Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

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11th Computer Science Guide Introduction to C++ Text Book Questions and Answers

Book Evaluation
Part I

Choose The Correct Answer

Question 1.
Who developed C++?
a) Charles Babbage
b) Bjarne Stroustrup
c) Bill Gates
d) Sundar Pichai
Answer:
b) Bjarne Stroustrup

Question 2.
What was the original name given to C++?
a) CPP
b) Advanced C
c) C with Classes
d) Class with C
Answer:
c) C with Classes

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Who coined C++?
a) Rick Mascitti
b) Rick Bjarne
c) Bill Gates
d) Dennis Ritchie
Answer:
a) Rick Mascitti

Question 4.
The smallest individual unit in a program is:
a) Program
b) Algorithm
c) Flowchart
d) Tokens
Answer:
d) Tokens

Question 5.
Which of the following operator is extraction operator of C++?
a) >>
b) <<
c) <>
d) AA
Answer:
a) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 6.
Which of the following statements is not true?
a) Keywords are the reserved words convey specific meaning to the C++ compiler.
b) Reserved words or keywords can be used as an identifier name.
c) An integer constant must have at least one digit without a decimal point.
d) Exponent form of real constants consists of two parts
Answer:
b) Reserved words or keywords can be used as an identifier name.

Question 7.
Which of the following is a valid string literal?
a) ‘A’
b) ‘Welcome’
c) 1232
d) “1232”
Answer:
d) “1232”

Question 8.
A program written in high level language is called as ………………………
a) Object code
b) Source code
e) Executable code
d) All the above
e) Executable code
Answer:
b) Source code

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 9.
Assume a=5, b=6; what will be result of a & b?
a) 4
b) 5
c) 1
d) 0
Answer:
a) 4

Question 10.
Which of the following is called as compile time operators?
a) size of
b) pointer
c) virtual
d) this
Answer:
a) sizeof

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Part – II

Very Short Answers

Question 1.
What is meant by a token? Name the token available in C++.
Answer:
C++ program statements are constructed by many different small elements such as commands, variables, constants, and many more symbols called operators and punctuators. Individual elements are collectively called Lexical units or Lexical elements or Tokens.
C++ has the following tokens:

  1. Keywords
  2. Identifiers
  3. Literals
  4. Operators
  5. Punctuators

Question 2.
What are keywords? Can keywords be used as identifiers?
Answer:
Keywords:
Keywords are the reserved words which convey specific meaning to the C++ compiler.
They are the essential elements to construct C++ programs.
Example:
int / float / auto / register Reserved words or keywords cannot be used as an identifier name.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
The following constants are of which type?

  1. 39
  2. 032
  3. OXCAFE
  4. 04.14

Answer:

  1. 39 – Decimal
  2. 032 – Octal
  3. OXCAFE – Hexadecimal
  4. 04.14 – Decimal

Question 4.
Write the following real constants into the exponent form:
i) 23.197
ii) 7.214
iii) 0.00005
iv) 0.319
Answer:
i) 23.197 : 0.23197 E2 (OR) 2.3197 E1 (OR) 23197E-3
ii) 7.214 : 0.7214 E1 (OR) 72.14 E-1 (OR) 721.4 E-2 (OR) 7214E-3
iii) 0.00005 : 5E-5
iv) 0.319 : 3.19 E-l (OR) 31.9 E-2 (OR) 319 E-3

Question 5.
Assume n=10; what will be result of n>>2;?
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 1

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 6.
Match the following:

A

B

(a) Modulus (1) Tokens
(b) Separators (2) Remainder of a division
(c) Stream extraction (3) Punctuators
(d) Lexical Units (4) get from

Answer:
a) 2
b) 3
c) 4
d) 1

Part – III

Short Answers

Question 1.
Describe the differences between keywords and identifiers.
Answer:
Keywords:

  • Keywords are the reserved words which convey specific meaning to the C++ compiler.
  • They are essential elements to construct C++ programs.
  • Most of the keywords are common to C, C++, and Java.

Identifiers:

  • Identifiers are the user-defined names given to different parts of the C++ program.
  • They are the fundamental building blocks of a program.
  • Every language has specific rules for naming the identifiers.

Question 2.
Is C++ case sensitive? What is meant by the term “case sensitive”?
Answer:
Yes. C++ is case sensitive as it treats upper and lower-case characters differently.
Example: NUM, Num, num are different in C++.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Differentiate “=” and “==”.
Answer:

  • ‘=’ is an assignment operator which is used to assign a value to a variable which is on the left hand side of an assignment statement.
  • ‘=’operator copies the value at the right side
    of the operator to the left side variable. Ex. num = 10; means 10 assign to the variable num.
  • ‘= =’ is a relational operator. It is used to compare both operands are same or not.
  • Ex. num = = 10 means it compare num value with 10 and returns true(l) if both are same or returns false(0)

Question 4.
Assume a=10, b=15; What will be the value of a∧b?
Answer:
Bitwise XOR (∧) will return 1 (True) if only one of the operand is having a value 1 (True). If both are True or both are False, it will return 0 (False).
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 2

Question 5.
What is the difference between “Run time error” and “Syntax error”?
Answer:
Run time Error:

  • A run time error occurs during the execution of a program. It occurs because of some illegal operation that takes place.
  • For example, if a program tries to open a file which does not exist, it results in a run-time error.

Syntax Error:

  • Syntax errors occur when grammatical rules of C++ are violated.
  • For example: if you type as follows, C++ will throw an error.
    cout << “Welcome to Programming in C++”

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 6.
What are the differences between “Logical error” and “Syntax error”?
Answer:

  • A Program has not produced the expected result even though the program is grammatically correct. It may be happened by the wrong use of variable/operator/order of execution etc. This means, the program is grammatically correct, but it contains some logical error. So, a Semantic error is also called a “Logic Error”.
  • Syntax errors occur when grammatical rules of C++ are violated.

Question 7.
What is the use of a header file?
Answer:
Header files contain definitions of Functions and Variables, which are imported or used into any C++ program by using the preprocessor #include statement. Header files have an extension “.h” which contains C++ function declaration and macro definition.
Example: #include

Question 8.
Why is main function special?
Answer:
Every C++ program must have a main function. The main() function is the starting point where all C++ programs begin their execution. Therefore, the executable statements should be inside the main() function.

Question 9.
Write two advantages of using include compiler directive.
Answer:

  1. The program is broken down into modules, thus making it more simplified.
  2. More library functions can be used, at the same time size of the program is retained.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Write the following in real constants.

  1. 15.223
  2. 211.05
  3. 0.00025

Answer:

  1. 15.223 → 1.5223E1 → 0.15223E2 → 15223E-3
  2. 211.05 → 2.1105E2 → 21105 E-2
  3. 0.00025 → 2.5E-4

Part – IV

Explain In Detail

Question 1.
Write about Binary operators used in C++.
Answer:
Binary Operators require two operands:
Arithmetic operators that perform simple arithmetic operations like addition, subtraction, multiplication, division (+, -, *, %, /), etc. are binary operators which require a minimum of two operands.

Relational operators are used to determining the relationship between its operands. The relational operators (<, >, >=, <=, ==, !=) are applied on two operands, hence they are binary operators. AND, OR (logical operator) both are binary operators. The assignment operator is also a binary operator (+=, – =, *=, /=, %=).

Question 2.
What are the types of Errors?
Answer:
COMMON TYPES OF ERRORS

Type of Error

Description

Syntax Error Syntax is a set of grammatical rules to construct a program. Every programming language has unique rules for constructing the sourcecode.
Syntax errors occur when grammatical rules of C++ are violated.
Example: if we type as follows, C++ will throw an error.
cout << “Welcome to C++”
As per grammatical rules of C++, every executable statement should terminate with a semicolon. But, this statement does not end with a semicolon.
Semantic error’ A Program has not produced expected result even though the program is grammatically correct. It may be happened by the wrong use of variable/operator/order of execution etc. This means, the program is grammatically correct, but it contains some logical error. So, Semantic error is also called a “Logic Error”.
Run­ time error  A run time error occurs during the execution of a program. It occurs because of some illegal operation that takes place.
For example, if a program tries to open a file which does not exist, it results in a run-time error.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Assume a=15, b=20; What will be the result of the following operations?
a) a&b
b) a|b
c) aAb
d)a>>3
e) (~b)
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 3

11th Computer Science Guide Introduction to C++ Additional Questions and Answers

Choose The Correct Answer

Question 1.
The latest standard version published in December 2017 as ISO/IEC …………….. which is informally known as C++ 17.
(a) 14882 : 1998
(b) 14883 : 2017
(c) 14882 : 2017
(d) 14882 : 2000
Answer:
(c) 14882 : 2017

Question 2.
C++ language was developed at ……………….
a) Microsoft
b) Borland International
c) AT & T Bell Lab
d) Apple Corporation
Answer:
c) AT & T Bell Lab

Question 3.
An integer constant is also called……………..
(a) fixed point constant
(b) floating-point constant
(c) real constants
(d) Boolean literals
Answer:
(a) fixed point constant

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
C++supports ………… programming paradigms.
a) Procedural
b) Object-Oriented
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 5.
…………….. relational operators are binary operators.
(a) 7
(b) 8
(c) 6
(d) 2
Answer:
(c) 6

Question 6.
C++ is a superset (extension) of …………….. language.
a) Ada
b) BCPL
c) Simula
d) C
Answer:
d) C

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7
…………….. used to label a statement.
(a) colon
(b) comma
(c) semicolon
(d) parenthesis
Answer:
(a) colon

Question 8.
The name C++ was coined by …………….
a) Lady Ada Lovelace
b) Rick Mascitti
c) Dennis Ritchie
d) Bill Gates
Answer:
b) Rick Mascitti

Question 9.
IDE stands for ……………..
(a) Integrated Development Environment
(b) International Development Environment
(c) Integrated Digital Environment
(d) None of the above
Answer:
(a) Integrated Development Environment

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Till 1983, C++ was referred to as …………………
a) New C
b) C with Classes
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 11.
…………….. data type signed more precision fractional value.
(a) char
(b) short
(c) long double
(d) signed doubles
Answer:
(c) long double

Question 12.
C# (C-Sharp), D, Java, and newer versions of C languages have been influenced by ………………. language.
a) Ada
b) BCPL
c) Simula
d) C++
Answer:
d) C++

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 13.
…………….. manipulator is the member of iomanip header file.
(a) setw
(b) setfill
(c) setf
(d) all the above
Answer:
(d) all the above

Question 14.
C++is ……………… language.
a) Structural
b) Procedural
c) Object-oriented
d) None of these
Answer:
c) Object-oriented

Question 15.
C++ includes ………………..
a) Classes and Inheritance
b) Polymorphism
c) Data abstraction and Encapsulation
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 16.
C language does not allow ………………
a) Exception handling
b) Inheritance
c) Function overloading
d) All the above
Answer:
d) All the above

Question 17.
……………. is a set of characters which are allowed to write a C++ program.
a) Character set
b) Tokens
c) Punctuators
d) None of these
Answer:
a) Character set

Question 18.
A character represents any …………………..
a) Alphabet
b) Number
c) Any other symbol (special characters)
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 19.
Most of the Character set, Tokens, and expressions are very common to C based programming languages like ………………..
a) C++
b) Java
c) PHP
d) All the above
Answer:
d) All the above

Question 20.
…………… is a white space.
a) Horizontal tab
b) Carriage return
c) Form feed
d) All the above
Answer:
d) All the above

Question 21.
C++ program statements are constructed by many different small elements called……………….
a) Lexical units
b) Lexical elements
c) Tokens
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 22.
…………… is a C++token.
a) Keywords
b) Identifiers and Literals
c) Punctuators and Operators
d) All the above
Answer:
d) All the above

Question 23.
The smallest individual unit in a program is known as a ……………
a) Token
b) Lexical unit
c) Token or a Lexical unit
d) None of these
Answer:
c) Token or a Lexical unit

Question 24.
………………… are the reserved words which convey specific meaning to the C++ compiler.
a) Keywords
b) Identifiers and Literals
c) Punctuators and Operators
d) All the above
Answer:
a) Keywords

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 25.
………………. are the essential elements to construct C++ programs.
a) Keywords
b) Identifiers and Literals
c) Punctuators and Operators
d)All the above
Answer:
a) Keywords

Question 26.
Most of the keywords are common to …………… languages.
a) C, and C++
b) C++ and Java
c) C, C++ and Java
d) None of these
Answer:
c) C, C++ and Java

Question 27.
……………. is a case sensitive programming language.
a) C, and C++
b) C++ and Java
c) C, C++ and Java
d) None of these
Answer:
c) C, C++ and Java

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 28.
In ……………. language, all the keywords must be in lowercase.
a) C, and C++
b) C++ and Java
c) C, C++ and Java
d) None of these
Answer:
c) C, C++ and Java

Question 29.
……………… is a new keyword in C++.
a) using
b) namespace
c) std
d) All the above
Answer:
d) All the above

Question 30.
…………….. is a new keyword in C++.
a) bal
b) static_cast
c) dynamic_cast
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 31.
………….. is a new keyword in C++.
a) true
b) false
c) Both A and B
c) None of these
Answer:
c) None of these

Question 32.
Identifiers are the user-defined names given to …………..
a) Variables and functions
b)Arrays
c) Classes
d) All the above
Answer:
d) All the above

Question 33.
Identifiers containing a …………. should be avoided by users.
a) Double underscore
b) Underscore
c) number
d) None of these
Answer:
a) Double underscore

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 34.
The first character of an identifier must be an ………….
a) Alphabet
b) Underscore (_)
c) Alphabet or Underscore (_)
c) None of these
Answer:
c) Alphabet or Underscore (_)

Question 35.
Only …………… is permitted for the variable name.
a) Alphabets
b) Digits
c) Underscore
d) All the above
Answer:
d) All the above

Question 36.
Identify the correct statement from the following.
a) C++ is case sensitive as it treats upper and lower-case characters differently.
b) Reserved words or keywords cannot be used as an identifier name.
c) As per ANSI standards, C++ places no limit on its length.
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 37.
ANSI stands for …………..
a) All National Standards Institute
b) Advanced National Standards Institute
c) American National Standards Institute
d) None of these
Answer:
c) American National Standards Institute

Question 38.
Identify the invalid variable name from the following.
a) num-add
b) this
c) 2myfile
d) All the above
Answer:
d) All the above

Question 39.
Identify the odd one from the following.
a) Int
b) _add
c) int
d) tail marks
Answer:
c) int

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 40.
……………… are data items whose values do not change during the execution of a program.
a) Literals
b) Constants
c) Identifiers
d) Both A and B
Answer:
d) Both A and B

Question 41.
…………. is a type constant in C++.
a) Boolean constant
b) Character constant
c) String constant
d) All the above
Answer:
d) All the above

Question 42.
…………… is a type of numeric constant.
a) Fixed point constant
b) Floating-point constant
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 43.
……………. are whole numbers without any fractions.
a) Integers
b) Real constant
c) Floating-point constant
d) None of these
Answer:
a) Integers

Question 44.
In C++, there are …………….. types of integer constants.
a) Two
b) Three
c) Four
d) Six
Answer:
b) Three

Question 45.
In C++, ……………… is a type of integer constant.
a) Decimal
b) Octal
c) Hexadecimal
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 46.
Any sequence of one or more digits (0 …. 9) is called ……………… integer constant.
a) Decimal
b) Octal
c) Hexadecimal
d) All the above
Answer:
a) Decimal

Question 47.
Any sequence of one or more octal values (0 …. 7) that begins with 0 is considered as a(n) …………… constant.
a) Decimal
b) Octal
c) Hexadecimal
d) All the above
Answer:
b) Octal

Question 48.
When you use a fractional number that begins with 0, C++ has consider the number as ………………..
a) An integer not an Octal
b) A floating-point not an Octal
c) An integer not a Hexadecimal
d) None of these
Answer:
a) An integer not an Octal

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 49.
Any sequence of one or more Hexadecimal values (0 …. 9, A …. F) that starts with Ox or OX is considered as a(n) ………….. constant.
a) Decimal
b) Octal
c) Hexadecimal
d) All the above
Answer:
c) Hexadecimal

Question 50.
Identify the invalid octal constant,
a) 05,600
b) 04.56
c) 0158
d) All the above
Answer:
d) All the above

Question 51.
Identify the valid hexa decimal constant
a) 0X1,A5
b) 0X.14E
c) CAFE
d) CPP
Answer:
c) CAFE

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 52.
The suffix ………….. added with any constant forces that to be represented as a long constant.
a) L or I
b) U or u
c) LO
d) Lg
Answer:
a) L or I

Question 53.
The suffix …………… added with any constant forces that to be represented as an unsigned constant.
a) L or I
b) U or u
c) US
d) us
Answer:
b) U or u

Question 54.
A _______ constant is a numeric constant having a fractional component.
a) Real
b) Floating point
c) Real or Floating point
d) None of these
Answer:
c) Real or Floating point

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 55.
……………. constants may be written in fractional form or in exponent form.
a) Real
b) String
c) Character
d) Integer constant
Answer:
a) Real

Question 56.
Exponent form of real constants consists of ………….. parts.
a) three
b) two
c) four
d) five
Answer:
b) two

Question 57.
Exponent form of real constants consists of ……………. part.
a) Mantissa
b) Exponent
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 58.
The mantissa must be a(n) …………. constant.
a) Integer
b) Real
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 59.
58.64 can be written as …………….
a) 5.864E1
b) 0.5864E2
c) 5864E-2
d) All the above
Answer:
d) All the above

Question 60.
Internally boolean true has value ……………
a) 0
b) 1
c) -1
d) None of these
Answer:
b) 1

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 61.
Internally boolean false has value …………….
a) 0
b) 1
c) -1
d) None of these
Answer:
a) 0

Question 62.
A character constant is any valid single character enclosed within ………….. quotes.
a) Double
b) Single
c) No
d) None of these
Answer:
b) Single

Question 63.
Identify the odd one from the following,
a) ‘A’
b) ‘2’
c) ‘$’
d) “A”
Answer:
d) “A”

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 64.
The value of a single character constant has an equivalent ………….. value.
a) BCD
b) ASCII
c) Nibble
d) None of these
Answer:
b) ASCII

Question 65.
The ASCII value of ‘A’ is …………..
a) 65
b) 97
c) 42
d) 75
Answer:
a) 65

Question 66.
The ASCII value of ‘a’ is ……………
a) 65
b) 97
c) 42
d) 75
Answer:
b) 97

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 67.
C++ allows certain non-printable characters represented as ……………. constants.
a) Integer
b) Real
c) Character
d) String
Answer:
c) Character

Question 68.
The non-printable characters can be represented by using …………………………
a) Escape sequences
b) String
c) Boolean
d) None of these
Answer:
a) Escape sequences

Question 69.
An escape sequence is represented by a backslash followed by …………. character(s).
a) One
b) Two
c) One or Two
d) None of these
Answer:
c) One or Two

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 70.
______ is escape sequence for audible or alert bell.
a) \a
b)\b
c) \n
d)\f
Answer:
a) \a

Question 71
_______ is escape sequence for backspace.
a)\a
b)\b
c) \n
d) \f
Answer:
b)\b

Question 72.
______ is escape sequence for form feed.
a)\a
b)\b .
c) \n
d) \f
Answer:
d) \f

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 73.
______ is escape sequence for new line or line feed.
a) a
b)\b
c)\n
d)\f
Answer:
c)\n

Question 74.
……………. is escape sequence for carriage return.
a)\r
b)\c
c)\n
d)\cr
Answer:
a)\r

Question 75.
______ is escape sequence for horizontal tab.’
a)\a ‘
b)\b
c)\t
d)\f
Answer:
c)\t

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 76.
_______ is escape sequence for vertical tab.
a)\v
b)\b
c) \t
d) \f
Answer:
a)\v

Question 77.
………………. is escape sequence for octal number.
a) \On
b) \xHn
c)\O
d)O
Answer:
a) \On

Question 78.
______ is escape sequence for hexadecimal number.
a) \On
b) \xHn
c)\O
d)\O
Answer:
b) \xHn

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 79
______ ¡s escape sequence for Null character.
a) \On
b) \xHn
c)\O
d)\n
Answer:
c)\O

Question 80.
______ ¡s escape sequence for Inserting?
a) \?
b) \\
c)\’
d)\”
Answer:
a) \?

Question 81.
______ is an escape sequence for inserting a single quote.
a)\?
b)\\
c) \‘
d) \“
Answer:
c) \‘

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 82.
______ is escape sequence for inserting double quote.
a)\?
b)\\
c) \‘
d) \“
Answer:
d) \“

Question 83.
______ is escape sequence for inserting
a)\?’
b)\\
c) \‘
d) \“
Answer:
b)\\

Question 84.
ASCII was first developed and published in 1963 by the …………. Committee, a part of the American Standards Association (ASA).
a) X3
b) A3
c) ASA
d) None of these
Answer:
a) X3

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 85.
Sequence of characters enclosed within ………. quotes are called as String literals,
a) Single
b) Double
c) No
d) None of these
Answer:
b) Double

Question 86.
By default, string literals are automatically added with a special character………..at the end.
a) ‘\0’ (Null)
b) ‘\S’
c) V
d) None of these
Answer:
a) ‘\0’ (Null)

Question 87.
Identify the valid string constant from the following.
a) “A”
b) “Welcome”
c) “1234”
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 88.
The symbols which are used, to do some mathematical or logical operations are called as ………………
b) Operands
d) None of these
a) Operators
c) Expressions
Answer:
a) Operators

Question 89.
The data items or values that the operators act upon are called as ……………
a) Operators
b) Operands
c) Expressions
d) None of these
Answer:
b) Operands

Question 90.
In C++, the operators are classified as ………… types on the basis of the number of operands,
a) two
b) three
c) four ,
d) five
Answer:
b) three

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 91.
………….. operators require only one operand.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
a) Unary

Question 92.
…………… operators require two operands.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
b) Binary

Question 93.
………… operators require three operands.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
c) Ternary

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 94.
C++ operators are classified as …………… types
a) 7
b) 3
c) 10
d) 4
Answer:
a) 7

Question 95.
…………… operators perform simple operations like addition, subtraction, multiplication, division etc.
a) Logical
b) Relational
c) Arithmetic
d) Bitwise
Answer:
c) Arithmetic

Question 96.
…………… operator is used to find the remainder of a division.
a) /
b) %
c) *
d) **
Answer:
b) %

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 97.
…………….. operator is called as Modulus operator.
a) /
b)%
c) *
d) **
Answer:
b)%

Question 98.
An increment or decrement operator acts upon a …………….. operand and returns a new value,
a) Single
b) Two
c) Three
d) None of these
Answer:
a) Single

Question 99.
………….. is a unary operator.
a) ++
b) —
c) Both ++ and —
d) None of these
Answer:
c) Both ++ and —

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 100.
The increment operator adds …………….. to its operand.
a) 1
b) 0\
c) -1
d) None of these
Answer:
a) 1

Question 101.
The decrement operator subtracts …………… from its operand.
a) 1
b) 0\
c) -1
d) None of these
Answer:
a) 1

Question 102.
The …………….. operators can be placed either as prefix (before) or as postfix (after) to a variable.
a) ++
b) –
c) ++or–
d) None of these
Answer:
c) ++or–

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 103.
With the prefix version, C++ performs the increment/decrement………….. using the operand.
a) Before
b) After
c) When required
d) None of these
Answer:
a) Before

Question 104.
With the postfix version, C++ performs the increment/decrement…………….. using the operand.
a) Before
b) After
c) When required
d) None of these
Answer:
b) After

Question 105.
With the postfix version, C++ uses the value of the operand in evaluating the expression …………… incrementing /decrementing its present value.
a) Before
b) After
c) When required
d) None of these
Answer:
a) Before

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 106.
……………… operators are used to determining the relationship between its operands.
a) Logical
b) Relational
c) Arithmetic
d) Bitwise
Answer:
b) Relational

Question 107.
When the relational operators are applied on two operands, the result will be a …………… value.
a) Boolean
b) Numeric
c) Character
d) String
Answer:
a) Boolean

Question 108.
C++ provides …………. relational operators.
a) Seven
b) six
c) Eight
d) Five
Answer:
b) six

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 109.
All six relational operators are ……………
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
b) Binary

Question 110.
A logical operator is used to evaluate …………… expressions.
a) Logical and Relational
b) Logical
c) Relational
d) None of these
Answer:
a) Logical and Relational

Question 111.
Which logical operator returns 1 (True), if both expressions are true, otherwise it returns 0 (false)?
a) AND
b) OR
c) NOT
d) All the above
Answer:
a) AND

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 112.
Which logical operator returns 1 (True) if either one of the expressions is true. It returns 0 (false) if both the expressions are false?
a) AND
b) OR
c) NOT
d) All the above
Answer:
b) OR

Question 113.
Which logical operator simply negates or inverts the true value?
a) AND
b) OR
c) NOT
d) All the above
Answer:
c) NOT

Question 114.
AND, OR both are ……………. operators.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
b) Binary

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 115.
NOT is a(n) …………… operator.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
a) Unary

Question 116.
Identify the correct statement from the following.
a) The logical operators act upon the operands that are themselves called logical expressions.
b) Bitwise operators work on each bit of data and perform the bit-by-bit operations.
c) There are two bitwise shift operators in C++, Shift left (<<) & Shift right (>>).
d) All the above
Answer:
d) All the above

Question 117.
In C++, there are …………… kinds of bitwise operator.
a) Three
b) Four
c) Two
d) Five
Answer:
a) Three

Question 118.
…………. is a type of bitwise operator.
a) Logical bitwise operators
b) Bitwise shift operators
c) One’s compliment operators
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 119.
______ will return 1 (True) if both the operands are having the value 1 (True); Otherwise, it will return 0 (False).
a) Bitwise AND (&) .
b) Bitwise OR (|)
c) Bitwise Exclusive OR(A)
d) None of these
Answer:
a) Bitwise AND (&)

Question 120.
………… will return 1 (True) if any one of the operands is having a value 1 (True); It returns 0 (False) if both the operands are having the value 0 (False)
a) Bitwise AND (&)
b) Bitwise OR (|)
c) Bitwise Exclusive OR(A)
d) None of these
Answer:
b) Bitwise OR (|)

Question 121.
…………. will return 1 (True) if only one of the operand is having a value 1 (True).If both are True or both are False, it will return 0 (False).
a) Bitwise AND (&)
b) Bitwise OR (|)
c) Bitwise Exclusive OR(A)
d) None of these
Answer:
c) Bitwise Exclusive OR(A)

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 122.
There are …………… bitwise shift operators in C++.
a) Three
b) Two
c) Four
d) Five
Answer:
b) Two

Question 123.
……………. is a type of * .wise shift operator in
C++.
a) Shift left
b) Shift right
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 124.
……………. is a type of bitwise shift left operator in C++.
a) <<
b) >>
c) &&
d) ||
Answer:
a) <<

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 125.
…………. is a type of bitwise shift right operator in C++.
a) <<
b) >>
c) &&
d) ||
Answer:
b) >>

Question 126.
The value of the left operand is moved to the left by the number of bits specified by the right operand using …………. operator.
a) <<
b) >>
c) &&
d) ||
Answer:
a) <<

Question 127.
The value of the left operand is moved to right by the number of bits specified by the right operand using …………. operator.
a) <<
b) >>
c) &&
d) ||
Answer:
b) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 128.
Right operand should be an unsigned integer for …………… operator.
a) Arithmetic
b) Relational
c) Bitwise Shift
d) None of these
Answer:
c) Bitwise Shift

Question 129.
………… is the bitwise one’s complement operator.
a) <<
b) >>
c) &&
d) ~
Answer:
d) ~

Question 130.
The bitwise ………….. operator inverts all the bits in a binary pattern, that is, all l’s become 0 and all 0’s become 1.
a) Shift left
b) Shift right
c) One’s complement
d) None of these
Answer:
c) One’s complement

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 131.
…………… is a unary operator.
a) Shift left
b) Shift right
c) Bitwise one’s complement
d) None of these
Answer:
c) Bitwise one’s complement

Question 132.
………….. operator is used to assigning a value to a variable which is on the left-hand side of an assignment statement.
a) Assignment
b) Logical
c) Bitwise
d) Conditional
Answer:
a) Assignment

Question 133.
…………. is commonly used as the assignment operator in all computer programming languages.
a) :=
b) ==
c) =
d) None of these
Answer:
c) =

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 134.
…………… operator copies the value at the right side of the operator to the left side variable,
a) Assignment
b) Logical
c) Bitwise
d) Conditional
Answer:
a) Assignment

Question 135.
The assignment operator is a(n) ……………….. operator.
a) Unary
b) Binary
c) Ternary
d) Conditional
Answer:
b) Binary

Question 136.
How many conditional operators are used in C++?
a) one
b) two
c) three
d) four
Answer:
a) one

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 137.
…………….. operator is a Ternary Operator.
a) Assignment
b) Logical
c) Bitwise
d) Conditional
Answer:
d) Conditional

Question 138.
…………… operator is used as an alternative to if … else control statement.
a) Assignment
b) Logical
c) Bitwise
d) Conditional
Answer:
d) Conditional

Question 139.
……………. is a pointer to a variable operator.
a) &
b) *
c) →
d) → *
Answer:
b) *

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 140.
…………… is an address operator.
a) &
b) *
c) →
d) →*
Answer:
a) &

Question 141.
……………. is a direct component selector operator.
a) .(dot)
b) *
c) →
d) →*
Answer:
a) .(dot)

Question 142.
…………… is an indirect component selector operator.
a) .(dot)
b) *
c) →
d) →*
Answer:
c) →

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 143.
…………. is a dereference operator.
a) . (dot)
b) .*
c) →
d) →*
Answer:
b) .*

Question 144.
……………… is a dereference pointer to class member operator.
a) . (dot)
b) .*
c) →
d) →*
Answer:
d) →*

Question 145.
……………. is a scope resolution operator.
a) .(dot)
b) .*
c) : :
d) →*
Answer:
c) : :

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 146.
The operands and the operators are grouped in a specific logical way for evaluation is called as………………
a) Operator precedence
b) Operator association
c) Hierarchy
d) None of these
Answer:
b) Operator association

Question 147.
Which operator is lower precedence?
a) Arithmetic
b) Logical
c) Relational
d) None of these
Answer:
b) Logical

Question 148.
Which operator is higher precedence?
a) Arithmetic
b) Logical
c) Relational
d) None of these
Answer:
a) Arithmetic

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 149.
Which operator is the lowest precedence?
a) Assignment
b) Comma
c) Conditional
d) Arithmetic
Answer:
b) Comma

Question 150.
In C++, asterisk ( * ) is used for ……………… purpose.
a) Multiplication
b) Pointer to a variable
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 151.
……………. punctuator indicates the start and the end of a block of code.
a) Curly bracket { }
b) Paranthesis ()
c) Sqaure bracket [ ]
d) Angle bracket < >
Answer:
a) Curly bracket { }

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 152.
……………. punctuator indicates function calls and function parameters.
a) Curly bracket { }
b) Paranthesis ()
c) Square bracket [ ]
d) Angle bracket < >
Answer:
b) Paranthesis ()

Question 153.
……………. punctuator indicates single and multidimensional arrays.
a) Curly bracket { }
b) Paranthesis ()
c) Square bracket [ ]
d) Angle bracket < >
Answer:
c) Square bracket [ ]

Question 154.
……………… punctuator is used as a separator in an expression.
a) Comma,
b) Semicolon;
c) Colon :
d) None of these
Answer:
a) Comma,

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 155.
Every executable statement in C++ should terminate with a ………..
a) Comma,
b) Semicolon;
c) Colon:
d) None of these
Answer:
b) Semicolon;

Question 156.
……………… punctuator is used to label a statement.
a) Comma,
b) Semicolon;
c) Colon:
d) None of these
Answer:
c) Colon:

Question 157.
………….. is a single line comment.
a) /I
b) /* ……..*/
c) \\
d) None of these
Answer:
a) /I

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 158.
……………… is a multi line comment.
a) //
b) /* */
c) \\
d) None of these
Answer:
b) /* */

Question 159.
C++ provides the operator to get input. .
a) >>
b) <<
c) ||
d) &&
Answer:
a) >>

Question 160.
…………….. operator extracts the value through the keyboard and assigns it to the variable on its right.
a) >>
b) <<
c) ||
d) &&
Answer:
a) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 161.
………………. operator is called as “Stream extraction” or “get from” operator.
a) >>
b) <<
c) ||
d) &&
Answer:
a) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 162.
Get from operator requires …………….. operands.
a) three
b) two
c) four
d) five
Answer:
b) two

Question 163.
…………….. is the operand of get from the operator.
a) Predefined identifier cin
b) Variable
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 164.
To receive or extract more than one value at a time ………… operator should be used for each variable.
a) >>
b) <<
c) ||
d) &&
Answer:
a) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 165.
……………. is called cascading of operator.
a) >>
b) <<
c) 11 .
d) Both A and B
Answer:
d) Both A and B

Question 166.
C++ provides …………… operator to perform output operation. a) >>
b) <<
c) ||
d) &&
Answer:
b) <<

Question 167.
The operator ………….. is called the “Stream insertion” or “put to” operator.
a) >>
b) <<
c) ||
d) &&
Answer:
b) <<

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 168.
…………… operator is used to send the strings or values of the variables on its right to the object on its left. a) >>
b) <<
c) ||
d) &&
Answer:
b) <<

Question 169.
The second operand of put to operator may be a …………….
a) Constant
b) Variable
c) Expression
d) Either A or B or C
Answer:
d) Either A or B or C

Question 170.
To send more than one value at a time …………… operator should be used for each constant/ variable/expression.
a) >>
b) <<
c) ||
d) &&
Answer:
a) >>

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 171.
The compiler ignores …………… statement.
a) Comment
b) Input
c) Output
d) Assignment
Answer:
a) Comment

Question 172.
Usually all C++ programs begin with include statements starting with a ……………… symbol.
a) $
b) #
c) {
d) %
Answer:
b) #

Question 173.
The symbol …………… is a directive for the preprocessor.
a) $
b) #
c) {
d) %
Answer:
b) #

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 174.
_____ means, statements are processed before the compilation process begins.
a) Preprocessor
b) Include
c) Header file
d) None of these
Answer:
a) Preprocessor

Question 175.
The header file ……………… should include in every C++ program to implement input/output functionalities.
a) iostream
b) stdio
c) conio
d) math
Answer:
a) iostream

Question 176.
………….. header file contains the definition of its member objects cin and cout.
a) iostream
b) stdio
c) conio
d) math
Answer:
a) iostream

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 177.
Namespace collects identifiers used for …………..
a) Class
b) Object
c) Variables
d) All the above
Answer:
d) All the above

Question 178.
…………….. provides a method of preventing name conflicts in large projects.
a) namespace
b) header files
c) include
d) None of these
Answer:
a) namespace

Question 179.
Every C++ program must have a …………… function.
a) user defined
b) main( )
c) Library .
d) None of these
Answer:
b) main( )

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 180.
The …………….. function is the starting point where all C++ programs begin their execution.
a) user-defined
b) main ( )
c) Library
d) None of these
Answer:
b) main ( )

Question 181.
The executable statements should be inside the ………… function.
a) user-defined
b) main ( )
c) Library
d) None of these
Answer:
b) main ( )

Question 182.
The statements between the …………… braces are executable statements.
a) Curly bracket { }
b) Paranthesis ()
c) Square bracket [ ]
d) Angle bracket < >
Answer:
a) Curly bracket { }

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 183.
For creating and executing a C++ program, one must follow ……………. important steps.
a) two
b) three
c) five
d) four
Answer:
d) four

Question 184.
For creating and executing a C++ program, one must follow ……….. step.
a) Creating source code and save with .cpp extension
b) Compilation
c) Execution
d) All the above
Answer:
d) All the above

Question 185.
………………. links the library files with the source code and verifies each and every line of code.
a) Interpreter
b) Compiler
c) Loader
d) Assembler
Answer:
b) Compiler

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 186.
If there are no errors in the source code, …………… translates the source code into a machine-readable object file.
a) Interpreter
b) Compiler
c) Loader
d) Assembler
Answer:
b) Compiler

Question 187.
The compiler translates the source code into machine-readable object file with an extension…………..
a) .cpp
b) .exe
c) .obj *
d) None of these
Answer:
c) .obj *

Question 188.
The object file becomes an executable file with extension ……………
a) .cpp
b) .exe
c) .obj
d) None of these
Answer:
b) .exe

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 189.
………….. files can run without the help of any compiler or IDE.
a) Source
b) Object
c) Executable
d) None of these
Answer:
c) Executable

Question 190.
…………… makes it easy to create, compile and execute a C++ program.
a) Editors
b) IDE
c) Compilers
d) None of these
Answer:
b) IDE

Question 191.
IDE stands for …………….
a) Integrated Development Environment
b) Integrated Design Environment
c) Instant Development Environment
d) Integral Development Environment
Answer:
a) Integrated Development Environment

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 192.
……………. is open-source C++ compiler.
a) Dev C++ / Geany / Sky IDE
b) Code Lite / Code::blocks / Eclipse
c) Ner Beans / Digital Mars
d) All the above
Answer:
d) All the above

Question 193.
Dev C++ is written in …………
a) Delphi
b) C++
c) C
d) Pascal
Answer:
a) Delphi

Question 194.
………….. error is possible in C++.
a) Syntax
b) Semantic
c) Run-time
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 195.
………….. error occurs because of some illegal operation that takes place.
a) Syntax
b) Semantic
c) Run-time
d) All the above
Answer:
c) Run-time

Question 196.
Semantic error is called as ……………error.
a) Syntax
b) Logic
c) Run-time
d) All the above
Answer:
b) Logic

Question 197.
If a program tries to open a file which does not exist, it results in a …………. error.
a) Syntax
b) Logic
c) Run-time
d) All the above
Answer:
c) Run-time

Question 198.
……………. errors occur when grammatical rules of C++are violated.
a) Syntax
b) Logic
c) Run-time
d) All the above
Answer:
a) Syntax

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Very Short Answers (2 Marks)

Question 1.
Mention any two benefits of C++.
Answer:

  1. C++ is a highly portable language and is often the language of choice for multi-device, multi-platform app development.
  2. C++ is an object-oriented programming language and includes classes, inheritance, polymorphism, data abstraction, and encapsulation.

Question 2.
What is a character?
Answer:
A character represents any alphabet, number, or any other symbol (special characters) mostly available in the keyboard.

Question 3.
What are the types of C++ operators based on the number of operands?
Answer:
The types of C++ operators based on the number of operands are:

  1. Unary Operators – Require only one operand
  2. Binary Operators – Require two operands
  3. Ternary Operators – Require three operands

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What are the recent keywords included in C++?
Answer:
The recent list of keywords includes: using, namespace, bal, static_cast, const_cast, dynamic_cast, true, false.

Question 5.
What is a stream extraction operator?
Answer:
C++ provides the operator >> to get input. It extracts the value through the keyboard and assigns it to the variable on its right; hence, it is called as “Stream extraction” or “get from” operator.

Question 6.
Why the following identifiers are invalid?
a) num-add
b) this
c) 2myfile
Answer:
a) num-add – It contains spedal character (-) which ¡s not permitted
b) this – It is a keyword in C++. Keyword can not be used as identifier
c) 2myflle – Name must begin with an alphabet or an underscore.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7.
What are the main types of C++ datatypes?
Answer:
In C++, the data types are classified into three main categories

  1. Fundamental data types
  2. User-defined data types
  3. Derived data types.

Question 8.
What are Boolean literals?
Answer:
Boolean literals are used to represent one of the Boolean values (True or false). Internally true has value 1 and false has value 0.

Question 9.
What are string literals?
Answer:
The sequence of characters enclosed within double quotes is called String literals. By default, string literals are automatically added with a special character ‘\0’ (Null) at the end. Valid string Literals: “A” “Welcome” “1234” Invalid String Literals : ‘Welcome’,’1234′

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Differentiate Operators and Operands.
Answer:
The symbols which are used to do some mathematical or logical operations are called as Operators.
The data items or values that the operators act upon are called as Operands.

Question 11.
What are the classifications of C++ operators based on operand requirements?
Answer:
In C++, the operators are classified on the basis of the number of operands as follows:
i) Unary Operators – Require only one operand
ii) Binary Operators – Require two operands
iii) Ternary Operators – Require three operands

Question 12.
List the C++ operators.
Answer:
C++ Operators are classified as:

  • Arithmetic Operators
  • Relational Operators
  • Logical Operators
  • Bitwise Operators
  • Assignment Operators
  • Conditional Operator
  • Other Operators ,

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 13.
Write note on increment and decrement operators.
Answer:
++ (Plus, Plus) Increment operator
– (Minus, Minus) Decrement operator
An increment or decrement operator acts upon a single operand and returns a new value. Thus, these operators are unary operators. The increment operator adds 1 to its operand and the decrement operator subtracts 1 from its operand.
Example:
x++ is the same as x = x+1; It adds 1 to the present value of x.
X– is the same as x = x—1; It subtracts 1 from the present value of x.

Question 14.
Write a note on bitwise operators.
Answer:
Bitwise operators work on each bit of data and perform the bit-by-bit operation.
In C++, there are three kinds of bitwise operators, which are:

  • Logical bitwise operators
  • Bitwise shift operators
  • One’s compliment operator

Question 15.
Write about bitwise one’s compliment operator.
Answer:
The Bitwise one’s compliment operator:
The bitwise One’s compliment operator ~(Tilde), inverts all the bits in a binary pattern, that is, all l’s become 0 and all 0’s become 1. This is a unary operator.
Example:
If a = 15; Equivalent binary values of a is 0000 1111
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 4

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 16.
Write about assignment operator.
Answer:
Assignment Operator:
The assignment operator is used to assigning a value to a variable which is on the left-hand side of an assignment statement. = (equal to) is commonly used as the assignment operator in all computer programming languages. This operator copies the value at the right side of the operator to the left side variable. It is also a binary operator.
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 5

Question 17.
What are the shorthand assignment operators? Give example.
Answer:

Operator

      Name of  Operator                                                         Example
+= Addition Assignment a = 10;
c = a+= 5;
(ie, a = a+5)
c = 15
-= Subtraction Assignment a = 10;
c = a-= 5;
(ie, a = a-5)
c = 5
* = Multiplication Assignment a = 10;
c = a*= 5;
(ie, a = a*5)
c = 50
/= Division Assignment a = 10;
c – a/= 5;
(ie, a = a/5)
c = 2
%= Modulus Assignment a = 10;
c = a%= 5;
(ie, a = a%5)
c = 0

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 18.
Write note on conditional or ternary operator.
Answer:
In C++, there is only one conditional operator is used. ?: is a conditional Operator. This is a Ternary Operator. This operator is used as an alternative to if… else control statement.

Question 19.
Write note on comma (, ) operator.
Answer:
The comma (,) is an operator in C++ used to bring together several expressions. The group of expressions separated by a comma is evaluated from left to right.

Question 20.
What are the pointer operators?
Answer:
* – Pointer to a variable operator
& – Address of operator

Question 21.
What are the component selection operators?
Answer:
. – Direct component selector operator
-> – Indirect component selector operator

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 22.
What are the class member operators?
Answer:
:: – Scope access / resolution operator
.* – Dereference operator
->* – Dereference pointer to class member operator

Question 23.
What is operator association?
Answer:
The operands and the operators are grouped in a specific logical way for evaluation. This logical grouping is called as an Association.

Question 24.
What are the cascading operators?
Answer:
Get from (>>) and Put to (<<) operators are cascading operators.

Question 25.
What are the popular C++ Compilers with IDE.
Answer:

Compiler

Availability

Dev C++ Open-source
Geany Open-source
Code:: blocks Open source
Code Lite Open-source
Net Beans Open-source
Digital Mars Open-source
Sky IDE Open-source
Eclipse Open-source

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Short Answers (3 Marks)

Question 1.
What are the benefits of C++?
Answer:
Benefits of learning C++:

  • c++ ¡s a highly portable language and ¡s often the language of choice for multi-device, multi- platform app development.
  • C++ is an object-oriented programming language and includes classes, inheritance, polymorphism, data abstraction and encapsulation.
  • C++ has a rich function library.
  • C++ allows exception handling, inheritance and function overloading which are not possible in C.
  • C++ is a powerful, efficient and fast language.

It finds a wide range of applications — from GUI applications to 3D graphics for games to real-time mathematical simulations.

Question 2.
What are the characters used In C++?
Answer:
C++ accepts the following characters:

Alphabets A …. Z, a…. z
Numeric 0 …. 9
Special Characters + – * / ~ ! @ # $ % A& [ ] ( ) {} = ><_\l?.,:'”;
White space Blank space, Horizontal tab (->), Carriage return (), Newline, Form feed
Other characters C++ can process any of the 256 ASCII characters as data.

Question 3.
What are Automatic conversion and Type promotion?
Answer:
Implicit type conversion is a conversion performed by the compiler automatically. So, the implicit conversion is also called “Automatic conversion”. This type of conversion is applied usually whenever different data types are intermixed in an expression. If the type of the operands differs, the compiler converts one of them to match with the other, using the rule that the “smaller” type is converted to the “wider” type, which is called “Type Promotion”.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What are the rules for naming an identifier/variable?
Answer:
Rules for naming an identifier:

  • The first character of an identifier must be an alphabet or an underscore (-).
  • Only alphabets, digits, and underscore are permitted. Other special characters are not allowed as part of an identifier.
  • c++ is case sensitive as it treats upper and lower-case characters differently.
  • Reserved words or keywords cannot be used as an identifier name.

