Category: 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 7 Mensuration Ex 7.1

9th Maths Mensuration Exercise 7.1 Question 1.
Using Heron’s formula, find the area of a triangle whose sides are
(i) 10 cm, 24 cm, 26 cm
(ii) 1.8 m, 8 m, 8.2 m
Solution:
(i) sides : 10 cm, 24 cm, 26 cm
Using Heron’s formula
Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) sq. units

9th Maths Mensuration Question 2.
The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and cost of leveling the ground at the rate of ₹ 20 per m².
Solution:

9th Maths Book Mensuration ex 7.1 Question 3.
The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot.
Solution:
s = 600 m
Side s are in the ratio 5 : 12 : 13
5x + 12x + 13x = 30x

∴ sides are 200 m, 480 m, 520 m.

9th Maths 7.1 Mensuration Question 4.
Find the area of an equilateral triangle whose perimeter is 180 cm.
Solution:
Perimeter of an equilateral triangle = 180 cm

9th Maths Exercise 7.1 Samacheer Kalvi Question 5.
An advertisement board is in the form of an isosceles triangle with perimeter 36m and each of the equal sides are 13 m. Find the cost painting it at ₹ 17.50 per square metre.
Solution:

Cost of painting 1 m² = ₹ 17.50
Cost of painting 60m² = 60 × 17.50 = ₹ 1050

9th Maths Mensuration Exercise 7.1 Question 6.
Find the area of the unshaded region.

Solution:

10th Maths Exercise 7.1 Samacheer Kalvi Question 7.
Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm
Solution:

9th Maths Exercise 7.1 Question 8.
A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park.
Solution:

Exercise 7.1 Class 10 Samacheer Kalvi Question 9.
A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the diagonal is 48 m. Find the area of the land.
Solution:
Perimeter of the rhombus land = 160 m

9th Maths 7.1 Question 10.
The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of the diagonal is 42 m. Find the area of Parallelogram.
Solution:

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 3.10

9th Maths Exercise 3.10 Solutions Question 1.
Draw the graph for the following
(i) y = 2x
(ii) y = 4x – 1
(iii) y = \(\left(\frac{3}{2}\right)\) x + 3
(iv) 3x + 2y = 14
Solution:
(i) Put x = -1, y = 2 × -1 = -2
When x = 0, y = 2 × 0 = 0
When x = 1, y = 2 × 1 = 2

The points (x, y) to be plotted: (-1, -2), (0, 0), (1, 2)

(ii) When x = -1 ⇒ y = 4 (-1) -1
y = – 4 – 1 = – 5
x = 0 ⇒ y = 4 × 0 – 1 = -1
x = 1 ⇒ y = 4 × 1 – 1 = 3

The points (x, y) to be plotted: (-1, -5), (0, -1), (1, 3)

The points to be plotted: (-2, 0), (0, 3), (2, 6)

The points to be plotted: (-2, 10), (0, 7), (2, 4)

9th Maths Graph Question 2.
Solve graphically
(i) x + y = 7; x – y = 3
(ii) 3x + 2y = 4; 9x + 6y – 12 =0
(iii) \(\frac{x}{2}+\frac{y}{4}=1 ; \frac{x}{2}+\frac{y}{4}=2\)
(iv) x – y = 0; y + 3 = 0
(v) y = 2x + 1; y + 3x – 6 = 0
(vi) x = -3, y = 3
Solution:
(i) We can find x and y intericepts and thus of the two points on the lines (1), (2)
x + y = 7 ……… (1), x – y = 3 …………. (2)
To draw the graph of (1)
Put x = 0 in (1)
0 + y = 7 ⇒ y = 7
Thus A (0, 7) is a point on the line
Put y = 0 in (1)
x + 0 = 7 ⇒ x = 7
Thus B (7, 0) is another point on the line
Plot A and B. Join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.

When x = 0,(2) ⇒ x – y = 3
0 – y = 3 ⇒ y = -3
P (0, -3) is a point on the line.
Put y = 0 in (2); x – 0 = 3
x = 3
∴ Q (3, 0) is another point on the line (2)
Plot P, Q
1 The point of intersection (5, 2) of lines (1), (2) is a solution

(ii) 3x + 2y = 4 ……. (1)
9x + 6y= 12 ………. (2)
To draw the graph of (1)
Put x = 0 in (1) ⇒ 3 (0) + 2y = 4
2y = 4
y =2
∴ A (0, 2) is a point on the line (1)
Put y = 0 in (1) ⇒ 3x + 2(0) = 4
3x = 4
x = \(\frac{4}{3}\) = 1.3
∴ B (1.3, 0) is another point on the line (1)
Plot the points A, B. Join them to produce the line (1)
To draw the graph of (2)
Put x = 0 in (2) ⇒ 9 (0) + 6y = 12
6y = 12
y = 2
∴ P (0, 2) is a point on the line (2)
Put y = 0 in (2) ⇒ 9x + 6(0) = 12
9x = 12
x = \(\frac{12}{9}=\frac{4}{3}\)
x = 1.3

Q (1.3, 0) is another point on the line (2)
Plot P, Q. Join them to produce the line (2).
The point of intersection of the lines (1), (2) is a solution.
[A, B],[P, Q] represent the same line.
∴ All the points on one line are also on the other.
This means we have an infinite number of solutions.

∴ Comparing (1), (2) we can conclude their slopes are equal
∴ The lines are parallel and will not meet at any point and hence no solution exists.
Let us draw the graphs of (1) & (2)
(1) ⇒ 2x + y = 4, Put x = 0 in (1) ⇒ y = 4
∴ A (0, 4) is a point on (1)
Put y = 0 in (1) ⇒ 2x = 4
x = 2
∴ B (2, 0) is another point on (1)
Plot A, B ; Join them to produce the line (1)
(2) ⇒ 2x + y = 8, Put x = 0 in (2)
2(0) + y = 8
y = 8, P (0,8) is a point
Put y = 0 in (2) ⇒ 2x + 0 = 8
2x = 8
x = 4, Q (4, 0) is another point on the line (2)

Plot P, Q: Join them to produce the line (2)

(iv) x – y = 0 ………….. (1) ⇒ -y = -x ⇒ y = x

Put x = 0 in (1), 0 – y = 0
-y = 0 ⇒ y = 0
A (0, 0) is a point on the line (1)
Put y = 0 in (1) ⇒ x – 0 = 0 ⇒ x = 0
B (-3, -3) is also the same point as A
(2) ⇒ y + 3 = 0
y = -3
∴ from (1) y = -3 = x
∴ B (-3, -3) is the solution

(v) y = 2x + 1 …………… (1)


The points of intersection (1, 3) of the lines (1) and (2) is a solution. The solution is the point that is common to both the lines.
∴ The solution is as x = 1, y = 3.

(vi) The point of intersection (-3, 3) is a solution.

x + y = 7 ……… (1), x – y = 3 …………. (2)
To draw the graph of (1)
Put x= 0 in (1)
0 + y = 7 ⇒ y =7
Thus A (0, 7) is a point on the line
Put y = 0 in (1)
x + 0 = 7 ⇒ x =7
Thus B (7, 0) is another point on the line
Plot A and B. Join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, …….. (2) ⇒ x – y = 3
0 – y = 3 ⇒ y = -3
P (0, -3) is a point on the line.
Put y = 0 in (2) ; x – 0 = 3
x = 3
∴ Q (3, 0) is another point on the line (2) Plot P, Q

9th Graph Exercise 3.10 Question 3.
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let x, y be the speed of the two cars. If the two cars travel towards each other they will meet in 1 hr. The distance between them d = 100; \(\frac{d}{s}\) = t
i.e., \(\frac{100}{x+y}\) = 1 ⇒ x + y = 100 ………. (1)
If the two cars travel in the same direction they will meet in 2 hrs.
x + y = \(\frac{100}{2}\) ⇒ x – y = 50 ………….. (2)

x + y = 10 ………….. (1)
Put x = 0 in (1), then 0 + y = 100 ⇒ y =100
A (0, 100) is a point on (1)
Put y = 0 in (1), then x + 0 = 100 ⇒ x =100
B (100, 0) is another point on (1)
Plot A & B. Join them to produce the line (1)
Similarly by x – y = 50
Put x = 0 in (2), then 0 – y = 50 ⇒ y = -50
P (0, -50) is a point on (2)
Put y = 0 in (2), then x – 0 = 50 ⇒ x = 50
Q (50, 0) is another point on (2)

Plot P & Q. Join them to produce the line (2)
The point of intersection (75, 25) of the two lines (1) & (2) is the solution.
∴ The solution i.e., the speed of the two cars x and y is given by x = 75 km and y = 25 km

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 Ex 1.6

Exercise 1.6 Class 9 Maths Samacheer Question 1.
(i) If n(A) = 25, n(B) = 40, n(A ∪ B) = 50 and n(B’) = 25 , find n(A ∩ B) and ii(U).
(ii) If n(A) = 300, n(A ∪ B) = 500, n(A ∩ B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
(i) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
n(A ∩ B) = 25 + 40 – 50 = 65 – 50 = 15
n(U) = n(B) + n(B’) = 40 + 25 = 65

(ii) n(U) = n(B) + n(B’)
n(A ∩ B) = n(A) + n(B) – n(A B)
n(B) = n(A ∪ B) + n(A ∩ B) – n(A) = 500 + 50 – 300 = 250
n(U) = 250 + 350 = 600

9th Maths Exercise 1.6 Question 2.
If U = {x : x ∈ N, x ≤ 10}, A = {2, 3, 4, 8, 10) and b = {1, 2, 5, 8, 10}, then verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Solution:
n(A) = 5, n(B) = 5
A ∪ B = {1, 2, 3, 4, 5, 8, 10}, A ∩ B= {2, 8, 10}
n(A ∪ B) = 7, n(A ∩ B) = 3
L.H.S n(A ∪ B) = 7
R.H.S = n(A) + n(B) – n(A ∩ B) = 5 + 5 – 3 = 7
∴ L.H.S = R.H.S proved.

9th Standard Maths Exercise 1.6 Question 3.
Verify n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
(ii) A = {1, 3, 5} B = {2, 3, 5, 6} and C = {1, 5, 6, 7}.
Solution:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)

(i) A = {a, c, e, f, h}, B = {c, d, e, f}, C = {a, b, c, f}
n (A) = 5, n (B) = 4, n (C) = 4
n( A ∩ B) =3
n(B ∩ C) = 2
n( A ∩ C) =3
n( A ∩ B ∩ C) = 2
A ∩ B = {c, e, f}
B ∩ C = {c, f}
A ∩ C = {a, c, f}
A ∩ B ∩ C = {c, f}
A ∪ B ∪ C = {a, c, d, e, f, b, h}
∴ n(A ∪ B ∪ C) = 7 ……………. (1)
n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8 = 7 ……………… (2)
∴ (1) =(2)
⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Hence it is verified.

(ii) A = {1, 3, 5}, B = {2, 3, 5, 6 }, C = {1, 5, 6,7} = 3, n (B) = 4, n (C) = 4
n(A ∩ B) = 2
n(B ∩ C) = 2
n(C ∩ A) = 2
n(A ∩ B ∩ C) = 1
n(A ∪ B ∪ C) = 6
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
6 = 3 + 4 + 4 – 2 – 2 – 2 + 1 = 12 – 6 = 6
Hence it is verified.

9th Maths 1.6 Question 4.
In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Solution:
Let the number of students take part in music is M.
Let the number of students take part in drama is D.
By using venn diagram

(i) The number of students take part in only music is 17.
(ii) The number of students take part in only drama is 22.
(iii) The total number of students in the class is 17 + 8 + 22 = 47.

Samacheer Kalvi Guru 9th Maths Question 5.
In a party of 45 people, each one likes tea or coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both tea and coffee.
(ii) do not like tea.
(iii) do not like coffee.
Solution:
Let the people who like tea be T.
Let the people who like coffee be C
By using formula
n( A ∪ B) = n(A) + n(B) – n(A ∩ B)

(i) n(T ∩ C) = n(T) + n(C) – n(T ∪ C) = 35 + 20 – 45 = 55 – 45 = 10
The number of people who like both coffee and tea = 10.

(ii) The number of people who do not like Tea
n(T) = n(U) – n(T) = 45 – 35 = 10

(iii) The number of people who do not like coffee
n(C’) = n(U) – n(C) = 45 – 20 = 25.

Samacheer Kalvi 9th Standard Maths Question 6.
In an examination 50% of the students passed in Mathematics and 70% of students passed in Science while 10% students failed in both subjects. 300 students passed in at least one subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let the students who appeared in the examination be 100%.
Let the percentage of students who failed in mathematics be M.
Let the percentage of students who failed in science be S.
Failed in Maths = 100 % – Pass% = 100% – 50% = 50%
Failed in Science% 100% – 70% = 30%
Failed in both% = 10%
n(M ∪ S) = n(M) + n(S) – n(M ∩ S)
= 50% + 30% – 10% = 70%
% of students failed in atleast one subject = 70%
∴ The % of students who have passed in atleast one subject = 100% – 70% = 30%
30% = 300
∴ \(100 \%=\frac{100 \times 300}{30}=1000\)
∴ The total number of students who appeared in the examination = 1000 students.

Samacheer Kalvi 9th Maths Solutions Pdf Question 7.
A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A ∩ B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B), calculate the value of x.
Solution:

n (A – B) = 32 +x
n(B – A) = 5x
n(A ∩ B) = x
n( A) = n(B)
32 + x + x = 5x + x
32 + 2x = 6x
4x = 32
x = 8

Samacheer Kalvi 9th Maths Question 8.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Solution:
n( A ∪ B) = n(A) + n(B) – n( A ∩ B)
n(A ∪ B) = 500 (given) …………. (1).
n( A) = 400
n(B) = 200
n( A ∩ B) =50
∴ n(A ∩ B) = 400 + 200 – 50 = 550 …………… (2)
1 ≠ 2
∴ This data is incorrect.

