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Materials: Metals and Non-Metals Class 8 Science NCERT Textbook Questions

Question 1.
Which of the following can be beaten into thin sheets?
(a) Zinc
(b) Phosphorus
(c) Sulphur
(d) Oxygen
Answer:
(a) Zinc

Question 2.
Which of the following statements is correct?
(a) All metals are ductile.
(b) All non-metals are ductile.
(c) Generally, metals are ductile.
(d) Some non-metals are ductile.
Answer:
(c) Generally, metals are ductile

Question 3.
Fill in the blanks.
(a) Phosphorus is a very ____ non-metal.
(b) Metals are _____ conductors of heat and _____
(c) Iron is ______ reactive than copper.
(d) Metals react with acids to produce ______ gas.
Answer:
(a) reactive
(b) good, electricity
(c) more
(d) hydrogen

Question 4.
Mark ‘T’ if the statement is true and ‘F’ if it is false.
(a) Generally, non-metals react with acids.
(b) Sodium is a very reactive metal.
(c) Copper displaces zinc from zinc sulphate solution.
(d) Coal can be drawn into wires.
Answer:
(a) False
(b) True
(c) False
(d) False

Question 5.
Some properties are listed in the following Table. Distinguish between metals and non-metals on the basis of these properties.

Properties	Metals	Non-metals
1. Appearance
2. Hardness
3. Malleability
4. Ductility
5. Heat Conduction
6. Conduction of Electricity

Answer:

Properties	Metals	Non-metals
1. Appearance	have metallic lustre	dull
2. Hardness	hard	soft
3. Malleability	malleable	non-malleable
4. Ductility	ductile	non-ductile
5. Heat Conduction	good conductors	bad conductors
6. Conduction of Electricity	good conductors	bad conductors/insulators

Question 6.
Give reasons for the following.
(a) Aluminium foils are used to wrap food items.
(b) Immersion rods for heating liquids are made up of metallic substances.
(c) Copper cannot displace zinc from its salt solution.
(d) Sodium and potassium are stored in kerosene.
Answer:
(a) Aluminium is highly malleable and it can be easily beaten in sheets to make its foil for wrapping purposes. It is also soft and does not react with food items. That is why aluminium foils are used . to wrap food items.
(b) Immersion rods made up of metallic substances because metals are good conductors of heat and electricity. They get hot very soon on the passage of electric current and warm the water.
(c) Copper is less reactive than zinc. So it cannot displace zinc from its solution.
(d) Sodium and potassium are highly reactive, so they are stored in kerosene.

Question 7.
Can you store the lemon pickle in an aluminium utensil? Explain.
Answer:
No, we cannot store lemon pickle in an aluminium utensil because aluminium is a metal and metals readily react with acids to produce hydrogen. When aluminium comes in contact with lemon, which is acidic, would react to give hydrogen and the pickles will be spoiled.

Question 8.
Match the substances given in column A with their uses given in column B.

A	B
Gold	Thermometers
Iron	Electric wire
Aluminium	Wrapping food
Carbon	Jewellery
Copper	Machinery
Mercury	Fuel

Answer:
(i) (d)
(ii) (e)
(iii) (c)
(iv) (f)
(v) (b)
(vi) (a)

Question 9.
What happens when
(a) Dilute sulphuric acid is poured on a copper plate?
(b) Iron nails are placed in a copper sulphate solution?
Write word equations of the reactions involved.
Answer:
(a) No reaction will take place because copper is very less reactive.
(b) Iron being more reactive than copper will replace copper from its solution and brown coating of copper is deposited on the iron nails. Also, the blue colour turns green.
Iron + Copper sulphate (solution) → Iron sulphate (solution) + Copper

Question 10.
Saloni took a piece of burning charcoal and collected the gas evolved in a test tube.
(a) How will she find the nature of the gas?
(b) Write down the word equations of all the reactions taking place in this process.
(a) She can find the nature of the gas by using a wet litmus paper. After bringing the litmus paper in contact with the gas, if it turns the blue litmus paper into red, it is acidic. Similarly, if it turn the red litmus into blue, it is basic.
(b) (i) Carbon + Oxygen → Carbon dioxide
(ii) Carbon dioxide + Lime water → Milky

Materials: Metals and Non-Metals Class 8 Science NCERT Intext Activities Solved

Activity 1

Take a small iron nail, a coal piece, a piece of thick aluminium wire and a pencil lead. Beat the iron nail with a hammer (Fig. 4.1).

(But take care that you don’t hurt yourself in the process). Try to hit hard. Hit hard the aluminium wire also. Then repeat the same kind of treatment on the coal piece and pencil lead. Record your observations in Table 4.1.
Table 4.1 Malleability of Materials

Object/Material	Change in Shape (Flattens/Breaks into Pieces)
Iron nail	Flattens
Coal piece	Breaks into pieces
Aluminium wire	Flattens
Pencil lead	Breaks into pieces

Solution: This activity shows that iron and aluminium are malleable while coal piece and pencil lead are brittle. Thus, metals are malleable and non-metals are non-malleable.

Activity 2

Recall how to make an electric circuit to test whether electricity can pass through an object or not (Fig. 4.2).

You might have performed the activity with various objects in Class VI. Now, repeat the activity with the materials mentioned in Table 4.2. Observe and group these materials into good conductors and poor conductors.

S. No.	Materials	Good Conductor/Poor Conductor
1.	Iron rod/nail	Good conductor
2.	Sulphur	Poor conductor
3.	Coal piece	Poor conductor
4.	Copper wire	Good conductor

Solution:
It shows that metals are good conductors of electricity and non-metals are poor conductors of electricity.

Activity 3

Let us check the nature of rust formed as a result of the reaction between iron, oxygen and water. Collect a spoonful of rust and dissolve it in a very little amount of water. You will find that the rust remains suspended in water. Shake the suspension well. Test the solution with red and blue litmus papers (Fig. 4.3).

What do you observe? Is the solution acidic or basic?
Solution:
We observed that the red litmus paper turns blue which shows that the nqjure of rust is basic. Blue litmus paper do not show any colour change with the solution.

Activity 4

(To be demonstrated by the teacher in the class)
Take a small amount of powdered sulphur in a deflagrating spoon and heat it. If deflagrating spoon is not available, you may take a metallic cap of any bottle and wrap a metallic wire around it and give it the shape shown in Fig. 4.4 (a).

As soon as sulphur starts burning, introduce the spoon into a gas jar/glass tumbler [Fig. 4.4(a)]. Cover the tumbler with a lid to ensure that the gas produced does not escape. Remove the spoon after some time. Add a small quantity of water into the tumbler and quickly replace the lid. Shake the tumbler well. Check the solution with red and blue litmus papers [Fig. 4.4. (b)].

Solution:
We observed that the solution of oxide turns the blue litmus red which shows that the solution is acidic in nature. This also shows that oxide of non-metals is acidic in nature.

Activity 5

Take a 250 mL beaker/glass tumbler. Fill half of it with water. Now carefully cut a small piece of sodium metal. Dry it using filter paper and wrap it in a small piece of cotton. Put the sodium piece wrapped in cotton into the beaker. Observe carefully.
When reaction stops, touch the beaker. What do you feel? Has the beaker become hot? Test the solution with red and blue litmus papers. Is the solution acidic or basic?

Solution:
On touching the beaker, it was felt hot. The solution turns the red litmus paper to blue which shows it is basic in nature. Blue litmus paper do not show any colour change with the solution.

Activity 6
Take samples of metals and non-metals given in Table 4.3 in separate test tubes and label them as A, B, C, D, E and F. With the help of a dropper add 5 mL of dilute hydrochloric acid to each test tube one by one. Observe the reactions carefully. If no reaction occurs in the cold solution, warm the test tube gently. Bring a burning matchstick near the mouth of each test tube.
Repeat the same activity using dilute sulphuric acid instead of dilute hydrochloric acid. Record your observations in Table 4.3.

Solution:
This activity shows that metals usually displace hydrogen from dilute acids whereas non-metals do not do so and no hydrogen gas is evolved.

Activity 7

Prepare a fresh solution of sodium hydroxide in a test tube by dissolving 3-4 pellets of it in 5 mL of water. Drop a piece of aluminium foil into it. Bring a burning matchstick near the mouth of the test tube. Observe carefully.
Solution:
We observed that a colourless gas is evolved which burns with a pop sound. This shows that aluminium react with bases on heating to produce hydrogen gas.

Activity 8
Take five 100 mL beakers and label them A, B, C, D and E. Take about 50 mL of water in each beaker. Dissolve in each beaker a teaspoonful of each substance as indicated in Fig. 4.6 (a), (i) Keep the beakers undisturbed for some time, (ii) Record you observations in your notebook.

Beaker A: Copper sulphate (CuSO4) + Zinc granule (Zn), Beaker B: Copper sulphate (CuSO4) + Iron nail (Fe)
Beaker C: Zinc sulphate (ZnSO4) + Copper turnings (Cu), Beaker D: Iron sulphate (FeSO4) + Copper turnings (Cu)
Beaker E: Zinc sulphate (ZnSO4) + Iron nail (Fe)
Solution:
In beaker ‘A’ zinc (Zn) replaces copper (Cu) from copper sulphate (CuS04) solution. That is why the blue colour of copper sulphate changes to colourless and a powdery red mass of copper is deposited at the bottom of the beaker. The reaction can be represented as follows:

In beaker B, iron replaces copper from its solution. That is why the blue colour of copper sulphate changes to green colour of ferrous sulphate.

In beaker C, D and E no change in colour or heat evolution is observed. This indicates that the metals are unable to displace the other metals from its solution.

NCERT Solutions for Class 8 Science Chapter 4 – 1 Mark Questions and Answers

Question 1.
Non-metals cannot be drawn into wires. Why ? [DAV2008]
Answer:
Non-metals are not ductile, therefore they cannot be drawn into wires.

Question 2.
Complete the following equation :
Zn + 2HCl ——-> __+ __   [MSE (Chandigarh) 2007]
Answer:
Zn + 2HCl ——-> ZnCl₂ + H₂

Question 3.
Which of the following can be beaten into thin sheets ? [NCERT]

• Zinc
• Phosphorus
• Sulphur
• Oxygen.

Answer:
Zinc.

Question 4.
The number of metals is much ………….. than non-metals.
Answer:
More.

Question 5.
……… are the good conductors of heat and electricity.
Answer:
Metals.

Question 6.
Examples of metals are ………., ……. and ………
Answer:
Iron, sodium and nickel.

Question 7.
Examples of non-metals are ………. , …….. and …….
Answer:
Sulphur, chlorine and oxygen.

Question 8.
Explain the term ‘metallurgy’.
Answer:
Metallurgy is the science of extracting metals from their ores and purifying them for various uses.

Question 9.
State general steps involved in metallurgy of a metal.
Answer:
The general steps of metallurgy are :

• Concentration of ore.
• Reduction of the metal compound.
• Refining of metal.

Question 10.
Metals are (softer/harder) than non-metals.
Answer:
Harder.

Question 11.
Most non-metals are (bad/good) conductors of heat.
Answer:
Bad.

Question 12.
The property that allows the metals to be hammered into thin sheets is called (ductility/ malleability).
Answer:
Malleability.

Question 13.
Melting point of most non-metals is (higher/lower) than metals.
Answer:
Lower.

Question 14.
(Metals/non-metals) display lustre.
Answer:
Metals.

Question 15.
Arrange the following metals in the order of their decreasing chemical activity : magnesium, potassium, iron, gold.
Answer:
Potassium, magnesium, iron, gold.

NCERT Solutions for Class 8 Science Chapter 4 – 2 Mark Questions and Answers

Question 1.
White phosphorous has to be kept in water. Why ? [NCT2007]
Answer:
Phosphorus is to be kept in water to prevent its contact with air because it is highly reactive.

Question 2.
Can you store lemon pickle in an aluminium utensils ? Explain. [NCERT]
Answer:
We cannot store acidic food stuffs in aluminium utensils because aluminium reacts with acids. The food gets spoilt.

Question 3.
One day Reeta went to a jeweller’s shop with her mother. Her mother gave an old gold jewellery to the goldsmith to polish. Next day when they brought the jewellery back, they found that there was a slight loss in its weight. Can you suggest a reason for the loss in weight ? [NCERT]
Answer:
The goldsmith must have used acid to clean the gold jewellery and some gold must have dissolved in it. Therefore, there was loss in weight of the jewellery.

Question 4.
Write short notes on

1. Metallurgical processes
2. Uses of common metals and non-metals
3. Noble metals

Answer:

1. Metallurgical process can be divided into following steps :
   • Concentration of the ore
   • Reduction of metal compound to get free metal
   • Refining of metal.
2. Uses of common metals and non-metals :
   Uses of metals – for making machinery, automobiles, industrial gadgets, building, bridges, cooking utensils, electrical gadgets, jewellery, sheets.
   Uses of non-metals – oxygen is used by plants and animals for their survival, nitrogen is used by plants for their growth, chlorine is used in water purification to kill germs, sulphur is used for making sulphuric acid, tincture iodine has antiseptic properties.
3. Noble metals – Gold, silver and platinum are noble metals. They occur free in nature . and maintain their lustre for a long time. Platinum, gold and silver are used for making jewellery as they do not tarnish.

Question 5.
Purity of gold is 15 carat. What is the percentage’of gold in the ornaments ?
Answer:
24 carat purity of gold =100
∴ 1 carat purity or gold = 100/24
15 carat purity of gold = (100*15)/24 = 62.5 %

Question 6.
Give two uses of sulphur in chemical industry.
Answer:

• It is used in the manufacture of sulphuric acid.
• It is used in the manufacture of carbon disulphide, which is used as an industrial solvent.

Question 7.
How is sulphur useful in agriculture ? How is sulphur useful in medicine ?
Answer:
Sulphur powder is an excellent insecticide and fungicide. It is used in spraying fruit trees.

• Sulphur is the main constituent of skin ointments.
• Metallic sulphides of sulphur are used in the preparation of Ayurvedic medicines.

Question 8.
Give two important uses of silver.
Answer:

• It is used for making coins.
• Silver salts (silver bromide and silver iodide) are used for making photographic films.

Question 9.
Give two uses of gold.
Answer:

• Gold is used for making ornaments.
• Gold foils are used in the preparation of Ayurvedic medicines.

Question 10.
Give two uses of platinum.
Answer:

• It is used as a catalyst in the manufacture of sulphuric and nitric acid.
• Platinum catalytic converters use platinum as catalytic agent.

Question 11.
Which of the following will form acidic oxide and why :
P, K, Na, Ca?
Answer:
P (Phosphorus) will form acidic oxide because it is a non-metal.

Question 12.
You are given two materials X and Y. On hammering X is flattened, but Y breaks. Which one is a metal ?
Answer:
X is a metal because it flattens, i.e., it is malleable.

Question 13.
There are four materials A, B, C and D. A and D are hard and shiny, but B and C are dull and not very hard. Identify the metals and non-metals from A, B, C and D.
Answer:
A and D are metals.
B and C are non-metals.

Question 14.
Gaurav knows that wires can be made from copper and aluminium. He tries to make wire . from sulphur and carbon. Will he succeed ? Give reason also.
Answer:
No, he will not succeed because sulphur and carbon are non-metals. Non-metals are not ductile, that is, they cannot be drawn into wires.

NCERT Solutions for Class 8 Science Chapter 4 – 3 Mark Questions and Answers

Question 1.
What happens when [KVS 2008]
(a) Hydrochloric acid is poured on aluminium foils ?
(b) Sodium is placed in water ?
(c) Sulphur dioxide is dissolved in water ?
(Write the chemical equation of the reaction involved)
Answer:

Question 2.

1. A copper spoon had fallen into a container containing dil.HCl. What would happen to it in three days time ? [DAV2008]
2. Give reasons for the following :
   • Metals are used for making bells.
   • We can’t use pure gold to make jewellery.
3. A metal ribbon bums in air with bright white light and forms a white powder.
   • Which metal is this ?
   • Give the equation of the reaction taking place.
   • The metallic oxide formed would be acidic or basic in nature ?

Answer:

1. Nothing will happen as copper does not react with hydrochloric acid.
   • Metals have the property of sonorosity so they are used for making bells.
   • Pure gold cannot be used for making jewellery because it is very soft.
   • Magnesium.
   • 2Mg + O₂ ——> 2MgO
   • Basic in nature.

Question 3.
Give reasons for the following : [KVS 2007]

1. Silver is used in making mirrors.
2. Aluminium is used to make electrical wire.
3. Iron is used in construction of bridges and houses.
4. Graphite is used as an electrode in the dry cell.
5. Iron sheets are galvanised before use.

Answer:

1. Silver has the ability to reflect light, therefore, it is used for making mirrors.
2. Aluminium is a good conductor of electricity, so, it is used for making electrical wires.
3. Iron is a strong metal, therefore, it is mixed with concrete to make bridges and houses.
4. Graphite is a good conductor of electricity, therefore, it is used as an electrode in dry cell.
5. Iron sheets are galvanised before use so that they do not corrode.

