Category: ICSE Solutions

Selina Concise Chemistry Class 8 ICSE Solutions – Carbon and Its Compounds

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 9 Carbon and Its Compounds. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 9 Carbon and Its Compounds

Points to Remember:

1. Carbon occurs in the earth’s crust in the free as well as in the combined state.
2. In the free state, it occurs as coal, diamond and graphite.
3. In the combined state, carbon occurs in atmosphere (CO₂) natural gas, food nutrients and carbonates.
4. Diamond is the hardest naturally occuring substance known.
5. Fullerenes are discovered only recently.

Exercise – I

Question 1.
Fill in the blanks.
(a) Carbon is present in both living and non-living things.
(b) The tendency of an element to exist in two or more forms but in the same physical state is called Allotropy.
(c) Crystalline and non-crystalline are the two major crystalline allotropes of carbon.
(d) Diamond is the hardest substance that occurs naturally.
(e) The name ‘carbon’ is derived from the Latin word carbo.

Question 2.
Choose the correct alternative.

(a) In a combined state, carbon occurs as
(i) coal
(ii) diamond
(iii) graphite
(iv) petroleum

(b) A crystalline form of carbon is
(i) lampblack
(ii) gas carbon
(iii) sugar
(iv) fullerene

(c) Graphite is not found in
(i) Bihar
(ii) Maharashtra
(iii) Orissa
(iv) Rajasthan

(d) Diamond is used for
(i) making the electrodes of electric furnaces.
(ii) making crucible for melting metals.
(iii) cutting and drilling rocks and glass.
(iv) making carbon brushes for electric motors.

(e) Carbon forms innumerable compounds because
(i) it has four electrons in its outermost shell.
(ii) it behaves as metal as well as non-metal.
(iii) carbon atoms can form long chains.
(iv) it combines with other elements to form covalent compounds.

Question 3.
Write ‘true’ or ‘false’ against the following statements.

(a) Carbon constitutes 0.03% of the earth’s crust. – True
(b) Graphite is the purest form of carbon. – False
(c) Coloured diamonds are costlier than colourless and transparent diamonds. – False
(d) Graphite has layers of hexagonal carbon bondings. – True
(e) Diamond is insoluble in all solvents. – True.

Question 4.
Define the following terms:
(a) Allotropy (b) Carat
(c) Crystal (d) Catenation
Answer:
(a) Allotropy: Allotropy is defined as the phenomenon due to which an element exists in two or more forms in the same physical state with identical chemical properties but with different physical properties.

(b) Carat – The weight of diamond is expressed in carats [ 1 carat = 0.2 g]

(c) Crystal – A crystal is a homogeneous solid in which particles (atoms, molecules, or ions) are arranged in the definite pattern due to which they have a definite geometrical shape with plane surfaces e.g. sugar and sodium chloride.

(d) Catenation – A large number of organic compounds is due to the ability of carbon atoms to form long chains with other carbon atoms through the sharing of electrons. This unique property of carbon is known as catenation.

Question 5.
State the terms:

(a) Substances whose atoms or molecules are arranged in a definite pattern. – Crystals.
(b) Different forms of an element found in the same physical state. – Allotropy.
(c) The property by which atoms of an element link together to form long chain or ring compounds. – Catenation

Question 6.
Name the following:

(a) The hardest naturally occurring substance. – Diamond.
(b) A greyish black non-metal that is a good conductor of electricity. – Graphite.
(c) The third crystalline form of carbon. – Fullerenes.

Question 7.
Answer the following questions:
(a) Why is graphite a good conductor of electricity but not diamond?
(b) Why is diamond very hard?
(c) What are fullerenes? Name the most common fullerenes.
(d) What impurity is present in black diamond?
(e) Explain the softness of graphite with reference to its structure.
Answer:
(a) In a graphite molecule, one valence electron of each carbon atom remains free, Thus making graphite a good conductor of electricity. Whereas in diamond, they have no free mobile electron. Thats why diamond are bad conductor electricity.

(b) A diamond is a giant molecule. The number of valence electrons in carbon atom is four. As such each carbon atom is linked with four neighboring carbon atoms. Thus forming a rigid tetrahedral structure. It is the strong bonding’that makes diamond the hardest substance.

(c) Fullerenes: Fullerenes are the third crystalline form of carbon.
Though they were discovered only recently. They have.been found to exist in interstellar dust as well as in the geological formations of the earth.
Common fullerenes are C – 32, C – 50, C – 70 and C – 76

(d) Black diamonds have copper oxide present in them as impurity.

(e) In a graphite molecule of each carbon atoms is linked with three neighboring carbon atoms. Thus forming a hexagonal arrangement of atoms. These hexagonal grouping of carbon atoms are arranged as layers or sheets piled one the top of other. The layers are held together by weak forces such that they can slide over one another. That is why graphite is soft.

Question 8.
Give two uses of (a) graphite (b) diamond.
Answer:

(a) Uses of graphite:

1. For making the electrodes of electric furnaces.
2. For making crucibles for melting metals due to its high melting points.

(b) Uses of Diamond:

1. Diamond is used in jewellery as a gem
2. It is used for cutting and drilling rocks, glass,

Question 9.
Write three differences between graphite and diamond.
Answer:
Difference between diamond and graphite.

Diamond

1. Pure diamond is colourless and transparent.
2. It is the hardest naturally occurring substance.
3. It has high density i.e. 3.5 g/cm³
4. It is bad conduct of electricity.
5. It bums in air at 900°C to form carbon dioxide.

Graphite

1. Graphite is greyish black opaque and shiny.
2. It is soft and greasy to touch.
3. It has low density i.e. 2.39 g / cm³
4. It is good conductor of electricity.
5. It bums in air at 700° C to form carbon dioxide.

Exercise – II

Question 1.
Fill in the blanks:

(a) Charcoal is formed when charcoal is burnt in a limited supply of air.
(b) Coal is a amorphous form of carbon.
(c) Peat is the most inferior form of coal.
(d) Wood charcoal is a bad conductor of heat and electricity.
(e) lampblack is used in making black shoe polish.

Question 2.
Choose the correct alternative

(a) Anthracite is
(i) an inferior type of coal
(ii) a superior type of coal
(iii) a cheapest form of coal
(iv) none of above

(b) Destructive distillation of coal yields
(i) coal tar
(ii) coal gas
(iii) coke
(iv) all of the above

(c) Lamp black is
(i) an amorphous form of carbon
(ii) a crystalline form of carbon
(iii) a pure form of carbon
(iv) a cluster of carbon atoms

(d) The process by which decayed plants slowly convert into coal is called.
(i) petrification
(ii) carbonisation
(ii) carbonification
(iv) fermentation

(e) The purest form of the amorphous carbon is
(i) wood charcoal
(ii) sugar charcoal
(iii) bone charcoal
(iv) lampblack

Question 3.
Write ‘true’ or ‘false’ against the following statements:

(a) Charcoal is a good adsorbent. True
(b) Coke is obtained by destructive distillation of sugar. False
(c) Activated charcoal is a good conductor of electricity. False
(d) Wood charcoal is an important constituent of gun powder. True
(e) Coal gas is used in the preparation of artificial ferilizers. False.

Question 4.
Define the following:
(a) Carbonization
(b) Adsorption
(c) Bone black
Answer: 
(a) Carbonization: The process of the slow conversion of vegetable matter into carbon-rich substances is called carbonization.
(b) Adsorption: Adsorption is the property due to which a substance absorbs gases, liquids and solids on its surface.
(c) Bone black: The Carbon content of bone charcoal is separated by treating the latter with hydrchloride acid, which dissolves the calcium phosphate. Carbon is then filtered out of the solution and in this form it is called bone black.

Question 5.
Name the following:

(a) Substances whose atoms or molecules are not arranged in a geometrical pattern. – Amorphous
(b) The best variety of coal. – Bituminous
(c) The purest form of amorphous carbon. – Anthracite
(d) An amorphous form of carbon that contains about 98% carbon. – Anthracite
(e) Mixture of carbon monoxide and hydrogen. – Water gas.

Question 7.
Answer the following questions:
(a) What is destructive distillation? What are the products formed due to the destructive distillation of coal?
(b) Why is wood charcoal used in water filters and gas masks?
(c) How is wood charcoal made locally? What other substances are formed in the process.
(d) How many carbon atoms are there in Buckminster fullerenes?
Answer: 
(a) Destructive Distillation: When a substances is heated in the absence of air. The process is called destructive distillation.
Products formed are: Coke, Coal tar, Coal gas and ammonia solution

(b) Due to its high adsorbing capacity, wood charcoal is used as gas masks to adsorb harmful gases. Wood charcoal is porous, that is why it is used to filter water.

(c) Wood charcoal is prepared when wood is heated in a limited supply of air. Locally wood charcoal is prepared by piling logs of wood one above the other with a gap in the centre of the pile. The pile is covered with wet clay to prevent the entry of air. A few holes are left at the bottom of the pile. The wood is set on fire. After some time when fire dies out, wood charcoal is left behind. The other substances are -wood tar, pyroligneous acid and wood gas.

(d) 60 carbon atoms are arranged in spherical structure in Buck minster fullerences.

Question 7.
(a) Descirbe the formation of coal,
(b) Name four types of coal with percentage of carbon present in each, with uses.
Answer:
(a) Formation of coal:- The formation of coal took millions of years. Coal was formed by the bacterial decomposition of ancient vegetable matter hurried under successive layers of the earth. Under in action of high temperature and pressure, and in the abcence of air, the decayed vegetable matter converted into coal.
(b) Types of Coal:

1. Peat: It is light brown in colour and contains only 50 – 60% carbon. It is the most inferior form of coal.
2. Lignite: it contains more than 60% carbon. It is brown in colour and harder than peat.
3. Bituminous: It has 90%, 80%, 70 – 75% carbon contents. Bituminous coal is the most common variety of coal and used as house hold coal.
4. Anthracite: It is the purest variety of coal. Its carbon contents vary between 92 – 98%. It is hard, dense and black, difficultto ignite.

Uses of coal:

1. Coal is used as both domestic and industrial fuel.
2. It is used to prepare coke, coal gas and coal tar.

Question 9.
Name the products formed when:
(a) wood is burnt in the absence of air.
(b) bone is heated in the absence of air.
(c) diamond is burnt in air at 900°C.
(d) graphite is subjected to high pressure and 3000°C temperature.
Answer:
(a) Wood charcoal is formed when wood is burnt in limited supply of air.
(b) Bone charcaol, bone oil and organic compound pyridine.
(c) Carbon dioxide.
(d) Artificial diamond.

Question 9.
Give two uses for each of the following:
(a) coal
(b) coke
(c) wood charcoal
(d) sugar charcoal
(e) bone charcoal
(f) lampblack
Answer: 
(a) Uses of coal

• It is used as both a domestic and an industrial fuel.
• It is used to prepare coke, coal gas and coal tar.

(b) Uses of coke

• Coke is used as a smokeless fuel, in smelting furnaces.
• It is used in the manufacturing of water and producer gas.

(c) Uses of wood charcoal:

• Wood charcoal is used as a fuel.
• It is an important constituent of gun powder.

(d) Sugar charcoal:

• Sugar charcoal is mostly used as a reducing agent.
• It is used to decolourise coloured solutions.

(e) Bone charcoal:

• It is extensively used to decolourise cane-sugar in the process of manufacturing sugar.
• It is also used in the manufacture of large number of phosphorous compounds.

(f) Uses of lamp black:

• It is used in making black shoe polish.
• It is used in the manufacture of tyres and gun powder.

Question 10.
Give balanced equations for the following chemical reactions:
(a) wood charcoal and cone, nitric acid
(b) coke and steam
(c) wood charcoal and lead monoxide.
Answer:

Exercise – III

Question 1.
(a) Name the chemicals required for the preparation of carbon dioxide in the laboratory.
(b) How will you collect the gas?
(c) Write the balanced chemical equation for the above reaction.
(d) Draw a labelled diagram for the preparation of CO₂ in the laboratory.
(e) Why is sulphuric acid not used for the preparation of carbon dioxide in the laboratory?
Answer:
(a) Calcium carbonate and dilute hydrochloric acid.
(b) By upward displacement of air.
(c) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.
(d) Laboratory preparation of carbon dioxide

(e) Dilute sulphuric acid reacts with calcium carbonate. But it is not used because the calcium sulphate which is formed during the reaction is insoluble in water. It covers the marble chips and stops the reaction.

Question 2.
Write the balanced chemical equations for the preparation of carbon dioxide by:
(a) heating calcium carbonate.
(b) the action of acetic acid on sodium bicarbonate.
(c) the action of dilute sulphuric acid on sodium bicarbonate.
Answer:

Question 3.
What happens when:
(a) a lit splinter is introduced into a jar containing carbon dioxide?
(b) moist blue litmus paper is placed in a jar containing carbon dioxide?
(c) carbon dioxide is passed through lime water first in small amounts and then in excess?
(d) a baking mixture containing baking powder is heated?
(e) a soda water bottle is opened?
Answer:
(a) Lit splinter extinguishes.
(b) Blue litmus paper turns red.
(c) When CO₂ is passed through lime water in small amount, it turns milky, when passed in excess milkiness disappears.
(d) Carbon dioxide is formed.
(e) When the pressure is released the bottled gas escapes with a bristling effervescence that ads fizz to the drink.

Question 4.
Give reasons for the following:
(a) An excess of carbon dioxide increases the temperature of the earth.
(b) Soda acid and foam types of fire extinguisher are not used for extinguishing electrical fires.
(c) Solid carbon dioxide is used for refrigeration of food.
Answer:
(a) Excess carbon dioxide increases the temperature of the earth. Due to the rise in temperature ice in the polar regions may melt causing floods in coastal regions island.
(b) In both of these fire extinguishers, the solutions are prepared in water, which conducts electricity. As a result, an electric shock might result, which might lead to short-circuiting and another fire.
(c) Solid carbon dioxide serves as a coolant and refrigeration for preserving food articles.

Question 5.
What is a fire extinguisher? What is the substance used in the modern type of fire extinguishers? How is it an improvement over the soda acid-type and the foam-type fire extinguishers?
Answer:
Fire Extinguisher— Fire extinguishers are a device in which carbon dioxide is produced in different forms for use as the extinguishing agent. It is a modem type of fire extinguisher in which liquid carbon dioxide is stored in a steel cylinder under pressure. Soda-acid and foam types of extinguisher cannot be used for extinguishing fire as they prepared in water, which conducts electricity and there can be short-circuiting, causing another fire.

Question 6.
Explain the term ‘greenhouse effect’. How can it be both beneficial and harmful for life on earth ?
Answer:
Green house effect— The trapping of the earth’s radiated energy by carbon dioxide present in air, so as to keep the earth warm, is called ‘green house effect’.
Green house is beneficial because this principle is applied to grow plants in colder regions.
Carbon dioxide increases the temperature of atmosphere. Due to rise in temperature; ice in the polar regions may melt, causing floods. So it is harmful for life on earth.

Question 7.
What steps should be taken to balance carbon dioxide in the atmosphere ?
Answer:
As global warming will cause an unbalanced ecological system, serious efforts should be made to balance the percentage of carbon dioxide in the atmosphere. Some of these steps are:

• Growing more trees and plants.
• Using smokeless sources of energy like solar energy, biogas, etc.
• Using filters in the chimneys of factories and power houses.

Question 8.
State three ways by which carbon dioxide gas is added into the atmosphere.
Answer:

1. By planting more trees.
2. By combustion of fuels
3. By decay of dead animals, plants and plants products.

Exercise – IV

Question 1.
Fill in the blanks:

(a) Carbon monoxide is formed when carbon is burnt in a limited supply of air or oxygen.
(b) Carbon monoxide bums in air with a pale blue flame to form carbon dioxide.
(c) Carbon monoxide is a products of incomplete combustion.
(d) A mixture of 95% oxygen and 5% carbon dioxide is called carbogen
(e) Carbon dioxide is used as a reducing agent in the extraction of pure metals from their corresponding ores.

Question 2.
Match the following.

Answer:

Question 3.
How is carbon monoxide gas formed?
Answer:
Mostly carbon monoxide is formed when a large amount of carbon or its compounds is burnt in a limited supply of air or oxygen.

Question 4.
State the poisonous nature of carbon monoxide?
Answer:
Carbon monoxide is a highly poisonous gas. If air containing 0.5% carbon monoxide by volume is inhaled, death can result This is because carbon monoxide combines with the haemoglobin present in the blood cells of our body to form a stable compound called carboxyl-haemoglobin. This does not allow to absorb of oxygen. Thus depriving our body cells of oxygen. This cause obstruction in respiration and causes death.

Question 5.
Give two uses of carbon monoxide.
Answer:
Uses of carbon monoxide:

• Carbon monoxide is a strong reducing agent.
• Carbon monoxide is used in the extraction of pure metals from their ores.

Question 6.
Why is carbon monoxide called a silent killer?
Answer:
Carbon monoxide is produced by burning coal or wood in a limited supply of air. Since the gas is colourless and a barely detectable smell, people do not feel it and it can be proved as a silent killer.

Question 7.
Explain the reducing action of carbon monoxide.
Answer:
Reducing action of carbon- monoxide: Carbon monoxide is a strong reducing agent. It reduces the oxides of the less active metals to their respective metals and itself gets oxidised to carbon dioxide.

Question 8.
Write two remedies for carbon monoxide poisoning.
Answer:

1. The victim should immediately be brought out into the open.
2. The victim should be given artificial respiration with carbogen.

