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

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 2 Number Systems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 2 Number Systems

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

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6

Inverse Function Calculator is an online tool that helps find the inverse value for the given function.

Choose the most suitable answer from the given four alternatives:

Question 1.
The value of sin-1(cos x), 0 ≤ x ≤ π is
(a) π – x
(b) x – \(\frac {π}{2}\)
(c) \(\frac {π}{2}\) – x
(d) x – π
Solution:
(c) \(\frac {π}{2}\) – x
Hint:
sin-1(cos x) = sin-1(sin(\(\frac {π}{2}\) – x)) = \(\frac {π}{2}\) – x

Question 2.
If sin-1 x + sin-1 y = \(\frac {2π}{3}\); then cos-1 x + cos-1 y is equal to
(a) \(\frac {2π}{3}\)
(b) \(\frac {π}{3}\)
(c) \(\frac {π}{6}\)
(d) π
Solution:
(b) \(\frac {π}{3}\)
Hint:
sin-1x + cos-1x + cos-1y + sin-1y = \(\frac {π}{2}\) + \(\frac {π}{2}\) = π
\(\frac {2π}{3}\) + cos-1x + cos-1y = π
cos-1x + cos-1y = π – \(\frac {2π}{3}\) = \(\frac {3π-2π}{3}\) = \(\frac {π}{3}\)

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 3.
sin-1\(\frac {3}{5}\) – cos-1\(\frac {12}{13}\) + sec-1\(\frac {5}{3}\) – cosec-1\(\frac {13}{12}\) is equal to
(a) 2π
(b) π
(c) 0
(d) tan-1\(\frac {12}{65}\)
Solution:
(c) 0
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 1

Question 4.
If sin-1 x = 2sin-1 α has a solution, then
(a) |α| ≤ \(\frac {1}{√2}\)
(b) |α| ≥ \(\frac {1}{√2}\)
(c) |α| < \(\frac {1}{√2}\)
(d) |α| > \(\frac {1}{√2}\)
Solution:
(a) |α| ≤ \(\frac {1}{√2}\)
Hint:
If sin-1 x = 2sin-1 α has a solution then
–\(\frac {π}{2}\) ≤ 2sin-1α ≤ \(\frac {π}{2}\)
–\(\frac {π}{4}\) ≤ sin-1α ≤ \(\frac {π}{4}\)
sin(\(\frac {-π}{4}\)) ≤ α ≤ sin\(\frac {π}{4}\)
–\(\frac {1}{√2}\) ≤ α ≤ \(\frac {1}{√2}\)
-|α| ≤ \(\frac {1}{√2}\)

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 5.
sin-1(cos x) = \(\frac {π}{2}\) – x is valid for
(a) -π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) –\(\frac {π}{2}\) ≤ x ≤ \(\frac {π}{2}\)
(d) –\(\frac {π}{4}\) ≤ x ≤ \(\frac {3π}{4}\)
Solution:
(b) 0 ≤ x ≤ π
Hint:
sin-1 (cosx) = \(\frac {π}{2}\) – x is valid for
cos x = sin (\(\frac {π}{2}\) – x)
cos x ∈ [0, π]
∴ 0 ≤ x ≤ π

Question 6.
If sin-1 x + sin-1 y + sin-1 z = \(\frac {3π}{2}\), the value of show that x2017 + y2018 + z2019 – \(\frac {9}{x^{101}+y^{101}+z^{101}}\) is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 2

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 7.
If cot-1 x = \(\frac {2π}{5}\) for some x ∈ R, the value of tan-1 x is
(a) –\(\frac {π}{10}\)
(b) \(\frac {π}{5}\)
(c) \(\frac {π}{10}\)
(d) –\(\frac {π}{5}\)
Solution:
(c) \(\frac {π}{10}\)
Hint:
tan-1 x + cos-1 \(\frac {π}{2}\)
tan-1x = \(\frac {π}{2}\) – cot-1 x = \(\frac {π}{2}\) – \(\frac {2π}{5}\)
= \(\frac {5π-4π}{10}\) = \(\frac {π}{10}\)

Question 8.
The domain of the function defined by f(x) = sin-1 \(\sqrt {x-1}\) is
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Solution:
(a) [1, 2]
Hint:
f(x) = sin-1 \(\sqrt {x-1}\)
\(\sqrt {x-1}\) ≥ 0
-1 ≤ \(\sqrt {x-1}\) ≤ 1
∴ 0 ≤ \(\sqrt {x-1}\) ≤ 1
0 ≤ x – 1 ≤ 1
1 ≤ x ≤ 2
x ∈ [1, 2]

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 9.
If x = \(\frac {1}{5}\) the value of cos(cos-1x + 2sin-1x) is
(a) –\(\sqrt{\frac {24}{25}}\)
(b) \(\sqrt{\frac {24}{25}}\)
(c) \(\frac {1}{5}\)
(d) –\(\frac {1}{5}\)
Solution:
(d) –\(\frac {1}{5}\)
Hint:
cos[cos-1x + sin-1x + sin-1x] = cos(\(\frac {π}{2}\) + sin-1x)
= -sin(sin-1x)
[∵ cos(90+θ) = -sin θ]
= -x = –\(\frac {1}{5}\)

Question 10.
tan-1(\(\frac {1}{4}\)) + tan-1(\(\frac {2}{9}\)) is equal to
(a) \(\frac {1}{2}\)cos-1(\(\frac {3}{5}\))
(b) \(\frac {1}{2}\)sins-1(\(\frac {3}{5}\))
(c) \(\frac {1}{2}\)tan-1(\(\frac {3}{5}\))
(d) tan-1(\(\frac {1}{2}\))
Solution:
(d) tan-1(\(\frac {1}{2}\))
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 3

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 11.
If the function f(x) = sin-1(x² – 3), then x belongs to
(a) [-1, 1]
(b) [√2, 2]
(c) [-2, -√2]∪[√2, 2]
(d) [-2, -√2]
Solution:
(c) [-2, -√2]∪[√2, 2]
Hint:
-1 ≤ x² – 3 ≤ 1
-1 + 3 ≤ x² ≤ 1 + 3
⇒ 2 ≤ x² ≤ 4
±√2 ≤ x ≤ ± 2
[-2, -√2]∪[√2, 2]

Question 12.
If cot-1 2 and cot-1 3 are two angles of a triangle, then the third angle is
(a) \(\frac {π}{4}\)
(b) \(\frac {3π}{4}\)
(c) \(\frac {π}{6}\)
(d) \(\frac {π}{3}\)
Solution:
(b) \(\frac {3π}{4}\)
Hint:
A + B + C = π (triangle)
cot-1 2 + cot-1 3 + C = π
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 4

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 13.
sin-1(tan\(\frac {π}{4}\)) – sin-1(\(\sqrt{\frac {3}{x}}\)) = \(\frac {π}{6}\). Then x is root of the equation
(a) x² – x – 6 = 0
(b) x² – x – 12 = 0
(c) x² + x – 12 = 0
(d) x² + x – 6 = 0
Solution:
(b) x² – x – 12 = 0
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 5

Question 14.
sin-1(2 cos²x – 1) + cos-1(1 – 2 sin²x) =
(a) \(\frac {π}{2}\)
(b) \(\frac {π}{3}\)
(c) \(\frac {π}{4}\)
(d) \(\frac {π}{6}\)
Solution:
(a) \(\frac {π}{2}\)
Hint:
sin-1(2 cos² x – 1) + cos-1(1 – 2 sin²x)
= sin-1 (2 cos² x – 1) + cos-1 (1 – sin² x – sin² x)
= sin-1(2 cos² x – 1) + cos-1(cos² x – (1 – cos²x))
= sin-1(2 cos² x – 1) + cos-1(cos² x – 1 + cos²x)
= sin-1(2 cos² x – 1) + cos-1(2 cos² x – 1)
= \(\frac {π}{2}\) [∵ sin-1 x + cos-1 x = \(\frac {π}{2}\)]

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 15.
If cot-1(\(\sqrt {sinα}\)) + tan-1(\(\sqrt {sinα}\)) = u, then cos 2u is equal to
(a) tan²α
(b) 0
(c) -1
(d) tan 2α
Solution:
(c) -1
Hint:
cot-1 x + tan-1 x = \(\frac {π}{2}\)
∴ u = \(\frac {π}{2}\)
cos 2u = cos 2(\(\frac {π}{2}\)) = cos π = -1

Question 16.
If |x| ≤ 1, then 2 tan-1 x – sin-1\(\frac {2x}{1+x²}\) is equal to
(a) tan-1x
(b) sin-1x
(c) 0
(d) π
Solution:
(c) 0
Hint:
sin-1\(\frac {2x}{1+x²}\) = 2 tan-1x
∴ 2 tan-1 x – 2 tan-1 x = 0

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 17.
The equation tan-1 x – cot-1 x = tan-1(\(\frac {1}{√3}\)) has
(a) no solution
(b) unique solution
(c) two solutions
(d) infinite number of solutions
Solution:
(b) unique solution
Hint:
tan-1 x – cot-1 x = tan-1(\(\frac {1}{√3}\)) …….. (1)
tan-1 x – cot-1 x = \(\frac {π}{2}\) ……… (2)
Add 1 and 2
2 tan-1 x = \(\frac {π}{6}\) + \(\frac {π}{2}\) = \(\frac {2π}{3}\)
tan-1 x = \(\frac {π}{3}\)
x = √3 which is uniqe solution.

Question 18.
If sin-1 x + cot-1(\(\frac {1}{2}\)) = \(\frac {π}{2}\), then x is equal to
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{√5}\)
(c) \(\frac {2}{√5}\)
(d) \(\frac {√3}{2}\)
Solution:
(b) \(\frac {1}{√5}\)
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 6

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Question 19.
If sin-1 \(\frac {x}{5}\) + cosec-1\(\frac {5}{4}\) = \(\frac {π}{2}\), then the value of x is
(a) 4
(b) 5
(c) 2
(d) 3
Solution:
(d) 3
Hint:
Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6 7

Question 20.
sin(tan-1 x), |x| < 1 is equal to
(a) \(\frac {x}{\sqrt{1-x^2}}\)
(b) \(\frac {1}{\sqrt{1-x^2}}\)
(c) \(\frac {1}{\sqrt{1+x^2}}\)
(d) \(\frac {x}{\sqrt{1+x^2}}\)
Solution:
(d) \(\frac {x}{\sqrt{1+x^2}}\)
Hint:
tan a = x
W.K.T 1 + tan² a = sec² a
1 + x² = sec² a
sec a = \(\sqrt{1+x^2}\)
\(\frac {1}{cosa}\) = \(\sqrt{1+x^2}\)
cos a= \(\frac {1}{\sqrt{1+x^2}}\)
sin a = \(\sqrt{1-cos^2a}\) = \(\sqrt{1-\frac {1}{1+x^2}}\)
\(\sqrt{\frac{1+x^2 -1}{1+x^2}}\) = \(\frac {x}{\sqrt{1+x^2}}\)

Samacheer Kalvi 12th Maths Guide Chapter 4 Inverse Trigonometric Functions Ex 4.6

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Having trouble working out with the Binomial theorem? You’ve come to the right place, our Binomial Expansion Calculator is here to save the day for you.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 1
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 2

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 3

(iii) (5 + x2)2/3
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 4

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 5
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 6

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 7

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 8
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 9
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 10
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 11

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 12
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 13

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 5.
Write the first 6 terms of the exponential series
(i) e5x
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 14

(ii) e-2x
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 15

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iii) ex/2
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 16
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 17

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 18

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(ii) log (1 – 2x)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 19

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 20

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 21
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 22

Question 7.
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 23
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 24

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 8.
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 25
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 26
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 27
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 28
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 29

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 9.
Find the coefficient of x4 in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 30

Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 10.
Find the value of Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 31
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 32
Samacheer Kalvi 11th Maths Guide Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 33

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 4 Work, Energy and Power Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

11th Physics Guide Work, Energy and Power Book Back Questions and Answers

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Part – I:
I. Multiple choice questions:

Question 1.
A uniform force of (2\(\hat{i}\) + \(\hat{j}\)) N acts on a particle of mass 1 kg. The particle displaces from position (3\(\hat{j}\) + \(\hat{k}\) )m to (5\(\hat{i}\) + 3\(\hat{j}\)) m. The work done by the force on the particle is _______. (AIPMT Model 2013)
a) 9 J
b) 6 J
c) 10 J
d) 12 J
Answer:
c) 10 J

Question 2.
A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of _______. (AIPMT model 2004)
a) \(\sqrt{2}\) : 1
b) 1 : \(\sqrt{2}\)
c) 2 : 1
d) 1 : 2
Answer:
d) 1 : 2

This online Velocity Calculator is used to find the velocity of water in a pipe with the flow rate and diameter of the pipe.

