Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Balbharti Maharashtra State Board Class 9 English Solutions Kumarbharati Chapter 1.1 Life Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

English Kumarbharati 9th Solutions Chapter 1.1 Life Textbook Questions and Answers

Warming Up:

1. There are many popular lines like:
‘life Is a game – play ft well’.
‘life Is a Journey – keep going ahead’.
Discuss and write down a few such metaphorical lines about ‘life’. You can make them up yourselves.
1. …………………………….. .
2. …………………………….. .
3. …………………………….. .
4. …………………………….. .
5. …………………………….. .

Question 1.
Discuss and write down a few metaphorical lines about life, like the two lines given on page 2 of the textbook.
Answer:
(Some examples)

  1. Life is a movie – enjoy it to the full.
  2. Life is a short picnic – relish it.
  3. Life is a see-saw – sometimes up, sometimes down.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

2. Prepare as many ‘acrostics’ using the word LIFE, as you can. Two examples are given below.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life
You can turn it into a game, using a time – limit. Make sure your acrostics are related to ‘life’.

Question 1.
Prepare as many acrostics as you can, using the word ‘LIFE’.
Answer:
(Some examples)
1. Love
Inspiration
Friendship
Enjoyment

2. Lively
Interesting
Fascinating
Enchanting

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

3. Listen carefully and write the word in the appropriate column.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life 2
Teacher: hopeful, lonely, happy, jealous, surprised, shy, loving, proud, cheerful, anxious, nervous, excited, embarrassed, scared, silly, comfortable, peaceful, depressed, enthusiastic, motivated, inspired, threatened, crushed, angry Note that it is possible to feel both kinds of emotions at the same time. Have you experienced it? Try to describe the situation in short.

Question 1.
Listen carefully to the words spoken by the teacher and write them in the appropriate columns :
(The answer is given directly.)
Answer:

Positive Feelings Negative Feelings
hopeful, excited, happy, surprised, loving, cheerful, comfortable, peaceful, enthusiastic, motivated, inspired. lonely, jealous, shy,  proud, anxious, nervous, embarrassed, scared, silly, depressed, angry, threatened, crushed.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

English Workshop :

1. Pick out from the extract two lines each that reflect an optimistic (positive) attitude and pessimistic (negative) attitude.

Optimism :
1. …………… .
2. ……………. .

Pessimism :
1. …………….. .
2. …………….. .

Question 1.
Pick out from the extract two lines each that reflect an optimistic (positive) attitude and pessimistic (negative) attitude.
Answer:
Optimism :

  1. Foretells a pleasant day.
  2. But these are transient all.

Pessimism :

  1. Oft a little morning rain.
  2. Sometimes there are clouds of gloom.

(Note: These lines cannot actually be called pessimistic. The poetess looks on problems as temporary and feels that any difficulties we face always foretell a happy future. There are no lines that are absolutely pessimistic.)

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

2. Give other ‘-ly’ adverbs of similar meaning for the following. Use a thesaurus if needed.

(a) rapidly → ……………… .
(b) merrily → ……………… .
(c) gratefully → ………………
(d) cheerily → ……………… .
(e) manfully → ……………… .
(f) fearlessly → ……………… .
(g) gloriously → ……………… .
(h) victoriously → ……………… .

Question 1.
Give other ‘-ly’ adverbs of similar meaning for the following. Use a thesaurus if needed :
(The answers are given directly.)
Answer:
(a) rapidly → swiftly
(b) merrily → cheerfully
(c) gratefully → thankfully
(d) cheerily → happily
(e) manfully → bravely
(f) fearlessly → courageously
(g) gloriously → wonderfully
(h) victoriously → triumphantly

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

3. Pick out three examples of interrogation (rhetorical questions) from the poem.
a. Explain in your own words the point that each one makes.
Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life 3

b. Explain the metaphor in the following lines :
1. Oft a little morning rain ………………………………….
Foretells a pleasant day ……………………………………
2. Sometimes there are clouds ………………………………….
3. … the shower will make the roses bloom ………………………………….
4. Life’s sunny hours flit by ………………………………….

c. Pick out four examples of personification. Write what is personified in each :
1. ………………………………….
2. ………………………………….
3. ………………………………….
4. ………………………………….

d. ‘And calls our Best away’ is a gentle way of expressing the unpleasant idea of a loved one dying. It is an example of euphemism. Think and write down 3 or 4 ways in which we can express the idea of ‘death’ in a tactful and gentle manner.
1. ………………………………….
2. ………………………………….
3. ………………………………….
4. ………………………………….

Question a.
Pick out two examples of interrogation (rhetorical questions) from the extract. Explain in your own words the point that each one makes.
Answer:

Interrogation Explanation
1. What though death at times steps in, And calls our Best away? If death, at times takes away the people we love most, we must accept it and hope for better times.
2. What though sorrow seems to win, O’er hope a heavy sway? It does not matter if we are overcome by sorrow and hope seems to be far away.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Question b.
Explain the metaphor in the following lines :

1. Oft a little morning rain ………………….
Foretells a pleasant day …………….
2. SometImes there are clouds……………..
3. … the shower wifi make the roses bloom …………..
4. life’s sunny hours flit by ……………

Question 1.
Oft a little morning rain
Foretells a pleasant day.
Answer:
If we have difficulties at some stage, there will be joyous times following it. Difficulties are implicitly compared to rain and joyous times to a pleasant day.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Question 2.
Sometimes there are clouds.
Answer:
Sometimes there are problems. Problems are implicitly compared to clouds.

Question 3.
……. the shower will make the roses bloom.
Answer:
The problems we face will make for better times ahead. Problems are implicitly compared to the showers of rain and better times Eire implicitly compared to roses blooming.

Question 4.
Life’s sunny hours flit by.
Answer:
The happy times of our life go by very quickly. Here, the happy times are implicitly compared to sunny hours.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Question c.
Pick out four examples of personification from the extract. Write what is personified in each.
Answer:

  1. ‘Death steps in.’ Death is personified.
  2. ‘Sorrow seems to win.’ Sorrow is personified.
  3. ‘Yet Hope again elastic springs, Unconquered though she fell.’ Hope is personified.
  4. ‘Can courage quell despair!’ Courage is personified.

Question d.
‘And calls our Best away’ is a gentle way of expressing the unpleasant idea of a loved one dying. It is an example of euphemism. Think and write down 3 or 4 ways in which we can express the idea of ‘death’ in a tactful and gentle manner.
Answer:

  1. The person has left for his heavenly abode.
  2. He/She has gone to meet his/her Maker.
  3. He/She has become the beloved of God.
  4. He/She has completed his work on this earth.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

4. ‘…………. and that’s how I realised that courage and hope can help me overcome any major mishap/problem in life.’ Write an episode/experience from your own life that leads to the above conclusion.

Question 1.
‘…… and that’s how I realized that courage and hope can help me overcome any major mishap/ problem in life.’ Write an episode/experience from your own life that leads to the above conclusion.
Answer:
I am a lover of sports. My favourite game is football. I wanted to play in my school team, and right from the age of eleven, I had been practising very sincerely.

Then one unfortunate day, when I was twelve, I fell down the stairs of my building and fractured my leg , badly. I was devastated. How could I play my favourite game? How could I make it to the team? Would my leg ever become completely alright again? Would I be I able to run around and kick the ball as before? I felt that my world had ended.

This was the time when my family gave me their full support. My grandparents helped me to while away the long hours; my parents made nutritious food for me and saw to the medical treatment. My sister, who is four years older than me, gave me sound advice and courage. She understood what I was going through.

She told me of various people who had suffered terrible disasters and had risen I above them. She brought me videos of umpteen cases where the mind had won over the body. She made me read biographies of achievers like Wilma Rudolph and Sudha Chandran. She made me do my exercises regularly and keep in touch with my sport. Slowly but surely, I regained courage and confidence.

Believe it or not, two years later I made it to my football team! My doctor and my family said that it was optimism and a positive attitude that had helped me recover completely. And that’s how I realized that courage and hope can help me (or anybody) overcome any major mishap/problem in life.

5. Read: ‘The Psalm of Life’ – a poem by H. W. Longfellow.

English Kumarbharati 9th Digest Chapter 1.1 Life Additional Important Questions and Answers

Warming Up:

Question 1.
It is possible to feel both kinds of emotions at the same time. Have you experienced it? Try to describe such a situation in short.
(An example is given below.)
Answer:
Yes, I have experienced it. Once, when I stood first in class and got excellent marks in Science and Maths, my parents gave a party. All my friends were also present. My father gave a speech praising me. At that time, I was happy as well as embarrassed.

Read the following extracts carefully and complete the activities :

Simple Factual Activity :

Question 1.
Say whether the following statements are True or False :
Answer:

  1. The sages say that life is a dark dream. True
  2. The clouds of gloom are permanent. False
  3. We must enjoy the happy hours of our life. True
  4. Life’s sunny hours flit by quickly. True

Complex Factual Activities :

Question 1.
What do the sages say?
Answer:
The sages say that life is a dark dream.

Question 2.
What does the rain often foretell?
Answer:
The rain often foretells a pleasant day.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Question 3.
What should we do during joyful times in life?
Answer:
We should enjoy the joyful times of our life gratefully and cheerfully.

Question 4.
Complete the following :
We should not be unhappy when we have difficulties because
Answer:
We should not be unhappy when we have difficulties because these difficulties are temporary. If we have difficulties at some stage in our lives, there will be joyous times following it later.

Activities based on Poetic Devices :

Question 1.
Pick out an example of interrogation (rhetorical question) from the extract. Explain j in your own words the point that it makes.
Answer:

Interrogation Explanation
If the shower will make the roses bloom, Oh, why lament its fall? If a little rain helps the roses to flower, we should not feel unhappy about it. This means that we must not express regret for the difficulties we face in life.

Simple Factual Activity :

Complete the following :
(The answers are given directly and underlined.)

1. One should bear the day of trial courageously and fearlessly.
2. The wings of hope are golden and buoyant, (strong, elastic)

Complex Factual Activity :

Question 1.
Does sorrow actually win over hope, according to the poet?
Answer:
No. Though hope is temporarily defeated, it springs up again and helps us to bear the days of trials.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Question 2.
Who does ‘Our Best’ refer to?
Answer:
‘Our Best’ refers to those whom we love and cherish the most.

Question 3.
Why are the wings of hope ‘golden’?
Answer:
Gold is the colour of optimism. The wings of hope are golden because hope helps us bear up with
our problems and suppress despair.

Activities based on Poetic Devices :

Question 1.
Write an appreciation of this poem. With the help of given points in paragraph format :
Note: The point format given for Appreciation of Poem is for easy understanding. However, it is to be written in I the form of a paragraph in the examination.
Answer:
Point Format
(for understanding)

  • Title: Life
  • Poet: Charlotte Bronte
  • Rhyme Scheme: abeb in the first stanza, and abab in all the remaining stanzas.
  • Figures of Speech: ‘What though death at times steps in’. The figure of speech is Personification, Death is given the human quality of ‘stepping in’.
  • The other important figures of speech are Metaphor and Interrogation (Rhetorical Questions).
  • Theme/Central idea: This is a motivating and optimistic poem. The poet says that the bad things of life are transient, and good things invariably follow them.
  • Hope will rescue us and help us to bear our trials, even during times of great adversity. We must be optimistic, and have the courage to overcome any mishaps or problems.

Maharashtra Board Class 9 English Kumarbharati Solutions Chapter 1.1 Life

Paragraph Format:

The poem ‘Life’ is by Charlotte Bronte. It is a motivating and optimistic poem. The Rhyme Scheme is abeb for the first stanza and abab in all remaining stanzas. There are many figures of speech e.g. Personification ‘What though death at times steps in’. Here, death is given the human quality of ‘stepping in’. The other important figures of speech are Metaphor and Interrogation (Rhetorical Questions).

The poetess says that the bad things of life are temporary. Good things invariably follow it. Even during times of great adversity, hope will rescue us and help us to bear our trials. We must look at the positive side of life, and have the courage to overcome any mishaps or problems. The poem makes one feel good and is easy to understand.

Maharashtra State Board Class 9 English Solutions

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 16 Heredity and Variation Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Class 9 Science Chapter 16 Heredity and Variation Textbook Questions and Answers

1. Complete the following sentences by choosing the appropriate words from the brackets.
(Inheritance, sexual reproduction, asexual reproduction, chromosomes, DNA, RNA, gene)

a. Hereditary characters are transferred from parents to offsprings by …………………………….., hence they are said to be structural and functional units of heredity.
b. Organisms produced by …………………………….. show minor variations.
c. The component which is in the nuclei of cells and carries the hereditary characteristics is called ……………………………..
d. Chromosomes are mainly made up of ……………………………..
e. Organisms produced through …………………………….. show major variations.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

2. Explain the following.

a. Explain Mendel’s monohybrid progeny with the help of any one cross.
Answer:

  • Mendel brought about a cross between two pea plants with only pair of contrasting characters. This type of cross is called a monohybrid cross.
  • Tall pea plants and dwarf pea plants were used in this cross. Hence this is parent generation (P1).
    Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 3
  • All the plants produced in F1 genration are tall, having genotype Tt. This indicates that the gene responsible for tallness in pea plants is dominant over the gene responsible for dwarfness.
  • When F1 plants are self pollinated they produce second filial generation (F2).
  • In F2 generation both tall and dwarf plants appeared in the ratio 3:1.
  • Thus, the genotypic ratio of F2 generation is 3 (Tall) : 1 (Dwarf) and the genotypic ratio is 1 TT : 2 Tt: 1 tt.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

b. Explain Mendel’s dihybrid ratio with the help of any one cross.
Answer:

  • In dihybrid cross, Mendel considered two pairs of contrasting characters.
  • He made a cross between a pea plant producing rounded and yellow couloured seeds and a pea plant with wrinkled and green coloured seeds.
    Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 4
  • All the plants produced in F1 generation had rounded yellow seeds. This is because in pea plants, round shape of seed is dominant over wrinkled shape and yellow colour of seed is dominant over green colour.
  • When F1 plants are self pollinated, they produce four types of gamates – RY, Ry, rY, ry.
  • F2 plants formed by the fusion of four types of male gametes and four types of female gametes, had phenotypes such as round yellow, wrinkled yellow, round green and wrinkled green.
  • Also, F2 generation showed nine different types of genotypes such as RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy, rryy.
  • Phenotypic ratio of dihybrid cross is
    Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 5
  • The genotypic ratio of dihybrid cross is
    Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 6

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

c. Distinguish between monohybrid and dihybrid cross.
Answer:

Monohybrid cross Dihybrid cross
(i) Cross involving a single pair of contrasting characters is called monohybrid cross. (i) Cross involving two pairs of contrasting characters is called a dihybrid cross.
(ii) F1 plants of monohybrid cross produce two types of gametes. (ii) F1 plants of dihybrid cross produce four types of gametes.
(iii) Monohybrid cross has a phenotypic ratio of 3 : 1 in F2 generation. (iii) Dihybrid cross has a phenotypic ratio of 9 : 3 : 3 : 1 in F2 generation.

d. Is it right to avoid living with a person suffering from a genetic disorder?
Answer:

  • No, it is not right to avoid living with a person suffering from a genetic disorder.
  • Genetic disorders are transmitted from parents to offsprings only and they are non-contagious, i.e., they do not spread from one person to another through contact.

3. Answers the following questions in your own words.

a. What is meant by ‘chromosome’. Explain its types.
Answer:

  • The structure in the nucleus of cells that carries the hereditary characteristics is called the chromosome.
  • It is made up mainly of nucleic acids and proteins.
  • Depending upon the position of the centromere, there are four types of chromosomes.
    Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 7

(a) Metacentric: The centromere is exactly at the mid-point in this chromosome, and therefore, it looks like the English letter ‘V’. The arms of this chromosome are equal in length.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

(b) Sub-metacentric: The centromere is somewhere near the mid-point in this chromosome which, therefore, looks like the English letter ‘U. One arm is slightly shorter than the other.

(c) Acrocentric: The centromere is near one end of this chromosome which therefore looks like the English letter One arm is much smaller than the other.

(d) Telocentric: The centromere is right at the end of this chromosome making the chromosome look like the English letter ‘i’. This chromosome consists of only one arm.

b. Describe the structure of the DNA molecule.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 8

  • In 1953, Watson and Crick proposed a model of the DNA molecule.
  • As per their model, two parallel threads (strands) of nucleotides are coiled around each other to form a double helix structure. This structure can be compared with a coiled and a flexible ladder.
  • Each strand of DNA is made up of many small molecules known as nucleotides.
  • Each nucleotide is made up of a molecule of nitrogen base and phosphoric acid joined to a molecule of sugar.
  • There are four types of nitrogen bases-adenine, guanine, cytosine and thymine. Adenine and guanine are called as purines while cytosine and thymine are called as pyrimidines.
  • Nucleotides are arranged like a chain in the DNA.
  • The two threads (strands) of the DNA are comparable to the two rails of the ladder and each rail is made up of alternately joined molecules of sugar and phosphoric acid.
  • Each rung of the ladder is a pair of nitrogenous bases joined by hydrogen bonds. Adenine always pairs with thymine and cytosine always pairs with guanine.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

c. Express your opinion about the use of DNA fingerprinting.
Answer:

  • DNA fingerprinting is the technique in which the sequence of the genes in the DNA of a person, i.e., the genome of the person is identified.
  • This technique is useful to identify the lineage and to identify criminals because it is unique to every person.
  • It is also useful to identify paternity and maternity disputes etc.
  • This technique was developed by Professor of genetics Sir Alec Jeffreys.
  • A common method of collecting a reference sample, is in the use of a buccal swab. If this is not available, blood or saliva or hair sample may be used.
  • Just like your actual fingerprint, your DNA fingerprint is something that you are born with. It is unique to you.
  • DNA fingerprint is very useful in forensic science.

d. Explain the structure, function and types of RNA.
Answer:

  • Ribonucleic acid (RNA) is an important nucleic acid of the cell.
  • RNA is made up of ribose sugar, phosphate molecules and four types of nitrogenous bases adenine, guanine, cytosine and uracil.
  • The nucleotide i.e., smallest unit of the chain of the RNA molecule is formed by the combination of a ribose sugar, phosphate molecule and one of the nitrogen bases.
  • Large numbers of nucleotides are bonded together to form the macromolecule of RNA.
  • RNA performs the function of protein synthesis.
  • According to function, there are three types of RNA:
    (a) Ribosomal RNA (rRNA): It is the component of cellular organelle ribosome. Ribosomes perform the function of protein synthesis.(b) Messenger RNA (mRNA): It carries the information for protein synthesis from genes (i.e. DNA segment in the cell nucleus) to ribosomes (in the cytoplasm) which produce the proteins.(c) Transfer RNA (tRNA): It carries the amino acid up to the ribosomes as per the message of the mRNA.

e. Why is it necessary for people to have their blood examined before marriage?
Answer:

  • If people have their blood examined before marriage, the partners will know about the possible genetic diseases that their children might inherit. So they may decide not to have children or not to get married.
  • Blood tests before marriage are also done to check for any contagious disease in the partners. This will help to protect the partners from contagious diseases like STDs.

4. Write a brief note on each.

a. Down syndrome
Answer:

  • Down syndrome is the disorder arising due to chromosomal abnormality.
  • This is the first discovered and described the chromosomal disorder in human beings.
  • This disorder is characterized by the presence of 47 chromosomes. It is described as the trisomy of the 21st pair.
  • Infants with this disorder have one extra chromosome with the 21st pair in every cell of the body. Therefore, they have 47 chromosomes instead of 46. Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation
  • Children suffering from Down syndrome are usually mentally retarded and have a short lifespan. Mental retardation is the most prominent characteristic.
  • Other symptoms include short height, short wide neck, flat nose, short fingers, scanty hair, single horizontal crease on palm and a life expectancy of about 16-20 years.

b. Monogenic disorders
Answer:

  • Disorders occurring due to mutation in any single gene into a defective one are called monogenic disorders.
  • Approximately 4000 disorders of this type are now known.
  • Due to abnormal genes, their products are either produced in insufficient quantity or not produced at all.
  • It causes abnormal metabolism and may lead to death at a tender age.
  • Examples of monogenic disorders are Hutchinson’s disease, Tay-Sachs disease, galactosaemia, phenylketonuria, sickle cell anaemia, cyctic fibrosis, albinism, haemophilia, night blindness etc.

c. Sickle cell anaemia: symptoms and treatment.
Answer:

  • Sickle-cell anaemia is a hereditary disease caused due to mutation in a single gene. It is a monogenic disorder.
  • Normal haemoglobin has glutamic acid as the 6th amino acid in its molecular structure. However, if it is replaced by valine, the shape/structure of the haemoglobin molecule, changes.
  • Due to this, the erythrocytes (RBCs) which are normally biconcave become sickle-shaped. This condition is called sickle-cell anaemia. The oxygen-carrying capacity of haemoglobin in such individuals is very low.
  • In this condition, clumping and thereby, destruction of erythrocytes occurs most often. As a result, blood vessels are obstructed and the circulatory system, brain, lungs, kidneys, etc. are damaged.
  • Symptoms of sickle-cell anaemia are swelling of legs and hands, pain in joints, severe general body aches, frequent cold and cough, constant low-grade fever, exhaustion, pale face, low haemoglobin count.
  • A person suffering from sickle-cell anaemia should take a tablet of folic acid daily.

5. How are the items in groups A, B and C inter-releated?

6. Filling the blanks based on the given relationship.
a. 44 + X : Turner syndrome : : 44 + XXY: – ……………………………..
b. 3:1 Monohybrid : : 9:3:3:1 : ……………………………..
c. Women : Turner syndrome : : Men : ……………………………..
Answer:
a. Klinefelter syndrome
b. Dihybrid
c. Klinefelter syndrome

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

7. Complete the tree diagram below based on types of hereditary disorders.
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 1
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 2

Class 9 Science Chapter 12 Study of Sound Intext Questions and Answers

Question 1.
How do specific traits or characteristics appear in organisms? (Can you tell; Textbook Page No. 180)
Answer:

  • Information necessary for synthesis of a particular protein is stored in the DNA.
  • The segment of DNA which contains all the information for synthesis of a particular protein is called a gene for that protein.
  • To understand how a specific trait is expressed, let us consider plant height as an example.
  • We know that there are growth hormones in plants. Increase in the height of plants depends upon the quantity of growth hormones.
  • The quantity of growth hormone produced by a plant depends upon the efficiency of the concerned enzyme.
  • Efficient enzymes produce a greater quantity of the hormone due to which the height of the plant increases.
  • However, if the enzymes are less efficient, a smaller quantity of hormone is produced leading to the stunting of the plant.
  • Thus, the expression of traits is controlled by the genes.

Question 2.
Show the monohybrid cross between (RR) and (rr) and write the phenotypic and genotypic ratio of F2 generation. (Use your brain power; Textbook Page No. 187)
Answer:
Parental Generation (P1)
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 9
Phenotypic ratio : 3 Round : 1 Wrinkled
Genotypic ratio : 1 RR : 2 Rr : 1 rr

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 3.
Why did the characteristic of the Rounded- Yellow seeds alone appear in the Fj generation but not the characteristic of the wrinkled- green seeds? (Use your brain power; Textbook Page No. 187)
Answer:

  • Rounded-Yellow seeds is a dominant characteristic whereas wrinkled-green seeds is a recessive characteristic.
  • Therefore only the characteristic of Rounded- Yellow seeds appeared in the F1 generation.

Question 4.
Do all boys and girls of your class look alike? (Think about it; Textbook Page No. 179)
Answer:

  • No, all the boys and girls of my class do not look alike.
  • There is a lot of variation among them.

Question 5.
Carefully observe your classmate’s earlobes. (Observe; Textbook Page No. 179)
Answer:

  • Most of the classmates have free earlobes while very few have attached ear lobes.
  • This shows that in humans free earlobes is a dominant characteristic whereas attached earlobe is a recessive characteristic.

Question 6.
Irrespective of all of us being humans, what difference do you notice in our skin colour? (Observe; Textbook Page No. 179)
Answer:

  • Irrespective of all of us being humans, there is a lot of variation in our skin colour. Some people are light-skinned while some are dark-skinned.
  • The difference in skin colour is due to the gene responsible for the production of the pigment melanin.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 7.
All of you are in std. IX. Why then are some students tall and some short? (Observe; Textbook Page No. 179)
Answer:
Our height is decided by gene. People who are tall have genes for tallness whereas people who are short have genes for shortness and hence the variation.

Class 9 Science Chapter 12 Study of Sound Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The similarities and differences are all the effect of
(a) Heredity
(b) Fertilization
(c) Evolution
(d) Natural selection
Answer:
(a) Heredity

Question 2.
Each chromosome appears midway during cell division.
(a) Circular
(b) Rod-shaped
(c) Dumbbell-shaped
(d) Bottle-shaped
Answer:
(c) dumbbell-shaped

Question 3.
The chromosome in which the centromere is exactly at the mid-point is called chromosome.
(a) sub-metacentric
(b) metacentric
(c) acrocentric
(d) telocentric
Answer:
(b) metacentric

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 4.
The chromosome in which the centromere is somewhere near the mid-point is called chromosome.
(a) metacentric
(b) acrocentric
(c) sub-metacentric
(d) telocentric
Answer:
(c) sub-metacentric

Question 5.
The chromosome in which the centromere is near one end of the chromosome is called chromosome.
(a) metacentric
(b) acrocentric
(c) sub-metacentric
(d) telocentric
Answer:
(d) telocentric

Question 6.
Sex chromosomes are called
(a) homologous chromosomes
(b) autosomes
(c) allosomes
(d) metacentric chromosomes
Answer:
(c) allosomes

Question 7.
Which of the following is absent in RNA?
(a) Adenine
(b) Uracil
(c) Cytosine
(d) Thymine
Answer:
(d) Thymine

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 8.
DNA was discovered by
(a) Watson and Crick
(b) Frederick Miescher
(c) Gregor Johann Mendel
(d) Robert Brown
Answer:
(b) Frederick Miescher

Question 9.
The double helix model of DNA was produced by
(a) Watson and Crick
(b) Frederick Miescher
(c) Gregor Johann Mendel
(d) Robert Hooke
Answer:
(a) Watson and Crick

Question 10.
The molecule of RNA which is a component of the ribosome organelle is called a
(a) mRNA
(b) tRNA
(c) rRNA
(d) DNA
Answer:
(c) rRNA

Question 11.
In a monohybrid cross, the phenotypic ratio of F, generation is
(a) 1 tall: 3 dwarf
(b) 2 tall: 2 dwarf
(c) 3 tall: 1 dwarf
(d) 3 tall: 2 dwarf
Answer:
(c) 3 talhl dwarf

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 12.
arises due to either inheritance of only X chromosome from parents or due to inactivation of the gender-related part of X-chromosomes.
(a) Down syndrome
(b) Turner syndrome
(c) Klinefelter syndrome
(d) Albinism
Answer:
(b) Turner syndrome

Question 13.
Progenies of normal man and sufferer woman for sickle-cell anaemia will be
(a) all normal
(b) 25% normal and 75% sufferer
(c) all carrier
(d) all sufferer
Answer:
(c) all carrier

Question 14.
is a mitochondrial disorder.
(a) Down syndrome
(b) Cleft palate
(c) Spina bifida
(d) Leber hereditary optic neuropathy
Answer:
(d) Leber hereditary optic neuropathy

Question 15.
is a monogenic disorder.
(a) Haemophilia
(b) Cleft palate
(c) Diabetes
(d) Spina bifida
Answer:
(a) Haemophilia

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 16.
is a recessive character of pea plant.
(a) Round shape of seeds
(b) White colour of flowers
(c) Green colour of pods
(d) Inflated shape of pods
Answer:
(b) White colour of flowers

Question 17.
is a dominant character of pea plant.
(a) Dwarf height
(b) Yellow colour of pod
(c) Yellow colour of seeds
(d) Terminal position of flower
Answer:
(c) Yellow colour of seeds

Question 18.
is a dominant character in human beings.
(a) Non-rolling tongue
(b) Attached ear lobe
(c) Absence of hair on arms
(d) Free ear lobe
Answer:
(d) Free ear lobe

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 19.
is a recessive character in human beings.
(a) Absence of hair on arms
(b) Black and curly hair
(c) Free earlobe
(d) Presence of hair on arms
Answer:
(a) Absence of hair on arms

Question 20.
If one parent is normal and one parent is carrier of sickle-cell anaemia, then the progenies will be
(a) all normal
(b) 50% normal and 50% carrier
(c) 50% carrier and 50% sufferer
(d) all carrier
Answer:
(b) 50% normal and 50% carrier

Question 21.
If one parent is carrier and one parent is a sufferer of sickle-cell anaemia, then the progenies will be
(a) 50% normal and 50% carrier
(b) all sufferers
(c) 50% carrier and 50% sufferer
(d) all carrier
Answer:
(c) 50% carrier and 50% sufferer

Find the odd man out:

Question 1.
Adenine, thymine, cytosine, uracil
Answer:
Adenine. It is a purine whereas the others are pyrimidines.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 2.
Axillary flower, green pod, green seed, inflated pod
Answer:
Green seed. It is a recessive character of pea plant whereas the others are dominant characters.

