Category: Class 9

Students can download Maths Chapter 1 Set Language Ex 1.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.7

I. Multiple Choice Questions.

Question 1.
Which of the following is correct?
(a) {7} ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(c) 7 ∉ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(d) {7} ⊄ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Question 2.
The set P = {x | x ∈ Z, -1 < x < 1} is a ……….
(a) Singleton set
(b) Power set
(c) Null set
(d) Subset
Solution:
(a) Singleton set
Hint: P = {0}

Question 3.
If U = {x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A’)’ is………..
{a) {1, 6, 7, 8, 9}
(b) {1, 2, 3, 4}
(c) {2, 3, 4, 5}
(d) { }
Solution:
(c) {2, 3, 4, 5}
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A= {2, 3, 4, 5}; A’ = {1, 6, 7, 8, 9}
(A’)’ = U – A
(A’)’ = {2, 3, 4, 5}

Question 4.
If B⊆A then n(A∩B) is………..
(a) n(A – B)
(b) n(B)
(c) n(B – A)
(d) n(A)
Solution:
(b) n(B)

Question 5.
If A = {x, y, z} then the number of non- empty subsets of A is…………..
(a) 8
(b) 5
(c) 6
(d) 7
Solution:
(d) 7
Hint:
n(A) = 3; P(A) = 2^m = 2³ = 8
Non-empty subsets of A = 8 – 1 = 7

Question 6.
Which of the following is correct?
(a) Ø ⊆ {a, b}
(b) Ø ∈ {a, b}
(c) {a} ∈ {a, b}
(d) a ⊆ {a, b}
Solution:
(a) Ø ⊆ {a, b}
Hint:
‘Q’ is a subset of every set.

Question 7.
If A∪B = A∩B then ……………
(a) A ≠ B
(b) A = B
(c) A ⊂ B
(d) B ⊂ A
Solution:
(b) A = B

Question 8.
If B – A is B, then A∩B is………….
(a) A
(b) B
(c) U
(d) Ø
Solution:
(d) Ø

Question 9.
From the adjacent diagram n[P(AΔB)] is………….

(a) 8
(b) 16
(c) 32
(d) 64
Solution:
(c) 32
Hint:
A – B = {60, 85, 75}
B – A = {90, 70}
AΔB = (A – B) ∪ (B – A)
= {60, 70, 75, 85, 90}
n[P(AΔB)] = 2⁵ = 32

Question 10.
If n(A) = 10 and n(B) = 15 then the minimum and maximum number of elements in A∩B is …………
(a) (10, 15)
(b) (15, 10)
(c) (10, 0)
(d) (0, 10)
Solution:
(d) (0, 10)

Question 11.
Let A = {Ø} and B = P(A) then A∩B is………..
(a) {Ø, {Ø}}
(b) {Ø}
(c) Ø
{d) {0}
Solution:
(b) {Ø}
Hint:
A = {Ø}, B = {{ }, {Ø}}
A∩B = {Ø}

Question 12.
In a class of 50 boys, 35 boys play carrom and 20 boys play chess then the number of boys play both games is……..
(a) 5
(b) 30
(c) 15
(d) 10
Solution:
(a) 5
Hint:
n(A∪B) = 50, n(A) = 35, n(B) = 20
n(A∩B) = n(A) + n(B) – n(A∪B)
= 35 + 20 – 50 = 5

Question 13.
If U = {x : x ∈ N and x < 10}, A = {1, 2, 3, 5, 8} and B = (2, 5, 6, 7, 9}, then n [(A∪B)’] is
(a) 1
(b) 2
(c) 4
(d) 8
Solution:
(a) 1
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A∪B ={1, 2, 3, 5, 8} ∪ {2, 5, 6, 7, 9}
= {1, 2, 3, 5, 6, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 5, 6, 7, 8, 9}
= {4}
n(A∪B)’ = 1

Question 14.
For any three sets P, Q and R, P – (Q∩R) is ………
(a) P – (Q∪R)
(i)(P∩Q) – R
(c) (P – Q) ∪ (P – R)
(d) (P – Q) ∩ (P – R)
Solution:
(c)(P – Q) ∪ (P – R)

Question 15.
Which of the following is true?
(a) A – B = A∩B
(b) A – B = B – A
(c) (A∪B)’ = A’∪B’
(d) (A∩B)’ = A’∪B’
Solution:
(d) (A∩B)’ = A’∪B’

Question 16.
If n(A∪B∪C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A∩B) = 20, n(B∩C) = 15 and n(A∩C) = 25, then n(A∩B∩C) = 10, then the value of x is……….
(a) 10
(b) 15
(c) 25
(d) 30
Solution:
(b) 15
Hint:
n(A∪B∪C) = n( A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
100 = 4x + 6x + 5x – 20 – 15 – 25 + 10
100 = 15x – 50 ⇒ 150 = 15x ⇒ x = \(\frac{150}{15}\) = 10

Question 17.
For any three sets A, B and C, (A – B) ∩ (B – C) is equal to
(a) A only
(b) B only
(c) C only
(d) \(\phi \)
Solution:
(d) \(\phi \)

Question 18.
If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of shapes with right angle, then J∩K∩L is………
(a) Set of isosceles triangles
(b) Set of equilateral triangles
(c) Set of isosceles right triangles
(d) Set of right angled triangles
Solution:
(c) Set of isosceles right triangles
J = Set of three sided shapes; It is a triangle
K = Set of shapes with two equal sides (Isosceles triangle)
L = Set of shapes with right angle (One angle is right angle)
∴ J∩K∩L = Set of isosceles right triangles

Question 19.
The shaded region in the Venn diagram is

(a) Z – (X∪Y)
(b) (X∪Y) ∩ Z
(c) Z – (X∩Y)
(d) Z ∪ (X∩Y)
Solution:
(c) Z – (X∩Y)

Question 20.
In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?
(a) 5
(b) 8
(c) 10
(d) 15
Solution:
(a) 5

Students can download Maths Chapter 1 Set Language Ex 1.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5

Question 1.
Using the adjacent Venn diagram, find the following sets:

(i) A – B
(ii) B – C
(iii) A’∪B’
(iv) A’∩B’
(v) (B∪C)’
(vi) A – (B∪C)
(vii) A – (B∩C)
Solution:
From the diagram we get
U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8},
A= {-2,-1, 3, 4, 6}, B = {-2,-1, 5, 7, 8}
C = {-3, -2, 0, 3, 8}
A’ = U – A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 3, 4, 6}
= {-3, 0, 1, 2, 5, 7, 8}
B’ = U – B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 5, 7, 8}
= {-3, 0, 1, 2, 3, 4, 6}
B∪C = {-2, -1, 5, 7, 8} ∪ {-3, -2, 0, 3, 8} = {-3, -2, -1, 0, 3, 5, 7, 8}
B∩C = {-2, -1, 5, 7, 8} ∩ {-3, -2, 0, 3, 8} = {-2, 8}

(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’∪B’= {-3, 0, 1, 2, 5, 7, 8} ∪ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A’∩B’ = {-3, 0, 1, 2, 5, 7, 8} ∩ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2}
(v) (B∪C)’ = U – (B∪C)= {-3,-2,-1,0, 1,2, 3,4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
= {1, 2, 4, 6}
(vi) A – (B∪C) = {-2, -1, 3, 4, 6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
(vii) A – (B∩C) = {-2,-1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

Question 2.
If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L) ∩ (K∪M)
(iv) (K∩L) ∪ (K∩M)
and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(i) K∪(L∩M)
(L∩M) = {b, c, d, g} ∩ [a, b, c, d, h}
= {b, c, d}
K∪(L∩M) = {a, b, d, e, f} ∪ {b, c, d}
= {a, b, c, d, e, f}

(ii) K∩(L∪M)
(L∪M) = {b, c, d, g} ∪ {a, b, c, d, h}
= {a, b, c, d, g, h}
K∩(L∪M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h}
= {a, b, d }

(iii) (K∪L) ∩ (K∪M)
(K∪L) = {a, b, d, e, f} ∪ {b, c, d, g}
= {a, b, c, d, e, f, g}
(K∪M) = {a, b, d, e, f} ∪ {a, b, c, d, h}
= {a, b, c, d, e, f, h}
(K∪L) ∩ (K∪M) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}

(iv) (K∩L) ∪ (K∩M)
(K∩L) = {a, b, d, e, f) ∩ {b, c, d, g}
= {b, d}
(K∩M) = {a, b, d, e, f} ∩ {a, b, c, d, h}
= {a, b, d}
(K∩L) ∪ (K∩M) = {b, d} ∪ [a, b, d}
= {a, b, d}
From (ii) & (iv) we get, K∩(L∪M) = (K∩L) ∪ (K∩M)
From (i) & (iii) we get, K∪(L∩M) = (K∪L) ∩ (K∪M)

Question 3.
For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}
verify A∪(B∩C) = (A∪B) ∩ (A∪C).
Solution:
A = {-1, 0, 1, 2, 3, 4}, B = {0, 1, 2, 3, 4, 5} and C = {-4, -1, 0, 2, 3, 4}
B∩C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4}
= {0, 2, 3, 4}
A∪(B∩C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(1)
A∪B = {-1, 0, 1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
= {-1, 0, 1, 2, 3, 4, 5}
A∪C = {-1, 0, 1, 2, 3, 4} ∪ {-4, -1, 0, 2, 3, 4}
= {-4, -1, 0, 1, 2, 3, 4}
(A∪B) ∩ (A∪C) = {-1, 0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 1, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(2)
From (1) and (2) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 4.
Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.
Solution:

From (ii) and (v) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h} ; B = {a, c, d, g, f}; C = {a, d, e, g, h}
B∩C = {a, c, d, g, i} ∩ {a, d, e, g, h}
= {a, d, g}
A – (B∩C) = {b, c, e, g, h} – {a, d, g}
= {b, c, e, h}…….(1)
A – B = {b, c, e, g, h} – {a, c, d, g, i}
= {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h}
= {b, c}
(A – B) ∪ (A – C) = {b, e, h} ∪ {b, c}
= {b, c, e, h)……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C)

Question 6.
If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and
C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)
Solution:
A = {0, 6, 12, 18, 24, 30}; B = {6, 8, 10, 12, 14, 16, 18}; C = {12, 15, 18, 21, 24, 27}
B∩C = {6, 8, 10, 12, 14, 16, 18} ∩ {12, 15, 18, 21, 24, 27}
= {12, 18}
A – (B∩C) = {0, 6, 12, 18, 24, 30} – {12, 18}
= {0, 6, 24, 30}………(1)
A – B = {0, 6, 12, 18, 24, 30} – {6, 8, 10, 12, 14, 16, 18}
= {0, 24, 30}
A – C = {0, 6, 12, 18, 24, 30} – {12, 15, 18, 21, 24, 27}
= {0, 6, 30}
(A – B) ∪ (A – C) = {0, 24, 30} ∪ {0, 6, 30}
= {0, 6, 24, 30}……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C).

Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that
A – (B∪C) = (A – B) ∩ (A – C).
Solution:
A= {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6}, C = {-1, 2, 5, 6, 7}
B∪C = {-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}
= {-1, 0, 2, 5, 6, 7}
A – (B∪C) = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6, 7}
= {-2, 1, 3} ………(1)
A – B = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6}
= {-2, 1, 3}
A – C = {-2, 0, 1, 3, 5}- {-1, 2, 5, 6, 7}
= {-2, 0, 1, 3}
(A- B) ∩ (A- C) = {-2, 1, 3} ∩ {-2, 0, 1, 3}
= {-2, 1, 3} ….(2)
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 8.
IF A = {y : y = \(\frac{a + 1}{2}\), a ∈ W and a ≤ 5}, B = {y : y = \(\frac{2n – 1}{2}\), n ∈ W and n < 5} and C = {-1, \(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).
Solution:

From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 9.
Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:


From (ii) and (v) we get A- (B∩C) = (A – B) ∪ (A – C).

Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U= {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {7, 8, 11, 12} ∪ {4, 8, 12, 15}
= {4, 7, 8, 11, 12, 15}
(A∪B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10,16} ………(1)
A’ = {4, 7, 8, 10, 11, 12, 15, 16} – {7, 8, 11, 12}
= {4, 10, 15, 16}
B’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 8, 12, 15}
= {7, 10, 11, 16}
A’∩B’ = {4, 10, 15, 16} ∩ {7, 10, 11, 16}
= {10,16} ………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’

(ii) A∩B = {7, 8, 11, 12} ∩ {4, 8, 12, 15}
= {8, 12}
(A∩B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {8, 12}
= {4, 7, 10, 11, 15, 16} ………(1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’∪B’ = {4, 10, 15, 16} ∪ {7, 10, 11, 16}
= {4, 7, 10, 11, 15, 16} ………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’

Question 11.
Verify (A∩B)’ = A∪B’ using Venn diagrams.
Solution:

From (ii) and (i) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 1 Set Language Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6

Question 1.
(i) If n(A) = 25, n(B) – 40, n(A∪B) = 50 and n(B’) = 25, find n(A∩B) and n(U).
Solution:
Given, n(A) = 25, n(B) = 40, n(A∪B) = 50 and n(B’) = 25 n(A∩B) = n(A) + n(B) – n(A∪B)
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
∴ n(A∩B) = 15 and n(U) = 65

(ii) If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
Given, n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350
n(A∪B) = n(A) + n(B) – n(A∩B)
500 = 300 + n(B) – 50
500 = 250 + n(B)
500 – 250 = n(B)
250 = n(B)
∴ n(B) = 250
n(U) = n(B) + n(B)’
250 + 350 = 600
∴ n(B) = 250 and n(U) = 600

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:
U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 3, 4, 8, 10}; B = {1, 2, 5, 8, 10}
n(U) = 10, n(A) = 5, n(B) = 5
(A∪B) = {2, 3, 4, 8, 10} ∪ {1, 2, 5, 8, 10}
= {1, 2, 3, 4, 5, 8, 10}
∴ n(A∪B) = 7 ……..(1)
(A∩B) = {2, 3, 4, 8, 10} ∩ {1, 2, 5, 8, 10}
= {2, 8, 10}
n(A∩B) = 3
n(A) + n(B) – n(A∩B) = 5 + 5 – 3
= 10 – 3
= 7 ……(2)
From (1) and (2) we get,
n(A∪B) = n(A) + n(B) – n(A∩B)

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}
Solution:
A∩B = {a, c, e, f, h} ∩ {c, d, e, f}
= {c, e, f}
B∩C = {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
A∩C = {a, c, e, f, h} ∩ {a, b, c, f}
= {c, f}
(A∩B∩C) = {a, c, e, f, h} ∩ {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
(A∪B∪C) = {a, c, e, f, h} ∪ {c, d, e, f} ∪ {a, b, c, f}
= {a, b, c, d, e, f, h}
n(A∩B) = 3, n(B∩C) = 2, n(A∩C) = 3, n(A∩B∩C) = 2
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)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

(ii) A= {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}
A∩B = {1, 3, 5} ∩ {2, 3, 5, 6}
= {3, 5}
B∩C = {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5, 6}
A∩C = {1, 3, 5} ∩ {1, 5, 6, 7}
= {1, 5}
A∩B∩C = {1, 3, 5} ∩ {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5}
A∪B∪C = {1, 3, 5} ∪ {2, 3, 5, 6} ∪ {1, 5, 6, 7}
= {1, 2, 3, 5, 6, 7}
n(A) = 3, n(B) = 4, n(C) = 4
n(A∩B) = 2, n(B∩C) = 2, n(A∩C) = 2
n(A∩B∩C) = 1
n(A∪B∪C) = 6……….(1)
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 – 2 – 2 – 2 + 1
= 12 – 6
= 6………(2)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

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 M be the set of all students take part in music.
Let D be the set of all students take part in drama.
n( M) = 25, n(D) = 30 and n(M∩D) = 8
By using venn-diagram
From the venn – diagram we get.
(i) Number of students take part in only music = 17
(ii) Number of students take part in only drama = 22
(iii) Total number of students in the class = 17 + 8 + 22 = 47

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’T’ be the set of people likes Tea
Let ‘C’ be the set of people likes Coffee
n(T∩C) = 45, n(T) = 35 and n(C) = 20
Let X be the number of people likes both Tea and Coffee.
By using venn diagram
From the venn – diagram we get.
35 – x + x + 20 – x = 45
55 – x = 45
55 – 45 = x
10 = x
(i) People like both tea and coffee = 10
(ii) People do not like tea = 20 – x
= 20 – 10 = 10
(iii) People do not like coffee = 35 – x
= 35 – 10 = 25

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 both the subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let M and S represent the student failed in Mathematics and Science.
Given: Number of students passed in Mathematics is 50%

∴ Number of students failed in Mathematics = 100 – 50% = 50%
n(M) = 50%
Number of students passed in Science is 70%
∴ Number of students failed in Science = 100 – 70% = 30%
n(S) = 30%
Number of students failed in both the subjects is 10%
n(M∩S) = 10%
n(M∪S)= n(M) + n(S) – n(M∩S)
= 50 + 30 – 10 = 80 – 10 = 70
Given: 70% of the students failed in atleast any one of the subject
∴ 30% of the students passed in atleast any one of the subjects.
30 students passed mean, the total number of students is 100.
∴ 300 students passed means, the total number of students = \(\frac{100 × 300}{30}\)
Total number of students appeared in the examination = 1000

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
From the Venn diagram:

Given n( A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x – 2x
32 = 4x
x = \(\frac{32}{4}\) = 8
The value of x = 8

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:
Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A∩B) = 50
n(A∪B) = 500………..(1)
n(A) + n(B) – n(A∩B) = 400 + 200 – 50
= 600 – 50
= 550………(2)
From (1) and (2) we get
n(A∪B) ≠ n(A) + n(B) – n(A∩B)
∴ The given data is not correct.

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:
Let T, E and H represent families buying Tamil newspaper, English newspaper and Hindi newspaper respectively.
n(T) = 275, n(E) = 150, n(H) = 45
n(T∩E) = 125, n(E∩H) = 17, n(T∩H) = 5
n(T∩E∩H) = 3
Let us represent the given data in Venn diagrams.

(i) Number of families buy only one news paper = 148 + 11 + 26
= 185
(ii) Number of families buy atleast two news paper = 122 + 2 + 14 + 3
= 141
(iii) Total number of families in the colony = 148 + 122 + 11 + 14 + 3 + 2 + 26
= 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:
Let P, R and C represent sets of farmers grew paddy, ragi and com respectively.
n(P∪R∪C) = 1000, n(P) = 600, n(R) = 350, n(C) = 280
n(P∩R) = 120, n(R∩C) = 100, w(P∩C) = 80 Let the number of farmers who grew all the three be “x”
n(P∪R∪C ) = n(P) + n( R) + n( C) – n(P∩R) – n(R∩C) – n(P∩C) + n(P∩R∩C )
1000 = 600 + 350 + 280 – 120 – 100 – 80 + x = 1230 – 300 + x.
1000 = 930 + x
1000 – 930 = x
70 = x
Number of farmers who grew all the three = 70.

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 and z = x + 10
n(U) = x + 4 + y + 17 + 3 + 6 + z + 5
125 = x + 4 + 2x + 17 + 3 + 6 + x + 10 + 5
125 = 4x + 45
125 – 45 = 4x
80 = 4x
x = 80/4 = 20
y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
∴ The value of x = 20, y = 40 and z = 30.

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:
Let A, B and C represent students play chess, carrom and table tennis.
n(A) = 22, n(B) = 21 , n(C) = 15
n(A∩C) = 10 , n(B∩C) = 8 , n(A∩B∩C) = 6
Let “x” represent student play chess and carrom but not table tennis.
Let us represent the data in Venn diagram.

From the Venn diagram we get,
Number of students play atleast one game = 35
12 – x + x + 13 – x + 2 + 6 + 4 + 3 = 35
40 – 35 = x
5 = x
(i) Number of students who play 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

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:
Let B, C and D represent students come to school by bus, bicycle and foot respectively.
n(B∪C∪D) = 50 , n(B) = 25 , n(C) = 20 , n(D) = 30, n(B∩C∩D) = 10
Let x, y and z represent the students come to school exactly by two modes of transport.
Let us represent the given data in Venn diagrams.
<
Total number of students in the class = 50
15 – x – z + x + 10 – x – y + y + 10 + z + 20 – z – y = 50
55 – x – y – z = 50
55 – 50 = x + y + z
5 = x + y + z
Number of students come to school exactly by two modes of transport = 5

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Samacheer Kalvi 9th English Book Solutions Prose

• Chapter 1 Learning the Game
• Chapter 2 I Can’t Climb Trees Anymore
• Chapter 3 Old Man River
• Chapter 4 Seventeen Oranges
• Chapter 5 Water – The Elixir of Life
• Chapter 6 From Zero to Infinity
• Chapter 7 A Birthday Letter

Samacheer Kalvi 9th English Book Solutions Poem

• Chapter 1 Stopping by Woods on a Snowy Evening
• Chapter 2 A Poison Tree
• Chapter 3 On Killing a Tree
• Chapter 4 The Spider and the Fly
• Chapter 5 The River
• Chapter 6 The Comet
• Chapter 7 The Stick-together families

Samacheer Kalvi 9th English Book Solutions Supplementary

• Chapter 1 The Envious Neighbour
• Chapter 2 The Fun They Had
• Chapter 3 Earthquake
• Chapter 4 The Cat and the Pain-killer
• Chapter 5 Little Cyclone: The Story of a Grizzly Cub
• Chapter 6 Mother’s Voice
• Chapter 7 The Christmas Truce

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Samacheer Kalvi 9th English Vocabulary / Grammar

• Homonyms, Homophones
• Prefix / Suffix
• Anagrams
• Preposition
• Prepositional Verbs
• Degrees of Comparison
• Clauses and Phrases
• Prepositional Phrases
• Non-finite verbs (Gerund, Infinitives, Participles)
• Auxiliaries Verbs
• Phrasal Verbs
• Idioms
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• Letter Writing (to a friend)
• Speech Writing (on a given topic)
• Article Writing
• Dialogue Writing
• Completion of Dialogues (Fill in type)
• Diary Writing
• Poster Making
• General Comprehension
• General Para Writing, Using Given Hints
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• Unknown Poem – Paraphrase
• Expressing Views on a Given Picture

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• Chapter 1 Evolution of Humans and Society – Prehistoric Period
• Chapter 2 Ancient Civilisations
• Chapter 3 Early Tamil Society and Culture
• Chapter 4 Intellectual Awakening and Socio-Political Changes
• Chapter 5 The Classical World
• Chapter 6 The Middle Ages
• Chapter 7 State and Society in Medieval India
• Chapter 8 The Beginning of the Modern Age
• Chapter 9 The Age of Revolutions
• Chapter 10 Industrial Revolution
• Chapter 11 Colonialism in Asia and Africa

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• Chapter 1 Lithosphere – I Endogenetic Processes
• Chapter 2 Lithosphere – II Exogenetic Processes
• Chapter 3 Atmosphere
• Chapter 4 Hydrosphere
• Chapter 5 Biosphere
• Chapter 6 Man and Environment
• Chapter 7 Mapping Skills
• Chapter 8 Disaster Management: Responding to Disasters

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• Chapter 1 Forms of Government and Democracy
• Chapter 2 Election, Political Parties and Pressure Groups
• Chapter 3 Human Rights
• Chapter 4 Forms of Government
• Chapter 5 Local Self Government
• Chapter 6 Road Safety

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• Chapter 1 Understanding Development: Perspectives, Measurement and Sustainability
• Chapter 2 Employment in India and Tamilnadu
• Chapter 3 Money and Credit
• Chapter 4 Tamilnadu Agriculture
• Chapter 5 Migration

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• Chapter 1 மனிதப் பரிணாம வளர்ச்சியும் சமூகமும்: வரலாற்றுக்கு முந்தைய காலம்
• Chapter 2 பண்டைய நாகரிகங்கள்
• Chapter 3 தொடக்ககாலத் தமிழ் சமூகமும் பண்பாடும்
• Chapter 4 அறிவு மலர்ச்சியும், சமூக – அரசியல் மாற்றங்களும்
• Chapter 5 செவ்வியல் உலகம்
• Chapter 6 இடைக்காலம்
• Chapter 7 இடைக்கால இந்தியாவில் அரசும் சமூகமும்
• Chapter 8 நவீன யுகத்தின் தொடக்கம்
• Chapter 9 புரட்சிகளின் காலம்
• Chapter 10 தொழிற்புரட்சி
• Chapter 11 ஆசிய ஆப்பிரிக்க நாடுகளில் காலனியாதிக்கம்

9th Std Social Science Book Back Answers Geography

• Chapter 1 நிலக்கோளம் – I புவி அகச்செயல்பாடுகள்
• Chapter 2 நிலக்கோளம் – II புவி புறச்செயல்பாடுகள்
• Chapter 3 வளிமண்டலம்
• Chapter 4 நீர்க்கோளம்
• Chapter 5 உயிர்க்கோளம்
• Chapter 6 மனிதனும் சுற்றுச் சூழலும்
• Chapter 7 நிலவரைபடத் திறன்கள்
• Chapter 8 பேரிடர் மேலாண்மை – பேரிடரை எதிர்கொள்ளுதல்

9th Std Social Science Guide Civics

• Chapter 1 அரசாங்க அமைப்புகள் மற்றும் மக்களாட்சி
• Chapter 2 தேர்தல், அரசியல் கட்சிகள் மற்றும் அழுத்தக் குழுக்கள்
• Chapter 3 மனித உரிமைகள்
• Chapter 4 அரசாங்கங்களின் வகைகள்
• Chapter 5 உள்ளாட்சி அமைப்புகள்
• Chapter 6 சாலை பாதுகாப்பு

9th Standard Social Science Guide Economics

• Chapter 1 மேம்பாட்டை அறிவோம்: தொலைநோக்கு, அளவீடு மற்றும் நிலைத் தன்மை
• Chapter 2 இந்தியா மற்றும் தமிழ்நாட்டில் வேலைவாய்ப்பு
• Chapter 3 பணம் மற்றும் கடன்
• Chapter 4 தமிழகத்தில் வேளாண்மை
• Chapter 5 இடம்பெயர்தல்

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• Chapter 4 வடிவியல் Ex 4.4
• Chapter 4 வடிவியல் Ex 4.5
• Chapter 4 வடிவியல் Ex 4.6
• Chapter 4 வடிவியல் Ex 4.7

9th Class Maths Guide State Syllabus Pdf ஆயத்தொலை வடிவியல்

• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.1
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.2
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.3
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.4
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.5
• Chapter 5 ஆயத்தொலை வடிவியல் Ex 5.6

9th Samacheer Maths Solution Book Chapter 6 முக்கோணவியல்

• Chapter 6 முக்கோணவியல் Ex 6.1
• Chapter 6 முக்கோணவியல் Ex 6.2
• Chapter 6 முக்கோணவியல் Ex 6.3
• Chapter 6 முக்கோணவியல் Ex 6.4
• Chapter 6 முக்கோணவியல் Ex 6.5

Samacheer 9th Maths Guide Tamil Medium Pdf Free Download Chapter 7 அளவியல்

• Chapter 7 அளவியல் Ex 7.1
• Chapter 7 அளவியல் Ex 7.2
• Chapter 7 அளவியல் Ex 7.3
• Chapter 7 அளவியல் Ex 7.4

