TN Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2

Question 1.
Define Truth Table.
Answer:
A truth table represents all the possible values of logical variable or statements along with all the possible results of given combination of truth values.

Question 2.
What are logical operators?
Answer:
The basic logical operations are AND, OR and NOT, which are symbolically represented by dot (.), plus (+), and by over bar / single apostrophe respectively. These symbols are also called as “Logical Operators”.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 3.
Define AND operator?
Answer:
The AND operator is defined in Boolean algebra by the use of the dot (.) operator. It is similar to multiplication in ordinary algebra. The AND operator combines two or more input variables. So that the output is true only if all the inputs are true. The truth table for a 2-input AND operator is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 5

The above 2-input AND operation is expressed as: Y = A . B

Question 4.
What is OR operator?
Answer:
The plus sign is used to indicate the OR operator. The OR operator combines two or more input variables. So that the output is true if at least one input is true. The truth table for a 2-input OR operator is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 6

The above 2-input OR operation is expressed as: Y = A + B

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 5.
What is NOT operator?
Answer:
The NOT operator has one input and one output. The input is either true or false, and the output is always the opposite, i.e., the NOT operator inverts the input. The truth table for a NOT operator where A is the input variable and Y is the output shown below:

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 7

The NOT operator is represented algebraically by the Boolean expression: Y = \(\overline{\mathrm{A}}\).

Question 6.
What is NAND operator?
Answer:
The NAND is the combination of NOT and AND. The NAND is generated by inverting the output of an AND operator. The algebraic expression of the NAND function is:
Y = \(\overline{\mathrm{A} \cdot \mathrm{B}}\)
The NAND function truth table is shown below:

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 8

A NAND B = NOT (A AND B)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 7.
What is called basic logic gates?
Answer:
A gate is a basic electronic circuit which operates on one or more signals to produce an output signal. There are three fundamental gates namely AND, OR and NOT.

Question 8.
Write the Distributive law.
Answer:
A . (B + C) = A . B + A . C
A + (B . C) = (A + B) . (A + C)

Question 9.
Complete the following:
(i) A + \(\overline{\mathrm{A}}\) =
(ii) A. \(\overline{\mathrm{A}}\) =
(iii) \(\overline{\overline{\mathrm{A}}}\) =
(iv) A + A =
(v) A. A =
(vi) A + 1 =
Answer:
(i) A+ \(\overline{\mathrm{A}}\) = 1
(ii) A. \(\overline{\mathrm{A}}\) = 0
(iii) \(\overline{\overline{\mathrm{A}}}\) = A
(iv) A + A = A
(v) A . A = A
(vi) A + 1 = 1

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 10.
Draw the logic circuit of Bubbled AND Gate and explain.
Answer:
The Logic Circuit of Bubbled AND Gate

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 9

In the above circuit, invertors on the input lines of the above circuit, invertors on the input lines of the AND gate gives the output as C = (\(\overline{\mathrm{A}}\) . \(\overline{\mathrm{B}}\))This circuit can be redrawn as the bubbles on the inputs, where the bubbles represent inversion.

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 10

It is referred as bubbled AND gate.

If A = 0 and B = 0 C = (\(\overline{0} . \overline{0}\)) = 1 . 1 = 1
If A = 0 and B = 1 C = (\(\overline{0} . \overline{1}\)) = 1 . 0 = 0
If A = 1 and B = 0 C = (\(\overline{1} . \overline{0}\)) = 0 . 1 = 0
If A = 1 and B = 1 C = (\(\overline{1} . \overline{1}\)) = 0 . 0 = 0

Here the truth table is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 11

A bubbled AND gate produces the same output as a NOR gate. So, replace each NOR gate by a bubbled AND gate. The circuits are interchangeable.

Therefore,
\((\overline{\mathrm{A}+\mathrm{B}})=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Which establishes the De Morgan’s second theorem.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 11.
Draw the logic circuit of Bubbled OR Gate and explain.
Answer:
The logic circuit of bubbled OR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 12

The output of this circuit can be written as
C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)
The above circuit can be redrawn as the bubbles on the input, where the bubbles represents the inversion.

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 13

It is referred as bubbled OR gate. The truth table for the bubbled OR is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 14

Comparing the truth tables of the bubbled OR gate with NAND gate, they are identical. So the circuits are interchangeable.

Therefore,

\((\overline{\mathrm{A} \cdot \mathrm{B}})=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

which establishes the De Morgan’s second theorem.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 12.
Draw the symbol of logic agtes and their corresponding Truth tables.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 15

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 13.
Explain XOR gate with truth table.
Answer:
The XOR (exclusive – OR) gate acts in the same way as the logical “either/or.” The output is “true” if either, but not both, of the inputs are “true”. The output is “false” if both inputs are “false” or if both inputs are “true”. The output is 1 if the inputs are different, but 0 if the inputs are the same. The logic circuit of XOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 16

The output of the XOR gate is
The truth table for XOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 17

In boolean algebra, exclusive – OR operator © or “encircled plus”.
Hence, C = A © B
The logical symbol of XOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 18

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 14.
What is Boolean Algebra?
Answer:
It is a mathematical discipline that is used for designing digital circuits in a digital computer. It describes the relation between inputs and outputs of a digital circuit.

Question 15.
Write a short note on NAND Gate.
Answer:
The NAND gate operates an AND gate followed by a NOT gate. It acts in the manner of the logical operation “AND” followed by inversion. The output is “false” if both inputs are “true”, otherwise, the output is “true”. The output of the NAND gate is 0 if and only if both the inputs are 1, otherwise the output is 1.

Question 16.
Draw the truth table for XOR gate.
Answer:
The truth table for XOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 19

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 17.
Write the associative laws?
Answer:
A + (B + C) = (A + B) + C
A. (B . C) = (A . B) . C

Question 18.
What are derived gates?
Answer:
Logic gates like NAND, NOR, XOR and XNOR are derived gates which are derived from the fundamental gates.