9th Maths Samacheer Kalvi Question 9.
In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Solution:
(i) Tamil Newspaper buyers n(A) = 275
English Newspaper buyers n(B) = 150
Hindi Newspaper buyers n(C) = 45
Tamil and English Newspaper buyers n(A ∩ B) = 125
English and Hindi Newspaper buyers n(B ∩ C) = 17
Hindi and Tamil Newspaper buyers n(C ∩ A) = 5
All the three Newspaper buyers n(A ∩ B ∩ C) = 3

(i) Number of families buy only one newspaper = 148 + 11 + 26 = 185
(ii) Number of families buy atleast two news papers = 122 + 14 + 2 + 3 = 141
(iii) Total number of families in the colony = 148 + 11 + 26 + 122 + 14 + 2 + 3 = 326

Question 10.
A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Solution:

a = 600 – (120 – x + x + 80 – x)
= 600 – (200 – x)
= 600 – 200 + x
= 400 + x
b = 350 – (120 – x + x + 100 – x)
= 350 – (220 – x)
= 350 – 230 + x
= 130 + x
c = 280 – (80 – x + x + 100 – x)
= 2800 – (180 – x) = 280 – 180 + x = 100 + x
Each farmer grew atleast one of the above three, the number of farmers who grew all the three is x.
= a + b + c + 120 – x + 100 – x + 80 – x + x = 1000
400 + x + 130 + x + 100 + x + 120 – x + 100 – x + 80 – x + x= 1000
∴ 930 + x = 1000
x = 1000 – 930 = 70
∴ 70 farmers grew all the three crops

9th Standard Maths Samacheer Kalvi Question 11.
In the adjacent diagram, if n(U) = 125,y is two times of x and z is 10 more than x, then find the value of x, y and z.

Solution:
n(U) = 125
y = 2x
z = x + 10
∴ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 – 45
4x = 80
x = 20
∴ y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
Hence x = 20 ; y = 40; z = 30

Samacheer Kalvi Guru Maths 9th Question 12.
Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Solution:

A – Chess
B – Carrom
C – Table Tennis
n(A) = 22
n(B) = 21
n(C) = 15
n(A ∩ C) = 10
n(B ∩ C) = 8
n(A ∩ B ∩ C) = 6
(i) y = 22 – (x + 6 + 4) = 22 – (x + 10)
= 22 – x – 10
= 12 – x
z = 21 – (x + 6 + 2) = 21 – (8 + x)
21 – 8 – x = 13 – x
y + z + 3 + x + 2 + 4 + 6 = 35
12 – x + 13 – x + 15 + x = 35
40 – x = 35
x = 40 – 35 = 5
(i) Number of students who pay only chess and Carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 – x = 12 – 5 = 7
(iii) Number of students who play only carrom = 13 – x = 13 – 5 = 8

9th Class Maths Exercise 1.6 Solution Question 13.
In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Solution:

A – by bus
B – by bicycle
C – on foot
n(A) = 25
n(B) = 20
n(C) = 30
n(A ∩ B ∩ C ) = 10
n(A ∪ B ∪ C ) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
50 = 25 + 20 + 30 – (10 + x) – (10 + y) – (10 + z) + 10
50 = 75 – 10 – x – 10 – y – 10 – z + 10
= 75 – 20 – (x + y + z)
= 55 – (x + y + z)
x + y + z = 55 – 50 = 5
∴ The number of students who come to school exactly by two modes of transport = 5

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 23 Economic Biology

Samacheer Kalvi 9th Science Economic Biology Textbook Exercises

I. Choose the correct answer.

Economic Biology 9th Class Question 1.
The production and management of fish is called ……………..
(a) Pisciculture
(b) Sericulture
(c) Aquaculture
(d) Monoculture
Answer:
(a) Pisciculture

Chapter 23 Economic Biology Question 2.
Which one of the following is not an exotic breed of cow?
(a) Jersey
(b) Holstein-Friesan
(c) Sahiwal
(d) Brown Swiss
Answer:
(c) Sahiwal

9th Science Economics Biology Question 3.
Which one of the following is an Italian species of honey bee?
(a) Apis mellifera
(b) Apis dorsata
(c) Apis florae
(d) Apis Cerana
Answer:
(a) Apis mellifera

Samacheer Kalvi 9th Science Question 4.
Which one of the following is not an Indian major carp?
(a) Rohu
(b) Catla
(c) Mrigal
(d) Singhara
Answer:
(d) Singhara

Samacheer Kalvi Guru 9th Science Question 5.
Drones in the honey bee colony are formed from
(a) unfertilized egg
(b) fertilized egg
(c) parthenogenesis
(d) both b and c
Answer:
(b) fertilized egg

Samacheer Kalvi 9th Science Solutions Question 6.
Which of the following is an high milk yielding variety of cow?
(a) Holstein- Friesan
(b) Dorset
(c) Sahiwal
(d) Red Sindhi
Answer:
(a) Holstein- Friesan

9th Science Samacheer Kalvi Question 7.
Which Indian variety of honey bee is commonly used for apiculture?
(a) Apis dorsata
(b) Apis florea
(c) Apis mellifera
(d) Apis indica
Answer:
(d) Apis indica

Samacheer Kalvi 9th Science Guide Question 8.
………………. is the method of growing plants without soil.
(a) Horticulture
(b) Hydroponics
(c) Pomology
(d) None of these
Answer:
(b) Hydroponics

Samacheer Kalvi 9th Science Solution Question 9.
The symbiotic association of fungi and vascular plants is …………………….
(a) Lichen
(b) Rhizobium
(c) Mycorhizae
(d) Azotobacter
Answer:
(c) Mycorhizae

9th Science Solutions Samacheer Kalvi Question 10.
The plant body of mushroom is …………………….
(a) Spawn
(b) Mycelium
(c) Leaf
(d) All of these
Answer:
(c) Leaf

II. Fill in the blanks.

Samacheer Kalvi Class 9 Science Solutions Question 11.
Quinine drug is obtained from ……………….
Answer:
Cinchona officinalis

Samacheer Kalvi 9th Biology Book Pdf Question 12.
Carica papaya leaf can cure …………….. disease.
Answer:
Apiculture

Samacheer Kalvi 9th Science Practical Question 13.
Vermicompost is a type of soil made by ………………. and microorganisms.
Answer:
earthworms

Samacheer Kalvi Biology Question 14
……………….. refers to the culture of prawns, pearl and edible oysters.
Answer:
Aquaculture

Samacheer Kalvi 9th Science Book Solutions Question 15.
The largest member in a honey bee haive is the …………………..
Answer:
F

Samacheer Kalvi 9th Standard Science Question 16
……………… is a preservative in honey.
Answer:
Formic acid

Science Solution Class 9 Samacheer Kalvi Question 17
…………………. is the method of culturing different variety of fish in a water body.
Answer:
Polyculture

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

1. Mycorrhiza is an algae – False.
   Correct Statement: Mycorrhiza is a fungi
2. Milch animals are used in agriculture and transport – False.
   Correct Statement: Milch animals are domesticated for obtaining only milk.
3. Apisflorea is a rock bee – False.
   Correct Statement: Apis Florea is a little bee
4. Ongole is an exotic breed of cattle – False.
   Correct Statement: Ongole is a dual-purpose breed of cattle
5. Sheep manure contains high nutrients than farmyard manure – True.

IV. Differentiate the following

9th Samacheer Kalvi Science Question 1.
Exotic breed and Indigenous breed.
Answer:

Exotic breed	Indigenous breed
Exotic breeds are imported from foreign countries	Indigenous breed are native to India
These foreign breeds are selected for long lactation periods.	These local breed show excellent resistance to diseases.
Example: Jersey, Brown Swiss and Holstein-Friesian	Example: Sahiwal, Red Sindhi, Deoni and Gir.

Samacheer Kalvi 9 Science Question 2.
Pollen and Nectar
Answer:

Pollen	Nectar
Pollen is a fine to a coarse powdery substance comprising pollen grains which are male microgametophytes of seed plants, which produce male gametes	It is a sweet viscous secretion secreted by the flower of plants.

Science 9th Samacheer Kalvi Question 3.
Shrimp and Prawn
Answer:

Shrimp	Prawn
Shrimp has lamellar gills.	Prawns have branching gills.
Shrimp have claws on two of their five pairs of legs.	Prawns have claws on three of their five pairs of legs.

Samacheer Kalvi Guru 9th Science Solutions Question 4.
Farmyard manure and Sheep manure
Answer:

Farmyard manure	Sheep manure
Well decomposed farm yard manure contains 0.5% Nitrogen, 0.2% available phosphate and 0.5% available potash.	It contains 3% Nitrogen, 1% phosphorus pentoxide and 2% potassium oxide.

V. Match the following.

	Column A	.           Column B
1.	Lobsters	(a) Marine fish
2.	Catla	(.b) Pearl
3.	Sea bass	(c) Shell fish
4.	Oysters	(d) Paddy
5	Pokkali	(.e) Fin fish
6	Pleurotus sps	(J) Psoriosis
7	Sarpagandha	(g) Oyster mushroom
8	Olericulture	(h) Reserpine
9	Wrighta tinctoria	(0 Vegetable farming

Answer:

1. (c) Shellfish
2. (e) Finfish
3. (a) Marine fish
4. (b) Pearl
5. (d) Paddy
6. (g) Oyster mushroom
7. (h) Reserpine
8. (i) Vegetable farming
9. (f) Psoriasis

VI. Answer in brief.

Samacheerkalvi.Guru 9th Science Question 1.
What are secondary metabolites?
Answer:
Most medicines are obtained either directly or indirectly from plants. All the major system of medicines such as Ayurveda, Yoga, Unani, Siddha, Homeopathy (AYUSH) use drugs obtained from plants and animals. These drugs from medicinal plants are called secondary metabolites.

Question 2.
What are the types of vegetable garden?
Answer:
Vegetable farming can be classified into:

1. Kitchen or Nutrition gardening,
2. Commercial gardening,
3. Vegetable forcing.

Question 3.
Mention any two mushroom preservation methods.
Answer:
Drying and Vacuum Cooling are some methods used to preserve mushroom.+

Question 4.
Enumerate the advantages of vermicompost over chemical fertiliser.
Answer:

• It is a rich source of nutrients essential for plant growth. It makes the soil fertile.
• It improves soil structure, texture, aeration and water holding capacity and helps to prevent soil erosion.
• It contains valuable vitamins, enzymes and growth regulator substances for increasing growth, vigour and yield of plants.
• It enhances decomposition of organic matter in soil.
• Vermicompost is free from pathogens and toxic elements.
• Vermicompost is rich in beneficial microflora.

Question 5.
What are the species of earthworm used for vermiculture?
Answer:
Among the vast community of earthworms only very few species can be used for vermicompost production. They are Perionyx excavatus (Indian blueworm), Eisenia fetida (Red worms), Eudrilus eugeniae (African night crawler).

Question 6.
List the medicinal importance of honey.
Answer:
Uses of Honey

• Honey has an antiseptic and antibacterial property. It is a blood purifier.
• It helps in building up of haemoglobin content in the blood.
• It is used in Ayurvedic and Unani system of medicines.
• It prevents cough, cold, fever and relieves sore throat.
• It is a remedy for ulcers of tongue, stomach and intestine.
• It enhances digestion and appetite.

VII. Answer in detail.

Question 1.
Enumerate the advantage of hydroponics.
Answer:
Hydroponics was demonstrated by a German Botanist Julius Von Sachs in 1980.
Advantages

• Crops can be grown in places where the land is limited, doesn’t exist, or is heavily contaminated.
• The climate – temperature, humidity, light intensification, the composition of the air can be monitored.
• Conservation of water and nutrients.
• Controlled plant growth.
• No intrusion by weeds.
• Fewer pests & diseases
• Minimal use of insecticide or herbicides
• In deserts and Arctic regions hydroponics can be an effective alternative method.
• Hydroponics is successfully employed for the commercial production of seedless cucumber and tomato.

Question 2.
Define Mushroom culture. Explain the mushroom cultivation methods.
Answer:
Mushroom is a fungi belonging to basidiomycetes. It is rich in proteins, fibres, vitamins and minerals. Mushroom culture is the process of producing food, medicine, and other products by the cultivation of mushrooms.
Mushroom can be cultivated either on paddy straw or on wood log.
Major stages of mushroom cultivation are;

• Composting:
  Compost is prepared by mixing paddy straw with number of organic materials like cow dung and inorganic fertilizers. It is kept at about 50°C for one week.
• Spawning:
  Spawn is the mushroom seed. It is prepared by growing fungal mycelium in grains under sterile conditions. Spawn is sown on compost.
• Casing:
  Compost is covered with a thin layer of soil. It gives support to the growing mushroom, provides humidity and helps regulate the temperature.
• Pinning:
  Mycelium starts to form little bud, which will develop into mushroom. Those little white buds are called pins.
• Harvesting:
  Mushroom grow better in 15°C – 23°C. They grow 3 cm in a week which is the normal size for harvesting. In the third week the first flush mushroom can be harvested.

Question 3.
What are the sources of organic resources for vermicomposting?
Answer:
Materials required for vermicomposting
Biologically degradable organic wastes are used as potential organic resources for vermicomposting. They are:

• Agricultural wastes (crop residue, vegetable waste, sugarcane trash)
• Crop residues (rice straw, tea wastes, cereal and pulse residues, rice husk, tobacco wastes, coir wastes)
• Leaf litter
• Fruit and vegetable wastes
• Animal wastes (cattle dung, poultry droppings, pig slurry, goat and sheep droppings)
• Biogas slurry

Question 4.
Give an account of different types of fish ponds used for rearing fishes.
Answer:
Types of ponds for fish culture
Fish farm requires different types of pond for the various developmental stages of fish growth. They are:

1. Breeding pond: Healthy and sexually mature male and female fishes are collected and introduced in this pond for breeding. The eggs released by the female are fertilized by the sperm and fertilized eggs float in water as frothy mass.
2. Hatchling pits: The fertilized eggs are transferred to hatching pits for hatching. Two types of hatching pits are hatcheries and hatching hapas.
3. Nursery ponds: The hatchlings are transferred from hatching pits after 2 to 7 days. The hatchlings grow into fry and are cultured in these ponds for about 60 days with proper feeding till they reach 2 -2.5 cm in length.
4. Rearing ponds: Rearing ponds are used to culture the fry. The fish fry are transferred from nursery pond to rearing ponds and are maintained for about three months till they reach 10 to 15 cm in length. In these rearing ponds the fry develops into fingerlings.
   e) Stocking pond: The stocking pond is also called a culture pond or production pond. These ponds are used to rear fingerlings upto the marketable size. Before releasing the fingerlings, the pond is manured with organic manure and inorganic fertilizers.

Question 5.
Classify the different breeds of the cattle with suitable examples.
Answer:
Cattle breeds
The Indian cattle include cows and buffaloes. They are domesticated for milk, meat, leather and transportation. They belong to two different species, Bos indicus (Indian cows and bulls) and Bos bubalis (buffaloes). These cattle animals are reared for milk and farm labour. They are classified into three types:

1. Dairy breeds,
2. Draught (or) Draft breeds,
3. Dual-purpose breeds

Dairy breeds:
Dairy animals are domesticated for obtaining milk. The cows (milk producing females) are high milk yielders (milch animals). The dairy breeds may be indigenous breeds (or) exotic breeds.