Samacheer Kalvi Class 9 Maths Solutions Chapter 8 புள்ளியியல்

• Chapter 8 புள்ளியியல் Ex 8.1
• Chapter 8 புள்ளியியல் Ex 8.2
• Chapter 8 புள்ளியியல் Ex 8.3
• Chapter 8 புள்ளியியல் Ex 8.4

9th Standard Samacheer Kalvi Maths Guide Chapter 9 நிகழ்தகவு

• Chapter 9 நிகழ்தகவு Ex 9.1
• Chapter 9 நிகழ்தகவு Ex 9.2
• Chapter 9 நிகழ்தகவு Ex 9.3

We hope these Tamilnadu State Board Samacheer Kalvi Class 9th Maths Book Solutions and Answers Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

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Samacheer Kalvi 9th Science Book Back Answers

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

Samacheer Kalvi 9th Science Physics Book Solutions

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

Samacheer Kalvi 9th Science Chemistry Book Solutions

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

Samacheer Kalvi 9th Science Biology Book Solutions

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

Samacheer Kalvi 9th Science Computer Science Book Solutions

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

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

Samacheer Kalvi 9th Science Physics Book Solutions

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

Samacheer Kalvi 9th Science Chemistry Book Solutions

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

Samacheer Kalvi 9th Science Biology Book Solutions

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

Samacheer Kalvi 9th Science Computer Science

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

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Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 18 Organization of Tissues Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 18 Organization of Tissues

9th Science Guide Organization of Tissues Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1.
The tissue composed of living thin walled polyhedral cell is
(a) parenchyma
(b) pollenchyma
(c) pclerenchyma
(d) none of above
Answer:
(d) none of above

Question 2.
The fibres consists of
(a) parenchyma
(b) sclerenchyma
(c) collenchyma
(d) none of above
Answer:
(b) sclerenchyma

Question 3.
Companion cells are closely associated with
(a) sieve elements
(b) vessel elements
(c) trichomes
(d) guard cells
Answer:
(a) sieve elements

Question 4.
Which of the following is a complex tissue?
(a) Parenchyma
(b) Collenchyma
(c) Xylem
(d) Sclerenchyma
Answer:
(c) Xylem

Question 5.
Aerenchyma is found in
(a) epiphytes
(b) hydrophytes
(c) halophytes
(d) xerophytes
Answer:
(b) hydrophytes

Question 6.
Smooth muscles occur in
(a) uterus
(b) artery
(c) vein
(d) all of the above
Answer:
(d) all of the above

Question 7.
Nerve cell does not contains
(a) axon
(b) nerve endings
(e) tendons
(d) dendrités
Answer:
(c) tendons

II. Match the following.

Question 1.

Sclereids	Chlorenchyma
Chloroplast	Sclerenchyma
Simple tissue	Collenchyma
Companion cell	Xylem
Tracheids	Phloem

Answer:

Sclereids	Sclerenchyma
Chloroplast	Chlorenchyma
Simple tissue	Collenchyma
Companion cell	Phloem
Tracheids	Xylem

III Fill in the blanks :

1. ……………. tissues provide mechanical support to organs.
Answer:
Compound epithelium

2. Parenchyma, Collenchyma, Sclerenchyma are ……………. type of tissue.
Answer:
simple

3. ……………. and ……………… are complex issues.
Answer:
Xylem, phloem

4. Epithelial cells with cilia are fóund in ……………….. of our body
Answer:
trachea or windpipe

5. Lining of the small intestine is made up of …………………..
Answer:
columnar epithelium

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

1. Epithelial tissue is protective tissue in an animal body.
Answer:
True.

2. Bone and cartilage are two types of areolar connective tissue.
Answer:
False.
Correct statement: Bone and cartilage are two types of supportive connective tissue.

3. Parenchyma is a simple tissue.
Answer:
True.

4. Phloem is made up of tracheids.
Answer:
False.
Correct statement: Phloem is made up of sieve tubes.

5. Vessels are found in collenchyma.
Answer:
False.
Correct statement: Vessels are found in the xylem.

V. Answer briefly :

Question 1.
What are intercalary meristems? How do they differ from other meristems?
Answer:
Intercalary meristem lies between the region of permanent tissues and is part of primary meristem which is detached due to formation of intermittent permanent tissues. It is found either at the base of leaf e.g. Pinus or at the base of intemodes e.g. grasses.

Question 2.
What is complex tissue? Name the various kinds of complex tissues.
Answer:

• Complex tissues are made of more than one type of cells that work together as a unit.
• Complex tissues consist of parenchyma and sclerenchyma cells. However, collenchymatous cells are not present in such tissues.
• Common examples are xylem and phloem.

Question 3,.
Mention the most abundant muscular tissue found in our body. State its function.
Answer:
Connective tissue is the most abundant and widely distributed tissue. It provides a structural framework and gives support to different tissues forming organs.

Question 4.
What is skeletal connective tissue? How is it helpful in the functioning of our body?
Answer:
The supporting or skeletal connective tissues forms the endoskeleton of the vertebrate body. They protect various organs and help in locomotion. The supportive tissues include cartilage and bone.

Question 5.
Why should gametes be produced by meiosis during sexual reproduction?
Answer:
Meiosis is important as it produces gametes i.e., male or female germ cells. During meiosis, a germ cell or gamete divides to make four new sex cells. As a result of fertilization two gamates join together to form an egg or zygote. Therefore only if gametes are produced, fertilization can take place.

Question 6.
In which stage of mitosis the chromosomes align in an equatorial plate? How?
Answer:
Metaphase (meta – after) The duplicated chromosomes arrange on the equatorial plane and form the metaphase plate. Each chromosome gets attached to a spindle fibre by its centromere. The centromere of each chromosome divides into two each being associated with a chromatid.

VI. Answer in detail:

Question 1.
What are the permanent tissues? Describe the different types of simple permanent tissues.
Answer:
Permanent tissues:
Permanent tissues are those in which, growth has stopped either completely or for the time being. At times, they become meristematic partially or wholly.

Different types of simple permanent tissue :
Simple tissue: Simple tissue is homogeneous-composed of structurally and functionally similar cells. Eg : Parenchyma, collenchyma, and sclerenchyma.

Parenchyma:

• Parenchyma are simple permanent tissue composed of living cells.
• Parenchyma cells are thin-walled, oval, rounded, or polygonal in shape with well-developed spaces among them.
• In aquatic plants, parenchyma possesses intercellular air spaces and is named as aerenchyma.
• When exposed to light, parenchyma cells may develop chloroplasts and are known as chlorenchyma.

Functions:

• Parenchyma may store water in many succulent and xerophytic plants.
• It also serves the functions of storage of food reserves, absorption, buoyancy, secretion, etc

Collenchyma:

• Collenchyma is a living tissue found beneath the epidermis.
• Cells are elongated with unevenly thickened non-lignified walls. Cells have rectangular oblique or tapering ends and persistent protoplast.
• They possess thick primary non-lignified walls.
  Functions: They provide mechanical support for growing organs.

Sclerenchymà:

• Scierenchyma consists of thick-walled cells which are often lignified.
• Scierenchyma cells do not possess living protoplasts at maturity. Scierenchyma cells are grouped into

(i) fibers and (ii) sclereids.

Fibres : Elongated scierenchymatous cells, usually with pointed ends. Their walls are lignified. Fibres are abundantly found in many plants. Eg. Jute.
Sclereids:

• Sclereids are widely distributed in plant body. They are usually broad, may occur in single or in groups.
• Sclereids are isodiametric, with lignified walls. Pits are prominent and seen along the walls.
• Lumen is filled with wall materials. Sclereids are also common in fruits and seeds.

Question 2.
Write about the elements of Xylem.
Answer:
Xylem is a conducting tissue which conducts water, mineral nutrients upward from root to leaves. Xylem is also meant for mechanical support to the plant body. Xylem is composed of different kinds of elements. They are

1. xylem tracheids
2. xylem fibres
3. xylem vessels and
4. xylem parenchyma.

(i) Xylem tracheids: They are elongated or tube-like dead cells with hard, thick and lignified walls. Their ends are tapering, blunt or chisel-like. These cells are devoid of protoplast. They have large lumen without any content. Their function is conduction of water and providing mechanical support to the plant.

(ii) Xylem fibers: These cells are elongated, lignified and pointed at both the ends. Xylem fibres help in the conduction of water and nutrients from root to the leaf and also provide mechanical support to the plant.

(iii) Xylem vessels: They are long cylindrical, tube-like structures with lignified walls and wide central lumen. These cells are dead as these do not have protoplast. They are arranged in longitudinal series in which the partitioned walls (transverse walls) are perforated, and so the entire structure looks-like a water pipe. Their main function is the transport of water and minerals from root to leaf, and also to provide mechanical strength.

(iv) Xylem parenchyma: Its cells are living and thin-walled. The main function of xylem parenchyma is to store starch and fatty substances.

Question 3.
List out the differences< between mitosis and meipsis.
Answer:

Mitosis	Meiosis
1. Occurs in somatic cells.	Occurs in reproductive cells
2. Involved in growth and occurs continuously throughout life.	Involved in gamete formation only during the reproductively active age.
3. Consists of single division.	Consists of two divisions.
4. Two diploid daughter cells are formed.	Four haploid daughter cells are formed.
5. The chromosome number in the daughter cell is similar to the parent cell (2n).	The chromosome number in the daughter cell is just half (n) of the parent cell.
6. Identical daughter cells are formed.	Daughter cells are not similar to the parent cell and are randomly assorted.

VII. Higher Order Thinking Skills :

Question 1.
What is the consequence that occurs if all blood platelets are removed from the blood?
Answer:
Blood platelets play a major role in the clotting of blood whenever there is a wound/injury. If blood platelets are removed from the blood, clotting of blood will not occur. In case of any injury/surgery etc., blood will be lost from the body in excess and may even prove to be fatal.

2. Which are not true cells in the blood? Why?
Answer:
Red blood cells or erythrocytes cannot be considered as true cells since they have a nucleus only in the early stages. A mature RBC lacks a nucleus which is the controlling centre of all living cells.

Intext Activities

ACTIVITY – 1
Rinse your mouth with water. Using a toothpick or ice-cream stick, scrap superficial cells from the inner side of the cheek and spread it on a clean glass slide. Dry the glass slide with the scrap cells taken from the inner side of the cheek. Add two drops of methylene blue stain. Identify the cells under low and high power of the microscope.
Solution:
1. Large irregularly shaped cells with cell walls.
2. Dark blue nucleus at the central part of each cell.
3. Lightly stained cytoplasm colour in each cell.

9th Science Guide Organization of Tissues Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
A meristematic tissue consists of
(a) immature cells that are capable of undergoing cell division.
(b) mature cells
(c) non-living cells
(d) sclerenchyma cells
Answer:
(a) immature cells that are capable of undergoing cell division

Question 2.
Two long bones of the hand are dislocated in a person who met with an accident. Which among the following may be the possible reason?
(a) Tendon injury
(b) Break of skeletal muscle
‘(c) Ligament tear
(d) Rupture of Areolar tissue
Answer:
(c) Ligament tear

Question 3.
Non-striated muscles are found in
(a) blood vessels
(b) gastric glands
(c) urinary bladder
(d) all of these
Answer:
(d) all of these

Question 4.
Which of the following is not found in a neuron?
(a) Sarcolemma
(b) Dendrite
(c) Myelin sheath
(d) Axon
Answer:
(a) Sarcolemma

Question 5.
Cylindrical, unbranched multinucleated cells are
(a) striated muscle cells
(b) smooth muscles
(c) cardiac muscles
(d) none of the above.
Answer:
(a) striated muscle cells

Question 6.
The matrix of the bone is rich in
(a) elastin
(b) reticular fibres
(c) collagen
(d) myosin
Answer:
(c) collagen

Question 7.
Which muscles act involuntarily?
(i) Striated muscles
(ii) Smooth muscles
(iii) Cardiac muscles
(iv) Skeletal muscles
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d)(i) and (iv)
Answer:
(b) (ii) and (iii)

Question 8.
Tendon connects
(a) cartilage with muscles
(b) bone with skeletal muscles
(c) ligament with muscles
(d) bone with bone
Answer:
(b) bone with skeletal muscles

Question 9.
In a certain type of cell division, the diploid number of chromosomes is reduced to half. This kind of division occurs in
(a) testis
(b) ovary
(c) both ovary and testis
(d) all body cells
Answer:
(c) both ovary and testis

Question 10.
……………… is derived from the ground meristem.
(a) Cortex
(b) Epidermis
(c) Xylem
(d) Cambium
Answer:
(a) Cortex

Question 11.
The function of phloem fibres is …………….to the plant body
(a) passage of food
(b) Store food
(c) mechanical strength
(d) preparation of food
Answer:
(c) mechanical strength

Question 12.
The …………….. epithelium is also known as pavement membrane.
(a) Ciliated
(b) Squamous
(c) Cuboidal
(d) Glandular
Answer:
(b) Squamous

Question 13.
Elastic structures that connect bones to bones are called ………………..
(a) muscles
(b) tendons
(c) ligaments
(d) areolar tissue
Answer:
(c) ligaments

Question 14.
………………… is seen in unicellular animals.
(a) Mitosis
(b) meiosis
(e) Amitosis
(d) none of the above
Answer:
(c) Amitosis

Question 15.
The disappearance of spindle fibres is seen in …………….
(a) metaphase
(b) prophase
(c) anaphase
(d) telophase
Answer:
(d) telophase

Question 16.
The ………………… is a single, long fiber like process that develops from the cyton.
(a) dendron
(b) axon
(c) dendrite
(d) neurilemma
Answer:
(b) axon

Question 17.
Bouquet stage refers to ……………..
(a) diakinesis
(b) leptotene
(c) zygotene
(d) pachytene
Answer:
(b) leptotene

Fill in the blanks :

1. The ……………. tissues are made up of more than one type of cells and these woks together as a unit.
Answer:
complex

2. The two types of skeletal connective tissues are ……………. and …………….
Answer:
bone, cartilage

3. Humans have 46 chromosomes. Their sperms and eggs will have ……………. chromosomes each.
Answer:
23

4. During pairing of chromosomes in meiosis, the …………….. chromosomes come to lie side by side.
Answer:
homologous

5. The word meristem is derived from a Greek word ………………
Answer:
Meristos

6. Cork cambium is an example of ……………… meristem.
Answer:
secondary

7. The meristem found at the base of intemodes is called ……………….
Answer:
intercalary meristem

8. In apple, paranchyma stores ………………
Answer:
sugar

9. Fibres are extensively longer ranging from 20 mm to 550 mm ……………….
Answer:
corchorus capsularis (jute)

10. During mieosis in pachytene, stages the paired chromosomes are called ………………….
Answer:
bivalents

11. Mitosis was discovered by …………………..
Answer:
Flemming

12. Both smooth and cardiac muscles are …………….. in nature.
Answer:
involuntary

13. ………………. is a non-flexible skeletal connective tissue.
Answer:
Bone

14. ……………. acts as a fat reservoir.
Answer:
Adipose tissue

15. …………….. epithelium is seen in sweat glands.
Answer:
Cuboidal

16. Genetic variations occur in meiosis because of …………….
Answer:
crossing over

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

1. Epithelial layer does not allow regulation of materials between the body and the external environment.
Answer:
False.
Correct statement: Epithelium is involved in the absorption and elimination of waste.

