Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 7 Different Ecosystems Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 7th Lesson Different Ecosystems

8th Class Biology 7th Lesson Different Ecosystems Textbook Questions and Answers

Improve Your Learning

Question 1.
Define an ecosystem. Explain it with a suitable example.
Answer:

1. An ecosystem can be defined as a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment.
2. For example, Mangroves are one of the most productive ecosystem on earth, deriving nourishment from terrestrial fresh water and tidal salt water.
3. Coringa mangrove is situated south of Kakinada Bay and is about 150 km south of Visakhapatnam.
4. It is named after the river coringa. Coringa mangroves receives fresh water from coringa and Gaderu rivers and salt waters from Kakinada Bay.
5. Biotic components in coringa:
   Producers: Mangrove, Spirogyra, Euglena, Oscilatoria, Blue Green Algae, Ulothrix etc.
   Consumers: Shrimp, crab, hydra, protozoans, mussel, snails, turtle, daphnia, brittle word, tube worm etc.
   Decomposers: Detritus feeding bacteria etc.
6. Abiotic components: Salt and fresh water, air, sunlight, soil, etc.

Question 2.
Explain how diversity of living organisms helps in enriching any ecosystem.
Answer:

1. The existence of the ecosystem depends on the continued survival of the organisms in the ecosystem.
2. All organisms require energy for growth, reproduction and survival.
3. This energy is obtained by the organisms from the food they consume.
4. Plants are the producers producing food in any ecosystem. The animals present in the ecosystem are consumers as they consume food from plants.
5. Some of the organisms in the ecosystem such as bacteria and fungi, obtained their nutritional requirements by decomposing the dead bodies of both producers and consumers.
6. They retain nutrients to the soil for the plants to use. As the cycle begins again.
7. Like this diversity of living organisms helps in enriching any ecosystem.

Question 3.
What happens when two animals having similar habits share one ecosystem?
Answer:
When two animals having similar habits, sharing one ecosystem , only the stronger and better equipped animal can survive, while the weaker one die or eliminated from the ecosystem. This is called ‘Survival of the fittest’.

Question 4.
What is the difference between habitat and ecosystem?
Answer:
Habitat is the natural living place of an organism or a group of organisms. Land and water are the major habitates.
An ecosystem is a Natural unit and has both Abiotic and biotic components, which interact and influence each other.

Question 5.
Who am I?
1. I am the base of food chain.
Answer:
Green plants.
2. I depend on plants for food.
Answer:
Consumers (Herbivorous Animals)
3. I break down the remains of dead plants and animals.
Answer:
Decomposers.
Ex: Bacteria, Fungi.

Question 6.
Which of the following is producer and why?
а) fox b) fungus c) chicken d) grass.
Answer:

1. Grass is the producer.
2. All green plants produce food materials with the help of carbondioxide and water in the presence of sunlight. So plants are called producers.
3. As grass is the green plant and produce food for other animals, it is called producer.

Question 7.
What do you understand by food web? Describe your own food web with the help of diagrammatic representation.
Answer:

1. A food web consists of several interlinked food chains and each organism in the food web will be a member of more than one food chain.
2. For example rats and insects eat seeds and other plant parts. As their food.
3. Insects are eaten by frogs and lizards.
4. Rats and frogs are eaten by snakes.
5. Lizards and snakes are eaten by birds.
6. Birds are eaten by fox, wolves. These are eaten by tigers and lions.
7. Thus a single plant or an animal may become food for more than one animal.
8. Similiarly an animal may consume more than one type of food depending on its taste and availability in the ecosystem.
9. Thus each organism in an ecosystem may be a member of more than one food chain.
10. When we looked at these relationships between various organisms for food in the ecosystem, it appears that several food chains are interlinked with each other forming a food web.

FOOD WEB

Question 8.
An ecosystem that had mice. What happens if more cats were added to it?
Answer:

1. In an ecosystem if the mice and the cats were existing equally, smooth balance would be maintained. When reproduction took place in these animals, generations would be continued then the ecosystem would be maintained healthy.
2. When more cats were added in that ecosystem all the mice would be eaten away by the cats it leads to the end of mice population.
3. Because of lack of food, the cats, either have to leave the ecosystem or they die.
4. If there is no continuity of the animals, the ecosystem would be destroyed.

Question 9.
List out producers (plants, bushes, trees), consumers (herbivores and carnivores) and decomposers that you observed in your agriculture field or school garden.
Answer:
Producers:
Plants – Grass plants, creepers like pumpkin, bottlegourd, etc.
Bushes – Rose, Jasmine, chrysanthemum, marigold.
Trees – Palm, coconut, mango, drumstick, lemon, sweet lemon, guava.
Consumers:
Herbivores – Goat, sheep, buffalo, ox, rats, butterflies, etc.
Carnivores – Crow, dogs, snakes, frogs, lizards.
Decomposers – Fungi (mushroom), Bacteria.

Question 10.
In grassland ecosystem, rabbit eats only plants. They eat plants faster than the plants can grow back. What must happen to bring the ecosystem into balance?
Answer:
The animals like fox, wolves, tigers, etc. which feed on rabbits will be introduced in that grass land ecosystem, then the rabbits will be controlled by them. Thus the ecosystem will comes into balance.

Question 11.
Plant, Tiger, Rabbit, Fox, Hawk.
Did you find any connection among the above list of things. If we remove Rabbit from the list what will happen?
Answer:

1. Plant, tiger, rabbit, fox, hawk these are the animals living in an ecosystem and are interdepending one on the other, and maintaining a food web – Plants → Rabbit → Hawk → Fox → Tiger.
2. A delicate balance is seen in nature between members of different species.
3. Any disturbance in this balance affects the organisms in a food web.
4. If we remove rabbit from the list the other animals like tiger, fox, hawk which are depending one on the other for food will die because of no food.
5. All the organisms, big or small, have right to live on this planet as man. We should respect this and allow other organisms to live and share the wealth of this planet.
6. “LIVE AND LET LIVE” should become our motto.

Question 12.
What do you understand by interdependency of animals and plants ? How do you appreciate ?
Answer:

1. An ecosystem is made up of groups of living things and their environments.
   
2. Living things like plants, animals and microorganisms are known as biotic components and others like soil, water, sunlight etc are called as abiotic components of the ecosystem.
3. All these organisms live together and interact with one another in many ways.
4. There is a feeding relationship between plants and animals. Along with this an interdependence between plants and animals for space, reproduction, shelter, etc.
5. All organisms in an ecosystem derive energy from food to live.
6. The sun is the main source of energy for all living things.
7. Plants being autotrophic, trap this energy through a process called photosynthesis and produce food to all living organisms. They are known as Primary producers.
8. Animals as they can not prepare food, they consume plants directly or indirectly and called consumers.
9. Living organisms like fungi and bacteria which are called decomposers, decay and decompose the dead animals of producers and consumers, and valuable nutrients to the soil for plants to use, as the cycle begins again.
10. A delicate balance is seen in nature between plants and animals by interdependence one to the other for thousands of years, which is unreachable to the human brain.

8th Class Biology 7th Lesson Different Ecosystems InText Questions and Answers

Question 1.
What is a Habitat?
Answer:

1. The dwelling place for plants and animals is called habitat.
2. One habitat shared by different types of plants and animals.
3. Try to add more such points to your list.
4. The natural home for plants and animals is called a habitat.
5. Habitat is the environment of an animal or plant.
6. Habitat is a suitable place for plants and animals to live.
7. Habitat is the origin for plants and animals.

Question 2.
Draw the diagram of Interdependence between the biotic components and answer the following questions.
Answer:
Interdependence between the biotic components:

1. What do the arrows in the figure indicate?
Answer:
The animals are depending one on the other for food.

2. Trace the path from grass to tiger. You may trace out other paths as well.
Answer:
Grass → grass hopper → frog → snake.
Carrot and grass → rabbit → fox → tiger
Plants → deer → bear → tiger
Seeds → squirrel → eagle → tiger

3. On how many organisms is rabbit dependent? Write their names.
Answer:
Carrot, grass.

4. How many organisms depend on rabbit? Write their names.
Answer:
Snake, fox, eagle, tiger.

5. Where do plants get their food from?
Answer:
Plants are autotrophs they can prepare their food from carbondioxide and water, in the presence of sunlight.

6. What other things do animals need for their survival?
Answer:
Abiotic components like soil, water, sunlight, etc.

8th Class Biology 7th Lesson Different Ecosystems Activities

Activity – 1

Lab Activity
Answer:
Aim: Study an ecosystem at your school/ home garden to understand it’s structure. Material Required : Measuring tape string, small sticks, hand lens, hand towel.
Procedure: To know about structure of the ecosystem we have to follow the following procedure.

1. Use the tape to measure a square area that is one meter long and one meter wide. It can be on grass, bare dirt or side wall.
2. Mark the edges of the square with the help of string/chalk.
3. Observe the study area (that has been marked). Look for the plants and animals that live there. Use the hand lens.
4. Record all the living organisms you see. You can even dig to go deeper to find out other living organisms that may be present there.

Observation / Findings: We find –

1. Plants like grass, herbs, shrubs, guava, neem and creepers.
2. Insects like ants, grass hoppers, butterflies, mosquitoes, houseflies, locusts, etc.
3. Animals like cat, dog, buffalo, frog, lizards, garden lizards, snake.
4. Mushroom, algae.
5. Deeper layers of soil we find earth worm, leech, rats, bandicoots, rabbits, etc.
6. Birds like crow, parrot, mynah, etc.

Discussion:
1. What living things did you find in your study area? Try to count them if possible.
Answer:
Grass, creepers, shrubs, herbs, trees, herbivores, carnivores, fungi.
2. Which kind of living thing was most common in your study area?
Answer:
Plants.
3. How was your study area different from those of other student groups?
Answer:
Living conditions, food, animals and plants are different.
4. Other than the living organisms what other things can you record from your study area?
Answer:
Soil, water, sunlight (temperature) are recorded.

Activity – 2

Question 2.
Observe the food web given below figure.

