Class 9

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x³ – x² – x + 1
Solution:
f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x⁴ -x³ +x² – x + 1
Solution:
f(- 1) = (- 1)⁴ – (- 1)³ + (- 1)² – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x⁴ + 2x³ + 2x² + x + 1
Solution:
f(- 1) = (-1)⁴ + 2 (- 1)³ + 2 (- 1)² + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x³ – x² – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)³ – (- 1)² – (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x³ + x² – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)³ + (- 1)² – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x³ + 3x² + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)³ + 3 (- 1)² + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x³ – 4x² + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 2³ – 4(2)² + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x³+ x² – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = \(x-\frac{2}{3}\) = x – a
∴ a = 2/3

v) f(x) = 4x³+ 20x²+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + \(\frac{3}{2}=\) = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
Given f(x) = x³ – 3x² – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 2³ – 3(2)² – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)³ – 3(- 3)²– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)³ – 3 (4)² – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 3³ – 6(3)² – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and \(\left(x-\frac{1}{2}\right)\) of px² + 5x + r, show that p = r.
Solution:
Let f(x) = px²+ 5x + r
As (x – 2) and \(\left(x-\frac{1}{2}\right)\) are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)² + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax⁴ + bx³ + cx² + dx + e
As (x – 1) is a factor of f(x) we have
x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x³ – 2x² – x + 2
Solution:
Let f(x) = x³ – 2x² – x + 2
By trial, we find f(l) = 1³ – 2(1)² – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x² – x – 2)
= (x – 1) [x² – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x³ – 3x² – 9x – 5
Solution:
Let f(x) = x³ – 3x² – 9x – 5By trial,
f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5)
But x²– 4x – 5 = x² – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x³ + 13x² + 32x + 20
Solution:
Let f(x) = x³ + 13x² + 32x + 20
Let f(- 1)
= (- 1)³ + 13 (- 1)² + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y³ + y² – y – 1
Let f(y) = y³ + y² – y – 1
f(1) = 1³+ 1²– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x² + 12x + 20)
But (x² + 12x + 20) = x²+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax² + bx + c and bx² + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax² + bx + c and g(x) = bx² + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)² + b(- 1) + c
= b(- 1)² + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x² – x – 6 and x² + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x² – x – 6 and
g(x) = x² + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a² – a – 6 = a² + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y³– 2y²– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y³– 2y² – 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y² + y – 6
= y² + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3

Question 1.
Find the remainder when
x³ + 3x² + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each
Solution:
Let f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is f (- 1)
f (- 1) = (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0

ii) \(x-\frac{1}{2}\)
f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is \(\mathrm{f}\left(\frac{1}{2}\right)\)

iii) x
Solution:
f(x) = x³ + 3x² + 3x + 1
The remainder is f(0)
∴ f(0) = 0³ + 3(0)² + 3(0) + 1 = 1

iv) x + π
Solution:
f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is f(- π)
f(- π) = (- π)³ + 3(-π)² + 3 (-π) + 1 .
= – π³ + 3π² – 3π + 1

v) 5 + 2x
f(x) = x³ + 3x² + 3x + 1
The remainder is \(\mathrm{f}\left(\frac{-5}{2}\right)\)

Question 2.
Find the remainder when x³ – px² + 6x – p is divided by x – p.
Solution:
Let f(x) = x³ – px² + 6x – p
(x – a) = x – p)
By Remainder theorem, the remainder is f(p)
∴ f(P) = P³ – P(P)² + 6p – p
= p³ – p³ + 5p = 5p

Question 3.
Find the remainder when 2x² – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.
Solution:
Let f(x) = 2x² – 3x + 5 and
x – a = 2x – 3 = x – \(\frac{3}{2}\)
By Remainder theorem f(x) when divided by (x – \(\frac{3}{2}\) ) leaves a remainder \(\mathrm{f}\left(\frac{3}{2}\right)\)

As the remainder is 5 we say that (2x – 3) is not a factor of f(x).

Question 4.
Find the remainder when 9x³ – 3x² + x – 5 is divided by x – \(\frac{2}{3}\)
Solution:
Let f(x) = 9x³ – 3x² + x – 5
x-a = x – \(\frac{2}{3}\)
Remainder theorem the remainder is \(\mathrm{f}\left(\frac{2}{3}\right)\)

Question 5.
If the polynomials 2x³ + ax² + 3x – 5 and x³ + x² – 4x + a leave the same remainder, when divided by x – 2, find the value of a.
Solution:
Let f(x) = 2x³ + ax² + 3x – 5
g(x) = x³ + x² – 4x + a
Given that f(x) and g(x) divided x – 2
give same remainder.
i e., f(2) = g(2)
By Remainder theorem.
But f(2) = 2(2)³ + a(2)² + 3(2) – 5
= 2 x 8 + 4a + 6 – 5
= 17 +4a
g(2) = 2³ + 2² – 4(2) + a .
= 8 + 4 – 8 + a = 4 + a
i.e., 4 + a = 17 + 4a
∴ a – 4a = 17 – 4
– 3a = 13
a = -13/3

Question 6.
If the polynomials x³ + ax² + 5 and x³ – 2x² + a are divided by (x + 2) leave the same remainder, find the value of a.
Solution:
Let f(x) = x³ + ax² + 5
g(x) = x³ – 2x² + a
Given that when f(x) and g(x) divided by (x + 2) leaves the same remainder.
i.e.,f(-2) = g(-2)
By Remainder theorem
f(- 2) = (- 2)³ + a(- 2)² + 5
= -8 + 4a + 5 = 4a – 3
g(- 2) = (- 2)³ – 2(- 2)² + a
= -8 – 8 + a = a – 16
By problem,
4a – 3 = a – 16
4a – a = – 16 + 3
⇒ 3a = – 13 ⇒ a = -13/3

Question 7.
Find the remainder when f(x) = x⁴ – 3x² + 4 is divided by g(x) = x – 2 and verify the result by actual division.
Solution:
Given f(x) = x⁴ – 3x² + 4
g(x) = x – 2
The remainder when f(x) is divided by g(x) is f(2).
f(2) = 2⁴ – 3(2)² + 4
= 16 – 12 + 4
= 8
Actual division

∴ The remainder either by Remainder theorem or by actual division is the same.

Question 8.
Find the remainder when p(x) = x³ – 6x² + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.
Solution:
Given p(x) = x³ – 6x² + 14x – 3
g(x) = 1 – 2x
By Remainder theorem when p(x) is divided by g(x) is p(1/2).

Question 9.
When a polynomial 2x³ + 3x² + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.
Solution:Let f(x) = 2x³ + 3x² + ax + b
The remainder when f(x) is divided by (x – 2) is 2.
i.e., f(2) = 2
⇒ 2(2)³ + 3(2)²+ a(2) + b = 2
⇒ 16 + 12 + 2a +b = 2
⇒ 2a + b = – 26 …………………..(1)

Also the remainder when f(x) is divided by (x + 2) is – 2.
i.e., f(- 2) = – 2
⇒ 2(- 2)³ + 3(- 2)² + a (- 2) + b = – 2
⇒ -16 + 12 – 2a + b = – 2
– 2a + b = 2 ………………..(2)
Solving (1) and (2),

b = – 12
and 2a – 12 = – 26
2a = -26+ 12
a = -14/2 = -7,
a = -7, b = – 12

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2

Question 1.
Find the value of the polynomial 4x² – 5x + 3, when
(i) x = 0
Solution:
The value at x = 0 is
4(0)² – 5(0) + 3
= 3

(ii) x = – 1
Solution:
The value at x = – 1 is
4 (- 1)² – 5 (- 1) + 3
= 4 + 5 + 3
= 12

iii) x = 2
Solution:
The value at x = 2 is
4(2)² – 5(2) + 3
= 16 -10 + 3
= 9

iv) x = \(\frac{1}{2}\)
Solution:

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials.
i) p(x) = x² – x + 1
Solution:
p(0) = 0² – 0 + 1 = 1
p(1) = 1² – 1 + 1 = 1
p(2) = 2² – 2 + 1 = 3

ii) P(y) = 2 + y + 2y² – y³
Solution:
p(0) = 2 + 0 + 2(0)² – 0³ = 2
p(1) = 2+ 1 + 2(1)² – 1³ = 4
p(2) = 2 + 2 + 2(2)² -2³ = 4 + 8- 8 = 4

iii) P(z) = z³
Solution:
p(0) = 0³ = 0
p(1) = 1³ = 1
p(2) = 2³ = 8

iv) p(t) = (t – 1)(t + 1) = t² – 1
Solution:
p(0) = (0 – 1) (0 + 1) = – 1
p(1) = t² – 1 = 1² – 1 = 0
p(2) = 2² – 1 = 4 – 1 = 3

v) p(x) = x² – 3x + 2
Solution:
p(0) = 0² – 3(0) + 2 = 2
p(1) = 1² – 3(1) + 2 = 1 – 3 + 2 = 0
p(2) = 2² – 3(2) + 2 = 4- 6 + 2 = 0

Question 3.
Verify whether the values of x given in each case are the zeroes of the polynomial or not ?
i) p(x) = 2x + 1; x = \(\frac{-1}{2}\)
Solution:
The value of p(x) at x = \(\frac{-1}{2}\) is
\(\mathrm{p}\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)+1\)
= -1 + 1 = 0
∴ x = \(\frac{-1}{2}\) is a zero of p(x).

