Class 11

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Chapter 4 Cyber Law Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Solutions Chapter 4 Cyber Law

1. Complete the following web.

Question 1.

Answer:

1. Use the Internet for the knowledge base.
2. Use the Internet to communicate messages.
3. Respect the privacy of other users on the Internet.
4. Download legitimate and authentic programs from the Internet.
5. Use licensed software on your computer.
6. Check to see if the web address begins with https:// whenever you sign in.
7. Report and flag content that is abusive or illegal.

2. Complete the following chart.

Question 1.

Answer:

3. Fill the following boxes with the appropriate cybercrime name.

Question 1.
Copyright violation of software created originally by an individual.

Answer:
Software Piracy

Question 2.
Gaining access without the user’s permission.

Answer:
Unauthorized Access

Question 3.
Extracting confidential information by email.

Answer:
Phishing

4. Read the following and answer the question.

Rahul lost his pen drive in his college computer lab. His classmate Madhav finds it. He carries it home to return it to him the next day. When Madhav opens it he finds his favourite game. He thinks of making a duplicate copy of the game software.

Question I.
Does Madhav think ethically?
Answer:
No

Question II.
What do you think should Madhav do?
Answer:
Madhav should return back the pen drive. He has done unauthorized access.

Question III.
If he makes a duplicate copy then which cybercrime will he commit?
Answer:
Software Piracy

5. Answer in brief.

Question 1.
What care should be taken by the user while doing online activities?
Answer:

1. Stick to copyright restrictions when downloading anything from the internet.
2. Use strong passwords with combinations of numbers. Uppercase and lowercase letters and special characters.
3. Always scan files before opening them.
4. Don’t talk to strangers.
5. Think before posting something on social media related to any sensitive topic like politics, religion, etc.
6. Report to the service provider if the account is hacked; if possible deactivate your account.
7. Whenever you sign in online check if the web address begins with https: //

Question 2.
Define the terms (1) Ethics (2) Moral
Answer:
Ethics:
Ethics also called moral philosophy is the discipline concerned with what is morally good and bad, right or wrong.

Moral:
The standards of behaviour; principles of right and wrong behaviour. Thus morals are dictated by society, culture, or religion while ethics are chosen by the person himself which governs his life.

Question 3.
Explain three examples related to unauthorized access?
Answer:

1. Hacking financial/bank account-related information.
2. Stealing organizational/intellectual information.
3. Illegal monitoring of information owned by other users.
4. Illegal use/break of login and password of other users.
5. Causing intentional irritation to other users by means of damaging software and important information.

Question 4.
Explain software piracy and Hacking.
Answer:
Software piracy:
Software piracy is the illegal copying, distribution, or use of the software. It 15 such a profitable “business” that it has caught the attention of organized crime groups in a number of countries. According to the Business Software Alliance (BSA), about 36% of all software in current use is stolen.

Hacking:
Hacking refers to unauthorized intrusion into a computer or a network. Hacker is a person intensely interested in the deep and hidden or concealed working of any computer operating system and programming language. They might discover loopholes within systems and reasons for such loopholes.

6. State true or false.

Question 1.
A firewall is used to encrypt the transmission of data.
Answer:
True

Question 2.
The standards of behaviour; the principle of right or wrong are referred to as moral.
Answer:
True

Question 3.
Hacking bank account-related information is an example of software piracy.
Answer:
False

Question 4.
Phishing is representing someone else’s work as own without permission.
Answer:
True

7. Match the following.

Question 1.

(A)	(B)
(1) Copying a game file	(a) Firewall
(2) Law related to internet	(b) Cyberthreat
(3) Network security	(c) Software piracy
(4) System crashes suddenly	(d) Cyber Law

Answer:

(A)	(B)
(1) Copying a game file	(c) Software piracy
(2) Law related to internet	(d) Cyber Law
(3) Network security	(a) Firewall
(4) System crashes suddenly	(b) Cyberthreat

8. Observe the following picture and give your opinion about it by responding to the following questions.

Question 1.
Is ‘B’s response correct or not?
Answer:
No B’s response is not correct.

Question 2.
What will be the consequences of ‘B’s reply?
Answer:
With all such crucial information like OTP in hand, the fraudster can easily carry out illegal financial transactions using B’s identity.

Question 3.
Which type of cybercrime does the picture depict?
Answer:
Fake Call Frauds

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Chapter 3 Impressive Web Designing Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Solutions Chapter 3 Impressive Web Designing

1. Answer the Following:

Question 1.
The data entry operator wants to insert.
1. Photograph
2. Write remarks about the photograph
3. Underline the heading.
He will use:
1. <Image>
2. <Text>
3. <TextArea>
4. <Img>
5. <UL>
6. <U>
Select the correct tags from the above and arrange them in the sequence.
Answer:
1. Photograph – <Img>
2. Write remarks about the photograph – <TextArea>
3. Underline the heading – <U>

Question 2.
Identify the logical operators in JavaScript.
1. OR
2. AND
3. ||
4. &
5. &&
6. ++
Answer:
| |, &&

2. Complete the following Activity:

Question 1.
State at least three attributes of <Input>

Answer:

Question 2.

Answer:

Question 3.
Group the Following.

Answer:

Question 4.
Write operator names with symbols in boxes.

Answer:

Question 5.
Complete following program to display multiplication of 6.40 and 300.
<!DOCTYPE html>
<html>
<head><title> Series </title></head>
<body>

</html>
Answer:
<!DOCTYPE html>
<html>
<head><title> Series </title></head>
<body>

</html>

3. Find out errors if any in the following javascript code.

Question 1.
var length, breadth;
length=4.5;
breadth=6;
area=1/2*length*breadth;
document.write(“Area of triangle is”area);
Answer:
Concatenate Operator ‘+’ is missing
document. write(“Area of triangle is” [+] area);

4. Solve the following puzzles.

Question A.
Fill the blocks

1) Across
2. The tag is used to create a table row.
4. Tag to create a form

2) Down
1. The attribute is used to specify the path of a linked document.
3. The tag used to display horizontal ruled line
Answer:

Across
2. <tr>
4. <form>
Down
1. href
3. <hr>

Question B.
Solve the puzzle by finding words with the help of the hint given below.

1. Boolean value.
2. Keyword used in conditional if statement.
3. Built-In function in JavaScript
4. Function to check given value is number or not
5. Keyword used to declare a variable
6. Function used to evaluate given expression
Answer:

1. False
2. if
3. confirm
4. nan
5. var
6. False

5. Trace the output of the following Html code.

Question 1.
<!DOCTYPE html>
<html>
<head>
<title>Heading tags</title></head>
<body>
<h1 align=left>Information Technology</h1>
<hr>
<h2 align=center>XI Standard</h2>
<h3 align=right>Division</h3>
</body>
</html>
Answer:

6. Discuss the following and answer the questions.

Question 1.
A Jr. web designer wants to design a webpage to enter the employee’s name, address. He needs a button to clear the form content and submit the data.
Write the different controls he will use to create the web page.
State the tags to be used for the controls.
Answer:
A Junior Web designer will use the following controls to design web as follows:
1. Text – to create a single-line textbox for employee name
2. Text or Textarea – If he wants to create a single-line address the text control is used otherwise to create a multiline address textarea is used.
3. Submit – It is used to submit form data to the server.
4. Reset – It is used to clear form data.

