Prasanna

Students can Download Chapter 4 Principles of Programming and Problem Solving Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 4 Principles of Programming and Problem Solving

Plus One Principles of Programming and Problem Solving One Mark Questions and Answers

Question 1.
The process of writing program is called _______.
Answer:
programming or coding

Question 2.
One who writes program is called _________
Answer:
Programmer

Question 3.
The step by step procedure to solve a problem is known as ________
Answer:
Algorithm

Question 4.
Diagrammatic representation of an algorithm is known as __________
Answer:
Flow Chart

Question 5.
Program errors are known as _________
Answer:
bugs

Question 6.
Process of detecting and correcting errors is called ________
Answer:
debugging

Question 7.
Mr. Ramu represents an algorithm by using some symbols. This representation is called _______.
Answer:
Flow Chart

Question 8.
Your computer teacher asked you that which symbol is used to indicate beginning or ending flow chart? What is your answer?
(a) Parallelogram
(b) Rectangle
(c) Oval
(d) Rhombus
Answer:
(c) Oval

Question 9.
You are suffering from stomach ache. The doctor prescribes you for a scanning. This process is related in a phase in programming. Which phase is this?
Answer:
Problem identification

Question 10.
Your friend asked you a doubt that to draw a flow chart parallelogram is used for what purpose?
Answer:
To input/output

Question 11.
Mr. Anil wants to perform a multiplication which symbol is used to represent in a flow chart.
Answer:
It is a processing so rectangle is used

Question 12.
Mr. George wants to check a number is greater than zero and to perform an operation while drawing a flow chart which symbol is used for this?
Answer:
Rhombus

Question 13.
ANSI means ________
Answer:
American National Standards Institute

Question 14.
To indicate the flow of an operation which symbol is used to draw a flow chart.
Answer:
Flow lines with arrow heads

Question 15.
Mr. Johnson is drawing a flow chart but it is not fit in a single page. Which symbol will help him to complete the flow chart?
Answer:
Connectors

Question 16.
Mr. Ravi developed a s/w student information system, He wants to protect the s/w from unauthorized copying. There is an act what is it?
Answer:
Copy right act

Question 17.
Odd man out.
(а) Parallelogram
(b) Oval
(c) Rectangle
(d) Star
Answer:
(d) Star, others are flowchart symbols

Question 18.
Odd man out
(a) Oval
(b) Rhombus
(c) Connector
(d) Triangle
Answer:
(d) Triangle, Others are flowchart symbols

Question 19.
Raju wrote a program and he wants to check the errors and correct if any? What process he has to do for this?
Answer:
debugging

Question 20.
Odd one out.
(a) Problem identification
(b) Translation
(c) debugging
(d) copyright
Answer:
(d) Copy right, others are phases in programming

Question 21.
Odd man out.
(a) syntax error
(b) logical error
(c) Runtime error
(d) printer error
Answer:
(d) Printer error, Others are different types of errors

Question 22.
A computerized system is not complete after the execution and testing phase? What is the next phase to complete the system?
Answer:
Documentation.

Question 23.
Mr. Sathian takes a movie DVD from a CD library and he copies this into another DVD without permission. This process is called _______________
Answer:
Piracy

Question 24.
Mr. Santhosh purchased a movie DVD and he takes several copies without permission. He is a _________
(a) Programmer
(b) Administrator
(c) Pirate
(d) Organizer
Answer:
(c) Pirate

Question 25.
The symbol used for copy right is a _______
(a) @
(b) Copy
(c) &
(d) ©
Answer:
(d) ©

Question 26.
Following are the advantages of flowcharts one among them is wrong. Find it.
(a) Better communication
(b) Effective analysis
(c) proper program documentation
(d) Modification easy
Answer:
(d) Modification easy. It is a disadvantage.

Question 27.
Which flow chart symbol has one entry flow and two exit flows?
Answer:
Diamond

Question 28.
Which flow chart symbol is always used in pair?
Answer:
connector

Question 29.
Program written in HLL is known as ____________
Answer:
Source code

Question 30.
Some of the components in the phases of programming are given below. Write them in order of their occurrence. (1)

1. Translation
2. Documentation
3. Problem identification
4. Coding of a program

Answer:
The chronological order is as follows

1. Problem Identification
2. Coding of a program
3. Translation
4. Documentation

Question 31.
_________ is the stage where programming errors are discovered and corrected.
Answer:
Debugging or compiling

Question 32.
Ramesh has written a C++ program. During compilation and execution there were no errors. But he got a wrong output. Name the type of error he faced.
Answer:
Logical Error

Question 33.
Pick out the software which rearranges the scattered files in the hard disk and improves the performance of the system.
(a) Backup software
(b) File compression software
(c) Disk defragmenter
(d) Antivirus software
Answer:
(c) Disk defragmenter

Question 34.
Some phases in programming are given below.

1. Source coding
2. Execution
3. Translation
4. Problem study

These phases should follow a proper order. Choose the correct order from the following:
(a) 4 → 2 → 3 → 1
(b) 1 → 3 → 2 → 4
(c) 1 → 3 → 4 → 2
(d) 4 → 1 → 3 → 2
Answer:
(d) 4 → 1 → 3 → 2

Question 35.
Which one of the following errors is identified at the time of compilation?
(a) Syntax error
(b) Logical error
(c) Run-time error
(d) All of these
Answer:
(a) Syntax error

Question 36.
Pick the odd one out and give a reason for your finding

Answer:
c. This has one entry flow and more than one exit flow.

OR

b. Used for both input and output.

Plus One Principles of Programming and Problem Solving Two Mark Questions and Answers

Question 1.
A debate on ‘Whether Free Software is to be promoted’ is planned in your class. You are asked to present points in support of Free Software. What would be your arguments, (at least three)?
Answer:
Freedom to use Comparatively cheap Freedom to modify and redistribute

Question 2.
Mr. Roy purchased a DVD of a movie and he found that on the cover there is a sentence copyright reserved and a mark ©. What is it? Briefly explain?
Answer:
It is under the act of copyright and the trademark is © copyright is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.

Question 3.
Can a person who knows only Malayalam talk to a person who knows only Sanskrit normally consider the corresponding situation in a computer program and justify your answer?
Answer:
Normally it is very difficult to communicate. But it is possible with the help of a translator. Translation is the process of converting programs written in High Level Language into Low Level Language (machine Language). The compiler or interpreter is used for this purpose. It is a program.

Question 4.
Define the term, debugging. Write the names of two phases that are included in debugging. (2)

OR

Define the different types of errors that are encountered during the compilation and running of a program.
Answer:
Debugging:
The program errors are called ‘bugs’ and the process of detecting and correcting errors is called debugging. Compilation and running are the two phases.

OR

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.

Question 5.
Write an algorithm to input the scores obtained in three unit tests and find the average score.

OR

Explain the flowchart and predict the output.
Answer:

• Step 1: Start
• Step 2: Read S1, S2, S3
• Step 3: avg = S1 + S2 + S3/3
• Step 4: Print avg
• Step 5: Stop

OR

This flowchart is used to print the numbers as 1,2, 3, ………, 10.

Question 6.
Differentiate between top down design and bottom up design in problem sloving.
Answer:
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.

Question 7.
Answer any one question from 5 (a) and 5 (b).
1. Draw a flowchart for the following algorithm.

• Step 1: Start
• Step 2: Input N
• Step 3: S = 0, K = 1
• Step 4: S = S + K
• Step 5: K = K + 1
• Step 6: If K < = N Then Go to Step 4
• Step 7: Print S
• Step 8: Stop

OR

2. Name the two stages in programming where debugging process is involved. What kinds of errors are removed in each of these stages?
Answer:
1.

2. The two stages are compile time and run time. In the debugging process can remove syntax error, logical error and runtime error.

Question 8.
Answer any one question from 7(1) and 7(2)
1. Observe the following portion of a flowchart. Fill in the blank symbols with proper instructions to get 321 as the output.

2. The following flowchart can be used to print the numbers from 1 to 100. Identify another problem that can be solved using this flowchart and write the required instructions in the symbols.

Answer:
1.

2. The following flowchart can be used to store another problem such as used to print odd numbers lessthan 200.

Question 9.
Write an algorithm to print the numbers upto 100 in reverse order, That is the output should be as 100, 99, 98, 97, …………., 1

OR

Draw a flow chart to check whether the given number is positive, negative or zero.
Answer:

• Step 1: Start
• Step 2: Set i ← 100
• Step 3: if i <= 0 then go to step 6
• Step 4: Print i
• Step 5: Set i ← i — 1 go to step 3
• Step 6: Stop

OR

Plus One Principles of Programming and Problem Solving Three Mark Questions and Answers

Question 1.
When you try to execute a program, there are chances of errors at various stages, Mention the types of errors and explain.
Answer:
1. Syntax error,
eg: 5 = x

2. Logic error:
If the programmer maks any logical mistakes, it is known as logical error.
eg: To find the sum of values A and B and store it in a variable C you have to write C = A + B. Instead of this if you write C = A × B, it is called logic error.

3. Runtime error:
An error occured at run time due to inappropriate data.
eg: To calculate A/B a person gives zero to B. There is an error called division by zero error during run time.

Question 2.
Following is a flow chart to find and display the largest among three numbers. Some steps are missing in the flowchart. Redraw the flow chart by adding necessary steps and specify its purpose. How can this flow chart be modified without using a fourth variable?
Answer:

Question 3.
A flow chart is given below.

1. What will be the output of the above flow chart?
2. How can you modify the above flow chart to display the even numbers upto 20, starting from 2.

Answer:
1. 1, 2, 3, 4, 5

2.

Question 4.
Write an algorithm to check whether the given number is even or odd.
Answer:

• Step 1: Start
• Step 2: Read a number to N
• Step 3: Divide the number by 2 and store the remainder in R.
• Step 4: If R = O Then go to Step 6
• Step 5: Print “N is odd” go to step 7
• Step 6: Print “N is even”
• Step 7: Stop

Question 5.
Write an algorithm to find the largest of 2 numbers?
Answer:

• Step 1: Start
• Step 2: Input the values of A, B Compare A and B.
• Step 3: If A > B then go to step 5
• Step 4: Print “B is largest” go to Step 6
• Step 5: Print “A is largest”
• Step 6: Stop

Question 6.
Write an algorithm to find the sum of n natural numbers and average?
Answer:

• step 1: Start
• Step 2: Set i ←1, S 0
• Step 3: Read a number and set to n
• Step 4: Computer i and n if i > n then go to step 7.
• Step 5: Set S ← S + i
• Step 6: i ← i + 1 go to step 4
• Step 7: avg ← S/n
• Step 8: Print “Sum = S and average = avg”
• Step 9: Stop

Question 7.
Write an algorithm to find the largest of 3 numbers.
Answer:

• Step 1: Start
• Step 2: Read 3 numbers and store in A, B, C
• Step 3: Compare A and B. lf A > Bthengotostep 6
• Step 4: Compare B and C if C > B then go to step 8
• Step 5: print “B is largest” go to step 9
• Step 6: Compare A and C if C > A then go to step 8
• Step 7: Print”A is largest” go to step 9
• Step 8: Print “C is largest”
• Step 9: Stop

Question 8.
Write an algorithm to calculate the simple interest (I =P × N × R/100)
Answer:

• Step 1: Start
• Step 2: Read 3 values for P, N, R
• Step 3: Calculate I ← P × N × R/100
• Step 4: Print “The simple interest = l”
• Step 5: Stop

Question 9.
Write an algorithm to calculate the compound interest (C.l = P × (1 + r/100)n – P)
Answer:

• Step 1: Start
• Step 2: Read 3 number for p, n, r
• Step 3: Calculate C.I = p × (1 + r/100)n – p
• Step 4: Print “The compound Interest = C.l”
• Step 5: Stop