Question 3.
Constricted pod, purple flower, axillary flower, yellow seeds.
Answer:
Constricted pod. It is a recessive character of pea plant whereas the others are dominant characters.

Question 4.
Green seeds, wrinkled seeds, terminal flower, green pod.
Answer:
Green pod. It is a dominant character of pea plant whereas the others are recessive characters.

Question 5.
Attached ear lobes, brown and straight hair, non-rolling tongue, presence of hair on arms.
Answer:
Presence of hair on arms. It is a dominant characteristic of human beings whereas the others are recessive characters.

Question 6.
Cystic fibrosis, albinism, spina bifida, sicklecell anaemia.
Answer:
Spina bifida. It is a polygenic disorder whereas the others are monogenic disorders.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 7.
Fiutchinson’s disease, phenylketonuria, nightblindness, leber hereditary optic neuropathy.
Answer:
Leber hereditary optic neuropathy. It is a mitochondrial disorder whereas the others are monogenic disorders.

Complete the analogy:

a. Tall plant: Phenotype :: Tt:
b. Dominant trait : Axial position of flower :: Recessive trait:
c. Women : 44 + XX :: Men :
d. Adenine and Guanine : Purine :: Cytosine and Thymine :
Answer:
a. Genotype
b. Terminal position of flower
c. 44 + XY
d. Pyrimidine

Match the columns

Question 1.

Column 1 Column 2 Column 3
(1) Leber hereditary optic neuropathy (a) 44 + XXY (i) Pale skin, white hairs.
(2) Diabetes (b) 45 + X (ii) Men are sterile.
(3) Albinism (c) Mitochondrial disorder (iii) Women are sterile.
(4) Turner syndrome (d) Polygenic disorder (iv) This disorder arises during development of zygote.
(5) Klinefelter Syndrome (e) Monogenic disorder (v) Effect on blood- glucose level.

Answer:
(1 – c – iv),
(2 – d – v),
(3 – e – i),
(4-b- Hi),
(5 – a – ii)

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 2.

Column ‘A’ Column ‘B’
(1) Yellow and rinkled (a) yyrr
(2) Green and round (b) YyRr
(3) Yellow and round (c) YYrr
(4) Green and wrinkled (d) yyRr

Answer:
(1 – c),
(2 – d),
(3 – b),
(4 – a)

Question 3.

Column ‘A’ Column ‘B’
(1) Tay-Sachs disease (a) Multifactorial disorder
(2) Diabetes (b) Destruction of erythrocytes
(3) Sickle-cell anaemia (c) Absence of melanin
(4) Albinism (d) Monogenic disorder

Answer:
(1 – d),
(2 – a),
(3 – b),
(4 – c)

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

State whether the following statements are true or false. Correct the false statement:

Question 1.
Offsprings produced through asexual reproduction show greater variations as compared to those produced through sexual reproduction.
Answer:
False. Offsprings produced through asexual reproduction show minor variations as compared to those produced through sexual reproduction.

Question 2.
Information necessary for protein synthesis is stored in the RNA.
Answer:
False. Information necessary for protein synthesis is stored in the DNA.

Question 3.
The quantity of growth hormone produced by a plant depends upon the efficiency of the concerned enzyme.
Answer:
True

Question 4.
The chromosome in which the centromere is exactly at the mid-point is called telocentric chromosome.
Answer:
False. The chromosome in which the centromere is exactly at the mid-point is called metacentric chromosome.

Question 5.
RNA molecules are called master molecules.
Answer:
False. DNA molecules are called master molecules.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 6.
The pair of sex chromosomes are called autosomes.
Answer:
False. The pair of sex chromosomes is called allosomes.

Question 7.
In DNA, Adenine always pairs with thymine and cytosine always pairs with guanine.
Answer:
True

Question 8.
In humans there are 23 pairs of autosomes and one pair of allosomes.
Answer:
False. In humans, there are 22 pairs of autosomes and one pair of allosomes.

Question 9.
The phenotypic and genotypic ratios are not same.
Answer:
True

Question 10.
Phenotype means the pairs of genes responsible for the visible characteristics of organisms.
Answer:
False. Phentotype means external appearance of visible characteristics of organisms.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 11.
During gamete formation, in Pj generation the pair of gametes separate independently.
Answer:
True

Question 12.
Down syndrome is caused due to monosomy of X chromosome.
Answer:
False. Down syndrome is caused due to trisomy of 21st chromosome.

Question 13.
In Klinefelter syndrome, women are sterile.
Answer:
False. In Klinefelter syndrome, men are sterile as this disorder arises in men due to abnormality in sex chromosome.

Question 14.
If the father and mother are both sufferers or carriers of sickle-cell anaemia, their offsprings are likely to suffer from this disease.
Answer:
True

Question 15.
During fertilization, mitochondria is contributed by the sperm cell and egg cell(ovum).
Answer:
False. During fertilization, mitochondria is contributed by the egg cell (ovum) alone.

Question 16.
Polygenic disorders strictly follow Mendel’s principles of heredity.
Answer:
False. Polygenic disorders do not strictly follow Mendel’s principles of heredity.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 17.
Genetic material is transferred in equal quantity from parents to progeny.
Answer:
True

Give scientific reasons:

Question 1.
DNA molecules are called as ‘Master molecules’.
Answer:

  • Molecules of DNA are present in all organisms from viruses and bacteria to human beings.
  • These molecules control the functioning, growth and division (reproduction) of the cell.
  • Genes present in the DNA are also responsible for transfer of hereditary characteristics from parents to offsprings.
  • Therefore, DNA molecules are called as ‘Master molecules’.

Question 2.
Phenotypic and genotypic ratios are different.
Answer:

  • Phenotype means external appearance or visible characteristics of organisms whereas the genotype is the pairs of genes responsible for the visible characteristics.
  • The genes responsible for any particular character are present in pairs.
  • Though, there are two genes, the phenotype
    depends on the presence of the dominant gene, e.g. Genotype for tall height of the plant is TT or Tt.
  • Therefore, phenotypic and genotypic ratios are different.

Question 3.
A carrier or sufferer of sickle-cell anaemia should avoid marriage with another carrier or sufferer.
Answer:

  • Sickle-cell anaemia is a hereditary disease caused due to mutation of a single gene.
  • It is a monogenic disorder that occurs due to changes in a gene during conception.
  • If father and mother both are sufferers or carriers of sickle-cell anaemia, their offsprings are likely to suffer from the disease.
  • Therefore, a carrier or sufferer of sickle-cell anaemia should avoid marriage with another carrier or sufferer.

Question 3.
Mitochondrial disorders are inherited from the mother only.
Answer:

  • Mitochondrial DNA becomes defective due to mutation.
  • During fertilization, mitochondria are contributed by the egg cell (ovum) alone.
  • Hence, mitochondrial disorders are inherited from the mother only.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 4.
Tobacco smoking causes cancer.
Answer:

  • Tobacco smoke contains harmful chemicals like pyridine, ammonia, aldehyde furfural, carbon monoxide, nicotine, sulphur dioxide etc.
  • They cause uncontrolled cell division.
  • Tobacco smoke is full of minute carbon particles which cause normal tissue lining of the lung to transform into thickened black tissue. This leads to cancer.
  • Therefore, tobacco smoking causes cancer.

Define the following terms:

Question 1.
Gene
Answer:
The segment of DNA which contains all the information for synthesis of a particular protein is called a ‘gene’.

Question 2.
Chromosome
Answer:
The structure in the nucleus of the cells that carries the hereditary characteristics is called chromosome.

Question 3.
Genetics
Answer:
The branch of biology which studies the transfer of characteristics of organism from one generation to the next and genes in particular, is called genetics.

Question 4.
Heredity
Answer:
Transfer of characteristics from parents to offsprings is called heredity.

Question 5.
Homologous chromosomes
Answer:
If the pair consists of chromosomes are similar in shape and organization, they are called homologous chromosomes.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 6.
Heterologous chromosomes
Answer:
If the pair which consists of chromosomes are not similar in shape and organization, they are called heterologous chromosomes.

Question 7.
Genetic disorders
Answer:
Diseases or disorders occuring due to abnormalities in chromosomes and mutations in genes are called genetic disorders.

Distinguish between:

Question 1.
DNA and RNA
Answer:

DNA RNA
(i) In DNA, the sugar present is deoxyribose. (i) In RNA, the sugar present is ribose.
(ii) In DNA, the nitrogen bases are adenine, guanine, cytosine and thymine. (ii) In RNA, the nitrogen base thymine is replaced by uracil.
(iii) DNA is double stranded. (iii) RNA is single-stranded.
(iv) DNA carries hereditary information (iv) RNA helps in protein synthesis.

Question 2.
Turner syndrome and Klinefelter syndrome.
Answer:

Turner syndrome Kline fater syndrome
(i) It is due to monosomy of sex chromosome. (i) It is due to felter of sex chromosome.
(ii) Disorder in sex chromosome results in 44 + X condition. (ii) Disorder in sex chromosome results is 44 + XY condition.
(iii) Seen in women. (iii) Seen in men
(iv) Women suffering from this syndrome are sexually sterile. (iv) Men suffering from this syndrome are sexually sterile.
(v) There is presence of total 45 chromosomes instead of 46. (v) There is presence of total 47 chromosomes instead of 46.

Write short notes on:

Question 1.
Albinism
Answer:

  • Albinism is a monogenic disorder.
  • Our eyes, skin and hair have colour due to the brown pigment melanin. In this disease, the body cannot produce melanin. Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation
  • The skin becomes pale, hair are white and eyes are usually pink due to absence of melanin pigment in the retina and sclera.

Question 2.
Polygenic disorders.
Answer:

  • Polygenic disorders are caused due to mutations in more than one gene.
  • In most such cases, their severity increases due to effects of environmental factors on the foetus.’
  • Common examples of such disorders are cleft lip, cleft palate, constricted stomach, spina bifida (a defect of the spinal cord), etc. Besides diabetes, blood pressure, heart disorders, asthma and obesity are also polygenic disorders.
  • Polygenic disorders do not strictly follow Mendel’s principles of heredity.
  • These disorders arise from a complex interaction between environment, life style and defects in several genes.

Question 3.
Turner syndrome.
Answer:

  • Turner syndrome is a disorder arising in women due to abnormality in sex chromosomes.
  • Turner syndrome arises due to either inheritance of only one X chromosome from parents or due to inactivation of the gender- related part of X-chromosomes.
  • Instead of the normal 44 + XX condition, women suffering from Turner syndrome show a 44 + X condition.
  • Such women are sterile i.e. unable to have children due to improper growth of the reproductive organ.

Answer the following questions:

Question 1.
What are the main objectives of National Health Mission?
Answer:
The main objectives of National Health Mission are:

  • Strengthening of the rural and urban health facilities.
  • Controlling various diseases and illnesses.
  • Increasing public awareness about health.
  • Offering financial assistance to patients through various schemes.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 2.
Which were the seven pairs of contrasting characteristics studied by Mendel in pea plant?
Answer:
The seven pairs of contrasting characters studied by Mendel in pea plant were as follows:

Characters Dominant Recessive
Shape of the seed Round(R) Wrinkled (r)
Colour of the seed Yellow (Y) Green (y)
Colour of the flower Purple (C) White (c)
Shape of pod Inflated (I) Constricted (i)
Colour of pod Green (G) Yellow (g)
Position of flower Axillary (A) Terminal (a)
Height of the plant Tall (T) Dwarf (t)

Question 3.
Name some dominant and recessive characteristics seen in human beings.
Answer:
Some dominant and recessive characteristics of human beings.

Dominant Recessive
Rolling tongue Non-rolling tongue
Presence of hair on arms Absence of hair on arms
Black and curly hair Brown and straight hair
Free earlobe Attached earlobe

Question 4.
What is Klinefelter syndrome?
Answer:

  • Klinefelter syndrome is a disorder arising in men due to abnormalities in sex chromosomes.
  • In this disorder, men have one extra X chromosome, hence their chromosomal condition becomes 44 + XXY.
  • Such men are usually sterile because their reproductive organs are not well developed.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 5.
How is the diagnosis for sickle-cell anaemia made?
Answer:

  • Under the National Health Mission scheme, the ‘Solubility Test’ for diagnosis of sickle-cell anaemia is available at all district hospitals.
  • Similarly, the confirmatory diagnostic test – ‘Electrophoresis’ is performed at rural and subdistrict hospitals.

Question 6.
Find out the Phenotypic ratio of the following:
(a) Round-Yellow
(b) Wrinkled-Yellow
(c) Round-Green
(d) Wrinkled-Green
Answer:
(a) Round-Yellow: 9
(b) Wrinkled-Yellow: 3
(c) Round-Green: 3
(d) Wrinkled-Green: 1
Phenotypic ratio: 9:3:3:1

Question 7.
Find out the Genotypic ratio of the following:
(a) RRYY
(b) RRYy
(c) RRyy
(d) RrYY
(e) RrYy
(f) Rryy
(g) rrYY
(h) rrYy
(i) rryy
Answer:
(a) RRYY -1
(b) RRYy-2
(c) RRyy-1
(d) RrYY-2
(e) RrYy-4
(f) Rryy-2
(g) rrYY – 1
(h) rrYy – 2
(i) rryy – 1
Genotypic ratio: 1:2:1:2:4:2:1:2:1

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Draw neat and labelled diagrams of the following:

Question 1.
Structure of chromosome
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 10

Question 2.
Types of RNA
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation 11

Answer the following questions:

Question 1.
What are the effects of tobacco consumption?
Answer:

  • Smoking of cigarettes and bidis adversely affects the process of digestion.
  • It causes a burning sensation in the throat and cough.
  • Excessive smoking causes instability and trembling of fingers.
  • It causes dry cough which leads to sleeplessness.
  • Tobacco consumption can lead to shortening of life span, chronic bronchitis, pericarditis, cancer of the lungs, mouth, larynx (voice box), pharynx, urinary bladder, lips or tongue.
  • The nicotine present in tobacco affects the central and peripheral nervous system. Arteries become hard i.e. it causes arteriosclerosis and hypertension.

Maharashtra Board Class 9 Science Solutions Chapter 16 Heredity and Variation

Question 2.
Complete the table for number of chromosomes in different organisms.
Answer:

Organism No. of Chromosomes
Crab 200
Maize 20
Frog 26
Roundworm 04
Potato 48
Human 46
Dog 78
Elephant 56
Fruit fly 08
Mango 40

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Class 9 Science Chapter 18 Observing Space: Telescopes Textbook Questions and Answers

1. Fill in the blanks with the proper words.

a. The wavelength of visible light is between …………………..….. and …………………..…. .
b. GMRT is used for …………………..….. waves.
c. A certain X-ray telescope is named after scientist …………………..…. .
d. The first scientist to use a telescope for space observations …………………..…. .
e. The biggest optical telescope in India is situated at …………………..…. .

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

2. Form pairs

Column ‘A’ Column ‘B’
(1) X-rays (a) GMRT
(2) Optical telescope (b) ISRO
(3) Indian radio telescope (c) Hubble
(4) Launching artificial-satellite (d) Chandra

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b),

3. What are the difficulties in using ground-based optical telescopes? How are they overcome?
Answer:
Difficulties:

  • The intensity of visible light coming from a heavenly body decreases due to absorption as it passes through the Earth’s atmosphere. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • Due to changes in atmospheric temperature and pressure light rays change their path.
  • Because of sunlight, we cannot use optical telescopes during the day.
  • During the night, city lights and cloudy weather cause difficulties.

Solution:

  • To reduce these problems, optical telescopes are situated on top of mountains, in uninhabited places.
  • To get rid of all the above problems completely, the telescope is placed above the earth’s atmosphere, in space.

4. Which type of telescopes can be made using a concave mirror, convex mirror, plane mirror, and a lens? Draw diagrams of these telescopes.
Answer:

  • A Newtonian telescope can be made using a concave mirror, plane mirror and a lens.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 4
  • A Cassegrain telescope can be made using a convex mirror, concave mirror and a lens.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 5
  • A refracting telescope can be made using two or more lenses.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 6

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

5. Study the figure and answer the following questions:
Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 3

a. What type of telescope is shown in the figure?
b. Label the main parts of the telescope.
c. Which type of mirror does the telescope use?
d. What other type of telescope uses a curved mirror?
e. Explain the working of the above telescope.
Answer:
(a) The Newtonian telescope (Reflecting telescope)
(b)
Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes 1
(c) Concave mirror
(d) The Cassegrain telescope
(e) Working.

  • Light rays coming from space are reflected by the concave mirror.
  • Before these reflected rays converge at the focus, they are deflected again by a small plane mirror.
  • As a result, they get focused at a point lying on the perpendicular to the axis of the telescope’s cylinder.
  • They pass through the eyepiece and we get a magnified image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

6. Answer the following questions.

a. Explain the construction of Galileo’s telescope.
Answer:

  • The Galileo’s telescope is made up of two or more convex lenses.
  • The lens facing the object is called the objective lens which is made as large as possible to collect the maximum amount of light coming from a heavenly object.
  • The lens closer to the eye is called the eyepiece which uses the light collected by the objective lens and produces a large image of the source.
  • Such a telescope is also called a refracting telescope.

b. Explain the construction of a radio telescope.
Answer:

  • The radio telescope is made from one or more dishes of a particular parabolic shape.
  • As in optical telescope, the incident radio waves are reflected by these dishes and converge at the focus.
    Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space Telescopes 2
  • A radio receiver is placed at the focal point.
  • The information gathered by this receiver is passed on to a computer which analyses it and constructs an image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

c. Why are optical telescopes located in uninhabited places on mountains?
Answer:

  • The visible light coming from a heavenly body has to pass through the earth’s atmosphere to reach the earth’s surface.
  • During this journey, some of the light is absorbed by the atmosphere and the intensity of the light reaching the earth’s surface decreases.
  • A second problem is caused by the changes in atmospheric pressure and temperature.
  • These changes cause turbulence in the atmosphere which in turn cause the light rays to change their path slightly and thereby shake the position of the image.
  • We cannot use optical telescope during the day. because of sunlight also.
  • During the night too, city lights and cloudy weather can cause difficulties in observing the heavenly bodies.
  • Hence, optical telescopes are located in uninhabited places on mountains.

d. Why can an X-ray telescope not be based on the earth?
Answer:

  • X-rays are emitted from distant heavenly bodies.
  • The rays coming towards the earth will be absorbed by the earth’s atmosphere and will go undetected.
  • Hence, X-ray telescopes are placed in an orbit outside the earth’s atmosphere and not on the earth.

Class 9 Science Chapter 18 Observing Space: Telescopes Intext Questions and Answers

Question 1.
What is meant by space observation? Why is it important?
Answer:
The observation of distant planets, galaxies and other astronomical objects is called as space observation.

  • Various space based satellites support our telecommunication networks by providing us television broadcasting.
  • Meteorology services such as weather forecasting help in disaster management.
  • Observation of positions of stars and other heavenly bodies has helped in the study of solar system, solar winds, pulsars, supernova, etc.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 2.
What is the difference between sky and space?
Answer:
Sky: It is a layer of gas that surrounds the earth in which sunlight is scattered making it visible from the surface of the earth.
Space: The vast emptiness beyond the sky is called as space.

Class 9 Science Chapter 18 Observing Space: Telescopes Additional Important Questions and Answers

Choose and write the correct options:

Question 1.
The first telescope was invented by
(a) Einstein
(b) Galileo
(c) Newton
(d) Mendel
Answer:
(b) Galileo

Question 2.
Eight is an wave.
(a) electric
(b) magnetic
(c) electromagnetic
(d) mechanical
Answer:
(c) electromagnetic

Question 3.
Visible radiation telescopes are also called as
(a) Gamma Ray telescopes
(b) Optical telescopes
(c) X-ray telescopes
(d) Radio telescopes
Answer:
(b) Optical telescope

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 4.
Reflecting telescopes are mainly of two types : ……………….. and ………………. .
(a) Newtonian and Galileoan
(b) Optical and Cassegrain
(c) Newtonian and Cassegrain
(d) Optical and Refracting
Answer:
(c) Newtonian and cassegrain

Question 5.
Giant Meterwave Radio Telescope (GMRT) has been erected at near Pune.
(a) Asangaon
(b) Talegaon
(c) Narayangaon
(d) Bhategaon
Answer:
(c) Narayangaon

Question 6.
Visible light coming from heavenly bodies have to pass through the to reach the earth surface.
(a) stratosphere
(b) troposphere
(c) atmosphere
(d) biosphere
Answer:
(c) atmosphere

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 7.
To collect the maximum amount of light coming from an object, the objective lens should be made as as possible.
(a) large
(b) small
(c) circular
(d) flat
Answer:
(a) large

Question 8.
The images formed by lenses have errors of colours. This is called
(a) chromatic disruption
(b) chromatic aberration
(c) chromatic reflection
(d) chromatic reaction
Answer:
(b) chromatic aberration

Question 9.
Radio telescope is made up of one or more dishes of shape.
(a) spherical
(b) hyperbolic
(c) parabolic
(d) hexagonal
Answer:
(c) parabolic

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) X-rays (a) GMRT
(2) Optical telescope (b) ISRO
(3) Indian radio telescope (c) Hubble
(4) Launching artificial satellite (d) Chandra
(5) Radio waves (e) telecommunication
(6) INSAT and GSAT (f) longer than 20cm

Answer:
(1 – d),
(2 – c),
(3 – a),
(4 – b),
(5 – f),
(6- e)

Question 2.

Column ‘A’ Column ‘B’
(1) Micro waves
(2) Infrared waves
(3) Visible light rays
(4) Ultraviolet rays
(5) X-rays
(a) 800 nm – 0.3mm
(b) 360 pm – 400 nm
(c) 3 pm – 300 pm
(d) 400 nm – 800nm
(e) 0.3 mm -20cm

Answer:
(1 – e),
(2 – a),
(3 – d),
(4 – b),
(5 – c)

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Define the following:

Question 1.
Visible radiation
Answer:
Light is an electromagnetic wave. Every wave has a characteristic wavelength. Our eyes can see only that light which has wavelengths between 400 nm to 800 nm. Such light is called visible radiation.

Question 2.
Reflecting Telescope
Answer:
When light falls on the mirror, it gets reflected. Telescopes that use concave mirrors are called reflecting telescope.

Question 3.
Optical telescopes
Answer:
Telescopes which are made from regular lenses and mirrors and used to see visible radiations coming from space are called optical telescopes.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 4.
Refracting Telescopes
Answer:
Light rays change their direction as they enter a lens from the atmosphere and again when they enter the atmosphere after passing through the lens. This is called refraction. The telescopes that use such lenses are called refracting telescopes.

Question 5.
Radio telescopes
Answer:
Many heavenly bodies emit radio waves in addition to visible radiation. A special type of telescope which is used to receive these rays are known as radio telescope.

Answer the following:

Question 1.
How do refracting telescopes work?
Answer:

  • Most refracting telescopes are made with two or more lenses.
  • The two lenses are called objective lens and an eyepiece lens.
  • The objective lens should be large so that maximum light can be collected from the source.
  • Using the light collected, the eyepiece lens which is small in size, produces a large image of the source.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 2.
What are the difficulties in using refracting telescopes?
Answer:

  • We require a large objective lens which is very difficult to make. Also large lenses are heavy and tend to get distorted.
  • As the objective and eyepiece lens are placed on the opposite sides of the telescope, the length of the telescope also increases with increase in the size of the lenses. This makes it difficult to manage a large telescope.
  • The images formed by lenses have errors of colours. This is called chromatic aberration.

Question 3.
How does a radio telescope work?
Answer:

  • A radio telescope is made from one or more dishes of a particular parabolic shape.
  • The incident radio waves are reflected by these dishes and converge at a focus.
  • A radio receiver is placed at focal point.
  • The information gathered by this receiver is then passed onto a computer which analyses and constructs the image of the source.

Question 4.
What are the two types of reflecting telescopes?
Answer:

  • The reflecting telescopes are mainly of two types: Newtonian and Cassegrain.
  • The newtonian telescope uses a plane mirror and a concave mirror.
  • The Cassegrain telescope uses a convex and a concave mirror.

Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes

Question 5.
What is the purpose of launching X-Ray telescope Chandra?
Answer:
Chandra was launched to study X-rays coming from heavenly objects. Special mirrors which . can reflect X-rays were used in this telescope.

Answer the following:

Question 1.
State some of the satellites and their uses.
Answer:

  • The INSAT and GSAT series of satellites support our telecommunication network, television broadcasting and meteorological services.
  • It is because of them that telephone, television and internet services are available everywhere in the country.
  • The EDUSAT satellite is used exclusively for education.
  • The IRS satellite series is used for the monitoring and management of natural resources as well as disaster management. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • Astrosat: It has ultraviolet and X-ray telescopes and detectors. It is a unique system having different kinds of telescopes on a single satellite.

Question 2.
What is the specialty of the artificial satellite Astrosat?
Answer:

  • This satellite was launched by Indian Space Research Organization (ISRO).
  • It has ultraviolet and X-ray telescopes and detectors. Most of the parts used in this satellite are made in India.
  • It is an unique system having different kinds of telescopes on a single satellite. Indian scientists are studying various aspects of the Universe using the data obtained with these telescopes.

