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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 7 Control Statements 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 7 Control Statements

Summary
These are classified into two decision making and iteration statements

Decision making statements:
if statement:
Syntax: if (condition)
{
Statement block;
}
First the condition is evaluated if it is true the statement block will be executed otherwise nothing will be happened.

if…else statement:
Syntax: if (condition)
{
Statement block1;
}
else
{
Statement block2;
}

Nested if:
An if statement contains another if statement completely then it is called nested if.
if (condition 1)
{
if (condition 2)
{
Statement block;
}
}
The statement block will be executed only if both the conditions evaluated are true.

The else if ladder:
The syntax will be given below
if (expression1)
{
statement block1;
}
else if (expression 2)
{
statement block 2;
}
else if (expression 3)
{
statement block 3;
}
……..
else
{
statement block n;
}
Here firstly, expression 1 will be evaluated if it is true only the statement blockl will be executed otherwise expression 2 will be evaluated if it is true only the statement block2 will be executed and so on. If all the expression evaluated is false then only statement block n will be executed

switch statement:
It is a multiple branch statement. Its syntax is given below.
switch(expression)
{
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
case value: statements;break;
………
default: statements;
}
First expression evaluated and selects the statements with matched case value. If all values are not matched the default statement will be executed.

Conditional operator:
It is a ternary operator hence it needs three operands. The operator is “?:”.
Syntax:
expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.

Iteration statements:
If we have to execute a block of statements more than once then iteration statements are used.

while statement:
It is an entry controlled loop. An entry controlled loop first checks the condition and execute(or enters in to) the body of loop only if it is true. The syntax is given below
Loop variable initialised
while(expression)
{
Body of the loop;
Update loop variable;
}
Here the loop variable must be initialised before the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.

for statement:
The syntax of for loop is
for(initialization; checking ; update loop variable)
{
Body of loop;
}
First part, initialization is executed once, then checking is carried out if it is true the body of the for loop is executed. Then loop variable is updated and again checking is carried out this process continues until the checking becomes false. It is an entry controlled loop.

do-while statement:
It is an exit controlled loop. Exit control loop first execute the body of the loop once even if the condition is false then check the condition.
do
{
Statements
} while(expression);
Here the body executes at least once even if the condition is false. After executing the body it checks the expression if it false it quits the body otherwise the process will be continue.

Students can Download Chapter 6 Data Types and Operators 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 6 Data Types and Operators

Summary
Concepts of data types:
The nature of data is different, data type specifies the nature of data we have to store.

C++ data types:

Fundamental data types:
It is also called built in data type. They are int, char, float, double and void
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory. i.e. 2³² numbers. That is 2³¹ negative numbers and 2³¹ positive numbers (0 is considered as +ve) So a total of 2³² numbers. We can store a number in between -2³¹ to + 2³¹-.

2. char data type:
Any symbol from the keyboard, eg: ‘A’, ‘?’, ‘9’ and so on. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 2⁸ = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory. eg: 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

1. User defined Data types: C++ allows programmers to define their own data type. They are Structure(struct), enumeration (enum), union, class, etc.
2. Derived data types: The data types derived from fundamental data types are called Derived data types. They are Arrays, pointers, functions, etc

Variables:
The named memory locations are called variable. A variable has three important things

1. variable name: A variable should have a name
2. Memory address: Each and every byte of memory has an address. It is also called location (L) value.
3. Content: The value stored in a variable is called content. lt is also called Read(R) value.

Operators:
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators
1. lnput(>>) and output(<<) operators are used to perform input and output operation.
eg: cin>>n;
cout<<n;

2. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction(-), division (/), multiplication(*) and modulus(%- gives the remainder) operations.
eg: If x = 10 and y = 3 then

x/y = 3, because both operands are integer. To get the floating point result one of the operand must be float.

3. Relational operator:
It is also a binary operator. It is used to perform comparison or relational operation between two values and it gives either true(1) or false(O). The operators are <, <=, >, >=, == (equality)and !=(not equal to)
eg: If x = 10 and y = 3 then

4. Logical operators:
Here AND(&&) , OR(||) are binary operators and NOT (!) is a unary operator. It is used to combine relational operations and it gives either true(1) orfalse(O). If x = 1 and y = 0 then

Both operands must be true to get a true value in the case of AND (&&) operation. If x = 1 and y = 0 then

Either one of the operands must be true to get a true value in the case of OR(||) operation. If x = 1 and y = 0 then

5. Conditional operator:
It is a ternary operator hence it needs three operands. The operator is”?:”.
Syntax:
expression ? value if true : value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.
eg: If x = 10 and y = 3 then x>y ? cout<<x : cout<<y;
Here the output is 10

6. sizeof():
This operator is used to find the size used by each data type. eg: sizeof(int) gives 2.

7. Increment and decrement operator:
These are unary operators.

• Increment operator (++): It is used to increment the value of a variable by one i.e., x++ is equivalent to x = x + 1.
• Decrement operator (- -): It is used to decrement the value of a variable by one i.e., x – – is equivalent to x = x – 1.

8. Assignment operator (=):
lt is used to assign the value of a right side to the left side variable.eg. x = 5; Here the value 5 is assigned to the variable x.

Expressions:
It is composed of operators and operands

Arithmetic expression:
It is composed of arithmetic operators and operands. In this expression the operands are integers then it is called Integer expression. If the operands are real numbers then it is called Floating point expression. If the operands are constants then it is called constant expression.

Relational expression:
It is composed of relational operators and operands

Logical expression:
It is composed of logical operators and operands

Statements:
Statements are smallest executable unit of a programming language. Each and every statement must be end with semicolon(;).

Declaration statement:
Each and every variable must be declared before using it. eg: int age;

Assignment statements:
Assignment operator is used to assign the value of RHS to LHS. eg: x = 100

Input statements:
lnput(>>) operator is used to perform input operation. eg: cin>>n;

Output statements:
output(<<) operator is used to perform output operation. eg: cout<<n;

Cascading of I/O operations:
The multiple use of input or output operators in a single statement is called cascading of i/o operators. eg: To take three numbers by using one statement is as follows
cin>>x>>y>>z;
To print three numbers by using one statement is as follows
cout<<x<<y<<z;

Students can Download Chapter 3 Classification of Elements and Periodicity in Properties 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 3 Classification of Elements and Periodicity in Properties

Introduction
The systematic classification of elements made the study of elements easy. In this unit, we will study the historical development of the periodic table and also learn how elements are classified.

Genesis Of Periodic Classification
While Dobereiner initiated the study of periodic relationship, it was Mendeleev who was responsible for publishing the Periodic Law for the first time. It states as follows:

The properties of the elements are a periodic function of their atomic weights.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing ‘ atomic weights. Elements with similar properties occupied the same vertical column or group. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. He ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

At the same time, keeping his primary aim of arranging the elements of similar properties in the same group, he proposed that some of the elements were still undiscovered and, therefore, left several gaps in the table. He left the gap under aluminium and a gap under silicon, and called these elements Eka-Aluminium and Eka-Silicon. Mendeleev predicted the existence of gallium and germanium, and their general physical properties. These elements were discovered later.

Modern Periodic Law And The Present Form Of The Periodic Table
Modem periodic law states that “The physical and chemical properties of the elements are periodic functions of their atomic numbers”. Atomic number is equal to the nuclear charge and the elements are arranged in the increasing order of atomic number.

The period number correspond to the highest principal quantum number (n) of the elements.

Nomenclature Of Elements With Atomic Number Greater Than 100
The names (IUPAC) are derived directly form the atomic number using numerical roots for zero and numbers 1 to 9. The roots are linked together in the order of digits and ‘ium’ is added at the end. The roots for 0,1, 2 9 are nil, un, bi, tri, quad, pent, hex, sept, oct and enn respectively. For example, the element with atomic number 110 will have the name Ununnilium (Un+ un+nil + ium), The element with atomic number 114 has the name Ununquadium (un + un + quad + ium) and the element with atomic number 120 will be Unbinilium (un + bi + nil + ium).

Electronic Configurations And Types Of Elements: s, p, d, f- Blocks

The s-Block Elements
The elements of Group 1 (alkali metals) and Group 2 (alkaline earth metals) which have ns1 and ns2 outermost electronic configuration belong to the s-Block Elements. They are all reactive metals with low ionization enthalpies.

They lose the outermost electron(s) readily to form 1+ ion (in the case of alkali metals) or 2+ ion (in the case of alkaline earth metals). The metallic character and the reactivity increase as we go down the group. Because of high reactivity they are never found pure in nature.

The compounds of the s-block elements, with the exception of those of lithium and beryllium are predominantly ionic.

The p-Block Elements
The p-Block Elements comprise those belonging to group 13 to 18 and these together with the s-Btock Elements are called the Representative Elements or Main Group Elements. The outermost electronic configuration varies from ns2np1 to ns2np6in each period. At the end of each period is a noble gas element with a closed valence shell ns2np6 configuration. All the orbitals in the valence shell of the noble gases are completely filled by electrons and it is very difficult to alter this stable arrangement by the addition or removal of electrons. The noble gases thus exhibit very low chemical reactivity. Preceding the noble gas family are two chemically important groups of non-metals. They are the halogens (Group 17) and the chalcogens (Group 16). These two groups of elements have high negative electron gain enthalpies and readily add one or two electrons respectively to attain the stable noble gas configuration. The non-metallic character increases as we move from left to right across a period and metallic character increases as we go down the group.

The d-Block Elements (Transition Elements)
These are the elements of group 3 to 12 in the centre of the Periodic Table. These are characterised by the filling of inner d orbitals by electrons and are therefore referred to as d-Block Elements. These elements have the general outer electronic configuration (n-1) d1-10ns^2. They are all metals. They mostly form coloured ions, exhibit variable valence (oxidation states), paramagnetism and oftenly used as catalysts. However, Zn, Cd and Hg which have the electronic configuration, (n-1) d10ns2 do not show most of the properties of transition elements. In a way, transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of groups 13 and 14 and thus take their familiar name “Transition Elements”.

The f-Block Elements (Inner-Transition Elements)
The two rows of elements at the bottom of the Periodic Table, called the Lanthanoids, Ce(Z = 58) -Lu(Z = 71) and actinoids, Th(Z = 90)-Lr(Z= 103) are characterised by the outer electronic configuration (n-2)f1-14 (n-1 )d°-1ns2. The last electron added to each element is filled in f- orbital. These two series of ‘ elements are hence called the Inner Transition Elements (f-Block Elements). They are all metals. Within each series, the properties of the elements are quite similar. The elements after Uranium are called Transuranium Elements.

Metals, Non-metals and Metalloids. In addition to displaying the classification of elements into s, p, d and f-blocks, they can be divided into Metals and Non-Metals. Metals usually have high melting and boiling points. They are good conductors of heat and electricity. They are malleable (can be flattened into thin sheets by hammering) and ductile (can be drawn into wires). In contrast, non-metals are located at the top right hand side of the Periodic Table.

In fact, in a horizontal row, the property of elements change from metallic on the left to non-metallic on the right. Non-metals are usually solids or gases at room temperature with low melting and boiling points (boron and carbon are exceptions). They are poor conductors of heat and electricity. Most nonmetallic solids are brittle and are neither malleable nor ductile. The elements become more metallic as we go down a group; the nonmetallic character increases as one goes from left to right across the Periodic Table. The elements (e.g., silicon, germanium, arsenic, antimony and tellurium) running diagonally across the Periodic Table show properties that are characteristic of both metals and nonmetals. These elements are called Semi-metals or Metalloids.

Periodic Trends In Properties Of Elements
Most of the properties such as atomic radius, ionic radius, Ionisation enthalpy, electron gain enthalpy and electron negativity are directly related to electronic configuration of their atoms. They show variation with change in atomic number within a period or a group.

Trends In Physical Properties

1. Atomic Radius :
lt is defined as the distance from the centre of the nucleus of an atom to the outermost shell of electrons. Electron cloud surrounding the atom does not have a sharp boundary since, the determination of the atomic size cannot be precise. Hence it is expressed in terms of different types of radii. Some of these are covalent radius and metallic radius. Covalent radius is defined as one half of the distance between the centres of nuclei of two similar atoms bonded by a single covalent bond. Metallic radius may be defined as half of the internuclear distance between two adjacent atoms in the metallic crystal.

2. Ionic Radius:
The removal of an electron from an atom results in the formation of a cation, whereas gain of an electron leads to an anion. The ionic radii can be estimated by measuring the distances between cations and anions in ionic crystals. When we find some atoms and ions which contain the same number of electrons, we call them isoelectronic species. For example, O2-, F~, Na+ and Mg2+ have the same number of electrons (10). Their radii would be different because of their different nuclear charges.

3. Ionization Enthalpy:
A quantitative measure of the tendency of an element to lose electron is given by its Ionization Enthalpy. It represents the energy required to remove an electron from an isolated gaseous atom (X) in its ground state. To understand the trends in ionization enthalpy, we have to consider two factors: (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of “shielding” or “screening” of the valence electron from the nucleus by the intervening core electrons.

