Selina Class 8 ICSE Solutions

Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants

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Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Biology Chapter 2 Reproduction in Plants. You can download the Selina Concise Biology ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Biology for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina Class 8 Biology ICSE SolutionsChemistryPhysicsMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Biology Chapter 2 Reproduction in Plants

REVIEW QUESTIONS

Multiple Choice Questions:

1. Put a tick mark (✓) against the correct alternative in the following statements

(a) Pollen is produced in the:

  1. Filament
  2. Style
  3. Pistil
  4. Anther

(b) Reproductive whorls of a flower are:

  1. Stamens and carpels
  2. Sepals and petals
  3. Sepals and stamens
  4. Petals and carpels

(c) Grafting is a method of:

  1. Artificial vegetative propagation
  2. Sexual reproduction
  3. Artificial pollination
  4. Cross-pollination

(d) Which one of the following is a false fruit ?

  1. Tomato
  2. Apple
  3. Potato
  4. Pea

Short Answer Questions:

Question 1.
Write two ways in which pollination may occur in plants.
Answer:
The two ways in which pollination may occur in plants are:
(a) Self-pollination.
(b) Cross-pollination.

Question 2.
Name the three agents of pollination.
Answer:
The three agents of pollination are:
(a) Insect
(b) Wind
(c) Water

Question 3.
Give two features of flowers which favour pollination by insects.
Answer:
Specialities of insect-pollinated flowers:
(a) These flowers are large with coloured petals to attract insects.
(b) These are scented so that insects locate the flowers by smell.

Question 4.
Name two characteristics of flowers in which pollination occur by wind.
Answer:
Special features of wind-pollinated flowers:
(a) They produce light pollen so that it is easily carried away.
(b) They produce a large quantity of pollen.

Question 5.
What is a “false fruit” ? Give one example:
Answer:
In false fruits the base of the flowers (thalamus) becomes the main fleshy part of the fruit, while the ovary remains a small central part containing seeds. Example: Apple and Pear.

Question 6.
Name any three agencies for dispersal of seeds.
Answer:
a. Wind
b. Water
c. Man and animals, birds, bats, squirrels.

Question 7.
Fill in the blanks by selecting suitable words:
(unisexual, fertilisation, fruit, stamen, anther, bisexual, pollination, seed, ovary)
Answer:
a. A flower that bears both the male and the female parts is known as bisexual flower.
b. A flower bearing only male or female parts is known as unisexual flower.
c. Transfer of pollen grains from the anther to the stigma is known as pollination.
d. Fusion of male cell with the female cell is called fertilisation.
e. The ovule develops into a seed.
f. The ovary of the flower develops into a fruit.

Long Answer Questions:

Question 1.
What is vegetative reproduction?
Answer:
Vegetative reproduction: In this method, new plants are produced by the vegetative parts of the plants. The vegetative part mean the leaf, stem and root. Potato, mint, ginger, banana etc. are reproduced by this method.
Vegetative propagation can occur by natural methods as well as by artificial methods.
Natural method includes reproduction by stem, by roots and by leaves.
Artificial method include reproduction by cutting, layering, grafting and tissue culture.

Question 2.
Briefly explain why a gardener prefers to grow certain plants vegetatively?
Answer:
Gardener prefer to grow certain plants by vegetative method. The advantages in doing so are as follows:

  1. Reproduction by vegetative parts takes place in a shorter time.
  2. New plants, thus produced, spread very fast in a small area.
  3. It is a surer method.
  4. All the characters of the mother plant are retained by the daughter plants.

Question 3.
Why is it disadvantageous to grow plants vegetatively?
Answer:
It is disadvantageous to grow plants vegetatively because of following:

  1. As all plants developed by vegetative propagation are identical, they are likely to be affected simultaneously if a disease spreads in the farm.
  2. Dispersal of plants does not take place on its own. Daughter plants, so developed, tend to remain nearby and are restricted to a particular area leading to competition for resources.

Question 4.
What is meant by pollination? Explain the structure of germinating pollen grain with the help of a labelled diagram.
Answer:
Pollination: Pollination is the process in which the pollen grains from the anthers are transferred to the stigma.
Structure of the pollen grain: After pollination pollen grains are deposited on the stigma of the carpel. Under suitable conditions pollen grains produce a tube known as pollen tube. This tube grows down through the stigma and style towards the ovary.
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 1
The nucleus of the pollen grains divides by mitosis and forms two male gametes. The tip of the pollen tube after entering ovary discharges the two male gametes into the embryo sac.
One of the male gametes fuses with the egg to form the zygote. This fusion is called fertilization. Another male gamete fuses with the diploid secondary nucleus and forms the endosperm, a food storing tissue.

Question 5.
Imagine all the seeds produced by a plant happen to fall under the same plant and sprout into new plants. Mention any two problems that will be faced by the new plants. –
Answer:
If all the seeds produced by a plant happen to fall under the same plant and sprout into new plants then in this situation plants will face the following problems:

  1. A large number of plants will grow in a small limited space. The water and the minerals available to them in the soil will be limited.
  2. The air surrounding them will not be enough and less sunshine will be available to them. As a result, most of these sprouted plants will die.

Question 6.
What is a flower ? Draw a neat labelled diagram showing the L.S. of a typical flower.
Answer:
A flower is a reproductive part of a plant. It helps in sexual reproduction as it has male parts and female parts. A fully opened flower has the following parts:
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 2

Stalk—A flower is attached to the shoot by means of stalk or pedicel stalk. The tip of the stalk is swollen or flattened. This is called torus or thalamus or receptacle.
The different parts of a flower are inserted on the thalamus. There are usually four whorls as

Calyx (Sepals)
Corolla (Petals)
Androecium (stamens)
Gynoecium (Carpels) Present on the thalamus

  1. Sepals (Calyx): These are the outermost part of the flower. These are leaf like and green in colour. This is the outer covering of the flower and form outer whorl in a flower. The Calyx (sepals) enclose the inner parts of the flower when it is a bud. It is protective in function.
  2. Corolla (Petals): Petals form the second whorl inner to the sepals. These are usually coloured, gaudy, or white in colour and scented and give sweet smell. The value of a flower is due to the attractive colour of the petals. These attract the insects for pollination.
  3. Stamens (Androecium): The third whorls inner to the petals are stamens. This third whorl is called Androecium. These are the male parts of the flower. Each stamen is formed of a long narrow, hair like structure called filament. On its tip it bears, a rounded broad sac like structure called anther. Each anther has two anther lobes. Each anther lobe has two pollen sacs which have powdery mass called pollen grains.
  4. Carpels (Gynoecium): Carpels are the inner most or fourth whorl in a flower. It is lodged on the thalamus and forms the female part of a flower. This whorl of carpels is called gynoecium. Each carpel or pistil has three parts.
    (a) The lower most, swollen part is ovary. It is attached to the thalamus.
    (b) The middle part is style which is narrow, thread like.
    (c) Stigma: The style ends in a knob like, rounded structure which is sticky in nature to receive the pollen grains.
    The ovaries contain ovules which later turn into seeds after fertilization and the ovary wall forms the fruit sometimes the thalamus also becomes a part of the fruit as in apple.

Question 7.
Write short notes on the following:
(a) Micro propagation
(b) Bryophyllum
(c) Vegetative reproduction
(d) Grafting
Answer:
(a) Micro propagation: This is the propagation of plants involving cell and tissue culture. If vegetative propagation is not possible in a crop, then buds, shoot apex or any other part of the plant can be used as explant for micropropagation.
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 3

  1. The explants are treated with sterilisation chemicals to prevent microbial growth, and then cultured in a particular nutrient medium.
  2. Cells grow and divide to form a cell mass called callus Some growth regulators (Plant hormones) are added.
  3. The callus differentiates into plant parts looking like plant (plantlet). After 4-6 weeks the plantlets transferred to the soil.

(b) Bryophyllum: It is a beautiful plant grown in gar xerophyte plant and can grow in any type soil or container. It requires sun and watering time to time. We can grow it vegetatively.
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 4
When a leaf is put on the soil or falls off accidentally from the parent plant. It starts giving out buds in the notches of the leaves. These buds start growing when in touch with soil having moisture. They give out adventitious roots which go into the soil and small aerial shoots which go into the air. So these adventitious buds form many plants from a single plant. Thus we can get many plants from a single leaf. We can grow these tiny plants into separate pots to get independent plants. This is one of the vegetative mode of reproduction.

(c) Vegetative reproduction: This is method of producing new plants from the vegetative parts of the plants. The vegetative part means the leaf, stem, root. As potato, mint, ginger, banana, sugar beet, gul-e-daudi, asparagus, sugar cane, are produced by this method.

(d) Grafting: In plants like mango, zizyphus (ber), guava apples, fruits, roses, a small bud is fixed on the stem. Many types of apples on a single plant thus a small orchard on a single plant. So we can have different types of roses and chrysanthemums on a single plant.
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 5

Question 8.
How artificial pollination is useful to plant breeders ? Discuss briefly.
Answer:
Artificial pollination means transfer of pollen to the stigma artificially. In ancient times, it was a common practice to sprinkle “male flowers of palms on the “female flowers”. However, in modem era, plant breeders use artificial pollination for developing new varieties. The breeders remove the anthers in young flowers and cover such flowers by plastic bags. Such flowers are then pollinated with pollen from the plants of the desired variety.

Question 9.
With the help of suitable diagrams, describe
(a) Binary fission in plants
(b) Budding in yeast cell
Answer:
(a) Binary fission: This is one of the asexual mean of reproduction. In lower plants like bacteria reproduction takes place by this process. In this process the nucleus of the cell divides into two. Then the cell wall splits across the middle of the cell. Thus each part has a nucleus. Thus each part is an independent bacterium.Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 6
Later on these two parts of the cell get separated from each other and form two independent individuals called daughter cells and lead independent life.

