NCERT Solutions Class 11 Computer Science Chapter 2 Encoding Schemes and Number System
NCERT Solutions Class 11 Computer Science Chapter 2 Encoding Schemes and Number System: National Council of Educational Research and Training Class 11 Computer Science Chapter 2 Solutions – Encoding Schemes and Number System. NCERT Solutions Class 11 Computer Science Chapter 2 PDF Download.
NCERT Solutions Class 11 Computer Science Chapter 2: Overview
Board |
NCERT |
Class |
11 |
Subject |
Computer Science |
Chapter |
2 |
Chapter Name |
Encoding Schemes and Number System |
Topic |
Exercise Solutions |
NCERT Solutions Class 11 Computer Science Chapter 2 Encoding Schemes and Number System
Question 1: Write base values of binary, octal and hexadecimal number system.
Answer:
NUMBER SYSTEM |
BASE VALUE |
Binary number system |
2 |
Octal number system |
8 |
hexadecimal number system |
16 |
Question 2: Give full form of ASCII and ISCII.
Answer:
- The full form of ASCII is American Standard Code for Information
- The full form of ISCII is Indian Script Code for Information Interchange.
Question 3: Try the following conversions.
(i) (514)8 = (?)10 (iv) (4D9)16 = (?)10
(ii) (220)8 = (?)2 (v) (11001010)2 = (?)10
(iii) (76F)16 = (?)10 (vi) (1010111)2 = (?)10
Answer:
- (514)8 = (?)10
Digits | 5 | 1 | 4 |
Position | 2 | 1 | 0 |
Weight | 82 | 81 | 80 |
Therefore,
Decimal number = 5×82 + 1×81 + 4×80
= 5×64 + 1×8 + 4× 1
= 320 + 8 + 4
= (332)10
- (220)8 = (?)2
Octal Digits | 2 | 2 | 0 |
Binary value
(3 bits) |
010 | 010 | 000 |
Therefore,
Binary number = (010010000)2
- (76F)16 = (?)10
Digits | 7 | 6 | F(15) |
Position | 2 | 1 | 0 |
Weight | 162 | 161 | 160 |
Therefore,
Decimal number = 7×162 + 6×161 + F×160
= 7×256 + 6×16 + F× 1
= 1792 + 96 + 15
= (1903)10
- (4D9)16 = (?)10
Digits | 4 | D | 9 |
Position | 2 | 1 | 0 |
Weight | 162 | 161 | 160 |
Therefore,
Decimal number = 4×162 + 13×161 + 9×160
= 4×256 + 13×16 + 9× 1
= 1024 + 208 + 9
= (1241)10
- (11001010)2 = (?)10
Digits | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
Position | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
Weight | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 |
Therefore,
Decimal number = 1×27+ 1×26+0×25 +0×24 +1×23 +0×22 +1×21 +0×20
= 128+64 +8 + 2
= (202)10
- (1010111)2 = (?)10
Digits | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
Position | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
Weight | 26 | 25 | 24 | 23 | 22 | 21 | 20 |
Therefore,
Decimal number = 1×26+ 0×25+1×24 +0×23 +1×22 +1×21 +1×20
= 64 +16 + 4+ 2 +1
= (87)10
Question 4: Do the following conversions from decimal number to other number systems.
(i) (54)10 = (?)2 (iv) (889)10 = (?)8
(ii) (120)10 = (?)2 (v) (789)10 = (?)16
(iii) (76)10 = (?)8 (vi) (108)10 = (?)16
Answer:
Question 5: Express the following octal numbers into their equivalent decimal numbers.
(i) 145
(ii) 6760
(iii) 455
(iv) 10.75
Answer:
(i) 145
Digits | 1 | 4 | 5 |
Position | 2 | 1 | 0 |
Weight | 82 | 81 | 80 |
Therefore,
Decimal number = 1×82 +4×81 + 5×80
=1×64 + 4×8 + 5× 1
=64 + 32 + 5
= (101)10
(ii) 6760
Digits | 6 | 7 | 6 | 0 |
Position | 3 | 2 | 1 | 0 |
Weight | 83 | 82 | 81 | 80 |
Therefore,
Decimal number = 6×83 +7×82 +6×81 + 0×80
=6×512 + 7×64 +6×8 + 0× 1
=3072 + 448 + 48+0
= (3568)10
(i) 455
Digits | 4 | 5 | 5 |
Position | 2 | 1 | 0 |
Weight | 82 | 81 | 80 |
Therefore,
Decimal number = 4×82 +5×81 + 5×80
=4×64 + 5×8 + 5× 1
=256 + 40 + 5
= (301)10
(iv) 10.75
Digits | 1 | 0 | 7 | 5 |
Position | 1 | 0 | -1 | -2 |
Weight | 81 | 80 | 8-1 | 8-2 |
Therefore,
Decimal number = 1×81+0×80+7×8-1+5×8-2
=1×8+0×1+7×0.125+5×0.015625
=8+0+0.875+0.078125
= (8.953125)10
Question 6: Express the following decimal numbers into hexadecimal numbers. (i) 548 (ii) 4052 (iii) 58 (iv) 100.25
No, 6 Ch 11 Computer ScienceQuestion 7: Express the following hexadecimal numbers into equivalent decimal numbers. (i) 4A2 (ii) 9E1A (iii) 6BD (iv) 6C.34
Answer:
(i) 4A2
Digits | 4 | A | 2 |
Position | 2 | 1 | 0 |
Weight | 162 | 161 | 160 |
Decimal number = 4 × 16 2 + A × 16 1+ 2 × 16 0
