Samacheer Kalvi 3rd Standard Tamil Guide Book Answers Solutions

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Samacheer Kalvi 3rd Standard Social Science Guide Book Answers Solutions

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Samacheer Kalvi 3rd Standard Maths Guide Book Back Answers Solutions

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Samacheer Kalvi 3rd Standard Books Solutions Guide

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Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q1

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q3
Answer:
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q4
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q4.1

Free online sample standard deviation calculator and variance calculator with steps.

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Answer:
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q5
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q5.1
Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Answer:
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q6
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q6.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q6.2
The standard deviation of bell strike in a day is 6.9

Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

Question 7.
Find the standard deviation of the first 21 natural numbers.
Answer:
Here n = 21
The standard deviation of the first ‘n’ natural numbers,
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q7
The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = \(\frac{3.6}{3}\) = 1.2
New Variance = (1.2)2 = 1.44 [∴ Variance = (S.D)2]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q10
Find its standard deviation.
Answer:
Assumed mean = 60
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q10.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q10.2

Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

Question 11.
In a study about viral fever, the number of people affected in a town were noted as
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q11
Find its standard deviation.
Answer:
Assumed mean = 35
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q11.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q11.2

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q12
Answer:
Assumed mean = 34.5
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q12.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q12.2

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q13
Answer:
Assumed mean = 11
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q13.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q13.2
Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
Number of candidates = 100
n = 100
Mean (\(\bar{x}\)) = 60
standard deviation (σ) = 15
Mean (\(\bar{x}\)) = \(\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}\)
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean (\(\bar{x}\)) = \(\frac{6050}{100}\) = 60.5
Given standard deviation = 15
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q14
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q14.1
Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q15
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q15.1
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q15.2
Samacheer Kalvi 10th Maths Guide Chapter 8 Statistics and Probability Ex 8.1 Q15.3

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

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11th Computer Science Guide Number Systems Text Book Questions and Answers

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Part – I

I. Choose The Correct Answer:

Question 1.
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 2.
How many bytes does 1 KiloByte contain?
a) 1000
b) 8
c) 4
d) 1024
Answer:
d) 1024

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 3.
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 4.
2^50 is referred as
a) Kilo
b) Tera
c) Peta
d) Zetta
Answer:
c) Peta

Question 5.
How many characters can be handled in Binary Coded Decimal System?.
a) 64
b) 255
c) 256
d) 128
Answer:
a) 64

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 6.
For 11012 whai is the Hexadecimal equivalent?
a) F
b) E
c) D
d) B
Answer:
c) D

Hex calculator for performing addition, subtraction, multiplication and division of hexadecimal numbers.

Question 7.
What is the 1’s complement of 00100110?
a) 00100110
b) 11011001
c) 11010001
d) 00101001
Answer:
b) 11011001

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 8.
Which amongst this is not an Octal number?
a) 645
b) 234
c) 876
d) 123
Answer:
c) 876

Part II

Very Short Answers.

Question 1.
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.
Example: Rajesh, 16, XI.

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

Example: Find 1’s complement for (-24)10
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 1

Question 3.
Convert (46)10 into a Binary number.
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 2

Question 4.
We cannot find l’s complement for (28)10 State reason.
Answer:
Since it is a positive number. 1’s complement will come only for negative numbers.

Question 5.
List the encoding systems for characters in memory.
Answer:
There are several encoding systems used for computers. They are

  • BCD – Binary Coded Decimal
  • EBCDIC – Extended Binary Coded Decimal Interchange Code
  • ASCII – American Standard Code for Information Interchange
  • Unicode
  • ISCII – Indian Standard Code for Information Interchange

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Part III

III. Very Short Answers

Question 1.
What is radix of a number system? Give example.
Answer:
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. Ex.

Number system Base / Radix
Binary 2
Octal 8
Decimal 10
Hexadecimal 16

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 2.
Write a note on the binary number system.
Answer:
There are only two digits in the Binary system, namely, 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 leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 3

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 3.
Convert (150)10 into Binary, then convert that Binary number to Octal.
Answer:
Decimal to Binary conversion 150
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 4

Binary to octal conversion
LSB to MSB divide the number into three-digit binary and write the equivalent octal digit
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 5

Question 4.
Write a short note on ISCII.
Answer:
ISCII – Indian Standard Code for Information Interchange (ISCII) is the system of handling the character of Indian local languages. This is an 8 – bit coding system. Therefore it can handle 256 (28) characters. It is recognized by the Bureau of Indian Standards (BIS). It is integrated with Unicode.

This system is formulated by the Department of Electronics in India in the year 1986-88 and recognized by the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Question 5.
Add : a) -2210 + 1510 b) 2010 + 2510.
Answer:
a) -2210 + 1510
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 6
Answer in 2’s complement form . 11111001 is 2’s complement of 7 which is the answer.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Part IV

IV. Detail Answers.

Question 1.
a) Write the procedure to convert fractional Decimal to Binary.
Answer:
Conversion of fractional Decimal to Binary
The method of repeated multiplication by 2 has to be used to convert such kinds 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 0’s and 1’s that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.
Integer part (last integer part obtained)
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 7
Write the integer parts from top to bottom to obtain the equivalent fractional binary number.
Hence
(0.2)10 = (0.00110011.,.)2 = (0.00110011)2.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

b) Convert (98.46)10 to Binary.
Convert (98.46)10 to Binary
Procedure: Conversion of an integral part:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 8
(98)10 = (1100010)2 Conversion of fractional part:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 9

Question 2.
Find l’s Complement and 2’s Complement for the following Decimal number.
a) -98
b) -135
Answer:
a) Conversion of (98)10 into binary
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 10

Conversion of (135)10 into binary
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 11

Question 3.
a) Add 11010102 + 101101)2
Answer:
a) Add 11010102 + 1011012
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 12

b) Subtract 11010112 – 1110102.
Subtract 11010112 – 1110102
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 13

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

11th Computer Science Guide Number Systems Additional Questions and Answers

Part I

Choose The Correct Answer.

Question 1.
The simplest method to represent a negative binary number is called ………………..
(a) signed magnitude
(b) sign bit or parity bit
(c) binary
(d) decimal
Answer:
(a) signed magnitude

Question 2.
Computer understand ________________language.
a) High level
b) Assembly
c) Machine
d) All the above
Answer:
c) Machine

Question 3.
Expansion for BCD ………………..
(a) Binary coded decimal
(b) binary complement decimal
(c) binary computer decimal
(d) binary convert decimal
Answer:
(a) Binary coded decimal

Question 4.
__________is the basic unit of data in computer.
a) BIT
b) BYTE
c) NIBBLE
d) WORD
Answer:
a) BIT

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 5.
The ……………….. operator is defined in boolean algebra by the use of the dot (.) operator.
(a) AND
(b) OR
(c) NOT
(d) NAND
Answer:
(a) AND

Question 6.
Binary digit means __________
a) 0
b) 1
c) either 0 or 1
d) None of these
Answer:
c) either 0 or 1

Question 7.
The convert (65)10 into its equivalent octal number ………………..
(a) (101)8
(b) (101)10
(c) (101)12
(d) (101)4
Answer:
(a) (101)8

Question 8.
A collection of 8 bits is called __________
a) BIT
b) BYTE
C) NIBBLE
d) WORD
Answer:
b) BYTE

Question 9.
……………….. is the general idea behind the positional numbering system.
(a) Radix
(b) Computer memory
(c) Binary number
(d) Decimal number
Answer:
(a) Radix

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 10.
__________refers to the number of bits processed by a computer’s CPU.
a) Word length
b) Nibble
c) Word size
d) None of these
Answer:
a) Word length

Question 11.
Bit means ………………..
(a) nibble
(b) byte
(c) word length
(d) binary digit
Answer:
(d) binary digit

Question 12.
__________is a valid word length of a computer.
a) 64
b) 32
c) 16
d) All the above
Answer:
d) All the above

Question 13.
The computer can understand ……………….. languages.
(a) computer
(b) machine
(c) post
(d) pre
Answer:
(b) machine

Question 14.
1 KiloByte equals to __________bytes.
a) 1024
b) 256
c) 1000
d) 128
Answer:
a) 1024

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 15.
How many bytes does 1 zettabyte contain?
(a) 290
(b) 280
(c) 270
(d) 260
Answer:
(c) 270

Question 16.
1024 MegaBytes equals to _________
a) 1 GigaByte
b) 1 TeraByte
c) 1 YottaByte
d) None of these
Answer:
a) 1 GigaByte

Question 18.
1-kilo byte represents ……………….. bytes.
(a) 512
(b) 256
(c) 1024
(d) 64
Answer:
(c) 1024

Question 18.
1Kb equals to _________bytes.
a) 210
b) 220
c) 230
d) 240
Answer:
a) 210

Question 19.
How many megabytes does 1 GB contain?
(a) 220
(b) 210
(c) 230
(d) 240
Answer:
(b) 210

Question 20.
1 GB equals to ________ bytes.
a) 210
b) 220
c) 230
d) 240
Answer:
c) 230

Question 21.
What is the 1’ s complement of 11001?
(a) 11100110
(b) 01010101
(c) 11110000
(d) 100100111
Answer:
(a) 11100110

Question 22.
1 PetaByte(PB) equals to _________bytes.
a) 250
b) 260
c) 270
d) 280
Answer:
a) 250

Question 23.
The hexadecimal equivalent of 15 is ………………..
(a) A
(b) B
(c) E
(d) F
Answer:
(d) F

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 24.
1 ZettaByte (1ZB) equals to _______ bytes.
a) 250
b) 260
c) 270
d) 280
Answer:
c) 270

Question 25.
The radix of a hexadecimal number is ………………..
(a) 2
(b) 8
(c) 16
(d) 10
Answer:
(c) 16

Question 26.
Computer memory is normally represented in terms of ________ bytes.
a) Kilo
b) Mega
c) Kilo or Mega
d) None of these
Answer:
c) Kilo or Mega

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 27.
The most commonly used number system is ………………..
(a) binary
(b) decimal
(c) octal
(d) hexadecimal
Answer:
(b) decimal

Question 28.
The most commonly used coding scheme to represent character set and the number is ________
a) BCD
b) ASCII
c) EBCDIC
d) All the above
Answer:
b) ASCII

Question 29.
What does MSB mean?
(a) Major sign bit
(b) Most sign bit
(c) Minor sign bit
(d) Most significant bit
Answer:
(d) Most significant bit

Question 30.
The ASCII value for blank space is _________
a) 43
b) 42
c) 32
d) 62
Answer:
c) 32

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 31.
The binary equivalent of hexadecimal number B is ………………..
(a) 1011
(b) 1100
(c) 1001
(d) 1010
Answer:
(a) 1011

Question 32.
The most commonly used numbering system in real life is the _________number system.
a) Hexadecimal
b) Octal
c) Binary
d) Decimal
Answer:
d) Decimal

Question 33.
What is the range of ASCII values for lower case alphabets?
(a) 65 to 90
(b) 65 to 122
(c) 97 to 122
(d) 98 to 122
Answer:
(c) 97 to 122

Question 34.
_________is the count of number of digits in each number system.
a) base
b) radix
c) base or radix
d) symbols
Answer:
c) base or radix

Question 35.
What is the ASCII value for blank space?
(a) 8
(b) 2
(c) 18
(d) 32
Answer:
(d) 32

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 36.
Identify the true statement from the following.
a) In the positional number system, each decimal digit is weighted relative to its position in the number.
b) A numbering system is a way of representing numbers.
c) The speed of a computer depends on the number of bits it can process at once.
d) All the above
Answer:
d) All the above

Question 37.
Which one of the following bits has the smallest positional weight?
(a) MSB
(b) LSB
(c) UPS
(d) USB
Answer:
(b) LSB

