Class 8

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SCERT Class 8 Maths Chapter 11 Solutions Parallelograms

Class 8 Kerala Syllabus Maths Solutions Chapter 11 Parallelograms Questions and Answers

Parallelograms Class 8 Questions and Answers Kerala Syllabus

Sides and Angles (Page 178)

Question 1.
The figures below are combinations of three, four, and five equal rhombuses. Draw each and colour it.

Answer:
1. Draw a circle of radius 2 cm (if any).
Divide the centre of the circle into angles of 60° each.
These lines meet the circle at the points A, B, C, D, E, and F.

Draw arcs of 2 cm (equal to radius) from A and B to get G.
Similarly, find H and I.
Draw the required parts and rub off unwanted parts.

2. Draw a circle of radius 2 cm (if any).
Mark the points A, B, C, D, E, F, G, and H on the circle by making 45° angles at the centre.
Draw an arc of 2 cm (if any), A and B to get I.
Similarly, find J, K, and L.
We get the required figure.

3. Draw AC, 4 cm long, and mark its midpoint B.
Since all are rhombuses, ABI is a equilateral triangle.
Its angles are 60° each.

In the rhombus BCDJ,
JBC = JDC = 60°
BCD = BJD = 120°
In the rhombus BJFI,
IBJ = IFJ = 60°
BJF = FIB =120°.
Draw each rhombus and complete the pattern.

Question 2.
This picture is made up of three equal parallelograms and an equilateral triangle. Draw and colour it.

Answer:
The figure is an optical illusion of a 3D cube or hexagon composed of three equal parallelograms and one equilateral triangle.

• Draw an equilateral triangle (all sides equal, all angles 60°).
• On each of the three sides of the triangle, draw a parallelogram.
• To make the figure symmetrical, the angles of the parallelograms meeting at the triangle’s vertices should be calculated such that they fit perfectly.
• Usually, the parallelograms are identical to the triangle in terms of side length.

Question 3.
Draw and colour this picture made up of four equal parallelograms and four equal right triangles:

Answer:
Construction:

• Draw a central square or rectangle (formed by the space or the meeting points).
• Attach right-angled triangles to the sides.
• Extend the pattern by attaching the parallelograms to the hypotenuse or legs of the triangles, depending on the specific pattern shown in your textbook.
• Without seeing the exact visual pattern in the book, the key is to ensure the right angles are exactly 90° and opposite sides of the parallelograms are equal.

Diagonals (Pages 182-183)

Question 1.
Draw a parallelogram with lengths of diagonals 8 centimetres and 6 centimetres and the angle between them 60°.
Answer:
Construction Steps:
Draw a line segment AC = 8 cm.
Mark the midpoint O of AC (so AO = 4 cm).
Through O, draw a line XY making an angle of 60° with AC.
On this line XY, mark points B and D on opposite sides of 0 such that OB = 3 cm and OD = 3 cm.
(since the second diagonal is 6 cm and diagonals bisect each other).
Join the points A, B, C, and D to form the parallelogram ABCD.

Question 2.
Prove that in a parallelogram, the line joining the midpoints of two opposite sides is parallel to the other two sides:

Answer:
(i) Let ABCD be a parallelogram.

Let P be the midpoint of side AB, and Q be the midpoint of side CD.
Since ABCD is a parallelogram, AB || CD and AB = CD.
Since P and Q are midpoints,
AP = \(\frac {1}{2}\)AB and DQ = \(\frac {1}{2}\)CD.
Therefore, AP = DQ and AP || DQ.
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
So, APQD is a parallelogram.
This implies PQ || AD.
Thus, the line joining the midpoints is parallel to the other two sides.

(ii) Let ABCD be a parallelogram.

Let P be the midpoint of side AD and Q be the midpoint of side BC.
Since ABCD is a parallelogram, AD || BC and AD = BC.
Since P and Q are midpoints,
AP = \(\frac {1}{2}\)AD and BQ = \(\frac {1}{2}\)BC.
Therefore, AP = BQ and AP || BQ.
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.
So, ABQP is a parallelogram.
This implies PQ || AB.
Thus, the line joining the midpoints is parallel to the other two sides.

Question 3.
The picture shows a parallelogram and the lines joining the midpoint of opposite sides:

(i) Prove that these lines bisect each other.
(ii) Prove that an angle between these lines is equal to an angle between the sides of the parallelogram.
Answer:
(i) Let the parallelogram be ABCD.
EF: joining midpoints of AB and CD
GH: joining midpoints of BC and AD

In parallelogram ABCD:
Opposite sides are parallel and equal:
AB || CD, BC || AD
E and F are midpoints,
AE = EB, CF = FD
Because AB || CD, the segment EF is a mid-segment between two parallel lines.
Similarly, GH is a mid-segment of the pair AD || BC.
The segment joining midpoints of one pair of opposite sides is parallel to the other pair of opposite sides.
Thus EF || AD || BC; GH || AB || CD
Because EF is parallel to BC and AD, and GH is parallel to AB and CD, the shape EFGH is a parallelogram (by opposite sides being parallel).
In every parallelogram, the diagonals bisect each other.
Therefore, X is the midpoint of both EF and GH.

(ii) We already saw:
EF || AD || BC
GH || AB || CD
So the angle between the lines EF and GH is equal to the angle between the pairs of parallel sides:
∠FXG = ∠HAE
Because corresponding angles between parallel lines are equal.
Thus, the angle between the mid-point joining lines equals an interior angle of the parallelogram.
Hence proved.

Question 4.
Draw a parallelogram with lengths of diagonals 6.5 centimetres and 4.5 centimetres and the angle between them 70°.
Answer:

Draw a line segment AC = 6.5 cm.
Mark its midpoint O (at 3.25 cm).
Through O, draw a line making an angle of 70° with AC.
On this new line, mark points B and D such that OB = 2.25 cm and OD = 2.25 cm (half of 4.5 cm).
Join A, B, C, and D.

Question 5.
Prove that if the diagonals of a parallelogram are of the same length, then it is a rectangle.
Answer:

Let ABCD be a parallelogram where diagonals AC = BD.
Consider the triangles ∆ABC and ∆BAD.
AB = AB (Common side).
BC = AD (Opposite sides of a parallelogram are equal).
AC = BD (Given).
By the SSS congruence rule,
∆ABC ≅ ∆BAD.
Therefore, their corresponding angles are equal:
∠ABC = ∠BAD
Since ABCD is a parallelogram, adjacent angles sum to 180° (∠ABC + ∠BAD = 180°).
Since they are equal,
2∠ABC = 180°
∴ ∠ABC = 90°.
A parallelogram with a right angle is a rectangle.

Question 6.
Draw a rectangle with a length of the diagonals of 6 centimetres and the angle between them 60°.
Answer:
Construction Steps:
Draw a line segment AC = 6 cm, and mark its midpoint O.
Through O, draw a line making an angle of 60° with AC.
On this line, mark points B and D such that OB = 3 cm and OD = 3 cm
Since the diagonals of a rectangle are equal and bisect each other, half of 6 cm is 3 cm.
Join A, B, C, and D.
You will get the required rectangle.

Area (Pages 187-188)

Question 1.
Draw a parallelogram with lengths of sides 5 centimetres and 6 centimetres, and an area of 25 square centimetres.
Answer:
Area of a parallelogram = one side × distance to the opposite side.
Since 5 × distance to the opposite side is 25 sq.cm.
Distance to the opposite side is 5 cm.
Draw a square ABCD of side 5 cm.
Draw an arc of radius 6 cm with centre A to cut DC at E.
Length of radius 6cm with centre B and arc of radius 5 cm with centre E meet at F.
Draw BC and FC.
ABFE is the required parallelogram.

Question 2.
Draw a parallelogram with area 25 square centimetres and perimeter 24 centimetres.
Answer:
Area of the parallelogram is to be 25 cm².
The product of one side and the distance to the opposite side is 25 cm².
One side and the distance to the opposite side can be 5 cm each.
The other side is 12 – 5 = 7 cm.
Draw a square with a side of 5cm and draw an arc of radius 7 cm with A as centre, such that it intersects the side CD.
Mark the point E and join AE.
Draw an arc of radius 7 cm with B as centre, and draw another arc of radius 5 cm with E as centre. Two arcs meet at the point P.
Join BF and CF.

Question 3.
The picture shows the parallelogram formed by the intersection of two pairs of parallel lines:

What is the area of the parallelogram? And the perimeter?
Answer:
One side of the parallelogram = 4 cm
Distance to the opposite side = 3 cm
Area of the parallelogram = 4 × 3 = 12 sq.cm

BC × DE = 12
BC × 2 = 12
BC = 6, AD = 6
Perimeter = 4 + 6 + 4 + 6 = 20 cm.

Question 4.
Two sides of a parallelogram are 12 centimetres and 10 centimetres, and the distance between the shorter sides is 6 centimetres.
(i) What is the area of the parallelogram?
(ii) What is the distance between the longer sides?
Answer:
Let the parallelogram have:
Longer side = 12 cm
Shorter side = 10 cm
Distance (height) between shorter sides = 6 cm
(i) Area of the parallelogram
Area = base × height
(If we take the shorter side (10 cm) as the base, its height is 6 cm)
Area = 10 × 6 = 60 cm²

(ii) Let the distance between the 12 cm sides be (h).
Since area = base × height for any pair of opposite sides:
60 = 12 × h
⇒ h = \(\frac {60}{12}\)
⇒ h = 5 cm

Rhombuses (Page 190)

Question 1.
Draw a square of area 4\(\frac {1}{2}\) square centimetres.
Answer:
\(\frac {1}{2}\)d² = 4\(\frac {1}{2}\)
⇒ d² = 2 × 4\(\frac {1}{2}\) = 9
⇒ d = 3 cm

Draw a circle of radius 3 cm.
Draw the diameter AC and construct the perpendicular bisector of AC, which meets the circle at B and D.
Join AC, AD, CD, and BD to get the square ABCD.

Question 2.
Draw a rhombus of area 9 square centimetres, which is not a square.
Answer:

Area = \(\frac {1}{2}\) × d₁ × d₂ = 9
d₁ × d₂ = 18
So, 1 × 18 = 18
2 × 9 = 18
3 × 6 = 18
Let d₁ = 1 and d₂ = 18
The product of the diagonals of the rhombus should be 18.
There are three pairs of natural numbers: 18, 1; 9, 2, and 6, 3 as the lengths of the diagonals.
You can find more pairs of fractions.
A rhombus with diagonals 6 cm and 3 cm is drawn here.
You draw the other two.

Draw AC, 6 cm long.
Draw its perpendicular bisector and mark points B and D on it such that the distance from AC is 1.5 cm each.
Join AB, BC, CD, and AD to complete the rhombus.

Question 3.
The picture shows a quadrilateral drawn by joining the midpoints of the diagonals of a rhombus:

(i) Is this quadrilateral also a rhombus? Why?
(ii) The area of the small quadrilateral is 3 square centimetres. What is the area of the large rhombus?
Answer:
(i) The diagonals of the rhombus intersect at O, and they bisect each other at right angles.

OA = OC
\(\frac{\mathrm{OA}}{2}=\frac{\mathrm{OC}}{2}\)
OQ = OS
Similarly, since OB = OD
i.e., OR = OP
The diagonals of PQRS bisect each other.
(Since OQ = OS and OR = OP)
Since the diagonal AC is perpendicular to BD, the diagonals of PQRS are also mutually perpendicular bisectors.
Therefore, PQRS is a rhombus.

(ii) Area of quadrilateral PQRS = 3 sq.cm
\(\frac {1}{2}\)d₁d₂ = 3 sq. cm
Area of rhombus ABCD = \(\frac {1}{2}\) × 2d₁ × 2d₂
= \(\frac {1}{2}\) × 4d₁d₂
= 4 × \(\frac {1}{2}\)d₁d₂
= 4 × 3
= 12 sq. cm

Question 4.
The sides of a rhombus are 10 centimetres long, and one of its diagonals is 16 centimetres long
(i) What is the length of the other diagonal?
(ii) What is the area of the rhombus?
(iii) What is the distance between the opposite sides?
Answer:
(i) d₁ = 16 cm

We have a rhombus with a side of 10 cm and a diagonal BD = 16 cm
We know that the diagonals of a rhombus bisect each other at 90°.
BO = OD = 8 cm
In ∆AOB, by Pythagoras theorem,
AO² + BO² = AB²
⇒ AO² + 8² = 10²
⇒ AO² = 100 – 64 = 36
⇒ AO = 6 cm [By above property]
Hence, AC = 6 + 6 = 12 cm

(ii) Area = \(\frac {1}{2}\)d1d2
= \(\frac {1}{2}\) × 16 × 12
= 96 cm²

(iii) The distance between opposite sides is the height.
Area = Base × Height (∵ Base is the side length)
96 = 10 × Height
Height = \(\frac {96}{10}\)
Height = 9.6 cm

Class 8 Maths Chapter 11 Kerala Syllabus Parallelograms Questions and Answers

Class 8 Maths Parallelograms Questions and Answers

Question 1.
Which of the following is a defining property of every parallelogram?
(A) Diagonals are equal
(B) All sides are equal
(C) Opposite sides are parallel
(D) All angles are 90°
Answer:
(C) Opposite sides are parallel
This is the basic definition of a parallelogram.

Question 2.
If the adjacent sides of a parallelogram are 8 cm and 6 cm, what is its perimeter?
(A) 14 cm
(B) 28 cm
(C) 48 cm
(D) 24 cm
Answer:
(B) 28 cm
Perimeter = 2(8 + 6)
= 2 × 14
= 28 cm

Question 3.
A parallelogram has a base of 10 cm, and the distance to the opposite side (height) is 6 cm. What is its area?
(A) 60 sq.cm
(B) 30 sq.cm
(C) 16 sq.cm
(D) 32 sq.cm
Answer:
(A) 60 sq. cm
Area = Base × Height
= 10 × 6
= 60

Question 4.
In a parallelogram, the diagonals:
(A) Are always equal
(B) Bisect each other
(C) Are always perpendicular
(D) Bisect the angles
Answer:
(B) Bisect each other
A key property of parallelogram diagonals.

Question 5.
If the diagonals of a parallelogram are equal in length, the figure is a:
(A) Rhombus
(B) Trapezium
(C) Rectangle
(D) Kite
Answer:
(C) Rectangle
If diagonals bisect each other and are equal, it’s a rectangle.

Question 6.
Calculate the area of a rhombus whose diagonals are 10 cm and 12 cm.
(A) 120 sq.cm
(B) 60 sq. cm
(C) 100 sq. cm
(D) 22 sq. cm
Answer:
(B) 60 sq. cm
Area of Rhombus = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 10 × 12
= 60

Question 7.
A square has a diagonal of length 8 cm. What is its area?
(A) 64 sq.cm
(B) 16 sq.cm
(C) 32 sq.cm
(D) 48 sq.cm
Answer:
(C) 32 sq.cm
Area of Square = \(\frac {1}{2}\) × diagonal²
= \(\frac {1}{2}\) × 8²
= \(\frac {1}{2}\) × 64
= 32

Question 8.
If one angle of a parallelogram is 70°, what is the measure of the angle opposite to it?
(A) 110°
(B) 70°
(C) 90°
(D) 20°
Answer:
(B) 70°
Opposite angles in a parallelogram are equal.

Question 9.
If one angle of a parallelogram is 60°, what is the measure of an adjacent angle?
(A) 60°
(B) 30°
(C) 120°
(D) 90°
Answer:
(C) 120°
Adjacent angles sum to 180°.
180° – 60° = 120°

Question 10.
A diagonal divides a parallelogram into two triangles. If the area of the parallelogram is 50 sq. cm, what is the area of one of these triangles?
(A) 50 sq.cm
(B) 100 sq.cm
(C) 25 sq.cm
(D) 10 sq.cm
Answer:
(C) 25 sq.cm
A diagonal splits a parallelogram into two equal triangles.
\(\frac {50}{2}\) = 25

Question 11.
Which quadrilateral has diagonals that are perpendicular bisectors of each other?
(A) Rectangle
(B) Parallelogram
(C) Rhombus
(D) Trapezium
Answer:
(C) Rhombus
Diagonals of a rhombus are perpendicular bisectors.

Question 12.
The area of a parallelogram is given by:
(A) Base × Height
(B) \(\frac {1}{2}\) × Base × Height
(C) Product of adjacent sides
(D) Product of diagonals
Answer:
(A) Base × Height

Question 13.
Read the following statements:
Statement I: All squares are rhombuses.
Statement II: All rhombuses are squares.
Choose the correct option:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(A) Statement I is true, Statement II is false
A square has all the properties of a rhombus (equal sides), but a rhombus doesn’t necessarily have 90° angles like a square.

Question 14.
Read the following statements:
Statement I: The area of a square is half the square of its diagonal \(\frac {1}{2}\)d².
Statement II: A square is a rhombus with equal diagonals.
Choose the correct option:
(A) Only I is true.
(B) Only II is true.
(C) Both are true, and II explains I.
(D) Both are true, but II does not explain I.
Answer:
(C) Both are true, and II explains I
Since a square is a rhombus with equal diagonals, we can use the rhombus formula \(\frac {1}{2}\) × d₁ × d₂, which becomes \(\frac {1}{2}\)d².

Question 15.
A quadrilateral where only one pair of opposite sides is parallel is called a:
(A) Parallelogram
(B) Rhombus
(C) Trapezium
(D) Rectangle
Answer:
(C) Trapezium

Question 16.
To construct a unique parallelogram, which of the following sets of measurements is sufficient?
(A) Lengths of two adjacent sides
(B) Lengths of two adjacent sides and the included angle
(C) Length of one side
(D) Measure of one angle
Answer:
(B) Lengths of two adjacent sides and the included angle
Just sides aren’t enough; the angle determines the shape/area.

Question 17.
The perimeter of a rhombus is 40 cm. What is the length of one side?
(A) 20 cm
(B) 5 cm
(C) 10 cm
(D) 8 cm
Answer:
(C) 10 cm
A rhombus has 4 equal sides.
\(\frac {40}{4}\) = 10

Question 18.
In the figure of a parallelogram, if the distance between the longer sides (10 cm) is 4 cm, what is the area?
(A) 14 sq. cm
(B) 20 sq. cm
(C) 40 sq. cm
(D) 80 sq. cm
Answer:
(C) 40 sq. cm
Area = Side × Distance
= 10 × 4
= 40

Question 19.
Read the following statements about a Rhombus:
Statement I: The diagonals are of equal length.
Statement II: The area is half the product of the diagonals.
Choose the correct option:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both are true.
(D) Both are false.
Answer:
(B) Statement I is false, Statement II is true
Diagonals of a rhombus are usually unequal (unless it’s a square).
The area formula is correct.

Question 20.
A parallelogram has sides 12 cm and 8 cm. The distance between the 12 cm sides is 4 cm. What is the distance between the 8 cm sides?
(A) 6 cm
(B) 4 cm
(C) 12 cm
(D) 8 cm
Answer:
(A) 6 cm
Area is constant.
12 × 4 = 48.
So, 8 × h = 48
∴ h = 6

Question 21.
Draw a rhombus of diagonals 5.5 cm and 3 cm in your notebook.
Answer:
Draw a line of length 5.5 cm, and find its midpoint by drawing the perpendicular bisector.
Mark the points on the upper and lower parts of the bisector line at a distance of 1.5 cm from the intersecting point of the first line and the perpendicular bisector.
Join these points to the end of the first line.

Question 22.
Draw these figures
1. Two equal rhombuses

2. Parallelograms on two sides of a square

Answer:
1. From the figure two sides of ∆BCD are equal, angles opposite these sides are also equal.
We can calculate them as 50° each.
In the same way, find other angles in the figure.

Draw a line BD vertically, 3 cm long.
At D, draw angles of 50° on both sides.
At B, also draw angles of 50° on both sides.
Then we get a rhombus ABCD.
Extend BC to G such that BC = CG, and extend DC to E such that DC = CE.
Draw GE. Draw an angle of 50° at G and E to find F.

2. Draw a square of side 3 cm.
And draw two parallelograms with sides 3 cm, 2 cm, and an angle between them of 45°, on the sides of the square.

Question 23.
In rhombus PQRS, PR = 7 cm and QS = 5 cm. Construct a rhombus PQRS.
Answer:

Question 24.
In the figure, ABCD is a parallelogram. D = 80°. Find all other angles?

Answer:
ABCD is a parallelogram
Opposite angles are equal.
B = 80°.
The sum of the angles on the same side is 180°.
∠A + ∠B = 180°.
∠A = 180° – 80° = 100°
And ∠C = 100° (opposite angles are equal).

Question 25.
In the figure, ABCD is a parallelogram. Find x, y, z.

Answer:
∠Y = 112° (opposite angles are equal)
In ADC, ∠x + ∠y + 40 = 180° (sum of angles in a triangle)
∠x + 112° + 40° = 180°
∠x = 180° – 152° = 28°
∠z = 28° (transversal alternate interior angles are equal).

Question 26.
In parallelogram ABCD, the diagonals AC and BD intersect at O. AC = 6.5 cm, BD = 7 cm, and ∠AOB = 100°. Construct the parallelogram.
Answer:

Question 27.
The diagonals of a rhombus are of lengths 16 cm and 12 cm. What is its perimeter?
Answer:

In right-angled AOB,
AO = 8 cm, BO = 6 cm, AOB = 90°
AB² = AO² + BO²
= 8² + 6²
= 64 + 36
= 100
Side, AB = 10 cm
Perimeter = 4 × 10 = 40 cm.

Question 28.
Draw the following patterns.
(a) 6 equal rhombuses

(b) 3 equal rhombuses

(c) Two equal rhombuses on the sides of a square

(d) 4 equal rhombuses

(e) Parallelograms on two sides of a rectangle of sides 6 cm and 3 cm.

Answer:
(a) Draw a circle of radius 2 cm with centre O.
Divide the centre of the circle into angles of 60° each.
These lines meet the circle at the points A, B, C, D, E, and F.
Draw an arc of 2 cm from A and B to get G.
In the same way, find H, I, J, K, L.
Draw needed part.

(b) Draw a square of side 4 cm.
Draw rhombuses with a side of 4 cm and an angle of 30° on the top and bottom sides of the square.
Complete the figure.

(c) Angles around the point at which three rhombuses joined together are 120° each.
Since one angle of the rhombus is 120°, another angle is 60°.
Draw three rhombuses with a side of 4 cm and an angle of 60°.
Complete the figure.

(d) Draw a semicircle of radius 2 cm with centre O.
Divide the centre of the circle into angles of 45° each.
These lines meet the circle at the points A, B, C, D, and E.
Draw an arc of 2 cm from A and B to get F.
In the same way, find G, H, and I.
Complete the figure.

(e) Draw a rectangle of length 6 cm and breadth 3 cm.
Draw two rhombuses on both sides of the rectangle, which makes angle 45° and 135° with length and breadth, respectively.

Question 29.
In a parallelogram ABCD, find x, y, z from the adjoining figure.

Answer:
ABCD is a parallelogram,
∠C = 45° (Opposite angles are equal)
∠C + Z = 180° (linear pair)
Z = 180° – 45° = 135°
45° + Y = 180° (Sum of the angles on the same side is 180°)
Y = 180° – 45° = 135°
Since Y = 135°
X = 135° (Opposite angles are equal)

Question 30.
Which of the following has a greater area: a parallelogram with one side 12 cm and a distance between parallel sides of 6 cm, or a square with diagonals of 12 cm each?
Answer:
Area of parallelogram = \(\frac {1}{2}\) × b × h
= \(\frac {1}{2}\) × 12 × 12
= 72 cm²
Area of square = \(\frac {1}{2}\) × d²
= \(\frac {1}{2}\) × 12²
= \(\frac {1}{2}\) × 144
= 72 cm²
Area of both are equal.

Question 31.
Find the area of a rhombus with one side of 6 cm and one diagonal of 6 cm.
Answer:
The diagonals of a rhombus bisect each other perpendicularly.

In ∆AOB, AO² = AB² – BO²
= 6² – 3²
= 36 – 9
= 25
AO = √25 = 5 cm
AC = 10 cm
Area of the rhombus = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 6 × 10
= 30 sq. cm

Question 32.
In the pictures given below, which one has more area?

Answer:
A rectangle has the maximum area among the parallelograms with the same sides.

Question 33.
The ratio of the two adjacent sides of a parallelogram is 3 : 2. The distance between the longer sides is 10 cm. If the area is 900 cm², find the sides of the parallelogram.
Answer:
Let the sides be 3x and 2x

3x × 10 = 900
⇒ 30x = 900
⇒ x = 30 cm
Sides are: 3 × 30 = 90 cm; 2 × 30 = 60 cm

Question 34.
If one side of a parallelogram is ‘a’ and the height of that side is h, prove that the area = ah.
Answer:
In the figure, ABCD is a parallelogram.
AB = a

The perpendicular distance from D to AB = h
By drawing the diagonal BD, we can divide the parallelogram into two equal triangles.
∆ABD and ∆BCD are equal triangles; their areas are equal.
i.e, the area of the parallelogram is two times of the area of ∆ABD.
Area of ∆ABD = \(\frac {1}{2}\) × a × h
Area of the parallelogram ABCD = 2 × \(\frac {1}{2}\) × ah = ah

Question 35.
PQRS is a rhombus. If the diagonals are 8 cm and 9 cm each, compute the area.