Question 4.
Which of the following statements is correct ? [NCERT]

• All metals are ductile.
• All non-metals are ductile.
• Generally, metals are ductile.
• Some non-metals are ductile.

Answer:
Generally, metals are ductile.

Question 5.
Fill in the blanks : [NCERT]

1. Phosphoms is a very ………… non-metal.
2. Metals are …….. conductors of heat and ………
3. Iron is ……… reactive than copper.
4. Metals react with acids to produce ………. gas.

Answer:

1. reactive
2. good, electricity
3. more
4. hydrogen.

Question 6.
Mark ‘T’ if the statement is true and ‘F’ if it is false. [NCERT]

1. Generally, non-metals react with acids. ( )
2. Sodium is a very reactive metal. ( )
3. Copper displaces zinc from zinc sulphate solution. ( )
4. Coal can be drawn into wires. ( )

Answer:

1. F
2. T
3. F
4. F

Question 7.
Some properties are listed in the following Table. Distinguish between metals and non-metals on the basis of these properties. [NCERT]

Properties	Metals	Non-metals
1. Appearance 2. Hardness 3. Malleability 4. Ductility 5. Heat conduction 6. Conduction of electricity

Answer:

Properties	Metals	Non-metals
1. Appearance	Solid at room temperature except mercury.	They are either solids or gases except bromine (liquid).
2. Hardness	They are hard	They are brittle.
3. Malleability	Malleable	Non-malleable
4. Ductility	Ductile	Non-ductile
5. Heat conduction	Good conductors	Bad conductors
6. Conduction of electricity	Good conductors	Bad conductors

Question 8.
Give reasons for the following : [NCT 2010]

1. Aluminium foils are used to wrap food items.
2. Immersion rods for heating liquids are made up of metallic substances.
3. Copper cannot displace zinc from its salt solution.
4. Sodium and potassium are stored in kerosene.

Answer:

1. Aluminium is a highly malleable metal and can be made into foils. So, it can be used to wrap food items.
2. Metals are good conductors of electricity, therefore, they are used for making immersion rods.
3. Copper is less reactive than zinc. Therefore, it cannot displace zinc from its salt solution.
4. Sodium and potassium are highly reactive metals. On exposure to air, they get oxidized. To avoid this they are stored in kerosene.

Question 9.
Match the substances given in Column I with their uses given in Column II. [NCERT]

Column I	Column II
(a) Gold	(i) Thermometers
(b) Iron	(ii) Electric wire
(c) Aluminium	(iii) Wrapping food
(d) Carbon	(iv) Jewellery
(e) Copper	(v) Machinery
(f) Mercury	(vi) Fuel

Answer:

Column I	Column II
(a) Gold (b) Iron (c) Aluminium (d) Carbon (e) Copper (f) Mercury	(iv) Jewellery (v) Machinery (iii) Wrapping food (vi) Fuel (ii) Electric wire (i) Thermometers

 Question 10.
What happens when [NCERT]

1. Dilute sulphuric acid is poured on copper plate ?
2. Iron nails are placed in copper sulphate solution ? Write word equations of the reactions involved.

Answer:

1. When sulphuric acid is poured on copper plate, copper sulphate and hydrogen gas are produced.
   Copper + Sulphuric acid ——-> Copper sulphate + Hydrogen (gas).
2. When iron nails are placed in copper sulphate solution, iron sulphate and copper are formed.
   Iron + Copper sulphate ——-> Iron sulphate + Copper

Question 11.
Saloni took a piece of burning charcoal and collected the gas evolved in a test tube. [NCERT]
(a) How will she find the nature of the gas ?
(b) Write down word equations of all the reactions taking place in this process.
Answer:
(a) The nature of gas can be found by passing it lime water, which will turn milky.

Question 12.
List different uses of metals that you come across in everyday life.
Answer:
Metals are used for making

• machinery
• automobiles, aeroplanes, trains, etc.
• pins, cooking utensils, electrical gadgets.
• electrical wires.
• thin sheets used for wrapping of food items, medicines, etc.

Question 13.
Choose appropriate words from the brackets and complete the statements.

1. Noble gases are found in (free state/compound forms).
2. Non-metals are generally (malleable/brittle).
3. Potassium after combustion will form (acidic oxide/basic oxide).
4. (Iodine/bromine) has antiseptic properties.
5. German silver has (copper/silver) as major constituent.

Answer:

1. Free state
2. Brittle
3. Basic oxide
4. Iodine
5. Copper

Question 14.
State whether the following statements are True or False :

1. Sodium is more reactive than magnesium.
2. Magnesium reacts with cold water.
3. All metals exist in solid form at room temperature.
4. Gallium has a low melting point.
5. Gold is alloyed with copper to make it hard.

Answer:

1. False
2. True
3. True
4. False
5. True

Question 15.
From among the set of metals — sodium, zinc, iron, copper, silver, select the following giving equations for each reaction :
(a) Two metals which will liberate hydrogen from water.
(b) One metal which is used to prepare hydrogen gas in the laboratory.
(c) One metal which will displace copper from copper sulphate solution.
(d) One metal which will not displace copper from copper sulphate solution.
Answer:

NCERT Solutions for Class 8 Science Chapter 4 MCQs

Question 1.
Which of the following properties is generally not shown by metals?
(a) Ductility
(b) Sonorous
(c) Dullness
(d) Electrical conduction
Answer:
(c)

Question 2.
The most abundant element in the universe is
(a) hydrogen
(b) oxygen
(c) helium
(d) carbon
Answer:
(a)

Question 3.
The ability of metals to be drawn into wires is known as
(a) ductility
(b) conductivity
(c) malleability
(d) sonorousity
Answer:
(a)

Question 4.
The most abundant element in the earth crust is
(a) iron
(b) oxygen
(c) silicon
(d) aluminium
Answer:
(b)

Question 5.
Galvanisation is a method qf protecting iron from rusting by coating with a thin layer of
(a) silver
(b) galium
(c) zinc
(d) aluminium
Answer:
(c)

Question 6.
The most abundant ihetal in earth crust is
(a) Cu
(b) Al
(c) Fe
(d) Zn
Answer:
(b)

Question 7.
An alloy is
(a) a compound
(b) a heterogeneous mixture
(c) a homogeneous mixture
(d) an element
Answer:
(c)

Question 8.
In extraction of copper, the flux used is
(a) FeO
(b) Si02
(c) CaO
(d) FeSi03
Answer:
(b)

Question 9.
Alloys are homogeneous mixtures of a metal with a metal or non-metal. Which among the following aljoys contain non-metal as one of its constituents?
(a) Amalgam
(b) Brass
(c) Bronze
(d) Steel
Answer:
(d)

Question 10.
Which of the following is purest form of carbon?
(a) Diamond
(b) Graph’ite
(c) Fullerenes
(d) Charcoal
Answer:
(c)

Question 11.
Which among the following alloys contain mercury as one of its constituents?
(a) Alnico
(b) Solder
(c) Stainless steel
(d) Zinc Amalgam
Answer:
(d)

Question 12.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying paint
(b) Applying grease
(c) Applying a coating of zinc
(d) all of these
Answer:
(c)

Question 13.
Generally, non-metals are not conductors of electricity, which of the following is a good conductor of electricity?
(a) Fullerenes
(b) Graphite
(c) Diamond
(d) Sulphur
Answer:
(b)

Question 14.
Food cans are coated with tin and not with zinc because
(a) zinc is costlier than tin
(b) zinc is less reactive than tin
(c) zinc is more reactive than tin
(d) zinc has a higher melting point than tin
Answer:
(a)

Question 15.
Electrical wires have a coating of an insulating materials. The material, generally used is
(a) sulphur
(b) graphite
(c) PVC
(d) none of these
Answer:
(c)

Synthetic Fibres and Plastics Class 8 Science NCERT Textbook Questions

Question 1.
Explain why some fibres are called synthetic.
Answer:
Some fibres are called synthetic fibres because they are made by man using chemicals.

Question 2.
Mark (✓) the correct answer.
Rayon is different from synthetic fibres because
(a) it has a silk-like appearance.
(b) it is obtained from wood pulp.
(c) its fibres can also be woven like those of natural fibres.
Answer:
(b) it is obtained from wood pulp.

Question 3.
Fill in the blanks with appropriate words.
(a) Synthetic fibres are also called ____ or ____ fibres.
(b) Synthetic fibres are synthesised from a raw material called _____
(c) Like synthetic fibres, plastic is also a _____
Answer:
(a) man-made, artificial fibres
(b) petrochemicals
(c) polymer

Question 4.
Give examples which indicate that nylon fibres are very strong.
Answer:
The following examples indicate that nylon fibres are very strong.
(i) They are used for making parachutes and ropes for rock climbing.
(ii) They are used in making seat-belts, fishing nets, tyre cord, a string for sports rackets and musical instruments.

Question 5.
Explain why plastic containers are favoured for storing food.
Answer:
Plastic containers are favoured for storing food because of the following reasons:
(i) the plastics do not react with the food stored in them.
(ii) the plastics are lightweight and are strong.
(iii) they are easy to handle and safe.

Question 6.
Explain the difference between thermoplastic and thermosetting plastics.
Answer:

Thermoplastics	Thermosetting plastics
(i) These plastics softened on heating and can be bent easily.	(i) These plastics when moulded once, can’t be softened again.
(ii) They do not lose their plasticity.	(ii) They lose their plasticity.
(iii) Examples are polyethene, PVC, etc.	(iii) Examples are bakelite and melamine.

Question 7.
Explain why the following are made of thermosetting plastics.
(a) Saucepan handles
(b) Electric plugs/switches/plugboards
Answer:
(a) Since, thermosetting plastics are a bad conductor of heat and do not get heated up while cooking, they are used for making saucepan handles.
(b) Since thermosetting plastics are a bad conductor of electricity and the electric current does not pass through such plastics, they are used for making electric plugs/switches/plugboards.

Question 8.
Categorise the materials of the following products into ‘can be recycled’ and ‘can not be recycled’.
Telephone instruments, plastic toys, cooker handles, carry bags, ballpoint pens, plastic bowls, plastic covering on electrical wires, plastic chairs, electrical switches.
Answer:

Can be recycled	Cannot be recycled
Plastic toys carry bags, plastic bowls, plastic covering on electrical wires, plastic chairs.	Telephone instruments, cooker handles, ballpoint pens, electrical switches.

Question 9.
Rana wants to buy shirts for summer. Should he buy cotton shirts or shirts made from synthetic material? Advise Rana, giving your reason.
Answer:
He should buy cotton shirts. This is because cotton has more capacity to hold moisture than synthetic clothes. In summers we have extensive sweating which is easily soaked by cotton shirts and hence, cotton clothes are much better than the clothes made from synthetic material.

Question 10.
Give examples to show that plastics are non-corrosive in nature.
Answer:
The literal meaning of non-corrosive is resistant to get destroyed by chemical action.
Following are the examples that show that plastics are non-corrosive in nature.

• Plastic containers do not react with items stored in it.
• They do not get rusted when exposed to moisture and air.
• They do not decompose when left in open for a long period.

Synthetic Fibres and Plastics Class 8 Science NCERT Intext Activities Solved

Activity 1
Take an iron stand with a clamp. Take a cotton thread of about 60 cm length. Tie it to the clamp so that it hangs freely from it as shown in Fig. 3.3.

At the free end suspend a pan so that weight can be placed in it. Add weight one by one till the thread breaks. Note down the total weight required to break the thread. This weight indicates the strength of the fibre. Repeat the same activity with threads of wool, polyester, silk and nylon. Tabulate the data as shown in Table 3.1. Arrange the threads in order of their increasing strength.
Observation Table 3.1

S. No.	Type of Thread/ Fibre	Total Weight required to break the thread
1.	Cotton	12 gm
2.	Wool	70 gm
3.	Silk	50 gm
4.	Nylon	100 gm

(Precaution: Note that all threads should be of the same length and most of the same thickness.)

Activity 2
Take two cloth pieces of the same size, roughly half a metre square each. One of these should be from natural fibre. The other could be a synthetic fibre. You can take the help of your parents in selecting these pieces. Soak the pieces in different mugs each containing the same amount of water. Take the pieces out of the containers after five minutes and spread them in the sun for a few minutes. Compare the volume of the water remaining in each container.
Solution:
It is observed that the volume of water of the container in which natural fibre is soaked contains less water as compared to the container in which synthetic fibre is soaked. Thus, natural fibre absorbs more water as compared to synthetic fibre. When both the fibres were spread in the sun, it was observed that synthetic fibre took less time to dry than natural fibre.

NCERT Solutions for Class 8 Science Chapter 3 – 1 Mark Questions and Answers

Question 1.
Explain why some fibres are called synthetic. [NCERT]
Answer:
Some fibres are called synthetic because they are made by human beings.

Question 2.
Mark the correct answer.
Rayon is different from synthetic fibres because [NCERT]

• It has a silk-like appearance.
• It is obtained from wood pulp.
• Its fibres can also be woven like those of natural fibres.

Answer:

• It is obtained from wood pulp.

Question 3.
Give examples which indicate that nylon fibres are very strong.
Answer:
Parachutes and nylon ropes are used for rock climbing.

Question 4.
Name the plastic whose sheets are used for packing liquids.
Answer:
Polythene.

Question 5.
Why is teflon used as tape for sealing purpose ?
Answer:
Becausee of its toughness, teflon is used as tape for sealing purpose.

Question 6.
Name the first fully synthetic fibre.
Answer:
Nylon.

Question 7.
Why nylon is used for making parachutes ?
Answer:
Nylon is used for making parachutes, because it is very strong, elastic and light.

Question 8.
Name the material used for making ropes for rock climbing.
Answer:
Nylon.

Question 9.
What is polyester ?
Answer:
Polyester is made up of the repeating units of an ester.

Question 10.
Name the plastic used for making polythene bags.
Answer:
Polythene bag is made of plastic which is a polymer of ethene.

Question 11.
What name is given to plastics which can be re-set a number of times ?
Answer:
Thermoplastics.

Question 12.
Classify the following into thermosetting and thermoplastics-combs, bakelite, melamine, toys.
Answer:
Thermosetting – bakelite, melamine
Thermoplastics – combs, toys

Question 13.
Why are the chemicals in the laboratory stored in plastic containers ?
Answer:
Plastics are non-reactive, so they can be used for storing chemicals.

Question 14.
Can you store pickles in plastic containers ? Why ?
Answer:
Plastics are non-reactive, so pickles can be stored in them.

Question 15.
Why are the electric wires covered with plastic ?
Answer:
Plastic is a poor conductor of electricity, therefore, the wires are covered with it.

Question 16.
Why do we use plastic cookwares in microwave ovens ?
Answer:
We use plastic cookware in microwave ovens, because they are not affected by heat but the food is cooked.

Question 17.
Tin takes about 100 years to degenerate. Is it biodegradable or non-biodegradable ?
Answer:
Tin is non-biodegradable.

Question 18.
Is it advisable to use recycled plastic containers for storing food ?
Answer:
No, we should not use recycled plastic containers for storing food because they contain colouring agents.

Question 19.
As a responsible citizen, what are the 4R’s we should remember ?
Answer:
Reduce, Reuse, Recycle and Recover

NCERT Solutions for Class 8 Science Chapter 3 – 2 Mark Questions and Answers

Question 1.
Aperson has to make a non-stick pan. He has three types of plastic-Bakelite, Teflon and PVC. Which plastic will he use for coating and why ?
Answer:
The person will use Teflon because it is not affected by heat and does not react chemically with other substances.

Question 2.
Explain the difference between thermoplastic and thermosetting plastics. [NCERT]
Answer:

Thermosetting	Thermoplastic
(i) These plastics can be moulded and (ii) e.g.,Bakelite and melamine.	(i) These plastics can be moulded and reset a number of times. (ii) e.g. Polythene and PVC.

 Question 3.
Differentiate between natural and synthetic fibres.
Answer:
Natural fibres are obtained from nature, e.g., cotton whereas synthetic fibres are made by man, e.g., nylon.

Question 4.
How is rayon made ? Give one advantage of using rayon.
Answer:
Rayon is obtained from wood pulp by treating it chemically. It is cheaper than silk, but can be woven like silk fibre.

Question 5.
Is nylon fibre so strong, that we can use it to make parachutes ?
Answer:
Yes, nylon is very strong and it can be used for making parachutes.

Question 6.
Give some uses of PET.
Answer:
PET is used for making bottles, utensils, fibres and wires.

Question 7.
Give the composition of Polycot and Polywool.
Answer:
Polycot is a mixture of polyester and cotton, Poly wool is a mixture of polyester and wool.