Question 3.
A body of mass 1 kg is thrown upwards with a velocity 20 m s-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (Take g= 10ms-2 ) (AIPMT 2009)
a) 20 J
b) 30 J
c) 40 J
d) 10 J
Answer:
a) 20 J

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 4.
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water of the jet. What is the rate at which kinetic energy is imparted to water? (AIPMT 2009)
a) \(\frac { 1 }{ 2 }\) mv3
b) mv3
c) \(\frac { 3 }{ 2 }\)mv²
d) \(\frac { 5 }{ 2 }\) mv²
Answer:
a) \(\frac { 1 }{ 2 }\) mv3

Question 5.
A body of mass 4 m is lying in xy-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed v. The total kinetic energy generated due to explosion is _______. (AIPMT 2014)
a) mv²
b) \(\frac { 3 }{ 2 }\)mv²
c) 2 mv²
d) 4 mv²
Answer:
b) \(\frac { 3 }{ 2 }\)mv²

Question 6.
The potential energy of a system increases, if work is done _______.
a) by the system against a conservative force
b) by the system against a non-conservative force
c) upon the system by a conservative force
d) upon the system by a non-conservative force
Answer:
a) by the system against a conservative force

Question 7.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
a) \(\sqrt{2gR}\)
b) \(\sqrt{3gR}\)
c) \(\sqrt{5gR}\)
d) \(\sqrt{gR}\)
Answer:
c) \(\sqrt{5gR}\)

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 8.
The work done by the conservative force for a closed path is _______.
a) always negative
b) zero
c) always positive
d) not defined
Answer:
b) zero

Question 9.
If the linear momentum of the object is increased by 0.1%, then the kinetic energy is increased by _______.
a) 0.1 %
b) 0.2%
c) 0.4%
d) 0.01%
Answer:
b) 0.2%

Question 10.
If the potential energy of the particle is α – \(\frac { β }{ 2 }\)x², then force experienced by the particle is _______.
a) F = \(\frac { β }{ 2 }\)x²
b) F = βx
c) F = – βx
d) F = – \(\frac { β }{ 2 }\)x²
Answer:
c) F = – βx

Question 11.
A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
a) v
b) v²
c) v3
d) v4
Answer:
c) v3

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 12.
Two equal masses m1 and m2 are moving along the same straight line with velocities 5ms-1 and -9ms-1 respectively. If the collision is elastic, then calculate the velocities after the collision of m1 and m2, respectively _______.
a) -4ms-1 and 10 ms-1
b) 10ms-1 and 0 ms-1
c) -9ms-1 and 5 ms-1
d) 5 ms-1 and 1 ms-1
Answer:
c) -9ms-1 and 5 ms-1

Question 13.
A particle is placed at the origin and a force F= kx is acting on it (where k is a positive constant). If U(0)=0, the graph of U(x) versus x will be (where U is the potential energy function) (IIT 2004)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 1
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 2

Question 14.
A particle which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = – kx + ax3. Here, k and a are positive constants. For x > 0, the functional form of the potential energy U(x) of the particle is _______. (IIT 2002)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 3
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 4

Question 15.
A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of _______.
a) \(\frac { 2 }{ 3 }\) k
b) \(\frac { 3 }{ 2 }\) k
c) 3k
d) 6k
Answer:
b) \(\frac { 3 }{ 2 }\) k

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

II. Short Answer Questions:

Question 1.
Explain how the definition of work in physics is different from general perception.
Answer:
The term work is used in diverse contexts in daily life. It refers to both physical as well as mental work. In fact, any activity can generally be called work. But in Physics, the term work is treated as a physical quantity with a precise definition. Work is said to be done by the force when the force applied to a body displaces it.

Question 2.
Write the various types of potential energy. Explain the formulate.
Answer:
The energy possessed by the body by virtue of its position is called potential energy.
The various types of potential energies are
1) Gravitational potential energy :
The energy possessed by the body due gravitational force gives gravitational potential energy
u = mgh.
where
u → Gravitational potential energy
m → Mass of the body
g → acceleration due to gravity
h → displacement produced

2) Elastic potential energy :
The energy due to spring force and other similar forces give rise to elastic potential energy
u = 1/2² Where
U → elastic potential energy
K → spring constant
x → elongation produced

3) Electrostatic potential energy
The energy due to electro static force on charges give rise to electrostatic potential energy
U = K \(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}\) where
K = \(\frac{1}{4 \pi \varepsilon_{0}}\) Constant
q1, q2 – magnitude of charges
d – displacement made by any one of the charges or by both charges.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 3.
Write the differences between conservative and non-conservative forces. Give two examples each.
Answer:
Conservative forces:
A Force is said to be a conservative force if work done by or against the force in moving the body depends only on the initial and final positions of the body and net depend on the nature of the path followed between the two positions
Examples:
Elastic spring force, electrostatic force, magnetic force, gravitational force.

Non-conservative force:
A force is said to be non-conservative if the work done by or against the force in moving a body depends on the path between initial and final positions. This means the value of work-done is different in different paths.

Examples:

  1. Frictional forces are non-conservative forces as the work done against friction depends on the length of the path moved by the body.
  2. The force due to air resistance, viscous forces are also non conservative forces because work done by or against these forces depends upon the velocity of motion.

Question 4.
Explain the characteristics of elastic and inelastic collision Elastic collision:
Answer:
In a collision, the total K.E of the bodies before collision is equal to the total K.E. of the bodies after collision, then it is an elastic collision.
Total K.E. before collision = Total K.E. after collision

Inelastic collision:
In a collision, the total K.E. of the bodies before collision is not equal to total K.E. after collision then it is called as inelastic collision Even though K.E. is not conserved but total energy is conserved. After collision of the two colliding bodies stick together such collision are called as perfectly inelastic a plastic collision.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 5.
Define the following
Answer:
a) Coefficient of restitution
b) Power
c) Law of conservation of energy
d) Loss of Kinetic energy in inelastic collision
Answer:
a) Coefficient of restitution :
Coefficient of restitution defined as the ratio of velocity of separation (after collision) to velocity of approach (before collision)
The coefficient of restitution = Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 5

b) Power :
Power is defined as the rate of work done or energy delivered
P = \(\frac { W }{ t }\) = \(\vec{F}\).\(\vec{V}\)

c) Law of conservation of energy :
Law states that “Energy can neither be created nor destroyed. It may be transformed from one to another but the total energy of an isolated system remains constant”

d) Loss of kinetic energy in inelastic collision
The difference in total K.E. before collision and total K.E. after collision’s is equal to loss of K.E. during collision.
∆Q = Total K.E. before collision – Total K.E. after collision.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

III. Long Answer Questions:

Question 1.
Explain with graphs the difference between work done by a constant force and by a variable force.
Answer:
a) Work done by a constant force:
When a constant force ‘F’ acts on a body the small work done (dw) by the force in producing a small displacement ‘dr’ is given by dw = (F cos θ) dr.
The total W.D in producing a displacement from initial position ri, to final position rf is,
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 6
The graphical representation of the W.D by constant force is shown below
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 7
The area under the graph shows the work done by constant force.

b) Work done by a variable force :
When the component of a variable force F acts on a body, the small work done (dw) by the force in producing a small displacement dr is given by dw = (F cos θ) dr [ Here F cos 0 is the component of variable forces]. where F and 0 one variables To total W.D for the displacement from initial position ri to final position rf is given by the relation.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 8
A graphical representation of the work done by a variable force is shown below. The area under the graph gives the W.D. by variable force.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 9

Question 2.
State and explain work energy principle Mention any three examples for it.
Answer:
Law:
Work done by a force on the body changes the kinetic energy of the body, ie change in K.E. = work done. This is called work energy theorem.
Proof:
Consider a body of mass m at rest on a frictionless horizontal surface. The work done (W) done by the constant force (F) for displacement (S) in the same direction is W = FS → (1)
The constant force is given by F = ma → (2)
We know that v² = u² + 2a
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 10
W = 1/2 mv² – 1/2 mu²
Here the term 1/2 mv² indicates K.E.
1/2 mv² – 1/2mu² = ∆K (change in K.E.)
∵ W = ∆K
Hence proved

Examples:

  1. A moving hammer drives a nail into the wood. Being in motion, it has K.E. or ability to do work.
  2. A fast moving stone can break a windowpane. The stone has K.E. due to its motion and so it can do work.
  3. The kinetic energy of air is used to run windmills.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 3.
Arrive at an expression for power and velocity. Give some examples for the same.
Answer:
Power is defined as the rate of work done or energy delivered. P = \(\frac { w }{ t }\)
Relation between power and velocity:
The work done by a force F for a displacement \(\overline{dr}\) is dw = \(\overline{F}\).\(\overline{dr}\)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 11

Examples:
1) A 100 Watt bulb consumes 100 joule of electrical energy in one second
2) Electrical motor supply enough power to bring water from a bore well.

Question 4.
Arrive at an expression for elastic collision in one dimension and discuss various cases.
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 12
Consider two elastic bodies of masses m1 and m2 moving in a straight line (along positive direction) on a frictionless horizontal surface. In order to have collision assume m1 moves faster than m2.

Let U1 and 1 be the initial velocities of m1 and m2 respectively. (u1 > u2). After collision let the masses m1 and m2 moves with velocities v1 and v2 respectively.
Incase of elastic collision both linear momentum and kinetic energies are conserved
∴ from law of conservation of linear momentum
Total momentum before (Pi)
collision = Total momentum afer collision (Pf)
m1u1 + m2u2 = m1v1 + m2v2 → (1)
m1(u1 – v1) = m2(v2 – u2 → (2)
Further
Total K.E. before collision (KEi) = Total K.E. after collision (KEf)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 13
rearranging
u1 – u2 = – (v1 – v2) → (5)
From this it is dear that for any elastic collision, relative speed of two elastic bodies after the collision has the same magnitude as before collision but in opposite direction.

Rewriting the above equation for v1 & v1
v1 = v2 + u2 – u1 → (6)
(or)
v2 = v1 + u1 – u2 → (7)
To find velocities of v1 & v2
Substituting (7) in (2)
m1(u1 – v1) = m2(v1 + u1 – u2 – u2)
m1(u1 – v1) = m2(v1 + u1 – 2u2)
m1u1 – m1v1 = m2v1 + m2u1 – 2m2u2
m1u1 – m2u1 + 2m2u2 = m1v1 + m2v1
u1(m1 – m2) + 2m2u2 = v1(m1 + m2)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 14

Case 1:
When bodies have same mass is m1 = m2 = m
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 15
The velocities get interchanged.

Case 2:
When both bodies have same mass m1 = m2 = m, but second body at rest in u2 = 0
v1 = 0, v2 = u1
After collision the first body comes to rest and the second body moves with the velocity of first body.

Case 3 :
The first body very much lighter than the second body in m1 << m2, \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\) << 1.
the ratio \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\) = 0 and also second body at rest, (u2 = 0)
Dividing numerator and denominator of equation 8 by m2
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 16
Similarly dividing numerator and denominator of equation 9 by m1
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 17

From this, the conclusion arrived is the first body which is lighter returns back (rebounds) in opposite direction with the same initial velocity as it has a negative sign. The second body since it has heavier mars continues to remain at rest even after the collision.

Case 4 :
The second body is very much lighter than the first body.
m2 << m1 then the ratio = 0 \(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\) and also if the target is at rest ie second body at rest (u2 = 0)
Dividing equation (8) both the numerator and denominator by m1
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 18
This shows that the first body which is heavier continues to move with same velocity and the second body which is lighter will move with twice the initial velocity of the first body, ie lighter body is thrown away from the point of collision.

Question 5.
What is inelastic collision? In which way it is different from elastic collision mention few examples in day to day life for inelastic collision.
Answer:
In a collision, the total K.E. on the bodies before collision is not equal to the total K.E. after collision then it is called as inelastic collision, i.e
Total K.E after collision ≠ Total K.E. before collision
Whereas in case of elastic collision Total K.E. before collison is equal to total K.E after collision.
Example: Collision between two vehicles, collision between a ball and floor.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

IV. Numerical Problems:

Question 1.
Calculate the work done by a force of 30N in lifting a load of 2kg to a height of 10m (g = 10ms-2)
Solution:
F = 30N
m = 2kg
s = 10 m
g = 10 ms-2
θ = 0
W.D = ?
W.D = \(\overline{F}\).\(\overline{S}\) = FS cos θ
W.D = 30 x 10
= 300 J

Question 2.
A ball with a velocity of 5 ms-1 impinges at an angle of 60° with the vertical on a smooth Horizontal plane. If the coefficient of restitution is 0.5 find the velocity and direction after the impact.
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 19
U1 = 5 ms-1
θ = 60°
e = 0.5
v = ?
Initial momentum = final momentum along the original line of m of con.
∵ Coefficient of restitution is 0.5 (less than 1) the collision is inelastic
Applying component of velocities. The x component of velocity is
u sin θ = v sin Φ → (1)
But the magnitude of y component is not same using coefficient of restitution
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 20

Question 3.
A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed center O as shown in the figure. What initial speed must be given to the object to reach the top of the circle?. (Hint: use law of conservation of energy). Is this speed less or greater than speed obtained in the section 4.2.9?
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 21
T.E. at lowest point (E1) = Total energy at (E2)
E1 = E2
At the lowest point potential energy = u1 = 0
Kinetic energy = KE1 = 1/2 mv1²
Total energy at lowest point = E1 = u1 + KE1
E1 = 0 + 1/2 mv² → (1)
∴ E1 = mv²1 … (1)
At the top of circle:
Potential energy = u2 = 2mgr
K.E energy at top = 1/2mv2²
Total energy at highest
point = 2mgr + 1/2 mv2² → (2)
According to law of conservation of energy E1 = E2
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 22

Question 4.
Two different unknown masses A and B collide. A is initially at rest when B has a speed V. After collision B has a speed V/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision.
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 23
Applying principle of conservation of momentum along x-axis
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 24
Applying principle of conservation of momentum along y-axis
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 25

Question 5.
A bullet of mass 20g strikes a pendulum of mass 5 kg. The centre of mass of the pendulum rises at a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.
Solution:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 26
mass of bullet m1 = 20 x 10-3kg
mass of pendulum m2 = 5kg
Let the speed of the bullet = u1
∵ The pendulum at rest u2 = 0
h = 10 x 10-2m
Let v be the common velocity after the bullet embedded inside the bob.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 27
This common velocity ‘v’ is the initial velocity of combined bullet and bob.
W.K.T
v² – u² = 2as
Here v = 6; u = 0, a = – g s = h
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 28

V. Conceptual Questions:

Question 1.
A spring which is initially in the unstretched condition is first stretched by a length x and again by a further length x. The work done in the first case w1 is one-third of the work done in the second case w2. True or false?
Answer:
True, W.D by the first case will be 1/3 of the second W.D
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 29

Question 2.
Which is conserved in inelastic collision? Total energy (or) kinetic energy?
Answer:
In inelastic collision total energy is only conserved but kinetic energy is not conserved. A part of kinetic energy is converted into some other form of energy such as sound, heat energy.
Note: The linear momentum is also conserved.