Answer:
Let the diagonals of the rhombus be d₁ and d₂
Area = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 8 × 9
= 36 cm²

Question 36.
The perimeter of a rhombus is 40 cm. If the length of one diagonal is 16 cm. What is the length of the other diagonal? Find the area?
Answer:
Perimeter = 40 cm
One side = 10 cm
∆POQ is a right-angled triangle
Since d₁ = 16 cm, OP = 8 cm
8² + OQ² = 10²
OQ² = 100 – 64 = 36

OQ = 6 cm
QS = 12 cm
d₁ = 16, d₂ = 12 cm
Area = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 16 × 12
= 8 × 12
= 96 sq.cm

Question 37.
What is the maximum area of a parallelogram with sides 8 cm and 5 cm? What is the speciality of the parallelogram of maximum area?
Answer:
The area will be maximum for a rectangle, and the maximum area is 40 cm².

Question 38.
The area of a rhombus is 112 sq. cm, and one of its diagonals is 16 cm long. Find the length of the other diagonal.
Answer:
Area of the rhombus = \(\frac {1}{2}\) × d₁ × d₂ = 112 sq. cm
⇒ \(\frac {1}{2}\) × d₂ × 16 = 112
⇒ d₂ = 14 cm

Class 8 Maths Chapter 11 Notes Kerala Syllabus Parallelograms

Parallelograms
A parallelogram is a quadrilateral where both pairs of opposite sides are parallel.
In any parallelogram, opposite sides are equal in length.

For example, ABCD is a parallelogram. Therefore, we can say,
AB and CD are parallel and equal
AD and BC are parallel and equal

Diagonals
The diagonals of a parallelogram bisect each other. This means they cut each other into two equal parts.
If the diagonals of a parallelogram are equal in length, the figure is a rectangle.

In the general parallelogram ABCD, we can relate the following properties:

• Two pairs of parallel sides: AB || DC and AD || BC
• Opposite sides are equal: AB = DC and AD = BC
• Equal opposite angles: ∠A = ∠C and ∠B = ∠D
• Diagonals bisect each other: AO = CO and DO = BO

In any parallelogram, the diagonals bisect each other; on the other hand, any quadrilateral in which the diagonals bisect each other is a parallelogram.

Drawing a Parallelogram
1. Sides and angles between them are given
Example:
Draw AB = 6 cm.
Draw lines of length 4 cm, 45° slanted at both ends of AB.
Join the ends of these lines.

2. Diagonals and the angle between them are given
Example:
Draw AC = 6 cm.
Mark the midpoint of AC.
Through the midpoint of AC, draw a line that makes an angle of 70° with the first line.
Mark B and D are on either side of the midpoint, which is 3 cm from it.
Complete the figure.

3. One diagonal and two sides are given.
Example:
Make a triangle with sides 7 cm, 4 cm, and 8 cm
Taking the 8 cm long side as one side, and the other two sides 7 cm and 4 cm, draw another triangle.
Complete the figure.

4. Two diagonals and one side are given.
Example:
Draw AB = 6 cm, mark ‘O’ which is 4 cm away from A and 3 cm away from B.
Complete AOB, extend line AO and BO at C and D respectively, OC = 4 cm and OD = 3 cm.
Complete the figure.

Area of a Parallelogram

The area of a parallelogram is the product of the length of one side and its distance to the opposite side.
Area = base × height = b × h

Rhombuses
A rhombus is a special parallelogram where all four sides are equal.

Area of Rhombuses
The area of any rhombus is half the product of the diagonals

The area of any square is half the square of its diagonal.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 18 Cells that Become Daughter Cells Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 18 Cells that Become Daughter Cells Question Answer Notes

Class 8 Basic Science Chapter 18 Notes Kerala Syllabus Cells that Become Daughter Cells Question Answer

Cells that Become Daughter Cells Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Complete the table by selecting the appropriate human reproductive organs and associated parts from the box.

Answer:

Details	Male reproductive system	Female reproductive system
1. Function	1. Produces sperm and sex hormones	5. Produces eggs and female sex hormones
2. Important parts	2. Testes, scrotum, penis	6. Ovary, fallopian tube, uterus, vagina
3. Where gamete production occurs	3. In the testes	4. In the ovary

Question 2.
Find the incorrect statements given below and rewrite them correctly.
a) The number of chromosomes in germ cells is 46.
b) The number of chromosomes in a human zygote is 92.
c) Budding is an example of asexual reproduction.
d) Four eggs are produced from a single germ cell of an ovary.
Answer:
a) Correct. (Germ cells in the reproductive organs, before meiosis, are diploid and have 46 chromosomes).

b) Incorrect.
Correction: The number of chromosomes in a human zygote is 46. (It is formed by the fusion of a sperm with 23 chromosomes and an egg with 23 chromosomes).

c) Correct. (As seen in Yeast).

d) Incorrect.
Correction: One egg (ovum) and three polar bodies are produced from a single germ cell of an ovary during meiosis.

Question 3.
Complete the flowchart and give an appropriate title.

Answer:

Basic Science Class 8 Chapter 18 Question Answer Kerala Syllabus

Textbook Page No : 288

Question 1.
Analyse the illustration and completethe given Table 18.1.?


Answer:

Number	Modes of reproduction	Examples
1	New plants grow from the leaf (Vegetative propagation from leaf)	Bryophyllum Sansevieria
2	New plants grow from the root	Curry leaf plant
3	New plants grow from the stem (Vegetative propagation from stem cutting)	Tapioca, Rose, Hibiscus, Sugarcane
4	Under favourable conditions, the cell divides and becomes two organisms.	Amoeba, Paramecium, Euglena
5	In unfavourable conditions, cells in organisms like plasmodium develop a thick outer covering.The cytoplasm and nuclear materials inside the cell divide into several parts. When the conditions become favourable, the outer covering of the cell breaks, open, releasing many tiny cells, each of which grows into a new organism.	Plasmodium
6	Regeneration: Body parts break off and regenerate into new organisms.	Planaria, Hydra
7	Budding: A small outgrowth (bud) forms on the parent cell, develops, and separates to form a new organism.	Yeast

Textbook Page No : 289

Question 2.
What is the fundamental difference between the reproductive methods we have discussed so far and the method of reproduction shown in the illustration below?

Answer:
Illustration 18.1 shows plants growing from seeds. This method is fundamentally different from the asexual methods discussed above because it involves the formation of seeds through sexual reproduction.

Question 3.
Observe the structure of flowers, which are the reproductive organs of plants and complete the illustration 18.2.

Answer:

Textbook Page No : 290

Question 4.
How do pollen grains from the anther reach the stigma?
Answer:
The process by which pollen grains are transferred from the anther (male part) to the stigma (female part) is called pollination. Since pollen grains cannot move on their own, they rely on external “agents” to carry them. Animals like insects and birds as well as wind and water, help in pollination

Question 5.
What happens to the pollen grain after pollination?
Answer:
After pollination, the pollen grain starts a journey to fertilize the egg inside the ovary. This process is called fertilization.

Question 6.
Shall we try a simple experiment with the help of the teacher? Prepare a note with diagram based on the experiment you conducted. [Illustration 18.3]

Answer:
Collect mature pollen grains from a flower’s anther.
Place them on a slide with a drop or two of sugar solution.
Observe under a microscope after 3 – 4 hours.
Observation: The pollen tube grows from the deposited pollen grain down towards the ovary. This tube carries the male gametes to fertilize the ovule inside the ovary.

Textbook Page No : 292
Indicators

Question 7.
Function of the tube nucleus …………………

Answer:

Controls the formation and growth of the pollen tube, then disintegrates.

Question 8.
Function of the pollen tube
Answer:
The pollen tube grows down through the style and enters the ovule, transporting the two male gametes into the ovary.

Question 9.
Formation of male gametes
Answer:
Generative nucleus divides to form two male gametes

Question 10.
Fertilization
Answer:
One of the male gametes that reaches the ovule through the pollen tube fuses with the egg cell to form a zygote. This process is known as fertilization.

Question 11.
Formation of endosperm
Answer:
The zygote grows and becomes an embryo. The embryo then grows into a seedling. The second male gamete fuses with the polar nuclei to form the endosperm. The endosperm stores the food needed for the growth of the embryo.

Textbook Page No : 294

Question 12.
Parts and functions of the reproductive system in human being.
Answer:

Parts	Functions
Male reproductive System
Vas Deferens	Transports male gametes from the testes to the urethra.
Prostate gland	It produces a fluid that contains the substances needed for the nourishment and movement of the sperms.
Testis	It produce sperm and male sex hormones (testosterone)
Penis	Through this duct, both urine and sperm are discharged. It helps to deliver the sperm into the vagina.
Female reproductive system
Uterus	The part that completes the growth of the embryo
Fallopian tube	It carries the ovum to the uterus. Fertilization takes place here.
Ovary	Ovary produces the ovum and female sex hormones, estrogen and progesterone.
Vagina	The part that opens the uterus to the outside. It is where the semen is deposited. During childbirth, the baby comes out through this part.
	The part that completes the growth of the embryo

Question 13.
Production of gametes
Answer:
Testis produces sperm in male. Ovary produces the ovum in females.

Question 14.
Sexual hormones
Answer:
Testis produces male sex hormones (testosterone). Ovary produces the female sex hormones, estrogen and progesterone.

Textbook Page No : 295

Question 15.
Complete the table 18.2

Answer:

Characteristics	Sperm	Ovum
Shape	Head, body, and tail	Spherical
Size	Smaller	Larger
Motility	Motile (using tail)	Non-motile

Textbook Page No : 297
Indicators

Question 16.
Site of fertilization.
Answer:
Fallopian Tube (specifically the ampulla region). This is where the sperm meets and fuses with the egg.

Question 17.
Embryo formation.
Answer:
Uterus (Womb). The fertilized egg divides to form an embryo while traveling, but it implants and develops into a baby in the wall of the uterus.

Question 18.
Menstruation.
Answer:
Uterus.Menstruation is the shedding of the inner lining of the uterus (endometrium), which flows out through the vagina.

Textbook Page No : 298
Let’s Find

Question 19.
What happens when cell division occurs in unicellular organisms?
Answer:
When a unicellular organism divides, it is reproducing, creating exact copies of itself, and the original parent cell ceases to exist as a single entity.

Textbook Page No : 299

Question 20.
Formation of chromosomes.
Answer:
Occurs during Prophase as the chromatin network condenses.

Question 21.
Stages of cell division
Answer:
Prophase, Metaphase, Anaphase, Telophase.

Question 22.
Number of chromosomes in the parent cell and daughter nuclei.
Answer:
The number of chromosomes in the daughter nuclei is the same as the number in the parent cell, (e.g., If the parent human cell had 46 chromosomes, each daughter nucleus will also have 46 chromosomes). Karyokinesis maintains the chromosome number.

Question 23.
Number of daughter nuclei formed in one division.
Answer:
Two daughter nuclei are formed in one division.

Textbook Page No : 301

Question 24.
Complete the illustration 18.13.

Answer:

Textbook Page No : 303
Indicators

Question 25.
Number of chromosomes in human germ cells
Answer:
23 chromosomes.
Germ cells (sperm and egg) are haploid (n). This is exactly half the number found in normal body cells (which have 46).

Question 26.
Characteristic of Meiosis Stage I
Answer:
It is a Reductional Division. The homologous chromosome pairs separate. This reduces the chromosome number by half (e.g., from 46 to 23), which is why the resulting cells are haploid.

Question 27.
Similarity between Meiosis Stage II and Mitosis
Answer:
Both involve the separation of sister chromatids. In both processes, the centromeres split and the individual chromatids are pulled to opposite poles. Because the chromosome number does not drop further, both are considered “equational divisions.”

Question 28.
Difference observed in the process of gamete formation in males and females
Answer:
The process is called Spermatogenesis in males and Oogenesis in females.

Spermatogenesis	Oogenesis
The cytoplasm is shared equally among daughter cells.	Most cytoplasm goes to one big egg; the others become tiny “polar bodies.”
One parent cell produces 4 active sperm.	One parent cell produces only 1 active egg.
Starts at puberty and is continuous.	Starts before birth (fetus stage), pauses, and resumes at puberty.

Question 29.
Complete the table 18.3.

Answer:

Details	Mitosis	Meiosis
In which cells does it occur?	In the normal cells of the body	In the germinal cells of the reproductive organs
Changes in chromosome number	No change (Remains the same)	Reduced to half
Number of daughter cells	2	4
Significance	Helps in the growth of the body and repair of tissues.	Germ cells are formed. In organisms that undergo sexual reproduction helps to maintain a constant number of chromosomes across generations.

Textbook Page No : 305

Question 30.
Complete the table 18.4.

Answer:

Details	Fraternal twins	Identical twins	Siamese twins
Formation method	Two eggs fertilized by two different sperm.	Single zygote divides completely into two.	Incomplete division of zygote OR fusion of separated cells.
Structural similarity	Will not be the same.	Will be the same.	Will be the same; bodies are joined.
Hereditary factors	Will not be the same.	Will be the same.	Will be the same.
Sexual characteristics	May be the same or different gender.	Will always be the same sex.	Will always be the same sex.

Textbook Page No : 305

Question 31.
Analyse illustrations 18.16, 18.17, and their descriptions, and write down your conclusions.


Answer:
Humans grow through distinct stages from birth to o|d age, including infancy, childhood, adolescence, adulthood, and old age. A particularly significant stage is adolescence. Adolescence (known as Kaurmaram in Malayalam) is the transitional phase leading from childhood to full maturity, generally spanning from ages 10 to 19. This period is characterized by rapid physical, mental, and emotional changes. Physically, individuals experience a growth spurt with a sudden increase in height and weight, alongside accelerated growth of the reproductive organs and increased activity of glands in the skin. Simultaneous with these physical changes, the brain continues to develop, resulting in significant shifts in thinking, emotions, and social interactions. It is also observed that the parts of the brain controlling these changes develop earlier in girls, leading to faster growth during puberty compared to boys.

Class 8 Basic Science Chapter 18 Question Answer Extended Activities

Question 1.
Prepare suitable questions about the human reproductive system and sexual hygiene, conduct an interview with a doctor and prepare a report.

Question 2.
Prepare a chart/slide presentation about diseases related to the reproductive organs.

Cells that Become Daughter Cells Class 8 Notes

Class 8 Basic Science Cells that Become Daughter Cells Notes Kerala Syllabus

Reproduction

• The biological process of producing new individual organisms (offspring), ensuring the continuation of a species.
• Types:
  1. Asexual Reproduction: From a single parent, offspring are genetically identical (e.g., Binary Fission, Budding, Regeneration, Spore Formation, Vegetative Propagation).
  2. Sexual Reproduction: Involves the fusion of male and female gametes (formed by meiosis) through fertilization to form a zygote. Offspring are genetically diverse.

ഒരു ജീവിയിൽ നിന്ന് പുതിയ തലമുറ ഉണ്ടാകുന്ന പ്രക്രിയയാണ് പ്രത്യുത്പാദനം. ഇത് വർഗ്ഗത്തിന്റെ (species) തുടർച്ച ഉറപ്പാക്കുന്നു. ഒരൊറ്റ ജീവിയിൽ നിന്ന് പുതിയ ജീവികളു ണ്ടാകുന്ന രീതിയാണ് അലൈംഗിക പ്രത്യുത്പാദ നം (Asexual Reproduction). ഉദാഹരണങ്ങൾ: ഇലമുളച്ചിയിൽ ഇലയിൽ നിന്നും, കപ്പയിൽ തണ്ടിൽ നിന്നും പുതിയ ചെടികളുണ്ടാകുന്നത് (കായിക പ്രജനനം / Vegetative Propagation); ബാക്ടീരിയ രണ്ടായി വിഭജിക്കുന്നത് (Binary Fission); പ്ലാനേറിയ മുറിഞ്ഞ ഭാഗങ്ങളിൽ നിന്ന് വളരുന്നത് (Regeneration); യീസ്റ്റിൽ മുകുള ങ്ങൾ ഉണ്ടാകുന്നത് (Budding).

Sexual Reproduction in Plants: Seeds
□ Plants growing from seeds. This method is fundamentally different from the asexual methods discussed above because it involves the formation of seeds through sexual reproduction.

Structure of a Flower
Main Parts:

• Petals (ദളങ്ങൾ): Often brightly coloured to attract pollinators.
• Sepals (വിതളങ്ങൾ): Usually green, leaf-like structures that protect the bud.
• Pedicel (പൂഞ്ഞട്ട്): The stalk of the flower.
• Thalamus: The part of the stalk where the floral organs are attached.

Reproductive Organs:

• Stamen (കേസരം): The Male reproductive organ.
• Anther (പരാഗി): The part that produces pollen grains.
• Filament: The stalk that holds the anther.
• Pistil / Carpel: The Female reproductive organ.
• Stigma: The receptive tip where pollen lands.
• Style: The stalk connecting the stigma to the ovary.
• Ovary: Contains the ovules, which develop into seeds after fertilization.

Pollination
Pollen grains containing the male gametes are formed in the anther (part of the stamen). The process of these pollen grains settling on the stigma (part of the pistil) is called pollination.

• Pollinating Agents: Animals like insects and birds, as well as wind and water, help in pollination.
• The polien tube grows from the deposited pollen grain down towards the ovary. This tube carries the male gametes to fertilize the ovule inside the ovary.
• After pollination, the pollen grain germinates on the stigma and the pollen tube grows towards the ovule inside the ovary.

Fertilization
One of the male gametes that reaches the ovule through the pollen tube fuses with the egg cell to form a zygote. This process is known as fertilization.

• The zygote grows and becomes an embryo. The embryo then grows into a seedling.
• The second male gamete fuses with the polar nuclei to form the endosperm. The endosperm stores the food needed for the growth of the embryo.

Endosperm vs. Cotyledons:

• In monocot plants (like paddy, maize, coconut), the endosperm itself is the primary stored food used during germination, and it’s often the edible part.
• In dicot plants, the food required for the embryo is stored in the cotyledon.

പരാഗണത്തിനുശേഷം (Pollination) പരാഗരേണു (pollen grain) മുളച്ച് പരാഗനാളി (pollen tube) അണ്ഡാശയത്തിലെ (Ovary) അണ്ഡത്തിലേക്ക് (Ovule) വളരുന്നു. പരാഗനാളിയിലൂടെ എത്തുന്ന രണ്ട് പുരുഷബീജങ്ങളിൽ (male gametes) ഒന്ന് അണ്ഡവുമായി (Ovum) ചേർന്ന് സിക്താണ്ഡം (Zygote) ഉണ്ടാകുന്നു. ഇതാണ് ബീജസങ്കലനം (Fertilization). സിക്താണ്ഡം വളർന്ന് ഭ്രൂണം (Embryo) ആകുന്നു. രണ്ടാമത്തെ പുരുഷബീജം അണ്ഡത്തിലെ പോളാർ ന്യൂക്ലിയസുമായി ചേർന്ന് എൻഡോസ്പേം (Endosperm) ഉണ്ടാകുന്നു. എൻഡോസ്പേം ആണ് ഭ്രൂണത്തിന്റെ വളർച്ചക്കാവ ശ്യമായ ആഹാരം സംഭരിക്കുന്നത്.

Fertilization
Like plants, animals also have specialized organ systems for reproduction. Reproduction in humans is sexual reproduction.
Hormones: Chemical substances produced by endocrine glands that control and coordinate various life activities, including reproduction.

Male reproductive System
In males, the testes are located in the scrotal sac just below the penis. For sperm production, a temperature 2 to 2.5 degrees Celsius lower than the normal body temperature is to be maintained. The contraction and relaxation of the scrotal sac helps in maintaining this temperature.

Female reproductive System

Gametes in Human Beings

Fertilization in Human Beings

• Semen: A combination of sperm cells formed in the testes and secretions from glands like the prostate.
• Ejaculation: The process by which semen is expelled through the penis.
• Sperm Transport: Sperm cells reach the vagina, pass through the uterus, and enter the fallopian tube.
• Fertilization: In the fallopian tube, one sperm unites with the egg (ovum) that has arrived there from the ovary. The process of the sperm uniting with the egg is called fertilization.
• Zygote Formation: Fertilization results in the formation of a zygote.
• Embryo Development: The zygote begins to divide and develops into an embryo. The embryo travels to the uterus and implants in the uterine wall (endometrium) to continue its growth.

പുരുഷബീജവും (Sperm) അണ്ഡവും (Ovum) സംയോജിക്കുന്ന പ്രക്രിയയാണ് ബീജസങ്കലനം (Fertilization). ഇത് സാധാരണയായി അണ്ഡവാ ഹിയിൽ (Fallopian tube) വെച്ചാണ് നടക്കുന്നത്. ബീജസങ്കലനത്തിന്റെ ഫലമായി സിക്താണ്ഡം (Zygote) രൂപപ്പെടുന്നു, അത് പിന്നീട് വളർന്ന് ഭ്രൂണമായി (Embryo) മാറുന്നു.

Ovulation & Menstruation

Illustration shows what happens after the ovary releases an egg.

• Ovulation: The process in which a fully matured ovum is released from the ovary.
• Preparation for Pregnancy: Along with ovulation, the thickness of the endometrium (inner lining of the uterus) increases, and more tissues and blood capillaries are formed in it, preparing to receive a fertilized egg.

Two Possibilities After Ovulation:

1. If Fertilization Occurs:

• A zygote is formed in the fallopian tube.
• The zygote develops into an embryo.
• The embryo implants in the thickened endometrium of the uterus and continues to grow. Pregnancy begins.

2. If Fertilization Does Not Occur:

• The released ovum disintegrates.
• The thickened endometrium (newly formed tissues and blood capillaries) is no longer needed.
• These tissues break down and are discharged through the vagina along with blood and mucus. This process is called Menstruation.

അണ്ഡാശയത്തിൽ നിന്ന് പൂർണ്ണവളർച്ചയെത്തിയ അണ്ഡം (Ovum) പുറത്തുവരുന്ന പ്രക്രിയയാണ് അണ്ഡവിസർജനം (Ovulation). ഇതോടൊപ്പം ഗർഭാശയത്തിന്റെ (Uterus) ആന്തരഭിത്തിയായ എൻഡോമെട്രിയം (Endometrium) കട്ടിയുള്ളതാ യി മാറുന്നു. ബീജസങ്കലനം (Fertilization) നടന്നാൽ, ഉണ്ടാകുന്ന ഭ്രൂണം (Embryo) ഈ ഭിത്തി യിൽ പറ്റിപ്പിടിച്ച് വളരും. ബീജസങ്കലനം നടന്നില്ലെ ങ്കിൽ, അണ്ഡം നശിച്ചുപോകുകയും, കട്ടിയുള്ള എൻഡോമെട്രിയം പാളി രക്തത്തോടൊപ്പം യോനി യിലൂടെ (Vagina) പുറന്തള്ളപ്പെടുകയും ചെയ്യുന്നു. ഇതാണ് ആർത്തവം (Menstruation).

Growth & Reproduction
Growth: From zygote to organism involves Cell Division, Cell Growth, and Cell Differentiation.

Cell Division Types:
1. Mitosis: For growth, repair, asexual reproduction. Produces two identical daughter cells with the same chromosome number as the parent. Stages: Prophase, Metaphase, Anaphase, Telophase. Cytokinesis differs in plant (cell plate) and animal (furrowing) cells.

• Mitosis takes place in two stages:
• Karyokinesis (division of the nucleus)
• Cytokinesis (division of the cytoplasm)

Karyokinesis: Each organism has a specific number of chromosomes. In human cells, the chromosome number is 46.

2. Meiosis: For gamete formation in sexual reproduction. Produces four daughter cells with half the chromosome number of the parent germ cell (e.g,, 46 → 23). In females, produces 1 ovum + 3 polar bodies. Maintains constant chromosome number across generations.

ശരീര വളർച്ചയ്ക്കും കേടുപാടുകൾ തീർക്കുന്നതി നും സഹായിക്കുന്ന കോശവിഭജനമാണ് ക്രമഭംഗം (Mitosis). ഇതിൽ ക്രോമസോം സംഖ്യയ്ക്ക് മാറ്റം വരുന്നില്ല. എന്നാൽ ലൈംഗിക പ്രത്യുത് പാദനത്തിനാവശ്യമായ ബീജകോശങ്ങൾ (Gametes) രൂപപ്പെടുന്നത് ഊനഭംഗം (Meiosis) വഴി യാണ്. ഇതിൽ ക്രോമസോം സംഖ്യ പകുതിയായി കുറയുന്നു (ഉദാ: 46 → 23).

Twins

• Identical Twins: Single zygote splits into two; genetically identical, same sex.
• Fraternal Twins: Two separate eggs fertilized by two different sperm; genetically like siblings, can be different sexes.
• Conjoined Twins: Incomplete splitting of a zygote; bodies joined, genetically identical, same sex.