Question 38.
The rightmost bit in the binary number is called as the __________
a) MSB
b) LSB
c) FSB
d) None of these
Answer:
b) LSB

Question 39.
Name the person who proposed the basic principles of Boolean Algebra?
(a) Wiliam Boole
(b) George Boole
(c) James Boole
(d) Boolean George
Answer:
(b) George Boole

Question 40.
_______ numbers are used as a shorthand form of a binary sequence.
a) Hexadecimal
b) Octal
c) Decimal
d) None of these
Answer:
a) Hexadecimal

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 41.
What is the other name for a logical statement?
(a) Truth values
(b) Truth functions
(c) Truth table
(d) Truth variables
Answer:
(b) Truth functions

Question 42.
In hexadecimal number system letter ‘E’ represents _______
a) 12
b) 13
c) 14
d) 15
Answer:
c) 14

Question 43.
The NOT operator is represented by the symbol.
(a) over bar
(b) single apostrophe
(c) a and b
(d) plus
Answer:
(c) a and b

Question 44.
_______is a method to convert decimal number to binary number.
a) Repeated division by 2
b) Sum of powers of 2
c) Repeated addition by 2
d) Either A or B
Answer:
d) Either A or B

Question 45.
The output for the AND operator is ………………..
(a) A + B
(b) –
(c) A.B
(d) AB + C
Answer:
(c) A.B

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 46.
Computer can handle _______ numbers.
a) signed
b) unsigned
c) signed and unsigned
d) None of these
Answer:
c) signed and unsigned

Question 47.
Which gate takes only one input?
(a) OR
(b) AND
(c) NOT
(d) XOR
Answer:
(c) NOT

Question 48.
In the signed magnitude method, the leftmost bit is called _______bit.
a) sign
b) parity
c) sign or parity
d) None of these
Answer:
c) sign or parity

Question 49.
Which is not a derived date?
(a) AND
(b) NAND
(c) NOR
(d) XOR
Answer:
(a) AND

Question 50.
The numbers are represented in computers in _______method.
a) Signed magnitude representation
b) 1’s complement
c) 2’s complement
d) All the above
Answer:
d) All the above

Question 51.
The statement “C equal the complement of A or B” means
(a) C = A + B
(b) C = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
(c) C = \(\overline{\mathrm{A}}\) + \(\overline{\mathrm{B}}\)
(d) C = \(\overline{\mathrm{A}\mathrm{B}}\)
Answer:
(a) C = A + B

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 52.
If the number has_______sign, it will be considered as negative in signed magnitude representation.
a) +
b) no
c) –
d) A or B
Answer:
c) –

Question 54.
What is the output of the XOR gate?
(a) C = A% B
(b) C = A \(\otimes\) A
(c) C = A \(\odot\) B
(d) C = A \(\oplus\) B
Answer:
(d) C = A \(\oplus\) B

Question 54.
2’s complement of (0001i000)2 is_______
a) 11100111
b) 00011001
c) 11101000
d) None of these
Answer:
c) 11101000

Question 55.
Find A + \(\overline{\mathrm{A}}\) .B = ………………..
(a) A + B
(b) A.B
(c) \(\overline{\mathrm{A}}\).B
(d) A.\(\overline{\mathrm{B}}\)
Answer:
(d) A.\(\overline{\mathrm{B}}\)

Question 56.
When two binary numbers are added _______will be the output.
a) sum
b) carry
c) sum and carry
d) None of these
Answer:
c) sum and carry

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 57.
When subtracting 1 from 0, borrow 1 from the next _______
a) LSB
b) MSB
c) either A or B
d) None of these
Answer:
b) MSB

Question 58.
Find the wrong pair from the following:
(a) Null element : A + 1 = 1
(b) Involution : \(\overset { = }{ A }\) = A
(c) Demorgan’s : \(\overline{\mathrm{A+B}}\) =\(\overline{\mathrm{A}}\) . \(\overline{\mathrm{A}}\)
(d) Commutative : A + B = B . A
Answer:
(d) Commutative : A + B = B . A

Question 59.
_______ is the character encoding system.
a) BCD and ISCII
b) EBCDIC
c) ASCII and Unicode
d) All the above
Answer:
d) All the above

Question 60.
With 2 inputs in the truth table, how many sets of values will be obtained.
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(a) 4

Question 61.
EBCDIC stands for _______
a) Extensive Binary Coded Decimal Interchange Code
b) Extended Binary Coded Decimal Interchange Code
c) Extended Binary Coded Digit Interchange Code
d) Extended Bit Coded Decimal Interchange Code.
Answer:
b) Extended Binary Coded Decimal Interchange Code

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 62.
ASCII stands for _______
a) Arithmetic Standard Code for Information Interchange
b) American Structured Code for Information Interchange
c) American Standard Code for Information Interchange
d) American Standard Code for Instant Interchange
Answer:
c) American Standard Code for Information Interchange

Question 63.
ISCII stands for_______
a) International Standard Code for Information Interchange
b) Indian Structured Code for Information Interchange
c) India’s Standard Code for Information Interchange
d) Indian Standard Code for Information Interchange
Answer:
d) Indian Standard Code for Information Interchange

Question 64.
BCD is _______bit code.
a) 6
b) 7
c) 8
d) None of these
Answer:
a) 6

Question 65.
EBCDIC is_______ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 66.
ASCII is________ bit code
a) 6
b) 7
c) 8
d) None of these
Answer:
b) 7

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 67.
Unicode is _______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
a) 16

Question 68.
ISCII is_______ bit code
a) 16
b) 7
c) 8
d) None of these
Answer:
c) 8

Question 69.
_______coding system is formulated by IBM.
a) BCD
b) EBCDIC
c) ISCII
d) None of these
Answer:
b) EBCDIC

Question 70.
IBM stands for_______
a) Indian Business Machine
b) International Basic Machine
c) International Business Method
d) International Business Machine
Answer:
d) International Business Machine

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 71.
_______is the system of handling the characters of Indian local languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
d) ISCII

Question 72.
ISCII system is formulated fay the _______ in India.
a) Department of Electronics
b) Department of Electricity
c) Department of E-commerce
d) Department of Economics
Answer:
a) Department of Electronics

Question 73.
SCO system can handle___________characterscharacters.
a) 64
b) 128
c) 256
d) 65536
Answer:
a) 64

Question 74.
EBCDIC system can handle _______ characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Question 75.
ASCII system can handle _______characters.
a) 64
b) 128
c) 256
d) None of these
Answer:
c) 256

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 76.
Unicode system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65536
Answer:
d) 65536

Question 77.
ISCII system can handle _______ characters.
a) 64
b) 128
c) 256
d) 65535
Answer:
c) 256

Question 78.
__________ language characters are not represented by ASCII.
a) Tamil
b) Malayalam
c) Telugu and Kannada
d) All the above
Answer:
d) All the above

Question 79.
Tamil, Malayalam, Telugu, and Kannada language characters are represented by _______ code.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 80.
_______scheme is denoted by hexadecimal numbers
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 81.
ISCII code was formulated in the year_______
a) 1986 – 88
b) 1984 – 86
c) 1988
d) 1987
Answer:
a) 1986 – 88

Question 82.
_______coding system is integrated with Unicode.
a) ASCII
b) EBCDIC
c) BCD
d) ISCII
Answer:
d) ISCII

Question 83.
_______was generated to handle all the coding system of Universal languages.
a) ASCII
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 84.
The popular coding scheme after ASCII is_______
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
b) Unicode

Question 85.
BCD system is_______bit encoding system.
a) 28
b) 216
c) 26
d) 24
Answer:
c) 26

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 86.
EBCDIC system is _______bit encoding system.
a) 28
b) 216
c) 26
d) 27
Answer:
a) 28

Question 87.
ASCII system is a bit encoding system.
a) 28
b) 216
c) 26
d) 27
Answer:
d) 27

Question 88.
Unicode system is _________bit encoding system.
a) 28
b) 216
c) 26
d) 27
Answer:
b) 216

Question 89.
ISCII svstem is _________bit encoding system.
a) 28
b) 216
c) 26
d) 27
Answer:
a) 28

Question 90.
The input code in ASCII can be converted into _________system.
a) EBCDIC
b) Unicode
c) BCD
d) ISCII
Answer:
a) EBCDIC

Question 91.
What is ASCII value for ‘A’ in a decimal number,
a) 97
b) 65
c) 98
d) 32
Answer:
b) 65

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 92.
What is the ASCII value for ‘A’ in a binary number?
a) 01100001
b) 01000001
c) 01100010
d) 00100000
Answer:
b) 01000001

Question 93.
What is the ASCII value for ‘A’ in an octal number?
a) 141
b) 101
c) 142
d) 40
Answer:
b) 101

Question 94.
What is the ASCII value for ‘A’ in hexadecimal numbers?
a) 61
b) 41
c) 62
d) 20
Answer:
b) 41

Question 95.
Find the false statement in the following.
a) Computers can handle positive and negative numbers.
b) MSB is called a sign bit
c) LSB is called a parity bit
d) All the above
Answer:
c) LSB is called a parity bit

Question 96.
Match the following.
a) 78 – (1) Binary number
b) linn – (2) Octal number
c) CAFE – (3) Decimal number
d) 71 – (4) Hexadecimal number

a) 3, 1, 4, 2
b) 4, 3, 2, 1
c) 1, 3, 2, 4
d) 3, 1, 2, 4
Answer:
a) 78 – (1) Binary number

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 97.
In signed magnitude representation,_________ in the sign bit represents negative number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
b) 1

Question 98.
In signed magnitude representation, __________in the sign bit represents positive number.
a) 0
b) 1
c) No symbol
d) None of these
Answer:
a) 0

Question 99.
The term data comes from the word __________
a) datum
b) date
c) fact
d) None of these
Answer:
a) datum

Part II

Very Short Answers.

Question 1.
What is nibble?
Answer:
Nibble is a collection of 4 bits. A nibble is half a byte.

Question 2.
Define information.
Answer:
Information is a processed fact and obtained from the computer as output. It conveys meaning.

Question 3.
What is radix?
Answer:
The base value of a number is also known as the radix.

Question 4.
Define Bit and Byte.
Answer:
Bit: A bit is the short form of a Binary digit which can be ‘0’ or ‘1’. It is the basic unit of data in computers.
Byte: A collection of 8 bits is called Byte. It is the basic unit of measuring the memory size in the computer.

Question 5.
Expand: BCD, EBCDIC, ASCII
Answer:
BCD – Binary Coded Decimal; EBCDIC – Extended Binary Coded Decimal Interchange Code; ASCII – American Standard Code for Information Interchange.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 6.
What are the different types of coding schemes to represent the character sets?
Answer:
The different coding schemes are

  • BCD – Binary Coded Decimal
  • EBCDIC – Extended Binary Coded Decimal Interchange Code
  • ASCII – American Standard Code for Information Interchange
  • Unicode
  • ISCII – Indian Standard Code for Information Interchange.

Question 7.
What are the methods of converting a number from decimal to binary?
Answer:

  1. Repeated division by two.
  2. Sum of powers of 2.

Question 8.
What does base or radix mean?
Answer:
Radix or base is the count of a number of digits in each number system. Radix or base is the general idea behind the positional numbering system.

Question 9.
What are the various ways for Binary representation of signed numbers?
Answer:

  1. Signed magnitude representation
  2. 1’s complement
  3. 2’s complement

Question 10.
Write a note on the 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.
Its base or radix is 10.

