AP 7th Class Science Important Questions 10th Lesson Changes Around Us

These AP 7th Class Science Important Questions 10th Lesson Changes Around Us will help students prepare well for the exams.

AP Board 7th Class Science 10th Lesson Important Questions and Answers Changes Around Us

Question 1.
What is a natural change?
Answer:
The change that is brought about by nature itself is called a natural change.

Question 2.
What are man made changes? Give two examples.
Answer:

  1. Changes which are taken place by the involvement of human beings are called man made changes.
  2. Examples for man made changes are cooking of rice, construction of buildings, preparation of laddu.

Question 3.
What are fast changes? Give examples.
Answer:

  1. Changes which occur in short duration of time are called fast changes.
  2. Burning of paper, cutting of piece of cake, firing of crackers, spinning of a top are the examples of fast changes.

Question 4.
What are the slow changes? Give examples.
Answer:

  1. Changes which take longer duration of time to happen are called slow changes.
  2. Examples for slow changes are growing of plant from seed to a tree, developmental changes in the child, rusting of iron, changes of season, formation of mountain etc.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

Question 5.
What are reversible changes? Give examples.
Answer:

  1. The changes in which the formed substance can be converted into their original substance are called reversible changes.
  2. Example for reversible changes are melting of wax, magnetizing a needle using bar magnet etc.

Question 6.
What are irreversible changes? Give examples.
Answer:

  1. Changes in which we cannot get the original substance by reversing the experimental conditions are called irreversible changes.
  2. Examples are burning of wood, burning of diwali crackers, ripening of fruits, rusting of iron etc.

Question 7.
What are periodic changes? Give examples.
Answer:

  1. The changes which are repeating at regular intervals of time are called as periodic changes.
  2. Formation of the full moon and new moon, occurrence of seasons in every year at regular intervals, the heart beats of human beings are examples for periodic changes.

Question 8.
What are non-periodic changes? Give examples.
Answer:

  1. Changes which do not occur at regular intervals of time and which can not be predicted are called non-periodic changes.
  2. Examples are pattern of rainfall, storms in seas, volcanic erruptions etc.

Question 9.
“In a physical change, the chemical properties of a substance do not change.” Do you agree with this statement? How?
Answer:

  1. Yes, I agree with this statement.
  2. Why because, when a piece of gold is melted, it’s chemical composition remains the same in the solid form and also in the liquid form.

Question 10.
“A physical change is usually temporary and reversible in nature” – Do you agree with this statement? State the reason.
Answer:

  1. I absolutely agree with this statement.
  2. It is because, when water is heated, water vapours are formed. Once water vapours are cooled water can be obtained again.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

Question 11.
“In a physical change, change in physical properties such as colour, shape and size of a substance may under go a change”. Do you support this statement? Why?
Answer:

  1. I support this statement.
  2. In cutting of vegetables changing physical properties, such as colour, shape and size of a substance.

Question 12.
What is galvanization?
Answer:
The process of deposition of a layer of zinc on iron is called “Galvanization.”

Question 13.
What is a chemical change? Give one example.
Answer:

  1. Changes that occur with the formation of new substance with different chemical composition or transformation of a substance into another substance with the evolution or absorption of heat or light energy are called as chemical changes.
  2. Example when Magnesium burns in the presence of oxygen, it forms magnesium oxide in the form of powdered ash.
    Magnesium + oxygen → Magnesium oxide

Question 14.
What is rusting?
Answer:

  1. When ron reacts with atmospheric oxygen and moisture it forms a new substance called Iron oxide as rust on iron articles made of iron.
  2. This process is known as rusting.
    Iron + oxygen (from air) + water → rust (iron oxide)

Question 15.
What are the reasons for global warming?
Answer:

  1. Global warming is due to drastic increase in the emission of carbondioxide by the burning of fossil fuels.
  2. Deforestation.

Question 16.
What are the reasons for acid rains?
Answer:

  1. When coal and oils are burnt, they release acidic gases like NO2 and SO2.
  2. They mix up with the water vapour and come down as acid rains.

Question 17.
The changes like chicks came out of eggs, blossoming flowers etc. are very pretty to see. List out such changes around you feel happy on observation.
Answer:

  1. Swarming of honeybees
  2. Rainbow formation
  3. Clouds passing across the mountains.
  4. Rows of birds flying in the sky.

Question 18.
Give examples for metals that do not rust.
Answer:

  1. Metals like gold, silver, zinc do not rust even though they are exposed to moist air.
  2. Zinc is used in Galvanization process to prevent rusting of iron articles.

Question 19.
How can you say that rusting is a chemical change?
Answer:

  1. When the metals are exposed to air, they form metal oxides.
  2. Hence, as new substances are formed, rusting is a chemical change.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

Question 20.
Observe the following changes and decode weather they are physical or chemical changes? Mention the reason, a) Preparation of tea b) Making dough for roti.
Answer:
a) This is a chemical change. Water + Milk + Tea powder + Sugar → Tea Here tea is a new substance. Hence it is a chemical change,

b) Making dough for roti is a physical change as there is no formation of new substance.

Question 21.
How is an iron gate prevented from rusting?
Answer:

  1. Iron gate when exposed to moisture and air gets rusted.
  2. To prevent rusting of Iron gates it should be painted with a paint.

7th Class Science 10th Lesson Changes Around Us Short Questions and Answers

Question 1.
What are the examples for natural changes?
Answer:

  1. Formation of day and night
  2. Changes of weather.
  3. Developmental changes of a child into an adult. ‘
  4. rowth of a seed into a plant.

Question 2.
How can we prevent browning of cut fruits and vegetables?
Answer:

  1. By keeping the cut vegetables in cold water, we can prevent browning of them.
  2. Cold water prevents the outer surface of the potato and brinjal from colouring.
  3. Small quantities of acids like vinegar or lemon juice in water will also prevent browning of vegetables.
  4. We can also rub the surface of cut fruits with juices of citrus fruits like lemon to avoid from browning.
  5. Ascorbic acid can also be used to prevent browning.

Question 3.
What is the reason for browning of vegetables in our daily life?
Answer:

  1. Some fruits and vegetables like potato and brinjal, when they cut react with oxygen in air.
  2. The process of reaction with oxygen is called oxidation.
  3. Due to this oxidation process, brown layer is formed on the surface of fruits and vegetables.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

Question 4.
What are the harmful physical and chemical changes in environment?
Answer:

  1. Plastic decomposition
  2. Acid rains
  3. Global warming
  4. Oil spills in seas and rivers
  5. Earthquakes and
  6. Floods are some of the harmful physical and chemical changes in environment.

Question 5.
What changes might have we noticed in our daily life?
Answer:

  1. The colour ,of tender leaf changing from red to green.
  2. Hard raw fruit becoming soft ripen fruit.
  3. Colour changes observed in slices of brinjal and apple after being cut.
  4. Change of milk into the curd.
  5. Ash produced on burning paper.
  6. Raw rice becoming soft after cooking etc.

Question 6.
What questions would you pose to your teacher about changes around us?
Answer:

  1. Why changes occur around us?
  2. Why would we depend on changes in the nature?
  3. What would happen if changes do not occur?
  4. How changes in the nature influence us?

Question 7.
AP 7th Class Science Important Questions 10th Lesson Changes Around Us 1
What type of change is it? Is itreversible? If so, how?
Answer:

  1. Ice slowly melts and becomes water.
  2. On further heating water changes to steam.
  3. This is a physical change.
  4. No new substance is formed.
  5. The change is reversible. If we reduce the temperature the water vapour changes back to water when the temperature is further reduced, it changes to ice.
    AP 7th Class Science Important Questions 10th Lesson Changes Around Us 2

Question 8.
What are the characteristics of a physical change?
Answer:

  1. No new substances are formed.
  2. It is usually temporary and reversible in nature.
  3. The chemical properties of a substance do not change.
  4. Change in physical properties such as colour, shape and size of a substance may undergo a change.

Question 9.
What are the characteristics of a chemical change?
Answer:

  1. During chemical change new substance are formed.
  2. It is a permanent change and irreversible in nature.
  3. Chemical composition of the substances changes.
  4. Heat, light may be released or absorbed.
  5. A colour change may take place and sound may be produced.

Question 10.
Where do you see the iron pilar that is not rusted for thousands of years? What are the reasons for that?
Answer:
AP 7th Class Science Important Questions 10th Lesson Changes Around Us 2

  1. We can see an iron pillar at the Qutub complex in Delhi which is not rusted. It is 1600 years old.
  2. It is made by 98% wrought iron, a special type of iron that has 1% of phosphorous.
  3. It don’t have Sulphur and Magnesium.
  4. More over, the pillar is covered by a thin layer of compound called Misawitea.
  5. So, rusting of this iron pillar is too slow and it will take thousands of years to get rusted.

Question 11.
What is so special about Lonar Lake?
Answer:

  1. The Lonar lake in Buldhana district of Maharashtra was created by plummeting meteor about 5200 years ago.
    AP 7th Class Science Important Questions 10th Lesson Changes Around Us
  2. The water of Lonar lake turned into pink colour recently due to Haloarchaea microbes present in the salty water.
  3. But the colour is not permanent.
  4. When the biomass of the microbes settled at the bottom, then the water again becomes transparent.

Questions 12.
Anurag appreciates the changes in ripe mango as “How nice its colour and taste are”? Give some examples of changes that makes you feel happy, or wonder.
Appreciate them in your own words.
Answer:

  1. When mango ripens, the contents present in the fruit gives good taste. The change in colour indicates that the fruit is ripend.
  2. Similar situations we experience with many fruits like banana, guava, papaya, pine apple … etc.
  3. In fact fruits are a gift to human beings as they give instantaneous energy when eaten.
  4. Nature is so kind to us to provide such ready made safe foods.

7th Class Science 10th Lesson Changes Around Us Long Questions and Answers

Question 1.
What is rust and rusting? Why does iron rust? What type of a change is this?
Answer:

  1. When iron nails, iron gates, iron benches or pieces of iron are left is the open ground for a long time, we find a brown layer on the surface of iron articles.
  2. This is called rust’ and process of forming this layer is called rusting.
  3. When iron is exposed to air for a long time, the Oxygen present in air reacts with it in the presence of moist air and forms a new substance called iron oxide as rust on iron articles. This process is known as rusting.
    Iron + Oxygen (from air) + Water → rust (Iron oxide)
  4. As a new substance is formed in this change, we call it a chemical change.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

Question 2.
Answer the following questions.
The gas we use in kitchen is in the form of liquid in the cylinder. When it comes out from the cylinder it becomes a gas (step-I), then it burns (step-II).
Choose the correct statement from the following.
a. Only step – I is a chemical change.
b. Only step – II is a chemical change.
c. Both steps – I & II are chemical changes.
d. Both steps – I & II are physical changes.
Answer:
a) Step -1 – is a physical change (not a chemical change)
b) Step – II – is a chemical change (It is correct)

c) Both steps I & II – are chemical changes.
This is not correct.
Step – I – Physical change and
Step – II – Chemical Change is correct.

d) Both steps I & II – are physical changes – This is not correct.
Step I – Physical change
Step II – Chemical change.

Question 3.
What are periodic changes? Explain briefly.
Answer:

S.No.Name of the Change
1.Change of day and night
2.Withering of leaves
3.Rising of the pole star
4.Change of Seasons
5.Change of Greenery in the fields of cultivation
6.Changes in lengths of shadows
7.Appearance of Full Moon
  1.  If we observe that every change mentioned is the table repeats after some period of time.
  2. Such changes are known as periodic changes.
  3. The events which repeat at regular intervals of time are called periodic events.

Question 4.
What is Galvanisation? Explain its importance.
Answer:

  1. Some articles made up of iron, don’t get rusted even they are exposed to air.
  2. To prevent iron articles from coming into contact with oxygen in air or water or both, we deposit a layer of a metal like Chromium or Zinc on them.
  3. This process of depositing a layer of zinc or Chromium on Iron is called Galvanisation.
  4. Generally we use Zinc for such type of coatings.
  5. We find in our house that water pipe lines are without rust on them.
  6. If we observe carefully, we notice that there is some metallic coating on these pipes to prevent rusting.
  7. They do not get rusted even after a long time since they are galvanized.
  8. The process of depositing one metal on another metal is called galvanisation.

Question 5.
Write the environmental issues due to physical and chemical changes.
Answer:
Physical and chemical changes occur all around us. These changes are essential parts of our daily lives. However a few of the physical and chemical changes that occur daily are harmful to the environment.

For example

  1. Plastic decomposition, global warming, effects of acid rain and oil spills, earthquakes, floods etc.
  2. Plastic waste is a widely recognised source of pollution. Most plastics are non- biodegradable. They take hundreds of years to decompose. Hence their disposal causes pollution.
  3. Global warming is due to drastic increase in the emission of carbon dioxide by the burning of fossil fuels. The climate change affects not only the atmosphere and living things on land but also creatures of ocean.
  4. Acidic gases are produced when fossil fuels such as coal and oil are burned in power station, factories and homes.
  5. Oil spills occur when liquid petroleum is released into the environment by human interference causing damage to creatures of ocean.
  6. Changes are quite common in nature. Human beings are misusing the natural re-sources for their needs.
  7. But we should take care that our needs should not bring drastic changes that causes harm to the nature and mankind.

AP Board 7th Class Science 10th Lesson 1 Mark Bits Questions and Answers Changes Around Us

I. Multiple Choice Questions

Choose the correct answer and write its letters in the brackets.

1. Change in the shape of balloon is done by bibwing air into it. This is a……
A) manmade change
B) natural change
C) chemical change
D) periodic change.
Answer:
A) manmade change

2. Which of the following is not a man made-change?
A) preparation of bricks
B) making of paper
C) weaving of clothes
D) growing of nails
Answer:
D) growing of nails

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

3. Which of the following is not a fast change?
A) burning of paper
B) firing of crackers
C) making of a cake
D) spinning of a top
Answer:
C) making of a cake

4. The change happens in less time is.
A) slow change
B) fast change
C) periodic change
D) non periodic change.
Answer:
B) fast change

5. Which of the following is not reversible?
A) Weight suspended from a spring
B) Water changed to water vapour
C) Inflating of a balloon
D) Burning of a coal
Answer:
D) Burning of a coal

6. On …………… ice converts to water
A) heating
B) cooling
C) freezing
D) evaporating
Answer:
A) heating

7. some changes we cannot regain the original substance, these are.
A) manmade change
B) natural change
C) irreversible change
D) reversible change
Answer:
C) irreversible change

8. Limewater
A) Calcium Hydroxide
B) Sodium hydroxide
C) Potassium hydroxide
D) Carbonic acid
Answer:
A) Calcium Hydroxide

9. Which of the following is an irreversible change?
A) burning of wood
B) burning of Diwali crackers
C) ripening of fruits
D) all
Answer:
D) all

10. Which gas is released when lemon juice reacts with baking soda?
A) oxygen
B) carbon dioxide
C) nitrogen
D)water vapour
Answer:
B) carbon dioxide

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

11. These changes are repeating at regular intervals of time.
A) slow change
B) fast change
C) periodic change
D) non periodic change.
Answer:
C) periodic change

12. Changes which do not occur at regular intervals of time and which cannot be predicted are called
A) slow change
B) fast change
C) periodic change
D) non periodic change
Answer:
D) non periodic change

13. Crystallization requires
A) heating
B) cooling
C) evaporating
D) A or C
Answer:
A) heating

14. In crystallization….
A) no new substances are formed.
B) new substances are formed.
C) no heating is required
D) none
Answer:
A) no new substances are formed.

15. Crystallization is
A) chemical change
B) physical change
C) periodic change
D) non periodic change
Answer:
B) physical change

16. Choose correct answer.
S: Crystallization is a Physical Change.
R: In crystallization no new substances are formed.
A) S and R are correct.
B) S and R are incorrect.
C) S is correct and R is incorrect.
D) S is incorrect and R is correct.
Answer:
A) S and R are correct.

17. Which change takes place when ice cube melts?
A) colour
B) phase
C) chemical
D) all
Answer:
B) phase

18. When a piece of gold is melted?
A) no new substances are formed.
B) new substances are formed.
C) chemical composition changes
D) none
Answer:
A) no new substances are formed.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

19. When a piece of gold is melted, its chemical composition in the Solid form and also in the liquid form.
A) changes
B) varies
C) remains same
D) different
Answer:
C) remains same

20. In physical change ……….. changes.
A) shape
B) colour
C) size
D) all
Answer:
D) all

21. It is not a characteristic of a physical change…
A) No new substances are formed
B) Temporary and reversible in nature.
C) The chemical properties of a substance do not change.
D) It is a periodic change.
Answer:
D) It is a periodic change.

22. In curdling of milk is
A) a physical change
B) reversible change
C) chemical change
D) temporary change
Answer:
C) chemical change

23. When magnesium burns in the presence of oxygen, it forms magnesium oxide
A) magnesium oxide
B) magnesium chloride
C) carbon dioxide
D) none
Answer:
A) magnesium oxide

24. Magnesium Hydroxide is…..
A) an acid
B) a base
C) a neutral
D) none
Answer:
B) a base

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

25. Characteristic of a chemical change
A) During chemical change new substances are formed.
B) It is a permanent change and irreversible in nature.
C) Chemical composition of the substance changes.
D) All
Answer:
D) All

26. In chemical change which is not happens.
A) Heat, light may be released or absorbed.
B) A colour change may take place and sound may be produced.
C) Original substances may be formed on reversing the process
D) None
Answer:
C) Original substances may be formed on reversing the process

27. Rusting of iron requires
A) moisture
B) air
C) both
D) none
Answer:
C) both

28. Rust is…
A) iron oxide
B) calcium chloride
C) iron peroxide
D) above all
Answer:
A) iron oxide

29. Oxidization is observed in
A) iron articles
B) apples
C) brinjal
D) none
Answer:
D) none

30. Which of the following is used to prevent browning of the outer surface of the potato and brinjal?
A) cold water
B) lemon juice
C) ascorbic acid
D) above all
Answer:
D) above all

31. Which of the following is used in galvanizing?
A) zinc
B) chromium
C) A & C
D) none
Answer:
C) A & C

32. This process of deposition of a layer of zinc on iron is called
A) oxidation
B) galvanization
C) crystallization
D) none
Answer:
B) galvanization

33. Due to this process of brown layer is formed on the surface of fruits and vegetables.
A) oxidation
B) galvanization
C) crystallization
D) none
Answer:
A) oxidation

34. Which of the following is a useful change?
A) global warming
B) acid rain
C) plastic decomposition
D) fermentation
Answer:
D) fermentation

35. Which of the following occurs due to drastic increase in the emission of carbon dioxide by the burning of fossil fuels
A) global warming
B) floods
C) earth quakes
D) fermentation
Answer:
A) global warming

36. Ice converting to water, water converting to steam are ………. changes.
A) reversible
B) chemical
C) periodic
D) all
Answer:
A) reversible

37. Ripening of fruits is ………….. change.
A) reversible
B) physical
C) periodic
D) irreversible
Answer:
D) irreversible

38. The change occurs only in Size, colour and shape of the substance and no change in chemical composition are called ………….. changes
A) chemical
B) physical
C) periodic
D) irreversible
Answer:
B) physical

39. ………. change occurs with the formation of new substance in different chemical composition.
A) Reversible
B) Physical
C) Periodic
D) Chemical
Answer:
D) Chemical

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

40. The process of depositihg zinc on iron metals is called
A) oxidation
B) galvanization
C) rusting
D) crystallization
Answer:
B) galvanization

II. Fill in the blanks.

1. The changes which were taken place by the involvement of human beings are called ………………. change.
2. Changes which occur in ………………. duration of time are called fast changes.
3. Changes which takes longer duration of time to happen are called ………………. Change.
4. Changing of vegetable to curry : slow reaction:: changing of acid into vapour ………………. .
5. On ………………. water converts to ice.
6. The changes in which the formed substance can be converted into their ………………. are called reversible changes.
7. ………………. changes Limewater into milky white.
8. Vinegar + Baking Soda → Sodium acetate + ………………. + water
9. ………………. + Lime water → Calcium carbonate + water
10. Changes in which we cannot get the original substance by reversing the experimental conditions are called ………………. Changes.
11. ………………. changes are repeating at regular intervals of time .
12. The process of separating a soluble solid from the solution by heating or evaporating is called ………………. .
13. A ………………. is usually temporary and reversible in nature.
14. The substances which undergo change in colour or state or size or shape are ………………. .
15. When a Magnesium ribbon burns it gives ………………. light leaving a powdery substance behind.
16. ………………. + Water → Magnesium Hydroxide
17. Changes that occur with the formation of new substance with different chemical composition or transformation of a substance into another substance with the evolution or absorption of heat or light energy are termed as ………………. .
18. Iron + Oxygen (from air) + Water → ……………….
19. Apply a coat of paint or grease on iron articles. Prevents ………………. .
20. To prevent iron articles from coming contact with oxygen in air and water, a layer of another metal like ………………. is coated on them.
21. The process of deposition of a layer of zinc on iron is called ………………. .
22. Browning is not only observed on iron articles but also on cut fruits and ………………. .
23. Rubbing of the surface of cut fruits with ………………. to avoid from browning.
24. The process of reaction with ………………. is called oxidation.
25. ………………. waste is a widely recognized source of pollution.
26. ………………. gases are produced when fossil fuels such as coal and oil are burned in power station, factories and homes.
27. Oil spills occur when ………………. is released into the environment.
28. Formation of day and night, occurrence of seasons are ………………. changes.
29. Curding of milk : useful changes :: ………………. : harmful change.
Answer:

  1. 1. man-made
  2. short
  3. slow
  4. fast reaction
  5. cooling
  6. original substance
  7. Carbon dioxide
  8. carbon dioxide
  9. Carbon dioxide
  10. irreversible
  11. periodic changes
  12. crystallization
  13. physical change
  14. physical changes
  15. brilliant white dazzling
  16. Magnesium Oxide
  17. chemical change
  18. rust (Iron oxide)
  19. rusting of iron
  20. chromium or zinc
  21. Galvanization
  22. vegetables
  23. juices of citrus fruits
  24. oxygen
  25. Plastic
  26. Acidic
  27. liquid petroleum
  28. periodic
  29. global warming

III. Match the following.

1.

Group – AGroup – B
1) Ripening of fruita) physical change
2) Burning of a dry leafb) chemical change
3) Melting of icec) periodic change
4) Day and nightsd) fast change

Answer:

Group – AGroup – B
1) Ripening of fruitb) chemical change
2) Burning of a dry leafd) fast change
3) Melting of icea) physical change
4) Day and nightsc) periodic change

2.

Group – AGroup – B
1) Carbon dioxidea) galvanizing
2) Oxygenb) crystallization
3) Zincc) global warming
4) sugard) oxidation

Answer:

Group – AGroup – B
1) Carbon dioxidec) global warming
2) Oxygend) oxidation
3) Zinca) galvanizing
4) sugarb) crystallization

3.

Group – AGroup – B
1) browning of vegetablesa) vinegar
2) browning of ironb) dazzling light
3) formation of crystalc) galvanization
4) burning of magnesiumd) crystallization

Answer:

Group – AGroup – B
1) browning of vegetablesa) vinegar
2) browning of ironc) galvanization
3) formation of crystald) crystallization
4) burning of magnesiumb) dazzling light

4.

Group – AGroup – B
1) more timea) physical change
2) less timeb) chemical change
3) time periodc) periodic change
4) reversibled) fast change
5) new substancese) slow change

Answer:

Group – AGroup – B
1) more timee) slow change
2) less timed) fast change
3) time periodb) chemical change
4) reversiblec) periodic change
5) new substancesa) physical change

5.

Group – AGroup – B
1) Zinca) Chemical changes
2) Formation of Magnesium oxideb) Natural changes
3) Belum Cavesc) Periodic changes
4) Changes in seasonsd) Oxidation
5) Photosynthesise) Galvanisation
f) Crystallization

Answer:

Group – AGroup – B
1) Zince) Galvanisation
2) Formation of Magnesium oxided) Oxidation
3) Belum Cavesb) Natural changes
4) Changes in seasonsc) Periodic changes
5) Photosynthesisa) Chemical changes

Do You Know?

→ Belum caves are naturally formed caves located near Kolimigundla mandal of Kurnool district. These are the second largest caves in Indian subcontinent after Meghalaya state caves. The name is derived from Bilum” Sanskrit word for caves. The caves reach its deepest point (120 feet from entrance level) at the point known as “Paathalaganga” only 1.5 km is open to tourists. In 1988 Andhra Pradesh Tourism , Development Corporation (APTDC) declared them protected and developed the caves as a tourist attraction in February 2002. Borra Caves are also famous natural caves located at Visakhapatnam District.

→ The Lonar lake in Buldhana district of Maharashtra was created by plummeting meteor about 5200 years ago. The water of Lonar lake turned into pink colour recently due to Haloarchaea microbes present in the salty water. But the colour is not permanent. When the biomass of the microbes settled at the bottom, then the water again becomes transparent.

AP 7th Class Science Important Questions 10th Lesson Changes Around Us

→ The Iron pillar at Delhi :
Amazingly there is an iron that did not rust! There is an iron pillar at the Qutub complex in Delhi which is more than 1600 years old. Even after such a long period, the iron pillar kept in open space has hot rusted at all. Do you know how? It is made by wrought iron which contains more amounts of phosphorus than usual. The main reason for rust resistance is due to the formation of Iron hydrogen phosphate on its surface. So its takes more time to rest. That’s why still the iron pillar at Delhi did not get rust.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

These AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate will help students prepare well for the exams.

AP Board 7th Class Science 9th Lesson Important Questions and Answers Heat, Temperature and Climate

Question 1.
What is heat?
Answer:
Heat is a form of energy which flows from hotter body (higher energy) to cooler body (lower energy).

Question 2.
How is heat measured?
Answer:
Heat is measured in Joules or calories with calorimeter.

Question 3.
What is temperature?
Answer:
The degree of hotness or coldness is called temperature.

Question 4.
How do we call the variations of hotness, coldness?
Answer:
The variations of hotness, coldness can be termed as degree of hotness and coldness.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 5.
What are the units of temperature?
Answer:
Temperature is measured in degrees of Celsius, degrees of Fahrenheit or Kelvin using a thermometer.

Question 6.
What is the SI unit of temperature?
Answer:
The SI unit of temperature is kelvin (K).

Question 7.
Write the different modes of transfer of heat?
Answer:

  1. Conduction of heat
  2. Convection of heat
  3. Radiation of heat.

Question 8.
What is conduction?
Answer:
The process of transfer of heat from hotter to colder end through the conductor is called conduction.

Question 9.
What is convection of heat?
Answer:
The process of transfer of heat from source of heat to surface by the motion of par¬ticles is called “convection of heat”.

Question 10.
What is radiation of heat?
Answer:
The process of transfer of heat in the form of waves is called radiation.

Question 11.
What is thermal contact?
Answer:
The contact which transfers heat by any mode is called Thermal contact.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 12.
What is the condition for conduction of heat?
(OR)
When does conduction take place?
Answer:
Conduction takes place, only when the conductor is in touch (contact) with the source of heat. The contact which transfers heat by any mode is called Thermal contact. Conduction doesn’t take place without thermal contact.

Question 13.
What are called convectional currents?
Answer:
Heat is transferred by means of currents called convectional currents.

Question 14.
How does heat transfer in liquids and gases?
Answer:
In liquids and gases heat is transmitted by mode of convection of heat.

Question 15.
What is a medium?
Answer:
The material which helps in transfer of heat from one place to another is called a medium. Solid, liquid and gaseous substances act as medium for transfer of heat.

Question 16.
How does heat transfer if there is no medium?
Answer:
Heat transfer, in the form of waves from one place to another. It does not require any medium.

Question 17.
Write one application of radiation of heat.
(OR)
How does a thermal scanner work?
Answer:
The thermal scanner receives the heat in the form of radiation to measure our body temperature.

Question 18.
What happens if we heat and cool a metal?
Answer:
The metal expands on heating and contracts on cooling.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 19.
Which device is used to measure the temperature?
Answer:
Thermometers are used to measure temperature. .

Question 20.
Based on which principle a Thermometer does work?
Answer:
Thermometer works on the expansion of liquids (mercury).

Question 21.
Where the thermometers are used?
Answer:
People use thermometers in homes, hospitals, automobiles, industries and restaurants etc., to measure temperatures of different objects and substances.

Question 22.
Which is essential for our healthy living? (OR) Why should we protect environment?
Answer:
A clean environment is essential for heaithy living. So, we want a concrete plan to protect environment.

Question 23.
Why does alcohol use in thermometer? (OR)How can a alcohol thermometer mea-sures very low temperatures?
Answer:
The freezing point of alcohol is less than -100°C. So, it can be used to measure very low temperatures.

Question 24.
Write the different types of thermometers?
Answer:
The different types of thermometers are Clinical Thermometer, Laboratory Thermometer, Digital Thermometer and Six’s makimum and minimum thermometer.

Question 25.
Which instrument is used to measure the maximum (highest) and minimum (lowest) temperatures of a place in a day?
Answer:
Six’s maximum and minimum thermometer is one of the Meteorological Instruments used to measure maximum (highest) and minimum (lowest) temperatures of a place in a day.

Question 26.
What is the normal temperature of the human body?
Answer:
The normal temperature,of the human body is 37°C or 98.4°F.

Question 27.
Why does smoke and hot air move up?
Answer:
Smoke and hot air moves up because it expands and becomes lighter. . .

Question 28.
What is the use of ventilators in our house?
Answer:
Smoke and hot air moves up because it expands and becomes lighter through ventila¬tors and exhaust fans on the upper parts of the wall.

Question 29.
What happens to the air on heating?
Answer:
On heating, the air expands, occupying more space and becomes lighter.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 30.
What happens when air is compressed?
Answer:
The air pressure becomes more when air is compressed.

7th Class Science 9th Lesson Heat, Temperature and Climate Short Questions and Answers

Question 1.
When do we feel hot and cool?
Answer:

  1. The heat when heat energy flows from hotter body to our body. We feel hot. Here we gain heat energy from the hotter body.
  2. We feel cool, when heat energy flows from our body to colder body. Here, we lose heat energy.

Question 2.
How does heat flow? How do we determine its direction?
Answer:

  1. Heat flows from a body of high temperature to a body of low temperature.
  2. This direction is determined by temperature. ‘

Question 3.
How do we written degree of Celsius, degree of Fahrenheit and Kelvin?
Answer:

  1. Degree Celsius: Celsius is written as °C and read as degree Celsius.
  2. Degree Fahrenheit: Fahrenheit is written as °F and read as degree fahrenheit.
  3. Kelvin is written as K and read as Kelvin.

Question 4.
What is Air pressure? How does it measure? What are the effects of the air pres-sure?
Answer:
Air pressure:
The force applied by air on any surface in contact is called “air pressure”.

Measuring of air pressure:
Air pressure is measured in height of mercury level in centimeters, and it is measured with a barometer. Now a days aneroid barometers are used to measure air pressure.

Effects of air pressure:
The air pressure becomes more when it is compressed. When air expands and rises up it creates low pressure, which drives the air high pressure from surrounding to move and occupy that place.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 5.
Why are cooking utensils made of metals while their handles are made of plastic or wood?
Answer:

  1. Some materials allow heat through them, this property is called conductivity.
  2. We use metals to make cooking vessels because they allow heat through them.
  3. We use material which does not allow heat to pass through as handles.
  4. So, cooking utensils made of metals while their handles are made of plastic or wood.

Question 6.
Write four applications of expansion of metals on heating.
Answer:

  1. The electric power lines (wires) are held loose on poles
  2. Rollers are kept under the beams of metal bridges
  3. Mercury is used in thermometers
  4. Small gaps are left between rails in railway tracks

Question 7.
Write the different modes of transfer of heat?
Answer:
Transfer of heat is in different types of materials. Heat gets transferred in three different modes, They are

  1. Conduction of heat
  2. Convection of heat
  3. Radiation of heat

1) Conduction of heat:
This process of transfer of heat from hotter to colder end through the conductor is called conduction. This mode of transfer of heat happens more in solid conductors.

2) Convection of heat:
This process of transfer of heat from source of heat to surface by the motion of particles is called “convection of heat”. In liquids and gases heat is transmitted by mode of convection of heat.

3) Radiation of heat:
This process of transfer of heat in the form of waves is called radiation. Radiation does not need any material medium. Sun’s heat transfers to earth in the form of radiation.

Question 8.
Write the differences between good conductors and poor conductors?
Answer:

Good conductorsPoor conductors
1) The materials which allow heat to pass through them are called conductors of heat.1) The materials which do not allow heat to pass through them easily are poor conductors of heat.
2) Example: Aluminum, iron and copper etc.2) Example : Water, air, clothes, glass, cork, plastic, wood etc.
3) Heat can pass.3) Heat cannot pass.
4) These are used to prepare cooking cooking vessels.4) These are used to prepare handles of vessels.

Question 9.
How does a thermos flask work?
(OR)
Describe working of a thermos flask.
Answer:

  1. Thermos flaSk has a two layered glass container and the air between these layers is removed to create vacuum.
  2. The inner silver coating protects the contents (tea, coffee, milk) poured in the flask from losing heat through radiation.
  3. As there is no medium between the walls of the flask. Neither conduction nor con-vection of heat takes place.
  4. As a result, heat is not transferred outside the flask so it is retained inside the flask for a few hours.

Question 10.
What is the reason for in the expansion of rails, mercury in thermometer, air inside the puri etc?
Answer:

  1. This is due to increase in the energy of particles present in these substances.
  2. Particles of substances occupy more space when they get heated.
  3. This is the reason for in the expansion of rails, mercury in thermometer, air inside the puri etc.

Question 11.
How does hot air balloon work?
Answer:

  1. Warm air is lighter than cold air.
  2. This property of air is used in flying of hot air balloons,
  3. It contains a bag which is called as envelope, and it is filled with heated air.
  4. There is a basket under the envelope which carries passengers and a source of heat.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 12.
Why mercury is used in thermometers?
Answer:
Mercury is used in thermometers, because it has the best properties to measure temperature.
Those are

  1. It’s expansion is uniform.
  2. It is bpaque and shining.
  3. It does not stick to the sides of glass tube.
  4. It is a good conductor of heat.
  5. It has a high boiling point (357°C) and a low freezing point (-39°C). Hence a wide range of temperatures can be measured using a mercury.

Question 13.
Why alcohol is used in thermometers?
Answer:
Alcohol is used in thermometers, because it has the best properties to measure temperature. Those are

  1. The freezing point of alcohol is less than -100°C.
  2. So, it can be used to measure very low temperatures.
  3. It’s expansion per degree Celsius rise in temperature is very large.
  4. It can be colored brightly and hence is easily visible.

Question 14.
What are the melting point of ice and boiling point of water in different scales?
Answer:
1) Celsius scale :
Melting point of ice is 0°C and boiling point of water is 100°C

2) Fahrenheit scale :
Melting point of ice is 32°F and boiling point of water is 212°F

3) Kelvin scale :
Melting point bf ice is 273K and boiling point of water is 373K

Question 15.
Write the Formulas for temperature conversion 1) Celsius to Fahrenheit 2) Celsius to Kelvin.
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 1

Question 16.
Explain two applications of transfer of heat.
Answer:
Thermal scanning:
The thermal scanner receives the heat in the form of radiation to measure our body temperature.

Thermos flask:
It has a two layered glass container and the air between these layers is removed to create vacuum. The inner silver coating protects the contents (tea, coffee, milk) poured in the flask from losing heat through radiation. As there is no medium between the walls of the flask. Neither conduction nor convection of heat takes place. As a result, heat is not transferred outside the flask so it is retained inside the flask for a few hours.

Question 17.
Fill in the table with suitable answers.
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 2
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 3

Question 18.
What is the use of a Clinical Thermometer? How it convenient to record the reading?
Answer:

  1. Clinical Thermometer is used in hospitals to measure the temperature of the human body.
  2. It has a kink that prevents the mercury from flowing back into the bulb when it was taken out of the patient’s mouth.
  3. This helps to record the temperature conveniently.

Question 19.
Where a Laboratory Thermometer is used? How it is better than the clinical ther-mometer?
(OR)
Describes laboratory Thermometer.
Answer:

  1. Laboratory Thermometer is used in school labs, industries etc. to measure temperature. It has no kink.
  2. It is made of a long stem and the bulb of mercury, so it can measure higher tempera-tures than clinical thermometer.

Question 20.
How does a Digital Thermometer work?
Answer:

  1. Digital Thermometer works without Mercury.
  2. It has a display which shows readings directly.
  3. It is very easy to use it, we can use it by just pressing on/off button present on it.
  4. Even children can also handle this one.

Question 21.
How do you use a clinical thermometer and a laboratory thermometer?
Answer:
Clinical thermometer

  1. Wash the clinical thermometer properly with an antiseptic solution.
  2. To lower the mercury level, hold the thermometer firmly and give some jerks.
  3. Ensure that it falls below 35°C. Now place the bulb of the thermometer under your tongue.
  4. After one or two minutes, take the thermometer out and note the reading.
  5. This is your body temperature.
  6. Don’t hold the thermometer by the bulb while reading it.

A Laboratory thermometer:

  1. Take cold or hot water in a bowl. Place the mercury bulb of the thermometer in the water so that the bulb immerses in it.
  2. Wait for some time till the mercury level shows a constant reading.
  3. Note down that reading.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 22.
How does sunstroke occur?
(OR)
What happens when humidity increases in the air?
Answer:

  1. Evaporation of sweat from our body makes us cool to maintain our body temperature.
  2. In summer, the humidity of air is high. Due to high humidity and temperature, it becomes difficult to evaporate the sweat from our body to cool It down.
  3. But still our body losses water.
  4. High temperatures, along with humidity sometimes may cause heat stroke or sun¬stroke.

Question 23.
What is the first aid for sunstroke?
Answer:
First Aid for sunstroke :

  1. Call 108 immediately. Move the victim to a cool ventilated place.
  2. Then loosen his dress and remove extra clothing. Put a cold wet towel on his head, back of the neck, on the grain and under the armpits.
  3. Fan air over the patient while wetting their skin with water from a sponge or wet cloth. Give plenty of cool, lightly salted water to drink often but in small sips.
  4. Help the person stay calm without any anxiety.

