Class 9

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Check whether p(x) is a multiple of g(x) or not.
p(x) = x³ – 5x² + 4x – 3, g(x) = x – 2
Solution:
p(x) = x³ – 5x² + 4x – 3; g(x) = x – 2
Let g(x) = 0
x – 2 = 0
x = 2
p(2) = 2³ – 5(2²) + 4(2) – 3
= 8 – 5 × 4 + 8 – 3 = 8 – 20 + 5 = -7 ≠ 0
⇒ p(x) is not a multiple of g(x)

Question 2.
By remainder theorem, find the remainder when, p(x) is divided by g(x) where,
(i) p(x) = x³ – 2x² – 4x – 1; g(x) = x + 1
(ii) p(x) = 4x³ – 12x² + 14x – 3; g(x) = 2x – 1
(iii) p(x) = x³ – 3x² + 4x + 50; g(x) = x – 3
Solution:
(i) p(x) = x³ – 2x² – 4x – 1; g(x) = x + 1
Let g(x) = x + 1
x + 1 = 0
x = -1
P(-1) = (-1)³ – 2(-1)² – 4(-1) – 1
= -1 -2 × 1 + 4 – 1
= -4 + 4 = 0
∴ Remainder = 0.

(ii) p(x) = 4x³ – 12x² + 14x – 3; g(x) = 2x – 1

(iii) p(x) = x³ – 3x² + 4x + 50 ; g(x) = x – 3
Let g(x) = x – 3
x – 3 = 0
x = 3
p(3) = 3³ – 3(3²) + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
∴ Remainder = 62.

Question 3.
Find the remainder when 3x³ – 4x² + 7x – 5 is divided by (x + 3)
Solution:
(3x³ – 4x² + 7x – 5) + (x + 3)

The remainder is -143.

Question 4.
What is the remainder when x²⁰¹⁸ + 2018 is divided by x – 1.
Solution:
x²⁰¹⁸ + 2018 is divided by x – 1
Let g(x) = x – 1 = 0
x = 1
p(x) = x²⁰¹⁸ + 2018
p(1)= 1²⁰¹⁸ + 2018
= 1 + 2018 = 2019

Remainder Theorem Calculator is a free online tool that displays the quotient and remainder of division for the given polynomial expressions.

Question 5.
For what value of k is the polynomial p(x) = 2x³ – kx² + 3x + 10 exactly divisible by (x – 2).
Solution:
Let g(x) = x – 2 = 0
x = 2
Since p(x) is exactly divisible by (x – 2)
p(2) = 2(2³) – k(2²) + 3(2)+ 10
= 16 – 4k + 6 + 10
= 32 – 4k = 0
= -k = -32
k= \(\frac{32}{4}\) = 8.

Question 6.
If two polynomials 2x³ + ax² + 4x – 12 and x³ + x² – 2x + a leave the same remainder when divided by (x – 3), find the value of a. and also find the remainder.
Solution:
Let f(x) = 2x³ + ax² + 4x – 12 and g(x) = x³ + x² – 2x + a
When f(x) is divided by x – 3, the remainder is f(3).
Now f(3) = 2(3)³ + a(3)² + 4(3) – 12 = 54 + 9a + 12 – 12
f(3) = 9a + 54 …………. (1)
When g(x) is divided by x – 3, the remainder is g(3).
Now g(3) = 3³ + 3² – 2(3) + a = 27 + 9 – 6 + a
g(3) = a + 30 ……….. (2)
Since, the remainder’s are same (1) = (2)
Given that f(3) = g(3)
That is 9a + 54 = a + 30
9a – a = 30 – 54 ⇒ 8a = -24 ∴ a = -3
Substituting a = -3 in f(3), we get
f(3) = 9(-3) + 54 = -27 + 54
f(3) = 27
∴ The remainder is 27.

Question 7.
Determine whether (x – 1) is a factor of the following polynomials:
(i) x³ + 5x² – 10x + 4
(ii) x⁴ + 5x² – 5x + 1
Solution:
(i) Let P(x) = x³ + 5x² – 10x + 4
By factor theorem (x – 1) is a factor of P(x), if P(1) = 0
P(1) = 1³ + 5(1²) – 10(1) + 4 = 1 + 5 – 10 + 4
P(1) = 0
∴ (x – 1) is a factor of x³ + 5x² – 10x + 4

(ii) Let P(x) = x⁴ + 5x² – 5x + 1
By/actor theorem, (x – 1) is a factor of P(x), if P( 1) = 0
P(1) = 1⁴ + 5 (1²) – 5(1) + 1 = 1 + 5 – 5 + 1 = 2 ≠ 0
∴ (x – 1) is not a factor of x⁴ + 5x² – 5x + 1

Question 8.
Using factor theorem, show that (x – 5) is a factor of the polynomial 2x³ – 5x² – 28x + 15
Solution:
LetP(x) = 2x³ – 5x² – 28x + 15
By factor theorem, (x – 5) is a factor of P(x), if P(5) = 0
P(5) = 2 (5)² – 5 (5)² – 28 (5) + 15
= 2 × 125 – 5 × 25 – 140 + 15
= 250 – 125 – 140 + 15 = 265 – 265 = 0
∴ (x – 5) is a factor of 2x³ – 5x² – 28x + 15

Question 9.
Determine the value of m, if (x + 3) is a factor of x³ – 3x² – mx + 24.
Solution:
Let P(x) = x³ – 3x² – mx + 24
By using factor theorem,
(x + 3) is a factor of P(x), then P (-3) = 0
P(-3) = (-3)³ -3 (-3)² – m (-3) + 24 = 0
⇒ -27 – 3 × 9 + 3m + 24 = 0 ⇒ 3m = 54 – 24
⇒ m = \(\frac{30}{3}\) = 10

Question 10.
If both (x – 2) and (x – \(\frac{1}{2}\)) are the factors of ax² + 5x + b, then show that a = b.
Solution:
Let P(x) = ax² + 5x + b
(x – 2) is a factor of P(x), if P(2) = 0
P(2) = a(2)² + 5(2) + b = 0
4a + 10 + b = 0
4 a + b = – 10 …………….. (1)
(x – \(\frac{1}{2}\)) is a factor of P(x), P(\(\frac{1}{2}\)) = 0

2a + 8b = -20
a + 4b = – 10 ……………… (2)
From (1) and (2)
4a + b = – 10 …………. (1)
a + 4b = – 10 ………… (2)
(1) and (2) ⇒ 4a + b = a + 4b
3a = 3b
∴ a = b. Hence it is proved.

Question 11.
If (x – 1)divides the polynomial kx³ – 2x² + 25x – 26 without remainder, then find the value of k.
Solution:
Let P(x) = kx³ – 2x² + 25x – 26
By factor theorem, (x – 1) divides P(x) without remainder, P (1) = 0
P(1) = k(1)³ -2 (1)² + 25 (1) – 26 = 0
k – 2 + 25 – 26 = 0
k – 3 =0
k = 3

Question 12.
Check if (x + 2) and (x – 4) are the sides of a rectangle whose area is x² – 2x – 8 by using factor theorem.
Solution:
Let P(x) = x² – 2x² – 8
By using factor theorem,(x + 2) is a factor of P(x), if P (-2) = 0
P(-2) = (-2)² – 2 (-2) – 8 = 4 + 4 – 8 = 0
and also (x -4) is a factor of P (x), if P (4) = 0
p (4) = 4² – 2(4) – 8 = 16 – 8 – 8 = 0
∴ (x + 2), (x – 4) are the sides of a rectangle whose area is x² – 2x – 8.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Additional Questions

Exercise 2.1

Question 1.
Find only two rational numbers between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 4 }\).
Solution:
A rational number between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 4 }\) = \(\frac { 1 }{ 2 }\) ( \(\frac { 1 }{ 4 }\) + \(\frac { 3 }{ 4 }\)) = \(\frac { 1 }{ 2 }\) (1) = \(\frac { 1 }{ 2 }\)
Another rational number between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\) = \(\frac { 1 }{ 2 }\) ( \(\frac { 1 }{ 2 }\) + \(\frac { 3 }{ 4 }\)) = \(\frac { 1 }{ 2 }\) ( \(\frac { 2+3 }{ 4 }\) = \(\frac { 31}{ 2 }\) × \(\frac { 5 }{ 4 }\)) = \(\frac { 5 }{ 8 }\)
The rational numbers \(\frac { 1 }{ 2 }\) and \(\frac { 5 }{ 8 }\) lies between \(\frac { 1 }{ 4 }\) and \(\frac { 3 }{ 2 }\) .

Question 2.
Is zero a rational numbers? Give reasons for your answer.
Solution:
Yes, since \(\frac { 0 }{ 2 }\) = 0, (i.e) it can be written in the form \(\frac { p }{ q }\) where q ≠ 0

Exercise 2.2

That’s literally all there is to it! 9/20 as a decimal is 0.45

Question 1.
Express the following decimal expansion is the form \(\frac { p }{ q }\) , where p and q are integers and q ≠ 0.
(i) 0.75
(ii) 0.625
(iii) 0.5625
(iv) 0.28
Solution:

Question 2.
Convert \(\overline { 0.9 }\) to a rational number.
Solution:
(i) Let x = \(0.\overline { 9 }\). Then x = 0.99999….
Multiplying by 10 on both sides, we get
10x = 9.99999….. = 9 + 0.9999….. = 9 + x
9x = 9
x = 1. That is, \(0.\overline { 9 }\) = 1 (∵ 1 is rational number).

Exercise 2.3

Question 1.
Classify the following number as rational or irrational.
(i) \(\sqrt { 11 }\)
(ii) \(\sqrt { 81 }\)
(iii) 0.0625
(iv) \(0.8\overline { 3 }\)
Solution:
(i) \(\sqrt { 11 }\) is an irrational number. (11 is not a perfect square number)
(ii) \(\sqrt { 81 }\) = 9 = \(\frac { 9 }{ 1 }\) , a rational number.
(iii) 0.0625 is a terminating decimal
∴ 0. 0625 is a rational number.
(iv) \(0.8\overline { 3 }\) = 0.8333
The decimal expansion is non-terminating and recurring.
∴ \(0.8\overline { 3 }\) is a rational number.

Question 2.
Find the decimal expansion of \(\sqrt { 3 }\).
Solution:

Question 3.
Find any 4 irrational numbers between \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 3 }\).
Solution:
\(\frac { 1 }{ 4 }\) = 0.25 and \(\frac { 1 }{ 3 }\) = 0.3333 = \(0.\overline { 3 }\)
In between 0.25 and \(0.\overline { 3 }\) there are infinitely many irrational numbers .
Four irrational numbers between 0.25 and \(0.\overline { 3 }\) are
0.2601001000100001 ……
0.2701001000100001 ……
0.2801001000100001 …..
0.3101001000100001 ……

Exercise 2.4

Question 1.
Visualise \(6.7\overline { 3 }\) on the number line, upto 4 decimal places.
Solution:
We locate 6.73 on the number line, by the process of successive magnification.

Step 1 : First we note that \(6.7\overline { 3 }\) lies between 6 and 7.
Step 2 : Divide the portion between 6 and 7 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.7 and 6.8.
Step 3 : Divide the portion between 6.7 and 6.8 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.73 and 6.74.
Step 4 : Divide the portion between 6.73 and 6.74 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.733 and 6.734.
Step 5 : Divide the portion between 6.733 and 6.734 into 10 equal parts and use a magnifying glass to visualise that \(6.7\overline { 3 }\) lies between 6.7332 and 6.7334.
We note that \(6.7\overline { 3 }\) is visualised closed to 6.7332 than to 6.7334.

Question 2.
Find whether x and y are rational or irrational in the following:
(i) a = 2 + \(\sqrt{3}\), b = 2 – \(\sqrt{3}\); x = a + b, y = a – b
(ii) a = \(\sqrt{2}\) + 7, b = x = a + b, y = a – b
Solution:
(i) Given that a = 2 + \(\sqrt{3}\), b = 2 – \(\sqrt{3}\)
x = a + b = (2+ \(\sqrt{3}\)) +(2 – \(\sqrt{3}\)) = 4, a rational number.
y = a – b = {2 + \(\sqrt{3}\)) – (2 – \(\sqrt{3}\)) = 2\(\sqrt{3}\) , an irrational number.