Question 11.
Write about the 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. Its base or radix is 8.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 12.
How will you convert decimal to hexadecimal?
Answer:
To convert Decimal to Hexadecimal, “Repeated division by 16” method can be used) In this method, we have to divide the given number by 16.
Example: Convert (31)10 into its equivalent hexadecimal number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 14

Question 13.
Give the procedure to Octal to Binary.
Answer:
Procedure: For each octal digit in the given number write its 3 digits binary equivalent using positional notation.
Example: Convert (6213)8 to equivalent Binary number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 15

Question 14.
How will you convert Hexadecimal to Binary?
Answer:
Procedure: Write 4 bits Binary equivalent for each Hexadecimal digit for the given number using the positional notation method.
Example:
Convert (8BC)16 into an equivalent Binary number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 16

Question 15.
Write short note on Binary Coded Decimal (BCD).
Answer:
This is 26 bit encoding system. This can handle 26 = 64 characters only. This encoding system is not
in the practice right now.

Question 16.
Write note on EBCDIC encoding system.
Answer:
Extended Binary Coded Decimal Interchange Code (EBCDIC) 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.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 17.
Write a note on the ISCII encoding system.
Answer:
ISCII is the system of handling the character of Indian local languages. This is an 8-bit coding system. Therefore 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 the Bureau of Indian Standards (BIS). Now, this coding system is integrated with Unicode.

Part III

III. Very Short Answers

Question 1.
Write about the binary number system.
Answer:
There are only two digits in the Binary system, namely, 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 leftmost bit in the binary number is called the Most Significant Bit (MSB) and it has the largest positional weight. The rightmost bit is the Least Significant Bit (LSB) and has the smallest positional weight.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 17

Question 2.
What is the octal number system?
Answer:
The 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.
Example: The Octal sequence (547)8 has the decimal equivalent:

Question 3.
Give the procedure to convert decimal to octal.
Answer:
To convert Decimal to Octal, “Repeated Division by 8” method can be used) In this method, we have to divide the given number by 8.
Example:
Convert (65)10 into its equivalent Octal number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 18

Question 4.
Give the procedure to convert Octal to Decimal
Answer:
To convert Octal to Decimal, we can use positional notation method)

  • Write down the Octal digits and list the powers of 8 from right to left (Positional Notation).
  • For each positional notation of the digit write the equivalent weight.
  • Multiply each digit with its corresponding weight.
  • Add all the values.

Example:
Convert (1265)8 to equivalent Decimal number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 19

(1265)8 = 512 x 1 + 64 x 2 + 8 x 6 + 1 x 5
= 512 + 128 + 48 + 5
(1265)8 = (693)10

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 5.
How will you convert Hexadecimal to Decimal?
Answer:
To convert Hexadecimal to Decimal we can use the positional notation method.

  • Write down the Hexadecimal digits and list the powers of 16 from right to left (Positional Notation)
  • For each positional notation written for the digit, now write the equivalent weight.
  • Multiply each digit with its corresponding weight
  • Add all the values to get one final value.

Example:
Convert (25F)16 into its equivalent Decimal number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 20
(25F)16 = 2 x 256 + 5 x 16 + 15 x 1
= 512 + 80 + 15 (25F)16
= (607)10

Question 6.
Write about binary representation for signed numbers.
Answer:
Computers can handle both positive (unsigned) and negative (signed) numbers. The simplest method to represent negative binary numbers is called Signed Magnitude. In signed magnitude method, the leftmost bit is the Most Significant Bit (MSB), which is called the sign bit or parity bit.
The numbers are represented in computers in different ways:

  • Signed Magnitude representation
  • 1’s Complement
  • 2’s Complement

Question 7.
Explain ASCII code in detail.
Answer:
This is the most popular encoding system recognized by the United States. Most of the computers use this system. Remember this encoding system can handle English characters only. This can handle 27 bit which means 128 characters.

In this system, each character has an individual number. The new edition ASCII -8, has 28 bits and can handle 256 characters are represented from 0 to 255 unique numbers.

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

Question 8.
Explain Unicode in detail.
Answer:
This coding system is used in most modern 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. Particularly there was a situation when the languages like Tamil, Malayalam, Kannada, and Telugu could not be represented by ASCII.

Hence, Unicode was generated to handle all the coding system of Universal languages. This is a 16-bit code and can handle 65536 characters. The unicode scheme is denoted by hexadecimal numbers.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Part IV

IV. Detail Answers.

Question 1.
Explain decimal to binary conversion using Repeated Division by 2 methods.
Answer:
To convert Decimal to Binary “Repeated Division by 2” method can be used. Any Decimal number divided by 2 will leave a remainder of 0 or 1. Repeated division by 2 will leave a sequence of 0s and Is that become the binary equivalent of the decimal number.

Suppose it is required to convert the decimal number N into binary form, dividing N by 2 in the decimal system, we will obtain a quotient N1 and a remainder Rl, where R1 can have a value of either 0 or 1. The process is repeated until the quotient becomes 0 or 1. When the quotient is ‘0’ or ‘1’, it is the final remainder value. Write the final answer starting from the final remainder value obtained to the first remainder value obtained.

Example:
Convert (65)10 into its equivalent binary number
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 21

Question 2.
Explain decimal to binary conversion using Sum of powers of 2 methods.
Answer:
A decimal number can be converted into a binary number by adding up the powers of 2 and then adding bits as needed to obtain the total value of the number.
a) Find the largest power of 2 that is smaller than or equal to 65.
6510 > 6410

b) Set the 64’s bit to 1 and subtract 64 from the original number
65 – 64 = 1

c) 32 is greater than the remaining total.
Therefore, set the 32’s bit to 0.

d) 16 is greater than the remaining total.
Therefore, set the 16’s bit to 0
.
e) 8 is greater than the remaining total.
Therefore, set the 8’s bit to 0.

f) 4 is greater than the remaining total.
Therefore, set the 4’s bit to 0.

g) 2 is greater than the remaining total.
Therefore, set the 2’s bit to 0.

h) As the remaining value is equivalent to l’s bit, set it to 1.
1 – 1 = 0
Conversion is complete 6510 = (1000001)2

Example:
The conversion steps can be given as follows:
Given Number: 65
Equivalent or value less than the power of 2 is: 64
(1) 65 – 64 = 1
(2) 1 – 1= 0
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 22
6510 = (1000001)2.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 3.
Explain the procedure to convert fractional decimal to Binary.
Answer:
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 Os and Is that becomes the binary equivalent of a decimal fraction.
Step 4: The final answer is to be written from the first integer part obtained till the last integer part obtained.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 23
Write the integer parts from top to bottom to obtain the equivalent fractional binary number. Hence (0.2)10= (0.00110011…)2 = 0.00110011)2

Question 4.
How will you convert Binary to Decimal?
Answer:
To convert Binary to Decimal we can use the 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.

Example:
Convert (111011)2 into its equivalent decimal number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 24
32 + 16 + 8 + 0 + 2 + 1 = (59)10
(111011)2 = (59)10

Question 5.
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 O to make a group of 3 bits if the leftmost group has less than 3 bits.
Step 3: Convert equivalent octal value using “2’s power positional weight method”

Example
Convert (11010110)2 into an octal equivalent number

Step 1: Group the given number into 3 bits from right to left.
011 010 110
The left-most groups have less than 3 bits, so 0 is added to its left to make a group of 3 bits.

Step-2: Find the Octal equivalent of each group.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 25

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 6.
Give the procedure to convert 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 leftmost group has less than 4 bits.
Step 3: Convert equivalent Hexadecimal value using “2’s power positional weight method”.

Example
Convert (1111010110)2 into Hexadecimal number
Step 1: Group the given number into 4 bits from right to left. 1
0011 1101 0110
0’s are added to the leftmost group to make it a group of 4 bits.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 26

Question 7.
Give the procedure to convert fractional Binary to Decimal equivalent.
Answer:
The steps to convert fractional Binary number to its decimal equivalent:
Step 1 : Convert an 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.

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

Example:
Convert the given Binary number (11.011)2 into its decimal equivalent Integer part (11)2 = 3
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 27

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 8.
Explain the method of representing signed binary numbers in the Signed Magnitude representation.
Answer:
The value of the whole numbers can be determined by the sign used before it. If the number has a ‘+’ sign or no sign it will be considered as positive. If the number has signed it will be considered negative.

Example:
+ 43 or 43 is a positive number
– 43 is a negative number

In signed binary representation, the leftmost bit is considered as a sign bit. If this bit is 0, it is a positive number and if it 1, it is a negative number. Therefore a signed binary number has 8 bits, only 7 bits used for storing values (magnitude), and 1 bit is used for signs.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 28

Question 9.
Explain the method of representing signed binary numbers in l’s complement representation.
Answer:
This is an easier approach to represent signed numbers. This is for negative numbers only i.e. the number whose MSB is 1.

The steps to be followed to find l’s complement of a number:

Step 1: Convert 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. Change 1 as 0 and 0 as 1)

Example: Find 1’s complement for (-24)10
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 29

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 10.
Explain the method of representing signed binary numbers in 2’s complement representation.
Answer:
The 2’s-complement method for the negative number is as follows:
a) Invert all the bits in the binary sequence (i.e., change every 0 to 1 and every 1 to 0 ie.,l’s complement)
b) Add 1 to the result to the Least Significant Bit (LSB).
Example: 2’s Complement represent of (-24)10

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 30

Question 11.
Explain binary addition with a suitable example.
Answer:
The following table is useful when adding two binary numbers.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 31
In 1 + 1 = 10, is considered as sum 0 and the 1 as carry bit. This carry bit is added with the previous position of the bit pattern.
Example: Add: 10112 + 10012
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 32

Example: Perform Binary addition for the following:
2310 + 1210
Step 1: Convert 23 and 12 into binary form
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 33

Step 2: Binary addition of 23 and 12:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 34

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 11.
Explain binary subtraction with a suitable example.
Answer:
The table for Binary Subtraction is as follows:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 35

When subtracting 1 from 0, borrow 1 from the next Most Significant Bit, when borrowing from the next Most Significant Bit, if it is 1, replace it with 0. If the next Most Significant Bit is 0, you must borrow from a more significant bit that contains 1 and replace it with 0 and 0s upto that point become Is.
Example : Subtract 10010102 — 101002.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 36
Example: Perform Binary addition for the
following:
(-21)10 + (5)10
Step 1: Change -21 and 5 into binary form
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 37 Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 38

Workshop

Question 1.
Identify the number system for the following numbers.
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 39

Question 2.
State whether the following numbers are valid or not. If invalid, give a reason.
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 40

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 3.
Convert the following Decimal numbers to their equivalent Binary, Octal, Hexadecimal.
i) 1920
ii) 255
iii) 126
Answer:
i) 1920
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 41

ii) 255
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 42

iii) 126
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 43

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 4.
Convert the given Binary number into its equivalent Decimal, Octal, and Hexadecimal numbers.
i)101110101
ii) 1011010
iii) 101011111
Answer:
i) 101110101
Binary to Decimal (Multiply by positional value and then add)
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 44

ii) 1011010
Binary to decimal
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 45 Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 46

iii) 101011111
Binary to decimal
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 47

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 5.
Convert the following Octal numbers into Binary numbers.
a) 472
b) 145
c) 347
d) 6247
e) 645
Answer:
Procedure: Write three digits binary number for every octal digit that will give the equivalent binary number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 48
Ans.
(472)8 = (100111010)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 49
Answer:
(145)8 = (001100101)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 50
Answer:
(347)8 = (011100111)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 51
Answer:
(6247)8 = (110010100111)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 52
Answer:
(645)8 = (110100101)2.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 6.
Convert the following Hexadecimal numbers to Binary numbers
a) A6
b) BE
c) 9BC8
d) BC9
Answer:
Procedure: Write four digits binary number for every Hexadecimal digit that will give the equivalent binary number.
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 53
Answer:
(A6)16 = (10100110)2.