(ii) p(x) = 5x – π ; x = \(\frac{-3}{2}\)
Solution:
The value of p(x) at x = \(\frac{-3}{2}\) is
\(\mathrm{p}\left(\frac{-3}{2}\right)=5\left(\frac{-3}{2}\right)-\pi=\frac{-15}{2}-\pi \neq 0\)
∴ x = \(\frac{-3}{2}\) is not a zero of p(x).

iii) p(x) = x² – 1; x = ±1
Solution:
The value of p(x) at x = 1 and – 1 is
p(1) = 1² – 1 = 0
p(-1) = (-1)² -1 = 0
∴ x = ±1 is a zero of p(x).

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2
Solution:
The value of p(x) at x = – 1 is
p(-1) = (-1 – 1) (-1 + 2)
=-2 x 1 =-2 ≠ 0
Hence x = – 1 is not a zero of p(x).
And the value of p(x) at x = – 2 is
p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0
Hence, x = – 2 is a zero of p(x).

v) p(y) = y²; y = o
Solution:
The value of p(y) at y = 0 is p(0) = 0² = 0
Hence y = 0 is a zero of p(y).

vi) p(x) = ax + b ; x = \(\frac{-\mathbf{b}}{\mathbf{a}}\)
Solution:
The value of p(x) at x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is
\(\mathrm{p}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)=\mathrm{a}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)+\mathrm{b}\)
= -b + b = 0
∴ x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is a zero of p(x).

vii) f(x) = 3x² – 1; x = \(\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
Solution:

viii) f(x) = 2x – 1; x = \(\frac{1}{2} ;-\frac{1}{2}\)
Solution:

Question 4.
Find the zero of the polynomial in each of the following cases.
i) f(x) = x + 2
Solution:
x + 2 = 0
x = – 2

ii) f(x) = x – 2
Solution:
x – 2 = 0
x = 2

iii) f(x) = 2x + 3
Solution:
2x + 3 = 0
2x = – 3
x = \(\frac{-3}{2}\)

iv) f(x) = 2x – 3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)

v) f(x) = x²
Solution:
x² = 0
x = 0

vi) f(x) = px, p ≠ 0
Solutin:
px = 0
x = 0

vii) f(x) = px + q; p ≠ 0; p, q are real numbers.
Solution:
px + q = 0
px = -q
x = \(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 5.
If 2 is a zero of the polynomial p(x) = 2x² – 3x + 7a, find the value of
a.
Solution:
Given that 2 is a zero of p(x) = 2x² – 3x + 7a
(i.e.) p(2) = 0
⇒ 2(2)² – 3(2) + 7a = 0
⇒ 8 – 6 + 7a = 0
⇒ 2 + 7a = 0
⇒ 7a = – 2
⇒ a = \(\frac{-2}{7}\)

Question 6.
If 0 and 1 are the zeroes of the polynomial f(x) = 2x³ – 3x² + ax + b, find the values of a and b.
Solution:
Given that f(0) = 0; f(1) = 0 and
f(x) = 2x³ – 3x² + ax + b
∴ f(0) = 2(0)³ – 3(0)² + a(0) + b
⇒ 0 = b
Also f(1) = 0
⇒ 2(1)³ – 3(1)² + a(1) + 0 = 0
⇒ 2 – 3 + a = 0 .
⇒ a = 1
Hence a = 1; b = 0

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.1

Question 1.
Find the degree of each of the polynomials given below,
i) x⁵ – x⁴ + 3
Solution:
Degree is 5.

ii) x² + x – 5
Solution:
Degree is 2.

iii) 5
Solution:
Degree is 0.

iv) 3x⁶ + 6y³ – 7
Solution:
Degree is 6.

v) 4 – y²
Solution:
Degree is 2.

vi) 5t – √3
Solution:
Degree is 1.

Question 2.
Which of the following expressions are polynomials in one variable and which are not ? Give reasons for your answer.
i) 3x² – 2x + 5
Solution:
Given expression is a polynomial in one variable.

ii) x² + √2
Solution:
Given expression is a polynomial in one variable.

iii) p² – 3p + q
Solution:
Given expression is not a polynomial in one variable. It involves two variables p and q.

iv) y + \(\frac{2}{\mathbf{y}}\)
Solution:
Given expression is not a polynomial. Since the second term contains the variable in its denominator.

v) \(5 \sqrt{x}+x \sqrt{5}\)
Solution:
Given expression is not a polynomial. Since the first term’s exponent is not an integer.

vi) x¹⁰⁰ + y¹⁰⁰
Solution:
Given expression has two variables. So it is not a polynomial in one variable.

Question 3.
Write the coefficient of x³ in each of the following.
i) x³ + x + 1
ii) 2 – x³+ x²
iii) \(\sqrt{2} x^{3}+5\)
iv) 2x³ + 5
v) \(\frac{\pi}{2} x^{3}+x\)
vi) \(-\frac{2}{3} x^{3}\)
vii) 2x² + 5
viii) 4
Solution:
i) x³ + x + 1 : co-efficient of x³ is 1.
ii) 2 – x³+ x² : co-efficient of x³ is – 1.
iii) \(\sqrt{2} x^{3}+5\) co-efficient of x³ is √2
iv) 2x³ + 5 : co-efficient of x³ is 2.
v) \(\frac{\pi}{2} x^{3}+x\) co-efficient of x³ is \(\frac{\pi}{2}\)
vi) \(-\frac{2}{3} x^{3}\) co-efficient of x³ is \(-\frac{2}{3}\)
vii) 2x² + 5 : co-efficient of x³ is ‘0’.
viii) 4 : co-efficient of x³ is ‘0’.

Question 4.
Classify the following as linear, quadratic and cubic polynomials.
i) 5x²+ x – 7 : degree 2 hence quadratic polynomial.
ii) x – x³ , : degree 3 hence cubic polynomial.
iii) x² + x + 4 : degree 2 hence quadratic polynomial.
iv) x – 1 : degree 1 hence linear polynomial.
v) 3p : degree 1 hence linear polynomial.
vi) πr² : degree 2 hence quadratic polynomial.

Question 5.
Write whether the following statements are True or False. Justify your answer.
i) A binomial can have at the most two terms
ii) Every polynomial is a binomial
iii) A binomial may have degree 3
iv) Degree of zero polynomial is zero
v) The degree of x² + 2xy + y² is 2
vi) πr² is monomial
Solution :
i) A binomial can have at the most two terms -True
ii) Every polynomial is a binomial – False
[∵ A polynomial can have more than two terms]
iii) A binomial may have degree 3 – True
iv) Degree of zero polynomial is zero – False
v) The degree of x² + 2xy + y² is 2 – True
vi) πr² is monomial – True

Question 6.
Give one example each of a monomial and trinomial of degree 10.
Solution :
– 7x¹⁰ is a monomial of degree 10.
3x²y⁸ + 7xy – 8 is a trinomial of degree 10.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 4 Movement of Materials Across the Cell Membrane Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 4th Lesson Movement of Materials Across the Cell Membrane

9th Class Biology 4th Lesson Movement of Materials Across the Cell Membrane Textbook Questions and Answers

Improve Your Learning

Question 1.
The structure which controls the entry and exit of the materials through the cell is
A) Cell wall
B) Cell membrane
C) Both
D) None of them
Answer:
Cell membrane.

Question 2.
Fill in the blanks.
a) The smell of flowers reaches us through the process of …………………..
Answer:
Diffusion

b) The MIC gas of Bhopal tragedy was spread throughout the city through the process of …………………
Answer:
Diffusion

c) Water enters the potato osmometer due to a process called ………………
Answer:
Osmosis

d) The fresh grape wrinkles, if kept in salt water because of …………………
Answer:
Osmosis

Question 3.
What do you mean by permeability of membrane? Explain with suitable example.
Answer:
Allowing only certain materials to pass through the membrane is called permeability.

Example :

1. The cell membrane is very much permeable to gases such as carbondioxide, oxygen, nitrogen and fat solvent compounds such as alcohol, ether and chloroform.
2. It is impermeable to polysaccharides, phospholipids and proteins.

Question 4.
If the dried vegetables are kept in water they become fresh. What is the reason?
Answer:

1. The dried vegetables have less water content and high salt concentration in cells.
2. When they are kept in water they absorb water and become fresh.
3. The water enter into the vegetables by a process known as osmosis.

Question 5.
Name the process by which we can get fresh water from sea water.
Answer:
Reverse Osmosis.

Question 6.
What will happen to a marine fish if kept in fresh water aquarium? Support your answer with reasons.
Answer:
The marine fish dies.
Reasons:

1. Usually marine fishes have high concentration of salts in their body.
2. When they are kept in fresh water, the water from the fresh water aquarium enters the body of fishes due to osmosis.
3. More amount of fresh water enters the cells of fish. This results in bursting of cells and fish dies.