(ii) Given that a = \(\sqrt{2}\) + 7,b = \(\sqrt{2}\) – 7
x = a + b = (\(\sqrt{2}\) + 7)+ (\(\sqrt{2}\) – 7) = 2\(\sqrt{2}\), an irrational number.
y = a – b = (\(\sqrt{2}\) + 7 ) – (\(\sqrt{2}\) – 7) = 14, a rational number.

Exercise 2.5

Question 1.
Evaluate :
(i) 10^-4
(ii) (\(\frac { 1 }{ 9 }\))^-3
(iii) (0.01)^-2
Solution:

Question 2.
Find the value of 625^{\(\frac { 3 }{ 4 }\)} :
Solution:

Question 3.
Find the value of 729^{\(\frac { -5 }{ 6 }\)} :
Solution:

Question 4.
Use a fractional index to write :
(i) (5\(\sqrt { 125 }\))⁷
(ii) \(\sqrt [ 3 ]{ 7 }\)
Solution:
(i) (5\(\sqrt { 125 }\))⁷ = 125^{\(\frac { 7 }{ 5 }\)}
(ii) \(\sqrt [ 3 ]{ 7 }\) = 7\(\frac { 1 }{ 3 }\)

Exercise 2.6

Question 1.
Can you reduce the following numbers to surds of same order.
(i) \(\sqrt{ 5 }\)
(ii) \(\sqrt [ 3 ]{ 5 }\)
(iii) \(\sqrt [ 4 ]{ 5 }\)
Solution:

Now the surds have same order

Question 2.
Express the following surds in its simplest form
(i) \(\sqrt { 27 }\)
(ii) \(\sqrt [ 3 ]{ 128 }\)
Solution:

Question 3.
Show that \(\sqrt [ 3 ]{ 2 }\) > \(\sqrt [ 5 ]{ 3 }\).
Solution:

Question 4.
Express the following surds in its simplest form \(\sqrt [ 4 ]{ 324 }\).
Solution:

order = 4 ; radicand = 4; Coefficient = 3

Question 5.
Simplify \(\sqrt { 63 }\) – \(\sqrt { 175 }\) + \(\sqrt { 28 }\)
Solution:

Question 6.
Arrange in ascending order: \(\sqrt [ 3 ]{ 2 }\), \(\sqrt [ 2 ]{ 4 }\), \(\sqrt [ 4 ]{ 3 }\)
Solution:
The order of the surds \(\sqrt [ 3 ]{ 2 }\), \(\sqrt [ 2 ]{ 4 }\), \(\sqrt [ 4 ]{ 3 }\) are 3, 2, 4
L.CM. of 3, 2, 4 = 12.

Exercise 2.7

Question 1.
Subtract 6\(\sqrt { 7 }\) from 9\(\sqrt { 7 }\). Is the answer rational or irrational?
Solution:
9\(\sqrt { 7 }\) – 6\(\sqrt { 7 }\) = (9 – 6) \(\sqrt { 7 }\) = 3\(\sqrt { 7 }\)
The answer is irrational.

Question 2.
Simplify,: \(\sqrt { 44 }\) + \(\sqrt { 99 }\) – \(\sqrt { 275 }\).
Solution:

Question 3.
Compute and give the answer in the simplest form : 3 \(\sqrt { 162 }\) x 7 \(\sqrt { 50 }\) x 6 \(\sqrt { 98 }\)
Solution:
3 \(\sqrt { 162 }\) × 7 \(\sqrt { 50 }\) × 6 \(\sqrt { 98 }\) = \((3 \times 9 \sqrt{2} \times 7 \times 5 \sqrt{2} \times 6 \times 7 \sqrt{2})\)
= \(3 \times 7 \times 6 \times 9 \times 5 \times 7 \times \sqrt{2} \times \sqrt{2} \times \sqrt{2}=79380 \sqrt{2}\)

Exercise 2.8

Question 1.
Write in scientific notation : (60000000)³
Solution:
(60000000)³ = (6.0 × 10⁷)⁴ = (6.0)⁴ × (10⁷)⁴
= 1296 × 10²⁸
= 1.296 × 10³ × 10²⁸ = 1.296 × 10³¹

Question 2.
Write in scientific notation : (0.00000004)3
Solution:
(0.00000004)³ = (4.0 × 10^-8)³ = (4.0)³ × (10^-8)³
= 64 × 10^-24 = 6.4 × 10 × 10^-24 = 6.4 × 10^-23

Question 3.
Write in scientific notation : (500000)⁵ × (3000)³
Solution:
(500000)⁵ × (3000)³ = (5.0 × 10⁵)³ × (3.0 × 10³)³
= (5.0)² × (10⁵)² × (3.0)³ × (10³)³
= 25 × 10¹⁰ × 27 × 10⁹ = 675 × 10¹⁹
= 675.0 × 10¹⁹ = 6.75 × 10² × 10¹⁹= 6.75 × 10²¹

Question 4.
Write in scientific notation : (6000000)³ ÷ (0.00003)²
Solution:
(6000000)³ + (0.00003)² = (6.0 × 10⁶)³ + (3.0 × 10^-5)²
= (6.0 × 10⁶)3 ÷ (3.0 × 10^-5)² = 216 × 10¹⁸ ÷ 9 × 10^-10
= \(\frac{216 \times 10^{9}}{9 \times 10^{-10}}\)
= 24 × 10¹⁸ × 10¹⁰ = 24 × 10²⁸
= 24.0 × 10²⁸ = 2.4 × 10 × 10²⁸ = 2.4 × 10²⁹

Exercise 2.9

Multiple Choice Questions :
Question 1.
A number having non-terminating and recurring decimal expansion is
(1) an integer
(2) a rational number
(3) an irrational number
(4) a whole number
Solution:
(2) a rational number
Hint:
Irrational number have nonterminating and non recurring decimal expansion.

Question 2.
If a number has a non-terminating and non-recurring decimal expansion, then it is
(1) a rational number
(2) a natural number
(3) an irrational number
(4) an integer
Solution:
(3) an irrational number
Hint: Rational number gave terminating or recurring and non-terminating decimal expansion.

Question 3.
Decimal form of \(\frac { -3 }{ 4 }\) is
(1) -0.75
(2) -0.50
(3) -0.25
(4) -0.125
Solution:
(1) -0.75
Hint:
\(\frac { 1 }{ 4 }\) = 0.25; \(\frac {1 }{ 2 }\) = 0.5; \(\frac { 3 }{ 4 }\) = 0.75

Question 4.
Which one of the following has a terminating decimal expansion?

Solution:
(1) \(\frac { 5 }{ 32 }\)
Hint:
32 = 2⁵ ⇒ \(\frac { 5 }{ 32 }\) has terminating decimal expansion

Question 5.
Which one of the following is an irrational number?
(1) π
(2) √9
(3) \(\frac { 1 }{ 4 }\)
(4) \(\frac { 1 }{ 5 }\)
Solution:
(1) π

Question 6.
Which one of the following are irrational numbers?

(a) (ii), (iii) and (iv)
(b) (i), (ii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
Hint:
\(\sqrt{4+\sqrt{25}}=\sqrt{9}=3 ; \sqrt{8-\sqrt[3]{8}}=\sqrt{8-2}=\sqrt{6}\)

Question 7.
Which of the following is not an irrational number?
(1) \(\sqrt {2}\)
(2) \(\sqrt {5}\)
(3) \(\sqrt {3}\)
(4) \(\sqrt {25}\)
Solution:
(4) \(\sqrt {25}\)

Question 8.
In simple form, \(\sqrt [ 3 ]{ 54 }\) is?
(1) 3 \(\sqrt [ 3 ]{ 2 }\)
(2) 3 \(\sqrt [ 3 ]{ 27 }\)
(3) 3 \(\sqrt [ 3 ]{ 2 }\)
(4) \(\sqrt { 3 }\)
Solution:
(1) 3 \(\sqrt [ 3 ]{ 2 }\)

Question 9.
\(\sqrt [ 3 ]{ 192 }\) + \(\sqrt [ 3 ]{ 24 }\)
(1) 3\(\sqrt [ 3 ]{ 6 }\)
(2) 6\(\sqrt [ 3 ]{ 3 }\)
(3) 3\(\sqrt [ 3 ]{ 216 }\)
(4) 3\(\sqrt [ 6 ]{ 216 }\)
Solution:
(2) 6\(\sqrt [ 3 ]{ 3 }\)

Question 10.
5\(\sqrt { 21 }\) × 6\(\sqrt { 10 }\)
(1) 30\(\sqrt { 210 }\)
(2) 30
(3) \(\sqrt { 210 }\)
(4) 210\(\sqrt { 30 }\)
Solution:
(1) 30\(\sqrt { 210 }\)

Text Book Activities

Activity – 1
Is it interesting to see this pattern? \(\sqrt{4 \frac{4}{15}}=4 \sqrt{\frac{4}{15}} \text { and } \sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}}\) Verify it. Can you frame 4 such new surds?
Solution:

Activity – 2
Take a graph sheet and mark O, A, B, C as follows:

Consider the following graphs:

Are they equal? Discuss. Can you verify the same by taking different squares of different lengths?
Solution:

Activity – 3
The following list shows the mean distance of the planets of the solar system from the Sun. Complete the following table. Then arrange in order of magnitude starting with the distance of the planet closest to the Sun.
Solution:

Arrange the planets in order of distance from the sun.
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Additional Questions

Text Book Activities

Question 1.
Discuss and give as many examples of collections from your daily life situations, which are sets and which are not sets.
Solution:
Which are sets

1. Collection of pen
2. Collection of dolls
3. Collection of books
4. Collection of red flower etc.

Which are not sets

1. Collection of good students in a class.
2. Collection of beautiful flowers in a garden etc.

Question 2.
Write the following sets in respective forms.
Solution:

Question 3.
Fill in the blanks with appropriate cardinal numbers.
Solution:

Additional Questions and Answers

Exercise 1.1

Question 1.
Let A = {0, 1, 2, 3, 4, 5}. Insert the appropriate symbol G or g in the blank spaces,
(i) 0 ___ A
(ii) 6 ___ A
(iii) 3 ___ A
(iv) 4 ____ A
(v) 7 ____ A
Solution:
(i) 0 ∈ A
(ii) 6 ∉ A
(iii) 3 ∈ A
(iv) 4 ∈ A
(v) 7 ∉ A

Question 2.
Write the following in Set-Builder form.
(i) The set of all positive even numbers.
(ii) The set of all whole numbers less than 20.
(iii) The set of all positive integers which are multiple of 3.
(iv) The set of all odd natural numbers less than 15.
(v) The set of all letters in the word ‘computer’.
Solution:
(i) A = {x : x is a positive even number}
(ii) B = {x : x is a whole number and x < 20}
(iii) C = {x : x is a positive integer and multiple of 3}
(iv) D = {x : x is an odd natural number and x < 15}
(v) E = {x : x is a letter in the word “Computer”}

Question 3.
Write the following sets in Roster form.
(i) A = {x : x ∈ N, 2< x < 10 }
(ii) B = {x : x ∈ Z, –\(\frac { 1 }{ 2 }\) < x < \(\frac { 11 }{ 2 }\) }
(iii) C = {x : x is a prime number and a divisor of 6 }
(iv) x = {x : x = 2ⁿ, it n ∈ N and n ≤ 5}
(v) M = {x : x = 2y – 1, y ≤ 5, j ∈ W}
Solution:
(i) A = {3,4, 5, 6, 7, 8, 9}
(ii) B = {0, 1, 2, 3, 4, 5}
(iii) C = {2, 3}
(iv) Given, x = 2ⁿ, n ∈ N and n ≤ 5.
Here n = 1, 2, 3, 4, 5
n = 1 ⇒ 21 = 2
n = 2 ⇒ 22 = 4
n = 3 ⇒ 23 = 8
n = 4 ⇒ 24 = 16
n = 5 ⇒ 25 = 32
X = {2, 4, 8, 16, 32}

(v) Given, x = 2y – 1, y ≤ 5 and y ∈ W Here j = 0, 1,2, 3, 4, 5
y = 0 ⇒ x = 2 (0) – 1 = -1
y = 1 ⇒ x = 2 (1) – 1 = 2 – 1 = 1
y = 2 ⇒ x = 2 (2) – 1 = 4 – 1 = 3
y = 3 ⇒ x = 2 (3) – 1 = 6 – 1 = 5
y = 4 ⇒ x = 2 (4) – 1 = 8 – 1 = 7
y = 5 ⇒ x = 2 (5) – 1 = 10 – 1= 9
M = {-1, 1, 3, 5, 7, 9}

Therefore, we conclude that 2 and 19 are prime factors of 38.