Question 9.
Complete the reactions and balance them.
(a) CuO + CO →
(b) Fe₂O₂ + CO →
Answer:

Selina Concise Physics Class 8 ICSE Solutions – Sound

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 7 Sound. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Physics Chapter 7 Sound

• SOUND: “Is energy which produces in us the sensation of hearing.” It is produced by vibration of body.
• Sound needs a medium for its propagation. Sound cannot travel in vacuum.
• Speed of sound is maximum in solids. 5000 ms^-1 in steel, in water 1500 ms^-1 and in air it is least 330 ms^-1 nearly.
• When a body vibrates, the particles of medium also start vibrating and K.E. of particles changes into potential energy and P.E. into
  K.E. This is why sound in energy.
•  Sound travels in a medium in the form of wave.
  
• Longitudinal wave : When the particles of medium move in the direction of motion of wave by forming compression and rarefaction.
  
• AMPLITUDE : “The maximum displacement of the particle of medium on either side of mean position.”
  
• TIME PERIOD: “The time taken by a particle of medium to complete its one vibration” “t”
•  FREQUENCY: “The number of vibrations made by a particle of
  the medium in one second. ƒ measured in Hertz (Hz)
•  FREQUENCY  ƒ = 1/ t or t = 1 / ƒ
•  WAVE LENGTH: “The distance travelled by the wave in one one time period of vibration of particle of medium.”
  Or
  “The distance between two consecutive compressions or between two consecutive rarefactions.” It is denoted by ‘ λ ’ and S.I. unit of wave length is metre (m).
  
•  CHARACTERISTICS OF SOUND :
  (i) Loudness.
  (ii) Pitch (or shrillness).
  (iii) Quality (or timbre or wave form).
  LOUDNESS : is the characteristic of sound by virtue of which a loud sound can be distinguished from a faint sound, both having same frequency and same wave form.
•  It depend on: (i) Amplitude of wave (ii) Surface area of vibrating body (ii) Distance from the source of sound (iv) Sensitivity of listener: Unit of loudness is (dB) decibel.
•  PITCH: It depends on number of vibrations per second or frequency : more frequency is high pitch shrilled sound and low frequency is flat sound.
•  QUALITY: is the characteristic which distinguishes two sounds’of the same pitch and same loudness. It depends on wave form.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) When sound propagates in air, it does not carry energy with it.
Answer. False.

(b) In a longitudinal wave, compression and rarefaction are formed.
Answer. True.

(c) The distance from one compression to nearest rarefaction is called wavelength.
Answer. False.

(d) The frequency is measured in second.
Answer. False.

(e) The quality of a sound depends on the amplitude of wave.
Answer. False.

(f) The pitch of sound depends on frequency.
Answer. True.

(g) Decibel is the unit of pitch of a sound.
Answer. False.

2. Fill in the blanks

(a) The time period of a wave is 2 s. Its frequency is 0.5 S^-1.
(b) The pitch of a stringed instrument is increased by increasing tension in string.
(c) The pitch of a flute is decreased by increasing length of air column.
(d) Smaller the membrane, higher is the pitch.
(e) If a drum is beaten hard, its loudness increases.
(f) A tuning fork produces sound of single frequency.

3. Match the following

4. Select the correct alternative

(a) Sound can not travel in

1.  solid
2.  liquid
3.  gas
4.  vacuum

(b) When sound travels in form of a wave

1. the particles of medium move from the source to the listener
2.  the particles of medium remains stationary
3.  the particles of medium start vibrating up and down
4.  the particles of medium transfer energy without leaving their mean positions.

(c) The safe limit of loudness of audible sound is

1.  0 to 80 dB
2.  above 80 dB
3.  120 dB
4.  above 120 dB

(d) The unit of loudness is

1.  cm
2.  second
3.  hertz
4.  decibel

(e) In a piano, pitch is decreased by

1.  using thicker string
2.  increasing tension
3.  reducing length of string
4.  striking it hard Ans.

Selina Concise Chemistry Class 8 ICSE Solutions – Water

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 8 Water. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 8 Water

Points to Remember:

• Water is the source of life for all living beings.
• Water occurs in all the three states of matter i.e. ice, liquid water and water vapours.
• Water found in nature (i) oceans and seas (ii) rivers and lakes (iii) springs and wells (iv) rain.
• Rainwater is the purest form of water. Sea water is very impure.
• Potable water should be free from suspended impurities and harmful germs.
• Water is a compound, with the molecular formula is H₂O.
• The boiling point of water is 100°C and the freezing point is 0°C.
• 0°C is also called tripple point, because water can exist in all its three states.
• The specific heat of water is higher than that of any other liquid. It is used as a cooling agent.
• Water has minimum volume and maximum density at 4°C. This is called anomalous behaviour of water.
• Water is an universal solvent. The gases dissolved in water have biological importance. They enable aquatic life to sustain itself.
• Water may be ‘hard or soft’. Hardness of water can be removed by boiling or by chemical treatment.
• Water pollution is a serious problem.
• Industrial and agriculture processes, nuclear and thermal plants pollute water.

Exercise – I

Question 1.
Name the four main sources of water.
Answer:

1. Oceans
2. Seas
3. Rivers
4. Lakes

Question 2.
State the importance of water cycle in nature.
Answer:
(i) Water cycle helps in regulating weather on the earth.
(ii) Water cycle makes water available in various forms on the earth.

Question 3.
Why is water very precious for all living beings?
Answer:
Water is one of the most precious substances for the existence of life. Since life on the earth began in the oceans, and since no living thing can survive without water, it is rightly called the source of life.
Water forms a large part of the body mass of all living organisms — 90% of human blood is water. Water has the ability to dissolve a number of substances. Therefore, it serve as the liquid medium in which all reactions within the living body take place.
Fruits and vegetables contain water in them. Even dry-looking substances like wood, peas, beans, grams, etc., contain some amount of water.

Question 4.
Name the two gases from which water is formed. What is the chemical composition of these two gases in water? Give the molecular formula of water?
Answer:
Oxygen and hydrogen
Chemical composition = H₂ and O₂ proportion 2:1
Molecular formula = H₂O

Question 5.
What is the effect on the boiling point of water when
(a) pressure is increased
(b) impurity is added
Answer:
(a) The boiling point of water increases with an increase, in pressure.
(b) Any impurity present in water lowers its freezing point and raises its boiling point.
For example, salt is added to ice to lower its melting point. Such a mixture is called a freezing mixture. The melting is called a freezing mixture is about -15° C.

Question 6.
Give reasons:
(a) Water is used as a cooling agent
(b) Water pipes burst in severe winters.
(c) It is difficult to cook in hills compared to plains.
(d) Ice floats on water.
(e) Seawater does not freeze at 0°C.
Answer:
(a) Water has high specific heat. Water neither heats up nor cools down quickly. This property makes water as an excellent cooling agent.
(b) Water pipes bursts in severe winter because the water inside I the pipes freezes and increases its volume.
(c) Water boils at a lower temperature in the hills, where the atmospheric pressure is lower than in the plains. This is why it takes a longer time to cook in hilly regions.
(d) Ice has a low density as compared to water. Water has a maximum density at 4°C. That is why ice floats on water.
(e) There are impurities dissolved in seawater which increases the freezing point. That is why seawater does not freeze at 0°C.

Question 7.
How does anomalous expansion of water help aquatic organisms in cold climates?
Answer:
The anomalous expansion of water helps in survivals of water animals in very cold climates. Initially when temperature of water falls, it becomes heavier and sinks down. This process continues till 4°C. Then after this expansion takes place. The surface layer of water gets freezed. Ice being bad conductor of heat does not allow loss of heat from the water below and results in survival of water animals.

Exercise – II

Question 1.
Explain the terms:
(a) Solution (b) Solute (c) Solvent.
Answer:
(a) Solution: “A homogeneous mixture of two or more substances can be varied is called a solution”.
(b) Solute: A substance which dissolves in a other substances to form a solution is called solute. Solute is smaller quantity in solution.
Or
“Substance which is dissolved in solvent.” is called Solute. Solute is smaller quantity in solution.
(c) Solvent: A solvent is a medium in which a solute dissolves. It is in large quantity in solution.
Solution = Solute + Solvent

Question 2.
What is meant by
(a) Unsaturated (b) Saturated and
(c) Supersaturated solutions.
Answer:
(a) Unsaturated solutions — A solution in which more of the solute can be dissolved at a given temperature is called an unsaturated solution.
(b) Saturated solutions — A solution that cannot dissolve any more of the solute at a given temperature is called a saturated solution.
(c) Supersaturated solutions — A solution that contains more solute than it can hold at room temperature is called supersaturated solution.

Question 3.
How do the solubility of a solid and a gas affected by –
(a) Increase in temperature
(b) Increase in pressure
Answer:
(a) Solubility of a solid solute generally increases with an increase in temperature. This makes it possible to prepare supersaturated solutions.
Solubility of a gas decreases with an increase in temperature.

(b) Pressure has practically no effect on the solubility of a solid (solute) in water.
In the case of gases, the amount of a gas dissolved in water increases with an increase in pressure.

Question 4.
Differentiate between:
(a) Solution and suspension
(b) Suspension and colloid
Answer:
(a) Solution and suspension

Solution

1. It is an example of homogeneous mixture.
2. Particle size less than 10^-10m
3. Transparent
4. Solute particles can not be filtered. Solution pass easily through filter paper.

Suspension

1. It is an example of heterogeneous mixture.
2. Particle size greater team 10^-7 m
3. Opaque
4. Particles of suspension do not pass through filter paper.

(b) Suspension and colloids

Suspension

1. Heterogeneous
2. Particle size greater than 10^-7 m.
3. Opaque.
4. Particles are visible.
5. Particles of suspension settle at the bottom of the container.
6. Particles of suspension do not pass through filter paper.

Colloid

1. Heterogeneous.
2. Particles size between 10^-10 to 10^-7m.
3. Translucent.
4. Particles can be seen with the help of a powerful microscope.
5. Particles of colloids do not settle.
6. Colloidal particles pass easily through ordinary filter paper but do not pass through ultra filters.

Question 5.
Define: ‘water of crystallisation’. Give two examples with formulae.
Answer:
The fixed amount of water which is in loose chemical combination with a salt in its crystal is called water of crystallisation. Examples:

Question 6.
Give two examples for each of the following:
(a) Hydrated substances
(b) Crystalline anhydrous substances
(c) Drying agents
(d) Deliquescent substances
(e) Efflorescent substances
(f) Colloids
(g) Solvents other than water.
Answer:
(a) Washing soda, Glauber’s salt (Na₂SO₄.10H₂O)
(b) Common salt (NaCl), potassium nitrate (KNO₃), sugar (C₁₂H₂₂O1₁₁) etc.
(c) Concentrated sulphuric acid (H₂SO₄), phosphorus pentoxide (P₂O₅), quicklime (CaO)
(d) Caustic soda (NaOH), crystalline-magnesium chloride (MgCl₂), calcium chloride (CaCl₂), Iron (III) chloride etc.
(e) Washing soda and glauber’s salt (Na₂SO₄.10 H₂O)
(f) Milk, blood, smoke, jellies, butter, ink etc.
(g) Acetone, ethanol, turpentive

Question 7.
What do you observe when:
(a) Blue vitriol is heated ?
(b) Washing soda is exposed to air ?
(c) Blue litmus solution is added to water ?
Answer:
(a) Blue vitriol is blue in colour as it contains 5 molecules of water of crystallisation (CuSO^HjO). When it is heated, it loses water of crystallisation and becomes an hydrous CuS04 which is grey-white in colour.
(b) Washing soda (Na₂CO₃.10H₂O) is a white crystalline substance and on exposure to air it gets changed to white powder.
(c) Pure water is neutral to litmus which means that no change in the colour of blue or red litmus solution is observed when 1 treated with water.

Question 8.
Give reason:
(a) Silica gel pouches are kept in unused water bottles.
(b) Table salt becomes moist during rainy season.
(c) On opening a bottle of a cold drink, a fizz sound is heard.
Answer:
(a) Silica get pouches are very commonly used to absorb moisture and to keep things dry. They are often kept in unused water bottles, with camera lenses etc. to keep them dry. These pouches are ideal to reuse throughout, in places at home where there is excess of moisture.

(b) On exposure to air, table salt (NaCl) turns moist and ultimately forms a solution especially during rainy season because it contains impurities like magnesium chloride and calcium chloride which are deliquescent. Sodium chloride is not deliquescent.

(c) The cold drink bottles contain carbon dioxide and are bottled under high pressure i.e. they contain a large amount of gas dissolved in them and on opening a bottle we hear a fizz sound, this is because of the solubility of CO₂ gas in it and pressure in it.

Question 9.
Give balanced chemical equations for the reaction of water with
(a) Sodium (b) Iron
(c) Carbon dioxide (d) Sodium oxide
Answer:

Question 10.
What is metal activity series ?
Answer:
The arrangement of metals in the decreasing order of their reactivity in the form of a series is called the activity or reactivity series of metals.

Question 11.
Name the gas produced when
(a) steam is passed over hot coke.
(b) chlorine is dissolved in water and exposed to sunlight
(c) a piece of calcium is added to water.
(d) when fossil fuel is burnt,
Answer:
(a) Water gas
(b) Oxygen
(c) Hydrogen
(d) Carbon dioxide

Exercise – III

Question 1.
Define:
(a) Soft water
(b) Hard water
Answer:
(a) Soft water: The water present in different natural sources has different substances dissolved in it. The water drawn from certain sources forms a lather with soap rather easily. Such water is called soft water.
(b) Hard water: Water obtained from various sources does not easily form a lather with soap, rather it forms a white sticky scum or a precipitate. This water is called hard water.

Question 2.
(a) Name the compounds responsible for
(i) temporary hardness
(ii) permanent hardness of water
(b) Suggest one method for the removal along with the reactions for
(i) temporary hardness
(ii) permanent hardness of water
Answer:
(a) (i) Temporary hard water— Water, which has bicarbonates of calcium and magnesium dissolved in it, is temporary hard water. This kind of hardness is easily removed by boiling.
(ii) Permanent hard water— Water, which has sulphates and chlorides of calcium and magnesium dissolved in it, is called permanent hard water. This hardness cannot be removed by boiling.

(b) Removal of the hardness of water:
(i) Temporary hardness—
By Boiling— This method helps to remove only the temporary hardness of water. When temporary hard water is boiled, the bicarbonates of calcium and magnesium break up to form their insoluble carbonates.
These can be filtered out so that water becomes soft.

(ii) Removal of hardness of permanent hardness of water—
By Adding sodium carbonate (washing soda)
Permanent hardness of water is removed when water is treated with a small quantity of sodium carbonate.
It reacts with the soluble chlorides and sulphates of calcium and magnesium to form their insoluble carbonates. These can be removed by filtration and then the water becomes soft. Sodium sulphate or sodium chloride formed after the reaction does not affect the soap.

Question 3.
Name three water-borne diseases.
Answer:
(i) Cholera (ii) Typhoid (iii) Hepatitis

Question 4.
What are the main causes of water pollution? How can it be controlled?
Answer:
Main causes of water pollution are:

• Chemical waste, industrial waste and agricultural processes.
• Thermal waste from nuclear and thermal power plants.
• Sewage and garbage.

Prevention of water pollution:

• Domestic sewage should be treated before being discharged into rivers.
• Using of non-biodegradable substances like detergents should be stopped.
• Trees and plants must be planted along the banks of rivers.
• Purification of water bodies should be carried out.
• The waste products of industries should be treated before they are discharged into rivers.

Question 5.
Give reasons:
(a) Alcohol is mixed with water and is used in car radiators.
(b) Icebergs float on ocean water.
(c) Carbonated drinks are bottled under high pressure.
Answer:
(a) Alcohol is mixed with the water used in car radiators to prevent it from freezing ki cold weather. Because it lowers the freezing point of water.
(b) Ice bergs float in ocean water because density of ice is less than water.
(c) Carbonated drinks are bottled under high pressure because the solubility of carbon dioxide increases with pressure.

OBJECTIVE TYPE QUESTIONS

1. Fill in the blanks:

(a) Water has maximum density and minimum volume at 4°C.
(b) Freezing mixture contains ice and salt.
(c) The solubility of a gas in water increases with rise in temperature and decreases with rise in pressure.
(d) Rain water is the purest form of natural water.
(e) Use of excessive fertilizers by farmers causes water pollution.
(f) Boiling removes the temporary hardness of water.
(g) Water turns the colour of anhydrous copper sulphate blue.
(h) Water turns the colour of anhydrous copper sulphate scum.

2. Give one word/words for the following statements:

(a) Water fit for human consumption potable water.
(b) The harmful substances dissolved in water impurities.
(c) The change of states of water from one form to another water cycle.
(d) The gaseous form of water found in air – water vapours.
(e) A mixture of common salt and ice – freezing mixture.
(f) A substance which does not contain water anhydrous substances.
(g) A property due to which a substance absorbs water without dissolving hygroscopic.
(h) Water molecules in loose chemical combination with other substances water of crystallisation.