Question 19.
Write the truth table of fundamental gates.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 20

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 21

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 20.
Write a short note on XNOR gate.
Answer:
The XNOR (exclusive – NOR) gate is a combination XOR gate followed by an inverter. Its output is “true” if the inputs are the same, and “false” if the inputs are different. In simple words, the output is 1 if the input are the same, otherwise the output is 0. The logic circuit of XNOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 22

The output of the XNOR is NOT of XOR
C = A ⨁ B
= \(\overline{\mathrm{A}}\) . B + A . \(\overline{\mathrm{B}}\)
= A B + \(\overline{\mathrm{A}}\) \(\overline{\mathrm{B}}\)
(Using De Morgan’s Theorem)
In boolean algebra, ⊙ or “included dot” stands for the XNOR.
Therefore, C = A ⊙ B
The logical symbol is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 23

Truth Table for XOR Gate

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 24

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 21.
Reason out why the NAND and NOR are called universal gates?
Answer:
NAND and NOR gates are called Universal gates, because the fundamental logic gates can be realized through them.

Question 22.
Give the truth table of XOR gate.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 25

Question 23.
Write the De Morgan’s law.
Answer:
De Morgan’s law \(\overline{\mathrm{A}+\mathrm{B}}\) = A . \(\overline{\mathrm{B}}\)
(\(\overline{\mathrm{A} \cdot \mathrm{B}}\)) = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 24.
Explain the fundamental gates with expression and truth table.
Answer:
The three fundamental gates namely AND, OR and NOT.
AND Gate:
The AND gate can have two or more input signals and produce an output signal. The output is “true” only when both inputs are “true”, otherwise, the output is “false”. The output will be 1 if and only if both inputs are 1; otherwise the output is 0. The output of the AND gate is represented by a variable say C, where A and B are two and if input boolean variables. In boolean algebra, a variable can take either of the values ‘0’ or ‘1’.

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 26

One way to symbolize the action of an AND gate is by writing the boolean function.
C = A AND B
In boolean algebra the multiplication sign stands for the AND operation. Therefore, the output of the AND gate is
C = A . B or simply C = AB
Read this as “C equals A AND B”. Since there are two input variables here, the truth table has four entries, because there are four possible inputs : 00, 01, 10 and 11.
For instance if both inputs are 0,
C = A .B = 0 . 0 = 0
The truth table for AND Gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 27

OR Gate:
The OR gate behaves like the logical inclusive “OR”. The output is “true” if either or both of the inputs are “true”. If both inputs are “false” then the output is “false”. The output will be 1 if and only if one or both inputs are 1; otherwise, the output is 0. The logical symbol of the OR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 28

The OR gate output is
C = A OR B
We use the + sign to denote the OR function. Therefore,
C = A + B
Read this as “C equals A OR B”.
For instance, if both the inputs are 1
C = A + B = 1 + 1 = 1
The truth table for OR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 29

NOT Gate:
The NOT gate, called a logical inverter, has only one input. It reverses the logical state. The output C is always the complement of the input. The logical symbol of the NOT gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 30

C = NOT A
In boolean algebra, the overbar stands for NOT operation. Therefore,
C = \(\overline{\mathrm{A}}\)
Read this as “C equals NOT A” or “C equals the complement of A”.
If A is 0,
C = \(\overline{1}\) = 0
On the otherhand, if A is 1,
C = \(\overline{1}\) = 0
The truth table for NOT gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 31

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 25.
How AND and OR can be realized using NAND and NOR gate.
Answer:
The output of NOR gate is
C = \(\overline{\mathrm{A}+\mathrm{B}}\)

The output of Bubbled AND gate is
C = \(\overline{\mathrm{A}}\) . \(\overline{\mathrm{B}}\)

The truth table for both Bubbled AND and NOR gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 32

From the above table, a bubbled AND gate produces the same output as a NOR gate. In this way we can realised AND gate using NOR gate.
The output of NAND gate is C = \(\overline{\mathrm{A} \cdot \mathrm{B}}\)
The output of Bubbled OR gate is C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)

The truth table for both bubbled OR gate and NAND gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 33

From the above table, a bubbled OR gate produces the same output as a NAND gate. In this way we can realised OR gate using NAND gate.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 26.
Explain the Derived gates with expression and truth table.
Answer:
The NAND gate operates an AND gate followed by a NOT gate. It acts in the manner of the logical operation “AND” followed by inversion. The output is “false” if both inputs are “true”, otherwise, the output is “true”. The output of the NAND gate is 0 if and only if both the inputs are 1, otherwise the output is 1. The logic circuit of NAND gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 34

The logical symbol of NAND gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 35

The output of the NAND gate is
C = (\(\overline{\mathrm{A} \cdot \mathrm{B}}\))

Read this as “C equals NOT of A AND B” or “C equals the complement of A and B”.

For example if both the inputs are 1

C = \((\overline{1.1})=\overline{1}\) = 0
The truth table for NAND gate is

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 36

TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 37

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Choose the correct answer:

Question 1.
The AND operator is defined in Boolean algebra by the use of the operator
(a) +
(b) –
(c) *
(d) .
Answer:
(d) .

Question 2.
The plus sign is used to indicate the ______ operator.
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(b) OR

Question 3.
The _______ is the combination of NOT and AND.
(a) NAND
(b) XOR
(c) XNOR
(d) NOT
Answer:
(a) NAND

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 4.
Which gate is the combination XOR gate followed by the inverter?
(a) XNOR
(b) XOR
(c) OR
(d) NOR
Answer:
(a) XNOR

Question 5.
How many fundamental logic gates?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 6.
A + 0 is:
(a) 1
(b) 0
(c) A
(d) \(\overline{\mathrm{A}}\)
Answer:
(c) A

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 7.
A. \(\overline{\mathrm{A}}\) is:
(a) A
(b) 0
(c) \(\overline{\mathrm{A}}\)
(d) 1
Answer:
(b) 0