Question 24.
What is weather of a place? What are its measuring components?
Answer:

  1. The day-to-day variations in the components like temperature, humidity, rainfall, wind speed are called weather.
  2. Maximum and minimum temperature of a day, air pressure, rainfall, wind speed and humidity are called measuring components of weather.

Question 25.
How do Meteorologists decide the climate of the place?
Answer:

  1. Meteorologists (scientists who study and work on weather) record weather every day. .
  2. These records of the weather have been preserved for the past several decades.
  3. They help us to determine the weather pattern of a place.
  4. The average weather pattern taken over a long period, say 25 years or more, is called the climate of the place.

Question 26.
Why does climate affect pur life style? What are its measuring components?
Answer:

  1. Climate of a place remains unchanged for a long period of time. So, it affects our lifestyle.
  2. The measuring components of climate are maximum and minimum temperature of a day, air pressure, rainfall, wind speed and humidity.

Question 27.
What are the differences between weather and climate?
Answer:

WeatherClimate
Weather keeps on changing rapidly.Climate of a place remains unchanged for a long period of time.
It affects our daily life.It affects our life style
Atmospheric conditions in a speciiic area and time.Atmospheric conditions of a place for a long period of time.
It is not constant.It remains constant for 25 years or more.

Question 28.
How weather and climate is effected our life?
Answer:

  1. Daily weather and long-term climatic conditions influence the life style of humans and other living things directly and indirectly.
  2. Some of them are diet, clothing, housing, occupations etc.

Question 29.
Doctor put the thermometer in the mouth of Karthikeya’s sister, Neha. By seeing this Karthikeya got some doubts and asked doctor. What could be those doubts?
Answer:

  1. What is in the thermometer?
  2. How does it show the temperature?
  3. How can you say Neha got the fever?
  4. How much time shall it place in the mouth?

Question 30.
Guess what will happen if we use the clinical thermometer to measure the temperature of the ice cubes or boiling water?
Answer:
We cannot measure too low and too high temperatures by using a clinical thermometer. It is made for clinical use only, so by using this we can measure from 35°C to 42°C only. If we measure temperature of boiling water or ice cubes it will break or it does not work.

Question 31.
Guess what will happen if a low air pressure is formed in your area?
Answer:
If a low air pressure is formed in our area, air blow from high air pressure area to occupy the low air pressure area. Sometimes this wind causes damage to roofs, trees etc.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 32.
Kalyan said that the climate was changed by being seen the dark clouds in the sky. Is the word climate correct in his statement?
Answer:
No. climate is different from weather. The changes occur during the period of some hours is called changing in weather. Climate is average weather conditions of a long period. So he had to use weather instead of climate in his statement.

Question 33.
What will happen, if there is no kink in the clinical thermometer?
Answer:
The kink prevents the mercury from flowing back into the bulb when it was taken out of the patient’s mouth. This helps to record the temperature conveniently. If there is no kink the level of the mercury will fall down in the gap of taking out from the patient and seeing the reading. So, it may show incorrect reading.

Question 34.
What is your observation in the activity of metal spoon with fixed pins and a candle and write the reason.
Answer:
I will observe the dropping of pins one after another from the flame end of the spoon. This is due to the transfer of heat from the end kept in the flame (hotter end) towards your hand (colder end) through a spoon (metal).

Question 35.
What are your observations in the activity of expansion of metal on heating?
Answer:
I will observe movement in the straw. The needle rolls on the second block of wood because of the expansion of the cycle spoke. If we remove the candles the needle rolls back causing movement in the straw opposite to the previous motion.

Question 36.
Read the graph and answer the following questions.
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 4
1. Which component was shown in the graph?
Answer:
Temperature

2. On which day the maximum temperatures recorded?
Answer:
On the dates, 1-12-2020 and 2-12-2020 the maximum temperature was recorded.

Question 37.
Take a glass and keep a postcard oh it. Wave your notebook above the post card to displace the air just above the postcard. What is your observation? What is the reason for it?
Answer:
The moving air creates low pressure. Hence the postcard lifts up due to the higher pressure on the card from air inside the glass.

Question 38.
Observe the figure and give answers to the following questions.
1. What mode of transfer of heat is observed in the spoon?
2. What is the source of heat?
3. What will happen after some time?
4. What is the medium here?
AP Board 7th Class Science Solutions 9th Lesson Heat, Temperature and Climate 9
Answer:
1. Conduction of heat is observed in the spoon
2. Candle is the source of heat.
3. All the pins will fall down.
4. Metal in the spoon is the medium (solid)

Question 39.
Fill in the table with suitable answers.

Component in the weatherUnitsDevice to measure
TemperatureKelvin or °C or °F
Air pressureMercury level in centimeters
MillimetersRain gauge
HumidityGrams per cubic meter.

Answer:

Component in the weatherUnitsDevice to measure
TemperatureKelvin or C or °FThermometer
Air pressureMercury level in centimetersBarometer
Rain fallMillimetersRain gauge.
HumidityGrams per cubic meter.Hygrometer

Question 40.
Fill in the table with suitable answers.

PhenomenonExamples
Convection1) Milk 2)…………………. 3) …………………….
Conduction1) Metal spoon 2)………. 3) ………………
Radiation1) Sun to earth 2)………. 3) ……………….
Insulators1) Rubber 2)…………….. 3) ………………
Conductors1) Iron spoke 2)…………. 3) ………………

Answer:

PhenomenonExamples
Convection1) Milk 2) Eater 3)oil
Conduction1) Metal spoon 2)copper vessel 3)nail
Radiation1) Sun to earth 2) flask inner to outer 3) sun to moon
Insulators1) Rubber 2) glass 3) water
Conductors1) Iron spoke 2) vessels 3) copper wire

Question 41.
Draw the diagram of mercury laboratory thermometer.
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 5

Question 42.
Draw the diagram of alcohol laboratory thermometer.
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 6

Question 43.
Draw the diagram of a thermos flask and label its inner parts.
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 7

44. Draw the diagram showing the arrangement of apparatus to show the expansion of liquids on heating?
Answer:
AP Board 7th Class Science Solutions 9th Lesson Heat, Temperature and Climate 12

45. Draw the diagram of Six minimum and maximum temperatures and label the parts.
Answer:
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 8

7th Class Science 9th Lesson Heat, Temperature and Climate Long Questions and Answers

Question 1.
Compare different modes of transfer of heat.
Answer:

ConductionConvectionRadiation
1) The process of transfer of heat from hotter to colder end through the conductor is called conduction of heat.The process of transfer of heat from source of heat to surface by the motion of particles is called convection of heat.The process of transfer of heat in the form of waves is called radiation.
2) In solids heat is transmitted by mode of conduction of heat.In liquids and gases heat is transmitted by mode of convection of heat.Radiation does not need any material medium
3) Ex: metalsEx: water, milkEx : Sun’s heat to earth
4) No movement in the particles.Movement in the particles.No movement in the particles
5) It needs thermal contact.It does not need thermal contact.It does not need thermal contact.

Question 2.
Write about different types of thermometers.
Answer:
Clinical thermometer:

  1. Wash the clinical thermometer properly with an antiseptic solution.
  2. To lower the mercury level, hold the thermometer firmly and give some jerks.
  3. Ensure that it falls below 35°C. Now place the bulb of the thermometer under your tongue.
  4. After one or two minutes, take the thermometer out and note the reading.
  5. This is your body temperature.
  6. Don’t hold the thermometer by the bulb while reading it.

Laboratory thermometer:

  1. Take cold or hot water in a bowl. Place the mercury bulb of the thermometer in the water so that the bulb immerses in it.
  2. Wait for some time till the mercury level shows a constant reading.
  3. Note down that reading.

Digital Thermometer:

  1. It works without mercury.
  2. It has a display which shows readings directly.
  3. It is very easy to use it.
  4. We can use it by just pressing on/off button present on it.
  5. Even children can also handle this one.

Six’s maximum and minimum thermometer:

  1. It is one of the Meteorological Instruments used to measure maximum (highest) and minimum (lowest) temperatures of a place in a day.
  2. It has a cylindrical ‘Bulb A’, and ‘Bulb B’ connected through a ‘U-shaped tube’ con¬taining mercury.
  3. Bulb A contains alcohol, and bulb B contains alcohol and its vapours.
  4. When the temperature increases, the alcohol in the bulb A expands and pushes the mercury in the U tube, this makes indicator(M) to move up.
  5. This indicates the maximum temperature of the day.
  6. When the temperature decreases, alcohol in the bulb A contracts and pulls the mercury back.
  7. This makes indicator (N) to move up.
  8. This indicates the minimum temperature of the day.
  9. After taking readings the indicators M and N are brought to their original places by using a magnet.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

Question 3.
What is humidity? How does it measure? How does it affect or body? What is the first aid of sunstroke?
Answer:
HumidityAll the evaporated water from different water bodies go into air. This water vapour present in the air is called humidity.

Measurement of humidity:
Hygrometer is used to measure humidity in air and it is expressed in grams per cubic meter.

Effects of humidity:
Evaporation of sweat from our body makes us cool to maintain our body temperature. In summer, the humidity of air is high. Due to high humidity and temperature, it becomes difficult to evaporate the sweat from our body to cool it down. But, still our body losses water. High temperatures, along with humidity sometimes may cause heat stroke or sunstroke.

First Aid for sunstroke:
Call 108 immediately. Move the person out of the heat. Fan while spraying water to cool him. Let the person drink water to rehydrate.

Question 4.
Read the following table and answer the questions.
AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 9
i) At what temperature, the ice melts? Express in Celsius.
ii) Which measurements are given in the table?
iii) Convert, 100°C into Fahrenheit and kelvin scale.
iv) What are the SI units of temperature?
Answer:
i) 0°C .
ii) Measurements of temperature are given.
iii) 100°C =212°F =373K
iv) Kelvin

Question 5.
How do you appreciate the weather reports?
Answer:

  1. The day-to-day variations in the components like temperature, humidity, rainfall, , wind speed are called weather.
  2. It keeps on changing and changes are very fast too.
  3. One day it may be dry and sunny and the next day it may rain.
  4. It gives the information about atmospheric conditions in a specific area and time.
  5. We can get details of weather from weather reports and you can see these symbols on television, newspapers and in weather forecasting.
  6. Weather forecasting is the application of science and technology to predict the conditions of the atmosphere using components of weather for a given location and time.
  7. Weather reports are very useful to farmers, fishermen etc.
  8. So I appreciate the weather reports.

Question 6.
How do you appreciate the properties of the mercury and alcohol as they are used in the thermometers?
Answer:
Properties of mercury:

  1. Its expansion is uniform.
  2. It is opaque and shining.
  3. It does not stick to the sides of glass tube.
  4. It is a good conductor of heat.
  5. It has a high boiling point (357°C) and a low freezing point (-39°C).

Hence a wide range of temperatures can be measured using mercury.

Properties of alcohol:

  1. The freezing point of alcohol is less than -100°C. So, it can be used to measure very low temperatures.
  2. It’s expansion per degree Celsius rise in temperature is very large.
  3. It can be coloured brightly and hence is easily visible.

Hence a wide range of temperatures can be measured using alcohol. Clinical ther-mometers, laboratory thermometers, six minimum and maximum thermometers etc. are made with alcohol or mercury or both.

Hence I appreciate the properties of the mercury and alcohol as they are used in the thermometers

AP Board 7th Class Science 9th Lesson 1 Mark Bits Questions and Answers Heat, Temperature and Climate

I. Multiple Choice Questions

Choose the correct answer and write its letters in the brackets.

1. It is a form of energy that flows from a hotter body to a cooler body.
A) Heat
B) Light
C) Sound
D) Electricity
Answer:
A) Heat

2. When heat energy flows from our body to Lassi. Here,
A) Lassi loss Heat energy
B) Body lose heat energy
C) Body gain heat energy
D) A and C
Answer:
B) Body lose heat energy

3. Joules are units of
A) Humidity
B) Climate
C) Heat
D) Air pressure
Answer:
C) Heat

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

4. The degree of hotness or coldness is called ………..
A) Humidity
B) Temperature
C) Heat
D) Air pressure
Answer:
D) Air pressure

5. Temperature is measured in…….
A) Degrees of Celsius
B) Degrees of Fahrenheit
C) Kelvin
D) All
Answer:
D) All

6. The SI unit of temperature is ………….
A) Degrees of Celsius
B) Degrees of Fahrenheit
C) Kelvin
D) All
Answer:
C) Kelvin

7. Kelvin is written as
A) °C
B) °F
C) °K
D) K
Answer:
D) K

8. Which is correct?
i) Heat is the degrees of hotness or coldness.
ii) Heat is measured in joules
A) i only
B) ii only
C) both i & ii
D) both are incorrect
Answer:
B) ii only

9. Cooking utensils are made of ………..
A) conductors
B) insulators
C) fuels
D) none of the above
Answer:
A) conductors

10. Handles of cooking vessels are made of …………….
A) conductors
B) insulators
C) fuels
D) none of the above
Answer:
B) insulators

11. A: Metals are used to make cooking vessels.
R: Metals allow heat through them.
Which is correct?
A) A & R are correct and R supports A
B) A & R are correct but R does not support A
C) A is correct but R is wrong
D) R is Correct but A is wrong
Answer:
A) A & R are correct and R supports A

12. The ability of a material to conduct heat is called …………..
A) Thermal conductivity
B) Thermal resistivity
C) Thermometer
D) None
Answer:
A) Thermal conductivity

13. It is not a good conductor.
A) Copper
B) Steel
C) Plastic
D) Cast iron
Answer:
C) Plastic

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

14. Insulator is….
A) Water
B) Glass
C) Plastic
D) All
Answer:
D) All

15. Heat cannot pass in the mode of
A) Conduction
B) Convection of heat
C) Radiation of heat
D) None
Answer:
D) None

16. In which heat is transferred from one end to another end by the mode of conduction.
A) Steel spoon
B) Water
C) A and B
D) Sun to earth
Answer:
A) Steel spoon

17. This mode of transfer of heat happens more in solid conductors
A) Conduction
B) Convection of heat
C) Radiation of heat
D) None
Answer:
A) Conduction

18. This process of transfer of heat from source of heat to surface by the motion of particles
A) Conduction
B) Convection of heat
C) Radiation of heat
D) all
Answer:
B) Convection of heat

19. This is not a medium of heat energy.
A) Solid
B) Liquid
C) Gas
D) None
Answer:
D) None

20. This mode of transfer of heat doesn’t require any media.
A) Conduction
B) Convection of heat
C) Radiation of heat
D) All
Answer:
C) Radiation of heat

21. The process of transfer of heat in the form of waves is called
A) Conduction
B) Convection of heat
C) Radiation of heat
D) All
Answer:
C) Radiation of heat

22. Which instrument can control the transfer of heat (loss of heat)?
A) Thermos flask
B) Thermometer
C) Thermal scanner
D) All
Answer:
A) Thermos flask

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

23. The flask retains hot inside the flask for
A) a few hours
B) ever
C) never
D) a few weeks.
Answer:
A) a few hours

24. Thermos flask was invented by
A) Sir James cook
B) Sir James Chadwick
C) Sir James Dewar
D) Sir James watts
Answer:
C) Sir James Dewar

25. Which of the following expands on heating and contracts on cooling?
A) Solid
B) Liquid
C) Gas
D) All
Answer:
D) All

26. Air on cooling
A) contracts
B) occupy less space.
C) expand
D) A and B
Answer:
D) A and B

27. Warm air is lighten than cold air. This property of air is used in
A) hot air balloons
B) kites
C) rockets
D) all
Answer:
A) hot air balloons

28. Thermometer contains bulb of
A) mercury
B) alcohol
C) A or B
D) silver
Answer:
C) A or B

29. Mercury is a in room temperature.
A) solid
B) liquid
C) gas
D) all
Answer:
A) solid

30. This is not a property of the mercury .
A) Its expansion is uniform.
B) It is a good conductor of heat.
C) It easily sticks to the sides of glass tube.
D) It has a high boiling point.
Answer:
C) It easily sticks to the sides of glass tube.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

31. The boiling point and a freezing point of the mercury are
A) -39°C and 357°C
B) 357°C and-39°C
C)100°C and 0°C
D) none
Answer:
B) 357°C and-39°C

32. A: The freezing point of alcohol is more than -100°C.
R: Alcohol can be used to measure very low temperatures.
A) A & R are correct and R supports A
B) A & R are correct but R does not support A
C) A is correct but R is wrong
D) R is correct but A is wrong
Answer:
D) R is correct but A is wrong

33. Number of divisions in Fahrenheit scale is
A) 100
B) 212
C) 180
D) 32
Answer:
C) 180

34. Number of divisions in Kelvin scale is
A) 100
B) 212
C) 180
D) 32
Answer:
A) 100

35. Number of divisions in Celsius scale is
A) 100
B) 212
C) 180
D) 32
Answer:
A) 100

36. Which prevents the mercury from flowing back into the bulb in clinical thermometer.
A) Capillary
B) Kink
C) Bulb
D) Magnet
Answer:
B) Kink

37. Laboratory thermometer can measure higher temperatures than clinical thermometer because, it has
A) long bulb
B) long stem
C) short bulb
D) short stem
Answer:
B) long stem

38. This thermometer can work without Mercury.
A) Six max and min thermometer
B) Digital Thermometer
C) Clinical thermometer
D)All
Answer:
B) Digital Thermometer

39. Who invented Six’s maximum and minimum thermometer in 1780?
A) Sir James cook
B) James watt
C) Sir James Dewar
D) James Six
Answer:
D) James Six

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

40. The correct order is
i) After one or two minutes, take the thermometer out and note the reading,
ii) Place the bulb of the thermometer under your friend’s tongue.
iii) Wash the clinical thermometer properly with an antiseptic solution.
iv) To lower the mercury level, hold the thermometer firmly and give some jerks.
A) iv, ii, i, iii
B) iv, iii, ii, i
C) iii, iv, ii, i
D) ii, iii, iv, i
Answer:
C) iii, iv, ii, i

41. The normal temperature of the human body is
A) 37°C
B) 98.4°F
C) 310K
D) all
Answer:
D) all

42. Ventilators are working on
A) Air expands on heating
B) Air contracts on heating
C) Metal expands on heating
D) Air contracts on cooling
Answer:
A) Air expands on heating

43. Which is used to measure the air pressure?
A) Hygrometer
B) Thermometer
C) Barometer
D) Rain gauge
Answer:
C) Barometer

44. A: If roofs were weak, they could be lifted and blown away.
R: The moving air creates high pressure.
A) A & R are correct and R supports A
B) A & R are correct but R does not support A
C) A is correct but R is wrong
D) R is correct but A is wrong
Answer:
C) A is correct but R is wrong

45. Rain fall is measured with a
A) hygrometer
B) thermometer
C) barometer
D) rain gauge
Answer:
D) rain gauge

46. This is not a measuring component of weather
A) humidity
B) temperature
C) wind speed
D) none
Answer:
D) none

47. Humidity is measured in
A) g/cubic meter
B) mm/cubic meter
C) °C
D) m/°C
Answer:
A) g/cubic meter

48. In summer, the humidity of air is
A) low
B) high
C) no change
D) zero
Answer:
B) high

49. High temperatures, along with humidity sometimes may cause
A) kwashiorkor
B) beriberi
C) corona
D) sunstroke
Answer:
D) sunstroke

50. It gives the information about atmospheric conditions in a specific area and time.
A) Weather
B) Climate
C) A and B
D) None
Answer:
A) Weather

51. This deals with a long period.
A) Weather
B) Climate
C) A and B
D) None
Answer:
B) Climate

52. It affects our lifestyle
A) Weather
B) Climate
C) A and B
D) None
Answer:
B) Climate

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

53. Which phenomenon measures atmospheric conditions in a specific area and time?
A) Weather
B) Climate
C) A and B
D) None
Answer:
A) Weather

II. Fill in the blanks

1. Heat is a form of energy that flows from a ………………. body to a ………………. body.
2. Heat is measured in ………………. with calorimeter
3. Heat flows from a body of high temperature to a body of low temperature. This direction is determined by ………………. .
4. The degree of ………………. is called ‘temperature’.
5. Degree of Celsius is written as ………………. .
6. Degree of Fahrenheit is written as ………………. .
7. Some materials allow heat through them, this property is called ………………. .
8. Water, air, clothes, glass, cork, plastic, wood etc. are some examples of ………………. .
9. This process of transfer of heat from hotter to colder end through the conductor’ is called ………………. .
10. The contact which transfers heat by any mode is called ………………. .
11. Water is a ………………. of heat.
12. Heat is transferred by means of ………………. called convectional currents.
13. The materials which help in transfer of heat from one place to another are called ………………. .
14. Sun’s heat transfers to earth in the form of ………………. .
15. The thermal scanner receives the heat in the form of ………………. to measure our body temperature.
16. The inner ………………. in thermos flask protects the contents (tea, coffee, milk) poured in the flask from losing heat through radiation.
17. As there is ………………. between the walls of the flask; neither conduction nor convection of heat takes place.
18. Particles of substances occupies ………………. space when they get heated.
19. Small gaps left between rails in railway tracks, because metal ………………. on heat
20. ………………. are used to measure temperature.
21. The principle involved in working of a thermometer is ………………. .
22. The melting point of ice is ………………. °C
23. The melting point of ice and boiling point of water in Fahrenheit are ………………. .
24. The melting point of ice and boiling point of water in SI units are ………………. .
25. The formula to convert Celsius to Kelvin is ………………. .
26. The formula to convert Celsius to Fahrenheit is ………………. .
27. ………………. is used in hospitals to measure the temperature of the human body.
28. Clinical Thermometer has a ………………. that prevents the mercury from flowing back into the bulb when it was taken out of the patient’s mouth.
29. After taking readings in six max and min thermometer the indicators I<sub>2</sub> and I<sub>1</sub> are brought to their original places by using ………………. .
30. The normal temperature of the human body is ………………. .
31. Smoke and hot air moves up because it ………………. on heating and becomes lighter.
32. The force applied by air oh any surface in contact is called ………………. .
33. The air pressure becomes ………………. when it is compressed.
34. When air expands and raises up it creates ………………. .
35. ………………. drives the air high pressure from surrounding to move and occupy that place.
36. Air pressure is measured in height of ………………. level in centimeters
37. ………………. is measured in millimeters by using a rain gauge.
38. The water vapour present in the air is called ………………. .
39. ………………. is used to measure humidity in air and it is expressed in grams per cubic meter.
40. Evaporation of ………………. from our body makes us cool to maintain our body temperature.
41. The day-to-day variations in the components like temperature, humidity, rainfall, wind speed are called ………………. .
42. ………………. forecasting is the application of science and technology to predict the conditions of the atmosphere using components of weather for a given location and time.
43. It is easier to know about weather with the help of ………………. than tables.
44. ………………. study and work on weather and record weather every day.
45. The average weather pattern taken over a long period, say 25 years or more, is called the ………………. of the place.
46. Expansion of IMD is ………………. .
47. The abnormal variation in the components of climate is called ………………. .
Answer:

  1. hotter, cooler
  2. Joules or calories
  3. temperature
  4. hotness or coldness
  5. °C
  6. °F
  7. conductivity
  8. insulators
  9. conduction
  10. Thermal contact
  11. poor conductor
  12. currents
  13. medium
  14. radiation
  15. radiation
  16. silver coating
  17. no medium or vacuum
  18. more
  19. expand
  20. Thermometers
  21. expansion of liquids on heating
  22. 0
  23. 32°F, 212°F
  24. 273K, 373K
  25. K °C + 273
  26. AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 10
  27. Clinical Thermometer
  28. kink
  29. a magnet
  30. 37°C or 98.4°F
  31. expands
  32. air pressure
  33. more
  34. low pressure
  35. Low pressure
  36. mercury
  37. Rainfall
  38. Humidity
  39. Hygrometer
  40. Sweat
  41. weather
  42. Weather
  43. graphs
  44. Meteorologists
  45. Climate
  46. Indian Meteorological Department
  47. climate change

III. Match the following

1.

Group – AGroup – B
1) heata) cm of mercury level
2) temperatureb) Kelvin
3) air pressurec) mm
4) rain falld) Joule

Answer:

Group – AGroup – B
1) heatd) Joule
2) temperatureb) Kelvin
3) air pressurea) cm of mercury level
4) rain fallc) mm

2.

Group – AGroup – B
1) humiditya) barometer
2) temperatureb) hygrometer
3) air pressurec) rain gauge
4) rainfalld) thermometer

Answer:

Group – AGroup – B
1) humidityb) hygrometer
2) temperatured) thermometer
3) air pressurea) barometer
4) rainfallc) rain gauge

3.

Group – AGroup – B
1) watera) radiation
2) heatb) convection
3) metalc) expansion
4) vacuumd) conduction

Answer:

Group – AGroup – B
1) waterb) convection
2) heatc) expansion
3) metald) conduction
4) vacuuma) radiation

4.

Group – AGroup – B
1) naila) clinical thermometer
2) clothb) lab thermometer
3) mercuryc) conductor
4) alcohold) insulator

Answer:

Group – AGroup – B
1) nailc) conductor
2) clothd) insulator
3) mercurya) clinical thermometer
4) alcoholb) lab thermometer

5.

Group – AGroup – B
1) weathera) low pressure
2) climateb) long period
3) windc) water vapour
4) humidityd) rapid change

Answer:

Group – AGroup – B
1) weatherd) rapid change
2) climateb) long period
3) winda) low pressure
4) humidityc) water vapour

6.

Group – AGroup – B
1) gas expansiona) alcohol thermometer
2) liquid expansionb) parachute
3) solid expansionc) railway rails
4) gas contractiond) LPG cylinder

Answer:

Group – AGroup – B
1) gas expansionb) parachute
2) liquid expansiona) alcohol thermometer
3) solid expansionc) railway rails
4) gas contractiond) LPG cylinder

7.

Group – AGroup – B
1) Kelvina) °F
2) Fahrenheitb) K
3) Celsiusc) J
4) Heatd) °c

Answer:

Group – AGroup – B
1) Kelvinb) K
2) Fahrenheita) °F
3) Celsiusd) °c
4) Heatc) J

Do You Know?

→ Heat is measured In Joules or Calories with calorimeter.

→ Warm air is lighter than cold air. This property of air is used in flying of hot air balloons. It contains a bag which is called as envelope, and it is filled with heated air. There is a basket under the envelope which carries passengers and a source of heat.

→ Formulas for temperature conversion

1) Celsius to Fahrenheit AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate 10
2) Celsius to Kelvin C=K – 273

→ Now a days cameras have been developed to detect heat. Those are called thermal- cameras.

AP 7th Class Science Important Questions 9th Lesson Heat, Temperature and Climate

→ The Indian Meteorological Department (IMD) studies climate of our country. 23rd March is celebrated as World Meteorological Day.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Andhra Pradesh BIEAP AP Inter 1st Year Botany Study Material 4th Lesson Plant Kingdom Textbook Questions and Answers.

AP Inter 1st Year Botany Study Material 4th Lesson Plant Kingdom

Very Short Answer Questions

Question 1.
What is the basis for the classification of Algae?
Answer:
The basis for the classification of Algae is pigmentation and type of stored food.

Question 2.
When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Answer:
In liver worts, reduction division occurs in sporophyte as a result spores are produced in the capsule.

In Mpsses, reduction division occurs in sporophyte in spore mother cells.

In ferns :
Reduction division occurs in Macro and Micro sporangia to produce Macrospores and Microspores.

In Gymnosperms – Reduction division occurs in Microsporangia and Megasporangia.

In Angiosperms – Reduction division occurs in Microspore Mother cells (Anther), Megaspore Mother cell (ovule).

Question 3.
Differentiate between syngamy and triple fusion.
Answer:

SyngamyTriple fusion
1. One of the Male gametes released in the embryosac fuses with the egg to form a zygote. This is called syngamy.1. In this, the 2nd male gamete fuses with the diploid secondary nucleus to produce primary endosperm nucleus. This is Triple fusion.
2. It was discovered by strasberger.2. It was discovered by Nawaschin.

Question 4.
Differentiate between antheridium and archegonium.
Answer:

AntheridiumArchegonium
1. It is the male sex organ.1. It is the female sex organ.
2. It is club shaped.2. It is flask shaped.
3. It produces biflagellete Antherozoids.3. It produces a single egg.

Question 5.
What are the two stages found in the gametophyte of mosses? Mention the structure from which these two stages develop?
Answer:
The gametophyte of Mosses consists of two stages namely
a) Juvenile stage, the Protonema and
b) Adult leafy stage, gametophore.

Protonema is developed directly from spore. Gametophore is developed from the protonema as a lateral adventitious bud.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 6.
Name the stored food materials found in Phaeophyceae and Rhodophyceae.
Answer:
In Phaeophyceae, the stored food material is Laminarin or Mannitol. The stored food material in Rhodophyceae is floridean starch.

Question 7.
Name the pigments responsible for brown colour of phaeophyceae and red colour of Rhodophyceae.
Answer:
“Fucoxanthin” pigment is responsible for brown colour of phaeophyceae and “r-phycoerythrin” is responsible for red colour of Rhodophyceae.

Question 8.
Name different methods of vegetative reproduction in Bryophytes. [A.P. Mar. 15]
Answer:
In Bryophytes vegetative Reproduction takes place by fragmentation or by Gemmae, or by budding in secondary protonema.

Question 9.
Name the integumehted megasporangium found in Gymnosperms. How many femals gametophytes are generally formed inside the megasporangium?
Answer:
The Integumented Megasporangium found in Gymnosperms is ovule. One multicellular female gametophyte develops inside the megasporangium which bears two or more archegonia.

Question 10.
Name the Gymnosperms which contain mycorrhiza and coralloid roots respectively.
Answer:
The Gynnosperm which contain Mycorrhiza is pinus, and which contain corralloid roots is cycas.

Question 11.
Mention the ploidy of any four of the following.
a. Protonemal cell of a moss.
b. Primary endosperm nucleus in a dicot.
c. Leaf cell of a moss.
d. Prothallus of a fern,
e. Gemma cell in Marchantia
f. Meristem cell of monocot
g. Ovum of a liverwort and
h. Zygote of a fern.
Answer:
a) Haploid
b) Triploid
c) Haploid
d) Haploid
e) Haploid
f) Diploid
g) Haploid
h) Diploid

Question 12.
Name the four classes of pteridophyta with one example each.
Answer:
The four classes of pteridophyta are :
i) Psilopsida Ex : Psilotum
ii) Lycopsida Ex : Lycopodium
iii) Sphenopsida Ex : Equisetum
iv) Pteropsida Ex : Pteris

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 13.
What are the first organisms to colonise rocks? Give the generic name of the moss which provides peat?
Answer:
The first organisms to colonise rocks are Mosses along with Lichens. Generic name is sphagnum.

Question 14.
Mention the fern characters found in Cycas.
Answer:
Some of the fern characters are :

  1. Circinate vernation of young leaves.
  2. Presence of ramenta.
  3. Multiciliated Male gametes.
  4. Presence of Archegonia.

Question 15.
Why are Bryophytes called the amphibians of the plant Kingdom?
Answer:
They live in moist soil and they depend on water for sexual reproduction. So they are called amphibians of plant Kingdom.

Question 16.
Name an algae which show
a) Haplo – diplontic and b) Diplontic types of life cycles.
Algae which show Haplo – diplontic life cycle is Ectocarpus. Algae that show Diplontic life cycle is Fucus.

Question 17.
Give examples for unicellular, colonial and filamentous algae.
Example are volvox, spirogyra and chara. These are the members of chlorophyceae.

Short Answer Type Questions

Question 1.
Differentiate between red algae and brown algae. [A.P. May. 18, Mar. 14]
Answer:

Red algaeBrown algae
1. Red algae belong to the class Rhodophyceae.1. Brown algae belong to the class Phaeo-phyceae.
2. Majority of them are marine and some are fresh water forms.2. They live in fresh waters, brackish and salt waters.
3. The thallus of Red algae are multicellular.3. The thallus range from simple branched filamentous form to profusely branched forms.
4. The major pigments are Chlorophyll a, d and r-phycoerythrin.4. The major pigments are chlorophyll a, c, carotexnoids and Xanthophylls (Fuco xanthin)
5. Flagella are absent.5. Flagella are 2, unequal lateral.
6. Cell wall is made up of Cellulose, Pectin and Polysulphate esters.6. Cell wall is made up of Cellulose and algin.
7. Food materials are stored in the form of floridean starch.7. Food materials are stored in the form of Mannitol and Laminarin.
8. Asdxual Reporduction is by non-motile spores.8. Asexual Reproduction is by biflagellate zoospores.
9. Sexual reproduction is by non-motile gametes.9. Sexual Reproduction is by motile gametes.
10. Red algae.
Ex : Polysiphonria, Porphyra Gracilaria, Gelidium.
10. Brown algae.
Ex : Ectocarpus, Laminaria, Sargassum, Focus.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 2.
Differentiate between liverworts and mosses.
Answer:

LiverwortsMosses
1. Plant body in liverworts is thalloid which is prostrate dorsiventral and closely appressed to the substrate.1. In Mosses the adult stage gametophore consists of upright, slender axis with spirally arranged leaves which gets attached to substratum by Rhizoids.
2. Antheridia (Male) and Archegonia (Female) are sex organs produced on the same or on different thalli.2. Male and female sex organs are produced at the apex of the leafy shoots.
3. Paraphyses are absent.3. Paraphyses are present.
4. Vegetative Reproduction is by fragmentation or by Gemmae.4. Vegetative Reproduction is by fragmen-tation or by Gemmae or by budding on the secondary protonema.
5. The sporophyte is small or reduced.5. The sporophyte in mosses is more elaborate.
6. Elaters are present in the capsule which help in spore dispersal.6. Peristomial teeth are present in capsule which help in spore dispersal.
7. Spores germinate to form free living gameto- phytes.7. Spores germinate to form creeping, green, branched protonema.
8. Ex : Marchantia.8. Ex : Funaria.

Question 3.
What is meant by Homosporous and Heterosporous pteridaphytes? Give two examples. [T.S. May. 18 A.P. Mar. 18, 15, 13]
Answer:
Pteridophytes which produce similar type of spores are called Homosporous pteridophytes.
Ex : Lycopodium, Pteris.

Pteridophytes which produce two types of spores are called Heterosporous Pteriodophytes.
Ex : Selaginella, Saivinia.

Question 4.
What is Heterospory? Briefly comment on its significance. Give two examples. [T.S. Mar, 15]
Answer:
Heterospory refers to the production of different types of spores.

Significance :

  1. Microspores formed from Microspore mother cells are small with 0.015 – 0.05 mp. Megaspores formed from Megaspore Mother cell are big and are with 1.5 mp.
  2. Microspores develop into Male gametophytes and Megaspores develop into female gametophytes which lead to unisexuality.
  3. The female gametophytes are retained on the parent sporophyte for variable periods.
  4. The development of zygote into young embryos takes place within the female gametophytes.
  5. The female gametophyte is with abundant food materials.
    Ex : Selagenella, Saivinia.

Question 5.
Write a note on economic importance of Algae and Bryophytes.
Answer:
Importance of Algae :

  1. At least a half of the carbon dioxide fixation on earth is carried out by Algae through photosynthesis and increases the level of oxygen in the environment.
  2. They are paramount importance as primary producers of energy rich compounds which form the basis of the food cycles of aquatic animals.
  3. Many species of Porphyra, Laminaria and sargassum are used as food.
  4. Some marine Brown and red algae produce large amounts of hydro carbons.
    Ex : A/gin and Carrageen.
  5. Iodine is extracted from kelps like Laminaria.
  6. Chlorella and Spirullina are used as food supplements even by space travellers.

Economic importance of Bryophytes :

  1. Some mosses provide food for herbaceous mammals, birds and other animals.
  2. Species of Sphagnum, a moss provide peat used as fuel and because of its capacity to hold water as packing material for trans – shipment of living material.
  3. Mosses along with lichens are the first organisms to Colonise rocks.
  4. They play significant role in plant succession.
  5. Mosses form dense mats on the soil, thus they reduce the impact of falling rain and prevent soil erosion.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 6.
How would you distinguish Monocots from Dicots.
Answer:

MonocotsDicots
1. Monocots contain only one cotyledon in their seeds.1. Dicots contain two cotyledons in their seeds.
2. Fibrous root system is present.2. Tap root system is present
3. Parallel venation is seen in leaves.3. Reticulate venation is seen in leaves.
4. Endosperm is absent.4. Endosperm is present.
5. During germination, seed produce only one leaf.5. During germination, the seed produce two leaves.

Question 7.
Give a brief account of Prothallus.
Answer:
In Pteridophytes, the spores germinate to give rise to Inconspicuous, small but multicellular free living, photosynthetic thalloid gametophytes called Prothalli. They require cool, damp, shady places to grow. Because of this specific requirement and water for fertilization, the spread of living pteridophytes is limited and restricted to narrow geographical regions. The gametophytes bear male and female sex organs, called antheridia and archegonia respectively. The sex organs are multicellular, jacketed and sessile.

Question 8.
Draw labelled diagrams of :
a) Female thallus and Male thallus of a liverwort.
b) Gametophyte and sporophyte of funaria.
Answer:
AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom 1

Long Answer Type Questions

Question 1.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Answer:
Bryophytes, Pteridophytes and Gymnosperms are the three ‘groups of plants that bear archegonia. The plant body in bryophytes is haploid. It produces gametes hence it is called gametophyte. The sex organs are multicellular, Jacketed and stalked. Antheredium, which is the male sex organ produces biflagellate antherozoids. Archegonium, the female sex organ, which is flask shaped, produces a small egg.

The antherozoids are released into water where they come in contact with archegonium. One antherozoid fuses with the egg to produce the zygote. This is called zooidogamous oogamy. Zygote produce a multicellular body called sporophyte. It is attached to the photosynthetic gametophyte and extracts nourishment from it.