Question 10.
Write an algorithm to find the cube of first n natural numbers (eg: 1, 8, 27, …., n3)
Answer:

• Step 1: Star
• Step 2: Set i ← 1
• Step 3: Read a number and store in n
• Step 4: Compare i and n if i > n then go to step 7
• Step 5: Print i × i × i
• Step 6: i ← i + 1 go to step 4
• Step 7: Stop

Question 11.
Write an algorithm to read a number and find its factorial (n ! = n × (n – 1) × (n – 2) ×………..3 × 2 × 1)
Answer:

• Step 1: Start
• Step 2: Fact ← 1
• Step 3: Read a number and store in n
• Step 4: If n = 0 then go to step 7
• Step 5: Fact ← Fact × n
• Step 6: n ← n – 1 go to step 4
• Step 7: Print “Factorial is fact”
• Step 8: Stop

Question 12.
Draw a flow chart to find the sum of n natural numbers and average.
Answer:

Question 13.
Draw a flow chart to find the largest of 3 numbers.
Answer:

Question 14.
Draw a flow chart to find the largest of 2 numbers.
Answer:

Question 15.
Draw a flow chart to check whether the given number is even or odd.
Answer:

Question 16.
Draw a flow chart to calculate simple interest.
Answer:

Question 17.
Draw a flow chart to calculate compound interest.
Answer:

Question 18.
Draw a flow chart to find the cube of n natural numbers.
Answer:

Question 19.
Draw a flow chart to read a number and find its factorial.
Answer:

Question 20.
Mr. Vimal wants to represent a problem by using a flowchart, which symbols are used for this. Explain.
Answer:
Flow chart symbols are explained below
1. Terminal (Oval)

It is used to indicate the beginning and ending of a problem

2. Input/Output (parallelogram)

It is used to take input or print output.

3. Processing (Rectangle)

It is used to represent processing. That means to represent arithmetic operation such as addition, subtraction, multiplication.

4. Decision (Rhombus)

It is used to represent decision making. One exit path will be executed at a time.

5. Flowlines (Arrows)

It is used to represent the flow of operation

6. Connector

These symbols will help us to complete the flow chart, which is not fit in a single page. A connector symbol is represented by a circle and a letter or digit is placed within the circle to indicate the link.

Question 21.
Jeena uses an algorithm to represent a problem while Neena uses flowchart which is better? Justify your answer?
Answer:
Flowchart is better. The advantages of flow chart is given below.
1. Better communication:
A flow chart is a pictorial representation while an algorithm is a step by step procedure to solve a program. A programmer can easily explain the program logic using a flow chart.

2. Effective analysis:
The program can be analyzed effectively through the flow chart.

3. Effective synthesis:
If a problem is big it can be divided into small modules and the solution for each module is represented in flowchart separately and can be joined together final system design.

4. Proper program documentation:
A flow chart will help to create a document that will help the company in the absence of a programmer.

5. Efficient coding:
With the help of a flowchart it is easy to write program by using a computer language.

Question 22.
A flow chart is a better method to represent a program. But it has some limitation what are they?
Answer:
The limitations are given below

• To draw a flowchart, it is time consuming and laborious work.
• If any change or modification in the logic we may have to redraw a new flow chart.
• No standards to determine how much detail can include in a flow chart.

Question 23.

1. Problem identification	a. Flowchart
2. Steps to obtain. the solution	b. Syntax Error
3. Coding	c. Runtime Error
4. Translation	d. COBOL
5. Debugging	e. X-ray
6. Execution & Testing	f. Compiler

Answer:
1 – e
2 – a
3 – d
4 – f
5 – b
6 – c

Question 24.
Alvis executes an error free program but he got an error. Explain different types of error in detail.
Answer:
There are two types of errors in a program before execution and testing phase.They are syntax error and logical error. When the programmer violates the rules or syntax of the programming language then the syntax error occurred.
eg: It involves incorrect punctuation.

Keywords are used for other purposes, violates the structure etc,… It detects the compiler and displays an error message that include the line number and give a clue of the nature of the error, When the programmer makes any mistakes in the logic, that types of errors are called logical error. It does not detect by the compiler but we will get a wrong output.

The program must be tested to check whether it is error free or not. The program must be tested by giving input test data and check whether it is right or wring with the known results. The third type of errors are Runtime errors.

This may be due to the in appropriate data while execution. For example consider B/C. If the end user gives a value zero for c, the execution will be interrupted because division by zero is not possible. These situation must be anticipated and must be handled.

Question 25.
The following are the phases in programming. The order is wrong rearrange them in correct order.

1. Debugging
2. Coding
3. Derive the steps to obtain the solution
4. Documentation
5. Translation
6. Problem identification
7. Execution and testing

Answer:
The correct order is given below.

1. Problem identification
2. Derive the steps to obtain the solution
3. Coding
4. Translation
5. Debugging
6. Execution and testing
7. Documentation

Question 26.
Draw a flow chart to input ten different numbers and find their average.
Answer:

Question 27.
Draw the flow chart to find the sum of first N natural numbers.
Answer:

Question 28.
Make a flow chart using the given labelled symbols, for finding the sum of all even numbers upto ‘N’

OR

Write an algorithm to accept an integer number and print the factors of it.

Answer:
Draw flowchart in any of the following order
e, c, d, f, i, h, a, b, g
e, d, c, f, i, h, a, b, g
e, c, d, a, f, i, h, b, g
e, d, c, a, f, i, h, b, g

• Step 1: Start
• Step 2: Input n
• Step 3: i = I
• Step 4: if i <= n/2 then repeat step 5 & 6
• step 5: if n % i == 0 print i
• Step 6: i = i + I
• Step 7: Stop

Question 29.
List the two approaches followed in problem solving or programming. How do they differ?
Answer:
Approaches in problem solving:
(a) 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

(b) 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.

Question 30.
Write an algorithm to find the sum of the squres of the digits of a number. (For example, if 235 is the input, the output should be 2² + 3² + 5² = 38)
Answer:

• Step 1: Start
• Step 2: Set S = O
• Step 3: Read N
• Step 4: if N > O repeat step 5, 6 and 7
• Step 5: Find the remainder. That is rem = N % 10
• Step 6 : S = S + rem × rem
• Step 7 : N = N/10
• Step 8 : Print S
• Step 9 : Stop

Question 31.
Consider the following algorithm and answer the following questions:

• Step 1: Start
• Step 2 : N = 2, S = 0
• Step 3: Repeat Step 4, Step 5 while N <= 10
• Step 4: S = S + N
• Step 5: N = N + 2
• Step 6: Print S
• Step 7: Stop

1. Predict the output of the above algorithm.
2. Draw a flowchart for the above algorithm

Answer:
1. The output is 30

2.

Question 32.
“It is better to give proper documentation within the program”. Give a reason.
Answer:
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.

Plus One Principles of Programming and Problem Solving Five Mark Questions and Answers

Question 1.
Mr. Arun wants to develop a program to computerize the functions of supermarket. Explain different phases he has to undergo is detail.

OR

Briefly explain different phases in programming.
Answer:
The different phases in programming is given below:
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 output 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 Ibu Musaa Al-Khowarizmi. The last part of his name Al-Khowafizmi was corrected to algorithm.

(b) Flowchart:
The pictorial or graphical representation of an algorithm is called flowchart.

3. Coding:
The dummy codes (algorithm)or flow chart 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.

Question 2.
Briefly explain the characteristic of an algorithm.
Answer:
The following are the characteristics of an algorithm

1. It must starts with ‘start’ statement and ends with ‘stop’ statement.
2. it should contains instructions to accept input and these are processed by the subsequent instructions.
3. Each and every instruction should be precise and must be clear. The instructions must be possible to carry out, for example consider the examples the instruction “Go to market” is precise and possible to carried out but the instruction “Go to hell” is also precise and can not be carried out.
4. Each instruction must be carried out in finite time by a person with paper and pencil.
5. The number of repetition of instructions must be finite.
6. The desired results must be obtained after the algorithm ends.

Students can Download Chapter 2 Data Representation and Boolean Algebra Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 2 Data Representation and Boolean Algebra

Plus One Data Representation and Boolean Algebra One Mark Questions and Answers

Question 1.
___________ is a collection of unorganized fact.
Answer:
Data

Question 2.
Data can be organized into useful ____________
Answer:
Information

Question 3.
___________ is used to help people to make decision.
Answer:
Information

Question 4.
Processing is a series of actions or operations that convert inputs into __________
Answer:
Output

Question 5.
The act of applying information in a particular context or situation is called ____________
Answer:
Knowledge

Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
(d) Adeline aged 17 years is in class 12. This is information. The others are data.

Question 8.
Raw facts and figures are known as _______
Answer:
data

Question 9.
Processed data is known as _______
Answer:
Information

Question 10.
Which of the following helps us to take decisions?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

Question 11.
Manipulation of data to get information is known as ___________
Answer:
Data processing

Question 12.
Arrange the following in proper order
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:

1. Data Capture
2. Input
3. Storage
4. Process
5. Output
6. Distribution

Question 13.
Pick the odd one out and give reason:
(a) Calculation
(b) Storage
(c) Comparison
(d) Categorization
Answer:
(b) Storage. It is one of the data processing stage the others are various operations in the stage Process

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.

1. (101)₂, (111)₂, (1001)₂, ……….
2. (1011)₂, (1110)₂, (10001)₂, ………

Answer:

1. 1011, 1101
2. 10101, 10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _______________
Answer:
Base

Question 18.
The base of decimal number system is ________
Answer:
10

Question 19.
MSD is ________
Answer:
Most Significant Digit

Question 20.
LSD is _________
Answer:
Least Significant Digit

Question 21.
Consider the number 627. Its MSD is _________
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is __________
Answer:
7

Question 23.
The base of Binary number system is ___________
Answer:
2

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series,
(101)₂, (111)₂, (1001)₂
Answer:
1011, 1101

Question 26.
State True or False. In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is ________
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1, 2, 3, 4, 5, 6, 7, __
Answer:
10

Question 29.
The base of Hexadecimal number system is ________
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexadecimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Question 32.
Convert (1110.01011)₂ to decimal.
Answer:
1110.01011 = 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 + 0 × 2 – 1 + 1 × 2 – 2 + 0 × 2 – 3 + 1 × 2 – 4 + 1 × 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)10

Question 33.
1 KB is bytes.
(a) 2⁵
(b) 2¹⁰
(c) 2¹⁵
(d) 2²⁰
Answer:
(b) 2¹⁰

Question 34.
The base of hexadecimal number system is ________.
Answer:
16

Question 35.
A computer has no _________
(a) Memory
(b) l/o device .
(c) CPU
(d) IQ
Answer:
(d) IQ

Question 36.
Pick the odd man out.
(AND, OR, NAND, NOT)
Answer:
NOT

Question 37.
Select the complement of X + YZ.
(a) \(\bar{x}+\bar{y}+\bar{z}\)
(b) \(\bar{x} .\bar{y}+\bar{z}\)
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)
(d) \(\bar{x}+\bar{y} \cdot \bar{z}\)
Answer:
(c) \(\bar{x} \cdot(\bar{y}+\bar{z})\)

Question 38.
Select the expression for absorption law.
(a) a + a = a
(b) 1 + a = 1
(c) o . a = 0
(d) a + a . b = a
Answer:
(a) a + a . b = a

Question 39.
What is the characteristic of logical expression?
Answer:
Logical expressions yield either true or false values

Question 40.
Name the table used to define the results of Boolean operations.
Answer:
Truth Table

Question 41.
According to ________ law, \(\bar{x}+\bar{y}=\overline{x y}\) and \(\overline{x y}=\bar{x}+\bar{y}\)
Answer:
De Morgan’s law

Question 42.
A NOR gate is ON only when all its inputs are
(a) ON
(b) Positive
(c) High
(d) OFF
Answer:
(d) OFF

Question 43.
The only function of a NOT gate is __________
Answer:
Invert an output signal

Question 44.
NOT gate is also known as _________
Answer:
Inverter

Question 45.
What is the relation between the following statements.
x + 0 = x and x . 1 = x
Answer:
One is the dual of the other expression.