Question 5.
List the kinds of literals in C++.
Answer:
C++ has several kinds of literals. They are:
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 6

Question 6.
What are the types of C++operators?
Answer:
C++ Operators are classified as:

  1. Arithmetic Operators
  2. Relational Operators
  3. Logical Operators
  4. Bitwise Operators
  5. Assignment Operators
  6. Conditional Operator
  7. Other Operators

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7.
Write a note on character constants.
Answer:
A character constant is any valid single character enclosed within single quotes. A character constant in C++ must contain one character and must be enclosed within a single quote.
Valid character constants : ‘A’, ‘2\ ‘$’
Invalid character constants : “A”
The value of a single character constant has an equivalent ASCII value. For example, the value of’A’ is 65.

Question 8.
What are escape sequences? Explain.
Answer:
Escape sequences (or) Non-graphic characters:
C++ allows certain non-printable characters represented as character constants. Non-printable characters are also called non-graphical characters. Non-printable characters are those characters that cannot be typed directly from a keyboard during the execution of a program in C++.
For example: backspace, tabs etc. These non-printable characters can be represented by using escape sequences. An escape sequence is represented by a backslash followed by one or two characters.
Example: \t \On \xHn

Question 9.
Tabulate the escape sequence characters.
Answer:

Escape sequence

Non-graphical character

\a Audible or alert bell
\b Backspace
\f Form feed
\n Newline or linefeed
\r Carriage return
\t Horizontal tab
\v Vertical tab
\\ Backslash
\’ Single quote
\” Double quote
\? Question Mark
\On Octal number
\xHn Hexadecimal number
\o Null

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Write a note on arithmetic operators.
Answer:
Arithmetic operators perform simple arithmetic operations like addition, subtraction, multiplication, division etc.

Operator

Operation

Example

+ Addition 10 + 5 = 15
Subtraction 10 – 5 = 5
* Multiplication 10 * 5 = 50
/ Division 10 / 5 = 2 (Quotient of the division)
% Modulus (To find the reminder of a division) 10 % 3 = 1 (Remainder of the division)

Question 11.
What are the relational operators in C++? Give examples.
Answer:
Relational operators are used to determining the relationship between its operands. When the relational operators are applied on two operands, the result will be a Boolean value i.e 1 or 0 to represents True or False respectively. C++ provides six relational operators. They are:

Operator Operation Example
> Greater than a > b
< Less than a < b
>= Greater than or equal to a >= b
<= Less than or equal to a <= b
== Equal to a == b
j= Not equal a != b
  • In the above examples, the operand ‘a’ is compared with ‘b’ and depending on the relation, the result will be either 1 or 0. i.e., 1 for true, 0 for false.
  • All six relational operators are binary operators.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 12.
What are the logical operators used in C++? Explain its operation.
Answer:
A logical operator is used to evaluate logical and relational expressions. The logical operators act upon the operands that are themselves called as logical expressions. C++ provides three logical operators.
table

Operator

Operation

Description

&& AND The logical AND combines two different relational expressions into one. It returns 1 (True), if both expressions are true, otherwise, it returns 0 (False).
II OR The logical OR combines two different relational expressions into one. It returns 1 (True) if either one of the expressions is true. It returns 0 (False) if both the expressions are false.
! NOT NOT works on a single expression/operand. It simply negates or inverts the truth value, i.e., if an operand/expression is 1 (True) then this operator returns 0 (False) and vice versa.

AND, OR both are binary operators where as NOT is a unary operator.
Example:
a = 5, b = 6, c = 7;

Expression

Result

(a<b) && (b<c) 1 (True)
(a>b) && (b<c) 0 (False)
(a<b) || (b>c) 1 (True)
!(a>b) 1 (True)

Question 13.
What are the logical bitwise operators? Explain its operation.
Answer:
Logical bitwise operators:

  • & Bitwise AND (Binary AND)
  • | Bitwise OR (Binary OR)
  • ∧ Bitwise Exclusive OR (Binary XOR)
  • Bitwise AND (&) will return 1 (True)if both the operands are having the value 1 (True); Otherwise, it will return 0 (False).
  • Bitwise OR (|) will return 1 (True) if any one of the operands is having a value 1 (True); It returns 0 (False) if both the operands are having the value 0 (False).
  • Bitwise XOR(A) will return 1 (True) if only one of the operand is having a value of 1 (True). If both are True or both are False, it will return 0 (False).

The truth table for bitwise operators (AND, OR, XOR)

A B A & B A | B A ∧ B
1 1 1 1 0
1 0 0 1 1
0 1 0 1 1
0 0 0 0 0

Example:
If a = 65, b=15
Equivalent binary values of 65 = 0100 0001; 15 = 0000 1111
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 7

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 14.
What are the bitwise shift operators? Explain its operation.
Answer:
The Bitwise shift operators:
There are two bitwise shift operators in C++, Shift left (<<) and Shift right (>>).

  1. Shift left (<<) – The value of the left operand is moved to the left by the number of bits specified by the right operand. The right operand should be an unsigned integer.
  2. Shift right (>>) – The value of the left operand is moved to the right by the number of bits specified by the right operand. The right operand should be an unsigned integer.

Example:
If a =15; the Equivalent binary value of a is 0000 1111
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 8

Question 15.
What is input operator in C++? Explain.
Answer:
C++ provides the operator >> to get input. It extracts the value through the keyboard and
assigns it to the variable on its right; hence, it is called as “Stream extraction” or “get from” operator.
It is a binary operator i.e., it requires two operands. The first operand is the pre-defined identifier cin that identifies keyboard as the input device. The second operand must be a variable.
Working process of cin
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 9
Example:
cin>>num; — Extracts num value
cin>>x>>y; — Extracts x and y values

Question 16.
What is an output operator in C++? Explain.
Answer:
C+ + provides << operator to perform output operation. The operator << is called the “Stream insertion” or “put to” operator. It is used to send the strings or values of the variables on its right to the object on its left. << is a binary operator.
The first operand is the pre-defined identifier cout that identifies monitor as the standard output object. The second operand may be a
constant, variable or an expression.
Working process of cout
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 10
Example:
cout<<“Welcome”; – Display Welcome on-screen cout<<“The Sum =”<

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Explain in Detail 5 Marks

Question 1.
Explain Integer Constants. (or) Fixed point constants In detail.
Answer:
Integers are whole numbers without any fractions. An integer constant must have at least one digit without a decimal point. It may be signed or unsigned. Signed integers are considered as negative, commas and blank spaces are not allowed as part of it.
In C++, there are three types of integer constants:

  • Decimal
  • Octal
  • Hexadecimal

i) Decimal
Any sequence of one or more digits (0 …. 9).

Valid

Invalid

725 7,500 (Comma is not allowed)
-27 66 5 (Blank space is not allowed)
4.56 9$ (Special Character not allowed)

If we assign 4.56 as an integer decimal constant, the compiler will accept only the integer portion of 4.56 ie. 4. It will simply ignore .56.

ii) Octal:
Any sequence of one or more octal values (0 ….7) that begins with 0 is considered as an Octal constant.

Valid

Invalid

012 05,600 (Comma is not allowed)
-027 04.56 (A decimal point is not allowed)**
+0231 0158 (8 is not a permissible digit in the octal system)

iii) Hexadecimal:
Any sequence of one or more Hexadecimal values (0 …. 9, A …. F) that starts with Ox or OX is considered as a Hexadecimal constant.

Valid

Invalid

0x123 0x1,A5 (Comma is not allowed)
0X568 0x.l4E (Decimal point is not allowed like this)

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 2.
Explain Real Constants. (or) Floating-point constants in detail.
Answer:
Real Constants (or) Floating-point constants:
A real or floating-point constant is a numeric constant having a fractional component. These constants may be written ¡n fractional form or ¡n exponent form.
The fractional form of a real constant is a signed or unsigned sequence of digits including a decimal point between the digits.
It must have at least one digit before and after a decimal point. It may have a prefix with the + or – sign.
A real constant without any sign will be considered positive.
Exponent form of real constants consists of two parts:

  1. Mantissa
  2. Exponent

The mantissa must be either an integer or a real constant. The mantissa followed by a letter E or e and the exponent. The exponent should also be an integer.
For Example:
58000000.00 may be written as 0.58 x 108 or 0. 58E8.

Mantissa (Before E)

Exponent (After E)

0.58 8

Example:
5.864 E1 → 5.864 x 101 → 58.64
5864 E-2 → 5864 x 10-2 → 58.64
0.5864 E2 → 0.5864 x 102 → 58.64

Question 3.
Explain the prefix and postfix operators’ working process with suitable examples.
Answer:
The ++ or – operators can be placed either as prefix (before) or as postfix (after) to a variable. With the prefix version, C++ performs the increment / decrement before using the operand.
For example: N1=10, N2=20;
S = ++N1 + ++N2;
The following Figure explains the working process of the above statement.
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 11
In the above example, the value of num is first incremented by 1, then the incremented value is assigned to the respective operand.
With the postfix version, C++ uses the value of the operand in evaluating the expression before incrementing /decrementing its present value.
For example: N1=10, N2=20;
S = N1++ + ++N2;
The following Figure explains the working process of the above statement.
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 12
In the above example, the value assigned to operand N1 is taken into consideration, first and then the value will be incremented by 1.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What are punctuators/separators? List the punctuators and their operations.
Answer:
Punctuators are symbols, which are used as delimiters while constructing a C++ program. They are also called “Separators”. The following punctuators are used in C++.

Curly braces { } Opening and closing curly braces indicate the start and the end of a block of code. A block of code containing more than one executable statement. These statements together are called as “compound statement”. int main ()
{int x=10,
y=20, sum;
sum = x + y;
cout << sum;
}
Parenthesis () Opening and closing parenthesis indicate function calls and function parameters. clrscr();
add (5, 6);
Square brackets [ ] It indicates single and multidimensional arrays. int num[5];
charname[50];
Comma (,) It is used as a separator in an expression. int x=10, y=20, sum;
Semicolon ; Every executable statement in C++ should terminate with a semicolon. int main ()

{
int x=10, y=20, sum; sum = x + y; cout << sum; }

Colon : It is used to label a statement. private:
Comments
///* *1
Any statement that begins with // are considered a comments. Comments are simply ignored by compilers, i.e., compiler does not execute any statement that begins
with a // // Single line comment
/*  ………….. /  Multiline comment
/* This is written By
myself to learn CPP */ int main ()
{
intx=10,
y=20, sum;  // to sum x  and y  sum = x + y;
cout << sum;
}

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Evaluate Yourself

Question 1.
What is meant by literals? How many types of integer literals available in C++?
Answer:
Literals are data items whose values do not change during the execution of a program. Literals are called as Constants.
In C++, there are three types of integer literals (constants). They are:

  1. Decimal
  2. Octal
  3. Hexadecimal

Question 2.
What kind of constants is following?
i) 26
ii) 015
iii) 0xF
iv) 014.9
Answer:
i) 26 : Decimal constant
ii) 015 : Octal constant
iii) 0xF : Hexadecimal constant
iv) 014.9 : Integer Constant. (A fractional number that begins with 0, C++ has consider the number as an integer not an Octal)

Question 3.
What is the character constant in C++?
Answer:
A character constant is any valid single character enclosed within single quotes. A character constant in C++ must contain one character and must be enclosed within a single quote.
Valid character constants: ‘A’, ‘2’, ‘$’
Invalid character constants: “A”

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
How are non-graphic characters represented in C++?
Answer:
Non-printable characters are also called as non-graphical characters.
Non-printable characters are those characters that cannot be typed directly from a keyboard during the execution of a program in C++, for example, backspace, tabs, etc. These non-printable characters can be represented by using escape sequences.

An escape sequence is represented by a backslash followed by one or two characters.

Escape Sequence

Non-graphical character

\a The audible or alert bell
\b Backspace
\f Form feed
\n Newline or linefeed
\r Carriage return
\t Horizontal tab
\v Vertical tab
\\ Backslash
V Single quote
\” Double quote
\On Octal number
\xHn Hexadecimal number
\0 Null

Even though an escape sequence contains two characters, they should be enclosed within single quotes because, C++ consider escape sequences as character constants and allocates one byte in ASCII representation.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 5.
Write the following real constants into exponent form:
i) 32.179
ii) 8.124
iii) 0.00007
Answer:
i) 32.179 → 3.2179E1 (OR) 0.32179E2 (OR) 32178E-3
ii) 8.124 → 0.8124E1 (OR) 8124E-3
iii) 0.00007 → 7E-5

Question 6.
Write the following real constants into fractional form:
i) 0.23E4
ii) 0.517E-3
iii) 0.5E-5
Answer:
i) 0.23E4 → 2300
ii) 0.517E-3 → 0.000517
iii) 0.5E-5 → 0.000005

Question 7.
What is the significance of the null (\0) character in a string?
Answer:
By default, string literals are automatically added with a special character ‘\0’ (Null) at the end. It is called as an end of string character.
The string “welcome” will actually be represented as “welcome\0” in memory and the size of this string is not 7 but 8 characters i.e., inclusive of the last character \0.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Evaluate Yourself

Question 1.
What is the use of operators?
Answer:
Operators are the symbols which are used to do some mathematical or logical operations on their operands.

Question 2.
What are binary operators? Give examples.
Answer:
Arithmetic binary operators.
Binary Operators – Require two operands.
The arithmetic operators addition(+), subtraction(-), multiplication(*), division(/) and Modulus(%) are binary operators which requires two operands.
Example:

Operator

Operation

Example

+ Addition 10 + 5 = 15
Subtraction 10 – 5 = 5 Slfil.r
* Multiplication 10 * 5 = 50
/ Division 10 / 5 = 2 (Quotient of the division)
% Modulus (To find the reminder of a division) 10 % 3 = 1 (Remainder of the division)

Question 3.
What does the modulus operator % do?
Answer:
The modulus operator is used to find the remainder of a division.
Example:
10%3 will return 1 which is the remainder of the division.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What will be the result of 8.5 % 2?
Answer:
The following error will appear while compiling the program.
Invalid operands of types ‘double’ and ‘int’ to binary ‘operator%’.
The reason is % operator operates on integer operands only.

Question 5.
Assume that R starts with a value of 35. What will be the value of S from the following expression? S=(R–)+(++R)
Answer:
S = 70

Question 6.
What will be the value of j = – – k + 2k. if k is ‘ 20 initially?
Answer:
The value of j will be 57 and k will be 19.
C++ Code;
#include
using namespace std;
int main()
{
int k=20,j;
j=–k+2*k;
cout<< “Vlaue of j=”<<j<< “\nVlaue of
k =”<<k;
return 0;
}
Out put
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 13

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7.
What will be the value of p = p * ++j where j is 22 and p = 3 initially?
Answer:
The value of p is 69 and j is 23.
C++ program:
#include
using namespace std;
int main()
{
int j=22, p=3;
P = P * ++j;
cout<< “Value of p =”<<p<< “\nValue of
j =”<<j;
return 0;
}
Out put
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 14

Question 8.
Give that i = 8, j = 10, k = 8, What will be result of the following expressions?
i) i < k
ii) i < j
iii) i > = k
iv) i = = j
v) j ! = k
Answer:

Expression

Result

i < k 0 (False)
i < j 1 (True)
i >= k 1 (True)
i == j 0 (False)
j != k 1 (True)

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 9.
What will be the order of evaluation for the following expressions?
i) i + 3 > = j – 9
ii) a +10 < p – 3 + 2 q
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 15

Question 10.
Write an expression involving a logical operator to test, if marks are 75 and grade is ‘A’.
Answer:
(marks == 75) && (grade == ‘A)

Hands On Practice

Type the following C++ Programs in Dev C++ IDE and execute, if the compiler shows any errors, try to rectify it and execute again and again till you get the expected result.
Question 1.
C++ Program to find the total marks of three
subjects.
#include
using namespace std;
int main()
{
int m1, m2, m3, sum;
cout << “\n Enter Mark 1:”; cin >> m1;
cout << ”\n Enter Mark 2: “; cin >> m2;
cout << “\n Enter Mark 3: “; cin >> m3;
sum = m1 + m2 + m3;
cout << “\n The sum = ” << sum;
}
Make changes in the above code to get the average of all the given marks.
Answer:
Modified Program:
#include
using namespace std;
int main()
{
int m1, m2, m3, sum;
float avg;
cout << “\n Enter Mark 1: “;
cin >> m1;
cout << “\n Enter Mark 2: “;
cin >> m2;
cout << “\n Enter Mark 3: “;
cin >> m3;
sum = m1 + m2 + m3;
avg = (float)sum / 3;
cout << “\n The sum = ” << sum;
cout << “\n The average = ” << avg;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 16

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 2.
C++ program to find the area of a circle.
#include
using namespace std;
int main()
{
int radius;
float area;
cout << “\n Enter Radius: “;
cin<< radius;
area = 3.14 * radius * radius;
cout << “\n The area of circle =“ <<area;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 17

Question 3.
Point out the errors in the following program:
Using namespace std;
int main( )
{
cout << “Enter a value”
cin << numl >> num2
num+num2=sum;
cout >> “\n The Sum= ” >> sum;
Answer:

Given code

Error

Using namespace std; The keyword must be in lowercase. So, Using should be written as using. Header file is missing.
int main() No Error
No Error
cout << “Enter a value “; Prompt should be “Enter two values” because cin contains two variables.
cin << numl >> num2 Variables are not declared. It should be declared first, cin must followed by Extraction operator(>>).

Semicolon is missing at the end of the statement.

num+num2=sum; Improper assignment statement and undefined variable name used. It should be replaced as sum=numl + num2;
cout >> “\n The Sum= ” >> sum; cout must followed by put to the operator.
Return 0; statement is missing. Close bracket} missing.

The correct program is given below:
using namespace std;
#include
int main()
{
int num1,num2,sum;
cout << “Enter two values”; cin >> numl >> num2;
sum= num+num2;
cout << “\n The Sum= “<< sum;
return 0;
}

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
Point out the type of error in the following program:
#include
using namespace std;
int main()
{
int h=10; w=12;
cout << “Area of rectangle ” << h+w; >
Answer:
Syntax error exists. Ie. int h=10;w=12; should written as int h=10,w=12;
There is also a logical error in the above program.
The formula for rectangle area is given wrong. This error will not indicate by the compiler.
MODIFIED PROGRAM:
#include
using namespace std;
int main()
{
int h=10, w=12;
cout << “Area of rectangle ” << h*w;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 18

DATATYPES, VARIABLES AND EXPRESSIONS
Book Evaluation
Part -I

Choose The Correct Answer

Question 1.
How many categories of data types available in C++?
a) 5
b) 4
c) 3
d) 2
Answer:
c) 3

Question 2.
Which of the following data types is not a fundamental type?
a) signed
b) int
c) float
d) char
Answer:
a) signed

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
What will be the result of following statement?
char ch= ‘B’;
cout << (int) ch;
a) B
b) b
c) 65
d) 66
Answer:
d) 66

Question 4.
Which of the character is used as suffix to indicate a floating point value?
a) F
b) C
c) L
d) D
Answer:
a) F

Question 5.
How many bytes of memory allocates for the following variable declaration if you are using Dev C++? short int x;
a) 2
b) 4
c) 6
d) 8
Answer:
a) 2

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 6.
What is the output of the following snippet?
char ch = ‘A’;
ch= ch + 1;
a) B
b) A1
c) F
d) 1A
Answer:
a) B

Question 7.
Which of the following Is not a data type modifier?
a) signed
b) int
c) long
d) short
Answer:
b) int

Question 8.
Which of the following operator returns the size of the data type?
a) size of( )
b) int ( )
c) long ( )
d) double ( )
Answer:
a) size of( )

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 9.
Which operator to be used to access a reference of a variable?
a) $
b) #
c) &
d) !
Answer:
c) &

Question 10.
This can be used as an alternate to end command:
a) \t
b) \b
c) \0
d) \n
Answer:
c) \0

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Part II

Very Short Answers

Question 1.
Write a short note const keyword with an example.
Answer:
const is the keyword used to declare a constant, const keyword modifies/restricts the accessibility of a variable. So, it is known as an Access modifier.
Example:
const int num =100; indicates that the variable num can not be modified. It remains constant.

Question 2.
What is the use of setw( ) format manipulator?
Answer:
setw ( ):
setw manipulator sets the width of the field assigned for the output. The field width determines the minimum number of characters to be written in the output.
Syntax:
setw(number of characters)
Example:
cout << setw(25) << “Net Pay : ” << setw(10)
<< np << endl;

Question 3.
Why is char often treated as an integer data type?
Answer:
Character data type is often said to be an integer type since all the characters are represented in memory by their associated ASCII Codes.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What is a reference variable? What is its use?
Answer:
A reference provides an alias for a previously defined variable. Declaration of a reference consists of base type and an & (ampersand) symbol; reference variable name is assigned the value of a previously declared variable.
Syntax:
<&reference_variable> =
Example:
#include
using namespace std;
int main( )
{
int num;
int &temp = num; //declaration of a reference variable temp
num = 100;
cout << “\n The value of num =” << num;
cout << “\n The value of temp =” << temp;
}
Output
The value of num = 100
The value of temp = 100

Question 5.
ConsIder the following C++ statement Are they equivalent?
char ch=67;
char ch=’C’;
Answer:
Yes. Both are equivalent. Both assignment statements will store character C’ in the variable ch.

Question 6.
what Is the difference between 561 and 56?
Answer:

  • 56 indIcate an Integer
  • 56L indicates Long Integer The suffix L indicates Long. So it stores a long integer

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7.
Determine which of the following are valid constant? And specify their type,
i) 0.5
ii) ‘Name’
iii) ‘\t’
iv) 27,822
Answer:
i) 0.5 : Valid. Floating-point constant
ii) ‘Name’ : Invalid. String constant must be enclosed within double-quotes.
iii) ‘\t’ : Valid. Character constant.
iv) 27,822 : Invalid. Comma not allowed with integer constant.

Question 8.
Suppose x and y are two double-type variables that you want add as integers and assign to an integer variable. Construct a C++ statement for doing so.
Answer:
double x, y;
int sum;
x = 12.64;
y = 13.56;
sum = (int) x + (int) y;
The variable sum will have the value of 25 due to explicit casting.

Question 9.
What will be the result of following if num=6 initially?
Answer:
a) cout << num;
6
b) cout << (num==5);
0

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Which of the following two statements are valid? Why? Also write their result, int a;
i) a=3,014;
ii) a =(3,014);
Answer:
i) a=3,014; – Invalid. Special character comma(,) not allowed.
ii) a=(3,014); – Valid. 014 is an octal constant. It will be converted into decimal and then stored in a. So, a will hold 12 as its value.

Part – III

Short Answers

Question 1.
What are arithmetic operators in C++? Differentiate unary and binary arithmetic operators. Give example for each of them.
Answer:
Arithmetic Operators:
Arithmetic operators perform simple arithmetic operations like addition, subtraction, multiplication, division, etc.

Operator Operation Example
+ Addition 10 + 5-15
 – Subtraction 10-5-5
* Multiplication 10 * 5 = 50
/ Division 10 / 5 = 2 (Quotient of the division)
% Modulus (To find the reminder of a division) 10 % 3 = 1 (Remainder of the division)

The above-mentioned arithmetic operators are binary operators which require a minimum of two operands.

Unary arithmetic operators:
– (Unary) – The unary minus operator changes the sign of its argument. A positive number ‘ becomes negative and negative number becomes
positive, int a = 5;
a = -a; // a becomes -5
+ (Unary) – The unary plus operator keeps the sign of its argument,
int a = -5;
a = +a; // a still have the value same value -5
(No change)
a = 5;
a = +a; // a stil have the same value 5
(No change)
Unary Minus is different from – (Binary) ie. subtraction operator requires two operands.
Unary Plus is different from + (Binary) ie. addition operator requires two operands.
#include
mt mamo
{
mt x=10;
intÿ= -10;
inta = -10;
‘nt b = 10;
y=+y;
a= -a;
b=+b;
cout«”\nx = “«X;
cout«”\ny = “«y;
cout«”\na = “«a;
cout«”\nb = “«b;
return 0;
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 19

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 2.
Evaluate x+= x + ++x; Let x=5;
Answer:
x=18
Code:
using namespace std;
#include
int main ()
{

int x;
x = 5;
x+= x + ++x;
cout «x;
return O;
}
Out put
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 20

Question 3.
How relational operators and logical operators related to one another?
Answer:
Both relational and logical operators will give the evaluation result as Boolean constant value 1 (True) or 0(False).

Question 4.
Evaluate the following C++ expressions where x, y, z are integers and m, n are floating
point numbers. The value of x = 5, y = 4 and
m=2.5;
i) n=x+y/x;
n=5
ii)z=m*x+y;
z=16
iii) z = (x++) * m + X;
z = 18
Code:
using namespace std;
#include
int main()
{
int x,y,z1,z2;
float m,n;
x=5;
y=4;
m=2.5;
n = x + y / X;
z1=m*x+y;
z2 = (x++) * m + X;
cout<<”\nN = “<<n;
còut<<”\nzl =<<z1;
cout<<”\nz2 =<<z2;
return 0;
}
output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 21

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

11th Computer Science Guide Introduction to C++ Additional Questions and Answers

Choose The Correct Answer 1 Mark

Question 1.
Every programming language has _____ fundamental element.
a) Data types
b) Variables
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 2.
c++ provides a predefined set of data types for handling the data item is known as …………….. data type.
a) Fundamental
b) Built-in data types
c) User-defined
d) Either A or B
Answer:
d) Either A or B

Question 3.
A programmer can create his own data types called as ______ data types.
a) Fundamental
b) Built-in data types
c) User defined
d) Either A or B
Answer:
c) User defined

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
In a programming language, fields are referred as ……………
a) Variables
b) Data
c) File
d) None of these
Answer:
a) Variables

Question 5.
In a programming language, values are referred to as ……………
a) Variables
b) Data
c) File
d) None of these
Answer:
b) Data

Question 6.
In C++, the data types are classified as ______ main categories.
a) three
b) four
c) two
d) five
Answer:
a) three

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 7.
In C++, ______ is a data type category.
a) Fundamental
b) Derived
c) User defined
d) All the above
Answer:
d) All the above

Question 8.
The ………….. are the named memory locations to hold values of specific data types.
a) Literals
b) Variables
c) Constants
d) None of these
Answer:
b) Variables

Question 9.
There are ………….. fundamental (atomic) data types in C++.
a) two
b) three
c) four
d) five
Answer:
d) five

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
…………… is an atomic data type in C++,
a) char / int
b) float / double
c) void
d) All the above
Answer:
d) All the above

Question 11.
………….. are whole numbers without any fraction.
a) Integers
b) Characters
c) Strings
d) None of these
Answer:
a) Integers

Question 12.
Identify the correct statement from the ; following.
a) Integers can be positive or negative.
b) If you try to store a fractional value in an int type variable it will accept only the integer portion of the fractional value.
c) If a variable is declared as an int, C++ compiler allows storing only integer values into it.
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 13.
……….. data type accepts and returns all valid \ ASCII characters.
a) Character
b) float
c) void
d) None of these
Answer:
a) Character

Question 14.
Character data type is often said to be an …………… type.
a) float
b) string
c) void
d) int
Answer:
d) int

Question 15.
……………… means significant numbers after decimal point.
a) Precision
b) Digit
c) Floating point
d) None of these
Answer:
a) Precision

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 16.
The _____ data type is larger and slower than type float.
a) Char
b) double
c) void
d) int
Answer:
b) double

Question 17.
The literal meaning for void is …………….
a) Empty space
b) Nothing
c) Blank
d) None of these
Answer:
a) Empty space

Question 18.
In C++, the ……………. data type specifies an empty set of values.
a) Char
b) double
c) void
d) int
Answer:
c) void

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 19.
……………. is used as a return type for functions that do not return any value.
a) Char
b) double
c) void
d) int
Answer:
c) void

Question 20.
Identify the correct statement from the following.
a) One of the most important reason for declaring a variable as a particular data type is to allocate appropriate space in memory.
b) As per the stored program concept, every data should be accommodated in the main memory before they are processed)
c) C++ compiler allocates specific memory space for each and every data handled according to the compiler’s standards.
d) All the above
Answer:
d) All the above

Question 21.
char data type needs ……………. bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
d) 1

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 22.
int data type needs ………….. bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
c) 2

Question 23.
float data type needs …………….. bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
b) 4

Question 24.
double data type needs ………… bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
a) 8

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 25.
The range of char data type is ………………
a) -127 to 128
b) -32,768 to 32,767
c) 3.4 x 10-38 to 3.4 x 1038 -1
d) 1.7 x 10-308 to 1.7 x 10308 -1
Answer:
a) -127 to 128

Question 26.
The range of int data type is …………..
a) -127 to 128
b) -32,768 to 32,767
c) 3.4 x HT38 to 3.4 x 1038-1
d) 1.7 x 10-308 to 1.7 x 10308 -1
Answer:
b) -32,768 to 32,767

Question 27.
The range of float data type is ……………
a) -127 to 128
b) -32,768 to 32,767
c) 3.4 x 10-38 to 3.4 x 1038 -1
d) 1.7 x lO”308 to 1.7 x 10308 -1
Answer:
c) 3.4 x 10-38 to 3.4 x 1038 -1

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 28.
The range of double data type is …………….
a) -127 to 128
b) -32,768 to 32,767
c) 3.4 x 10″38 to 3.4 x 1038-1
d) 1.7 x 10-308 to 1.7 x 10308 -1
Answer:
d) 1.7 x 10-308 to 1.7 x 10308 -1

Question 29.
…………….. can be used to expand or reduce the memory allocation of any fundamental data type.
a) Modifiers
b) Access specifiers
c) Promoters
d) None of these
Answer:
a) Modifiers

Question 30.
………….. are called as Qualifiers.
a) Modifiers
b) Access specifiers
c) Promoters
d) None of these
Answer:
a) Modifiers

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 31.
There are …………….modifiers used in C++.
a) five
b) four
c) three
d) two
Answer:
b) four

Question 32.
……………… is a modifier in C++,
a) signed / unsigned
b) long
c) short
d) All the above
Answer:
d) All the above

Question 33.
short data type needs …………. bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
c) 2

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 34.
unsigned short data type needs ………….. bytes of memory. .
a) 8
b) 4
c) 2
d) 1
Answer:
c) 2

Question 35.
signed short data type needs…………… bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
c) 2

Question 36.
signed long data type needs ……………. bytes of memory.
a) 8
b) 4
c) 2
d) 1
Answer:
b) 4

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 37.
The range of unsigned short is …………..
a) -32,768 to 32768
b) 0 to 65535
c) -2,147,483,648 to 2,147,483,647
d) 0 to 4,294,967,295
Answer:
b) 0 to 65535

Question 38.
The range of unsigned long is ……………..
a) -32,768 to 32768
b) 0 to 65535
c) -2,147,483,648 to 2,147,483,647
d) 0 to 4,294,967,295
Answer:
d) 0 to 4,294,967,295

Question 39.
The range of signed long is ……………..
a) -32,768 to 32768
b) 0 to 65535
c) -2,147,483,648 to 2,147,483,647
d) 0 to 4,294,967,295
Answer:
c) -2,147,483,648 to 2,147,483,647

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 40.
The range of unsigned char is …………….
a) -32,768 to 32768
b) 0 to 65535
c) 0 to 255
d) 0 to 4,294,967,295
Answer:
c) 0 to 255

Question 41.
The range of long double data type is ……………..
a) -127 to 128
b) -32,768 to 32,767
c) 3.4 x 10^932 to 1.1 x 104932 -1
d) 1.7 x 10~308 to 1.7 x 10308 -1
Answer:
c) 3.4 x 10^932 to 1.1 x 104932 -1

Question 42.
int data type needs …………. bytes of memory in Dec C++.
a) 8
b) 4
c) 2
d) 1
Answer:
b) 4

Question 43.
unsigned int data type needs ……………… bytes of memory in Dec C++.
a) 8
b) 4
c) 2
d) 1
Answer:
b) 4

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 44.
signed int data type needs …………… bytes of memory in Dec C++.
a) 8
b) 4
c) 2
d) 1
Answer:
b) 4

Question 45.
long double data type needs…………… memory in Dec C++.
a) 10
b) 8
c) 2
d) 12
Answer:
d) 12

Question 46.
long double data type needs…………… memory in Turbo C++.
a) 10
b) 8
c) 2
d) 12
Answer:
a) 10

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 47.
…………….. is an operator which gives the size of a data type.
a) sizeof()
b) byteof()
c) datatype()
d) None of these
Answer:
a) sizeof()

Question 48.
The suffix……………. is used for floating point values
a) U
b) L
C) F
d) None of these
Answer:
C) F

Question 49.
The suffix ……………… is used for long int values.
a) U
b) L
C) F
d) None of these
Answer:
b) L

Question 50.
The suffix ………….. is used for unsigned int
a) U
b) L
C) F
d) None of these
Answer:
a) U

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 51.
……………… are user-defined names assigned to specific memory locations in which the values are stored.
a) Literals
b) Variables
c) Operators
d) None of these
Answer:
b) Variables

Question 52.
There are ………….. values associated with a symbolic variable
a) two
b) three
c) four
d) five
Answer:
a) two

Question 53.
…………… is data stored in a memory location.
a) L-value
b) R-value
c) T-value
d) B-value
Answer:
b) R-value

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 54.
…………… is the memory address in which the R-value is stored.
a) L-value
b) R-value
c) T-value
d) B-value
Answer:
a) L-value

Question 55.
The memory addresses are in the form of ……………. values.
a) Binary
b) Octal
c) Decimal
d) Hexadecimal
Answer:
d) Hexadecimal

Question 56.
Every …………. should be declared before they are actually used in a program.
a) Variable
b) Operand
c) Literals
d) None of these
Answer:
a) Variable

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 57.
If we declare a variable without any initial value, the memory space allocated to that variable will be occupied with some unknown value is called as ………….. values.
a) Junk
b) Garbage
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 58.
A variable can be initialized during the execution of a program is known as ……………..
a) Dynamic initialization
b) Static initialization
c) Random initialization
d) None of these
Answer:
a) Dynamic initialization

Question 59.
………….. is the keyword used to declare a constant.
a) constant
b) const
c) cons
d) None of these
Answer:
b) const

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 60.
…………… keyword modifies / restricts the accessibility of a variable.
a) constant
b) const
c) cons
d) None of these
Answer:
b) const

Question 61.
……………….. is known as Access modifier of a variable.
a) constant
b) const
c) cons
d) None of these
Answer:
b) const

Question 62.
Declaration of a reference consists of ……………..
a) Base type
b) An 8i (ampersand) symbol
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 63.
……………. are used to format the output of any
C++ program,
a) Qualifiers
b) Modifiers ‘
c) Manipulators
d) None of these
Answer:
c) Manipulators

Question 64.
ManIpulators are functions specifically designed to use with the ______ operators.
a) Insertion («)
b) Extraction(»)
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 65.
Commonly used manipulator is …………..
a) endl and setw
b) setfill
c) setprecision and setf
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 66.
endl manipulator is a member of  ………………….. header file.
a) iomanip
b) iostream
c) manip
d) conio
Answer:
b) iostream

Question 67.
setw, setfihl, setprecision and setf manipulators are members of______ header file.
a) iomanip
b) iostream
c) manip
d) conio
Answer:
a) iomanip

Question 68.
______ is used asa line feeder in C++.
a) setw
b) setfill
c) endi
d) setf
Answer:
c) endi

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 69.
______ can be used as an alternate to ‘sn’.
a) setw
b) setfill
c) endi
d) setf
Answer:
c) endi

Question 70.
______ inserts a new line and flushes the buffer.
a) setw
b) setfill
c) endi
d) setf
Answer:
c) endi

Question 71.
______ manipulator sets the width of the field assigned for the output.
a) setw
b) setñhl
c) endi
d) setf
i.) IIUI
U) SeIT
Answer:
a) setw

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 72.
_______ manipulator is usually used after setw.
a) endl
b) setfill
c) endl
d) setf
Answer:
b) setfill

Question 73.
______ is used to display numbers with fractions In specific number of digits.
a) ‘endI
b) setfill i1
c) endl
d) setprecision
Answer:
d) setprecision

Question 74.
setf() manipulator may be used in form…………..
a) Fixed
b) Scientific
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 75.
An expression ¡s a combination of ……………………… arranged as per the rules of C++.
a) Operators
b) Constants
c) Variables
d) All the above
Answer:
d) All the above

Question 76.
InC++,there are ………………………….. types of expressions used.
a) four
b) five
c) seven
d) two
Answer:
c) seven

Question 77.
The process of converting one fundamental type into another is called as …………………………
a) Type Conversion
b) Compiling
c) Inverting
d) None of these
Answer:
a) Type Conversion

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 78.
c++ provides …………………… types of conversions
a) three
b) two
c) four
d) five
Answer:
b) two

Question 79.
C++ provides _____types of conversion.
a) Implicit
b) Explicit
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 80.
A(n) ………………………. type conversion is a conversion performed by the compiler automatically.
a) Implicit
b) Explicit
c) BothAandB
d) None of these
Answer:
a) Implicit

Question 81.
______conversion is also called as Automatic conversion.
a) Implicit ‘
c) BothAandB
b) Explicit
d) None of these
Answer:
a) Implicit ‘

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 82.
Data of smaller type  converted to the wider type, which is called is as ………………
a) Type Promotion
c) Type extended
b) Type upgrade
d) None of these
Answer:
a) Type Promotion

Question 83.
c++ allows explicit conversion of variables or expressions from one data type to another specific data type by the programmer Is called as ………….
a) Type Promotion
b) Type upgrade
c) Type casting
d) None of these
Answer:
c) Type casting

Very Short Answers 2 Marks

Question 1.
What are the classification of data types?
Answer:
In C++, the data types are classified as three main categories.

  1. Fundamental data types
  2. User-defined data types and
  3. Derived data types.

Question 2.
Define: Variable.
Answer:
The variables are the named memory locations to hold values of specific data types.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Give the syntax for declaring a variable with an example.
Answer:
Syntax for declaring a variable:
Example:
int num1;
int num1, num2, sum;

Question 4.
What a the fundamental/atomic data types in C++?
Answer:
Fundamental (atomic) data types are predefined data types available with C++. There are five fundamental data types in C++: char, int, float, double and void.

Question 5.
Write about int data type.
Answer:
Integers are whole numbers without any fraction. Integers can be positive or negative. Integer data type accepts and returns only integer numbers.
If a variable is declared as an int, C++ compiler allows storing only integer values into it.
If we try to store a fractional value in an int type variable it will accept only the integer portion of the fractional value.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 6.
What are the advantages of using float data type?
Answer:
There are two advantages of using float data types.

  1. They can represent values between the integers.
  2. They can represent a much greater range of values.

Question 7.
What is the disadvantage of using float data type?
Answer:
The floating point operations takes more time to execute compared to the integer type ie., floating point operations are slower than integer operations. This is a disadvantage of floating point operation.

Question 8.
What do you mean by precision?
Answer:
Precision means significant numbers after decimal point of floating point number.

Question 9.
Tabulate the memory allocation for fundamental data types?
Answer:
Memory allocation for fundamental data types

Data type

Space in memory

in terms of bytes in terms of bits
char 1 byte 8 bits
int 2 bytes 16 bits
float 4 bytes 32 bits
double 8 bytes’ 64 bits

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 10.
Tabulate the range of value for fundamental data types?
Answer:

Data type

Range of value

char -127 to 128
int -32,768 to 32,767
float 3.4×10--38 to 3.4×1038 -1
double 1.7×10--308 to 1.7xl0308-1

Question 11.
What is modifier?
Answer:
Modifiers can be used to modify the memory allocation of any fundamental data type. They are also called as Qualifiers.

Question 12.
What are the modifiers in C++?
Answer:
There are four modifiers used in C++.
They are;

  1. signed
  2. unsigned
  3. long
  4. short

Question 13.
What are the two values associated with a variable?
Answer:
There are two values associated with a symbolic variable; they are R-value and L-va!ue.

  • R-value is data stored in a memory location.
  • L-value is the memory address in which the R-value is stored.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 14.
How memory addresses are represented?
Answer:
The memory addresses are in the form of Hexadecimal values.
Example: .
0x134e represents a memory address.

Question 15.
What are garbage or junk values?
Answer:
If we declare a variable without any initial value, the memory space allocated to that variable will be occupied with some unknown value. These unknown values are called as “Junk” or “Garbage” values.

Question 16.
What do you mean by dynamic ¡nitialization
Answer:
A variable can be initialized during the execution of a program. It is known as “Dynamic
initiaIization’
For example: .
int sum = num1+num2;
This initializes sum using the known values of num1 and num2 during the execution.

Question 17.
What is the difference between reference and pointer variable?
Answer:
A reference is an alias for another variable whereas a pointer holds the memory address of a variable.

Question 18.
What is the purpose of manipulators in C++?
Answer:
Manipulators are used to format the output of any C++ program. Manipulators are functions specifically designed to use with the insertion (<<) and extraction>>) operators.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 19.
What are the manipulators used in C++?
Answer:
Commonly used manipulators are:
endl, setw, setfill, setprecision and setf.

Question 20.
Write about the endl manipulator.
Answer:
endl (End the Line)
endl is used as a line feeder in C++. It can be used as an alternate to ‘\n’. In other words, endl inserts a new line and then makes the cursor to point to the beginning of the next line.
Example:
cout << “The value of num = ” << num <<endl;

Question 21.
What is the difference between endl and \n.
Answer:
There is a difference between endl and ‘\n’, even though they are performing similar tasks.
endl – Inserts a new line and flushes the buffer (Flush means – clean)
‘\n’ – Inserts only a new line.

Question 22.
Write about setw( ) manipulator.
Answer:
setw ( ) :
setw manipulator sets the width of the field assigned for the output. The field width determines the minimum number of characters to be written in output.
Syntax: ‘
setw(number of characters)
Example:
cout<< setw(25) << “Basic Pay :”<< setw(10)<< basic<< endl;

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 23.
What is the use of setfill( ) manipulator? setfill ():
Answer:
This manipulator is usually used after setw. If the presented value does not entirely fill the given width, then the specified character in the setfill argument is used for filling the empty fields.
Syntax:
setfill (character);
Example:
cout<<“\n H.R.A :”<<setw(10)< For example, if we assign 1200 to hra, setw accommodates 1200 in a field of width 10 from right to left and setfill fills p in the remaining 6 spaces that are in the beginning. The output will be, 0000001200.

Question 24.
What is the purpose of setprecision() manipulator?
Answer:
setprecision ( ):
This is used to display numbers with fractions in specific number of digits.
Syntax:
setprecision (number of digits);
Example:
float hra = 1200.123;
cout << setprecision (5) << hra;
In the above code, the given value 1200.123 will be displayed in 5 digits including fractions. So, the output will be 1200.1
setprecision ( ) prints the values from left to right. For the above code, first, it will take 4 digits and then prints one digit from fractional portion.

Question 25.
WhIch of the following statements are valid? Why? Also write their result
Inta; .
i) a – (014,3);
ii) a = (5,017)
iii) a = (3,018)
Answer:
i) a = (014,3); – Valid. A will hold 3. (The second value)
ii) a = (5,017) – Valid. 014 ¡s an octal constant. It will be converted into decimal and then stored in a. So, a will hold 15 as its value.
iii) a = (3,018) – Invalid. Because 8 is not an octal digit. (A number starts with 0 is considered as an octal)

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 26.
Which of the following statements are valid? Why? Also write their result.
inta;
i) a = (3,0xA);
ii) a = (5,0xCAFE)
iii) a = (OXCAFE,0XF)
iv) a = (0XCAFE,0XG)
Answer:
i) a = (3,0xA):
Valid. OxA is a hexadecimal constant. It will be converted into decimal and then stored in a. So, a will hold 10 as its value.
ii) a = (5,0xCAFE):
Valid. OxCAFE is a hexadecimal constant. It will be converted into decimal and then stored in a. So, a will hold 51966 as its value.
iii) a = (OXCAFE,0XF):
Valid. 0XF is a hexadecimal constant. It will be converted into decimal and then stored in a. So, a will hold 15 as its value.
iv) a= (OXCAFE,0XGAB):
Invalid. Because G is not a hexadecimal digit in OXGAB.\

Short Answer 3 Marks

Question 1.
LIstthedattypesinC++.
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 22

Question 2.
Write about character data type.
Answer:
Character data type accepts and returns all valid ASCII characters. Character data type is often said to be an integer type, since ail the characters are represented in memory by their associated ASCII Codes.
If a variable is declared as char, C++ allows storing either a character or an integer value.
Example:
char c=65;
char ch = ‘A’; Both statements will assign ‘A’ to c and ch.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Write note on double data type.
Answer:
double data type:
This data type is for double precision floating point numbers. The double is also used for handling floating point numbers. But, this type occupies double the space than float type. This means, more fractions can be accommodated in double than in float type.
The double is larger and slower than type float. The double is used in a similar way as that of float data type.