(b) Budding in yeast cell: Budding is the most common method in yeast. The Parent cell produces an outgrowth called a bud. The bud grows, and then gets detached from the parent body to lead an independent life.
Selina Concise Biology Class 8 ICSE Solutions – Reproduction in Plants 7

Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid

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Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

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

Selina Class 8 Biology ICSE SolutionsChemistryPhysicsMathsGeographyHistory & Civics

Selina Concise ICSE Solutions for Class 8 Biology Chapter 9 Diseases and First Aid

REVIEW QUESTIONS

Multiple Choice questions:

1. Put a tick mark (✓) against the correct alternative in the following statements:

(a) A mosquito is a vector for

  1. Typhoid
  2. Cholera
  3. Malaria
  4. Jaundice

(b) Dengue is caused by a

  1. Protozoan
  2. Virus
  3. Worm
  4. Fungus

(c) The idea of vaccination was conceived by

  1. Charles Darwin
  2. Alexander Flemming
  3. Issac Newton
  4. Edward Jenner

(d) Which one of the following is not a psychotropic drug ?

  1. Morphine
  2. Cocaine
  3. Heroin
  4. Penicillin

(e) Which one of the following is a communicable disease ?

  1. Measles
  2. Cancer
  3. Heart stroke
  4. Allergy

(f) Cataract is a disease of: –

  1. Ears
  2. Nose
  3. Eyes
  4. Throat

(g) Infectious diseases can be prevented by:

  1. Medicines
  2. Proper food
  3. Immunisation
  4. Exercise

(h) Which one of the following is a genetic disease ?

  1. Scurvy
  2. Leukemia
  3. Goitre
  4. Haemophilia

(i) Which one of the following is a degenerative disease ?

  1. Thalassemia
  2. Beri-beri
  3. Cataract
  4. Diabetes

(j) Pellagra is one disease caused by the deficiency of:

  1. Vit. B3
  2. Vit. B1
  3. Vit. C
  4. Vit. D

(k) Hay fever and asthma are

  1. Deficiency diseases
  2. Genetic diseases
  3. Organic diseases
  4. Allergy diseases

(I) Which one of the following vitamin deficiency diseases can be cured by eating a diet which includes carrot, yellow fruits, vegetables, butter, milk, fish ?

  1. Beri-beri
  2. Dermatitis
  3. Night blindness
  4. Scurvy

Short Answer Questions:

Question 1(a).
What is a non-communicable disease ?
Answer:
The diseases which are caused due to improper functioning of the body organs e.g. diabetes, heart attack. They are not caused by germs and not transmitted from one to another.

Question 1(b).
What are communicable diseases ?
Answer:
Those disease which spread from one person to another by the entry of microorganisms are known as communicable diseases.

Question 1(c).
How can we control spreading of diseases by mosquitoes and houseflies ?
Answer:
We can control spreading of diseases by mosquitoes and houseflies by using repellants, throwing garbage in covered bins, avoiding stagnation of water and checking breeding of these insects.

Question 1(d).
Public hygiene is equally important as personal hygiene. Give reasons.
Answer:
Keeping oneself clean is personal hygiene. But public hygiene is equally important as dirty surrounding may lead to more of houseflies, mosquitoes and other insects which are major vectors of many communicable diseases. Clean surrounding also provides good mental health as it soothes our mind.

Question 1(e).
What is a deficiency disease ?
Answer:
These disease are caused by lack of nutrients, vitamins, minerals as a anaemia, goitre.

Question 1(f).
Biting nails should be strictly avoided. Give reason.
Answer:
Nail biting may cause many disease as the dirt has many bacteria causing diseases. Nails should be cut from time to time to save from diseases.

Question 1(g).
Regular exercise and proper rest is a must. Give reason.
Answer:
Regular exercise keeps our body strong and immune to many diseases, rest refreshes our body.

Question 1(h).
Children eating more of fast food tend to suffer from obesity (overweight). Comment.
Answer:
Fast food like pizza, burger, patty, oily foods etc. have much carbohydrates and fats. Children eating these become more and more fat and gain weight soon as they do not do much of physical work.

2. Name the following:

Question 2(a).
A viral disease caused due to unhealthy sexual contact
Answer:
AIDS.

Question 2(b).
A disease caused due to Plasmodium
Answer:
Malaria.

Question 2(c).
A disease caused by the bite of female Anopheles mosquito
Answer:
Malaria.

Question 2(d).
Two viral diseases caused by mosquito bites
Answer:
Dengu, Chikungunya.

Question 2(e).
Any droplet — borne disease.
Answer:
Amoebiasis, Cholera, Hepatitis A.

Question 2(f).
A viral disease caused by the bite of a dog
Answer:
Rabies/Hydrophobia

Question 2(g).
A disease due to choking of coronary artery
Answer:
Atheraosclerosis.

Question 2(h).
Two diseases caused due to deficiency of protein in the diet of a child.
Answer:
Kwashiorkor and marasmus.

3. Write short (2-3 sentences) notes on the following:
Disease, immunisation, pathogen, AIDS, vaccination, vector.

Answer:

Disease: Disease is a departure from normal health due to structural or functional disorder of the body. Disease may be due to deficiency of nutrients or malfunctioning of organs or 1 genetic disorders, improper metabolic activity, or allergies, or cancer and mental illness as diabetes, haemophilia, leukemia, schizophrenia.

Immunisation: It means, we make the body immune to certain diseases by introducing respective weakened germs into the body. Thus we develop resistance to the concerned disease this process is called immunisation. The germs or the material introduced into the body to make it resistant to the concerned disease is called vaccine. This produces antibodies in the body of the person and the person can be saved by these antibodies. The vaccine can be given by the injection or orally as polio drops, tap vaccine for typhoid, BCG vaccine for tuberculosis.

Pathogens: The germs that cause diseases to human beings and to other animals and plants are called pathogens. They spread the diseases from person to person or through the air or through the articles of the diseased persons. Pathogens may be different kinds of bacteria, viruses, fungi, protozoans or worms.

AID’S (Acquired Immune Deficiency Syndrome): It is a viral disease caused by the virus called HIV (Human immuno deficiency virus) This virus makes the defence mechanism of the human body very weak. The immune system in the body as W.B.C. becomes weak. Thus the person catches the infectious diseases very easily. This disease spreads through sexual contact as one of the partner may be carrier of the disease. It may spread through the blood transfusion and infected syringes, blades of the barbers, it may infect the developing baby through the blood by the mother. It is very deadly disease.

Vaccination: Vaccination is a method of making the body immune to a particular disease by injecting killed or weakened disease causing microbe into a body to stimulate the formation of antibodies and develop immunity to that disease causing microbe.

Vector: A vector is an organism that carries disease causing microbes (pathogens) from one host to another. They are the carriers of infection. Example: Mosquito, housefly, etc.

P.Q. Fill in the blanks by selecting suitable words given below:
(clotting, goitre, insuline, rickets, iron, proteins)

(a) Anaemia is caused due to the deficiency of iron.
(b) Deficiency of Vit. D causes rickets in children.
(c) Deficiency of iodine in the diet may cause goitre.
(d) Diabetes is caused due to undersecretion of insulin.
(e) Kwashiorkor is caused due to the deficiency of proteins.
(f) Haemophilia is a disease caused due to slow clotting of the blood.

P.Q. Find the odd one out:

(a) Typhoid, cholera, jaundice, tuberculosis, tetanus.
Jaundice is odd one
(b) Cold, AIDS, plague, malaria, measles.
Malaria is odd one.
(c) Scurvy, rickets, haemophilia, pellagra, night blindness.
Haemophilia is odd one.
(d) Proteins, carbohydrates, fats, minerals, cancer.
Cancer is odd one.

P.Q. Fill in the blank in the following table:
Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid 1
Answer:
Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid 2
Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid 3

Long Answer Questions:

Question 1.
What is vaccination? Mention the four ways in which vaccine’s are prepared, giving the name of one disease for which each type of vaccine is used.
Answer:
For developing resistance in the body we introduce germs or germ substances in the body to develop resistance in the body against a particular disease. The material introduced into the body is called vaccine, this practice is called prophylaxis.
The germ or the germ substance is put into the body orally as polio drops or it is introduced by injection as TAB vaccine. Vaccine or vaccination was attached with small pox, but it is now used in a general sense.

Preparation:

  1. Killed germs are introduced into the body These act as vaccine for TAB, vaccine for typhoid, Salk’s vaccine for poliomyelitis. Rabies vaccine for dog bite.
  2. Living weakened germs: The living germs are treated in such a way that they become very weak and as such, they cannot cause the disease. They can induce antibody formation such as the vaccine for measles and the freezed dried BCG vaccine for tuberculosis.
  3. Living fully virulent germs: These virulent germs in small doses are introduced into the body as vaccine and these produce antibodies in the body and these do not allow the germs of particular type to cause that disease: In this vaccination the person is inoculated with cowpox virus. It is very similar to small pox virus.
  4. Toxoids: Toxoids are prepared from the extracts., of toxins secreted by bacteria. These toxins are poisons and these are made harmless by adding formalin into them. They retain their capacity and as a result when introduced into the body they produce antibodies into the body and do not allow the germs to grow in the body as vaccines for diphtheria and tetanus.

Question 2.
Burns can be superficial burns, deep burns or chemical burns. What emergency care you would suggest in each case.
Answer:
When one gets burnt up accidentaly one should not pull clothes from the burnt area and one should not cut the blisters. One should apply butter, vaseline or ointment.

  • Superficial burns: When there are superficial bums, put cold water on it and apply some oily substance, bumol, castor oil, vaseline, butter etc.
  • Deep burns: When there are deep bums do not put cold water on the burnt area. Seek the help of hospital and one should be kept under the treatment of some expert doctor of skin.
  • Chemical burns: In case of chemical burns i.e., due to acid and other chemicals, wash with running water for 10 minutes and then cover with dressing.