Adolescence

• Age: ~10 to 19 years; transition from childhood to maturity.
• Changes: Growth spurt, reproductive organ growth, brain development, skin changes (acne due to sebum), mental/ emotional changes.
• Puberty: Physical changes for reproductive capability.
• Health: Importance of balanced nutrition, avoiding harmful influences (drugs, abuse), wise use of media.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 17 The Beauty of Diversity Important Questions ensure conceptual clarity.

The Beauty of Diversity Extra Questions and Answers Class 8 Basic Science Chapter 17 Kerala Syllabus

The Beauty of Diversity Class 8 Important Questions

Question 1.
Regarding the levels of biodiversity, choose the correct statements.
i. Genetic diversity refers to the diversity at the ecosystem level.
ii. Species diversity refers to the variation within a species.
iii. Ecosystem diversity refers to the variety of different habitats in a region.
iv. Different varieties of mangoes in India represent genetic diversity.
(a) i, ii correct
(b) iii, iv correct
(c) i, iii correct
(d) ii, iv correct
Answer:
(b) iii, iv correct

Question 2.
Choose the correct statements about decomposers.
i. Green plants are examples of decomposers.
ii. Decomposers break down complex molecules in dead remains into simpler ones.
iii. Bacteria and Fungi are examples of decomposers.
iv. Decomposers are placed at the top of the food chain.
a) i, iv correct
b) ii, iii correct
c) i, iii correct
d) All are correct
Answer:
(b) ii, iii correct

Question 3.
Select the correct statements regarding ecological interactions.
i. In mutualism, both organisms benefit.
ii. In parasitism, both organisms are harmed.
iii. In commensalism, one benefits, and the other is neither harmed nor benefited.
iv. Predation is beneficial to both the predator and the prey.
a) i, iii correct
b) ii, iv correct
c) i, ii, iii correct
d) All are correct
Answer:
(a) i, iii correct

Question 4.
Complete the food chain by placing the organisms in the correct sequence: Snake, Eagle, Paddy, Rat.
a) _________ → (b) _________ → (c) _________ → (d) _________
Answer:
a) Paddy
b) Rat
c) Snake
d) Eagle

Question 5.
Fill in the blanks representing the trophic levels in the energy pyramid shown in Illustration 17.4.

Trophic level 4: (a) _________Consumers
Trophic level 3: Secondary Consumers
Trophic level 2: (b) _________Consumers
Trophic level 1: (c) _________
Answer:
(a) Tertiary
(b) Primary
(c) Producers

Question 6.
Complete the diagram illustrating the Carbon Cycle by filling in the processes (A), (B), and (C) based on Illustration 17.7.

Atmospheric CO₂ → (A) Plants → (B) Animals → (C) Atmospheric CO₂
Answer:
A) Photosynthesis
B) Consumption / Eating
C) Respiration

Question 7.
Define the following terms:
a) Habitat
b) Ecosystem
Answer:
a) Habitat: The natural surroundings where each living being lives.
b) Ecosystem: The community of plants, animals, and microorganisms in an area interacting with each other and with abiotic factors.

Question 8.
Give one example for each of the following ecological interactions:
a) Mutualism
b) Parasitism
Answer:
a) Mutualism: Flower and Butterfly (pollination for nectar) OR Lichens (Algae and Fungus).
b) Parasitism: Flea on a Dog OR Lice on humans.

Question 9.
What is the Red Data Book, and which organisation publishes it?
Answer:
The Red Data Book is a document containing information about rare and endangered animals and plants. It is published by the IUCN (International Union for Conservation of Nature).

Question 10.
Observe Figure 17.3 showing different habitats. List two abiotic (non-living) factors essential for the survival of organisms in the Desert habitat shown.

Answer:
Two abiotic factors essential in a desert are:
Sunlight (provides energy, but also contributes to high temperature).
Water (scarce, crucial for survival). (Other possible answers: Air, Soil type/ Sand).

Question 11.
Illustration 17.5 shows a Vanda plant growing on a tree.

a) Identify this type of ecological interaction.
b) Explain how each organism is affected.
Answer:
a) Commensalism.
b) The Vanda (orchid) benefits by getting support and access to sunlight without harming the tree. The Tree is neither benefited nor harmed.

Question 12.
Figure 17.7 shows deforestation. List two negative impacts of deforestation on biodiversity shown or implied in the comparison with Figure 17.6.


Answer:
Two negative impacts are:
Habitat Loss: The removal of trees destroys the homes and living spaces for many animals and plants seen in Fig 17.6.
Loss of Food Sources: Animals that depend on the trees or other forest plants for food lose their sustenance.
(Other possible answers: Soil erosion, disruption of food webs).

Question 13.
Explain why decomposers like bacteria and fungi are essential for the balance of an ecosystem.
Answer:
Decomposers break down dead organisms and waste products. This process is crucial because it returns essential nutrients (like carbon, nitrogen) back into the soil, water, and air, making them available for producers (plants) to use again. Without decomposers, nutrients would remain locked in dead matter, and the ecosystem would eventually run out of usable nutrients.

Question 14.
Why does the amount of energy available decrease at each higher trophic level in a food chain?
Answer:
Energy transfer between trophic levels is inefficient. When one organism consumes another, a large portion of the energy obtained is used by the consumer for its own life processes (like respiration, movement, maintaining body temperature) and is ultimately lost as heat to the environment. Only a small fraction of the energy is stored in the consumer’s body and becomes available to the next trophic level.

Question 15.
Explain the importance of biodiversity in pollination and seed dispersal.
Answer:
Pollination: Many plants rely on animals (like bees, butterflies, birds – part of biodiversity) to transfer pollen, which is necessary for fertilization and fruit/seed production. Without these pollinators, many plant species, including important food crops, could not reproduce.
Seed Dispersal: Animals (like birds, mammals – part of biodiversity) often eat fruits and disperse the seeds in their droppings far from the parent plant. This helps plants colonize new areas, increasing their chances of survival and maintaining genetic diversity across landscapes.

Question 16.
Evaluate the statement: “Protecting only the top predators like tigers is enough for biodiversity conservation.” Is this sufficient? Explain why or why not.
Answer:
No, protecting only top predators is not sufficient. While important, they are just one part of the complex ecosystem. Biodiversity conservation requires protecting all levels of the food web and their habitats. Producers (plants) form the base, herbivores control plant populations, and decomposers recycle nutrients. Protecting the entire ecosystem, including lower trophic levels and habitats, ensures the long-term survival of top predators and maintains overall ecological balance.

Question 17.
Explain the difference between In-situ and Ex-situ conservation methods, giving one example for each.
Answer:
In-situ conservation: Involves protecting species within their natural habitats. Example: National Parks (like Eravikulam) or Wildlife Sanctuaries (like Periyar).

Ex-situ conservation: Involves protecting species outside their natural habitats, often in controlled environments. Example: Zoological gardens (Zoos) or Botanical gardens.

Question 18.
Evaluate the importance of the People’s Biodiversity Register (PBR) in conservation efforts.
Answer:
The PBR is important because it is a local-level documentation of the biodiversity (microorganisms, plants, animals) present in a specific area, prepared with community participation. This detailed, localized information helps:
Understand the unique characteristics and availability of local species.
Identify species that may need conservation focus.
Involve the local community, making conservation efforts more effective and sustainable.
Provide baseline data for monitoring changes in biodiversity over time.

Question 19.
Classify the following as producers and consumers.
Lizard, Planktons, Paddy, Calotes, Carrot, Grasshopper, Tortoise, Algae, Snake
Answer:

Producers	Consumers
Paddy	Lizard
Carrot	Calotes
Planktons	Grasshopper
Algae	Tortoise
	Snake

Question 20.
Find out suitable example for the animal relations mentioned.
i. Parasitism ___________
ii. Mutualism ___________
iii. Commensalism ___________
Crops × Weeds
Mango tree × Vanda
Mango tree × Loranthus
Fish × Heron
Hermit crab × sea anemone
Answer:
i. Mango tree and Loranthus
ii. Hermit Crab – Sea anemone
iii. Mango tree and Vanda.

Question 21.
Which are the different types of conservation of biodiversity?
Answer:
These are mainly two types.

1. In-situ conservation in which organisms are conserved within their natural environment
2. Ex-situ conservation in which animals are protected out their natural environment.

Question 22.
Complete the illustration Suitably

Answer:
a. Environmental/Ecological services
b. Cultural services
c. Food, Medicine
d. Nutrient Cycle, Pollination

Question 23.
Classify the following into Exsitu and In-situ.
(Zoological Gardens, Sacred Groves, Gene banks, Biosphere Reserves, Botanical Gardens, National parks)
Answer:

In-situ	Ex-situ
National parks	Zoological Gardens
Sacred Groves	Gene Banks
Biosphere Reserves	Botanical Gardens.

Question 24.
Expand the following terms.
Answer:
JNTBGRI – Jawaharlal Nehru Tropical Botanic Garden and Research Institute.
MBG – Malabar Botanical Garden
RGCB – Rajiv Gandhi Centre for Biotechnology.

Question 25.
Why do all the food chains start from green plants?
Answer:
The basis of all food chain is the plants. They are the producers. These are eaten by the herbivores which in turn are eaten by the carnivores.

Question 26.
The members at the successive higher levels are lesser in number and larger in size in a food chain. What about their body weight? What would be the reason for that?
Answer:
In the successive higher levels in the food chain the number of consumers decreases and the size of their body increases. The number of producers will be very large. The number of animals which feed on them is less in number. But their body size increases.

Question 27.
Hay → Horse
Paddy → FowlFox
Phytoplankton → Tadpole → Fish → Man
Grass → Grasshopper → Frog → Snake → Vulture
Examine the food chain given above and classify them as primary consumers, secondary and tertiary consumers.
Answer:
Producers
Hay, Paddy, Phytoplankton, Grass

Primary consumer
Horse, Fowl, Tadpole, Grass hopper

Secondary consumers
Fox, Fish, Frog, Snake

Tertiary consumers
Vulture, Man

Question 28.
What will happen if the number of herbivores increase?
Answer:
If the number of herbivores increase they will eat away all the grass and shrubs and they will have to face shortage of food. The destruction of grass and shrubs will cause soil erosion and the top fertile soil will be washed away.

Question 29.
Animals which are facing extinction.
Answer:

• Wild goat
• Musk deer
• Indian wild Ass
• Lion tailed Monkey
• Lion
• Rhinoceros
• The large Indian Bustard
• Tiger
• Kashmir deer
• Himalayan Tiger
• Silver owl
• Panda

Question 30.
Write down the name of any two organisations that work for the protection of nature
Answer:
IUCN (International Union for
Conservation of Nature)
WWF (World Wide Fund For Nature)

Question 31.
Choose the appropriate ones from the box and complete the table

(a) Nilgiris	i) Wild life Sanctuary
(b) Periyar	ii) National Parks
(c) Silent Valley	iii) Biosphere Reserves
(d) Kadalundi	iv) Community Reserve
(e) Malabar Botanical Garden

Answer:
i. Wildlife Sanctuaries – b) Periyar
ii. National Parks – c) Silent Valley
iii. Biosphere reserves – a) Nilgiris
iv. Community reserves – d) Kadalundi

Question 32.
Ecological interactions and their examples are given in the boxes. Make pairs according to the model given below.
Model – Predation: Deer and tiger

Ecological interactions	Examples
Predation	Mango tree and loranthus
Mutualism	Deer and tiger
Parasitism	Flower and butterfly
Commensalism	Mango tree and vanda

Answer:
Predation – Deer and tiger
Mutualism – Flower and butterfly
Parasitism – Mango tree and loranthus
Commensalism – Mango tree and vanda

Question 33.
Observe the pictures
i) Identify and name the organisms given above.
ii) Mention their specific importance related to biodiversity conservation.


Answer:
(i) A. Lion – tailed macaque
(Scientic name: Macaca silenus)
B. i. Nilgiri tahr (Scientific name: Nilgiritragushylocrius)
ii) These organisms on the verge of extinction due to several reasons.

Question 34.
Analyse the given statement and answer the questions.
There exist diverse ecosystems in earth, each differs in richness of biodiversity.
i. Is it necessary to conserve all kinds of ecosystems? Why?
ii. Write any four services of biodiversity.
Answer:
a) Yes
Ecosystems are places where there is rich biodiversity. Destruction of an ecosystem means destruction of many living organisms. So, in order to protect living organisms, we have to protect natural ecosystems.

b) Availability of essential materials

• Food
• Medicine
• Fuels
• Construction materials
  Cultural Services
• Aesthetics
• Recreation
• Study Practice of rituals
• Ecological services
• Soil Formation
• Prevention of soil erosion.
  O₂, CO₂ balance
• Availability of freshwater
• Flood control
• Climate control
  Auxiliary services
• Nutrient cycling, Pollination, Biological Control. Seed dispersal

Question 35.
Give reason.
It is not possible to illustrate a food chain using the following organisms
Snake, Rat, Eagle
Answer:
Eagle feed on both rats and snakes. So we cannot illustrate food chain

Question 36.
Complete the table given below:

A	B	C
Rat	Primary consumer	(a)  ………………….
Paddy	(b) ………………….	first trophic level
snake	(c) …………………….	Third trophic level
Eagle	Tertiary consumer	(c)  ………………….

Answer:
a) Second trophic level
b) Producers
c) Secondary consumers
d) Fourth trophic level

Question 37.
Pickout the organisms that are related to commensalism from the following and write as a pair.
Mango tree, Loranthus, Vanda, Butterfly, Snail, Flower
Answer:
Mango tree – Vanda

Question 38.
Correct mistakes if any, in the underlined part of the following statements.
• Dodo is an extinct organism.
• Palode Tropical Botanical Garden is an example for In-Situ conservation.
• Indo-Burma Region is an Ecological hotspot.
Answer:
Palode Tropical Botanical Garden is an example for Ex-situ conservation

Question 39.
Analyse the given food chain and an-swer the following.
A: Paddy → Hen → Fox → Lion
B: Grass → Deer → Tiger
a) In these food chains A and Lion and Tiger belong to which trophic levels.
b) What is the role of these organisms in maintaining ecological balance?
c) Write two messages for the conservation of these organisms, for preparing a poster.
Answer:
a) Lion – Fourth trophical level
Tiger – Third trophical level

b)They control the number of organisms that they eat. If they become extinct, the number of organism in
the lower trophic levels will be increasing in number beyond a certain limit. It will cause imbalance to the environment.

c) Earth without wild life is life without beautiful things.
If you like to shoot animals, do it with a camera, not a gun.

Question 40.
Identify the examples of given ecological interactions and answer the questions.
(i) Mango tree and Loranthus
(ii) Mango tree and Vanda
a. Name the ecological interactions (i) and (ii)
b. Write the name and peculiarity of another ecological interaction.
Answer:
a. i) Parasitism
ii) Commensalism
b. Predators – Eg: Tiger and deer

The Beauty of Diversity Class 8 Questions and Answers Notes Basic Science Chapter 16 Kerala Syllabus

Question 41.
Which trophic level has the maximum amount of energy available?
a) Primary Consumers
b) Producers
c) Secondary Consumers
d) Decomposers
Answer:
(b) Producers

Question 42.
Which interaction is characterized by the relationship where one organism benefits and the other is harmed?
(a) Mutualism
(b) Commensalism
(c) Predation
(d) Parasitism
Answer:
(d) Parasitism (Note: Predation also involves harm, but parasitism is a longterm, dependent relationship, e.g., Flea and Cat).

Question 43.
The organization responsible for publishing the Red Data Book is:
(a) WWF
(b) BMC
(c) IUCN
(d) PBR
Answer:
(c) IUCN (International Union for Conservation of Nature)

Question 44.
A Purple Frog (Mahabali Frog) is often called a ‘living fossil’ primarily because:
(a) It is found only in the Himalayas.
(b) It comes out only once a year to mate.
(c) It has undergone very little evolutionary change over millions of years.
(d) It lives underground.
Answer:
(c) It has undergone very little evolutionary change over millions of years.

Question 45.
Define Biodiversity Hotspot
Answer:
Biodiversity Hotspot: A geographically rich area that harbors a large number of endemic species and is under threat from human activities.

Question 46.
Differentiate between In-situ and Ex-situ conservation.
Answer:
In-situ Conservation: Protecting organisms within their natural habitats (e.g., National Park). Ex-situ Conservation: Protecting organisms outside their natural habitats in controlled conditions (e.g., Zoological Garden).

Question 47.
What is the main role of the Fungus component in the symbiotic relationship of a Lichen?
Answer:
The Fungus provides structure, protection, and helps in the absorption of water for the lichen organism.

Question 48.
Why is the amount of available energy found to decrease at successive trophic levels?
Answer:
Energy is lost as heat at each transfer step between trophic levels as organisms use energy for respiration and metabolic processes.

Question 49.
Explain the concept of Ecosystem Diversity with an example.
Answer:
Ecosystem Diversity: This refers to the variety of different habitats, ecosystems, and ecological processes found in an area.
Example: The diversity of hills, rivers, mangrove forests, and paddy fields found within a single region.

Question 50.
Differentiate between Genetic Diversity and Species Diversity. (2 points each)
Answer:
Genetic Diversity:
Variations within the individuals of a single species.
Example: Different varieties of rice or mango. Species Diversity:
Species Diversity.
The variety and number of different species in a particular area.
Example: The difference between a forest containing elephants, tigers, and bamboo plants.

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 10 Decimal Forms Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 10 Solutions Decimal Forms

Class 8 Kerala Syllabus Maths Solutions Chapter 10 Decimal Forms Questions and Answers

Decimal Forms Class 8 Questions and Answers Kerala Syllabus

Earlier Forms (Page 164)

Question 1.
Write in fractional form, the numbers given in decimal form below:
(i) 0.1
(ii) 0.01
(iii) 0.11
(iv) 0.101
(v) 0.0101
Answer:
(i) 0.1 = \(\frac {1}{10}\)
(ii) 0.01 = \(\frac {1}{100}\)
(iii) 0.11 = \(\frac {11}{100}\)
(iv) 0.101 = \(\frac {101}{1000}\)
(v) 0.0101 = \(\frac {101}{10000}\)

Some Other Fractions (Page 166)

Question 1.
Write the fractions below in decimal form:
(i) \(\frac {3}{20}\)
(ii) \(\frac {3}{40}\)
(iii) \(\frac {13}{40}\)
(iv) \(\frac {7}{80}\)
(v) \(\frac {5}{16}\)
Answer:
(i) \(\frac{3}{20}=\frac{3 \times 5}{20 \times 5}=\frac{15}{100}\) = 0.15

(ii) 40 = 2 × 2 × (2 × 5)
Multiplied by 5 × 5
(2 × 5) × (2 × 5) × (2 × 5) = 1000
It is a multiple of 10.
\(\frac{3}{40}=\frac{3 \times 5 \times 5}{40 \times 5 \times 5}=\frac{75}{1000}\) = 0.075

(iii) 40 = 2 × 2 × (2 × 5)
Multiplied by 5 × 5
(2 × 5) × (2 × 5) × (2 × 5) = 1000
It is a multiple of 10.
\(\frac{13}{40}=\frac{13 \times 5 \times 5}{40 \times 5 \times 5}=\frac{325}{1000}\) = 0.325

(iv) 80 = 2 × 2 × 2 × (2 × 5)
Multiplied by 5 × 5 × 5
(2 × 5) × (2 × 5) × (2 × 5) × (2 × 5) = 10000
It is a multiple of 10.
\(\frac{7}{80}=\frac{7 \times 5 \times 5 \times 5}{80 \times 5 \times 5 \times 5}=\frac{875}{10000}\) = 0.0875

(v) 16 = 2 × 2 × 2 × 2
Multiplied by 5 × 5 × 5 × 5
(2 × 5) × (2 × 5) × (2 × 5) × (2 × 5) = 10000
It is a multiple of 10.
\(\frac{5}{16}=\frac{5 \times 5 \times 5 \times 5 \times 5}{16 \times 5 \times 5 \times 5 \times 5}=\frac{3125}{10000}\) = 0.3125

Question 2.
Find the decimal form of the sums below:
(i) \(\frac{1}{5}+\frac{1}{25}+\frac{1}{125}\)
(ii) \(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\)
(iii) \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\)
Answer:

Question 3.
A two-digit number divided by another two-digit number gives 5.875. What are the numbers?
Answer:
5.875 = \(\frac {5875}{1000}\)
5875 = 5 × 5 × 5 × 47
1000 = 5 × 5 × 5 × 8

5.875 = \(\frac{5875}{1000}=\frac{5 \times 5 \times 5 \times 47}{5 \times 5 \times 5 \times 8}=\frac{47}{8}\)
But 8 is not a two-digit number.
So we multiplied both numerator and denominator by 2, and we get
\(\frac{47}{8}=\frac{94}{16}\)
5.875 = \(\frac {94}{16}\)

New forms (Page 172)

Question 1.
Find fractions with powers of 10 as denominators that get closer and closer to each of the fractions below, and write the decimal form of each:
(i) \(\frac {5}{6}\)
(ii) \(\frac {1}{9}\)
(iii) \(\frac {1}{11}\)
Answer:

Question 2.
Prove that the numbers \(\frac{1}{10}, \frac{11}{100}, \frac{111}{1000}, \ldots\) get closer and closer to \(\frac {1}{9}\).
(ii) Find the decimal forms of \(\frac{1}{9}, \frac{2}{9}, \frac{3}{9}, \frac{4}{9}, \frac{5}{9}, \frac{6}{9}, \frac{7}{9}, \frac{8}{9}\)
(iii) What are the decimal forms of \(\frac {2}{3}\)?
Answer:

Question 3.
Compute the following and write answers in decimal form:
(i) 0.111… + 0.222…
(ii) 0.333… + 0.777…
(iii) 0.333… × 0.666…
Answer:
(i) 0.111 = \(\frac {1}{9}\)

Question 4.
For each of the fractions below, write the decimal form of a fraction with a difference less than \(\frac {1}{1000}\)
(i) \(\frac {1}{3}\)
(ii) \(\frac {1}{6}\)
(iii) \(\frac {2}{3}\)
(iv) \(\frac {5}{6}\)
(v) \(\frac {1}{7}\)
Answer:
(i) \(\frac {1}{3}\) = 0.333
(ii) \(\frac {1}{6}\) = 0.166 (or 0.167)
(iii) \(\frac {2}{3}\) = 0.666 (or 0.667)
(iv) \(\frac {5}{6}\) = 0.833
(v) \(\frac {1}{7}\) = 0.142

Class 8 Maths Chapter 10 Kerala Syllabus Decimal Forms Questions and Answers

Class 8 Maths Decimal Forms Questions and Answers

Question 1.
What is the decimal form of the fraction \(\frac {1}{8}\)?
(A) 0.18
(B) 0.125
(C) 0.8
(D) 0.0125
Answer:
(B) 0.125

Question 2.
Which of the following fractions has the decimal form 0.04?
(A) \(\frac {1}{4}\)
(B) \(\frac {4}{10}\)
(C) \(\frac {1}{25}\)
(D) \(\frac {2}{5}\)
Answer:
(C) \(\frac {1}{25}\)

Question 3.
Read the given statements.
Statement I: The fraction \(\frac {1}{3}\) cannot be written as a fraction with a denominator that is a power of 10 (like 10, 100, 1000).
Statement II: The decimal form of \(\frac {1}{3}\) is the recurring decimal 0.333….
Choose the correct answer:
(A) Statement I is true, and Statement II is false.
(B) Statement I is false, and Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(C) Both statements are true

Question 4.
Read the given statements.
Statement I: 0.03 = \(\frac {3}{100}\)
Statement II: 0.03 can be split as \(\frac {0}{10}\) + \(\frac {3}{100}\).
Choose the correct answer:
(A) Statement I is true, and Statement II is false.
(B) Statement I is false, and Statement II is true.
(C) Both statements are true, and II explains I.
(D) Both statements are false.
Answer:
(C) Both statements are true, and II explains I.