The first ionization enthalpy of boron (Z = 5) is slightly less than that of beryllium (Z = 4) even though the former has a greater nuclear charge. It is because, s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron Compared to the removal of a 2s-electron from beryllium.

Thus, boron has a smaller first ionization enthalpy than beryllium. Another “anomaly” is the smaller first ionization enthalpy of oxygen compared to nitrogen. This arises because in the nitrogen atom, three 2p-electrons reside in different atomic orbitals (Hund’s rule) whereas, in the oxygen atom, two of the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion. Consequently, it is easier to remove the fourth 2p-electron from oxygen than it is, to remove one of the three 2p-electrons from nitrogen.

4. Electron Gain Enthalpy :
When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change accom-panying the process is defined as the Electron Gain Enthalpy (∆_eg H).

5. Electronegativity:
A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity. Unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity. However, a number of numerical scales of electronegativity of elements viz., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale have been developed.

Trends In Chemical Properties
1. Oxidation State :
The atomic property, valency is better explained in terms of oxidation state. It is the charge which an atom of element has or appears to have when present in the combined state. Electronegative elements generally acquire negative oxidation states while electropositive elements acquire positive oxidation states.

2. Anomalous properties of second-period elements:
The first element of each group in s and p block differs in many respects from the remaining members of the respective groups. This is due to their small size, high charge/ radius ratio, high electronegativity and availability of less valence orbitals. The first member has only 4 valence orbitals (2s, 2p) whereas the second member of the same group will have nine valence orbitals (3s, 3p, 3d) for bonding. B can form only (BF₄)^– while Al forms (AlF₆)^3-

In group 1 only Li forms covalent compounds and in many respects, Li resembles Mg of group 2. Similarly, Be resembles Al of group 13. This type of similarity in properties is known as diagonal relationship.

Chemical Reactivity
Across a period ionisation enthalpy increases and electron gain enthalpy becomes more negative. Thus elements at the extreme left show lower ionisation enthalpies (more electropositive nature) and those at the right (excluding nobel gases) show larger negative electron gain enthalpies (more electronegative). Therefore high chemical reactivity is found with elements at the two extremes compared to those at the centre. Electropositivity leads to metallic behaviour and electronegativity leads to non-metallic behaviour.

Students can Download Chapter 5 Introduction to C++ Programming 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 5 Introduction to C++ Programming

Summary
It is developed by Bjarne Stroustrup. It is an extension of C Language.

Character set:
To study a language first we have to familiarize the character set. For example to study English language first we have to study the alphabets. Similarly here the character set includes letters(A to Z & a to z), digits(0 to 9), special characters(+, -, ?, *, /, …..) white spaces(non printable) etc..

Token:
It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens
1. Keywords:
These are reserved words for the compiler. We can’t use for any other purposes eg: float is used to declare variable to store numbers with decimal point. We can’t use this for any other purpose

2. Identifier:
These are user defined words. Eg: variable name, function name, class name, object name, etc…

3. Literals (Constants):
Its value does not change during execution
(a) Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
eg:

• For decimal 100, 150, etc
• For octal 0100, 0240, etc
• For hexadecimal 0x100, 0x1A, etc

(b)Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals. eg: 3.14157, 79.78, etc.

(c) Character literal-: A valid C++ character enclosed in single quotes, its value does not change during execution. eg: ‘m’, ‘f ’ etc

(d) String literal:
One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’)
eg: “Mary’s”, ’’India”, etc.

4. Punctuators:
In English or Malayalam language punctuation mark are used to increase the readability but here it is used to separate the tokens. eg: {,}, (,).

5. Operators:
These are symbols used to perform an operation(Arithmetic, relational, logical, etc…)

Integrated Development Environment(IDE):
It is used for developing programs

1. It helps to write as well as editing the program.
2. It helps to compile the program and linking it to other (header files and other user) programs
3. It helps to run the program

Turbo C++ IDE:
Following is an C++ IDE

(a) Opening the edit window:
Method I: File → Click the menu item New
Method II: Press Alt and F simultaneously then press N

(b) Saving the program:
Click File → Save or Press Function key F2 or Alt + F and then press S. Then give a file name and press ok.

(c) Running/executing the program:
Press Alt + R then press R OR Click Run → press R, OR Press Ctrl + F9

(d) Viewing the output:
Press Alt + F5

(e) Closing Turbo C++ IDE:
Click File → then press Quit menu Or Press Alt + X

Geany IDE

Step 1: Take Geany Editor and type the program (source code)
Step 2: Save the file with extension .cpp
Step 3: Compile the program by Click the Compile Option
Step 4: After successful compilation, Click Build option
Step 5: Then click on Execute option

Students can Download Chapter 4 Principles of Programming and Problem Solving 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 4 Principles of Programming and Problem Solving

Summary
Problem solving using computers:
It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
They can perform tasks based upon the instructions given by the humans (programmers) .

Approaches in problem solving:
Top down design:
Larger programs are divided into smaller ones and solve each tasks by performing simpler activities. This concept is known as top down design in problem solving

Bottom up design:
Here also larger programs are divided into smaller ones and the smaller ones are again subdivided until the lowest level of detail has been reached. We start solving from the lowest module onwards. This approach is called Bottom up design.

Phases in Programming:
1. Problem identification:
This is the first phase in programming. The problem must be identified then only it can be solved, for this we may have to answer some questions.

During this phase we have to identify the data, its type, quantity and formula to be used as well as what activities are involved to get the desired out put is also identified for example if you are suffering from stomach ache and consult a Doctor.

To diagnose the disease the Doctor may ask you some question regarding the diet, duration of pain, previous occurrences etc, and examine some parts of your body by using stethoscope X-ray, scanning etc.

2. Deriving the steps to obtain the solution:
There are two methods, Algorithm and flowchart, are used for this.
(a) Algorithm:
The step-by-step procedure to solve a problem is known as algorithm. It comes from the name of a famous Arab mathematician Abu Jafer Mohammed Ibn Musaa Al-Khowarizmi, The last part of his name Al-Khowarizmi was corrected to algorithm.

(b) Flowchart:
The pictorial or graphical representation of an algorithm is called flowchart.
Flow chart symbols are explained below
(i) Terminal (Oval):

It is used to indicate the beginning and ending of a problem.
(ii) Input/Output (parallelogram):

It is used to take input or print output.
(iii) Processing (Rectangle):

It is used to represent processing. That means to represent arithmetic operation such an addition, subtraction,multiplication and, etc.
(iv) Decision (Rhombus):

It is used to represent decision making. It has one entry flow and two exit flows but one exit path will be executed at a time.
(v) Flow lines (Arrows):

It is used to represent the flow of operation
(vi) Connector:

3. Coding:
The dummy codes (algorithm) or flowchart is converted into program by using a computer language such s Cobol, Pascal, C++, VB, Java, etc.

4. Translation:
The computer only knows machine language. It does not know HLL, but the human beings HLL is very easy to write programs. Therefore a translation’ is needed to convert a program written in HLL into machine code (object code).

During this step, the syntax errors of the program will be displayed. These errors are to be corrected and this process will be continued till we get “No errors” message. Then it is ready for execution.

5. Debugging:
The program errors are called ‘bugs’ and the process of detecting and correcting errors is called debugging. In general there are two types of errors syntax errors and logical errors. When the rules or syntax of the language are not followed then syntax errors occurred and it is displayed after compilation.

When the logic of a program is wrong then logical errors occurred and it is not displayed after compilation but it is displayed in the execution and testing phase.

6. Execution and Testing:
In this phase the program will be executed and give test data for testing the purpose of this is to determine whether the result produced by the program is correct or not. There is a chance of another type of error, Run time error, this may be due to inappropriate data.

7. Documentation:
It is the last phase in programming. A computerized system must be documented properly and it is an ongoing process that starts in the first phase and continues till its implementation. It is helpful for the modification of the program later.

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.

Students can Download Chapter 1 Physical World 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 1 Physical World

Summary
What is Physics?

a. Science And Scientific Method
Science is exploring, experimenting and predicting from what we see around us. ie. It is basically an systematic attempt to understand natural phenomena.

b. Physics
Two approaches are used in Physics: unification and reduction. In unification diverse physical phenomena are explained in terms of a few concepts or laws. For example all electric and magnetic phenomena can be explained by laws of electromagnetism (Maxwell’s equations).

In reduction, we derive properties of complex (bigger) system from properties and interactions of constituent parts. For example, the temperature of system is related to average kinetic energy of molecule of system.

Scope And Excitement Of Physics
The different subdisciplines of physics belongs to two domains: microscope domain and macroscopic domain. The macroscopic domain includes phenomena at laboratory, terrestrial and astronomic scales.

The microscopic domain of physics deals with constitution and structure of matter and their interaction with elementary particles like electron, proton, photon etc.

Physics covers a wide range of magnitude of physical quantities like length, time, mass, energy, etc. Physics includes phenomena involving elementary particles like electron, proton etc. whose range is 10^-14m.

It also deals with astronomical phenomena at the scale of even the entire universe (10⁺²⁶m). The range of time extends from 10^-22 s to 10¹⁸s. The range of mass goes from 10^-30 kg (mass of electron) to 10⁵⁵kg (mass of entire universe).

Physics, Technology And Society
The relation between Physics, technology and society can be seen in many examples. The steam engine has an important role in the Industrial Revolution in England in eighteenth century. The discovery of basic laws of electricity and magnetism contributed wireless communication technology.

Fundamental Forces Of Nature
There occur four fundamental forces in nature. They are gravitational force, electromagnetic force, strong nuclear force and weak nuclear force.

1. Gravitational Force:
It is a universal force. Gravitational force is the attractive force existing between any two bodies by virtue of its mass.

2. Electromagnetic Force:
The electromagnetic force exist between charged bodies. The electrostatic force of attraction or repulsion exist between charges at rest. A moving charge has magnetic effect in addition to electric effect. The electric and magnetic effects are inseparable and hence force experienced by charge is called electromagnetic force.

3. Strong Nuclear Force:
The strong nuclear force binds the nucleons (protons and neutrons) inside the nucleus. It is the strongest of all fundamental forces. The range of nuclear force is 1o^-15m (fermi) and it is charge independent.

4. Weak Nuclear Force:
The range of weak nuclear force is 10^-16m. This force exists only in few nuclear, reactions like b-decay.

5. Towards Unification of Forces:
Isac Newton unified terrestrial and celestial domain by applying law of gravitation in two domains. Oersted and Faraday showed that electric and magnetic phenomena are inseparable. Maxwell unified electromagnetism and optics by showing light is an electromagnetic wave.

1.5 Nature Of Physical Laws
The physical quantity that remains unchanged in process is called conserved quantity. Some of the conservation laws in nature are laws of conservation of mass, energy, linear momentum, angular momentum, charge etc.

Conservation laws have a deep connection with symmetries of nature. The symmetry of nature w.r.t. translation in time is equivalent to conservation of energy. Similarly the symmetry of nature w.r.t. translation in space is equivalent to conservation of linear momentum.

Symmetries of space and time and other types of symmetry play an important role n modern theories of fundamental forces in nature.

Students can Download Chapter 15 Waves 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 15 Waves

Plus One Physics Waves One Mark Questions and Answers

Question 1.
Velocity of sound in vacuum is
(a) 330ms^-1
(b) 165ms^-1
(c) zero
(d) 660ms^-1
Answer:
(c) zero
Sound requires a material medium for propation. Hence, velocity of sound in vacuum is zero.

Question 2.
The physical quantity that remains unchanged when a sound wave goes from one medum to another is
(a) amplitude
(b) speed
(c) wavelength
(d) frequency
Answer:
(d) frequency
When a sound wave goes from one medium to another the frequency of the wave remains unchanged.

Question 3.
What is the range of frequency of audible sound?
Answer:
20Hz to 20KHz.

Question 4.
Why does sound travel faster in iron than in air?
Answer:
Because solids are more elastic compared to air.

Question 5.
What kind of waves help the bats to find their way in dark?
Answer:
Ultrasonic wave

Question 6.
In which gas, hydrogen and oxygen will the sound have greater velocity?
Answer:
As velocity, v α \(\sqrt{1 / ρ}\), velocity of sound will be greater in hydrogen gas.

Question 7.
Why transverse waves can not setup in gas?
Answer:
The rigidity modulus of gas is zero.

Question 8.
What is the effect of pressure on the velocity of sound waves?
Answer:
No effect

Question 9.
Why bells are made up of metal and not wood?
Answer:
The wood causes high damping.

Question 10.
What is the velocity of sound in perfect rigid body?
Answer:
The velocity is infinite cause young’s modulus of perfect rigid body is infinite.

Plus One Physics Waves Two Mark Questions and Answers

Question 1.
If a tuning fork is held above a resonance column, then maximum sound can be heard at certain height of the air column.