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 54
Answer:
(BE)16 = (1011 1110)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 55
Answer:
(9BC8)16 = (1001101111001000)2

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 56
Answer:
(BC9)16 = (101111001001)2

Question 7.
Write the l’s complement number and 2’s complement number for the following decimal numbers:
Perform the following binary computations:
a) -22
b) -13
c) -65
d) -46
Answer:
a) -22
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 57

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 58
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 59

Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems

Question 8.
a) 1010 + 1510
b) – 1210 + 510
c) 1410 – 1210
d) (-2)10 – (-6)10
Answer:
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 60
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 61
Samacheer Kalvi 11th Computer Science Guide Chapter 2 Number Systems 62

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 2 Kinematics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 2 Kinematics

11th Physics Guide Kinematics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 1.
Which one of the following Cartesian coordinate systems is not followed in physics?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 1
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 2

Question 2.
Identify the unit vector in the following _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 3
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 4

Question 3.
Which one of the following quantities cannot be represented by a scalar?
a) Mass
b) length
c) momentum
d) magnitude of the acceleration
Answer:
c) momentum

Question 4.
Two objects of masses m1 and m2 fall from the heights h1 and h2 respectively. The ratio of the magnitude of their momenta when they hit the ground is _______. (AIPMT 2012)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 5
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 6

Question 5.
If a particle has negative velocity and negative acceleration, it speeds _______.
a) increases
b) decreases
c) remains the same
d) zero
Answer:
a) increases

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 6.
If the velocity is \(\overline{V}\) = \(2 \hat{i}+t^{2} \hat{j}-9 \hat{k}\) then the magnitude of acceleration at t = 0.5s is _______.
a) 1ms-2
b) 2 ms-2
c) zero
d) -1ms-2
Answer:
a) 1ms-2

Question 7.
If an object is dropped from the top of a building and it reaches the ground at t = 4s, then the height of the building is (ignoring air resistance) (g = 9.8ms-2)
a) 77.3m
b) 78.4m
c) 80.5
d) 79.2m
Answer:
b) 78.4m

Question 8.
A ball is projected vertically upwards with a velocity v. It comes back to the ground in time t. Which v-t graph shows the motion correctly? (NSEP 00-01)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 7
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 8

Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant t is _______.
a) 1
b) 2
c) 4
d) 0.5
Answer:
a) 1

Question 10.
A bail is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 9
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 10

Question 11.
If a particle executes uniform circular motion in the XY plane in a clockwise direction, then the angular velocity is in _______.
a) +y direction
b) +z direction
c) -z direction
d) -x direction
Answer:
c) -z direction

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 12.
If a particle executes uniform circular motion, choose the correct statement _______. (NEET 2016)
a) The velocity and speed are constant.
b) The acceleration and speed are constant.
c) The velocity and acceleration are constant.
d) The speed and magnitude of acceleration are constant
Answer:
d) The speed and magnitude of acceleration are constant

Question 13.
If an object is thrown vertically up with the initial speed u from the ground, then the time taken by the object to return back to the ground is _______.
(a) \(\frac{u^{2}}{2 g}\)
(b) \(\frac{u^{2}}{g}\)
(c) \(\frac { u }{ 2g }\)
(d) \(\frac { 2u }{ g }\)
Answer:
(d) \(\frac { 2u }{ g }\)

Question 14.
Two objects are projected at angles 30° and 60° respectively with respect to the horizontal direction. The range of two objects are denoted as R30° and R60° Choose the correct relation from the following
a) R30° = R60°
b) R30° = 4R60°
c) R30° = R\(\frac { 60° }{ 2 }\)
d) R30° = 2R60°
Answer:
a) R30° = R60°

Question 15.
An object is dropped in an unknown planet from a height of 50m, it reaches the ground in 2s. The acceleration due to gravity in this unknown planet is _______.
a) g = 20ms-2
b) g = 25ms-2
c) g = 15ms-2
d) g = 30ms-2
Answer:
b) g = 25ms-2

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

II. Short Answer Questions:

Question 1.
Explain what is meant by the Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates (x,y,z) is called “Cartesian coordinate system”.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 11
If x, y, and z axes are drawn in an anticlockwise direction, then the coordinate system is called a right-handed Cartesian coordinate system.

Question 2.
Define a vector. Give Example.
Answer:
Vector is a quantity which is described by both magnitude and direction. Geometrically a vector is a directed line segment.
Example – force, velocity, displacement.

Question 3.
Define a Scalar. Give Examples.
Answer:
Scalar is a property of a physical quantity which can be described only by magnitude.
Example: Distance, Mass, Temperature, Speed, Energy, etc.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 4.
Write short note on the scalar product between two vectors.
Answer:
The Scalar product of two vectors (dot product) is defined as the product of the magnitudes of both the vectors and the cosine of angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vectors having an angle θ between them, then their scalar or dot product is
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 12
Example: W = \(\vec{F}\).\(\vec{dr}\). Work done is a scalar product of force \(\vec{F}\) and \(\vec{r}\)

Question 5.
Write a Short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if\(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, then their vector product is written as \(\overrightarrow{\mathrm{A}}\) × \(\overrightarrow{\mathrm{B}}\) which is a vector C defined by \(\overrightarrow{\mathrm{c}}\) = \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) = (AB sin 0) \(\hat{n}\)
The direction \(\hat{n}\) of \(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) , i.e., \(\overrightarrow{\mathrm{c}}\) is perpendicular to the plane containing the vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\).

Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If the vector product of the two given vectors is having maximum magnitude.
i.e sinθ = 90°, [ (\(\vec{A}\) x \(\vec{B}\))Max = AB\(\hat{n}\) ] then the two vectors are said to be perpendicular.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 7.
Define Displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.

Question 8.
Define velocity and speed.
Answer:
Velocity – Velocity is defined as the rate of change of position vector with respect to time (or) defined as the rate of change of displacement. It Is a vector quantity.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 13

Speed – Speed is defined as the rate of change of distance. It is a scalar quantity.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 14

Question 9.
Define acceleration.
Answer:
Acceleration is defined as the rate of change of velocity.
Acceleration \(\vec{a}\) = \(\frac{d \vec{v}}{d t}\)
Acceleration is a vector quantity.
Unit – ms-2
Dimensional formula-[LT-2]

Question 10.
What is the difference between velocity and average velocity?
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 15

Question 11.
Define a radian.
Radian is defined as ratio of length of the arc to radians of the arc. One radian is the angle subtended at the center of the circle by an arc that is equal to in length to the radius of the circle.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 16

Question 12.
Define angular displacement and angular velocity.
Answer:

  1. Angular displacement: The angle described by the particle about the axis of rotation in a given time is called angular displacement.
  2. Angular velocity: The rate of change of angular displacement is called angular velocity.

Question 13.
What is non-uniform circular motion?
Answer:
When an object is moving in a circular path with variable speed, it covers unequal distances in equal intervals of time. Then the motion of the object is said to be a non-uniform circular motion. Here both speed and direction during circular motion change.

Question 14.
Write down the kinematic equations for angular motion.
Answer:
The Kinematic equations for angular motion are ω = ω0 + αt
θ = ω0t + \(\frac { 1 }{ 2 }\)αt²
ω² = ω0² + 2αθ
θ = \(\left(\frac{\omega_{0}+\omega}{2}\right)\) x t
ω0 → initial angular velocity
ω → final angular velocity
α → angular acceleration
θ → angular displacement
t → time interval

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non-uniform circular motion.
Answer:
In the case of non-uniform circular motion, the particle will have both centripetal and tangential acceleration. The resultant acceleration is obtained as the vector sum of both centripetal and tangential acceleration.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 17
This resultant acceleration makes an angle 6 with a radius vector, which is given by
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 18

III. Long Answer Questions:

Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vector \(\vec{A}\) and \(\vec{B}\) as shown In fig.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 19
Law: To find the resultant of two vectors, the triangular law of addition can be applied as follows.
A and B are represented as the two adjacent sides of a triangle taken in the same order. The resultant is given by the third side of the triangle taken in reverse order.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 20
Magnitude of the resultant vector:
from figure
Let θ be the angle between two vectors.
from ∆ ABN, Sin θ = \(\frac { BN }{ AB }\) ⇒ ∴ BN = B sinθ
Cos θ = \(\frac { AN }{ AB }\) ⇒ ∴ AN = B Cos θ
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 21
Which is the magnitude of the resultant \(\vec{A}\) and \(\vec{B}\).

The direction of the resultant vector:
If \(\vec{R}\) makes an angle α with \(\vec{A}\) then
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 22

Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product:
formula : \(\vec{A}\).\(\vec{B}\) = ABCosθ

1. The product quantity \(\overline{A}\).\(\overline{B}\) is always a scalar. It is positive if the angle between the vectors is acute (θ< 90°) and negative if angle between them is obtuse (90 < θ < 180)

2. The scalar product is commutative \(\overline{A}\).\(\overline{B}\) = \(\overline{B}\).\(\overline{A}\)

3. The scalar product obey distributive law. \(\overline{A}\).( \(\overline{B}\) + \(\overline{C}\) ) = \(\overline{A}\).\(\overline{B}\) + \(\overline{A}\).\(\overline{C}\)

4. The angle between the vector is θ = Cos-1\(\frac{\bar{A} \cdot \bar{B}}{A B}\)

5. The scalar product of two vectors will be maximum when cos θ = 1 i.e θ = 0 ie when they are parallel.
[ ( \(\overline{A}\).\(\overline{B}\) ) max = AB.]

6. The scalar product of two vectors will be minimum when cos θ = -1 ie θ = 180°
( \(\overline{A}\).\(\overline{B}\))mm = – AB [the vector are anti-parallel]

7. If two vector \(\overline{A}\) & \(\overline{B}\) are perpendicular to each other then \(\overline{A}\).\(\overline{B}\) = O. Because cos 90 = 0. Then vectors A & B are mutually orthogonal.

8. The scalar product of a vector with it self is termed as self or dot product and is given by
( \(\overline{A}\) )² = \(\overline{A}\).\(\overline{A}\) = AA cos θ = A²
Here 0=0
The magnitude or norm of the vector \(\overline{A}\) is
|A| = A = \(\sqrt{\bar{A} \cdot \bar{A}}\) = A.

9. Incase of orthogonal unit vectors
\(\hat{n}\).\(\hat{n}\) = 1 x 1cos0 = 1
for eg \(\hat{i}\).\(\hat{i}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 1

10. Incase of orthogonal unit vectors \(\hat{i}\), \(\hat{f}\), \(\hat{k}\) then \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{k}\) = \(\hat{k}\).\(\hat{j}\) = 1.1 cos 90 = 0.

11. In terms of components the scalar product of A and B can be written as
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 23

Properties of cross product:
Formula \(\vec{A}\) x \(\vec{B}\) = ABsinθ

1. The vector product of any two vectors is always an another vector whose direction perpendicular to the plane containing these two vectors, ie. Orthogonal to \(\overline{A}\) & \(\overline{B}\) even though \(\overline{A}\) & \(\overline{B}\) may not be mutually orthogonal.

2. Vector product is not commutative
\(\overline{A}\) x \(\overline{B}\) = – \(\overline{B}\).\(\overline{A}\)
\(\overline{A}\) x \(\overline{B}\) ≠ \(\vec{B}\) x \(\vec{A}\)
Here magnitude | \(\overline{A}\) x \(\overline{B}\) | = | \(\overline{B}\).\(\overline{A}\) | are equal but opposite direction.

3. The vector product of two vector is maximum when sine = 1, ie θ = 90°
ie. when \(\overline{A}\) and \(\overline{B}\) are orthogonal to each other.
( \(\overline{A}\) x \(\overline{B}\) ) max = AB \(\hat{n}\).