Question 7.
Why do the doctors administer saline (salt solution) only, but not the distilled water?
Answer:

• Distilled water causes cells to lyse, so injecting distilled water into a vein will cause some degree of haemolysis.
• Haemolysis is the rupture of red blood cells.
• Large amount of distilled water would cause much more damage not just limited to haemolysis and also cause brain damage or cardiac arrest and death.
• That is why fluids are administered to patients as saline (which include appropriate amount of salt)

Question 8.
What will happen if 50% glucose solution (dextrose) is injected intravenously (into vein)?
Answer:

• 50% glucose solution (dextrose) is used for reduction of increased cerebrospinal pressure and cerebral edema.
• If 50% glucose solution is injected intravenously it may produce allergic reactions in sensitive persons.
• The allergic reactions include nervous excitement infection at the joint site, tissues necrosis, venous thrombosis extending from the site of injection etc.
• Hence concentrated dextrose (glucose) should be administered via central vein only after suitable dilution.

Question 9.
What will happen if cells do not have ability of permeability?
Answer:

1. If the cells do not have ability of permeability they would not be able to carryout any of their fundamental life functions.
2. Oxygen, glucose, fats, proteins and vitamins are needed by cells to perform life process.
3. Mature cells become impermeable to any molecules or atoms it would die of toxicity and it would not be able to remove its wastes.

Question 10.
Draw the flow chart showing different stages in doing the experiment with egg.
Answer:

Question 11.
You have purchased a coconut in the market. By shaking it you found there is less water in coconut. Can you fill the coconut with water without making a hole to the coconut?
Answer:

• No, it is not possible to fill the coconut with water without making a hole.
• The husk of coconut is mostly made up of sclerenchymatous cells which are dead.
• Osmosis do not takes place in dead cells.
• It is not possible to fill the coconut with water without making a hole.

Question 12.
What are your observations in experiments to know about diffusion?
Answer:
Observations in experiments to know about diffusion are :

1. Materials kept in medium (water/air) get dissolves in the medium.
2. These dissolved molecules gradually move randomly in all directions. (from center to periphery)
3. They move from higher concentration to lower concentration.
4. This movement occures till these molecules spread equally throughout the medium.

Question 13.
Discuss with your friends and write the list of incidences where diffusion occurs.
Answer:

• A sugar cube in a glass of milk/water diffuses throughout it and make it sweet.
• The smell of cookies diffuses through the house as they bake.
• Tea leaf pigments diffuse through the tea bag into the water to give it colour and taste.
• Air freshner/deodorent molecules diffuse into the air when put on so we can smell it.
• If the cooking gas is leaked it spreads all over the house through diffusion.
• CO₂ bubbles in soft drink diffuses out of soda leaving the soda flat.
• Robbin Blue drops diffuses in water, making the water blue.
• Agarbatti, mosquito repellents work on the principle of diffusion.

Question 14.
How diffusion is useful in everyday life?
Answer:

• A wilted carrot made firm again by soaking in water.
• Cigarette smoke. It diffuses into air and spreads through the room.
• A sugar cube in a glass of water that is not stirred will dissolve slowly and the sugar molecules will distribute over the water by diffusion.
• The smell of cookies diffuses through the house as they bake.
• Tea leaf pigments diffuse through the tea bag to give the water its colour and taste of tea.
• Air freshner / deodorant molecules diffuse into the air when put on. So we can smell it.
• If the cooking gas is leaked, it spreads all over the house through diffusion.
• CO₂ bubbles in soft drink diffuses out of our soda leaving our soda flat.
• Air freshners, agarbatti, mosquito repellents work on the principle of diffusion.

Question 15.
Give examples of three daily life activities in which osmosis is involved?
Answer:

• Water enters into the roots through osmosis.
• In our body waste materials are filtered from the blood.
• Osmosis helps in the opening and closing of stomata.

9th Class Biology 4th Lesson Movement of Materials Across the Cell Membrane Activities

Activity – 1

Question 1.
Look at the substances in the table identify the (✓) substances that should go into the cell and should go out of the cell?
Answer:

Substance	Should go into the cell	Should go out of the cell
Oxygen	✓
Glucose	✓
Proteins	✓
Fats	✓
Vitamins	✓
Minerals	✓
Carbondioxide		✓
Wastes		✓

Procedure :

• Keep the raw eggs in dil HCl / toilet cleaning acid for 4 to 5 hours.
• Take out the egg with the help of table spoon.
• Wash the eggs under tap water.

• Measure the circumference of each egg with long strip of paper, as its widest place, and mark on the paper with pen or pencil.
• Prepare a concentrated salt solution in a beaker.
• Place one egg in the beaker with tap water and place the other in the salt water.

• Leave the beakers for 2 to 4 hours.
• Take the eggs out, wipe them and measure the circumference with the same strip of paper. Mark on the paper with pen or pencil.

Observation :
The egg placed in salt water shrinks, the egg placed in the tap water swells.

Result:

• Shrinking of egg placed in the salt water is due to exosmosis in which water molecules leave the cell.
• Swelling of egg placed in the tap water is due to endosmosis in which water molecules enter the cell.

Lab Activity – 3

Question 2.
Prepare semi-permeable membranes and conduct an experiment to prove osmosis with it.
Answer:
Preparing semi-permeable membranes.

• Take two raw eggs.
• Keep the two eggs in dil. HCl for 4 to 5 hours.
• The shells which are made of calcium carbonate (CaCO₃) are dissolved.
• Wash the eggs under tap water.
• Carefully pierce a pencil sized hole in the egg membrane and drain the contents.
• Wash the membrane with fresh water. Now the semi-permeable membrane is ready for use.

Experiment of osmosis with egg membranes :

Aim :
To prove osmosis through semi- permeable membrane of an egg.

Materials required :
Two egg membranes, three beakers, sugar, water, thread, measuring jar, disposable syringe.

Procedure :

1. Take one egg membrane and fill it with 10 ml of saturated sugar solution with a syringe.
2. Tie its mouth with a thread.
3. Measure 100 ml of tap water in a beaker.
4. Keep the egg membrane in fresh water beaker.
5. Leave it for overnight.
6. Take the second egg membrane and fill it with 10 ml of tap water with the syringe.
7. Prepare 100 ml of saturated sugar solution and keep the egg membrane in it.
8. Leave it for overnight.
9. Measure the contents of the egg membranes and beakers.

Observations:

1. Water entered into the egg membrane in which sugar solution is filled. So size of the membrane increased.
2. Water left from the egg membrane in which water is filled. The size of the membrane decreased.
   Result: Water move across membranes from solutions of one concentration to the other through a process called osmosis.

Activity – 4

Question 3.
How do you observe the diffusion of coffee powder in water? Write your findings.
Answer:

1. Take half bowl water.
2. Prepare a small ball of coffee powder.
3. Slowly put in water and observe.

Observations:

1. The ball of coffee powder starts dissolving in water.
2. The water around the coffee powder will appear dark in colour.
3. As time progresses, all the water in the beaker becomes coloured.
4. Initially pale in colour and slowly all the water in the beaker becomes uniformly coloured. Coffee powder molecules diffuse into the water forming uniform colour.

Activity – 5

Question 4.
Observe the diffusion of potassium permanganate in water. Write your findings.
Answer:

• Keep a crystal of KMNO₄ (Potassium permanganate) in the centre of the petridish with the help of a forceps.
• Carefully fill the petridish with water.
• Observe the movement of pink colour in the petridish every minute.
• Also observe the spreading of colour from centre to periphery.

Observations:

1. Potassium permanganate crystal starts dissolving in water.
2. The water around the crystal will appear in pink colour.
3. As time progresses all the water in the beaker becomes coloured.
4. Initially pale in colour and slowly all the water in the beaker become uniformly pink coloured.

Diffusion :
The permanganate molecules moves from higher concentration to lower concentration in water through diffusion.

Activity – 6

Question 5.
How do you observe the diffusion of copper sulphate in water? Write your findings.
Answer:

• Keep a small crystal of copper sulphate in the center of the petridish with the help of a forceps.
• Carefully fill the petridish with water.
• Observe the movement of blue colour in the petridish every minute.
• Also observe the spreading of colour from centre to periphery.

Observations :

1. Copper sulphate crystal starts dissolving in water.
2. The water around the crystal will appear in blue color.
3. As time progresses, all the water in the beaker becomes coloured.
4. Initially pale blue in colour and slowly all the water in the beaker becomes uniformly blue in color.

Diffusion :
The copper sulphate molecules move from higher concentration to lower concentration in water through diffusion.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 11 Bio Geo Chemical Cycles Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 11th Lesson Bio Geo Chemical Cycles

9th Class Biology 11th Lesson Bio Geo Chemical Cycles Textbook Questions and Answers

Improve Your Learning

Question 1.
What is the importance of different biogeochemical cycles in the nature? (AS 1)
Answer:

• Biogeochemical cycles enables the transformation of matter from one ecosystem to another.
• Biogeochemical cycles enable the transfer of molecules from one locality to another.
• Some elements such as nitrogen are highly concentrated in the atmosphere, but some of the atmospheric nitrogen is transfer it to soil through the nitrogen cycle.
• Biogeochemical cycles facilitates the storage of elements.
• Biogeochemical cycles assists in functioning of ecosystem.
• Biogeochemical cycles link living organisms with living organisms, living organisms with non-living organisms and non-living organisms with non-living organisms.
• Biogeochemicals regulate the flow of substances.

Question 2.
What do you understand by Ozone layer? Write an essay to participate in elocution competition on importance of ozone layer. (AS 6)
Answer:
Ozone is concentrated in a layer in the stratosphere, about 15-30 kilometres above the earth’s surface. Ozone is a molecule containing three oxygen atoms. It is blue in colour and has a strong odour.