Answer in detail:

Question 1.
Explain any two space observation telescope.
Answer:
Hubble Telescope:

  • In 1990, the National Aeronautics and Space Administration launched into space an optical telescope called the Hubble telescope.
  • It has a mirror of diameter 94 inches and is orbiting the earth at a height of 589 km from it.
  • This telescope is still working and has helped to make important discoveries.

Chandra Telescope:

  • In 1999, the National Aeronautics and Space Administration launched an X-ray telescope named Chandra, in space, to study X-rays coming from heavenly objects.
  • Special mirrors which can reflect X-rays were used in this telescope.
  • Chandra has given us very useful information about stars and galaxies.
  • The telescope is named after the famous Indian scientist Subramanian Chandrashekhar.

Question 2.
Write short note : GMRT
Answer:

  • Giant Meterwave Radio Telescope (GMRT) is a large radio telescope erected at Narayangaon near Pune.
  • It uses radio waves having wavelengths of about a metre, coming from heavenly bodies to study those bodies. This telescope is actually a collection of 30 dishes, each having a diameter of 45 m.
  • It is called a giant telescope as the arrangement of the 30 dishes over an area which measures up to 25 km across. It works as a single dish having a diameter of 25 km. Maharashtra Board Class 9 Science Solutions Chapter 18 Observing Space: Telescopes
  • GMRT gives the same data that we would have got from a telescope having a single dish of 25 km diameter.
  • Scientists study the solar system, solar winds, pulsars, supernova, interstellar hydrogen clouds, etc. with the help of the GMRT.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Class 9 Science Chapter 17 Introduction to Biotechnology Textbook Questions and Answers

1. Each of the following statements is wrong. Rewrite them correctly by changing either one or two words.

a. Simple squamous epithelium is present in the respiratory tract.
b. Glandular epithelium is present in kidneys.
c. Chlorenchyma helps the plant to float in water.
d. Striated muscles are also called involuntary muscles.
e. Chloroplast is present in permanent tissue.
Answer:
a. False. Ciliated columnar epithelium is present in respiratory tract.
b. False. Cuboidal epithelium is present in kidneys.
c. False. Aerenchyma helps the plant to float in water.
d. False. Striated muscles are also called voluntary muscles.
e. False. Chlorenchyma is present in permanent tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

2. Identify the odd word and explain why it is odd.

a. Xylem, phloem, permanent tissue, meristematic tissue.
Answer:
Meristematic tissue. It has the ability to divide whereas the others have lost the ability to divide.

b. Epithelium, Muscle fibre, nerve fibre, the epidermis.
Answer:
Epidermis. It is a plant tissue whereas the others are animal tissues.

c. Cartilage, bone, tendon, cardiac muscle.
Answer:
Cardiac muscle. It is a muscular tissue whereas the others are connective tissues.

3. Write the names of the following tissues.

a. Tissue lining inner surface of mouth.
Answer:
Squamous epithelium.

b. Tissue joining muscles and bones.
Answer:
Tendon

c. Tissue responsible for increasing height of plants.
Answer:
Apical meristem.

d. Tissue responsible for increasing girth of stem.
Answer:
Lateral meristem.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

4. Write the differences.
Simple tissue and complex tissues in plants.
Answer:

Simple Tissue in plants Complex Tissues in plants
(i) They are made up of only one type of cells. (i) They are made up of more than one type of cells.
(ii) They are found in all parts of the plant. (ii) They are found in the vascular regions of the plant.
(iii) They perform different functions like storage (iii) They mainly perform the function of
of food, support, giving strength etc. conduction of water and food.
(iv) Examples – Parenchyma, Collenchyma, Sclerenchyma. (iv) Example – Xylem and phloem.

5. Write short notes.

a. Meristematic tissue.
Answer:

  • Meristematic tissue is present in specific parts of a plant where growth takes place.
  • Cells of meristematic tissue contain thick cytoplasm, a conspicuous nucleus and a thin cell wall and are compactly packed together.
  • Vacuoles are usually absent in these cells and they are highly active.
  • The main function of meristematic tissue is to bring about plant growth.
  • According to the location, meristematic tissues are of three types: Apical meristem, intercalary meristem and lateral meristem.

b. Xylem
Answer:

  • Xylem is a complex permanent tissue in plants.
  • It consists of thick-walled dead cells.
  • The type of cells in xylem are trachieds, vessels, xylem fibres (dead cells) and xylem parenchyma (living cells).
  • Its structure is like interconnected tubes conduct water and minerals only in upward direction.

c. Striated muscles.
Answer:

  • The cells of striated muscles are long, cylindrical, multinucleate and have no branches.
  • These are alternate dark and light bands on these muscles.
  • As they are attached to bones, they are also called skeletal muscles.
  • They move as per our will, hence they are called voluntary muscles.
  • Striated muscles bring about movements of arms and legs, running, speaking etc.

d. Agro-complementary business.
Answer:
(i) The business that are complementary to agriculture and generate supplementary income for the farmers are called agro complementary business.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

(ii) These include:

  • Animal Husbandry: It is practiced for milk production and for using the cattle as help in farming operations, e.g. cows and buffaloes are raised for milk whereas bulls and male buffaloes for pulling heavy loads.
  • Poultry farming: It is the rearing of egg and meat yielding chickens. Chickens raised for eggs are called layers while those raised for meat are called broilers.
  • Sericulture: It is the rearing of silkworms (moths) for silk production. The silk fibres obtained are processed, reeled and then woven into fabric.

e. Genetic engineering
Answer:

  • Genetic engineering is the deliberate modification of the characteristics of an organism by manipulating its genetic material.
  • An organism that is generated through genetic engineering is called a genetically modified organism (GMO).
  • Genetic engineering is applied in many fields like research, agriculture, industrial biotechnology and medicine.
  • In agriculture, genetic engineering is used in the production of cash crops, improvement in varieties of cash crops, increase in ability of plants to withstand environmental stresses.
  • In medicine, genetic engineering is used for vaccine production, early diagnosis of congenital disease, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

f. Sericulture
Answer:

  • Sericulture is the rearing of silkworms (moths) for production of silk.
  • Bombyx mori is the most commonly used variety of silkworm for this purpose.
  • The life cycle of silkworm consist of four stages namely egg, larva, pupa and adult.
  • Thousands of eggs deposited by female moths are incubated artificially to shorten the incubation period.
  • Larvae hatching out of eggs are released on mulberry plants.
  • Larvae are nourished by feeding on mulberry leaves.
  • After feeding for 3-4 days, larvae move to branches of mulberry plant.
  • The silk thread is formed from the secretion of their salivary glands.
  • Larvae spin this thread around themselves to form a cocoon. The cocoon may be spherical in nature.
  • Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
  • Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
  • These fibres are unwound, processed and reeled. Various kinds of fabric is woven from silk threads.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

6. Explain the meaning of biotechnology and its impact on agricultural management with suitable examples.
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called biotechnology. Impact of biotechnology on agricultural management:

  • Genetically modified crops (GM crops) are being produced by introducing changes in DNA of natural crops.
  • Normally such varieties are not found in nature.
  • Thus, new varieties are produced artificially Different useful characters are introduced in such varieties.
  • Some naturally occurring varieties cannot withstand environmental stress like frequently changing temperature.
  • Wet and dry famines, changing climates etc. However, GM crops can grow in any of such adverse conditions.
  • As GM crops are resistant to insect pests, pathogens, chemical weedicides, etc. the use of harmful chemicals like pesticides can be avoided.
  • Due to use of seeds of GM crops, there is improvement in nutritive value and decrease in loss of crops.

Draw neat and labelled diagrams of the following:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 2
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 3

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

7. Which two main techniques are used in biotechnology? Why?
Answer:

  • The two main techniques used in biotechnology are – Genetic engineering and tissue culture.
  • These techniques are used to bring about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings.
  • Its uses are in the production of cash crops, improvement in varieties of cash crops, early diagnosis of congenital diseases, organ transplant, cancer research, production of artificial skin, cartilage etc. in laboratories.

8. Discuss ‘Agritourism’ in the class and write a project on an agrotourism centre nearby. Present it in the class in groups.
Answer:

  • In agritourism, plantlets of flowering, medicinal, ornamental, vegetable plants and fruit trees are produced on a large scale by tissue culture technique.
  • By growing some of the plants fully, an agritourism centre can be developed.
  • If sufficient land is available, the emerging field of agritoursim would be a good business.
  • An agritourism centre consists of following:
    (a) Mango, chikoo (sapota), guava, coconut, custard apple and some other regional fruit trees.
    (b) Shade giving local or exotic attractive plants.
    (c) Ornamental and flowering plants.
    (d) Butterfly garden.
    (e) Medicinal plant garden.
    (f) Organic vegetables and fruits.
  • People visit places with such attraction in large numbers.
  • Selling plantlets/seedlings, fruits, vegetables at such places can be quite profitable.

9. Define the term tissue and explain the concept of tissue culture.
Answer:
A group of cells having the same origin, same structure and same function is called tissue. Concept of tissue culture:

  • Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium is called tissue culture.
  • Nowadays, a complete organism can be developed from a single cell or from tissue with the help of the tissue culture technique. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • A liquid, solid or gel-like medium prepared from agar, which supplies nutrients and energy necessary for tissue culture is used in this technique.
  • Tissue culture can be used to grow plants on a large scale, which bear flowers, fruits of excellent quality, in shorter durations and are disease free.
  • The various processes involved in tissue culture are:

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 7

10. ‘Rearing of sheep is a livestock’. Justify this statement.
Answer:

  • The term livestock refers to animals reared for profit or for use.
  • Sheep provides us with wool, skin, meat and milk.
  • Therefore, rearing of sheep is a livestock.

Class 9 Science Chapter 17 Introduction to Biotechnology Intext Questions and Answers

Question 1.
Which components bring about important processes in the living organisms?
Answer:
Cells, tissues, organs, organ systems bring out important processes in living organisms.

Question 2.
Which is the smallest structural and functional unit of the body of living organisms?
Answer:
Cell is the smallest structural and functional unit of the body of living organisms.

Question 3.
Which type of muscle is the diaphragm of the respiratory system?
Answer:
Diaphragm is a skeletal muscle. It is an involuntary muscle that regulates breathing, although some voluntary control can be achieved.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
What is the main difference between the growth of animals and plants?
Answer:

  • Growth in animals is uniform whereas the growth in plants occurs in specific parts of the body.
  • Growth in animals occurs for a certain period of life whereas the growth in plants occur throughout their life.

Question 5.
Why does the growth of a plant occur only at specific parts of the plant body?
Answer:
Growth of a plant occurs only at specific parts of the body due to the presence of meristematic tissues which contain dividing cells.

Question 6.
Suppose you want to grow a garden like the one shown in the picture, around your home or school. What would you do to achieve that? By which methods will you cultivate the seedlings?
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 4
Answer:

  • To grow a garden around the school or home, one must cultivate seedlings of different plants and those seedlings must be watered regularly and nurtured properly.
  • Seedlings can be cultivated by the following methods:
    (a) Seed sowing: Seeds are sown, watered and allowed to germinate.
    (b) Transplantation: Seeds are sown somewhere else and then the seedlings are transplanted in the garden.

Question 7.
You must have seen flowers of same variety but of 2 or 3 different colours borne by same plant. How is this possible?
Answer:

  1. The different coloured flowers borne by the same plant are due to the pigments like anthocyanins, carotenoids etc. according to the genetic makeup of the plant.
  2. Also, by using the latest techniques of biotechnology and tissue culture, it is possible to manipulate the genes for flower colour and get the desired flower colour.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
What keeps the various organs and organ systems separate from each other? Why?
Answer:
The epithelial tissue keeps the various organs and organ systems separate from each other by forming a layer on the organs.

Question 9.
Why are epithelial tissues said to be simple tissues?
Answer:
Epithelial tissues are said to be simple tissues as they are made up of only one type of cells.

Question 10.
Why do slim persons feel more cold in winter than those who are obese?
Answer:

  • Below the skin, there is a tissue called adipose tissue.
  • The cells of these tissue are filled with fat droplets.
  • Due to storage of fat, it acts as an insulator and helps to retain heat in the body.
  • Obese people have more fat deposited in the adipose tissue as compared to slim persons. Therefore, slim persons feel more cold in winter than those who are obese.

Question 11.
Why can bones not be folded?
Answer:

  • Bone cells called osteocytes are embedded in solid ground substance made up of calcium phosphate.
  • This makes the bone hard, rigid and non- flexible. Therefore, bones cannot be folded.

Question 12.
Which other industries can be started as an extension of the plant nursery business?
Answer:
Businesses like Agritourism, Ecotourism, forest resorts and organic fruit gardens, yoga and meditation centres can be started as an extension of the plant nursery business.

Question 13.
To which places do people choose to go on vacation in order to relax when they are tired of crowds and stressful life?
Answer:
People choose to go to hill stations, beaches, forest resorts and places where there is lot of greenery to relax when they are tired of crowds and stressful life.

Question 14.
What is the inter-relationship between the two questions (7 and 8) above?
Answer:
The inter-relationship between the above two questions is that businesses like agritourism, ecotourism, forest resorts etc. can fetch a huge profit as people look for these kind of places to relax, to be away from the hustle and bustle of city life and feel close to nature.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 15.
Why are the cocoons transferred to boiling water before the pupa develops into an adult?
Answer:

  • Once the pupa develops into an adult, it will secrete a fluid to dissolve the silk and emerge out.
  • Hence the cocoons are transferred to boiling water before the pupa develops in to an adult to kill the silkworms before transformation is complete.

Question 16.
Why we cannot see our organs like the heart, blood vessels and intestines?
Answer:
We cannot see our organs like the heart, blood vessels and intestines as they are located inside the body.

Question 17.
What is meant by white revolution? Who was its pioneer? What benefits did it bring?
Answer:

  • White revolution was the programme launched by the National Dairy Development Board to increase the milk production in India.
  • Dr. Verghese Kurien was the pioneer of white revolution.
  • It transformed India from a milk-deficient nation to the largest producer of milk. It also helped the dairy farmers in directing their own development and empowering them.

It also helped to reduce the malpractices carried out by milk traders and merchants.

Answer the following questions:

Question 1.
Collect information about GM varieties of crops in your area and make a note of them. Also find out if there are adverse effects of GM crops on human beings and environment.
Answer:

  • Some of the GM crops are:
    (a) Maize: MON 810, MON 863
    (b) Rice: Golden rice
    (c) Brinjal: BT brinjal
    (d) Cotton: BT cotton
  • Effects of GM Crops on human beings: GM crops may cause toxicity and allergic reactions in humAnswer:
  • Effects of GM Crops on environment:
    (a) GM crops may be toxic to pollinators and non-target species like butterflies.
    (b) Many GM crops may be a threat to soil ecosystem as they secrete their toxins into the soil.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
As shown in the figure, place an onion on each gas jar in such a way that its base (roots) will remain dipped in water. Measure and record the length of the roots of both onions on the first, second and third day. On the fourth day, cut off 1 cm of the roots of the onion in flask B Measure the length of the roots of both onions for the next five days and record your observations in the following table.
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 6

Length (cm) Day 1 Day 2 Day 3 Day 4 Day 5
Flask A
Flask B

Answer:
(Students are expected to record their observation in the given table).

(a) Which onion has longer roots? Why?
Answer:
The onion in jar Ahas longer roots as it continues to grow due to the presence of meristematic tissue at root tip which had dividing cells.

(b) Why did the roots of the onion in jar B stop growing?
Answer:
The roots of onion in jar B stopped growing as the meristematic tissue present in the root tips were cut off.

Question 3.
Take a fresh and fleshy leaf of Rhoeo, lily or any other plant. Pull and press it, tearing it obliquely in such a way that its transparent epidermis will be visible at the cut margins. Take the transparent epidermis with the forceps and keep it in dilute safranin solution for 1 minute. Spread it on a slide, cover it with a cover-slip and observe it under a compound microscope.
Answer:

  • The cells of the epidermis are flat and polygonal and there are no intercellular spaces between them.
  • A single continuous layer is formed.
  • The epidermis is covered by a waxy cuticle.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 4.
Establish your own plant nursery near your school or home. Prepare the seedlings of flowering plants, fruit plants, and ornamental plants being grown in your area. Can you start a business in the future with the help of this activity? Think it over.
Answer:
(Students are expected to do this activity on their own.)

Yes, we can start a business of plant nursey in future. If sufficient land is available, the emerging field of agritourism would be a good business.

Question 5.
Observe the skin of the back of your hand with the help of a magnifying lens. Do you see the closely attached squarish and pentagonal shapes?
Answer:
Yes, we can see the squarish and pentagonal shapes. This is the stratified squamous epithelium present on the outer layer of skin.

Question 6.
Observe a permanent slide of blood smear under a compound microscope. What did you see?
Answer:

  • A permanent slide of blood smear under a compound microscope shows different types of cells like RBCs, WBCs and platelets.
  • The different types of WBCs that can be seen are eosin mphocytes.

Question 7.
Visit a modern cowshed nearby and record the following points – The number of cattle, their variety, total milk production, cleanliness in cattle-shed, arrangements for health care of cattle.
Answer:
Students are expected to do this activity on their own.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 8.
Collect more information about animal husbandry.
Answer:

  • The branch of agriculture which deals with the feeding, shelter, health and breeding of domestic animals is called animal husbandry.
  • It is a scientific and systematic management of livestock.
  • Animal husbandry practices include:
    (a) Proper food and clean drinking water.
    (b) Proper shelter
    (c) Proper methods of breeding
    (d) Prevention and cure of disease.
  • Animal husbandary practices serve as an alternative income for the farmers and help to satisfy the need of food for man.

Question 9.
Find out from the internet the average daily milk yield from local and exotic varieties of cow.
Answer:

  • Milk yield from local cow : 12-15 litres/day.
  • Milk yield from exotic cow: 15-30 litres/day.

Question 10.
Are the structure and functions of the bodies of plants and animals the same?
No. The structure and functions of the bodies of plants and animals are not the same. Some differences between plants and animals:
Answer:

Plants Animals
Plants continue to grow throughout their life. Animals grow only for a specific period of life.
Growth in plants in not uniform. Growth in animals is uniform.
Plants have dividing and non-dividing tissues. Animals do not have different dividing and non­dividing tissues.
Plants are sedentary. Animals generally move from place to place in search of food, shelter and partners.
Energy needs of plants are less. Energy needs of animals are greater.
Plants can prepare their own food by photosynthesis. Animals cannot prepare their own food. They depend on plants and other animals for their food.

Class 9 Science Chapter 17 Introduction to Biotechnology Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
Protective coverings in the animal body are called tissues.
(a) meristematic
(b) muscular
(c) epithelial
(d) bone
Answer:
(c) epithelial

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Epithelial tissue is present in
(a) skin
(b) inner surface of blood vessels
(c) walls of the alveoli
(d) all of these
Answer:
(d) all of these

Question 3.
helps in selective transport of substances.
(a) Squamous epithelium
(b) Stratified epithelium
(c) Cuboidal epithelium
(d) Columnar epithelium
Answer:
(a) Squamous epithelium

Question 4.
helps in secretion of digestive juice, absorption of nutrients.
(a) Glandular epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Ciliated epithelium
Answer:
(b) Columnar epithelium

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
supports internal organs
(a) Areolar tissue
(b) Adipose tissue
(c) Lymph
(d) Tendon
Answer:
(a) Areolar tissue

Question 6.
tissue helps in insulation, supply of
Answer:
(b) Adipose tissue

Question 7.
connect muscles to bones.
(a) Ligaments
(b) Tendons
(c) Cartilages
(d) Nerves
Answer:
(b) Tendons

Question 8.
join two bones to each other.
(a) Cartilages
(b) Tendons
(c) Ligaments
(d) Muscles
Answer:
(c) Ligaments

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 9.
meristem increases the length of root and stem.
(a) Intercalary
(b) Apical
(c) Lateral
(d) Permanent
Answer:
(b) Apical

Question 10.
showed that cells and tissues can be grown ex vivo.
(a) Frederick Campion Steward
(b) Gregor Johann Mendel
(c) Frederick Miescher
(d) Robert Brown
Answer:
(a) Frederick Campion Steward

Question 11.
is an example of genetically modified maize.
(a) Vaishali
(b) Vistive Gold
(c) MON 810
(d) Amflora
Answer:
(c) MON 810

Question 12.
is an exotic variety of cow.
(a) Holstein
(b) Gir
(c) Sahiwal
(d) Dangi
Answer:
(a) Holstein

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 13.
is a layer chicken.
(a) Brahma
(b) Leghorn
(c) Cochin
(d) Aseel
Answer:
(b) Leghorn

Question 14.
is a broiler chicken.
(a) Leghorn
(b) Lehman
(c) Aseel
(d) Minorca
Answer:
(c) Aseel

Question 15.
is a broiler chicken.
(a) Minroca
(b) Ancona
(c) Leghorn
(d) Cochin
Answer:
(d) Cochin

Question 16.
Cartilage is found
(a) all around the cells in body
(b) in nose, ear, larynx, trachea
(c) between the muscles and skin
(d) around the blood vessels
Answer:
(b) in nose, ear, larynx, trachea

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 17.
Which of the following is an exotic variety of COW?
(a) Plymouth Rock
(b) New Hampshire
(c) Black Rock
(d) Brown Swiss
Answer:
(d) Brown Swiss

Question 18.
epithelium is present in the inner surface of respiratory tract.
(a) Stratified
(b) Columnar
(c) Ciliated
(d) Cuboidal
Answer:
(c) Ciliated

Question 19.
Bones consist of osteocytes embedded in solid ground substance made up of
(a) calcium bicarbonate
(b) calcium carbonate
(c) calcium phosphate
(d) calcium sulphate
Answer:
(c) calcium phosphate

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 20.
helps in growth of branches, formation of leaves and flowers.
(a) Lateral meristem
(b) Apical meristem
(c) Intercalary meristem
(d) Xylem
Answer:
(c) Intercalary meristem

Question 21.
tissue is present in the hard coat of seeds, outer covering of coconut.
(a) Parenchyma
(b) Xylem
(c) Collenchyma
(d) Sclerenchyma
Answer:
(d) Sclerenchyma

Question 22.
The cells of tissue are dead.
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) All of these
Answer:
(c) Sclerenchyma

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 23.
Cuboidal epithelium is found in
(a) inner surface of mouth, blood vessels
(b) inner layer of skin
(c) tubules of kidney, salivary gland
(d) inner surface of respiratory tract
Answer:
(c) tubules of kidney, salivary gland

Find the odd man out:

Question 1.
Cell body, muscle fibre, axon, dendrites.
Answer:
Muscle fibre. It is a muscle cell whereas the others are parts of nerve cell.

Question 2.
Tracheids, vessels, companion cells, xylem fibres.
Answer:
Companion cells. It is an element of phloem whereas the others are elements of xylem.

Question 3.
Sieve tubes, tracheids, phloem parenchyma, phloem fibres.
Answer:
Tracheids. It is an element of xylem whereas the others are elements of phloem.

Question 4.
Holstein, Brown swiss, Bombyx mori, Jersey.
Answer:
Bombyx mori. It is a variety of silkworm whereas the others are varieties of cows.

Complete the analogy:

(1) Inner surface of mouth : Squamous epithelium :: Inner surface of intestine :
(2) Glandular epithelium : Secretion of sweat, oil :: Cuboidal epithelium:
(3) Respiratory tract: Ciliated columnar epithelium :: Kidney tubules:
(4) Outer layer of skin : Stratified epithelium :: Inner layer of skin :
(5) Muscular tissue : Movement :: Nervous tissue :
(6) Tendons : Join muscles to bones :: Ligaments :
Answer:
(1) Columnar epithelium
(2) Secretion of saliva
(3) Cuboidal epithelium
(4) Glandular epithelium
(5) Conduction of excitation
(6) Join two bones to each other

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Squamous epithelium (a) Secretion of digestive juice
(2) Cuboidal epithelium (b) Selective transport of substances
(3) Columnar epithelium (c) Protection of organs
(4) Stratified epithelium (d) Secretion of saliva

Answer:
(1 – b),
(2 – d),
(3 – a),
(4 – c)

Question 2.

Column ‘A’ Column ‘B’
(1) Brahma (a) Local variety of cow
(2) Jersey (b) Layer chicken
(3) Devin (c) Exotic variety of cow
(4) Lehman (d) Broiler chicken

Answer:
(1 – d),
(2 – C),
(3 – a),
(4 – h)

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 3.

Column ‘A’ Column ‘B’
(1) Potato (a) Vistive Gold
(2) Maize (b) Amflora
(3) Soybean (c) Vaishali
(4) Tomato (d) MON 863

Answer:
(1-b),
(2-d),
(3-a),
(4-c)

Answer the following:

Question 1.
Explain the different types of muscular tissues.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 5

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Explain the types of complex permanent tissues.
Answer:

Name of tissue Xylem Phloem
Characteristics Consists of thick-walled dead cells Consists of cytoplasm containing living cells.
Types of cells Tracheids, vessels and xylem fibres – dead cells. Xylem parenchyma – living cells. Sieve tubes, companion cells, phloem parenchyma – living cells. Phloem fibres – dead cells.
Function A structure like interconnected tubes, conduct water and minerals only in upward direction. Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in upward and downward direction

State whether the following statements are true or false. Correct the false statement:

Question 1.
Lateral meristem increase girth(diameter) of the root and stem.
Answer:
True

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Due to use of seeds of GM crops, there is decrease in nutritive value and increase in loss of crops.
Answer:
False. Due to use of seeds of GM crops there is improvement in nutritive value and decrease in loss of crops.

Question 3.
In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
Answer:
True

Question 4.
Rare and endangered plants can be grown by tissue culture and can be protected from extinction.
Answer:
True

Question 5.
The silk thread is formed from the secretion of the sweat glands of silkworm.
Answer:
False. The silk thread is formed from the secretion of the salivary glands of silkworm.

Question 6.
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
Answer:
True

Question 7.
Phloem consists of thick-walled dead cells.
Answer:
False. Phloem consists of living cells containing cytoplasm.

Question 8.
Stratified squamous epithelium is present in inner surface of mouth, oesophagus, blood vessels and alveoli.
Answer:
False. Stratified squamous epithelium is present in the outer layer of skin.

Question 9.
Broiler chickens are raised for eggs.
Answer:
False. Broiler chickens are raised for meat.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 10.
Non-striated muscles bring about contraction and relaxation of heart.
Answer:
False. Cardiac muscles bring about contraction and relaxation of heart.

Name the following:

Question 1.
Tissue responsible for growth of branches, formation of leaves and flowers.
Answer:
Intercalary meristem.

Question 2.
Tissue that helps in insulation, supply of energy, storage of fats.
Answer:
Adipose tissue.

Question 3.
Tissue present in outer layer of skin.
Answer:
Glandular epithelium.

Question 4.
Tissue present in nose, ear, larynx, trachea.
Answer:
Cartilage.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Dead cells in xylem.
Answer:
Tracheids, vessels and xylem fibres.

Question 6.
Living cells in xylem.
Answer:
Xylem parenchyma.

Question 7.
Living cells in phloem.
Answer:
Sieve tubes, companion cells, phloem parenchyma.