4. The vector product of two non zero vectors is minimum if |sinθ| = 0. ie. θ = 0 or 180°
( \(\overline{A}\) x \(\overline{B}\) ) m in = 0
Vector product of two non zero vectors is equal to zero if they either parallel or anti parallel

5. The self cross product ie product of a vector with itself is a null vector \(\overline{A}\) x \(\overline{B}\) = AA sinθ = 0

6. The self-vector product of the unit vector is zero
i.e. \(\hat{i}\).\(\hat{j}\) = \(\hat{j}\).\(\hat{j}\) = \(\hat{k}\).\(\hat{k}\) = 0

7. In case of orthogonal unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) in accordance with right hand cork screw rule \(\hat{i}\).\(\hat{j}\) =\(\hat{k}\), \(\hat{i}\).\(\hat{k}\) = \(\hat{i}\), \(\hat{k}\).\(\hat{i}\) = \(\hat{j}\) also since cross product is not commutative
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 24

9. If two vectors \(\overline{A}\) & \(\overline{B}\) form adjacent sides of a parallelogram then the magnitude of |\(\overline{A}\) x \(\overline{B}\)| will give area 0f parallelogram.

10. Since one can divide a parallelogram into two equal triangles, the area of the triangle is \(\frac { 1 }{ 2 }\) |\(\overline{A}\) x \(\overline{B}\)|.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 3.
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the initial velocity at t = 0, and v be the final velocity after a time of t seconds

(i) Velocity time relation:
The acceleration of the body at any instant is given by first derivative of the velocity with time
a = \(\frac { dv }{ dt }\)
dv = adt
integrating both sides
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 25

Displacement time relation:

(ii) The velocity of the body is given by the first derivative of the displacement with respect to time
But v = ds/dt
∴ dv = v dt
v = u + at
ds = (u + at)dt
ds = udt + atdt
Integrating both sides
 Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 26

Velocity-displacement relation:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 27
also we can derive from the relation v = u + at
v – u = at
Substituting in equation s = ut + \(\frac { 1 }{ 2 }\)at²
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 28

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
For a body falling vertically from a height ‘h’:
Consider an object of mass m falling from height h.
Neglecting air resistance, the downward direction as the positive y-axis.
The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the earth.
In kinematic equations of motion \(\vec{a}\) = g\(\hat{i}\)
By comparing the components ax = 0, ag = 0, ay= g
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 29
Case – 1
If the particle is thrown with initial velocity ‘u’ downward then
v = u + gt
y = ut + 1/2gt²
v² – u² = 2gy

Case – 2
Suppose the particle starts from rest,
u = 0
v = gt
y = 1/2gt²
v² = 2gy
For a body projected vertically: Consider an object of mass m thrown vertically upwards with an initial velocity u. Ne-glect air friction. The vertical direction as positive y axis then the acceleration,
a = – g
The kinematic equation of motion are v = u – gt
v = u – gt
s = ut – 1/2 gt²
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 30

Projectile motion calculator solving for vertical velocity at time given initial vertical velocity, acceleration of gravity and time.

Question 5.
Derive the equations of motion, range, and maximum height reached by a particle thrown at an oblique angle θ with respect to the horizontal direction.
Answer:
Consider an object thrown with an initial velocity u at an angle θ with horizontal.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 31
Then initial velocity is resolved into two components
ux = u cos θ horizontally and
uy = u sin θ vertically
At maximum height uy = 0 (since acceleration due to gravity is opposite to the direction of the vertical component).
The Horizontal component of velocity
ux = u cos θ remains constant throughout its motion.
hence after the time t the velocity along the horizontal motion
Vx = Ux + axt
= ux = cos θ
The horizontal distance travelled by the projectile in a time ‘t’ is Sx = uxt + 1/2 axt².
Here Sx = x ux = u cos θ
ax = 0
∴ x = u cos θt ____ (1)
∴ t = \(\frac { x }{ u cos θ }\) ____ (2)
For vertical motion
Vy = uy + ayt
Here vy = vy
uy = u sin θ
ay = – g
vy = u sin θ – gt
The vertical distance travelled by the projectile in the same time ‘t’ is
Sy = Uy t + ay
Sy = y, Uy = u sin θ ay = – g
y = u sin θ t – 1/2 gt² ____ (4)
Substituting the value of t in (4) we get equation:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 32
Which indicates the path followed by the projectile is an inverted parabola.

Expression for Maximum height:
The maximum vertical distance travelled by the projectile during its motion is called maximum height.
We know that
vy² = uy² + 2ays
Here uy = u sin 0, ay = – g, s = hmax
vy = 0
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 33

Expression for horizontal range:
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range.
Horizontal range = Horizontal component of velocity x time of flight
R = u cos θ x tf → (1)
Time of flight (tf) is the time taken by the projectile from point of projection to point the projectile hits the ground again
w.k.t = Sy = uy tf + 1/2 ayf)
Here Sy = 0 uy = u sin θ, ay = – g
0 = u sin θ tf – 1/2g t²f
1/2 gt t²f = u sin θ tf
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 34

Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously with out changing its magnitude. ie speed remains constant and direction changes. Even though the velocity is tangential to every point is a circle, the acceleration it acting towards the centre of the circle along the radius. This is called centripetal acceleration

Expression:
The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors. Let the directions of position and velocity vectors shift through same angle θ in a small time interval ∆t
For uniform circular motion r = \(\left|\bar{r}_{1}\right|\) = \(\left|\bar{r}_{2}\right|\)
and v = \(\left|\bar{v}_{1}\right|\) = \(\left|\bar{v}_{2}\right|\)
If the particle moves from position vector \(\bar{r}_{1}\) to \(\bar{r}_{2}\) the displacement is given by \(\overrightarrow{\Delta r}\) = \(\bar{r}_{2}\) – \(\bar{r}_{1}\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 35
and change in velocity from \(\bar{v}_{1}\) to \(\bar{v}_{2}\) is given ∆\(\bar { v }\) = \(\bar{v}_{2}\) – \(\bar{v}_{1}\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 36
The magnitudes of the displacement ∆r and ∆v satisfy the following relation
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 37
Here negative sign indicates that ∆v points radially inwards, towards the centre of the circle
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 38
For uniform circular motion v=rω where ω is the angular velocity of the particle about the center
The centripetal acceleration a = ω²r.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 7.
Derive the expression for total acceleration in the non-uniform circular motion.
Answer:
If the velocity changes both in speed and direction during circular motion, then we get non-uniform circular motion. Whenever the speed is not the same in a circular motion then the particle will have both centripetal and tangential acceleration.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 39
The resultant acceleration is obtained by the vector sum of centripetal and tangential acceleration
Let the tangential acceleration be at.
Centripetal acceleration is v²/r.
The magnitude of the resultant acceleration is aR = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

IV. Exercises:

Question 1.
The position vector particle has a length of 1m and makes 30° with the x-axis what are the lengths of x and y components of the position vector?
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 40

Question 2.
A particle has its position moved from \(\left|\bar{r}_{1}\right|\) = 3\(\hat{i}\) + 4\(\hat{j}\) to r\(\left|\bar{r}_{2}\right|\) = \(\hat{i}\)+ 2\(\hat{j}\) calculate the displacement vector ( ∆ \(\vec{r}\) ) and draw the \(\left|\bar{r}_{1}\right|\), \(\left|\bar{r}_{2}\right|\) and ( ∆ \(\vec{r}\) ) vector in a two dimensional Cartesian co-ordinate system.
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 41

Question 3.
Calculate the average velocity of the particle whose position vector changes from \(\left|\bar{r}_{1}\right|\) = 5\(\hat{i}\) + 6\(\hat{j}\) to \(\left|\bar{r}_{2}\right|\) = 2\(\hat{i}\) + 3\(\hat{j}\) in a time 5 seconds.
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 42

Question 4.
Convert the vector \(\overline{r}\) = 3\(\hat{i}\) + 2\(\hat{j}\) into a unit vector.
Solution:
A vector divided by its magnitude is a unit vector
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 43

Question 5.
What are the resultants of the vector product of two given vector given by
\(\overline{A}\) = 4\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) and \(\overline{B}\) = 5\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\)?
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 44

Question 6.
An object at an angle such that the horizontal range is 4 times the maximum height. What is the angle of projection of the object?
Solution:
Incase of obliging projection
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 45

Question 7.
The following graphs represent velocity-time graph. Identify what kind of motion a particle undergoes in each graph.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 46
Solution:
(a) When the body starts from rest and moves with uniform acceleration is constant
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 47
(b) This graph represents, for a body moving with a uniform velocity or constant velocity. The zero slope of curve indicates zero acceleration.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 48
(c) This v-t graph is a straight line not passing through origin indicates the body has a constant acceleration but greater than fig(i) as slope is more than the first one (more steeper)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 49
(d) Greater changes in velocity (velocity variations are taking place in equal as travels of time. The graph indicates increasing acceleration.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 50

Question 8.
The following velocity-time graph represents a particle moving in the positive x-direction. Analyse its motion from o to 7s calculate the displacement covered and distance traveled by the particle from 0 to 2s.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 51
Solution:
From o to A(o to Is):
At t = os the particle has a zero velocity at t > 0 the particle has a negative velocity and moves in positive x-direction the slope dr/dt is negative. The particle is decelerating. Thus the velocity decreases during this time interval.

From A to B (Is to 2s):
From time Is to 2s the velocity increase and slope dv/dt becomes positive. The particle is accelerating. The velocity increases in this time interval.

From B to C (2s to 5s):
From 2s to 5s the velocity stays constant at 1 m/s. The acceleration is zero.

From C to D (6s to 7s):
From 5s to 6s the velocity decreases. Slope dv/dt is negative. The particle is decelerating. The velocity decreases to zero. The body comes to rest at 6s.

From D to E (6s to 7s)
The particle is at rest during this time interval.

Displacement: in 0 – 2s:
The total area under the curve from 0 to 2s displacement = 1/2bh + 1/2bh
=1/2 x 1.5 x (- 2) + 0.5 x 1
= – 1.5 + 0.25
= – 1.25 m

Distance: is 0 – 2s
The distance covered is = 1.5 + 0.25 = 1.75 m

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 9.
A particle is projected at an angle of θ with respect to the horizontal direction. Match the following for the above motion.
(a) vx – decreases and increases
(b) Vy – remains constant
(c) Acceleration – varies
(d) Position vector – remains downwards
Solution:
(a) Vx – remains constant
(b) Vy – decreases and increases
(c) Acceleration (a) – remains downwards
(d) Position vector (r) – varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountains is V. Calculate the total area around the fountain that gets wet.
Solution :
Speed of water = V
(Range)max = radius = u2/g = v²/g
This range becomes the radius = (v²/g) of the circle where water sprinkled.
Area covered = Area of circle
= πr² = π\(\left(\frac{v^{2}}{g}\right)\)²
= π \(v^{4} / g^{2}\)

Question 11.
The following table gives the range of the particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity (g value)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 52
Solution:
R = – (sin 2θ)
∵ the initial velocity and angle of projection are constants
R ∝ \(\frac { 1 }{ g }\)
g ∝ \(\frac { 1 }{ R }\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 53
According to acceleration, due to gravity In ascending order, the solution is. Mercury, Mars, Earth, Jupiter

Question 12.
The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B. Then the angle between A and B is
(a) 30°
(b) 45°
(c) 150°
(d)120°
Solution:
Let two vectors be A & B
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 54
Magnitude of B = B
Magnitude of A = A
∝ = 90°
Given:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 55