Exercise 1.2

Question 1.
Find the number of subsets and number of proper subsets of a set X = {a, b, c, x, y, z}.
Solution:
Given X = {a, b, c, x, y, z}.
Then, n(X) = 6
The number of subsets = n[P(X)] = 2⁶ = 64
The number of proper subsets = n[P(X)] – 1 = 2⁶ – 1 = 64 – 1 = 63

Question 2.
Find the cardinal number of the following sets.
(i) A = {x : x is a prime factor of 12}.
(ii) B = {x : x ∈ W, x ≤ 5}.
(iii) X = {x : x is an even prime number}
Solution:
(i) Factors of 12 are 1, 2, 3, 4, 6, 12. So, the prime factors of 12 are 2,3.
We write the set A in roster form as A = {2, 3} and hence n(A) = 2.
(ii) In Tabular form B = {0, 1, 2, 3, 4, 5}
The set B has six elements and hence n(B) = 6
(iii) X = {2} [2 is the only even prime number]
∴ n (X) = 1

Question 3.
State whether the following sets are finite or infinite.
(i) A = {x : x is a multiple of 5, x ∈ N}.
(ii) B = {0,1, 2, 3, 4, 75}.
(iii) The set of all positive integers greater than 50.
Solution:
(i) A = {5, 10, 15, 20, …… } ∴A is an infinite set
(ii) Finite
(iii) Let X be the set of all positive integers greater than 50
Then X= (51, 52, 53, ….. }
∴ X is an infinite set.

Question 4.
Which of the following sets are equal?
(i) A = (1, 2, 3, 4}, B = {4, 3, 2, 1}
(ii) A = (4, 8, 12, 16}, B = (8, 4, 16, 18}
(iii) X ={2, 4, 6, 8}
Y = {x : x is a positive even integer and 0 < x < 10}
Solution:
(i) Since A and B contain exactly the same elements, A and B are equal sets.
(ii) A and B has different elements.
∴ A and B are not equal sets.
(iii) X = {2, 4, 6, 8}, Y = {2, 4, 6, 8}
∴ X and Y are equal sets.

Question 5.
Write ⊆ or ⊈ in each blank to make a true statement.
(i) {4, 5, 6, 7} ____ {4, 5, 6, 7, 8}
(ii) {a, b, c} ____ {b, e, f, g}
Solution:
(i) {4, 5, 6, 7} ⊈ {4, 5, 6, 7, 8}
(ii) {a, b, c} ⊈ {b, e, f, g}

Question 6.
Write down the power set of A= {3, {4, 5}}.
Solution:
The subsets of A are {Ø, {3}, {4, 5}, {3,{4, 5}}
P(A) = {Ø, {3}, {4,5}, {3{4,5}}

Exercise 1.3

Question 1.
Find the union of the following sets.
(i) A = {1, 2, 3, 5, 6} and B = {4, 5, 6, 7, 8}
(ii) X = {3, 4, 5} and Y = Ø
Solution:
(i) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}
(ii) X ∪ Y = {3, 4, 5}

Question 2.
Find A ∩ B if (i) A = {10, 11, 12, 13}, B = {12, 13, 14, 15}, (ii) A = {5, 9, 11}, B = Ø.
Solution:
(i) A ∩ B = {12, 13}
(ii) A ∩ B = Ø

Question 3.
Given the sets A = {4, 5, 6, 7} and B = {1, 3, 8, 9}, find A ∩ B.
Solution:
A ∩ B = Ø

Question 4.
If A= {-2, -1, 0, 3, 4}, B = {-1, 3, 5}, find (i) A – B, (ii) B – A.
Solution:
(i) A – B = {-2, 0, 4}
(ii) B – A = {5}

Question 5.
If A = {2, 3, 5, 7,11} and B = {5, 7, 9, 11, 13}, find A ∆ B.
Solution:
A ∆ B= {2, 3, 9, 13}

Question 6.
Draw a venn diagram similar to one at the side and shade the regions representing the following sets (i) A’, (ii) B’, (iii) A’ ∪ B’, (iv) (A ∪ B)’, (v) A’ ∩ B’
Solution:
(i) A’

(ii) B’

(iii) A’ ∪ B’

(iv) (A ∪ B)’

(v) A’ ∩ B’

Question 7.
State which of the following sets are disjoint.
(i) A = {2, 4, 6, 8}, B = {x : x is an even number < 10, x ∈ N}
(ii) X = {1, 3, 5, 7, 9}, Y = {0, 2, 4, 6, 8, 10}
(iii) R = {a, b, c, d, e}, S = {d, e, b, c, a}
Solution:
(i) A = {2, 4, 6, 8}, B = {2, 4, 6, 8}
A ∩ B = {2, 4, 6, 8} ≠ Ø
∴ A and B are not disjoint sets.
(ii) X ∩ Y = { } = Φ, X and Y are disjoint sets.
(iii) R ∩ S = {a, b, c, d, e} ≠ Ø
∴ R and S are not disjoint sets.

Question 8.
If A = {a, b, c, d, e}and B = {a, e, i, o, u} find AB.
Solution:
A ∩ B = {a, b, c, d, e} ∩ {a, e, i, o, u} = {a, e}

Exercise 1.4

Question 1.
If A and B are two sets containing 13 and 16 elements respectively, then find the minimum and maximum number of elements in A ∪B?
Solution:

n(A) = 13 ; n(B) = 16
Minimum n(A ∪ B) = 16
Maximum n(A ∪ B) = 13 + 16 = 29

Question 2.
If n(U) = 38, n(A) = 16, n(A ∩ B) = 12, n(B’) = 20, find n(A ∪ B).
Solution:

n(U) = 38
n(A) = 16
N(A ∩ B) = 12
n(B’) = 20
n(A ∪ B) = ?
Hint: n(B) = n(U) – n(B)’
n(B) = 38 – 20
n(B) = 18
n(A ∪ B) = n(A) + n(B) – n (A ∩ B)
n(A ∪ B) = 16 – 12 + 18
n(A ∪ B) = 4 + 18 = 22.

Question 3.
Let A = {b, d, e, g, h} and B = {a, e, c, h} verify that n(A – B) = n(A) – n(A ∩ B)
Solution:
A = {b, d, e, g, h),
B = {a, e, c, h}
A ∩ B = {b, d, g}
n(A ∩ B) = 3 ………….. (1)
A ∩ B = {e, h}
n(A ∩ B) = 2, n(A) = 5
n(A) – n(A ∩ B) = 5 – 2 = 3 ……………… (2)
Form (1) and (2) we get
n(A – B) = n(A) – n(A ∩ B)

Question 4.
If A = {2, 5, 6, 7} and B = {3, 5, 7, 8}, then verify the commutative property of
(i) union of sets
(ii) intersection of sets
Solution:
Given, A = {2, 5, 6, 7} and B = {3, 5, 7, 8}
(i) A ∪ B = {2, 3, 5, 6, 7, 8} ……………. (1)
B ∪ A = {2, 3, 5, 6, 7, 8} …………… (2)
From (1) and (2) we have A ∪ B = B ∪ A
It is verified that union of sets is commutative.

(ii) A n B = {5, 7} …………… (3)
B n A = {5, 7} ……………. (4)
From (3) and (4) we get, A ∩ B = B ∩ A
It is verified that intersection of sets is commutative.

Question 5.
If A = {b, c, d, e} and B = {b, c, e, g} and C = {a, c, e}, then verify A ∪ (B ∪ C) = (A ∪ B) ∪ C.
Solution:
Given, A = {b, c, d, e} and B = {b, c, e, g} and C = {a, c, e}
Now B ∪ C = {a, b, c, e, g}
Au(B ∪C) = {a, b, c, d, e, g} ……………. (1)
Then, A ∪ B = {b, c, d, e, g}
(A ∪ B) ∪ C = {a, b, c, d, e, g} ……………… (2)
From (1) and (2) it is verified that
A ∪ (B ∪ C) = (A ∪ B) ∪ C

Exercise 1.5

Question 1.
If A = {1, 3, 5, 7, 9}, B = {x : x is a composite number and x < 12} and C = {x : x ∈ N and 6 < x < 10} then verify A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Solution:
Given, A = {1, 3, 5, 7, 9} and B = {4, 6, 8, 9, 10}and C = {6, 7, 8, 9}
B ∩ C = {4, 6, 8, 9, 10} n {6, 7, 8, 9} = {6, 8, 9}
A ∪ (B ∩ C) = {1, 3, 5, 6, 7, 8, 9} ……………. (1)
Then (A ∪ B) = {1, 3, 5, 7, 9} ∪ {4, 6, 8, 9, 10} = {1, 3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ C) = {1, 3, 5, 7, 9} ∪ {6, 7, 8, 9} = {1,3, 5, 6, 7, 8, 9}
(A ∪ B) ∩ (A ∪ C) = {1, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {1, 3, 5, 6, 7, 8, 9}
= {1, 3, 5, 6, 7, 8, 9} …………….. (2)
From (1) and (2), it is verified that
A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪ C)

Question 2.
If A, B and C are overlapping sets, draw venn diagram for : A ∩ B
Solution:

Question 3.
Draw Venn diagram for A ∩ B ∩ C
Solution:

Question 4.
If P = {x : x ∈ N and 1 < x < 11}, Q = {x : x = 2n, n ∈ N and it < 6} and R = {4, 6, 8, 9, 10, 12}, then verify P – (Q ∩ R) = (P – Q) ∪ (P – R).
Solution:
The roster form of sets P, Q and R are P = {2, 3, 4, 5, 6, 7, 8, 9, 10}, Q = {2, 4, 6, 8, 10} and R = {4, 6, 8, 9, 10, 12}
First, we find Q ∩ R = {4, 6, 8, 10}
Then, P – (Q ∩ R) = {2, 3, 5, 7, 9} ………….. (1)
Next, P – Q = {3, 5, 7, 9}
and P – R = {2, 3, 5, 7}
and so, (P – Q) ∪ (P – Q) = {2, 3, 5, 7, 9} ……………. (2)
Hence from (1) and (2), it verified that P – (Q ∩ R) = (P – Q) ∪ (P – R)
Finding the elements of set Q
Given, x = 2 n
n = 1 → x = 2 (1) = 2
n = 2 → x = 2(2) = 4
n = 3 → x = 2 (3) = 6
n = 4 → x = 2(4) = 8
n = 5 → x = 2(5) = 10
Therefore, x takes values such as 2, 4, 6, 8, 10

Question 5.
If U = {x : x ∈ Z, -3 < x ≤ 9}, A = {x : x = 2P + 1, P ∈ Z , -2 ≤ P ≤ 3}, B = {x : x = q + l, q ∈ Z, 0 ≤ q ≤ 3}, verify De Morgan’s laws for complementation.
Solution:
Given, U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {-3, -1, 1, 3, 5, 7} and B = {1, 2, 3, 4}
Law (i) (A ∪ B)’ = A’ ∩ B’
Now, A ∪ B = {-3, -1, 1, 2, 3, 4, 5, 7}
(A ∪ B)’ = {-2, 0, 6, 8, 9} ………….. (1)
Then, A’ = {-2, 0, 2, 4, 6, 8, 9) and
B’ = {-3, -2, -1, 0, 5, 6, 7, 8, 9}
A’ ∩ B’ = {-2, 0, 6, 8, 9} ……………… (2)
From (1) and (2) it is verified that
(A ∪ B)’ = A’ ∩ B’
Law (ii) (A ∩ B)’ = A’ ∪ B’
Now, A ∩ B = {1, 3}
(A ∩ B)’ = {-3, -2, -1, 0, 2, 4, 5, 6, 7, 8, 9} ………….. (3)
Then, A’ ∪ B’ = {-3, -2, -1, 0, 2, 4, 5, 6, 7, 8, 9} ………………. (4)
From (3) and (4) it is verified that
(A ∩ B)’ = A’ ∪ B’

Exercise 1.6

Question 1.
From the given venn diagram. Find (i) A, (ii) B, (iii) A ∪ B (iv) A ∩ B also verify that n(A ∪B) = n(A) + n(B) – n(A ∩ B).