Question 5.
If \(\frac {1}{9}\) = 0.111…, then what is the decimal form of \(\frac {5}{9}\)?
(A) 0.5
(B) 0.0505…
(C) 0.555…
(D) 0.9595…
Answer:
(C) 0.555…

Question 6.
What is the missing number in the expansion of 0.72?
0.72 = \(\frac{7}{10}+\frac{\cdots .}{100}\)
(A) 7
(B) 2
(C) 72
(D) 0
Answer:
(B) 2

Question 7.
Write in decimals
(a) \(\frac {1}{9}\)
(b) \(\frac {2}{9}\)
(c) \(\frac {1}{7}\)
(d) \(\frac {1}{11}\)
(e) \(\frac {2}{11}\)
(f) \(\frac {1}{12}\)
Answer:
(a) 0.111…
(b) 0.222…
(c) 0.14285…
(d) 0.090909…
(e) 0.181818…
(f) 0.08333…

Question 8.
Find the fraction of the denominator that is a power of 10 equal to each of the fractions below, and then write their decimal forms:
(i) \(\frac {1}{50}\)
(ii) \(\frac {3}{40}\)
(iii) \(\frac {5}{16}\)
(iv) \(\frac {12}{625}\)
Answer:

Question 9.
Find the fraction of the denominator that is a power of 10, getting closer and closer to each of the fractions below, and then write their decimal forms.
(i) \(\frac {5}{6}\)
(ii) \(\frac {3}{11}\)
(iii) \(\frac {23}{11}\)
(iv) \(\frac {1}{13}\)
Answer:

Question 10.
(i) Explain using algebra that the fractions \(\frac{1}{10}, \frac{11}{100}, \frac{111}{1000}, \ldots\) get closer and closer to \(\frac {1}{9}\).
(ii) Using the general principle above on single digit numbers, find the decimal forms of \(\frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\) (Why \(\frac {3}{9}\) and \(\frac {6}{9}\) left out in this?)
(iii) What can we say in general about those decimal forms in which a single digit repeats?
Answer:
(i) Let x be the number
\(\frac{x}{9}-\frac{x}{10}=\frac{x}{90}=\frac{x}{9}-\frac{11 x}{100}=\frac{x}{900}\)



(\(\frac {3}{9}\), \(\frac {6}{9}\). These fractions having common factor in the numerator and the denominator.
(iii) Repeated decimals.

Question 11.
(a) Find the decimal form of \(\frac {1}{4}\).
(b) Write the decimal form of \(\frac{7}{10}+\frac{3}{100}+\frac{4}{1000}\).
Answer:

Question 12.
(a) Write the decimal forms of \(\frac {1}{3}\) and \(\frac {1}{9}\).
(b) What is the decimal form of (0.33333….)²?
Answer:

Question 13.
Write the decimal forms of \(\frac {3}{25}\) and \(\frac {1}{8}\).
Answer:
\(\frac{3}{25}=\frac{3 \times 4}{25 \times 4}=\frac{12}{100}\) = 0.12
\(\frac{1}{8}=\frac{1 \times 125}{8 \times 125}=\frac{125}{1000}\) = 0.125

Question 14.
(a) Write the decimal form of the fractions \(\frac {1}{2}\) and \(\frac {2}{5}\).
(b) If \(\frac {4}{k}\) is a fraction between \(\frac {1}{2}\) and \(\frac {2}{5}\). What is k?
(c) Write the decimal form of the fraction \(\frac {4}{k}\).
Answer:

If \(\frac {4}{k}\) is between \(\frac {1}{2}\) and \(\frac {2}{5}\), then \(\frac {4}{k}\) is between \(\frac {4}{8}\) and \(\frac {4}{10}\).
Then the number is \(\frac {4}{9}\)
∴ k = 9
(c) \(\frac {4}{9}\) = 0.444…

Question 15.
(i) Find the decimal form of \(\frac {1}{11}\).
(ii) Find the decimal forms of \(\frac {2}{11}\), \(\frac {3}{11}\).
(iii) What is the decimal form of \(\frac {10}{11}\)?
Answer:

Question 16.
Write the results of the operations below as decimals:
(i) (0.333….)²
(ii) \(\sqrt{0.444 \ldots \ldots}\)
Answer:

Question 17.
Write the decimal form of \(\frac {1}{6}\)
Answer:
\(\frac {1}{6}\) = 0.1666…..

Class 8 Maths Chapter 10 Notes Kerala Syllabus Decimal Forms

Fractions with Powers of 10 as Denominators
Fractions with powers of 10 as denominators can be written in decimal form.
\(\frac {3}{10}\) = 0.3
\(\frac {54}{10}\) = 5.4
\(\frac {29}{100}\) = 0.29
\(\frac {54}{100}\) = 0.54
\(\frac {347}{1000}\) = 0.347
\(\frac {54}{1000}\) = 0.054

Any number in decimal form can be written using fractions with powers of 10 as denominators.
0.7 = \(\frac {7}{10}\)
0.91 = \(\frac {91}{100}\)
0.673 = \(\frac {673}{1000}\)

Decimal numbers can be splitted according to place value.

Some Other Fractions
Convention of some fractions, whose denominators are not powers of 10.

New forms
Conversion of some fractions, whose denominators were not powers of 10.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 17 The Beauty of Diversity Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 17 The Beauty of Diversity Question Answer Notes

Class 8 Basic Science Chapter 17 Notes Kerala Syllabus The Beauty of Diversity Question Answer

The Beauty of Diversity Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Based on Box A, identify and write the appropriate items from Boxes B and C.

A	B	C
1. Parasitism	a. At first, harm to both; later, benefit for the winner.	a. Butterfly and the flower.
2. Predation	b. Beneficial to one, neither beneficial nor harmful to another.	b. Vanda and mango tree.
3. Commensalism	c. Beneficial to both organisms.	c. Tiger and deer.
4. Mutualism	d. Beneficial to one, harmful to another. The prey becomes food for the predator.	d. Weed and crop.
5. Competition	e. One benefits, the other is harmed.	e. Cat and flea.

Answer:

A (Interaction)	B (Description)	C (Example)
1. Parasitism	e. One benefits, the other is harmed.	Cat and flea.
2. Predation	d. Beneficial to one, harmful to another. The prey becomes food for the predator.	Tiger and deer.
3. Commensalism	b. Beneficial to one, neither beneficial nor harmful to another.	Vanda and mango tree.
4. Mutualism	c. Beneficial to both organisms.	Butterfly and the flower.
5. Competition	a. At first, harm to both; later, benefit for the winner.	Weed and crop.

Question 2.
Examine the illustration and answer the following questions.

a) Rearrange the given food chain in the correct order.
b) Identify and write the producer, primary consumer, secondary consumer and tertiary consumer in the food chain.
c) Which organism belongs to the third trophic level in this food chain?
d) If the fish in this food chain completely disappear, how would it affect the ecosystem?
Answer:
a) Algae → Tadpole → Fish → Kingfisher
(Reasoning: Algae are producers (plants). Tadpoles eat algae. Fish eat tadpoles. Kingfishers eat fish.)

b)

• Producer: Algae (Makes its own food).
• Primary Consumer: Tadpole (Eats the producer).
• Secondary Consumer: Fish (Eats the primary consumer).
• Tertiary Consumer: Kingfisher (Eats the secondary consumer).

c) Fish
(Reasoning: Trophic Level 1 = Producer (Algae); Trophic Level 2 = Primary Consumer (Tadpole); Trophic Level 3 = Secondary Consumer (Fish)).

d)

• The Kingfisher population might decrease or move elsewhere due to a lack of food (fish).
• The Tadpole population might increase significantly because their main predator (fish) is gone.
• An increase in tadpoles could lead to a decrease in the Algae population due to overgrazing.
• This disrupts the balance of the ecosystem.

Question 3.
A food chain created by a student is given below. Examine it and answer the following questions.

a) Is the food chain created by the student correct? What is your opinion? What is the justification for it?
b) What would happen if decomposers like bacteria did not exist?
c) Write an example of a decomposer other than bacteria.
Answer:
a) No, the food chain is incorrect.
Justification: A food chain shows the flow of energy from producers to consumers. Bacteria are decomposers, not consumers. Decomposers break down dead organic matter from all trophic levels (including dead paddy plants), but they are not typically shown as the next step after a producer in a simple feeding chain. A correct chain would be Paddy Plant Herbivore (e.g., Rat).

b) If decomposers did not exist, dead plants and animals would not break down effectively. Nutrients locked within this dead organic matter would not be returned to the soil, water, and air. Over time, producers (plants) would run out of essential nutrients, and the entire ecosystem would eventually collapse. Dead bodies would accumulate everywhere.

c) Fungi (e.g., mushroo1ms, mould).

Basic Science Class 8 Chapter 17 Question Answer Kerala Syllabus

Texbook Page No : 269

Question 1.
Which are the living organisms that you see in the picture (Biodiversity Park)?
Answer:
Based on a typical biodiversity park setting, one would see trees, flowering plants, students (humans), birds, butterflies, and potentially insects on the ground.

Question 2.
In what different ways can you classify the organisms you identified?
Answer:

• Plants vs. Animals: Trees and bushes vs. Birds and Humans.
• Vertebrates vs. Invertebrates: Birds / Humans vs. Insects.
• Terrestrial vs. Aquatic: If there is a pond, organisms can be classified by where they live.:

Question 3.
Apart from the organisms you identified in the picture, what other organisms can be seen in a biodiversity park?
Answer:
Soil organisms (earthworms, millipedes), microorganisms (bacteria, fungi), reptiles (lizards, garden snakes), and various insects (ants, beetles).

Question 4.
Why is this place called a biodiversity park?
Answer:
Because it is a protected area designed to conserve a wide variety of plant and animal species (biodiversity) in one location, allowing them to thrive in their natural habitat.

Textbook Page No : 270 & 271

Question 5.
What conclusions can we arrive about the biodiversity?
Answer:
Biodiversity is vast and not yet fully discovered. Even today, hundreds of new species are being found, indicating that our forests and ecosystems hold many secrets and hidden species.

Question 6.
What are the organisms found only in these habitats?
Answer:
Organisms have specific adaptations that limit them to particular environments.

• Coral Reef: Corals, sea anemones, clownfish, starfishes.
• Mangrove Forest: Mangrove trees (plants with breathing roots), mud-skippers (fish that can move on mud), crabs, water birds.
• Desert: Cactus plants (modified leaves to save water), camels, rattle snakes, scorpions.
• Wetlands: Water lilies, lotus, frogs, cranes, dragonflies.

Question 7.
What are the non-living (abiotic) factors that help the survival of organisms in these habitats?
Answer:
Abiotic factors are the non-living parts of the environment that influence living organisms.

• Coral Reef: Saline (salty) water, sunlight penetration, steady water temperature. dissolved oxygen.
• Mangrove Forest: Brackish water (mix of salt and fresh water), clay-rich/muddy soil, tides,
• Desert: High temperature, intense sunlight, scarcity of water, sandy soil.
• Wetlands: Abundance of fresh water, marshy soil, moderate temperature.

Let’s Find

Question 8.
“Some organisms are able to live only in certain habitats. Why?”
Answer:
Organisms have specific adaptations (physical features or behaviors) evolved over huge periods of time that allow them to survive only in specific conditions (e.g., gills for breathing in water, thick fur for warmth in polar regions).

Question 9.
Have you noticed that a single organism can exist in different forms as shown in the pictures?
Answer:
Yes. Some organisms undergo a process called metamorphosis, where they change their form completely during their life cycle.
butterfly (Egg → Larva/Caterpillar → Pupa → Adult) or a frog (Egg → Tadpole → Frog).
Example: A frog starts as an egg, hatches into a tadpole (which lives in water and has a tail), and eventually grows legs and loses its tail to become an adult frog. This allows the organism to live in different ecosystems (water vs. land) at different stages of its life.

Textbook Page No : 273 & 274

Question 10.
Identify and write down which level of biodiversity that each of the following represents. (Based on Illustration 17.1: Ecosystem diversity, Species diversity, Genetic diversity).


Answer:

Question 11.
What does the poster (Illustration 17.2) indicate?

Answer:
It highlights the “International Day for Biodiversity” (May 22) and the theme “Our biodiversity, our food, our health,” emphasizing that our survival depends on nature.

Question 12.
How does biodiversity benefit each living organism?
Answer:
It provides food, shelter, oxygen, and maintains the ecological balance necessary for survival.

Question 13.
Shall we collect information on these topics and organize a seminar in the class on the topic “Biodiversity and the Survival of Living Beings”.
Seminar subtopics
• Food, clothing, shelter, fuel, medicines and biodiversity.
• Biodiversity and the stability of soil, water and air.
• Biodiversity in nutrient cycling, pollination, biological control and seed dispersal.
• Biodiversity and aesthetic values.
Answer:
□ Food, Clothing, Shelter, Fuel, Medicines, and Biodiversity
This section highlights the direct utilization of biodiversity by humans.

□ Biodiversity and the Stability of Soil, Water, and Air
This section focuses on the regulatory and life-support services of ecosystems.
Soil Stability:

• Plant Roots bind the soil together, preventing soil erosion by wind and water.
• Decomposers (like earthworms, bacteria, and fungi) enrich the soil by breaking down organic matter, increasing its fertility and water retention capacity.
• Water Stability (Hydrological Cycle):
  • Forests (especially rainforests and hill forests) act as natural sponges, regulating the flow of water. They absorb rainfall, reduce surface runoff, and slowly release water into streams, preventing floods and maintaining water tables.
  • Wetlands filter pollutants from water.
• Air Quality and Stability:
  • Plants (Producers) are responsible for taking in atmospheric Carbon Dioxide (CO₂) and releasing Oxygen (O₂) during photosynthesis, maintaining breathable air quality. Large forest ecosystems help stabilize the global climate by acting as carbon sinks (absorbing CO₂).

□ Biodiversity in Nutrient Cycling, Pollination, Biological Control, and Seed Dispersal These are the essential functional roles (ecosystem services) provided by living organisms.

• Nutrient Cycling (e.g., Carbon Cycle):
  • Decomposers (bacteria and fungi) break down dead organisms, releasing essential nutrients (like nitrogen, phosphorus, and carbon) back into the soil and atmosphere for reuse by plants. This process is crucial for life to continue.
• Pollination:
  • Insects (like bees, butterflies), birds, and bats transfer pollen between plants, enabling reproduction in flowering plants. This is vital for the production of fruits, vegetables, and seeds.
• Biological Control:
  • Nature uses natural predators and parasites to keep pest populations in check, reducing the need for harmful chemical pesticides.
  • Example: Ladybugs eat aphids (plant pests).
• Seed Dispersal:
  • Animals (e.g., birds, squirrels, elephants) eat fruits and carry the seeds elsewhere, often depositing them in fertile soil via their droppings, which helps plants colonize new areas. This is crucial for forest regeneration.

□ Biodiversity and Aesthetic Values
This section discusses the intangible, non-material benefits of biodiversity.

• Aesthetic Appeal: The beauty of nature, including colorful flowers, diverse wildlife, majestic landscapes, and vibrant coral reefs, provides profound sensory and emotional satisfaction.
• Recreation and Ecotourism: Natural areas (national parks, sanctuaries, beaches) are destinations for relaxation, learning, and tourism. This supports local economies.
• Cultural and Spiritual Value: Many indigenous communities and cultures have deep, spiritual connections to specific plants, animals, or natural sites (e.g., sacred groves).
• Education and Research: Biodiversity provides a living laboratory for scientific research, helping us understand genetics, ecology, and evolution.

Textbook Page No : 275
Indicators

Question 14.
From whom does the food chain begin?
Answer:
The food chain begins with Producers (Paddy green plants/algae).

Question 15.
Why are green plants called producers?
Answer:
They are the only organisms capable of producing their own food using solar energy and simple substances through photosynthesis.

Question 16.
Who are the consumers?
Answer:
The consumers are organisms that feed on producers or other consumers. In the example: Rat (Primary), Snake (Secondary), and Eagle (Tertiary).

Question 17.
What is the characteristic of primary consumers?
Answer:
They are herbivores; they feed directly on the producers (green plants).

Question 18.
What is the characteristic of second ary consumer s?
Answer:
They are carnivores or omnivores; they feed on the primary consumers (herbivores).

Question 19.
Who are the tertiary consumers?
Answer:
They are carnivores; they feed on the secondary consumers.

Question 20.
What is the role of decomposers in the food chain?
Answer:
Decomposers (like Bacteria and Fungus) break down the complex organic matter in dead organisms and remains into simpler molecules, returning essential substances and elements to nature. They play a vital role in the flow of matter.

Question 21.
Create a food web using the given organisms

Answer:
The food web starts with the Producers (Trophic Level 1) and shows how energy flows up through the different Consumers.

1. Producers (Trophic Level 1)
Plant: Produces its own food (often including Grass/Paddy/other generic plants as assumed for the ecosystem).

2. Primary Consumers (Trophic Level 2 – Herbivores)
These organisms feed directly on the Plant:

• Deer
• Rabbit
• Rat (Also an Omnivore, but eats seeds/plants)
• Grasshopper
• Squirrel

3. Secondary Consumers (Trophic Level 3 – Primary Carnivores/Omnivores)
These organisms feed on the Primary Consumers:

• Snake: Eats Rat, Frog.
• Fox: Eats Rabbit, Rat, Squirrel, Grasshopper, Frog.
• Frog: Eats Grasshopper.
• Calotes (Lizard): Eats Grasshopper.
• Vulture: Eats dead or dying animals (Scavenger, but occupies this level when eating small, newly dead herbivores).

4. Tertiary Consumers (Trophic Level 4 – Secondary Carnivores)
These organisms feed on Secondary Consumers:

• Tiger: Eats Deer (Primary Consumer) and Fox (Secondary A Tertiary Consumer).
• Fox: Eats Snake, Frog, Calotes.
• Vulture: Eats dead or dying animals (Scavenger, but occupies this level when eating larger, newly dead carnivores like snakes/foxes).

Text Book Page No : 276
Indicators

Question 22.
Trophic level with the maximum number of organisms:
Answer:
Trophic Level 1 (Producers). The base of the pyramid is always the largest.

Question 23.
Trophic level with the least number of organisms.
Answer:
The highest trophic level (e.g., Trophic Level 4, Tertiary Consumers). Energy is lost at each step, supporting fewer organisms at higher levels.

Question 24.
How does the disappearance of top-level organisms affect balance?
Answer:
The disappearance of top predators can lead to an overpopulation of organisms in the lower trophic levels (e.g., secondary consumers). This imbalance can disrupt the entire food web and ecosystem stability (e.g., herbivores might overgraze if their predators disappear).

Text Book Page No : 277

Question 25.
Types of Interactions (Illustration 17.5):

Answer:
Commensalism:
Example: Vanda (orchid) growing on a Tree. (Vanda gets support, tree is unaffected)

Mutualism:
Example: Tree providing nectar to a Butterfly, which helps in pollination. (Lichens are another example).

Competition:
Example: Crop plants and Weeds competing for nutrients and space.

Parasitism:
Example: Flea living on a Dog. (Other examples: intestinal worms, lice).

Predation:
Example: Eagle hunting a Chicken. (Other examples: Tiger hunting deer, Snake eating rat).

Text Book Page No : 278 & 279

Question 26.
Are such relationships necessary to maintain the balance and stability of an ecosystem? Discuss.
Answer:
Yes, ecological relationships are absolutely necessary to maintain the balance and stability of an ecosystem.
They ensure:

Population Control (Predation & Competition): Relationships like predation prevent overpopulation of certain species (prey), which stops them from destroying producers (plants).
Competition limits resource dominance, ensuring diversity.

Energy Flow and Cycling (Food Webs & Decomposers): They facilitate the continuous flow of energy from producers to consumers and the crucial recycling of nutrients (like carbon and nitrogen) back into the soil and atmosphere by decomposers.

Survival and Reproduction (Mutualism): Relationships like mutualism (e.g., pollination) are vital for the reproduction and survival of many plant and animal species.
Essentially, these interdependencies create a resilient web of life; if one part is lost, the entire system can collapse, making these connections essential for ecological balance.

Indicators
Question 27.
The ultimate source of energy in the food chain.
Answer:
The Sun

Question 28.
Storage of solar energy in producers.
Answer:
Plants absorb light and convert it into chemical energy via photosynthesis.

Question 29.
Energy transfer and energy loss.
Answer:
Energy is transferred from one level to the next (Plant → Herbivore → Carnivore). However, a portion is lost as heat at each stage for bodily functions.

Question 30.
Energy availability at different trophic levels.
Answer:
The amount of available energy decreases as you move up the trophic levels.

Question 31.
Which are the important elements and substances among them?
Answer:
The most important elements and substances are the basic building blocks and energy sources of life:
Elements: Carbon (C), Hydrogen (H), Oxygen (O), Nitrogen (N), and Phosphorus (P).
Substances: Water (H₂O), Proteins, Carbohydrates, and Fats (Lipids).

Text Book Page No : 280
Indicators

Question 32.
In what form is carbon present in the atmosphere?
Answer:
As Carbon dioxide (CO₂).

Question 33.
Why do plants use carbon dioxide?
Answer:
To perform photosynthesis and produce food.

Question 34.
How does carbon dioxide return to the atmosphere?
Answer:
Through the respiration of living organisms, decomposition of dead matter, and combustion (burning) of fossil fuels.

Question 35.
What situations lead to an increase in atmospheric carbon dioxide?
Answer:
Excessive burning of fossil fuels (factories, vehicles) and deforestation (cutting down trees).

Question 36.
How does biodiversity conservation help in maintaining carbon balance?
Answer:
Plants/Forests absorb large amounts of CO₂. Protecting them ensures that CO₂ levels in the atmosphere are regulated.

Question 37.
Collect information and illustrate other material cycles.
Water cycle Nitrogen cycle
Answer:
Water Cycle (Hydrologic Cycle)
Water constantly moves between the Earth and the atmosphere through changes in state.
Evaporation/Transpiration:Water turns into vapor and rises to the atmosphere.
Condensation: Vapor cools and forms clouds.
Precipitation: Water falls back as rain, snow, etc.
Runoff/Infiltration: Water flows over or soaks into the ground.
Nitrogen Cycle
Nitrogen is converted into usable forms by bacteria for living things.
Nitrogen Fixation: Bacteria convert atmospheric Nitrogen gas (N₂) into Ammonia (usable form).
Assimilation: Plants absorb this nitrogen to make proteins/DNA.
Nitrification: Bacteria convert Ammonia to Nitrates (preferred plant food).
Denitrification: Other bacteria return nitrogen compounds back to Nitrogen gas (N₂) in the atmosphere.

Text Book Page No : 281
Indicators

Question 38.
What would happen if decomposers disappeared?

Answer:
Fig 17.6 shows a dense, healthy forest. Fig 17.7 shows a developed city area where the forest has been destroyed.

Question 39.
How does deforestation affect other living beings?
Answer:
It destroys habitats, causing food scarcity and death for wild animals, leading to species extinction.

Question 40.
Are all the organisms that once lived in your area still seen today?
Answer:
No. Many organisms have disappeared (become locally extinct or extirpated) due mainly to habitat loss and pollution.

Question 41.
Is biodiversity in our area increasing or decreasing?
Answer:
Decreasing. In most regions, the expansion of human settlements and activities leads to biodiversity depletion (loss of species and ecosystems).

Question 42.
Is this kind of biodiversity loss happening only in our region?
Answer:
No. It is a global phenomenon that is happening all over the world. The crisis is tracked globally by organizations like IUCN.

Text Book Page No : 282 & 283

Question 43.
Can we protect all species at all times?
Answer:
No. While the goal is protection, limited resources and changing environments mean conservation efforts often prioritize rare and endangered species (like those in the Red Data Book).

Question 44.
Is it possible to fulfill human needs and conserve biodiversity at the same time?
Answer:
Yes. It requires careful planning and implementation of conservation strategies, such as In-situ conservation (protecting natural habitats) and Ex-situ conservation (protecting species outside their habitat).

Question 45.
What are the reasons for wild animals to enter human settlements?
Answer:
The main reasons are the depletion of resources and loss of habitat within the forest. For example, intensifying summer heat can drive animals like elephants into settlements in search of food and water.

Question 46.
Can you suggest any solutions to this problem?
Answer:
Using Al technology to generate sounds to drive wild animals back into the forest.
Preparing special sanctuaries or shelters outside the forest to house animals like tigers.
Providing adequate compensation to farmers for crop damage.

Question 47.
Why is it necessary to conserve biodiversity?
Answer:
Biodiversity is necessary because it ensures human survival by providing:

• Food (fruits, vegetables, fish, etc.).
• Medicine (many modern medicines are made from plants and other organisms).
• Ecosystem Services (creatures like bees help in pollination).
• Clean Air and overall environmental health.

Question 48.
What are the organizations in our country that work for nature conservation?
Answer:
There are various government institutions, schemes, and local bodies that work for conservation:

• Kerala State Biodiversity Board
• Biodiversity Management Committees (BMC) at the local self-government level, which prepare the People’s Biodiversity Register (PBR).
• (Internationally, WWF and IUCN are mentioned as examples).

Class 8 Basic Science Chapter 17 Question Answer Extended Activities

Question 1.
Examine the People’s Biodiversity Register (PBR) of the local self-government body (Panchayat/Municipality / Corporation) where your school is located. List the species found under each category and their current status (increasing/decreasing). Present your findings in the class.

Question 2.
Plant at least 10 different types of plants at your home and make a small garden. Before making your garden, record the organisms that were already present and the composition of the soil. After one month of planting, observe and record the changes that occurred. Present your findings and conclusions in the class.

The Beauty of Diversity Class 8 Notes

Class 8 Basic Science The Beauty of Diversity Notes Kerala Syllabus

Introduction: Biodiversity

• Biodiversity Park: A place (like in a school) that showcases a variety of living organisms, both plants and animals, in their natural setting. It’s called a biodiversity park because it contains diverse life forms.
• Biodiversity: Refers to the vast variety of life forms found on Earth, including plants, animals, fungi, microorganisms, etc.

