AP State Important Questions

These AP 8th Class Biology Important Questions 9th Lesson Production and Management of Food From Animals will help students prepare well for the exams.

AP State Syllabus 8th Class Biology 9th Lesson Important Questions and Answers Production and Management of Food From Animals

8th Class Biology 9th Lesson Production and Management of Food From Animals 1 Mark Important Questions and Answers

Question 1.
From where do we obtain food?
Answer:
We obtain food from plants and animals.

Question 2.
What are the food items that are obtained from animals?
Answer:
We obtain milk, meat and eggs from animals.

Question 3.
What is animal husbandry?
Answer:
Providing food, shelter, protection and breeding of animals is called animal husbandry.

Question 4.
How did man use animals since long time?
Answer:
Since long time, man used animals not only for obtaining food but also for agriculture, transportation etc.

Question 5.
Why did early man domesticate only some of the animals?
Answer:
The early man domesticated only some of the animals which provide him food, clothing and the animals which are helpful for agriculture and transportation.

Question 6.
In what way the animals are useful to us?
Answer:
Buffalo and cow give us milk, hens give us eggs and meat, goats and sheep provide us meat, ox, bulls, donkeys are useful for agriculture and transportation.

Question 7.
What do farmers believe in our country?
Answer:
In our country farmers believe that animal husbandry is part and parcel of agriculture.

Question 8.
Name the animals that domesticate by the people living in rural areas.
Answer:
People living in rural areas domesticate animals like cows, buffaloes, bullocks, goats, sheep, pigs, hens, etc.

Question 9.
What is the important issue in Animal husbandry?
Answer:
Supplying of nutritious food, accommodating clear and hygienic shelters for animals is important issue in animal husbandry.

Question 10.
Where do people rear their cattle in the village?
Answer:
Generally villagers send their cattle to rear at the places where grass is easily available.

Question 11.
Where do farmers keep their cattle?
Answer:
Farmers keep their cattle in the sheds.

Question 12.
Name the agricultural practices done by using bullocks and he-buffaloes.
Answer:
Ploughing and levelling the field farmers use bullocks and he-buffaloes.

Question 13.
How do cattle rearers protect their cattle?
Answer:
Cattle rearers make fences in the fields at off crop seasons.

Question 14.
How is the milk production effected?
Answer:
Milk production is effected by viral and bacterial diseases.

Question 15.
In rainy season how are cattle protected from mosquitoes?
Answer:
Cattle can be protected by covering mosquito nets.

Question 16.
Who will provide treatment and health care for cattle?
Answer:
Veterinary doctor will provide treatment and health care for cattle.

Question 17.
Name the dangerous disease in cows and buffaloes.
Answer:
Galikuntu is a common and dangerous disease in cows and buffaloes.

Question 18.
From which disease sheep and goats will suffer?
Answer:
Sheep and goats will suffer from worm infections (Nattala Vyadhi)

Question 19.
How do our government treats milk production?
Answer:
Our Government treats milk production as an industry.

Question 20.
How much milk is given by traditional species of cows?
Answer:
2 to 5 litres of milk per day.

Question 21.
How much milk is given by murra species?
Answer:
Murra species give up to 8 litres of milk per day.

Question 22.
Name the traditional varieties of cows.
Answer:
Haryana, Jaferabad, Nagapuri are the traditional varieties which give good quantity of milk.

Question 23.
Name the foreign varieties of cows.
Answer:
Jersy (England) and Holstein (Denmark).

Question 24.
What is the milk yield from foreign varieties?
Answer:
They give 8 to 20 litres of milk.

Question 25.
What is pasteurization?
Answer:
The destruction of disease producing organisms present in the milk. In this process milk is heated at 62° temperature and cooled below 10°C.

Question 26.
In which months milk production is high?
Answer:
In the months of October and November milk production is high.

Question 27.
Who is the father of white revolution?
Answer:
Prof. J.K. Korian is the father of white revolution in India.

Question 28.
How is milk secreted?
Answer:
Milk is secreted from the mammary glands of animals.

Question 29.
Why do people in our country decorate their cattle during festivals?
Answer:
People believe that cattle are part and parcel of our culture. They treat them as their family members. So they decorate their cattle.

Question 30.
Why do some persons collect bones of dead animals?
Answer:
because bones are used in fertilizer industry.

Question 31.
What is the use of leather of animals?
Answer:
Leather of cattle is used in the leather industry.

Question 32.
What is Biogas? How is it produced?
Answer:
Biogas is produced from the wastes from cattle, home etc. As this is produced biologically, this gas is called biogas used for domestic purposes.

Question 33.
What are Broilers and layers?
Answer:
Broilers are the hens reared for meat and layers are the hens reared for eggs in poultry.

Question 34.
Name some local variety of poultry varieties.
Answer:
Aseel, Kadaknath, Chittagang, Longshan, Bursa are the pure local varieties.

Question 35.
What is a poultry?
Answer:
Production and rearing of hens on a large scale is called poultry.

Question 36.
In which place India is occupied in production of eggs?
Answer:
India achieved 4th position in the world by producing 41.06 million eggs per annum.

Question 37.
Name the foreign varieties of hens giving meat.
Answer:
New Hampshire, White plymouth, Rhode island red, while leg horn and Anoka.

Question 38.
What are the major practices in food production?
Answer:
Animal husbandry, poultry, fish culture, bee culture etc., are the major practices in food production.

Question 39.
In which months the egg prices are high?
Answer:
During January to April, egg prices are high.

Question 40.
Why egg prices are more in January to April?
Answer:
This is because of most of the eggs are used for hatching.

Question 41.
What is the temperature for hatchability of eggs?
Answer:
Hatchability of eggs is generally influenced by 37°C to 38°C temperature.

Question 42.
Write a slogan on nutritional Egg.
Answer:
“If you want to be healthy person eat egg every day”.

Question 43.
Name the traditional variety meant for fighting.
Answer:
Aseel (Berisa Kodi) is the Indian traditional variety meant for fighting.

Question 44.
How is hatching done in our rural areas?
Answer:
Our rural areas, the practitioner hatch eggs by placing them under broody hen.

Question 45.
What is the weight of Emu bird?
Answer:
The weight of Emu bird is nearly 50 kg.

Question 46.
What is the life span of natural wild varieties?
Answer:
Natural wild varieties grow fully in 5 to 6 years.

Question 47.
What is the life span of broilers?
Answer:
Broilers grow fully in just 6 to 8 weeks.

Question 48.
What is Apiculture?
Answer:
Culture of honey bees (apis) is called Apiculture.

Question 49.
What is the use of honey bee production?
Answer:
Development of apiculture is not only for honey production but also very much useful for crop pollination.

Question 50.
Which insects are the best pollinators?
Answer:
Honey bees are the best pollinators of many agricultural crops.

Question 51.
How much honey is produced by Indian honey bee?
Answer:
Indian honey bee produces 3 – 10 kg of honey.

Question 52.
Which honey bee is produces more honey?
Answer:
European honey bee produces 25 – 30 kgs of honey per annum.

Question 53.
Name the honey bees present in honey bee colony.
Answer:
A honey bee colony consists of one queen bee, several thousands of workers and few hundreds of drones.

Question 54.
What is the primary function of a queen bee?
Answer:
The primary function of a queen bee is to lay eggs.

Question 55.
What is the life span of a queen bee?
Answer:
The life span of queen bee is two-three years.

Question 56.
What are worker bees?
Answer:
The sterile females are called worker bees in the hive.

Question 57.
What is the function of worker bees?
Answer:
They attend to indoor duties during first three weeks of their lives such as secretion of royal jelly feeding of the brood, collecting nectar, pollen and water.

Question 58.
Name the plants that are sources of nectar.
Answer:
Fruit trees like citrus, apple, guava, tamarind, cultivated fields, crops like mustard, gingelly, wheat, cotton, sunflower.

Question 59.
What are the other products of Apiculture?
Answer:
Bee venom and bee wax are the other products of Apiculture.

Question 60.
How is bee venom used?
Answer:
Bee venom is used for the preparation of “Apistincture” used in homeopathic treatment.

Question 61.
What are the uses of bee wax?
Answer:
The uses of bee wax are production of polish cream, nail polish etc.

Question 62.
Name the parts present in a artificial bee hive.
Answer:
It consists of floor board, brood chamber, super chamber, top cover, inner cover, frames and entrance rod.

Question 63.
Name the pests and predators that attack honey bee colonies.
Answer:
Wax moths, wasps, rubber flies, dragon flies attack honey colonies. King crow, Bee eater are more harmful.

Question 64.
How are fish important for us?
Answer:
Fish constitute an important and rich sources of high quality animal protein.

Question 65.
What are the inland water areas?
Answer:
Rivers, fresh water and brackish water lakes, reservoirs, tanks, ponds, swamps etc., are the inland water areas.

Question 66.
Name the crustacean fishery.
Answer:
Prawns, lobsters and crabs together constitute the crustacean fishery.

Question 67.
Name the local varieties offish.
Answer:
Murrel (Korramenu), Katla (jalla), Katrana (bochalu), rohu (mosu), seer (vanjram) are the local varieties.

Question 68.
Name the marine fish varieties.
Answer:
Macerel, tuna, saradines are the marine fish varieties.

Question 69.
What are mechanized fishing?
Answer:
Fishermen catch fish by using machines is called mechanized fishing.

Question 70.
Name some marine fishes which are of high economic value.
Answer:
Mullets, Bhetki and peral spots, shel fishes such as prawns mussels and oysters as well as sea weed.

Question 71.
What are estuaries?
Answer:
Brackish water resources where sea water and fresh water mix together are called estuaries.

Question 72.
What is the reason for growing fish in paddy fields?
Answer:
The reason for this is increasing use of inorganic fertilizers and insecticides in paddy fields.

8th Class Biology 9th Lesson Production and Management of Food From Animals 2 Marks Important Questions and Answers

Question 1.
How do you appreciate the role of poultry to meet the food needs of the present day increasing population ?
Answer:

1. The world is facing the problem of over growth in population.
2. To meet the increasing population food needs, poultry is the best suitable option.
3. So billion hens are reared world wide for eggs and chicken.
4. India achieved 4th position in the world by producing 41.06 million eggs per annum.
5. India is placed in 5th position in the production of 1000 million kgs of chicken per year.
6. Hence, I appreciate the poultry, in fulfilling the food needs of the society.

Question 2.
“Fish has high nutrition values” said Shravya. Do you support Shravya? Why?
Answer:

1. The meat of fish contains 15-25% of proteins. Hence they are the rich sources of proteins.
2. The meat of fish has vitamins like A and D.
3. Fishes are the sources for very important fatty acids and unsaturated fatty acids which are very vital in our body’s metabolic activities.
4. Meat of fish can be easily digested. It helps in the growth and development.

Question 3.
Name the three Indian Major fresh water crops grown in fish culture.
Answer:

1. Rohu – Labeo rohita:
2. Boche – Catla catla
3. Erramosu – Cirrhinus mrigala are the three Indian major crop fishes.

8th Class Biology 9th Lesson Production and Management of Food From Animals 4 Marks Important Questions and Answers

Question 1.
This fish is called Tuna. Collect information about availability catching and marketing of this particular fish.

Answer:
Availability:
English: Big eye Tuna
Latin: Thunnus obesus
size + weight: Average today about 40 -180 c.m. about 1, 4-130 Kg.
Biggest Angled Fish: 197, 3 kg, 236 cm, Peru, 1957 Russel Lee
Maturity: Size 105 cm, weight 25 Kg., age 3 – 4 years
Maximum: Size 230 c.m., Weight 210 Kg., Age 15 years
Catching: 21% Eastern Pacific
38% Western Pacific
22% Indian Ocean
19% Atlantic Ocean
Catching Methods: Longlining, Purse seining and pole-and-line (by-catch)
Share of all Tuna: 2011 about 10% – 398.000 m.t
Main Production: Thailand, Philippines, Indonesia, Mexico, Venezuela, Ecuador, Colombia, Spain, Italy.
Life cyle: About 7-8
Major Markets: Japan (Sashimi)
Populr Product: Fresh (Whole fish)
Forms: Fresh Fillets (Sashimi)

Question 2.
Write a note on different methods used to store the fish in your area.
Answer:

1. The fish caught should be preserved in chilled conditions.
2. After catching the fish, gut is removed. Before transportation, fishes are thoroughly washed in clean and chlorinated waters.
3. They are perceived in insulated ice boxes with alternate layers of crushed ice and fishes in 1 : 1 proportion.
4. Preservation can also be done by
   a) Drying b) Salting c) Smoking d) Canning methods
5. Before canning, harmful bacteria like clostridum botulinum must be destroyed.
6. Cold storage facility and vans with refrigerated chamber have improved the storage and transport of fish.

Question 3.
Visit a poultry farm of your village. Collect the diseases of poultry and their preventive control measures. Prepare them in theform of a table.
Answer:

Diseases in Poultry birds	Casual organism	Precautionary control measures
Fowl cholera	Bacteria	Vaccination for fowl cholera is must.
Salmonellasis	Bacteria	Antibiotics with sulphur and vaccines.
Coryza	Bacteria	Sulfa drugs and antibiotics are necessary.
Fowl fox	Virus	They should be separated from the farm.
Ranikhet	Virus	Fowls should be separated from the farm and treatment should be given separately.

Question 4.
Write some diseases of tamed animals.
Answer:

1. Galikuntu is a common and dangerous diseases occur in cows and buffaloes.
2. Sheep and goats suffer from worm infections (Nattala vyadhi).
3. Some parasitic diseases cause damage to liver and intestine of tamed animals.
4. Viral and bacterial diseases also effect on milk production of cows and buffaloes.
   These are some examples for diseases in domestic animals.

8th Class Biology 9th Lesson Production and Management of Food From Animals Important Questions and Answers

Question 1.
How are the microorganisms like bacteria killed and how can the milk be preserved at milk chilling centres?
Answer:

1. Pasteurization of milk ensures the destruction of disease producing organisms present in milk.
2. In this process, milk is heated at 72°C for 30 minutes.
3. Prior to this, the milk was being cooled to below 10°C.
4. This process was invented by Louis Pasteur.

Question 2.
Dung is used as biofuel. You know that it is an accessory product. Write about such accessory products produced in animal husbandry.
Answer:
1. The supplimentary products produced from animal husbandry are meat, dung, leather, bones, horns, etc,
2. Animal husbandry is also contributing to different industries like

1. Dairy industry: Milk and milk products are produced.
2. Slaughter houses: Production of meat.
3. Leather industry: Tanning of leather and making shoes, belts and suitcases, etc.
4. Fertilizer industry : Bones of cattle are used in the production of chemical fertilizers.
5. Bio gas industry: Cattle dung is used in the production of biogas.
6. Toys industry: The horns of cattle are used to make toys and ornamental items.

These AP 8th Class Social Studies Important Questions 1st Lesson Reading and Analysis of Maps will help students prepare well for the exams.

AP State Syllabus 8th Class Social 1st Lesson Important Questions and Answers Reading and Analysis of Maps

Question 1.
Do you think the free access to maps is a good thing?
Answer:
No, I do not think so. Any government has to maintain secrecy in access to maps. Otherwise, it helps the enemies. But at present satellite images reveal every place on the earth.

Question 2.
What is meant by the Relief feature?
Answer:
Relief feature means the high and low places on the surface of the earth.

Question 3.
Who were the Portuguese explorers?
Answer:
Magellan, Vasco-da-Gama and Bartholomeu Dias were the Portuguese explorers.

Question 4.
What do you know about Marcopolo?
Answer:
Marcopolo was an Italian explorer. He explored China and Asia.

Question 5.
Why do you think maps are useful to armies in times of war?
Answer:
Maps were also in great demand during times of war as armies and airforces needed them.
They use strategy maps to illustrate an organization’s vision, mission, overarching strategies and key goals and initiatives.

Question 6.
Why were the Greeks and the Romans greatly interested in making maps?
Answer:
The Greeks and after them the Romans were greatly interested in making maps and knowing about places near and far. They wanted to conquer the world, build colonies in far off places and trade with them. e.g.: Alexander, the Greek king.

Question 7.
What is aerial photography?
Answer:

Aerial photography is the technique of taking of photographs of the ground from an elevated position, using aircrafts, helicopters, hot air balloons etc.
Aerial photographs are not maps.

Question 8.
What are satellite imagery?
Answer:
Satellite imageries are the photographs of the earth taken by the artificial satellites launched into the space. These are used in many ways: map making, planning, meteorology, forestry, warfare etc.

Question 9.
Write the difference between maps and photographs.
Answer:
Maps:
A map is used by geographers to show features that we consider important.
(Or)
A map actually is a model of a place giving those features that the map maker considers to be of importance.
Photographs:
A photograph may not be able to show you the features of a place like a map.

Question 10.
Name the early map makers.
Answer:
The Sumerians, the Babylonians, the Greeks, the Arabs, the Chinese, the Europeans were the early map makers.

Question 11.
How can we read Thematic maps?
Answer:

1. A map which focuses on only one aspect is called “Thematic Map”,
   e.g.: Political Maps, Physical Maps etc.
2. We should have the knowledge of different kinds of symbols, colours and patterns which are used on thematic maps.
   e.g.: Dark Grey – Mountain Black – Boundary.
3. Then only we can read Thematic maps.

Question 12.
What are main relief features on the earth?
Answer:

1. Relief feature means the high and low places on the surface of the earth.
2. The main relief features are: hills, valleys, plateaus, plains, river basins, rocky and sandy places.

Question 13.
What are called Isolines?
Answer:
The lines which join the places with some common features are called Isolines or Contour lines.

Question 14.
What are the uses of contour lines?
Answer:
Contour lines give an indication of the slope of the land as well as the elevation above sea level.

Question 15.
How did Mt. Everest get its name?
Answer:
In 1802 William Lambton began one of the most important geographical surveys in the world starting from Chennai in the south and culminating in the Himalayas to determine the length of longitude and also the heights of various places. This survey was completed by Sir George Everest. It is this survey that established that Mt. Everest is the highest peak in the world. It was named after George Everest.

Question 16.
What are conventional symbols?
Answer:
The symbols which are used by map makers conventionally are called conventional symbols.

Question 17.
Observe the given picture and write your comment.

Answer:
This was a model of the world according to the Bible. It is surrounded by oceans and is divided into three continents – Asia, Europe and Africa.
Of these, Asia was considered the largest and the most important as it had Jerusalem which was the birthplace of Jesus Christ. It is therefore also shown on the top.

Question 18.
Statement:
“All the important features can be shown on a map. So people make different kinds of maps”.
* Read this statement and prepare a question on it.
Answer:
“Why do people make different kinds of maps?”

Question 19.
Why were the Greeks and the Romans greatly interested in making maps?
Answer:
The Greeks and after them the Romans were greatly interested in making maps and knowing about places near and far. They wanted to conquer the world, build colonies in far off places and trade with them.
e.g.: Alexander, the Greek king.

Question 20.
Why do you think that in Idrisi’s map the top side points to the south while in the maps prepared by the Greeks it pointed to the north?
Answer:
In many issues Islamic traditions are different with the rest of the world’s, e.g.: Script.
In the similar way the top side points to the south in Idrisi’s map.
(Or)
While facing towards east, he would have given importance to the right hand side (south) and showed it on the top side of the map.

Question 21.
Appreciate the Babylonian clay tablet world map.
Answer:
The Babylonian clay tablet was dated from the Persian period. It was flat and round. The inner circle had all the places they knew about. The city of Babylon was shown in the middle. Beyond the inner circle was ‘Bitter river, or ‘Salt water ocean’ in which were seven triangular islands.
Their thought, imagination, map making ability were really appreciable.

Question 22.
Who invented the latitudes and longitudes and the grid system?
Answer:
Hipparchus (190-120 B.C.) was perhaps the greatest of the Greek astronomers. He devised a method of locating geographical positions by means of latitudes and longitudes. Ptolemy also was a Greek astronomer and mathematician. He lived and worked in Egypt. His work which employed a system of latitudes and longitudes, influenced map makers for hundreds of years. He evolved the science of map-making.

Question 23.
Write the biography of Al Idrisi. (in a few lines)
Answer:

Ash – Sharif Al Idrisi was born in 1099. He was a Muslim geographer, cartographer, Egyptologist and traveller. He lived in Sicily, at the Court of King Roger – II. He spent much of his early life travelling through North Africa and Spain.
Al Idrisi incorporated the knowledge of Africa, the Indian Ocean and the far east gathered by Islamic merchants and explorers and recorded on Islamic maps. He wrote a book ‘A Diversion for the Man Longing to Travel to Far Off Places’. The Tabula Rogeriana’ was drawn by him in 1154 for the Norman King Roger II of Sicily. He died in Sicily in 1165/1166.

Question 24.
Observe the following paragraph and answer the given questions.
Relief feature means the high and low places on the surface of the earth. The main relief features are: hills, valleys, plateaus, plains, river basins, rocky and sandy places. Since the maps are flat we cannot show the height on them. We, therefore, use a special symbol for this called contours or contour lines. Contours are lines on the map joining places of same height – measured from the sea level. In other words, all places on a contour line will have the same height from the sea level. Contour lines are also called isolines – lines joining places with some common features.
1. What is meant by Relief feature?
Answer:
Relief feature means the high and low places on the surface of the earth.

2. What are the main relief features?
Answer:
The main relief features are hills, valleys, plateaus, plains, river basins, rocky and sandy places.

3. Why can we not show the heights of the maps?
Answer:
Since the maps are flat we cannot show the height on them.

4. Special symbols are called contour lines.

5. Contours are lines on map joining places of same height measured from the sea level.

6. Contour lines are also called isolines.

7. What are isolines?
Answer:
Isolines are the lines which join the places with some common features.

Question 25.
Observe the given information and answer the following questions.

Names of Famous Explorers	Their Nationality	Their Lifespan	Their Voyage of Exploration
1) Marcopolo	Italian	1254- 1324	China and Asia
2) Ferdinand Magellan	Portuguese	1480- 1521	First voyage around the world.
3) Christopher Columbus	Italian / Spanish	1451 – 1506	America, the new world.
4) Vasco-da-Gama	Portuguese	1469-1524	Ocean route from Portugal to east.
5) Bartholomeu Dias	Portuguese	1457- 1500	Voyage around the Cape of Good Hope on the southern most tip of Africa.

1. Who were the Protuguese explorers?
Answer:
Magellan, Vasco-da-Gama and Bartholomeu Dias were the Portuguese explorers.

2. What do you know about Marcopolo?
Answer:
Marcopolo was an Italian explorer. He explored China and Asia.

3. Who discovered America?
Answer:
Christopher Columbus discovered America.

4. What was the lifespan of Magellan?
Answer:
His lifespan of Magellan was 1480 – 1521. (41 years)

5. Who made his first voyage around the world?
Answer:
Ferdinand Magellan.

Question 26.
Locate the following on the given world map.

1. Present place of Babylonians / Sumarians
2. Greece
3. Sicily
4. Libya
5. Asia
6. Europe
7. Arabia
8. China
9. North America
10. South America

Answer:

Question 27.
Observe the given map.

Now answer the following questions.
1. What are the islands in the Arabian sea
Answer:
Lakshadweep.