Question 8.
A + A is:
(a) 2A
(b) A2
(c) A
(d) +1
Answer:
(c) A

Question 9.
A + 1 is:
(a) 1A
(b) A
(c) 1
(d) 2A
Answer:
(c) 1

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 10.
________ Gate is called logic inverter.
(a) OR
(b) NOT
(c) AND
(d) None
Answer:
(b) NOT

Question 11.
Who proposed the basic principles of Algebra?
(a) George Boole
(b) George William
(c) George Kingston
(d) George Barsley
Answer:
(a) George Boole

Question 12.
The variable which can stored truth values are called:
(a) Logical variable
(b) Binary valued variables
(c) Boolean variable
(d) All the above
Answer:
(d) All the above

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 13.
The basic logical operations symbol are also called:
(a) Arithmetic operators
(b) Logical operators
(c) Conditional operators
(d) Relational operators
Answer:
(b) Logical operators

Question 14.
Which logical operator has one input and one output?
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(c) NOT

Question 15.
Which operator is generated by inverting the output of an AND operator?
(a) NAND
(b) NOT
(c) OR
(d) NOR
Answer:
(a) NAND

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 16.
Which operator is the combination of NOT and OR?
(a) NAND
(b) NOT
(c) OR
(d) NOR
Answer:
(d) NOR

Question 17.
Which operator is generated by inverting the output of an OR operator?
(a) NOR
(b) NAND
(c) NOT
(d) OR
Answer:
(a) NOR

Question 18.
Which operator uses multiplication sign in boolean algebra?
(a) NOR
(b) NOT
(c) AND
(d) OR
Answer:
(c) AND

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 19.
Which gate gets its name from its behaviour like the logical inclusive?
(a) OR
(b) AND
(c) NOT
(d) NAND
Answer:
(a) OR

Question 20.
Which gate circuit is an OR gate followed by an inverter?
(a) NAND
(b) NOR
(c) NOT
(d) AND
Answer:
(b) NOR

Question 21.
⊙ – (included dot) stands for the:
(a) XNOR
(b) XOR
(c) NAND
(d) NOR
Answer:
(a) XNOR

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 22.
Which of the logic gates are found in digital integrated circuits?
(a) gates
(b) arrays
(c) operation
(d) physical space
Answer:
(b) arrays

Question 23.
A + \(\overline{\mathrm{A}}\) is:
(a) 0
(b) 1
(c) A
(d) \(\overline{\mathrm{A}}\)
Answer:
(b) 1

Question 24.
A.A is:
(a) 0
(b) 1
(c) A
(d) \(\overline{\mathrm{A}}\)
Answer:
(c) A

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 25.
A.0 is:
(a) 0
(b) 1
(c) A
(d) \(\overline{\mathrm{A}}\)
Answer:
(a) 0

Question 26.
\((\overline{\overline{\mathrm{A}})}\) is:
(a) 0
(b) 1
(c) A
(d) \(\overline{\mathrm{A}}\)
Answer:
(c) A

Question 27.
The symbol of NOT gate is:
(a) TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 1

(b) TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 2

(c) TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 3

(d) TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 4

Answer:

(a) TN State Board 11th Computer Science Important Questions Chapter 2 Boolean Algebra Part 2 1

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 28.
Choose the odd man out:
(a) F
(b) E
(c) D
(d) H
Answer:
(d) H

Question 29.
Choose the odd man out:
(a) AND
(b) OR
(c) NOR
(d) NOT
Answer:
(c) NOR

Question 30.
Match the following:

(i) AND Operator  (a) Y = A + B
(ii) OR Operator  (b) Y= \(\overline{\mathrm{A}}\)
(iii) NOT Operator  (c) Y = \(\overline{\mathrm{A} \cdot \mathrm{B}}\)
(iv) NAND Operator  (d) Y = A . B

Answer:

(a) (i) – (a); (ii) – (b); (iii) – (c); (iv) – (d)
(b) (i) – (b); (ii) – (a); (iii) – (c); (iv) – (d)
(c) (i) – (c); (ii) – (b); (iii) – (d); (iv) – (a)
(d) (i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)
Answer:
(d) (i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 31.
Match the following:

(i) A + 0 (a) A + B
(ii) A . \(\overline{\mathrm{A}}\) (b) 1
(iii) A + 1 (c) 0
(iv) A . \(\overline{\mathrm{A}}\) . B (d) A

(a) (i) – (d); (ii) – (c); (iii) – (b); (iv) – (a)
(b) (i) – (c); (ii) – (b); (iii) – (a); (iv) – (d)
(c) (i) – (d); (ii) – (b); (iii) – (a); (iv) – (c)
(d) (i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)
Answer:
(a) (i) – (d); (ii) – (c); (iii) – (b); (iv) – (a)

Question 32.
Choose the incorrect pair:

Column I

 Column II

(a) A + B  B + A
(b) A + (A . B)  A
(c) A + \(\overline{\mathrm{A}}\) . B  A . B
(d) \(\overline{\mathrm{A}+\mathrm{B}}\)  \(\overline{\mathrm{A}}\) . \(\overline{\mathrm{B}}\)

Answer:
(c)

Question 33.
Assertion (A):
The NAND gate operates on OR gate followed by a NOT gate.
Reason (R):
It acts in the manner of the logical operation ‘AND’ followed by inversion.
(a) Both A and R are true, and R is the correct explanation for A.
(b) Both A and R are true, but R is not the correct explanation for A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(d) A is false, but R is true.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 34.
Assertion (A):
The XOR(exclusive-OR) gate acts in the same way as the logical ‘either/or’.
Reason (R):
The output is ‘true’ if either, but not both of the inputs are ‘true’.
(a) Both A and R are true, and R is the correct explanation for A.
(b) Both A and R are true, but R is not the correct explanation for A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation for A.