= 4 × 256 + 10 × 16 + 2 × 1
= (1186)10
In case you are missed :- Previous Chapter Solution
(ii) 9E1A
Digits | 9 | E | 1 | A |
Position | 3 | 2 | 1 | 0 |
Weight | 163 | 162 | 161 | 160 |
Decimal number = 9 × 163 + E × 162 + 1 × 161 + A × 160
= 9 × 4096 + 14 × 256 + 1 × 16 + 10 × 1
= (40474)10
(iii) 6BD
Digits | 6 | B | D |
Position | 2 | 1 | 0 |
Weight | 162 | 161 | 160 |
Decimal number = 6 × 162 + B × 161 + D × 160
= 6 × 256 + 11 × 16 + 13 × 1
= (1725)10
(iv) 6C.34
Digits | 6 | C | 3 | 4 |
Position | 1 | 0 | -1 | -2 |
Weight | 161 | 160 | 16-1 | 16-2 |
Decimal number = 6 × 16^1 + C × 16^0 + 3 × 16^-1 + 4 × 16^-2
= 96 + 12 × 1 + 0.1875 + 0.015625 [as C = 12 in base 16]
= (108.203125)10
Question 8: Convert the following binary numbers into octal and hexadecimal numbers. (i) 1110001000 (ii) 110110101 (iii) 1010100 (iv) 1010.1001
8 computer science Class 11Question 9: Write binary equivalent of the following octal numbers. (i) 2306 (ii) 5610 (iii) 742 (iv) 65.203
Answer:
(i) 2306
OCTAL | 2 | 3 | 0 | 6 |
BINARY | 010 | 011 | 000 | 110 |
Binary Number = (10011000110)2
(ii) 5610
OCTAL | 5 | 6 | 1 | 0 |
BINARY | 101 | 110 | 001 | 000 |
Binary Number = (101110001000)2
(iii) 742
OCTAL | 7 | 4 | 2 |
BINARY | 111 | 100 | 010 |
Binary Number = (111100010)2
(iv) 65.203
Digits | 6 | 5 | 2 | 0 | 3 |
Weight | 110 | 101 | 010 | 000 | 011 |
Binary Number = (110101.010000011)2
Question 10: Write binary representation of the following hexadecimal numbers. (i) 4026 (ii) BCA1 (iii) 98E (iv) 132.45
Answer:
(i) 4026
HEXADECIMAL | 4 | 0 | 2 | 6 |
BINARY | 0100 | 0000 | 0010 | 0110 |
Binary Number = (100000000100110)2
(ii) BCA1
HEXADECIMAL | B | C | A | 1 |
BINARY | 1011 | 1100 | 1010 | 0001 |
Binary Number = (1011110010100001)2
(iii) 98E
HEXADECIMAL | 9 | 8 | E |
BINARY | 1001 | 1000 | 1110 |
Binary Number = (100110001110)2
(iv) 132.45
HEXADECIMAL | 1 | 3 | 2 | 4 | 5 |
BINARY | 0001 | 0011 | 0010 | 0100 | 0101 |
Binary Number = (100110010.01000101)2
Question 11: How does computer understand the following text? (hint: 7 bit ASCII code). (i) HOTS (ii) Main (iii) CaSe
Answer:
(i) HOTS: 072 078 084 083
H | 072 |
O | 078 |
T | 084 |
S | 83 |
(ii) Main: 077 097 105 110
M | 077 |
a | 097 |
i | 105 |
n | 110 |
(iii) CaSe: 067 097 083 101
C | 067 |
a | 097 |
S | 083 |
e | 101 |
Question 12: The hexadecimal number system uses 16 literals (0–9, A–F). Write down its base value.
Answer:
The base value of hexadecimal system is 16.
Question 13: Let X be a number system having B symbols only. Write down the base value of this number system.
Answer:
The base value of t number system X will be B. Because base value is the number of bits the number system contain.
Question 14: Write the equivalent hexadecimal and binary values for each character of the phrase given below. ‘‘ हम सब एक”
Answer:
Hexadecimal value:
Binary value:
Question 15: What is the advantage of preparing a digital content in Indian language using UNICODE font?
Answer:
UNICODE language is universally acceptable .there is no need to install any special software to recognize unlike Hindi, Marathi. UNICODE provide special value for each character. Hence, digital content in Indian language are prepared using UNICODE language.
Question 16: Explore and list the steps required to type in an Indian language using UNICODE.
Answer:
Step 1: write the characters, words, sentence in Indian language.
Step 2: Unicode provide unique value for each character which is readable by machine.
Step 3: system converts our Unicode to binary language/ machine language.
Question 17: Encode the word ‘COMPUTER’ using ASCII and convert the encode value into binary values.
Answer:
WORD | ASCII | BINARY VALUE |
C | 067 | 01000011 |
O | 079 | 01001111 |
M | 077 | 01001101 |
P | 080 | 01010000 |
U | 085 | 01010101 |
T | 084 | 01010100 |
E | 069 | 01000101 |
R | 082 | 01010010 |
ASCII:
067 079 077 080 085 084 069 082
BINARY VALUE:
01000011 01001111 01001101 01010000 01010101 01010100 01000101 01010010 |
In case you are missed :- Next Chapter Solution