2. This indicates
Answer:
External boundaries.

3. What is the scale of the map?
Answer:
1 cm = 200 kms.

4. Name any state on east coast.
Answer:
Odisha, Andhra Pradesh.
(you can write any one)

5. Name any state on the west coast.
Answer:
Gujarat.

Question 28.
Observe the given picture and write down your opinion on Mercator projection.

Answer:

1. Gerardus Mercator was a Dutch geographer and cartographer.
2. The Mercator projection distorts the size and the shape of large objects, as the scale increases from the equator to the poles, where it becomes infinite,
   e.g.: 1) Greenland takes as much space on the map as Africa. In reality Africa’s area is 14 times greater than that of Greenland. Greenland is comparable to Algeria only.
   2) Alaska – Brazil
3. Finland – India.

 

These AP 10th Class Maths Chapter Wise Important Questions Chapter 9 Tangents and Secants to a Circle will help students prepare well for the exams.

AP State Syllabus 10th Class Maths 9th Lesson Important Questions and Answers Tangents and Secants to a Circle

Question 1.
What do we call the part a and b in the below circle ?

Solution:
‘a’ is minor segment and ‘b’ is major segment.

Question 2.
Find the length of the tangent to a circle of radius 7 cm at a point from a distance 25 cm from the centre.
Solution:
Given OA = 25 cm, OB = r = 7 cm
In ΔAOB, ∠B = 90°

OA² = OB² + AB²
⇒ AB² = OA² – OB² = 25² – 7²
⇒AB = \(\sqrt{25^{2}-7^{2}}=\sqrt{625-49}\)
= \(\sqrt{576}\)
= 24 cm

Question 3.
Find the area of a sector of a circle whose radius is 7 cm and angle at the centre is 60°.
Solution:
Radius = 7 cm, Angle at centre = 60°
Area of the sector = \(\frac{\mathrm{x}}{360}\) x πr²
= \(\frac{60 \times \frac{22}{7} \times 7 \times 7}{360}=\frac{154}{6}\) = 25.66 cm²

Question 4.
A tangent is drawn to a circle of radius 4 cm. from a point that lies at a distance of 5 cm. from the centre. Find the measure of length of the tangent.
Solution:

ΔOAB is a right triangle.
OA² = 0B² + AB²
5² = OB² + 4²
OB² = 25 -16 = 9
OB = \(\sqrt{9}\) = 3 cm.

Question 5.
Find the area of the shaded part in the given figure.

Solution:
Area of shaded part
= Area of semi-circle – Area of triangle

Question 1.
Find the length of the tangent from a point 13 cm away from the centre of the circle of radius 5 cm.
Solution:
Radius of the circle = 5 cm
Length of the tangent = x cm

Distance between centre to point be = 13 cm
13² = 5² + x²
x² = 169 – 25 = 144 ⇒ x = 12 cm

Question 2.
If tangents PA and PB from a point P to a circle with centre ‘O’ are inclined to each other at angle of 80°, then find the measured of ∠POA.
Solution:

Between ΔA0P and ΔB0P
∠OAP = ∠OBP = 90° (∵ tangent and line from origin meet at 90° to each other)
0A=0B = radius of the circle = r (say)
and OP is the common side
Hence, we can say ΔAOP = ΔBOP
Therefore, we can say
∠OPA = ∠OPB = \(\frac { 1 }{ 2 }\)(∠APB)
= \(\frac { 1 }{ 2 }\)(80°) = 40°
(∵ given that tangents PA and PB are inclined to each other by 80°)
Now, in ΔAOP,
∠POA + ∠OPA + ∠A = 180°
⇒ ∠POA + 40° + 90° = 180°
⇒ ∠POA = 50°

Question 3.
Draw a circle of radius 3 cm, mark a point ‘P’ on the circle and draw a tan gent at ‘P’.
Solution:
Steps of construction:

1) Draw a circle of radius ‘3’ cm from the centre “O” and pick a point ‘P’ on the circle. Join \(\overline{\mathrm{OP}}\)
2) Now draw a perpendicular at the point ‘P’ to the line segment \(\overline{\mathrm{OP}}\)
such that XY ⊥ \(\overline{\mathrm{OP}}\).
3) Then \(\overline{\mathrm{XY}}\) is the desired tangent at ‘P’ to the given circle.

Question 4.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding minor segment (use π = 3.14),
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.

Area of sector OACB = \(\frac{\mathrm{x}}{360}\) × πr²
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of Δ AOB = \(\frac { 1 }{ 2 }\) × 10 × 10
= 50 cm²
Area of the Minor Segment = Area of sector OACB – Area of Δ OAB
= 78.5 – 50.0 = 28.5 cm²

Question 1.
A chord of circle of radius 10 cm sub-tends a right angle at the centre. Find the area of the corresponding :
i) Minor segment ii) Major segment (use π = 3. 14)
Solution:
Radius of circle (r) = 10 cm
Sector angle (x) = 90°
Radius of sector (r) = 10 cm.

Area of sector OACB = \(\frac{\mathrm{x}}{360}\) × πr²
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10
= 78.5 Sq. cm
Area of ∆ AOB = \(\frac { 1 }{ 2 }\) × 10 × 10
= 50 cm²
Area of the Minor Segment = Area of sector OACB – Area of ∆ OAB
= 78.5 – 50.0 = 28.5 cm²
Area of Major Segment = Area of the circle – Area of the Minor Segment
= (3.14 × 10 × 10) – 28.5
= 314-28.5 = 285.5 cm²

Question 2.
Draw a circle of radius 3 cm. Take a point ’P’ at a distance of 5 cm from the centre of the circle. From P, draw 2 tangents to the circle.
Solution:
Award marks for construction as follows:

i) To draw a circle with radius 3 cm
ii) To plot a point P such that OP = 5 cm.
iii) To Bisect QP at M and draw circle with radius OM br MP
iv) To draw tangents from intersecting points of two circles.

Question 3.
Draw a Circle of radius 4 cm. From a point 7.5 cm away from its centre, construct the pair of tangents to the circle.
Solution:
1) Draw a circle with:radius 4 cm with centre O.
2) Locate a point P such that OP 7.5 cm
3) Bisect OP and draw circle with radius MO or MP with centre M.
4) Draw tangents PA and PB from external point P to the given circle.

Question 4.
Draw a circle of radius 5 cm. From a point 8 cm away from its centre, con-struct a pair of tangents to the circle. Find the lengths of tangents.
Solution:
Steps of construction :
1) Construct a circle with a radius of 5 cm.
2) Trace the point ‘p’ in the exterior of the circle which is at a distance of ‘8’ cm from its centre.
3) Construct a perpendicular bisector to OP which meets at M.
4) The draw a circle with a radius of MP or MO from the point M. This circle cuts the previous circle drawn from the centre ‘O’ at the points A and B.
5) Now join the points PA and then PB.
6) PA, PB are the required tangents which are measured 6.2 cm long.

OA = 5 cm ; OP = 8 cm
AP = PB = 6.2 cm.

Question 5.
Draw a circle of radius 4 cm and draw a pair of tangents to the circle, which are intersecting each other 6 cm away from the centre.
Solution:

Steps of Construction:
1) Draw a circle with centre ‘O’ and radius 4 cm.
Take a point ‘P’ outside the circle such that OP = 6 cm. Join OP.
Draw the perpendicular bisector to OP which bisects it at M.
4) Taking M as centre and PM or MO as radius draw a circle.
Let the circle intersects the given circle at ’A’ and ‘B’.
Join P to A and B.
PA and PB are the required tangents of lengths.

Question 6.
Two tangents TP and TQ are drawn to a circle with centre ‘O’ from an external point T, then prove that ∠PTQ = 2. ∠OPQ.
Solution:
Given a circle with centre 0.
Two tangents TP, TQ are drawn to the circle from an external point T.
We need to prove ∠PTQ = 2∠OPQ
Let ∠PTQ = θ

TP = TQ (The lengths of tangents drawn from an external point to a circle are equal)
So ΔTPQ is an isosceles triangle
∴ ∠TPQ + ∠TQP + ∠PTQ = 180°
(Sum of three angles in a triangle)
∠TPQ = ∠TQP = \(\frac { 1 }{ 2 }\)(180° – θ)
= 90°- \(\frac{\theta}{2}\)
∠OPQ = ∠OPT – ∠TPQ
= 90° – 0(90° – \(\frac{\theta}{2}\)) = \(\frac{\theta}{2}\)
= \(\frac { 1 }{ 2 }\) ∠PTQ
∴∠PTQ = 2∠OPQ

Question 7.
Find the area of the segment shaded in the figure in which PQ = 12 cm, PR = 5 cm and QR is the diameter of the circle with centre ‘O’. (Take π = \(\frac{22}{7}\))

Solution:
To find the area of the segment shaded in the given figure.
Here ‘PQ’ = 12 cm; ‘PR’ = 5 cm; ‘QR’ is diameter
Now PQOR is a semicircle then angle in a semicircle is 90°.
then ∠QPR = 90°
∴ ΔPQR is a right angled triangle
∴ Area of ΔPQR = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) × PQ × PR
= \(\frac { 1 }{ 2 }\) × 12 × 5 =30 cm² ……………(1)
Now the area of shaded part = area of semicircle – area of ΔPQR
= \(\frac { 1 }{ 2 }\)πrcm² – 30 cm² …………….(2)
In ΔPQR, QR² = PQ² + PR²
(from Pythagoras theorem)
QR² = 12² + 5²
= 144 + 25 = 169 = 13²
∴ QR =13 then
Radius of the circle (r) = QO = \(\frac{\mathrm{Q} R}{2}\)
= \(\frac{13}{2}\) = 6.5 cm
then area of semicircle
= \(\frac { 1 }{ 2 }\)πrcm²
= \(\frac{1}{2} \times \frac{22}{7} \times \frac{13}{2} \times \frac{13}{2}\) = 66.39 cm² …………….(3)
Now putting the values of (1) and (3) in (2) we get
Area of shaded part = (66.39 – 30)
= 36.39 cm².

Question 8.
Draw two tangents to a circle of radius 2.5 cm, from a point ‘P’ at a distance of 7 cm from its centre.
Solution:
1. Draw a line segment PO = 7 cm.
2. From the point O, draw a circle of radius = 2.5 cm.
3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.
4. Taking M as centre and OM as radius, draw a circle.
5. Let this circle intersects the given circle at the point Q and R.
6. Join PQ and PR.

Question 1.
As shown in the figure, radius of the given circle is 21 cm and ∠AOB = 120°. Then find the area of segment AYB.

Solution:

Given radius of area = OA = OB = 21 cm
Now3 the angle at centre for the sector \(\widehat{\mathrm{OAB}}\) = 120°.
Formula for area of sector = \(\frac{x}{360}\) x πrcm²
= \(\frac{120}{360} \times \frac{22}{7}\) × 21 × 21
= 22 × 21 = 462 cm² ……………………… (1)

Now area of segment \(\widehat{\mathrm{AYB}}\)
= Area of sector – area of Δ OAB.
Let \(\overline{\mathrm{OD}}\) is perpendicular to AB, then
∠AOB = \(\frac{120}{2}\) = 60°
∴ sin 60° = \(\frac{\mathrm{AD}}{\mathrm{OA}}\)

∴ area of ΔOAB = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) × AB × OD = AD × OD
= \(\) = 190.95 cm² ………….. (2)
∴ area of segment \(\widehat{\mathrm{AYB}}\)
= 462 – 190.95
= 271.05 cm²
= 271.05 cm²

Question 2.
In a wall clock, length of minutes needle is 7 cm. Then find the area covered by it in 10 minutes of time.
Solution:
Length of minutes needle =(r) = 7 cm
We need to calculate the area covered by in 10 minutes of time = Area of sector.

Now the angle covered by it in 60 minutes = 360°
∴ In 10 minutes = \(\frac{360}{60} \) × 10 = 60°
Area of sector = \(\) × 7 × 7
= \(\frac{154}{6}=\frac{77}{3}\)
∴ Area covered by it in 10 minutes of time = \(\frac{77}{3}\) cm² = 25.66 cm²

Question 3.
Find the area of a right hexagon inscribed in a circle having 14 cm of radius.
Solution:
Radius of circle = OA = OB = OC = OD = OE = OF = 14 cm

∠AOB = \(\frac{360}{6} \) = 60°
∴ In ΔAOB AO = BO = 14 cm
∠AOB = 60°
then ∠OAB = ∠OBA
(∵ Opposite to equal sides)
And ∠OAB + ∠OBA + 60 = 180
=> ∠OAB = ∠OBA = ∠AOB = 60°
Hence it is an equilateral triangle.
∴ OA = OB = AB = 14 cm
∴ Area of hexagon = 6 (Area of ΔAOB)
= 6. \(\frac{\sqrt{3}}{4}\) a²
= 6. \(\frac{\sqrt{3}}{4}\) × 14 × 14
Area of hexagon = 294\(\sqrt{3}\) cm²

Question 4.
Four carrorp board pans are arranged as shown in figure. Radius of the pan is 3 cm each. Then find the area in between of them.
Solution:
Area in between 4 pans = Area of square ABCD formed by joining their centres – 4 (area of sector)

Now side of square ABCD = 3 + 3 = 6
Then area of square ABCD
= 6 × 6
= 36 cm²
and now area of sector = \(\frac{x}{360}\) × πr²
Area of 4 sectors
= 4 × \(\frac{90}{360} \times \frac{22}{7}\) × 3 × 3
= \(\frac{198}{7}\) = 28.3 cm²
∴ Area in between 4 pans i.e., shaded
= 36 – 28.3 = 7.7 cm²

These AP 9th Class Physical Science Important Questions 12th Lesson Units and Graphs will help students prepare well for the exams.

AP State Syllabus 9th Class Physical Science 12th Lesson Important Questions and Answers Units and Graphs

9th Class Physical Science 12th Lesson Units and Graphs Textbook Questions and Answers

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9th Class Physical Science 12th Lesson Units and Graphs 1 Mark Questions and Answers

Question 1.
What is Unit?
Answer:
A unit is a standard measure used for comparing measurements.

Question 2.
Why do we use different units for different items?
Answer:
Depending on the quantities of material the units may be expressed.

Question 3.
Write fundamental quantities in MKS system.
Answer:
Mass (m), Length (l) and Time (t).

Question 4.
What are fundamental quantities?
Answer:
Fundamental quantities are those physical quantities that cannot be expressed in terms of other quantities.

Question 5.
What is fundamental unit?
Answer:
Units used to express fundamental quantities are called fundamental units.

Question 6.
Write fundamental units in MKS system.
Answer:
Meter, Kilogram and Second.

Question 7.
What is CGS system?
Centimeter, Grams and Seconds.

Question 8.
What is SI system? When was it introduced?
Answer:
SI system means International System of units. It was introduced in 1971.

Question 9.
List the fundamental quantities that are added to SI system.
Answer:
Electricity, light intensity, quantity of substance, temperature and plane angle.

Question 10.
What is difference between MKS and SI systems?
Answer:

• MKS system has only three fundamental quantities. Those are length, mass and time.
• SI system has eight fundamental quantities. Those are length, mass, time, electricity, light intensity, quantity of substance, temperature, plane angle.

Question 11.
What is derived quantity?
Answer:
The quantities that are obtained by combining fundamental quantities either by multiplication or division or both operations are called derived quantities.

Question 12.
What are derived units?
Answer:
The derived units of measurement derived from the fundamental units.

Question 13.
Write derived units of area and volume.
Derived units of area = m²
Derived units of volume = m³

Question 14.
What are the fundamental units are used to express the units of Force?
Answer:

Hence fundamental units used to express force are kg, m, s.

Question 15.
Which conversion factor is used to convert kilometers into meter?
Answer:
1000.

Question 16.
Which conversion factor is used to convert meters into kilometers?
Answer:
10^-3.

Question 17.
Write conversion factor to convert km/hr into m/s.
Answer:
\(\frac{5}{18}\)

Question 18.
Write some units of measurements expressed in the name of scientists.
Answer:
Newton (N), Pascal (Pa), Joule (J), Watt (W), etc.

Question 19.
What is graph?
Answer:
A pictorial form of representation that shows the relation between two quantities can be called a graph.

Question 20.
What are independent and dependent variables?
Answer:
Independent variables are controlled by us. Whereas dependent variables are changed due to the change in dependent variable.

Question 21.
Which variable is taken on x – axis?
Answer:
Independent variable.

Question 22.
What are grids on graph paper?
Answer:
On a graph paper, there are thick as well as fine vertical and horizontal lines. These intersecting lines form squares or grids.

Question 23.
What is Range?
Answer:
Range = highest value – lowest value.

Question 24.
What is scaling?
Answer:
Pointing of the values in the table on x – axis and y – axis is called scaling.

Question 25.
What is scale?
Answer:
The interval taken pointing values on axis is called scale.

Question 26.
How do you find x – axis scale?
Answer:

Question 27.
What is curved line graph?
Answer:
If a graph is in curved shape, then it is curved line graph.

Question 28.
What is Hooke’s law?
Answer:
The elongation of the spring is proportional to the applied mass within the elastic limit of the spring.

Question 29.
What is slope of the straight line graph?
Answer:

Question 30.
What is slope of a graph?
Answer:
The tan value of the angle at x – axis is the slope of a graph.
tan θ = \(\frac{\Delta \mathrm{y} }{\Delta \mathrm{x}}\)

Question 31.
What is area of a graph?
Answer:
The product of the physical quantity of y – axis and x – axis may explain another physical quantity. Hence, it is said to be area of the graph.

Question 32.
What does slope indicate to the graph of displacement and time?
Answer:
Velocity.

Question 33.
What does area of the graph of acceleration Vs time indicate?
Answer:

Question 34.
Predict shape of the graph if you plot a graph by taking time on x – axis and velocity of a free fall body on y – axis.
Answer:
Straight line.

Question 35.
Guess the shape of the graph if you plot a graph of time and distance covered by a bus.
Answer:
Curved graph.

Question 36.
Rithvik wrote ‘3 kgs of sugar’ on the black board. Correct him by asking a question.
Answer:
Is it correct to mention kg in plural form while writing units?

Question 37.
Karthik got a doubt by seeing ‘3 newtons’ on the black board. What would be it?
Answer:
Which quantity is measured in newtons ?

Question 38.
What are the materials / apparatus required to prove Hooke’s law?
Answer:

1. Spring,
2. Weights,
3. Scale,
4. Stand.

Question 39.
Area, volume, density, mass
a) Write any one fundamental quantity from the above data.
b) Write any one derived quantity from the above data.
Answer:
a) mass.
b) volume.

Question 40.
‘Dheeraj counted distance travelled by him for every 10 min.’
From the above data write independent and dependent variables.
Answer:
Time (min) is independent variable distance is dependent variable.

Question 41.

Plot the data points (6, 4) and (0, 0) in the above graph.
Answer:

Question 42.
How do you remember scientists through units of measurements?
Answer:
Some units are expressed in the names of the scientists like newton, joule, etc. When we use these units, scientists are remember once again.

Question 43.
What is the use of graph?
Answer:

• We can solve so many problems by using graphs.
• We can know the relation between two physical quantities through graphs.

Question 44.
Calculate the Range of 5, 7, 3, 8, 18, 4, 2, 6.
Answer:
Range = highest value – lowest value
= 18-2 = 16.

Question 45.
If the range is 10 and no. of grids horizontally is 20. How can you take scale X-axis?
Answer:

2) I will take 0.5,1,1.5, 2, 2.5,3,3.5, 4, 4.5, 10 on the X-axis. As a scale on x-axis.

Question 46.
How slope of displacement Vs time graph is useful?
Answer:
We can calculate velocity by calculating slope of line in displacement Vs time graph.

Question 47.
How area of velocity Vs time graph is useful?
Answer:
Area of velocity Vs time graph is useful to find the position of the object.

Question 48.

What is the acceleration at ‘B’?
Answer:

Acceleration is 1 m/s².

Question 49.

Find the acceleration at ‘A’.
Answer:

Question 50.

How much distance is travelled by the object from A to B in the graph?
Answer:
Zero.

9th Class Physical Science 12th Lesson Units and Graphs 2 Marks Questions and Answers

Question 1.
What is a unit? Where it be placed?
Answer:

• A unit is a standard measure used for comparing measurements.
• Unit should be placed right side of the magnitude of every physical quantity. For eg : 2 kg, 7 m.

Question 2.
Differentiate fundamental quantities and derived quantities.
Answer:

Fundamental quantity	Derived quantity
1) Fundamental quantities are those physical quantities that cannot be expressed in terms of other quantities.	1) The quantities that are obtained by combining fundamental quantities either by multiplication or division or both are called derived quantities.
2) These are independent.	2) These are dependent.
3) Eg : Mass, length, time.	3) Eg : Area, density, force.

Question 3.
Derive units of density.
Answer:

Question 4.
Write different system of measurements.
Answer:

1. FPS : Foot, Pound, Second.
2. CGS : Centimeter, Gram, Second.
3. MKS : Meter, Kilogram, Second.
4. SI : Meter, Kilogram, Second, Ampere, Candela, Mole, Kelvin, Radian.

Question 5.
Derive units to the given physical quantities.
Answer:

Question 6.
Which of the following are correct? Why?
Answer:
A) m/s
B) m/s/s
C) m-s²
D) 10.kgs
E) Pascal
F) joule
Answer:
A) and F) only are correct
B) two solidus are used – wrong.
C) line break is used – wrong.
D) ‘•’ and plural of kg are used – wrong.
E) capital letter used (P) – wrong.

Question 7.
Write correct form of units in the following.
(j) DB
(ii) mhz
(iii) 20 – m
(iv) kg.m-s^-2
Answer:
(i) dB
(ii) MHz
(iii) 20 m
(iv) Kg.m.s^-2

Question 8.
How do you expect the shape of the given graphs?
(a) a ∝ b
(b) a ∝ \(\frac{1}{\mathbf{b}}\)
Answer:
a) It may be straight line graph.
b) It may be curved line graph.

Question 9.
From the graph,
a) What is indicated by slopes of the lines OA, OB, OC?
b) Which line shows high speed than other lines? Why?

Answer:
a) Slopes of the lines indicate speeds of the objects.
b) OA shows high speed than other lines.
Because it has high slope than others.

Question 10.

a) Which is independent variable?
b) Which is dependent variable?
c) Predict how the graph is.
d) Find the range on Y – axis.
Answer:
a) Mass
b) Weight
c) It may be in straight line.
d) On Y – axis, values are 98, 196, 294, 392, 490.
Range = 490 – 98 = 392

Question 11.
Plot the given data points on the given graph paper and join the dots.
Data points : (8, 10) (20, 15) (40, 22.5) (48, 0)

Answer:

Question 12.
Draw a graph roughly to show uniform velocity of a body.
Answer:

Question 13.
Draw a graph that shows a body without movement after travelling sometime.
Answer:

Question 14.
How do you appreciate the role of graphs in solving problems?
Answer:

• Graphs can be used to compare different physical quantities like time- displacement, time-velocity, pressure-volume, etc. and helps in solve the problems.
• Graphs are very useful to find the path of an object which is in motion and it is easy to calculate velocity, acceleration, position, etc. at particular time and particular point of the object.
• Graphs are very useful minimum, maximum values of some physical quantities. For example to find minimum deviation of a prism graphs are very useful.
• In this way graphs are appreciable for their role in solving problems.

