Class 8

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

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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:

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:

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.

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’.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 12 Factorisation Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 12th Lesson Factorisation Exercise 12.1

Question 1.
Find the common factors of the given terms in each.

(i) 8x, 24
(ii) 3a, 2lab
(iii) 7xy, 35x²y³
(iv) 4m², 6m², 8m³
(v) 15p, 20qr, 25rp
(vi) 4x², 6xy, 8y²x
(vii) 12 x²y, 18xy²
Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x²y³
7xy = 7 × x × y
35x²y³ = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x²y³
= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m², 6m², 8m³
4m² = 2 × 2 × m × m
6m² = 2 × 3 × m × m
8m³ = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m² , 6m² , 8m³
= 2, m, m², 2m, 2m².

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x², 6xy, 8y²x
4x² = 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y²x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x², 6xy, 8xy² = 2, x, 2x.

vii) 12x²y, 18xy²
12x²2y = 2 × 2 × 3 × x × x × y
18xy² = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x²y, 18xy²
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2.
Factorise the following expressions
(i) 5x² – 25xy
(ii) 9a² – 6ax
(iii) 7p² + 49pq
(iv) 36 a²b – 60 a²bc
(v) 3a²bc + 6ab²c + 9abc²
(vi) 4p² + 5pq – 6pq²
(vii) ut + at²
Solution:
(i) 5x² – 25xy
= 5 x × x × – 5 × 5 × x × y
= 5 × x [x – 5 × y]
= 5x [x – 5y]

ii) 9a² – 6ax
= 3 × 3 × a × a – 2 × 3 × a × x
= 3a [3a – 2x]

iii) 7p² + 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a²b – 60a²bc
= 2 × 2 × 3 × 3 × a × a × b – 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 – 5c]
= 12a²b [3 – 5c]

v) 3a²bc + 6ab²c + 9abc²
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p² + 5pq – 6pq²
= 2 × 2 × p × p + 5 × p × q – 2 × 3 × p × q × q
= p [4p + 5q – 6q²]

vii) ut + at²
= u × t + a × t × t = t [u + at]

Question 3.
Factorise the following:
(i) 3ax – 6xy + 8by – 4bx
(ii) x³ + 2x² + 5x + 10
(iii) m² – mn + 4m – 4n
(iv) a³ – a²b² – ab + b³
(v) p²q – pr² – pq + r²
Solution:
i) 3ax – 6xy + 8by – 4ab
= (3ax – 6xy) – (4ab – 8by)
= (3 × a × x – 2 × 3 × x × y)
– (4 ×a × b – 4 × 2 × b × y)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y)(3x – 4b)

ii) x³ + 2x² + 5x + 10
= (x³ + 2x²) + (5x +10)
= (x² × x + 2 × x²) + (5 × x + 5 × 2)
= x²(x + 2) + 5(x + 2)
= (x + 2) (x² + 5)

iii) m² – mn + 4m – 4n
= (m² – mn) + (4m – 4n)
= (m × m – m × n) + (4 × m – 4 × n)
= m(m – n) + 4(m – n)
= (m – n) (m + 4)

iv) a³ – a²b² – ab + b³
= (a³ – a²b²) – (ab – b³)
= (a² × a – a² × b²) – (a × b – b × b²)
= a²(a – b²) – b(a – b²)
= (a – b²) (a² – b)

v) p²1 – pr² – pq + r²
= (p²q – pr²) – (pq – r²)
= (p × p × q – p × r × r) – (pq – r²)
= p(pq – r²) – (pq – r²) × 1
= (p – 1) (pq – r²)

AP State Syllabus AP Board 8th Class Biology Solutions Chapter 10 Not For Drinking-Not For Breathing Textbook Questions and Answers.

AP State Syllabus 8th Class Biology Solutions 10th Lesson Not For Drinking-Not For Breathing

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing Textbook Questions and Answers

Improve Your Learning

Question 1.
How does air pollution lead to water pollution?
Answer:

1. Air and water are not separate problem. There is a close link between the health of air and the health of water.
2. Nitrogen and chemical contaminants are two types of pollutants that harm both the air and water.
3. Up to l/3rd of the nitrogen that pollutes the bay and it’s rivers comes from the air.
4. Air pollution from a very large geographic area can eventually wind up in the Bay.
5. Use of fertilizers and pesticides in agriculture pollutes not only air but also land and water.
6. Sources of air pollution includes vehicles, industries, power plants and farm operations which lead to water pollution.

Question 2.
What steps can be taken up to control air pollution and water pollution?
Answer:
Some of the methods for controlling air pollutions are:

1. Tall chimneys should be installed in all factories to reduce air pollution at the ground level.
2. Better designed fuel burning equipment should be used in homes and industries so that fuel burnt completely.
3. Install electrostatic precipitators in the chimneys of industries.
4. Reduce vehicular emissions by using non polluting fuels like CNG (compressed natural gas)
5. Use LPG for domestic use (Liquify Petroleum Gas)
6. Improve the quality of fuel in automobiles and use catalytic converters in them.
7. Make use of renewable alternative source of energy like solar energy, wind energy and hydro energy.
8. All motor vehicles should be maintained properly so that they comply with pollution norms.
9. Use unleaded petrol.
10. Plant and grow, more and more trees, we can protect plants and trees. Vanamahotsav should be continued every year in July month.