ജീവവൈവിധ്യം (Biodiversity) എന്നാൽ ഭൂമിയി ലുള്ള വിവിധ തരത്തിലുള്ള ജീവജാലങ്ങൾ (സസ്യ ങ്ങൾ, ജന്തുക്കൾ, സൂക്ഷ്മജീവികൾ) എന്നാണ് അർത്ഥമാക്കുന്നത്. ഒരു ജൈവവൈവിധ്യ പാർക്ക് (Biodiversity Park) എന്നാൽ ഇത്തരത്തിലുള്ള വിവിധ ജീവജാലങ്ങളെ ഒരുമിച്ച് കാണാൻ കഴി യുന്ന ഒരിടമാണ്.

Habitat(ആവാസവ്യവസ്ഥ)

• The natural surroundings where each living being lives is called its habitat.
• Examples of Habitats (Fig 17.3):

• • Coral reef (പവിഴപ്പുറ്റ്)
  • Mangrove Forest (കണ്ടൽക്കാട്)
  • Desert (മരുഭൂമി)
  • Wetlands (തണ്ണീർത്തടം)
• Habitat Examples (Fig 17.2):

• • Hibiscus Plant: Land
  • Fish: Water (Ocean/Sea)
  • Frog: Land and Water (Amphibious)
  • Polar Bear: Polar regions (Snow/Ice)

ഓരോ ജീവിയും സ്വാഭാവികമായി ജീവിക്കുന്ന ചുറ്റുപാടിനെയാണ് അതിന്റെ ആവാസവ്യവസ്ഥ (Habitat) എന്ന് പറയുന്നത്. കര, ജലം, മരുഭൂമി, ധ്രുവപ്രദേശങ്ങൾ എന്നിവയെല്ലാം വിവിധതരം ആവാസവ്യവസ്ഥകളാണ്.

Lichens (ലൈക്കനുകൾ)
Lichens are an example of a symbiotic relationship and contribute to biodiversity.

• Structure: Some algae or cyanobacteria live inside certain fungi.
• Symbiosis (സഹജീവനം): This is a mutually beneficial relationship.
  • The fungus provides structure and protection.
  • The algae or cyanobacteria produce food for both organisms through photosynthesis.
• The symbiotic organisation of these organisms is known as lichens.

ലൈക്കനുകൾ എന്നത് ഫംഗസുകളും ആൽഗകളും (അല്ലെങ്കിൽ സയനോബാക്ടീരിയകളും ഒരുമിച്ച് ജീവിക്കുന്ന ഒരു കൂട്ടായ്മയാണ്. ഫംഗസ് സംര ക്ഷണവും ഘടനയും നൽകുമ്പോൾ, ആൽഗകൾ പ്രകാശസംശ്ലേഷണം (photsoynthesis) വഴി രണ്ടുപേർക്കും വേണ്ട ആഹാരം നിർമ്മിക്കുന്നു. ഇത് സഹജീവനത്തിന് (symbiosis) ഒരു ഉദാഹ രണമാണ്.

Biodiversity (ജീവവൈവിധ്യം)
Biodiversity refers to all the diverse living organisms on Earth, including plants, animals, and microorganisms, along with their habitats.

• Operational Definition: “All the diverse living organisms on Earth, including plants, animals, and microorganisms, along with their habitats, together form biodiversity.”
• Variation: Even within a single species, variations can exist (e.g., different colours of hibiscus flowers, different types of ants – Fig 17.4).
  There are over 15,000 types of ants discovered, adapted to various habitats like deserts, rainforests, and mountains.

Levels of Biodiversity (ജീവവൈവിധ്യത്തിന്റെ തലങ്ങൾ)
Biodiversity can be studied at different levels (Illustration 17.1):

• Genetic Diversity (ജനിതക വൈവിധ്യം):
  • Variation within a species.
  • Example: Different varieties of rice (over 50,000 in India) or mangoes (over 1,000 varieties in India) show genetic diversity. Different breeds of dogs or different colours of hibiscus flowers also represent this.
• Species Diversity (സ്പീഷീസ് വൈവിധ്യം):
  • Species-level diversity; the variety of different species within a particular area or habitat.
  • Example: In a forest, there are elephants, monkeys, birds, snakes, grasses. bacteria, etc. – all different species living together.
• Ecosystem Diversity (ആവാസവ്യവസ്ഥാ വൈവിധ്യം):
  • Diversity at the ecosystem level; the variety of different habitats or ecosystems within a larger region.
  • Example: In a village, there might be paddy fields, coconut groves, hills, rivers, and marshes – each representing a different ecosystem. Coral reefs, mangrove forests, deserts, and wetlands are also examples of different ecosystems.

ജീവവൈവിധ്യത്തെ മൂന്ന് തലങ്ങളിൽ പഠിക്കാം:

• ജനിതക വൈവിധ്യം: ഒരു സ്പീഷീസിനുള്ളി ലെ വ്യതിയാനങ്ങൾ (ഉദാ: വിവിധയിനം നെല്ലുകൾ, മാവുകൾ).
• സ്പീഷീസ് വൈവിധ്യം: ഒരു പ്രദേശത്തുള്ള വ്യത്യസ്ത സ്പീഷീസുകളുടെ എണ്ണം (ഉദാ: കാട്ടിലെ ആന, കുരങ്ങ്, പാമ്പ് തുടങ്ങിയവ).
• ആവാസവ്യവസ്ഥാ വൈവിധ്യം: ഒരു വലിയ പ്രദേശത്തെ വിവിധതരം ആവാസവ്യവസ്ഥ കൾ (ഉദാ: വയൽ, കാവ്, പുഴ, കുളം എന്നിവ ചേർന്ന ഒരു ഗ്രാമം).

Importance of Biodiversity (ജീവവൈവിധ്യത്തിന്റെ പ്രാധാന്യം)
Biodiversity plays a crucial role in the survival of all living beings, including humans.
The theme “Our biodiversity, our food, our health” highlights this connection.

• Benefits / Roles of Biodiversity:
  • Provisioning Services: Provides essential resources like food, clothing, shelter, fuel, and medicines.
  • Regulating Services: Helps maintain the stability of soil, water, and air; plays a role in nutrient cycling (e.g., decomposition), pollination (പരാഗണം), biological control (ജൈവീക കീടനിയന്ത്രണം), and seed dispersal (വിത്ത് വിതരണം).
  • Supporting Services: Essential for processes like photosynthesis and soil formation.
  • Cultural & Aesthetic Values: Provides recreational opportunities, beauty, and inspiration.
• May 22 is celebrated as the International Day for Biodiversity.

Biodiversity and the Balance of the Eco system (ജീവവൈവിധ്യവും ആവാസവ്യവസ്ഥയുടെ സന്തുലനവും)
The interactions between different organisms are crucial for maintaining the balance and stability of an ecosystem. One key interaction is the flow of energy through feeding relationships, represented by food chains and food webs.

• Food Chain (ഭക്ഷ്യ ശൃംഖല)
  • A representation of how energy flows through an ecosystem, showing who eats whom.
  • Example (Illustration 17.3): Paddy → Rat → Snake → Eagle.

• • Components of a Food Chain (based on Indicators):
• Producers (ഉൽപ്പാദകർ):
  • Food chains begin with Producers.
  • Green plants are called producers because they produce their own food using sunlight (photosynthesis). They form the base of most food chains, (e.g., Paddy).
• Consumers (ഉപഭോക്താക്കൾ):
  • Organisms that obtain energy by feeding on other organisms.
  • Primary Consumers (പ്രാഥമിക ഉപഭോ ക്താക്കൾ): Herbivores that feed directly on producers (plants), (e.g., Rat eating paddy).
• Secondary Consumers ((ദ്വിതീയ ഉപഭോ ക്താക്കൾ):) Carnivores or omnivores that feed on primary consumers, (e.g., Snake eating rat).
• Tertiary Consumers (തൃതീയ ഉപഭോക്താക്കൾ): Carnivores or omnivores that feed on secondary consumers, (e.g., Eagle eating snake).
• Decomposers (വിഘാടകർ)::
  • Bacteria, fungi, and other organisms that break down complex molecules in dead bodies and other organic remains into simpler molecules.
  • Role: They return essential nutrients back to nature (soil, water, air), making them available for producers again, (e.g., Bacteria, Fungus).
• Food Web (ഭക്ഷ്യ ശൃംഖലാജാലം)
  • A network of interconnected food chains found in an ecosystem. Most organisms eat more than one type of food and are eaten by more than one type of predator.

Trophic Level (പോഷണ തലം)

• The position an organism occupies in a food chain is called a trophic level.

Trophic Level 1 : Producers (Plants).
Trophic Level 2 : Primary Consumers (Herbivores feeding directly on plants).
Trophic Level 3 : Secondary Consumers (Carnivores/Omnivores feeding on herbivores).
Trophic Level 4 : Tertiary Consumers (Carnivores/Omnivores feeding on secondary consumers).

• Complexity: As food webs become complex, a single organism can belong to different trophic levels depending on what it eats.

Ecological Interactions (പാരിസ്ഥിതിക ബന്ധങ്ങൾ)
Organisms in an ecosystem interact in various ways besides just eating each other. These relationships are necessary to maintain the balance and stability of the ecosystem.

Types of Interactions (Illustration 17.5):

• Commensalism:
  • Beneficial to one organism, neither beneficial nor harmful to the other.
  • Example: Vanda (orchid) growing on a Tree. (Vanda gets support, tree is unaffected).
• Mutualism:
  • Beneficial to both organisms.
  • Example: Tree providing nectar to a Butterfly, which helps in pollination. (Lichens are another example).
• Competition:
  • At first, harmful to both (as they compete for the same limited resources like sunlight, water, food); later, benefit for the winner.
  • Example: Crop plants and Weeds competing for nutrients and space.
• Parasitism:
  • Beneficial to one (the parasite), harmful to the other (the host). The parasite depends on the host for food.
  • Example: Flea living on a Dog. (Other examples: intestinal worms, lice).
• Predation:
  • Beneficial to one (the predator), harmful to the other (the prey). The prey becomes food for the predator.
  • Example: Eagle hunting a Chicken. (Other examples: Tiger hunting deer, Snake eating rat).

Energy Flow in an Ecosystem (ഊർജ്ജപ്രവാഹം ആവാസവ്യവസ്ഥയിൽ)
Energy is needed for all bodily functions and is transferred from one trophic level to the next in a food chain.

Role of Decomposers: Decomposers break down dead organic matter, releasing energy, but primarily returning nutrients (matter) to the ecosystem.

Flow of Matter in an Ecosystem (പദാർത്ഥങ്ങളുടെ ചാക്രിക പ്രവാഹം)
Various substances and elements are needed for life and must be cycled through the ecosystem. For ecosystems and biodiversity to survive, the flow of matter and energy must take place effectively. Substances absorbed from the non-living environment circulate through living organisms and return to where they came from. Food chains, biodiversity, and decomposers play vital roles in this.

Carbon Cycle (കാർബൺ ചകം)
Plants use atmospheric carbon dioxide for photosynthesis. Carbon returns to the atmosphere through respiration, decomposition, and combustion of fossil fuels.

• Atmospheric Carbon: Carbon exists in the atmosphere primarily as Carbon dioxide (CO₂).
• Photosynthesis: Plants absorb CO₂ and convert it into chemical energy (organic compounds).
• Respiration & Decomposition: Carbon returns to the atmosphere through the respiration of living organisms and the decomposition of dead matter.
• Combustion: Burning fossil fuels also releases CO₂ back into the atmosphere.
• Conservation: Conserving biodiversity helps maintain the carbon balance in ecosystems.

Biodiversity Depletion (ജീവവൈവിധ്യ ശോഷണം)
Human activities can severely damage eco-systems and lead to a loss of biodiversity.
Habitat Loss:

• Deforestation and human activities destroy natural habitats, forcing wild animals into human settlements.
• Pictures in the text contrast a natural forest with an industrialized/deforested area to highlight habitat destruction.
  • Human-Wildlife Conflict:
    Wild animals (like elephants and tigers) enter farmlands and settlements due to food scarcity and habitat loss, causing agricultural loss and danger to humans.
  • Technological Solution: Al technology is being used to detect animals via cameras and generate specific sounds to drive them back into the forest.
  • Unique Species:
    Purple Frog (Mahabali Frog): A rare frog that lives underground and surfaces only once a year to mate. It is called a “living fossil” because it has undergone very little evolutionary change since the time of the dinosaurs.

The Red Data Book:

• Publisher: IUCN (International Union for Conservation of Nature).
• Purpose: It contains information about rare and endangered plants and animals.
• Listed Species Examples: Malabar civet cat, Ganges Shark, Kashmir stag, Rameswaram Parachute spider, Lion-tailed macaque, and Nilgiri tahr.

• Importance of Conservation:
  • Biodiversity is essential for food, clean air, pollination, and medicines.
  • Medicinal Plants: Many modern medicines are derived from plants. Even plants considered “weeds” may have medicinal value.
• Conservation Organizations:
  • International: IUCN, WWF (World Wide Fund for Nature).

• • Local: Kerala State Biodiversity Board.

Methods of Biodiversity Conservation
Conservation is classified into two main types:

• In-situ Conservation
  Protection of organisms within their natural habitats.
  • National Parks: e.g., Eravikulam, Silent Valley.
  • Wildlife Sanctuaries: e.g., Periyar.
  • Biosphere Reserves: e.g., Nilgiri.
  • Community Reserves: e.g., Kadalundi.
  • Ecological Flotspots: e.g., Western Ghats, Himalayas.
• □ Ex-situ Conservation
  Protection of organisms outside their natural habitats in controlled environments.
  • Examples: Botanical gardens, Zoological gardens, Gene banks, Aquariums.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 16 Tree of Life Important Questions ensure conceptual clarity.

Tree of Life Extra Questions and Answers Class 8 Basic Science Chapter 16 Kerala Syllabus

Tree of Life Class 8 Important Questions

Question 1.
Characteristics of Kingdom Fungi are given below. Choose the correct option.
i. They are Prokaryotes.
ii. They have a cell wall made of chitin.
iii. They are autotrophs containing chlorophyll.
iv. They are saprophytes.
(a) i, iii correct
(b) ii, iv correct
(c) i, ii, iv correct
(d) All are correct
Answer:
(b) ii, iv correct

Question 2.
Rules of Binomial Nomenclature developed by Carl Linnaeus are given. Choose the correct statements.
i. The scientific name has two parts: species name first, genus name second.
ii. The first letter of the Genus name should be capital.
iii. The first letter of the Species name should be capital.
iv. The scientific name should be printed in italics or underlined if handwritten.
(a) i, ii, iii correct
(b) ii, iv correct
(c) i, iv correct
(d) All are correct
Answer:
(b) ii, iv correct

Question 3.
Which of the following statements correctly define a ‘Species’ according to the biological concept?
i. It is the most basic level of classification.
ii. Organisms within a species look exactly identical.
iii. Members of a species can naturally reproduce to form fertile offspring.
iv. It is the level just below Kingdom in classification.
(a) i, iii correct
(b) i, ii, iii correct
(c) iii, iv correct
(d) i only correct
Answer:
(a) i, iii correct

Question 4.
Complete the missing levels in the hierarchical classification system shown below.
Kingdom → Phylum → (a) _______ → Order → (b) ________ → Genus → (c) ________
Answer:
a) Class b) Family c) Species

Question 5.
Complete the following table summarizing characteristics of the Five Kingdoms.

Kingdom	Nuclear Type (Prokaryotic/Eukaryotic)	Mode of Nutrition (Autotroph/Heterotroph/ Saprotroph	Example
Monera	(a) ………………….	Autotroph or Heterotroph	Bacteria
Fungi	Eukaryotic	(b) ………………….	Yeast
Plantae	Eukaryotic	(c) ………………….	Coconut Tree

Answer:

Kingdom	Nuclear Type (Prokaryotic/Eukaryotic)	Mode of Nutrition (Autotroph/Heterotroph/ Saprotroph	Example
Monera	(a) Prokaryotic	Autotroph or Heterotroph	Bacteria
Fungi	Eukaryotic	(b) Saprotroph	Yeast
Plantae	Eukaryotic	(c) Autotroph	Coconut Tree

Question 6.
Fill in the missing classification levels for the Tiger (Panthera tigris).

Level	Taxon
Kingdom	Animalia
Phylum	Chordata
Class	(a) …………………….
Order	Carnivora
Family	(b)  …………………….
Genus	Panthera
Species	(c)  …………………….

Answer:
a) Mammalia,
b) Felidae,
(c) tigris

Question 7.
Who introduced the Five-Kingdom classification system, and what were the five kingdoms proposed?
Answer:
Robert H. Whittaker introduced the Five-Kingdom classification. The five kingdoms are Monera, Protista, Fungi, Plantae, and Animalia.

Question 8.
What is Taxonomy?
Answer:
Taxonomy is the branch of science that deals with classifying organisms by comparing their characteristics, identifying similarities and differences, and giving them scientific names.

Question 9.
What contribution did John Ray make to the field of classification?
Answer:
John Ray initiated the scientific method of classification and was the first to use the term ‘species’.

Question 10.
Explain the need for a universally accepted scientific naming system like Binomial Nomenclature.
Answer:
The same organism often has different common names in different regions or languages (e.g., Papaya). This creates confusion and makes it difficult to accurately identify and study organisms globally. Binomial Nomenclature provides a unique, two-part scientific name (Genus and species) that is universally accepted, overcoming language barriers and ensuring clarity in scientific communication.

Question 11.
Why was Kingdom Monera split into two separate kingdoms (Archaebacteria and Eubacteria) in the Six-Kingdom classification?
Answer:
The split was proposed by Carl Woese based on modern scientific observations, particularly differences at the molecular level (like genetics and cell wall composition) which indicated that the organisms originally grouped in Monera actually represented two very distinct and ancient lineages of prokaryotes.

Question 12.
Explain the main limitation of classifying organisms based only on their external structure, as was done in early classification systems.
Answer:
Classifying based only on external structure can be misleading because organisms that look similar externally may not be closely related evolutionarily or genetically (like Wolffia and Salvinia). Conversely, organisms that look

different might be closely related. Modern classification uses evolutionary and genetic data along with morphology for a more accurate system.

Question 13.
Identify the organism that belongs to Kingdom Monera from the following list: Amoeba, Yeast, Paramecium, Bacteria.
Answer:
Bacteria. (Amoeba and Paramecium are Protista; Yeast is Fungi).

Question 14.
Find the odd one out based on the Five-Kingdom classification: Mushroom, Mould, Yeast, Euglena.
Answer:
Euglena.
Reason: Mushroom, Mould, and Yeast belong to Kingdom Fungi. Euglena belongs to Kingdom Protista.

Question 15.
Which of the following is NOT an example of a species? Human beings, Cats, Mammalia, Coconut Tree.
Answer:
Mammalia.
Reason: Human beings (Homo sapiens), Cats (Felis catus), and Coconut Tree (Cocos nucifera) represent specific species. Mammalia is a Class, a higher level of classification that includes many different species.

Question 16.
Analyse the statement and answer the questions.
The scientific name of dog is Canis familiaris
a. How is this method of giving scientific names to organisms known as?
b. What does the words Canis and familiaris in the scientific name indicate?
Answer:
a. Binomial nomenclature
b. Canis indicates the name of the genus and the ‘familiaris’ indicates the name of the species.

Question 17.
Cocos nucifera is the scientific name of coconut. Identify the level of classification indicated by the underlined part.
Answer:
Genus

Question 18.
The various levels of classification of human beings based on the six kingdom classification are given below. Complete the table suitably.

Domain	Eukarya
Kingdom	a. ________
Phylum	Chordata
Class	b. ________
Order	Primates
Family	Hominidae
Genus	c. ________
Species	d. ________

Answer:
a. Animalia
b. Mammalia
c. Homo
d. Sapiens

Question 19.
Find the correct pair from the following.
i. Robert Hooke-Discovered the centre of the cell
ii. Robert H Whittaker – 5 Kingdom classification
iii. M. J. Schleiden – Found animal body is made up of cells
Answer:
ii. Robert. H. Whittaker – 5 kingdom classification

Question 20.
Analyse the statement and answer the following questions.
According to the Binomial nomenclature, scientific name of human being is Homo sapiens.
a) What does the First word Homo indicate
b) Who proposed the Binomial nomenclature?
c) What is the merit of Binomial nomenclature?
Answer:
a) Genus

b) Carl Linnaeus

c) Each organism is known by different names in different places and in different languages. So a commonly acepted name is needed for identifying organisms.

Question 21.
Identify the relation between the words of first pair and complete the second pair suitably
a) Mushroom: Fungi
Amoeba: …………………….
b) Cat: Felis domesticus
Crow: ……………………
Answer:
a. Protista
b. Corvus splendens

Question 22.
Rewrite the following sentences correctly. If there is any error in the underlined part
a. Aristotle is known as the father of Botany
b. Monera includes unicellular organisms with distinct nucleus
c. Five Kingdom classification was introduced by Robert H Whitaker
Answer:
a. Theophrastus is known as the father of Botany
b. Protista includes unicellular organisms with distinct nucleus

Question 23.
Name of two scientists related to taxonomy is given below. Write their contributions in this field.
i) Charakan
ii) John Ray
Answer:
a. Charakan Wrote the treatise ‘Charakasamhita1 including around 200 plants and animals.
b. John Ray Recorded more than 18000 plants in his book named Historia generalis plantarum’ Used the term ‘species’ for the first time.

Question 24.
The scientific name of Tiger is Panthera Tigris.
a. What do the words ’Panthera’ and ’Tigris’ indicate ?
b. What are the advantages of giving scientific name to organisms?
c. Name the scientist who proposed the scientific naming of organisms?
Answer:
a. Panthera indicates Genus and Tigris indicates Species

b. By giving scientific name, organisms can be easily identified all over the world and it helps to solve the difficulties arising out of an organisms known by various names in different languages and regions.

c. Carl Linnaeus

Question 25.
Write the taxonomic hierarchy of coconut tree by choosing appropriate terms from the box
Plantae cocos Series
Nucifera Dicotyledonae
Genus Monocotyledonae

Domain	Eukarya
Kingdom	i. ……………………..
Phylum	Angiospermophyta
Class	ii. ……………………
iii. ……………………	Calysinae
Family	Aracasiae
iv. ………………………	v. ……………………
Species	vi. ………………….

Answer:
i. Plantae
ii. Monocotyledonae
iii. Order
iv. Genus
v. Cocos
vi. Nucifera

Question 26.
Correct the mistakes if any in the underlined part.
a. Taxonomic keys are the scientific indicators used for identifying and classifying plants and animals.
b. Genus is the basic level of classification.
c. The first word of the scientific name indicates the genus and the second word indicates the species.
Answer:
Species is the basic level of classification

Question 27.
Name of scientists who contributed in the field of taxonomy are given below. Identify their contribution which is given in the box and make pairs.
A) Carl Woese
B) John Ray
C) Theophrastus
d) R.H. Whittaker
i. Classified the organisms as those with red blood cells and those without red blood cells.
ii. Classified plants into annuals, biennials and perennials.
iii. Six kingdom classification.
iv. Five kingdom classification.
v. Used the term species for the first time
Answer:
i. Aristotle
ii. Theophrastus
iii. Cari woese
iv. Robert H Whittaker
v). John ray

Question 28.
Identify the word pair relation and fill the blanks.
a. 2 Kingdom classification:
Carls Linnaeus :: 5 kingdom
Classification: ………………………….
b. Mushrooms: Fungi::
Bacteria: ………………………….
c. Aristotle – Father of Biology
Carls Linnaeus – ………………………….
d. Golden shower : Cassia fistula
:: ………………………….: Corvus splendens
e. Charaka: Father of Ayurveda
:: ………………………….: Father of Botany
Answer:
a. R H Whittaker
b. Monera
c. Father of Taxonomy
d. Crow
e. Theophrastus

Question 29.
Identify the odd one and write the characteristic features of others.
a. Lion, Tiger, Rabbit, Cat.
b. Genus, Order, Carl Linnaeus, Phylum.
Answer:
a. Rabbit – Others including order Carnivora
b. Carl Linnaeus – He is the Father of Modern Taxonomy. Others are different levels of classification.

Question 30.
Which organism is most suitable for the following indicators (amoeba, bacteria, virus, Fungus)
• Lives only in living cell
• Pathogen
• Only genetic material and a protein covering.
Answer:
Virus

Question 31.
Find out the scientists suitable to the statements given.
i. Author of Charaka Samhitha
ii. Author of Historia Generalis Plantarum
iii. Father of Modern Taxonomy
iv. Father of Biology
Answer:
i. Charaka
ii. John ray
iii. Carl Linnaeus
iv. Aristotle

Question 32.
Complete the table
Organism Scientific Name
Elephant ……………………..
Peacock ……………………..
Dog ……………………..
Hibiscus ……………………..
Neem ……………………..
Paddy
……………………..
Answer:
Elephas Maximus / Elephas Indicus
• Pavo Cristatus
• Canis Familiaris
• Hibiscus Rosasinensis
• Azadirachta Indica
• Oryza sativa

Question 33.
What are limitations in the system of classification of Carl Linnaeus.
Answer:
Some lower organisms share the characters of both animals and plants. So it is difficult to recognize them as plants or animals. Linnaeus considered only plants and animals for his classification. Microscopic organisms like bacteria, fungus, protozoa etc were not included either in plant kingdom or in animal kingdom. Certain animals which were included in the classification of Linnaeus shows characters of both the animal and plant kingdoms. Eg: Euglena protozoan shows locomotory movements like animals, but it contains chlorophyll like plants.

