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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.1

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.1 Textbook Questions and Answers

Question 1.
A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?
Answer:
Let the height of the tower = h m
Distance of the point of observation from the foot of the tower =15 cm.
Angle of elevation of the top of the tower = 45°
From the figure tan θ = \(\frac{\text { opp. side }}{\text { adj. side }}\)
tan 45° = \(\frac{h}{15}\)
⇒ 1 = \(\frac{h}{15}\)
∴ h = 1 × 15 = 15 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
Answer:
Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.
Let the length of the remaining part be = h m.
Let the length of the broken part be = x m.
Angle made by the broken part with the ground = 30°.
From the figure
tan 30° = \(\frac{h}{6}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{6}\)
∴ h = \(\frac{6}{\sqrt{3}}=\frac{3 \times 2}{\sqrt{3}}\) = 2√3 m
Also cos 30° = \(\frac{6}{x}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}\) = \(\frac{3 \times 2 \times 2}{\sqrt{3}}\) = 4√3
∴ Height of the tree = broken part + remaining part
= x + h
= 2√3 + 4√3 = 6√3 m
= 6 × 1.732
≃ 10.392 m.

Question 3.
A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2 m and by making an angle of 30° with the ground. What should be the length of the slide?
Answer:
Height of slide = 2 m
Let the length of the slide = x m.
Angle made by the slide with the ground = 30°
From the figure
sin 30° = \(\frac{2}{x}\)
⇒ \(\frac{1}{2}\) = \(\frac{2}{x}\)
⇒ x = 2 × 2 = 4 m
Length of the slide = 4 m.

Question 4.
Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
Answer:
Height of the pole = 15 m
Length of the shadow = 5√3 m
Let the angle of elevation be ‘θ’.
Then from the figure
tan θ = \(\frac{15}{5 \sqrt{3}}=\frac{5 \times \sqrt{3} \times \sqrt{3}}{5 \times \sqrt{3}}\) = √3
tan θ = √3 = tan 60°
∴ θ = 60°
∴ Angle of elevation of Sun rays with the ground = 60°.

Question 5.
You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30° with the pole. What should be the length of the rope?
Answer:
Height of the pole = 10 m
Let the length of each rope = x
Angle made by the rope with the pole = 30°

From the figure
cos 30° = \(\frac{10}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{10}{x}\)
⇒ x = \(\frac{10 \times 2}{\sqrt{3}}=\frac{20}{\sqrt{3}}\)
∴ Length of each rope = \(\frac{20}{\sqrt{3}}\)m
= 11.546 m.

∴ Total length of the rope = 3 × \(\frac{20}{\sqrt{3}}\)
= 20√3
= 20 × 1.732
≃ 34.64 m.

Question 6.
Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between you and the object?
Answer:
Height of the building = 6 m
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{6}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}=\frac{2 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\) = 4√3
∴ Distance = 4√3 m or
4 × 1.732 = 6.928 m.

Question 7.
An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?
Answer:
Height of the pole = 9m
Height of the point from the ground where he reaches the pole = 9 – 1.8 = 7.2 m
Angle of elevation = 60°
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{7.2}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{7.2}{x}\)
⇒ x = \(\frac{7.2 \times 2}{\sqrt{3}}=\frac{3 \times 2.4 \times 2}{\sqrt{3}}=\frac{\sqrt{3} \times \sqrt{3} \times 4.8}{\sqrt{3}}\)
⇒ x = 1.732 × 4.8
≃ 8.31 m
Also tan 60° = \(\frac{7.2}{d}\)
√3 = \(\frac{7.2}{d}\)
⇒ d = \(\frac{7.2}{\sqrt{3}}=\frac{2.4 \times 3}{\sqrt{3}}\) = 2.4 × √3 = 2.4 × 1.732
∴ d ≃ 4.1568 m

Question 8.
A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?
Answer:
Let the width of the river = AB = x m
Angle made by the boat = 60°
Distance travelled = AC = 600 m
From the figure
cos 60° = \(\frac{x}{600}\)
\(\frac{1}{2}\) = \(\frac{x}{600}\)
⇒ x = \(\frac{600}{2}\) = 300 m.
In the figure
A = Boat’s place
C = Reach place of another side (or) Point of observation.
AC = Travelling distance of the boat ∠AC = 60°
AB = width of the river AB
In △ABC, sin 60° = \(\frac{AB}{AC}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{AB}{600}\)
⇒ AB = 600 × \(\frac{\sqrt{3}}{2}\) = 300√3

Question 9.
An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of the palm tree?
Answer:
Height of the observer = 1.8 m
Distance of the observer from the palm tree = 13.2 m
From the figure
tan 45° = \(\frac{x}{13.2}\)
⇒ 1 = \(\frac{x}{13.2}\)
⇒ x = 13.2 m
∴ Height of the palm tree = 13.2 + 1.8 = 15 m.

Question 10.
In the given figure, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle.
Answer:
Draw a perpendicular BD to AC
∴ BD ⊥ AC
Now let AD = 6 – x and DC = x
Given AB = 5 cm and ∠BAD = 30° then in △ABD
sin 30° = \(\frac{BD}{AB}\) = \(\frac{BD}{5}\) = \(\frac{1}{2}\)
⇒ BD = \(\frac{5}{2}\) = 2.5 cm
and cos 30° = \(\frac{AD}{AB}\) = \(\frac{6-x}{5}\) = \(\frac{\sqrt{3}}{2}\)
⇒ 6 – x = \(\frac{5 \sqrt{3}}{2}\)
⇒ x = 6 – \(\frac{5 \sqrt{3}}{2}\) = 6 – \(\frac{5(1.732)}{2}\)
∴ x = 1.67
∴ Area of △ABC = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) × AC × BD
= \(\frac{1}{2}\) × 6 × 2.5
= 7.5 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3

10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer:

Sum of the frequencies = 68
∴ \(\frac{n}{2}\) = \(\frac{68}{2}\) = 34
Hence median class = 125 – 145
Lower boundary of the median class, l = 125
cf – cumulative frequency of the class preceding the median class = 22
f – frequency of the median class = 20
h = class size = 20
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 125 + \(\frac{[34-22]}{20}\) × 20
∴ Median = 125 + 12 = 137
Maximum number of consumers lie in the class 125 – 145
Modal class is 125 -145
l – lower limit of the modal class =125
f₁ – frequency of the modal class = 20
f₀ – frequency of the class preceding the modal class =13
f₂ – frequency of the class succeeding the modal class =14
h – size of the class = 20
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
Mode (Z) = 125 + \(\frac{20-13}{(20-13)+(20-14)} \times 20\)
= 125 + \(\frac{7}{7+6}\) × 20
= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
∴ Mode = 135.769
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 135
∴ \(\overline{\mathbf{x}}\) = 135 + \(\frac{7}{68}\)
= 135 + 0.102941
≃ 135.1
Mean, Median and Mode are approximately same in this case.

Question 2.
If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
It is given that ∑f = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y = 15 ….. (1)
The median is 28.5 which lies be-tween 20 and 30.
Median class = 20 – 30
Lower boundary of the median class ‘l’ = 20
\(\frac{N}{2}\) = \(\frac{60}{2}\) = 30
cf – cumulative frequency = 5 + x
h = 10
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
⇒ 28.5 = 20 + \(\frac{30-5-x}{20}\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
\(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25- 17 = 8
also from (1); x + y = 15
8 + y = 15
y = 7
∴ x = 8; y = 7.

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.

Answer:

The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
Observe that from the given distribution 2 persons with age less than 20.
i.e., frequency of the class below 20 is 2.
Now there are 6 persons with age less than 25 and 2 persons with age less than 20.
∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
Similarly, the frequencies can be calculated as shown in table.
Number of observations = 100
n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50, which lies in the class 35-40
∴ 35 – 40 is the median class and lower boundary l = 35
cf = 45;
h = 5;
f = 33
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 35 + \(\frac{50-45}{33}\) × 5
= 35 + \(\frac{5}{33}\) × 5
= 35 + 0.7575
= 35.7575
∴ Median ≃ 35.76

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…, 171.5 – 180.5.)
Answer:
Since the formula, Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\) assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
The classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.

∑f_i = n = 40
\(\frac{n}{2}\) = \(\frac{40}{2}\) = 20
\(\frac{n}{2}\)th observation lie in the class 144.5- 153.5
∴ Median class = 144.5 – 153.5
Lower boundary, l = 144.5
Frequency of the median class, f = 12
c.f. = 17
h = 9
∴ Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 144.5 + \(\frac{20-17}{12}\) × 9
= 144.5 + \(\frac{3}{12}\) × 9
= 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
∴ Median length = 146.75 mm.

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps.

Find the median life-time of a lamp.
Answer:

Total observations are n = 400
\(\frac{n}{2}\)th observation i.e \(\frac{400}{2}\) = 200
200 lies in the class 3000 – 3500
∴ Median class = 3000 – 3500
Lower boundary l = 3000
frequency of the median class f = 86
c.f = 130
Class size, h = 500
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 3000 + \(\frac{200-130}{86}\) × 500
= 3000 + \(\frac{70}{86}\) × 500
= 3000 + 406.977
= 3406.98
∴ Median life ≃ 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows.

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Answer:
Number of letters in the surnames.
Also find the modal size of the surnames.

Total observations are n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50
50 lies in the class 7 – 10
∴ Median class = 7 – 10
l – lower boundary = 7
f – frequency of the median class = 40
cf = 36
Class size h = 3
Median:
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 7 + \(\frac{50-36}{40}\) × 3
= 7 + \(\frac{14}{40}\) × 3
= 7 + \(\frac{42}{40}\)
= 7 + 1.05
= 8.05
∴ Median = 8.05.

Mean:
Assumed mean, a = 8.5
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 8.5 + \(\frac{(-18)}{100}\)
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32.

Mode:
Maximum number of surnames = 40
∴ Modal class = 7-10
l – lower boundary of the modal class = 7
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
l = 7; f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode (Z) = 7 + \(\frac{40-30}{(40-30)+(40-16)}\) × 3
= 7 + \(\frac{10}{10+24}\) × 3
= 7 + \(\frac{30}{34}\)
= 7 + 0.882
= 7.882

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Number of observations (n) = ∑f_i
\(\frac{n}{2}\) = \(\frac{30}{2}\) = 15
15 lies in the class 50 – 55
∴ Median class = 50-55
l – lower boundary of the median class = 55
f – frequency of the median class = 8
c.f = 5
Class size h = 6
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 50 + \(\frac{15-5}{8}\) × 6
= 50 + 7.5
= 57.5
= 50 + 7.5 = 57.5
∴ Median weight = 57.5 kg.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.1

10th Class Maths 14th Lesson Statistics Ex 14.1 Textbook Questions and Answers

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Answer:

Since f_i and x_i are of small values we use direct method.
∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{162}{20}\)
= 8.1

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:

Here the x_i are of large numerical values.
So we use deviation method then,
\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
Here the assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 275 + \(\frac{1900}{50}\)
= 275 + 38
= 313.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Answer:

\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
x_i = 18 (given)
\(\Rightarrow 18=\frac{752+20 \mathrm{f}}{(44+\mathrm{f})}\)
18 (44 + f) = 752 + 20 f
⇒ 20f- 18f= 792-752
⇒ 2f = 40
∴ f = \(\frac{40}{20}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Answer:

\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
75.5 + \(\frac{12}{30}\)
= 75.5 + 0.4
= 75.9.

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?
Answer:

Here we use step deviation method where a = 135, h = 5,a multiple of all d_i
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 22 + \(\frac{25}{400}\) × 5
= 22 + 0.31
= 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Answer:

Here a = 125, h = 50, ∑f_iu_i = 43
Now
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 125 + \(\frac{43}{25}\) × 50
= 125 + (43 × 2)
= 125 + 86
= 211.
NOTE: If we consider first value as “a” then we dont get negative values in u_i, f_iu_i columns. Then it becomes easy for calculation.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Answer:

∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\)
= 0.00986666…….
≃ 0.099

Question 8.
A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Answer:

Here, a = 51.5
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
= 51.5 – \(\frac{99}{40}\)
= 51.5 – 2.475
= 49.025
≃ 49 days

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

a = 70; h = 10
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}} \times \mathbf{h}\)
\(\Rightarrow \bar{x}=70-\frac{2}{35} \times 10\)
= 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\)
= 70 – 0.57142
= 69.4285
≃ 69.43%

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Answer:
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)
cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)
tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)
∴ sin A = \(\frac{15}{17}\);
cos A = \(\frac{8}{17}\)
tan A = \(\frac{15}{8}\)

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.
Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR² = PQ² + PR²
(25)² = (7)² + PR²
PR² = (25)² – (7)² = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);
tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)
∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{(24)^{2}-(7)^{2}}{168}\)
= \(\frac{576-49}{168}\)
= \(\frac{527}{168}\)

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Answer:
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC² = AB² + BC²
(25)² = AB² + (24)²
AB² = 25² – 24² = 625 – 576
AB² = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)
tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)
Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A.
Answer:
From the identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\left(\frac{12}{13}\right)^{2}\)
= 1 – \(\frac{144}{169}\)
= \(\frac{169-144}{169}\)
= \(\frac{25}{169}\)
∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Answer:
Given 3 tan A = 4
⇒ tan A = \(\frac{4}{3}\)
From the identify sec² A – tan² A = 1
⇒ 1 + tan² A = sec² A

If cos A = \(\frac{3}{5}\) then from
sin² A + cos² A = 1
We can write sin²A = 1 – cos²A
= 1 – \(\left(\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}\)
⇒ sin² A = \(\frac{16}{25}\)
⇒ sin A = \(\frac{4}{5}\)
∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.
Answer:

In the given triangle, cos A = cos X
⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot² θ = (cot θ)²
= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ
So cot θ = \(\frac{7}{8}\)
⇒ tan θ = \(\frac{8}{7}\)
⇒ tan² θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)
From sec² θ – tan² θ = 1
⇒ 1 + tan² θ = sec² θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:
Given, tan A = \(\frac{\sqrt{3}}{1}\)
Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC² = AB² + BC²
(By Pythagoras theorem)
⇒ AC² = (1k)² + (√3k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k²
∴ AC = \(\sqrt{4 k^{2}}\) = 2k

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.2

10th Class Maths 13th Lesson Probability Ex 13.2 Textbook Questions and Answers

Question 1.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red?
Answer:
i) Total number of balls in the bag = 3 red + 5 black = 8 balls.
Number of total outcomes when a ball is drawn at random = 3 + 5 = 8
Now, number of favourable outcomes of red ball = 3.
∴ Probability of getting a red ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{3}{8}\)
ii) If P( E) is the probability of drawing no red ball, then P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E)= 1 – \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 2.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Answer:
Total number of marbles in the box = 5 red + 8 white + 4 green = 5 + 8 + 4= 17
Number of total outcomes in drawing a marble at random from the box =17.
i) Number of red marbles = 5
Number of favourable outcomes in drawing a red ball = 5
∴ Probability of getting a red ball P(R) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(R) = \(\frac{5}{17}\)

ii) Number of white marbles = 8
Number of favourable outcomes in drawing a white marble = 8
∴ Probability of getting a white marble
P(W) = \(\frac{8}{17}\)

iii) Number of ‘non-green’ marbles = 5 red + 8 white = 5 + 8 = 13
Number of outcomes favourable to drawing a non-green marble =13.
∴ Probability of getting a non- green marble
P(non – green) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(non – green) = \(\frac{13}{17}\)
Probability of getting a green ball = \(\frac{\text { No. of green balls }}{\text { Total no. of balls }}\) = \(\frac{4}{17}\)
Now P(G) = \(\frac{4}{17}\)
and P(G) + P(\(\overline{\mathrm{ G}}\)) = 1
∴ P(\(\overline{\mathrm{G}}\)) = 1 – P(G)
= 1 – \(\frac{4}{17}\)
= \(\frac{13}{17}\)

Question 3.
A Kiddy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Answer:
i) Number of 50 p coins = 100
Number of Rs. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
Number of total outcomes for a coin to fall down = 180.
Number of outcomes favourable to 50 p coins to fall down = 100.
∴ Probability of a 50 p coin to fall down = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{100}{180}\) = \(\frac{5}{9}\)

ii) Let P(E) be the probability for a Rs. 5 coin to fall down.
Number of outcomes favourable to Rs. 5 coin = 10.
∴ Probability for a Rs. 5 coin to fall down = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{10}{180}\) = \(\frac{1}{18}\)
Then P(\(\overline{\mathrm{E}}\)) is the probability of a coin which fall down is not a Rs. 5 coin.
Again P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\))= l-P(E)
= 1 – \(\frac{1}{18}\)
= \(\frac{17}{18}\).

Question 4.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (See figure). What is the probability that the fish taken out is a male fish?
Answer:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 m + 8 f
= 13 fishes.
∴ Number of total outcomes in taking a fish at random from the aquarium =13.
Number of male fish = 5
∴ Number of outcomes favourable to male fish = 5.
∴ The probability of taking a male fish = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{5}{13}\)
= 0.38

Question 5.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer:
Number of total outcomes are (1,2,……….., 8) = 8

i) Number of outcomes favourable to 8 = 1.
∴ P(8) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{8}\)

ii) Number of ‘odd numbers’ on the spinning wheel = (1, 3, 5, 7) = 4
∴ Number of outcomes favourable to an odd number.
∴ Probability of getting an odd number = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{8}\) = \(\frac{1}{2}\)

iii) Number greater than 2 are (3, 4, 5, 6, 7, 8)
Number of outcomes favourable to ‘greater than 2’ are = 6.
Probability of pointing a number greater than 2
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

iv) Number less than 9 are: (1,2, 3, 4, 5, 6, 7, 8 …… 8)
∴ Number of outcomes favourable to pointing a number less than 9 = 8.
∴ Probability of a number less than 9
P(E) = \(\frac{\text { No. of outcomes favourable to less than } 9}{\text { No.of total outcomes }}\)
= \(\frac{8}{8}\) = 1
Note : This is a sure event and hence probability is 1.

Question 6.
One card is drawn from, a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Answer:
Total number of cards = 52.
∴ Number of all possible outcomes in drawing a card at random = 52.
i) Number of outcomes favourable to the king of red colour = 2(♥ K, ♦ K)
∴ Probability of getting the king of red colour
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Number of face cards in a deck of cards = 4 × 3 = 12 (K, Q, J)
Number of outcomes favourable to select a face card = 12.
∴ Probability of getting a face card
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) Number of red face cards = 2 × 3 = 6.
∴ Number of outcomes favourable to select a red face card = 6.
∴ Probability of getting a red face
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{52}\) = \(\frac{3}{26}\)

iv) Number of outcomes favourable to the jack of hearts = 1.
∴ Probability of getting jack of hearts
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

v) Number of spade cards = 13
∴ Number of outcomes favourable to ‘a spade card’ = 13.
∴ Probability of drawing a spade
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

vi) Number of outcomes favourable to the queen of diamonds = 1.
∴ Probability of drawing the queen of diamonds
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

Question 7.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked is (a) an ace? (b) a queen?
Answer:
Total number of cards = 5.
∴ Number of total outcomes in picking up a card at random = 5.
i) Number of outcomes favourable to queen = 1.
∴ Probability of getting the queen
= \(\frac{\text { No.of outcomes favourable to the ‘Q’ }}{\text { No.of total outcomes }}\)
= \(\frac{1}{5}\)

ii) When queen is drawn and put aside, remaining cards are four.
∴ Number of total outcomes in drawing a card at random = 4.
a) Number of favourable outcomes to ace 1
Probability of getting an ace
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{4}\)

b) Number of favourable outcomes to ‘Q’ = 0 (as it was already drawn and put aside)
∴ Probability that the card is Q = \(\frac{0}{4}\) = 0
After putting queen aside, selecting the queen from the rest is an impossible event and hence the probability is zero.