Question 15.
From the given graph find the slope and area at point B.
Answer:

Question 16.
Convert 3 m/s into km/hr.
Answer:

Question 17.
A ball of 100 g mass is thrown with initial velocity of 0.01 km/s. What is its momentum at that time?
Answer:
Momentum = mass × velocity
Here, mass of the ball = 100 gr = 100 × \(\frac{1}{100}\) kg ⇒ m = 0.1 kg
Velocity of the ball = 0.01 km/s = 0.01 × 1000 m/s ⇒ v = 10 m/s
Momentum of the ball = mxv = 0.1 × 10 kg.m/s = 1 kg.m/s

Question 18.

Find the temperature on Day – 4 by using graph.
Answer:

From the graph the temperature on day – 4 is 43°C.

Question 19.
Observe table and answer the following.

i) 12 cm = …………. m
ii) 2 nm = …………. cm
Answer:
i) 12 cm = 12 × 10^-2 m
ii) 2 nm = 2 × 10^-7 cm
[From the table 1 nanometer = 10_-9 m; 1 centimeter = 10^-2 m.
1 nanometer to centimeter conversion factor = 10^-9-(-2) = 10^-7.]

Question 20.
Find the conversion factors
a) Picometers to Meters
b) Gigabytes to Kilobytes.
Answwr:
From the above table,
a) 1 picometer = 10^-12 m; 1 meter = 10° m
Conversion factor to convert picometers to meters = 10^-12-(-0) = 10^-12 m

b) 1 GB = 10⁹ bytes ; 1 KB = 10³ bytes
Conversion factor to convert GB to KB = 10^9-3 = 10⁶ bytes.

Question 21.
Find the conversion factors
a) Megawatts to Kilowatts
b) Kilowatts to Megawatts.
Answer:
From the above table,
a) 1 Megawatt = 10⁶ watts; 1 Kilowatt = 10³ watts
Conversion factor MW to KW = 10^6-3 = 10³ W

b) 1 KW = 10³ W and 1 MW = 10⁶ W
Conversion factor KW to MW = 10^3-6 = 10^-3 W

9th Class Physical Science 12th Lesson Units and Graphs 4 Marks Questions and Answers

Question 1.
Write any four rules of writing units of measurements.
Answer:
1) Values are written as a number followed by a space and a unit symbol.
eg : 24 kg, 2 cm, 5 s.

2) Symbols for derived units formed by multiplication are joined with a dot or a space.
eg : 30 N.m, 30 N.m

3) Symbols are mathematical entities, not abbreviations. So, do not have an appended period / full stop.
eg : 4 kg is correct, but 4 k.g. is incorrect.

4) All symbols for units are written in lowercase.
eg : m, s, kg, etc.

5) Symbols derived from the name of a person use uppercase letter for shortform.
eg : Pa, N, W, etc.

Question 2.
Write steps in the construction of a graph, to the given data in a tabular form.
Answer:

• Take a graph paper. Draw x – axis and y – axis on it.
• Identify independent variable and dependent variable from the data given in the table. Generally independent variables are decided by us, whereas dependent variables are changed due to the change in independent variable. Take independent variable on x – axis and dependent variable on y – axis.
• Calculate range for values taken on x – axis as well as y – axis, by using the formula. Range = highest value – lowest value.
• Determine the scale to be taken on x – axis and y-axis by calculating, range/no. of grids on axis. And mark the scale on the axes.
• Write names of the physical quantity which is taken on the axis. For eg : time (s), distance (m).
• Write the data points from the data given in the table such as (x₁ y₁) (x₂ y₂)
• Label the data points in the graph paper by put a dot where the corresponding horizontal and vertical line intersects of a data point.
• Joint the dots in the graph paper.

Question 3.
Fill the given table with suitable answers.

Answer:
a) N/m²
b) Pa [pascal]
c) s^-1
d) Force
e) m.kg.s^-2
f) m² . kg . s^-2
g) J/s (or) \(\frac{N.m}{s}\)
h) W [watt]

Question 4.
Fill the table with suitable answers.

Derived quantity	Relation with base and derived quantities	Unit
Area	 
Volume
Density
Speed
Velocity
Acceleration
Work
Momentum

Answer:

Derived quantity	Relation with base and derived quantities	Unit
Area	length × breadth	m2
Volume	length × width × height	m3
Density	mass / volume	kg m-3
Speed	distance / time	m s-1
Velocity	displacement / time	m s-1
Acceleration	change in velocity / time	m s-2
Work	force × distance	kg m2 s-2
Momentum	mass × velocity	kg.m.s-1

Question 5.

1) Who took rest while travelling?
2) After 100 min., what is the positions of Padma and Bhavani?
3) Whose speed is uniform?
4) What is the speed of Bhavani?
Answer:

1. Padma
2. Padma at 700 meters and Bhavani at 600 meters from the beginning point.
3. Speed of Bhavani is uniform.
4. \(\frac{600}{100}\)m/min. = 6 m/min.

Question 6.

a) Which physical quantity is taken as independent variable?
b) What is the scale on x – axis?
c) What is the scale on y – axis?
d) How the graph is?
Answer:
a) Distance
b) 1 cm = 100 m
c) 1 cm = 2 min.
d) It is a curved line graph.

Question 7.
Draw a graph to the given data.

Answer:
1) Draw x – axis and y – axis on a graph paper.
2) Here time is independent variable and displacement is dependent variable. So, take time on x – axis and displacement on y – axis.

3) Calculate the ranges of values on x – axis and y – axis.
Range on x – axis = 24 – 4 = 20
Range on y – axis = 15-2 = 13

4) Estimate scale on x – axis and y – axis.

So, scale on x – axis 1 cm = 2 min.
Scale on y – axis 1 cm = 1 km.
Mark the scales on x and y – axis.

5) Write time (min) at x – axis and displacement (km) at y – axis.

6) Write data points or data values as (4, 2) (8, 3) (12, 7) (16, 11) (20, 13) (24, 15).

7) Plot the above data points in the graph paper with dots.

8) Join the data points.

Question 8.
Draw velocity – time graph to the given table.

Answer:
1) Draw x – axis and y – axis on a graph paper.
2) Here time is independent variable and velocity is dependent variable. So, take time on x – axis and velocity on y – axis.

3) Calculate the ranges of values on x – axis and y – axis.
Range on x – axis = 180 – 20 = 160
Range on y – axis = 20 – 5 = 15

4) Estimate scale on x – axis and y – axis.

Take scale as 1 cm = 8 s on x – axis.
1 cm = 1 m/s on y – axis.

5) Write time (s) at x – axis, velocity (m/s) on y – axis.

6) Write data points as (20, 5), (40, 10), (60, 15) (80, 20) (100, 20), (120, 20) (140, 20) (160, 20) (180, 15).

7) Plot the data points in the graph paper with dots.

8) Join the dots.

Question 9.
Show that i) v = u + at, ii) s = ut + ½ at² in graphical method.
Answer:

1) Graph is plotted with time on x – axis and velocity on y – axis.

2) According to the above graph,
Velocity of the object at “t” value ”0″ = u
Velocity of the object at “t” value “t” = v

3) Difference between the time intervals = v – u

4) From the graph, x₁ = 0, x₂ = t, y₁ = u, y₂ = v

5) We know that slope refers acceleration to the velocity – time graph.
Hence acceleration (a) = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) ⇒ v – u = at ⇒ v = u + at ………… (1)

6) The area between the two straight lines drawn at u, v gives displacement of the object. The area of graph is in the shape of trapezium. It has a rectangle and a triangle.

7) Area of the graph = area of rectangle (ABCD) + area of triangle (DCE)

8) We know that area of the graph of velocity – time gives displacement(s).

Question 10.
Observe the following graph and calculate displacement of the object in 12 sec.

Answer:
1) Displacement (s) = Area of the graph = Area of ABDO rectangle + BDC triangle.
2) Area of the rectangle ABDO = 8 × 20 = 160.
3) Area of the triangle BDC = ½ × 20 × (12 – 8)
= ½ × 20 × 4 = 40.
4) Displacement (s) = 160 + 40 = 200 m.
5) Hence, displacement of the object in 12 sec. = 200 m.

Question 11.
Find the velocity from the graph at A.

Answer:

1. We know that area of the graph of acceleration – time gives velocity (v).
2. Join the line AB.
3. Velocity (v) = Area of ΔABO = ½ × 60 × 120 = 3600 m/s.

Question 12.
Find the force at point ‘B’.

Answer:

We know that
Force= Area of trapezium ABDO.
= Area of triangle AEB + Area of rectangle EODB

Question 13.
Using the following data. Draw displacement time graph.

Using above graph, what is the average velocity in First “4” Seconds
Answer:
Average velocity in first “4” seconds

9th Class Physical Science 12th Lesson Units and Graphs InText Questions and Answers

9th Class Physical Science Textbook Page No. 205

Question 1.
Which is bigger, 2 kg or 100 gr?
Answer:
2 kg.

Question 2.
What would be Rasheeda’s answer?
Answer:
2 kg.

Question 3.
If Rasheeda asked the shop keeper to give 2 sugar, 100 tea powder, can he be able to weigh the items? Why?
Answer:
No, the shop keeper cannot weigh. Because there is no measuring terms (units) to weigh.

Question 4.
Can you name the units of time? What are they?
Answer:
Second, minutes, hours, etc.

Question 5.
Why are they (different units to measure a physical quantity) required?
Answer:
Depending upon the quantity, different units to measure are required.

Question 6.
Can we express the mass of chalk in kilogram?
Answer:
It is difficult to measure directly in kilograms.

9th Class Physical Science Textbook Page No. 206

Question 7.
Why do we use different units for different items?
Answer:
Depending on the quantities of material the units may be expressed.

Question 8.
Why are these called fundamental quantities?
Answer:
These (length, mass, time, etc.) are independent quantities, cannot be expressed in terms of other quantities.

Question 9.
What fundamental quantities are there in the table?
Answer:
Length, mass, time electricity, light intensity, quantity of substance, temperature, plane angle.

Question 10.
What similarities have you noticed in MKS and SI systems?
Answer:
Length, mass, time are common in MKS and SI systems.

Question 11.
How do you find the area of the book?
Answer:
By multiplying length and breadth of the book.

Question 12.
What measurements are required to express area?
Answer:
Length and breadth.

9th Class Physical Science Textbook Page No. 208

Question 13.
How can you convert kilometers into meters?
Answer:
We can convert kilometers into meters by multiplying meters with conversion factor 1000.

9th Class Physical Science Textbook Page No. 211

Question 14.
What is the length of each side (of square grid)?
Answer:
1 cm.

Question 15.
What is the distance between the adjacent thin lines?
Answer:
1 mm.

Question 16.
Which axis is time plotted on and which axis represents distance?
Answer:
Time is plotted on x – axis, y – axis represents distance.

Question 17.
How are these (numbers on x – axis) interpreted?
Answer:
By calculating range of the values of data on x – axis.

Question 18.
What is the difference between the highest and the lowest values?
Answer:
40-5 = 35.

9th Class Physical Science Textbook Page No. 212

Question 19.
What is the advantage of knowing the range of values?
Answer:
If the graph is plotted with the help of calculated values of range, the graph is uniformly distributed through the graph paper and has a correct shape. Otherwise, it will be either too big or too small and will not be any help for analysis.

Question 20.
How to identify values on x-axis and y-axis?
Answer:
By determine the scale.

Question 21.
How to determine scale?
Answer:

Question 22.
Which values of distance in the table are equal to the values marked on y-axis?
Answer:
3, 12, 24, 24, 30, 33.

9th Class Physical Science Textbook Page No. 213

Question 23.
What is the shape of the line in the graph?
Answer:
It is curved line.

Question 24.
Did you notice any elongation of the spring?
Answer:
Yes, I noticed elongation of the spring.

Question 25.
How much was the elongation?
Answer:
Some extend. It can be calculated by difference in the lengths of the spring.

9th Class Physical Science Textbook Page No. 214

Question 26.
Can you identify the independent and the dependent variables?
Answer:
Mass is independent variable and elongation of the spring is dependent variable.

9th Class Physical Science Textbook Page No. 215

Question 27.
At what point plunger cannot move forward.
Answer:
It is depending upon conditions of the experiment.

Question 28.
Do you experience any pressure with air?
Answer:
Yes.

9th Class Physical Science Textbook Page No. 217

Question 29.
Let us say what is the slope of x – axis?
Answer:
Slope of the x – axis is zero.

9th Class Physical Science Textbook Page No. 218

Question 30.
What does the ratio explains the physical quantities on the y – axis and x – axis?
Answer:

The ratio explains the physical quantities on the y – axis and x – axis is velocity.

Question 31.
What is the velocity of the object at 1.5 seconds of time?
Answer:

9th Class Physical Science Textbook Page No. 207

Question 32.
Is litre a fundamental unit or a derived unit?
Answer:
One litre is the volume of a cube with 10 cm sides. (10 cm × 10 cm × 10 cm)
(or) 1 L = 10^-3 m³
Hence, litre is a derived unit.

9th Class Physical Science 12th Lesson Units and Graphs Activities

Activity – 1

Question 1.
You have noticed that different units are used to measure different materials. Based on your past experience fill in the table with suitable units of measurements used in general, put a tick mark.

Activity – 2 Derived Quantities – Derived Units

Question 2.
1) Let us measure the surface area of your book using a scale.
2) Measure the length and breadth of the book.
3) Area of the book can be obtained by multiplying length and breadth.
4) Length = ……………………..
5) Breadth = ……………………
Is area of fundamental quantity?
Answer:
No.

Which fundamental quantity is used to derive area?
Answer:
Length.

From the activity, we can say area is a derived quantity, which is obtained by multiplication of length and breadth which are fundamental quantities.

Activity 3

Question 3.
Let us observe the given table and try to find out fundamental and derived units for the given information.

1) Which derived units are obtained by the multiplication of fundamental units?
Answer:
Area, volume

2) Which units are derived from only the fundamental quantity length?
Answer:
Area, volume

3) Which derived quantity has no units?
Answer:
Relative density

4) What are the fundamental units of volume?
Answer:
Length

5) What quantities are derived from time?
Answer:

1. Velocity
2. Acceleration
3. Force
4. Pressure

Activity – 4

Question 4.
Draw a curved line graph to the given table.
Answer:
1) Anitha travels to her grandmother’s village by car.
2) She observed the Odometer (the instrument that measures the distance travelled) fitted in the car.
3) She recorded the reading shown by the odometer once every 5 minutes for 40 minutes.
4) Look at the table prepared by Anitha.

5) ‘We show the relation between two quantities using a graph.
6) Here time is independent variable and distance is dependent variable.
7) The steps in the construction of a graph

i) Draw x – axis and y – axis on a graph paper.
Plot time on x – axis as it is independent variable whereas distance on y – axis as it is dependent variable.

ii) Find the range on x – axis
Range on x – axis = highest value – lowest value = 40 – 5 = 35 Find the range on y – axis
Range on y – axis = highest value – lowest value = 33 – 3 = 30

iii) Determine the scale

iv) Names on the axis
Write time (min.) on x – axis.
Write distance (km) on y – axis.

v) Write the data points from the table.
(0, 0), (5, 3), (10, 8), (15, 12), (20, 19), (25, 24), (30, 24), (35, 30), (40, 33).

vi) Labelling of data points :
Label above data points with dots where vertical and horizontal lines are intersect to the respective data points.

vii) Join the dots.

What is the shape of the line graph?
Answer:
It is curved in shape.

Activity 5

Question 5.
Draw a straight line graph to the given table.

Answer:

• Take a spring / rubber band.
• Measure the length of the spring with a scale.
• Fix the spring to the stand.
• Suspend the weight to the second end of the spring.
• Now, again measure the length of the spring.
• Calculate the elongation.
•  This way, keep on changing the weights and measure the elongation of the spring with scale.

Record the readings in the table.

9) Here mass (gm) is independent variable and elongation is dependent variable.
10) Plot the graph.

1. Which quantity is taken on the x-axis ? In what units it was expressed? Why was it taken on the x-axis?
Answer:

1. Mass is taken on the x – axis.
2. It was expressed in grams.
3. Mass is independent variable, so it was taken on the x – axis.

2. Which quantity is taken on the y-axis? In what units it was expressed? Can this be called an independent variable?
Answer:

1. Elongation of the spring was taken on the y – axis.
2. It was expressed in millimeters.
3. It cannot be called as independent variable, because the values of elongation are obtained in the experiment.

3. The value of range on the x – axis
Answer:
Range = 50 – 0 = 50

4. The value of range on the y-axis
Answer:
Rang = 10 – 0 = 10

5. Scale on the x-axis
Answer:
10

6. Scale on the y-axis
Answer:
2

7. Data values on the x-axis
Answer:
0, 10, 20, 30, 40, 50.

8. Data values on the y-axis
Answer:
0, 2, 4, 6, 8, 10.

9. What is the shape of the line that joined the intersecting points?
Answer:
Straight line.

10. What does the slope of a curve tell us?
Answer:
Relation between mass and extension of the spring.

11. What is the reason for the elongation of the spring?
Answer:
Due to mass suspension to the spring.

12. What relation did you notice between the mass and elongation of spring?
Answer:
Elongation is directly proportional to mass.

13. Wliat could be the relation between the two quantities when the graph is a straight line?
Answer:
It could be directly proportional.

14. Based on the graph, which of the following sentences are correct?
a) If mass increases the elongation of spring increases.
b) If mass decreases the elongation of spring increases.
c) Even if mass increases, there will be no change in the elongation of the spring.
Answer:
a), b) are correct.

Activity – 6

Question 6.
Plot a graph showing inversely proportional measurements.
Answer:

1. Take a 50 ml syringe.
2. Fill the syringe with air by pulling a plunger.
3. To avoid escape of air from syringe close the nozzle of syringe with your finger.
4. Now push (apply the force) the plunger slowly.
5. While you are pushing the plunger you may experience more pressure with the decrease of volume of air in syringe, hence you need to apply more force on plunger.
6. Let us find out the relation between the pressure and volume of air.
7. Observe the following data table.

Physical Quantity on x – axis
Answer:
Pressure

Physical Quantity on y – axis
Answer:
Volume

What is the independent variable?
Answer:
Pressure

What is the dependent variable?
Answer:
Volume

What is the range of values on vertical axis?
Answer:
Range = 50 – 18.7 = 31.3

What is range of values on horizontal axis?
Answer:
Range = 3.2 – 1.2 = 2

What is the shape of the graph?
Answer:
Curved in shape (Parabolic).

What relation you noticed between two physical quantities based on data points in the graph?
Answer:
I have noticed that there is decrease in volume with increase of pressure. Hence these two quantities said to be inversely proportional.

Activity – 7

Question 7.
Observe the following graphs and write your analysis in the given table.

These AP 9th Class Social Studies Important Questions 8th Lesson Service Activities in India will help students prepare well for the exams.

AP State Syllabus 9th Class Social 8th Lesson Important Questions and Answers Service Activities in India

9th Class Social 8th Lesson Service Activities in India 4 Marks Important Questions and Answers

Question 1.
‘Service sector is beyond the other two sectors in providing employment’.
“Ramesh is a self employee who established an internet centre after completion of B.Tech. He also provided employment for two more persons”.
‘Service sector plays key role in providing employment’ comment on it. (SA-III : 2015-16)
Answer:
Yes. Services sector plays key role in providing employment by the following ways.

1. There has been tremendous change in the communication technology which has opened a wide range of jobs for educated people.
2. Business Process Outsourcing (BPO) has brought in new kinds of employment opportunities.
3. They provide services to people located across the globe using telecommunication links.
4. The entertainment industry creates jobs to work in various print media firms, cable television channels.
5. We see number of internet cafes and telephone booths in most of the cities and towns.
6. The advertising industry has also brought new job opportunities.
7. Many new jobs are available in Banking and insurance sectors.

Hence service sector can reduce educated unemployment in our country.

Question 2.
The table below shows the number of workers (in lakhs) employed in different service activities in large government enterprises in 1991 and 2010. Study the table and answer the question given below. (SA-II : 2018-19)

Service sector activities	Government jobs
	1991	2010
Wholesale and retail trade	1.5	1.7
Transport, storage and Communications	30.3	25.3
Finance, insurance, real estate etc.	11.9	14.1
Community, social and personal services	92.3	90.5

a) Which service activity provided least employment?
b) Write any two jobs related to community, social and personal services.
c) What kind of service activity jobs have shown maximum decrease in 2010?
d) What kind of service activity jobs has government-generated most in 2010?
Answer:
a) Wholesale and retail trade

b)

1. Washing and drycleaning work
2. Beauty treatment
3. Tailoring work
4. Barbour work (Hair cutting)
5. Who runs the photo studio.

c) Transport, storage and communications.

d) Finance, insurance, real estate etc.

9th Class Social 8th Lesson Service Activities in India Important Questions and Answers

Question 1.
How is service sector different from other sectors?
Answer:

• Other sectors, agriculture and industry produce goods that are tangible.
• But service activities do not produce any tangible good.
• But service activities fulfill the needs cater the requirements of people or organisations.
• They are also essential for agriculture and industrial activities.
• Non-production of commodity or good is the essential nature of service sector.

Question 2.
Is service sector, the heaven of dream jobs?
Answer:

• No, I don’t think so.
• Dream jobs in service sector are at a rate of one in tens of thousands of jobs.
• Remaining, especially uneducated and less educated do jobs with meagre wages.
• Those who have dreams are also under constant pressure and job stress.

Question 3.
Read the following case history and answer the following questions.

My lifestyle has improved due to IT. In college, I had just two trousers and two shirts. Now, I have many shirts. I buy good quality clothes and I don’t have to think about how much I spend. So definitely, working in the software industry has added comfort to my life. Earlier, there were a lot of financial problems in the family – my brothers and sisters were not able to study properly due to this, and I am very happy that I have been able to solve this problem. But it is a stressful and mechanical life; there is no social life. In my hometown, people were always there around me – neighbours and relatives. There was always someone to go and talk to but here, it’s difficult to make friends.

1) What was the position of the family before getting a job?
Answer:
Earlier there were a lot of financial problems in the family – his brothers and sisters were not able to study properly due to this.

2) What is his family’s financial position now?
Answer:
Working in software has added comfort to his life, and his brothers and sisters are able to study properly now.

3) Is the individual happy with his lifestyle?
Answer:

• He is earning good salary and that added comfort to his life.
• But at the same time the job is very stressful and his life is mechanical.
• He is missing his village, neighbours and relatives.

Question 4.
What is the work going on here in the given picture? Describe it in five sentences.
Answer:

1. It is a call centre job.
2. A call centre is an office where people work answering or making telephone calls for a particular company or customer regarding certain information.
3. The information may be regarding customer’s bank deposits or about their hospital records.
4. There will be three shifts and employees may have to work during nights too.
5. Salaries are high and at the same time lots and lots of stress in the job can be experienced.

Question 5.
What services, in your opinion, are desired today?
Answer:

• India faces an acute shortage of skilled human resources in the health sector. So health services are desired today.
• More infrastructure service is the need of the hour. So roads, hospitals and educational institutions are to be set up.
• We need good storage facilities as well as credit facilities.
• New economic policies in India shall emphasize the expansion of service activities.

Question 6.
Why do you think more people are engaged in low income service activities today?
Answer:

• With lack of job opportunities many people today are employed in low income services.
• Closure of factories forced workers to work as small vendors, rickshaw pullers and other low income service professions.
• People from rural areas are not getting opportunities in urban areas.
• They shift to many unskilled low earning employment opportunities in services.