MULTIPLE CHOICE QUESTIONS

1. Two gases found dissolved in natural water are
(a) oxygen and carbon dioxide
(b) hydrogen and oxygen
(c) sulphur dioxide and hydrogen
(d) chlorine and ammonia

2. Temporary hardness of water can be removed by
(a) filtering
(b) boiling
(c) loading
(d) none of the above

3. The ultimate source of all water on the earth is
(a) oceans and seas
(b) spring and wells
(c) rivers and lakes
(d) rain

4. Colloids have the particle size range between
(a) 10^-7 to 10^-10 m
(b) 10^-10 to 10^-12 m
(c) 10^-7 to 10^-5 m
(b) 10^-12 to 10^-15 m

Selina Concise Physics Class 8 ICSE Solutions – Matter

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 1 Matter. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Physics Chapter 1 Matter

•  Matter Every substance living and non-living that we see is made up of matter and MATTER “is something which has mass, occupies space and can be perceived by our senses.” e.g. Hydrogen, milk, oxygen, pen, table, water, iron, air, oil, sugar etc.
• Matter is composed of tiny particles called molecules, which are in constant motion, have spaces between them and have inter-molecular attraction.
• Every molecule can exist freely in nature and has all the properties of matter.
• A molecule is composed of ATOMS, but atom cannot exist free in nature.
•  INTER-MOLECULAR FORCE ‘The molecules of a matter exert a force of attraction on each other – The force of attraction is called INTER-
• MOLECULAR FORCE This force in solid is very strong and we cannot break a solid easily. In liquids this force is less strong and in molecules of gas it is very less. –
• FORCE OF COHESION “The inter-molecular force of attraction between the molecules of same substance is called FORCE OF COHESION.” i.e. between water and water.
• FORCE OF ADHESION “The force of attraction between the molecules of two different substances is called FORCE OF ADHESION” i.e. between glue and paper.
  Matter is composed of tiny particles and molecules of matter have spaces between them can be proved by experiment.
  Take 50 c.c of water in a measuring cylinder. Add a small quantity of salt in it. Salt gets dissolved in water and still level remains at 50 c.c. Where has salt gone?
  The salt molecules enter into spaces of water and water molecules into spaces of salt molecules. This experiment show that particles of matter are very minute and cannot be seen by naked eye and there are spaces between molecules.
  

• The molecules of matter are in constant motion can be seen by opening a bottle of perfume in a comer of room, it reaches the other parts of the room.
•  SUBLIMATION Change of solid directly into vapours on absorbing heat.
• DEPOSITION “The change of vapours directly into solid on cooling.”
• MELTING “Change of solid in liquid at fixed temperature on heating.”
• FUSION or FREEZING “Change of liquid to solid state on cooling at a fixed temperature.”
• FUSION or MELTING “Change of a solid to liquid state at a fixed temperature on absorbing heat.”
• EVAPORATION Surface phenomenon i.e. only takes place at surface “Is change of liquid to vapours.” Evaporation has cooling effect. Takes place at all temperatures.
• VAPORIZATION “Change of liquid to vapour state on heating at constant temperature.”
  It is fast process and produces hotness.

Test Yourself

A. Objective Questions

1. Write true or false for each statement

(a) The temperature of a substance remains unaffected during its change of state.
Answer: True.

(b) Ice melts at 100°C.
Answer: False. The ice melts at 0° by absorption of heat.

(c) Water at 100°C has more heat than the steam at 100°C.
Answer: False.

(d) Evaporation of a liquid causes cooling.
Answer: True.

(e) Water evaporates only at 100°C.
Answer: False.

(f) Boiling takes place at all temperatures.
Answer: False.

(g) Evaporation takes place over the entire mass of the liquid.
Answer: False.

(h) The process of a gas converting directly into solid is called vaporization.
Answer: False.
The process of a liquid converting directly into gas is called vaporization.

(i) At high altitudes water boils above 100° C.
Answer: False.

(j) The melting point of ice is 0°C.
Answer: True.

 

2. Fill in the blanks

(a) Evaporation takes place at all temperature.
(b) Freezing process is just reverse of melting.
(c) Sublimation is a process that involves direct conversion of a solid into its vapour on heating.
(d) The temperature at which a solid converts into a liquid is called its melting point.
(e) The smallest unit of matter that exists freely in nature is called molecule.
(f) Molecules of a substance are always in a state of motion and so they possess kinetic energy.
(g) Intermolecular space is maximum in gases less in liquids and the least in solids.
(h) Intermolecular force of attraction is maxiumum in solids, less in liquids and the least in gases.

3. Match the following

4. Select the correct alternative

(a) The inter-molecular force is maximum in

1. solids
2. gases
3. liquids
4. none of the above

(b) The inter-molecular space is maximum in

1. liquids
2. solids
3. gases
4. none of the above

(c) The molecules can move freely anywhere in

1. gases
2. liquids
3. solids
4. none of the above

(d) The molecules move only within the boundary of

1.  liquids
2. gases
3. solids
4. none of the above

(e) The temperature at which a liquid gets converted into its vapour state is called its

1. melting point
2. boiling point
3. dewpoint
4. freezing point.

(f) Rapid conversion of water into steam is an example of

1. evaporation
2. freezing
3. melting
4. vapourization.

(g) Evaporation takes place from the

1. surface of liquid
2. throughout the liquid
3. mid-portion of the liquid
4. bottom of liquid.

(h) Boiling takes place from the

1. the surface of the liquid
2. throughout the liquid
3. mid-portion of liquid
4. none of the above.

Short/Long Answer Questions

Question 1.
Define the term matter. What is it composed of ?
Answer:
Anything which occupies space and has mass is called matter. Matter is composed of tiny particles called MOLECULES.

Question 2.
State three properties of molecules of a matter.
Answer:

1. They are very small in size.
2. They have spaces between them.
3. They are in constant motion and they posses kinetic energy.

Question 3.
What do you mean by the inter-molecular spaces ? How do they vary in different states of matter ?
Answer:
INTER-MOLECULAR SPACES “The spacing between the molecules of matter is called Inter-molecular spaces.”
The inter-molecular spaces is less in solids more in liquids and still more in gases.
Explanation of inter-molecular space : Take water in a measuring cylinder say upto 80 ml. mark. Add 10 gm of salt to it. The volume in cylinder should increase. On dissolving salt we find volume remains same i.e. upto 80 ml mark. This is because there are spaces in water molecules and salt molecules occupy these spaces and volume remains the same.

Question 4.
What is meant by the inter-molecular forces of attraction ?
Answer:
How do they vary in solids, liquids and gases ?
INTER-MOLECULAR FORCES OF ATTRACTION : “The forces of attraction between the molecules of matter is called the inter-molecular force of attraction.”
This inter molecular force is maximum in solids, less in liquids and least in gases.

Question 5.
Which of the following are correct ?
Answer:
(a) Solids have definite shape and definite volume.
True.
Reason As the molecules here have negligible intermolecular distance between them and have maximum intermolecular force of attraction.
(b) Liquids have definite volume but do not have definite shape.
True.
(c) Gases have definite volume but no definite shape.
False.
Correct Gases have neither definite volume nor a definite shape.
(d) Liquids have definite shape and definite volume.
False.
Correct Liquids have a definite volume but not definite shape.

Question 6.
Discuss the three states of matter solid, liquids and gas on the basis of molecular model.
Answer:
Solids

Here the molecules are very tightly packed that there is no or very less intermolecular space and there is high intermolecular force of attraction (force of cohesion). The molecules do not move about their mean position and thus solids have a definite shape and volume.
Liquids :

Here the molecules are less tightly packed as compared to solids and also there is lesser force of intermolecular attraction. The intermolecular distance is greater than that in the solids. Thus, they do not have a definite shape but acquire the shape of the vessel in which they are contained but have a definite volume at a given temperature.
Gases :

Here the molecules are far apart from each other i.e. have the greatest intermolecular distance which result into the weakest intermolecular forces of attraction. The molecules as are not bound by any strong force, move about freely and thus gases do not have a definite shape and’hlso do not have any definite volume.

Question 7.
What do you mean by the change of state ? Write the flow chart showing the complete cycle of change of state.
Answer:
CHANGE OF STATE: “The process of change from one state(form) to another state (form) either by absorption or rejection of heat at a constant temperature is called the CHANGE OF STATE.”
COMPLETE CYCLE OF CHANGE OF STATE : On heating a solid changes to liquid and liquid on heating changes to vapours. On cooling vapours condense to LIQUID, LIQUIDS on cooling freeze to SOLIDS. Some SOLIDS on heating change to vapours. On rejection of heat vapours solidify.

This cycle can be shown by diagram

 

Question 8.
Differentiate between melting point and boiling point, giving atleast one example of each.
Answer:
MELTING POINT:
The temperature at which a solid starts changing into LIQUID without further increase in temperature is called MELTING POINT.” Or The constant temperature at which a solid changes into liquid.”
Example : Ice (solid) melts at Q?C into water (liquid) when heated.
BOILING POINT : “The temperature at which a LIQUID start changing in vapour without further rise in temperature.
Or
‘The constant temperature at which a LIQUID starts changing into GAS (vapours)
Example : Boiling point of water (liquid) is 100°C.

Question 9.
Describe the process of condensation and sublimation with examples.
Answer:
CONDENSATION :
“The change of vapours on cooling at fixed temperature to liquid is called condensation.”
Example: When water vapours at 100°C are cooled they change into water (liquid).
SUBLIMATION : “The process of change of solid directly into vapours on heating is called sublimation.”

Question 10.
Explain the term melting and melting point.
Answer:
Melting — The change from the solid state to the liquid state on heating at a fixed temperature is called melting.
Melting Point — It can be defined as the fixed temperature at which a solid starts changing to its liquid state.

Question 11.
Describe an experiment to demonstrate that a substance absorbs heat during melting without change in its temperature.
Answer:
MELTING POINT OF SOLID (WAX): Put some wax in a test tube. Insert a thermometer in solid wax, so that bulb of thermometer remains in wax and does not touch the sides. Clamp the test tube along with thermometer in hot bath i.e. in water contained in the beaker and set up the apparatus as shown. Note the temperature Heat the beaker over the flame of burner and record the temperature after every minute. First temperature rises and then reaches 55 °C and wax shines in the test tube. Temperature remains constant for nearly 5 minutes i.e. at 55 °C. This means Wax is melting and temperature remains constant till whole of wax is melted. Then temperature rises again every minute till it reaches
Conclusion : The temperature remains constant at 55°C while changing from solid to liquid. This means 55°C is the melting point and heat is absorbed without change in temperature. This heat is absorbed at constant temperature till whole of wax is melted.

Question 12.
Explain the terms vaporization and boiling point.
Answer:
VAPORIZATION: “Change of liquid to vapours (gas) on heating at constant temperature is called VAPORIZATION.”
When we heat a liquid temperature starts rising till it starts changing into vapours and then temperature remains constant for sometime, through we are supplying heat. This heat supplied is being used to change every molecule of liquid into vapours and temperature does not rise till the whole of liquid is changed into vapours.
BOILING POINT : “The temperature at which a liquid starts changing into vapours or gas at constant temperature is called its BOILING POINT.”

Question 13.
A liquid can change into vapour state
(a) at a fixed temperature, and
(b) at all temperatures
Name the processes involved in two cases.
Answer:
(a) is Boiling point
(b) is Evaporation.
The process involved in two’cases is vaporization or boiling.

Question 14.
Some ice is taken in a beaker and its temperature is recorded after each one minute. The observations are listed below

From the above observations what conclusion do you draw about the melting point of ice ?
Answer:
From the above observations we conclude that ice melts at 0°C during which heat is supplied but temperature does not rise shows that heat supplied is used to change every molecule of ice into water and when whole of ice is melted, temperature starts rising.

Question 15.
Describe an experiment to demonstrate that water absorbs heat during boiling at a constant temperature.
Answer:
BOILING POINT OF WATER AT CONSTANT TEMPERATURE:
Take some water in a beaker. Suspend and clamp a thermometer in beaker in water so that bulb of thermometer remains in water without touching bottom and sides of beaker. Supply heat by burner and note the temperature at room temperature (20°C nearly)

Record the temperature after evefy minute. Temperature rises and as it reaches 100°C water starts boiling. Though heat is being supplied temperature does not rise i.e. it remains constant at 100°C and bubles formed are seen. Thus, boiling point of water is 100°C and at boiling point heat supplied is absorbed by water at constant temperature. Because this heat is being used to change every molecule of water into vapours

Question 16.
State (a) the melting point of ice, and (b) the boiling point of water.
Answer:
(a) MELTING POINT OF ICE: “Is the constant temperature at which it starts (melting) changing from ice to water.”
It is 0°C for ice.
(b) BOILING POINT OF WATER : “Is that constant temperature at which water starts (BOILING) changing from water to steam (vapours)”.
It is 100°C for water.

Question 17.
What is evaporation ?
Answer:
EVAPORATION : “The change of state of a liquid to vapour at all temperatures from the surface of liquid is called evaporation.”

Question 18.
State three factors which affect the rate of evaporation of a liquid.
Answer:
Three factors on which affect the rate of evaporation of a liquid:
(i) AREA OF EXPOSED SURFACE.
(ii) TEMPERATURE OF LIQUID.
(iii) NATURE OF THE LIQUID.
(iv) PRESENCE OF HUMIDITY.

Question 19.
Wet clothes dry more quickly on a warm dry day than on a cold humid day. Explain.
Answer:
Rate of evaporation is directly proportional to temperature. Thus, rate of evaporation is higher on warm day i.e. hot day than cold day having low temperature and clothes dry soon on warm day.

Question 20.
Water in a dish evaporates fasterjhan in a bottle. Give reason.
Answer:
Rate of evaporation is more when the area of exposed surface is more. As area exposed in a dish is more, evaporation is also more.

Question 21.
Why are volatile liquids such as alcohol and spirit stored in tightly closed bottles ?
Answer:
Rate of evaporation depends on NATURE OF LIQUID i.e. more volatile liquids like ALCOHOL and SPIRIT evaporate easily, hence these are stored in tightly closed bottles to avoid their evaporation.

Question 22.
A certain quantity of water is heated from 20°C to 100°C. Its temperature is recorded after each 1 minute. The observations are:

What conclusion do you draw from the above table about the boiling point of water ? Explain.
Answer:
From the table given above we note that as thermometer shows 100°C, it becomes constant and through heat is being supplied. This means boiling point of water is 100°C and heat supplied is being used to convert every molecule of water into vapours (steam) till whole of the water gets boiled off.

Question 23.
Why is cooling produced on evaporation of a liquid ?
Answer:
For evaporation of a liquid it requires HEAT. This heat is taken from the surroundings like body or palm or fore-head or finger and its temperature falls and we feel cool.

Question 24.
Explain with an example to demonstrates that when a liquid
evaporates, it takes heat from its surroundings.
Answer:
If some spirit is poured on cotton wrapped around the bulb of a thermometer, the reading of the thermometer falls. This shows that cooling is produced when a liquid evaporates taking heat from surroundings.

Question 25.
Give two applications of evaporation.
Answer:
Two APPLICATIONS OF EVAPORATION:
(i) When we sprinkle water on the roads in summer evening, water evaporates by taking heat from the road and produces coolness in the surroundings and it becomes pleasant.
(ii) After taking a bath in summer when we come out of water, water evaporates taking heat from our body. The temperature of body falls and we feel refreshed.

Question 26.
Explain why in hot summer days water remains cool in earthen pots.
Answer:
Water seeps out through the pores in the earthen pot and it evaporates. The latent heat required for evaporation is taken, from water inside the~pot which gets cooled.

Question 27.
A patient suffering from high fever is advised to put wet clot strips on his forehead. Why ?
Answer:
Water in wet’ strips evaporates taking latent heat required for evaporation from the forehead. The temperature of forehead (body of the patient) falls and he feels relieved.

Question 28.
What do you mean by sublimation ? Explain with an example.
Answer:
SUBLIMATION : “Change of solid state of matter directly on heating to vapour state (without becoming liquid) and on cooling vapours to solid is called sublimation

Question 29.
Why does the size of naphthalene balls decrease when left open ?
Answer:
When naphthalene balls are left open, due to sublimation they change to vapours and their size decreases.

Question 30.
Describe an experiment to demonstrate the process of sublimation.
Answer:
Experiment: Take some Ammonium chloride powder in a china dish. Cover the china dish with inverted funnel and put a cotton plug in end of funnel so that vapours do not escape. Set up the apparatus as shown. Heat the dish with burner. Solid ammonium chloride changes into vapours. Which when come in contact of walls of funnel get cooled and change to solid and get deposited there.