Answer:
The diagram showing food web.
Now answer the following Questions.
1. Which are the producers in the food web?
Answer:
Grass, rice plants, maize, bushes.
2. Which are consumers?
Answer:
Fish, frog, birds, rats, rabbit, deer, tadpole, larva ,sheep, cat, fox, tiger, crane, eagle, snake, owl, peacock, insects, lion.
3. Where does the food web start from?
Answer:
Food web starts from green plants.
4. Name the organism where the food web ends.
Answer:
Crane, eagle, owl, peacock, lion.
5. What happens when plants and animals die in a food web?
Answer:
When plants and animals die, they are decayed and decomposed by Decomposers like bacteria and fungi. They return nutrients to the soil for plants to use, as the cycle begin again. This is the reason ‘Decomposers are also called as recyclers.

Activity – 3

Question 3.
Collect the information forests of Andhra Pradesh and write the flora and fauna and fill up the following table:
Answer:
Forests of Andhra Pradesh

Name of the Forest – Kondapalli Reserve Forest

Investigations:

1. Do all the forest have same type of vegetation?
Answer:
No, there are mainly trees that show much species diversity and greater degrees stratification.
2. Are producers afforest ecosystem higher than its consumers?
Answer:
Trees are higher than consumers, besides trees shrubs and ground vegetation also there.
3. Do all the forests have same type of animals?
Answer:
No, the availability of food and environment different type of animals are present in different forest.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.4

Question 1.
Check whether 25110 is divisible by 45.
Solution:
The given number = 25110
If 25110 is divisible by 45 then it should be divisible by 5 and 9.

∴ The number 25110 is divisible by 45

Question 2.
Check whether 61479 is divisible by 81.
Solution:
If 61479 is divisible by 81 then it is divisible by 9.
If the sum of the digits of a number is dívisible by 9 then the entire number is divisible by 9.
∴ 61479 → 6 + 1 + 4 + 7 + 9 → \(\frac { 27 }{ 9 }\) (R = 0)
∴ 61479 is divisible by 81. [∵ 9 is factor of 81]

Question 3.
Check whether 864 is divisible by 36? Verif,’ whether 864 is divisible by all the factors of 36 ?
Solution:
864 is divisible by 2 and 3.
∴ 864 is divisible by 6.
∴ 864 is divisible by 36 [ ∵ 6 is the factor of 36]
∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18. 36.


∴ 864 is divisible by all the factor of 36.

Question 4.
Check whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?
Solution:
756 is divisible by 2 and 3.
∴ 756 is divisible by 6.
2a + 3b + c = 2 x 7 + 3 x 5 + 6 = 14 + 15 + 6 → \(\frac { 35 }{ 7 }\) (R = 0)
∴ 756 is divisible by 7.
∴ 756 is divisible by 42. [ ∵ 6, 7 are the factors of 42]
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42.


∴ 756 is divisible by all the factor of 42.

Question 5.
Check whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by product of 11 and 7?
Solution:

Question 6.
Check whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?
Solution:

Question 7.
Check whether 456 and 618 are divisible by 6’? Also check whether 6 divides the sum of 456 and 618 ‘?
Solution:

Question 8.
Check whether 876 and 345 are divisible by 3. Also check whether 3 divides the difference of 876 and 345?
Solution:

Number	Divisible by 3	Y/N	Difference is divisible by 3 Y/N
876	8 + 7 + 6 → \(\frac { 21 }{ 3 }\) (R = 0)	Yes	876 – 345 = 531
345	3 + 4 + 5 → \(\frac { 12 }{ 3 }\) (R = 0)	Yes	The difference of 876, 345 is divisible by 3.
531	5 + 3 + 1 → \(\frac { 9 }{ 3 }\) (R = 0)	Yes

Question 9.
Check whether 2² + 2³+2⁴ is divisible by 2 or 4 or by both 2 and 4’?
Solution:

∴ 2² + 2³+2⁴ is divisible by both 2 and 4.

Question 10.
Check whether 32² is divisible by 4 or 8 or by both 4 and 8’?
Solution:

32² is divisible by 4 and 8

Question 11.
If A679B is a 5-dit number is divisible by 72 find ‘A’ and ‘B”?
Solution:
If A679B is divisible by 72 then it should be divisible by 8 and 9.
[ ∵ 8, 9 are the factors of 72]
A679B is divisible by 9 then
A + 6 + 7 + 9 + B = A + B + 22 = 27 (= 9 x 3)
=A + B = 5 ……………. (1)
A679B → \(\frac{79 \mathrm{~B}}{8}\) [From B (2,4,6,8) we take B = 2]
= \(\frac{792}{8}\) (R = 0)
∴ B = 2
From (1) ⇒ A + 2 = 5
∴ A = 3, B = 2

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 3 Story of Microorganisms 1 Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 3rd Lesson Story of Microorganisms 1

8th Class Biology 3rd Lesson Story of Microorganisms 1 Textbook Questions and Answers

Improve Your Learning

Question 1.
Which organisms are interlinked between living and non-living organisms? Why do you think so?
Answer:
Viruses are an interesting type of microorganisms. They usually made up of crystalized proteins. They behave like nonliving things when they are outside of a living cell. But they behave like living organisms when they are inside host living cells and reproduce just like bacteria. Hence viruses can also call as connecting links between living and nonliving things.

Question 2.
What are microorganisms? Where do you find them?
Answer:
We can see several organisms in our surroundings but we cannot see many of them with our unaided eyes. They can be seen only with the help of microscope only. They are called microorganisms. They can found in air, water, soil and all living organisms.

Question 3.
What type of microorganisms we can observe in pond water?
Answer:
Usually pond water contains bacteria, phytoplanktons, algal members, fungi, rotifers, hydra etc. Collect some pond water with greenish scrapings on a slide and we can observe different algal members like Spirogyra, Chara and Chlamydomonas through the microscope.

Question 4.
Whether microorganisms are useful or harmful. How? Explain.
Answer:

1. Microorganisms are useful and some are harmful.
2. Some microorganisms are useful in formation process, medicine preparation and increase soil fertility.
3. Some microorganisms are harmful by causing diseases and spoling food items.

Question 5.
How are the human actions causing the death of useful bacteria and fungi? What will happen if it continuous?
Answer:

1. Soil is highly rich in microorganisms such a bacteria, fungi, protozoa, micro arthropods.
2. The top eight inches of soil of one acre many contain as much as five and half tons of fungi and bacteria.
3. This is very much useful for growing crops.
4. Excess use of pesticides kills these bacteria.
5. Thus human actions causing death of useful bacteria and fungi.
6. If it happens continue, then it causes to ecological imbalance.

Question 6.
Why the cooked food spoil soon but not uncooked food ? Give your reasons.
Answer:

1. Cooking of food items makes the proteins in the food materials coagulate.
2. It also degrades the protective surface of the food.
3. Thus the cooked foods can be easily inhabited by microoganisms.
4. So they can be spoiled in less time than the uncooked food.

Question 7.
What questions would you like to ask your teacher to know about different shapes of bacteria ?
Answer:

1. Where can we find bacteria?
2. How can we see bacteria?
3. What do we call the round shaped bacteria?
4. What do we call the spring shaped bacteria?
5. What is the name of coma shape bacteria?
6. How many types of bacteria do we find in nature?
7. What is the shape of Lactobacillus bacteria? How is it useful?
8. How is septicemia bacteria harmful?
9. Which type of bacteria is responsible for food poisoning?
10. Which bacteria is present in root nodules of leguminous plants? How do they useful?
11. Name the bacteria that causes leprosy.
12. Which type of bacteria is responsible for tuberculosis?
13. What is the shape of Bacillus thuringiensis bacteria how is it useful to plants?
14. What is the shape of staphylococci bacteria? In what way it affect the health of people?

Question 8.
What will happen if you add buttermilk to chilled milk?
Answer:

1. Lacto bacillus bacteria is responsible for the formation of curd.
2. When we add buttermilk to luke warm milk it takes 2 or 3 hours time to form curd.
3. But if we add buttermilk to chilled milk it takes more time or curd would not form.
4. Curdling indicates that the increase in number of bacteria in milk.
5. In chilled condition the number of bacteria do not increase in number there by curd would not be formed.

Question 9.
How do you observe Lactobacillus bacterium?
Answer:
Take one or two drops of buttermilk on a slide and spread it. Heat the slide slightly on a lamp (3-4 sec¬onds). Add a few drops of crystal violet. Leave it for 30-60 seconds and wash the slide gently with water.
Observe the slide under the Compound Microscope to see the Lactobacillus bacterium.

Question 10.
Visit any bakery or milk chilling center near your school with the help of your teacher or parents. Learn about some techniques to culture and usage of some Microorganisms and prepare a note on them.
Answer:
The Milk Collection Station is a specially designed, integrated unit, which combines the several functions of a milk collection centre. It measures the weight, fat content and gives the price of the milk brought in by the each producer. The equipment is particularly useful for the milk cooperatives / milk collection centres as it can also maintain a summary of milk supplied. This state of the art equipment operates both on battery and mains and is able to process and record 120-150 milk collection per hour. An Electronic Milk weighing Unit, the Electronic Milk Tester and Data Processor Unit are main components of the system. The membership code of individual mem¬bers is entered automatically by member identity card / manually by an electronic key-board.

Question 11.
Observe some permanent slides of microorganisms in your school lab with the help of microscope. Draw its picture.
Answer:

Question 12.
Prepare a model of any microorganism. And write a note on them.
Answer:

Question 13.
Why should we clean our hands with soap before eating ?
Answer:

1. We touch the objects.
2. Microbes are present on them.
3. When we touch them, they will inhabit our hands.
4. Washing our hands with soap kills all the microbes.
5. And makes our hands clean and hygenic.
6. So we should clean our hands with soap before eating.

8th Class Biology 3rd Lesson Story of Microorganisms 1 Activities

Activity – 2

Question 1.
Identify the fungi present, in rotten vegetables.
Answer:
Take some rotten part of vegetable or black spoiled part of bread or coconut with help of a needle on a slide, Put a drop of water. Place a cover slip on it and we can see the following microorganisms through microscope.

Activity – 6

Question 2.
Observe different soil microorganisms through microscope and draw rough sketches.
Answer:
Collect some soil from the field in a beaker or in a glass. Add some water to it and stir it. Wait for some time to allow the soil particles to settle down. Take a drop of water on a slide and we can observe the following microorganisms.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers.