Question 3.
Describe the ways in which communicable diseases are transmitted through various indirect methods.
Answer:
Indirect methods of transmission of communicable diseases:

  1. Using items used by the infected persons: The healthy persons may be infected by using the articles like towel, hankey, utensils, bedding used by the patient infected by the communicable diseases. Diseases like tuberculosis, ring worm; common cold, influenza are transmitted by this method.
  2. Contaminated food and water: Diseases like dysentery, cholera spread through the contaminated food and water. Flies sitting on the food, if taken by a healthy person may be infected by the germs which may cause vomiting and loose motions. Similarly water and food infected by entamoeba may cause dysentery to persons who may take contaminated food.
  3. Vectors or carriers: Organisms like mosquitoes and house flies, ticks carry germs from the source of infection and pass on the germs to the normal persons and thus they are infected by diseases like malaria, cholera, plague. These organisms which carry the disease are called vectors and are not infected themselves.
    Mosquitoes suck blood and carry the disease causing protozoans from infected persons to healthy persons.
    Similarly houseflies carry the germs from garbage and sewage to the food. If this food is taken by the persons they become prey to typhoid and other diseases.
  4. Air: One sneeze from a person infected by cold may give billions of germs which are carried by air and may infect the healthy person. Tuberculosis passes from one person to other by coughing or sneezing of the infected person.
    These germs remain suspended in the air and persons may be infected by these spores or germs. Common cold, measles, diptheria, chicken pox.

Question 4.
Given a crossword puzzle. Read the clues across and clues downward, and fill up the blank squares. Check up your performance with the correct solution given at the end.

Clues across

  1. Category of pathogen that causes diseases, like common cold and mumps.
  2. These may readily grow in your hair, if you do not wash it regularly.
  3. This is the vaccine for preventing tuberculosis.
  4. Cover this part of your body by a handkerchief while sneezing to prevent droplet infection to others.
  5. The disease pertussis is popularly known as whooping
  6. One of the most common insects which visits our exposed foods and contaminates them.
  7. A disease that weakens body’s defence system
    Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid 4
  8. Germ or germ – substance introduced into the body to prevent occurrence of an infectious disease.
  9. A disease in which the eyes, the skin and the urine turn yellow.
  10. An organ usually affected by tuberculosis.
  11. A disease caused by the bite of an infected dog, and which affects the central nervous system.

Answer:
Selina Concise Biology Class 8 ICSE Solutions – Diseases and First Aid 5

Selina Concise Chemistry Class 8 ICSE Solutions – Hydrogen

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Selina Concise Chemistry Class 8 ICSE Solutions – Hydrogen

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 7 Hydrogen. 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 7 Hydrogen

Points to Remember:

  1. Hydrogen is the most abundant element found in the universe.
  2. Hydrogen is much more common in the form of compounds. The most important compound of hydrogen is water.
  3. The chief sources of hydrogen are water, acids and alkalies.
  4. Hydrogen is prepared by the action of water, acids or alkalies on active metals.
  5. Electrolysis of water results in the formation of hydrogen and oxygen.
  6. Hydrogen is lighter than air.
  7. Hydrogen bums in air with pop sound.
  8. Hydrogen acts as a reducing agent.
  9. Hydrogen is used to produce oxyhydrogen flame and in weather forecast balloons.

Exercise

1. Fill in the blanks:

(a) Hydrogen is lighter than air.
(b) Hydrogen is sparingly soluble in water.
(c) Hydrogen bums with a oxyhydrogen pale blue flame and pop sound is heard.
(d) A metal sodium hydrogen in the reactivity series gives hydrogen with water.
(e) Hydrogen reacts with metal oxides to form metal and water.
(f) Oxidation is the removal of hydrogen and addition of oxygen.
(g) In redox reaction oxidation and reduction occur simultaneously.

2. Indicate which of the following statements are true and which are false:

(a) Hydrogen molecule is monovalent.
(b) The removal of hydrogen from a substance is called reduction.
(c) Nitric acid can not be used to prepare hydrogen by its action on active metals ?
(d) The reaction between hydrogen and nitrogen to form ammonia is reversible.
(e) Zinc can liberate hydrogen from water, acid and alkali solution.
(f) Hydrogen is combustible as well as a supporter of combustion.
(g) Hydrogen gas is easily liquefiable.
Answer:
(a) False
(b) True
(c) False. Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide.
(d) True
(e) True
(f) False
(g) False

3. Complete and balance the following equations:

selina-concise-chemistry-class-8-icse-solutions-hydrogen-1
selina-concise-chemistry-class-8-icse-solutions-hydrogen-2

4. Give reasons for the following:

(a) Hydrogen be used as a fuel?
(b) Though hydrogen is lighter than air it cannot be collected by downward displacement of air.
(c) A pop sound produced when hydrogen is burnt?
(d) Helium replaced hydrogen in weather observation balloons?
(e) Nitric acid not used for the preparation of hydrogen gas?
(a) Because of its high heat of combustion, it is used as a fuel.
Answer:
(a) Coal gas, water gas and liquid hydrogen are some significant fuel.
(b) Since hydrogen is lighter than air. it is possible to collect the gas by downward displacement of air. But it is not safe to do so since a mixture of hydrogen and air can lead to an explosion.
(c) Impure hydrogen gas bums in air with a pop sound. This is because of the presence of impurities in it.
(d) If there is small leakage of hydrogen in a balloon, it forms a mixture with air that can explode. So helium has replaced hydrogen.
(e) Hydrogen cannot be prepared by the action of nitric acid on metals because it also releases nitrous oxide and nitric oxide and oxides the hydrogen to form water.

5. Name the following:
(a) Two metals which give hydrogen with cold water.
(b) A metal which liberates hydrogen only when steam is passed over red hot metal.
(c) The process in which oxygen is added or hydrogen is removed.
(d) A metallic oxide which can be reduced into metal by hydrogen.
Answer:
(a) Sodium (Na) and Potassium (K) give hydrogen with cold water.
(b) Iron
(c) Oxidation
(d) Copper oxide (CuO)

6. (a) Name the chemicals required to prepare hydrogen
gas in the laboratory.
(b) Give a balanced chemical equation for the reaction.
(c) Draw a neat and well-labelled diagram for the laboratory preparation ofhydrogen.
(d) How is hydrogen gas collected?
Answer:
(a) Granulated Zinc and dil. Hydrochloric acid.
(b) Zn + 2 HCl → ZnCl2 + H2 (g)
(c)
selina-concise-chemistry-class-8-icse-solutions-hydrogen-3
(d) Hydrogen gas is collected by the down-ward displacement of water.

7. How would you show that hydrogen:
(a) is a non-supporter of combustion?
(b) is lighter than air?
Answer:
(a) Hold a hydrogen gas filled jar with its mouth downwards.
Place a lighted candle inside the jar. The candle gets extinguished but the gas bums with a pop sound. This shows that hydrogen is non-supporter of combustion.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-4

(b) Take a delivery tube and place one of its ends in a soap solution kept in a trough and the other one in a flat bottom jar as shown in the figure. The soap bubbles containing hydrogen rise upward the air. The rising soap bubbles prove that hydrogen is lighter than air.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-5
Hydrogen-filled soap bubbles rising upward in the soap solution and into the air shows that hydrogen is lighter than air.

8. Hydrogen is a good reducing agent: What do you understand by the above statement? Explain with the help of copper oxide as an example.
Answer:
Hydrogen acts as a good reducing agent means, when hydrogen gas is passed over hot metallic oxides of copper, lead, iron, etc. it removes oxygen from them and thus reduces them to their corresponding metal.
Let us consider the following example, in each of which metallic oxide react with hydrogen. Metallic oxide act as oxidising agents and hydrogen acts as a reducing agents.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-6

9. (a) Name a process by which hydrogen gas is manufactured.
(b) Give equations for the reactions.
(c) How is hydrogen separated from carbon dioxide and carbon monoxide?
Answer:
(a) Commercially, hydrogen is prepared by Bosch process.

(b) (i) Steam is passed over hot coke at 1000°C in a furnace called converters. As a result water gas is produced which is a mixture of carbon monoxide and hydrogen gases.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-7
This reaction is endothermic in nature.
(ii) Water gas is mixed with excess of steam and passed over a catalyst ferric oxide (Fe2O3) and a promotor chromium trioxide (Cr2O3).
selina-concise-chemistry-class-8-icse-solutions-hydrogen-8
This reaction is exothermic in nature

(c) (i) The products are hydrogen, carbon dioxide and some
unreacted carbon monoxide. Hydrogen is separated from carbon dioxide by passing the mixture through water under pressure, in which carbon dioxide gets dissolved leaving behind hydrogen. Carbon dioxide can also be separated by passing it through caustic potash (KOH) solution.
2KOH + CO2 → K2CO3 + H2O
(ii) To separate carbon monoxide the gaseous mixture is passed through ammoniacal cuprous chloride in which carbon monoxide dissolves leaving behind hydrogen.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-9
Thus hydrogen gas is obtained.

10. Match the statements in Column A with those in Column B.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-10

11. State four uses of hydrogen:
Answer:

  1. Hydrogen with oxygen produce oxy-hydrogen flame which is used for cutting and welding.
  2. Hydrogen gas is used as a fuel.
  3. Hydrogen is used for hydrogenation of vegetable oil.
  4. Hydrogen gas is used extensively in the manufacture of ammonia gas, which is used to produce fertilizers.

12. Define:
(a) catalytic hydrogenation (b) oxidation
(c) reduction (d) redox reaction
Answer:
(a) Catalytic hydrogenation: catalytic hydrogenation is a process by which hydrogen gas is passed through vegetable oils in the presence of catalyst like Ni, Pt or Pd to convert them into solid vanaspati ghee.

