:

1. Serial Port: It transmits data one bit at a time(eg: 101000001010 ). Its transmission speed is low but it is cheaper. It is suitable to transmit data over long distance.
2. Parallel port: It can transmit data more than one bit at a time. It is faster and used to connect printer.
3. USB (Universal Serial Bus) port: It has high bandwidth hence it is faster. Nowadays it is used to connect all the devices like keyboard,mouse,printer, scanner, pen drive, digital camera, mobile phones, dongle etc.
4. LAN port: By using this port we can connect our computer to another network by a cable.
5. PS/2(Personal System/2) port: It is introduced by IBM for connecting keyboard and mouse earlier.
6. Audio ports: It is used to connect audio devices like speakers, mic etc.
7. Video Graphics Array(VGA) port: It is introduced by IBM to connect monitor or LCD projector to a computer.
8. High Definition Multimedia Interface(HDMI): Through this port we can connect high definition quality video and multi-channel audio over a single cable.

Memory:
Storage Unit(Memory Unit): A computer has huge storage capacity. It is used to store data and instructions before starts the processing. Secondly it stores the intermediate results and thirdly it stores information(processed data), that is the final results before send to the output unit(Visual Display Unit, Printer, etc). Memory measuring units are given below.

• 1 bit = 1 or 0(Binary Digit)
• 4 bits = 1 Nibble
• 8 bits = 1 Byte
• 1024 Bytes = 1 KB(Kilo Byte)
• 1024 KB = 1 MB(Mega Byte)
• 1024 MB = 1 GB(Giga Byte)
• 1024 GB = 1 TB(Tera Byte)
• 1024 TB = 1 PB(Peta Byte)

1. Primary Storage alias Main Memory:
It is further be classified into Two – Random Access Memory (FRAM) and Read Only Memory(ROM). The one and only memory that the CPU can directly access is the main memory at a very high speed. It is expensive hence storage capacity is less.

RAM is volatile(when the power is switched off the content will be erased) in nature but ROM is non volatile(lt is permanent). In ROM a “boot up” program called BIOS(Basic Input Output System) is stored to “boots up” the computer when it switched on. Some ROMs are given below.

• PROM(Programmable ROM): It is programmed at the time of manufacturing and cannot be erased.
• EPROM (Erasable PROM): It can be erased and can be reprogrammed using special electronic circuit.
• EEPROM (Electrically EPROM): It can be erased and rewritten electrically

Cache Memory:
The processor is a very high speed memory but comparatively RAM is slower than Processor. So there is a speed mismatch between the RAM and Processor, to resolve this a high speed memory is placed in between these two this memory is called cache memory. Commonly used cache memories are Level(L1) Cache(128 KB), L2(1 MB),L3(8 MB), L4(128MB).

2. Secondary Storage alias Auxiliary Memory:
Because of limited storage capacity of primary memory its need arises. When a user saves a file, it will be stored in this memory hence it is permanent in nature and its capacity is huge. Eg: Hard Disc Drive(HDD), Compact Disc(CD), DVD, Pen Drive, Blu Ray Disc etc.
(i) Magnetic storage device:
It uses plastic tape or metal/plastic discs coated with magnetic material. .
(a) Hard Disk:
Instead of flexible or soft disk it uses rigid material hence the name hard disk. Its storage capacity and data transfer rate are high and low access time. These are more lasting and less error prone. The accessing mechanism and storage media are combined together in a single unit and connect to the motherboard via cable.

(ii) Optical storage device:
(a) Optical Disk:
The high power laser uses a concentrated, narrow beam of light, which is focuses and directed with lenses, prisms and mirrors for recording data. This beams burns very very small spots in master disk, which is used for making molds and these molds are used for making copies on plastic disks.

A thin layer of aluminium followed by a transparent plastic layer is deposited on it. The holes made by the laser beam are called pits, interpreted as bit 0 and unburned areas are called lands interpreted as bit 1. Lower power laser beam is used to retrieve the data.