8th Class Maths 14th Lesson Surface Areas and Volume (Cube-Cuboid) InText Questions and Answers

Do this

Question 1.
Find the total surface area of the following cuboid.      [Page No. 298]

Answer:
i) l = 4 cm, b = 4 cm, h = 10 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (4 × 4 + 4 × 10 + 4 × 10) = 2(16 + 40 + 40)
= 2 × 96
= 192 Sq. cms.

ii) l = 6 cm, b = 4 cm, h = 2 cm.
The total surface area of a cuboid = 2 (lb + bh + lh)
= 2(6 × 4 + 4 × 2 + 6 × 2)
= 2 (24 + 8 + 12)
= 2 × 44
= 88 sq. cms.

Question 2.
Let us find the volume of a cuboid whose length, breadth and height are 6 cm, 4 cm and 5 cm respectively.      [Page No. 287]

Let place 1 cubic centimeter blocks along the length of the cuboid. How many blocks can we place along the length? 6 blocks, as the length of the cuboid is 6 cm.
How many blocks can we place along its breadth? 4 blocks, as the breadth of the cuboid is 4 cm. So there are 6 × 4 blocks can be placed in a layer.
How many layers of blocks can be placed in the cuboid? 5 layers, as the height of the cuboid is 5 cm. Each layer has 6 × 4 blocks. So, all the 5 layers will have 6 × 4 × 5 blocks i.e. length × breadth × height.
This discussion leads us to the formula for the volume of a cuboid.
Volume of a cuboid = length × breadth × height      [Page No. 305]
Answer:
The dimensions of a cuboid are 6 cm, 4 cm, 5 cm respectively.
∴ Volume (V) = lbh
= 6 × 4 × 5.
= 120 cm³

Question 3.
Arrange 64 unit cubes in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid cuboid of same volume have same surface area? [Page No. 306]
Answer:
No. of cuboids are formed using 64 unit cubes
64 = 1 × 64 ……. (1)
= 2 × 32 …….. (2)
= 4 × 16 …….. (3)
1) l = 64 cm, b = 1 cm, h = 1 cm.
The total surface area of a cuboid, A = 2 (lb + bh + lh)
= 2 (64 × 1 + 1 × 1 + 1 × 64)
= 2 (64 + 1 + 64)
= 2 × 129
= 258 Sq. cm.

2) l = 32 cm, b = 2 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (32 × 2 + 2 × 1 + 32 × 1)
= 2 (64 + 2 + 32)
= 2 × 98 = 196 Sq. cm.

3) l = 16 cm, b = 4 cm, h = 1 cm.
A = 2 (lb + bh + lh)
= 2 (16 × 4 + 4 × 1 + 16 × 1)
= 2 (64 + 4 + 16)
= 2 × 84 = 168 Sq. cm.

No, the volume of a cuboid is not same as the surface area of a cuboid.

Try These

Question 1.
Find the surface area of cube ‘A’ and lateral surface area of cube ‘B’.      [Page No. 300]

Answer:
a = 10 cm.
The total surface area of a figure ‘A’ = 6a²
= 6 × (10)²
= 6 × 100 = 600 Sq. cm.
Lateral surface area of a figure ‘B’ – 4a²
= 4 × (8)² [∵ a = 8 cm.]
= 4 × 64 = 256 Sq. cm.

Question 2.
Two cubes each with side ‘b’ are joined to form a cuboid as shown in the given fig. What is the total surface area of this cuboid?      [Page No. 300]

Answer:

Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 (2b × b + b × b + 2b × b)
= 2 (2b² + b² + 2b²)
= 2(5b²) = 10b² Sq. cm.

Question 3.
How will you arrange 12 cubes of equal lengths to form a cuboid of smallest surface area?      [Page No. 300]

Answer:
We can’t obtain the least total surface area by arranging 12 cubes by side by side.

∴ A = 2 (lb + bh + lh)
= 2 (12 × 1 + 1 × 1 + 12 × 1)
= 2 (12 + 1 + 12)
= 2 × 25 = 50 Sq. cm.
We can obtain the least total surface area by arranging 3 cubes by 4 cubes.
∴ A = 2 (lb + bh + lh)
= 2 (3 × 1 + 1 × 4 + 3 × 4) (∵ l = 3; b = 1; h = 4)
= 2 (3 + 4 + 12)
= 2 × 19
= 38 Sq. cm.

Question 4.
The surface area of a cube of 4 × 4 × 4 dimensions is painted. The cube is cut into 64 equal cubes. How many cubes have
(a) 1 face painted? (b) 2 faces painted? (c) 3 faces painted? (d) no face painted?       [Page No. 300]
Answer:
If the 4 × 4 × 4 cube is divided into 64 equal cubes then the length of its each side = 1 unit.

[∵ \(\frac{4 \times 4 \times 4}{64}\) = 1]
a) No.of cubes (a = 4) have painted 1 face = 6(a – 2)² = 6(4 – 2)² = 6 × 4 = 24
b) No.of cubes have painted 2 faces = 12(a – 2) = 12(4 – 2) = 24
c) No.of cubes have painted 3 faces = 4 × a = 4 × 2 = 8
d) No.of cubes have painted no faces = (a – 2)³ = (4 – 2)³ = (2)³ = 8

Think, Discuss and Write

Question 1.
Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base.      [Page No. 299]
Answer:
Total surface area of a cuboid = L.S.A + 2 × Area of base
= 2h (l + b) + 2 × lb
= 2lh + 2bh + 2lb
= 2 (lb + bh + lh)
We can conclude that total surface area of a cuboid = L.S.A + 2 × Area of base

Question 2.
If we change the position of cuboid from Fig. (i) to Fig. (ii) do the lateral surface areas become equal?     [Page No. 299]

Answer:
There will be no change in the L.S.A of a cuboid if its positions are changed.

Question 3.
Draw a figure of cuboid whose dimensions are l, b, h are equal. Derive the formula for LSA and TSA.       [Page No. 299]
Answer:

Lateral surface area of a cuboid
= 4 × (areas of 4 faces)
= 2 (l × h) + 2 × (b × h) (1 + 2 + 3 + 4 faces)
= 2h(l + b) sq.units (1 = 3, 4 = 2)
∴ Total surface area of a cuboid
= 4 × (Area of 4 faces) + (Areas of upper & lower faces)
= 2h (l + b) + 2 (lb)
= 2lh + 2bh + 27b
= 2 (lb + bh + lh) sq.units.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.3

Question 1.
Check whether the given numbers are divisible by ‘6’ or not?
(a) 273432
(b) 100533
(c) 784076
(d) 24684
Solution:
if a number is divisible by ‘6’, it has to be divisible by 2 and 3.

Question 2.
Check whether the given numbers are divisible by ‘4’ or not?
(a) 3024
(b) 1000
(c) 412
(d) 56240
Solution:

Number	Divisible by 4	Yes/No
a) 3024	3024 → \(\frac { 24 }{ 4 }\) (R = 0)	Yes
b) 1000	1000 → \(\frac { 0 }{ 4 }\) (R = 0)	Yes
c) 412	412 →  \(\frac { 12 }{ 4 }\) (R = 0)	Yes
d) 56240	56240 →  \(\frac { 40 }{ 4 }\) (R = 0)	Yes

Question 3.
Check whether the given numbers are divisible by ‘8’ or not?
(a) 4808
(b) 1324
(c) 1000
(d) 76728
Solution:

Number	Divisible by 4	Yes/No
a) 4808	4808 → \(\frac { 808 }{ 8 }\) (R = 0)	Yes
b) 1324	1324 → \(\frac { 324 }{ 8 }\) (R ≠ 0)	No
c) 1000	1000 →  \(\frac { 0 }{ 8 }\) (R = 0)	Yes
d) 76728	76728 →  \(\frac { 728 }{ 8 }\) (R = 0)	Yes

Question 4.
Check whether the given numbers are divisible by ‘7’ or not?
(a) 427
(b) 3514
(e) 861
(d) 4676
Solution:

Question 5.
Check whether the given numbers are divisible by ‘11’ or not?
(a) 786764
(b) 536393
(c) 110011
(d) 1210121
(e) 758043
(f) 8338472
(g) 54678
(h) 13431
(i) 423423
(j) 168861
Solution:

Question 6.
If a number is divisible by ‘8’, then it also divisible by ‘4’. also Explain?
Solution:
If a number is divisible by 8 it ¡s also divisible by 4.
∴ If a number is divisible by 8, then it ¡s also divisible by the factors of 8.
Factors of 8 = 1, 2, 4, 8.
∴ The number which is divisible 8, is also divisible by 4.

Question 7.
A 3-digit number 4A3 is added to another 3-digit number 984 to give four digit number 13B7, which is divisible by 11. Find (A + B).
Solution:
The given 3 – digited numbers are = 4A3, 984
∴ 4A3 + 984 = 13B7. If It is divisible by 11 then,
⇒ 1 3 B 7
(1 + B) – (3 + 7)
⇒ (B+1) – 10 = 0 ⇒ B – 9 = 0
∴ B = 9

⇒ A + 8 = 9 ⇒ A = 9 – 8 = 1
∴ A = 1
A + B= 1+9
∴ A + B = 10

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 6 Biodiversity and its Conservation Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 6th Lesson Biodiversity and its Conservation

8th Class Biology 6th Lesson Biodiversity and its Conservation Textbook Questions and Answers

Improve Your Learning

Question 1.
Read this and answer the following questions.
Answer:
Biodiversity – 2050.
A news item on Biodiversity discussed by Conference of Parties (CoP) -2012- Hyderabad says in the next four decades the earth’s natural resources will be limited to grass lands, mountains, ice and arid and semi arid plains.
By 2050 the loss of Biodiversity will lead to unprecedented. Climate change would be the key factor. Nearly 1.3 million natural ecosystems will be without any original species.
(The coloured areas are indicators of biodiversity loss. The red areas show maximum biodiversity loss.)

a) What does the areas with colour codes indicate?
Answer:
The coloured areas are the indicators of Biodiversity loss.

b) Which areas show maximum biodiversity loss?
Answer:
The red areas show maximum biodiversity loss.

c) Which areas show minimum biodiversity loss?
Answer:
The blue areas show minimum biodiversity loss.

d) From 2010 – 2050 what difference do you find in the state of biodiversity?
Answer:
In the few decades earth’s natural areas will be limited to grass lands, mountains, ice and arid and semi arid plans. Nearly 1.3 million natural ecosystems will be without any original species.

e) So what steps would you suggest to conserve our biodiversity?
Answer:
Utilizing the forests resources judiciously without effecting the ecosystems. So that we can have a sustainable development in the forests and the biodiversity can be conserved for future generations.