Some cells of the sporophyte, called spore mother cells undergo reduction division to produce Haploid spores. These spores germinate to produce gametophyte. Bryophytes show alternation of generations (because garhetophytic and sporophytic bodies are different) and life cycle is Haplo-diplontic type.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 2.
Describe the important characteristics of Gymnosperms.
Answer:

  1. Gymnosperms are embryophytic, trachaeophytic, archegoniate phanerogams.
  2. They include medium size trees or tall trees and shrubs.
  3. The root system is tap root system. In some genera roots have fungal association in the form of mycorrhiza (Pinus) and in some Cycas roots have coralloid roots which are associated with Nitrogen fixing Cyanobacteria (Nostbc and Anabaena)
  4. The stems are unbranched (Cycas) or branched (Pinus Cedrus).
  5. The leaves may be simple or compound. In Cycas, the pinnate leaves persists for a years in Cycas.
  6. Anatomically the stem shows eustele. The vascular bundles are conjoint, collateral and open.
  7. Vessels are generally absent in xylem and companion cells are absent in phloem.
  8. Secondary growth occurs in stem and roots.
  9. The Gymnosperms are heterosporous, they produce haploid microspores and megaspores.
  10. The two types of spores are produced in sporangia that are borne on sporophylls which are arranged spirally an aixs to form compact strobili.
  11. The male strobili consists of microsporophylls and Microsporangia which produce Micro-spores.
  12. Microspores or pollen grains develop into Male gametophyte.
  13. The strobili bearing Megasporophylls with ovules are called female strobili.
  14. Micro and Megasporphylls may be borne on the same tree (Pinus) or on different trees (Cycas).
  15. Megaspore develop into female gametophyte. The pollination is direct and anemophilous.
  16. Gymnosperms are divided into three classes namely Cycadopsida, Coniferopsida and Gnetopsida.

Question 3.
Give the sailent features of Pteridophytes.
Answer:

  1. Pteridophytes are used for medicinal purposes and as soil-binders.
  2. These are the first terrestrial plants to possess vascular tissues.
  3. They are embryophytic, archegoniate vascular cryptogams.
  4. They prefer cool, damp and shady places.
  5. The plant body is a sporophyte which is differentiated into true roots, stem and leaves.
  6. The root system is adventitious.
  7. The stele may be protostele or siphonostele or solenostele or’dictyostele.
  8. The leaves are small (selagenella) or large as in ferns.
  9. The sporophytels bear sporangia that are subtended by leaf like sporophylls.
  10. Most of the pteridophytes are homosporous but selaginella and Salvinia shows Heterosporous.
  11. The spores germinate to give rise to Prothalli.
  12. The gametophytes bear male and female sex organs called antheridia and archegonia.
  13. The sex organs are multicellular, jacketed and sessile.
  14. Fusion of Male gamete with the egg present in the archegonium results in the formation of zygote.
  15. Zygote develops into young embryo which produces a multicellular sporophyte.

Question 4.
Give an account of plant life Cycles and alternation of Generations.
Answer:
In plant, both haploid and diploid cells can divide by mitosis which leads to the formation of different plant bodies, haploid and diploid. The haploid plant body produces gametes by mitosis and is called gametophyte. It is followed by fertilization, which results in the formation of zygote which also divides by mitosis to form diploid sporophytic plant body. In this, Meiosis occurs, results in the formation of spores. These spores again divide by mitosis to form a Haploid plant body. Thus during life cycle there is a alternation of generations between gamete producing haploid gametophyte and spore producing diploid sporophyte.

Different plants show different life cycles. For Ex :
1) Many Algae such as Vo/vox, Spirogyra and some species of Chlamydomomas shows Haplontic life cycle. In this, zygote represents the sporophytic stage which divides by meiosis results in the formation of Haploid spores. These spores divide mitotically and form the gametophyte.

2) In some species, the diploid sporophyte is the dominant photosynthetic. Independent phase of the paint. The haploid phase is represented by gametes only. So this lifecycle is called diplontic type. In some pteridophytes, the gametophyte is represented by few celled stage so called diplo-haplontic type. Other exmples are polysiphonia.

3) In Bryophytes, both phases are multicellular with dominant gametophytic phases and dependent sporophytic phage. So this life cycle is called haplo-diplontic type. Other examples for this are Ectocarpus, Laminaria.

AP Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Question 5.
Both Gymnospoerms and Angiosperms bear seeds then why are they classified separately?
Answer:

GymnospermsAngiosperms
1. Herbs are absent. Mainly trees.1. Most of the Angiosperms are herbs.
2. Reproductive parts are cones.2. Reproductive parts are flowers.
3. Cones are unisexual.3. Flowers are uni or Bisexual.
4. Ovules are Naked. So called naked seeded plants.4. Ovules are hidden within the Ovary. Seeds are present in the fruit.
5. Pollen grains reach the ovules directly.5. Pollen grains reach the ‘stigma’.
6. Male gametophyte consists of prothallial cells.6. Prothallial cells are absent.
7. Male gametes are Multiciliated.7. Cilia are absent on Male gametes.
8. Fertilization occurs only once.8. Fertilization occurs twice.
9. Archegonia are present.9. Archegonia are absent.
10. Female gametophyte acts on endosperm formed before fertilization which is Haploid.10. Endosperm is formed after fertilization, and is Triploid.
11. During Embryogenesis, free nuclear divisions occurs.11. Free nuclear divisions are absent.
12. Xylem vessels and companion cells are absent.12. Xylem vessels and Companion Cells are present.
13. Vegetative reproduction is rare.13. Vegetative reproduction is common.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

These AP 7th Class Science Important Questions 8th Lesson Wonders of Light will help students prepare well for the exams.

AP Board 7th Class Science 8th Lesson Important Questions and Answers Wonders of Light

Question 1.
What is the source of light? Give examples.
Answer:
Light comes from different objects called sources, of light. Ex: Sun, candle, tube light.

Question 2.
What are Parallel beams of light rays?
Answer:
Light rays which travel parallel to each other are called Parallel beams of light rays.

Question 3.
What are converging beam of light, rays?
Answer:
Light rays which travel from different directions to meet at a point are called as con¬verging beam of light rays.

Question 4.
What is reflection of light?
Answer:
The process of bouncing back of light rays into the same medium after falling on a smooth or rough surface from the light source is called “reflection of light”.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 5.
How objects are visible?
Answer:
Objects are visible only when light falls on the objects and bounces back to the eye.

Question 6.
What is angle of incidence?
Answer:
The angle made by incident ray with the normal is called angle of incidence (i).

Question7.
What is angle of reflection?
Answer:
The angle made by the reflected ray with the normal is called the angle of reflection (r).

Question 8.
Write the first law of reflection.
Answer:
The angle of incidence is always equal to angle of reflection.
Angle of incidence = Angle of reflection
This is the first law of reflection.

Question 9.
Write the second law of reflection.
Answer:
The incident ray reflected ray and normal to the surface are present in the same plane and incident, reflected rays are on either side of normal. This is the second law of reflection.

Question 10.
How is the object distance and image distance of a plane mirror?
Answer:
The distance of the object from the mirror is equal to the distance of image from the mirror.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 11.
What is object distance?
Answer:
The distance of the object from the mirror is called object distance.

Question 12.
What is image distance?
Answer:
The distance of the image from the mirror is called image distance.

Question 13.
We are able to see our image in the mirror. Can we get pur image on screen?
Answer:
We cannot get the image formed by a plane mirror on the screen.

Question 14.
What is a real image?
Answer:
The image which can get on a screen is called real image.

Question 15.
What is a virtual image?
A. The image which cannot get it on screen is called virtual image.

Question 16.
How is the nature of the image formed by a plane mirror?
Answer:
The nature of the image formed by a plane mirror is virtual and erect image.

Question 17.
What is Periscope?
Answer:
Periscope is an instrument used in submarines to see the objects or persons above the water level.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 18.
What is the use of a Periscope?
Answer:
We can use this to see the objects outside the room through the window while hiding ourself in the room.

Question 19.
What is a convex mirror?
Answer:
The spherical mirror which has reflecting surface bent outward is called convex mirror.

Question 20.
What is a concave mirror?
Answer:
The spherical mirror which has reflecting surface bent inward is called concave mirror.

Question 21.
What is a lens?
Answer:
A piece of glass or any other transparent material with curved sides is called a lens.

Question 22.
What is a convex lens?
Answer:
Lens which is thick in center and thin at the edges is called convex lens.

Question 23.
What is a concave lens?
Answer:
Lens which is thin in the Centre and thick at the edges is called concave lens.

Question 24.
What basic principle is involved in a periscope?
Answer:
Periscope is prepared based on the principle of reflection of light from plane mirrors.

Question 25.
What are the uses of a concave mirror?
Answer:
Concave mirrors are used by dentists, ophthalmologists, ENT doctors and also in head lights of vehicles.

Question 26.
What are the uses of a convex mirror?
Answer:
Convex mirrors are used as rearview mirrors in vehicles and safety mirrors at curved roads.

Question 27.
How white light is composed?
Answer:
White light is composed of seven colours.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 28.
Why do dentists use a concave mirror?
Answer:
Dentists also use concave mirrors to get a bigger image of the teeth.

7th Class Science 8th Lesson Wonders of Light Short Questions and Answers

Question 1.
What is a beam of light? How many types of beam of lights are there?
Answer:
The bundle of light rays are called beam of light rays. These are three types.

  1. Parallel beam of light rays
  2. Converging beam of light rays
  3. Diverging beam of light rays.

Question 2.
Write the differences between natural and artificial light sources.
Answer:

Natural sources of lightArtificial sources of light
1. Objects that emit light on their own are known as natural sources of light.1. Objects that release light artificially are called man made sources of light or artificial sources of light.
2. Sources like sun, stars are natural sources of light.2. Sources like bulb, torch light, candle artificial sources of light.
3. These emit light on their own.3. These don’t emit light on their own.

Question 3.
What is a ray of light? How does it represent?
Answer:

  1. The direction or path along which light travels is called a ray of light.
  2. It is denoted by a straight line with an AP 7th Class Science Important Questions Chapter 8 కాంతితో అద్భుతాలు 1 arrow mark.
  3. The straight line indicates the path of light and arrow mark indicates the direction of light from the source.

Question 4.
Differentiate between regular and irregular reflections.
Answer:

Regular reflectionIrregular reflection
1. Reflection from a smooth and shiny surface is called regular reflection.1. Reflection from an irregular or uneven surface is called irregular reflection or diffused reflection.
2. Clear images are formed in case of regular reflection.2. Images are not clear or sometimes cannot form the images at all in case of irregular reflection.
3. Plane mirrors, stable water surfaces, polished metals and stones can give regular reflections.3. Rough surfaces, unstable water surfaces, unpolished metals and stones can give irrregular reflections.

Question 5.
What are the characteristics of image by plane mirror?
Answer:
Characteristics of image by plane mirror:

  1. Object distance is equal to image distance.
  2. Size of the object is equal to size of the image.
  3. The image formed is always virtual and erect.
  4. Laterally inverted image is formed, (left and right alternates)

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 6.
When you are eating food with right hand it appears to be eating with left hand in a mirror. Why?
Answer:
When you are eating food with right hand it appears to be eating with left hand in a mirror. Such a shift of lateral side of images in opposite direction is called lateral inversion. Images formed by plane mirrors undergo lateral inversion.

Question 7.
Write the differences between real and virtual images.
Answer:

Real imagesVirtual images
1. The image which can get on a screen is called real image.1. The image which cannot get it on screen is called virtual image.
2. Real image is always inverted.2. Virtual image is always erected.
3. Concave mirrors form real images.3. Plane mirrors form virtual images.
4. We cannot see directly in the mirror.4. We can see directly in the mirror.

Question 8.
Find the number of images will be formed, if the two mirrors are kept at an angle 50° between them?
Answer:
If the two mirrors are kept at an angle 50s between them, then the number of images between them are,
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 1

Seven is the next integer after 6.2.
So, the number of images formed will be seven

Question 9.
Write the differences between a convex mirror and a concave mirror.
Answer:

Convex mirrorConcave mirror
1. The spherical mirror which has Reflecting surface bent outward is called convex mirror.1. The spherical mirror which has Reflecting surface bent inward is called concave mirror.
2. The light is diverged by the convex mirror.2. The light is converged by the concave mirror.
3. It is used in vehicles as rear view mirrors.3. It is used by dentists and ophthal-mologists.
4. A convex mirror always forms virtual, erect, smaller image irrespective of the position of the object.4. A concave mirror forms real and virtual images, erect and inverted images smaller, same size and bigger images depending on the position of object in front of it.

Question 10.
Write the characteristics of image formed by a concave mirror and a convex mirror.
Answer:
1) Characteristics of image formed by a concave mirror :
A concave mirror forms real and virtual images, erect and inverted images, smaller, same size and bigger images depending on the position of object infront of it.

2) Characteristics of image formed by a convex mirror :
A convex mirror always forms virtual, erect, smaller image irrespective of the position of the object.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 11.
How the ENT doctors are using a concave mirror?
Answer:

  1. Concave mirrors are used by ENT doctors as Head mirrors.
  2. They have this mirror strapped on their head.
  3. Light from the bulb is made to fall on the mirror gets reflect from it.
  4. This reflected light is focused into the throat or ear of the patient. So that the doctor can examine inner parts clearly.

Question 12.
How the ophthalmologists are using a concave mirror?
Answer:

  1. Ophthalmologists use a special instrument called Ophthalmoscope.
  2. It is fitted with a.concave mirror having a small hole near its center.
  3. The concave mirror helps the doctor to direct a beam of light into the patient’s eye and see his/her retina clearly through the hole in the mirror.

Question 13.
Why the surface of reflection in a torch light or a headlight of vehicle is made with a concave mirror, behind the bulb?
Answer:

  1. The surface of reflection in a torch light or a headlight of vehicle is made with a concave mirror, behind the bulb.
  2. Light released from the bulb falls on the concave surface.
  3. After reflection from concave surface, all the light rays travel parallel to each other.
  4. So that we can observe focused light at a distance.

Question 14.
Why do convex mirrors use beside the drivers in vehicles?
Answer:

  1. The convex mirrors which are used beside the drivers in vehicles are called rear view mirrors. .
  2. These mirrors form images of objects spread over a large area.
  3. So, these help the driver to see the traffic behind them.

Question 15.
Write the uses of convex mirrors.
Answer:

  1. Large convex mirrors kept at the corners of curved roads to avoid accidents.
  2. Convex mirrors form images of vehicles on both sides of curved road.
  3. These mirrors are also used at the junctions of roads.
  4. These mirrors used as rear view mirrors in vehicles.

Question 16.
Which colour of light is better to see? Why?
Answer:

  1. Yellow light is better to see.
  2. Yellow light has been proven effective in protecting retina than blue light which causes damage to retina.

Question 17.
What is the rule of 20 – 20 – 20?
Answer:
While watching TV or computer a simple rule 20 – 20 – 20 helps us in protecting eyesight. After 20 minutes of watching TV or computer screen, take 20 seconds break and watch the TV or computer around 20 feet away.

Question 18.
Akshaya got many doubts while her brother is burning a paper with a magnifier. What would be those doubts?
Answer:

  1. How does a magnifier burn the paper?
  2. Why are the sun rays passing through the magnifier, without reflecting back?
  3. What is the shape of the magnifier?
  4. What happens to the sun rays after passing through the magnifier?

Question 19.
Guess, what will happen if a concave mirror or a plane mirror is used as a rear view mirror?
Answer:

  1. Generally convex mirror is used in rear view mirror.
  2. Because these mirrors form images of objects spread over a large area.
  3. The images’formed by the convex mirror have the characteristics like erect, virtual and smaller images.
  4. So, these help the driver to see the traffic behind them.
  5. If we U9e a concave mirror instead of a convex mirror it does not form any images of the rear objects.
  6. If we use a plane mirror instead of a convex mirror we cannot see the images of the rear objects totally.

Question 20.
Ramesh is not able to identify different spherical mirrors. Explain him, the differ-ences between concave and convex mirrors by asking some questions.
Answer:

  1. How the shining surface of the mirrors are bent?
  2. What do we call the mirror, if it has reflecting surface bent inwards?
  3. What do we call the mirror, if it has reflecting surface bent outwards?
  4. How is your image in the both the mirrors?
  5. Which mirror forms smaller image?
  6. Which mirror can form your image on the wall?

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 21.
Guess, why does we cannot see our face on a wall clearly?
Answer:
Light rays coming from the face fall on the wall and suffer irregular reflection, because wall has rough surface. Irregular reflection causes formation of no image on the wall or sometimes not cleared images.

Question 22.
How do you show the parallel light by an activity?
Answer:

  1. Take a plank and cardboard.
  2. Make small slits on cardboard.
  3. Keep the cardboard on the plank perpendicular to it. Keep it in sunlight during afternoon.
  4. The light rays from the sun fall on the cardboard and passed through the slits. We can observe that the light rays are travelling parallel to each other.

Question 23.
How do make your own spherical mirrors?
Answer:

  1. Take a silver paper used for decoration.
  2. Paste the silver paper without folds on a postcard size chart paper.
  3. Allow it to dry by keeping it under some heavy books.
  4. Bend the sheet slightly forward as its shiny surface comes inwards.
  5. It works as a concave mirror.
  6. Now bend the sheet slightly backward as its shiny surface comes out wards.
  7. It works as a convex mirror.

Question 24.
Fill in the table with suitable answer.
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 2
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 3

Question 25.
How do you appreciate the role of the spherical mirrors in our daily life?
Answer:

  1. Concave and convex mirrors are spherical mirrors.
  2. Concave mirrors are used by dentists, ophthalmologists, ENT doctors and also in head lights of vehicles.
  3. Convex mirrors are used as rear view mirrors in vehicles and safety mirrors at curved roads.
  4. Hence, I appreciate the role of the spherical mirrors in our daily life.

Question 26.
What precautions do you take to watch a TV?
Answer:
While watching TV or computer a simple rule 20 – 20 – 20 helps us in protecting eyesight. After 20 minutes of watching TV or computer screen, take 20 seconds break and watch the TV or computer around 20 feet away.

Question 27.
Draw the diagrams of regular reflection and diffused reflection.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 1

Question 28.
Draw the diagram of a source of light.
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 4

Question 29.
Draw the diagram showing the reflection of an incident ray.
(OR)
Draw the diagram showing an incident ray, reflected ray and a normal.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 9

Question 30.
Draw the diagram of a periscope and lable its inner parts.
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 5

Question 31.
Draw the diagram of a convex mirror and concave mirror.
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 6

Question 32.
Draw the diagram of a light ray.
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 7

Question 33.
Draw the diagrams of a convex lens and a concave lens.
Answer:
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 8

Question 34.
Draw the diagram of arrangement of the apparatus in the activity of verifi¬cation of the laws of reflection.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 8

7th Class Science 8th Lesson Wonders of Light Long Questions and Answers

Question 1.
a) What is reflection of light?
b) How many types of reflection are there? What are they?
c) Write the laws of reflection.
Answer:
a) The process of bouncing back of light rays into the same medium after falling on a smooth or rough surface from the light source is called “reflection of light”.

b) Reflections are of two types :
1. Regular reflection
2. Diffused reflection.

Reflection from a smooth and shiny surface is called regular reflection. Reflection from an irregular or uneven surface is called irregular reflection or diffused reflection.

c) Laws of reflection :
There are two laws of reflection.
1. Angle of incidence is equal to angle of reflection.
2. The incident ray reflected ray and normal to the surface are present in the same plane. Incident and reflected rays are on either side of normal.

Question 2.
Draw the reflected ray to the given incident ray.
AP 7th Class Science Important Questions 8th Lesson Wonders of Light 9
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 9

Question 3.
Answer the following.
1) What is light?
2) What is source of light?
3) How do you classify the light sources?
4) What are the different types of light beams? Explain.
Answer:
1) Light is a form of energy.

2) Light energy releases from different objects called sources of light.

3) Sources which release light energy on their own are called natural sources.
Ex: sun, stars.
Sources which need the human involvement to release light energy are called manmade sources of light.
Ex: torch light, candle, bulb etc.

4) Actually, light is not a single ray, but a bundle of rays. This bundle of light rays are called beam of light rays. These are three types :

i. Parallel beam of light rays :
Light rays which travel parallel to each other are called Parallel beam of light rays.

ii. Converging beam of light rays :
“Light rays which travel from different directions to meet at a point are called as Converging beam of light rays”.

iii. Diverging beam of light rays :
“Light rays which travel from a source moving in different’ directions are called as Diverging beam of light rays.”

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

Question 4.
What precautions do you take while watching TV or modern gadgets? Why?
Answer:

  1. Light plays an important role in the sensation of vision.
  2. Now a day’s people, due to the usage of modern gadgets, like mobile phone, com-puters, televisions through which light enters the eyes.
  3. Watching TV too much or sitting very close to it may make your eyes-tired cause dryness of eyes and gradually lead to headache.
  4. While using computer or TV ensures that your room is well lighted.
  5. While watching TV or computer a simple rule 20 – 20 – 20 helps us in protecting eyesight.
  6. After 20 minutes of watching TV or computer screen, take 20 seconds break and watch the TV or computer around 20 feet away.
  7. Reduce blue light in cell phones and computers are switched on power saving mode during night.
  8. Yellow light has been proven effective in protecting retina than blue light which causes damage to retina.

Question 5.
How do you appreciate the role of the reflection of light in our daily life?
Answer:

  1. The process of bouncing back of light rays into the same medium after falling on a smooth or rough surface from the light source is called “reflection of light”.
  2. When light falls on an object it reflects back. Reflected ray when reaches to our eyes, causes sensation of vision.
  3. If there is no phenomenon of reflection, we cannot see the colourful world.
  4. Periscope is prepared based, on the principle of reflection of light from plane mirrors.
  5. Kaleidoscope is prepared based on the principle of multiple reflection of light from plane mirrors.
  6. The dentists, ophthalmologists, ENT doctors etc. give us treatment by using reflection of light through the concave mirrors.
  7. The drivers of the vehicles observe the rear objects of his vehicles by using reflection of light through the convex mirrors.
  8. Hence, I appreciate the role of the reflection of light in our daily life.

AP Board 7th Class Science 8th Lesson 1 Mark Bits Questions and Answers Wonders of Light

I. Multiple Choice Questions

Choose the correct answer and write its letters in the brackets.

1. This is not a natural sources of light.
A) sun
B) stars
C) moon
D) none
Answer:
C) moon

2. Example to human involvement is needed to release light
A) sun
B) stars
C) candle
D) none
Answer:
C) candle

3. Light rays which travel from a source moving in different directions are called as of light rays.
A) parallel
B) converge
C) diverge
D) B and C
Answer:
C) diverge

4. Which of the following does not give regular reflection?
A) mirror
B) cloth
C) new steel plate
D) polished marble
Answer:
B) cloth

5. In case of irregular reflection images cure
A) not clear
B) cannot form
C) clear
D) A & B
Answer:
D) A & B

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

6. These are in the same plane.
i. The incident ray
ii. normal to the surface
iii. The reflected ray
A) i,ii
B) i,iii
C) ii, iii
D) i,ii,iii
Answer:
D) i,ii,iii

7. Angle between normal and the incident ray
A) angle of incidence
B) angle of reflection
C) angle of vision
D) A & B
Answer:
A) angle of incidence

8. Which of the following is not a characteristic of the image of the plane mirror?
A) Object distance is equal to image distance.
B) Size of the object is equal to size of the image.
C) The image formed is always real and erect.
D) Laterally inverted image is formed.
Answer:
C) The image formed is always real and erect.

9. The distance of the object from the mirror is …………….. the distance of image from the mirror.
A) is equal to
B) is greater than to
C) is less than to
D) A & C
Answer:
A) is equal to

10. The size of image is not equal to that of the object in any situation.
A) plane mirror
B) convex mirror
C) concave mirror
D) none
Answer:
B) convex mirror

11. The distance of the image from the mirror is called
A) Image distance
B) Object distance
C) Normal
D) None
Answer:
A) Image distance

12. This type of image can catch on the screen
A) real
B) virtual
C) both
D) none
Answer:
A) real

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

13. Image formed by a plane mirror
A) erect & real
B) inverted & virtual
C) erect & virtual
D) inverted & real
Answer:
C) erect & virtual

14. No. of images formed when the angle between two plane mirrors is 60°
A) 4
B) 5
C) 6
D) 7
Answer:
B) 5

15. No. of images formed when the angle between two plane mirrors is 90°
A) 4
B) 2
C) 1
D) 3
Answer:
D) 3

16. No. of images formed when the angle between two plane mirrors is 0°
A) infinity
B) 5
C) 6
D) 0
Answer:
A) infinity

17. No. of images formed is 1, then the angle between two plane mirrors is
A) 0°
B) 90°
C) 180°
D) 360°
Answer:
C) 180°

18. The instrument which is used in submarines to see the objects or persons above the water level.
A) microscope
B) telescope
C) kaleidoscope
D) periscope
Answer:
D) periscope

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

19. Angle between two plane mirrors in periscope is
A) 0°
B) 90°
C) 180°
D) 360°
Answer:
A) 0°

20. The steel spoon can act as …
A) plane mirror
B) convex mirror
C) concave mirror
D) B & C
Answer:
D) B & C

21. The mirror forms all types of images
A) plane mirror
B) convex mirror
C) concave mirror
D) B & C
Answer:
C) concave mirror

22. The mirror is used as rear view mirror
A) plane mirror
B) convex mirror
C) concave mirror
D) B & C
Answer:
B) convex mirror

23. Which rule we should follow to protect eyes while watching TV or computer?
A) 20-20-20
B) 40-4-40
C) 10-10-10
D) 12-12-12
Answer:
A) 20-20-20

24. Which light is not dangerous to our eye?
A) yellow
B) red
C) blue
D) violet
Answer:
A) yellow

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

25. Which colour of light is composition of all colours?
A) yellow
B) red
C) blue
D) white
Answer:
D) white

II. Fill in the blanks

1. Light is a form of ………………. that helps us to see all the things around us.
2. Light comes from different objects called ………………. .
3. Objects that emit light are known as ………………. .
4. Sources that release light artificially [with human involvement] are called ………………. sources of light or artificial sources of light.
5. ………………. source emit light on their own.
6. The direction or path along which light travels is called ………………. .
7. The straight line indicates the of light and arrow mark indicates the ………………. of light from the source.
8. A bundle of light rays are called ………………. of light rays.
9. Light rays which travel parallel to each other are called ………………. of light rays.
10. Light rays which travel from different directions to meet at a point are called as ………………. of light rays.
11. Light rays which travel from a source moving in different directions are called as ………………. of light rays.
12. The process of bouncing back of light rays into the same medium after falling on a smooth or rotigh surface from the light source is called ………………. .
13. Objects are visible only when ………………. falls on the objects and bounces back to the eye.
14. Light rays from the sun fall on the mirror and bounced back and formed a spot of light on the wall. It is the image of ………………. .
15. The light rays that fall on the objects are called ………………. .
16. The light rays that bounce back from the objects are called ………………. .
17. Reflection from a ………………. surface is called regular reflection.
18. Reflection from an irregular or uneven surface is called ………………. reflection.
19. Clear images are formed in case of ………………. reflection.
20. Angle of ………………. is equal to Angle of reflection.
21. Incident and reflected rays are on either side of ………………. .
22. The angle made by incident ray with the normal is called ………………. .
23. The angle made by the reflected ray with the ………………. is called the angle of reflection.
24. The distance of the object from the mirror is called ………………. .
25. The distance of the image from the is called image distance.
26. When you are eating food with right hand it appears to be eating with left hand. This characteristic is called ………………. .
27. The image which cannot catch on the screen is called ………………. .
28. Plane mirror forms ………………. image.
29. The relation between number of images formed and the angle between two plane mirrors is ………………. .
30 . ………………. is an instrument used in submarines to see the objects or persons above the water level.
31. ………………. mirror reflect the light in different directions.
32. ………………. mirror reflect the light to one point.
33. A ………………. mirror forms real and virtual images, erect and inverted images, smaller, same size and bigger images depending on the position of object infront of it.
34. Ophthalmoscope is fitted with a ………………. mirror having a small hole near its center.
35. Dentists also use ………………. mirrors to get a bigger image of the teeth.
36. ………………. mirrors are used in head lights of the vehicles.
37. A piece of glass or any other transparent material with curved sides is called ………………. .
38. Lens which is thick in center and thin at the edges is called ………………. lens.
39. Lens which is thin in the centre and thick at the edges is called ………………. lens.
40. ………………. light has been proven effective in protecting retina than blue light which causes damage to retina.
41. Splitting of white light into seven colours is called ………………. .
Answer:

  1. energy
  2. sources of light
  3. sources of light
  4. man made
  5. natural
  6. a ray of light
  7. path, direction
  8. beam
  9. parallel beam
  10. Converging beam
  11. Diverging beam
  12. reflection of light
  13. Light
  14. sun
  15. incident rays
  16. reflected rays
  17. smooth and shiny
  18. irregular or diffused
  19. regular
  20. incidence
  21. normal
  22. angle of incidence
  23. normal
  24. object distance
  25. mirror
  26. lateral inversion
  27. virtual image
  28. virtual
  29. 360°/θ -1
  30. periscope
  31. convex
  32. concave
  33. concave
  34. concave
  35. concave
  36. concave
  37. a lens
  38. convex
  39. concave
  40. Yellow
  41. dispersion

III. Match the following

1.

Group – AGroup – B
1) Natural sourcea. cloth
2) Artificial sourceb. steel plate
3) Regular reflectionc. star
4) Irregular reflectiond. candle

Answer:

Group – AGroup – B
1) Natural sourcec. star
2) Artificial sourced. candle
3) Regular reflectionb. steel plate
4) Irregular reflectiona. cloth

2.

Group – AGroup – B
1) Convex mirrora. make up
2) Concave mirrorb. ENT doctor
3) Plane mirrorc. magnifier
4) Convex lensd. rear view

Answer:

Group – AGroup – B
1) Convex mirrord. rear view
2) Concave mirrorb. ENT doctor
3) Plane mirrora. make up
4) Convex lensc. magnifier

3.

Group – AGroup – B
1) Periscopea. concave mirror
2) Ophthalmoscopeb. convex mirror
3) Road safetyc. convex lens
4) Telescoped. plane mirror

Answer:

Group – AGroup – B
1) Periscoped. plane mirror
2) Ophthalmoscopea. concave mirror
3) Road safetyb. convex mirror
4) Telescopec. convex lens

4.

Group – AGroup – B
1) Convex mirrora. inwards curve
2) Concave mirrorb. outwards curve
3) Plane mirrorc. thick at middle
4) Convex lensd. plane surface
5) Concave lense. thin at middle

Answer:

Group – AGroup – B
1) Convex mirrorb. outwards curve
2) Concave mirrora. inwards curve
3) Plane mirrord. plane surface
4) Convex lensc. thick at middle
5) Concave lense. thin at middle

5.

Group – AGroup – B
1) Virtual imagea. pass through one point
2) Real imageb. bundle of rays
3) Rayc. caught on the screen
4) Beamd. seen in the mirror
5) Converginge. path of light

Answer:

Group – AGroup – B
1) Virtual imaged. seen in the mirror
2) Real imagec. caught on the screen
3) Raye. path of light
4) Beamb. bundle of rays
5) Converginga. pass through one point

6.

Group – AGroup – B
1) Incident anglea. perpendicular line to surface
2) Reflecting angleb. angle of incidence angle of reflection
3) Normalc. angle between normal and reflected ray
4) First law of reflectiond. normal, incident ray, reflected ray are in one plane
5) Second law of reflectione. angle between normal and incident ray

Answer:

Group – AGroup – B
1) Incident anglee. angle between normal and incident ray
2) Reflecting anglec. angle between normal and reflected ray
3) Normala. perpendicular line to surface
4) First law of reflectionb. angle of incidence angle of reflection
5) Second law of reflectiond. normal, incident ray, reflected ray are in one plane

7.

Group – AGroup – B
1) Convex mirrora. converges
2) Concave mirrorb. diverges
3) Plane mirrorc. parallel rays
4) Sund. irregular reflection
5) Thermocol sheete. regular reflection

Answer:

Group – AGroup – B
1) Convex mirrorb. diverges
2) Concave mirrora. converges
3) Plane mirrore. regular reflection
4) Sunc. parallel rays
5) Thermocol sheetd. irregular reflection

Do You Know?

→ We cannot see the real image with our naked eye. But we can catch it on screen. Whereas we can see the virtual image in the mirror with our naked eye. But we cannot catch it on screen.

AP 7th Class Science Important Questions 8th Lesson Wonders of Light

→ In olden days the concave mirrors can also be used as weapons. Archimedes, a Greek scientist used this mirror as weapon, 2000 years ago. When the Romans attacked Syracus, a coastal City – state in Greece, Archimedes arranged concave mirrors. The mirrors could be moved in any direction. They were positioned such that they reflected the sunlight on the Roman soldiers. The soldiers were dazzled by the sunlight. They did not know what was happening. Roman soldiers got confused and returned back.

→ Convex mirrors are used in Automatic Teller Machine (ATM) for security purpose to get a wider background look. This is to avoid others from over looking at your password.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

These AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants will help students prepare well for the exams.

AP Board 7th Class Science 7th Lesson Important Questions and Answers Reproduction in Plants

Question 1.
What is reproduction?
Answer:
The process of giving rise to new ones of their own kind is called Reproduction.

Question 2.
What are the two types of reproduction?
Answer:
Generally, plants reproduce in two ways. They are 1) Sexual Reproduction and 2) Asexual Reproduction.

Question 3.
What is Sexual reproduction?
Answer:
Reproduction in plants with seeds is called Sexual Reproduction.

Question 4.
What is Asexual reproduction?
Answer:
Reproduction in plants without seeds is called Asexual Reproduction.

Question 5.
Give examples for plants reproduce through seeds.
Answer:
Tamarind, drumsticks, mango, neem etc. reproduce through seeds.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 6.
Give examples for plants reproduce without seeds.
Answer:
Jasmine, banana, rose etc. reproduce without seeds.

Question 7.
Give examples for plants reproduce by both with seeds and without seeds.
Answer:
Curry leaf, Coriander etc. can reproduce by both means.

Question 8.
Mention some asexual reproduction methods.
Answer:
Binary fission, budding, fragmentation, spore formation etc.

Question 9.
What are incomplete flowers?
Answer:
Flowers which do not have any one of the four whorls are called Incomplete flowers.

Question 10.
Give examples for complete flowers.
Answer:
Hibiscus, Datura

Question 11.
Give examples for incomplete flowers.
Answer:
Pumpkin, Ridge guard

Question 12.
What are unisexual flowers?
Answer:
In some flowers either androecium or gynoecium is present. Such type of flowers are called unisexual flowers.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 13.
What are bisexual flowers?
Answer:
Generally, both androecium and gynoecium are present in the same flower in most of the plants. This type of flowers are called as bisexual flowers.

Question 14.
What are male flowers?
Answer:
Unisexual flowers that contain only androecium are called male flowers.

Question 15.
What are female flowers?
Answer:
Unisexual flowers that contain only gynoecium are called female flower.

Question 16.
What is the third whorl of a flower?
Answer:
The third whorl of a flower is androecium. This is the male reproductive part of the flower.

Question 17.
What is the 4th whorl of a flower?
Answer:
The fourth whorl of a flower is gynoecium. This is the female reproductive part of the flower.

Question 18.
What are the reproductive parts of a flower?
Answer:
Androecium and Gynoecium are the reproductive parts of a flower.

Question 19.
What is pollination?
Answer:
The process of transferring pollen grains from anther to stigma is called as Pollination.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 20.
What is self-pollination?
Answer:
If the pollen grains are transferred from the anther of a flower to stigma of the same flower, is known as SELF POLLINATION.

Question 21.
What is cross pollination?
Answer:
If the transfer of pollen grains takes place from the anther of one flower to the stigma of another flower, it is called CROSS POLLINATION.

Question 22.
Mention some pollinating agents.
Answer:
Insects, birds, animals, air and water are the pollinating agents.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 23.
Why do plants try to spread their seeds to distant places?
Answer:
All the plants try to spread their seeds to distant places to increase the chance of survival and propagation.

7th Class Science 7th Lesson Reproduction in Plants Short Questions and Answers

Question 1.
What is reproduction? How is it useful to the plants?
Answer:

  1. The process of giving rise to new ones of their own kind is called Reproduction.
  2. Reproduction helps to increase their number and continue their existence.
  3. All plants produce new once like them.
  4. There is no existence for plants without reproduction.

Question 2.
Write the differences between sexual and asexual reproductions in plants.
Answer:

Sexual reproductionAsexual reproduction
1. Reproduction in plants with seeds is called Sexual Reproduction.Reproduction in plants without seeds is called Asexual Reproduction.
2. Pollination takes placePollination does not takes place
3. Fertilization occursThere is no fertilization
4. Occurs in most of the plants like Datura, Mango etc.Occurs in few plants like Bryophyllum, banana etc.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 3.
Write the differences between calyx and corolla.
Answer:

CalyxCorolla
1. First whorl of the flower.Second whorl of the flower.
2. They consisting of sepals.They consisting of petals.
3. They are green in colour.They are in different colours in different plants.
4. They give protection to the inner parts during bud condition.They attract the insects and helps in pollination.

Question 4.
Write the differences between bisexual flowers and unisexual flowers?
Answer:

Bisexual flowersUnisexual flowers
Both androecium and gynoecium are present.Either androecium or gynoecium is present.
Generally, they have four whorls.Generally, they have only three whorls.
They are complete flowers.
Ex: Hibiscus, datura
They are incomplete flowers.
Ex : Ridge gourd, bottle gourd

Question 5.
Differentiate male and female flowers.
Answer:

  1. Unisexual flowers that contain only androecium are called male flowers.
  2. Unisexual flowers that contain only gynoecium are called female flowers.

Question 6.
How can you identify a female flower?
Answer:

  1. Female flowers are not difficult to identify.
  2. Notice the flowers on the variety of gourd creepers we grow around our home.
  3. Commonly below the petals of the flower are just a stalk found in the male flowers, while in the female flowers, a small berry-like form is seen below the flower which is ready to bear fruit in further days.
  4. By observing the presence of that berry-like form, we can identify the female flowers.

Question 7.
Write the differences between self-pollination and cross pollination.
Answer:

Self-pollinationCross pollination
If the pollen grains are transferred from the anther of a flower to stigma of the same flower, is known as SELF POLLINATION.If the transfer of pollen grains takes place from the anther of one flower to the stigma of another flower, it is called CROSS POLLI – NATION.
It is possible only in bisexual flowersIt happens in both unisexual and bisexual flowers.
It may happen even without blooming of the flowerIt happens only after blooming of the flower
Pollinating agents such as insects, air, water etc. may not necessary.Pollinating agents such as insects, air; water etc. are necessary.