Question 8.
What properties of plastics make it useful for many things ?
Answer:
Plastics can be recycled, reused, coloured, melted, rolled into sheets or made into wires, so they are very useful.

Question 9.
Why is melamine used for making kitchenware ?
Answer:
Melamine resists fire and can tolerate heat better than other plastics. So, it is used for making kitchenware.

Question 10.
Buckets made up of plastics are better. Why ?
Answer:
Buckets made up of plastic are better because they are light, strong and durable.

NCERT Solutions for Class 8 Science Chapter 3 – 3 Mark Questions and Answers

Question 1.
Give three advantages of polythene over natural materials. [MSE (Chandigarh)]
Answer:
Three advantages of polythene over natural materials are :

• It is strong but flexible.
• Can be rolled into sheets.
• Water resistant.

Question 2.
Give one use each of bakelite, nylon and acrylic.
Answer:
Bakelite is used for making plugs and switches. Nylon is used in textile industry to produce clothes. Acrylic is used as a substitute of natural wool for knitting sweaters, blankets etc.

Question 3.
Fill in the blanks with appropriate words. [NCERT]

1. Synthetic fibres are also called ……….. or ……….
2. Synthetic fibres are synthesised from raw material called …………
3. Like synthetic fibres, plastic is also a …………..

Answer:

1. artificial, man-made
2. mono or petrochemicals
3. polymers.

Question 4.
Explain why plastic containers are favoured for storing food. [NCERT]
Answer:
Food can be stored in plastic containers because

• Plastics are non-reactive.
• Plastics are light and strong.
• Plastics are durable.

Question 5.
Explain why the following are made of thermosetting plastics : [NCERT]

1. Saucepan handles
2. Electric plugs/switches/plug boards.

Answer:

1. Saucepan handles are made of thermosetting plastics because they are bad conductors of heat.
2. Electric plugs/switches/plug boards are made of thermosetting plastics because they are bad conductors of electricity.

Question 6.
Rana wants to buy shirts for summer. Should he buy cotton shirts or shirts made from synthetic material ? Advice Rana, giving your reason.
Answer:
Rana should buy cotton shirts. Cotton clothes are preferred in summer because they absorb sweat and the person feels cool whereas synthetic materials do not absorb sweat and the person feels uncomfortable.

Question 7.
Give examples to show that plastics are noncorrosive in nature. [NCERT]
Answer:

• Buckets made of iron get rusted but those made of plastic are not rusted.
• Food can be stored in plastic containers, as they are resistant to chemicals in the food
• Teflon coated pans are used instead of metallic as it doesnot react with chemicals in the food.

Question 8.
Give three advantages of rayon.
Answer:
Advantages of rayon are :

• It is cheap.
• It can be dyed in many colours.
• It can be woven like silk.

Question 9.
What are the advantages of nylon ?
Answer:
Nylon fibre is strong, elastic and light. It is lustrous and easy to wash.

Question 10.
Give some uses of nylon.
Answer:
Nylon can be used for making socks, ropes, tents, toothbrushes, sleeping bags, curtains, parachutes and ropes for rock climbing

NCERT Solutions for Class 8 Science Chapter 3 – 5 Mark Questions and Answers

Question 1.
Categorise the materials of the following products into ‘can be recycled’ and ‘cannot be recycled’. [NCERT]
Telephone instruments, plastic toys, cooker handles, carry bags, ball point pens, plastic bowls, plastic covering on electrical wires, plastic chairs, electrical switches.
Answer:

Can be recycled	Non-biodegradable Materials
Toys, carry bags, plastic bowls, plastic covering on electrical wires, plastic chairs.	Telephone instruments, cooker handles, electric switches, ball point pens, electrical switches.

 Question 2.
Match the terms of Column A correctly with the phrases given in Column B. [NCERT]

Column A	Column B
(a) Polyester (b) Teflon (c) Rayon (d) Nylon	(i) Prepared by using wood pulp. (ii)Used for making parachutes and stockings. (iii) Used to make non-stick cookwares. (v) Fabrics do not wrinkle easily.

Answer:

Column A	Column B
(a) Polyester	(iv) Fabrics do not wrinkle easily.
(b) Teflon	(iii) Used to make non-stick cookwares.
(c) Rayon	(i) Prepared by using wood pulp.
(d) Nylon	(ii) Used for making parachutes and stockings.

 Question 3.
‘Manufacturing synthetic fibres is actually helping conservation of forests’. Comment. [NCERT]
Answer:
The said statement is correct to a certain extent. The forests would be conserved if synthetic fibres are used, but other effects of synthetic fibres are more harmful. Disposal of synthetic fibres causes lot of environmental pollution. When synthetic fibres bum, lot of smoke is produced.

Question 4.
‘Avoid plastics as far as possible’. Comment on this advice. [NCERT]
Answer:
Plastics are very useful, but it causes serious environmental and health concern :

• Plastics are non-biodegradable.
• Careless disposal of plastic bags, chokes, drains and blocks the soil.
• If eaten by cows, it can kill them.
• Plastic bags can also contaminate foodstuffs due to poisonous dyes getting absorbed into food

Question 5.
How is plastic useful in healthcare industry ?
Answer:
In healthcare, plastics are used

• for packaging of tablets
• as threads for stitching wounds
• in syringes
• doctor gloves
• for making a number of medical instruments.

Question 6.
“Even though plastics are very useful, they are not environment friendly.” Justify the statement.
Answer:
Plastics are very useful, but their disposal is a big problem. Plastics are non-biodegradable and take many years to decompose. They release a lot of poisonous fumes, when burnt so they cause environmental pollution.

Question 7.
Give methods by which pollution due to plastics can be solved.
Answer:
Pollution due to plastics can be solved by following methods :

• Avoid the use of plastics as far as possible.
• Use bags, made of jute or cloth.
• Biodegradable and non-biodegradable wastes should be separated and disposed off separately.
• Plastics should be recycled and reused.

NCERT Solutions for Class 8 Science Chapter 3 MCQs

Question 1.
Electrical switches are made of
(a) nylon
(b) bakelite
(c) polythene
(d) melamine
Answer:
(b)

Question 2.
Out of the following, which is not biodegradable ?
(a) Vegetable peels
(b) Plastic bags
(c) Cotton
(d) Jute
Answer:
(b)

Question 3.
Clothes made of which fabric are best suited for hot climate ?
(a) Cotton
(b) Nylon
(c) Acrylic
(d) Polycot
Answer:
(a)

Question 4.
Out of the following, which is not a natural fibre ?
(a) Cotton
(b) Silk
(c) Jute
(d) Rayon
Answer:
(d)

Question 5.
Naturally occurring po/ymer is
(a) cellulose
(b) polyester
(c) nylon
(d) PVC
Answer:
(a)

Question 6.
Plastic used for coating non-stick pans is
(a) PVC
(b) ester
(c) bakelite
(d) melamine
Answer:
(d)

Question 7.
Pickles are kept in plastic containers because plastic containers are
(a) non-corrosive
(b) light
(c) colourful
(d) cheap
Answer:
(a)

Question 8.
Which of the following is a thermoplastic
(a) Bakelite
(b) Melamine
(c) Polythene
(d) Jute
Answer:
(c)

Question 9.
Which of the following can be recycled?
(a) Plastic bowls
(b) Ballpoint pens
(c) Telephone instruments
(d) Electrical switches
Answer:
(a)

Question 10.
Which material is best suited for covering electric wires?
(a) Plastic bowls
(b) PVC
(c) Polystyrene
(d) Nylon
Answer:
(b)

Microorganisms: Friend and Foe Class 8 Science NCERT Textbook Questions

Question 1.
Fill in the blanks.
(a) Microorganisms can be seen with the help of a _____
(b) Blue-green algae fix ______ directly from the air to enhance the fertility of soil.
(c) Alcohol is produced with the help of _____
(d) Cholera is caused by ______
Answer:
(a) microscope
(b) nitrogen
(c) yeast
(d) bacteria

Question 2.
Tick the correct answer.
(a) Yeast is used in the production of
(i) sugar
(ii) alcohol
(iii) hydrochloric acid
(iv) oxygen
Answer:
(ii) alcohol

(b) The following is an antibiotic:
(i) sodium bicarbonate
(ii) streptomycin
(iii) alcohol
(iv) yeast
Answer:
(ii) streptomycin

(c) Carrier of malaria-causing protozoan is:
(i) female Anopheles mosquito
(ii) cockroach
(iii) housefly
(iv) butterfly
Answer:
(i) female Anopheles mosquito

(d) The most common carrier of communicable diseases is
(i) ant
(ii) housefly
(iii) dragonfly
(iv) spider
Answer:
(ii) housefly

(e) The bread or idli dough rises because of:
(i) heat
(ii) grinding
(iii) growth of yeast cells
(iv) kneading
Answer:
(iii) growth of yeast cells

(f) The process of conversion of sugar into alcohol is called
(i) nitrogen fixation
(ii) moulding
(iii) fermentation
(iv) infection
Answer:
(iii) fermentation

Question 3.
Match the organisms in column A with their action in column B.

A	B
(i) Bacteria	(a) Fixing nitrogen
(ii) Rhizobium	(b) Setting of curd
(iii) Lactobacillus	(c) Baking of bread
(iv) Yeast	(d) Causing malaria
(v) A protozoan	(e) Causing cholera
(vi) A virus	(f) Causing AIDS
	(g) Producing antibodies

Answer:
(i) (e)
(ii) (a)
(iii) (b)
(iv) (c)
(v) (d)
(vi) (f)

Question 4.
Can microorganisms be seen with the naked eye? If not, how can they be seen?
Answer:
The microorganisms cannot be seen with our naked eyes because they are very small in size. Some of these, such as fungus growing on bread, can be seen with a magnifying glass. Others cannot be seen without the help of a microscope.

Question 5.
What are the major groups of microorganisms?
Answer:
Microorganisms are classified on the basis of their size into four major groups. These groups are:
(a) Bacteria
(b) Fungi
(c) Protozoa
(d) Some algae

Question 6.
Name the microorganisms which can fix atmospheric nitrogen in the soil.
Answer:
Rhizobium, Clostridium and Azotobacter.

Question 7.
Write 10 lines on the usefulness of microorganisms in our lives.
Answer:
Microorganisms are useful to us in many ways. For example,

• Bacteria like Lactobacillus convert milk into curd.
• Bacteria are also involved in the making of cheese.
• Acetobacter aceti is used for producing acetic acid from alcohol.
• Yeast is used in the commercial production of alcohol, wine and bakery products.
• Some specific microorganisms are helpful in manufacturing of antibiotics.
• Microorganisms act as cleansing agents and decompose the waste products into manure.
• Dead or weakened microbes are used in the preparation of vaccines.
• Some bacteria fix atmospheric nitrogen and increase soil fertility.
• Algae, yeast, fungi or bacteria may be used as an ingredient or a substitute for protein-rich foods that are suitable for human or animal consumption.
• Some microorganisms are taken as probiotics, that are believed to provide health benefits when consumed.

Question 8.
Write a short paragraph on the harms caused by microorganisms.
Answer:
Microorganisms are harmful to us in many ways. For example, microorganisms, called pathogens cause disease in humans, plants and animals. Pathogens or germs enter a healthy body through air, water, contaminated food and infected person by direct or indirect contact or by the carrier. Common ailments like cold, influenza (flu), cough, polio, chicken pox are caused by viruses. Foot and mouth diseases in the cattle are also caused by viruses. Typhoid, tuberculosis (TB) are caused by bacteria. Anthrax a dangerous human and cattle diseases is also caused by bacteria.

Diseases like dysentery and malaria are caused by protozoa. Ringworm is caused by fungi. Several microbes causes diseases in plants and thus reduces the yield. Citrus canker, a bacterial disease, affects trees of citrus fruit and is spread by air. Bhendi yellow vein mosaic disease is caused by a virus and is spread by insects in lady fingers. Rust of wheat is a fungal disease spread through air. Microorganisms that grow on our food sometimes produce toxic substances. These make the food poisonous causing serious illness and even death. This food-borne illness is called food poisoning.

Question 9.
What are antibiotics? What precautions must be taken while taking antibiotics?
Answer:
Antibiotics are the medicines which kill or stop the growth of the disease-causing microbes. They are manufactured by growing specific microorganisms. They are used to cure a variety of diseases.

It is important to take antibiotic only on the advice of a qualified doctor. One must finish the course prescribed by the doctor to make the drug more effective. Antibiotics must not be taken unnecessarily because it may kill beneficial bacteria also. Antibiotics are, however, not effective against cold and flu as they are caused by viruses.

Microorganisms: Friend and Foe Class 8 Science NCERT Intext Activities Solved

Activity 1
Collect some moist soil from the field in a beaker and add water to it. After the soil particles have settled down, observe a drop of water from the beaker under a microscope. What do you see?
Solution:
It is observed that some tiny organisms are moving around.

Activity 2
Take a few drops of water from a pond. Spread on a glass slide and observe through a microscope.
Solution:
It is observed that some tiny organisms are moving around.

Activity 3
Jake 1/2 kg flour (atta or maida), add some sugar and mix with warm water. Add a small amount of yeast powder and knead to make a soft dough. What do you observe after two hours? Did you find the dough rising?

Solution:
It is observed that the dough begins to rise up in volume. Yeast reproduces rapidly and releases C02 during respiration. Etubble of this gas fill the dough and increase its volume.

Activity 4

Take 500 ml. beaker filled upto 3/4 with water. Dissolve 2-3 teaspoons of sugars in it. Add half a spoon of yeast powder to the sugar solution. Keep it covered in a warm place for 4-5 hours. Now smell the solution. Can you get a smell?
Solution:
It is observed that the solution smell like alcohol. This process of conversion of sugar in alcohol is known as fermentation.

Activity 5

Take two pots and fill each pot half with soil. Mark them A and B. Put plant waste in pot A and things like polythene bags, empty glass bottles and broken plastic toys in pot B. Put the pots aside. Observe them after 3-4 weeks.
Solution:
It is observed that plant waste in pot A has been decomposed, whereas the polythene bags, empty glass bottles and broken plastic toys in pot B did not undergo such changes.

NCERT Solutions for Class 8 Science Chapter 2 – 1 Mark Questions and Answers

Question 1.
Name the bacteria responsible for the disease typhoid. [MSE (Chandigarh) 2008]
Answer:
Salmonella typhi.

Question 2.
Chicken pox is caused by ………. virus. [KVS 2008]
Answer:
Varicella zoster.

Question 3.
………….. (disease) is caused by mycoplasma. [KVS 2008]
Answer:
Pleuropneumonia.

Question 4.
Name the toxin released by T.B. bacteria.[MSE (Chandigarh) 2007]
Answer:
Tuberculin toxin is released by T.B. bacteria.

Question 5.
Which microorganism is the cause of malaria ? [NCT 2006]
Answer:
A Protozoan, Plasmodium.

Question 6.
Write two diseases caused by bacteria. [NCT2005]
Answer:
Two diseases caused by bacteria are tuberculosis and diphtheria.

Question 7.
Name any two items that are prepared by using yeast. [KVS 2005]
Answer:
Yeast is used to prepare bread and alcohol.

Question 8.
Fill in the blanks : [NCERT]

1. Microorganisms can be seen with the help of a …………..
2. Blue-green algae fix ………… directly from air to enhance fertility of soil.
3. Alcohol is produced with the help of ………….
4. Cholera is caused by ……………

Answer:

1. Microscope
2. Nitrogen
3. Yeast
4. Bacteria.

Question 9.
Tick the correct answer : [NCERT]

• (a) Yeast is used in the production of
  (i) sugar
  (ii) alcohol
  (iii) hydrochloric acid
  (iv) oxygen.
• (b) The following is an antibiotic
  (i) Sodium bicarbonate
  (ii) Streptomycin
  (iii) Alcohol
  (iv) Yeast.
• (c) Carrier of malaria-causing protozoan is
  (i) female anopheles mosquito
  (ii) cockroach
  (iii) housefly
  (iv) Butterfly.
• (d) The most common carrier of communicable diseases is
  (i) ant
  (ii) housefly
  (iii) dragonfly
  (iv) spider.
• (e) The bread or idli dough rises because of
  (i) heat
  (ii) grinding
  (iii) growth of yeast cells
  (iv) kneading.
• (f) The process of conversion of sugar into alcohol is called
  (i) nitrogen fixation
  (ii) moulding
  (iii) fermentation
  (iv) infection.