Question 3.
Is there any network done by external forces on a car moving at a constant speed along a straight road?
Answer:
Since the car is moving at a constant speed along a straight line, displacement is caused. So work in done by the force.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 4.
A car starts from rest and moves on a surface with uniform acceleration. Draw the graph of kinetic energy versus displacement. What information you can get from that graph?
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 30
According to work-energy theorem change in K.E = W.D
K.E is Kept constant slope is constant

Question 5.
A charged particle moves towards another charged particle under what conditions the total momentum and the total energy of the system conserved?
Answer:
Coulomb force is acting in between the charged particles Internal force is a conservative force. If no external forces act or the work done by external forces is zero then the mechanical energy of the system and also total linear momentum also remains constant.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

11th Physics Guide Work, Energy and Power Additional Important Questions and Answers

I. Multiple choice questions:

Question 1.
Thrust and linear momentum
(a) Thrust and linear momentum
(b) Work and energy
(c) Work and power
(d) Power and energy
Answer:
(b) Work and energy

Question 2.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 31 For the surface μ = 0.4, The work done by applied force, frictional force and net force are
a) 50J, -40J, 10J
b) 50J, -20J, 10J
c) 10J, -50J, 40J
d) 50J, -40J, 20J
Answer:
a) 50J, -40J, 10J

Question 3.
The system is released from rest. Find the work done by the force if gravity during first 2 seconds of motion.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 32
a) 80J
b) 20J
c) 40J
d) 100J
Answer:
c) 40J

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 4.
Dimensional formula for work done is
(a) MLT-1
(b) ML2T2
(c) M-1L-1T2
(d) ML2T-2
Answer:
(d) ML2T-2

Question 5.
The variation of force acting on a particle along x ax is is shown. The W.D by the force during the displacement x = 0 to x = 25m is _______.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 33
a) 100J
b) 115J
c) 130J
d) 125J
Answer:
b) 115J

Question 6.
The amount of work done by centripetal force on the object moving in a circular path is
(a) zero
(b) infinity
(c) positive
(d) negative
Answer:
(a) zero

Question 7.
A ball of mass 200g is attached to a string 50 mm and a force F is applied as shown. The W.D by this force if the string makes an angle 60° with vertical is? [at initial and final positions speed of the ball is zero.]
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 34
a) 1J
b) 0.5J
c) 0.05J
d) 0.25J
Answer:
b) 0.5J

Question 8.
An object of mass 4kg falls from rest through a vertical distance of 20m and reaches with velocity 10ms-1 on ground. The work done by air friction is .
a) 800J
b) – 800J
c) 600J
d) – 600J
Answer:
d) – 600J

Question 9.
If the force and displacement are perpendicular to each other, then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(c) zero

Question 10.
What is the power of an engine which can lift 600kg of coal per minute from a mine 20m deep?
a) 2000 w
b) 100 w
c) 1000 w
d) 200 w
Answer:
a) 2000 w

Question 11.
If w1, w2 and w3 represent the work done in moving a particle from A to B along 3 different paths 1, 2 and 3 respectively as shown in figure, in the gravitational field of a point mass m. Find the correct relation an between w1, w2 and w3 _______.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 35
a) w1 > w2 > w3
b) w1 = w2 = w3
c) w1 < w2 < w3
d) w2 > w1 > w3
Answer:
b) w1 = w2 = w3

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 12.
A position – depends force F = 7 – 2x + 3x² newton acts on a small body of mass 2kg and displacement if from x = 0 to x = 5m. The work done in Joule is _______.
a) 70
b) 270
c) 35
d) 135
Answer:
d) 135

Question 13.
The energy possessed by a body due to its motion is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(b) kinetic energy

Question 14.
According to the work-energy theorem, the work done by the net force on a particle is equal to change in its _______.
a) kinetic energy
b) potential energy
c) linear momentum
d) angular momentum
Answer:
a) kinetic energy

Question 15.
A block of 2kg is resting on a smooth surface. At what angle a force of 10N be acting on the block so that it will acquire a K.E of 10J after moving 2m _______.
a) 30°
b) 60°
c) 45°
d) 90°
Answer:
b) 60°

Question 16.
The K.E acquired by a body of mass m in travelling a certain distance starting from rest under a constant force is _______.
a) directly proportional to m
b) directly proportional to \(\sqrt{m}\)
c) inversely proportional to \(\sqrt{m}\)
d) independent of m
Answer:
b) directly proportional to \(\sqrt{m}\)

Question 17.
1-kilowatt hour is equivalent to
(a) 10-7 J
(b) 1.6 × 10-19 J
(c) 4.186 J
(d) 3.6 × 10-6 J
Answer:
(d) 3.6 × 10-6 J

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 18.
A particle moves in a straight line with retardation proportional to displacement. Its loss of K.E in any displacement x is proportional to _______.
a) x²
b) ex
c) x
d) logex
Answer:
a) x²

Question 19.
Which one of the following is not a conservative force?
a) gravitational force
b) electrostatic force between the charges
c) Magnetic force between two magnetic dipoles
d) Frictional force
Answer:
d) Frictional force

Question 20.
A spring of force constant 800 N/M has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is _______.
a) 163
b) 8J
c) 32J
d) 24J
Answer:
b) 8J

Question 21.
The kinetic energy of the body is always
(a) zero
(b) infinity
(c) negative
(d) positive
Answer:
(d) positive

Question 22.
A bullet moving with a speed of 150 m/s strikes a wooden plank. After passing through the plank its speed becomes 125ms-1. Another bullet of the same mass and size strikes the plank with a speed of 90m/s. Its speed after passing through the plank would be _______.
a) 25 m/s
b) 35 m/s
c) 50 m/s
d) 70 m/s
Answer:
a) 25 m/s

Question 23.
A 2kg block slides on a horizontal floor with a speed of 4 m/s. It strikes on uncompressed spring and compresses it fill the block is motionless. The kinetic friction force is 15N and spring constant is 104N/m. The spring compress by _______.
a) 5.5 cm
b) 2.5 cm
c) 11 cm
d) 8.5 cm
Answer:
a) 5.5 cm

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 24.
If two objects of masses m1 and m2 (m1 > m2) are moving with the same momentum then the kinetic energy will be greater for
(a) m1
(b) m2
(c) m1 or m2
(d) both will have equal kinetic energy
Answer:
(b) m2

Question 25.
The power of a pump which can pump 200 kg of water to a height if 200m in 10s is _______. (g = 10ms-2)
a) 40 kW
b) 80 Kw
c) 400 Kw
d) 960 Kw
Answer:
a) 40 Kw

Question 26.
A body of mass m accelerates uniformly from rest to velocity v0 in a time t0. The instantaneous power delivered to the body at any time t is _______.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 36
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 37

Question 27.
particle to mass ‘m’ starting from rest undergoes uniform acceleration. If the speed acquired in time T is V, the power delivered to the particle is _______.
a) \(\frac{M V^{2}}{T}\)
b)\(\frac { 1 }{ 2 }\)\(\frac{M V^{2}}{T^{2}}\)
c) \(\frac{M V^{2}}{T^{2}}\)
d) \(\frac { 1 }{ 2 }\)\(\frac{M V^{2}}{T}\)
Answer:
d) \(\frac { 1 }{ 2 }\)\(\frac{M V^{2}}{T}\)

Question 28.
A particle is placed at the origin and a force F = kx is acting on it (where k is the positive constant). If U (o) = 0, the graph u(x) verses x will be _______. (u=potential energy function)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 38
Answer:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 39

Question 29.
The force acting on a body moving along x-axis varies with the position of the particle as in fig _______.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 40
The body in stable equilibrium at
a) x = x1
b) x = x2
c) both x1, and x2
d) neither x1 nor x2
Answer:
b) x = x2

Question 30.
Non-conservative force is
(a) frictional force
(b) viscous force
(c) air resistance
(d) all the above
Answer:
(d) all the above

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 31.
If the K.E of the body is increased by 300% then for the percentage change in momentum will be _______.
a) 100%
b) 150%
c) 265%
d) 73.2%
Answer:
a) 100%

Question 32.
A particle is projected making an angle 45’ with horizontal having kinetic energy k. The K.E at the highest point will be _______.
a) K/\(\sqrt{2}\)
b) K/2
c) 2K
d) K
Answer:
b) K/2

Question 33.
The coefficient of restitution e for a perfectly elastic collision is _______.
a) 1
b) 0
c) ∞
d) -1
Answer:
a) 1

Question 34.
A ball moving with a certain velocity hits another identical ball at rest. If the plane is frictionless and collision is elastic, the angle between the directions in which the balls move after collision will be _______.
a) 30°
b) 60°
c) 90°
d) 120°
Answer:
c) 90°

Question 35.
Which of the following is zero at the highest point in vertical circular motion?
(a) velocity of the particle
(b) tension of the spring
(c) potential energy
(d) none
Answer:
(a) velocity of the particle

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

II. Additional Questions:

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative.
(a) Work done by man lifting a bucket out of the well by means of a rope tied to the bucket
Solution:
Work done is positive, because the bucket moves in the direction of force applied by the man.

(b) work done by gravitational of force in the above case.
Solution:
work done by gravitational force in negative because the bucket is moving upwards against gravitational force.

(c) Work done by friction on a body sliding down an inclined plane.
Solution:
Work done is negative, as frictional force is opposite to direction of motion.

(d) work done by the applied force on a body moving on a rough horizontal plane with uniform velocity.
Solution:
Work done is positive because the applied force acts in the direction of motion of the body

(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution:
Work done is negative, because the resistive force of air acts in a direction opposite to the direction of motion of the vibration pendulum.

Question 2.
Comets move around the sun in highly elliptical orbits. The gravitational force an the comet due to sun is not normal to the comet velocity in general. Yet the W.D by the gravitational force over every complete orbit of the comet is zero why?
Solution:
The gravitational force acting on the comet is a conservative force. The work done by the conservative force over any path is equal to the negative of change in potention energy. Over a complete orbit of any shape, there is no change in P.E of the comet. Hence no work is done by the gravitational force on the comet.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 3.
A body is moving unidirectionally under the influence of a source of constant power. How displacement (S) and time (t) are related?
Answer:
By work energy theorem
W = Pxt = 1/2 mv²
v² = \(\frac { 2pt }{ m }\)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 41

Question 4.
A molecule in a gas container hits a horizontal wall with speed 200 ms-1 and angle 30° with normal and rebounds with same speed. Is momentum conserved in collision? Is the conserved is elastic or in elastic.
Solution:
Momentum is always conserved, whether collision is elastic or in elastic
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 42
Let m be the mars of the molecule and M be the mass of the wall. As the wall is heavy the molecule rebounding with same velocity does not produce any velocity to wall.
K.E. before collision = Ki = 1/2 m(200)² + 0
= 2 x 104 x 1/2m – 3
K.E. after collision = Kf = 1/2 m(200)² + 0
= 1/2 x m x 2 x w4
∵ Ki = Kf
Collision is elastic in nature

Question 5.
An elastic spring of force constant K is compressed by a amount x. Show that its potential energy is 1/2 kx².
Answer:
Consider the plot of spring force Fs versus displacement x of the block attached to free end of a spring.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 43
According to Hookes law Fs = – k xm
The work done by the spring force for an extension xm is
Ws = Area ∆OBA = 1/2 AB x OB
= 1/2 Fs Xm
= 1/2 (-kxm) xm²
Ws = – 1/2 kxm²
In order stretch the spring slowly an external force F equal to and opposite to Fs has to be applied. So work done by the external force F is
W = – Ws = + 1/2 kxm²
This work done is stored as P.E of the spring
u = 1/2 kx²

Question 6.
Obtain the relation between momentum and K.E
Answer:
Consider an object of mass on moving with velocity v. Then its linear momentum is \(\overline{p}\) = \(\overline{mv}\) and its kinetic energy
k.E = 1/2 mv²
multiply and divide by m both numerator and denominator
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 44

Question 7.
Explain motion of a body in vertical circle.
Answer:
Consider a body of mass m attached to one end of marsless and inextensible string executes circular motion is a vertical plane with other end fixed at 0. The length of the string equal to radius.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 45
There are two forces acting an mass m.
i) Gravitational force (mg) which acts downwards
ii) Tension along the string (T)
Applying Newtons II law
mg sin θ = m at
mg sin θ = – m\(\frac { dv }{ dt }\) → (1)
Where at = tangential acceleration
In the radial direction
T – mg cos θ = m ar
T – mg cos θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
Where ar = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = centripetal or radial acceleration from the above 2 equations four important facts to get understood are.