1. Name the type of wave produced in the air col¬umn.
2. What do you mean by beats?

Answer:

1. Longitudinal
2. Periodic variation of intensity of sound is called beats.

Question 2.
Why bells of colleges and temple are of large size?
Answer:
Larger, the area of source of sound more is the energy transmitted into the medium. Hence intensity of sound is more and loud sound is heard.

Plus One Physics Waves Three Mark Questions and Answers

Question 1.
A sound wave of frequency 400 Hz is travelling in air at a speed of 320m/s.

1. The speed of sound wave in vaccum is______
   • 320m/s
   • more than 320m/s
   • less than320m/s
   • Zero
2. What is the wavelength of the above wave?
3. Calculate the diffference in phase between two points on the wave 0.2m apart in the direction of travel.

Answer:
1. Zero

2. v = fλ
λ = \(\frac{v}{f}=\frac{320}{400}\) = 0.8m

3. At 0.2 m apart, the phase differnce’is given by
∆Φ = \(\frac{0.2}{0.8} \times 2 \pi=\frac{\pi}{2} \mathrm{rad}\).

Question 2.
A string fixed one end is suddenly brought in to up and down motion.

1. What is the nature of the wane produced in the string and name the wave.
2. A brass wire 1 m long has a mass 6 × 10^-3 kg. If it is kept at a tension 60N, What is the speed of the wave on the wire.

Answer:
1. Transverse wave

2.

Plus One Physics Waves Four Mark Questions and Answers

Question 1.
A sonometer wire of length 30cm vibrates in the second overtone

1. Represent it pictorially
2. What is the distance between two points in the string which has a phase difference of n
3. A violin string resonates in its fundamental frequency of 196hz. Where along the string must you place your finger so that the fundamental frequency becomes 440Hz, If the length of violin string is 40cm.

Answer:
1.

2. 0.30 = \(\frac{3}{2}\)λ
λ = \(\frac{2 \times 0.3}{3}\) = 0.2
We know 2π = λ
ie. π radian = \(\frac{λ}{2}\) wave length
= \(\frac{0.2}{2}\) = 0.1m
∴ The distance between two points is 0.1 m (for a a phase difference of π radian).

3. For fundamental mode of vibration.

The finger must be placed 37.5 from one end.

Question 2.
A horizontal metal wire is fixed on a state of tension between two vertical supports when plucked it gives a fundamental frequency f_O.

1. Obtain a mathematical expression for f_O
2. A 5.5 m wire has a mass of 0.035 kg. If the tension of the string is 77N, the speed of wave on the string is
   • 110 ms^-1
   • 11\(\sqrt{10}\)ms^-1
   • 77 ms^-1
   • 11 ms^-1
   • 102 ms^-1
3. What change, if any, will be observed in the fundamental frequency if the wire is now immersed in water and plucked again?

Answer:
We know velocity on a string, V = \(\sqrt{T / m}\)
But V = λf

2. Mass per unit length,m = \(\frac{M}{\ell}=\frac{0.035}{5.5}\)
= 6.36 × 10^-3kg/m
T = 77N

3. Frequency does not change.

Question 3.
When a pebble is dropped to the surface of water, certain waves are formed on the water surface.

1. What type of wave is it?
2. Is it a progressive wave? Explain?
3. The equation for such a wave is y = 4sinπ (2t – 0.01x). where ‘y’ and ‘x’ are in cm. and ‘t’ in sec. find its
   • Amplitude
   • Wavelength
   • initial phase
   • Frequency

Answer:
1. Transverse wave.

2. It is a progressive wave. It moves from one point to another point.

3. y = 4sinπ(2t – 0.01x)
y = 4 sin (2πt-0.01πx)
= -4 sin (0.01 πx – 2πt)
Comparing with standard wave equation, y = A sin (kx – ωt), we get

• Amplitude A = 4 m
• Kx = 0.01 × πx
  \(\frac{2 \pi}{\lambda}\)x = 0.01 πx, λ = 200m
• Initial phase = 0
• ωt = 2 πt, 2 π f t = 2 πt, f = 1 Hz.

Question 4.
The speed of a wave along a stretched string depends only on the tension and the linear mass density of the string and does not dependent on the frequency of the wave.

1. Give the equation of speed of transverse wave along a stretched string.
2. Why the speed does not depend on the frequency of the wave.
3. A steel wire 0.72 long has a mass of 5 × 10^-3Kg. If the wire is under a tension 60N, what is the speed of transverse wave on the wire?

Answer:

1. v = \(\sqrt{\frac{T}{m}}\) Where T is the tension and m is the mass per unit length.
2. The frequency of the wave is determined by the source that generates the wave.
3. 

Question 5.
While discussing the propagation of sound through atmospheric air, one argued that the velocity of sound is 280 ms-1 and said that he calculated it using Newton’s formula. But another learner argued that velocity of sound is 330 ms1. He justified his argument by saying that he has applied Laplace corrected formula.

1. Write the formula used by the second learner.
2. Using the above relation, show that velocity depends on temperature and humidity while is independent of pressure.
3. “Sound can be heard over longer distance on rainy days.” Justify.

Answer:
1. a = \(\sqrt{\frac{2 \lambda P}{\rho}}\)

2.

If temperature remain same
PV = constant
∴ v is independent of p.

3. During rainy day as ρ decreases. Hence v increase and sound propagate longer distance.

Question 6.
When a stone is dropped in to the river, certain waves are formed on its surface.

1. What type of wave it is?
2. Is it a progressive wave? Explain.
3. If yes, derive a mathematical expression for the above wave.

Answer:
1. Transverse wave.

2. Yes, because each particles of the medium vi¬brates simple harmonically.

3. Consider a harmonic wave travelling along the +ve x-direction with a speed V. Let ‘0’ be the particle in the medium. Its displacement at any instant of time may be written as y = A sin ωt

Consider another particle ‘p’ at a distance x from ‘0’ to its right. The displacement of p at any instant.
y = A sin (ωt – α) ______(1)
α is the phase difference between 0 and P. Here
α = \(\frac{2 \pi}{\lambda}\)x equation (1) becomes

Plus One Physics Waves Five Mark Questions and Answers

Question 1.
A boy standing near a railway track found that the pitch of the siren of a train increases as it approaches him

1. State the phenomenon behind it?
2. List any two applications of the same Phenomenon
3. Obtain an expression for the apparent frequency of the siren as heard by the boy.

Answer:
1. Doppler effect.

2. Doppler effect can be used to find the speed of moving object. Dopplar effect in light is used to find speed of galaxies.

3. The apparent change in the frequency of sound wave due to the relative motion of source or listener or both is called Doppler effect. It was proposed by John Christian Doppler and it was experimentally tested by Buys Ballot.

Considers source is producing sound of frequency v. Let V be the velocity of sound in the medium and λ the wavelength of sound when the source and the listener are at rest.

The frequency of sound heard by the listener is ν = \(\frac{v}{\lambda}\). Let the source and listener be moving with velocities v_s and v_l in the direction of propogation of sound from source to listener. (The direction S to L is taken as positive).

The relative velocity of sound wave with respect to the source = V – V_s
Apparent wavelength of sound,
λ¹ = \(\frac{V-V_{s}}{v}\) _____(1)
Since the listener is moving with velocity v^, the relative velocity of sound with respect to the listener,
V¹ = V – V_l ______(2)
Apparent frequency of sound as heard by the listener is given by
ν = \(\frac{v^{1}}{\lambda^{1}}\) ______(3)
Sub (1) and (2) in eq.(3) we get

Question 2.

1. Waves are means of transferring energy from one point to another. Distinguish between longitudinal and transverse waves.
2. What is a plane progressive wave? Arrive at an expression for the displacement of a particle on the path of the wave, advancing in the positive x-direction.
3. The velocity of sound is greater in solids than in gases. Explain.

Answer:
1.

2. Out of syllabus

3. Solids are highly elastic as compared to liquids and gases.

Question 3.

1. Sound produced by an open pipe contains:
   • Fundamental component only
   • Odd harmonics only
   • All the harmonics
   • Even harmonics only
2. A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly exerted by a 1.1 kHz source?
3. Will resonance with the same source be observed if one end of the pipe is closed? (Take the speed
   of sound in air to be 330 ms^-1)

Answer:
1. All the harmonics.

2. We know frequency of oscillation in the Pipe, f = \(\frac{n V}{2 L}\)
Substituting the values v, L and f we get

n = 2 means that the oscillation is second harmonics.

3. The condition for second harmonics in closed pipe is

The frequency required for resonance in a closed pipe is 825 Hz. Hence we do not get resonance at 1.1 Khz.

Question 4.
A boy plucks at the centre of a stretched string of length 1 m and abserves a wave pattern.

1. Which type of wave is produced on the string?
2. What are the conditions for the formation of the above mentioned wave?
3. The distance between consecutive nodes is
   1. λ
   2. λ/2
   3. 2λ
   4. λ/4
4. A steel rod100 cm long is clamped at its middle. The fundamental frequency of longitudiral vibrations of the rod is given to be 2.5kHz. What is the speed of sound in steel?

Answer:
1. Standing wave (or) stationary wave

2. Same frequency, same amplitude, travelling in opposite direction.

3. λ/2

4. Length of the rod l = \(\frac{\lambda_{1}}{2}\)
λ₁ = 2l =2m
v = ν₁λ₁
v = 2500 × 2 = 5000 m/s.

Plus One Physics Waves NCERT Questions and Answers

Question 1.
A string of mass 2.50kg is under a tension of200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Tension T = 200N
Length I = 20.0m; Mass M = 2.50kg

Question 2.
A stone droped from the top of a tower of height 300m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top. Given that the speed of sound in air is 340ms^-1? (g = 9.8ms^-2)
Answer:
Time after which the splash is heard at the top is equal to the sum of the time t₁ taken by the stone to fall down and the time t₂ taken by the sound to travel from bottom to top.
Using S = ut + \(\frac{1}{2}\) at², we 9et S = \(\frac{1}{2}\)gt₁²
(∵ u = 0 and a =g)

Question 3.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7kms^-1? The operating frequency of the scanner is 4.2MHz.
Answer:
λ = ? u = 1.7kms^-1 = 1700ms^-1
v = 4.2 × 10⁶ Hz, u = vλ = or λ = \(\frac{u}{v}\)
or λ = \(\frac{1700}{4.2 \times 10^{6}}\) m = 4.05 × 10^-4 m = 0.405mm.

Question 4.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of the wire is 3.5 × 10^-2kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. the speed of a transverse wave on the string and
2. the tension in the string?

Answer:
mass of wire M = 3.5 × 10^-2kg
Linear density µ = mass / length
= M/l = 4.0 × 10^-2kg
∴ Length of wire l = \(\frac{M}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\) m = 0.875m
In the fundamental mode,
λ =2l = 2 × 0.875m = 1.75m

1. Speed of transverse vyaves u = v λ
= 45 × 1.75ms^-1 = 78.75ms^-1

2. u= \(\sqrt{\frac{T}{\mu}}\) or T = µu² = 4 × 10^-2(78.75)²N
= 278.06N.

Question 5.
A steel rod 100cm long is clamped at its middle. The fundemental frequency of longitudinal vibrations of the rod are given to be 2.53kHz. What is the speed ‘ of sound in steel?
Answer:
l = 1m, v = 2.53 × 10³HZ, \(\frac{\lambda}{2}\) = I or λ = 2m
u = v λ = 2.53 × 10³ × 2ms^3-1 = 5.06 × 10³ ms^-1
= 5.06kms^-1

Students can Download Chapter 2 Data Representation and Boolean Algebra 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 2 Data Representation and Boolean Algebra

Number System:
It is a systematic way to represent numbers in different ways. Each number system has its own Base, that is a number and that number of symbols or digits used.

1. Most Significant Digit (MSD): The digit with most weight is called MSD. MSD is also called Left Most Digit(LMD)
2. Least Significant Digit (LSD): The digit with least weight is called LSD. LSD is also called Right Most Digit(RMD)
   • eg: 106 : Here MSD : 1 and LSD : 6
   • 345.78: Here MSD : 3 and LSD : 8
3. A Binary Digit is also called a bit.
4. The weight of each digit of a number can be represented by the power of its base.

Number conversions:
In general, to convert a decimal number into another number system(binary or octal or hexadecimal) do the following. Divide the number successively by the base of the number system do you want to convert and write down the remainders from bottom to top.

To convert a decimal fraction into another number system .multiply the number by the base of the number system do you want to convert then integer part and fractional part are separated again multiply the fractional part by the base and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Binary:
Divide the number by the base 2 successively and write down the remainders from bottom to top.

Decimal fraction to binary:
multiply the number by the base 2 then integer part and fractional part are separated again multiply the fractional part by the base 2 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Decimal to Octal:
Divide the number by the base 8 successively and write down the remainders from bottom to top.