(b) DVD(Digital Versatile Disc):
It is similar to CD but its storage capacity is much higher. The capacity of a DVD starts from 4.7 GB

(c) Blu-ray Disc:
It is used to read and write High Definition video data as well as to store very huge amount of data. While Cd and DVD uses red laser to read and write but it uses Blue-Violet laser, hence the name Blu ray disc. The blue violet laser has shorter wavelength than a red laser so it can pack more data tightly.

(iii) Semiconductor storage (Flash memory):
It uses EEPROM chips. It is faster and long lasting.

• USB flash drive: It is also called thumb drive or pen drive. Its capacity varies from 2 GB to 32 GB.
• Flash memory cards: It is used in Camera, Mobile phones, tablets etc to store all types of data.

Input / Output devices:
It is used to supply: data to the computer for processing
1. Keyboard:
It is the most widely used device to input information in the form of words, numbers etc. There are 101 keys on a standard keyboard. The keys on the keyboard are often classified into alpha numeric keys (A to Z, 0 to 9), function keys (F1 to F12), special purpose keys (Special characters), cursor movement keys (arrow keys). While pressing a key, the corresponding code’s signal is transmitted to the computer.

2. Mouse:
It is a pointing device, that controls the movement of the cursor, or pointer as a display screen. A mouse has two or three buttons, it is often used in GUI oriented computers.

Under the mouse there is a ball, when the mouse moves o.n a flat surface this ball also moves. This mechanical motion is converted into digital values that represents x and y values of the mouse movement.

3. Light Pen:
It is an input device that use a lightsensitive detector to select objects directly on a display screen using a pen. Light pen has a photocell placed in a small tube. By using light pen, we can locate the exact position on the screen.

4. Touch screen:
It allows the user to enter data by simply touching on the display screen. This technology is applied in tablets, cell phones, computers etc.

5. Graphic tablet:
It consists of an electronic writing area. We can create graphical images by using a special pen.

6. Touchpad:
It is a pointing device found on the portable computers(lap top). Just like a mouse it consists of two buttons below the touch surface to do the operations like left click and right click. By using our fingers we can easily operate.

7. Joy Stick:
It is a device that lets the user move an object quickly on the screen. It has a liver that moves in all directions and controls the pointer or object. It is used for computer games and CAD/CAM systems.

8. Microphone:
By using this device we can convert voice signals into digital form.

9. Scanner:
It is used to read text or pictures printed on paper and translate the information into computer usable form. It is just like a photostat machine but it gives information to the computer.

10. Optical Mark Reader (OMR):
This device identifies the presence or absence of a pen or pencil mark. It is used to evaluate objective type exams. In this method special preprinted forms are designed with circles can be marked with dark pencil or ink.

A high intensity beam in the OMR converts this into computer usable form and detects the number and location ofthe pencil marks. By using this we can evaluate easily and reduce the errors.

11. Bar code/Quick Response (QR) code reader:
Light and dark bars are used to record item name, code and price is called Bar Code. This information can be read and input into a computer quickly without errors using Bar Code Readers.

It consists of a photo electric scanner and it is used in super market, jewellery, textiles etc. QR codes are similar to barcodes but it uses two dimensional instead of single dimensional used in Barcode.

12. Biometric sensor:
It is used to read unique human physical features like finger prints, retina, iris pattern, facial expressions etc. Most of you give these data to the Government for Aadhaar.

13. Smart card reader:
A plastic card(may be like your ATM card) stores and transmit data with the help of a reader.

14. Digital Camera:
By using digital camera, we can take photographs and store in a computer. Therefore we can reduce the use of film. Hence it is economical.

Output devices:
After the data processing the result is displayed as soft copy(soft copy can view, only by using a device) or hard copy(lt can read easily).
1. Visual Display Unit:
a. Cathode Ray Tube (CRT):
There are two types of CRT’s, monochrome (Black and white) and colour. Monochrome CRT consists of one electron gun but colour CRT consists of 3 electron guns (Red, Green and Blue) at one end and the other end coated with phosphor. It is a vacuum tube. The phosphor coated screen can glow when electron beams produced by electron guns hit.