Prevention and controlling of water pollution

1. Toxic industrial wastes should be treated chemically to neutralize the harmful substances present in it before discharging into rivers and lakes.
2. The sewage should not be dumped in to the rivers directly.
3. The use of excessive fertilizers and pesticides should be avoided.
4. The use of synthetic detergent should be minimized or biodegradable detergents should be used.
5. Dead bodies of human beings and animals should not be thrown into rivers.
6. The excreta and other garbage should be treated in a biogas plant to get fuel as well as manure.
7. The water of rivers, streams, ponds and lakes should be purified or cleaned. For example Ganga action plait launched by the Indian Government.
8. Trees and shrubs should bp planted along the banks of the rivers.
9. There should be general Rareness among the people regarding the harmful of water pollution and the ways of prevention.
10. Waste papers, plastic, waste fbod materials and rotten food and vegetables should not be thrown into open drains.
11. Go for the alternate energy resources that can replenish themselves without affecting our environment.
12. Reuse the materials for secondary purpose. Recycling is the next stage of reuse.

Question 3.
Why does the increased level of nutrients in the water affect the survival of aquatic organisms?
Answer:
Plants nutrients:

1. Phosphates and nitrates – chemical fertilizers from agriculture run – off due to rain and industrial wastes enter into water through sewage and pollute the water.
2. It helps algae to bloom, weeds to grow and bacteria is spread. As a result water turn green and cloudy and smell bad.
3. Decomposing plants use up the oxygen in water, disrupting aquatic life, reducing biodiversity and even killing aquatic life.
4. Thus, this enrichment of water by nutrients leading to excessive plant growth and depletion of oxygen is known as ‘Eutrophication’. This affects aquatic life badly.

Question 4.
Road side plants cannot grow properly. Find your own reasons and explain with your argument.
Answer:

1. The plantation along the roads mainly includes neem, peepal, banyan, almond etc.
2. It was observed that vegetation at roadside with heavy traffic and markets was much affected by vehicular emissions.
3. Significant decrease in total chlorophyll and protein content was observed with reduced leaf area.
4. It is concluded that plants can be uses as indicators for urban air pollution and there is need to protect the road side plants from air pollution.
5. Biomonitoring of plants is an important tool to evaluate the impact of air pollution on plants.
6. A study suggests that plants have the potential to serve as excellent quantitative and qualitative indices of pollution levels.
7. So plants should be grown and protected.

Question 5.
Sudheer is a traffic constable. What do you think about his health? Give some suggestions to protect his health during duty period.
Answer:

1. Environmental pollution place a significance role in the development of various respiratory diseases. Different particles and gases from vehicular emissions like carbondioxide, carbon monoxide, sulphur, benzene, lead, nitrogen dioxide and black smoke are the root of the problem.
2. Traffic police are increasingly becoming victims of a diabetes and allergies at a younger age. Irregular work schedule were posing a challenge to the health of the police. Besides physical strain, mental stress and asymmetrical food habits are also contributing the problem.
3. The traffic police men who work at busy intersections are at the highest risk of developing asthma, bronchitis, shortness of breath, sore throat, chest pain, lung cancer, eye irritation, skin ailments. Impaired hearing, excessive carboxy haemoglobin and annoyance with noise also high blood pressure and cardiovascular problems.
4. On the basis of the study, pollution masks should be used by the traffic police men on duty at higher polluted junctions.
5. He should be recommended better and special medical care for his protection. Various duty places for them need to be scientifically evaluated for their exposure risk.
6. To protect ears from deafening noises, they can put small cotton balls in the ear and also wear ear masks for this purpose.
7. Ecofriendly solar traffic booths at traffic intersections will be provided for the traffic police men. These booths contains ionisers which supress the suspended particulate matter and provide a healthy environment for a police man seated inside.

Question 6.
Write a short note on the effects of water pollution in your village. Suggest precautions.
Answer:

1. Patancheru is a suburban mandal headquarters in Medak district, located about 25 km from Hyderabad. It is a major industrial hub of the state.
2. It is one of the most polluted areas in India. Nearly 14 villages were badly affected by pollution related diseases like cancer, respiratory diseases and heart diseases caused by number of poisons in air, water and on land.
3. The presence of pharmaceutical and chemical industries, pesticide units, steel rolling industries, distilleries releasing the pollutants like chlorine, Hydrogen sulphide, which are entering into the atmosphere.
4. Most of the agricultural lands became barren. The lives of people there depend on agriculture and animal husbandry. They became helpless. Most of the people converted themselves as workers in the factories.

Question 7.
Visit a pollution check centre. Observe the process of conducting a pollution check and record your findings. You may consider the following areas your record.
Average number of vehicles checked in a certain time period, Time taken to check each vehicle, Pollutants checked for, The process of testing, Permissible limits of emission of various pollutants, Measures taken if the emitted gases are above the permissible limits.