• Indigenous breeds are native of India. They include Sahiwal, Red Sindhi, Deoni and Gir. These cattle are well built with strong limbs, prominent hump and loose skin. These local breed animals show excellent resistant to diseases.
• The exotic breeds (Bos taurus) are imported from foreign countries. They include Jersey, Brown Swiss and Holstein-Friesian etc. These foreign breeds are selected for long lactation periods. The Indian (local) breeds and foreign breeds can be cross bred to produce animals with both desired qualities.

Draught (or) Draft breeds:
They are used for agricultural work, such as tilling, irrigation and carting. These include Amritmahal, Kangayam, Umblachery, Malvi, Siri and Hallikar breeds. Bullocks are good draft animals while the cows are poor milk yielders.

Dual-purpose breeds:
These breeds provide milk and they are useful for farm work. In India these breeds are favoured by farmers as the cows are fairly good milk yielders and bullocks are good for draught work. They includes Haryana, Ongole, Kankrej and Tharparkar.

Buffalo breeds:
In India, buffaloes are domesticated in great number. They are the main milk producers. The milk production of buffaloes is more than that of cows. Murrah, Mehsana and Surti are indigenous buffalo breeds which are good milk yielders.

VIII. Higher Order Thinking Skills.

Question 1.
Biomanuring plays an important role in agriculture. Justify
Answer:
Biomanure also knwon as organic manures, are predominantly derived from plant debris, animal faeces and microbes. They make the soil fertile by adding nutrients like nitrogen. They are eco-friendly. Biomanure is easy to generate and very economical. Some examples of biomanure are Animal manure, Vermicompost and Green Manure.

Question 2.
Each bee hive consists of hexagonal cells. Name the material in which the cell is formed and mention the significance of the hexagonal cells.
Answer:
The cell is formed in a sheet of wax. The hexagonal shape allows to hold the queen bee’s eggs and store the pollen and honey the worker bees bring to the hive.

Activity

Question 1.
Discuss in your class room about the importance of crop insurance to farmers.
Answer:
In agriculture there are risks beyond one’s control. Hence precautionary measures are to be considered to control damage faced by farmers. Farmers who take crop insurance protect their crop and families from unforeseen setbacks.
The advantages of crop insurance are,

1. Stability in Income
2. Minimal Debts
3. Farmers can safely invest on new technological advancements to improve their crop production.
4. Protection against loss of crops
5. Provides Awareness on natural calamities and the preventive measures to be taken

Question 2.
Collect at least five medicinal plants from your locality. Identify the plant and try to find out its medicinal value.
Answer:
Students can perform this activity under the guidance of the class teacher.

Question 3.
Visit a fish farm during the breeding season near your locality and collect information.
Answer:
Students can perform this activity under the guidance of the class teacher.

Samacheer Kalvi 9th Science Economic Biology Additional Questions

I. Choose the correct answer.

Question 1.
………………. is a branch of agriculture that deals with cultivation of fruits, vegetables and
ornamental plants.
(a) Floriculture
(b) Horticulture
(c) Olericulture
(d) Moriculture
Answer:
(b) Horticulture

Question 2.
……………….. is growing of vegetables in small scale in households.
(a) Vegetable farming
(b) Flower farming
(c) Kitchen gardening
(d) Vegetable forcing
Answer:
(c) Kitchen gardening

Question 3.
Which one of the following is an African species of Honey bee?
(a) Apis florae
(b) Apis indica
(c) Apis mellifera
(d) Apis adamsoni
Answer:
(d) Apis adamsoni

Question 4.
Which one of the following is not an indegenous of cow?
(a) Sahiwal
(b) Jersey
(c) Red Sindhi
(d) Deoni
Answer:
(b) Jersey

Question 5.
……………… involves rising of cattle for milk production.
(a) Dairy farming
(b) Drying
(c) Freezing
(d) Canning
Answer:
(a) Dairy farming

Question 6.
………………. are low in fibre and contain high level of carbohydrates, protein and other nutrients.
(a) Cattle feed
(b) Roughage
(c) Concentrates
(d) Feed management
Answer:
(c) Concentrates

Question 7.
The ………………. contain the young stages of the honey bees and they are built in the centre
and lower part of the comb.
(a) brood cells
(b) storage cells
(c) drone chamber
(d) queen chamber
Answer:
(a) brood cells

Question 8.
………………….. feeds on the organic wastes.
(a) Bees
(b) Earthworms
(c) Prawns
(d) Cattle
Answer:
(b) Earthworms

II. Fill in the blanks.

1. ………………. is the science of growing vegetables.
2. Compost is a ……………….. as well as a fertilizer which is rich in nutrients.
3. …………….. is a small bacterium that colonize the roots of leguminous plants to form root nodules.
4. Application of ……………… has been found to increase yield of wheat, rice, maize and sorghum.
5. Mycelium starts to form little bud which develops into a ……………..
6. Compost for mushroom cultivation is prepared by mixing ………………… with number of organic materials like cow dung and inorganic fertilizers.

Answer:

1. Olericulture
2. soil conditioner
3. Rhizobium
4. Azotobacter
5. mushroom
6. paddy straw

III. State whether the following statements are true or false. If false, write the correct statement.

1. Vermicomposting is the rearing of earthworms for the production of vermicompost – True
2. Pasturage is the production of fruits – False.
   Correct Statement: Pasturage is the availability of flowers to bees for nectar and pollen collection.
3. Binomial name of Nilavembu is Leucas aspera – False.
   Correct Statement: Binomial name of Nilavembu is Andrographis paniculata.
4. Mariculture is the culture of fishes and another aquatic organism in marine water near the sea coast – True
5. Operation flood programme is based on dairy commodity to increase milk supply in urban areas – True.

IV. Match column A with column B.

S.No.	Column A		Column C
1.	Tulsi	a.	Wrightia tinctoria
2.	Nannari	b	Cathyranthus roseus
3.	Vepalai	c	Eucalyptus globulus
4	Cinjona maram	d	Hemidesmus indicus
5	Nithya kalyani	e	Cinchona officinalis
6	Thaila maram	f	Ocimum sanctum

Answer:

S.No.	Column A		Column C
1.	Tulsi	a.	Ocimum sanctum
2.	Nannari	b	Hemidesmus indicus
3.	Vepalai	c	Wrightia tinctoria
4	Cinjona maram	d	Cinchona officinalis
5	Nithya kalyani	e	Cathyranthus roseus
6	Thaila maram	f	Eucalyptus globulus

V. Differentiate the following.

Question 1.
Marine water prawn culture and Freshwater prawn culture
Answer:

Marine water prawn culture	Freshwater prawn culture
The rearing of marine penaied prawn is called marine prawn culture or shrimp culture.	The rearing of freshwater prawn is called fresh water prawn culture.

Question 2.
Extensive fish culture and Intensive fish culture
Answer:

Extensive fish culture	Intensive fish culture
Culture of fishes in large areas with low stocking density and natural feeding.	Culture of fishes in small areas with high stocking density and providing artificial feed to increase production.

Question 3.
Storage cells and Brood cells
Answer:

Storage cells	Brood cells
The storage cells contain honey and pollen.	The brood cells contain the young stages of the honey bees.
They are built in the margin and at the top of the comb.	They are built in the centre and the lower part of the comb.

VI. Answer in brief.

Question 1.
Who are the Father of Indian medicines.
Answer:

Ayurveda	Charaka Samhita
Yoga	Patanjali
Unani	Hippocrates
Siddha	Agasthya
Homeopathy	Samuel Hahnemann

Question 2.
What is spawning?
Answer:
Spawn is the mushroom seed. It is prepared by growing fungal mycelium in grains under sterile conditions. Spawning is the sowing or planting of spawn on compost.

Question 3.
What are the types of aquaculture?
Answer:
Aquaculture is classified into;

1. Freshwater aquaculture
2. Brackish water aquaculture
3. Marine water aquaculture (Mariculture)

Question 4.
What is aquaponics?
Answer:
Aquaponics is a system of a combination of conventional aquaculture with hydroponics in a symbiotic environment in which plants are fed with the aquatic animals excreta or wastes.

Question 5.
What is green manure? How is it beneficial?
Answer:
Green manure is obtained by collection and decomposition of green leaves, twigs of trees, shrubs and herbs growing in wastelands, field bunds etc. Green manure improves soil structure, increases water holding capacity and decreases soil loss by erosion. It also helps in reclamation of alkaline soils and reduces weed proliferation. It is a manure obtained from undecomposed green material derived from leguminous plants e.g. Sunhemp, Dhaincha, etc.

Question 6.
Write a note on nutritional value of fishes.
Answer:
Cultivable freshwater and marine food fishes are highly nutritious, rich source of animal proteins and are easily digestible. They are rich in essential amino acids such as lysine and methionine, minerals like calcium, phosphorus, iron, sodium, potassium and magnesium. Fat soluble vitamins A, D and water soluble B-complex vitamins like pyridoxine, cyanocobalamine and niacin ate found in fishes. Polyunsaturated fatty acid (PUFA) which are helpful in regulation of cholesterol are present in plenty in fishes and thus promote cardiac health.

VII. Answer in detail.

Question 1.
Explain hydroponics and give its importance.
Answer:
Hydroponics is the method of growing plants without soil, using mineral nutrient solutions in water. The containers are made of glass, metal or plastic. They range in size from small pots for individual plants to huge tank for large scale growing. It was demonstrated by a German Botanist Julius Von Sachs in 1980. Hydroponics is successfully employed for the commercial production of seedless cucumber and tomato.

Plants are suspended with their roots submerged in water that contain plant nutrients. The roots absorb water and nutrients, but do not perform the anchoring function. Therefore, the plants must be mechanically supported from above.
Importance of hydroponics

• Conservation of water and nutrients.
• Controlled plant growth.
• In deserts and Arctic regions hydroponics can be an effective alternative method.

Question 2.
What are the types of honey bees found in a colony?
Answer:
There are three types of individuals in a colony namely the Queen bee, the drones and the worker bees.

1. Queen Bee: The queen is the largest member and the fertile female of the colony. They are formed from fertile eggs. The queen is responsible for laying eggs in a colony. The life span of the queen bee is 3-4 years.
2. Drones: They are the fertile males. They develop from unfertilized eggs. They are larger than the workers and smaller than the queens. Their main function is to fertilize the eggs produced by the queen.
3. Worker Bees: They are sterile female bees and are the smallest members of the colony. Their function is to collect honey, look after the young ones, clean the comb, defend the hive and maintain the temperature of the bee hive.

Question 3.
Explain the methods used for vermicomposting.
Answer:
Vermicomposting methods can range from a wormbin in the kitchen for household scraps to large mechanized systems, which can be able to accommodate tons of organic material. In general these methods are of the following types:

• Bin (or) Container method
• Vermicomposting of organic wastes in field pits
• Vermicomposting of organic wastes on ground heaps

Bin method:
Vermicomposting by bin method is the rearing of earthworms in a container or bin. The container is half filled with bedding materials such as shredded cardboard, leaves, paddy husk, chopped straw, saw dust and manure. Small quantity of soil and sand is added to provide necessary grit for the worms. The bedding material should be moistened by adding water that enables free movements of the worms. The worms are gently placed and spread evenly on the bedding.

Organic wastes (kitchen wastes, vegetable and fruit wastes) are added which are fed by the earthworms. The bin is covered with coconut leaves or gunny bags to conserve moisture, provide darkness and keep out of pests. After a period of 60 days the wastes are completely transformed into nutrient-rich materials that are excreted by earthworms known as worm castings. These castings are harvested and used as organic manure.

Question 4.
What are concentrates? Why should they be given to cattle?
Answer:
Concentrates are low in fibre and contain high level of carbohydrates, protein and other nutrients. A variety of raw materials such as cholam (jowar), kambu (pearl millet), ragi (finger millet), rice bran, wheat bran, cotton seed cake, mustard cake, linseed cake, groundnut cake, mango seed, neem cake and yellu (sesame) cake can be used to make concentrate feed. The concentrates are fed at the time of milking. This helps in ‘let down’ of milk.
The daily average feed ratio of a milking cow is:

1. 15-25 kg of roughage (dry grass and green fodder)
2. 4-5 kg of grain mixture
3. 100-150 litres of water

For a cow that gives above 2.5 kg milk yield per day, 1 kg of concentrate feed should be given for every additional milk yield.

You can Download From Zero to Infinity Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 9th English Book Solutions Guide Pdf Prose Chapter 6 help you to revise complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th English Solutions Prose Chapter 6 From Zero to Infinity

From Zero To Infinity 9th Standard English Warm Up:

Answer:

From Zero To Infinity Summary Question 1.
Did you enjoy solving this?
Answer:
Yes.

From Zero To Infinity Prose Summary Question 2.
Was it easy or hard to solve?
Answer:
It was very easy to solve.

From Zero To Infinity Question 3.
Do you like Mathematics? Give reasons.
Answer:
I like mathematics because it is fun to solve and find the right solutions. It is very useful in our daily lives.
OR
I do not like mathematics because it is complicated and too hard to understand.

From Zero to Infinity Intext Questions

From Zero To Infinity 9th Standard English Question 1.
What was the reaction of the classmates to Ramanujan’s question?
Answer:
The classmates laughed at Ramanujan’s question.

From Zero To Infinity Questions And Answers Question 2.
What did the Indian mathematician Bhaskara prove?
Answer:
The Indian mathematician Bhaskara proved that zero divided by zero is infinity.

From Zero To Infinity Book Back Answers Question 3.
Where did Ramanujan get S.L. Loney’s book on Trigonometry?
Answer:
Ramanujan got Loney’s “Trigonometry” book from a college library.

Zero To Infinity Lesson Summary Question 4.
Where did Ramanujan do his mathematical problems?
Answer:
Ramanujan did his mathematical problems on loose sheets of paper or on a slate.

From Zero To Infinity Lesson Plan Question 5.
What were the subjects neglected by Ramanujan in college?
Answer:
History, English, Physiology were the subjects neglected by Ramanujan in college.

From Zero To Infinity Lesson Summary Question 6.
Which University granted him a fellowship of H75 a month?
Answer:
University of Madras granted him a fellowship of? 75 a month.

From Zero To Infinity Biography Of Ramanujan Question 7.
What did Ramanujan send to G.H. Hardy?
Answer:
Ramanujan sent a letter in which he set out 120 theorems and formulae to G.H Hardy.

From Zero To Infinity Summary In English Question 8.
Who discovered a rare mathematical genius in Ramanujan?
Answer:
G.H Hardy and his colleague J.E. Littlewood discovered a rare mathematical genius in Ramanujan.

From Zero to Infinity Textual Questions

A. Answer the following questions in a sentence or two.

From Zero To Infinity Essay Question 1.
Why did the students laugh at Ramanujan?
Answer:
The students laughed at Ramanujan because he asked if no banana was distributed among no one, would every one get one banana.