2. Striated and non-striated tissues are types of epithelial tissues.
Answer:
False.
Correct statement: They are types of muscular tissues.

3. Spindle formation occur in amitosis.
Answer:
False.
Correct statement: Spindle formation occur in mitosis.

4. Movement of food in the alimentary canal is because of cardiac muscles.
Answer:
False.
Correct statement: Movement of food is alimentary canal by the rhythmic con t no-tom and relaxation of the muscular nails of the alimentary canal.
Correct statement: Movement of food in the alimentary canal.

5. A mature RBC lacks a nucleus.
Answer:
True

6. Excessive pulling of bones causes a sprain.
Answer:
False.
Correct statement: Sprain is caused by excessive pulling of ligaments.

7. Glandular epithelium gives a stratified appearance.
Answer:
False.
Correct statement: Compound epithelium given a stratified appearance.

8. Sieve cells have no companion cells.
Answer:
True.

9. Conduction can be bidirectional in phloem tissue.
Answer:
True.

10. White blood corpuscles contain respiratory pigment hemoglobin.
Answer:
False.
Correct statement: Red blood corpuscles contain respiratory pigment hemoglobin.

IV. Assertion and Reason type:

Direction: In each of the following questions, a statement of Assertion is given and a corresponding statement of Reason is given just below it. Of statements, given below, mark the correct answer as
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true that Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.

Question 1.
Assertion: Non-striated muscles are said to be voluntary in nature.
Reason: Non-striated muscles are under the control of our will.
Answer:
(d) Both Assertion and Reason are false

Question 2.
Assertion: Materials are exchanged between epithelial and connective tissues by diffusion.
Reason: Blood vessels are absent in epithelial tissue.
Answer:
(a) Assertion and Reason are true and Reason is the correct explanation of Assertion

V. Answer in one or two sentences :

Question 1.
Name the two types of sclerenchyma cells.
Answer:
Sclerenchyma cells are grouped into

1. fibres and
2. sclereids.

Question 2.
Name the components of xylem and phloem.
Answer:
Xylem is composed of :

• Xylem tracheids
• Xylem fibres
• Xylem vessels
• Xylem parenchyma

Phloem is composed of:

• Sieve elements
• Companion cells
• Phloem fibres
• Phloem parenchyma

Question 3.
Name the tissue that connects muscle to bone in humans.
Answer:
Tendons join skeletal muscles to bones in our body.

Question 4.
Name the tissue that stores excess fat in our body.
Answer:
Adipose tissue.

Question 5.
Name the connective tissue with a fluid matrix.
Answer:
Blood and lymph

Question 6.
Name the tissue present in the brain.
Answer:
Nervous tissue.

Question 7.
What is plate meristem?
Answer:
These cells divide into two planes resulting in an increase in the area of an organ.
Eg: Leaf formation.

Question 8.
Differentiate collenchyma and sclerenchyma.
Answer:

Collenchyma	Sclerenchyma
It consists of living cells	It consists of dead cells
Cells contain protoplasm	Cells do not possess living protoplast
Cell walls are non-lignified	Cell walls are lignified

Question 9.
Mention the type of epithelium seen in alveoli of lungs.,
Answer:
Squamous epithelium.

Question 10.
Name the supportive connective tissues.
Answer:
Cartilage and bone

Question 11.
Name the cartilage cells present in the matrix.
Answer:
Chondrocytes.

Question 12.
What is the role of RBC?
Answer:
RBC contains a respiratory pigment called hemoglobin which is involved in the transport of oxygen to tissues.

Question 13.
Mention the stages of meiotic Prophase -I.
Answer:
Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis.

Question 14.
What is the significance of Meiosis?
Answer:
The constant number of chromosomes in a given species is maintained by meiotic division.

Question 15.
Draw a shoot apex and label the meristem’s parts.
Answer:

VI. Short Answer Type :

Question 1.
How would you differentiate meristematic and permanent tissue?
Answer:

Meristematic tissue	Permanent tissue
Cytoplasm is dense, and vacuoles are nearly absent.	Usually large central vacuole present in living permanent cells.
Intercellular spaces absent.	Intercellular spaces present.
Component cells are small, spherical or polygonal and undifferentiated.	Component cells are large, differentiated with different shapes.
Cell wall is thin and elastic.	Cell wall is thick.
Nucleus is large and prominent.	Nucleus is less conspicuous.
Cells grow and divide regularly.	Cells do not normally divide.
Provides mechanical support and elasticity to the plant body.	Provides only mechanical support.

Question 2.
Differentiate fibres from sclereids.
Answer:

Sclereids	Fibres
Usually broad	Elongated narrow thread-like
End walls blunt	Usually with pointed ends.
Occur singly	Occur in bundles
Deep pits	Narrow pits

Question 3.
Which tissue is the main component of tendons and ligaments? How do they differ in function?
Answer:
Dense Connective Tissue is a fibrous connective tissue densely packed with fibres and fibroblasts. It is the principal component of tendons and ligaments.

a. Tendons: Cord-like, strong, structures that join skeletal muscles to bones. Tendons have great strength and limited flexibility. They consist of parallel bundles of collagen fibres, between which are present rows of fibroblasts.

b. Ligaments: They are highly elastic structures and have great strength which connects bones to bones. They contain very little matrix. They strengthen the joints and allow normal movement.

Question 4.
What are the fibres present in the connective tissue proper?
Answer:
Connective tissue proper: Connective tissue proper consists of collagen fibres, elastin fibres and fibroblast cells.

Areolar tissue: It has cells and fibres loosely arranged in a semi-fluid ground substance called matrix. It takes the form of fine threads crossing each other in every direction leaving small spaces called areolae. It joins skin to muscles, fills space inside organs and is found around muscles, blood vessels and nerves. It helps in repair of tissues after injury and fixes skin to underlying muscles.

Adipose Tissue: Adipose tissue is the aggregation of fat cells or adipocytes spherical or oval in shape. It serves as fat reservoir. The matrix consists of collagen fibres, elastin fibres and fibroblast cells.

Question 5.
How are collagen fibres organized in dense connective tissues?
Answer:

• Dense connective tissue is a fibrous connective tissue densely packed with fibres and fibroblasts. It is the principal component of tendon and ligaments.
• Tendons consist of parallel bundles of collagen fibres, between which are present rows of fibroblasts.
• Ligaments are highly elastic structures and contain very little matrix.

Question 6.
Write one point of difference between
a) Bone and cartilage.
b) Simple and compound epithelial tissue.
Answer:
a)

Bone	Cartilage
It is solid, rigid, and strong, non-flexible skeletal connective tissue.	It is a soft, semi-rigid, flexible skeletal connective tissue.
The matrix of the bone is in the form of concentric rings called lamellae	The matrix is composed of large cartilage cells called chondrocytes

b)

Simple epithelium tissue	Çompound epithelium tissue
It is composed of a single layer of cells resting ón a basement membrane.	It is composed of several layers of cells. Only the cells of the deepest layer rest on the basement membrane.

Question 7.
Why is blood considered to be a fluid connective tissue?
Answer:

• The blood and the lymph are the fluid connective tissues which link different parts of the body. The cells of the connective tissue are loosely spaced and are embedded in an intercellular matrix.
• Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes), and platelets. In this fluid connective tissue, blood cells move in a fluid matrix called plasma. The plasma contains inorganic salts and organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

Question 8.
Give the sequence of the events occurring during prophase of mitosis.
Answer:

• During this stage, chromosomes become short and thick and are clearly visible inside the nucleus.
• Centrosome splits into two daughter centrioles and occupies opposite poles of the cell.
• Each centriole is surrounded by aster rays. Spindle fibres appear between the two centrioles.
• The nuclear membrane and nucleolus disappear gradually.

Question 9.
Why is meiosis called reductional division and mitosis as equational division?
Answer:

• In mitosis one parent cell divides into two identical daughter cells, each with a nucleus having the same amount of DNA, same number of chromosomes and genes as the parent cells. It is therefore called an equational division.
• Meiosis is called reduction division because the chromosome number is reduced to haploid (n) from diploid (2n) in the daughter cells.

Question 10.
What is terminalization?
Answer:
In the stage of diplotene of meiotic prophase I, chiasmata begin to move along the length of the chromosome from the centromere towards the end resulting in terminalization.

Question 11.
What is a tetrad?
Answer:
The chromosomes are visible as long paired twisted threads. The pairs so formed are called bivalents. Each bivalent now contains four chromatids (tetrad stage) in pachytene of mieotic prophase I. The condition of bivalent containing four chromatids are called tetrad stage.

Question 12.
What is crossing over?
During pachytene of mieotic prophase I, the chromatids break and the broken segments are interchanged between homologous chromosomes. The points of exchange are the chiasmata. This is called .crossing over.

Question 13.
What is bouquet stage?
Answer:
During leptotene of mieotic prophase I, the chromosomes become uncoiled and assume long thread-like structures and take up a specific orientation inside the nucleus. They form a bouquet stage.

Question 14.
What is zygotene?
Answer:
It is one of the stages of mieoticphophase I. Two homologous chromosomes approach each other and begin to pair. The pairing of homologous chromosomes is called synapsis.

Question 15.
Explain amitosis.
Answer:
It is the simplest model of cell division and occurs in unicellular animals, aging cells and in foetal membranes. During amitosis, the nucleus elongates first, and a constriction appears in it which deepens and divides the nucleus into two, followed by this cytoplasm divides resulting in the formation of two daughter cells.

Question 16.
Write the salient features of the compound epithelium.
Answer:

• It consists of more than one layer of cells and gives a stratified appearance. Hence, they are also known as stratified epithelial cells.
• The main function of this epithelium is to give protection to the underlying tissues against mechanical and chemical stress.
• They also cover the dry surface of the skin, the moist surface of the buccal cavity, and the pharynx.

Question 17.
Write a note the significance of mitosis.
Answer:

• This equational division results in the production of diploid daughter cells (2n) with equal distribution of genetic material (DNA).
• In multicellular organisms growth, organ development and an increase in body size are accomplished through the process of mitosis.
• Mitosis helps in the repair of damaged and wounded tissues by the renewal of the lost cells.

Question 18.
Draw a neuron and label the parts.
Answer:

VII. Answer in detail:

Question 1.
What are meristems? Describe the distribution and functions of various types of meristems.
Answer:
Meristematic tissues are group of immature cells that are capable of undergoing cell division. In plants, the meristem is found in zones where growth can take place. Example: apex of stem, root, leaf primordia, vascular cambium, cork cambium, etc.,

Types of Meristems based on position:
On the basis of their position in the plant, meristems are of three types: Apical meristem, Intercalary meristem, and Lateral meristem.

Apical meristem: These are found at the apices or growing points of root and shoot and bring about an increase in length.

Intercalary meristem: It lies between the region of permanent tissues and is part of the primary meristem. It is found either at the base of leaf e.g. pinus or at the base of internodes e.g. grasses.

Lateral Meristem: These are arranged parallel and causes the thickness of the plant part.

Functions: Meristems are actively dividing tissues of the plant, that are responsible for primary (elongation) and secondary (thickness) growth of the plant.

Question 2.
Give one reason for the following:
a. Blood is fluid connective tissue.
b. Skeletal muscles are voluntary muscles.
c. Heart muscles are involuntary in nature.
Answer:
The blood and the lymph are the fluid connective tissues which link different parts of the body. The cells of the connective tissue are loosely spaced and are embedded in an intercellular matrix.

(a) Blood: Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes), and platelets. In this fluid connective tissue, blood cells move in a fluid matrix called plasma. The plasma contains inorganic salts and
organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

(b) They work under our control and are also known as voluntary muscles. They are not under the control of our will and so are called involuntary muscles.

(c) Cardiac muscle: It is a special contractile tissue present in the heart. The muscle fibers are cylindrical, branched, and uninucleate The branches join to form a network called as itercalated disc which are unique distinguishing features of the
cardiac muscles. The contraction of cardiac muscle is involuntary and rhythmic.

Question 3.
Explain simple epithelium and its types.
Answer:
Simple Epithelium :
1. It is formed of a single layer of cells. It forms a lining for the body cavities and ducts.
2. Simple epithelium is further divided into the following types.

• Squamous epithelium
• Cuboidal epithelium
• Columnar epithelium
• Ciliated epithelium
• Glandular epithelium

(i) Squamous epithelium :

• It is made up of thin, flat cells with prominent nuclei. These cells have irregular boundaries and bind with neighbouring cells.
• The squamous epithelium is also known as pavement membrane, which forms the delicate lining of the buccal cavity, alveoli of lungs, proximal tubule of kidneys, blood vessels etc.
• It protects the body from mechanical injury, drying and invasion of germs.
  