It is possible to create all the colours using Red, Green and Blue. The images produced by this is refreshed at the rate of 50 or 60 times each second. Its disadvantage is it is heavy and bulky. It consumes more power and emits heat. But it is cheap. Nowadays its production is stopped by the company.

b. Liquid Crystal Display (LCD):
It consists of two electrically conducting plates filled with liquid crystal. The front plate has transparent electrodes and the back plate is a mirror.

By applying proper electrical signals across the plates, the liquid crystals either transmit or block the light and then reflecting it back from the mirror to the viewer and hence produce images. It is used in where small sized displays are required.

c. Light Emitting Diode(LED):
It uses LED behind the liquid crystals in order to light up the screen. It gives a better quality and clear image with wider viewing angle. Its power consumption is less.

d. Plasma Panels:
It consists of two glass plates filled with neon gas. Each plate has several parallel electrodes, right angles to each other. When low voltage is applied between two electrodes, one on each plate, a small portion of gas is glow and hence produce images.

Plasma displays provide high resolution but are expensive. It is used in, where quality and size is a matter of concern.

e. Organic Light Emitting Diode(OLED) Monitors:
It is made up of millions of tiny LEDs. OLED monitors are thinner and lighter than LCDs and LEDs. It consumes less power and produce better quality images but it is very expensive.

• LCD projector: It is used to display video, images or data from a computer on a large screen. Its main component is a high intensity light producing bulb and a lens.

2. Printer:
There are two types of printers impact and non impact printers. Printers are used to produce hard copy.
Impact Printers:
There is a mechanical contact between print head and the paper.
(i) Dot Matrix Printer:
Here characters are formed by using dots. The printing head contains a vertical array of pins. The letters are formed by using 6 dot rows and 7 dot columns. Such a pattern is called 5 × 7 matrix.

This head moves across the paper, the selected pins fire against an inked ribbon to form characters by dot. They are capable of faster printing, but their quality is not good.

Non-impact Printers:
There is no mechanical contact between print head and paper so carbon copies cannot be possible to take. They are inkjet, laser, thermal printers etc.
(a) Inkjet Printer:
It works in the same fashion as dot matrix printers, but the dots are formed with tiny droplets of ink to be fired from a bottle through a nozzle. These droplets are deflected by an electric field using horizontal and vertical deflection plates to form characters and images.

It is possible to generate colour output. They produce less noise and produce high quality printing output. The printing cost is higher. Here liquid ink is used.

(b)Laser Printer:
It uses photo copying technology. Here instead of liquid ink dry ink powder called toner is used. A drum coated vyith positively charged photo conductive material is scanned by a laser beam.

The positive charges that are illuminated by the beam are dissipated. The drum is then rolled through a reservoir of negatively charged toner which is picked up by the charged portions of the drum.

It adheres to the positive charges and hence creating a page image on the drum. Monochrome laser printer uses a single toner whereas the colour, laser printer uses four toners. Its print quality is good less noise and printing cost is higher.

(c) Thermal Printers:
It is same as dot matrix printer but it needs heat sensitive paper. It produces images by pushing electrically heated pins to the special paper. It does not make an impact on the paper so we cannot produce carbon copies. It produce less noise, low quality print and inexpensive. It is used in fax machine.

3. Plotter:
A plotter is a device that draws pictures or diagrams on paper based on commands from a computer. Plotters draw lines using.a pen. Pen plotters generally use drum or flat bed paper holders. In a drum plotter the paper is mounted on the surface of a drum.

Here the paper is rotated. But in a flat bed plotter the paper does not move and the pen holding mechanism provides the motion that draws pictures. Plotters are used in engineering applications where precision is needed.