Question 35.
Which is a basic electronic circuit which operates on one or more signals?
(a) Boolean algebra
(b) Gate
(c) Fundamental gates
(d) Derived gates
Answer:
(b) Gate

Question 36.
Which gate is called as the logical inverter?
(a) AND
(b) OR
(c) NOT
(d) XNOR
Answer:
(c) NOT

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 37.
A + A = ?
(a) A
(b) O
(c) 1
(d) A
Answer:
(a) A

Question 38.
NOR is a combination of ?
(a) NOT(OR)
(b) NOT(AND)
(c) NOT(NOT)
(d) NOT(NOR)
Answer:
(a) NOT(OR)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Boolean Algebra Part 2

Question 39.
NAND is called as ________ Gate.
(a) Fundamental Gate
(b) Derived Gate
(c) Logical Gate
(d) Electronic Gate
Answer:
(b) Derived Gate

TN Board 11th Computer Science Important Questions

TN Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1

Question 1.
Define word length.
Answer:
Word length refers to the number of bits processed by a Computer’s CPU. A word length can have 8 bits, 16 bits, 32 bits and 64 bits (Present Computers use 32 bits or 64 bits).

Question 2.
Define byte.
Answer:
A collection of 8 bits is called Byte. A byte is considered as the basic unit of measuring the memory size in the computer.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 3.
What is a bit?
Answer:
A bit is the short form of Binary digit which can be ‘0’ or ‘ 1 ’. It is the basic unit of data in computers.

Question 4.
What is Decimal Number System?
Answer:
It consists of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits). It is the oldest and most popular number system used in our day-to-day life. In the positional number system, each decimal digit is weighted relative to its position in the number. It means that each digit in the number is multiplied by 10 raised to a power corresponding to that digit’s position.

Question 5.
What is meant by Octal Number System?
Answer:
Octal number system uses digits 0, 1, 2, 3, 4, 5, 6 and 7 (8 digits). Each octal digit has its own positional value or weight as a power of 8.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 6.
Convert (547)<sub>8</sub> to its decimal equivalent?
Answer:
The Octal sequence (547)<sub>8</sub> has the decimal equivalent:
(547)8 = 5 × 82 + 4 × 81 + 7 × 80
= 5 × 64 + 4 × 8 + 7 × 1
= 320 + 32 + 7 = (359)10

Question 7.
(11.011)2 convert into decimal number.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 1

= 3 + (0 × 0.5 + 1 × 0.25 + 1 × 0.125)
= 3 + 0 + 0.25 + 0.125 = (3.375)10

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 8.
(6213)8 convert into binary number.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 2

Question 9.
What is a character set?
Answer:
In general, 26 uppercase letters, 26 lowercase letters, 0 to 9 digits and special characters are used in a computer is called character set. All these character set are denoted through numbers only.

Question 10.
Convert (65)10 into its equivalent binary number.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 3

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 11.
How will you convert decimal to octal? Give example.
Answer:
To convert Decimal to Octal, “Repeated Division by 8” method can be used. Here, we have to divide the given number by 8.
Eg: Convert (65)10 into its equivalent Octal Number.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 4

Question 12.
Write octal numbers and their binary equivalent.
Answer:

Octal  Binary Equivalent
0  000
1  001
2  010
3  011
4  100
5  101
6  110
7  111

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 13.
In what ways the numbers are represented in computers?
Answer:
The numbers are represented in computers in different ways:
(i) Signed Magnitude representation.
(ii) 1 ’s Complement.
(iii) 2’s Complement.

Question 14.
Convert (11010110)2 into octal equivalent.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 5

Question 15.
Convert the following binary numbers to decimal
(i) 11101,
(ii) 1011010.
Answer:
(i) 11101

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 6

(ii) 1011010

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 7

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 16.
Convert (3EF)16 to decimal.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 8

Question 17.
Convert (111011)2 into its equivalent decimal number.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 9

Question 18.
Write short note on hexadecimal number system.
Answer:
A hexadecimal number is represented using base 16. Hexadecimal or Hex numbers are used as a shorthand form of binary sequence. This system is used to represent data in a more compact manner. Since 16 symbols are used, 0 to F, the notation is called hexadecimal. The first 10 symbols are the same as in the decimal system, 0 to 9 and the remaining 6 symbols are taken from the first 6 letters of the alphabet sequence, A to F, where A represents 10, B is 11, C is 12, D is 13, E is 14 and F is 15.
Eg: The hexadecimal sequence (25)16 has the decimal equivalent:
(25)<sub>16</sub> = 2 × 16<sup>1</sup> + 5 × 16<sup>0</sup>
= 32 + 5 = (37)<sub>10</sub>

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 19.
Write the hierarchy of data representation.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 10

Question 20.
Write the steps for binary to decimal conversion.
Answer:
To convert Binary to Decimal we can use
positional notation method.
Step 1: Write down the Binary digits and list the powers of 2 from right to left (Positional Notation)
Step 2: For each positional notation written for the digit, now write the equivalent weight.
Step 3: Multiply each digit with its corresponding weight.
Step 4: Add all the values.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 11

Question 21.
How will you convert binary to octal?
Answer:
Step 1: Group the given binary number into 3 bits from right to left.
Step 2: You can add preceding 0 to make a group of 3 bits if the left most group has less than 3 bits.
Step 3: Convert equivalent octal value using “2’s power positional weight method”.
Eg: Convert (11010110)2 to its Octal equivalent.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 12

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 13

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 22.
Write the steps for converting binary to hexadecimal.
Answer:
Step 1: Group the given number into 4 bits from right to left.
Step 2: You can add preceding 0’s to make a group of 4 bits if the left most group has less than 4 bits.
Step 3: Convert equivalent Hexadecimal value using “2’s power positional weight method”.
Eg: Convert (111 1010110)2 to Hexadecimal.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 14

[Note: 0’s are added to the left most group to make it a group of 4 bits]

Question 23.
Write the steps for octal to decimal conversion.
Answer:
To convert octal to decimal, we can use positional notation method.
(i) Write down the octal digits and list the powers of 8 from right to left(Positional Notation).
(ii) For each positional notation of the digit write the equivalent weight.
(iii) Multiply each digit with its corresponding weight.
(iv) Add all the values.
Eg: Convert (1265)8 to equivalent Decimal number.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 15