(ii) Cuboidal epithelium:

• It is composed of single layer of cubical cells.
• The nucleus is round and lies in the centre.
• This tissue is present in the thyroid vesicles, salivary glands, sweat glands, exocrine pancreas.
• It is also found in the intestine and tubular part of the nephron (kidney tubules) as microvilli that increase the absorptive surface area.
• Their main function is secretion and absorption.
  

(iii) Columnar epithelium:

• It is composed of a single layer of slender, elongated and pillar like cells.
• Their nuclei are located at the base.
• It is found lining the stomach, gall bladder, bile duct, small intestine, colon, oviducts and forms a mucous membrane. ‘
• They are mainly involved in secretion and absorption.
  

(iv) Ciliated epithelium :

• Certain columnar cells bear numerous delicate hair like out growths called ilia and are called ciliated epithelium.
• Their function is to move particles or mucus in a specific direction over the epithelium.
• It is seen in the trachea of wind-pipe, bronchioles of respiratory tract, kidney tubules and fallopian tubes of oviducts.
  

(v) Glandular epithelium :

• Epithelial cells are often modified to form specialized gland cells which secrete chemical substances at the epithelial surface.
• This lines the gastric glands, pancreatic tubules and intestinal glands.
  

Question 4.
Explain the components of phloem tissue.
Answer:
Phloem is a complex tissue and consists of the following elements :
(i) Sieve elements
(ii) Companion cells
(iii) Phloem fibres
(iv) Phloem parenchyma

(i) Sieve elements :

• The conducting elements of phloem are collectively called as Sieve elements.
• Sieve tubes are elongated, tube-like slender cells placed end to end. The transverse walls at the ends are perforated and are known as sieve plates.
• The main function of sieve tubes is translocation of food, from leaves to the storage organs of the plants.

(ii) Companion cells : These are elongated cells attached to the lateral wall of the sieve tubes. A companion cell may be equal in length to the accompanying sieve tube element or the mother cell may be divided transversely forming a series of companion cells.

(iii) Phloem parenchyma : The phloem parenchyma are living cells which have cytoplasm and nucleus. Their function is to store food materials.

(iv) Phloem fibers : Sclerenchymatous cells associated with primary and secondary phloem are commonly called phloem fibers. These cells are elongated, lignified and provide mechanical strength to the plant body.

Question 5.
Write a note on blood and its components.
Answer:
Blood is a fluid connective tissue.
Blood contains corpuscles which are red blood cells (erythrocytes), white blood cells (leucocytes) and platelets. In this fluid connective tissue, the blood cells move in a fluid matrix called. The plasma contains inorganic salts and organic substances. It is a main circulating fluid that helps in the transport of nutrient substances.

Red blood corpuscles (Erythrocytes):

• The red blood corpuscles are oval shaped, circular, biconcave disc-like and lack nucleus when mature (mammalian RBC).
• They contain a respiratory pigment called haemoglobin which is involved in the transport of oxygen to tissues.

White blood corpuscles (Leucocytes): They are larger in size, contain distinct nucleus and are colourless. They are capable of amoeboid movement and play an important ‘ role in body’s defense mechanism. WBC’s are of two types :
(i) Granulocytes.
(ii) Agranulocytes.

(i) Granulocytes have irregular shaped nuclei and cytoplasmic granules. They include the neutrophils, basophils and eosinophils. Agranulocytes lack cytoplasmic granules and include the lymphocytes and monocytes.

Blood platelets : They are minute, anucleated, fragile fragments of giant bone marrow called mega karyocytes They play an important role in blood clotting mechanism.

VIII. Higher Order Thinking Skills :

Question 1.
Identify the figure given below

(a) Label the parts A, B and C.
(b) What is the chemical composition of the tissue?
Answer:
(a) T.S. of Bone
(A) Lamellae
(B) Lacunae
(C) Central (Haversian canal)
(b) The matrix of the bone is rich in calcium salts and collagen fibres which gives the bone its strength.
(c) C – Haversian canal

Question 2.
Identify figures A and B.

(a) …………… epithelium forms the outer lining of the buccal cavity.
(b) ………………. epithelium consist of ceils that are tall and pillar-like.
(c) Which one allows diffusion of substances?
(d) Which is called pavement epithelium?
(e) Which epithelium lines the gastrointestinal tract and epiglottis?
Answer:
Figure A – Squamous Epithelium
Figure B. – Glandular epithelium
(a) Squamous
(b) Columnar
(c) Columnar epithelium
(d) Squamous epithelium
(e) Columnar epithelium

Question 3.
If cell (A) has undergone one mitotic division and another cell (B) has completed its meiotic division. The number of cells produced in A and B would be
Cell A: Cell B :
Answer:
Cell A : 2 daughter cells.
Cell B : 4 daughter cells.

Question 4.
Identify the stage of mitosis from the below picture. List the chromosomal events in this stage.

Answer:
Mitotic anaphase
(i) The centromeres attaching the two chromatids divide and the two daughter chromatids of each chromosome separate and migrate towards the two opposite poles.
(ii) The migration of the daughter chromosomes is achieved by the contraction of spindle fibres.

Question 5.
Identify the following relationship
Cuboidal : Epithelial
Cardiac : ………..
Granulocytes : …………
Osteocytes : ………….
Answer:
Cardiac : Muscular
Granulocytes : Blood cells
Osteocytes : Bone cells

Question 6.
Umbilical cord blood is collected at the time of child birth and stored in stem cell banks? Reason out.
Answer:

• Umbilical cord blood consists of stem cells, they are undifferentiated cells which undergo unlimited divisions and give rise to one or more different types of cells. – Embryonic stem cells differentiate into different tissues and organs.
• Stem cells can be used in the treatment of certain degenerative diseases in future.

Question 7.
How do WBC help in defence?
Answer:
They are capable of amoeboid movement and play an important role. They engulf or destroy foreign bodies.

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 17 Animal Kingdom Text Book Back Questions and Answers, Notes.

Samacheer Kalvi 9th Science Solutions Chapter 17 Animal Kingdom

9th Science Guide Animal Kingdom Text Book Back Questions and Answers

I. Choose the correct answer :

Question 1.
Find the group having only marine members.
(a) Mollusca
(b) Coelenterata
(c) Echinodermata
(d) Porifera
Answer :
(d) Echinodermata

Question 2.
Mesoglea is present in
(a) Porifera
(b) Coelenterata
(c) Annelida
(d) Arthropoda
Answer:
(b) Coelenterata

Question 3.
Which one of the following pairs is not a poikilothermic animal
(a) Fishes and Amphibians
(b) Amphibians and Aves
(c) Aves and Mammals
(d) Reptiles and Mammals
Answer:
(c) Aves and Mammals

Question 4
Identify the animal having four chambered heart.
(a) Lizard
(b) Snake
(c) Crocodile
(d) Calotes
Answer:
(c) Crocodile

Question 5
The animal without skull is
(a) Acrania
(b) Acephalia
(c) Apteria
(d) Acoelomate
Answer:
(a) Acrania

Question 6.
Hermaphrodite organisms are
(a) Hydra, Tape worm, Earthworm, Amphioxus
(b) Hydra, Tape worm, Earthworm, Ascidian
(c) Hydra, Tape worm, Earthworm, Balanoglossus
(d) Hydra, Tape worm, Ascaris, Earthworm
Answer:
b) Hydra, Tap worm, Earthworm, Ascidian

Question 7.
Poikilothermic organisms are
(a) Fish, Frog, Lizard, Man
(b) Fish, Frog, Lizard, Cow
(c) Fish, Frog, Lizard, Snake
(d) Fish, Frog, Lizard, Crow
Answer:
(c) Fish, Frog, Lizard, Snake

Question 8.
Air sacs and Pneumatic bones are seen in
(a) fish
(b) frog
(c) bird
(d) bat
Answer:
(c) Bird

Question 9.
Excretory organ of tape worm is
(a) flame cells
(b) nephridia
(c) body surface
(d) solenocytes
Answer:
(a) Flame cells

Question 10.
Water vascular system is found in
(a) Hydra
(b) Earth worm
(c) Star fish
(d) Ascaris
Answer:
(c) Starfish

II. Fill in the blanks :

1. The skeletal framework of Porifera is ……………..
Answer:
spicules

2. Ctenidia are respiratory organs in ……………….
Answer:
phylum mollusca

3. Skates are ………………. fishes.
Answer:
cartilaginous

4. The larvae of an amphibian is …………………..
Answer:
bilaterally symmetrical

5. ………………. are jawless vertebrates.
Answer:
Cyclostomes

6. …………………… is the unique characteristic feature of mammal.
Answer:
Placenta

7. Spiny anteater is an example for …………….. mammal.
Answer:
egg-laying

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

1. Canal system is seen in coelenterates.
Answer:
False.
Correct statement: Canal system is seen in porifera

2. Hermaphrodite animals have both male and female sex organs.
Answer:
True,

3. Trachea are the respiratory organ of Annelida.
Answer:
False.
Correct statement: Trachea are the respiratory organ of Arthopoda.

4. Bipinnaria is the larva of Mollusca.
Answer:
False.
Correct statement: Bipinnaria is the larva of star fish.

5. Balanoglossus is a ciliary feeder.
Answer:
True.

6. Fishes have two chambered heart.
Answer:
True.

7. Skin of reptilians are smooth and moist.
Answer:
False.
Correct statement: Skin of cyclostomata are smooth and moist.

8. Wings of birds are the modified forelimbs.
Answer:
True

9. Female mammals have mammary glands.
Answer:
True.

IV. Match the following :

PHYLUM	EXAMPLES
(A) Coelenterata	(i) Snail
(B) Platyhelminthes	(ii) Starfish
(C) Echinodermata	(iii) Tapeworm
(D) Mollusca	(iv) Hydra

Answer:
A – iv, B – iii, C – ii, D – i

V. Answer very briefly :

Question 1.
Define taxonomy.
Answer:
The theoretical study of classification includes its basic principles, procedures, and rules.

Question 2.
What is nematocyst?
Answer:
The tentacles of organisms belonging to phylum Coelenterata bear stinging cells called cnidoblast or nematocyst.

Question 3.
Why coelenterates are called diploblastic animals?
Answer:
The animals in phylum coelenterates have two layers the outer ectoderm and the inner endoderm in the body wall. So they are called diploblastic animals.

Question 4.
List the respiratory organs of amphibians.
Answer:
Respiration is through by gills, skin, buccopharynx and lungs.

Question 5.
How does locomotion take place in starfish?
Answer:
Locomotion is affected by tube feet.

Question 6.
Are jellyfish and starfish similar to fishes? If no justify the answer.
Answer:

• Jellyfish is a coelenterate. Their bodies are made of calcium carbonate.
• Starfish fish is an echinoderm. Their bodies are made of calcium carbonate.
• Catfish is a fish species.
• Jellyfish and starfish are invertebrates.
• Fishes are vertebrates.

Question 7.
Why are frogs said to be amphibians?
Answer:
The frogs have dual adaptation in land and aquatic environments. So they are called amphibians.

VI. Answer briefly:

Question 1.
Give an account on phylum Annelida.
Answer:

• These are bilaterally symmetrical, triploblastic, first true coelomate animals with an organ-system grade of organization.
• Body is externally divided into segments called metameres joined by ring-like structures called annuli.
• It is covered by a moist thin cuticle.
• Setae and parapodia are locomotor organs.
• Sexes may be separate or ünited (hermaphrodites).
  e.g- Nereis, Earthworm, Leech.

Question 2.
Differentiate between flatworms and roundworms?
Answer:

Flatworms	Roundworms
a) They belong to Phylum Platyhelminthes.	They belong to Phylum Aschelminthes.
b) They are mostly parasitic.	Exist as free-living soil forms or as parasites.
c) Mostly hermaphrodites.	Sexual dimorphism is seen.
d) They are acoelomate organisms.	They are pseudocoelomate organisms.
e) The alimentary canal is absent or simple. Eg: Tapeworm.	The alimentary canal is a straight tube. Eg: Roundworm.

Question3.
Outline the flow charts of Phylum Chordata.
Answer:
Classification of phylum Chordata:

Question 4.
List five characteristic features of fishes.
Answer:

• Fishes are poikilothermic, whose internal temperature varies, considerably.
• The body has a covering of scales.
• Body muscles are arranged into segments called myotomes.
• The body is differentiated into head, trunk and tail.
• Respiration is done by 5 to7 pairs of gills, which are covered by an operculum or sometimes maybe naked.

Question 5.
Comment on the aquatic and terrestrial habits of amphibians.
Answer:
Aquatic habits of amphibians:

• The larva of amphibians (tadpole) lives in water and breathes with gills.
• External fertilization occurs in frog with water as a medium of fertilization.
• The adult frog has webbed feet to swim in water.
• The skin is moist and glandular which helps in Respiration.

Terrestrial habits of amphibians:

• The adults live on land and breathe through the lungs. Bucco-pharynx also helps in Respiration.
• The forelimbs are short and help to hop on land.

Question 6.
How are the limbs of the birds adapted for avian life?
Answer:

1. Birds have a spindle-shaped body.
2. The forelimbs are modified as wings for aerial locomotion.
3. The air sacs present in the birds, make the bird lightweight.
4. The body is covered with feathers.

VII. Answer in detail :

Question 1.
Describe the characteristic features of different Prochordates.
Answer:
Prochordata :

• The urochordates are considered as the forerunner of vertebrates.
• Based on the nature of the notochord, protochordate is classified into subphylum Urochordata and subphylum Cephalochordata.
  Phylum Hemichordata :
  
• Hemichordates are marine organisms with soft, vermiform and unsegmented body.
• They are bilaterally symmetrical, coelomate animals with non-chordate and chordate features.
• They have gill slits but do not have notochord.
• They are ciliary feeders and mostly remain as tubiculous forms, e.g- Balanoglossus (Acorn worms).
  Subphylum Urochordata:
  
• Notochord is present only in the tail region of free-living larva, e.g. Ascidian
• Adults are sessile forms and mostly degenerate.
• The body is covered with a tunic or test.