Question 8.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
Number of good pens = 132
Number of defective pens = 12
Total number of pens = 132 + 12 = 144
∴ Total number of outcomes in taking a pen at random = 144.
No. of favourable outcomes in taking a good pen = 132.
∴ Probability of taking a good pen
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 9.
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
Given :
4 out of 20 bulbs are defective
(i.e.) Number of defective bulbs = 4
Number of non-defective bulbs = 20 – 4 = 16
If a bulb is drawn at random, the total outcomes are = 20
Number of outcomes favourable to ‘defective bulb’ = 4
∴ Probability of getting a defective bulb
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{20}\) = \(\frac{1}{5}\)
Suppose a non-defective bulb is drawn and not replaced, then the bulbs remaining are = 19
∴ Total outcomes in drawing a bulb from the rest = 19
Number of favourable outcomes in drawing non-defective bulb from the rest = 16 – 1 = 15
∴ Probability of getting a non-defective bulb in the second draw
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{15}{19}\)

Question 10.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Answer:
Total number of discs in the box = 90
∴ Number of total outcomes in drawing a disc at random from the box = 90.

i) Number of 2-digit numbers in the box (10, 11,….., 90) = 81
i.e., Number of favourable outcomes in drawing a 2 – digit numbers = 81
∴ Probability of selecting a disc bearing a 2 – digit number
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{81}{90}\) = \(\frac{9}{10}\) = 0.9

ii) Number of perfect squares in the box (1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64 and 9² = 81) = 9
i.e., Number of favourable out-comes in drawning a disc bearing a perfect square = 9
∴ Probability of drawning a disc with a perfect square
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{9}{90}\) = \(\frac{1}{10}\)

iii) Number of multiples of 5 from 1 to 90 are (5, 10, 15, ……….., 90) = 18
i.e., Number of favourable outcomes in drawing a disc with a multiple of 5 = 18
∴ Probability of drawing a disc bearing a number divisible by 5
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{18}{90}\) = \(\frac{1}{5}\)

Question 11.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Answer:
3 m.
Length of the given rectangle = 3 m.
and its breadth = 2 m.
Area of the rectangle
= length × breadth = 3 × 2 = 6 m²
∴ Total area of the region for landing = 6 m².
Diameter of the given circle = 1 m.
Area of the circle = \(\frac{\pi \mathrm{d}^{2}}{4}\)
= \(\frac{22}{7} \times \frac{1 \times 1}{4}\left[\text { or } \pi r^{2}=\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}\right]\)
= \(\frac{22}{28}\)
∴ Probability of the coin to land on the circle
= \(\frac{\frac{22}{28}}{6}\)
= \(\frac{22}{28×6}\)
= \(\frac{11}{28×3}\)
= \(\frac{11}{84}\)

Question 12.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it?
Answer:
Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20
no. of good ball pens = 144 – 20 = 124
∴ Total outcomes in drawing a ball pen at random = 144.

i) Sudha buys it if it is not defective / a good one.
No. of outcomes favourable to a good pen = 124.
∴ Probability of buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{124}{144}\) = \(\frac{31}{36}\)

ii) Sudha will not buy it-if it is a defective pen
No. of outcomes favourable to a defective pen = 20
∴ Probability of not buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{20}{144}\) = \(\frac{5}{36}\)

!! (not buying) = 1 – P (buying)
= 1 – \(\frac{31}{36}\) = \(\frac{5}{36}\)

Question 13.
Two dice are rolled simultaneously and counts are added
(i) Complete the table given below:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Answer:
When two dice are rolled, total number of outcomes = 36 (see the given table).
(i)
(ii) The above (given) argument is wrong [from the above table].
The sum 2, 3, 4, ………… and 12 have different no. of favourable outcomes, moreover total number of outcomes are 36.

Question 14.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer:
When a coin is tossed for n – times, the total number of outcomes = 2ⁿ.
∴ If a coin is tossed for 3 – times, then the total number of outcomes = 2³ = 8
Note the following :
TTT
TTH
THT
HTT
HHT
HTH
THH
HHH
Of the above, no. of outcomes with different results = 6.
Probability of losing the game
= \(\frac{\text { No. of favourable outcomes to lose }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 15.
A dice is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up atleast once? [Hint : Throwing a dice twice and throwing two dice simultaneously are treated as the same experiment].
Answer:
If a dice is thrown n-times or n-dice are thrown simultaneously then the total
number of outcomes = 6×6×6….×6
(n – times) = 6ⁿ.
No. of total outcomes in throwing a dice for two times = 6² = 36.
i) Let E be the event that 5 will not come up either time, then the favourable outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3/2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) = 25.
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{25}{36}\)

ii) Let E be the event that 5 will come up atleast once.
Then the favourable outcomes are (1,5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) = 11 No. of favourable outcomes = 11
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{11}{36}\)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.2

10th Class Maths 14th Lesson Statistics Ex 14.2 Textbook Questions and Answers

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
Maximum number of patients joined in the age group 35 – 45.
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
Class size h = 10
Frequency of modal class, f₁ = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeeding the modal class f₂ = 14

∴ Mode = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(\begin{array}{l}
=35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times 10 \\
=35+\left(\frac{2}{46-35}\right) \times 10
\end{array}\)
= 35 + \(\frac{2}{11}\) × 10
= 35 + 1.81818……
= 36.8 years.
Mean x = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2830}{80}\)
= 35.37 years.
Interpretation: Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are of the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the Mean.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Answer:

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f₁ = 61.
Frequency of the class preceding the modal class f₀ = 52
Frequency of the class succeeding the modal class f₂ = 38
Lower boundary of the modal class, l = 60
Height of the class, h = 20
∴ Mode (Z) = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(=60+\left[\frac{61-52}{2 \times 61-(52+38)}\right] \times 20\)
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625
= 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Answer:

Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of the modal class (l) = 1500
Frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class f₀ = 24
Frequency of the class succeeding the modal class f₂ = 33
Height of the class, h = 500
Hence, modal monthly income = Rs. 1847.83.
Assumed mean (a) = 3250
∑f_i = 200, ∑u_if_i = -235
Mean monthly income = \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= Rs. 2662.50

Question 4.
The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Answer:

Since the maximum number of states ‘10’ lies in the class interval 30-35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f₁ = 10
Frequency of the class preceding the modal class = f₀ = 9
Frequency of the class succeeding the modal class = f₂ = 3
Height of the class, h = 5
∴ Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
\(=30+\frac{10-9}{(10-9)+(10-3)} \times 5\)
= 30 + \(\frac{1×5}{1+7}\)
= 30 + \(\frac{5}{8}\)
= 30 + 0.625
= 30.625
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 32.5; h = height of the class = 5
∴ x = 32.5 – \(\frac{23}{35}\) × 5
= 32.5 – 3.28
= 29.22
Mean = 30.625
Mode = 29.22
Mode states have a students – teacher ratio 29.22 and on an average this ratio is 30.625.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Answer:

Maximum number of batsmen are in the class 4000 – 5000.
∴ Modal class is 4000 – 5000.
Lower boundary of the modal class ‘l’ = 4000
Frequency of the modal class, f₁ = 18
Frequency of the class preceding the modal class, f₀ = 4
Frequency of the class succeeding the modal class, f₂ = 9
Height of the class, h = 1000
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(4000+\frac{18-4}{(18-4)+(18-9)} \times 1000\)
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\)
= 4000 + 608.695
= 4608.69
≃ 4608.7 runs

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

Find the mode of the data.
Answer:

Since the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class ‘l’ = 40
Frequency of the modal class, f₁ = 20
Frequency of the class preceding the modal class, f₀ = 12
Frequency of the class succeeding the modal class, f₂ = 11
Height of the class, h = 10;
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(40+\frac{(20-12)}{(20-12)+(20-11)} \times 10\)
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≃ 44.7 cars

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.
Answer:

Let the height of the tower = h mts say
Width of the road be = x m.
Distance between two points of observation = 10 cm.
Angles of elevation from the two points = 60° and 30°.
From the figure
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
⇒ h = √3x …….(1)
Also tan 30° = \(\frac{h}{10+x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)
⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)
From equations (1) and (2) h
h = √3x = \(\frac{10+x}{\sqrt{3}}\)
∴ √3x = \(\frac{10+x}{\sqrt{3}}\)
√3 × √3x = 10 + x
⇒ 3x – x = 10
⇒ 2x = 10
⇒ x = \(\frac{10}{2}\) = 5m
∴ Width of the road = 5 m
Now Height of the tower = √3x = 5√3 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.
Answer:

Height of the temple = 30 m
Height of the man = 1.5 m
Initial distance between the man and temple = d m. say
Let the distance walked = x m.
From the figure
tan 30° = \(\frac{30-1.5}{d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)
∴ d = 28.5 × √3m ………(1)
Also tan 60° = \(\frac{28.5}{d-x}\)
⇒ √3 = \(\frac{28.5}{d-x}\)
⇒ √3(d-x) = 28.5
⇒ √3(28.5 × √3-x) = 28.5
⇒ 28.5 × 3 – √3x = 28.5
⇒ √3x = 3 × 28.5-28.5
⇒ √3x = 2 × 28.5 = 57
∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3
= 19 × 1.732
= 32.908 m.
∴ Distance walked = 32.908 m.

Question 3.
A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.
Answer:

Height of the pedestal = 2 m.
Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.
Angle of elevation of top of the pedestal = 45°.
Let the distance between the point of observation and foot of the pedestal = x m.
From the figure
tan 45° = \(\frac{2}{x}\)
1 = \(\frac{2}{x}\)
∴ x = 2 m.
Also tan 60° = \(\frac{2+h}{x}\)
⇒ √3 = \(\frac{2+h}{x}\)
⇒ 2√3 = 2 + h
⇒ h = 2√3 – 2
= 2(√3-1)
= 2(1.732 – 1)
= 2 × 0.732
= 1.464 m.

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.
Answer:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Distance between tower and building = 7 m.
Let the height of the building = x m and tower = y m.
From the figure
tan 45° = \(\frac{x}{7}\)
1 = \(\frac{x}{7}\)
∴ x = 7 m.
Also tan 60° = \(\frac{y-x}{7}\)
⇒ √3 = \(\frac{y-x}{7}\)
⇒ 7√3 = y – 7
∴ y = 7 + 7√3
= 7 (√3 + 1)
= 7(1.732 + 1)
= 2.732 × 7
= 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?
Answer:

Length of the wire = 18 m
Let the length of the wire removed = x
Height of the pole be = h
From the figure
sin 30° = \(\frac{h}{18}\)
⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)
⇒ h = \(\frac{18}{2}\) = 9 m
Also sin 60° = \(\frac{h}{18-x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)
√3(18-x) = 9 × 2
18√3 – √3x = 18
√3x = 18√3 – 18
√3x = 18(√3-1)
x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)
= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)
= 6√3(√3-1)
= 6(3-√3)
= 18 – 6√3
= 18 – 10.392
= 7.608 m.

Question 6.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
Answer:

Height of the tower = 30 m
Angle of elevation of the top of the tower = 60°.
Angle of elevation of the top of the building = 30°.
Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
From the figure
tan 60° = \(\frac{30}{d}\)
√3 = \(\frac{30}{d}\)
⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m
Also tan 30° = \(\frac{x}{d}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)
⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m
∴ Height of the building = 10 m

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:

Width of the road = 120 f.
Angle of elevation of the top of the 1st tower = 60°.
Angle of elevation of the top of the 2 tower = 30°.
Let the distance of the point from the 1st pole = x.
Then the distance of the point from
the 2nd pole = 120 – x.
and height of each pole = h say.
From the figure
tan 60° = \(\frac{h}{x}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x ……..(1)
Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ h = \(\frac{120-x}{\sqrt{3}}\)
From (1) and (2)
√3x = \(\frac{120-x}{\sqrt{3}}\)
⇒ √3.√3x = 120-x
⇒ 3x = 120 – x
⇒ 3x + x = 120
⇒ 4x = 120
⇒ x = \(\frac{120}{4}\) = 30 ft
Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet
∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
Height of each pole = 51.96 ft.

Question 8.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
Answer:

Let the height of the tower = h m.
Angles of elevation of the top of the tower from two points = x° and (90° – x)
From the figure
tan x = \(\frac{h}{4}\) ……. (1)
Also tan (90° – x) = \(\frac{h}{9}\)
⇒ cot x = \(\frac{h}{9}\)
⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)
∴ tan x = \(\frac{9}{h}\) …….. (2)
From (1) and (2)
tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)
∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)
h × h = 9 × 4
⇒ h² = 36
⇒ h = 6 m

Question 9.
The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)
Answer:

Height of the plane from the ground PM = RN = 1500√3 m.
Angle of elevation are 30° and 60°.
From the figure
tan 60° = \(\frac{PM}{QM}\)
√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)
QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
Also tan 30° = \(\frac{RN}{QN}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)
QM + MN = 1500√3 × √3
1500 + MN = 1500 × 3
MN = 4500 – 1500
MN = 3000 mts.
∴ Distance travelled in 15 seconds = 3000 mts.
∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s
= 200 × \(\frac{18}{5}\) kmph
= 720 kmph
Speed = 200 m/sec. or 720 kmph.

AP SSC 10th Class Maths Textbook Solutions

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment.

AP State Syllabus SSC 10th Class Biology Important Questions 9th Lesson Our Environment

10th Class Biology 9th Lesson Our Environment 1 Mark Important Questions and Answers

Question 1.
Suggest one alternative method in place of pesticides to protect crops?
Answer:
Alternative methods for using pesticides to save the crops from pests:

1. Rotation of crops
2. Studying the life histories of pests
3. Biological Control
4. Sterility
5. Genetic Strains

Question 2.
By taking two plants of your surroundings as examples, explain how they protect themselves against the animals which eat them.
Answer:

1. Neem Tree: Neem leaves contain an alkaloid Nimbin to protect themselves from the animals which eat them.
2. Cactus: They have thorns to protect themselves.
3. Datura: Datura leaves gives bad odour.

Question 3.
Identify one food chain from your surroundings. Name the producers and different levels of consumers in that food Chain.
Answer:
Grass → Insects → Frog → Snake.
Producers – Grass.
Primary Consumers – Insects.
Secondary Consumers – Frog
Tertiary consumers – Snake.

Aquatic Plants → Insects → Fish → Crane.
Producers – Aquatic Plant.
Primary Consumers – Insects
Secondary Consumers – Fish
Tertiary consumers – Crane

Question 4.
Write the names of producers and consumers in the food chain, you have observed.
Answer:
Producers – Plants, Green Algae
Consumers – All Animals.

Question 5.
Write any two slogans to promote awareness among the people about Ecofriendly programs.
Answer:
a) Lets go green to get global clean.
b) If you disturb the nature, the nature will disturb you.
c) The best solution to arrest pollution is plantation.
d) Reduce the usage of plastic and reduce the pollution.

Question 6.
What happens if decomposers are removed from the food web?
Answer:

1. If decomposers are removed from the food web then the biological cycles are not completed.
2. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.

Question 7.
Observe the following given below. Draw the pyramid of numbers.
Grass → Goat → Man
Answer:

Question 8.
We can’t expect the world without sparrows. So how should be our concern towards their conservation?
Answer:

1. Sparrows are useful to control harmful insects like locust which damage food grains.
2. Chemical pesticides are the cause for destruction of sparrows and useful insects.
3. By using biological methods we can conserve the sparrow population.

Question 9.
Human being is modifying agriculture lands and lakes into residential areas. What is its effect on Bio-diversity?
Answer:

1. The shelter may not be provided for migratory birds.
2. Food chain get disturbed.
3. Decrease in the ground water level.

Question 10.
How do you protect the plants, which were planted in “Haritha Haaram” programme in your school?
Answer:
We protect the plants:

1. Watering of plants at regular intervals.
2. Fencing or gaurding of plants.
3. Adoption of plants.
4. Providing organic manure.

Question 11.
The figure given below represents a food pyramid. Study it and answer the following questions.

i) Which trophic level has maximum energy?
Answer:
T₁ (or) Primary producers (green plants)
ii) Give one example for T₄ trophic level.
Answer:
Lion, tiger, hawk, etc.

Question 12.
“We can’t imagine the world without insects and birds, conserve them.
Answer:
Methods to conserve insects and birds:

1. Avoid indiscriminate usage of pesticides
2. Protect the natural habitats of insects and birds.
3. Development of bird sanctuaries
4. Everyone should follow environmental ethics.

Question 13.
What is environment?
Answer:
The sum of physical and biological factors along with their chemical interactions that affect an organism is called environment.

Question 14.
What is biosphere?
Answer:
The world of living things is called biosphere.
(OR)
The life supporting zone on the earth is called biosphere.

Question 15.
What are the physical or abiotic factors in a biosphere?
Answer:
Land, air, water, sunlight, humidity etc. are the physical or abiotic factors in a biosphere.

Question 16.
What does a food chain show?
Answer:
Food chain shows that how the energy is passed from one organism to another.

Question 17.
How the terrestrial ecosystems are determined?
Answer:
The terrestrial ecosystems are determined largely by the variations in climatic conditions between the poles and equator.

Question 18.
Where did the Kilimanjaro mountain located?
Answer:
The Kilimanjaro mountain is located in equatorial Africa (present in Tanzania, Africa).

Question 19.
What is the main source of energy for all the organisms in an ecosystem?
Answer:
Sunlight is the main source of energy for all the organisms in an ecosystem.

Question 20.
What is food web?
Answer:
The network of a large number of food chains existing in an ecosystem is called a food web.

Question 21.
What does a food web indicate?
Answer:
A food web indicates that the number of possible links for food in an ecosystem and reflects the fact that the whole community is a complex inter-connected unit.

Question 22.
What is ‘niche’?
Answer:
The position of organisms in a food web.

Question 23.
What does the word ‘niche’ denotes?
Answer:
‘Niche’ is the term used to describe not only the animals position in the food web and what it eats but also its mode of life.

Question 24.
What is an ecological pyramid?
Answer:
The graphic representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.

Question 25.
Who was the first one to introduce “Ecological pyramid”?
Answer:
Ecological pyramid was first introduced by a British Ecologist Charles Elton in 1927.

Question 26.
In ecological pyramids the producers are represented at?
Answer:
The producers are represented at the base of the ecological pyramids.

Question 27.
What is pyramid of number?
Answer:
A graphical representation designed to show the number of organisms at each tropic level in a given ecosystem is called “pyramid of number”.

Question 28.
What does the pyramid of number show?
Answer:
Pyramid of number shows the population of organisms at each tropic level in a food chain.

Question 29.
What does each bar represent in a pyramid of number?
Answer:
In a pyramid of number, each bar represents the number of individuals at each tropic level in a food chain.

Question 30.
When does the pyramid of number not look like a pyramid at all?
Answer:
If the producer is a large plant such as tree or if one of the organisms at any tropic level is very small, then the pyramid of number does not look like a pyramid.

Question 31.
What is biomass?
Answer:
Any type of plant or animal material that can be converted into energy is called biomass.

Question 32.
What is biofuels?
Answer:
The materials which are used for energy production are known as biofuels.

Question 33.
What is Pyramid of biomass?
Answer:
A graphical representation designed to show the quantity of living matter at each tropic level in a given ecosystem is called “Pyramid of biomass”.

Question 34.
Why Pyramid of biomass inverted in case of aquatic ecosystem?
Answer:

1. In an aquatic ecosystem, the biomass of phytoplankton is quite negligible as compared to that of the crustaceans and small herbivorous fish that feed on these producers.
2. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.

Question 35.
How much percentage of the biomass is transferred from one tropic level to the next in a food chain?
Answer:
The percentage of the biomass transferred from one tropic level to the next level in food chain is nearly 10 – 20%.

Question 36.
When does the species at the top of the pyramid get more energy?
Answer:
The species at the top of the pyramid get, more energy when the steps in a food chain are fewer.

Question 37.
What are bio-geochemical cycles?
Answer:
Flow of materials between organisms and their environment is called cycling of materials or bio-geochemical cycles.

Question 38.
What is ecological efficiency?
Answer:
The ratio between energy flows at different tropic levels among the food chain expressed as percentage is called ecological efficiency.

Question 39.
What is ten per cent law?
Answer:
During the transfer of energy from one tropic level to the next, only about ten per cent of the energy from organic matter is stored as flesh. This is called “Ten per cent law”.

Question 40.
Where do Kolleru one of the largest fresh water lakes in India exists?
Answer:
Kolleru is one of the largest fresh water lakes in India exists between West Godavari and Krishna districts of Andhra Pradesh.

Question 41.
What is the catchment area of the lake Kolleru?
Answer:
A catchment area of the Kolleru lake extends up to 6121 Km2.