(1265)8 = 512 × 1 + 64 × 2 + 8 × 6 + 1 × 5
= 512 + 128+ 18 + 5
(1265)8 = (693)10

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 24.
Write the steps for converting hexadecimal to decimal.
Answer:
To convert Hexadecimal to Decimal we canuse positional notation method.
(i) Write down the Hexadecimal digits and list the powers of 16 from right to left (Positional Notation).
(ii) For each positional notation written for the. digit, now write the equivalent weight.
(iii) Multiply each digit with its corresponding weight.
(iv) Add all the values to get one final value.
Eg: Convert (25F)16 into its equivalent Decimal number.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 16

(25F)16 = 2 × 256 + 5 × 16 + 15 × 1
= 512 + 80 + 15
(25F)16 = (607)10

Question 25.
Write short note on signed magnitude representation.
Answer:
The value of the whole numbers can be determined by the sign used before it. If the number has ‘+‘ sign or no sign it will be considered as positive. If the number has ‘—‘ sign it will be considered as negative.
Eg: +43 or 43 is a positive number.
– 43 is a negative number.

In signed binary representation, the left most bit is considered as sign bit. If this bit is O, it is a positive number and if it is 1, it is a negative number. Therefore a signed binary number has 8 bits, only 7 bits used for storing values (magnitude) and the 1 bit is used for sign.
+ 43 is represented in memory as follows:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 17

– 43 is represented in memory as follows:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 18

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 26.
Write 2’s complement procedure with example.
Answer:
The 2’s complement method for negative number is as follows:
(i) Invert all the bits in the binary sequence
(i.e., change every 0 to 1 and every 1 to 0 i.e.,1‘s complement).
(ii) Add I to the result to the Least Significant Bit (LSB).
Eg: 2’s Complement represent of (-24)10
Binary equivalent of +24 = 11000
8bit format = 00011000
1’s complement = 11100111
Add 1 to LSB = +1
2’s complement of -24 = 11101000

Question 27.
Write short note unicode.
Answer:
Unicode is used in most of the modem computers. The popular coding scheme after ASCII is Unicode. ASCII can represent only 256 characters. ‘English and European Languages alone can be handled by ASCII.
Languages like Tamil, Malayalam, Kannada and Telugu could not be represented by ASCII. Hence, the Unicode was generated to handle all the coding system of Universal languages. Unicode is 16 bit code and can handle 65536 characters.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 28.
Give the steps to convert fractional binary to decimal equivalent with example.
Answer:
Step 1:
Convert integral part of Binary to Decimal equivalent using positional notation method.

Step 2:
To convert, the fractional part of binary to its decimal equivalent.

Step 2.1:
Write down the Binary digits in the fractional part.

Step 2.2:
For all the digits write powers of 2 from left to right starting from 2-1, 2-2, 2-3, ………….2-n, now write the equivalent weight.

Step 2.3:
Multiply each digit with its corresponding weight.

Step 2.4:
Add all the values which you obtained in Step 2.3.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 19

Step 3:
To get final answer write the integral part (after conversion), followed by a decimal point(.) and the answer arrived at

Step 2.4.
Eg: Convert the given Binaxy number(11.011)2 into its decimal equivalent Integer part
(11)2 = 3.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 20

3 + (0 × 0.5 + 1 × 0.25 + 1 × 0.125) = 3.375
3 + 0.75 = 3.75
(11.011)2 = (3.375)10

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 29.
Explain in detail the different encoding systems used for computer.
Answer:
There are several encoding systems used for computer. They are –
(i)Binary Coded Decimal (BCD):
It is 26 bit encoding system. It can handle 26 = 64 characters only. This encoding system is not in the practice right now.

(ii)American Standard Code for information Interchange (ASCII):
This is the most popular encoding system recognized by United States. Most of the computers use this system. This encoding system can handle English characters only. This can handle 2<sup>7</sup> bit which means 128 characters. Here each character has individual number.

The new edition (version) ASCII – 8, has 2<sup>8</sup> bits and can handle 256 characters represented from 0 to 255 unique umbers.

The ASCII code equivalent to the uppercase letter ‘A’ is 65. The binary representation of ASCII (7 bit) value is 1000001. Also 01000001 in ASCII-8 bit.

(iii)Extended Binary Coded Decimal Interchange Code (EBCDIC):
This is similar to ASCII Code with 8 bit representation. This coding system is formulated by International Business Machine(IBM). The coding system can handle 256 characters. The input code in ASCII can be converted to EBCDIC system and vice – versa.

(iv)Indian Standard Code for Information Interchange (ISCII):
ISCII is the system of handling the character of Indian local languages. This is a 8-bit coding system and can handle 256 (2<sup>8</sup>) characters. It is formulated by the department of Electronics in India in the year 1986- 88 and recognized by Bureau of Indian Standards (BIS). Now this coding system is integrated with Unicode.

(v) Unicode:
This coding system is used in most of the modem computers. The popular coding scheme after ASCII is Unicode. ASCII can represent only 256 characters. Therefore English and European Languages alone can be
handled by ASCII. Languages like Tamil, Malayalam, Kannada and Telugu could not be represented by ASCII. Hence, the Unicode was generated to handle all the coding system of Universal languages. This is 16 bit code and can handle 65536 characters.

Question 30.
Identify the number system for the following numbers.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 21

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 31.
State whether the following numbers are valid or not. If invalid, give reason.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 22

Question 32.
Convert the following Decimal numbers to its equivalent Binary, Octal, Hexadecimal.
(i) 1920,
(ii) 255,
(iii) 126.
Answer:
(i) 1920

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 23

(ii) 255

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 24

(iii) 126

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 25

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 33.
Convert the given Binary number into its equivalent Decimal, Octal and Hexadecimal number.
(i) 101110101, (ii) 1011010, (iii) 101011111.

(i) 101110101 into Decimal number
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 26

(ii) 1011010 to Decimal
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 27

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 28

(iii) 101011111 to Decimal
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 29

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 34.
Convert the foltowing Octal numbers into Binary numbers.
(a) 472, (b) 145, (c) 347, (d) 6247, (e) 645.