Question 8.
Dead cells in phloem.
Answer:
Phloem fibres.

Question 9.
GM crops.
Answer:
BT cotton, Amflora, Golden Rice.

Question 10.
Local Indian varieties of cow.
Answer:
Sahiwal, Sindhi, Lai kandhari, Devni, Khillari, Dangi.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 11.
Exotic varieties of cow.
Answer:
Jersey, Brown swiss, Holstein.

Question 12.
Layers.
Answer:
Leghorn, Minorca, Ancona, Lehman.

Question 13.
Broilers.
Answer:
Brahma, Long, Cochin, Aseel.

Question 14.
Chickens reared for both eggs and meat.
Answer:
Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock.

Give scientific reason.

Question 1.
Blood is a complex tissue.
Answer:

  • A complex tissue is made up of more than one type of cells.
  • In blood, cells of different types, colour and shapes are mixed together.
  • Blood contains erythrocytes (RBCs), leucocytes (WBCs) and platelets in a liquid plasma.
  • Therefore, blood is a complex tissue.

Define the following:

Question 1.
Tissue
Answer:
A group of cells having the same origin, same structure and same function is called tissue.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Differentiation
Answer:
When cells acquire a specific structure, shape and location and perform a specific function, it is called differentiation.

Question 3.
Biotechnology
Answer:
The techniques of bringing about improvements in living organisms by artificial genetic changes and by hybridization for the welfare of human beings, are together called ‘Biotechnology’.

Question 4.
Tissue culture
Answer:
Ex vivo growth of cells or tissues in an aseptic and nutrient-rich medium’ is called tissue culture.

Write short notes

Question 1.
Animal husbandry
Answer:

  • In India, animal husbandry is practised for milk production and for using the cattle as help in farming operations.
  • Example – Cows and buffaloes are raised for milk and bulls and male buffaloes for pulling the heavy loads.
  • Local Indian varieties of cows like Sahiwal, Sindhi, Gir, Lai kandhari, Devni, Khillari, Dangi, etc. and exotic varieties like Jersey, Brown swiss, Holstein, etc. are kept for their milk.
  • Proper care of cattle is necessary for a clean and high yield of milk which includes:
    (a) A balanced diet, i.e. all constituents of food should be given to cattle. It must include fibre- rich coarse food, fodder, and sufficient water.
    (b) The cattle-shed should be clean and dry with proper ventilation and a roof.
    (c) Cattle should be regularly vaccinated.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Poultry farming.
Answer:

  • Rearing of egg and meat yielding chickens is called poultry farming.
  • Chickens raised for laying eggs are called layers . e.g. Leghorn, Minorca, Ancona, Lehman.
  • Chickens raised for meat are called broilers, e.g. Brahma, Long, Cochin, Aseel.
  • Rhode Island Red, New Hampshire, Plymouth Rock, Black Rock are varieties of chicken reared for both eggs as well as meat.
  • The objectives behind development of new hybrid varieties from a cross between Indian varieties like Aseel and exotic varieties like Leghorn are to produce good quality chickens in large numbers.
  • To develop the ability to withstand high temperature, to use by-products of agriculture as poultry feed, etc.

Distinguish between:

Question 1.
Xylem and Phloem
Answer:

Xylem Phloem
(i) Consists of mostly thick-walled dead cells. (i) Consists of mostly living cells containing cytoplasm.
(ii) The types of cells include dead cells- tracheids, vessels and xylem fibres and living cells-Xylem parenchyma. (ii) The types of cells include living cells – Sieve tubes, companion cells, phloem parenchyma and dead cells – phloem fibres.
(iii) Structure like interconnected tubes, conduct water and minerals only in upward direction. (iii) Tubes joined to each other, conduct sugar and amino acids from leaves to various parts in both upward and downward direction.

Question 2.
Striated muscles and Non-striated muscles.
Answer:

Striated muscles Non-striated muscles
(i) Muscle cells are long, cylindrical, multinucleate and have no branches. (i) Muscle cells are short, spindle-shaped, uninucleate and have no branches.
(ii) There are alternate dark and light bands on these muscles and they are attached to bones. (ii) Dark and light bands are absent. Not attached to bones.
(iii) They move as per our will, hence they are called voluntary muscles. (iii) They are not under the control of our will, hence they are called involuntary muscles.
(iv) These muscles bring about movements of arms and legs, running, speaking etc. (iv) These muscles bring about movement of eyelids, passage of food through alimentary canal, contraction and relaxation of blood vessels etc.

Question 3.
Cartilage and Bone
Answer:

Cartilage Bone
(i) They are present in nose, ear, larynx, trachea. (i) They form the skeleton of the body.
(ii) They contain cells supported by fibrous, flexible, jelly-like ground substance. (ii) They contain osteocytes embedded in solid ground substance made up of calcium phosphate.
(iii) Lubricates the surface of bones, gives support and shape to organs. (iii) Supports and protects different organs, helps in movement.

Answer the following questions:

Question 1.
Close your eyes and try to identify different objects by feeling them with your hand. Why is it possible for you to identify things like a note-book, text-book, bench, compass-box, etc. only by touching them?
Answer:

  • We can identify objects just by touching them because of the memory that we retain in our brain.
  • The nervous tissue enables us to respond to the stimuli of touch.

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Answer in detail:

Question 1.
Write the location, structure and function of following epithelial tissues:
(a) Squamous epithelium
(b) Stratified epithelium
(c) Glandular epithelium
(d) Columnar epithelium
(e) Ciliated epithelium
(f) Cuboidal epithelium
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 8

Question 2.
Write the structure and function of the following connective tissues:
(a) Blood
(b) Lymph
(c) Areolar tissue
(d) Adipose tissue
(e) Cartilage
(f) Bones
(g) Tendons and ligaments
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 9

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 2.
Describe the structure of nervous tissue with the help of a neat labelled diagram.
Answer:

  • Nervous tissue enables us to respond to stimuli like touch, sound, odour, colour, etc.
  • Cells of the nervous tissue are specifically made to become excited and conduct the excitation from one part of the body to another.
  • The main part of the nerve cell is the cell body which contains the cytoplasm and the nucleus.
  • Numerous, small, branched fibres called dendrites arise from the cell body.
  • One of the fibres, is extremely long and is called
    the axon.
  • The length of the nerve cell may be up to one metre.
  • Many nerve cells are bound together with the help of connective tissue to form a nerve.
  • Nervous tissue is present in the brain, spinal cord and the network of nerves spread all throughout the body.
  • In most animals, action in response to a stimulus occurs due to the integrated functioning of the nervous tissue and muscular tissue.
    Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology 1

Question 3.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Question 4.
Explain types, location and function of Meristematic tissue in tabular form:
Answer:

Types Apical Intercalary Lateral
Location At tip of the root and stem At the base of the petiole of leaves and branches At sides of root and stem
Function Increase the length of the root and stem Growth of branches, the formation of leaves and flowers Increases diameter of the root and stem

Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology

Question 5.
Explain the types of simple permanent tissues in plants.
Answer:
Types of simple permanent tissues.

Name of Tissue Parenchyma Collenchyma Sclerenchyma
Struc­ture of cells Living cells with thin cell wall and intercellular spaces. Elongated living cells with thickened cell wall at corners due to cellulose and pectin. Dead and fibrous cells with tapering ends, cell wall contains lignin.
Loca­tion All parts like roots, stem, leaves, flowers and seeds. At the base of leaf petiole, branches and stem Stem, veins of leaves, hard coats of seeds, outer covering of coconut.
Func­tions Support, storage of food and filling vacant spaces. Support and flexibility to various parts. Give strength and rigidity to parts of the plants.
Sub­ types Chlorenchyma: Leaves, perform photosynthesis. Aerenchyma: Helps aquatic plants, leaves and stem to float.

Question 6.
Write down the applications of biotechnology (tissue culture) in floriculture, nurseries and forestry.
Answer:
Applications of biotechnology (tissue culture) in floriculture, nurseries and foresty:

  • Tissue culture can be used to grow those plants on a large scale which bear flowers, fruits of excellent quality.
  • Fully grown plants can be produced in shorter durations.
  • Plants can be grown on a large scale even if means of pollination or germinating seeds are not available.
  • For example, orchids or pitcher plants do not germinate but these plants can easily be produced by means of tissue culture.
  • In a bioreactor, cells can be grown in a more nutritive medium and protected from pathogens.
  • Bioreactors are useful for producing plantlets on a very large scale.
  • A large number of seedlings/plantlets can be produced in a short time using minimum resources and materials.
  • Usually, plants produced by tissue culture and genetic modification techniques are disease-free.
  • Plantlets produced by tissue culture technique of the meristem are virus-free.
  • Embryos produced using conventional hybridization technique between two or more varieties may not grow fully for some reasons. Maharashtra Board Class 9 Science Solutions Chapter 17 Introduction to Biotechnology
  • However, embryos produced by tissue culture technique always complete their growth.
  • Rare and endangered plants can be grown using tissue culture technique and can thus be protected from extinction.
  • Similarly, various parts and seeds of such plants can be preserved by tissue culture and those varieties can be protected.

Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 14 Substances in Common Use Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 14 Substances in Common Use

Class 9 Science Chapter 14 Substances in Common Use Textbook Questions and Answers

1. Fill in the blanks.

a. The number of molecules of water of crystallization in washing soda is …………….…. .
b. The chemical name of baking soda is …………….…. .
c. …………….…. is used in treatment of hyperthyroidism.
d. The chemical name of Teflon is …………….…. .
Answer:
a. 10
b. Sodium bicarbonate
c. Iodine -123
d. Polytetra fluoroethylene (C2F4)n

Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use

2. Match the pairs

Group A  Group B
1. Saturated brine  a. sodium metal freed
2. Fused salt  b. basic salt
3. CaOCl2  c. crystallization of salt
4. NaHCO3  d. oxidation of colour

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 – b)

3. Write answers to the following

a. What is meant by radioactivity?
Answer:

  • Elements with a high atomic number such as uranium, thorium, radium have a property of spontaneously emitting invisible, highly penetrating and high energy radiation.
  • This property is called radioactivity.
  • A substance having this property is called a radioactive substance.

b. When is said to be the nucleus unstable?
Answer:

  • It is the balance of protons and neutrons in a nucleus which determines whether a nucleus will be stable or unstable.
  • Too many neutrons or protons upset this balance disrupting the binding energy from the strong nuclear forces making the nucleus unstable.

c. Which diseases are caused by artificial food colours?
Answer:
Diseases like ADHD (Attention Deficit Hyperactivity Disorder) can affect children due to excessive consumption of foods with added food colours.

d. Where in the industrial field is radioactivity used?
Answer:
Industrial field Radiography

  • Internal cracks and voids in cast iron articles and iron solder can be detected with the help of gamma rays.
  • For this purpose, isotopes like cobalt-60, iridium-192 are used in the radiography camera.
  • This technique is used for detecting flaws in metal work.

Measurement of thickness, density and level

  • It is necessary to maintain the required thickness in the manufacture of aluminium, plastic, iron sheets of differing thickness.
  • In the manufacturing process, a radioactive substance is placed on one side and an instrument to measure radiation on the other.
  • The radiation read by the measuring instrument varies with the thickness of the sheet.
  • Material inside a packing can also be examined by the same technique.

Luminescent paint and radio luminescence:

  • The radioactive substances radium, promethium, tritium with some phosphor are used to make certain objects visible in the dark, for example, the hands of a clock, and certain other objects.
  • Krypton-85 is used in HID (High-Intensity Discharge) lamps while promethium – 147 is used in portable X-ray units as the source of beta rays.

Use in Ceramic articles:

  • Luminous colours are used to decorate ceramic tiles, utensils, plates, etc.
  • Earlier uranium oxide was used in these paints.

e. Write down properties of teflon.
Answer:
Properties of teflon :

  • The atmosphere and chemical substances have no effect on Teflon.
  • Neither water nor oil will stick to Teflon coated articles.
  • High temperatures do not affect Teflon as its melting point is 327 °C.
  • Teflon coated articles are easy to clean.

f. What type of colours will you use to celebrate ecofriendly Rang Panchami? Why?
Answer:

  • We regularly use artificial colours on Rang Panchami.
  • The red colour used on Rang Panchami is very dangerous. It contains a high proportion of mercury in it.
  • This poses risks like blindness, skin cancer, asthma, itching of the skin, permanent blocking of sweat pores, etc.
  • Therefore, it is necessary to use eco-friendly colours.
  • We will prepare colours for Rang Panchami from natural resources such as beet root, flowers of flame of forest, spinach, flame tree (gulmohar) and protect your health by using these.

g. Why has the use of methods like Teflon coating become more common?
Answer:
The use of methods like Teflon coating became more common because of following properties of teflon:

  • The atmosphere, rain water and chemical substances have no effect on Teflon.
  • Neither water nor oil will stick to Teflon coated articles.
  • High temperatures do not affect Teflon as its melting point is 327 °C.
  • Teflon coated articles are easy to clean.
  • Teflon is a poor conductor of electricity.

4. Give a scientific explanation

a. Bleaching powder has the odour of chlorine.
Answer:

  • Bleaching powder undergoes slow decomposition due to the carbon dioxide in air and chlorine gas is released.
  • Bleaching powder gets its property because of this release of chorine gas.
    Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 7

b. The hard water of a well becomes soft on adding washing soda to it.
Answer:

  • The hard water from a well or a bore-well becomes soft on adding washing soda and we come to know this from the lather formed on it.
  • The hardness of water is due to the presence of chlorides and sulphates of calcium and magnesium in it.
  • Na2CO3 is added to it to soften such water and make it suitable for use.
  • The reaction with Na2CO3 causes the formation of insoluble carbonate salts of magnesium and calcium.
    MgCl2(aq) + Na2CO3(s) → MgCO3(s) + 2 NaCl(s)

c. Soap forms a precipitate in hard water.
Answer:

  • When soap is mixed with hard water, calcium and magnesium salts of fatty acids are formed.
  • These being water insoluble, they form a precipitate and that is why lather is not formed.

d. The particles of powder are given an electric charge while spraying them to form the powder coating.
Answer:

  • Powder coating is a method of applying a layer harder than paint on the surface of an iron object to prevent rusting.
  • In this method, a polymer resin, a pigment and some other ingredients are melt, mixed, cooled and ground into a uniform powder.
  • This powder is sprayed on the polished metal surface by electrostatic spray deposition (ESD).
  • In this method, the particles of the powder are given an electrostatic charge due to which a uniform layer of the powder sticks to the metal surface.
  • Then the object is heated in the oven along with the coating.
  • A chemical reaction occurs in the layer, resulting in the formation of long cross-linked polymeric chains.

e. The aluminium article is used as an anode in the anodising process.
Answer:

  • A protective layer is formed naturally on the surface of aluminium metal by reaction with oxygen in air.
  • In the anodizing process, this layer is made of the desired thickness. Anodizing is done by electrolysis.
  • Dilute Sulphuric acid is taken in the electrolytic cell and the aluminium article is dipped in it as the anode.
  • When an electric current is passed, hydrogen gas is released at the cathode and oxygen gas at the anode.
  • A reaction with oxygen occurs and a layer of hydrated aluminium oxide is formed on the anode. Therefore, aluminium article is used as an anode in the anodizing process.

f. When the radiation coming out from certain radioactive substance is passed through an electric field, marks are found at three places on the photographic plate placed in its path.
Answer:
(i) When the radiation coming out from certain radioactive substance is passed through an electric field, marks are found at three places on the photographic plate placed in its path. This is because the radiation coming out from radioactive substance are of three types Alpha rays, Beta rays and Gamma rays.

(ii) The rays which get deviated slightly towards negatively charged plate are called alpha rays. They get deviated towards negatively charged plate because they are made of positively charged particles, called as alpha particles (He++).

(iii) The rays which get deviated substantially towards the positively charged plate are called beta rays. They get deviated towards a positively charged plate because they are made of negatively charged particles called as beta particles (e-).

(iv) The rays which do not deviate at all are called gamma rays. They are uncharged electromagnetic radiation. Hence, marks are found at three places on the photographic plate. (For diagram refer Fig. 14 .2 from Q. 4(3))

g. A certain type of ceramic tiles are fixed on the outer layer of a space shuttle.
Answer:
(i) Each shuttle is covered by more than 24,000 of the six- by six-inch advance ceramic tiles. This tiles withstand high temperatures without decomposing. They are brittle, water-resistant and electrical insulator. Most of the tiles are made of silica fibres, which are produced from high-grade sand (SiO2).

(ii) Silica is an excellent insulator because it transports heat slowly. When the outer portion of a tile gets hot, the heat takes a long time to work its way down through the rest of the tile to the shuttle’s skin. The tiles keep the orbiter’s aluminium skin at 350 degrees or less.

(iii) The silica fibers are mixed with water and chemicals, and the mixture is poured into molds, which are zapped in microwave ovens at 2,350°C to fuse the silica fibres. Tiles are too brittle to attach to the orbiter directly.

(iv) The shuttle’s skin contracts slightly while in orbit, then expands during re-entry.

(v) In addition, the stresses of launch and re-entry cause the skin to flex and bend. Such motions could easily crack the tiles or shake them off. To keep them in place, workers glue the tiles to flexible felt-like pads, then glue the pads to the orbiter.

5. Write answers to the following

a. Write about artificial food colours, the substances used in them and their harmful effects.
Answer:
(a) Artificial food colour and substances in them.

  • Food colours are mixed in most soft drinks and foodstuffs available in the market.
  • These food colours are in the form of powders, gels and pastes.
  • Food colours are used in domestic as well as commercial products.
  • Certain colours and essences are added to ice cream, ice candies, sauce, fruit juices, cold drinks, pickles, jams and jelly.
  • Food colours are often found to be added to packaged meat (chicken, mutton), chilli powder, turmeric, sweets and other similar substances so as to give them a good colour.
  • Tetrazene, sunset yellow are artificial food colours used extensively.

(b) Harmful effects of artificial food colours

  1. Food colours added to pickles, jam and sauce contain small quantities of lead and mercury. These can be harmful for those who consume these products on a regular basis.
  2. Diseases like ADHD (Attention Deficit Hyperactivity Disorder) can affect children due to excessive consumption of foods with added food colours.

b. What is meant by water of crystallization? Give examples of salts with water of crystallization, and their uses.
Answer:
The exact number of water molecules which are chemically bonded to a molecule of a salt within a hydrated crystalline compound is called as water of crystallization are:

Some substances in our daily use which contain water of crystallization are:

  • Alum (Potash alum – K2SO4.Al2(SO4)3.24H2O)
  • Borax (Na2B4O7.10H2O)
  • Epsom salt (Magnesium sulphate MgSO4.7H2O)
  • Barium chloride (BaCl2.2H2O)
  • Sodium sulphate (Glauber’s salt Na2SO4.10 H2O)
  • Blue vitriol (Copper Sulphate – CuSO4.5H2O)

Uses of these salts are as given below:
(i) Alum (Potash alum – K2SO4.Al2(SO4)3.24H2O)

  • Alum is used in the process of water purification.
  • Because of the property of coagulation, the solid impurities in water come together, become heavy and settle to the bottom. As a result, the impure water or muddy above becomes clear.
  • Alum powder, found in the spice section of many grocery stores, may be used in pickling recipes as a preservative to maintain fruit and vegetable crispness.
  • Alum is used as the acidic component of some commercial baking powders.
  • Alum has been used as an after shave treatment.

(ii) Borax (Na2B4O7.10H2O)

  • Borax today is used for many cleaning purposes.
  • It is used to make homemade laundry detergent.
  • It is added as a cleaning boost to any other detergent.
  • It even keeps ants and other pests away.

(iii) Epsom salt (Magnesium sulphate MgSO4.7H2O)

  • Epsom salt is used as a relaxing magnesium bath soak.
  • To grow better vegetables – Add a tablespoon of Epsom salt to the soil below a vegetable plant to boost growth.

(iv) Barium chloride (BaCl2.2H2O)

  • In industry, Barium chloride is mainly used in the purification of brine solution in caustic chlorine plants.
  • It is also used in the manufacture of heat treatment salts.
  • It is used in hardening of steel.
  • It is used in the manufacture of pigments.

(v) Sodium sulphate (Glauber’s salt Na2SO4.10H2O)

  • In the laboratory, anhydrous Sodium sulphate is widely used as an inert drying agent, for removing traces of water from organic solutions.
  • Glauber’s salt, the decahydrate, is used as a laxative.

(vi) Blue vitriol (Copper sulphate CuSO4.5H2O)

  • Blue vitriol is used in the blood test for diagnosing anaemia.
  • Slaked Time is used with blue vitriol in the Bordeaux mixture which is used as a fungicide on fruits like grapes and musk melon.

c. Write briefly about the three methods of electrolysis of sodium chloride.
Answer:

  • When an electric current is passed through a saturated solution of sodium chloride (brine) it is electrolysed and hydrogen gas is released at the cathode while chlorine gas is released at the anode.
  • This method is used for production of chlorine gas. In this method an important basic compound NaOH is formed in the cell.
  • Chemical reaction
    2NaCl + 2H2O → 2NaOH + Cl2 ↑ + H2
  • When salt is heated to a high temperature (about 800 °C), it melts. This is called the fused state of the salt.

6. Write the uses.

a. Anodizing
b. Powder coating
c. Radioactive substances
d. Ceramic
Answer:
(a) Anodizing : Anodizing is done on aluminum cooking utensils like griddles and cookers.

(b) Powder coating : Powder coating is done on iron objects to prevent rusting. Also, on Plastic and Medium density fibre (MDF) board, to make them highly durable, hard and attractive.

(c) Radioactive substances:
(I) Industrial field:
(i) Industrial Radiography : Internal cracks and voids in cast iron articles and iron solder can be. detected with the help of gamma rays. For this purpose, isotopes like cobalt-60, iridium-192 are used in the radiography camera. This technique is used for detecting flaws in metal work.

(ii) Measurement of thickness, density and level: It is necessary to maintain the required thickness in the manufacture of aluminium, plastic, iron sheets of differing thickness. In the manufacturing process, a radioactive substance is placed on one side and an instrument to measure radiation on the other. The radiation read by the measuring instrument varies with the thickness of the sheet. Material inside a packing can also be examined by the same technique.

(iii) Luminescent paint and radioluminescence:
The radioactive substances radium, promethium, tritium with some phosphor are used to make certain objects visible in the dark, for example, the hands of a clock, and certain other objects. Krypton-85 is used in HID (High Intensity Discharge) lamps while promethium-147 is used in portable X-ray units as the source of beta rays.

Use in Ceramic articles – Luminous colours are used to decorate ceramic tiles, utensils, plates, etc. Earlier uranium oxide was used in these paints.

(II) Field of agriculture :

  • The genes and chromosomes that give seeds properties like fast growth, higher productivity, etc. can be modified by means of radiation.
  • The radioactive isotope cobalt-60 is used for food preservation.
  • Onions, potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.
  • Strontium-90 is used as a tracer in the research on various crops.

(III) Medical science:

  • Polycythemia : The red blood cell count increases in the disease polycythemia. Phosphorus- 32 is used in its treatment.
  • Bone cancer : Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.
  • Hyperthyroidism : Enlargement of thyroid gland, weight loss in spite of appetite, insomnia are the symptoms of hyperthyroidism.
  • It occurs due to overproduction of hormones by the thyroid gland. Iodine-123 is used in the treatment of hyperthyroidism.
  • Tumour detection : Boron-10, iodine-131, cobalt-60 are used in treatment of brain tumour, while arsenic-74 is used in detection of small tumours in the body.

(d) Ceramic : Pots made by a potter, Mangalore roofing tiles, construction bricks, pottery, terracotta articles are some examples of common ceramic articles that we see around.

7. Write the harmful effects

a. Artificial dye
b. Artificial food colour
c. Radioactive substances
d. Deodorant
Answer:
(a) Artificial dye:

  • Dyeing hair can have adverse effects like hair fall, damage to hair texture, burning of skin, adverse effect on eyes, etc.
  • Lipstick contains a dye named carmine. It does not affect lips but causes stomach disorders.
  • Excessive use of plants for making natural dyes results in deterioration of the environment.

(b) Artificial food colour:

  • Food colours added to pickles, jams and sauces contain small quantities of lead and mercury. These can be harmful for those who consume these products on a regular basis.
  • Diseases like ADHD (Attention Deficit Hyperactivity Disorder) can affect children due to excessive consumption of foods with added food colours.

(c) Radioactive substances :

  • The central nervous system is affected by radioactive radiations.
  • Hereditary defects are generated by bombardment of radiation on D.N.A in the body.
  • Radioactive radiation can penetrate the skin, and causes diseases like skin cancer, leukemia.
  • The radioactive pollutants created due to explosions enter the body through air and it is difficult to control them.
  • The radioactive pollutants released in the sea enter the bodies of fishes and through them enter the human body.
  • The radioactive paint on the watch can cause cancer.
  • The radioactive isotopes strontium-90 can enter the body through plants, fruits, flowers, cereals, milk, etc. and cause diseases like bone cancer, leukemia.

(d) Deodorant:

  • Aluminium – Zirconium compounds are the most harmful chemicals in the deodorant. Disorders like headache, asthma, respiratory disorders, heart disease are likely to occur without our knowledge.
  • There is a possibility of various skin disorders and also skin cancer due to the aluminium chlorohydrates.

8. Write the chemical formula Bleaching powder, common salt, baking soda, washing soda
Answer:

Common name Chemical name
Bleaching powder CaOCl2
Common salt NaCl
Baking soda NaHCOs
Washing soda Na2CO3.10 H2O

9. Explain what you see in the following picture
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 19
Powder coating
Answer:

  1. This picture shows powder coating of the given object. Powder coating is a method of applying a layer harder than paint on the surface of an iron object to prevent rusting.
  2. In this method, a polymer resin, a pigment and some other ingredients are melt mixed, cooled and ground into a uniform powder.
  3. This powder is sprayed on the polished metal surface by electrostatic spray deposition (ESD).
  4. In this method, the particles of the powder are given an electrostatic charge due to which a uniform layer of the powder sticks to the metal surface.
  5. Then the object is heated in the oven along with the coating.
  6. A chemical reaction occurs in the layer, resulting in the formation of long cross-linked polymeric chains.
  7. This powder coating is highly durable, hard and attractive. Powder coating can be done on plastic and medium density fibre (MDF) board in day to day use as well.

Class 9 Science Chapter 14 Substances in Common Use Intext Questions and Answers

Can you recall?

Question 1.
What are detergents?
Answer:

  • Detergent is a water-soluble cleansing agent which combines with impurities and dirt to make them more soluble, and differs from soap in not forming a scum with the salts in hard water.
  • Today, detergents are more likely to be a mixture of synthetic chemicals and additives cooked up in a huge chemical plant and, unlike traditional soap, they’re generally liquids rather than solids.
  • Detergents are used in everything from hair shampoo and clothes, washing powder to shaving foam and stain removers.
  • The most important ingredients in detergents are chemicals called surfactants – a word made from bits of the words surface-active agents.

Question 2.
What are the important substances that we use in day to day life? For what purposes do we use them?
Answer:
Some of the substances we use in our day to day life and the purpose of using them are as follows:

Toothpaste: for cleaning teeth and maintaining hygienic conditions of mouth.