Question 46.
The algebra used to solve problems in digital systems is called __________
Answer:
Boolean Algebra

Question 47.
Pick the one which is not a Basic Gate.
(AND, OR, XOR, NOT)
Answer:
XOR

Question 48.
Select the universal gates from the list. (NAND, NOR, NOT, XOR)
Answer:
NAND, NOR

Question 49.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 50.
Fill up the missing digit.
(41)₈ = ( )₁₆
Answer:

• Step 1: Divide the number into one each and write down the 3 bits equivalent.
• Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.

Answer:
So the answer is 21.

Question 51.
Real numbers can be represented in memory by using __________
Answer:
Exponent and Mantissa

Question 52.
Consider the number 0.53421 x 10^-8 Write down the mantissa and exponent.
Answer:
Mantissa: 0.53421
Exponent: -8

Question 53.
Characters can be represented in memory by using _________
Answer:
ASCII Code

Question 54.
ASCII Code of A’ is __________
Answer:
(100 0001)₂ = 65

Question 55.
ASCII Code of’a’ is __________
Answer:
(110 0001)₂ = 97

Question 56.
Define the term ‘bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Question 57.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 58.
ASCII stands for __________.
Answer:
American Standard Code for Information Interchange

Question 59.
List any two image file formats.
Answer:
BMP, GIF

Question 60.
Name the operator which performs logical multiplication.
Answer:
AND

Question 61.
Name a gate which is ON when all its inputs are OFF .
Answer:
NAND or NOR

Question 62.
Specify the laws applied in the following cases.

1. a (b + c) = ab + ac
2. (a + b) + c = a + (b + c)

Answer:

1. Distribution law
2. Associative law

Question 63.
Pick the correct Boolean expression from the following.
(a) \(A +\bar{A}=163.\)
(b) \(\text { A. } \bar{A}=1\)
(c) \(A \cdot \overline{A B}=A + B\)
(d) A + AB = A
Answer:
(a) & (d)

Question 64.
1’s complement of the binary number 110111 is _________
Answer:
Insert 2 zeroes in the left hand side to make the binary number in the 8 bit form 00110111
To find the 1’s complement, change all zeroes to one and all ones to zero. Hence the answer is 11001000

Plus One Data Representation and Boolean Algebra Two Mark Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get information as a result that must be stored in the computer for future references.

Question 2.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 3.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee. Here data is the ingredients – water, sugar, milk and coffee powder
Information is the final product i.e. Coffee Processing is the series of steps to convert the ingredients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Question 4.
ASCII is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because it is a 7 bit code so it can represent 2⁷ = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 5.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact matvh

A	B
(12)8	(1110)2
F16	25
(19)16	10
(11)8	(13)16
	(17)8
	9

Answer:

A	B
12	10
F	(17)8
(19)16	25
(11)8	9

Question 6.

1. Name various number systems commonly used in computers.
2. Include each of the following numbers into all possible number systems
   123 569 1101

Answer:

1. The number system are binary, octal, decimal and hexa decimal
2. All possible number systems are
   • 123 Octal, decimal and hexa decimal
   • 569 Decimal, hexa decimal
   • 1101 Binary, Octal, Decimal, Hexa decimal

Question 7.
Fill up the missing digit. (Score 2)
If (220)_a = (90)_b then (451)_a = (  )₁₀
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 10. The values of b can be 10 or 16. L.H.S > R.H.S. a < b and a b also.
The possible values of a and b are given below.

Question 8.
Convert (106)₁₀ = (  )₂?
Answer:

Question 9.
Convert (106)₁₀ = (  )₈?
Answer:

Question 10.
(106)₁₀ = (  )₁₆?
Answer:

Question 11.
Convert (55.625)₁₀ = (  )₂?
Answer:
First convert 55, for this do the following

Write down the remainders from bottom to top.
(55)₁₀ = (110111 )₂
Next convert 0.625, for this do the following.

Write down the remainder from top to bottom. So the answer is
(55.625)₁₀ = (110111.101)₂

Question 12.
Convert (55.140625)₁₀ = (  )₈?
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
(55)₁₀ = (67)₈
Next convert 0.140625, for this do the following.

Write down the remainders from top to bottom. So the answer is
(55.140625)₁₀ = (67.11 )₈

Question 13.
Convert (55.515625)₁₀ = (  )₁₆
Answer:
First convert 55, for this do the following.

Write down the remainders from bottom to top.
ie. (55)₁₀ = (37)₁₆
Next convert .515625

So the answer is
(55.515625)₁₀ = (37.84)₁₆

Question 14.
Convert (101.101)₂ = ( )₁₀?
Answer:
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

Question 15.
Convert (71.24)₈ = (  )₁₀?
Answer:
71.24 = 7 × 8¹ + 1 × 8⁰ + 2 × 8^-1 + 4 × 8⁼²
= 56 + 1 + 2/8 + 4/82
= 57 + 0.25 + 0.0625 (71.24)8
(71.24)₈ = (57.3125)₁₀

Question 16.
Convert (AB.88)₁₆ = (  )₁₀?
Answer:

= 160 + 11 + 0.5 + 0.03125
(AB.88)₁₆ = (171.53125)₁₀

Question 17.
Convert (1011)₂ = (  )₈?
Answer:
Step I: First divide the number into groups of 3 bits starting from the right side and insert necessary zeroes in the left side.
0 0 1 | 0 1 1

Step II: Next write down the octal equivalent.

So the answer is (1011)₂ = (13)₈.

Question 18.
Convert (110100)₂ = (  )₁₆
Answer:

• Step I: First divide the number into groups of 4 bits starting from the right side and insert necessary zeroes in the left side.
• Step II: Next write down the hexadecimal equivalent.

So the answer is (110100)₂ = (34)₁₆.

Question 19.
(72)₈ = ( )₂?
Answer:
Write down the 3 bits equivalent of each digit.

So the answer is (72)₈ = (111010)₂.

Question 20.
Convert (AO)₁₆ = (  )₂ ?
Answer:
Write down the 4 bits equivalent of each digit

So the answer is (AO)₁₆ = (1010 0000)₂.

Question 21.
Convert (67)₈ = (  )₁₆?
Answer:
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

Question 22.
Convert (A1)₁₆ = (  )₈?
Answer:
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.

Question 23.
What is the use of the ASCII Code?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the keyboard is represented in memory by using ASCII Code.
eg: A’s ASCII Code is 65 (1000001), a’s ASCII Code is 97 (1100001)

Question 24.
Pick invalid numbers from the following.

1. (10101)₈
2. (123)₄
3. (768)₈
4. (ABC)₁₆

Answer:

1. (10101)₈ – Valid
2. (123)₄ – Valid
3. (768)₈ – Invalid. Octal number system does not contain the symbol 8
4. (ABC)₁₆ – Valid

Question 25.
Convert the decimal number 31 to binary.
Answer:

(31)₁₀ = (11111)₂

Question 26.
Find decimal equivalent of (10001 )₂
Answer:

= 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20
= 16 + 0 + 0 + 0 + 1
= (17)₁₀

Question 27.
If (X)₈ =(101011 )₂ then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.

X = 53

Question 28.
Fill the blanks:
(a) (………..)₂ = (AB)₁₆
(b) (——D—–)₁₆ = (1010 1000)₂
(c) 0.25₁₀ = (—–)₂

Answer:
Write down the 4bit equivalent of each digit
(a) (………..)₂ = (AB)₁₆

= (10101011)₂
(b) (——D—–)₁₆ = (1010 1000)₂
(A D 8)₁₆ =(1010 1101 1000)₂

(c) 0.25₁₀ = (—–)₂

0.25₁₀ = (0.01)₂

Question 29.
Which is the MSB of representation of -80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 30.
Write 28.756 in Mantissa exponent form.
Answer:
28.756 = .28756 × 100
= .28756 × 102
= .28756 E + 2

Question 31.
Represent -60 in 1’s complement form.
Answer:
+60 = 00111100
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
-60 is in 1’s complement is 11000011

Question 32.
Define Unicode.
Answer:
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.

Question 33.
Substract 1111 from 10101 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here the subtrahend is 1111 and minuend is 5 bits. So insert a zero. So subtrahend is 01111. First take the 1’s complement of subtrahend and add 1 to it

Here is a carry. So ignore the carry and the result is +ve.
So the answer is 110

Question 34.
You purchased a soap worth Rs. (10010)₂ and you gave Rs. (10100)₂ and how much rupees will you get back in binary.
Answer:
Substract (10010)₂ from (10100)₂

You will get rupees (10)₂

Question 35.
Draw the logic circuit diagram for the following Boolean expression.
Answer:

Question 36.
Simplify the expression using basic postulates and laws of Boolean algebra.

1. \(\bar{x}+x \cdot \bar{y}\)
2. \(x(y+y . z)+y(\bar{x}+x z)\)

Answer:

1. \(\bar{x}+\bar{y}\)
2. y

Question 37.
Show \(A(\bar{B}+C)\) using NOR gates only.
Answer:

Question 38.
The following statement Demorgan’s theorem of Boolean algebra. Identify and state ‘Break the line, change the sign’.
Answer:
Demorgan’s theorems,
Demorgan’s first theorem,
\(\overline{x + y}\) = \(\bar{x} . \bar{y}\)
Demorgan’s second theorem,
\(\overline{x – y}\) = \(\bar{x} + \bar{y}\)

Question 39.
Prove algebraically that
\(x \cdot y + x \cdot \bar{y} \cdot z\) = x . y + x . z
Answer:
\(x \cdot y + x \cdot \bar{y} \cdot z\) = \(x(y+\bar{y} . z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 40.
State which of the following statements are logical statements.
(a) AND is a logical operator
(b) ADD 3 to y
(c) Go to class
(d) Sun rises in the west.
(e) Why are you so late?
Answer:
(a) and (d) are logical statements because these statements have a truth value which may be true or false.

Question 41.
Express the integer number -39 in sign and magnitude represnetation.
Answer:
First find the binary equivalent of 39 for this do the following

In sign and magnitude representation -39 in 8 bit form is (10100111)₂.

Question 42.

1. Which logic gate does the Boolean expression \(\overline{\mathrm{AB}}\) represent?
2. Some NAND gates are given. How can we construct AND gate, OR gate and NOT gate using them?

Answer:
1. NAND

2. AND gate

Question 43.
Perform the following number conversions.

1. (110111011.11011)₂ = (….)₈
2. (128.25)₁₀= (…..)₈

Answer:
1. 110111011.11011
Step 1: Insert a zero in the right side of the above number and divide the number into groups of 3 bits as follows
110 111 011 . 110 110

Step 2: Write down the corresponding 3 bit binary equivalent of each group
6 7 3 .6 6
Hence the result is (673.66)₈

2. It consists of 2 steps.
Step 1: First convert 128 into octal number for this do the following

Write down the remainders from bottom to top.
(128)₁₀ = (200)₈

Step 2: Then convert .25 into octal number for this do the following

(0.25)₁₀ = (0.2)₈.
Combine the above two will be the result.
(128.25)₁₀= (200.2)₈

Question 44.
Represent -38 in 2’s complement form.
Answer:
+38 = 00100110
First take the 1 ’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of -38 is (11011010)₈.