Question 34.
Hints about some organisms are given below. Name the kingdom to which these organisms belong.
a. Multicelluar heterotropic organisms with a nucleus and capacity for locomotion.
b. Multicelluar heterotropic, nonmotile organisms with a nucleus.
c. Unicelluar organisms with a nucelus.
d. Multicelluar, autotropic, nonmotile organisms with a nucleus.
Answer:
a. Animalia
b. Fungi
c. Protista
d. Plantae

Question 35.
Write the scientific name of following animals and plants.
Coconut, paddy, wheat, crow, mango, grapes
Answer:
Coconut – Cocos nucifera
Paddy – Oryza sativa
Wheat – Triticum aestivum
Crow – Corvus splendens
Mango – Mangifera indica
Grapes – Vitis vinifera

Question 36.
What are two important characters of species?
Answer:

1. Species is a group of organisms that can freely interbreed to produce fertile offsprings.
2. A group of organisms that closely resemble each other in structure biochemical make up and external charecteristics but which are genetically different. In one species there may be sub species.

Question 37.
“The method of classification adopted by Whittakar is much better than the method adopted by Carl Linnaeus the father of the science of classification. What is your response to this statement?
Answer:
The classification of Linnaeus had only two kingdom, plants and animals. He did not consider bacteria, protozoa, fungus etc. Certain characters of animals considered by Linnaeus for classification are found in both the kingdoms, eg. photosynthesis found in plants is seen in some animals like Euglena.

Certain characters of animals can be seen in plants also, eg certain types of algea. But Whittakar adopted the method of having five kingdoms including protozoa and bacteria.

1. Monera,
2. protista
3. fungi
4. plantae
5. animalia

Question 38.
Bacteria does not have a well defined Nucleus. Viruses are also like Bacteria. Why is it not possible to include viruses under Monera.
Answer:
Viruses exhibit living nature only when they enter the host cells. On other occasions they do not exhibit living nature. But organisms in Monera are not like that.

Question 39.
How do the levels of classification of plants made by Carl Linnaeus differ from the levels of classification of Animals?
Answer:
Not much differences are there. In the place of ‘Order’ in animal classification, plant classification has series. And in the place of ‘Phylum’ in animal classification. plant classification has ‘Division’.

Question 40.
Cell is the smallest unit of life. But there are certain organisms that live without cell too. Analyse these statement.
Answer:
Life is not possible without a cell. Viruses do not have cells. As it is so, it does not have life when it outside a living cell. As it enters a living cell, it will show the features of life forms. It makes use of the components of the host cell and continues to live. Though it does not have a cell, it can continue its life only after entering a host cell and by making use of its components.

Question 41.
Observe the given statement, and write correct answer if you find false statements.
a. Aristotle is the Father of Biology.
b. John Ray is the Father of Botany.
c. Carl Linnaeus used the term ‘species’ for the first time
d. Charakan proposed binomial nomenclature.
Answer:
a. True
b. False, Theophrastus
c. False, John Ray
d. False, Carl Linnaeus.

Question 42.
Read the following statements and choose the correct option:
Statement I: John Ray was the first scientist to use the term ‘species’.
Statement II: Aristotle classified organisms into five kingdoms.
A) Both Statements are true.
B) Statement I is true, but Statement II is false.
C) Statement I is false, but Statement II is true.
D) Both Statements are false.
Answer:
B) Statement I is true, but Statement II is false. : Explanation: John Ray used the term species (True). Aristotle classified animals by red blood, not 5 kingdoms 9. (Whittaker did that).

Question 43.
Read the following statements regarding classification levels:
Statement I: The level just below Kingdom is Phylum.
Statement II: Species is the basic and smallest unit of classification.
A) Both Statements are correct.
B) Only Statement I is correct.
C) Only Statement II is correct.
D) Both Statements are incorrect.
Answer:
A) Both Statements are correct. : Explanation: Phylum is below Kingdom, and Species is the basic unit. Both are correct.

Question 44.
Assertion (A): According to modern classification, Wolffia and Anthurium are included in the same family, Araceae.
Reason (R): Wolffia and Anthurium look exactly the same in their external structure.
A) Both A and R are true, and R is the correct explanation of A.
B) Both A and R are true, but R is NOT the correct explanation of A.
C) A is true, but R is false.
D) A is false, but R is true.
Answer:
C) A is true, but R is false. : Explanation: They are in the same family due to genetic/evolutionary relationships, despite having different external structures.

Question 45.
Assertion (A): Fungi like mushrooms cannot perform photosynthesis.
Reason (R): Fungi cells lack chlorophyll and are saprophytes.
A) Both A and R are true, and R is the correct explanation of A.
B) Both A and R are true, but R is NOT the correct explanation of A.
C) A is true, but R is false.
D) A is false, but R is true.
Answer:
A) Both A and R are true, and R is the correct explanation of A. : Explanation: Fungi are saprophytes because they lack chlorophyll.

Question 46.
Who introduced the method of Binomial Nomenclature?
A) Robert H. Whittaker
B) Carl Woese
C) Carl Linnaeus
D) Ernst Haeckel
Answer:
(C) Carl Linnaeus

Question 47.
Which of the following organisms belongs to Kingdom Monera?
A) Amoeba
B) Bacteria
C) Mushroom
D) Coconut tree
Answer:
(B) Bacteria

Question 48.
Write the scientific names of the following organisms:
a) Human
b) Tiger
Answer:
a) Human: Homo sapiens
b) Tiger: Panthera tigris

Question 49.
List any two rules that must be followed while writing scientific names (Binomial Nomenclature).
Answer:
Rules of Binomial Nomenclature:
1. The first word is the Genus (starts with a Capital letter).
2. The second word is the Species (starts with a small letter).
(Optional: The name should be in italics).

Question 50.
Why was the Two-Kingdom classification system proposed by Carl Linnaeus considered insufficient? Give one reason.
Answer:
It was insufficient because it did not classify organisms like bacteria, fungi, and some unicellular organisms correctly. For example, Fungi are neither plants (no photosynthesis) nor animals, yet they had to be forced into one of the two groups.

Question 51.
Differentiate between Kingdom Monera and Kingdom Protista based on their cell structure (nucleus) and give one example for each.
Answer:
Monera: Prokaryotes (do not have a distinct nucleus). Example: Bacteria. Protista: Eukaryotes (have a distinct nucleus). Example: Amoeba.

Question 52.
Complete the table below regarding the classification of the Tiger:

Level	Name
Kingdom	Animalia
Phylum	(a) ……………….
Class	Mammalia
Order	(b) ……………….
Family	Felidae
Genus	(c) ……………….

Answer:
(a) Chordata
(b) Carnivora
(c) Panthera

Question 53.
Explain the Six-Kingdom Classification system. Who proposed it, and how was the Kingdom Monera divided in this system?
Answer:
Proposed by: Carl Woese.

Explanation: He analyzed the Kingdom Monera and found significant differences among its members. He divided Monera into two separate kingdoms: Archaebacteria and Eubacteria. This, along with Protista, Fungi, Plantae, and Animalia, makes up the six kingdoms.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 16 Chemistry of Matter Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 16 Chemistry of Matter Question Answer Notes

Class 8 Basic Science Chapter 16 Notes Kerala Syllabus Chemistry of Matter Question Answer

Chemistry of Matter Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Examine the following passage and correct the errors if any, found in the underlined parts.
Classifying living beings into larger groups based on similarities and further into smaller groups based on differences is the basis of the cell theory. The various levels of classification were developed by Theophrastus. He is known as the Father of Taxonomy. The smallest unit in classification is phylum. The method of scientifically naming organisms was developed by Carl Linnaeus. The practice of identifying an organism by using both its genus and species name is known as taxon. According to this, when writing the scientific name of an organism, the species name should be written first.
Answer:

Original Underlined Text	Correction and Reason
…is the basis of the cell theory	Correction: Taxonomy or Classification. Classification is the basis for grouping organisms.
The various levels of classification were developed by Theophrastus	Correction: Carl Linnaeus. Linnaeus ‘ introduced the different levels of classification.
The smallest unit in classification is phylum	Correction: Species. Species is the most basic/smallest level of classification.
..practice of identifying an organism by using both its genus and species name is known as taxon	Correction: Binomial Nomenclature. Taxon refers to the classification levels themselves (like Kingdom, Phylum, etc.).
…the species name should be written first	Correction: Genus name. The genus name comes first, and the species name comes second in Binomial Nomenclature.

Question 2.
Match the following.

A	B
1. John Ray	a) Living things were classified into 5 kingdoms
2. Carl Linnaeus	b) Plants were classified as trees, shrubs, and herbs
3. Ernst Haeckel	c) Living things were classified into 3 kingdoms
4. Robert H. Whittaker	d) Father of taxonomy
5. Aristotle	e) Initiated the scientific method of classification
	f) Animals were classified into those with red blood and those without red blood.

Example: Aristotle classified animals into two groups – those with red blood and those without red blood.
Answer:
1. John Ray – e) Initiated the scientific method of classification
2. Carl Linnaeus – d) Father of taxonomy
3. Ernst Haeckel – c) Living things were classified into 3 kingdoms
4. Robert H. Whittaker – a) Living things were classified into 5 kingdoms
5. Aristotle – f) Animals were classified into those with red blood and those without red blood

Question 3.
Which of the following does not belong to the group? What is the common feature of the others?
A. Housefly, Lizard, Butterfly, Grasshopper
B. Cat, Dog, Tiger, Snake
C. Arecanut tree, Coconut tree, Jack tree, Palm tree
Answer:
A. Housefly, Lizard, Butterfly, Grasshopper
Does not belong: Lizard Common feature of others: They are all insects (belonging to Class Insecta,
Phylum Arthropoda). Lizards are reptiles (Class Reptilia, Phylum Chordata).

B. Cat, Dog, Tiger, Snake Does not belong: Snake Common feature of others: They are all mammals (belonging to Class Mammalia). Snakes are reptiles (Class Reptilia).

C. Arecanut tree, Coconut tree, Jack tree, Palm tree
Does not belong: Jack tree
Common feature of others: They are all types of palm trees (belonging to the Family Arecaceae, Class Liliopsida – Monocots). Jack tree belongs to a different family (Moraceae) and is a dicot.

Basic Science Class 8 Chapter 16 Question Answer Kerala Syllabus

Textbook Page No : 255

Question 1.
What is the main idea conveyed by the poster?
Answer:
The main idea conveyed by the poster is the vastness of biodiversity on Earth, by showing the large and varied number of identified species in major groups of living organisms (Animals, Plants, Fungus, Protista, and Bacteria/Archaea).

Question 2.
Can you suggest a suitable title for the poster?
Answer:
A suitable title for the poster could be:

• The World of Life: A Glimpse of Biodiversity
• Biodiversity on Earth: Identified Species Count
• Tree of Life: Species Numbers

Question 3.
How is it possible to identify and study so many living organisms?
Answer:
It is possible to identify and study so many living organisms through the branch of science called Taxonomy.

Taxonomy deals with:

• Classifying organisms by comparing their characteristics, and identifying similarities and differences.
• Giving them scientific names (Nomenclature).
• This systematic approach, using different levels of classification (Taxa), allows scientists to organize and understand the relationships between the vast number of organisms (biodiversity).

Textbook Page No : 257 & 258

Question 4.
What are the classification levels between species and phylum?
Answer:
Referring to the levels of classification, the levels between species and phylum are:

• Genus
• Family
• Order
• Class

Question 5.
Complete Table 16.1 for the classification levels of the Tiger:

Answer:

Levels of classification	Group and characteristics	Organisms (Example/ Group)
Kingdom:	Animalia (Animals) – Multicellular, eukaryotic, heterotrophic organisms.	Tiger, Cat, Human, Fish etc.
Phylum:	Chordata (Vertebrates) – Presence of vertebrae and vertebral column.	Tiger, Cat, Human, Fish
Class:	Mammalia (Mammals) – Warmblooded, body covered with hair, nourish young ones with milk.	Tiger, Cat, Human
Order:	Carnivora (Carnivores) – Sharp teeth and claws adapted for hunting and eating meat.	Tiger, Cat
Family:	Felidae – Retractable claws, excellent night vision.	Tiger, Cat
Genus:	Panthera (Big cats) – Ability to roar, large and powerful body structure.	Tiger, Lion, Leopard
Species:	Panthera tigris (Tiger) – Orange fur with black stripes and strong large body.	Tiger

Textbook Page No : 260

Question 6.
Can we include Bacteria, Paramecium, Mushroom, and Yeast in the Plant or Animal kingdoms? Why?
Answer:
No, we cannot include them in the Plant or Animal kingdoms.
Reason: The Two-Kingdom classification (Plantae and Animalia) was insufficient because it did not account for characteristics of these organisms.

• Bacteria are prokaryotes (no distinct nucleus).
• Paramecium is a unicellular eukaryote.
• Fungi (Mushroom/Yeast) are saprophytes and do not have chlorophyll to perform photosynthesis like plants.

Textbook Page No : 262

Question 7.
Complete Table 16.3: Five Kingdom Classification

Answer:

Kingdom	Characteristics	Examples
Monera	Prokaryotes, do not have a distinct nucleus.	Bacteria
Protista	Mostly unicellular eukaryotes. Can be autotrophs or heterotrophs. Move via flagella, cilia, or pseudopodia.	Amoeba, Paramecium, Euglena
Fungi	Eukaryotic saprophytes. Do not have chlorophyll.	Mould, Yeast, Mushroom Plantae
Plantae	Multicellular eukaryotes that produce their own food (photosynthesis).	Plants (Coconut tree, etc.)
Animalia	Multicellular eukaryotes and heterotrophs. Have locomotory organs.	Sponges, Humans, Tiger

Textbook Page No : 264

Question 8.
Complete Table 16.4 for Scientific Names:

Answer:

Organism	Scientific Name
Tiger	Panthera tigris
Coconut tree	Cocos nucifera
Dog	Canis familiaris (The text provides no specific information, but this is the correct scientific name)

Textbook Page No : 266
Indicators

Question 9.
Limitations of classification based only on external features
Answer:
Classification based solely on external appearance is often inaccurate because it can separate organisms that are actually closely related biologically.

• The text uses the example of Wolffia (a tiny floating plant) and Anthurium (a larger flowering plant). Despite looking very different externally, they belong to the same family (Araceae) because they are genetically related and share a common ancestor.
• Conversely, Salvinia and Wolffia might look somewhat similar (both small aquatic plants), but Salvinia is a non-flowering fern (Family Salviniaceae) while Wolffia is a flowering plant.

Therefore, relying only on external structure can lead to errors in grouping.

Question 10.
Changes in the criteria of classification
Answer:
The criteria for classification have evolved from simple observations to complex scientific analysis.

• Past Criteria: In early days, classification was based merely on external structure and utility (usefulness).
• Modern Criteria: Today, classification considers evolutionary (phylogenetic) relationships and genetic characteristics.
• Result: This shift led to the splitting of Kingdom Monera into Archaebacteria and Eubacteria based on molecular studies.

Textbook Page No : 267

Question 11.
Do all humans belong to the same species?
Answer:
Yes.
Although humans living in different parts of the world show variations in external appearance (due to local geography and environmental factors), there are no significant genetic or other fundamental differences among them. All humans around the world share a common ancestor and can interbreed to produce fertile offspring. That is why all humans are classified under the same species, sapiens (within the genus Homo).

Class 8 Basic Science Chapter 16 Question Answer Extended Activities

Question 1.
Prepare a magazine including pictures and contributions of scientists related to the history of classification.

Question 2.
List the plants found in the school surroundings and biodiversity parks, find their scientific names, and display them.

Tree of Life Class 8 Notes

Class 8 Basic Science Tree of Life Notes Kerala Syllabus

Introduction: Biodiversity
The Earth is home to a vast number of diverse living organisms, including plants, animals, fungi, bacteria, and protists. How is it possible to identify and study so many different living things?. Attempts to classify and study them began a long time ago.

Early Attempts at Classification

• Aristotle (അരിസ്റ്റോട്ടിൽ): Classified living beings based on whether they had red blood or not.
• Theophrastus (തിയോഫാസ്റ്റസ്): Classified plants into trees, shrubs, and herbs.

• Carl Linnaeus (കാൾലിനേയസ്): Introduced the different levels of classification and provided a scientific foundation for it. He is considered the Father of Taxonomy.

• John Ray (ജോൺ റേ): Initiated the scientific method of classification and was the first to use the term ‘species’.

• Ernst Haeckel (ഏണസ്റ്റ് ഹെക്കൽ): Classified living organisms into three kingdoms: Animalia, Plantae, and Protista.

Taxonomy (വർഗ്ഗീകരണശാസ്ത്രം)
The branch of science that deals with classifying organisms by comparing characteristics, identifying similarities/differences, and giving scientific names.
Importance: Helps understand biodiversity and relationships between organisms.

Father of Taxonomy: Carl Linnaeus.
ജീവികളെ തിരിച്ചറിയുകയും, സാമ്യതകളുടെയും വ്യത്യാസങ്ങളുടെയും അടിസ്ഥാനത്തിൽ തരംതി രിക്കുകയും, ശാസ്ത്രീയ നാമം നൽകുകയും ചെ യ്യുന്ന ശാസ്ത്രശാഖയാണ് ടാക്സോണമി. ഇതി ലൂടെ ജീവവൈവിധ്യത്തെക്കുറിച്ചും ജീവികൾ ത മ്മിലുള്ള ബന്ധങ്ങളെക്കുറിച്ചും മനസ്സിലാക്കാം.

Levels of Classification (വർഗ്ഗീകരണ തലങ്ങൾ)
Linnaeus proposed a hierarchical system for classifying organisms, starting from broad categories and becoming more specific.

The Main Steps in Classification:

1. Identification (തിരിച്ചറിയൽ)
2. Classification (വർഗ്ഗീകരണം)
3. Naming (നാമകരണം)
   The Hierarchy (Illustration 16.1):
   The levels, from broadest to most specific, are:

Species

• A species is a group of organisms that can naturally reproduce to form fertile offspring (പ്രത്യുത്പാദന ശേഷിയുള്ള തലമു).
• This is the most basic level of classification.
• Examples: Human beings, cats, dogs, and cashew trees are examples of different species. All humans belong to the same species, sapiens.

Classification Systems (വർഗ്ഗീകരണ രീതികൾ)

• The classification levels developed by Carl Linnaeus are still used today. These levels (Kingdom, Phylum, Class, etc.) are called Taxa. Plants can also be classified using these levels, from species to kingdom.
• Two-Kingdom Classification System (ദ്വികിങ്ഡം വർഗീകരണം)
  • Proposed by Carl Linnaeus.
  • Divided all living organisms into two kingdoms:
    1. Kingdom Animalia (Animals)
    2. Kingdom Plantae (Plants).
• Classification level of the coconut tree

Kingdom	Plantae (Plants) – Multicellular, eukaryotic, autotrophic organisms.
Division	Magnoliophyta (Flowering plants) – Plants with seeds enclosed inside fruits.
Class	Liliopsida (Monocotyledons) – Plants with parallel leaf venation and fibrous roots.
Order	Arecales – Plants with long and hard stems.
Family	Arecaceae – Plants with large compound leaves and unbranched stems.
Genus	Cocos – Produces large, single-seeded fruits.
Species	Cocos nucifera (Coconut) – Produces fruits with a hard shell, fibrous husk, with a liquid and solid endosperm, inside.

Limitations of the Two-Kingdom System

• Many organisms (like Bacteria, Paramecium, Mushroom, Yeast – see Fig 16.8) do not fit neatly into either the Plant or Animal kingdom.
• Bacteria: Unicellular, Prokaryote. (Plants/Animal are Eukaryotes).
  • Paramecium: Unicellular, Eukaryote, Heterotroph, Mobile. (Plants are autotrophs; some animals are multicellular).

• • Mushroom/Yeast: Eukaryote, Heterotroph (saprophyte), No mobility. (Plants are autotrophs; Animals are usually mobile).

• This showed the need for a more comprehensive system.

Five-Kingdom Classification System (പഞ്ചകിങ്ഡം വർഗീകരണം)

• Proposed by Robert H. Whittaker to overcome the limitations of the two-kingdom system.
• Classified organisms into five kingdoms:

Kingdom	Cell Type	Multicellularity	Mode of Nutrition	Key Characteristics & Examples
Monera	Prokaryotes	Unicellular	Autotrophic or Heteroirophic	Do not have a distinct nucleus. Simplest and most ancient organisms. Eg: Bacteria.
Protista	Eukaryotes	Unicellular	Autotrophic or Heterotrophic	Mostly aquatic. Have locomotory structures like flagella, cilia, and pseudopodia. Eg: Amoeba, Paramecium, Euglena.
Fungi	Eukaryotes	Mostly Multicellular	Heterotrophs (Saprophytes)	Do not have chlorophyll. Absorb nutrients from dead and decaying matter. Cell wall made of chitin. Eg: Mould, Yeast, Mushroom.
Plantae	Eukaryotes	Multicellular	Autotrophs (Photosynthesis)	Primary producers. Have roots, stems, and leaves.
Animalia	Eukaryotes	Multicellular	Heterotrophs	Have locomotory organs and definite receptors. Eg: Sponges to humans.

• Advantage of Five-Kingdom Classification: It provides a more accurate and detailed way to group organisms, especially microorganisms like bacteria, protists, and fungi, which didn’t fit well into the older two-kingdom system.

Closer Look at Some Kingdoms

• Kingdom Monera
  • Characteristics: The simplest and most ancient organisms. They are all unicellular and prokaryotic (lack a distinct nucleus).
  • Main Group: Bacteria.
  • Occurrence: Found everywhere – soil, water, air, food, inside our bodies.
  • Role: Important in nutrient cycles (like nitrogen fixation). Most are useful, but some cause diseases.
• Kingdom Protista

• • Characteristics: Unicellular eukaryotes. Most live in aquatic habitats.
  • Examples: Paramecium, Amoeba, Euglena.
  • Disease-causing Protists: Plasmodium (causes malaria), Entamoeba histolytica (causes amoebic dysentery).
• Kingdom Fungi
  • Characteristics: Eukaryotes. Mostly multicellular (except for unicellular yeast). Have a cell wall made of chitin. Do not have chlorophyll and cannot perform photosynthesis. They are saprophytes, absorbing nutrients from dead and decaying matter.
  • Examples: Mould on bread, mushrooms, yeast.

• • Uses: Edible mushrooms, yeast (used in baking wine alcohol production) candidiasis and ringworm.

The section discusses:

• The variety of local names for a plant (like Papaya).
• The difficulty in identifying and studying the same organism when it is known by different names.
• The solution: giving names that are universally accepted beyond language barriers.
• This solution is Binomial Nomenclature, a scientific method developed by Carl Linnaeus, where the name consists of the genus and species name.
• The names are often in Latin or transformed to the Latin style to ensure they are the same everywhere in the world.
• For example, the scientific name of a human is Homo sapiens.

Binomial Nomenclature (ദ്വിനാമ പദ്ധതി)

• Developed by: Carl Linnaeus.
• A universally accepted scientific method of naming organisms using two parts: the Genus name (first, capitalized) and the Species name (second, lowercase).
• Format: Written in italics or underlined.
• Importance: Provides a unique, universal name for each organism, avoiding confusion from common names.
• Example: Homo sapiens (Human), Panthera tigris (Tiger).

ജീവികൾക്ക് ലോകമെമ്പാടും അംഗീകരിക്കപ്പെട്ട ഒരു ശാസ്ത്രീയനാമം നൽകുന്ന രീതിയാണ് ദ്വിനാ മ പദ്ധതി. ഇതിൽ ആദ്യപദം ജീനസിനെയും (Genus) രണ്ടാം പദം സ്പീഷീസിനെയും (Species) കുറിക്കുന്നു. ഉദാ: Homo sapiens (മനുഷ്യൻ).

New Paths in Classification (വർഗ്ഗീകരണത്തിലെ പുതിയ രീതികൾ)
Classification methods are not static; they are constantly being revised based on new discoveries and understanding.

Limitations of Early Methods:

• Early classification was often based only on external features (morphology) and utility.
• Problem: Organisms that look similar externally might not be closely related, and organisms that look different might actually be closely related.
• Example (Fig 16.16): Wolffia (a tiny flowering plant) and Salvinia (a non-flowering fern-like plant) look some what similar externally but belong to completely different families based on evolutionary and genetic characteristics. Wolffia is in the family Araceae (like Anthurium), while Salvinia is in Salviniaceae (like Azolla).