Question 4.
Compare memory allocation by Turbo C++ and Dev C++.
Answer:

Data type Memory size in bytes
Turbo C++ Dev C++
short 2 2
unsigned short 2 2
signed short 2 2
int 2 4
unsigned int 2 4
signed int 2 4
long 4 4
unsigned long 4 4
signed long 4 4
char 1 1
unsigned char 1 1
signed char 1 1
float 4 4
double 8 8
long double 10 12

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 5.
What is the purpose of number suffix in C++?
Answer:
There are different suffixes for integer and floating point numbers. Suffix can be used to assign the same value as a different type.
For example, if we want to store 45 in an int, long, unsigned int and unsigned long int, you can use suffix letter L or U (either case) with 45 i.e. 45L or 45U.
This type of declaration instructs the compiler to store the given values as long and unsigned.
‘F’ can be used for floating point values, example: 3.14F

Question 6.
How setprecision( ) is used to set the number of decimal places?
Answer:
setprecision can also be used to set the number of decimal places to be displayed. In order to do this task, we will have to set an ios flag within setf( ) manipulator.
This may be used in two forms:
(i) fixed and
(ii) scientific.
These two forms are used when the keywords fixed or scientific are appropriately used before the setprecision manipulator.
Example:
#include
#indude
using namespace std;
int main()
{
cout.setf(ios::fixed);
cout << setprecision(2)<<0.1;
}
In the above program, ios flag is set to fixed type; it prints the floating point number in fixed notation. So, the output will be, 0.10
cout.setf(ios: scientific); cout << setprecision(2) << 0.1;
In the above statements, ios flag is set to scientific type; it will print the floating point number in scientific notation. So, the output wiil
be, 1.00e-001

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Explain In Detail 5 Marks

Question 1.
What is an expression? Explain its types with suitable example.
Answer:
An expression is a combination of operators, constants and variables arranged as per the rules of C++.
An expression may consist of one or more operands, and zero or more operators to produce a value. In C++, there are seven types of expressions as given below.

Expression

Description

Example

1.Constant Expression Constant expression consist only constant values int num=100;
2. Integer Expression The combination of integer and character values and/or variables with simple arithmetic operators to produce integer results. sum = num1 +

num2;

avg=sum/5;

3. Float Expression The combination of floating point values and/or variables with simple                 arithmetic operators to produce floating point results. Area=3.14*r*r;
4. Relational Expression The combination of values and/or variables with relational operators to produce bool(true means 1 or false means 0) values as results. x>y;

a+b==c+d;

5. Logical Expression The combination of values and/or variables with Logical operators to produce bool values as results. (a>b)&& (c= = 10);
6. Bitwise Expression The combination of values and/or variables with Bitwise operators. x>>3;

a<<2;

7. Pointer Expression A Pointer is a variable that holds a memory address. Pointer                declaration statements. int *ptr;

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 2.
Explain type conversion in detail.
Answer:
The process of converting one fundamental type into another is called as “Type Conversion”. C++ provides two types of conversions.

  1. Implicit type conversion
  2. Explicit type conversion

Implicit type conversion:
An Implicit type conversion is a conversion performed by the compiler automatically. So, implicit conversion is also called as “Automatic
conversion “.
This type of conversion is applied usually whenever different data types are intermixed in an expression. If the type of the operands differs, the compiler converts one of them to match with the other, using the rule that the “smaller” type is
converted to the “wider” type, which is called as “Type Promotion”
Example:
#include<iostream>
using namespace std;
int main()
{
int a=6;
float b =3.14;
cout << a+b
}
In the above program, operand ‘a’ ¡s an mt type and ‘b’ is a float type. During the execution of the program, ‘mt’ is converted into a ‘float because a float is wider than mt.
Hence, the output of the above program will be: 9.14
Explicit type conversion:
C++ allows explicit conversion of variables
or expressions from one data type to another
specific data type by the programmer. It is called
as “type casting”.
Syntax:
(type-name) expression;
Where type-name is a valid C++ data type to which the conversion is to be performed.
Example:
#include<iostream>
using namespace std;
int main ()
{
float varf=78.685;
cout << (int) varf;
}
In the above program, variable varf is declared as a float with an initial value 78.685. The value of varf is explicitly converted to an int type in cout statement. Thus, the final output will be 78.
During explicit conversion, if we assign a value to a type with a greater range, it does not cause any problem. But, assigning a value of a larger type to a smaller type may result in loosing or loss of precision values.
Example:
#include <iostream>
using namespace std;
int main()
{
double varf= 178.25255685;
cout << (float) varf < < endl;
cout << (int) varf << endl;
}
Output
178.253
178

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Evaluate Yourself

Question 1.
What do you mean by fundamental data types?
Answer:
C++ provides a predefined set of data types for handling the data items. Such data types are known as fundamental or built-in data types.

Question 2.
The data type char is used to represent characters. Then why is it often termed as an integer type?
Answer:
Character data type is often said to be an integer type, since all the characters are represented in memory by their associated ASCII Codes.

Question 3.
What is the advantage of floating point numbers over integers?
Answer:
Advantages of using float data types.

  • They can represent values between the integers.
  • They can represent a much greater range of values.

Question 4.
The data type double is another floating point type. Then why is it treated as a distinct data type?
Answer:
The double is also used for handling floating point numbers. But, this type occupies double the space than float type. This means, more fractions can be accommodated in double than in float type.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 5.
What is the use of void data type?
Answer:
The literal meaning for void is ’empty space’. In C++, the void data type specifies an empty set of values. It is used as a return type for functions that do not return any value.

Evaluate Yourself

Question 1.
What is modifiers? What is the use of modifiers?
Answer:
Modifiers are used to modify the storing capacity of a fundamental data type except void type.
For example, int data type can store only two bytes of data. In reality, some integer data may have more length and may need more space in memory. In this situation, we should modify the memory space to accommodate large integer values. .
Modifiers can be used to modify the memory allocation of any fundamental data type. They are also called as Qualifiers.

Question 2.
What is wrong with the following C++ statement?
Answer:
long float x;
The modifier long must be associated with only double.

Question 3.
What Is variable? Why a variable called symbolic variable?
Answer:
Variables are user-defined names assigned to specific memory locations in which the values are stored. Variables are also identifiers; and hence,
the rules for naming the identifiers should be followed while naming a variable.
These are called as symbolic variables because these are named locations.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 4.
What do you mean by dynamic initialization of a variable? Give an example.
Answer:
A variable can be initialized during the execution of a program. It is known as Dynamic
initialization.
For example:
int sum = num1+num2;
This initializes sum using the known values of num1 and num2 during the execution.

Question 5.
What is wrong with the following statement?
Answer:
const int x ;
In the above statement x must be initialized. It is missing. It may rewritten as
const mt x =10;

Evaluate Yourself

Question 1.
What is meant by type conversion?
Answer:
The process of converting one fundamental type into another is called as “Type Conversion”. C++ provides two types of conversions.

  1. Implicit type conversion
  2. Explicit type conversion

Question 2.
How implicit conversion different from explicit conversion?
Answer:
An Implicit type conversion is a conversion performed by the compiler automatically. Implicit conversion is also called as “Automatic conversion”.
An explicit conversion of variables or expressions from one data type to another specific data type is by the programmer. It is called as “type casting”.

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
What is difference between endl and \n?
Answer:
There is a difference between endl and ‘\n’ even though they are performing similar tasks.
endl – Inserts a new line and flushes the buffer (Flush means – dean)
‘\n’ – Inserts only a new line.

Question 4.
What is the use of references?
Answer:
A reference provides an alias for a previously defined variable. Declaration of a reference consists of base type and an & (ampersand) symbol; reference variable name is assigned the value of a previously declared variable.
Syntax:
<&reference_variable>=

Question 5.
What is the use of setprecision ( )?
Answer:
setprecision ( ):
This is used to display numbers with fractions in specific number of digits.
Syntax:
setprecision (number of digits);
Example:
float hra = 1200.123;
cout << setprecision (5) << hra;
The given value 1200.123 will be displayed in 5 digits including fractions. The output will be 1200.1

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Hands On Practice

Question 1.
Write C++ programs to interchange the values of two variables.
a) Using with third variable C++PROGRAM:
#include
using namespace std;
int main()
{
intx,y,t;
cout<<“\nEnter two numbers”; cin>>x>>y;
cout<<“\nValues before interchage
x=”<<x<<“\ty=”<<y;
t=x;
x=y;
y=t;
cout<<“\nValues after interchage x=”<<x<<“\ty=”<<y;
return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 23

b) Without using third variable C++ PI OGRAM:
#include
using namespace std;
int main ()
{
int x,y;
cout<<“\nEnter two numbers”; cin>>x>>y;
cout<<“\nValues before interchage
x=”<<x<<“\ty=”<<y;
//interchage process without using third
variable
x=x+y;
y=x-y;
x=x-y;
cout<<“\nValues after interchage x=”<<x<<“\ty=”<<y;
return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 24

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 2.
Write C++ programs to do the following:
a) To find the perimeter and area of a quadrant.
C++ PROGRAM:
#include
using namespace std;
int mainQ()
{
float ppm,area;
cout<<“\nEnter radius cin>>r;
area = 3.14 * r * r / 4;
pm = 3.14 * r / 2;
cout<<“\nQuadrant Area = “<<area;
cout<<“\n\nQuadrant Perimeter = “<<pm;
return 0;
}
OutPut
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 25

b) To find the area of triangle.
C++ PROGRAM 1:
(Area of triangle when b and h values are known)
#include
using namespace std;
int main()
{
float b,h,area;
cout<<“\nEnter b and h value of triangle cin>>b>>h;
// Area of triangle when b and h values are known
area = b * h / 2;
cout<<“‘\nBase value = “<<b<<“Height = “<<h<<“Triangle Area = “<<area;
return 0;
}
OutPut
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 26
C++ PROGRAM 2:
(Area of triangle when three sides are known)
#inciude
#include using namespace std;
int main()
{
float a,b,c,area,s;
cout<<“\nEnter three sides of triangle cin>>a>>b>>c;
//Area of triangle when three sides are known
s = (a+b+c)/2;
area = sqrt(s*(s-a)*(s-b)*(s-c));
cout<<“\n Sidel value = “<<a<<“Side2
value = “<<b<<” Side3 value =”<<c;
cout<<“\nTriangie Area = “<<area;
return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 27
c. To convert the temperature from Celsius to Fahrenheit.
C++ PROGRAM:
//Convertion of the temperature from Celsius to Fahrenheit
#include< iostream>
using namespace std;
int main()
{
float c,f;
cout<<”\nEnter Celsius value “; cin>>c;
f=9*c/5+32;
cout«”\nTemperature in Celsius = “<<C;
cout < <“\nTemperature in Fahrenheit =
“<<f;
return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 28

Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++

Question 3.
Write a C++ to find the total and percentage of marks you secured from 10th Standard Public Exam. Display all the marks one-by- one along with total and percentage. Apply formatting functions.
C++ PROGRAM:
#include
#include
using namespace std;
int rnain()
{
int tarn,enm,mam,som,scm,total,avg; char name[30];
cout<<“\nEnter name of the student cin>>name;
cout<<“\nEnterTamil mark”; cin>>tam;
cout<<“\nEnter English mark”; cin>>enm;
cout<<“\nEnter Maths mark “; cin>>mam;
cout<<“\nEnter Science mark “; cin>>scm;
cout<<“\nEnter Social Science mark”; cin>>som; .
total = tarn + enm + mam + scm + som; avg = total / 5;
cout<<“\n\t\tlOth Standard Public Exam Mark”<<end!<<endl;
cout<<setw(30)<<“Name of the student «name<<endk<endl;
cout<<setw(30)<<setfill(“)<<“Tamil mark
< <setw(3)< <setfill(‘0’)< <tam< cout<<setw(30)<<setfillC ‘)<<“English
mark “<setw(3)<<setfill(‘0’)<<enm< cout<<setw(30)<<setfillC ‘)<<“Maths mark
< <setw(3)< <setfillCO’)< <mam< <endl< <endl;
cout<<setw(30)<<setfillC ‘)<<“Science
mark :”<<setw(3)<<setfill(‘0’)<<scm< cout< < setw(3G) < < setfi 11C ‘) < < “Soda I
Science mark :”<<setw(3)<<setfillC0’)< cout<<setw(30)<<setfillC ‘)«’Total Marks <<setw(3)<<setfillC0’)<<totak<endk<endl;
cout<<setw(30)«setfillC ‘)«”Average mark:”
<<setw(3)<<setfill(‘0’)< return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 9 Introduction to C++ 29

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 10 Flow of Control Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 10 Flow of Control

11th Computer Science Guide Flow of Control Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer

Question 1.
What is the alternate name of null statement?
a) No statement
b) Empty statement
c) Void statement
d) Zero statement
Answer:
b) Empty statement

Question 2.
In C++, the group of statements should enclosed within:
a) { }
b) [ ]
c)()
d)<>
Answer:
a) { }

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 3.
The set of statements that are executed again and again in iteration is called as:
a) condition
b) loop
c) statement
d) body of loop
Answer:
d) body of loop

Question 4.
The multi way branching statement:
a) if
b) if… else
c) switch
d) for
Answer:
c) switch

Question 5.
How many types of iteration statements?
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 6.
How many times the following loop will execute?
for (int i=0; i<10; i++)
a) 0
b) 10
c) 9
d) 11
Answer:
b) 10

Question 7.
Which of the following is the exit control loop?
a) for
b) while
c) do…while
d) if…else
Answer:
c) do…while

Question 8.
Identify the odd one from the keywords of jump statements:
a) break
b) switch
c) go to
d) continue
Answer:
b) switch

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 9.
Which of the following is the exit control loop?
a) do-while
b) for
c) while
d) if-else
Answer:
a) do-while

Question 10.
A loop that contains another loop inside its body:
a) Nested loop
b) Inner loop
c) In line loop
d) Nesting of loop
Answer:
a) Nested loop

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Part – II

Short Answers

Question 1.
What is a null statement and compound statement?
Answer:
null statement; The “null or empty statement” is a statement containing only a semicolon. It takes the flowing form: ;
// it is a null statement
compound Statement:
In C++, a group of statements enclosed by pair of braces {} is called as a compound statement or a block.
The general format of compound statement is:
{
statement1;
statement2;
statement3;
}

Question 2.
What is a selection statement? Write its types?
Answer:
The selection statement means the statements are executed depends upon a condition. If a condition is true, a true block (a set of statements) is executed, otherwise, a false block is executed. This statement is also called a decision statement or selection statement because it helps in making a decision about which set of statements are to be executed.
Types:

  1. Two-way branching
  2. Multiway branching

Question 3.
Correct the following code segment:
Answer:
if (x=1)
p= 100;
else
P = 10;
Correct code: (Equal to operator is wrongly given)
if (x==1)
p= 100;
else
p = 10;

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 4.
What will be the output of the following code:
int year;
cin >> year;
if (year % 100 == 0)
if (year % 400 == 0)
cout << “Leap”;
else
cout << “Not Leap year”;

If the input is given is
(i) 2000
(ii) 2003
(iii) 2010
Answer:
i) 2000 AS INPUT
Output
Leap
ii) 2003AS INPUT
Output
Not Leap year
iii) 2010 AS INPUT
Output
Not Leap year

Question 5.
What is the output of the following code?
for (int i=2; i<=10 ; i+=2)
cout << i;
Answer:
Output
2 4 6 8 10

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 6.
Write a for loop that displays the number from 21 to 30.
Answer:
# include
using namespace std;
int main()
{
for (int = 21; i < 31; i++)
cout << “value of i:” << i << endl;
return 0;
}
Output:
Value of i: 21
Value of i: 22
Value of i: 23
Value of i: 24
Value of i: 25
Value of i: 26
Value of i: 27
Value of i: 28
Value of i: 29
Value of i: 30

Question 7.
Write a while loop that displays numbers 2,4, 6, 8…….20.
Answer:
While loop to
i = 2;
while(i<=20)
{
cout<< i <<“,”;
i = i+2;
}

Question 8.
Compare an if and a ? : operator.
Answer:
The if statement evaluates a condition, if the condition is true then a true-block is executed, otherwise false-block will be executed! Block may consists of one or more statements. The conditional operator (or Ternary operator) . is an alternative for ‘if else statement’. Here if the condition is true one statement otherwise another statement will be executed.

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Part – III

Short Answers

Question 1.
Convert the following if-else to a single conditional statement:
Answer:
if (x >= 10)
a = m + 5;
else
a = m;
a = (x >= 10)? m + 5 : m ;

Question 2.
Rewrite the following code so that it is functional:
Answer:
v = 5;
do;
{
total += v;
cout << total;
while v <=10
CORRECT CODE:
int v = 5;
do
{
total += v;
v++;
} while (v <= 10); cout << total;

Question 3.
Write a C++ program to print the multiplication table of a given number.
Answer:

# include
using namespace std;
int main ()
{

int num;
cout << “Enter Number to find its multiplication table”; cin >> num;
for (int a = 1; a < = 10; a++)
{
cout << num << “*” << a << “=” << num*a << endl;
}
return( );
}

Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 1

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 4.
Write the syntax and purpose of switch statement.
Answer:
The syntax of the switch statement is:
switch(expression)
{
case constant 1:
statement(s);
break;
case constant 2:
statement(s);
break;
.
.
.
.
default:
statement(s);
}
Purpose of switch statement:
The switch statement is a multi-way branch statement. It provides an easy way to dispatch execution to different parts of code based on the value of the expression.

Question 5.
Write a short program to print following series: a) 1 4 7 10…… 40
Answer:
PROGRAM
using namespace std;
#inciude
int main()
{
for(int i=1;i<=40;i=i+3)
cout<<i<<” “;
return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 2

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Part – IV

Explain In Detail

Question 1.
Explain the control statement with a suitable example.
Answer:
Control statements are statements that alter the sequence of flow of instructions.
In a program, statements may be executed sequentially, selectively, or iteratively. Every programming language provides statements to support sequence, selection (branching), and iteration.
Selection statement:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 3
The selection statement means the statement(s) are executed depends upon a condition. If a condition is true, a true block ¡s executed otherwise a false block is executed. This statement is also called a decision statement or selection statement because it helps in making decisions about which set of statements are to be executed.
Iteration statement:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 4
The iteration statement is a set of the statement are repetitively executed depends upon conditions. If a condition evaluates to true, the set of statements (true block) is executed again and again. As soon as the condition becomes false, the repetition stops. This is also known as a looping statement or iteration statement.

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 2.
What entry control loop? Explain any one of the entry control loops with a suitable example.
Answer:
In an entry-controlled loop, the test- expression is evaluated before entering into a loop, while and for statements are called as entry controlled loop.
Working of while loop:
A while loop is a control flow statement that allows the loop statements to be executed as long as the condition is true. The while loop ¡s an entry-controlled loop because the test-expression is evaluated before the entering into a loop.
The while loop syntax is:
while (Test expression )
{
Body of the loop;
}
Statement-x;
In a while loop, the test expression is evaluated and if the test expression result is true, then the body of the loop is executed and again the control is transferred to the while loop. When the test expression result is false the control is transferred to statement-x.
The control flow and flow chart of the while loop is shown below.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 5
Example:
#include <iostream>
using namespace std;
int main ()
{
int i=1,sum=0;
while(i<=10)
{
sum=sum+i;
i++;
}
cout<<“The sum of 1 to 10 is “<<sum; return 0;
}
Output
The sum of 1 to 10 is 55

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 3.
Write a program to find the LCM and GCD of two numbers.
Answer:
PROGRAM:
using namespace std;
#include <iostream>
int main()
{
int n1, n2, i, gcd;
cout<<“Enter two integers:”;
cin>>n1>>n2;
for(i=1; i <= n1 && i <= n2; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0)
gcd = i;
}
int lcm = n1*n2 /gcd;
cout<<“\nG.C.M.OF”<<nl<<“and”<<n2<<”
is”<<gcd;
cout< < “\nL.C. M. OF “< < n 1 < < ” and ” < < n2< < ” is “<<lcm; return 0;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 6
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 7

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 4.
Write programs to find the sum of the following series:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 8
Answer:
PROGRAM
using namespace std;
#include
int main()
{
int i,x,n,f=1,sign=1;
float sum=0,t;
cout<<“\nEnter N value”;
cin>>n;
cout<<“\nEnter x value …”;
cin>>x;
t=x;
for(i=1;i<=n;i++)
{
f = f * i;
sum = sum + sign * t/f;
t = t * x;
sign = -sign;
}
cout<<”SUM OF THE SERIES = “<<sum;
return O;
}
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 9
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 10
PROGRAM
using namespace std;
#include
int main()
{
int i,x,n;
float sum=0,t;
cout<<“\nEnter N value”;
cin>>n;
cout<<“\nEnter x value …”;
cin>>x;
t=x;
for(i=l;i<=n;i++)
{
sum = sum + t/i;
t = t * x;
}
cout<<“SUM OF THE SERIES = “<<sum;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 11

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 5.
Write a program to find sum of the series s=1 +x +x2 + ……..+xn
Answer:
PROGRAM
using namespace std;
#include
int main()
{
int sum=1,x,i,t,n;
cout<<“\nEnter N value”;
cin>>n;
cout<<“\nEnter x value … “;
cin>>x;
t=x;
for(i=l;i<=n;i++)
{
sum = sum + t;
t = t * x;
}
cout<<“SUM = “<<sum;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 12

11th Computer Science Guide Flow of Control Additional Questions and Answers

Choose The Correct Answer 1 Mark

Question 1.
The empty statement is otherwise called as ………………..
(a) Control statement
(b) Zero statement
(c) Null statement
(d) Block statement
Answer:
(c) Null statement

Question 2.
The basics of control structure are……………. statement.
a) Selection
b) Iteration
c) Jump
d) All the above
Answer:
d) All the above

Question 3.
Iteration statement is called as ………………..
(a) Null statement
(b) Block statement
(c) Selection statement
(d) Looping statement
Answer:
(d) Looping statement

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 4.
In a program, the action may be ……………..
a) Variable declarations
b) Expression evaluations
c) Assignment operations
d) All the above
Answer:
d) All the above

Question 5.
……………….. is a multi-path decision-making statement.
(a) if
(b) if-else
(c) else – if
(d) if-else ladder
Answer:
(d) if-else ladder

Question 6.
The ……………. is a statement containing only a semicolon.
a) Null
b) Empty statement
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 7.
……………….. is more efficient than the if-else statement.
(a) Control statement
(b) Switch statement
(c) Empty statement
(d) Null statement
Answer:
(b) Switch statement

Question 8.
C++ allows a group of statements enclosed by pair of ……………..braces.
a) ()
b) { }
c) [ ]
d) < >
Answer:
b) { }

Question 9.
C++ supports types of iteration statements.
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(a) 3

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 10.
……………. statements alter the sequence of flow of instructions.
a) Control
b) Null
c) Compound
d) None of these
Answer:
a) Control

Question 11.
……………….. is used to transfer the control from one place to another place without any condition in a program.
(a) Break statement
(b) Continue statement
(c) goto statement
(d) All the above
Answer:
(c) goto statement

Question 12.
The ……………. statement are the statements, that are executed one after another only once from top to bottom.
a) Sequential
b) Selective
c) Iterative
d) None of these
Answer:
a) Sequential

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 13.
…………….. statements do not alter the flow of execution.
a) Sequential
b) Selective
c) Iterative
d) None of these
Answer:
a) Sequential

Question 14.
…………….. statements always end with a semicolon (;).
a) Sequential
b) Selective
c) Iterative
d) None of these
Answer:
a) Sequential

Question 15.
The ……………. statement means the statement (s) are executed depends upon a condition.
a) Sequential
b) Selective
c) Iterative
d) None of these
Answer:
b) Selective

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 16.
If a condition is true, a true block (a set of statements) is executed otherwise a false block is executed is called …………… statement.
a) Decision
b) Selection
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 17.
The ……………. statement is a set of the statement are repetitively executed depends upon conditions.
a) Sequential
b) Selective
c) Iterative
d) None of these
Answer:
c) Iterative

Question 18.
The condition on which the execution or exit from the loop is called ………………
a) Exit-condition
b) Test-condition
c) Either A or B
d) None of these
Answer:
c) Either A or B

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 19.
In C++, any non-zero is treated as …………… including negative numbers
a) False
b) True
c) Complement
d) None of these
Answer:
b) True

Question 20.
In C++, zero is treated as …………..
a) False
b) True
c) Complement
d) None of these
Answer:
a) False

Question 21.
Decisions in C++ are made using …………….. statement.
a) if..else
b) switch., case
c) Either A or B
d) None of these
Answer:
c) Either A or B

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 22.
……………. statement which chooses between two alternatives.
a) if..else
b) switch., case
c) Either A or B
d) None of these
Answer:
a) if..else

Question 23.
………………. creates branches for multiple alternatives sections of code, depending on the value of a single variable.
a) if..else
b) switch., case
c) Either A or B
d) None of these
Answer:
b) switch., case

Question 24.
…………. is a two-way branching statement.
a) if..else
b) switch., case
c) Either A or B
d) None of these
Answer:
a) if..else

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 25.
……………….. is a multiple branching statement.
a) if..else
b) switch., case
c) Either A or B
d) None of these
Answer:
b) switch., case

Question 26.
An if statement contains another if the statement is called …………….
a) Nested if
b) Compound if
c) Block if
d) Group if
Answer:
a) Nested if

Question 27.
The nested if can have ……………. forms.
a) four
b) two
c) three
d) five
Answer:
c) three

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 28.
……………. is nested if form.
a) If nested inside if part
b) If nested inside both if part and else part
c) If nested inside else part
d) All the above
Answer:
d) All the above

Question 29.
The ………….. is a multi-path decision-making statement.
a) if-else ladder
b) Simple if
c) if… else
d) None of these
Answer:
a) if-else ladder

Question 30.
In ……………..type of statement ‘if’ is followed by one or more else if statements and finally end with an else statement.
a) if-else ladder
b) Simple if
c) if… else
d) None of these
Answer:
a) if-else ladder

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 31.
The……………. operator is an alternative for the if-else statement.
a) Conditional
b) Ternary
c) Continue
d) Either A or B
Answer:
d) Either A or B

Question 32.
The conditional operator consists of …………….. symbols.
a) two
b) three
c) four
d) five
Answer:
a) two

Question 33.
The conditional operator takes …………….. arguments.
a) two
b) three
c) four
d) five
Answer:
b) three

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 34.
The switch statement is a …………… statement.
a) Multi-way branch
b) Two-way branch
c) Jump
d) None of these
Answer:
a) Multi-way branch

Question 35.
………….. statement provides an easy way to dispatch execution to different parts of code based on the value of the expression.
a) if..else
b) switch
c) goto
d) None of these
Answer:
b) switch

Question 36.
The ………………statement replaces multiple if-else sequences.
a) if..else
b) switch
c) go to
d) None of these
Answer:
b) switch

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 37.
The expression provided in the switch should result in a…………… value.
a) Constant
b) Variant
c) Null
d) None of these
Answer:
a) Constant

Question 38.
In switch statement …………… statement is optional.
a) default
b) break
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 39.
The …………… statement is used inside the switch to terminate a statement sequence.
a) default
b) break
c) continue
d) None of these
Answer:
b) break

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 40.
In a switch statement, when a ………….. statement is reached, the flow of control jumps to the next line following the switch statement.
a) go to
b) break
c) continue
d) None of these
Answer:
b) break

Question 41.
……………. statement checks only for equality.
a) go to
b) if,.else
c) continue
d) switch
Answer:
d) switch

Question 42.
………….. statement checks for equality as well as for logical expression.
a) go to
b) if..else
c) continue
d) switch
Answer:
b) if..else

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 43.
………….. statement uses a single expression for multiple choices.
a) go to
b) if..else
c) continue
d) switch
Answer:
d) switch

Question 44.
A(n) ………….. statement uses multiple statements for multiple choices.
a) go to
b) if..else
c) continue
d) switch
Answer:
b) if..else

Question 45.
The ……………. statement evaluates integer, character, pointer or floating-point type or Boolean type.
a) go to
b) if..else
c) continue
d) switch
Answer:
b) if..else

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 46.
……….. statement evaluates only character or an integer data type.
a) go to
b) if..else
c) continue
d) switch
Answer:
d) switch

Question 47.
If the expression inside the switch statement turns out to be false then statements are executed.
a) default
b) break
c) else block
d) None of these
Answer:
a) default

Question 48.
If the expression inside if turns out to be false, statement inside ………….. will be executed.
a) default
b) true block
c) else block
d) None of these
Answer:
c) else block

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 49.
The ………….. statement is more flexible.
a) switch
b) if
c) continue
d) None of these
Answer:
b) if

Question 50.
The …………… statement is more efficient than the if-else statement.
a) switch
b) go to
c) continue
d) None of these
Answer:
a) switch

Question 51.
Which is true related to the switch statement?
a) A switch statement can only work for the quality of comparisons.
b) No two case labels in the same switch can have identical values.
c) If character constants are used in the switch statement, they are automatically converted to their equivalent ASCII codes.
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 52.
When a switch is a part of the statement sequence of another switch, then it is called as …………….. switch statement.
a) Iterative
b) Sequential
c) Nested
d) None of these
Answer:
c) Nested

Question 53.
A(n) ………….. is a sequence of one or more statements that are repeatedly executed until a condition is satisfied.
a) Iteration
b) Looping
c) Iteration or Looping
d) None of these
Answer:
c) Iteration or Looping

Question 54.
………….. is used to reduce the length of code, to reduce time, to execute the program and takes less memory space.
a) Iteration
b) Looping
c) Iteration or Looping
d) None of these
Answer:
c) Iteration or Looping

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 55.
C++supports ………….. types of iteration statements.
a) five
b) four
c) three
d) two
Answer:
c) three

Question 56.
C++ supports ………….. type of iteration statement.
a) for
b) while
c) do..while
d) All the above
Answer:
d) All the above

Question 57.
Every loop has …………….. elements.
a) five
b) four
c) three
d) two
Answer:
b) four

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 58.
Every loop has ………….. element.
a) Initialization expression
b) Test expression / Update expression
c) The body of the loop
d) All the above
Answer:
d) All the above

Question 59.
The control variable(s) must be initialized …………. the control enters into a loop.
a) after
b) before
c) Either A or B
d) None of these
Answer:
b) before

Question 60.
Whose value decides whether the loop-body will be executed or not?
a) Test expression
b) Update expression
c) Initialization expression
d) None of these
Answer:
a) Test expression

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 61.
In an ……….. loop, the test-expression is evaluated before entering into a loop,
a) Entry-controlled
b) Exit-controlled
c) Both A and B
d) None of these
Answer:
a) Entry-controlled

Question 62.
In an ………….. loop, the test-expression is evaluated before exiting from the loop.
a) Entry-controlled
b) Exit-controlled
c) Both A and B
d) None of these
Answer:
b) Exit-controlled

Question 63.
……………. is used to change the value of the loop variable.
a) Test expression
b) Update expression
c) Initialization expression
d) None of these
Answer:
b) Update expression

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 64.
……………. statement is executed at the end of the loop after the body of the loop is executed.
a) Test expression
b) Update expression
c) Initialization expression
d) None of these
Answer:
b) Update expression

Question 65.
In an …………. loop, first, the test expression is evaluated and if it is nonzero, the body of the loop is executed otherwise the loop is terminated.
a) Entry-controlled
b) Exit-controlled
c) Both A and B
d) None of these
Answer:
a) Entry-controlled

Question 66.
In an …………. loop, the body of the loop is executed first then the test-expression is evaluated.
a) Entry-controlled
b) Exit-controlled
c) Both A and B
d) None of these
Answer:
b) Exit-controlled

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 67.
………….. loop statement contains three different statements.
a) for
b) while
c) do..while
d) All the above
Answer:
a) for

Question 68.
The for statement contains ……………. statement
a) Test expression
b) Update expression
c) Initialization expression
d) All the above
Answer:
d) All the above

Question 69.
The for statement sections are separated by ………………
a) Semicolons
b) Colon
c) Comma
d) None of these
Answer:
a) Semicolons

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 70.
The initialization part of the loop is used to initialize variables or declare a variable which is executed …………… time(s).
a) Loop repeated
b) Only one
c) two
d) None of these
Answer:
b) Only one

Question 71.
How many times the following loop will be executed?
for (i=0; i <= 10; i++);
a) 10
b) 9
c) 11
c) None of these
Answer:
c) 11

Question 72.
If the body of the loop contains a ……………. statement(s), need not use curly braces.
a) Single
b) More than one
c) Either A or B
d) None of these
Answer:
a) Single

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 73.
In for loop, multiple initializations and multiple update expressions are separated by ……………
a) Semicolons
b) Colons
c) Commas
d) None of these
Answer:
c) Commas

Question 74.
In a for loop, which expression is optional?
a) Test expression
b) Update expression
c) Initialization expression
d) All the above
Answer:
d) All the above

Question 75.
What will be formed if the following loop is executed?
for( i=0; ++i);
a) Finite loop
b) Indefinite loop
c) No iteration
d) None of these
Answer:
b) Indefinite loop

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 76.
A loop that has no statement in its body is called an ……………
a) Empty loop
b) Indefinite loop
c) Finite loop
d) None of these
Answer:
a) Empty loop

Question 77.
Identify the empty loop from the following.
a) for( i+0 ; i<=5; +=i);
b) for( i+0 ; i<=5; +=i); {cout<<“C++”;>
c) for( i+0 ; i<=5; +=i) { }
d) All the above
Answer:
d) All the above

Question 78.
A ………….. loop is useful for pausing the program for some time.
a) Time delay
b) Infinite loop
c) Finite loop
d) None of these
Answer:
a) Time delay

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 79.
……………. is a time delay loop.
a) for(i=l; i<=100; i++);
b) int i=l;
while( ++i < 10)
{
}
c) int i=l;
do
{
} while( ++i < 10);
d) All the above
Answer:
d) All the above

Question 80.
How many times the following loop will be executed?
int i=l;
while( ++i < 10)
cout<< “The value of i is “<<i;
a) 10
b) 9
c) 8
d) Infinite times
Answer:
c) 8

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 81.
How many times the following loop will be executed?
int i=l;
while( i++< 10)
cout<< “The value of i is “<<i;
a) 10
b) 9
c) 8
d) Infinite times
Answer:
b) 9

Question 82.
In ………… loop, the body of the loop will be executed atleast once.
a) while
b) do..while
c) for
d) All the above
Answer:
b) do..while

Question 83.
In ……………….. loop, the body of the loop is executed at least once, even when the condition evaluates false during the first iteration.
a) while
b) do..while
c) for
d) All the above
Answer:
b) do..while

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 84.
What will be the output of the following program?
using namespace std;
int main ()
{
int n = 10;
do
{
cout<<n<<“”; n-; }while (n>0);
}
a) 109 8 7 6 5 43 2 1
b) 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
c) 1 2 3 4 5 6 7 8 9 10
d) 10, 9, 8, 7, 6, 5, 4, 3, 2, 1,0
Answer:
b) 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Question 85.
……………. statements are used to interrupt the normal flow of the program.
a) Jump
b) Loop
c) Decision making
d) None of these
Answer:
a) Jump

Question 86.
……………. is a jump statement.
a) go to statement
b) break statement
c) continue statement
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 87.
The ………….. statement is a control statement which is used to transfer the control from one place to another place without any condition in a program.
a) go to
b) break
c) continue
d) All the above
Answer:
a) go to

Question 88.
A …………… statement is a jump statement which terminates the execution of the loop and the control is transferred to resume normal execution after the body of the loop.
a) go to
b) break
c) continue
d) All the above
Answer:
b) break

Question 89.
…………….. statement forces the loop to continue or execute the next iteration,
a) go to
b) break
c) continue
d) All the above statement is executed in
Answer:
c) continue

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 90.
When the …………….statement is executed in the loop, the code inside the loop following the …………… statement will be skipped and the next iteration of the loop will begin.
a) go to
b) break
c) continue
d) All the above
Answer:
c) continue

Question 91.
……………. statement breaks the iteration.
a) continue
b) break
c) Both A and B
d) None of these
Answer:
b) break

Question 92.
…………… statement skips the iteration.
a) continue
b) break
c) Both A and B
d) None of these
Answer:
a) continue

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 93.
…………….. statement is used with loops as well as switch case.
a) continue
b) break
c) Both A and B
d) None of these
Answer:
b) break

Question 94.
………….. statement is only used in loops.
a) continue
b) break
c) Both A and B
d) None of these
Answer:
a) continue

Very Short Answers (2 Marks)

Question 1.
What is an if statement? Write its syntax.
Answer:
The if statement evaluates a condition, if the condition is true then a true – block (a statement or set of statements) is executed, otherwise the true – block is skipped. The general syntax of the if the statement is:
if (expression)
true – block;
statement – x;

Question 2.
What are the basic control structures?
Answer:
The basics of control structures are:

  • Selection statement
  • Iteration statement
  • Jump statement

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 3.
What is a computer program?
Answer:
A computer program is a set of statements or instructions to perform a specific task.

Question 4.
What is nested if? Mention its types.
Answer:
An if statement contains another if the statement is called nested if. The nested can have, one of the following three forms.

  1. If nested inside if part
  2. If nested inside else part
  3. If nested inside both if part and else part

Question 5.
What are control statements?
Answer:
Control statements are statements that alter the sequence of flow of instructions.
In a program, statements may be executed sequentially, selectively, or iteratively. Every programming language provides statements to support sequence, selection (branching), and iteration.

Question 6.
What is the exit condition of a loop?
Answer:
The condition on which the execution or exit from the loop is called exit-condition or test-condition.

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 7.
What is an iteration or loop?
Answer:
An iteration or loop is a sequence of one or more statements that are repeatedly executed until a condition is satisfied. These statements are also called control flow statements.

Question 8.
What are the advantages of a loop or an iteration?
Answer:
It is used to reduce the length of code, to reduce time, to execute the program, and takes less memory space

Question 9.
What are loop elements?
Answer:
Every loop has four elements that are used for different purposes. These elements are

  1. Initialization expression
  2. Test expression
  3. Update expression
  4. The body of the loop

Question 10.
What are the parts of a loop?
Answer:
Every loop has four elements that are used for different purposes. These elements are:

  • Initialization expression
  • Test expression
  • Update expression
  • The body of the loop

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 11.
Write a note on the declaration of a variable in a for a loop.
Answer:
Declaration of a variable in a for loop:
In C++, the variables can also be declared within a for loop.
For instance,
for(int i=0; i<=5; ++i)
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 13
A variable declared inside the block of main() can be accessed anywhere inside main() i.e., the scope of variable in main().

Question 12.
Write a program to display numbers from 1 to 10 except 6 using a continue statement.
Answer:
C++ program to display numbers from 1 to 10 except 6 using continue statement
#include
using namespace std;
int main()
{
if (i = 6)
continue;
else
cout << i << ” ”
}
return 0;
}
Output:
1 2 3 4 5 7 8 9 10

Question 13.
Write about the go-to statement.
Answer:
go to the statement: The go-to statement is a control statement which is used to transfer the control from one place to another place without any condition in a program.
The syntax of the go-to statement is;
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 14
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 16
In the syntax above, the label is an identifier. When go-to label; is encountered, the control of the program jumps to label: and executes the code below ¡t.

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 14.
Write a note on the break statement.
Answer:
break statement: A break statement is a jump statement which terminates the execution of the loop and the control is transferred to resume normal execution after the body of the loop. The following figure shows the working of break statement with looping statements;
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 15

Question 15.
Write a note on the continue statement.
Answer:
continue statement:
The continue statement works quite similar to the break statement. Instead of terminating the loop, the continue statement forces the loop to continue or execute the next iteration. When the continue statement is executed in the loop, the code inside the loop following the continue statement will be skipped and the next iteration of the loop will begin.
The following figure describes the working flow of the continue statement
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 17

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Short Answers 3 Marks

Question 1.
Write a note on the sequence statement.
Answer:
sequence statement:
The sequential statement is the statements, that are executed one after another only once from top to bottom. These statements do not alter the flow of execution. These statements are called sequential flow statements. They always end with a semicolon (;).
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 18

Question 2.
What is a selection statement? Mention its types.
Answer:
Selection statement: The selection statement means the statement (s) are executed depends upon a condition. If a condition is true, a true block is executed otherwise a false block is executed. This statement is also called a decision statement or selection statement because it helps in making a decision about which set of statements are to be executed.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 19

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 3.
Explain the Iteration statement.
Answer:
Iteration statement: The iteration statement is a set of the statement are repetitively executed depends upon conditions. If a condition evaluates to true, the set of statements (true block) is executed again and again. As soon as the condition becomes false, the repetition stops. This is also known as a looping statement or iteration statement.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 20

Question 4.
Explain if statement with syntax and an example.
Answer:
The if statement evaluates a condition, if the condition is true then a true-block is executed, otherwise, the true-block is skipped.
Syntax: if (expression) true-block; statement-x;
In the above syntax, if is a keyword that should contain expression or condition which is enclosed within parentheses. Working of if statement: If the expression is true then the true-block is executed and followed by statement-x is also executed, otherwise, the control passes to statement-x. It is implemented in the following flowchart.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 21
Example 1:
if ( qty> 500)
dis=10;
Example 2:
if(age>=18)
cout<< “\n You are eligible for voting ….”;

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 5.
Explain if..else statement with syntax and an example.
Answer:
The syntax of the if..else statement:
if (expression)
{
True-block;
}
else
{
False-block;
}
Statement-x
Working of if..else statement:
In if..else statement, first the expression or condition is evaluated as either true or false. If the result is true, then the statements inside the true-block are executed and the false-block is skipped. If the result is false, then the statement inside the false-block is executed i.e., the true-block is skipped. It is implemented in the following flowchart.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 22
Example:
if(num%2==0)
cout<< “\n The given number is Even”;
else
cout<< “\n The given number is Odd”;

Question 6.
What is a conditional or ternary operator? Explain.
Answer:
The conditional operator (or Ternary operator) is an alternative for the ‘if-else statement’. The conditional operator consists of two symbols (?:). It takes three arguments. The control flow of the conditional operator is shown below.
The syntax of the conditional operator is:
expression 1? expression 2: expression 3
In the above syntax, expression 1 is a condition which is evaluated, if the condition is true (Non-zero), then the control is transferred to expression 2, otherwise, the control passes to expression 3.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 23
Example:
#include
using namespace std;
int main()
{
int a, b, largest;
cout << “\n Enter any two numbers:
cin >> a >> b;
largest = (a>b)? a : b;
cout << “\n Largest number: ” << largest;
return 0;
}
Output
Enter any two numbers: 12 98
Largest number: 98

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 7.
Explain the nested switch statement.
Answer:
Nested switch:
When a switch is a part of the statement sequence of another switch, then it is called a nested switch statement. The inner switch and the outer switch constant may or may not be the same.
The syntax of the nested switch statement is: switch (expression)
{
case constant 1:
statement(s);
break;
switch(expression)
{
case constant 1:
statement(s);
break;
case constant 2:
statement(s);
break;
.
.
.
default: statement(s);
}
case constant 2:
statement(s);
break;
.
.
.
default:
statement(s);
}
Example:
#include
using namespace std;
int main()
{
int a = 8;
cout<<“The Number is : ” < switch (a)
{
case 0 :
cout<<“The number is zero” <<endl;
break;
default:
cout<<“The number is a non-zero
integer” <<endl;
int b = a % 2;
switch (b)
{
case 0:
cout<<“The number is even” <<endl;
break;
case 1:
cout<<“The number is odd” <<endl;
break;
}
}
return 0;
}
Output
The Number is: 8
The number is a non-zero integer
The number is even

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 8.
Explain variations of for loop?
Answer:
Variations of for loop:
Variations increase the flexibility and applicability of for loop.
Multiple initializations and multiple update expressions:
Multiple statements can be used in the initialization and update expressions of for loop.
This multiple initializations and multiple update expressions are separated by commas.
For example:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 24
Output
The value of i is 0 The value of j is 10
The value of i is 1 The value of j is 9
The value of i is 2 The value of j is 8
The value of i is 3 The value of j is 7
The value of i is 4 The value of j is 6
In the above example, the initialization part contains two variables i and j, and the update expression contains i++ and j++. These two variables are separated by commas which are executed in sequential order i.e., during initialization firstly i-0 followed by j=10. Similarly, in update expression, firstly i++ is evaluated followed by j++ is evaluated.

Question 9.
Explain for loop optional expression with suitable examples.
Answer:
Optional expressions;
The for loop contains three parts, i.e., initialization expressions, test expressions and
update expressions. These three expressions are optional in a for loop.
Case 1:
#include
using namespace std;
int main ()
{
int i, sum=0, n;
cout<<“\n Enter The value of n”;
cin>>n;
i =1;
for (; i<=10;i++)
{
sum += i;
}
cout<<“\n The sum of 1 to” <<n<<“is “<<sum;
return 0;
}
Output
Enter the value of n 5
The sum of 1 to 5 is 15
In the above example, variable i is declared and the sum is initialized at the time of variable declaration.
The variable i is assigned to 0 before the for loop but still, the semicolon is necessary before test expression.
In a for loop, if the initialization expression is absent then the control is transferred to the test expression/conditional part.
Case 2:
#include
using namespace std;
int main ()
{
int i, sum=0, n;
cout<<“\n Enter The value of n”; cin>>n;
i =1;
for (; i<=10; )
{
sum += i;
++i;
cout<<“\n The sum of 1 to” <<n<<“is “<<sum;
return 0;
}
Output
Enter the value of n 5
The sum of 1 to 5 is 15
In the above code, the update expression is hot done, but a semicolon is necessary before the update expression.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 25
In the above code, neither the initialization nor the update expression is done in the for a loop. If both or any one of the expressions are absent then the control is transferred to the conditional part.
Case 3:
An infinite loop will be formed if a test- expression is absent in a for a loop.
For example:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 26

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 10.
What is an empty loop? Give a suitable example.
Answer:
Empty loop:
An empty loop means a loop that has no statement in its body is called an empty loop. Following for loop is an empty loop:
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 27
In the above code, the for loop contains a null statement, it is an empty loop.
Similarly, the following for loop also forms an empty loop.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 28
In the above code, the body of a for loop enclosed by braces is not executed at all because a semicolon is ended after the for a loop.