Solution:
A = {a, b, d, e, g, h}
B = {b, c, e, f, h, i, j}
A ∪ B = {a, b, c, d, e, f, g, h, i, j}
A ∩ B = {b, e, h}
So, n(A) = 6, n(B) = 7, n(A ∪B) = 10, n(A ∩ B) = 3
Now, n(A) + n(B) – n(A ∩ B) = 6 + 7 – 3 = 10
Hence, n(A) + n(B) – n(A ∩ B) = n(A ∪ B)

Question 2.
If n(A) = 12, n(B) = 17 and n(A ∪ B) = 21, find n(A ∩B).
Solution:
Given that n(A) = 12, n(B) = 17 and n(A ∪ B) =21
By using the formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∩ B) = 12 + 17 – 21 = 8

Question 3.
In a school, 80 students like Maths, 90 students like Science, 82 students like History, 21 like both Maths and Science, 19 like both Science and History 20 like both Maths and History and 8 liked all the three subjects. If each student like atleast one subject, then find (i) the number of students in the school (ii) the number of students who like only one subject.
Solution:
Let M, S and H represent sets of students who like Maths, Science and History respectively.
Then, n(M) = 80, n(S) = 90, n(H) = 82, n(M ∩ S) = 21, n(S ∩ H) = 19, n(M ∩ H) = 20, n(M ∩ S ∩ H) = 8
Let us represents the given data in a venn diagram.

(i) The number of student in the school = 52 + 59 + 55 + 12 + 11 + 8 + 8 = 205
(ii) The number of students who like only one subject = 52 + 59 + 55 = 166

Question 4.
State the formula to find n(A ∪ B ∪ C).
Solution:
n( A ∪ B ∪ C) = n(A) + n(B) +n(C) – n(A ∩ B) – (B ∩ C) – n(A ∩ C) + (A ∩ B ∩ B)

Question 5.
Verify n (A ∪ B ∪ C) = n (A) + n(B) + n (C) – n(A ∩ B) – (B ∩ C) – n(A ∩ C) + (A ∩ B ∩ C) for the following sets A = {1, 3, 5, 6, 8}, B = {3, 4, 5, 6} and C = {1, 2, 3, 6}
Solution:
(A ∪ B ∪ C) = {1,2, 3, 4, 5, 6, 8}
n (A ∪ B ∪ C) = 7
Also, n (A) = 5, n (B) = 4, n (C) = 4,
Further, A ∩ B = {3, 5, 6} ⇒ n(A ∩ B) = 3
B ∩ C = {3, 6} ⇒ n(B ∩ C) = 2
A ∩ C = {3, 5, 6} ⇒ n(A ∩ C) = 3
Also, A ∩ B ∩ C = {3, 6} ⇒ n(A ∩ B ∩ C) = 2
Now n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n( B ∩ C) – n (A ∩ C) + n(A ∩ B ∩ C)
7 = 5 + 4 + 4 – 3 – 2 – 3 + 2
7 = 13 – 8 + 2
7 = 5 + 2
7 = 7
Thus verified

Exercise 1.7

Multiple Choice Questions
Question 1.
If A = {5, {5, 6}, 7} which of the following is correct?
(1) {5, 6} ∈ A
(2) {5} ∈ A
(3) {7} ∈ A
(4) {6} ∈ A
Solution:
(1) {5, 6} ∈ A
Hint: {5, 6} is an element of A.

Question 2.
If x = {a, {b, c}, d}, which of the following is a subset of X?
(1) {a, b}
(2) {b, c}
(3) {c, d}
(4) {a, d}
Solution:
(4) {a, d}
Hint: b is not an element of X. Similarly c.

Question 3.
If a finite set A has m elements, then the number of non-empty proper subset of A is
(1) 2^m
(2) 2^m – 1
(3) 2^m-1
(4) 2(2^m-1 – 1)
Solution:
(4) 2(2^m-1 – 1)
Hint: P(A) = 2m Proper non empty subset = 2m – 2 = 2 (2^m-1 – 1)

Question 4.
For any three A, B and C, A – (B ∪ C) is
(1) (A – B) ∪ (A – C)
(2) (A – B) ∩ (A ∪ C)
(3) (A – B) ∪C
(4) A ∪ (B – C)
Solution:
(2) (A – B) ∩ (A ∪ C)

Question 5.
Which of the following is true?
(1) (A ∪ B) = B ∪ A
(2) (A ∪ B)’ = A’ – B’
(3) (A ∩ B)’ = A’ ∩ B’
(4) A – (B ∩ C) = (A – B) ∩ (A – C)
Solution:
(1) (A ∪ B) = B ∪ A

Question 6.
The shaded region in the venn diagram is
(1) A ∪ B
(2) A ∩ B
(3) (A ∩ B)’
(4) (A – B) ∪ (B – A)
Solution:
(4) (A – B) ∪ (B – A)
Hint:

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science Geography Solutions Chapter 2 Lithosphere – II Exogenetic Processes

Lithosphere – II Exogenetic Processes Textual Exercise

I. Choose the correct answer.

Weathering the Storm in Ersama MCQ Questions Class 9 English with Answers.

Question 1.
The disintegration or decomposition of rocks is generally called as …………………
(a) weathering
(b) erosion
(c) transportation
(d) deposition
Answer:
(a) weathering

Question 2.
The process of the levelling up of land by means of natural agents.
(a) aggradation
(b) degradation
(c) gradation
(d) none
Answer:
(c) gradation

Question 3.
……………. is seen in the lower course of the river.
(a) Rapids
(b) Alluvial fan
(c) Delta
(d) Gorges
Answer:
(c) Delta

Question 4.
Karst topography is formed due to the action of ………………
(a) Glacier
(b) Wind
(c) Sea waves
(d) Ground water
Answer:
(d) Ground water

Question 5.
Which one of the following is not a depositional feature of a glacier?
(a) cirque
(b) Moraines
(c) Drurrtlins
(d) Eskers
Answer:
(a) cirque

Question 6.
Deposits of fine silt blown by wind is called as …………….
(a) Loess
(b) Barchans
(c) Hamada
(d) Ripples
Answer:
(a) Loess

Question 7.
Stacks are formed by …………….
(a) Wave erosion
(b) River erosion
(c) Glacial erosion
(d) Wind deposion
Answer:
(a) Wave erosion

Question 8.
……………… erosion is responsible for the formation of cirque.
(a) wind
(b) glacial
(c) river
(d) underground water
Answer:
(b) glacial

Question 9.
Which one of the following is a second order land form?
(a) Asia
(b) Deccan Plateau
(c) Kulu valley
(d) Marina Beach
Answer:
(b) Deccan Plateau

II. Match the following.

Answer:
1. (c)
2. (d)
3. (a)
4. (e)
5. (b)

III. Answer in brief.

Question 1.
Define weathering.
Answer:
Weathering is the disintegration and decomposition of the material of the earth’s crust by its exposure to the atmosphere.

Question 2.
What do you mean by biological weathering?
Answer:
Biological weathering occurs due to the penetration and expansion of plant roofs, earthworms, burrowing animals (rabbits, rats), and same human activities.

Question 3.
Mention the three courses of a river with any two landforms associated with each course.
Answer:

1. The course of a river is divided into
   • The Upper course
   • The Middle course
   • The Lower course.
2. The land features carved by a river in its upper course are V-shaped valleys, gorges, canyons, rapids, potholes, spurs, and waterfalls.
3. In the middle course of a river some typical landforms like alluvial fans, flood plains, meanders, oxbow lakes, etc., are formed.
4. The lower course of a river develops typical landforms like delta and estuary.

Question 4.
What are the Ox-bow lakes?
Answer:

1. Meanders in due course of time become almost a complete circle with narrow necks.
2. This in turn gets abandoned and forms a lake. This is called an Ox-bow lake.

Question 5.
How does a cave differ from a sea arch?
Answer:

Sea cave	Sea Arch
A prolonged wave attack on the base of a cliff erodes rock materials which results in the formation of caves.	When two caves approach one another from either side of a headland and Unite, they form an arch, e.g., Neil Island, Andaman & Nicobar.

Question 6.
List out any four karst topographical areas found in India.
Answer:
Karts areas in Indian are given below:

1. Guptadham caves – Western Bihar
2. Robert cave and Tapkeshwar Temple – Uttarakhand.
3. Pandav caves – Pachmarhi, Madhya Pradesh.
4. Kutumsar – Bastar District in Chattisgarh
5. Borra caves of Visakhapatnam – Andhra Pradesh.

Question 7.
What do you mean by a hanging valley?
Answer:
These are valleys eroded by the tributary glacier and that hangs over the main valley.

Question 8.
Define:
(a) Moraine
(b) Drumlin
(c) Esker.
Answer:
(a) Moraine:
Landforms formed by the glacial deposits of the valley or continental glaciers are termed as Moraines. They are of various shapes and sizes, like the ground, terminal, and lateral moraine, etc.

(b) Drumlins:
Drumlins are deposits of glacial moraines that resemble giant inverted teaspoons or half cut eggs.

(c) Eskers:
Long Narrow ridges composed of boulders gravel and sand deposited by streams of melting water which run parallel to a glacier are called Eskers.

Question 9.
Mention the various features formed by wind erosion.
Answer:

• The erosional landforms of wind: Mushroom rocks, Inselbergs, and Yardangs
• The Depositional Landforms of wind: Sand dunes, Barchans, and loess.

Question 10.
What are the wave-cut platforms?
Answer:

1. Flat surface found at the foot of sea cliffs are called wave platform
2. Wave cut platform is also referred to as beach, shelf, terrace, and plain.

IV. Distinguish between:

Question 1.
Physical and chemical weathering
Answer:

S.No.	Physical weathering	Chemical weathering
(i)	It is the breakdown of rocks without changing their chemical composition through the action of physical forces.	Disintegration and. decomposition of rocks due to chemical reactions is called Chemical weathering.
(ii)	Cracks are formed and disintegration occurs eventually.	Chemical weathering takes place through the processes of oxidation, carbonation solution and hydration.
(iii)	Exfoliation, block disintegration, granular disintegration, etc., are the different types of weathering.	The agents of Chemical weathering are oxygen, CO2 and Hydrogen.

Question 2.
Delta and Estuary
Answer:

S.No.	Delta	Estuary
(i)	A triangular-shaped low lying area formed by the river at its mouth is called Delta.	Deltas have fine deposits of sediments enriched with minerals.
(ii)	Estuary is formed where the rivers meet the sea.	Deposition of silt by the river is not possible in the estuaries like Delta as if the waves keep on eroding the deposits e.g. River Narmada, River Tapti.

Question 3.
Stalactite and stalagmite
Answer:

Stalactite	Stalagmite
When the water containing dissolved calcite gradually drips from the ceiling of the caves, water evaporates and the remaining calcite hangs from the ceiling and thus Stalactites are formed.	When the calcite deposits rises upward like a pillar Stalagmites are formed.

Question 4.
Longitudinal and Transverse sand dunes
Answer:

Longitudinal	Transverse sand dunes.
Longitudinal dunes are long narrow ridges of sand, which extend in a direction parallel to the prevailing winds. These dunes are called Seifs in Sahara.	Transverse dunes are asymmetrical in shape. They are formed by alternate slow and fast winds that blow from the same direction.