(b) Oxidation: A reaction in which a substance combine with oxygen or in which hydrogen is removed is called oxidation reaction.
Example: H2S + Cl → 2HCl + S

(c) Reduction: Those reactions in which hydrogen combines with a substance or oxygen is removed from a substance, are known as reduction reactions.
selina-concise-chemistry-class-8-icse-solutions-hydrogen-11

(d) Redox reaction: Redox reactions are those in which reduction and oxidation both takes place simultaneously i.e. one substance is reduced while the other gets oxidised.

13. Multiple Choice Questions

(a) Equal volumes of hydrogen and chlorine are exposed to diffused sunlight to prepare

  1. hydrogen chloride
  2. water
  3. sodium hydroxide
  4. hydrochloric acid

(b) The metal which reacts with cold water to produce hydrogen is

  1. magnesium
  2. aluminium
  3. calcium
  4. iron

(c) In metal activity series the more reactive metals are at

  1. top
  2. bottom
  3. middle
  4. none

(d) Hydrogen is responsible for producing

  1. heat and light
  2. hydrogenated oil
  3. fertilizers
  4. all of the above

(e) Hydrogen is

  1. combustible
  2. non-combustible
  3. supporter of combustion
  4. neither supporter nor combustible

(f) Water gas is a mixture of

  1. carbon monoxide and oxygen
  2. carbon monoxide and hydrogen
  3. hydrogen and oxygen
  4. hydrogen and nitrogen.

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

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

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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.
    Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 1

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 2

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 3

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 4

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”.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 5

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 6

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 7

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)

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 8
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 9

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.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 10

 

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:
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 11

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
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 12
∴ 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 ?

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 13Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 14
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.

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 15

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
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 16
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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 17
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

Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 18
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) = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 19
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) =?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 20
(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
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 21

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) = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 22
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
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 23
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
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 24
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 = ?
Selina Concise Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 25
(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 Physics Class 8 ICSE Solutions Chapter 2 Physical Quantities and Measurement 26

Selina Concise Chemistry Class 8 ICSE Solutions – Chemical Reactions

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Selina Concise Chemistry Class 8 ICSE Solutions – Chemical Reactions

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions. 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.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions

Exercise – I

Question 1.
(a) Define a chemical reaction.
(b) What happens during a chemical reaction ?
(c) What do you understand by a chemical bond ?
Answer:
(a) Any chemical change in matter which involves transformation into one or more substances with entirely different properties is called a chemical reaction.
(b) A chemical reaction involves breaking of chemical bonds between the atoms or groups of atoms of reacting substances and rearrangement of atoms making new bonds to form new substances.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-1
(c) A chemical bond is the attractive force that holds the atoms of a molecule together, in a compound.

Question 2.
Give one example each of which illustrates the following characteristics of a chemical reaction:
(a) evolution of a gas
(b) change of colour
(c) change in state
Answer:
(a) When Zinc reacts with dil. sulphuric acid. Hydrogen gas is evolved, with an effervescence
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-2

(b) When blue coloured copper sulphate reacts with hydrogen sulphide gas, a black coloured substance copper sulphide is formed.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-3

(c) The reaction between hydrogen sulphide and chlorine (both gases) produces sulphur (solid) and hydrogen chloride (gas).
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-4

Question 3.
How do the following help in bringing about a chemical change?
(a) pressure (b) light
(c) catalyst (d) heat.
Answer:
(a) Some chemical reactions take place when reactants are subjected to high pressure.
e.g: Nitrogen and hydrogen when subjected to high pressure produce ammonia gas.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-5

(b) Some chemical reactions can take place in the presence of light. Ex. Photosynthesis.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-6

(c) A catalyst can either increases or decreases the rate of chemical reaction and some chemical reactions need a catalyst to change the rate of the reaction, in case it is too slow or too fast.

  1. Positive catalyst: When a catalyst increase the rate of reaction finely divided iron is used as a positive catalyst in the manufacturing of ammonia from hydrogen and nitrogen.
    selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-7
  2. Negative Catalyst: When a catalyst decreases the rate of reaction.
    Ex. Phosphoric acid act as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide.

(d) Some chemical reactions take place only in the presence of heat.
e.g. When lead nitrate is heated, it breaks into lead monoxide, nitrogen dioxide and oxygen.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-8

Question 4.
(a) Define catalyst.
(b) What are (i) positive catalysts and (ii) negative catalysts? Support your answer with one example for each of them.
(c) Name three biochemical catalysts found in the human body.
Answer:
(a) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change during the reaction.

(b) (i) Positive catalyst: When a catalyst increases the rate of chemical reaction, it is called positive catalyst.
e.g. when potassium chlorate heated to 700°C decomposes to evolve oxygen gas, when MnO2 is added the decomposition takes place at 300°C
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-9
(ii) Negative catalyst: When a catalyst decreases the rate of chemical reaction it is called negative catalyst.
Example. Phosphoric acid acts as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide. Alcohol too acts as a negative catalyst in certain chemical reactions.

(c) Biochemical catalysts found in human body:

  1. Pepsin
  2. Tryspin
  3. lipase.

Question 5.
What do you observe when
(a) dilute sulphuric acid is added to granulated zinc?
(b) a few pieces of iron are dropped in a blue solution of copper sulphate?
(c) silver nitrate is added to a solution of sodium chloride?
(d) ferrous sulphate solution is added to an aqueous solution of sodium hydroxide.
(e) solid lead nitrate is heated?
(f) when dilute sulphuric acid is added to barium chloride solution ?
Answer:
(a) When Zinc reacts with dilute sulphuric acid, hydrogen gas is evolved with an effervesence.
Zn + dil. H2SO4 → Zn SO4 + H2.

(b) When a few pieces of iron are dropped into a blue coloured copper sulphate solution, the blue colour of the solution fades and eventually turns into green.

(c) When a solution of silver nitrate is added to a solution of sodium chloride, white insoluble ppt. of silver chloride is formed.
AgNO3 (aq) + NaCl (aq) → AgCl (ppt) + NaNO3 (aq)

(d) When ferrous sulphate solution is added to sodium hydroxide solution, a dirty green ppt. of ferrous hydroxide is formed.
FeSO4 (aq) + 2NaOH (aq) → Fe(OH)2 ↓ + Na2SO4(aq)

(e) When solid lead nitrate is heated, it decomposes to produce light yellow solid lead monoxide, reddish brown nitrogen dioxide gas and colourless oxygen gas.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-42

(f) When few drops of dilute sulphuric acid is added to barium chloride solution, a white precipitate of barium sulphate is formed.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-43

Question 6.
Complete and balance the following chemical equations:
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-12
Answer:
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-13
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-14

Exercise – II

Question 1.
1. Fill in the blanks.
(a) A reaction in which two or more substances combine to form a single substance is called a combination reaction.
(b) A catalyst is a substance which changes the rate of a chemical reaction without undergoing a chemical change.
(c) The formation of gas bubbles in a liquid during a reaction is called effervesence
(d) The reaction between an acid and a base is called neutralization reaction.
(e) Soluble bases are called alkalis.
(f) The chemical change involving iron and hydrochloric acid illustrates a displacement reaction.
(g) In the type of reaction called double decomposition reaction, ions two compounds exchange their positive and negative radicals ions respectively.
(h) A catalyst either increases or decreases the rate of a chemical change but itself remains unchanged at the end of the reaction.
(i) The chemical reaction between hydrogen and chlorine is a combinaton reaction
(j) When a piece of copper is added to silver nitrate solution, it turns blue in colour.

Question 2.
Classify the following reactions as combination, decomposition, displacement, precipitation and neutralization. Also balance the equations.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-15
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-16

Question 3.
Define:
(a) precipitation (b) neutralization (c) catalyst
Answer:
(a) Precipitation: A chemical reaction in which two compounds in their aqueous state react to form an insoluble salt as one of the product.
Acid + Base → Salt + Water
Example.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-44

(b) Neutralization: A chemical reaction in which a base or an alkali reacts, with an acid to produce a salt and water only.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-18

(c) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-19
here iron act as a catalyst and increases the rate of chemical reaction.

Question 4.
Explain the following types of chemical reactions giving two examples for each of them.
(a) combination reaction
(b) decomposition reaction
(c) displacement reaction
(d) double decomposition reaction
Answer:
(a) Combination reaction: A reaction in which two or more substances combine to form a single substance is called combination reaction.
A + B → AB
e.g (i) When iron and sulphur are heated together, they combine to form iron sulphide.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-20
(ii) When carbon bums in oxygen to form a gaseous compound called carbon dioxide.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-21
(b) Decomposition reaction: A reaction in which a compound breaks up due to the application of heat into two or more simple substances is called decomposition reaction.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-22
e.g. (i) Mercuric oxide when heated, decomposes to form two elements mercury and oxygen
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-23
(ii) CaCO3 when heated decomposes to calcium oxide and carbon dioxide.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-24

(c) Displacement reaction: A reaction in which a more active element displaces a less active element from a compound is called displacement reaction.
AB + C → CB + A
e.g. (i) Zinc, displaces copper from copper sulphate solution.
Zn + CuSO4 (aq) → ZnSO4 (aq) + Cu
(ii) Iron piece when added to copper sulphate solution, copper is displaced.
Fe + CuSO4 → FeSO4 + Cu.

(d) Double decomposition reaction: A chemical reaction in which two compounds in their aqueous state exchange their ions to form new compounds is called a double decomposition reaction.
AB + CD → CB + AD
e.g. (i) AgNO3 + HCl  AgCl + HNO3(aq)
(ii) NaOH (aq) + HCl (aq)  NaCl (aq) + H2O.