Question 2.
How can you say that forests are biosphere reserves? Give reasons.
Answer:

1. Forests are the natural habitates for many types of plants (flora) and animals (fauna).
2. Plants are primary producers as they provide food for entire human population and all other living organisms on earth.
3. Every part of the plant is used by man and animals in their daily life, and also some of the exudates are used by man.
4. We get cereals, pulses, oil seeds, sugars, spices, drugs, timber, fibres and coir from plants.
5. In addition to this products like rubber, resins, fruits, vegetables, dyes, etc. are obtained from plants.
6. Variety of animals live in the forests. Major products obtained from animals are meat, milk hair and skin.
7. Primitive man obtained his food requirements primarily by hunting of animals in the forests.
8. The skin of animals like tiger, lion, leopard, deer, snakes and ivory from elephants are very valuable.
9. Thus we can say that forests are biosphere reserves.

Question 3.
What do you understand about the terms (a) extinct (b) endangered (c) endemic?
Give examples.
Answer:
a) Extinct: When animals vanish for ever from the earth it is said that the species has become extinct. E.g.: Sparrow, vulture.
b) Endangered : It is a warming signal about the organisms whose numbers have declined rapidly and the species might be wiped off from the earth in near future. Such organisms are called endangered species. E.g.: Lion, red fox, loris, wild cat, vulture.
c) Endemic: Plants or animal species found restricted to a particular area of a country are called endemic species. E.g.: Tiger, peacock, kangaroo, kiwi.

Question 4.
a) Extinct: When animals vanish for ever from the earth it is said that the species has become extinct. E.g.: Sparrow, vulture.
b) Endangered: It is a warming signal about the organisms whose numbers have declined rapidly and the species might be wiped off from the earth in near future. Such organisms are called endangered species. E.g.: Lion, red fox, loris, wild cat, vulture.
c) Endemic: Plants or animal species found restricted to a particular area of a country are called endemic species. E.g.: Tiger, peacock, kangaroo, kiwi.
Answer:

1. Some birds live in the same habitate throughout the year. Other birds which don’t have permanent nestlings join into small flocks and move from one region to the other for food and shelter called as ‘migration’ and such birds are called migratory birds.
2. Primary motivation for migration appears to be food. Also longer days of northern summer provide extended time per breeding birds to feed their young ones.

Question 5.
Identify the endemic and endangered species and write them below the pictures.
Answer:

Question 6.
What is the need of conducting biodiversity meet ? Collect information about these meetings when and where it was conducted and its agenda also.
Answer:
The need of conducting biodiversity meet is

1. Managing biodiversity in transboundary landscapes in Hindukush Mountains.
2. Conserving high altitudes wet lands of the Hindukush Himalayans.
3. Operationalizing nagoya protocall in South Asia.

AGENDA

1. Conservation of Biodiversity.
2. Sustainable use of components of Biological diversity.
3. Fair and equitable sharing of benefits arising out of the utilization of genetic resources.

33 Decisions were taken at cop – 11, Hyderabad among them some are given below.

1. Status of Nagoya protocol an axis to genetic resources and equitable sharing of benefits.
2. Review of progress in implementation of strategic plan for Biodiversity – 2011 – 20 and Aichi biodiversity targets.
3. Review of implementation of the strategy for resource mobilization including establishments of targets.
4. Financial mechanism.
5. Cooperation with other conventions International organization and initiatives.
6. Business and Biodiversity.
7. Engagement of other state holders major groups and sub – National authorities.
8. Progress report on gender main streaming
9. Periodicity of meetings.

Question 7.
Nowadays we find animals like leopards and bears intruding into our living places. What may be the reason for this?
Answer:

1. Forests are the living places for wild animals. Animals can get plenty of food, shelter from the shade of trees and they feel secure in the forests.
2. People are so greedy that they cut down the forest areas for logging of wood, to increase agriculture, and for human habitations.
3. They clear the forests to construct, thermal power stations, industries and many buildings which leads to destruction of forests.
4. Due to deforestation, nowadays the animals like leopard and bears, lost their food and shelter, to satisfy their hunger and to keep themselves alive, they intruding into our living places.
5. To escape from hunters and from climatic conditions because of deforestation, is may be one more reason, for the animals coming into our living places.

Question 8.
Make a list of animals/birds seen now and 30 years ago. Take the help of your elders. Write few reasons for their disappearance.
Answer:

1. List of Animals and Birds: Lion, Tiger, Jackel, Fox, Wolves, Deer, Monkeys, Hyena, Squirrel, jungle foul etc. Birds like Crow, Pigeon, Peacock, Koel, Parrot, Flemingo, Mynah, King Fisher, Emu, Migratory shore birds etc.
2. The endangered species of plants and animals of India are:
   Lion, Red fox, Single horned (Rhinoceros), Vulture, Spotted chital deer, Loris, Black spider monkey, wild cat, cycas, Rauvolfia serpentine, Nepenthes, Sandle wood tree.
3. These species include mammals – Indian cheetah, Japan Rhinoceros and Sumatran Rhinoceros. Some species of birds of gone extinct in recent times – including pink headed duck (Rhodonessa caryophyllacca) and Himalayan quail (Ophrysia superciliosia)
4. Warbler (Acrocephalus orinus) – Rampur in H.P was rediscovered after 139 years in Thailand.
5. Based on the case study we find that many animals that were found earlier are not found now.
6. The reason for this is exploitation of land and forest resources by humans, along with hunting and trapping for food and sport has led to the extinction. Feeding of Diclofenac treated cattle is the reason for disappearance of vulture.

Question 9.
Select an area in your locality. Observe the animals (living and visiting) for a day. Prepare a list and plot a graph.
Answer:
The animals in our locality:

1. Dog
2. Cat
3. Rat
4. Mouse
5. Bandicoots
6. Squirrels
7. Frogs
8. Lizards
9. Garden Lizard
10. Monkeys
11. Buffaloes
12. Goats
13. Donkeys
14. Mongoose
15. Snake
16. Sheep
17. Fish
18. Tortoise
19. Rabbit
20. Parrot
21. Crow
22. Hens
23. Koel
24. Pigeon
25. Butterflies
26. Houseflies
27. Dragonflies
28. Mosquitoes
29. Honey bee
30. Cockroach

Question 10.
When tree is considered as an ecosystem, record the flora and fauna connected with it.
Answer:
Flora and Fauna of a tree.
Flora:

1. Grass,
2. Trida (shrub),
3. Datura (Herbs),
4. Creppers,
5. Mosses,
6. Fungi.

Fauna: Squirrels, Butterflies, Dragonfly, Mosquitoes, Birds, Snakes, Ants, Catterpillers, Beetles, Buffaloes, Goats, Human beings, Mouse, Lizard.

Question 11.
Browse through the internet or books on wild life and gather information on birds sanctuaries in India. Prepare a list of birds migrating to India.
Answer:

Birds migration to India:

1. Siberian cranes
2. Greater flamingos
3. Ruff
4. Black winged stilt
5. Common teal
6. Common Green Shark
7. Northern Pintail
8. Yellow wag tail
9. White wag tail

Question 12.
Visit local forest office and collect the data of local flora and fauna.
Answer:
Horticultural Resources: Guava, Mango, Papaya, Sapota, Banana, Coconut, Citrus.
Major oil seeds: Groundnut, Sunflower, Oil palm.
Major food crops: Paddy, Jowar, Ragi, Bazra, Maize, Green gram, Red gram, Black gram.
Commercial crops: Sugarcane, Jute, Chillies, Cotton, Turmeric.
Forest based Resources: Plantation of Eucalyptus Trees, Ponuku wood, Casuarina,
Subabul, Jatropa, Pongamia (bio – diesel plantation).
And also:
Vegetables, flowers, plantation crops, spices and medicinal plants, aromatic crops.
They occupy 6517, 177, 4440, 14315 and 369 hectors respectively in Khammam District.
Local Fauna:
Life Stock Resources: Poultry, Dairy form.
Marine Resources: Fish, Prawn.
Animal Husbandary: Plough animals, Dairy animals like Cow, Buffaloes, Sheep, Goat, Pig.
Forest resources:

1. Panther, Hyena, Jungle cats, Foxes, Bears, and Carnivores, Mammalian are found.
2. Deer, Spotted deer, Sambar, Black buck and other Herbivorous animals found in inland forests.
3. The district has a large number of murrah buffaloes and cows.
4. Migrant grey billed pelican is a protected bird in Kolleru lake and Pulicat lake.

Question 13.
Where do you find most of the biodiversity on the earth? Draw A.P map showing maximum biodiversity areas.
Answer:

1. In areas with sufficient amount of water, a wide variety of plants ranging from grasses to tall trees are seen.
2. Most of the forests are seen in these areas.
3. As there is sufficient vegetation, there will be a large number of herbivorous animals. Carnivorous animals which feed on the herbivores are also found in these regions.
   
4. Generally there is a increase in biodiversity from poles to trophies. Thus localities at lower lattitudes have more species than localities at higher lattitudes.
5. Ultimate factor behind many of the other factors is greater mean temperature at the equator.

Question 14.
What do you understand by biodiversity? How can you say variations are present in them?
Answer:
The word biodiversity is a contraction of biological diversity. It is commonly used to describe the number variety and variability of living organisms. This very broad usage embracing many different parameters, is essentially a synonym of LIFE ON EARTH.

1. The whole world has wide variety of living organisms we can see both invisible (microbial) and the visible world around us are diverse.
2. Different microorganisms like algae, fungi, bacteria, viruses etc., and also the micro-arthropods. There is diverse among microbes.
3. There are different variety of plants like grass, herbs, shrubs, creepers, trees etc. Among the individuals also there are variations like height, colour and size of flower and fruits.
4. We find variations in animals even though they are similar kind. We find differences in colour of fur, nails, claws or hoofs etc.
5. Among birds we can see variations in their feathers, feet, crown, tail etc.
6. As all the humans belong same genus but there is variation in their hands, fingers, toes, nails and hair, height and shape. We can even see variation in the texture of skin dry, oily, smooth or rough. Whether they are twins also we can find variations among them.