Question 7.
Many unskilled have low earning employment opportunities in service sector. State your opinion.
Answer:

• People from rural areas come to work in the city but there are not enough employment opportunities in urban areas.
• Hence most of the unskilled are earning their livelihood as small vendors, rickshaw pullers, etc.
• They are unable to make both ends meet.

Question 8.
New job opportunities are depressing the employees. Comment on it (especially in IT).
Answer:

• New jobs are providing good salaries but the employees are also expected to work very long hours.
• The life is very stressful and mechanical.
• There is no social life.
• These jobs definitely depress the employees.

Question 9.
What measures are taken by companies to reduce cost and avoid paying benefits to workers?
Answer:

• In order to reduce cost and avoid paying health, pension and provident fund benefits to workers, industries ‘outsource’ security services to security agencies.
• Many companies outsource their work for research and development, accounting, legal services, customer service, public relations etc.

Question 10.
Many people work in service jobs. Few people do voluntary service. What measures would you take to distinct them?
Answer:

• Voluntary service means that is done free of cost or out of love and devotion.
• It doesn’t contribute to their livelihood.
• They have other occupation but do this as charity.
• On the otherhand, the people who work in service sector earn livelihood through it.
• They do specialised services required for people or organisations.

These AP 9th Class Social Studies Important Questions 19th Lesson Expansion of Democracy will help students prepare well for the exams.

AP State Syllabus 9th Class Social 19th Lesson Important Questions and Answers Expansion of Democracy

9th Class Social 19th Lesson Expansion of Democracy 1 Mark Important Questions and Answers

Question 1.
The map below shows the democratic countries that existed between 1900 and 1950. (SA-II : 2018-19)

Answer:
Canada, Great Britain, U.S.A., Alaska, Peru, Chile, etc

Question 2.
Name the country which is marked by “A”. (SA-II : 2016-17)

Answer:
Russia.

9th Class Social 19th Lesson Expansion of Democracy 2 Marks Important Questions and Answers

Question 1.
Do you think our country has a liberal democratic political system? Give your reasons? (SA-II : 2016-17)
Answer:
Yes, our country has a liberal democratic political system by following methods.

1. Free and fair regular elections.
2. Written constitution.
3. Equality before law.
4. Universal Adult Franchise etc.

9th Class Social 19th Lesson Expansion of Democracy Important Questions and Answers

Question 1.
Compare the struggles for democracy in Libya and Burma on the following aspects
A) Leadership
B) Nature of struggle
C) Process of transition.
Answer:

Libya	Burma
A. Leadership	A. Leadership
1. Monarchy was established in 1951 under the leadership of king Idris.	1. Democracy was established. Three elec­tions were held in 1951, 56 and 1960.
2. Monarchy was overthrown by “Revolutionary Command Council” under the leadership of Muammar Gaddafi in 1969.	2. In 1962, the elected government was overthrown by a coup and army. General “Ne Win” took charge of the country.

B. Nature of Struggle	B. Nature of Struggle
1. People took the help of mobiles and internet.	1. People took the lead and thousands were brutally killed.
2. They protested against the military regime.	2. Suu Kyi was a central figure in the for establishing democracy in Burma.
3. Soon there broke out a civil war.	3. In 1990 Suu Kyi’s ‘NLD’ won the majority but was not allowed to form government and was kept under house arrest.
4. Gaddafi’s army crushed the protests and thousands were killed.

C. Process of transfer	C. Process of transfer
1. “UNO” intervened and declared Libya as “No fly zone”	1. Economic sanctions were imposed on Burma.
2. France, US, Britain etc., jointly used their force to bombard Libyan government.	2. There were a world wide pressure on the military government to extent civil liberty.
3. Libyans succeeded in overthrowing Gaddafi regime.	3. Elections were held in 2010 but due to the allegations of malpractices, the military junta was dissolved.
4. A new government was sworn in on14th Nov 2012.	4. Thein Sein became the president of Burma.
5. The story is still incomplete.	5. The struggle is still incomplete.

Question 2.
Read the following information and answer the questions given below.
Events and year of occurrence

1) Which of the above two countries first became independent? And when did it happen?
Answer:
Burma was the first country to get independence. And she got it in 1948.

2) Who was the dictator that ruled over Libya?
Answer:
“Gaddafi” ruled over Libya.

3) Who was Aung San?
Answer:
Aung San was the leader of Nationalist movement in Burma.

4) Which of the two countries became democratic first?
Answer:
Burma in 2010.

5) Where are the two countries located?
Answer:
Libya is in Africa and Burma is in Asia.

Question 3.
What are “economic sanctions” ? How do they work on any country’s economy?
Answer:

• Economic sanctions are restrictions imposed on any country regarding imports and exports.
• Import of goods from that country will be banned.
• Export of goods to that country will be stopped.
• Due to that the country’s economy will generally be effected.

Question 4.
What are the steps taken by “the UNO” with regard to Libya or Burma?
Answer:

• The UNO intervened and supported rebels by declaring Libya to be a “No fly zone”.
• Later with the help of US, Britain and France it bombarded the government positions in Libya.
• As a result a new government was formed in on 14th Nov. 2012,
• “Economic Sanctions” were imposed to put pressure on rulers.
• An election was held under the observation of the UNO in 2010.

These AP 10th Class Biology Important Questions and Answers 2nd Lesson Respiration will help students prepare well for the exams.

AP State Syllabus 10th Class Biology 2nd Lesson Important Questions and Answers Respiration

10th Class Biology 2nd Lesson Respiration 1 Mark Important Questions and Answers

Question 1.
What are the end products of Aerobic and Anaerobic Respirations?
Answer:
End products of aerobic respiration: Carbon dioxide, Water, Energy
End products of anaerobic respiration: Ethanol / Lactic acid, Carbon dioxide, Energy

Question 2.
In which organisms, blood does not supply the Oxygen?
Answer:
Arthropoda organisms (or) Insects (OR) Tracheal respiratory Organisms.

Question 3.
Hari said that stem also respires along with leaves. How do you support him?
Answer:
Lenticels on stem also help in gaseous exchange in some woody plants along with stomata.

Question 4.
Arrange the apparatus as above and heat the glucose. What will happen to lime water when glucose burns?
Answer:
Lime water turns milky due to carbon dioxide (CO₂).

Question 5.
What is the role of mitochondria in anaerobic respiration?
Answer:
The release of energy from glucose in the presence of oxygen occurs in mitochondria. In anaerobic respiration, as oxygen is absent, mitochondria have no role in respiration.

Question 6.
Fermented idli, dosa produce smell. Name the microorganism responsible for producing such smell.
Answer:
Yeast is responsible for producing such smell in fermented idli, dosa.

Question 7.
In what compound, the energy released during the breakdown of glucose is stored?
Answer:
“ATP” (Adenosine Triphosphate).

Question 8.
Label a and b in the given diagram.

Answer:
(a) Matrix, (b) Cristae.

Question 9.
Name chemical substance produced in human muscles during Anaerobic respiration.
Answer:
Lactic acid is produced in human muscles during Anaerobic respiration.

Question 10.
Why is Diazene Green solution added to the Glucose solution in anaerobic respiration experiment?
Answer:
Diazene Green solution is added to the Glucose solution in anaerobic respiration experiment to check the presence of oxygen in glucose solution.

Question 11.
Name the food material on which trypsin acts and name the end products.
Answer:
i) protein ii) end products – peptones.

Question 12.
“Respiration is the energy releasing process.” Write your opinion on this statement.
Answer:
The given statement is absolutely correct. We respire to use the oxygen to oxidise our food and release energy. This is similar like burning but a slower process. With the help of respiratory enzymes, energy released can be stored in the form of ATP for later use.

Question 13.
Identify the figure.

Answer:
Aerial roots in Mangrove plants.

Question 14.
Can we say that combustion and respiration are almost same actions? What evidences do you have for this?
Answer:

1. In both these processes sugar is converted to carbon dioxide and water.
2. Both these processes require oxygen.
3. Both combustion and respiration releases energy.

Question 15.
What is the role of epiglottis in respiration and swallowing food?
Answer:
The epiglottis is a flexible flap at the superior end of the pharynx in the throat. Epiglot¬tis acts as a lid over glottis and prevents food from entering into larynx. Air from pharynx enters the larynx while food enters into oesophagus.

Question 16.
What is the function of haemoglobin?
Answer:
During respiration haemoglobin carries oxygen to the cells and CO, from cells to lungs.

Question 17.
What is respiration?
Answer:
Respiration is the process by which food is broken down to release energy.

Question 18.
What does the word respiration mean in Latin?
Answer:
In Latin the word respiration means “to breathe”.

Question 19.
Who did comprehensive work on properties of gases, their exchange and respiration?
Answer:
Lavoisier and Priestly.

Question 20.
What was the gas liberated on heating powdered charcoal in a bell jar?
Answer:
It was fixed air. In those days carbon dioxide was known as fixed air.

Question 21.
What is oxygen debt?
Answer:
It is the inadequate supply of oxygen when we undertake strenuous exercise.

Question 22.
What is vitiated air?
Answer:
It is the term used then to show air from which the component needed for burning had been removed.

Question 23.
What is the total lung capacity of human being?
Answer:
The total lung capacity of human being is nearly 5800 ml.

Question 24.
Who was the renowned chemist who wrote a textbook of Human Physiology?
Answer:
John Daper was the renowned chemist who wrote a textbook of Human Physiology.

Question 25.
What happens when air passes through nasal cavities?
Answer:

1. Air is filtered in nasal cavity by mucus lining and the hairs growing from its sides, remove some of the tiny particles of dirt in the air.
2. The temperature of the air is brought close to that of the body.

Question 26.
What is the function of epiglottis?
Answer:
Epiglottis controls the movement of air and food towards their respective passages.

Question 27.
What is breathing?
Answer:

1. Breathing is the process of inhaling and exhaling.
2. The mechanism by which organisms obtain oxygen from the environment and release CO₂ is called breathing.

Question 28.
What are pleura?
Answer:
Pleura are the two membranes that protect lungs from injury.

Question 29.
What is the concentration of oxygen at a height of 13 km from the sea level?
Answer:
At a height of 13 km above sea level the concentration of oxygen is much lower about one-fifth as great as at sea level.

Question 30.
What is cellular respiration?
Answer:
Oxidation of glucose or fatty acids takes place in the cells releasing energy. Hence this process is known as cellular respiration.

Question 31.
Where does aerobic respiration occur in eukaryotic cells?
Answer:
Aerobic respiration occur in cytoplasm and mitochondria of eukaryotic cells.

Question 32.
What is Glycolysis?
Answer:
It is the first stage of respiration. In this breakdown of glucose molecule into two molecules of 3 carbon compound called pyruvic acid or pyruvate releasing energy.

Question 33.
What is the fate of pyruvate in the absence of oxygen in animals?
Answer:
In the absence of oxygen pyruvate will be converted to lactic acid and release small amount of energy in animals.

Question 34.
In which type of respiration pyruvate is converted into carbon dioxide and water?
Answer:
In aerobic respiration pyruvate is converted into carbon dioxide and water.

Question 35.
What is the main reason for feeling pain in muscles after strenuous exercise?
Answer:
Due to the anaerobic respiration in muscles large amounts of lactic acid is accumulated and this results in muscular pains.

Question 36.
What is fermentation?
Answer:
In the absence of oxygen, yeast cells convert pyruvic acid to ethanol. This process is called fermentation.

Question 37.
What is the method used to separate ethanol from the yeast glucose mixture in anaerobic respiration?
Answer:
The method used to separate ethanol from the yeast glucose mixture in anaerobic respiration is fractional distillation.

Question 38.
In which organisms does exchange of gases take place through diffusion?
Answer:
In Amoeba, hydra and planarians exchange of gases takes place through diffusion.

Question 39.
In tracheal respiratory system which carry air directly to the cells in the tissues?
Answer:
Trachioles, the fine branches of trachea carry air directly to the cells in the tissues.

Question 40.
What are the respiratory organs in fishes?
Answer:
Gills or bronchiae are the respiratory organs in fishes.

Question 41.
What is cutaneous respiration?
Answer:
If the respiration occurs through skin, it is known as cutaneous respiration, e.g : Leech, Earthworm and Frog.

Question 42.
What are the other areas on the plant body through which gaseous exchange take place?
Answer:
The areas on the plant body through which geseous exchange take place are the surface of roots, lenticels on the stem.

Question 43.
What is the full form of ATP? How is it formed?
Answer:
I) ATP stands for Adenosine triphosphate.
2) ATP is used to supply energy in the cells for the carrying all the metabolic processes.

Question 44.
What are the factors that control respiration?
Answer:
Oxygen and temperature are the two important factors that control the process of respiration.

Question 45.
What are the substances that are used for the production of energy in all living organisms?
Answer:
Glucose and fatty acids are used for the production of energy in all living organisms.

Question 46.
How many types of respiration are present? What are they?
Answer:
There are two types of respiration. They are :

1. Aerobic respiration and
2. Anaerobic respiration.

Question 47.
Where is energy stored in ATP?
Answer:
Energy is stored in the terminal phosphate bond in ATP which is having three phosphates attached to a molecule of Adenosine.

Question 48.
What are the power houses of the cell?
Answer:
Mitochondria are the power houses of the cell.

Question 49.
What is the main difference between respiration and combustion?
Answer:
In respiration several intermediates are produced and in combustion, there are no such intermediates are produced.

Question 50.
What is the equation that represents respiration?
Answer:
The equation that represents respiration is

Question 51.

.
What are the sites of cellular respiration?
Answer:
Mitochondria are the sites of cellular respiration.

Question 52.
What are cristae in mitochondria?
Answer:
The inner membrane of mitochondria is thrown into several folds called cristae.

Question 53.
What is the net gain of ATP molecules in Glycolysis?
Answer:

1. Four ATP molecules are produced when one molecule of glucose is converted to two molecules of pyruvate but two are consumed.
2. The remaining two ATP molecules are net gain in glycolysis.

Question 54.
How many ATP molecules are produced when one glucose molecule is completely oxidised?
Answer:
A net gain of 38 ATP molecules are formed from the total oxidation of one glucose molecule.

Question 55.
What are the three stages present in complete oxidation of glucose molecule?
Answer:
The three stages present in complete oxidation of glucose molecule are

1. Glycolysis
2. Kreb’s cycle and
3. Electron transport.

Question 56.
Why does oxidation of fatty acids give more energy?
Answer:
Oxidation of fatty acids give more energy due to the presence of more carbon atoms in them.

Question 57.
What are aquatic and terrestrial animals?
Answer:
Animals that live in water are called aquatic animals and that live on land are known as terrestrial animals.

Question 58.
Why is the rate of breathing in aquatic organisms much faster than terrestrial organisms?
Answer:

1. The amount of oxygen dissolved in water is low when compared to the amount of oxygen present in air.
2. Therefore the rate of breathing in aquatic animals is much faster than in terrestrial animals.

Question 59.
Which part of the roots is involved in the exchange of respiratory gases?
Answer:
The part of roots that are involved in the exchange of respiratory gases are root hairs.

Question 60.
What is the average breathing rate in an adult mem at rest?
Answer:
The average breathing rate in an adult man at rest is about 15 to 18 times per minute.

Question 61.
Why is the trachea prevented from collapsing?
Answer:
The walls of the trachea are supported by several ‘C’ shaped cartillagenous rings. They prevent the trachea from collapsing and closing.

Question 62.
Why deos the percentage of carbon dioxide increase in exhaled air?
Answer:
During oxidation of glucose carbon dioxide is produced as waste product. Hence the concentration of carbon dioxide increases in exhaled air.

Question 63.
How does breathing take place in mangrove plants?
Answer:
In mangrove plants breathing takes place through specialised structures called breath¬ing roots or pneumatophores.

Question 64.
How does respiration take place in plants where roots are present in wet places?
Answer:
The plants which have their roots in very wet places have much larger air spaces, connect the stems with the roots, making diffusion from upper parts.

Question 65.
Which form a continuous network all over the plant?
Answer:
The stomatal openings lead to a series of spaces between the cells inside the plant which form a continuous network all over the plant.

Question 66.
What are the reasons for the animals to develop different types of respiratory organs?
Answer:
Body size, availability of water, habitat in which they live and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.

Question 67.
Why do fishes die when taken out of water?
Answer:
Fishes do not have lungs to utilise oxygen for breathing. They have gills which can utilize only dissolved oxygen from water.

Question 68.
What would be the consequences of deficiency of haemoglobin in our bodies?
Answer:
Deficiency of haemoglobin in blood can affect the oxygen supplying capacity of blood to body cells. It can also lead to a disease called Anaemia.

Question 69.
What are the stages of respiration in man?
Answer:
Respiration in man occurs in two stages 1) Inhalation (or) Inspiration 2) Exhalation (or) Expiration.

Question 70.
Which part plays major role in respiration of man?
Answer:
Diaphragm plays a major role in respiration in man.

Question 71.
Which part plays major role in respiration of woman?
Answer:
In woman ribs play a major role in respiration.

Question 72.
How are lungs protected?
Answer:
Lungs are protected by two membranes called pleura. A fluid between these membranes protects the lungs from injury.

Question 73.
What is the composition of exhaled air?
Answer:
Exhaled air contains 16% of oxygen, 4% of carbon dioxide and 79% of nitrogen.

Question 74.
Why are red blood cells red in colour?
Answer:
Red blood cells are red in colour due to the presence of haemoglobin in their cytoplasm.

Question 75.
How is haemoglobin made up of?
Answer:
Haemoglobin is made up of a protein called globin, Iron (Hearn) and organic molecule called porphyrin.

10th Class Biology 2nd Lesson Respiration 2 Marks Important Questions and Answers

Question 1.
(a) Which gas turns lime water milky in this experiment?
Answer:
Carbondioxide (or) CO₂

(b) Which gas do you think might be present in less quantities in the air we breath out as compared to air around us?
Answer:
Oxygen (or) O₂

Question 2.
Balu said that, “Plants perform Photosynthesis during day time. They respire during night time”.
Do you agree with Bain? Why? Why not?
Answer:

1. I do not agree with Balu’s statement.
2. Photosynthesis depends on light for energy but respiration does not depend on light.
3. Hence, photosynthesis takes place during day time only whereas respiration takes place both day and night.

Question 3.
The sportsman who participated in 100 mtr race get more muscle pains. But the sportsman who participates in 5 km’s race get less muscle pains. What is the reason?
Answer:

1. Accumulation of lactic acid results in muscular pain.
2. During 100 m race a well trained athlete can hold his breath and afterwards he pants.
3. In this case, the muscles are using energy released during the anaerobic break down of glucose, lactic acid is produced.
4. The presence of lactic acid in the blood is the main cause of muscle fatigue. Whether it is 100 mtr race or 5 km race.
5. If the body is rested long enough the tiredness goes.

Question 4.
What happens if there is no epiglottis in human beings?
Answer:

1. Food may enters into the larynx.
2. Food may enters into the lungs leading to the death.
3. May not speak properly.
4. Entry of food and air may not be regulated properly.

Question 5.
Write two chemicals and two materials required to conduct the experiment “Heat and Carbon dioxide are evolved during anaerobic respiration”.
Materials required: Thermosflask, splitted corks, thermometer, wash bottle, glass tubes.
Chemicals required: Liquid paraffin, glucose solution, bicarbonate solution, Janus green B and Yeast cells.

Question 6.
Observe the below diagram.
A) To which biosystem is this picture related?
Answer:
Respiratory system.

B) Write the names of the parts of A, B.
Answer:
A – alveolus; B – blood capillary network

C) To which system are they linked with?
Answer:
Respiratory system; circulatory system.

D) Which process is happening here? What happens as a result of it?
Answer:
Gaseous exchange between alveolus of lungs and blood capillaries. Due to this the CO₂, present in blood capillaries enter alveolus and oxygen present in alveolus en¬ter blood capillaries.

Question 7.
A person reached a specific distance once on foot and once by running. In which situation his legs pain? Why?
Answer:

1. When a person runs to reach a specific distance gets pain in his legs.
2. This is due to the production of lactic acid in the muscles.
3. Due to the Anaerobic respiration glucose in muscles converts into lactic Acid.
4. Accumulation of lactic acid causes pain in leg muscles.

Question 8.
What is the advantage of the wet and warm passage of air from the nostrils to capillaries?
Answer:
When the air passes in nasal cavity and in the pharynx some changes take place.

1. The mucus layer and hair in the nasal cavity removes the dust particles in the air.
2. The temperature of the air brought to the body temperature.
3. Moistening the air.

Question 9.
In the experiment of anaerobic respiration with yeast
i) Why was liquid paraffin poured on glucose?
ii) What did you understood about anaerobic respiration?
Answer:
i) The supply of oxygen from the air can be stopped by pouring liquid paraffin on glucose.
ii) Anaerobic respiration takes place in the absence of oxygen. In this glucose molecule is incompletely oxidised. The end products of anaerobic respiration are ethyl alcohol or lactic acid and CO₂.
During anaerobic respiration small amount of energy is liberated (2ATP). Anaero¬bic respiration occurs in many anaerobic bacteria and human muscles cells. The anaero¬bic respiration can be represented as:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂+ 56 K.Cal.

Question 10.
See the below table. Write what you know from it.

Gas	% in inhaled air	% of exhaled air
Oxygen	21	16
Carbon dioxide	0.04	4
Nitrogen	79	79

Answer:

1. The inhaled air consists of 21% of oxygen whereas the exhaled air contains 16% of oxygen only. This is due to utlilisation of oxygen during cellular respiration in the body. Hence the difference occurs.
2. Inhaled air contains 0.04% of carbondioxide whereas exhale air contains 4% of carbondioxide.
   The concentration of CO₂ is increased a lot due to the release of CO₂ during cellular respiration in the body.
3. Both inhale and exhale air contains 79% of nitrogen because nitrogen has no role to play in cellular respiration.

Question 11.
What is the pathway of air from nostril to alveolus?
Answer:
Draw a flow chart of Respiratory passage of Humans.

Question 12.
What happens when a baker prepares a dough by mixing yeast in it?
Answer:

1. The yeast is commonly used for fermenting bread is saccharomyces cerevisiae.
2. Baker’s yeast has the advantage of producing uniform, quick, and reliable results because it is obtained from pure culture.
3. Water is mixed with flour, salt and the fermenting agent.
4. The mixed dough is then allowed to rise one or more times.
5. Then loaves are formed and the bread is baked in air oven.

Question 13.
How does respiration in amoeba and hydra occur through diffusion? (OR)
What are the similarities in respiration of amoeba and hydra?
Answer:

1. Amoeba and hydra are aquatic organisms.
2. Respiration in them occurs through diffusion.
3. As oxygen is used by these organisms in respiration, its concentration is reduced in cytoplasm. Hence oxygen diffuses into cytoplasm from surrounding water.
4. During respiration CO₂ is continuously produced, its concentration increases in the cytoplasm, hence it diffuses into surrounding water.

Question 14.
Write a short note on ATP. (OR) Expand ATP.
Answer:

1. From the break down of glucose the energy is released and stored up in a special compound known as ATP (Adenosine Triphosphate).
2. It is a small parcel of chemical energy. The energy currency of these cells is ATP an energy rich compound that is capable of supplying energy whenever needed within the cell.
3. Each ATP molecule gives 7200 calories of energy. This energy is stored in the form of phosphate bonds.
4. If the bond is broken, the stored energy is released.