Question 42.
Through which The lake Kolleru discharges its excess water into Bay of Bengal?
Answer:
The lake Kolleru discharges its excess water into Bay of Bengal through the twisty channel called Upputeru which is about 65 km long.

Question 43.
When did Government of Andhra Pradesh had declared the lake as Bird Sanctuary?
Answer:
In November 1999, Government of Andhra Pradesh had declared the lake as Bird Sanctuary.

Question 44.
What is the number of species of birds being hosted by the Kolleru?
Answer:
The Kolleru lake is hosting 193 species of birds.

Question 45.
What are the major sources of pollution in Kolleru lake?
Answer:
The major sources of pollution are agricultural runoff containing residues of several agrochemicals, fertilizers, fish tank discharges, industrial effluents containing chemical residues and different types of organic substances, municipal and domestic sewage.

Question 46.
What is the objective of “Operation Kolleru” by the ministry of environment and forest, Government of India?
Answer:
The objective of operation Kolleru by the ministry of environment and forest, Government of India is to bring back the ecological balance of Kolleru lake which is a gift of nature.

Question 47.
What is Bioaccumulation?
Answer:
The process of entering of pollutants in a food chain is known as Bioaccumulation.

Question 48.
What is Biomagnification?
Answer:
The tendency of pollutants to concentrate as they move from one tropic level to the next is known as Biomagnification.

Question 49.
What are pesticides?
Answer:
The chemical materials used to control pests that attack crop plants or live as parasites on the body of farm animals are called pesticides.

Question 50.
What is a perfect pesticide?
Answer:
The perfect pesticide is one which destroys a particular pest and is completely harmless to every other form of life.

Question 51.
Why did the egg breaking among the peregrines increase?
Answer:
Egg breaking among the peregrines increased due to their disturbed behaviour caused by the nerve poisons that entered into their tissues through food chain.

Question 52.
Why did the aquatic biota is being contaminated?
Answer:
The aquatic biota is being contaminated with heavy metals due to industrialization and anthropogenic activities.

Question 53.
Why fish are considered to be the bioindicators of metal contamination?
Answer:
Fish are considered to be the bioindicators of metal contamination in environmental monitoring because fish species are strongly respond to stress conditions.

Question 54.
Where did Edulabad water reservoir located?
Answer:
Edulabad water reservoir is located in urban areas of Ranga Reddy district of Telangana.

Question 55.
Which fish species is grown in Edulabad water reservoir?
Answer:
Cyprinus carpio (common scale carp) is the fish species grown in Edulabad water reservoir.

Question 56.
What are the effects of bioaccumulation of metals in human beings that eat cyprinus carpio?
Answer:
The bioaccumulation of various metals in cyprinus carpio cause disorders. Such as hypertensions, sporadic fever, renal damage, nausea, etc.

Question 57.
In which country sparrows were hunted extensively in 1958?
Answer:
In China sparrows were hunted extensively in 1958.

Question 58.
In your opinion what are the effective methods to control pests?
Answer:
Rotation of crops, biological control, developing genetically modified plants are the effective methods to control pests in my opinon.

Question 59.
Why the temperatures are very high during the day and cold during the nights in deserts?
Answer:
In deserts, the rainfall and humidity are very low, so the sun’s rays easily penetrate the atmosphere making ground temperatures very high during the day. But the nights are often cold as the earth loses heat rapidly.

Question 60.
How can we draw a food chain?
Answer:
We can draw a food chain by connecting the pictures or names of organisms by putting arrows between them. These arrows should always point from food to the feeder.

Question 61.
How many types of ecological pyramids are there in practice? Name them.
Answer:
There are three types of ecological pyramids. They are:

1. Pyramid of number
2. Pyramid of biomass and
3. Pyramid of energy.

Question 62.
Which process helps to convert the solar energy into suitable form of energy for animals to consume?
Answer:
Photosynthesis helps to convert the solar energy into suitable form of energy (food) for animals to consume.

10th Class Biology 9th Lesson Our Environment 2 Marks Important Questions and Answers

Question 1.
Grass → Grasshopper → Frog → Snake → Hawk
What will happen if we remove Frog from the above food chain? Explain.
Answer:

1. Frog is secondary consumer in this food chain.
2. If we remove frog from the food chain, the number of grasshopper will increase on other hand the number of snakes which depend on frogs will decrease.
3. Hence, ecological balance may be damaged.

Question 2.
Observe the diagram and answer the following.
i) Write any two food chains from the diagram.
ii) What are the secondary consumers in the food chain that are written by you?
Answer:
i) a) Plants → Goat → Tiger
b) Plants → Rabbit → Wolf / Fox
ii) Tiger, Wolf / Fox

Question 3.
Study the given paragraph and answer the questions.

Solar energy from sun enters into the producers of an ecosystem. No organisms except green plants and photosynthetic bacteria can absorb solar energy and convert it into chemical energy.

A) What are the producers mentioned in the given paragraph?
B) What form is energy converted into photosynthesis? In Photosynthesis, which form is energy converted into?
Answer:
A) Green plants and photosynthetic bacteria.
B) In photosynthesis, the light (or) solar energy is converted into chemical energy.

Question 4.
Explain the flow chart given below.
Answer:
It is the pyramid of biomass.

1. In this pyramid 10% of the food will reach to the next trophical level and so on at each level.
2. It would take 1000 kg of phytoplankton to provide 100 kg of zooplankton and to form 1 kg of human tissue, 10 kg of frog is needed.
3. The fewer the steps in the food chain, the more energy will be for the species at the top.

Question 5.
Observe the pyramid of number which is given below and answer the questions.
i) As per the number of organisms in the tropic level, which group of organisms
are more in number and which are less in number?
ii) What happens if Secondary consumers disappear?
Answer:
i) If producers are more in number, then tertiary consumers are less in number,
ii) If secondary consumers disappear the primary consumers increase in number and the tertiary consumers found no food to live. It leads to death.

Question 6.
Explain in brief about the alternate methods to be followed to prevent the harmful effects of over usage of pesticides.
(OR)
Mention any four effective methods of controlling pests, which are less harmful on environment based on biological principles.
Answer:
Some of alternative pest control methods are

1. Rotation of Crop: Growing different crops on a particular piece of land in successive years.
2. Studying the life histories of the pests: When this is done it is sometimes possible to sow the crops at a time when least damage will be caused.
3. Biological Control: Introducing Natural predator or parasite of the pest.
4. Sterility: Rendering the males of a pest species sterile.
5. Genetic Strains: The development of genetic strains (genetically modified plants) which are resistant to certain pest.
6. Environmental ethics: People need to know besides laws regarding environment there are some basic ethics what is right and what is wrong in view of environment.

Question 7.
Write any 4 slogans on the necessity of forests and on their conservation.
Answer:

1. Save the trees, save the earth. We are the guardians of nature’s birth.
2. Don’t destroy the greenary and don’t spoil the scenery.
3. Don’t make trees rare, we should keep them with care.
4. To live for future in rest, saving forest is the best.

Question 8.
How does the given below concepts differs?
(a) Bioaccumulation b) Biomagnification
Answer:
a) Bioaccumulation: The process of entry of pollutants into a food chain is known as bioaccumulation.
b) Biomagnification: It is the tendency of pollutants to concentrate as they move from one tropic level to the next is known as biomagnification.

Question 9.
The biomass of a producer in an ecosystem is calculated as 3500 kgs. Calculate the biomass of primary, secondary, tertiary consumers.
Answer:
In a food chain roughly 90% of the food is lost at each step. So if the biomass of a producer in an ecosystem is calculated as 3500 kgs. the biomass of primary consumer as will be 350 kgs. and of secondary consumer is 35 kgs and biomass of tertiary consumer is 3.5 kgs.

Question 10.
Write a short note on food chain and food web.
Answer:

1. Food chain is a pathway along which food is transferred from one tropic level to another tropic level beginning with producers.
2. It shows who eats what in a particular habitat.
3. The arrows between each item in the chain always point from the food to the feeder.
4. For example
   Grass → Rabbit → Snake → Hawk
5. The elaborate interconnected feeding relationships in an ecosystem is said to be food web.
6. Many of the food chains in an ecosystem are crosslinked to form food web.
7. For example,
   
8. Food chain and food web help us to understand the food relations among living things.

Question 11.
Write a short notes on ecological pyramids.
Answer:

1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.
2. It was first introduced by a British Ecologist Charles Elton in 1927.
3. In the ecological pyramid, the producers (First tropic level) are represented at the base, and the successive tropic levels (primary, secondary and tertiary consumers) are represented one above the other with top carnivores at the tip.
4. There are three types of pyramids.
   i) Pyramid of number ii) Pyramid of biomass and iii) Pyramid of energy.
5. Pyramid of number shows the population of organisms at each tropic level in a food chain.
6. Pyramid of biomass represents the available food as a source of energy at each tropic level in the food chain.
7. Pyramid of energy represents the available energy at each tropic level in food chain.

Question 12.
Write a short notes on pyramid of number.
Answer:

1. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
2. The shape of this pyramid varies from eco-system to ecosystem.
   
3. In aquatic and grassland ecosystems, numerous small autotrophs support lesser herbivores which support further small number of carnivores and hence the pyramid structure is upright.
   
4. In forest ecosystem, less number of producers support greater number of herbivores who in turn support a fewer number of carnivores. Hence the pyramid structure is partly upright.
5. In parasitic food chain, one primary producer supports numerous parasites which support still more hyperparasites. Hence the pyramid structure is inverted.
   

Question 13.
Write a short notes on pyramid of biomass.
Answer:

1. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (biomass) at each tropic level in a given ecosystem.
2. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores. Hence the pyramid structure is upright.
   
3. In an aquatic ecosystem, the biomass of phytoplankton (producers) is quite negligible as compared to that of crustaceans and small herbivorous fish that feed on these producers. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.
   

Question 14.
How do pesticides cause Bioaccumulation and Biomagnification?
(OR)
What are the effects of pesticides on environment?
Answer:

1. Pesticides are the toxic chemicals used to destroy pest and insects which damage our crops and stored foods.
2. These pesticides vary in their length of life as toxic materials.
3. Some of the pesticides are degradable that can be broken down into harmless substances in a comparatively short time and others are non-degradable.
4. Non-degradable pesticides accumulate in the bodies of animal and pass right through food web.
5. Thus the pesticides cause bioaccumulation.
6. These accumulated pesticides concentrate as they move from one tropic level to the next, thus leads to biomagnification.

Question 15.
List out some human activities which altered the communities of plants and animals in their natural ecosystem.
Answer:

1. Industrialization
2. Damming rivers
3. Draining marshes
4. Re-claiming land from the sea
5. Cutting down forests
6. Using chemical fertilisers and pesticides
7. Building towns, cities, canals and motor ways.

Question 16.
What kind of changes may come in 2m ecosystem due to development of a large town?
Answer:
The following changes are expected due to development of a large town.

1. Some plants and animal species will die out.
2. Some will adapt to the new conditions sufficiently to survive in reduced numbers.
3. Some will benefit by the new conditions and will increase in numbers.

Question 17.
Write a comparative note on pyramid of number and pyramid of biomass.
(OR)
Write the differences between pyramid of number and pyramid of biomass.
Answer:

Question 18.
Write a comparative note on pyramid of biomass and pyramid of energy.
(OR)
What are the differences between pyramid of biomass and pyramid of energy?
Answer:

Question 19.
Write a comparative note on pyramid of number and pyramid of energy?
(OR)
What are the differences between pyramid of number and pyramid of energy?
Answer:

Question 20.
What is ecological efficiency? Write a short notes on Ten per cent law?
Answer:
Ecological efficiency: The ratio between energy flows at different tropic levels along the food chain expressed as percentage is called “ecological efficiency”.
Ten per cent law:

1. The amount of energy transferred decreases with successive tropic levels.
2. Slobodkin (1959) suggested that the transfer of energy from one tropic level to the next is of the order of 10% and this is called “Gross ecological efficiency”.

Question 21.
Geetha said “a given species may occupy more than one tropic level in the same ecosystem at the same time”. Do you support her or not? Explain your answer with example.
Answer:
I support her for the following reason.

1. A snake eating a mouse in a field or lawn occupies the third tropic level.
   Plant → Mouse → Snake
2. When the snake eats a frog in the same field, it occupies the fourth tropic level in a food chain. It is because the frog feed on some of the insects that depend on the plants.
   Plant → Insect → Frog → Snake
3. Thus, a given species may occupy more than one tropic level in the same ecosystem at the same time.
4. This is to satisfy its food needs, as it cannot do so by occupying one tropic level.

Question 22.
If we introduce a man into a forest ecosystem, at which level of food chain we will place him? Explain your answer.
Answer:

1. If we introduce a man into a forest ecosystem, he can fit for any level of consumers of food chain.
2. He may feed on plant parts such as fruits. Then we can place him at primary consumer level.
   Plant → Man
3. He may feed on some of the herbivorous organisms such as rabbit, then we can place him at secondary consumers level.
   Plant → Rabbit → Man
4. He may also feed on some of the carnivorous organisms such as insectivorous birds then we can place him at tertiary consumers level.
   Plant → Insect → Bird → Man
5. This is possible to place him at any level of consumers, as he is an omnivore, who feed on both plant originated and animal originated foods.

Question 23.
Draw the ecological pyramids for the given food chain.
Banyan Tree → Herbivorous birds → Carnivorous birds.
Answer:
1) Pyramid of number

2) Pyramid of biomass

3) Pyramid of energy

Question 24.
Write briefly about Minamata disease.
Answer:

1. Minamata disease was first discovered in Minamata city in Kumamoto prefecture, Japan, in 1956.
2. It was caused by the release of methyl mercury in the industrial waste water from the Chisso corporation’s chemical factory, which continued from 1932 to 1968.
3. This highly toxic chemical bioaccumulated in shellfish and fish in Minimata Bay and the Shiranui Sea, which, when eaten by the local populace, resulted in mercury poisoning.
4. While cat, dog, pig and humans death continued for 36 years.

Question 25.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:

1. Trophic levels is the feeding position in a food chain.
2. It is the functional level occupied by an organism in a food chain.
3. Examples of trophic levels include ‘herbivores’ and ‘decomposers’
4. An example of food chain depicting various trophic levels is as follows:
   

Question 26.
What is the role of decomposers in the ecosystem? (OR)
How decomposers help in cleaning the environment?
Answer:

1. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.
2. Eventually the world would run out of carbon dioxide or nitrate or phosphate or other inorganisms material essential for life.
3. The decomposers breakdown the organic waste products and dead remains of organisms into the inorganic substances needed by the producers.
4. Most decomposition is carried out by saprophytic fungi, by bacteria and by invertebrates.

10th Class Biology 9th Lesson Our Environment 4 Marks Important Questions and Answers

Question 1.
Prepare some slogans about ‘Vanam – Manam’ programme to display in your school rally?
Answer:

1. Save paper – Save trees
2. Plant a tree – Plant a life
3. Saving trees is our duty
4. Think green – Go green
5. If we protect plants – they protects us
6. Conserve plants – Conserve life
7. Plant a tree – get the air free
8. Plant a tree – Reduce the pollution.
9. Tree on – Global warming gone.
10. If cut a tree – It kills a life.

Question 2.
Read the information about Kolleru lake in the given table and answer the following questions.

a) In which year, lake water spread area is more?
b) Why do you think weeds are more in the lake?
c) Guess the reasons for decrease in the lake area.
d) What measures are to be taken to control pollution in the lake?
Answer:
a) In the year 1967
b) Addition of excessive nutrients from aquaculture ponds and rice fields.
c) Aquaculture ponds, Ricefields and Encroachment are the reasons for decrease in lake area.
d) Anthropogenic activities are to be controlled in the lake catchment area. (Or)
Fish ponds are to be removed in the lake catchment area. (Or)
Agricultural practices in the lake area should be minimised as per the norms of government.

Question 3.
Observe the following pyramid of biomass and answer the following questions.
a) This pyramid shows a decrease in the biomass as we move up, why the biomass is decreasing?
Answer:
The pyramid of biomass for the given food chain, at each step 90% of the food is lost. That means 1000 kg of phytoplankton to produce 100 kg of Zooplankton to form 10 kg of fish to produce 1kg of human tissues. The fewer the steps in the food chain the more energy will be for the species at the top.

b) Give some examples of producers and primary consumers.
Answer:
Examples for producers: Plants, Grass, Diatoms.
Examples for primary consumers: Grasshopper, Rabbit, Deer,

c) Where do producers get the energy from?
Answer:
From the sun.

d) How much biomass is lost at each step?
Answer:
90%

Question 4.
Observe the following diagram and answer the following questions.
i) Name the primary producers in the given food web.
ii) Prepare any one food – chain from the diagram.
iii) What are the tertiary consumers?
iv) Write names of any two herbivores.
Answer:
i) Plants, Grass, Trees Phytoplanktons etc.
ii) Grass → Rabbit → Fox → Tiger
iii) Tiger, Vulture, Crane, Owl, Peacock etc.
(OR)
The animals which are at 4th trophic level in a food chain are called as Tertiary Consumers.
iv) Rabbit, Deer, Goat, Cow

Question 5.
What is number pyramid? What does it indicate?
Answer:

1. The number of organisms in a food chain can be represented graphically in a pyramid of number.
2. Each bar represents the number of individuals at each tropic level in a food chain.
3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size but decrease in number.
4. For example in a wood, the aphids are very small and occur in astronomical numbers.
5. The lady birds which feed on them are distinctly larger and not so numerous.
6. The insectivorous birds which feed on the lady birds are larger still and are only present in a small number and there may only be a single pair of hawks of much larger size than the insectivorous birds on which they prey.

Question 6.
Draw the diagram of number pyramid keeping foxes as third consumers. What are the consequences if their number increases?
Answer:

1. If the number of foxes increases, then the competition for food will be very severe and less amount of food will be available for them.
   
2. As a result some of the foxes may not get enough food and die due to starvation.
3. This reduces the population of foxes and very few foxes will be left in the forest.
4. This increases the chances of survival of secondary consumers birds, hence their number increases.
5. This increases the availability of food for foxes. Very soon a balance will be established between the number of secondary consumers and foxes.

Question 7.
What reasons are responsible for decrease in number of top carnivores and biomass starting production in a food chain?
(OR)
Why the number of organisms get decreased as we move from producers to consumer levels?
Answer:

1. In a food chain as we move from producers to different levels of consumers the energy available will decrease gradually.
2. Only ten percent of the energy present in one tropic level transfer to another tropic level.
3. Biomass also decreases gradually as only 10 – 20% of the biomass is transferred from one tropic level to the next in a food chain.
4. As there is less energy of less biomass available at top levels, number of organisms also less generally.
5. So the number of organisms get decreased as we move from producers to different level of consumer.

Question 8.
Show food chain of different organisms, number of pyramid of your school.
Answer:
Food chain of different organisms in our school:
Plant → aphids → spiders thirds.

1. The pyramid of organisms in a food chain can be represented graphically in a pyramid of number.
2. Each bar represents the number of individuals at each tropic level in a food chain.
3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size, but decrease in number.

Question 9.
What determines the terrestrial ecosystems on the earth?
Answer:

1. The terrestrial ecosystems on the earth are being determined largely by the variations in climatic conditions between the poles and equator.
2. The main climatic influences which determine these ecosystems are rainfall, temperature and availability of light from the sun.
3. For instance, forests are usually associated with high rainfall, but the type Is influenced by temperature and light.
4. The same applies to deserts which occur in regions where rainfall is extremely low.
5. Thus, the climatic conditions along the horizontal climatic regions determined the terrestrial ecosystems on the earth.
6. If we move from equatorial region to the polar region, we can come across tropical rain forests, savannah, deciduous forest, coniferous forests and then tundras respectively.
7. Similarly altitude of the place is also a determining factor.
8. If we climb a mountain such as Kilimanjaro in equatorial Africa, we can go through a comparable system of ecosystems, starting with tropical rain forest at the base and ending with perpetual snow and ice at the summit.