Question 15.
Most of our biodiversity is being lost due to human activities. Suggest few ways to protect them.
Answer:
Most of our biodiversity is being lost due to human activities like logging of wood, increased agriculture, increased human habitation and pollution etc. Man has realised this mistake before it was too late. Government of India also realised the importance of wild life and initiated several programmes to preserve wild life in the country and the wild life act was passed in 1972.
Efforts towards conservation:

1. Activities leading to deforestation have been declared as illegal activities and severe punishments have been imposed.
2. Pouching of birds and hunting their eggs are prohibited.
3. Usage of pesticides should be minimised. Usage of biological control methods of pests should be maintained.
4. Efforts to be made to substitude chemical fertilizers with more (natural) bio fertilizers.
5. Pollution from the industries should be reduced.
6. Automobiles should be designed to reduce pollution.
7. Reforestation programmes will be conducted.
8. Gardens, parks, lakes and zoos should be developed.
9. National parks, wild life sanctuaries, where wild life is protected, have been created.
10. Collection, marketing and selling of forest products such as sandal wood, ivory by private parties is banned and is taken up by the Government.
11. Construction of cell phone towers which produce radiation, should be in greater height, so that they should not be reachable to the birds.

Question 16.
When you see a park, sanctuary or a zoo with many kinds of plants and animals, how would you express your happiness? Write a few lines on them.
Answer:

1. Imagine a forest with a carpet of wet leaves littering the ground, the flowers on the trees, we can hear the water drops, sounds of insects, birds chirping and perhaps the distant screech of a monkey – the place picturing is a park, a sanctuary or a zoo. Which gives pleasure to us.
2. These are the homes for many plants and wild animals – and also decorate the world. Any of them are airy and shadowy places.
3. These dwelling places of plants and animals give us happiness. When we are in distress. They give relax – when we feel tired. They give us enjoyment when playing with our friends and they give good health – when we fall sick as they give fresh air and are the lungs of the world.
4. They maintain ecological balance in the environment where we live.
5. We notice the pet dog licks our feet, wags it’s tail, sits near us and walks with us we feel the affection, which gives pleasure mentally.
6. Like this we can experience many situations plants and animals as they are the partners of our environment. So be kind towards them and protect the environment. By maintaining eco – friendly activities.

Question 17.
Prepare an essay to give a talk on biodiversity and conservation.
Answer:

1. The existence of biodiversity in nature teaches us that every plant and animal whether useful or not has right to exist on earth.
2. Every organism is a part of our ecosystem. Loss of any organism endemic or otherwise effects the food chain and food web of that ecosystem, which has impact on the world biodiversity.
3. Hence if we want to protect the biodiversity on our planet, first we must be a part of conservation and then make other aware of it because today we see extinction of some species tomorrow it could be our species.
4. Conserving the biodiversity in a wider prospective is utilizing the forest resources judiciously without affecting the ecosystems so that we can have a sustainable development in the forests and the biodiversity can be conserved for future generations.
5. Nature is for human’s need, not for his greed. If we protect nature, it protects us.

Question 18.
Rani said conservation, of biodiversity starts from our home. Is she correct? How do you support Her? What will be your action for this?
Answer:

1. We live in houses that protect us from heat, cold and rain etc.
2. We keep some animals and birds as pets in our houses. We also grow some plants which give us fruits and vegetables.
3. Thus we can say that our house is also a habitat. Several animals like dogs, cats, goats, cows, birds, hens, ducks, pigeons, spiders, ants, cockroaches live with us.
4. Plants like money plant and some crotons are also kept inside our house.
5. We know that every bit of effort towards conservation helps. If we take due care of plants growing around us, we may not be adding a forest, but adding to greenery around us which is essential for our own existence.
6. So Rani is correct. Conservation of biodiversity starts from our house.

Question 19.
When we take steps towards conserving the tiger, what are the other things that have to be conserved?
Answer:

1. When we take steps towards conserving the tiger we have to conserve the other flora and fauna related to the tiger.
2. If a tiger has to be saved it’s food web should be protected.
3. The tiger depends for food on deer and many other herbivores.
4. If the tiger disappears, the deer and other herbivores population will increase and that would affect the flora of the area.
5. All organisms in nature influence each other in some way or the other. So we need to protect all of them.

Question 20.
Prepare some slogans or a pamphlet to make aware of people about conservation of biodiversity. (OR)
Prepare two slogans to conserve biodiversity in your area.
Answer:

1. Save tree – Save other lives too.
2. Save the lungs of the earth.
3. Be kind towards biodiversity.
4. Reduce pollution.
5. Protect Nature, it protects us.
6. Nature is for human’s need not for his greed.
7. Hunting of wild life is a crime.
8. Forest is our life.

8th Class Biology 6th Lesson Biodiversity and its Conservation InText Questions and Answers

Question 1.
Rampachodavaram: East Godavari District, 60-70 years back Rampachodavaram had dense forest with a rich heritage of wild life. These forest extended to border areas of Aswaraopet of West Godavari district. It was an abode for wild animals like tigers, leopard, deers, hyenas (kondrigallu), foxes, wild boars (adavi pandi), bears, pythons, cobras, porqupines (mulla pandhi), owls, hares, monitor lizard (udumu) scorpions, geremandals (like the desert spider) etc.
After the erection of mines (colour soil) and other industries, human activities increased. Then many buildings, roads and stone quarries have come into exist¬ence. Forest area was cleared and so several organisms started disappearing.
Though an area near Maredumilly, Addateegala (very close to Eleswaram) was once known as Tiger valley, shows no signs of tigers now. Animals like foxen, deers are also not seen these days.
Now there are several human settlements in the area. Some areas of less dense forests with animals like pythons, cobras, deers, scorpions etc., are commonly seen. Bears are rarely found. Peacocks have been sighted recently.
The above case study explains you the need of conservation of biodiversity.
a. What is the difference between the situation regarding types of animals present 70 years ago and now?
Answer:
70 years ago Rampachodavaram had dense forest with a rich heritage of wild life. Now no animals are not seen. Now there are several human settlements in that area.

b. What might have happened to tigers of Rampachodavaram?
Answer:
After the erection of Mines (colour soil) and other industries, human activities increased. Many buildings, roads and stone quarries have come into existence. Forest areas were cleared and so several organisms started disappearing.

c. Do we find tigers any where else in our country?
Answer:
Tigers are found in other parts of our country and the world as well.

d. Peacocks love eating snakes. Can you guess why they dwell in this place?
Answer:
We can find snakes in desert areas. Peacocks love eating snakes. So they dwell in this place where it finds it’s food (snakes) plenty.

Question 2.
Is there any extinct species in your area ? Name them and write a note on them.
Answer:
Sparrow, Vulture.
i) Over use of pesticides and radiation from cell phone towers led to extinction of sparrow.
ii) By feeding of diclofenac – treated cattle led to extinction of vulture.

Question 3.
Give your reasons as for why the organisms become so extinct?
Answer:

1. Either knowingly or unknowingly, man has destroyed the wild life.
2. Hunting of animals either for food or for pleasure, cutting of trees and clearing of forests for fuel, timber and for human settlements, construction of dams and reservoirs has resulted in large scale destruction of forests. This has destroyed the wild life.

Question 4.
How biodiversity is depleting in your area? How to improve it?
Answer:
Biodiversity depleting: Reduction of plant and animal species is called biodiversity depletion. Causes:
1) It may be caused by natural causes which include floods, earthquakes, land slides, diseases etc.
2) Man made causes are called ‘Anthropocentric’ causes. These are

1. Urbanization
2. Expansion of agriculture
3. Deforestation
4. Pollution

In my area: Lot of animal species are in danger of depletion in may areas due to human activities. Sparrow’s, voltures, become rare ones. Their population decreased rapidly.
How to improve?

1. Give importance to plantation
2. Avoid deforestation
3. Installing bird boxes
4. Protect the native species
5. Provide wild life corridors
6. Use organic manures
7. Utilise existing green space connections
8. Be mercy with other creatures.

Question 5.
Observe the pictures and identify the animals. Also try to find out where these can be found.

Answer:
1. Peacock – India (It is our National bird)
2. Tiger – India (It is our National animal)
3. Kiwi – Newzealand

Question 6.
Name an Endemic species of our state.
Answer:
Indian lion, Leopard.

Question 7.
Why should we conserve a small insect like a bee or butterfly?
Answer:
The insects like bee and butterfly, suck nectar from the flowers. By this pollination takes place in flowers.

Question 8.
What will happen if these insects become extinct?
Answer:
Insects help in pollination of flowers. By this pollination fertilization of flowers takes place and seeds will form which helps in the growth of next generation of plants. If insects become extinct – no pollination – no fertilization and no future generations of plants, there by extinction of plants takes place.

Question 9.
What can be done to save these insects?
Answer:
Spraying of pesticides will be minimised. Biological Methods will be used to control pests. (The animals which feed on pests will be used in the agricultural lands.)

8th Class Biology 6th Lesson Biodiversity and its Conservation Activities

Activity – 1 & 2

Question 1.
How many different colours could you mark on your sheet?

Answer:
Seven different colours. (Refer textbook page 85 for colours)
i) What does the colours indicate?
Answer: The colours indicate the existence of plants, animals, insects, humans, and birds,
ii) What does your total colour code count indicate?
Answer: Plants, animals, birds, insects live in our surroundings.
iii) What are the things that attract you very much in the nature?
Answer: Bird’s nests, cobwebs, worms, leaves, insects, mosses etc attract us.
iv) Write your feelings without any hesitation.
Answer: Enjoyment, happiness and pleasure.

Activity – 3

Question 2.
Variations in plants.
Answer:

Similarities:

1. Paddy and Maize belong to grass plants.
2. Both of them have same root system (fibrous R.S.)
3. They are green colour.
4. Seeds are enclosed.
5. They produce cereals.

Variations in Animals:
a) Do you find any differences between animals?
Answer:
They show difference in colour of fur, nails, claws, hoofs etc.

b) Do you find any differences among birds?
Answer:
Birds have differences in their feathers, feet, crown, tail etc.

Variations in Human beings:
a) Observe two students of your class. Do they appear similar?
Answer:
No. Human beings show differences in their height & shape, hands, fingers, toes, nails and hair.
The texture of skin also may be dry, oily, smooth or rough.

b) Suppose two of your classmates happen to be twins, will they look same in structure and shape.
Answer:
If they are twins also we find little differences.