Tags used for the controls are-
1. Employee Name :<Input type-”text”>
2. Address :<Input type=”text”> or
Address: <textarea rows=”4” cols-”30”><textarea>
3. <Input type=”submit” value=”submit”>
4. <Input type-”reset” value=”reset”>

Question 2.
A teacher has asked a student to create a web page to accept numbers and check whether it is between 50 to 100.
List the variable, operators to be used.
Specify the built-in function used and structure used.
Answer:
Program:
<!DOCTYPE html>
<head>
</head>
<body>
<script language =”Javascript”>
//To check whether the given no is between 50 to 100
var n;
n=prompt(“Enter any number”);
if(n<50 | | n>100)
{
alert(“it is less than 50, Please Re-Enter”);
}
else
{
alert(“it is greater than 50. Please Re-Enter”);
}
</script>
</body>
</html>

Variables used ‘n’
Operators used Not is equals | | less then < grater then > etc
Function is Prompt(), Alert() if then Else

7. Create web pages for the following.

Question 1.
Write a program using HTML to design your class Time Table.
Answer:
Program <!Doctype Html>
<html>
<head><title> Time Table of Class XI</title>
</head>
<body bgcolor=yellow>
<h2>Time Table of Class XI</h2>
<table border=”2” bgcolor=skyblue>
<tr>
<th>Period No</th>
<th>Monday</th>
<th>Tuesday</th>
<th>Wednesday</th>
<th>Thursday</th>
<th>Friday</th>
<th>Saturday</th>
</tr>
<tr>
<td>1</td>
<td>Book Keeping</td>
<td>Economics</td>
<td>Maths</td>
<td>Maths</td>
<td>English</td>
<td>English</td>
</tr>
<tr>
<td>2</td>
<td>Maths</td>
<td>Maths</td>
<td>Book Keeping</td>
<td>Economics</td>
<td>English</td>
<td>IT</td>
</tr>
<tr>
<td>3</td>
<td>Maths</td>
<td>Maths</td>
<td>Book Keeping</td>
<td>Economics</td>
<td>IT</td>
<td>IT</td>
</tr>
<tr>
<th colspan=”7”>”L U N C H_B R E A K”</th>
</tr>
<tr>
<td>4</td>
<td>Organization of Commerce</td>
<td>Secretarial Practice</td>
<td>EVS</td>
<td>IT</td>
<td>Economics</td>
<td>English</td>
</tr>
<tr>
<td>5</td>
<td>Secretarial Practice</td>
<td>EVS</td>
<td>EVS</td>
<td>IT</td>
<td>Economics</td>
<td>English</td>
<tr>
<th colspan=”7”>”S H O R T_B R E A K”</th>
</tr>
<tr>
<td>6</td>
<td>Organization of Commerce</td>
<td>Secretarial Practice</td>
<td>Maths</td>
<td>Maths</td>
<td>English</td>
<td>English</td>
</tr>
<tr>
<td>7</td>
<td>Economics</td>
<td>Maths</td>
<td>Book Keeping</td>
<td>Book Keeping</td>
<td>English</td>
<td>IT</td>
</tr>
</table>
</body>
</html>

Question 2.
Write a program using HTML to create a form and submit it with personal data like name, address, and standard.
Answer:
Program:
<!DOCTYPE html>
<head>
<title> Form</title>
</head>
<body bgcolor=”#FF0100”>
<h1>Personal Data</h1><br><br><br>
Name :<input type=text name=t1><br><br>
Address:<Textarea> Address Here </textarea><p><br>
Class XI <input type=radio name=c1 value=xi>
Class XII <input type=radio name=c1 value=xii><br>
<input type=Submit name=b1 value=Submit>
<input type=reset name=b2 value=Reset>
</body>
</html>

Output:

Question 3.
Write a javascript program to accept two numbers and perform the addition of two numbers by using the mouseover event.
Answer:
<html>
<head>
<title> JavaScript Program </title>
<script language = “JavaScript”>
function iload()
{
var c=0
a=parseInt(prompt(“Enter 1st no.”));
b=parseInt(prompt(“Enter 2nd no.”));
c=a+b
document.write(“Addition is”+c);
}
</script>
</head>
<body><center><br><br><br><br><br>
<a onMouseover=”iload();”>Addition</a></center>
</body>
</html>

8. Complete the following.

Question 1.

Answer:

9. Write HTML Code for the following table.

Question 1.

Answer:
<!DOCTYPE html>
<head>
<title>Html Program </title>
</head>
<BODY>
<table border=”1”>
<tr bgcolor=pink>
<th>Place</th>
<th>State</td>
<th>Max <br>Temperature <br>in C</th>
</tr>
<tr align=center bgcolor=skyblue>
<td>Wardha</td>
<th rowspan=”2”>Maharashtra</th>
<td>47.5</td>
</tr>
<tr align=center bgcolor=skyblue>
<td>Akola</td><td>46.4</td>
</tr>
<tr align=center bgcolor=skyblue>
<td>Khajuraho</td>
<th rowspan=”2”>Madhya Pradesh</th>
<td>46.4</td>
</tr>
<tr align=center bgcolor=skyblue>
<td>Sagar</td><td>46.2</td>
</tr>
</table>
</body>
</html>

10. Multiple choice one correct answer.

Question 1.
The default method of submitting form data is __________
(a) Post
(b) Get
(c) Submit
(d) Reset
Answer:
(b) Get

Question 2.
In JavaScript the post increment operator is __________
(a) x++
(b) x–
(c) –x
(d) ++x
Answer:
(a) x++

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Chapter 2 Introduction to DBMS Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Solutions Chapter 2 Introduction to DBMS

1. Complete the following activity.

Question 1.
Tick whichever box is not valid.

Answer:

Question 2.
The student wants to create a field Pincode in a table, which data type he will choose?

Answer:
Int

Question 3.
Tick the appropriate box.

Answer:
Uniquely identifies a record

Question 4.
Tick the appropriate circle.

Answer:
One or Many Tables

2. Observe the field names of a database given below in ‘Column A’ related to Bus reservation. Write suitable data types for each field in front of the respective field in ‘Column B’.

Question 1.

Answer:

Column A (Field Name)	Column B (DataType)
Passenger Name	Varchar
Age	Int
Gender	Char
Mobile Number	Int

3. Write the use of the following SQL command.

Question 1.

Answer:

4. Create a table for the information given below by choosing appropriate data types. Specify proper primary key for the table (1) Movie (2) Actor

Question 1.
Movie (Registeration_no, movie_name, Realease_Date)
Answer:
CREATE TABLE Movie (Registeration_no int(20) PRIMARY KEY, movie_name VARCHAR(20), Realease_Date DATE);

Question 2.
Actor (actor_id, Actor_name, birth_date )
Answer:
CREATE TABLE Actor (actor_id int(20) PRIMARY KEY, Actor_name VARCHAR(20), Birth_Date DATE);

5. Consider the following table Stationary. Write SQL commands for the following statements.

Table: Stationary

Question 1.
Write SQL command to create above Table.
Answer:
CREATE TABLE Stationary (S_ID int(20) PRIMARY KEY, S_Name VARCHAR(20), C_Name VARCHAR(20), Price int(20), Quantity int(20));

Question 2.
Write SQL command to insert the above-mentioned record in the table.
Answer:
INSERT INTO Stationary Values (001, ‘Note Book’, ‘ABC’, 20,50);
INSERT INTO Stationary Values (002, ‘Note Book’, ‘XYZ’, 10,80);
INSERT INTO Stationary Values (003, ‘Note Book’, ‘PQR’, 600,02);

Question 3.
To delete the above table.
Answer:
DROP TABLE Stationery;

6. Answer the following questions.

Question 1.
What is a database?
Answer:
A DBMS is a collection of programs (computer-based system) that enables the user to create and maintain a database it is used to define, construct and manipulate the data in the database for various applications. It provides information storage, organization, and retrieval capabilities. The DBMS also enforces necessary access restrictions and security measures in order to protect the database.

Various types of control systems within the DBMS make sure that the database continues to function properly. They Include Integrity system Security system Concurrency control system Recovery control system Some DBMS enables us to define “views” of the database. A view is how the database appears to the user. This enables us to show only the relevant information to different types of users and it increases security, as certain users will not be able to see data which they are not meant to see.

Question 2.
What are the advantages of a DBMS?
Answer:
Advantages of a DBMS:
1. Redundancy is controlled: In File Processing System, duplicate data is created in many places because all the programs have their own files. This creates data redundancy which in turn wastes labor and space. ‘In Database Management System, all the files are integrated into a single database. The whole data is stored only once in a single place so there is no chance of duplicate data.

2. Sharing of Data: In a database, the users of the database can share the data among themselves. There are various levels of authorization to access the data. And consequently, the data can only be shared based on the correct authorization protocols being followed.

3. Data Security: Data Security is a vital concept in a database. Only authorized users should be allowed to access the database and their identity should be authenticated using a username and password. Unauthorized users should not be allowed to access the database under any circumstances as it violates the integrity constraints.

4. Enforces integrity constraints: Constraints are used to store accurate data because there are many users who feed data in the database. Data stored in the database should always be correct and accurate. DBMS provides the capability to enforce these constraints on the database.

5. Provides backup and recovery of data: Data loss is a very big problem for all organizations. In a traditional tile processing system, a user needs to back up the database after a regular interval of time that wastes lots of time and resources. If the volume of data is large then this process may take a very long time.

Question 3.
What do you understand by Data Model?
Answer:
A Database model defines the logical design and structure of a database and defines how data will be stored, accessed, and updated in a database management system. While the Relational Model is the most widely used database model.