Question 8.
How do insects and some birds help in pollination?
Answer:

  1. Insects like butterflies, honeybees, bumble bees, humming birds, bats, ants visit the flowers in search of nectar.
  2. When insects come in contact with a flower, the pollen grains stick to their legs and wings.
  3. When insects visit another flower, the pollen fall on its slimy stigma.
  4. Thus they help in pollination.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 9.
How do fertilization takes place in plants?
Answer:

  1. The pollen grains which fall on the stigma of a flower germinates.
  2. A pollen tube is formed from the pollen grains.
  3. The pollen tube travels from stigma to the ovules in the ovary.
  4. Fertilization and zygote formation occur in ovary.
  5. The zygote develops into an embryo.

Question 10.
What changes will come in the flower after fertilization?
Answer:

  1. After fertilization the ovary ripens and turns into fruit.
  2. The remaining floral parts will fall off.
  3. The ovules become the seeds.

Question 11.
What is seed dispersal? Mention different means of seed dispersal.
Answer:

  1. Spreading of seeds from one place to another place is called seed dispersal.
  2. Seed dispersal may happen through wind, water, animals, birds, man and even through some mechanical processes such as bursting of fruits.

7th Class Science 7th Lesson Reproduction in Plants Long Questions and Answers

Question 1.
What is vegetative propagation? Mention different vegetative propagation methods and their examples.
Answer:
Reproduction in plants through vegetative parts such as stem, root, and leaves is called vegetative propagation.

Vegetative propagation methodPart involvedExample
1. Through suckersStemBanana, Chrysanthemum
2. Through nodesStemMint plant, Sugarcane
3. Through eyes (buds)StemPotato
4. Through Stem cuttingStemSugarcane, Rose, Hibiscus
5. Through root cuttingRootCarrot, Beetroot
6. Leaf budsLeafBryophyllum( Ranapala )
7. LayeringStemJasmine, Jaji, Bougainvillea, Strawberry
8. GraftingStemMango, Rose

Question 2.
Explain the process of ground layering.
Answer:
AP 7th Class Science Important Questions Chapter 7 మొక్కలలో ప్రత్యుత్పత్తి 1
Ground Layering

  1. Layering is done with branches nearer to the ground.
  2. Remove the bark of the stem at one place.
  3. Bend the branch into the ground and cover it with soil.
  4. Within one month roots will develop from that branch.
  5. Separate it from the mother plant and plant it where ever you want.
  6. This method is useful to propagate Jasmine, Jaji, Bougainvillea, Strawberry etc. Plants with creeping stem.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 3.
What is grafting? Explain the method of grafting.
Answer:
Grafting is a technique where two plants are joined so as to continue their growth together. The upper part of the combined plant is called the scion, while the lower part is called the stock.
AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 1a

Procedure:

  1. Remove the bark of the Stock and Scion plants in such a way that removed portions face each other.
  2. The parts where bark is removed are joined tightly with the help of a jute twine.
  3. Wrap and tie with a polythene paper.
  4. After a month, cut the upper part of Stock and lower part in Scion.
  5. Within an another month the Scion attaches to the Stock and grows.
  6. Now remove the new branches which grow on Stock to facilitate the growth of Scion.

Question 4.
Describe the structure of a flower.
Answer:
Flower is the sexual part of a plant. The green part which connects the flower to stem is called the stalk. The stalk bears a slightly swollen head called Thalamus.

Over the thalamus floral parts are arranged in four whorls.

Calyx:
The green cup shaped leaf like, fused structures that covers the internal parts are Sepals. They are collectively called as Calyx. (Whorl 1)

Corolla:
The white or brightly coloured structures are called Petals. They are collec¬tively called as Corolla. (Whorl 2) .

Androecium:
The soft elongated structures attached to the petals are called Stamens. All the stamens are collectively called Androecium (Whorl 3). This is the male reproductive part of the flower. Each stamen has a swollen structure at the top called Anther.

Gynoecium:
A bulged structure seated on the thalamus is called Ovary which continues into a fine tube like structure Style with a slimy bead like tip called Stigma. These are collectively called the pistil Gynoecium (Whorl 4). This is the female reproductive part of the flower’
AP Board 7th Class Science Solutions 7th Lesson Reproduction in Plants 2

Question 5.
Write the differences between Androecium and Gynoecium.
Answer:

AndroeciumGynoecium
This is the third whorl of the flower.This is the fourth and inner most whorl of the flower.
This is male reproductive part of the flower.This is female reproductive part of the flower.
It consisting of stamens.It consisting of ovary, style and stigma.
Number of stamens are more.In general gynoecium consists of single ovary.
They produce pollen grains.It produce eggs or ova.
Pollen grains are released from stamens for pollinationFertilization occurs in ovary.
Generally they drop off after, fertilizationOvary develop into fruit after fertilization.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

Question 6.
Draw the different whorls of a flower.
Answer:
AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 2

Question 7.
Do you find any relationship between the weight of the seeds and the dispersal mechanism? Discuss with suitable examples.
Answer:

  1. I find relationship between the weight of the seeds and the dispersal mechanism.
  2. For example : Seeds of calotropis.
    a) These are light and have hairy structure at one end.
    b) So they travel with wind and settle at a suitable place to germinate.
    c) Milk weed, cotton seeds, etc. are all dispersed by wind.
  3. Another example : Coconut fruit
  4. a) The outer covering of the seed has empty spaces filled with air and are fibrous,
    b) The coconut fruit floats on water and reaches the ground some where. There it germinates. Seeds of lotus is another example.
  5. Another example : Neem fruit.
    a) Birds like Bulbuls, Mynahs, crows eat the outer fleshy part gets digested in the food canal and the seed coats of them become tender.
    b) They are then dispersed to other places as bird droppings.
  6. Similarly animals, Human beings carry the seeds to other places.

AP Board 7th Class Science 7th Lesson 1 Mark Bits Questions and Answers Reproduction in Plants

I. Multiple Choice Questions

Choose the correct answer and write its letters in the brackets.

1. Seeds are related to
A) stem
B) root
C) flower
D) leaf
Answer:
C) flower

2. Sugarcane propagates through
A) nodes
B) leaf
C) root
D)suckers
Answer:
A) nodes

3. Ground layering is seen in
A) Jasmine
B) Bougainvillea
C) Strawberry
D) All
Answer:
D) All

4. Which one is formed from layering branch?
A) Roots
B) Leaves
C) Nodes
D) Fruits
Answer:
A) Roots

5. The rooted plant in grafting is called as
A) Scion
B) Stock
C) Node
D) Cutting
Answer:
B) Stock

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

6. In grafting which have desirable characters
A) Scion
B) Stock
C) Root
D) Leaf
AnsA) Scionwer:

7. AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 3 This symbol represents
A) Female
B) Male
C) Both
D) Flower
Answer:
B) Male

8. AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 4 This symbol represents
A) Male
B) Female
C) Both
D) Pollination
Answer:
B) Female

9. AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 5 This diagram indicates
A) Pistil
B) Corolla
C) Stamens
D) Sepals
Answer:
A) Pistil

10. How many whorls that the complete flower has
A) 2
B) 3
C) 4
D) 5
Answer:
C) 4

11. Complete flowers also called as
A) Male flower
B) Female flower
C) Unisexual flower
D) Bisexual flower
Answer:
D) Bisexual flower

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

12. Ipomea have
A) Unisexual flower
B) Bisexual flower
C) Male flower
D) Female flower
Answer:
B) Bisexual flower

13. The zygote develops into
A) Embryo
B) Fruit
C) Flower
D) Stem
Answer:
A) Embryo

14. Which of the flower colour change after fertilization?
A) Mango
B) Rose
C) Cotton
D) Neam
Answer:
C) Cotton

15. Seeds compitect for
A) Soil
B) water
C) light
D) All
Answer:
D) All

16. Which process makes life immortal?
A) Reproduction
B) Digesion
C) Respiration
D) Circulation
Answer:
A) Reproduction

17. Transfer of pollen to stigma is
A) Pollination
B) Fertilization
C) Germination
D) Growth
Answer:
A) Pollination

18. Assertion (A) : A stamen has 2 parts.
Reason (R) : Filament and anther are parts of a stamen.
A) Both ’A’ and R’ are true and R’ is the correct explanation of ‘A’.
B) Both A’ and ‘R’ are true but ‘R’ is not the correct explanation of ‘A‘.
C) ‘A’ is true but R’ is false.
D) A’ is false but ‘R’ is true.
Answer:
A) Both ’A’ and R’ are true and R’ is the correct explanation of ‘A’.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

19. Assertion (A) : Parts of pistil are anther and filament.
Reason (R) : Bisexual flowers are those which contain both stamen ahd pistil.
A) Both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation of ‘A’.
B) Both ‘A’ and ‘R’ are true but ‘R’ is. not the correct explanation of ‘A’.
C) ‘A’ is true but ‘R’ is false.
D) ‘A’ is false but ‘R’ is true.
Answer:
D) ‘A’ is false but ‘R’ is true.

20. Assertion (A) : A flower may either have a male or a female part.
Reason (R) : A flower having both male and female parts is called bisexual flower.
A) Both A’ and ‘R’ are true and ‘R’ is the correct explanation of ‘A’.
B) Both ‘A’ and ‘R’ are true but ‘R’ is not the correct explanation of ‘A’.
C) ‘A’ is true but ‘R’ is false.
D) ‘A’ is false but ‘R’ is true.
Answer:
B) Both ‘A’ and ‘R’ are true but ‘R’ is not the correct explanation of ‘A’.

21. Statement (I): After fertilization the ovary becomes the fruit and ovules the seeds.
Statement (ii) : Study of pollen grains is called polynology.
A) Statement (i) is correct while statement (ii) is incorrect.
B) Statement (ii is correct while statement (i) is correct.
C) Both statements are correct.
D) Both statements are incorrect.
Answer:
C) Both statements are correct.

22. It is the seat on which the parts of a flower are present.
A) Corolla
B) Thalamus
C) Calyx
D) Androecium
Answer:
B) Thalamus

23. It has funnel shape.
A) Corolla
B) Sepals
C) Calyx
D) Thalamus
Answer:
A) Corolla

24. This is the male part of the flower.
A) Gynoecium
B) Androecium
C) Petals
D) Sepals
Answer:
B) Androecium

25. This is the female part of a flower.
A) Petals
B) Calyx
C) Gynoecium
D) Androecium
Answer:
C) Gynoecium

26. An example for unisexual flower.
A) Bittergourd
B) Datura
C) Ipomea
D) Hibiscus
Answer:
A) Bittergourd

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

27. This is a bisexual flower.
A) Papaya
B) Cucumber
C) Bottlegourd
D) Hibiscus
Answer:
D) Hibiscus

28. The fleshy base to which all the floral parts are attached to is
A) Pedicel
B) Sepals
C) Thalamus
D) Petals
Answer:
C) Thalamus

29. The pollen grains are found in
A) Ovule
B) Anthers
C) Ovary
D) Stigma
Answer:
B) Anthers

30. The number of stamens present in Datura :
A) 5
B) 3
C) 6
D) 10
Answer:
A) 5

31. Ovules develop into :
A) flowers
B) plants
C) leaf
D) seeds
Answer:
D) seeds

32. The pollen grains contain
A) male garnets
B) female garnets
C) filaments
D) Stigma
Answer:
A) male garnets

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

33. It is a …………… AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants 6
A) Unisexual flower
B) Bisexual flower
C) Female flower
D) Male flower
Answer:
C) Female flower

34. In which type of flower this part is present?
A) Male flower
B) Bisexual flower
C) Both A & B
D) None of these
Answer:
C) Both A & B

35. Identify the part of flower
A) Stamen
B) Pistil
C) Petals
D) Sepals
Answer:
A) Stamen

36. Identify the part of flower.
A) Stamen
B) Pistil
C) Petals
D) Sepals
Answer:
B) Pistil

37. It shows that
A) T.S of flower
B) L.S of flower
C) T.S of stem
D) T.S of root
Answer:
B) L.S of flower

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

38. Pollen grains are formed from
A) Stamen
B) Pistil
C) Ovary
D) Sepals
Answer:
A) Stamen

39. In the pollination what is transfer from anther to stigma?
A) Ovule
B) Zygote
C) Seed
D) Pollengrain
Answer:
D) Pollengrain

40. Agents of pollination
A) Birds
B)Insects
C) Air
D) All of these
Answer:
D) All of these

41. New plants grow from the following part of a plant is not a vegetative reproduction
A) Root
B) Stem
C) Leaf
D) Flower
Answer:
D) Flower

42. This develops as fruit.
A) Ovary
B) Ovule
C) Petals
D) Flower
Answer:
A) Ovary

43. Ovules develop into
A) fruit
B) seed
C) flower
D) plant
Answer:
B) seed

44. Dispersal of this seed is by water.
A) Coconut
B) Soap nut
C) Neem
D) Milk weed
Answer:
A) Coconut

45. Dispersal of calotropis is through
A) Water
B) Animals
C) Wind
D) Humans
Answer:
C) Wind

46. These seeds are dispersed by animals.
A) Fleshy fruits
B) Coconut
C) Neem
D) Balsam
Answer:
A) Fleshy fruits

47. Neem seed is dispersed to other places by
A) Animals
B) Human beings
C) Birds
D) Water
Answer:
C) Birds

48. Dispersal by bursting of these fruits is done.
A) Neem
B) Coconut
C) Fleshy fruits
D) Bhendi
Answer:
D) Bhendi

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

49. People transferred these seeds across the globe.
A) Coconut
B) Neem
C) Soap nut
D) Sugarcane
Answer:
D) Sugarcane

50. Among the following dispersal of seed occurs through wind in
A) Coconut
B) Lotus
C) Milkweed
D) Mango
Answer:
C) Milkweed

51. Seeds of Lotus travel by
A) Birds
B) Insects
C) Animals
D) Water
Answer:
D) Water

52. These are carried by birds
A) Maple
B) Milkweed
C) Lotus
D) Neem
Answer:
D) Neem

53. On drying the pod explodes releases seeds with great force in
A) Mustard
B) Sugarcane
C) Pears
D) Tomato
Answer:
A) Mustard

54. These seeds are dispersal through
A) Water
B) Wind
C) Animals
D) Birds
Answer:
B) Wind

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

55. These seeds are dispersal through
A) Wind
B) Water
C) Animals
D) Birds
Answer:
B) Water

56. Seeds of sweet and fleshy fruits can dispersal through
A) Human beings
B) Birds
C) Animals
D) All of these
Answer:
C) Animals

57. Seed dispersal is essential for survival of
A) Plants
B) Animals
C) Birds
D) Human beings
Answer:
A) Plants

58. Cotton seeds have the following structures
A) Wing like outer growths
B) Hooks
C) Hairy parts
D) Thorns
Answer:
C) Hairy parts

59. The following seeds dispersal through human beings
A) Bhendi
B) Coconut
C) Tomato
D) Cotton
Answer:
C) Tomato

60. Identify the wrong statement.
A) All the seeds of a fruit should be able to germinate
B) Some seeds germinate but plants die before maturation
C) Some seeds never germinate
D) All seeds do not germinate
Answer:
A) All the seeds of a fruit should be able to germinate

61. This seed is dispersed by
A) Water
B) Air
C) Animals
D) Bursting
Answer:
C) Animals

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

62. Dispersal of seeds by bursting of fruits happens in
A) Sugarcane
B) Coconut
C) Lotus
D) Ladies finger
Answer:
D) Ladies finger

II. Fill in the blanks.

1. The propagation of Banana is through ………………………. .
2. The propagation of sugarcane is done by planting ………………………. .
3. Mint is propagated through ………………………. .
4. In potatoes, ………………………. are used for propagation.
5. In onions ………………………. are used for propagation.
6. Example for rhizome is ………………………. .
7. Example for corm is ………………………. .
8. Example for stem cutting is ………………………. .
9. Examples for ground layering are ………………………. .
10. Artificial propagation method used for getting plants with desired characters is ………………………. .
11. Sepals of a flower collectively called as ………………………. .
12. Petals of a flower are are collectively known as ………………………. .
13. Calyx is present in ………………………. whorl.
14. Corolla is present in ………………………. whori.
15. The male reproductive part of the flower is ………………………. .
16. The female reproductive part of the flower is ………………………. .
17. Androecium consists of ………………………. .
18. Gynoecium consists of ………………………. .
19. Androecium is present in ………………………. whorl of the flower.
20. Gynoecium is present in ………………………. whorl of the flower.
21. The smooth yellow coloured powder in anther is called ………………………. .
22. Transferring pollen from anther to stigma of the same flower is
23. Transferring of pollen grains from anther of one flower to stigma of another flower is known as ……………….. .
24. When pollen grains fall on stigma is ………………………. formed.
25. The fusion of pollen grains with ovules in ovary is known as ………………………. .
26. Fertilization results in the formation of ………………………. .
27. After fertilization, ovary ripes and turn into a ………………………. .
28. After fertilization, ovules turn into ………………………. .
29. Seeds with wings are carried away by ………………………. .
30. Example for seed dispersal through water is ………………………. .
31. Seeds of calotropis dispersed with the help of ………………………. .
32. Seeds of fleshy fruits are dispersed through ………………………. .
33. Seeds having hooks are dispersed with the help of ………………………. .
34. Neem seeds are dispersed with the help of ………………………. .
35. Seeds having hairs dispersed through ………………………. .
36. Dispersal of seed take place by bursting of fruits can be seen in ………………………. .
37. Lotus seeds are dispersed through ………………………. .
38. In Mustard, seeds dispersal takes place through ………………………. .
39. Seeds dispersed through aeroplanes and ships are ………………………. .
40. Flower shown in the picture is ………………………. .
41. Grafting technique is used in the plants like ………………………. for getting desired characters.
Answer:

  1. suckers
  2. nodes of sugarcane
  3. cut the twigs with nodes and sow them
  4. eyes
  5. Bulb
  6. Ginger
  7. colocasia
  8. rose
  9. Jasmine, Jaji
  10. grafting
  11. Calyx
  12. Corolla
  13. 1st
  14. 2nd
  15. Androecium
  16. Gynoecium
  17. Stamens
  18. Ovary with style and stigma
  19. 3rd
  20. 4th
  21. pollen grains
  22. Self – pollination
  23. cross – pollination
  24. pollen tube
  25. fertilization
  26. zygote
  27. fruit
  28. seeds
  29. wind
  30. coconut, louts
  31. wind
  32. animals, birds and man
  33. animals
  34. birds
  35. wind
  36. bhendi, balsam
  37. water
  38. bursting mechanism
  39. pulses, wheat, paddy
  40. Unisexual flower (or) incomplete flower
  41. mango, guava etc.

III. Match the following

1.

Group – AGroup – B
1. Daturaa) Bursting Mechanism
2. Cucumberb) Gynoecium
3. Papayac) Androecium
4. Stamend) Male and female flowers are on different plants.
5. Pistile) Male and female flowers are on same plant.
f) Complete flower

Answer:

Group – AGroup – B
1. Daturaf) Complete flower
2. Cucumbere) Male and female flowers are on same plant.
3. Papayad) Male and female flowers are on different plants.
4. Stamenc) Androecium
5. Pistilb) Gynoecium

2.

Group – AGroup – B
1. Fertilizationa) Birds
2. Calotropisb) Water dispersal
3. Coconutc) Animals
4. Seeds of grassd) Wind dispersal
5. Neeme) Zygote
f) Bursting mechanism

Answer:

Group – AGroup – B
1. Fertilizatione) Zygote
2. Calotropisd) Wind dispersal
3. Coconutb) Water dispersal
4. Seeds of grassc) Animals
5. Neema) Birds

3.

Group – AGroup – B
1. Ovarya) Wind dispersal
2. Ovulesb) Bursting mechanism
3. Antherc) Nodes
4. Mapled) Pollengrain
5. Balsame) Seeds
f) Fruit

Answer:

Group – AGroup – B
1. Ovaryf) Fruit
2. Ovulese) Seeds
3. Antherd) Pollengrain
4. Maplea) Wind dispersal
5. Balsamb) Bursting mechanism

4.

Group – AGroup – B
1. Suckersa) Nodes
2. Rhizomeb) Colocasia
3. Bulbc) Potato
4. Eyesd) Onion
5. Corme) Ginger
f) Banana

Answer:

Group – AGroup – B
1. Suckersf) Banana
2. Rhizomee) Ginger
3. Bulbd) Onion
4. Eyesc) Potato
5. Cormb) Colocasia

5.

Group – AGroup – B
1. Sugar canea) Suckers
2. Seeds of Drumsticksb) Bursting mechanism
3. Seeds of Xanthiumc) Man
4. Seeds of Custard appled) Animals
5. Seeds of Lady’s fingere) Wind
f) Nodes

Answer:

Group – AGroup – B
1. Sugar canef) Nodes
2. Seeds of Drumstickse) Wind
3. Seeds of Xanthiumd) Animals
4. Seeds of Custard applec) Man
5. Seeds of Lady’s fingerb) Bursting mechanism

6.

Group – AGroup – B
1. Rosea) Buds on leaf
2. Jasmineb) Rhizome
3. Mangoc) Suckers
4. Chrysanthemumd) Grafting
5. Turmerice) Layering
f) Stem cutting

Answer:

Group – AGroup – B
1. Rosef) Stem cutting
2. Jasminee) Layering
3. Mangod) Grafting
4. Chrysanthemumc) Suckers
5. Turmericb) Rhizome

Do You Know?

→ You will be amazed that banana contains seeds. The rose plant also has red fruits with seeds. Do you know that Nandivardhanam and Hibiscus also have seeds* There are nuts and seeds in the rose and banana plants that grow wild without human interference in forests. The rose and banana plants that are found in our surroundings have no seeds. Do you know the reason? Our ancestors grew these plants with seeds of the plants with seeds of the plants that were found in the forest.

With the efforts made over many generations to propagate them as plants with favourable characteristics, they became seedless plants.

AP 7th Class Science Important Questions 7th Lesson Reproduction in Plants

→ Can you imagine a single plant giving 40 types of fruits such as peaches, apricots, plums, cherries, nectarines and so on? Such types of plants can be produced by Grafting.

AP 7th Class Science Important Questions 6th Lesson Electricity

These AP 7th Class Science Important Questions 6th Lesson Electricity will help students prepare well for the exams.

AP Board 7th Class Science 6th Lesson Important Questions and Answers Electricity

Question 1.
How does a bulb glow?
Answer:
When we switched ON the bulb, electricity flows through the wire and makes the bulb glow.

Question 2.
What is a cell?
Answer:
A cell is a device used to generate electricity.

Question 3.
What are the anpde and cathode in a cell?
Answer:

  1. Anode and cathode are the two electrodes in a cell.
  2. The positive electrode is called anode and the negative electrode is called cathode.

Question 4.
What is the electrolyte in a cell?
Answer:
The electrolyte is a chemical component that conducts the electricity.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 5.
Which parts of the battery function as an anode and a cathode?
Answer:

  1. The container which is made of zinc metal works as the cathode (negative terminal).
  2. A carbon rod with a metal cap works as an anode (positive terminal).

Question 6.
Name the common chemicals present in the dry cell.
Answer:

  1. Carbon powder
  2. Ammonium chloride.

Question 7.
What Is a battery?
Answer:
A group of cells Is called a battery.

Question 8.
Write different types of cells.
Answer:
1) Dry cell, 2) Lithium cell, 3) Button ceils, 4)Alkaline cell

Question 9.
Write one difference between fuse and MCB.
Answer:
There is no need to change the fuse wire in MCB. But fuse wire has to be changed whenever it melts in fuse device.

Question 10.
What precantion do you take while connecting copper wire in the circuit?
Answer:
While connecting copper wire in the circuit, the copper coating at the two ends erf the copper wire should be cleaned or scratched.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 11.
What is a simple electric circuit?
Answer:
A simple electric circuit consisting of a battery, a bulb, a switch, and connecting wires.

Question 12.
What are circuit diagrams?
Answer:
The diagrams which show the arrangement of electric components in a circuit are called circuit diagrams.

Question 13.
Write the common sequence of components in electric circuit diagram.
Answer:
Positive terminal of the cell → wire → bulb → wire → switch → wire → negative terminal of the cell.

Question 14.
What is the tearing effect of electricity?
Answer:
The production of heat due to the flow of electricity through a wire is called heating effect of electricity.

Question 15.
Name the metal used as a filament in the iron box and buld?
Answer:

  1. Generally, the filament in an iron box is made up of Niehrome.
  2. The filament in a bulb is made of Tungsten.

Question 16.
What is magnetic effect of electricity?
Answer:
The production of magnetic force due to the flow of electricity thrombi a wire is called Magnetic effect of electricity.

Question 17.
What are electromagnets?
Answer:
The devices which act as magnets when electricity passes through them are called electromagnets.

Question18.
On what principle do electric cranes work?
Answer:
Electric cranes are worked on the principle of magnetic effects of electricity.

Question 19.
Answer:

  1. By using repulsion property of a magpet, a magnetic object can levitate.
  2. Electromagnetic train runs on this principle.

Question 20.
How does electricity bill calculated?
Answer:
Generally, electricity bills are calculated on the basis of number of units of electricity consumed.

Question 21.
What is wattage of a bulb? What does it mean?
Answer:

  1. 1) The wattage measures how powerful the electric device is.
  2. The brighter the bulb, the higher its wattage and more the electricity used by it.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 22.
What is meant by KWH?
Answer:

  1. Wiser? I kilo watt is used per one hour, it Is considered as 1 kilo watt hour (KWH)
  2. 1 KWH = 1 UNIT of electricity.

Question 23.
What is the unit of electricity consumption?
Answer:
Kilo watt hour (or) UNIT.

Question 24.
From where electricity soppiied to owr booses?
Answer:
Electricity supplied to our houses comes ban the power stations and substations.

Question 25.
Where do we use lithium-ion cells and why?
Answer:
Lithium ion cells are rechargeable and are used in mobile phones, laptops etc.

Question 26.
Where do we use button cells?
Answer:
Button cells are used in wrist watches, laser lights. thermometers (electronic), computers etc.

Question 27.
Where is the use of a switch?
Answer:
Switch is used to open or dose a circuit.

Question 28.
Write different types of Im21» which areasedht omr daily life.
Answer:
In our daily life we use different types of bulbs. Some of them are Incandescent bulb, fluorescent both (tube light). CFL (Compact Fluorescent Lamp), LED bulb etc.

Question 29.
What is the use of parallel connection of cells in a circuit?
Answer:
The parallel connection of cells, increases the battery life.

Question 30.
Guess, what happens if one of bulbs is burnt in a series connection?
Answer:
It one of the bulbs is burnt, remaining bulbs also not glow because of circuit is opened.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 31.
What are the material required to make a cell with lemons?
Answer:
Zinc plate, copper plate, a small bulb, or LED, connecting wires, crocodile dips, and lemons.

Question 32.
Where do you find series and parallel connections of bulbs in your real life?
Answer:

  1. Parallel connections of bulbs are found in houseshold connections.
  2. Series connections of bulbs are found in decorative purpose connections.

Question 33.
What are the risks of electric shock?
Answer:
Explosure to electricity may result in no injury at all or may result in major damage or sometimes lead to death also.

7th Class Science 6th Lesson Electricity Short Questions and Answers

Question 1.
Answer:

  1. A cell is device used to generate electricity.
  2. It contains two elect rides and a chemical.
  3. When electrodes come in contact with electrolyte, a chemical reaction takes place inside the cell to produce electricity.
  4. It converts the chemical energy into electrical energy.
  5. When electrodes and connected to the wires in a circuit, electricity passes through them.

Question 2.
What are the main components of a cell? Explain.
Answer:

  1. A cell is a device used to generate electricity.
  2. It mainly consists of two components.
    i) Electrolyte : This is a chemical component that conducts the electricity.
    ii) Electrodes : There are two electrodes in a cell. One is positive electrode called anode and the other is negative electrode called cathode.

Question 3.
Why and how do we save electricity?
Answer:

  1. Our country faces shortage of electricity.
  2. So, we should use electricity careful! and only when it is needed.

Question 4.
How long a dry cell works?
Answer:

  1. The cell can supply electricity in a circuit for a period of time.
  2. After that, chemicals present in it get exhausted and it cannot produce electricity.

Question 5.
Why should we replace an ordihary bulb with a LED bulb?
Answer:

  1. The ordinary bulb we use gives us light and also heat.
  2. This is not desirable.
  3. This results in the wastage of electricity.
  4. Hence, we should replace an ordinary bulb with a LED bulb.
  5. LED bulbs consume less electricity than ordinary bulb.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 6.
How do you find an electrical device which consumers less electricity?
Answer:

  1. Now-a-day every electrical device is marked with a power saving guide which contains star symbols.
  2. The number of stars on them indicates the energy efficiency of that device.
  3. The electrical appliance with more stars consumes less electricity.
  4. Hence, it is advised to select electrical appliances with more stars.

Question 7.
How does an electric fuse protect home appliances?
(OR)
How do you appreciate the electric fuse?
Answer:

  1. Electric appliances may get damaged when excess of electricity flows through them.
  2. To protect them from such accidents, a safety device called electric fuse is used.
  3. Electric fuse has a body made of ceramic.
  4. It has two points for connecting the fuse wire.
  5. If there is an overload of electricity in the circuit, the fuse wire melts because of low melting point.
  6. This breaks the circuit and helps in preventing damage to electrical appliances.

Question 8.
How does a MCB work?
(OR)
How do you appreciate the role of MCB in the protection of electrical home appliances?
Answer:

  1. MCB means Miniature Circuit Breaker.
  2. It is being used in place of electric fuses.
  3. There is a switch which automatically turns off when electricity in circuit exceeds the safe limit.
  4. So, the circuit is opened.
  5. We just need to switch ON to resume electric supply.
  6. At present MCB is available with automatically resumed built-in mechanism.
  7. MCB is better than electric fuses.
  8. There is no need to change it often.
  9. But fuse wire has to be changed whenever it melts.
  10. Hence, I appreciate the MCB for protecting home appliances.

Question 9.
What is series connection of devices? What happens if any one of the devices is removed?
Answer:

  1. If the second terminal of the first device is connected to the first end of the second device, such type of connection is called series connection.
  2. In series connection electricity has only one path.
  3. If any of them is removed or not functioning properly, it becomes an open circuit.

Question 10.
What is parallel connection of devices ? What is its advantage?
Answer:

  1. If all the first terminals of all the devices are connected to one point and all the second terminals are connected to another point such type of connection is called parallel connection.
  2. In parallel connection electricity has more than one path.
  3. Even if any one of them is removed the circuit remains closed.

Question 11.
How does the brightness of the bulb changes, if a number of cells are connected 0 in series ii) in parallel?
Answer:
i) If a number of cells are connected in series the brightness of the bulb is enhanced,
ii) If a number of cells are connected in parallel the brightness of the bulb remains same (or) does not change.

Question 12.
Why the batteries in a torch light are connected in series?
Answer:

  1. If we connect batteries/cells in series it enhance the brightness of the bulb, the batteries in a torch light are connected in series.
  2. So, to enhance the brightness of the bulb.

Question 13.
How does the brightness of the bulb change when a number of bulbs are connected in i) series ii) parallel?
Answer:
i) If a number of bulbs are connected in series the brightness of the bulb is reduced.

ii) If the number of bulbs are connected in parallel, the brightness of the bulb remains same.

Question 14.
How does an electric-iron box work?
Answer:

  1. The electric iron box contain a coil of wire, which is generally made up of Nichrofne.
  2. This coil of wire is called a filament, which produce heat, when current is passed through it.

Question 15.
How does a bulb with filament glow?
Answer:

  1. When electric current is passed through the filament (made with Tungsten), it becomes red hot and further becomes bright white.
  2. This bright white glow of the filament gives light.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 16.
What happens if we pass current through the copper coil around a nail?
Answer:
The nail acts as a magnet, when current is passed through the coil around it.

Question 17.
Are electro-magnets permanent magnets? Justify your answer.
Answer:

  1. Electromagnets are not permanent magnets.
  2. Because, they lose the property of magnetism when the circuit is open.

Question 18.
Write some situations which lead to an electric shock.
Answer:

  1. Operating a switch with a wet hand.
  2. Removing plug pins when the switch is on.
  3. Using wires without insulation.
  4. Changing bulb when the switch is on.
  5. Holding a person with a current shock.

Question 19.
What immediate action should we take to save a person who gets electric shock?
Answer:

  1. When anybody gets an electric shock.
  2. First, cut off the power supply.
  3. If it is not possible, push away the person with a dry stick.
  4. If the shock is severe and the person is unconscious, give mouth to mouth artificial respiration.
  5. Sometimes heart beat ceases, at that time keep your palms on his chest and press down and release until the heartbeat is recovered. This is called Cardio Pulmonary Resuscitation (CPR). Immediately take him to the hospital.

Question 20.
What is ISI? What is its use?
Answer:

  1. ISI means Indian Standards Institution.
  2. It can be observed on the electrical devices.
  3. The ISI mark ensures the quality of appliances and safety of users.

Question 21.
When the power to house cut off, the Saritha noticed that her father was changing the fuse wire. Then she had many doubts. What would be those doubts?
Answer:

  1. What is fuse?
  2. How does it work?
  3. What happens to the fuse, when current is cut off?
  4. What is the use of fuse?

Question 22.
Filament bulbs are hot, but some other bulbs like LED, CFL are cold. Guess, what is the reason?
Answer:

  1. Filament bulbs are made up of Tungsten filaments.
  2. These are working on the principle of heating effect of the electricity.
  3. So, they are hot.
  4. But, other bulbs like CFL and LED are not working on this principle. Hence they are cool.

Question 23.
Guess, what would happen if home appliances were connected in a series connection?
Answer:

  1. If one device is stopped or failed to work, all other devices will also stop.
  2. If one device is switched on the other devices will also works.

Question 24.
Write the procedure of making of our own cell.
Answer:

  1. Take a lemon and insert a zinc plate and a copper plate into it.
  2. These plates acts as electrodes and the juice inside the fruit acts as electrolyte.
  3. Connect two terminals of the bulb to zinc and copper plates using connecting wires.
  4. The bulb will glow.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 25.
Write the required materials to do an activity to know about heating effect of electric current.
Answer:

  1. Electric cell,
  2. Iron nails – 2,
  3. Wooden board,
  4. Connecting wires,
  5. 10 cm of Nichrome wire etc.

Question 26.
Write the material required to make a electromagnet.
Answer:
Battery, Switch, Iron nail, Insulated copper wire, and small pins.

Question 27.
What happens if the switch is open in the experiment of given picture.
AP Board 7th Class Science Solutions 6th Lesson Electricity 3
Answer:
When we switched off the circuit, immediately all the pins fall from the iron nail.

Question 28.
Observe the diagram and write answers to the given questions.
AP 7th Class Science Important Questions 6th Lesson Electricity 1
1) Which bulb is more power saver?
2) Which bulb has less life span?
3) Which bulb do you suggest for your home?
4) Which bulb works on the principle of heating effect of electricity?
Answer:

  1. LED
  2. Incandescent
  3. LED
  4. Incandescent.

Question 29.
Fill in the table with suitable answers.

Name of the DeviceUse
To give light
Switch
To protect from high voltage
Cell

Answer:

Name of the DeviceUse
BulbTo give light
SwitchTo open or close a circuit
FuseTo protect from high voltage
CellTo given electricity

Question 30.
Observe the following circuit diagrams and answer the question.
AP Board 7th Class Science Solutions 6th Lesson Electricity 17
What difference do you find between diagrams A and B?
Answer:
We observe that the bulb in circuit A is not glowing and in circuit B it is glowing.

We can also see the switch is in OFF mode in circuit diagram A and it is in ON mode in circuit diagram B. Circuit diagram A represents open circuit, where as circuit diagram B represents closed circuit.

Question 31.
Observe the figure and give answers to the questions.
1) How are the cells connected?
2) If one of the cells is disconnected what will happen?
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 10
Answer:

  1. Cells are connected in parallel.
  2. The bulb will glow as usual if one of the cells is disconnected.

Question 32.
Draw the given diagram by using symbols instead of realistic components.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 2

Question 33.
Draw the diagram of setup of material in make a cell with lemon.
Answer:
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 8

Question 34.
Draw the diagram of a simple circuit.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 3

Question 35.
Draw a circuit diagram showing the parallel connection of the cells.
Answer:
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 10

Question 36.
Draw a circuit diagram showing the series connection of the cells.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 4

37. Draw a circuit diagram showing the parallel connection of bulbs.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 5

38. Draw a circuit diagram showing the series connection of bulbs.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 6

39. Draw a device alignment diagram of equipment in the experiment of heating effect of electric current.
Answer:
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 17

40. Draw the diagram showing the arrangement of circuit to make an electromagnet.
Answer:
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 2

41. Draw the symbol of ISI mark.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 7

42. Where do we use dry cells? What are its advantages?
Answer:

  1. In our daily life we use dry cells in torch lights, wall clocks, radios etc.
  2. Dry cell consists of an electrolyte in dried form.
  3. So, it is easy to move anywhere.

Question 43.
How does a switch work in a circuit?
Answer:

  1. We know that switch Is used to open or dose a circuit
  2. When the switch is ‘off’ position, bulb does not glow, because the circuit Is opened.
  3. If we keep the switch in ‘on’ mode, the circuit is closed and the bulb glows.

Question 44.
How do circuit diagrams help to electricians?
Answer:

  1. A circuit diagrams is a graphic representation of an electric circuit
  2. It shows how the electric components are connected together.
  3. Electricians and engineers draw circuit diagrams to help them design the actual circuits.

Question 45.
What is the first aM few electric shock?
(OR)
How do you perform CPR?
Answer:

  1. It the electric shock is severe and the person is unconscious, given mouth to mouth artificial respiration.
  2. Sometimes heartbeat ceases, at that time keep your palms on his chest and press down and release until the heartbeat is recovered, this is called Cardio Pulmonary Resuscitalion (CPR)

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 46.
How do you check for the good electric appliances?
Answer:
While selecting electric appliances:

  1. Check ISI (Indian Standard institution) mark on all electrical appliances before you buy.
  2. Check for more star rating.

7th Class Science 6th Lesson Electricity Long Questions and Answers

Question 1.
What is a cell? How does it work?
Answer:
A cell is a device used to generate electricity. It mainly consists of two components.

  1. Electrolyte – This is a chemical component that conducts the electricity.
  2. Electrodes – There are two electrodes in a ceiL One is positive electrode called anode and the other is negative electrode called cathode.

When electrodes come in contact with electrolyte, a chemical reaction takes place inside the cell to produce electricity. Thus, it converts the chemical energy into electrical energy When electrodes are connected to the wires in a circuit, electricity passes through them.