Question 15.
How do Dolphin and Crocodile respire?
Answer:

1. The aquatic animals like dolphin and crocodile respire with the help of lungs.
2. They come out of the water for air.
3. These two animals were lived on land initially.
4. Later they lived in water and developed several adaptations to live in water.

Question 16.
Why are Mitochondria called “Power houses of cell”? (QR)
What is the energy producing organ in a cell? How does it produce energy?
Answer:

1. Cellular respiration in prokaryotic cells like that of bacteria occurs within the cytoplasm.
2. In eukaryotic cells cytoplasm and mitochondria are the sites of reaction.
3. The produced energy is stored in mitochandria in the form of ATP.
4. Hence, mitochondria are called “Power houses of cell”.

Question 17.
Write the rate of respiration in different age groups of human beings.
Answer:

1. Newborn child: 32 times per minute
2. Children of 5 years: 26 times per minute
3. Man of 25 years: 15 times per minute
4. Man of 50 years: 18 times per minute

10th Class Biology 2nd Lesson Respiration 4 Marks Important Questions and Answers

Question 1.
Write about respiration in mangroves that grow in marshy lands.
Answer:

1. Mangroves grown near the marshy places respire through aerial roots or respiratory roots.
2. The root hairs exchange the gases from their surface.
3. They obtain oxygen from the airspaces present between the soil particles.
4. The plants grown in marshy places are adapted to develop aerial roots above the soil surface which helps in gaseous exchange.

Question 2.
a) What is the aim of this experiment?
Answer:
Heat is liberated during respiration.

b) What change do you observe in thermometer readings?
Answer:
Reading increases in the thermometer.

c) In your opinion, where did this heat come from?
Answer:
The heat comes from the germinating seeds which respire and releasing heat.

d) What precaution should we take, while doing this experiment?
Answer:
The bulb of the thermometer should be dip in the germinating seeds (or) sprouts.

Question 3.
You have conducted this experiment in your classroom. Now answer the following questions.
a) What do you prove by conducting this experiment?
Answer:
To test the production of heat and carbon dioxide during anaerobic respiration.

b) Why do you heat glucose solution?
Answer:
To remove the dissolved oxygen in the glucose solution.

c) How do you confirm that glucose solution is free from oxygen after heating it?
Answer:
By adding diazine green (Janus green B) solution to glucose solution, it turns to pink.

d) What are the changes you notice in the lime water?
Answer:
Lime water turns milky white.

Question 4.

i) What change did you observe in the thermometer in the given experiment?
Answer:
Raise in the temperature

ii) Where does the heat come from?
Answer:
From the germinating seeds during respiration

iii) What result you will get, if you perform this experiment with dry seeds?
Answer:
No change of temperature in thermometre.

iv) What are the apparatus used in this experiment?
Answer:
Glass jar, germinating seeds, cork, thermometer.

Question 5.
Observe the set of apparatus and answer the following questions.
i) Which process do we know with the help of this experiment?
Answer:
Combustion.

ii) How does this process differ with respiration?
Answer:
Respiration occurs in the presence of water.
Combustion occurs in the absence of water.

iii) What are the similarities between this process and respiration?
Answer:
In both processes energy is released.

iv) Which gas turns lime – water milky?
Answer:
Carbon-di-oxide (CO₂)

Question 6.
Look at the following experiment. Answer the questions.

a) What is the aim of the experiment?
Answer:
The aim of the experiment is CO₂ is released during anaerobic respiration.

b) Which agent is used to find the presence of oxygen?
What changes do you observe when oxygen is present in Glucose solution?
Answer:
To find the presence of oxygen diazine green (Janus Green B) solution is used. The blue diazine green solution turns pink when oxygen is present in the glucose solution.

c) Why is liquid paraffin poured on glucose solution?
Answer:
By pouring liquid paraffin on glucose solution, the supply of oxygen from the air can be cut off.

d) Which gas is released during the experiment? How can you prove it?
Answer:
Carbon dioxide is released.
The released CO₂ passes into lime water it turns milky.

Question 7.
Observe the following diagram and answer the following questions.

1. What do we call the membranes that cover the lungs?
2. What is the functional unit of lungs ?
3. Which part produces the sound ?
4. What does ‘X’ denote ?

Answer:

1. Pleura
2. Alveoli
3. Larynx
4. Trachea

Question 8.
Observe the diagram and answer the following questions.

a) What does the given diagram indicate?
b) What is the part ‘X’ in the diagram?
c) What is the function of the given picture?
d) To which system the given picture belongs to?
Answer:
a) The given diagram indicates mitochondria.
b) Matrix
c) Performing cellular respiration and releasing energy in the form of ATP.
d) Respiratory system.

Question 9.
Observe the experimental setup and answer the given questions.
A) What is the aim of this experiment?
B) What are the apparatus required for this experiment?
C) What changes do you observe in thermometer during this WKm experiment?
D) What will happen, if dry seeds are taken instead of germinating seeds in this experiment?
Answer:
A) Heat is liberated during respiration.
B) Glass jar, Germinating seeds, Cork and Thermometer.
C) We can notice the raise in temperature after observing the thermometer readings.
D) There will be no change of temperature in the thermometer. We can’t prove the aim of the experiment.

Question 10.
Observe the below diagram and answer the following questions:
i) What does the above setting (diagram) indicate?
Answer:
The above setting (diagram) indicates to prove that carbon dioxide and heat are liberated during anaerobic respiration by yeast cells.

ii) Why is boiled and cooled glucose covered with paraffin?
Answer:
To prevent supply of air, boiled and cooled glucose is covered with paraffin.

iii) What is the use of adding diazine green to glucose solution? What change you notice in glucose solution?
Answer:
Diazine green is added to glucose solution to know whether oxygen is present or not in glucose solution. When the availability of oxygen is less the diazine green changes to pink colour.

iv) Why is lime water used in this experiment?
Answer:
To know whether carbon dioxide is released or not in this experiment lime water is used. Carbon dioxide changes lime water to milky white.

v) Why is bulb of thermometer dipped in the glucose water?
Answer:
To know the rise in temperature of glucose solution when heated, the bulb of thermometer is dipped in the glucose water.

Question 11.
Explain with the help of a flow chart, the path way of air in humans.
Answer:

Question 12.
Study the graph and answer the following questions :
Graph showing effects of vigorous excercise on the concentration of lactic acid in blood.
i) What was the concentration of lactic acid in blood to start with?
ii) What was the greatest concentration of lactic acid reached during the experiment?
iii) What is the concentration of lactic acid after 25 minutes of exercise?
iv) What is the relationship between lactic acid and muscle pain?
Answer:
i) 20 mg/cm³
ii) 20 minutes (Or) at “B” point,
iii) 101 mg/cm³
iv) If concentration of lactic acid increases, muscle pains also increases.

Question 13.
Observe the following :

Write the answers to the following questions:
i) How many Pyruvic acid molecules form from one Glucose?
Answer:
2 Pyruvic acid molecules.

ii) What conditions influence Pyruvic acid to participate in Aerobic and Anaerobic respiration?
Answer:
Presence of oxygen

iii) In which we get more energy in both Aerobic and Anaerobic respirations?
Answer:
Aerobic respiration

iv) The chemical that is formed in human muscles during Anaerobic respiration.
Answer:
Lactic acid

Question 14.
Why does the exchange of gases happen only in alveoli, though arteries are present in pharynx, trachea and bronchus?
Answer:

1. Alveoli are tiny air sacs in the lungs surrounded by capillaries
2. They are numerous and only single cell thickness
3. They increase the efficiency of gas exchange.
4. Due to the difference in a gradient of O₂ oxygen diffuse from alveoli to blood capillaries.

Question 15.
What are the events or steps in respiration?
Answer:
The following are the events or steps in respiration.

1. Breathing: Air moves into lungs and out of lungs.
2. Gaseous exchange in lungs: Exchange of gases between alveoli and blood.
3. Gas transport by blood: Transport of oxygen from blood capillaries of alveoli to body cells and return of carbon dioxide.
4. Gaseous exchange in cells: Exchanging oxygen from blood into the cells and carbon dioxide from cells into the blood.
5. Cellular respiration: Using oxygen in cell processes to produce carbon dioxide and water, releasing energy to be used for life processes.

Question 16.
What will happen if the respiratory tract is not moist? (OR)
Why respiratory tract should be moist?
Answer:

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

Question 17.
Explain the process of transportation of gases through the blood.
Answer:

1. The relative amount of gases and their combining capacity with haemoglobin and other substances in blood determine their transport via blood in the body.
2. When oxygen present in the air is within normal limits (around 21%) then almost all of it is carried in the blood by binding to haemoglobin, a protein present in the red blood cells.
3. As oxygen is diffused in the blood, it rapidly combines with the haemoglobin to form oxyhaemoglobin.
4. Not only can haemoglobin combine with oxygen, but it can easily broken into haemoglobin and oxygen.
5. Carbon dioxide is usually transported as bicarbonate, while some amount of it combines with haemoglobin and rest is dissolved in blood plasma.
   

Question 18.
Why is human life impossible at higher altitudes without a supplementary supply of oxygen? (OR)
The concentration of oxygen in air decreases as we go up from sea level. Explain briefly.
Answer:

1. If haemoglobin is exposed to air at sea level, every molecule in air combines with oxygen to form oxyhaemoglobin.
2. At a height of 13 km above sea level, the concentration of oxygen is much lower about l/5th of a sea level.
3. Under these conditions about half as many molecules of oxygen combine with haemoglobin to form oxyhaemoglobin.
4. Blood cannot carry enough oxygen to the tissues.
5. Hence human life is impossible at such a high altitude without a supplementary supply of oxygen.
6. Provision for such a supply is built into modern aircraft which have pressurized cabins that maintain an enriched air supply.

Question 19.
What are the different ways in which glucose is oxidised to provide energy in various organisms? Give one example of each.
How does oxidation of glucose occur in various organisms?
Answer:

1. Glucose is the most commonly used sugar for deriving energy in plants, animals and in microorganisms.
2. In all these organisms glucose is oxidized in two stages.
3. The first stage is known as Glycolysis. It occurs in cytoplasm.
4. During glycolysis glucose is converted to two molecules of pyruvic acid.
5. In the second stage if oxygen is available pyruvic acid is converted to C02 and water, large amount of energy is released. This is known as aerobic respiration. It occurs in most of the plant and animal cells.
6. If oxygen is inadequate or not available, pyruvic acid is converted into ethanol and carbon dioxide. This is anaerobic respiration taking place in yeast cells that is called fermentation.
7. If oxygen is not available in muscle cells, the pyruvic acid is converted into lactic acid.
   

Question 20.
Write the adaptations seen in plants living in water logged conditions.
(OR)
What are the adaptations seen in magrove plants?
Answer:

1. Most plants can aerate their roots by taking in the oxygen through lenticels or through the surface of their root hairs.
2. But plants which have their roots in very wet places, are unable to do this.
3. They are adapted to these water logged conditions by having much larger air spaces which connect the stems with the roots, making diffusion from the upper parts much more efficiently.
4. The problem of air transportion is more difficult for trees and may not survive with their roots permanently in water.
5. To overcome this problem the mangrove tree of the tropics which raise up aerial roots above the surface and takes in oxygen.

Question 21.
Describe the mechanism of branchial or gill respiration in fishes.
(OR)
Briefly explain the process of exchange of gases in fishes during respiration.
Answer:

1. Some aquatic animals like fishes have developed special organs for respiration which are known as gills or branchiae.
2. Blood is supplied to gills through capillaries which have thin walls where gases are exchanged. Gills are present in the gill pouches or branchial pouches.
   
3. Gills are provided with leaf-like folds called gill lamellae.
4. Fish keeps its mouth open and lowers the floor of the oral cavity. As a result water from outside will be drawn into the oral cavity.
5. Now the mouth is closed and the floor of the oral cavity is raised.
6. Water is pushed into the pharynx and is forced to gill pouches through internal branchia apertures.
7. When water passes through gill lamellae exchange of gases takes place, that is oxygen diffuses from water to blood and CO₂ from blood into water.
8. Then water flows through external branchia aperture.

Question 22.
Explain briefly about Pranayama- the art of breathing. (OR)
How can the capacity of lungs be improved by yoga?
Answer:

1. To improve breathing capacity the saint Patanjali developed Yogabyasa.
2. The art of breathing in Yogabyasa is called Pranayama. Prana means gas, ayama means journey.
3. In Pranayama practice air is allowed to enter three lobes of lungs in order to in¬crease the amount of oxygen to diffuse into blood.
4. More amount of oxygen available to brain and tissues the body will be more active.
5. It is very important to practise Pranayama regularly to make our life healthy and active.
6. All people irrespective of age and sex should practise Pranayama under the guidance of well trained Yoga Teacher to improve the working capacity of lungs.

Question 23.
What are the experiments carried out by Lavoisier to understand the property of gases?
Answer:

1. In his early experiments Lavoisier thought that the gas liberated on heating powdered charcoal in a bell jar kept over water in a trough was like fixed air i.e., carbon dioxide.
2. The next series of experiments deals with the combustion of phosphorous in a bell jar. From this he showed that whatever it was in the atmospheric air which combined with the phosphorous was not water vapour.
3. This was respirable air, a component of air that also helped in burning.
4. The air that we breathe out precipitated lime water while that after heating metal did not.
5. From this, he concluded that there were two processes involved in respiration.
6. Lavoisier carried out another experiment by which he showed that about one sixth of the volume of ‘vitiated air’ consists of chalky acid gas (fixed air).
7. Either eminently respirable air is changed in the lungs to chalky acid air; or an exchange takes place, the eminently respirable air being absorbed, and an almost equal volume of chalky acid air being given up to the air from the lungs.
8. Lavoisier had to admit that there were strong grounds for believing that eminently respirable air did combine with the blood to produce the red colour.

Question 24.
Explain the evolutionary changes in energy-releasing system.
(OR)
What are the different respiratory systems in animal groups?
Answer:
Exchange of gases is a common life process in all living organisms, but it is not same in all.

1. Diffusion:
   1. Single-celled organisms like amoeba or multicellular organisms like hydra and planarians obtain oxygen and expel carbon dioxide directly from the body by the process of diffusion.
   2. In multicellular animals special organs are evolved.
   3. Body size, availability of water and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.
2. Tracheal respiratory system : In insects tracheal respiratory system is present in which small branches of trachea called trachioles carry air directly to the cells in the tissues.
3. Bronchial respiration : In fishes gills are utilised for the exchanges of gases. Blood is supplied to gills through capillaries which have thin walls for exchange of gases. This is called bronchial respiration.
4. Cutaneous respiration: 0 Respiration through skin is called cutaneous respiration.
   Eg: i) Earth worms and leeches.
   ii) Frog, an amphibian can respire through lungs and skin.
5. Pulmonary respiration : Most of the higher animals respire with the help of lungs. This type of respiration is known as pulmonary respiration. Eg: Mammals.

Question 25.
Describe the structure of mitochondria with the help of a diagram. (OR)
Which cell organelle is called energy currency or power house of cell? What do you know about its construction?
Answer:
Mitochondria is known as energy currency or power house of cell.
Structure of mitochondria:

1. Mitochondria are sac-like structures present in the cytoplasm of the cells.
2. Mitochondria have two compartments-an inner compartment and an outer compartment. The substance in the inner compartment is called matrix.
3. The matrix is surrounded by a membrane called inner membrane of mitochondria.
4. The inner membrane is thrown into several folds called cristae. The cristae extended into the matrix.
5. The space between the folds is continuous with the outer compartment.
6. On the inner membrane, projecting into the matrix are a large number of particles called elementary particles.
7. These particles have a spherical head and a stalk. They are attached to the inner membrane by their stalk and the head portion of the particle is in the matrix.
8. The outer compartment is surrounded by another membrane – the outer membrane. The outer membrane is smooth and has no projections.
9. The inner membrane, the matrix and the elementary particles in the mitochondria have large number of enzymes and other required proteins for the respiration and energy production.

Question 26.
Draw and label mitochondria. Why should we call it cell of power ?
Answer:

Oxidation of glucose molecule occurs in the mitochondria, ot cell. This is known as cellular respiration. The energy produced during cellular respiration stored in the form of ATP molecule. Energy producing cellular respiration occurs in mitochondria hence we call it cell of power or power house of the ceil.

Question 27.
Describe how oxygen enters the blood in lungs with the help of block diagram.
(OR)
How does gaseous exchange occur in lungs?
Answer:

1. Gaseous exchange takes place within the lungs by diffusion from the alveoli to blood capillaries and vice versa. Alveoli in lungs are numerous and only one cell thick.
   
2. Alveoli are surrounded by capillaries that are also one cell thick.
3. Blood, dark red in colour flows from the heart through these capillaries and collects oxygen from the alveoli.
   At the same time, carbon dioxide passes out of the capillaries and into the alveoli.
4. When we breathe out, we get rid of carbon dioxide.
5. The bright red, oxygen rich blood is returned to the heart and pumped out to all parts of the body.

Question 28.
What is the role of diaphragm and ribs in respiration? Are both active in man and woman?
Answer:
Diaphragm:

1. Diaphragm is a muscular dome shaped tissue present at the floor of the chest cavity separating abdomen from respiratory system.
2. Diaphragm expands downwards into the abdomen thus increasing chest cavity. This allows the lungs to expand as we inhale.
3. As the diaphragm contracts upwards thus decreasing the chest cavity, it allows the air to expel from the lungs.
   Ribs:
4. The ribs protect the lungs and expand as we inhale to facilitate space for the lungs to expand. The ribs then contract expelling the air from the lungs.
5. The intercostal muscles present between the ribs help in contraction and relaxation of ribs.
6. In man, diaphragm plays a major role in the respiration, while in woman, the ribs play a major role.

Question 29.
Why are alveoli so small and uncountable in number? (OR)
How do alveoli increase the area for exchange of gases?
Answer:

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

Question 30.
Write a brief note on respiration in plants. (OR)
Does respiration occur in plants? Explain briefly about it.
Answer:

1. In most plants exchange of gases takes place through stomata.
2. There are other areas on the plant body like surface of roots, lenticels on stem, etc. the gaseous exchange takes place.
   
3. Some plants have specialized structures like breathing roots of mangrove plants as well as the tissue in orchids.
4. Breathing roots and tissue in orchids help plants to take oxygen to produce energy and release carbon dioxide.
5. Inside the plants openings lead to a series of spaces between the cells which form a continuous network all over the plant.
6. The whole system works by diffusion.
7. As the oxygen is used up by the cells a gradient develops between the cells and the air in the spaces.
8. So oxygen passes in between the air spaces and the air outside stomata and lenticels.
9. In the same way, as more carbon dioxide is given out by the cells, a gradient occurs in the reverse direction and it passes out.

Question 31.
Write a brief note on tracheal respiration in insects.
Answer:

1. In insects blood do not contain haemoglobin, and blood is white in colour. Hence it cannot carry oxygen.
2. For respiration insects adopt a special system called tracheal system.
   
3. This system consists of a series of tubes called trachea.
4. These trachea open out through small apertures called spiracles on either side of the body.
5. All tracheal tubes of each side join and form a longitudinal tracheal trunk.
6. Trachea divide into a number of branches called tracheoles which carry air directly to the tissues.
7. As the air moves in and out of the trachea, oxygen present in the air diffuses into the cells and CO₂ diffuses into the air from the cells.

Question 32.
Write about the mechanism of respiration in human beings. (OR)
How does exchange of gases take place in human beings?
Answer:

1. Respiration in man occurs in two stages. They are inspiration and expiration.
2. During inspiration air from outside enters into the lungs by increasing the chest cavity.
   
3. Increase in the chest cavity is made by pulling the diaphragm down and pushing the ribs forward.
4. As the air pressure in the lungs is reduced, air from outside enters the lungs through external nostrils, nasal cavities, internal nares, pharynx, epiglottis, larynx, trachea, bronchi and branchioles and finally reach the alveoli where exchange of gases takes place.
5. During expiration the diaphragm and ribs come back to original positions.
   
6. This reduces the volume of chest cavity.
7. So the volume of lungs is decreased and air under pressure comes out of the lungs.

Question 33.
Study the graph given below and analyse the reasons for accumulation of lactic acid in blood after strenuous exercise.

Answer:

1. This graph shows the relation between time accumulation of lactic acid in the muscles.
2. At the beginning, the amount of lactic acid in the blood is very less.
3. Gradually it is increased by vigorous exercise.
4. Within 15 minutes it goes to maximum level which causes muscle pain.
5. Then the lactic acid is removed from muscles in an hour.
6. Muscles produce energy by anaerobic respiration.
   C₆H₁₂O₆ → lactic acid + CO₂ + energy
7. In the vigorous exercise, muscle work rapidly and produce more lactic acid.
8. That’s why lactic acid concentration is increased in muscle after strenuous exercise.

Question 34.
Observe the above graph of lactic acid accumulation in the muscles of an athlete and answer the following questions.
a) What was the concentration of lactic acid in the blood to start with?
Answer:
It is 20 mg/km³.

b) What was the greatest concentration reached during the experiment?
Answer:
101 mg/cm³.

c) If the trend between points C and D were to continue at the same rate, how long might it take for the original lactic acid level to be reached once again?
Answer:
55 minutes.

d) What does high level of lactic acid indicate about the condition of respiration?
Answer:
It indicates the accumulation of lactic acid in muscles through anaerobic respiration. The presence of lactic acid in the blood is the main cause of muscular pain and fatigue.

Question 35.
Describe the structure of human lungs with the help of a diagram.

Answer:

1. A pair of lungs is present in the chest cavity one on either side of the heart.
2. Lungs are spongy and elastic. They are enclosed by two membranes called pleura.
3. Space between the two membranes of pleura is filled with fluid. Pleura protects the lungs from injury.
4. Right lung is larger than the left lung.
5. Right lung is made of three lobes while the left lung has only two lobes.
6. Lung has several thousands of alveoli which are supplied with blood capillaries.
7. Pulmonary artery brings deoxygenated blood from heart to lungs.
8. After entering the lung, this artery divides into several arterioles and capillaries and supplies deoxygenated blood to alveoli.
9. Gas exchange occurs in the alveoli.
10. Oxygenated blood is carried from the lung to heart by the pulmonary vein.

Project work
Question 1.
Observe and analyse the questions in the table given below.

	Newly borned	(Children)	(Children)	Children	Youth/Adults	Athletics
	(0-3 months)	(3-6 months)	(6-12 months)	(1-10 years)
Heart beat	100 -150	90-120	80 -120	70-130	60-100	40-60

A) In which age group rate of heart beat is more?
B) In which age group rate of heart beat is less?
C) Why heart beat in Athletics is less?
D) What are reasons for more rate of heart beats differences between the newly born and children?
Answer:
A) In newly borned babies which are in 0 – 3 months of age group rate of heart beat is more i.e., 100 to 150 times.
B) In athletics the rate of heart beat is less i.e., 40 – 60 times / minute.
C) The heart of athlete pump more blood per beat due to increased cardio-vascular fitness in the structure of the heart. The muscles in the heart wall thicken and the heart pumps more blood with each beat.
D)

1. Mothers who have special medical conditions such as thyroid diseases or diabetes may give birth to new borns who are temporarily tachscardic from altered hormone and glucose levels. Tachycardia is a medical term for a very rapid heart beat.
2. Some infants are born with accessory electrical tissue in the heart causes epi¬sodes of rapid heart rate.
3. In wolf – parkinson syndrome – white syndrome there are extra cells and an ac-cessory path way, causing additional heart beats.