From Zero To Infinity Summary In Tamil Question 2.
Why did the teacher compliment Ramanujan?
Answer:
The teacher complimented Ramanuj for asking a question that took centuries for mathematicians to answer.

Summary Of From Zero To Infinity Question 3.
Question What did Ramanujan do after reading the book on Trigonometry?
Answer:
After reading the book on Trigonometry, Ramanujan began his own research. He came forth with many mathematical theorems and formulae not given in the book

From Zero To Infinity Pdf Question 4.
What disappointed Ramanujan’s father?
Answer:
Ramanujan failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. This disappointed his father.

From Zero To Infinity Mind Map Question 5.
How did Ramanujan manage his paper crisis?
Answer:
Ramanujan needed about 2,000 sheets of paper every month. He started using even scraps of paper he found lying on the streets. Sometimes he used a red pen to write over what was written in blue ink.

Question 6.
Why were Ramanujan’s application for jobs rejected?
Answer:
Ramanujan would show his frayed notebooks to every officers. But no one could understand what was written in the notebooks. So, his applications for jobs were rejected.

Question 7.
Why was Ramanujan sent back to India?
Answer:
While Ramanujan continued his research work, Tuberculosis, then an incurable disease, was devouring him. So, he was sent back to India.

Additional Questions:

Question 1.
Who asked the intriguing question to the arithmetic teacher? What do you know about him?
Answer:
The intriguing question was asked by Srinivasa Ramanujan. He was a native of Kumbakonam. Both during his school and research work at Cambridge, he was always ahead of his mathematics teachers.

Question 2.
Where and when was Ramanujan born? What do you know of his father?
Answer:
Ramanujan was born in Erode in Tamil Nadu on December 22, 1887. His father was a petty clerk in a cloth shop.

Question 3.
What was the most significant turn in Ramanujan’s life?
Answer:
The most significant turn came when one of Ramanujan’s senior friends showed him Synopsis of Elementary> Results in Pure Applied Mathematics by George Shoobridge Carr. Ramanujan was delighted than intimidated. This book triggered the mathematical genius in him and he began to work on the problems given in it.

Question 4.
What was Hardy’s opinion of Ramanujan?
Answer:
According to Hardy, Ramanujan was an unsystematic mathematician, similar to one who knows the Pythagorus theorem but does not know what a congruent triangle means. He felt that many discrepancies in his research could be due to his lack of formal education.

Question 5.
How can you say that Ramanujan was multi-talented?
Answer:
Besides Ramanujan being a mathematician, he was a reputed astrologer. He was also an excellent orator and many were eager to listen to his practical and intellectual talks. He used to give lectures on subjects like “God, Zero and Infinity”, topics that no ordinary man can easily indulge in. Hence we can surely say he was multi-talented.

B. Answer the following questions in about 80 -100 words.

Question 1.
Describe the life of Srinivasa Ramanujan in India.
Answer:
Ramanujan was born in Erode in Tamilnadu on December 22,1887. From early childhood, it was evident that he was a prodigy. Senior students used to get his assistance in solving math problems. At the age of 13, he began his own research on Trigonometry. The book “Elementary Results in Pure Applied Mathematics” by George Shoobridge Carr triggered the genius in Ramanujan. He used to do problems on loose sheets and enter the results in notebooks which are now famous as “Ramanujan’s Frayed Notebooks”.

Although Ramanujan secured a first class in Mathematics in the matriculation examination and was awarded the Subramanyan Scholarship, he failed twice in his first year arts examination in college as he neglected other subjects such as History, English and Physiology. He searched for job for food and papers to do calculations. The Director of Madras Port Trust gave a clerical job to Ramanujan on a monthly salary of Rupees 25.

Question 2.
Narrate the association of Ramanujan with G.H. Hardy.
Answer:
Ramanujan sent a letter to the great Mathematician G.H. Hardy of Cambridge University, in which he set out 120 theorems and formulae which included the Reimann Series. Hardy and his colleague Littlewood realized that they had discovered a rare mathematical genius.

They invited him to Britain.Despite the cold weather and food, Ramanujan continued his research with determination in the company of Hardy and Littlewood. Hardy found an unsystematic mathematician in Ramanujan due to his lack of formal education. Ramanujan’s achievements include the Hardy-Ramanujan-Littlewood circle method in number theory.

Additional Questions:

Question 1.
What prompted Ramanujan to ask an intriguing question?
Answer:
The Mathematics teacher was teaching and solving concepts in division. She drew three bananas on the blackboard and she pointed to three boys who were there and asked them how many each would get. A smart student quickly answered that each would get one. After appreciating the students’ answer, the teacher introduced a similar instance of 1,000 bananas distributed among 1,000 boys where each would again get one. While the teacher was explaining, a boy seated in one comer who was none other than Ramanujan asked an intriguing question wherein if no banana was distributed among no one, would everyone still get one banana?

Question 2.
Mention the achievements of Srinivasa Ramanujan.
Answer:
Ramanujan was elected Fellow of the Royal Society on February 28, 1918. In October, 1918 he became the first Indian to be elected Fellow of Trinity College, Cambridge. His achievements at Cambridge include the Hardy-Ramanujan-Littlewood circle method in number theory.

He is also popular with Roger-Ramanujan’s identities in partition of integers. A long list of the highest composite numbers, besides work on the number theory and the algebra of inequalities are also his noted achievements. In algebra his work on continued fractions is considered on par with great mathematicians like Leonard Euler and Jacobi.

Question 3.
What did Ramanujan do when his mind was flooded with ideas?
Answer:
Mathematical ideas flooded Ramanujan’s mind. He was not able to write all of them down. ‘He solved problems on loose sheets of paper, slate and jotted the results down in notebooks. Before he went abroad he had filled three notebooks, later known as Ramanujan’s Frayed Notebooks.

His father who found him scribbling, mistook him to be mad. He had to find money for food and papers to do calculations. Every month, he needed at least 2,000 sheets of paper. He started using scraps of paper found lying on the streets. Sometimes, he even wrote using red ink over used papers found on the streets.

C. Match the words with correct Synonym and Antonym from the table.

Answer:

Listening:

D. Listen to the anecdote “Two Geniuses” and narrate it in your own words.
Answer:

Narration of “Two Geniuses”

There’s a story about how Dr. Albert Einstein was travelling to Universities in his car, delivering lectures on his theory of relativity. During one tired journey, his driver Hans remarked “Dr. Einstein, I have heard you deliver that lecture about 30 times. ,t I know it by heart and bet I could give it myself.”

“Well, I’ll give you the chance”, said the Dr. “They don’t know me at the next University, so when we get there, I’ll put on your cap, and you introduce yourself as Dr. Einstein and give the lecture.”

The driver delivered Einstein’s lecture without any mistakes. When he finished, he started to leave, but one of the professor stopped him and asked a complex question filled with mathematical equations and formulae. The driver thought fast. “The answer to that problem is so simple,” he said,” I’m surprised you have to ask me. In fact, to show you just how simple it is, I’m going to ask my driver Hans to come up here and answer your question”.

Speaking:

E. Divide the students into groups of five and conduct a group discussion on the topic “Importance of Mathematics in Our Everyday Life” The teacher will act as a moderator.
Answer:

Group Discussion on Importance of Mathematics

Teacher: Good morning students! We have just learnt the life of the great mathematician Ramanujan. Now let’s have a group discussion on “Importance of Mathematics in Our Everyday Life”. Divide yourselves into groups of five.

Harsha (Group A) : The importance of maths in everyday life. Mathematics is a methodical application of matter. It is so said because the subject makes a man methodical or systematic. Mathematics makes our life orderly and prevents chaos.

Varsha (Group B) : In Hebrew, it’s root is “thinking.” They tell us that mathematics gives us the critical ability to learn and think logically in any field of endeavor. The skills of learning today are more important than knowledge, which is so readily available on the Internet.

Yusuf (Group C) : Math is an important part of our lives, because in the future you will get a job that deals with math. Math is pretty much in everything you do, really. Math is important because it is the most widely used subject in the world. Every career uses some sort of math.

Adhira (Group D) : Maths improves problem-solving abilities. Teaches clearer logical reasoning. Sharpens concentration and observance. Develops confidence and self-esteem.

Danny (Group E) : Knowing basic math principles keeps you from having to carry around a calculator because good use of math allows you to do many calculations in your head.

Reading:

F. Answer the following questions based on the given passage.

Question 1.
What made John Shepherd-Barron to come up with the idea of ATM?
Answer:
It was then John’s habit to withdraw money on a Saturday, but on this particular weekend he had arrived one minute late and found the bank doors locked against him. This made him to come up with idea of ATM.

Question 2.
When and where was the first ATM installed?
Answer:
The first ATM was installed at a branch in the North London suburb of Enfield on June 27, 1967.

Question 3.
Who was the first person to withdraw cash from the ATM?
Answer:
The first person to withdraw cash from the ATM was Reg Varney, a celebrity resident of Enfield known for his part in the number of popular television series.

Question 4.
Why did Shepherd-Barron reduce the PIN number from six digits to four?
Answer:
Shepherd-Barron’s wife said that she could only remember four figures, because of her, four figures became the world standard.

Question 5.
Which theory of Ramanujan helps the ATMs to dispense cash?
Answer:
Ramanujan’s Partition theory helps the ATMs to dispense cash.

Writing:

G. Paragraph Writing

Question 1.
Write a paragraph of 100-120 words about a memorable anecdote / incident of your life.
Answer:
A memorable anecdote/incident in my life:
I was then a student of class four. One day, I was left at home with my grandmother. It was in the afternoon, my grandmother was taking a nap. I was a very restless one. The toys soon bored me and I looked around for something new. The unique thing which caught my attention was my Grandma’s spectacles.

I put it on my nose just in the style of my Grandma and looked around. Soon my eyes got tired. As I felt pain in my eyes, I removed the specs and threw them away. They struck the wall and landed on the ground broken. Now I got worried and afraid. I started trying to repair it. As I was holding these glass pieces I felt a severe pain in the middle finger of my right hand, I looked at it .

Blood was trickling down from a deep cut in my finger. I started crying loudly. On hearing my loud wailing my Grandma woke up. She hurriedly came out of her room, took a quick glance at my adventure and detecting the source of my trouble, she pressed her hand on my cut finger for some time and then she took me to the doctor for bandaging. I was very much afraid of punishment but my Grandma forgave me although she had to suffer difficulty in seeing until the glasses were repaired. However I was naturally punished as I could neither eat my meals nor do my homework for three days.

Question 2.
Write a paragraph of 100-120 words about your favorite personality.
Answer:
My favorite personality:
There are many people all around the world who are very famous and celebrities. But my favourite personality is my father. My father is my hero. He is kind, polite and really friendly to everyone. He is a teacher by profession and is very good in teaching. He is always ready to help and support the needy and helpless. He is a God fearing person and always teaches us to remember the God’s gifts and God’s love for the world.

I am so proud to have a father like him. He is a simple man with kind rules. He is handsome, my favourite and my ideal man. He is my friend and always ready to encourage, appreciate me for success and always ready to help me wherever I need a friend or a support of my father. I am proud of my father and wish him good health forever.

Grammar:

A. Complete the following sentences using appropriate Connectors from the box.

1. She felt cold _________ she was wearing a winter coat.
Answer:
although

2. This restaurant has some of the best chefs in the town.__________ their service is excellent.
Answer:
Moreover

3. I’m not going to the party tonight __________ I didn’t get an invitation.
Answer:
because

4. You can set the table. __________, I’ll start making dinner.
Answer:
Meanwhile

5. I can play quite a few instruments __________ , the flute, the guitar and the piano.
Answer:
For instance

6. The store was out of chocolate chips; __________ they would need to make a different type of cookies.
Answer:
therefore

7. The stores are open daily __________ Sundays.
Answer:
except

8. I’ll stay __________ you need me.
Answer:
as long as

9. This detergent is highly concentrated and __________ you will need to dilute it.
Answer:
thus

10. It was the thing he prized __________ .
Answer:
above all

Active Voice and Passive Voice:

B. Convert the following active sentences into passive sentences by supplying an appropriate passive verb form.

Question 1.
She will not recognize us. / We__________ by her.
(a) will not recognize
(b) will not being recognized
(c) will not be recognized
Answer:
(c) will not be recognized

Question 2.
They didn’t invite me, but I went anyway. /I __________ but I went anyway.
(a) wasn’t invited
(b) wasn’t being invited
(c) wasn’t inviting
Answer:
(a) wasn’t invited

Question 3.
They broke up the table for firewood. / The table __________ up for firewood.
(a) broke
(b) had broken
(c) was broken
Answer:
(c) was broken

Question 4.
She has won the first prize. / The first prize __________ by her.
(a) has won
(b) has been won
(c) had been won
Answer:
(b) has been won

Question 5.
A friend of mine is repairing the car. / The car __________ by a friend of mine.
(a) is repairing
(b) is repaired
(c) is being repaired
Answer:
(c) is being repaired

Question 6.
Begin the work tomorrow. / Let the work __________ tomorrow.
(a) be begun
(b) begin
(c) is beginning
Answer:
(a) be begun

Question 7.
They speak English in New Zealand. / English __________ in New Zealand.
(a) is speaking
(b) is spoken
(c) is being spoken
Answer:
(b) is spoken

Question 8.
His attitude shocked me. / I __________ by his attitude.
(a) had shocked
(b) had been shocked
(c) was shocked
Answer:
(c) was shocked

Question 9.
She had already sent the parcel. / The parcel __________ by her.
(a) has already been sent
(b) had already been sent
(c) was already sent
Answer:
(b) had already been sent

Question 10.
Her silence worries me / I __________ her silence.
(a) am worrying by
(b) am worried by
(c) have worried by
Answer:
(b) am worried by

C. Match the following Active voice sentences with Passive voice.

Answers:

D. Change the following into passive voice.

Question 1.
Stanley will inform you later.
Answer:
You will be informed by Stanley later.

Question 2.
People speak Portuguese in Brazil.
Answer:
Portuguese is spoken by people in Brazil.

Question 3.
My grandfather built this house in 1943.
Answer:
This house was built by my grandfather in 1943.

Question 4.
Do not hurt the animals.
Answer:
You are warned not to hurt the animals.

Question 5.
You must not drop litter in the streets.
Answer:
You are warned not to drop litter in the streets.

Question 6.
Carry it home.
Answer:
Let it be carried to home.

Question 7.
They are decorating the wall.
Answer:
The wall is being decorated by them.

Question 8.
He has already mended the TV set.
Answer:
The TV set has already been mended by him.

E. Make a scrapbook of’Famous Biographies’ by collecting at least five biographies of famous scientists, mathematicians, inventors, artists etc., of your choice. You may also collect the pictures related to their achievements, inventions etc.
Answer :
Sir Isaac Newton (Scientist):
Sir Isaac Newton was born on Christmas day, 1642. He was an English physicist and mathematician, who was the culminating figure of the scientific revolution of the 17th century. Born in the hamlet of Woolsthorpe, Newton was the only son of a local yeoman. Newton would eventually pick up his idea of a mathematical science of motion and bring his work to full fruition.