Answer:
(a) —> (ii)
(b) —> (ii)
(c) —> (i)
(d) —> (ii)
(e) —> (iii)
(f) —> (iii)

Question 10.
Match the organisms in Column I with their action in Column II. [NCT2010, NCERT]

Column I	Column II
(a) Bacteria (b) Rhizobium (c) Lactobacillus (d) Yeast (e) A protozoan (f) A virus	(i) Fixing nitrogen (ii) Setting of curd (iii) Baking of bread (iv) Causing malaria (v) Causing cholera (vi) Causing AIDS

Answer:

Column I	Column II
(a) Bacteria (b) Rhizobium (c) Lactobacillus (d) Yeast (e) A protozoan (f) A virus	(i) Causing cholera (ii) Fixing nitrogen (iii) Setting of curd (iv) Baking of bread (v) Causing malaria (vi) Causing AIDS

Question 11.
Name the microorganisms which can fix atmospheric nitrogen in the soil. [NCERT]
Answer:
Rhizobium, Clostridium md Azotobacter fix atmospheric nitrogen.

Question 12.
Can microorganisms be seen with the naked eye ? If not, how can they be seen ? [NCERT]
Answer:
No, microorganisms cannot be seen with the naked eye. They can only be seen with the help of microscope.

Question 13.
What are the major groups of microorganisms ? [NCERT]
Answer:

• Bacteria
• Fungi
• Protozoa
• Algae.

Question 14.
What are microorganisms ?
Answer:
Microorganisms are organisms that are so small that they can only be seen through a microscope.

Question 15.
Name two diseases caused by viruses.
Answer:
Influenza and cough.

NCERT Solutions for Class 8 Science Chapter 2 – 2 Mark Questions and Answers

Question 1.
Describe the role of blue-green algae in fertility of soil. [KVS 2006, MSE (Chandigarh) 2008]
Answer:
Blue-green, algae, also called cyanobacteria, can fix atmospheric nitrogen into usable , compounds. These are then used as fertilizers.

Question 2.
Name three habitats of microorganisms. [MSE (Chandigarh) 2008, 2006]
Answer:
Microorganisms are present in soil, water, outer space and inside the body of animals.

Question 3.
Name the bacterium found in the roots of pea plant. How is this bacterium useful for human beings ? [NCT2007]
Answer:
Rhizobium is the bacterium found in the roots of pea plant. The. bacterium absorbs the atmospheric nitrogen and converts it to nitrates.

Question 4.
What is a vaccine ? Why is it important to vaccinate small children ? [DAV2006]
Answer:
Vaccine is prepared from weak or dead disease-causing microbe. Vaccine is given to healthy persons to prevent occurrence of disease. It is important to vaccinate small children because it creates antibodies in blood.

Question 5.

1. Give full form of ORS.
2. What is vaccination ? [DAV2007]

Answer:

1. ORS — Oral Rehydration Solution.
2. Vaccination is a method to immunise the body against diseases by making the body’s immune system produce antibodies against the disease-causing microbe in the vaccine.

Question 6.
Mention two important uses of fungi. [KVS 2006]
Answer:
Uses of fungi:

• They convert dead organic matter into simple soluble minerals and gases, which can be used again by plants.
• Fungi like yeast is used in bread, beer and wine.

Question 7.
How is pasteurised milk obtained ? [NCT 2010]
Answer:
In pasteurisation, the milk is heated to 62.5°C for 30 minutes or to 71.5°C for 15 seconds. It is then rapidly cooled to 10°C and packed in airtight containers.

Question 8.
Draw a neat and well-labelled diagram of Spirogyra.
Answer:

Question 9.
Draw a labelled diagram of Rhizopus.
Answer:

Question 10.
Draw a labelled diagram of Chlamydomonas. [DAV 1997]
Answer:

Question 11.
Draw a labelled diagram of Amoeba.
Answer:

Question 12.
How are viruses different from other microorganisms ?
Answer:
Viruses reproduce only inside the host organisms that is, bacterial, plant or animal cell.

Question 13.
Give two examples where microorganisms are useful at home.
Answer:
At home microorganisms are used for preparation of curd and cakes.

Question 14.
How are microorganisms useful commercially ?
Answer:
Microorganisms are used for the large scale production of alcohol, wine and acetic acid (vinegar).

Question 15.
How do microorganisms clean the environment ?
Answer:
Microorganisms degrade the harmful substances and clean the environment.

Question 16.
How can we prevent a person from getting Hepatitis B ?
Answer:

• By giving boiled water for drinking.
• By vaccination.

Question 17.
Name one disease caused by bacteria and one disease caused by virus in cattle.
Answer:

• Bacteria – Anthrax.
• Virus – Foot and mouth disease.

NCERT Solutions for Class 8 Science Chapter 2 – 3 Mark Questions and Answers

Question 1.

• Tuberculosis is a highly infectious disease. Justify the statement.
• Which vitamin helps in the prevention of common cold ? [MSE (Chandigarh) 2008]

Answer:

• Tuberculosis is a highly infectious disease. TB is transmitted through minute droplets of infected sputum on phelgm, by drinking milk of an infected animal.
• Common cold is prevented by taking vitamin C.

Question 2.
Draw a labelled diagram of virus. [KVS 2008, DAV 2006]
Answer:

Question 3.
Can you store pickles in iron containers ? Why ? [KVS 2007]
Answer:
We can not store pickles in iron containers because the acid present in the pickles reacts with iron. This can cause food poisoning, if consumed.

Question 4.
What precautions must be taken while taking antibiotics ?
Answer:
Antibiotics should be taken after consulting a doctor. The complete dose of the antibiotics should be taken as prescribed. They should not be taken unnecessarly because they will kill the useful bacteria.

Question 5.

1. Write the causal organism of cholera.
2. Write any two symptoms of this disease.
3. Why ORS should be given to the patients suffering from cholera ? [MSE (iChandigarh) 2007]

Answer:

1. Cholera is caused by Vibrio cholerae.
2. Two symptoms of cholera are :
   • profuse and painless watery diarrhoea.
   • muscular cramps.
3. ORS should be given to the patients suffering from cholera to avoid excessive loss of body fluids.

Question 6.
Why are viruses considered to be on the borderline between living organisms and non-living things ? [DAV2007]
Answer:
Viruses cannot reproduce, respond to changes or use energy to grow. Since viruses reproduce in the host cell, scientist regard viruses as a link between living and non-living.

Question 7.
How are bacteria beneficial for us ? [NCT 2006]
Answer:
Bacteria are useful to us as :

• Nitrogen fixing bacteria increase the fertility of the soil.
• They can be used to form curd, alcohol, etc.
• They can decompose organic matter.

Question 8.
Write three types of bacteria on the basis of their shape. Give one example of each. [MSE (Chandigarh) 2005]
Answer:
Three types of bacteria are :

1. Bacillus or rod-shaped, e.g., Lactobacillus.
2. Coccus or spherical, e.g., Streptococcus.
3. Spirillum or spiral, e.g., Vibrio.

Question 9.
(a) Are bacteria plants or animals ? Give reasons in support of your answer.
(b) Write two differences between autotrophic and heterotrophic bacteria. [DAV2005]
Answer:
(a) Bacteria is considered as plant due to the presence of a rigid cell wall in it.
(b)

Autotrophic bacteria	Heterotrophic bacteria
(i) Chlorophyll is present. (ii) Synthesize their own food.	(i) Chlorophyll is absent. (ii) Depend on readymade food from other sources.

Question 10.
How does a housefly transmit diseases ? [MSE (Chandigarh) 1999]
Answer:
A housefly gets attracted towards garbage and excreta. The harmful microorganisms present in excreta and garbage easily stick to its fine array of body hair and are thus, transferred to food stuffs whenever they sit on them and a result food gets poisoned.

Question 11.

1. What are antibiotics ?
2. How are antibiotics manufactured ?
3. Name two important antibiotics.

Answer:

1. Medicines which kill or stop the growth of the disease causing microorganisms are called antibiotics.
2. The antibiotics are manufactured by growing specific microorganisms.
3. Two important antibiotics are Tetracycline and Bacitracin.

Question 12.
What will happen if you take antibiotics when not needed ?
Answer:
If you take antibiotics when not needed, you help bacteria in your body to develop resistance to them. Next time when you fall ill and need these antibiotics, they would be less effective.

Question 13.
How does a vaccine act ?
Answer:
The vaccine acts by making the body’s immune system to produce antibodies against the disease-causing microbe in the vaccine. The antibodies attack and destroy the weakened microbe as it enters the body.

Question 14.
How is common cold spread ?
Answer:
When a person suffering from common cold sneezes, fine droplets of moisture carrying thousands of viruses causing common cold are spread in the air. The viruses may enter the body of a healthy person while breathing.

Question 15.
How can you prevent the spread of communicable diseases ?
Answer:
It is better to avoid contact with the infected person. We should keep distance from infected persons.

NCERT Solutions for Class 8 Science Chapter 2 – 5 Mark Questions and Answers

Question 1.
Describe the nitrogen cycle with the help of a neat and labelled diagram. [MSE (Chandigarh) 2005, 2007]
Answer:

Question 2.
Write 10 lines on the usefulness of microorganisms in our lives. [NCERT]
Answer:
Microorganisms are useful in our lives in the following ways :

• Bacteria cause the decay of dead plants and animals.
• Nitrogen fixing bacteria convert the atmospheric nitrogen into nitrates.
• Bacteria are used in the production of vinegar, curd, cheese, etc.
• Bacteria are used in sewage disposal plants.
• Scientists have discovered ways to change bacteria so that they produce valuable medical, agricultural and industrial products.

Question 3.
Write a short paragraph on the harms caused by microorganisms. [NCERT]
Answer:

• Microorganisms cause diseases in human beings, plants and animals.
• Microorganisms spoil food and cause food poisoning.

Question 4.
How can we control the spread of malaria or dengue ?
Answer:
All mosquitoes breed in water. Hence, by keeping the surroundings clean and dry, we can prevent mosquitoes from breeding. We should not allow water to collect in coolers, flower pots, etc. We should wear full sleeves shirts in the evenings and at night. We can also use mosquito repellant creams and mosquito nets.

NCERT Solutions for Class 8 Science Chapter 2 MCQs

Question 1.
Microorganisms exhibiting the characteristics of living and non-living organisms are known as
(a) bacteria
(b) virus
(c) algae
(d) fungi
Answer:
(b)

Question 2.
A person bitten by dog gets
(a) malaria
(b) tetanus
(c) typhoid
(d) rabies
Answer:
(d)

Question 3.
Amoebic dysentery is transmitted by
(a) sneezing
(b) using contaminated water
(c) direct contact
(d) wound
Answer:
(b)

Question 4.
Anthrax in cattle is caused by
(a) fungi
(b) virus
(c) bacteria
(d) algae
Answer:
(c)

Question 5.
Smut of rice is caused by
(a) virus
(b) protozoa
(c) fungi
(d) bacteria
Answer:
(c)

Question 6.
Wheat, maize and oats are preserved by
(a) using sugar and salt
(b) pasteurisation
(c) drying
(d) using chemical preservatives
Answer:
(c)

Question 7.
Jams and jellies are preserved by
(a) drying
(b) using chemical preservatives
(c) using acetic acid
(d) using heat and cold treatments
Answer:
(b)

Question 8.
Pasteurisation is the method used for preserving
(a) milk
(b) jams
(c) pickles
(d) grains
Answer:
(a)

Question 9.
Fermentation is the process used for making
(a) curd
(b) cheese
(c) idlis
(d) chappatis
Answer:
(c)

Crop Production and Management Class 8 Science NCERT Textbook Questions

Question 1.
Select the correct word from the following list and fill in the blanks.
float, water, crop, nutrients, preparation
(a) The same kind of plants grown and cultivated on a large scale at a place is called _____
(b) The first step before growing crops is _______ of the soil.
(c) Damaged seeds would ______ on top of the water.
(d) For growing a crop, sufficient sunlight and ______ and ______ from the soil are essential.
Answer:
(a) crop
(b) preparation
(c) float
(d) water, nutrients

Question 2.
Match items in column A with those in column B.

A	B
(i) Kharif crops	(a) Food for cattle
(ii) Rabi crops	(b) Urea and superphosphate
(iii) Chemical fertilisers	(c) Animal excreta, cow dung, urine and plant waste
(iv) Organic manure	(d) Wheat, gram, pea
	(e) Paddy and maize

Answer:
(i) (e)
(ii) (d)
(iii) (b)
(iv) (c)

Question 3.
Give two examples of each.
(a) Kharif crop
(b) Rabi crop
Answer:
(a) Kharif crop: Paddy and maize
(b) Rabi crop: Wheat and gram

Question 4.
Write a paragraph in your own words on each of the following.
(a) Preparation of soil
(b) Sowing
(c) Weeding
(d) Threshing
Answer:
(a) Preparation of soil: Soil preparation is necessary before growing a crop. It involves tilling and loosening the soil. This allows the roots to penetrate deep in the soil and to breath easily even when they are deep.

(b) Sowing: The process of putting seeds into the soil is called sowing. The tool used traditionally for sowing seeds is funnel-shaped. Nowadays a seed drill is used for sowing with the help of tractors. This tool sows the seed uniformly at a proper distance and depth.

(c) Weeding: Some undesirable plants grow along with crop and these unwanted plants are called weeds. The process of removing these unwanted plants is called weeding.

(d) Threshing: The process of separating the grain seeds from the chaff is called threshing.

Question 5.
Explain how fertilisers are different from manure.
Answer:

Fertilisers	Manures
(i) A fertiliser is an inorganic salt.	(i) Manure is a natural substance obtained by the decomposition of cattle dung, human waste and plant residues.
(ii) A fertiliser is prepared in factories.	(ii) Manure can be prepared in the fields.
(iii) A fertiliser does not provide any humus to the soil.	(iii) Manure provides a lot of humus to the soil.
(iv) Fertilisers are very rich in plant nutrients like nitrogen, phosphorus and potassium.	(iv) Manure is relatively less rich in plant nutrients.

Question 6.
What is irrigation? Describe two methods of irrigation which conserve water.
Answer:
The artificial method of watering the plants for assisting in their growth is called irrigation. Main sources of irrigation are wells, tube-wells, ponds, lakes, rivers.
Two methods which help us to conserve water are:
(i) Sprinkler irrigation system: This irrigation system has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.

(ii) Drip irrigation system: This irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.

Question 7.
If wheat is sown in the kharif season, what would happen? Discuss.
Answer:
Wheat crop is sown from November/December to March/April. It is grown in winter and requires less water. If wheat is sown in Kharif season, its production will be decreased considerably.

Question 8.
Explain how soil gets affected by the continuous plantation of crops in a field.
Answer:
Continuous plantation of crops makes the soil poorer in certain nutrients as the crops take up nutrients from the soil. The soil becomes infertile. It does not get enough time to replenish the nutrients.

Question 9.
What are the weeds? How can we control them?
Answer:
The undesirable and unwanted plants which grow naturally along with the crop are called weeds. The growth of weeds can be controlled by adopting many ways. Tilling before sowing of crops helps in the uprooting and killing of weeds, which may then dry up and get mixed with the soil. Weeds are also controlled by using certain chemicals, called weedicides. Weedicides are sprayed in the fields to kill the weeds.

Question 10.
Arrange the following boxes in the proper order to make a flow chart of sugarcane crop production.

Answer:

Question 11.
Complete the following word puzzle with the help of clues given below.
Down
1. Providing water to the crops.
2. Keeping crop grains for a long time under proper conditions.
5. Certain plants of the same kind grown on a large scale.
Across
3. A machine used for cutting the matured crop.
4. A rabi crop that is also one of the pulses.
6. A process of separating the grain from the chaff.
Answer:

Crop Production and Management Class 8 Science NCERT Intext Activities Solved

Activity 1

Take a beaker and fill half of it with water. Put a handful of wheat seeds and stir well. Wait for some time.
Solution:
We observe that most of the seeds sink while some float on water. Damaged seeds become hollow and lighter so they float. In this way, we can separate damage seeds from the healthier ones.

Activity 2

Jake moong or gram seeds and germinate them. Select three equal sized seedlings out of these. Now take three empty glasses or similar vessels. Mark them A, B and C.To glass A add little amount of soil mixed with a little cow dung manure. In glass B put the same amount of soil mixed with a little urea. Take the same amount of soil in glass C without adding anything [Fig. 1.2(a)]. Now pour the same amount of water in each glass and plant the seedlings in them. Keep them in a safe place and water them daily. After 7 to 10 days observe their growth [Fig. 1.2(b)].

Solution:
After 7-10 days we observed that the growth was fastest in glass B followed by glass A. Glass C showed mini-mum growth.

Activity 3
Make the following Table in your notebook and complete it.