Soap : for bathing and maintaining hygienic conditions of body.

Natural and artificial fibers : we wear them as clothes to protect our body and many other purposes.

Washing powder: for cleaning clothes. Vegetables, fruits and food grains : they provide us with necessary vitamins, minerals and carbohydrates required for proper functioning of body.

Water : very important for our survival and many other functions of day to day life.

Metals : for making buildings, bridges, roads, vehicles, trains, ships, airplanes, utensils etc.

Acids, bases and salts : used for various purposes, for example, Sodium chloride which is used as common salt, a very necessary ingredient of our daily food without which our food is tasteless.

Question 3.
How are the various substances in day to day use classified from the scientific point of view?
Answer:
The various substances in day to day use are classified from the scientific point of view as elements, compounds, mixtures, metals, non-metals, acids, bases and salts.

Question 4.
Which chemicals and apparatus will you use in the laboratory for making soap?
Answer:
For making soap in laboratory following chemicals and apparatus are used :
Chemicals : Vegetable oil (coconut oil, castor oil, soya bean oil etc.)
Sodium hydroxide (NaOH)
Salt (Sodium chloride NaCl)

Apparatus :
500 ml glass beaker
Burner
Flat glass tray
Knife

Soaps are essential to personal and public health. They safely remove germs, soils and other contaminants and help us to stay healthy and make our surroundings more pleasant. Soaps are made from fats and oils or their fatty acids.

Take about 20 ml of any vegetable oil (coconut oil, castor oil, soybean oil etc.) in a beaker. Put about 30 – 35 ml of NaOH in it and heat and stir the mixture till a paste is formed. Now add a little salt. Stir again and allow the mixture to cool in a flat glass tray, on cooling it solidifies. Take a knife and cut it in soap bars as per the size required.

Question 5.
What are salts?
Answer:

  • Inorganic substances occur naturally in the form of salts rather than acids or bases.
  • The ionic compounds which do not contain H+ and OH ions and contain only one kind of cation and anion are called simple salts.
  • For example, Na2SO4, K3PO4, CaCl2.
  • The sea is said to be a rich source of salts. In fact, the sea is a rich source of several salts of various elements such as chlorine, sodium, magnesium, potassium, calcium, bromine.
  • The important salts found in sea water are sodium chloride, magnesium chloride, magnesium sulphate, potassium chloride, calcium carbonate and magnesium bromide.

Answer the following questions:

Question 1.
How is it determined whether a substance is acidic, basic or neutral?
Answer:

  • In the pH scale, the pH of solutions varies from 0 to 14 in accordance with the strength of the acids or base.
  • The pH scale shows different colours at different values of pH. Colour from dark red to yellow or from pH value 0 to 6 indicates the substances are acidic.
  • Colour from light blue to violet or from pH value 8 to 14 indicates the substances are basic.
  • Green colour or pH value 7 indicates the substances are neutral.

Question 2.
A sweets shop looks attractive because of the colourful sweets displayed there. Which colours are used in these substances?
Answer:

  • Food colours and essences are used in sweets to make them colourful and tasty.
  • These food colour are in the form of powders, gels and pastes.
  • Food colours are natural as well as artificial.
  • The food colours prepared from seeds, beetroot, flowers and fruit concentrate are natural.
  • Tetrazene, sunset yellow are artificial food colours used extensively.

Question 3.
Make a list of substances in day to day use in accordance with their pH value (O to 14).
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 18

Question 4.
A doctor gives you medicines when you fall ill. What are the medicines made from?
Answer:

  • Long before pure chemicals were manufactured in labs, people used plants for medicine.
  • There are over a hundred active ingredients derived from plants for use as drugs and medicines.
  • The emergence of synthetic chemistry in the 1800s opened up new avenues for scientific research into drugs.
  • Many chemists tried to locate medically active ingredients in plants (for instance the important malaria remedy quinine), and subsequently tried to make those substances in the laboratory in order to become independent of plant supplies.
  • At the same time, a new industry developed as scientists developed processes to synthesise new chemicals on a large scale.
  • New production methods developed in the late 1800s and early 1900s made it easier to standardise, package and transport medicines.

Question 5.
What problems do you get after playing colours on Rang Panchami? Why?
Answer:

  • After playing colours on Rang Panchami we may get problems such as risks of blindness, skin cancer, asthma, itching of the skin, permanent blocking of sweat pores etc.
  • This is because the colours used in rang panchami specially the red colour is very dangerous.
  • It contains a high proportion of mercury in it.

Question 6.
Which colours will you use to prevent the occurrence of these problems?
Answer:

  • To prevent the occurrence of these problem, instead of using artificial colours, natural colours should be used.
  • Natural colours are prepared from natural resouroes such as beet root, flowers of flame of forest, spinach, flame tree (gulmohar).

Question 7.
What problem do you have on painting the house and furniture?
Answer:

  • On painting the house and furniture with artificial colours which are dangerous, we may get problems such as risks of blindness, skin cancer, asthma, itching of the skin, permanent blocking of sweat pores, etc.
  • This is because the colours contain a high proportion of mercury in it.

Question 8.
What is the property of Teflon because of which it is used in non-stickware?
Answer:
Following are the properties of teflon because of which it is used in nonstick cookware.

  • The atmosphere and chemical substances have no effect on Teflon.
  • Neither water nor oil will stick to Teflon coated non-stickware.
  • High temperatures do not affect Teflon as its melting point is 327 °C.
  • Teflon coated non-stickware are easy to clean.

Answer the following questions:

Question 1.
What is the strip shown below? What is it used for?
Answer:
The strip shown is pH metre scale. It is used to determined pH of solutions.

Question 2.
Prepare saturated solutions of given salts and put 2-3 drops of the universal indicator in them and note your observations in the table below:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 1

Class 9 Science Chapter 14 Substances in Common Use Additional Important Questions and Answers

Select the correct option:

Question 1.
Baking soda is ………………………… .
(a) non-metallic
(b) metallic
(c) acidic
(d) basic
Answer:
(d) basic

Question 2.
During electrolysis of NaCl in the fused state ………………………… is formed at cathode.
(a) sodium
(b) oxygen
(c) oxalic acid
(d) butyric acid
Answer:
(a) sodium

Question 3.
………………………… is known as chloride of lime.
(a) Calcium carbonate
(b) Sodium bicarbonate
(c) Sodium chloride
(d) Bleaching powder
Answer:
(d) Bleaching powder

Question 4.
Salt obtained from certain type of rock is called ………………………… .
(a) hard salt
(b) rock salt
(c) stone salt
(d) pure salt
Answer:
(b) rock salt

Question 5.
The ………………………… % aqueous solution of salt is called brine.
(a) 25
(b) 60
(c) 30
(d) 50
Answer:
(a) 25

Question 6.
The molecular formula for baking soda is ………………………… .
(a) Na2CO3
(b) NaHCO3
(c) Na2SO4
(d) NaNO3
Answer:
(b) NaHCO3

Question 7.
Bleaching powder is also called ………………………… .
(a) Calcium oxygen chlorine
(b) Calcium oxychloride
(c) Calcium hydroxide
(d) Calcium oxalate
Answer:
(b) Calcium oxychloride

Question 8.
NaHCO3 is also called as ………………………… .
(a) Sodium carbonate
(b) Sodium bicarbonate
(c) Sodium carbon oxide
(d) Sodium chloride
Answer:
(b) sodium bicarbonate

Question 9.
Molecular formula for Magnesium bromide is ………………………… .
(a) MgBr
(b) Mg2Br
(c) Mg2Br2
(d) MgBr2
Answer:
(d) MgBr2

Question 10.
The molten state of NaCl is also called as ………………………… .
(a) solid state
(b) semi-solid
(c) gaseous state
(d) fused state
Answer:
(d) fused state

Question 11.
Substance used to make hard water soft is ………………………… .
(a) Na2CO3.10H2O
(b) Na2CO3.H2O
(c) NaCO3.10H2O
(d) NaCO3.H2O
Answer:
(a) Na2CO3.10H2O

Question 12.
The fixed number of molecules of water in alum is ………………………… .
(a) 24
(b) 10
(c) 7
(d) 2
Answer:
(a) 24

Question 13.
Chlorine gas is a strong ………………………… agent.
(a) reducing
(b) oxidising
(c) neutral
(d) acidic
Answer:
(b) oxidising

Question 14.
Bleaching powder is ………………………… in colour.
(a) brownish red
(b) yellowish-white
(c) pinkish brown
(d) greenish blue
Answer:
(b) yellowish-white

Question 15.
Soap is ………………………… in nature.
(a) acidic
(b) basic
(c) neutral
(d) fuming
Answer:
(b) basic

Question 16.
Melting point of NaCl is ………………………… .
(a) 400°C
(b) 100°C
(c) 800°C
(d) 600°C
Answer:
(c) 800°C

Question 17.
Rock salt is ………………………… in colour.
(a) black
(b) white
(c) brown
(d) blue
Answer:
(c) brown

Question 18.
Strong odour in swimming pool is due to ………………………… .
(a) chlorine gas
(b) impurities
(c) calcium carbonate
(d) calcium hydroxide
Answer:
(a) chlorine gas

Question 19.
Molecular formula of sodium bicarbonate is ………………………… .
(a) NaHCO3
(b) Na2HCO3
(c) Na2HCO2
(d) NaHCO2
Answer:
(a) NaHCO3

Question 20.
Soap is ………………………… in nature.
(a) radioactive
(b) natural
(c) acidic
(d) basic
Answer:
(d) basic

Question 21.
The molecular formula for Magnesium bromide is ………………………… .
(а) MgBr
(b) MgBr2
(c) Mg2Br2
(d) Mg2Br
Answer:
(b) MgBr2

Fill in the blanks and rewrite the complete statements:

(1) 25% of NaCl is known as
(2) gas is released when bleaching powder is added to water.
(3) Washing soda is as it absorbs moisture if left exposed to air.
(4) is used to make hard water soft.
(5) Molecular formula for Alum is
(6) is used in blood test for diagnosing anaemia.
(7) High grade of and are used as raw material to manufacture bathing soap.
(8) Pickles, jams and sauce contain small quantities of and
(9) Excessive consumption of foods with added food colours leads to diseases like
(10) Melting point of Teflon is
(11) Anodizing is done by
(12) The red blood cell count increases in the disease
(13) Magnesium sulphate (MgS04.7H20) is also called
Answer:
(1) Brine
(2) Chlorine
(3) hygroscopic
(4) Na2CO3 (Washing soda)
(5) K2SO4.Al2(SO)3.24H2O
(6) Blue Vitriol
(7) fats, oils
(8) lead, mercury
(9) ADHD (Attention Deficit Hyperactivity Disorder)
(10) 327°C
(11) Electrolysis
(12) Polycythemia
(13) Epsom salt
(14) Soap

State whether the following statements are true or false. Correct the false statements.
(1) Bleaching powder is used in preparation of chloroform which is used as an anaesthetic.
(2) On addition of universal indicator the colour of table salt changes to pink.
(3) Borax is Na2B4O7.9H2O.
(4) Boron -10 is used to cure bone cancer.
(5) Clinical deo contains 20 to 25% of aluminium.
(6) NaHCO3 is sodium carbonate.
(7) Soap is acidic in nature.
(8) Baking powder is neutral in nature.
(9) POP is crystalline in nature.
(10) pH value of universal indicator is 0 – 7.
(11) Sodium chloride is formed by neutralization reaction between sodium hydroxide and calcium carbonate.
(12) Boiling point of NaCl is 800 °C.
(13) Molten state of NaCl is also called as solid state.
(14) Mineral halite is an example of Rock Salt.
(15) Sodium bicarbonate is used to reduce acidity.
(16) Bleaching powder is also called as chloride of lime.
(17) Bleaching powder is obtained by reaction of chlorine gas with slaked lime.
(18) Bleaching powder available in market is classified on the basis of colour.
(19) The hard water becomes soft on adding baking soda (NaHCO3)
Answer:
(1) True
(2) False. On addition of universal indicator the colour of table salt changes to green.
(3) False. Borax is Na2B4O7.9H2O
(4) False. Boron – 10 is used in treatment of brain tumour.
(5) True
(6) False. NaHCO3 is sodium bicarbonate.
(7) False. Soap is basic in nature.
(8) False. Baking powder is basic in nature.
(9) False. POP is amorphous in nature.
(10) False. pH value of universal indicator is 0 -14.
(11) False. Sodium chloride is formed by a neutralization reaction between sodium hydroxide and hydrochloric acid.
(12) False. Melting point of NaCl is 800°C.
(13) False. Molten state of NaCl is also called as fused state.
(14) True
(15) True
(16) True
(17) True
(18) False. Bleaching powder available in market is classified on the basis of percentage of chlorine present in it.
(19) False. Hard water becomes soft on adding washing soda (Na2CO3.10H2O)

Name the following:

Question 1.
Important salts found in sea water.
Answer:
Magnesium chloride, Potassium chloride

Question 2.
Chemical used in preparation of breads and cakes to make them soft.
Answer:
Sodium bicarbonate

Question 3.
Gas liberated when bleaching powder is added to CO2 gas.
Answer:
Chlorine (Cl2) gas.

Question 4.
Salt used in refining petroleum
Answer:
Sodium carbonate (Na2CO3)

Question 5.
Chemical formula of bleaching powder
Answer:
CaOCl2

Question 6.
The chemical formula of Sodium hydrogen carbonate.
Answer:
NaHCO3

Question 7.
Nature of soap.
Answer:
Basic

Question 8.
Molecular formula for Magnesium bromide
Answer:
MgBr2

Question 9.
Melting point of NaCl
Answer:
800 °C

Question 10.
Two examples of Rock Salt
Answer:
Mineral halite and Himalayan rock salt.

Question 11.
Uses of sodium bicarbonate
Answer:
To prepare breads, cakes and dhokla

Question 12.
Constituents of baking powder.
Answer:
Baking soda and mild acid (tartaric acid)

Question 13.
Substance used to make hard water soft.
Answer:
Washing soda i.e. Na2CO3.10H2O.

Question 14.
Radioactive isotopes used in treatment of bone cancer.
Answer:
Strontium – 89, Strontium – 90, Samarium -153, Radium – 223

Question 15.
Use of Strontium 89 and strontium 90.
Answer:
Treatment of bone cancer.

Select the odd man out:

Question 1.
Sodium sulphate, Barium chloride, Magnesium sulphate, Bleaching powder.
Answer:
Bleaching powder, all others are crystalline salt with water of crystallization while bleaching powder is not.

Question 2.
Strontium – 89, Strontium – 90, Radium – 223, Iodine -123.
Answer:
Iodine -123, it is used to cure hyperthyroidism while others are used to cure bone cancer.

Question 3.
Soap, Toothpaste, Baking soda, Curd.
Answer:
Curd, it is acidic in nature while others are basic in nature.

Question 4.
Na2CO3.10H2O, Na2SO4.10H2O, BaCl2.2H2O, CaOCl2
Answer:
CaOCl2, it is amorphous in nature while rest are crystalline in nature.

Question 5.
Boron-10, Iodine – 131, Cobalt – 60, Iodine – 123.
Answer:
Iodine -123, it is used in hyperthyroidism while rest are used in treatment of brain tumour.

Question 6.
Na2SO4, K3PO4, MgBr2, HCl.
Answer:
HCl, rest all are salts but HCl is an acid.

Write the correlated terms:

(1) Washing Powder : Basic :: Milk : ……………………………
(2) Na2SO4: Sodium sulphate :: K3PO4 : ……………………………
(3) CaOCl2: Calcium Oxychloride : : NaHCO3 : ……………………………
(4) Borax : 10H2O : : Alum : ……………………………
(5) Sodium Sulphate : Na2SO4.10H2O : : Magnesium Sulphate : ……………………………
(6) Bathing soap : High grades fats and oils :: Washing soap : ……………………………
(7) Radium – 23 : Bone cancer :: Phosphorus – 32 : ……………………………
(8) Iodine -123 : Hyperthyroidism :: Iodine -131 : ……………………………
(9) Clinical deo : 20-25% of aluminium : : Antiperspirant deo : of aluminium chlorohydrate : ……………………………
(10) Sodium carbonate : Na2CO3 : : Sodium bicarbonate : ……………………………
(11) Curd : Acidic : : Soap : ……………………………
(12) Calcium Carbonate : CaCO3 : : Magnesium bromide : ……………………………
(13) Baking Soda : NaHCO3 : : Washing Soda : ……………………………
(14) Baking Soda : Breads and Cakes : : Washing Soda : ……………………………
(15) Borax : Na2B4O7.10H2O : : Alum : ……………………………
Answer:
(1) Acidic
(2) Potassium phosphate
(3) Sodium hydrogen carbonate
(4) 24H2O
(5) MgSO4.7H2O
(6) Low-grade fats and oils
(7) Polycythemia
(8) Tumour detection
(9) 15%
(10) NaHCO.
(11) Basic
(12) MgBr2
(13) Na2CO3.10H2O
(14) Hard water soft
(15) K2SO4.Al2(SO4)3.24H2O

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Barium chloride (a) 24H2O
(2) Sodium sulphate (b) 7H2O
(3) Magnesium sulphate (c) 2H2O
(4) Alum (d) 10H2O

Answer:
(1 – c),
(2 – d),
(3 – b),
(4 – a)

Question 2.

Column ‘A’ Column ‘B’
(1) Strontium – 90 (a) Brain tumour
(2) Cobalt – 60 (b) Hyperthyroidism
(3) Iodine – 131 (c) Polycythemia
(4) Phosphorus – 32 (d) Bone cancer

Answer:
(1 – d),
(2 – a),
(3 – b),
(4 -c)

Give scientific reasons:

Question 1.
Baking soda is used while making cakes.
Answer:

  • Baking soda is chemically known as sodium bicarbonate.
  • On heating, sodium bicarbonate decomposes to form sodium carbonate, water and carbon dioxide (CO,).
  • The carbon dioxide produced is released due to which the cake becomes soft and spongy.
  • Hence, baking soda is used to make cake spongy and soft.

Question 2.
Baking soda is used to reduce acidity in the stomach.
Answer:

  • Baking soda is used as a medicine (antacid). When it is taken, it undergoes hydrolysis to give sodium hydroxide in the stomach. Thus, sodium hydroxide neutralizes the hydrochloric acid produced by gastric juice and gives relief to the patient from acidity.
    Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 4

Question 3.
Tap water and water in swimming pool has a typical strong odour in the rainy season.
Answer:

  • Tap water and water in swimming pool has a typical strong odour in the rainy season, it is the odour of the chlorine gas that is used to destroy the microbes in the water.
  • Chlorine gas is a strong oxidizing agent and therefore, it has a strong disinfecting as well as bleaching action.
  • Chlorine is inconvenient to handle because of its gaseous state. Instead, the solid bleaching powder which has the same effect is more convenient to use.
  • Bleaching powder undergoes slow decomposition due to the carbon dioxide in air and chlorine gas is released. Bleaching powder gets its property because of this release of chlorine gas.
    Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 5

Question 4.
Sodium carbonate is added to hard water to make it soft.
Answer:

  • The hardness of water is due to the presence of chlorides and sulphates of calcium and magnesium in it.
  • Na2CO3 is added to it to soften such water and make it suitable for use.
  • The reaction with Na2CO3 causes the formation of insoluble carbonate salts of magnesium and calcium thus making the water soft.
    Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 6

Question 5.
Alum is used in the process of water purification.
Answer:

  • Alum is used in the process of water purification because of property of coagulation.
  • The solid impurities in water come together, become heavy and settle to the bottom.
  • As a result, the water above becomes clear.

Question 6.
Overconsumption of artificial food colours should be avoided?
Answer:

  • The overconsumption of artificial food colours should be avoided because they can be detrimental to health.
  • Food colours added to pickles, jam and sauce contain small quantities of lead and mercury. These can be harmful for those who consume these products on a regular basis.
  • Diseases like ADHD (Attention Deficit Hyperactivity Disorder) can affect children due to excessive consumption of foods with added food colours.

Question 7.
Artificial colours in Rang Panchami should be used cautiously.
Answer:

  • We regularly use artificial colours on Rang Panchami.
  • The red colour used on Rang Panchami is very dangerous. It contains a high proportion of mercury in it.
  • This poses risks like blindness, skin cancer, asthma, itching of the skin, permanent blocking of sweat pores etc.
  • Therefore, it is necessary that artificial colours are used cautiously.

Question 8.
There are various brands of bleaching powder in the market.
Answer:

  • When bleaching powder comes in contact with air, it reacts with carbon dioxide to form calcium carbonate and releases chlorine gas. CaOCl2 + CO2 → CaCO3 + Cl2 ↑ (from air)
  • This generated chlorine is known as “available chlorine.”
  • On the basis of the percentage of “available chlorine” various brands of bleaching powder are available in the market.

Question 9.
We use anodized cooking utensils like griddles and cookers.
Answer:

  • Anodizing is process of forming a protective rayer of hydrated aluminium oxide on utensils.
  • Due to anodizing, the surface of the cooking utensils become non-standing, non-toxic and non-reactive with the ingredients of the foot.
  • These utensils get heated faster and withstand high heat.
  • Also the amount of butter, ghee or oil used to cook food in these type of utensils is neglible.
  • These utensils are tough, durable and resistant to corrosion.
  • Hence, we use anodized cooking utensils like gridders and cookers.

Explain the following chemical reactions with the help of balanced equations :

Question 1.
Sodium hydroxide reacts with dilute Hydrochloric acid.
Answer:
When Sodium hydroxide reacts with dilute Hydrochloric acid, it forms sodium chloride and water.
This is a neutralization reaction.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 8

Question 2.
When an electric current is passed through a saturated solution of sodium chloride (brine)
Answer:
When an electric current is passed through a saturated solution of sodium chloride (brine) it is electrolysed and hydrogen gas is released at the cathode while chlorine gas is released at the anode. Also, an important basic compound NaOH is formed in the cell.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 9

Question 3.
Baking soda (Sodium hydrogen carbonate) reacts with dilute hydrochloric acid.
Answer:
When Sodium bicarbonate (Sodium hydrogen carbonate) reacts with dilute hydrochloric acid, it gives sodium chloride, water and carbon dioxide gas is liberated.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 10

Question 4.
Bleaching powder is exposed to air.
Answer:
When bleaching powder is exposed to air, it undergoes slow decomposition due to the carbon dioxide in air and chlorine gas is released.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 11

Question 5.
When dry slaked lime reacts with chlorine gas.
Answer:
When dry slaked lime reacts with chlorine gas, it gives bleaching powder.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 12

Question 6.
Bleaching powder reacts with dilute sulphuric acid.
Answer:
Dilute sulphuric acid reacts rapidly with bleaching powder to form calcium sulphate and release chlorine gas completely.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 13

Question 7.
Bleaching powder reacts with dilute hydrochloric acid.
Answer:
Dilute hydrochloric acid reacts rapidly with bleaching powder to form Calcium chloride and release chlorine gas completely.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 16

Question 8.
Sodium carbonate reacts with magnesium chloride.
Answer:
When Sodium carbonate reacts with magnesium chloride, it forms insoluble magnesium carbonate and sodium chloride. This reaction is used to make hard water soft.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 15

Question 9.
Crystalline sodium carbonate is exposed to air.
Answer:
When crystalline sodium carbonate is exposed to air it loses its water of crystallization readily and a white powder is obtained. This powder is called washing soda.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 14

Question 10.
Sodium carbonate reacts with sulphuric acid.
Answer:
When sodium carbonate reacts with sulphuric acid it forms sodium sulphate, water and carbon dioxide gas is liberated.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 17

Distinguish between;

Question 1.
Washing soda and Baking soda
Answer:

Washing soda Baking soda
(i) Chemical name is sodium carbonate. (i)    Chemical name is sodium bicarbonate or sodium hydrogen carbonate.
(ii) Its molecular formula is Na2CO3.10H2O (ii) Its molecular formula is NaHCO3
(iii) It is a crystalline substance. (iii) It is an amorphous powder.
(iv) It is used in manufacturing soaps and detergents. (iv) It is used in bakery for making cakes and bread lighter and spongy.

Question 2.
Bathing soap and Washing soap
Answer:

Bathing soap Washing soap
(i) High-grade fats and oils are used as the raw material. (i) Low-grade fats and oils are used as the raw material.
(ii) Expensive perfumes added. (ii) Cheaper perfumes added.
(iii) No free alkali content present to prevent injuries to skin. (iii) Free alkali present for cleaning action.

Answer the following:

Question 1.
Give the properties and uses of Sodium bicarbonate.
Answer:
Properties and uses of sodium bicarbonate

  • (i) NaHCO2 reacts with moist litmus paper and red litmus turns blue which means that it is basic in nature.
  • (ii) It is used to make bread, cake, dhokla.
  • (iii) Being basic in nature it is used to reduce acidity in the stomach.
  • (iv) NaHCO3 is used to make the active substance CO2 in the fire extinguisher.
  • (v) Baking soda is used to clean an oven.

Question 2.
Give the properties and uses of washing soda.
Answer:
Properties and uses of washing soda :

  • Washing soda is a whitish and odourless powder at room temperature.
  • Litmus has a blue colour in its aqueous solution.
  • It is hygroscopic, that is, it absorbs moisture if left exposed to air.
  • It is used mainly for washing clothes.
  • Sodium carbonate is used in the glass and paper industry and also in refining of petrol.

Question 3.
Name some substances in our daily use which contain water of crystallization.
Answer:

  • Alum (Potash alum – K2SO4.Al2(SO4)3.24H2O)
  • Borax (Na2B4O7.10H2O)
  • Epsom salt (Magnesium sulphate MgSO4.7H2O)
  • Barium chloride (BaCl2.2H2O)
  • Sodium sulphate (Glauber’s salt Na2SO4.10H2O)

Question 4.
Give the uses of radioactive isotopes in the measurement of thickness, density and level.
Answer:
Measurement of thickness, density and level:

  • It is necessary to maintain the required thickness in the manufacture of aluminium, plastic, iron sheets of differing thickness.
  • In the manufacturing process, a radioactive substance is placed on one side and an instrument to measure radiation on the other.
  • The radiation read by the measuring instrument varies with the thickness of the sheet.
  • Material inside a packing can also be examined by the same technique.

Question 5.
Give the uses of radioactive isotopes in Luminescent paint and radioluminescence.
Answer:
Luminescent paint and radioluminescence:

  • The radioactive substances radium, promethium, tritium with some phosphour are used to make certain objects visible in the dark, for example, the hands of a clock, and certain other objects.
  • Krypton-85 is used in HID (High Intensity Discharge) lamps while promethium-147 is used in portable X-ray units as the source of beta rays.

Question 6.
Give the uses of radioactive isotopes in Field of agriculture.
Answer:
Field of agriculture:

  • The genes and chromosomes that give seeds properties like fast growth, higher productivity, etc. can be modified by means of radiation.
  • The radioactive isotope cobalt-60 is used for food preservation.
  • Onions, potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.
  • Strontium-90 is used as a tracer in the research on various crops.