A tiny and weak baby, Newton was not expected to survive his first day of life, much less 84 years. Deprived of a father before birth, he soon lost his mother as well, for within two years she married a second time. He was left with his grandmother and moved to a neighbouring village.

For nine years, Isaac was effectively separated from his mother, and his pronounced psychotic tendencies have been ascribed to this traumatic event. Like thousands of other undergraduates, Newton began his higher education by immersing himself in Aristotle’s work. Even though the new philosophy was not in the curriculum, it was in the air.

Sometime during his undergraduate career, Newton discovered the works of the French natural philosopher Rene Descartes and the other mechanical philosophers. Newton had also begun his mathematical studies. Within little more than a year, he had mastered the literature; and, pursuing his own line of analysis, he began to move into new territory.

Despite the. fact that only a handful of savants were even aware of Newton’s existence, he had arrived at the point where he had become the leading mathematician in Europe.

Leonardo da Vinci (Artist):
Leonardo da Vinci was a leading artist and intellectual of the Italian Renaissance who’s known for his enduring works ‘ The Last Supper’ and the ‘Mona Lisa’. Leonardo da Vinci was born on April 15, 1452, in a farmhouse nestled amid the undulating hills of Tuscany outside the village of Anchiano, in present-day Italy. Born out of wedlock to respected Florentine notary Serpiero and a young peasant woman named Caterina.

Leonardo da Vinci was raised by his father and his stepmother. At the age of five, he moved to his father’s family estate in nearby Vinci, the Tuscan town from which the surname associated with Leonardo derives, and lived with his uncle and grandparents. Young Leonardo received little formal education beyond basic reading, writing and mathematics instruction, but his artistic talents were evident from an early age.

Around the age of 14, da Vinci began a lengthy apprenticeship with the noted artist Andrea del Verrocchio in Florence. He learned a wide breadth of technical skills including metalworking, leather arts, carpentry, drawing, painting and sculpting. His earliest known dated work, a pen-and-ink drawing of a landscape – the Amo Valley – was sketched in 1473. With a curious mind and keen intellect, da Vinci studied the laws of science and nature, which greatly improved his work. His ideas and body of work have influenced countless artists and made da Vinci a leading light of the Italian Renaissance.

Pythagoras (Mathematician):
Pythagoras is often known as the first pure mathematician. born on the island of Samos, Greece in 569 BC, his father, Mnesarchus, was a gem merchant. His mother’s name was Pythais and Pythagoras lived with his two or three brothers. Pythagoras was well educated, and played the lyre. He knew poetry and recited Homer. He was interested in mathematics, philosophy, astronomy and music.
Pythagoras believed:

1. The sum of the angles of a triangle is equal to two right angles.
2. The theorem of Pythagoras – for a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides.
3. Constructing figures of a given area and geometrical algebra. For example they solved various equations by geometrical means.
4. Pythagoras taught that Earth was a sphere in the center of the Universe, that the planets, stars, and the universe were spherical because the sphere was the most perfect solid figure.
5. Pythagoras recognized that the morning star was the same as the evening star, Venus.

The Pythagorean Theorem is a cornerstone of mathematics, and continues to be so interesting to mathematicians that there are more than 400 different proofs of the theorem, including an original proof by President Garfield.

Michael Joseph Jackson (Singer and Dancer):

Michael Joseph Jackson was an American singer, songwriter and dancer. Dubbed the “King of Pop”, he is regarded as one of the most significant cultural icons of the 20th century and is also regarded as one of the greatest entertainers of all time. Jackson’s contributions to music, dance, and fashion, along with his publicized personal life, made him a global figure in popular culture for over four decades.

The eighth child of the Jackson family, Michael made his professional debut in 1964 with his elder brothers Jackie, Tito, Jermaine and Marlon as a member of the Jackson 5. He began his solo career in 1971 while at Motown Records.

In the early 1980s, Jackson became a dominant figure in popular music. His music videos, including “Beat It”, “Billie Jean”, and “Thriller” from his 1982 album Thriller, are all time favourites. Through stage and video performances, Jackson popularized a number of complicated dance techniques, such as the robot and the moonwalk, to which he gave the name.

His distinctive sound and style has influenced numerous artists of various genres. While preparing for his comeback concert series, This Is It, Jackson died of acute propofol and benzodiazepine intoxication in 2009, after suffering from cardiac arrest.

Kapil Dev (Cricketer):

Dev was born as Kapil Dev Nikhanj to Ram Lai Nikhanj, a prominent timber merchant and his wife Raj Kumari Ram Lai Nikhanj in Chandigarh on 6 January 1959. His mother was born in Pakpattan, in the town of the Sufi Saint Baba Farid. His father was from Dipalpur. They lived in Shah Yakka which is now in Okara district, Pakistan. His four sisters were born there before partition and his two brothers in Fazilka, where they moved after partition. His father spent his early life after the partition in Fazilka. They moved to the capital city Chandigarh.

Dev was a student at D.A.V. School and joined Desh Prem Azad in 1971. Dev captained the Indian cricket team that won the 1983 Cricket World Cup. He was India’s national cricket coach between October 1999 and August 2000. He retired in 1994, holding the world record for the most number of wickets taken in Test cricket, a record subsequently broken by Courtney Walsh in 2000.

He is the first player to take 200 ODI wickets. He is the only player in the history of cricket to have taken more than 400 wickets (434 wickets) and scored more than 5000 runs in Tests, making him one of the greatest all-rounders to have played the game. On 11 March 2010, Dev was inducted into the ICC Cricket Hall of Fame.

From Zero to Infinity Lesson Summary By Biography Of Srinivasa Ramanujan

Ramanujan was born in Erode in Tamil Nadu on December 22, 1887. His father was a petty clerk in a cloth shop. From early childhood it was evident that he was a prodigy. Mathematical ideas flooded in his mind for which he did not find enough papers to note it down. Hence he started writing them in loose sheets which was later known as Ramanujan’s Frayed Notebook. At the age of 13, he was not only able to master the Loney’s Trigonometiy, but also started his own research and came up with many mathematical theorems and formulae.

At the age of 15, the book Synopsis of Elementary Results in Pure Applied Mathematics by George Shoobridge Carr given by his senior friends triggered the mathematical genius in him. Though Ramanujan was a mathematical genius and was awarded the Subramanyan Scholarship, he failed twice in his first-year arts examination in college, which disappointed his father.

Then Ramanujan started looking for a job as he needed money not only for food but also for papers to do his calculations. His applications were rejected as no one could understand what he had scribbled in his notebooks. Luckily, Director of Madras Port Trust, Francis Spring, understood his capability and he gave Ramanujan a clerical job on a monthly salary of ?25. Later on, some teachers and educationists helped him to get a research fellowship. Thus, on 1st May 1913, the University of Madras granted him a fellowship of ?75 a month, though he had no qualifying degree.

Meanwhile, he had sent a letter to the great mathematician G.H. Hardy of Cambridge University, in which he set out 120 theorems and formulae. Among them was what is known as the Reimann Series, a topic in the definite integral of Calculus. These letters made G.H. Hardy and his colleague J.E. Littlewood realize that they bad discovered a rare mathematical genius. They made quick arrangements for his passage and stay at Cambridge University. Ramanujan sailed to Britain on March 17, 1914.

Even though he found it difficult to adapt to the new environment, he continued his research in Mathematics with determination. He forgot his hardships in the company of Hardy and Littlewood. Ramanujan was elected Fellow of the Royal Society on February 28, 1918. He was the youngest Indian to receive this distinguished fellowship. In October that year, he became the first Indian to be elected Fellow of Trinity College, Cambridge.

Ramanujan continued his works even though Tuberculosis, then an incurable disease, was devouring him. When his friends found him pale, exhausted and emaciated, they sent him back to India. He continued to play with numbers until his death. Apart from a mathematician, Ramanujan was an astrologer of repute and a good speaker.

From Zero to Infinity Glossary:

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 Civics Solutions Chapter 2 Election, Political Parties and Pressure Groups

Election, Political Parties and Pressure Groups Textual Exercise

I. Choose the correct answer.

Election Political Parties And Pressure Groups Question 1.
India has adapted the electoral system followed in the
(a) USA
(b) United Kingdom
(c) Canada
(d) Russia
Answer:
(b) United Kingdom

Election Political Parties And Pressure Groups Questions And Answers Question 2.
The Election Commission of India is a/an
(a) Independent body
(b) Statutory body
(c) Private body
(d) Public corporation
Answer:
(a) Independent body

Class 9 Civics Chapter 2 Question 3.
Which Article of the Constitution provides for an Election Commission?
(a) Article 280
(b) Article 315
(c) Article 324
(d) Article 325
Answer:
(c) Article 324

Political Parties And Pressure Groups Question 4.
Which part of the constitution of India says about the election commission?
(a) Part III
(b) Part XV
(c) Part XX
(d) Part XXII
Answer:
(b) Part XV

Chapter 2 Civics Class 9 Question 5.
Who accords recognition to various political parties as national or regional parties?
(a) The President
(b) The Election Commission
(c) The Parliament
(d) The President in consultation with the Election Commission
Answer:
(b) The Election Commission

Question 6.
Assertion (A): Indian Constitution provides for an independent Election Commission
Reason (R): To ensure free and fair elections in the country.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 7.
NOT A was introduced in the year ………..
(a) 2012
(b) 2013
(c) 2014
(d) 2015
Answer:
(c) 2014

Question 8.
The term pressure groups originated in …….
(a) USA
(b) UK
(c) USSR
(d) India
Answer:
(a) USA

Question 9.
Assertion (A): A large number of pressure groups exist in India.
Reason (R): Pressure Groups are not developed in India to the same extent as in the USA.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

II. Fill in the blanks.

1. The Election Commission of India is a body of ……… members.
2. National Voters day has been celebrated on ………
3. In India ……… party system is followed.
4. In 2017, there were ……. recognised national parties.
5. Narmada Bachao Andolan is a …………
Answers:
1. 3
2. 25^th January
3. Multi
4. Seven
5. Pressure group
III. Match the following.

Answers:
1. (d)
2. (c)
3. (b)
4. (a)

IV. Give short answers.

Question 1.
Explain the electoral system in India.
Answer:

1. The electoral system in India has been adapted from the system followed in the United Kingdom.
2. India is a socialist, secular, democratic, republic and the largest democracy in the world.
3. The modem Indian nation state came into existence on 15th August 1947.

Question 2.
Give the meaning of a political party.
Answer:
A political party is an organisation formed by a group of people with a certain ideology and agenda to contest elections and hold power in the government.

A political party has three components: a leader, active members and the followers.

Question 3.
Distinguish between two-party system and the multi-party system.
Answer:

Two party system	Multi-party system
Two party system in which only two major parties exist, for example, USA, UK.	Multi-party system in which there are more than two political parties, for example India, Srilanka, France and Italy.

Question 4.
What is a pressure group?
Answer:

1. A pressure group is a group of people who are organised actively for promoting and defending their common interest. It is so called as it attempts to bring a change in the public policy by exerting pressure on the government.
2. The pressure groups are also called ‘interest groups’ or vested groups.
3. They are different from the political parties in that they neither contest elections nor try to capture political power.

V. Answer in detail.

Question 1.
Discuss merits and demerits of direct elections?
Answer:
Merits:

1. As the voters elect their representatives directly, direct elections are considered to be a more democratic method of election.
2. It educates people regarding the government activities and helps in choosing the appropriate candidates. Also, it encourages people to play an active role in politics.
3. It empowers people and makes the rulers accountable for their actions.

Demerits:

1. Direct elections are very expensive.
2. Illiterate voters sometimes get misguided by false propaganda and sometimes campaigning based on caste, religious and various other sectarian consideration spose serious challenges.
3. Since conducting direct elections is a massive exercise, ensuring free and fair elections at every polling station is a major challenge to the Election Commission.
4. There are instances of some political candidates influencing the voters through payments in the form of cash, goods or services.
5. Election campaigns sometimes results in violence, tension, law and order problems and
   affects the day-to-day life of people.

Question 2.
What are the functions of political parties?
Answer:

1. Parties contest elections. In most democracies, elections are fought mainly among the candidates put up by political parties.
2. Parties put forward their policies and programmes before the electorate to consider and choose.
3. Parties play a decisive role in making laws for a country. Formally, laws are debated and passed in the legislature.
4. Parties form and run the governments.
5. Those parties that lose in the elections play the role of the Opposition to the party or a group of coalition parties in power, by voicing different views and criticising the government for its failures or wrong policies.
6. Parties shape public opinion. They raise and highlight issues of importance.
7. Parties function as the useful link between people and the government machinery.

Question 3.
What are the functions of Pressure groups in India?
Answer:
Pressure groups are the interest groups that work to secure certain interest by influencing the public policy. They are non-aligned with any political party and work as an indirect yet powerful group to influence the policy decisions. Pressure groups carry out a range of functions including representation, political participation, education, policy formulation and policy implementation.

Political Participation: Pressure groups can be called the informal face of politics. They exert influence precisely by mobilising popular support through activities such as petitions, marches, demonstrations and other forms of political protest. Such forms of political participation have been particularly attractive to young people.

Education: Many pressure groups devote significant resources by carrying out research, maintaining websites, commenting on government policy and using high-profile academics, scientists and even celebrities to get their views across, with an emphasis to cultivate expert authority.

Policy Formulation: Though the pressure groups themselves are not policy-makers, yet it does not prevent many of them from participating in the policy-making process. Many pressure groups are vital sources of information and render advice to the government and therefore they are regularly consulted in the process of policy formulation.

VI. Project and Activity

Question 1.
Compare the policies, programmes and achievements of a national party and a state party.
Answer:

1. Refer the National policies, programmes and achievements from the Internet and library books.
2. The students are instructed to compare the policies programmes and achievements.
3. This is a group Activity.

VII. HOTS

Question 1.
“Elections are considered essential for any representative democracy”. Why?
Answer:
“A democracy requires a mechanism by which people can choose their representatives at regular intervals and change them if they wish to do so. Therefore, elections are considered essential for any representative democracy. In an election the voters make many choices.

1. This helps the public in choosing the development course which they want and want to adopt.
2. This directly provide the public the opportunity to select their representatives and these representatives decision can be legitimacy.
3. This also ensures the transparency as well as the accountability as the public representatives are chosen directly.

Question 2.
What is the principle of universal adult franchise? What is its importance?
Answer:
Principle: Universal Adult Franchise means that the right to vote should be given to all adult citizens without the discrimination of caste, class, colour, religion (or) gender. It is based on equality, which is a basic principle of democracy.