Activity – 4

Question 3.
Collect and paste some pictures of your favourite cricket players belonging to countries like West Indies, Australia, India, etc. in your note book.
Write the differences and similarities that you have noticed in them. What diversity you observed?
Display your finding of above activity in the class and discuss the following questions.
a) Are there any two organisms with 100% similarities between them?
Answer:
No, there are no organisms with 100% similarities.

b) Why do they differ from each other?
Answer:
Because they belong to different species.

c) What will happen if all plants are creepers?
Answer:
If all plants are creepers there will be no shelter for many birds and animals.

d) Hen and goat both have legs. What diversity do you find between them?
Answer:
Hen is a bird and goat is an animal. So hen has two legs and goat has 4 legs.

e) Are all the nests of birds similar why?
Answer:
Because of their living conditions and food habits, the nests of birds are not similar.

f) Do animals all around the world have similar organs and functions? What is the diversity behind them?
Though they look similar, upon careful observation we find differences or variations between them that leads to Biodiversity.

There is no mononamy or uniqueness in structure and functions of nature.
Diversity is the nature’s way.

Activity – 5 (Project work)

Question 4.
Studying migration and its effect on biodiversity of an area.

Read the following ways to conserve biodiversity try to enrich this list in your own way.
1. Look at the sky in the morning and evening. Do you observe birds flying in groups ?
Answer:
Yes. We can see birds flying in the sky in the morning and evening in groups.

2. Did you get the same number and types of birds every day?
Answer:
No. Some times more and some times less in number.

3. Was there any sudden variation in a particular season?
Answer:
Particularly in winter season we can see large number of birds flying in the sky.

4. Did you notice any new type of bird population in any season?
Answer:
During rainy season most of the birds from far away places migrate to Kolleru and Pulikot lakes of our states.

5. Why do these birds move from one place to another?
Answer:
Birds move from one place to another for food and shelter (nestling habits)

6. Sometimes at night we see birds flying in groups. Where do you think they fly to?
Answer:
Sometimes to protect themselves from climatic conditions, for food, for reproduction and to escape from hunting and also due to deforestation, we see birds flying in groups.

7. Perhaps the most important value of biodiversity, particularly in a country like India. Is that it meets the basic survival needs of a vast number of people.
Answer:
Cereals, pulses, oil seeds, sugars, spices, drugs, fibres, coir, timber, resins, gums, fruits, vegetables, dyes are obtained from plants. Meat, skin and hair are obtained from animals.
Like this the biodiversity in our country meets the basic survival needs of a vast number of people.
Flora and Fauna are renewable resources and are to be use judiciously.

Activity – 6 (Project work)

Question 5.
How to make recycled newspaper from waste newspapers? (OR)
Write the procedure of preparation of recycled paper which you did in your school lab.
Answer:
Materials:
2 plastic tubs, wooden spoon, water, clean cotton cloth, old news paper, wire screen, measuring cup, plastic wrap, blender, heavy books / roller.
Procedure:

1. Add cut news papers strips in a tub full of water and soak it for a day.
2. Put two cups of soaked paper and six cups of water in a blender. Blend till the mixture turns into a pulp (like nanny oat meal).Pour it in a clean tub.
3. Fill the tub with one fourth of blended paper pulp.
4. Lay a cloth on a flat, waterproof surface. Slide the wire screen under the wet paper. Remove the screen gently. Press the news paper pulp to squeeze out any extra water.
5. Carefully flip the screen onto the cloth. Press it down firmly. Remove the screen.
6. Lay another cloth on top of the mixture. Cover the cloth with a plastic wrap and stack the books on the wrap.
7. After several hours remove the books on the cloth and let the paper dry.
8. You can even use a hair dryer to blow the paper dry.
9. By adding few drops of edible colours to the pulp you can make your paper colourful. Iron the new made paper with a iron box and cut it to your required size and shape.
10. Beautiful greeting cards, file covers, bags etc., can be made using recycled paper.

Question 6.
How is a compressed cardboard prepared?
Answer:
Materials: bits of wood, saw dust and chemicals sulphate.
Producer:

1. The pulp is made by using bits of wood.
2. It is spead evenly as layers.
3. The saw dust is sandwiched between the two layers.
4. This is compressed and dried.
5. It becomes hard and strong as wooden board.
6. Hence there is no need to cut down the whole tree. This helps in reducing deforestation.

AP State Syllabus 8th Class Maths Solutions 4th Lesson Exponents and Powers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 4 Exponents and Powers InText Questions and Answers.

8th Class Maths 4th Lesson Exponents and Powers InText Questions and Answers

Do this

Question 1.
Simplify the following.   (Page No. 81)
i) 3⁷ × 3³
ii) 4 × 4 × 4 × 4 × 4
iii) 3⁴ × 4³
Answer:
(i) 3⁷ × 3³ = 3⁷ + 3³ = 3¹⁰       [∵ a^m × aⁿ = a^m+n]
(ii) 4 × 4 × 4 × 4 × 4 = 4⁵      [∵ a × a × a × ……. m times = a^m]
(iii) 3⁴ × 4³ = 3⁴⁺³ = 3⁷      [∵ a^m × aⁿ = a^m+n]

Question 2.
The distance between Hyderabad and Delhi is 1674.9 km by rail. How would you express this in centimetres? Also express this in the scientific form.     (Page No. 81)
Answer:
Distance from Hyderabad to Delhi is
= 1674.9 km = 1674.9 × 1000 m = 1674900 mts
= 1674900 × 100 cm
= 167490000 cm
= 16749 × 10⁴ cm

Question 3.
What is 10^-10 equal to?     (Page No. 83)
Answer:
10^-10 = \(\frac{1}{10^{10}}\)      [∵ a^-n = \(\frac{1}{a^{n}}\)]

Question 4.
Find the multiplicative inverse of the following. (Page No. 83)
Answer:

Question 5.
Expand the following numbers using exponents. (Page No. 84)
Answer:
i) 543.67
= (5 × 100) + (4 × 10) + (3 × 10⁰) + \(\left(\frac{6}{10}\right)\) + \(\left(\frac{7}{10^{2}}\right)\)
= (5 × 10²) + (4 × 10) + (3 × 10⁰) + (6 × 10^-1) + (7 × 10^-2)   [∵ aⁿ = a^-n]

ii) 7054.243
= (7 × 1000) + (0 × 100) + (5 × 10) + (4 × 10⁰) + \(\left(\frac{2}{10}\right)\) + \(\left(\frac{4}{100}\right)\) + \(\left(\frac{3}{1000}\right)\)
= (7 × 10³) + (0 × 10²) + (5 × 10¹) + (4 × 10⁰) + (2 × 10^-1) + (4 × 10^-2) + (3 × 10^-3)

iii) 6540.305
= (6 × 1000) + (5 × 100) + (4 × 10) + (0 × 10⁰) + \(\left(\frac{3}{10}\right)\) + \(\left(\frac{0}{100}\right)\) + \(\left(\frac{5}{1000}\right)\)
= (6 × 10³) + (5 × 10²) + (4 × 10¹) + (0 × 10⁰) + (3 × 10^-1) + (0 × 10^-2) + (5 × 10^-3)

iv) 6523.450
= (6 × 1000) + (5 × 100) + (2 × 10) + (3 × 10⁰) + \(\left(\frac{4}{10}\right)\) + \(\left(\frac{5}{100}\right)\) + \(\left(\frac{0}{1000}\right)\)
= (6 × 10³) + (5 × 10²) + (2 × 10¹) + (3 × 10⁰) + (4 × 10^-1) + (5 × 10^-2) + (0 × 10^-3)

Question 6.
Simplify and express the following as single exponent.    (Page No. 85)
(i) 2^-3 × 2^-2
(ii) 7^-2 × 7⁵
(iii) 3⁴ × 3^-5
(iv) 7⁵ × 7^-4 × 7^-6
(v) m⁵ × m^-10
(vi) (-5)^-3 × (-5)^-4
Answer:
(i) 2^-3 × 2^-2 = 2^(-3)+(-2) = 2^-5 = \(=\frac{1}{2^{5}}\) = \(\frac{1}{2 \times 2 \times 2 \times 2 \times 2}\) = \(\frac{1}{32}\) [∵ a^m × aⁿ = a^m+n]
(ii) 7^-2 × 7⁵ = 7^-2+5 = 7³ = 343
(iii) 3⁴ × 3^-5 = 3^4+(-5) = 3^-1 = \(\frac{1}{3}\) [∵ a^-n = \(\frac{1}{a^{n}}\)]
(iv) 7⁵ × 7^-4 × 7^-6 = 7^5+(-4)+(-6) = 7^5-10 = 7^-5 = \(=\frac{1}{7^{5}}\)
(v) m⁵ × m^-10 = m^5+(-10) = m^-5 = \(=\frac{1}{m^{5}}\)
(vi) (-5)^-3 × (-5)^-4 = (-5)^(-3)+(-4) = (-5)^-7 = \(\frac{1}{(-5)^{7}}\) = –\(\frac{1}{5^{7}}\)

Question 7.
Change the numbers into standard form and rewrite the statements.      (Page No. 93)
i) The distance from the Sun to Earth is 149,600,000,000 m
Answer:
149,600,000,000 m = 1496 × 10⁸ m

ii) The average radius of the Sun is 695000 km
Answer:
695000 km = 695 × 10³ km

iii) The thickness of human hair is in the range of 0.005 to 0.001 cm.
Answer:
0.005 to 0.001 cm
= \(\frac{5}{1000}\) to \(\frac{1}{1000}\) cm = 5 × 10^-3 to 1 × 10^-3 cm

iv) The height of Mount Everest is 8848 m
Answer:
8848 m, itself is a standard form.

Question 8.
Write the following numbers in the standard form.      (Page No. 93)
The standard form of the following numbers are
Answer:
(i) 0.0000456 = \(\frac{456}{10000000}\) = 456 × 10^-7
(ii) 0.000000529 = \(\frac{529}{1000000000}\) = 529 × 10⁹
(iii) 0.0000000085 = \(\frac{85}{10000000000}\) = 85 × 10¹⁰
(iv) 6020000000 = 602 × 10000000 = 602 × 10⁷
(v) 35400000000 = 354 × 100000000 = 354 × 10⁸
(vi) 0.000437 × 10⁴ = \(\frac{437}{1000000}\) × 10⁴
= 437 × 10^-6 × 10⁴
= 437 × 10^(-6)+4
= 437 × 10^-2

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 2 Cell: The Basic Unit of Life Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 2nd Lesson Cell: The Basic Unit of Life

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Textbook Questions and Answers

Improve Your Learning

Question 1.
Who discovered the cell for the first time?
Answer:
It was the year 1665 Robert Hooke, a British scientist observed thin slices of cork under a simple magnifying device which he had made himself. He observed that the cork resembled the structure of a honey comb consisting of many empty spaces or empty box like structures. He thought it was made up of very small cavities, Robert Hooke called these cavities “Cell”.