(a) 472
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 30

(b) 145
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 31

(c) 347
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 32

(d) 6247
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 33

(e) 645
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 34

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 35.
Convert the following Hexadecimal numbers to Binary numbers:
(a) A6, (b) BE, (c) 9BC8, (d) BC9.

(a) A6
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 35

(b) BE
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 36

(c) 9BC8
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 37

(d) BC9
Answer:
B → 11 C → 12

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 38

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 36.
Write the l’s complement number and 2’s complement number for the following decimal numbers:
(a) 22 (b) -13 (c) -65 (d) -46

(a) – 22 → 101102
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 39

(b) -13 → 1101
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 40

(c) -65 → 1000001
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 41

(d) -46 → 1011102
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 42

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 37.
Perform the following binary computations:
(a) 1010 + 1510
(b) – 1210 + 510
(c) 1410 – 1210
(d) (-2)10 – (- 610).

(a) 1010 + 1510
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 43

(b) – 1210 + 510
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 44

(c) 1410 – 1210
Answer:
14 = 1110 = 00001110
– 12 = 1100 = 00001100

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 45

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 46

[The final carry bit is lost because it cannot be accommodated in the 8 – bit register]

(d) (-2)10 – (- 610)
Answer:
-2 = 10
8 bit from = 00000010
1’s complement = 11111101

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 47

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 38.
What is data?
Answer:
The term data comes from the word datum, which means a raw fact. The data is a fact about people, places or some objects.
Eg: Let ‘Name’, ‘Age’, ‘Class’, ‘Marks’ and ‘Subject’ be some defined variables. Assign a value to each of these variables.
Name = Rajesh
Age = 16
Class = XI
Mark = 65
Subject = Computer Science
Here, the values assigned to the five different variables are called data.

Question 39.
Write the l’s complement procedure.
Answer:
Step 1: Convert the given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits , if less add 0 at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e.,s Change 1 as 0 and 0 as 1)

Question 40.
Convert (46)10 into Binary number.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 48

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 41.
We cannot find l’s complement for (28)10. State reason.
Answer:
1 ’s complement representation is an easier approach to represent signed numbers. This is for negative numbers only. This (28)10 this whole numbers cannot be determined by negative number because the number whose MSB is 1.

Question 42.
List the encoding systems for characters in memory.
Answer:
There are several encoding systems used for computer. They are
(i) BCD: Binary Coded Decimal.
(ii) EBCDIC: Extended Binary Coded Decimal Interchange Code.
(iii) ASCII: American Standard Code for Information Interchange.
(iv) Unicode
(v) ISCII: Indian Standard Code for Information Interchange.

Question 43.
What is radix of a number system? Give example.
Answer:
A numbering system is a way of representing numbers. The most commonly used numbering system in real life is Decimal number system. Other number systems are Binary, Octal, Hexadecimal number system. Each number system is uniquely identified by its base value or radix. Radix or base is the count of number of digits in each number system. Radix or base is the general idea behind positional numbering system.
Eg:(123)10, (547)8, (1001)2, (25)16.
In this, 10, 8, 2, 16 are radix or base value.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 44.
Write note on binary number system.
Answer:
There are only two digits in the Binary system, 0 and 1. The numbers in the binary system are represented to the base 2 and the positional multipliers are the powers of 2. The left most bit in the binary number is called as the Most Significant Bit (MSB) and it has the largest positional weight. The right most bit is the Least Significant Bit (LSB) and has the smallest positional weight.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 49

Eg: The binary sequence (1101)2 has the decimal equivalent:
(1101)2 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20
= 8 + 4 + 0 + 1 =(13)10

Question 45.
Convert (150)10 into Binary, then convert that Binary number to Octal.
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 50

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 51

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 46.
Write short note on ISCII.
Answer:
ISCII is the system of handling the character of Indian local languages. This is a 8-bit coding system. So it can handle 256 (28) characters. This system is formulated by the department of Electronics in India in the year 1986-88 and recognized by Bureau of Indian Standards (BIS). Now this coding system is integrated with Unicode.

Question 47.
Add: (i)-2210 + 1510;
(ii) 2010 + 2510.

(i) -2210 + 1510
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 52

(ii) 20 + 25
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 53

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 48.
(a) Write the procedure to convert fractional Decimal to Binary.
(b) Convert (98.46)10 to Binary.
Answer:
(a) The method of repeated multiplication by 2 has to be used to convert such kind of decimal fractions.
The steps involved in the method of repeated multiplication by 2:

Step 1: Multiply the decimal fraction by 2 and note the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of 0s and Is that become the binary equivalent of decimal fraction.
Step 4: The final answer is to be written from first integer part obtained till the last integer part obtained.

(b) (98.46)10
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 54

Question 49.
Find 1’s Complement and 2’s Complement for the following Decimal number: (a) – 98,
(b) -135.
(a) – 98
Answer:
First, convert given decimal number into binary.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 55

Binary number = 11000102
Second, check binary number as 8 bits, If less add 0 as the left most bit, 01100010
Third, Invert all bits (change 1 as 0 and 0 as 1) 1 ’s complement for 10011101.
2’s complement:
Binary equivalent of + 98 = 1100010
8 bit format = 01100010
1 ’s complement = 10011101
Add 1 to LSB = 1
10011110
2’s complement of – 98 = 100111102

(b) – 135
Answer:
First, convert given decimal number into Binary.

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 56

Second, check binary number as 8 bits, If less add 0 at the left most bit. It has 8 bits, 10000111.
Third, Invert all bits (change 1 as 0 and 0 as 1) 1 ’s complement for 01111000.