Question 7.
Give the uses of dyes.
Answer:
Uses of dyes :

  • They are used for colouring cloth and hair.
  • Fluorescent colours are used to make street boards that are visible at night.
  • Dyes are used to polish leather shoes, purses and chappals.

Question 8.
What are the adverse effects of dyes?
Answer:
Adverse effects dyes :

  • Dyeing hair can have adverse effects like hair fall, damage to hair texture, burning of skin, adverse effect on eyes, etc.
  • Lipstick contains a dye named carmine. It does not affect lips but causes stomach disorders.
  • Excessive use of plants for making natural dyes results in deterioration of the environment.

Question 9.
Give the uses of teflon.
Answer:
Uses of teflon :

  • Teflon is a poor conductor of electricity. Therefore, Teflon clad wires and parts are used in high technology electronic instruments.
  • It is used for making non-stick kitchenware.
  • The coloured metal sheets of two-wheelers and four-wheelers are given a Teflon coating to protect them from damage due to high temperature and rain.

Question 10.
What is ceramic? Give examples.
Answer:

  • Ceramic is a heat resistant substance formed by kneading an inorganic substance in water and then shaping it and hardening it by heating.
  • Pots made by a potter, Mangalore roofing tiles, construction bricks, pottery, terracotta articles are some examples of common ceramic articles that we see around.

Question 11.
Give the uses of ceramics.
Answer:

  • Ceramics can withstand high temperatures without decomposing. Ceramic is brittle, water-resistant and an electrical insulator.
  • Therefore, it is used in electrical instruments, for coating the interior of a kiln, the outer surfaces of ships and blades of jet engines.
  • A certain type of ceramic tiles are fixed on the outer layer of a space shuttle. Some types of ceramics are used as superconductors.

Question 12.
What is brine? What happens when ^th of this solution is evaporated?
Answer:

  • The 25% aqueous solution of salt is called saturated brine.
  • When th of this solution is evaporated the dissolved salt gets crystallized and salt gets separated from the solution.

Question 13.
What is baking soda? Give its chemical name and molecular formula.
Answer:

  • Baking soda is a white non-crystalline powder.
  • Its chemical name is sodium hydrogen carbonate or sodium bicarbonate and its molecular formula is NaHC03.

Question 14.
Give two uses of Blue vitriol.
Answer:

  • Blue vitriol is used in the blood test for diagnosing anaemia.
  • Bordeaux mixture which is a mixture of slaked lime and blue vitriol is used as a fungicide on fruits like grapes and musk melon.

Question 15.
What is radioactivity and what are radioactive substances?
Answer:

  • Elements with a high atomic number such as uranium, thorium, radium have a property of spontaneously emitting invisible, highly penetrating and high energy radiation. This property is called radioactivity.
  • A substance having this property is called a radioactive substance.

Question 16.
What can you say about the uses of radioactive isotopes?
Answer:

  • It is a misconception that radioactive elements are used only for making an atom bomb.
  • Radioactive isotopes are used in various fields such as scientific research, agriculture, industry, medicine, etc.
  • Radioactive substances are used in two ways.
  • By using the radiation alone.
  • By using the radioactive element itself.

Question 17.
Give the uses of radioactive isotopes in Radiography.
Answer:

  • Internal cracks and voids in cast iron articles and iron solder can be detected with the help of gamma rays.
  • For this purpose, isotopes like cobalt-60, iridium-192 are used in the radiography camera.
  • This technique is used for detecting flaws in metalwork.

Question 18.
Give the uses of radioactive isotopes in ceramic articles.
Answer:

  • Luminous colours are used to decorate ceramic tiles, utensils, plates, etc.
  • Earlier uranium oxide was used in these paints.

Question 19.
What is a dye?
Answer:
The coloured substance which on applying to an article, imparts that colour to the article, is called a dye

Question 20.
What is done to fix a dye on the cloth ?
Answer:

  • A mordant is to be used to fix the colour after dyeing a cloth.
  • A mordant is a chemical binding agent that adheres well to both the fibers and the dye.

Question 21.
What is Teflon? Give its chemical name.
Answer:

  • Teflon is the polymer of tetrafluoroethylene.
  • Its chemical name is polytetrafluoroethylene (C2F4)n

Question 22.
What is powder coating? Why is it done?
Answer:

  • Powder coating is a method of applying a layer harder than paint on the surface of plastic, medium density fibre (MDF).
  • Board and iron object to make them attractive, durable and to prevent rustihg respectively.

Question 23.
Give the properties of ceramics.
Answer:

  • Ceramics can withstand high temperatures without decomposing.
  • Ceramic is brittle, water resistant and an electrical insulator.

Question 24.
How is bone china made? How is it different from porcelain?
Answer:

  • Bone china is made by adding some ash of animal bones in the mixture of china clay, feldspar and fine silica while making porcelain.
  • This ceramic is harder than porcelain.

Question 25.
What compounds are used instead of clay for making advanced ceramic?
Answer:
Oxides like Alumina (Al2O3), Zirconia (ZrO2) Silica (SiO2) and some other compounds like silicon carbide (SiC), boron carbide (B4C) are used instead of clay for making advanced ceramic.

Question 26.

What is sintering?
Answer:

  • Advanced ceramics requires a temperature of 1600 to 1800 °C and an oxygen-free atmosphere for firing.
  • This process is called sintering.

Write short notes on :

Question 1.
Chernobyl disaster.
Answer:

  • On 26th April 1986, the graphite reactor in the Chernobyl atomic power plant exploded, and suddenly the radioactive isotopes and radiation came out.
  • Due to this episode, radioactive isotopes entered the human body through water and land and caused genetic disorders.
  • These got carried further into the next generation.
  • Thyroid disorders increased in children as well as adults.
  • As a result, the incidence of throat diseases is greater there than in other places.

Question 2.
Food colours.
Answer:

  • Food colours are mixed in most soft drinks and foodstuffs available in the market.
  • These food colours are in the form of powders, gels and pastes.
  • Food colours are used in domestic as well as commercial products.
  • Certain colours and essences are added to ice cream, ice candies, sauces, fruit juices, cold drinks, pickles, jams and jellies.
  • Food colours are often found to be added to packaged meat (chicken, mutton), chilli powder, turmeric, sweets and other similar substances so as to give them a good colour.
  • Food colours are natural as well as artificial.
  • The food colours prepared from seeds, beetroot, flowers and fruit concentrate are natural.
  • Tetrazene, sunset yellow are artificial food colours used extensively.
  • However, over-consumption of artificial food colours can be detrimental to health. Therefore, usage of natural food colours is always good.

Question 3.
Anodizing.
Answer:

  • A protective layer is formed naturally on the surface of aluminium metal by reaction with oxygen in air.
  • In the anodizing process, this layer can be made of the desired thickness.
  • Anodizing is done by electrolysis.
  • Dilute sulphuric acid is taken in the electrolytic cell and the aluminium article is dipped in it as the anode.
  • When an electric current is passed, hydrogen gas is released at the cathode and oxygen gas at the anode.
  • A reaction with oxygen occurs and a layer of hydrated aluminium oxide is formed on the anode, i.e. the aluminium article.
  • This layer can be made attractive by adding colour in the cell during electrolysis.

Answer the following questions:

Question 1.
What are salts? Give their source.
Answer:

  • The ionic compounds which do not contain H+ and OH ions and contain only one kind of cation and anion are called simple salts.
  • For example, Na2SO4, K3PO4, CaCl2.
  • The sea is said to be a rich source of salts of various elements such as chlorine, sodium, magnesium, potassium, calcium, bromine.
  • The important salts found in sea water are:
    (a) Sodium chloride
    (b) Magnesium chloride
    (c) Magnesium sulphate
    (d) Potassium chloride
    (e) Calcium carbonate
    (f) Magnesium bromide

Question 2.
Which are the three kinds of salts depending on their pH value? How are they formed?
Answer:
The three kinds of salts depending on their pH value are:

Neutral salts:
Salt is neutral when its pH value is 7. Such a salt is made from a strong acid and a strong base.

Acidic salts:
The pH value of a salt made from a strong acid and a weak base is less than 7 and it is acidic.

Basic salts:
The pH value of a salt made from a weak acid and strong base is more than 7 and it is basic.

Question 3.
Which is the most used salt? What is its chemical name? How is it formed?
Answer:

  • Table salt, or common salt, which gives a salty taste to food, is the most used of all salts.
  • Its chemical name is Sodium chloride.
  • Sodium chloride is formed by a neutralization reaction between sodium hydroxide and hydrochloric acid.
  • Chemical reaction
  • This is a neutral salt and the pH value of its aqueous solution is 7.

Question 4.
How is bleaching powder obtained?
Answer:

  • Bleaching powder is obtained by the reaction of chlorine gas with slaked lime.
  • Chemical reaction

Question 5.
How is hard water from well converted to soft water?
Answer:

  • The hard water from a well or a bore-well becomes soft on adding washing soda and we come to know this from the lather formed on it.
  • The hardness of water is due to the presence of chlorides and sulphates of calcium and magnesium in it.
  • Na2CO3 is added to it to soften such water and make it suitable for use.
  • The reaction with Na2CO3 causes the formation of insoluble carbonate salts of magnesium and calcium.
  • Chemical reaction
    MgCl2(aq) + Na2CO3(s) → MgCO3(s) + 2NaCl(s) Magnesium Sodium Magnesium Sodium chloride carbonate carbonate chloride

Question 6.
How is washing soda obtained from sodium carbonate?
Answer:

  • Sodium carbonate is a water soluble salt of sodium.
  • Crystalline sodium carbonate, on keeping, loses its water of crystallization readily and a white powder is obtained.
  • This powder is called washing soda.
  • Chemical reaction

Question 7.
How is soap prepared? Why it does not form lather with hard water?
Answer:

  • When oil or animal fat is boiled with an aqueous solution of sodium or potassium hydroxide, sodium or potassium salts of carboxylic acids (fatty acids) are formed.
  • These salts are called soap.
  • When soap is mixed with hard water, calcium and magnesium salts of fatty acids are formed.
  • These being water insoluble, they form a precipitate and that is why lather is not formed.

Question 8.
What are the harmful effects of artificial food colours?
Answer:

  • Food colours added to pickles, jams and sauces contain small quantities of lead and mercury. These can be harmful for those who consume these products on a regular basis.
  • Diseases like ADHD (Attention Deficit Hyperactivity Disorder) can affect children due to excessive consumption of foods with added food colours.

Question 9.
Which is the main source of colour for preparing a natural dye?
Answer:

  • Plants are the main source of colour for preparing a natural dye.
  • Roots, leaves, flowers, bark, fruits, seeds, fungus and pistils are used for making dyes.
  • In Kashmir a very good dye is made from saffron, which is used to dye fibers from which saris, shawls and dresses are made.
  • These are very costly. The use of henna leaves to colour hair is safe for health.

Question 10.
How are artificial colours harmful to us?
Answer:

  • We regularly use artificial colours on Rang Panchami.
  • The red colour used on Rang Panchami is very dangerous.
  • It contains a high proportion of mercury in it.
  • This poses risks like blindness, skin cancer, asthma, itching of the skin, permanent blocking of sweat pores, etc.
  • Therefore, it is necessary that artificial colours are used cautiously.

Question 11.
What are the harmful effects of deodorants?
Answer:

  • Aluminium – Zirconium compounds are the most harmful chemicals in deodorants. Disorders like headache, asthma, respiratory disorders, heart disease are likely to occur without our knowledge.
  • There is a possibility of various skin disorders and also skin cancer due to the aluminium chlorohydrates.

Question 12.
How is a ceramic article made?
Answer:

  • When clay is kneaded in water, shaped and then fired in a kiln at a temperature of 1000 to 1150 °C, a porous ceramic is formed.
  • To overcome the porosity, the fired object is covered with finely ground glass powder suspended in water (glaze) and is then fired again. As a result, the surface of the ceramic becomes shiny and its porosity disappears.

Question 13.
What is porcelain? How is it made?
Answer:

  1. Porcelain is a hard, translucent and white coloured ceramic. It is made by using the white clay called kaolin, found in China. Glass, granite and the mineral feldspar is mixed with kaolin and kneaded with water.
  2. The resulting mixture is shaped and fired in a kiln at a temperature of 1200 to 1450 °C. On firing again after glazing, beautiful articles of porcelain are obtained.

Answer the following questions:

Question 1.
What are the constituents of baking powder? Where is the baking powder used?
Answer:

  • Baking powder consist of 30% sodium bicarbonate, 5-12% monocalcium phosphate and 21-26% sodium aluminium sulphate.
  • Baking powder is a leavening agent and is used for increasing the volume and lightening the texture of baked goods.

Question 2.
Take a piece of coloured cloth. Put some saturated solution of bleaching powder on a small part and observe what changes take place in the colour of the cloth.
Answer:
When bleaching powder is added to a piece of coloured cloth, the fabric looses its colour due to the oxidising reaction of the chlorine present in the bleaching powder.

Answer in detail:

Question 1.
Give the properties and uses of Sodium chloride.
Answer:
Properties and uses of sodium chloride:

  • Common salt is a colourless and crystalline ionic compound. There is no water of crystallization in its crystalline structure.
  • It is a neutral salt, salty in taste.
  • This compound is used for the production of salts like Na2CO3, NaHCO3.
  • When an electric current is passed through a saturated solution of sodium chloride (brine), it is electrolysed and hydrogen gas is released at the cathode while chlorine gas is released at the anode.
  • This method is used for production of chlorine gas. In this method an important basic compound NaOH is formed in the cell.
    2NaCl + 2H2O → 2NaOH + Cl2↑ + H2
  • When salt is heated to a high temperature (about 800 °C), it melts. This is called the fused state of the salt.
  • When fused salt is electrolysed, chlorine gas is released at the anode and liquid sodium metal, at the cathode.

Question 2.
Give the properties and uses of Bleaching Powder.
Answer:
Properties and uses bleaching powder:

  • Bleaching powder is a yellowish-white coloured solid substance.
  • Its chemical name is Calcium oxychloride.
  • It has a strong odour of chlorine gas.
  • It is used for disinfection of drinking water at the water works and the water in the swimming pool.
  • It is used for bleaching of cloth.
  • It is used for disinfection by the road side and garbage sites.
  • Dilute sulphuric acid and dilute hydrochloric acid react rapidly with bleaching powder to release chlorine gas completely.
    CaOCl2 + H2SO4 → CaSO4 + Cl2 ↑ + H2O CaOCl2 + 2HCl → CaCl2 + Cl2 ↑ + H2O
  • Calcium oxychloride reacts slowly with carbon dioxide to form calcium carbonate and chlorine.
    CaOCl2 + CO2 → CaCOs + Cl2

Question 3.
Explain with neat labelled diagram the nature of radioactive radiation.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 2
Answer:

  • When radioactive radiations are allowed to pass through two oppositely charged plates they get separated. This method was introduced by Rutherford in 1902.
  • Rutherford and Willard studied the radiation emitted by radioactive substances.
  • For this purpose, the rays were allowed to pass through an electrical field and a photographic plate was held in their path.
  • It was found that the radiation was divided into three types.
  • One type of radiation deviated slightly towards the negatively charged plate, while the second type of radiation deviated substantially towards the positively charged plate.
  • However, the third type of radiation did not deviate at all in the electrical field.
  • The rays which deviated slightly toward negatively charged plate are called alpha rays.
  • Those which deviate substantially towards the positively charged plate are called beta rays and those which did not deviate at all are called gamma rays.

Question 4.
Give the characteristics of alpha, beta and gamma rays.
Answer:
Characteristics of alpha, beta and gamma rays are as given in the table.
Maharashtra Board Class 9 Science Solutions Chapter 14 Substances in Common Use 3

Question 5.
Give the uses of radioactive isotopes in Medical science.
Answer:
Medical science:

  • Polycythemia: The red blood cell count increases in the disease polycythemia. Phosphorus- 32 is used in its treatment.
  • Bone cancer: Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.
  • Hyperthyroidism: Enlargement of thyroid gland, weight loss in spite of appetite, insomnia are the symptoms of hyperthyroidism.
  • It occurs due to overproduction of hormones by the thyroid gland. Iodine-123 is used in the treatment of hyperthyroidism.
  • Tumour detection: Boron-10, iodine-131, cobalt-60 are used in the treatment of brain tumour, while arsenic-74 is used in detection of small tumours in the body.

Question 6.
What are the hazards of radioactive substances and radiation?
Answer:

  • The central nervous system is affected by radioactive radiations.
  • Hereditary defects are generated by bombardment of radiation on D.N.A in the body.
  • Radioactive radiation can penetrate the skin, and causes diseases like skin cancer, leukemia.
  • The radiative pollutants created due to explosions enter the body through air and it is difficult to control them.
  • The radioactive pollutants released in the sea enter the bodies of fishes and through them enter the human body.
  • The radioactive paint on the watch can cause cancer.
  • The radioactive isotopes strontium-90 can enter the body through plants, fruits, flowers, cereals, milk, etc. and cause diseases like bone cancer, leukemia.

Question 7.
Why are deodorant used? Give their types and content.
Answer:

  • Body odour is caused by the bacterial decomposition of the sweat. A deodorant is used to prevent this odour.
  • Deodorants contain parabens (methyl, ethyl, propyl, benzyl and butyl) and also alcohol in large proportions. Aluminium compounds and silica are also used.
  • Types of deodorants are :
    (a) Ordinary deo: It contains a smaller proportion of aluminium. It decreases the odour of the sweat.
    (b) Antiperspirant deo : This decreases the extent of sweating. It contains about 15% of aluminium chlorohydrate. It clogs the sweat pores on the skin.
    (c) Clinical deo : Some people sweat heavily and it has harmful effects on the skin. Clinical deo is meant for such people. It contains 20 to 25% aluminium. It is used during the night.

Question 8.
Explain how powder coating is done.
Answer:

  • Powder coating is a method of applying a layer harder than paint on the surface of an iron object to prevent rusting.
  • In this method, a polymer resin, a pigment and some other ingredients are melt, mixed, cooled and ground into a uniform powder.
  • This powder is sprayed on the polished metal surface by electrostatic spray deposition (ESD).
  • In this method, the particles of the powder are given an electrostatic charge due to which a uniform layer of the powder sticks to the metal surface.
  • Then the object is heated in the oven along with the coating.
  • A chemical reaction occurs in the layer, resulting in the formation of long cross-linked polymeric chains.
  • This powder coating is highly durable, hard and attractive.
  • Powder coating can be done on plastic and medium density fibre (MDF) board in day to day use as well.

 

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Class 9 Science Chapter 13 Carbon: An Important Element Textbook Questions and Answers

1. Select the proper option and complete the statements
(single, all, double, ionic, carbon, give and take, hydrogen, multiple, share, most, covalent)

a. A carbon atom forms a ….……………. bond with other atoms. In this bond the two atoms ….electrons.
Answer:
covalent, share

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. All the carbon bonds in a saturated hydrocarbon ….……………. electrons.
Answer:
share

c. At least one carbon bond in an unsaturated hydrocarbon is ….…………… .
Answer:
multiple

d. ….……………. is the essential element in all the organic compounds.
Answer:
Carbon

e. The element hydrogen is present in ….. organic compound.
Answer:
all

2. Answer the following questions

a. Why are carbon and its compounds used as fuels?

Answer:
(i) The name ‘carbon’ is derived from Latin word ‘carbo’meaning coal. In the earth’s crust, carbon is present to an extent of approximately 0.27% in the form of carbonate, coal and petroleum. One of the non-crystalline and amorphous form of carbon is coal. Coal is a fossil fuel.

(ii) Peat, lignite, bituminous and anthracite are the four types of coal in the increasing order of their carbon content and heat produced respectively. Charcoal and coke are the other amorphous forms of carbon used as fuel.

(iii) Compounds of carbon such as hydrocarbons consist of carbon and hydrogen and they are easily combustible. For example, methane (CH4) which occurs in natural gas is highly inflammable. It bums by reacting with oxygen to give a bluish flame. In this reaction, 213 Kcal/mol of heat is given out. Methane bums completely.

Chemical reaction:
CH4 + 2O2 → CO2 + 2H2O + Heat

(iv) Thus when hydrocarbons are burnt in air, large amount of heat is evolved with formation of carbon dioxide (CO2) and water (H2O). Due to evolution of heat on combustion, carbon and its compounds are used as fuels.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. In which compound forms does carbon occur?
Answer:
Carbon in its combined state exists as various compounds such as:

  • Carbon dioxide and in the form of carbonates such as calcium carbonate, marble, calamine (ZnCO3).
  • Fossil fuel – coal, petroleum, natural gas.
  • Carbonaceous nutrients – carbohydrates, proteins, fats.
  • Natural fibres – cotton, wool, silk.
  • Hydrocarbons – compound of carbon and hydrogen.

c. Write the uses of the diamond.
Answer:
Uses of diamonds are:

  • Diamonds are used in glass cutting and rock drilling machines.
  • Diamonds are used in ornaments.
  • Diamond knives are used in the eye surgery.
  • Diamond dust is used for polishing other diamonds.
  • Diamond is used to make windows giving protection from radiation in space and in artificial satellites.

3. Explain the difference:

a. Diamond and graphite.

Answer:

Diamond  Graphite
(i) Diamond is a brilliant, hard and crystalline allotrope of carbon. (i) Graphite is a black, soft, brittle and slippery crystalline allotrope of carbon.
(ii) In diamonds, every carbon atom is bonded to four neighbouring atoms by covalent bonds forming tetragonal three dimensional structure which makes it very hard. (ii) In graphite, every carbon atom is bonded to three other carbon atoms by covalent bonds in such a way that a hexagonal layered structure is formed. A graphite crystal is made of many such layers of carbon atoms. These layers slip over each other on applying pressure.
(iii) Density of diamond is 3.5 g/cm3. (iii) Density of graphite is 1.9 to 2.3 g/cm3.
(iv) Diamond is a bad conductor of electricity as it does not have free electrons. (iv) Inside each layer of graphite, free electrons move continuously within the entire layer. Hence, graphite is a good conductor of electricity.

b. Crystalline and non-crystalline forms of carbon.
Answer:

Crystalline forms of carbon Non-crystalline forms of carbon
(i) A crystalline form has a regular and definite arrangement of atoms.
(ii) They have high melting points and boiling points.
(iii) A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.
(iv) Diamond, graphite and fullerene are different crystalline forms of carbon.
(i) A non-crystalline form does not have a regular and definite arrangement of atoms.
(ii) They have low melting points and boiling points.
(iii) They are amorphous, hence, they do not have definite geometrical shape.
(iv) Coal, charcoal and coke are different non­crystalline/amorphous forms of carbon.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

4. Write scientific reasons

a. Graphite is a conductor of electricity.
Answer:

  • In graphite, each carbon is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • Due to this structure, graphite has free electrons available.
  • These free electrons move continuously within the entire layer.
  • Hence, graphite is a good conductor of electricity.

b. Graphite is not used in ornaments.
Answer:

  • Graphite is a black, soft, brittle and dull form of carbon.
  • It is neither malleable nor ductile.
  • These properties of graphite make it unsuitable for making of ornaments.
  • Hence, graphite is not used for making ornaments.

c. Limewater turns milky when CO2 is passed through it.
Answer:

  • Limewater traditionally means a weak solution of the alkali calcium hydroxide Ca(OH)2.
  • When CO2 is passed through limewater, it reacts with calcium hydroxide to form insoluble particulates (precipitate) of Calcium carbonate (CaCO3).
  • Calcium carbonate is weak basic salt and this gives a milky white precipitate.
  • Hence, lime water turns milky when CO2 gas is passed through it.

d. Biogas is an eco-friendly fuel.
Answer:

  • Biogas is formed by the decomposition of animal dung, dry leaves, wet garbage in a biogas plant.
  • This produces methane gas also called biogas.
  • Biogas is a very cheap fuel option which meets the demand for cooking gas.
  • Biogas is eco-friendly as it contains about 55% to 60% of methane and rest is carbon dioxide, hence, on combustion it does not produce harmful gases which cause pollution.
  • Biogas is a fuel which is convenient to use and in addition to this it produces a very good manure as a side product of the process.
  • Hence, biogas is an eco-friendly fuel.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

5. Explain the following.

a. Diamond, graphite and fullerenes are crystalline forms of carbon.
Answer:

  • Carbon exhibits a property of allotropy in which an element exists in more than one form in nature.
  • The chemical properties of these different forms are the same but their physical properties are different.
  • Carbon exists in crystalline as well as non-crystalline (amorphous form).
  • Crystalline form has a regular and definite arrangement of atoms. They have high melting points and boiling points.
  • A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.
  • Carbon has three crystalline allotropes such as diamond, graphite and fullerene.
  • In the structure of diamond, every carbon atom is bonded to four neighbouring atoms by covalent bonds.
  • Therefore, diamond has a tetragonal three dimensional structure which makes it very hard.
  • Brilliant and pure diamond is the hardest natural substance.
  •  In the structure of graphite, every carbon atom in graphite is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • A graphite crystal is made of many sheets or layers of carbon atoms.
  • These layers slip over each other on applying pressure. One layer of graphite is called graphene.
  • Fullerene is rarely found in nature. It is found in soot and in interstellar space.
  • The first example of fullerene is Buckminster fullerene (C60).
  • This allotrope of carbon is named fullerene after the architect.
  • Richard Buckminster Fuller because the structure of C60 resembles the structure of the geodesic dome he designed.
  • (xi) Molecules of fullerenes are found in the form of buckyballs and buckytubes.
  • (xii) There are 30 to 900 carbon atoms in one molecule of a fullerene. , C60, C70, C76, C82 and C86 are other examples of fullerene. Their molecules occur in small numbers in soot.

b. Methane is called marsh gas.
Answer:

  • Methane is formed by the decomposition of plant and animal matter in swamps or marshy areas.
  • As methane gas bubbles out from marshy area, it is called as marsh gas.

c. Petrol, diesel, coal are fossil fuels.
Answer:
(i) A fossil fuel is a fuel formed by natural processes, such as anaerobic decomposition of buried dead organisms. Fossil fuels contain high percentage of carbon. The word carbon is derived from the Latin word ‘Carbo’ meaning coal.

(ii) Coal is formed from the remains of trees and other vegetation. Approximately 350 million years ago, these remains were trapped on the bottom of swamps, accumulating layer after layer and creating a dense material called peat. As this peat was buried under more and more ground, the high temperature and pressure transformed it into coal.

(iii) Petrol and diesel are obtained from mineral oil. Mineral oil also called as crude oil or petroleum oil is formed from the remains of plants and animals that lived in the seas millions of years ago. This plant and animal matter has been drawn down and subjected to extremes of temperature and pressure over millions of years ago.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(iv) Mineral oil is commonly formed in rocks under the sea bed. The word petroleum is derived from Latin word ‘Rock oil’. Petrol and Diesel are obtained from mineral oil by the process called as fractional distillation.
Thus petrol, diesel and coal are fossil fuels.

d. Uses of various allotropes of carbon.

e. Use of CO2 in fire extinguisher.
Answer:

  • CO2 based fire extinguishers do not cause corrosion and are non-conductors of electricity.
  • It is beyond their capacity to extinguish a big fire.
  • Therefore these fire extinguishers are used to extinguish small scale fire of electrical and electronic equipments.

f. Practical uses of CO2.
Answer:
Practical uses of CO2 are:

  • CO2 is used to make aerated drinks.
  • CO2 obtained by chemical reaction or kept under pressure is used in fire extinguishers. Liquified CO2 is used to remove caffeine from coffee.
  • Liquid CO2 is used as solvent in modem eco-friendly dry cleaning.
  • Solid carbon dioxide is used in cold storage and to keep milk and milk products and frozen substances cool during transport. It is also used for getting special effects of a mist in dramas and movies.