• % Average number of vehicles checked in a certain time period.
  Answer:
  10-15
• Time taken to check each vehicle.
  Answer:
  5-7 minutes
• Pollutants checked for
  Answer:
  Carbon dioxide, carbon monoxide, nitrogen, nitrogen oxide, methane, hydro carbons, sulphur dioxide, particular matter, trace elements, water vapour etc.
• The process of testing:
  1. The test shall be carried out with the engine mounted on a test bench and connected to a dynamometer.
  2. The gases emission from the exhaust of the engine include hydrocarbons, carbondioxide, carbon monoxide and oxides of nitrogen.
  3. During prescribed sequence of warmed up engine operating conditions, the amount of the above gases in the exhaust, shall be examined continuously.
  4. The prescribed sequence of operations consists of a number of speed and power modes which span the typical operating range of engine.
  5. During each mode, the concentration of each pollutant, exhaust flow and power out put shall be determined and the measured values, weighed and used to calculate the grams of each pollutant emitted for kilowatt hour.
• Permissible limits of emission of various pollutants.
  Pefrol engine
  Carbondioxide                                  14%
  Carbon monoxide                            1 to 2%
  Nitrogen oxide less than                  0.5%
  Hydrocarbons                                   0.5%
  Sulphur dioxide                                possible traces
  Particular matter less than                0.5%
  Trace elements less than                   0.5%
  Nitrogen                                            71%
  Water vapour                                    12%
  Diesel Engine
  Nitrogen                                                   67%
  Carbondioxide                                          13%
  Water vapour                                            11%
  Carbon monoxide                             less than 0.045%
  Nitrogen oxide                                 less than 0 to 1.5%
  Sulphur dioxide                                less than 0.03%
  Hydro carbons                                  less than 0.43%
  Particulate Pollutants                       less than 0.045%
  Trace elements                                           0.3%
• Measures taken if the emitted gases are above the permissible limits.
  Answer:
  The carbon particles (soot) deposited in the engine head will be checked and cleaned or the vehicle will be ceased by .R.T.O.

Question 8.
Organize a field visit to a pond/lake/river present in or near to your village with the help of your teachers.
Observations followed by discussion could focus on…. The history of the pond or lake or river, Water resources available other than that river/pond/or lake, Cultural traditions, Pollution concerns, Source of pollution, Effects of pollution on the people living by the river side as well as those living far away.
Answer:
Kolleru lake: It is a 2nd largest fresh water lake located in Andhra Pradesh located between Krishna and Godavari delta.
History: Two copper plates of the early Pallava dynasty have been found in the lake, tracing it’s history to Langula Narasimha Deva an Ganga Vanshi Odisha King. According to legend, the Gajapathi fort was located at Kolleti kota on one of the eastern islands of the lake. The enemy general “Muhammadan” general escamped at “Chiguru kota” located on the shores. In some ways the lake protected the Oriya forces. The enemy finally tried to excavate a channel, the modern day Upputeru. So that the water of the lake would empty into the sea and the level would fall so that they could attack the Gajapathi fort. The royal oriya army general sacrificed his own daughter to propitiate Gods and ensure his success against Muhammadan and her name was “Perantala Kanama”. Therefore the channel Was called perantala Kanama. “Sri Peddinti Ammavari Temple” is one of the oldest and famous temples found in Kolleru.
Water sources available other than Kolleru: Wells, taps. The lake is fed directly water from seasonal Budameru and Tammileru streams and connected to the Krishna and Godavari systems by channels.
Cultural traditions: The vast majority of the district is rural in nature. Thus the culture of the kolleru lake people is mostly conservative and traditional. The joint family system, the arranged marriages are the norms. Telugu language is spoken in this place. Pollution concerns: Kolleru lake is suffering from the unsatisfied greed of people and selfish interests of mankind who exploit the lake’s integrity. Thousands of fish tanks were dug up effectively converting the lake into a mere drain. This had great impact in terms of pollution, leading to difficulty in getting drinking water for the local people.
Source of pollution: Satellite images taken on February 9, 2001 by the Indian remote sensing satellite found that approximately 42% of the 245 km² lake was occupied by aquaculture. While agriculture had encroached another 8.5% they were mostly rice paddies. Surprisingly no clear water could be found in the satellite image. The rest of the lake is being diminished by weeds like elephant grass and water hyacinth.
Effects of pollution on people: Thousands of fish tanks were dug up effectly leading to difficulty in getting drinking water for the local people. An adverse effect on the thousands of acres of crop in the upper reaches of water flow into the sea because of obstruction by bunds of the fish tanks that appeared illegally.

Question 9.
What is air pollution? Make a flow-chart to describe its causes and effects.
Answer:
If solid, liquid and gaseous substances are present in higher volumes than required in the air, it is harmful to air. It is called air pollution.
Table shows causes and effects of air pollution:

COMMON POLLUTANTS AND THEIR SOURCES

Question 10.
Clear and transparent water is always suitable for drinking. Comment.
Answer:

1. No. Clear and transparent water is not always suitable for drinking.
2. Water might appears clean but it may contain some disease causing micro-organisms and other dissolved impurities.
3. Hence it is advised to purify water before drinking.
4. Purifying can be done by water purifying systems or by boiling the water.

Question 11.
If our monument like Tajmahal is effected by air pollution, what is your advise to protect it?
Answer:
The Taj Mahal one of the seven wonders of the world is located in Agra. It is made of white marbles. The effect of pollutants like SO₂, NO₂, smoke, dust, soot, etc. on it has turned the marble from white to yellow.
Precautions to protect Taj Mahal:

1. Switch over to cleaner fuels like CNG (Compressed Natural Gas) and LPG (Liquify Petroleum Gas)
2. Use unleaded petrol in vicinity of Taj Mahal.
3. Shift polluting industries to the outside of Agra city.
4. Industrial pollution should be banned around the Taj Mahal.
5. Limited vehicular use in a specific radius.
6. An electronic board should be installed in the Taj Mahal premises, that compares current pollution levels, to the levels that deemed safe.

Question 12.
Reshma going to talk about controlling measures of soil pollution. Prepare write up for her.
Answer:
Controlling measures of soil pollution:

1. Limit the use of fertilizers and pesticides.
2. Awareness about biological control methods and their implementation.
3. The grazing of cattle must be controlled and forest management should be done properly.
4. The afforestation and reforestation must take place.
5. Proper preventing methods like shields should be used in areas of wind erosion .and wind breakes.
6. Remember to carry paper bags and minimising plastic bags.
7. The soil binding grass must be planted and the large trees must be plant along the banks.
8. Industrial waste must be dumped in the low lying areas.
9. There should be a definite technic of cropping which does not allow weeds to settle on the fields.
10. The mining waste must be improved along with the transportation.
11. The are must not be left barren and dry.