Selina Concise Physics Class 8 ICSE Solutions – Energy

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 4 Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions For Class 8 Physics Chapter – 4 – Energy

• When we are pushing a wall we are not doing any work as the position of wall is not change i.e. wall has not moved in the direction of force.
• WORK: “is said to be done if on applying force on a body, the body moves (or changes it position) from it place in the direction of force. W = F × d
  Or
  “Work is said to be done by a force applied on a body, if it changes its size or shape.”
•  FACTORS AFFECTING THE AMOUNT OF WORK DONE : W = F × d
  (i) Magnitude of force applied.
  (ii) Distance moved by the body in the direction of force.
  UNIT OF WORK : W = F × d
  s.i  unit W= 1N × 1m = Nm = joule (J)
  1 kgf = 9.8 N is force on 1 kg ∴ F = mg
  Work done – 1 kgf × m = 9.8 N m = 9.8 J = 10 J nearly
• A cooli standing with a box on his head, does no work as distance moved is zero. ~
•  A cooli with a box on his head and walking is doing no work as force is acting vertically downward and direction of motion is at right angle.
•  ENERGY: “is capacity of doing work.”
  Or
  “The work done on a body in changing its state is called energy.”
  s.1. unit of energy = S.I. unit of work = (J)
•  JOULE: “A body is said to possess a energy of one joule. If a force of 1 Newton moves the body by a distance of 1 metre in the direction of force.”
• MECHANICAL ENERGY: “The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
•  Potential energy and kinetic energy are mechanical energies.
•  POTENTIAL Energy (P.E.) : “Is energy possessed by body due to its state of rest or position.” P.E. = mgh
•  KINETIC Energy (K.E.) : “Is energy possessed by body due to its motion.”
  K.E. = 1/2 M V²
• GRAVITATIONAL POTENTIAL ENERGY: “When a stone or water is raised (lifted) from ground to a height, work is done against the force of gravity. This work is stored in the stone or water in the form of GRAVITATIONAL POTENTIAL ENERGY.”
• A stretched bow, due to change in position possesses potential energy. When stretched bow is relreased the arrow comes in motion and due to motion possesses the kinetic energy and hits the body on which it strikes.
•  When a body at a hight, it possess P.E. = mgh. When it falls, height decreases and speed increases
  ∴ its P.E. decreases and K.E. increases.
• Powder : “Rate of doing work”. P = W/t
  

Test your self

A.Objective Questions

1. Write true or false for each statement

(a) A coolie does no work against the force of gravity while carrying a luggage on a road.
Answer. True.

(b) The energy stored in water of a dam is the kinetic energy.
Answer. False.
The energy stored in water of a dam is the potential energy.

(c) The energy of a flying kite is kinetic energy.
Answer. True.

(d) Work done by a boy depends on the time in which he does work.
Answer. False.

(e) Power spent by a body depends on the time for which it does work.
Answer. True.

2. Fill in the blanks

(a) Work is said to be done by a forte only when the body moves.
(b) Work done = Force × distance moved in direction of force.
(c) The energy of a body is its capacity to do work.
(d) The S.I. unit of energy is joule.
(e) The potential energy of a body is due to its state of rest or position and kinetic energy of body is due to its state of motion.
(f) Gravitational potential energy U = mass × force of gravity on unit mass × height.
(g) Kinetic energy = 1/2 × mass × (speed)²
(h) Power P=work done/time taken.
(i) The S . i.  unit of power is watt
(j) I H.P. = 746 W

3. Match the following

4. Select the correct alternative

(a) The S.I. unit of work is

1. second
2. metre
3. joule
4. newton

Answer:
joule

(b) No work is done by a force if the body

1. moves in direction of force
2. does not move
3. moves in opposite direction
4. none of the these

Answer:
does not move

(c) Two coolies A and B do some work in time 1 minute and 2 minute respectively. The power spent is

1. same by both coolies
2. is more by coolie A than by B
3. is less by coolie A than by B
4. nothing can be said.

Answer:
is more by coolie A than by B

(d) The expression of power P is

1. P = mgh
2. P = P = 1/2 Mv²
3. P = F × d
4. P = F × d/t

Answer:
P = F × d/t

(e) I H.P. ¡s equal to

1. 1 W
2.  1 J
3.  746 J
4.  746 W

Answer:
 746 W

(f) When a boy doubles his speed, his kinetic energy becomes

1. half
2. double
3. four times
4. no change

Answer:
four times

(g) A boy lifts a luggage from height 2 m to 4 m. The potential energy will become

1. half
2. double
3. one-third
4. one-fourth

Answer:
double

Selina Concise Physics Class 8 ICSE Solutions – Heat Transfer

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 6 Heat Transfer. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Physics Chapter 6 Heat Transfer

•  Heat is a form of energy. When two bodies are in contact heat flows from body at higher temperature to body at lower temperature till the lower temperature of both is same.
•  When a body is heated, its molecules move faster about their means position and kinetic energy increases and with fall in temperature their K.E. decreases.
•  When a substance is heated
  (i) It expands i.e. a change in size takes place
  (ii) Change in temperature takes place.
  (iii) Change in state takes place.
• CHANGE OF STATE : “The process of change from one state to another at a constant temperature is called change of state.”
• Solid on heating changes into LIQUID. LIQUID on absorbing heat changes to VAPOURS some SOLIDS on heating DIRECTLY change in vapours called SUBLIMATION. Substance is called SUBLIMATE.
  SOLIDIFICATION on cooling when a vapours change into SOLID. GAS OR VAPOURS on cooling \(\xrightarrow { Condensation }\) changes to LIQUID also called LIQUIFACTION.
•  MELTING: Change of solid into liquid at constant temperature. FUSION ⇒ FREEZING is change of LIQUID into SOLID at constant temperature and change of solid into liquid at a constant _ temperature is called FUSION.
•  EVAPORATION: “Change liquid to gas at ALLTEMPERATURES” It is surface phenomenon. “
• VAPOURIZATION : “Change of liquid into vapours at fixed temperature”.
•  METING POINT: “Is the temperature at which a solid starts melting and remains constant till the whole of solid melts.”
  M.P. is same as freezing point.
  M.P. of ice is 0°C or freezing point, of water is 0°C.
•  BOILING POINT: “Is the temperature of a liquid at which it start, boiling i.e. change into vapours or gaseous state.”
  B .P. of pure water is 100°C.
• ABSOLUTE ZERO: “The temperature at which molecular motion completely ceases.”
•  FACTORS EFFECTING THE RATE OF EVAPORATION :
  (i) Temperature: Increases with increase in temperature
  (ii) S.A.: Increases with increase in S.A.
  (iii) BLOWING AIR—Renewal of air increases evaporation.
  (iv) NATURE—Some liquids like spirit, alcohol, petrol evaporate easily.
•  EVAPOURATION → produces coolness, BOILING produces Hotness.
•  LINEAR EXPANSION: When a solid rod (metal) is heated change in length takes place, which depends upon
  (i) original length (L₀)
  (ii) Increase in temperature
  (iii) Material of rod.
  Let L0 be the original length at 0°C, when heated to T°C final length becomes L
  Increase in length (L_t – L₀) a L₀ (T – 0)
  Or
  Coefficient of linear expension a which depends upon material of rod.
  L_t– L₀ = L₀ α T
  α = L_t – L₀ / L₀ T = increase in length / original length × Rise in temperature
•  When a metal plate is heated, change in area takes place and the expansion is called SUPERFICIAL expansion.
•  When a solid of volume v₀ is heated change in volume called cubical expansion takes place.
• α : β : γ = 1 : 2 : 3

Test your self

A. Objective Questions

1. Write true or false for each statement

(a) Evaporation is rapid on a wet day.
Answer. False.

(b) Evaporation takes place only from the surface of liquid.
Answer. True.

(c) All molecules of a liquid take part in the process of evaporation.
Answer. False.

(d) Temperature of a liquid rises during boiling or vaporization
Answer. False.

(e) All molecules of a liquid take part in boiling.
Answer. True.

(f) Boiling is a rapid phenomenon.
Answer. True.

(g) All solids expand by the same amount when heated to the same rise in temperature.
Answer. False.

(h) Telephone wires are kept tight between the two poles in winter.
Answer. True.

(i) Equal volumes of different liquids expand by the different amount when they are heated to the same rise in temperature.
Answer. True.

(j) Solids expand the least and gases expand the most on being heated.
Answer. True.

(k) A mercury thermometer makes use of the property of expansion of liquids on heating.
Answer. True.

(l) Kerosene contracts on heating.
Answer. False.

2. Fill in the blanks 

(a) Boiling occurs at a fixed temperature.
(b) Evaporation takes place at all temperature.
(c) The molecules of liquid absorb heat from surroundings in evaporation.
(d) Heat is absorbed during boiling.
(e) Cooling is produced in evaporation.
(f) A longer rod expands more than a shorter rod on being heated to the same temperature.
(g) Liquids expand more than the solids.
(h) Gases expand more than the liquids.
(i) Alcohol expands more than water.
(j) Iron expands less than copper.

3. Match the following

4. Select the correct alternative 

(a) In evaporation

1. all molecules of liquid begin to escape out
2.  only the molecules at the surface escape out
3.  the temperature of liquid rises by absorbing heat from surroundings.
4.  the molecules get attracted within the liquid.

(b) The rate of evaporation of a liquid increases when :

1.  temperature of liquid falls
2.  liquid is poured in a vessel of less surface area
3.  air is blown above the surface of liquid
4.  humidity increases.

(c) During boiling or vaporization

1.  all molecules take part
2.  temperature rises
3.  no heat is absorbed
4.  the average kinetic energy of molecules increases.

(d) The boiling point of a liquid is increased by

1.  increasing the volume of liquid
2.  increasing the pressure, on liquid
3.  adding ice to the liquid
4.  decreasing pressure on liquid.

(e) Two rods A and B of the same metal, but of length 1 m and 2 m respectively, are heated from 0°C to 100°C. Then

1.  both the rods A and B elongate the same
2.  the rod A elongates more than the rod B
3.  the rod B elongates more than the rod A
4.  the rod A elongates, but the rod B contracts.

(f) Two rods A and B of the same metal, same length, but one solid and the other hollow, are heated to the same rise in temperature.
Then

1.  the solid rod A expands more than the hollow rod B
2.  the hollow rod B expands more than the solid rod A
3.  the hollow rod B contracts, but the solid rod A expands
4.  both the rods A and B expand the same.

(g) A given volume of alcohol and the same volume of water are heated from the room temperature to the same temperature then.

1.  alcohol contracts, but water expands
2.  water contracts, but alcohol expands
3.  water expands more than alcohol
4.  alcohol expands more than water.

(h) The increase in length of a metal rod depends on

1.  the initial length of the rod only
2.  the rise in temperature only
3.  the material of rod only
4.  all the above three factors.

(i) The correct statement is

1.  Iron rims are cooled before they are placed on the cart wheels.
2.  A glass stopper gets tighten on warming the neck of the bottle.
3.  Telephone wires sag in winter, but become tight in summer.
4.  A little space is left between two rails on a railway track.

Selina Concise Physics Class 8 ICSE Solutions – Electricity

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 8 Electricity. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Physics Chapter 8  Electricity

•  ELECTRICITY: “Is the rate of flow of electrons”. i = \(\overset { q }{ \underset { t }{ – } } \)
•  To keep electrons move, potential difference is needed. This is done by a cell or battery.
•  Potential difference “is the amount of work done in moving a unit positive charge from one point to other.”
  Potential difference V = Work done (W) / charge moved (Q)
  Or W = QV but charge = it
  Hence, W = VIt or Electrical energy = VIt
•  Power is “Rate of doing work”
  P = W/t = VIt/t = VI
  Power is measured in watt or J S^-1
  1 Watt = 1 Volt × 1 Ampere
• S.I. unit of charge is coulomb (C).
  S.I. unit of current is Ampere (A).
  S.I. unit of P.D. is volt.
  S.I. unit of electrical energy is Joule (J) and of power is watt (W)
  1 kWh-3600000J = 3.6 x 10⁶ J
•  ELECTRIC power is generated at the GENERATING STATION at 11000 volt, or 11 kV as these stations are at very far off place from areas where it is to be used. The voltage (A.C.) is of 50 HZ frequency.
• AT GRID SUB-STATION this alternating current (A.C) voltage is stepped up from 11 kV to 132 kV to minise the loss of energy in transmission line wires.
• At MAIN-SUB-STATION this voltage is stepped down from 132 kV to 33 kV and transmitted to city SUB-STATION.
•  At CITY SUB STATION, it is further stepped down from 33 kV to 220 V for supply to hourses for consumers.
•  Colour coding: Live wire — Red or Brown
  Neutral—Black or light blue
  Earth wire — Green or yellow
•  1 kWh = 1 unit: Power Rating on an appliance 100 W – 220 V means the appliance when worked on a 220 V will consume 100 W electricity power
• OVER LOADING: is the condition of Electric circuit, when it draws more current than it is designed for i. e. when a number of appliances are switched on at a time i.e. geyser, A.C. Electric motor etc. or a large number of plugs are put in the same socket.
•  EARTHING: is done in a house near the kWh meter. Earthing is a safety device which puts the appliance at zero potential.
•  SHORT CIRCUITING: If the insulation on the wire of cable used f in the wiring (or used with an appliance) breaks. The LIVE WIRE
  COMES IN CONTACT WITH THE NEUTRAL WIRE, this result in SHORT CIRCUITING
•  FUSE: “Is a device used to limit the current in an electric circuit”. The use of fuse protects the appliance in circuit from being damaged Fuse is always connected in live wire. A fuse wire should have
  (i) High resistance
  (ii) Low melting point.
  These days miniature circuit breakers (MCB) are used. It is AUTOMATIC breaker, when current flowing excess.
•  Appliances in a house are connected in parallel.

Test yourself

A. Objective Questions

1. Write true or false for each statement:

(a) A fuse wire has a high melting point.
Answer. False.

(b) Flow of protons constitutes electric current.
Answer. False.

(c) Silver is an insulator of electricity.
Answer. False.

(d) S.I. unit and commercial unit of electrical energy are same.
Answer. False.

(e) Overloading of electric current in circuits can lead to short circuiting.
Answer. True.

(f) Our body can pass electricity through it.
Answer. True.

(g) All metals are insulators of electricity.
Answer. False.

(h) The earth wire protects us from an electric shock.
Answer. True.

(i) A switch should not be touched with wet hands.
Answer. True.

(j) AH electrical appliances in a household circuit work at the same voltage.
Answer. True.

(k) In a cable, the green wire is the live wire.
Answer. False.

(l) A fuse is connected to the live wire.
Answer. True.

(m) A switch is connected to the neutral wire.
Answer. False.

2. Fill in the blanks

(a) The unit in which we pay the cost of electricity is kWh.
(b) The electrical energy consumed in a house is measured by kWh meter.
(c) In a household electrical circuit, the appliance are connected in parallel with the mains.
(d) A switch is connected to the live wire.
(e) The red colour insulated wire in a cable is the live wire.
(f) One kilowatt hour is equal to 3.6 x 106 joule.
(g) A fuse wire should have low melting point.

3. Match the following

4. Select the correct alternative

(a) All wires used in electric circuits should be covered with

1.  colouring material
2.  conducting material
3.  an insulating material
4.  none of the above

(b) Electric work done per unit time is

1.  electrical energy
2.  electric current
3.  electric voltage
4.  electrical power

(c) One kilowatt ¡s equal to

1.  100 watt
2.  1000 watt
3.  10 watt
4.  none of these

(d) Fuse wire is an alloy of

1.  tin-lead
2.  copper-lead
3.  tin-copper
4.  lead-silver

(e) A fuse wire should have

1.  a low melting point
2.  high melting point
3.  very high melting point
4.  none of the above

(f) When switch of an electric appliance is put off, it disconnects

1.  the live wire
2.  the neutral wire
3.  the earth wire
4.  the live and the neutral wire

(g) The purpose of an electric meter in a house is

1.  to give the cost of electricity directly
2.  to give the consumption of electrical energy
3.  to safeguard the circuit from short circuiting
4.  to put on or off the mains.

(h) If out of the two lighted bulbs in a room, one bulb suddenly fuses, then

1.  other bulb will glow more
2.  other bulb will glow less
3.  other bulb will also fuse
4.  other bulb will remain lighted unaffected.

Selina Concise Physics Class 8 ICSE Solutions – Force and Pressure

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 3 Force and Pressure. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Physics ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions For Class 8 Physics Chapter – 3 – Force and Pressure

• FORCE : “Is the cause which changes the state of a body (rest or
  state of motion) or it changes the size or shape of a body”.

Weight of a body → The force with which a body is attracted towards the centre of earth mg = force of gravity.

• A force does not change the mass of the body that is why mass of a
  body on earth and moon in same but weight → force exerted on
  body is different.
•  Force cannot be seen but it is felt.
• → represents force, length of arrow gives magnitude and arrow points the direction.
•  S.I. unit of force is Newton (N).
• Newton: “Is that much force, which when acting on a body of mass 1 kg produces in it (increases) a speed of 1 M s^-1 in the direction of its motion.
• 1 kgf = 9.8 N = 10 N (nearly)
• RIGID body: “When a force is applied on a body and inter-spacing between its constituent particles do not change is called RIGID body” force can cause only the motion in it.
•  NON-RIGID body: “When force applied changes inter-spacing.” Force causes both change in its size (shape) and the motion in body.
• TURNING EFFECT: “When force is applied on a pivoted (at a point) body, it can turn it and turning of body about point of rotation is called TURNING EFFECT.” or Moment of force.
  This is measured as:
  TURNING EFFECT = MOMENT OF FORCE
  Force x perpendicular distance from point of rotation.
  
  Moment of force = F x OP
  S.I. unit of moment of force=N × m=Nm
• THRUST: “Force acting normally on a surface.” Smaller the area of surface, larger is thrust.
• PRESSURE : “Thrust per unit area.p = Thrust/area = F/A S.I unit of area A pressure is Nm^-2 or pascal (Pa)
•  If Thrust is measured in kgf and area in Cm², then pressure is expressed as kgf Cm^-2.
  ATMOSPHERIC Pressure: 1 atm = 76 cm of mercury column 1 atm = 1.013 x 10⁵ Pa
•  FACTORS AFFECTING THE PRESSURE : P = F/A
  (i) Area: Greater the area, lesser is pressure and lesser area, greater is pressure.
  (ii) Magnitude of thrust acting: greater thrust, greater pressure.
• Factors Affecting LIQUID PRESSURE = hdg
  (i) High of liquid column: increases with height
  (ii) Density of liquid: increases with density of liquid.
  (iii) Gravity constant.
• ATMOSPHERIC PRESSURE : “Pressure exerted by the air of atmosphere around us.”