Importance: Under this system a government is elected that is accountable to the people it governs. Because every vote counts, issues in a society receive their appropriate weight in terms of importance and urgency.

Question 3.
Discuss merits and demerits of democracy.
Answer:

	Merits		Demerits
1.	Safeguards the interests of the people.	1.	More emphasis on quantity than on quality.
2.	Based on the principle of equality.	2.	Rule of the incompetent.
3.	Stability and responsibility in administration.	3.	Based on unnatural equality.                •
4.	Political education to the people.	4.	Voters do not take interest in election.
5.	Little chance of revolution.	5.	Lowers the moral standard.
6.	Stable government.	6.	Democracy is a government of the rich.
7.	Helps in making people good citizens.	7.	Misuse of public funds and time.
8.	Based on public opinion.	8.	No stable government.
		9.	Dictatorship of majority.
		10.	Bad influence of political parties.

Question 4.
Discuss the multi-party system.
Answer:

1. A multi-party system is a system where multiple political parties that have ideas participate in the national elections.
2. A lot of countries that use this system have a coalition government, meaning many parties are in control, and they all work together to make laws.
3. Countries with a multi-party political system tend to have greater voter participation.
4. No democracy can survive without multi-party system.

VIII. Life Skill

Question 1.
Conduct a mock poll in your classroom.
Answer:

1. Help the students to understand the process of electing officials and the power of vote by holding a mock election.
2. These are great activities to enjoy during the presidential election.
3. Students explain the steps taken from party formation to National election.
4. The students will act out the campaigning and voting process by stimulating a real election in their own class room.

Steps set for a mock-poll in the class room:

1. Setting up the political parties.
2. Preparing a manifesto.
3. Running a campaign.
4. Holding the class room election.

Election, Political Parties and Pressure Groups Additional Questions

I. Choose the correct answer.

Question 1.
India is the democracy in the world.
(a) largest
(b) smallest
(c) strongest
(d) None of the above
Answer:
(a) largest

Question 2.
Kudavolai was the system of voting followed during the ……. period in Tamil Nadu.
(a) Chera
(b) Chola
(c) Pandya
(d) Pallava
Answer:
(b) Chola

Question 3.
Which country has single party system?
(a) USA
(b) UK
(c) Cuba
(d) France
Answer:
(c) Cuba

Question 4.
Assertion (A): Parties shape public opinion.
Reason (R): They raise and highlight issues of importance.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A).
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 5.
India is the ………… th country in the world to introduce NOTA.
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 6.
The ……. is elected by members of the Lok Sabha.
(a) Prime Minister
(b) President
(c) Governor
(d) Cabinet Minister
Answer:
(a) Prime Minister

II. Fill in the blanks.
1. ……. in elections are the best way to make your ‘voice’ heard.
2. Indirect elections are less ……..
3. ……… parties are an essential part of Democracy.
4. …… treats all the parties equally.
5. The pressure groups are also called ……. groups.
6. A political party has three components: a ………. and the ……….
Answers:
1. Voting
2. expensive
3. Political
4. Election Commission
5. Interest
6. a leader, acting members, followers

III. Match the following.

Answers:
1. (d)
2. (a)
3. (b)
4. (c)

IV. Give short answers.

Question 1.
What do you know about Voters Verified Paper Audit Trail?
Answer:

1. Voters Verified Paper Audit Trail (WPAT) is the way forward to enhance credibility and transparency of the election process.
2. This system was first introduced in 2014 General election.

Question 2.
Mention the merits and demerits of Indirect elections.
Answer:
Merits:

1. Indirect elections are less expensive.
2. It is more suited to elections in large countries.

Demerits:

1. If the number of voters is very small, there exists the possibility of corruption, bribery, horse trading and other unfair activities.
2. It is less democratic because people do not have a direct opportunity to elect, but they instead do it through their representatives. So, this may not reflect the true will of the people.

Question 3.
Write a short note on “State Parties”.
Answer:
Other than the seven national parties, most of the major parties of the country are classified by the Election Commission as ‘state parties’. These are commonly referred to as regional parties. A party is recognised as a state party by the Election Commission of India based on certain percentage of votes secured or a certain number of seats won in the Assembly or Lok Sabha elections.

Question 4.
Classify pressure groups in India.
Answer:
The pressure groups in India can be broadly classified into the following categories:

1. Business groups
2. Trade unions
3. Agrarian groups
4. Professional associations
5. Student organisations
6. Religious organisations
7. Tribal organisations
8. Linguistic groups
9. Ideology-based groups
10. Environmental protection groups.

V. Answer in detail.

Question 1.
Give an account of Mobilisation and Democratic Participation.
Answer:
Mobilising people towards socially productive activities that lead to the overall betterment of people’s lives is essential. Sometimes earthquakes, tsunamis, floods and other such natural disasters on a massive scale occur and people’s immediate mobilisation for evacuation and emergency relief becomes most essential.

Democratic Participation: Democracy can-succeed only when smaller local groups and, in fact, every citizen can take action that supports the tax and revenue collection systems, observance of national norms in environmental protection, cleanliness, health and hygiene, sanitary drives and immunisation programmes like pulse polio.

However, we must keep in mind that there is no better form of government than Democratic government. To create a better society and nation, the people of India along with the union and state governments should come together to fight against the miseries of human life.

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 Ex 1.2

9th Maths Exercise 1.2 Samacheer Kalvi Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {v : v = \(\frac { 4 }{ 3n }\) ,n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integers, x ∈ Z and -5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n(M) = 6
(ii) W = {0, 1, 2, 3, ……. }
if n = 0, x = 3(0) + 2 = 2
if n = 1, x = 3(1) + 2 = 5
if n = 2, x = 3(2) + 2 = 8
if n = 3, x = 3(3)+ 2 =11
if n = 4, x = 3(4) + 2=14
∴ P= {2, 5, 8, 11, 14}
n(P) = 5

(iii) N = {1,2, 3, 4, …..}
n ∈ {3, 4, 5}

n(Q) = 3

(iv) x ∈ z
R = {-5, – 4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R)= 10.

(v) S = {1884, 1888, 1892, 1896, 1904}
n (S) = 5.

9th Maths Set Language Exercise 1.2 Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite set
(ii) Infinite set
(iii) A = { ……. , -2, -1, 0, 1, 2, 3, 4}
∴ Infinite set

(iv) x² – 5x + 6 = 0
(x – 3) (x – 2) = 0
B = {3, 2}
∴ Finite set.

9th Maths Exercise 1.2 Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = A = { x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) A = {a, e, i, o, u}
B = {V, O,W, E, L}
The sets A and B contain the same number of elements.
∴ Equivalent sets

(ii) C ={2, 3, 4, 5}
D = {2, 3, 4}
∴ Unequal sets

(iii) X = {L, I, F, E}
Y = {F, I, L, E}
The sets X and Y contain the exactly the same elements.
∴ Equal sets.

(iv) G = {5, 7, 11, 13, 17, 19}
H = {1, 2, 3, 6, 9, 18}
∴ Equivalent sets.

9th Maths Set Language Exercise 1.2 Solutions Question 4.
Identify the following sets as null set or singleton set.
(i) A = (x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2.
(iii) C = {0}
(iv) D = The set of all triangles having four sides.
Solution:
(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.
∴ Null set

(ii) B = { } ∵ All even natural numbers are divisible by 2.
∴ B is Null set

(iii) C = {0}
∴ Singleton set

(iv) D = { }
∵ No triangle has four sides.
∴ D is a Null set.

9th Maths Exercise 1.2 In Tamil Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s)
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x: x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n,f h, s} = {f, a, s}
Since A ∩ B ≠ ϕ , A and B are overlapping sets.

(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = Ø, C and D are disjoint sets.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ ϕ, E and F are overlapping sets.

9th Standard Maths Exercise 1.2 In Tamil Question 6.
If S = {square,rectangle,circle,rhombus,triangle}, list the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°
(iv) The set of shapes which have 5 sides.
Solution:
(i) {Square, Rhombus}
(ii) {Circle}
(iii) {Triangle}
(iv) Null set.

9th Standard Maths Exercise 1.2 Question 7.
If A = {a, {a, b}}, write all the subsets of A.
Solution:
A= {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.

9th Std Maths Exercise 1.2 Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = Ø
Solution:
(i) The subsets of A are Ø, {a}, {b}, {a, b}
The power set of A
P(A ) = {Ø, {a}, {b}, {a,b}}

(ii) The subsets of B are ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
The power set of B
P(B) = {Ø, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

(iii) The subset of D are Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s},{p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}}
The power set of D
P(D) = {Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}

(iv) The power set of E
P(E) = { }.

9th Maths 1.2 In Tamil Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red,blue, yellow}
(ii) X = { x² : x ∈ N, x² ≤ 100}.
Solution:
(i) Given W = {red, blue, yellow}
Then n(W) = 3
The number of subsets = n[P(W)] = 2³ = 8
The number of proper subsets = n[P(W)] – 1 = 2³ – 1 = 8 – 1 = 7

(ii) Given X ={1,2,3, }
X² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(X) = 10
The Number of subsets = n[P(X)] = 2¹⁰ = 1024
The Number of proper subsets = n[P(X)] – 1 = 2¹⁰ – 1 = 1024 – 1 = 1023.

9th Maths 1.2 Exercise Question 10.
(i) If n(A) = 4, find n[P(A)].
(ii) If n(A) = 0, find n[P(A)].
(iii) If n[P(A)] = 256, find n(A).
Solution:
(i) n( A) = 4
n[ P(A)] = 2n = 2⁴ = 16
(ii) n(A) = 0
n[P(A)] = 2⁰ = 1
(iii) n[P(A)] = 256

n[P(A)] = 28
∴ n(A) = 8.

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 6 Light

Samacheer Kalvi 9th Science Light Textbook Exercises

I. Choose the correct answer.

Light Samacheer Kalvi Question 1.
A ray of light passes from one medium to another medium. Refraction takes place when angle of incidence is …………..
(a) 0°
(b) 45°
(c) 90°
Answer:
(b) 45°

Light Book Back Answers Question 2.
…………. is used as reflectors in torchlight.
(a) Concave mirror
(b) Plane mirror
(c) Convex mirror
Answer:
(a) Concave mirror

Light 9th Class Question 3.
We can create enlarged, virtual images with ………………
(a) Concave mirror
(b) Plane mirror
(c) Convex mirror
Answer:
(a) Concave mirror

9th Science Light Question 4.
When the reflecting surface is curved outwards the mirror formed will be
(a) concave mirror
(b) convex mirror
(c) plane mirror
Answer:
(b) convex mirror

Chapter 6 Light Question 5.
When a beam of white light passes through a prism it gets
(a) reflected
(b) only deviated
(c) deviated and dispersed
Answer:
(a) Reflected

Unit 6 Light Question 6.
The speed of light is maximum in
(a) vacuum
(b) glass
(c) diamond
Answer:
(a) vacuum

II. True or False – If false give the correct answer.

1. The angle of deviation depends on the refractive index of the glass – True.
2. If a ray of light passes obliquely from one medium to another, it does not suffer any deviation – False.
   Correct Statement: When light travels from one medium to another, it suffers deviation.
3. The convex mirror always produces a virtual, diminished and erect image of the object – True.
4. When an object is at the centre of curvature of concave mirror the image formed will be virtual and erect – False.
   Correct Statement: The image formed is real, inverted and same size of the object.
5. The reason for brilliance of diamonds is total internal reflection of light – True.

III. Fill in the blanks.

1.  In going from a rarer to denser medium, the ray of light bends …………….
2. The mirror used in search light is ………………
3. The angle of deviation of light ray in a prism depends on the angle of ……………..
4. The radius of curvature of a concave mirror whose focal length is 5 cm is …………….
5. Large ………… mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Answer:

1. towards normal
2. concave mirror
3. prism and angle of incident
4. 10 cm
5. concave

IV. Match the following.

1. Ratio of height of image to height of object	(a) Concave Mirror
2. Used in hairpin bends in mountains	(b) Total Internal Reflection
3. Coin inside water appearing slightly raised	(c) Magnification
4. Mirage	(d) convex Mirror
5. Used as Dentist’s mirror	(e) Refraction

Answer:

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

V. Assertion & Reason.

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

Light Lesson Class 9 Question 1.
Assertion: For observing the traffic at a hairpin bend in mountain paths a plane mirror is
preferred over convex mirror and concave mirror.
Reason : A convex mirror has a much larger field of view than a plane mirror or a concave mirror.
Answer:
(d) If assertion is false but reason is true.

9th Science Reflection Of Light Exercise Question 2.
Assertion : Incident ray is directed towards the centre of curvature of spherical mirror.
After reflection it retraces its path.
Reason : Angle of incidence i = Angle of reflection r = 0°.
Answer:
(b) If both assertion and reason are true and reason is not the correct explanation.

VI. Answer very briefly.

9th Science Reflection Of Light Question 1.
According to cartesian sign convention, which mirror and which lens has negative focal length?
Answer:
Concave mirror is having a negative focal length.

Light Chapter Of Class 6 Pdf Question 2.
Name the mirror(s) that can give (i) an erect and enlarged image, (II) same sized, inverted image.
Answer:
Concave mirror

Science Solution Class 9 Samacheer Kalvi Question 3.
If an object is placed at the focus of a concave mirror, where is the image formed? Image will be formed at infinity as real and inverted.

Light Class 9 Question 4.
Why does a ray of light bend when it travels from one medium to another?
Answer:
The bending of light rays when they pass obliquely from one medium to another medium is called refraction of light.

Light rays get deviated from their original path while entering from one transparent medium to another medium of different optical density. This deviation (change in direction) in the path of light is due to the change in velocity of light in the different medium. The velocity of light depends on the nature of the medium in which it travels.

9th Science Samacheer Kalvi Question 5.
What is speed of light in vacuum?
Answer:
The speed of light in vacuum is known to be almost exactly 300,000 km per second. In 1665 the Danish astronomer Ole Roemer first estimated the speed of light by observing one of the twelve moons of the planet Jupiter.

Samacheerkalvi.Guru 9th Science Question 6.
Concave mirrors are used by dentists to examine teeth. Why?
Answer:
As a dentist’s head mirror: You would have seen a circular mirror attached to a band tied to the forehead of the dentist/ENT specialist. A parallel beam of light is made to fall on the concave mirror; this mirror focuses the light beam on a small area of the body (such as teeth, throat etc.).

VII. Answer briefly.

9th Science Chapter 11 Reflection Of Light Exercise Question 1.
(a) Complete the diagram to show how a concave mirror forms the image of the object.
(b) What is the nature of the image?

Solution:
(a)
(b) magnified, real and inverted.