Question 5.
Write the missing reactants and products and balance the equations.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-25
Answer:
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-26

Question 6.
How will you obtain?
(a) Magnesium oxide from magnesium.
(b) Silver chloride from silver nitrate.
(c) Nitrogen dioxide from lead nitrate.
(d) Zinc chloride from zinc.
(e) Ammonia from nitrogen.
Also give balanced equations for the reactions
Answer:
(a) Magnesium when burnt in air (oxygen) Magnesium oxide is formed
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-27

(b) When silver nitrate solution reacts with sodium chloride, silver chloride is formed.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-28

(c) Lead nitrate when heated nitrogen oxide is obtained
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-29

(d) Zinc when reacts with hydrochloric acid zinc chloride and hydrogen (g) is formed.
Zn + 2HCl → ZnCl2 + H2

(e) Nitrogen when reacts with hydrogen at 450°C and under 200 atm, ammonia is formed.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-30

Question 7.
What do you observe when
(a) Iron nail is kept in copper sulphate solution for sometime.
(b) Phenolphthalein is added to sodium hydroxide solution.
(c) Blue litmus paper is dipped in dilute hydrochloric acid.
(d) Lead nitrate is heated.
(e) Magnesium ribbon is burnt in oxygen.
(f) Ammonia is brought in contact of hydrogen chloride.gas.
Answer:
(a) A brown layer of copper gets deposited on iron nail. This is due to chemical reaction.
Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
(b) Solution turns pink.
(c) Blue litmus turns red in an acid solution.
(d) The pale yellow solid is lead monoxide, the reddish brown gas is nitrogen dioxide and the colourless gas is oxygen.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-31
(e) Magnesium ribbon bums with a dazzling white light and produces a white powder which is magnesium oxide.
The reaction can be represented as
2Mg + O2 → 2MgO (white powder)
(f) Ammonia and hydrogen chloride, both compounds, combine to form a compound, ammonium chloride.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-32

Question 8.
Give reason:
(a) A person suffering from acidity is advised to take an antacid.
(b) Acidic soil is treated with quick lime.
(c) Wasp sting is treated with vinegar.
Answer:
(a) An antacid neutralizes stomach acidity.
(b) If the soil is acidic it can be treated with base like quick lime, to make it neutral.
(c) Wasp stings are alkaline and they can be neutralized by vinegar which is a weak acid.

Question 9.
What is meant by the metal reactivity series ? State its importance, (any two points).
Answer:
A list in which the metals are arranged in the decreasing order of their chemical reactivity is called the metal reactivity series.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-33
Special features of the activity series:

  1. The ease with which a metal in solution loses electron(s) and forms a positive ion decreases down the series, i.e. from potassium to gold.
  2. Hydrogen is included in the activity series because, like metals do, it too loses an electron and becomes positively charged (H+) in most chemical reactions.
  3. The series facilitates the comparative study of metals in terms of the degree of their reactivity.
  4. The compounds of the metals (oxides, carbonates, nitrates and hydroxides) too can be easily compared.

Question 10.
What are oxides ? Give two examples of each of the following oxides.
(a) Basic oxide (b) Acidic oxide
(c) Amphoteric oxide (d) Neutral oxide
Answer:
An oxide is a compound which essentially contains oxygen in its molecule, chemically combined with a metal or a non-metal.
selina-concise-chemistry-class-8-icse-solutions-chemical-reactions-34

Question 11.
Define exothermic and endothermic reactions. Give two examples of each.
Answer:
Exothermic reactions: The chemical reaction in which heat is given out is called exothermic reactions. It causes rise in temperature. .
e.g. (i) When carbon bums in oxygen to form carbon dioxide, a lot of heat is produced.
C + O2 → CO2 + heat.
When water is added to quicklime a lot of heat is produced which boils the water.
CaO + H2O → Ca (OH)2 + Heat.

Endothermic reaction: A chemical reaction in which heat is absorbed is called endothermic reaction. It causes fall in temperature.
e.g. (i) When nitrogen and oxygen together are heated to a temperature of about 3000°C, nitric oxide gas is formed.
N2 + O2 + heat → 2NO (g)
(ii) Decomposition of calcium carbonate into carbon dioxide and calcium oxide when heated to a 1000°C.
CaCO3 + Heat → CaO (s) + CO2 (g)

Question 12.
State the effect of:
(a) an endothermic reaction
(b) an exothermic reaction on the surroundings.
Answer:
(a) Carbon dioxide present in the atmosphere is trapped by infrared radiations, gives rise to temperature which is exothermic reaction.
(b) The melting of glaciers by global warming.

Question 13.
What do you observe when
(a) an acid is added to a basic solution.
(b) ammonium chloride is dissolved in water.
Answer:
(a) A chemical reaction in which a base or an alkali reacts with an acid to produce a salt and water.
Acid + Base → Salt + Water
(b) Dissolution of ammonium chloride in water is an endothemic reaction in which heat energy is absorbed.

Selina Concise Chemistry Class 8 ICSE Solutions – Water

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Selina Concise Chemistry Class 8 ICSE Solutions – Water

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.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:
(i) Oceans (ii) Seas
(iii) Rivers (iv) 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 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) Sea water 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 a 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 low density as compared to water. Water has maximum density at 4°C. That is why ice floats on water.
(e) There are impurities dissolved in sea water which increases the freezing point. That is why sea water 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:
selina-concise-chemistry-class-8-icse-solutions-water-6

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:
selina-concise-chemistry-class-8-icse-solutions-water-1

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.
selina-concise-chemistry-class-8-icse-solutions-water-2

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 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
selina-concise-chemistry-class-8-icse-solutions-water-3
(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.
selina-concise-chemistry-class-8-icse-solutions-water-4

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. Match the statements in column A with that in column B.

selina-concise-chemistry-class-8-icse-solutions-water-7

3. 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

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Selina Concise Physics Class 8 ICSE Solutions – Matter

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APIusTopper.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.Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 1
    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.
    Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 2
  • 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

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 3

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.

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 4

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 5

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 :

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 6
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 :

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 7
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

 

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 8

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.”

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 9

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

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 10

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 11

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)

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 12

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:

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 13
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

Selina Concise Physics Class 8 ICSE Solutions Chapter 1 Matter 14

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 Chapter 1 Matter 15

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 19 Representing 3-D in 2-D. 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.

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Representing 3-D in 2-D Exercise 19 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
If a polyhedron has 8 faces and 8 vertices, find the number of edges in it.
Solution:
Faces = 8
Vertices = 8
using Eulers formula,
F + V – E = 2
8 + 8 – E = 2
-E = 2 – 16
E= 14

Question 2.
If a polyhedron has 10 vertices and 7 faces, find the number of edges in it.
Solution:
Vertices = 10
Faces = 7
Using Eulers formula,
F + V – E = 2
7 + 10 – E = 2
-E = -15
E = 15

Question 3.
State, the number of faces, number of vertices and number of edges of:
(i) a pentagonal pyramid
(ii) a hexagonal prism
Solution:
(i) A pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10

(ii) A hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18

Question 4.
Verily Euler’s formula for the following three dimensional figures:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 1
Solution:
(i) Number of vertices = 6
Number of faces = 8
Number of edges = 12
Using Euler formula,
F + V – E = 2
8 + 6 – 12 = 2
2 = 2 Hence proved.

(ii) Number of vertices = 9
Number of faces = 8
Number of edges = 15
Using, Euler’s formula,
F + V – E = 2
9 + 8 – 15 = 2
2 = 2 Hence proved.

(iii) Number of vertices = 9
Number of faces = 5
Number of edges = 12
Using, Euler’s formula,
F + V – E = 2
9 + 5 – 12 = 2
2 = 2 Hence proved.

Question 5.
Can a polyhedron have 8 faces, 26 edges and 16 vertices?
Solution:
Number of faces = 8
Number of vertices = 16
Number of edges = 26
Using Euler’s formula
F + V – E
⇒ 8 + 16 – 26 ≠ -2
⇒ -2 ≠ 2
No, a polyhedron cannot have 8 faces, 26 edges and 16 vertices.

Question 6.
Can a polyhedron have:
(i) 3 triangles only ?
(ii) 4 triangles only ?
(iii) a square and four triangles ?
Solution:
(i) No.
(ii) Yes.
(iii) Yes.

Question 7.
Using Euler’s formula, find the values of x, y, z.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 2
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 3

Question 8.
What is the least number of planes that can enclose a solid? What is the name of the solid.
Solution:
The least number of planes that can enclose a solid is 4.
The name of the solid is Tetrahedron.

Question 9.
Is a square prism same as a cube?
Solution:
Yes, a square prism is same as a cube.

Question 10.
A cubical box is 6 cm x 4 cm x 2 cm. Draw two different nets of it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 4

Question 11.
Dice are cubes where the sum of the numbers on the opposite faces is 7. Find the missing numbers a, b and c.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 5
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 6

Question 12.
Name the polyhedron that can be made by folding each of the following nets:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 7
Solution:
(i) Triangular prism. It has 3 rectangles and 2 triangles.
(ii) Triangular prism. It has 3 rectangles and 2 triangles.
(iii) Hexagonal pyramid as it has a hexagonal base and 6 triangles.

Question 13.
Draw nets for the following polyhedrons:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 8
Solution:
Net of hexagonal prism:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 19 Representing 3-D in 2-D - 9
Net of pentagonal pyramid:

Selina Concise Chemistry Class 8 ICSE Solutions – Physical and Chemical Changes

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Selina Concise Chemistry Class 8 ICSE Solutions – Physical and Chemical Changes

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes. 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.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes

Exercise

Question 1.
Define:
(a) a physical change, (b) a chemical change.
Answer:
(a) Physical Change: A physical change is a temporary change in which no new substance is formed and the chemical composition of the original substance remains the same, even though its physical properties like colour, state, shape, size etc. might change.
(b) Chemical Change: A chemical change is permanent change in which new substances are formed whose chemical composition and physical and chemical properties are different from those of in original substance.