STANDARD ATMOSPHERIC PRESSURE 1 Atm = 76 cm of Hg column = 1.013 x 10⁵ Pa

 

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) The S.I. unit of force is kgf.
Answer. False.
The S.I. unit of force is newton.

(b) A force always produces both the linear and turning motions.
Answer. False.

(c) Moment of force = force × perpendicular distance of force –  from the pivoted point.
Answer. True.

(d) Less force is needed when applied at a farther distance from the pivoted point.
Answer. True.

(e) For a given thrust, pressure is more on a surface of large j area.
Answer. False.
For a given thrust, pressure is less on a surface of large area.

(f) The pressure on a surface increases with an increase in the thrust on the surface.
Answer. True.

(g) A man exerts same pressure on the ground whether he is standing or he is lying.
Answer. False.
A man exerts different pressure on the ground whether he is  standing or he is lying.

(h) It is easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.
Answer. False.
It is not easier to hammer a blunt nail into a piece of wood than a sharply pointed nail.

(i) The S.I. unit of pressure is pascal.
Answer. True.

(j) Water in a lake exerts pressure only at its bottom.
Answer. False.

(k) A liquid exerts pressure in all directions.
Answer. True.

(l) Gases exert pressure in all directions.
Answer. True.

(m) The atmospheric pressure is nearly 10⁵ Pa.
Answer. True.

(n) Higher we go, greater is the air pressure.
Answer. False.

2. Fill in the blanks

(a) 1 kgf = 10 N (nearly).
(b) Moment of force = force × distance of force from the point of turning
(c) In a door, handle is provided farthest from the hinges.
(d) The unit of thrust is newton .
(e) Thrust is the normal force acting on a surface.
(f) Pressure is the thrust acting on a surface of unit area.
(g) The unit of pressure is pascal
(h) Pressure is reduced if area of surface increases.
(i) Pressure in a liquid increases with the depth.
(j) The atmospheric pressure on earth surface is nearly 10⁵ Pa.

3. Match the following

4. Select the correct alternative

(a) SI. unit of moment of force ¡s

1. N
2.  N cm
3.  kgfm
4.  N m

(b) To obtain a given moment of force for turning a body, the force needed can be decreased by

1. applying the force at the pivoted point
2.  applying the force very close to the pivoted point
3.  applying the force farthest from the pivoted point
4.  none of the above

(c) The unit of thrust is

1.  kgf
2.  kg
3.  g
4. m s^-1

(d) The unit of pressure is

1.  N × m
2.  kgf
3.  N m^-2
4.  kgf m²

(e) The pressure and thrust are related as

1.  Pressure = Thrust
2.  Pressure = Thrust x Area
3.  Pressure = Thrust / Area,
4.  Pressure = Area / Thrust

(f) A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a thrust on the surface equal to

1.  50 kgf
2.  5 kgf
3.  50 kgf  m^-2
4.  5 kgf  m^-2

P.Q. A body weighing 5 kgf, placed on a surface of area 0.1 m2, exerts a pressure on the surface equal to

1.  50 kgf
2.  5 kgf
3.  50 kgf m^-2
4.  5 kgf m^-2

(g) The feet of lizards act like

1.  moving pads
2.  drilling pads
3.  suction pads
4.  none of the above

(h) Pressure exerted by a liquid is due to its

1.  weight
2.  mass
3. volume
4.  area

(i) Pressure inside a liquid increases with :

1.  increase in depth
2.  decrease in depth
3.  decrease in density
4.  none of the above

(j) The atmospheric pressure at sea level is nearly

1.  10 Pa
2.  100,000 Pa
3.  100 Pa
4.  10,000 Pa

(k) Nose bleeding may occur at a high altitude because

1.  the atmospheric pressure decreases
2.  the oxygen content of atmosphere decreases
3.  the atmospheric pressure increasess
4.  there are strong air currents at the high altitude

Selina Concise Chemistry Class 8 ICSE Solutions – Language of Chemistry

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 5 Language of Chemistry. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 5 Language of Chemistry

Points to Remember:

• The valency of an element is the number of electrons donated or accepted by its ‘atom’ during chemical combination.
• There are some elements with more than one valency e.g., iron, copper, tin, lead.
• Two or more different non-metals that collectively accept or donate one or more electrons and become negatively or positively charged in the process are called radicals.
• A chemical reaction involves the transformation of original substance into an altogether new substance(s).
• A chemical reaction can be represented with the help of the symbols or the formulae of the elements and the compounds taking part in that reaction. This gives a chemical equation.
• Certain necessary conditions for a chemical reaction to happen are — close contact, solution form, heat, light and catalyst.
• Characteristics of chemical reactions are — change of colour, evolution of a gas, formation of a precipitate, change of state, change of smell and evolution/absorption of heat.
• A complete chemical equation symbolically represents the reactants, products and their physical states.
• The substances that react with each other are called reactants and they are represented on the left hand side of the equation. The substances that are formed as a result of the reaction are called products. They are represented on the right hand side of the equation.
• A chemical equation needs to be balanced to make it follow the law of the conservation of mass.
• The law of conservation of mass states that mass can be neither created nor destroyed, it can only be transformed from one form to another.
• A chemical equation gives both qualitative and quantitative information about the reactants and products.

ACTIVITY 1
Write the names and symbols of the first twenty elements that you have studied in class VI & VII.
Answer:

ACTIVITY 2
Write the molecular formulae of:

1. Copper oxide
2. Iron (III) chloride
3. Sodium hydroxide
4. Iron (II) sulphide
5. Lead (II) oxide
6. Hydrogen nitrate (nitric acid)
7. Hydrogen sulphate (sulphuric acid)
8. Calcium hydroxide
9. Magnesium carbonate
10. Ammonium carbonate

Answer:

1. Copper oxide – CuO
2. Iron (III) chloride – FeCl₃
3. Sodium hydroxide – NaOH
4. Iron (II) sulphide – FeS
5. Lead (II) oxide – PbO
6. Hydrogen nitrate (nitric acid) – HNO₃
7. Hydrogen sulphate (sulphuric acid) – H₂SO₄
8. Calcium hydroxide – Ca(OH)₂
9. Magnesium carbonate – MgCO₃
10. Ammonium carbonate – (NH₄)₂CO₃

ACTIVITY 3
Write the molecular formula for each of the following compounds:

1. Sulphur trioxide
2. Iron (II) sulphide and
3. Ammonia

Find the number and names of elements present in them and calculate their molecular masses.
Answer:
1. Sulphur trioxide

1. A molecule of sulphur trioxide is represented by the formula SO₃.
2. The elements present in it are sulphur dioxide and oxygen.
3. One molecule of sulphur trioxide has one atom of sulphur and three atoms of oxygen.
4. Molecular mass of sulphur trioxide (SO₃)
   = 32 + 3 x 16
   = 32 + 48 = 80 amu.

2. Iron (II) sulphide

1. A molecule of iron (II) sulphide is represented by the formula FeS.
2. The elements present in it are iron and sulphur.
3. One molecule of iron (II) sulphide has one atom of iron and one atom of sulphur.
4. Molecular mass of iron (II) sulphide (FeS)
   = 55.5 + 32
   = 87.5 amu.

3. Ammonia

1. A molecule of ammonia is represented by the formula NH₃.
2. The elements present in it are nitrogen and hydrogen.
3. One molecule of ammonia has one atom of nitrogen and three atoms of hydrogen.
4. Molecular mass of ammonia (NH₃)
   = 14 + 3 x 1
   = 14 + 3
   = 17 amu.

Exercise

Question 1.
Define:
(a) Radical
(b) Valency
(c) Molecular formula
Answer:
(a) Radical: A radical is an atom of an element or a group of atoms of different elements that behaves as a single unit with a positive or negative charge on it.
(b) Valency: It is the number of electrons donated or accepted by the valence shell of an atom during chemical combination.
(c) Molecular formula: It is a symbolic representation of a molecule. It shows the number of atoms of each element present in it. These atoms combine in the whole numbers to form the molecule.

Question 2.
Give the symbols and valencies of the following radicals:
(a) Hydroxide (b) Chloride
(c) Carbonate (d) ammonium
(e) Nitrate
Answer:

Question 3.
Write the molecular formula for the oxide and sulphide of the following elements.
(a) Sodium (b) Calcium
(c) Hydrogen
Answer:
(a) Sodium oxide Na₂O
Sodium sulphide Na₂S
(b) Calcium oxide CaO
Calcium sulphide CaS
(c) Hydrogen oxide H₂O
Hydrogen sulphide H₂S

Question 4.
Write the molecular formulae for the following compounds and name the elements present.
(a) Baking soda (b) Common salt
(c) Sulphuric acid (d) Nitric acid
Answer:
(a) Baking soda — NaHCO₃
Elements present in Baking soda are sodium, hydrogen, oxygen and carbon.
(b) Common salt — NaCl
Element present are: Sodium and chlorine.
(c) Sulphuric acid — H₂SO₄
Element present are: Hydrogen, sulphur and oxygen.
(d) Nitric acid — HNO₃
Elements present are: Hydrogen, nitrogen and oxygen.

Question 5.
The valency of aluminium is 3. Write the valency of other radicals present in the following compounds.
(a) Aluminium chloride
(b) Aluminium oxide
(c) Aluminium nitride
(d) Aluminium sulphate
Answer:
(a) Aluminium chloride — (AlCl₃) here valency of Al is 3.
Other radical – Chloride (Cl^–)
Valency of chloride = 1
(b) Aluminium oxide — (Al₂O₃)
Here valency of Al is 3
Other radical presents = oxide (O^2-)
Valency of O^2- = 2
(c) Aluminium nitride — (Al N)
Here valency of aluminium = 3
Another radical = Nitride (N^3-)
Valency of nitride (N^3-) = 3
(d) Aluminium sulphate — Al₂(SO₄)₃
Here valency of aluminium is 3
Another radical = Sulphate (SO₄^2-)
Valency of (SO₄2-) = 2

Question 6.
What is variable valency? Give two examples of elements showing variable valency.
Answer:
Certain elements exhibit more than one valency, which means they show variable valency.
Ferrous is written as Iron (II) and Ferric is written as Iron (III).

Metal	Radicals	Valency
Iron	Ferrous [Iron (II)] Ferric [Iron (III)]	2 3
Copper	Cuprous [Copper (I)] Cupric [Copper (II)]	1 2

Question 7.
(a) What is a chemical equation?
(b) Why it is necessary to balance a chemical equation?
(c) What are the limitations of a chemical equation?
Answer:
(a) Chemical Equation— A chemical equation is the symbolic representation of a chemical reaction using the symbols and the formulae of the substances involved in the reaction.
(b) A chemical equation needs to be balanced so as to make the number of the atoms of the reactants equal to the number of the atoms of the products.
(c)

1. It does not inform about the physical states of the reactants and the product i.e. whether they are solids, liquids, and gases.
2. It does not inform about the concentration of reactants and products.
3. It does not inform about the time taken for the completion of the reaction.
4. It does not inform about the rate at which a reaction proceeds.
5. It does not inform about the heat changes during the reaction i. e. whether the heat is given out or absorbed.
6. It does not inform about the conditions such as temperature, pressure, catalyst, etc. which affect the reaction.
7. It does not inform about the nature of the reaction i.e. whether it is reversible or irreversible.

Question 8.
What are the ways by which a chemical equation can be made more informative?
Answer:
A chemical equation can give more information in the following ways:

1. The physical state of the reactants and products can be indicated by putting (s) for solid, (l) for liquid, (g) for gas and (aq) for aqueous state.
2. Evolution or absorption of heat during the reaction can be denoted by adding or subtracting a heat term on the product side.
3. Temperature, pressure and catalyst can be indicated above the arrow (→ or =) separating the reactants and products.
4. Concentration of reactants and products are indicated by adding word (dil) for dilute and (cone) for concentrated before their formulae.
5. By the sign → or \(\rightleftharpoons \) information about irreversible and reversible reactions can be obtained.

Question 9.
State the law of conservation of mass.
Answer:
Law of conservation of mass: It states that mass can neither be created nor destroyed in a chemical reaction. During any change (physical or chemical), matter is neither created nor destroyed. However it may change from one form to another.

Experimental Verification of Law of Conservation of Mass

Requirements: H-shaped tube called Landolt’s tube, Sodium chloride solution, silver nitrate solution, etc.
Procedure: A specially designed H-shaped tube is taken. Sodium chloride solution is taken in one limb of the tube and silver nitrate solution in the other limb as shown in the figure.
Both the limbs are now sealed and weighed. Now the tubes is averted so that the solutions can mix up together and react chemically. The reaction takes place and a white precipitate of silver chloride is obtained.

The tube is weighed again. The mass of the tube is found to be exactly the same as the mass obtained before inverting the tube.
Thus, this experiment clearly verifies the law of conservation of mass.

Question 10.
Differentiate between:
(a) Reactants and products
(b) A balanced and an unbalanced chemical equation
Answer:
(a) Reactants and products

Reactants

1. The substances that react with one another are called reactants.
2. Reactants are written on the left-hand side of the equation.

Products

1. The new substances formed are called products.
2. Products are written on the right-hand side of the equation.

(b) A balanced and an unbalanced chemical equation

Balanced chemical

1. A balanced chemical reaction is one in which the number of atoms of each element on the reactant side is equal to the number of atoms of that element on the product side.
2. Ex- H₂ + Cl₂ → HCl

Unbalanced chemical

1. The number of elements on the reactant side is not equal to the number of elements on the product side.
2. Ex- H₂ + Cl₂ → 2HCl

Question 11.
Balance the following equations:

Answer:

Question 12.
12. Write balanced chemical equations for the following word equations:
(a) Iron + Chlorine → Iron (III) chloride
(b) Magnesium + dil sulphuric acid → Magnesium sulphate + water
(c) Magnesium + oxygen → Magnesium oxide
(d) Calcium oxide + water → Calcium hydroxide
(e) Sodium + chlorine → Sodium chloride
Answer:
(a) Iron + Chlorine → Iron (III) chloride
4Fe + 3Cl₂ → 2F₂Cl₃
(b) Magnesium + dil sulphuric acid → Magnesium sulphate + water
2Mg + 2H₂SO₄ → 2MgSO₄ + 2H₂
(c) Magnesium + oxygen → Magnesium oxide
2Mg + O₂ → 2MgO
(d) Calcium oxide + water → Calcium hydroxide
CaO + H₂O → Ca(OH)₂
(e) Sodium + chlorine → Sodium chloride
2Na + Cl₂ → 2NaCl

Question 13.
What information do you get from the following chemical equation:
Zn(s) + 2HCl (dil) → ZnCl₂ (aq) + H₂(g)
Answer:
This gives zinc chloride and hydrogen. The word equation is:
Zinc + Hydrochloric acid → Zinc chloride + Hydrogen
Formulae for the products are ZnCl₂ and H₂

Selina Concise Chemistry Class 8 ICSE Solutions – Atomic Structure

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 4 Atomic Structure. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Chemistry ICSE SolutionsPhysicsBiologyMathsGeographyHistory & Civics

Exercise

1. Fill in the blanks.

(a) Dalton said that atoms could not be divided
(b) An ion which has a positive charge is called a cation.
(c) The outermost shell of an atom is known as valence shell.
(d) The nucleus of an atom is very hard and dense.
(e) Neutrons are neutral particles having mass equal to that of protons.
(f) Isotopes are the atoms of an element having the same atomic number but a different mass number.

2. Write ‘true’ or ‘false’ for the following statements:

(a) An atom on the whole has a positive charge.
false
(b) The maximum number of electrons in the first shell can be 8.
false
(c) The central pad of the atom is called nucleus.
True.

3. Give the following a suitable word/phrase.

(a) The sub-atomic particle with negative charge and negligible mass.
(b) Protons and neutrons present in the nucleus.
(c) The electrons present in the outermost shell.
(d) Arrangement of electrons in the shells of an atom.
(e) The number of protons present in the nucleus of an atom.
(f) The sum of the number of protons and neutrons of an atom.
(g) Atoms of same element with same atomic number but a different mass number.
(h) The smallest unit of an element which takes part in a chemical reaction.

Answer:

(a) Neutron
(b) Mass number
(c) Valency
(d) Orbits or Valence shells
(e) Atomic number
(f) Mass number
(g) Isotopes
(h) Atom

4. Multiple Choice Questions

(a) The outermost shell of an atom is known as

1. valency
2. valence electrons
3. nucleus
4. valence shell

(b) The number of valence electrons present in magnesium is

1. two
2. three
3. four
4. five

(c) The sub atomic particle with negative charge is

1. proton
2. neutron
3. electron
4. nucleon

(d) If the atomic number of an atom is 17 and mass number is 35 then number of neutron will be

1. 35
2. 17
3. 18
4. 52

(e) The number of electrons in an atom is equal to number of

1. protons in a neutral atom
2. neutrons in a neutral atom
3. nucleons in a neutral atom
4. none of the above

(f) The sum of number of protons and number of neutrons present in the nucleus of an atom is called its

1. mass number
2. atomic number
3. number of electrons
4. all of the above

Question 5.
Name three fundamental particles of the atom. Give the symbol with charge, on each particle.
Answer:
The fundamental particles of the atom are: electrons, protons and neutrons.