Question 13.
To conduct a quiz program on air and water pollution, prepare five thought provoking questions.
Answer:

1. What is environmental pollution?
2. What are the natural disasters of pollution?
3. What are the human activities that leads to pollution?
4. What is the unforgettable industrial tragedy took place in Bhopal on second December
5. Who is responsible for thepollution in Patancheru ? What are the interim orders, released by supreme court of India for the sake of people and environment?
6. What is the sad story of River Musi?
7. Can we save Taj Mahal from pollution?
   What is the present situation of Taj Mahal?
8. Natural resources are the devine gift for us by nature. Can we keep these resources clean and healthy for the future generations? What is your role and response towards this?

Question 14.
“Use Bicycle – Avoid motor bikes and cars”. This slogan is prepared by Sravani. Prepare some more slogans on pollution.
Answer:

1. Slogans towards biodiversity:
2. Segregate the reusable waste.
3. Don’t produce lot of waste.
4. Don’t use plastic covers.
5. Compost the wet waste.
6. Turn the waste into compost.
7. Make Compost out of fallen leaves.
8. Never burn the fallen leaves of trees.
9. Use bicycles if the destination is manageable.
10. Plant trees in vacant places and take care to ensure their growth.
11. Grow plants and protect them for fresh air and ventilation.
12. Use cattle dung, organic fertilizers.
13. Practice eco-friendly methods.
14. Strictly follow environmental policies and laws.
15. Limit the use of fire wood and use bio fuels for cooking.
16. Plant a sapling on your birthday water it every day.
17. Wash your hands, feet close to the trees and plants.

Question 15.
If you are a general manager of a chemical industry, what precautions would you take to control air and water pollution?
Answer:
Precautions to control air and water pollution :

1. Stoppage of effluent flowing into air and water bodies immediately.
2. Installing tall chimneys in factory to reduce air pollution at the ground level.
3. Installing electrostatic precipitators in the chimney of industry to reduce.
4. Plant and grow, more and more. Trees in the industry surroundings.
5. Protecting plants and trees.
6. Toxic industrial wastes should be treated chemically to neutralize the harmful substances present in it before discharging into water bodies.
7. Waste water from the industry is to be filtered first to remove particulate material and stored in shallow tanks where bacterial degradation of organic compounds takes place.

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing InText Questions and Answers

Question 1.
Observe this certificate try to find out answers for the following questions.
Answer:
The pollution under control certificate

a) Which department issues the pollution under control certificate?
Answer:
The pollution check up centre issues the certificate.

b) For how much time is it valid?
Answer:
It is valid for six months.

c) For which type of vehicle has it been issued?
Answer:
Motor bike, scooters, cars, bus, lorry all type of vehicles.

d) What is emission test ? What components are tested ini the pollution check up centre?
Answer:
The test conducting of the gases releasing from the vehicle is called emission test. Components tested in the pollution check up center are carbondioxide, carbon monoxide, nitrogen, hydrocarbons, sulphur dioxide etc.

e) What will happen if carbon monoxide (CO) and hydro carbons (HQ readings are higher that the permissible limits reading ?
Answer:
If the above said gases are higher than the permissible limits reading it leads to pollution is harmful to the living organisms and hurt the health and well of living organisms.

f) Think of why there is a peed of “pollution under control certificate”?
Answer:
With a rapid increase in Ifie number of vehicles the problem of automobile pollution has assumed greater significance. The emission of smoke from motor vehicles is a major source of air pollution.

Question 2.
What will happen if harmful organisms or substances enter your body? How do you feel?
Answer:
If harmful organisms enter the body, the normal functioning of the body will be disrupted or disturbed. We feel sick.

What is air pollution ?
Question 3.
List out the gases that you know present in the air.
Answer:
The gases present in the air are nitrogen, oxygen, carbondioxide, inert gases mainly argon, and water vapour.

Question 4.
What are the four major gases in the air?
Answer:
The four major gases are nitrogen oxygen, argon and carbondioxide.

Question 5.
Draw a neat diagram showing the composition of air in the atmosphere arid write the percentage of gases.
Answer:

Question 6.
Collect the pictures of natural activities and human activities which leads to pollution and paste them in your record book.
Answer:

Question 7.
If a person burnt out types or dried leaves at a particular place. Where shall go the smoke and ash goes?
Answer:
The smoke and ash raise up, mixes with gases in the atmosphere which leads to pollution.

Water Pollution
Question 8.
Let us read the following news paper clipping.

Answer the following questions based on your understanding of the paper clipping.
a. What do you understand after reading the news paper clippings?
Answer:
Some areas in Nalgonda district, the water got polluted because of the chemicals dissolved in it. These chemicals are released Into the water by the oil factory, textiles, pesticide factory and steel factory.

b. What are the issues discussed in this news paper clipping?
Answer:
Total dissolved solids (T.D.S.) in water should be maximum 500. Due to pollution it is estimated that T.D.S. has reached to 10,000. The pollution in water is higher which is dangerous for use.

c. What are its causes and effects?
Answer:
The causes for the TDS: The chemical pollutants released by the factories into the water. Effects: The water becomes unfit for drinking. By using the water the crops are beeing dead.

d. How does the problem arise?
Answer:
Large amounts of chemicals discharged into the ground water leads to water pollution. Due to water pollution the problem arise in those areas.

e. Are you also facing this type of problems in your area? Can you explain reasons behind?
Answer:
Yes, I am facing this type problems in our area. Because industries are developed in our area. So population is increasing. That’s why water is polluted in our area.