Subphylum Cephalochordata :

• Cephalochordates are small fish like marine chordates with unpaired dorsal fins.
• The notochord extends throughout the entire length of the body. e.g. Amphioxus

Question 2.
Give an account on phylum Arthropoda.
Answer:

• Arthropoda is the largest phylum.
• The organisms have jointed legs.
• The body is segmented into head, thorax and abdomen.
• The exoskeleton is made up of chitin.
• The coelomic cavity is filled with haemolymph (blood).
• They do not have defined blood vessels. This is called open circulatory system.
• The insects shed the exoskeleton and this process is called moulting.
• Small Arthropods absorb oxygen through the body and larger aquatic species breathe through book gills.
• Land Arthropods breathe through a system of tiny body tubes called tracheae.
• Excretion occurs through malphigian tubules and through green glands in crabs and prawns. .
• Insects, spiders, crabs, shrimps, butterflies, millipedes, centipedes, and scorpions are some arthropods.

Intext activities

ACTIVITY – 1

Observation :

• Scorpion
• Arachnid
• Cockroach
• Crab
• Crustacean
  [End of the activity]

9th Science Guide Animal Kingdom Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
Which is not an insect?
(a) Housefly
(b) Bedbug
(c) Mosquito
(d) Spider
Answer:
d) Spider

Question 2.
Which is not a feature of chordates?
(a) Green glands
(b) Sweat glands
(c) Sebaceous gland
(d) Mammary gland
Answer:
(a) Green glands

Question 3.
Choose the correct terms related to Hemichordate
(a) Vermiform, unsegmented, bilaterally symmetrical, ciliary feeders
(b) Vermiform, segmented, triploblastic, ciliary feeders
(c) Vermiform, unsegmented, diploblastic, ciliary feeders
(d) Vermiform, unsegmented; triploblastic, filter feeders
Answer:
a) Vermiform, unsegmented, bilaterally symmetrical, ciliary feeders

Question 4.
The …………. can swim faster than a cheetah.
(a) Sail fish
(b) Catfish
(c) Jelly fish
(d) Sucker fish
Answer:
a) Sail fish

Question 5.
The biggest vertebrate animal is the ……………….
(a) Blue whale
(b) Dolphin
(c) Elephant
(d) Rhinoceros
Answer:
(a) Blue whale

Question 6.
Who introduced binomial nomenclature?
(a) Huxley
(b) Carolus Linnaeus
(c) Nicolaus Copernicus
(d) Charles Darwin
Answer:
(b) Carolus Linnaeus

Question 7.
What is the binomial name of tapeworm?
(a) Ascaris lumbricoides
(b) Taenia solium
(c) Lampito mauritii
(d) Hirudinaria granulosa
Answer:
(b) Taenia solium

Question 8.
Which is the state bird of Tamil Nadu?
(a) Peacock
(b) Parrot
(c) Emerald dove
(d) Eagle
Answer:
(c) Emerald dove

Question 9. ;
……………. is an example of bony fish.
(a) Mullets
(b) Sharks
(c) Skates
(d) Hagfish
Answer:
(a) Mullets

Question 10.
The notochord extends throughout the entire length of the body.
(a) Acorn worms
(b) Lamprey
(c) Ascidian
(d) Amphioxus
Answer:
(d) Amphioxus

Question 11.
Animals which possess notochord are called as …………………
(a) invertebrates
(b) non-chordates
(c) chordates
(d) all the above
Answer:
(c) chordates

Question 12.
What ¡s the name of the bone which is filled with air?
(a) Flat bones
(b) Irregular bones
(c) Short bones
(d) Pneumatic bones
Answer:
d) Pneumatic bones

Question 13.
_________ ¡s the unique characteristic feature of mammals.
(a) Placenta
(b) Egg
(c) Tail
(d) Ear
Answer:
(a) Placenta

Question 14.
________ have gill slits but do not have notochord.
(a) Phylum Hemichordata
(b) Phylum Echirodermata
(c) Phylum Mollusca
(d) Phylum Annelida
Answer:
Phylum Hemichordata

II. Fill In the blanks:

1. The binomial name of our National Bird is …………………………
Answer:
pavocristatus

2. In mammals testis are enclosed by ………………….
Answer:
scrotal sacs

3. In birds, the air sacs communicate with …………………
Answer:
lungs

4. The animals belonging to phylunm …………….. are said to be soft-bodied.
Answer:
mollusca

5. The name ………….. means “thousand legs”.
Answer:
millipede

6. The skeletal framework of phylum Porifera is made up of ……………..
Answer:
spicules

7. A true body cavity is located within ………………
Answer:
mesoderm

8………………. separates the digestive tract from the body wall.
Answer:
Coelom

9. ……………….. formed during embryonic development.
Answer:
Notochord

10. …………….. is the largest phylum of the animal kingdom.
Answer:
Arthropoda

11. ………………… is a fluid-filled body cavity.
Answer:
Coelom

12. The body cavity of the phylum Aschelminthes is …………………….
Answer:
pseudocoelom

13. The most common diseases caused by nematodes in human beings are …………and ……………
Answer:
elephantiasis, ascariasis

14. Cyclostmes are ………………….
Answer:
jawless vertebrates

15. Malphigian tubules also called as ……………….
Answer:
green glands

16. Locomotion of phylum Echinodermata is affected by .
Answer:
tube feet

III. Analogy:

Question 1.
Echinodermata: ________ : : Soft Bodies Animals: ________
Answer:
Spiny Skinned Animals; Mollusca.

Question2.
Birds: _______ : : _________ : Cold-blooded.
Answer:
Warm-blooded; Fishes.

IV. State whether true or false. If false write the correct statement:

1. Order is a number of related families having common characters are placed in an order.
Answer:
True.

2. Tissue organization can be seen in phylum Porifera.
Answer:
False.
Correct statement: Cellular grade of organization can be seen in phylum Porifera.

3. Aschelminthes is the largest phylum of the animal kingdom.
Answer:
False.
Correct statement: Arthropods is the largest phylum of the animal kingdom.

4. In the animals belonging to phylum Echinodermata, the body is divided into head, muscular food and visceral mass………….
Answer:
False.
Correct statement: In phylum mollusca, the body is divided into head, muscular food and visceral mass.

5. Setae and parapodia are locomotor organs of earthworms.
Answer:
True.

6. Mammals have four-chambered Hearts.
Answer:
True,

V. Match the following:

Question 1.

(A) Cartilaginous	(i) Mullets
(B) Bony fishes	(ii) Sharks
(C) Pneumatic bones	(iii) Reptiles
(D) Three chambered heart	(iv) Birds

Answer:
A – i, B – ii, C – iv, D – iii

VI. Assertion and Reason type :

Question 1.
Assertion (A) : The Arthropods have an open circulatory system. .
Reason (R) : The cells and tissues are bathed in blood.
(a) A is right and R is wrong.
(b) A is the wrong R is right.
(c) A and R are wrong.
(d) A is right R explains A.
Answer:
(d) A is right R explains A

Question 2.
Assertion (A) : Mammals give birth to their young ones.
Reason (R) : Platypus is an egg-laying mammal.
(a) A and R are correct.
(b) A and R are wrong.
(c) R explains A.
(d) A is the wrong R is right.
Answer:
(a) A and R are correct

VII. Answer in one or two sentences :

Question 1.
Which organism is called Friends of farmers? Why?
Answer:
Earthworms are called ‘Friends of the farmers’ because after digesting organic matter, earthworms excrete a nutrient-rich waste product called castings which are used as Vermicompost.

Question 2.
What are the sub-phylum of prochordates?
Answer:
Sub phylum of prochordates are
Subphylum Cephalochordata
Subphylum Urochordata

Question 3.
Name the three germ layers.,
Answer:
Ectoderm, mesoderm and endoderm

Question 4.
Name the locomotary organs seen in phylum protozoa.
Answer:
Pseudopodia, cilia, or flagella.

Question 5.
Mention two special features of mammals.
Answer:
(i) External ear or pinna is seen in most the mammals.
(ii) Presence of the placenta is a unique characteristic feature.

Question 6.
Write the binomials of a) Man b) Tapeworm
Answer:
a) Homo sapiens
b) Taenia solium.

Question 7.
What is the binomial name?
Answer:
Carolus Linnaeus introduced the method of naming the animals with two names. This is known as binomial name and is followed universally.
Eg: Binomial of frog is

VIII. Short answer questions :

Question 1.
How is the body wall of coelenterates arranged?
Answer:
Body wall of coelenterate is disloblastic with two layers, namely the outer ectoderm and the inner endoderm, are separated by non-cellular jelly-like substance called mesoglea. Due to the presence of two layers in the body wall, they are said to be diploblastic animals.

Question 2.
What is polymorphism?
Answer:
Many coelenterates exhibit polymorphism, which is the variation in the structure and function of the individuals of the same species. They reproduce both asexually and sexually, e.g. Hydra, Jellyfish.

Question 3.
What is the unique feature of chordates?
Answer:
Presence of notochord is the unique feature of chordates.

Question 4.
On the basis of position of notochord, classify the different prochordates. Justify your answer.
Answer:

Subphyla of prochordates	Position of notochord
b) Cephalochordata	The notochord is present only in the tail region of the free-living larva. e.g. Ascidian
c) Urochordata	The notochord extends throughout the entire length of the body. e.g. Amphioxus

Question 5.
List the integumentary glands of mammals.
Answer:
Integumentary glands are seen in mammals:

• Sweat glands
• Sebaceous glands
• Scent glands
• Mammary glands are modified
• Integumentary glands

Question 6.
How is the reproductive characters of mammals different from those of aves.

Mammals	Aves
(i) Mammals have paired sex organs, eggs are small with little or no yolk.	(i) They are oviparous and lay eggs.
(ii) They are viviparous and give birth to young one.	(ii) Formation of placenta is not seen. The embryo is nourished,by egg yolk. Egg has an abundant quantity of yolk.
(iii) Formation of placenta is a special feature. The tissue connects the foetus to maternal tissues.

Question 7.
Mention two characteristic features of phylum Echinodermata.
Answer:
They have a water vascular system which is a unique feature. Locomotion is affected by tube feet.

Question 8.
Mention two features seen in Aves to aid in flying.
Answer:

• The forelimbs are modified into wings for flight.
• Bones are filled with air (pneumatic bones), which reduces body weight.

Question 9.
Why is crocodile unique among reptiles?
Answer:
It has a four-chambered heart whereas the reptiles have a three-chambered heart.

Question 10
Why are eggs of reptiles covered with shell unlike eggs of amphibians?
Answer:
First-class of vertebrates adapted for terrestrial life and lay their eggs with tough outer Shell e.g Calotes, Lizard, Snake, Tortoise, Turtle. Hence the eggs are covered with shells to protect them. Whereas amphibians eggs are laid in water.

Question 11.
What are pneumatic bones?
Answer:
The bones are filled with air. So they are called pneumatic bones and are seen in aves.

IX. Long answer questions :

Question 1.
Write a note on mammals.
Answer:

• Mammals are -warm-blooded animals.
• The skin is covered with hair. It also bears sweat and sebaceous (oil) glands.
• The body is divisible into head, neck, trunk and tail.
• Females have mammary glands, which secrete milk for feeding the young ones.
• The external ear or pinnae is present.
• Heart is four-chambered and they breathe through lungs.
• Except egg-laying mammals (Platypus, and Spiny anteater), all other mammals give birth to their young ones (viviparous).
• Placenta is the unique characteristic feature of mammals, e.g Rat, Rabbit, Man.

Question 2.
Write a note on aves.
Answer:

• Birds are homeothermic (warm-blooded) animals with several adaptations to fly.
• The spindle or boat-shaped body is divisible into head, neck, trunk and tail.
• The body is covered with feathers. Forelimbs are modified into wings for flight.
• Hindlimbs are adapted for walking, perching or swimming.
• The respiration is through lungs, which have air sacs.
• Bones are filled with air (pneumatic bones), which reduces body weight.
• They lay large yolk laden eggs.
• They are covered by hard calcareous shell, e.g. Parrot, Crow, Eagle, Pigeon, Ostrich.