Question 13.
Compare the components for the following vector equations.
(a) T\(\hat{j}\) – mg\(\hat{j}\) = ma\(\hat{j}\)
(b) \(\overline{T}\) + \(\overline{F}\) = \(\overline{A}\) + \(\overline{B}\)
(c) \(\overline{T}\) – \(\overline{F}\) = \(\overline{A}\) – \(\overline{B}\)
(d) T\(\hat{j}\) + mg\(\hat{j}\) = ma\(\hat{j}\)
Solution:
We can resolve all vectors in x, y, z components w.r.t. Cartesian co-ordinate system. After resolving the components separately equate x components on both sides y components on both sides and z components on both side we get.
(a) T\(\hat{j}\) – mg\(\hat{j}\) = ma\(\hat{j}\)
T – mg = ma

(b) \(\overline{T}\) + \(\overline{F}\) = \(\overline{A}\) + \(\overline{B}\)
Tx + Fx = Ax + Bx

(c) \(\overline{T}\) – \(\overline{F}\) = \(\overline{A}\) – \(\overline{B}\)
Tx – Fx = Ax – Bx

(d) T\(\hat{j}\) + mg\(\hat{j}\) = ma\(\hat{j}\)
T + mg = ma

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors \(\overline{A}\) = 5\(\hat{i}\) – 3\(\hat{j}\) \(\overline{B}\) = 4\(\hat{i}\) + 6\(\hat{j}\).
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 56

Question 15.
If the earth completes one revolution in 24 hours, what is the angular displacement made by the earth in one hour? Express your answer in both radian and degree.
Solution:
ω = θ/t θ = wt
In 24 hours, angular displacement made
θ = 360° (or) 2π rad
In 1 hours, angular displacement
θ = \(\frac { 360° }{ 24 }\)
θ = 15°
In radian θ = \(\frac { 2π }{ 24 }\) = \(\frac { π }{ 12 }\) radians.
θ = \(\frac { π }{ 12 }\) rad.

Question 16.
An object is thrown with initial speed of 5ms-1 with an angle of projection of 30°. What is the height and range reached by the particle?
Solution:
u = 5 m/s
θ = 30°
hmax = ?
R = ?
Height reached
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 57

Range:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 58

Question 17.
A football player hits the ball with a speed 20m/s with angle 30° with respect to as shown in the figure horizontal directions. The goal post is at a distance of 40 m from him. Find out whether the ball reaches the goal post.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 59
Solution :
In order to find whether the ball is reaching the goal post the range should be equal to 40m so range
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 60
= \(\frac { 692.8 }{ 19.6 }\)
= 35.35 m.
Which is less than the distance of the goal post which is 40 m away so the ball won’t reach the goal post.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 18.
If an object is thrown horizontally with an initial speed 10 ms-1 from the top of a building of height 100 m. What is the horizontal distance covered by the particle?
Solution:
u = 10 m/s
h = 100 m
x = ?
x = u x T
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 61
x = 45.18 m.

Question 19.
An object is executing uniform circular motion with an angular speed of π/12 radians per second. At t = 0 the object starts at an angle θ = 0. What is the angular displacement of the particle after 4s?
Solution :
ω = π/12 rad/s
ω = θ/t
θ = w x t = π/12 x 4
θ = π/3 radian
θ = \(\frac { 180° }{ 3 }\)
= 60°

Question 20.
Consider the x-axis as representing east, the y-axis as north, and the z-axis as vertically upwards. Give the vector representing each of the following points.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 62
(a) 5m northeast and 2m up.
(b) 4m southeast and 3m up.
(c) 2m northwest and 4m up.
Solution:
5m northeast and 2m up.
(a) The vector representation of 5m N-E and 2m up is (5i + 5j) Cos 45° + 2\(\hat{k}\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 63

(b) 4m south east and 3m up.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 64
The vector representing 4m south east and 3m up is
(4i – 4j) cos 45 + 3\(\hat{k}\)
\(\frac{4(i-j)}{\sqrt{2}}\) + 3\(\hat{k}\)

(c) 2m north west and 4m up.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 65
The vector representing 2m northwest and 4m up
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 66

Question 21.
The moon is orbiting the earth approximately once in 27 days. What is the angle transversed by the moon per day?
Solution :
Angle described in 27 days = 2π rad = 360° days
Angie described in one day = 2π/27 radian
= \(\frac { 360° }{ 27 }\)
θ = 13.3°

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 22.
An object of mass m has an angular acceleration ∝ = 0.2 rad/s². What is the angular displacement covered by the object after 3 seconds? (Assume that the object started with angle zero with zero angular velocity)
Solution:
∝ = 0.2 rad/s²
θ = ? t = 3s.
w0 = 0
w.k.T θ = ω0t + 1/2 ∝ t²
θ = 0 + 1/2 x 0.2 x 9
θ = 0.9 rad
θ = 0 = 0.9 x 57.295° = 51°
The magnitude of the resultant vector R is given by
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 67

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

11th Physics Guide Kinematics Additional Important Questions and Answers

I. Multiple choice questions:

Question 1.
A particle moves in a circle of radius R from A to B as in the figure.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 68
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 69

Question 2.
The branch of mechanics which deals with the motion of objects without taking force into account is –
(a) kinetics
(b) dynamics
(c) kinematics
(d) statics
Answer:
(c) kinematics

Question 3.
A particle moves in a straight line from A to B with speed v1 and then from B to A with speed v2. The average velocity and average speed are _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 70
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 71

Question 4.
A particle is moving in a straight line under constant acceleration. It travels 15m in the 3rd second and 31m in the 7th second. The initial velocity and acceleration are _______.
a) 5 m/s, 4 m/s²
b) 4 m/s, 5 m/s²
c) 4 m/s, 4 m/s²
d) 5 m/s, 5 m/s²
Answer:
a) 5 m/s, 4 m/s²

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 5.
If an object is moving in a straight line then the motion is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(a) linear motion

Question 6.
A car is moving at a constant speed of 15 m/s. Suddenly the driver sees an obstacle on the road and takes 0.4 s to apply the brake. The brake causes a deceleration of 5 m/s². The distance traveled by car before it stops _______.
a) 6 m
b) 22.5 m
c) 28.5 m
d) 16.2 m
Answer:
c) 28.5 m

Question 7.
A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate (3 to come to rest. If the total time lapses in ‘t’ seconds, then the maximum velocity reached is _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 72
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 73

Question 8.
Spinning of the earth about its own axis is known as –
(a) linear motion
(b) circular motion
(c) curvilinear motion
(d) rotational motion
Answer:
(d) rotational motion

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 9.
A particle is thrown vertically up with a speed of 40m/s, The velocity at half of the maximum height _______.
a) 20 m/s
b) 20\(\sqrt{2}\)m/s
c) 10 m/s
d) 10\(\sqrt{2}\)m/s
Answer:
b) 20\(\sqrt{2}\)m/s

Question 10.
The ratio of the numerical values of the average velocity and the average speed of the body is always _______.
a) unity
b) unity or less
c) unity or more
d) less than unity
Answer:
b) unity or less

Question 11.
The motion of a satellite around the earth is an example for –
(a) circular motion
(b) rotational motion
(c) elliptical motion
(d) spinning
Answer:
(a) circular motion

Question 12.
One car moving on a straight road covers one-third of the distance with 20 km/h and the rest with 60 km/h. The average speed is _______.
a) 40 km/h
b) 80km/h
c) 46\(\frac { 2 }{ 3 }\) km/hr
d) 36 km/h
Answer:
d) 36 km/h

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 13.
A 150m long train is moving with a uniform velocity of 45 km/h. The time taken by the train to cross a bridge of length 850m is _______.
a) 56s
b) 68s
c) 80s
d) 92s
Answer:
c) 80s

Question 14.
A particle moves in a straight line with constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance of 135 m in ‘t’ seconds. The value of t is _______.
a) 12s
b) 9s
c) 10s
d) 1.8s
Answer:
b) 9s

Question 15.
If a ball is thrown vertically upwards with a speed u the distance covered during the last ‘t’ seconds of its ascent is _______.
a) 1/2 gt²
b) ut – 1/2gt²
c) (u – gt)t
d) ut
Answer:
a) 1/2 gt²

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 16.
A particle moves along a straight line such that its displacement ‘s’ at any time ‘t’ is given by s = t3 – 6t² + 3t + 4 meters, t being in second. The velocity when acceleration is zero is _______.
a) 3 m/s
b) -12m/s
c) 42 m/s
d) -9 m/s
Answer:
d) – 9 m/s

Question 17.
Which of the following is not a scalar?
(a) Volume
(b) angular momentum
(c) Relative density
(d) time
Answer:
(b) angular momentum

Question 18.
Vector is having –
(a) only magnitude
(b) only direction
(c) bot magnitude and direction
(d) either magnitude or direction
Answer:
(c) both magnitude and direction

Question 19.
The displacement-time graph of a moving particle is shown below.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 74
The instant velocity of the particle is negative at the point
a) D
b) F
c) C
d) E
Answer:
d) E

Question 20.
If two vectors are having equal magnitude and the same direction is known as –
(a) equal vectors
(b) col-linear vectors
(c) parallel vectors
(d) on it vector
Answer:
(a) equal vectors

Question 21.
The velocity-time graph of a body moving in a straight line is shown below _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 75
Which are of the following represents its acceleration-time graph?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 76
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 77

Question 22.
Indicate which of the following graph represents the one-dimensional motion of particle?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 78
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 79

Question 23.
The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in 4s is _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 80
a) 60m
b) 55m
c) 25m
d) 30m
Answer:
b) 55m

Question 24.
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s), velocity (v) graph of this object is _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 81
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 82

Question 25.
A unit vector is used to specify –
(a) only magnitude
(b) only direction
(c) either magnitude (or) direction
(d) absolute value
Answer:
(b) only direction

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 26.
A vector is not changed if _______.
a) It is rotated through an arbitrary angle
b) It is multiplied by an arbitrary scalar
c) It is cross multiplied by a unit vector
d) It is parallel to itself.
Answer:
d) It is parallel to itself.