Question 10.
“All the energy in the ecosystem is ultimately derived from sunlight.” Justify.
Answer:

1. All the organisms in an ecosystem derive energy from food.
2. The food by its nature is the chemical energy and by in its stored form, it is the potential energy.
3. In an ecosystem, all the consumers at any level depend upon producers for their food either directly or indirectly.
4. The producers in any ecosystem are nothing but photosynthetic organisms such as plants, phytoplanktons and photosynthetic bacteria.
5. Energy enters the producers in the ecosystem from the sun in the form of solar energy during photosynthesis.
6. From the producers, the chemical energy passes to the consumers from one tropic level to the next through food.
7. For example, in a grassland ecosystem, grass traps the solar energy and stores in its body.
8. When this grass is eaten and assimilated by insects this stored energy enters into the body of insects.
9. From the insects it will pass to frog, from them to snake and so on to eagle.
10. Thus, all the energy in the ecosystem is ultimately derived from sunlight.

Question 11.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:

1. The tendency of pollutants to concentrate as they move from one trophic level to the next trophic level is known as Biomagnification.
2. Plants absorb pesticides, heavy metals from the soil.
3. The primary consumers when eat these plants the remaining of pesticides and heavy metals enter their bodies.
4. As these chemicals are not degradable, they accumulate in the bodies of organisms of all trophic levels in the food chain.
5. Most of the plants products which we eat are grown in fields in which pesticides and fertilisers have been used.
6. These are absorbed by the plants and cannot be removed by washing or other means.
7. Human beings are at the top level of the food chain these chemicals get accumulated in our bodies and cause various disorders.
8. Levels of biological magnification would increase as the trophic level increases.

Question 12.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels ? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

1. If we remove producers from ecosystem, herbivores will not survive and the entire ecosystem collapse.
2. Removing herbivores result in increase number of producers and carnivores would not get food.
3. Removing carnivores result in increase of herbivores to unsustainable levels.
4. If we remove decomposers from ecosystem waste material and animal dead remains would pile up and nutrients would not be available to the producers.
5. Some or the other damage would be caused to the ecosystem if the organisms of any trophic level is removed.
6. However impact of removing producers or decomposers would be serve as the whole ecosystem would collapse.
7. Without plants sun’s energy cannot be converted to chemical energy which is the basis of life on earth.
8. Without decomposers the nutrients cannot be recycled and made available to producers.

Question 13.
Every organism has got the right to live on this planet. Write slogans to motivate the people on preservation of biodiversity.
Answer:

1. Live and let live.
2. Conserve nature – conserve life.
3. Clean the environment, live happily.
4. Think eco-friendly and live eco-friendly.
5. If we protect the environment, it protect us.
6. Reduce pollution – conserve the biodiversity.

Question 14.
Write some friendly ecosystem activities you will conduct in your school.
Answer:

1. Forming eco-clubs: These clubs consists of student representatives from each class. They will take up the eco-friendly activities and encourage the people of that village to follow environment friendly activities.
2. Setting up garden at school: This ensures the school and its premises green through planting of flowering plants, vegetables and fruit trees. It is a symbol of biodiversity because various plants and animals inhabit the garden.
3. Electricity conservation programme: To save energy the school implements certain hours to be switched off habit. This switching off programme for one hour from 3.30 p.m. to 4.30 p.m. help conserve electricity in every classroom.
4. Pollution prevention programme: A ‘no burning of trash policy should be implemented in the school. Waste materials are recycled and properly disposed to ensure a clean, waste-free environment.
5. Making compost by organic wastes: By digging a pit at one corner of the school and throwing the organic waste particularly of mid day meal waste into pit and covering with soil layers prepares compost which can be used as manure for plants. This creates a clean environment in the school.
6. Using cloth bags instead of polythene bags by pupil.
7. Collection of solid waste materials and proper management of its helps in reducing soil pollution.
8. Children should be encouraged to follow ‘3R’ system i.e. Reduce, Re use and Recycle different substances.

Question 15.
What is Ecological pyramid? Describe different types of Ecological pyramids.
Answer:

1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called ecological pyramid.
   
2. There are three types of ecological pyramids. They are
   1) Pyramid of number, 2) Pyramid of biomass and 3) Pyramid of energy.
3. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
4. The shape of this pyramid varies from ecosystem to ecosystem.
5. In forest ecosystem the pyramid structure is partly upright and in parasitic food chain is inverted.
6. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (bio mass) at each trophic level in a given ecosystem.
7. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores hence the pyramid structure is upright whereas in aquatic ecosystem it is inverted.
8. Pyramid of energy is a graphical representation designed to show the quantity of energy present at each tropic level in a given ecosystem. The pyramid of energy is always upright.

Question 16.
Collect information regarding pesticides commonly used in your area and prepare a chart showing pesticide and common name and on which crop and pest it is commonly used.
Answer:

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 10 Natural Resources.

AP State Syllabus SSC 10th Class Biology Important Questions 10th Lesson Our Environment

10th Class Biology 10th Lesson Natural Resources 1 Mark Important Questions and Answers

Question 1.
Recently a new programme was launched in our state known as “Vanam – Manain”. Prepare any two slogans to promote the programme.
Answer:
a) Save forest, forest will save you.
b) Grow the plants and get the fresh air.

Question 2.
Suggest any two practices suitable to farmers with less water resources.
Answer:

1. Construction of percolation tanks (or) Soak pits
2. Irrigation techniques like drip irrigation and usage of sprinklers.

Question 3.
Why should we conserve forests? Give two reasons.
Answer:
a) Forests serve as lungs for the world. They purify the air and protect the earth from greenhouse effect and global warming,
b) Forests are rich habitats for plants

Question 4.
Ravi observed symbol on the plastic water bottle purchased by him. What does this symbol indicate? and animals.
Answer:
The symbol on the plastic bottle indicates that the bottle is made from recycled plastic and after its use it can be recycled.

Question 5.
Write any two suggestions for the conservation of biodiversity at your village.
Answer:

1. Protecting and preserving the natural habitats of birds and animals.
2. Replace the wood products with alternative sources.
3. Using Recycled products and following the 4’R Principle in day to day life.

Question 6.
Suggest any two activities to check soil erosion in your school.
Answer:

1. Observe the school ground after the rain.
2. Conduct a field project on soil erosion.

Question 7.
To create awareness on “Water conservation” in your locality, what slogan you will suggest?
Prepare two slogans on ‘Save Water’ propaganda.
Answer:
“Don’t Waste Water”.
“Save every drop”.
“Water is life”.

Question 8.
The symbol is there on the item you bought. What it instructs? (OR)
What does the given logo indicate? What does it mean?

Answer:
It is the Recycle logo. It indicates that the item we bought is prepared from recycled materials or the item can be recycled after use.

Question 9.
What happens if the forest area decreases rapidly?
Answer:
If the forest area decreases

1. It destroys wildlife habitat.
2. It increases soil erosion.
3. It releases green house gases into the atmosphere contributing to global warming.
4. It also harms people who relay on forest for their survival, hunting and gathering, harvesting forest products or using timber and firewood.

Question 10.
Write two activities which you are performing to save electricity.
(OR)
Write any two measures vou take in your home to reduce the consumption of electricity.
Answer:

1. We can reduce the consumption of electricity by putting off the fans and lights when there is no need.
2. We can use LED (Lighting Emitting Device) bulbs to save electricity.
3. To shut down laptops and computers when they are not in use.

Question 11.
Prepare two slogans on protecting non-renewable resources.
Answer:

1. Use Biofuel – Reduce Fossil Fuel.
2. Use alternative resources – Save the environment.

Question 12.
Write two examples for non-renewable resources.
Answer:
Examples for non-renewable resources are coal, petroleum and natural gas.

Question 13.
What is sustainable development? Is it needful for us?
Answer:
When we use the environment In ways that ensure we have resources for the future, It Is called sustainable development. It Is needed because development and conservation can coexist In harmony.

Question 14.
What are examples for natural resources?
Answer:
Examples for natural resources are water, soil, forests, flora, fauna, etc.

Question 15.
What are percolation tanks?
Answer:
Percolation tanks are normally earthen dams with masonry structures where water may overflow.

Question 16.
What are Kharif crops?
Answer:
Crops grown In the rainy season are termed as Kharif crops, e.g: Paddy, maize, millet and cotton crops.

Question 17.
What are Rabi crops?
Answer:
The crops that are grown only in winter season are generally called Rabi crops, e.g.: Wheat, Gram and Mustard.

Question 18.
What is the average fall of ground water level in Andhra Pradesh state during the period of 1998 – 2002?
Answer:
The average fall of ground water level In Andhra Pradesh state during the period of 1998 – 2002 Is 3 meters.

Question 19.
Which agency in villages of Warangal district helped in recharging wells that were being dried up?
Answer:
Centre for water solidarity (Secundrabad, T.S.) helped In recharging wells that were drying up In the villages.

Question 20.
Give examples for micro irrigation techniques.
Answer:
Drip irrigation, sprinklers are the examples for micro irrigation techniques.

Question 21.
Mow did the boundaries between the villages were fixed in ancient times?
Answer:
In ancient times village boundaries were decided upon a water shed (Land between water sources usually of two rivers or streams) basis fixed at the common point of the drainage system In between two villages by the expert farmers In the village.

Question 22.
Expand ICRISAT.
Answer:
International Crop Research Institute for Semi-Arid Tropics.

Question 23.
What is the other name for Sri Rama Sagar Project?
Answer:
Sri Rama Sagar Project also known as the Pochampadu project on the Godavari river,

Question 24.
What is the use qf planting Gliricidia on field bunds?
Answer:
Planting Gliricidia on field bunds help In strengthen them and make the soli nitrogen-rich.

Question 25.
What is the micro irrigation system that can reduce water consumption by 70%?
Answer:
Drip irrigation can reduce water consumption by 70%.

Question 26.
Who predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity ?
Answer:
The Food and Agriculture Organisation (FAO) of the united nations has predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity.

Question 27.
What happens if we use resources wisely?
Answer:
If resources are used wisely and efficiently they will last much longer. Through conservation people can reduce waste and manage natural resources wisely.

Question 28.
Give an example of country where restrictions on water usage were imposed.
Answer:
In Australia restrictions were imposed on activities like, watering lawns by using sprinkler systems, washing vehicles, using house pipes to clean paved areas, and refilling swimming pools.

Question 29.
Why are the natural resources used up quickly?
Answer:
The population of human beings has grown enormously in the past two centuries. Billions of people use up resources quickly as they eat food, build houses, produce goods and burn fuel for transportation and electricity.

Question 30.
What happens if we damage a forest resource?
Answer:
Harm to animals that may be forced to find new habitats. If we damage a forest resource indiscriminately the depletion of resources occur and we may have to face problem for water and timber in future.

Question 31.
What are the results of deforestation?
Answer:
Deforestation destroys wild life habitats and increases soil erosion and also releases green house gases into atmosphere, contributing to global warming.

Question 32.
How the people in China and Mexico recycle paper? (OR)
Give an example of recycling paper by the people. What is the use of recycling paper?
Answer:
People in China and Mexico reuse much of their waste paper, including writing paper, wrapping paper and card board.

Question 33.
How the soil is important for us ? How the soil is importane for us?
Answer:
Soil is vital to food production and also important to plants that grow in the wild.

Question 34.
What are the reasons for depletion of nutrients in soil?
Answer:
Poor farming methods, such as repeatedly planting the same type of crop in the same place cause depletion of nutrients in the soil.

Question 35.
What is biodiversity?
Answer:
Biodiversity is the variety of living things that populate the Earth.

Question 36.
How are people speeding up the loss of biodiversity?
Answer:
Through hunting, pollution, habitat destruction, people are speeding up the loss of biodiversity.

Question 37.
How many plant species are being used by us for medicines world wide?
Answer:
We use between 50,000 to 70,000 plant species for medicines world wide.

Question 38.
What is selective harvesting?
Answer:
The practice or removing individual plants or small groups of plants leaving other plants standing to anchor the soil is called selective harvesting.

Question 39.
How are fossil fuels produced?
Answer:
The fuels that are produced from the remains of ancient plants and animals are called fossil fuels. They include coal, petroleum and natural gas.

Question 40.
What are the alternate sources of energy?
Answer:
The alternate sources of energy are sun, wind and water.

Question 41.
What are the other products made from petroleum?
Answer:
Plastic, synthetic, rubber, fabrics like nylon, medicines, cosmetics, waxes, cleaning products, medical devices, etc., are the other products made from petroleum.

Question 42.
Which plant’s seeds are used for the production of bio-fuel?
Answer:
Seeds from the Jatropa Curcas plant are used for the production of bio-fuel.

Question 43.
How does the mining method, Mountain Top Removal mining (MTR) devastate the environment?
Answer:
The mining method Mountain Top Removal mining devastate the environment. They destroy soil, plants and animal habitats.

Question 44.
In which country car manufacturers recycle many raw materials used in making automobiles?
Answer:
In Japan car manufacturers recycle many raw materials used in making automobiles.

Question 45.
In which country nearly one third of the iron produced comes from recycled automobiles?
A. In the United States, nearly one-third of the iron produced comes from recycled automobiles.

Question 46.
What does the Indian tradition teach us?
Answer:
The Indian tradition teaches us that all forms of life – human, animal and plant are so closely inter linked that disturbance of one gives rise to imbalance in the other.

Question 47.
Expand IUCN.
Answer:
IUCN stands for International Union for the Conservation of Nature.

Question 48.
How is IUCN planning to protect wild life and habitats?
Answer:
IUCN monitors the status of endangered wild life, threatened national parks and preserves.

Question 49.
What are the four R’s to save the environment?
Answer:
Reduce, Reuse, Recycle, Recover are the four R’s to save the environment.

Question 50.
How did Amritha Devi and her daughters protest against the king’s order?
Answer:

1. Amritha Devi and her daughters, followed by villagers, who clung to trees in the forest surrounding their village and laid down their lives to save them.
2. They protested against the king’s order to collect wood for the construction of his palace.

Question 51.
Write a method of soil conservation.
Answer:
One soil conservating method is called contour strip cropping. Several crops such as corn, wheat and clover are planted to alternating strips across a slope or across the path of the prevailing wind.

Question 52.
What is the rate of extinction by the estimation of the scientists?
Answer:

1. Scientists estimate that the current rate of extinction is 1,000 times the natural rate through hunting, pollution, habitat destruction.
2. Based on various estimates of the number of species on Earth, we could be losing anywhere from 200 to 1,00,000 species each year.

Question 53.
What is the need to protect biodiversity?
Answer:
We need to protect biodiversity to ensure plentiful and varied food sources. Biodiversity is important for more than just food because many plant species are being used for medicines.

Question 54.
Mention two ways in which water harvesting can be undertaken?
Answer:
The two ways by which water harvesting can be undertaken are

1. Capturing run off water from, rooftops.
2. Capturing run off water from local catchments.

Question 55.
On the basis of the issues raised in the chapter management of natural resources, what changes you in corporate in your lifestyle in a move towards a sustainable use of our resources?
Answer:
I would incorporate the maximum of four R’s i.e., reduce, recycle, reuse and recover in my lifestyle in a move towards a sustainable use of our resources.

10th Class Biology 10th Lesson Natural Resources 2 Marks Important Questions and Answers

Question 1.
Rahul remarked that different human activities are responsible for global warming.
What might be the reasons for his statement?
Answer:

1. Deforestation.
2. Industrialisation and urbanization.
3. Conversion of agriculture lands into residential areas.
4. Home appliances like A/C, refrigerators, vehicle pollution.
5. Population explosion.

Question 2.
What steps do you take to improve natural resources?
Answer:

1. Motivate the people to conserve water.
2. Try to avoid wastage of water whenever possible.
3. Plantation in the vacant lands.
4. Educating the farmer regarding proper utilization of water for irrigation.
5. Encourage the people to recycle the water wherever possible.

Question 3.
Proper utilisation of natural resources is the way to show gratitude to our nation.
Can you support this statement? Give your argument.
Answer:

1. Natural resources of a nation influence its economical and social development.
2. Natural resources are freely available in nature and help in many activities and development of people.
3. The generation of natural resources take a lot of time.
4. They disappear by indiscriminate usage.
5. So proper utilization of natural resources is the way to show gratitude to our nation.

Question 4.
The humans are utilising natural resources indiscriminately. These resources are decreasing more rapidly. Guess what will be the consequences in future?
Answer:
Indiscriminate usage of natural resources causes the following consequences.

1. Reduction in rainfall
2. Drought will occur.
3. Atmospheric temperature becomes increase.
4. The rare species become extinct.

Question 5.
Write any four slogans on the conservation of natural resources.
Answer:
Slogans:

1. Waste water today – live in desert tomorrow
2. Practice eco-friendly methods.
3. Use natural resources judiciously.
4. Save nature – Save future.

Question 6.
There is an increase in the atmospheric temperature year by year. If it continues, guess and write what would be the consequences?
Answer:
If the temperature on earth increases, the consequences would be as follows.
a) All the glaciers and the frozen ice in the polar region start melting leading to rise in the sea water levels.
b) It results in the submergence of low lying coastal areas throughout the world. Millions of people of those areas would lost their homes.
c) Changes in rainfall patterns take place and it result in the occurance of droughts and decrease in crop production.
d) Global warming results in climate change which cause the breakout of climatic sensitive diseases like Malaria, Dengue, Diarrhoea, etc.

Question 7.
There is water scarcity in Ravi’s village during summer. He wants to conduct a rally to create awareness regarding conservation of water. Write any four slogans required to conduct this rally.
Answer:

1. Water is life.
2. Save water – Save a life.
3. Today’s rain water is tomorrow’s life saver.
4. No matter your occupation, water conservation is your obligation.

Question 8.
What steps you take to conserve the biofuels in your daily life?
Answer:

1. Development and usage of alternative energy resources in place of bio-fuels.
2. Minimise the usage of bio-fuels whenever possible.
3. Use public transport, ride by bicycle and walking regularly.
4. Use and purchase energy efficient appliances to save bio-fuels.

Question 9.
Why do we use fossil fuels judiciously?
Answer:

1. Fossil fuels were produced from the remains of ancient plants and animals.
2. They include coal, petroleum (oil) and natural gas.
3. We need to use fossil fuels judiciously because they are non – renewable resources.
4. We need to conserve fossil fuels so we don’t run out of them.
5. The pollution caused by them when burnt, to limit our fossil fuel use.
6. Future generations may not get these resources.
7. Balance in the nature will be disturbed.
8. Electricity production will be stopped.
9. Vehicles running with fossil fuels become useless.

Question 10.
Write two suggestions to create awareness on groundwater conservation.
Answer:

1. We need to adapt different methods to Improve the quality and increase the quantity of groundwater.
2. We should dig water harvesting pits for every house.
3. We should clean the silt, mud fill in tanks and ponds.
4. We should prohibit the establishment of borewells for extraction of groundwater for agricultural and Industrial use.
5. These measures will improve quality and quantity of groundwater.

Question 11.
What is the importance of 4R’s in achieving the goal of “Swachh Bharat”?
Answer:

1. Reduce the production of garbage.
2. Reuse the garbage for the production of manure and electricity.
3. Recycle the garbage by separating It as dry and wet garbage.
4. Recover the plants.

Question 12.
Suggest four measures to conserve fossil fuels.
Answer:
Measures to conserve fossil fuels:

1. Usage of alternatives to fossil fuel.
2. Minimise the usage of fossil fuel.
3. Walk, ride by bicycle and use public transportation whenever possible.
4. Purchase energy efficient appliances.
5. Turn off light and other electronics when you are not using them.

Question 13.
The indiscriminate digging of Borewells may result in what type of consequences in future?
Answer:

1. Due to over drilling of borewells and pulling out water by electric motors, the ground water level Is decreasing day by day.
2. It Is goes on without recharging, ground water becomes scarce.
3. It shows impact on agriculture and the productivity will decrease.
4. Fluorine levels In ground water will increase.
5. Sometimes, saline water may intrude Into the interior places of land and water becomes unfit for consumption.
6. Farmers have to drill the bore wells to more depths which Increase the losses for them.

Question 14.
Ramaiah made broad bed furrow around his field under employment guarantee scheme. Guess the reasons for if. If all the farmers of your village work together, will their water scarcity meet?
Answer:
The reason for Ramaiah making broad bed furrow around his field was, it is useful to conserve soil and water, fertilizer application weeding operations. It also conserves rain water.

The farmers are over coming the water scarcity by sharing water available in the village. They formed groups of farmer including large and small ones who would use the same water resource. Farmers were also motivated to use irrigation techniques like drip irrigation.