Significance of ozone layer :
Even the small amount of ozone plays a key role in the atmosphere. The ozone layer absorbs a portion of the radiation from the sun, preventing it from reaching the planets.
Most important of all it absorbs the portion of ultra violet light which causes many harmful effects including various types of skin cancer and harm to some crops, certain materials and some forms of marine life.

Ozone depletion :
Certain industrial processes and consumer products results in the emission of ozone depleting substances to the atmosphere. Chlorofluoro carbons used in almost all refrigeration and air conditioning systems destroy ozone layer. The ozone hole is not really a hole, but it was observed that there is less ozone in Antarctica than in arctic region.

Conservation of ozone layer :
The discovery of an ozone hole over Antarctica prompted action to control the use of gases which have a destructive effect on the ozone layer. From this concern emerged the Montreal protocol on substances that deplete the ozone layer signed by 24 countries in 1987.

Question 3.
What emissions from human activities lead to ozone depletion? And what are the principal steps in stratospheric ozone depletion caused by human activities? (AS 1)
(OR)
Which human activities emit gases that lead to Ozone depletion. What measures you suggest to control the emission of these gases?
Answer:

• Certain industrial processes and consumer products result in emission of ozone depletion substances to the atmosphere.
• These gases bring chlorine and flourine atoms to the atmosphere when they destroy ozone in chemical reactions.
• Important emissions from human activities are chlorofluoro carbons used in all most all refrigeration and air conditioning system.
• Most of these gases accumulate in the lower atmosphere because they are unreactive and do not dissolve readily in rain or snow.
• Natural air motions transport these accumulated gases to the stratosphere, where they are converted to make reactive gases.
• Some of these gases then participate in reactions that destroy ozone.

Measures to control these gases :

• We should control and phase out the production and supply of ozone depleting chemicals specifically CFCs and their derivatives.
• We should control and phase out of Halons, which destroy the growing plants in waste lands and starting reforestation works.

Question 4.
Why could we say that biogeochemical cycles are in “balance”? (AS 1)
Answer:

• We can say that biogeochemical cycles are in balance because the composition of various gases present in atmosphere does not change.
• And also even the substances of the biogeochemical cycles change from one ecosystem to the other, their percentage in soil, water and atmosphere remain same.
• By this, we can say that the biogeochemical cycles are in balance.

Question 5.
What role does carbon dioxide play in plant life processes? (AS 7)
Answer:

• The fixing of carbon in biological form takes place within plant and other organisms known as producers – in a process called photosynthesis, by which energy from sunlight is converted into chemical form.
• In photosynthesis, light energy helps to combine carbon dioxide and water to create the simplest of sugars, the carbohydrate molecules known as glucose (C₆H₁₂O₆).
• The carbohydrates then become the source of chemical energy that fuel living cells in all plants and animals.
• In plants, some carbon remains as simple glucose for short term energy use, while some are converted to large complex molecules such as starch for longer term energy storage.

Question 6.
If all the vegetation in the pond died, what effects would it have on the animals? Why? (AS 2)
Answer:

• If all the vegetation in the pond dies, the animals which are herbivores also die due to the lack of food materials.
• So, the herbivores depend on vegetation for their food, dies immediately.

Question 7.
Burning of fossil fuels a concern for scientists and environmentalists. Why? (AS 6)
Answer:

• There are two problems associated with the use of fossil fuels.
• The first problem is that they are non – renewable resources.
• In other words as we use these fuels, their supply gets exhausted.
• It is estimated that the available supply of fossil fuels will get exhausted in another 50 to 100 years.
• The second problem with the use of fossil fuels is pollution.
• When these fuels are burnt various gases are produced.
• These are carbon dioxide, carbon monoxide, sulphur dioxide etc.
• Carbon dioxode is responsible for green house effect in the environment.
• As its concentration increases, more heat is retained in the atmosphere and the temperature all over the world increases and this is called global warming.
• Global warming causes floods in some areas and droughts in some areas.
• Sulphur dioxide released by the industries in to the atmosphere mixes with water vapour forming sulphuric acid and sulphurous acids. These are known as acid rains.

Question 8.
How human activities caused an imbalance in biogeochemical cycles? (AS 7)
Answer:

• In recent years human activities have directly or indirectly affected the biogeochemical cycles that determine climatic conditions of earth.
• Use of fertilizers mainly has affected the phosphorous and nitrogen cycles.
• Plants may not be able to utilize all of the phosphate fertilizer as a consequence, much of it lost from the land through the water run off. This result in pollution of water bodies.
• Humans have interfered with carbon cycle where fossil fuels have removed from the earth crust.
• Additionally, clearing of vegetation that serve as carbon sinks has increased the concentration of CO₂ in the atmosphere.
• Human impact on the sulphur cycle is primarily in the production of sulphur dioxide from industry.
• Sulphur dioxide can precipitate on to surfaces where it can be oxidized to sulphate in the soil, reduced to sulphide in atmosphere, or oxides to sulphate in the atmosphere as sulphuric acid.
• As a result of extensive cultivation of legumes, creation of chemical fertilizers, and pollution emitted by vehicles and industrial plants, human beings have more than doubled the annual transfer of nitrogen in to biologically available form.

Question 9.
List three ways we, as humans, have affected the water cycle. (AS 7)
Answer:

• The earth’s water supply stays the same but humans can alter the cycle. As population increases, and living standards rise this can increase the demand for water.
• Human impact the water cycle by polluting the water in rivers, streams, reservoirs etc.
• We are polluting it with harmful chemicals and disgusting substances. Technically we cannot alter the water cycle, however we can mess it up by dumping waste in to the ocean.

Question 10.
Describe interdependence of biotic and abiotic components by taking Nitrogen cycle as an example. Draw Nitrogen cycle. (AS 5)
Answer:
Interdependence of biotic and abiotic components in nitrogen cycle :

• Atmospheric nitrogen is present in inert form.
• From abiotic atmosphere nitrogen fixing bacteria abiotic component fixes nitrogen and uses it and stores in the body cells.
• Nitrates can also be converted to ammonia by the denitrifying bacteria in the soil.
• From soil plants take up nitrates as well as ammonium ions from the soil to convert them to proteins and nucleic acids.
• When animals and plants die, the nitrogen in the organic matter reenters the soil and water bodies.
• There the decomposing bacteria releases ammonia into soil and water.
• From abiotic soil component nitrogen makes its way back into atmosphere through a process called denitrification in which soil nitrate is converted back to gaseous nitrogen.

Question 11.
Go to a nearby pond observe organisms living in the pond and biodegradable substances mixing in water. How they effect on those organisms? Write your observation. (AS 4)
Answer:

• Biodegradable pollutants could have serious environmental consequences if large quantities are released in a small area.
• For example, dumping of biodegradable waste in to a small pond will deplete the •pond’s oxygen supply.
• Microorganisms in the ponds uses oxygen for degrading biological wastes.
• More amount of oxygen will be utilised by microorganisms for degradation.
• Left with no oxygen the aquatic organisms like fish die.
• Thus biodegradable substances become pollutants.

Question 12.
Prepare an article for newspaper on the item “How human activities effects the environment”. (AS 7)
Answer:
When the human population was smaller, people lived in small communities, so the effects of their activities were small and localised. A rapid increase in the human population and increase in the standard of living have lead to wide spread damage of the environment.

Question 13.
Write an experiment to prove Green house effect on temperature.
Answer:
Aim :
To prove the green house effect on temperature.

Appratus :
Two glass bottles, two corks, two thermometers, vinegar, baking soda, high voltage lamp

Procedure:

1. Take 100ml of vinegar and a table spoon of baking soda in one bottle and close its mouth with cork.
2. Insert the thermometer into the bottle through cork such that the bulb of the thermometer should not touch the material in the bottle.
3. Insert another thermometer into the empty bottle through the cork.
4. Keep these two bottles opposite to a high voltage bulb such that both bottles receives the same amount of temperature.
5. Note down the initial temperatures and record the temperatures for an hour.

Observation:

1. We can observe that the vinegar and baking soda in the first bottle react with each other to produce CO₂.
2. This CO₂ absorbs and retains the more heat from the bulb than the normal air in the second bottle.
   Inference : This proves the green house effect (green house gases such as CO₂) increases the temperature of the earth.

9th Class Biology 11th Lesson Bio Geo Chemical Cycles Activities

Lab Activity – 1

Question 1.
Aim :
Test the effect of a green house on temperature.

Materials required :
Plastic bottle, nail, 2 thermometers, notebook and pencil.

Procedure:
1) Make a hole near the top of the plastic bottle with the nail.
2) Insert the first thermometer into the hole.
3) Place the second thermometer next to the bottle.
4) Make sure that the same amount of sunlight reaches both thermometers.
5) After 10 minutes, note temperature values from both thermometers.
6) Record the data in the notebook.
7) Take the temperature records again after another 10 minutes and repeat it for 2 – 3 times more.

Answer the following questions :
1) Do both thermometers record the same temperature?
Answer:
No.

2) If not, which one is higher?
Answer:
The thermometer kept in the plastic bottle shows higher temperature.