1) The mass is having tangential acceleration (g sin θ) for all values of 0 (expect θ =0°). It is clear that vertical circular motion is not a uniform circular motion

2) From (1) & (2) as the magnitude of velocity is not constant in the course of motion, the Tension in the string is also not constant.

3) From equ(2) T = mg cos θ + \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\), indicates that in section A and D (for – π/2 < θ < π/2) cosθ is five, ie the term mg cos θ is always > 0.
Hence Tension can not vanish even when the velocity vanishes.

4) The equation (2) \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = T – mg cos θ further indicates that in section B and C
(for π/2< θ < 3π/2) cosθ is negative.
(-mg con θ) is always greater than zero. Hence velocity cannot vanish even tension vanishes.
Consider two positions, lowest position (1) and the highest position 2.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 46
Let v1 be the velocity at the lowest point 1.
Let v2 be the velocity at the lowest point 2.
The direction of velocity is always tangential to the circular path at all points.
Let T1 and T2 be the Tensions at lowest end highest point respectively. Tension at each point act towards the centre. Tension and Similarly velocity can be determined by applying the law of conservation of energy.
For the lowest point (1)
when the body is at its lowest point (1) the gravitational force mg which acts vertically downwards and Tension (T1) vertically upwards.
 Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 47
For the highest point:
At the highest point (2) both gravitational force mg and tension T2 act vertically downwards.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 48
Here the term can (v1² – v2²) be found by applying law of conservation of energy at point 1 and 2.
Total energy at 1 = Total energy at 2
E1 = E2
P.E at 1, u1 = 0
K.E at 1, KE1 = 1/2 mv1²
T.E at 1 = 1/2 0 + mv1² = 1/2 mv1² → (8)
Similarly
P.E at 2, u2 = mg(2r)
K.E at 2, K.E2 = 1/2 mv²
T.E at 2 = mg(2r) + 1/2 mv22 → (9)
equating (8) and (9)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 49
(v1² – v2²) = 4gr → (10)
Substituting (10) in (7)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 50
Minimum speed at the highest point (2)
In order to loop the circle, the body must have a minimum speed at point 2 – To find minimum speed at 2 consider 2.
Consider T2 = 0 in equation … (6)
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 51
At highest point 2 the body should have a minimum velocity v2 = \(\sqrt{gr}\) to stay in circular path.
Minimum speed at the lowest point:
From equation 10
v1² – v2² = 4gr
v1² – gr = 4gr
v1² = \(\sqrt{5gr}\)
The body must have a speed of \(\sqrt{5gr}\) at the lowest point 1.
ie v1 ≥ \(\sqrt{5gr}\) so that the mars can stay in circular path.
Comparing values of v1 & v2 it is clear that minimum speed at the lowest point should be \(\sqrt{5}\) times more than the minimum speed at the highest point, so that the body loops in a vertical circle.

Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power

Question 8.
What is perfect in elastic collision?
Answer:
Obtain an expression for velocity after collision. A perfect or complete in elastic collision is the are in which the object stick permanently after collision such that they move with a common velocity.

Expression for common velocity:
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 52
Let two bodies of masses m1 and m2 move with initial velocities u1 and u2 respectively before collision. After perfect inelastic collision both two objects move together with a common velocity v.

Since linear momentum is conserved during collisions.
m1u1 + m2u2 = (m1 + m2)v
V = \(\frac{\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\)
This gives the expression for common velocity.

Question 9.
Obtain an expression for loss of K.E. in perfect elastic collision.
Answer:
Let KEi be the initial total kinetic energy before collision and KEf be the total final kinetic energy after collision.
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 53
Samacheer Kalvi 11th Physics Guide Chapter 4 Work, Energy and Power 54

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 2 Motion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 2 Motion

9th Science Guide Motion Text Book Back Questions and Answers

I. Choose the correct answer:

Online Average Acceleration calculator can be used to calculate Acceleration,​initial and final velocity and time taken.

Question 1.
The area under velocity – time graph represents the
(a) velocity of the moving object.
(b) displacement covered by the moving object,
(c) speed of the moving object.
(d) acceleration of the moving object.
Answer:
(b) displacement covered by the moving object

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 2.
Which one of the following is most likely not a case of uniform circular motion?
(a) Motion of the Earth around the Sim.
(b) Motion of a toy train on a circular track.
(c) Motion of a racing car on a circular track.
(d) Motion of hours’ hand on the dial of the clock.
Answer:
(c) Motion of a racing car on a circular track

Question 3.
Which of the following graph represents uniform motion of a moving particle?
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 1
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 2
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 3

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 4.
The centrifugal force is
(a) a real force.
(b) the force of reaction of centripetal force.
(c) a virtual force.
(d) directed towards the centre of the circular path.
Answer:
(c) a virtual force

II. Fill in the blanks :

1. Speed is a …………………….quantity whereas velocity is a …………….quantity.
Answer:
Scalar, Vector

2. The slope of the distance – time graph at any point gives ………………
Answer:
Speed

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

3. Negative acceleration is called ……………………
Answer:
retardation (or) deceleration

Displacement calculator is really a nice calculator that will help you solve a lot of your problems related to displacement.

4. Area under velocity – time graph shows ……………………………….
Answer:
displacement

III. State whether true or false. If false, correct the statement:

1. The motion of a city bus in a heavy traffic road is an example for uniform motion.
Answer:
False.
Correct statement: The motion of a city bus in a heavy traffic road is an example for non-uniform motion.

2. Acceleration can get negative value also.
Answer:
True.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

3. Distance covered by a particle never becomes zero but displacement becomes zero.
Answer:
True.

4. The velocity – time graph of a particle falling freely under gravity would be a straight line parallel to the x axis. .
Answer:
False.
Correct statement: The velocity – time graph of aparticle moving at uniform do dry. would be straight line parallel to the x axis.

5. If the velocity – time graph of a particle is a straight line inclined to X-axis then its displacement – time graph will be a straight line.
Answer:
True.

IV. Assertion and Reason Type Questions :

Mark the correct choice as:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 1.
Assertion : The accelerated motion of an object may be due to change in magnitude of velocity or direction or both of them..
Reason : Acceleration can be produced only by change in magnitude of the velocity. It does not depend the direction.
Answer:
(c) If assertion is true but reason is false.

Use Average Velocity Calculator to find the average velocity, initial velocity, and final velocity with average velocity formula (vavg=(v+u)/2).

Question 2.
Assertion : The Speedometer of a car or a motor-cycle measures its average speed.
Reason : Average velocity is equal to total displacement divided by total time taken.
Answer:
(d) Assertion is false but reason is true

Question 3.
Assertion : Displacement of a body may be zero when distance travelled by it is not zero.
Reason : The displacement is the shortest distance between initial and final position.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

V. Match the Following :

Samacheer Kalvi 9th Science Guide Chapter 2 Motion 4
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 5

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

VI. Answer briefly :

Question 1.
Define velocity.
Answer:
Velocity is the rate of change of displacement. It. is the displacement in unit time.

Question 2.
Distinguish distance and displacement.
Answer:
Table
Distance

  1. The actual length of the path travelled by a moving body irrespective of the direction
  2. It is a Scalar quantity

Displacement:

  1. The change in position of a moving body in a particular direction
  2. It is a Vector quantity

Question 3.
What do you mean by uniform-motion?
Answer:
An object is said to be in uniform motion if it covers equal distances in equal intervals of time how so ever big or small these time intervals may be.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 4.
Compare speed and velocity.
Answer:

Speed Velocity
1. It is the rate of change of distance with respect to time It is the rate of change of displacement with respect to time
2. It is a scalar quantity having magnitude only It is a vector quantity having both magnitude and direction
3. Speed is velocity without a particular direction Velocity is speed in a particular direction
4. It is measured in ms-1 in SI system It is also measured in ms-1 in a particular direction in SI system
5. Speed in any direction would be a positive quantify, since the distance in any direction is a positive quantity. Velocity can get both positive and negative values. If velocity in one direction is assumed to be positive then the velocity in the opposite direction would be a negative quantity. Velocity can get zero value also.

Question 5.
What do you understand about negative acceleration?
Answer:
If v < u, i.e. if final velocity is less than initial velocity, the velocity decreases with time and the value of acceleration is negative. It is called negative acceleration. It is also called as retardation (or) deceleration.

Question 6.
Is the uniform circular motion accelerated? Give reasons for your answer.
Answer: When an object is moving with a constant speed along a circular path, the change in velocity is only due to the change in direction. Hence it is accelerated motion.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 7.
What is meant by uniform circular motion? Give two examples of uniform circular motion.
Answer:
When an object moves with constant speed along a circular path, the motion is called uniform circular motion.
Example :

  • The earth moves around the sun in the uniform circular motion.
  • The moon moves in uniform circular motion around the earth.

VII. Answer in detail :

Question 1.
Derive the equations of motion by graphical method.
Answer:
Equations of motion from velocity-time graph:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 6

Graph shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t it reaches the point B on the graph.
The initial velocity of the object = u = OD = EA
The final velocity of the object = v = OC = EB
Time = t = OE = DA
Also from the graph we know that, AB = DC

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

1. First equation of motion :

By definition, acceleration = change in velocity / time
= (final velocity – initial velocity)/time
= (OC – OD) / OE
= DC / OE
a = DC /t
DC = AB = at
From the graph EB = EA + AB
v = u + at…………..(1)
This is first equation of motion. .

2. Second equation on of motion :
From the graph the distance covered by the object during time t is given by the area of quadrangle DOEB

s = area of the quadrangle DOEB
= area of the rectangle DOEA + area of the triangle DAB
= (AE x OE) + (1/2 × AB × DA) .
s = ut + 1/2 (at2) …………(2)
This is the second equation of motion.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

3. Third equation of motion :
From the graph the distance covered by the object during time, t is given by the area of the quadrangle DOEB. Here DOEB is a trapezium. Then,
s = area of trapezium DOEB
= 1/2 × sum of length of parallel side × distance between parallel sides
= 1/2 × (OD + BE) × OE
S = 1/2 ×× (u + v) × t
since a = (v – u) /1 or t = (v – u) / a
Therefore s = 1/2 × (v + u) x (v – u) / a
2as = v2 – u2
v2 = u2 + 2 as …………….. (3)
This is the third equation of motion.

Question 2.
Explain different types of motion.
Answer:
In physics, motion can be classified as below.

1. Linear motion: Motion along a straight line.

2. Circular motion: Motion along a circular path.

3. Oscillatory motion: Repetitive to and fro motion of an object at regular intervals of time. Random motion: Motion of the object which does not fall in any of the above categories.

Uniform and Non-uniform motion:

Uniform motion: Consider a car that covers 60 km in the first hour, 60 km in the second horn-, and another 60 km in the third hour, and so on. The car covers equal distance at equal intervals of time. We can say that the motion of the car is uniform. An object is said to be in uniform motion if it covers equal distances in equal intervals of time howsoever big or small these time intervals may be.

Non-uniform motion: Now, consider a bus starting from one stop. It proceeds slowly when it passes through a crowded area of the road. Suppose, it manages to travel merely 100 m in 5 minutes due to heavy traffic and is able to travel about 2 km in 5 minutes when the road is clear. Hence, the motion of the bus is non-uniform i.e. it travels unequal distances in equal intervals of time.

VIII. Exercise Problems:

Question 1.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Here we have
Initial velocity, u = 0
Distance, s = 20 m
Acceleration, a = 10 m/s2
Final velocity, v = ?
Time, t = ?

a) Calculation of final velocity, v
We know that, v2 = u2 + 2as
v2 = 0 + 2 × 10m/s2 × 20m
v2 = 400m2/s2
\(=\sqrt{400 m^{2} / s^{2}}\)
v = 20m/s

b) Calculation of time, t
We know that, v = u + at
20 m/s = 0 + 10m/s2 × t
t = \(\frac{20 m / s^{2}}{20 m / s}=2 s\)
∴ Ball will strike the ground at a velocity of 20 ms-1
Time taken to reach the ground = 2s.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 2.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 m and 20 s?
Answer:
Here we have
Diameter = 200. m
∴ Radius = 200 m/2 = 100 m
Time of one rotation = 40 s
Time after 2m 20 s = 2 × 60 s +20 s = 140 s
Distance after 140 s = ?
Displacement after 140 s = ?
Circular track with a diameter of 200m
We know that, velocity
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 7
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 8

a) Distance after 140 s
We know that, distance = velocity × time
= Distance 15.7 rn/s × 140 s
= 2198m

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

b) Displacement after 2 min 20 s i.e, in 140 s
We know that, distance = velocity × time
Since, rotation in 40 s = 1
∴ Rotation in 1 s = \(\frac{1}{40}\)
∴ Rotation in 140 s = \(\frac{1}{40} \times 140=3.5\)
∴ In 3.5 rotation athlete will be just at the opposite side of the circular track,
i.e. at a distance equal to the diameter of the circular track which is equal to 200m
∴ Distance covered in 2min 20 s = 2198 m
Displacement after 2min 20 5 = 200 m.