Question 2.
Classify the following as a physical or a chemical change.
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(c) Butter getting rancid
(d) Boiling of water
(e) Burning of paper
(f) Melting of wax
(g) Burning of coal
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb
(k) Curdling of milk
Answer:
Physical change
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(d) Boiling of water
(f) Melting of wax
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb.
Chemical change
(c) Butter getting rancid
(e) Burning of paper
(g) Burning of coal
(k) Curdling of milk

Question 3.
Fill in the blanks.
Answer:
(a) The process of a liquid changing into a solid is called freezing.
(b) A change, which alters the composition of a substances, is known as a chemical change.
(c) There is no change in the composition of the substance during a physical change.
(d) The reaction in which energy is evolved is called exothermic reaction.

Question 4.
Given reason:
(a) Freezing of water to ice and evaporation of water are physical changes.
(b) Burning of a candle is both a physical and chemical change.
(e) Burning of paper is a chemical change.
(d) Cutting of a cloth piece is a physical change, though it cannot be reversed.
Answer:
(a) Freezing of water to ice and evaporation of water are physical change because water can be brought back to its original (liquid) form by

  1. We can heat the ice to bring it back to water.
  2. We can cool down the vapours to bring it back to water.

(b) When a candle is lighted, some of the solid wax first melts and turns into liquid, then it turns into vapours to produce a flame. New substances CO2 and H2O vapours are formed alongwith the evolution of light and heat energy. This shows a chemical change. When some of the molten wax drops to the floor, it again solidifies. Which shows a physical change. Thus the melting of candle wax is a physical change and the production of CO2 and H2O represents chemical change.

(c) When a piece of paper is burnt a new substance ash is produced. Even when the burning is stopped, the ash cannot be changed back into paper. This shows that the formation of the ash from paper is a permanent and irreversible change.

(d) Because it does not change chemical composition of cloth and the change is only in the state, size, shape, colour, texture or the smell of some or all of the substances that undergo physical change.

Question 5.
Give four difference between physical and chemical changes.
Answer:
The differences are Physical and Chemical Changes:
Physical change

  1. In a physical change no new substance is formed and the chemical composition of substance remains same. There are changes only in physical properties and state.
  2. Temporaiy change which can be reversed by simple physical methods.
  3. Weight of original substance doesn’t change
  4. Energy like heat, light etc. may or may not be absorbed or released

Chemical change

  1. In a chemical change new substance with entirely different chemical composition and properties is formed.
  2. Permanent change and irreversible
  3. Weight of original substances may increase or decrease
  4. Energy like heat, light etc. are given out or absorbed.

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions

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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions)

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 11 Algebraic Expressions (Including Operations on Algebraic Expressions). 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

Algebraic Expressions Exercise 11A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Separate the constants and variables from the following :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 1
Solution:
Clearly constants are : -7, √5, 8 – 5
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 2

Question 2.
Write the number of terms in each of the following polynomials.
(i) 5x2 + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 3

Question 3.
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 4
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 5

Question 4.
Write the degree of each polynomial given below :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 6
Solution:
(i) degree = 2 (Polynomial is xy + 7z)
(ii) degree = 3 (Polynomial is x2 – 6x3 + 8 y)
(iii) degree = 8 (Polynomial is y – 6y2 + 5y8)
(iv) degree = 3 (Polynomial is xyz – 3)
(v) degree = 4 (Polynomial is xy + yz2 – xz3)
(vi) degree = 12 (Polynomial is x5y7 – 8x3y8 + 10x4, y4z4)

Question 5.
Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x2 in 5x2 – 5x ;
(iv) 8 in a2 – 8ax + a ;
(v) 4xy in x2 – 4xy + y2.
Solution:
(i) The coefficient of ab in 7abx = 7x
(ii) The coefficient of 7a in 7abx = bx
(iii) The coefficient of 5x2 in 5x2 – 5x = 1
(iv) The coefficient of 8 in a2 – 8ax + a = – ax
(v) The coefficient of 4xy in x2 – 4xy + y2 = -1

Question 6.
In \(\frac { 5 }{ 7 }\) xy2z3, write the coefficient of
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 7
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 8

Question 7.
In each polynomial, given below, separate the like terms :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 9
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 10

Algebraic Expressions Exercise 11B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 11
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 12

Question 2.
Add :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 13
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 14
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 15

Question 3.
Find the total savings of a boy who saves ₹ (4x – 6y) ; ₹ (6x + 2y) ; ₹ (4y – x) and ₹ (y – 2x) for four consecutive weeks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 16

Question 4.
Subtract :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 17
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 18

Question 5.
(i) Take away – 3x3 + 4x2 – 5x+ 6 from 3x3 – 4x2 + 5x – 6
(ii) Take m2 + m + 4 from -m2 + 3m + 6 and the result from m2 + m + 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 19

Question 6.
Subtract the sum of 5y2 + y – 3 and y2 – 3y + 7 from 6y2 + y – 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 20

Question 7.
What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 21

Question 8.
(i) How much more than 2x2 + 4xy + 2y2 is 5x2 + 10xy – y2 ?
(ii) How much less 2a2 + 1 is than 3a2 – 6 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 22

Question 9.
If x = 6a + 86 + 9c ; y = 2b – 3a – 6c and z = c – b + 3a ; find
(i) x + y + z
(ii) x – y + z
(iii) 2x – y – 3z
(iv) 3y – 2z – 5x
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 23
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 24

Question 10.
The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 25

Question 11.
The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 26

Question 12.
The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 27

Question 13.
What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 28
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 29

Question 14.
How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?
Solution:
The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)
= 22x – 20y – 13z + 26 – 15x + 18y – 19z
= 7x – 2y – 32z + 26

Question 15.
How much bigger is 15x2y2 – 18xy2 – 10x2y than -5x2 + 6x2y – 7xy ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 30

Algebraic Expressions Exercise 11C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Multiply :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 31
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 32
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 33
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 34
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 35
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 36

Question 2.
Multiply :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 37
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 38
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 39

Question 3.
Simplify :
(i) (7x – 8) (3x + 2)
(ii) (px – q) (px + q)
(iii) (5a + 5b – c) (2b – 3c)
(iv) (4x – 5y) (5x – 4y)
(v) (3y + 4z) (3y – 4z) + (2y + 7z) (y + z)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 40

Question 4.
The adjacent sides of a rectangle are x2 – 4xy + 7y2 and x3 – 5xy2. Find its area.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 41

Question 5.
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Solution:
Reqd. Area = \(\frac { 1 }{ 2 }\) (base) x (altitude)
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 42

Question 6.
Multiply -4xy3 and 6x2y and verify your result for x = 2 and y= 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 43

Question 7.
Find the value of (3x3) x (-5xy2) x (2x2yz3) for x = 1, y = 2 and z = 3.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 44

Question 8.
Evaluate (3x4y2) (2x2y3) for x = 1 and y = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 45

Question 9.
Evaluate (x5) x (3x2) x (-2x) for x = 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 46

Question 10.
If x = 2 and y = 1; find the value of (-4x2y3) x (-5x2y5).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 47

Question 11.
Evaluate:
(i) (3x – 2)(x + 5) for x = 2.
(ii) (2x – 5y)(2x + 3y) for x = 2 and y = 3.
(iii) xz (x2 + y2) for x = 2, y = 1 and z= 1.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 48

Question 12.
Evaluate:
(i) x(x – 5) + 2 for x = 1.
(ii) xy2(x – 5y) + 1 for x = 2 and y = 1.
(iii) 2x(3x – 5) – 5(x – 2) – 18 for x = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 49
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 50

Question 13.
Multiply and then verify :
-3x2y2 and (x – 2y) for x = 1 and y = 2.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 51

Question 14.
Multiply:
(i) 2x2 – 4x + 5 by x2 + 3x – 7
(ii) (ab – 1)(3 – 2ab)
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 52
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 53

Question 15.
Simplify : (5 – x)(6 – 5x)(2 -x).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 54

Algebraic Expressions Exercise 11D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Divide :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 55
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 56
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 57
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 58
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 59
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 60
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 61
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 62
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 63
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 64

Question 2.
Find the quotient and the remainder (if any) when :
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 65
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 66
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 67
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 68
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 69
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 70

Question 3.
The area of a rectangle is x3 – 8x2 + 7 and one of its sides is x – 1. Find the length of the adjacent side.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 71

Question 4.
The product of two numbers-is 16x4 – 1. If one number is 2x – 1, find the other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 72

Question 5.
Divide x6 – y6 by the product of x2 + xy + y2 and x – y.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 73

Simplification
(Using removal of brackets)
The signs for different types of brackets are :

  1. ____ ; Vinculum or bar brackets,
  2. ( ); Parenthesis or small brackets,
  3. { }; Curly brackets or middle brackets,
  4. [ ]; Square brackets or big brackets.
    In a combined operation, the brackets must be removed in the same order as written above:

Algebraic Expressions Exercise 11E – Selina Concise Mathematics Class 8 ICSE Solutions

Simplify :
Question 1.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 74
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 75

Question 2.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 76
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 77

Question 3.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 78
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 79

Question 4.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 80
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 81

Question 5.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 82
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 83
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 84

Question 6.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 85
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 86

Question 7.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 87
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 88

Question 8.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 89
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 90

Question 9.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 91
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 92

Question 10.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 93
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 94

Question 11.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 95
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 96

Question 12.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 105
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 98

Question 13.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 99
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 100

Question 14.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 101
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 102

Question 15.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 103
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 11 Algebraic Expressions image - 104

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

by

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

Selina Class 8 Maths SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 7 Percent and Percentage. 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.