Relational Model:
It is the most popular data model in DBMS. The relational model is the primary data model. Which is widely used for data processing. This model has all properties required to Process data with storage efficiency.

Question 4.
What is a primary key?
Answer:
Primary Key: A column in the table that uniquely identifies each row in that table is called a primary key.

Question 5.
What is DDL (Data Definition Language)
Answer:
DDL statements or commands are used to define and modify the database structure of your tables or schema. When you execute a DDL statement, it takes effect immediately.
Some examples of DDL commands are CREATE, ALTER, and DROP.

7. In a company the data is stored in a table under the following fields Employee number, Last name, Date of birth, Address. Which data type will you use for the above field?

Question 1.

Answer:

Field Name	Data Type
Employee Name	Char
Last Name	Char
Address	Varchar
Date of Birth	Date

8. Multiple choice select three correct answers.

Question 1.
Valid relationships in RDBMS are
(a) one to one
(b) one to two
(c) one to many
(d) many to two
(e) many to many
(f) one to three
Answer:
(a) one to one, (c) one to many, (e) many to many

9. Complete the following.

Question 1.

Answer:

Use	Command
To remove access rights or privileges from the database	Revoke
Extracts data from a table	Insert into Select

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Chapter 1 Basics of Information Technology Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Solutions Chapter 1 Basics of Information Technology

1. Complete the following activity.

Question 1.

Answer:

Question 2.
Tick the appropriate box.
Internet is a ____________ network connecting millions of computer.

Answer:
Global

Question 3.

Answer:

Question 4.
Tick the appropriate box.

Answer:

Question 5.

Answer:

2. Divide the following list of devices into appropriate categories.

Question 1.
Monitor, Barcode reader, Printer, Keyboard. Optical character reader, Speaker

Answer:

Categories	Names
Input Devices	Barcode reader, Keyboard, Optical Character reader
Output Devices	Monitor, Printer, Speaker

3. Multiple choice two correct answers.

Question 1.
The primary memory consists of ____________ and ____________
(a) Pendrive
(b) Hard Disk
(c) RAM
(d) Scanner
(e) ROM
Answer:
(c) RAM, (e) ROM

Question 2.
The network architectures which are widely used are ____________
(a) Server
(b) Client
(c) Peer to peer
(d) Client-server
(e) Internet
Answer:
(c) Peer to peer, (d) Client-server

4. Match the following.

Question 1.

1. IS	(a) change directory
2. FTP	(b) Translates Network Address
3. CD	(c) List of Directory
4. DNS	(d) To transfer file on interent

Answer:

1. IS	(c) List of Directory
2. FTP	(d) To transfer file on interent
3. CD	(a) change directory
4. DNS	(b) Translates Network Address

5. Name the following and complete the diagram.

Question 1.

Answer:

6. Complete the following with Linux commands with their use.

Question 1.

Answer:

7. Complete the list of the following protocols.

Question 1.

Answer:

8. Complete the following Long form.

Question 1.

Answer:

9. Identify the following activity.

Question 1.
You are typing a letter using a computer and suddenly there is a power failure.
Which type of Memory does this activity deal with?
Answer:
Random Access Memory

10. Answer the following.

Question 1.
What are Data and Information? Give examples of data and information.
Answer:
Data can be any character, text, word, number, or raw facts.
Example of Data:
Mumbai, 1234, Aditya, MG Road, Maharashtra, 9444444441, 411004

Information is data formatted in a manner that allows it to be utilized by human beings in some significant way.
Example of Information:
Aditya, 1234, MG Road, Mumbai 400004, Maharashtra, 944444444114.

Question 2.
Explain functional units of a computer system.
Answer:
The computer system has the following three basic components:
Input Unit:
An input device is any hardware device that sends data to a computer, allowing you to interacts with and controls it. Data can be in the form of words, symbols, numbers, etc. The function of the input device is to direct commands and data into the computer.
For example keyboard, mouse, scanners, digital cameras, joysticks, and microphones.

Central Processing Unit:
After receiving data and commands from the user, a computer system has to process the instructions provided using Central Processing Unit (CPU). It has three elements:

(a) Arithmetic and Logic Unit: An arithmetic logic unit (ALU) is a major component of the central processing unit of a computer system. It does all processes related to arithmetic and logic operations like add, subtract, multiply, etc.

(b) Control Unit: The control unit (CU) is a component of a computer’s central processing unit (CPU) that directs the operation of the processor. It tells the computer’s memory, arithmetic, and logic unit, and input and output devices how to respond to the instructions that have been sent to the processor.

(c) Memory Unit: A memory unit is the amount of data that can be stored in the storage unit. Once the data has been entered using input devices, the system stores the data in the memory unit.

Types of Memory: Primary Memory & Secondary Memory.

• Primary Memory: It has 18 internal memory of the computer also known as main memory. It is of two types RAM and ROM.
• RAM (Random Access Memory): RAM stands for Random Access Memory also known as reading/write memory. Information stored in this memory is lost as the power supply to the computer is switched off; it is also called “Volatile Memory”.
• ROM(Read Only Memory): ROM stands for Read-Only Memory. ROM is a permanent type of memory. The contents are not lost as the power supply to the computer is switched off. ROM cannot be overwritten by the computer. It is also called “Non Volatile Memory”.
• Secondary Memory: It is the external memory of the computer which is used to store a large amount of data. The secondary storage devices are a Hard disk, Pen drive, CD, DVD, etc.

Output Unit:
An output device is any device used to send data from a computer to another device or user. Most computer data output that is ‘meant for humans is in the form of audio or video. Thus, most output devices used by humans are in these categories. Examples include monitors, projectors, speakers.

Question 3.
What is a storage unit? Explain types of primary memory storage.
Answer:
When a user enters data using input devices, the computer system stores this data in a memory unit i.e. storage unit. The storage unit uses a set of pre-programmed instructions to further transmit this data to other parts of the CPU. There are two types of memory

• Primary Memory
• Secondary Memory

RAM: It stands for Random Access Memory. RAM is known as reading/write memory. It is the main memory of the computer system. The information stored in this memory is lost as the power supply to the computer is switched off, so it is also called as “Volatile Memory”.

ROM: It stands for Read-Only Memory. ROM is permanent memory. The content is not lost when the power supply is switched off. ROM cannot be overwritten by the computer, so it is also called “Non-Volatile Memory”.

Question 4.
Explain how Linux is different from Windows.
Answer:

• Linux is open sources system whereas the window operating system is commercial.
• Linux has access to source code and alters the code as per user need whereas a window does not have access to source code.
• Linux distribution doesn’t collect user data” whereas Windows collects all the user details which leads to privacy concerns.
• As the software is open to the public, it constantly updates, improves, and expands as more people can work on its improvement.

Question 5.
Write down the difference between LAN, MAN, and WAN.
Answer:

LAN (Local Area Network)	MAN (Metropolitan Area Network)	WAN (Wide Area Network)
LAN Stands for Local Area Network.	MAN stands for Metropolitan Area Network.	WAN stands for Wide Area Network.
A LAN is a network of connected devices that exist within a specific location.	A public or private network is used to connect various locations including suburbs in metropolitan cities.	A WAN is any network that crosses metropolitan, regional, or national boundaries.
LANs may be found in homes, offices, educational institutions, or other areas.	A MAN is a network, which covers an entire city, but uses LAN topology.	Most networking professionals define a WAN as any network that uses routers and public network links (e.g. Telephone lines).
LAN is easy to set up.	MANs are formed by connecting multiple LANs.	Due to long-distance transmission, the noise and errors are more in WAN.
Data transmits at a very fast rate.	Examples of a MAN are the cable TV network in a city.	The best example of WAN is the Internet.

Maharashtra State Board Information Technology 11th Std Textbook Solutions Digest

Information Technology 11th Std Commerce | Information Technology Class 11 Pdf Maharashtra Board

• Chapter 1 Basics of Information Technology
• Chapter 2 Introduction to DBMS
• Chapter 3 Impressive Web Designing
• Chapter 4 Cyber Law

Information Technology 11th Std Skill Oriented Practicals (SOP)

(Select Any Four Skill Oriented Practicals from the following list)

• Skill Set 1 Daily Computing (Libre Office)
• Skill Set 2 Web Designing (HTML – 5)
• Skill Set 3 Client Side Scripting (JavaScript)
• Skill Set 4 Accounting Package (GNUKhata)
• Skill Set 5 Digital Content Creation (GIMP, Inkscape)
• Skill Set 6 DBMS (PostgreSQL)

Maharashtra State Board Class 11 Textbook Solutions

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 8 Sound Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 8 Sound

1. Choose the correct alternatives

Question 1.
A sound carried by air from a sitar to a listener is a wave of following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationary
(D) Longitudinal progressive
Answer:
(D) Longitudinal progressive

Question 2.
When sound waves travel from air to water, which of these remains constant ?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer:
(B) Frequency

Question 3.
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer:
(C) propagate adiabatically

Question 4.
Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer:
(D) solid

Question 5.
The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer:
(D) Absorb sound

2. Answer briefly.

Question 1.
Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.