Question 27.
Two forces each of magnitude ‘F’ have a resultant of the same magnitude. The angle between two forces
a) 45°
b) 120°
c) 150°
d) 60°
Answer:
b) 120°

Question 28.
The magnitude of a vector can not be-
(a) positive
(b) negative
(e) zero
(cl) 90
Answer:
(b) negative

Question 29.
Six vectors \(\vec{a}\) through \(\vec{f}\) have magnitudes and directions as indicated in figure. Which of the following statement is true?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 83
a) \(\overline{b}\) + \(\overline{e}\) = \(\overline{f}\)
b) 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) \(\hat{b}\) + \(\hat{c}\) = \(\hat{f}\)
c) \(\hat{d}\) + \(\hat{c}\) = \(\hat{f}\)
d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)
Answer:
d) \(\hat{d}\) + \(\hat{e}\) = \(\hat{f}\)

Question 30.
A force of 3 N and 4 N are acting perpendicular to an object, the resultant force is-
(a) 9 N
(b) 16 N
(c) 5 N
(d) 7 N
Answer:
(c) 5 N

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 31.
The figure shows ABCDEF as regular hexagon. What is the value of
\(\overline{AB}\) + \(\overline{AC}\) + \(\overline{AD}\) + \(\overline{AE}\) + \(\overline{AF}\)?
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 84
a) \(\overline{A0}\)
b) 2 \(\overline{A0}\)
c) 4 \(\overline{A0}\)
d) 6 \(\overline{A0}\)
Answer:
d) 6 \(\overline{A0}\)

Question 32.
One of the two rectangular components of a force is 20N. And it makes an angle of 30° with the force. The magnitude of the other component is _______.
a) 20/\(\sqrt{3}\)
b) 10/\(\sqrt{3}\)
c) 15/V\(\sqrt{3}\)
d) 40\(\sqrt{3}\)
Answer:
a) 20/\(\sqrt{3}\)

Question 33.
The angle between (\(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)) and (\(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)) can be –
(a) only 0°
(b) only 90°
(c) between 0° and 90°
(d) between 0° and 180°
Answer:
(d) between 0° and 180°

Question 34.
If the sum of two unit vectors is a unit vector the magnitude of the difference is _______.
a) \(\sqrt{2}\)
b) \(\sqrt{3}\)
c) 1/\(\sqrt{2}\)
d) \(\sqrt{5}\)
Answer:
b) \(\sqrt{3}\)

Question 35.
If P = mV then the direction of P along-
(a) m
(b) v
(c) both (a) and (b)
(d) neither m nor v
Answer:
(b) v

Question 36.
If \(\overline{A}\) = 2\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\), \(\overline{B}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) and \(\overline{C}\) = 6\(\hat{i}\) – 2\(\hat{j}\) – 6\(\hat{k}\) then angle between \(\overline{A}\) + \(\overline{B}\) and \(\overline{C}\) will be _______.
a) 30°
b) 45°
c) 60°
d) 90°
Answer:
d) 90°

Question 37.
The scalar product \(\overrightarrow{\mathrm{A}}\).\(\overrightarrow{\mathrm{B}}\)is equal to-
(a) \(\overrightarrow{\mathrm{A}}\) +\(\overrightarrow{\mathrm{B}}\)
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)
(c) AB sin θ
(d) (\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\)
Answer:
(b) \(\overrightarrow{\mathrm{A}}\). \(\overrightarrow{\mathrm{B}}\)

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 38.
If \(\overline{A}\) x \(\overline{B}\) = \(\overline{C}\) then which of the following statement is wrong?
a) \(\overline{C}\) ⊥\(\overline{A}\)
b) \(\overline{B}\) ⊥\(\overline{B}\)
c) \(\overline{C}\) ± ( \(\overline{A}\) + \(\overline{B}\) )
d) \(\overline{C}\) ± ( \(\overline{A}\) x \(\overline{B}\) )
Answer:
d) \(\overline{C}\) ± ( \(\overline{A}\) x \(\overline{B}\) )

Question 39.
The scalar product of two vectors will be minimum. When θ is equal to –
(a) 0°
(b) 45°
(c) 180°
(d) 60°
Answer:
(c) 180°

Question 40.
If | \(\overline{A}\) x \(\overline{B}\) |, then value of | \(\overline{A}\) x \(\overline{B}\) | is _______.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 85
Answer:
d) (A² + B² + AB)\(\frac { 1 }{ 2 }\)

Question 41.
The angle between vectors \(\overline{A}\) and \(\overline{B}\) is A. The value of the triple product \(\overline{A}\) ( \(\overline{A}\) x \(\overline{B}\) ) is _______.
a) A² B
b) zero
c) A² B sinθ
d) A² B cos θ
Answer:
d) A² B cos θ

Question 42.
Two adjacent sides of a parallelogram are represented by the two vectors \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) and 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\). The area parallelogram _______.
a) 8
b) 8\(\sqrt{3}\)
c) 3\(\sqrt{8}\)
d) 192
Answer:
b) 8\(\sqrt{3}\)

Question 43.
If \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) are two vectors, which are acting along x, y respectively, then \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) lies along-
(a) x
(b) y
(c) z
(d) none
Answer:
(c) z

Question 44.
Galileo writes that for angles of the projectile (45 + θ) and (45 – θ) the horizontal ranges described by the projectile are in the ratio of (if θ ≤ 45)
a) 2:1
b) 1:2
c) 1:1
d) 2:3
Answer:
c) 1:1

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 45.
A projectile is thrown into the air so as to have the minimum possible range equal to 200. Taking the projection point as the origin the Coordinates of the point where the velocity of the projectile is minimum are _______.
a) 200,50
b) 100,50
c) 100,150
d) 100,100
Answer:
b) 100,50

Question 46.
\(\overrightarrow{\mathrm{A}}\) x \(\overrightarrow{\mathrm{B}}\) isequal to –
(a) \(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(b) \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)
(d) \(\overrightarrow{\mathrm{A}}\) – \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) –\(\overrightarrow{\mathrm{B}}\) x \(\overrightarrow{\mathrm{A}}\)

Question 47.
The vector product of any two vectors gives a –
(a) vector
(b) scalar
(e) tensor
(d) col-linear
Answer:
(a) vector

Question 48.
A 150 m long train is moving the north at a speed of 10 m/s. A parrot flying towards the south with a speed of 5 m/s crosses the train. The time taken would be _______.
a) 30s
b) 15s
c) 8s
d) 10s
Answer:
d) 10s

Question 49.
A boat is moving with a velocity of 3i+4j with respect to the ground. The water in the river is moving with a velocity of -3i-4j with respect to the ground. The relative velocity of the boat with respect to water _______.
a) 8j
b) -6i -8j
c) 6i + 8j
d) 5\(\sqrt{2}\)
Answer:
c) 6i + 8j

Question 50.
The vector product of two non-zero vectors will be minimum when O is equal to-
(a) 0°
(b) 180°
(e) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(e) both (a) and (b)

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

II. Long Answer Questions:

Question 1.
What are the different types of motion? State one example for each & explain.
Answer:
The different types of motions are:
a) Linear motion: An object is said to be in linear motion if it moves in a straight line.
Example: An athlete running on a straight tack.

b) Circular motion: It is defined as a motion described by an object traveling a circular path.
Example: The motion of a satellite around the earth

C) Rotational motion: If any object moves in a rotational motion about an axis the motion is rotational motion. During rotation, every point in the object traverses a circular path about an axis.
Example: Spiring of earth about its own axis

D) Vibratory motion: If an object or a particle executes to and fro motion about the fixed point it is said to be in vibratory motion. Sometimes called oscillatory motion.
Example: Vibration of a string on a Guitar.

Question 2.
How will you differentiate motion in one dimension, two dimensions, and in three dimensions?
Answer:
Motion in one dimension: One-dimensional motion is the motion of a particle moving along a straight line.
Example: An object falling freely under gravity close to the earth.

Motion in two dimensions: If a particle is moving along a curved path in-plane, then it is said to be in two-dimensional motion.
Example: Motion of a coin in a carom board.

Motion in three dimensions: A particle moving in usual three-dimensional space has three-dimensional motion.
Example: A bird flying in the sky.

Question 3.
State and define different types of vectors.
Answer:
The different types of vectors are:
1. Equal vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be equal when they have equal magnitude and same direction and represent the same physical quantity.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 86

(a) Coilinear vectors: Collinear vectors are those which act along the same line. The angle between them can be 0° or 180°

(i) Parallel vectors – If two vectors \(\vec{A}\) & \(\vec{B}\) act in the same direction along the same line or in parallel lines. Angle between them is equal to zero
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 87

(ii) Antiparallel vectors:
Two vectors \(\vec{A}\) & \(\vec{B}\) are said to be antiparallel when they are in opposite direction along the same line or in parallel lines. The angle between them is 180°
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 88

2. Unit vector:
A vector divided by its own magnitude is a unit vector.
The unit vector of \(\vec{A}\) is represented as \(\hat{A}\)
Its magnitude is equal to 1 or unity
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 89

3. Orthogonal unit vectors:
Let \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) be three unit vectors which specify the direction along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directly perpendicular to each other
The angle between any two of them is 90°. Then \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 90

Question 4.
Explain how two vectors are subtracted when they are inclined to an angle θ.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 91
Let \(\overline{A}\) and \(\overline{B}\) be non zero vectors inclined at an angle θ.
The difference \(\overline{A}\) – \(\overline{B}\) can be obtained as follows.
First obtain – \(\overline{B}\)
The angle between \(\overline{A}\) – \(\overline{B}\)
= 180 – θ.
The difference \(\overline{A}\) – \(\overline{B}\) is the same as the resultant of \(\overline{A}\) – – \(\overline{B}\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 92
[∵ cos 180 – θ = – cos θ]
(cos 180 – θ = – cos θ)
The gives the resultant magnitude. The resultant is inclined by an angle α2 to \(\overline{A}\)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 93
This gives the direction of the resultant. \(\vec{A}\) – \(\vec{B}\)

Question 5.
Write short notes on relative velocity.
Answer:
When two objects A and B are moving with uniform velocities then the velocity of one object A with respect to another object B is called the relative velocity of A with respect to B.

Case 1:
Consider two objects A and B moving with uniform velocities \(\overline{V}\)A and \(\overline{V}\)B along straight line in same direction with respect to ground.
The relative velocity of object A with respect to object B is \(\vec{V}\)AB = \(\vec{V}\)A – \(\vec{V}\)B
The relative velocity of object B with respect to object A is \(\vec{V}\)BA = \(\vec{V}\)B –\(\overline{V}\)A
Thus, if two objects are moving it’s the same direction the magnitude of the relative velocity of one object with respect to another is equal to the difference in magnitude of the two velocities.

Case 2:
Consider two objects A and B moving with uniform velocities VA and VB along the same track in the opposite direction
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 94
The relative velocity of object A with respect to B is
\(\overline{V}\)AB = \(\overline{V}\)A – ( – \(\overline{V}\)B) = \(\overline{V}\)A + \(\overline{V}\)B
The relative velocity of object B with respect to A is
\(\vec{V}\)BA = – \(\vec{V}\)B – \(\vec{V}\)A) = – ( \(\vec{V}\)A + \(\vec{V}\)B )
Thus if two objects are moving in opposite directions the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitudes of their velocities.

Case 3:
Consider two objects A&B moving with velocities VA and VB at an angle 0 between their directions, then the relative velocity of A with respect to B
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 95
tan θ = (β is the angle between \(\overline{V}\)AB and VB)

Special cases:
(i) When θ = 0, the bodies move along parallel straight lines in the same direction.
VAB = (VA – VB) in the direction of VA.
VBA = (VB – VA) in the direction of VB

(ii) When θ = 180° the bodies move along parallel straight lines in opposite direction.
VAB = VA – (- VB) = (VA + VB) in the direction of VA
VBA = ( VB + VA) in the direction of VB

(iii) If the two bodies are moving at right angles to each other, then θ = 90°
VAB = \(\sqrt{V_{A}^{2}+V_{B}^{2}}\)

(iv) Consider a person moving horizontally with velocity \(\vec{V}\)m Let the rain fall vertically with velocity \(\overline{V}\)R.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 96
An umbrella is held to avoid the rain.
Then relative velocity \(\overline{V}\)M of rain with respect to man is
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 97

Question 6.
Explain Horizontal projection. Derive the equation for its motion, horizontal range & time of flight.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 100
Consider an object thrown horizontally with an initial velocity u, from atop of a tower of height h. The horizontal velocity remains constant throughout its motion and the vertical component of velocity go on increases. The constant acceleration acting along the downward direction is g. The horizontal distance travelled is x(t) = x and the vertical distance travelled is y(t)=y. since the motion is two-dimensional the velocity will have both horizontal (ux) and vertical (uy) components.

Motion along horizontal direction:
The particle has zero acceleration along the x-direction and so initial velocity ux remains constant throughout its motion.
The distance travelled by projectile in a time’t’ is given by
x = ut+1/2 at²
x = uxt → (1)
Motion along vertical direction
Here uy =0, a = g, s = y
S = ut + \(\frac { 1 }{ 2 }\) at²
y = \(\frac { 1 }{ 2 }\) gt² → (2)
from (1) t = x/ux sub in equation (2)
y = \(\frac { 1 }{ 2 }\) g (x/ux)²
y = k x² Where k = \(\frac{g}{2 u_{x}^{2}}\) .x²
This equation resemble the equation of a parabola. Thus the path followed by the projectile is a parabola.