Question 3.
A racing car has a uniform acceleration of 4ms-2. What distance it covers in 10s after the start?
Answer:
Here we have
Acceleration, a = 4 m/s2.
Initial velocity u = 0
Time t = 10 s
Distance (s) covered =?
We know that, s = ut + \(\frac { 1 }{ 2 }\) at2
s = (0 × 10s) + [\(\frac { 1 }{ 2 }\) × 4 m/s2 × (10 s)2]
= \(\frac { 1 }{ 2 }\) × 4 m/s2 × 100 s2
= 2 × 100 m = 200 m

Thus, racing car will cover a distance of 200 m after start in 10s with given acceleration.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Intex Activities

ACTIVITY – 1

Look around you. You can see many things: a row of houses, large trees, small plants, flying birds, running cars and many more. List the objects which remain fixed at their position and the objects which keep on changing their position.
Solution:

  1. Row of houses, large trees, small plants are the examples, of immovable objects.
  2. Flying birds, running cars and buses are the examples of movable objects.

Activity to be done by the students themselves

ACTIVITY – 2

Tabulate the distance covered by a bus in a heavy traffic road in equal intervals of time and do the same for a train which is not in an accelerated motion. From your table what do you understand?
The bus covers unequal distance in equal intervals of time but the train covers equal distances in equal intervals of time.
Solution:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 9

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

ACTIVITY – 3

Observe the motion of a car as shown in the figure and answer the following questions:
Compare the distance covered by the car through the path ABC and AC. What do you observe? Which path gives the shortest distance to reach D from A? Is it the path ABCD or the path ACD or the path AD?
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 10
Solution:

  1. Distance covered by the car through the path ABC = 4m + 3m = 7 m. and AC = 5 m. The distance covered by the car through the path ABC is large compared to AC.
  2. The shortest distance to resell D from A is path AD = 3m.
  3. The total distance covered by the car ABCD A = 14 m. It finally reaches to A.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

ACTIVITY – 4

Take a large stone and a small eraser. Stand on the top fit a table and drop them simultaneously from the same height? What do you observe? Now, take a small eraser and a sheet of paper. Drop them simultaneously from the same height. What do you observe? This time, take two sheets of paper having same mass and crumple one of the sheets into a ball. Now, drop the sheet and the ball from the same height. What do you observe?
Solution :
Both the stone and the eraser have reached the surface of the earth almost at the same time.
The eraser reaches first and the sheet of paper reaches later.
The paper crumpled into a ball reaches ground first and plain sheet of paper reaches later, although they have equal mass. It is because of air resistance. The magnitude of air resistance despends on the area of object exposed to air. So the sheet of paper reaches later.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

9th Science Guide Motion Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
A particle is moving in a circular pattern of radius r. The displacement after half a circle would be
(a) zero
(b) πr
(c) 2r
(d) 2πr
Answer:
(c) 2r

Question 2.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving in the straight road.
(b) If the car is moving in a circular road.
(c) The earth is revolving around the sun.
(d) The pendulum is moving to and fro
Answer:
(a) If the car is moving in the straight road

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 3.
A body is thrown vertically upward with velocity, the greatest height h to which it will rise is
(a) u2/2g
(b) u2/g
(c) u/g
(d) u/2g
Answer:
(a) u2/2g

Question 4.
If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
Answer:
(b) uniform acceleration

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 5.
From the given v-t graph, u can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration
Answer:
(a) in uniform motion

Question 6.
The area under v-t graph represents a physical quantity which has the unit.
(a) m2
(b) m
(c) m3
(d) ms-1
Answer:
(b) m

Question 7.
m/s2 is the unit of
(a) distance
(b) displacement
(c) velocity
(d) acceleration
Answer:
(d) acceleration

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 8.
The rate of change of displacement
(a) speed
(b) velocity
(c) acceleration
(d) retardation
Answer:
(b) velocity

Question 9.
A scalar quantity has T
(a) magnitude only
(b) direction only
(c) both
(d) none
Answer:
(a) magnitude only

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 10.
When an object undergoes acceleration
(a) there is always an increase in its velocity
(b) there is always an increase in its speed
(c) a force always acting on it.
(d) all of the above
Answer:
(c) a force always acting on it

Question 11.
A body is projected up with an initial velocity u m/s. It goes up to a height, ‘h’ metres in seconds time. Then it comes back at the point of projection. Considering negligible air resistance, which of the following statement is true?
(a) the acceleration is zero
(b) the displacement is zero
(c) the average velocity is 2hit
(d) the final velocity is 2u when body reaches projection point
Answer:
(b) the displacement is zero

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 12.
A car accelerates at 1.5m/s2in a straight road. How much is the increase in velocity in 4s.
(a) 6 m/s
(b) 4 m/s
(c) 3 m/s
(d) 2.66 m/s
Answer:
(a) 6 m/s

Question 13.
The slope of the distance-time curve is steeper / greater is the
(a) velocity
(b) acceleration
(c) displacement
(d) speed
Answer:
(d) speed

Question 14.
The given graph represents motion with …………….speed.
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 11
(a) uniform
(b) non-uniform
(c) constant
(d) none
Answer:
(b) non-uniform

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 15.
The relation between displacemeñt and time is given by the equation of
(a) v2 = ut + at
(b) s = ut + \(\frac { 1 }{ 2 }\) at2
(e) c = s/t
(d) v2 = u2 + 2as
Answer:
(b) s = ut + \(\frac { 1 }{ 2 }\) at2

Question 16.
A body moves in a uniform circular motion
(a) It is moving with constant velocity
(b) its acceleration is zero
(c) the body has an acceleration y
(d) none of the above
Answer:
(a) It is moving with constant velocity

Question 17.
Speed of the body in particular direction can be called
(a) acceleration
(b) displacement
(c) velocity
(d) distance
Answer:
(c) velocity

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 18.
Statement A: Uniform circular motion is a case of accelerated motion
Statement B: In third equation of motion we do not have the term time
(a) Statement B is true, A is false
(b) Statement A is true, B is false
(c) neither statement A nor B is true
(d) both are true
Answer:
(d) both are true

Question 19.
Which of the following is correct about uniform circular motion
(i) direction of motion is continuously changed
(ii) direction of motion is not changed
(iii) speed and direction both remain constant
(iv) speed is constant but direction is changing
(a) ii & iii are correct
(b) i, ii & iii are correct
(c) i & iv are correct
(d) all of these
Answer:
(c) i & iv are correct

Question 20.
Which of the quantities have the same SI unit?
(a) speed, velocity
(b) acceleration, time
(c) velocity, time
(d) velocity, acceleration
Answer:
(a) speed, velocity

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 21.
Rest and motion of body are
(a) non-relative
(b) not related
(c) relative
(d) none
Answer:
(c) relative

Question 22.
An ant moves from one corner of a room diagonally to the opposite corner. If the dimensions of the hall are 8m x 6m, the displacement of the ant is
(a) 10m
(b) 14m
(c) 28m
(d) 2m
Answer:
(a) 10m

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 23.
The displacement covered by a second hand of radius V in a clock after one revolution is
(a) 360°
(b) 0
(c) 3r
(d) 2r
Answer:
(b) 0

Question 24.
A man leaves his house at 6.30 a.m. for a morning walk and returns back at 7.30 a.m. after covering 4 km. Displacement covered by him is …………….
(a) 2 km
(b) zero
(c) 8 km
(d) 4 km
Answer:
(b) zero

Question 25.
A body is said to be in non uniform motion if it travels
(a) equal distance in unequal interval of time
(b) equal distance in equal interval of time
(c) unequal distance in unequal interval of time
(d) unequal distance in equal interval of time.
Answer:
(d) unequal distance in equal interval of time

Question 26.
A quantity which has both magnitude and direction is
(a) scalar
(b) distant
(c) vector
(d) moving body
Answer:
(c) vector

Question 27.
A bus accelerating with 4ms-2 changes its speed from 60ms_1 to a certain value in 5s. The final speed is
(a) 40 m/s
(b) 25 ms-1
(c) 60 ms-1
(d) 30 ms-1
Answer:
(a) 40 m/s

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 28.
A quantity has a value of 16ms-2. It is the
(a) acceleration of an object
(b) velocity of an object
(c) retardation of an object
(d) speed of an object
Answer:
(c) retardation of an object

Question 29.
A boy throws a ball up and catches it when the ball falls back. In which part of the motion the ball is accelerating?
(a) during downward motion
(b) when the ball comes to rest
(c) during upward motion
(d) when the boy catches the ball.
Answer:
(a) during downward motion]

Question 30.
Choose the correct option.
(a) distance is a scalar, velocity is a vector, acceleration is a vector
(b) distance is a vector, velocity is a scalar, acceleration is a vector
(c) distance is a vector, velocity is a vector, acceleration is a vector
(d) distance is a scalar, velocity is a vector, acceleration is scalar
Answer:
(a) distance is a scalar, velocity is a vector, acceleration is a vector

Question 31.
If a moving body comes to rest, then its acceleration is
(a) positive
(b) negative
(c) zero
(d) all of these depending upon initial velocity.
Answer:
(b) negative

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 32.
If the velocity of a body changes uniformly from u to v in time t, the sum of average velocity and acceleration is
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 12
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 13
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 14

Question 33.
Acceleration is defined as the rate of change of
(a) distance
(b) velocity
(c) speed
(d) displacement
Answer:
(b) velocity

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 34.
When an object undergoes acceleration
(a) there is always an increase in its velocity ‘
(b) there is always an increase in its speed
(c) a force always acting on it.
(d) all the above
Answer:
(c) a force always acting on it

Question 35.
The equation v = u + at gives information as
(a) velocity is a function of time
(b) velocity is a function of position
(c) position is a function of time
(d) position is a function of time and velocity
Answer:
(a) velocity is a function of time

Question 36.
Which of the following can determine the acceleration of a moving object.
(a) area of the velocity-time graph
(b) the slope of the velocity-time graph
(c) area of a distance-time graph
(d) the slope of a distance-time graph
Answer:
(b) slope of the velocity-time graph

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 37.
What is the slope of the body when it moves with uniform velocity?
(a) positive
(b) negative
(c) zero
(d) may be positive or negative
Answer:
(c) zero]

Question 38.
If a body starts from rest, what can be said about the acceleration of the body?
(a) positively accelerated
(b) negative accelerated
(c) uniform accelerated
(d) none of the above
Answer:
(a) positively accelerated

Question 39.
When a body moves uniformly along the circle then
(a) its velocity changes but speed remain the same
(b) its speed changes but velocity remains the same
(c) both speed and velocity changes
(d) both speed and velocity remains same
Answer:
(a) its velocity changes but speed remains the same

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 40.
Distance travelled by a freely falling body is proportional to
(a) mass of the body
(b) square of the acceleration due to gravity
(c) square of the time of fall
(d) time of fall
Answer:
(c) square of the time of fall

Question 41.
If the displacement-time graph of a particle is parallel to the time axis, then velocity of the particle is.
(a) infinity
(b) unity
(c) equal to acceleration
(d) zero
Answer:
(d) zero

Question 42.
In the velocity-time graph, AB shows that the body has
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 15
(a) uniform acceleration
(b) non-uniform retardation
(c) uniform speed
(d) initial velocity OA & is moving with uniform retardation.
Answer:
(d) initial velocity OA & is moving with uniform retardation

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 43.
The magnitude of the centripetal force is given by (F= ….)
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 16
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 17
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 18

Question 44.
A body moving with an initial velocity 5ms-1 and accelerates at 2ms -2. Its velocity after 10s is
(a) 20ms-1
(b) 25ms-1
(c) 5ms-1
(d) 22.55ms-1
Answer:
(b) 25ms-1

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 45.
In a 100m race, the winner takes 10s to reach the finishing point. The average speed of the winner is
(a) 5ms-1
b) 20ms-1
c) 40ms-1
d) 10ms-1
Answer:
(d) 10ms-2

Question 46.
The area under the velocity-time graph represents
(a) the velocity of the moving object
(b) displacement covered by the moving object
(c) speed of the moving object.
(d) acceleration of the moving object
Answer:
(b) displacement covered by the moving object

Question 47.
A car is being driven at a speed of 20ms-1 when brakes are applied to bring it to rest in 5 s. The deceleration produced in this case will be
(a) +4ms-2
(b) -4ms-2
(c) -0.25ms-2
(d) +0.25ms-2
Answer:
(b) -4ms-2

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 48.
Unit of acceleration is
(a) ms-1
(b) ms-2
(c) ms
(d) ms2
Answer:
(b) ms-2

Question 49.
The force responsible for drying clothes in a washing machine is …………….
(a) Centripetal force
(b) Centrifugal force
(c) Gravitational force
(d) Electro static force
Answer:
(b) Centrifugal force

II. Fill in the blanks:

1. If a body does not change its position, then it is said to be at …………………
Answer:
rest

2. The back and forth motion of a swing is an ………………… motion.
Answer:
Oscillatory

3. In uniform motion an object travels equal ………………… in ………………… interval of time.
Answer:
distances, equal

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

4. The actual path covered by a body is called …………………
Answer:
distance

5. Displacement is the ………………… distance covered by a body.
Answer:
shortest

6. The motion of the bus is ………………… motion.
Answer:
non-uniform

7. Rate of change of displacement is …………………
Answer:
velocity

8. Speed is a ………………… quantity whereas velocity is a …………………
Answer:
scalar, vector

9. If final velocity is less than initial velocity the acceleration is ………………….
Answer:
negative

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

10. If final velocity is equal to initial velocity the value of acceleration is ………………….
Answer:
zero

11. The slope of the distance-time graph becomes steeper & steeper the speed ………………….
Answer:
increases

12. A straight line parallel to the x-axis in the velocity-time graph, represents the object moves in…………………
Answer:
uniform velocity