4. Three Dimensional (3D) printer:
This device is used to print 3D objects.

5. Audio output devices:
Speakers are used to produce sound by the backward and forward movement of the diaphragm in the speaker according to the electrical signals from the computer through the audio port.

e-Waste(electronic waste):
It refers to the mal functioning electronic products such as faulty computers, mobile phones, tv sets, toys, CFL etc.

Why should we concern about e Waste:
Itcontains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.

What happens to the e Waste:
A small amount is recycled. Due to this our natural resources are contaminated(poisoned). Some of them can recycle properly. But it is a very big problem in front of the Government to collect segregate, recycle and disposal of e-Waste.

e-Waste disposal methods:

1. Reuse: Reusability has an important role of e-Waste management and can reduce the volume of e-Waste
2. Incineration: It is the process of burning e Waste at high temperature in a chimney
3. Recycling of e-Waste: It is the process of making new products from this e-Waste.
4. Land filling: It is used to level pits and cover by thick layer of soil.

Students’ role in e-Waste disposal:

• Stop buying unnecessary electronic equipments
• Repair Faulty electronic equipments instead of buying a new one.
• Give electronic equipments to recycle
• Buy durable, efficient, quality, toxic free, good warranty products
• check the website or call the dealer if there is any exchange scheme
• Buy rechargeable battery products

Green computing or Green IT:
It is the study and practice of eco friendly computing or IT such as designing, manufacturing, using and disposal of. computers and components (monitors, printers, storage devices etc.)
Following are some steps to follow to reduce the adverse impact on the global environment

• Turn off computer and other devices when not in use
• Use power saver mode
• Use laptops instead of desktops
• Avoid print outs if not needed
• Use LCD s instead of CRT s to save power
• Use Energy Star rated H/W or S/w and Solar energy(Hybrid Energy)
• Dispose e Waste properly as per norms

Following are the steps to promote green computing:

1. Green design: Design energy efficient and eco friendly devices
2. Green manufacturing: reduce non eco friendly parts while manufacturing
3. Green use: Use energy saver devices
4. Green disposal: Use easily disposable devices

Software:
The set of instructions that tell the hardware how to perform a task is called software. Without software computer cannot do anything. Two types System s/w and Application s/w
A. System software: It is a collection of programs used to manage system resources and control its operations. It is further classified into two.

1. Operating System
2. Language Processor

1. Operating System:
It is collection of programs which acts as an interface between user and computer. Without an operating system computer cannot do anything. Its main function is make the computer usable and use hardware in an efficient manner, eg:- Windows XP, Windows Vista, Linux, Windows 7, etc.
Major functions of an operating System

1. Process management: It includes allocation and de allocation of processes(program in execution) as well as scheduling system resources in efficient manner
2. Memory management: It.takes care of allocation and de allocation of memory in efficient manner
3. File management: This includes organizing, naming, storing, retrieving, sharing , protecting and recovery of files.
4. Device management: Many devices are connected to a computer so it must be handled efficiently.

2. Language Processes:
We know that a program is a set of instructions. The instructions to the computer are written in different languages. They are high level language (HLL) and low level language. In HLL English like statements are used to write programs. They are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

Low level language are classified into Assembly Language and Machine Language. In assembly language mnemonics (codes) are used to write programs

In Machine Language 0’s and 1 ’s are used to write program. It is very difficult but this is the only language which is understood by the computer. Usually programmers prefer HLL to write programs because of its simplicity.

But computer understands only machine language. So there is a translation needed. The program which perform this job are language processors. The different language processors are given below:

1. Assembler: This converts programs written in assembly language into machine language.
2. Interpreter: This converts a HLL program into machine language by converting and executing it line by line. The first line is converted if there is no error it will be executed otherwise you have to correct it and the second line and so on.
3. Compiler: It is same as interpreter but there is a difference it translate HLL program into machine language by converting all the lines at a time. If there is no error then only it will executed.

B. Application Software:
Programs developed to serve a particular application is known as application software. eg: MS Office, Compression Utility, Tally etc. Application software can further be sub divided into three categories.