Modern Classification Methodology:
Today, classification considers not only physical characteristics but also:

• Evolutionary relationships (Phylogenetics)
• Genetic characteristics

Six-Kingdom Classification System (ആറ് കിങ്ഡം വർഗ്ഗീകരണം)
As scientific understanding, particularly in molecular studies, progressed, the Five-Kingdom system was further refined.

• Proposer: The scientist Carl Woese introduced the Six-Kingdom classification system.
• Change: Based on modern scientific observations, the Kingdom Monera (which contained all prokaryotes) was divided into two distinct kingdoms.
• The Split: Monera was split into:
  • Kingdom Archaebacteria ( കീരിയ ആർക്കിബാ)
  • Kingdom Eubacteria (യുബാക്ടീരിയ).

ശാസ്ത്രീയ നിരീക്ഷണങ്ങളുടെ അടിസ്ഥാനത്തിൽ, കാൾ ഷൂസ് എന്ന ശാസ്ത്രജ്ഞൻ വർഗ്ഗീകരണ രീതി പരിഷ്കരിച്ചു. അദ്ദേഹം മൊണീറ എന്ന കി ങ്ഡത്തെ ആർക്കിബാക്ടീരിയ, യൂബാക്ടീരിയ എ ന്നിങ്ങനെ രണ്ടായി വിഭജിച്ചു. അങ്ങനെ മൊത്തം ആറ് കിങ്ഡങ്ങൾ നിലവിൽ വന്നു.

Do all Humans Belong to the Same Species?
Despite differences in external appearance among humans living in different parts of the Earth, which may be due to the influence of local geography and environmental factors, there is no significant genetic or other fundamental difference among them.

• All humans around the world share a common ancestor.
• Therefore, all humans are classified under the same species: sapiens.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 15 Chemistry of Matter Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 15 Chemistry of Matter Question Answer Notes

Class 8 Basic Science Chapter 15 Notes Kerala Syllabus Chemistry of Matter Question Answer

Chemistry of Matter Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Which of the following is made of a synthetic material?
a. Iron nail
b. Aluminium vessel
c. Glass test tube
d. Gold ornament
Answer:
W c. Glass test tube.
Reason: Glass is a synthetic material made by reacting silica (sand) with metal salts at high temperatures. Iron, Aluminium, and Gold are metals derived from natural sources.

Question 2.
Which of the following is not a polymer?
a. Bakelite
b. PVC
c. Rubber
d. Copper
Answer:
d. Copper.
Reason: Bakelite, PVC, and Rubber (natural or synthetic) are polymers, which are large molecules formed from repeating smaller units called monomers. Copper is a metallic element.

Question 3.
What is the difference between thermoplastics and thermosets?
Answer:
The main difference lies in their response to heat after moulding:
Thermoplastics (like polythene) can be reheated, melted, and reshaped into new products multiple times.
Thermosetting plastics (or thermosets, like Bakelite) cannot be reheated and melted to make new reshaped products once they are set.

Question 4.
Write one reason why synthetic rubber is used in the manufacture of tyres.
Answer:
Synthetic rubber shows less wear and tear compared to natural rubber, making it more durable for use in tyres.

Question 5.
What is the advantage of blending natural fibres along with synthetic fibres?
Answer:
Blending is done to overcome the limitations of both types of fibres. For example, blending cotton (natural) with polyester (synthetic) can create a fabric that has the comfort and breathability of cotton but the durability, wrinkle resistance, and quick-drying properties of polyester.

Question 6.
“Material chemistry provides us with both advantages and disadvantages” Do you agree with this statement? Why?
Answer:
B Yes, I agree with the statement. Advantages: Material chemistry has transformed various natural substances into countless useful products that have improved our quality of life, such as glass for various applications, durable plastics, elastic rubbers, versatile fibres, medicines, soaps, paints, etc.. It plays a crucial role in human progress by creating new materials with suitable properties.

Disadvantages: The production and uncontrolled use of some synthetic materials, particularly plastics, lead to significant environmental problems like plastic pollution. This pollution affects soil, water, and air, harming plants, animals, and ecosystems. Disposal of these materials is often difficult as many are not easily biodegradable.

Basic Science Class 8 Chapter 15 Question Answer Kerala Syllabus

Textbook Page No : 245 & 246

Question 1.
Why does the panchayat give importance to the conservation of water resources? (Discussion)?

Answer:

Items made from natural materials	Items made from synthetic materials
Door – Made of wood	Jar – Made of glass
Chair (wicker) – Made of cane/ rattan	Chair (plastic) – Made of plastic
Rope (jute/coir) – Made of natural fibers	Tyre – Made of synthetic rubber/ plastic
Hammer handle – Made of wood	Bucket – Made of plastic
Vessels (steel) – Made from processed metals	Balloons – Made of rubber/latex (can be natural or synthetic)
Tiffin carrier (steel) – Made from processed metals	Rope (nylon) – Made of synthetic fiber

Question 2.
Try writing down the items made of glass:

Answer:
Mirror

1. Glass jar
2. Windows/Panes
3. Car windshields
4. Beakers and laboratory equipment
5. Bottles
6. Light bulbs
7. Spectacle lenses
8. Lenses
9. Prisms

Textbook Page No : 247

Question 3.
Are the glasses used to make the objects shown in the picture of the same type?
Answer:
No, they are not of the same type. Different types of glass are manufactured by modifying the chemical composition to achieve specific properties for various applications (e.g., thermal resistance for ovenware, or specific refractive index for lenses).

Question 4.
Why can water be used for so many purposes?
Answer:
Water can be used for so many purposes primarily because of its special property as a universal solvent. Water can dissolve a wide variety of substances , and the nature of the water changes depending on what is dissolved in it, allowing it to be used to make different types of solutions. This solubility is one of the main reasons for its varied uses.

Question 5.
Haven’t you seen the shattered glasses of vehicles met with accidents? What is special about them?
Answer:
The glasses used in most vehicle wind-screens (front windshields) are usually Laminated glass, and sometimes side/ rear windows are Tempered glass.

• Laminated glass is special because it is made of two layers of glass bonded by a plastic inner layer (PVB). When it shatters, the fragments stick to the plastic, preventing sharp pieces from flying and causing injury.
• Tempered glass shatters into small, blunt, granular fragments instead of large, sharp shards, which reduces the risk of injury.

Question 6.
Do you know the name of glass used in the display of mobile phones?
Answer:
display of most modern mobile phones is covered with a type of chemically strengthened Aluminosilicate glass (e.g., Gorilla Glass or Dragontrail Glass). It is chosen for its superior scratch resistance and toughness.

Question 7.
Table Completion (Table 15.2)

Answer:

Types of Glasses	Uses	Major Components
Soda-lime glass/Soft glass	To make mirrors, window glass, bottles, etc.	Silica, Sodium oxide, Calcium oxide
Borosilicate Glass	To make laboratory equipment, cookwares, etc.	Silica, Boric acid (or Boric oxide), Sodium oxide, Aluminium oxide
Flint Glass /Optical Glass	Used in the manufacture of lenses, prisms, etc.	Silica, Lead oxide, Potassium oxide
Quartz glass	Apparatus used in chemical laboratories, electric bulbs.	Silica

Question 8.
Which is the component present in ail types of glass?
Answer:
Silica (SiO₂). It is the primary glass-forming material in almost all commercial types of glass.

Question 9.
Find out the other components present in each type of glass.
Answer:
The main components besides Silica are:
Soda-lime glass: Sodium oxide (Na₂O) and Calcium oxide (CaO).
Borosilicate glass: Boric acid/Boric oxide (B₂O₃), Sodium oxide (Na₂O), and Aluminium oxide (Al₂O₃).
Flint Glass/Optical Glass: Lead oxide (PbO) and Potassium oxide (K₂O). (In modern, lead-free flint glasses, lead oxide is sometimes replaced by oxides of Titanium or Zirconium.)

Textbook Page No : 248 & 249

Question 10.
Try writing down the common properties of glass:
Answer:
Based on the nature and uses of glass, the common properties are:

1. Transparency
2. Amorphous (non-crystalline solid)
3. Brittle (breaks suddenly)
4. Chemically resistant (to most common chemicals)
5. Poor conductor of heat and electricity

Question 11.
Haven’t you seen glasses of different colours? How do they get their colour?
Answer:
Glasses get a variety of colors when different chemicals (usually metal oxides) are added during the manufacturing process. These chemicals mix into the molten glass and impart a specific hue.

Question 12.
Why does the molecular mass increase when monomers participate in polymerisation? Can you observe and explain that from the above picture?

Answer:
The molecular mass increases significantly because monomers (small molecules) chemically combine under high pressure and temperature to form a single, very large molecule called a polymer.
Explanation from the picture (Fig 15.6): The diagram illustrates that the Dimer is formed by combining two Monomers, and the Polymer is formed by joining many Monomers in a long chain. The final polymer molecule is simply the accumulated mass of all the individual monomers, resulting in a much higher molecular mass compared to the initial molecules.

Monomer and Polymer

Question 13.
Plastic tables, chairs, etc. can be manufactured easier than wooden ones. Why?
Answer:
Plastic items are easier to manufacture because plastics are polymers that are generally thermoplastics. This means they can be easily melted and molded into complex shapes and designs using industrial processes (like injection molding) in a continuous and rapid manner. Manufacturing wooden objects requires more complex processes like cutting, carving, and joining.

Textbook Page No : 250 & 251

Question 14.
Plastics are widely used due to their unique properties compared to natural materials:
Answer:

1. Can be reshaped (miss)
2. Lightweight
3. Durable
4. Often resistant to chemicals
5. Can be electrical insulators
6. Can be moulded into various shapes easily

Question 15.
Are most of the objects we use daily, natural or synthetic? Discuss.
Answer:
Most of the objects we use daily are made of synthetic materials (polymers like plastics) or contain major synthetic components. While natural materials (like wood, metal, cotton) are still essential, synthetic polymers are favored for massproduced items because they are:

• Less expensive to produce.
• Lightweight and durable.
• Easily molded into various shapes (versatility).
• Resistant to moisture and decay.

Question 16.
How can we identify different types of plastics? What do they indicate?
Answer:
We identify different types of plastics using the Recycling Symbols. These are symbols consisting of a triangle made of chasing arrows, with a number (1 – 7) inside and an abbreviation below

These symbols indicate:

1. The specific type of polymer the item is made from (e.g., PET, PVC, HDPE).
2. The recyclability of the material and how it should be sorted for recycling.

Textbook Page No : 252 & 253

Question 17.
Analysis and Debate: Plastic – Friend or Foe?
Answer:
The discussion revolves around the necessary regulation of plastics in addition to the 4R method (Refuse, Reduce, Reuse, Recycle).

Aspect	Argument (Plastic)	Justification
Friend (Advantages)	Plastic is essential and unavoidable.	Used for diverse benefits. Essential for modern life (e.g., packaging, cables, pipes, bottles.
Foe (Disadvantages)	Uncontrolled use causes environmental damage.	Accumulates and causes water and soil pollution. Alters the natural chemical composition of soil, air, and water, affecting plant and organism growth.
Regulation	Regulation is necessary.	The 4R method alone may not completely avoid pollution; production and distribution of single-use disposable products must also be regulated. The newspaper report highlights strict government actions, fines, and bans as part of this regulation.

Question 18.
Write one reason why synthetic rubber is used in the manufacture of tyres.
Answer:
Synthetic rubbers are used in the manufacture of tyres because they show less wear and tear compared to natural rubber, making them more durable.

Question 19.
Try pulling and stretching a jute rope and a rubber band with force. What happens? Why?
Answer:

Material	What Happens	Why?
Jute Rope	It will stretch very little and may snap or break if enough force is applied.	Jute is a natural fibre, and fibres are polymers that are designed to be strong with low elasticity.
Rubber Band	It stretches significantly with less force and recoils when released.	Rubber is a polymer with an inherent elastic nature.

Question 20.
Comparison of Natural and Synthetic Fibres (Completing Table 15.3)

Answer:

Characteristics	Natural Fibres	Synthetic
Air circulation	More	Fibres
Water absorption	More	Less
Durability	Lower	Less
Weight	Higher	Higher

Question 21.
What is the advantage of blending natural fibres along with synthetic fibres?
Answer:
The advantage is to overcome the limitations of both types. Blending combines the best properties of both, such as the comfort and good water absorption of natural fibres with the durability, strength, and wrinkle resistance of synthetic fibres.

Question 22.
Write down the uses of fibres.
Answer:
Apart from clothing, fibres are used for:

• Making ropes, mats, and nets.
• Making fishing nets (e.g., Nylon fibres).
• Used to prevent soil erosion and to make geotextiles for agricultural use.

Question 23.
Complete the table based on polymer characteristics (Completing Table 15.8).

Answer:

Characteristics	Polymer
Suitable for making strong threads	Nylon (or Polyester)
Polymer that can be moulded into various shapes	Plastic
Polymer with elastic nature	Rubber (Natural or Synthetic)

Textbook Page No : 254

Question 24.
List the other materials around us that are made with the help of chemistry.
Answer:

1. Dyes
2. Perfumes
3. Soaps
4. Medicines
5. Fertilizers
6. Pesticides
7. Cosmetics

Class 8 Basic Science Chapter 15 Question Answer Extended Activities

Question 1.
Organize an exhibition on the theme “The Discovery of Glass and the Growth of Science,” including descriptions and pictures of the history of glass making and different types of glasses.

Question 2.
Prepare a seminar paper on the topic “Microplastic Pollution”

Chemistry of Matter Class 8 Notes

Class 8 Basic Science Chemistry of Matter Notes Kerala Syllabus

Natural vs. Synthetic Materials
We use diverse objects made from both natural substances and synthetic materials. Material Chemistry is the branch of science that helps transform various substances into useful products for humans.

• Natural Materials: Obtained directly from nature (e.g., wood, cotton, jute).
• Synthetic Materials: Man-made materials, often produced through chemical processes (e.g., glass, plastic).

Glass
Glass is a synthetic material and an amorphous (non-crystalline) transparent substance.

Manufacturing and Composition

• Glass is formed when silicon dioxide in sand undergoes a chemical reaction with metal salts at high temperatures.
• The primary component present in all types of glass is Silica (SiO₂).
• The accidental discovery of glass is attributed to Phoenician mariners who used blocks of washing soda (Sodium carbonate) to build a hearth on a sandy bank.

□ Properties and Colouring of Glass
Common Properties of Glass:

• Transparency (സുതാര്യത)
• Brittleness (എളുപ്പത്തിൽ പൊട്ടുന്നത്)
• Hardness
• Resistance to heat (varies by type)
• Usually an electrical insulator

□ Colouring Glass:
Glasses get different colours when specific chemicals are added during the manufacturing process.

□ Chemicals and Glass Colours

Chemicals Added	Colour of the Glass Obtained
Cadmium sulphide	Yellow
Gold chloride	Ruby red
Chromium oxide	Green
Cobalt oxide	Blue

□ Types:
Different types of glass are made by varying the components added to silica.

• Soda lime glass: Bottles, window glass.
• Borosilicate glass: Lab equipment, cookware.
• Flint glass: Lenses, prisms.
• Common Component: Silica (SiO₂) is present in all types.

ഗ്ലാസ് ഒരു കൃത്രിമ പദാർത്ഥമാണ്. മണലിലെ സിലിക്കയും (SiO₂) ലോഹ ലവണങ്ങളും (metal salts) ഉയർന്ന താപനിലയിൽ പ്രവർത്തിച്ചാണ് ഗ്ലാസ് ഉണ്ടാക്കുന്നത്. വിവിധ രാസവസ്തുക്കൾ ചേർത്ത് ഗ്ലാസിന് നിറം നൽകാം.

Polymers
Many synthetic materials like plastics, rubber, and synthetic fibres belong to a group called polymers.

Polymerisation (പോളിമറൈസേഷൻ):
Polymerisation is the process in which many small molecules combine under high pressure and temperature to form large molecules.

Monomers (മോണോമറുകൾ): The small molecules used in polymerisation are called monomers.
Polymers (പോളിമറുകൾ): The large molecules formed by the combination of a large number of monomers are called polymers.

Key Points:

• The study of polymers is known as polymer chemistry.
• Hermann Staudinger is known as the Father of Polymer Chemistry.
• Bakelite is the first synthetic polymer.
• Polymers have much higher molecular mass than ordinary molecules, leading to different properties.

അനേകം ചെറിയ തന്മാത്രകൾ (മോണോമറുകൾ) ഉയർന്ന മർദ്ദത്തിലും താപനിലയിലും സംയോജി ച്ച് വലിയ തന്മാത്രകൾ (പോളിമറുകൾ) ഉണ്ടാകു ന്ന പ്രിയയാണ് പോളിമറൈസേഷൻ. പ്ലാസ്റ്റിക്, റബ്ബർ, നാരുകൾ എന്നിവയെല്ലാം പോളിമറുകൾ ക്ക് ഉദാഹരണമാണ്.

□ Classification of Polymers
Polymers can be classified in different ways:
1. Based on Physical Properties: They can be grouped into plastics, rubber, and fibres.

□ Based on Origin:
Natural Polymers: Found in nature.
Synthetic Polymers: Man-made polymers.

Natural Polymers	Synthetic Polymers
Cellulose	Polythene
Starch	Nylon
Natural rubber	Bakelite
	PVC, Synthetic rubbers

□ Common Synthetic Polymers and Their Uses

Polymer	Characteristics	Uses
Polyethylene (Polythene)	Inert nature (രാസപ്രവർത്ത നത്തിൽ ഏർപ്പെടാത്ത സ്വഭാവം)	Food containers, packaging
Polytetrafluoroethylene (PTFE / Teflon)	Heat resistance (താപം പ്രതിരോധിക്കാനുള്ള കഴിവ്)	Cookwares used at high temperatures (Non-stick)
Polyvinyl chloride (PVC)	Electrical resistance (വൈദ്യുതി പ്രതിരോധിക്കാനുള്ള കഴിവ്)	Cables, Pipes
Polyethylene terephthalate (PET)	Prevents gas permeation (വാതകങ്ങളെ കടത്തിവിടില്ല)	Used in bottle manufacturing

Plastics
Plastics are highly demanded due to their unique properties, such as being

• Can be reshaped (രൂപമാറ്റം വരുത്താം)
• Lightweight
• Durable
• Often resistant to chemicals
• Can be electrical insulators
• Can be moulded into various shapes easily

□ Classification of Plastics
Plastics can be further classified based on their behaviour when heated:
1. Thermoplastics:
These plastics can be reheated, melted, and reshaped into new products multiple times. They soften on heating and harden on cooling.
Example: The plastic carry bags we use (Polythene), PVC pipes.

2. Thermosetting Plastics:
These plastics cannot be reheated and melted to reshape them once they have been set. They undergo a permanent chemical change during moulding.
Example: Items like the handle of a pressure cooker (e.g., Bakelite).
ചൂടാക്കുമ്പോൾ മൃദുവാകുകയും തണുപ്പിക്കു മ്പോൾ ഉറയ്ക്കുകയും ചെയ്യുന്ന പ്ലാസ്റ്റിക്കുകളാണ് തെർമോപ്ലാസ്റ്റിക്കുകൾ. ഇവയെ വീണ്ടും ചൂടാക്കി രൂപമാറ്റം വരുത്താൻ കഴിയും. എന്നാൽ, നിർമ്മാ ണ വേളയിൽ ചൂടാക്കുമ്പോൾ ഉറയ്ക്കുകയും, പി ന്നീട് ചൂടാക്കിയാൽ മൃദുവാകാത്തതുമായ പ്ലാസ്റ്റി ക്കുകളാണ് തെർമോസെറ്റിംഗ് പ്ലാസ്റ്റിക്കുകൾ. ഇ വയെ വീണ്ടും രൂപമാറ്റം വരുത്താൻ സാധ്യമല്ല.

□ Identifying Different Types of Plastics
Different types of plastics have specific properties and recycling potential. They are often identified using Resin Identification Codes (the number inside the chasing arrows symbol).

Plastic Pollution ((പ്ലാസ്റ്റിക് മലിനീകരണം)
While plastics have many benefits, their uncontrolled use leads to significant environmental problems.

• Problem: Plastics accumulate in the environment, causing water and soil pollution. They alter the natural composition of soil, air, and water, affecting the growth of plants and soil organisms.
• Solution (The 4R Approach): To reduce plastic pollution, the 4R approach is recommended:
  • Refuse (ഒഴിവാക്കുക): Avoid giving or accepting plastic products whenever possible.
  • Reduce (ഉപയോഗം കുറയ്ക്കുക): Minimise the use of plastic products.
  • Reuse (പുനരുപയോഗിക്കുക): Reuse plastic products multiple times.
  • Recycle (പുനഃചംക്രമണം ചെയ്യുക): Convert plastic products into new items through physical and chemical processes.
• Regulation: Governments are also taking action by banning certain single-use plastics and imposing fines for littering.

Rubbers (റബ്ബറുകൾ)
Rubbers are polymers known for their elastic nature.

• Natural Rubber:
  • An elastic natural polymer.
  • Obtained from the latex collected from the rubber tree.
• Synthetic Rubber:
  • Produced to meet increased demand and overcome limitations of natural rubber.
  • Shows less wear and tear compared to natural rubber.
  • Use: Widely used in the manufacture of tyres.

Fibres (നാരുകൾ)

• Fibres are polymers that are strong and long.
• For a polymer to be considered a fibre, its length must be at least a hundred times its diameter.
• Uses: Apart from clothing, fibres are used to make ropes, mats, nets, geotextiles, etc..
• Synthetic Fibres:
  • Developed to overcome the limited availability and quality issues of natural fibres.
  • Nylon is the first synthetic fibre made by humans.
• Blending Fibres: To overcome the limitations of both types, natural fibres are often blended with synthetic fibres.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 15 Chemistry of Matter Important Questions ensure conceptual clarity.

Chemistry of Matter Extra Questions and Answers Class 8 Basic Science Chapter 15 Kerala Syllabus

Chemistry of Matter Class 8 Important Questions

Question 1.
Match the polymer type in Column A with its characteristic in Column B.

Column A	Column B
(i) Thermoplastic	(a) Cannot be remelted and reshaped.
(ii) Thermosetting Plastic	(b) Can be reheated and reshaped multiple times.
(iii) Fibre	(c) Polymer with elastic nature.
(iv) Rubber	(d) Polymer suitable for making strong threads.

Answer:
(i) – b, (ii) – a, (iii) – d, (iv) – c

Column A	Column B
(i) Thermoplastic	(b) Can be reheated and reshaped multiple times.
(ii) Thermosetting Plastic	(a) Cannot be remelted and reshaped.
(iii) Fibre	(d) Polymer suitable for making strong threads.
(iv) Rubber	(c) Polymer with elastic nature.

Question 2.
Match the type of glass in Column A with its major use in Column B.

Column A	Column B
(i) Soda lime glass	(a) Lenses, prisms
(ii) Borosilicate glass	(b) Window glass, bottles
(iii) Flint glass	(c) Laboratory equipment, cookware

Answer:
(i) – b, (ii) – c, (iii) – a

Column A	Column B
(i) Soda lime glass	(b) Window glass, bottles
(ii) Borosilicate glass	(c) Laboratory equipment, cookware
(iii) Flint glass	(a) Lenses, prisms

Question 3.
Match the chemical added to glass in Column A with the colour obtained in Column B.

Column A	Column B
(i) Cobalt oxide	(a) Yellow
(ii) Chromium oxide	(b) Ruby red
(iii) Cadmium sulphide	(c) Blue
(iv) Gold chloride	(d) Green

Answer:
(i) – c, (ii) – d, (iii) – a, (iv) – b

Column A	Column B
(i) Cobalt oxide	(c) Blue
(ii) Chromium oxide	(d) Green
(iii) Cadmium sulphide	(a) Yellow
(iv) Gold chloride	(b) Ruby red

Question 4.
Statement (i): Thermoplastics can be reheated and reshaped.
Statement (ii): Thermosetting plastics can also be reheated and reshaped.
(a) Both statements (i) and (ii) are correct.
(b) Statement (i) is correct, but statement (ii) is incorrect.
(c) Statement (ii) is correct, but statemeht (i) is incorrect.
(d) Both statements (i) and (ii) are incorrect.
Answer:
(b) Statement (i) is correct, but statement (ii) is incorrect. (Thermosets cannot be reshaped after setting).

Question 5.
Statement (i): Nylon is the first synthetic fibre made by humans.
Statement (ii): Natural rubber is obtained from the latex of the rubber tree.
(a) Both statements (i) and (ii) are correct.
(b) Statement (i) is correct, but statement (ii) is incorrect.
(c) Statement (ii) is correct, but statement (i) is incorrect.
(d) Both statements (i) and (ii) are incorrect.
Answer:
(a) Both statements (i) and (ii) are correct.