Question 2.
What is a both? Write different types of bain How does an Incandescent bulb work? Does if save Are electrical energy?
Answer:

  1. Bulb is an electric device which converts electrical energy into light energy.
  2. There are different types at bulbs. They are 1) incandescent bulb, 2) Fluorescent bulb (Tube light), 3) CFL (Compact Fluorescent Lamp), 4) LED bulb, etc.
  3. Incandescent bulb works on the principle of heating effect of electricity.
  4. Incandescent bulb doesn’t save electrical energy. It consumes more electrical energy compared to CFL and LED lamps.

Question 3.
Write the differences between series and parallel connections of cells.
Answer:

Series connectionParrllel connection
I) If the second terminal of the first device is connected to the first end of the second device such type of  connection is called series connection.1) If all the first terminals of all the devices are connected to one point and all the second terminals are cennected to another point, such type of connection is called of parallel connection.
2) In series connection, electricity has only one path.2) In parallel connection electricity has more than one path.
3) If any one of the cells is removed or not functioning properly it becomes an open circuit.3) Even if anyone of them is removed the circuit remains closed.

 

4) In series brightness of the bulb is enhanced.4) In parallel connection bulb will glow as usual.
5) We can use series connection of cells in torch lights, toys, remotes, wall clocks etc.5) We can use parallel connection of cells for a long lasting battery life.

AP 7th Class Science Important Questions 6th Lesson Electricity

Question 4.
Write the differences between series and parallel connections of bulbs.
Answer:

Series connection of bulbsParallel connections of bulbs
1) In series connection electricity has only one path.1) In parallel connection electricity has more than one path.
2) If one of the bulbs is removed or broken, the circuit becomes open, and the otter bulbs will not glow.2) If one of tie bulbs is removed or broken, the circuit remains same and the busib will j glow as usual.
3) The brightness of the bciib is reduced in series connection.3) The brightness erf the bulb remains same in parallel connection.
4) Series eettnectioo of btsifes is used for decorative purpose.4) FaraEei coaraeetkm of bulbs is used in household connection.

Question 5.
Explain the following :
a) Heating effects of electric current.
b) Magnetic effects of electric current.
Answer:
a) Heating effects of electric current:

  1. The production of heart due to the flow of electricity through a wire is called heating effects of electricity.
  2. Electric devices like electric stoves, room heaters, iron boxes, electric kettles, electric cookers, heaters, geysers, hair dryers work on the principle of heating effects of electric current.
  3. All these appliances contain a heating element is called a filament.
  4. Generally, a filament is made up of Nichrome in iron boxes and the filaments of an electric bulb is made of Tungsten.

b) Magnetic effects of electric current:

  1. The production of magnetic force due to the flow of electricity through a wire is called magnetic effect of electricity.
  2. The devices which act as magnets when electricity passes through them are called electromagnets.
  3. The electric devices like loudspeakers, fan, bell, motor, mixer grinders, MRI machines, generators, metal detectors, mobile phones work on the principle of magnetic effect of electricity.
  4. Magnetic effects produced in a coil which is wound around a soft iron due to the passing of electricity in the coil.

Question 6.
1) What is an electric shock?
2) Which situations lead to an electric shock?
3) What is the first aid for efectric shock?
4) How does CPR help to the person?
Answer:
1) An electric shock occurs when a person comes into contact with an electrical source. Electricity flows through a portion of the body causing a shock It leads to major damage, sometimes it may cause death.

2) i) Operating a switch with a wet hand.
ii) Removing plug pins when the switch is on.
iii) Using wires without insulation.
iv) Changing bulb when the switch is on.

3) i) When anybody gets an electric shock.
ii) First, cut off the power supply.
iii) If it is not possible, push away the person with a dry stick.
iv) If the shock is severe and the person is unconscious, give mouth-to-mouth artificial respiration.

4) Sometimes heart beat ceases, at that time keep your palms on his chest and press down and release until the heartbeat is recovered. This is called Cardio Pulmonary Resuscitation (CPR). Immediately take him to the hospital.

Question 7.
Fill in the table with suitable answers.
AP 7th Class Science Important Questions 6th Lesson Electricity 8
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 9

Question 8.
Observe the given figures and answer the following questions.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 10
1) In which circuit bulbs are connected in parallel?
2) In which circuit electricity passes through only one path?
3) Which circuit represents household connections?
4) What happens if one bulb is damaged in circuit A?
5) In which circuit glow of bulb is enhanced?
Answer:

  1. Circuit B
  2. Circuit A
  3. Circuit B
  4. Circuit will be opened
  5. Circuit A

Question 9.
Fill in the table.

Name of the deviceMagnetic effect of electricityHeating effect of electricity
1. Electric stove
2. Electric iron box
3. Electric fan
4. Electric bell
5. Metal detector
6. Loud speaker
7. Filament bulb
8. Electric kettle
9. Geyser

Answer:

Name of the deviceMagnetic effect of electricityHeating effect of electricity
1. Electric stove
2. Electric iron box
3. Electric fan
4. Electric bell
5. Metal detector
6. Loud speaker
7. Filament bulb
8. Electric kettle
9. Geyser

Question 10.
Draw the diagram of a dry cell and label the parts.
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 9

Question 11.
Draw the symbols of given electrical components.
1) Cell 2) Battery 3) Bulb 4) Glowing bulb 5) Switch in ‘ON’ mode 6) Switch in ‘OFF’ mode 7) Fuse 8) Connecting wires.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 11

Question 12.
Here a circuit with real components are given. Draw the circuit by using symbols of the components.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 12
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 13

Question 13.
Kalyan uses 5 LED bulbs of 9W for 10 hours, 3 fans of 60 W for 8 hours, 1 TV of 100 W for 6 hours in his house per day. Calculate the cost of his consumed electricity for a month at the rate of ₹ 10 per unit.
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 14

Problems Given In The Lesson

Exercise – 1

The meter reading in Chandu’s house for the month of December 2020 is 29171 units. The meter reading in the previous month is 29062 units. Calculate how much he has to pay towards electricity bill in December. The unit cost is Rs.3.16.
Answer:

Reading in December29171 units
Reading in previous month29062 units
No. of units of electricity consumed109 units
Cost per unitRs.3.16
Amount to be paid109 x 3.16= Rs.344.44 ( Rs.344 only)

AP 7th Class Science Important Questions 6th Lesson Electricity 15

Exercise – 2

Suppose in a house there are 5 bulbs of 100W each, five of 60 W each, five of 40W each. All of them are used for five hours everyday. How many units of electricity is used in the month of February 2021? How much will they have to pay at Rs.2.80/- per unit?
Answer:
AP 7th Class Science Important Questions 6th Lesson Electricity 16

AP Board 7th Class Science 6th Lesson 1 Mark Bits Questions and Answers Electricity

I. Multiple Choice Questions

1. The positive electrode in a cell is called
A) Electrolyte
B) Anode
C) Cathode
D) Filament
Answer:
B) Anode

2. Which of the following converts chemical energy into electrical energy?
A) Cell
B) Bulb
C) Fan
D) Switch
Answer:
A) Cell

3. The container of a dry cell works as
A) Anode
B) Cathode
C) Electrolyte
D) None
Answer:
B) Cathode

AP 7th Class Science Important Questions 6th Lesson Electricity

4. A carbon rod with a metal cap in a cell works as
A) Anode
B) Cathode
C) Electrolyte
D) None
Answer:
A) Anode

5. Generally the chemical used in a dry cell is
A) Copper sulphate
B) Sodium chloride
C) Ammonium chloride
D) Potassium hydride
Answer:
C) Ammonium chloride

6. The lemon juice in a cell is acts as a
A) Anode
B) Cathode
C) Electrolyte
D) None
Answer:
C) Electrolyte

7. Dry cell is not used in
A) Torch light
B) Wall clock
C) Radio
D) Mobiles
Answer:
D) Mobiles

8. Assertion (A) : Dry cell is easy to move any where.
Reason (R) : Dry cell consists of also electrolyte in dried form.
A) A and R are correct and R is a correct reason to A
B) A and R are correct but R is not a correct reason to A
C) A is correct but R is incorrect
D) A is incorrect but R is correct
Answer:
A) A and R are correct and R is a correct reason to A

9. Generally, button cells are used in
A) Torch lights
B) Mobile phones
C) Wrist watches
D) Vehicles
Answer:
C) Wrist watches

10. 1) When switch is in ‘off position the circuit is a closed.
2) When switch is in’off position the bulb does not glow.
A) 1 is correct
B) 2 is correct
C) 1 and 2 are correct
D) 1 and 2 are wrong
Answer:
B) 2 is correct

AP 7th Class Science Important Questions 6th Lesson Electricity

11. Which of the following consumes more power than others?
A) Incandescent bulb
B) Fluorescent bulb
C) CFL
D) LED
Answer:
A) Incandescent bulb

12. Fuse wire is made of
A) Low melting point wire
B) High melting point wire
C) No melting point wire
D) A or B
Answer:
A) Low melting point wire

13. Which of the follow device prodect the electrical home appliances?
A) Fuse
B) MCB
C) A and B
D) Filament
Answer:
C) A and B

14. A siaifsie decftfc circuit amwBlmg of
A) a battery
B) a bulb
C) a switch
D) above all
Answer:
D) above all

15. AP 7th Class Science Important Questions 6th Lesson Electricity 17 This is the symbol of
A) a cell
B) a battery
C) two switches
D) fuse
Answer:
B) a battery

16. AP Board 7th Class Science Solutions 6th Lesson Electricity 1 This symbol represents
A) a cell
B) an open switch
C) a closed switch
D) an open cell
Answer:
B) an open switch

17. WMADftefofloif mg represents a glowing bulb?
AP 7th Class Science Important Questions 6th Lesson Electricity 18
Answer:
A

18. On series connection has
A) Only two paths
B) Only one path
C) More than one path
D) None
Answer:
B) Only one path

AP 7th Class Science Important Questions 6th Lesson Electricity

19. If a number of cells are connected in series the brightness of the bulb is
A) Enhanced
B) Reduced
C) As usual
D) A or B
Answer:
A) Enhanced

20. If a number of cells are connected in parallel the brightness of the bulb is
A) Enhanced
B) Reduced
C) As usual
D) A or B
Answer:
C) As usual

21. In which connection one of the cells are removed the circuit does not open?
A) Series
B) Parallel
C) A and B
D) None
Answer:
B) Parallel

AP 7th Class Science Important Questions 6th Lesson Electricity

22. If a number of bulbs are connected in series the brightness of the bulb is
A) Enhanced
B) Reduced
C) Sanafe issmre
D) Aor B
Answer:
B) Reduced

23. If a number of bulbs are connected in parallel the brightness of the bulb is
A) Enhanced
B) Reduced
C) Remains same
D)A or B
Answer:
C) Remains same

24. Household devices are connected in
A) Series
B) Parallel
C) A or B
D) None
Answer:
B) Parallel

25. Which of the following gives light by glowing in Bulbs?
A) Nichrome
B) Tungsten
C) Alnico
D) Copper
Answer:
B) Tungsten

26. Generally filament in iron boxes are made up of
A) Nichrome
B) Tungsten
C) Alnico
D) Copper
Answer:
A) Nichrome

27. The devices which acts as magnets when electricity passes through them
A) Electric kettles
B) Electric heaters
C) Electromagnets
D) MCB
Answer:
C) Electromagnets

AP 7th Class Science Important Questions 6th Lesson Electricity

28. Which of the following works on principle of magnetic effect of electricity?
A) Electric rice cooker
B) Thermometer
C) Electric bell
D) Electric bulb
Answer:
C) Electric bell

29. Which of the following is required to make an electromagnet?
A) Coiled copper wire
B) Nail
C) Electricity
D) Above all
Answer:
D) Above all

30. Which of the following property can levitate a magnetic object?
A) Heat effect of electricity
B) light effect of electricity
C) Chemical effect of electricity
D) Magnetic effect erf etedtricity
Answer:
D) Magnetic effect erf etedtricity

31. Electromagnets are
A) permanent magnets
B) temporary msassets
C) not the magnets
D) bad conductors
Answer:
A) permanent magnets

32. 1 kilo watt =
A) 1000 watts
B) 100 watts
C) 1/1000 watts
D) 1/100 watts
Answer:
A) 1000 watts

33. 1 unit of electricity means
A) 1 WH
B) 1 KWH
C) 1000 KWH
D) 1/1000 KWH
Answer:
B) 1 KWH

34. Immediate action to be tacken when an electric shock occurs is
A) pouring erf water
B) danger with the device
C) Cut off power supply
D) Expiry date of the device
Answer:
C) Cut off power supply

AP 7th Class Science Important Questions 6th Lesson Electricity

35. AP 7th Class Science Important Questions 6th Lesson Electricity 7 Indicates
A) Power saving erf the device
B) Danger with the device
C) Quality erf the device
D) Expiry date of the device
Answer:
C) Quality erf the device

II. Fill in the blanks

1. …………………. is a device used to generate electricity
2. The chemical componesrf that conducts electricity is called …………………. .
3. The negative electrode is called …………………. .
4. The …………………. electrode is called anode.
5. A cell converts the …………………. energy into …………………. energy.
6. The positive and negative terminals are called …………………. .
7. A group of …………………. is called a battery.
8. When switch is in ‘ON’ mode, the circuit is …………………. .
9. An ordinary bulb gives …………………. and …………………. .
10. CFL means …………………. .
11. LED means …………………. .
12. The electrical appliance with more stars consumes …………………. electricity.
13. …………………. protects home appliances from the excess of electricity flows through them.
14. Generally, fuse wire made of …………………. melting point.
15. MCB means …………………. .
16. …………………. is better, when compared to electric fuse.
17. A …………………. is a graphical representation of an electrical circuit.
18. The electrical symbol of fuse is …………………. .
19. In …………………. connection of cells life of battery increases.
20. Nichrome wire gets …………………. when electricity passed through it.
21. The production of …………………. due to the flow of electricity through a wire is called heating effect of electricity.
22. Household electricity is measured in …………………. .
23. 1 …………………. = 1000 watts.
24. KWH means …………………. .
25. A …………………. occurs when a person comes into contact with an electrical source.
26. Electricity flows through a portion of the body causing a
27. CPR means …………………. .
28. ISI means …………………. .
29. ISI mark ensures …………………. of the appliances.
Answer:

  1. A cell
  2. Electrolyte
  3. Cathode
  4. positive
  5. chemical, electrical
  6. electrodes
  7. cells
  8. closed
  9. heat, light
  10. Compact Fluorescent Lamp
  11. Light Emitting Diode
  12. less
  13. Fuse
  14. Low
  15. Miniature Circuit Breaker
  16. MCB
  17. circuit diagram
  18. AP 7th Class Science Important Questions 6th Lesson Electricity 19
  19. parallel
  20. heated up
  21. heat
  22. KWH or units
  23. kilo watt
  24. kilo watt hour
  25. Electric
  26. shock
  27. Cardio Pulmonary Resuscitation
  28. Indian Standards Institutions
  29. quality

III. Match the following

1.

Group – AGroup – B
A) Electrolyte1) Copper
B) Electrode2) Nichrome
C) Filament3) Ammonium chloride
D) Connecting wire4) Carbon rod
E) Cell5) Lithium – ion

Answer:

Group – AGroup – B
A) Electrolyte3) Ammonium chloride
B) Electrode4) Carbon rod
C) Filament2) Nichrome
D) Connecting wire1) Copper
E) Cell5) Lithium – ion

2.

Group – AGroup – B
A) Dry cell1) Bikes
B) Lithium cell2) Wrist watches
C) Button cell3) Torches
D) Alkaline cell4) Mobiles

Answer:

Group – AGroup – B
A) Dry cell3) Torches
B) Lithium cell4) Mobiles
C) Button cell2) Wrist watches
D) Alkaline cell1) Bikes

3.

Group – AGroup – B
A) Cell1) Protects home appliances
B) Bulb2) Opens or closes circuits
C) Fuse3) Produces light
D) Switch4) Produces electricity
E) Iron box5) Produces heat
F) Cranes6) Produce magnetism

Answer:

Group – AGroup – B
A) Cell4) Produces electricity
B) Bulb3) Produces light
C) Fuse1) Protects home appliances
D) Switch2) Opens or closes circuits
E) Iron box5) Produces heat
F) Cranes6) Produce magnetism

4.
AP 7th Class Science Important Questions 6th Lesson Electricity 20
Answer:
A – 7, B – 3, C – 1, D – 5, E – 6, F – 2, G – 4.

Do You Know?

→ In 1752, Franklin conducted the kite experiment to show that lightning was also electricity. He flew a kite during a thunderstorm. He tied a metal key to the string of the kite to conduct the electricity.

→ LED (Light Emitting Device) has two legs. Long leg indicates positive(+ve) and short leg indicates negative (-ve). Early LEDs produced only red light, but modern LEDs can produce several different colors, including red, green and blue light. LEDs are used as indicators in mobiles, laptops, T.V remotes etc.

→ A switch can be placed anywhere in the circuit and a cell can be connected in any direction in the electric circuit.

AP 7th Class Science Important Questions 6th Lesson Electricity

→ The scientist Oersted discovered that a magnetic field is produced around a current carrying conductor. Electromagnets are made by using the relationship between electricity and magnetism.

→ Michael Faraday made a magnet to revolve around a coil of wire which produced electricity. He invented electric generator or dynamo and transformer.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 4th Lesson States of Matter: Gases and Liquids Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 4th Lesson States of Matter: Gases and Liquids

Very Short Answer Questions

Question 1.
Name the different intermolecular forces experienced by the molecules of a gas.
Answer:
The different intermolecular forces experienced by the molecules of a gas are London (or) dispersion forces, Dipole-Dipole forces, Dipole- induced dipole forces, and hydrogen bond.

Question 2.
State Boyle’s law. Give its mathematical expression.
Answer:
At constant temperature, the pressure of a given mass (fixed amount) of gas varies inversely with it’s volume. This is Boyle’s law.

  • Mathematically it can be written as
    P ∝ \(\frac{1}{v}\) (At constant T and no.of moles (n))
    ⇒ Pv = \(\frac{k}{v}\) (constant).

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 3.
State Charle’s law. Give its mathematical expression.
Answer:
At constant pressure the volume of a fixed mass of a gas is directly proportional to it’s absolute temperature. This is charle’s law.

  • Mathematically it can be written as
    V ∝ T (At constant P and no.of moles (n))
    ⇒ V = kT
    ⇒ \(\frac{V}{T}\) = k (constant).

Question 4.
What are Isotherms?
Answer:
At constant temperature the curves which shows the relationship between variation of volume of a given mass of gas and pressure are called isotherms.

Question 5.
What is Absolute Temperature?
Answer:
It is also called thermodynamic temperature (or) Kelvin temperature. It is a temperature on the absolute (or) kelvin scale in which zero lies at – 273.16°C.
T = (t° C + 273.16) K

Question 6.
What are Isobars?
Answer:
The curves (or) graphs that can be drawn at constant pressure are called Isobars.
Eg : Graph drawn between volume and temperature.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 7.
What is Absolute Zero?
Answer:
It is the lowest temperature theoretically possible at which volume of a perfect gas is zero.

Question 8.
State Avogadro’s law.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contains equal number of molecules
V ∝ n (mathematically)
v = kn

Question 9.
What are Isochores ?
Answer:
At constant volume a line on a graph showing the variation of temperature of a gas with its pressure is called Isochores.

  • It is also called Isoplere.

Question 10.
What are S T P Conditions ?
Answer:
STP means Standard Temperature and Pressure conditions.

  • Standard temperature is 0° C = 273 K
  • Standard pressure is 1 atmosphere = 76 cm = 760 mm. of Hg.

At S.T.P. one mole of any gas occupy 22.4 lit. of volume.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 11.
What is Gram molar Volume ?
Answer:
The volume occupied by one gram molecular weight (or) one gram mole of an element (or) compound in the gaseous state is called gram molar volume.
(or)

  • At STP one mole of any gas occupy 22.4 lit. of volume This is known as gram molar volume.

Question 12.
What is an Ideal gas ?
Answer:
A gas which obeys gas laws i.e. Boyle’s law, charle’s law and avagadro’s law exactly at all temperatures is called an ideal gas.

Question 13.
Why the gas constant ‘R’ is called Universal gas constant ?
Answer:
Gas constant ‘R’ is called universal gas constant because the value of ‘R‘ is same for all gases.

Question 14.
Why Ideal gas equation is called Equation of State ?
Answer:
Ideal gas equation is a relation between four variables (p, v, n, T) and it describes the state of any gas. Hence it is called equation of state.

Question 15.
Give the values of gas constant in different units.
Answer:
Gas constant ‘R’ has values in different units as follows.
R = 0.0821 lit. atm. k-1 mol-1
= 8.314 J. k-1 mol-1
= 1.987 (or) 2 cal. k-1 mol-1
= 8.314 × 107 ergs. k-1 mol-1.

Question 16.
How are the density and molar mass of a gas related?
Answer:
Pv = n RT
Pv = \(\frac{w}{m}\) RT m
P = \(\left(\frac{w}{v}\right) \frac{R T}{M}\)
Molar mass M = \(\frac{\mathrm{dRT}}{\mathrm{P}}\) [∴ \(\frac{w}{v}\) = density(d)]
P = Pressure of gas
R = Universal gas constant
T = Temperature of gas in kelvins scale.

Question 17.
State Graham’s law of diffusion. (A.P. Mar. ‘16, ’14)
Answer:
The rate of diffusion of a given mass of gas at a given pressure and temperature is inversely proportional to the square root of its density
rate of diffusion r ∝ \(\frac{1}{\sqrt{d}}\).

Question 18.
Which of the gases diffuses faster among N2, O2 and CH4? Why? (T.S. Mar. ‘15)
Answer:
CH4 gas diffuse faster among N2, O2 and CH4.
Reason : CH4 (16) has low molecular weight than N2 (28) and O2 (32).

Question 19.
How many times methane diffuses faster than sulphurdioxide?
Answer:
According to Graham’s law of diffusion.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 1
Hence methane gas diffuses 2 times faster than SO2.

Question 20.
State Dalton’s law of Partial pressures. (Mar. ‘14)
Answer:
The total pressure exerted by a mixture of chemically non – reacting gases at given temperature and volume, is equal to the sum of partial pressures of the component gases.
P = P1 + P2 + P3.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 21.
Give the relation between the partial pressure of a gas and its mole fraction.
Answer:
Partial pressure of a gas = mole fraction of the gas × Total pressure of the mixture of gases
Eg : Consider A and B in a container which are chemically non reaction.
∴ Partial pressure of A (PA) = XA × PT
Partial pressure of B (PB) = XB × PT
XA = \(\frac{n_A}{n_A+n_B}\), XB = \(\frac{n_B}{n_A+n_B}\)
XA, XB are mole fractions
PT = Total pressure.

Question 22.
What is aqueous tension?
Answer:
The pressure exerted by the water vapour which is equilibrium with liquid water is called aqueous tension.
(or)
The pressure exerted by the saturated water vapour is called aqueous tension.

Question 23.
Give the two assumptions of Kinetic molecular theory of gases that do not hold good in explaining the deviation of real gases from ideal behaviour.
Answer:
The two assumptions of kinetic molecular theory of gases that do not hold good in explaining the deviation of real gases from ideal behaviour are

  1. There is no force of attraction between the molecules of a gas.
  2. Volume of the gas molecules is negligible when compared to the space occupied by the gas.

Question 24.
Give the Kinetic gas equation and write the terms in it.
Answer:
Kinetic gas equation is PV = \(\frac{1}{3} \mathrm{mnu}_{\mathrm{rms}}^2\)
P = Pressure of the gas
V = Volume of the gas
m = Mass of 1 mole of the gas
urms = RMS speed of the gas molecules.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 25.
Give an equation to calculate the kinetic energy of gas molecules.
Answer:
Kinetic energy for ‘n1 moles of gas is given by
K.E. = \(\frac{3}{2} \mathrm{nRT}\)
R = Universal gas constant
T = absolute temperature.

Question 26.
What is Boltzman’s constant ? Give its value.
Answer:
Boltzman’s constant is the gas constant per molecule.
Boltzman’s constant K = \(\frac{R}{N}\)
= 1.38 × 10-16 erg/k. molecule
= 1.38 × 10-23 J/k. molecule.

Question 27.
What is RMS speed ?
Answer:
The square root of mean of the squares of the speeds of all molecules of a gas is known as RMS speed (uRMS)
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 2

Question 28.
What is Average speed ?
Answer:
The arithematic mean of speeds of gas molecules is known as average speed (uav).
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 3

Question 29.
What is Most probable speed ?
Answer:
The speed possessed by the maximum number of molecules of the gas is known as most probable sPeed (ump).

Question 30.
What is the effect of temperature on the speeds of the gas molecules ?
Answer:
Temperature and speeds of the gases are directly related.
∴ By the rise of temperature the speeds of the gas molecules also increases.

Question 31.
What is the effect of temperature on the kinetic energy of the gas molecules ?
Answer:
According to the postulates of kinetic molecular theory of gases.
The kinetic energy of gas molecules is directly proportional to the absolute temperature.
K.E. ∝ Tabs

Question 32.
Give the ratio of RMS average and most probable speeds of gas molecules.
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 4

Question 33.
Why RMS speed is taken in the derivation of Kinetic gas equation ?
Answer:
RMS speed is the mean of squares of speeds of all molecules of gas. Hence RMS speed, is taken into the derivation of kinetic gas equation.
PV = \(\frac{1}{3} m n u_{r m s}^2\)

Question 34.
What is Compressibility factor ?
Answer:
The ratio of the actual molar volume of a gas to the molar volume of a perfect gas under the same conditions is called compressibility factor.
Compressibility factor Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)
For a perfect gas Z = 1.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 35.
What is Boyle Temperature?
Answer:
The temperatue at which a real gas exibits ideal behaviour for a considerable range of pressure is called Boyle’s temperature.

Question 36.
What is critical temperature ? Give its value for CO2.
Answer:
The temperature above which no gas can be liquified how ever high the pressure may be applied is called critical temperature.

  • Critical temperature of CO2 gas is 31.98° C.

Question 37.
What is critical Volume ?
Answer:
The volume occupied by one mole of gas at critical temperature and critical pressure is known as critical volume.

Question 38.
What is critical Pressure ?
Answer:
The pressure required to liquify a gas at critical temperature is known as critical pressure.

Question 39.
What are critical constants ?
Answer:
Critical temperatue (TC), critical volume (VC) and critical pressure (PC) are called as critical constants.

Question 40.
Define vapour Pressure of a liquid.
Answer:
The pressure exerted by the vapour on the liquid surface. When it is in equilibrium with the liquid at a given temperature is known as vapour pressure of the liquid.

Question 41.
What are normal and standard boiling points ? Give their values for H2O.
Answer:

  • The boiling points at 1 atm. pressure are called normal boiling points.
  • The boiling points at 1 bar pressure are called standard boiling points.
  • For water normal boiling point is 100° C.
  • For water standard boiling point is 99.6° C.

Question 42.
Why pressure Cooker is used for cooking food on hills ?
Answer:
At hill areas pressure cooker is used for cooking food because low atmospheric pressure is observed at high altitudes. At high altitudes liquids boil at low temperature. So water boils at low temperature on hills.

Question 43.
What is surface tension ?
Answer:
The force acting at right angles to the surface of the liquid along unit length of surface is called surface tension.

  • Units : dynes / cm.

Question 44.
What is laminar flow of a liquid ?
Answer:
In liquids a regular gradation of velocity for layers in passing from one layer to the next observed. This flow of liquid is called Laminar flow.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 45.
What is coefficient of Viscosity ? Give its units.
Answer:
The force of friction required to maintain velocity difference of 1 cm. sec-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1cm2 is called coefficient of viscosity.

  • It is denoted by η
  • F = η A \(\frac{\mathrm{du}}{\mathrm{dx}}\)
  • Units : Poise : In CGS system 1 poise = 1g. cm-1 sec-1.

Short Answer Questions

Question 1.
State and explain Boyle’s law.
Answer:
Boyle’s law : At constant temperature, the volume of a given mass of gas is inversely proportional to pressure of the gas.
If ‘V’ is the volume of a given mass of the gas and its pressure, then the law can be written as
V ∝ \(\frac{1}{P}\) at constant temperature
V = \(\frac{k}{P}\)
∴ PV = k
or P1V1 = P2V2 = k.
Boyle’s law may also be stated as, “at constant temperature the product of the’ pressure and volume of a given mass of gas is constant.”

Question 2.
State and explain Charle’s law.
Answer:
Charles’ law : At constant pressure, the volume of a given mass of gas is directly proportional to absolute temperature.
V ∝ T at constant pressure
V = kT
\(\frac{V}{T}\) = k
Where V is the volume and T is the absolute temperature. V1 and V2 are the initial and final volumes of a given mass of the gas at the absolute temperatures T1 and T2 respectively at constant pressure.
\(\frac{V_1}{T_1}\) = k; \(\frac{V_2}{T_2}\) = k
or \(\frac{v_1}{T_1}\) = \(\frac{V_2}{T_2}\) = k

Question 3.
Derive Ideal gas equation. (T.S. Mar. ’16)
Answer:
Ideal gas equation : The combination of the gas laws leads to the development of an equation which relates to the four parameters volume, pressure, absolute temperature and number of moles. This equation is known as ideal gas equation.
In this Boyle’s law and Charles’ law combined together and an equation obtained is called the gas equation.
V ∝ \(\frac{1}{p}\) (Boyle’s law)
V ∝ T (Charles law)
V ∝ n (Avogadro’s law)
Combining above three laws, we can write
V ∝ \(\frac{1}{p}\) ∝ T ∝ n (or) V = R × \(\frac{1}{P}\) × T × n
(Or) PV = nRT
Where V = volume of the gas
P = pressure of the gas
n = no. of moles of gas
T = absolute temperature
R = Universal gas constant.

Question 4.
State and explain Graham’s law of Diffusion. (A.P. Mar.’16) (Mar.’13)
Answer:
Graham’s law of diffusion : At a given temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of density, vapour density or molecular weight.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 5
If r1 and r2 are the rates of diffusion of two gases d1 and d2 are their densities respectively, then
\(\frac{r_1}{r_2}\) = \(\sqrt{\frac{d_2}{d_1}}\)
This eqaution can be written as:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 6
Comparison of the volumes of the gases that diffuse in the same time. Let V1 and V2 are the volumes of two gases that diffuse in the same time ‘t’.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 7
When time of flow is same then : \(\frac{\mathrm{r}_1}{\mathrm{r}_2}\) = \(\frac{v_1}{v_2}\)
When volume is the same then : \(\frac{r_1}{r_2}\) = \(\frac{t_2}{t_1}\).

Applications:

  • This principle is used in the separation of isotopes like U235 and U238.
  • Molar mass of unknown gas can be determined by comparing the rate of diffusion of a known gas molecular mass.
  • Ansil’s alarms which are used in coal mines to detect the explosive maršh gas works on the principle of diffusion.

Question 5.
State and explain Dalton’s law of Partial pressures.
Answer:
Dalton’s law of Partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.
Explanation: Consider a mixture of three gases ¡n a vessel. Let P1, P2, P3 be the partial pressures of the three gases in the mixture. According to Dalton’s law of partial pressures, the pressure (P) of the gaseous mixture is at the same temperature.
P = P1 + P2 + P3

Let n1, n2, n3 be the number of moles of three gases respectively in the mixture. Let ‘V’ be the volume of the mixture of gases at T K temperature.
According to ideal gas equation,
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 8
∴ P1 = x1 P
In a similar way P2 = x2 P and P3 = x3P
∴ Partial pressure = mole fraction × total pressure

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 6.
Deduce
(a) Boyle’s law and
(b) Charle’s law from Kinetic gas equation.
Answer:
a) Deduction of Boyle’s law:
Kinetic gas equation is PV = \(\frac{1}{3} m n u^2\)
PV = \(\frac{1}{3}\) mnu2
= \(\frac{1}{3}\) × \(\frac{2}{2}\) mnu2
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu2
PV = \(\frac{2}{3}\) [Kinetic energy (KE)] [∵ KE‘n; moles = \(\frac{1}{2}\) mnu2]
PV = \(\frac{2}{3}\) KT [∵ KE = KT According to kinetic theory]
PV = \(\frac{2}{3}\)KT
According to Boyle’s law T is constant
∴ PV = \(\frac{2}{3}\)(constant)
∴ PV = constant
(or)
P ∝ \(\frac{1}{V}\)
Hence Boyle’s law proved from kinetic gas equation.

b) Deduction of Charle’s law:
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu2
PV = \(\frac{1}{3}\) mnu2
= \(\frac{2}{2}\) × \(\frac{1}{3}\) mnu2
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu2
= \(\frac{2}{3}\)(KE) [Kinetic energy (KE) = \(\frac{1}{2}\) mnu2]
PV = \(\frac{2}{3}\)KT [According to kinetic theory KE = KT]
\(\frac{V}{T}\) = \(\frac{2}{3}\) × \(\frac{K}{P}\)
According to Charles law ‘P is constant
∴ \(\frac{V}{T}\) = constant (or) V ∝ T
Hence Charle’s law proved from kinetic gas equation.

Question 7.
Deduce
(a) Graham’s law and
(b) Dalton’s law from Kinetic gas equation.
Answer:
a) Graham’s law: At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density, r ∝ \(\frac{1}{\sqrt{d}}\)
Deduction : Kinetic gas equation is
PV = \(\frac{1}{3}\) mnu2 = \(\frac{1}{3}\) mu2
u2 = \(\frac{3 P V}{M}\) = \(\frac{3}{d}\)
∴ u = \(\sqrt{\frac{3 \mathrm{P}}{d}}\)
At constant pressure u = k. \(\frac{1}{\sqrt{d}}\)
k = constant; or u ∝ \(\frac{1}{\sqrt{d}}\)
∴ The rate of diffusion of gases depends upon the velocity of the gas molecules.
So r ∝ \(\frac{1}{\sqrt{d}}\).
This is Graham’s law.

b) Dalton’s law of partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.
Deduction :
Consider a gas present in vessel of volume = V
no. of molecules = n1
mass of each molecule = m1
RMS velocity = u1
According to kinetic gas equation the pressure of the gas. P1 = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{v}\)
When this gas is replaced by another gas in the same vessel, P2 = \(\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\)
When these two gases are mixed in the same vessel, the total pressure of the mixture is P = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{V}+\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\) = P1 + P2
∴ P = P1 + P2.
This is Dalton’s law of partial pressures.

Question 8.
Derive an expression for Kinetic Energy of gas molecules.
Answer:
Kinetic gas equation is PV = \(\frac{1}{3}\)mnu2
For one mole of gas ‘n’ the no.of molecules will be equal to Avagadro’s number ‘N’.
∴ m × N = ‘M’ (gram molar mass of the gas)
∴ PV = \(\frac{1}{3}\) Mu2
= \(\frac{2}{2}\) × \(\frac{1}{3}\) Mu2
= \(\frac{2}{3}\) × \(\frac{1}{2}\) Mu2
= \(\frac{2}{3}\) (K.E)
Ideal gas equation for 1 mole of gas is PV = RT
∴ \(\frac{2}{3}\) KE = RT
⇒ KE = \(\frac{3}{2}\) RT
For ‘n’ moles KE = \(\frac{3}{2}\) nRT

Question 9.
Define
(a) RMS
(b) average and
(c) most probable speeds of gas molecules. Give their interrelationship.
Answer:
a) RMS speed:
The square root of mean of the squares of the speeds of all molecules of a gas is known as RMS speed (uRMS)
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 9
b) Average speed:
The arthematic mean of speeds of gas molecules is known as average speed (uav).
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 10
c) Most probable speed:
The speed possessed by the maximum number of molecules of the gas is known as most probable speed (ump).
\(u_{m p} \sqrt{\frac{2 R T}{M}}\) = \(\sqrt{\frac{2 P V}{M}}\) = \(\sqrt{\frac{2 P}{d}}\)

Inter relationships : –

  • ump : uav : urms = \(\sqrt{\frac{2 R T}{M}}\) : \(\sqrt{\frac{8 R T}{\pi M}}\) : \(\sqrt{\frac{3 R T}{M}}\)
    = 1 : 1.128 : 1.224.
  • uav = 0.9213 × urms
  • ump = 0.8166 × urms

Question 10.
Explain the physical significance of Vander Waals paramaters.
Answer:
Vander Waals equation : [P + \(\frac{a n^2}{\mathrm{~V}^2}\)] [V – nb] = nRT
Where P = Pressure of the gas
n = Number of moles of the gas
a, b = Vander Waals parameters (or) empirical parameters
V = Volume of the container
R = Gas constant
T = Absolute temperature
Units of ’a’: – bar lit-2 mole-2
Units of ’b’: – lit. mol-1

Significance : –

  • ‘a’ is the measure of magnitude of inter molecular forces (attractive) with in the gas and is independent of temperature and pressure. If ‘a’ value is high the gas can be easily liquified.
  • ‘b’ is the effective volume of the gas molecule. It indicates the effective size of the gas molecules. If the value of ‘b’ is constant over a long range of temperature and pressure then the gas cannot be compressed easily.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 11.
What is Surface Tension of liquids ? Explain the effect of temperature on the surface tension of liquids.
Answer:
Surface tension property (γ): ‘It is defined as the force acting along the surface of a liquid at right angles to any line of 1 unit length.”
It is numerically and dimensionally equal to surface energy. It has dimensions kg. s-2 and in SI unit Nm-1.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 11
If we consider a molecule in the bulk of the liquid it experiences equal intermolecular forces in all directions. Hence there is no net force acting on it. But a molecule at the surface has intermolecular forces from inside only. Therefore there is a net attractive force on it towards the interior of the liquid. Due to this the surface area of the liquid tends to minimise. That is the molecules experience a downward force and have more energy than the molecules in the bulk.
Surface tension decreases with increase of temperature because of increase in K,E. of molecules and decrease in intermolecular forces,
e.g.:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 12

Question 12.
What is Vapour Pressure of liquids? How the Vapour Pressure of a liquid is related to its boiling point ?
Answer:
The pressure exerted by vapour of a liquid when it is in equilibrium with liquid is known as vapour pressure.
Effect of temperature : When the temperature of a liquid increased the average kinetic energy of molecules increases. This increase in kinetic energy overcomes the attractive forces between the liquid molecules so that liquid molecules rapidly escape into air. Thus rise in temperature raises the escaping tendency of molecules. Hence the vapour pressure of a liquid increases with increase in temperature.