Questions On Reflection Of Light Class 9 Question 2.
Pick out the concave and convex mirrors from the following and tabulate them
Answer:
Rear-view mirror, Dentist’s mirror, Torch-light mirror, Mirrors in shopping malls, Make-up mirror.

Concave mirror	Convex mirror
Dentist’s mirror	Rear view mirror
Torch light mirror	Mirrors in shopping malls
Make-up mirror

Question 3.
State the direction of incident ray which after reflection from a spherical mirror retraces its path. Give reason for your answer.
Answer:
When incident ray is directed towards the centre of curvature, at all the points of spherical mirror, the ray is always normal. Therefore, angle of incidence i = Angle of reflection r = 0°.

Question 4.
What is meant by magnification? Write its expression. What is its sign for real image and virtual image?
Answer:
Magnification produced by a spherical mirror gives how many times the image of an object is magnified with respect to the object size.
It can be defined as the ratio of the height of the image (h_i) to the height of the object (h_o).
Magnification = m = \(\frac{h_{i}}{h_{o}}=\frac{\text { height of the image }}{\text { height of the object }}\)

• for real image it is negative,
• for virtual image it is positive.

Question 5.
Write the spherical mirror formula and explain the meaning of each symbol used in it.
Answer:
The expression relating the distance of the object u, distance of image v and focal length/of a spherical mirror is called the mirror equation. It is given as:
Mirror formula: \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Here, f – focal length of spherical mirror; u – distance of the objective; v – distance of the image.

VIII. Answer in detail.

Question 1.
(a) Draw ray diagrams to show how the image is formed, using a concave mirror when the position of object is

1.  at C
2.  between C and F
3.  between F and P of the mirror.

Answer:
(1) At the centre of curvature C

(2) Between C and F

(3) Between the focus F and the Pole P of the mirror.

(b) Mention the position and nature of image in each case.
Answer:

Position of object	Position of Image	Nature of Image
At the centre of curvature C	AtC	Real and Inverted
Between C and F	Beyond C	Real and inverted
Between F and P	Behind the mirror	Virtual and Erect

Question 2.
Explain with diagrams how refraction of incident light takes place from
(a) rarer to denser medium
(b) denser to rarer medium
(c) normal to the surface separating the two media.
Answer:
(a) rarer to denser medium:
When a ray of light travels from optically rarer medium to optically denser medium, it bends towards the normal.

(b) denser to rarer medium:
When a ray of light travels from an optically denser medium to an optically rarer medium it bends away from the normal.

(c) normal to the surface separating the two media:
A ray of light incident normally on a denser medium, goes without any deviation.

IX. Numerical problems.

Question 1.
A concave mirror produces three times magnified real image of an object placed at 7 cm in front of it. Where is the image located? (21 cm in front of the mirror)
Answer:
Given: h_i = 3h_o
u = 7 cm
Solution:

1. m = \(\frac{h_{i}}{h_{O}} \Rightarrow \frac{3 h o}{h_{O}}\) ∴ m = 3.
2. m = – \(\frac{v}{u}\)
   v = – m × u
   = – 3 × 7 cm = – 21 cm.
   Hence, Real, inverted and magnified image will be formed at 21 cm in front of the mirror.

Question 2.
Light enters from air into a glass plate having refractive index 1.5. What is the speed of light in glass? (2 × 10⁸ ms^{– 1})
Given: Refractive index (µ) = 1.5
Speed of light in vacuum (c) = 2 × 10⁸ms^{– 1}
Speed of light in glass (υ) = ?
Solution:

∴ Speed of light in glass = 1.3 × 10⁸ ms< sup>- 1

Question 3.
The speed of light in water is 2.25 × 10⁸ms^{– 1}. If the speed of light in vacuum is 3 × 10⁸ms^{– 1}, calculate the refractive index of water.
Given:
Speed of light in water (υ) = 2.25 × 10⁸ms^{– 1}
Speed of light in vacuum (c) = 3 × 10⁸ms^{– 1}
Solution:
μ = \(\frac{c}{υ}\) ⇒ μ = \(\frac{3 \times 10^{8} \mathrm{ms}^{-1}}{2.25 \times 10^{8} \mathrm{ms}^{-1}}\)
∴ μ = 1.33

X. HOTS.

Question 1.
Light ray emerges from water into air. Draw a ray diagram indicating the change in its path in water.
Answer:

When a ray of light travels from dense medium to rarer medium [from water medium to air medium], light ray moves away from the normal.
∴ Angle of incidence < Angle of refraction.

Question 2.
When a ray of light passes from air into glass, is the angle of refraction greater than or less than the angle of incidence?
Answer:

When a ray of light travels from rare (air) medium to dense [glass] medium, light ray moves towards the normal.
∴ Angle of refraction < Angle of incidence.

Question 3.
What do you conclude about the speed of light in diamond if the refractive index of diamond is 2.41? way from the mirror. Does the image become smaller or larger? What do you observe?
Answer:
µ = 2.41; Ca = 3 × 10⁸ms [Velocity of light]
µ = \(\frac{\text { Speed of light in air }}{\text { Speed of light in diamond }}=\frac{\mathrm{C}_{\mathrm{a}}}{\mathrm{C}_{\mathrm{d}}}\)
C_d = \(\frac{C_{a}}{\mu}=\frac{3 \times 10^{8}}{2.41}\) = 1. 245 × 10⁸ m/s
∴ Speed of light decreases when the light ray travels from air to diamond.

ACTIVITY

Question 1.
Stand before the mirror in your dressing table or the mirror fixed in a steel almirah. Do you see your whole body?
Answer:
To see your entire body in a mirror, the mirror should be atleast half of your height.
Height of the mirror = Your height/2.

Question 2.
Hold a concave mirror in your hand (or place it in a stand). Direct its reflecting surface towards the sun. Direct the light reflected by the mirror onto a sheet of paper held not very far from the mirror. Move the sheet of paper back and forth gradually until you find a bright, sharp spot of light on the paper. Position the mirror and the paper at the same location for few moments. What do you observe? Why does the paper catches fire?
Answer:
The concave mirror focus the sunlight at one particular point. At that particular point, amount of sunlight is converged and heat is trapped at that point. This is because, it bum the paper on which the image is focused.

Question 3.
Take a convex mirror. Hold it in one hand. Hold a pencil close to the mirror in the upright position in the other hand. Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged? Move the pencil slowly away from the mirror. Does the image become smaller or larger? What do you observe?
Answer:
When we hold a pencil in the upright position in front of a convex mirror we observe the image of the pencil at the back of the mirror. The image is erect, virtual and smaller in size than the object. As the pencil is moved away from the mirror, the image becomes smaller. When the object is moved away from the mirror, the image would move closer to the focus of the mirror.

Question 4.
Refraction of light at air – water interface
Put a straight pencil into a tank of water or beaker of water at an angle c 45° and look at it from one side and above. How does the pencil look now?
Answer:

The pencil appears to be bent at the surface of water.

Samacheer Kalvi 9th Science Light In Text Problems

Question 1.
Find the size, nature and position of the image formed when an object of size 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm.
Solution:
Object distance, u = – 15 cm (to the left of mirror)
Image distance, v = ?
Focal length,f = – 10 cm (concave mirror)
Using mirror formula,

Thus, image distance, v = – 30 cm (negative sign indicates that the image is on the left side of the mirror).
∴ Position of image is 30 cm in front of the mirror. Since the image is in front of the mirror, it is real and inverted.
To find the size of the image, we have to calculate the magnification.
m = \(\frac{-v}{u}=\frac{-(-30)}{(-15)}\) = – 2
We know that, m = \(\frac{h_{2}}{h_{1}}\)
Here, height of the object h₁ = 1 cm
– 2 = \(\frac{h_{2}}{1}p.\)
h₂ = – 2 × 1 = – 2 cm
The height of image is 2 cm (negative sign shows that the image is formed below the principal axis).

Question 2.
An object 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a real image 3 cm high. Find the position of the image.
Solution:
Height of object h₁ = 2 cm
Height of real image h₂ = – 3 cm

The position of image is 24 cm in front of the mirror (negative sign indicates that the image is on the left side of the mirror).

Question 3.
A car is fitted with a convex mirror of focal length 20 cm. Another car is 6 m away from the first car. Find the position of the second car as seen in the mirror of the first. What is the size of the image if the second car is 2 m broad and 1.6 m high?
Solution:
Focal length = 20 cm (convex mirror)
Object distance = – 6m = – 600 cm
Image distance v =?

Height of image = \(\frac{1}{31}\) × 160 cm = 5.16 cm

Question 4.
The speed of light in air is 3 × 10⁸ ms^{– 1} and in glass it is 2 × 10⁸ ms^{– 1}. What is the
refractive index of glass?
Solution:
\(_{a} \mu_{g}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}\) = 1.5

Question 5.
Light travels from a rarer medium to a denser medium. The angles of incidence and refraction are respectively 45° and 30°. Calculate the refractive index of the second medium with respect to the first medium.
Solution:
Angle of incidence i = 45° ; Angle of refraction r = 30°
\(_{1} \mu_{2}=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1 / \sqrt{2}}{1 / 2}=\sqrt{2}\) = 1.414

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 Ex 2.2

9th Maths Exercise 2.2 Samacheer Kalvi Question 1.
Express the following rational numbers into decimal and state the kind of decimal expansion.
(i) \(\frac { 2 }{ 7 }\)
(ii) \(-5 \frac{3}{11}\)
(iii) \(\frac { 22 }{ 3 }\)
(iv) \(\frac { 327 }{ 200 }\)
Solution:
(i) \(\frac { 2 }{ 7 }\)

\(\frac{2}{7}=0 . \overline{285714}\)
Nen-terminating and recurring

(ii) \(-5 \frac{3}{11}\)

\(-5 \frac{3}{11}=-5 . \overline{27}\)
Nen-terminating and recurring

(iii) \(\frac { 22 }{ 3 }\)

\(\frac{22}{3}=7 . \overline{3}\)
Nen-terminating and recurring

(iv) \(\frac { 327 }{ 200 }\)

\(\frac { 327 }{ 200 }\) = 1.635, Terminating.

9th Maths Exercise 2.2 In Tamil Question 2.
Express \(\frac { 1 }{ 13 }\) in decimal form. Find the length of the period of decimals.
Solution:

\(\frac{1}{13}=0 . \overline{076923}\) has the length of the period of decimals = 6.

9th Standard Maths Exercise 2.2 Question 3.
Express the rational number \(\frac { 1 }{ 13 }\) in recurring decimal form by using the recurring decimal expansion of \(\frac { 1 }{ 11 }\) . Hence write \(\frac { 71 }{ 33 }\) in recurring decimal form.
Solution:
The recurring decimal expansion of \(\frac { 1 }{ 11 }\) = 0.09090909…. = \(0.\overline { 09 }\)

9th Maths Exercise 2.2 Question 4.
Express the following decimal expression into rational numbers.
(i) \(0.\overline { 24 }\)
(ii) \(2.\overline { 327 }\)
(iii) -5.132
(iv) \(3.1\overline { 7 }\)
(v) \(17.\overline { 215 }\)
(vi) \(-21.213\overline { 7 }\)
Solution:
(i) \(0.\overline { 24 }\)
Let x = \(0.\overline { 24 }\) = 0.24242424……… ….(1)
(Here period of decimal is 2, multiply equation (1) by 100)
100x = 24.242424 ………. ….(2)
(2) – (1)
100x – x = 24.242424…. – 0.242424….
99x = 24
x = \(\frac { 24 }{ 99 }\)

(ii) \(2.\overline { 327 }\)
Let x = 2.327327327…… …………. (1)
(Here period of decimal is 3, multiply equation (1) by 1000)
1000x = 2327.327… ……………. (2)
(2) – (1)
1000x – x = 2327.327327… – 2.327327….
999x = 2325
x = \(\frac { 2325 }{ 999 }\)

(iii) -5.132
\(x=-5.132=\frac{-5132}{1000}=\frac{-1283}{250}\)

(iv) \(3.1\overline { 7 }\)
Let x = 3.1777 ……. ………… (1)
(Here the repeating decimal digit is 7, which is the second digit after the decimal point, multiply equation (1) by 10)
10x = 31.7777 …….. …………. (2)
(Now period of decimal is 1, multiply equation (2) by 10)
100x = 317.7777…….. …………….. (3)
(3) – (2)
100x – 10x = 317.777…. – 31.777….
90x = 286
\(x=\frac{286}{90}=\frac{143}{45}\)

(v) \(17.\overline { 215 }\)
Let x = 17.215215 ……. ………. (1)
1000x = 17215.215215…… …………. (2)
(2) – (1)
1000x – x = 17215.215215… – 17.215…
999x = 17198
x = \(\frac { 17198 }{ 999 }\)

(vi) \(-21.213\overline { 7 }\)
Let x = -21.2137777… ……….. (1)
10x = -212.137777…… ……….. (2)
100x = -2121.37777…… ………… (3)
1000x = -21213.77777…. ……….. (4)
10000x = 212137.77777….. ………… (5)
(Now period of decimal is 1, multiply equation (4) it by 10)
(5) – (4)
10000x – 1000x = (-212137.7777…) – (-21213.7777…)
9000x = -190924
x = –\(\frac { 190924 }{ 9000 }\)

9th Maths 2.2 Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expansion.
(i) \(\frac { 7 }{ 128 }\)
(ii) \(\frac { 21 }{ 15 }\)
(iii) 4\(\frac { 9 }{ 35 }\)
(iv) \(\frac { 219 }{ 2200 }\)
Solution:
(i) \(\frac { 7 }{ 128 }\)

So \(\frac{7}{128}=\frac{7}{2^{7} 5^{0}}\)
This of the form 4m, n ∈ W
So \(\frac { 7 }{ 128 }\) has a terminating decimal expansion.

(ii) \(\frac { 21 }{ 15 }\)

So \(\frac { 21 }{ 15 }\) has a terminating decimal expansion.

(iii) 4\(\frac { 9 }{ 35 }\) = \(\frac { 149 }{ 35 }\)

\(\frac{49}{35}=\frac{149}{5^{1} 7^{1}}\)
∴ This is not of the form \(\frac{p}{5^{1} 7^{1}}\)
So 4\(\frac { 9 }{ 35 }\) has a non-terminating recurring decimal expansion.

(iv) \(\frac { 219 }{ 2200 }\)

\(\frac{219}{2200}=\frac{219}{2^{3} 5^{2} 11^{1}}\)
∴ This is not of the form \(\frac{p}{2^{m} 5^{n}}\)
So \(\frac { 219 }{ 2200 }\) has a non-terminating recurring decimal expansion.