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

13. From v-t graph ………………… can be calculated.
Answer:
displacement

14. ………………… measures the instantaneous speed of the automobile.
Answer:
Speedometer

15. Slope of the velocity-time graph gives…………………
Answer:
acceleration

16. The value of acceleration for a body at rest is…………………
Answer:
zero

17. At the highest point, when a body is thrown vertically upwards, the velocity is ………………….
Answer:
zero

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

18. A body moves in a circular pattern the ………………… of velocity does not change but changes.
Answer:
magnitude, direction

19. When a body moves in a circular pattern ………………… acceleration is directed radially towards the centre of the circle.
Answer:
centripetal

20. The separation of cream from milk-hn example for the application of ………………….
Answer:
centrifugal

21. Consider an object is rest at position x = 20m. Then its displacement – time graph will be straight-line ………………… to the time axis.
Answer:
Parallel

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

III. State whether true or false. If false, correct the statement:

1. Displacement can be zero but distance never.
Answer:
True.

2. Time is a vector quantity.
Answer:
False.
Correct statement: Time is a scalar quantity.

3. Displacement magnitude can be greater than the distance travelled by the object.
Answer:
True.

4. If the velocity of the body decreases with time the acceleration is negative and the motion is called decelerated motion.
Answer:
True.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

5. Acceleration is a scalar.
Answer:
False.
Correct statement: Acceleration is a vector.

6. The area of the velocity-time graph gives the displacement of the body.
Answer:
True.

7. Motion & rest are relative terms.
Answer:
True.

8. An object can be moving with uniform speed but variable acceleration.
Answer:
True.

9. Slope of the distance-time graph indicates the speed.
Answer:
True.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

10. It is possible to have object moving with uniform velocity but non-uniform acceleration.
Answer:
True.

11. It is possible to have object moving with uniform speed but variable acceleration.
Answer:
False.
Correct statement : It is possible to have object moving with uniform speed but constant acceleration.

12. The force experienced by a boy in the merry-go-round is a centripetal force.
Answer:
False.
Correct statement: The force experienced by a boy in the merry-go-round is a centrifugal force.

13. The initial velocity of a freely falling object is zero as it is released from rest.
Answer:
True.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

IV. Assertion and Reason Type Questions :

(a) If both assertion & reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion & reason are true but the reason is not correct explanation of the assertion.
(c) If the assertion is true but the reason is false.
(d) If assertion & reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion : A body can have acceleration even if its velocity is zero at a given instant of time.
Reason: A body is momentarily at rest when it reverses its direction of motion.
Answer:
(a) Both assertion & reason are true and the reason is the correct explanation of the assertion

Question 2.
Assertion : If the displacement of the body is zero, the distance covered by it may not be zero.
Reason: Displacement is a vector & distance is a scalar quantity.
Answer:
(a) Both assertion & reason are true and the reason is the correct explanation of the assertion

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 3.
Assertion: An object can have constant speed but the variable velocity
Reason: Speed is a scalar but velocity is a vector.
Answer:
(a) Both assertion & reason are true and the reason is the correct explanation of the assertion

Question 4.
Assertion: The speed of a body can be Negative.
Reason : If the body is moving in the opposite direction of positive motion, then its speed is Negative.
Answer:
(d) Assertion & reason both are false

Question 5.
Assertion : The position-time graph of a uniform motion in one dimension of a body can have Negative slope
Reason : When the speed of body decreases with time then, position-time graph of the moving body has Negative slope.
[Answer:
(c) Assertion is true but the reason is false

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 6.
Assertion: A positive acceleration of a body can be associated with slowing down of the body.
Reason: Acceleration is a vector.
[Answer: (b) Both assertion & reason are true but the reason is not correct explanation of the assertion]

Question 7.
Assertion :A negative acceleration of a body can be associated with speeding up of the body.
Reason: Increase in speed of a moving body is independent of its direction of motion.
Answer:
(b) Both assertion & reason are true but the reason is not the correct explanation of the assertion

Question 8.
Assertion When a body is subjected to a uniform acceleration, it is always moving in a straight line.
Reason: Motion may be straight-line motion or circular motion.
Answer:
(e) Assertion is false but the reason is true

Question 9.
Assertion : Position-time graph of a stationary object is a straight line parallel to time axis. ’
Reason : For a stationary object, position does not change with time.
Answer:
(a) Both assertion & reason are true and the reason is the correct explanation of the assertion

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 10.
Assertion : The slope of distance-time graph of a body moving with high speed is steeper than slope of distance-time graph of a body with low velocity.
Reason: Slope of distance-time graph = speed of the body.
Answer:
(a) Both assertion & reason are true and the reason is the correct explanation of the assertion

V. Answer briefly :

Question 1.
What are centripetal acceleration and centripetal force?
Answer:
When a body moves in a circular pattern the acceleration is directed radially towards the centre of the circle.
The force causing this acceleration is also directed towards the centre of the circle and it is called centripetal force.

Question 2.
Find the magnitude of centripetal force.
Answer:
Consider an object of mass m, moving along a circular path of radius r, with a velocity v, its centripetal acceleration is given by a = v2 / r
Hence, the magnitude of centripetal force is given by,
F = mass x centripetal acceleration
F = mv2 / r

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 3.
What is centrifugal force? Give examples.
Answer:
Force acting on a body away from the centre of circular path is called centrifugal force. Thus centrifugal force is in a direction opposite to the direction of centripetal force. Its magnitude is same as that of centripetal force.
Example : Spin dryer of a washing machine, ride on a merry-go-round.

Question 4.
When an object is thrown upwards, what is true of velocity and acceleration at the highest point of motion of the object?
Answer:

  • Velocity becomes zero
  • Acceleration remains same as g.

Question 5.
Name the two quantities, the slope of whose graph gives (i) speed (ii) acceleration.
Answer:
(i) Distance – Time
(ii) Speed – Time

Question 6.
Define Average speed.
Answer:
It is the total distance travelled divided by the total time taken to cover this distance.
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 19

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 7.
What do you infer if
(i) Distance – time graph is straight line.
(ii) Velocity time graph is curved.
(iii) Displacement time is zig zag.
Answer:
(i) Speed is constant.
(ii) Acceleration is not uniform.
(iii) Non uniform velocity.

Question 8.
Give the formula for each.
(i) Relation between initial, final velocity, acceleration and displacement in a uniformly accelerated straight line motion.
(ii) Relation between initial, final velocity, acceleration & time in a uniformly accelerated straight line motion.
(iii) Relation between initial velocity, acceleration, displacement and time.
Answer:
(i) Relation between initial, final Velocity, acceleration & displacement
in a uniformly accelerated straight line motion. v2 = u2 + 2as
(ii) Relation between initial, final velocity, acceleration & time
in a uniformly accelerated straight line motion. v = u + at
(iii) Relation between initial velocity, acceleration, displacement and time. s = ut+ 1/2 at2

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 9.
What is the difference between uniform acceleration and non – uniform acceleration?
Answer:
Uniform Acceleration

  1. It is the acceleration in which the object changes its velocity with equal intervals of time.
  2. eg. The motion of a ball rolling down.

Non – Uniform Acceleration

  1. It is the acceleration in which the object changes its velocity with unequal intervals of time.
  2. A car travels 2 km in 1st hour, 3 km in 2nd hour and 3.5 km in 3rd horn.

Question 11.
Define Acceleration.
Answer:
Acceleration is the rate of change of velocity with respect to time or it is the rate of change of velocity in unit time. It is a vector quantity. The SI unit of acceleration is

VI. Paragraph Questions :

Question 1.
Define acceleration and state its SI unit for motion along a straight line, when do we consider the acceleration to be (i) positive (ii) negative? Give an example of a body in uniform acceleration.
Answer:
Answer: Acceleration is the rate of change of velocity with respect to time or it is the rate of change of velocity in unit time. It is a vector quantity. The SI unit of acceleration is ms-2
Acceleration = Change in velocity/time
= (Final velocity – initial velocity)/time
a = \(\frac{v-u}{t}\)

If v > u, then ‘a’ is positive. If final velocity is greater than initial velocity, the velocity increase with time, the value of acceleration is positive.

If v < u, then a is negative. If final velocity is less than initial velocity
Example : The motion of a freely falling body and vertically thrown up body are the examples of uniform acceleration.
The motion of ball rolling down on an inclined plane is another example.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 2.
Distinguish between uniform motion and non
Answer:
uniform motion.

  1. An object is said to be in uniform motion if it covers equkl distances in equal intervals of time.
  2. example of uniform motion ‘train’

non-uniform motion.

  1. If a body covers unequal distances in equal interval of time (or) equal distances in a different interval of time
  2. example of non – uniform motion ‘bus’

Question 3.
Define uniform circular motion and give an example of it. Why is it called accelerated motion?
Answer:
When an object is moving with a constant speed along a circular path, the change in velocity is only due to the change in direction. Hence it is accelerated motion. Example:

  1. The earth moves around the sun in a uniform circular motion.
  2. The moon moves in uniform circular motion around the earth.

Question 4.
When a body is said to be in (i) uniform acceleration (ii) non – uniform acceleration?
Answer:
(i) A body is said to be in uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal time intervals.
(ii) A body is said to be in non-uniform acceleration if the rate of change of its velocity is not constant i.e. differs in different time intervals.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 5.
What remains constant in uniform circular motion? And what changes continuously in uniform circular motion?
Answer:

  • Speed remains constant in a uniform circular motion.
  • Velocity changes continuously in a uniform circular motion.

Problems

Question 1.
A bus speed decreases from 50 km/h to 40 km/h in 3s, find the acceleration of the bus.
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 20

Question 2.
A car starting from rest moves with uniform acceleration of 0.2 ms-2 for 3 min. Fine the (a) speed acquired (b) the distance travelled.
Answer:
Initial speed (u) = 0 m/s
Acceleration (a) = 0.2 ms-2
Time taken (t) = 3 min = 3 × 60 = 180 s
Final velocity (v) = ?
Distance covered(s) = ?
v = u + at = 0 + 0.2 × 180 = 36 m/s
v = 36m/s
s = ut + 1/2 at2 = 0 + 1/2 × 0.2 × (180)2
= 0.1 × 32400 = 3240 m
s = 3240m

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 3.
A train is travelling at a speed of 90 kmh-1. Brakes are applied so as to produce a uniform acceleration of-0.5 ms-2, find how far the train will go before it is brought to rest.
Answer:
Initial velocity of train (u) = 90 km/h = \(\frac{90,000 \mathrm{~m}}{3,600 \mathrm{sec}}=25 \mathrm{~ms}^{-1}\)
Final velocity (v) = 0 ms-1
Acceleration (a) = – 0.5 ms-2
v2 = u2+ 2as
s = (v2 – u2) / 2a = (02 – 252) / -(2 × 0.5)
s = -625/-1 = 625m
s = 625m

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 4.
In a long-distance race the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 300m,
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 21
(i) What is the total distance to be covered by the athletes?
What is the total displacement of the athletes when they touch the finish line?
(iii) Is the motion of the athletes uniform or non- D .^starting point uniform?
(iv) Is the displacement & distance moved by an athlete at the end of the race equal?
Answer:
(i) Total distance covered = 4 × 300 = 1200 m
(ii) Displacement = 0 [final position – initial position]
(iii) Non – uniform.
∵ the direction of motion is changing while running on the track.
(iv) Both are not equal.

Question 5.
Ram swims in a 80m long swimming pool. He covers 160m in 1 min by swimming from one end to the other and back along the same straight pattern. Find the average speed and average velocity.
Answer:
Total distance = 160m
Total displacement = 0
Time taken (t) = 1 min = 60s
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 22

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 6.
Abus from Chennai travels to Trichy passes loo km, 160 km at 10.15 am, 11.15 am respectively. Find the average speed of the bus during 10.15 – 11.15 am.
Answer:
The distance coveredbetween 10.l5am& 11.15 am = 160 – 100
= 60km
The time interval = 1 h
Average speed = 60/1
= 60km/h

Question 7.
In a distance-time graph of two objects A & B, which object is moving with greater speed when both are moving?
Answer:
Object B makes a longer angle with the time – axis. Its slope is greater than the slope of object A. Thus the speed of B is greater than that of A.
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 23

Question 8.
Find the distance covered by a particle during the time interval which the speed-time graph is-shown in figure.
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 24
Distance covered in the. time interval 0 to 20s is equal to the area of the triangle OAB.
Area of A OAB. \(\frac { 1 }{ 2 }\) × base × height
\(\frac { 1 }{ 2 }\) × 20 x 20 = 200 ms-1

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 9.
A car moves 30 km in 30 min and the next 30 km in 40 min. Calculate the average speed for the entire journey.
Answer:
Answer:
Total time taken = 30 + 40 = 70 min. = \(\frac { 70 }{ 60 }\) hour
Total distance = 30 + 30 = 60 km
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 25

Question 10.
A boy travels a distance of 3m due east and then 4m due north.
(a) How much is the total distance covered?
(b) What is the magnitude of the displacement?
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 26
(a) Total distance covered = 3 + 4 = 7m
(b) Net displacement: OB2 = OA2 + AB2
= 32 + 42
OB2 = 25m2
∴OB = 5m 0 3m
Net displacement = 5m

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 11.
During an experiment, a signal from a spaceship reached the ground station in five seconds. What was the distance of the spaceship from the ground station? The signal
travels at the speed of light that is 3 × 108 ms-1
Answer:
Time taken = 5 seconds.
Speed of signal u = 3 × 108 m/s ?
Distance = ?
Speed = Distance / Time
∴ Distance = Speed × Time
Distance = 3 × 108 × 5 = 15 × 108 m.