The vapour pressure of a liquid increases with increase of temperature. This goes on until the critical temperature of the liquid is reached. Above the critical temperature liquid state does not exist when the vapour pressure of the liquid becomes equal to the external (atmospheric) pressure the liquid is said to be boiling and the temperature at which this happens is known as boiling point.

For water the boiling point is 100°C at 1. atm pressure. If the external pressure is reduced, the liquid boils at lower temperature. The boiling point of a liquid can be increased by increasing the external pressure.

Question 13.
Define Viscosity and Coefficient of Viscosity. How does the Viscosity of liquids varies with temperature.
Answer:
Viscosity : Viscosity is a measure of resistance to flow of liquids. This arises due to internal friction between layers of fluids as they slip past one another while liquid flows.

Coefficient of Viscosity:

The force of friction required to maintain velocity difference of 1 cm. sec-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1 cm2 is called coefficient of viscosity.

  • It is denoted by η
  • F = ηA\(\frac{d u}{d x}\)
  • Units : Poise : In CGS system 1 poise = 1g. cm-1 sec-1.
  • Viscosity of liquids decrease with increase of temperature due to high kinetic energy of molecules that over come the inter molecular forces.

Long Answer Questions

Question 1.
Write notes on Intermolecular Forces.
Answer:
Intermolecular forces :
a) Ion – Dipole forces : Ion dipole forces are mainly important in aqueous solutions of ionic substances such as NaCl in which dipolar water molecules surround the ions.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 13
Water molecules are polar and in them hydrogen atoms possess partial positive charges and oxygen atoms possess partial negative charges due to electronegativity difference between hydrogen and oxygen atoms. When ionic compounds like NaCl dissolve in water, they dissociate into component ions like Na+ and Cl. Now the water molecules orient in the presence of ions in such a way that the positive end of the dipole is near an anion and the negative end of the dipole is near a cation.

b) Dipole-Dipole forces : Neutral but polar molecules experience dipole-dipole forces. These are due to the electrical interactions among dipoles on neighbouring molecules. These forces are again attractive between unlike poles and repulsive between like poles and depend on the orientation of the molecules. The net force in a large collection of molecules results from many individual interactions of both types. The forces are generally weak and are significant only when the molecules are in close contact.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 14

c) London dispersion forces : These forces result from the motion of electrons around atoms. Take, for example, atoms of helium. The electron distribution around a helium atom is for averaged over time spherically symmetrical. However, at a given instant the electron distribution in an atom may be unsymmetrical giving the atom a short – lived dipole moment. This instantaneous dipole on one atom can affect the electron distribution is neighbouring atoms and induce temporary dipoles in those neighbours. As a result, weak attractive forces develop known as London forces or dispersion forces. London forces are generally small. Their energies are in the range 1 – 10k J mol-1.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 15

d) Dipole – Induced Dipole forces : These forces are between polar molecules with permanent dipole moments and the molecules with no permanent dipole moment. Permanent dipole of the polar molecule induces dipole on the electrically neutral molecule by deforming into electronic cloud.
Magnitude of these forces depends on the magnitude of the dipole moment of permanent dipole and polarisatricity of neutral molecule. This interaction is proportional to \(\left(\frac{1}{r^2}\right)\), where r = distance between molecules.

AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter: Gases and Liquids

Question 2.
State Boyle’s law, Charle’s law and Avogadro’s law and derive Ideal gas equation.
Answer:
Boyle’s law : At constant temperature, the volume of a given mass of gas is inversely proportional to pressure of the gas.
If ‘V is the volume of a given mass of the gas and ‘P’ its pressure, then the law can be written as V ∝ \(\frac{1}{P}\) at constant temperature
V = \(\frac{k}{P}\)
∴ PV = k or P1V1 = P2V2 = k

Boyle’s law may also be stated as, “at constant temperature the product of the pressure and volume of a given mass of gas is constant.”

Charles’ law: At constant pressure, the volume of a given mass of gas is directly proportional to absolute temperature.
V ∝ T at constant pressure
V = kT
\(\frac{\mathrm{T}}{\mathrm{T}}\) = k

Where V is the volume and T is the absolute temperature. V1 and V2 are the initial and final volumes of a given mass of the gas at the absolute temperatures T1 and T2 respectively at constant pressure.
\(\frac{V_1}{T_1}\) = k ; \(\frac{V_2}{T_2}\) = k
or \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) = k
Avogadro’s law : Equal volumes of all gases contain equal number of moles at constant temperature and pressure.
V ∝ n (pressure and temperature are constant).

Ideal gas equation : The combination of the above gas laws leads to the development of an equation which relates to the four parameters volume, pressure, absolute temperature and number of moles. This equation is known as ideal gas equation.

In this Boyle’s law and Charles’ law combined together and an equation obtained called the gas equation.
V ∝ \(\frac{1}{P}\) (Boyle’s law) ‘
V ∝ T (Charles law)
V ∝ n (Avogadro’s law)

Combining above three laws, we can write
V ∝ \(\frac{1}{P}\) ∝ T ∝ n (or) V = R × \(\frac{1}{P}\) × T × n
(Or) PV = nRT
Where V = volume of the gas
P = pressure of the gas
n = no. of moles of gas
T = absolute temperature
R = Universal gas constant

Question 3.
Write notes on diffusion of Gases.
Answer:
Diffusion : The property of gases to spread and occupy the available space is known as diffusion.

  • It is a non – directional phenomenon.
    Effusion : The escape of a gas from high pressure region into space through a fine hole is called effusion.
  • It is uni directional phenomenon.

Rate of diffusion : No. of molecules diffused per unit time is called rate of diffusion.

Graham’s law of diffusion : At a given temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of density, vapour density or molecular weight.
r ∝ \(\frac{1}{\sqrt{d}}\) ; r ∝ \(\frac{1}{\sqrt{V D}}\) ; r ∝ \(\frac{1}{\sqrt{M}}\)
If r1 and r2 are the rates of diffusion of two gases d1 and d2 are their densities respectively, then
\(\frac{r_1}{r_2}\) = \(\sqrt{\frac{\mathrm{d}_2}{\mathrm{~d}_1}}\)
This equation can be written as :
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 16
Comparison of the volumes of the gases that diffuse in the same time. Let V1, and V2 are the volumes of two gases that diffuse in the same time t’.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 17
When time of flow is same then \(\frac{r_1}{r_2}\) = \(\frac{V_1}{V_2}\)
When volume is the same then : \(\frac{r_1}{r_2}\) = \(\frac{t_2}{t_1}\)

Applications :

  • This principle is used in the separation of isotopes like U235 and U238.
  • Molar mass of unknown gas can be determined by comparing the rate of diffusion of a known gas molecular mass.
  • Ansil’s alarms which are used in coal mines to detect the explosive marsh gas works on the principle of diffusion.

Question 4.
State and explain Dalton’s law of Partial Pressures.
Answer:
Dalton’s law of Partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.

Explanation: Consider a mixture of three gases in a vessel. Let P1, P2, P3 be the partial pressures of the three gases in the mixture. According to Dalton’s law of partial pressures, the pressure (P) of the gaseous mixture is at the same temperature.
P = P1 + P2 + P3
Let n1, n2, n3 be the number of moles of three gases respectively in the mixture. Let V be the volume of the mixture of gases at T K temperature.
According to ideal gas equation,
P1 = \(\frac{n_1 R T}{V}\) ; P2 = \(\frac{\mathrm{n}_2 \mathrm{RT}}{\mathrm{V}}\) ; P3 = \(\frac{\mathrm{n}_3 \mathrm{RT}}{\mathrm{V}}\)
∴ Total pressure of the mixture P = P1 + P2 + P3
P = \(\frac{n_1 R T}{V}\) + \(\frac{n_2 R T}{V}\) + \(\frac{n_3 R T}{V}\)
P = \(\frac{R T}{V}\)(n1 + n2 + n3)
Since nn1 + n2n2 + nn3 = n
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 18
∴ P1 = x1 P
In a similar way P2 = x2 P and P3 = x3P
∴ Partial pressure = mole fraction × total pressure

Question 5.
Write the postulates of Kinetic Molecular Theory of Gases.
Answer:
Assumptions:

  1. Gases are composed of minute particles called molecules. All the molecules of a gas are identical.
  2. Gaseous molecules are always, at a random movement. The molecules are moving in all possible directions in straight lines with very high velocities. They keep on colliding against each other and against the walls of the vessel at very small intervals of time.
  3. The actual volume occupied by the molecules is negligible when compared to the total volume occupied by the gas.
  4. There is no appreciable attraction or repulsion between the molecules.
  5. There is no loss of kinetic energy when the molecules collide with each other or with the wall of vessel. This is because the molecules are spherical and perfectly elastic in nature.
  6. The pressure exerted by the gas is due to the bombardment of the molecules of the gas on the walls of the vessel.
  7. The average kinetic energy of the molecules of the gas is directly proportional to the absolute temperature, Average K.E. ∝ T.
  8. The force of gravity has no effect on the speed of gas molecules.

Boyle’s law : According to kinetic theory of gases, the pressure of a gas is due to collisions of gas molecules on the walls of the vessel. At a particular temperature the molecules make definite number of collisions with the walls of the vessel; When the volume of the vessel is reduced the molecules have to travel lesser distance only before making collisions on the walls. As a result the number of collisions per unit increases. The pressure then increases, i.e., the pressure increases when the volume is reduced at constant temperature. This explains Boyle’s law.

Charles’ law : According to kinetic theory of gases, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas.
K.E. ∝ T
but K.E. = \(\frac{1}{2}\) mc2

As temperature increases, the velocity of the molecules also increases. As a result the molecules make more number of collisions against the walls of the vessel. This results in an increase of pressure if the volume is kept constant. If the volume is allowed to increase the number of collisions decrease due to the increased distance between the molecules and the walls of the vessel. The pressure then decreases. In other words, with rise of temperature, the volume should increase in order to keep the pressure constant.
V ∝ T at constant pressure.
This is Charles’ law.

Question 6.
Deduce gas laws from Kinetic gas equation.
Answer:
a) Deduction of Boyle’s law :
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu2
PV = \(\frac{1}{3}\) mnu2
= \(\frac{1}{3}\) × \(\frac{2}{2}\) mnu2
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu2
PV = \(\frac{2}{3}\)[Kinetic energy (KE)] [∵ KE‘n’ moles = \(\frac{1}{2}\) mnu2]
PV = \(\frac{2}{3}\)KT [∵ KE = KT According to kinetic theory]
PV = \(\frac{2}{3}\)KT
According to Boyle’s law T is constant
∴ PV = \(\frac{2}{3}\)(constant)
∴ PV = constant
(or)
P ∝ \(\frac{1}{V}\)
Hence Boyle’s law proved from kinetic gas equation.

b) Deduction of Chartes law:
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu2
PV = \(\frac{1}{3}\) mnu2
= \(\frac{2}{2}\) × \(\frac{1}{3}\) mnu2
= \(\frac{2}{3}\) × \(\frac{1}{2}\) mnu2
= \(\frac{2}{3}\) (KE) [Kinetic Energy (KE) = \(\frac{1}{2}\) mnu2]
PV = \(\frac{2}{3}\)KT [According to kinetic theory KE = KT]
\(\frac{V}{T}\) = \(\frac{2}{3}\) × \(\frac{K}{P}\)
According to Charles law P’ is constant
∴ \(\frac{V}{T}\) = constant (or) V ∝ T
Hence Charles law proved from kinetic gas equation.

c) Graham’s law: At constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density, r ∝ \(\frac{1}{\sqrt{d}}\)
Deduction: Kinetic gas equation is
Pv = \(\frac{1}{3}\) mnu2 = \(\frac{1}{3}\) Mu2
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 19
At constant pressure u = k. \(\frac{1}{\sqrt{d}}\)
k = constant; or u ∝ \(\frac{1}{\sqrt{d}}\)
∵ The rate of diffusion of gases depends upon the velocity of the gas molecules.
So r ∝ \(\frac{1}{\sqrt{d}}\)
This is Graham’s law.

d) Dalton’s law of partial pressures : The total pressure exerted by a mixture of chemically non-reacting gases at given temperature and volume, is equal to the sum of the partial pressures of the component gases.

Deduction:
Consider a gas present in vessel of volume = V
no. of molecules = n1
mass of each molecule = m1
RMS velocity = u1

According to kinetic gas equation the pressure of the gas. P1 = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{V}\)
When this gas is replaced by another gas in the same vessel, p2 = \(\frac{1}{3} \frac{m_2 n_2 \cdot u_2^2}{V}\)
When these two gases are mixed in the same vessel, the total pressure of the mixture is P = \(\frac{1}{3} \frac{m_1 n_1 u_1^2}{v}\) + \(\frac{1}{3} \frac{m_2 n_2 u_2^2}{V}\) = P1 + P2
∴ P = P1 + P2
This is Dalton’s law of partial pressures.

Question 7.
Explain Maxwell-Boltzmann distribution curves of molecular speeds and give the important conclusions. Discuss the effect of temperature on the distribution of molecular speeds.
Answer:
According to kinetic gas equation it was assumed that all the molecules in a gas have the same velocity. But it is not correct. When any two molecules collide exchange of energy takes place and hence their velocities keep on changing. At any instant few molecules may have zero velocity, a few molecules may be at high velocities and some may be with low velocities.

The distribution of speeds between different molecules were worked out by Maxwell by applying probability considerations.

If one plots a graph between fraction of molecules \(\frac{\Delta \mathrm{N}}{\mathrm{N}}\) vs velocity one gets distribution curve of the type.
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 20
These curves are shown at different temperatures T1 T2 (T1 < T2)

The graph reveals that

  1. There are no molecules with zero velocity and only very few molecules possess the highest velocity.
  2. The velocities of most of the molecules lie near a mean value.
  3. As the temperature of the gas is increased, the curve becomes more flattened and shifts towards higher velocity. It means that at higher temperature the number of molecules possessing higher velocities is more than at lower temperature.

The peak point corresponds to the most probable velocity. It is the velocity possessed by maximum number of molecules.
The average velocity of the molecules is slightly higher than the most probable velocity. The RMS velocity is slightly higher than the average velocity.

Question 8.
Write notes on the behaviour of real gases and their deviation from ideal behavior.
Answer:
Real gases are also called non – ideal gases; A gas which does not obey ideal gas equation PV = nRT is called Real gas.

  • Real gases show ideal behaviour at low pressure and high temperature.
    The deviation of real gas from ideal behaviour can be measured in terms compressibility factor (Z), which is the ratio of product PV and nRT. (i.e.,) Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)

For ideal gas, Z = 1 at all temperatures and pressures because PV = nRT. The graph of Z Vs P will be a straight line parallel to pressure axis. For gases which deviate from ideality, value of Z deviates from unity. At very low pressures all gases shown Z = 1 and behave as ideal gas. At low pressures, inter molecular forces are negligible hence show ideal behaviour. At high pressures all the gases have Z > 1. These are more difficult to compress. At intermediate pressures, most gases have Z < 1.

Question 9.
Derive the Vander Waals equation of state. Explain the importance of Vander Waal’s gas eqaution.
Answer:
Vander Waal’s equation of state : Vander Waal’s proposed an approximate equation of state which involves the intermolecular interactions that contribute to the deviations of a gas from perfect gas law. It may be explained as follows. The repulsive interactions between two molecules cannot allow them to come closer than a certain distance. Therefore, for the gas molecules the available volume for free travel is not the volume of the container V but reduced to an extent proportional to the number of molecules present and the volume of each exclude.

Therefore, in the perfect gas equation a volume correction is made by changing v to (v – nb). Here, ‘b’ is the proportionality constant between the reduction in volume and the amount of molecules present in the container. P = \(\frac{n R T}{V-n b}\) If pressure is low, the volume is large compared with the volume excluded by the molecules (V > > nb). The nb can be neglected in the denominator and the equation reduces to the perfect gas equation of state.

The effect of attractive interactions between molecules is to reduce the pressure that the gas exerts. The attraction experienced by a given molecule is proportional to the concentration n/V of molecules in that container. As the attractions slow down the molecules, the molecules strike the waals less frequently and strike with a weaker impact. Therefore, we can expect the reduction in pressure to be proportional to the square of the molar-concentration, one factor of n/V showing the reduction in frequency of collisions and the other factor the reduction in the strength of their impulse.

Reduction in pressure ∝ \(\left(\frac{n}{V}\right)^2\)
Reduction in pressure = a. \(\left(\frac{n}{v}\right)^2\),
Where a = the proportionality constant.
Vander Waals equation is
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 21
The equation is called Vander Waals equation of state.

The constants ‘a’ and ‘b’ known as Vander Waals parameters (or) empirical parameters. They depend on the nature of the gas independent of temperature.

Significance : –

  • ‘a’ is the measure of magnitude of inter molecular forces (attractive) with in the gas and is independent of temperature and pressure. If ‘a’ value is high the gas can be easily liquified.
  • ‘b’ is the effective volume of the gas molecule. It indicates the effective size of the gas molecules. If the value of ‘b’ is constant over a long range of temperature and pressure then the gas cannot be compressed easily.

Question 10.
Explain the principle underlying the liquefacation of gases.
Answer:
Liquifacation of gases can be done by decreasing the temperature and increasing the pressure.

Liquefaction of gases: Any gas, if it to be liquefied, it must be cooled below its critical temperature. A gas liquefies if it is cooled below its boiling point at given pressure. For example, chlorine at room pressure say 1 atmosphere can be liquefied by cooling it to – 34.0°C in a dry ice bath. For N2 and O2 that have very low boiling points -196°C and -183°C. Such simple technique is not possible. Then, to liquify such type of gases the technique based on intermolecular forces is used. It is as follows. If the velocities of molecules are reduced to such lower values that neighbours can attract each other by their interaction or intermolecular attractions, then the cooled gas will condense to a liquid.

For this, the molecules are allowed to expand into available volume without supplying any heat from outside. In this, the molecules have to overcome the attractions of their neighbours and in doing so, the molecules convert some of their kinetic energy into potential energy and now travel slowly. The average velocity decreases and therefore the temperature of the gas decreases and the gas cools down compared to its temperature before its expansion. For this the gas is allowed to expand through a narrow opening called throttle. This way of cooling of gas by expansion from high pressure side to low pressure is called Joule – Thomson effect.

Question 11.
Write notes on the following properties of liquids
(a) Vapour Pressure
(b) Surface Tension
(c) Viscosity.
Answer:
(a) Vapour Pressure : The pressure exerted by vapour of a liquid when it is in equilibrium with liquid is known as vapour pressure.
Effect of temperature : When the temperature of a liquid increased the average kinetic energy of molecules increases. This increase in kinetic energy overcomes the attractive forces between the liquid molecules so that liquid molecules rapidly escape into air. Thus rise in temperature raises the escaping tendency of molecules. Hence the vapour pressure of a liquid increases with increase in temperature.

The vapour pressure of a liquid increases with increase of temperature. This goes on until the critical temperature of the liquid is reached. Above the critical temperature liquid state does not exist when the vapour pressure of the liquid becomes equal to the external (atmospheric) pressure the liquid is said to be boiling and the temperature at which this happens is known as boiling point.

For water the boiling point is 100°C at 1. aim pressure. If the external pressure is reduced, the liquid boils at lower temperature. The boiling point of a liquid can be increased by increasing the external pressure.

b) Surface tension property (γ) : “It is defined as the force acting along the surface of a liquid at right angles to any line of 1 unit length.”
It is numerically and dimensionally equal to surface energy. It has dimensions kg. s-2 and in SI unit Nm-1
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 22

If we consider a molecule in the bulk of the liquid it experiences equal intermolecular forces in all directions. Hence there is no net force acting on it. But a molecule at the surface has intermolecular forces from inside only. Therefore there is a net attractive force on it towards the interior of the liquid. Due to this the surface area of the liquid tends to minimise. That is the molecules experience a downward force and have more energy than the molecules in the bulk.

Surface tension decreases with increase of temperature because of increase in K, E. of molecules and decrease in intermolecular forces.
e.g.:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 23

c) Viscosity : Viscosity is a measure of resistance to flow of liquids. This arises due to internal friction between layers of fluids as they slip past one another while liquid flows.
Coefficient of Viscosity:
The force of friction required to maintain velocity difference of 1cm. sec-1 between two parallel layers of a liquid 1 cm. apart and each layer having an area 1cm2 is called coefficient of viscosity.

  • It is denoted by η
  • F = η A \(\frac{\mathrm{du}}{\mathrm{dx}}\)
  • Units : Poise : In CGS system 1 poise = 1g. cm-1 sec-1.
  • Viscosity of liquids decrease with increase of temperature due to high kinetic energy of molecules that over come the inter molecular forces.

Solved Problems

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?
Solution:
Formula:
P1y1 = P2y2
P1 = 1 bar
V1 = 500 dm3
V2 = 200 dm3
P2 = ?
1 × 500 = P2 × 200
P2 = \(\frac{5}{2}\) = 2.5 bar.

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure ?
Solution:
Formula:
P1V1 = P2V2
P1 = 1.2 bar
V1 = 120 ml
V2 = 180 ml
P2 = ?
1.2 × 120 = P2 × 180
P2 = \(\frac{1.2 \times 12}{18}\)
= \(\frac{2.4}{3}\) = 0.8 bar

Question 3.
Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Solution:
Consider the equation of state .
PV = nRT
PV = \(\frac{w}{M} R T\)
P = \(\frac{W}{V} \times \frac{R T}{M}\)
P = \(\frac{\mathrm{dRT}}{\mathrm{M}}\) (∵ d = \(\frac{w}{V}\))
From the above relation
P ∝ d

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution:

  • Given two gases one is unknown oxide and another one is dinitrogen.
  • Density of two gases is same
    AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 24

Question 5.
Pressure of 1 gm. of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Solution:
Given
Weight of gas A = 1 gm
Weight of gas B = 2 gms
Molecular mass of A = MA
Molecular mass of B = MB
Pressure of A = PA = 2 bar
Given Total pressure = 3 bar (PA + PB)
∴ PB = 3 – 2 = 1 bar
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 25

Question 6.
The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Solution:
Chemical equation is
2Al + 2 NaOH + 2H2O → 2 NaAlO2 + 3H2
From the above equation
2 gram atom of Al liberates 3 moles of H2 at NTP
2 × 27 gms Al liberates 3 × 22.4 lit.
0.15 gms of Al liberates?
= \(\frac{0.15 \times 3 \times 22.4}{2 \times 27}\)
= 0.1866 li.t = 186.6 ml
P1 = 1.013 bar P2 = 1 bar
V1 = 186.6 ml V2 = ?
T1 = 273 K T2 = 20° C = 293 K
Formulae:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 26

Question 7.
What will be the pressure extracted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °c?
Solution:
Formula:
Given 3.2 gmš of CH4
no.of moles of CH4 = \(\frac{w t}{\text { GMW }}\) = \(\frac{3.2}{16}\) = 0.2
no.of moles of CO2 = \(\frac{4.4}{44}\) = 0.1
∴ n = \(\mathrm{n}_{\mathrm{CH}_4}\) + \(\mathrm{n}_{\mathrm{CO}_2}\)
= 0.2 + 0.1 = 0.3
R = 8.314
T = 27°C = 300 K
V = 9 dm3
PV = nRT
P = \(\frac{n R T}{V}\)
= \(\frac{0.3 \times 8.314 \times 300}{9}\) = 83.14
= 83.14 × 103 pa
= 83.14 × 104 pa
∴ P = 8.314 × 104 pa

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution:
Case – I
Hydrogen gas
P1 = 0.8 bar
P2 = ?
V1 = 0.5 lit
V2 = 1.0 lit
P1y1 = P2V2
P2 = \(\frac{0.8 \times 0.5}{1}\)
P2 = 0.4 bar
Partial pressure of H2 = 0.4 bar. \(\left[\mathrm{P}_{\mathrm{H}_2}\right]\)

Case-II:
Oxygen gas
P1 = 0.7 bar
V1 = 2 lit
V2 = 1.0 lit
P2 = ?
P1V1 = P2V2
P2 = \(\frac{P_1 V_1}{V_2}=\frac{0.7 \times 2}{1}\)
= 1.4 bar
Partial pressure of O2 = 1.4 bar. \(\left[\mathrm{P}_{\mathrm{O}_2}\right]\)
∴ Total pressure = \(P_{\mathrm{H}_2}\) + \(P_{\mathrm{O}_2}\)
= 0.4 + 1.4 = 1.8 bar

Question 9.
Density of a gas is found to be 5.46 g/dm3 at 27 °c at 2 bar pressure. What will be its density at STP?
Solution:
d1 = 5.46 gm/dm3
T1 = 27° C = 300 K
P1 = 2 bar
P2 = 1.013 bar (STP)
T2 = 273 K(STP)
d2 = ?
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 27

Question 10.
34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °c and 0.1 bar pressure. What is the molar mass of phosphorus ?
Solution:
P = 0.1 bar
W = 0.0625 gms
R = 0.083 bar dm3 k-1 mol-1
V = 34.05 × 10-3 lit
T = 546°C = 819 K
Formula:
PV = nRT
PV = \(\frac{w}{M} R T\)
0.1 × 34.05 × 10-3 = \(\frac{0.0625}{M}\) × 0.083 × 819
M = \(\frac{0.0625 \times 0.083 \times 819}{0.1 \times 34.05 \times 10^{-3}}\)
= \(\frac{0.0625 \times 83 \times 819}{34.05}\)
= 124.77 gm/mole.

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out ?
Solution:
Formula:
T1 = 27° C – 300 K
T2 = 477° C = 750 K
\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)
\(\frac{V_1}{300}\) = \(\frac{V_2}{750}\)
V2 = \(\frac{750 \times \mathrm{V}_1}{300}\)
V2 = 2.5 V1
The volume of air expelled = V2 – V1
= 2.5V1 – V1
= 1.5V1
Fraction of air expelled out
= \(\frac{1.5 \mathrm{~V}_1}{2.5 \mathrm{~V}_1}\) = \(\frac{1.5}{2.5}\) = \(\frac{15}{25}\) = \(\frac{3}{5}\)

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R = 0.083 bar dm3 k-1 mol-1)
Solution:
Formulae: —
P = 3.32 bar
V = 5 dm3
R = 0.083 bar dm3 k-1 mol-1
n = 4 moles
PV = nRT
T = \(\frac{\mathrm{PV}}{\mathrm{nR}}\)
= \(\frac{3.32 \times 5}{4 \times 0.083}\)
= \(\frac{16.6}{0.332}\)
= 50
∴ T = 50 k

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution:
14 gms of N2 gas contains
6.023 × 1023 atoms
1.4 gms of N2 gas contains
6.023 × 1022 atoms
Each ‘N’ atom contains 7 electrons.
∴ Number of electrons present in 1.4 gms of Nitrogen
= 6.023 × 1022 × 7
= 42.161 × 1022
= 4.2161 × 1023 electrons.

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains are distributed each second ?
Solution:
Given that
1010 grains are distributed in each second i.e., one second Avagodro number means 6.023 × 1023
6.023 × 1023 grains distributed in ?
X seconds
x = \(\frac{6.023 \times 10^{23}}{10^{10}}\) = 6.023 × 1013 seconds
The time taken to distribute the one Avagadro number of grains
= \(\frac{6.023 \times 10^{13}}{60 \times 60 \times 24 \times 365}\)
= \(\frac{6.023 \times 10^{13}}{3.153 \times 10^7}\) = 1.909 × 106 years.

Question 15.
Ammonia gas diffuses through a fine hole at the rate 0.5 lit min-1. Under thé same conditions find the rate of diffusion of chlorine gas.
Solution:
Rate of diffusion of ammonia (r1)
= 0.5 lit min-1
Molecular weight of ammonia (M1) = 17
Rate of diffusion of chlorine (r2) = ?
Molecular weight of chlorine (M2) = 71
According to Graham’s law of diffusion
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 28
∴ Rate of diffusion of Cl2 = 0.245 lit/min.

Question 16.
Find the relative rates of diffusion of CO2 and Cl2 gases.
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 29

Question 17.
If 150 mL carbon monoxide effused in 25 seconds, what volume of methane would diffuse in same time?
Solution:
Rate of diffusion of CO (r1)
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 30
Molecular weight of CO (M1) = 28
Rate of diffusion of methane (r2)
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 31
Molecular weight of methane (M2) = 16
According to Graham’s law of diffusion
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 32

Question 18.
Hydrogen chloride gas is sent into a 1oo metre tube from one end ‘A’ and ammonia gas from the other end ‘B’, under similar conditions. At what distance from ‘A’ will be the two gases meet?
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 33
The two gases HCl and NH3 diffuse into the pipe from the ends A and B respectively to meet at a point O as indicated by formation of white ring of NH4Cl. If the distance AO is x meters, the distance OB will be (100 – x) metres.
According to Graham’s law of diffusion. Ration of rates of diffusion of HCl and NH3 gas is given by
\(\frac{735 \mathrm{~mm} \times 101.3 \mathrm{kPa}}{1760 \mathrm{~mm}}\) = 98 k Pa
It means that the two gases meet at the point O such that the ratio of the distances from the end A to O and B to O is 0.68 : 1.00
∴ \(\frac{0.68}{1}\) = \(\frac{x}{(100-x)}\)
or 0.68 (100 – x) = x or 68 – 0.68 x = x
or 68 = x + 0.68 x or 68 = x (1 + 0.68)
= 1.68 x
x = \(\frac{68}{1.68}\) = 40.48 metres.
Hence, the two gases meet at a distance of 40.48 metres from the end ‘A’.

Question 19.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K-1 mol-1.
Solution:
Formula:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 34

Question 20.
Calculate the total pressure in a mixture of 3.5g of dinitrogen 3.0g of dihydrogen and 8.0g dioxygen confined in vessel of 5 dm3 at 27°C (R = 0. 083 bar dm3 k-1 mol-1)
Solution:
Formula:
V = 5 dm3
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 35

Question 21.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m-3 and R = 0.083 bar dm3 k-1 mol-1).
Solution:
Formula :
r = 10 m
m = 100 kg
T = 27° C = 300 K
d = 1.22 kg/m3
Volume of the ballon = \(\frac{4}{3} \pi r^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times 10^3\)
= 4190.5 m3
P = 1.66 bar
T = 300 K
V = 4190.5 m3
R = 0.083 bar dm3 k-1 mol-1
PV = nRT
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}\) = \(\frac{1.66 \times 4190.5}{0.083 \times 10^{-3} \times 300}\)
= 2793.70 moles
= 0.083 × 10-3 bar m3k-1mol-1
∴ Weight of 279370 moles of He
= 279390 × \(\frac{4}{1000}\)
= 1117.48 kg
Total weight of balloon = 100 + 1117.48
= 1217.48 kg
Maximum weight of He = V × d
=4190.5 × 1.2
= 5028.6 kg
∴ Payload = 5028.6 – 1219.48
= 3811.12 kg.

Question 22.
Calculate the volume occupied by 8.8 g of CO2 at 31 .1°C and 1 bar pressure. R = 0.083 bar L K-1 mol-1.
Solution:
Formula:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 36

Question 23.
2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Solution:
Given unknown gas and dihydrogen
For unknown gas
V1 = V
n1 = \(\frac{2.9}{\mathrm{~m}}\)
T1 = 95° C = 368 K
P1V1 = n1RT1
P1 = \(\frac{n_1 R T_1}{V_1}\)
= \(\frac{2.9}{m} \times \frac{R \times 368}{V}\)
For dihydrogen
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 37

Question 24.
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Solution:
Given 20% by wt of dihydrogen so 80% oxygen remained for dihydrogen
n = \(\frac{w t}{\text { GMwt }}\) = \(\frac{0.2}{2}\) = 0.1
For dioxygen
n = \(\frac{w t}{\text { GMwt }}\) = \(\frac{0.8}{32}\) = 0.025
mole fraction of H2
= \(\frac{0.1}{0.1+0.025}\) = \(\frac{0.1}{0.125}\) = 0.8
Partial pressure of dihydrogen
= mole fraction of H2 × Ptotal
= 0.8 × 1 = 0.8 bar

Question 25.
What would be the SI unit for the quantity \(\frac{\mathrm{PV}^2 \mathrm{~T}^2}{\mathrm{n}}\)?
Solution:
Given quantity \(\frac{\mathrm{PV}^2 \mathrm{~T}^2}{\mathrm{n}}\) = \(\frac{\mathrm{N} / \mathrm{m}^2\left(\mathrm{~m}^3\right)^2(\mathrm{~K})^2}{\text { mole }}\)
= N × mole-1m4k2
∴ The given quantity has SI units Nm4k2 mole-1.

Question 26.
In terms of Charles’ law explain why — 273°C is the lowest possible temperature.
Solution:
According to Charles law if we put the value of t = -273°C
in the equation Vt = V0 \(\left[\frac{273.15+t}{273.15}\right]\).
In this case the volume of the gas becomes zero.
V0 = Volume at 0° C
Vt = Volume at t° C

  • This means the gas will not exist
  • In fact all gases liquified before this temperature.

Question 27.
Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has
stronger intermolecular forces and why?
Solution:
Given Critical temperatures of CO2, CH4
TC (CO2) = 31.1°C
TC (CH4) = -81.9°C

  • The gas with highest critical temperature value can be easily liquified because of high inter molecular forces.
    ∴ TC(CO2) is very high.
    So CO2 gas liquified easily.
  • ‘He’ gas has low T value so it is highly difficult to liquify.

Question 28.
Air is cooled form 25°C to 0°C. Calculate the decrease in rms speed of the molecules.
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 40

Question 29.
Find the rms, most probable and average speeds of SO2 at 27°c.
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 41

Question 30.
Find the RMS. average and most probable speeds of O2 at 27°c.
Solution:
AP Inter 1st Year Chemistry Study Material Chapter 4 States of Matter Gases and Liquids 42
uaverage = 0.9213 × urms
= 0.9213 × 4.835 × 104
= 4.455 × 104 cm/sec.
ump = 0.8166 × urms
= 3.948 × 104 cm/sec.

Question 31.
Give the values of Gas constant ‘R’ in different units.
Answer:
R = 0.0821 lit. atm. K-1 .mol-1
= 8.314 J.K-1. mole-1
= 1.987 (or) 2 cal.K-1.mol-1
= 8.314 × 107 erg.K-1.mol-1

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties

Very Short Answer Questions

Question 1.
What is the difference in the approach between Mendeleev’s periodic law and the modern periodic law?
Answer:

  • According to Mendeleev, elements’ physical and chemical properties are periodic functions of their atomic weights.
  • According to modem periodic law, elements’ physical and chemical properties are periodic functions of their atomic numbers.

Question 2.
In terms of period and group, where would you locate the element with Z = 114?
Answer:
Element Z = 114 is present in 7th period and IVA group (Group – 14)

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 3.
Write the atomic number of the element, present in the third period and seven¬teenth group of the periodic table.
Answer:
The Element present in 3rd period and Group – 17 (VIIA group) is chlorine (Cl). It’s atomic number is 17.

Question 4.
Which element do you think would have been named by
a) Lawrence Berkeley Laboratory .
b) Seaborg’s group
Answer:
a) Lawrence Berekeley Laboratory – Lanthanide
b) Seaborg’s group – Actinide (Transuranic element).

Question 5.
Why do elements in the same group have similar physical and chemical properties ?
Answer:
Elements in the same group have same no. of valency shell electrons and have similar outer elec¬tronic configuration so these have similar physical and chemical properties.

Question 6.
What are representative elements ? Give their valence shell configuration.
Answer:

  • Representative elements are s and p-block elements except zero group.
  • These have general electronic configuration ns1-2np1-5.

Question 7.
Justify the position of f-block elements in the periodic table.
Answer:
The two series of elements lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table, though they belong to the sixth and seventh periods of third group (III B).

The justification for assigning one place to these elements has been given on the basis of their similar properties. The properties are so similar that elements from Ce to Lu can be considered as equivalent to one element. In case these elements are assigned different positions (i.e.,) arranged in order of their increasing atomic numbers, the symmetry of the whole arrangement would be disrupted. The same explanation can be given in the case of actinides.

Question 8.
An element ‘X’ has atomic number 34. Give its position in the periodic table.
Answer:
The element X with atomic no (z) = 34 has electronic configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4. Hence the element is present in 4,h period and 16th group (VIA group).

Question 9.
What factors impart characteristic properties to the transition elements ?
Answer:
Transition elements exhibits characteristic properties because

  • The differentiating electron enters into penultimate d-subshell
  • These elements have small size.
  • These possess high effective nuclear charge.

Question 10.
Give the outer shells configuration of d-block and f-block elements.
Answer:

  • The outer shell electronic configuration of d-block – elements is ns1-2 (n-1)d1-10
  • The outer shell electronic configuration of f-block – elements is ns2(n-1)d0 (or) 1 (n – 2) f1-14

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 11.
State and give one example for Dobereiner’s law of triads and Newland’s law of octaves.
Answer:
1) Dobereiner law :

  • According to Dobereiner in a traid (3 – elements) the atomic weight of the middle element is the arithmatic mean of the other two elements.
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 1

2) Newland’s law of octaves : According to Newlands, the elements arranged in the increasing order of atomic weights noticed that every eight element had properties similar to the first element. This relationship was just like every eight note that resembles the first in octaves of music.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 2

Question 12.
Name the anomalous pairs of elements in the Mendaleev’s periodic table.
Answer:
In Mendeleev’s periodic table anamalous pairs are the elements whose atomic weights increasing order is reversed.
Eg:
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 3

Question 13.
How does atomic radius vary in a period and in a group ? How do you explain the variation ?
Answer:
In a period : Atomic radius decreases generally from left to right in a period.
Reason : In periods electrons are entered into same subshells.
In a group : Atomic radius increases generally from top to bottom in a group.
Reason : In groups electrons are entered into new subshells.

Question 14.
Among N-3; O-2, F, Na+, Mg+2 and Al+3
a. What is common in them ?
b. Arrange them in the increasing ionic radii.
Answer:
Given ions are
N-3, O-2, F, Na+, Mg+2 and Al+3.
a) The above ions have same number of electrons (All have 10 electrons). So these are called iso electronic species.
b) The increasing order of ionic radii among above ions is
Al+3 < Mg+2 < Na+ < F < O-2 < N-3
Reason : – In case of iso electronic species as the nuclear charge increases ionic radii decreases.