2’s complement:
Binary equivalent of + 135 = 10000111
8 bit format = 10000111
1 ’s complement = 01111000
Add 1 to LSB = 1
01111001
2’s complement of- 135 = 011110012

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 50.
(a) Add 11010102+ 1011012
(b) Subtract 11010112 – 1110102

(a) Add 11010102+ 1011012
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 57

(b) Subtract 11010112 – 1110102
Answer:

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 58

11010112 – 1110102

TN State Board 11th Computer Science Important Questions Chapter 2 Number Systems Part 1 59

11010112 – 1110102 = 1100012

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Choose the correct answer:

Question 1.
A bit is the short form of:
(a) binary data
(b) binary digit
(c) binary decimal
(d) big digit
Answer:
(b) binary digit

Question 2.
A ________ is a collection of 4 bits.
(a) bit
(b) byte
(c) nibble
(d) word length
Answer:
(c) nibble

Question 3.
A collection of 8 bits is called:
(a) bit
(b) byte
(c) word length
(d) nibble
Answer:
(b) byte

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 4.
2^40 is referred as:
(a) mega
(b) giga
(c) peta
(d) tera
Answer:
(d) tera

Question 5.
2^70 is referred as:
(a) peta
(b) exa
(c) zetta
(d) yotta
Answer:
(c) zetta

Question 6.
1024 GB is referred as :
(a) kilo byte
(b) mega byte
(c) giga byte
(d) tera byte
Answer:
(d) tera byte

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 7.
____________ are used to represent characters in a text.
(a) Bits
(b) Bytes
(c) Nibble
(d) Wordlength
Answer:
(b) Bytes

Question 8.
The ASCII value for blank space is:
(a) 32
(b) 91
(c) 48
(d) 65
Answer:
(a) 32

Question 9.
The ASCII value for numeric 0 is:
(a) 32
(b) 97
(c) 48
(d) 65
Answer:
(c) 48

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 10.
Each number system is uniquely identified by its:
(a) decimal value
(b) binary value
(c) base value or radix
(d) octal value
Answer:
(c) base value or radix

Question 11.
The range of ASCII values for lower case alphabets is from:
(a) 97 to 122
(b) 65 to 90
(c) 98 to 122
(d) 97 to 123
Answer:
(a) 97 to 122

Question 12.
The range of ASCII values for upper case alphabets is from:
(a) 97 to 122
(b) 65 to 90
(c) 66 to 90
(d) 65 to 97
Answer:
(b) 65 to 90

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 13.
The expansion of MSB is:
(a) Most Significant Bit
(b) Most Signed Bit
(c) Must Significant Bit
(d) Must Signed Bit
Answer:
(a) Most Significant Bit

Question 14.
The expansion of LSB is:
(a) Lower Significant Bit
(b) Least Significant Bit
(c) Lower Signed Bit
(d) Least Signed Bit
Answer:
(b) Least Significant Bit

Question 15.
Radix of octal number is:
(a) 2
(b) 10
(c) 16
(d) 8
Answer:
(d) 8

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 16.
The binary sequence (1101)2 has the decimal equivalent:
(a) (25)10
(b) (15)10
(c) (17)10
(d) (13)10
Answer:
(d) (13)10

Question 17.
In hexadecimal C represents:
(a) 11
(b) 10
(c) 12
(d) 15
Answer:
(c) 12

Question 18.
The simplest method to represent negative binary numbers is called:
(a) signed magnitude
(b) unsigned magnitude
(c) magnitude bit
(d) unmagnitude bit
Answer:
(a) signed magnitude

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 19.
EBDIC coding system can handle characters.
(a) 64
(b) 255
(c) 256
(d) 128
Answer:
(c) 256

Question 20.
The most commonly used coding scheme is the:
(a) Binary Coded Decimal
(b) Extended Binary Coded Decimal Interchange Code
(c) Indian Standard Code for Information Interchange
(d) American Standard Code for Information Interchange
Answer:
(d) American Standard Code for Information Interchange

Question 21.
Which is the basic unit of measuring the memory size in the computer?
(a) bit
(b) byte
(c) nibble
(d) word length
Answer:
(b) byte

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 22.
What is the most commonly used numbering system in real life?
(a) Binary
(b) Octal
(c) Decimal
(d) Hexadecimal
Answer:
(c) Decimal

Question 23.
Which of the following is octal number?
(a) 1101
(b) 1178
(c) 778
(d) 8005
Answer:
(a) 1101

Question 24.
Which of the following is not a binary number?
(a) 2101
(b) 1000
(c) 1111
(d) 1001
Answer:
(a) 2101

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 25.
Which of the following is a Hexadecimal number?
(a) 1234
(b) ABCD
(c) 1001
(d) All the above
Answer:
(d) All the above

Question 26.
Unicode is a ______ bit code.
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(b) 16

Question 27.
What is the 1 ’s complement for (-24)10?
(a) 00011000
(b) 11100111
(c) 11110011
(d) 11100001
Answer:
(b) 11100111

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 28.
Who formulated ASCII coding system?
(a) Microsoft
(b) Oracle
(c) IBM
(d) SVN
Answer:
(c) IBM

Question 29.
What is the ASCII value of A in decimal number?
(a) 65
(b) 66
(c) 67
(d) 68
Answer:
(a) 65

Question 30.
ISCII coding system is formulated by:
(a) Electronics in India
(b) IBM
(c) Oracle corporation
(d) Sun micro systems
Answer:
(a) Electronics in India

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 31.
Which coding system is used in most of the modern computers?
(a) ASCII
(b) BCD
(c) ISCII
(d) Unicode
Answer:
(d) Unicode

Question 32.
Unicode scheme is denoted by:
(a) Binary numbers
(b) Octal numbers
(c) Hexadecimal numbers
(d) Decimal numbers
Answer:
(c) Hexadecimal numbers

Question 33.
What is the ASCII value of a in decimal number?
(a) 95
(b) 96
(c) 97
(d) 98
Answer:
(c) 97

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 34.
What is the ASCII value of * symbol in Hexadecimal number?
(a) 2B
(b) 2A
(c) 2C
(d) 2D
Answer:
(b) 2A

Question 35.
What is the ASCII value of @ symbol in octal number?
(a) 111
(b) 571
(c) 100
(d) 123
Answer:
(c) 100