6. Write two physical properties each.

a. Diamond
Answer:
Properties of diamond are:

  • Brilliant and pure diamond is the hardest natural substance.
  • The density of diamond is 3.5 g/cm3.
  • The melting point of diamond is 3500 °C.
  • When a diamond is heated at 800 °C in the presence of oxygen, CO2 is given away. In this process no other product besides CO2 is formed.
  • Diamond does not dissolve in any solvent.
  • Acids/bases have no effect on diamond.
  • Diamond is a bad conductor of electricity as it does not have free electrons.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. Charcoal
Answer:

  • The charcoal that is made from animals is made from their bones, horns, etc.
  • On the other hand, the charcoal made from plants is formed by combustion of wood in an insufficient supply of air.

c. Fullerene
Answer:
Properties of fullerenes are:

  • Molecules of fullerenes are found in the form of buckyballs and buckytubes.
  • There are 30 to 900 carbon atoms in one molecule of a fullerene.
  • Fullerenes are soluble in organic solvents such as carbon disulphide, chlorobenzene.

7. Complete the following Chemical reactions.

1. ………………..+………………..→ CO + 2H2O + Heat
2. ………………..+………………..→ HCl + Cl + HCl
3. 2 NaOH + CO2 →………………..+………………..
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 1

8. Write answers to the following in detail.

a. What are the different types of coal? What are their uses?
Answer:
Coal is a fossil fuel. It contains carbon, hydrogen and oxygen. It also contains nitrogen, phosphorus and sulphur. It occurs in the solid state. It is of four types.

  • Peat: Formation of peat is the first step in the formation of coal. It contains a high proportion
    of water and less than 60% of carbon. Therefore, not much heat can be obtained from peat.
  • Lignite: Peat was transformed into Lignite due to increased pressure and temperature inside the earth. It contains 60 to 70% of carbon.- Lignite is the second step of the formation of coal.
  • Bituminous coal: Bituminous coal was formed as the third step of formation of coal. It contains 70 to 90% of carbon.
  • Anthracite: Anthracite is known as the pure form of coal. This coal is hard and contains about 95% of carbon.

Uses of coal:

  • Coal is used as fuel in factories and homes.
  • Coal is used to obtain coke, coal gas and coal tar.
  • Coal is used in thermal power plants for generation of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

b. How will you prove experimentally that graphite is good conductor of electricity?
Answer:
Apparatus required: Lead pendi, electrical wires, battery/cell, small bulb, etc.

Step-I: Remove the lead from a pencil and assemble the apparatus as shown in the diagram.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 28

Step-II:

  • Start the electric current in the circuit, the moment the electric current is passed through the circuit, the bulb glows.
  • This experiment proves that graphite is a good conductor of electricity as graphite has free electrons moving continuously within the entire layer and these free electrons conduct electricity in the lead of the pencil.

c. Explain the properties of carbon.
Answer:
Allotropic nature of Carbon: Some elements occur in nature in more than one form. The chemical properties of these different forms are the same but their physical properties are different. This property of elements is called allotropy. Carbon shows the property of allotropy.

Carbon allotropes are of two types:
(A) Crystalline forms:
Carbon has three crystalline allotropes: Diamond, Graphite and Fullerene. Properties of crystalline forms of carbon are as follows:

  • A crystalline form has a regular and definite arrangement of atoms.
  • They are made up of only carbon atoms.
  • They have high melting points and boiling points.
  • A crystalline form has a definite geometrical shape, sharp edges and plane surfaces.

(B) Amorphous forms or non-crystalline forms: Coal, charcoal, coke are the non-crystalline forms of carbon.
Properties of non-crystalline forms of carbon are as follows:

  • The arrangement of carbon atoms in this form is not regular.
  • Apart from carbon atoms, they also contain hydrogen, oxygen, nitrogen, phosphorus and sulfur atoms.
  • Compared to a crystalline form, they have low melting and boiling points.
  • Most of them are used as fuels.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

d. Classify carbon.
Answer:
Carbon is classified as follows:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 2

9. How will you verify the properties of carbon dioxide?
Answer:
Properties of carbon dioxide can be verified in the following ways:

  • When a burning candle is placed in a gas jar of carbon dioxide, it extinguishes indicating that carbon dioxide is a non-combustible gas and does not support combustion.
  • When carbon dioxide gas is passed through lime water, it turns lime water milky due to the formation of insoluble calcium carbonate.
  • Moist blue litmus turns red in a gas jar of carbon dioxide indicating, it is acidic in nature.
  • Carbon dioxide is fairly soluble in water, it dissolves in water forming carbonic acid.

Class 9 Science Chapter 13 Carbon: An Important Element Intext Questions and Answers

Can you recall?

Question 1.
Whatremains behind on complete combustion of any organic compound?
Answer:
A black colour substance ‘Carbon’ remains behind on complete combustion of any organic compound.

Question 2.
What type of element is carbon? Give some information about it.
Answer:

  • Carbon is a non-metallic element. It is available ‘ abundantly in the nature and occurs in free as well as combined state.
  • Carbon in free state is found as diamond and graphite and in combined state it is present to the extent of approximately 0.27% in the form of carbonate, coal, petroleum.
  • In atmosphere, the proportion of carbon in the form of carbon dioxide is approximately 0.03%.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.
What is an element? What are the different types of elements.
Answer:

  • A substance which cannot be decomposed into simple substances by any physical or simple chemical method is called as an Element.
  • An element is composed of atoms of only one kind.
  • The different types of elements are:
    (a) Metals – Examples: Gold, Silver, etc.
    (b) Non-metals – Examples: Carbon, Sulphur, etc.
    (c) Metalloids – Examples: Silicon, Antimony, etc.

Answer the following questions:

Question 1.
Does an electric charge form on atoms when a covalent bond is formed between them? Why is a single bond between two carbon atoms strong and stable?
Answer:

  • No, electric charge is not formed on atoms when a covalent bond is formed between them. This is because covalent bond is formed by sharing of electrons.
  • Therefore, there is no change in number of electrons and protons in these atoms and they remain electrically neutral.
  • Covalent bonds occur when electrons are shared between two atoms. A single covalent bond is formed when only one pair of electrons is shared between atoms.
  • In this sharing, the atomic orbitals directly overlap between the nuclei of two atoms forming the strongest type of covalent bond called as sigma bond.
  • Hence, a single covalent bond between two carbon atoms is strong and stable.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 2.
In which of the solvents – water, kerosene and cooking oil does the coal powder dissolve?
Answer:
Solvents such as water, kerosene and cooking oil do not dissolve coal powder in them.

Question 3.
What inference will you draw about the solubility of carbon?
Answer:
Carbon is insoluble in water, kerosene and cooking oil.

Question 4.
Is the density of CO2 more or less than that of air?
Answer:

  1. Density is defined as mass per unit volume of a substance, expressed as kilograms per cubic meter.
  2. At standard temperature and pressure, the density of air is 1.29 kg/m3. While that of carbon dioxide (CO2) is 1.79 kg/m3. Hence, density of CO2 is more than that of air.

Question 5.
What is a compound? How are compounds formed?
Answer:

  • A compound is a substance formed when two or more elements are chemically bonded with each other in definite proportion by weight.
  • When atoms or two or more different elements chemically react with each other in a definite proportion by weight, a compound is formed. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • The properties of a compound are altogether different from its constituent elements.
  • Example: Pure water is a compound made up of two elements Hydrogen and Oxygen. Hydrogen (H) atoms and oxygen (O) atoms have chemically reacted with each other in definite proportion to form a compound-water (H2O). The proportion of hydrogen and oxygen in water by volume is 2 : 1 and by weight is 1 :8 respectively.
  • The properties of water are altogether different from the properties of its constituent elements, hydrogen and oxygen.

Class 9 Science Chapter 13 Carbon: An Important Element Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The organic compound having double or triple bond in them is termed as
(a) unsaturated
(b) inorganic
(c) saturated
(d) complete
Answer:
(a) Unsaturated hydrocarbon

Question 2.
Covalent compounds are generally soluble in solvents.
(a) inorganic
(b) organic
(c) mixed
(d) mineral
Answer:
(b) organic

Question 3.
Methane is also called as
(a) maha gas
(b) mar’s gas
(c) Anthracite
(d) marsh gas
Answer:
(d) Marsh gas

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 4.
is an alio trope of carbon used to make ornaments.
Answer:
Diamond

Question 5.
The number of valence electrons in carbon is
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
(a) 4

Question 6.
In saturated hydrocarbons, two carbon atoms are linked by
(a) double bond
(b) triple bond
(c) multiple bond
(d) single bond
Answer:
(d) single bond

Question 7.
Carbon has 4 valence electrons in its outermost shell, hence, it is
(a) divalent
(b) tetravalent
(c) trivalent
(d) pentavalent
Answer:
(b) tetravalent

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 8.
Methanogenic bacteria act on the organic acids to produce
(a) oxygen gas
(b) nitrogen gas
(c) methane gas
(d) carbon dioxide gas
Answer:
(c) Methane gas

Question 9.
is used to remove caffeine from coffee.
(a) liquefied O2
(b) liquefied N2
(c) liquefied CO2
(d) liquefied CH4
Answer:
(c) Liquified CO2

Question 10.
Exhaled air contains about CO2.
(a) 4%
(b) 3%
(c) 5%
(d) 6%
Answer:
(a) 4%

Question 11.
Electronic configuration of carbon is
(a) (2, 2)
(b) (2,4)
(c) (2, 5,4)
(d) (2, 6)
Answer:
(b) 2, 4

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 12.
Molecular formula of ethane is
(a) C3H4
(b) C2H4
(c) C2H6
(d) C2H2
Answer:
(c) C2H6

Question 13.
Molecular mass of carbon dioxide is
(a) 28
(b) 22
(c) 56
(d) 44
Answer:
(d) 44

Question 14.
On heating a mixture of hydrogen and carbon monoxide gases at 300°C in the presence of nickel (catalyst) is formed.
(a) carbon dioxide gas
(b) oxygen gas
(c) methane gas
(d) ethylene gas
Answer:
(c) methane gas

Question 15.
Ethane with molecular formula C2H6 has covalent bonds.
(a) six
(b) seven
(c) eight
(d) nine
Answer:
(b) seven

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 15.
Weight of Kohinoor Diamond was
(a) 186 carats
(b) 27 carats
(c) 252 carats
(d) 23 carats
Answer:
(a) 186 carats

Question 16.
Covalent compounds have
(a) high melting point
(b) low melting point
(c) moderate melting point
(d) very high melting point
Answer:
(b) low melting point

Question 17.
Methane is
(a) C2H6
(b) C3H8
(C) CH2
(d) CH4
Answer:
(d) CH4

Question 18.
Wohler, a German scientist synthesized the compound from an inorganic compound ammonium cyanate.
(a) methane
(b) ethylene
(c) urea
(d) acetic acid
Answer:
(c) urea

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 19.
Molecular mass of methane is
(a) 19
(b) 16
(c) 17
(d) 18
Answer:
(b) 16

Question 20.
coal contains highest percentage of carbon.
(a) Bituminous
(b) Peat
(c) Anthracite
(d) Lignite
Answer:
(c) Anthracite

Question 21.
H2C = CH2 is ……………………… .
(a) Propane
(b) Ethyne
(c) Ethane
(d) Ethene
Answer:
(d) Ethene

Question 22.
CH3 – C = CH is ……………………… .
(a) Propene
(b) Propyne
(c) Ethyne
(d) Ethene
Answer:
(b) Propyne

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 23.
Carbon dioxide gas is not used in
(a) photosynthesis
(b) aerated drinks
(c) glass cutting
(d) fire extinguishers
Answer:
(c) glass cutting

Question 24.
Melting point of diamond is
(a) 3700 °C
(b) 3500 °C
(c) 4000 °C
(d) 2500 °C
Answer:
(b) 3500 °C

Question 25.
Melting point of C02 is
(a) 26 °C
(b) 56 °C
(c)-56.6°C
(d)-98°C
Answer:
(c) -56.6 °C

Question 26.
Melting point o* methane is
(a) -182.5 °C
(b) -161.5 °C
(c) 182.5 °C
(d) 161.5 °C
Answer:
(a) -182.5 °C

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 27.
Biogas contains about methane.
(a) 55% to 60%
(b) 20% to 25%
(c) 90% to 95%
(d) 40% to 45%
Answer:
(a) 55% to 60%

Question 28.
is not a property of carbon dioxide gas.
(a) Supporting combustion
(b) Odourless
(c) Colourless
(d) Turns blue litmus red
Answer:
(a) Supporting combustion

Name the following:

Question 1.
Industries that use methane in the form of natural gas.
Answer:
Fabric mills, paper mills, food processing industry, petrol purification.

Question 2.
Organic compounds prepared from methane.
Answer:
Ethanol, methyl chloride, methylene chloride, acetylene.

Question 3.
Chemicals used in regular fire extinguisher.
Answer:
Sodium bicarbonate and dilute sulphuric acid.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 4.
Components of biogas
Answer:
Methane, carbon dioxide

Question 5.
Organic solvents is which fullerenes are soluble.
Answer:
Carbon disulfide, chlorobenzene.

Question 6.
Use of carbon dioxide is dramas and movies
Answer:
The special effect of mist.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) Diamond
(2) Fullerenes
(3) Graphite
(a) Hexagonal layered structure
(b) Tetragonal three-dimensional structure
(c) Geodesic dome

Answer:
(1-b),
(2- c),
(3 – a)

Question 2.

Column ‘A’ Column ‘B’
(1) Peat
(2) Lignite
(3) Bituminous
(4) Anthracite
(a) 60-70% of Carbon
(b) 95% of Carbon
(c) less than 60% of Carbon
(d) 70 – 90% of Carbon

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.

Column ‘A’ Column ‘B’
(1) Propyne
(2) Propene
(3) Propane
(a) CH3 – CH2 – CH3
(b) CH3 – C = CH
(c) CH3 – CH = CH2

Answer:
(1 – b),
(2 – c),
(3 – a)

Question 4.

Column ‘A’ Column ‘B’
(1) Fullerene
(2) Diamond
(3) Graphite
(a) Lubricants
(b) Insulator
(c) Ornaments

Answer:
(1 – b),
(2 – c),
(3 – a)

Question 5.

Column ‘A Column ‘B’
(1) Water gas
(2) Methane gas
(3) Producer gas
(4) Carbon dioxide gas
(a) CH4
(b) CO + H2
(c) CO2
(d) CO + H2 + CO2 + N2

Answer:
(1 – b),
(2 – a),
(3 – d),
(4 – c)

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

State whether the following statements are true or false. Correct the false statement.

(1) Study of organic compounds is called as organic chemistry.
(2) Hydrocarbons with double bonds are called as saturated hydrocarbons.
(3) Ethene is saturated hydrocarbon.
(4) Covalent compounds are good conductor of electricity.
(5) Methane is a covalent compound.
(6) Covalent compounds are soluble in organic solvents.
(7) Graphite is used in making lubricants and lead pencils.
(8) The density of diamond is 9.8 g/cm3.
(9) Diamond knives are used in eye surgery.
(10) Fullerenes are used as insulators.
(11) Coke is used in production of water gas i.e. (CO2 + H2O).
(12) Structural formula of propyne is CH3 – C = CH.
(13) COz is used to make aerated drinks.
(14) Methane gas is black in colour.
(15) Production of biogas is aerobic process.
Answer:
(1) True
(2) False. Hydrocarbons with double bonds are called as Unsaturated hydrocarbons.
(3) False. Ethene is Unsaturated hydrocarbon.
(4) False. Covalent compounds are bad conductor of electricity.
(5) True
(6) True
(7) True
(8) False: The density of diamond is 3.5 g/cm3.
(9) True
(10) True
(11) False. Coke is used in production of water gas i.e. (CO + H2).
(12) False. Structural formula of propyne is CH3 – C = CH.
(13) True
(14) False. Methane gas is colourless.
(15) False. Production of biogas is anaerobic process.

Find the odd man out:

Question 1.
Propane, Methane, Ethene, Pentane.
Answer:
Ethene. It is an unsaturated hydrocarbon with double bond between two carbon atoms while rest are saturated hydrocarbons with single bond between two carbon atoms.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 2.
CH4, C2H6, C3H8, CaCO3.
Answer:
CaCO3. It is a salt which is an inorganic’ compound, while rest are hydrocarbon compounds, i.e. organic compounds.

Question 3.
C2H2, C3H8, C2H6, CH4.
Answer:
C2H2. It is an unsaturated hydrocarbon with triple bonds while rest are saturated hydrocarbons with single bonds between two carbon atoms.

Question 4.
Diamond, Fullerene, Graphite, Methane.
Answer:
Methane. It is a marsh gas while rest are allotropes of carbon.

Question 5.
Coal, Petroleum, Natural gas, Cotton
Answer:
Cotton. Cotton is a natural fibre while rest all are fossil fuels.

Question 6.
Cotton, Silk, Proteins, Wool
Answer:
Proteins. Proteins are carbonaceous nutrients while rest all are natural fibres.

Question 7.
Carbohydrates, Coal, Proteins, Fats
Answer:
Coal. Coal is a fossil fuel while rest all are carbonaceous nutrients.

Question 8.
Peat, Charcoal, Lignite, Bituminous
Answer:
Charcoal. Charcoal is a non-crystalline form of carbon while rest all are types of coal.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 9.
Lubricants, Electrodes, Ornaments, Arc Lamps
Answer:
Ornaments. Ornaments are made from diamonds while rest all are made from graphite.

Write the correlated terms:

(1) Propene : Double bond :: Propyne : ……………………. .
(2) Ethane : CH3 – CH3 :: Ethene : ……………………. .
(3) Hydrogen : Monovalent :: Carbon : ……………………. .
(4) Methane : Low melting point i.e. -182.5°C Diamond : ……………………. .
(5) Graphite : Hexagonal structure :: Diamond : ……………………. .
(6) Density of Diamond : 3.5 g/cm3 :: Density of Graphite : ……………………. .
(7) Peat: 60% of carbon :: Anthracite : ……………………. .
(8) Melting point of CO2: -56.6°C :: Melting point of CH4: ……………………. .
Answer:
(1) Triple Bond
(2) CH2 = CH2
(3) Tetravalent
(4) High melting point i.e. 3500°C
(5) Tetragonal structure
(6) 1.9 to 2.3 g/cm3
(7) 95% of carbon
(8) -182.5 °C

Give scientific reasons:

Question 1.
Covalent compounds have low melting and boiling points.
Answer:

  • When substances melt or boil, bonds between the molecules are broken due to heat supplied.
  • On covalent compounds, the intermolecular forces of attraction are weak.
  • Hence, intermolecular forces in covalent compounds are broken easily due to which they have low melting and boiling points.

Question 2.
Graphite is a good conductor of electricity whereas diamond is a non-conductor of electricity.
Answer:

  • In diamond, each carbon atom uses all its four electronic to get bonded to another carbon atom, whereas in graphite each carbon uses there out of four electrons during bonding.
  • Hence in diamond no free electrons are left whereas in graphite free electrons are available.
  • Due to the presence of electron which are free to flow graphite is a good conductor of electricity, whereas due to the absence of free electrons diamond is a non-conductor of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 3.
Graphite is used as a lubricant.
Answer:

  • Lubricants are used to reduce friction and wear and tear of mechanical parts.
  • Graphite is smooth and slippery and hence is useful in reducing friction. Hence, graphite is used as a lubricant.

Question 4.
Methane is used as a domestic fuel.
Answer:

  • Methane is highly inflammable.
  • It burns by reacting with oxygen to give a bluish flame.
  • It burns completely and producers 213 Kcal/ mol of heat.
  • Being the smallest hydrocarbon, the proportion of CO2 released in the combustion of methane is small.
  • Therefore methane is used as a domestic fuel.

Answer in short:

Question 1.
What was the contribution of chemist Wohier in organic chemistry?
Answer:

  • The German chemist Wohier synthesized an organic compound urca from an inorganic compound ammonium cyanate,
  • Ever since then, many organic compounds have been made from inorganic compounds.
  • Carbon was found to he the main element in all these compounds.
  • Hence, organic chemistry is also referred to as chem is try of carbon corn pounds.

Question 2.
With neat diagram explain the structure of diamond.
Ans.

  • In diamond, every carbon atom is bonded to four neighbouring atoms by covalent bonds.
  • Therefore, diamond has a tetragonal three dimensional structure which makes it very hard.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 29

Question 3.
With neat diagram explain the structure of graphite?
Answer:

  • Every carbon atom in graphite is bonded to three other carbon atoms in such a way that a hexagonal layered structure is formed.
  • A graphite crystal is made of many sheets or layers of carbon atoms.
  • These layers slip over each other on applying pressure.
  • One layer of graphite is called graphene.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 30

Question 4.
Give the properties of graphite.
Answer:
Properties of graphite are:

  • Graphite found in nature is black, soft, brittle and slippery.
  • Inside each layer of graphite, free electrons move continuously within the entire layer. That is why graphite is a good conductor of electricity. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • Due to the layered structure graphite can be used for writing on paper.
  • The density of graphite is 1.9 to 2.3 g/cm3.
  • Graphite does not dissolve in most solvents.

Question 5.
Give the uses of graphite.
Answer:
Uses of graphite are:

  • Graphite is used for making lubricants.
  • Graphite is used for making carbon electrodes.
  • Graphite is used in pencils for writing.
  • Graphite is used in paints and polish.
  • Graphite is used in arc lamps which gives a very bright light.

Question 6.
Give the uses of fullerenes.
Answer:
Uses of Fullerenes are:

  • Fullerenes are used as insulators.
  • Fullerenes are used as a catalyst in water purification.
  • At a certain temperature, fullerene exhibits superconductivity.

Question 7.
What is Coal?
Answer:

  • Coal is a fossil fuel. It contains carbon, hydrogen and oxygen.
  • It also contains nitrogen, phosphorus and sulphur.
  • It occurs in the solid state. It is of four types – peat, lignite, bituminous and anthracite.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 8.
What is Coke?
Answer:
The pure coal that remains when coal gas has been taken away from coal, is called coke.

Question 9.
Give the uses of Coal.
Answer:
Uses of Coal are:

  • Coal is used as fuel in factories and homes.
  • Coal is used to obtain coke, coal gas and coal tar.
  • Coal is used in thermal power plants for generation of electricity.

Question 10.
Give the uses of Coke.
Answer:
Uses of Coke are:

  • Used as domestic fuel.
  • Coke is used as a reducing agent.
  • Coke is used in production of water gas (CO + H2) and producer gas (CO + H2 + CO2 + N2)

Question 11.
Name the types of coal and give the differences in them.
Answer:
The four types of coal are Peat, Lignite, Bituminous and Anthracite.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 3

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 12.
Give the properties of covalent compounds.
Answer:
Properties of covalent compounds:

  • Covalent compounds have low melting points and boiling points.
  • Generally they are insoluble in water and soluble in organic solvents.
  • They are poor conductors of heat and electricity.

Question 13.
What are saturated hydrocarbons? Give examples.
Answer:

  • The hydrocarbons having only single bonds between carbon atoms are called saturated hydrocarbons.
  • For example, ethane (C2H6) which is (CH3 – CH3), propane (C3H8) which is (CH3 – CH2 – CH3).

Question 14.
What are unsaturated hydrocarbons? Give examples.
Answer:

  • Some hydrocarbons have a multiple bond between two carbon atoms.
  • A multiple bond can be a double bond or a triple bond.
  • Hydrocarbons having at least one multiple bond are called unsaturated hydrocarbons.
  • For example, ethene (H2C = CH2), ethyne (HC = CH), propene (CH3 – CH = CH2), propyne (CH3 – C = CH).

Question 15.
Give the following information of carbon dioxide: Molecular formula, Molecular mass, Melting point, Percentage occurrence in air.
Answer:
Molecular formula – CO2, Molecular mass – 44, Melting point -56.6 °C Percentage occurrence in air – 0.03%.

Question 16.
Give the physical and chemical properties of carbon dioxide.
Answer:
Physical properties of carbon dioxide are:

  • It is an odourless gas.
  • It is a colourless gas.

Chemical properties of carbon dioxide are:

  • It is non-combustible and does not support combustion.
  • It turns lime water milky.
  • It is fairly soluble in water and dissolves in water forming carbonic acid.
  • It turns blue litmus red indicating it is acidic in nature.
  • The colour of universal indicator turns orange/ yellow in C02 indicating its pH value between 4 and 6, i.e. acidic in nature.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 17.
Give the uses of Carbon dioxide.
Answer:
Uses of Carbon dioxide are:

  • CO2 is used to make aerated drinks
  • Solid carbon dioxide is used in cold storage and also to keep milk and milk products and frozen substances cool during transport. It is also used for getting special effects of a mist in dramas and movies.
  • CO2 obtained by chemical reaction or kept under pressure is used in fire extinguishers.
  • Liquified CO2 is used to remove caffeine from coffee.
  • Liquid CO2 is used as solvent in modern eco-friendly dry cleaning.
  • Plants use CO2 in air for photosynthesis.

Question 18.
Give the occurrence of methane.
Answer:
The occurrence of methane is as follows:

  • Methane occurs in natural gas to the extent of 87%.
  • Decomposition of organic matter in the absence of air (anaerobic) produces methane.
  • Methane is present in biogas.
  • Methane is found in coal mines.
  • Methane is found at the surface of marshy places which is why it is also called marsh gas.
  • On heating a mixture of hydrogen and carbon monoxide gases at 300 °C in the presence of nickel (catalyst), methane gas is formed.
  • Fractional distillation of natural gas gives methane in pure form.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 19.
Give the physical properties of methane.
Answer:
Physical properties of methane are:

  • Melting point of methane is (-182.5 °C).
  • Boiling point of methane is (-161.5 °C).
  • It is a colourless gas.
  • The density of liquid methane is less than that of water.
  • Methane is sparingly soluble in water. It is highly soluble in organic solvents like gasoline, ether and alcohol.
  • Methane is in gaseous state at room temperature.

Question 20.
Give the uses of methane.
Answer:
Uses of methane are:

  • Methane in the form of natural gas is used in industries such as fabric mills, paper mills, food processing industry, petrol purification.
  • Being the smallest hydrocarbon, the proportion of CO2 released in the combustion of methane is small and, therefore, it is used as a domestic fuel.
  • Methane is used for production of organic compounds such as ethanol, methyl chloride, methylene chloride and acetylene.