Question 11.
What do you mean by the nesting of loops?
Answer:
Nesting of loops:
A loop which contains another loop is called a nested loop.
The syntax is given below:
for (initialization(s); test-expression; update expression(s))
{
for (initialization(s); test-expression; update expression(s)
{
statement(s);
}
statement(s);
{
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 29
Example:
#include
using namespace std;
int main(void)
{
cout<< “A multiplication table:”
< <<endl<< “” <<endl;
for(int c = 1; c < 10; C++)
{
cout<< c << “l”;
for(int i = 1; i< 10; i++)
{
cout< }
cout<<endl;
}
return 0;
}

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 12.
What are the Differences between the Break and Continue statement?
Answer:

Break

Continue

The break is used to terminate the execution of the loop. Continue is not used to terminate the execution of the loop.
It breaks the iteration. It skips the iteration.
When this statement is executed, control will come out from the loop and executes the statement immediately after the loop. When this statement is executed, it will not come out of the loop but moves/jumps to the next iteration of the loop.
The break is used with loops as well as switch cases. Continue is only used in loops, it is not used in switch cases.

Book Evaluation

Hands-On Practice
Write a C++ program to solve the following problems:
Question 1.
Temperature – conversion program that gives the user the option of converting Fahrenheit to Celsius or Celsius to Fahrenheit and depending upon the user’s choice.
Answer:
PROGRAM :
using namespace std;
#include<iostream>
int main()
{
float f,c;
int choice;
cout<<“\nl, FahrenheiT to Celsius”;
cout<<“\n2. Celsius to FahrenheiT”;
cout<<“\n3.Exit”;
cout<<“\nENTER YOUR CHOICE “;
cin>>choice;
switch(choice)
{
case 1: cout<<“\nENTER TEMPERATURE IN FAHRENHEIT “;
cin>>f;
c=5.0/9.0 *(f-32);
cout<<“\nTEMPERATURE IN FAHRENHEIT “<<f;
cout«”\nTEMPERATURE IN CELSIUS “<<c;
break;
case 2: cout<<“\nENTER TEMPERATURE IN CELSIUS “;
cin>>c;
f = 9*c/5 + 32;
cout< <“\nTEMPERATURE IN FAHRENHEIT “<<f;
cout< < “\nTEMPERATURE IN CELSIUS “<<c;
break;
default: cout< <“\nPROCESS COMPLETED”; > .
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 30

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 2.
The program requires the user to enter two numbers and an operator. It then carries out the specified arithmetical operation: addition, subtraction, multiplication, or division of the two numbers. Finally, it displays the result.
using namespace std;
#include <iostream>
int main()
{
float num1,num2,result;
char op;
coutczcz”\nl. Enter two numbers”;
cin>>num1>>num2;
cout<<”\nEnter operaotr + or – or * or / : “; cin>>op;
switch(op)
{
case’+’:
result = num1 + num2;
cout<<“\nAdded value “<<result;
break;
case’-‘:
result = num1 – num2;
cout<<“\nSubtracted value “<<result;
break;
case ‘*’:
result = num1 * num2;
cout<<“\Multiplied value “<<result;
break;
case’/’:
result = num1 / num2;
cout<<“\nDivided value “<<result;
break;
default:cout<<“\nPROCESS COMPLETED”;
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 31
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 32

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 3.
Program to input a character and to print whether a given character is an alphabet, digit or any other character.
Answer:
PROGRAM
using namespace std;
#include <iostream>
#include
int main()
{
char ch;
cout<<“\nl. Enter a character :”; cin>>ch;
if(isdigit(ch))
cout<<“\nThe given character “<<ch<<” is a digit”;
else if(isalpha(ch))
cout<<“\nThe given character “<<ch<<” is an alphabet”;
else
cout<<“\nThe given character “<<ch<<” is other character”;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 33

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 4.
Program to. print whether a given character Is an uppercase or a lowercase character or a digit or any other character. Use ASCII codes for it. The ASCII codes are as given below:
Answer:
Characters -ASCII Range
‘0’ – ‘9’ – 48 – 57
‘A’ – ‘Z’ – 65 – 90
‘a’ – ‘z’ – 97 – 122
other characters 0 – 255 excluding the above-mentioned codes.
PROGRAM
using namespace std;
#include
int main()
{
char ch;
cout<<“\nl. Enter a character: cin>>ch;
int av = ch;
if(av>47 m av<58)
cout<<“\nThe given character “<<ch<<” is a digit”; else if(av>64 && av<91)
cout<<“\nThe given character “<<ch<<” is an Uppercase character”; else if(av>96 && av < 123)
cout<<“\nThe given character “<<ch<<” is n Lowercase character”;
else
cout<<“\nThe given character “<<ch<<” is an Other character”;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 34
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 35

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 5.
Program to calculate the factorial of an integer.
Answer:
PROGRAM
using namespace std;
#include
int main()
{
int i,n;
long int fact=l;
cout<<“\nEnter a number… cin>>n; ‘
for(i=l; i<=n; i++)
fact = fact * i;
cout<<“\nFactorial of “<<n<<” is “<<fact;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 36

Question 6.
Program that print 1 2 4 8 16 32 64 128.
Answer:
PROGRAM
using namespace std;
#include
int main()
{
int i;
for(i=l; i< =128; i=i*2)
cout<<i<<“\t”;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 37

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 7.
Program to generate divisors of an integer.
Answer:
PROGRAM
using namespace std;
#include
int main() {
int n,d;
cout<<“\n Enter a number…”; cin>>n;
cout<<“\nThe divisors for “< for(d=l; d<=n/2;d++)
if(n%d==0)
cout<<d<<“\t”;
cout<<n;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 38

Question 8.
Program to print fibonacci series i.e., 0112 3 5 8
Answer:
PROGRAM
using namespace std;
#include
int main()
{
int n,fl,f2,f3;
fl=-l; ,
f2=l;
cout«”\nHOW MANY TERMS … “; .
cin>>n;
cout<<,r\nThe FIBONACCI TERMS ARE \n”; for(int i=l; i<=h;i++)
{
f3=fl+f2;
cout<<f3<<“\n”;
fl-f2;
f2=f3;
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 39

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 9.
Programs to produces the following design using nested loops.
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 41
Answer:
9. A)
PROGRAM
using namespace std;
#include
int main()
{
int i,j;
for(i=l; i<=6; i++)
{
char c = ‘A’; for(j=l;j<=i;j++)
{
cout<<c<<“\t”;
C++;
}
cout<<“\n\n”;
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 42
9. B)
PROGRAM
using namespace std;
#include
int main()
{
int i,j;
for(i=l; i<=5; i++) { for(j=5;j>=i;j–)
cout<<j<<“\t”;
cout<<“\n\n”;
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 43
9.c)
PROGRAM
using namespace std;
#include
int main()
{
int i,j,k=0,m;
for(i=7; i> = l; i=i-2)
{
for(j=l;j<=i;j++)
cout<<“#\t”;
cout<<“\n\n”;
k++;
for(m=l;m<=k;m++)
cout<<“\t”;
}
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 44

Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control

Question 10.
Program to check whether the square root of a number is a prime or not.
Answer:
PROGRAM
using namespace std;
#include
#include int main()
{
int n,srn,d,p=0;
cout<<“\nENTER A NUMBER … cin>>n;
srn=sqrt(n);
cout<<“\nTHE GIVEN NUMBER IS “<<n;
cout<<“\nTHE SQUARE ROOT OF”<<n<<” IS”<<srn;
for(d=2; d<=srn/2;d++)
{
if(srn%d==0)
{
p=1;
break;
}
}
if(p==0)
cout <<“\nTHE SQUARE ROOT OF THE GIVEN NUMBER “<<n<<” IS A PRIME”; else
cout <<“\nTHE SQUARE ROOT OF THE GIVEN NUMBER “<<n<<” IS NOT A PRIME”;
}
Output
Samacheer Kalvi 11th Computer Science Guide Chapter 10 Flow of Control 45

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 4 Theoretical Concepts of Operating System Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 4 Theoretical Concepts of Operating System

11th Computer Science Guide Theoretical Concepts of Operating System Text Book Questions and Answers

Part I

Choose the correct answer.

Question 1.
Operating system is a
a) Application Software
b) Hardware
c) System Software
d) Component
Answer:
c) System Software

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 2.
Identify the usage of Operating Systems
a) Easy interaction between the human and computer
b) Controlling input & output Devices
c) Managing use of main memory
d) All the above
Answer:
d) All the above

Question 3.
Which of the following is not a function of an Operating System?
a) Process Management
b) Memory Management
c) Security management
d) Complier Environment
Answer:
d) Complier Environment

Question 4.
Which of the following OS is a Commercially licensed Operating system?
a) Windows
b) UBUNTU
C) FEDORA
d) REDHAT
Answer:
a) Windows

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 5.
Which of the following Operating systems support Mobile Devices?
a) Windows 7
b) Linux
c) BOSS
d) iOS
Answer:
a) Windows 7

Question 6.
File Management manages
a) Files
b) Folders
c) Directory systems
d) All the Above
Answer:
d) All the Above

Question 7.
Interactive Operating System provides
a) Graphics User Interface (GUI)
b) Data Distribution
c) Security Management
d) Real Time Processing
Answer:
a) Graphics User Interface (GUI)

Question 8.
Android is a
a) Mobile Operating system
b) Open Source
c) Developed by Google
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 9.
Which of the following refers to Android operating system’s version?
a) JELLYBEAN
b) UBUNTU
c) OS/2
d) MITTIKA
Answer:
a) JELLYBEAN

Part II

Short Answers

Question 1.
What are the advantages of management in Operating System?
Answer:

  1. Allocating memory is easy and cheap
  2. Any free page is ok, OS can take the first one out of the list it keeps
  3. Eliminates external fragmentation
  4. Data (page frames) can be scattered all over PM
  5. Pages are mapped appropriately anyway
  6. Allows demand paging and pre – paging
  7. More efficient swapping
  8. No need for considerations about fragmentation
  9. Just swap out the page least likely to be used

Question 2.
What is the multi-user Operating system?
Answer:
It is used in computers and laptops that allow same data and applications to be accessed by multiple users at the same time.
The users can also communicate with each other. Windows, Linux and UNIX are examples for multi-user Operating System.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 3.
What is a GUI?
Answer:
GUI – Graphical User Interface – It allows the use of icons or other visual indicators to interact with electronic devices, rather than using only text via the command line. For example, all versions of Microsoft Windows utilize a GUI, whereas MS-DOS does not.

Question 4.
List out different distributions of Linux operating system.
Answer:
Linux distributors are Ubuntu, Mint, Fedora, RedHat, Debian, Google’s Android, Chrome OS, and Chromium OS.

Question 5.
What are the security management features available in Operating System?
Answer:
Security Management features:

  1. File access level security
  2. System-level security
  3. Network-level security

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 6.
What is multi-processing?
Answer:
This is a one of the features of Operating System. It has two or more processors for a single running process .

Question 7.
What are the different Operating Systems used in computer?
Answer:
Different operating system used:

  1. Single user, single Task Operating system
  2. Multi-user operating system
  3. Multiprocessing operating system
  4. Distributed Operating system

Part III

Explain in Brief

Question 1.
What are the advantages and disadvantages of Time-sharing features?
Answer:

Advantages Disadvantages
Many applications can run at the same time. it consumes much resources.
The CPU is not idle. Constraint on security and integrity of data.
Provides the advantage of quick response. Reliability constraint exists.

Question 2.
Explain and List examples of mobile operating system.
Answer:
A Mobile Operating System (or mobile OS) is an operating system that is specifically designed to run on mobile devices such as phones, tablets, smartwatches, etc.
Example: Apple IOS

Google Android: Android is a mobile OS developed by Google, based on Linux and designed for smartphones and tabs. iOS was developed by Apple.
Example: Android, ColorOS, LGUX, MIUI.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 3.
What are the differences between Windows and Linux Operating systems?
Answer:

WINDOWS LINUX
It is a licensed OS Linux is an open-source operating system
Vulnerable to viruses and malware attacks More secure
Windows must boot from the primary- partition It can be booted from an either primary or logical partition
Windows file name are not case sensitive In Linux, file names are case sensitive
Windows uses the microkernel which takes less space Linux uses the monolithic kernel which consumes more running space

Question 4.
Explain the process management algorithms in Operating System.
Answer:
The following algorithms are mainly used to allocate the job (process) to the processor: FIFO, SJF, Round Robin, based on priority.

  1. First In First Out (FIFO) Scheduling – This algorithm is based on queuing technique. Technically, the process that enters the queue first is executed first by the CPU, followed by the next, and so on. The processes are executed in the order of the queue.
  2. Shortest Job First (SJF) Scheduling – This algorithm works based on the size of the job being executed by the CPU.
  3. Round Robin Scheduling – Round Robin (RR) Scheduling algorithm is designed especially -for time-sharing systems, jobs are assigned, and processor time in a circular method.
  4. Based on priority – The given job (process) is assigned on a priority. The job which has higher priority is more important than ether jobs.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Part IV

Explain in detail

Question 1.
Explain the concept of a Distributed Operating System.
Answer:
The data and application that are stored and processed on multiple physical locations across the world over the digital network (internet/intranet). The Distributed Operating System is used to access shared data and files that reside in any machine around the world. The user can handle the data from different locations. The users can access it as if it is available on their own computer.

The advantages of distributed Operating System are as follows:

  • A user at one location can make use of all the resources available at another location over the network.
  • Many computer resources can be added easily in the network
  • Improves the interaction with the customers and clients.
  • Reduces the load on the host computer.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 2.
Explain the main purpose of an operating system.
Answer:
The main use of the Operating System is:

  • To ensure that a computer can be used to extract what the user wants it to do.
  • Easy interaction between the users and computers.
  • Starting computer operation automatically when power is turned on (Booting).
  • Controlling Input and Output Devices.
  • Manage the utilization of main memory.
  • Providing security to user programs.

Question 3.
Explain the advantages and disadvantages of open-source operating systems.
Answer:
The benefits of open source are tremendous and have gained huge popularity in the IT field in recent years. They are as follows:

  1. Open-source (OS) is free to use, distribute and modify.
  2. Open source is independent of the company as an author who originally created it.
  3. It is accessible to everyone. Anyone can debug the coding.
  4. It doesn’t have the problem of incompatible formats that exist in proprietary software.
  5. It is easy to customize as per our needs.
  6. Excellent support will be provided by programmers who will assist in making solutions.

Some of the disadvantages are:

  1. The latest hardware is incompatible, i.e. lack of device drivers.
  2. It is less user-friendly and not as easy to use.
  3. There may be some indirect costs involved such as paying for external support.
  4. Malicious users can potentially view it and exploit any vulnerability.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

11th Computer Science Guide Theoretical Concepts of Operating System Additional Questions and Answers

Part I

Choose the correct answer

Question 1.
Software is classified into ………………. types.
(a) five
(b) two
(c) four
(d) six
Answer:
(b) two

Question 2.
_________ interacts basically with the hardware to generate the desired output.
a) hardware
b) freeware
c) software
d) None of these
Answer:
c) software

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 3.
Which one of the following is not an algorithm?
(a) NTFS
(b) FIFO
(c) SJE
(d) Round Robin
Answer:
(a) NTFS

Question 4.
_________is a software classification.
a) application software
b) system software
c) both A and B
d) None of these
Answer:
c) both A and B

Question 5.
Which one of the following is not a prominent operating system?
(a) UNIX
(b) IOS
(c) GUI
(d) Android
Answer:
(c) GUI

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 6.
_________is an application software.
a) MS-Word
b) VLC player
c) MS-Excel
d) All the above
Answer:
d) All the above

Question 7.
Which one of the following comes under proprietary license?
(a) Apple Mac OS
(b) Google’s Android
(c) UNIX
(d) LINUX
Answer:
(a) Apple Mac OS

Question 8.
_________is an application software to play audio, video files.
a) MS-Word
b) VLC player
c) MS-Excel
d) All the above
Answer:
b) VLC player

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 9.
………………. is the second most popular mobile operating system globally after Android.
(a) Microsoft Windows
(b) iOS
(c) UNIX
(d) LINUX
Answer:
(b) iOS

Question 10.
_________is a system software.
a) Operating System
b) Language Processor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 11.
Which one of the following is System software?
(a) Operating System
(b) Language Processor
(c) Both a & b
(d) none of these
Answer:
(c) Both a & b

Question 12.
_________controls input, output of all other peripheral devices.
a) Operating System
b) Language Processor
c) VLC player
d) MS-Word
Answer:
a) Operating System

Question 13.
Hardware and software are managed by ……………….
(a) GUI
(b) OS
(c) Bootstrap
(d) keyboard
Answer:
(b) OS

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 14.
Without a(n) _________, a computer cannot effectively manage all the resources,
a) Operating System
b) Language
c) Both A and B
d) None of these
Answer:
a) Operating System

Question 15.
An OS that allows only a single user to perform a task at a time is called……………….
(a) Single user os
(b) Single task os
(c) Both a & b
(d) Multi-tasking os
Answer:
(c) Both a & b

Question 16.
A user cannot communicate directly computer _________
a) software
b) hardware
c) Both A and B
c) None of these
Answer:
b) hardware

Question 17.
Identify the multi-user OS?
(a) Windows
(b) Linux
(c) UNIX
(d) All of these
Answer:
(d) All of these

Question 18.
The mobile devices mostly use _________as mobile OS.
a) Android
b) iOS
c) Either A or B
d) None of th
Answer:
c) Either A or B

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 19.
GUI stands for ……………….
(a) Geo User Interact
(b) Global User InterChange
(c) Graphical User Interface
(d) Global User Interface
Answer:
(c) Graphical User Interface

Question 20.
Operating System is basically – an interface between the _________
a) user and hardware
b) user and memory
c) programmer and hardware
d) None of these
Answer:
a) user and hardware

Question 21.
The operating system processes are executed by ……………….
(a) User code
(b) System code
(c) Task
(d) Program
Answer:
(b) System code

Question 22.
_________translates the user request into machine language.
a) Operating System
b) Language Processor
c) Application program
d) None of these
Answer:
a) Operating System

Question 23.
NTFS is a ……………….
(a) game
(b) file management technique
(c) OS
(d) System-level security
Answer:
(b) file management technique

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 24.
The main use of Operating 5ystem is _________
a) Controlling Input and Output Devices
b) Manage the utilization of main memory
c) Providing security to user programs
d) All the above
Answer:
d) All the above

Question 25.
Unix was developed in the year ……………….
(a) 1970
(b) 1980
(c) 1990
(d) 1960
Answer:
(a) 1970

Question 26.
A user task is a _________function.
a) printing a document
b) writing a file to disk
c) editing a file or downloading a file
d) All the above
Answer:
d) All the above

Question 27.
………………. is a windows alternative open-source operating system.
(a) React OS
(b) Boss
(c) Redhat
(d) Fedora
Answer:
(a) React OS

Question 28.
______ OS is used in computers and laptops.
a) Multi-user
b) Multitask
c) Either A or B
d) None of these
Answer:
a) Multi-user

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 29.
Which among the following is not an android mobile open source version?
(a) Donut
(b) Froyo
(c) Nougat
(d) Alpha
Answer:
(a) Donut

Question 30.
In _________OS users can also communicate with each other.
a) Multi-user
b) Multitask
c) Single user
d) MS-DOS
Answer:
a) Multi-user

Question 31.
_____ is an example of a multi-user Operating System.
a) Windows
b) Linux
c) UNIX
d) All the above
Answer:
d) All the above

Question 32.
A cheap computer can be build with _________
a) raspbion OS
b) Raspberry Pi
c) Both A and B
d) None of these
Answer:
c) Both A and B

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 33.
_________is a platform that’s designed to teach how to build a computer.
a) raspbion OS
b) Raspberry Pi
c) Both A and B
d) None of these
Answer:
a) raspbion OS

Question 34
______ is a key feature of an OS.
a) User interface
b) Fault tolerance
c) Memory management
d) All the above
Answer:
d) All the above

Question 35.
_________is the only way that a user can make interaction with a computer.
a) User interface
b) Fault tolerance
c) Memory management
d) All the above
Answer:
a) User interface

Question 36.
_________is the main reason for the key success of GUI.
a) User friendly
b) Fault tolerance
c) Robust
d) None of these
Answer:
a) User friendly

Question 37.
GUI means _________
a) Graphical User Interlink
b) Good User Interface
c) Graph User Interface
d) Graphical User Interface
Answer:
d) Graphical User Interface

Question 38.
The _________is a window based system.
a) command mode
b) GUI
c) both A and B
d) None of these
Answer:
b) GUI

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 39.
_________are playing vital role of the particular application.
a) Icons
b) Commands
c) Navigations
d) None of these
Answer:
a) Icons

Question 40.
The___________should satisfy the customer based on their needs.
a) user interface
b) fault tolerance
c) memory management
d) all the above
Answer:
a) user interface

Question 41.
The ________ should save the user’s precious time.
a) user interface
b) fault tolerance
c) memory management
d) all the above
Answer:
a) user interface

Question 42.
The ________is to satisfy the customer.
a) user interface
b) fault tolerance
c) memory management
d) all the above
Answer:
a) user interface

Question 43.
The ________should reduce the number of errors committed by the user.
a) process management
b) fault tolerance
c) memory management
d) None of these
Answer:
d) None of these

Question 44.
_____ is the process of controlling and coordinating the computer’s main memory.
a) process management
b) fault tolerance
c) memory/ management
d) None of these
Answer:
c) memory/ management

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 45.
________is the process of assigning memory space to various running programs to optimize overall computer performance.
a) user interface
b) memory management
c) fault tolerance
d) all the above
Answer:
b) memory management

Question 46.
________involves the allocation of specific memory blocks to individual programs based on user demands.
a) user interface
b) memory management
c) fault tolerance
d) all the above
Answer:
b) memory management

Question 47.
________ensures the availability of adequate memory for each running program at all times.
a) process management
b) fault tolerance
c) memory management
d) None of these
Answer:
c) memory management

Question 48.
The objective of the Memory Management process is to improve ________
a) the utilization of the CPU
b) the speed of the computer’s response
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 49.
The Operating System is responsible for the________
a) Keeping track of which portion of memory is currently being used and who is using them.
b) Determining which processes and data to move in and out of memory.
c) Allocation and de-allocation of memory blocks as needed by the program in main memory.
d) All the above
Answer:
d) All the above

Question 50.
________ is function that includes creating and deleting processes.
a) process management
b) fault tolerance
c) memory management
d) None of these
Answer:
a) process management

Question 51.
________ providing mechanisms for processes to communicate and synchronize with each other.
a) process management
b) fault tolerance
c) memory management
d) None of these
Answer:
a) process management

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 52.
A(n)____i_____ s the unit of work in a computer.
a) task
b) process
c) operation
d) None of these
Answer:
b) process

Question 53.
A word-processor program being run by an individual user on a computer is a________
a) task
b) process
c) operation
d) None of these
Answer:
b) process

Question 54.
A system task, such as sending output to a printer or screen, can also be called as a ________
a) task
b) process
c) operation
d) none of these
Answer:
b) process

Question 55.
A computer processes are classified as _________ categories.
a) 5
b) 4
c) 3
d) 2
Answer:
d) 2

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 56.
A category of computer process is _________
a) operating system process
b) user process
c) both A and B
d) None of these
Answer:
c) both A and B

Question 57.
________process is executed by system code.
a) operating system process
b) user process
c) both A and B
d) None of these
Answer:
a) operating system process

Question 58.
________process is execute by user code.
a) operating system process
b) user process
c) both A and B
d) None of these
Answer:
b) user process

Question 59.
All the computer processes can potentially execute concurrently on a ________CPU,
a) single
b) parallel
c) linear
d) None of these
Answer:
a) single

Question 60.
A process needs certain resources including _________ to finish its task.
a) CPU time
b) Memory
c) Files and I/O devices
d) All the above
Answer:
d) All the above

Question 61.
The Operating System is responsible for the_______activity.
a) Scheduling processes and threads on the CPUs
b) Suspending and resuming processes
c) Providing mechanisms for process synchronization
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 62.
The ________ algorithm is mainly used to allocate the job to the processor.
a) FIFO or SJF
b) Round Robin
c) Based on Priority
d) All the above
Answer:
d) All the above

Question 63.
FIFO means ________
a) First In Fast Out
b) Fast In First Out
c) First In First Out
d) None of these
Answer:
c) First In First Out

Question 64.
________is based on queuing technique.
a) FIFO
b) Round Robin
c) Based on Priority
d) All the above
Answer:
a) FIFO

Question 65.
________algorithm works based on the size of the job being executed by the CPU.
a) FIFO
b) Round Robin
c) Based on Priority
d) SJF
Answer:
d) SJF

Question 66.
________algorithm is designed especially for time-sharing systems.
a) FIFO
b) Round Robin
c) Based on Priority
d) SJF
Answer:
b) Round Robin

Question 67.
In the _______algorithm jobs are assigned and processor time in a circular method.
a) FIFO
b) Round Robin
c) Based on Priority
d) SJF
Answer:
b) Round Robin

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 68.
The challenge in the computer and software industry is to protect user’s ________from hackers.
a) data
b) operation
c) hardware
d) all the above
Answer:
a) data

Question 69.
The Operating System provides ________levels of securities to the user end.
a) four
b) three
c) two
d) many
Answer:
b) three

Question 70.
The Operating System security is ________
a) File access level
b) System-level
c) Network level
d) All the above
Answer:
d) All the above

Question 71.
In order to access the files created by other people, you should have the ________
a) user name and password
b) ogin id
c) email id
d) access permission
Answer:
d) access permission

Question 72.
File access permissions can be granted by the__________
a) creator of the file
b) administrator of the system
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 73.
________level security is offered by the password in a multi-user environment.
a) File access
b) System
c) Network
d) All the above
Answer:
b) System

Question 74.
________offers the password facility.
a) Windows
b) Linux
c) Windows and Linux
d) None of these
Answer:
c) Windows and Linux

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 75.
________security is an indefinable one.
a) File access level
b) System-level
c) Network level
d) All the above
Answer:
c) Network level

Question 76.
The people from all over the world try to provide ________security.
a) File access level
b) System-level
c) Network level
d) All the above
Answer:
c) Network level

Question 77.
The Operating Systems should be ________
a) patience
b) error-free
c) robust
d) None of these
Answer:
c) robust

Question 78.
When there is a fault, the ________should not crash.
a) application program
b) operating system
c) data
d) None of these
Answer:
b) operating system

Question 79.
The operating system manages the ________on a computer.
a) files
b) folders
c) directory systems
d) all the above
Answer:
d) all the above

Question 80.
Any type of data in a computer is stored in the form of ________
a) blocks
b) files
c) archives
d) None of these
Answer:
b) files

Question 81.
FAT stands for ________
a) File Allocation Task
b) File Authentication Table
c) Fixed Allocation Table
d) File Allocation Table
Answer:
d) File Allocation Table

Question 82.
The FAT stores general information about files like_________
a) filename and type
b) size
c) starting address and access mode
d) all the above
Answer:
d) all the above

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 83.
______ is the file access mode.
a) sequential
b) indexed / indexed-sequential
c) direct/relative
d) all the above
Answer:
d) all the above

Question 84.
The ________ of the operating system helps to create, edit, copy, allocate memory to the files, and also updates the FAT.
a) system agent
b) file agent
c) file supervisor
d) file manager
Answer:
d) file manager

Question 85.
ext2 stands for _______________
a) secondary extended file system
b) second extended folder system
c) second extended file scheme
d) second extended file system
Answer:
d) second extended file system

Question 86.
ext2 used in_________
a) Linux
b) MSDOS
c) Unix
d) None of these
Answer:
a) Linux

Question 87.
NTFS stands for_______
a) New Technology Focus System
b) New Technology File System
c) New Technology Filter System
d) New Trend File system
Answer:
b) New Technology File System

Question 88.
NTFS developed by________
a) Apple
b) IBM
c) Intel
d) Microsoft
Answer:
d) Microsoft

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 89.
______ has two or more processors for a single running process.
a) Mega processing
b) Micro processing
c) Multi-processing
d) Mixed-processing
Answer:
c) Multi-processing

Question 90.
Processing takes place in parallel is known as __________ processing.
a) parallel processing
b) distributed processing
c) parent
d) None of these
Answer:
a) parallel processing

Question 91.
______feature is used for high-speed execution which increases the power of computing.
a) parallel processing
b) distributed processing
c) multi-processing
d) None of these
Answer:
a) parallel processing

Question 92.
________allows execution of multiple tasks or processes concurrently.
a) Extended processing
b) Time sharing
c) Batch processing
d) None of these
Answer:
b) Time sharing

Question 93.
In _______ each task a fixed time is allocated.
a) Extended processing
b) Time sharing
c) Batch processing
d) None of these
Answer:
b) Time sharing

Question 94.
In_________the processor switches rapidly between various processes after a time is elapsed or the process is completed.
a) Extended processing
b) Time sharing
c) Batch processing
d) None of these
Answer:
b) Time sharing

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 95.
________feature takes care of the data and application that are stored and processed on multiple physical locations across the world over the digital network.
a) Multi-user OS
b) Distributed OS
c) Singe user OS
d) None of these
Answer:
b) Distributed OS

Question 96.
_________ is used to access shared data and files that reside in any machine around the world.
a) Multi-user OS
b) Distributed OS
c) Singe user OS
d) None of these
Answer:
b) Distributed OS

Question 97.
In_________ OS the user can handle the data from different locations.
a) Multiuser OS
b) Distributed OS
c) Singe user OS
d) None of these
Answer:
b) Distributed OS

Question 98.
________is the advantage of distributed Operating System.
a) A user at one location can make use of all the resources available at another location over the network.
b) Many computer resources can be added easily to the network
c) Improves the interaction with the customers and clients.
d) All the above
Answer:
d) All the above

Question 99.
_______reduces the load on the host computer.
a) Multi-user OS
b) Distributed OS
c) Singe user OS
d) None of these
Answer:
b) Distributed OS

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 100.
Prominent OS is________
a) UNIX and Windows
b) Linux
c) iOS and Android
d) All the above
Answer:
d) All the above

Question 101.
Modem operating systems use a
a) command mode interaction
b) GUI
c) visual
d) None of these
Answer:
b) GUI

Question 102.
OS can be
a) Proprietary with a commercial license
b) Open source
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 103.
_______is a proprietary with a commercial license OS.
a) Microsoft Windows
b) Apple Mac OS
c) Apple iOS
d) All the above
Answer:
d) All the above

Question 104.
________ is a open source free license OS.
a) Unix
b) Linux
c) Google’s Android
d) All the above
Answer:
d) All the above

Question 105.
Unix derive originally from _______
a) Borland International
b) AT&T Bell Labs
c) Intel
d) None of these
Answer:
b) AT&T Bell Labs

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 106.
The development of the Unix began in the year__________
a) 1960
b) 1966
c) 1970
d) 1976
Answer:
c) 1970

Question 107.
The Unix was developed by _______
a) Ken Thompson
b) Dennis Ritchie
c) Both A and B
d) Bjarne Stroustrup
Answer:
c) Both A and B

Question 108.
_______ OS can be modified and distributed by anyone around the world.
a) MS-DOS
b) Windows
c) Linux
d) None of these
Answer:
c) Linux

Question 109.
Most of the servers run on Linux because_______
a) it is easy to customize
b) it is rigid
c) it is not case sensitive
d) None of these
Answer:
a) it is easy to customize

Question 110.
_______is Linux distributor.
a) Ubuntu
b) Mint
c) Fedora
d) All the above
Answer:
d) All the above

Question 111.
_______is Linux distributor.
a) RedHat
b) Debian
c) Google’s Android
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 112.
_______is Linux distributor.
a) Chrome OS
b) Chromium OS
c) Both A and B
d) MS-DOS
Answer:
c) Both A and B

Question 113.
The Linux operating system was originated in the year _________
a) 1990
b) 1991
c) 1890
d) 1980
Answer:
d) 1980

Question 114.
The Linux operating system was developed by________
a) Ken Thompson
b) Dennis Ritchie
c) Linus Torvalds
d) Bjarne Stroustrup
Answer:
c) Linus Torvalds

Question 115.
Linux is similar to the________operating system.
a) Windows
b) UNIX
c) MS-DOS
d) None of these
Answer:
b) UNIX

Question 116.
Unix and the C programming language were developed by
a) Borland International
b) AT&T Bell Labs
c) Intel
d) None of these
Answer:
b) AT&T Bell Labs

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 117.
_______OS is primarily targeted to Intel and AMD architecture based computers.
a) Windows
b) UNIX
c) MS-DOS
d) None of these
Answer:
a) Windows

Question 118.
_______is a Windows-alternative open-source operating system.
a) RearOS
b) ReachOS
c) ReactOS
d) None of these
Answer:
c) ReactOS

Question 119.
________is a mobile device.
a) phones
b) tablets
c) MP3 players
d) All the above
Answer:
d) All the above

Question 120.
Android is a mobile operating system developed by
a) Borland International
b) AT&T Bell Labs
c) Intel
d) Google
Answer:
d) Google

Question 121.
_________ OS is designed primarily for touch screens mobile devices such as smartphones and tablets.
a) Windows
b) UNIX
c) MS-DOS
d) Android
Answer:
d) Android

Question 122.
Google has developed _______for televisions.
a) Android Wear
b) Android Car
c) Android TV
d) None of these
Answer:
c) Android TV

Question 123.
Google has developed _______ for cars.
a) Android Auto
b) Android Car
c) Android TV
d) None of these
Answer:
a) Android Auto

Question 124.
Google has developed_________for wrist watches.
a) Android Wear
b) Android Car
c) Android TV
d) None of these
Answer:
a) Android Wear

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 125.
Google has developed separate Android for_________
a) game consoles
b) digital cameras
c) PCs
d) All the above
Answer:
d) All the above

Question 126.
________ is an Android version.
a) Alpha and Beta
b) Cupcake and Donut
c) Eclair and Froyo
d) All the above
Answer:
d) All the above

Question 127.
_________ is an Android version.
a) Gingerbread
b) Honeycomb
c) Icecream Sandwich
d) All the above
Answer:
d) All the above

Question 128.
___________ is an Android version.
a) Jelly Bean
b) Kitkat and Lollipop
c) Marshmallow and Nought
d) All the above
Answer:
d) All the above

Question 129.
________ is a mobile Operating System created and developed by Apple Inc.
a) Android
b) iOS
c) Unix
d) None of these
Answer:
b) iOS

Question 130.
__________ is a mobile Operating System created and developed only for hardware of the Apple iPhone.
a) Android
b) iOS
c) Unix
d) None of these
Answer:
b) iOS

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 131.
________is the second most popular mobile Operating System globally.
a) Android
b) iOS
c) Unix
d) None of these
Answer:
b) iOS

Question 132.
________ is the top most popular mobile Operating System globally.
a) Android
b) iOS
c) Unix
d) None of these
Answer:
a) Android

Question 133.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 1
is the logo of _______OS.
a) Windows 7
b) Windows 8
c) Mac OS
d) None of these
Answer:
b) Windows 8

Question 134.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 2
is the logo of ______OS
a) Windows 7
b) Windows 8
c) Mac OS
d) None of these
Answer:
c) Mac OS

Question 135.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 3
is the logo of______OS
a) Linux
b) Apple iOS
c) Mac OS
d) None of these
Answer:
a) Linux

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 136.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 4
is the logo of _______OS
a) Windows 7
b) Apple iOS
c) Mac OS
d) None of these
Answer:
b) Apple iOS

Question 137.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 5
is the logo of _______OS
a) Android
b) Apple iOS
c) Mac OS
d) None of these
Answer:
a) Android

Question 138.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 6
is the logo of ______OS
a) Android
b) Apple iOS
c) Mac OS
d) Unix
Answer:
d) Unix

Part II

Short Answers

Question 1.
What is an operating system?
Answer:
An operating system is software which serves as the interface between a user and a computer.

Question 2.
What are the classifications of software?
Answer:
Software is classified into two types:

  • Application Software.
  • System Software.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 3.
What is a Real-Time operating system?
Answer:
It is multi-tasking and multi-user operating system designed for real-time-based applications such as robotics, weather, and climate prediction software, etc.

Question 4.
What is system software?
Answer:
The system software is a type of computer program that is designed to run the computer’s hardware and application programs. Example: Operating System.

Question 5.
What are the advantages of Distributed Operating system?
Answer:
Resources can be used in different locations. Improves interaction with customers and clients. Reduces load on host computers. The data can be exchanged via email and chat.

Question 6.
List any 4 system software.
Answer:

  1. Operating System.
  2. Language Processor.
  3. Compiler.
  4. Loader.

Question 7.
Explain Round Robin Scheduling.
Answer:
This type of scheduling is also known as the Time-sharing scheduling process. In this, each program is given a fixed amount of time to execute.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 8.
What are the functions of an operating system?
Answer:
The functions of an Operating System include file management, memory management, process management and device management and many more.

Question 9.
Mention different management techniques?
Answer:
Single continuous allocation, Partitioned allocation, Paged memory management, Segmented memory management.

Question 10.
What are the popular operating systems used in mobile devices?
Answer:
The mobile devices mostly use Android and iOS as mobile OS.

Question 11.
What is an Android?
Answer:
Android is a mobile operating system developed by Google, based on Linux, and designed primarily for touch screens mobile devices such as smartphones and tablets.

Question 12.
How OS cars be developed and released?
Answer:
OS can be either proprietary with a commercial license or can be open source.

Question 13.
What is Process Management?
Answer:
Process management is a function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other.

Question 14.
Write note on Microsoft Windows.
Answer:
Microsoft Windows is a family of proprietary operating systems designed by Microsoft Corporation and primarily targeted to Intel and AMD architecture-based computers.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 15.
What are the 3 levels of security?
Answer:

  1. File Access Level
  2. System Level
  3. Network Level.

Question 16.
Write about raspbion OS.
Answer:
Raspbion OS is a platform that’s designed to teach how to build a computer, what every part of a circuit board does, and finally how to code apps or games. The platform is available in pre-designed kits.

Question 17.
Write not on User Interface.
Answer:
User interface is one of the significant features in Operating System. The only way that users can make interact with a computer. If the computer interface is not user-friendly, the user slowly reduces the computer usage from their normal life.

Question 18.
What is the objective of memory management?
Answer:
The objective of the Memory Management process is to improve both the utilization of the CPU and the speed of the computer’s response to its users via main memory.

Question 19.
What is process management?
Answer:
Process management is a function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other.

Question 20.
What is the process? Give an example.
Answer:
A process is the unit of work in a computer. A word¬processing program being run by an individual user on a computer is a process.

Question 21.
Give any two examples for a process. Examples:
Answer:
A word-processing program being run by an individual user on a computer.
System task, such as sending output to a printer or screen.

Question 22.
What is the classification of a process?
Answer:
A computer consists of a collection of processes. They are classified as two categories:

  1. Operating System processes are executed by system code.
  2. User Processes which is executed by user code.

Question 23.
What are the requirements of a process?
Answer:
A process needs certain resources including CPU time, memory, files, and I/O devices to finish its task.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 24.
What is the major challenge in the computer and software industry?
Answer:
The major challenge in the computer and software industry is to protect user’s legitimate data from hackers.

Question 25.
Explain File Access level security.
Answer:
In order to access the files created by other people, you should have access permission. Permissions can either be granted by the creator of the file or by the administrator of the system.

Question 26.
How will you offer system-level security?
Answer:
System-level security is offered by the password in a multi-user environment.

Question 27.
Which OS offers system-level security?
Answer:
Both Windows and Linux offer the password facility to enable system-level security.

Question 28.
Write a note on network-level security.
Answer:
Network security is an indefinable one. So people from all over the world try to provide such security.

Question 29.
Write a note on Fault Tolerance.
Answer:
Fault Tolerance: The Operating Systems should be robust. When there is a fault, the Operating System should not crash, instead, the Operating System has fault tolerance capabilities and retains the existing state of the system.

Question 30.
How data is stored in a computer?
Answer:
Any type of data in a computer is stored in the form of files and directories/folders through File Allocation Table (FAT).

Question 31.
What will be stored in FAT?
Answer:
The FAT stores general information about files like filename, type (text or binary), size, starting address, and access mode.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 32.
What are the various file access modes?
Answer:
File access modes are:

  • Sequential.
  • Indexed.
  • Indexed-sequential.
  • Direct.
  • Relative.

Question 33.
Write note on file manager.
Answer:
The file manager of the operating system helps to create, edit, copy, allocate memory to the files and also updates the FAT,

Question 34.
What is multi-process?
Answer:
This is one of the features of Operating System, It has two or more processors for a single running process. Each processor works on different parts of the same task or on two or more different tasks. This feature is used for high-speed execution which increases the power of computing.

Question 35.
Write note on parallel processing.
Answer:
Processing takes place in parallel is known as parallel processing.

Question 36.
Write a note on Time Sharing,
Answer:
It allows the execution of multiple tasks or processes concurrently. For each task, a fixed time is allocated. This division of time is called Time- sharing.

Question 37.
Why distributed operating system is used?
Answer:
The Distributed Operating System is used to access shared data and files that reside in any machine around the world.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 38.
What are the advantages of GUI?
Answer:
A GUI lets you use your mouse to click icons, buttons, menus, and everything is clearly displayed on the screen using a combination of graphics and text elements.

Question 39.
What are the key features of OS?
Answer:
Features of the Operating System:
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 7

Question 40.
What devices are controlled by OS?
Answer:
OS controls input, output and other peripheral devices such as disk drives, printers and electronic gadgets

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Part III

Explain in Brief

Question 1.
Write a short note on Android.
Answer:
Android:
Android is a mobile operating system developed by Google, based on Linux, and designed primarily for touch screens mobile devices such as smartphones and tablets. Google has further developed Android TV for televisions, Android Auto for cars, and Android Wear for wrist watches, each with a specialized user interface. Variants of Android are also used on game consoles, digital cameras, PCs, and other electronic gadgets.

Question 2.
List various OS with their symbol.
Answer:
Various Operating Systems:
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 8

Question 3.
Explain the classification of Operating Systems according to availability.
Answer:
Classification of Operating Systems according to availability.
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 9

Question 4.
Write a note on iOS – iPhone OS.
Answer:
iOS (formerly iPhone OS) is a mobile Operating System created and developed by Apple Inc., exclusively for its hardware. It is the Operating System that presently powers many of the company’s mobile devices, including the iPhone, iPad, and iPod Touch. It is the second most popular mobile Operating System globally after Android.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 5.
List the various Linux distributions.
Answer:
Linux Distributions:
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 10

Question 6.
Write a note on single-user OS?
Answer:
An os allows only a single user to perform a task at a time. It is called a single user and single task os.
Example: MS-DOS.

Question 7.
Explain Android OS.
Answer:
Android is a mobile operating system developed by Google, based on Linux and designed primarily for touch screen mobile devices such as smartphones and tablets.

Google has further developed Android TV for televisions, Android Auto for cars and Android Wear for wrist watches, each with a specialized user interface.

Variants of Android are also used on game consoles, digital cameras, PCs and other electronic gadgets.

Question 8.
What are the various Android versions?
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 11

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 9.
Define Process?
Answer:

  1. A process is a unit of work (program) in a computer.
  2. A word processing program being run by an individual user on a computer is a process.
  3. A system task, such as sending output to a printer or screen can also be called a process.

Question 10.
What are the points are considered when User Interface is designed for an application.
Answer:
The following points are considered when User
The interface is designed for an application.

  • The user interface should enable the user to retain this expertise for a longer time.
  • The user interface should also satisfy the customer based on their needs.
  • The user interface should save user’s precious time. Create graphical elements like Menus,Window,Tabs, Icons and reduce typing work will be an added advantage of the Operating System.
  • The ultimate aim of any product is to satisfy the customer. The User Interface is also to satisfy the customer.
  • The user interface should reduce number of errors committed by the user with a little practice the user should be in a position to avoid errors.

Question 11.
Name the activities done by os related to the process management?
Answer:

  1. Scheduling processes and threads on the CPU.
  2. Creating and deleting both user and system processes.
  3. Suspending and resuming processes.
  4. Providing mechanisms for process synchronization.
  5. Providing mechanisms for process communication.

Question 12.
What are the responsibilities of the Operating System in connection with memory management?
Answer:
The Operating System is responsible for the following activities in connection with memory management:

  • Keeping track of which portion of memory are currently being used and who is using them.
  • Determining which processes and data to move in and out of memory.
  • Allocation and de-allocation of memory blocks as needed by the program in main memory.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 13.
Write a note on the File Allocations Table (FAT).
Answer:

  1. Any type of data in a computer is stored in the form of flies and directories/folders through File Allocation Table (FAT).
  2. The FAT stores general information about files like file name, type (text or binary), size, starting address and access mode (sequential / indexed / indexed – sequential / direct / relative).

Question 14.
Explain File Management.
Answer:
File Management:
File management is an important function of OS which handles the data storage techniques. The operating system manages the files, folders, and directory systems on a computer.