8th Class Biology 10th Lesson Not For Drinking-Not For Breathing Activities

Activity – 1

Question 1.
NATURAL DISASTERS POLLUTION
Collect information from your school library for the following natural disasters in the world.
Answer:
Volcanic eruptions:
Deep under the earth’s surface, it’s so hot that even rock melts. Sometimes this molten rock, called ‘magma’, is pushed up to the surface. At this point it is referred to as lava. And the opening or vent that lets the lava out is a volcano. ‘
A volcano may explode violently throwing out rocks for miles around. It releases various gases and ash into the atmosphere.

Forest fires:
Forest fire is a moving combustion reaction spreading outwards in a band from it’s ignition point leaving burnt forest behind it and also called as wild fire.
It can be large uncontrolled disasters that burnt through hundred to hundred thousand acres.
Causes : Natural cause, lightening, volcanic eruptions, sparks from rocks falls and spontaneous combustion.
Pollution :
Forest fires release carbon particles (ash) into the air and pollute the air.

Sand storms and Tsunamis:
A sand storm or a dust storm is a meteorological phenomenon common in arid and semi arid regions. They arise when a gust front or other strong wind blows, loose sand and dirt from a dry surface. Particles are transported by saltation and suspension. The Sahara where sand is more prevalent, soil type than dirt or rock.
Causes:
As a force of wind passing over loosly held particles increased, the particles of sand first start to vibrate, then to saltate. As they repeatedly strike the ground, they loosen and break off smaller particles of dust, which then begin to travel in suspension.

Tsunamis:

1. Tsunami is also called as a ‘seismic sea wave’ is a series of waves in a water body caused by the displacement of a large volume of water.
2. Tsunami impact is limited to coastal areas. But their destructive power can be enormous, and they can effect entire ocean basin.
3. The 2004 Indian Ocean tsunami was among the deadliest natural disasters in human history with, atleast 2,30,000 people killed or missing in 14 countries bordering the Indian Ocean.
4. Tsunami causes much damage by two mechanisms. The smashing force of water travelling at a high speed and destructive power of a large volume of water draining of the land and carrying away large amount of debris with it.
5. About 80% of tsunamis occur in the Pacific ocean, and there is possibility for tsunamis wherever there are large bodies of water including lakes.

Activity – 2

Question 2.
A) OIL PAPER EXPERIMENT
Answer:
Take three square pieces of white paper of 5 x 5 cm size dipped in oil. Hang these oil dipped paper at three different locations, say your backyard, your school, near a park or parking lot etc. Let it be there for 30 minutes. Observe and compare all three papers.
a) What did you found on those papers dipped in oil ?
Answer:
Some dust particles sticking to the oil paper.

b) Is there any difference observed for all the three locations ?
Answer:
Some differences are observed. The soot and ash dust particles are seen on the oil paper hung at back yard which comes from burning of wood.
More dust particles seen on the oil paper at the school and less dust particles on the oil paper which was hung near a park.

c) Try to find out the answer why this difference occurred.
Answer:
When different types of fuels are burnt they release different types of particles into air. These particles are different according to different locations.

d) Do you know where the dust particles could have come from ?
Answer:
The dust particles could have come from the effect of air pollution caused by man made resources largely affects nature.

B) HUMAN ACTIVITIES:
Name of the fuels we burnt in our daily activities including rural and urban areas.
Answer:
Petrol, diesel, wood, charcoal, tyres etc.

Activity – 3

Question 3.
POWER GENERATION PLANTS.
Go to your school library and collect information to make a list of these power generation plants and where they are located.
Answer:
THERMAL POWER PLANTS

NUCLEAR POWER PLANTS

Discuss about the adverse effects of Global warming.
Answer:
Effects of Global Warming: Global warming is having measurable effects on the planet right now. They are

1. Ice is melting in both polar ice caps and mountain glaciers.
2. Lakes around the world, including lake superior are warming rapidly.
3. Animals are changing migration patterns.
4. Plants are changing the dates of activity. Ex : Leaf flash.
5. Global warming increase in temperature around the world.
6. The average global temperature increased 0.8°C over the post 100 years.
7. Weather is changed by globed warming, some places become more hot and some more cool.
8. Global warming increases the sea level. So the costal areas will sink into the sea.
9. Ice is melted which increases the sea level that leeds to ocean acidification.

Ask your teacher about secondary pollutants why they are called so?
Answer:
Oxides of nitrogen NO, NO₂ (NOx), peroxy acetyl nitrate, formaldehyde, ozone, etc., are the secondary pollutants.
Pollutants are defined as primary pollutants resulting from combustion of fuels and industrial operations and secondary pollutants, those which are produced due to reaction of primary pollutants in the atmosphere.

Activity – 4

Question 4.
FIELD VISIT:
Visit nearby factory, industry (boiled rice mill, brick making kiln, oil mill food processing mill etc.,) present in your area and observe.

• How are they polluting air and water ?
  Answer:
  They are releasing smoke, smooth dust particles into air and releasing waste material into water sources and polluting air and water.
• Is there any green belt around the factory ? Name the trees they are growing.
  Answer:
  Ashoka, Gulmohar, Neem, Eucalyptus are growing around the factory.
•  What precautions are they taking to prevent pollution ?
  Answer:
  Suction devices known as vacuum pans are used to collect the pollutants from the water.
  To control air pollution, ventury type wet scrubber and meeting the norms described by A.P. Pollution board -have arranged.