Decimal fraction to octal:
multiply the number by the base 8 then integer part and fractional part are separated again multiply the fractional part by the base 8 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.
eg: (55)₁₀ = ()₈

(0.140625)₁₀ = (0.11)₈

Decimal to Hexadecimal:
Divide the number by the base 16 successively and write down the remainders from bottom to top.

Decimal fraction to hexadecimal:
multiply the number by the base 16 then integer part and fractional part are separated again multiply the fractional part by the base 16 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Converting a number from any number system into decimal: For this multiply each digit by its corresponding weight and sum it up.

Binary to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 2.

Converting binary fraction to decimal

101.101 = 1 × 2² + 0 × 2¹ + 1 × 2⁰ + 1 × 2^-1 + 0 × 2^-2 + 1 × 2^-3
= 4 + 0 + 1 + 1/2 + 0 + 1/8
= 5 + 0.5 + 0.125
(101.101)₂ = (5.625)₁₀

Octal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 8.
Eg: (1007)₈ =()₁₀?

1 × 8³ + 0 × 8² + 0 × 8¹ + 7 × 8⁰
= 512 + 0 + 0 + 7
=(519)₁₀
Converting octal fraction to decimal (600.005)₈ =()₁₀?

= 6 × 8² + 0 × 8¹ + 0 × 8⁰ + 0 × 8^-1 + 0 × 8^-2 + 5 × 8^-3
= 384 + 0 + 0 + 0 + 0 + 0.009765625
= (384.009765625)₁₀

Hexadecimal to decimal conversion:
For this multiply each bit by its corresponding weight and sum it up. The weights are power of 16.
Eg: (100)₁₆ = ()₁₀?

= 1 × 16² + 0 × 16¹ + 0 × 16⁰
= 256 + 0 + 0
= (256)₁₀
Converting Hexadecimal fraction to decimal (60A.4)₈ =()₁₀?

= 6 x 16² + 0 x 16¹ + 10 x 16⁰ + 4 x 16^-1
= 1536 + 0 + 0 + .25
= (1536.25)₁₀

Octal to binary conversion:
Convert each octal digit into its 3 bit binary equivalent. Consider the following table

Hexadecimal to binary conversion:
Convert each Hexadecimal digit into its 4 bit binary equivalent. Consider the following table

1010 1011 1100 (ABC)₁₆=(101010111100)₂

Binary to octal conversion
Divide the binary number into groups of 3 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding octal equivalent.
eg: (10100110)₂= ()₈?
Insert one zero in the left side to form 3 bits group

(10100110)₂= (246)₈

Binary to Hexadecimal conversion:
Divide the binary number into groups of 4 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding Hexadecimal equivalent.
eg: (100100110)₂ = ()₁₆?
Insert 3 zeros in the left side to form 4 bits group

(100100110)₂ = (126)₁₆

Octal to Hexadecimal conversion:
First convert octal number into binary(see 1.6.7), then convert this binary into hexadecimal(also see 1.6.10)
eg: Convert (67)₈ = ( )₁₆
Step I: First convert this number into binary equivalent for this do the following:

Step II: Next convert this number into hexadecimal equivalent for this do the following.

So the answer is (67)₈ = ( 37)₁₆

Hexadecimal to octal conversion:
First convert Hexadecimal to binary(see 1.6.8), then covert this binary into octal(also see 1.6.9)
eg: Convert (A1)₁₆ = ( )₈?
Step I: First convert this number into binary equivalent. For this do the following

Step II. Next convert this number into octal equivalent. For this do the following.
So the answer is (A1)₁₆ = (241)₈

Data representation:
The data stored in the computer memory is in the form of binary.

Representation of integers
There are three ways to represent integers in computer. They are as follows:

1. Sign and Magnitude Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1. SMR:
Normally a number has two parts sign and magnitude, eg: Consider a number+5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is -ve. eg: If a computer has word size is 1 byte then

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can , represent 2⁷ = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2. 1’s Complement Representation:
To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1 ’s complement,
eg:
(i) To find the 1’s complement of -21
+21 = 00010101
To get the 1’s complement change all 0 to 1 and.all 1 to 0.
-21 = 11101010
1’s complement of-21 is 11101010

(ii) Find the 1’s complement of +21. Positive numbers are represented by using SMR.
+21 = 00010101 (No need to take the 1’s complement)

3. 2’s Complement Representation:
To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg: To find the 2’s complement of -21
+21 = 00010101
First take the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -21 is 1110 1011

Representation of floating point numbers:
A real number consists of an integer part and fractional part and represent by using Exponent and Mantissa method. This method is also used to represent too big numbers as well as too small numbers.
Eg: .0000000000000000000000001 can be represented easily as 1 × 10^-25. Here T is the Mantissa and -25 is the exponent.

A computer with 32 bit word length is used 24 bits for mantissa and the remaining 8 bits used to store exponent.

Representation of characters:
1. ASCII(American Standard Code for Information Interchange):
It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by U.S. government. Each character in the keyboard has a unique number.

Eg: ASCII code of ‘a’ is 97, when you press ‘a’ in the keyboard , a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 2⁷ = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2. EBCDIC(Extended Binary Coded Decimal Interchange Code):
It is an 8 bit code introduced by IBM(lnternational Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(lndian Standard Code for Information Interchange):
It uses 8 bits to represent data and introduced by standardization committee and adopted by Bureau of Indian Standards(BIS).

4. Unicode:
The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 2¹⁶ = 65536 characters (i.e, world’s all written language characters) can store by using this.

Binary arithmetic:
Binary addition:
The rules for adding two bits

eg: Find the sum of binary numbers 110011 and 100001.

Binary subtraction:
The rules for subtracting a binary digit from another digit.

Subtraction using 1’s complement:
The steps are given below:

Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 1’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive then add the carry 1 to the result.
eg: Subtract 1101 from 111100 using 1’s complement method.
Step 1: Insert two Os to the left of 1101. Hence the subtrahend is 001101.
Step 2: 1’s complement of 001101 is 110010
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive and add add the carry 1 to 101110

Hence the result is 101111.

Subtraction using 2’s complement:
The steps are given below :
Step 1: Add 0s to the left of the subtrahend, to make two numbers with same number of bits.
Step 2: Find 2’s complement of subtrahend.
Step 3: Add the complement with minuend.
Step 4: If there is a carry, ignore the carry, the result is positive.
eg: Subtract 1101 from 111100 using 2’s complement method.
Step 1: Insert two 0s to the left of 1101. Hence the subtrahend is 001101.
Step 2: Find the 2’s complement of 001101.
1’s complement is 110010.
2’s complement is 110010 + 1 = 110011
Step 3: Add this to the minuend.

Step 4: Ignore the carry the result is positive. Hence the result is 101111.

Introduction to Boolean algebra:
The name Boolean Algebra is given to honour the British mathematician George Boole. Boolean algebra deals with two states true or false otherwise Yes or No and numerically either 0 or 1.

Binary valued quantities:
A logical decision which gives YES or No values is a binary decision, A statement which gives YES or NO values(TRUE or FALSE) is a logical statement or truth function. A variable which can assign TRUE or FALSE (1 or 0) values is a logical variable

Boolean operators and logic gates:
Logical Operators are AND, OR and NOT. A logical gate is a physical device (electronic circuit)that can perform logical operations on one or more logical inputs and produce a single logical output. A table represents the set f all possible values and the corresponding results in a statement is called truth table.
1. The OR operator and OR gate:
The OR operator gives a 1 either one of the operands is 1. If both operands are 0, it produces 0. The truth table of X OR Y is

The logical OR gate is given below.

The truth table and the gate for the Boolean expression Y = A + B + C

2. The AND operator and AND gate:
The AND operator gives a 1 if and only if both operands are 1. If either one of the operands is 0, it produces 0 The truth table of X AND Y is

The logical AND gate is given below.

The truth table and the gate for the Boolean expression Y = A . B . C

3. The NOT operator and NOT gate:
It produces the vice versa. NOT gate is also called inverter. It is a unary operator that means it has only one input and one output. The truth table of NOT X is

Basic postulates of Boolean algebra:
Boolean algebra consists of some fundamental laws. These laws are called postulates.
Postulate 1: Principles of 0 and 1
If A ≠ 0 , then A = 1 and A 1, then A = 0

Principle of Duality:
When changing the OR(+) to AND(.), AND (.) to OR(+), 0 to 1 and 1 to 0 in a Boolean expression we will get another Boolean relation which is the dual of the first, this is the principle of duality.

Basic theorems of Boolean algebra:
There are some standard and accepted rules in every theory, these rules are known as axioms of the theory.

Identity law:
If X is a Boolean variable, the law states that

1. 0 + X = X
2. 1 + X = 1 (these are additive identity law)
3. 0 . X = 0
4. 1 . X = X (these are multiplicative identity law)

Following are the truth tables

Idempotent law:
This law states that

1. X + X = X
2. X . X = X

Involution law:
This states that
\(\overline{\overline{\mathrm{X}}}=\mathrm{x}\)
The compliment of compliment of a number is the number itself.

Complimentary law:
This law states that

1. \(x+\bar{x}=1\)
2. \(x \cdot \bar{x}=0\)

The truth table is given below

Commutative law:
This law allows to change the position of variable in OR and AND

1. X + Y = Y + X
2. X . Y = Y . X

The truth table is given below

Associative law:
It allows grouping of variables differently

1. X + (Y + Z) = (X + Y) + Z
2. X . (Y . Z) = (X . Y) . Z

The truth table is given below

Distributive law:
This law allows expansion of multiplication over addition and also allows addition operation over multiplication.

1. X . (Y + Z) = X . Y + X . Z
2. X + Y . Z = (X + Y) . (X + Z)

The truth table is given below

Absorption law:
It is a kind of distributive law in which two variables are used and result will be one of them

1. X + (X . Y) = X
2. X . (X + Y) = X

The truth table is given below

De Morgan’s theorem:
Demorgan’s first theorem states that
\(\overline{\mathrm{X}+\mathrm{Y}}=\overline{\mathrm{X}} \cdot \overline{\mathrm{Y}}\)
ie. the compliment of sum of two variables equals product of their compliments.

The second theorem states that
\(\overline{\mathrm{X} . {\mathrm{Y}}}=\overline{\mathrm{X}}+\overline{\mathrm{Y}}\)
ie. The compliment of the product of two variables equals the sum of the compliment of that variables.

Circuit designing for simple Boolean expressions:
By using basic gates such as AND, OR and NOT gates we can create logic circuits.

Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.

NAND gate:
The output of AND gate is inverted by NOT gate is the NAND gate

NOR gate:
The output of OR gate is inverted by NOT gate is the NOR gate.

Students can Download Chapter 1 Some Basic Concepts of Chemistry 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 1 Some Basic Concepts of Chemistry

INTRODUCTION
Chemistry is the branch of science which deals with the composition, properties and transformation of matter. These aspects can be best understood in terms of basic constituents of matter: atoms and molecules. That is why chemistry is called the sci-ence of atoms and molecules.

IMPORTANCE OF CHEMISTRY
Chemistry plays an important role in almost all walks of life. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesised.

NATURE OF MATTER
Matter is anything which has mass and occupies space. Matter can exist in three physical states: solid, liquid and gas.

At the macroscopic or bulk level, matter can be clas-sified as mixtures or pure substances.
A material containing only one substance is called a pure substance. Materials containing more than one substance are called mixtures. Pure substances are further classified into two types: elements and com¬pounds. Mixtures are also of two types: homoge¬neous mixtures and heterogeneous mixtures. A mix¬ture is said to be homogeneous if it has same com¬position throughout. Some examples of homoge¬neous mixtures are air, gasoline, kerosene, milk, alloys, etc. Heterogeneous mixtures are the mixtures which have different composition in different parts. Some examples of heterogeneous mixtures are iron and sulphur, muddy water, etc.

PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substances has characteristic properties. These are classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the iden¬tity or the composition of the substance. Some ex¬amples of physical properties are colour, odour, melt¬ing point, boiling point, density, etc. chemical properties are characteristic reactions of different substances; these include acidity or basic¬ity, combustibility, etc.

THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS)
A unit may be defined as the standard of reference chosen to measure any physical quantity. There are many different systems of units.
The improved metric system of units accepted inter-nationally is called International System of Units or SI units. The SI system has seven base units from which all other units are derived.

UNCERTAINTY IN MEASUREMENT
Scientific measurements involving some measuring devices have some degree of uncertainty. The magnitude of uncertainty depends on the accuracy of the measuring device and also on the skill of its operator.
The closeness of a set of values obtained from identical measurements of a quantity is known as precision of the measurement.
The term accuracy is defined as the closeness of a measurement or a set of measurements to its true value.

SIGNIFICANT FIGURES
The total number of digits in a measurement is called the number of significant figures. It includes the num¬ber of figures that are known with certainty plus the last uncertain digit, beginning with the first non-zero digit.
1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example,0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000

LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
It states that matter can neither be created nor destroyed.
This law was put forth by Antoine Lavoisier in 1789.

Law of Definite Proportions
This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. It is sometimes also referred to as Law of definite composition.

Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Avogadro Law
In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.

DALTON’S ATOMIC THEORY
The main postulates of the theory are:-

1. Matter is made up of extremely small, indivisible particles called atoms.
2. Atoms of the same element are identical in all respects i.e. size and mass.
3. Atoms of different elements are different, i.e., they possess different sizes, shapes, masses, and chemical properties.
4. Atoms of different elements may combine with each other in a simple whole number ratio to form compound atoms or molecules.
5. Atoms can neither be created nor destroyed, i.e., atoms are indestructible.

ATOMIC AND MOLECULAR MASSES
ATOMIC MASS
The mass of an atom is extremely small. These are very inconvenient for calculations. This difficulty was overcome by expressing atomic masses as relative masses, i.e., with respect to the mass of an atom of a standard substance. The scale in which the relative masses are expressed is called atomic mass unit scale or amu scale. One atomic mass unit (amu) is equal to one-twelfth (1/ 12) the mass of an atom of carbon-12. Recently the symbol ‘u’ (unified mass) is used in place of amu.

AVERAGE ATOMIC MASS
The atomic masses of many elements have fractional values because they exist as mixture of isotopes.
In the case of such elements, the atomic mass is taken as the average of the atomic masses of the various isotopes. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they are present in the ration 3:1. Therefore,
Atomic mass of Chlorine = \(\frac{35 \times 3+37 \times 1}{3+1}\) = 35.5u

MOLECULAR MASS
Molecular mass is the sum of atomic masses of the elements present in a molecule.
For example, molecular mass of water
= 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 × (1.008u) + 1 × 16.00 u
= 18.02 u

FORMULA MASS
In ionic compounds, we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the atomic masses of all atoms in a formula unit of the compound.

MOLE CONCEPT AND MOLAR MASSES
‘Mole’ was introduced as the seventh base quantity for the amount of substance in SI system. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the ¹²C isotope. This number is known as ‘Avogadro constant’ (N_A = 6.022 × 10²³). The mass of one mole of a substance in grams is called its molar mass. The molar mass is numerically equal to atomic/ molecular/ formula mass in u.

PERCENTAGE COMPOSITION
One can check the purity of a sample by analysing its mass percentage
Mass % of an element in a compound =

EMPIRICAL FORMULA FOR MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Problem 1.2
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?
Solution:
Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.

Step 3. Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH₂Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
ForCH₂Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass

c) Multiply empirical formula by ‘n’ obtained above to get the molecular formula
Empirical formula = CH₂Cl, n = 2.
Molecular formula = 2 × [CH₂Cl]=C₂H₄Cl₂

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
‘Stoichiometry’ deals with the calculation of masses (sometimes volumes also) of the reactants and prod¬ucts involved in a chemical reaction. The coefficients of reactants and products in a balanced chemical equation is called the stoichiometric coefficients.

LIMITING REAGENT
The amount of the product obtained in a chemical reaction is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is com¬pletely consumed in a reaction containing two or more reactants.

REACTIONS IN SOLUTIONS
1. Mass per cent :
It is obtained by the following relation:

2.Mole Fraction :
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as

3. Molarity :
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution.

4. Molality :
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by ‘m’.

Note :
Molarity of a solution changes with temeprature. But molality of a solution does not change with temperature since mass remains unaffected with temperature.

Students can Download Chapter 14 Oscillations 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 14 Oscillations

Plus One Physics Oscillations One Mark Questions and Answers

Question 1.
Fill in the blanks :
A girl is swinging on a swing in a sitting position. When she stands up, the period of the swing will______.
Answer:
Decreases

Question 2.
A particle executes a simple harmonic motion with a frequency f. What is the frequency with which its kinetic energy oscillates?
Answer:
Frequency of oscillation of kinetic energy is 2f.

Question 3.

Answer:
a. 7/2R
b. ω = \(\sqrt{k / m}\)

Question 4.
Can a simple pendulum vibrate at centre of earth?
Answer:
No. Because ‘g’ at centre of earth is zero.

Question 5.
A glass window may be broken by a distant explosion. Why?
Answer:
The sound waves can cause forced vibrations in glass due to difference between frequency of sound wave and natural frequency of glass. This can break the glass window.

Plus One Physics Oscillations Two Mark Questions and Answers

Question 1.
A simple pendulum is transferred from earth to moon. Will it go faster or slower?
Answer:
The value of g at moon is low compared to earth. The decrease in g will increase time period of simple pendulum. Hence pendulum will vibrate slower.

Question 2.
Here five examples of accelerated motion are given in first column. Match each examples given in the second column.

Answer:

Question 3.
A girl is swinging a swing in sitting position. What shall be the effect of frequency of oscillation if

1. if she stands up
2. if another girl sits gently by her side

Answer:
1. If the system is considered as simple pendulum, length of pendulum is reduced as girl stands up.

So frequency of oscillation is increased.

2. The time period and hence frequency of simple pendulum is independent of mass. Hence there is no change in frequency.

Plus One Physics Oscillations Three Mark Questions and Answers

Question 1.
A student is advised to study the variation of period of oscillation with the length of a simple pendulum in the laboratory. According he recorded the period of oscillation for different lengths of the pendulum.

1. If he plots a graph between the length and period of oscillation, what will be the shape of the graph?
2. How would you determine the value of acceleration due to gravity using l – T² graph?

Answer:
1.

This is the equation parabola. Hence the shape of graph between period(T) and length (l) will be parabola.

2. Find slope of l – T² graph. The acceleration due to gravity can be found using formula g = 4P² × slope of (l – T²) graph.

Question 2.
A simple pendulum has a bob of mass m is suspended from the ceiling of a lift which is lying at the ground floor of a multistoried building.

1. Find the period of oscillation of pendulum when the lift is stationary.
2. What is the tension of the string of the pendulum when it is ascending with an acceleration ‘a’?
3. What is the period of oscillation of the pendulum while the lift is ascending?

Answer:

1. T = \(2 \pi \sqrt{l /g}\)
2. Tension, T= m (g + a)
3. T = 2π\(\sqrt{\frac{\ell}{g+a}}\)

Question 3.
A body tied a spherical pot with a string and suspended it on a clamp. He then filled it with water. Length of the string if 90 cm and diameter of the pot is 20 cm. The pot is slightly displaced to one side and leave it to oscillate. Considering the above example as a simple pendulum (g = 9.8 ms^-2)

1. What is the length of the Pendulum
2. Calculate the period of oscillation of the pendulum.

Answer:
1. Length of pendulum l = 90 + 10 = 100 cm

2.

Question 4.

1. Motion repeated at regular intervals of time is called periodic. Explain the simple harmonic motion with a figure.
2. A particle executes a simple harmonic motion with a period 2 seconds, starting from its equilibrium at time t = 0. Find the minimum time in which it is displaced by half the amplitude.

Answer:
1. A periodic motion in which acceleration is directly proportional to displacement but opposite in direction is called SHM.

The graphical variation of simple harmonic motion with time given above.

2. y = a sin wt

Sin π/6 = sin wt
wt = π/6
\(\frac{2 \pi}{T}\)t = π/6
But T = 2s. Hence we get

Question 5.
A spring of spring constant ‘k’ is used to suspend a mass ‘m’ at its free end while the other end of the spring is rigidly fixed.

1. If the mass is slightly depressed and released, then name the motion of the mass.
2. Write down the expression for the period of oscillation of the mass.
3. If this system is taken into outer space then what happens to its period? Why?

Answer:

1. Simple Harmonic Motion
2. T = \(2 \pi \sqrt{m / k}\)
3. Period of oscillation does not change.

Plus One Physics Oscillations Four Mark Questions and Answers

Question 1.
A simple harmonic motion is represented by x(t) = Acosωt.

1. What do you mean by simple harmonic motion.
2. An SHM has amplitude A and time period T, What is the time taken to travel from x = A to x = A/2

Answer:
1. Simple harmonic motion is the simplest form of oscillatory motion. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the partide from the origin varies with time as

2. x = Acosωt

Question 2.

1. Define simple harmonic motion for a particle moving in a straight line.
2. Use your definition to explain how simple harmonic motion can be represented by the equation.
3. Show that the above equation is dimensionally correct.
4. A mechanical system is known to perform simple harmonic motion. What quantity must be measured in order to determine frequency for the system?

Answer:

1. A periodic motion in which acceleration is directly proportional to displacement and opposite in direction is called simple harmonic oscillation.
2. Mathematically, a simple harmonic oscillation can be expressed as
   y = a sin wt (or) y = a cos wt
3. y = a sin w t
   Sin wt has no dimension. Hence we need to consider dimension of ‘a’ only, ie, y = a, L = L
4. Its period is determined.

Question 3.
A particle executes simple harmonic motion according to the equation x = 5sin\(\left(\frac{2 \pi}{3} t\right)\)

1. find the period of the oscillation
2. What is the minimum time required for the particle to move between two points 2.5cm on either side of the mean position?

Answer:
1. x = 5sin\(\left(\frac{2 \pi}{3} t\right)\), when we compare this equation
with standard S.H.M, x = a sin wt.
We get wt = \(\frac{2 \pi}{3} t\)

2. y = a sin wt
2.5 = 5 sin w × t

Time taken to travel 2.5 from the mean position is 0.25 sec. Hence time taken to travel 2.5 cm on either side of the mean position is 0.5 sec.

Question 4.
A mass m is suspended at one end of a spring and the other end of the spring is firmly fixed on the ceiling. If the mass is slightly depressed and released it will execute oscillation.

1. Write down the expression for the frequency of oscillation of the mass.
2. If the spring is cut into two equal halves and one half of the spring is used to suspend the same mass then obtain an expression for the ratio of periods of oscillation in two cases.
3. If this system is completely immersed in water then what happens to the oscillation?

Answer:
1. f = \(\frac{1}{2 \pi} \sqrt{k / m}\)

2. f₁ = [atex]\frac{1}{2 \pi} \sqrt{k / m}[/latex] ____(1)

3. When spring is cut in to half,spring constant becomes 2k. Hence frequency,

4. If this system is immersed in water, the amplitude of oscillation decrease quickly. Hence system comes to rest quickly.

Question 5.
Starting from the origin, a body oscillates simple harmonically with an amplitude of 2m and a period of 2s.

1. What do you mean simple harmonic motion.
2. Draw the variation of displacement with time for the above motion.
3. After what time, will its kinetic energy be 75% of the total energy?

Answer:
1. The oscillatory motion is said to be simple harmonic motion if the displacement ‘x’ of the particle from the origin varies with time as

where
x(t) = displacement x as a function of time t
A = amplitude
ω = angular frequency
(ω t + Φ) = phase (time-dependent)
Φ = phase constant or initial phase

2.

3. y = a sin ω t
ν = aω cos ωt
Kinetic energy,

Question 6.
A body of mass 16 kg is oscillating on a spring of force constant 100 N/m.

1. What do you mean by spring constant.
2. Derive a general expression for period of oscillating spring.

Answer:
1. Spring constant is the force required for unit extension.

2.

Consider a body of mass m attached to a massless spring of spring constant K. The other end of spring is connected to a rigid support as shown in figure. The body is placed on a frictionless horizontal surface.

If the body be displaced towards right through a small distance Y, a restoring force will be developed.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0
We get ω² = k/m

Question 7.
A particle execute simple harmonic motion according to the equation, x = 5sin \(\left(\frac{2 \pi}{3}\right) t\)

1. What is the period of the oscillation?
2. Write an expression for velocity and acceleration of the above particle.

Answer:
1. When x = 5sin\(\left(\frac{2 \pi}{3}\right) t\) with standard equation.
x = asin ω t we get

2. Velocity V = \(\frac{d}{d t}\) 5sin\(\left(\frac{2 \pi}{3}\right) t\)

Question 8.

1. Arrive the differential equation of SHM.
2. What do you mean by seconds pendulum.

Answer:
1. The force acting simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.

Comparing this equation with standard differential equation \(\frac{d^{2} x}{d t^{2}}\) + ω²x = 0,
We get ω² = k/m.

2. A pendulum having period 2 sec is called seconds pendulum.

Question 9.
SHM is a type of motion in which both speed and acceleration change continuously.

1. Which of the following condition is sufficient for SHM?
   • a = ky,
   • a = ky
   • a = ky²
2. Draw a graph of SHM between
   • displacement-time
   • speed – time
   • acceleration -time

Answer:
1. a = ky

2. Variation of displacement Y with time t

Variation of velocity Y with time t

Variation of acceleration with time

Question 10.

1. Is oscillation of a mass suspended by a spring is simple harmonic?
2. Write period of oscillation of the spring?
3. There are two springs, one delicate and another stout. For which spring, the frequency of oscillation will be more?
4. Two un equal springs of same material are loaded with same load, which one will have large value of time period?