Question 2.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

Question 3.
Describe a transverse wave.
Answer:
Transverse wave:
A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave.
Example: Ripples on the surface of water, light waves.

Characteristics of transverse waves:

1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength ‘λ’ of the wave.
4. Crests and troughs advance in the medium and are responsible for transfer of energy.
5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
7. Transverse wave can be polarised.
8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.

Question 4.
Define a longitudinal wave.
Answer:
A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.

Question 5.
State Newton’s formula for velocity of sound.
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Question 6.
What is the effect of pressure on velocity of sound?
Answer:
Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is ρ.

ii. Using Laplace’s formula, we can write,
v = \(\sqrt{\frac {γP}{ρ}}\) ….(1)

iii. If V be the volume of a gas having mass M then, ρ = \(\frac {M}{V}\)

iv. Substituting ρ in equation (1), we get,
v = \(\sqrt{\frac {γPV}{M}}\) ….(2)

v. But according to Boyle’s law,
PV = constant (at constant temperature)
Also, M and γ are constant.
∴ v = constant

vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
v = \(\sqrt{\frac {γnRT}{M}}\)

viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

Question 7.
What is the effect of humidity of air on velocity of sound?
Answer:
Effect of humidity:
i. Let v_m and v_d be the velocities of sound in moist air and dry air respectively.

ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density ρ_m of moist air is less than the density ρ_d of dry air i.e., ρ_m < ρ_d.

iii. Thus \(\frac {v_m}{v_d}\) > 1
∴ v_m > v_d

iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.

Question 8.
What do you mean by an echo?
Answer:
An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.

Question 9.
State any two applications of acoustics.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.

Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.

Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 10.
Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave.
SI unit: (m)

ii. Wavelength (λ): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave.
SI unit: (m)

Question 11.
Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.
Answer:

i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

Question 12.
Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (λ) during one period (T).
Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = \(\frac {Distance covered}{Corresponding time}\)

ii. v = \(\frac {22}{7}\) i.e., v = λ × (\(\frac {1}{T}\)) …………….. (1)

iii. But reciprocal of the period is equal to the frequency (n) of the waves.
∴ \(\frac {1}{T}\) = n …………… (2)

iv. From equations (1) and (2), we get
v = nλ
i.e., wave velocity = frequency × wavelength.

Question 13.
State and explain principle of superposition of waves.
Answer:
Principle:
As waves don’t repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.

Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.

ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
y₁ = displacement due to first wave.
y₂ = displacement due to second wave.

iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.
∴\(\vec{y}\) = \(\vec{y_1}\) + \(\vec{y_2}\)

Question 14.
State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

ii. Apparent frequency heard by the listener is given by,
n = n₀(\(\frac {v±v_L}{v±v_s}\))
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.

iii. If source and listener are moving towards each other, then apparent frequency is given by,
n = n₀(\(\frac {v+v_L}{v-v_s}\)) i.e., apparent frequency increases.

iv. If source and listener are moving away from each other, then apparent frequency is given by,
n = n₀(\(\frac {v-v_L}{v+v_s}\)) i.e., apparent frequency decreases.

Question 15.
State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer:
Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

i. If listener is moving towards source then apparent frequency is given by,
n = n₀(\(\frac {v+v_L}{v}\)) i.e., apparent frequency increases.

ii. If listener is receding away from source then apparent frequency is given by,
n = n₀(\(\frac {v-v_L}{v}\)) i.e., apparent frequency decreases.

Question 16.
State the expression for apparent frequency when listener is stationary and source is.

(i) moving towards the listener
(ii) moving away from the listener
Answer:
Let,
n = actual frequency of the source.
n₀ = apparent frequency of the source,
v = velocity of sound in air.
v_s = velocity of the source.
v_l = velocity of the listener.

i. If source is moving towards observer then apparent frequency is given by,
n = n₀(\(\frac {v}{v-v_s}\)) i.e., apparent frequency increases.

ii. If source is receding away from observer then apparent frequency is given by,
n = n₀(\(\frac {v}{v+v_s}\)) i.e., apparent frequency decreases.

Question 17.
Explain what is meant by phase of a wave.
Answer:

i. The state of oscillation of a particle is called the phase of the particle.

ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.

iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.

iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.

v. Separation between two particles which are in phase is wavelength (λ).

vi. The two successive particles differ by ‘1’ in their oscillation number i.e., if particle v is at its n^th oscillation, particle p will be at its (n + 1)^th oscillation as the wave is travelling along + X direction.

vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle v is \(\frac {π^c}{2}\) = 90° at this instant.

viii. Phase angles of particles u and q are π^c (180°) and 2rcc (360°) respectively.

ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.
∴ phase angle = 2π^c + \(\frac {π^c}{2}\)
= \(\frac {5π^c}{2}\)

x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by 2π^c.

Question 18.
Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.

Question 19.
Distinguish between traverse waves and longitudinal waves.
Answer:

Longitudinal wave	Transverse wave
1. The particles of the medium vibrate along the direction of propagation of the wave.	1. The particles of the medium vibrate perpendicular to the direction of propagation of the wave.
2. Alternate compressions and rarefactions are formed.	2. Alternate crests and troughs are formed.
3. Periodic compressions and rarefactions, in space and time, produce periodic pressure and density variations in the medium.	There are no pressure and density, variations in the medium.
4. For propagation of a longitudinal wave, the medium must be able to resist changes in volume.	For propagation of a transverse wave, the medium must be able to resist shear or change in shape.
5. It can propagate through any material medium (solid, liquid or gas).	It can propagate only through solids.
6. These waves cannot be polarised.	These waves can be polarised.
7. eg.: Sound waves	eg.: Light waves

Question 20.
Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton’s formula and the experimental value.

2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.

3. Solve the following problems.

Question 1.
A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = \(\frac {v}{λ}\)
ii. T = \(\frac {1}{n}\)
Calculation: From formula, (i)
n = \(\frac {340}{1.7}\)
∴ n = 200 Hz
From formula, (ii)
T = \(\frac {1}{n}\) = \(\frac {1}{2×10^2}\)
= 5 × 10^-3
…….. (using reciprocal Table)
∴ T = 0.005 s

Question 2.
A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 × 2
∴ v = 340 m/s

Question 3.
An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = \(\frac {distance}{time}\)
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
v = \(\frac {2 × d}{t}\)
∴ d = \(\frac {1500 × 0.45}{2}\) = 337.5 m

Question 4.
A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.

ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 5.
Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: T_A = 0.015 s, T_B = 0.025 s
To find: greater frequency (n)
Formula: n = \(\frac {1}{T}\)
Calculation: From formula,
n_A = \(\frac {1}{T_A}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ n_A = 66.67
…. (using reciprocal table)
n_B = \(\frac {1}{T_B}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ n_B = 40 Hz
…. (using reciprocal table)
∴ n_A > n_B

Question 6.
At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
v_air = 1.75 V_S.T.P = \(\frac {7}{4}\) v_S.T.P
T_S.T.P = 273 K
To find: temperature T_air
Formula: v ∝ √T
Calculation: From formula,

Question 7.
A man standing between 2 parallel eliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = \(\frac {distence}{time}\)
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t₁
= \(\frac {3}{2}\) = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t₂
= \(\frac {5}{2}\) = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
v = \(\frac {1360}{4}\)
∴v = 340 m/s

Question 8.
If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v₁ and v₂ be the velocity of sound on day 1 and day 2 respectively.
\(\frac {v_1}{v_2}\) = \(\frac {1}{1.1}\)
We know, v ∝ \(\frac {1}{√ρ}\)
Let ρ₁ and ρ₂ be the density of air on day 1 and day 2 respectively.
∴ \(\sqrt{\frac {ρ_2}{ρ_1}}\) = \(\frac {1}{1.1}\)
∴ \(\frac {ρ_2}{ρ_1}\) = (\(\frac {1}{1.1}\))²
∴ ρ₁ = 1.1² ρ₂ = 1.21 ρ²
From above equation, we can conclude,
ρ₁ > ρ₂
∴ v₂ > v₁ i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (v_m) is greater than speed of sound in dry air (v_d).
∴ We can conclude, air is moist on second day and dry on the first day.