Expression for time of flight:
The time taken for the projectile to complete its trajectory is called the time of flight.
Let h be the height of the tower or the vertical distance traversed.
Let T be the time of flight w.k. S = ut + 1/2 at²
here s = y = h, u = uy, t = T, a = g
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 98
T depends on height of tower or vertical distance & independent of Horizontal velocity.

Expression for Horizontal Range:
The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground it called horizontal range.
w. k. t, S = ut + \(\frac { 1 }{ 2 }\) at²
Here,
t = T, a = 0, S = x = R, u = ux
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 99
Hence R ∝ u ∝ & R ∝ \(\frac{1}{\sqrt{g}}\)

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 7.
Obtain an expression for resultant velocity and the speed of the projectile when it hits the ground in case of a horizontal projection.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 100
At any instant t, the projectile has velocity components along both the x and y-axis.
The velocity component at any time t along with horizontal component Vx = u → (1)
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 101
Speed of the projectile when it hits the ground:
When the projectile hits the ground after thrown horizontally from top of tower of height h, the time of flight is T = \(\sqrt{\frac{2 h}{g}}\)
The horizontal component of velocity Vx = u
The vertical component of velocity
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 102

Conceptual Questions:

Question 1.
Can a body have a constant speed and still have varying velocity?
Answer:
Yes, a particle in uniform circular motion has a constant speed but varying velocity because of the change in its direction of motion at every point.

Question 2.
When an observer is standing on earth appear the trees and houses appear stationary to him. However, when he is sitting in a moving bus or a train all objects appears to move in a backward direction why?
Answer:
For a stationary observer, the relative velocity of trees and houses is zero. For the observer sitting in the moving train, the relative velocity of houses and trees are negative. So these objects appear to move in the backward direction.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 3.
Draw position-time graphs for two objects having zero relative velocity?
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 103
As relative velocity is zero the two bodies A and B have equal velocities. Hence their position-time graphs are parallel straight lines, equally inclined to the time axis.

Question 4.
Can a body be at rest as well as in motion at the same time? Explain.
(OR)
Rest and motion are relative terms. Explain.
Answer:
Yes, the object may be at rest relative to one object and at the same time if maybe in motion relative to another object.

For example, a passenger sitting in a moving train is at rest with respect to his fellow passengers but he is in motion with respect to the objects outside the train. Hence rest and motion are relative terms.

Question 5.
Use integration technique to prove that the distance travelled in-the nth second of motion is Sth =u + \(\frac { a }{ 2 }\) (2n – 1)
Answer:
By definition of velocity v = \(\frac { ds }{ dt }\)
ds = Vdt = (u + at) dt → (1)
when t = (n – 1) second, let distance travelled = Sn-1
when t = n, second, let distance travelled = Sn
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 104

Question 6.
An old lady holding a purse in her hand was crossing the road. She was feeling difficulty in walking. A pickpocket snatched the purse from her and started running away. Can seeing this incident Suresh decided to help that old lady. He informed the police inspector who was standing nearby the inspector chased the pickpocketed and caught hold of him. He collected the purse from the pickpocket and gave back the purse to the old lady.
a) What were the values displayed by Suresh?
b) A police jeep is chasing with a velocity of 45 km/h. A thief in another jeep is moving at 155 km/hr. Police fire a bullet strike the jeep of the thief?
Answer:
The values displayed by Suresh are the presence of mind, helping tendency, and also a sense of social responsibility.
Relative velocity of the bullet with respect to thief’s Jeep = (Vb + Vp)-Vt.
= 180 m/s + 45 km/hr – 155 km/hr
= 180 m/s – 110 x 5/18 m/s
= 180 – 30.5
= 149.5 m/s.

Question 7.
A stone is thrown vertically upwards and then it returns to the thrower. Is it projective?
Answer:
No. It is not a projectile. A projectile should have two-component velocities in two mutually perpendicular directions. But in this case, body has a velocity in only one direction.

Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics

Question 8.
Can two non-zero vectors give zero resultant when they multiply with each other?
Answer:
If yes condition for the same. Yes. for example, the cross product of two non-zero vectors will be zero when θ = 0 or θ = 180°.

Question 9.
Justify that a uniform motion is an accelerated motion.
Answer:
In a uniform circular motion, the speed of the body remains the same but the direction of motion changes at every point.
Samacheer Kalvi 11th Physics Guide Chapter 2 Kinematics 105
Fig. shows the different velocity vectors at different positions of the particle. At each position, the velocity vector V is perpendicular to the radius vector. Thus the velocity of the body changes continuously due to the continuous change in the direction of motion of the body. As the rate of change is of velocity is acceleration a uniform circular motion is an accelerated motion.

Question 10.
State polygon law of vector addition.
Answer:
If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in the same order then their resultant is represented both in magnitude arid direction by the closing side of the polygon taken in the opposite order.

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.1

Question 1.
Fill in the blanks with the correct term from the given list.
(in proportion, similar, corresponding, congruent, shape, area, equal)
(i) Corresponding sides of similar triangles are _________ .
Answer:
in proportion

(ii) Similar triangles have the same ________ but not necessarily the same size.
Answer:
Shape

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

(iii) In any triangle _______ sides are opposite to equal angles.
Answer:
equal

(iv) The symbol is used to represent _________ triangles.
Answer:
congruent

(v) The symbol ~ is used to represent _________ triangles.
Answer:
similar

Question 2.
In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 1
Answer:

Statements Reasons
1. CI = CO ∵ CIP ≡ COP, by CPCTC
2. IP =  OP By CPCTC
3. CP = CP By CPCTC
4. Also HI = HO CPCTC ∆HIP ≡ HOP given
5. IP = OP By CPCTC and (4)
6. ∴ IP ≡ OP By (2) and (4)

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 3.
In the given figure, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆ BCF ≡ ∆ EDF.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 2
Answer:

 Statements Reasons
1. ∠BCF = ∠EFD Vertically opposite angles
2. ∠CBD = ∠DEC Angles on the same base given
3. ∠BCF = ∠EDF Remaining angles of ∆BCF and ∆EDF
4. ∆BCF ≡ ∆EDF By (1) and (2) AAA criteria

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 4.
In the given figure, △ BCD is isosceles with base BD and ∠BAE ≡∠DEA. Prove that AB ≡ ED.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 3
Answer:

Statements Reasons
1. ∠BAE ≡ ∠DEA Given
2. AC = EC By (1) sides opposite to equal angles are equal
3. BC = DC Given BCD is isosceles with base BD
4. AC – BC = EC – DC 2 – 3
5. AB ≡ ED By 4

Midpoint Calculator is a free online tool that displays the midpoint of the line segment.

Question 5.
In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that △ODC ≡ △EDC
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 4
Answer:

Statements Reasons
1. OD = ED D is the midpoint OE (given)
 2. DC = DC Common side
3. ∠CDE = ∠CDO = 90° Linear pair and given ∠CDE = 90°
4. △ODC ≡ △EDC By RHS criteria

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 6.
Is △PRQ ≡ △QSP? Why?
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 5
Answer:
In △PRQ and △PSQ
∠PRQ = ∠PSQ = 90° given
PR = QS = 3 cm given
PQ = PQ = 5 cm common
It satisfies RHS criteria
∴ △PRQ congruent to △QSP.

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 7.
From the given figure, prove that △ABC ~ △EDF
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 6
Answer:
From the △ABC, AB = AC
It is an isosceles triangle
Angles opposite to equal sides are equal
∴ ∠B = ∠C = 65°
∴ ∠B + ∠C = 65° + 65°
= 130°
We know that sum of three angles is a triangle = 180°
∠A + ∠B + ∠C = 180°
∠A + 130° = 180°
∠A = 180° – 130°
∠A = 50°
From △EDF, ∠E = 50°
∴ Sum of Remaining angles = 180° – 50° = 130°
DE = FD
∴ ∠D = ∠F
From △ABC and △EDF ∴ △D = \(\frac{130}{2}\) = 65°
∠A = ∠E = 50°
∠B = ∠D = 65°
∠C = ∠F = 65°
∴ By AAA criteria △EDF ~ △ABC

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 8.
In the given figure YH || TE. Prove that △WHY ~ △WET and also find HE and TE.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 7
Answer:

Statements Reasons
1. ∠EWT = ∠HWY Common angle
2.  ∠ETW = ∠HYW Since YH || TE, corresponding angles
3. ∠WET = ∠WHY Since YH || TE corresponding angles
4. △WHY ~ △WET By AAA criteria

Also △ WHY ~ △WET
∴ Corresponding sides are proportionated
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 8
⇒ 6+HE = \(\frac{6}{4}\) × 16
⇒ 6 + HE = 24
∴ HE = 24 – 6
HE = 18
Again \(\frac{4}{\mathrm{ET}}=\frac{4}{16}\)
ET = \(\frac{4}{4}\)
ET = 16

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 9.
In the given figure, if △EAT ~ △BUN, find the measure of all angles.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 9
Answer:
Given △EAT ≡ △BUN
∴ Corresponding angles are equal
∴ ∠E = ∠B …… (1)
∠A = ∠U …… (2)
∠T = ∠N ……. (3)
∠E = x°
∠A = 2x°
Sum of three angles of a triangle = 180°
In △ EAT, x + 2x + ∠T = 180°
∠T = 180° – (x° + 2x°)
∠T = 180° – 3x°
Also in △BUN
(x + 40)° + x° + ∠U = 180°
x + 40° + x + ∠U = 180°
2x° + 40° + ∠U = 180°
∠U = 180° – 2x – 40° = 140° – 2x°
Now by (2)
∠A = ∠U
2x = 140° – 2x°
2x + 2x = 140°
4x = 140°
x = \(\frac{140}{4}\) = 35°
∠A = 2x° = 2 × 35° = 70°
∠N = x + 40° = 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠8 = 35°
∠A = ∠U = 70°

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 10.
In the given figure, UB || AT and CU ≡ CB Prove that △CUB ~ △CAT and hence △CAT is isosceles.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 10
Answer:

Statements Reasons
1. ∠CUB = ∠CBU ∵ In △CUB, CU = CB
2. ∠CUB = ∠CAB ∵ UB || AT, Corresponding angle if CA is the transversal.
3. ∠CBU = ∠CTA CT is transversal UB || AT,

Corresponding angle commom angle.

4. ∠UCB = ∠ACT Common angle
5. △CUB ~ △CAT By AAA criteria
6. CA = CT ∵ ∠CAT = ∠CTA
7. Also △CAT  is isoceles By 1, 2 and 3 and sides opposite to equal angles are equal.

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Objective Type Questions

Question 11.
Two similar triangles will always have _______ angles
(A) acute
(B) obtuse
(C) right
(D) matching
Answer:
(D) matching

Question 12.
If in triangles PQR and XYZ, \(\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{ZX}}\) then they will be similar if
(A) ∠Q = ∠Y
(B) ∠P = ∠X
(C) ∠Q = ∠X
(D) ∠P ≡∠Z
Ans:
(C) ∠Q = ∠X

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 13.
A flag pole 15 m high casts a shadow of 3m at 10 a.m. The shadow cast by a building at the same time is 18.6m. The height of the building is
(A) 90 m
(B) 91 m
(C) 92 m
(D) 93 m
Answer:
(D) 93 m

Question 14.
If ∆ABC – ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer:
(A) 50°

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.1

Question 15.
In the figure, which of the following statements is true?
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.1 11
(A) AB = BD
(B) BD < CD
(C) AC = CD
(D) BC = CD
Answer:
(C) AC = CD