S. No.	Food	Sources
1.	Milk	Cow, Buffalo, She-goat, She-camel
2.	Meat	Goat, Hen, Pig, Duck, Sheep
3.	Egg	Hen, Duck, Goose
4.	Honey	Honey bee

NCERT Solutions for Class 8 Science Chapter 1 – 1 Mark Questions and Answers

Question 1.
Select the correct word from the following list and fill in the blanks : [NCERT]
float, water, crop, nutrients, preparation
(a) The same kind of plants grown on a large scale at a place is called …………..
(b) The first step before growing crops is …………. of the soil.
(c) Damaged seeds would …………. on top of water.
(d) For growing a crop sufficient sunlight, …………….. and …….. the soil are essential.
Solution:
(a) crop
(b) preparation
(c) float
(d) nutrients, water

Question 2.
Match items in column ‘A’ with those in column ‘B’ [NCERT]

(A)	(B)
1. Kharif crops 2. Rabi crops 3.Chemicalfertilisers 4. Organic manure	(a) Food for cattle (b) Urea and super phosphate (c) Animal excreta,cowdung, urine and plant waste. (d) Wheat, gram, pea. (e) Paddy and maize

Solution:

(A)	(B)
1. Kharif crops 2. Rabi crops 3. Chemical fertilisers 4. Organic manure	(e) Paddy and maize (d) Wheat, gram, pea (b) Urea and super phosphate (c) Animal excreta, cow dung, urine and plant waste.

Question 3.
What are crops?
Solution:
Crops are plants of the same kind grown in large quantities for food.

Question 4.
What is the basis of classification of crops in our country?
Solution:
In our country, crops are classified on the basis of the season in which they grow.

Question 5.
What are kharif crops?
Solution:
The crops which are sown in the rainy season and harvested in September/October are called kharif crops.

Question 6.
What are summer crops?
Solution:
The crops which are grown in the summer season and harvested before rainy season are called summer crops or zayed crops.

Question 7.
Name two summer season crops.
Solution:
Moong and muskmelon are summer season crops.

Question 8.
What are rabi crops?
Solution:
The crops which are grown in the winter season and harvested in March/April are called rabi crops.

Question 9.
Why paddy cannot be grown in the summer season?
Solution:
Paddy requires a lot of water, so it can only be grown during rainy season.

Question 10.
What is meant by agricultural practices?
Solution:
The activities undertaken by farmers over a period of time for cultivation of crops are known as agricultural practices.

Question 11.
Write a paragraph in your own words on preparation of soil.
Solution:
Soil is prepared by tilling i.e., loosening and turning of soil.

Question 12.
Name the tool used for tilling of soil.
Solution:
A plough is used for tilling of soil.

Question 13.
What are crumbs?
Solution:
A ploughed field may have big pieces of soil called crumbs.

Question 14.
How are crumbs broken?
Solution:
The crumbs are broken with the help of a plank.

Question 15.
Why should loose soil be levelled?
Solution:
Loose soil be levelled for sowing and irrigation.

Question 16.
How is levelling of soil done?
Solution:
Levelling of soil is done with the help of a leveller.

Question 17.
How is ploughing done these days?
Solution:
Now a days ploughing is done by tractor having a multipronged plough.

Question 18.
What is meant by sowing?
Solution:
Sowing is the process of putting seeds in the soil.

Question 19.
What is meant by good quality seeds?
Solution:
The good quality seeds means clean and healthy seeds of a good variety free from diseases.

Question 20.
What is the advantage of sowing seeds with a seed drill?
Solution:
The advantage of sowing seeds with a seed drill is that the seeds are sown at a proper depth under the soil and the distance between them is uniform.

Question 21.
What are manure and fertilisers?
Solution:
The substances which are added to the soil in the form of nutrients for the healthy growth of plants are called manure and fertilisers.

Question 22.
What is organic manure?
Solution:
Manure obtained from animal or plant waste such as cattle dung, droppings is called organic manure.

Question 23.
What is meant by crop rotation?
Solution:
Growing crops alternatively to prevent depletion of any one nutrient from soil is called crop rotation.

Question 24.
What is meant by irrigation?
Solution:
The supply of water to crops in the fields at different intervals is called irrigation.

Question 25.
What is the drip system of irrigation?
Solution:
Falling of water drop by drop at the roots of the plant is called drip irrigation.

Question 26.
Why should weeds be removed?
Solution:
Weeds compete with the crop plants for water, nutrients, space and light and thus affect the growth of the crop. So, they should be removed.

Question 27.
Name the process of removal of weeds.
Solution:
Weeding is the process of removal of weeds.

Question 28.
How is harvesting done in our country?
Solution:
Harvesting in our country is done either manually by sickle or by a machine called harvester.

Question 29.
Name the farm machine used for harvesting and threshing both.
Solution:
Combine.

Question 30.
What is meant by winnowing?
Solution:
After threshing, the grain is separated from the chaff, with the help of wind. This is known as winnowing.

Question 31.
How are food grains stored?
Solution:
Food grains are dried in the sun to remove the excess moisture and then stored.

Question 32.
Why should grains be dried before storage?
Solution:
Grains should be dried before storage to remove the excess moisture in them, so that microbes are not able to attack the grains.

Question 33.
What is meant by animal husbandry?
Solution:
The study of the care of animals is known as animal husbandry.

Question 34.
Name some animals from whom milk can be obtained.
Solution:
Milk can be obtained from cow, buffalo, she goat and she camel.

Question 35.
How is fish useful for us?
Solution:
Fish is highly nutritious and easily digestible food. Cod liver oil from fish is also a rich source of vitamin D.

Question 36.
Name some animals which are reared for their meat.
Solution:
Sheep, goat, pigs, chicken and fish are reared for their meat.

Question 37.
Why is honey so useful?
Solution:
Honey is an antiseptic and its enzymes help in digestion. It is also used for making several ayurvedic medicines.

NCERT Solutions for Class 8 Science Chapter 1 – 2 Mark Questions and Answers

Question 1.
If wheat is sown in the kharif season, what would happen? Discuss. [NCERT]
Solution:
The farmer will not get a good crop because wheat should be sown in winter season.

Question 2.
Explain how soil gets affected by the continuous plantation of crops in a field. [NCERT]
Solution:
When the crops are planted continuously in a field, the soil becomes deficient in nutrients.

Question 3.
What are weeds? How can we control them? [NCERT]
Solution:
Weeds are unwanted plants in the fields. It can be controlled by

• During tilling they are removed.
• By removing them manually.
• By using weedicides.

Question 4.

(a) Give two examples of each : [NCT 2011, NCERT]
(i) Kharif crop
(ii) Rahi crop
(b) Can you explain why most crops have a particular season in which they grow?
Solution:
(a) (i) Groundnut and cotton.
(ii) Pea and mustard.
(b) Most crops have a particular season in which they grow because different crops need different temperature, humidity and rainfall.

Question 5.
What is a seed drill?
Solution:
A seed drill is used for sowing seeds. It has a funnel shaped opening leading to long tubes attached to a plough. Seeds are put into the funnel. As the plough makes furrows in the soil, the seeds are deposited in the soil by the drill.

Question 6.
Give two reasons why seeds should be sown at correct distance.
Solution:
Seeds should be sown at a correct distance. The reasons are:

• If the seeds are too close, they will not get enough water, sunlight and nutrients.
• If the seeds are too far apart, there is wastage of field space.

Question 7.
Give two methods by which threshing can be done.
Solution:
Threshing can be done manually by making oxen or buffaloes trample over the cut crop or by a machine called thresher.

Question 8.
A farmer grow moong during the rainy season. Will he get a good crop?
Solution:
The farmer will not get a good crop of moong because moong should be grown during summer season.

Question 9.
Why does loosening of soil allow roots to breathe easily?
Solution:
When soil is loosened, the roots can breathe easily because there is more air present in the soil.

Question 10.
A farmer never leaves his field fallow. Will he get a good crop?
Solution:
If the field is never left fallow, the soil will become deficient in certain nutrients and the farmer will not get a good crop.

NCERT Solutions for Class 8 Science Chapter 1 – 3 Mark Questions and Answers

Question 1.
Define lodging. How does it happen? [MSE (Chandigarh) 2008, 2006]
Solution:
Lodging is the falling of crop plants at the grain maturation stage. It happens due to untimely rains and strong winds.

Question 2.

1. What is harvesting?
2. Mention two important uses of tilling the soil. [MSE (Chandigarh) 2008]

Solution:

1. Cutting and gathering of crops after maturation is known as harvesting.
2. Uses of tilling the soil are :
   • It improves air circulation.
   • Roots can penetrate deeper into the soil.

Question 3.

1. Define crop.
2. Name two categories of crops based on season. [KVS 2005]
3. What are pesticides? [KVS 2005]

Solution:

1. The plants of same kind grown at a place is referred to as crop.
2. Two categories of crop based on season are kharif and rabi crops.
3. Pesticides are chemicals used to protect crops from harmful organisms called pests.

Question 4.

• Earthworms are nature’s ploughmen. How?
• How is soil a resource for all living organisms? Give any two points. [DAV2006]

Solution:

• They make burrows in soil and bring lower fertile layer above the ground.
  • It is habitat for many living organisms.
  • Plants grow in soil which provide food, shelter, material for clothes.
  • Plants grows in soil and provide O2.
  • Decomposition occurs in soil.

Question 5.
What is irrigation? Describe two methods of irrigation which conserve water. [NCERT]
Solution:
Supply of water to crops at appropriate intervals is called Irrigation.
Two methods of irrigation are :

• Sprinkler system – Where water is sprinkled on the crops as if it is raining.
• Drip system – In this system, the water falls drop by drop just at the position of the roots.

Question 6.
Name three natural methods of replenishing the nutrients of the soil. Are these natural methods sufficient to maintain the fertility of the soil?
Solution:
Field fallow, crop rotation and mixed cropping are three natural methods of replenishing the nutrients of the soil. These natural methods are not enough and farmers have to resort to manures and fertilisers.

Question 7.
Give three reasons, why soil should be turned and loosened?
Solution:

• It allows the roots to penetrate deep in the soil.
• It helps the growth of earthworms and microbes in the soil.
• Various nutrients held in the dead organisms are released back in the soil.

Question 8.
What are the three steps involved in the preparation of soil?
Solution:

• Ploughing the soil to loosen it.
• Levelling the soil to ensure uniform distribution of water and nutrients.
• Manuring to provide nutrients to the soil.

Question 9.
Name the three tools used for ploughing and give the function of each.
Solution:

• Plough – It is used for tilling of soil, adding fertilisers to the crop, removing the weeds, scrapingof soil.
• Hoe – It is used for removing weeds and for loosening the soil.
• Cultivator – Used for ploughing.

Question 10.
How is organic manure obtained?
Solution:
Farmers dump plant and animal waste in pits at open places and allow it to be decomposed by microorganisms. This decomposed material is converted into organic manure.

NCERT Solutions for Class 8 Science Chapter 1 – 5 Mark Questions and Answers

Question 1.

1. What are the two ways of sowing the seeds? [DAV2007]
2. Give any three advantages of ploughing.

Solution:

1. Seeds can be sown manually or by seed drills.
2. Ploughing has following advantages :
   • Loose soil has lot of air trapped which allows roots to breathe.
   • Loose soil allows the roots to penetrate deeper.
   • Loose soil helps the growth of microbes and other organisms.

Question 2.

1. What will happen if field is not ploughed before sowing the seeds? Give any two disadvantages. [DAV2006]
2. How are perishable food items stored on a commercial scale?
3. Why should excessive supply of water be avoided?
4. What are weeds?

Solution:

1. Disadvantages:
   • Seeds cannot be sown at proper depth.
   • Water and air holding capacity of soil will be poor.
2. Perishable food items can be stored in deep freezers and cold storages.
3. Roots get damaged and the plants die because of excessive supply of water.
4. They are unwanted plants which grow along with the cultivated plants.

Question 3.
(a) Why does farmer rotate crops in the field?
(b) Differentiate between manure and fertiliser. [DAV2005]
Solution:
(a) To replenish the nutrients of the soil.
(b)

Manure	Fertiliser
(i) It is natural. (ii) It is organic. (iii) It adds humus to the soil. (iv) It is not nutrient specific. (v) It is cheap. (vi) It is  prepared in the fields.	(i) It is man-made. (ii) It is inorganic. (iii) It does not add humus. (iv) It is nutrient specific. (v) It is costly. (vi) It is prepared in factories.

or
(a) (i) While spraying pesticides, nose and mouth should be closed.
(ii) Position of the farmer should be such that the wind blows away from his face.
(iii) Immediately after spraying is done, hands should be washed, face and mouth should be cleaned with water.

Question 4.
Arrange the following boxes in proper order to make a flow chart of sugarcane crop production: [NCERT]

Solution:

Question 5.
List the steps involved in crop production in a flow chart.
Solution:

Question 6.
Explain the method used these days to sow seeds.
Solution:
These days the seed drill is used for sowing seeds with the help of a tractor. This sows the seed uniformly at a proper distance and depth. Seeds get covered by soil during sowing with a drill. This prevents damage caused by birds. Sowing by using seed drill saves time and labour.

Question 7.
What are the advantages of organic manure?
Solution:
The advantages of organic manure are :

• It enhances the water holding capacity of the soil.
• It makes the soil porous due to which exchange of gases takes place.
• The number of friendly microbes is increased.
• The organic manure improves the texture of the soil.

Question 8.
How does a plough work?
Solution:
A plough is an agricultural implement used for tilling and loosening of soil. It has a triangular iron strip called plough share. Plough shaft is the main part made of a long wooden log. The other end is hung on the bull’s necks. The plough can be with two bulls and one person.

Question 9.
Write a paragraph in your own words on each of the following : [NCERT]

1. Preparation of soil
2. Sowing
3. Weeding
4. Threshing

Solution:

1. Preparation of soil : Preparation of the soil involves loosening and turning the soil.
   This process, known as ploughing, is done by using a wooden or iron plough which is pulled either by an animal or by tractors. Loose soil is then levelled by using a wooden or iron leveller.
2. Sowing : Seeds are sown after preparation of the soil. Seeds can be sown manually or by seed drills by the process called broadcasting. Seeds should be sown at the correct depth and at correct distance.
3. Weeding : The process of removing weeds from a field is called weeding. Weeding can be done manually by pulling the weeds out by hand or by using a harrow to uproot them. Weeding can also be done by spraying special chemicals called weedicides.
4. Threshing: Threshing is the process of separating the grain from the cut crop. Threshing can be done manually by making oxen or buffaloes trample over the cut crop or by using a machine called thresher.

NCERT Solutions for Class 8 Science Chapter 1 – Multiple Choice Questions

Question 1.
Paddy can be grown in
(a) summer season
(b) autumn
(c) rainy season
(d) winter
Solution:
(c)

Question 2.
Wheat can be grown in
(a) winter
(b) rainy season
(c) spring
(d) summer
Solution:
(a)

Question 3.
Which of the following is a rabi crop?
(a) Paddy
(b) Maize
(c) Mustard
(d) Soyabean
Solution:
(c)

Question 4.
Loosening and turning of the soil is known as
(a) tilling
(b) broadcasting
(c) transplantation
(d) manuring
Solution:
(a)

Question 5.
Sowing the seeds manually is known as
(a) ploughing
(b) broadcasting
(c) tilling
(d) transplantation
Solution:
(b)

Question 6.
The method of transferring seedlings from nursery to field is known as
(a) broadcasting
(b) transplantation
(c) crop rotation
(d) harvesting
Solution:
(b)

Question 7.
Leaving the agricultural land uncultivated for one or more seasons is known as
(a) field fallow
(b) crop rotation
(c) manuring
(d) threshing
Solution:
(a)

Question 8.
Chemical substances rich in specific nutrients are called
(a) manures
(b) fertilisers
(c) pesticides
(d) weedicides
Solution:
(b)

Question 9.
Chemicals which kill weeds are known as
(a) fertilizers
(b) pesticides
(c) weedicides
(d) none of these
Solution:
(c)

Question 10.
Cutting and gathering of crops after maturation is known as
(a) harvesting
(b) threshing
(c) broadcasting
(d) tilling
Solution:
(a)

Exercise 16.1 | Class 8th Mathematics

1. Find the values of the letters in each of the following and give reasons for the steps involved.
   
   
   Solution:
   

2.

Solution:

3.

Solution:

4.

Solution:

5.

Solution:
3 × B = B
⇒ B = D
3 × A = CA
⇒ 3 × 5 = 15
Thus A = 5 and C = 1
Hence A = 5, B = 0 and C = 1

6.

Solution:
5 × B = B
⇒ B = 0 or 5
5 × A = CA
5 × 5 = 25
Only possible when B = 0
Thus A = 5 and C = 2
Hence A = 5, B = 0 and C = 2

7.

Solution:
B × 6 = B
6 × 4 = 24 → B = 4 and 2 is carried to
6 × A = BB
⇒ 6 × 7 = 42 + 2 (carried on) = 44
Thus B = 7
Hence A = 7 and B = 4

Solution:
1 + B = 0
1 + 9 = 10 → unit digit is 0 and 1 is carried to A
+ 1 +1 (carried on) = B = 9
A + 2 = 9 ⇒ A = 9 – 2 = 7
Hence A = 7 and B = 9

9.