Question 9.
A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: v_s = 15 m/s, n0 = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n₀(\(\frac {v}{v-v_s}\))
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 (\(\frac {340}{340-15}\)) = \(\frac {3400}{13}\)
∴ n = 261.54 Hz

Question 10.
The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: v_s = 30 m/s, n₀ = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (n_A – n’_A)
Formulae:
i. When the ambulance moves towards he stationary observer then n_A = n₀(\(\frac {v}{v-v_s}\))

ii. When the ambulance moves away from the stationary observer then, n’_A = n₀(\(\frac {v}{v+v_s}\))

Calculation:
From formula (i), icon’ 340
n_A = 1500(\(\frac {340}{340-30}\))
∴ n_A = 1645 Hz
From (ii)
n’_A = 1500(\(\frac {340}{340+30}\))
∴ n_A = 1378 Hz
Difference between n_A and n’_A
= n_A – n’_A = 1645 – 1378 = 267 Hz

11th Physics Digest Chapter 8 Sound Intext Questions and Answers

Can you recall? (Textbook page no. 142)

i. What type of wave is a sound wave?
ii. Can sound travel in vacuum?
iii. What are reverberation and echo?
iv. What is meant by pitch of a sound?
Answer:
i. Sound wave is a longitudinal wave.

ii. Sound cannot travel in vacuum.

iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

Activity (Textbook page no. 144)

i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.

Answer:

Wave	Wavelength (λ)	Amplitude (A)
A	4 m	2 m
B	2 m	4 m

Wave	Wavelength (λ)	Amplitude (A)
P	6 units	3 units
Q	4 units	2 units

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• Chapter 1 प्रेरणा
• Chapter 2 लघु कथाएँ
• Chapter 3 पंद्रह अगस्त
• Chapter 4 मेरा भला करने वालों से बचाएँ
• Chapter 5.1 मध्ययुगीन काव्य (अ) भक्ति महिमा
• Chapter 5.2 मध्ययुगीन काव्य (आ) बाल लीला
• Chapter 6 कलम का सिपाही
• Chapter 7 स्वागत है !
• Chapter 8 तत्सत
• Chapter 9 गजलें
• Chapter 10 महत्त्वाकांक्षा और लोभ
• Chapter 11 भारती का सपूत
• Chapter 12 सहर्ष स्वीकारा है

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• Chapter 17 ई – अध्ययन : नई दृष्टि

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• भावार्थ : भक्ति महिमा और बाल लीला
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Maharashtra State Board Class 11 Textbook Solutions

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 11 Study of Animal Type: Cockroach Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 11 Study of Animal Type: Cockroach

Question (A)
Chemical nature of chitin is ____________ .
(A) protein
(B) carbohydrate
(C) lipid
(D) glycoprotein
Answer:
(B) carbohydrate & (D) glycoprotein

Question (B)
Cockroach has ___________ type of mouthparts.
(A) sponging
(B) chewing and biting
(C) piercing and sucking
(D) lapping
Answer:
(B) chewing and biting

Question (C)
Spiracle is a part of ________ system of cockroach.
(A) circulatory
(B) respiration
(C) reproductive
(D) nervous
Answer:
(B) respiration

Question (D)
________ is a part of digestive system.
(A) Trachea
(B) Hypopharynx
(C) Haemocyte
(D) Seminal vesicle
Answer:
(B) Hypopharynx

Question (E)
_________ is also called as brain of cockroach.
(A) Supra-oesophageal ganglion
(B) Sub-oesophageal ganglion
(C) Hypo-cerebral ganglion
(D) Thoracic ganglion
Answer:
(A) Supra-oesophageal ganglion

2. Answer the following questions

Question (A)
Describe the digestive system of cockroach.
OR
With the help of neat and labelled diagram, describe the digestive system of cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.

(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.

(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.

(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.

(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is divisible into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (B)
Give an account on tracheal system of cockroach.
Answer:
1. Cockroach has an internal respiratory system of air tubes called tracheal system by which the air is brought into the body and is in contact with every part of the body. It allows the exchange of gases directly between the air and tissues without the need of blood.
These air tubes of internal respiratory system begin at the opening on body surface called spiracles.

2. Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

3. Trachea: The trachea form a definite pattern of branching tubes arranged transversely as well as longitudinally. They are about 1mm thick and have spiral or annular thickening of chitin. The inner lining of chitin prevents the trachea from collapsing. Each trachea further branches into smaller tubes called tracheoles.

4. Tracheoles: These are fine intracellular tubes that penetrate deep into tissues. They are thin and not lined by chitin. They end blindly in the cells. Each tracheole at the blind end is filled with a watery fluid through which exchange of gases takes place. The content of this fluid keeps changing. At high muscular activity, part of fluid part is drawn into the tissues to enable more and rapid oxygen intake.

Question (C)
Describe the nervous system of cockroach.
Answer:
Nervous system in cockroach:
Nervous system of cockroach is ventral, solid and ganglionated. It consists of central nervous system (CNS), peripheral nervous system (PNS) and autonomous nervous system (ANS).
Central nervous system (CNS): Central nerv ous system consists of nerve ring and ventral nerve cord.
1. Nerve ring consists of:
a. a pair of supra-oesophageal ganglia
b. a pair of circum-oesophageal connectives
c. a pair of sub-oesophageal ganglia
(a) Supra-oesophageal ganglia or cerebral ganglia: A pair of supra-oesophageal ganglia is collectively known as the brain. Brain is present in head, above the oesophagus and between antennal bases. Each supra-oesophageal ganglion is formed by the fusion of three small ganglia – protocerebram, deutocerebrum and tritocerebrum.
(b) Circum-oesophageal connectives: Supra-oesophageal ganglia are connected to sub-oesophageal ganglion by a pair of lateral nerves called as circum-oesophageal connectives. Connectives arise from supra-oesophagial ganglia.
(c) Sub-oesophageal ganglia: It is a bilobed and present below the oesophagus, in head. It is also formed by the fusion of three pairs of ganglia.

2. Ventral nerve cord:
a. It arises from the sub-oesophageal ganglion. It is present along mid-ventral position, in perineural sinus.
b. It is double ventral nerve cord and consists of nine segmental, paired ganglia.
c. First three pairs of segmental ganglia are large and known as thoracic ganglia. The other six pairs of segmental ganglia are in abdomen (abdominal ganglia).
d. 6th abdominal ganglion is the largest and it is present in 7th abdominal segment.
e. There is no ganglion in 6th segment.

Peripheral nervous system (PNS):

1. The peripheral nervous system comprises of nerves that arise from various ganglia of CNS.
2. Six pairs of nerves arise from the supra-oesophageal ganglia.They supply to the eyes, antenna and labrum.
3. Nerves arising from the sub-oesophageal ganglion supply to the mandibles, maxillae and labium.
4. Nerves arising from the thoracic ganglia supply to the wings, legs and internal thoracic organs.
5. Nerves from abdominal ganglia go to the abdominal organs of respective abdominal segments.
6. Autonomic nervous system (ANS): It consists of four ganglia and a retrocerebral complex.

The ganglia are as follows:

1. Frontal ganglion: It is present above the pharynx and in front of brain.
2. Hypocerebral ganglion: It is present on the anterior region of oesophagus.
3. Ingluvial ganglion: It is present on crop. It is also called as visceral ganglion.
4. Ventricular ganglion: It is present on gizzard.

Question (D)
With the help of neat labelled diagram, describe female reproductive system of cockroach.
Answer:

1. Female reproductive system of cockroach consists a pair of ovaries, a pair of oviducts, vagina, spermatheca and accessory glands.
2. Ovaries are primary reproductive organs. They are paired and lie lateral in position in 2nd – 6lh abdominal segments. Each ovary is formed of a group of 8 ovarian tubules or ovarioles, containing a chain of developing ova. All ovarioles of an ovary open in lateral oviduct of respective side.
3. The lateral oviducts unite to form a common oviduct or vagina.
   Common oviduct or vagina opens into the Bursa copulatrix (genital chamber), the female organ of copulation.
4. Spermatheca, is a sperm storing structure present in the 6th segment opens into genital chamber. It receives the sperms during copulation and store them for fertilization.
5. Collaterial glands are accessory paired glands that open in genital chamber.
6. Female gonapophyses consists of six chitinous plates surrounding the genital pore.
7. In males, genital pouch or genital chamber lies at the hind end of abdomen which is bounded dorsally by 9th and 10th terga and ventrally b; male genital pore and gonapophysis.