Lab Activity 

Question 5.
POLLUTANTS:
Aim: Observation of pollutants in local available water samples.
Material: Glass tumblers, water samples from tap, pond, river, well, lake, red, blue litmus papers, soap.
Procedure: Collect water samples from a tap, pond, river, well and lake. Pour each into separate glass containers. Compare these for smell, colour pH and hardness.

• pH of water samples can be determined by using litmus paper. If blue litmus paper turns to the red colour, that water sample is acidic in nature and if red litmus turns to blue, water sample is basic in nature.
• Hardness of water can be determined using soap. If water produces lesser foam it is referred as hard water.

Observations and findings:
Record your observation in the following table.

Activity – 5

Question 6.
Visit your nearby pond/lake or river and find out the material being discharged in it. Prepare a biography on it.
Answer:
As Hyderabad has grown in size and is emerging as a global mega city, its growing water requirements have been met by under taking long distance water projects over the years. These projects are dependent on Musi River. Thousands of people depend on it for their daily needs and livelihood. The Moosi has been polluted for many years.
The people living near the Musi river, throw large quantities of garbage, untreated sewage, industrial waste, dead bodies, polythene bags, hot water and statues of deities and many other materials directly in to the river. –
The ‘Musi reservoir action plan project’ was undertaken to reduce the pollution level in the river. Pollution control activities include under the project are:

• Solid waste management.
• Installation of sewage treatment plant.
• Provision of low cost sanitary facilities.
• Development of river front.
• Efforts to develop public awareness.

Ask your teacher about aerobic bacteria and write a note on it with some examples.
Answer:
An aerobic bacteria is the one that can survive and grow in an oxygenated environment. Bacillus, nocardic and pseudomonas space aeruginosa, myobacterium tuberculosis are some of the examples of aerobic bacteria.

Do you know oil slick on sea water ?In what way it is dangerous to aquatic life?
Answer:
Several kinds of plants and animals live in oceans. Oceans maintain equilibrium in nature. We transport several kinds of oils and fuels over seas. The spillage of these oils and fuels by accident creates a layer of oil over the surface of the sea water for hundreds of kilometers. This is called oil spill or oil slog. When it happens, air and light cannot enter the sea water and several marine creatures like fish, tortoise and other marine life forms die of asphyxiation.

Think and Discuss

Question 1.
When we go on a busy road in the evening a lot of smoke is spread in the surroundings. We get cough and feel uneasy even when we close the nose with napkins.
a) Why this type of symptoms we observe? Think about it.
Answer:
Carbon monoxide is poisonous gas combines with haemoglobin of our blood and forms carboxy haemoglobin. Due to this haemoglobin is unable to carry oxygen to various parts. This leads to respiratory problems. It causes suffd&Stion and may cause even death.

b) If these symptoms will continue, what happens?
Answer:
Air pollution is like a slow poison. The effect of air pollution are not seen immediately. But over a long period of time the pollutants present in air damage our health and property.

Question 2.
Do you find any relation between pH and hardness of water?
Answer:
No relation between pH and hardness of water. pH is scale to measure acid, base or neutral hardness is the percentage of salts dissolved in the water.

Question 3.
Which water sample is colourless?
Answer:
Tap water.

Question 4.
Which water sample is suitable for drinking and why?
Answer:
Tap water is suitable for drinking because the tap water is cleaned and chlorinated and safe for drinking.

Question 5.
Do you find any change in colour and smell of water in some water samples ? What are your reasons ?
Answer:
The colour and smell of water is due to the nature of the soil and the plants grown in the water.

AP State Syllabus 8th Class Maths Solutions 13th Lesson Visualizing 3-D in 2-D InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 13 Visualizing 3-D in 2-D InText Questions and Answers.

8th Class Maths 13th Lesson Visualizing 3-D in 2-D InText Questions and Answers

Do This

Question 1.
Name some 3 – Dimensional objects.   [Page No. 282]
Answer:

1. Cube
2. Cylinder
3. Sphere
4. Cuboid
5. Cone

Question 2.
Give some examples of 2 – Dimensional objects.     [Page No. 282]
Answer:

1. Square
2. Rectangle
3. Line segment
4. Circle
5. Triangle

Question 3.
Draw a kite in your notebook. Is it 2 – D or 3 – D object?      [Page No. 282]
Answer:

Kite is a 2 – D object.

Question 4.
Identify some objects which are in cube or cuboid shape.      [Page No. 282]
Answer:
Shapes of Cube                           Shapes of Cuboid
a) Chalk piece box                       a) Duster
b) Dice                                         b) Cell phone (Cuboidal in shape)
c) Cube shaped cake                   c) Plasma T.V.

Question 5.
How many dimensions that a circle and sphere have?      [Page No. 282]
Answer:
Circle has 2 dimensions.
Sphere has 3 dimensions.

Question 6.
Identify the faces, edges and vertices of given figures.      [Page No. 288]

Answer:

Question 7.
Write the names of the prisms given below:     [Page No. 290]

Answer:
(i) Cube
(ii) Triangular prism
(iii) Pentagonal prism
(iv) Hexagonal prism
(v) Rectangular prism

Question 8.
Write the names of the pyramids given below:       [Page No. 290]

Answer:
(i) Square pyramid
(ii) Pentagonal pyramid
(iii) Hexagonal pyramid

Question 9.
Fill the table:      [Page No. 290]

Answer:

Question 10.
Explain the difference between prism and pyramid.      [Page No. 290]
Answer:
Upper and lower sides of a prism are equal in number. But, in a pyramid the base is a plane and all the edges are coincide in a single point on the top.