Solution:
B + 1 = 8 ⇒ B = 8 – 1 = 7
A + B = a number with unit digit 1
A + B = 11
⇒ A + 7 = 11
⇒ A = 11 – 7 = 4 (1 Carried to)
Now 1 carried on + 2 + A = B
3 + A = 7
⇒ A = 7 – 3 = 4
Hence A = 4, B = 7

10.

Solution:
9 = A + B
9 = 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5 or 8 + 1 or 7 + 2 or 6 + 3 or 5 + 4 or 0 + 9 or 9 + 0
Now 0 is required at unit place
2 + A = 10
⇒ A = 10 – 2 = 8
B = 9 – 8 = 1
1 + 6 + 1 (carried on) = A = 8
Hence A = 8 and B = 1

Exercise 16.2 | Class 8th Mathematics

1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
   Solution:
   A number is divisible by 9 if the sum of its digits is also divisible by 9.
   Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y
   (8 + y) ÷ 9 = 1
   ⇒ 8 + y = 9
   ⇒ y = 9 – 8 = 1
   Hence, the required value of y = 1.

2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
Solution:
A number is a multiple of 9 when the sum of its digits is also divisible by 9.
Sum of the digits of 31z5 = 3 + 1 + z + 5
3 + 1 + 2 + 5 = 9k where k is an integer.
For k = 1,
3 + 1 + z + 5 = 9
⇒ z = 9 – 9 = 0
For k = 2,
3 + 1 + z + 5 = 18
⇒ z = 18 – 9 = 9
k = 3 is not possible because 3 + 1 + z + 5 = 27
⇒ z = 27 – 9 = 18 which is not a digit.
Hence the required value of z is 0 or 9

3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
Solution:
Since 24x is a multiple of 3, the sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers; 0, 3, 6, 12, 15, 18, ……..
6 + x = 3k where k is any integer.
For k = 0,
6 + x = 3 × 0
⇒ 6 + x = 0
x = -6. Not possible
For k = 1,
6 + x = 3 × 1
⇒ 6 + x = 3
⇒ x = 3 – 6 = -3. Not possible
For k = 2,
6 + x = 3 × 2
⇒ 6 + x = 6
⇒ x = 6 – 6 = 0
2 + 4 + 0 = 6 multiple of 3
For k = 3,
6 + x = 3 × 3
⇒ x = 9 – 6 = 3
2 + 4 + 3 = 9 multiple of 3
For k = 4,
6 + x = 3 × 4
⇒ 6 + x = 12
⇒ x = 12 – 6 = 6
2 + 4 + 6 = 12 which is multiple of 3
For k = 5,
6 + x = 3 × 5
⇒ x = 15 – 6 = 9
2 + 4 + 9 = 15 which is multiple of 3
For k = 6,
6 + x = 3 × 6
⇒ x = 18 – 6 = 12 not possible as x is digit
Hence the required values of x are 0, 3, 6 or 9.

4. If 31z5 is a multiple of 3, where z is a digit, what might be the value of z?
Solution:
A number is a multiple of 3 if the sum of its digits is divisible by 3.
3 + 1 + z + 5 = 3k where k is an integer
⇒ 9 + z = 3k
⇒ z = 3k – 9
Here, k = 0, 1, 2 is not possible as z is a digit of the number.
For k = 3,
z = 3 × 3 – 9 = 9 – 9 = 0
9 + 0 = 9 multiple of 3
For k = 4,
z = 3 × 4 – 9 = 12 – 9 = 3
9 + 3 = 12 multiple of 3
For k = 5,
z = 3 × 5 – 9 = 15 – 9 = 6
9 + 6 = 15 multiple of 3
For k = 6,
z = 3 × 6 – 9 = 18 – 9 = 9
9 + 9 = 18 multiple of 3
For k = 7,
z = 3 × 7 – 9 = 21 – 9 = 12 not possible as z is a digit
Hence, the required values of 2 are 0, 3, 6 and 9.

Extra Questions | Class 8th Mathematics

Exercise 15.1 | Class 8th Mathematics

1. The following graph shows the temperature of a patient in a hospital, recorded every hour.
   
   (a) What was the patient’s temperature at 1 pm?
   (b) When was the patient’s temperature 38.5°C?
   (c) The patient’s temperature was the same two times during the period given. What were the two times?
   (d) What was the temperature at 1:30 pm? How did you arrive at your answer?
   (e) During which periods did the patient’s temperature show an upward trend?
   Solution:
   (a) The patient’s temperature at 1 pm was 36.5°C
   (b) The patient’s temperature was 38.5°C at 12:00 noon.
   (c) The patients temperature was 36.5°C at 1 pm and 2 pm.
   (d) The temperature at 1:30 pm was 36.5°C. We have taken the mid value of 1 pm. and 2 pm, i.e., 1:30 p.m and proceed perpendicularly upwards to meet the horizontal line showing 36.5°C.
   (e) During 9 am to 10 am and 10 am to 11 am, the temperature showed upwards trend.

2. The following line graph shows the yearly sales figures for a manufacturing company.

(a) What were the sales in (i) 2002 (ii) 2006?
(b) What were the sales in (i) 2003 (ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?
Solution:
(a) The sales in the year 2002 was ₹ 4 crore and in the year 2006 was ₹ 8 crore.
(b) The sales in the year 2003 was ₹ 7 crore and in 2005 was ₹ 10 crore.
(c) Sales in 2002 = ₹ 4 crore
Sales in 2006 = ₹ 8 crores
Difference = ₹ (8 – 4) crore = ₹ 4 crore.
(d) The greatest difference between the sales was in the year 2005, as compared to previous year.

3. For an experiment in Botany, two different plants, plant A and plant B, were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.

(a) How high was plant A after (i) 2 weeks (ii) 3 weeks
(b) How high was plant B after (i) 2 weeks (ii) 3 weeks
(c) How much did plant A grow during the 3rd week?
(d) How much did plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did plant A grow most?
(f) During which week did plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.
Solution:
(a) The height of plant A (i) after 2 weeks was 7 cm (ii) after 3 weeks was 9 cm.
(b) The height of plant B (i) after 2 weeks was 7 cm (ii) after 3 weeks was 10 cm.
(c) Plant A grew 7 cm to 9 cm i.e, 2 cm.
(d) Plant B grew 7 cm to 10 cm i.e., 3 cm.
(e) Plant A grew most in a 2nd week i.e., 5 cm.
(f) Plant B grew the least in a first week i.e., 1 cm.
(g) Yes, the two plants grew the same height at the end of 2nd week i.e., 7 cm.

4. The following graph shows the temperature forecast and the actual temperature for each day of a week.

(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?
Solution:
(a) On Tuesday, Friday and Sunday, the forecast temperature was the same as the actual temperature i.e., 20°C, 15°C and 35°C respectively.
(b) The forecast maximum temperature during the week was 35°C
(c) The minimum actual temperature during the week was 17.5°C
(d) On Thursday, the actual temperature differed the most from the forecast temperature
i.e., 22.5°C – 15°C = 7.5°C.

5. Use the tables below to draw linear graphs.
(a) The number of days a hillside city recovered show in different years.

Year	2003	2004	2005	2006
Days	8	10	5	12

(b) Population (in thousands) of men and women in a village is different years.

Year	2003	2004	2005	2006	2007
Number of men	12	12.5	13	13.2	13.5
Number of women	11.3	11.9	13	13.6	12.8

Solution:

6. A courier-person cycle from a town to a neighbouring suburban area to deliver a parcel to the merchant. His distance from the town at different times is shown by the following graph.

(a) What is the scale taken for the time axis?
(b) How much time did the person take for travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain?
(e) During which period did he ride fastest?
Solution:
(a) Scale taken on time-axis i.e., the x-axis is 4 units = 1 hour
(b) Total time taken by the person for the total journey is 3 hours 30 minutes.

(c) The place of the merchant is at a distant of 22 km from the town.

7. Can there be a time-temperature graph as follows? Justify your answer.

Solution:
(i) It represents a time-temperature graph where temperature increases as the time increase.
(ii) It shows a time-temperature graph where temperature decreases as the time increases.
(iii) It does not represent a time-temperature graph. Here the temperature is increasing at a constant time which is not possible.
(iv) It represents a time-temperature graph where the temperature remains constant when the time is increasing.

Exercise 15.2 | Class 8th Mathematics

1. Plot the following points on a graph sheet. Verify if they lie on a line.
   (a) A (4, 0), B(4, 2), C(4, 6), D(4, 2.5)
   (b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
   (c) K(2, 3), L(6, 3), M(5, 5), N (2, 5)
   Solution:
   
   Yes, all the coordinate points lie on a line.
   
   Yes, all the coordinate points lie on a line.
   
   No, the coordinate points do not lie on a line.

2. Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.

Solution:
CD is the required line passing through the points A(2, 3), B (3, 2) which meets x-axis at C(5, 0) and y-axis at D(0, 5).

3. Write the coordinates of the vertices of each of these adjoining figures.

Solution:
Required coordinates are as follows:
Coordinates of rectangle OABC are:
O(0, 0), A(2, 0), B(2, 3), C(0, 3)
Coordinates of parallelogram PQRS are:
P(4, 3), Q(6, 1), R(6, 5), S(4, 7)
Coordinates of triangle KLM are:
K(10, 5), L(7, 7), M (10, 8).

4. State whether True or False. Correct that is false.
(i) A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y-coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
Solution:
(i) True
(ii) False, the point will lie on the x-axis with coordinates (5, 0).
(iii) True

Exercise 15.3 | Class 8th Mathematics

1. Draw the graphs for the following tables of values, with suitable scales on the axes.
   (a) Cost of apples
   
   (b) Distance travelled by car.
   
   (i) How much distance did the car cover during the period 7:30 am to 8 am?
   (ii) What was the time when the car had covered a distance of 100 km since its start?
   (c) Interest on deposits for a year.
   
   (i) Does the graph pass through the origin?
   (ii) Use the graph to find the interest on ? 2500 for a year.
   (iii) To get an interest of t 280 per year, how much money should be deposited?
   Solution:
   
   (b) (i) The distance covered by the car during the period 7:30 am to 8 am is (120 km – 100 km) = 20 km.
   (ii) At 7:30 am, the car had covered a distance of 100 km.
   
   (c) (i) Yes, the graph passes through the origin,
   (ii The interest on ₹ 2500 is ₹ 200 for 1 year.
   (iii) ₹ 3500 should be invested to earn the interest of ₹ 280.
   

2. Draw the graph for the following:

Solution:

Extra Questions | Class 8th Mathematics

Exercise 14.1 | Class 8th Mathematics

1. Find the common factors of the given terms.
   (i) 12x, 36
   (ii) 2y, 22xy
   (iii) 14pq, 28p²q²
   (iv) 2x, 3x², 4
   (v) 6abc, 24ab², 12a²b
   (vi) 16x³, -4x², 32x
   (vii) 10pq, 20qr, 30rp
   (viii) 3x²y³, 10x³y², 6x²y²z
   Solution:
   (i) 12x, 36
   (2 × 2 × 3 × x) and (2 × 2 × 3 × 3)
   Common factors are 2 × 2 × 3 = 12
   Hence, the common factor = 12

(ii) 2y, 22xy
= (2 × y) and (2 × 11 × x × y)
Common factors are 2 × y = 2y
Hence, the common factor = 2y

(iii) 14pq, 28p²q²
= (2 × 7 × p × q) and (2 × 2 × 7 × p × p × q × q)
Common factors are 2 × 7 × p × q = 14pq
Hence, the common factor = 14pq

(iv) 2x, 3x², 4
= (2 × x), (3 × x × x) and (2 × 2)
Common factor is 1
Hence, the common factor = 1 [∵ 1 is a factor of every number]

(v) 6abc, 24ab², 12a²b
= (2 × 3 × a × b × c), (2 × 2 × 2 × 3 × a × b × b) and (2 × 2 × 3 × a × a × b)
Common factors are 2 × 3 × a × b = 6ab
Hence, the common factor = 6ab

(vi) 16x³, -4x², 32x
= (2 × 2 × 2 × 2 × x × x × x), -(2 × 2 × x × x), (2 × 2 × 2 × 2 × 2 × x)
Common factors are 2 × 2 × x = 4x
Hence, the common factor = 4x

(vii) 10pq, 20qr, 30rp
= (2 × 5 × p × q), (2 × 2 × 5 × q × r), (2 × 3 × 5 × r × p)
Common factors are 2 × 5 = 10
Hence, the common factor = 10

(viii) 3x²y², 10x³y², 6x²y²z
= (3 × x × x × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)
Common factors are x × x × y × y = x²y²
Hence, the common factor = x²y².

2. Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a² + 14a
(iv) -16z + 20z³
(v) 20l²m + 30alm
(vi) 5x²y – 15xy²
(vii) 10a² – 15b² + 20c²
(viii) -4a² + 4ab – 4ca
(ix) x²yz + xy²z + xyz²
(x) ax²y + bxy² + cxyz
Solution:
(i) 7x – 42 = 7(x – 6)
(ii) 6p – 12q = 6(p – 2q)
(iii) 7a² + 14a = 7a(a + 2)
(iv) -16z + 20z³ = 4z(-4 + 5z²)
(v) 20l²m + 30alm = 10lm(2l + 3a)
(vi) 5x²y – 15xy² = 5xy(x – 3y)
(vii) 10a² – 15b² + 20c² = 5(2a² – 3b² + 4c²)
(viii) -4a² + 4ab – 4ca = 4a(-a + b – c)
(ix) x²yz + xy²z + xyz² = xyz(x + y + z)
(x) ax²y + bxy² + cxyz = xy(ax + by + cz)

3. Factorise:
(i) x² + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Solution:
(i) x² + xy + 8x + 8y
Grouping the terms, we have
x² + xy + 8x + 8y
= x(x + y) + 8(x + y)
= (x + y)(x + 8)
Hence, the required factors = (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2
Grouping the terms, we have
(15xy – 6x) + (5y – 2)
= 3x(5y – 2) + (5y – 2)
= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by
Grouping the terms, we have
= (ax – ay) + (bx – by)
= a(x – y) + b(x – y)
= (x – y)(a + b)
Hence, the required factors = (x – y)(a + b)

(iv) 15pq + 15 + 9q + 25p
Grouping the terms, we have
= (15pq + 25p) + (9q + 15)
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5) (5p + 3)
Hence, the required factors = (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz
Grouping the terms, we have
= (-xyz + 7xy) + (z – 7)
= -xy(z – 7) + 1 (z – 7)
= (-xy + 1) (z – 1)
Hence the required factor = -(1 – xy) (z – 7)

Exercise 14.2 | Class 8th Mathematics

1. Factorise the following expressions.
   (i) a² + 8a +16
   (ii) p² – 10p + 25
   (iii) 25m² + 30m + 9
   (iv) 49y² + 84yz + 36z²
   (v) 4x² – 8x + 4
   (vi) 121b² – 88bc + 16c²
   (vii) (l + m)² – 4lm. (Hint: Expand (l + m)² first)
   (viii) a⁴ + 2a²b² + b⁴
   Solution:
   (i) a² + 8a + 16
   Here, 4 + 4 = 8 and 4 × 4 = 16
   a² + 8a +16
   = a² + 4a + 4a + 4 × 4
   = (a² + 4a) + (4a + 16)
   = a(a + 4) + 4(a + 4)
   = (a + 4) (a + 4)
   = (a + 4)²

(ii) p² – 10p + 25
Here, 5 + 5 = 10 and 5 × 5 = 25
p² – 10p + 25
= p² – 5p – 5p + 5 × 5
= (p² – 5p) + (-5p + 25)
= p(p – 5) – 5(p – 5)
= (p – 5) (p – 5)
= (p – 5)²

(iii) 25m² + 30m + 9
Here, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225
25m² + 30m + 9
= 25m² + 15m + 15m + 9
= (25m² + 15m) + (15m + 9)
= 5m(5m + 3) + 3(5m + 3)
= (5m + 3) (5m + 3)
= (5m + 3)²

(iv) 49y² + 84yz + 36z²
Here, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764
49y² + 84yz + 36z²
= 49y² + 42yz + 42yz + 36z²
= 7y(7y + 6z) +6z(7y + 6z)
= (7y + 6z) (7y + 6z)
= (7y + 6z)²

(v) 4x² – 8x + 4
= 4(x² – 2x + 1) [Taking 4 common]
= 4(x² – x – x + 1)
= 4[x(x – 1) -1(x – 1)]
= 4(x – 1)(x – 1)
= 4(x – 1)²

(vi) 121b² – 88bc + 16c²
Here, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936
121b² – 88bc + 16c²
= 121b² – 44bc – 44bc + 16c²
= 11b(11b – 4c) – 4c(11b – 4c)
= (11b – 4c) (11b – 4c)
= (11b – 4c)²

(vii) (l + m)² – 4lm
Expanding (l + m)², we get
l² + 2lm + m² – 4lm
= l² – 2lm + m²
= l² – Im – lm + m²
= l(l – m) – m(l – m)
= (l – m) (l – m)
= (l – m)²