Question (E)
Draw a labelled diagram of digestive system of a cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.
(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.
(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.
(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.
(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food,

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (F)
A student observed that the cockroaches are killed for dissection by simply putting them in soap water. He inquired whether soap is so poisonous. Teacher said it is due to its peculiar respiratory system. How?
Answer:
Cockroaches when put in soap solution, the solution enters into their body through the small respiratory openings called spiracles. The spiracles lead to trachea which further branches into smaller tubes called tracheoles. Each of these tracheoles has body fluid which acts as a stationary medium for diffusion. The soap solution rapidly diffuses through the entire respiratory system which may result in suffocation and eventually lead to the death of cockroach.

Question (G)
Describe the circulatory system of cockroach.
Answer:

1. Haemolymph: Haemolymph is colourless as it is without any pigment. It consists of plasma and seven types of blood cells/haemocytes. Plasma consists of water with some dissolved organic and inorganic solutes. It is rich in nutrients and nitrogenous wastes like uric acid.Cockroach has open circulatory system. It consists of colourless blood (haemolymph), a dorsal blood vessel (heart and dorsal aorta) and haemocoel.

2. Haemocoel: The body cavity of cockroach (haemocoel) can be divided into three sinuses due to two diaphragms i.e. dorsal and ventral diaphragm. These diaphragms are thin, fibromuscular septa (sing.septum)
which remain attached to terga along lateral sides at intermittent points.
(a) Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.
(b) Ventral diaphragm: It is flat and present just above the ventral nerve cord. Laterally, it is attached to sterna at intermittent points.
(e) Sinuses: The coelom of cockroach is divided into three sinuses – pericardial sinus, perivisceral sinus and perineural sinus.

1. Pericardial sinus: It is dorsal, very small and contains dorsal vessel.
2. Perivisceral sinus: It is middle and largest sinus. It contains fat bodies and almost all major visceral organs of alimentary canal and reproductive system.
3. Perineural sinus: It is ventral, small and contains ventral nerve cord. It is continuous into legs. All the three sinuses communicate with each other through the pores present between two successive points of attachments of diaphragms.
4. Dorsal blood vessel: This is present in pericardial sinus, just below the tergum. It is divisible into posterior heart and anterior aorta (dorsal aorta/cephalic vessel).
(a) Heart: It is about 2.5 cm long, narrow, muscular tube that is open anteriorly and closed posteriorly. It starts from 9th abdominal segment and extends anteriorly upto 1st thoracic segment. Heart of cockroach is 13 chambered, out of which 10 chambers are in abdominal region and 3 chambers are in thoracic region. Each chamber has a pair of vertical slit-like incurrent aperture or opening called ostium (plural: ostia). Ostia are present along lateral side in the posterior region of first 12 chambers. Each ostium has lip-like valves that allow the flow of blood from sinus to heart only.
(b) Anterior aorta: Heart is continued by a short, thin-walled vessel called dorsal aorta. It lies in head region and opens in haemocoel.

3. Answer the following questions

Question (A)
How will you identify male or female cockroach?
Answer:
Male and female cockroach can be identified with the help of following differences:

Male cockroach	Female cockroach
1. Abdomen is relatively long and narrow.	Abdomen is short and broad.
2. 7th tergum covers 8,h tergum.	7th tergum covers 8th and 9th terga.
3. Antennae are longer in size.	Antennae are shorter in size.
4. Anal styles are present.	Anal styles are absent.
5. Brood pouch is absent.	Brood pouch is present.
6. All 9 sterna visible.	Only 7 sterna visible.

Question (B)
Write a note on: Gizzard of cockroach.
Answer:
Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question (C)
Give the systematic position of cockroach.
Answer:
Systematic position of cockroach:

Classi	fication	Reasons
Kingdom	Animalia	Cell wall absent, heterotrophic nutrition.
Phylum	Arthropoda	They have jointed appendages. Body is chitinous and segmented.
Class	Insecta	They possess two pairs of wings and three pairs of walking legs.
Genus	Periplaneta	Straight wings and nocturnal.
Species	americana	Originated in the continent of America.

Question (D)
What would have happened if cockroach did not have gizzard?
Answer:
1. The gizzard in cockroach is a spherical organ which has chitinous teeth and bristles.
2. The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.
3. If the cockroach did not have gizzard, the food will not be crushed into small particles and unfiltered food will enter the hindgut. Thus, digestion will be affected in the absence of gizzard.

Question (E)
What is the functional difference between eyes of cockroach and human being?
Answer:
1. Cockroaches have compound eyes whereas humans have simple eyes.
2. Eyes of cockroach possess several ommatidia that collectively form an image and help them to detect even the slightest movement of its predator. They provide mosaic or hazy vision.
3. Human eyes contain single lens and a clear image is formed on the retina. Humans have binocular vision which provides an improved perception of depth and gives a three-dimensional image of their surroundings.

Question (F)
What is the functional difference between respiratory systems of cockroach and human being?
Answer:
The functional difference between the respiratory systems of cockroach and human being is that in respiratory system of cockroach transport of gases does not occur via. blood whereas in human respiratory system transport of gases takes place via blood. In cockroach, the circulatory system has no role in respiratory process whereas in humans, circulatory system plays an important in respiratory process.

4. Explain the following in short.

Question (A)
What are anal cerci?
Answer:
1. Anal cerci are a pair of appendages at the end of the abdomen that arise from the 10th segment of the body of both male and female cockroach.
2. They are sensitive to wind movements and detect vibrations.

Question (B)
What is ganglion?
Answer:
1. Ganglion is a group of nerve cell bodies.
2. It represents the brain in advanced invertebrates.

Question (C)
Write a short note on hypopharynx.
Answer:
Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

Question (D)
What is mesentron?
Answer:
Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

Question (E)
Location of tergum.
Answer:
1. Tergum is a chitinous plate located in the abdomen of cockroach.
2. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.

Question (F)
What is ootheca?
Answer:
1. The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.
2. It is about 8 mm long and ranges from dark reddish to blackish brown.
3. Ootheca contains 14 to 16 fertilized eggs in two rows.
4. They are dropped or glued to a suitable surface, like a crack or crevice with good humidity near a food source.
5. A female cockroach on an average, produces 9 to 10 oothecae during its lifespan.

Question (G)
How many chambers are present in heart of a cockroach?
Answer:
13 chambers are present in heart of a cockroach, out of which 10 chambers are in abdominal region and 3 are in thoracic region.

Practical/Project:

Question 1.
Visit to nearest sericulture farm and study the life cycle of silk worm.
Answer:

1. The life cycle of the silk moth consists of four stages namely, egg, larva, pupa and adult.
2. Thousands of eggs deposited by female moths are incubated artificially to reduce the incubation period.
3. Larvae hatching out of eggs are released on mulberry plants to obtain nourishment from mulberry leaves.
4. After feeding for 3 – 4 weeks, larvae move to branches of mulberry plant.
5. The silk thread is formed from the secretion of salivary glands of larvae.
6. Larvae spin this thread around themselves forming a cocoon, which may be spherical in shape.
7. Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
8. Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
9. These fibres are then unwound, processed and reeled.
10. Different kinds of fabric are woven from silk threads.

[The life cycle of silkworm is given for reference. Students are expected to visit the nearest sericulture farm and attempt this activity on their own.]

11th Biology Digest Chapter 11 Study of Animal Type: Cockroach Intext Questions and Answers

Can you recall? (Textbook Page No. 127)

How many different types of animals are present around us?
Answer:
Animals on earth show great diversity. The different types of animals present around us are;
a. Unicellular and multicellular
b. Prokaryotic and eukaryotic
c. Vertebrates and invertebrates
d. Unisexual and hermaphrodite
e. Aquatic, terrestrial, amphibian, reptilian, aerial, etc.

Can a person do a complete detailed study of each of those animals?
Answer:
Yes, a person can do a complete detailed study of each of those animals. Classification of animals based on characteristics into various groups has made it easier to study them.

Which phylum is most diverse and populous?
Answer:
Phylum Arthropoda is most diverse and populous.