Particle	Symbol	Charge
electron	e–	-1 or 1.602 x 10-19 C. Where -1 represent its one unit negative electrical charge
proton	p+	+ 1 or 1.602 x 10-19 C. Where +1 represents one unit +ve electrical charge.
neutron	no	0

Question 6.
Define the following terms:
(a) Atomic number
(b) Mass number
(c) Nucleons
(d) Valence shell
Answer:
(a) Atomic number: Atomic number refers to the number of protons present in an atom. It is denoted by Z. Example: An atom of oxygen contains 8 proton Therefore its atomic number is 8.
(b) Mass number: Mass number refers to the sum of the number of protons and neutrons present in the nucleus of an atom and denoted by A Mass number = Number of protons + Number of neutrons.
(c) Nucleons: The protons and neutrons collectively are known as nucleons.
(d) Valence Shell: The outermost shell of an atom is known as its valence shell.

Question 7.
Mention briefly the salient features of Dalton’s atomic theory (five points).
Answer:
Salient features of Dalton’s atomic theory:

1. Matter consists of very small and indivisible particles called atoms, which can neither be created nor can be destroyed.
2. The atoms of an element are alike in all respects i.e. size, mass, density, chemical properties but they differ from the atoms of other elements.
3. Atoms of an element combine in small numbers to form molecules of the element.
4. Atoms of one element combine with atoms of another element in simple whole number ratio to form molecules of compounds.
5. Atoms are the smallest units of matter that can take part in a chemical reaction during which only rearrangement of atoms takes place.

Question 8.
(a) What are the two main features of Rutherford’s atomic model?
(b) State its one drawback.
Answer:
(a) According to Rutherford’s model an atom consists of:

1. The centrally located nucleus: The nucleus is a centrally located positively charged mass. The entire mass of the atom is concentrated in it. It is the densest part of the atom. Its size is very small as compare to the atom as a whole.
2. The outer circular orbits: Electrons revolve in circular orbits (shell) in the space available around the nucleus. An atom is electrically neutral i.e., number of protons and electrons present in an atom are equal.

(b) Rutherford’s atomic model could not explain the stability of the atom as it is like a solar system, the sun is at the centre and the planets revolve around it, in an atom the electrons revolve around the centrally located nucleus containing protons.

Question 9.
What are the observations of the experiment done by Rutherford in order to determine the structure of an atom?
Answer:
Following were the observations made by Rutherford:

1. Most of the alpha particle passed straight through the foil without any deflection from their path.
2. A small fraction of them were deflected from their original path by small angles.
3. Only a few particles bounced back.

Question 10.
State the mass number, the atomic number, number of neutrons and electronic configuration of the following atoms.

Also, draw atomic diagrams for them.
Answer:

Question 11.
What is variable valency? Name two elements having variable valency and state their valencies.
Answer:
Variable valency: Some elements exhibit more than one valency. They are said to have variable valency, e.g. Iron, copper, tin, lead.
Iron         Fe       Fe²⁺ or Fe³⁺
Copper   cu        cu⁺ or cu²⁺

Question 12.
The atomic number and the mass number of sodium are 11 and 23 respectively. What information is conveyed by this statement?
Answer:
Atomic number = 11; No of protons = 11
Mass number = 23 = Number of protons + Number of neutrons.
No of neutrons = 23-11 = 12.

Question 13.
Draw the diagrams representing the atomic structures of the following:
(a) Nitrogen (b) Neon
Answer:

Question 14.
Explain the rule with example according to which electrons are filled in various energy levels,
Answer:
The maximum number of electrons that can be present in any shell or orbit of an atom is given by the formula 2n², where n is the serial number of the shell.
Therefore:
K shell, n = 1, no. of electrons = 2 x 1² = 2
L shell, n = 2, no. of electrons = 2 x 2² = 8
M shell, n = 3, no. of electrons = 2 x 3² = 18
N shell, n = 4, no. of electrons = 2 x 4²= 32
Electrons are not accommodated in a given shell, unless the inner shells are filled.
That is, the shells are filled in a stepwise manner.
Example:

Question 15.
The atom of an element is made up of 4 protons, 5 neutrons and 4 electrons. What is its atomic number and mass number?
Answer:
Protons = 4, neutrons = 5, electrons = 4
Atomic number = 4,
Mass number = 4 + 5 = 9

Question 16.
(a) What are the two main parts of which an atom is made of?
(b) Where is the nucleus of an atom situated ?
(c) What are orbits or shells of an atom ?
Answer:
(a)

1. The centrally located nucleus
2. The outer circular orbits.

(b) The nucleus is a centrally located positively charged mass.
(c) The circular orbits (shell present) in the space available around the nucleus on which electrons revolve are called orbits or shells of an atom.

Question 17.
What are isotopes? How does the existence of isotopes contradict Dalton’s atomic theory?
Answer:
Atoms of an element must have the same atomic number, but their mass number can be different due to the presence of a different number of neutrons. These atoms of an element having a different number of neutrons are called groups.
According to Dalton’s theory, all atoms of an element are similar to all respects, for example, they have the same shape, size etc. and have similar physical and chemical properties like mass, density and reactivity. Whereas isotopes of an element have atoms that are similar as they have same number of protons and electrons but differ in the number of neutrons. So, the isotopes have atoms that are not similar in all aspects.

Question 18.
Complete the table below by identifying A, B, C, D, E and F.
Answer:

Selina Concise Physics Class 8 ICSE Solutions – Physical Quantities and Measurement

ICSE Solutions  Selina ICSE Solutions  ML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 2 Physical Quantities and Measurement. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Concise physics Class 8 ICSE SolutionsChemistryBiologyMathsGeographyHistory & Civics

Selina Concise ICSE Solutions For class 8 Physics chapter 2 – Physical Quantities and Measurement

• MASS Is the quantity of matter contained in a body.
  VOLUME is the space occupied by body.
• Equal mass of IRON and cotton, iron will have less volume than cotton.
• Equal volume of Iron and cotton, the mass of iron is more than mass of cotton, because iron denser than cotton.
• DENSITY “Is ratio of mass of substance to volume of substance”
  D = M/V = KG/M³ The SI. unit of density is kg M^-3
• Density of a substance does not change with change in shape or size.
• When a substance is heated it expands and volume increases. Hence density decreases.
  Water has maximum density at 4°C i.e. density of water increases from 0°C to 4°C and decreases above 4°C.
• Volume of substance is measured by formula V = L × B × H or 4/3 or by measuring cylinder.
• Mass is measured by beam balance or spring balance.
• RELATIVE DENSITY of substance is the density compare with water i.e. How many times the substance is DENSER than water. Since density of water is 1  Gcm^-3, so density of a substance in Gcm^-3 = relative density of substance.
  S.I. unit of R.D. > has no units — since it is the ratio of same
  quantities.
• If a substance has density more than liquid it SINKS in the liquid and if the density of substance is LESS than liquid it floats on liquid.
• BUOYANT FORCE “The force exerted by liquid acting vertically
  upward on a body and is equal to the weight of liquid displaced by its immersed part.”
• Weight of body Acting vertically downward. This force has the tendency to sink the body.
•  LAW OF FLOATATION “When a body floats in a liquid, the weight of the liquid displaced by its immersed part is equal to the total weight of the body.” While floating
  wt. of floating body W=wt. of liquid displaced by its immersed part FB i.e. Apparent wt. of floating body is zero.
  

Density of body is greater than density of liquid. The body sinks.

Density of body is equal to the density of liquid. The body float where ever it is left in liquid.

Density of body is less than density of liquid. The body rises to the surface and floats.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) Equal volumes of the two different substances have equal masses.
Answer False.
Equal volumes of the two different substances have different masses.

(b) The density of a piece of brass will change by changing its size or shape.
Answer False.

(c) The density of a liquid decreases with increase in its temperature.
Answer True.

(d) Relative density of water is 1.0.
Answer True.

(e) Relative density of a substance is expressed in g cm-3.
Answer False.
Relative density of a substance has no units.

(f) When a body is immersed in a liquid, the buoyant force experienced by the body is equal to the volume of the liquid displaced by it.
Answer False.
The buoyant force is equal to the weight of the liquid displaced by the immersed part of body.

(g) A body experiences the same buoyant force while floating in watr or alcohol.
Answer True.

(h) A body experiences the same buoyant force when it floats or sinks in water.
Answer False.

(i) A body floats in a liquid when its weight becomes equal to the weight of the liquid displaced by its submerged part. .
Answer True.

(j) A body while floating, sinks deeper in a liquid of low density than in a liquid of high density.
Answer True.

2. Fill in the blanks

(a) 1 kg is the mass of 1000 ml of water at 4°C.
(b) Mass = density x volume.
(c) The S.I. unit of density is Kg m^-3
(d) Density of water is 1000 Kg m^-3.
(e) 1 g cm^-3 = 1000 Kg m^-3.
(f) The density of a body which sinks in water is more than 1000 Kg m^-3.
(g) Abody sinks in a liquid A, butt floats in a liquid B. The density of liquid A is less than the density of liquid B.
(h) A body X sinks in water, but a body Y floats on water. The density of the body X is more than the density of body Y.
(i) The buoyant force experienced by a body when floating in salt¬water is equal to or same that of when floating in pure water.
(j) The weight of a body floating in a liquid is zero.

3. Match the following

4. Select the correct alternative 

(a) The correct relation is

1. Density = Mass x Volume
2. Mass = Density x Volume
3.  Volume = Density x Mass
4. Density = Mass + Volume

(b) The relative density of alcohol is 0.8. Its density is

1.  0.8
2. 800 kg nr3
3.  800 g cm-3
4. 0.8 kg m-3

(c) A block of wood of density 0.8 g cm-3 has a volume of 60 cm3. The mass of block is

1.  60.8 g
2.  75 g
3. 48 g
4. 0.013 g

(d) The density of aluminium is 2.7 g cm-3 and that of brass 8.4 g cm’3. The correct statement is

1.  Equal masses of aluminium and brass have equal volumes
2. The mass of a certain volume of brass is more than the mass of equal volume of aluminium.
3.  The volume of a certain mass of brass is more than the volume of equal mass of aluminium.
4.  Equal volumes of aluminium and brass have equal masses.

(e) A density bottle has a marking 25 mL on it. It means that:

1.  the mass of density bottle is 25 g
2. the density bottle will store 25 ml of any liquid in it
3.  the density bottle will store 25 ml of water, but more volume of liquid denser than water.
4.  the density bottle will store 25 ml of water, but more volume of a liquid lighter than water.

(f) The correct statement is

1.  The buoyant force on a body is equal to the volume of the liquid displaced by it ‘
2. The buoyant force on a body is equal to the volume of the body
3. The buoyant force on a body is equal to the weight of the liquid displaced by it
4.  The buoyant force on a body is always equal to the weight of the body.

(g) A piece of wood floats on water. The buoyant force on wood will be

1.  zero
2.  more than the weight of the wood piece
3. equal to the weight of the wood piece
4. less than the weight of the wood piece.

(h) The weight of a body is more than the buoyant force experienced by it, due to a liquid. The body will

1. sink
2.  float with its some part outside the liquid
3.  float just below the surface of liquid
4. float with whole of its volume above the surface of liquid.

B. Short/Long Ans Questions 

Question 1.
Define the term density of a substance.
Answer:
Density of a substance is defined as “Mass per Unit volume”.

Question 2.
Name the S.I. unit of density. How is it related to g Cm^-3 ?
Answer:
S.I. unit of density is kg M^-3 In C.GS. system unit of mass is g and unit of volume is Cm³, so CGS unit of density is g Cm^-3 (gram per cubic centimetre)
Relationship between S.I. and C.GS. units

Question 3.
The density of brass is 8.4 g cm-3. What do you mean by this statement ?
Answer:
This statement meAns one cubic centimetre volume of brass has mass of 8.4 g.

Question 4.
Arrange the following substances in order of their increasing density:
Iron, Cork, Brass, Water, Mercury.
Answer:
Cork, Water, Iron, Brass, Mercury.

Question 5.
How does the density of a liquid (or gas) vary with temperature?
Answer:
Most of the liquids increase in volume with increase in temperature, but water shows anomalous behaviour. Water has maximum volume at 4°C and maximum density at 4°C.
Actually, when volume increases density decreases and when volume decreases the density increases.
But water when cooled from a high temperature, contracts upto 4°C because volume decreases and expands when cooled further below 4°C and hence density of water increases when it is cooled upto 4°C while decreases when cooled further below 4°C. In other words, the density of water is maximum at 4°C equal to 1 g Cm^-3 or lOOO kg m^-3.

Question 6.
A given quantity of a liquid is heated. Which of the following quantity will vary and how ?
(a) mass, (b) volume and (c) density
Answer:
When a given quantity of liquid is heated
(a) Mass : does not change.
(b) Volume: changes and increases with rise in temperature.
(c) Density : Changes and decreases.
Density = Mass / volume

Question 7.
Describe an experiment to determine the density of the material of a coin.
Answer:
Density = Mass / volume
To find the density of the material of a coin, we need to find its (i) mass—by common beam balance and (ii) Its volume by measuring cylinder.
Measure the mass of coin.
EXPERIMENT – Let the mass of coin shown by beam balance = M (gram) = 50 g (ray)
Measure the vol. of coin.
Initial volume of water = V₁ = 40 ml (say)
Final volume of water
When coin is added in the cylinder=V₂ = 50 ml (say)
Then vol. of coin = V₂ – V₁ = 50 – 40 = 10 ml

Question 8.
Describe an experiment to determine the density of a liquid.
Answer:
To determine the density of a liquid D = M / V
We need to find (i) the vol. of liquid say milk, (ii) mass of liquid.
EXPERIMENT:
(i) To find the mass of milk:
wt. of empty 100 c.c beaker = M₁ g = 70 g (say)
Fill the beaker (half) with milk and weigh again=M₂ g = 116 g (say)
(ii) To find the vol. of milk:
TrAnsfer this milk into measuring cylinder and note the volume V = 40 c.c (say)

Question 9.
What is a density bottle ? How is it used to find the density of a liquid ?
Answer:
DENSITY bottle is a small glass bottle having a glass stopper at its neck. The bottle can store a fixed volume of a liquid. Generally the volume of bottle is 25 ml or 50 ml. Stopper has a narrow hole through it. When bottle is filled with liquid and stopper is inserted, THE EXCESS LIQUID RISES THROUGH THE HOLE and drains out. Thus the bottle will contain the same volume of liquid each time when it is filled. It is used to determine the density of a liquid.

 

Question 10.
Define the term relative density of a substance.
Answer:
RELATIVE DENSITY: “is the ratio of density of a substance to the density of water at 4° C.”
Or
RELATIVE DENSITY “is theratio of mass of the substance to the mass of an equal volume of water at 4° C.”

Question 11.
What is the unit of relative density ?
Answer:
UNIT OF RELATIVE DENSITY: No units since it is a pure ratio.

Question 12.
Distinguish between density and relative density.
Answer:

Question 13.
Explain the meaning of the statement ‘relative density of aluminium is 2.7’ ?
Answer:
The statement ‘Relative density of aluminium is 2.7’ meAns .
A piece of aluminium of any volume has mass 2.7 times that of an equal volume of water.
i.e. Aluminium is 2.7 times heavier than water.

Question 14.
How does the density of a body and that of a liquid determine whether the body will float or sink into that liquid ?
Answer:
If the density of a body is LESS than the density of LIQUID, the body will FLOAT on the surface of liquid.
If the density of a body is MORE than the density of liquid, the body will SINK in a liquid.

Question 15.
A cork piece floats on water surface while an iron nail sinks in it. Explain the reason.
Answer:
CORK floats on water meAns density of cork is LESS than density of water.
IRON nail: Sinks in water meAns density of iron nail is MORE than density of water.

Question 16.
Which of the following will sink or float on water ? (Densityof water = 1 g Cm^-3)
(a) body A having density 500 kg m^-3
(b) body B having density 2520 kg m^-3
(c) body C having density 1100 kg m^-3
(d) body D having density 0.85 g m^-3
Answer:
Density of water = 1 g Cm^-3
(a) Density of body A = 500 kg m^-3 = 500 × = 0.5 = 0.5 g Cm^-3
Density of body A ¡s less than density of water hence A will float on water
(b) Density of body B = 2520 kg m^-3 = 2520 × 1/1000 = 2.52 g Cm^-3
Density of body B is more than density of water and hence B will SiNK in water
(c) Density of body C = 1100kg m^-3 = 1100 × 1/1000 = 1.1 g Cm^-3
is greater than water.
Hence, body C will sink in water.
(d) Density of body D = 0.85 g Cm^-3 < 1.0 g Cm^-3
Density of body D is less than the density of water hence body D will FLOAT on water

Question 17.
What is the iaw of floatation ?
Answer:
When a body floats in a liquid, the weight of the liquid displaced by its immersed part is equal to the total weight of the body. This is the law of floatation, i.e. while floating. Weight of the floating body = Weight of the liquid displaced by its immersed part.