Plus One Data Representation and Boolean Algebra Three Mark Questions and Answers

Question 1.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.

In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to input data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 2.
Complete the series.

1. 3248, 3278 ,3328, …., ….
2. 5678, 5768, 605s, ……, …..

Answer:
1. (324)₈ = 3 × 82 + 2 × 81 + 4 × 80
= 3 × 64 + 2 × 8 + 4 × 1
= 192 + 16 + 4 = (212)10

(327)₈ = 3 × 82 + 2 × 81 + 7 × 80
= 192 + 16 + 7 = (215)₁₀

(332)₈= 3 × 82 + 3 × 81 +2 × 80
= 192 + 24 + 2 = (218)₁₀
So the missing terms are (221)₁₀, (224)₁₀ we have to convert this into octal.

So the missing terms are (335)₈, (340)₈

2. (567)₈ = 5 x 82 + 6 x 81 + 7 x 80
= 5 x 64 + 6 x 8 + 7 x 1
= 320 + 48 + 7 = (375)₁₀

(576)₈ = 5 x 82 + 7 x 81 + 6 x 80
= 320 + 56 + 6 = (382)₁₀

(605)₈ = 6 x 82 + 0 x 81 + 5 x 80
= 6 + 64 + 0 + 5
= 384 + 0 + 5 = (389)₁₀
So the missing terms are (396)₁₀, (403)₁₀ we have to convert this into octal.

So the missing terms are (614)₈, (623)₈

Question 3.
Fill up the missing digits.

1. (4……)₈ = (……110)₂
2. (…….7……)₈ = (100…….110)₂

Consider the following

Answer:
1. 4…….100 and 110……….6
So (46)₈ = (100 110)₂

2. 100…….4
7……111
110………6
So (476)₈ = (100 111 110)₂

Question 4.
Fill up the missing numbers.

1. (A…….)₁₆ = (……..1001)₂
2. (…….B…….)₁₆ = (1000………1111)₂

Consider the following:


Answer:
1. A……..1010
1001………9
So (A9)₁₆ = (1010 1001)₂

2. B…….1011
1000………8
1111………F
So (8BF)₁₆ = (1000 1011 1111)₂

Question 5.
Complete the Series.

1. 6ADD, 6ADF, 6AE1, ……., …….
2. 14A9, 14AF, 14B5, …….., ……

Answer:
1. Consider the sequence
6ADD, 6ADF, 6AE1, ………….
Here the ‘numbers’ are
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11,——–
The difference between 6ADD & 6ADF is 2

Similarly 6ADF & 6AE1 is 2

So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5

2. Consider the sequence.
14A9, 14AF, 14B5, ———
The difference between 14A9 and 14AF is 6.
The normal sequence is

The difference between 14AF and 14B5 is also 6.

The normal sequence is

Similarly the next 6 terms in the sequence are given below.

Similarly the next 6 terms are

So the missing terms are 14BB and 14C1

Question 6.
Find the octal numbers corresponding to the following numbers using shorthand method.

1. (ADD)₁₆
2. (DEAD)₁₆

Answer:
1. (ADD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(ADD)(ADD)₁₆ = (5335)(ADD)₈

2. (DEAD)₁₆
Step 1: Write down the 4 bit binary equivalent of each digit.

Step 2: Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

(DEAD)₁₆ = (157255)₈

Question 7.
If (126)x = (56)y, then find x and y.
Answer:
L.H.S contains 2 & 6, so x ≠ 2
R.H.S contains 5 & 6, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also The possible values of x and y are given below.

Case I:
Let x = 8 then y = 10

It is grater than (56)₁₀
so when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

Question 8.
If (102)x = (42)y then (154)x = (  )y.
Answer:
L.H.S contains 2, so x ≠ 2
R.H.S contains 5 & 4, so y ≠ 2
L.H.S > R.H.S
So x < y and x ≠ y also
The possible values of x and y are given below.

case I:
let x = 8 and y = 10

So when x = 8 then y ≠ 10

case II:
let x = 8 and y = 16

So x = 8 and y = 16
then we have fo find the hexadecimal equivalent of (154)₈ For this first convert this into binary thus again convert it into hexadecimal. First write down the 3 bit equivalent of 154.

Then divide this number into groups of 4 bits starting from the right and write down the hexa decimal equivalent.

so the result is (154)₈ = (6C)₁₆

Question 9.
Fill up the missing digit.
If (121)_a = (441)_b then (121)_b = (  )₁₀
Answer:
L.H.S. contains 2, so a ≠ 2
R.H.S. contains 4, so b ≠ 2
L.H.S. < R.H.S. So a > b and a b also.
Hence the values of a can be 10 or 16.
The values of b can be 8 or 10.
The possible values of a and b are given below.

Case I:
Let a = 16 and b = 10
(121 )₁₆ = (289)_10, so b ≠ 10

Case II:
Let a = 16 and b = 8
(121)₁₆ = (289)₁₀
(441)₈ = 4 × 8² + 4 × 8¹ + 1 × 8⁰
= 256 +32 + 1
= (289)₁₀.
So a = 16 and b = 8.
Then (121)₈ = 1 × 82 + 2 × 81 + 1 × 80
= 64 + 16 + 1 = (81)₁₀

Question 10.
Fill up the missing digit. (Score 3)
If (128)_a = (450)_b then (16)_a = (  )₁₀
Answer:
L.H.S. contains 2 & 8, so a 2 and a ≠ 8.
R.H.S. contains 4 and 5, so b ≠ 2.
L.H.S. < R.H.S. so a > b and a ≠ b also.
The possible values of a and b are given below.

Case I:
a = 16 and b = 8
(128)₁₆ = (296)₁₀
(450)₈ = (296)₁₀ So a = 16 and b = 8.
Then (16)₁₆ = 1 × 16 + 6 × 16⁰ = (22)₁₀

Question 11.
Fill up the missing digit.
(3A.6D)₁₆ = (  )₈
Answer:
Step I: Write down the 4 bits equivalent of each digit.

Step II: Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010

Step III: Write the octal equivalent of each group. SO we will get. (72.332)₈.
(3A.6D)₁₆ = (72.332)₈

Question 12.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:

1. Sign 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: 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

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

Question 13.
Write short notes about Unicode (3)
Answer:
It is like ASCII Code. By using ASCII, we can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
eg: Malayalam, Hindi, Sanskrit .

Question 14.
Match the following.

1. (106)10	a. (171.53125)10
2. (71.24)8	b. (6a)16
3. (AB.88)16	c. (20)8
4. (10)16	d. (10000000)2
5. (128)10	e. (10)16
6. (16)10	f. (57.3125)10

Answer:
1 – b, 2 – f, 3 – a, 4 – c, 5 – d, 6 – e

Question 15.
Find the largest number in the list.

1. (1001)₂
2. (A)₁₆
3. (10)₈
4. (11)₁₀

Answer:
Convert all numbers into decimal
1. (1001)₂ = 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 0 + 0 + 1
= (9)₁₀

2.) (A)₁₆ = (10)₁₀

3. (10)₈ = 1 × 8¹+0 × 8⁰
= (8)₁₀
So the largest number is 4 – (11)₁₀

Question 16.
Subtract 10101 from 1111 by using 2’s complement method.
Answer:
To subtract a number from another number find the 2’s complement of the subtrahend and add it with the minuend. Here subtrahend is 10101 and minuend is 1111 First take the 1’s complement of subtrahend and add 1 to it.
1’s complement of 10101 is 01010 add 1 to it

Here is no carry. So the result, is -ve and take the 2’s complement of 11010 and put a -ve symbol. So 1’s complement of 11010 is 00101 add 1 to this

So the result is -00110

Question 17.
Mr. Geo purchased (10)₂ kg sugar @Rs. (110 10)₂ and (1010)₂ kg Rice @Rs. (10100)₂. So how much rupees he has to pay in decimal.
Answer:
Convert each into decimal number system multiply and sum it up.
(10)₂ = (2)₁₀

(11010)₂ = 1 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2¹
= 16 +8 + 0 +2 + 0
= (26)₁₀

(1010)₂ = 1 × 2³ + 0 × 2² + 1 × 2¹ +0 × 2⁰
= 8 + 0 + 2 + 0
= (10)₂

(10100)₂ = 1 × 2⁴ + 0 × 2³ + 1 × 2² +0 × 2¹ + 0 × 2⁰
= 16 + 0 + 4 + 0 + 0
= (20)₂
therfore 2 × 26 + 10 × 20
= 52 + 200
= 252
So Mr. Geo has to pay Rs. 252/-

Question 18.
Mr. Vimal purchased a pencil @ Rs. (101)2, a pen @ Rs. (1010)2 and a rubber @ Rs. (10)2. So how much rupees he has to pay in decimal.
Answer:
Add 101 + 1010 + 10

then convert (10001)₂ into decimal
(10001)₂ = 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 0 + 0 + 0 + 1
= (17)₁₀
So Mr. Vimal has to pay Rs. 17/-

Question 19.
Mr. Antony purchased 3 books worth Rs. a total of (1100100)₂. Atlast he returned a book worth Rs. (11001)₂. So how much amount he has to pay for the remaining two books in decimal number sys¬tem.
Answer:

then convert (1001011 )₂ into decimal
(1001011)₂ = 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=64 + 0 + 0 + 8 + 0 + 2 + 1
= (75)₁₀
So he has to pay Rs. 75/-

Question 20.
Mr. Leones brought two products from a super market a total of Rs. (11010010)₂ and he got a dicount of Rs. (1111)₂ So how much he has to pay for this products in decimal number system.
Answer:
Substract (1111)₂ from (11010010)₂

then convert (11000011)₂ into decimal
(11000011)₂ = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 1 × 2⁰
=128 + 64 + 0 + 0 + 0 + 0 + 2 + 1
= (195)₁₀

Question 21.
A textile showroom sells shirts with a discount of Rs. (110010)₂ on all barads. Mr. Raju wants to buy a shirt worth Rs. (11111000)₂. So after discount how much amount he has to pay in decimal.
Answer:
Substract (110010)₂ from (11 111 000)₂

then convert (11000110)₂ into decimal
(11000110)2 = 1 × 2⁷ + 1 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 0 × 2⁰
= 128 + 64 + 0 + 0 + 0 + 4 + 2 + 0
= (198)₁₀

Question 22.
Mr. Lijo purchased a product worth Rs. (1110011)₂ and he has to pay VAT @ Rs. (1100)₂. Then calculate the total amount he has to pay in decimal.
Answer:
Add (1110011)₂ and (1100)₂

then convert (1111111)₂ into decimal
(1111111)₂ = 1 × 2⁶ + 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= (127)₁₀

Question 23.
By using truth table, prove the following laws of Boolean Algebra.

1. Idempotent law
2. Involution law

Answer:
1. A + A = A
A = A = A

2. (A¹)¹ = A

Question 24.
Consider the logical gate diagram.

1. Find the logical expression for the circuit given.
2. Find the compliment of the logical expression.
3. Draw the circuit diagram representing the compliment.

Answer:
1. \((x+\bar{y}) \cdot z\)

2. \((\bar{x} . y)+\bar{z}\)
3.

Question 25.
Draw the logic circuit diagram for the following Boolean expression.
\(A \cdot(\bar{B} + \bar{C})+\bar{A} \bar{B} \bar{C}\)
Answer:

Question 26.
Consider a bulb with three switches x, y and z. Write the Boolean expression representing the following states.

1. All the switches x, y and z are ON
2. x is ON and y is OFF or Z is OFF
3. Exactly one switch is ON.

Answer:

1. xy . z
2. \(x \bar{y}+\bar{z}\)
3. \(x . \bar{y} . \bar{z}+\bar{x} . y . \bar{z}+\bar{x} . \bar{y} . \bar{z}\)

Question 27.
Match the following.