Try These

Question 1.
Name three things which are the examples of polyhedron. [Page No. 287]
Answer:

Question 2.
Name three things which are the examples of non-polyhedron. [Page No. 287]
Answer:

Think, Discuss and Write

Question 1.
How to find area and perimeter of top view and bottom view of the given figure.       [Page No. 283]

Answer:
Let the side of each face be ‘1’ unit say.
Shapes of different positions (I)                      Their areas (II)
1. Front view                                                    A = (1 × 1) + (1 × 1) + (1 × 1) = 3 Sq. Units
2. Top view                                                      A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
3. Bottom view                                                A = (1 + 1 + 1) × (1 + 1) = 3 × 2 = 6 Sq. Units
Perimeters (III)
1. —————>                                             1 + 1 + 1 = 3 Units
2. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units
3. —————>                                             2(l + b) = 2 (3 + 2) = 2 × 5 = 10 Units

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.1

Question 1.
Using divisibility rules, fmd which of the following numbers are divisible by 2,5,10 ( say
yes or no ) in the given table. What do you observe?

Solution:

Question 2.
Using divisibility tests, determine which of following numbers are divisible by 2
(a) 2144 (b) 1258 (c) 4336 (d) 633 (e) 1352
Solution:
If a number is divisible by 2 then the units digit of the number be 0, 2, 4, 6, 8.
∴ a) 2144, b) 1258, c) 4336 e) 1352 are divisible by ‘2’.

Question 3.
Using divisibility tests, determine which of the following numbers are divisible by 5
(a) 438750 (b) 179015 (c) 125 (d) 639210 (e) 17852
Solution:
If a number is divisible by 5 its units digit be either ‘0’ or 5.
∴ a) 438750, b) 179015, c) 125 d) 639210 are divisible by 5.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 10:
(a) 54450 (b) 10800 (c) 7138965 (d) 7016930 (e) 10101010
Solution:
If a number is divisible by 10 then its units digit must be 0’.
a) 54450, b) 10800, d) 7016930, e) 1010100 are divisible by 10.

Question 5.
Write the number of factors of the following’?
(a) 18 (b) 24 (e) 45 (d) 90 (e) 105
Solution:

Question 6.
Write any 5 numbers which are divisible by 2,5 and 10.
Solution:
10, 20, 30, 40. are divisible by 2, 5 and 10
[∵ The L.C.M. of 2, 5, 10 is 10]

Question 7.
A number 34A is exactly divisible by 2 and leaves a remainder 1, when divided by 5, find A.
Solution:
If 34A Is divisible by 2 then the remainder should be equal to 0.
∴ A should be equal to 0, 2, 4, 6, 8.
∴ 340, 342, 344, 346, 348 are divisible by 2 and gives the remainder ‘0’.
Among these 346 is divisible by 5 and gives the remainder 1.
∴ 346 → \(\frac{6}{5}\) (R = 1)
∴ The value of A = 6.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volumes Ex 14.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volumes Exercise 14.2

Question 1.
Find the volume of the cuboid whose dimensions are given below.

Solution:

Question 2.
Find the capacity of the tanks with the following internal dimensions. Express the capacity in cubic meters and litres for each tank.

Solution:

Question 3.
What will happen to the volume of a cube if the length of its edge is reduced to half? Is the volume get reduced? If yes, how much?
Solution:
Volume of a cube of side (s) is V₁ = a³
If the length of the side is reduced by half then
s = \(\frac{\mathrm{a}}{2}\)
∴ Volume of cube (V₂ ) = s³

∴ V₂ = \(\frac { 1 }{ 8 }\) × V₁
∴ V₁ = 8V₂

Question 4.
Find the volume of each of the cube whose sides are.
(i) 6.4 cm
(ii) 1.3 m
(iii) 1.6 m.
Solution:
Volume of a cube(V) = a³

i) a = 6.4 cm
ii) a = 1.3 m
iii) a = 1.6 m

V = (6.4)³
Volume of a cube (V) = a³
= 6.4 × 6.4 × 6.4
= 262.144 cm³

V = (1.3)³
= 1.3 × 1.3 × 1.3
= 2.197 m³

V = (1.6)³
= 1.6 × 1.6 × 1.6
= 4.096 m³

Question 5.
How many bricks will be required to build a wall of 8 m long, 6m height and 22.5 cm thici if each brick measures 25 cm by 11.25 cm by 6 cm?
Solution:
The volume of a wall of measures
8 m × 22.5 cm × 6 m
(V₁) = l₁b₁h₁
= 8 m × 22.5 cm × 6 m
= 800 cm × 22.5 cm × 600 cm
The volume of a brick each measures
25 cm × 11.25 cm × 6 cm
(V₂) = l₂b₂h₂
= 25 × 11.25 × 6 cm³
∴ The no.of bricks will be required

= 32 × 2 × 100 = 6400

Question 6.
A cuboid is 25 cm long, 15 cm broad, and 8 cm high . How much of its volume will differ from that of a cube with the edge of 16 cm’?
Solution:
Volume of a cuboid (V₁) of measures
= 25 cm, b = 15 cm, h = 8 cm.
V₁ = 25 × 15 × 8 = 3000 cm³
Volume of a cube of measure side (s) = 16 cm is
V₂ = (s)³ = (16)³ = 16 × 16 × 16
= 4096 cm³
The difference between their volumes
= V₂ – V₁
= 4096 – 3000
= 1096 cm³