Question 15.
What is the significance of the term isolated gaseous atom while defining the ionization enthalpy.
Hint: Requirement for comparison.
Answer:

  • Isolated gaseous atom’s ionisation enthalpy is taken as reference value and it is required to compare this values to various ions of this elements and to compare this values with various elements.

Question 16.
Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.
Answer:
Given that the energy of the electron in the ground state for hydrogen atom = – 2.18 × 10-18 J.
For 1 mole of atoms is given by – 2.18 × 10-18 J × 6.023 × 1023
= -13.13 × 105 J/Mole
∴ Ionisation enthalpy of hydrogen atom = 13.13 × 105 J/Mole.

Question 17.
Ionization enthalpy, (IE1) of O is less than that of N — explain.
Answer:

  • Oxygen has electronic configuration 1s2 2s2 2p4
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 4
  • Nitrogen has electronic configuration 1s2 2s2 2p3
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 5
  • Nitrogen has half filled shell and is stable so more amount of energy is required to remove an electron, than in oxygen.
    Hence IE, of ‘O’ is less than that of ‘N’.

Question 18.
Which in each pair of elements has a more negative electron gain enthalpy?
a. O or Fb. F or Cl
Answer:
a)
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 6
has more negative electron gain enthalpy than that of
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 7
b)
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 8
has more negative electron gain enthalpy than that of
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 9

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 19.
What are the major differences between metals and non-metals ?
Answer:
Metals

  • These are generally in solid form (Except Hg)
  • These are good conductors of heat and electricity.
  • These have high m.pts and b.pts.
  • Generally these are electropositive.
  • These forms more ionic compounds.

Nón metals

  • These may be solids (or) gases (or) liquids.
  • These are not good conductors of heat and electricity.
  • These have low m.pts and b.pts.
  • Generally these are electronegative.
  • These forms more covalent compounds.

Question 20.
Use the periodic table to identify elements.
a. With 5 electrons in the outer subshell
b. Would tend to lose two electrons
c. Would tend to gain two electrons.
Answer:
a) The elements possessing 5 electrons in the outer most shell are group 15 (VA) elements.

  • General outer electronic configuration is ns2 np3 Eg. : N, P, As………

b) The elements tend to lose two electrons are Group — II elements.

  • General outer electronic configuration is ns2 Eg: Mg, Ca, Sr etc.

c) The elements tend to gain two electrons are Group – VIA elements (16th group).

  • General outer electronic configuration is ns2 np4 Eg : O, S, Se ………

Question 21.
Give the outer electronic configuration of s, p, d and f – block elements.
Answer:
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 10

Question 22.
Write the increasing order of the metallic character among the elements B, Al, Mg and K.
Answer:
Given elements are B, Al, Mg and K
The increasing order of metallic character is
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 11

Question 23.
Write the correct increasing order of non – metallic character for B, C, N, F and Si.
Answer:
Given elements are B, C, N. F and Si
The increasing order of non- metallic character is
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 12

Question 24.
Write the correct increasing order of chemical reactivity in terms of oxidizing property for N, O, Fand Cl.
Answer:
The correct increasing order of chemical reactivity in terms of oxidizing property for N, O, F and Cl is F > 0 > Cl > N.

Question 25.
What is electronegativity ? How is this useful in understanding the nature of elements?
Answer:
Electronegativity : The tendency of an element (or) atom to attract the shared pair of electrons towards itself in a molecule is called electronegativity.

  • On the basis of electronegativity values nature of elements can be predicted. Higher electro-negativity values indicates that element is non metal and lower values indicates that the element is a metal.
  • On the basis of electronegativity values bond nature also predicted (Ionic/covalent).

Question 26.
What is screening effect? How is it related to lE?
Answer:
The decrease of nuclear attraction on outer most shell electrons due to presence of inner energy electrons is called screening effect.

  • As the screening effect increases LE. values decreases.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 27.
How are electronegativity and metallic & non-metallic characters related?
Answer:

  • Greater the electronegativity values of an element indicates that non metallic nature and low metallic nature of that element.
  • Lower the electronegativity value of an element indicates that low non metallic nature and high metallic nature of that element.

Electronegativity ∝ Non metallic nature
Electronegativity ∝ \(\frac{1}{\text { Metallic Nature }}\)

Question 28.
What is the valency possible to Arsenic with respect to oxygen and hydrogen?
Answer:

  • The valency of Arsenic with respect to hydrogen is ‘3’
    Eg : AsH3
  • The valency of Arsenic with respect to oxygen is ‘5’
    Eg : As2O5

Question 29.
What is an amphoteric oxide? Give the formula of an amphoteric oxide formed by an element of group – 13.
Answer:
The oxide which contains both acidic as well s basic nature is called amphoteric oxide.

  • The oxides reacts with both acids and bases and forms salts.
    Eq : Al2O3 is one of the amphoteric oxide formed by the Group — 13 element Aluminium.

Question 30.
Name the most electronegative element. Is it also having the highest electron gain enthalpy? Why or Why not?
Answer:
The most electronegative element is fluorine (F).

  • It doesnot have high electron gain énthalpy.

Reasons : —

  • Due to small size
  • Due to high inter electronic repulsions.
  • Chlorine has high electron gain enthalpý.

Question 31.
What is diagonal relation? Give one pair of elements, that have this relation.
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship. e.g. : (Li, Mg); (Be, Al); (B, Si)
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 13

Question 32.
How does the nature of oxides vary in the third period?
Answer:
In 3rd period from left to right the oxide nature varies from high basic nature to high acidic nature.

  • Basic nature gradually decreases and acidic nature gradually increases.
  • AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 14

Question 33.
Radii of iron atom and its ions follow Fe > Fe2+ > Fe3+ – explain.
Answer:
When the positive charge on the ion increases, the effective nuclear charge on the outer electrons increases.
Hence the ionic size decreases in the order Fe > F2+ > F3+.

Question 34.
IE2 > IE1 for a given element — why?
Answer:
IE2 > IE1 for a given element

Reason : —

  • IE1 means minimum amount of energy required to remove an electron from isolated neutral
    gaseous atom.
  • IE2 means minimum amount of energy required to remove an electron from uni positive ion.
  • In case of unipositive ion nuclear attraction increases on outer most electrons than in isolated gaseous atom. So more amount of energy needed to remove an electron from unipositive ion.
    Hence IE2 > → IE1.

Question 35.
What is Ianthanide contraction? Give one of its consequences.
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

  1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
  2. Due to this, it becomes difficult to separate lanthanides from a mixture.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 36.
What is the atomic number of the element, having maximum number of unpaired 2p electrons ? To which group does it belong ?
Answer:

  • The atomic number of the element having maximum no.of unpaired 2p electrons is 7′ (Z = 7)
  • Element is nitrogen.
  • Electronic configuration is 1s2 2s2 2p3 (3 unpaired 2p electrons) AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 15

Question 37.
Sodium is strongly metallic, while chlorine is strongly non-metallic – explain.
Answer:
Sodium is an alkali metal and it is present in group – I and it has the ability to lose the valency electron readily.

  • It has high electropositive nature. So it has metallic nature.
  • Chlorine is a halogen and it is placed in Group – 17 and it has the ability to gain the electron readily.
  • It has high electronegative nature. So it has non metallic nature.

Question 38.
Why are zero group elements called noble gases or inert gases ?
Answer:

  • Zero group elements has general outer electronic configuration ns2 np6 (except for He).
  • These contains stable octet configuration. So these are stable and chemically inert. Hence these are called inert gases.
  • These elements neither lose nor gain electrons. Hence these are called ‘noble gases’.

Question 39.
Select in each pair, the one having lower ionization energy and explain the reason,
a. I and I
b. Br and K
c. Li and Li+
d. Ba and Sr
e. O and S
f. Be and B
g. N and O
Answer:
a) I has lower ionisation energy than I because of increase of size \(\mathrm{I}^{\ominus}\) ion than ‘I’.
b) K has lower ionisation energy than ‘Br’ because of low electronegative value of K (0.8) than ‘Br’ (2.8).
c) ‘Li” has lower ionisation energy than Li+ because of large size of ‘Li’ than Li+.
d) ‘S’ has lower ionisation energy than ‘O’ because of large size of ‘S’ than ‘O’.
e) ‘B’ has lower ionisation energy than ‘Be’ because ‘Be’ has completely filled electronic configuration (1s2 2s2).
f) ‘O’ has lower ionisation energy than ‘N’ because ‘N’ has half filled electronic configuration (1s2 2s2 2p3).

Question 40.
IE1 of O < IE1 of N but IE2 of O > IE2 of N – Explain.
Answer:

  • ‘N’ has half filled electronic configuration (1s2 2s2 2p3)
    So IE1 of O < IE1 of ‘N’.
  • O+ ion has half filled electronic configuration (1s2 2s2 2p3)
    So IE2 of O > IE1 of N.

Question 41.
Na+ has higher value of ionization energy than Ne, though both have same electronic configuration – Explain.
Answer:
Na+ has higher value of I.E. than Ne, though both have same electronic configuration.

Reason : –

  • Both have electronic configuration 1s2 2s2 2p6
  • In case of Na+ ion effective nuclear charge increases and size decreases than in ‘Ne’.

Question 42.
Which in each pair of elements has a more electronegative gain enthalpy ? Explain.
a. N or O
b. F or Cl
Answer:
a) Oxygen has high electronegative gain enthalpy than Nitrogen because ‘N’ has stable half filled electron configuration.
b) Chlorine (- 349 KJ /mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

Question 43.
Electron affinity of chlorine is more than that of fluorine – explain.
Answer:
Chlorine (- 349 KJ / mole) has high electronegative gain enthalpy than Fluorine (- 328 KJ/mole) because ‘F’ has small size and more inter electronic repulsions.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 44.
Which in each has higher electron affinity ?
a. F or Cl
b. O or O-
c. Na+ or F
d. F or F
Answer:
a) Fluorine has high electron affinity than Cl ion because of inert gas configuration of Cl ion.
b) Oxygen has high electron affinity than O because O has positive of 2nd electron affinity.
c) F has high electron affinity than Na+ because Na+ has inert gas configuration.
d) F has high electron affinity than F because F has inert gas configuration.

Question 45.
Arrange the following in order of increasing ionic radius :
a. Cl, P-3, S-2, F
b. Al+3, Mg+2, Na+, O-2, F
c. Na+, Mg+2, K+
Answer:
a) The increasing order of ionic radius is F < Cl < S-2 < P-3
b) The increasing order of ionic radius is Al+3 < Mg+2 < Na+ < F < O-2
c) The increasing order of ionic radius is Mg+2 < Na+ < K+

Question 46.
Mg+2 is smaller than O-2 in size, though both have same electronic configuration –
explain.
Answer:
Mg+2 and O-2 ions are iso electronic species.
In case of iso electronic species nuclear charge increases size of ion decreases. So Mg+2 has small size than O-2.

Question 47.
Among the elements B, Al, C and Si
a. Which has the highest first ionization enthalpy ?
b. Which has the most negative electron gain enthalpy ?
c. Which has the largest atomic radius ?
d. Which has the most metallic character ?
Answer:
a) Highest I.E. is possessed by the element carbon
b) Most negative gain enthalpy is for carbon (- 122 KJ/mole)
c) Large atomic radius is for Al (1.43 A)
d) Most metallic nature having element is ‘Al’.

Question 48.
Consider the elements N, P, O and S and arrange them in order of ;
a. Increasing first ionization enthalpy
b. Increasing negative electron gain enthalpy
c. Increasing non-metallic character
Answer:
a) Increasing first Ionisation energy order is S < P < O < N.
b) Increasing negative electron gain enthalpy order is N < P < O < S. .
c) Increasing non metallic nature order is P < N < S < O.

Question 49.
Arrange in given order :
a. Increasing EA :O, S and Se
b. Increasing IE1 : Na, K and Rb
c. Increasing radius : I, I+ and I
d. Increasing electronegativity : F, Cl, Br, I
e. Increasing EA : F, Cl, Br, I
f. Increasing radius : Fe, Fe+2, Fe+3
Answer:
a) Increasing order of electron affinity is O < Se < S.
b) Increasing order of IE1 is Rb < K < Na.
c) Increasing order of radius is I+ < I < I
d) Increasing order of electronegativity is I < Br < F < Cl
e) Increasing order of electron affinity is I < Br < F < Cl
f) Increasing order of radius is Fe+3 < Fe+2 < Fe

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 50.
a. Name the element with highest ionization enthalpy
b. Name the family with highest value of ionization enthalpy.
c. Which element possesses highest electron affinity ?
d. Name unknown elements at the time of Mendeleef
e. Name any two typical elements.
Answer:
a) Highest I.E1 possessing element is ‘Helium’.
b) The family that possess highest values of I.E is Noble gases (or) inert gases.
c) Highest electron affinity element is ‘Chlorine’.
d) Unknown elements at the time of mendeleef are Germanium (Eka silicon). Scandium (Eka Alu-minium), Gallium (Eka Boron).
e) 3rd period elements are called typical elements
Eg : Al, Si, P, Na, Mg etc

Question 51.
a. Name any two bridge elements.
b. Name two pairs showing diagonal relationship.
c. Name two transition elements.
d. Name two rare earths.
e. Name two transuranic elements.
Answer:
a) Bridge elements 2nd period elements are called Bridge elements Eg : Be, B.
b) i) ‘Li’ diagonally relates with ‘Mg’,
ii) ‘Be diagonally relates with ‘Al’.
c) Scandium, Titanium, Vanadium, Chromium, etc., are examples of transition elements.
d) Lanthanides are called rare earths
Eg : Cerium (Ce), Prasodimium (Pr), Promethium (Pm).
e) Neptunium (Np), Californium (Cf), Fermium (Fm) are examples of transuranic elements.

Question 52.
On the basis of quantum numbers, justify that the 6th period of the periodic table should have 32 elements.
Answer:
6th period contains the subshells 6s, 4f, 5d, 6p
6s can accomodate two electrons (2 elements)
4f can accomodate 14 electrons (14 elements)
5d can accomodate 10 electrons (10 elements)
6p can accomodate 6 electrons (6 elements)

Total no.of electrons can accomodation 6th period are 2 + 14 + 10 + 6 = 32
∴ 6th period of periodic table contains 32 elements.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 53.
How did Moseley’s work on atomic numbers show that atomic number is a fundamental property better than atomic weight ?
Answer:
Mosley’s equation is
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 16
where v = frequency, Z = atomic number a, b = constants
A plot of \(\sqrt{v}\) against ‘Z’ gives a straight line.
However, no such relationship was obtained when the plot was drawn between frequency and the atomic mass. The atomic number of the elements, according to Mosley, stands for serial numbers of the elements in the periodic table. As the atomic number of the elements increase, the wavelengths of characteristic X – rays decrease. Mosley concluded that there is a fundamental quantity in an atom which increases in regular steps with increasing atomic number. The correlation between X – ray spectra and atomic number indicated that an element is characterized by its atomic number and not by atomic mass.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 17

Question 54.
State modern periodic law. How many groups and periods are present in the long form of the periodic table ?
Answer:
Modern periodic law : – The physical and chemical properties of elements are periodic functions of their atomic numbers.

  • In modern periodic table 7 periods and 18 groups are present.

Question 55.
Why f-block elements are placed below the main table ?
Answer:
In Lanthanides 4f – orbitals and in Actinides 5f – orbitals are filled. Since these elements have the same electronic configuration in the ultimate and penultimate shells they have similar properties. Hence they were placed at the bottom of the periodic table though they belongs to the sixth and seventh periods of IIIB groups.

Question 56.
Mention the number of elements present in each of the periods in the long form periodic table.
Answer:
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 18

Question 57.
Give the outer orbit general electronic configuration of
a. Noble gases
b. Representative elements
c. Transition elements
d. Inner transition elements
Answer:
Type of elements – General electronic configurations
a) Noble gases – ns2 np6 (except ‘He’ which has 1s2)
b) Representative elements – ns1-2 np0-5
c) Transition elements – (n – 1)d1-10ns1-2
d) Inner transition elements – (n – 2) f1-14(n – 1)d0, 1 ns2

Question 58.
Give any four characteristic properties of transition elements.
Answer:
Characteristic properties of elements:
a) They exhibit more than one oxidation state.
b) Most of the elements and their ions exhibit colour.
c) These elements and their compounds are good catalysis for various chemical processes.
d) They and their ions exhibit paramagnetic properties.
e) They form useful alloys.

Question 59.
What are rare earths and transuranic elements?
Answer:

  1. Lanthanides are rare earths. In these elements the differentiating electron enters into 4f – orbital.
  2. The elements present after Uranium are called Transurariic elements. All of these are radioactive and synthetic elements.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 60.
What is isoelectronic series? Name a series that will be isoelectronic with each of the following atoms or ions.
a. F
b. Ar
c. He
d. Rb+
Answer:
The species containing same no.of electrons are called Iso electronic species and this series is called Isoelectronic series.
a) F relating series
N-3, O-2, F, Ne, Na+, Mg+2, Al+3
b) Ar relating series P-3, S-2, Cl, Ar, K+, Ca+2
c) ‘He’ relating series H, He, Li+, Be+2
d) Rb+ relating series
As-3, Se-2, Br, Kr, Rb+, Sr+2

Question 61.
Explain why cation is smaller and anion is larger in radii than their parent atoms.
Answer:

  • Cation means a positively charged species which is formed by an atom (or) element when an electron is lost.
    M → M+ + \(\mathrm{e}^\Theta\)
  • Cation has high effective nuclear charge and decrease in size observed.
    Hence cation has smaller radii.
  • Anion means a negatively charged species which is formed by an atom (or) element when an electron is gained.
    M + e → \(\mathrm{M}^{\Theta}\).
  • Anion has very low effective nuclear charge and increase in size observed.
    Hence anion has larger radii.

Question 62.
Arrange the second period elements in the increasing order of their first ionization enthalpies. Explain why Be has higher IE1 than B.
Answer:
The increasing order of the I.E.s of 2nd period elements are as follows.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 19

  • Due to presence of incompletely filled p – orbitals in Boron, its IE value is less.
  • Due to presence of completely filled s2 configuration in ‘Be’, it has higher IE value.

Question 63.
IE1 of Na is less than that of Mg but IE2 of Na is higher than that of Mg – explain.
Answer:
IE1 of Na is less than that of ‘Mg’
Reason :-
Na — has electronic configuration [Ne] 31
Mg — has electronic configuration [Ne] 3s2
Mg has completely filled configuration so Mg has more IE1 than Na.
IE2 of Na is higher than that of Mg

  • AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 20
  • Na+ has stable inert gas configuration so IE2 of Na is very high
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 21
  • By the lose of one electron from Mg+ ion forms Mg+2 ion which is more stable so low amount of energy is required.
    ∴ IE2 of Na is higher than Mg.

Question 64.
What are the various factors due to which the IE of the main group elements tends to decrease down a group ?
Answer:
The factors influencing on IE are

  1. Atomic radius
  2. Nuclear charge
  3. Screening effect
  4. Half filled, completely filled electronic configurations
  5. peretrating power
    • In main group elements in a group IE decrease from top to bottom.

Reason: –
In groups from top to bottom size of elements increases hence IE values decreases.

Question 65.
The first ionization enthalpy values (in KJ mol-1) of group 13 elements are :
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 22
How do you explain this deviation from the general trend ?
Answer:
The given IE, values (in KJ / mole) of group 13 are as follow
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 23

  • In genaral in a group IE values decrease down a group but from the above values we observe that there is no smooth decrease down the group.
  • The decrease from B to Al is due to increase of size.
  • The observed discontinuity in the IE values of Al and Ga, and between In and Tl are due to inability of d- and f – electrons, which have low screening effect to compensate the increase in nuclear charge.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 66.
Would you expect the second electron gain enthalpy of oxygen as positive, more negative or less negative than the first ? Justify.
Answer:
2nd gain enthalpy means energy released when an electron is added to uni negative ion.
O(g) + e → \(\mathrm{O}_{(g)}^{-}\) + 141 KJ/mole
O(g) + e → \(O_{(g)}^{-(2)}\) – 780 KJ/mole

  • 2nd gain enthalpy of oxygen is positive because O ion doesnot accept an electron readily and the entering electron have to over come the repulsive force.

Question 67.
What is the basic difference between the electron gain enthalpy and electropositivity?
Answer:

  • Electron gain enthalpy means energy released when an electron is added to isolated neutral gaseous atom.
  • The tendency to lose the electrons by an element is called electropositivity.
  • Electron gain enthalpy is the measure of electronegativity.
  • Electron gain enthalpy and electropositivity are inversely related.

Question 68.
Would you expect IE1 for two isotopes of the same element to be the same or different ? Justify.
Answer:

  • Isotopes means existance of same elements with different mass no.s
  • The isotope with higher mass no. have low I.E value than the normal isotope.
  • This is due to the less nuclear attraction on valency electrons in case of heavier nuclide.
  • But overall, the IE values of the isotopes are nearly same and the difference in IE values is negligible.

Question 69.
Increasing order of reactivity among group-1 elements is Li < Na < K < Rb < Cs, where as among group-17 elements it is F > Cl > Br > I- explain.
Answer:
a) Increasing order of reactivity among group -1 elements is Li < Na < K < Rb < Cs

Explanation: –

  • Group -1 elements are Alkali metals.
  • Group -1 elements have the tendency to lose the electrons
  • Group -1 elements forms ionic bonds readily by losing electrons
  • Group – 1 elements good reducing agents.
  • Electro positive character increases from top to bottom this is due to increase of size.

b) Increasing order of reactivity among group -17 elements is F > Cl > Br > I

Explanation: –

  • These are halogens (Group – 17)
  • These have high electronegativity due to small size.
  • These have the tendency to gain electrons.
  • In a group from top to bottom electronegativity decrease, due to increase of size.
  • These are oxidising agents and forms ionic bonds by gaining electrons.

Question 70.
Assign the position of the element having outer electronic configuration.
a. ns2np4 for n = 3
b. (n – 1)d2ns2 for n = 4
Answer:
a) ns2np4 for n = 3 .

  • 3s23p4 → element is sulphur
  • Sulphur belongs to VIA group (Group – 16) and 3rd period in periodic table,

b) (n – 1 )d2 ns2 for n = 4
3d2 4s2 → element is Titanium

  • Titanium belongs to IVB group (Group – 4) and 4th period in the periodic table.

Question 71.
Predict the formulae of the stable binary compounds that would be formed by the combination of the following pairs of elements.
a. Li and O
b. Mg and N
c. Al and I
d. Si and O
e. p and Cl,
f. Element with atomic number 30 and Cl
Answer:
a) Stable binary compound formed between Li and O is Li2O (Lithiumoxide).
b) Stable binary compound formed between Mg and N is Mg3N2 (Magnesium nitride).
c) Stable binary compound formed between Al and I is AlI3 (Aluminium Iodide).
d) Stable binary compound formed between Si and I is SiO2 (Silicondioxide).
e) Stable binary compound formed between P and Cl is PCl3 and PCl5 (Phosphorous trichloride and phosphorous pentain chloride).
f) Stable binary compound formed between element with At. NO – 30 and Cl is ZnCl2 (Zinc chloride) [At. No – 30 – (Zn)].

Question 72.
Write a note on the variation of metallic nature in a group or in a period.
Answer:
Metals shows electropositive nature (i.e.,) loss of electrons and form positive ions.
Non – metals shows electronegative nature (i.e.,) gain of electrons and form negative ions.
Periodicity:
a) Down the group : Going down a group of the periodic table, the tendency to form positive ions, increases. That means there is an increase in metallic nature as the size of atom increases down the group.
b) Along a period : From left to right in a period, size of atom decreases. So there is decrease in metallic nature.

Question 73.
How does the covalent radius increase in group- 7 ?
Answer:

  • Covalent radius increases in a group from top to bottom
  • The increase of covalent radius in VIIA group elements as follows.
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 24

Question 74.
Which element of 3rd period has the highest IE1 ? Explain the variation of IE1 in this period.
Answer:

  • In periods IE values increase from left to right
  • Among 3rd period elements Argon (Ar) [Z = 18] possess highest IE,sub>1 value.
  • IE1 values of 3rd period elements given below.
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 25

Exceptions:

  • IE, of ‘Mg’ is higher than ‘Al’ because ‘Mg’ has completely filled ‘s’ subshells.
  • IE, of ‘p’ is higher than ‘s’ because ‘p’ has half filled ‘p’ subshells.

Question 75.
What is valency of an element ? How does it vary with respect to hydrogen in the third period?
Answer:
Valency : The combining capacity of an element with another element is called valency.

The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens = no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule.
e.g.:
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 26
Periodicity of valency:
1) Each period starts with valency ‘1’ and ends in ‘0’.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 27

2) In a group valency is either equal to the group number (i.e., upto 4th group) or is equal to (8 – group number) (i.e., from 5th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 76.
What is diagonal relationship ? Give a pair of elements having diagonal relationship. Why do they show this relation ?
Answer:
On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship, e.g. : (Li, Mg); (Be, Al); (B, Si)
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 28
Diagonal relationship is due to similar sizes of atoms (or of ions) and similar electronegativities of the representative elements.
Diagonally similar elements posses the same polarizing power.
Polarizing power = \(\frac{\text { (ionic charge) }}{\text { (ionic radius) }^2}\)

Question 77.
What is lanthanide contraction ? What are its consequences ?
Answer:
Lanthanide contraction : Slow decrease in size of the atoms or ions among the lanthanides is known as lanthanide contraction.

Consequences due to lanthanide contraction:

  1. Due to lanthanide contraction, the crystall structure and other properties of the elements become very closely similar.
  2. Due to this, it becomes difficult to separate lanthanides from a mixture.

Question 78.
The first IP of lithium is 5.41 eV and electron affinity of Cl is – 3.61 eV. Calculate ΔH in kJ mol-1 for the reaction : Li(g) + Cl(g) → \(\mathrm{Li}_{(\mathrm{g})}{ }^{+}\) + \(\mathrm{Cl}{ }_{(g)}^{-}\)
Answer:
Given reaction is
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 29
ΔH = ΔH1 + ΔH2
= 5.41 – 3.61
= 1.8 ev
= 1.8 × 9.65 × 104 J/mole
= 17.37 × 104 J/Mole
= 173.7 KJ/mole

Question 79.
How many Cl atoms can you ionize in the process Cl → Cl+ + e by the energy liberated for the process Cl + e → Cl for one Avogadro number of atoms. Given IP = 13.0 eV and EA = 3.60 eV. Avogadro number = 6 × 1023
Answer:
Given
Cl(g) + e → \(\mathrm{Cl}_{(\mathrm{g})}^{-}\) ΔH = – 3.6ev
1 – atom → Electron affinity = 3.6 ev
6.23 × 1023 atoms → ?
6.23 × 1023 × 3.6 = 21.6828 × 1023
∴ For one avagadro no.of Cl atoms electron affinity = 21.6828 × 1023 eV
Given 13 eV can ionise 1 atom of Cl
21.6828 × 1023 eV ionise -?
\(\frac{21.6828 \times 10^{23}}{13}\) = 1.667 × 1023 eV

Question 80.
The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2g of chlorine atoms is completely converted to Cl ions in the gaseous state ? (1 e V = 23.06 kcal)
Answer:
Given electron affinity of Cl = 3.7 ev
Cl + e → Cl ΔH = – 3.7ev
35.5 gms of CZ contains 6.023 × 1023 atoms
2 gms of Cl contains ?
= \(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atoms
6.23 × 1023 atoms can liberate 3.7 eV
\(\frac{2 \times 6.023 \times 10^{23}}{35.5}\) atom can liberate?
= \(\frac{2 \times 6.023 \times 10^{23} \times 3.7}{35.5 \times 6.023 \times 10^{23}}\) = \(\frac{2 \times 3.7}{35.5}\) = \(\frac{7.4}{35.5}\)
= 0.2084 eV
= 0.2084 × 23.06 kcal/mole
= 4.81 k.cal/mole

Long Answer Questions

Question 1.
Discuss the classification of elements by Mendaleev’
Answer:
The periodic classification of elements based on “atomic weights” was done by “Lothar Meyer” and “Mendeleev” independently.
Mendeleev’s periodic law : “The physical and chemical properties of elements and their compounds are a periodic function of their atomic weights”.

Mendeleev arranged the 65 elements in a periodic table. He did not blindly follow the atomic weight but gave more importance to their chemical properties in arranging them in the table.
Explanation of the periodic law : When the elements are arranged in the increasing order of their atomic weights, elements with similar properties appear again and again, at regular intervals. This is called, periodicity of properties.

Mendeleev’s table : Mendeleev introduced a periodic table containing the known 65 elements. In this table, while arranging the elements, he gave importance only to their atomic weights, but also to their physical and chemical properties. This table was defective in some responses. Then he introduced another table, after rectifying the defects of that table. lt is called, “short form of periodic table”. He named the horizontal rows as ‘periods’ and the vertical columns, as ‘groups’. It has in all ‘9’ groups, I to VIII and a ‘O’ group. The first ‘7’ groups were divided into A and B sub groups. There are ‘7’ periods in the table. The VIII group contains three triods, namely, (Fe, Co, Ni), (Ru, Rh, Pd) and (Os, Ir, Pt).

Mendeleev’s observations :

  1. When the elements are arranged according to their atomic weights, they exhibit periodicity of properties.
  2. Elements with similar chemical properties have nearly equal atomic weights: Iron (55.85),. Cobalt (58.94) and Nickel (58.69).
  3. The group number corresponds to the valency of element in that group.
  4. Most widely distributed elements like H,C, O, N, Si, S etc., have relatively low atomic weights.
  5. The atomic weight of an element may be corrected if the atomic weights of the adjacent elements are known. The properties of an element are the average properties of the neighbouring elements.
  6. The atomic weights of beryllium. Indium, Uranium etc. were corrected, based on this observation.

Merits of Mendeleev’s table:

  1. Actually it formed the basis for the development of other modern periodic tables.
  2. Mendeleeffs left some vacant spaces in his periodic table, for the unknown elements. But the predicted the properties of those elements. Later on, when these elements were discovered, they exactly fitted into those vacant places having properties, predicted by Mendeleev.
    Ex : Eka – boron (Scandium), EKa – Silicon (germanium)
    EKa – aluminium (gallium) etc.
  3. ‘O’ group elements were not known at the time of Mendeleeff. Later when they were discovered, they found a proper place in that table under ‘0’ group of elements. Similarly, the radioactive elements.
  4. In case of these pairs of elements Tellurium – Iodine Argon – Potassium and Cobalt – Nickel, there is a reversal of the trend. The first element has higher atomic weight than the second one. These are called anomalous pairs.
    However, based on their atomic numbers, and chemical properties, this arrangement proves quite justified.

Draw-backs of Mendeleev’s periodic table :

  1. Dissimilar elements were placed in the same group.
    Ex : The coinage mentals Cu, Ag and Au are placed along with the alkali metals K, Rb, Cs etc. in the I group. The only common property among them is that they are all univalent (Valency = 1).
  2. The 14 rare earths having different atomic weights are kept in the same place.
  3. Hydrogen could not be given a proper place, as it resembles alkali metals and halogens in its properties.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 2.
From a study of properties of neighbouring elements, the properties of an unknown element can be predicted – Justify with an example.
Answer:
From a study of adjacent elements and their compounds, it is possible to predict the characteristics of certain elements. These predictions were found to be very accurate. These predicted properties helped the future scientists in the discovery of unknown elements, e.g. : EKa Aluminium (EKa Al) (now known as Gallium); EKa Silicon (EKa Si) (now known as Germanium); EKa Boron (EKa B) (now known as Scandium).

Illustration : Following table shows a comparison of the properties predicted by’Mendeleeff’ for the elements and those found experimentally after their discovery.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 30

Question 3.
Discuss the construction of long form periodic table.
Answer:
The elements are arranged in the long form of the periodic table in the increasing order of atomic numbers. ‘Neils Bohr’ constructed the long form of the periodic table based on electronic configuration of elements.
The important features of the long form of the periodic table are :
It consists seven horizontal rows called periods and 18 vertical columns which are classified into 16 groups only.

Periods : Every period starts with an alkali metal and ends with an inert gas. The first period consists of two elements only (H, He) and is called very short period. Second period consists 8 elements (Li to Ne) and is called first short period. The third period consists (Na to Ar) 8 elements and is called second short period.

Fourth period contains 18 elements (K to Kr) and is called first long period. Fifth period is the second long period with 18 elements (Rb to Xe).
Sixth period is the longest period with 32 elements which starts with Cs and ends with Rn. This period includes 14 lanthanides.
Seventh period is an incomplete period with 20 radioactive elements.
Groups : There are 16 groups in the long form of the periodic table (in transition elements three vertical columns are fused and designated as VIII group). These groups are IA, IIA, NIB, IVB, VB, VIB, VIIB, VIII, IB, MB, IMA, IVA, VA, VIA, VIIA and zero group.

The elements of IA, IIA, IIIA, IVA, VA, VIA, VIIA groups are called representative elements or normal elements. Elements of IB, MB, NIB, IVB, VB, VIB, VIIB and VIII groups have their ultimate and penultimate shell incomplete. These are called transition elements. MB elements have (n – 1) d10 ns2 outermost electronic configuration.

Zero group elements have stable electronic configuration. These elements are called inert gases, noble gases. These elements have been grouped at the extreme right of the periodic table.
In this long periods have been expanded and short periods are broken to accommodate the transitional elements in the middle of the long period.

Lanthanides and actinides have been grouped separately and placed at the bottom of the periodic table.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 31

Question 4.
Discuss the relation between the number of electrons filled into the sub energy levels of an orbit and the maximum number of elements present in a period.
Answer:
Elements have been accommodated in these periods according to the following scheme.
1st period : The first main energy shell (K – shell) is completed. As the maximum capacity of K- shell is of 2 electrons, it consists of only two elements, hydrogen (1s1)and helium (1s2).
2nd period : The second main energy shell is completed (i.e.,) 2s and 2p are completed. It includes eight elements from Li (2s1) to Ne (2s22p6).
3rd period : The 3s and 3p energy shells are completed. It includes also eight elements from Na(3s1) to Ar (3s2 3p6).
4th period : The 4s, 3d and 3p energy shells are completed. It includes 18 elements from K(4s1) to Kr (3d10 4s24p6). It includes two s – block elements, ten d – block elements and six p – block – elements.
5th period : The 5s, 4d and 5p energy shells are completed. It includes 18 elements from Rb (5s) to Xe (4d10 5s1 5p6). 1st also includes two s – block elements, ten d – block elements and six p – block elements.
6th period : The 6s, 4f, 5d and 6p energy shells are completed, (i.e.,) it includes 32 elements from Cs (6s1) to Rn (4f14 5d10 6s26p6). It consists of two s – block elements, ten d – block elements, six p – block elements and fourteen f – block elements.
As this period can accommodate only 18 elements in the table, 14 members of 4f – series are separately accommodated in a horizontal row below the periodic table.
7th period : This period at present consists 26 elements. The 7s, 5f and 6d are completed (i.e.,) twenty six elements. Seven of the elements from atomic numbers 106 to 112 have recently been reported.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 5.
Write an essay on s, p, d and f block elements .
Answer:
According to the electronic configuration of elements, the elements have been classified into four blocks. The basis for this classification is the entry of the differentiating electron into the subshell. They are classified into s, p, d and f blocks.
‘s’ block elements : If the differentiating electron enters into ‘s’ orbital, the elements belongs to ‘s’ block.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 32
In every group there are two ‘s’ block elements. As an ‘s’ orbital can have a maximum of two electrons, ‘s’ block has two groups IA and IIA.
‘p‘ block elements : If the differentiating electron enters into ‘p’ orbital, the elements belongs to ‘p’ block.

‘p’ block contains six elements in each period. They are IIIA to VIIA and zero group elements. The electronic configuration of ‘p’ block elements varies from ns2np1 to ns2np6.

‘d‘ block elements : It the differentiating electron enters into (n – 1) d – orbitals the elements belongs to’d’ block. These elements are in between ‘s’ and ‘p’ blocks. These elements are also known as transition elements. In these elements n and (n – 1) shells are incompletely filled. The general electronic configuration of’d’ block elements is (n – 1) d1-10 ns1-2. This block consists of IIIB to VIIB, VIII, IB and IIB groups.

‘f’ block elements : If the differentiating electron enters into ‘f’ orbitals of antipenultimate shell (n – 2) of atoms of the elements belongs to ‘f block. They are in sixth and seventh periods in the form of two series with 14 elements each. They are known as lanthanides and actinides and are arranged at the bottom of the periodic table. The general electronic configuration is (n – 2)f1-14 (n – 1)d0 – 1 ns2.

In these shells the last three shells (ultimate, penultimate and anti penultimate) are incompletely filled. Lanthanides belongs to 4f series. It contains Ce to Lu. Actinides belong to 5f series. It contains Th to Lr.

Advantages of this kind of classification :

As a result of this classification of elements were placed in correct positions in the periodic table. It shows a gradual gradation in physical and chemical properties of elements. The metallic nature gradually decreases and non – metallic nature gradually increases from’s1′ block to ‘p1 block. This classification gave a special place for radioactive elements.

Question 6.
Relate the electronic configuration of elements and their properties in the Classification of elements.
Answer:
The chemical properties of all elements depends upon the electronic configuration. Upon the basis of complete and incomplete electron shells and chemical properties, the elements are classified into four types. (Type -1, Type – II, Type – III and Type – IV).

Type – I(inert gas elements): All the elements with an electronic configuration ns2 np6 including He belongs to this type, nth shell of those elements are completely filled.
The elements show chemical inertness due to completely filled shells and hence they have extra stability. Because of their stability, they are chemically inactive.
e.g. : He, Ne, Ar, Kr, Xe and Rn.

Type – II (Representative elements) : Except inert gases, the remaining elements of s and p – blocks are called representative elements. All the elements with an electronic configuration ns1 to ns2 np5 excluding. He comes under this type. These are atoms in which all except the outermost shells (nth) are complete. Elements of this type enter into chemical combination by loosing, gaining or sharing electrons to get stable inert gas configuration. Many of the metals, all non – metals and metalloids come under this type. Chemically these elements are reactive.