Question 3.
Outline the flow chart of invertebrate phyla.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 9th Science Guide Pdf Chapter 19 Plant Physiology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 19 Plant Physiology

9th Science Guide Plant Physiology Text Book Back Questions and Answers

I. Choose the correct answer :

Question 1.
The tropic movement that helps the climbing vines to find a suitable support is …………….
(a) phototropism
(c) thigmotropism
(b) geotropism
(d) chemotropism
Answer:
(c) thigmotropism

Question 2.
The chemical reaction occurs during photosynthesis is .
(a) CO₂ is reduced and water is oxidized
(b) water is reduced and CO₂ is oxidized
(c) both CO₂ and water are oxidized
(d) both CO₂ and water are produced
Answer:
(a) CO₂ is reduced and water is oxidized

Question 3.
The bending of root of a plant in response to water is called …………………….
(a) Thigmonasty
(b) Phototropism
(c) Hydrotropism
(d) Photonasty
Answer:
(c) Hydrotropism

Question 4.
A growing seedling is kept in the dark room. A burning candle is placed near it for a few days. The tip part of the seedling bends towards the burning candle. This is an example of ……………
(a) Chemotropism
(b) Geotropism
(c) Phototropism
(d) Thigmotropism
Answer:
(c) Phototropism

Question 5.
The root of the plant is …………………
i) positively phototropic but-negatively geotropic
ii) positively geotropic but negatively phototropic
iii) negatively phototropic but positively hydrotropic
iv) negatively hydrotropic but.positively phototropic
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) (ii) and (iii)

Question 6.
The non-directional movement of a plant part in response to temperature is called………………..
(a) Thermotropism
(b) Thermonasty
(c) Chemotropism
(d) Thigmonasty
Answer:
(b) Thermonasty

Question 7.
Chlorophyllin a leaf is required for …………………
(a) photosynthesis
(b) tropic movement
(c) transpiration
(d) nastic movement
Answer:
(a) photosynthesis

Question 8.
Transpiration takes place through ………………..
(a) fruit
(b) seed
(c) flower
(d) stomata
Answer:
(d) stomata

II. Fill in the blanks :

1. The shoot system grows upward in response to ………………..
Answer:
light

2. ………………..is positively hydrotropic as well as positively geotropic.
Answer:
root

3. The green pigment present in the plant is………………..
Answer:
chlorophyll

4. The solar tracking of sunflower in accordance with the path of sun is due to …………………
Answer:
phototropism:

5. The response of a plant part towards gravity is ………………..
Answer:
geotropism

6. Plants take in carbon dioxide for photosynthesis but need ……………….. for their living.
Answer:
oxygen

III. Match column A with column B

Column A	Column B
1. Roots growing downwards into soil.	Positive phototropism
2. Shoots growing towards the light.	Negative geotropism
3. Shoots growing upward.	Negative phototropism
4. Roots growing downwards away from light.	Positive geotropism

Answer:

Column A	Column B
1. Roots growing downwards into soil.	Positive geotropism
2. Shoots growing towards the light.	Positive phototropism
3. Shoots growing upward.	Negative geotropism
4. Roots growing downwards away from light.	Negative phototropism

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

1. The response of a part of plant to the chemical stimulus is called phototropism.
Answer:
False.
Correct statement : The response of a plant part to the chemical stimulus is called Itemoiropisin.

2. Shoot is positively phototropic and negatively geotropic.
Answer:
True.

3. When the weather is hot, water evaporates lesser which is due to opening of stomata.
Answer:
False.
Correct statement: When the weather is hot, water evaporates more but stomata begin to close to reduce evaporation.

4. Photosynthesis produces glucose and carbon dioxide.
Answer:
False.
Correct statement: Photosynthesis produces glucose and oxygen.

5. Photosynthesis is important in releasing oxygen to keep the atmosphere in balance.
Answer:
True

6. Plants lose water when the stomata on leaves are closed.
Answer:
False.
Correct statement: Plants lose water when the stomata on leaves are open.

V. Answer very briefly :

Question 1.
What is nastic movement?
Answer:
Non-directional response to the stimulus is called Nastic movements.

Question 2.
Name the plant part
a) Which bends in the direction of gravity but away from the light.
b) Which bends towards the light but away from the force of gravity.
Answer:
a) Root system
b) Shoot system.

Question 3.
Differentiate phototropism from photo nasty.
Answer:
Phototropism:

1. Movement of a plant part towards the light.
2. Eg: Shoot of a plant.

Photonasty:

1. Movement of a plant part is a response to light.
2. Eg: Moonflower, Taraxacum Officinale.

Question 4.
Photosynthesis converts energy X into energy Y.
a) What are X and Y?
b) Green plants are autotrophic in their mode of nutrition. Why?
Answer:
a) X → light energy,
Y → Chemical energy.
b) Green plants are autotrophic in their mode of nutrition because they prepare then- own food materials through photosynthesis.

Question 5.
Define transpiration.
Answer:
The loss of water in the form of water vapour from the aerial parts of the plant body is called Transpiration.

Question 6.
Name the cell that surrounds the stoma.
Answer:
Each stomata is surrounded by guard cells.

VI. Answer briefly :

Question 1.
Give the technical terms for the following :
a) Growth dependent movement in plants.
b) Growth independent movement in plants.
Answer:
a) Tropic movements
b) Nastic movements

Question 2.
Explain the movement seen in Pneumatophores of Avicennia.
Answer:

• Pneumato[hores are specialized roots that can involve in the respiration of plants,
• This type of roots intakes the gas through its lenticel. a small hole in their both.

Question 3.
Fill in the blanks:

Answer:

Question 4.
What is chlorophyll?
Answer:
Chlorophyll is a green pigment present in all green plants with is responsible for the absorption of light to provide energy for photosynthesis.

Question 5.
Name the part of plant which shows positive geotropism. Why?
Answer:
Root shows positive geotropism because of the unidirectional movement in response to gravity.

Question 6.
What is the difference between the movement of flowers in the sunflower plant and the closing of the leaves in the Mimosa pudica?
Answer:
Movement of flower in sunflower plant:
In sunflower plant the stem tip follows the path of the sun from dawn to dusk (East to est) and in night it moves from West to East. This is a growth movement and takes place in response to the stimulus ‘light’. It is an example of tropic movement in response to light and is called phototropism.

Closing of the leaves in the Mimosa pudica :
The closing of leaves in Mimosa pudica occurs in response to touch. It is not a growth movement and occurs independently of the direction of stimulus. This is nastic movement . and is called Thigmonasty.

Question 7.
Suppose you have a rose plant growing in a pot, how will you demonstrate transpiration in it?
Answer:
Tie a plastic bag over leaves of rose plant and place in sunlight. After a few hours, we see water condensing inside the plastic bag. This is due to loss of water in the form of water vapour, which condenses into water. This is due to transpiration.

Question 8.
Mention the differences between stomatal and lenticular transpiration.
Answer:
Stomatal transpiration :

1. Loss of water from plants through stomata.
2. 90-95% of transpiration in a plant takes place through stomata only.

Lenticular transpiration :

1. Loss of water from plants as vapour through the lenticels.
2. A very small percentage of water is lost by through plants lenticular transpiration.

Question 9.
To which directional stimuli do (a) roots respond (b) shoots respond?
Answer:
(a) Roots respond positively geostrophic and negatively phototrophic.
(b) Shoots respond negatively geostrophic and positively phototrophic.

VII. Answer in detail :

Question 1.
Differentiate between tropic and nastic movements
Answer:
Tropic movements

1. Unidirectional response to the stimulus.
2. Growth dependent movements.
3. More or less permanent and irreversible.
4. Found in all plants.
5. Slow action.

Nastic movements :

1. Non-directional response to the stimulus
2. Growth independent movements. –
3. Temporary and reversible.
4. Found only in a few specialized plants.
5. Immediate action.

Question 2.
How will you differentiate the different types of transpiration?
Answer:
There are three types of transpiration:
Stomatal transpiration: Loss of water from plants through stomata. It accounts for 90- 95% of the water transpired from leaves.
Cuticular transpiration: Loss of water in plants through the cuticle.
Lenticular transpiration: Loss of water from plants as vapour through the lenticels. The lenticels are tiny openings that protrude from the barks in woody stems and twigs as well as in other plant organs.
But transpiration is necessary for the following reasons.

1. It creates a pull in leaf and stem.
2. It creates an absorption force in roots.
3. It is necessary for a continuous supply of minerals.
4. It regulates the temperature of the plant.

VIII. Higher Order Thinking Skills:

Question 1.
There are 3 plants A, B and C. The flowers of A open their petals in bright light during the day but closes when it gets dark at night. On the other hand, the flowers f of plant B open their petals at night but closes during the day when there is a bright light. The leaves of plant C fold up and droop when touched with fingers or any other solid object.
(a) Name the phenomenon shown by the flowers of plants A and B.
(b) Name one plant each which behaves like the flowers of plant A and B.
(c) Name the phenomenon exhibited by the leaves of plant C.
Answer:
(a) Photonasty
(b) Plant A – Common Dandelion – Taraxacum officinale
Plant B – Moonflower – Ipomoea alba
(c) Nyctinasty

Question 2.
Imagine that student A studied the importance of certain factors in photosynthesis.
He took a potted plant and kept it in dark for 24 hours. In the early hours of the next
morning, he covered one of the leaves with dark paper in the centre only. Then he placed the plant in sunlight for a few hours and tested the leaf which was covered
with black paper for starch.
(a) What aspect of photosynthesis was being investigated?
(b) Why was the plant kept in the dark before the experiment?
(c) How will you prove that starch is present in the leaves?
Answer:
(a) Necessity of light as a factor for Photosynthesis.
(b) To make the plant starch-free before starting the experiment. After the experiment, if starch is found in the leaf it will be the starch prepared by leaves during the experiment only.
(c) Presence of starch in the leaves can be proved by doing the starch test.

• The potted plant is kept in bright sunlight for 4 to 6 hours.
• The selected covered leaf is plucked and the black paper is removed.
• The leaf is immersed in boiling water for a few minutes and then in alcohol to remove chlorophyll.
• The leaf is now tested with iodine solution for the presence of starch.
• The covered part of the leaf does not turn blue-black whereas the uncovered part of the leaf turns blue-black colour.
• The covered part of the leaf which did not receive the sunlight was unable to synthesize starch.
• Hence it does not turn blue-black colour.
• But the uncovered part of the leaf which received sunlight was able to synthesize starch and so it turns blue-black in colour.

Intext Activities

ACTIVITY – 1

Take a glass trough and fill it with sand. Keep a flower pot containing water, plugged at the bottom at the centre of the glass trough. Place some soaked pea or bean seeds around the pot in the sand. What do you observe after 6 or 7 days? Record your observation.

Aim : To demonstrate hydrotropism.
Materials required : Glass trough, sand, flower pot, plugged at the bottom, pea or bean seeds and water.

Procedure :

1. A glass trough is taken and is filled with sand. A flowerpot containing water plugged at the bottom is kept at the centre of the glass trough.
2. Soaked pea or bean seeds are placed around the pot in the sand. It is observed after 6 or 7 days.

Observation: Observed that radicle has grown towards the pot and moisture instead of growing vertically downward.
Inference: It proves that primary root is positively hydrotropic.
Result: Hydrotropism has been demonstrated by showing positive hydrotropism in roots.

[End of the activity]

ACTIVITY – 2

Take pea seeds soaked in water overnight. Wait for the pea seeds to germinate. Once the seedling has grown put it in a box with an opening for light on one side. After few hours, you can clearly see how the stem has bent and grown towards the light.

Aim : To demonstrate phototropism.
Materials required : Box, water, light and pea seeds.

Procedure:

1. Take pea seeds soaked in water overnight. Wait for the pea seeds to germinate.
2. Once the seedling has grown put it in a box with an opening for light on one side.
3. After few hours, you can clearly see how the stem has bent and grown towards the light.

Observation : Observed that movement of a stem of a plant (pea seeds) towards light moist condition. This is the way a new plant develops from a seed.
Inference : Positive phototropism of a plant has been demonstrated.

[End of the activity]

ACTIVITY – 3

Pluck a variegated leaf from Coleus plant kept in sunlight. De- starch it by keeping in dark room for 24 hours. Draw the picture of this leaf and mark the patches of cholorphyll on the leaf. Immerse the leaf in boiling water followed by alcohol and test it for starch using iodine solution. Record your observation.

Aim : To show that chlorophyll is essential for photosynthesis.
Materials required : Plant with variegated leaves, boiling water, alcohol and iodine solution.

Procedure:
Variegated leaf is plucked from Coleus plant kept in sunlight. It isde-starched by keeping it in dark room for 24 hours. The picture of this leaf is drawn and the patches of cholorphyll on the leaf are marked. After immersing the leaf in boiling water followed by alcohol it is tested for starch with iodine solution.

Observation : The patches of the leaf with chlorophyll turn blue-black. The other portions remain colourless.
Inference: The chlorophyll is essential for photosynthesis.

ACTIVITY – 4

Place a potted plant in a dark room for about 2 days to de-starch its leaves. Cover one of its leaves with the thin strip of black paper as shown in the picture, make sure that the leaf is covered on both sides. Keep the potted plant in bright sunlight for 4 to 6 hours. Pluck the selected covered leaf and remove the black paper.
Immerse the leaf in boiling water for a few minutes and then in alcohol to remove chlorophyll. Test the leaf now with iodine solution for the presence of starch. The covered part of the leaf does not turn blue-black whereas the uncovered part of the leaf turns blue-black colour. Why are the changes in colour noted in the covered and uncovered part of the leaf?

Aim : To show that light is essential for photosynthesis.
Materials required : Covered leaf, boiling water, alcohol and iodine.

Procedure:

1. A potted plant is placed in a dark room for about 2 days to de-starch its leaves.
2. One of its leaves is covered with a thin strip of black paper as shown in the picture.
3. Make sure that the leaf is covered on both sides.
4. The potted plant is kept in bright sunlight for 4 to 6 hours.
5. The selected covered leaf is plucked and the black paper is removed.
6. Th e leaf is immersed in boiling water for a few minutes and then in alcohol to remove chlorophyll.
7. The leaf is now tested with iodine solution for the presence of starch.

Observation : The covered part of the leaf does not turn blue-black whereas the uncovered part of the leaf turns blue black colour. The covered part of the leaf which did not receive sunlight was unable to synthesize starch.
Inference : The light is essential for photosynthesis.