Answer:
1. Yes

2. T = \(2 \pi \sqrt{\frac{m}{k}}\)

3. When a stout spring loaded with mass ‘m’, the extension (l) produced is large.
∴ T is large, because T = \(2 \pi \sqrt{\frac{m}{k}}\),
T is small, i.e, frequency is large. Stout spring oscillate with larger frequency.

4. When a longer spring is locked with weight mg, the extension T is more
∴ T is large, because T = \(2 \pi \sqrt{\frac{l}{g}}\)
So longer spring will have a large value of period.

Question 11.
A simple pendulum consists of a metallic bob suspended from a long straight thread whose one
end is fixed to a rigid support.

1. What is the time period of second’s pendulum?
2. Derive an expression for period of simple pendulum.

Answer:
1. 2 sec

2.

Consider a mass m suspended from one end of a string of length L fixed at the other end as shown in figure. Suppose P is the instantaneous position of the pendulum. At this instant its string makes an angle θ with the vertical.

The forces acting on the bob are (1) weight of bob F_g(mg) acting vertically downward. (2) Tension T in the string.

The gravitational force F_g can be divided into a radial component F_gCosθ and tangential component F_gSinθ. The tangential component F_gSinθ produces a restoring torque.
Restoring torque τ = – F_g sinθ . L
τ = – m_g sinθ . L _____(1)
-ve sign shown that the torque and angular displacement θ are oppositely directed. For rotational motion of bob,
τ = Iα ______(2)
Where I is moment of inertia about the point of suspension and α is angular acceleration. From eq (1) and eq (2).
Iα = – mgsinθ . L
If we assume that the displacement θ is small, sinθ ≈ θ
∴ Iα = -mgθ . L
Iα + mgθ L = 0

Comparirig eq (3) with standard differential equation


∴ period of simple pendulum,

for simple pendulum I = mL²
Substituting I = mL² we get

The above equation gives that T is indepent of mass.

Plus One Physics Oscillations Five Mark Questions and Answers

Question 1.

1. Which of the following condition is sufficient for the simple harmonic motion?
   • a = ky
   • a = ky²
   • a = -ky
   • a = -ky²
     Where ‘a’ – acceleration, y – displacement
2. Prove that simple harmonic motion is the projection of uniform circular motion on any diameter of the circle.
3. Represent graphically the variations of potential energy, kinetic energy and total energy as a function of position ‘x’ for a linear harmonic oscillator. Explain the graph.

Answer:
1. a = -ky

2.

Consider a particle moving along the circumference of a circle of radius ‘a’ and centre O, with uniform angular velocity w. AB and CD are two mutually perpendicular diameters along X and Y axis. At time t = 0.

let the particle be at P₀ so that ∠P₀OB = Φ. After time ‘t’ second, let the particle reach P so that ∠POP₀ = ω t. N is the foot of the perpendicular drawn from P on the diameter CD.

Similarly M is the foot of the perpendicular drawn from P to the diameter AB. When the particle moves along the circumference of the circle, the foot of the perpendicular executes to and fro motion along the diameter CD or AB with O as the mean position.

From the right angle triangle O MP, we get
Cos (ωt + Φ) = \(\frac{O M}{O P}\)
∴ OM = OPcos(ωt + Φ)
X= a cos (ωt + Φ) _______(1)
Similarly, we get
Sin (ωt + Φ) = \(\frac{y}{a}\) (or)
Y = a sin (ωt + Φ) _______(2)
Equation (1) and (2) are similar to equations of S.H.M. The equation(1) and (2) shows that the projection of uniform circular motion on any diameter is S.H.M.

3. KE = PE
\(\frac{1}{2}\)mω²(a² – x²) = \(\frac{1}{2}\) = mω²x²
Solving we get, x = \(\frac{a}{\sqrt{2}}\)
where a is the amplitude of oscillation.

Question 2.
The spring has a scale that reads from zero to 30 kg. The length of the scale is 30cm.

1. Calculate the force constant K
2. If the period of oscillation is 1 sec. Calculate mass of the body attached to the spring.
3. If the spring is cut into two halves, What is the force constant of each half?

Answer:
1. 30 cm = 30kg
1 cm = \(\frac{30}{30}\) = 1kg
∴ Spring constant K = 1 kg/cm = 10N/cm =1000N/m.

2.

3. If spring of spring constant K is cut in to half, spring constant of each half became, 2K

Plus One Physics Oscillations NCERT Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow.
Answer:
(b) A freely suspended bar magnet displaced from its N-S direction and released and
(c) A hydrogen molecule rotating about its centre of mass

Question 2.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = 200x²
(c) a = 10x
(d) a = 100x²
Answer:
Only (c) ie. a = -10x represent SHM. This is because acceleration is proportional and opposite to displacement (x).

Question 3.
A spring having spring constant 1200Nm^-1 is mounted on a horizontal table as shown in figure. A mass of 3 kg is attached to the free end of the spring. The mass is the pulled sideways to a distance of 2.0cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass and
3. the maximum speed of the mass.

Answer:
k = 1200Nm^-1, m = 3kg, a = 2.0cm = 2 × 10^-2m
1.

2. Maximum acceleration = ω² a = \(\frac{k}{m}\)a

3. Maximum speed = aω = a\(\sqrt{\frac{K}{m}}\)

Question 4.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m. If the position moves with simple harmonic motion with an angular frequency of 200 rev/min, what is its maximum speed?
Answer:
a = \(\frac{1}{2}\)m, ω = 200 rev/min
U_max = aω = \(\frac{1}{2}\) × 200m/min = 100 m/min.

Question 5.
A circular disc of mass 10kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -αθ, where J is the restoring couple and θ the angle of twist).
Answer:

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Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Plus One Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
The greatest affinity for haemoglobin is shown by
a) NO
b) CO
c) O₂
d) CO₂
Answer:
b) CO

Question 2.
London smog is ___________ in nature.
Answer:
reducing

Question 3.
Addition of phosphate fertilizers into water leads to
a) Increased growth of decomposers
b) Reduced algal growth
c) Increased algal growth
d) Nutrient enrichment
Answer:
d) Nutrient enrichment

Question 4.
Which of the following is a viable particulate?
a) Algae
b) Smoke
c) Mist
d) Fumes
Answer:
a) Algae

Question 5.
Which of the following is not a greenhouse gas?
a) CO₂
b) CH₄
c) O₂
d) Water vapour
Answer:
c) O₂

Question 6.
Methyl isocyanate is prepared by the action of ___________ on methyl amine
Answer:
COCl₂

Question 7.
In a photo chemical smog the gas that causes eye irritation is ___________ .
a) CO₂
b) CH₄
c) PAN
d) Acrolein
e) both (c) and (d)
Answer:
PAN

Question 8.
Super sonic jet planes can contribute to zone depletion by introduction of the gas straight to the stratosphere
Answer:
NO

Question 9.
The difference between the amounts of dissolved oxygen in a sample of water saturated with oxygen and that after incubation for a period of 5 days is known as.
Answer:
BOD

Question 10.
Excess of nitrate ions in drinking water causes.
Answer:
Methenoglobinemia

Plus One Chemistry Environmental Chemistry Two Mark Questions and Answers

Question 1.
Write the equation for the combustion of ethane.
a) Complete combustion
b) Incomplete combustion
Answer:
a) 2C₂H₆(g) +7O₂ → 4CO₂(g) + 6H₂O(v)
b) 2C₂H₆(g)+5O₂ → 4CO(g) + 6H₂O(v)

Question 2.
Write the three common pollutants.
Answer:
1. Gases such as CO and SO₂
2. Compounds of metals like lead, mercury, zinc.
3. Radioactive substance

Question 3.
Three equations are given below. After studying it, write about Acid rain.
(1) CO₂(g) + H₂O(l) → 2H+ (aq) + CO₃^2-(aq)
(2) 2SO₂(g) + O₂(g) +2H₂O(l) → 2H₂SO₄(aq)
(3) 4NO₂(g) + O₂(g) + 2H₂O(l) → 2HNO₃ (aq)
Answer:
Oxides of Sulphur and Nitrogen, mist of HCI and Phosphoric acid, etc. present in polluted airdissolve in rain water making it more acidic. This is known as acidic rain. SO₂ and NO₂ present in polluted air are the major contributors of acid rain.
2SO₂(g) +O₂(g) + 2H₂O(l) → 2H₂SO₄(aq)
4NO₂(g) +O₂(g) + 2H₂O(l) → 2HNO₃(aq)

Question 4.
Marble of Taj Mahal reacts with acid rain. What is its result?
Answer:
The marble with which Taj Mahal is made is continuously eaten away by acid rain. It is due to the presence of chemical factories in Agra. The acids present in acid rain react with marble and making the marble dull and rough.
CaCO₃ + H₂SO₄ → CaSO₄ +H₂O + CO₂

Question 5.
Why usage of chlorofluorocarbons being discouraged? or Explain ozone layer depletion?
Answer:
The decomposition of CFC’s destroying ozone. CF₂Cl₂(g) + hv → Cl(h) + CF₂Cl(g)
The reactive Cl atom reacts with O₃ to form ClO radical. Cl(g) + O₃ → ClO(g) + O₂(g)
Thus each Cl atom produced, can destroy many O₃ molecules. This leads to ozone depletion.

Question 6.
What is meant by BOD?
Answer:
It is biochemical oxygen demand. It is the amount of dissolved oxygen required by micro-organisms to oxidise organic and inorganic matter present in polluted water.

Question 7.
1. How we can control air pollution?
2. How does the soil pollution occur?
Answer:
1. Air pollution can be controlled by the following ways.
• Exhaust to oxidize CO to CO₂.
• CO₂ level can be maintained by planting trees.
• Hydrogen gas is looked upon as pollution less future fuel.
• The large amount of nitrogen oxides emitted from power plant can be removed by scrubbing it with H₂SO₄.
2. It occurs due to
• Indiscriminate use of fertilizers, pesticides etc.
• Dumping of waste materials.
• Deforestation

Question 8.
What do you mean by greenhouse effect?
Answer:
Global warming is caused by the excess amount of CO₂ in the atmosphere. When CO₂ accumulate due to the decrease of trees it will increase the temperature of earth. This leads to melting of ice. So the sea level rises.

Question 9.
What do you mean by global warming?
Answer:
As more and more infrared radiations are trapped, the atmosphere becomes hotter and the global temperature rises up. This is known as global warming. There has been a marked increase in the levels of CO₂ in the atmosphere due to severe deforestation and burning of fossil fuels.

Question 10.
Give two examples in which green chemistry has been applied.
Answer:
1) Oxidative cracking process in the formation of ethylene is a significant step.
2) Fuel cells for cellular phones have been developed. This cell last for the full life time of the phone.

Question 11.
Oxygen play a key role in the troposphere while ozone in the stratosphere.
i) How is ozone formed in the atmosphere?
ii) What are the causes of depletion of ozone layer?
Answer:
i) By electric discharge of O₂ during lightning.
ii) UV radiations enter to the earth surface and it causes skin diseases.

Question 12.
No new industries are allowed in thickly populated cities by Government order. Why such an order is issued?
Answer:
The city will be destroyed due to acid rain caused by industrial pollution. The Govt, order was to protect the environment (orthe city).

Question 13.
Discuss the importance of dissolved oxygen in water.
Answer:
It helps living organisms in water. They inhale dissolved oxygen in water.

Question 14.
Define environmental chemistry.
Answer:
Environmental chemistry is defined as the branch of science which deals with the chemical processes occurring in the environment. It involves the study of origin, transport, reactions, effects and the fates of chemical species in the environment.

Question 15.
Explain tropospheric pollution.
Answer:
Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.
• Gaseous air pollutants. These include mainly oxides of sulphur (SO₂, SO₃), oxides of nitrogen (NO, NO₂) and oxides of carbon (CO, CO₂) in addition , to hydrogen sulphide (H₂S), Hydrocarbons, ozone and other oxidants.
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Question 16.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
The main gas responsible for greenhouse effect is CO₂. Other greenhouse gases are methane, nitrous oxide, water vapours, chlorofluorocarbons (CFC’s) and ozone.

Question 17.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process they consume the oxygen dissolved in water. Water has already limited dissolved oxygen (≈10 ppm) which gets is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead in water.

Question 18.
How can domestic waste be used as mannure?
Answer:
Domestic waste comprises two types of materials, biodegradable such as leaves, rotten food, etc, and non-biodegradable such as plastics, glass, metal scrap, etc. The non-biodegradable waste is sent to industry for recycling. The biodegradable waste should be deposited in the land fills. With the passage of time, it is converted into compost manure.

Question 19.
For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Plus One Chemistry Environmental Chemistry Three Mark Questions and Answers

Question 1.
Fish grow in cold water as well as in warm water.
1. Do you agree? What is the reason?
2. pH of rain water is 5.6. Is it true? What is the reason?
Answer:
1. No. Fish grow in cold water. Warm water contains less dissolved O₂ than cold water,

2. Normally rain water contains dissolved CO₂ and hence it shows acidic pH of 5.6.
H₂O + CO₂ \(\rightleftarrows \) H₂CO₃ or 2H⁺ + CO₂
When pH of rain water became below 5.6, it becomes acid rain.