A	B
i. Idem potent law	a. x + (y + z)=(x + y)+z
ii. Involution law	b. x + xy = x
iii. Complementarity law	c. x + y = y + x
iv. Commutative law	d. xx- 0
v. Absorption law	e. x = x
vi. Associative law	f. x + x = x

Answer:
i – f, ii – e, iii – d, iv – c, v – b, vi – a

Question 28.
Explain the principle of duality.
Answer:
It states that, starting with a Boolean relation, another Boolean relation can be derived by

1. Changing each OR sign (+) to a AND sign (.)
2. Changing each AND sign (.) to an OR sign (+)
3. Replacing each 0 by 1 and each 1 by 0.

The relation derived using the duality principle is called the dual of the original expression,
eg: x + 0 = x is the dual of x . 1 = x

Question 29.
Draw the circuit diagram for \(F=A \bar{B} C+\bar{C} B\) using NAND gate only.
Answer:
\(F=A \bar{B} C+\bar{C} B\)
= (A NAND (NOT B) NAND C) NAND ((NOT C) NAND B)

Question 30.
Draw a logic diagram for the function f = YZ + XZ using NAND gates only.
Answer:
f = YZ + XZ
= (Y NAND Z) NAND (X NAND Z)

Question 31.
How do you make various basic logic gates using NAND gates.
Answer:
1. AND operation using NAND gate,
A.B = (A NAND B) NAND (A NAND B)

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)

3. NOT operation using NAND gate,
NOT A = (A NAND A)

Question 32.
Which of the following Boolean expressions are correct? Write the correct forms of the incorrect ones.

1. A + A¹ = 1
2. A + 0 = A
3. A . 1 = A
4. A . A¹ = 1
5. A + A . B = A
6. A . (A + B) = A
7. A + 1=1
8. \((\overline{\mathrm{A} . \mathrm{B}})=\overline{\mathrm{A}} . \overline{\mathrm{B}}\)
9. A + A¹B = A + B
10. A + A = A
11. A + B . C = (A+B) . (B+C)

Answer:

1. Correct
2. Correct
3. Correct
4. Wrong, A . A¹ = 0
5. Correct
6. Correct
7. Correct
8. Wrong \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
9. Correct
10. Correct
11. Wrong, A + B . C = (A + B) . (A + C)

Question 33.
Prove algebraically that (x + y)’ . (x’ + y’) = x’ . y’
Answer:
LHS = (x + y)’ . (x’ + y’)
= (x’ . y’) . (x’ . y’)
= x’ . y’ . x’ + x’ . y’ . y’
= x’ . y’ + x’ . y’
= x ‘. y’ = RHS
Hence proved.

Question 34.
Give the complement of the following Boolean Expression.

1. (A + B) . (C + D)
2. (P + Q) + (Q + R) . (R + P)
3. (B + D’) . (A + C’)

Answer:
1. ((A+B) . (C+D))1 = (A+B)’ + (C+D)’
= A’ . B’ + C’ . D’

2. ((P+Q) + (Q+R) . (R.P))’ = (P+Q) ‘. ((Q+R) . (R+P))’
= P’ . Q’ . (Q+R)’ + (R+P)’
= P’ . Q’ . (Q’ . R’ + R’ . P’),

3. ((B+D’).(A+C’))’ = (B+D’)’0 + (A+C’)’
= B’ . D” + A’ . C”
= B’ . D + A’ . C

Question 35.
State and prove the idempotent law using truth table. Idempotent law
Answer;
Idempotent law states that

1. A + A = Aand
2. A . A = A Proof

1. A + A = A
Truth table is as follows:

ie. A + A = A as it is true for both values of A. Hence proved.

2. A . A = A
Truth table is as follows:

ie. A . A = A itself. It is true for both values of A. Hence proved.

Question 36.
State the Absorption laws of Boolean algebra with the help of truth tables.
Answer:
Absorption law states that
A + A . B = A and A . (A + B) = A
Proof:
The Truth table of the expression A + A . B=A is as follows.

Here both columns A and A + A . B are identical. Hence proved.
For A . (A + B) = A, the truth table is as follows:

Both columns A & A . (A + B) are identical. Hence proved

Question 37.
State Demorgen’s laws. Prove anyone with truth table method.
Answer:
Demorgan’s first theorem states that (A + B)’ = A’ . B’
ie. the complement of sum of two variables equals product of their complements,

The second theorem states that (A . B)’ = A’ + B’
ie. The complement of the product of two variables equals the sum of the complement of that variables.
Proof:
Truth table of first one is as follows:

From the truth table the columns of both (A + B)’ and A’ . B’ are identical. Hence proved.

Question 38.
Fill in the blanks:

1. (0.625)₁₀ = (……….)₂
2. (380)₁₀ = (……..)₁₆
3. (437)₈ = (………)₂

Answer:

1. (0.101)₂
2. (17C)₁₆
3. (100 011 111)₂

Question 39.
What do you mean by universal gates? Which gates are called Universal gates? Draw their symbols.

OR

Construct a logical circuit for the Boolean expression \(\bar{a} \cdot b+a \cdot \bar{b}\). Also write the truth table.
Answer:
Universal gates:
By using NAND and NOR gates only we can create other gate hence these gates are called Universal gate.
NAND gate:

NOR gate:

Truth table:

Logical circute:

Question 40.
Computers uses a fixed number of bits to respresent data which could be a number, a character, image, sound, video etc. Explain the various methods used to represent characters in memory.
Answer:
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 the 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(International Business Machine). 2⁸ = 256 characters can be represented by using this.

3. ISCII(Indian 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.

Question 41.
Draw the logic circuit for the function
\(f(a, b, c)=a . b . c+\bar{a} . b+a . \bar{b}+a . b . \bar{c}\)

OR

Prove algebrically.
\(x . y+x . \bar{y} . z=x . y .+x . z\)
Answer:

OR

\(x \cdot y+x \cdot \bar{y} \cdot z=x \cdot(y+\bar{y} \cdot z)\)
= x . (y + z) = x . y + x . z
Hence proved.

Question 42.
Following are the numbers in various number systems. Two of the numbers are same. Identify them:

1. (310)₈
2. (1010010)₂
3. (C8)₁₆
4. (201)₁₀

OR

Consider the following Boolean expression:
(B’ + A)’ = B . A’
Identify the law behind the above expression and prove it using algebriac method.
Answer:
1. (310)₈ = 3 * 8² + 1 * 8¹ + 0 * 8⁰
= 192 + 8 + 0
= (200)₁₀

2. (1010010)₂ = 1 × 2⁶+ 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰
= 64 + 0 + 16 + 0 + 0 + 2 + 0
= (82)₁₀

3. (C8)₁₆ = C × 16 + 8 × 16⁰
= 12 × 16 + 8 × 1
= 192 + 8
= (200)₁₀
Here (a) (310)₈ and (C8)₁₆ are same

OR

This is De Morgan’s law (B’ + A’) = (B’)’ . A’
= B . A’
Hence it is proved

Question 43.
Find the decimal equivalent of hexadecimal number (2D)₁₆. Represent this decimal number in 2’s complement form using 8 bit word length.
Answer:
Convert (2D)₁₆ to binary number for this write down the 4 bit binary equivalent of each number

(2D)₁₆ = (00101101 )₂
First find the 1’s complement of (00101101 )₂ and add 1 to it

Hence 2’s complement is (11010011)₂

Question 44.
Answer any one question from 15(a) and 15(b).
1. Draw the logic circuit for the Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

2. Using algebraic method prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:
1.

OR

2. L.H.S. = \(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y Z+Y\)
= \(\bar{y} \cdot(\bar{z}+z)+y \cdot(z+1)\)
= \(\bar{y}. 1+\bar{y} \cdot 1=y \cdot y=1\)

Question 45.
With the help of a neat circuit diagram, prove that NAND gate is a universal gate.
Answer:
1. AND operation using NAND gate,
A . B = (A NAND B) NAND (A NAND B)

2. OR operation using NAND gate,
A + B = (A NAND A) NAND (B NAND B)

3. NOT operation using NAND gate,
NOT A = (A NAND A)

Question 46.
Boolean expression:
\((A+\overline{B C})+\overline{A B}\)

OR

Using algebraic method, prove that
\(\bar{Y} \cdot \bar{Z}+\bar{Y} \cdot Z+Y \cdot Z+Y=1\)
Answer:

OR

= Y . Z + Y . Z + Y . Z + Y
= Y . (Z + Z) + Y . (Z + 1)
= Y . 1 + Y. 1
= Y + Y
= 1
Hence the result.

Plus One Data Representation and Boolean Algebra Five Mark Questions and Answers

Question 1.
Explain the components of Data processing.
Answer:
Data processing consists of the techniques of sorting, relating, interpreting and computing items of data in orderto convert meaningful information. The components of data processing are given below.

1. Capturing data: In this step acquire or collect data from the user to input into the computer.
2. Input: It is the next step. In this step appropriate data is extracted and feed into the computer.
3. Storage: The data entered into the computer must be stored before starting the processing.
4. Processing/Manipulating data: It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
5. Output of information: In this stage we will get the results as information after processing the data.
6. Distribution of information: In this phase the information(result) will be given to the concerned persons/computers.

Question 2.
Define computer. What are the characteristics?
Answer:
A computer is an electronic device used to perform operations at very high speed and accuracy.
Following are the characteristics of the computer.

1. Speed: It can perform operations at a high speed.
2. Accuracy: It produces result at a high degree of accuracy.
3. Diligence: Unlike human beings, a computer is free from monotony, tiredness, lack of concentration etc. We know that it is an electronic ma chine. Hence it can work four hours without making any errors.
4. Versatility: It is capable of performing many tasks. It is useful in many fields.
5. Power of Remembering: A computer consists of huge amount of memory. So it can store and recall any amount of information. Unlike human beings, it can store huge amount of data and can be retrieved when needed.

Disadvantages of computer:

1. No. IQ: It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
2. No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers)

Students can Download Chapter 6 Work, Energy and Power 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 6 Work, Energy and Power

Plus One Physics Work, Energy and Power One Mark Questions and Answers

Question 1.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Question 2.
What is the work done by the tension in the string of simple pendulum?
Answer:
Zero

Question 3.
When is the exchange of energy is maximum during an elastic collision?
Answer:
When mass of two colliding bodies are same, there will be maximum exchange of energy.

Question 4.
In atom, an electron is revolving around the nucleus. What is the work done?
Answer:
Work done is zero because work done by centripetal force is zero.

Question 5.
What is the type of collision when macroscopic particles collide?
Answer:
Perfectly inelastic collision.

Question 6.
Name the parameter which is a measure of degree of elasticity of a body.
Answer:
Coefficient of restitution.

Question 7.
What is the source of kinetic energy for falling rain drops?
Answer:
Gravitational potential energy.

Plus One Physics Work, Energy and Power Two Mark Questions and Answers

Question 1.
The law of conversation of energy states that energy can neither be created nor be destroyed but can only change from one form into another. A bus and a car, moving with the same kinetic energy are brought to rest by applying an equal retardation force by the breaking systems. Which one will come to rest at a shorter distance? Give the reason behind your answer.
Answer:
Change in K.E. = Force × Displacement
1/2 mv² = F × S
ie. KE α s
KE_car α S_car _____(1)
KE_bus α S_bus ______(2)

ie. S_car = S_bus
Both will travel equal distance.

Question 2.
A body constrained to move alomg the Z-axis of a co-ordinate system is subjected to a constant force \(\bar{F}=(\hat{i}+2 \hat{\jmath}+3 \hat{k}) N\)

1. What is the magnitude of force along z direction.
2. What is the work done by this force in moving the body over a distance of 4m along z-axis.