Any type of data in a computer is stored in the form of files and directories/folders through File Allocation Table (FAT). The FAT stores general information about files like filename, type (text or binary), size, starting address and access mode (sequential/indexed/indexed-sequential/direct/ relative).

The file manager of the operating system helps to create, edit, copy, allocate memory to the files, and also updates the FAT. The OS also takes care of the files that are opened with proper access rights to read or edit them.

There are few other file management techniques available like Next-Generation File System (NTFS) and ext2 (Linux).

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 15.
Write a note on React OS.
React os is a window – alternative open-source os which is being developed on the principles of windows – without using any of the Microsoft code.

PART – IV

Explain in detail.

Question 1.
What is the Need for Operating System? Explain in detail.
Answer:
Operating System has become essential to enable the users to design applications without the knowledge of the computer’s internal structure of hardware. Operating System manages all the Software and Hardware.

Most of the time there are many different computer programmes running at the same time, they all need to access the Computers, CPU, Memory, and Storage. The need for an Operating System is basically – an interface between the user and hardware.

Interaction of Operating system and user
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 12

Operating System works as translator, while it translates the user request into machine language (Binary language), processes it and then sends it back to Operating System. Operating System converts processed information into the user-readable form.

Question 2.
What are the uses of the Operating System?
Answer:
The main use of the Operating System is:

  • To ensure that a computer can be used to extract what the user wants it to do.
  • Easy interaction between the users and computers.
  • Starting computer operation automatically when power is turned on (Booting).
  • Controlling Input and Output Devices.
  • Manage the utilization of main memory.
  • Providing security to user programs.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Question 3.
Explain User Interface?
Answer:
User Interface:
The user interface is one of the significant features of the Operating System. The only way that a user can make interacting with a computer. If the computer interface is not user-friendly, the user slowly reduces the computer usage from their normal life. This is the main reason for the key success of GUI (Graphical User Interface) based Operating System. The GUI is a window-based system with a pointing device to direct I/O, choose from menus, make selections, and a keyboard to enter text. Its vibrant colours attract the user very easily. Beginners are impressed by the help and pop-up window message boxes. Icons are playing a vital role in the particular application.

Now Linux distribution is also available as a GUI-based Operating System. The following points are considered when User Interface is designed for an application.

  1. The user interface should enable the user to retain this expertise for a longer time.
  2. The user interface should also satisfy the customer based on their needs.
  3. The user interface should save the user’s precious time. Create graphical elements like Menus, Window, Tabs, Icons and reduce typing work will be an added advantage of the Operating System.
  4. The ultimate aim of any product is to satisfy the customer. The User Interface is also designed to satisfy the customer.
  5. The user interface should reduce the number of errors committed by the user with a little practice the user should be in a position to avoid errors (Error Log File)

Question 4.
Explain distributed operating system.
Answer:
This feature takes care of the data and applications that are stored and processed on multiple physical locations across the world over the digital network. The Distributed Operating System is used to access shared data and files that reside in any machine around the world. The user can handle the data from different locations. The users can access as if it is available on their own computer.

The advantages of distributed Operating System are as follows:

  • A user at one location can make use of all the resources available at another location over the network.
  • Many computer resources can be added easily to the network.
  • Improves the interaction with the customers and clients.
  • Reduces the load on the host computer.

Question 5.
Explain process Management.
Answer:
Process Management:
Process management is a function that includes creating and deleting processes and providing mechanisms for processes to communicate and synchronize with each other. A process is a unit of work (program) in a computer. A word processing program being run by an individual user on a computer is a process. A system task, such as sending output to a printer or screen, can also be called a Process.

A computer consists of a collection of processes, they are classified into two categories:

  1. Operating System processes which is executed by system code.
  2. User Processes which is executed by user code.

All these processes can potentially execute concurrently on a single CPU. A process needs certain resources including CPU time, memory, files, and I/O devices to finish its task.

The Operating System is responsible for the following activities associated with the process management:

  1. Scheduling processes and threads on the CPUs.
  2. Creating and deleting both user and system processes.
  3. Suspending and resuming processes.
  4. Providing mechanisms for process synchronization.
  5. Providing mechanisms for process communication.

The following algorithms are mainly used to allocate the job (process) to the processor.

  1. FIFO
  2. SJF
  3. Round Robin
  4. Based on Priority

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

ACTIVITY

Question 1.
Draw a line between the operating system logo and the correct description.
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System 13

Question 2.
Discuss and provide suitable answers to the questions below.
Answer:
One of the functions of an Operating System is multi-tasking.
i) Explain one reason why multi-tasking is needed in an operating system.
Reasons:

  • CPU is used most of the time and never becomes idle.
  • The system looks fast as all the tasks run in parallel.
  • Short-time jobs are completed faster than long-time jobs.
  • Resources are used nicely.
  • Total read time is taken to execute program/job decreases.
  • Response time is shorter.

ii) State two other functions of an Operating System.
Disk Management:
The operating system manages the disk space. It manages the stored files and folders in a proper way.

Device Controlling:
The operating system also controls all devices attached to the computer. The hardware devices are controlled with the help of small software called a device driver.

Samacheer Kalvi 11th Computer Science Guide Chapter 4 Theoretical Concepts of Operating System

Print controlling:
The operating system also controls the printing function. If a user issues two print commands at a time, it does not mix data of these files and prints them separately.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 8 Heat and Thermodynamics Text Book Back Questions and Answers, Notes.

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11th Physics Guide Heat and Thermodynamics Book Back Questions and Answers

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Part – I:
I. Multiple choice questions:

Question 1.
In hot summer after a bath, the body’s:
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer:
(a) internal energy decreases

Question 2.
The graph between volume and temperature in Charle’s law is:
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer:
(c) a straight line

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer:
(b) adiabatic

Question 4.
An ideal gas passes from one equilibrium state (P1, V1, T1, N) to another equilibrium state (2P1, 3V1, T2, N). Then:
(a) T1 = T2
(b) T1 = \(\frac{\mathrm{T}_{2}}{6}\)
(c) T1 = 6T2
(d) T1 = 3T2
Answer:
(b) T1 = \(\frac{\mathrm{T}_{2}}{6}\)

Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase:
(a) mass
(b) weight
(c) centre of mass
(d) moment of inertia
Answer:
(d) moment of inertia

Hint:
M.I. = MK²
K – radius of gyration.
During heating K would be increased. Hence moment of inertia will increase.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process:
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0
Answer:
(a) Q > 0, W > 0

Hint:
Q > 0; W > 0
∆Q = ∆U1 + ∆W.
∴ ∆Q ∝ ∆W
When ∆Q > 0
∆W > 0.

Question 7.
When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U< 0, W < 0
(d) ∆U = 0, W > 0
Answer:
(b) ∆U < 0, W > 0

Hint:
∆Q = ∆U + ∆W
When exercise is performed, internal energy is utilised (∆U < 0). So it will decrease.
But as internal energy is utilised, the exercise (∆W) will be more. So work will be more.
∴ ∆W > 0
∆U < 0 W > 0

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0
(c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
Answer:
(c) ∆U > 0, Q > 0

Hint: During the thermal equilibrium with surrounding as heat energy (Q) is increased, internal energy will be increased.
∴ ∆U > 0 Q > 0

Question 9.
An ideal gas is taken from (Pi, Vi) to (Pf, Vf) in three different ways. Identify the process in which the work done on the gas the most.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 1
(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer:
(b) Process B

Hint:
When work is done on the gas, compression takes place. Final pressure Pf will be increased and remains constant. It is shown by process B.

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A → B → is shown below. (Processes D → A and B → C are adiabatic)
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 2
The corresponding PV diagram for the process is (all figures are schematic)
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 3
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 4

Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is:
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer:
(a) 8280 K

Hint:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 5

Question 12.
Identify the state variables given here.
(a) Q, T, W
(b) P,T,U
(c) Q, W
(d) P,T,Q
Answer:
(b) P,T,U

Question 13.
In an isochoric process, we have:
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
Answer:
(a) W = 0

Hint:
Work done in an isochoric process is zero.

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018)
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(b) 20%

Hint:
T2 = 0° C = 0 + 273 = 273 K
T1 = 100° C = 100 + 273 = 373 K
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 6

Question 15.
An ideal refrigerator has a freezer at a temperature of -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is:
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer:
(c) 40.2°C

Hint:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 7

I. Short Answer Questions:

Question 1.
‘An object contains more heat’- is it a right statement? If not why?
Answer:
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at a higher temperature to an object at lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.

Question 2.
Obtain an ideal gas law from Boyle’s and Charles’law.
Answer:
According to Boyle’s law, Pressure ∝ \(\frac { 1 }{ 2 }\)
i.e., P ∝ \(\frac { 1 }{ V }\)
According to Charle’s law,
Volume ∝ Temperature, i.e., V ∝ T
By combining these two laws, we get
PV= CT … (1)
Where C is a positive constant
But C = k x Number of particles (N)
C = kN
Where k – Boltzmann’s constant.
∴ Equation (1) becomes
PV = NkT

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 3.
Define one mole.
Answer:
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).

Question 4.
Define specific heat capacity and give its unit.
Answer:
Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or 1 °C.
The SI unit for specific heat capacity is J kg-1 K-1.

Question 5.
Define molar specific heat capacity.
Answer:
Heat energy required to increase the temperature of one mole of substance by IK or 1°C.

Question 6.
What is a thermal expansion?
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Question 7.
Give the expressions for linear, area and volume thermal expansions.
Answer:
(i) Linear expansion:
\(\frac { ∆L }{ L }\) = α ∆T ⇒ αL = \(\frac { ∆L }{ L∆T }\)

(ii) Area expansion:
\(\frac { ∆A }{ A }\) = 2α ∆T ⇒ αA = \(\frac { ∆A }{ A∆T }\)

(iii) Volume expansion:
\(\frac { ∆V }{ V }\) = 3α ∆T ⇒ αv = \(\frac { ∆V }{ V∆T }\)

Question 8.
Define latent heat capacity. Give its unit.
Answer:
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material. The SI unit for latent heat capacity is J kg-1.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 9.
State Stefan-Boltzmann law.
Answer:
The total amount of heat energy radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.

Question 10.
What is Wien’s law?
Answer:
The wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.

Question 11.
Define thermal conductivity. Give its unit.
Answer:
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady-state conditions is known as the thermal conductivity of a material.

Question 12.
What is a black body?
Answer:
A black body is one which neither reflects nor transmits but absorbs whole of the radiation incident on it.

Question 13.
What is a thermodynamic system? Give examples.
Answer:
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of a large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V), and Temperature (T). The remaining part of the universe is called the surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas, and radiation.

Question 14.
What are the different types of thermodynamic systems?
Answer:
(i) Open system can exchange both matter and energy with the environment.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 8
(ii) Closed system exchange energy but not matter with the environment.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 9
(iii) Isolated system can exchange neither energy nor matter with the environment.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 10

Question 15.
What is meant by ‘thermal equilibrium’?
Answer:
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

Question 16.
What is mean by state variable? Give example.
Answer:
The quantities that are used to describe the equilibrium states of a Thermodynamic system. .
Example: Pressure, Volume, Temperature.

Question 17.
What are intensive and extensive variables? Give examples.
Answer:
Extensive variable depends on the size or mass of the system.
Example: Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.

Question 18.
What is an equation of state? Give an example.
Answer:
The equation that relates the state variables in a specific manner is called equation of state.
Examples:
(i) Gas equation
PV = NkT

(ii) Vander Waal’s equation, .
(p +\(\frac{a}{v^{2}}\))(v – b) = RT

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 19.
State Zeroth law of thermodynamics,
Answer:
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Question 20.
Define the internal energy of the system.
Answer:
The sum of kinetic and potential energies of all the molecules of the system with respect to the centre of mass of the system.

Question 21.
Are internal energy and heat energy the same? Explain.
Answer:
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”

Question 22.
Define one calorie.
Answer:
To raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat energy was measured in calorie.
1 cal = 4.186J

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
In Joule’s experiment, when the masses fall, the paddle wheel turns. The frictional force between paddle wheel and water causes a rise in temperature of water. It is due to the work done by the masses. This infers that gravitational potential energy is transformed into internal energy of water. Thus Joule converted mechanical energy into heat energy.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 24.
State the first law of thermodynamics.
Answer:
‘Change in internal energy (AU) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.

Question 25.
Can we measure the temperature of the object by touching it?
Answer:
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.

Question 26.
Give the sign convention for Q and W.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 11

Question 27.
Define the quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical, and chemical equilibrium with its surroundings throughout.

Question 28.
Give the expression for work done by the gas.
Answer:
In general, the work done by the gas by increasing the volume from Vi to Vf is given by
W = \(\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}\)

Question 29.
What is PV diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer:
When increasing the temperature of a gas at constant pressure, gas expands and consume some heat to do work. It is not happened, while increasing the temperature of a gas at constant volume. Hence less heat energy is required to increase the temperature of the gas at constant volume. Hence CP is always greater than Cv.

Question 31.
State the equation of state for an isothermal process.
Answer:
The equation of state for isothermal process is given by,
PV = constant

Question 32.
Give an expression for work done in an isothermal process.
Answer:
W = μRTln \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\)

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at the constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT

Question 34.
Apply first law for (i) an isothermal (ii) adiabatic (Hi) isobaric processes.
Answer:
(i) For an isothermal process temperature is maintained as constant. Hence,
∆V= 0
∴ W = P∆V = 0
According to first law of thermodynamics
∆U = Q – W
= Q – 0
= Q
∴ ∆U = Q

(ii) For adiabatic process, Q = 0
By applying first law of thermodynamics
∆U = Q – W
∆U = |- W| = W
∆U = W = \(\frac{\mu R}{\gamma-1}\) = (Ti – Tf)

(iii) (a) For isobaric expansion Q > O
According to first law of thermodynamics
∆U = – Q – W
∆U = Q – P∆V

(b) For isobaric compression Q < O.
According to first law of thermodynamics
∆U = – Q – W
= – Q – P∆V

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic proc ess is given by,
PVγ = Constant

Question 36.
State an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by,
P = (\(\frac { μR }{ V }\))T

Question 37.
if the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? ff not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 38.
Draw the PV diagram for,
(a) isothermal process
(b) Adiabatic process
(c) lsobaric process
(d) lsochoric process
Answer:
(a) isothermal process:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 12
(b) Adiabatic process:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 13
(c) lsobaric process:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 14
(d) lsochoric process:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 15

Question 39.
What is a cyclic process?
Answer:
It is a process in which the thermodynamic system returns to its initial state after undergoing a series of changes.

Question 40.
What is meant by a reversible and irreversible processes?
Answer:
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.

Question 41.
State Clausius form of the second law of thermodynamics.
Answer:
“Heat energy always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.

Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
Answer:
It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work.

Question 43.
Define heat engine.
Answer:
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Question 44.
What are processes involves in a Carnot engine?
Answer:

  1. Isothermal expansion
  2. Adiabatic expansion
  3. Isothermal compression
  4. Adiabatic compression.

Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
Answer:
No.
According to second law it can’t be completely converted. It is not possible to convert heat completely converted into work, in a cyclic process. But, in Non-cyclic process, heat can be completely converted into work.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 46.
State the second law of thermodynamics in terms of entropy.
Answer:
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural processes should occur.

Question 47.
Why does heat flow from a hot object to a cold object?
Answer:
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will be decreased. So second law thermodynamics is violated.

Question 48.
Define the coefficient of performance.
Answer:
The ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = ß = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}}\).

III. Long Answer Questions:

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
The spontaneous flow of energy from the object at higher temperature to the one at lower temperature. When they are contact. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating.

Example: ‘A hot cup of coffee has more heat’. This statement is correct or wrong?

When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and that flows from a body at higher temperature to an other body at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat’ is wrong, instead ‘coffee is hot’ will be appropriate.

Meaning of Work: If a body undergoes a displacement then only work is said to be done by the body. Work is also the transfer of energy by other means such as moving a piston of a cylinder containing gas.

Example: When you rub your hands against each other the temperature of the hand is increased. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now when you place your hands on the chin, the temperature of the chin increases.

This is because the hands are at higher temperature than the chin. In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.

The temperature in the system will increase and sometimes may not by doing work on the system. Like heat, work is also not a quantity and through the work energy is transferred to the system.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 2.
Discuss the ideal gas laws.
Answer:
(i) Boyles law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.
P ∝ \(\frac { 1 }{ v }\)

(ii) Charles law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T

(iii) By combining these two equations we have PV= CT … (1)

(iv) Let the constant C as k times the number of particles N. Here k is the Boltzmann constant and it is found to be a universal constant. So the ideal gas law can be stated as follows,
PV = NkT … (2)
If the gas consists of p mole of particles then total number of particles is
N = μNA … (3)
Where NA is Avogadro’s number.
Substituting the value of N in the equation (2) we get
PV = μNAkT
Here NAk = R
So, the ideal gas law can be written for μ mole of gas is
PV= pRT
It is called equation of state for an ideal gas.

Question 3.
Explain in detail the thermal expansion.
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. At that time, relative change in the size of solids is small.

Liquids expand more than solids because, they have less intermolecular forces than solids. In the case of gas molecules, the intermolecular forces are almost negligible and so they expand much more than solids.

Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\frac { ∆L }{ L }\) is directly proportional to ∆T.
\(\frac { ∆L }{ L }\) = αL∆T
Therefore, αL = \(\frac { ∆L }{ L∆T }\)

Area Expansion: For an infinitesimal change in temperature ∆T the fractional change in area of a substance is directly proportional to ∆T and it can be written as
\(\frac { ∆A }{ A }\) = αA∆T
Therefore, αA = \(\frac { ∆A }{ A∆T }\)
Volume Expansion: For a very small change in temperature ∆T the fractional change in volume \(\frac { ∆V }{ V }\) of a substance is directly proportional to AT.
\(\frac { ∆V }{ V }\) = αV∆T
Therefore, αV = \(\frac { ∆L }{ V∆T }\)

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 4.
Describe the anomalous expansion of water. How is it helpful in our lives?
Answer:
On heating liquids expand and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as . it is cooled from room temperature, until it reach 4°C . Below 4°C the volume increases and so the density decreases. It means that the water has a maximum density at 4°C. This behaviour of water is called anomalous expansion of water.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 16
During winter in cold countries, the surface of the lakes would be at lower temperature than the bottom. Since the solid water (ice) has lower density than its liquid form,’below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). It is due to the anomalous expansion of water. The water in lakes and ponds freeze only at the top. So, the species living in the lakes will be safe at the bottom.

Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, due to transformation of heat. The heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be mathematically expressed as
Qgain = – Qlost
Qgain + Qlost = 0
Heat gained or lost can be measured with a calorimeter.
When a sample is heated at high temperature (T1) and immersed into water at room temperature (T2) in the calorimeter. After some time both sample and water reach a final equilibrium temperature Tf. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water.
Qgain = – Qlost
As per the sign convention, the heat energy lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity,
Qgain= m2s2(Tf – T2)
Qlost = m1s1(Tf – T1)
Here s1 and s2 specific heat capacity of hot sample and water respectively.
So we can write,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 17

Question 6.
Discuss various modes of heat transfer.
Answer:
There are three modes of heat transfer: Conduction, Convection and Radiation.

Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one.

Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Example: Boiling of water in a pot.

Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example: Solar energy from the Sun.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 7.
Explain in detail Newton’s law of cooling.
Answer:
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac { dQ }{ dt }\) ∝ – (T – Ts) … (1)
The negative sign implies that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T = Temperature of the surrounding
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 18
From the above graph, it is clear that the rate of cooling is high initially and decreases with falling temperature.

Consider a body of mass m and specific heat capacity s at temperature T. Let T be the temperature of the surroundings. If the temperature falls by a small amount dl in time dt, then the amount of heat lost is,
dQ = msdT
Dividing both sides of equation (2) by dt
\(\frac { dQ }{ dt }\) ∝ \(\frac { msdT }{ dt }\) … (3)
From Newton’s law of cooling
\(\frac { dQ }{ dt }\) ∝ (T – Ts)
\(\frac { dQ }{ dt }\) ∝ (T – Ts) … (4)
Where a is some positive constant.
From equation (3) and (4)
– a(T – Ts) = ms\(\frac { dT }{ dt }\)
\(\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{s}}\) = – \(\frac { a }{ ms }\)dt … (5)
Integrating equation (5) on both sides,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 19
Where b1 is the constant of integration. Taking exponential both sides, we get
T = Ts + b2 \(e^{-\frac{a}{m s} t}\)
here b2 = \(e^{b_{1}}\) = constant.

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Answer:
Wien’s law states that, ‘the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body’.
λm ∝ \(\frac { 1 }{ T }\)
(or)
λm = \(\frac { b }{ T }\) … (1)
It is implied that if temperature of the body increases, maximal intensity wavelength (λm) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum.

Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)?

The Sun is approximately considered as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation (1),
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 20
It is the wavelength at which maximum intensity is 508nm. Since the Sun’s temperature is around 5700K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum.

The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 21

Question 9.
Discuss the,
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium
(iv) Thermodynamic equilibrium.
Answer:
(i) Thermal equilibrium: Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

(ii) Mechanical equilibrium: A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermodynamic system or on the surrounding by thermodynamic system.

(iii) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.

(iv) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Answer:
Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy which is equal to 2mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.

This causes a rise in temperature of the water. This confirms that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat energy. He found that to raise lg of an object by 1°C , 4.186J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 22
This is called Joule’s mechanical equivalent of heat.

Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Answer:
Let us consider a gas contained in the cylinder fitted with a movable piston. When the gas is expanded quasi-statically by pushing the piston by a small distance dx as shown in Figure. Because the expansion occurs quasi- statically. The pressure, temperature and internal energy will have unique values at every instant.
The small work done by the gas on the piston
dW = F dx … (1)
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 23
The force exerted by the gas on the piston F – PA.
A – area of the piston and
P – pressure exerted by the gas on the piston.
Equation (1) can be rewritten as
dW = PA dx … (2)
But A dx = dV change in volume during this expansion process.
Hence the small work done by the gas during the expansion is given by
dW = dV … (3)
It is noted that is positive since the volume is increased. In general the work done by the gas by increasing the volume from Vi to Vf is given by
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathbf{P} d \mathrm{~V}\)
Suppose if the work is done on the system, then Vi > Vf.
Then, W is negative.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 12.
Derive Meyer’s relation for an ideal gas.
Answer:
Let us consider μ mole of an ideal gas in a container with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dT. Since no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If Cv is the molar specific heat capacity at constant volume,.
dU= μCvdT … (1)
When the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = μCpdT … (2)
If W is the workdone by the gas in this process, then
W = P dV … (3)
But from the first law of thermodynamics,
Q = dU + W … (4)
Substituting equations (1), (2) and (3) in (4), we get,
μCpdT = μCvdT + P dV … (5)
For mole of ideal gas, the equation of state is given by,
PV = μRT
⇒ PdV + VdP = μRdT … (6)
Since the pressure is constant, dP = 0
∴ CpdT = CvdT + RdT
∴ Cp = Cv + R
(or) Cp – Cv = R … (7)
This relation is called Meyer’s relation.

Question 13.
Explain in detail the isothermal process.
Answer:
It is a process in which both the pressure and volume of a thermodynamic system will change at constant temperature. The ideal gas equation is
PV = μRT
Here, T is constant for this process.
So the equation of state for isothermal process is given by,
PV= constant … (1)
It is implied that if the gas goes from one equilibrium state (P1,V1) to another equilibrium state (P2,V2) the following relation holds for this process
P1,V1 = P2,V2 … (2)
Since PV = constant, P is inversely proportional to V(P ∝ \(\frac { 1 }{ V }\)). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.
The following figure shows the PV diagram for quasi-static isothermal expansion and quasi-static isothermal compression. For an isothermal process since temperature is constant, the internal energy is also constant. It implies that dU or ∆U – 0.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 24
For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W … (3)
From equation (3), it implies that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat energy to the system, the temperature will increase. For isothermal process it is not true. When the piston of the cylinder is pushed, the isothermal compression takes place. This will increase the internal energy which will flow out of the system through thermal contact.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 14.
Derive the work done in an isothermal process.
Answer:
Let us consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (Pi,Vi) to the final state (Pf, Vf. Let us calculate the work done by the gas during this process.
gas, W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid.
P = \(\frac { μRT }{ V }\) … (2)
Substituting equation (2) and (1) we get,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 25
In equation (3), μRT is taken out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get
W = μRTln\(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) … (4)
Since we have an isothermal expansion,
\(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) > 1, so \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) > 0. As a result the work done by the gas during an isothermal expansion is positive.
Equation (4) is true for isothermal compression also. But in an isothermal compression \(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) < 1. So \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) < 0.
Hence the work done on the gas in an isothermal compression is negative. In the PV diagram the work done during the isothermal expansion is equal to the area under the graph as shown in figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 25
Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas that turns out to be the area with a negative sign.

Question 15.
Explain in detail an adiabatic process.
Answer:
It is a process in which no heat energy flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. Hence the pressure, volume and temperature of the system may change in an adiabatic process.

From the first law for an adiabatic process, ∆U = W. This implies that the work is done by the gas at the expense of internal energy or work is done on the system that increases its internal energy. The adiabatic process can be achieved by the following methods

(i) Thermally insulating the system from surroundings. So that no heat energy flows into or out of the system. For instance, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 27
(ii) If the process occurs so quickly so that there is no time to exchange heat with surroundings even though there is no thermal insulation

The equation (1). Implies that if the gas goes from an equilibrium state (Pi, Vi) to another equilibrium state (Pf,Vf) adiabatically then it satisfies the relation,
PVγ = constant … (1)
The equation (1) Here γ is called adiabatic exponent ( γ = \(\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}\)) which depends on the nature of the gas.
The equation (1). Implies that if the gas goes from an equilibrium state (Pi,Vi) to another equilibrium state (Pf, Vf) adiabatically then it satisfies the relation,
PiViγ = PfVfγ … (2)
The PV diagram of an adiabatic expansion and adiabatic compression process.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 28
The PV diagram for an adiabatic process is also called adiabatic. It is noted that the PV diagram for isothermal and adiabatic processes look similar. But actually the adiabatic curve is steeper than isothermal curse.
Let us rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac { μRT }{ V }\). Substituting this equation in the equation (1), we have
\(\frac { μRT }{ V }\)Vγ = constant
(or) \(\frac { T }{ V }\)Vγ = \(\frac { constant }{ μR }\)
Note here that is another constant. So it can
be written as
TVγ-1 = constant … (3)
From the equation (3) it is known that if the gas goes from an initial equilibrium state (Ti,Vi) to final equilibrium state (Tf, Vf) adiabatically then it satisfies the relation
\(\mathrm{T}_{i} \mathrm{~V}_{i}^{\gamma-1}=\mathrm{T}_{f} \mathrm{~V}_{f}^{\gamma-1}\) … (4)
The equation of state for adiabatic process can also be written in terms of T arid P as
\(\mathrm{T}^{\gamma} \mathrm{P}^{1-\gamma}\) = constant

Question 16.
Derive the work done in an adiabatic process.
Answer:
Let us consider p moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston A is fitted in the cylinder as shown in Figure. It’s area of cross-section be A.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 29
Let W be the work done when the system goes from the initial state (PiViTi) to the final state (PfVfTf) adiabatically.
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
It is assumed that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PVγ = constant (or) P = \(\frac{\text { constant }}{\mathrm{V}^{\gamma}}\) can be substituted in the equation (1), we get
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 30
From ideal gas law,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 31
In adiabatic expansion, work is done by the gas. i.e., Wadia is positive. As Ti > Tf the gas cools during adiabatic expansion.

In adiabatic compression, work is done on the gas. i.e., Wadia is negative. As Ti < Tf hence the temperature of the gas increases during adiabatic compression.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 17.
Explain the isobaric process and derive the work done in this process.
Answer:
Isobaric process is a thermodynamic process that occurs at constant pressure. Inspite of the pressure consistency of this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
V = (\(\frac { μR }{ P }\))T … (1)
Here, \(\frac { μR }{ P }\) = constant
In an isobaric process the temperature is directly proportional to volume.
V ∝ T(Isobaric process) … (2)
It is implied that for a isobaric process, the V-T graph is a straight line passing through the origin.
Work done by the gas
W= \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\)
In an isobaric process, the pressure is constant, So P comes out of the integral,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 32
∆V – change in the volume. If AV is negative, W is also negative. It is implied that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas. From the ideal gas equation.
equation (4) can be rewritten as
PV = μRT and V = \(\frac { μRT }{ P }\)
Substituting this in equation (4) we get,
W = μRTf(1 – \(\frac{\mathrm{T}_{i}}{\mathrm{~T}_{f}}\)) … (5)
In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the following Figure is equal to the work done by the gas.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 33
The first law of thermodynamics for isobaric process is given by,
∆U = Q – P∆V

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables. The pressure-volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 34
The equation of state for an isochoric process is given by,
P = \(\frac { μR }{ V }\) … (1)
Here \(\frac { μR }{ V }\) = constant
∴ P ∝ T
It is inferred that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (Pi, Ti) to (Pf, Tf) at constant volume, then the system satisfies the following equation,
\(\frac{P_{i}}{T_{i}}=\frac{P_{f}}{T_{f}}\)
For an isochoric processes, AV=0 and W-0. Then the first law of thermodynamics becomes
∆U = Q … (3)
It is implied that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
The first law of thermodynamics explains well the inter convertibility of heat and work. But is not indicated the direction of change.
Example:
(i) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law of thermodynamics, it is possible for the energy to flow from hot object to cold object and viceversa. But in nature the direction of heat flow is always from higher temperature to lower temperature.

(ii) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted into the kinetic energy of the car. Hence the first law is not sufficient to explain many of natural phenomena.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 20.
Explain the heat engine and obtain its efficiency.
Answer:
Heat engine is a device which takes heat as input and converts this heat into work by undergoing a cyclic process.
A heat engine has three parts:
(i) Hot reservoir
(ii) Working substance
(iii) Cold reservoir
A Schematic diagram for heat engine is given below in the figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 35

  • Hot reservoir (or) Source: It supplies heat to the engine. Which is always maintained at a high temperature TH.
  • Working substance: It is a substance like gas or water, that converts the heat supplied into work.
  • Cold reservoir (or) Sink: The heat engine ejects some amount of heat (QL) into cold reservoir after it doing work. It is always maintained at a low temperature TL.

The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.

The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process. Let the working substance absorb heat QH units from the source and reject QL units to the sink after doing work W units.
We can write,
Input heat = Work done + ejected heat
QH = W + QL
W = QH – QL
Then the efficiency of heat engine
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 36

Question 21.
Explain in detail Carnot heat engine.
Answer:
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine consists of four parts that are given below.

  1. Source: It is the source of heat maintained at constant high temperature TH. Without changing its temperature any amount of heat can be extracted from it.
  2. Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
  3. Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
  4. Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.

The four parts are shown in the following Figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 37
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 38
Carnot’s cycle: In camot’s cycle, the working substance is subjected to four successive reversible processes.
Let the initial pressure, volume of the working substance be P1, V1.
When the cylinder is placed on the source, the heat (QH) flows from source to the working substance (ideal gas) through the bottom of the cylinder.
The input heat increases the volume of the gas.So the piston is allowed to move out very’ slowly.
W1 is the work done by the gas it expands from volume V1 to volume V2 with a decrease of pressure from P1 to P2.
Then the work done by the gas (working substance) is given by
∴ QH = QA→B \(\int_{V_{1}}^{\mathrm{V}_{2}} \mathrm{P} d \mathrm{~V}\)
When the cylinder is placed on the insulating stand and the piston is allowed to move out, the gas expands adiabatically from volume V2 to volume V3 the pressure falls from P2 to P3. The temperature falls to TLK.
The work done by the gas in an adiabatic expansion is given by,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 39
Area under the curve BC.
Isothermal compression from (P3, V3, TL) to (P4, V4, TL).
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 40
Let the cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P4 and V4 respectively.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 41
Let the cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P1, volume V2 and temperature TH, Which is shown by the curve DA in the P – V diagram.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 42
= – Area under the curve DA
Let ‘W’ be the net work done by the working substance in one cycle
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 43
The net work done by the Carnot engine in one cycle is given by,
W = WA→B – WC→D
Net work done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 44

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 45
Let us omit the negative sign. Since we are interested in only the amount of heat (QL) ejected into the sink, we have
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 46
By applying adiabatic conditions, we get,
TH V2γ-1 = TL V3γ-1
TH V1γ-1 = TL V4γ-1
By dividing the above two equations, we get
\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)γ-1 = \(\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\right)\)γ-1
Which implies that
\(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\) … (5)
Substituting equation (5) and (4), we get
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
∴ The efficiency
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) … (1)
According to efficiency of Carnot’s engine equation, quantity \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\).
The quantity \(\frac { Q }{ T }\) is called entropy. It is a very important thermodynamic property of a System, which is also a state variable.
\(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) is the entropy received by the Carnot engine from hot reservoir and \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\) is entropy given out by the Carnot engineLto the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. It is proved in equation (1).
But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\) > \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\).
In fact the second law of thermodynamics can be reformulated as follows “For all the processes that occur in nature(irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process must occur.

Question 24.
Explain in detail the working of a refrigerator.
Answer:
A refrigerator is a Carnot’s engine working in the reverse order which is shown in the figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 47
8.17 Schematic diagram of a refrigerator The working substance (gas) absorbs a quantity of heat QL from the cold body (sink) at a lower temperature TL. A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) i.e., the atmosphere at TH. When we stand beneath of the refrigerator, we can feel warmth air.
From the first law of thermodynamics, we have QL + W = QH
Hence the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

IV. Numerical problems:

Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 48
The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa.
The pressure inside the balloon P = 1.8 x 105 P
Room temperature
T = 273 + 30 = 303K.
Let the radius of the balloon be R = 10 x 10-2m.
Volume of the balloon
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 49

Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of
the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 50
= 0.9 x 10³Pa
PV = NkT
Since volume is unity V = 1
Equation (1) becomes
P = NkT
Where k is Boltzmann constant
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 51

Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V1 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
In first chamber let the pressure be P1
In second chamber let the pressure be P2
In first chamber let the volume be V1
In second chamber let the volume be V2
In first chamber let the temperature be T1
In second chamber let the temperature be T2
According to conservation of energy,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 52

Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion αL is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its centre and perpendicular to an axis of the rod.
Answer:
Moment of inertia of uniform rod of mass M, and length L about axis passing through its centre and perpendicular to an axis of rod is given by
I = \(\frac{\mathrm{ML}^{2}}{12}\)
The increase in length of the rod
∆l = LαL∆T
αL – Coefficient of linear expansion.
∆T – change in temperature
After the rod is heated,
Moment of inertia,
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 53

Question 5.
Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) isobaric process.
Answer:
(a) For isochoric process:
V = V0 = constant
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 54
The points a and b are represented as a = (P1V0T0) b = (P2, V0, T2)

(b) For isothermal process:
T = T0 = constant
PV = nRT0
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 55
P(T) = multivalue
P(T) = \(\frac{n \mathrm{RT}_{0}}{\mathrm{~V}}\)
The points a and b are represented as
a = (P1V1,T) b = (P2V0T0)

(c) For isobaric process:
P = P0 = constant
T(V) = \(\frac{\mathrm{P}_{0} \mathrm{~V}}{n^{2}}\)
P0V = nRT
P(T) = P0
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 56

Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 57

Question 7.
The temperature of a normal human body is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body)
Answer:
Normal temperature of human body = 98.6°F
Conversion of F° to C°
C = \(\frac { (F-32) }{ 1.8 }\)
= \(\frac { (98.6-32 }{ 1.8 }\)
= \(\frac { 66.6 }{ 1.8 }\)
= 37°C
Conversion of C° to Kelvin
T = 37 + 273 = 310
Peak wavelength at 98.6°F is
λm = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{-3}}{310}\)
= 0.009348 x 10-3 m
= 9348 x 10-9
= 9348 nm
Final temperature of the human body = 104°F
Conversion of F° to C°
C = \(\frac { F-32 }{ 1.8 }\)
= \(\frac { 104-32 }{ 1.8 }\)
= \(\frac { 72) }{ 1.8 }\)
Conversion of 40° to Kelvin
T = 40+ 273 = 313 K
Peak wavelength at 104°F is
λm = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{3}}{313}\)
= 0.009258 x 10-3 m
= 9258 x 10-9 = 9258 nm
λmax = 9348 nm at 98.6°F
λmax = 9258 nm at 104°F

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Answer:
Percentage of increased volume = 4%
\(\frac { ∆V }{ V }\) x 100 = 4%
γ = 1.4
In adiabatic process
PVγ = Constant
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 58

Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
Answer:
P1 = 1 atmospheric pressure, V1 = V, V2 = \(\frac { V }{ 8 }\)
T1 = 20 + 273 – 293K, T2 – ?, γ = 1.4.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 59
∴ The temperature of the compressed air = 400°C

Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 60
Answer:
Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x direction and V is taken along y-direction. Analyze the nature of heat exchange in each process.

Process 1 to 2: increase in volume. So heat must be added.

Process 2 to 3: Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.

Process 3 to 1: Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 61
Explanation: In the graph, during the process (1 to 2), the gas undergoes isothermal expansion. It receives certain amount of heat from the surroundings. It uses this heat in doing the work. Hence internal energy of the gas remains unchanged.

During the process represented by (2 to 3) the gas is heated at constant volume. Since no work is done and volume does not change, the process is isochoric process.

Since heat is transferred to the gas from the surroundings, the internal energy of the gas is increased. During the process represented by (3 to 1) the gas is compressed isobarically. Work is done on the gas. Since temperature drops internal energy is reached. Hence the gas gives up heat to the surroundings.

Question 11.
An ideal gas is taken in a cyclic process as shown in the figure. Calculate
(a) work done by the gas.
(b) work done on the gas.
(c) Net work done in the process
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 62
Answer:
(a) Let the work done by the gas along AB be W. From the fig pressure,
P = 600 N/m²
From the fig change in volume,
∆V = 3
∴Work done,
W = PAV
= 600 x 3 = 1800J
= 1.8 kJ

(b) Isobaric compression take place and work is done on the gas along BC.
W = – PAV
∆V = (6 – 3) = 3
P = +400N/m²
∴ W= – 400 x 3
= – 1200
= – 1.2 kJ

(c)
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 64

Question 12.
For a given ideal gas 6 x 105 J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m² at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change in internal energy of the gas
(c) graph this process in PV and TV diagram.
Answer:
Heat energy supplied to the gas
Q = 6 x 105J

(b) Change in internal energy
∆U = ∆Q – AW
= ∆Q – P∆V
= 6 x 105 – 2.026 x 105
= 3,974 x 103
∆U = 3.974 kJ

(c)
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 63

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency.
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) By increasing the temperature of the hot reservoir from 300°C tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Temperature of cold reservoir
T1 = 100°C
= 100 + 273 – 373 K
Temperature of hot reservoir
TH = 300 + 273 = 573 K
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
η = 1 – \(\frac { 373 }{ 573 }\)
η = 1 – \(\frac { 573 – 373 }{ 573 }\)
= \(\frac { 200 }{ 753 }\)
= 0.349 x 100
Initial efficiency = 34.9%

(a) By decreasing the temperature of the cold reservoir from 100°C to 50PC.
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 67

(b) By increasing the temperature of the hot reservoir from 300°C to 350°C,
T1 = 350 + 273 = 623K
T1 = 100 + 273 =373K
∴ Efficiency,
η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
η = 1 – \(\frac { 373 }{ 623 }\)
η = \(\frac { 623 – 373 }{ 623 }\)
η = \(\frac { 250 }{ 623 }\)
η = 0.4012
η = 0.4012 x 100 = 40.12%
∴ Method (a) is more efficient than method (b).

Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained atatemperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Temperature of a source,
T1 = 327°C = 327 + 273 = 600 k
η = \(\frac { 45 }{ 100 }\) = 0.45
Efficiency η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
∴ 0.45 = 1 – \(\frac{\mathrm{T}_{2}}{600}\)
\(\frac{\mathrm{T}_{2}}{600}\) = 1 – 0.45 = 0.55
∴ T2 = 0.55 x 600 = 330 K
η = 60% = \(\frac { 60 }{ 100 }\) = 0.6
0.6 = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\) = 1 – \(\frac{330}{\mathrm{~T}_{1}}\)
\(\frac{330}{\mathrm{~T}_{1}}\) = 1 – 0.6 = 0.4
In take temperature
T1 = \(\frac { 330 }{ 0.4 }\) = 825K
= 825 – 273
= 552°C
The intake temperature = 552°C

Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 8 Heat and Thermodynamics 68

Samacheer Kalvi 11th Bio Botany Guide Chapter 3 Vegetative Morphology

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 3 Vegetative Morphology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

11th Bio Botany Guide Vegetative Morphology Text Book Back Questions and Answers

Part-I

I. Choose the Right Answer:

Question 1.
Roots are
a. Descending-negatively geostrophic positively phototrophic
b. Descending-positively geostrophic negatively phototrophic
c. Ascending, positively geostrophic negatively phototrophic
d. Ascending, negatively geostrophic negatively phototrophic
Answer:
b. Descending-positively geostrophic negatively phototrophic

Question 2.
When the root is thick and fleshy but does not take a definite shape is said to be
a. Nodulose root
b.Tuberous root
c. Moniliform root
d. Fasciculated root
Answer:
b. Tuberous root

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 3.
Example for negatively geotrophic roots
a. Ipomoea, Dahlia
b. Asparagus, Ruellia
c. Vitis, Portulaca
d. Avicennia, Rhizophora
Answer:
d. Avicenniarhizophora

Question 4.
Cureumaamada curcuma domestica Asparagus maranta are examples of
a. Tuberous root
b. Beaded root
c. Moniliform root
d. Nodulose root
Answer:
d. Nodulose root

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 5.
Bryophyllum and Dioscorea are examples for
a. Foliar bud, apical bud
b.Foliar bud, cauline bud
c. Cauline bud, apical bud
d.Cauline bud, fohar bud
Answer:
b. Foliar bud, cauline bud

Two Marks

Question 6.
Why lateral roots are endogenous?
Answer:
Lateral roots arise from the pericycle, part Eg. Inner part – so it is known as endogenous in origin.

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 7.
Write the similarities and differences between

  1. Avicennia & trapa
  2. Banyan & silk cotton
  3. Fusiform and Napiform root

Answer:
I. Avicennia & trapa

 Avicennia

Trapa (water chestnut)

Live in marshy leaves Live in aquatic habitat
Has negatively geotrophic root known as respiratory roots-with pneumatophores help in exchange of gases Has photosynthetic or assimilatory roots – help in photosynthesis.

II. Banyan & silk cotton

Banyan

Silk cotton

Has pillar roots- grow vertically downward from the lateral branches to soil -to give additional support. Has broad plant-like outgrowths develop obliquely towards the base all around the trunk – to give support.

III. Fusiform and Napiform root

Fusiform

Napiform

Roots are swollen in the middle and tapering towards both ends (like a spindle-shaped Eg. Raphanus sativus Roots broad and suddenly tapers like a tall at the apex (top-shaped) Eg. Beta vulgaris

Question 8.
How root climbers differ from stem climbers?
Answer:

Root Climbers

Stem Climbers

Climb with the help of adventitious root( arise from the node) Eg: Piperbetal, piper nigrum No special climbing structures – stem itself coils around the support Eg: Ipomoeaellittoria

Question 9.
Compare sympodial branching with monopodial branching.
Answer:

Sympodial

Monopodial

Determinate — growth
Terminal bud ceases -to grow-and further growth by lateral buds- Eg. Cycas
Indeterminate growth
TerminaI bud — grows uninterrupted and produce several lateral branches — Eg. Polyalthi

Question 10.
Compare pinnate unicostate and palmate multicoastate venation?
Answer:

Pinnately reticulate (unicoastate)

Palmately reticulate (multicoastate)

One mid rib in the centre with many laterlal braches
Eg: Mangifera indica
Several veins arise from the help of peticole & run parallel & unite at the apex – 2 types
Divergent Eg: Borassussflabellifereg,
Covergant Eg. Paddy

Part – II

11th Bio Botany Guide Vegetative Morphology Additional Important Questions and Answers

Choose the Correct Answer:

Question 1.
The study about external features of an organism is known as …………… .
(a) morphology
(b) anatomy
(c) physiology
(d) taxonomy
Answer:
(a) morphology

Question 2.
Onion lettuce, fennel, radish, cabbage are examples of
a. perennial
b. annual.
c. centennial
d. biennial
Answer:
d.biennial

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 3.
The branch of science that deals with the classification of organisms is called as …………… .
(a) taxonomy
(b) morphology
(c) physiology
(d) anatomy
Answer:
(a) taxonomy

Question 4.
Palmately reticulate, convergent venation is seen in
a. zizipus
b. mango
c. cucurbita
d. carica papaya
Answer:
a. zizipus

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 5.
Rolling or folding of individual leaves may be as follows
a. pteryix
b. ptyxis
c. typxis
d. xyptes
Answer:
b. ptyxis

Question 6.
The general form of a plant is referred to as …………… .
(a) habitat
(b) structure
(c) habit
(d) shape and size
Answer:
(c) habit

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 7.
These are examples for shrubs
a. coconut and Palmyra
b. mango and bamboo
c. hibiscus and castor
d. cotton and bougainvillea
Answer:
c. Hibiscus and Castor

Question 8.
Angiosperms are also known as
a. Bryophytes
b. pteridophytes
c. Magnoliophytes
d. Tracheophytes
Answer:
c.Magnoliophytes

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 9.
Climbers are also called as …………… .
(a) herbs
(b) trees
(c) vines
(d) shrubs
Answer:
(c) vines

Question 10.
The phyllotaxy seen in Nerium is known as
a. whorled
b. opposite
c. alternate
d. ternate
Answer:
d.Ternate

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 11.
…………… is an example for xerophytes.
(a) Lichens
(b) Euphorbia
(c) Ficus
(d) Ipomoea
Answer:
(b) Euphorbia

II. FILL UP THE BLANKS

Question 1.