Question 5.
Inselbergs and yardangs
Answer:

S.No.	Inselbergs	Yardangs
(i)	Certain hard rocks like igneous rocks are more resistant to wind action.	In arid regions, certain rocks .have hard and soft layers arranged vertically.
(ii)	Isolated residual hills rising abruptly from their surroundings are termed as inselbergs. e.g., Uluru (or) Ayers Rock – Australia.	When winds blow over these rocks, the soft layers get eroded leaving irregular crests. These are called Yardangs.

Question 6.
Spit and bar
Answer:

Spit	Bar
A spit is a ridge (or) embankment of sediment, attached to the land on one end and terminating in open water on the other end. Spits are common at the mouth of estuaries, e.g., Kakinada Spit.	A bar is an elongated deposit of sand, shingle (or) mud found in the sea almost parallel to the shoreline.

V. Give Reasons.

Question 1.
Chemical weathering is predominant in hot and humid zones.
Answer:

1. Chemical is predominant in hot and humid zones because the warm temperature and rainfall increases the chemical weathering.
2. It encourages the decomposition of plant matter to produce chemicals such as humic acids and CO₂
   These chemicals increase the rate of weathering.

Question 2.
Slit deposits are less at estuaries than deltas.
Answer:

1. Deltas form at the mouths of large rivers when sediments and silt accumulate.
2. As sediments continue to accumulate, the course of the river may even be changed.
3. Estuary is formed where the rivers meet the sea.
4. Deposition of silt by the river is not possible here in the estuaries like delta as if the waves keep on eroding the deposits.

Question 3.
The snow line is at the sea level in Polar regions.
Answer:
The snow line is at the sea level in Polar regions because the higher the latitude lowers the snow line from sea level. –

Question 4.
Wind can possibly erode the rocks from aO sides.
Answer:

• Wind erosion can occur in any area where the Soil (or) Sand is not compacted (or) if it is finely granulated in nature.
• Wind can loosen the materials and send them in all directions.

Question 5.
In limestone regions, surface drainage is rarely found.
Answer:

1. Groundwater percolating through cracks removes the soluble rock while leaving an enlarged channel for the further flow of water.
2. If there is a thick cover of soil above the soluble rock, surface streams may flow above the subterranean karst drainage system.
3. But most commonly, dissolution features occur at the surface.
4. Therefore there are a few continuous surface streams.

VI. Answer in Paragraph.

Question 1.
Write a note on weathering classify and explain.
Answer:
Weathering is the disintegration and decomposition of materials of the earth’s crust by their exposure to the atmosphere. There are three types of weathering,
(a) Physical weathering
(b) Chemical weathering
(c) Biological weathering

Physical weathering: It is the breakdown of rocks without changing their chemical composition, through the action of physical forces. The constant freezing and thawing of rocks during the night and day leads to the expansion and contraction of rocks. Cracks are formed and disintegration occurs eventually. Exfoliation, block disintegration, granular disintegration, etc., are the different types of weathering.

Chemical weathering: Disintegration and decomposition of rocks due to chemical reactions are called Chemical Weathering. This is predominantly high in the hot and humid regions such as the equatorial, tropical, and subtropical zones. Chemical weathering takes place through the processes of oxidation, carbonation, solution, and hydration. The agents of Chemical weathering are Oxygen, Carbon-dioxide, and Hydrogen.

Biological weathering: Biological weathering occurs due to the penetration and expansion of plant roots, earthworms, burrowing animals (rabbits, rats), and some human activities.

Question 2.
Explain the erosional landforms formed by underground water.
Answer:
(a) Erosional landforms formed by underground water.

• Most of the erosions take place due to the process of solution.
• When rainwater mixes with carbon-di-oxide and enters a limestone region, it dissolves and destroys much of the limestone.
• As a result, landforms such as Terra Rossa, Lappies, sinkholes, swallow holes, dolines, uvulas, poljes, caves, and caverns are formed.

(b) Terra Rossa: The deposition of red clay soil on the surface of the Earth is due to the dissolution of limestone content in rocks.

(c) When the joints of limestone rocks are corrugated by groundwater, long furrows are formed. These are called Lappies.

(d) A funnel-shaped depressions formed due to the dissolution of limestone rock is called sinkholes.

(e) Caves are hollows that are formed by the dissolution of limestone rocks when carbon-di-oxide in the air turns into carbonic acid. Caverns are caves with irregular floors.

Question 3.
What is a glacier? Explain its types.
Answer:
A Glacier is a large mass of ice that moves slowly over the land, from its place of accumulation. It is also known as the ‘River of ice’. The place of accumulation is called a snowfield. The height above which there is a permanent snow cover in the higher altitude or latitude is called snowline. The higher the latitude, the lower the snowline from sea level.

The gradual transformation of snow into granular ice is called ‘firn’ or ‘ neve’ and finally, it becomes solid glacial ice.
Movement of Glacier: The large mass of ice creates pressure at its bottom and generates heat. Due to this, the glacier melts a little and starts to move The rate of movement of a glacier varies from a few centimeters to several hundred meters a day. The movement of glaciers depends on slope, the volume of the glacier, thickness, roughness at the bottom (friction), etc., and Temperature. Like the rivers, glaciers also carry out erosion, transportation, and deposition.

Types of Glacier: Glaciers are broadly divided into two types based on the place of occurrences, such as Continental glacier and valley glacier.

Question 4.
Describe the depositional work of winds.
Answer:
(a) Depositional landforms of wind:

• Deposition occurs when the speed of wind is reduced by the presence of obstacles like bushes, forests and rock structures.
• The sediments carried by wind get deposited on both the windward and leeward sides of these obstacles.

(b) Sand dunes: In deserts, during sandstorms, wind carries loads of sand. When the speed of wind decreases, huge amount of sand gets deposited. These mounds or hills of sand are called sand dunes.

(c) Barchans: Barchans are isolated, crescent-shaped sand dunes. They have gentle slopes on the windward side and steep slopes on the leeward side.

(d) Transverse dunes: They are asymmetrical in shape. They are formed by alternate slow and fast winds that blow from the same direction.

(e) Longitudinal dunes: Longitudinal dunes are long narrow ridges of sand, which extend in a direction parallel to prevailing winds.

(f) Loess:

1. The term loess refers to the deposits of fine silt and porous sand over a vast region.
2. Extensive loess deposits are found in Northern and Western China, the Pampas of Argentina, in Ukraine, and in the Mississippi valley of the United States.

Question 5.
Give a detailed account of the three orders of landforms.
Answer:
Major land forms:
(i) First order landforms : Continents & Oceans
(ii) Second-order landforms: Mountains, Plateaus, and plains minor land forests
(iii) Third-order landforms: Deltas, Fjords coasts, Sand dimes, Beaches, Valleys, Cirques, Mushroom rocks, Limestone rocks.

First-order land forms:

1. Continents: (i) It is a very large area of land.
   (ii) One of the seven large landmasses on the earth’s surface, surrounded by sea.
   Asia, Africa, Europe, North America, South America, Australia, and Antarctica.
2. Oceans: A very large expanse of sea, Atlantic ocean, Arctic ocean, Pacific ocean, Indian ocean, and Antarctic ocean.

Second-order landforms: Mountains, Plateaus, and Plains.

1. Mountains: A large natural elevation of the earth’s surface, rising abruptly from the surrounding level, e.g., the Himalayas.
2. Plateaus: An area of fairly level high ground, e.g., Tibetan plateau.
3. Plains: A large area of flat land, e.g., Coastal plains.

Third-order landforms: Deltas, Fjords, Sand dunes, Beaches, Valleys, Cirques, Mushroom rocks, Limestone rocks.

1. Deltas: A triangular-shaped low lying area formed by the river at its mouth is called Delta. Fjords: These are glacial valleys that are partly submerged in the sea.
2. Sand Dunes: In deserts, a huge amount of sand gets deposited. These mounds (or) hills of sand are called sand dunes.
3. Beaches: Sand and gravel are moved and deposited by waves along the shore to form Beaches. Valleys: A low area of land between hills (or) mountains typically with a river (or) stream flowing through it.
4. Cirques: The glacier erodes the steep sidewalls of the mountain and farms bowl shaped armchair. It is termed as a cirque.
5. Mushroom rocks: By the constant wearing down action of wind the bottom of the rock gets eroded away to form a mushroom-like structure. This is called Mushroom rock (or) Pedestal rock.
6. Limestone rocks: The underground water creates distinct landforms in limestone regions called Karst Topography. It consists of calcite, aragonite.

VII. Consider the given statements and choose the right option given below.

Question (i).
1. ‘I’ shaped valley is an erosional feature of the river.
2. ‘U’ shaped valley is an erosional feature of the glacier.
3. ‘V’ shaped valley is an erosional feature of the glacier.
(a) (i), (ii) and (iii) are right
(b) (i) and (ii) are right
(c) (i) and (iii) are right
(d) only (i) is right
Answer:
(d) only (ii) is right

Question (ii).
Statement I: Running water is an important agent of gradation.
Statement II: The work of the river depends on the slope of land on which-it flows.
(a) Statement I is false II is true
(b) Statement I and II are false
(c) Statement I is true II is false
(d) Statement I and II are true
Answer:
(a) Statement I is false II is true

Question (iii).
Statement: Limestone regions have less underground water.
Reason : Water does not percolate through limestone.
(a) The statement is right reason is wrong.
(b) The statement is wrong Reason is right.
(c) The statement and reason are wrong.
(d) The statement and reason are right.
Answer:
(d) The statement and reason are right.

VIII. HOTS

Question 1.
Is wind the only gradational agent in the desert?
Answer:
Yes, the wind is the only gradational agent in the desert.
e.g., Erosional activity: Yardung
Depositional activity: Sand Dimes.

Question 2.
Underground water is more common in limestone areas than surface runoff. Why?
Answer:
The chief constituent of limestone is calcium carbonate which is soluble in pure water and easily soluble in carbonated water.

Question 3.
The river channels in the lower course are wider than the upper course.
Answer:
The reasons are,

1. The river splits into a number of channels called distributaries.
2. The river brings downloads of debris from its upper and middle.
3. The river deposits and develop typical landforms like Delta and Estuary.

In-text HOTs Questions

Question 1.
Is weathering a pre-requisite in the formation of soil?
Answer:

1. Yes, weathering a pre-requisite in the formation of soil.
2. The rock materials in due course of time are weathered further to form soil.
3. Soil is a mixture of disintegrated rock material.

Question 2.
The snowline of the Alps’is 2700 metre whereas the snowline of Greenland is just 600 mts. Find out the reason.
Answer:
On tropical mountains, the snowline may be as high as 500 mts, but when traced poleward it descends to 2700 mts in the European Alps to 600 meters in Greenland and just to se-a level near the poles. –

IX. Map Skill.

Question 1.
On the given outline map of the world, mark the following.
1. Any two deltas
2. A Karst region
3. Any two hot and cold deserts
Answer:
1. Any two deltas – Euphrates & Tigris Delta and Amazon river Delta

2. A Karst region – China

3. Any two hot and cold desets

X. Give geographical terms for the following:

Question 1.
(a) Chemical alteration of carbonate rocks on limestone region.
(b) Flat surfaces near cliffs.
(c) Erosion + Transportation + Deposition =
(d) The bottom line of a snowfield.
Answers:
(a) Carbonation
(b) Plateau
(c) Gradation
(d) The snowline

Lithosphere – I Endogenetic Processes Additional Questions

I. Choose the correct answer.

Question 1.
The process of dissolution of rock substances in water is …………….
(a) oxidation
(b) solution
(c) gradation
(d) hydration
Answer:
(b) solution

Question 2.
This generally originate from mountains.
(a) lake
(b) sea
(c) river
(d) ocean
Answer:
(c) river

Question 3.
The cylindrical holes drilled vertically in the river bed are …………….
(a) Potholes
(b) canyons
(c) rapid
(d) Gorge
Answer:
(a) Potholes

Question 4.
The largest Delta in the world is …………….
(a) The Nile River Delta
(b) The Ganga – Brahmaputra Delta
(c) The Yellow river Delta
(d) The Indus Delta
Answer:
(b) The Ganga – Brahmaputra Delta

Question 5.
The redness of the red clay soil is due to. the presence of ……………
(a) iron oxide
(b) carbon
(c) copper
(d) magnesium
Answer:
(a) iron oxide

Question 6.
The most powerful agents of gradation are ……………
(a) Rivers
(b) Glaciers
(c) Sea waves
(d) Streams
Answer:
(c) Sea waves

II. Match the following.

Answer:
1. (e)
2. (a)
3. (d)
4. (b)
5. (a)

III. Answer in brief.

Question 1.
Define Granular Disintegration.
Answer:
Granular disintegration takes place in crystalline rocks where the grains of the rocks become loose and fall out. This is due to the action of temperature & frost.