Question 2.
Write the mechanism of formation of photochemical smog.
Answer:
At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in airto form NO₂ which absorbs sunlight and form NO and free O atom.

The O atoms being reactive and combine with O₂ to form O₃.
O₂ (g) + O(g) → O₃(g)
The O₃ react with NO formed by the photochemical decomposition of NO₂.
NO(g) + O₃(g) → NO₂(g) + O₂(g)
NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted airto form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Question 3.
1. What are the major pollutants of water?
2. What is meant by eutrophication?
Answer:
1. Micro-organism present in domestic sewages, organic wastes, plant nutrients, toxic metals, sediments, pesticides and radioactive substances,

2. Addition of P to water as PO₄^3- ion encourages the formation of algae which reduces the dissolved oxygen content of water. This is called eutrophication.

Question 4.
1. Acid rain causes extensive damage to vegetation and aquatic life. )
i) What do you mean by acid rain?
ii) Name the chemicals responsible for acid rain.
2. List gases which are responsible for greenhouse effect.
Answer:
1. i) When the pH of the rain water drops below 5.6, it is called acid rain.
ii) Oxides of nitrogen and sulphur mist of hydrochloric acid and phosphoric acid etc.

2. Such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFC).

Question 5.
Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer:
The statues of monuments in India are affected by acid rain. For example, the air in the vicinity of Taj Mahal contains very high levels of oxides of sulphur and nitrogen. This results in acid rain which reacts with marble of Taj Mahal causing pitting.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)2 +H₂O + CO₃.
As a result, the monument is being slowly eaten away and the marble is getting decolourised and lustreless. Thus, acid rain is considered as a threat to Taj Mahal.

Question 6.
1. Writethree major consequences of air pollution.
2. Write any two suitable methods to control air pollution.
Answer:
1. We cannot get pure oxygen for inspire.
Air pollution causes global warming.
It leads to acid rain.
It leads to diseases.

2. Plant trees.
Reduce the use of motor vehicles.
Do not burn plastics.

Question 7.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
CO combines with haemoglobin to forms a complex entity, carboxyhaemoglobin which is about 300 times more stable than oxy-haemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches 3 -4%, the oxygen-carrying capacity of the blood is significantly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eyesight etc.

Plus One Chemistry Environmental Chemistry Four Mark Questions and Answers

Question 1.
Classify the following pollutants:
(a) Carbon monoxide (CO)
(b) Detergents
(c) Plastic
(d) DDT
(e) Sewage
(f) Cigarette smoke
Answer:

Question 2.
As a result of stratospheric pollution, the ozone layer is destructed.
a) What is ozone layer?
b) How is it useful to us?
c) Write a harmful effect of ozone layer depletion.
Answer:
a) The layer of ozone seen in stratosphere is called ozone layer.
b) Ozone layer plays a significant role in protecting earth from UV rays.
c) Due to ozone layer depletion agricultural crops were found to give reduced yields. Small aquatic organisms which are very sensitive are destroyed due to the increase in the level of UV radiation.

Question 3.
a) What is smog?
b) What are the adverse effects of photochemical smog?
c) Write any two methods to control photochemical smog.
Answer:
a) Smog is a mixture of smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog:
1) Classical smog
2) Photochemical smog

b) Adverse effects of photochemical smog:
• Eye irritants
• Nose irritation
• Head ache
• Chest pain
• Dryness of throat
• Cough
• Difficulty in breathing

c) 1) Use catalytic converters in automobiles.
2) Plant certain plants (e.g. Pinus) which can metabolise nitrogen oxide.

Students can Download Chapter 13 Hydrocarbons 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 13 Hydrocarbons

Plus One Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
Which of the following cannot be prepared by Wurtz reaction?
a) CH₄
b) C₂H₆
c) C₃H₈
d) C₄H₈
Answer:
a) CH₄

Question 2.
The cyclic polymerization of propane produces __________ .
Answer:
1, 3, 5-trimethylbenzene

Question 3.
The reaction

a) Hydration
b) Dehydration
c) Dehydrogenation
d) Dehalogenation
Answer:
b) Dehydration

Question 4.
Say TRUE or FALSE.
Calcium carbide on hydrolysis gives ethylene.
Answer:
False

Question 5.
3-Hexyne reacts with Na/liquid NH₃ to produce
a) cis-3-Hexene
b) trans-3-Hexene
c) 3-Hexylamine
d) mm2-Hexylamine
Answer:
b) trans-3-Hexene

Question 6.
Fill in the blanks after finding the correct relationship
CH₃ – O – CH₃ : Ether, CH₃-CH₂-OH: …………….
Answer:
Alcohol

Question 7.
Choose the correct answer from the brackets given below:
1) General formula of alkene (C_nH_2n, C_nH_4n-2)
2) The 2 different forms of elemental carbon (Bitumen, Diamond, Charcoal, Coke, Led, Graphite)
3) Find the odd man out. Give reason (C₂H₄, C₃H₆, C₄H₈, C₂H₆)
Answer:
1) C_nH_2n
2) Diamond, graphite
3) C₂H₆. It is an alkane. All others are alkenes.

Question 8.
The structural formula of a compound and the name given by a student to it is given below:

Answer:
The name given by the student is wrong. The correct name of the compound is 3-Ethyl -2-methyl nonane.

Question 9.
Which is not the isomer of CH₃-CH₂-O-CH₂-CH₃? (1)
a) CH₃-O-CH₂-CH₂-CH₃
b) CH₃-CH₂-CH₂-CH₂-OH
c) CH₃-CH₂-CO-CH₃
d) CH₃-CH₂(OH)-CH-CH₃
Answer:
c) CH₃CH₂COCH₃

Question 10.
Gammexane has the formula
Answer:
C₆H₆Cl₆

Question 11.
Bayer’s reagent is __________ .
Answer:
dil. alkaline KMnO₄

Question 12.
The electrophile attacking benzene during nitration is __________ .
Answer:
NO₂⁺

Question 13.
The compound that is least readily nitrated is __________ .
a) phenol
b) Toluene
c) Ethylbenzene
d) Benzoic acid
e) Xylene
Answer:
d) Benzoic acid

Question 14.
The hydrocarbon formed when Beryllium carbide is treated with water is __________ .
Answer:
Methane

Plus One Chemistry Hydrocarbons Two Mark Questions and Answers

Question 1.
The lUPAC names of 2 compounds with their structural formulae is given below. Make out the errors and correct them.
A. 2, 2-Dimethyl-3-hexyne

B. 1-Butyne
CH₃-CH₂-CH=CH₃
Answer:
A is wrong. The strucutral formula of 2,2-Dimethyl -3- hexyne is

Question 2.
Is there an organic compound named 2-Ethylpentane? Why? If no, write the correct answer.
Answer:
No. When an ethyl group comes with the second carbon atom, the longest chain will have 6 carbon atoms. Hence its correct name will be 3-Methyl hexane.

Question 3.
Write down 2 similarities and 2 difference of the hydrocarbons given below:

Answer:

Question 4.
An equation for combined Chemical reaction of acetylene is given below:

Find the products X’ and ‘Y’?
Answer:

Question 5.
In a Chemistry class, teacher asked students to write the geometrical form of dicarboxylic acid with the formula C₄H₄O₄. For this question, one student wrote

Both of them argued fortheir answers. Hearing this argument teacher told that both of them are right and she also explained the reason for it. Can you write the answer given by the teacher?
Answer:
Two geometrical isomers are possible for dicarboxylic acid. They are cis and transform. In the cis form, similar groups are present on the same side of the double bond whereas in transform, identical groups are present on different side of the double bond.

Question 6.
Match the following:

Answer:

Question 7.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect. Anti markonikov’s rule.

Plus One Chemistry Hydrocarbons Three Mark Questions and Answers

Question 1.
Reactions
CH₃-C ≡ C-CH₃ + H₂
Reaction: 2
X + HCl → Y
a) What type of reaction is reaction 1 ?
b) Find X and Y.
c) Which are the different products obtained from the reaction of X with oxygen?
Answer:
a) Addition reaction

Question 2.

Here are some functional groups. You are supposed to form 3 structures of hydrocarbons using 3 different functional groups and try to find their names.
Answer:

Question 3
CH₃ – CH₂ – CH₂ – CH₂ – OH, CH₃ – CH₂ – CH – CH₃
i)For which isomerism can the example above be considered?
ii) Define it.
Answer:
i)Position isomerism
ii) Position isomerism arises as a result of the difference in the position of double bond, triple bond or functional group.

Question 4.
Explain the following with necessary chemicals equations.
i) Wurtz reaction
ii) Kolbe’s reaction
iii) Ozonolysisof alkenes
Answer:
i) Wurtz reaction – when alkyl halide is allowed to react with metallic sodium in presence of dry ether an alkane is obtained.

ii) Kolbes reaction – When a solution of sodium acetate is electrolized, ethane is obtained.

iii) Ozonolysis ofalkene: When an alkene is allowed to react with ozone, an ozonide is obtained. This on hydrolysis gives Aldehyde or ketone. The whole process is called ozonolysis.

Question 5.
b) Complete the reaction:

Answer:

Question 6
1. Explain the reaction between sodium metal and bromoethane in dry ether. (3)
2. Draw Sawhorse and Newman’s projections of the different conformers of ethane.
Answer:
1. 2CH₃CH₂Br + 2Na → CH₃-CH₂-CH₂-CH₃ + 2NaBr
2.

Question 7.
Analyse the given reactions, give the major products. Justify your answer.
a) HBris added to 1-Butene two products are obtained.
b) Action of excess chlorine with benzene in dark.
c) Addition of chlorine to benzene in uv light.
Answer:

c) When Benzene is allowed to react with chlorine in presence of sunlight. Benzene hexa chloride (BHC) is obtained.

Question 8.
Predict the product in the following reactions and identify the rules:-

Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti Markownikoffs rule of addition).

Question 9.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2)π electrons.

Plus One Chemistry Hydrocarbons Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
Fill in the blanks:

Answer:

Question 3.
Given below are the structures of some hydrocarbons. Pick out the correct IUPAC names for them from the box.

Answer:
1) 4 – Ethyl -2, 3-dimethyl heptane
2) 3, 4-Dimethyl hex – 3-ene
3) 4, 5, 5- triethyl 3 methyl 2 heptyne
4) 3ethyl-2, 5, 5, trimethylheptane

Question 4.
a) I am an unsaturated hydrocarbon.
b) My wordroot is pent.
c) My suffix is ene. The double bond lies between 2nd and 3rd carbon atoms.
d) l have a branch of methyl group on my second carbon atom. Who am I?
Answer:

Question 5.
From the given table find out the isomer pairs and which type of isomerism they have?

Answer:
Isomer pairs
a – ii – functional group isomerism
b – i – chain isomerism
c-iv-position isomerism
d – iii – metamerism Question 6

Qn 6.
Given below is the structural formula of a compound written by a student.

i) Draw all the possible conformers of the compound?
ii) Arrange them in the order of stability?
Answer:
i)

ii) Eclipsed

Question 7.
A cross word puzzle.
Down
1. Two or more compounds having the same
molecular formula but different physical or chemical properties are known as isomers and the phenomenon is known as
2. Hydrocarbons having the general formula C_nH_2n.
3. IUPAC name of C₆H₅CH₃.
4. ………….. is added to the word root to show whether the hydrocarbon is saturated or unsaturated.
5. Hydrocarbons contain carbon-carbon triple bond.

Cross
6. From where does the inorganic compounds mainly originate from?
7. Name the alkene with the moelcular formula C₁₀H₂₀
8. Prefix of the functional group carboxylic acid.
9. Compounds with two rings.
10. Compounds having same molecular formula but different arrangements of carbon atoms on either side of the same functional group are called ……..

Answer:

Question 8.
1. Name the product obtained when HBr is added to propene. Why?
2. Acetylene is more acidic than ethylene or ethane. Why?
Answer:
1. When propene is treated with HBr, both 2-bromopnopane and 1-bromopropane are formed as products. The major product is2-Bromo-propane. This is due to the rule known as Markonikoffs rule of addition. It states that when a hydrogen halide is added to an unsymmetrical alkene, the halogen atom will goes to the doubly bonded carbon containing lesser number of hydrogen atoms.

2. In Acetylene, the hybridisation of carbon is sp and hence 50% s-character is present.

Question 9.
a) How will you prepare butane in the laboratory using ethyl bromide (CH₃CH₂Br) as one of the raw materials. Write relevant equation.

Identify the product ‘X’. Statethe law that explains the formation of X.
Answer:

The law of Anti-Markownikoff’s rule of addition explains the formation of 1 -Bromopropane.