1.

Definition

Term used

Example

a. The occurrence of 2 kinds of leaves (a)
b. When leaf directly vertically upwards (b) Limnophyllaheterophylla
c. One leaflet articulated to the petiole unifoliolate Grass (c)
d. Lower leaves with longer petioles, Successive leaves with shorter petioles Mosaic (d)

Answer:
a. Heterophylly
b. isobilateral leaf
c. citrus
d. acalypha

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

III. Identify the Diagram

Question 1.
Identify The Diagram and Label ABCD
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 1

Question 2.
IDENTIFY THE DIAGRAM and Label ABCD from the diagram
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 2

Question 3.
Identify the Diagram and Label ABCD from the diagram
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 3

IV. Read the following Assertion and Reason Find the correct answer

Question 1.
Assertion: Rootstock laek root cap and root hairs but they possess terminal but which is a characteristic of stem
Reason: Rootstocks also known as underground stem
a. Assertion and Reason are correct reason explaining stem
b. Assertion and Reason are correct but the reason is not explaining assertion
c. Assertion is true, but Reason wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct -Reason is explaining assertion

Question 2.
Assertion: Avieennia develop special kinds of the root (negatively-geotropic) known as respiratory roots
Reason: They are mangrove plants
a. Assertion and Reason are correct Reason is explaining assertion
b. Assertion and Reason are correct but, the reason is not explaining assertion
c. Assertion is true, but Reason is wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct, but the reason is not explaining assertion.

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

V. Find out the Wrong answer

Question 1.
Buttress roots are not traced in
a. Terminalia arjuna
b. Delonixregia
c. Bombax spp
d. Piper betel
Answer:
d. Piper betel

Question 2.
Among the given which one doesn’t have foliar roots
a. Bryophyllum
b. Begnonia
c. Zamiaculeas
d. Ranunculus
Answer:
d. Ranunculus

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 3.
Among the given, Find out the odd man with reference to the fibrous root system.
a. Eleusineeoracana
b. Pennisetumamericanum
c. Zingiferaoffieinalis
d. Ficusbenahaliensis
Answer:
d. Ficusbenahaliensis

VI. Form the match and Find the Wrong Pair

Question 1.
(1) Tendril as stem modification – Passiflora
(2) Tendril as leaf modification – Lathyrus
(3) Tendril as stipule modification- Smilax
(4) Tendril as a modification of petiole of the leaf – Nepenthes
Answer:
(4) Tendril as a modification of petiole of the leaf – Nepenthes

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 2.
(1) Zingifereffienalis – Rhizome
(2) Eolehicum – Eorm
(3) Allium cepia – Tunicatedbulle
(4) Tuber – Tulipa.spp
Answer:
(4) Tuber – Tulipa .spp

Question 3.
(1) Leaf base – Hypopodium
(2) petiole – Mesopodium
(3) Midrib – Endopodium
(4) Lamenia – Epipodium
Answer:
(3) Midrib – Endopodium

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 4.
(1) Flattened – Cynodon dactyla
(2) Cylindrical cladode – Asparagus
(3) Flattened phylloclade – Opuntia
(4) Cylindrical – Euphorbia antiquorum
Answer:
(3) Flattened phylloclade- Opuntia

Question 5.
(1) Additional outgrowth between leafe base & lamina – Ligute
(2) Sheathing leaf base – Mesopodium
(3) Stiples occur in – Fabaceae
(4) Stiples absent in – Monocots
Answer:

VII. Match And Find The Correct Answer

Question 1.
(1) Unipennate – Eaesalpinia A
(2) Bipinnate – Eoriandrumsativum B
(3) Tripinnate – Neem C
(4) Decompound – Moringa D
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 4
Answer:
b)C-A-D-B

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 6.
Tabulate the Aerial Stem modification
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 5

Question 7.
Draw the structure of prop roots
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 6

Question 8.
Differentiate between Excurrent and decurrent types of stem.
Answer:

Excurrent

Decurrent

Main axis – continuous growth lateral branches shorter and tapper towards tip conical appearance Ex: Polyalthia Lateral branches more vigorous growth- giving rounded spreading appearance Ex: Mangifera

Question 9.
Differentiate between Runner and Sucker:
Answer:

Runner

Sucker

Prostrate branch of aerial stem creeping on the ground and rooting at nodes.
Ex: Centellacynodondaetylon
The underground stem grows obliquely upwards give rise to a new plant.
Ex: Chrysanthemum Bambusa, Musa

Question 10.
Differentiate between Ternate and Whorled Phyilotaxy
Answer:

Ternate

Whorled

At each node, three leaves are attached
Ex: Nerium
At each node more than 3 leaves form a whorl
Ex: Allamanda

Question 11.
What is plant morphology?
Answer:
Plant morphology is also known as external morphology deals with the study of shape, size, and structure of plants and their parts like (roots, stems, leaves, flowers, fruits, and seeds).

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 12.
Name any 2 brace roots and write down their botanical name
Answer:

  1. Sugarcane – Saccharum officinarum
  2. Maize- Zea mays

Question 13.
Draw the Regions of the root tip and label the parts
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 7

Question 14.
What is meant by ‘Eye’ of a potato?
Answer:
The axillary bud ensheathes by the scale appears as eye-like on the potato surface each and every eye can develop into a potato plant.
‘S’ Scale Leaf Auxiliary bud
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 8

Question 15.
Draw the structure of a typical leaf and label the parts
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 9

Give Short Answer

Question 1.
The morphological study is important in Taxonomy. Why?
Answer:
Morphological features are important in determining the productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 2.
Classify leaves on the basis of duration
Answer:

  1. Cauducuous (fagaceous)- falling off soon after formation – Opuntia
  2. Deciduous – Falling at the end of the growing season (winter& summer-leaf less)- Erythrina indica
  3. Evergreen- persistent throughout the year tree never remain leafless Mimusops
  4. Marcescent- no falling-but withering on the plants – Fagaceae

Question 3.
Classify compound leaf types
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 10

Question 4.
Give the diagrammatic representation of leaf modifications
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 12

Question 5.
Give a brief account on the tap root system.
Answer:
Primary root is the direct prolongation of the radicle. When the primary root persists and continues to grow as in dicotyledons, it forms the main root of the plant and is called the tap root. Tap root produces lateral roots that further branch into finer roots. Lateral roots along with their branches together called secondary roots.

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 6.
Classify venation
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 13
Question 7.
Notes on Heterophylly.
Answer:
Definition:
Morphologically 2 different kinds of leaves in the same plant is called heterophylly.
Types-2

  1. Structural
  2. Developmental

1. Structural – In Limnophyllaheterophylla, aquatic plant half of its plant body is submerged and half is above water level. Here aerial leaves are normal & the submerged leaves are highly dissected.

2. Developmental – In Sterculiavillosa Varying structure during different developmental stages- Young leaves – lobed or dissected Mature leaves – entire

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 8.
How the leaf hooks helps the Bignonia plant?
Answer:
In cat’s nail (Bignonia unguiscati) an elegant climber, the terminal leaflets become modified into three, very sharp, stiff, and curved hooks, very much like the nails of a cat. These hooks cling to the bark of a tree and act as organs of support for climbing.

Phyllode- It is a winged leaf petiole or stalk or rachis Eg. Nepenthus – modified to perform the function as leaf
Acacia auriculiformis Leaf- petiole modification to do photosynthesis

Essay Question – Five marks

Question 1.
Classify terrestrial habitats
Answer:

Types

Examples

Mesophytes (soil-&with sufficient water) Azadurachitaindica
Xerophytes (in dry habitals) Opuntia .euphorbia
Psammophytes (on sand) Spinifex littoralis
Lithophytes (on rocks) Liehensficusspp

Question 2.
Classify aquatic habitat.
Answer:

Types

Examples

1. Free-floating Eichhomia, pistia
2. Submerged Hydrilla,vallisneria
3. Emergent Limnophytes, typha
4. Floating leaves but submerged Nelumbo, nymphaea
5. Mangroves (marshy plants) Avicennia, Rhizophora

Question 3.
Draw the structure of a typical plant and neatly label the parts
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 14

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 4.
What are the various types of root modification
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology 15

Question 5.
Give a clear-cut distinction of horks Spines & prickles.
Answer:

Horks

Spines

Prickles

Organ of climbing Protective & adaptive to xeric condition Protective & adaptive to xeric condition
Leaf modification terminal 3 leaflets – modified into sharp sliff -curved hooks (like nails of eat)- bignonia unguiseatr Leaf modification – opuntia Leaf & stipule modification-  Euphorbia Leaf surface or margin of leaf – Argemone mexicana Our growth from epidermal cells of stem or leaves – Rosa spp

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Question 6.
List out various types of Phyllotaxy.
Answer:

Type

Definition

Examples

1. Alternate Spiral Only one leaf — each node successive nodes have alternate – in a spiral manner Hibiscus
2. Alternate bifarous Leaves in 2 rows – alternatively Polyalthia
3. Opposite superposed 2 leaves in each node opposite in same direction in successive nodes Eg. Guava
4. Opposite deeussate One pair of leaves at right angles to the next lower pair Eg.Calotropis
5. Temate 3 leaves at each node Eg.Nerium
6. Whorled or verticulate More than 3 leaves at each node Eg.Allamanda
7. Leaf Mosaic Upper leaves with short petiole lower leaves have long petioles Eg.Aealypha

Question 7.
Compare & Contrast pitcher from bladderwort.
Answer:

Pitcher plant

Bladder plant

Grow in terrestrial habitat – where there is scarcity of nitrogen in the soil Root less free-floating or slightly submerged aquatic plant
All parts of leaf modified, specially the leaf lamina- as pitcher with lid to trap insects Eg. Nepenthus Leaf highly segmented and some segments of leaf modified into the bladder with trap door to trap aquatic animalcules Eg. Utricularia

Question 8.
Define Ptyxis or Vernation list out the various types
Answer:

Types Definition Examples
Reclinate Upper half of leaf blade bent upon lower leaf Eriobotry a japonica
Consolidate Lengthwise folding along mid rib Guava, potato Bauhenia
Plicate Repeatedly folded longitudinally along ribs – zigzag manner Borassus
Cricinate Leaf rolled apex to base Ferns
Convolute Leaf rolled from margin from one to other Musa members of Arecaceae
Involute Two margins rolled on the upper surface of leaf towards mid rib. Lotus lily
Crumpled Irregularly folded Cabbage

Samacheer Kalvi 11th Bio Botany Chapter 3 Vegetative Morphology

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 33 Indirect Taxation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 33 Indirect Taxation

11th Commerce Guide Indirect Taxation Text Book Back Questions and Answers

I. Choose the Right Answer:

Question 1.
Who is the chairman of the GST council?
a) RBI Governor
b) Finance Minister
c) Prime Minister
d) President of India
Answer:
b) Finance Minister

Question 2.
GST Stands for ………………
a) Goods and Supply Tax
b) Government Sales Tax
c) Goods and Services Tax
d) General Sales Tax
Answer:
c) Goods and Services Tax

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 3.
What kind of Tax the GST is?
a) Direct Tax
b) Indirect Tax
c) Dependence on the Type of Goods and Services
d) All Business Organisations
Answer:
b) Indirect Tax

Question 4.
What is IGST?
(a) Integrated Goods and Service Tax
(b) Indian Goods and Service Tax ‘
(c) Initial Goods and Service Tax
(d) All the Above
Answer:
(a) Integrated Goods and Service Tax

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 5.
In India, GST became effective from?
a) 1st April 2017
b) 1st January 2017
c) 1 st July 2017
d) 1 st March 2017.
Answer:
d) 1 st March 2017.

II. Very Short Answer Questions:

Question 1.
Define Indirect tax.
Answer:
Indirect Tax is levied on goods and services. It is collected from the buyers by the sellers and paid by the sellers to the Government. Since it is indirectly imposed on the buyers it is called an indirect tax.

Question 2.
List out any four types of indirect taxes levied in India.
Answer:
Goods and Services Tax, Excise Duty

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 3.
What do you mean by Goods and Services Taxes?
Answer:
Goods and Services Tax (GST) is the tax imposed on the supply (consumption) of goods and services. It is a destination-based consumption tax and collected on those value-added items at each stage of the supply chain.

Question 4.
Write a note on SGST.
Answer:
State Goods and Services Tax – imposed and collected by the State Governments under State GST Act. (Tamil Nadu GST Act 2017 passed by Tamil Nadu Govt.) SGST means State goods and service tax, replace the existing tax like sales tax, luxury tax, entry tax, etc. and it is levied by the state government.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 5.
What is CGST?
Answer:
CGST – Central Goods and Services Tax – is imposed and collected by the Central Government on all supply of goods within a state (intrastate) under CGST Act 2017.

III. Short Answer Questions

Question 1.
Write any two differences between direct taxes and indirect taxes.
Answer:

Basis of Difference Direct Taxes Indirect Taxes
1. Meaning If a tax levied on the income or wealth of a person is paid by that person directly to the Government it is called direct tax If tax is levied on the goods or services of a person is collected from the buyers by another person(seller) and paid by him to the Government it is called an indirect tax
2. Evasion Tax evasion is possible Tax evasion is more difficult

Question 2.
What are the objectives of GST?
Answer:

  1. To create a common market with uniform tax rate in India. (One Nation, One Tax, One Market)
  2. To eliminate the cascading effect of taxes, GST allows set-off of prior taxes for the same transactions as input tax credit.
  3. To boost Indian exports, the GST already collected on the inputs will be refunded and thus there will be no tax on all exports.
  4. To increase the tax base by bringing more number of tax payers and increase tax revenue.
  5. To simplify tax return procedures through common forms and avoidance of visiting tax departments.
  6. To provide online facilities for payment of taxes and submission of forms.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 3.
Briefly explain the functions of GST council.
Answer:
The GST Council will oversee the implementation of the GST. But the Central Board of Excise and Customs is responsible for administration of the CGST and IGST Acts. The Council makes recommendations on rate of GST, apportionment of IGST, exemptions, model GST laws, etc. The Chairman of the Council is the Union Finance Minister.

The Minister of State in the Finance Ministry and all Finance Ministers of the State Governments shall be its members.The Central Government shall have l/3rd voting power and all State Governments shall have 2/3rd voting powers. All decisions of the Council can be passed only with 3/4h of the total votes. Each state has one vote, irrespective of its size or population. Twenty four council meetings were held until 2017.

Question 4.
Explain IGST with an example.
Answer:
IGST – Inter-State Goods and Services Tax is imposed and collected by the Central Government and the revenue is shared with States under IGST Act 2017.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 5.
Write any three demerits of UGST.
Answer:
The demerits of GST are :
Several Economists says that GST in India would impact negatively on the real estate market. It would add up to 8 percent to the cost of new homes and reduce demand by about 12 percent. Another criticism is that CGST, SGST are nothing but new names for Central Excise/Service Tax, VAT and CST.

Hence, there is no major reduction in the number of tax layers. A number of retail products currently have only four percent tax on them. After GST, garments and clothes could become more expensive.

IV. Long Answer Questions

Question 1.
Distinguish between direct taxes and indirect taxes.
Answer:

Bases of Difference

Direct Taxes

Indirect Taxes

1. Meaning If a tax levied on the income or wealth of. a person is paid by that person (or his office) directly to the Government, it is called direct tax. If tax is levied on the goods or services of a person is collected from the buyers by another person (seller) and paid by him to the Government it is called indirect tax
2. Incidence and Impact Falls on the same person. Imposed on the income of a person and paid by the same person. Falls on different persons. Imposed on the sellers but collected from the consumers and paid by sellers.
3.Burden More income attracts more income tax. Tax burden is progressive on people. Rate of tax is flat on all individuals. Therefore more income individuals pay less and lesser portion of their income as tax. Tax burden is regressive
4.Evasion Tax evasion is possible. Tax evasion is more difficult
5.Inflation Direct Tax helps in reducing the inflation Indirect tax contributes to inflation
6. Shift ability Cannot be shifted to others Can be shifted to others
7.Examples Income Tax,Wealth Tax, Capital Gains Tax, Securities Transaction Tax, Perquisites Tax GST, Excise Duty

Question 2.
Discuss the different kinds of GST.
Answer:
GST is of three kinds: CGST, SGST/UGST, and IGST.
1. CGST – Central Goods and Services Tax – imposed and collected by the Central Government on all supply of goods within a state (intra-state) under CGST Act 2017.

2. SGST – State Goods and Services Tax – imposed and collected by the State Governments under State GST Act. (Tamil Nadu GST Act 2017 passed by Tamil Nadu Govt.)

3. UGST – Union Territory Goods and Services Tax – imposed and collected by the five Union Territory Administrations in India under UGST Act 2017.

4. IGST – Inter-State Goods and Services Tax – imposed and collected by the Central Government and the revenue shared with States under IGST Act 2017.

5. IGST on exports – All exports are treated as Inter-State supply under GST. Since exports are zero-rated, GST is not imposed on all goods and services exported from India. Any input credit paid already on exports will be refunded.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 3.
Elucidate the merits of GST.
Answer:
A. To the Society and country:

  1. A unified common national market will attract more foreign investment. GST has integrated the economy of all States and Union Territories.
  2. It brings parity in taxation among imported goods and Indian manufactured goods. All imported goods will be charged with IGST which will be more or less equivalent to the total of CGST and SGST levied on manufactured goods. Removal of several taxes will make the price of Indian products more competitive in the world market.
  3. It will boost manufacturing, export, GDP leading to economic growth through an increase in economic activity.
  4. The creation of more employment opportunities will result in poverty eradication.
  5. It will bring more tax compliance (more taxpayers) and increase revenue to the Governments.
  6. It is transparent and will improve India’s ranking in the Ease of Doing Business in the world.
  7. Uniform rates of tax will reduce tax evasion and rate arbitrage between States.

B. To Business Community:

  1. Simpler Tax System with fewer exemptions. 17 taxes were abolished and one tax exists today.
  2. The input tax credit will reduce cascading effect of taxes. Reduction in average tax burden will encourage manufacturers and help the “Make in India” campaign and make India a manufacturing hub.
  3. Common procedures, common classification of goods and services, and timelines will lend greater certainty to the taxation system.
  4. GSTN facility will reduce multiple record-keeping, lesser investment in manpower and resources and improve efficiency.
  5. All interactions will be through a common GSTN portal and will ensure corruption-free administration
  6. Uniform prices throughout the country. Expansion of business to all states is made easy.

C. To Consumers:

  1. The input tax credit allowed will lower the prices to the consumers.
  2. All small retailers will get exemptions and purchases from them will cost less for the consumers.

Question 4.
Compare CGST, SGST, and IGST.
Answer:

Basis of Difference CGST SGST IGST
1.Meaning CGST means Central goods and service tax to replace the existing tax like service tax, excise, etc. and It is levied by the central government SGST means State goods and service tax, replace the existing tax like sales tax, luxury tax, entry tax,-etc. and it is levied by the state government IGST refers to the Integrated Goods and Services Tax and it is a combined form of CGST and IGST and it is levied by the central government
2. Collection of Tax Central Government State Government Central Government
3. Applicability Intra-State supply Intra-state supply Inter-State supply
4.Registration No registration till the turnover crosses 20 lakhs (10 lakhs for northeastern states) No registration till the turnover crosses 20 lakhs (10 lakhs for northeastern states) Registration is mandatory
5.Composition The dealer can use the benefit upto 75 lakhs under the composition scheme The dealer can use the benefit up to 75 lakhs under the composition scheme The composition scheme is not applicable in interstate supply

11th Commerce Guide Indirect Taxation Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
GST is a …………….. based tax on consumption of goods and services.
a. duration
b. destination
c. dividend
d. development
Answer:
b. destination

Question 2.
India GST model has …………….. rate structure.
a. 3
b. 5
c.4
d. 6
Answer:
c. 4

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 3.
What is the maximum rate prescribed under GST?
a. 12
b. 20
c. 28
d. 18
Answer:
b. 20

Question 4.
When does liability to pay GST arise in the case of a supply of goods?
a. on raising of invoice
b. At the time of supply of goods
c. On receipt of payment
d. Earliest of any of above
Answer:
d. Earliest of any of above

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 5.
Special provisions are there in the GST Act for the …………….. north-eastern states.
a. 8
b. 9
c. 6
d. 18
Answer:
a. 8

II. Very Short Answer Questions:

Question 1.
What do you mean by UGST?
Answer:
Union Territory Goods and Services Tax is formed to impose and collect tax from the five union territory administrations in India under UGST Act 2017.

III. Long Answer Questions

Question 1.
Write a note on GST Council:
Answer:
The GST Council will oversee the implementation of the GST. But the Central Board of Excise and Customs is responsible for the administration of the CGST and IGST Acts. The Council makes recommendations on the rate of GST, apportionment of IGST, exemptions, model GST laws, etc. The Chairman of the Council is the Union Finance Minister.

The Minister of State in the Finance Ministry and all Finance Ministers of the State Governments shall be its members. The Central Government shall have l/3rd voting power and all State Governments shall have 2/3rd voting powers. All decisions of the Council can be passed only with 3/4h of the total votes. Each state has one vote, irrespective of its size or population. Twenty four council meetings were held until 2017.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Question 2.
Who are all the officials involved in GST Secretariat?
Answer:
The following are the officials involved in GST officials:

  • The Secretary (Revenue) will be appointed as the Ex-officio Secretary to the GST Council.
  • The Chairperson, Central Board of Excise and Customs (CBEC), will be a permanent invitee (non-voting).
  • One post of Additional Secretary to the GST, and
  • Four posts of Commissioner in the GST Council Secretariat will also be created.

Question 3.
What are all the limitations of GST?
Answer:
The limitations/disadvantages of GST are stated below:

  • Several Economists say that GST in India would impact negatively the real estate market. It would add up to 8 percent to the cost of new homes and reduce demand by about 12 percent.
  • Another criticism is that CGST, SGST are nothing but new names for Central Excise/ Service Tax, VAT, and CST. Hence, there is no major reduction in the number of tax layers.
  • A number of retail products currently have only four percent tax on them. After GST, garments, and clothes could become more expensive.
  • The aviation industry would be affected. Service taxes on airfares currently range from six to nine percent. With GST, this rate will surpass fifteen percent and effectively double the tax rate.
  • Adoption and migration to the new GST system would involve teething troubles and learn for the entire ecosystem.

Samacheer Kalvi 11th Commerce Guide Chapter 33 Indirect Taxation

Samacheer Kalvi 11th Bio Botany Guide Chapter 1 Living World

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 1 Living World Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 1 Living World

11th Bio Botany Guide Living World Text Book Back Questions and Answers

Part-I

Choose the Right Answer: 

Question 1.
Which one of the following statements about viruses is correct?
a. Possess their own metabolic system
b. They are the facultative parasites
c. They contain DNA or RNA
d. Enzymes are present
Answer:
b. They are the facultative parasites

Question 2.
Identify the Archaebacterium
a. Acetobacteria
b. Erwinia
c. Treponema
d. Methanobacterium
Answer:
d. Methanobacterium
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
Identify the correctly matched pair
a. Actinomycete – a) Late blight
b. Mycoplasma  – b) lumpy jaw
c. Bacteria – c) crown gall
d. Fungi – d) sandal spike
Answer:
a. Actinomycete – Lumpy jaw
b. Mycoplasma – sandal spike
c. Bacteria – crown gall
d. Fungi – late blight

Question 4.
Identify the incorrect statement about the gram-positive bacteria.
a. Teichoic acid absent
b. A high percentage of peptidoglycan is found in the cell wall.
c. Cell wall is single-layered
d. Lipopolysaccharide is present in the cell wall.
Answer:
a. Teichoic acid absent
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 5.
The correct statement regarding Blue Green Algae is
a. Lack of motile structures
b. Presence of cellulose in cell wall
c. Absence of mucilage around the thallus
d. Presence of Floridian starch
Answer:
a. lack of motile structure

Question 6.
Differentiate Homoiomerous and Heteromerous lichens.
Answer:

Homoiomerous

Heteromerous

Here algae cells evenly distributed in the thallus Heteromerous-a distinct layer of alga and fungi present.

Question 7.
Write the distinguishing features of Monera.
Answer:
Distinguishing Features of Monera:

  1. This kingdom includes all prokaryotic organisms. Example: Mycoplasma, bacteria, actinomycetes, and cyanobacteria.
  2. These are microscopic. They do not have a true nucleus and membrane-bound organelles.
  3. Many other bacteria like Rhizobium, Azotobacter, and Clostridium can fix atmospheric nitrogen into ammonia.
  4. Some bacteria are parasites and others live as symbionts.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 8.
Why do farmers plant leguminous crops in crop rotations/mixed cropping?
Answer:
Rhizobium- Nitrogen-fixing bacteria,
Living in the root modules of leguminous plants has a symbiotic association with it, fix atmospheric nitrogen, and convert it into nitrates, thereby increases the fertility of the soil. Growing legumes alternatively with paddy can help paddy to give high yield- This method of growing paddy, alternatively with leguminous plants is known as crop rotation.

Mixed cropping:
Amidst, other crops. The leguminous crop is also raised as a mixed crop – so that it enriches the soil and increases the yield by fixing atmospheric nitrogen.

Question 9.
Briefly discuss the 5 kingdom system of classification. Add a note on their merits and demerits.
Answer:
a. Proposed by R.H. Whittaker (American taxonomist)
b. Criteria considered – cell structure, Thallus Organization, Mode of Nutrition, Reproduction, and Phytogenitic Relations.
5 kingdom classifications include: –
a. Monera b. Protista c. Fungi d. Plantae e. Animalia

S. No

Merits:

Demerits:

1. Based on the complexity of cell structure & organization of thallus Monera & Protista – Include both autotrophic & heterotrophic organisms
2. Based on mode of nutrition. Include cell wall lacking & cell wall
3. Fungi-kept in a separate category from plants Bearing organisms.
4. It shows the phylogeny of the organisms So the group is more heterogeneous

Question 10.
Give a general account of lichens
Answer:
a. Definition: A symbolic association of algae and Fungi helping each other & living together known as lichens.
b. Partners: Algal partner known as Phycobiont & Fungal partner known as Mycobiont
c. Role of Algal partner – Autotrophic prepare food – give nutrition to fungal partner also
d. Role of fungal partner – gives protection- helps in fixing to the substratum by rhizines.
Classification:

Character

Phycobiont

Mycobiont

1 Asexual reproduction Akinetes, hormogonia, Aplanospore, etc. fragmentation soredia, and isidia
2 Sexual reproduction absent sexual reproduction by ascocarp & ascospores

Character

Classification of lichens

1. Habitat Corticolous – growing on the bark
Lichnicolous – growing on wood
Saxicolous – growing on rock
Terricolous – growing on the ground
Marine – siliceous rock sea
Freshwater – siliceous rocks (freshwater habitat).
2. Morphology of thallus Leprose – distinct fungal layer absent Crustose – crust like Foliose – Leaf-like Fruticose-branched pendulous shrub-like
3. On the basis of the distribution of algae cells Homoiomerous – Algae cells evenly distributed Heteromerous – A distinct layer of Algae and Fungi present
4. On the basis of the fungal partner If it is Ascomycetes – Ascolichen If it is basidiomycetes-Basidiolichen

Economic importance:

Secretion of acids of lichens

Uses

1 Oxalic acid Weathering of rocks Pioneers in xerosere
2 Usnic acid Antibacterial

II. a. Pollution Indicators – Lichens sensitive to air pollutants- (pollution indicators)
b. Rocella Montagne – Produces a dye used in litmus paper (acid-base indicator)
c. Cladonia rangiferina – Food for animals in tundra regions

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Part – II.

11th Bio Botany Guide Living World Additional Important Questions and Answers

Choose The Right Answer:

Question 1.
Earth has formed around billion years ago ……………
(a) 3.3
(b) 5.6
(c) 4.6
(d) 5.9
Answer:
(c) 4.6

Question 2.
The organism that is reproductively sterile is
a. Wasp
b. Worker bees
c. Housefly
d. Drosophila
Answer:
c. Worker bees

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
Which of the following is NOT a prokaryote?
(a) Bacteria
(b) Blue-green algae
(c) Oedogonium
(d) Nostoc
Answer:
(c) Oedogonium

Question 4.
Recombination is the result of
a. Binary fission
b. Asexual reproduction
c. Sexual reproduction.
d. Vegetative propagation
Answer:
c. sexual reproduction

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 5.
Vaccination for smallpox was discovered by …………….
(a) W.M. Stanley
(b) Adolf Mayer
(c) Robert Koch
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 6.
Blister-like pustules occur due to
a. Chickenpox
b. Rust
c. Smut
d. Mumps
Answer:
a. Chickenpox

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 7.
Expand Bt-toxin
a. Biotechnology
b. Biotoxin
c. Beta-toxin
d. Bacillus thuringiensis
Answer:
d. Bacillus thuringiensis

Question 8.
One nanometer equals to metres …………….
(a) 10-9
(b) 10-6
(c) 10-5
(d) 10-12
Answer:
(a) 10-9

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 9.
Saprophytic angiosperm with mycorrhiza
a. Clostridium
b. Azolla
c. Monotropa
d. Viscum
Answer:
c. Monotropa

Question 10.
The famous roqueforti cheese is produced by employing
a. Aspergillus roquefortic
b. Penicillium camemberti
c. Penicillium notatum
d. Aspergillus terreus
Answer:
b.Penicillium camémberti

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 11.
Identify the criteria not used in classifying viruses by Baltimore …………….
(a) ss (or) ds
(b) use of RT
(c) capsid
(d) sense or antisense
Answer:
(c) capsid

Question 12.
Both viruses and bacteria contain
a. Plasma membrane
b. Protein wat
c. Peptidoglycan
d. Nucleic acids
Answer:
d. Nucleic acids

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 13.
Lactobacillus bulgaricus is responsible for the formation of
a. Lactic acid
b. Cheese
c. Yogurt
d. Curd
Answer:
C. Yoghurt

Question 14.
Parvo viruses have …………….
(a) ssDNA
(b) dsDNA
(c) ssRNA
(d) dsRNA
Answer:
(a) ssDNA

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 15.
Bacterial chlorophyll is also known as
a. Chlorophyll
b. Bilirubin
c. Chromatium
d. Chioridin .
Answer:
Chromatium

Question 16.
This drug is also known as wonder drug.
a. Streptomycin
b. Aureomycin
c. Bacitracin
d. Pencillin
Answer:
Pencillin

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 17.
The empty protein coat left outside after penetration is …………….
(a) host
(b) ghost
(c) capsid
(d) capsomeres
Answer:
(b) ghost

Question 18.
Among the given 4 – one is not viral diseases- find it out.
a. Cucumber mosaic
b. Citrus canker
c. Ricetungro
d. Potato leaf roll
Answer:
b. citrus canker

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 19.
Bacillus thuringiensis is an
a. Biofertilizer
b. Bio-fuel
c. Bio-pesticide
d. Bio-medicine
Answer:
c. Bio-pesticide

Question 20.
Mad cow disease is caused by …………….
(a) viroids
(b) virusoids
(c) prions
(d) viruses
Answer:
(c) prions

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 21.
Bacteria that grow in high salinity condition is known as
a. Methane bacterium
b. Halobacterium
c. Thermos aquaticus
d. Agrobacterium
Answer:
b. Halobacterium

Question 22.
Budding is an unique feature of
a. Schizo saccharomyces
b. Aspergillus
c. Penicillium
d. Neurospora
Answer:
Schizo saccharomyces

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 23.
Mycophages infect …………….
(a) blue-green algae
(b) bacteria
(c) fungi
(d) cyanobacteria
Answer:
(c) fungi

Question 24.
The colourless cell in Nostoc in the intercalary position is responsible for nitrogen fixation is
a. Holoblast
b. Heterozygote
c. Homocyst
d. Heterocyst
Answer:
d. Heterocyst

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 25.
Genetic trait carried in the bacterial
a. Cell wall
b. Chromosome
c. Plasmid
d. Cell membrane
Answer:
c. Plasmid

Question 26.
Three kingdom classification was proposed by …………….
(a) Copeland
(b) Theophrastus
(c) Linnaeus
(d) Haeckel
Answer:
(d) Haeckel

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 27.
Developing a vaccine for SARS is difficult because
a. It spreads through nucleic acid
b. It is an enveloped virus
c. It has RNA
d. It constantly changes its form
Answer:
d. It constantly changes its form

Question 28.
Arrange correctly the following viruses according to the given shape and symmetry, Cuboidal, spherical, helical and complex respectively
I. a. Vaccinovirus b. Influenza, c. HIV, d. Herpes
II. a. Influenza b. HIV c, Herpes d. Vaccinovirus
III. a. Herpes b. HIV, c. Influenza d. Vaccinovirus
IV. a. HIV b. Herpes c. Vaccinio Virus d. Influenza
Answer:
III. a. Herpes, b, HIV, c, Influenza, d, Vaccinovirus

II.Match the following and find the correct answer.

Question 1.
I. Five kingdom system of classification
II. Three kingdom system of classification
III. Four kingdom system of classification
IV. Two kingdom system of classification
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 1
Answer:
a. C-D-B-A

Question 2.
I. TMV Discovered by world
II. Bacterium word coined by
III. Father of Mycology
IV. Classification virus given by
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 2
Answer:
b. B-D-A-C

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
I. Genophore
II. Bacteria
III. Extra Chromosomal DNA
IV. Fimbrial
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 2
Answer:
b. D-A-B-C

Question 4.
I. Plasmid
II. Heterocyst
III. Glycocalyx
IV. Mesosome
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 4
Answer:
c. B-A-D-C

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

III.

Question 1.
Which one of the following is a false statement regarding Prions
a. Prions were discovered by B. Prusiner in 1982
b. They are infectious particles of lipo protein
c. They cause about a dozen fatal degenerative disorders of CNS
d. Creutzfeldt-Jakob disease(cjD) and bovine spongiform encephalopathy (BSE) are some commonly known diseases
Answer:
b. They are infectious particles of lipoprotein

Question 2.
Which one of the following is a false statement regarding Ribosomes
a. Ribosomes are the sites of protein synthesis
b. The number of ribosomes per cell varies from 1000 to 1500.
c. The ribosomes are 70s type and consists of a 2 subunits (50s and 30s)
d. The nbosomes are held together by mRNA and form polyribosomes or polysomes.
Answer:
b. The number of ribosomes per cell varies from 1000 to 1500.

IV. Find out the True and False statements from the following and on that basis find the correct answer:

Question 1.
(i) Poly-B hydroxybutysate is a microbial plastic which is biodegradable
(ii) Transfer of DNA from one bacterium to another is known as transduction
(iii) Micrococcus must have oxygen to survive-known as an obligate aerobe
(iv) Spirulina is rich in carbohydrates so treated as an alternative food.
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 5
Answer:
c. True False True False

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 2.
(i) Toad stools are known as an edible mushroom
(ii) Volvariella volvaceae and Agaricus bisporous are known for their high poisonous nature
(iii) Claviceps purpurea produces ergot-used as vasoconstrictor
(iv) Aspergillus flavus infest dried foods and produce carcinogenic toxin called -aflatoxin
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 6
Answer:
B. False False True True

Question 3.
Which one of the following is a correct statement regarding TMV
a. David Baltimore in 1971 discovered TMV
b.First visible symptom of TMV is visible discoloration of leaves along the veins-but with typical yellow and green symptom molting-(mosaic symptom)
c. The plant grow abnormally at the nodal point
d. Infection spread by house flies and mosquitoes
Answer:
b. First visible symptom of TMV is discoloration of leaves along with the veins-but visible with typical yellow and green symptom molting-(mosaic symptom)

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

V.

Question 1.
Which one of the following is a correct statement regarding bacterial antibiotic
a. Chloromycetin got from Streptomyces venezuelae cure T.B
b. Bactracin is got from bacillus mycoides- used to treat UTI
c. Aurecomycin got from Streptomyces aureofaciens is used to treat whooping cough and eye infections
d. Streptomycin got from Streptmyces griseus cure typhoid fever
Answer:
c. Aurecomycin got from streptomyces aureofaciens is used to treat whooping cough and eye infections

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 2.
Which one of the following is a correct statement regarding mycoplasma
a.  Mycoplasm are very small (0.1-0.5mm) pleomorphic gram negative micro organisms
b.  The have whole body appear like boiled egg-like structure in culture
c.  Little leaf of tomato Witches broom of solanum.
d.  Mycoplasma is also known as mollicutes
Answer:
Mycoplasm arevery small (0.1- 0.5mm) pleomorphic gram negative micro organisms

VI.

Question 1.
Sac fungi & club fungi are common names of
a.  Ascomycetes
b.  Basidiomycetes
c.  Deuteromycetes
d.  Phycomycetes
(i) a & b
(ii) b & c
(iii) a & c
(iv) c & d
Answer:
(i) a & b

Question 2.
Recurrence of fungal skin disease is due to
a. Resistance to antibiotics
b. Dormant spores become active at the onset of favourable condition
c. Non-availability of specific drugs.
d. Moisture favor fungal mycelium to spread,
(i) a & C
(ii) b & c
(iii) b & d
(iv) c & d
Answer:
(i) b & d

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
Find out the symbiotic associations from the given options.
a. Nitrogen fixing bacteria on the leguminous plants.
b. Rhizoids of Neprolepis
c. Lichens on rocks
d. Mycorrhizal roots.
(I) ac & d
(II) ab & c
(III) ab & d
(IV) a & b
Answer:

VII. Find out the wrong statement

Question 1.
Which one of the following is a wrong statement regarding ehemo lithotrophs
a. Sulphur bacteria
b. Iron bacteria
c. Methane bacteria
d. Hydrogen bacteria
Answer:
c. Methane bacteria

Question 2.
Among the following, which one is not viral?
a. Cucumber mosaic
b. Citrus canker
c. Rice tungro
d. Potato leaf roll
Answer:
b. Citrus canker

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
Which one of the following is not a Ribovirus?
a. Tobacco mosaic virus
b. Cauliflower mosaic virus
c. Human immune deficiency virus.
d. Wound tumour virus d. Pilobolus
Answer:
d. Wound tumour virus

Question 4.
Find out from the given, which one is not a Zygomycetes fungi.
a. Mucor
b. Rhizopus
c. Yeast
d. Pilobolus
Answer:
c. yeast

VIII. Read the following Assertion A and Reason R. Find the correct Answer

Question 1.
Assertion ‘A’: Viruses have genetic material but cannot divided on its own. They also don’t have in built metabolic machinery Reason ‘R’: Virus kept between living and non living
(a) A & R correct. R is explaining A
(b) A & R correct R is not explaining A
(c) A is true but R is wrong
(d) A is true but R is not explaining A
Answer:
a. A & R correct R is explaining A.

Question 2.
Assertion ‘A : Some bacteria have the capacity to retain gramstain after treatment with acid alcohol.
Reason ‘R’: Known as gram +ve as attracted towards positive pole under the influence of electric current.
Answer:
c. A is true but R is wrong.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 3.
Assertion’A’: Aflatoxin produced by Aspergillus flavus.
Reason ’R’: These toxin are useful to mankind to cure few disease
Answer:
c. A is true but R is wrong

Question 4.
Assertion ’A’: In septal mycelium the septa complete the partition walls between cells Reason
‘R’: There is no cytoplasmic connection between adjacent cells.
Answer:
d. A is true but R is not explaining A.

I. Additional 2 Marks

Question 1.
Define Growth.
Answer:
Growth is an intrinsic property of all living organisms through which they can increase cells both in number and mass.

Question 2.
Tabulate Milestones in Virology
Answer:

Year

Name of the Scientist

Achievement

1796 Edward Jenner Vaccination for smallpox
1886 Adolf Mayer Proved infectious nature of- TMV- from mosaic leaves sap
1892 Dimitry Ivanowsky Viruses are smaller than bacteria

Question 3.
Draw the structure of TMV and label the parts and explain in a word or two
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 7
Answer:
a. Nucleic acid – It is ss RNA(Single Standed)
b. Capsomere – Protein Units
c. Capsid – Protein Coat.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 4.
Define reproduction and Mention its types.
Answer:
Reproduction is the tendency of a living organism to perpetuate its own species. There are two types of reproduction namely asexual and sexual.

Question 5.
Bacteria is a indeed friend- discuss
Answer:
Even though Bacteria cause many diseases to plants animals and human beings they are beneficial to day-to-day life also. Ex: Milk (lactobacillus acidophobus)/(lactobaci llus lacti)
a. Curd b. Butter c. Cheese d. Yogurt These are a few of the beneficial activities.

Question 6.
Draw the ultra structure of bacterial cell.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 8

Question 7.
Name the four types of ascocarps produced by ascomycetes.
Answer:
Four Types Of Ascocarps Produced By Ascomycetes:

  1. Cleistothecium
  2. Perithecium
  3. Apothecium and
  4. Pseudothecium.

Question 8.
Why is it essential to do classification?
Answer:

  • To relate on the basis of common features
  • To define, on the basis of salient features
  • To know the relationship among different groups.
  • To understand evolutionary relationship.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 9.
Ruggerio etal’s recent classification-Explain.
Answer:

  • Ruggerio etal in 2015-published-7-kingdom system of classification
  • It is an extension of Cavalier’s 6 – Kingdom Scheme’Include 2 superkingdom
    Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 9

Question 10.
What is Prophage?
Answer:
a. In the lysogenic cycle, the injected phage DNA become circular and integrates into bacterial chromosome by recombination.
b. The integrated DNA of phage and bacteria is known as Prophase.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 11.
Distinguish between Cyanophage and Mycophage
Answer:

Cyanophage Mycophage
1 Vims infecting blue green Algae are known as
Cyanophage
Vims allacking fungi are called Mycovimses or
Mycophage
2 1st reported by Safferman and Mores(1963) 1st reported by Holling(1962)
3 Eg. Lyngbuya, Plectonema Eg. Myc ovims attacking Mushrooms

Question 12.
Differentiate between flagella of Prokaryotes and Eukaryotes
Answer:

1 Size 20-30mm in diameter 15 pm in length Bigger in size
2 Structure Simple made up of single fibril In C.S flagella contain 9+2 microtubules
3 No of position Many types few types
4 Function Locomotion Locomotion

Question 13.
Differentiate between Photolithotrophs and Photo organotrophs
Answer:

Photolithotrophs

Photo organotrophs

1 .Hydrogen donor is an in organic substance 2 types – Green sulphur bacteria Hydrogen donor H2S Posses bacterioviridin.
E.g chlorobium Purple sulphur bacteria
Here hydrogen donor is thiosulphate
pigment bacterial chlorophyll in chlorosomes
E.g chromatium
2. The hydrogen donor is an organic acid or alcohol
e.g purple non sulphur bacteria Rhodospirillum
E.g

  • Purple non sulphur bacteria
  • Rhodospirillum

Question 14.
If you think endospore formation not a reproduction method then justify you answer
Answer:
Yes Endospores are formed not during reproduction, but during unfavorable season the thick walled endospores are resting spores when favorable condition comes, they germinate and form bacteria.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 15.
What are the 3 different methods by which gene recombination occur in bacteria ? Or write about Sexual reproduction in bacteria?
Answer:
Sexual reproduction is so simple formation and fusion of gametes is absent. However by  Three different methods gene recombination can occur

  1. Conjugation
  2. Transduction
  3. Transformation

2 Marks

Question 16.
Define chemolithotrophs – give examples
Answer:

  • The type of bacteria oxidise in organic compound to release energy
  • Eg Sulphur bacteria – Thiobacillus thio oxidants
  • Iron bacteria – Ferrobacillus ferro oxidants
  • Hydrogen bacteria – Hydrogenomonas

Question 17.
Label the given diagram properly
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 10
Answer:
1. F- plasmid
2. Conjugation pilus
3. Chromosome
4. F+ cell

Question 18.
Write any 2 vitamin yielding bacteria
Answer:

Escherichia coli Live in human intestine produce large quantities of vitamin K&B – complex
Clostrdiumacctobutylieum Vitamin B2 is prepared by the fermentation of sugar

Question 19.
Name any 2 bacteria diseases affecting Potato
Answer:

Name of disease Causative organism
1. Ringrot_____________ Clavi bacter michiganensis sub sp sepedonicus
2. Scab_____________ Streptomyees scabies

Question 20.
What is meant by Probiotics
Answer:

  • Microorganism such as lactobacillus bifidobacterium when consume as a dietary supplement help to maintain or restores beneficial bacteria to the digestive tract. They are called friendly or good bacteria. They keep our gut healthy
  • They help to increase the immunity of the body
  • Eg: Probiotic Yoghurt
  • Probiotie tooth paste.