3) Can you explain why these two temperature records are not the same?
Answer:
a) The plastic bottle traps the sun’s rays and keeps the heat from escaping.
b) That is why it is warm inside the bottle.
c) The higher temperature in thermometer kept inside the bottle is due to the warmness inside the bottle.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 10 Soil Pollution Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 10th Lesson Soil Pollution

9th Class Biology 10th Lesson Soil Pollution Textbook Questions and Answers

Improve Your Learning

Question 1.
Define soil pollution. (AS 1)
Answer:
Soil or land pollution can be defined as the buildup in soils of persistent toxic compounds, chemicals, salts, radioactive materials, or disease causing agents, which have adverse effects on plant growth and animal health.

Question 2.
Why are plastic bags a big environmental nuisance? (AS 6)
Answer:

• Plastics are so versatile in use that their impact on environment are extremely wide ranging.
• Careless disposal of plastic bags chokes drains, blocks the porosity of the soil, and causes problems for ground water recharge.
• Plastic disturbs the soil microbe activity, and once ingested can kill animals.
• Plastic bags can also contaminate food stuffs due to leaching of toxic dyes and transfer of pathogens.
• Plastic bags remains strewn on the ground, or in unmanaged garbage dumps.
• Though small percentage lies strewn, it is this portion that is of concern as it causes extensive damage to the environment.

Question 3.
Describe an environmental friendly method to profitably dispose of human waste and cattle waste. (AS 1)
Answer:

• In recent years, an alternate and better method is used to obtain energy from not only from cattle waste but also from human waste.
• This is by anaerobic fermentation of the wastes to produce a gas which can be used as fuel.
• As this gas is produced from biological waste, this is called biogas.
• Biogas is a mixture of several gases : methane, carbondioxide, and small amounts of hydrogen, nitrogen, and hydrogen sulphide.

Question 4.
Chemical fertilizers are useful to crops. In which way they cause environmental pollution? (AS 1)
(OR)
We often read in newspapers that environmentalists often show deep concern of threats posed by chemical fertilisers and pesticides, etc. What are those threats to environment?
Answer:

• Fertilisers contaminate the soil with impurities, which come from the raw materials used for their manufacture.
• Due to excessive use of phosphate fertilizers soil becomes an indestructible poison for crops.
• Excessive use of fertilizers can endup polluting lakes, rivers and streams.
• This leads to promote the growth of algae in water bodies and is called eutropication.
• This abundant uncontrolled growth of plants blocks the flow of water and reduces oxygen content in the water.
• Other organisms living in the water do not get sufficient water, oxygen and ultimately die.
• Nitrogen fertilizer contribute to air pollution when it enters the atmosphere as ammonia and nitrogen oxide.
• This inturn cause acid rain and city smog associated health and environmental problems such as respiratory illness.

Question 5.
What steps can be taken to reduce pollution due to particulate matter from industries?
Answer:

• Industrial wastes can be treated physically, chemically and biologically until they are less hazardous.
• Acidic and alkaline wastes should be first neutralized; the insoluble material if biodegradable should be allowed to degrade under controlled conditions before being disposed.
• Electrostatic precipitators are used to reduce the particulate matter in the factory smoke.

Question 6.
What is a medical waste? Why it is called hazardous waste? What is the safe way to dispose medical waste? (AS 1)
Answer:

• Waste that is produced from hospitals is known as medical waste.
• Medical waste include needles, syringes, saline bottles, instruments used in surgeries, bandages soaked with blood and pus, used medicines, human excreta etc.
• Medical waste is called as hazardous waste because it containing toxic substances.
• Burying the medical waste in locations situated away from residential areas is the simplest method to dispose medical waste.

Question 7.
Prepare a flow chart to describe soil pollution, causes and methods of control. (AS 5)
(OR)
Prepare a pamphlet of your own to create awareness on soil pollution among the people in your area.
Answer:

Question 8.
What soil problems do you find in your area? Prepare a list of those problems and suggest a method for each of them to control those problems. (AS 7)
Answer:
The soil problems identified by me in our area :

Question 9.
What farm practices impact soil? Do they impact soil in a positive or a negative way?
Answer:

• Indiscriminate use of fertilizers, pesticides, insecticides, herbicides, no-till farming and growing same crop in all seasons are the farm practices impact soil.
• These farm practices may show positive or negative impact on the soil.
• By using chemical fertilizers we can get high yielding for only 20 to 30 years.
• After that soil becomes reluctant to plant growth. These chemicals damage fertility.
• Due to the extensive use of pesticides, insecticides, herbicides the salinity of the soil increases and it is not suitable for growing crops.
• Notill farming is a way of growing crops without disturbing the soil through tillage.
• Tillage activity can lead to compaction of soil, loss of organic matter in soil, loss of native vegetation, and death of the organisms in the soil.
• Growing the same crop in all seasons decreases the fertility.

Question 10.
Rank the negative impact practices in your area in the order in which you think they should be eliminated. (AS 1)
Answer:
Negative impact practices in our area :

1. Using chemical fertilizers
2. Using pesticides
3. Using insecticides
4. Using herbicides
5. Till farming
6. Deforestation
7. Using weedicides
8. Growing same crop in all seasons
9. Using locally prepared seeds

Question 11.
Rank the positive impact practices in order in which you think they should be used for the most benefit on your farm. (AS 1)
Answer:

1. Hybridised seeds
2. Organic manures
3. Organic weedicides
4. Predatory insects
5. No-till farming
6. Maintaining suitable pH value
7. Crop rotation
8. Salinity management
9. Soil organisms

Question 12.
Ravi said soil health is important. How can you support him? (AS 7)
Answer:

1. I support Ravi’s statement.
2. Healthy soil is fundamental to the quality of food it produces and to the health of those who eat the food produced from it.
3. When the soil components are present in appropriate percentage, the productivity is high.

Question 13.
How would soil texture affect the nutrients in soil? What would be its impact on crop production? (AS 2)
Answer:

• Soil with loose pores will allow water to collect and roots to expand. Loose soil is better than hard compact soil.
• Finer particles like clay increase surface area of the soil which allow nutrients to stay in the soil.
• Very porous soil, such as sand will allow nutrients to be leached more easily which can make less nutrients available to plants.
• Generally, a loose, airy soil structure is best for most plants.
• This can be accomplished by digging the bed and mixing together coarse and finer textures such as tilling compost into clay soil.

Question 14.
What are the three main physical properties of soil? What effects do this have on the plants? (AS 1)
Answer:

1. Colour texture, structure and porosity are the three main physical properties of soil.
2. These properties regulate and affect air and water movement in the soil and thus, soil ability to function.

Question 15.
What is pH? What is its range? What are the negative impacts if the pH of soil is too low or too high? (AS 1)
Answer:

1. The term pH is used to indicate the level of acidity or alkalinity of a soil.
2. The range of pH values of a good soil live from 5.5 to 7.5.
3. Below pH 7 the soils are termed as acidic and above pH 7 alkaline.

Negative impacts of low pH value :

1. The concentration of soluble metals especially aluminium and manganese may be toxic.
2. Calcium may be deficient.
3. Soil organisms responsible to transform N, S and P to plant available forms may be reduced.
4. Symbiotic nitrogen fixation in legume crops is greatly impaired.
5. Soils will be having low organic matter.
6. The availability of mineral elements to plants may be effected.

Negative impacts of high pH value :

1. If the pH is beyond 7, nutrient absorption and microbial activity will be affected which can be poisonous to plants.
2. pH extremes are unhealthy for most plants because they close or open membranes of plant cells too much.
3. This affects plant structure and their ability to uptake nutrients.
4. pH extremes make minerals and nutrients either too available or not available enough.

Question 16.
What is soil fertility? What are the sources of soil fertility? (AS 1)
Answer:

• Fertility of soil is closely associated with the properties of soil and is defined by its capacity to hold water and nutrients and supply them to plants when they need them, independent of direct application of nutrients.
• Soil organisms contribute to buildup soil organic matter, including humus, the soils most important nutrient reservoir.
• A major part of the soil microbial biomass is composed of fungi.
• Soil fertility is a complex process that involves the constant cycling of nutrients between organic and inorganic forms.
• As plant material and animal wastes decompose they release nutrients to the soil solution.
• Soil pH, its acidity or alkalinity is highly relevant to how readily nutrients become available in the soil.

Question 17.
Name 5 living things that live in soil. What do these things do to affect the soil?
Answer:

1. Viruses, earthworms, rats, ground squirrels, bacteria, fungi, algae, protozoa, dung beetle and different types of worms live in the soil.
2. These organisms feed on plant residues burrow the soil and help in aeration and percolation of water.
3. Soild microbes convert organic forms of elements to their inorganic forms.
4. Soil bacteria also control the forms of ions in which these nutrients occurs.

Question 18.
What is organic matter? Why it is important to plants? (AS 1)
Answer:

• Organic matter is the organic component of soil which includes the residues of dead plants and animals.
• Organic matter consists of nutrients necessary for plant growth such as nitrogen, phosphorus, and potassium.
• Soils which contain 30% or more organic matter are considered organic soil, all other soils are identified as mineral soils.
• Organic matter in soil improves water in filteration, decreases evaporation, and increases the water holding capacity.
• And also where there is organic matter, there will be numerous organisms present helping to convert it back to nutrients and these organisms help to create small pieces of nutrients, ideal for cultivation.