1. Packages
2. Utilities
3. Customized Software

1. Packages:
Application software that makes the computer useful, for people to do every task. Packages are used to do general purpose application.
They are given below:
(i) Word Processes:
This is used for creation and modification of text document. That means a word processor helps the people to create, edit and format a textual data with less effort and maximum efficiency.

By using word processor we can change font and font size of character, change alignment (left, right, center and justify), check spelling and grammar of the whole document etc. eg: MS Word.

(ii) Spread Sheets:
It contains data or information in rows and columns and can perform calculation (Arithmetic, Relational and logical operation). It helps to calculate results of a particular formula and the formula can apply different cells (A cell is the intersection of a row and column. Each column carries an alphabet for its name and row is numbered).

It is used to prepare budgets, balance sheets, P & L account, Pay roll etc. We can easily prepare graphs and charts using data entered in a worksheet. A file is a work book that contains one or more work sheets, eg: MS Excel is a spread sheet software.

(iii) Presentation and Graphics:
You can present your idea with sound and visual effects with the help of presentation software by preparing slides. The application software that manipulate visual images is known as graphics software. Eg: MS Power Point is a presentation package.

(iv) Data base package:
Data base is a collection of large volume of data. DBMS is a set of programs that manages the datas are for the centralized control of data such that creating new records to the database, deleting, records whenever not wanted from the database and modification of the existing database. Example for a DBMS is MS Access.

DTP Packages:
DTP means Desk Top Publishing. By using this we can create books, periodicals, magazines etc. easily and fastly. Now DTP packages are used to create in Malayalam also, eg: PageMaker.

2. Utilities:
Utilities are programs which are designed to assist computer for its smooth functioning. The utilities are given below.

1. Text editor: It is used for creating and editing text files.
2. Backup utility: Creating a copy of files in another location to protect them against loss, if your hard disk fails or you accidentally overwrite or delete data.
3. Compression Utility: It is used to reduce the size of a file by using a program and can be restored to its original form when needed.
4. Disk Defragmenter: It is used to speeds up disk access by rearranging the files that are stored in different locations as fragments to contiguous memory and free space is consolidated in one contiguous block.
5. Virus Scanner: It is a program called antivirus software scans the disk for viruses and removes them if any virus is found.

3. Specific purpose software (Customized software):
It is collection of programs which are developed to meet user needs to serve a particular application. It is also called tailor made software.

Free and open source software:
Here “free” means there is no copyright or licensing. That is we can take copies of the s/w or modify the source code without legal permission of its vendor (creator) we can use and distribute its copy to our friends without permission. That is Freedom to use to modify and redistribute.
The Four freedoms are:

1. Freedom 0: To run program for any purpose
2. Freedom 1: To study how it works and allows you to adapt according to your needs. Also allows to access source code.
3. Freedom 2: Allows to take copies and distribute
4. Freedom 3: Allows you to change source code and release the program .

Examples for Free and open source software are given below
(i) Linux:
it is a free s/w. It was written by Linux Trovalds at the University of Helsinki. It is a GUI and multi-user, multi-tasking O.S. with many more other features. It is also independent of the hardware. It is used by ISP’s, programmers who uses Java to write program, etc. The main Linux distributors are. Open Linux Red hat, Debian, Gnu Linux, etc.

(ii) GNU/Linux:
It was organized by Richard Stallman in 1983

(iii) GIMP (GNU Image Manipulation Program):
It is a very helpful software to perform all the activities to an image.

(iv) Mozilla Firefox:
This web browser helps the users to browse safely.

(v) OpenOffice.org:
This package contains different s/w s that helps the users to draft letters neatly by using “Writer”, perform calculations by using “Calc” and prepare presentations by using “Impress”. It is platform independent (That means it works on both Linux and Windows platforms.

Freeware:
A s/w with Copy right is available free of cost for unlimited use.
Shareware: It is an introductory pack distributed on a trial basis with limited functionality and period.