Question 7.
A closed box is made up of wood which is 1cm thick .The outer dimensions of the box is 5 cm × 4 cm × 7 cm. Find the volume of the wood used.
Solution:
The volume of a box formed with outer measures 5 cm × 4 cm × 7 cm
V₁ = l × b × h
= 5 × 4 × 7
∴ V₁ = 140 cm³
Inner measures
= l – 2w, b – 2w, h – 2w
= (5 – 2 × 1), (4 – 2 × 1), (7 – 2 × 1)
= (5 – 2), (4 – 2), (7 – 2)
= 3 cm, 2 cm, 5 cm
∴ Volume of a box formed with inner measures
V₂ = (l – 2w)(b – 2w)(h – 2w)
= 3 × 2 × 5 = 30 cm³
∴ The volume of wood used = V₁ – V₂
= 140 – 30 = 110 cm³

Question 8.
How many cubes of edge 4cm, each can be cut out from cuboid whose length, breadth and height are 20 cm, 18 cm and 16 cm respectively
Solution:
The volume of a cuboid formed with the measures 20 cm × 18 cm × 16 cm
(V₁) = l₁b₁h₁ = 20 × 18 × 16
Volume of a cube (V₂) = s³
s = 4 cm (given)
∴ V₂ = (s)³ = (4)³= 4 × 4 × 4 cm³
∴ No.of cubes are required
= \(\frac{V_{1}}{V_{2}}=\frac{20 \times 18 \times 16}{4 \times 4 \times 4}\)
= 90

Question 9.
How many cuboids of size 4 cm × 3 cm × 2 cm can be made from a cuboid of size 12 cm x 9cm x 6cm?
Solution:
Volume of a cuboid of measures 12 cm × 9 cm × 6 cm
V₁ = l × b × h = 12 × 9 × 6
Volume of the smaller cuboid of measures 4 cm × 3 cm × 2 cm
V₂ = l₂b₂h₂ = 4 × 3 × 2
∴ No.of cuboids are made
= \(\frac{V_{1}}{V_{i}}=\frac{12 \times 9 \times 6}{4 \times 3 \times 2}\) = 27

Question 10.
A vessel in the shape of a cuboid is 30 cm long and 25 cm wide. What should be its height to hold 4.5 litres of water ?
Solution:
Length of a cuboidal vessel (l) = 30 cm
breadth (b) = 25 cm
height (h) = ?
The volume of water m a cuboidal vessel = 4.5 Lts.
= 4.5 × 1000 cm³
= 4500 cm³
∴ l × b ×h = 4500
⇒ 30 × 25 × h = 4500
⇒ h = \(\frac{4500}{30 \times 25}\)
∴ h = 6 cm
∴ Height of the vessel (h) = 6 cm

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a – b)² ≡ a² – 2ab+ b² geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)² = a² – 2ab + b² [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)² = 4 sq. units
[∵ (3 – 1 )² = 2² = 4]

ii)

∴ (a – b)² = a² – 2ab + b²
Area of ABCD + Area of CYZS = a² – 2ab + b²
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a² – b² geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a² – b² = 9 – 4= 5sq . units
[ ∵ 3² – 2² = 9 – 4 = 5]

ii)

a² – b² = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a² – b² = 3 sq. units
[∵ (2² – 1²) = 4 – 1 = 3]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 14 Surface Areas and Volumes Ex 14.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 14th Lesson Surface Areas and Volumes Exercise 14.1

Question 1.
There are two cuboidal boxes as shown in the given figure. Which box requires the less amount of material to make?

Solution:
Volume of a cuboid V₁ = lbh
= 60 × 40 × 50
V₁ = 1,20,000 cubic units.
Volume of a cube V₂ = (a)³
= (50)³ = 50 × 50 × 50
V₂ = 1,25,000 cubic units.
∴ The cuboidal box requires less amount of material.
∴ V₁ < V₂

Question 2.
Find the side of a cube whose surface area is 600 cm²
Solution:
Total surface area of a cube = 6a²
⇒ 6a² = 600
⇒ a² = \([latex]\frac { 600 }{ 6 }\)[/latex] = 100
⇒ a² = 100
⇒ a = √100 = 10
∴ The side of a cube (a) = 10 cm.

Question 3.
Prameela painted the outer surface of a cabinet of measures 1m × 2m × 1 .5m. Find the surface area she cover if she painted all except the bottom of the cabinet?
Solution:
The area of outer surface of a cabinet except the bottom of the cabinet will be equal to its lateral surface area.
I = lm,b = 2m, h = 1.5m.
A = 2h(l + b)
= 2 × 1.5(1 + 2)
= 3 × 3 = 9 m².

Question 4.
Find the cost of painting a cuboid of dimensions 20cm × 15 cm × 12 cm at the rate of 5 paisa per square centimeter.
Solution:
l = 20cm, b = 15cm, h = 12cm.
∴ Total surface area of a cuboid
A = 2 (lb + bh + lh)
=2(20 × 15 + 15 × 12 + 20 × 12)
= 2 (300 + 180 + 240)
= 2 × 720
= 1440 sq.cm.
The cost of painting a cuboid at the rate of 5 paisa per sq. cm for 1440 sq.cm.
= 1440 × 5 paisa
= 7200 paise
= ₹ \(\frac { 7200 }{ 100 }\)
= ₹ 72