Percent and Percentage Exercise 7A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Evaluate :
(i) 55% of 160 + 24% of 50 – 36% of 150
(ii) 9.3% of 500 – 4.8% of 250 – 2.5% of 240
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 1

Question 2.
(i) A number is increased from 125 to 150 ; find the percentage increase.
(ii) A number is decreased from 125 to 100 ; find the percentage decrease.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 2

Question 3.
Find :
(i) 45 is what percent of 54 ?
(ii) 2.7 is what percent of 18 ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 3

Question 4.
(i) 252 is 35% of a certain number, find the number.
(ii) If 14% of a number is 315 ; find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 4
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 5

Question 5.
Find the percentage change, when a number is changed from :
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 6

Question 6.
An auctioneer charges 8% for selling a house. If a house is sold for Rs.2, 30, 500; find the charges of the auctioneer.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 7

Question 7.
Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Solution:
Total number of oranges = 800
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 8

Question 8.
A cistern contains 5 thousand litres of water. If 6% water is leaked. Find how many litres of water are left in the cistern.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 9

Question 9.
A man spends 87% of his salary. If he saves Rs.325 ; find his salary.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 10

Question 10.
(i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number 5.78 x 103 is wrongly written as 5.87 x 103; find the percentage error
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 11

Question 11.
In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18, 336 votes. Find the total number of votes polled and the votes secured by each candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 12
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 13

Question 12.
In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12, 366 votes. Find the total votes polled and die votes secured by the winning candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 14
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 15

Question 13.
The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is
₹ 8,000; find its cost:
(i) after one year
(ii) after 2 years
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 16

Question 14.
In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks ; find the maximum marks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 17

Question 15.
In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35% ; find the maximum marks.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 18
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 19

Question 16.
In an objective type paper of 150 questions; John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
(ii) What percent is Mohan’s correct answers to John’s correct answers ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 20

Question 17.
The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 21

Question 18.
The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 22

Question 19.
The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 23

Question 20.
The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 25

Percent and Percentage Exercise 7B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A man bought a certain number of oranges ; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still has 522 oranges left. Find how many had he bought?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 26
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 27

Question 2.
5% pupil in a town died due to some diseases and 3% of the remaining left the town. If 2, 76, 450 pupil are still in the town; find the original number of pupil in the town.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 28
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 29

Question 3.
In a combined test in English and Physics ; 36% candidates failed in English ; 28% failed in Physics and 12% in both ; find:
(i) the percentage of passed candidates
(ii) the total number of candidates appeared, if 208 candidates have failed.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 30

Question 4.
In a combined test in Maths and Chemistry; 84% candidates passsed in Maths; 76% in Chemistry and 8% failed in both. Find :
(i) the percentage of failed candidates ;
(ii) if 340 candidates passed in the test ; then how many appeared ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 31

Question 5.
A’s income is 25% more than B’s. Find, B’s income is how much percent less than A’s.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 32
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 33

Question 6.
Mona is 20% younger than Neetu. How much percent is Neetu older than Mona ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 34

Question 7.
If the price of sugar is increased by 25% today; by what percent should it be decreased tomorrow to bring the price back to the original ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 35
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 36

Question 8.
A number increased by 15% becomes 391. Find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 37

Question 9.
A number decreased by 23 % becomes 539. Find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 38
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 39

Question 10.
Two numbers are respectively 20 percent and 50 percent more than a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 40

Question 11.
Two numbers are respectively 20 percent and 50 percent of a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 41
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 42

Question 12.
Two numbers are respectively 30 percent and 40 percent less than a third number. What percent is the second of the first ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 43

Percent and Percentage Exercise 7C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 44

Question 2.
Mohan gets Rs. 1, 350 from Geeta and Rs. 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit. what percent does he get from Rohit ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 45

Question 3.
The monthly income of a man is Rs. 16, 000. 15 percent of it is paid as income-tax and 75% of the remainder is spent on rent, food, clothing, etc. How much money is still left with the man?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 46

Question 4.
A number is first increased by 20% and the resulting number is then decreased by 10%. Find the overall change in the number as percent.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 47

Question 5.
A number is increased by 10% and the resulting number is again increased by 20%. What is the overall percentage increase in the number ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 48

Question 6.
During 2003, the production of a factory decreased by 25%. But, during 2004, it (production) increased by 40% of what it was at the beginning of2004. Calculate the resulting change (increase or decrease) in production during these two years.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 49

Question 7.
Last year, oranges were available at Rs. 24 per dozen ; but this year, they are available at Rs. 50 per score. Find the percentage change in the price of oranges.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 50

Question 8.
In an examination, Kavita scored 120 out of 150 in Maths, 136 out of 200 in English and 108 out of 150 in Science. Find her percentage score in each subject and also on the whole (aggregate).
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 51

Question 9.
A is 25% older than B. By what percent is B younger than A ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 52
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 53

Question 10.
(i) Increase 180 by 25%.
(ii) Decrease 140 by 18%.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 54

Question 11.
In an election, three candidates contested and secured 29200, 58800 and 72000 votes. Find the percentage of votes scored by winning candidate.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 55

Question 12.
(i) A number when increased by 23% becomes 861 ; find the number.
(ii) A number when decreased by 16% becomes 798 ; find the number.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 56
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 57

Question 13.
The price of sugar is increased by 20%. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 7 Percent and Percentage image - 58

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

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Special Types of Quadrilaterals Exercise 17 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 1
Let ∠B = x
∠A = 3 ∠B = 3x
AD||BC
∠A + ∠B = 180°
3x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
∠B = 45°
∠A = 3x = 3 x 45 = 135°
and ∠B = ∠D = 45°
opposite angles of || gm are equal.
∠A = ∠C = 135°
opposite sides of //gm are equal.
AB = CD
5x – 7 = 3x + 1
⇒ 5x – 3x = 1+7
⇒ 2x = 8
⇒ x = 4
CD = 3 x 4+1 = 13
Hence 135°, 45°, 135° and 45° ; 13

Question 2.
In parallelogram PQRS, ∠Q = (4x – 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.
Solution:
In parallelogram PQRS,
∠Q = (4x – 5)° and ∠S = (3x + 10)°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 2
opposite ∠s of //gm are equal.
∠Q = ∠S
4x – 5 = 3x + 10
4x – 3x = 10+5
x = 15
∠Q = 4x – 5 =4 x 15 – 5 = 55°
Also ∠Q + ∠R = 180°
55° + ∠R = 180°
∠R = 180°-55° = 125°
∠Q = 55° ; ∠R = 125°

Question 3.
In rhombus ABCD ;
(i) if ∠A = 74° ; find ∠B and ∠C.
(ii) if AD = 7.5 cm ; find BC and CD.
Solution:
AD || BC
∠A + ∠B = 180°
74° + ∠B = 180°
∠B =180° – 74°= 106°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 3
opposite angles of Rhombus are equal.
∠A = ∠C = 74°
Sides of Rhombus are equal.
BC = CD = AD = 7.5 cm
(i) ∠B = 106° ; ∠C = 74°
(ii) BC = 7.5 cm and CD = 7.5 cm Ans.

Question 4.
In square PQRS :
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS
(ii) if PR = 5x and QR = 9x – 8. Find QS
Solution:
(i) sides of square are equal.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 4
PQ = QR
=> 3x – 7 = x + 3
=> 3x – x = 3 + 7
=> 2x = 10
x = 5
PS=PQ = 3x – 7 = 3 x 5 – 7 =8
(ii) PR = 5x and QS = 9x – 8
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 5
As diagonals of square are equal.
PR = QS
5x = 9x – 8
=> 5x – 9x = -8
=> -4x = -8
=> x = 2
QS = 9x – 8 = 9 x 2 – 8 =10

Question 5.
ABCD is a rectangle, if ∠BPC = 124°
Calculate : (i) ∠BAP (ii) ∠ADP
selina-concise-mathematics-class-8-icse-solutions-special-types-of-quadrilaterals-5
Solution:
Diagonals of rectangle are equal and bisect each other.
∠PBC = ∠PCB = x (say)
But ∠BPC + ∠PBC + ∠PCB = 180°
124° + x + x = 180°
2x = 180° – 124°
2x = 56°
=> x = 28°
∠PBC = 28°
But ∠PBC = ∠ADP [Alternate ∠s]
∠ADP = 28°
Again ∠APB = 180° – 124° = 56°
Also PA = PB
∠BAP = \(\frac { 1 }{ 2 }\) (180° – ∠APB)
= \(\frac { 1 }{ 2 }\) x (180°- 56°) = \(\frac { 1 }{ 2 }\) x 124° = 62°
Hence (i) ∠BAP = 62° (ii) ∠ADP =28°

Question 6.
ABCD is a rhombus. If ∠BAC = 38°, find :
(i) ∠ACB
(ii) ∠DAC
(iii) ∠ADC.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 7
Solution:
ABCD is Rhombus (Given)
AB = BC
∠BAC = ∠ACB (∠s opp. to equal sides)
But ∠BAC = 38° (Given)
∠ACB = 38°
In ∆ABC,
∠ABC + ∠BAC + ∠ACB = 180°
∠ABC + 38°+ 38° = 180°
∠ABC = 180° – 76° = 104°
But ∠ABC = ∠ADC (opp. ∠s of rhombus)
∠ADC = 104°
∠DAC = ∠DCA ( AD = CD)
∠DAC = \(\frac { 1 }{ 2 }\) [180° – 104°]
∠DAC = \(\frac { 1 }{ 2 }\) x 76° = 38°
Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104° Ans.