Question 19.
What are the factors affecting organic matter levels in soil? How this level of organic matter can be increased? (AS 1)
Answer:
1. Temperature, rainfall, natural vegetation, texture, drainage, cropping and tillage and crop rotation are the factors affecting organic matter levels in soil.

2. Temperature :
The decomposition of organic matter is accelerated in warm climates as compared to cooler climates.

3. For each 10°C decline in mean annual temperature the total organic matter and nutrients increases by two to three times.

4. Rainfall:
There is an increase in organic matter with an increase in rainfall.

5. Natural vegetation :
The total organic matter is higher in soils developed under grasslands than those under forests.

6. Texture :
Fine textured soils are generally higher in organic matter than coarse textured soils.

7. Drainage:
Poorly drained soils because of their high moisture content and relatively poor aeration are much higher in organic matter and nutrients than well drained soils.

8. Cropping and Tillage :
The cropped lands have much low nutrients and organic matter than comparable virgin soils.

9. Crop rotation :
Crop rotation of cereals with legumes results in higher soil organic matter.

Question 20.
What is solid waste? Explain best practices for solid waste management. (AS 1)
Answer:
Solid waste:
Solid waste may be defined as the organic and inorganic waste produced by various activities of the society which have lost their value to the first user.

Best practices for solid waste management:

1. By practicising four R’s : Reduce, Reuse, Recycle and Recover we would get less solid waste.
2. Materials such as glass containers, plastic bags, paper, cloth, etc., can be reused at domestic levels rather than being disposed, reducing solid waste pollution.
3. Solid waste management involves activities including collection, transfer, and transport to suitable sites and safe disposal of wastes by methods which are environmentally friendly methods.
4. Burying the waste in locations situated away from residential areas is the simplest and most widely used technique of solid waste management.
5. Solid waste management can also be done by methods such as sanitary landfill, composting and incineration, etc.

Question 21.
What is bioremediation? How it helps in controlling soil pollution? (AS 1)
Answer:

1. Bioremediation means to use a biological remedy to reduce or clean up contamination.
2. Microbes are often used to remedy environ¬mental problems found in soil, water and sediments.
3. Plants have also been used to assist bio¬remediation processes. This is called phytoremediation.
4. Biological processes have been used for some inorganic materials, like metals to lower radioactivity and to remediate organic contaminants.

Question 22.
Why soil conservation is important to us? What will happen if no preventive measures would be taken? (AS 2)
Answer:

• Soil conservation is important to us because it forms the basis for habitats and plants, which act as source of food to both humans and animals.
• Soil conservation is also important because with the erosion of the top soil layer, valuable nutrients are lost and crop yield diminish, which means very less food is produced per acre.
• We have to conserve soil because it has organic material that is good for plant growth.
• If no preventive measures are taken for soil conservation, soil erosion takes place.
• And also soil will be over used and it has more chemicals leading to unproductive soil.
• Amount of nutrients present in the soil decreases.

Question 23.
Look at the following symbol, what does it mean?

Answer:
1. It is the symbol of bioremediation.
2. Plants have been used to assist bioremediation.

9th Class Biology 10th Lesson Soil Pollution InText Questions and Answers

9th Class Biology Textbook Page No. 155

Question 1.
Today what are the pollutants produced from your school. How many of these are non-degradables?
Answer:
Wastes produced from our school :
Peels of fruits, vegetables, rice, glass materials, pens, polythene bags, biscuit and chocolate covers, icecream sticks, rubber, plastic tea glasses, paper leaves twigs etc.

Non-degradable pollutants :
Glass materials, pens, polythene bags, biscuit and chocolate wrappers, rubber, plastic glasses.

9th Class Biology 10th Lesson Soil Pollution Activities

Activity – 1

Question 1.
Answer:
1. During interval time Venu was eating a fruit.
2. He was about to throw the peel in corner of verandah.
3. His friend Ramu stopped him.
4. Ramu said you should not throw waste in the verandah. Drop it in the bin/basket given.
5. Prepare a list of waste materials we throw out in a day from morning to evening.
6. Classifying them as wet wastes and dry wastes with the help of the example given in the table.

Wet waste	Dry waste
Vegetable peels	Biscuit wrapper
Banana peels	Polythene covers
Food materials	Used papers
Fruit peels	Plastic materials
Dung	Glass materials
Hay	Card board

7. Weigh the wet wastes, which you have listed in the table for one day.
8. Divide the weight by number of people in your home.
9. The result will be the per capita wet waste we are producing in one day.
10. Suppose if a family containing four members throws 400 gms of wet wastes per day,

Multiply it by 30 = 100 × 30 = 3000 gms per month
Multiply it by 365 = 3000 × 365 = 10,95,000 gms = 1095 kgs per year.

Activity – 2

Question 2.
Dumping and decomposing,
Answer:
1. Take a polythene bag / plastic bucket / or any container.
2. Fill half of it with soil.
3. Keep wet wastes and other wastes in it.
4. Wastes should include vegetable peels, rubber, plastic etc.
5. Add some more soil and sprinkle water regularly on it.
6. Dig it and observe in 15 days intervals.
7. Note your observations in the table.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 9 Adaptations in Different Ecosystems Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 9th Lesson Adaptations in Different Ecosystems

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Textbook Questions and Answers

Improve Your Learning

Question 1.
What do you understand by adaptations in organisms and why do they adapt? (AS 1)
Answer:

• The ways and means that organisms adapt or develop over a certain period of time in different conditions for better survival are adaptation of organisms.
• Adaptation is a feature that is common in any population because it provides some improvement for better survival.

Question 2.
With the help of two examples, explain how these organisms have adapted themselves in the ecosystem? (AS 1)
Answer:

• Mangroves grow in a wet and salty place.
• They have evolved to have curious looking projections from their roots called pneu- matophores or knees.
• These pneumatophores develop from the lateral roots that are growing near the surface, and protrude upto 12 inches out of the soil.
• Pneumatophores aid the plants in maintaining adequate root respiration in a watery environment.
• We don’t find such structures in plants growing around us.
• Another example is in kaiabanda, the leaves are reduced to spines so that there is little transpiration loss and water is stored in the tissues of the stem (succulent stems)
• This helps the plant to live in conditions of water scarcity as we come across in deserts.
• With the above two examples, we can say that these organisms have adapted them-selves in the ecosystem.

Question 3.
Collect some aquatic plants- cut the leaves and stems. Observe them under microscope and record your observations like presence air /absence of air spaces etc. and answer the below. (AS 3)
a) Are there any other reasons for their floating?
Answer:
The bodies of aquatic plants are delicate with more than 80% of their weight consisting of water.

b) Draw a diagram of what you have observed under microscope.

Question 4.
What special adaptations can be seen in the following organisms? (AS 1)
a) mangrove trees
b) camel
c) fish
d) dolphins
e) planktons

a) Mangrove trees :
Answer:

1. Mangroves grow in a wet and salty place near the sea shore.
2. From their roots arise pneumatophores or knees.
3. These pneumatophores develop from the lateral roots that are growing near the surface and protrude upto 12 inches out of the soil.
4. Pneumatophores. aid the plants in maintaining adequate root respiration in a watery environment.

b) Camel:
Answer:

1. In camel hump stores fat fordater use.
2. Long eyelashes protects eye from sand.
3. Nostrils closes voluntarily to protect from blowing sand.
4. Long legs keeps the body away from hot ground.

c) Fish :
Answer:’

1. The body is covered by scales.
2. Fishes bear specialised structures to swim like fins.
3. Fishes have floaters in their body (special structures of their digestive canals) to be able to inhabit particu¬lar levels in the water body.
4. Fishes respire with gills.

d) Dolphins:
Answer:
Dolphins have adapted to their environment in the following ways :
Fins shape – A dolphins tail goes up and down to help it dive up to get the air. The shape of their fins also help to propel them through the water.

To help dolphins save oxygen while they dive under water, their heart beat slower during a dive and their blood is diverted from other parts of their body to their heart, lungs and brain. They also save oxygen via muscles, which have a protein called myo-globin which in turn stores oxygen.

They have a blubber or fat which provides insulation helping the dolphin stay warmer under cold water.

They have a body covering of skin. The upper most layer of skin produces an oil which forms a film that cover the dolphin’s body.

Being mammal dolphin breathe with lungs rather than gills. So they breathe from a blow hole which closes before the dolphin goes into the water. The long nose helps the dolphin to fight sharks and their teeth help them to catch fish.

They have well developed echo location by which they locate other animals and also communicate with each other.

e) Planktons:
Answer:
Microscopic photosynthetic organisms like planktons have droplets of oil in their cells that keeps them float.

Question 5.
If an animal of euphotic zone has to survive in abyssal zone, what adaptations are required to survive there? (AS 1)
Answer:
Adaptations required to survive in abyssal zone are :

1. The animals should have wide mouths and huge curved teeth which prevent escape of any prey.
2. Absence of skeleton, flattened bodies are required.
3. Animals should have special structures that produce light on their bellies, around their eyes, and at the sides of their bodies.
4. The animals should show bioluminescence in the dark waters.