Question 15.
What are renewable sources and non-renewable resources?
Answer:
Renewable resources: Resources that can be replaced after they are used are called renewable resources.
Ex: Air, water and soil.
Non-renewable resources: Some other resources, cannot be replaced at all: Once they are used up they are gone forever and are called non renewable resources.
Ex: Coal, Petroleum, Natural gas (fossil fuels).

Question 16.
How do people waste natural resources?
Answer:

1. People often waste natural resources.
2. Animals are over hunted, forests are cleared, exposing land to wind and water damage.
3. Fertile soil is exhausted and lost to erosion because of poor farming practices.
4. Fuel supplies are depleted.
5. Water and air are polluted.
6. Water resources is indiscriminately used for crop growth.

Question 17.
How do people use the forest resources differently?
Answer:

1. The need to conserve resources often conflicts with other needs.
2. For some people, a forest area may be a good place to put a farm.
3. A timber company may want to harvest the area’s trees for construction materials.
4. A business company may want to build a factory or a shopping mall on the land.

Question 18.
What are die effects of deforestation?
Answer:

1. Deforestation destroys wild life habitats and increases soil erosion.
2. It also releases green house gases into the atmosphere, contributing to global warming,
3. Deforestation accounts for 15 per cent of the world’s green house gas emissions.
4. Deforestation also harms the people who rely on forests for their survival, hunting and gathering, harvesting forest products, or using the timber for firewood.

Question 19.
In your opinion What are the causes for soil erosion?
Answer:

1. Soil erosion is caused by poor farming methods such as repeatedly planting the same type of crop in the same place.
2. These methods deplete nutrients in the soil.
3. Soil erosion Is also caused by water and wind currents.
4. When farmers plough up and down hills, soil erosion occurs.
5. Overgrazing by cattle also causes soil erosion.
6. Natural floods causes the extensive damage to the top layer of the soil.

Question 20.
What is Biodiversity? Explain.
(OR)
What is the importance of biodiversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. The products and benefits we get from nature rely on biodiversity.
3. We need to protect biodiversity to ensure plentiful and varied food sources.
4. Biodiversity is important for more than just food. For instance we use between 50,000 to 70,000 plant species for medicines world wide.

Question 21.
How can we use the fossil fuels carefully?
Answer:
We can use the fossil fuels carefully by taking the following measures.

1. Turn off lights and other electronics when we are not using them.
2. Purchase energy-efficient appliances.
3. Walk, ride a bicycle, if the distance is less.
4. Use public transportation whenever possible.
5. It is better to prefer public transport system like bus or train, instead of travel in personal vehicles.

Question 22.
Why the prices of aluminium and iron are expensive?
Answer:
Earth’s supply of raw material resources is in danger. Many mineral deposits that have been located and mapped have been depleted. As the ores for minerals like aluminium and iron become harder to find and extract, their prices go up.
This makes tools and machinery more expensive to purchase and operate.

Question 23.
What are the effects of mining?
Answer:

1. Many mining methods such as Mountain Top Removal mining (MTR) devastate the environment.
2. They destroy soil, plants and animal habitats.
3. Many mining methods also pollute water and air, as toxic chemicals leak into the surrounding ecosystem.

Question 24.
What did Smt. Indira Gandhi said, while launching the world conservation strategy in India on 6th March 1980?
Answer:
“The interest in conservation is not a sentimental one but the discovery of a truth well known to our ancient stages. The Indian tradition teaches us that all forms of life- human, animal and plant – are so closely inter-linked that disturbance in one gives rise to imbalance in the other” said by Smt. Indira Gandhi.

Question 25.
What are the steps taken by the government to conserve resources?
Answer:

1. Government enacts laws defining how land should be used and which areas should be set aside as parks and wild life preserves.
2. The government enforces laws designed to protect the environment from pollution, such as requiring factories to install pollution control devices and also provide incentives for conserving resources.

Question 26.
What is the necessity of sustainable management of natural resources? Out of the two methods reuse and recycle which one would you suggest to practice and why?
Answer:

1. Sustainable management of natural resources is necessary to Increase the over all life of natural resources specially non renewable resources and also to control the environmental pollution.
2. Both reuse and recycle are the good choice.
3. Reuse: If we reuse something then the cost of recycle will be saved.
4. Recycle: It is not necessary that each and everything can be reused, so after getting recycled the life of the resource will be enhanced.

Question 27.
“Burning fossil fuels is a cause of global warming”. Justify this statement?
Answer:

1. Fossil fuels are composed of carbon, hydrogen, nitrogen and sulphur.
2. When these are burnt they produce CO₂, H₂O, Oxides of Nitrogen and Sulphur.
3. Incomplete combustion of fossil fuels produces green house gases such as CO₂,
4. If huge amount of fossil fuels are burnt, It would produce high amount of CO₂ resulting intense global warming.

Question 28.
Can you suggest some changes in your school which would make it environment friendly?
Answer:
The changes that would make my school environment friendly are

1. Save energy by turning off lights that we are not using.
2. I will suggest to buy recycled paper for decoration and other purposes.
3. Use writing paper on both the sides.
4. Growing trees and plants all around the play ground.

Question 29.
What is the necessity of replenishment of forest? State four reasons.
Answer:
The replenishment of forest is necessary because of the following reasons.

1. It is used to conserve soil.
2. It provides shelter to wild animals.
3. It reduces atmospheric pollution.
4. It controls flood and increases frequency of rainfall.

10th Class Biology 10th Lesson Natural Resources 4 Marks Important Questions and Answers

Question 1.
Forest is renewable resource. But, each year, the Earth loses about 36 million acres of forest. In this type of situation, what suggestions do you give to save forests from turning into non-renewable resources ?
Answer:

1. Forests are the lungs of the world. So I will suggest the following measures to save forests from turning into non-renewable resources.
2. Sustainable forestry practices for ensuing resources into the future.
3. Low impact logging practices, harvesting with natural regeneration in mind. Prevention of removing all the high value trees or all the largest trees from the forests. Recycling methods should be adopted.
4. Replace wood products with alternative sources.
5. Preventing forest fires.
6. Implementing methods like agro forestry, social forestry crop rotation, green plantation, etc. are essential.

Question 2.
What are four R’s? Explain how they help to conserve the environment?
(OR)
Write about the 4 ‘R’s needed for the protection and conservation of environment.
Answer:
By pursuing the maximum of four R’s i.e., Reduce, Reuse, Recycle and Recover, we can save the environment in an effective way.

1. Reduce : It means to use less, I would save electricity by switching off unnecessary lights and fans, prefer walking or cycling than using a vechicle, turn off the engine of car at red light, repair leaky taps and would not waste food.
2. Recycle: It means to collect used things like plastic, paper, galss and metal items and recycle these materials to make required things instead of synthesising or extracting fresh plastic, paper, glass or metal.
3. Reuse: It refers to use things again and again. For example instead of throwing away used envelops, they can be used by pasting new labels.
4. Recover: We should implement ‘recover’ to prevent environmental threat. For example when we cut trees to construct industries or roads for transportation, it is important to grow trees in another areas.

Question 3.
What steps you would like to follow on your part to conserve bio-diversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. To conserve biodiversity we should avoid hunting.
3. Sustainable forest conservation methods should be followed.
4. I will actively participate Vana Mahosthavam programmes.
5. I will educate and encourage people and make them participate in conservation programmes.
6. Create awareness programmes in and around school.
7. Writing slogans and also make some posters about conservation of biodiversity
8. Judicious use of electricity wherever possible.
9. Finding out of various alternative sources of energy.
10. Plant the saplings in the habitat.
11. Encouraging of social forestry.

Question 4.
Observe the pie diagram showing water resources available in our state for agriculture and answer the given questions.
A) Which water resource is using more for agriculture?
B) What are the consequences of excess utilization of underground water?
C) Which water resource should be utilized for agriculture?
D) What are the alternative ways to increase the underground water resources?

Answer:
A) Ground water.
B) Underground water table will be depleted and scarcity of drinking water will arise.
C)

1. Tanks should be constructed to harvest with rain water.
2. Projects should be constructed across the rivers to store water that can be utilized for agriculture.

D)

1. Construction of rain water storage structures on large scale.
2. Constructing soaking or percolation pits.
3. Contour field bunding.
4. Recharge of wells by building dykes or barriers in the nalla.
5. Plantation in waste lands.
6. Adapting micro irrigation techniques.
   (Any two points you can write)

Question 5.
Forests are renewable resource. Write four sentences supporting this.
(OR)
“Forest is a renewable resource”. Do you agree? Justify.
Answer:

1. Forests are rich habitat for plants and animals. Forests serve as lungs for the world and a bed of nutrients for new fife to prosper.
2. Forest’s pure air protects the earth from green house effect by removing carbon dioxide from the atmosphere and converts it into oxygen.
3. Many fruits, medicines, dyes, sandle wood and bamboo is obtained from forest by local people.
4. Forest provide employment to large number of people and also help in generating revenue.

Question 6.
Observe the above table and answer the following questions.

1. Which crop is most suitable to cultivate for small farmer in both the villages?
Answer:
Cotton, paddy

2. If you are a large farmer, which crop do you select to cultivate?
Answer:
Cotton, paddy, mirchi

3. What similarities you have identified in village A and village B?
Answer:
Small and large fanners cultivated same type of crop in both villages. Large farmer gets more income per acre on crops than small farmer in both the villages.

4. Which is the lowest income crop ?
Answer:
Mirchi.

5. Is there any relationship between production of crops and income ? How ?
Answer:
Commercial crops are good for income. Income may or may not related to production of crop. It depends upon demand of the market.

Question 7.
Read the given information and answer the following questions.

A survey was conducted in two villages, Vanaparthy and Vaddicherla of Warangal district in Telangana State. The first with no scarcity and the second with scarce groundwater. Well census was carried out in the villages in order to get a complete picture of well irrigation and its status as well as availability of water. There are no alternative sources of supply as against wells in Vaddicherla, where there is an existing tank that has been converted into a percolation tank, so that the water situation is much better in Vanaparthy.

i) Why did they conduct survey?
Answer:
A compartive study on available water resources irigation method in the Vaddicharla and Wanaparthi of Warangal Dist of Telangana State.

ii) What are irrigation resources in Telangana State?
Answer:
Lakes, wells, canals and ground water etc…,

iii) In which village, do you suggest drip irrigation?
Answer:
Vaddicherla.

iv) Why the water situation is much better in Vanaparthy village compared to Vaddicherla?
Answer:
Existing tank has been converted into a percolation tank.

Question 8.
Observe the Pie diagram and answer the following questions.
i) Identify the fossil fuels from the above diagram.
Answer:
Coal, natural gas, oils are fossil fuels.

ii) Why wastes should be considered as primary energy source in future?
Answer:
The fossil fuels may be exhausted in future. So we may be considered that wastes are primary alternative energy resources.

iii) Why can’t we depend on fossil fuels forever?
Answer:
We can’t depend on fossil fuels forever because fossil fuels are non-renewable resources.

iv) What are the alternatives for fossil fuels?
Answer:
Solar energy, wind energy, tide energy, nuclear energy, energy from waste materials.

Question 9.
Explain the importance and implementation of community based interventions and farmer based interventions for water management.
Answer:
Community based interventions:

1. For water harvesting, there is an urgent need to construct earthen and masonry dams. They help us to store rain water during rainy seasons. They are help in increasing the ground water table.
2. Construction of percolation pits and field bunding are very helpful in the harvesting every rain drop.
3. Open dry wells near nalla canal were recharged by building dykes or barriers in the nalla and maintaining the run – off rain water. The ground water is recharged by these community based interventions.
   Farmer based interventions:
4. Broad Bed Furrow (BBF) land form and contour planting methods are very useful to conserve soil, water and fertilizer application and weeding operations.
5. Planting Gliricidia, a leguminous plant adapted to grow in dry areas on field bunds to strengthen them and make the soil nitrogen rich.
6. Farmers were encouraged to use water resource jointly and irrigate land using micro irrigation methods like sprinklers and drip irrigation.

Question 10.
Explain the farmer based and community based interventions to conserve soil and water resources.
Answer:

Question 11.
“The humans who were developed by using the natural resources, today has become the reason for destroying them”. Explain analytically.
Answer:
“The humans who were developed by using the natural resources, today has become the reason for destroying them” – This statement is absolutely true.

1. Primitive man lived in forests and hills. He used the natural resources for his livelyhood. He worshipped nature and used them wisely for his development.
2. After his development, he becomes greedy and using the natural resources indiscriminately and held responsible for their destruction.
3. To meet the needs of growing population, industrialization, urbanization, and huge constructive activities, man utilised natural resources Indiscriminately. At the same time, he did not planned for their revival.
4. But now he realised the importance of natural resources and taken up steps for their conservation. The concept of “Sustainable development” is being implemented in natural resource management.
5. He focussed on development of alternatives for fossil fuels, conservation of water and soil at community level and farmer based interventions.
6. Now he is so keen on conserving forests, wild life and biodiversity.
7. He is so cautious in minimising the utilization of natural resources by following 4’R principle in the day to day life [R – Reduce, R – Reuse, R – Recycle, R – Recover]
8. Now, he is adopting micro-irrigation methods like sprinklers and drip Irrigation to minimise the water usage in low water available areas.
9. He is very interested in following eco-friendly techniques, natural farming methods, using biofertilizers, vermicompost and natural pest control methods in place of toxic chemical pesticides.

Question 12.
The wells and tanks in your village become dry. Ground water levels decreased. Assume the causes for this. Will there be no water scarcity if all the farmers of your village work collectively?
Answer:
Causes for decreasing ground water levels:

1. Varying monsoon behaviour in recent years, there is a pressure on ground water utilization.
2. Indiscriminate tapping of ground water in our village by too much drilling and construction of deep tube wells and bore wells have resulted in over exploitation and depletion of ground water resources.
3. There will be no water scarcity if all the farmers of our village work collectively. Farmers in our village were encouraged to use water resource jointly and irrigate land using micro irrigation techniques. By using micro irrigation techniques farmers in our village obtained more crop yield. Farmers in our village follow the micro irrigation method i.e. drip irrigation and can reduce water consumption by 70% in our village.

Question 13.
Whom do you meet to collect the information of the methods of farmer based, community based water management? Prepare information table to note down your observation.
Answer:
I will meet officials of International Crop Research Institute for Semi – Arid Tropics (ICRISAT) located at Hyderabad to collect information of the methods of farmer based and community based water management.
I also collect information from Central Research Institute for Dry Land Agriculture (CRIDA), National Remote Sensing Agency (NRSA), District Water Management Agency (DWMA) and M Venkatarangaiah Foundation (MVF) and NGO.
The information I gathered from these institutions is summarised below.
Information table:
For Information table See Q.No. 10 in 4 Marks.

Question 14.
Think that there is much scarcity of water for drinking and cultivation in your village. What advice do you give to prevent this?
(OR)
How do you overcome the problem of water scarcity in your village?
Answer:

1. Motivate the people to conserve water.
2. I will educate the people to avoid wastage of water whenever possible.
3. Construction of recharge pits in the house, school and in the open areas to increase the underground water level.
4. Planting trees wherever possible in the village particularly in the vacant lands.
5. Educate the farmers about the micro irrigation system like drip irrigation, sprin¬klers, etc.
6. Encourage the farmers to form groups to share available water among themselves.
7. Construction of percolation tanks in the low lying areas of the village.

Question 15.
What type of fossil fuels are used in your house? What measures do you take to conserve them?
Answer:
Fossil fuels are sources of energy for cooking, heating and burning in our households. Petrol and diesel are being used in our house for transport and running generators and water pumps.
Measures to be taken to conserve fossil fuels in my house :

1. I will put the food material to be cooked on the stove only after arranging all the things which are necessary for cooking.
2. By using pressure cookers 20% gas on rice and 41.5% on meat would be saved when compared to Other cooking means.
3. We must reduce the flame as soon as the boiling process starts in a pressure cooker. This process saves nearly 35% of fuel.
4. I will soak the food material before cooking. It saves 22% of fuel.
5. I will cook food in broad and low depth vessel.
6. I will keep lid on the cooking vessel. If not, it takes more time to cook.
7. For short distances to travel I will go by walk to save fuel for longer distance. I use public transport.
8. Encourage people to use solar water heater and solar cooker.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 1 Nutrition – Food Supplying System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 1st Lesson Nutrition – Food Supplying System

10th Class Biology 1st Lesson Nutrition – Food Supplying System Textbook Questions and Answers

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Question 1.
Write differences between
(a) Autotrophic nutrition – Heterotrophic nutrition:
Answer:

(b) Ingestion – Digestion :
Answer:

(c) Light reaction – Dark reaction :
(OR)
Differentiate the reactions that take place in presence of light and the reactions which do not require light in photosynthesis.
Answer:

(d) Chlorophyll – Chloroplast:
Answer:

Question 2.
Give reasons.
a) Why photosynthesis is considered as the basic energy source for most of living world?
(OR)
Why can we say that photosynthesis is the basic energy source for the living world?
Answer:

1. All living organisms constantly need energy to be alive.
2. They get energy from the food they eat.
3. The food directly or indirectly comes from the green plants through photosynthesis.
4. Hence photosynthesis can be considered as the basic energy source for most of living world.

b) Why is it better to call the dark phase of photosynthesis as a light independent phase?
Answer:

1. The term dark reaction or light independent does not mean that they occur when it is dark at night.
2. It only means that the reactions are not depend on light.
3. Hence we call the dark phase of photosynthesis as a light independent phase.

c) Why is it necessary to destarch a plant before performing any experiment on photosynthesis?
Answer:
1) To get better results, it is necessary to destarch a plant before performing any experiment on photosynthesis.
2) Because if starch is present it may interfere with the result of the experiment.

d) Why is it not possible to demonstrate respiration in green plants kept in sunlight?
Answer:

1. We cannot demonstrate an experiment of respiration in green plants kept in sunlight.
2. Because if sunlight is present, the C02 produced in respiration will be used in photosynthesis.
3. So we must conduct an experiment on respiration in green plants in a dark room.

Question 3.
Give examples.
a) Digestive enzymes
Answer:
The digestive enzymes are:

1. Salivary Amylase (Ptyalin),
2. Pepsin,
3. Trypsin,
4. Lipase,
5. Peptidases,
6. Sucrase,
7. Amylase (Pancreatic juice)

b) Organisms having heterotrophic nutrition is seen in organisms like:
Answer:
Heterotrophic nutrition is seen in organisms like:

1. All animals and human beings.
2. Some protozoans Ex: Amoeba.
3. Some parasitic plants Ex: Cuscuta
4. Saprophytes Ex: Bread moulds, yeast, mushrooms, etc.

c) Vitamins
Answer:
Water soluble vitamins:
B complex (B₁) Thiamine, (B₂) Riboflavin, (B₃) Niacin, (B₆) Pyridoxine,
(B₁₂) Cyanocobalamine, Folic acid, Pantothenic acid, Biotin, (C) Ascorbic Acid.
Fat soluble: (A) Retinol, (D) Calciferol, (E) Tocoferol, (K) Phylloquinine.

d) Nutritional deficiency diseases
Answer:
Eg: Kwashiorkor, Marasmus etc.

Question 4.
Where do plants get each of the raw materials required for photosynthesis?
Answer:

Question 5.
Explain the necessary conditions for autotrophic nutrition and what are its by products.
Answer:
A. Necessary conditions:

1. Autotrophic nutrition takes place through the process of photosynthesis.
2. Carbon dioxide, water, chlorophyll pigment and sunlight are the necessary conditions required for autotrophic nutrition.
3. The rate of photosynthesis depends on availability of sunlight.

B. By products:

1. Photosynthesis is the main process for autotrophic nutrition.
2. Carbohydrates and oxygen are the by products of photosynthesis.

Question 6.
With the help of chemical equation explain the process of photosynthesis In detail with the help of a flow chart.
Answer:
Process of photosynthesis:

1. The chemical equation representing the process of photosynthesis is
   
2. Definition: Photosynthesis is a photochemical reaction during carbohydrates are formed using carbon dioxide and water in the chloroplasts of the green plants in the presence of sunlight.
3. CO₂ water, sunlight and chlorophyll are the requirements of photosynthesis.
4. Glucose, O and water are the end products of the reaction.
5. Photosynthesis have two phases.
   1) Light reaction 2) Dark reaction
6. Light reaction have three steps, i) Oxidation of chlorophyll ii) Photolysis iii) Formation of ATP, NADPH and O₂
7. In dark reaction CO₂ is utilized and finally glucose is formed which is converted and stored as starch.