Question 2.
Name the factors on which shape of the cells depend.
Answer:
The shape and size of the cells vary considerably but all of these cells ultimately determined by the specific function of the cells.
e.g.: Amoeba is changing its shape for specific functions like collection of food and locomotion.
The shape of the cell may vary for giving definite structure to the organism, e.g,: Epidermal cells.

Question 3.
Distinguish between unicellular and multi cellular organisms.
Answer:

Unicellular	Multicellular
1) An organism composed of just one cell.	1) An organism composed of more than cell.
2) Many of nature’s simplest creatures called “unicellular organisms”.	2) More biologically advanced creatures, called “multicellular organisms”.
3) Cell is an individual form no gathering to perform tasks, but they live together.	3) Different kinds of cells are joined together to perform specialized tasks.
4) The single cell a unicellular organism possesses, the smaller its body, e.g.: Amoeba, Chlamydomonas.	4) The more cells a multicellular organism possesses, the larger its body, e.g.: Fish, Neem tree.

Question 4.
How will you prepare slide without drying quickly?
Answer:
Preparation of slide is a technique to observe microscopic structures. Microscopic slide is prepared on a 2 mm thick. Thin flat plant material directly placed in a drop of water on the glass slide. A drop of glycerin is added to the water to keep the material for longer time. Glycerine saves the material from drying quickly.

Question 5.
Deekshith said that, “we can’t see cells with unaided eye.” Is the statement true or false? Explain.
Answer:
We can’t see cell with naked eye is true. All living things formed by microscopic cells which are visible through microscope only but the egg of birds is visible without microscope. The size of the cells in living organism may be as small as the millionth of a meter.
Most of the cells either in unicellular and multicellular are small in size to perform all the life processes perfectly. The smallest cell 0.1 to 0.5 micrometers found in bacteria. Some of the cells can be seen with naked eye. The largest cell 17 cm x 18 cm egg of Ostrich.

Question 6.
Correct the statement and if necessary rewrite. (OR)
What is cell wall and what are its functions?
a) Cell wall is essential in plant cells.
b) Nucleus controls cell activity.
c) Unicellular organisms perform all life processes like respiration, excretion, growth and reproduction.
d) To observe nucleus and organelles clearly, staining is not necessary.
Answer:
a) Cell wall is essential in plant cells.
Plant cell walls are essential for plant life and also have numerous industrial applications, ranging from wood to nutraceuticals.
The cell wall is the tough, usually flexible but sometimes fairly rigid layer that surrounds some types of cells. It is located outside the cell membrane and provides these cells with structural support and protection, in addition to acting as a filtering mechanism. A major function of the cell wall is to act as a pressure vessel, preventing over-expansion when water enters the cell. Cell wall is found in plants, bacteria, fungi, algae, archaea. Animals and protozoa do not have cell walls. “Plant cell wall is an essential component of biotic stress response mechanisms.”

b) Nucleus controls cell activity.
By containing the instructions for protein products in the DNA of the nucleus. All “control” work in the cell is carried out by proteins, such as enzymes, though DNA codes for other structural material, only protein has metabolic and behavioural control in the organism’s cells. Thus, the nucleus is the cell’s control center.

c) Unicellular organism perform all life processes like respiration, excretion, growth and reproduction.
All living organisms perform some basic life processes like respiration, excretion, etc., to sustain its life and improve its race. Unicellular organisms also perform all life processes.

d) To observe nucleus and organelles clearly, staining is not necessary.
To observe nucleus and organelles clearly, staining is necessary. Staining is a technique to get attached color to different parts of a cell. This helps to highlight particular areas in the cell.

Question 7.
Describe the structure of Nucleus.
Answer:
The nucleus is the largest cellular organelle in animals. In mammalian cells, the average diameter of the nucleus is approximately 6 micrometers (pm), which occupies about 10% of the total cell volume.

Question 8.
Explain the functions of Nucleus.
Answer:
Functions:

1. The main function of the cell nucleus is to control gene expression and mediate the replication of DNA during the cell cycle.
2. The nucleus provides a site for genetic transcription that is segregated from the location of translation in the cytoplasm, allowing levels of gene regulation that are not available to prokaryotes.

Question 9.
What is difference between cells in onion peel and cells in Spinach?
Answer:
Cells in onion peel arranged systematically with prominent nucleus. Cells in spinach are in different sizes and shapes without nucleus to perform nutrition.

Question 10.
Label parts of diagrammes given below. And identify which one is plant cell and which one is animal cell.

Answer:
A. Nucleus
B. Cytoplasm
C. Cell membrane
D. Vacuole
E. Nucleus
F. Cell wall
G. Cell membrane
H. Vocuole
I. Vacuole

Question 11.
What questions will you pose to know diversity in cells?
Answer:

1. Are all the cells similar in shape and size?
2. Do you find nuclei in all the cells?
3. How many different types of cells could you see?
4. What are the different shapes of the cells?
5. Do all the cells one of the same in length?

Question 12.
If you want to know about unicellular and multicellular organisms, what questions will you pose?
Answer:

1. What do you mean by unicelluar organism?
2. What do you mean by multicellular organism?
3. Give examples for unicellular and multicellular organisms.
4. What are the differences between unicellular and multicellular organism?

Question 13.
Get some floating slime from a puddle, pick a very small amount of slime and put it on a slide. Separate out one fiber and look at it through the microscope. Draw the diagram what you have observed.
Answer:

Question 14.
Collect different kinds of leaves from your surroundings and observe the shapes of the epidermal cells under microscope. Make a table which contains serial number, name of the leaf, shape of the leaf, shape of the epidermal cells. Do not forget to write specific findings below the table.
Answer:

Question 15.
Make sketches of animal and plant cells which you observed under microscope.
Answer:

Question 16.
Ameer said “Bigger onion has larger cells when compared to the cells of smaller onions”! Do you agree with his statement or not? Explain why.
Answer:
The sizes of the cells in living organisms are too small to be seen with naked eye. The size of the cell is related to its function. The cell has to perform similar function in all living organisms.
The size of the onion depends upon the number of cells and not the size of the cell. Cells are of different shape, size and number.
Hence, I don’t agree with Ameer. “Bigger onions have larger cells when compared to the cells of smaller onions”.

Question 17.
How do you appreciate the fact that a huge elephant, man and trees are made of cells, which are very small and we can look at them through microscope?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell both in man and animals are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number.

Question 18.
Deepak said, “A plant can’t stand erect without cell wall”. Support this statement.
Answer:
Deepak said, “A plant can’t stand erect without cell wall.” We support this statement with the following reasons.
i) Plant cells differ from those of animals in having an additional layer around the cell membrane.
ii) We called if as ‘cell wall’.
iii) Cell wall gives strength and rigidity to planks.
So a plant can’t stand erect without cell wall.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life InText Questions and Answers

Question 1.
Make different questions to know cells and cell organelles.
Answer:

1. What are the structures present in the cells?
2. Why cells are considered to be structural and functional unit of life?
3. What is the need of cell wall in plant cells?
4. Did we see the cells with naked eye?

Question 2.
Prepare different questions to know the discovery of cell.
Answer:

1. In which year cell discovered?
2. Name the scientist who observed cells.
3. What type techniques used to observe the cell?
4. Is there any special devices used to study the cell?
5. Name the different devices used for discovery of cell.
6. Can we see living cells under the microscope?

Question 3.
Prepare permanent slides of Onion cell, cheek cell and compare practically.
Answer:
Take inner layer of onion and cheek cells. Stain them with saffranin or methylene blue and keep coverslip.
Observe both the slides under microscope. Cell membrane and cell wall is present in both the cells. Dark stained nucleus is presented in the centre of the cell.
Comparison between onion and cheek cells:

Onion cells	Cheek cells
Cells arranged compactly	Cells arranged loosely
Boundary of onion is cell membrane	Boundary of cheek cell is cell wall
Cytoplasm and nucleus is present	Cytoplasm surrounds the nucleus
Cell organelles present in cytoplasm	Cell organelles present in cytoplasm.

Question 4.
Explain diversity in leaf cells practically.
Answer:
Take a section of Neem leaf on the slide and put a drop of water, cover it with a coverslip and observe it under the microscope. We can see different types of tissues present in the leaf. The section of leaf shows the following features.
Three groups of tissues arranged in leaf.
The first group epidermis where the cells are barrel shaped and covered by waxy material for protection.
The second group mesophyll tissue where the cells contain chloroplasts for nutrition. Cells arranged loosely with air spaces and stomata for exchange of gases.
The third group vascular tissue where the cells are thick walled to transport water and food.

Question 5.
Make a sketch of blood cells.
Answer:

Question 6.
Draw a neat diagram of Clilamydomonas.
Answer:

Question 7.
Make a sketch of Amoeba. (OR)
a) Draw a labelled diagram of Amoeba.
b) What are pseudopodia?
Answer:
a)

b) The projections on the body of amoeba which help in locomotion and collecting food.

Question 8.
Have you listened to the words of the cell? Guess how big a cell is? Is the number and sizes of cells in both man and elephant the same? Are the cells of an elephant bigger than that of a man?
Answer:
The size of the cells in living organisms may be as small as the millionth of a meter (micron) or may be as large as a few centimeters. Majority of the cells are too small to be seen with naked eye.
The size of the cell is related to its function. For example, nerve cell in both man and elephant are long and branched. They perform the same functions that of transferring message.
The size of the organism is depending upon the number of cells and not the size of the cell. Cells are of different shape, size and number. Hence size of cells in both elephant and man are same. The number of cells are more in elephant than man.

8th Class Biology 2nd Lesson Cell: The Basic Unit of Life Activities

Activity – 2

Question 1.
Prepare a slide of an onion peel and find out the special characters.
Answer:
Peel an onion and cut out a small fleshy portion from the bulb. Break this into two small parts and try to separate them. We notice a thin film like material holding the pieces together. Take out small portion and spread it evenly on a slide. Cover it with a cover slip and observe it under microscope.
Draw the figure what you have observed.
Cells are arranged side by side without any gaps.