Curiosity box: (Textbook Page No. 127)

Why do insects need moulting?
Answer:
a. Insects undergo metamorphosis (change of form or structure in an individual after hatching or birth). Each time an insect enters the next growth stage it has to molt.
b. Moulting is the process in which formation of new chitinous exoskeleton and subsequent shedding of the old one occurs.
c. The insects need moulting as their exoskeleton is rigid unlike the skin and does not allow the body to grow.

What is the difference between simple and compound eyes?
Answer:

Simple eyes	Compound eyes
1. Simple eyes contain single lens and several sensory cells.	Compound eyes contain several lenses (around 2000) called ommatidia (sing. Ommatidium).
2. Single lens collect light and focuses onto retina to form a single image	Each ommatidium forms an image of an object thereby forming several images of an ob ject.
3. Simple eye does not form a complex image but can detect movement of the object.	Compound eye forms a complex image of an object 1 and detects even a slightest movement of the object.

Use your brainpower. (Textbook Page No. 131)

Why do body cavity of cockroach is called as haemocoel?
Answer:
The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.

Internet my friend. (Textbook Page No. 136)

Collect the information about techniques and objectives of rearing the cockroaches in countries like China and make a Powerpoint presentation including video clips.
Answer:
1. Cockroach rearing industry is a booming industry in China. Cockroaches are reared in more than hundred farms in China. A giant farm in China produces around 6 billion cockroaches.
2. It is believed that cockroaches can be used to prepare a medicine that can prevent stomach cancer. They are also used to treat compost waste.
[Students can search on internet for more information about the techniques and objectives of rearing the cockroaches]

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 5 Chemical Bonding

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order ?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. Energy levels in an atom
b. the shapes of molecules and ions.
c. the electrone getivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

	C=O bond	CO2 molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer:

	C=O bond	CO2 molecule
A	polar	non-polar

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

• formation of the excited state and
• mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

• Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
• Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

Balbharti Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System Textbook Exercise Questions and Answers.

Maharashtra State Board Bookkeeping and Accountancy 11th Solutions Chapter 10 Single Entry System

1. Answer in One sentence only.

Question 1.
What do you mean by a Single Entry System?
Answer:
A system of bookkeeping in which an accountant or businessman records only one aspect of business transaction (either debit or credit and ignores the other aspect is called ‘Single Entry System’.

Question 2.
What is a Statement of Affairs?
Answer:
A list of all assets and liabilities prepared under a single entry system to find out capital balance is called a statement of affairs.

Question 3.
Which type of accounts are normally not kept under the Single Entry System?
Answer:
Under a single entry system, records of impersonal accounts i.e. real accounts and nominal accounts are not maintained.

Question 4.
Which statement is prepared under the Single Entry system to ascertain the capital balances?
Answer:
A statement of Affairs is prepared under a single entry system to ascertain capital balances.

Question 5.
How Opening Capital is calculated under the Single Entry System?
Answer:
Under a single entry system, opening capital is ascertained by preparing the opening statement of affairs.

Question 6.
Which types of accounts are maintained under the Single Entry System?
Answer:
Under a single entry system, all personal accounts and cash accounts are maintained.

Question 7.
Can a Trial Balance be prepared under a Single Entry System?
Answer:
A trial balance cannot be prepared under a single entry system.

Question 8.
Which type of organizations generally follow the Single Entry System?
Answer:
Organizations having small sizes of business such as sole trading concerns and partnership firms follow a single entry system.

2. Write a word, term, or phrase which can substitute each of the following statements.

Question 1.
A statement that is similar to the Balance Sheet.
Answer:
Statement of Affairs

Question 2.
The system of Accounting is normally suitable for small business organizations.
Answer:
Single Entry System

Question 3.
A statement similar to the Balance Sheet is prepared to find out the amount of opening capital.
Answer:
Opening Statement of Affairs

Question 4.
An excess of assets over liabilities.
Answer:
Capital

Question 5.
Excess of closing capital over opening capital of proprietor under Single Entry System.
Answer:
Profit

Question 6.
Name of the method of accounting suitable to firms having limited transactions.
Answer:
Single Entry System

Question 7.
A System of accounting that is unscientific.
Answer:
Single Entry System

Question 8.
Further capital introduced by the proprietor in the business concern over and above his existing capital.
Answer:
Additional Capital

3. Select the most appropriate answer from the alternatives given below and rewrite the sentence.

Question 1.
The capital balances are ascertained by preparing _______________
(a) Statement of Affairs
(b) Cash Account
(c) Drawings Accounts
(d) Debtors Accounts
Answer:
(a) Statement of Affairs

Question 2.
Under Single Entry System, Opening Capital = Opening Assets less _______________
(a) Opening Liabilities
(b) Closing Liabilities
(c) Debtors Account
(d) Creditors Account
Answer:
(a) Opening Liabilities

Question 3.
Additional Capital introduced during the year is _______________ from closing capital in order to find out the correct profit.
(a) Added
(b) Deducted
(c) Divided
(d) Ignored
Answer:
(b) Deducted

Question 4.
Single Entry System may be useful for _______________
(a) Sole traders
(b) Company
(c) Government
(d) None of these
Answer:
(a) Sole traders

Question 5.
In order to find out the correct profit, drawings is _______________ from closing capital.
(a) Multiplies
(b) Divided
(c) Deducted
(d) Added
Answer:
(d) Added

Question 6.
The difference between assets and liabilities is called _______________
(a) Capital
(b) Drawings
(c) Income
(d) Expenses
Answer:
(a) Capital

Question 7.
When Closing Capital is greater than the Opening Capital, the difference is _______________
(a) Profit
(b) Loss
(c) Assets
(d) Liabilities
Answer:
(a) Profit

Question 8.
Opening Capital is ₹ 30,000; Closing Capital is ₹ 60,000; Withdrawals are ₹ 5,000; and further capital brought in is ₹ 3,000; Profit is _______________
(a) ₹ 45,000
(b) ₹ 35,000
(c) ₹ 32,000
(d) ₹ 22,000
Answer:
(c) ₹ 32,000

4. State True or False with reasons:

Question 1.
The double Entry System of Book-keeping is a scientific method of books of accounts.
Answer:
This statement is True.
In the double-entry system of book-keeping, there are two-fold effects. Both the effects are recorded simultaneously with an equal amount. This system also follows principles and rules of debit and credit. Due to this, there are very fewer chances of mistakes. So double entry system of Book-Keeping is a scientific method of the book of accounts.

Question 2.
Preparation of Trial Balance is not possible under the Single Entry System.
Answer:
This statement is True.
Under the single entry system, only cash and personal accounts of debtors and creditors are open. So it is not possible to prepare. Trail balance under single entry system as it has incomplete information of Accounting.

Question 3.
Statement of Affairs and Balance Sheet are one and the same.
Answer:
This statement is False.
There is a difference between a statement of Affairs and the Balance sheet. Statement of Affair shows estimated values of assets and liabilities.

Question 4.
The single Entry System is not useful for large organizations.
Answer:
This statement is True.
Under the Single Entry System, only the cash book and personal account of Debtor and Ciygditor are maintained. Real and Nominal accounts are not maintained. It has no proper set of rules to be followed. It is useful for small organisations and not for a large organisations.

Question 5.
Only Cash and Personal accounts are maintained under the Single Entry System.
Answer:
This statement is True.
The single Entry System is an ancient and unscientific method of recording business transactions. This system maintains minimum accounts so it is easy for traders to write books of accounts. This system does not follow any accounting rules. To know the cash collections and amount payable or receivable only cash and personal accounts are maintained under a single entry system.

5. Do you agree with the following statements?

Question 1.
Further capital introduced during the year increases profit.
Answer:
Disagree

Question 2.
Interest in Drawings decreases the amount of profit under the Single Entry System.
Answer:
Disagree

Question 3.
Real and Nominal accounts are not maintained under the Single Entry System.
Answer:
Agree

Question 4.
The single Entry System is based on certain rules and principles.
Answer:
Disagree

Question 5.
Statement of Profit is just like Profit and Loss Account.
Answer:
Disagree

6. Fill in the Blanks.

Question 1.
Statement of Affairs is just like _______________
Answer:
Balance Sheet

Question 2.
Under Single Entry System, Profit = Closing Capital Less _______________
Answer:
Opening Capital

Question 3.
In order to find out the correct profit, drawings are _______________ to the closing capital.
Answer:
Added

Question 4.
In _______________ Book Keeping System, in every business transactions we find two effects.
Answer:
Double Entry System

Question 5.
The difference between Assets and Liabilities is called _______________
Answer:
Capital

Question 6.
Single Entry System is more popular for _______________
Answer:
Sole Trader

Question 7.
Additional Capital introduced during the year is _______________ from Closing Capital in order to find out the correct profit.
Answer:
Deducted

Question 8.
Single Entry System is Suitable for _______________ business.
Answer:
Small

7. Find the odd one:

Question 1.
Interest on Drawings, Outstanding Expenses, Undervaluation of Assets, Prepaid Expenses.
Answer:
Outstanding Expenses

Question 2.
Interest on Capital, Interest on Loan, Overvaluation of Liabilities, Depreciation on Assets.
Answer:
Overvaluation of Liabilities

Question 3.
Creditors, Bills Payable, Bank Overdraft, Stock in Trade.
Answer:
Stock in Trade

8. Complete the following table:

Question 1.