FLOW CHART:

Question 7.
Name the three end products of photosynthesis.
Answer:
Glucose, oxygen and water are the three end products of photosynthesis.

Question 8.
What is the connecting substance between light reaction and dark reaction?
Answer:
The hydrogen of NADPH present in the stroma is the connecting substance between light reaction and dark reaction.

Question 9.
Most leaves have the upper surface is more green and shiny than the lower ones. Why?
(OR)
In most of leaves the upper surface will be more green and shiny than the lower surface. Why?
Answer:

1. The upper surface comprising of the palisade parenchyma.
2. The lower surface comprising of the spongy parenchyma.
3. Palisade parenchyma contains more number of chloroplasts than the spongy parenchyma.
4. Thus the upper surface is more green and shiny than the lower ones.

Question 10.
Explain the structure of chloroplast with a neatly labelled sketch.
(OR)
Explain the structure of a chloroplast with the help of a rough diagram.
Answer:

1. Chloroplast is a membranous structure consisting of 3 membranes.
2. The third layer forms stacked sac like structures called granum.
3. The intermediatery fluid filled colourless portion is called stroma.
4. It is responsible for enzymatic reaction leading to the synthesis of glucose in plants.
5. Substances found in chloroplast, capture sunlight are called photosynthetic pigments.
6. Chlorophyll pigment contain one atom of magnesium.
7. Two major kinds of chlorophylls are associated with thyakoid membranes.
8. Chlorophyll-a is blue-green in colour and chlorophyll-b is yellow-green colour.
9.  Around 250-400 pigments molecules are grouped as light harvesting complex units in granum.
10. Some of the events occur in chloroplast are :
    a) Conversion of light energy to chemical energy.
    b) Splitting of water molecule.
    c) Reduction of carbondioxide to carbohydrates.

Question 11.
What is the role of acid in stomach?
Answer:

1. The internal walls of stomach has number of gastric glands. They secret gastric juice.
2. It contains HCl and enzymes.
3. HCl kills the bacteria present in the food and protects us from their harmful effects.
4. And also denatures the proteins so that enzymes can act easily on them.

Question 12.
What is the function of digestive enzyme?
Answer:

1. The function of digestive enzyme is to increase the process of breaking up of complex molecules into simpler and absorb molecules.
2. This makes easy for the body to absorb food.

Question 13.
How is the small intestine designed to absorb digested food? Explain.
(OR)
How is food absorbed by villi in small intestine?
Answer:

1. Small intestine is the largest part in digestive system.
2. Absorption is its main function including last stage of digestion.
3. The inner surface of small intestine has millions of tiny finger-like projections called villi.
4. Due to the presence of villi, the absorbing surface area of small intestine increases.
5. And the large surface area of small intestine helps in the rapid absorption of digested food.
6. The digested food which is absorbed through the walls of the small intestine goes into our blood.
7. Long and folding structure increase the ability of small intestine.

Question 14.
How are fats digested in our bodies? Where does this process take place?
What is emulsification? How it helps in digestion of fats? (OR)
How are fats digested? Where do they get digested?
Answer:

1. Bile juice and lipase enzymes helps in fat digestion.
2. Bile juice is secreted by liver.
3. Fats are digested by converting them into small globules like forms by the help of the bile juice.
4. This process is called emulsification.
5. Lipase enzyme is secreted by pancreas.
6. It converts emulsified fats into fatty acids and glycerol.
7. 
8. This process takes place in duodenum and small intestine.

Question 15.
What is the role of saliva in the digestion of food ?
(OR)
How does saliva digest food ?
Answer:

1. Saliva is secreted by three pairs of salivary glands present in the mouth.
2. Human saliva contains an enzyme called amylase (ptyalin).
3. It converts starch into maltose (a sugar).
4. The food is mixed thoroughly with saliva and becomes wet and slippery.
5. Saliva helps in the smooth passage of food in the food pipe.

Question 16.
What will happen to protein digestion as the medium of intestine is gradually rendered alkaline ?
Answer:

1. The food coming from the stomach to intestine is acidic in nature.
2. Bile and pancreatic juices render the internal condition of the intestine gradually to a basic or alkaline one.
3. Protein digestion continues even if the medium of intestine is gradually changed to alkaline.
4. In the alkaline medium pancreatic enzyme trypsin can act on the food and digests the proteins.
   
5. The enzymes present in the intestinal juice like peptidases complete the digestion of proteins into amino acids.
   

Question 17.
What is the role of roughages in the alimentary tract?
Answer:

1. Roughages are the fibres of either carbohydrates or proteins.
2. Plenty of roughages in the diet avoid constipation.
3. Roughages help in the easy movement of faeces in the large intestine.
4. They help in the easy digestion of food and keep the alimentary canal clean and healthy.

Question 18.
What is malnutrition? Explain some nutrition deficiency diseases.
Answer:
Malnutrition: Eating of food that does not have one or more than one nutrients in required amount is known as malnutrition.
Malnutrition is of three types:

1. Calorie malnutrition,
2. Protein malnutrition,
3. Protein calorie malnutrition.

Nutrition deficiency diseases:

1. Kwashiorkor disease: This is due to protein deficiency in diet.
   Symptoms:
   i) Body parts becomes swollen due to accumulation of water in the intercellular spaces,
   ii) Very poor muscle development,
   iii) Swollen legs,
   iv) Fluffy face,
   v) difficult to eat,
   vi) diarrhoea,
   vii) Dry skin.
2. Marasmus: This is due to deficiency of both protein and calories. Generally this disease occurs when there is an immediate pregnancy or repeated child births.
   Symptoms:
   i) Lean and weak,
   ii) Swelling in joints of limbs,
   iii) Less developed muscles,
   iv) Dry skin,
   v) diarrhoea.

Question 19.
How do non-green plants such as fungi and bacteria obtain their nourishment?
Answer:

1. Bacteria and fungi are non-green plants. So they cannot prepare their own food materials.
2. They are saprophytes which feed on dead and decaying plant and animal bodies.
3. The fungi and bacteria breakdown the complex organic molecules present in dead and decaying matter by releasing chemical substances into simple substances out¬side the body.
4. These simpler substances are then absorbed by fungi and bacteria as their food.

Question 20.
If we keep on increasing CO₂ concentration in the air, what will be the rate of photosynthesis?
Answer:

1. If the CO₂ concentration in the air increases, the rate of photosynthesis also increases.
2. If the CO₂ concentration raises above 5% then the rate of photosynthesis reduces.
3. At certain CO₂ concentration the rate of photosynthesis is constant.
4. Here a rise in CO₂ levels has no affect on the rate of photosynthesis as the other factors such as light intensity become limited.

Question 21.
What happens to plant if the rate of respiration becomes more than the rate of photosynthesis ?
Answer:

1. Respiration is a catabolic (destructive) process and photosynthesis is an anabolic (constructive) process.
2. If the rate of respiration becomes more than the rate of photosynthesis, the amount of food oxidised will be more than the food produced.
3. This affects the growth and development of plants and may even results in the death of the plant.

Question 22.
Why do you think that carbohydrates are not digested in the stomach?
(OR)
Where are carbohydrates digested in alimentary canal?
Answer:

1. For the digestion of carbohydrates enzyme ptyalin or amylase are required.
2. The gastric juice produced by stomach do not contain the enzyme ptyalin or amylase, it contains only pepsin which digests proteins.
3. Hence carbohydrates are not digested in the stomach.
4. Carbohydrates are partially digested in the mouth and completely in small intestine.

Question 23.
What process do you follow in your laboratory to study the presence of starch in leaves?
(OR)
(Activity – 1)
How do you test the presence of starch in leaves ? (OR)
Mention the materials required and explain the experiment to prove the presence of starch in leaves. What inference do you draw from this experiment?
Answer:
Aim: To study the presence of starch in leaves.
Apparatus: Beaker, test tube, bunsen burner, tripod stand, asbestos gauze, ethanol, leaf, petridish, iodine solution.
Procedure:

1. Select a leaf of a potted plant with soft thin leaves.
2. Boil the leaf in methylated spirit over a water bath till it becomes pale white due to the removal of chlorophyll.
3. Observe the leaf.
   
4. Spread the leaf in a dish and add a few drops of tincture iodine / betadine solution on it. Again observe the leaf.

Observation: The presence of starch will be indicated by a blue-black colour in leaf. Result: The experiment proves that starch is present in leaves. It is formed by Photo-synthesis.
Precautions:

1. Do not boil the methylated spirit test tube directly on flame.
2. Boil the water bath with low flame.

Question 24.
How would you demonstrate that green plants release oxygen when exposed to light? (OR) (Lab Activity)
Write the experimental procedure to prove that oxygen is produced during photosynthesis in the presence of light. (OR)
What materials are required to prove that oxygen is produced during photosynthesis in the presence of light? What procedure we need to follow to perform the above experiment?
We have conducted experiment that prove the release of oxygen when photosynthesis happens?
i) What are the plants used for this experiment? Where do they grow?
ii) How did you conduct the above experiment? In which context large number of air bubbles released? Do you noticed?
Answer:
i) Hydrilla plants are used for this experiment they grow in water.
ii) Experiment to demonstrate the release of oxygen during photosynthesis.
Aim: To prove that oxygen is produced during photosynthesis by hydrilla funnel experiment.
Apparatus: Beaker with water, test tube, funnel, hydrilla twigs, glowing splinter.
Procedure:

1. Arrange the apparatus as shown in the figure.
2. Place some water plant hydrilla in a beaker containing pond water, and cover these by a short stemmed funnel.
3. Invert a test – tube full of water over the stem of the funnel.
4. Ensure that the level of water in the beaker is above the level of stem of the inverted funnel.
5. Place the apparatus in the sun for at least 2 or 3 hours.
6. After sometime it is observed that gas bubbles come from the hydrilla plant. These bubbles are collected at the end of the test tube pushing the water into the beaker.
7. After sufficient gas is collected test – tube is taken out of the beaker carefully by closing it with thumb.

Observation: Test the gas in the test – tube by inserting a glowing incense stick which would burst into flames. This shows the presence of oxygen.
Result: This shows that oxygen is produced during photosynthesis.
Precautions:

1. Funnel should be smaller than the beaker.
2. Necessary care is to be taken while removing the test tube from the stem of the funnel.

Question 25.
Collect information from your primary health centre of malnutrition child at various ages and make a table your own and display in the classroom.
Answer:

Question 26.
If there were no green plants, all life on the earth would come to an end ! Comment.
(OR)
The survival of organisms would become difficult, if there are no green plants on the earth. How do you support?
Answer:

1. Plants play the most important part in the cycle of nature.
2. Without plants there could be no life on earth.
3. Plants are the only organisms that can make their own food and all other living beings directly or indirectly depend on plants for their food.
4. Moreover plants release oxygen into the atmosphere through photosynthesis.
5. Oxygen is essential for the organisms to respire.
6. Hence without green plants, all life on the earth would come to an end.

Question 27.
Draw a neat labelled diagram of chloroplast found in leaf, and its role in photosynthesis.
Answer:
Role of Chloroplast in photosynthesis:

1. Chloroplasts trap solar energy.
2. They convert that solar energy into chemical energy.
3. They help in the formation of glucose.

Question 28.
Draw the label diagram of human digestive system. List out the parts where peristalsis takes place.
Answer:
Parts where peristalsis takes place: Oesophagus, stomach, small intestine and large intestine.

Question 29.
Raheem prepared a model showing the passage of the food through different parts of the alimentary canal. Observe this and label its parts.
Answer:

Question 30.
Observe the following diagram and write a note on light dependent, light independent reactions.
Answer:
Note on light dependent reactions:

1. Light dependent reactions are also called as photochemical phase.
2. The light dependent reaction takes place in chlorophyll containing thylakoids called grana of chloroplasts.
3. Several steps occur in the light dependent reaction.
4. Step – 1: The chlorophyll on exposure to light energy becomes activated by absorbing photons.
5. Step – II: The energy is used in splitting the water molecule into two component ions named hydrogen (H⁺), hydroxyl ion (OH^–). This reaction is known as photolysis.
6. Step – III: OH^– ions through a series of steps produce water (H₂O) and O₂.
7. The end products of light reaction are ATP, NADPH and O₂.

Note on dark reaction or light independent reaction:

1. In light independent phase the hydrogen of the NADPH is used to combine it with CO₂, by utilizing ATP energy and to produce glucose.
2. This synthesis occurs in a number of steps using certain special intermediate compounds (mainly RUBP – Ribulose hi phosphate) and enzymes. Finally glucose is converted to starch.
3. All these reactions occur in the stroma region of the chloroplast.

Question 31.
Almost all the living world depends on plants for food material. How do you appreciate the process of making food by the green plants?
(OR)
What facts about the green plants do you appreciate?
Answer:
Leaf is a wonderful machine to synthesize food:

1. The leaf is the important site of photosynthesis and is called as food factory of the plant.
2. This plant organ can be treated as a wonderful natural machine which converts solar energy into useful chemical energy.
3. With all his scientific knowledge and technical skills, man has not produced anything similar leaf for utilization of solar energy without polluting the atmosphere.
4. This machine provides food and supports the life by providing oxygen for all the organisms including man on this planet.
5. Nature has given us such a wonderful machine free !!

Question 32.
Even a hard solid food also becomes smooth slurry in the digestive system by the enzymes released at a particular time. This mechanism is an amazing fact. Prepare a cartoon on it.
Answer:

Question 33.
What are good food habits?
Answer:
The food habits I am going to follow after reading this chapter are:

1. I take balanced diet which contains proper amounts of carbohydrates, proteins, fats, vitamins and minerals.
2. I avoid taking food containing high proportion of fat.
3. I eat food as much required by my body. I do not over eat.
4. I will not eat rich meals over several days.
5. I eat simple balanced meals, eat it leisurely and thoroughly masticating the food.
6. I avoid doing violent exercise soon after eating food.
7. I empty the bowels regularly avoiding constipation.
8. I will see to have plenty of roughages in the diet.

(OR)
After reading the chapter nutrition, I would like to follow the following food habits.

1. Having simple, well balanced meals.
2. Eating them in a leisurely manner.
3. Thoroughly masticating the food.
4. Avoiding strenuous exercise soon after eating food.
5. Drinking plenty of water and having regular bowel movement.
6. Decreasing consumption of coffee or tea per day.
7. Taking leafy vegetables at least 3 times a week and taking of fruits and vegetables plenty.
8. Maintaining regular timings for daily food consumption.

Fill in the blanks.

1. The food synthesized by the plant is stored as ———–.
2. ———– are the sites of photosynthesis.
3. Pancreatic juice contains enzymes for carrying the process of digestion of ———– and ———–.
4. The finger-like projections which increases the surface area in small intestine are called ———–.
5. The gastric juice contains ———– acid.
6. ———– vitamin is synthesized by bacteria present in intestine.

Answer:

1. carbohydrates
2. Chloroplasts
3. proteins, fats
4. Villi
5. HCl
6. Cyanocobalamin

Choose the correct answer.

1. Which of the following plant take the food by parasitic nutrition? [ ]
   A) Yeast
   B) Mushrooms
   C) Cuscuta
   D) Leeches
   Answer: C & D
2. The rate of photosynthesis is not affected by [ ]
   A) Light intensity
   B) Humidity
   C) Temperature
   D) Carbon dioxide concentration
   Answer: B
3. A plant is kept in dark cupboard for about forty eight hours before conducting any experiment on photosynthesis in order to [ ]
   A) Remove chlorophyll from leaves
   B) Remove starch from leaves
   C) Ensure that no photosynthesis occurred
   D) Ensure that leaves are free from the starch
   Answer: B
4. The digestive juice without enzyme is [ ]
   A) Bile
   B) Gastric juice
   C) Pancreatic juice
   D) Saliva
   Answer: A
5. In single-celled animals, the food is taken by [ ]
   A) the entire body surface
   B) mouth
   C) teeth
   D) vacuoles
   Answer: A
6. Which part of the plant takes in carbon dioxide from the air for photosynthesis? [ ]
   A) Root hair
   B) Stomata
   C) Leaf veins
   D) Sepals
   Answer: B

10th Class Biology 1st Lesson Nutrition – Food Supplying System Activities

Activity – 1

How do you prove experimentally that carbon dioxide is necessary for photosynthesis by Mohl’s half leaf experiment?
(OR)
List out the materials required and the procedure to be followed to prove that ‘carbon dioxide’ is essential for photosynthesis.
(OR)
You know that the factors like CO₂, Light and Chlorophyll are essential for photosynthesis. Write any one of experiment related to the factors essential for photosynthesis.
Answer:
Aim:
To prove-that carbon dioxide is essential for photosynthesis by Mohl’s half leaf experiment.
Apparatus:
Wide mouthed transparent bottle, KOH solution, potted plant, vertically split cork, Iodine solution.
Procedure:
Arrange the apparatus as shown in the figure.

1. Take a healthy potted plant and keep it in the dark for nearly a week for the removal of starch from the leaves.
2. A wide mouthed transparent bottle is taken.
3. Put potassium hydroxide pellets or potassium hydroxide solution (KOH) in the bottle.
4. This KOH absorbs CO₂ present in the bottle.
5. Insert splitted cork in the mouth of the bottle.
6. Insert one of the leaves of destarched plant through a split cork into transparent bottle.
7. Arrange half of the leaf is inside bottle and the remaining half outside.
8. Leave the plant in the sunlight for 2-3 hours.
9. After a few hours, test this leaf and other leaf of this plant for starch.

Observation :

1. The part of the leaf outside the bottle turns blue-black because starch is formed in this part due to photosynthesis.
2. The part of the leaf inside the bottle does not turn blue-black because the carbon dioxide present inside the bottle is absorbed by potassium hydroxide solution.
3. All the other factors water, sunlight and chlorophyll are available but not CO₂. Hence starch is not formed in the leaf part which is inside the bottle.

Result: This experiment proves that CO₂ is necessary for photosynthesis. Precautions:

1. The part of the leaf kept inside the bottle should not touch potassium hydroxide solution.
2. The apparatus should be kept air tight by applying grease or vaseline.

Activity – 2

Sunlight is necessary to form starch in green leaves.
(OR)
Write the materials required and the procedure to prove that light is essential for Photosynthesis.
(OR)
Write the procedure, precautions and observations in the lab activity, “Sunlight is necessary for photosynthesis”.
Answer:
Aim:
To prove that light is necessary for photosynthesis to form starch.
Apparatus:
Potted plant, light screen, iodine solution.

1. Keep potted plant in dark for one week to remove starch.
2. Take one black paper and cut it with your own design.
3. Keep design paper properly on the both sides with the help of clips.
4. Ensure that light does not pass through the covered area with black paper.
5. Keep the arranged apparatus at sunlight available area.
6. After few hours of exposure to bright sunlight detach the leaf.
7. Boil the leaf in methylated spirit over water bath. It becomes pale white due to the removal of chlorophyll. Take the leaf from test tube and spread the leaf in a petridish.
8. Add few drops of Iodine on leaf. The parts of the leaf, which could get light through the cut out design, turns blue-black colour.
9. The parts of the leaf which could not get light are not turned into blue – black colour.
   

Observation:
It is observed that only the parts of the leaf, which could get light through the cut out design, turn blue black, showing the presence of starch.

Result:
This experiment proves that light is necessary to form starch in the process of photosynthesis.

Activity – 3

Demonstrate litmus paper test on salivary amylase in the mouth.
Answer:

1. Before taking food into the mouth, take a litmus indicator paper and touch it to the tongue.
2. We observe no colour change in litmus paper.
3. Perform the litmus test again after chewing the food and swallowing it.
4. The red litmus paper turns to blue colour.
5. The blue litmus paper do not turns to red colour.
6. This demonstrates that amylase converts complex carbohydrates to simple sugar.
7. Amylase is alkaline in nature. This turns litmus paper blue when touches glucose at the second time.
   

Activity – 4

Observe different digestive enzymes and their role in digesting food in a tabular form.
Answer:

10th Class Biology 1st Lesson Nutrition – Food Supplying System InText Questions and Answers

10th Class Biology Textbook Page No. 2

Question 1.
Can you think of some raw materials needed for photosynthesis?
Answer:
Yes. Photosynthesis needs the following raw materials.