Question 10.
1. Addition of HBrto propene yields 2-Bromopropane, while in the presence of Benzoyl peroxide. The same reaction yields 1-Bromopropane. Give reason. Justify your answer.
2. Three compounds are given. Benzene, m- dinitrobenzene and toluene. Identify the compound which will undergo nitration most easily and why?
Answer:
1. When propene is allowed to react with HBr in the presence of “peroxide” 2-Bromopropane is obtained as the minor product. (Peroxide effect or Kharach effect or Anti-Markownikoff’s role of addition) Peroxide effect is applicable only in the case of HBr.

2. Toluene. This is because the -CH₃ group, being an activating group activates the benzene ring towards electrophilic substitution in toluene.

Question 11.
a) Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) Howwill you convert:
i) Benzene to toluene
ii) Benzene to nitrobenzene
Answer:

Question 12.
The equations for two chemical reactions are given below:
i) CH ≡ CH + HCl → A → B
ii) OH₄ + O₂ → C + D
Which are the products A, B, C, and D?
Answer:

Question 13.
The 2 conformations shown below belongs to the compound cyclohexane. Infinite number of conformations are possible for cyclo hexane. But these 2 conformations given below has a peculiarity. Try to find it. Also, define the terms conformers and the phenomenon conformation?

Answer:
Infinite number of conformations are possible for cyclohexane. Out of it chair conformation is the most stable one and the boat conformation is the least stable form.

The different arrangement of a compound which arises as a result of rotation about carbon single bond are called conformers and the phenomenon is called conformation.

Plus One Chemistry Hydrocarbons NCERT Questions and Answers

Question 1.
How do you account for formation of ethane during chlorination of methane? (3)
Answer:
Chlorination of methane takes place through a free radical chain mechanism as given below:

From the above mechanism, it is evident that during chain propagation step, CH3free radicals areproduced. In the chain termination step, the two free CH₃ radicals may combine together to form ethane (CH₃-CH₃) molecule.

Question 2.
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (4)
a) C₄H₈ (one double bond)
b) C₃H₈ (one triple bond)
Answer:
a) Isomers of C₄H₈ having one double bond are:

Question 3.
What are the necessary conditions for any compound to show aromaticaly? (3)
Answer:
The conditions for a compound to show aromaticity are:
i) The molecule must be cyclic
ii) It must have a conjugated system of (4n+2) π- electrones
iii) The molecule must be planar so that delocalization of π-electrones can take place.

Question 4
Explain why the following system are not aromatic?

Answer:

It dose not contain all the π- electrons in the ring. Therefore, it is not an aromatic compound.

It contains only four electrons, therefore, the system is not aromatic because it dose not contain (4n + 2) π- electrones.

Cyclootatetraene is a conjugated system having 8 π-electrons.
Therefore, the molecule is not contain (4n+2) π-electrons.

Question 5.
In the alkane, H₂CCH₂C(CH₃)₂ CH₂CH(CH₃)₂, identify 1°, 2°, 3° carbon atoms and give the total number of atoms bonded to each one of these.
Answer:

1° Carban atoms = 5
Hydrogen atoms attached to 1° carbon atoms = 15 2° Carbon atoms = 2
Hydrogen atoms attached to 2° carbon atoms = 4 3° Carbon atom = 1
Hydrogen atoms attached to 3° carbon atom = 1

Question 6.
What effect does branching of an alkane chain has on its melting point?
Answer:
As the branching increases melting point increases.

Question 7.
Why does benzene undergo electrophilic subsitution easily and nucleophilic substitutions with difficulty? (2)
Answer:
Benzene molecule has two n cloud rings, one above and the other below the plane of atoms. Therefore, it is likely to be attached by electrophiles which subsequently brings about substitution.

The nucleophiles would be repelled by the π-electron rings and hence benzene reacts with nucleophiles with difficulty.

Students can Download Chapter 1 The Discipline of 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 1 The Discipline of Computing

Summary
Computing milestones and machine evolution:
People used pebbles and stones for counting earlier days. They draw lines to record information. eg: 1 line for one, 2 lines for two, 3 lines for three, etc. In this number system the value will not change if the lines are interchange. This type of number system is called non positional number system.

Counting and the evolution of the positional number system:
In positional number system, each and every number has a weight. Earlier sticks are used to count items such as animals or objects. Around 3000 BC the Egyptians use number systems with radix 10(base-the number of symbols or digits used in the number system) and they write from right to left.

Later Sumerian/Babylonian use number system with largest base 60 and were written from left to right. They use space for zero instead of a symbol, 0. In 2500 BC, the Chinese use simple and efficient number system with base 10 very close to number system used in nowadays.

In 500 BC, the Greek number system known as Ionian, it is a decimal number system and used no symbols for zero. The Roman numerals consists of 7 letters such as l, V, X, L, C, D, M. The Mayans used number system with base 20 because of the sum of the number of fingers and toes is 10 + 10 = 20.

It is called vigesimal positional number system. The numerals are made up of three symbols; zero (shell shape, with the plastron uppermost), one (a dot) and five (a bar or a horizontal line). To represent 1 they used one dot, two dots for 2, and so on

The Hindu – Arabic number system had a symbol(0)for zero originated in India 1500 years ago. Consider the table to compare the number system

Evolution of the computing machine:
(a) Abacus:
In 3000 BC Mesopotamians introduced this and it means calculating board or frame. It is considered as the first computer for basic arithmetical calculations and consists of beads on movable rods divided into two parts. The Chinese improved the Abacus with seven beads on each wire. Different Abacus are given below.

(b) Napier’s bones:
A Mathematician John Napier introduced this in AD 1617.

(c) Pascaline:
A French mathematician Blaise Pascal developed this machine that can perform arithmetical operations.

(d) Leibniz’s calculator:
In 1673, a German mathematician and Philosopher Gottfried Wilhelm Von Leibniz introduced this calculating machine.

(e) Jacquard’s loom:
In 1801, Joseph Marie Jacquard invented a mechanical loom that simplifies the process of manufacturing textiles with complex pattern. A stored program in punched cards was used to control the machine with the help of human labour. This punched card concept was adopted by Charles Babbage to control his Analytical engine and later by Hollerith.

(f) Difference engine:
The intervention of human beings was eliminated by Charles Babbage in calculations by using Difference engine in 1822. It could perform arithmetic operations and print results automatically

(g) Analytical engine:
In 1833. Charles Babbage introduced this. Charles Babbage is considered as the “Father of computer It is considered as the predecessor of today’s computer. This engine was controlled by programs stored in punched cards. These programs were written by Babbage’s assistant, Augusta Ada King, who was considered as the first programmer in the World.

(h) Hollerith’s machine:
In 1887, Herman Hollerith an American made first electromechanical punched cards with instructions for input and output. The.card contained holes in a particular pattern with special meaning. The Us Census Bureau had large amount of data to tabulate, that will take nearly 10 years.

By this machine this work was completed in one year. In 1896, Hollerith started a company Tabulating Machine Corporation. Now it is called International Business Machines(IBM).

(i) Mark-1:
In 1944 Howard Aiken manufactured automatic electromechanical computer in collaboration with engineers at IBM that handled 23 decimal place numbers and can perform addition, subtraction, multiplication and subtraction.

Generations of computers:
There are five generations of computers from 16^th century to till date.

First generation computers (1940 – 1956):
Vacuum tubes were used in first generation computers. The input was based on punched cards and paper tapes and output was displayed on printouts. The Electronic Numerical Integrator and Calculator(ENIAC) belongs to first generation was the first general purpose programmable electronic computer built by J. Presper Eckert and John V. Mauchly.

It was 30-50 feet long, weight 30 tons, 18,000 vacuum tubes, 70,000 registers, 10,000 capacitors and required 1,50,000 watts of electricity. It requires Air Conditioner. They later developed the first commercially successful computer, the Universal Automatic Computer(UNIVAC) in 1952. Von Neumann architecture

The mathematician John Von Neumann designed a computer structure that structure is in use nowadays. Von Neumann structure consists of a central processing unit(CPU), Memory unit, Input and Output unit. The CPU consists of arithmetic logical unit(ALU) and control unit(CU).

The instructions are stored in the memory and follows the “Stored Program Concept”. Colossus is the secret code breaking computer developed by a British engineer Tommy Flowers in 1943 to decode German messages.

Second generation computers (1956 -1963):
Transistors, instead of Vacuum tubes, were used in 2^nd generation computers hence size became smaller, less expensive, less electricity consumption and heat emission and more powerful and faster.

A team contained John Bardeen, Walter Brattain and William Shockley developed this computer at Bell Laboratories. In this generation onwards the concept of programming language was developed and used magnetic core (primary) memory and magnetic disk(secondary) memory.

These computers used high level languages(high level language means English like statements are used)like FORTRAN (Formula translation) and COBOL(Common Business Oriented Language). The popular computers were IBM 1401 and 1620.

Third generation computers (1964 – 1971):
Integrated Circuits(IC’s) were used. IC’s or silicon chips were developed by Jack Kilby, an engineer in Texas Instruments. It reduced the size again and increased the speed and efficiency. The high level language BASIC(Beginners All purpose Symbolic Instruction Code) was developed during this period.

The popular computers were IBM 360 and 370. Due to its simplicity and cheapness more people were used. The number of transistors on IC’s doubles approximately every two years. This law is called Moore’s Law, it is named after Gordon E Moore. It is an observation and not a physical or natural law.

Fourth generation computers (1971 onwards):
Microprocessors are used hence computers are called microcomputers. Microprocessor is a single chip which contains Large Scale of IC’s(LSI) like transistors, capacitors, resistors,etc due to this a CPU can place on a single chip. Later LSI were replaced by Very Large Scale Integrated Circuits(VLSI). The popular computers are IBM PC and Apple II.

Fifth generation computers (future):
Fifth generation computers are based on Artificial Intelligence(AI). Al is the ability to act as human intelligence like speech recognition, face recognition, robotic vision and movement etc. The most common Al programming language are LISP and Prolog.

Evolution of computing:
Computing machines are used for processing or calculating data, storing and displaying information. In 1940’s computer were used only for single tasks like a calculator. But nowadays computer is capable of doing multiple tasks at a time.

The “Stored Program Concept” is the revolutionary innovation by John Von Neumann helped storing data and information in memory. A program is a collection of instructions for executing a specific job or task.

Augusta Ada Lowelace: She was the Countess of Lowelace and she was also a mathematician and writer. She is considered as the first lady computer programmer.

Programming languages:
The instructions to the computer are written in different languages. They are Low Level Language(Machine language), Assembly Language(Middle level language) and High Level Language(HLL).

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. In assembly language mnemonics (codes) are used to write programs

Electronic Delay Storage Automatic Calculator(EDSAC) built during 1949 was the first to use assembly language. In HLL English like statements are used to write programs. A-0 programming language developed by Dr. Grace Hopper, in 1952, for UNIVAC-I is the first HLL.

A team lead by John Backus developed FORTRAN @IBM for IBM 704 computer and ‘Lisp’ developed by Tim Hart and Mike Levin at Massachusetts Institute of Technology. The other HLLs are C, C++, COBOL, PASCAL, VB, Java etc. HLL is very easy and can be easily understood by the human being.

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.

Algorithm and computer programs:
The step-by-step procedure to solve a problem is known as algorithm. It comes from the name of a famous Arab mathematician Abu Jafer Mohammed Ibn Musaa Al-Khowarizmi, The last part of his name Al-Khowarizmi was corrected to algorithm.

Theory of computing:
It deals with how efficiently problems can be solved by algorithm and computation. The study of the effectiveness of computation is based upon a mathematical abstraction of computers is called a model of computation, the most commonly used model is Turing Machine named after the famous computer scientist Alan Turing.

1. Contribution of Alan Turing:
He was a British mathematician, logician, cryptographer and computer scientist. He introduced the concept of algorithm and computing with the help of his invention Turing Machine.

He asked the question Can machines think’ led the foundation for the studies related to the computing machinery and intelligence. Because of these contributions he is considered as the Father of Modern Computer Science as well as Artificial Intelligence.

2. Turing Machine:
In 1936 Alan Turing introduced a machine, called Turing Machine. A Turing machine is a hypothetical device that manipulates symbols on a strip of tape according to a table of rules. This tape acts like the memory in a computer. The tape contains cells which starts with blank and may contain 0 or 1.

So it is called a 3 Symbol Turing Machine. The machine can read and write, one cell at a time, using a tape head and move the tape left or right by one cell so that the machine can read and edit the symbol in the neighbouring cells. The action of a Turing machine is determined by

1. the current state of the machine
2. the symbol in the cell currently being scanned by the head and
3. a table of transition rules, which acts as the program.

3. Turing Test:
The Turing test is a test of a machine’s ability to exhibit intelligent behaviour equivalent to, or indistinguishable from, that of a human. The test involves a human judge engages in natural language conversations with a human and a machine designed to generate performance indistinguishable from that of a human being.

All participants are separated from one another. If the judge cannot reliably tell the machine from the human, the machine is said to have passed the test. The test does not check the ability to give the correct answer to questions; it checks how closely the answer resembles typical human answers. Turing predicted that by 2000 computer would pass the test.