Question 6.
Statement (i): Silica (SiO₂) is the major component present in all common types of glass.
Statement (ii): Glass is considered a natural material.
(a) Both statements (i) and (ii) are correct.
(b) Statement (i) is correct, but statement (ii) is incorrect.
(c) Statement (ii) is correct, but statement (i) is incorrect.
(d) Both statements (i) and (ii) are incorrect.
Answer:
(b) Statement (i) is correct, but statement (ii) is incorrect. (Glass is a synthetic material).

Question 7.
Define Polymerisation, Monomer, and Polymer.
Answer:
Polymerisation: The process in which many small molecules combine under high pressure and temperature to form large molecules.
Monomers: The small molecules used in polymerisation.
Polymers: The large molecules formed by the combination of a large number of monomers.

Question 8.
Explain the ‘4R’ approach recommended to reduce plastic pollution.
Answer:
The 4R approach consists of:
Refuse: Avoid giving or accepting plastic products.
Reduce: Minimise the use of plastic products.
Reuse: Reuse plastic products multiple times.
Recycle: Convert plastic products into new items.

Question 9.
Why are synthetic rubbers often preferred over natural rubber for making tyres?
Answer:
Synthetic rubbers generally show less wear and tear compared to natural rubber, making them more durable and suitable for the demanding conditions tyres face.

Question 10.
How can you identify the type of plastic used in a container using the identification code?
Answer:
Look for the recycling symbol (chasing arrows) on the plastic item. Inside the symbol, there will be a number from 1 to 7. This number corresponds to a specific type of plastic resin (e.g., 1 is PET, 2 is HDPE, 3 is PVC, etc.). You can then refer to a chart (like Table 15.6) to find the abbreviation and name of the plastic.

Question 11.
Describe the observation when a thermoplastic material (like polythene) and a thermosetting material (like Bakelite) are heated strongly after being initially shaped.
Answer:
Thermoplastic: When heated strongly, the thermoplastic material will soften, melt, and can be reshaped.

Thermosetting plastic: When heated strongly, the thermosetting material will not soften or melt. It may char or decompose at very high temperatures but cannot be reshaped.

Question 12.
A student observes the characteristics of natural cotton fibre and synthetic polyester fibre. What differences would they likely note regarding water absorption and durability?
Answer:
Water absorption: The natural cotton fibre would show more water absorption ability compared to the synthetic, polyester fibre.

Durability: The synthetic polyester fibre would likely show more durability (resistance to wear and tear) compared to the natural cotton fibre.

Question 13.
What is the main chemical compound in sand that reacts with metal salts at high temperatures to form glass?
Answer:
Silicon dioxide (SiO₂).

Question 14.
Identify the polymer formed when many ethylene monomers undergo polymerisation.
Answer:
Polyethylene (or Polythene). (The name implies it’s formed from ethylene).

Question 15.
What type of substance is generally formed when silicon dioxide (SiO₂) reacts with metal salts (like sodium carbonate) at high temperature?
Answer:
Glass.

Question 16.
Write the chemical formula for Silica, the main component of glass.
Answer:
SiO₂

Question 17.
Write the full name for the plastic represented by the abbreviation PVC.
Answer:
Polyvinyl chloride.

Question 18.
Observe Figure 15.6, which shows monomers combining to form a polymer. Explain why the molecular mass of a polymer is much higher than that of its monomer.

Answer:
A polymer is formed by chemically combining a large number of individual monomer molecules into a long chain. Therefore, the total mass of the polymer molecule is essentially the sum of the masses of all the monomer units that joined together, making it significantly higher than the mass of a single monomer.

Question 19.
Differentiate between thermoplastic and thermosetting plastic.
Answer:
The plastic that get softened on heating and hardened on cooling is thermoplastics. When heating physical change occurs.

The plastic which remains soft when heated and gets hardened permanently on cooling is thermosetting plastics. Chemical change occurs when heating.

Question 20.
Given some occasions of using plastic. Find the peculiarity of plastic used and fill the table.

Occasion	Peculiarity
As the covering of conductor
To make the handles of cooking vessels
To keep chemicals
To make water tanks
To make house hold materials

Answer:

Occasion	Peculiarity
As the covering of conductor	Plastics are insulators
To make the handles of cooking vessels	Not conducting heat
To keep chemicals	Do not react with chemical
To make water tanks	No rusting. Less weight
To make house hold materials	Easy touse. Less weight

Question 21.
Write 4 occasions which plastic is harmful to daily life.
Answer:

1. Enviromental pollution when it is thrown without any control.
2. When burning air pollution.
3. Hindrance in drainages
4. The water absorption property of soil decreases when plastics are dumped in soil.

Question 22.
What are the uses of plastic in the field of human health?
Answer:

• To produce I V tubes, bottles
• To produce heart valves
• To produce packets

Question 23.
What are the uses of plastic in the field of production of house building.
Answer:
To produce roof materials, doors, plumping and wiring materials.

Question 24.
List the peculiarities of plastic.
Answer:
Can mould in any shape, longlasting, Insulator, do not conduct heats, not reacting with chemicals and water, will burn.

Question 25.
Separate natural and artificial polymer from the list Rubber, wool, p.v.c, Bakelite, nylon, rayon cellulose, silk, polythene, polyester
Answer:

Natural	Artificial
Rubber	p.v.c
Wool	polythene
Cellulose	nylon
Silk	rayon
	Polyester

Question 26.
List the merits of natural and artificial polymer.
Answer:

Merits	Demerits
Comfortable to wear	Less availability
More aeration	wrinkle easily
Absorb water, sweat	high cost
Not easily burns	not durable

Question 27.
If we heat polyethene cover can we convert into earlier stage? Justify.
Answer:
No. Undergoes chemical change because it is thermosetting plastic.

Question 28.
The primary component present in all common types of glass is:
a. Calcium oxide (CaO)
b. Lead oxide (PbO)
c. Silica (SiO₂)
d. Sodium oxide (Na₂O)
Answer:
c. Silica (SiO₂)

Question 29.
The small molecules that combine in a process called polymerisation to form a very large molecule are known as:
a. Isomers
b. Polymers
c. Monomers
d. Dimers
Answer:
c. Monomers

Question 30.
Which of the following is a Thermosetting Plastic?
a. Polythene
b. PVC
c. Bakelite
d. PET
Answer:
c. Bakelite

Question 31.
The plastic recycling code ‘2’ with the abbreviation HDPE is used for:
a Polyvinyl Chloride
b. Low-Density Polyethylene
c. High-Density Polyethylene
d. Polystyrene
Answer:
c. High-Density Polyethylene

Question 32.
Which chemical is typically added to molten glass to achieve a Ruby Red colour?
a. Cobalt oxide
b. Chromium oxide
c. Cadmium sulphide
d. Gold chloride
Answer:
d. Gold chloride

Question 33.
Explain why Borosilicate Glass is the preferred material for making laboratory equipment like beakers and flasks, rather than Soda-lime glass.
Answer:
Borosilicate Glass is preferred because it contains Boric acid/Boric oxide. This gives it a low coefficient of thermal expansion which makes it highly resistant to heat and thermal shock, preventing it from cracking when heated or cooled rapidly.

Question 34.
Differentiate between Thermoplastics and Thermosetting Plastics based on their ability to be reshaped and provide one example of an object made from each.
Answer:
Thermoplastics: Can be reheated, melted, and reshaped.
Example: Polythene/PVC.
Thermosetting Plastics: Cannot be reheated and melted/reshaped.
Example: Bakelite/Pressure cooker handle.

Question 35.
State and briefly explain the 4R approach that is essential for resolving the problem of plastic pollution in our environment.
Answer:
The 4R Approach is:

1. Refuse: Avoid single-use disposableplastic products.
2. Reduce: Minimise overall plastic usage.
3. Reuse: Find new ways to use plastic products instead of discarding them.
4. Recycle: Convert waste plastic into new usable items through suitable processes.

Question 36.
Analyse the properties of polymers by completing the following table and answering the question below.

Category/Type	Example Polymer	Key Property or Characteristic
Natural Polymer	(i) ____________	A component of wood and paper. A component of wood and paper.
Fibre Polymer	Nylon	(iii)  ____________ (excluding its use in clothing)
Glass Type	Flint Glass	(iii) ____________
Thermoset Plastic	(iv)  ____________	Key Property or Characteristic

Cannot be reheated and molded into a new shape.
What is the primary advantage gained by blending a natural fibre like Cotton with a synthetic fibre like Nylon?
Answer:
(i) Cellulose or Natural Rubber
(ii) Strong and long threads/Low wear and tear/High durability (Any one appropriate property of a fibre)
(iii) High refractive index (Used for lenses and prisms)
(iv) Bakelite or Melamine (Any one appropriate thermoset)
The advantage is to combine the best properties of both. Specifically, it provides the comfort and water absorption of cotton with the strength, durability, or wrinkle resistance of nylon.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 14 Water Important Questions ensure conceptual clarity.

Water Extra Questions and Answers Class 8 Basic Science Chapter 14 Kerala Syllabus

Water Class 8 Important Questions

Question 1.
Match the property of water in Column A with its definition or consequence in Column B.

Column A	Column B
(i) Surface Tension	(a) Allows many substances to dissolve in it.
(ii) High Heat Capacity	(b) Causes ice to float on water.
(iii) Universal Solvent	(c) Allows water to absorb much heat without a large temperature rise.
(iv) Anomalous Expansion	(d) Force on the water surface minimizing its area.

Answer:
(i) – d, (ii) – c, (iii) – a, (iv) – b

Column A	Column B
(i) Surface Tension	(d) Force on the water surface minimizing its area.
(ii) High Heat Capacity	(c) Allows water to absorb much heat without a large temperature rise.
(iii) Universal Solvent	(a) Allows many substances to dissolve in it.
(iv) Anomalous Expansion	(b) Causes ice to float on water.

Question 2.
Match the condition with its effect on the boiling point of water.

Condition	Effect on Boiling Point
(i) Increased Pressure (e.g., pressure cooker)	(a) Boiling point remains 100°C.
(ii) Decreased Pressure (e.g., high altitude)	(b) Boiling point increases (above 100°C).
(iii) Normal Pressure(sea level, 1 atm)	(c) Boiling point decreases (below 100°C).

Answer:
(i) – b, (ii) – c, (iii) – a

Condition	Effect on Boiling Point
(i) Increased Pressure (e.g., pressure cooker)	(b) Boiling point increases (above 100°C).
(ii) Decreased Pressure (e.g., high altitude)	(c) Boiling point decreases (below 100°C).
(iii) Normal Pressure(sea level, 1 atm)	(a) Boiling point remains 100°C.

Question 3.
Match the type of water hardness with its cause and removal method.

Type of Hardness	Cause / Removal
(i) Soft Water	(a) Caused by chlorides/sulphates; cannot be removed by boiling.
(ii) Temporary Hardness	(b) Soap lathers well; contains few dissolved salts.
(iii) Permanent Hardness	(c) Caused by bi-carbonates; can be removed by boiling.

Answer:
(i) – b, (ii) – c, (iii) – a

Type of Hardness	Cause / Removal
(i) Soft Water	(b) Soap lathers well; contains few dissolved salts.
(ii) Temporary Hardness	(c) Caused by bi-carbonates; can be removed by boiling.
(iii) Permanent Hardness	(a) Caused by chlorides/sulphates; cannot be removed by boiling.

Question 4.
Statement (i): Water has a high surface tension, which helps it wet clothes easily.
Statement (ii): Adding soap or heating water decreases its surface tension.
a) Both statements (i) and (ii) are correct.
b) Statement (i) is correct, but statement (ii) is incorrect.
c) Statement (ii) is correct, but statement (i) is incorrect.
d) Both statements (i) and (ii) are incorrect.
Answer:
(c) Statement (ii) is correct, but statement (i) is incorrect. (High surface tension prevents water from wetting clothes easily).

Question 5.
Statement (i): When water freezes to form ice, its volume increases and its density decreases.
Statement (ii): Because ice is denser than water, it floats on water.
a) Both statements (i) and (ii) are correct.
b) Statement (i) is correct, but statement (ii) is incorrect.
c) Statement (ii) is correct, but statement (i) is incorrect.
d) Both statements (i) and (ii) are incorrect.
Answer:
(b) Statement (i) is correct, but statement (ii) is incorrect. (Ice floats because it is less dense than water).

Question 6.
Statement (i): The boiling point of water increases as atmospheric pressure decreases.
Statement (ii): Food cooks faster in a pressure cooker because the increased pressure raises the boiling point of water.
a) Both statements (i) and (ii) are correct.
b) Statement (i) is correct, but statement (ii) is incorrect.
c) Statement (ii) is correct, but statement (i) is incorrect.
d) Both statements (i) and (ii) are incorrect.
Answer:
(c) Statement (ii) is correct, but statement (i) is incorrect. (Boiling point decreases as pressure decreases).

Question 7.
Why is water called the “universal solvent”?
Answer:
Water is called the universal solvent because it has the ability to dissolve a wide variety of substances, more than many other liquids.

Question 8.
Why does steam at 100°C cause more severe burns than boiling water at 100°C?
Answer:
When water boils and turns into steam at 100°C, it absorbs extra heat energy called latent heat without changing its temperature. This latent heat is used for the change of state. Steam at 100°C contains this additional latent heat energy compared to water at 100°C, so it transfers more heat upon contact, causing more severe burns.

Question 9.
Why does ice float on water, and why is this important for aquatic life?
Answer:
When water freezes, its volume increases (anomalous expansion), causing its density to decrease. Because ice is less dense than liquid water, it floats. This is important because lakes freeze from the top down, forming an insulating layer of ice that prevents the entire water body from freezing, allowing aquatic life to survive underneath.

Question 10.
Describe a simple experiment to distinguish between hard water and soft water using soap.
Answer:
Take two separate flasks or bottles.

• Pour equal amounts of the water samples (one hard, one soft) into each flask.
• Add equal-sized pieces of soap to both flasks.
• Shake both flasks well.
• Observation: The flask containing soft water will produce a good amount of lather (foam). The flask containing hard water will produce very little lather or scum.

Question 11.
How can you demonstrate that the boiling point of water decreases when pressure is reduced?
Answer:

• Fill a boiling tube half with water and boil it using a flame.
• After turning off the flame, immediately close the tube with a rubber cork fitted with a glass tube connected to a syringe.
• Pull back the piston of the syringe.
• Observation: The water inside the tube will start boiling again, even though it has cooled slightly.
• Conclusion: Pulling the piston reduces the pressure inside the tube, which lowers the boiling point, causing the water to boil at the lower temperature.

Question 12.
Describe an activity to demonstrate surface tension using a razor blade or needle.
Answer:
Fill a glass or tumbler completely with still water, perhaps even slightly overfull so the surface bulges.

• Carefully and gently place a clean, dry razor blade or needle flat onto the surface of the water.
• Observation: The blade or needle will float on the surface without sinking, supported by the “skin” created by surface tension.

Question 13.
Water (H₂O) can be decomposed by electrolysis. Write the balanced chemical equation for this reaction.
Answer:

Question 14.
During the electrolysis of water using a Hoffmann voltameter, identify the products formed at the positive electrode (anode) and the negative electrode (cathode).
Answer:
Positive electrode (anode): Oxygen gas (O₂).
Negative electrode (cathode): Hydrogen gas (H₂).

Question 15.
When quicklime (CaO) is added to water (H₂O), slaked lime (Ca(OH)₂) is formed. Write the balanced chemical equation for this reaction.
Answer:
CaO + H₂O → Ca(OH)₂.

Question 16.
What is the chemical formula for a molecule of water, and what does it indicate about the atoms present?
Answer:
The chemical formula is H₂O. It indicates that one molecule of water contains two hydrogen atoms and one oxygen atom.

Question 17.
During the electrolysis of water, twice the volume of hydrogen gas is produced compared to oxygen gas. Based on the formula of water (H₂O), explain why this ratio occurs.
Answer:
The formula H₂O shows that each water molecule contains two hydrogen atoms for every one oxygen atom. When water is split during electrolysis, these atoms combine to form H₂ and O₂ molecules. Since there are twice as many hydrogen atoms available, they form twice the volume (or number of molecules) of hydrogen gas (H₂) compared to oxygen gas (O₂).

Question 18.
What are the chemical formulas for the salts typically responsible for:
a) Temporary hardness (as bicarbonates)
b) Permanent hardness (as chlorides)
Answer:
(a) Calcium bicarbonate: Ca(HCO₃)₂;
Magnesium bicarbonate: Mg(HCO₃)₂.

(b) Calcium chloride: CaCl₂;
Magnesium chloride: MgCl₂.

Question 19.
a.What are the gases liberated in the water voltameter arranged for decomposition of water?
b. Which gas is liberated at the negative pole of the battery?
c. What will be the ratio of the volume of the gases?
Answer:
a. Hydrogen and oxygen
b. Hydrogen
c. Ratio of hydrogen and oxygen – 2 : 1

Question 20.
Heat equal quantity of water and coconut oil in two vessels.
a. The temperature of which liquid will be raised rapidly?
b. Which has more heat capacity?
Answer:
a.Coconut oil
b.water

Question 21.
Take water in a vessel and put a needle on the surface of water carefully.
a. Will the needle be immersed in water. Justify your answer.
b. Why do the water drops assume spherical shape?
Answer:
a. No. Because of the surface tension of water.
b. The surface tension reduces the surface area.

Question 22.
Name two metals which react with cold water.
Answer:
Potassium, sodium

Question 23.
Which is the metal react with hot water? Which is the gas produced? Write the chemical equation.
Answer:
Magnesium; Hydrogen.
Mg + H₂O MgO + H₂

Question 24.
Why does water is known as universal solvent?
Answer:
Since water can dissolves various substances and widely used for preparing solutions it is a universal solvent.

Question 25.
What are hard water and soft water?
Answer:
The water in which soap does not
lather easily is called hard water. Water in which soap gives lather readily is called soft water.

Question 26.
Give the reason for temporary hardness of water.
Answer:
The hardness of water is due to the presence of calcium or magnesium bicarbonate. This type of hardness is called temporary hardness.

Question 27.
Hardness of water is not removed even after it is boiled. What may be the reason for this?
Answer:
This is due to chloride and sulphate of magnesium and calcium are dissolved in water.

Question 28.
Why is air continuously introduced in to the water in an aquarium?
Answer:
Aquatic plants and animals make use of oxygen dissolved in water. Water get polluted as the amount of oxygen in it decreases. Air is introduced in to water to make up the quantity of oxygen.

Question 29.
Write two remedies to prevent water pollution.
Answer:

1. Stop dumping of waste in water resources.
2. Reduce the use of fertilisers and pesticides.

Question 30.
Write two methods to reduce the surface tension of water
Answer:

1. Add soap to water
2. Boil the water.

Question 31.
Complete the given chemical equation.
Mg + H₂O → MgO + …………………
Answer:
H₂

Question 32.
Identify the property of water shown in each of the following situations.
a. Water is used in radiators of automobiles.
b. Small water drops assume spherical shape.
Answer:
a. Heat capacity of water
b.Surface tencsion of water

Question 33.
Two samples of water are given.
Sample 1 – Soap gives lather easily.
Sample 2 – Soap does not give lather easily.
a. Which sample represents hard water?
b. Which among the following salts causes hardness of water?
(Sodium carbonate, Magnesium chloride, Zinc hydroxide, Ammonium chloride)
c. Which type of hardness of water can be removed by boiling?
Answer:
a. Sample 2
b. Magnesium Chloride (MgCl₂)
c. Temporary hardness of water

Question 34.
Water pollution is a burning issue now-a-day.
a. How do aquatic animals and plants get oxygen for respiration?
b. The wastes dumped into rivers and ponds result in the destruction of aquatic life. Explain.
c. Write any other activity that causes water pollution.
Answer:
a. Aquatic animals and plants make use of the oxygen and carbon dioxide dissolved in water.

b. The Wastes dumped in the rivers and Ponds decay using the oxygen dissolved in them. This depletes the amount of oxygen and results in the destruction of aquatic life.

c. Excessive use of detergents Rampant use of fertilizers.

Question 35.
Boling point of water is …………………..
Answer:
100°c.

Question 36.
Following are the observations
when a student added soap solution to two samples of water collected from two different sources.

Sample	Observation
Water collected from source 1	Lathers easily
Water collected from source 2	Does not lather easily

a. Which one of these is hard water?
b. Which salts are responsible for the hardness of water?
Answer:
a. 2.
b. Ca or Mg.

Question 37.
Water is used to cool hot objects. Which property of water is utilised here?
Answer:
Heat capacity of water, (or Specific heat capacity)

Question 38.
Water pollution is a major issue we are facing nowadays. Write any two activities that cause water pollution.
Answer:
Dumping of wastes in water resources Rampant use of fertilizers

Question 39.
Water taken in a round bottomed flask is heated continuously and the temperature is noted.
a. What is the boiling point of water?
b. Eventhough we are heating the flask continuously the temperature remains constant at 100° C. Why?
c. Steam causes more severe burns than boiling water. Why?
Answer:
a. The temperature at which water boils at normal atmospheric pressure is called boiling point of water -100°C

b. Once boiling starts the temperature will not change because all the heat supplied will be used for the change of state.

c. When water starts boiling all the heat supplied will be used for the change of state. Hence all heat supplied will be contained in the steam at the same temperature. So steam causes more severe burns.

Question 40.
Soap is added to two samples of water taken in two test tubes and shaken well.
a. Soap lathers easily in water taken in one test tube. Which type of water is it?
b. Soap does not lather easily in water taken in the second test tube Write the name of a salt which causes this behaviour.
c. Suggest any one method to remove this behaviour of water.
Answer:
a. Soft water
b. Bicarbonates, chlorides and sulphates of calcium and magnesium
c. Temporary hardness is removed by boiling and permanent hardness by any chemical methods.

Question 41.
The volume of water When it solidifies
(increases, decreases, does not change)
Answer:
Increases

Question 42.
Water pollution is a major issue we face today
a. Write down any two situations in which water in your locality gets polluted.
b. Suggest a method to prevent water pollution.
Answer:
a. Plastic thrown into the soil
b. Chemical Pesticides and Chemical fertilizers.

Question 43.
In water from certain sources soap lathers easily. But in water from some other sources soap does lather easily
a. What is the name of water in which soap doesn’t lather easily?
b. Give an example of any salt responsible for this defect of water.
c. Write a method to remove this defect.
Answer:
a. Hard water
b. Bicarbonate, Chloride or Sulphate of Calcium and Magnesium
c. Boil water / add suitable Chemicals

Question 44.
Water is known as the ……………………. solvent.
a) Polar
b) Universal
c) Common
d) Basic
Answer:
b) Universal

Question 45.
Hardness caused by sulphates of calcium and magnesium is known as:
a) Temporary hardness
b) Basic hardness
c) Permanent hardness
d) Chemical hardness
Answer:
c) Permanent hardness

Question 46.
The special force due to the mutual attraction between water molecules that causes the surface to behave like a stretched film is called:
a) Cohesion
b) Adhesion
c) Surface Tension
d) Capillary action
Answer:
c) Surface Tension

Question 47.
Water has its maximum density at ……………………. °C.
a) 0
b) 100
c) 4
d) 37
Answer:
c) 4

Question 48.
When a water solution is tested with litmus paper, it turns blue litmus red. The solution is:
a) Neutral
b) Basic
c) Acidic
d) Alkaline
Answer:
c) Acidic

Question 49.
State any two uses of water’s high heat capacity in daily life.
Answer:
Two uses of high heat capacity:

1. Used as a coolant in car radiators,
2. Helps in maintaining the body temperature within a limit.

Question 50.
How is temporary hardness different from permanent hardness in water?
Answer:
Temporary Hardness is caused by bicarbonates of Calcium/Magnesium and can be removed by boiling. Permanent Hardness is caused by sulphates or chlorides of Calcium/Magnesium and cannot be removed by boiling.

Question 51.
Explain why food takes a longer time to cook in high mountain regions compared to sea level.
Answer:
In high mountain regions, the atmospheric pressure is lower. Since the boiling point is related to pressure, water boils at a temperature lower than 100°C. Cooking takes longer because the cooking medium (water) is at a lower maximum temperature.

Question 52.
Explain why steam causes a more severe burn than boiling water at 100°C.
Answer:
Steam contains more energy than boiling water at the same temperature (100°C). This extra heat is called latent heat of vaporization. When steam condenses on the skin, it releases this large amount of latent heat, causing a much more severe burn.

Question 53.
List three problems caused by water pollution, as mentioned in the chapter.
Answer:
Three problems caused by water pollution: (Any three of the following):

1. Destruction of aquatic plants,
2. Damage to the food chain,
3. Soil pollution,
4. Spread of water-borne diseases (leading to sickness and death).

Question 54.
Explain the Density Anomaly of water and its importance for the survival of aquatic life in cold regions.
Answer:
The maximum density of water is at 4°C. When water is cooled below 4°C and freezes into ice, its volume increases, which means its density decreases. Because ice is less dense than water, it floats on the surface. This floating ice forms an insulating layer that prevents the water underneath from freezing completely, allowing fish and other aquatic life to survive below the surface.