Question 12.
A train travelling at a speed of 90kmph. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest?
Answer:
Here we have
Initial velocity , u = 90km/h
\(=\frac{90 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}=25 \mathrm{~m} / \mathrm{s}\)
Final velocity, v = 0
Acceleration, a = -0.5m/s2
Thus, distance travelled = ?
We know that, v2 = u2 + 2as
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 27

Question 13.
The adjacent diagram shows the velocity-time graph of the body.
a) During what time interval is the motion of the body accelerated?
Answer:
At 0 to 4 second

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

b) Find the acceleration in the time Interval mentioned in part ‘a’.
Answer:
\(\mathrm{a}=\frac{v-u}{t}=\frac{30-0}{4}=7.5 \mathrm{~m} / \mathrm{s}^{2}\)

c) What is the distance travelled by the body in the time interval mentioned ¡n part ‘a’?
Answer:Distance travelled Area under the graph
= Area of the triangle = 1/2 bh
= 1/2 × 4 × 30 = 60m

Question 14.
The following graph shows the motion of a car. What do you infer from the graph along with OA and AB? What is the speed of the car along with AB and what time it reached this speed?
a) What do you infer from the graph along OÄ and AB
Answer:
Graph along with OA: The car travels with uniform acceleration and uniform motion.
Graph along with AB : The car travels with constant speed and unaccelerated motion.
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 28

b) What is the speed of the car along AB?
Answer:
Along AB : The speed of the car is constant.
From the graph, it seems the speed along AB is 72 km/hr.

c) What time it reached this speed
Answer:
It reaches this speed after 3.2 hours, that is, 3 hours, 12 minutes.

Samacheer Kalvi 9th Science Guide Chapter 2 Motion

Question 15.
From the following table, check the shape of the graph.
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 29
Answer:
Samacheer Kalvi 9th Science Guide Chapter 2 Motion 30

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Tamilnadu State Board New Syllabus Samacheer Kalvi 5th Science Guide Pdf Term 1 Chapter 3 Energy Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 5th Science Solutions Term 1 Chapter 3 Energy

Samacheer Kalvi 5th Science Guide Energy Text Book Back Questions and Answers

Evaluation
I. Choose the correct Answer:

Question 1.
When diesel is burnt chemical energy is converted into _______.
a) wind energy
b) heat energy
c) solar energy
d) sound energy
Answer:
b) heat energy

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Kinetic Energy calculator – online mechanical engineering tool to calculate the observed energy of a body due to its motion, in both US customary & metric (SI)

Question 2.
Running water possesses _______.
a) potential energy
b) chemical energy
c) kinetic energy
d) sound energy
Answer:
c) kinetic energy

Question 3.
Unit of energy is _______.
a) Kilogram
b) Newton
c) Kelvin
d) Joule
Answer:
d) Joule

Question 4.
Which one of the following requires wind energy?
a) Bicycle
b) Photosynthesis
c) Parachute
d) Automobiles
Answer:
c) Parachute

Question 5.
Cow dung possesses _______ .
a) kinetic energy
b) chemical energy
c) solar energy
d) heat energy
Answer:
b) chemical energy

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

II. Find out the energy conversion that takes place in the following:

Question 1.
Iron box: _______.
Answer:
Chemical energy to Heat energy

Question 2.
Electric Iron box: _______.
Answer:
Electric energy to Heat energy

Question 3.
Electric fan:
Answer:
Electric energy to Mechanical energy

Question 4.
Speaker:
Answer:
Electric energy to Sound energy

Question 5.
Generator: _______.
Answer:
Mechanical energy to Electrical energy

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

III. Find out form of energy possessed by the following things:

Question 1.
A rock on the top of a hill.
Answer:
Pontential Energy

Question 2.
A rolling ball.
Answer:
Kinetic Energy

Question 3.
Charcoal.
Answer:
Heat Energy

Question 4.
Water falls.
Answer:
Kinetic Energy

Question 5.
Battery.
Answer:
Chemical Energy

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

IV. Match the following:

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy 1
Answer:

  1. e
  2. d
  3. a
  4. c
  5. b

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

V. Say True or False:

Question 1.
An apple falling from a tree is an example for kinetic energy.
Answer:
True

Question 2.
Electrical energy is used to run electric trains.
Answer:
True

Question 3.
Heat energy cannot be produced by friction.
Answer:
False

Question 4.
Potential energy and heat energy are the two forms of mechanical energy.
Answer:
False

Question 5.
The unit of energy is joule.
Answer:
True

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

VI. Answer in brief:

Question 1.
What is energy?
Answer:
Energy is defined as capacity to do work. S.I unit of work is Joule.

Question 2.
What are the different forms of energy?
Answer:
There are different forms of energy like mechanical energy, heat energy, light energy, wind energy, solar energy, electrical energy, and chemical energy.

Question 3.
What are the uses of mechanical energy?
Answer:

  • In hydroelectric plants, kinetic energy of water is converted into electrical energy.
  • Windmills convert kinetic energy of winds into electrical energy.
  • Mechanical energy of the hammer is used to apply a force on a nail.
  • Mechanical energy can bring a moving body to rest and make a body at rest to move.

Question 4.
State the Law of conservation of energy.
Answer:
Law of conservation of energy states that energy Can neither be created nor be destroyed. One form of energy is converted into another form of energy.

Question 5.
Give the uses of Light energy.
Answer:

  • We are able to see objects with the help of light energy.
  • Plants use light energy to synthesis their food.
  • With the help of light energy, our skin is able to synthesis Vitamin D.
  • Electricity can be produced with the help of light energy.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

VII. Answer in detail:

Question 1.
Explain the types of Mechanical energy.
Answer:
Energy possessed by an object due its position is called mechanical energy.
Mechanical energy can be classified into two;

  • Kinetic energy
  • Potential energy

a) Kinetic energy: Energy possessed by a moving objectis known as kinetic energy. It is also known as energy of motion.
Eg: Moving car, cricket ball bowled by a player, bullet coming out of a gun.

b) Potential energy: Energy possessed by an object which is at rest is known as potential energy. It is also known as stores energy of position.
Eg: Object lifted above, stone in the stretched rubber, water in the dam.

Question 2.
Explain Conservation of energy.
Answer:
Energy cannot be created cannot be destroyed also. It is changed from one form to another form or transferred from one object to another object. Examples for conservation of energy in our daily life are :

1. Water dam :

  • Water stored in water dams posseses potential energy.
  • When water falls down, potential energy of water is converted, into kinetic energy.
  • Kinetic energy of water rotates the turbines and electric energy is generated.

2. Electrical appliances:

  • Electrical energy is used in many domestic appliances such as electric stove, iron box and fan.
  • Electric energy flows into the coil in the devices.
  • As current flows, it heats up the coil.
  • With the help of this heat energy, we do many useful works.
  • Thus, electrical energy is converted into heat energy.
  • Electical energy is converted to mechanical energy in fan, light enrgy in bulb and sound energy in computer.

3. Driving a car:

  • We use fuel in the form of petrol or diesel or gas to run vehicles.
  • When this fuel burns in the engine, chemical energy is converted into heat energy.
  • Burning fuel produces hot gases which pushes the piston in the engine to move the vehicle.
  • Thus heat energy is converted into mechanical energy.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Samacheer Kalvi 5th Science Guide Energy Additional Questions and Answers

I. Choose the correct Answer:

Question 1.
___________ is the only form of energy visible to human eye.
a) Heat energy
b) Light energy
c) Electrical energy
d) Chemical energy
Answer:
b) Light energy

Question 2.
Light travels at a speed of _________.
a) 3,00,000 km/s
b) 4,00,000 km/s
c) 5,00,000 km/s
d) 6,00,000 km/s
Answer:
a) 3,00,000 km/s

Question 3.
Sunlight takes _________ to reach earth.
a) 5 minutes
b) 6 minutes
c) 7 minutes
d) 8 minutes
Answer:
a) 5 minutes

Question 4.
Which of the following device converts chemical energy into electrical energy?
a) Battery
b) Loud speaker
c) Solar cell
d) Electrical motor
Answer:
a) Battery

Question 5.
___________ stands first in generating electricity from wind mill.
a) Kerala
b) Tamil Nadu
c) Karnataka
d) Andhra pradesh
Answer:
b) Tamil Nadu

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Question 6.
The energy, possessed by a object due to its position is called _______.
a) Kinetic energy
b) Potential energy
c) Mechanical energy
d) Electrical energy
Answer:
b) Potential energy

Question 7.
The energy possessed by a cricket ball bowled by a player is _______.
a) Potential energy
b) Kinetic energy
c) Mechanical energy
d) Electrical energy.
Answer:
b) Mechanical energy

Question 8.
Electric eel is a _______.
a) fish
b) snake
c) frog
d) none of the above
Answer:
a) fish

Question 9.
Which of the following is the example potential energy?
a) Object lifted above
b) Bullet coming out of a sun
c) Moving car
d) Cricket ball bowled by a player
Answer:
c) Moving car

Question 10.
What energy is possessed by stretched rubber.
a) Nuclear
b) Potential
c) Kinetic
d) Thermal
Answer:
b) Potential

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

II. Fill in the blanks:

Question 1.
_______ is a measure of heat in a body.
Answer:
Temperature

Question 2.
________ use wind energy to generate electricity.
Answer:
Windmills

Question 3.
The form of energy that produces feeling of hotness is called as _______.
Answer:
heat

Question 4.
Light energy helps our skin to synthesis _________.
Answer:
vitamin – b

Question 5.
When stored water falls down, potential energy of water is converted into _______.
Answer:
Kinetic energy

Question 6.
Electric eel can generate _______.
Answer:
electricity

Question 7.
Law of conversation of energy given by _________.
Answer:
Julius Robert Mayar

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

III. Say True or False:

Question 1.
Light does not require any medium to travel.
Answer:
True

Question 2.
Mechanical energy can be used for pumping water.
Answer:
False

Question 3.
Heat energy is used to run vehicles.
Answer:
True

Question 4.
Heat is measured in Joule.
Answer:
True

Question 5.
Water stored in water dams possesses kinetic energy.
Answer:
False

Question 6.
Photosynthesis changes solar energy into chemical energy.
Answer:
True

Question 7.
Wind mills convert kinetic energy of winds into electrical energy.
Answer:
True

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

IV. Circle the odd one:

Question 1.
a) Proton
b) Dentron
c) Electron
d) Neutron
Answer:
b) Dentron

Question 2.
a) Nuclear power plants
b) Hydroelectric plants
c) Water heaters
d) Windmills
Answer:
c) Water heaters

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

V. Match the following:

Question 1:

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy 2
Answer:

  1. d
  2. a
  3. b
  4. c

Question 2.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy 3
Answer:

  1. d
  2. c
  3. b
  4. a

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

VI. Answer briefly:

Question 1.
What is mechanical energy?
Answer:

  • Energy possessed by an object due to its position is called mechanical energy.
  • Mechanical energy can be classified into two.
      • Kinetic energy
      • Potential energy

Question 2.
What is energy of motion? Give examples.
Answer:
Energy possessed by a moving object is known as kinetic energy. It is also known as energy of motion.
Examples: Moving car, Cricket ball bowled by a player, Bullet coming out of a gun.

Question 3.
Write any two uses to mechanical energy.
Answer:

  • Windmills convert kinetic energy of winds into electrical energy.
  • Mechanical energy of the hammer is used to apply a force on a Class nail.

Question 4.
What is wind energy and write its uses?
Answer:
Energy possessed by the wind is known as wind energy.
Uses of wind energy

  • Wind mills use wind energy to generate electricity.
  • Ships sail by the power of wind.
  • Sports like wind surfing, sailing, kite surfing use wind energy.
  • Wind energy can be used for pumping water.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Question 5.
Name the particles of atom?
Answer:
Atoms possess particles like protons, electrons, and neutrons.

Question 6.
What is electric energy?
Answer:
Movement of electron in the objects causes energy. This energy is called electric energy.

Question 7.
What are the different ways electricity can be generated?
Answer:
Electric energy is also generated from nuclear power plants, hydroelectric plants, and windmills. It is also generated from solar energy.

Question 8.
Write the uses of Electric energy.
Answer:

  • Electric energy is needed for the working of fan, light, television, washing machine, refrigerator, etc.
  • Electric iron box, electric stove, and electric water heater work by electrical energy.
  • It is used to run cars and trains.
  • It is used in factories to produce materials.

Question 9.
What is law of conservation of energy?
Answer:
Law of conservation of energy states that energy can neither be created nor be destroyed. One form of energy is converted into another form of energy. This law was given by Julius Robert Mayor.

Question 10.
What energy transfer happen in a car engine?
Answer:
When Petrol or diesel or gas burns in the engine, chemical energy is converted into heat energy. Burning fuel produces hot gases which pushes the piston in the engine to move the vehicle. Thus heat energy is converted into mechanical energy.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

VII. Answer in detail:

Question 1.
Write the uses of heat energy and light energy.
Answer:
Uses of heat energy:

  • Heat energy obtained from power stations is used to generate electricity.
  • Heat energy obtained from petrol and diesel is used to run vehicles.
  • We cook food with the help of heat. Heat energy renders the food material soft and easy to digest.
  • Hard substances like iron are heated to mold them into different shapes.
  • Heat is used to dry clothes and other wet substances.