(viii) a⁴ + 2a²b² + b⁴
= a⁴ + a²b² + a²b² + b⁴
= a²(a² + b²) + b²(a² + b²)
= (a² + b²)(a² + b²)
= (a² + b²)²

2. Factorise.
(i) 4p² – 9q²
(ii) 63a² – 112b²
(iii) 49x² – 36
(iv) 16x⁵ – 144x³
(v) (l + m)² – (l – m)²
(vi) 9x²y² – 16
(vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²
Solution:
(i) 4p² – 9q²
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)
[∵ a² – b² = (a + b)(a – b)]

(ii) 63a² – 112b²
= 7(9a² – 16b²)
= 7 [(3a)² – (4b)²]
= 7(3a – 4b)(3a + 4b)
[∵ a² – b² = (a + b)(a – b)]

(iii) 49x² – 36 = (7x)² – (6)²
= (7x – 6) (7x + 6)
[∵ a² – b² = (a + b)(a – b)]

(iv) 16x⁵ – 144x³ = 16x³ (x² – 9)
= 16x³ [(x)² – (3)²]
= 16x³(x – 3)(x + 3)
[∵ a² – b² = (a + b)(a – b)]

(v) (l + m)² – (l – m)²
= (l + m) – (l – m)] [(l + m) + (l – m)]
[∵ a² – b² = (a + b)(a – b)]
= (l + m – l + m)(l + m + l – m)
= (2m) (2l)
= 4ml

(vi) 9x²y² – 16 = (3xy)² – (4)²
= (3xy – 4)(3xy + 4)
[∵ a² – b² = (a + b)(a – b)]

(vii) (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y – z) (x – y + z)
[∵ a² – b² = (a + b)(a – b)]

(viii) 25a² – 4b² + 28bc – 49c²
= 25a² – (4b² – 28bc + 49c²)
= (5a)² – (2b – 7c)²
= [5a – (2b – 7c)] [5a + (2b – 7c)]
= (5a – 2b + 7c)(5a + 2b – 7c)

3. Factorise the expressions.
(i) ax² + bx
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Solution:
(i) ax² + bx = x(ax + 5)

(ii) 7p² + 21q² = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an²
= m² (a + b) + n²(a + b)
= (a + b)(m² + n²)

(v) (lm + l) + m + 1
= l(m + 1) + (m + 1)
= (m + 1) (l + 1)

(vi) y(y + z) + 9(y + z) = (y + z)(y + 9)

(vii) 5y² – 20y – 8z + 2yz
= 5y² – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1(5b + 2)
= (5b + 2)(2a + 1)

(ix) 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y(3x – 2) – 3(3x – 2)
= (3x – 2) (2y – 3)

4. Factorise.
(i) a⁴ – b⁴
(ii) p⁴ – 81
(iii) x⁴ – (y + z)⁴
(iv) x⁴ – (x – z)⁴
(v) a⁴ – 2a²b² + b⁴
Solution:
(i) a⁴ – b⁴ – (a²)² – (b²)²
[∵ a² – b² = (a – b)(a + b)]
= (a² – b²) (a² + b²)
= (a – b) (a + b) (a² + b²)

(ii) p⁴ – 81 = (p²)² – (9)²
= (p² – 9) (p² + 9)
[∵ a² – b² = (a – b)(a + b)]
= (p – 3)(p + 3) (p² + 9)

(iii) x⁴ – (y + z)⁴ = (x²)² – [(y + z)²]²
[∵ a² – b² = (a – b)(a + b)]
= [x² – (y + z)²] [x² + (y + z)²]
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²]
= (x – y – z) (x + y + z) [x² + (y + z)²]

(iv) x⁴ – (x – z)⁴ = (x²)² – [(y – z)²]²
= [x² – (y – z)²] [x² + (y – z)²]
= (x – y + z) (x + y – z) (x² + (y – z)²]

(v) a⁴ – 2a²b² + b⁴
= a⁴ – a²b² – a²b² + b⁴
= a²(a² – b²) – b²(a² – b²)
= (a² – b²)(a² – b²)
= (a² – b²)²
= [(a – b) (a + b)]²
= (a – b)² (a + b)²

5. Factorise the following expressions.
(i) p² + 6p + 8
(ii) q² – 10q + 21
(iii) p² + 6p – 16
Solution:
(i) p² + 6p + 8
Here, 2 + 4 = 6 and 2 × 4 = 8
p² + 6p + 8
= p² + 2p + 4p + 8
= p (p + 2) + 4(p + 2)
= (p + 2) (p + 4)

(ii) q² – 10q + 21
Here, 3 + 7 = 10 and 3 × 7 = 21
q² – 10q + 21
= q² – 3q – 7q + 21
= q(q – 3) – 7(q – 3)
= (q – 3) (q – 7)

(iii) p² + 6p – 16
Here, 8 – 2 = 6 and 8 × 2 = 16
p² + 6p – 16
= p² + 8p – 2p – 16
= p(p + 8) – 2(p + 8)
= (p + 8) (p – 2)

Exercise 14.3 | Class 8th mathematics

1. Carry out the following divisions.
   (i) 28x⁴ ÷ 56x
   (ii) -36y³ ÷ 9y²
   (iii) 66pq²r³ ÷ 11qr²
   (iv) 34x³y³z³ ÷ 51xy²z³
   (v) 12a⁸b⁸ ÷ (-6a⁶b⁴)
   Solution:
   

2. Divide the following polynomial by the given monomial.
(i) (5x² – 6x) ÷ 3x
(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
(iii) 8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
(iv) (x³ + 2x² + 3x) ÷ 2x
(v) (p³q⁶ – p⁶q³) ÷ p³q³
Solution:

3. Work out the following divisions.
(i) (10x – 25) ÷ 5
(ii) (10x – 25) ÷ (2x – 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7)
(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)
(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4)(b – 6)
Solution:

4. Divide as directed.
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
(ii) 26xy (x + 5)(y – 4) ÷ 13x(y – 4)
(iii) 52pqr(p + q) (q + r) (r + p) ÷ 104pq(q + r)(r + p)
(iv) 20(y + 4)(y² + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
Solution:

5. Factorise the expressions and divide them as directed.
(i) (y² + 7y + 10) ÷ (y + 5)
(ii) (m² – 14m – 32) ÷ (m + 2)
(iii) (5p² – 25p + 20) ÷ (p – 1)
(iv) 4yz(z² + 6z – 16) ÷ 2y(z + 8)
(v) 5pq(p² – q²) ÷ 2p(p + q)
(vi) 12xy(9x² – 16y²) ÷ 4xy(3x + 4y)
(vii) 39y³(50y² – 98) ÷ 26y²(5y + 7)
Solution:

Extra Questions | Class 8th Mathematics

Exercise 13.1 | Class 8th Mathematics

1. Following are the car parking charges near a railway station up to.
   4 hours – ₹ 60
   8 hours – ₹ 100
   12 hours – ₹ 140
   24 hours – ₹ 180
   Check if the parking charges are in direct proportions to the parking time.
   
   Solution:
   We have the ratio of time period and the parking charge.
   
   Hence the given quantities are not directly proportional.

2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Solution:
Let the number to be filled in the blanks be a, b, c and d respectively.

3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Solution:
Let the required red pigment be x part.

Hence, the required amount of red pigment = 24 parts.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the required number of bottles be x.

Hence the required number of bottles = 700.

5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Solution:
Let the actual length be x cm.

6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Solution:
Let the required length of the model ship be x m.

7. Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution:
Let x be the number of sugar crystals needed.

8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road of 72 km. What would be her distance covered in the map?
Solution:
Let the required distance be x km.

Hence the distance covered in the map = 4 cm.

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high,
(ii) the height of a pole which casts a shadow 5 m long.
Solution:
(i) Let the required length of shadow be x m.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Let the required distance be x km.

Hence the required distance = 168 km.

Exercise 13.2 | Class 8th Mathematics

1. Which of the following are in inverse proportion?
   (i) The number of workers on a job and the time to complete the job.
   (ii) The time taken for a journey and the distance travelled in a uniform speed.
   (iii) Area of cultivated land and the crop harvested.
   (iv) The time taken for a fixed journey and the speed of the vehicle.
   (v) The population of a country and the area of land per person.
   Solution:
   (i) As the number of workers increase, the job will take less time to complete. Hence, they are inversely proportional.
   (ii) For more time, more distance to travel. Hence, they are not inversely proportional.
   (iii) More area of land cultivated, more crop to harvest. Hence, they are not inversely proportional.
   (iv) If speed is increased, it will take less time to complete the fixed journey. Hence, they are inversely proportional.
   (v) If the population of a country increases, then the area of land per person will be decreased. Hence, they are inversely proportional.

2. In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners	1	2	4	5	8	10	20
The prize for each winner (in ₹)	1,00,000	50,000	–	–	–	–	–

Solution:
Let, the blank spaces be denoted by a, b, c, d and e.
So, we observe that 1 × 100,000 = 2 × 50,000
⇒ 1,00,000 = 1,00,000
Hence they are inversely proportional.
2 × 50,000 = 4 × a

Number of winners	1	2	4	5	8	10	20
The prize for each winner (in ₹)	1,00,000	50,000	25,000	20,000	12,500	10,000	5,000

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.

Number of spokes	4	6	8	10	12
The angle between a pair of consecutive spokes	90°	60°	–	–	–

(i) Are the number of spokes and the angle formed between the pairs of consecutive spokes in inverse proportion.
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
From the above table, we observe that
4 × 90° = 6 × 60°
360° = 360°
Thus the two quantities are inversely proportional.
Let the blank spaces be denoted by a, b, and c.
4 × 90° = 8 × a

Hence, the required table is

Number of spokes	4	6	8	10	12
The angle between a pair of consecutive spokes	90°	60°	45°	36°	30°

(i) Yes, they are in inverse proportion
(ii) If the number of spokes is 15, then
4 × 90° = 15 × x
x = 4×9015 = 24°
(iii) If the angle between two consecutive spokes is 40°, then
4 × 90° = y × 40°
y = 4×9040 = 9 spokes.
Thus the required number of spokes = 9.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is decreased by 4?
Solution:

Number of children	Number of Sweets
24	5
(24 – 4) or 20	a

We observe that on increasing the number of children, number of sweets got by each will be less. So, they are in inverse proportion.
x₁y₁ = x₂y₂
where x₁ = 24, y₁ = 5, x₂ = 20
and y₂ = a(let)
24 × 5 = 20 × a
a = 6
Hence, the required number of sweets = 6.

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:
If the number of animals increases, then it will take fewer days to last.
Then the two quantities are in inverse proportions.

Number of animals	Number of days
20	6
(20 + 10) or 30	P

Let the required number of days be p.
x₁y₁ = x₂y₂
where x₁ = 20, y₁ = 6, x₂ = 3
and y₂ = p (let)
20 × 6 = 30 × p
p = 4
Hence the required number of days = 4.

6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:
If the number of persons is increased, it will take less number of days to complete the job.
Thus, the two quantities are in inverse proportion.

Number of persons	Number of days
3	4
4	k

Let the required number of days be k.
x₁y₁ = x₂y₂
3 × 4 = 4 × k
k = 3 days.
Hence, the required number of days = 3.

7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:
If the number of bottles is increased then the required number of boxes will be decreased. Thus the two quantities are in inverse proportion.

Number of boxes	Number of bottles per box
25	12
x	20

Let the required number of boxes be x.
x₁y₁ = x₂y₂
25 × 12 = x × 20
x = 15
Hence, the required number of boxes = 15.

8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:
If the number of machines is increased then less number of days would be required to produced the same number of articles.
Thus, the two quantities are in inverse proportion.

Number of machines	Number of days
42	63
x	54

Let the required number of machines be x.
x₁y₁ = x₂y₂
42 × 63 = x × 54
x = 49
Hence, the required number of machines is 49.

9. A car takes 2 hours to reach a destination by traveling at a speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:
On increasing the speed, it will take less time to travel a distance.
Thus the two quantities are in inverse proportions.

Speed in km/h	Time in hour
60	2
80	x

Let the required times be x hours.
x₁y₁ = x₂y₂
60 × 2 = 80 × x
x = 32 hours = 112 hrs.
Hence, the required time = 112 hours.

10. Two persons could fit new windows in a house in 3 days.
(i) One of the people fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution:
On increasing the number of persons, less time will be required to complete a job.
Thus, the quantities are in inverse proportion.

Number of persons	Number of days
2	3
(i) 1(2 – 1)	x
(ii) y	1

(i) Let the required number of days be x.
x₁y₁ = x₂y₂
2 × 3 = 1 × x
x = 6
Hence, the required number of days = 6
(ii) Let the required number of persons be y.
x₁y₁ = x₂y₂
2 × 3 = y × 1
y = 6
Hence, the required number of persons = 6.

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:
On increasing the duration of periods, the number of periods will be reduced.
Thus, the two quantities are in inverse proportion.

Number of periods	Duration of periods in minutes
8	45
9	x

Let the required duration of each period be x.
x₁y₁ = x₂y₂
8 × 45 = 9 × x
x = 40 minutes
Hence, the required duration of period = 40 minutes.

Extra Questions | Class 8th Mathematics

Direct and Inverse Proportions Class 8 Extra Questions Very Short Answer Type

1. A train is moving at a uniform speed of 100 km/h. How far will it travel in 20 minutes?
Solution:
Let the distance travelled by train in 20 minutes be x km.

Since the speed is uniform, the distance travelled will be directly proportional to time.

Hence, the required distance is 3313 km.

2. Complete the table if x and y vary directly.

Solution:
Let the blank spaces be filled with a, b and c.


Hence, the required values are a = 7, b = 15 and c = 7.5.

3. If the cost of 20 books is ₹ 180, how much will 15 books cost?
Solution:
Let the required cost be ₹ x.
Here, the two quantities vary directly.

4. If x₁ = 5, y₁ = 7.5, x₂ = 7.5 then find y₂ if x and y vary directly.
Solution:
Since x and y vary directly.

Hence, the required value is 11.25.

5. If 3 kg of sugar contains 9 × 10⁸ crystals. How many sugar crystals are there in 4 kg of sugar?
Solution:
Let the required number of crystals be x.

Hence, the required number of crystals = 1.2 × 10⁹.

6. If 15 men can do a work in 12 days, how many men will do the same work in 6 days?
Solution:
Let the required number of men be x.
Less days → more men.
Thus the two quantities are inversely proportional to each other.

Hence, the required number of days = 30.

7. A train travels 112 km in 1 hour 30 minutes with a certain speed. How many kilometres it will travel in 4 hours 45 minutes with the same speed?
Solution:
Let the required distance be x km.
More distance → more time
Thus, the two quantities are directly proportional.


Hence, the required distance = 354.6 km.

8. The scale of a map is given as 1 : 50,000. Two villages are 5 cm apart on the map. Find the actual distance between them.
Solution:
Let the map distance be x cm and actual distance be y.
1 : 50,000 = x : y

Hence, the required distance = 250 km.

9. 8 pipes are required to fill a tank in 1 hr 20 min. How long will it take if only 6 pipes of the same type are used?
Solution:
Let the required time be ‘t’ hours.

10. 15 men can build a wall in 42 hours, how many workers will be required for the same work in 30 hours?
Solution:
Let the required number of workers be x.
The number of workers, faster will they do the work.
So, the two quantities are inversely proportional.

x₁y₁ = x₂y₂
⇒ 42 × 15 = 30 × x
⇒ x = 21
Hence, the required number of men = 21

Direct and Inverse Proportions Class 8 Extra Questions Short Answer Type

11. The volume of a gas V varies inversely as the pressure P for a given mass of the gas. Fill in the blank spaces in the following table:

Solution:
Since volume and pressure are inversely proportional.
PV = K

12. The cost of 5 metres of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.
Solution:
Let the length of the cloth be x m and its cost be ₹ y. We have the following table.

13. Six pumps working together empty a tank in 28 minutes. How long will it take to empty the tank if 4 such pumps are working together?
Solution:
Let the required time be t minutes.

Less pump → More time
Since there is an inverse variation.
x₁y₁ = x₂y₂
6 × 28 = 4 × x
x = 42
Hence, the required time = 42 minutes.

14. Mohit deposited a sum of ₹ 12000 in a Bank at a certain rate of interest for 2 years and earns an interest of ₹ 900. How much interest would be earned for a deposit of ₹ 15000 for the same period and at the same rate of interest?
Solution:
Let the required amount of interest be ₹ x.


Hence, the required amount of interest = ₹ 1125.

15. A garrison of 120 men has provisions for 30 days. At the end of 5 days, 5 more men joined them. How many days can they sustain on the remaining provision?
Solution:
Let the number of days be x.