Question 2.
Observe the table given below and analyse the questions.

Name of the animal	Weight of the body	Weight of the heart	No. of beats/min
Blue whale	1,30,000 kg	750 kg	7
Elephant	3000 kg	12-21 kg	46
Man	60 – 70 kg	300 gm	76
Coaltit (Bird)	8 gm	0.15 gm	1200

A) Why heart beat is less in animals with more weight?
B) Why heart beat is more in animals with less weight?
C) What is the relationship between weight of the body and rate of heart beat?
D) Why the weight of heart is less than body weight?
Answer:
A) The animals with more weight usually have weighted hearts. In one heart beat the large-sized hearts sends high amounts of blood to circulatory system. It takes time for the fulfilment of heart. Hence heart beat is less in animals with more body weight.
B) Usually the heart is very small in less weight animals. When the animal shrinks or contracts , its heart actually decrease the volume of blood proportionately. It can compensate for the reduced volume by increasing the rate at which it can supply blood to all body parts.
C) As the weight of the body of the animal increases the rate of heart beat per minute decreases. And also as the weight of the body decrease the rate of heart beat increases.
D) Usually the body of an organism is made by number of organs which makes the body functional. As all the body parts constitute the whole organism, the heart one of the organ is usually has less weight than body weight of an animal.

These AP 8th Class Biology Important Questions 1st Lesson What is Science will help students prepare well for the exams.

AP State Syllabus 8th Class Biology 1st Lesson Important Questions and Answers What is Science

8th Class Biology 1st Lesson What is Science Textbook Questions and Answers

Improve Your Learning

Question 1.
What is Science?
Answer:
Science:

1. Science is the concerted human effort to understand the history of the natural world how the natural world works, with observable physical evidence.
2. Science is an organized study of knowledge which is based on experimentation.
3. Science is a tool for searching truths of nature.
4. Science is the way of exploring the world.

Question 2.
What are scientific methods ? Write the steps involved in a scientist’s work.
Answer:
Scientific Methods:

1. Scientists solve a problem or answer a question by using organized ways. They are called “Scientific Methods”.
2. The following are the steps involved in a scientific work.
   a) Step -1 : Ask questions.
   b) Step – 2 : Form hypothesis.
   c) Step – 3 : Plan experiments.
   d) Step – 4 : Conduct experiments.
   e) Step – 5 : Draw conclusions.

Question 3.
What is the way to find out solutions for the problems in scientific way?
Answer:
To find out solutions for the problems in scientific way we need to follow a sequential order. So we go through the following.

1. Identifying problem : Let us identify any problems from your surroundings.
   Ex: The bulb did not lit in the room.
2. Making hypothesis: List out different solutions which your think for the identifying problem.
   Ex: De filament, fuse failure, switch problem, wire problem.
3. Collecting information: To solve the identifying problem collect material, apparatus, Information, persons.
   Ex: Collect material like tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
4. Data analysis: Arrange the collected data or information to conduct experiment.
5. Experimentation: To prove selecting hypothesis conduct experiment.
   Ex: Observe filament of the bulb.
6. Result analysis: Analyzing the results to find out the solution for the problem based on the results you need to select another hypothesis to prove.
   Ex: Filament of the bulb is good in condition so we need to observe fuse.
7. Generalisation: Based on the experiment and its results explain the solution for the problem.
   Ex: Fuse is damaged so the bulb not glow, so we need to replace the fuse.

Question 4.
During investigation in science, name some rules to be followed for the safety.
Answer:

1. Think ahead: Study the steps of the investigation, so you know what to expect.
2. Be neat: Keep your work area clean.
3. Oops!: If you should spill or break something or get cut, tell your teacher right away.
4. Watch your eyes: Wear safety goggles anytime you are directed to do so.
5. Yuck!: Never eat or drink anything during a scientific activity.
6. Protect yourself from shocks: Be especially careful if an electric appliance is used. Be sure that electric cords are in a safe place where you can’t trip over them. Don’t ever pull a plug out an outlet by pulling on the cord.
7. Keep it clean: Always clean up when you have finished.

Question 5.
If you plan a test to find out how much water different brands of paper towels absorb, write down the steps of the experiment.
Answer:

1. Pour 1 liter of water into each of three beakers.
2. Put a towel from each of the three brands into a different beaker for 10 seconds.
3. Pull the towel out of the water, and let it drain back into the beaker for 5 seconds.
4. Measure the amount of water left in each beaker.
5. The towel put in the beaker in which less amount of water is left absorbs more water.

Question 6.
Collect information about scientists and their works and prepare a chart and paste it in your classroom.
Answer:

Question 7.
What are process skills ?
Answer:
Observe, compare, classify are the process skills.

1. Observe – Use the senses to learn about objects and events.
2. Compare – Identify characteristics of things or events to find out how they are alike and different.
3. Classify – Group or organize objects or events in categories based on specific characteristics.

Question 8.
Read the following.
Answer:
Endemic Species:

You may find that these animals are specifically found in certain regions of the world.
You are also aware of the fact that many plants and animals are widely distributed throughout the world. But some species of plants and animals are found restricted to some areas only. Plants or animal species found restricted to a particular area of a country are called Endemic Species.
Now answer the following questions.
a) Name an Endemic Species of our State.
Answer:
Tiger, Peacock and Crane.

b) You may notice that Kangaroo is endemic to Australia and Kiwi to New Zealand. Can you tell which among the above picures represent an endemic species of India ?
Answer:
Peacock and Tiger.

c) Name some other endemic species of India.
Answer:
Indian Lion, Indian Wolf, Great Indian bustard bill.

Question 9.
Reading the following poster.

Answer:
Biotic Components:
Producers – Mangrove, spirogyra, euglena, oscillatoria, blue green algae, ulothrix, etc.
Consumers – Shrimp, crab, hydra, protozoans, mussel, snails, turtle, daphnia, brittle word, tube worm, etc.
Decomposers – Detritus feeding bacteria, etc.
Abiotic components – Salt and fresh water, air, sunlight, soil, etc.

Question 10.
What is Generalisation? Give an example.
Answer:
Based on the experiment and its results explaining the solution for the problem is called Generalisation.
Example:

a) The bulb did not light in the room.
b) Identify the problem that may be defilament, fuse failure, switch problem, wire problem.
c) Then take tester, screwdriver, wooden scale, wires, insulation tape, table and blade.
d) Observe the filament of the bulb.
e) If the filament of the bulb is good, then observe fuse.
f) As the fuse is damaged, we need to replace the fuse.

Based on this, the bulb did not lit in the room because the fuse is damaged. This is the generalisation in the above experiment.

Question 11.
What are the different types of writings used by scientists to describe what they are doing or learning?
Answer:

1. Informative writing, Narrative writing, Expressive writing, Persuasive writing are used by the Scientists.
2. In informative writing, scientists describe the observation, inferences and their conclusion.
3. In narrative writing the scientists describe about something, give examples or tell a story.
4. In expressive writing, they may write letters, poems, or songs.
5. In persuasive writing they write letters about important issues in science and also write about what they have learned about science which helps others to understand about their thinking.

Question 12.
How do scientists use numbers in their investigations?
Answer:

1. Measuring, interpreting data, using number sense are few methods used by scientists in their investigations.
2. Scientists make accurate measurements by using different measuring instruments like thermometer clocks, timers, rules, a spring scale, beakers, balance and other containers to measure liquids.
3. Scientists collect, organize, display and interpret data by using tables, charts and graphs.
4. Scientists compare and order numbers, compute with numbers shown on graphs and read the scales on thermometers, measuring cups, beakers and other tools.
5. Good scientists apply their maths skills to help them display and interpret the data they collect.

Question 13.
What is Hypothesis? What are variables?
Answer:

1. Making a statement about an expected outcome is called Hypothesis.
2. Variables are factors that can affect the outcome of the investigation.

Question 14.
What do the following persons do?
a) A geologist
b) A chemist
c) A biologist
d) An ecologist
e) A climatologist
Answer:
a) A geologist examines the distribution of fossils and makes observations to find patterns in natural phenomena.
b) A chemist observes the rate of a chemical reaction at a variety of temperatures.
c) A biologist observes the reaction of a particular tissue to various stimulants.
d) An ecologist observes the territorial behaviours of different animals and birds.
e) A climatologist collects data from weather balloons and makes observations basing on it.

These AP 10th Class Physical Science Chapter Wise Important Questions 8th Lesson Structure of Atom will help students prepare well for the exams.

AP State Syllabus 10th Class Chemistry 8th Lesson Important Questions and Answers Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom 1 Mark Important Questions and Answers

Question 1.
Write the electronic configuration of chromium. (AP June 2016)
Answrr:
The electronic configuration of chromium is
1s²2s²2p⁶3s²3p⁶4s¹3d⁵ or [Ar] 4s¹3d⁵

Question 2.
Out of 3d and 4s, which has more (n + l) value? Explain. (AP June 2017)
Answer:
1) 3d ⇒ n + l ⇒ 3 + 2 ⇒ 5 (energy)
4s ⇒ n + l ⇒ 4 + 0 ⇒ 4 (energy)

2) Hence, ‘3d’ has more (n + l) value than ‘4s’.

Question 3.
Prepare a question on nl^x method. (AP SA-I:2018-19)
Answer:

• How is nl^x method useful.
• Explain the nl^x method with an example.

Question 4.
Which colours do you observe when an iron rod is gradually heated to higher tem-peratures? (TS June 2015)
Answer:
First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).

Question 5.
Which principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p⁴? Give reasons. (TS June 2015)
Answer:
1) Principle :
Aufbau principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p⁴.

2) Reasons :
i) According to Aufbau principle electron enters into orbital of lowest energy.
ii) Between 2s and 2p, 2s has least energy. So 2s must be filled before the electron has to enter 2p.

Question 6.
Write the symbol of the outermost shell of magnesium (Z = 12) atom. How many electrons are present in the outermost shell of magnesium? (TS June 2017)
Answer:
Symbol of the outermost shell of magnesium (3rd shell) = M
No. of electrons in outermost shell of Magnesium = 2.

Question 7.
The four quantum number values of the 21st electrons of scandium (Sc) are given in the following table. (TS March 2017)

Write the values of the four quantum numbers for the 20th electron of scandium (Sc) in the form of the table.
Answer:

Question 8.
If n = 3, mention the orbitals present in the shell and write maximum number of electrons in the shell. (TS March 2018)
(OR)
Write the maximum number of electrons and number of orbitals in the shell, when n = 3.
Answer:

1. When n = 3, number of subshells = 3 (3s₁ 3p₁ 3d)
2. Number of orbitals = 9 (3s(1) 3p(3) 3d(5))
3. Maximum number of electrons (3s² 3p⁶ 3d¹⁰) = 18

Question 9.
What is dispersion?
Answer:
The splitting of light into different colours is called dispersion.

Question 10.
What is an electromagnetic wave?
Answer:
When electric field and magnetic fields are perpendicular to each other and at right angles to the direction of propagation of wave is formed. Such a wave is called electromagnetic wave.

Question 11.
What is a Zeeman effect?
Answer:
The splitting of spectral lines in the presence of magnetic field is called Zeeman effect.

Question 12.
What is a spectrum?
Answer:
Group of wavelengths is called spectrum (or) A collection of dispersed light giving its wavelength composition is called a spectrum.

Question 13.
What is speed of electromagnetic wave?
Answer:
It is equal to speed of light, i.e. 3 x 10⁸ ms^-1

Question 14.
Which colour has highest wavelength and which colour has least wavelength on visible spectrum?
Answer:
The colour that has highest wavelength in visible spectrum is red and least wavelength is violet.

Question 15.
If n = 5, then what is the maximum value for l?
Answer:
The maximum value for l is 4.

Question 16.
If l = 4, what is the number erf values for m_l?
Answer:
m_l = 2l + 1 = 2(4) +1=9.

Question 17.
What are the values of m_s?
Answer:
½ or – ½

Question 18.
What is electronic configuration?
Answer:
Distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.

Question 19.
What is Heisenberg’s principle of uncertainty?
Answer:
It is not possible to find the exact position and velocity of electron simultaneously.

Question 20.
Give ascending order of various atomic orbitals according to Moeller diagram.
Answer:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4d < 5d < 6p < 7s < 5f < 6d < 7p < 8s.

Question 21.
What is Hund’s Rule?
Answer:
Electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

Question 22.
What is Planck’s equation?
Answer:
Planck’s equation is E = hv.
E = Energy of the radiation
h = Planck’s constant = 6.625 × 10^-34 J
v = Frequency of radiation.

Question 23.
What is electromagnetic spectrum?
Answer:
Electromagnetic waves can have a wide variety of wavelengths. The entire range of wavelengths is known as the electromagnetic spectrum.

Question 24.
What is wavelength?
Answer:
The distance from one wave peak to the next is called wavelength (λ).

Question 25.
What is frequency?
Answer:
The number of wave peaks that pass by a given point per unit time is called frequency.

Question 26.
When are electromagnetic waves produced?
AnElectromagnetic waves are produced when an electric charge vibrates.

Question 27.
Which is the example for line spectrum?
Answer:
The atomic spectrum of hydrogen atom.

Question 28.
Which model explains fine spectrum of atom?
Answer:
Bohr – Sommerfeld model.

Question 29.
How are wavelength and velocity of light related?
Answer:
c = vλ
where
c = velocity of light,
v = frequency of light,
λ = wavelength of light.

Question 30.
Give the equation which gives electromagnetic energy (light) that can have only Certain discrete energy values.
Answer:
E = hv
E = Energy of light
h = Planck’s constant = 6.62 5 × 10^-27 erg sec or 6.625 × 10^-34 Joule-sec
v = Frequency of radiation
This equation is called Planck’s equation.

Question 31.
Which group elements are called Noble gases?
Answer:
VIII A group or 18th group elements are called inert gases (or) Noble gases.

Question 32.
Which elements are highly stable?
Answer:
Noble gases are highly stable.

Question 33.
Write the set of quantum numbers for the electrons in a 3p_z orbital.
Answer:

Question 34.
What is the difference between an orbit and orbital?
Answer:
An orbit is a well defined path of electron that revolves around the nucleus.

An orbital is the space around the nucleus, where the probability of finding electrons is maximum.

Question 35.
What are the factors which influence electromagnetic energy?
Answer:
Electromagnetic energy depends on two factors

1. wavelength
2. frequency.

Question 36.
What is a wave?
Answer:
The disturbance occurred in a medium is called wave.

Question 37.
When cupric chloride is kept in non-luminous flame then what is the colour of flame?
Answer:
Green colour.

Question 38.
If the colours gradually changes there are no sharp boundaries in between them, then what is the name given to that type of spectrum?
Answer:
Continuous spectrum of emission.

Question 39.
What is the information given by magnetic orbital quantum numbers?
Answer:
Orientation of orbitals in space.

Question 40.
How many orbitals are present in a sub-shell?
Answer:
The number of orbitals are present in a sub-shell is n² (where n is principal quantum number).

Question 41.
What happens when an object is suitably excited by heating?
Answer:
Light is emitted by the object.

Question 42.
What is meant by Aufbau?
Answer:
The German word Aufbau means building up.

Question 43.
Which elements are examples for Noble gases?
Answer:
Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn) are examples for Noble gases.

Question 44.
Which element has duplet configuration?
Answer:
Helium (1s²)

Question 45.
Write the set of quantum number for the added electron of oxygen atom.
Answer:
Configuration of oxygen is 1s² 2s² 2p⁴.
The added electron is 4th in the 2p.
The set of quantum numbers (2, 1, -1, -½)

Question 46.
Can we apply c = vλ, to sound waves?
Answer:
Yes. It is a universal relationship and applies to all waves.

Question 47.
What is the value of Planck’s constant?
Answer:
The value of Planck’s constant is 6.626 × 10^-34 Js.

Question 48.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:’
The light emitted by different kind of atoms is different because the excited states electrons will go are different.

Question 49.
How many values can ‘l’ have for n = 4?
Answer:
If n = 4, l can take values 0, 1, 2, 3. So there are four values.

Question 50.
Write the four quantum numbers for the differentiating electrons of lithium (Li) atom.
Answer:
The electronic configuration of lithium is 1s² 2s¹. So differentiating electron enters into 2s. The values of four quantum numbers are as given below.

Question 51.
Write four quantum numbers for 2p¹ electrons.
Answer:
The four quantum numbers for 2p are

Question 52.
Which rule is violated in the following electronic configuration?

Answer:
The rule violated is Hund’s rule.

Question 53.
How many maximum number of electrons that can be accommodated in N principle energy shell?
Answer:
For N shell n = 4.
The maximum number of electrons accommodated in a shell is 2n².
∴ A maximum of 32 electrons can be filled in N shell.

Question 54.
How many maximum number of electrons that can be accommodated in a ‘l’ sub-shell?
Answer:
l sub-shell has 3 orbitals. Each orbital accommodates 2 electrons. So 6 electrons can be filled in l sub-shell.

Question 55.
How many maximum number of electrons can be accommodated in ‘d’ sub-shell?
Answer:
d sub-shell has 5 orbitals. So 10 electrons can be filled in d sub-shell.

Question 56.
How many sub-shells present in a ‘M’ principal energy shell?
Answer:
For M shell n = 3
The number of sub-shells in M shell is 3.

Question 57.
How many spin orientations are possible for an electron in s-orbital?
Answer:
The spin quantum number values for electrons are ½ or -½. So 2 spin orientations are possible.

Question 58.
Write valence electronic configuration of element which has the following set of quantum numbers.
Answer:

n = 3 indicates 3rd orbit and l = 1 indicates p orbital and there is one electron in p orbital. So the valence electron configuration is 3p¹.

Question 59.
How many unpaired electrons are present in chromium?
Answer:
The electronic configuration of chromium 1s² 2s² 2p⁶ 3s² 3p6 4s¹ 3d⁵.

The number of unpaired electrons = 6.

Question 60.
Find the four quantum number values of 3^rd and 4^th electrons of Beryllium.
The electronic configuration of Beryllium is 1s² 2s².

Question 61.
What is the n + l value of 4f orbital?
Answer:
For 4f orbital n = 4 and f orbital l = 3.
∴ n + l = 4 + 3 = 7

Question 62.
When you heat iron rod first it turns red. Why?
Answer:
Iron turns into red because red has higher wavelength. So it has lower energy which is emitted by iron.

Question 63.
What is the significance of Planck’s proposal?
Answer:
Electromagnetic energy can be gained or lost in discrete values and not in a continuous manner.

Question 64.
When do you see an emission line?
The energy emitted by an electron is seen in the form of an electromagnetic energy and when the wavelength is. in the visible region it is visible as an emission line.

Question 65.
How many elliptical orbits are there in 4^th orbit of Sommerfeld?
Answer:
The number of elliptical orbits in 4^th orbit of Sommerfeld is 3.

Question 66.
Why is spin quantum number introduced?
Answer:
When we observe spectrum of yellow light by using high resolution spectroscope it has very closely spaced doublet. Similar patterns are shown by Alkali and Alkaline earth metals. In order to account this spin quantum number is introduced.

Question 67.
Which of the following magnetic quantum number values is not possible for 3d orbital?
a) – 2
b) – 1
c) 0
d) 4
Answer:
For d orbital the possible m; values – 2, – 1, 0, 1, 2. So the value 4 is not possible.

Question 68.
If an element has 11 electrons in its M shell, then what is the name of element and its atomic number?
Answer:
The electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³.
[M shell electrons = 2 + 6 + 3=11]

So the element is Titanium.

Question 69.
The wave length of a wave is 100 nm. Find its frequency.
Answer:

10th Class Chemistry 8th Lesson Structure of Atom 2 Marks Important Questions and Answers

Question 1.
Explain Hund’s Rule with an example. (TS March 2016) (AP SA-1:2018-19)
Answer:
Hund’s Rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

(1) Ex :

1. The configuration of carbon atom (Z = 6) is 1s² 2s² 2p².
2. The first four electrons go into the 1s and 2s orbitals.
3. The next two electrons go into 2p_x and 2p_y orbitals.
   
4. But, they do not pair in 2p_x orbital.

(2) Ex :

1. In oxygen atom (₈O), distribution of electrons is given below.
2. 
3. Here, pairing of electrons in 2px orbital takes place after, 2p_x, 2p_y and 2p_z orbitals are filled with a single electron.

Question 2.
The electronic configuration of Sodium is 1s² 2s² 2p⁶ 3s¹. (AP March 2017)
What information that it gives?
Answer:

1. Its atomic number is 11
2. It is s-block element
3. It is in 3rd period ’
4. It is in 1st group
5. Its valency is 1
6. Number of valency electrons are 1
7. It can form uni positive ion
8. It can form ionic bond, 9. It is metal.

Question 3.
Explain the principle which describes the arrangement of electrons in degenerate orbitals.
Answer:
According to Hund’s rule the degenerate orbitals are occupied with one electron each before pairing of electron starts.
Ex : Electronic configuration of carbon is 1s² 2s² 2p²

the last two electrons will enter into separate 2p orbitals.

Question 4.
Name the principle, which says an orbital can hold only 2 electrons and explain. (AP March 2018)
Answer:

• Name of the principle : Pauli’s exclusion principle.
• No, two electrons in an orbital can have all four quantum numbers same.
• It says there is a chance to hold only 2 electrons, one rotates in clockwise direction (+½)other rotates in anticlock wise direction (-½).

Question 5.
For a better understanding about the electronic configuration in an atom, the teacher wrote shorthand notation nl^x on the blackboard.
Looking at this notation, what could be the probable questions that generate in the student’s mind? Write any two of them. (TS March 2015)
Answer:

1. What n, l, x indicates related to atoms?
2. How nl^x indicates the position of the electrons in the atom?

Question 6.
Write the ‘Octet Rule’. How does Mg (12) get stability while reacting with chlorine as per this rule? (TS June 2017)
Answer:
Octet Rule :
The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer – shell electrons.

Magnesium atom looses 2 electrons and get 8 electrons in its outermost shell as Neon. So that it gets stability.

Question 7.
Write the electronic configuration of the atom of an element having atomic number 11. Write the names of the rules and the laws followed by you in writing this electronic configuration. (TS March 2017)
Answer:
1s²2s²2p⁶3s¹.
(OR)
Principles followed :

1. Aufbau principle.
2. Hund’s Rule.
3. Pauli Exclusion Principle

Question 8.
The electron enters into 4s orbital after filling 3p orbital but not into 3d. Explain the reason (TS March 2018)
Answer:
Based on (n+l) values energy value of 3d orbitals is 3 + 2 = 5, energy value of 4s orbitals is 4 + 0 = 4
The energy level of 4s orbital is less than the 3d orbital according to the Aufbau principle electron enters into lower energy orbital first.
Thats why electrons enters into 4s orbital after filling 3p, but not into 3d.

Question 9.
Write the electronic configuration of Na⁺ and Cl^–.
Answer:
Electronic Configuration of Na⁺ is 1s² 2s² 2p⁶ and Cl^– is 1s² 2s² 2p⁶ 3s² 3p⁶.