Question 2.
Mention the land features carved by a river in its upper course.
Answer:
‘V’ shaped valleys, Gorges, Canyons, rapids, potholes, spurs and waterfalls.

Question 3.
What are “Pot Holes”?
Answer:
Due to the river action, cylindrical holes are drilled vertically in the river bed with varying depth and diametre.

Question 4.
State the other erosional features of Karst regions in other parts of the world.
Answer:
Swallow Holes, Uvalas, Dolines, Poljis are the other erosional features.

Question 5.
What are Transverse Dunes?
Answer:

1. Transverse Dunes are asymmetrical in shape.
2. They are formed by alternate slow and fast winds that blow from the same direction.

Question 6.
What are the wave-cut Platforms?
Answer:
Flat surface found at the foot of the sea cliffs are called a wave Cut platform. It is also referred to as Beach, shelf, terrace, and plain.

IV. Distinguish between.

Question 1.
Oxidation and Carbonation.
Answer:

S.No.	Oxidation	Carbonation
(i)	Oxygen in the atmosphere reacts with the Iron found in rocks thus leading to the formation of Iron oxide. This process is known as oxidation.	Carbonation is the mixing of water with atmospheric CO2 forming carbonic acid.
(ii)	Oxidation weakens the rocks.	It is important in the formation of caves, in the limestone region.

Question 2.
Alluvial Plain and Flood Plain.
Answer:

S.No.	Alluvial Plain	Flood Plain
(i)	A fan-shaped deposition made by the river at the foothills is called an alluvial plain.	Fine sediments are deposited on river banks when a river floods and is called flood plain.
(ii)	These deposits are rich and fertile useful for cultivation.	These sediments make the region rich and fertile.

Question 3.
Arete and Matterhorn.
Answer:

Arete	Matterhorn
Aretes are narrow ridges formed when two cirque walls joined together back to back and forms narrow knife-like ridges.	The pyramidal peaks formed when three (or) more cirques meet together are referred as matterhoms.

Question 4.
Sea Cave and Sea Arch.
Answer:

Sea Cave	Arch
Prolonged wave attack on the base of a cliff erodes rock materials which result in the formation of caves.	When two caves approach one another from either side of a headland and Unite, they form an arch, e.g., Neil Island, Andaman Nicobar.

V. Give reasons.

Question 1.
Why do Biological weathering occur?
Answer:
Biological weathering occurs due to the penetration and expansion of plant roots, earthworms, burrowing animals (rabbits and rats), and some human activities.

Question 2.
Why is Karst Topography formed?
Answer:
Karst Topography is formed due to the dissolution of soluble rocks such as limestone, dolomite, and Gypsum.

Question 3.
Why do the Pedestal rock look like mushroom?
Answer:
By the constant wearing down action of wind, the bottom gets eroded away to form a mushroom-like structure. So the Pedestal rock looks like a mushroom.

VI. Answer in a Paragraph.

Question 1.
Explain the origin of the river and its course.
Answer:
Rivers generally originate from mountains and end in a sea or lake. The whole path that a river flows through is called its course. The course of a river is divided into:
(i) The upper course
(ii) The middle course and
(iii) The lower course

(i) The Upper Course: Erosion is the most dominant faction of the river in the upper course. In this course, a river usually tumbles down the steep mountain slopes. The steep gradient increases the velocity and the river channel performs erosion with great force to widen and deepen its valley. The land features carved by a river in its upper course are V-shaped valleys, gorges, canyons, rapids, potholes, spurs, and waterfalls.

(ii) The Middle Course: The river enters the plain in its middle course. The volume of water increases with the confluence of many tributaries and thus increases the load of the river. Thus, the predominant action of a river is transportation. The deposition also occurs due to the sudden decrease in velocity. The river in the middle course develops some typical landforms like alluvial fans, flood plains, meanders, ox-bow lakes, etc.,

(iii) The Lower course: The river, moving downstream across a broad, level plain is loaded with debris, brought down from its upper and middle courses. Large deposits of sediments are found at the level bed and the river splits into a number of channels called distributaries. The main work of the river here is a deposition and it develops typical landforms like delta and estuary.

Question 2.
Describe the Erosional landforms of Sea.
Answer:
Some of the erosional landforms of sea waves are sea cliff, sea cave, arch, stack, beach, bar and spit and wave-cut platform.

• Sea Cave: Prolonged wave attack on the base of a cliff erodes rock materials, which results in the formation of caves.
• Sea Arch: When two caves approach one another from either side of a headland and unite, they form an arch, e.g., Neil Island, Andaman, and Nicobar.
• Sea Stack: Further erosion by waves ultimately leads to the total collapse of the arch. The seaward portion of the headland will remain as a pillar of rock known as a stack. Eg the Old man of Hoy in Scotland.
• Sea Cliffs: Sea cliffs are steep rock faces formed when sea waves dash against them. The rocks get eroded to form steep vertical walls.
• WaveCut Platforms: Flat surface found at the foot of sea cliffs are called wave-cut platforms. Wave cut platform is also referred to as beach, shelf, terrace, and plain.

VII. Consider the given statements and choose the right option given below.

Question 1.
(i) The nature and magnitude of weathering differ from place to place and region to region.
(ii) Granular disintegration takes place due to the action of volcanoes.
(iii) Weathering is a pre-requisite in the formation of soil.
Which of the above statement is/are the right statement.
(a) (i), (ii) and (iii) are right
(b) (i) & (ii) are right
(c) (i) & (iii) are right
(d) only (i) is right.
Answer:
(c) is right

Question 2.
(i) Small streams that join the main river is tributary.
(ii) River Gangas is a tributary.
Which of the above statement is/are the right statement.
(a) The statement is the right reason is wrong
(b) The statement is the wrong reason is right
(c) The statement & reason are wrong
(d) The statement & reason are right
Answer:
(a) is right.

VIII. Map Skill.

Question 1.
Indus and Ganga Brahmaputra Delta
Answer:

2. Sri Lanka & Myanmar

You can Download Samacheer Kalvi 9th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 2 Motion

Samacheer Kalvi 9th Science Motion Textbook Exercises

I. Choose the correct answer.

This is free online Displacement calculator. enter Initial and final value then click on calculate and result will be instant displayed

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:
(d) acceleration of the moving object

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

Question 3.
Which of the following graph represents uniform motion of a moving particle?

Answer:
(b)

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

The Velocity Calculator is a free tool to predict how much work a team will complete during upcoming iterations.

II. Fill in the blanks.

1. Speed is a ____________ quantity whereas velocity is a _____________ quantity.
2. The slope of the distance-time graph at any point gives ____________
3. Negative acceleration is called ____________
4. Area under velocity-time graph shows ___________

Answer:

1. scalar, vector
2. speed
3. retardation (or) deceleration
4. displacement

This displacement calculator finds the displacement (distance traveled) by an object using its initial and final velocities as well as the time traveled.

III. True or False.

1. The motion of a city bus in a heavy traffic road is an example of uniform motion.
2. Acceleration can get a negative value also.
3. Distance covered by a particle never becomes zero but displacement becomes zero.
4. The velocity-time graph of a particle falling freely under gravity would be a 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:

1. False
2. True
3. True
4. False
5. True

IV. Assertion and Reason Type Question.

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

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.

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 the total time taken.
Answer:
(d) If 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) If both assertion and reason are true and reason is the correct explanation of assertion.

V. Match the following.

Answer:
1. (D)
2. (C)
3. (A)
4. (B)

VI. Answer briefly.

Question 1.
Define velocity.
Answer:

• Velocity is the rate of change of displacement. It is the displacement with unit time. It is a vector quantity. The SI unit of velocity is ms^-1
• Thus, Velocity = Displacement/time taken.

Question 2.
Distinguish distance and displacement.
Answer:

S.No.	Distance	Displacement
1.	The actual length of the path traveled by a moving body irrespective of the direction	The change in position of a moving body in a particular direction
2.	Scalar quantity	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 howsoever big or small these time intervals may be.

Question 4.
Compare Speed and Velocity.
Answer:

S.No.	Speed	Velocity
1.	The rate of change of distance	The rate of change of displacement
2.	Scalar quantity	Vector quantity
3.	Speed = \(\frac{\text { Distance travelled }}{\text { time taken }}\)	Velocity = \(\frac{\text { Displacement }}{\text { time taken }}\)

Question 5.
What do you understand about negative acceleration?
Answer:
If velocity decreases with time the value of acceleration is negative.
Note: Negative acceleration is called 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 velocity changes due to the change in direction. Hence it is an accelerated 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.
Ex. (i) Revolution of Earth around the Sun
(ii) Revolution of Moon around the Earth.

VII. Answer in detail.

Question 1.
Derive equations of motion by graphical method.
Answer:
An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement S.
Let us try to derive these equations by graphical method. Equations of motion from velocity – time graph:

Graph shows the change in velocity with time for an 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

For 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.

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

For 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) / t or t = (v – u)/a
Therefore = 1/2 × (v + u) × (v – u)/a .
2as = v² = u² + 2as
v² = u² + 2as ………(3)
This is third equation of motion.

Question 2.
Explain different types of motion.
Answer:
Different types of motion:

1. Linear motion: The motion of an object along a straight line is known as linear motion.
   Ex: Car moving on a straight road.
2. Circular motion: The motion of an object in a circular path is known as circular motion.
   Ex: Earth revolving around the sun.
3. Oscillatory motion: Repetitive to and fro motion of an object at regular interval of time is called oscillatory motion. Ex: Motion of pendulum of a clock.
4. Random motion: The disordered or irregular motion of a body is called random motion.
   Ex: Movement of fish underwater.

VIII. Exercise problems.

Question 1.
A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms^{– 2}. With what velocity will it strike the ground? After what time will it strike the ground?
Given: height = 20 m
acceleration = 10 ms^{– 2}
Formula: For free falling body,

1. v = gt
2. s = 1/2 gt²
   Solution:
   time taken to strike the ground
   s = 1/2 gt²
   20m = 1/2 × 10 ms ^{– 2} × t²
   t² = \(\frac{40 m}{10 m s^{-2}}\) = 4s ²
   ∴ t = 2s
3. Velocity of the ball when it strikes ground v = gt
   v = 10ms^{– 2} × 2s
   v = 20ms^{– 1}

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?
Given: Diameter of circular track = 200m
time to complete = 40s
Formula: Circumference of circular track = d. π m
speed = distance/time
Solution:
Circumference of the track = d.π
= 200 × 3.14 = 628m
speed = \(\frac{628 m}{40 s}\) = 15.7 ms^{– 1}
Distance covered in 2 min 20 s = speed × time
In the given time athlete covers 31/2 rounds = 2198m

The final position of athlete is as shown in figure (A – initial position, B – final position)
∴ Displacement = Distance × time
= 200 m

Question 3.
A racing car has a uniform acceleration of 4 ms^{– 2}. What distance it covers in 10 s after the start?
Given: acceleration = 4ms^{– 2}.
time = 10s and initial velocity = 0
Formula: s = ut + 1/2 at²
Solution:
distance covered s = 0 × t + 1/2 × 4 ms^{– 2} × (10 s)².
s = 1/2 × 4 ms^{– 2} × 100 s²
= 1/2 × 400m
= 200m

Solved Examples.