Answer:

1. 3N
2. Work done = Force × Displacement = 3 × 4 = 12J.

Question 3.
Match the following

Answer:
Collision of two balls – inelastic – TE and momentum Collision of two molecules – elastic – KE, TE, and momentum.

Plus One Physics Work, Energy and Power Three Mark Questions and Answers

Question 1.
A car of mass 1000kg moving with a speed 18mk/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25 × 10³N/m

1. What do you mean by Spring constant.
2. What is the maximum compression of the spring?

Answer:
1. Spring constant is the force required to stretch the spring by a unit distance.

2. \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)kx², 18km/h = 5m/s.

x = 2m.

Question 2.
A man tries to lift a mass 200kg with a force 100N

1. Is he doing work? Explain.
2. If yes, find the amount of work done If No, find the force required to lift it.
3. If it is lifted to 2m in 10 seconds, find his power.

Answer:

1. No work is done, as there is no displacement, 100N force is insufficient to raise 200kg.
2. Force required to lift 200 kg = 200 × 9.8 = 1960N
3. Power = \(\frac{m g h}{t}=\frac{200 \times 9.8 \times 2}{10}\) = 392W.

Question 3.
Two cricket balls are colliding each other.

1. Name the collision
2. Say whether law of conservation of Kinetic Energy hold good in this case. Why?
3. State and prove the other conservation law applicable here.

Answer:

1. Inelastic collision
2. No, Total KE before collision is not equal to total KE after collision.
3. Proof and statement of law of conservation of momentum.

Plus One Physics Work, Energy and Power Four Mark Questions and Answers

Question 1.
Two cars A and B travelling with speeds 20m/s and 10m/s respectively applies breaks,.so that A comes to rest in 15 second and B in 7.5s

1. From the graph determine which of the two cars travelled further after brakes were applied and by how much distance it travelled?
2. Draw the velocity time graph of A and B in the same graph.
3. In the above process ,the wear and tear of which the car gets affected more ?

Answer:
1. The area of velocity time graph gives displacement distance travelled by the car A, S_A = 1/2 × 20 × 15 = 150 m
distance travelled by the car B, S_B = 1/2 × 10 × 7.5 = 37.5.

2.

3. Wear and tear gets affected more for the car A.

Question 2.
A sphere of mass m is moving with a velocity u and makes a head on collision with another identical mass which is at rest. It is observed that the stationary mass starts moving with a lesser velocity than u, after the collision.

1. Which physical quantity is conserved here?
2. Define coefficient of restitution.
3. Determine the ratio of the velocities of the two spheres after elastic collision if ‘e’ is the coefficient of restitution.

Answer:
1. conservation of linear momentum.

2. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

3. Coefficient of restitution,
e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
in this cos u₁ = u, u₂ = 0, v₁ = 0, v₂ = u
∴ e = \(\frac{u-0}{u-0}\)
e = 1.

Question 3.
From the table given below

1. Draw the force displacement curve
2. Analyse the graph & find the type of force involved
3. Estimate the workdone

Answer:
1.

2. Workdone by a variable force.

3. Workdone = Area of the graph
= \(\frac{1}{2}\)bh
= \(\frac{1}{2}\)5 × 10 = 25J.

Question 4.
Raju increased the speed of moving mass ‘50 kg’ from 2 m/s to 4m/s.

1. How much force will be required, if velocity change takes place with in 0.2 sec?
2. How much work is done by Raju?

Answer:
1. F = mass × acceleration
= 50 × \(\frac{(4-2)}{0.2}\)
= 500N.

2. w = \(\frac{1}{2}\)mv² – \(\frac{1}{2}\)mu².
=\(\frac{1}{2}\)50 (4² – 2²)
= 300 J.

Plus One Physics Work, Energy and Power Five Mark Questions and Answers

Question 1.
A car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. (Assume that the truck is heavier than the car)

1. Which one will have greater momentum?
2. Write the relationship between kinetic energy and linear momentum.
3. If the mass of truck is 100 times greater than that of the car, find the ratio between their velocities.

Answer:
1. Kinetic energy, of car, K.E_c = \(\frac{P_{c}^{2}}{2 m_{c}}\)
Kinetic energy of truck,

Since m_c < m_t
Hence P_t > P_c
∴ momentum of truck is greater than car.

2. KE = P²/2m.

3. KE_c = KE_t
1/2m_c V_c² = 1/2 x m_tV_t²
But m_t = 100m_c
\(V_{t}^{2}=\frac{V_{c}^{2}}{100}\)
Velocity of truck, Vt = \(\frac{V_{c}}{10}\)
ratio of velocity, 10V_t = V_c
10:1.

Question 2.
Raju dropped a rubber ball of mass m from a height h to the ground. He observed that the ball rebounds vertically and along the same line to a height h₁, which is less than h.

1. Is it an elastic or inelastic collision?
2. Find the velocity with which it strikes the ground?
3. If it is replaced by a solid aluminium ball, then what happens to the height of rebound?
4. If the rubber ball is allowed to fall on a spring placed on the ground then what change will Raju notice in the height of rebound?

Answer:
1. Inelastic collision.

2. The velocity with which ball strikes the ground,
v² = u² + 2as
v² = 0 + 2g × h
v = \(\sqrt{2 g h}\).

3. Height of rebound decreases.

4. Height of rebound depends on the state of potential energy stored in the spring. If ball falls on a compressed spring, the height of rebound increases due to potential energy given by the spring to bail.

Question 3.
A man tries to lift a mass 200kg with a force 100N.

1. Is he doing work? Explain.
2. If it is lifted to 2m in 10s, find the power.
3. Show that total mechanical energy is conserved fora freely falling body.

Answer:
1. No. Force required to lift the body is 2000N (w = mg = 200 × 10). But the applied force is 100N. Hence there is no displacement due to this applied force.

2. Power P = \(\frac{w}{t}=\frac{m g h}{t}=\frac{200 \times 10 \times 2}{10}\) = 400 watt.

3.

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
Potential energy at A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2}\) mv² = 0
(since the body at rest, v = 0).
∴ Total mechanical energy = PE + KE
= mgh + 0 = mgh
Total energy at the point B
The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
v² = u² + 2as
v² = 0 + 2 gx

∴ KE at B, = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) m2gx
= mgx
P.E. at B, = mg(h – x)
Total mechanical energy = PE + KE
= mg(h – x) + mgx
= mgh
Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2 gh
∴ KE at C, = \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE = 0 + mgh = mgh.

Question 4.
An elevator of total mass 1800kg is moving up with a constant speed of 2m/s. A frictional force of 400N acts on this motion.

1. The direction of frictional force is______
   • Opposite to direction of motion
   • In the direction of motion
2. What is the work done by gravitational force.
3. What is the total work done by the elevator?

Answer:
1. Opposite to direction of motion

2. w = F × V
w = mg × 2
= 1800 × 10 × 2
w = 36000J

3. P = P_total × V
= (mg + F_fricti0n) × V
= (1800 × 10 + 4000) × 2
= (18000 + 4000) × 2
P = 44000w.

Question 5.
A stone of mass ‘m’ is to be thrown to a height h

1. What is the acceleration of the stone?
2. With what minimum velocity should it be thrown.
3. At what height does the KE and PE become equal?
4. Find the velocity at that height

Answer:
1. ^–g or ^–9.8m/s.

2. v = 0, a = -g, S = h
Substitute this values in
V² = u² + 2as we get
0 = u² – 2gh
u = \(\sqrt{2 g h}\)

3. at \(\frac{h}{2}\)., KE and PE are equal.

4. V² = U² + 2aS
= U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^{2}}{2}\),
V² = \(\frac{U^{2}}{2}\),
V = \(\frac{U}{\sqrt{2}}\).

Question 6.
A toy gun, with a spring compresser 3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 7.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Question 8.
Atoy gun, with aspring compresser3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 9.

A graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N/m]

1. The energy possessed by the stone due to its height is called_______
2. If this stone falls on the spring, find the length of mark that produced on the graph paper due to pencil [The change in P E of stone due to compression may be negleted]
3. What will happen to the length of mark, if spring having smaller spring constant is used? Justify.

Answer:
1. Potential energy.

2. \(\frac{1}{2}\)kx² = mgh
\(\frac{1}{2}\) × 98 × x² = 50 × 10^-3 × 9.8 × 1
x² = 100 × 10⁴
x = 10cm.

3. The length of mark will be decreased. Compression of spring depends on spring constant.

Plus One Physics Work, Energy and Power NCERT Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. work done by gravitational force in the above case.
3. work done by friction on a body sliding down an inclined plane.
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
5. work done by the resistive force of air on vibrating pendulum in bringing it to rest.

Answer:

1. +ve
2. -ve
3. -ve
4. +ve
5. -ve

Question 2.
The potential energy function fora particle executing linear simple harmonic motion is given by
V(x) = \(\frac{k x^{2}}{2}\), where k is the force constant of the oscillator, For k = 0.5N nm^-1, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2m.
Answer:
We know that maximum potential energy = total energy
∴ (\(\frac{1}{2}\)kx²) max = 1 joule or \(\frac{1}{2}\) × 0.5 × (x²)_max = 1
or (x²)_max = 4 or (x)_max = ± 2m.

Question 3.
Choose the correct alternative:

1. When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/remains unaltered.
2. Work done by a body against friction always results in a loss of its kinetic/potential energy.
3. The rate of change of total momentum of a many particle system is proportional to the external force/sum of the internal forces on the system.
4. In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/ total energy of the system of two bodies.

Answer:

1. decreases
2. kinetic energy
3. external force
4. total linear momentum and also total energy (if the system of two bodies is isolated).

Question 4.
State if each of the following statements is true or false.

1. In an elastic collision of two bodies, the momentum and energy of each body is conserved.
2. Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False
2. False
3. False
4. False (true usually but not always).

Question 5.
A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) untill at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10ms^-1?
Answer:
r = 2 × 10^-3m,
volume = \(\frac{4}{3} \times \frac{22}{7}\) (2 × 10^-3)³ m³
p = 1000kgm^-3,
h = 250m
W= \(\frac{4}{3} \times \frac{22}{7}\) × 8 × 10^-9 × 1000 × 9.8 × 250J = 0.082J
Data reamains unchanged in the next half.

Question 6.
A bullet of mass 0.012kg and horizontal speed 70ms-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
If V be the velocity of the block after collision, the using law of conservation of momentum, we get
0.012 × 70 + 0 = (0.012 + 0.4)V
or V = \(\frac{0.012 \times 70}{0.412}\) ms^-1 = 2.04ms^-1
If h be the height through which block rises, then
(M + m) gh = \(\frac{1}{2}\) (M + m)V²
or h = \(\frac{v^{2}}{2 g}\) or
h = \(\frac{2.04 \times 2.04}{2 \times 9.8}\)m = 0.212m = 21.2 cm
Amount of heat produced in the block = loss of K.E.
= \(\frac{1}{2}\) × 0.012 × 70 × 70 – \(\frac{1}{2}\) × 0.412 × 2.04 × 2.04
= 29.4J – 0.857J = 28.543J.

Students can Download Chapter 8 Gravitation 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 8 Gravitation

Plus One Physics Gravitation One Mark Questions and Answers

Question 1.
If a satellite of mass m is revolving around the earth with distance rfrom centre, then total energy is

Answer:
(c) \(-\frac{G M m}{2 r}\)
The satellite revolving around the earth has two types of energies.
1. Gravitational potential energy U due to position of the satellite is given by
U = \(-\frac{\mathrm{GMm}}{\mathrm{r}}\).