Question 10.
Observe the given table and answer the following questions. (TS March 2019)

Sl.No.	Electron Configuration
1.	1s2 2s2 2p6 3s2 3p3
2.	1s2 2s2 2p6 3s2 3p6 4s2
3.	1s2 2s2 2p6 3s2 3p6

1) Mention the divalent element name.
2) Name the element belongs to 3^rd period and VA Group.
Answer:

1. Name of the divalent element in the table is Calcium.
2. Name of the element which belongs to 3^rd period and VA Group is Phosphorous.

Question 11.
Your friend is unable to understand nl^x. What questions will you ask him to understand nl^x method? (AP SCERT: 2019-20)
Answer:

1. What is nl^x method?
2. Where does it use ?
3. What is meant by ‘n’, 7′ and ‘x’?
4. How can we use nl^x method in the writing of electronic configuration?

Question 12.
Why do valency electrons involve in bond formation, than electrons of inner shells? (AP SCERT: 2019-20)
Answer:

1. When two atoms come sufficiently close together the valence electrons of each atom experience the attractive force of the nucleus in the other atom.
2. The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
3. The electrons in outer most shell of an atom get affected.
4. Thus electrons in valence shell are responsible for the formation of bond between atoms.

Question 13.
Explain Pauli’s exclusion principle with an example. (AP SA-I:2019-20)
Answer:
Pauli’s exclusion principle :
No two electrons of the same atom can have all four quantum numbers the same. If n, l and m_l are same for two electrons, then m_s must be different.
Suppose take the example of Helium atom.
The four quantum numbers for two electrons in the Helium atom given below.

We can observe from the table that three of the quantum numbers are same but fourth quantum number is different. The electronic configuration of Helium atom is² ↑↓. So the maximum number of electrons filled in an orbital is 2.

Question 14.
Explain Aufbau principle. (AP SA-I:2019-20)
Answer:
Aufbau principle :

1. In the ground state the electronic configuration can be built up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to atomic number.
2. Thus orbitals are filled in the order of increasing energy.
3. Electrons are assigned to orbitals in order of increasing value of n + l.
4. For sub-shells with the same value of n + l, electrons are assigned first to the subshell with lower n.

Question 15.
The electronic configuration of an atom is as follows 1s² 2s² 2p².
a) Which element’s atom is it?
b) Which orbital is the last electron in?
c) When excited what could be the number of lone / single electrons in this atom?
d) What is the value of principal quantum numbers of two electrons in the first box?
Answer:
Given electronic configuration of atom is 1s² 2s² 2p².
a) The element is carbon.
b) The valence electron enters 2p orbital.
c) In excited state the electron in 2s orbital enters 2p orbital. So it has 4 unpaired electrons.
d) The value of principal quantum number is 1.

Question 16.
Draw the table which gives the information about the quantum numbers and the number of the quantum states.
Answer:

Question 17.
Explain briefly about spin quantum number.
Answer:

• This gives spin of the electrons about their own axes. It is denoted by ms.
• This quantum number refers to the two possible orientations of the spin of an electron, one clockwise and the other anti-clockwise spin.
• These are represented by +½ -½ and .

Question 18.
Write electronic configurations of following elements,
a) Hydrogen
b) Helium
c) Lithium
d) Beryllium
e) Boron
Answer:

Question 19.
What does a line spectrum tell us about the structure of an atom?
Answer:
The electrons in ground state i.e. lowest, energy state absorb energy and move into excited state where they are unable to stay for long periods so lose the energy and come back to the ground state. The emitted radiation appears as line in line spectrum.

Question 20.
What are the spins of electrons in Helium atom?
Answer:
The quantum numbers for two electrons of Helium are given below as per Pauli’s exclusive principle.

Three of quantum numbers are same. So fourth must be different so the two electrons have anti-parallel spins.

Question 21.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ is the electronic configuration of Cu (Z = 29). Which rule is violated while writing this configuration? What might be the reason for writing this configuration?
Answer:
The rule violated is Aufbau principle. The elements which have half filled or completely filled orbitals have greater stability. So copper can get stability by transferring one electron from 4s to 3d (their energies are close to each other).

So the electronic configuration of copper is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹, not 1s² 2s² 2p 3s² 3p⁶ 3d⁹ 4s².

Question 22.
Why are chromium and copper exceptions to electronic configuration?
Answer:
Elements which have half-filled or completely filled orbitals have greater stability. So in chromium and copper the electrons in 4s and 3d redistribute their energies to attain stability by acquiring half-filled and completely filled d orbitals.

Hence the actual electronic configurations of chromium and copper are as follows.

Question 23.
Distinguish between emission and absorption spectrum.
Answer:

Emission spectrum	Absorption spectrum
1) The spectrum produced by emitted radiation is called emission spectrum.	1) The spectrum produced by absorption of radiation is called absorption spectrum.
2) The emission spectrum contains bright lines on the dark back-ground.	2) The absorption spectrum contains dark lines on the bright back-ground.
3) The emission spectrum corresponds the radiation emitted when an excited electron comes back to the ground state.	3) The absorption spectrum corresponds the radiation absorbed in exciting an electron from lower to higher energy levels.

Question 24.
Draw a table which will give the relationship between l values and number of orbitals and name of sub-shell and maximum number of electrons.
Answer:

Question 25.
Distinguish between line and band spectrum.
Answer:

Line spectrum	Band spectrum
1) The spectrum has sharp distinct lines.	1) The spectrum has many closed, spaced lines.
2) The spectrum is characteristic of atoms and is also called atomic spectrum.	2) The spectrum is characteristic of molecules is also called molecular spectrum.
3) The spectrum is given by inert gases, metal vapours and atomised non-metals.	3) The spectrum is given by hot metals and molecular non-metals.

Question 26.
What is Dispersion of light? Explain a natural example of dispersion of light.
Answer:
Dispersion:
Splitting of white light into colours (VIBGYOR) is called Dispersion of light.

The natural example for dispersion of light is formation of Rainbow. It is caused by dispersion of sunlight by tiny water droplets present in atmosphere which act as small prisms.

Question 27.
Wien is an electromagnetic wave produced? Write about characteristics of electromagnetic wave.
Answer:
Electromagnetic wave is produced when an electric charge vibrates (moves back and fortn).

Characteristics of electromagnetic waves :

1. Efectric field and magnetic fields are perpendicular to each other and at right angles to direction of propagation of wave.
2. It travels with speed of light i.e., 3 × 10⁸ ms^-1.
3. Electromagnetic energy is characterized by wavelength (λ) and frequency (v). The relation is given by c = vλ.

Question 28.
The valence electron configuration of element is given as 4s¹. Then give the following information.
1) What is the name of that element?
2) What is the outermost orbit of element?
3) What is ‘l’ value of outermost sub-shell?
4) What is the atomic number of element?
Answer:

1. Potassium.
2. N
3. The outermost sub-shell is 4s its l value is ‘O’.
4. Its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. So its atomic number is 19.

Question 29.
Write all the quantum numbers for valance electron of sodium.
Answer:

1. The electronic configuration of sodium is 1s² 2s² 2p⁶ 3s¹.
2. The valance orbital is 3s.
3. The quantum numbers for this orbital is
   

Question 30.
Give electronic configurations of following elements.
a) Sodium
b) Phosphorous
Answer:
Sodium – 1s² 2s² 2p⁶ 3s¹
Phosphorous – 1s² 2s² 2p⁶ 3s² 3p³

Question 31.
Why does nitrogen has more chemical stability when compared with oxygen?
Answer:
The electronic configurations of Nitrogen and Oxygen are as follows.

Nitrogen has half filled 2p³ configuration. So it has greater chemical stability when compared with Oxygen.

Question 32.
Ramu gave electronic configuration of potassium as 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ whereas Ravi expressed the configuration as 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. Who gave the correct configuration? Why?
Answer:
Ravi gave the correct configuration because according to Aufbau principle after completion of 3p orbital electron may enter either 4s or 3d.
Their n + l values are given below.

Orbital	n + l Values
4s	4 + 0 = 4
3d	3 + 2 = 5

So 4s orbital has lower n + l value when compared with 3d orbital. So electron enters into 4s.

Question 33.
Given the valence electron configuration of an element is 4s¹. Then what are its quantum number values. Which element does it represent?
Answer:
Given valence electronic configuration is 4s¹. So n = 4 for ‘s’ sub shell l = 0 and if l is ‘O’ then m_l is also zero. m_s takes only two values that is +½ or -½ for convenience we can take m_s as +½.
∴ The quantum number values are like this

The element is potassium.

Question 34.
The atomic number of an element is 17, then calculate the total number of electrons present in its s and p orbitals.
Answer:
The element with atomic number 17 is chlorine.
Its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
So, the total number of electrons present in s orbitals = 2 + 2 + 2 = 6.
The total number of electrons present in p orbitals = 6 + 5 = 11.

Question 35.
Based on Aufbau’s principle, in which of the three 4d, 5p and 5s orbitals the electrons will be filled first? Why?
Answer:
According to Aufbau’s principle, the electron enters the orbital having lower n + l value. If both orbitals have same n + l values, electron enters the orbital with lower ‘n’ value.

Oprbital	n + l value
4d	4 + 3 = 7
5p	5 + 1 = 6
5s	5 + 0 = 5

So, 3s has least n + l value. Therefore the electron enters 5s orbital first.

Question 36.
Find the following.
1) Number of orbitals present in M orbit.
Answer:
Number of orbitals present in an orbit = n² For M orbit n = 3.
∴ So the number of orbitals = 3² = 9.

2) The maximum and minimum possible m_l values for 4f orbital.
Answer:
For f orbital ‘l’ value is 3. If l is 3, then m_l takes values from – 3 to + 3.
So the minimum value for m_l is – 3 and maximum value is + 3.

3) The possible values of l if n = 4.
Answer:
If n = 4, then l take values from 0 to 3 i.e., 0, 1, 2, 3.

4) The maximum number of electrons that can be filled in ‘N’ energy level.
Answer:
For N orbit n = 4. The maximum number of electrons present in a shell = n².
∴ The maximum number of electrons filled in N shell = 4² = 16.

Question 37.
How do the vibrating electric and magnetic fields around the charge become a wave that travel through space?
Answer:

• A vibrating electric charge creates a change in the electric field.
• The changing electric field creates a changing magnetic field.
• This process continues, with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.

Question 38.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nuclear where the electrons are found and also their energies.

Question 39.
An electron in an atom has the following set of four quantum numbers. Which orbital does it belong to?

n	l	ml	ms
3	0	0	+½

Answer:
The quantum numbers of an atom is given below.

n	l	ml	ms
3	0	0	+½

By using nl^x method n = 3, if l = 0 then the sub-shell is s.
So the electron belongs to 3s.

Question 40.
Look at the following table and answer the following questions.

a) What is the law that is voilated in the above table? State that law.
b) Write the correct table using that law.
Answer:
a) The law violated is Pauli’s Exclusive Principle. Pauli’s law states that no two electrons of the same atom can have all the four quantum numbers same.
b)

10th Class Chemistry 8th Lesson Structure of Atom 4 Marks Important Questions and Answers

Question 1.
Draw Moeller chart of filling order of atomic orbitals.
(OR)
Draw a diagram showing the increasing value of (n + l) of orbitals. (AP June 2017)
Answer:

The filling order of atomic orbitals (Moeller Chart)

Question 2.
Complete the following table based on quantum numbers related to atomic orbitals and electron of an atom. (AP SA-I-2018-19)

Answer:

Question 3.
Based on the information given in the table, answer the questions given below.

i) For the 4th main shell, how many values are there for m_l? What are they? (TS June 2016)
Answer:
1) There are 16 m( values in the 4th main shell.
2) They are

Orbital	ml values	Total
4s	0	1
4p	-1, 0, + 1	3
4d	-2, -1, 0, + 1, +2	5
4f	-3, -2, -1, 0, +1, +2, +3	7
Total		16

ii) For sub-shell with n = 3, l = 1, write the m, values.
Answer:
For sub – shell with n = 3, l = 1 the m, values are -1, 0, 1.

iii) Write the principal quantum number value for ‘N’ shell. How many sub-shells are there in the main shell?
Answer:

1. The principal quantum number value for ‘N’ shell is 4.
2. The number of sub-shells is 4. They are 4s, 4p, 4d, 4f.

iv) In the above table m_l and l values are given. Write a formula that gives the relationship between m_l and l based on those values.
Answer:
m_l (No. of values) = 2l + 1.

Question 4.
Observe the information provided in the table about quantum numbers. Then answer the questions given below it. (TS June 2017)

n	l	ml
1	0	0
2	0	0
	1	-1, 0, +1
3	0	0
	1	-1, 0, +1
	2	-2, -1, 0, +1, +2

i) Write the ‘l’ value and symbol of the spherical shaped sub-shell.
ii) How many values that ‘m_l‘ takes for 1 = 2? What are they?
iii) Write the symbols of the orbitals for l = 1 sub-shell.
iv) What is the shape of the sub-shell for l = 2? What is the maximum number of electrons that can occupy this sub-shell?
Answer:
i) Spherical shaped sub-shell “l” value is zero and symbol is ‘s’.
ii) Number of m; values for 1 = 2 is 5, those are -2, -1, 0, 1, 2.
iii) Symbols of the orbitals for l = 1 sub-shell are p_x, p_y, p_z.
iv) Shape of the subshell l = 2 is double dumbel.
The maximum number of electrons that can occupy in this sub-shell is 10.

Question 5.
Write postulates and limitations of Bohr Hydrogen atomic model. (TS June 2018)
Answer:
Postulates :

1. Niels Bohr proposed that electrons in an atom occupy ‘stationary’ orbits of fixed energy at different distances from the nucleus.
2. When an electron jumps from a lower energy state (ground state) to higher energy state it absorbs energy or emits energy when such a jump occurs from a higher energy state.
3. The energies of an electron in an atom can have only certain values E₁, E₂, E₅…. that is, the energy is quantized.

Limitations :

1. Bohr’s model failed to account for splitting of line spectra of hydrogen atom into liner lines.
2. Bohr theory could not explain the Zeeman and Stark effect.

Question 6.
Explain Bohr’s model of hydrogen atom and its limitations. (March 2019)
Answer:
Niels Bohr proposed that,
a) electrons in an atom occupy stationary orbits of fixed energy (K, L, M, N,…) at different distances from the nucleus.

b) when an electron jumps from a lower energy state to higher energy state, it ab¬sorbs energy or emits energy when such a jump occurs from a higher energy state to lower energy state.

c) the energies of an electron in an atom can have only certain values E₁, E₂, E₃, ……… i.e. the energy is quantized. The states corresponding to these energies are called stationary states and the possible values of the energy are called energy levels.

d) the angular momentum of electron is multiple integral of \(\frac{L}{2 \pi}\).
∴ L = mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
m = mass of electron ;
v = velocity of electron ;
r = radius of circular path;
h = plank constant

Limitations :

1. Bohr’s model failed to account for splitting of line spectra (Zeeman effect).
2. This model failed to account for the atomic spectra of atoms of more than one electron.
3. Bohr theory was not explained the quantisation of angular momentum of an electron.
4. It was not explained the formation of chemical bonds.

Question 7.
Explain four quantum numbers with an example. (AP SA-I; 2019-20)
Answer:
Each electron iii an atom is described by a set of three numbers called Quantum numbers.

1) Principal quantum number (n) :
It is used to know the size and energy of the main shell The values of ‘n’ are 1, 2, 3 …..
energy of the shell, n = 1 < energy of the Shell n = 2.

2) Angular momentum of quantum number (l) :
It is used to know the shape of a particular sub shell.
The values of ‘l’ are 0, 1, 2, 3
l = 0 = s orbital = spherical in shape
l = 1 = p orbital = dumbel in shape
l = 2 = d orbital = double dumbel Shape

3) i) Magnetic quantum number (m_l) :
It is used to describe the orientation of the orbital in space relative to the other orbitals in the atom.
The values of mt are – ‘l’ to ’+l’ including zero,

ii) Spin quantum number (m_s) :
It is used to know the orientation of the spins of electrons.
The values of ms are +½ and – ½.

Question 8.
How does Hund’s rule helps in writing electronic configuration of an atom? Explain with a suitable example. (TS June 2019)
Answer:
Hund’s rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

This Hund’s rule helps in writing of electronic configuration of an element.

Example :

1. The electronic configuration of carbon (C) atom (Z = 6) is 1s² 2s² 2p⁶.
2. The first four electrons go into the Is and 2s orbitals.
3. The next two electrons go into separate 2p orbitals, with both electrons having the same spin.
   

Another example :

1. The electronic configuration of oxygen (₈O) is 1s² 2s² 2p⁴.
2. The first four electrons go into the 1s, 2s orbitals.
3. The next four electrons go into 2p orbits as 2 in 2p_x, 1 in 2p_y and 1 in 2p_z orbital.
   
4. Here pairing of electron in 2px starts after filling of electron in each 2p_x, 2p_y, 2p_z orbitals.
5. But electrons do not occupy like this
   

Question 9.
Your father asked you to go to the market and purchase an electric lamp. The shop-keeper displayed two lamps – one is violet and another is red. Which coloured lamp do you purchase to put in your bedroom? Support your choice of solution.
Answer:
Red has highest wavelength and violet has lowest wavelength. We know that relationship between energy as follows.
E = hv = h\(\frac{c}{\lambda}\)

h and c are remains constant.
∴ E ∝ \(\frac{1}{\lambda}\)

Energy is inversely proportional to wavelength. Red coloured light has lower energy and least intensity. So red coloured lamp is preferable as bed light.

Question 10.
Heisenberg contradicts Neils Bohr. Explain in what way he contradicts.
Answer:

• According to Bohr, electrons revolve around nucleus in definite paths or orbits. So the exact position of the electron at various times will be known to us.
• In order to explain Bohr’s postulate we have to know the velocity and exact position of electron.
• In order to find the position of electron we have to take the help of suitable light to find the position. As the electrons are very small, light of very short wavelength is required for this task.
• This short wavelength light interacts with the electron and disturbs the motion of the electron.
• Hence, it is not possible to find the exact position and velocity of electron simultaneously. This was stated by Heisenberg which is called Heisenberg’s principle of uncertainty.
• In this way Heisenberg contradicts Neils Bohr.

Question 11.
Explain Bohr-Sommerfeld model of an atom. What is the merit of this model? What are its limitations?
Answer:

• In an attempt to account for the structure of line spectra, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.
• While retaining the first of Bohr’s circular orbit as such, he added one elliptical orbit to Bohr’s second orbit, two elliptical orbits to Bohr’s third orbit, etc.
• Nucleus of the atom is one of the principal foci of these elliptical orbits because periodic motion under the influence of a central force will lead to elliptical orbits with the force situated at one of the foci.

Merit:
Bohr-Sommerfeld model is successful in accounting for the fine line structure of hydrogen atomic spectra.

Limitations :

1. This model failed to account for the atomic spectra of atoms of more than one electron.
2. It did not explain Zeeman and Stark effects.

Question 12.
In an atom the number of electrons in N-shell is equal to the number of electrons in K, L and M shells. Answer the following questions.
i) Which is the outermost shell?
Answer:
The outermost shell is p(n = 6).

ii) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

iii) What is the atomic number?
Answer:
Its atomic number is 56.

iv) Write the electronic configuration of the elements.
Answer:
The electronic configuration of element is
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s².

Question 13.
Explain the following electron configurations by using nl^x method.
a) 2p¹
b) 3d⁵
c) 4f⁹
d) 6s²
Answer:
In nl^x method n is the principle quantum number and l is the angular momentum quantum number and x is number of electrons. Now let us explain the following configurations
a) 2p¹ – It indicates that there is one electron in ‘p’ sub-shell of second orbit or shell.
b) 3d⁵ – It indicates that there are five electrons in ‘d’ sub-shell of third orbit or shell.
c) 4f⁹ – It indicates that there are nine electrons in ‘f’ sub-shell of fourth orbit or shell.
d) 6s² – It indicates that there are two electrons in ‘s’ sub-shell of sixth orbit or shell.

Question 14.
In an atom the number of electrons in L shell is equal to three times of K shell. Then answer the following.
1) Which is the outermost shell?
2) How many electrons are there in outermost shell?
3) What is the atomic number of element?
4) Write electronic configuration of element.
5) Write name of element.
Given that the number of electrons in L shell is three times of K shell.
Answer:
We know that number of electrons in K shell is 2.
Therefore number of electrons in L shell = 3 × 2 = 6

1. So the outermost shell is L.
2. The number of electrons in outermost shell is 6.
3. The atomic number of element is 8.
4. The electronic configuration of element is 1s² 2s² 2p⁴.
5. The element is oxygen.

Question 15.
We know that the electron configuration of copper is [Ar] 4s¹ 3d¹⁰. Is it against to Aufbau principle or not. If so, why is the configuration violated?
Answer:

• The atomic number of copper is 29. So its electron configuration should be [Ar] 4s² 3d⁴.
• But if one electron from 4s orbital jumps into 3d orbital, then copper gets half filled ‘d’ orbitals which gives stability to the atom.
• The energy difference between 4s and 3d is very less. So one electron can easily jump from 4s to 3d which gives half-filled 3d⁵ configuration.
• So, in order to get additional stability, Aufbau principle is violated i.e., electron enters the orbital of higher energy before the completion ortrital of lower energy.

Question 16.
Electronic configurations of following elements are written wrongly. Correct those configurations with proper explanation.

Answer:
1) The electron configuration of oxygen is 1s² 2s² 2p⁴ because the maximum number of electrons that can be filled in s orbital is 2 and so the extra electron should be entered in 2p.

2) The correct electron configuration of nitrogen is

The reason is that the pairing of electrons does not take place until each degenerate orbital is filled with one electron each (Hund’s principle).

3) The correct electronic configuration of scandium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ because after completion of 3p orbital electron enters in 4s because the energy of 4s orbital is less than 3d (Aufbau principle).

4) The correct electronic configuration of chromium is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵. Because atoms having half filled or completely filled orbitals are more stable. So by transferring one electron from 4s to 3d the atom gets extra stability.

Question 17.
Here is set of quantum numbers. Do they form correct values of quantum numbers or not. If not, give reason.

The values l, m_l, m_s are not correct values for given ‘n’.
Answer:

• The maximum value for l is n – 1. If n = 3, then l takes values from 0 to 2. That is 0, 1, 2.
• m_l values depend on ‘l’. m_l take values from – l to + l including zero. So, the possible values for m7 may be from – 2 to + 2.
• Spin quantum number has only two values, i.e. and –\(\frac{1}{2}\) and \(\frac{1}{2}\). So \(\frac{1}{4}\) value is not possible.

Question 18.
Answer the following questions.
a) If n = 4, then what energy level does it represent?
Answer:
If n = 4, then it represents N energy level.

b) If n = 5, then what is the maximum value for l and why?
Answer:
If n = 5, then the maximum value of l for 4 because the maximum value for l is n- 1.

c) If l = 3, then what are the maximum possible values for m_l?
Answer:
Given l = 3.
Then possible values for m_l is 2l + 1.
∴ Maximum possible values = 2(3) +1 = 7

d) What is the number of electrons present in M energy level?
Answer:
For M energy level n = 3.
The maximum number of electrons in an orbit = 2n² = 2 × 3² = 2 × 9 = 18

Question 19.
Draw the shapes of s, p and d orbitals.
Answer:
s – orbital (Spherical) :

p – orbital (Dumbell) :

All P:

d – orbital (Double Dumbell):

Question 20.
Draw electromagnetic wave and label its parts.
Answer:

Question 21.
Draw the diagram of electromagnetic spectrum.
Answer:

These AP 9th Class Maths Important Questions 11th Lesson Areas will help students prepare well for the exams.