Question 18.
The density of water is 1.0 g Cm^-3. The density of iron is 7.8 × 10″3 g Cm^-3. The density of mercury is 13.6 g Cm^-3.
Ans the following:
(a) Will a piece of iron float or sink in water ?
(b) Will a piece of iron float or sink in mercury ?
Answer:
Density of water 1.0 g Cm^-3
(a) Density of piece of iron = 7.8 × 10^-3 g Cm^-3

∴ Density of piece of iron is LESS than density of water.
Hence, piece of iron will FLOAT in water.
(b) Density of piece of iron = 7.8 × 10^-3
Density of mercury is 13.6 × 10^-3 g Cm^-3
Since 7.8 × 10^-3 < 13.6 × 10^-3
∴ Density of piece of iron is LESS than density of mercury
∴ Piece of iron will FLOAT in mercury

Question 19.
The diagram given below show a body floating in three different liquids. A, B and C at different levels.
(a) In which liquid does the body experience the greatest buoyant force ?
(b) Which liquid has the least density ?
(c) Which liquid has the highest density ?

Answer:
(a) Buoyant force is same in each case as the wt. of body is same in each case and Buoyant force is equal to the weight of liquid displaced by the immersed part of body which balances the wt. of body.
(b) The liquid A has the least density as body immerses the maximum.
(c) Liquid C has the highest density as the body immerses the least.

Question 20.
For a floating body, how is its weight related to the buoyant force ?
Answer:
When a body floats in a liquid. The weight of the liquid displaced by its immersed part is equal to the total weight of the body.

Question 21.
Why does a piece of ice float on water ?
Answer:
FLOATATION OF ICE ON WATER : Density of 0.9 g Cm^-3 is less than density of water 1 g Cm^-3. Hence, ice floats on water.

Question 22.
Explain why an iron needle sinks in water, but a ship made of iron floats on water.
Answer:
Density of iron is more than density of water, ∴ weight of iron nail is more than wt. of water displaced by it and nail SINKS. While shape of iron ship is made in such a way that it displaces MORE WEIGHT OF WATER than its own weight. Secondly the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF SHIP LESS THAN THAT OF WATER and hence ship floats on water.

Question 23.
It is easier to swim in sea water than in river water. Explain the reason.
Answer:
Density of sea water is greater than density of river water, [because of impurities]
(i) In each case the weight of water displaced will be equal to the weight of the man.
∴ Ratio of weight of sea water and river water displaced by man is 1: 1.
(ii) With smaller portion of man’s body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body. While to displace the same weight of river water, a larger portion of the body will have to be submerged ¡n water.
∴ It is easier for man to swim in sea water.

Question 24.
Icebergs floating on sea water are dangerous for ships. Explain the reason.
Answer:
ICEBERGS are very dangerous for ships as ICEBERGS are huge masses of ice floating in sea [density of ice being 0.917 g Cm^-3]
with about 9/10 portion below water and only 1/10 portion of it above surface of water.

Question 25.
Explain why it is easier to lift a stone under water than in air.
Answer:
In water, the stone experience a buoyant force which counter balances the weight of the stone acting downward and this makes the stone lighter and thus easier to lift the stone in water.

Question 26.
What is a submarine ? How can it be made to’dive in water and come to the surface of water.
Answer:
SUBMARINE: Submarine is a water-tight boat which can travel under water like a ship. It is providgd with water tanks. When submarine is to dive, water is filled in water tanks and it is made heavier and average density of submarine becomes greater than the density of sea water and it sinks. To make the submarine rise to the surface of water, water tanks are emptied and average density.of submarine becomes less than the density of sea water and it rises to surface of water.

While submarine is underwater soldiers can see the enemy activities through periscope.

Question 27.
A balloon filled with hydrogen rises in air. Explain the reason.
Answer:
A balloon filled with hydrogen rises to a certain height as it displaces more wt. of air than wt. of balloon but as it rises higher density of air DECREASES there and upthrust becomes less and ultimately upthrust becomes equal to the weight of balloon and balloon stops rising further.

C. Numericals

Question 1.
The density of air is 1.28 g/Iitre. Express it in:
(a) g cm³ (b) kg m
Answer:
(a) The density of air is I .28g/litre

Question 2.
The dimensions of a hail are 10 m × 7 m × 5 m. If the density of air is 1.11 kg m^-3, find the mãss of air in the hail.
Answer:
The dimensions of hall 10m × 7m × 5m
i.e. V350 m³
Density of air(D)= 1.11 kg m^-3
M = V × D 350 ×  1.11 =388.5 kg

Question 3.
The density of aluminium is 2.7 g cm3. Express it in kg m^-3
Answer:
Density of aluminium = 2.7 g/Cm³

Question 4.
The density of alcohol is 600 kg m^-3. Express it in g Cm^-3.
Answer:
Density of alcohol is = 600 kg/m^-3

Question 5.
A piece of zinc of mass 438.6 g has a volume of 86 Cm³. Calculate the density of zinc.
Answer:

Mass of Zinc (M) = 438.6 g
Volume V = 86 Cm³
Density (D) = ?

Question 6.
A piece of wood of mass 150 g has a volume of 200 Cm³. Find the density of wood ¡n
(a) C.GS. unit, (b) S.l. unit
Answer:
(a) Mass of wood (M) = 150 g
Volume of wood (V) = 200 Cm³
Density (D) =?

(b) In S.I. system = 0.75 × 1000 750 kg/ m³

Question 7.
Calculate the volume of wood of mass 6000 kg if the density of wood is 0.8 g Cm^-3
Answer:
Volume of wood (V) = ?
Mass of wood (M) = 6000 kg
Density of wood D = 0.8 g/ Cm³
D=O.8g/Cm³=o.8 × IOOO = 800kg /m³

Question 8.
Calculate the density of solid from the following data :
(a) Mass of solid = 72 g
(b) Initial volume of water in measuring cylinder = 24 ml
(c) Final volume of water when solid is completely immersed in water = 42 ml
Answer:
Mass of solid (M) = 72 g
Intial volume of water V₁ = 24 ml
Final volume of water V₂ = 42 ml
Volume of solid (V) = V₂ – V₁ = 42 – 24 = 18 Cm³
Density of solid (D) = ?

Question 9.
The mass of an empty density bottle is 21.8 g, when filled completely with water is 41.8 g and when filled completely with liquid it is 40.6 g. Find :
(a) the volume of density bottle
(b) the relative density of liquid
Answer:
Density of water is 1 g Cm³
∴ Volume of density bottle = weight of water in grams completely filling the bottle
(a) Volume of density bottle:
Mass of empty density bottle = M₁ =21.8 g
Mass of bottle + water = M₂  41.8 g
∴ Mass of water completely fih1ig the density bottle = M₂ — M₁
=41.8 —21.8
20g
But 1 g of water has volume = 1 cc
∴ Volume of bottle (density bottle) = volume of water =20 c.c. =20 ml
(b) The relative density of liquid:
Mass of 20 c.c. of liquid = (mass of density bottle + mass of 20 c.c of liquid- mass of density bottle)
= 40.6—21.8
= 18.8 g
Mass of 20 C.C of water = 20g
Relative density of liquid

Question 10.
From the following observations, calculate the density and relative density of a brine solution. Mass of empty density
bottle = 22 g
Mass of bottle + water = 50 g
Mass of bottle + brine solution = 54 g

Answer:
Mass of empty bottle, M₁ = 22 g
Mass of bottle + water, M₂ =50 g
Mass of bottle + brine solution, M₃ =54 g
Mass of water = M₂ — M₁ =50—22=28 g
Mass of brine solution = M₃ — M₁ 54—22 = 32 g
Density of brine solution = Mass of brine solution / Mass of water

Question 11.
The mass of an empty density bottlfe is 30 g, it is 75 g when filled completely with water and 65 g when filled completely with a liquid. Find :
(a) volume of density bottle,
(b) density of liquid, and
(c) relative density of liquid.
Answer:
Mass of empty density bottle (M₁) =30 g
Mass of bottle + Water (M₂) 75 g
Mass of liquid + Liquidx (M₃)= 65 g
Mass ofwater=M₂—M₁=75—30=45 g
(a) Volume of density bottle = Mass of water 45 g
(b) Density of Iiquid x = ?

(c) Mass of water in the density bottle =75 — 30 = 45 g
∴ Volume of water in density bottle = 45 cc
and mass of equal volume of liquid in density bottle 65—30 = 35g

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 21 Surface Area, Volume and Capacity (Cuboid, Cube and Cylinder). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Surface Area, Volume and Capacity Exercise 21A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the volume and the total surface area of a cuboid, whose :
(i) length = 15 cm, breadth = 10 cm and height = 8 cm.
(ii) l = 3.5 m, b = 2.6 m and h = 90 cm,
Solution:
(i) Length =15 cm, Breadth = 10 cm, Height = 8 cm.
Volume of a cuboid = Length x Breadth x Height = 15 x 10 x 8 =1200 cm³.
Total surface area of a cuboid 2 (l x b + b x h + h x l) = 2 (15 x 10 + 10 x 8 + 8 x 15) = 2(150 + 80 +120) = 2 x 350 = 700 cm²
(ii) Length = 3.5 m Breadth = 2.6 m, Height = 90 cm = \(\frac { 90 }{ 100 }\) m = 0.9 m.
Volume of a cuboid = l x b x h = 3.5 x 2.6 x 0.9 = 8.19 m³
Total surface area of a cuboid = 2(l x b + b x h + h x l)
= 2 (3.5 x 2.6 + 2.6 x 0.9 + 0.9 x 3.5) = 2 (910 + 2.34 + 3.15) = 2(14.59)= 29.18 m²

Question 2.
(i) The volume of a cuboid is 3456 cm³. If its length = 24 cm and breadth = 18 cm ; find its height.
(ii) The volume of a cuboid is 7.68 m³. If its length = 3.2 m and height = 1.0 m; find its breadth.
(iii) The breadth and height of a rectangular solid are 1.20 m and 80 cm respectively. If the volume of the cuboid is 1.92 m³; find its length.
Solution:
(i) Volume of the given cuboid = 3456 cm³.
Length of the given cuboid = 24 cm.
Breadth of the given cuboid = 18 cm
We know,
Length x Breadth x Height = Volume of a cuboid
⇒ 24 x 18 x Height = 3456
⇒ Height = \(\frac { 3456 }{ { 24 }\times { 18 } }\)
⇒ Height = \(\frac { 3456 }{ 432 }\)
⇒ Height = 8 cm
(ii) Volume of a cuboid = 7.68 m³
Length of a cuboid = 3.2 m
Height of a cuboid = 10m
We know
Length x Breadth x Height = Volume of a cuboid
3.2 x Breadth x 1.0 = 7.68
⇒ Breadth = \(\frac { 7.68 }{ { 3.2 }\times { 1.0 } }\)
⇒ Breadth = \(\frac { 7.68 }{ 3.2 }\)
⇒ Breadth = 2.4 m
(iii) Volume of a rectangular solid = 1.92 m³
Breadth of a rectangular solid = 1.20 m
Height of a rectangular solid = 80 cm = 0.8 m
We know
Length x Breadth x Height = Volume of a rectangular solid (cubical)
Length x 1.20 x 0.8 = 1.92
Length x 0.96 = 1.92
⇒ Length = \(\frac { 1.92 }{ 0.96 }\)
⇒ Length = \(\frac { 192 }{ 96 }\)
⇒ Length = 2 m

Question 3.
The length, breadth and height of a cuboid are in the ratio 5 : 3 : 2. If its volume is 240 cm³; find its dimensions. (Dimensions means : its length, breadth and height). Also find the total surface area of the cuboid.
Solution:
Let length of the given cuboid = 5x
Breadth of the given cuboid = 3x
Height of the given cuboid = 2x
Volume of the given cuboid = Length x Breadth x Height
= 5x x 3x x 2x = 30x³
But we are given volume = 240 cm³
30x³ = 240 cm³
⇒ x³ = \(\frac { 240 }{ 30 }\)
⇒ x³ = 8
⇒ x = \({ 8 }^{ \frac { 1 }{ 3 } }\)
⇒ x = \(\left( { 2 }\times { 2 }\times { 2 } \right) ^{ \frac { 1 }{ 3 } }\)
⇒ x = 2 cm
Length of the given cube = 5x = 5 x 2 = 10 cm
Breadth of the given cube = 3x = 3 x 2 = 6 cm
Height of the given cube = 2x = 2 x 2 = 4cm
Total surface area of the given cuboid = 2(l x b + b x h + h x l)
= 2(10 x 6 + 6 x 4 + 4 x 10) = 2(60 + 24 + 40) = 2 x 124 = 248 cm²

Question 4.
The length, breadth and height of a cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm2; find its dimensions. Also, find the volume of the cuboid.
Solution:
Let length of the cuboid = 6x
Breadth of the cuboid = 5x
Height of the cuboid = 3x
Total surface area of the given cuboid = 2 (I x b + b x h + h x l)
= 2(6x x 5x + 5x x 3x + 3x x 6x) = 2(30×2 + 15×2 + 18×2)
= 2 x 63×2 = 126x²
But we are given total surface area of the given cuboid = 504 cm²
126x² = 504 cm²
=> x² = \(\frac { 504 }{ 126 }\)
=> x² = 4
=> x = √4
=> x = 2 cm.
Length of the cuboid = 6x = 6 x 2 = 12 cm
Breadth of the cuboid = 5x = 5 x 2 = 10cm
Height of the cuboid = 3x = 3 x 2 = 6 cm
Volume of the cuboid = l x b x h = 12 x 10 x 6 = 720 cm³

Question 5.
Find the volume and total surface area of a cube whose each edge is :
(i) 8 cm
(ii) 2 m 40 cm.
Solution:
(i) Edge of the given cube = 8 cm
Volume of the given cube = (Edge)³ = (8)³ = 8 x 8 x 8 = 512 cm³
Total surface area of a cube = 6(Edge)² = 6 x (8)² = 384 cm²
(ii) Edge of the given cube = 2 m 40 cm = 2.40 m
Volume of a cube = (Edge)³
Volume of the given cube = (2.40)³ = 2.40 x 2.40 x 2.40 = 13.824 m²
Total surface area of the given cube = 6 x 2.4 x 2.4 = 34.56 m²

Question 6.
Find the length of each edge of a cube, if its volume is :
(i) 216 cm³
(ii) 1.728 m³
Solution:

Question 7.
The total surface area of a cube is 216 cm2. Find its volume.
Solution:
6(Edge)² = Total surface area of a cube
6(Edge)² = 216 cm²
=> (Edge)² = \(\frac { 216 }{ 6 }\)
=> (Edge)² = 36
=> Edge = √36
=> Edge = 6 cm
Volume of the given cube = (Edge)³ = (6)³ = 6 x 6 x 6 = 216 cm³

Question 8.
A solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm. Find its volume.
Solution:
Length of the cuboid = 24 cm
Breadth of the cuboid = 18 cm
Height of the cuboid = 4 cm
Volume of the cuboid = l x b x h = 24 x 18 x 4 = 1728 cm³

Question 9.
A wall 9 m long, 6 m high and 20 cm thick, is to be constructed using bricks of dimensions 30 cm, 15 cm and 10 cm. How many bricks will be required.
Solution:
Length of the wall = 9 m = 9 x 100 cm = 900 cm
Height of the wall = 6 m = 6 x 100 cm = 600 cm
Breadth of the wall = 20 cm
Volume of the wall = 900 x 600 x 20 cm3 = 10800000 cm³
Volume of one Brick = 30 x 15 x 10 cm³ = 4500 cm³
Number of bricks required to construct the wall = \(\frac { Volume\quad of\quad wall }{ Volume\quad \quad of\quad one\quad brick }\)
= \(\frac { 10800000 }{ 4500 }\)
= 2400

Question 10.
A solid cube of edge 14 cm is melted down and recasted into smaller and equal cubes each of edge 2 cm; find the number of smaller cubes obtained.
Solution:
Edge of the big solid cube = 14 cm
Volume of the big solid cube = 14 x 14 x 14 cm3 = 2744 cm³
Edge of the small cube = 2 cm
Volume of one small cube = 2 x 2 x 2 cm³ = 8 cm³
Number of smaller cubes obtained = \(\frac { Volume\quad of\quad big\quad cube }{ Volume\quad of\quad one\quad small\quad cube }\)
= \(\frac { 2744 }{ 8 }\) = 343

Question 11.
A closed box is cuboid in shape with length = 40 cm, breadth = 30 cm and height = 50 cm. It is made of thin metal sheet. Find the cost of metal sheet required to make 20 such boxes, if 1 m² of metal sheet costs Rs. 45.
Solution:
Length of closed box (l) = 40 cm
Breadth (b) = 30 cm
and height (h) = 50 cm

Total surface area = 2 (lb + bh + hl)
= 2 (40 x 30 + 30 x 50 + 50 x 40) cm²
= 2 (1200 + 1500 + 2000) cm²
= 2 x 4700 = 9400 cm²
Surface area of sheet used for 20 such boxes = 9400 x 20 = 188000 cm²
Cost of 1 m² sheet = Rs. 45
Total cost = \(\frac { { 18000 }\times { 45 } }{ { 100 }\times { 100 }\times { 100 } }\)
= Rs.846

Question 12.
Four cubes, each of edge 9 cm, are joined as shown below :

Write the dimensions of the resulting cuboid obtained. Also, find the total surface area and the volume of the resulting cuboid.
Solution:
Edge of each cube = 9 cm
(i) Length of the cuboid fonned by 4 cubes (l) = 9 x 4 = 36 cm
Breadth (b) = 9 cm and height (h) = 9 cm
(ii) Total surface area of the cuboid = 2(lb + bh + hl)
= 2 (36 x 9 + 9 x 9 + 9 x 36) cm²
= 2 (324 + 81 + 324) cm²
= 2 x 729 cm²
= 1458 cm²
(iii) Volume = l x b x h = 36 x 9 x 9 cm² = 2916 cm³

Surface Area, Volume and Capacity Exercise 21B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
How many persons can be accommodated in a big-hall of dimensions 40 m, 25 m and 15 m ; assuming that each person requires 5 m³ of air?
Solution:

Question 2.
The dimension of a class-room are; length = 15 m, breadth = 12 m and height = 7.5 m. Find, how many children can be accommodated in this class-room ; assuming 3.6 m³ of air is needed for each child.
Solution:
Length of the room = 15 m
Breadth of the room = 12 m
Height of the room = 7.5 m
Volume of the room = L x B x H = 15 x 12 x 7.5 m3 = 1350 m³
Volume of air required for each child = 3.6 m³
No. of children who can be accommodated in the class room.