2. Kinetic energy K due to orbital velocity of satellite is given by

Question 2.
The dimensions of universal gravitational constant are
(a) [M^-2L³T^-2]
(b) [M^-2L²T^-1]
(c) [M^-1L³T^-2]
(d) [ML²T^-1]
Answer:
(c) [M^-1L³T^-2]
According to Newton’s law of gravitation

Question 3.
When a radius of earth is reduced by 1% without changing the mass, then change in the acceleration due to gravity will be
(a) increased by 2%
(b) decreased by 1.5%
(c) increased by 1 %
(d) decreased by 1 %
Answer:
(a) increased by 2%

Question 4.
The kinetic energy of a satellite is 2MJ. What is the total energy of the satellite?
(a) -2M J
(b) -1MJ
(c) -1/2MJ
(d) -4MJ
Answer:
(a) -2MJ
Total energy of satellite – Kinetic energy of satellite.

Question 5.
If total energy of this stone is negative, can it escape from the earth’s surface? Justify.
Answer:
It will not escape as the force is attractive.

Question 6.
Why a man can jump higher on the moon than on earth?
Answer:
The value of g on the moon is one sixth of that on earth.

Question 7.
The gravitational potential energy of a body of mass ‘m’ is -10⁷J. What energy is required to project the body out of gravitational field of earth?
Answer:
10⁷J

Question 8.
How much energy is required by a satellite to keep it orbiting? Neglect air resistance.
Answer:
No energy is required because work done by centripetal force is zero.

Question 9.
What would happen to an artificial satellites, if its orbital velocity is slightly decreased due to some defects in it?
Answer:
It will fall on to earth.

Question 10.
Why a tennis ball bounces higher on hills than in plain?
Answer:
As height increases ‘g’ decreases. The value of g on hills is less than that on the surface of earth. Hence tennis ball bounces higher on hills than in plains.

Question 11.
The orbiting velocity of an earth-satellite is 8 km/s. What will be the escape velocity?
Answer:
V_e = \(\sqrt{2}\) V₀
V_e = \(\sqrt{2}\) × 8 = 11.3 kms^-1.

Plus One Physics Gravitation Two Mark Questions and Answers

Question 1.
1. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant

• linear speed
• total energy
• angular momentum

2. What are the consequences if the angular momentum is conserved?
Answer:
1. Highly elliptical orbit:

• No
• Yes
• Yes

2. consequences if the angular momentum:

• Areal velocity of planet becomes constant
• Angular velocity changes according to the position of planet around sun.

Question 2.
Fill in the blanks
R – Radius of earth, g – Acceleration due to gravity h – Height from surface of earth, d – Depth from surface of earth

Answer:
a) g(h) = g(1 – \(\frac{2 h}{R}\))
b) Variation of g with depth
c) g Decreases with depth
d) (Change with radius (polar, equitorial org is maximum on the earth’s surface)

Plus One Physics Gravitation Three Mark Questions and Answers

Question 1.
The Earth moving round the Sun in a circular orbit is acted upon by a force.

1. Name the force acting on the Earth.
2. Obtain a mathematical expression for the force.
3. What is the work done by the force?

Answer:

1. Centripetal force
2. F_e = ma
   F_c = m \(\frac{v^{2}}{r}\)
3. zero. (Displacement is always perpendicular to centripetal force).

Question 2.
Escape velocity of objects in a planet depends on the mass and size of the planet.

1. Write down the expression for the escape velocity on the surface of the earth.
2. What is its value on the surface of our planet?
3. Give the reason why the moon does not have an atmosphere around it.

Answer:

1. Escape velocity Ve = \(\sqrt{\frac{2 G M}{R}}\)
2. 11.2km/sec
3. The escape velocity of moon is very low. Hence all the gas molecules can escape from the moon’s surface.

Question 3.
The acceleration due to gravity is measured using ticker timer and is found to be 9.8ms^-2

1. What does the value 9.8 for the acceleration implies?
2. One of your friends argues that the acceleration due to gravity at the centre of earth is infinity. Do you agree with it?
3. Justify your answer with mathematical support.

Answer:
1. An acceleration of 9.8m² implies that velocity change by 9.8ms¹ in every second.

2. No, at centre g = 0

3. At centre d = R
g_d = g × 0 = 0

Plus One Physics Gravitation Four Mark Questions and Answers

Question 1.
A body of mass ‘m’ falls freely under gravity, near the surface of earth.

1. Will the acceleration of the body change if a part of the mass is thrown away from it?
2. What will be the free fall acceleration if it is falling from a height equal to R, the radius of earth?
3. If the mass is taken to the moon’s surface, will the free fall acceleration increase or decrease?

Answer:
1. No. The acceleration of body does not depend on the mass of falling body.

2. We know acceleration due to gravity

3. Decrease.

Question 2.
Geo Stationary satellites are commonly used for communication purpose.

1. Name one geostationary satellite of earth.
2. What are the requirements of a geostationary satellite for its orbital motion.
3. Explain the phenomenon of ‘weightlessness’ in orbiting satellites.
4. Distinguish between gravitational mass and inertial mass.

Answer:

1. INSAT.
2. The period of this satellite must be same as the period of the rotation of earth about its own axis. Its direction of rotation must be the same as that of the earth (from west to east).
3. The weight of satellite is used for providing centripetal force required for rotation. Hence anybody inside the satellite appears to be weightlessness.
4. The mass related with newtons second law of motion is called inertial mass. But the mass related with newtons law of gravitation is called gravitational mass.

Question 3.

1. What is acceleration due to gravity?
2. Does a body have the same weight at the equator and at the poles? Explain.
3. If the value of gravitational constant is 6.6 × 10^-11 Nm²kg^-2 and g = 9.8 m/s², find the mass of the earth. Given, radius is 6.4 × 10⁶m.

Answer:
1. 9.8 m/se²

2. No. The weight of the body depends on the acceleration due to gravity at a place. The magnitude of acceleration due to gravity at equator and poles are different.

3.

Question 4.
Two identical satellites are orbiting in circular orbits around the earth at heights R and 3R respectively from the surface of the earth. The radius of the earth is R.

1. Define orbital velocity.
2. What is its value on the surface of the earth?
3. How do you compare the periods of revolution of these two satellites?

Answer:
1. The minimum velocity required for a satellite to revolve around earth in a stable orbit is called orbital velocity.

2.

3.

Question 5.
The uniform acceleration produced in a freely falling body due to gravitational pull of the earth is known as acceleration due to gravity.

1. What is the value of acceleration due to gravity on the surface of earth.
2. Obtain an expression for acceleration due to gravity at a depth ‘d’ from the surface of earth
3. What is the value of gat the centre of earth?

Answer:
1. 9.8m/s^-2

2.

If we assume the earth as a sphere of radius R with uniform density?
mass of earth = volume × density

Substituting eq(1) in eq(2), we get
g = \(\frac{4}{3}\) πGR_ρ ……(3)
Therefore the acceleration due to gravity at a depth d is given by
g_d = \(\frac{4}{3}\) πG(R – d)ρ ……(4)
eq(4)/eq(3) and solving we get

g_d = g(1 – d/R)
The above equation shows that, when depth increases g decreases.

3. Increase.

Plus One Physics Gravitation Five Mark Questions and Answers

Question 1.
Earth can be treated as a sphere of radius R and mass M. A is a point at a height h above the Earth’s surface and B is another point at a depth h1 below the Earth’s surface. The acceleration due to gravity at Earth’s surface is g.

1. Obtain a formula to evaluate the acceleration due to gravity at A.
2. What is the value of acceleration due to gravity at B?
3. If we move from equator to pole, the value of g.
   • increases
   • decreases
   • remains the same
   • first increases and then decreases
   • first decreases and then increases
4. What is the value of g at the centre of Earth? Explain.

Answer:
1. The acceleration due to gravity on the surface of earth,
g = \(\frac{G m}{R^{2}}\) …..(1)
At a height h, the acceleration due to gravity can be written as,

eq(1)/eq(2) and solving we get
g_h = g(1 – \(\frac{2 h}{R}\))
This equation shows that acceleration due to gravity decreases as height increases. The above equation is valid when h << R.

2.

3. Increase

4. Zero. At centre, the force acting on the body is zero.

Question 2.
Fora particle to leave from earth’s gravitational field it should be projected with a minimum velocity.

1. Name the velocity.
2. Derive an expression for this velocity.
3. What is the magnitude of this velocity when this particle is projected from another planet whose mass and radius is twice that of the earth?

Answer:
1. Escape velocity

2. Expression for escape speed:
Force on a mass m at a distance r from the centre of earth = \(\frac{G M m}{r^{2}}\)
Work done in taking the body to infinity from surface of earth,

This energy is given in the form of K.E. = \(\frac{1}{2}\)mv_e²,
where v_e is the escape speed.

This escape velocity \(\sqrt{2 \mathrm{Rg}}\) is estimated to be 11.2 km/son the earth.

3. Escape velocity of planet,

Question 3.
Moon is the only satellite of earth. Mass of moon is very much lower than that of earth.

1. Is there any difference in gravitational force of moon and earth? Explain.
2. Deduce an equation for gravitational potential energy.
3. A girl argues that even today if we visit moon, we can see the foot steps of Neil Armstrong. What is your opinion?

Answer:
1. Yes. Gravitational force of earth longer than gravitational force of moon.

2. Expression for gravitational P.

Consider the earth as a uniform sphere of radius R and mass M. Considera point A at distance ‘r’ from the centre of earth. P is another point at a distance ‘x’ from O. Q lies at distance dx from P.

By definition, the gravitational potential energy of the body at point A, is the work done in bringing the body of mass ‘m’ from infinity to that point A The gravitational force on the body at the point P.
is given by F = \(\frac{G M m}{x^{2}}\).
If the body is displaced from P to Q Work done, dw = F.dx
= \(\frac{G M m}{x^{2}}\) Therefore, workdone in bringing the body from infinity to the point A,

Since this workdone is stored inside the body as its gravitational potential energy, the gravitational potential energy (U) of a body of mass m at distance rfrom the centre of the earth is given by,

The gravitational potential energy of a body at a point is defined as the amount of workdone in bringing the body from infinity to that point without acceleration.

3. There is no atmosphere on the moon. Hence we can see the foot steps of Neil Armstrong.

Plus One Physics Gravitation NCERT Questions and Answers

Question 1.
Suppose there existed a planent that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer:
T_e = 1 year, T_p = \(\frac{1}{2}\) year, R_p = ?, R_e = 1 AU
Using Kepler’s third law

Question 2.
lo, one of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸m. Show that the mass of the Jupiter is about one-thousandth that of the Sun.
Answer:

Substituting the given data for Jupiter
Mass of Jupiter

Substituting the known data forthe revolution of earth around Sun,

Question 3.
Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete on revolution? Take the diameter of the Milky way to be 10⁵ly.
Answer:

Question 4.
A rocket is fired from the earth towards the Sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 × 10³⁰ kg, Mass of the earth = 6 × 10²⁴kg. Neglect the effect of other planets etc. (Orbital radius = 1.5 × 10¹¹m).
Answer:
Mass of Sun, M = 2 × 10³⁰ kg
Mass of Earth m = 6 × 10²⁴kg.
Distance between Sun and Earth, r= 1.5 × 10¹¹m
Suppose, at the point P, the gravitational force on the rocket due to earth = gravitational force on the rocket due to Sun
Let x be the distance of the point P from the earth

Question 5.
How will you “weigh the Sun’, that is estimate its mass? The mean orbital radius of the earth around the Sun is 1.5 × 10⁸km.
Answer:
For the revolution of earth around sun, the gravitational force between the sun and the earth provides the necessary centripetal force.