Question 1.
An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled by the object =16m + 16m = 32m
Total time taken = 4s + 2s = 6s
Average speed = \(\frac{\text { Total distance travelled }}{\text { total time taken }}\) = \(\frac{32}{6}\) = 5.33 ms^{– 1}
Therefore, the average speed of the object is 5.33 ms^{– 1}.

Question 2.
A sound is heard 5 s later than the lightning is seen in the sky on a rainy day. Find the distance of location of lightning? Given the speed of sound = 346 ms^{– 1}
Solution:
Speed = \(\frac{\text { Distance }}{\text { time }}\)
Distance = speed x time = 346 x5 = 1730 m
Thus, the distance of location of lightning = 1730 m

Question 3.
The brakes applied to a car produce an acceleration of 6 ms^{– 2} in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance traveled during this time.
Solution:
We have been given a = – 6 ms^{– 2}, t = 2s and v = 0
From the equation of motion,
v = u + at
0 = u + (- 6 × 2) .
0 = u – 12 ∴ u = 12 ms^{– 1}
s = ut + 1/2 at²
= (12 × 2) + 1/2 (- 6 × 2 × 2) .
= 24 – 12 = 12m
Thus, the car will move 12 m before it stops after the application of brakes.

Question 4.
A 900 kg car moving at 10 ms^{– 1} takes a turn around a circle with a radius of 25 m.
Determine the acceleration and the net force acting upon the car.
Solution:
When the car turns around circle, it experiences centripetal acceleration a = \(\frac{v^{2}}{r}\)
a = \(\frac{(10)^{2}}{25}=\frac{100}{25}\) ∴ a = 4ms^{– 2}
Net force acting upon the car,
F = ma = 900 × 4 = 3600N

ACTIVITY

Question 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.
Answer:

1. The objects which remain fixed at their position, and do not change their position are a row of houses, large trees and small plants.
2. The objects which keep on changing their position are flying birds, running cars and buses.

Question 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?
Answer:
The bus covers unequal distance in equal intervals of time but the train covers equal distances in equal intervals of time.
(The students can do this activity by themselves)

Question 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?
Answer:

1. AB + BC
   ABC = 4m + 3m=7m
   AC = 5 m
2. AB + BC + CD
   ABCD = 4m + 3m + 4m = 11m
   ACD = AC + CD = 5m + 4m = 9m
   AD = 3 m
   The shortest distance to reach D from A will be AD, that is 3 m.

Question 4.
Take a large stone and a small eraser. Stand on the top of 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?
Answer:
Both the stone and the eraser have reached the surface of the Earth almost at the same time in the absence of air medium (vacuum). But in air medium, due to friction, air offers resistance to the motion of free falling objects. The eraser reaches the first, the sheet of paper reaches later. The air resistance exerted on the sheet of paper is much higher than that of the eraser. The paper crumpled into a ball reaches ground first and plain sheet of paper reaches later, although they have equal mass. The air resistance offer to the plain sheet of paper is much higher than that offered to the paper ball. This is because the magnitude of air resistance depends on the area of objects exposed to air.

Question 5.
Take a piece of thread and tie a small piece of stone at one of its ends. Rotate the stone to describe a circular path with constant speed by holding the thread at the other end. Now, release the thread and let the stone go. Can you tell the direction in which the stone moves after it is released?
Answer:
The stone moves along the straight line tangential to the circular path. This is because once the stone releases, it continues to move along the direction it has been moving at that instant.

Question 6.
Take a piece of rope and tie a small stone at one end. Hold the other end of the rope and rotate it such that the stone follows a circular path.
Do you experience any pull or push in your hand?

In this activity, a pulling force that acts away from the centre is experienced. This is called as centrifugal force.

Samacheer Kalvi 9th Science Motion Additional Questions:

I. Choose the best answer.

Question 1.
The area under velocity time graph represents ………….
(a) Velocity of the moving object
(b) Displacement covered by the moving object
(c) Speed of the moving object
Answer:
(b) Displacement covered by the moving object

Question 2.
Unit of acceleration is ………………
(a) ms^{– 1}
(b) ms^{– 2}
(c) ms
(d) ms²
Answer:
(b) ms^{– 2}

Question 3.
When a body starts from rest, the acceleration of the body after 2 second in ………………. of its displacement.
(a) Half
(b) Twice
(c) Four times
(d) One fourth
Answer:
(a) Half

II. Short Answer Questions.

Question 1.
A bus travels, a distance of 20 km from Chennai central to airport in 45 minutes. What is the average speed?
Given: Distance = 20 km = 20,000 m
Time = 45 min = 2700 s
Formula: Average speed = \(\frac{\text { Total Distance }}{\text { Total Time taken }}\)
Solution:
Average speed = \(\frac{20 \mathrm{km}}{45 \mathrm{min}}=\frac{20,000 \mathrm{m}}{2700 \mathrm{s}}\)
= \(\frac{200 m}{27 s}\) = 7.4 ms ^{– 1}

Question 2.
Why did the actual speed differ from average, speed?
Answer:
Actual speed gives instantaneous speed of a body at any instant but average speed is the total distance covered by total time taken.

Question 3.
Mention the uses of velocity-time graph.
Answer:

1. Area covered under velocity – time graph gives us the displacement
2. Slope of the velocity – time graph gives us the acceleration.

Question 4.
The speed of a particle is constant. Will it have acceleration? Justify with an example.
Answer:

1. When a particle is moving with a constant speed along a straight line, it has no acceleration.
2. When a particle is moving with a constant speed along a circular path, the velocity changes due to the change in direction. Hence it has a acceleration.
   Ex. Revolution of Earth around the Sun.

Question 5.
Distinguish distance and displacement of a moving object.
Answer:

S.No.	Distance	Displacement
1.	The actual length of the path traveled by a moving body irrespective of the direction	The change in position of a moving body in a particular direction
2.	Scalar quantity	Vector quantity

III. Answer the following Question briefly.

Question 1.
Derive the three equations of motion by graphical method.
An object is in motion with initial velocity u attains a final velocity v in time t due to acceleration a, with displacement s.
Let us try to derive these equations by graphical method. Equations of motion from velocity-time graph:

Graph shows the change in velocity with time for an 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

For 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.

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

For 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) / t or t = (v – u)/a
Therefore s = 1/2 × (v + u) × (v – u)/a
2as = v² – u²
v² = u² + 2 as ………..(3)
This is third equation of motion.

I. Choose the best answer.

Question 1.
In a 100 m race, the winner takes 10s to reach the finishing point. The average speed of the winner is ……….. ms^{– 1}.
Answer:
(b) 10

II. Choose correct statement.

Question 1.
(a) Action and reaction forces act on same object
(b) Action and reaction forces act on different objects
Both (a) and (b) are possible
Neither (a) nor (b) is correct
Answer:
(a) is wrong
(b) is correct.

III. Short Answer Questions.

Question 1.
A motorcycle travelling at 20 ms^{– 1} has an acceleration of 4 ms^{– 2}. What does it explains about the velocity of the motorcycle?
Answer:

It explains that the motorcycle travels with uniform velocity.

Question 2.
Complete the following sentences.
(a) The acceleration of the body that moves with a uniform velocity will be …………….. .
Answer:
constant

(b) A train travels from A to station B with a velocity of 100 km/h and returns from station B to station A with a velocity of 80 km/h. Its average velocity during the whole journey in ……………. and its average speed is ……………… .
Answer:
zero, 90 km/h

Question 3.
Distinguish speed and velocity.
Answer:

S.No.	Speed	Velocity
1.	The rate of change of distance	The rate of change of displacement
2.	Scalar quantity	Vector quantity
3.	Speed = \(\frac{\text { Distance travelled }}{\text { time taken }}\)	Velocity = \(\frac{\text { Displacement }}{\text { time taken }}\)

Question 4.
What is meant by 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.

IV. Answer the following Question.

Question 1.
A boy moves along the path ABCD. What is the total distance covered by the boy? What is his net displacement?
(a) Total distance covered by a boy = 110 m
(b) Net displacement = \(\sqrt{30^{2}+40^{2}}\)
= \(\sqrt{2500}\)
= 50 m

Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board Samacheer Kalvi 9th Tamil Book Answers and Solutions Guide Pdf Free Download are part of Samacheer Kalvi 9th Books Solutions. Here we have given TN State Board New Syllabus Samacheer Kalvi 9th Std Tamil Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes.

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Tamilnadu State Board Samacheer Kalvi 9th Tamil Book Back Answers Solutions Guide Term 1, 2, 3.

Samacheer Kalvi 9th Tamil Book Back Answers

• Chapter 1.1 திராவிட மொழிக்குடும்பம்
• Chapter 1.2 தமிழோவியம்
• Chapter 1.3 தமிழ்விடு தூது
• Chapter 1.4 வளரும் செல்வம்
• Chapter 1.5 தொடர் இலக்கணம்
• Chapter 2.1 நீரின்றி அமையாது உலகு
• Chapter 2.2 பட்டமரம்
• Chapter 2.3 பெரியபுராணம்
• Chapter 2.4 புறநானூறு
• Chapter 2.5 தண்ணீர்
• Chapter 2.6 துணைவினைகள்
• Chapter 3.1 ஏறு தழுவுதல்
• Chapter 3.2 மணிமேகலை
• Chapter 3.3 அகழாய்வுகள்
• Chapter 3.4 வல்லினம் மிகும் இடங்கள்
• Chapter 3.5 திருக்குறள்
• Chapter 4.1 இயந்திரங்களும் இணையவழிப் பயன்பாடும்
• Chapter 4.2 ஓ, என் சமகாலத் தோழர்களே
• Chapter 4.3 உயிர்வகை
• Chapter 4.4 விண்ணையும் சாடுவோம்
• Chapter 4.5 வல்லினம் மிகா இடங்கள்
• Chapter 5.1 கல்வியில் சிறந்த பெண்கள்
• Chapter 5.2 குடும்ப விளக்கு
• Chapter 5.3 சிறுபஞ்சமூலம்
• Chapter 5.4 விட்டிற்கோர் புத்தகசாலை
• Chapter 5.5 இடைச்சொல் – உரிச்சொல்
• Chapter 6.1 சிற்பக்கலை
• Chapter 6.2 இராவண காவியம்
• Chapter 6.3 நாச்சியார் திருமொழி
• Chapter 6.4 செய்தி
• Chapter 6.5 புணர்ச்சி
• Chapter 6.6 திருக்குறள்
• Chapter 7.1 இந்திய தேசிய இராணுவத்தில் தமிழர் பங்கு
• Chapter 7.2 சீவக சிந்தாமணி
• Chapter 7.3 முத்தொள்ளாயிரம்
• Chapter 7.4 மதுரைக்காஞ்சி
• Chapter 7.5 சந்தை
• Chapter 7.6 ஆகுபெயர்
• Chapter 8.1 பெரியாரின் சிந்தனைகள்
• Chapter 8.2 ஒளியின் அழைப்பு
• Chapter 8.3 தாவோ தொ ஜிங்
• Chapter 8.4 யசோதர காவியம்
• Chapter 8.5 மகனுக்கு எழுதிய கடிதம்
• Chapter 8.6 யாப்பிலக்கணம்
• Chapter 9.1 விரிவாகும் ஆளுமை
• Chapter 9.2 அக்கறை
• Chapter 9.3 குறுந்தொகை
• Chapter 9.4 தாய்மைக்கு வறட்சி இல்லை
• Chapter 9.5 அணியிலக்கணம்

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Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board Samacheer Kalvi 9th Books Answers and Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi Books Solutions. Here we have given TN State Board New Syllabus Samacheer Kalvi 9th Std Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, Formulas.

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TN Samacheer Kalvi 9th Book Back Answers Solutions Guide

• Samacheer Kalvi 9th English Grammar
• Samacheer Kalvi 9th Maths Guide
• Samacheer Kalvi 9th Science Guide
• Samacheer Kalvi 9th Social Science Guide
• Samacheer Kalvi 9th English Guide
• Samacheer Kalvi 9th Tamil Guide

We hope the given Tamilnadu State Board Samacheer Kalvi Class 9th Std Books Solutions and Answers Pdf Free Download in English Medium and Tamil Medium will help you. If you have any queries regarding TN State Board New Syllabus Samacheer Kalvi 9th Standard Guides Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, Formulas, drop a comment below and we will get back to you at the earliest.

Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board Samacheer Kalvi 9th Science Book Answers and Solutions Guide Pdf Free Download in English Medium and Tamil Medium are part of Samacheer Kalvi 9th Books Solutions. Here we have given TN State Board New Syllabus Samacheer Kalvi 9th Std Science Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes.

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Samacheer Kalvi 9th Science Book Back Answers

Samacheer Kalvi 9th Science Physics Book Solutions

• Chapter 1 Measurement
• Chapter 2 Motion
• Chapter 3 Fluids
• Chapter 4 Electric Charge and Electric Current
• Chapter 5 Magnetism and Electromagnetism
• Chapter 6 Light
• Chapter 7 Heat
• Chapter 8 Sound
• Chapter 9 Universe

Samacheer Kalvi 9th Science Chemistry Book Solutions

• Chapter 10 Matter Around Us
• Chapter 11 Atomic Structure
• Chapter 12 Periodic Classification of Elements
• Chapter 13 Chemical Bonding
• Chapter 14 Acids, Bases and Salts
• Chapter 15 Carbon and its Compounds
• Chapter 16 Applied Chemistry

Samacheer Kalvi 9th Science Biology Book Solutions

• Chapter 17 Animal Kingdom
• Chapter 18 Organization of Tissues
• Chapter 19 Plant Physiology
• Chapter 20 Organ Systems in Animals
• Chapter 21 Nutrition and Health
• Chapter 22 World of Microbes
• Chapter 23 Economic Biology
• Chapter 24 Environmental Science

Samacheer Kalvi 9th Science Computer Science Book Solutions

• Chapter 25 Computer – An Introduction
• Chapter 26 Parts of Computer
• Chapter 27 Hardware and Software

9th New Science Book Back Questions Tamil Medium with Answers

9th Science Book Back Questions Tamil Medium with Answers, 9th Std Science Guide Pdf Download 2021 Tamil Medium, 9th New Science Book Back Answers in Tamil 2021-2022.

Samacheer Kalvi 9th Science Physics Book Solutions

• Chapter 1 அளவீடு
• Chapter 2 இயக்கம்
• Chapter 3 பாய்மங்கள்
• Chapter 4 மின்னூட்டமும் மின்னோட்டமும்
• Chapter 5 காந்தவியல் மற்றும் மின்காந்தவியல்
• Chapter 6 ஒளி
• Chapter 7 வெப்பம்
• Chapter 8 ஒலி
• Chapter 9 அண்டம்

Samacheer Kalvi 9th Science Chemistry Book Solutions

• Chapter 10 நம்மைச் சுற்றியுள்ள பொருட்கள்
• Chapter 11 அணு அமைப்பு
• Chapter 12 தனிமங்களின் வகைப்பாட்டு அட்டவணை
• Chapter 13 வேதிப்பிணைப்பு
• Chapter 14 அமிலங்கள், காரங்கள் மற்றும் உப்புகள்
• Chapter 15 கார்பனும் அவற்றின் சேர்மங்களும்
• Chapter 16 பயன்பாட்டு வேதியியல்

Samacheer Kalvi 9th Science Biology Book Solutions

• Chapter 17 விலங்குலகம்
• Chapter 18 திசுக்களின் அமைப்பு
• Chapter 19 தாவர உலகம் – தாவர செயலியல்
• Chapter 20 விலங்குகளின் உறுப்பு மண்டலங்கள்
• Chapter 21 ஊட்டச்சத்து மற்றும் ஆரோக்கியம்
• Chapter 22 நுண்ணுயிரிகளின் உலகம்
• Chapter 23 பொருளாதார உயிரியல்
• Chapter 24 சூழ்நிலை அறிவியல்

Samacheer Kalvi 9th Science Computer Science

• Chapter 25 கணினி – ஓர் அறிமுகம்
• Chapter 26 கணினியின் பாகங்கள்
• Chapter 27 வன்பொருளும் மென்பொருளும்

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You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Factorise each of the following polynomials using synthetic division:
(i) x³ – 3x² – 10x + 24
(ii) 2x³ – 3x² – 3x + 2
(iii) -7x + 3 + 4x³
(iv) x³ + x² – 14x – 24
(v) x³ – 7x + 6
(vi) x³ – 10x² – x + 10
Solution:
(i) x³ – 3x² – 10x + 24
Let p(x) = x³ – 3x² – 10x + 24
Sum of all the co-efficients = 1 – 3 – 10 + 24 = 25 – 13 = 12 ≠ 0
Hence (x – 1) is not a factor.
Sum of co-efficient of even powers with constant = -3 + 24 = 21
Sum of co-efficients of odd powers = 1 – 10 = – 9
21 ≠ -9
Hence (x + 1) is not a factor.
p (2) = 2³ – 3 (2²) – 10 × 2 + 24 = 8 – 12 – 20 + 24
= 32 – 32 = 0 ∴ (x – 2) is a factor.
Now we use synthetic division to find other factor

Thqs (x – 2) (x + 3) (x – 4) are the factors.
∴ x³ – 3x² – 10x + 24 = (x – 2) (x + 3) (x – 4)

The polynomial division calculator allows you to divide two polynomials to find the quotient and the remainder of the division.

(ii) 2x² – 3x² – 3x + 2
Let p (x) = 2x³ – 3x² – 3x + 2
Sum of all the co-efficients are
2 – 3 – 3 + 2 = 4 – 6 = -2 ≠ 0
∴ (x – 1) is not a factor
Sum of co-efficients of even powers of x with constant = -3 + 2 = – 1
Sum of co-efficients of odd powers of x = 2- 3= -1
(-1) = (-1)
∴ (x + 1) is a factor
Let us find the other factors using synthetic division

Quotient is 2x² – 5x + 2 = 2x – 4x – x + 2 = 2x (x – 2) – 1 (x – 2)
= (x – 2) (2x – 1)
∴ 2x³ – 3x² – 3x + 2 = (x + 1) (x – 2) (2x – 1)

(iii) -7x + 3 + 4x³
Let p(x) = 4x³ + 0x² – 7x + 3
Sum of the co-efficients are = 4 + 0 – 7 + 3
= 7 – 7 = 0
∴ (x- 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 3 = 3
Sum of co-efficients of odd powers of x with constant = 4 – 7 = -3
-3 ≠ -3
∴ (x + 1) is not a factor
Using synthetic division, let us find the other factors.


Quotient is 4x² + 4x – 3
= 4x² + 6x – 2x – 3
= 2x (2x + 3) – 1 (2x + 3)
= (2x + 3) (2x – 1)
∴ The factors are (x – 1), (2x + 3) and (2x – 1)
∴ -7x + 3 + 4x³ = (x + 1) (2x + 3) (2x – 1)

(iv) x³ + x² – 14x – 24
Let p (x) = x³ + x² – 14x – 24
Sum of the co-efficients are = 1 + 1 – 14 – 24 = -36 ≠ 0
∴ (x – 1) is not a factor
Sum of co-efficients of even powers of x with constant = 1 – 24 = -23
Sum of co-efficients of odd powers of x = 1 – 14 = -3
-23 ≠ -13
∴ (x + 1) is also not a factor
p(2) = 2³ + 2² – 14 (2) – 24 = 8 + 4 – 28 – 24
= 12 – 52 ≠ 0, (x – 2) is a not a factor
p (-2) = (-2)³ + (-2)² – 14 (-2) – 24
= -8 + 4 + 28 – 24 = 32 – 32 = 0
∴ (x + 2) is a factor
To find the other factors let us use synthetic division.

∴ The factors are (x + 2), (x + 3), (x + 4)
∴ x³ + x² – 14x – 24 = (x + 2) (x + 3) (x – 4)

(v) x³ – 7x + 6
Let p (x) = x³ + 0x² – 7x + 6
Sum of the co-efficients are = 1 + 0 – 7 + 6 = 7 – 7 = 0
∴(x- 1) is a factor
Sum of co-efficients of even powers of x with constant = 0 + 6 = 6
Sum of coefficient of odd powers of x = 1 – 7 = -7
6 ≠ -7
∴ (x + 1) is not a factor
To find the other factors, let us use synthetic division.

∴ The factors are (x – 1), (x – 2), (x + 3)
∴ x³ + 0x² – 7x + 6 = (x – 1) (x – 2) (x + 3)

(vi) x³ – 10x² – x + 10
Let p (x) = x³ – 10x² – x + 10
Sum of the co-efficients = 1 – 0 – 1 + 10
= 11 – 11 = 0
∴ (x – 1) is a factor
Sum of co-efficients of even powers of x with constant = -10 + 10 = 0
Sum of co-efficients of odd powers of = 1 – 1 = 0
∴(x + 1) is a factor
Synthetic division

∴ x³ + 10x² – x + 10 = (x – 1) (x + 1) (x – 10)

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

Exercise 8.1

Question 1.
The following data gives the number of residents in an area based on their age. Find the average age of the residents.

Solution:


Question 2.
Find the mean for the following frequency table :

Solution:
Let Assumed mean A = 170

Median Calculator Instructions. This calculator computes the median from a data set: To calculate the median from a set of values, enter the observed values.

Question 3.
Find the mean for the following distribution using step Deviation Method.

Solution:
Let Assumed mean A = 28, Class width C = 8

Exercise 8.2

Question 1.
For the following up grouped data 8, 15, 14, 19, 11, 16, 10, 8, 17, 20. Find the median.
Solution:
Arrange the values in ascending order 8, 8, 10, 11, 14, 15, 16, 17, 19, 20
The number of values = 10

Question 2.
The following table gives the weekly expenditure of 200 families. Find the median of the weekly expenditure.

Solution:


Question 3.
The median of the following data is 24. Find the value of x.

Solution:

Question 4.
The following are the scores obtained by 11 players in a cricket match 7, 21, 45, 12, 56, 35, 25, 0, 58, 66, 29. Find the median score.
Solution:
Let us arrange the values in ascending order 0, 7, 12, 21, 25, 29, 35, 45, 56, 58, 66
The number of values = 11 which is odd.

Exercise 8.3

Question 1.
Find the mode of the given data : 65, 65, 71, 71, 72, 75, 82, 72, 47, 72.
Solution:
In the given data 72 occurs thrice. Hence the mode is 72.

Question 2.
Find the mode:

Solution:
7 has the maximum frequency 21. Therefore 7 is the mode.

Question 3.
Find the mode for the following data.

Solution:

Question 4.
In a distribution, the mean and mode are 46 and 40 respectively. Calculate the median.
Solution:
Given, Mean = 46 and mode = 40
Using mode ≈ 3 median – 2 mean ,
40 ≈ 3 Median – 2 (46)
3 Median ≈ 40 + 92
Therefore, Median ≈ \(\frac{132}{3}\) = 44

Exercise 8.4

Multiple Choice Questions :

Question 1.
The mean of first 10 natural numbers.
(1) 25
(2) 55
(3) 5.5
(4) 2.5

Solution:
(3) 5.5

Question 2.
The mean of a distribution is 23, the median is 24 and the mode is 25.5. It is most likely that this distribution is :
(1) Positively skewed
(2) Symmetrical
(3) Asymptotic
(4) Negatively skewed
Hint: For Negatively skewed means is likely to be less than mode and median
Solution:
(4) Negatively skewed

Question 3.
The middle value of an ordered array of numbers is the
(1) Mode
(2) Mean
(3) Median
(4) Mid point
Solution:
(3) Median

Question 4.
The weights of students in a school is a :
(1) Discrete variable
(2) Continuous variable
(3) Qualitative variable
(4) None of these
Solution:
(2) Continuous variable

Question 5.
The first hand and unorganized form data is called
(1) Secondary data
(2) Organised data
(3) Primary data
(4) None of these
Solution:
(3) Primary data