Question 21.
What is the meant by Ray fungi? Give example
Answer:

  • Actinomycetes are called as ray fungi due to their mycelia like growth
  • They are anaerobic or facultative anaerobic
  • They are gram + ve
  • Don’t produce aerial mycelium
  • DNA contain high guanine and cytosine content Eg strepotmyces

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 22.
What is this structure?
Answer:
The figure is the structure of mycoplasma
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 11
Answer:
1. Cell membrane
2. Ribosome
3. DNA strain
4. Cytoplasm

Question 23.
Name few Renowed mycologists?
Answer:
A. Arthur H.R. Buller, John Webster D.L. Hawksworth, G.C Ainsworth
B. B mundkur, K.C. Meta, C.V. Subramanian and T.S. Sadasivam & Father of Indian Mycology -E.J. Butler-

Question 24.
Identify the diagram and label any three parts.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 12
Conidia formation-Penicillium

  1. Conidiophores
  2. Ramus
  3. Metula
  4. Sterigma
  5. Conidium or Conidiospore

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 25.
Define Homeostasis. Why it is essential?
Answer:
Property of self-regulation and tendency to maintain a steady-state within an external environment which is liable to change is called homeostasis. It is essential for the living organism to maintain internal conditions to survive in the environment.

Question 26.
Name 4 fungi, from which we derive organic acids.
Answer:

  1. Citric acid& Gluconic acid – Aspergillus niger
  2. Itaconicacid – Aspergillus terreus
  3. Kojicacid – Aspergillus oryzae

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 27.
Name 4 common basidiomycetes
Answer:

  1. Puffballs,
  2. Toadstools,
  3. Birds nest’s fungi,
  4. Bracket fungi,
  5. Smuts
  6. Rusts,
  7. Smuts.

Question 28.
Define aflatoxin.
Answer:
Aspergillus, Polyporus, Mucor and Penicillium are involved in spoilage of food material Aspergillus flavus infest dried food & produce caranogenic toxin known as aflatoxin.

Question 29.
Name 3 Dermatophytes
Answer:

  1. Trichophyton
  2. Tinea
  3. Microsporum
  4. Epidermophyton are some fungi causing skin problems

3 Marks Additional Questions

Question 1.
State the living and Non-living character of the Virus.
Answer:
(i) Living characters:

  • Presence of nucleic acid & protein
  • Capable of mutation
  • Ability to multiply with living cells
  • Able to infect and cause diseases
  • Show irritability and host-specific.

(ii) Non-living characters:

  • Can be crystallized
  • Don’t have metabolic machinery or functional autonomy
  • In active outside the host
  • Energy producing enzyme system is absent

Question 2.
Draw the ultra structure of bacterial cell. Image Parts:
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 13
Question 3.
What are the economic importance of Cyanophyceae
Answer:

SNO

NAME OF THE ORGANISM

ECONOMIC IMPORTANCE

1 Microcystis aeruginosa anabaeria Anabaena Flos aquae Waterbloom-release toxins-affect aqualic organisms
2 Nostoc, Anabaena Fix atmospheric nitrogen(bio fertilizer)
3 Spirulina Used to prepare SCP

Question 4.
Explain any 3 asexual method of reproduction in fungi
Answer:

1. Zoospores Flagellate structures-produced zoosporangia e.g. chytrids
2. Conidia Spores produced on conidiosphores e.g. penicillillium, Aspergillus
3. Budding A small out growth of parental cell, gets detached E.g. Saccharomyces.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 5.
Tabulate animal diseases caused by bacteria
Answer:

S.NO Name of the animal Name of the diseases Name of the pathogen
1. Sheep Anthrax Bacillus anthracis
2. Cattle Brucellosis Bacillus abortus
3. Cattle Bovine Tuberculosis Mycobacterium bovis
4. Cattle Black leg Clostridium chanvei

Question 6.
Distinguish between Ammonification & Nitrification.
Answer:

Ammonification Nitrification
1. Convert protein of dead plant animal bodies

Ammonia

Ammonium salt
After Ammonification the ammonium salts converted into Nitrites & nitrates
2. The bacteria bringing forth this conversion is known as  Ammonifying Bacteria.  This conversion of Ammonia into nitrites & nitrates is known as Nitrification.

Question 7.
By drawing diagrams, classify bacteria on the basis of flagellatin
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 14

Question 8.
What are the prominent symptoms of TMV-affected tobacco plants?
Answer:
The first visible symptom of TMV is discoloration of leaf colour along the veins and show typical yellow and green mottling which is the mosaic symptom. The downward curling and distortion of young apical leaves occurs, plant becomes stunted and yield is affected.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 9.
Define Transduction and explain
Answer:
The three types of Transduction

  1. Phage mediated DNAtransfer is Transduction
  2. Zinder and Lederberg (1952) discovered it in Salmonella typhimurum
  3. 2 types – Generalised and Specalised

I. Generalised Transduction:
a. Ability of bacteriophage to carry the genetic material of any region of bacteria DNA is called generalized transduction.

II. Specialised Transduction:
a. Ability of bacteriophage to carry only a specific region of the bacterial DNA is called Specialized or Restricted Transduction.

Question 10.
Identify from the diagram – & table correctly
Answer:
The given diagram is Basidiocarp of Agaricus
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 15
Answer:
1. Rhizoids
2. Stipe
3. Pileus
4. Gills

Question 11.
Identify from the diagram & label correctly.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 16
The given diagrams sporangium of mucor
1. Rhizoids
2. Sporangiophore
3. Zygosporangium
4. Zygospores

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 12.
Define biopesticides and give 2 examples from fungi.
Answer:

  1. The substances derived from microbes and plants can be used to kill or eradicate pests weeds and diseases causing germs of crops.
  2. This is known Bio-pesticide. They are ecofriendly, non hazardous, non phytotoxic, e.g. Beauveria bassiana Metarhizium anisopliae

Question 13.
Write down any 4 uses of Mycorrhiza.
Answer:

  • Nutrition – Saprophytic Angiosperm cant prepare food-due to absence of green leaves & it get nutrition via mycorrhiza e.g. Monotropa
  • Availability of water and minerals: improve availability of water & minerals
  • Protection-help plant to resist drought & attack of plant pathogen

Question 14.
Progametangium is formed at the tip.
Answer:
The fusion occurs & Zygosporangium is formed then undergo meiosis & zygospores These zygospores germinate are formed, during favourable condition into fungal hyphae.
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 17

Question 15.
Explain briefly the characteristics of Oomycetes
Answer:

Mycelium
Cell wall
Asexual reproduction
Sexual reproduction
Branched & Coenocytic
(multinucleate made up of Glucan & Cellulose)
Heterokont with one whiplash & one tinsel flagellum Oogamous-in nature E.g. Albugo

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 16.
Explain briefly the characteristics of zygomycetes
Answer:

  • Nutrition cell wall
  • Cellwall
  •  Mycelium
  • Asexual reproduction
  • Sexual reproduction
Mostly saprophytic
Chitin & Cellulose
Branched & Coenocylic
Zygospore produced in Zygosporangia
fusion of gametangia result in zygospores

Question 17.
Planogametic copulation in fungi has 3 types Explain.
Answer:
Planogametic copulation means fusion of motile gametes it has 3 types

  1. Isogamy
  2. Anisogamy
  3. Oogamy
    Isogamy — Morphologically smilar gametes Anisogamy Morphologically dissimilar gametes
    Oogamy – But in Oogamy it is highly advanced anisogamy

Question 18.
Explain gametangial copulation in Rhizopus with the help of diagrams.
Answer:

  • In rhizopus & in Mucor there occur heterothallism- there are 2 strains of the hyphae.
  • In a sexual copulation only the 2 opposite strains +ve and -ve strains come together.
    Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 18

Question 19.
The tea or tobacco is not a result of mere drying process. Explain the value addition in these things?
Answer:
Yes tea & tobacco they are not mere dried leaves of Camellia sinensis (tea) & Nicotiana tobaccum so tea and tobacco to reach the utility stage, they have to be subjected to biological process known as curing, where specific bacteria are added & curing occur by a process of fermentation specific flavor and aroma of tea and tobacco are due to this fermentation process.

Tea – Mycococcus candisans
Tobacco – Bacillus megatherium

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 20.
Write about the harmful activities of fungi.
Answer:

Amanita phalloides
Amanita yema
Boletus satanus
known as toad stools
Poisonous – toxins are produced
Aspergillus
Rhizopus
Mucor
Penicillum
Cause food spoilage
Aspergillus flavus infest dried foods produce carcinogenic toxin called aflatoxin
Patutin ochratoxin A arc other toxins produced by fungi
Diseases Various diseases are caused to plants animals & human beings.

Question 21.
Mycorrhiza is known as bio fertilizer. Explain
Answer:

  • Symbolic association between fungal mycelium and roots of higher plants is known as mycorrhiza.
  • Fungi absorbs nutrition from root of higher plant, and intum fungal hyphae helps the paint to absorb water and minerals nutrients from soil.
  • Any nutrient of biological orgin is biofertilizer & This is Bio fertilizer it is non hazardous, Non phytotoxic and ecofriendly.

Question 22.
Name the Antibiotics derived from fungi.
Answer:

Organism Antibiotics Uses
Penicillium notatum Penicillin To treat Pneumonia and throat infections
Penicillium griseofulvum Grisofulvin To treat ring worm athletes foot & fungal infections of scalp.
Acremorium chrysogenum Cephatosporin To treat respiratory tract infections skin infections & UTI (urinary tract infections)
Claviceps purpurea Ergotamine To treat migraine head aches induce uterus contraction at the time of child birth.

5 Marks

Question 1.
Compare the five-kingdom system of classification
Answer:
Kingdom
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 28

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 2.
Ultrastructure of a typical bacterial cell.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 19

Question 3.
Explain industrial uses of bacteria.
Answer:
Industrial Uses
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 29

Question 4.
List out bacterial diseases caused to plants, animals & human brings.
Answer:
Plant diseases caused by bacteria

SNo

Name of the host Name of the disease

Name of the pathogen

1 Rice Bacterial blight Xanthomonas oryzae
2 Apple Fire blight Erwinia amylovora
3 Carrot Soft rot Erwinia caratovora
4 Citrus Citrus canker Xanthomonas citri
5 Cotton Angular leaf spot Xanthomonas malvacearum
6 Potato Ring rot Clavibacter michiganesis subsp sepdonicus
7 Potato scab Sterptomyces scabies

Animal diseases caused by bacteria

S.No

Name of the animal Name of the diseases

Name of the pathogen

1 Sheep anthrax Bacillus anthracis
2 Cattle brucellosis Brucella abortus
3 Cattle Bovine tuberculosis Mycobacterium bovis
4 Cattle Black leg Clostridium chanvei

Human diseases caused by bacteria

S.NO

Name of the disease

Name of the pathogen

1. Cholera Vibrio cholerae
2. Typhoid Salmonella typhi
3. Tuberculosis Mycobacterium tuberculosis
4. Leprosy Mycobacterium leprae
5. Pneumonia Diplococcus pneumonie
6. Plague Yersinia pestis
7. Diphtheria Corynebacterium diptheriae
8. Tetanus Clostridium tetani
9. Food poisoning Clostridium botulinum
10. Syphilis Treponema pallidum

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 5.
Tabulate the salient features of Cyanophyceae?
Answer:

Thallus Unicellular- Eg. Chroococcus
Colonial- Eg. Gleocapsa
Filamentous trichome- Eg. Nostoc
Movement Gliding movement – Eg. Oscillatoria
Protoplasm Central – centroplasm
Peripheral – chromoplast
Photosynthetic pigments c. phyco cyanin &
c. phyco erythrin
Myxo xanthum
Myxo xanthophyll

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 20

Question 6.
Give an account of Mycoplasma as Mollicutes.
Answer:

  • Small-(0.1-0.5gm)
  • Pieomorphic- gram-negative
  • 1 st isolation by Nocard & co in 1898-from pleural fluid of cattle affected with bovine pleuropneumonia.
  • Cell wall-absent
  • In culture appear as fried egg.
  • DNA contains low guanine and cytosine than true bacteria
  • Cause disease in Animals / Plants
  • Little leaf of brinjal – Witches broom of legumes Phyllody of cloves Sandal spike, -plant diseases
  • Pleuropneumonia – animal disease caused by Mycoplasma mycoide.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 7.
Differentiate between Gram-positive and Gram-Negative bacteria.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 30

Question 8.
Explain transformation in bacteria.
Answer:
Definition: Transfer of DNA from one bacteria to another is called transformation.

  • 1928 Frederick Griffth demonstrated it in mice using Diplococcus pneumonia.
  • 2 strains
    • Smooth colonies – virulent type (S)
    • Rough colonies – Avirulent type (R)
  • S type injected – mouse died because it is virulent
  • R type injected – mouse lived because R type is Avirulent
  • Heat killed S type injected-mouse lived
  • Heat killed S type + R type called when injected mixedly} mouse died

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 21
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 22

 

 

Question 9.
Classify the nitrogen-fixing biological systems.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 23
I. Symbiolic Angiosperm:
Leguminous Rhizobium root nodules of Ground nut Peas etc. (soil bacterium) Non-Leguminous (Alnus casuarinas) Frankia actinomycetes (root nodules)
II. Symbiolic gymnosperm Cycas unidentified spices of Blule green algae (corolloid roots)
III. Symbiotic ferns Azolla- Anbaena azollae – leaf pockets
IV. Symbiotic (fungi & Algae) Lichen (phycobiont mycobiont mutually benefited).

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 10.
Tabulate the human disease caused by bacteria.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 31

Question 11.
Explain General characteristics of fungi
Answer:

  • Plant body- the plant body is filamentous & branched hyphae No of hyphae interwoven to form – my celium
  • Cell wall – made up of chitin (polymer of N acetyl glucose)
  • Mycelium 2 types septate cross wall between cells Aseptate hyphae (multi nucleus coenocytic mycelium) E.g Albugo
  • Mycelium of higher fungi. Septum present between cells if the hyphae Eg. Fusarium.
    Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 24

Types of mycelium
Mycelium organized to compactly interwoven tissue Pletenchyma – 2 Types

  • Prosenchyma-hyphae loosely arranged
  • Pseudoparenchyma hyphae compactly arranged Holocarpic form- Entire thallus- become reproductive structure.
  • In Eucarpic form Some region of thallus reproductive & some other region vegetative Reproduction there are 3 types
    • Anamorph-Asexual
    • Teleomorph -Sexual
    • Holomorph-both sexual & Asexual.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 25

Question 12.
Tabulate various types of Mycorrhizae
Answer:
Mycorrhizae – 3 Types

Ectotropic mycorrhizae

Endotropicmycorrhizae

Ectrndotrophicmycorrhizae

The fungal mycelium forms a dense sheath around the root called mantle. The hypha network penetrate the intercellular spaces of the epidermis, and cortex to form Hartignet. Example pisolithus tinctorius. The hyphae grows mainly inside the roots, penetrate the outer cortical cells of the plant root. A small portion of the mycelium is found outside the root. This form is also called Vesicular Arbuscular Mycorrhizal fungi (VAM fungi) due to the presence of vesicle or arbuscle like haustoria.

1. Arbuscular mycorrhizae(VAM)- Example Gigaspora
2. Ericoid mycorrhizae- Example liriodendron
3. Orchid mycorrhizae- Example Rhizoctonia

The fungi form both mantle and also penetrates the cortical cells.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Question 13.
List out diseases caused by fungi
Answer:
Diseases caused by fungi

S no

Name of the disease

Causative organism

                                               Plant diseases
1 Blast of paddy Magnoporthegriesea
2 Red rot of sugarcane Colletotrichum falcatum
3 Anthracnose of beans Colletotrichum lindemuthianum
4 White rust of crucifers Albugo Candida
5 Peach leaf curl Taphrina deformans
6 Rust of wheat Puccinia graminis tritici

Human diseases

1 Athlete’s foot Epidermophyton floccosum
2 Candidiasis Candida albicans
3 Coccidioidomycosis Coccidiosis immitis
4 Aspergillosis Aspergillus fumigatus

Question 14.
Fruit bodies or Ascocarps 4types
Answer:

  • Cleistothecium
  • Ascothecium(flask shape)
  • Apothecium(cup shape)
  • Psedothecium
    1. Image structure of cleistothecium
    2. Structure of PeritheciumV.S
    3. Structure of Apothecium

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 26
Samacheer Kalvi 11th Bio Botany Chapter 1 Living World 27

Question 15.
Give an account of Basidiomycetes (club fungi).
Answer:

Fungi included Puffballs, toadstools, bird nest’s fungi. Bracket fungi- stinkhorns, rusts smuts etc.
Habitat Mostly terrestrial Saprophytic Parasitic mode of life
Mycelium Well-developed, septate with a bracket like 3 types Monokaryotic mycelium Dikaryotic mycelium (Primary & Secondary) Tertiary mycelium
Sexual reproduction steps Plasmogamy Karyogamy &Meiosis Occur – but sex organs absent Somatogamy or – Spermatisation result in Plasmogamy Karyogamy delayed & dikaryotic phase is prolonged followed by meiosis
image 4 nuclei become basidiospores bom on sterigmata -(exogenous) Club-shaped basidium bear basidiospores so-known as club fungi.

Samacheer Kalvi 11th Bio Botany Chapter 1 Living World

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 32 Direct Taxes Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 32 Direct Taxes

11th Commerce Guide Direct Taxes Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
Income Tax is ………………
a) a business tax
b) a direct tax
c) an indirect tax
d) none of these
Answer:
b) a direct tax

Question 2.
Period of assessment year is …………
a) 1 st April to 31 st March
b) 1st March to 28th Feb
c) 1st July to 30th June
d) 1st Jan. to 31st Dec
Answer:
a) 1 st April to 31 st March

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 3.
The year in which income is earned is known as …………………
a) Assessment Year
b) Previous Year
c) Light Year
d) Calendar Year
Answer:
b) Previous Year

Question 4.
The aggregate income under five heads is termed as ……………….
(a) Gross Total Income
(b) Total Income
(c) Salary Income
(d) Business Income
Answer:
(b) Total Income

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 5.
Agricultural income earned in India is ……………
a) Fully Taxable
b) Fully Exempted
c) Not Considered for Income
d) None of the above
Answer:
b) Fully Exempted

II. Very Short Answer Questions

Question 1.
What is Income tax?
Answer:
Income tax is a direct tax under which tax is calculated on the income, gains, or profits earned by a person such as individuals and. other artificial entities (a partnership firm, company, etc.).

Question 2.
What is meant by the previous year?
Answer:
The year in which income is earned is called the previous year. It is also called as financial year.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 3.
Define the term person?
Answer:
The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu, Undivided Family, Firm, Company, local authority, Association of Person or Body of Individual or any other artificial juridical persons.

Question 4.
Define the term assessed?
Answer:
As per S. 2(7) of the Income Tax Act, 1961, the term “assessee” means a person by whom any tax or any other sum of money is payable under this Act. Assess includes individual, HUF, Firm, Company, Local authority, AOP, BOJ or any other artificial juridical persons.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 5.
What is an assessment year?
Answer:
The term has been defined under section 2(9). The year in which tax is paid is called the assessment year. It normally consists of a period of 12 months commencing on 1st April every year and ending on 31st March of the following year.

III. Short Answer Questions

Question 1.
What is Gross Total Income?
Answer:
According to section 80 B (5) Income computed under the following heads shall be aggregated after adjusting past and present losses and the total so arrived is known as ‘Gross Total income’.

  • Income from Salaries
  • Income from House Property
  • Income from Business or Profession
  • Income from Capital Gain
  • Income from Other Sources

Question 2.
List out the five heads of Income.
Answer:
The five heads of income are:

  1. Income from‘Salaries’ [Sections 15 – 17];
  2. Income from ‘House Property’ [Sections 22 – 27];
  3. Income from ‘Profits and Gains of Business or Profession’ [Sections 28 – 44];
  4. Income from ‘Capital Gains’ [Sections 45 – 55]; and
  5. Income from other Sources’ [Sections 56 – 59].

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 3.
Write a note on Agricultural Income.
Answer:
According to Section 2(1 A) of the Income Tax Act 1961, Agricultural Income includes, “Any rent or revenue derived from land which is situated in India and is used for agriculture purposes”.

Question 4.
What do you mean by Total income?
Answer:
Out of Gross Total Income, Income Tax Act 1961 allows certain deductions under section 80. After allowing these deductions the figure which we arrive at is called ‘Total Income’ and on this figure tax liability is computed at the prescribed rates.

  1. Gross Total Income
  2. Less: Deductions (Sec. 80C to 80U)
  3. Total Income (T.I.)

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 5.
Write short notes on:
Answer:
a. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, it is called direct tax e.g. Income-Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax. Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs.50,000 and above -from 2005), etc. In India all direct taxes are levied and administered by Central Board of Direct Taxes.

b. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from the buyers and is paid by seller to the Government. It is called indirect tax. e.g. GST.

IV. Long Answer Questions

Question 1.
Elucidate any five features of Income Tax.
Answer:
Features of Income Tax in India:
1. Levied as Per the Constitution Income tax is levied in India by virtue of entry No. 82 of List I (Union List) of Seventh Schedule to Article 246 of the Constitution of India.

2. Levied by Central Government Income tax is charged by the Central Government on all incomes other than agricultural income. However, the power to charge income tax on agricultural income has been vested with the State Government as per entry 46 of List II, i.e., State List.

3. Direct Tax Income tax is a direct tax. It is because the liability to deposit and ultimate burden are on the same person. The person earning income is liable to pay income tax out of his own pocket and cannot pass on the burden of tax to another person.

4. Annual Tax Income tax is an annual tax because it is the income of a particular year which is chargeable to tax.

5. Tax on Person It is a tax on income earned by a person. The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu Undivided Family, Firm, Company, local authority, Association of Person or Body of Individual or any other artificial juridical persons. The persons who are covered under Income-tax Act are called ‘assessees’.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 2.
Define Tax. Explain the term direct tax and indirect tax with an example.
Answer:
A tax is a compulsory financial contribution imposed by a government to raise revenue, levied on the income or property of persons or organizations, on the production costs or sales prices of goods and services etc.

a. Direct Tax:
A direct tax is paid directly by an individual or organization to an imposing entity. A taxpayer, for example pays direct taxes to the government for different purposes, including real property tax, personal property tax, income tax or taxes on assets.

b. Indirect Tax:
If tax is levied on the goods or services of a person(seller), it is known as indirect tax. It is , collected from the buyers and is paid by the seller to the government. It is paid to the government by one entity in the supply chain. Example: GST.

Question 3.
List out any ten kinds of incomes chargeable under the head income tax.
Answer:

  1. Profits and gains of business or profession.
  2. Dividend
  3. Voluntary contribution received by a charitable / religious trust or university/education institution or hospital/electoral trust[ w.e.f. 01.04.2010]
  4. Value of perquisite or profit in lieu of salary taxable u/s 17 and social allowance or benefit specifically granted either to meet personal expenses or for performance Of duties of an office or employment of profit.
  5. Export incentives, like duty drawback, cash compensatory support, sale of licenses, etc.
  6. Interest, salary, bonus, commission, or remuneration earned by a partner of a firm from such firm.
  7. Capital gains chargeable u/s 45.
  8. Profits and gains from the business of banking carried on by a cooperative society with its members.
  9. Winnings from lotteries, crossword puzzles, races including horse races, card games, and other games of any sort or from gambling or betting of any form or nature whatsoever.
  10. Deemed income u/s 41 or 59.

Question 4.
Discuss the various kinds of assessments.
Answer:
The following are the different types of assesses:

  • Individual
  • Partnership firm
  • Hindu Undivided Family
  • Companies
  • Association of Persons
  • Body of Individual.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

11th Commerce Guide Direct Taxes Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
How many heads of income are there to compute Gross total income?
a. Six
b. Five
c. Four
d. Three
Answer:
b. Five

Question 2.
Income Tax Act came into force on ……………
a.1.4.1932
b. 1.4.1962
c. 1.4.1947
d. 1.4.1954
Answer:
b. 1.4.1962

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 3.
The compensation received for loss of trading asset is a ……………….
a. Capital receipt
b. Revenue receipt
c. a casual receipt
d. None of the above
Answer:
a. Capital receipt

Question 4.
The legislative powers of the Union Government and the State Governments are given in the …………………….. of the Indian Constitution.
a. Article 246 (VII schedule)
b. Article 246 (VI schedule)
c. Article 264 (VII schedule)
d. Article 446 (VII schedule)
Answer:
a. Article 246 (VII schedule)

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 5.
Tax charged on Long Term Capital Gain is ……………..
a. 20%
b. 15%
c. 25%
d. 30%
Answer:
a. 20%

II. Very Short Answer Questions:

Question 1.
What do you mean by Tax?
Answer:
Tax is a compulsory contribution to state revenue by the Government. It is levied on the income or profits from the business of individuals and institutions.

Question 2.
What is the reasoñ for collecting tax?
Answer:
The revenue earned through tax is utilized for the expenses of civil ädministration, internal and external security, building infrastructure, etc.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

III. Short Answer Questions

Question 1.
Who do income tax is treated as annual tax?
Answer:
Income tax is an annual tax because it is the income of a particular year which is chargéable to tax.

Question 2.
What are all the tax rates prescribed for LTCG, STCG, and lottery income?
Answer:
The following tax rates have been prescribed under Income Tax Act. ‘

  • Tax on long term capital gain @ 20% (Section 112);
  • Tax on short term capital gain on shares covered under STT @15% (Section 111A).
  • Tax on lottery income @ 30% (Section 11 5BB)

Question 3.
What do you mean by the previous year?
Answer:
According to Section (3) of the Income Tax Act 1961, “The year in which income is earned is called the previous year”. It is also normally consisting of a period of 12 months commencing on 1 st April every year and ending on 31 st March of the following year. It is also called as financial year’ immediately following the assessment year.

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

IV. Long Answer Questions

Question 1.
Write a note on the structure of the Indian Taxation system:
Answer:
The Indian taxation system is one of the largest systems in the world. The authority to levy tax is derived from the Indian constitution and is well-structured. The tax administration has a clear demarcation between Central Government and State Governments and then between state Governments and Local Bodies. Article 246 (Seventh Schedule) of the Indian constitution contains the legislative powers (including taxation) of the Union government and the State Governments.

Question 2.
Write a note on Heads of Income under Income Tax Act:
Answer:
Section 14 of the Income Tax Act 1961 provides for the computation of the total income of an assessee which is divided into five heads of income. Each head of income has its own method of computation. These five heads are;

  1. Income from ‘Salaries’ [Sections 15-17]
  2. Income from ‘House Property’ [Sections 22-27]
  3. Income from ‘Profits and Gains of Business or Profession’ [Sections 28- 44]
  4. Income from ‘Capital Gains’ [Sections 45-55] and
  5. Income from ‘Other Sources’ [Sections 56-59].

Samacheer Kalvi 11th Commerce Guide Chapter 32 Direct Taxes

Question 3.
Write a note on slab raite of Income-tax charged on Individual:
Answer:
According to the Assessment year 2018-2019 the following rates will be charged:

Total Income (Rs)

Income Tax Rate

Up to 2,50,000 Nil
2,50,001 – 5,00,000  5%
5,00,000 – 10,00,000 20%
Above 10,00,000 30%

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 3 Periodic Classification of Elements Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

11th Chemistry Guide Periodic Classification of Elements Text Book Back Questions and Answers

Textual Questions:

I. Choose the best Answer:

Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

Question 2.
The electronic configuration of the elements A and B are 1s2, 2s2, 2p6, 3s2 and 1s2, 2s2, 2p5 respectively. The formula of the ionic compound that can be formed between these elements is
(a) AB
(b) AB2
(c) A2B
(d) none of the above
Answer:
(b) AB2

Question 3.
The group of elements in which the differentiating electron enters the anti penultimate shell of atoms are called
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it?
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al3+ < Mg2+ < Na+ < F (increasing ionic size)
(d) B < C < O < N (increasing first ionisation enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

Question 5.
Which of the following elements will have the highest electronegativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 6.
Various successive ionisation enthalpies (in kjmol 1) of an element are given below.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 1

The element is
(a) phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 7.
In the third period the first ionization potential is of the order.
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na < Al < Mg < P < Si
Answer:
(b) Na < Al < Mg < Si < P

Question 8.
Identify the wrong statement.
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Amongst isoelectric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius

Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy
(a) Al < O < C < Ca < F
(b) Al < Ca < O < C < F
(c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53 respectively is
(a) I > Br > Cl > F
(b) F > Cl > Br > I
(c) Cl > F > Br > I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 11.
Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen

Question 12.
The element with positive electron gain enthalpy is
(a) Hydrogen
(b) Sodium
(c) Argon
(d) Fluorine
Answer:
(c) Argon

Question 13.
The correct order of decreasing electronegativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y > A > Z
Answer:
(a) Y > Z > X > A

Question 14.
Assertion:
Helium has the highest value of ionisation energy among all the elements known
Reason:
Helium has the highest value of electron affinity among all the elements known
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Question 15.
The electronic configuration of the atom having | maximum difference in first and second ionisation j energies is
(a) 1s2, 2s2, 2p6, 3s1
(b) 1s2, 2s2, 2p6, 3S2
(c) 1s2, 2s2, 2p6, 3s2, 3s2, 3p6, 4s1
(d) 1s2, 2s2, 2p6, 3s2, 3p1
Answer:
(a) 1s2, 2s2, 2p6, 3s1

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 16.
Which of the following is second most electronegative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Question 17.
IE1 and IE2 of Mg are 179 and 348 kcal mol-1 respectively. The energy required for the reaction Mg → Mg2+ + 2e is
(a) + 169 kcal mol-1
(b) -169 kcal mol-1
(c) +527 kcalmol-1
(d) -527 kcal mol-1
Answer:
(c) +527 kcalmol-1

Question 18.
In a given shell the order of screening effect is
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Question 19.
Which of the following orders of ionic radii is correct?
(a) H > H+ > H
(b) Na+ > F > O2-
(c) F > O2- > Na+
(d) None of these
Answer:
(d) None of these

Question 20.
The First ionisation potential of Na, Mg and Si are 496, 737 and 786 kJ mol-1 respectively. The ionisation potential of Al will be closer to
(a) 760 kJ mol-1
(b) 575 kJ mol-1
(c) 801 kJ mol-1
(d) 419 kJ mol-1
Answer:
(b) 575 kJ mol-1

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(b) Increases in a period and decreases in a group

Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases

Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

II. Write brief answer to the following questions:

Question 24.
Define modern periodic law.
Answer:
The modem periodic law states that “The physical and chemical properties of the elements are a periodic function of their atomic numbers.”

Question 25.
What are isoelectronic ions? Give examples.
Answer:
Two ions having the same number of electrons are called isoelectronic ions.
Example: Na+ (1s2 2s2 2p6) and F (1s2 2s2 2p6). Both these ions contain eight electrons.

Question 26.
What is an effective nuclear charge?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Zeff = Z – S
Where Z = Atomic number
S = Screening constant calculated by using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
Answer:
“Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
The given statement is not correct. Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg+, Mg2+, and Mg3 ions. Which step will have the highest ionization energy and why?
Answer:
Magnesium loses electrons successively in the following steps,
Step 1:
Mg(g) + IE1 → Mg+(g) + 1e, Ionisation energy = I.E1
Step 2:
Mg+(g)+ IE2 → Mg2+(g) + 1e, Ionisation energy = I.E2
Step 3:
Mg2+(g) + IE3 → Mg3+(g) + 1e, Ionisation energy = I.E3
The total number of electrons is less in the cation than the neutral atom while the nuclear charge remains the same. Therefore, the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus, the successive ionization energies, always increase in the following order I.E1 < I.E2 < I.E3. Thus, Step-3 will have the ionization energy.

Question 29.
Define electronegativity.
Answer:
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionisation potential is always higher than the first ionisation potential?
Answer:

  • The second ionization potential is always higher than the first ionization potential.
  • Removal of one electron from the valence orbit of a neutral gaseous atom is easy so first ionization energy is less. But from a uni positive ion, removal of one more electron becomes difficult due to the more forces of attraction between the excess of protons and less number of electrons.
  • Due to greater nuclear attraction, second ionization energy is higher than first ionization energy.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 31.
The energy of an electron in the ground state of the hydrogen atom is -2.8 × 10-8 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol-1
Answer:
Energy of an electron in the ground state of the hydrogen atom is – 2.8 × 10-8.
The ionization energy of atomic hydrogen is 2.8 × 10-18 × 6.023 × 1023 J/mol
= 16.86 × 105 J/mol
= 1686 kJ/mol.

Question 32.
The electronic configuration of an atom is one of the important factors which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:

  • The electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
  • Half-filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
  • For e.g. Beryllium (Z = 4) 1s2 2s2 (completely filled electronic configuration)
    Nitrogen (Z = 7) 1s2  2s2  2px1  2py1 2pz1 (half-filled electronic configuration) Both beryllium and nitrogen have high ionization energy due to more stable nature.
  • In the case of beryllium (1s2 2s2), nitrogen (1s2 2s2 2p3) the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.
  • Noble gases have stable ns2 np6 configuration and the addition of further electrons is unfavorable and they have zero electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element Ununoctium (Oganesson, Z – 118) present in the 7th period and 18th group of the periodic table.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
The fifth period of the periodic table has 18 elements. 5th period starts from Rb to Xe (18 elements).
5th period starts with principal quantum number n = 5 and l = 0, 1, 2, 3 and 4.
When n = 5, the number of orbitals = 9.
1 for 5s
5 for 4d
3 for 5p
The total number of orbitals = 9.
Total number of electrons that can be accommodated in 9 orbitals = 9 x 2 = 18.
Hence the number of elements in the 5th period is 18.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 35.
Elements a, b, c and d have the following electronic configurations:
a: 1s2, 2s2, 2p6
b: 1s2, 2s2, 2p6, 3s2, 3p1
c: 1s2, 2s2, 2p6, 3s2, 3p6
d: 1s2, 2s2, 2p1
Which elements among these will belong to the same group of the periodic table?
Answer:
The elements ‘a’ [Ne (Z= 10), 1s2, 2s2, 2p6] and ‘c’ [Ar (Z = 18), 1s2, 2s2, 2p6, 3s2, 3p6] have the same valence electronic configuration and hence, belongs to the same group, i.e., group 18 of the periodic table . Similarly, the elements ‘b’ [ Al (Z = 13), 1s2, 2s2, 2p6, 3s2, 3p1] and ‘d’ [B (Z= 5), 1s2, 2s2, 2p1] have the same valence electronic configuration and hence, belongs to the same group, i.e., group 13 of the periodic table.

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:

  • The electronic configuration of lanthanides is 4f1-14 5d0-16s2.
  • The electronic configuration of actinides is 5f1-14 6d0-1 7s2.

Question 37.
Why do halogens act as oxidizing agents?
Answer:
Halogens are having the general electronic configuration of ns2, np5 and readily accept an electron to get the stable noble gas electronic configuration. Therefore, halogens have high electron affinity. Hence, halogens act as oxidizing agents.

Question 38.
Mention any two anomalous properties of second-period elements.
Answer:

  • In the 1st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
  • In the 2nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

Question 39.
Explain the Pauling method for the determination of ionic radius.
Answer:
Ionic radius is defined as the distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud of the ion. The ionic radius of a uni-univalent crystal can be calculated using Pauling’s method from the interionic distance between the nuclei of the cation and anion. Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other and therefore,
d = rc+ + rA- …………..(1)
where ‘d’ is the distance between the centre of the nucleus of the cation C+ and A. r c+ and rA- are the radius of the cation and anion respectively.

Pauling also assumed that the radius of the ion having noble gas electronic configuration (Na+ and Cl having 1s2, 2s2, 2p6 configuration) is inversely
proportional to the effective nuclear charge felt at the periphery of the ion.
i.e.,   rc+ ∝ \(\frac{1}{\left(Z_{e f f}\right)^{C+}}\) ………(2)

and rA- ∝ \(\frac{1}{\left(Z_{e f f}\right)^{A-}}\) ………….(3)

where Zeff is the effective nuclear charge.
Zeff = Z – S.
Dividing the equation (2) by (3)
\(\frac{r_{c^{+}}}{r_{A^{-}}}=\frac{\left(Z_{e f f}\right)^{A-}}{\left(Z_{e f f}\right)^{C+}}\)

On solving the equations (1) and (4), the ionic radius of cation and anion are calculated.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 40.
Explain the periodic trend of ionisation potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called ionization energy.
(b) Variation in a period:
Ionization energy is a periodic property. On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

  • Increase of nuclear charge in a period
  • Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

  • A gradual increase in atomic size
  • Increase of screening effect on the outermost electrons due to the increase of the number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 41.
Explain the diagonal relationship.
Answer:
On moving diagonally across the periodic table, the second and third-period elements show certain similarities. Even though the similarity is not the same as we see in a group, it is quite pronounced in the following pair of elements.
Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 2
The similarity in properties existing between the diagonally placed elements is called ‘diagonal relationship’.

Question 42.
Why the first Ionisation enthalpy of sodium is lower than that of magnesium while its second ionisation enthalpy Is higher than that of magnesium?
Answer:
The 1st ionization enthalpy of magnesium is higher than that of Na due to the higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of the first electron, Na+ formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to a very high second ionization enthalpy for sodium. On the other hand, Mg+ formed after losing the first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 43.
By using Pauling’s method calculate the ionic radii of K+ and Cl ions in the potassium chloride crystal. Given that dK + – Cl = 3.14 Å.
Answer:
d = rK+ + rCl- = 3.14 Å
\(\frac{r_{K+}}{r_{C l-}}=\frac{\left(Z_{e f f}\right)^{C l-}}{\left(Z_{e f f}\right)^{K+}}\)

(Zeff)Cl- = Z – S = 17 – 10.9 = 6.1
(Zeff)K+ = Z – S = 19 – 16.8 = 2.2
\(\frac{r_{K+}}{r_{C l-}}=\frac{6.1}{2.2}\) = 2.77

rK+ = 2.77 rCl-
2.77 rCl- + rCl- = 3.17 Å
3.77 rCl- = 3.17 Å
rCl- = 0.83 Å
rK+ = (3.14 – 0.83) Å = 2.31 Å
The ionic radius of the K+ ion is 2.31 Å and Cl ion is 0.83 Å.

Question 44.
Explain the following, give appropriate reasons.
(i) Ionisation potential of N is greater than that of O:
Answer:
Nitrogen with 1s2, 2s2, 2p3 electronic configuration has higher ionization energy than oxygen. Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from the 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

(ii) First ionisation potential of the C-atom is greater than that of the B atom, whereas the reverse is true is for the second ionisation potential.
Answer:
The first ionization potential of the C atom and B-atom are as follows:
C(1s2, 2s2, 2p2) + IE1 → C+ (1s2, 2s2, 2p1)

B(1s2, 2s2, 2p1) + IE1 → B+ (1s2, 2s2)

The ionization energy usually increases along a period. Hence, the first ionization energy of carbon is greater than that of Boron.
The second ionization potential of the C atom and B atom is as follows:
C+ (1s2, 2s2, 2p1) + IE2 → C+(1s2, 2s2)
B+ (1s2, 2s2) + IE2 → B+ (1s2, 2s1)

B+ has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of the C+ atom. Hence, the second ionization energy of Boron is greater than that of carbon.

(iii) The electron affinity values of Be, Mg, and noble gases are zero, and those of N (0.02 eV) and P (0.80 eV) are very low.
Answer:
Be, Mg and noble gases have completely filled stable configuration and the addition of further electron is unfavourable and requires energy. The addition of extra electrons will disturb their stable electronic configuration and hence, they have almost zero electron affinity.

Nitrogen and Phosphorus have a half-filled stable configurations and the addition of further electrons is unfavourable and requires energy. The addition of extra electrons will disturb their stable electronic configuration and hence, they have very low zero electron affinity.

(iv) The formation of from F(g)  from F(g) is exothermic while that of O2-(g) from O(g) is endothermic.
Answer:
The sizes of oxygen and fluorine atoms are comparatively small and they have high electron density. The extra electron added to fluorine has to accommodate in the 2p orbital which is relatively compact. Hence, the formation of F- from F is exothermic. In the case of oxygen, the formation of O2- from O is endothermic due to extra stability of the completely filled 2p orbital of O2- formation.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 45.
What is the screening effect?
Answer:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect (or) screening effect.

Question 46.
Briefly give the basis for Pauling’s scale of electronegativity.
Answer:
Pauling’s scale:

  • Electronegativity is the relative tendency of an element present in a covalently bonded molecule to attract the shared pair of electrons towards itself.
  • Pauling assigned arbitrary values of electronegativities for hydrogen and fluorine as 2.2 and 4, respectively.
  • Based on this the electronegativity values for other elements can be calculated using the following expression.
    (XA-XB) = 0.182 √EAB – (EAA EBB)
    Where EAB , EAA, and EBB are the bond dissociation energies of AB, A2, and B2 molecules respectively.
    XA and XB are electronegativity values of A and B.

Question 47.
State the trends in the variation of electronegativity in groups and periods.
Answer:
Variation of Electronegativity in a period:
The electronegativity generally increases across a period from left to right. As discussed earlier, the atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases. Hence, the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.

Variation of Electronegativity in a group:
The electronegativity generally decreases down a group. As we move down a group, the atomic radius increases, and the nuclear attractive force on the valence electron decreases. Hence, the electronegativity decreases. Noble gases are assigned zero electronegativity. The electronegativity values of the elements of 5-block show the expected decreasing order in a group. Except for 13th and 14th groups, all other p-block elements follow the expected decreasing trend in electronegativity.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

11th Chemistry Guide Periodic Classification of Elements Additional Questions and Answers

I. Very Short Question and Answers (2 Marks):

Question 1.
State periodic law.
Answer:
Periodic law states that ‘‘the properties of the elements are the periodic functions of their atomic weights”.

Question 2.
Write a note about Chancourtois classification.
Answer:
In this system, elements that differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line. Elements lying directly under each other showed a definite similarity. This was the first periodic law.

Question 3.
What is Lavoiser’s classification of elements?
Answer:
Lavoiser classified the substances into four groups of elements namely acid-making elements, gas-like elements, metallic elements, and earthy elements.

Question 4.
State Mendeleev’s periodic law.
Answer:
This law states that “The physical and chemical properties of elements are a periodic function of their atomic weights.”

Question 5.
What are groups and periods?
Answer:
All the elements are arranged in the modem periodic table which contains 18 vertical columns and 7 horizontal rows. In the periodic table, the vertical columns are called groups, and horizontal rows are called periods.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 6.
State modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 7.
What are p-block elements? Give their general electronic configuration.
Answer:
The elements of groups 13 to 18 are called p-block elements or representative elements and have a general electronic configuration ns2, np1 – 6.

Question 8.
Mention the names of the elements with atomic numbers 101, 102, 109, and 110.
Answer:
Z = 101  IUPAC  name : Mendelevium
Z = 102  IUPAC  name : Nobelium
Z = 109  IUPAC  name : Meitnerium
Z = 110  IUPAC  name : Darmstadtium

Question 9.
What are f-block elements? Give their properties.
Answer:
The lanthanides and the actinides are called f-block elements. These elements are metallic in nature and have high melting points. Their compounds are mostly coloured. These elements also show variable oxidation states.

Question 10.
Give the name and electronic configuration of elements of group and 2 groups.
Answer:

  • Elements of 1st group are called alkali metals. Their electronic configuration is ns1.
  • Elements of 2nd group are called alkaline earth metals. Their electronic configuration is ns2.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 11.
Define Atomic radius.
Answer:
The atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Question 12.
Write any two characteristic properties of alkaline earth metals.
Answer:

  • Alkaline earth metals readily lose their outermost electrons to form a +2 ion.
  • As we go down the group. their metallic character and reactivity are increased.

Question 13.
Why is the covalent radius shorter than the actual atomic radius?
Answer:
The formation of a covalent bond involves the overlapping of atomic orbitals and it reduces the expected internuclear distance. Therefore, the covalent radius is always shorter than the actual atomic radius.

Question 14.
Define metallic radius.
Answer:
Metallic radius is defined as one-half of the distance between two adjacent metal atoms in the closely packed metallic crystal lattice.

Question 15.
Halogens and chalcogens have highly negative electron gain enthalpies. Why?
Answer:

  • Group 16 (chalcogens) and Group 17 (halogens) are interested to add two or one electrons respectively to attain a stable noble gas configuration.
  • Because of this interest, these elements have highly negative electron gain enthalpies.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 16.
What is the effective nuclear charge? How is it approximated?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the below-mentioned equation Zeff = Z – S.
where Z is the atomic number and S is the screening constant which can be calculated using Slater’s rules.

Question 17.
Elements Zn, Cd, and Hg with electronic configuration (n-1)d10 ns2 do not show most of the transition elements properties. Give reason.
Answer:

  • Zn, Cd, and Hg are having completely filled d-orbitais (d10 electronic configuration).
  • They do not have partially filled d-orbitais Like other transition elements. So they do not show much of the transition elements properties.

Question 18.
Define Ionisation energy.
Answer:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
Why d-block elements are called transition elements?
Answer:
d-block elements form a bridge between the chemically active metals of s-block elements and the less active elements of groups of 13th and 14th and thus take their familiar name transition elements.

Question 20.
Beryllium has higher ionization energy than Boron. Give reason.
Answer:
Beryllium has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of Boron (2s2 – 2p1). Hence, Beryllium has higher ionization energy than Boron.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 21.
Write the electronic configuration of lanthanides and actinides?
Answer:

  • The electronic configuration of lanthanides is 4f1-144 5d0-11 6s2.
  • The electronic configuration of actinides is 5f1-14 6d0-17s2.