Type – III (Transition elements): All the elements with an electronic configuration (n – 1) d1 – 9 ns1or2 belongs to this type. Atoms in which the two outermost shells are incomplete. These elements show variable oxidation states, form complex ions and coloured ions. The electronic configuration of d – block elements is (n – 1) d1 – 10 ns1 – 2. Small size, high nuclear charge and unpaired’d’ orbitals impart characteristic properties to be transition elements.

Type- IV (Inner transition elements): All the elements with an electronic configuration (n – 2) f1 – 14 (n -1) d0, 1 ns2 belongs to this type. Atoms in which three outermost shells are incomplete. Lanthanides and Actinides belong to this type.

Question 7.
What is a periodic property ? How the following properties vary in a group and in a period ? Explain
a) Atomic radius
b) Electron gain enthalpy.
Answer:
Recurrence of similar properties of elements at definite regular intervals with increasing atomic number i.e., according to their electronic configurations is known as periodicity. Any property which is periodic in nature is called periodic property.
a) Atomic radius : The atomic radius decreases from left to right in a period. With an increase in the atomic number in a period the nuclear charge increases. As a result the effective nuclear charge over the outermost electrons increases, due to this the orbitals are pulled closer to the nucleus causing in a decrease in the atomic radius.

The atomic radius increases from top to the bottom in a group because – with an increase in the atomic number the electrons are added to new shells resulting an increase in the number of inner shells. Hence atomic radius increases from top to bottom in a group.

b) Electron gain enthalpy : Electron gain enthalpy increases in a period from left to right because the size of the atom decreases and the nature of the element changes from metallic to non – metallic nature when we move from left to right in a period.

Electron gain enthalpy decreases from top to bottom in a group because there is an increase in the atomic size. But the second element has greater electron gain enthalpy than the first element.
e.g. : Chlorine has more electron affinity value (- 348 kJ mol-1) than Fluorine (-333 kJ mol-1). It is because fluorine atom is smaller in size than chlorine atom. There is repulsion between the incoming electron and electrons already present in fluorine atom i.e., due to stronger inter electronic repulsions.

Question 8.
What is a periodic property ? How the following properties vary in a group and in a period ?
Explain
a. IE
b. EN
Answer:
a) IE : In groups and periods of the periodic table the ionization enthalpy values are periodically change depend upon the electronic configuration and size of elements.
In a group of elements ionization energy decreases from top to bottom because atomic radius increases.
In general, in a period the atomic size decreases. Because of this, the ionization energy increases across a period.

b) Electro negativity : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases.
In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 9.
Write a note on
a. Atomic radius
b. Metallic radius
c. Covalent radius
Answer:
a) Crystal Radius (Atomic radius or Metallic radius): The term is used for metal atoms. A metallic crystal contains metal atoms in close packing. These metal atoms are considered spherical. They are supposed to touch each other in the crystal.
The crystal radius is half the internuclear distance between two adjacent atoms.
e.g.: The internuclear distance between two adjacent sodium atoms in a crystal of sodium metal is 3.72 A. So crystal radius of sodium is \(\frac{3.72}{2}\) = 1.86 A
For potassium it is 2.31 A.

b) Covalent radius : It is used generally for non – metals. Covalent radius is half the equilibrium distance between the nuclei of two atoms with a covalent bond.
Covalent radii of two atoms can be added to give internuclear distance between them.
e.g. : Covalent radius of H is 0.37A and for chlorine it is 0.99A. Hence internulear distance between H and Cl in HCl is 1.36A.

c) Vander Waals radius (Collision radius): The Vander Waals radius is half the equilibrium distance between the nuclei of two atoms bound by Vander Waals forces.
It is used for molecular substances in solid state and for inert gases.
e.g. : The Vander Waals radius for hydrogen is 1.2 A and that of chlorine is 1.80 A.
The Vander Waals radius of an atom is 40% larger than its covalent radius.

Question 10.
Define IE1 and IE2. Why is IE2 > IE1 for a given atom? Discuss the factors that effect IE of an element. (A.P. Mar. ’16)(T.S. Mar. ’13)
Answer:
1) Ionization energy is the amount of energy required to remove the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom.
It is denoted as I1, and is expressed in electron volts per atom, kilo calories (or) kilo joules per mole.
M(g) + I1 → \(M_{(g)^{+}}\) + e
I1 is first ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I2.
\(M_{(g)}^{+}\) + I2 → \(\mathrm{M}_{(\mathrm{g})}{ }^{2+}\) + e

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential.
For sodium, I1 is 5.1 eV and I2 is 47.3 eV.
I1 < I2 < I3 ….. In

Factors affecting ionization potential:

1. Atomic radius : As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2. Nuclear charge : As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3. Screening effect or shielding effect: In multielectron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. The electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.

(Magnitude of screening effect) ∝ \(\frac{1}{(\text { Ionization enthalpy) }}\)
TREND IN A GROUP : The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.
TREND IN A PERIOD : In a period from left to right I.P. value increases as the size of the elements decreases along the period.

Question 11.
How do the following properties change in group-1 and in the third period ? Explain with example.
a. Atomic radius
b. IE
c. EA
d. Nature of oxides
Answer:
a) Atomic radius

  • In Group – 1 atomic radius from Li to Cs increases
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 33
    → In 3rd period from Na to Cl atomic radius decreases.
    AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 34

b) Ionization energy : In a group ionization energy values decreases with an increase in the size of the atom. In IA group Li is the element with highest ionization potential and Cs is the element with lowest ionization potential.

In third period the ionization potential increases from Na to Mg and then decreases at Al and increases upto P and decreases in case of S and then increases upto argon, i.e., Al has a lower value than Mg and S has a lower value than P, due to stable electronic configurations of Mg and P Among these elements argon has highest ionization potential.

c) Electron affinity :

  • In 3rd period E.A > from ‘Si’ to ‘P’ decreases and P to Cl increases.
    ‘Mg’ and ‘Ar’ has positive values.
  • In 1st group from Li to Cs electron affinity decreases due to increase of size.

d) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes fed litmus blue.
e.g. : Na2O, CaO , MgO etc.
Na2O + H2O → 2 NaOH
CaO + H2O → Ca(OH)2
MgO + H2O → Mg (OH)2
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 35

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 12.
Define electron gain enthalpy. How it varies in a group and in a period? Why is the electron gain enthalpy of O or F is less negative than that of the succeeding element in the group ?
Answer:

  1. Electron affinity is the amount of energy released when an electron is added to a neutral gaseous atom in its ground state. It is known as first Electron affinity EA,sub>1. It has -ve value.
    X(g) + e → \(\mathrm{X}_{(\mathrm{g})}{ }^{-}\) + EA1
  2. Energy can be absorbed when an other electron is added to uni negative ion. It is because to overcome the repulsion between negative ion and electron. Hence second electron affinity, EA2 has +ve value.
    \(X_{(g)}{ }^{-}\) + e → \(X_{(g)}^{2-}\) + EA2
  3. It can be indirectly determined from Born —Haber cycle. Its units are kJ/mole.
  4. Trend in a group : Generally EA decreases down the group. In a group the second element has higher EA value than the first member.
    e.g. : EA of Cl is more than that of F
    Fluorine possess lower EA value than chlorine. It is due to inner electron repulsions in Fluorine.
  5. Trend in a period : Generally EA value increases across a period, but some irregularities can be observed.

a) I A group elements possess low EA values than the corresponding III A elements, e.g. : Be has EA value zero, it is due to completely filled 2s2 orbital.
b) VA elements have low EA values than that of VI A elements.
e.g. : Due to presence of half filled p – orbitals [2s2 2p3], nitrogen has lower EA value than that of oxygen.
c) For inert gases EA value is zero.
d) The element with the highest EA value is chlorine.

  • The electron gain enthalpy of O and F is less negative than the succeeding element in the group because. These have small size and inter electronic repulsions are high in these elements
    O → 141 KJ / mole and S → 200 kJ/Mole
    F → 328 KJ / mole and Cl → 349 kJ/Mole

Question 13.
a. What is electronegativity ?
b. How does it vary in a group and in a period ?
Answer:
a) Electronegativity : ‘The tendency of the atom of an element to attract the shared electron pair(s) more towards itself in a hetronuclear diatomic molecule or in a polar covalent bond.” Measuring electronegativity Pauling scale : Pauling scale is based on the values of bond energy. The bond energy of a compound A – B is the average of bond energies of A – A and B – B molecules.
EA – B = \(\frac{1}{2}\left(E_{A-A}+E_{B-B}\right)\)
But the experimental value of EA – B is found to exceed the theoretical value. The difference is ∆.
∴ ∆ = E’A – B – EA – B
∆ indicates the polarity of the covalent bond. It is measured in k. cal. mol-1.
Pauling gave the relation XA – XB = 0.208 × \(\sqrt{\Delta}\)
In S.l. units XA – XB = 0.1017 \(\sqrt{\Delta}\) where ∆ is measured in kJ/mole
XA and XB are the electronegativities of A and B. Pauling arbitrarily fixed 2.1 as the electronegativity value of Flydrogen and calculated the electronegativities of other elements. On Pauling scale Fluorine has the highest EN value 4.0.

From the values of the electronegativity of elements, the nature of the chemical bond formed can be understood. If two bonded atoms differ by 1.70 or more ion their EN values, the bond between them would be either 50% or more than 50% ionic in nature. Similarly if the difference in the EN values of the atoms is less than 1.70, the bond formed is more than 50% covalent in nature.

b) Variation in a group and period : Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases. In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 14.
Explain the following
a. Valency
b. Diagonal relation
c. Variation of nature of oxides in the Group -1
Answer:
a) Valency : The combining capacity of an element with another element is called valency. The number of hydrogen atoms (or) chlorine atoms (or) double the number of oxygen atoms, with which one atom of the element combine is also called valency.
∴ Valency = no. of hydrogens
= no. of chlorine atoms
= 2 × no. of oxygen atoms present in the molecule
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 36

Periodicity of valency :
1) Each period starts with valency’11 and ends in ‘O’.
e.g.:
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 37

2) In a group valency is either equal to the group number (i.e., upto 4th group) or is equal to (8 – group number) (i.e., from 5th group onwards).
Significance :
Valency of an element is useful in writing the formulae of compounds.

b) Diagonal relation : On moving diagonally across the periodic table the elements show certain similarities. An element of group in 2nd period has similar properties with second element of the next higher group in the 3rd period. This type of resemblance is called diagonal relationship,
e.g. : (Li, Mg); (Be, Al); (B, Si)
AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 38
c) Nature of oxides of elements : All IA group elements are alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions which changes red litmus blue.
e.g. : Na2O, CaO , MgO etc.
CaO + H2O → Ca(OH)2
MgO + H2O → Mg (OH)2
The basic nature of these oxides increases from top to bottom in the group. In a period basic properties decreases and acidic properties increases.

Solved Problems

Question 1.
What would be the IUPAC name and symbol for the element with atomic number 120?
Solution:
The roots for 1, 2 and 0 are un, bi and nil, respectively. Hence, the symbol and the name respectively are Ubn and unbinilium.

Question 2.
How would you justify the presence of 18 elements in the 5t” period of the Periodic Table ?
Solution:
When n = 5, l = 0, 1, 2, 3. The order in which the energy of the available orbitals 4d, 5s and 5p increases is 5s < 4d < 5p. The total number of orbitals available are 9. The maximum number of electrons that can be accommodated is 18; and therefore 18 elements are there in the 5th period.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 3.
The elements Z = 117 and 120 have not yet been discovered. In which family / group would you place these elements and also give the electronic configuration in each case.
Solution:
we see from that element with Z = 117, would belong to the halogen family (Group 17) and the electronic configuration would be [Rn] 5f146d107s27p5. The element with Z = 120, will be placed in Group 2 (alkaline earth metals), and will have the electronic configuration [Uuo]8s2

Question 4.
Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.
Solution:
Metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is : P < Si < Be < Mg < Na.

Question 5.
Which of the following species will have the largest and the smallest size ?
Mg, Mg2+, Al, Al3+.
Solution:
Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius.
Hence the largest species is Mg; the smallest one is Al3+.

Question 6.
The first ionization enthalpy (∆iH) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol-1. Predict whether the first ∆iH value for Al will be more close to 575 or 760 kJ mol-1 ? Justify your answer.
Solution:
It will be more close to 575 kJ mol-1. The value for Al should be lower than that of Mg because of effective shielding of 3p electrons from the nucleus by 3s-electrons.

Question 7.
Which of the following will have the most negative electron gain enthalpy and which the least negative ? P, S, Cl, F. Explain your answer.
Solution:
Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Hence the element with most negative electron gain enthalpy is chlorine; the one with the least negative electron gain enthalpy is phosphorous.

Question 8.
Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements;
(a) silicon and bromine
(b) aluminium and sulphur.
Solution:
(a) Silicon is group 14 element with a valence of 4; bromine belongs to the halogen family with a valency of 1. Hence the formula of the compound formed would be SiBr4.
(b) Aluminium belongs to group 13 with a valence of 3; sulphur belongs to group 16 elements with a valence of 2. Hence, the formula of the compound formed would be Al2S2.

AP Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 9.
Are the oxidation state and covalency of Al in [AICI(H2O)5]2+ same ?
Solution:
No. The oxidation state of Al is +3 and the covalency is 6.

Question 10.
Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 is an acidic oxide.
Solution:
Na2O with water forms a strong base Whereas Cl2O7 forms strong acid.
Na2O + H2O → 2NaOH
Cl2O7 + H2O → 2HClO4
Their basic or acidic nature can be qualitatively tested with litmus paper.

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets InText Questions and Answers.

10th Class Maths 2nd Lesson Sets InText Questions and Answers

Question 1.
List the teeth under each of the following type (Page No. 25)
AP SSC 10th Class Maths Chapter 2 Sets InText Questions 1
i) Incisors
Answer:
Central incisors = 4
Lateral incisors = 4
Total incisors = 8
ii) Canines
Answer:
Total canines = 4
iii) Pre-molars
Answer:
First premolars = 4
Second premolars = 4
Total premolars = 8
iv) Molars
Answer:
First molars = 4
Second molars = 4
Third molars = 4
Total molars = 12

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
Identify and write the “common property” of the following collections. (Page No. 26)
1) 2, 4, 6, 8, …….
Answer:
All are even numbers {x : x is even}
2) 2, 3, 5, 7, 11, …….
Answer:
All are prime numbers (x : x is a prime}
3) 1, 4, 9, 16, …….
Answer:
All are perfect squares
(x : x is a perfect square}
4) January, February, March, April,…
Answer:
All are English months
{x : x is a month of the year}
5) Thumb, index finger, middle finger, ring finger, pinky.
Answer:
All are fingers of a hand
{x : x is a finger of a hand}

Question 3.
Write the following sets. (Page No. 27)
1) Set of the first five positive integers.
Answer:
{11, 2, 3, 4, 5}
2) Set of multiples of 5 which are more than 100 and less than 125.
Answer:
{105, 110, 115, 120}
3) Set of first five cubic numbers.
Answer:
{13, 23, 33, 43, 53}
{1, 8, 27, 64, 125}
4) Set of digits in the Ramanujan number.
Answer:
Ramanujan’s number is 1729
{1, 2, 7, 9}

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 4.
Some numbers are given below. Decide the numbers to which number sets they belong to and does not belong to and express with correct symbols. (Page No. 28)
i) 1
Answer:
1 ∈ N
ii) 0
Answer:
0 ∈ W and 0 ∉ N
iii) -4
Answer:
– 4 ∈ I and – 4 ∉ N
iv) \(\frac{5}{6}\)
Answer:
\(\frac{5}{6}\) ∈ Q and \(\frac{5}{6}\) ∉ Z
v) \(1 . \overline{3}\)
Answer:
\(1 . \overline{3}\) ∉ N and \(1 . \overline{3}\) ∉ Z
vi) √2
Answer:
√2 ∈ S and √2 ∉ Q
vii) log 2
Answer:
log 2 ∉ N
viii) 0.03
Answer:
0.03 ∉ Q
ix) π
Answer:
π ∉ Z
x) \(\sqrt{-4}\)
Answer:
\(\sqrt{-4}\) ∉ Q and \(\sqrt{-4}\) ∈ C

Question 5.
List the elements of the following sets. (Page No. 29)
i) G = {all the factors of 20}
ii) F = {the multiples of 4 between 17 and 61 which are divisible by 7}
iii) S = {x : x is a letter in the word ‘MADAM’}
iv) P = {x : x is a whole number between 3.5 and 6.7}
Answer:
i) G = {1, 2, 4, 5, 10, 20}
ii) Multiples of 4 between 17 and 61
x = {20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60}
F = {28, 56}
iii) S = {M, D, A}
iv) P = {4, 5, 6}

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 6.
Write the following sets in the roster form.   (Page No. 29)
i) B is the set of all months in a year having 30 days.
ii) P is the set of all prime numbers smaller than 10.
iii) X is the set of the colours of the rainbow.
Answer:
i) B = {April, June, September, November}
ii) P = {2, 3, 5, 7}
iii) X = {Violet, Indigo, Blue, Green, Yellow, Orange, Red}

Question 7.
A is the set of factors of 12. Which one of the following is not a member of A?   (Page No. 29)
A) 1
B) 4
C) 5
D) 12
Answer:
[C]

Think & Discuss

Question 1.
Observe the following collections and prepare as many as generalized statements you can describing their more properties. (Page No. 26)
i) 2, 4, 6, 8,….
Answer:
a) All even natural numbers
b) All positive even integers
c) Multiples of 2

ii) 1, 4, 9, 16, …..
Answer:
a) Squares of natural numbers
b) All perfect square numbers

Question 2.
Can you write set of rational numbers listing elements in it? (Page No. 28)
Answer:
We can’t list all elements in the set of rational numbers. We know that rational numbers are infinite.

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Try this

Question 1.
Write some sets of your choice, involving algebraic and geometrical ideas. (Page No. 29)
Answer:
The set of all natural numbers ‘x’ such that 4x + 9 < 50,
ii) A = {x : x is an integer and -3 ≤ x ≤ 7}
iii) B = {Equilateral triangle, Right angled triangle, Scalene triangle, Obtuse angled triangle, Acute angled triangle)
iv) C = {Rectangle, Square, Parallelogram, Rhombus, Trapezium}

Question 2.
Match roster forms with the set builder form. (Page No. 29)
AP SSC 10th Class Maths Chapter 2 Sets InText Questions 2
Answer:
i) d
ii) c
iii) a
iv) b

Do this

Question 1.
A = {1, 2, 3, 4},
B = {2, 4},
C = {1, 2, 3, 4, 7}, ∅ = { }.
Fill in the blanks with ⊂ or ⊄.  (Page No. 33)
i) A …. B
ii) C …. A
iii) B …. A
iv)A …. C
v) B …. C
vi) ∅ …. B
Answer:
i) A ⊄ B
ii) C ⊄ A
iii) B ⊆ A
iv) A ⊆ C
v) B ⊆ C
vi) ∅ ⊆ B

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
State which of the following statements are true.    (Page No. 33)
i) { } = ∅
ii) ∅ = 0
iii) 0 = { 0 }
Answer:
i) True (T)
ii) False (F)
iii) False (F)

Question 3.
Let A = {1, 3, 7, 8} and B = [2, 4, 7, 9}.
Find A ∩ B.     (Page No. 37)
Answer:
Given sets A = (1, 3, 7, 8} and B = {2,4, 7,9}
A ∩ B = {1, 3, 7, 8} ∩ (2, 4, 7, 9} = {7}

Question 4.
If A = {6,9,11 }; ∅ = {}, find A ∪ ∅, A ∩ ∅).  (Page No. 37)
Answer:
Given sets
A = {6, 9, 11} and ∅ = { }
A ∪ ∅ = {6, 9, 11} ∪ { }
= {6, 9, 11} = A
∴ A ∪ ∅ = A
A ∩ ∅ = {6,9,11} ∩ { } = { } = ∅
∴ A ∩ ∅ = ∅

Question 5.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B = {2, 3, 5, 7}. Find A ∩ B and show that A ∩ B = B.    (Page No. 37)
Answer:
Given sets
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7} = B
∴ A ∩ B = B

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 6.
If A = {4, 5, 6}; B = {7, 8}, then show that A ∪ B = B ∪ A.  (Page No. 37)
Answer:
Given sets are
A = {4, 5, 6} and B = {7, 8}
A ∪ B = {4, 5, 6} ∪ {7, 8}
= {4, 5, 6, 7, 8}.
B ∪ A = {7, 8} ∪ {4, 5, 6}
= {4, 5, 6, 7, 8}
∴ A ∪ B = B ∪ A.

Question 7.
If A = {1, 2, 3, 4, 5 }; B = {4, 5, 6, 7}, then find A – B and B – A. Are they equal?  (Page No. 38)
Answer:
Given sets are
A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}
A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7}
= {1, 2, 3}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
= {6, 7}
∴ No, A – B ≠ B – A.

Question 8.
If V = {a, e, i, o, u} and B = {a, i, k, u}, find V – B and B – V.    (Page No. 38)
Answer:
Given sets are
V = {a, e, i, o, u} and B = {a, i, k, u}
V – B = {a, e, i, o, u} – {a, i, k, u}
= {e, o}
B – V = {a, i, k, u} – {a, e, i, o, u}
= {k}.

Try this

Question 1.
A = {set of quadrilaterals},
B = {square, rectangle, trapezium, rhombus}.
State whether A ⊂ B or B ⊂ A. Justify your answer.     (Page No. 33)
Answer:
A = {set of quadrilaterals} means A = {square, rectangle/trapezium, rhombus, parallelogram}
B = {square, rectangle, trapezium, rhombus}
So, B ⊂ A because A’ is having elements more than ‘B’.

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
If A = {a, b, c, d}. How many subsets does the set A have?    (Page No. 33)
A) 5 B) 6 C) 16 D) 65
Answer:
Given A = {a, b, c, d}
n(A) = 4
Number of subsets for a set, which is having ‘n’ elements is 2n.
So n(A) = 4
Number of subsets for A is 24 = 16.
Answer is [C].

Question 3.
P is the set of factors of 5, Q is the set of factors of 25 and R is the set of factors of 125. Which one of the following is false?    (Page No. 33)
A) P ⊂ Q
B) Q ⊂ R
C) R ⊂ P
D) P ⊂ R
Answer: [C]

Question 4.
A is the set of prime numbers smaller than 10, B is the set of odd numbers < 10 and C is the set of even numbers < 10. How many of the following statements are true?    (Page No. 33)
i) A ⊂ B
ii) B ⊂ A
iii) A ⊂ C
iv) C ⊂ A
v) B ⊂ C
vi) C ⊂ B
Answer:
All the statements are false.

Question 5.
List out some sets A and B and choose their elements such that A and B are disjoint.  (Page No. 37)
Answer:
Consider the disjoint sets
A = {1, 2, 3, 4} and B = {a, b, c}

Question 6.
If A = {2, 3, 5}, find A ∪ ∅ and ∅ ∪ A and compare.    (Page No. 37)
Answer:
Given sets A = {2, 3, 5} and ∅ = { }
A ∪ ∅ = {2,3,5} ∪ { } = {2,3,5}
∅ ∪ A = { } ∪ {2, 3, 5} = {2,3,5}
A ∪ v = ∅ ∪ A = A

Question 7.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 4, 5, 6, 7, 8}, then find A ∪ B, A ∩ B. What do you notice about the result?   (Page No. 37)
Answer:
Given sets are
A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7, 8}
A ∪ B = {1, 2, 3, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8} = B
A ∩ B = {1, 2, 3, 4} ∩ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4} = A
If A ⊂ B, then A ∪ B = B and A ∩ B = A

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 8.
A = {1, 2, 3, 4, 5, 6}; B = {2, 4, 6, 8, 10}. Find the intersection of A and B.     (Page No. 37)
Answer:
Given sets are
A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10}
A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6, 8, 10} = {2, 4, 6}

Think & Discuss

Question 1.
Is empty set subset to every set?    (Page No. 34)
Answer:
‘Yes’. Empty set is subset to every set
Justify: If A ⊂ B, it means all the elements of set ‘A’ belong to set ‘B’.
In other words we can say no element of ‘A’ missed in set B.
now empty set means which has no elements, now no element of empty set can be missed in any set. So we can write empty set is subset of every set.

Question 2.
Is any set subset to itself?    (Page No. 34)
Answer:
‘Yes’. Every set is subset to itself.
Let ‘A’ is any set.
Now every element of ‘A’ definitely belongs to ‘A’.
So A ⊂ A
Hence every set is a subset to it.

Question 3.
You are given two sets such that a set is not a subset of the other. If you have to prove this, how do you prove?    (Page No. 34)
Answer:
Let the given sets are ‘A’ and ‘B’.
To prove are set (A) is not subset of other (B).
We check if all elements of ‘A’ belong to the set ‘B’ or not.
If any of the element doesn’t belong to ‘B’ then we can say ‘A’ is not subset of ‘B’. So we have to prove at least one element of ‘A’ does not belong to ‘B’.
Hence ‘A’ is not subset of ‘B’.

Question 4.
The intersection of any two disjoint sets is a null set. Justify your answer.    (Page No. 37)
Answer:
Let A and B be any two disjoint sets,
i.e., A and B have no elements in common.
∴ A ∩ B is a null set. (∵ A ∩ B is the set of all elements which are common to both A and B)

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 5.
The sets A – B, B – A and A ∩ B are mutually disjoint sets. Use examples to observe if this is true.    (Page No. 38)
Answer:
Let the sets are A = {1, 2, 3, 4} and B = {5, 6, 7, 8}
A – B = {1, 2, 3, 4} – {5, 6, 7, 8} = {1, 2, 3, 4}
B – A = {5, 6, 7, 8} – {1, 2, 3, 4} = {5, 6, 7, 8}
A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7, 8} = { } = ∅
∴ A – B, B – A and A ∩ B are disjoint sets.

Do these

Question 1.
Which of the following are empty sets? Justify your answer.    (Page No. 44)
i) Set of integers which lie between 2 and 3 .
ii) Set of natural numbers that are smaller than 1.
iii) Set of odd numbers that leave remainder zero, when divided by 2.
Answer:
i) This is null set. We know that there is no integer that lie between 2 and 3.
ii) This is also a null set. We know that there is’ no natural number less than ‘1’.
iii) This is a null set. We know that odd numbers do not leave remainder zero when divided by 2.

Question 2.
State which of the following sets are finite and which are infinite. Give reasons for your answer.    (Page No. 44)
i) A = {x : x e N and x < 100}
ii) B = {x : x e N and x ≤ 5}
iii) C = {12 , 22, 32, ……}
iv) D = {1, 2, 3, 4}
v) {x : x is a day of the week}
Answer:
i) A = (1, 2, 3, 4, , 98, 99}
This set is finite, because there are 99 numbers possible to count.
ii) B = {1, 2, 3,  4, 5}
This set is finite because there are 5 numbers possible to count.
iii) C = {12 , 22, 32, ……}
This set is infinite because there are infinite numbers.
iv) D – {1, 2, 3, 4}
This set is finite because there are 4 numbers that are possible to count.
v) E = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
This set is finite, because there are 7 days in a week possible to count.

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
Tick the set which is infinite.      (Page No. 44)
A) The set of whole numbers < 10
B) The set of prime numbers < 10
C) The set of integers < 10
D) The set of factors of 10
Answer:
[C]
The set of integers < 10
{….., -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Try this

Question 1.
Which of the following sets are empty sets? Justify your answer.    (Page No. 44)
i) A = {x : x2 = 4 and 3x = 9}.
ii) The set of all triangles in a plane having the sum of their three angles less than 180.
Answer:
i) x2 = 4 ⇒ x = ± 2
3x = 9 ⇒ x = 3
The value of ‘x’ is not same in both cases, so this is a null set.
ii) This is a null set, because the sum of the three angles of a triangle is equal to 180°.

Question 2.
B = {x : x + 5 = 5} is not an empty set. Why?   (Page No. 44)
Answer:
B = {x: x + 5 = 5} is not an empty set
let x ∈ Z or x ∈ W
then for x = 0 ⇒ x + 5 = 0 + 5 = 5
So if x ∈ W, or x ∈ Z then for x = 0,
x + 5 = 5 is true.
Then the set B = {0} which is not an empty set.
Note: But if x ∈ N
We will have no ‘x’ such that x + 5 = 5
then ‘B’ will be an empty set.
But in the textbook it is not given whether x ∈ N (or) x ∈ W (or) x ∈ Z.
Hence we consider first one.

AP SSC 10th Class Maths Solutions Chapter 2 Sets InText Questions

Think & Discuss

Question 1.
An empty set is a finite set. Is this statement true or false? Why?   (Page No. 44)
Answer:
Yes, it is a finite set because there is finite number i.e., ‘0’ elements it consists.

Think & Discuss

Question 1.
What is the relation between n(A), n(B), n(A ∩ B) and n(A ∪ B)?   (Page No. 45)
Answer:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B). This is called Fundamental theorem of sets.

Question 2.
If A and B are disjoint sets, then how can you find n(A ∪ B)?    (Page No. 45)
Answer:
If A and B are disjoint then A ∩ B is a null set.
∴ n(A ∩ B) = 0 and it gives us n(A ∪ B) = n (A) + n(B).

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.4

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.4

10th Class Maths 2nd Lesson Sets Ex 2.4 Textbook Questions and Answers

Question 1.
State which of the following sets are empty and which are not?
i) The set of lines passing through a given point.
ii) Set of odd natural numbers divisible by 2.
iii) {x : x is a natural number, x < 5 and x > 7}
iv) {x: x is a common point to any two parallel lines}
v) Set of even prime numbers.
Answer:
i) Not empty
ii) Empty
iii) Empty
iv) Empty
v) Not empty

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.4

Question 2.
Which of the following sets are finite or infinite?
i) The set of months in a year.
ii) {1, 2, 3, …, 99, 100}
iii) The set of prime numbers smaller than 99.
Answer:
i) Finite
ii) Finite
iii) Finite

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.4

Question 3.
State whether each of the following sets is finite or infinite.
i) The set of letters in the English alphabet.
ii) The set of lines which are parallel to the X-axis.
iii) The set of numbers which are multiples of 5.
iv) The set of circles passing through the origin (0, 0).
Answer:
i) Finite
ii) Infinite
iii) Infinite
iv) Infinite

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.3

10th Class Maths 2nd Lesson Sets Ex 2.3 Textbook Questions and Answers

Question 1.
Which of the following sets are equal?
A = {x : x is a letter in the word FOLLOW}
ii) B = {x : x is a letter in the word FLOW}
iii) C = {x : x is a letter in the word WOLF}
Answer:
i) Elements in set A are {F, L, O, W}
ii) Elements in set B are {F, L, O, W}
iii) Elements in set C are {F, L, O, W} Sets A, B and C have same elements, Hence, they are equal sets.

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

Question 2.
Consider the following sets and fill up the blank in the statement given below with = or ≠ so as to make the statement true.
A = {1, 2, 3};
B = {The first three natural numbers};
C = {a, b, c, d};
D = {d, c, a, b};
E = {a, e, i, o, u};
F = {Set of vowels in English Alphabet}
i) A …. B
ii) A …. E
iii) C …. D
iv) D …. F
v) F …. A
vi) D …. E
vii) F …. B
Answer:
i) A = B
ii) A ≠ E
iii) C = D
iv) D ≠ F
v) F ≠ A
vi) D ≠ E
vii) F ≠ B

Question 3.
In each of the following, state whether A = B or not.
i) A = {a, b, c, d} ; B = {d, c, a, b}
ii) A = {4, 8, 12, 16} ; B = {8, 4, 16, 18}
iii) A = {2, 4, 6, 8, 10}; B = {x : x is a positive even integer and x ≤ 10}
iv) A = {x : x is a multiple of 10}; B = {10, 15, 20, 25, 30, …}
Answer:
i) A = B
ii) A ≠ B
iii) A ≠ B
iv) A ≠ B

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

Question 4.
State the reasons for the following :
i) {1, 2, 3, …., 10} ≠ {x : x ∈ N and 1 < x < 10}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n+1 and x ∈ N}
iii) {5, 15, 30, 45} ≠ {x : x is a multiple of 15}
iv) {2, 3, 5, 7, 9} ≠ {x : x is a prime number}
Answer:
i) In R.H.S ‘x’ is greater than 1 and less than 10 but L.H.S is having both 1 and 10.
ii) L.H.S ≠ R.H.S
R.H.S: x = 2n + 1 is definition of odd numbers.
L.H.S: Given set is even numbers set.
iii) x is a multiple of 15.
So 5 does not exist.
iv) x is a prime number but 9 is not a prime number.

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

Question 5.
List all the subsets of the following sets.
i) B = {p, q}
ii) C = {x, y, z}
iii) D = {a, b, c, d}
iv) E = {1, 4, 9, 16}
v) F = {10, 100, 1000}
Answer:
i) Subsets of ‘B’ are {p}, {q}, {p, q}, φ
ii) Subsets of ‘C’ are {x}, {y} {z}, {x, y}, {y, z}, {z, x}, {x, y, z} and φ (23 = 8)
iii) Subsets of ‘D’ are {a}, {b}, {c}, {d}, {a,b}, {b,c}, {c, d}, {a, c}, {a, d}, {b, d}, {a, b, c}, {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d} and φ
iv) Subsets of ‘E’ are
φ, {1}, {4}, {9}, {16}, {1,4}, {1,9}, {1, 16}, {4, 9}, {4, 16}, (9, 16}, {1, 4, 9}, {1, 9, 16}, {4, 9, 16}, {1, 4, 16}, {1, 4, 9, 16}
v) Subsets of ‘F’ are
φ, {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000}, {10, 100, 1000}.

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.2

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets Ex 2.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets Exercise 2.2

10th Class Maths 2nd Lesson Sets Ex 2.2 Textbook Questions and Answers

Question 1.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 5, 6} then find A ∩ B and B ∩ A. Are they equal ?
Answer:
Given sets are A = {1, 2, 3, 4} and B = {1,2,3, 5,6}
A ∩ B = {1,2, 3,4} ∩ {1,2, 3, 5, 6}
= {1,2,3} …… (1)
B ∩ A = {1, 2, 3, 5, 6} ∩ {1, 2, 3, 4}
= {1,2,3} …….(2)
From (1) and (2)
A ∩ B and B ∩ A are same.

Question 2.
A = {0, 2, 4}, find A ∩ φ and A ∩ A. Comment.
Answer:
Given set A = {0, 2, 4} and φ is a null set.
A ∩ φ = {0, 2, 4} ∩ { }
= { } ……. (1)
A ∩ A = {0, 2, 4} ∩ {0, 2, 4}
= {0, 2,4} …….. (2)
From (1) and (2),
We conclude that A ∩ φ = φ and A ∩ A = A

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.2

Question 3.
If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A – B and B – A.
Answer:
Given sets are
A {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}
A – B = {2, 4, 6, 8, 10} – {3, 6, 9, 12, 15}
= {2, 4, 8, 10} …… (1)
B – A = {3, 6, 9, 12, 15} – {2, 4, 6, 8, 10}
= {3, 9, 12, 15} …… (2)
From (1) and (2), A – B ≠ B – A

Question 4.
If A and B are two sets such that A ⊂ B then, what is A ∪ B?
Answer:
Let us consider A ⊂ B
Set A = {1, 2, 3} and
Set B = {1, 2, 3, 4, 5}
Now A ∪ B = {1, 2, 3} ∪ {1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5} = B
∴ A ∪ B = B

Question 5.
If A = {x : x is a natural number},
B = {x : x is an even natural number},
C = {x : x is an odd natural number} and
D = {x : x is a prime number}
Find A ∩ B, A ∩ C, A ∩ D, B ∩ C, B ∩ D, C ∩ D.
Answer:
Given sets are
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ……}
B = {2, 4, 6, 8, 10, …….}
C = {1, 3, 5, 7, 9, …….}
D = {2, 3, 5, 7, 11, …….}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …….} ∩ {2, 4, 6, 8, 10, ……}
= {2, 4, 6, 8, 10, ……}
A ∩ C = {1, 2, 3,4, 5, 6, 7, 8, 9, 10, …} ∩ {1, 3, 5, 7, 9 }
= {1, 3, 5, 7, 9, ……}
A ∩ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …} ∩ {2, 3, -5, 7, 11,….}
= {2, 3, 5, 7, 11, ……}
B ∩ C = {2, 4, 6, 8, 10, ……} ∩ {1, 3, 5, 7, 9, …….}
= { } = φ
B ∩ D = {2, 4, 6, 8, 10, ……} ∩ {2, 3, 5, 7, 11, ……}
= {2}
C ∩ D = {1, 3, 5, 7, 9, ……} ∩ {2, 3, 5, 7, 11, 13, ……}
= {3, 5, 7, …..}

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.2

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21}; B = {4, 8, 12, 16, 20}; C = {2, 4, 6, 8, 10, 12, 14, 16}; D = {5, 10, 15, 20} find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
Answer:
Given sets are A = {3, 6, 9, 12, 15, 18, 21}
B = {4, 8, 12, 16, 20}
C = {2, 4, 6, 8, 10, 12, 14, 16} and
D = {5, 10, 15, 20}
i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20} = {3, 6, 9, 15, 18, 21}
ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16} = {3,9,15,18,21}
iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20} = {3,6,9,12,18,21}
iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21} = {4, 8, 16, 20}
v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
vi) D – A = {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
vii) B – C = {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
viii) B – D = {4, 8, 12, 16,20} – {5, 10, 15, 20} = {4, 8, 12, 16}
ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20} = {2, 6, 10, 14}
x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20} = {5, 10, 15}

AP SSC 10th Class Maths Solutions Chapter 2 Sets Ex 2.2

Question 7.
State whether each of the following statement is true or false. Justify your answers.
i) {2,3,4,5} and {3,6} are disjoint sets.
ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
Answer:
i) Rule: If two sets are disjoint their intersection is null set.
= {2, 3, 4, 5} n {3, 6} = { 3 } ≠ φ
∴ Given statement is False.

ii) Given sets are
{a, e, i, o, u} and {a, b, c, d}
= {a, e, i, o, u} ∩ {a, b, c, d}
= { a } ≠ φ
∴ Given statement is False.

iii) Given sets are
{2, 6, 10, 14} and {3, 7, 11, 15}
= {2, 6, 10, 14} ∩ {3, 7, 11, 15}
= { }
∴ Given statement is True.

iv) Given sets are
{2, 6, 10} and {3, 7, 11}
= {2, 6, 10} ∩ {3, 7, 11} = { }
∴ Given statement is True.