Answer:
₹ 5,000

Question 2.

Answer:
₹ 30,000

Question 3.

Answer:
₹ 5,000

Question 4.

Answer:
₹ 25,000, ₹ 20,000

Question 5.

Answer:
₹ 19,000

9. Complete the following table. Put Proper mark in Box.

Question 1.

Answer:

1. Add
2. Add
3. Add
4. Less
5. Add
6. Less
7. Add
8. Less
9. Less
10. Less

Practical Problems

Question 1.
Mr. Poonawala keeps his books under the Single Entry System and gives the following information.
Capital as of 31.3.2017 – ₹ 60,000
Capital as on 31.3.2018 – ₹ 1,00,000
Drawings made during the year ₹ 2,000
Additional capital introduced during the year ₹ 12,000
Calculate Profit or Loss during the year.
Solution:
In the books of Mr. Poonawala
Statement of Profit or Loss for the year ended 31st March 2018

Question 2.
Sujit a small trader provides you with the following details of his business.

Additional information:
1. Sujit withdraws ₹ 5,000 for his personal use, on 1st Oct. 2017.
2. He had also withdrawn ₹ 30,000 for rent of his residential flat.
3. Depreciation Furniture by 10% p.a. and writes off ₹ 1,000 from Motor Van.
4. Charge interest on Drawings ₹ 3,000.
5. 10% Govt. Bonds were purchased on 1st Oct. 2017.
6. Allow interest on capital at 10% p.a.
7. ₹ 1,000 is written off as bad debts and provides 5% p.a. R.D.D. on Debtors.
Prepare Opening Statement of Affairs, Closing Statement of Affairs, and Statement of Profit or Loss for the year ending 31st March 2018.
Solution:
In the books of Sujit
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 3.
Anjali keeps her books by the Single Entry System. Her position on 1.4.2016 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 1,000, Stock ₹ 6,000; Sundry Debtors ₹ 8,400, Plant and Machinery ₹ 7,500, Bill Receivable ₹ 2,600, Creditors ₹ 3500; Bills Payable ₹ 4,000
On 31.3.2017 her position was as follows; cash at Bank ₹ 3,900, Cash in Hand ₹ 2,000. Stock ₹ 9000, Sundry Debtors, ₹ 7,500; Plant and Machinery ₹ 7,500; Bills Payable ₹ 2,200, Bills Receivable ₹ 3,400; Creditors ₹ 1,500.
During the year Anjali introduced further Capital of ₹ 1,500 and she spent ₹ 700 per month for her personal use.
Depreciation Plant and Machinery by 5% p.a. and create Reserve for Doubtful debts @ 5% p.a. on the debtor. Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Anjali
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 4.
Mr. Vijay is dealing in the business of fruits. He maintains an accounting record with a single entry. The following figures are taken from his record.

Additional information:
1. Mr. Vijay introduced ₹ 7,000 as fresh capital.
2. He spent ₹ 40,000 from his business for his daughter’s marriage.
3. Depreciate Building by ₹ 6,000.
4. Create a 5% reserve for doubtful debts on Sundry Debtor.
Prepare:
1. Opening Statement of Affairs.
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Vijay
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 5.
Miss. Fiza keeps her books on the Single Entry System and disclosed the following information about her business.

Additional information:
1. Miss. Fiza transferred ₹ 2,000 per month during the first half-year and ₹ 1000 per month for the second half-year from a business account to her personal account.
2. She sold her private asset for ₹ 40,000 and brought the proceeds into her business.
3. She also took goods worth ₹ 12,000 for private use.
4. Plant and Machinery is to be depreciated by 10% p.a.
5. Provide R.D.D. on debtors at 5% p.a.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018
Solution:
In the books miss Fiza
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 6.
Miss. Sanika keeps her books on the Single Entry System. The statement of affairs is given on 31st March 2018.

On 31st March 2018 their Assets and Liabilities were as follows:
Plant and Machinery ₹ 42,000, Stock ₹ 38,000, Cash in Hand ₹ 10,000, Creditors ₹ 7,000, Debtors ₹ 25,000, Bills Payable ₹ 6,000
Drawings during the year were ₹ 5,500, Plant and Machinery were found Overvalued by 5% p.a. and Stock was found Undervalued by 20% p.a., R.D.D. was to be created at 10% p.a. on Debtors, Interest on Capital was allowed at 10% p.a.
Prepare:
1. Closing Statement of Affairs.
2. Statement of Profit or Loss for the year ended 31st March 2018.
Solution:
In the books of miss Sanika
Closing statement of Affairs as on 31.03.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 7.
Mr. Suhas commenced his business with the Capital of ₹ 1,50,000 on 1st April 2017. His financial position was as follows as on 31st March 2018, Cash ₹ 20,000, Stock ₹ 15,000, Debtors ₹ 30,000, Premises ₹ 90,000, Vehicles ₹ 45,000, Creditors ₹ 18,500, Bills Payable ₹ 10,000.
Additional information:
1. He brought additional capital ₹ 10,000 on 30th Sept. 2017, Interest on capital is to be provided at 5% p.a.
2. He withdrew ₹ 15,000 for personal use on which interest is to be charged at 5% p.a.
3. Write off Bad debts ₹ 500.
Prepare:
1. Closing Statement of Affairs
2. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Mr. Suhas
Closing statement of Affairs as on 31.3.2018

Statement of Profit or Loss for the year ended 31st March 2018.

Question 8.
Ganesh keeps his books by the Single Entry Method. Following are the details of his business:

During the year he has withdrawn ₹ 25,000 for his private purpose and goods of ₹ 3,000 for household use. On 1st Oct. 2016. He sold his household furniture for ₹ 4,000 and deposited the same amount in a business Bank Account.
Provide Depreciation on Plant and Machinery at 10% p.a. (assuming additions were made on 1st Oct. 2016) and Furniture at 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2017.
Solution:
In the books of Ganesh
Opening and Closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2017

Question 9.
Peter keeps his books on the Single Entry System. From the following particulars, Prepare Opening and Closing Statement of Affairs and Statement of Profit or Loss for the year ending 31st March 2018.

Additional Information:
1. Peter has withdrawn ₹ 15,000 from the business for his personal use.
2. He has introduced additional capital of ₹ 10,000 in the business on 1st January 2018.
3. Depreciate furniture @ 10% p.a.
4. Maintain reserve for doubtful debts @ 5% on Sundry Debtors.
5. Closing Stock is overvalued by 25% in the books.
Solution:
In the books of Peter
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018

Question 10.
Suresh keeps his books by the Single Entry System. His position on 1.4.2017 was as follows.
Cash at Bank ₹ 4,000, Cash in Hand ₹ 3,000; Stock ₹ 8,000; Sundry Debtors ₹ 9,000; Plant & Machinery ₹ 10,000; Bills Receivable ₹ 3000; Creditors ₹ 1500; Bills Payable ₹ 2000.
On 31st March 2018, his position was as follows:
Cash at bank ₹ 6,400; Cash in Hand ₹ 1,800; Stock ₹ 10000; Sundry and Debtors ₹ 8,000; Plant & Machinery ₹ 10,000; Bills Payable ₹ 4,000; Bills Receivable ₹ 5,200; Creditors ₹ 2,000 During the year Suresh introduced further capital of ₹ 3,000 and his drawings were ₹ 700 per months. Depreciate Plant & Machinery by 5% and create a reserve for bad doubtful debts @ 5%.
Prepare:
1. Opening Statement of Affairs
2. Closing Statement of Affairs
3. Statement of Profit or Loss for the year ended 31.3.2018.
Solution:
In the books of Suresh
Opening and closing statement of Affairs as on _______________

Statement of Profit or Loss for the year ended 31st March 2018