Question 7.
ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.
Solution:
Given : Rhombus ABCD in which ∠BCA = 35°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 8
To find : ∠ADC
Proof : AD || BC
∠DAC = ∠BCA (Alternate ∠s)
But ∠BCA = 35° (Given)
∠DAC = 35°
But ∠DAC = ∠ACD ( AD = CD) & ∠DAC +∠ACD + ∠ADC = 180°
35°+ 35° + ∠ADC = 180°
∠ADC = 180° – 70° = 110°
Hence ∠ADC = 110°

Question 8.
PQRS is a parallelogram whose diagonals intersect at M.
If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find :
(i) ∠RPS
(ii) ∠PRS
(iii) ∠PSR.
Solution:
Given : ||gm PQRS in which diagonals PR & QS intersect at M.
∠PMS = 54° ; ∠QSR = 25° and ∠SQR=30°
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 9
To find : (i) ∠RPS (ii) ∠PRS (iii) ∠PSR
Proof : QR || PS
=> ∠PSQ = ∠SQR (Alternate ∠s)
But ∠SQR = 30° (Given)
∠PSQ = 30°
In ∆SMP,
∠PMS + ∠ PSM +∠MPS = 180° or 54° + 30° + ∠RPS = 180°
∠RPS = 180°- 84° = 96°
Now ∠PRS + ∠RSQ = ∠PMS
∠PRS + 25° =54°
∠PRS = 54° – 25° = 29°
∠PSR = ∠PSQ + ∠RSQ = 30°+25° = 55°
Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

Question 9.
Given : Parallelogram ABCD in which diagonals AC and BD intersect at M.
Prove : M is mid-point of LN.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 10
Proof : Diagonals of //gm bisect each other.
MD = MB
Also ∠ADB = ∠DBN (Alternate ∠s)
& ∠DML = ∠BMN (Vert. opp. ∠s)
∆DML = ∆BMN
LM = MN
M is mid-point of LN.
Hence proved.

Question 10.
In an Isosceles-trapezium, show that the opposite angles are supplementary.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 11
Given : ABCD is isosceles trapezium in which AD = BC
To Prove : (i) ∠A + ∠C = 180°
(ii) ∠B + ∠D = 180°
Proof : AB || CD.
=> ∠A + ∠D = 180°
But ∠A = ∠B [Trapezium is isosceles)]
∠B + ∠D = 180°
Similarly ∠A + ∠C = 180°
Hence the result.

Question 11.
ABCD is a parallelogram. What kind of quadrilateral is it if :
(i) AC = BD and AC is perpendicular to BD?
(ii) AC is perpendicular to BD but is not equal to it ?
(iii) AC = BD but AC is not perpendicular to BD ?
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 12
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 13

Question 12.
Prove that the diagonals of a parallelogram bisect each other.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 14
Given : ||gm ABCD in which diagonals AC and BD bisect each other.
To Prove : OA = OC and OB = OD
Proof : AB || CD (Given)
∠1 = ∠2 (alternate ∠s)
∠3 = ∠4 = (alternate ∠s)
and AB = CD (opposite sides of //gm)
∆COD = ∆AOB (A.S.A. rule)
OA = OC and OB = OD
Hence the result.

Question 13.
If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 15
Given : //gm ABCD in which AC = BD
To Prove : ABCD is rectangle.
Proof : In ∆ABC and ∆ABD
AB = AB (Common)
AC = BD (Given)
BC = AD (opposite sides of ||gm)
∆ABC = ∆ABD (S.S.S. Rule)
∠A = ∠B
But AD // BC (opp. sides of ||gm are ||)
∠A + ∠B = 180°
∠A = ∠B = 90°
Similarly ∠D = ∠C = 90°
Hence ABCD is a rectangle.

Question 14.
In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 16
Given : //gm ABCD in which E and F are mid-points of AD and BC respectively.
To Prove : BFDE is a ||gm.
Proof : E is mid-point of AD. (Given)
DE = \(\frac { 1 }{ 2 }\) AD
Also F is mid-point of BC (Given)
BF = \(\frac { 1 }{ 2 }\) BC
But AD = BC (opp. sides of ||gm)
BF = DE
Again AD || BC
=> DE || BF
Now DE || BF and DE = BF
Hence BFDE is a ||gm.

Question 15.
In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :
(i) AE = AD,
(ii) DE bisects and ∠ADC and
(iii) Angle DEC is a right angle.
Solution:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 17
Given : ||gm ABCD in which E is mid-point of AB and CE bisects ZBCD.
To Prove : (i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
Const. Join DE
Proof : (i) AB || CD (Given)
and CE bisects it.
∠1 = ∠3 (alternate ∠s) ……… (i)
But ∠1 = ∠2 (Given) …………. (ii)
From (i) & (ii)
∠2 = ∠3
BC = BE (sides opp. to equal angles)
But BC = AD (opp. sides of ||gm)
and BE = AE (Given)
AD = AE
∠4 = ∠5 (∠s opp. to equal sides)
But ∠5 = ∠6 (alternate ∠s)
=> ∠4 = ∠6
DE bisects ∠ADC.
Now AD // BC
=> ∠D + ∠C = 180°
2∠6+2∠1 = 180°
DE and CE are bisectors.
∠6 + ∠1 = \(\frac { { 180 }^{ 0 } }{ 2 }\)
∠6 + ∠1 = 90°
But ∠DEC + ∠6 + ∠1 = 180°
∠DEC + 90° = 180°
∠DEC = 180° – 90°
∠DEC = 90°
Hence the result.

Question 16.
In the following diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 18
Show that:
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 90°
(vi) ABCD is a rectangle
Thus, the bisectors of the angles of a parallelogram enclose a rectangle.
Solution:
Given : In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.
To prove :
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 9°
(vi) ABCD is a rectangle
Proof : In parallelogram PQRS,
PS || QR (opposite sides)
∠P +∠Q = 180°
and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram
∠APQ + ∠AQP = \(\frac { 1 }{ 2 }\) x 180° = 90°
But in ∆APQ,
∠A + ∠APQ + ∠AQP = 180° (Angles of a triangle)
∠A + 90° = 180°
∠A = 180° – 90°
(v) ∠A = 90°
Similarly PQ || SR
∠PSB + SPB = 90°
(ii) and ∠PBS = 90°
But, ∠ABC = ∠PBS (Vertically opposite angles)
(iii) ∠ABC = 90°
Similarly we can prove that
(iv) ∠ADC = 90° and ∠C = 90°
(vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)
Hence proved.

Question 17.
In parallelogram ABCD, X and Y are midpoints of opposite sides AB and DC respectively. Prove that:
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram.
Solution:
Given : In parallelogram ABCD, X and Y are the mid-points of sides AB and DC respectively AY and CX are joined
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 19
To prove :
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram
Proof : AB || DC and X and Y are the mid-points of the sides AB and DC respectively
(i) AX = YC ( \(\frac { 1 }{ 2 }\) of opposite sides of a parallelogram)
(ii) and AX || YC
(iii) AXCY is a parallelogram (A pair of opposite sides are equal and parallel)
Hence proved.

Question 18.
The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 20
Prove that:
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM CN
(iii) ANCM is a parallelogram.
Solution:
Given : In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BN
AN, NC, CM and MA are joined
To prove :
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM = CN
(iii) ANCM is a parallelogram
Proof :
(i) In ∆DMC and ∆BNA.
CD = AB (opposite sides of ||gm ABCD)
DM = BN (given)
∠CDM = ∠ABN (alternate angles)
∆DMC = ∆BNA (SAS axiom)
CM =AN (c.p.c.t.)
Similarly, in ∆AMD and ∆CNB
AD = BC (opposite sides of ||gm)
DM = BN (given)
∠ADM = ∠CBN – (alternate angles)
∆AMD = ∆CNB (SAS axiom)
AM = CN (c.p.c.t.)
(iii) CM = AN and AM = CN (proved)
ANCM is a parallelogram (opposite sides are equal)
Hence proved.

Question 19.
The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 21
Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at 90°
∠BCD = 80°
Diagonals bisect the opposite angles also ∠BCD = ∠BAD (Opposite angles of rhombus)
∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°
Diagonals bisect opposite angles
∠OCB or ∠PCB = \(\frac { { 80 }^{ 0 } }{ 2 }\) = 40°
In ∆PCM,
Ext. CPD = ∠OCB + ∠PMC
110° = 40° + x
=> x = 110° – 40° = 70°
and ∠ADO = \(\frac { 1 }{ 2 }\) ∠ADC = \(\frac { 1 }{ 2 }\) x 100° = 50°
Hence x = 70° and y = 50°

Question 20.
Use the information given in the alongside diagram to find the value of x, y and z.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 22
Solution:
ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle
Opposite sides of a parallelogram are equal
3x + 14 = 2x + 25
=> 3x – 2x = 25 – 14
=> x = 11
∴ x = 11 cm
∠DCA = ∠CAB (Alternate angles)
y + 9° = 24
y = 24° – 9° = 15°
∠DAB = 3y° + 5° + 24° = 3 x 15 + 5 + 24° = 50° + 24° = 74°
∠ABC =180°- ∠DAB = 180° – 74° = 106°
z = 106°
Hence x = 11 cm, y = 15°, z = 106°

Question 21.
The following figure is a rectangle in which x : y = 3 : 7; find the values of x and y.
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 23
Solution:
ABCD is a rectangle,
x : y = 3 : 1
In ∆BCE, ∠B = 90°
x + y = 90°
But x : y = 3 : 7
Sum of ratios = 3 + 7=10
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 24
Hence x = 27°, y = 63°

Question 22.
In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:
Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals image - 25
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram.
Solution:
ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC
BA is produced to D
AE bisects ∠DAC
To prove:
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram
Proof:
(i) In ∆ABC and ∆ZAEC
AC=AC (common)
AB = CE (given)
∠BAC = ∠ACE (Alternate angle)
∆ABC = ∆AEC (SAS Axiom)
(ii) ∠BCA = ∠CAE (c.p.c.t.)
But these are alternate angles
AE || BC
But AB || EC (given)
∴ ABCD is a parallelogram