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 History Solutions Chapter 5 The Classical World

The Classical World Textual Exercise

I. Choose the correct answer.

The Classical World 9th Class Question 1.
…………….. is the Greek city-state which resisted the Persians to the end.
(a) Acropolis
(b) Sparta
(c) Athens
(d) Rome
Answer:
(c) Athens

Attempt An Account Of Slavery In Rome Question 2.
The other name for Greeks was …………….
(a) Hellenists
(b) Hellenes
(c) Phoenicians
(d) Spartans
Answer:
(b) Hellenes

Samacheer Kalvi 9th Books Social Science Question 3.
The founder of Han dynasty was …………..
(a) Wu Ti
(b) Hung Chao
(c) Liu Pang
(d) Mangu Khan
Answer:
(c) Liu Pang

Question 4.
…………… was the Roman Governor responsible for the crucifixion of Jesus.
(a) Innocent I
(b) Hildebrand
(c) Leo I
(d) Pontius Pilate
Answer:
(d) Pontius Pilate

Question 5.
The Peloponnesian War was fought between ……………… and ……………
(a) Greeks and Persians
(b) Plebeians and Patricians
(c) Spartans and Athenians
(d) Greeks and Romans
Answer:
(c) Spartans and Athenians

II. Find out the correct statement.

Question 1.
(i) First Persian attack on Greece failed.
(ii) The downfall of Roman Empire is attributed to Julius Caesar.
(iii) The Barbarians who invaded Rome were considered to be culturally advanced.
(iv) Buddhism weakened the Roman Empire.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(a) (i) is correct

Question 2.
(i) Euclid developed a model for the motion of planets and stars.
(ii) Romans established a republic after overthrowing Etruscans.
(iii) Acropolis became a famous slave market.
(iv) Rome and Carthage united to drive out the Greeks.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iv) are correct
(d) (iv) is correct
Answer:
(c) (ii) and (iv) are correct

Question 3.
(i) Silk road was closed during the Han dynasty.
(ii) Peasant uprisings posed threats to Athenian democracy.
(iii) Virgil’s Aeneid glorified Roman imperialism.
(iv) Spartacus killed Julius Caesar. f
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iv) are correct
(d) (iii) is correct
Answer:
(d) (iii) is correct

Question 4.
(i) Roman Emperor Marcus Aurelius was a tyrant.
(ii) Romulus Aurelius was the most admired ruler in Roman History.
(iii) Fabius was a famous Carthaginian General.
(iv) Tacitus is respected more than Livy as a historian.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(d) (iv) is correct

Question 5.
(i) Buddhism went to China from Japan
(ii) After crucifixion of Jesus, St Thomas spread the Christian doctrine
(iii) St Sophia Cathedral was the most magnificent building in Europe
(iv) Trajan was one of the worst dictators Rome had.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(c) (iii) is correct

III. Match the following:

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

IV. Fill in the blanks.

1. Greeks defeated the Persians at ………………
2. …………… stood in favour of poor peasants in Roman republic.
3. Buddhism came to China from India during the reign of …………… dynasty.
4. The most magnificent building in Europe was ……………
5. ……….. and …………… were Magistrates in Rome.
Answers:
1. Marathon in 490 B.C.
2. Tiberius Gracchus and Garius do Gracchus
3. Han
4. St. Sophia Cathedral
5. Marius and Sulla

V. Answer all questions given under each heading.

Question 1.
Emergence of Rome as an empire
(a) Who were the Gracchus brothers?
Answer:
Tiberius Gracchus and Garius do Gracchus were Patricians. They voiced their opinion in favour of the poor peasants.

(b) What role did they play?
Answer:
They voiced their opinion in favour of the poor peasants.

(c) What was the outcome of their martyrdom?
Answer:
The Martyrdom of the Gracchus brothers played a decisive role in the transformation of the Roman Republic into the Roman Empire.

(d) Who was the first Roman Emperor?
Answer:
Marus Aureillus was the first Roman Emperor.

Question 2.
Han Dynasty
(a) Who was the founder of Han Empire?
Answer:
Han Dynasty was founded by Liu Pang.

(b) What was the capital of Han Empire?
Answer:
The capital of Han Empire was Chang-an.

(c) Where did they have their new capital?
Answer:
They had their new capital at Xuchang.

(d) Who was the powerful ruler of the Han dynasty?
Answer:
The most popular and powerful ruler was Wu Ti.

VI. Answer the following briefly.

Question 1.
Attempt an account of slavery in Rome.
Answer:

• A major source of revenue to the Roman state was slave trade.
• The island of Delos became a great slave market.
• There were more slave revolts in Rome than in Greece.
• The revolt of spartacus was the most famous.

Question 2.
Highlight the main contribution of Constantine.
Answer:
The main contribution of Constantine was, conversion to Christianity. He himself became a Christian and Christianity became the official religion of the Empire.

Question 3.
What do you know of the Carthaginian leader Hannibal?
Answer:

• Hannibal was a Carthaginian General who defeated the Roman Army and made a great part of Italy a desert in the Punic War.
• In the Second Punic War Hannibal was defeated in the Battle of Zama.
• Pursued by the Roman army, Hannibal ended his life by poisoning himself.

Question 4.
What were the reasons for the prosperity of Han Empire?
Answer:

1. The Han Empire threw open the silk road for trade.
2. A large export trade mainly in silk reached as far as the Roman Empire.
3. In the North artisans and herders of rival “barbarian” dynasties brought in new techniques like the methods harnessing horses, use of saddle and stirrup, techniques of building bridges and mountain roads and seafaring.
4. Such innovations made Han Empire prosperous.

Question 5.
Write about St. Sophia Cathedral.
Answer:

• St. Sophia Cathedral was built in mid-sixth century AD (CE) The most magnificent building in Europe at that time, it was known for its innovative architectural techniques.
• This Cathedral was turned into a mosque by the Ottoman Turks when they captured Constantinople.

VII. Answer the following in detail:

Question 1.
Discuss the rise and growth of Athens, pointing out its glorious legacy.
Answer:
In Athens, the pressure from below resulted in the replacement of both oligarchy’and tyranny by “democracy”. The law-making power in Athens was vested in an assembly open to all freemen. Judges and lower officials were chosen by lots. This arrangement was resented by ‘ the upper classes who considered democracy to be the rule of the mob.

The Persian danger had united the Greeks. When this danger was removed, they started quarrelling again. The history of many Greek city-states was one of continual struggles by the rich landowners against “democracy”. The only exception was Athens, where “democracy’ survived for about 200 years.

Question 2.
Highlight the contributions of Rome to World Civilization.
Answer:
The Byzantine emperors, who ruled from the city of Constantinople for about 1,000 years,
called themselves Romans. But their language was Greek. The splendor of Constantinople with its luxurious royal palaces, its libraries, its scholars familiar with the writings of Greeks 1 and Romans and its fascinating St. Sophia Cathedral are the legacies they have left behind.

However, in terms of the development of science and technology, there was no progress during this I period. The economies of the Empire’s provinces were in the hands of large local landowners. The small peasants always lived on the edge of poverty. The fundamental weakness of
Byzantine Civilization stood exposed when the participants of Fourth Crusade pillaged it and I ruled it.

Student Activities

Question 1.
In an outline map of Europe, the students are to sketch the extent of Western and Eastern Roman Empire.
Answer:

The Classical World Additional Questions

I. Choose the correct answer.

Question 1.
………….. a fortified city of ancient Greeks on a hill in Athens is an illustrative example of their advancement.
(a) Acropolis
(b) Athens
(c) Sparta
(d) None of the above
Answer:
(a) Acropolis

Question 2.
The word …………….. literally means “rule of the people”.
(a) Autocracy
(b) Communism
(c) Socialism
(d) Democracy
Answer:
(d) Democracy

Question 3.
Aristotle was the disciple of ……………
(a) Socrates
(b) Plato
(c) Pericles
(d) Augustus
Answer:
(b) Plato

Question 4.
“Natural History” was completed by ………….
(a) Pliny the Elder
(b) Antoninus Pius
(c) Marcus Aurelius
(d) Tacitus
Answer:
(a) Pliny the Elder

Question 5.
……………. came to China from India during the reign of Han Dynasty.
(a) Jainism
(b) Sikhism
(c) Buddhism
(d) Christianity
Answer:
(c) Buddhism

Question 6.
……………. started spreading the Christian doctrine after the Crucifixion of Jesus.
(a) St. Paul
(b) St. Thomas
(c) St. Antony
(d) St. John
Answer:
(a) St. Paul

II. Find out the correct statement.

Question 1.
(i) Until 8th Century B.C. (BCE) Greece was different from the rest of the world.
(ii) Democracy literally means “rule of the people”.
(iii) The Greek city-states have an elaborate bureaucracy.
(iv) The entire period of Alexander’s reign was not spent on wars.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(b) (ii) is correct

Question 2.
(i) Prisoners of war were enslaved in Rome.
(ii) Rome developed into a normal town.
(iii) In the beginning Rome was a society of Business men.
(iv) A major source of revenue to the Roman state was slave trade.
(a) (i) is correct
(b) (ii) is correct
(c) (i) and (iv) are correct
(d) (iii) is correct
Answer:
(c) (i) and (iv) are correct

Question 3.
(i) The revolt of spartacus was the most famous.
(ii) The Island of Delos became a great slave market.
(iii) Catalina’s victory led to mob violence.
(iv) Livy was a Poetist.
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(b) (i) and (ii) are correct

Question 4.
(i) Han Empire once again threw open the silk road for trade.
(ii) Buddhism came to China from Indonesia.
(iii) With Buddhism came the influence of Indian art to Korea.
(iv) The period after Han rule witnessed political stability across the country.
(a) (i) is correct
(b) (i) and (iii) are correct
(c) (ii) is correct
(d) (iii) and (iv) are correct
Answer:
(a) (i) is correct

Question 5.
(i) Jewish had no faith on Jesus.
(ii) They hoped Messiah would not arrive.
(iii) Jesus was against the rich and the hypocrites.
(iv) Christianity did not spread in Europe.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(c) (iii) is correct

III. Match the following:

Answer:
1. (f)
2. (g)
3. (a)
4. (b)
5. (c)
6. (d)
7. (e)

IV. Fill in the blanks.

1. Under ……….. the new rich exploited the smaller landholders.
2. The Persian danger had united the ……………..
3. Cultural development that took place rapidly after Alexander’s death. 323 BC is called ………….. civilization.
4. A major source of revenue to the Roman state was ……………
5. The most distinguished writers of the ……………… Age brought glory to the empire.
6. The most popular and powerful ruler of Han Dynasty was …………….
7. Some of the Buddhist art of the time show the impact of …………… styles.
8. One of the Roman emperors …………….. himself became a Christian.
9. Internal Crisis and invasion of Franks, Goths and Vandals ended the …………….
10. Christianity became a state religion of …………….. and began to spread in Europe.
Answers:
1. Oligarchy
2. Greeks
3. Hellenistic
4. Slave trade
5. Augustus
6. Wu Ti
7. Hellenistic
8. Constantine
9. Roman Empire
10. Byzantium

V. Answer all questions given under each heading.

Question 1.
Athenian Democracy.
(a) What replaced the oligarchy and tyranny in Athens?
Answer:
Democracy.

(b) With whom the law-making power was vested in Athens?
Answer:
The law making power in Athens was vested in an assembly open to all freemen.

(c) What united the Greeks?
Answer:
The Persian danger had united the Greeks.

(d) How many years were Democracy survived in Athens?
Answer:
Democracy survived in Athens for about 200 years.

Question 2.
Hellenistic Civilization.
(a) Who established the kingdom in Macedonia?
Answer:
Alexander the Great.

(b) How did this kingdom succeed?
Answer:
This kingdom succeeded in annexing two historic empires of Egypt and the Middle East.

(c) How did Alexander spend his life?
Answer:
The entire period of Alexander’s reign was spent on wars.

(d) When did the cultural development take place? What is it called?
Answer:
After the death of 323 B.C. and is called Hellenistic civilization.

VI. Answer the following briefly.

Question 1.
What do you know about the classical world?
Answer:

1. Greece, Rome and China represented the classical world which ended with the fall of Western Roman Empire.
2. Until the end of 5th Century AD (CE) Christianity was confined to the Roman Empire.

Question 2.
Why did the first Persian attack on Greece failed?
Answer:
The first Persian attack on Greece failed due to the fact that Persian army suffered from disease and lack of food during its March.

Question 3.
Comment on the ‘Age of Pericles’.
Answer:

1. Athens, despite hostility and disturbance from sparta, became a noble city with magnificent buildings.
2. There were great artists and great thinkers.
3. Historians therefore call this the Age of Pericles.

Question 4.
What do you know about the Socrates?
Answer:

1. Socrates was the greatest of the thinkers of the Pericles Era.
2. As a philosopher he discussed difficult problems with his friends to bring out the truth out of discussions.

Question 5.
Who were the capable rulers in Rome after the death of Augustus?
Answer:
Trajan (98-117), Antoninus Pius (138-161) and Marcus Aurelius (161-180) were the capable enlightened rulers.

Question 6.
Who were Barbarians?
Answer:
A group of people from a very different country (or) culture that is considered to be less culturally advanced and more violent than their own were Barbarians.

Question 7.
What is known as Silk Road (or) Silk Route?
Answer:
The trade route from China to Asia Minor and India, known as the Silk Road or Silk Route, linked China with the West. Goods and ideas between the two great civilizations of Rome and China were exchanged through this route. Silk went westward, and wools, gold, and silver went east. China received Buddhism from India via the Silk Road.

Question 8.
Who spreaded Christianity after Jesus’s crucifixion?
Answer:

1. St. Paul started spreading the Christian doctrine.
2. Paul succeeded in his effort and Christianity gradually spread.

VII. Answer the following in detail.

Question 1.
Explain the “Rise of Christianity”.
Answer:
(i) After a brief period of glory in the days of David and Solomon, the Jewish people had a great fall and experienced extreme hardship.

(ii) While spreading out all over the Roman Empire and elsewhere, they hoped that a Messiah would arrive to restore their pristine glory.

(iii) Initially they laid much hopes on Jesus. Jesus was against the rich and the hypocrites, and condemned certain observances and ceremonials.

(iv) This was not to the liking of the priests, who turned against Jesus and handed him over to the Roman Governor Pontius Pilate. Looked upon as a political rebel by the Roman authorities, Jesus was tried and crucified.

(v) After Jesus’s crucifixion, St Paul started spreading the Christian doctrine. Paul succeeded in his effort and Christianity gradually spread.

(vi) Romans were prepared to tolerate Christianity. But the refusal of the Christians to pay respect to the Emperor’s image was viewed as political treason.

(vii) It led to the persecution of Christians. Their property was confiscated and they were thrown to the lions.

(viii) Yet the Roman Empire did not succeed in suppressing Christianity. One of the Roman emperors Constantine himself became a Christian. Christianity thus became the official religion of the Empire.