Question 6.
A geostationary satellite orbits the earth at a height of nearly 36,000km from the surface of the earth. What is the potentional due to earth’s gravity at the site of this sattelite? (Take the potential energy at infinity to be zero). Mass of the earth = 6 × 10²⁴ kg., radius of earth = 6,400 km.
Answer:
Required potential

Question 7.
A satum year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸km away from the sun?
Answer:
We know that T² α R³

where subscripts s and e refer to the satum and earth respectively.
Now \(\frac{T_{s}}{T_{e}}\) = 29.5 (given); R_e = 1.50 × 10⁸km

R_s = R_e × [(29.5)²]^1/3 = 1.50 × 10⁸ × (870.25)^1/3km
= 1.43 × 10⁹km = 1.43 × 10¹²m.

Question 8.
A body weighs 63N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:
Let g_h be the acceleration due to gravity at a height equal to half the radius of the earth (h = R/2) and g its value on earth’s surface. Let the body have mass m, we know that

Let W be the weight of body on the surface of earth and W_h the weight of the body at height h.
Then,

Students can Download Chapter 12 Thermodynamics 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 12 Thermodynamics

Summary
Thermo dynamics is the branch of Physics that deals with concept of heat and temperature, their inter-conversions and other forms of energy. In thermo dynamics state of system is specified by macroscopic variables such as pressure, temperature, volume, mass, composition… etc.

Thermo Equilibrium
Thermo dynamic equilibrium:
A system is said to be in thermodynamic equilibrium if the macroscopic variables used to describe the system does not change with time.
Eg: Agas enclosed in a rigid container (characterized by P, V, T, m, and composition) The equilibrium of a thermodynamic system depends on surroundings and nature of wall that separates the system from surroundings. The wall can be

• Adiabatic wall – It does not allow flow of heat (energy).
• Diathermic wall – It allows the flow of heat (so that it can comes to thermal equilibrium).

Zeroth Law Of Thermodynamics
Zeroth law of thermodynamics gives concept of temperature. R.H Fowler formulated this law in 1931. According to Zeroth law of thermodynamics, the systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other.

Let A & B are in thermal equilibrium with each other. If they are separately in thermal equilibrium with C, then the three systems are in thermal equilibrium
(T_A = T_B = T_C).

Heat, Internal Energy And Work
Internal Energy (U):
internal energy of a gas is the sum of potential energy and kinetic energy (translational kinetic energy, rotational kinetic energy and vibrational energy) Internal energy is a thermodynamic variable and hence it depends on state of the thermodynamic system.

Heat and Work:
The internal energy of a system can be changed either by heat or by work. Heat is energy transfer between a system and surroundings due to temperature difference. Work is mode of energy transfer brought about by means like moving piston (it is not due to temperature difference)

First Law Of Thermodynamics
According to first law of thermodynamics, heat supplied to a system is used to increase its internal energy and to do work.
If ∆Q is heat supplied to the system, ∆W is work done by the system and Au is change in internal energy, then ∆Q = ∆U + AW
Note:

• If the entire heat supplied to the system is used to do work, then ∆Q = ∆W
• The work done against constant pressure, ∆W = P∆V, ∆Q = ∆U + P∆V.

Specific Heat Capacity
Molar specific heat capacity of solid:
Consider a solid consisting of N atoms. Each atom have an average energy 3K_BT, where K_B is Botlzman constant. Total energy for one,mole
U = 3K_BT × N_A
N_A is avagadro number.
At constant pressure, ∆Q = ∆U + P∆V
For solid ∆V is negligible ∆Q = ∆U
Molar specific heat capacity

Thus molar specific heat capacity of solids is found to be 3R. But at low temperature molar specific heat capacity is not a constant.
Specific heat capacity of water:
The specific heat capacity of water is 4186 J Kg-‘K1. But the specific heat capacity of water changes with temperature as shown.

Mayor’s Relation; C_p – C_v = R
If molar specific heat capacity of constant pressure is C_p and that at constant volume is C_v then C_p – C_v = R, for an ideal gas.
Proof:
According to 1^st law of thermodynamics
∆Q = ∆U + P∆V
If ∆Q heat is absorbed at constant volume (∆V = 0)

From ideal gas equation for one mole PV = RT Differentiating w.r.t. temperature (at constant pressure)

Substituting in equation (2)

Equation (4) – Equation (1), we get C_p – C_v = R.

Thermodynamic State Variable And Equation Of State
Extensive and Intensive variables:
Thermodynamic variables can be extensive or intensive. Extensive variables depend on size of the system.
Eg: mass, internal energy, volume.
Intensive variables are independent of size of system.
Eg: Pressure, temperature, density.
Equation of state:
The relation between state variables of a system is called equation of state. For an ideal gas equation of state is PV = µRT

Thermodynamic process
1. Quasi-static process:
The thermodynamic process in which thermo dynamic variables (P, V, T… etc) changes so slowly that the system remains in thermal and mechanical equilibrium is called quasi static (nearly static) process. In quasi-static process change in temperature or pressure will be infinitesimally small. The different types of quasi static process are listed in the above table.

2. Work done in isothermal process:
For an isothermal process, the equation of state is
PV = constant = µRT
P = \(\frac{\mu \mathrm{RT}}{\mathrm{V}}\)
Let a system undergoes an isothermal process at temperature T, from the state (P₁, V₁) to (P₂, V₂). Let ‘DV’ be a small charge in volume due to pressure P.
Then work done (for ∆V)
∆W = P∆V

3. Work done by adiabatic process:
Let an ideal gas undergoes adiabatic charge from (P₁, V₁) to (P₂, V₂). The equation for adiabatic charge is PV^γ = constant = k

from equation (a) P₁V₁^γ = P₂V₂^γ = k

Substituting ideal gas equation.

Note:

1. The graph connecting P and V of isothermal process is called isotherm.
2. In adiabatic process
   • T₁ < T₂, then work is done on gas (w < 0)
   • T₁ > T₂, then work is done by gas (w > 0)
3. For Isothermal process
   • V₁ < V₂, w > 0 work is done by gas
   • V₁ > V₂, w < 0 work is done on gas.
4. In isochoric process no work is done on or by gas because volume is constant.

Cyclic process:
In cyclic process, the system returns to its initial state such that change in internal energy is zero. The P – V diagram for cyclic process will be closed loop and area of this loop gives work done or heat absorbed by system.

Heat Engines
Heat engines converts heat energy into mechanical energy. Heat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work.
Heat engines consists of:

1. Working substance (the system which undergoes cyclic process) eg: mixture of fuel vapor and air in diesel engine, steam in steam engine.
2. An external reservoir at a high temperature (T₁) – it is the source of heat.
3. An external reservoir at low temperature (T₂) or sink

Working:
The working substances absorbs an energy Q₁ from source reservoir at a temperature T₁. It undergoes cyclic process and releases heat Q₂ to cold reservoir. The change in heat (Q₁ – Q₂) is converted in to work (mechanical energy).
Efficiency of heat engine (η):
The efficiency of heat engine is the ratio of work done to input heat.

Note:

• Q₂ = 0, η = 1. When entire heat input is converted into work heat engine is 100% efficient. But practically 100% efficiency cannot be achieved. It is limited by second law of thermodynamics.
• Heat engine can be external combustion engine or internal combustion engine.

In external combustion engine, the fuel (system) is heated by external furnace.
Eg.: Steam engine.
In internal combustion engine, fuel is heated internally by exothermic chemical reactions.
Eg: Diesel engine, Patrol engine.

Refrigerators And Heat Pumps
Refrigerator is reverse of heat engine, the device used to cool a portion of space (inside a chamber) is refrigerator. The device used to pump heat into a portion of space (to warm-up room) is called heat pump.

In both devices, the working substance absorbs heat Q₂ from cold reservoir at temperature T₂. Some external work (by compression of gas by electric means) is done on it and heat Q₁ is supplied to hot reservoir at T₁.

The working cycle of refrigerator:
In refrigerator the working substance is a gas (freon)

Step1: The gaseous working substance is converted into vapor- liquid mixture at lower temperature (T₂)
Step 2: The cold fluid absorbs heat from region to be cooled (cold reservoir) and convert it into vapor.
Step 3: The vapor is heated by external work.
Step 4: The vapor release heat to surroundings and then comes to initial temperature T₂.

The coefficient of performance (α):
The coefficient of performance of refrigerator is defined as
α = \(\frac{Q_{2}}{W}\)
where Q₂ is heat extracted from cold reservoir and W is work done on system.
Note:

• The working substance in refrigerator is termed as refrigerant.
• For heat engine η can not exceed 1. But α can be greater than one.

Second law of Thermo dynamics
Kelvin – Plank statement:
No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of heat into work.
Clausius statement:
No process is possible whose sole result is the transfer of heat from a colder object to hotter object.

Reversible And Irreversible Processes
A themodynamic process is said to be reversible if the process can bring both system and surrounding back to the original state without any change.

A reversible thermodynamic process is ideal case. Most thermodynamic processes are irreversible because the process involves dissipative effects like friction, viscous force etc. and during such process, system passes through non-equilibrium states.
Note: Condition for a thermodynamic process to be reversible is

1. Quasi static
2. non-dissipative

Question 1.
What is the importance of reversibility in thermo dynamics?
Answer:
The main concern of thermodynamics is efficiency with which heat can be converted into work. According to second law of thermodynamics, the efficiency can not be 100%. The heat engine can have highest efficiency only if the cyclic process is reversible.

Carnot’S Engine
Carnot’s theorem:
Sadi carnot proposed Carnot’s theorem. According to Carnot’s theorem

1. No engine operating between two temperature can have efficiency more than that of Carnot’s engine,
2. The efficiency of Carnot’s engine is independent of nature of working substance.

Carnot’s engine:
A reversible heat engine operating between two temperatures is called Carnot’s engine.
Carnot’s cycle:
The Carnot cycle consists of two isothermal processes and two adiabatic processes.

The cylindar is placed on the source. The gas expands and temperature increases. This is the first stroke of the heat engine. The expansion is an isothermal expansion.

During this expansion the working substance originally at the state A (P₁, V₁, T₁) has new variable (P₂, V₂, T₁). The variation A to B is shown byAB in v-p graph.

During the second stroke, the cylinder is placed on the stand and the gas is allowed to expand further and reaches the state C. The new Co-ordinate are (P₃, V₃, T₂).

This expansion is adiabatic expansion. The third stroke is carried out when cylinder is placed on the zink. The cylinder undergoes for isothermal compression and coordinate becomes (P₄, V₄, T₂).

In the fourth and final stroke, the cylinder is placed on the non conducting stand. The gas is compressed back to state A. This is adiabatic compression.

1. The work done by gas in one Carnot cycle
Step 1: Isothermal expansion : The gas absorbs heat Q, from hot reserviorand undergoes Isothermal expansion.
[(P₁, V₁, T₁) → (P₂, V₂, T₂)]

Step 2: Adiabatic expansion
[(P₂, V₂, T₁) → (P₃, V₃, T₂)]

Step 3: Isothermal compression: The gas releases heat Q₂ to cold reservoir at T₂.
[(P₃, V₃, T₂) → (P₄, V₄, T₂)]

Step 4: Adiabatic compression
[(P₄, V₄, T₂) → (P₁, V₁, T₁)]

2. Efficiency of Carnot’s engine:

In adiabatic expansion from
(P₂, V₂, T₁) to (P₃, V₃, T₂)

In adiabatic compression
(P₄, V₄, T₂) to (P₁, V₁, T₁)

Thus equation (b) becomes η = 1 – \(\frac{T_{2}}{T_{1}}\) ……..(3)
Comparing this with equation (a)

Note:

• The equation
• Shows that efficiency of heat engine is independent of nature of working substance.