Question 36.
Unicode can handle characters.
(a) 65536
(b) 75536
(c) 66535
(d) 85536
Answer:
(a) 65536

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 37.
BCD can handle _______ characters only.
(a) 65
(b) 75
(c) 64
(d) 94
Answer:
(c) 64

Question 38.
ASCII can handle characters.
(a) 365
(b) 128
(c) 256
(d) 255
Answer:
(b) 128

Question 39.
________ can handle all the coding system of universal languages.
(a) Unicode
(b) ISCII
(c) EBCDIC
(d) BCD
Answer:
(a) Unicode

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 40.
Choose the odd man out:
(a) Bit
(b) Byte
(c) Nibble
(d) IBM
Answer:
(d) IBM

Question 41.
Choose the odd man out:
(a) Octal variable
(b) Logical variable
(c) Binary valued variable
(d) Boolean variable
Answer:
(a) Octal variable

Question 42.
Match the following:

(i) (1010)2  (a) Octal
(ii) (989)16  (b) Binary
(iii) (750)8  (c) Decimal
(iv) (926)10  (d) Hexadecimal

(a) (i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)
(b) (i) – (b); (ii) – (d); (iii) – (a); (iv) – (c)
(c) (i) – (b); (ii) – (a); (iii) – (d); (iv) – (c)
(d) (i) – (d); (ii) – (b); (iii) – (a); (iv) – (c)
Answer:
(b) (i) – (b); (ii) – (d); (iii) – (a); (iv) – (c)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 43.
Match the following:

Hexadecimal

 Binary

(i) A  (a) 1101
(ii) B  (b) 1010
(iii) D  (c) 1110
(iv) E  (d) 1011

(a) (i) – (d); (ii) – (b); (iii) – (a); (iv) – (c)
(b) (i) – (c); (ii) – (d); (iii) – (a); (iv) – (b)
(c) (i) – (b); (ii) – (d); (iii) – (a); (iv) – (c)
(d) (i) – (b); (ii) – (a); (iii) – (d); (iv) – (c)
Answer:
(c) (i) – (b); (ii) – (d); (iii) – (a); (iv) – (c)

Question 44.
Match the following:

(i) BCD  (a) 0 – 15
(ii) EBCDIC  (b) 65536 characters
(in) Unicode  (c) 256 characters
(iv) Hexadecimal  (d) 64 characters

(a) (i) – (d); (ii) – (c); (iii) – (b); (iv) – (a)
(b) (i) – (d); (ii) – (a); (iii) – (b); (iv) – (c)
(c) (i) – (a); (ii) – (b); (iii) – (d); (iv) – (c)
(d) (i) – (c); (ii) – (a); (iii) – (d); (iv) – (b)
Answer:
(a) (i) – (d); (ii) – (c); (iii) – (b); (iv) – (a)

Question 45.
Choose the correct pair.

Column – I

 Column – II

(a) 2  011
(b) 4  100
(c) 5  110
(d) 7  010

Answer:
(b)

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 46.
Choose the incorrect pair.

Column – I

 Column – II

(a) Kilo byte  2^10
(b) Mega byte  2^20
(c) Peta byte  2^50
(d) Exa  2^70

Answer:
(d)

Question 47.
Assertion (A):
The computer can understand only Machine language.
Reason (R):
The Machine language is 0 and 1.
(a) Both A and R are true, and R is the correct explanation for A.
(b) Both A and R are true, but R is not the correct explanation for A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation for A.

Question 48.
Assertion (A):
A numbering system is a way of representing numbers.
Reason (R):
The most commonly used numbering system in real life is Binary number system.
(a) Both A and R are true, and R is the correct explanation for A.
(b) Both A and R are true, but R is not the correct explanation for A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(c) A is true, but R is false.

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 49.
Assertion (A):
The popular coding scheme after ASCII is Unicode.
Reason (R):
The Unicode was generated to handle all the coding system of universal language.
(a) Both A and R are true, and R is the correct explanation for A.
(b) Both A and R are true, but R is not the correct explanation for A
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
(a) Both A and R are true, and R is the correct explanation for A.

Question 50.
A collection of 8 bits is called:
(a) bit
(b) byte
(c) word length
(d) nibble
Answer:
(b) byte

Question 51.
The ASCII value for numeric 0 is:
(a) 32
(b) 97
(c) 48
(d) 65
Answer:
(c) 48

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 52.
The expansion of LSB is:
(a) Lower Significant Bit
(b) Least Significant Bit
(c) Lower Significant Bit
(d) Least Signed Bit
Answer:
(b) Least Significant Bit

Question 53.
Which refers to the number of bits processed by a computer’s CPU?
(a) Byte
(b) Nibble
(c) Word length
(d) Bit
Answer:
(c) Word length

Question 54.
How many bytes does 1 KiloByte contain?
(a) 1000
(b) 8
(c) 4
(d) 1024
Answer:
(d) 1024

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 55.
Expansion for ASCII:
(a) American School Code for Information Interchange
(b) American Standard Code for Information Interchange
(c) All Standard Code for Information Interchange
(d) American Society Code for Information Interchange
Answer:
(b) American Standard Code for Information Interchange

Question 56.
2^50 is referred as:
(a) kilo
(b) tera
(c) peta
(d) zetta
Answer:
(c) peta

Question 57.
How many characters can be handled in Binary Coded Decimal System?
(a) 64
(b) 255
(c) 256
(d) 128
Answer:
(a) 64

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 58.
For 11012 what is the Hexadecimal equivalent?
(a) F
(b) E
(c) D
(d) B
Answer:
(c) D

Question 59.
What is the 1 ’s complement of 00100110?
(a) 00100110
(b) 11011001
(c) 11010001
(d) 00101001
Answer:
(b) 11011001

Samacheer Kalvi TN State Board 11th Computer Applications Important Questions Chapter 2 Number Systems Part 1

Question 60.
Which amongst this is not an Octal number?
(a) 645
(b) 234
(c) 876
(d) 123
Answer:
(c) 876

TN Board 11th Computer Science Important Questions