1. Sunlight, CO₂ water are external factors.
2. Chlorophyll and enzymes are internal factors.

Question 2.
What could be the end products of the process of photosynthesis?
Answer:
Glucose, water and oxygen are the end products of photosynthesis.

10th Class Biology Textbook Page No. 4

Question 3.
Do you think solar energy transforms into chemical energy by the process of photosynthesis?
Answer:
Yes, solar energy transforms into chemical energy by the process of photosynthesis.

Question 4.
What are the materials that you think would be essential for the synthesis of carbohydrates in the process of photosynthesis?
Answer:
The materials essential for the synthesis of carbohydrates in the process of photosynthesis are carbon dioxide, water, sunlight and chlorophyll.

Question 5.
Do you think the equation tells us about all the materials involved?
Answer:
Yes, the materials which are essential for photosynthesis and the products formed are involved in the equation.

10th Class Biology Textbook Page No. 5

Question 6.
What had Priestly done to introduce the mint plant without disturbing the experimental setup?
Answer:
Priestly should have tilted the bell jar to one side and introduced the mint plant without disturbing the experimental set up.

Question 7.
How did Priestly light the candle from outside?
Answer:
Priestly might have used convex lens through which beam of sun rays can light the candle from outside or he might have used long burning stick, to light the candle by lifting the jar partially.

Question 8.
Do you find any relationship between candle, rat, mint plant? Discuss.
Answer:
Priestly’s experiment confirmed that gaseous exchange was going on and plants were giving out a gas that supported burning and was essential for the survival of animals.
By combustion process candle releases carbondioxide. By respiration process rat also releases carbondioxide. During photosynthesis process mint plant uses this carbondioxide and releases oxygen. This oxygen will be used by rat to stay alive and for the candle to burn.
So there is a relationship between respiration and photosynthesis by candle, rat and mint plant.

10th Class Biology Textbook Page No. 6

Question 9.
Why was the plant kept in dark and then in sunlight?
Answer:

1. The plant is kept in the dark for nearly a week to remove the starch from the leaves.
2. Then only we can understand that the starch is formed in the leaves or not after the experiment when the plant is kept in the sunlight.

Question 10.
Why did we study two leaves in the Mohl’s half leaf experiment?
Answer:

1. To test CO₂ is essential or not for photosynthesis, two leaves are used in the experiment.
2. One leaf with the plant and another one used in the experiment.
3. The leaf which is exposed to the atmospheric air becomes bluish-black. It proves that starch is prepared in the leaf by using CO₂ from atmosphere.
4. The leaf inside the flask containing potassium hydroxide, which absorbs CO₂ present in the bottle does not become bluish black. It shows that CO₂ is necessary for photosynthesis.

10th Class Biology Textbook Page No. 7

Question 11.
What precautions do you need while removing test tube from the beaker? Discuss with your teacher.
Answer:

1. When sufficient gas is collected lift the test tube carefully from the beaker by closing its mouth with the thumb.
2. Because of that the gas present in the test tube cannot escape into the atmosphere.

10th Class Biology Textbook Page No. 8

Question 12.
Which part of leaf turns blue black? What about the remaining part?
Answer:

1. The part of the leaf, which could get light through the cut design turns to blue black showing the presence of starch.
2. The remaining part of the leaf which did not get light, do not turn blue, indicating that starch is not prepared.

Question 13.
Observe the colour of the leaf stained with iodine. Can you tell why it is stained differently?
Answer:

1. Some parts of the leaf prepared starch.
2. Some parts of the leaf does not prepared starch.
3. So it is stained differently.

Question 14.
What about plants having coloured leaves?
Answer:
Plants having coloured leaves also carry out photosynthesis. The coloured leaves containing pigments pass on the energy of sunlight trapped by them to chlorophyll.

Question 15.
How is that new leaves which look dark red in colour in several plants turn green?
Answer:

1. The new leaves which look dark red in colour contain coloured chromoplasts.
2. As the leaf grows the chromoplasts turns to chloroplasts and the leaf appears green in colour.

Question 16.
Do plants having reddish or yellowish leaves also carry out photosynthesis?
Answer:

1. Yes. Plants having reddish or yellowish leaves also carry out photosynthesis.
2. The pigments present in reddish or yellowish leaves pass on the energy of sunlight trapped by them to chlorophyll.

Question 17.
What made plants carry out photosynthesis while even green coloured animals (like some birds) could not?
Answer:

1. Chlorophyll and other pigment molecules trap (harvest) solar energy, convert it into chemical energy in the thylakoid membranes of the chloroplast.
2. But animals having green colour on their body cannot trap solar energy and cannot perform photosynthesis. Photosynthesis is possible only in plants but not in animals.

10th Class Biology Textbook Page No. 9

Question 18.
Where is chlorophyll and other pigments present in the plant?
Answer:
Chlorophyll and other pigments are present in the grana thylakoids of chloroplast in leaf.

Question 19.
Do you think the new reddish leaves of plants also carry out photosynthesis? What could be the role of their colour?
Answer:

1. Yes. New reddish leaves of plants also carry out photosynthesis.
2. Chromoplasts are responsible for the reddish colour of leaves.
3. They also pass on the energy of sunlight which they trap to the photosystems.

10th Class Biology Textbook Page No. 10

Question 20.
What makes chloroplast appear completely different from other cell organelles?
Answer:

1. Substances found in chloroplast, which capture sunlight are called photosynthetic pigments.
2. Two major kinds of chlorophyll are associated with thylakoid membranes.
3. Chlorophyll – a (blue – green in colour) and chlorophyll – b (yellow – green); around 250 to 400 pigment molecules are grouped as light harvesting units in each granum.
4. Such innumerable units in chloroplasts make them appear completely different from other organelles.

10th Class Biology Textbook Page No. 13

Question 21.
What happens to the food once it enters our body?
Answer:

1. The food once enters our body it gets digested by various enzymes in different parts of alimentary canal.
2. Digestion starts in the mouth and it completes in the small intestine.
3. Finally it absorbed in the small intestine into the circulatory system.

10th Class Biology Textbook Page No. 15

Question 22.
Name the enzymes which act on carbohydrates.
Answer:
Ptyalin (salivary amylase), amylase and sucrose are the enzymes that act on carbohydrates.

10th Class Biology Textbook Page No. 16

Question 23.
What are the end products of fats?
Answer:
The end products of fats are fatty acids and glycerol.

Question 24.
What are the enzymes that act on proteins?
Answer:
Pepsin, Trypsin and Peptidases are the enzymes that act on proteins.

Question 25.
Which digestive juice contains no enzymes?
Answer:
Bile juice produced by liver contains no enzymes.

Question 26.
What do you think about the process of digestion?
Answer:

1. The process of digestion occurs in the alimentary canal or digestive system.
2. During the process of digestion large complex macro molecules present in the food are converted to simple and small molecules.
3. Digestion provides the food material properly absorbed by the body.

Question 27.
What are the major steps of digestion?
Answer:
The major steps of digestion are

1. Ingestion
2. Digestion
3. Absorption and
4. Defecation.

10th Class Biology Textbook Page No. 18

Question 28.
Collect information about pellagra and discuss with your teacher.
Answer:

1. Pellagra is a vitamin-deficient disease.
2. Niacin (Vitamin – B₃) is essential for the metabolism of carbohydrates, proteins and fats.
3. The resources for vitamin – B₃ are kidney, liver, meat, egg, fish, oil seeds and legumes.
4. Deficiency of niacin results in a disease called PELLAGRA.

Symptoms: Pellagra is described by 3 Ds – Diarrhoea, Dermatitis, Dementia.
A more comprehensive list of symptoms include

1. Sensitivity to sunlight
2. Aggression
3. Dermatitis
4. Alopecia (hair loss)
5. Edema (swelling)

Smooth, beefy red, glossitis, red skin lesions, insomnia (sleepless), weakness, mental confusion, nerve damage are the symptoms of this pellagra.

Prevention: By taking of yeast, meat, fish, milk, eggs, green vegetables, beans and cereal grains, we can prevent this disease.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 2 Respiration – The Energy Releasing System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 2nd Lesson Respiration – The Energy Releasing System

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System Textbook Questions and Answers

Improve your learning

Question 1.
Distinguish between
(a) Inspiration and Expiration
(b) Aerobic and Anaerobic respiration
(c) Respiration and Combustion
(d) Photosynthesis and Respiration
Answer:
(a) Inspiration and Expiration:

(b) Aerobic respiration and Anaerobic respiration:
(OR)
Respiration is energy-producing process in the organisms. It takes place both in the presence and absence of oxygen. Laxmi said there are some differences between the two processes. How do you support her?
Answer:

Aerobic respiration	Anaerobic respiration
1. It takes place in the presence of oxygen.	1. It takes place in the absence of oxygen.
2. In aerobic respiration, complete oxidation of glucose takes place.	2. In anaerobic respiration, the glucose molecule is incompletely oxidised.
3. End products are CO2 and water.	3. End products are either ethyl alcohol or lactic acid and CO2.
4. Lot of energy is liberated (38 ATP).	4. Relatively small energy is liberated (2 ATP).
5. It occurs in plant’s and animal’s cells.	5. Occurs in many anaerobic bacteria and human muscle cells.
6.	6.
7. It has two stages – Glycolysis and Krebs cycle.	7. It has two stages – Glycolysis and Fermentation.

(c) Respiration and Combustion:
(OR)
Even though both are oxidation processes, combustion and respiration are different in many aspects. Explain those differences.
(OR)
Combustion and respiration are oxidative processes but still there are differences between them. What are they?
(OR)
Write the differences between respiration and combustion.
Answer:

(d) Photosynthesis and Respiration:

Photosynthesis	Respiration
1. Occurs only in all plants and photo­synthetic bacteria.	1. Occurs in all living organisms.
2. Takes place in the presence of sunlight.	2. Takes place throughout day and night.
3. A plant can survive without performing photosynthesis for few days.	3. No organism can survive without respiration for few minutes even.
4. In plants, only few cells perform photosynthesis.	4. All living cells in an organism perform this process.
5. Raw materials are C02 and water.	5. Uses carbohydrates and oxygen.
6. Oxygen is liberated.	6. Carbon dioxide is released.
7. It occurs in chloroplast.	7. It takes place in cytoplasm and mitochondria.
8. Adds weight to the organism.	8. Decrease weight of the organism.
9. It is an anabolic process.	9. It is a catabolic process.
10.	10.

Question 2.
State two similarities between aerobic and anaerobic respiration.
Answer:
Similarities between aerobic and anaerobic respiration:

1. Both are catabolic processes.
2. Both aerobic and anaerobic respiration takes place in all cells.
3. Energy is released in both the processes.
4. CO₂ is the end product of both processes.
5. First stage of both respiration is glycolysis.
6. Respiratory substances in both processes are glucose, fatty acids and amino acids.

Question 3.
Food sometimes enters the wind pipe and causes choking. How does it happen?
Answer:

1. Pharynx is the common passage of both air and food.
2. From here air enters into trachea and food enters into oesophagus.
3. Pharynx is connected to larynx through glottis a slit like opening.
4. A cartilagenous flap called EPIGLOTTIS act as a lid over glottis and prevents food from entering into trachea during swallowing.
5. Any food particles enters the trachea it causes chocking.
6. Sometimes the food particles are forced back by cough.

Question 4.
Why does the rate of breathing increase while walking uphill at a normal pace in the mountains? Give two reasons.
Answer:
The rate of breathing increases while walking uphill at a normal pace in the mountains.

1. It is because as we go up the hill above sea level the concentration of oxygen is greatly reduced. So we have to breathe more to get required amount of oxygen.
2. While walking uphill a lot of oxygen is used by our body to release energy from glucose.
3. This leads to lack of oxygen in the cells.
4. We take in oxygen when we breathe.
5. Hence to increase the amount of oxygen intake there is an increase in breathing rate during walking uphill.

Question 5.
Air leaves the tiny sacs in the lungs to pass into capillaries. What modification is needed in the statement?
Answer:

1. Gaseous exchange takes place within the lungs by diffusion from the alveoli to blood capillaries and vice versa.
2. The carbondioxide in the blood is exchanged for oxygen in alveoli.
3. This sentence may be modified as “Air that contains oxygen reaches the tiny sacs in the lungs to pass into capillaries.

Question 6.
Plants photosynthesize during daytime and respire during the night. Do you agree to this statement? Why? Why not?
Answer:

1. No, I do not agree with this statement. Plants photosynthesize during daytime only and respire during the daytime as well as night time also.
2. During daytime when photosynthesis occurs in the presence of sunlight. Oxygen is produced. The leaves use some of this oxygen for respiration and the rest diffuses into air.
3. During daytime CO₂ produced by respiration is all used up in photosynthesis by leaves.
4. At night time no photosynthesis occurs and oxygen diffuses into leaves to carryout respiration.

Question 7.
Why does a deep sea diver carry oxygen cylinder on his/her back?
Answer:

1. When we go deep into the sea, the oxygen level decreases. Oxygen is in dissolved state in water.
2. Humans are adapted for utilizing oxygen in gaseous state. They cannot use dissolved oxygen for breathing.
3. Only aquatic animals like fish can utilize the dissolved oxygen for breathing using gills.
4. Human beings have lungs for respiration. Therefore, sea divers have to carry oxygen cylinders in their back so as to receive oxygen.
5. If they do not carry them, they do not get oxygen and there is a chance even to die.

Question 8.
How are alveoli designed to maximise the exchange of gases?
Answer:
The human lungs have been designed to maximise the exchange of gases as follows.

1. The interior of lung is divided into millions of small chambers called alveoli.
2. The presence of millions of alveoli in the lungs provide a very large surface area.
3. If all alveoli of our lungs are spread out they will cover an area of nearly 160 m².
4. Availability of large surface area maximises the exchange of gases.

Question 9.
Where will the release of energy from the glucose in respiration take place? Mala writes lungs, while Jiya writes muscles. Who is correct and why?
Answer:

1. Respiration is the process of releasing energy from the breakdown of glucose.
2. Respiration takes place in every living cell, all the time.
3. All cells need to respire in order to produce the energy that they require.
4. During respiration the release of energy from the glucose takes place in muscles but not in lungs.
5. So Jiya is correct. The energy is released from the muscle cells during respiration. Only gaseous exchange takes place in lungs.

Question 10.
What is the role of epiglottis and diaphragm in respiration?
Answer:
Epiglottis:

1. Epiglottis, a flap like muscular valve controls movement of air and food towards their respective passages.
2. Epiglottis is partly closed when we swallow food and it opens more widely when we take a breath and air enters the lungs.
3. Epiglottis allows air pass through the larynx and the respiratory system.
   

Diaphragm :

1. The diaphragm in the respiratory system is the dome shaped sheet of muscle that separates the chest from the abdomen.
2. When the diaphragm contracts during inhalation it flattens out a bit. This results in the enlargement of the volume of the chest cavity.
3. This reduces the pressure in the lungs and air enters into lungs from outside the body.
4. During exhalation, the diaphragm relaxes and assumes its dome shape. This change increases the pressure on the lungs and squeezes the air through the nose to the atmosphere.
   

Question 11.
How does gaseous exchange take place at blood level?
Answer:

1. Lungs are made up of several thousands of small chambers called alveoli.
2. Within the alveoli, exchange of gases take place between the gases inside the alveoli and blood.
3. Blood arriving in the alveoli has higher CO₂ concentration which is produced during respiration by the body cells.
4. At the same time air in the alveoli has a much lower concentration of CO₂ and this allows the diffusion of CO₂ out of the blood and to alveolar air.
5. Similarly blood arriving in the alveoli has a lower oxygen concentration while air in the alveoli has a higher oxygen concentration.
6. Therefore oxygen moves into the blood by diffusion.
   

Question 12.
Explain the mechanism of gaseous exchange at bronchiole level.
Answer:

1. the trachea (wind pipe) is divided into two tubes called BRONCHI. As there are two lungs each bronchus (singular) enters the lungs on the same side.
2. In the lung, the bronchus divides into smaller and smaller branches called BRONCHIOLES which enters into each alveoli.
3. When oxygen from outside reaches the alveoli through bronchioles and the carbondioxide from alveoli moves out.
4. The inhaled air from outside enters into bronchioles through nostrils → nasal cavities → pharynx → larynx → trachea → bronchus.
5. The exhaled air from alveoli enters bronchioles → pharynx → nasal cavities → nostrils → outside.
   

Question 13.
After a vigorous exercise or work we feel pain in muscles. What is the relationship between pain and respiration?
Answer:

1. We obtain energy by oxidation of glucose molecule.
2. In the absence of oxygen (anaerobic respiration) glucose is converted to latic acid.
3. During vigorous exercise oxygen gets used up faster in the muscle cells that can be supplied by blood.
4. When oxygen supply is inadequate the muscles use energy released during the anaerobic breakdown of glucose.
5. The anaerobic respiration by muscles bring about partial breakdown of glucose to form lactic acid.
6. The accumulation of lactic acid in the muscles causes muscular pains or cramps.

Question 14.
Raju said, “Stems also respire along with leaves in plants”. Can you support this statement? Give your reasons.
Answer:
Yes. I support the statement of Raju that stems also respire along with leaves in plants.
The reasons are

1. The stems of herbaceous plants have stomata.
2. So the exchange of respiratory gases in the stems of herbaceous plants takes place through stomata.
3. The oxygen from air diffuses into the stem through stomata and reaches all the cells for respiration.
4. The carbon dioxide released during respiration diffuses out into the air through the stomata.
5. In woody stems the bark has lenticels for gaseous exchange. Through lenticels, oxygen diffuses in and carbon dioxide diffuses out into the air.
   

Question 15.
What happens if diaphragm is not there in the body?
Answer:

1. The lungs cannot draw in air or push it out by themselves. The chest wall muscles and the diaphragm helps the lungs in moving air into and out of them.
2. If diaphragm is not there in the body, we would not be able to breathe.
3. The diaphragm is the major muscle in the process of respiration.
4. It separates the thoracic and abdominal cavities.
5. In the absence of diaphragm, the relaxation and contraction of the chest wall muscles do not take place and thereby inspiration and expiration become difficult that leads to death of the person.

Question 16.
If you have a chance to meet pulmonologist, what questions are you going to ask about pulmonary respiration?
Answer:
If I have a chance to meet pulmonologist, I would like to ask the following questions:

1. What is pulmonary respiration?
2. What is the organ involved in pulmonary respiration?
3. What is the name of blood vessel that brings deoxygenated blood to lungs?
4. What is the name of the blood vessel that carries oxygenated blood from the lungs to the heart?
5. Out of the two lungs which one is larger than the other?
6. What type of diagnostic test will be performed to assess the function of lungs?
7. What is pulmonary edema? How does it occur?
8. Can all the diseases of the lungs be cured permanently?

Question 17.
What procedure do you follow to understand anaerobic respiration in your school laboratory?
(OR) (Lab Activity)
Write the procedure and observations of the experiment which you have conducted in your laboratory to prove that CO₂ and heat are evolved during anaerobic respiration by using yeast.
(OR)
How do you prove that carbon dioxide is released during anaerobic respiration?
(OR)
How do yeast cells convert glucose solution to CO₂ and ethyl alcohol?
Answer:
Aim : To prove that CO₂ is released during anaerobic respiration.
Apparatus: Thermos flask, splitted corks, thermometer, wash bottle, glass tubes, liquid
paraffin, glucose solution, yeast cells, bicarbonate solution.
Procedure:

1. Remove dissolved oxygen from glucose solution by boiling it for a minute and then cooling it without shaking.
2. Now add some yeast to the glucose solution and fix a two-holed rubber stopper to the flask.
3. The supply of oxygen from the air can be cut off by pouring a 1cm layer of liquid paraffin into the mixture through the holes.
   
4. Insert one end of the thermometer into the thermos flask. See the mercury bulb of thermometer keep inside the solution.
5. Arrange for any gas produced by the yeast to escape through a wash bottle containing bicarbonate solution or lime water as shown in the figure.
6. Add a few drops of diazine green (Janus Green B) solution to the yeast suspension before you pour liquid paraffin over it.
7. The blue diazine green solution turns pink when oxygen is in short supply around it.
8. Warm the apparatus to about 37° F in order to speed up the test.
9. Keep the apparatus undisturbed for one or two days.