Question 3.
The length, breadth and height of a room are 6 m, 5.4 m and 4 m respectively. Find the area of :
(i) its four-walls
(ii) its roof.
Solution:
Length of the room = 6 m
Breadth of the room = 5.4 m
Height of the room = 4 m
(i) Area of four walls = 2(L+B) x H
= 2(6 + 5.4) x 4 = 2 x 11.4 x 4 = 91.2 m²
(ii) Area of the roof = L x B = 6 x 5.4 = 32.4 m²

Question 4.
A room 5 m long, 4.5 m wide and 3.6 m high has one door 1.5 m by 2.4 m and two windows, each 1 m by 0.75 m. Find :
(i) the area of its walls, excluding door and windows ;
(ii) the cost of distempering its walls at the rate of Rs.4.50 per m².
(iii) the cost of painting its roof at the rate of Rs.9 per m².
Solution:
Length of the room = 5 m
Breadth of the room = 4.5 m
Height of the room = 3.6 m

Question 5.
The dining-hall of a hotel is 75 m long ; 60 m broad and 16 m high. It has five – doors 4 m by 3 m each and four windows 3 m by 1.6 m each. Find the cost of :
(i) papering its walls at the rate of Rs.12 per m²;
(ii) carpetting its floor at the rate of Rs.25 per m².
Solution:
Length of the dining hall of a hotel = 75 m
Breadth of the dining hall of a hotel = 60 m
Height of the dining hall of a hotel = 16 m
(i) Area of four walls of the dining hall = 2[L+B) x H = 2(75 + 60) x 16

Question 6.
Find the volume of wood required to make a closed box of external dimensions 80 cm, 75 cm and 60 cm, the thickness of walls of the box being 2 cm throughout.
Solution:
External length of the closed box = 80 cm
External Breadth of the closed box = 75 cm
External Height of the closed box = 60 cm
External volume of the closed box = 80 x 75 x 60 = 360000 cm³
Internal length of the closed box = 80 – 4 = 76 cm
Internal Breadth of the closed box = 75 – 4 = 71 cm
Internal Height of the closed box = 60 – 4=56 cm
Internal volume of the closed box = 76 x 71 x 56 cm = 302176 cm³
Volume of wood required to make the closed box = 360000 – 302176 = 57824 cm³

Question 7.
A closed box measures 66 cm, 36 cm and 21 cm from outside. If its walls are made of metal-sheet, 0.5 cm thick ; find :
(i) the capacity of the box ;
(ii) volume of metal-sheet and
(iii) weight of the box, if 1 cm³ of metal weights 3.6 gm.
Solution:
External length of the closed box = 66cm.
External breadth of the closed box = 36 cm
External height of the closed box =21 cm
External volume of the closed box= 66 x 36 x 21 = 49896 cm³
Internal length of the box =(66 – 2 x 0.5) = 66 – 1 = 65 cm
Internal breadth of the box =(36 – 2 x 0.5) = 36 – 1 = 35 cm
Internal height of the box = (21 – 2 x 0.5) = 21 – 1 = 20 cm
Internal Volume of the box = 65 x 35 x 20 = 45500 cm³
(i) Capacity of the box = 45500 cm³
(ii) Volume of metal sheet of the box = External volume – Internal volume
= 49896 – 45500 = 4396 cm³
(iii) 1 cm3 of metal weigh 3.6 grams.
Weight of the box = 4396 x 3.6 gm = 15825.6 gm

Question 8.
The internal length, breadth and height of a closed box are 1 m, 80 cm and 25 cm. respectively. If its sides are made of 2.5 cm thick wood ; find :
(i) the capacity of the box
(ii) the volume of wood used to make the box.
Solution:
Internal length of the closed box = 1m = 100 cm

Question 9.
Find the area of metal-sheet required to make an open tank of length = 10 m, breadth = 7.5 m and depth = 3.8 m.
Solution:
Length of the tank = 10 m
Breadth of the tank = 7.5 m
Depth of the tank = 3.8 m
Area of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8
= 2 x 17.5 x 3.8 = 35 x 3.8 = 133 m²
Area of the floor = L x B = 10 x 7.5 = 75 m
Area of metal sheet required to make the tank =Area of four walls + Area of floor = 133 m² + 75 m² = 208 m²

Question 10.
A tank 30 m long, 24 m wide and 4.5 m deep is to be made. It is open from the top. Find the cost of iron-sheet required, at the rate of ₹ 65 per m², to make the tank.
Solution:
Length of the tank = 30 m
Width of the tank = 24 m
Depth of the tank = 4.5 m
Area of four walls of the tank = 2[L+B] x H = 2(30 + 24) x 4.5 = 2 x 54 x 4.5 m² = 486 m²
Area of the floor of the tank = L x B = 30 x 24 = 720 m²
Area of Iron sheet required to make the tank = Area of four walls + Area of floor = 486 + 720 = 1206 m²
Cost of iron sheet required @ ₹ 65 per m² = 1206 x 65 = ₹ 78390

Surface Area, Volume and Capacity Exercise 21C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The edges of three solid cubes are 6 cm, 8 cm and 10 cm. These cubes are melted and recast into a single cube. Find the edge of the resulting cube.
Solution:
Edge of first solid cube = 6 cm
Volume = (6)³ = 216 cm³
Edge of second cube = 8 cm
Volume = (8)³ = 512 cm³
Edge of third cube = 10 cm
Volume = (10)³ = 1000 cm³
Sum of volumes of three cubes = 216 + 512 + 1000= 1728 cm³
Let a be the edge of so formed cube volume = a³
a³ = 1728 = (12)³
a = 12 cm

Question 2.
Three solid cubes of edges 6 cm, 10 cm and x cm are melted to form a single cube of edge 12 cm, find the value of x.
Solution:
Edge of first cube = 6 cm
Volume = (6)³ = 216 cm³
Edge of second cube = 10 cm
Volume = (10)³ = 1000 cm³
Edge of third cube = x
Volume =x³
Edge of resulting cube = 12 cm
Volume = (12)³ = 1728 cm³
216 + 1000 + x³ = 1728
x³ = 1728 – 216 – 1000 = 512 = (8)³
x = 8
Edge of third cube = 8 cm

Question 3.
The length of the diagonals of a cube is 8√3 cm.
Find its:
(i) edge
(ii) total surface area
(iii) Volume
Solution:
(i) Length of diagonal of a cube = 8√3 cm
Length of edge = \(\frac { 8\surd 3 }{ \surd 3 }\) = 8 cm
(ii) Total surface area = 6a² = 6 x 8² = 6 x 64 cm² = 384 cm²
(iii) Volume = a³ = (8)³= 512 cm³

Question 4.
A cube of edge 6 cm and a cuboid with dimensions 4 cm x x cm x 15 cm are equal in volume. Find:
(i) the value of x.
(ii) total surface area of the cuboid.
(iii) total surface area of the cube.
(iv) which of these two has greater surface and by how much?
Solution:
Edge of a cube = 6 cm
Volume = a³ = (6)³ = 216 cm³
Dimensions of a cuboid = 4 cm x x cm x 15 cm
Volume = 60x cm³
Volume of both is equal

Question 5.
The capacity of a rectangular tank is 5.2 m³ and the area of its base is 2.6 x 104 cm²; find its height (depth).
Solution:
Capacity of a tank = 5.2 m³
and area of its base = 2.6 x 10⁴ cm²

Question 6.
The height of a rectangular solid is 5 times its width and its length is 8 times its height. If the volume of the wall is 102.4 cm³, find its length.
Solution:
Height of rectangular solid = 5 x width
and length = 8 x height = 8 x 5 x width = 40 x width
Volume = 102.4 cm³
Let width = w
Then height = 40w
and height = 5w

Question 7.
The ratio between the lengths of the edges of two cubes are in the ratio 3 : 2. Find the ratio between their:
(i) total surface area
(ii) volume.
Solution:
Ratio in edges of two cubes = 3:2
Let edge of first cube = 3x
Then edge of second cube = 2x
(i) Now total surface area of first cube = 6 x (3x)² = 6 x 9x² = 54x²
and of surface area of second cube = 6 x (2x)² = 6 x 4x² = 24x²
Ratio = 54x²: 24x² = 9:4
(ii) Volume of first cube = (3x)³ = 27x³
and second cube = (2x)³ = 8x³
Ratio = 27x³: 8x³ = 27 :8

Question 8.
The length, breadth and height of a cuboid (rectangular solid) are 4 : 3 : 2.
(i) If its surface are is 2548 cm², find its volume.
(ii) If its volume is 3000 m³, find its surface area.
Solution:
Surface area of cuboid = 2548 cm²
Ratio in length, breadth and height of a cuboid = 4 : 3 : 2
Let length = 4x, Breadth = 3x and height = 2x

Surface Area, Volume and Capacity Exercise 21D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
The height of a circular cylinder is 20 cm and the diameter of its base is 14 cm. Find:
(i) the volume
(ii) the total surface area.
Solution:
Height of cylinder (h) = 20 cm
and diameter of its base (d)= 14 cm
and radius of its base (r)= 14/2 = 7 cm
(i) Volume = πr²h
= 22/7 x 7 x 7 x 20 cm³ = 3080 cm³

(ii) Total surface area = 2πr(h + r)
= 2 x 22/7 x 7 (20 + 7) cm² = 44 x 27 = 1188 cm²

Question 2.
Find the curved surface area and the total surface area of a right circular cylinder whose height is 15 cm and the diameter of the cross-section is 14 cm.
Solution:
Diameter of the base of cylinder = 14 cm
Radius (r) = 14/2 cm = 7 cm
Height (h) = 15 cm

Curved surface area = 2πrh
= 2 x 22/7 x 7 x 15 = 660 cm²
Total surface area = 2πr (h + r)
= 2 x 22/7 x 7(15 + 7)
= 2 x 22/7 x 7 x 22 = 968 cm²

Question 3.
Find the height of the cylinder whose radius is 7 cm and the total surface area is 1100 cm².
Solution:
Total surface area =1100 cm²
Radius = 7 cm
Let height of the cylinder = h
Then, total surface area = 2πr(h + r)

Question 4.
The curved surface area of a cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.
Solution:
Height (h) = 14 cm
Curved surface area (2πrh) = 88 cm²
Then, 2πrh = 88 cm²

Question 5.
The ratio between the curved surface area and the total surface area of a cylinder is 1 : 2. Find the ratio between the height and the radius of the cylinder.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr² = 2πr(h + r)
But their ratio is 1 : 2

Question 6.
Find the capacity of a cylindrical container with internal diameter 28 cm and height 20 cm.
Solution:
Diameter = 28 cm
Radius = 28/2 cm = 14 cm
Height = 20 cm
Volume = πr²h = 22/7 x 14 x 14 x 20
Volume = 12320 cm³

Question 7.
The total surface area of a cylinder is 6512 cm² and the circumference of its bases is 88 cm. Find:
(i) its radius
(ii) its volume
Solution:
Let r be the radius and h be the height of the given cylinder.
Circumference = 2πr = 88 cm (Given)

Question 8.
The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.
Solution:
Let r and h be the radius and height of the solid cylinder respectively.
Given, r + h = 37 cm
Total surface area of the cylinder = 1628 cm² (Given)

Question 9.
A cylindrical pillar has radius 21 cm and height 4 m. Find :
(i) the curved surface area of the pillar
(ii) cost of polishing 36 such cylindrical pillars at the rate of ₹ 12 per m².
Solution:
Radius of the cylinder = 21 cm

Question 10.
If radii of two cylinders are in the ratio 4 : 3 and their heights are in the ratio 5 : 6, find the ratio of their curved surfaces.
Solution:
Ratio in radii of two cylinders = 4 : 3
and ratio in their heights = 5 : 6

Surface Area, Volume and Capacity Exercise 21E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A cuboid is 8 m long, 12 m broad and 3.5 high, Find its
(i) total surface area
(ii) lateral surface area
Solution:
Length of a cuboid = 8 m
Breadth of a cuboid = 12 m
Height of a cuboid = 3.5 m
(i) Total surface area = 2(lb + bh + hl)
= 2(8 x 12 + 12 x 3.5 + 3.5 x 8)
= 2(96 + 42 + 28)
= 2 x 166 = 332 m²
(ii) Lateral surface area = 2h(l + b)
= 2 x 3.5(8 + 12) = 7 x 20= 140 m²

Question 2.
How many bricks will be required for constructing a wall which is 16 m long, 3 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm ?
Solution:
Length of the wall = 16 m = 16 x 100 cm = 1600 cm
Height of the wall = 3 m = 3 x 100 cm = 300 cm
Breadth of the wall = 22.5 cm
Volume of the wall = 1600 x 300 x 22.5 cm3 = 1,08,00,000 cm³
Volume of one brick = 25 x 11.25 x 6 cm3 = 1687.5 cm³

Question 3.
The length, breadth and height of cuboid are in the ratio 6 : 5 : 3. If its total surface area is 504 cm², find its volume.
Solution:
Let length of the given cuboid = 6x
Breadth of the given cuboid = 5x
Height of the given cuboid = 3x
Total surface area of the given cuboid

Question 4.
The external dimensions of an open wooden box are 65 cm, 34 cm and 25 cm. If the box is made up of wood 2 cm thick, find the capacity of the box and the volume of wood used to make it.
Solution:
External length of the open box = 65 cm
External breadth of the open box = 34 cm
External height of the open box = 25 cm
External volume of the open box = 65 x 34 x 25 cm³ = 55250 cm³
Internal length of open box = 65 – (2 x 2) cm = 61 cm
Internal breadth of a open box = 34 – (2 x 2) cm = 30 cm
Internal height of open box = 25 – 2 cm = 23 cm
Internal volume of open box or capacity of the box = 61 x 30 x 23 cm³ = 42090 cm³
Volume of wood required to make the closed box = 55250 – 42090 cm³ = 13160 cm³

Question 5.
The curved surface area and the volume of a toy, cylindrical in shape, are 132 cm² and 462 cm³ respectively. Find, its diameter and its length.
Solution:
Let the radius of a toy = r and
height of the toy = h
Curved surface area of a toy = 132 cm²
=> 2πrh = 132 cm²

Question 6.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m² is ₹ 15,000, find the height of the hall.
Solution:
Let length, breadth and height of the rectangular hall be l m, b m and h m respectively.

Area of four walls = Area of cuboid – Area of floor – Area of top
= 2 (lb + bh + hl) – (l x b) – (l x b)
= 2(lb) + 2 (bh) + 2(hl) – 2lb = 2 lh + 2 bh
= 2h(l + b)
= 2h x 125 [From (i)]
= 250h m²
Area of four walls = 250h m²
Cost of painting 1 m² area = ₹ 10
Cost of painting 250h m² area = ₹ 10 x 250h = 2500h
15000 = 2500h
h = 15000/2500
The height of the hall is 6 m.

Question 7.
The length of a hall is double its breadth. Its height is 3 m. The area of its four walls (including doors and windows) is 108 m², find its volume.
Solution:
Let the breadth be x
and the length be 2x
Height = 3 m
Area of four walls = 108 m²

Question 8.
A solid cube of side 12 cm is cut into 8 identical cubes. What will be the side of the new cube? Also, find the ratio between the surface area of the original cube and the total surface area of all the small cubes formed.
Solution:
Here, cube of side 12 cm is divided into 8 cubes of side 9 cm.

Given that,
Their volumes are equal.
Volume of big cube of 12 cm = Volume of 8 cubes of side a cm
(Side of big cube)³ = 8 x (Side of small cube)³
(12)³ = 8 x a³

Question 9.
The diameter of a garden roller is 1.4 m and it 2 m long. Find the maximum area covered by it 50 revolutions?
Solution:
Diameter of the roller = 1.4 m
Radius (r) = 1.4/2 = 0.7 m
and length (h) = 2m
Curved surface area = 2πrh = 2 x 22/7 x 0.7 x 2 cm² = 8.8 m²
Area covered in 50 complete revolutions = 8.8 x 50 m² = 440 m²
Area of the playground = 440 m²

Question 10.
In a building, there are 24 cylindrical pillars. For each pillar, radius is 28 m and height is 4 m. Find the total cost of painting the curved surface area of the pillars at the rate of ₹ 8 per m².
Solution:
Radius (r) of each pillar = 28 m
Height (h) = 4 m
Curved surface area of each pillar = 2πrh
= 2 x 22/7 x 28 x 4 m² = 704 m²
Surface area of 24 pillars = 704 x 24 m² = 16,896 m²
Cost of cleaning = ₹ 8 per m²
Total cost = ₹ 16,896 x 8 = ₹ 1, 35, 168