Question 6.
Marine water fishes drink more water than fresh water fishes. Do you agree? Justify.
Answer:

• Yes, marine water fishes drink more water than fresh water fishes.
• Because several marine fishes have a lower internal salt concentration than that of the water they swim in.
• So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amount of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 7.
Visit a nearby pond or lake. Record the organisms you have observed and their adaptations. (AS 4)
Answer:

• Nearby pond or lake consists of three zones namely littoral zone, limnetic zone, and profundal zone.
• In the topmost littoral zone, the edge of a water body is home to snails, insects, several crustaceans, fishes and amphibians, and the eggs and larvae of dragonflies.
• Predators present are tortoise, snakes, and ducks.
• Adaptations : Several organisms have well developed sight, usually have dull and greyish bodies, and are fast swimmers.
• Limnetic zones contains fresh water fish, crustaceans like daphnia, cyclops, and small shrimps are present.
• Floating- plants like water hyacinth, wolfia, pistia along with algae are present.
• Adaptations seen in the plants this zone are presence of air space, leaves covered with wax, etc.
• In the profundal zone scavengers and predators for example crustaceans, crabs, fishes like eels and snails, turtles are present.
• They adapt themselves by feeding on dead animals that settle down.

Question 8.
Draw a lake showing different zones. Why are they called so? (AS 5)
Answer:
Zones of Lake :

1) Littoral zone :
The zone close to shore. They reaches all the way. Plants living in this zone perform photosynthesis.

2) Limnetic zone :
Sunlit part of the lake surrounded by the littoral zone. This zone extends at a depth where sunlight penetrates.

3) Profundal zone :
It is much colder and denser than previous zones.

Question 9.
Collect information of one lake from internet and prepare a table of organisms adapted at different zones.
Answer:
Different zones in lakes and types of organisms present:
1) The littoral zone :
a) The topmost and warmest zone at the edge of a water body is home to snails, clams, insects, several crustaceans, fishes and amphibians and eggs and larvae of dragonflies.
b) Plants like mosses, water lily, vallisneria, hydrilla etc. are found along with several types of algae.
c) Predators of this zone are tortoise, snakes and ducks.

2) The limnetic zone :
a) This zone contains variety of fresh water fish with bright shiny scales.
b) Transparent or whitish bodied crustaceans like daphnia, cyclops, small shrimps are also found in this zone.
c) There are different types of floating plants like water hyacinth wolfia, pistia along with a variety of algae.

3) The profundal zone :
a) Mostly heterotrophs are present.
b) Scavengers and predators like crustaceans, crabs, fishes like eels and glossogobius (isika dondu), snails, turtles etc are present.
c) Many kinds of bacteria are also present in this zone that help in decomposition.

Question 10.
Write the effect of temperature on the organisms adapted in a lake and pond in a tabular form. (AS 1)
Answer:

• In deeper lakes during summer only the surface water is heated up while the deeper layer remain cold. During summer the ponds dry up.
• In tropical regions water gets heated up and evaporates in lakes. During average temperatures the water in the pond heated up and evaporates.
• The requirements necessary to the organisms like oxygen and nutrients gets decreased in the lake.
• The salinity of the water increases, concentration of oxygen decreases and availability of food decreases in pond during average temperatures.
• In the cold regions upper layers of the lake gets frozen during winter and lower layers does not.
• The entire pond gets frozen during winter.
• Aquatic animals in tropics undergo aestivation or hybernation to overcome extreme cold or hot seasons.

Question 11.
Amphibians are wonderful creatures on the earth. How do you appreciate their adaptation? (AS 6)
Answer:

• Amphibian body has small waist, no neck. Streamlined body shape helps in swimming.
• Skin is thin and moist allows gaseous exchange in cutaneous respiration.
• Front legs used to keep the front portion of the body off the ground.
• Hind legs able to jump great distances and change direction quickly.
• Eyes are positioned on top of head gives the frog a wide angled visual field.
• Mouth is very large and broad can able to catch and eat large prey.
• Tongue attached at front of mouth enables it stick the prey when caught.
• Frogs start their lives as aquatic tadpoles with gills to breathe. As tadpole grows into frogs lungs replace the gills and allows frog to breathe on land.

Question 12.
Some animals and plants survive only in certain conditions. Nowadays human activities cause damage to these conditions. What do you think about this? (AS 7)
Answer:

• Human activities are causing lot of damage to biodiversity.
• Human activities such as deforestation, overgrazing, conversion of forest land to agricultural land, hunting and indiscriminate killing of animals for their products, and pollution can endanger the plant and animal species.
• If proper care is not taken plants and animals may disappear totally from the surface of the earth.

Question 13.
In the chapter on ecosystem, we had studied about the mangrove ecosystems. What kind of abiotic conditions did you study in them? (AS 1)
Answer:
Kinds of abiotic conditions in mangrove ecosystems are soil, pH, oxygen, nutrients, winds and currents, light, temperature, humidity, tides, salinity.

Question 15.
Are there any rivers meeting in the Bay of Bengal in the Coringa ecosystem? Collect information and make a note on them.
Answer:

• Coringa mangrove is situated South of Kakinada Bay and is about 150 km South of Visakhapatnam.
• Coringa is named after the river Coringa.
• Coringa mangroves receive fresh water from Coringa and Gaderu rivers, distributors of Gautami, Godavari rivers, and neritic waters from Kakinada Bay.
• Numerous creeks and canal traverse this coringa ecosystem.

Question 17.
The Murrel (Korramatta) and Rohu are fishes found in rivers. Will they be able to live in the coringa ecosystem ? Give reasons for your answer.
Answer:

• Yes, Murrel and Rohu be able to live in the coringa ecosystem.
• Because coringa ecosystem gets fresh water from rivers coringa, Gaderu and distributories of Gautami, Godavari rivers.
• If the salinity of the water in the coringa ecosystem increases, the water enters the body of fresh water fishes.
• The water can be excreted in the form of urine, but to maintain a suitable salt bal¬ance fresh water fish need to reabsorb salt through the kidneys and salt collecting cells in gills.

Question 18.
How the frogs got protected themselves from cold and heat?
Answer:

• Frogs are cold blooded animals so they can’t tolerate extreme cold or heat conditions.
• They protect themselves from extreme cold conditions by a process called hibernation (winter sleep) and from extreme heat conditions by Aestivation (summer sleep)
• During these processes they burrow deep in the ground and remain motionless until the conditions are favourable.
• During this period the rate of metabolic activities slow down and the animal goes into a nearly unconscious sleepy condition.

Question 19.
How do you appreciate the processing protection pebble plants from the enemies?
Answer:

• Pebble plants are also called living stones.
• They protect themselves from their enemies by adapting themselves to their habitat.
• They survive by living partly underground.
• They avoid being eaten by blending in with surrounding rocks.
• Leaves of these plants are not green as in almost all higher plants, but various shades of cream, grey and brown, patterned with darker windowed areas, dots and red lines.
• The markings on the top surface disguise the plant in its surroundings (camouflage)
• Thus, they adopt wonderfully to their habitats and protect themselves from their enemies.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems InText Questions and Answers

9th Class Biology Textbook Page No. 131

Question 1.
What is a habitat?
Answer:
Habitat is the immediate environment occupied by an organism or the living place of an organism.

Question 2.
Is a tree habitat only for a crow?
Answer:’
No. Tree is a habitat for variety of birds and insects.

Question 3.
In what way an ecosystem is different from habitat?
Answer:
In ecosystem biotic and abiotic components are present. Habitat is the place where organisms live in an ecosystem.

9th Class Biology Textbook Page No. 134

Question 4.
You may know animals that live in water. Do you find in them any suitable characters adapted to live in water? Write a note on them in your notebook.
Answer:

• Structural adaptations in the bodies like presence of special air spaces.
• Such air spaces help them to swim and float in water.
• The aquatic organisms bear specialized structures to swim like flippers as in turtles and fins in fishes.
• Fishes, dolphins have floaters in their body to be able to inhabit particular levels in the water body.

9th Class Biology Textbook Page No. 135

Question 5.
In what way flexible stem is useful to the aquatic plants?
Answer:

• In aquatic plants flexible stem contains a parenchymatous tissue known as arenchyma.
• Arenchyma consists of number of air filled spaces.
• These air spaces help the plant to float on water.

9th Class Biology Textbook Page No. 137

Question 6.
Observe the table and answer the following questions.
Answer:

a) How many zones can you see in the figure basis of light penetration? Name them.
Answer:
Three zones are present. They are eu- photic zone, bathyal zone and abyssal zone.

b) What types of abiotic conditions do you find as per the given table?
Answer:
Light, temperature and depth.

c) What will effect adaptation to marine life other than the conditions shown in the table and figure?
Answer:
Salinity, oxygen, rainfall, regular windflow, soil, pH, nutrients, humid-ity, oceanic currents effect adaptation to marine life.

d) What happens to the temperature and pressure as depth increases?
Answer:
As depth increases temperature decreases and pressure increases.

e) Which zone has more animals? Guess why.
Answer:

1. Bathyal zone has more animals. Because the conditions in this zone are suitable for the organisms to grow.
2. Red and brown kelps are the primary producers. They provide food to other organisms in that zone.

9th Class Biology Textbook Page No. 139

Question 7.
Does Pulikat lake of Nellore come under fresh water ecosystem or not? Why?
Answer:

• Pulikat lake of Nellore comes under marine or salt water ecosystem.
• Because the salinity of water in the lake is 3.5%.
• Main salts present in the Pulikat lake are sodium and potassium.