Activity – 4

Question 2.
Observation of the Nucleus in onion peel cells. (OR)
Sahitya is trying to observe the nucleus in the cells of onion peel. Explain the procedure to be followed for the experiment.
Answer:
Peel a membrane an onion now keep this membrane on a slide and add 1-2 drops of the stain (saffranin, methylene blue or red ink). Cover this with a cover slip and leave it for about five minutes. Then add water drop-wise from one side of the cover slip while soaking the extra water with a filter paper from the other side. This will help in washing away the extra stain. Now observe this slide under a microscope. The blue spot observed within the cell is the nucleus.

Activity – 5

Question 3.
Observe the following pictures and answer the questions given below.

i) What are the structures present in the cells ?
Answer:
Cell wall in onion cell and cell membrane in cheek cell.
Cytoplasm and nucleus are common in both the cells.

ii) Did you see a tiny dark stained thing in all the cells ?
Answer:
Yes there is a tiny dark stained nucleus is common in both the cells.

iii) Are they located in the center of the cell in both the cells ?
Answer:
Nucleus located in center of both the cells.

iv) What is the difference between boundary of onion cell and cheek cell ?
Answer:
Cell wall is the boundary of onion and cell membrane is the boundary of cheek cell.

Activity – 6

Question 4.
Collect leaves stems and roots of different plants from the field and take sections to study different types of cells and tissues present in leaf and stems practically.
Answer:
Take a section of grass leaf on the slide and put a drop of water, cover it with a cover slip and observe it under the microscope. We can see different types of tissues present in the leaf.
The section of root or stem shows the following features.
Four groups of cells can be observed.
First group is outermost layer called epidermis.
It protects stem or root externally.
Major portion of stem or root has second group of cells.
This group synthesizes food and preserve food.
Third group of cells transport water and food.
Fourth group of cells placed centrally.

Question 5.
a) Observe the following cells and collect permanent slides.

b) Fill the following table with help of your teacher.

Name of the cell	Shape of the cell	Name of the parts observed in it
RBC	Biconcave	Blood tissue
Smooth Muscle Cell	Rod	Muscles
Nerve Cell	Tree	Brain, spinal cord and nerves
Bone Cell	Star	All bones
White Blood cell	Amoeboid	Blood tissue

i) Are there any similarities in shape of the cells?
Answer:
No similarity is found in the shape and size of the above cells.

ii) Do you find nuclei in all the cells?
Answer:
In human RBC nucleus is absent. Muscle, Nerve, Bone and White Blood Cells consist nucleus.

iii) Do you know, which cell is the longest in all animals?
Answer:
Nerve cell is the longest in all animals.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures InText Questions and Answers.

8th Class Maths 8th Lesson Exploring Geometrical Figures InText Questions and Answers

Do this

Question 1.
Identify which of the following pairs of figures are congruent.     [Page No. 184]

Answer:
The congruent figures are (1, 10), (2, 6, 8), (3, 7), (12, 14), (9, 11), (4, 13).

Question 2.
Look at the following pairs of figures and find whether they are congruent. Give reasons. Name them.    [Page No. 185]

Answer:
i) △ABC, △PQR
∠A = ∠Q Angle
There is no information about other angles (or) sides.
But if we overlap each other, they coincide.
∴ △ABC ≅ △PQR

ii) From △PLM, △QNM
PL = QN (S)
LM = MN (S)
PM = QM (S)
By S.S.S congruency, these two triangles are congruent.
∴ △PLM ≅ △QNM

iii) From △LMN, △PQR
NL ≠ PQ,LM ≠ QR, NM ≠ RP [∵ The corresponding angles are not given]
∴ △LMN ≆ △PQR

iv) From fig. ABCD is a parallelogram and LMNO is a rectangle.
In any case a rectangle and a parallelogram are not congruent.
∴ ▱ ABCD ≆ □ DLMNO

v) Both the circles are having same radii,
i.e., r₁ = r₂ = 2 units
∴ The given circles are congruent to each other.

Question 3.
Identify the out line figures which are similar to those given first.    [Page No. 186]

Answer:
The similar figures are (a) (ii), (b) (ii).

Question 4.
Draw a triangle on a graph sheet and draw its dilation with scale factor 3. Are those two figures are similar?      [Page No. 191]
Answer:
Step – 1: Draw a △ PQR and choose the center of dilation C which is not on the triangle. Join every vertex of the triangle from C and produce.

Step – 2: By using compasses, mark three points P’, Q’ and R’ on the projections
so that
CP’ = k(CP) = 3CP
CQ’ = 3 CQ
CR’ = 3 CR

Step- 3: Join P’Q’,Q’R’and R’P’.
Notice that △P’Q’R’ ~ △PQR

Question 5.
Try to extend the projection for any other diagram and draw squares with scale factor 4, 5. What do you observe? [Page No. 191]
Answer:
Sometimes we need to enlarge 10 the figures say for example while making cutouts, and sometimes we reduce the figures during designing. Here in every case the figures must be similar to the original. This means we need to draw enlarged or reduced similar figures in daily life. This method of drawing enlarged or reduced similar figure is called ‘Dilation’.

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
Every vertex A, B, C, D are joined from the sign ‘O’ and produced to 4 times the length upto A, B, C and D respectively. Then A, B, C, Dare joined to form a square which 4 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and
\(\frac{OA’}{OA}\) = \(\frac{4}{1}\) = 4 is called scale factor.

Question 6.
Draw all possible lines of symmetry for the following figures.     [Page No. 193]

Answer:

Try these

Question 1.
Stretch your hand, holding a scale in your hand vertically and try to cover your school building by the scale (Adjust your distance from the building). Draw the figure and estimate height of the school building.      [Page No. 189]
Answer:
Illustration: A girl stretched her arm towards a school building, holding a scale vertically in her arm by standing at a certain distance from the school building. She found that the scale exactly covers the school building as in figure. If we compare this illustration with the previous example, we can say that

By measuring the length of the scale, length of her arm and distance of the school building, we can estimate the height of the school building.

Question 2.
Identify which of the following have point symmetry.     [Page No. 196]
1.

2. Which of the above figures are having symmetry ?
3. What can you say about the relation between line symmetry and point symmetry?
Answer:
1. The figures which have point symmetry are (i), (ii), (iii), (v).
2. (i), (iii), (v).
3. Number of lines of symmetry = Order of point symmetry.

Think, discuss and write

Question 1.
What is the relation between order of rotation and number of axes of symmetry of a geometrical figure?     [Page No. 195]
Answer:
The line which cuts symmetric figures exactly into two halves is called line of symmetry. The figure is rotated around its central point so that it appears two or more times as original. The number of times for which it appears the same is called the order of rotation.

From the above table number of lines of symmetry = Number of order of rotation.

Question 2.
How many axes of symmetry does a regular polygon has? Is there any relation between number of sides and order of rotation of a regular polygon?      [Page No. 195]
Answer:
Number of sides of a regular polygon are n. Then its lines of symmetry are also n.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.2

Question 1.
If 345 A 7 is divisible by 3,supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
∴ 345A7 ⇒ 3 + 4 + 5 + A + 7 = 19 + A
19 + A = 3 x 7
⇒ A = 21 – 19 = 2 ⇒ A = 24 – 19 = 5

A + 19 = 3 x 8
⇒ A = 24 – 19 = 5

A + 19 = 3 x 9
⇒ A = 27 – 19 = 8

∴ A = {2,5,8}

Question 2.
If 2791 A,is divisible by 9, supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
∴ 2791A = 2 + 7 + 9 + 1 + A = 9 x 3
⇒ 19 + A = 9 x 3 = 27
⇒ A = 27 – 19 = 8
∴ A = 8

Question 3.
Write some numbers which are divisible by 2,3,5,9 and 10 also.
Solution:
90, 180, 270. are divisible by 2, 3, 5, 9 and 10.
[∵ The L.C.M. of 2, 3, 5, 9, 10 is 90]

Question 4.
2A8 is a number divisible by 2, what might be the value of A’?
Solution:
If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.
∴ 2A8 is divisible by 2 for any value of A.
∴ A = (0, 1, 2 ………………….9)

Question 5.
50B is a number divisible by 5, what might be the value of B?
Solution:
Given number is 50B.
The units digit of a number ¡s either ‘0’ or 5, then it is divisible by 5.
∴ 500 → \(\frac { 0 }{ 5 }\) (R = 0)
505 → \(\frac { 5 }{ 5 }\) (R = 0)
∴ B = {0, 5}

Question 6.
2P is a number which is divisible by 2 and 3, what is the value of P
Solution:
The given number is 2P.
If 2P is divisible by 2, 3 then 2P should be a multiple of 6. [ ∵ L.C.M. of 2, 3 is 6]
∴ 2P = 24, 30 ………….
24 → 2 + 4 → \(\frac { 6 }{ 3 }\) (R = 0)
∴ P = 4

Question 7.
54Z leaves remainder 2 when divided by 5 , and leaves remainder 1 when divided by 3, what is the value of Z’?
Solution:
If 54Z is divisible by 3 then the sum of the digits of the number is divisible by 3.
According to problem 54Z is divisible by 3 and leaves remainder 1’.
∴ 5 + 4 + Z = (3 x 4) + 1
= 9 + Z = 13
∴ Z = 4(or)
9 + Z = (3 x 5) + 1
9 + Z = 16
Z = 7
If 54Z is divisible by 5 then Z should be equal to either ‘0’ or ‘5’.
∴ 54(0 + 2) = 542 (Z = 2)
54(0 + 7) = 547 (Z = 7)
∴ From the above two cases
Z = 7
∵ 547 → \(\frac{7}{5}\)(R = 2)

Question 8.
27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?
Solution:
27Q is divided by 5 gives the remainder 3
Le.,27Q = 27 (0 + 3) = 273(Z = 3)(T)
= 27 (0 + 8) = 278 (Z = 8)
27Q is divided by 2 gives the remainder 1.
i.e., 27Q = 27(0 + 1) = 271 (Z = 1)
27Q = 27 (0 + 3) = 273 (Z = 3) (T)
∴ From above situations Z = 3
∴ 27Q = 273→ 2 + 7 + 3 → \(\frac{12}{3}\)(R = 0)
∴ 273 is divisible by 3 and gives the remainder 0’.