Question 22.
What is the effect of shielding on ionization energy?
Answer:
As we move down a group, the number of inner-shell electrons increases which in turn increases the repulsive force exerted by them on the valence electrons, i.e., the increased shielding effect caused by the inner electrons decreases the attractive force acting the valence electron by the nucleus. Therefore, the ionization energy decreases.

Question 23.
Define electron affinity.
Answer:
Electron affinity is defined as the amount of energy released when an electron is added to the valence shell of an isolated neutral gaseous atom in its ground state to form its anion.

Question 24.
What are periodic properties? Give example.
Answer:
The term periodicity of properties indicates that the elements with similar properties reappear at certain regular intervals of atomic number in the periodic table.
Example:

  • Atomic radii
  • Ionization energy
  • Electron affinity
  • Electronegativity.

Question 25.
Why is the electron affinity of Nitrogen almost zero?
Answer:
Nitrogen has half-filled stable (1s2, 2s2, 2p3) electronic configuration. The addition of extra electrons will disturb their stable electronic configuration and it has almost zero electron affinity.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 26.
The cationic radius is smaller than its corresponding neutral atom. Justify this statement.
Answer:

  • When a neutral atom loses one or more electrons it forms cation.
    Na → Na+ + e
  • The radius of this cation (rNa+)is decreased than its parent atom (rNa).
  • When an atom is charged to cation, the number of nuclear charges becomes greater than the number of orbital electrons. Hence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Question 27.
Define electronegativity.
Answer:
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 28.
What are isoelectronic ions? Give example.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example: Na+ Mg2+, Al3+, F, O2-, N3-

Question 29.
What is the variation of electronegativity in a group?
Answer:
The electronegativity generally decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence, the electronegativity decreases.

Question 30.
The ionization energy of beryllium is greater than the ionization energy of boron. Why?
Answer:
Be (Z= 4) 1s2 2s2. it has completely filled valence electrons, which requires high IE1.
B (Z =5) 1s2 2s2 2p1. It has incompletely filled valence electrons, which requires comparatively
less IE1 Hence I.E1 Be > I.E1 B.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 31.
What is the diagonal relationship?
Answer:
On moving diagonally across the periodic table, the second and third-period elements show certain similarities. The similarity in properties existing between the diagonally placed elements is called the ‘diagonal relationship’.

Question 32.
Define electron gain enthalpy or electron affinity. Give its unit.
Answer:
The electron gain enthalpy of an element is the amount of energy released when an electron is added to the neutral gaseous atom.
A + electron → A + energy (E.A)
Unit of electron affinity is KJ mole.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

II. Short Question and Answers (3 Marks):

Question 1.
How Moseley determined the atomic number of an element using X-rays?
Answer:

  • Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
  • He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding a clement with the high energy of electrons.
  • Moseley correlated the frequency of the X-ray emitted by an equation as,
    \(\sqrt{v}\) = a (Z – b)
    Where υ = Frequency of the X-rays emitted by the elements.
    a and b = Constants.
  • From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

Question 2.
Write notes on Newlands classification of elements.
Answer:
Newland made an attempt to classify the elements and proposed the law of octaves. On arranging the elements in the increasing order of atomic weights, he observed that the properties of every eighth element are similar to the properties of the first element. This law holds good for lighter elements up to calcium.

Question 3.
Describe Mendeleev’s periodic classification of elements.
Answer:
Dmitriev Mendeleev proposed that “the properties of the elements are the periodic functions of their atomic weights” and this is called periodic law. Mendeleev listed the 70 known elements at that time in several vertical columns in order of increasing atomic weights. Thus, Mendeleev constructed the first periodic table based on the periodic law.

In the periodic table, he left some blank spaces since there were no known elements with the appropriate properties at that time. He and others predicted the physical and chemical properties of the missing elements. Eventually, these missing elements were discovered and found to have the predicted properties. For example, Gallium of group-III and Germanium of group- IV were unknown at that time. But Mendeleev predicted their existence and properties. After the discovery of the actual elements, their properties were found to match closely to their predicted by Mendeleev.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 4.
Write notes on Moseley’s work.
Answer:
Henry Mosley studied the characteristic X-rays spectra of several elements by bombarding them with high energy electrons and observed a linear correlation between atomic number and the frequency of X-rays emitted which is given by the following expression, √υ = a(Z – b)
where, υ is the frequency of the X-rays emitted by the element with atomic number ‘Z’; ‘a’ and ‘b’ are constants and have same values for all the elements. The plot of √υ against Z gives a straight line. Using this relationship, we can determine the atomic number of an unknown element from the frequency of X-rays emitted.

Question 5.
What are the reasons behind Moseley’s attempt in finding an atomic number?
Answer:

  • The number of electrons increases by the same number as the increase in the atomic number.
  • As the number of electrons increases, the electronic structure of the atom changes.
  • Electrons in the out can not shell of an atom (valence shell electrons) determine the chemical properties of the elements.

Question 6.
What are s-block elements? Give their properties.
Answer:
The elements of group-1 and group-2 are called s-block elements, since the last valence electron enters the ns orbital.

  • The group-1 elements are called alkali metals while group-2 elements are called alkaline earth metals.
  • These are soft metals and possess low melting and boiling points with low ionization enthalpies.
  • They are highly reactive and form ionic compounds.
  • They are highly electropositive in nature and most of the elements imparts colour to the flame.

Question 7.
Explain the classification of elements based on electronic configuration.
Answer:

  • The distribution of electrons into orbitais, s, p, d, and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, l, m, and s. of these the principal quantum number (n) defines the main energy level known as shells.
  • The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
  • The electronic configuration of elements in the periodic table can be studied along with the periods and groups separately for the best classification of elements.
  • Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
  • A vertical column of the periodic table is called a group. A group consists of a series of elements having a similar configuration to the outermost shell. There are 18 groups in the periodic table.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 8.
What are d-block elements? Give their properties.
Answer:
The elements of the groups 3 to 12 are called d-block elements or transition elements with general valence shell electronic configuration ns1 – 2, (n – 1 )d1 – 10.

  • These elements show more than one oxidation state and form ionic, covalent, and coordination compounds.
  • They can form interstitial compounds and alloys which can also act as catalysts.
  • These elements have high melting points and are good conductors of heat and electricity.

Question 9.
What is a covalent radius? Explain.
Answer:
Covalent radius is one-half of the internuclear distance between two identical atoms linked together by a single covalent bond. The internuclear distance can be determined using X-ray diffraction studies.
For example, the experimental intemuclear distance in Cl2 molecule is 1.98 Å. The covalent radius of chlorine is calculated as follows,
dcl – cl = rcl + rcl
rcl = \(\frac{d_{c l-d t}}{2}=\frac{1.98}{2}\) = 0.99 Å .

The formation of a covalent bond involves the overlapping of atomic orbitals and it reduces the expected internuclear distance. Therefore, the covalent radius is always shorter than the actual atomic radius.

Question 10.
How is the covalent radius of an individual atom can be calculated?
Answer:
The covalent radius of an individual atom can be calculated using the internuclear distance (dA-B) between two different atoms A and B. The simplest method proposed by Schomaker and Stevenson is as follows.
dA – B = rA + rB – 0.09(χA – χB)
where χA and χB are the electronegativities of A and B respectively in Pauling units. Here, χA > χB and radius is in Å.

Question 11.
Write about the electronic configuration of the 1st and 2nd periods.
Answer:
Electronic configuration of t period:
In 1 period only two elements are present. This period starts with the filling of electrons in the first energy level, n1. This level has only one orbital as is. Therefore it can accommodate two electrons maximum.

Electronic configuration of 2nd period:
In the 2’ period 8 elements are present. This period starts with the filling of electrons in the second energy level, n = 2. In this level four orbitais (one 2s and three 2p) are present. Hence the second energy level can accommodate 8 electrons. Thus, the second period has eight elements.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 12.
Discuss the variation of electron affinity in the period.
Answer:
As we move from alkali metals to halogens in a period, generally electron affinity increases, i.e., the amount of energy released will be more. This is due to an increase in the nuclear charge and a decrease in the size of the atoms. However, in the case of elements such as beryllium and nitrogen the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have stable ns2, np6 configurations, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns2, np5 readily accept an electron to get the stable noble gas electronic configuration (ns2, np6), and therefore, in each period the halogen has high electron affinity.

Question 13.
What are the two exceptions of block division in the periodic table?
Answer:
1. Helium has two electrons. Its electronic configuration is 1s2. As per the configuration, it is supposed to be placed in ‘s’ block but actually placed in s group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in the 18th group. It also resembles 18th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. it has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17th group elements). Because of these assumptions, the position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

III. Long Question and Answers (5 Marks):

Question 1.
Describe Mosley’s work and Modern Periodic Law.
Answer:
Henry Mosley studied the characteristic X-rays spectra of several elements by bombarding them with high energy electrons and observed a linear correlation between atomic number and the frequency of X-rays emitted which is given by the following expression,
√υ = a(Z – b)
where, υ is the frequency of the X-rays emitted by the element with atomic number ‘Z’; ‘a’ and ‘b’ are constants and have the same values for all the elements. The plot of √υ against Z gives a straight line. Using this relationship, we can determine the atomic number of an unknown element from the frequency of X-rays emitted.

Based on his work, the modern periodic law was developed which states that, “the physical and chemical properties of the elements are periodic functions of their atomic numbers”. Based on this law, the elements were arranged in order of their increasing atomic numbers. This mode of arrangement reveals an important truth that the elements with similar properties recur after regular intervals. The repetition of physical and chemical properties at regular intervals is called periodicity.

Question 2.
The first ionization enthalpy of magnesium is higher than that of sodium. On the other hand, the second ionization enthalpy of sodium is very much higher than that of magnesium. Explain.
Answer:
The 1st ionization enthalpy of magnesium is higher than that of Na+ due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of the first electron, N& formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand. Mg+ formed after losing first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 3.
By using Pauling’s method calculate the ionic radii of Na+ and F ions in the potassium chloride crystal. Given that dNa+ – F = 231 pm.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 4
Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 5

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 4.
Discuss the variation of electronic configuration along the periods.
Answer:
Each period starts with the element having general outer electronic configuration, ns1 and ends with ns2, np6 where n is the period number. The first period starts with the filling of valence electrons in is orbital, which can accommodate only two electrons. Hence, the first period has two elements, namely hydrogen and helium.

The second period starts with the filling of valence electrons in 2s orbital followed by three 2p orbitals with eight elements from lithium to neon. The third period starts with filling of valence electrons in the 3s orbital followed by 3p orbitals. The fourth period starts with filling of valence electrons from 4s orbital followed by 3d and 4p orbitals in accordance with Aufbau principle. Similarly, we can explain the electronic configuration of elements in the subsequent periods.

In the fourth period, the filling of 3d orbitals starts with scandium and ends with zinc. These 10 elements are called first transition series. Similarly, 4d, 5d and 6d orbitals are filled in successive periods and the corresponding series of elements are called second third and fourth transition series respectively.

In the sixth period the filling of valence electrons starts with 6s orbital followed by 4f, 5d and 6p orbitals. The filling up of 4f orbitals begins with Cerium (Z = 58) and ends at Lutetium (Z = 71). These 14 elements constitute the first inner-transition series called Lanthanides. Similarly, in the seventh period 5f orbitals are filled, and it’s -14 elements constitute the second inner-transition series called Actinides. These two series are placed separately at the bottom of the modem periodic table.

Question 5.
Describe the nomenclature of elements with Atomic Number greater than 100.
Answer:
Usually, when a new element is discovered, the discoverer suggests a name following IUPAC guidelines which will be approved after a public opinion. In the meantime, the new element will be called by a temporary name coined using the following IUPAC rules until the IUPAC recognizes the new name.
1. The name was derived directly from the atomic number of the new element using the following numerical
roots.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 6

2. The numerical roots corresponding to the atomic number are put together and ‘ium’ is added as suffix.
3. The final ‘n’ of ‘enn is omitted when it is written before ‘nil’ similarly the final ‘i’ of ‘bi’ and ‘tri’ is omitted when it written before ‘ium’.
4. The symbol of the new element is derived from the first letter of the numerical roots.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 7

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 6.
Explain the merits of Moseley’s long form of the periodic table.
Answer:
Merits of Moseley’s long form of the periodic table:

  •  As this classification is based on atomic number, it relates the position of an element to its electronic configuration.
  • The elements having similar electronic configuration fall in a group. They also have similar physical and chemical properties.
  • The completion of each periõd is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached.
  • The position of zero group is also justified in the table as group 18.
  • The table completely separates metals and non-metals.
  • The table separates two subgroups. lanthanides and actinides, dissimilar elements do not fall together.
  • The greatest advantage of this periodic table is that this can be divided into four blocks namely s, p. d, and f-block elements.
  • This arrangement of elements is easier to remember, understand and reproduce.

Question 7.
Explain the variation of the atomic radius in periods and groups.
Answer:
Variation in a period:
Atomic radius tends to decrease in a period. As we move from left to right along a period, the valence electrons are added to the same shell. The simultaneous addition of protons to the nucleus increases the nuclear charge, as well as the electrostatic attractive force between the valence electrons and the nucleus. Therefore atomic radius decreases along a period.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 8

Variation in a group:
In the periodic table, the atomic radius of elements increases down the group. As we move down a group, new shells are opened to accommodate the newly added valence electrons. As a result, the distance between the centre of the nucleus and the outermost shell containing the valence electron increases. Hence, the atomic radius increases.

Question 8.
What is an Effective nuclear charge? How is it calculated using slater’s rule?
Answer:
In addition to the electrostatic forces of attraction j between the nucleus and the electrons, there | exists repulsive forces among the electrons. The repulsive force between the inner shell electrons j and the valence electrons leads to a decrease in the j electrostatic attractive forces acting on the valence 1 electrons by the nucleus. Thus, the inner shell | electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective 1 nuclear charge. It is approximated by the below-mentioned equation.
Zeff = Z – S
Where Z is the atomic number and ‘S’ is the screening constant which can be calculated using Slater’s rules as described below.
Step 1:
Write the electronic configuration of the atom and rearrange it by grouping ns and np orbitals together and others separately in the following form.
(1s)(2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) ……

Step 2:
Identify the group in which the electron of interest is present. The electron present right to this ! group does not contribute to the shielding effect. Each of the electrons within the identified group (denoted by ‘n’) shields to, an extent of 0.35 unit of nuclear charge. However, it is 0.30 unit for 1s electron.

Step 3:
Shielding of inner-shell electrons. If the electron of interest belongs to either s or p orbital,
(i) each electron within the (n – 1) group shields to an extent of 0.85 unit of nuclear charge, and
(ii) each electron within the (n – 2) group (or) even lesser group (n – 3), (n – 4) etc…
completely shields i.e. to an extent of 1.00 unit of nuclear charge.

If the electron of interest belongs to d or f orbital, then each of electron left of the group of electron of interest shields to an extent of 1.00 unit of nuclear charge.

Step 4:
Summation of the shielding effect of all the electrons gives the shielding constant ‘S’.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 9.
Explain the periodic variation of ionization energy in a period.
Answer:
The ionization energy usually increases along a period with few exceptions. When we move from left to right along a period, the valence electrons are added to the same shell, at the same time protons are added to the nucleus, This successive increase of nuclear charge increases the electrostatic attractive force on the valence electron and more energy is required to remove the valence electron resulting in high ionization energy. Consider the variation in ionization energy of second period elements. The plot of atomic number vs ionisation energy is given below.

In the graph, there are two deviations in the trends of ionisation energy. It is expected that boron has higher ionisation energy than beryllium since it has higher nuclear charge. However, the actual ionisation energies of beryllium and boron are 899 and 800kJ mol-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s2, 2p1).

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements 9

The electronic configuration of beryllium (Z = 4) in its ground state is 1s2, 2s2 and that of boron (Z = 5) 1s2, 2s2, 2p1. Similarly, nitrogen with 1s2, 2s2, 2p1 electronic configuration has higher ionisation energy (l402 kJmol-1) than oxygen (1314 kJmol-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

Question 10.
Explain the periodic variation of electron affinity in a group.
Answer:
Electron affinity is defined as the amount of energy released (required in the case of noble gases) when an electron is added to the valence shell of an isolated ^neutral gaseous atom in its ground state to form its anion. It is expressed in kJmol-1
A + 1e → A + EA

Variation of Electron Affinity in a period:
As we move from alkali metals to halogens in a period, generally electron affinity increases, i.e., the amount of energy released will be more. This is due to an increase in the nuclear charge and a decrease in size of the atoms. However, in case of elements such as beryllium (1s2, 2s2), nitrogen (1s2, 2s2, 2p3) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have stable ns2, np6 configurations, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns2, np5 readily accept an electron to get the stable noble gas electronic configuration (ns2, np6), and therefore in each period the halogen has high electron affinity, (high negative values)

Question 11.
Define electron affinity. How does it vary along with the group?
Answer:
Electron affinity is defined as the amount of energy released (required in the case of noble gases) when an electron is added to the valence shell of an isolated neutral gaseous atom in its ground state to form its anion. It is expressed in kJmoF1
A + 1e → A + EA

Variation of Electron affinity in a group:
As we move down a group, generally the electron affinity decreases. It is due to an increase in atomic size and the shielding effect of inner-shell electrons. However, oxygen and fluorine have lower affinity than sulphur and chlorine respectively. The sizes of oxygen and fluorine atoms are comparatively, small and they have high electron density.

Moreover, the extra electron, added to oxygen and fluorine has to be accommodated in the 2p orbital which is relatively compact compared to the 3p orbital of sulphur and chlorine so, oxygen and fluorine have lower electron affinity than their respective group elements sulphur and chlorine.

Samacheer Kalvi 11th Chemistry Guide Chapter 3 Periodic Classification of Elements

Question 12.
Explain the salient features of groups.
Answer:
1. Number of electrons in the outermost shell:
The number of electrons present in the outermost shells does not change on moving down in a group, i.e remains the same. Hence, the valency also remains the same within a group.

2. Number of shells:
In going down a group the number of shells increases by one at each step and ultimately becomes equal to the period number to which the element belongs.

3. Valency:
The valencies of all the elements of the same group are the same. The valency of an element with respect to oxygen is same in a group.

4. Metallic character:
The metallic character of the elements increases in moving from top to bottom in a group.

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 2 Plant Kingdom
Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

11th Bio Botany Guide Plant Kingdom Text Book Back Questions and Answers

Part-I

Choose the Right Answer:

Question 1.
Which of the plant group has gametophyte as a dominant phase ?
a. Pteridophytes
b. Bryophytes
c. Gymnosperm
d. Angiosperm
Answer:
b. Bryophytes

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 2.
Which of the following represent gametophytic generation in Pteridophytes?
a. Prothallus
b. Thallus
c. Cone
d. Rhizophore
Answer:
a. Prothallus

Question 3.
The haploid number of chromosome for an Angiosperm is 14, the number of chromosome in its endosperm would be
a. 7
b. 14
c. 42
d. 28
Answer:
c. 42

Question 4.
Endosperm is gymnosperm is formed
a. At the time of fertilization
b. Before fertilization
c. After fertilization
d. Along with the development of embryo
Answer:
b. Before fertilization

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 5.
Differentiate Haplontic and Diplontic life cycle.
Answer:

Haplontic life cycle Diplontic life cycle .
Gametophyte phase Sporophytic phase Dominant photosynthetic independent Recessive represented by the zygote.Zygote undergoes meiosis to restore haploidy e.g. Volvox.Spirogyra Recessive represented by single to few celled stage Zygote develop in to dominant sporophyte e.g.Fucus, Gymnosperm, Angiosperm.

Question 6.
What is plectostele – Give example.
Answer:
Plectostele: Xylem plates alternates with phloem plates. Example: Lycopodium clavatum.

Question 7.
What do we infer from the term Pucnoxylic?
Answer:

  • Secondary growth is also traced in gymnosperms, E.g. Cycas and Pinus.
  • The wood may be compact with narrow medullary ray this condition known as Pycnoxlic seen in Pinus.
  • It is opposite to Manoxylic condition which is seen in Cycas.

Question 8.
Mention two characters shared by Gymnosperms and Angiosperms.
Answer:

Gymnosperms

Angiosperms

1 Seed bearing plants Seed bearing plants
2 Plant body is a sporophyte and it is dominant phase. Plant body is a sporophyte and it is also the dominant phase
3 Stem show secondary growth. Stem show secondary growth only in dicots not in monocots.
4 Alternation of generation is present. Alternation of generation is present.

Question 9.
Do you think shape of chloroplast is unique for algae. Justify your answer.
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup-shaped (chlamydomonas). Discoid ((Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 10.
Do you agree with the statement that Bryophytes need water for fertilization? Justify your answer.
Answer:
Being amphibians of plant kingdom they are simplest land inhabiting cryptogams and restricted to moist shady habitats. They need water for the male gametes to reach Archegonia to effect fertilization. So water is needed for completing their fertilization and their life cycle.

Part – II

11th Bio Botany Guide Plant Kingdom Additional Important Questions and Answers

I. Choose the correct option.

Question 1.
Gametophytic phase is …………… .
(a) triploid
(b) tetraploid
(c) haploid
(d) diploid
Answer:
(c) haploid

Question 2.
The numbers of known species of Angiosperms in the world is
a. 268600
b. 286600
c. 224400
d. 274832
Answer:
a. 268600

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
Which algae leads an endozoic life in Hydra?
(a) Chlorella
(b) Gracilaria
(c) Ulothrix
(d) Chlamydomonas
Answer:
(a) Chlorella

Question 4.
The protein bodies found in chromatophores & assist in the synthesis and storage of starch is
a. Leucoplasts
b. Floridean starch
c. Pyrenoids
d. Mannitol
Answer:
c. Pyrenoids

Question 5.
Postelia palmaeformis is commonly known as
a. Sea kelp
b. Sea shell
c. Sea palm
d. Sea worth
Answer:
c. Sea palm

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 6.
In Chara, thallus is encrusted with …………… .
(a) calcium carbonate
(b) hydrogen sulphate
(c) silica
(d) ammonium carbonate
Answer:
(a) calcium carbonate

Question 7.
Gemmae formation is not traced in which three of the given four options
a. Marchanlia
b. Riella
c. Ricciocarpus
d. Anthoceros
(i) ab & c
(ii) be & d
(iii) ab & d
(iv) ac & d
Answer:
(ii) be & d

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 8.
Find out the aquatic bryophytes of the following.
a. Riella
b. Ricciocarpus
c. Riccia
d. Bryopteris
(i) a&c
(ii) b&c
(iii) c&d
(iv) a&b
Answer:
(iv) a&b

Question 9.
…………… are thin-walled non – motile spores.
(a) Zoospores
(b) Akinetes
(c) Aplanospores
(d) Genunae
Answer:
(c) Aplanospores

Question 10.
which one of the following is a terrestrial chiorophycea
a. Chara
b. Zygnema
c. Trentipohlia
d. Ulva
Answer:
c. Trentipohlia

Question 11.
Thick walled spores meant for perrennation are known as
a. Aplanospores
b. Akinetes
c. Endospores
d. Zoospores
Answer:
b. Akinetes

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 12.
The photosynthetic part of the Phaeophyceae thallus Is called as
(a) holdfast
(b) stipes
(c) lamina
(d) fronds
Answer:
(d) fronds

Question 13.
The female sex organ of red algae is known as
a. Archegonium
b. Spermatogonium
c. Carpogonium
d. Oogonium
Answer:
c. Carpogonium

Question 14.
Endosperm is triploid and haploid in
a. Pteridophyta & Gymnosperm
b. Angiosperm & Gymnosperm
c. Gymnosperm & monocot
d. Gymnosperm & dicot
Answer:
b. Angiosperm & Gymnosperm

Question 15.
Gelidium belongs to …………… members.
(a) Rhodophyccae
(b) Phaeophyceae
(c) Cyanophyccae
(d) Dinophyceae
Answer:
(a) Rhodophyceae

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 16.
……………….. is known as leather leaf fern
a. Marsilea
b. Azolla
c. Selaginella
d. Rumohra adiantiformis
Answer:
d. Rumohra adiantiformis

Question 17.
Stele includes
a. Xylem, phloem & cambium
b. Xylem, phloem & medulla
c. Xylem, phloem & pericycle
d. Xylem, phloem, pericycle & medulla
Answer:
d. Xylem, phloem, pericycle & medulla

Question 18.
Marchantia vegetatively propagates by …………… .
(a) tubers
(b) gemmae
(c) buds
(d) brood bodies
Answer:
(b) gemmae

Question 19.
The organ in bryophytes that help to attach the thallus to the substratum is
a. Hold fast
b. Rhizoids
c. Rhizopore
d.Roots
Answer:
a. Holdfast

Question 20.
Coralloid roots of cycas have a symbiotic association with …………….
(a) Blue-green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(a) Blue-green algae

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

II. Match the following & find the correct answer.

Question 1.
(i) Spores look similar to parental cell – Zoospore (A)
(ii) Thick walled aplanospores – Autospore (B)
(iii) Thin walled non-motile spores – Hypnospore (C)
(iv) Thin walled motile spores – Aplanospore (D)
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 1
Answer:
(b) B-C-D-A

Question 2.
(i) Helianthusannum – Amphiboloicstele (A)
(ii) Lycopodium serratum – Eustele(B)
(iii) Zeamays – Actinostele (C)
(iv) Adiantumpedatum – Atactostele (D)
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 2
Answer:
(a) B-C-D-A

Question 3.
(i) Fossil bryophyte – Lepidodendron, Williamson
(ii) Fossil Algae – Calamites Baragwanthia
(iii) Fossil pteridophyte – Naiadita, Hepaticites
(iv) Fossil Gymnosperm – Palaeoporella,Dimorphosiphon
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 3
Answer:
(d) C-D-B-A

Question 4.
(i) Abies balsamea – Drug for cancer treatment (A)
(ii)Taxus brevifolia – Wood for making door, boat & railway sleepers (B)
(iii) Cedrus deodara – Treatment for asthama & bronchitis(C)
(iv) Ephedra gerardiana – Slide mounting medium(D)
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 4
Answer:
(c) D-A-B-C

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 5.
Find the chromosome number of the following by choosing the correct option.
(i) Embryo ofblyophyta
(ii) Embryo ofAngiosperm
(iii) Endosperm ofAngiosperm
(iv) Sporophyte ofpteridophyta
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 5
Answer:
(a) (I) n (II) 2n (III) 3n (IV) 2n

III. Choose the wrong statement.

Question 1.
The following statement is not applicable to which one of the following options
The sporophyte is dominant, photosynthetic, and independent. The gametophytic phase is represented by a single to few celled gametophyte
a. Fucus
b. Mango
c. Pinus
d. Marchantia
Answer:
d. Marchantia

Question 2.
One of the following is not a Marine Algae.
a. Gracilaria
b. Sargassum
c. Oedogonium
d. Cladophora
Answer:
c. Oedogonium

Question 3.
Which one of the following is not a Vascular cryptogam
a. Lycopodium
b. Anthoceros
c. Equisetum
d. Selaginella
Answer:
b. Anthoceros

Question 4.
Which one among the given four doesn’t belong to Chlorophyceae
a. Chiorella
b. Volvox
c. Chara
d. Sargassum
Answer:
d. Sargassum

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 5.
Pollination is not entomophilous in
a. Hibiscus
b. Mangifera
c. Chrysanthemum
d. Cycas
Answer :
d. Cycas

Question 6.
The following one is not a monoecious plant
a. Pinus
b. Cycas
c. Alnus
d. Ginkgo
Answer:
b. Cycas

Question 7.
Which one of the following is not the correct statement regarding Algae?
a. The study of Algae is known as Phycology
b. A wide range of thallus organization is found in Algae
c. Algae are eukaryotic except Blue Green Algae
d. They are the simplest plant group with root stem and leaves
Answer:
d. They are the simplest plant group with root stem and leaves

IV. Find out the true or false statements from the following and on that basis find the correct answer.

Question 1.
(i) Chara thallus is encrusted with calcium carbonate
(ii) Siliceous wall occurs in the cell wall of Diatom
(iii) Soil inhabiting algae – Fritshchiella
(iv) Cladophora crispate is growing now
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 6
Answer:
c. (I) True (II) True (III) False (IV) True

Question 2.
(i)  Prothallus develop into a sporophyte
(ii)  Algae growing on snow is known as cryptophytes
(iii)  The common name of Postelia Palmaeformis is known as sea palm.
(iv)  Endosperm is triploid in pinus.
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 7
Answer:
a. (I) False (II) True (III) True (IV) False

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
(i)  Apogamy and Apospory is common in pteridophytes
(ii) Spore bearing leaves in pteridophytes are known as a sorus
(iii) Branches of limited growth and branches of unlimited growth are seen in gymnosperm
(iv) Cambium occur in gymnosperm as in dicots
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 8
Answer:
d. (I) True (II) False (III) False (IV) True

Question 4.
(i) Fungi play important role in soil conservation
(ii) Vascular cryptogams were predominant in the paleozoic era
(iii) Gymnosperms were dominant in the early cretaceous period.
(iv) Angiosperms appeared during the Jurassic period
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 9
Answer:
b (I) False (II) True (III) False (IV) False

Question 5.
(i) Polyembryony is traced in Pteridophyta
(ii) Vessels are present in Gnetum and ephedra
(iii) Heterosporus condition is seen in Lycopodium
(iv) Corolloid root occur in Cycas
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 10
Answer:
a. (I) False (II) True (III) False (IV) True

Question 6.
Which one of the following is the correct statement regarding Phaeophyta
a.  It is commonly known as Red Algae
b. The plant body has fronds, stipe & hold fast
c. The reserve material is Floridian starch
d. Sexual reproduction is isogamous
Answer:
b. The plant body has fronds, stipe & holdfast

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 7.
Choose the right statement regarding leaves of Pteridiumsp.
a. It is used as food
b. Green dye is derived from it.
c. It is used as bio-fertilizer
d. It is an ornamental foliage plant
Answer:
b.Green dye is derived from it

Question 8.
Which one of the following is a correct statement regarding Bryophyta
a. Mostly terrestrial plants so water is not essential for reproduction
b. The gametophyte is dominant but the sporophyte is independent
c. They have well-developed xylem and phloem tissues
d. They are the simplest land inhabiting cryptogams lacking vascular tissues.
Answer:
d.They are the simplest land inhabiting cryptogams lacking vascular tissues

Question 9.
Choose the correct statement regarding Gymnosperm
a. The spores are generally homosporous
b. The leaves are dimorphic, foliage and scale leaves are present
c. The stem is aerial erect and unbranched in conifers
d. Xylem mostly consists of only vessels.
Answer:
b. The leaves are dimorphic, foliage and scale leaves are present.

Question 10.
Choose the correct statement regarding the common characters of Gymnosperm and Angiosperm only
a. Pollen tube help in the transfer of male nucleus & fertilization is Siphonogamous
b. Heterospory is of common occurrence
c. Vessels are the chief water-conducting elements
d. Pollination is by Anemophilous method only
Answer:
a. Pollen tubes help in the transfer of the male nucleus & fertilization is Siphonogamous.

Question 11.
Look at the picture and find out the correct answer.
a. Gemmae of Marchantia
b. Thallus of Riccia
c. The gametophyte of Anthoceos
d. The tubers of Anthoceos
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 11
Answer:
d.The tubers of Anthoceos

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 12.
Look at the picture and find out the correct answer
a. The Asexual reproduction in Ultrix by the formation of zoospores
b. Akinetes formation in Pithophora
c. Scalariform conjugation in Zygnema
d. Zygospore formation in spirogyra
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 12
Answer:
c. Scalariform conjugation in zygoma.

V.

Question 1.
Which one of the following is a wrong pair?
a. National wood fossil park – Thuruvakkarai
b. Shiwalik fossil park – Arunachala Pradesh
c. Mandal fossil park – Madhya Pradesh
d. Raj mahal hill – Jharkhand
Answer:
b. Shiwalik fossil park – Arunachala Pradesh

Question 2.
Which one of the following is a wrong pair?
a. Halophytic Algae – Dunaliella salina
b. Epiphytic Algae – Rhodymenia
c. Endophytic algae – Cladophora crisp ala
d. Endozoic Algae – Chlorella
Answer:
c. Endophytic Algae – Cladophora crisp ala

Question 3.
Which one of the following is a wrong pair?
a. Father of Indian Bryology- Prof. Shiv. Ram Kashyap
b. Father of Indian Phycology – F.E. Fritsch
c. The classification of gymnosperm – Sporn
d. The father of Indian Paleobotany – Prof. Birbal Sahni
Answer:
b.Father of Indian Phycology F.E.Fritsch

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

VI. Read the following assertion & reason. Find the correct answer.

Question 1.
Assertion: Chlorophyceae is known as green algae.
Reason: These plants have chlorophyll & chlorophyll as their major photosynthetic pigments.
(a) Assertion and Reason are correct. The reason is explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and Reason are correct. The reason is explaining Assertion.

Question 2.
Assertion: In Bryophytes haploid gametophyte, alternate with diploid sporophyte phase.
Reason: Bryophytes lack vascular tissue and hence called non-vascular cryptogams.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
Assertion: Gnetum has flowers and it also has vessels as conducting elements like angiosperm.
Reason: Gnetum is a primitive gymnosperm.
(a) Assertion and Reason are correct, Reason explaining assertion.
(b) Assertion and Reason are correct, but Reason not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(c) Assertion is true, but Reason is wrong.

Question 4.
Assertion: The endosperm is haploid and develops before fertilization in a gymnosperm.
Reason: The endosperm is triploid and develops after fertilization in angiosperm.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.

Question 5.
Assertion: The embryogeny is endoscopic in Bryophytes.
Reason: The first division of the zygote is transverse & the apex of the embryo develops from the outer cell.
(a) Assertion and Reason are correct, Reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and reason are correct, Reason explaining Assertion.

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

VII. In the following diagram what are the parts (A) (B) (C) (D) representing?

Question 1.
(a) (A) Epidermis (B) Xylem
(C) Phloem (D) Cambium
(b) (A) Epidermis (B) Phloem
(C) Cambium (D) Xylem
(c) (A) Epidermis (B) Cambium
(C) Xylem (D) Phloem
(d) (A) Phloem (B) Xylem
(C) Cambium (D) Epidermis
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 13
Ans: (b) (A) Epidermis-(B) Phloem-(C) Cambium-(D) Xylem

Additional Questions – 2 Marks

Question 1.
Define alternation of generation.
Answer:
Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 2.
Chlorella – structure label the parts.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 14
Label (1) nucleus
(2) cup-shaped chloroplast

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
Name any two freshwater algae.
Answer:
Two freshwater algae:

  1. Oedogonium and
  2. Ulothrix

Question 4.
What is the use of Diatomaceous earth?
Answer:
Diatomaceous earth is got from the siliceous frustules (Diatom). It belongs to bacillario phyta. The transparent cell walls of a diatom are made up of hydrated silica. Generally known as Diatomaceous earth.
Use:

  • It is used in water filters as an insultation material.
  • Reinforcing agent in concrete and rubber.

Question 5.
What is the use of chlorella in sewage treatment?
Answer:

  • Chlorella, Scenedesmus, Chlamydomonas are used in the sewage treatment plants. For their photosynthetic activity, they utilize the carbon dioxide from sewage and release oxygen.
  • The aerobic bacteria, by utilising this oxygen degrade & decompose organic materials in the sewage. Thus play a vital role in sewage treatment plants. (STPs)

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 6.
Identify and label the given diagram.
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 15
Answer:
The given figure represents Marchantia

  1. Apical notch
  2. Sex organs
  3. Gametophyte of thallus
  4. Rhizoids
  5. Gemma

Question 7.
Differentiate between Sorus, Sporangia, Sporophyll.
Answer:

Sporangia

Sorus

Sporophyll

Spore producing organ Group of Sporangia known as Sorus The leaf-bearing Sorus in Neprolepis is known as Sporophyll

Question 8.
Bryophytes are amphibians of plant kingdom – Justify.
Answer:
Bryophytes are called ‘amphibians of plant kingdom’ because they need water for completing their life cycle.

Question 9.
Label the given diagram.
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 16
Answer:
The given diagram is the sporophyte of Funaria parts

  1. Calyptra
  2. Capsule
  3. Leaves
  4. Rhizoids

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 10.
What are the aspects that helped Pteridophtes to evolve into terrestrial habitats?
Answer:

  • Pteridophytes are the first to acquire vascular tissues, xylem and phloem, and a well-developed root system. So known as vascular cryptogam.
  • This aspect had helped pteridophytes to evolve into terrestrial habitats.

Question 11.
Assign few plants in the group Pteridophytes.
Answer:
Type Botanical name
Club moss – Lycopodium
Horsetails – Equisetum
Quill worts – Isoetes
Water ferns – Salviniales
Tree ferns – Dryopteris

Question 12.
Write down Reimer’s classification of Pteridophytes.
Answer:
5 subdivisions – 19 orders – 48 families

  1. Psilophytopsida
  2. Psilotopsida
  3. Lycopsida
  4. Sphenopsida
  5. Pteropsida

Question 13.
What Is amber? Which group of plants produce amber?
Answer:
Amber is a plant secretion that is an efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 14.
What is meant by Siphonostele?
Answer:
In siphonostele, the xylem is surrounded by phloem with pith in the centre- eg. Marsilea. It includes ectophloic, siphonoslele, Amphiphloic-siphonoslele, soleonslete, eustele, Atactostele, and polycyclic stele thus altogether 6 types.

Question 15.
What is Amber? Where have you noticed it?
Answer:

  • In Spielberg’s Jurassic park-movie (A mosquito embedded in a transparent substance called amber is
    mentioned)
  • Amber is thus a plant secretion, used as an efficient preservative, that doesn’t get degraded.
  • Pinites succinifera a Gymnosperm plant produces Amber.

Question 16.
What is the word Jurassic – denote?
Answer:

  • ‘Jurassic’ is a specific period of the dinosaurs. It comes under the Mesozoic era.
  • Gymnosperms were also dominant in that period.

3 Marks

Question 1.
Give the widely accepted outline classification for plants.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 17

Question 2.
Give the total number of plant groups in the world and India.
Answer:

Total Number of plant groups in the world and India
N Number of known species
Plant group World India
Algae 40000 7357
Bryophytes 16236 2748
Pteridophytes 12000 1289
Gymnosperms 1012 79
Angiosperms 268600 18386

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
Name the 3 types of life cycles seen in plants?
Answer:
The 3 types of life cycles seen in the plant:

  1. Haplontic life cycle
  2. Diplontic life cycle
  3. Haplodiplontic life cycle

Question 4.
More than half of the total productivity of the world is done by Marine Algae -Justify.
Answer:

  • Yes two-third of earths surface is covered by oceans and seas.
  • Of this the photosynthetic plants called algae are the major primary producers Nearly 1/2 of total productivity of the world is done by marine Algae. All other marine organisms depend upon them for the very existence.

Question 5.
Classify Algae according to F.E.Fritsch.
Answer:
F.E. Fritsch in his the structure and reproduction of Algae (1935)-classified Algae into 11 classes.

  1. Chlorophyceae
  2. Xanthophyceae
  3. Chrysophyceae
  4.  Bacillariophyceae
  5. Cryptophyceae
  6. Dinophyceae
  7. Chloromondineae
  8. Euglenophyceae
  9. Phaeophyceae
  10. Rhodophyceae
  11. Cyanophyceae

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 6.
Write about Reproduction in Chlorophyceae.
Answer:

1. Vegetative Reproduction Fragmentation-E.g. spirogyra
2. Asexual Reproduction Zoospores, Aplanospores, Akinetes
3. Sexual Reproduction Isogamous, Anisogamous, Oogamous

Question 7.
Write down the economic importance of Bryophyte?
Answer:

Name of the Bryophyte

Use

Sphagnum A large amount of dead thallus compressed & hardened to form-peat
Peat Northern Europe (Netherlands) peat is a commercial fuel.
Sphagnumt peat Nitrates, brown dye, tanning materials are derived from the peat used in horticulture as packing material.
Marechantia polymorpha Cure pulmonary tuberculosis
Sphagnum, Bryum polytrichum Used as food
All bryophytes From major role in soil formation through succession & help in soil conservation

Question 8.
List down the economic importance of Pteridophytes.
Answer:

Pteridophyte

Uses

Rumohra adiantiformis(leather leaf fem) Cut flower arrangements
Marsilea Food
Azolla Biofertilizer
Dryopteris filix-mas Treatment for tapeworm.
Pteris vittata Removal of heavy metals from soils-(Bioremediation)
Pteridium sp. Leaves yield a green dye
Equisetum sp. Stems for scouring
Psilotum, Lycopodium, Selaginella, Angiopteris, Marattia Ornamental plants

Question 9.
Name the three classes of Bryophytes, according to Proskauer.
Answer:
Three Classes of Bryophytes, According to Proskauer:

  1. Hepaticopsida
  2. Anthocerotopsida and
  3. Bryopsida.

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 10.
List down the salient features of Angiosperm.
Answer:

  • Vascular system – Xylem and Phloem well developed
  • Flowers are produced (instead of cones)
  • Ovules (embryosac and seeds) – remain enclosed in ovary/fruit Pollen tube -Help in fertilization water no necessary
  • Double fertilization & triple fusion present is one of the unique features.

Question 11.
Differentiate between Dicotyledons & Monocotyledons.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 18

5 Marks

Question 1.
Classify Algae on the basis of their habitats.
Answer:

S.no

Habitat

Name of Algae

1. Aquatic / Marine Gracilaria & Sargassum
2. Freshwater Oedogonium & ulothrix
3. Soil Fritschella & Vaucheria
4. Endozoic life in (Hydra & sponges) Chlorella
5. Epizoic (on the shells of mollusks) Cladophora crispata
6. Salt pans – (Halophytic algae) Dunaliella salina
7. Growing in snow converted mountains – (Cryophytic) Chlamydomonas nivulis (give red colour to snow / red Snow)
8. Epiphytic on the surface of aquatic plants Coleochaete, Rhodymenia

Question 2.
Give an account of General Characteristics of Algae?
Answer:
Criteria I

Thallus Organization

Unicellular motile – Chlamydomonas

Unicellular non-motile – Chlorella
Colonial motile – Volvox
Colonial non-motile – Hydrodictyon
Siphonous – Vaucheria
Filamentous (unbranched)- Spirogyra
Filamentous (branched) – Cladophora
Discoid – Coleochaete
Heterotrichous – Frlschiella Sc macrocyslis
Foliaceous – Ulva
Giant kelps – Laminaria, Macrocystis

Criteria II

Eukaryotic Cell wall

Almost all except Blue-green alga (prokaryotic)

Cellulose & Hemicelluloses Most algae
Siliceous walls Diatoms
Cell wall-encrusted with Calcium Carbonate Chara thallus
Algin, Polysulphate (Agar Agar)
Ester of Polysaccharide Carrageenin Gracillaria Chondrus crispus Gellidella

Criteria III

Pigmentation Reserve food material & flagellation A lot of difference exists
Reproduction Vegetative Asexual & sexual
Vegetative reproduction Fission, fragmentation budding, bulbils, akinetes, tubers etc
Asexual reproduction Zoo spores- ulolhrix Aplano spores- Vaucheria Auto spores- Chlorella Hypno spores – Chlamydomonas nivalis Tetra spores etc polysiphonia
Sexual reproduction Isogamy – ulothrix Anisogamy- Pandorina Oogamy- Sargassum

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 3.
Give an account of Phaeophyceae.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 19
Question 4.
Give an account of Rhodophyceae (red algae) criteria.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 20
Question 5.
Economic importance of Algae.
Answer:

Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 22

Question 6.
General characteristic features of Bryophytes.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 22
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 23

Samacheer Kalvi 11th Bio Botany Guide Chapter 2 Plant Kingdom

Question 7.
Write down differences between Gymnosperm & Angiosperm.
Answer:

Gymnosperms

Angiosperms

Vessels are absent [except Gnetales] Vessels are present
Phloem lacks companion cells Companion cells are present
Ovules are naked Ovules are enclosed within the ovary
Wind pollination only Insects, wind, water, animals etc., act as pollinating agents
Double fertilization is absent Double fertilization is present
The endosperm is haploid (Pre-Fertilisation) The endosperm is triploid (Post-Fertilisation)
Fruit formation is absent Fruit formation is present
Flowers absent Flowers present

Question 8.
Write down the Economic Importance of Gymnosperm.
Answer:

Plants

Products

Uses

1. Cycas circinalis, Cycas revolute Sago Starch used as a food
2. Pinus gerardiana Roasted seed Used as food
3. Abies balsamea Resin (Canada balsam) Used as a mounting medium in permanent slide preparation
4. Pinus insularis, Pinus roxburghii Rosin and Tupertine Paper sizing and varnishes
5. Araucaria (monkey’s puzzle), Picea and Phyllocladus Tannins Bark yield tannins and is used in leather industries
6. Taxus brevifolia Taxol Drug used for cancer treatment
7. Ephedra gerardiana Ephedrine For the treatment of asthma, bronchititis
8. Pinus roxburghii, Oleo resin Used to make soap, varnishes, and printing ink
9. Pinus roxburghii, Picea smithiana Wood pulp Used to make papers
10. Cedrus deodara Wood Used to make doors, boats, and railway sleepers
11. Cedrus atlantica Oil Used in perfumery
12. Thuja, Cupressus, Araucaria, & Cryptomeria Decorative Ornamental plants

Question 9.
Give an account of Fossil plants.
Answer:
Samacheer Kalvi 11th Bio Botany Chapter 2 Plant Kingdom 24