Uses of light energy:

  • We are able to see objects with the help of light energy.
  • Plants use light energy to synthesis their food.
  • With the help of light energy, our skin is able to synthesis Vitamin-D.
  • Electricity can be produced with the help of light energy.

Question 2.
Explain about chemical energy and its uses.
Answer:
Chemical energy:
Chemical energy is stored in substances when atoms join together to form chemical compounds. When two or more chemical substances react with each other, this energy is released.

Uses of chemical energy:

  • The food we eat contains chemical energy.
  • Chemical energy in wood provides heat energy which helps us to cook food.
  • Chemical energy in coal is used to generate electricity.
  • Batteries we use in our daily life contain chemical energy.
  • Fuels like petrol and diesel possess chemical energy which is used to run vehicles.

Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy

Question 3.
Draw the diagram to show how the electric energy generated from water.
Answer:
Samacheer Kalvi 5th Science Guide Term 1 Chapter 3 Energy 4

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Applications Guide Pdf Chapter 18 Tamil Computing Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Applications Solutions Chapter 18 Tamil Computing

11th Computer Applications Guide Tamil Computing Text Book Questions and Answers

Part – I

I. Very Short Answers

Question 1.
List of the search engines supporting Tamil.
Answer:
Google and Bing provide searching facilities in Tamil, which means we can search everything through Tamil.

Question 2.
What are the keyboard layouts used in Android?
Answer:
Sellinam and Ponmadal – are familiar Tamil keyboard layouts that works on Android operating system in Smart phone using phonetics.

Question 3
Write a short note about Tamil Programming Language.
Answer:
Based on Python programming language, the first Tamil programming language “Ezhil” (sr^leo) is designed. With the help of this programming language, we can write simple programs in Tamil.

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 4.
What TSCII?
Answer:
TSCII (Tamil Script Code for Information Interchange) is the first coding system to handle our Tamil language in an analysis of an encoding scheme that is easily handled in electronic devices, including non-English computers. This encoding scheme was registered in IANA (Internet Assigned Numbers Authority) unit of ICANN.

Question 5.
Write a short note on Tamil Virtual Academy
Answer:
Tamil Virtual Academy:
With the objectives of spreading Tamil to the entire world through internet, Tamil Virtual University was established on 17th February 2001 by the Govt, of Tamilnadu.

Now, this organisation functioning with the name “Tamil Virtual Academy”. This organisation offers different courses regarding Tamil language, Culture, heritage etc., from kindergarten to under graduation level.

Website: http://www.tamilvu.org/index.php

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

11th Computer Applications Guide Tamil Computing Additional Important Questions and Answers

Part – I

I. Choose The Correct Answers

Question 1.
In India, _______ are using the Internet in Tamil based on Google and KPMG report.
a) 42%
b) 50%
c) 22%
d) 40%
Answer:
a) 42%

Question 2.
________ internet users consider local language digital content to be more reliable than English.
a) 74%
b) 68%
c) 42%
d) 28%
Answer:
b) 68%

Question 3.
Currently, ________ has the highest Internet adoption levels followed by the Indian language users.
a) Hindi
b) Kannada
c) Tamil
d) Telugu
Answer:
c) Tamil

Question 4
In 2021 onwards, ________ of people in India will access internet using Tamil.
a) 74%
b) 68%
c) 42%
d) 28%
Answer:
a) 74%

Question 5.
The _______ are used to search any information from the cyber space,
a) Search Engines
b) Browsers
c) Cookies
d) None of these
Answer:
a) Search Engines

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 6.
In the top ten search engines, ________ takes first place.
a) Google
b) Bing
c) Yahoo
d) None of these
Answer:
a) Google

Question 7.
In the top ten search engines, _______ takes second place.
a) Google
b) Bing
c) Yahoo
d) None of these
Answer:
b) Bing

Question 8.
In the top ten search engines, takes third place.
a) Google
b) Bing
c) Yahoo
d) None of these
Answer:
c) Yahoo

Question 9.
_______ Search engine provides searching facilities in Tamil.
a) Google
b) Bing
c) Yahoo
d) Google and Bing
Answer:
d) Google and Bing

Question 10.
A _______ search engine gives you an inbuilt Tamil virtual keyboard.
a) Google
b) Bing
c) Yahoo
d) Google and Bing
Answer:
a) Google

Question 11.
Getting Government services through internet is known as _______.
a) e-Services
b) e-Governance
c) Web Governance
d) None of these
Answer:
b) e-Governance

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 12.
Outside India, Government of _______ provides all their services through the official website in Tamil.
a) Srilanka
b) Canada
c) Nepal
d) None of these
Answer:
a) Srilanka

Question 13.
_______ are portal or website of collection of e-books.
a) E-Libraries
b) E-Learning
c) E-Content
d) None of these
Answer:
a) E-Libraries

Question 14.
_______ is a familiar Tamil keyboard interfaces software that is used for Tamil typing which works on Tamil Unicode, using phonetics.
a) NHM Writer
b) E-Kalappai
c) Lippikar
d) All the above
Answer:
d) All the above

Question 15.
________ is a familiar Tamil keyboard layouts that works on Android operating system in Smart phone using phonetics.
a) Sellinam
b) Ponmadal
c) Sellinam and Ponmadal
d) None of these
Answer:
c) Sellinam and Ponmadal

Question 16.
________ office automation software provides complete Tamil interface facility.
a) Microsoft Office
b) Open Office
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 17.
_______ is office automation software working exclusively for Tamil.
a) Azhagi Unicode Editor
b) Ponmozhi & Menthamiz
c) Kamban and Vani
d) All the above
Answer:
d) All the above

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 18.
________ is a Tamil translation application having more than 30000 Tamil words equivalent to English words.
a) Thamizpori
b) Sellinam
c) Ponmadal
d) None of these
Answer:
a) Thamizpori

Question 19.
The first Tamil programming language is _______.
a) Thamizpori
b) Ezhil
c) Kamban
d) None of these
Answer:
b) Ezhil

Question 20.
With the help of _______ programming language, you can write simple programs in Tamil.
a) Thamizpori
b) Ezhil
c) Kamban
d) None of these
Answer:
b) Ezhil

Question 21.
Expand TSCII _______.
a) Telugu Script Code for Information Interchange
b) Total Script Code for Information Interchange
c) Tamil Script Code for Information Interchange
d) Technical Script Code for Information Interchange
Answer:
c) Tamil Script Code for Information Interchange

Question 22.
Computers are handle data and information as _______ system.
a) Binary
b) Decimal
c) Ocal
d) Hexa decimal
Answer:
a) Binary

Question 23.
The ASCII encoding system is applicable only for handling _______ language.
a) English
b) Hindi
c) French
d) Tamil
Answer:
a) English

Question 24.
________ is the first coding System to handle our Tamil language in an analysis of an encoding scheme that is easily handled in electronic devices, including non-English computers.
a) ASCII
b) TSCII
c) EBCDIC
d) None of these
Answer:
b) TSCII

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 25.
IANA means _______.
a) Internet Assigned Numbers Authority
b) Internet Assigned Names Authority
c) Internet Access Numbers Authority
d) Intranet Assigned Numbers Authority
Answer:
a) Internet Assigned Numbers Authority

Question 26.
Expand ISCII
a) International Script Code for Information Interchange
b) Internet Script Code for Information Interchange
c) Indian Script Code for Information Interchange
d) Instant Script Code for Information Interchange
Answer:
c) Indian Script Code for Information Interchange

Question 27.
________ encoding schemes specially designed for Indian languages including Tamil.
a) ASCII
b) EBCDIC
c) BCD
d) ISCII
Answer:
d) ISCII

Question 28.
________ is an encoding system, designed to handle various world languages, including Tamil.
a) ASCII
b) EBCDIC
c) UNICODE
d) ISCII
Answer:
c) UNICODE

Question 29.
Unicode first version 1.0.0 was introduced on _______.
a) October 1991
b) October 2001
c) October 1993
d) October 1999
Answer:
a) October 1991

Question 30.
A(n) _______ is needed to access electronic systems such as computer and smart phone.
a) Browser
b) Operating system
c) Search Engine
d) None of these
Answer:
b) Operating system

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 31.
Identify the correct statement from the following.
a) An operating system should be easy to work and its environment should be in understandable form.
b) Windows Tamil Environment interface shows all windows elements such as Taskbar, desktop elements, names of icons, commands in Tamil.
c) Among the various encoding scheme, Unicode is the suitable to handle Tamil.
d) All the above
Answer:
d) All the above

Question 32.
Tamil Virtual University was established oh _____ by the Govt, of Tamilnadu.
a) 17th February 1991 V
b) 17th February 2011
c) 17th February 2001
d) 27th February 2001
Answer:
c) 17th February 2001

Question 33.
________ organisation offers different courses regarding Tamil language, Culture, heritage etc., from kindergarten to under graduation level.
a) Tamil Virtual Institute
b) Tamil Virtual Academy
c) Tamil Virtual Association
d) Tamil Virtual Organisation
Answer:
b) Tamil Virtual Academy

Question 34.
________ is an open and voluntary initiative to collect and publish free electronic editions of ancient tamil literary classics.
a) Project Madurai
b) Project Tamilnadu
c) Project Madras
d) Project Nellai
Answer:
a) Project Madurai

Question 35.
In 1998, Project Madurai released in Tamil script form as per ________ encoding.
a) ASCII
b) TSCII
c) BCD
d) None of these
Answer:
b) TSCII

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Part – II

II. Very Short Answers

Question 1.
Write note on search engines.
Answer:
The “Search Engines” are used to search any information from the cyber space. Although there are many search engines, but only a few of them are frequently in use. In the top ten search engines, Google, Bing and Yahoo are takes first three places respectively.

Question 2.
Which search engine provide inbuilt Tamil virtual keyboard.
Answer:
The Google search engine gives you an inbuilt Tamil virtual keyboard.

Question 3.
Write note on E-Library.
Answer:
E-Libraries are portal or website of collection of e-books. Tamil e-Library services provide thousands of Tamil Books as ebooks mostly at free of cost. It is the most useful service to Tamil people who live far away from their home land.

Question 4.
Write about Tamil translation application.
Answer:
Thamizpori is a Tamil translation application having more than 30000 Tamil words equalent to English words. Using this application, we can transaite small English sentences into Tamil.

Google also gives an online translation facility, using this online facility we can translate from Tamil to any other language vice versa.

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Question 5.
Write about Project Madurai.
Answer:
Project Madurai:
Project Madurai is an open and voluntary initiative to collect and publish free electronic editions of ancient tamil literary classics. This means either typing-in or scanning old books and archiving the text in one of the most readily accessible formats for use on all popular computer platforms. Since its launch in 1998, Project Madurai released in Tamil script form as per TSCII encoding. Since 2004 they started releasing ebooks in Tamil Unicode as well.

Part – III

III. Short Answers

Question 1.
Write note on e-Governance,
Answer:
Getting Government services through internet is known as e-Governance. Government of Tamilnadu has been giving its services through Internet. One can communicate with Government of Tamilnadu from any corner of the state. One can get important announcements, government orders, and government welfare schemes from the web portal of Government Of Tamilnadu.

Question 2.
List the website address of popular e-Libraries.
Answer:
Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing 1

Question 3.
List the familiar Tamil keyboard interfaces and Tamil keyboard layouts.
Answer:

  1. NHM Writer, E-Kalappai and Lippikar – are familiar Tamil keyboard interfaces software that is used for Tamil typing which works on Tamil Unicode, using phonetics.
  2. Sellinam and Ponrpadal – are familiar Tamil keyboard layouts ,that works on Android operating system in Smart phone using phonetics.

Question 4.
Write note on Unicode.
Answer:
Unicode is an encoding system, designed to handle various world languages, including Tamil. Its first version 1.0.0 was introduced on October 1991. While introduction of this scheme, can be able to handle nearly 23 languages including Tamil. Among the various encoding scheme, Unicode is the suitable to handle Tamil.

Question 5.
Write note on Tamil operating system.
Answer:
An operating system should be easy to work and its environment should be in understandable form. Thus, all operating systems used in computers and smart phones offered environment in Tamil.

Windows Tamil Environment interface should be downloading and install from internet. It shows all windows elements such as Taskbar, desktop elements, names of icons, commands in Tamil.

Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing

Part – IV

IV. Explain In Detail

Question 1.
Write about Tamil Office Automation Applications.
Answer:
Famous Office automation software like Microsoft Office, Open Office etc., provides complete Tamil interface facility. These softwares are downloadable and installed in your computer. After installation, our office automation software environment will completely changed to Tamil. Menu bars, names of icons, dialog boxes will be shown in Tamil. Moreover, you can save files with Tamil names and create folders with Tamil names.

Libra Office Writer Environment in Tamil:
Samacheer Kalvi 11th Computer Applications Guide Chapter 18 Tamil Computing 2
Apart from that Tamil Libra Office, Tamil Open Office, Azhagi Unicode Editor, Ponmozhi, Menthamiz, Kamban, Vani are office automation software working exclusively for Tamil. These applications are designed to work completely in Tamil.