AP State Syllabus 9th Class Maths 11th Lesson Important Questions and Answers Areas

Question 1.
Fill in the following table with suitable units:

Solution:

Question 2.
Find the distance between the parallel lines of a parallelogram whose base is given 12.5 cm. and area is 75 cm².
Solution:
Base of parallelogram = b = 12.5 cm
Area of parallelogram = bh = 75 sq.cm.
Height of parallelogram = \(\frac{75}{12.5}\) = 6 cm

Question 3.
The following statements belong to which quadrilateral area and express them in symbolic form.
i) Half of the product of base and height.
ii) Half of the product of it’s diagonals.
Solution:
i) Area of triangle = 1/2 × base × height
ii) Area of Rhombus = 1/2 × diagonal × diagonal₂

Question 4.
In the given figure \(\overline{\mathbf{B A}}\) ⊥ \(\overline{\mathbf{E A}}\), \(\overline{\mathbf{B E}}\) ⊥ \(\overline{\mathbf{E D}}\) and \(\overline{\mathbf{B D}}\) ⊥ \(\overline{\mathbf{C D}}\).

Find the area of the pentagon ABCDE.
Solution:
Given \(\overline{\mathbf{B A}}\) ⊥ \(\overline{\mathbf{E A}}\), \(\overline{\mathbf{B E}}\) ⊥ \(\overline{\mathbf{E D }}\), \(\overline{\mathbf{B D}}\) ⊥ \(\overline{\mathbf{C D }}\)
Area of Pentagon = area of ΔABE + area of ΔBED + area of ΔBDC
= \(\frac{1}{2}\) × AB × AE + \(\frac{1}{2}\) × BE × DE + \(\frac{1}{2}\) × BD × CD
= \(\frac{1}{2}\) × 4 × 3 + \(\frac{1}{2}\) × 5 × 12 + \(\frac{1}{2}\) × 13 × 24
= 6 cm² + 30 cm² 156 cm²
= 192 cm²

Question 5.
□ABCD is a parallelogram where \(\overline{\mathbf{A B}}\) || \(\overline{\mathbf{C D}}\). E is a point on \(\overline{\mathbf{C D}}\) If the area of ΔABE = 14.2 cm², find the area of the parallelogram ABCD. Support your answer with a proper reason.
Solution:
Area of ΔABE from figure = 14.2 cm

From theorem.
Area of ΔADE + area of ΔBCE – area of ΔAEB
∴ Area of ΔADE = area of ΔBCE,
x + x = area of ΔAEB
2x = 14.2 cm
x = \(\frac{14.2}{2}\) = 7.1 cm
∴Area of ΔADE = area of ΔBEC

Question 6.
Give any two examples for each from your daily life to find
i) Area of triangle
ii) Area of rectangle.
Solution:
i) Uses of Area of triangle :
a) To payment the white wash of a triangular wall
b) To prepare canvas of a boat,

ii) Uses of area of rectangle :
a) To payment the white wash of a rect-angular wall.
b) To measure length of a rectangular plot.

Question 7.
ABCD is a parallelogram. ABEM is a rectangle on the same base AB as shown in the figure. If AB = 10 cm, DM = 3cm and BE = 7 cm, find the area of the quadrilateral ABED.

Solution:
Area of rectangle ABEM = AB × BE
=10 × 7 = 70 sq.cm.
Area of ΔAMD = \(\frac{1}{2}\) × AM × MD
= \(\frac{1}{2}\) × 7 × 3 = 10.5 sq.crn
Area of quadrilateral ABED
= Area of ABEM – Area of ΔAMD
= 70 – 10.5 = 59.5 sq.cm.

Question 8.
PQRS is a rhombus. Find the ratio of area of rhombus PQRS to the area of triangle PQR.
Solution:

Area of Rhombus = \(\frac{1}{2}\) × PR × QS
Area of ΔPQR = \(\frac{1}{2}\) × PR × QO
Ratio = \(\frac{1}{2}\) × PR × QS : \(\frac{1}{2}\) × PR × QO
= QS : QO
= 2 (QO) : QO = 2 : 1

Question 9.
In ΔABC, AB = 12 cm and BC = 6.5 cm, ∠ABC = 90°. If D is the mid-point of \(\overline{\mathbf{A C}}\), find the area of ΔABD and ΔBCD. What did you observe?
Solution:
In ΔABC, ∠B = 90°, AB = 12 cm and BC = 6.5 cm

Area of ΔABC = \(\frac{1}{2}\) × BC × AB
= \(\frac{1}{2}\) × 6.5 × 12
= 39 sq. cm.
From theorem, D is a mid point of AC.
So, areas of ΔABD and ABDC are same.
∴ ΔABD + ΔBDC = ΔABC

These AP 8th Class Biology Important Questions 3rd Lesson Story of Microorganisms 2 will help students prepare well for the exams.

AP State Syllabus 8th Class Biology 3rd Lesson Important Questions and Answers Story of Microorganisms 2

8th Class Biology 3rd Lesson Story of Microorganisms 2 1 Mark Important Questions and Answers

Question 1.
What are antibodies?
Answer:
Whenever a disease causing microorganisms enter in our body, it produces some de¬fenders to fight against them, these are called antibodies.

Question 2.
Which one is better to take antibiotics or vaccines?
Answer:
Vaccines are better than antibiotics when the disease is prevented by vaccine.
Ex: Tuberculosis, Tetanus, etc.

Question 3.
What questions do you ask on role of mosquitoes and houseflies in spreading the diseases like Malaria and Swine flu?
Answer:

1. How can we be safe, without affected by malaria and typhoid?
2. Which organisms spread diseases like malaria and typhoid?

Question 4.
What is ‘Triple Antigen Vaccine’?
Answer:
D.P.T vaccine which prevents

1. Diptheria
2. Pertusis (Wooping cough)
3. Tetanus

Question 5.
What is Tet – vac?
Answer:
Tet – vac is the vaccine that is used to prevent Tetanus.

Question 6.
What is M.M.R vaccine?
Answer:
M.M.R is the vaccine that prevents the

1. Measles
2. Mumps
3. Rubella

Question 7.
Name the vaccine that prevents T.B.
Answer:
B.C.G (Bacillus Cardio Gram)

Question 8.
What is the mode of transmission of Dengue disease?
Answer:
Aedes Mosquitoes spread Dengue disease.

8th Class Biology 3rd Lesson Story of Microorganisms 2 2 Marks Important Questions and Answers

Question 1.
Mention different questions to know about antibiotics.
Answer:

1. What are antibiotics?
2. In which year penicillin was announced as antibiotic?
3. Can antibiotics destroy the viruses?
4. Write different types of antibiotics.

Question 2.
Prepare different questions to know the importance of vaccines.
Answer:

1. Name the scientist who invented vaccines first.
2. What type of materials used for preparation of vaccines?
3. Name the vaccine which is used to prevent rabies.
4. Name the disease which is eradicated by vaccine.

Question 3.
Ask different questions about the significance of microbes.
Answer:

1. Name different nitrogen fixing microbes.
2. Can all microbes useful to mankind?
3. What will happen if vaccines are not invented?
4. Write about significance of microbes.

Question 4.
Make a sketch of Nostoc.
Answer:

Question 5.
How is preservation of food helps us?
Answer:

1. Preservation prevents spoilage of food.
2. It helps us to store food for a longer time without spoilage.
3. Variety of food items may be available in off seasons also.

8th Class Biology 3rd Lesson Story of Microorganisms 2 4 Marks Important Questions and Answers

Question 1.
Which one is better to take antibiotics or vaccines? Why?
Answer:
Antibiotics are small molecules or compounds that are effective in treating infections caused by organisms such as bacteria, fungi and protozoa. Some antibiotics act by killing the bacteria while others prevent the bacteria from multiplying, leaving your own immune system to ‘mop them up’.
Vaccines are dead or inactivated organisms or compounds that are used to provide immunity to a particular infection or disease.
Vaccines are used to prevent infection, particularly viral infections. Vaccines are produced from portions of viruses or viruses that are rendered harmless.
The killed or modified virus is introduced into the body where it stimulates part of the body’s natural defenses against infection without causing the illness itself.
Because of this, if the body is exposed to the specific virus in the future, it will recognize it and can fight the infection off much more quickly and effectively, meaning that you don’t succumb to the illness.
Hence vaccines are better than antibiotics when the disease is prevented by vaccine. Ex: Tuberculosis, Tetanus, etc.

Question 2.
Refrigerator helps us in saving health and money. Can you explain?
Answer:
Food and drink, being organic in origin, does not last forever. Shortly after food is bought or prepared it will begin to spoil, quickly becoming inedible through the effects of bacteria. By lowering the temperature of the food through refrigeration, you can slow or even temporarily halt the effects of bacteria, making the food stay edible longer.we can also transport fruits vegetables and other food items up to long distances by using refrigerators. Refrigerator is more useful to keep vaccines and medicines for long time. Hence we can save money and as well as health.

Question 3.
Take root nodules of any pulse or leguminous plants crush on a slide, and draw a rough sketches of what you observe under microscope.
Answer:

Question 4.
Can you give some important key points you noticed that scientists followed in their inventions and discoveries?
Answer:
Invention of microscope is most important one to find out microorganisms. Basing on this several inventions were made, which break the unsolved problems since 400 years. This credit goes to Antonie van Leeuwenhoek who invented single lens powerful microscope, which could magnify the object 300 times. His curiosity and skill of making powerful lenses were the secrets of this invention of powerful microscope.
Dr. Alexander Flemming observed some fungi were preventing the growth of bacteria in his experiments. He separated the substances released by the fungus and tested it on some other disease causing bacteria. This substance also killed many other diseases causing bacteria and named it as Pencillin. The discovery of pencillin paved the way to the discovery of many antibiotics like streptomycin, erythromycin etc.
After several attempts Ross suddenly realized that he used the wrong species of mosquitoes in his experiments. His assistant brought larvae, which hatched to several dappled winged mosquitoes. Delighted with this capture Ross identified Anopheles mosquito is the host for malarial parasite.

Question 5.
Which invention is most important one related to microorganisms? Can you explain the reasons why it is so important?
Answer:
Invention of microscope is most important one to find out microorganisms. Basing on this several inventions were made, which break the unsolved problems since 400 years. This credit goes to Antonie van Leeuwenhoek who invented single lens powerful microscope, which could magnify the object 300 times. His curiosity and skill of making powerful lenses were the secrets of this invention of powerful microscope. Later several microbes were identified by using microscopes. This provides a good source to recognize several diseases and treatment by vaccines and antibiotics.

Question 6.
What are antibodies? When do they develop? How they help us?
Answer:
Whenever a disease causing microorganisms enter in our body, it produces some defenders to fight against them, these are called antibodies. The lymphocytes of white blood cells develop antibodies against to the antigen of pathogen. Antibodies fight the disease causing microorganisms and make resistant to particular strains. Antibodies are used by the immune system to identify and neutralize foreign objects such as viruses and bacteria.

Question 7.
If there are no microorganisms on the earth what will happen?
Answer:
Microorganisms are present everywhere, they are to be found in air, water, soil within the bodies of animals and plants. Some microorganisms are very useful and help us in many ways while some of them are harmful. Microorganisms are most essential to maintain inorganic and organic cycles, to give freshness to the nature by degrading waste organic matter and as well as to fight against harmful microorganisms. Hence no life will be sustained in the nature if there are no microorganisms on the earth.

Question 8.
Prepare a table showing some of plant diseases, causative microorganisms and mode of transmission.
Answer:

Name of the plant disease	Causative microorganisms	Mode of transmission
1. Citrus canker	Bacteria, Xanthomonoss citri	Air
2. Red rot of sugarcane	Fungi	Air, seedlings
3. Tikka disease of groundnut	Fungi	Air, seeds
4. Tobacco mosaic	Virus	Insects
5. Smut disease of rice	Fungus	Air

Question 9.
If there are no microorganisms on the earth what will happen?
Answer:

1. If micro organisms are not present on earth the earth will be full of dead organisms and plants.
2. The plants can not fix the atmospheric nitrogen in the soil.
3. We can not get fermented products.
4. The life becomes impossible on the earth.

8th Class Biology 3rd Lesson Story of Microorganisms 2 Important Questions and Answers

Question 1.
Many people to. a colony are suffering from ‘Cholera’. What do you think could be the reason?
Answer:

1. The main reasons for the spread of cholera is consumption of contaminated food and water. House flies rapidly spread this disease among people.
2. I would imagine that people were not following personal hygienic practices before consuming food and water.
3. Perhaps people might not be keeping their surroundings clean and hygienic.
4. Perhaps the people might have poor sanitary conditions.

Question 2.
Suggest any two methods of fish preservation.
Answer:

1. Drying
2. Smoking
3. Canning
4. Chilling

Question 3.
What is symbiosis? Give two examples for it.
Answer:

1. In Greek language, symbiosis means “living together”.
2. Symbiosis is any type of a close and long term biological interaction between two different biological organisms in which both are benefited.
3. Examples for symbiotic relationship is Rhizobium. Rhizobium bacteria is living in the root nodules of legume plants.
4. Rhizobium fixes the atmospheric nitrogen in the form of nitrates which are very essential for the growth of plants. Inturn legume plants provide shelter for Rhizobium. Here, both are benefited.
5. Another example for symbiosis is Lichens. Here, we can see the symbiotic relationship between Algae and Fungi.

Question 4.
In an experiment if we keep the dosa dough in Bacillus free chamber, what changes can you observe in the dosa dough after one day?
Answer:

1. Fermentation of dosa dough will not take place.
2. The volume of the dosa dough will not increase and it will not turn into spongy texture.

Question 5.
Classify the given below microorganisms into useful and harmful microorganisms.
Plasmodium, Lactobacillus, Rhizobium, Pencillium, Yeast, Virus.
Answer:
Useful microorganisms:

1. Lacto bacillus
2. Rhizobium
3. Pencillium
4. Yeast

Harmful microorganisms:

1. Plasmodium
2. Virus.

Question 6.
What will happen if we add butter milk to chilled milk?
Answer:
If we add a few drops of butter milk to chilled milk 1) the milk will not turn into curd. It is because, cold milk won’t provide the conditions to grow lactobacillus. Hence, curd will not be formed.

Question 7.
If you are going to conduct a quiz on micro-organisms, what type of questions do you prepare for the quiz programme?
Answer:

1. Can you name the scientist who discovered the pencillin?
2. Name the bacteria that converts the milk into curd.
3. Can you name the food items that can be preserved through pasteurisation?
4. Can you name the microorganism which is very helpful in the preparation of alcohol?

Question 8.
What is the name of the nitrogen fixation bacteria in the root nodules?
Answer:
Rhizobium

Question 9.
What precautions do you take to avoid diseases?
Answer:

1. We should consume boiled, purified water and healthy food daily.
2. Keep our surroundings neat and clean.
3. We should take vaccines to prevent different diseases.
4. We should practice personal hygiene and good sanitary habits.

Question 10.
Defecation in open areas spreads micro-organisms easily. Write some slogans to conduct a rally to protect us from micro-organisms.
Answer:

1. Keep your country clean by keeping your area clean.
2. Stop being so open….. else everything is broken.
3. Use a toilet. Don’t let anything spoil it,
4. Use a toilet – to achieve Swach Bharath.

Question 11.
Draw a table of microorganisms and its diseases.
Microorganisms : Virus, Bacteria, Protozoa, Arthropods.
Disease: Scabies, Malaria, Polio, Typhoid.
Answer:

Microorganism	Disease
1. Virus	Polio
2. Bacteria	Typhoid
3. Protozoa	Malaria
4. Arthropods	Scabies

Question 12.
Why is the idly or dosa batter prepared one day before?
Answer:
If we prepare idly or dosa batter one day before, fermentation occurs in it, carbondioxide gas fill the batter increasing it’s volume and makes it spongy.

Question 13.
Read the following passage and answer the following questions.
When an infected person sneezes or coughs, the pathogens enter the air. When this air containing pathogens enters into the body of a healthy person, it may cause cold. These type of diseases which spread from infected people to healthy ones are known as “communicable diseases”. These get spread through air, water, food or through physical contact of infected person or through insects like houseflies and mosquitoes. Common cold, conjunctivitis, typhoid, smallpox, chickenpox, swine flu, tuberculosis are some examples of communicable diseases. Some insects and animals carry diseases causing microorganisms. They are called as ‘vectors’. The microorganism ‘plasmodium’causes malaria. The female Anopheles mosquito carries plasmodium and thus is a vector. Mosquitoes are vectors for other diseases as well. By controlling mosquitoes, we can prevent dis¬eases caused by them. Mosquitoes breed in stagnant water. We should be careful not to let water collect anywhere in our surroundings, including left over waste pots, waste flower pots, tyres, bowls etc.
a) What are communicable diseases? Give examples.
b) What are vectors?
c) Name the vector which spreads malaria.
d) How do you prevent spread of malaria?
Answer:
a) The diseases which spread from infected people to healthy ones are known as “communicable diseases”.
Eg: Conjuctivitis, Common cold, Smallpox, Rubella, Typhoid, etc…
b) The insects and animals which carry disease causing microorganisms are called as “vectors”.
c) Female Anopheles mosquito.
d) By controlling mosquito population.

These AP 10th Class Maths Chapter Wise Important Questions Chapter 5 Quadratic Equations will help students prepare well for the exams.

AP State Syllabus 10th Class Maths 5th Lesson Important Questions and Answers Quadratic Equations

Question 1.
If b² – 4ac ≥ 0, then write the roots of a quadratic equation ax² + bx + c = 0
Solution:
When b² – 4ac ≥ 0 then the roots of given quadratic equation ax² + bx + c = 0 are
\(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\) and \(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\)

Question 2.
Find the Quadratic polynomial with zeroes -2 and \(\frac{1}{3}\)
Solution:
Let α = -2; β = \(\frac{1}{3}\)
= -2 + \(\frac{1}{3}\) = \(\frac{-6+1}{3}=\frac{-5}{3}\)
α.β = -2. \(\frac{1}{3}\) = \(\frac{-2}{3}\)
Quadratic polynomial is [x² – (α + β )x + αβ ] = [x² + \(\frac{5}{3}\)x – \(\frac{-2}{3}\)
the quadratic polynomial will be 3x² + 5x – 2

Question 3.
Two angles are complementary and one angle is 18° more than tne other, then find angles.
Solution:
Let smaller angle be x°
bigger angle be y°
Since these two angles are comple-mentary
x + y = 90° ………… (1)
Since bigger angle is more than smaller angle
by 18°, y – x = 18° ………….(2)
By solving (1) & (2), we get x = 36°, y = 54°

Question 4.
Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0.
Solution:
Discriminant of standard quadratic equation
ax² + bx + c = 0 is b² – 4ac
Now comparing the given quadratic equation 2x² – 4x + 3 = 0 with stan¬dard form of quadratic equation.
We get a = 2, b = -4, c = 3 then its discriminant = b² – 4ac
= (-4)² – 4(2) (3)
= 16 – 24 = -8
∴ Discriminant = -8.

Question 1.
Find the roots of x + \(\frac { 6 }{ x }\) = 7, x ≠ 0
Solution:
x + \(\frac { 6 }{ x }\) = 7 ⇒ \(\frac{x^{2}+6}{x}\) = 7
⇒ x² – 7x + 6 =0
⇒ (x – 6) (x – 1) = 0
⇒ x = 6 or 1
Roots = 6, 1

Question 2.
Length of a rectangle is 2 units greater than its breadth. If the area of the rect-angle is 120 sq. units then find its length.
Solution:
Let breadth of the rectangle = x
Length = x + 2 ,
Area = 120 sq. units
x(x+2) = 120
x2 + 2x- 120 = 0
(x + 12) (x- 10) = 0
x = – 12 or 10
Breadth cannot be negative
∴ Breadth of the rectangle = x
= 10 units
∴ Length = x + 2 = 10 + 2 = 12 units

Question 3.
Find the zeroes of the quadratic poly-nomial x² – x – 30 and verify the rela¬tion between the zeroes and its co¬efficients.
Solution:
Given quadratic polynomial = x² – x – 30
⇒ x² – x – 30 = 0 = 0
x² – 6x + 5x – 30 = 0
⇒ x(x – 6) + 5(x – 6) = 0
⇒ (x – 6) (x + 5) = 0
⇒ x – 6 = 0
x = 6
x + 5 = 0 x = -5
∴ Zeroes are α = 6 and β = – 5
Sum of zeroes = α + β = \(\frac{-\mathrm{b}}{\mathrm{a}}\)
⇒ 6 – 5 = \(\frac{-(-1)}{1}\)
⇒ 1 = 1
Product of zeroes α + β = 6(-5) = \(\frac{\mathrm{c}}{\mathrm{a}}\)
= -30 = \(\frac{-30}{1}\)
Hence the relation was verified.

Question 4.
Find the roots of the quadratic equa¬tion (3x – 2)² – 4(3x – 2) + 3 = 0.
Solution:
(3x – 2)² – 4(3x – 2) + 3 = 0
9x² + 4 – 12x – 12x + 8 + 3 = 0
9x² – 24x +15 = 0
3x² – 8x + 5 = 0
3x² – 3x – 5x + 5 = 0
3x(x – 1) – 5 (x – 1) = 0
(x- 1) (3x – 5) = 0
x = 1 (or) x = \(\frac{5}{3}\)
Roots of quadratic equation are 1, \(\frac{5}{3}\).

Question 5.
Two numbers differ by 4 and their product is 192. Find the numbers.
Solution:
Let the larger number be ‘x’
Since the difference of two numbers is 4,
Then the smaller number is (x – 4)
Their product = x(x – 4)
Given that product = 192
∴ x(x – 4) = 192
⇒ x² – 4x – 192 = 0
⇒ x² – 16x + 12x- 192 = 0
⇒ x(x – 16) + 12(x – 16) = 0
⇒ (x – 16)(x + 12) = 0
⇒ x = 16 or x = -12
If x = 16, then x – 4 = 16 – 4 = 12
then the numbers are 16 and 12.
If x = – 12, thenx-4 = – 12 – 4 = – 16
then the numbers are – 12 and – 16.

Question 1.
Find the roots of the equation 5x² – 7x – 6 = 0 by the method of completing the square.
Solution:
Given that 5x² – 7x – 6 = 0
5x² – 7x = 6
x² – \(\frac{7}{5}\)x = \(\frac{6}{5}\)
⇒ x².2.\(\frac{7}{10}\)x = \(\frac{6}{5}\)
adding \(\frac{49}{100}\) on the both sides

The roots of the quadratic equation are 2 and \(\frac{-3}{5}\)

Question 2.
Find the roots of the quadratic equation 3x² + 11x + 10 = 0 by method of – completing the Square.
Solution:
Given : 3x² + 11x + 10 = 0
Dividing both sides by 3.

Question 3.
Solve the Quadratic equation 9x² – 9x + 2 = 0 by the method of com-pleting the square.
Solution:
Given : 9x² – 9x + 2 = 0
⇒ x² – x + \(\frac{2}{9}\) = 0
⇒ x² – x = –\(\frac{2}{9}\)