9th Science Guide Plant Physiology in Animals Additional Important Questions and Answers

I. Choose the correct answer :

Question 1.
The bending of root of a plant in response to water is called …………………..
(a) thigmonasty
(b) phototropism
(c) hydrotropism
(d) photonasty
Answer:
(c) hydrotropism

Question 2.
A growing seedling is kept in the dark room. A burning candle is placed near it for a few days. The top part of the seedling bends towards the burning candle. This is an example of ……………………
(a) chemotropism
(b) thigmotropism
(c) phototropism
(d) geotropism
Answer:
(c) phototropism

Question 3.
The root of the plant is ……………………
(i) positively phototropic but negatively geotropic.
(ii) positively geotropic but negatively phototropic.
(iii) negatively phototropic but positively hydrotropic.
(iv) negatively hydrotropic but positively phototropic.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) (ii) and (iii)

Question 4.
The plant part which exhibits negative geotropism is ……………………
(a) root
(b) stem
(c) branch
(d) leaves
Answer:
(b) stem

Question 5.
The non-directional movement of a plant part in response to temperature is called…………………..
(a) thermotropism
(b) thermonasty
(c) chemotropism
(d) thigmonasty
Answer:
(b) thermonasty

Question 6.
Sunflowers open the petals in bright light during the day time but close the petals in dark at night. This response of sunflowers is called ……………………
(a) geonasty
(b) thigmonasty
(c) chemonasty
(d) photonasty
Answer:
(d) photonasty

Question 7.
During photosynthesis plants exhale ……………………
(a) Carbon dioxide
(b) oxygen
(c) hydrogen
(d) helium
Answer:
(b) oxygen

Question 8.
Chlorophyll in a leaf is required for ……………………
(a) photosynthesis
(b) transpiration
(c) tropic movement
(d) nastic movement
Answer:
(a) photosynthesis

Question 9.
A plant is kept in a dark room for about 24 hours before conducting any experiment on photosynthesis in order to ……………………
(a) remove chlorophyll from the leaf
(b) remove starch from the leaves
(c) ensure that photosynthesis occurred
(d) to prove transpiration
Answer:
(b) remove starch from the leaves

Question 10.
Transpiration takes place through ……………………
(a) fruit
(b) seed
(c) flower
(d) stomata
Answer:
(d) stomata

Question 11.
Thigmonasty can be seen in ……………………
(a) Mimosa pudica
(b) Taraxacum officinale
(c) Ipomea alba
(d) Rhizophora
Answer:
(a) Mimosa pudica

Question 12.
The venus flytrap plants are examples for …………………..
(a) chemotropism
(b) thigmonasty
(c) hydrotropism
(d) thigmotropism
Answer:
(b) thigmonasty

Question 13.
………………….. flowers bloom as the temperature increases.
(a) Rose
(b) Jasmine
(c) Lilly
(d) Tulip
Answer:
(d) Tulip

Question 14.
Only green plants have the unique capacity to release ………………….. into the atmosphere.
(a) nitrogen
(b) oxygen
(c) carbon dioxide
(d) ozone
Answer:
(b) oxygen

Question 15.
The end product of photosynthesis is ……………………
(a) glucose
(b) sucrose
(c) fructose
(d) none of the above
Answer:
(a) glucose

II. Fill in the blanks :

1. The minerals like nitrogen, potassium and phosphorus, are required in substantial quantity by the plants are called …………………..
Answer:
macronutrients

2. The solar tracking of sunflower in accordance with the path of sun is due to ………………………
Answer:
phototropism

3. The response of a plant part towards gravity is ………………………
Answer:
geotropism

4. When the leaves of a sensitive plant are touched with a finger, they fold up and when light fades at dusk the petals of a sunflower. These two plants show ……………………..and ……………………..movements.
Answer:
thigmonastic, photonastic

5. Opening and closing of Moon flower is not a tropism because the movement in this is……………………..
Answer:
independent of stimulus

6. The raw materials for photosynthesis are …………………….. and ………………………
Answer:
carbon dioxide, water

7. When iodine solution is added for testing starch, part of the leaf with …………………….. turn blue- black colour.
Answer:
starch

8. In leaves, the food is stored in the form of……………………..
Answer:
starch

9. Plants may inhale carbon dioxide for photosynthesis but need ……………………..for their living.
Answer:
oxygen

10. Plants utilize only ……………………..% of the absorbed water for photosynthesis and the other activities.
Answer:
1

11 . Plants inhale and exhale continuously through the . ……………………..
Answer:
stamata

12. The bending of shoot towards light is due to the hormone ……………………..
Answer:
auxin

13. Heliotropism is a kind of ……………………..
Answer:
phototropism

14. The ……………………..plant exhibits one of the fastest known nastic movements.
Answer:
venus flytrap

15. Chlorophyll differs from haemoglobin by the presence of its central molecule ……………………..
Answer:
magnesium

16. Plants exchange gases ……………………..continuously through these stomata.
Answer:
CO₂ to O₂

17. Guard cells help in regulating the ……………………..
Answer:
rate of transpiration

III. Analogy :

Question 1.
Towards a stimulus : ……………….:: Away from the stimulus : Negative tropism
Answer:
Positive tropism

Question 2.
Hydrotropism : Response towards water :: Phototropism : …………..
Answer:
Response towards light

Question 3.
Photosynthesis : ……………….. :: Transpiration : Stomata
Answer:
Green leaves

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

1. Scientific term used to represent the bending of roots towards water is called geotropism.
Answer:
False.
Correct statement: Scientific term used to represent the bending of roots towards water is called hydrotropism.

2. When the leaves of Mimosa pudica plant are touched with the finger, they fold up quickly.
Answer:
True.
Correct statement: This is an example of thigmonasty.

3. The petals of moon flower open up in morning and closes in the evening. This is called photonasty.
Answer:
False.
Correct statement :The petals of moon flower dost ‘ in the morning and opens up in the evening. This is called photonasty.

4. Photosynthesis produces glucose and carbon dioxide.
Answer:
False.
Correct statement: Photosynthesis produces glucose and oxygen.

5. Photosynthesis is important in releasing oxygen to keep the atmosphere in balance.
Answer:
True.

6. Plants lose water when the stomata on leaves are closed.
Answer:
False.
Correct statement: Plants will not lose water when the stomata on leaves are closed.

V. Assertion and reason :

Question 1.
Assertion (A) : If the plant part moves in the direction of gravity, it is called positive geotropism.
Reason (R) : Stem shows positive geotropism.
(a) A and R are incorrect
(b) A is incorrect, R is correct
(c) A is correct, R is incorrect
(d) Both A and R are correct
Answer:
(c) A is correct, R is incorrect

Question 2.
Assertion (A) : The loss of excess water from the aerial parts of the plant in the form of water vapour is known as transpiration.
Reason (R) : Stomata of the leaf perform transpiration.
(a) A and R are incorrect
(b) A is incorrect, R is correct
(c) A is correct, R is incorrect
(d) Both A and R are correct
Answer:
(d) Both A and R are correct

VI. Short answer questions :

Question 1.
Write the scientific terms used to represent the following:
(a) Growing of roots towards the gravity.
(b) Bending of roots towards the water.

Answer:
(a) Positive geotropism
(b) Positive hydrotropism.

Question 2.
Observe the given picture.
(a) Identify this plant. What type of special movement is shown by this plant?
(b) What are the other movements seen in this plant?
Answer:
(a) Mimosa pudica.
Special movement shown by the plant: Just a casual touch is enough to make the leaves of Mimosa pudica (Touch-me- not plant) fold up and droop. This is described as seismonasty or thigmonasty.
Movements which are not directed towards stimuli but are independent of the stimulus direction are called as nastie movements. They may or may not be growth movement.

(b) The foliage of Mimosa closes during darkness and reopens in the presence of light. Note : The ‘rapid movement of in leaflets in Mimosa is rare in the plant kingdom and is related to changes in turgor pressure.

Question 3.
What is the end product of photosynthesis?
Answer:
The end product of photosynthesis is glucose which is converted into starch and stored in the plant parts.

Question 4.
Name the minute openings seen on the lower surface of the leaf.
Answer:
Stomata are the minute openings seen on the lower surface of the leaf.

Question 5.
Give the scientific terms for the following:
(a) Growth dependent movement in plants.
(b) Growth independent movement in plants.
Answer:
(a) Tropic movements
(b) Nastie movements.

Question 6.
Study the pictures below and then complete the table by putting a plus (+) if the shoot or root grows towards the stimulus and a minus (-) if it grows away from it.


Answer:

Question 7.
What is the other name for thigmonasty?
Answer:
Seismonasty.

Question 8.
Which flowering plant shows photonasty just opposite to that of Dandelion?
Answer:
Moon flower.

Question 9.
Give an example for negative hydrotropism.
Answer:
Growth of certain fungal hyphae away from the source of water/moisture.
Note : No clear evidences of negative hydrotropism.

Question 10.
Which gas is evolved during photosynthesis?
Answer:
Oxygen.

Question 11.
Give an example for micronutrients.
Answer:
Zinc.

Question 12.
What does a Mimosapudica plant do in response to touch? What is the phenomenon known as?
Answer:
In response to touch, the leaves of the plant Mimosa pudica fold up and droop. The movement is known as seismonasty or thigmonasty.

Question 13,
(i) What happens to the dandelion flower
(a) During the daytime?
(b) At night?
(ii) What is the phenomenon known as?
Answer:
(i) (a) The dandelion flower unfolds.
(b) The dandelion flower closes.
(ii) The phenomenon is known as photonasty.

Question 14.
Define photosynthesis.
Answer:
‘Photo’ means Tight’ and ‘synthesis’ means ‘to build’ thus photosynthesis literally means ‘building up with the help of light’. During this process, the light energy is converted into chemical energy. Green plants are autotrophic in their mode of nutrition because they prepare their food materials through a process called photosynthesis.\

Question 15.
Draw the structure of stomata and label the parts.
Answer:

Question 16.
Complete the following table with the different types of tropism:

Answer:

Question 17.
Label the diagram with the raw materials and products of photosynthesis.

Answer:
A. Carbon dioxide
B. Water
C. Glucose
D. Oxygen

Question 18.
Give an example for the movement of plant part which is very quick and can be observed easily.
Answer:
Just a casual touch is enough to make the leaves of Mimosa pudica fold up and droop. This takes place quickly and can be easily observed.

Question 19.
Name the cell that surrounds the stoma.
Answer:
Guard cells.

Question 20.
Give an example for chemotropism.
Answer:
During fertilization, pollen tube grows down the style in response to the sugars in the style is an example of chemotropism.

Question 21.
Give the overall equation for photosynthesis.
Answer:

Question 22.
Guard cells of stomata are green but cannot photosynthesize? Give reason.
Answer:
The enzymes needed for photosynthetic reactions are absent and hence guard cells cannot photosynthesize. _ .

Question 23.
Why do roots of halophytes show negative geotropism.
Answer:
The roots turn 180° upright for respiration.

Question 24.
Why do stems show postive phototropism?
Answer:

• Stem growing up and towards light is more likely to get sunlight for photosynthesis and display its flowers prominently for pollinators to arrive.
• It also has better chance of spreading its seeds.

Question 25.
Mention four factors required for photosynthesis?
Answer:
Chlorophyll, Water, Carbon dioxide and Light.

Question 26.
Why is transpiration called as necessary evil?
Answer:
Water is lost by transpiration but still it is a necessary process since it gives the following advantages to the plant.

• Creates an absorption force in roots to suck more water form the soil.
• Regulates the temperature of the plant. Hence transpiration is said to be a necessary evil.

Question 27.
What are stomata?
Answer:
The leaves have tiny holes, t ailed slomaia, through which the exchange of air takes place. The leaves have tiny, microscopic pores called stomata. Water evaporates through these stomata. Each stomata is surrounded by guard cells. These guard cells help in regulating the rate of transpiration by opening and closing of stomata.

Question 28.
Write the scientific terms used to represent.
Answer:
Leaves fold up and droop in touch-me-not plant Mimosa pudica Ans. Thigmonasty or Seismonasty.
Example: Brunnichia ovata and Mimosa pudica.

VII. Long answer questions :

Question 1.
Write a note on Transpiration.
Answer:

• The loss of water in the form of wate r vapour from the aerial parts of the plant body is called as transpiration.
• The leaves have tiny, microscopic pores called stomata.
• Water evaporates through these stomata. Each stomata is surrounded by guard cells.
• These guard cells help in regulating the rate of transpiration by opening and closing of stomata.

Question 2.
Write the types of tropism.
Answer:
Types of Tropism :
Based on the nature of stimuli, tropism can be classifi ed as follows.
Phototropism : Movement of a plant part towards light, e.g. shoot of a plant.
Geotropism : Movement of a plant in response to gravity, e.g. root of a plant.
Hydrotropism : Movement of a plant or part of a plant towards water, e.g root of a plant.
Thigmotropism : Movement of a plant part due to touch, e.g. climbing vines.
Chemotropism : Movement of a part of plant in response to chemicals, e.g growth of a pollen tube in response to sugar present on the stigma.

Tropism is generally termed positive if growth is towards the signal and negative if it is away from the signal. Shoot of a plant moves towards the light, the roots move away. Th us the shoots are positively phototropic.

Usually shoot system of a plant is positively phototropic and negatively geotropic and root system is negatively phototropic and positively geotropic.

VIII. Higher Order Thinking Skills :

Question 1.
While conducting experiments to study the effects of various stimuli on the plants, it was observed that the roots of a plant X grow and bend towards two stimuli A and B but bend away from a third stimulus C. The stem of the plant X however bends ” away from stimulus A and B but bends towards the stimulus C. The stimulus B is known to act on the roots due to factors related with Earth. Keeping these points in mind, answer the following questions:
(a) What could be stimulus A ?
(b) Name the stimulus seen in B.
(c) What could be stimulus-C?
Answer:
(a) Stimulus A – Water
(b) Stimulus B – Gravity
(c) Stimulus C – Light

Question 2.
An organism A which cannot move from one place to another makes a simple food B from the substances C and D available in the environment. This food is made in the presence of green coloured substance E present in organs F in the presence of light energy in a process called G. Some of the simple food B also gets converted into a complex food H for storage purposes. This food gives blue-black colour with iodine solution?
(a) What is (i) organism A (ii) food B and food H?
(b) What are C and D?
(c) Name (i) green coloured substance E and organ F.
(d) What is the process G?
Answer:
(a) (i) Organism A refers to Green plants.
(ii) Food B – Glucose
Food H – Starch

(b) C – Carbon dioxide
D – Water

(c) Green coloured substance E – Chlorophyll
Organ F – Leaves

(d) G – Photosynthesis process.