9th Class Biology Textbook Page No. 140

Question 8.
‘Think, how webbed feet helps ducks?
Answer:

1. Webbed feet of birds help them to adapt conditions on land as well as in water.
2. Webbed feet have enabled them to be good swimmers.

Question 9.
Why cranes have long legs and long beaks?
Answer:

1. Cranes have long, thin legs wander through the mud shallows searching for insects.
2. Long beak help them in searching of insects in the mud.

9th Class Biology Textbook Page No. 141

Question 10.
How are marine ecosystems different from fresh water ones?
Answer:

1. The saliny of water in marine ecosystem is 3.5% whereas it is 1.8% in fresh water.
2. Marine ecosystems are huge and they make up about three-fourths of the earths surface.
3. The number of organisms present in marine ecosystems are more when compared to fresh water ecosystem.

Question 11.
Write two types of adaptations you find in marine ecosystems, different from fresh water ecosystems.
Answer:

• Many marine animals have blubber fur insulation from the cold and some fish have an antifreeze like substance in their blood to keep it flowing.
• Marine animals must regulate the interaction of fresh water and salt water in their bodies.
• Specially developed kidneys, gills and body functions help to maintain salt concentrations across members through osmosis.

Question 12.
What are the similarities in adaptation on the basis of light penetration in the two aquatic ecosystems?
Answer:

• In both the aquatic ecosystems, light penetrates upto a depth of zoom only.
• The light intensity is sufficient to perform photosynthesis.
• In the low light intensities below 200 mts depth is sufficent to perform photosynthesis by some kelps.
• Due to the lack of light in abyssal and profundal zones, usually scavengers and predators exists.

9th Class Biology Textbook Page No. 142

Question 13.
Which zone do you think, when compared to marine ecosystems, is absent in fresh water ecosystem?
Answer:
Benthic zone is absent in fresh water ecosystem when compared to marine ecosystem.

Question 14.
What would be a major factor leading to different types of adaptations in marine, fresh water ecosystems?
Answer:
Light would be a major factor leading to different types of adaptation in marine, fresh water ecosystems.

Question 15.
Do all plants shed their leaves at same time in a year throughout the world?
Answer:

1. No. Some plants in temperate regions shed their leaves before the winter starts.
2. In tropical regions some plants shed their leaves before the start of summer.

9th Class Biology Textbook Page No. 143

Question 16.
Are thorny leaves also an adaptation to temperature?
Answer:

1. No. They are not adaptation to temperature.
2. They are adaptation to protect themselves from the animals who eat them.

Question 17.
If the trees have broad leaves at the time of snow fall season what will happen?
Answer:
If the trees have broad leaves at the time of snow fall season, the branches of tree can break due to the weight of snow gathered on each leaf and branch during snow fall.

Question 18.
Why polar bear has thick fur on its body?
Answer:

1. Polar bear has thick fur coat or hair covering on their bodies.
2. The fur act as insulator preventing heat loss from its body.

Question 19.
In what way thick skin helps the seal to protect from cold weather?
Answer:

1. In the thick layer of skin, fat is deposited in seals.
2. The thick layer of fat deposited under their act as insulators preventing heat loss from its body.
3. The fat not only insulates the body but helps in producing heat and energy.

9th Class Biology Textbook Page No. 132

Question 20.
Can you give some examples of fleshy leaf plants?
Answer:
Yes. Bryophyllum, Aloe, and Agave are the examples for fleshy leaved plants.

Question 21.
Why xerophytic plants do not have broad leaves?
Answer:
To prevent the excessive loss of water through respiration xerophytic plants do not have broad leaves.

Question 22.
You may see Kittanara, a xeric plant, grown as fence around crop fields in some areas in our state. Actually those places are not desert. How can they grow there?
Answer:
They grow there because this plant shows adaptations in that places.

9th Class Biology Textbook Page No. 133

Question 23.
Do all animals living in desert conditions show adaptations?
Answer:
Yes, all animals living in desert conditions show adaptations.

Question 24.
Why some animals have scales on their body?
Answer:

1. Scales mainly protect the animals from environment.
2. In desert animals scales allow them to retain moisture by preventing the evaporation of water through the skin.
3. This allows the animal to become dehydrated and animal requires small amount of water to survive.

Question 25.
Why the animals that lives in burrows usually comeout during night time only?
Answer:
To protect themselves from extreme hot conditions, animals that live in burrows usually comeout during night time only.

9th Class Biology Textbook Page No. 139

Question 26.
Which organism among jelly fishes and decomposers present in euphotic zone?
Answer:
Jelly fishes are present in euphotic zone.

Question 27.
What kinds of adaptations can be seen in the organisms of the euphotic zone?
Answer:

1. The organisms living in this zone are mostly floaten and swimmen.
2. Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface are transparent.
3. These usually have sharp vision.

Question 28.
What kind of adaptations can be seen in the organisms of abyssal zone?
Answer:

• The larger animals in abyssal zone have wide mouths and huge curved teeth which prevent escape of any prey.
• Absence of skeleton, flattened bodies are some other characteristics observed.
• Some animals also have special structures that produce light on their bellies, around their eyes and at the sides of their bodies.
• Some animals shows bioluminiscence in the dark waters.

Question 29.
What differences can you find in the animals of bathyal zone when compared to animals of euphotic and abyssal zones?
Answer:

• Most of the plants found in this zone are the red and brown kelps, sponges, corals even animals with tubular bodies like squids and large animals like whales, etc.
• Some of the animals in the bathyal zone have a flat body like the ray fishes.
• Big eyes sensitive to very dimlight may present in bathyal zone animals.

Question 30.
How organisms of different zones of marine ecosystem are adapted?
Answer:

• The animals of euphotic zone are mostly floaters and swimmers.
• Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface.
• Animals of euphotic zone have very sharp vision.
• Some of the animals in bathyal zone have a flat body like the ray fishes.
• The animals may have big eyes sensitive to very dim light in bathyal zone.
• Absence of skeleton, flattened bodies are some adaptations found in animals of abyssal zone.
• Some animals in abyssal zone may have special structures that produce light on – their bellies, around their eyes and at the sides of their bodies.
• Some animals in abyssal zone shows bioluminiscence in the dark waters.

9th Class Biology Textbook Page No. 141

Question 31.
Organisms of the oceans have a lesser salt content in their bodies than the seawater around 3.5%. The fluid could drain out of the body of the organisms into the sea. This could be dangerous and fatal to the organism. How do they survive under such conditions?
Answer:

• Several marine species have a lower internal salt concentration than that of the water they swim in. So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amounts of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 32.
Can fish in estuarine ecosystem survive in river as well as in sea?
Answer:

• Yes, fish in estuarine ecosystem survive in river as well as in sea.
• Two of the main challenges of estuarine life are the variability in salinity and sedimentation.
• Many species of fish living in estuarine have various methods of control to the salt shifts.
• They regulate the salt concentrations using osmoregulaters.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Activities

Activity – 1

Question 1.
i) Take a Kalabanda (Aloevera) and a Balsam plant in two separate pots.
ii) Water each of them with two tablespoons of water.
iii) Do not water them for a week.
iv) Observe the condition of the plants after a week.

Observations :
a) Which plant showed growth?
Answer:
Kalabanda plant showed growth.

b) Which plant dried first? Why?
Answer:
Balsam plant dried first. Because Balsam plants are not watered regularly. They need water to grow.

Activity – 2

Question 2.
i) Collect an aquatic plant out of a water body (e.g. Duck weed, Hydrilla, Vallisneria etc.) ii) Carry it back home and plant it in a pot and water it.
Observations :
a) From the above activity we see that some plants dry up without water very quickly, while other can grow even with very little water.
b) Each of these plants are adapted to the conditions in their surroundings on the basis of need of water.

Activity – 3

Question 3.
You know some of the animals that reside in and around lake or pond. Make a list of those animals and the characteristics of their body.
List of animals and reside in and around lake or pond :
Insects : Dragonfly, Damsefly, Mayfix, Stonefly, Dobsofly, Caddisfly, Cranefly, Water bugs, Beetles, etc.

Crustaceans	Cray fish, Scuds, Shrimps
Molluscs	Snails
Annelids	Leeches
Fish	Blugill, Bass, Catfish, Sculpin, Minnow
Reptiles	Snakes, turtles
Amphibia	Frogs

Characteristics of the body of animais living in and around lake:

Animals	Characteristics
1) Mosquito	The body is segmented and it is a carrier of diseases.
2) Shrimps	‘ These are small, bottom dwelling crustaceans with a trans­lucent exoskeleton.
3) Snails	A soft bodied animal with a hard protective shell.
4) Swan	Swans are long necked water birds, webbed feet are present.
5) Crayfish	Fresh water crustaceans with four pairs of walking legs. Body is segmented with head and thorax united.
6) Dragonfly	it is a flying insect with a long abdomen. Body is elongated with two pairs of transparent wings.
7) Earthworm	It is a little animal with a long, soft body and no legs.
8) Fish	It lives in the water and breathe with gills.
9) Goldfish	It is a type of crap that makes a nice pet, kept in aquariums and swims with fins.
10) Toads	The skin is dry and leathery. Toads are amphibians with poison glands, short legs and snout like parotid glands. Drier skin. Webbed feet helps in walking and swimming.
11) Leech	The body is segmented. It sucks blood of other animals.