[∵ Remaining days = 30 – 5 = 25]
[Total men = 120 + 5 = 125]
Since there is an inverse variation.
x₁y₁ = x₂y₂
120 × 25 = 125 × x
x = 24
Hence, the required number of days be 24.

16. In a scout camp, there is food provision for 300 cadets for 42 days. If 50 more persons join the camp, for how many days will the provision last? (NCERT Exemplar)
Solution:
More the persons, the sooner would be the provision exhausted. So, this is a case of inverse proportion.
Let the required number of days be x.
Hence, 300 × 42 = (300 + 50) × x
300 × 42 = 350 × x
x = 36

17. If two cardboard boxes occupy 500 cubic centimetres space, then how much space is required to keep 200 such boxes? (NCERT Exemplar)
Solution:
As the number of boxes increases, the space required to keep them also increases.
So, this is a case of direct proportion.

Thus, the required space is 50,000 cubic centimetres.

18. Under the condition that the temperature remains constant, the volume of gas is inversely proportional to its pressure. If the volume of gas is 630 cubic centimetres at a pressure of 360 mm of mercury, then what will be the pressure of the gas if its volume is 720 cubic centimetres at the same temperature? (NCERT Exemplar)
Solution:
Given that, at constant temperature pressure and volume of a gas are inversely proportional.
Let the required pressure be x.

Therefore, the required pressure is 315 mm of mercury.

Exercise 12.1 | Class 8th Mathematics

1. Evaluate:
   (i) 3^-2
   (ii) (-4)^-2
   (iii) (12)^-5
   Solution:
   

2. Simplify and express the result in power notation with a positive exponent.

Solution:

3. Find the value of

Solution:

4.

Solution:

5. Find the value of m for which 5^m ÷ 5^-3 = 5⁵.
Solution:
5^m ÷ 5^-3 = 5⁵
⇒ 5^m-(-3) = 5⁵ [∵ a^m ÷ aⁿ = a^m-n]
⇒ 5^m+3 = 5⁵
Comparing the powers of equal bases, we have
m + 3 = 5
m = 5 – 3 = 2, i.e., m = 2

6. Evaluate:

Solution:

7. Simplify:

Solution:

Exercise 12.2 | Class 8th Mathematics

1. Express the following numbers in standard form:
   (i) 0.0000000000085
   (ii) 0.00000000000942
   (iii) 6020000000000000
   (iv) 0.00000000837
   (v) 31860000000
   Solution:
   

(v) 31860000000
= 3.186 × 10000000000
= 3.186 × 10¹⁰
Hence, the required standard from = 3.186 × 10¹⁰

2. Express the following numbers in usual form.
(i) 3.02 × 10^-6
(ii) 4.5 × 10⁴
(iii) 3 × 10^-8
(iv) 1.0001 × 10⁹
(v) 5.8 × 10¹²
(vi) 3.61492 × 10⁶
Solution:

3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 11000000 m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm.
Solution:

4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Solution:
Thickness of books = 5 × 20 = 100 mm
Thickness of 5 paper sheets = 5 × 0.016 mm = 0.080 mm.
Total thickness of the stack = 100 mm + 0.080 mm = 100.080 mm = 1.0008 × 10² mm

Extra Questions | Class 8th Mathametics

1. Find the multiplicative inverse of:
   (i) 3^-3
   (ii) 10^-10
   Solution:
   

2. Expand the following using exponents.
(i) 0.0523
(ii) 32.005
Solution:

3. Simplify and write in exponential form.

Solution:

4. Simplify the following and write in exponential form.

Solution:

5. Express 8^-4 as a power with the base 2.
Solution:
We have 8 = 2 × 2 × 2 = 2³
8^-4 = (2³)^-4 = 2^3×(-4) = 2^-12

6. Simplify the following and write in exponential form.
(i) (3⁶ ÷ 3⁸)⁴ × 3^-4
(ii) 127 × 3^-3
Solution:

7. Find the value of k if (-2)^k+1 × (-2)³ = (-2)⁷
Solution:
(-2)^k+1 × (-2)³ = (-2)⁷
⇒ (-2)^k+1+3 = (-2)⁷
⇒ (-2)^k+4 = (-2)⁷
⇒ k + 4 = 7
⇒ k = 3
Hence, k = 3.

8. Simplify the following:

Solution:

9. Find the value of [(−34)−2]2
Solution:

10. Write the following in standard form
(i) 0 0035
(ii) 365.05
Solution:

Exponents and Powers Class 8 Extra Questions Short Answer Type

11. Find the value of P if

Solution:

12.

Solution:

13. Find the value of x if

Solution:

⇒ 3 + x = 18 [Equating the powers of same base]
x = 18 – 3 = 15

14. Solve the following: (81)^-4 ÷ (729)^2-x = 9^4x
Solution:

15.

Solution:

16.

Solution:

17. Find x so that (-5)^x+1 × (-5)⁵ = (-5)⁷ (NCERT Exemplar)
Solution:
(-5)^x+1 × (-5)⁵ = (-5)⁷
(-5)^x+1+5 = (-5)⁷ {a^m × aⁿ = a^m+n}
(-5)^x+6 = (-5)⁷
On both sides, powers have the same base, so their exponents must be equal.
Therefore, x + 6 = 7
x = 7 – 6 = 1
x = 1.

Exercise 11.1 | Class 8th Mathematics

1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
   Solution:
   Perimeter of figure (a) = 4 × side = 4 × 60 = 240 m
   
   Perimeter of figure (b) = 2 [l + b]
   Perimeter of figure (b) = Perimeter of figure (a)
   2[l + b] = 240
   ⇒ 2 [80 + b] = 240
   ⇒ 80 + b = 120
   ⇒ b = 120 – 80 = 40 m
   Area of figure (a) = (side)² = 60 × 60 = 3600 m²
   Area of figure (b) = l × b = 80 × 40 = 3200 m²
   So, area of figure (a) is longer than the area of figure (b)

2. Mrs Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Solution:
Area of the plot = side × side = 25 m × 25 m = 625 m²
Area of the house = l × b = 20 m × 15 m = 300 m²
Area of the garden to be developed = Area of the plot – Area of the house = 625 m² – 300 m² = 325 m²
Cost of developing the garden = ₹ 325 × 55 = ₹ 17875

3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres]

Solution:
Length of the rectangle = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Area of the rectangle = l × b = 13 × 7 = 91 m2
Area of two circular ends = 2(12 πr²)
= πr²
= 227 × 72 × 72
= 772 m²
= 38.5 m²
Total area = Area of the rectangle + Area of two ends = 91 m² + 38.5 m² = 129.5 m²
Total perimeter = Perimeter of the rectangle + Perimeter of two ends
= 2 (l + b) + 2 × (πr) – 2(2r)
= 2 (13 + 7) + 2(227 × 72) – 4 × 72
= 2 × 20 + 22 – 14
= 40 + 22 – 14
= 48 m

4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners).
Solution:
Area of the floor = 1080 m² = 1080 × 10000 cm² = 10800000 cm² [∵ 1 m² = 10000 cm²]
Area of 1 tile = 1 × base × height = 1 × 24 × 10 = 240 cm²
Number of tiles required

= 45000 tiles

5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression C = 2πr, where r is the radius of the circle.

Solution:

Exercise 11.2 | Class 8th Mathematics

1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
   
   Solution:
   Area of the trapezium = 12 × (a + b) × h
   = 12 × (1.2 + 1) × 0.8
   = 12 × 2.2 × 0.8
   = 0.88 m²
   Hence, the required area = 0.88 m²

2. The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel sides.
Solution:
Given: Area of trapezium = 34 cm²
Length of one of the parallel sides a = 10 cm
height h = 4 cm
Area of the trapezium = 12 × (a + b) × h
34 = 12 × (10 + b) × 4
⇒ 34 = (10 + b) × 2
⇒ 17 = 10 + b
⇒ b = 17 – 10 = 7 cm
Hence, the required length = 7 cm.

3. Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Solution:
Given:
AB + BC + CD + DA = 120 m .
BC = 48 m, CD = 17 m, AD = 40 m
AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m
Area of the trapezium ABCD = 12 × (BC + AD) × AB
= 12 × (48 + 40) × 15
= 12 × 88 × 15
= 44 × 15 = 660 m².
Hence, the required area = 660 m²

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
Area of the field = area of ∆ABD + area of ∆BCD

= 12 × b × h + 12 × b × h
= 12 × 24 × 13 + 12 × 24 × 8
= 12 × 13 + 12 × 8
= 12 × (13 + 8)
= 12 × 21
= 252 m²
Hence, the required area of the field = 252 m².

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Here, d1 = 7.5 cm, d2 = 12 cm
Area of the rhombus = 12 × d1 × d2
= 12 × 7.5 × 12
= 7.5 × 6
= 45 cm²
Hence, area of the rhombus = 45 cm².

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Given: Side = 5 cm
Altitude = 4.8 cm
Length of one diagonal = 8 cm
Area of the rhombus = Side × Altitude = 5 × 4.8 = 24 cm²
Area of the rhombus = 12 × d₁ × d₂
24 = 12 × d₁ × d₂
24 = 4d₂
d₂ = 6 cm
Hence, the length of other diagonal = 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.
Solution:
Given: Number of tiles = 3000
Length of the two diagonals of a tile = 45 cm and 30 cm
Area of one tile = 12 × d₁ × d₂
= 12 × 45 × 30
= 45 × 15
= 675 cm²
Area covered by 3000 tiles = 3000 × 675 cm² = 2025000 cm² = 202.5 m²
Cost of polishing the floor = 202.5 × 4 = ₹ 810
Hence, the required cost = ₹ 810.

8. Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:
Let the side of the trapezium (roadside) be x cm.
The opposite parallel side = 2x m
h = 100 m
Area = 10500 m²
Area of trapezium = 12 (a + b) × h
10500 = 12 (2x + x) × 100
2 × 10500 = 3x × 100
21000 = 300x
x = 70 m
So, AB = 2x = 2 × 70 = 140 m
Hence, the required length = 140 m.

9. The top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Solution:
Area of the octagonal surface = area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF
Area of trapezium ABCH = Area of trapezium GDEF
= 12 (a + b) × h
= 12 (11 + 5) × 4
= 12 × 16 × 4
= 32 m²
Area of rectangle HCDG = l × b = 11 m × 5 m = 55 m²
Area of the octagonal surface = 32 m² + 55 m² + 32 m² = 119 m²
Hence, the required area = 119 m².

10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
(i) From Jyoti’s diagram:

Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF
= 2 × Area of trapezium ABCD
= 2 × 12 (a + b) × h
= (15 + 30) × 7.5
= 45 × 7.5
= 337.5 m²
(ii) From Kavita’s diagram:

Area of the pentagonal shape = Area of ∆ABE + Area of square BCDE
= 12 × b × h + 15 × 15
= 12 × 15 × 15 + 225
= 112.5 + 225
= 337.5 m²

Yes, we can also find the other way to calculate the area of the given pentagonal shape.
Join CE to divide the figure into two parts, i.e., trapezium ABCE and right triangle EDC.
Area of ABCDE = Area of ∆EDC + Area of square ABCE

11. Diagram of the picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is the same.

Solution:

Hence, the areas of the four parts A, B, C, and D are 80 cm², 96 cm², 80 cm^2, and 96 cm² respectively.

Exercise 11.3 | Class 8th Mathematics

1. There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?
   
   Solution:
   (a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 120000 cm³
   (b) Volume of cube = (Side)³ = (50)³ = 50 × 50 × 50 = 125000 cm³
   Cuboidal box (a) requires lesser amount of material.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of the suitcase = 80 cm × 48 cm × 24 cm
l = 80 cm, b = 48 cm and h = 24 cm
Total surface area of the suitcase = 2[lb + bh + hl]
= 2 [80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]
= 2 × 6912
= 13824 cm²
Area of tarpaulin = length × breadth = l × 96 = 96l cm²
Area of tarpaulin = Area of 100 suitcase
96l = 100 × 13824
l = 100 × 144 = 14400 cm = 144 m
Hence, the required length of the cloth = 144 m.

3. Find the side of a cube whose surface area is 600 cm²?
Solution:
Total surface area of a cube = 6l²
6l² = 600
l² = 100
l = √100 = 10 cm
Hence, the required length of side = 10 cm.

4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution:
l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box
= 2 [lb + bh + hl] – lb
= 2[2 × 1.5 + 1.5 × 1 + 1 × 2] – 2 × 1
= 2[3 + 1.5 + 2] – 2
= 2[6.5] – 2
= 13 – 2
= 11 m²
Hence, the required area = 11 m².

5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of the area is painted. How many cans of paint will she need to paint the room?
Solution:
Surface area of a cuboidal hall without bottom = Total surface area – Area of base
= 2 [lb + bh + hl] – lb
= 2 [15 × 10 + 10 × 7 + 7 × 15] – 15 × 10
= 2[150 + 70 + 105] – 150
= 2 [325] – 150
= 650 – 150
= 500 m²
Area of the paint in one can = 100 m²
Number of cans required = 500100 = 5 cans.

6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Solution:
The two figures given are cylinder and cube.
Both figures are alike in respect of their same height.
Cylinder: d = 1 cm, h = 7 cm
Cube: Length of each side a = 7 cm
Both of the figures are different in respect of their shapes.
Lateral surface of cylinder = 2πrh
= 2 × 227 × 72 × 7 = 154 cm²
Lateral surface of the cube = 4l² = 4 × (7)² = 4 × 49 = 196
So, cube has the larger lateral surface = 196 cm².

7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How many sheets of metal is required?
Solution:
Area of metal sheet required = Total surface area of the cylindrical tank = 2πr(h + r)
= 2 × 227 × 7(3 + 7)
= 2 × 227 × 7 × 10
= 440 m²
Hence, the required area of sheet = 440 m².

8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.
Solution:
Width of the rectangular sheet = Circumference of the cylinder

h = 128 cm
l = 128 cm, b = 33 cm
Perimeter of the sheet = 2(l + b) = 2(128 + 33) = 2 × 161 = 322 cm
Hence, the required perimeter = 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

Solution:
The lateral surface area of the road roller = 2πrh

Hence, the area of road = 1980 m2

10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Solution:
Here, r = 142 = 7 cm
Height of the cylindrical label = 20 – (2 + 2) = 16 cm
Surface area of the cylindrical shaped label = 2πrh
= 2 × 227 × 7 × 16
= 704 cm²
Hence, the required area of label = 704 cm².

Exercise 11.4 | Class 8th Mathematics

1. Given a cylindrical tank, in which situation will you find the surface area and in which situation volume.
   
   (a) To find how much it can hold.
   (b) Number of cement bags required to plaster it.
   (c) To find the number of smaller tanks that can be filled with water from it.
   Solution:
   (a) In this situation, we can find the volume.
   (b) In this situation, we can find the surface area.
   (c) In this situation, we can find the volume.

2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Solution:
Cylinder B has a greater volume.
Verification:
Volume of cylinder A = πr²h

3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.
Solution:
Given: Area of base = lb = 180 cm²
V = 900 cm³
Volume of the cuboid = l × b × h
900 = 180 × h
h = 5 cm
Hence, the required height = 5 cm.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:
Volume of the cuboid = l × b × h = 60 cm × 54 cm × 30 cm = 97200 cm³
Volume of the cube = (Side)³ = (6)³ = 216 cm³
Number of the cubes from the cuboid

Hence, the required number of cubes = 450.

5. Find the height of the cylinder whose volume is 1.54 m³ and the diameter of the base is 140 cm.
Solution:
V = 1.54 m³, d = 140 cm = 1.40 m
Volume of the cylinder = πr2h

Hence, the height of cylinder = 1 m.

6. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank.

Solution:
Here, r = 1.5 m
h = 7 m
.’. Volume of the milk tank = πr²h
= 227 × 1.5 × 1.5 × 7
= 22 × 2.25
= 49.50 m³
Volume of milk in litres = 49.50 × 1000 L (∵ 1 m³ = 1000 litres)
= 49500 L
Hence, the required volume = 49500 L.

7. If each edge of a cube is doubled,
(i) how many times will it be surface area increase?
(ii) how many times will its volume increase?
Solution:
Let the edge of the cube = x cm
If the edge is doubled, then the new edge = 2x cm
(i) Original surface area = 6x² cm²
New surface area = 6(2x)² = 6 × 4x² = 24x²
Ratio = 6x² : 24x² = 1 : 4
Hence, the new surface area will be four times the original surface area.

(ii) Original volume of the cube = x³ cm³
New volume of the cube = (2x)³ = 8x³ cm³
Ratio = x³ : 8x³ = 1 : 8
Hence, the new volume will be eight times the original volume.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the reservoir = 108 m³ = 108000 L [∵1 m³ = 1000 L]
Volume of water flowing into the reservoir in 1 minute = 60 L
Time taken to fill the reservoir

Hence, the required hour to fill the reservoir = 30 hours.