Observations :

1. After two days it can be observed that lime-water of the wash bottle turns into milky white precipitate.
2. Increase in temperature in thermometer.
3. Alcohol smell given off from the flask.

Result: These observations indicate that yeast cells respire anaerobically converting glucose solution into CO₂, ethyl alcohol and release heat energy.

Question 18.
What are your observations in the combustion of sugar activity?
Answer:
Observations in the combustion of sugar:

1. When sugar is heated, first it melts, chars and later burns producing flames.
2. When sugar is combusted, carbon dioxide and water are produced.
3. Energy is also released in the form of heat and it released at once.
4. We cannot control the combustion of sugar.
5. Intermediate products are not formed.
6. We can combust sugar in the absence of water and enzymes.
7. When combustion of sugar, heat energy is released into the atmosphere and we cannot store it for further use.

Question 19.
Collect information about cutaneous respiration in frog. Prepare a note and display them in your classroom.
(OR)
How does frog respire with the help of skin?
Answer:

1. Respiration through skin is called cutaneous respiration.
2. In frog, skin is an additional or secondary or accessory respiratory organ.
   
3. Skin is a very important respiratory organ in both on land and water.
4. One-third of the total oxygen taken up by frog is through the skin.
5. Frog also keeps its skin moist. Frog skin has a large number of mucous glands which secrete mucous onto the surface of the skin.
6. The mucous layer retains water and reduces evaporation of water from body.
7. To keep the skin wet and moist frogs jump into water very frequently.
8. Frog skin is supplied with a large number of blood vessels which help in absorbing oxygen from the water.
9. The carbon dioxide produced during to respiration, diffuses out into the water through the blood vessels present in the skin of the frog.

Question 20.
Collect information about respiratory diseases (because of pollution, tobacco) .and discuss with your classmates.
Answer:
Respiratory diseases because of pollution:

1. Irritation of eyes, nose, mouth and throat.
2. Headaches and dizziness.
3. Respiratory symptoms such as coughing and running nose.
4. Respiratory and lung diseases including
   a) Asthma attacks
   b) Chronic obstructive pulmonary disease (COPD)
   c) Reduced lung function
   d) Pulmonary cancer caused by a series of carcinogen chemicals that through inhalation
   e) Mesothelioma: A particular type of lung cancer, usually associated with expo¬sure to asbestos (it usually occurs 20 – 30 years after the initial exposure)
   f) Pneumonia: Infection of lungs caused by bacteria.
   g) Bronchitis: It is inflammation or swelling of bronchial tubes.
   h) Emphysema: It is a lung condition in which tiny air sacs in lungs alveoli fill up with water.

Respiratory diseases due to tobacco :

1. Chronic bronchitis: A long term inflammation of the bronchi is characterized by coughing.
2. Lung cancer: An abnormal continuous multiplication of cells that can result in tumors in the lining of the bronchi.
3. Emphysema: A chronic lung condition that affects the air sacs in the lungs characterized by shortness of breath, coughing, fatigue, sleep and heart problems.

Question 21.
What is the pathway taken by air in the respiratory system? Illustrate with a labelled diagram.
Answer:
The path way taken by air in the respiratory system:
Nostrils → Nasal cavity → Pharynx → Larynx → Trachea → Bronchus → Bronchioles → Alveolus → Blood.

1. Nostrils : Air enters the body through the nostrils.
2. Nasal cavity: Air is filtered and its temperature is also brought close to that of the body.
3. Pharynx: It is the junction of respiratory and digestive system. Epiglottis – a flap like muscular valve controls movement of air and food towards their respective passages.
4. Larynx: Also called voice box. This stiff box contains vocal cords. When air passes out of the lungs and over vocal cords, it causes them to vibrate. This produces sounds on the basis of our speech, song etc.
5. Trachea: This is also called wind pipe. It channels air to lungs.
6. Bronchi: Trachea at it’s lower end divides into two bronchi one leading to each lung.
7. Bronchioles: The bronchi further divided into smaller and smaller branches called bronchioles.
8. Alveoli: Clusters of air sacs called alveoli in the lungs which are very small and numerous. The gaseous exchange takes place here as blood capillaries take up oxygen and expel CO₂.
9. Blood: It carries oxygen, to each and every cell of the body and collects CO₂ from them.
   

Question 22.
Draw a block diagram showing events in respiration. Write what you understood about cellular respiration.
Answer:
Events in respiration :

Cellular respiration :

1. All living cells must carry out cellular respiration.
2. Oxidation of glucose of fatty acids releasing energy takes place in cells, hence it is called cellular respiration.
3. It can be in the presence of oxygen that is aerobic respiration or in its absence that is anaerobic respiration (fermentation).
4. Cellular respiration in prokaryotic cells like that of bacteria occurs within the cytoplasm.
5. In Eukaryotic cells cytoplasm and mitochondria are the site of cellular respiration.
6. The energy released in cellular respiration is stored in a special compound called ATP.
7. ATP is utilised for carrying out other functions in the cell.

Question 23.
How do you appreciate the mechanism of respiration in our body?
Answer:

1. Respiration is essential for life because it provides energy for carrying out all the life processes which are necessary to keep the organism alive.
2. The energy that is obtained from respiration is used to build the organism by way of cell growth, reproduction and cell repair, etc.
3. All systems in living beings need energy to survive.
4. Respiration helps to expel out the toxic carbon dioxide out of the cells. This CO₂ will be utilised by the plants to produce food materials through the process of photosynthesis.
5. The respiratory system goes into operation from the movement of our birth and works without ever stopping as long as we live our breath continues.
6. During exhalation, the vocal cords in the larynx vibrate to produce sounds and help in speaking as we like.

Question 24.
Prepare an article on anaerobic respiration to present school symposium.
Answer:
Anaerobic respiration :

1. Respiration that occurs without oxygen is known as anaerobic respiration.
2. It is present in primitive organisms and muscular cells in higher animals.
3. Alcohol, CO₂ and H₂O are end products in this process.
4. 

Merits:

1. In the absence of oxygen it is good process.
2. Mechanism is simple.
3. Suitable to microorganisms

Demerits:

1. Provides less energy
2. Not suitable to higher animals

Question 25.
Prepare a cartoon on discussion between haemoglobin and chlorophyll about respiration.
Answer:
Discussion between haemoglobin and chlorophyll about respiration:

Haemoglobin: Hello good morning chlorophyll. How are you?

Chlorophyll: Very good morning haemoglobin. I am fine.

Haemoglobin: I am also doing well. Let me know something about you.

Chlorophyll: I am a green coloured pigment present in leaves of plants.

Haemoglobin: How many types of chlorophylls are there ?

Chlorophyll: We are four types i.e., chlorophyll – a, chlorophyll – b, chlorophyll – c and chlorophyll – d.

Haemoglobin: May I know your job in leaves?

Chlorophyll: Yes. Why not? I am an essential factor required to prepare food through the process of photosynthesis in plants.

Haemoglobin: Oh! You are participating in the process of preparing food materials by green plants.

Chlorophyll: Now tell me about your presence.

Haemoglobin: I am present only in animal cells. That too in the red blood cells of the blood. The red colour of the blood is due to my presence.

Chlorophyll: Then tell me about your function in respiration?

Haemoglobin: During respiration, I carry oxygen to the cells in the body tissues.

Chlorophyll: How are you able to do this?

Haemoglobin: I have an oxygen binding element iron. It binds oxygen on four corners of it. I form oxy-haemoglobin with oxygen in the lungs.

Chlorophyll: What happens to the digested food in the cells?

Haemoglobin: In cells, oxygen breakdown the glucose molecule into C02 and H20 releasing large amount of energy. Around 38 ATP molecules are produced.

Chlorophyll: What is the use of this energy?

Haemoglobin: This energy is utilised by the cell to carry other functions.

Chlorophyll: Thank you haemoglobin. You have taught me everything about respiration. In our next meeting we will discuss another topic.

Haemoglobin: Thank you chlorophyll for your interest and patience.

Fill in the blanks.

1. Exhaled air contains ———– and ———–.
2. A flap like muscular valve controls movement of air and food is ———–.
3. Energy currency of the cell is called ———–.
4. Lenticels are the respiratory organs that exist in ———– part of the plant.
5. Mangrove trees respire with their ———–.

Answer:

1. carbon dioxide, water vapour
2. Epiglottis
3. ATP (Adenosine Tri Phosphate)
4. stem
5. aerial roots

Choose the correct answer.

6. We will find vocal cords in [ ]
A) Larynx
B) Pharynx
C) Nasal cavity
D) Trachea
Answer: A

7. Cluster of air sacs in lungs are called [ ]
A) Alveoli
B) Bronchi
Answer: A

8. Which of the following is correct ? [ ]
i) The diaphragm contracts – volume of chest cavity increased
ii) The diaphragm contracts – volume of chest cavity decreased
iii) The diaphragm expands – volume of chest cavity increased
iv) The diaphragm expands – volume of chest cavity decreased
A) i B) i and ii C) ii and iii D) iv
Answer: A

9. Respiration is a catabolic process because of [ ]
A) Breakdown of complex food molecules
B) Conversion of light energy
C) Synthesis of chemical energy
D) Energy storage
Answer: A

10. Energy is stored in [ ]
A) Nucleus
B) Mitochondria
C) Ribosomes
D) Cell wall
Answer: B

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System

Activity – 1

How do you test the presence of water vapour and heat in the exhaled air?
Answer:

1. Keep your palm around an inch away from your nose.
2. Feel you breathing out.
3. Do not remove your palm until you have finished the activity.
4. Breathe steadily for 1 – 2 minutes.
5. Now take a piece of any fruit.
6. Chew and before swallowing it keep the fingers of the other palm on your neck, now swallow it.

Questions:

1. What did you notice? What happens to your breath as you try to swallow?
   Answer:
   We cannot swallow while breathing. We usually stop breathing when we swallow food.
2. What is helping you to swallow without deflecting it to the windpipe?
   Answer:
   Epiglottis is helping me to swallow without deflecting food to the windpipe.

Activity – 2
Write an experiment to observe changes during combustion of sugar.
(OR)
What are your observations in combustion of sugar activity.
Answer:
Aim: To observe changes during combustion of sugar.
Apparatus: Wooden stand, test tubes, rubber stopper, delivery tube, glucose or sucrose powder, lime water, spirit lamp.
Procedure:

1. Take a small amount of glucose in a small test tube.
2. Arrange the apparatus as shown in the figure.
3. Heat the test tube until the glucose catches fire.
   

Observations :

1. In the initial stage the glucose becomes liquid.
2. Later it turns to black colour after catching fire.
3. In this process carbon dioxide and water are produced.
4. Energy is released as heat.
5. The carbon dioxide released changes lime water to milky white.

Conclusion:

1. From this experiment, we can conclude that carbon dioxide, water and heat are produced during combustion of glucose in the laboratory.
2. The carbon dioxide changes lime water to milky white in nature.

Activity – 3

How can you prove that carbon dioxide is evolved during respiration?
(OR)
Write the experimental procedure and draw the arrangement of apparatus to show that CO₂ is evoloved in respiration.
To understand that CO₂ is evolved during respiration, what experiment you have performed in your laboratory? Explain the procedure.
(OR)
Write an experiment to prove that CO₂ is released during respiration.
Answer:
Aim: To prove that CO₂ is released during aerobic respiration.
Apparatus: Two wide mouthed plastic or glass bottles, germinating seeds, dry seeds, two small injection bottles or beakers with lime water.

1. Take two wide mouthed glass bottles.
2. Keep germinating bengal gram seeds in one bottle and dry seeds in another bottle.
3. Keep two small beakers with lime water in each glass bottles.
4. Close the glass bottles tightly.
5. Keep both the sets undisturbed for one or two days.
   

Observation:

1. After two days it can be observed that lime water of the beaker placed in the bottle containing germinating seeds turns into milky white.
2. And the lime water kept in the glass bottle containing dry seeds do not change its colour.
   Result: It indicates that germinating seeds liberated carbon dioxide which turns lime
   water into milky white.

Activity – 4

Explain the procedure you have adopted in your school to prove that heat is liberated during respiration. What result we will get, if you perform this experiment with dry seeds?
(OR)
Write the procedure you have followed to observe “heat is evolved during respiration” in your laboratory. What precautions did you take during the activity?
Answer:
Aim: To prove that heat is liberated during respiration.
Apparatus: Two Thermos flasks, two thermometers, two rubber corks, dry seeds, germinating seeds.
Procedure:

1. Take a handful of moong or bazra seeds.
2. Soak the seeds in water a day before experiment.
3. Keep these soaked seeds in a cloth pouch and tie with a string tightly.
4. Next day collect the sprouts / germinated seeds from the pouch in a thermos flask and take dry seeds in another thermos flask.
5. Remove the lid and prepare a cork through which you can bore a hole to insert thermometers into two flasks in such a way that the bulb of the thermometer should dip into the germinating and dry seeds.
6. Close the thermos flasks with tight fitting rubber corks.
7. Record the initial temperature in both the flasks and record it for every two hours for at least 24 hours.
   

Observation: Constant increase in the temperature is observed in thermometer placed in the germinating seeds.
Result: Hence it is proved that germinated seeds respire and liberate heat which is responsible for increase in the temperature.
Questions:

1. Make a graph by using your observations.
   Answer:
   
2. Is there any increase in temperature?
   Answer:
   Yes, there is increase in temperature.
3. Does the temperature increase steadily or does it abruptly increase at a time of the day?
   Answer:
   The temperature in the thermometer increases steadily.
4. Where does the heat come from?
   Answer:
   The sprouting or germinating seeds respire and liberate heat.

10th Class Biology 2nd Lesson Respiration – The Energy Releasing System InText Questions and Answers

10th Class Biology Textbook Page No. 25

Question 1.
Can it be said that Priestly’s experiment helped us to find out more about composition of air? How?
Answer:
Yes, Priestly’s experiments helped to find out the composition of air when burning of charcoal, carbon dioxide is produced which is the one of the composition of air done by Lavoisier.
Another experiment with phosphorus done by Lavoisier was cleared that a gas which is the respirable air that is helped in burning was oxygen also a component of air.
Lavoisier proved experimentally that carbon dioxide and oxygen were the components of air. Lavoisier confirmed the experiments of Priestly about the gases present in the air.

Question 2.
What gas was produced by combustion according to Lavoisier?
Answer:
The gas produced by combustion is carbon dioxide.

Question 3.
What did Lavoisier find out about air from his experiments?
Answer:
A fixed air carbon dioxide and respirable air oxygen which helped in burning were liberated during his experiments.

Question 4.
What conclusion can be drawn from Lavoisier’s experiments?
A. Oxygen and carbon dioxide are the compositions of air.

10th Class Biology Textbook Page No. 26

Question 5.
Which gas do you think is Lavoisier talking about when he says chalky acid gas?
Answer:
Carbon dioxide.

Question 6.
Which gas according to Lavoisier is respirable air?
Answer:
Oxygen.

Question 7.
What steps in the process of respiration does Lavoisier mention as an inference of his experiments?
Answer:
Lavoisier mentioned that there were two steps in the respiration.
1. Inspiration: Breathing oxygen.
2. Expiration: Eliminating carbon dioxide from lungs.

Question 8.
It is a common observation that our breath is warmer than the air around us ; does respiration have anything to do with this?
Answer:
Our exhaled air is warmer than the air around us because heat is liberated during respiration.

10th Class Biology Textbook Page No. 27

Question 9.
What does this experiment indicate?
Answer:
This experiment indicates that carbon dioxide is liberated during respiration.

Question 10.
Which gas turns lime water milky?
Answer:
Carbon dioxide.

Question 11.
Which gas do you think might be present in greater quantities in the air we breathe out as compared to air around us?
Answer:
Nitrogen (78%) is present in greater quantities in the air.

Question 12.
We are also aware of the fact that water vapour deposits on a mirror if we breathe out on it; where does this water vapour come from in Exhaled air?
Answer:
Water vapour is liberated during respiration.

10th Class Biology Textbook Page No. 29

Question 13.
Why are we advised not to talk while eating food?
Answer:

1. When we are eating food, epiglottis helps to avoid food entering into trachea.
2. If we talk while we eat food, there is a chance of food entering into trachea and causes choking.
3. As a result irritation and inflammation takes place in the respiratory tract.
4. So we are advised not to talk while eating.

10th Class Biology Textbook Page No. 30

Question 14.
What is the role of diaphragm and ribs in respiration? Are both active in man and woman?
Answer:

1. Diaphragm and ribs are helpful the chest cavity to increase or decrease the volume for inspiration and expiration in respiration.
2. Diaphragm plays a major role in men and ribs play a major role in mechanism of respiration.

10th Class Biology Textbook Page No. 31

Question 15.
What can be concluded from this?
Answer:
All movements of breathing is controlled by nerves leading from the brain.

Question 16.
What happens during the process of breathing?
Answer:
During the process of breathing, the patterns of breathing show a great range for they are coordinated with moment by moment needs of the body for supply of oxygen and removal of carbon dioxide.

Question 17.
Which gas needs to be removed from our body during exhalation? Where does the extra amount of gas come from?
Answer:
Carbon dioxide needs to be removed during exhalation. The extra amount of gas comes from the breakdown of glucose to release energy in the mitochondria. Carbon dioxide gas is released here.

Question 18.
What is the composition of inhaled air?
Answer:
Inhaled air contains oxygen -21%, C09 – 0.03%, Nitrogen – 78%.

Question 19.
When exhaled air is compared with inhaled air, is there any difference in composition?
Answer:
Yes, there is a difference between inhaled air and exhaled air.
The difference is

10th Class Biology Textbook Page No. 32

Question 20.
Why does the amount of oxygen vary between exhaled and inhaled air?
Answer:
Because some amount of oxygen will be utilised during cellular respiration in the body. Hence the difference in amount of oxygen occurs.

Question 21.
What has raised the percentage of carbon dioxide in exhaled air?
Answer:
CO₂ is released from all the cells in the body in respiration and is added to the exhaled air.

10th Class Biology Textbook Page No. 34

Question 22.
Do cells of alveoli or lungs also require oxygen to carry out cellular respiration? Why / Why not?
Answer:

1. Alveoli are made of squamous epithelium tissue which is very thin and elastic
2. Alveoli are so thin that oxygen can pass from air-filled alveoli to R.B.C inside the vessels.
3. Simple squamous epithelial cells function as mediators of filtration and diffusion. As these cells are living tissue they also need oxygen.
4. This is done through the exchange of gases in the alveoli.

Question 23.
After undergoing strenuous exercise we feel pain in muscles, does adequate oxygen reach the muscles?
Answer:
No. Adequate oxygen does not reach the muscles.

Question 24.
What is being formed in the muscles?
Answer:
Lactic acid.

10th Class Biology Textbook Page No. 42

Question 25.
In which set does the colour change faster? Why?
Answer:
In the set which has germinating seeds the colour changes faster. Because CO₂ is formed faster in aerobic respiration.

Think and Discuss

10th Class Biology Textbook Page No. 29

Question 1.
What will happen if the respiratory tract is not moist?
Answer:

1. If the respiratory tract is not moist the dirt particles in the inhaled air will not be removed from air in the nasal cavities and reaches lungs and create problems to lungs.
2. The temperature of the inhaled air is brought close to that of the body for the smooth passage in the respiratory tract. If it is dry, it is not possible.
3. If the surface dries out, gas exchange will happen at a very reduced rate since fast moving gaseous oxygen molecules do not efficiently cross the alveoli membrane.
4. The reduced gas exchange is most likely not enough to support blood oxygenation for vital functions of the body.
5. Hence respiratory tract should be moist for smooth exchange of gases.

Question 2.
Are both lungs similar in size?
Answer:
No. Right lung is slightly bigger than left lung.

Question 3.
Why are alveoli so small and uncountable in number?
Answer:

1. The pouch-like air sacs at the ends of the smallest bronchioles are called alveoli.
2. The walls of the alveolus are very thin and they are surrounded by very thin blood capillaries.
3. It is in the alveoli that gaseous exchange takes place.
4. There are millions of alveoli in the lungs. The presence of millions of alveoli in the lungs provides a very large area for the exchange of gases.
5. And the availability of large surface area maximises the exchanges of gases.