Question 21.
How is methane formed? Give structural formula and electron dot model of methane.
Answer:

  • Atomic number of carbon is 6. The electronic configuration of carbon is (2, 4). The valency of carbon is 4.
  • Carbon atom can form four covalent bonds with other carbon atoms or atoms of different elements.
  • When a carbon atom shares one electron each with four hydrogen atoms and forms four C-H bonds, a methane (CH4) molecule is formed.
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 4

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 22.
What are organic and inorganic compounds?
Answer:
Compounds obtained directly or indirectly from plants and animals are called organic compounds and compounds obtained from minerals are called inorganic compounds.

Question 23.
What is allotropy?
Answer:

  • Allotropy – Some elements occur in nature in more than one form.
  • The chemical properties of these different forms are the same but their physical properties are different.
  • This property of elements is called allotropy.

Question 24.
What are basic organic compounds? What are they also called as?
Answer:

  • The compounds formed from only carbon and hydrogen are called basic organic compounds.
  • These are also called hydrocarbons.

Write a balanced chemical equation and explain the following chemical reactions.

Question 1.
Preparation of urea from Ammonium cyanate:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 5
Organic compound Urea can be synthesized from an inorganic compound Ammonium cyanate.

Question 2.
Coal when burnt in air
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 6
When coal is burnt in air, the carbon present in coal combines with oxygen present in air to form carbon dioxide gas.

Question 3.
Calcium carbonate reacts with dilute hydrochloric acid:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 7
When Calcium carbonate reacts with dilute hydrochloric add it forms Caldum chloride, water and Carbon dioxide gas is evolved.

Question 4.
Carbon dioxide is passed through an aqueous solution of Sodium hydroxide:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 8
When Carbon dioxide gas is passed through an aqueous solution of Sodium hydroxide it forms Sodium carbonate and Water.

Question 5.
Carbon dioxide is passed through an aqueous solution of Sodium carbonate:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 9
When carbon dioxide is passed through an aqueous solution of Sodium carbonate, it forms Sodium bicarbonate.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 6.
Carbon dioxide is passed through freshly prepared lime water for a short duration and then for longer duration:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 10
When Carbon dioxide is passed through freshly prepared lime water, it forms water and white precipitate of Calcium carbonate because of which lime water turns milky.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 11
When Carbon dioxide is continuously passed through milky lime water, it forms Calcium bicarbonate which is soluble in water and therefore, water once again turns colourless.

Question 7.
Carbon dioxide gas is dissolved in water.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 12
Carboh dioxide gas is fairly soluble in water, it dissolves in water under pressure to form Carbonic acid.

Question 8.
Sodium bicarbonate reacts with dilute Sulphuric acid, (reaction in fire extinguisher):
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 13
When Sodium bicarbonate reacts with dilute Sulphuric acid it forms Sodium sulphate, water and Carbon dioxide gas.

Question 9.
Methane gas is burnt in air:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 14
Methane is highly inflammable, it burns in air with a bluish flame and combines with oxygen present in air to form Carbon dioxide and water. In these reaction 213 kcal/mol of heat is given out.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Question 10.
Methane and Chlorine gases react with each other.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 15
Methane and chlorine gases react with each other at the temperature of 250 °C to 400 °C in presence of ultra voilet light and form mainly Methyl chloride (Chloromethane and ‘ Hydrogen chloride)
This reaction is called Chlorination of methane.

Question 11.
Production of methane gas in biogas plant.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 16
In biogas plant, microbes act on the bio-degradable complex organic compounds and produce organic acids.

The methanogenic bacteria act on the organic acids to produce Methane gas and Carbon dioxide gas.

Question 12.
Production of water gas:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 17
When super heated steam is passed over red hot coke at high temperature, it forms carbon monoxide gas and hydrogen gas.

This gaseous product of carbon monoxide gas and hydrogen gas together is called as water gas.

With the help of ne abelled diagram explain the following:

Question 1.
Regular fire extinguisher.
Answer:

  • A fire extinguisher contains Sodium bicarbonate powder. There is also dilute sulphuric acid placed in a glass capsule.
  • On pressing the knob, the capsule breaks and sulphuric acid comes in contact with the sodium bicarbonate. These two react chemically to release CO2 which comes out.
  • CO2 based fire extinguishers do not cause corrosion and are non-conductors of electricity. Therefore, they are used when electrical or electronic equipment catches fire. Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element
  • CO2 based fire extinguishers are used to extinguish small scale fire. It is beyond their capacity to extinguish a big fire.
  • In modern fire extinguishers liquid and solid CO2 is filled under pressure. On reducing the pressure, it becomes gaseous and comes out forcefully through the horn-like hose pipe.
  • Chemical reaction:
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 18

Question 2.
Biogas plant.
Answer:

  • Animal dung, dry leaves, wet garbage get decomposed by anaerobic microbes in a biogas plant.
  • This produces methane gas, also called biogas.
  • Biogas is a very cheap fuel option which meets the demand for cooking gas. It is also used for the production of electricity.
  • Biogas contains about 55% to 60% methane and the rest is carbon dioxide.
  • Biogas is a fuel which is convenient to use and, in addition to this, a very good manure is also produced as a side product of the process.
  • Biogas production process – Production of biogas is an anaerobic process. It takes place in two stages.
    (a) Production of acids – The microbes act on the biodegradable complex organic compound and produce organic acids.
    (b) Methane gas production – The methanogenic bacteria act on the organic acids to produce methane gas.

Chemical reaction:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 19

Answer the following

Question 1.
What happens when substances like milk, sugar, wool, dry leaves, hair and seeds are heated strongly in hard glass test tube?
Answer:
When milk, sugar, wool, dry leaves, hair and seeds are heated strongly in hard glass test tube, they get charred and a black substance is left behind. This black substance is carbon.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Answer the following questions:

Question 1.
Take some milk in an evaporating dish. Heat the evaporating dish on a bunsen burner. What remains behind at the bottom of an evaporating dish on complete evaporation of the milk?
Answer:

  • On complete evaporation of milk, a black residue is left behind.
  • This residue is of carbon.

Question 2.
Take small samples of sugar, wool, dry leaves, hair, seeds, split pulses and plastic in separate test tubes. Heat each test tube and observe the changes taking place in the substances. What does the black substance in each test tube indicate?
Answer:

  • On heating the above samples, they get charred and a black substance is left behind.
  • The black substance is carbon.

Question 3.
Apparatus : Coal, match box, moist litmus paper, etc.

Procedure : Ignite the coal. Hold the moist blue litmus paper over the gas released on igniting the coal. Note the observation.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 20

(a) With which gas in the air does the coal react on igniting?
Answer:
Coal reacts with oxygen gas present in the air, on igniting.

(b) What is the substance formed?
Answer:
The substance formed is carbon dioxide.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(c) What change takes place in the litmus paper?
Answer:
The moist blue litmus paper turns red.

(d) Write down the chemical reaction taking place in the above procedure.
Answer:
(i) Carbon combines with oxygen to form Carbon dioxide gas.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 21
(ii) This C02 combines with water present on moist blue litmus paper to form Carbonic acid which turns blue litmus to red.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 22

Question 4.
Apparatus : Pencil, electrical wires, battery/ cell, small bulb, water, kerosene, test tube, lead pencils, etc.

Procedure : Remove the lead from a pencil and arrange the apparatus as shown in the above diagram. Note your observations for the following.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 23

(a) What is the colour of lead in the pencil? .
Answer:
The colour of the lead in the pencil is black as it is made from graphite which is an allotrophic form of carbon.

(b) Try to break lead with your hand.
Answer:
The lead breaks easily as it is made up of graphite which is brittle in nature.

(c) Start the electric current in the circuit and observe. What did you find?
Answer:
When we start the electric current in the circuit, the bulb in the circuit glows, indicating that the lead in the pencil is a good conductor of electricity.

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

(d) Take some water in a test tube. Take some kerosene in another test tube. Put lead dust in both the test tube. What did you observe?
Answer:
(i) Lead dust does not dissolve in water nor in kerosene.
(ii) It remains insoluble in both the test tubes.

Question 5.
Apparatus : Test tube, straw, lime water, etc.

Procedure: Take freshly prepared lime water in a test tube. Blow air in it for some time through the straw and observe the lime water. Note the observation.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 24

(a) What did you see?
Answer:
When we blow air through the straw in freshly prepared lime water, it turns milky.

(b) What might be the reason behind the change?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 25
Freshly prepared lime water is obtained by dissolving lime (CaO) in water (H2O).

When we blow air through the straw in lime water, carbon dioxide (CO2) present in the air reacts with lime water to form white precipitate (insoluble substance) of Calcium Carbonate (CaCO3) due to which lime water turns milky.
Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 26

Question 6.
With the help of a neat labelled diagram explain the laboratory preparation of Carbon dioxide gas.
Answer:
The laboratory preparation of Carbon dioxide gas is as follows:

Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon: An Important Element

Apparatus : Retort stand, round bottom flask, thistle funnel, gas delivery tube, gas jars.
Chemicals : Calcium carbonate (Pieces of Shahabad tiles/marble pieces/limestone), dilute hydrochloric acid.

Procedure:

  • Assemble the apparatus as shown in the figure. While assembling, place CaCO3 in the round bottom flask.
  • Add dilute HCl in the flask through thistle funnel. See to it that the end of the funnel dips in the acid.
  • CO2 is formed as a result of the reaction between CaCO3 and HCl. Collect this gas in four to five gas jars. The chemical equation of the above reaction is as follows. CaCO3 + 2HCl → CaCl2 + H2O + CO2
    Maharashtra Board Class 9 Science Solutions Chapter 13 Carbon An Important Element 27

 

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Class 9 Science Chapter 10 Information Communication Technology (ICT) Textbook Questions and Answers

1. Fill in the blanks to complete the statements. Justify the statements.

a. While working with a computer we can read the information stored in its memory and perform other actions in ………………………. memory.
b. While presenting pictures and videos about the works of scientists, we can use ……………………… .
c. To draw graphs based on the quantitative information obtained in an experiment, one uses ……………………… .
d. The first generation computers used to shut down because of ……………………… .
e. A computer will not work unless ………………………. is supplied to it.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

2. Answer the following questions.

a. Explain the role and importance of information communication in science and technology.
Answer:

  • ICT plays a key role in creating, displaying, collecting, processing and communicating information in the field of science and technology.
  • Following is the importance of ICT in science and technology:
    (a) Access to wide range of information
    (b) Storing of Data
    (c) Processing of Data
    (d) Securing work files
    (e) Proper representation of data

b. Which application software in the computer system did you find useful while studying science, and how?
Answer:

  • Microsoft word: To write down the information collected and making a document for further evaluation.
  • Microsoft excel : To draw graph based on the obtained numerical information from the experiment.
  • Internet explorer: To search for information in finding out the solution or solving the queries by reading the available information.

c. How does a computer work?
Answer:

Input unit Processor Output Unit
All types of information/ data is entered into the computer through this unit. Generally, a keyboard is used to enter data or information Processing Unit
(1) Memory unit
(2) Control unit
(3) ALU unit
The result/solution/ answer is eventually sent to the output unit. Generally, a screen/monitor or printer is used as an output unit.

d. What precautions should be taken while using various types of software on the computer?
Answer:

  • Antivirus must be installed.
  • Software should be legal and from a trusted place.
  • Application should be scanned before using.
  • Pirated Software should not be used Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)
  • Provide all necessary data to obtain the best possible results.

e. Which are the various devices used in information communication? How are they used in the context of science?
Answer:

  • Various devices used in information communication are: Computers, Laptops, Mobiles, Radios, Television, etc.
  • Computers, Laptops and Mobiles: Help in accessing, collecting, processing, communicating, sharing and storing of information. It helps in determining the appropriate conclusions in all fields, including the field of science.
  • Television: Help in getting information about the new and innovative technology.

3. Using a spreadsheet, draw graphs between distance and time, using the information about the movements of Amar, Akbar and Anthony given in the table on page 4, in the lesson on Laws of Motion. What precautions will you take while drawing the graph?
Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 3
Answer:
Precautions to be taken while drawing a graph:

  • The data should be kept in tabular form.
  • Whenever there is ‘drag and fill’ option used, ‘smart tag’ option should be used after ‘drag data’ to fill data as required. Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)
  • Entered data should be formatted in the manner required.
  • Various types of graphs can be created by using the same data, so appropriate graph should be selected.
  • Chart titles and axes titles should be updated as per the data.

4. Explain the differences between the different generations of computers. How did science contribute to these developments?
Answer:
Generation: 1st
Time Period: 1946 – 1956
Development: Vacuum Tubes Characteristics:

  • Huge in size
  • Expensive
  • Lot of electricity consumption
  • Heat generation

Generation: 2nd
Time Period: 1956 – 1963
Development: Transistors
Characteristics:

  • Frequent shutdowns
  • Superior to 1st Generation
  • Small in size and fast
  • Cheaper as compared to 1st Generation
  • Less consumption of electricity

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Generation: 3rd
Time Period: 1963 -1971
Development: IC
Characteristics:

  • Keyboards and monitors
  • OS
  • Smaller and still cheaper

Generation: 4th
Time Period: 1971 – 2010
Development: Microprocessor Characteristics:

  • Use of Internet
  • GUI
  • Introduction of portable devices like mobiles, laptops, etc.

Generation: 5th
Time Period: 2010 – Till Date
Development: Artificial Intelligence (AI) Characteristics:

  • Voice recognition
  • Sensors
  • Nano technology

1st Generation computers occupied the entire room, but due to advancement in science and technology, today’s computer fits into our pockets.

Initially computers needed a specific language to interact but today we use voice recognition for the same.

In these ways, science has contributed in making the computers faster, smaller, cheaper and much more useful.

5. What devices will you use to share with others the knowledge that you have?
Answer:
Devices like radios, televisions, pendrives, computers, laptops, mobiles, landlines, hard drives, CDs, memory cards help us in sharing our knowledge with others.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

6. Using information communication technology, prepare powerpoint presentations on at least three topics in your textbook. Make a flowchart of the steps you used while making these presentations.
Answer:
Steps for preparation of PowerPoint presentations:
Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 1

7. Which technical difficulties did you face while using the computer? What did you do to overcome them?
Answer:

  • Lagging: Due to lot of applications running at the same time, the computer starts lagging and becomes slow. Closing a few applications helped solve the problem of lagging.
  • Viruses and Bugs: Cybercrimes are rising daily, even from single mail the computer can be attacked by viruses. Installing a valid antivirus helps solve the problem of viruses and bugs.
  • Breach of Privacy: Confidential information being accessed by anyone is the breach of privacy. Putting privacy setting in place helps solve the problem. Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)
  • Physical Damage: Hardware over a period of time might get physically damaged. Taking precautions while using will help to solve the problem.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Intext Questions and Answers

Question 1.
Make a list of various hardware and software items of a computer,
Answer:
Hardware: Mouse, Keyboard, Pendrive, Monitor and other parts of computer.
Software: Operating Systems, Application Programs, Antivirus, etc.

Answer the following questions:

Question 2.
Which devices do we directly or indirectly use for collecting, sharing, processing and communicating information?
Answer:

  • Computers
  • Laptops
  • Mobiles
  • Memory Cards
  • Pendrives
  • Landlines
  • Hard disks etc.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 3.
How is information communication technology important for dealing with explosion of information?
Answer:

  • Information explosion means a situation whère information is available in abundance, in other words, too much information.
  • Devices like computers, laptops help us in easier accessment of information that we need from all the data.

Class 9 Science Chapter 10 Information Communication Technology (ICT) Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
…………………….. includes communication devices and the use of those devices as well as the services provided with their help.
(a) Operating System
(b) Office
(c) Computers
(d) Information Communication Technology
Answer:
(d) Information Communication Technology

Question 2.
Computers have gone through …………………….. generations.
(a) 5
(b) 7
(c) 10
(d) 8
Answer:
(a) 5

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 3.
First generation of Computers were considered to be present in the period of ……………………. .
(a) 2000 – 2001
(b) 1901 – 2001
(c) 1946 -1959,
(d) 1996 – 2001
Answer:
(c) 1946-1959

Question 4.
Full form of RAM is ……………………. .
(a) Roaming Application Memory
(b) Random Accessible Media
(c) Random Access Memory
(d) None of the above
Answer:
(c) Random Access Memory

Question 5.
Full form of ROM is ……………………. .
(a) Roaming Only Memory
(b) Random Output Media
(c) Read Only Memory
(d) None of the above
Answer:
(c) Read Only Memory

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 6.
…………………….. is raw information.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(c) Data

Question 7.
…………………….. are used for sharing information.
(a) Telephones
(b) Hard disks
(c) RAM
(d) ROM
Answer:
(a) Telephones

Question 8.
Hard disks are used for …………………….. information.
(a) storing
(b) communicating
(c) sharing
(d) all of the above
Answer:
(a) storing

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 9.
Computers are used for …………………….. information.
(a) storing
(b) managing
(c) sharing
(d) all of the above
Answer:
(d) All of the above

Question 10.
RAM and ROM are 2 types of …………………….. memory.
(a) external
(b) internal
(c) physical
(d) garbage
Answer:
(b) internal

Question 11.
The information stored in ROM is only …………………….., changes cannot be made.
(a) external memory
(b) readable
(c) accessible
(d) physical
Answer:
(b) readable

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 12.
…………………….. is a group of commands to be given to the computer.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(a) Program

Question 13.
…………………….. communicates between the computer and the person working on it.
(a) Program
(b) Memory
(c) Data
(d) Operating System
Answer:
(d) Operating System

Match the columns:

Column ‘A’ Column ‘B’
(1) Antivirus
(2) OS
(3) CPU
(4) Printer
(5) Mouse
a) Output Device
b) Software
c) Input Device
(d) Operating System
(e) Brain of the computer

Answer:
(1 – b),
(2 – d),
(3 – e),
(4 – a),
(5 – c)

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

State whether the following statements are true or false and if false, correct the false statements:

(1) RAM and ROM are the types of external memory.
(2) ICT includes communication devices and the use of those devices as well as services provided with their help.
(3) A computer cannot be used without its operating system.
(4) Microsoft Excel is used to make PowerPoint.
(5) Software refers to the set of commands given to the computer.
Answer:
(1) False. RAM and ROM are the types of internal memory.
(2) True
(3) True
(4) False. Microsoft Excel is used to make spreadsheets.
(5) True

Answer the following in one sentence:

Question 1.
Name the computer which was made between 1946 -1959.
Answer:
The ENIAC computer was made in the period of 1946-1959.

Question 2.
Give one example of Input Unit.
Answer:
Keyboard.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 3.
Name the 3 major parts of the processing units.
Answer:

  • Memory unit
  • Control unit
  • ALU unit

Question 4.
What precautions need to the taken care while entering formula into the excel?
Answer:
While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Question 5.
What is Internet Explorer?
Answer:
This is a kind of Search Engine. It helps to find the information we want from all the information available on the internet.

Question 6.
What is a PDF?
Answer:
A PDF or Portable Document Format file can be used to view the file to print it or to handle files.

Question 7.
What is C-DAC?
Answer:
C-D AC, is a well-known Centre for Development of Advanced Computing, situated in Pune.

Write the Full forms of the following abbreviation:

Question 1.
ICT
Answer:
Information Communication Technology

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 2.
OS
Answer:
Operating System

Question 3.
RAM
Answer:
Random Access Memory

Question 4.
ROM
Answer:
Read Only Memory

Question 5.
CPU
Answer:
Central Processing Unit

Question 6.
DOS
Answer:
Disk Operating System

Question 7.
PDF
Answer:
Portable Document Format

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 8.
ALU
Answer:
Arithmetic Logical Unit

Question 9.
GUI
Answer:
Graphical User Interface

Question 10.
C-DAC
Answer:
Centre for Development of Advanced Computing

Question 11.
ISCII
Answer:
Indian Script Code for Information Interchange

Define the following:

Question 1.
Memory
Answer:
Memory is the place for storing data obtained from the input and also the generated solution or answer by the computer.

Question 2.
RAM
Answer:
RAM is created from electronic components and can function only as long as it is supplied with electricity.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 3.
ROM
Answer:
Information stored in ROM can only be read and changes cannot be made to the information originally stored here.

Question 4.
Operating System
Answer:
It is a program which provides a means of communication between the computer and the person working on it. It is called the DOS (Disk Operating System).

Question 5.
Program
Answer:
A program is a group of commands to be given to a computer.

Question 6.
Data and Information
Answer:
Data is information in its raw (unprocessed) form.

Question 7.
Hardware
Answer:
Hardware consists of all the electronic and mechanical parts used in computers.

Question 8.
Software:
Answer:
Software refers to the commands given to the computer, information supplied to it (input) and the results obtained from the computer after analysis (output).

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Give scientific reasons:

Question 1.
Computer cannot function without its operating system.
Answer:

  • Operating system is like a link between the computer and the person working on it.
  • Operating system manages all the activities performed by the computer.
  • Without the operating system, the user won’t be able to input any data or run any program. Thus, a computer cannot run without an operating system.

Question 2.
ROM is a Read Only Memory.
Answer:

  • ROM also known as Read Only Memory is a part of internal memory of a computer where the information stored can only be read.
  • ROM helps store data permanently for a long period of time and the information stored cannot be deleted.
  • Thus, data in a ROM can only be read and cannot be altered or modified and hence, it is called as Read Only Memory.

Complete the table:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 2

Answer the following questions:

Question 1.
What precautions will you take when entering data?
Answer:

  1. As far as possible, the data should be kept in tabular form. Different types of data should be entered in different cells. Data should be entered neatly and in one ‘flow’. Unnecessary space and special characters should not be used.
  2. Many times we ‘drag and fill’ data. At such times, the ‘smart tag’ can be used after ‘drag data’ to fill any data in any manner as required. Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)
  3. Once the data has been entered, it can be formatted in different ways. Similarly, we can perform different types of calculations, using different formulae.
  4. While using a formula, the ‘=’ sign should be typed first. Similarly, no space should be inserted while typing any formula.

Answer in detail:

Question 1.
Write in short about the opportunities in the field of ICT.
Answer:
(i) Software Field: This is an important field. Having accepted the challenge of creating software, many companies have entered this field. The opportunities in the software field can be classified as follows – application program development, software package development, operating systems and utility development, special purpose scientific applications.

(ii) Hardware Field: Today, there are several companies in our country too, which make computers. They sell computers that they have themselves made. Others sell computers brought from outside as well as repair them and take maintenance contracts to keep computers in big companies working efficiently without a break. Plenty of jobs are available here. There are job opportunities in hardware designing, hardware production, hardware assembly and testing, hardware maintenance, servicing and repairs, etc.

(iii) Marketing: There are many establishments which make and sell computers and related accessories. They need good sales personnel who are experienced in the working of computers as well as skilled in marketing.

(iv) Training: The training of new entrants for various jobs is a vast field. It is very important to have dedicated teachers who are competent in the field of computers.

Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)

Question 2.
Write in short about the industries conducting research in the field of computers.
Answer:

  • C-DAC, the well-known Centre for Development of Advanced Computing, situated in Pune, is the leading institute in India that conducts research in the field of computers.
  • The first Indian supercomputer was made with help from this institute. Valuable guidance for making this computer (the Param computer) was received from the senior scientist Vijay Bhatkar. Param means the supreme.
  • This computer can perform one billion calculations per second. It is used in many fields like space research, movements in the interior of the earth, research in oil deposits, medicine, meteorology, engineering, military etc.
  • C-DAC has also participated in developing the ISCII code for writing different language scripts. (Indian Script Code for Information Interchange).

Question 3.
Use Microsoft Word to create a document and write equations.
Answer:

  • Click on the Microsoft word 2010 icon on the desktop.
  • Select the ‘New option in the ‘File’ tab, and then select the ‘Blank document’ option.
  • Type your material on the blank page on the screen using the keyboard. Use the language, font size, bold, etc. options in the Home tab to make the typed material attractive.
  • To type equations in the text, select the ‘Equation’ option in the ‘Insert’ tab. Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT)
    Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 4 Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 5
  • Select the proper equation and type it using mathematical symbols.
    Maharashtra Board Class 9 Science Solutions Chapter 10 Information Communication Technology (ICT) 5

 

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Class 9 Science Chapter 5 Acids, Bases and Salts Textbook Questions and Answers

1. Identify the odd one out and justify.

(a) Chloride, nitrate, hydride, ammonium
Answer:
Ammonium is the odd one out as it is a basic radical and rest all are acidic radicals. Generally, basic radicals are formed by the removal of electrons from the atom of metals such as Na+, Cu2+. But there are some exceptions, such as NH4+.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(b) Hydrogen chloride, sodium hydroxide, calcium oxide, ammonia
Answer:
Hydrogen chloride is the odd one out. It is acidic and rest all are basic.

(c) Acetic acid, carbonic acid, hydrochloric acid, nitric acid
Answer:
Carbonic acid is the odd one out. It is a dibasic acid and rest are all monobasic acids.

(d) Ammonium chloride, sodium chloride, potassium nitrate, sodium sulphate
Answer:
Ammonium chloride is the odd one out, as it is made up of a strong acid and weak base and rest all are formed from strong acid and strong base.

(e) Sodium nitrate, sodium carbonate, sodium sulphate, sodium chloride
Answer:
Sodium carbonate is the odd one out, as it is made up of a weak acid and strong base, and rest all are formed from strong acid and strong base.

(f) Calcium oxide, magnesium oxide, zinc oxide, sodium oxide.
Answer:
Zinc oxide is the odd one out, as it is an amphoteric oxide, and rest all are basic oxides.

(g) Crystalline blue vitriol, crystalline common salt, crystalline ferrous sulphate, crystalline sodium carbonate.
Answer:
Crystalline common salt is the odd one out, as it does not contain water of crystallisation. It is an ionic compound and ionic compounds are crystalline in nature and rest all have their crystalline structure because of their water of crystallization.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(h) Sodium chloride, potassium hydroxide, acetic acid, sodium acetate.
Answer:
Acetic acid is the odd one out. It is an acid, the rest are all salts.

2. Write down the changes that will be seen in each instance and explain the reason behind it.

(a) 50ml water is added to 50ml solution of copper sulphate.

Answer:

  • Copper sulphate solution is blue. It is a concentrated solution.
  • When 50 ml of water is added to this concentrated solution, it becomes a diluted solution.
  • The intensity of the blue colour is now different in this homogenous mixture.

(b) Two drops of the indicator phenolphthalein were added to 10ml solution of sodium hydroxide.

Answer:

  • Sodiumhy droxide is a base and phenolphthalein is a synthetic indicator.
  • Sodium hydroxide solution will turn pink if phenolphthalein is added to it.
  • It is a test for identifying bases.

(c) Two or three filings of copper were added to 10ml dilute nitric acid and stirred.
Answer:
When copper metal reacts with dilute nitric acid, the metal does not displace hydrogen from the acid like reaction with other metals. Instead the reaction produces nitric oxide, (NO).
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 39

(d) A litmus paper was dropped into 2ml dilute HCl. Then 2ml concentrated NaOH was added to it and stirred.
Answer:
Blue litmus Paper:

  • HCl is hydrochloric acid, so the blue litmus turns red.
  • When equal amount of NaOH is added the colour again changes to blue and remains the same.

Red litmus paper:

  • Red litmus paper shows no colour change in hydrochloric acid.
  • When some amount of NaOH is added the colour changes to blue initially but when the amount of NaOH is sufficient the blue colour dissappears.
  • Equal amounts of HC1 and NaOH results in the formation of NaCl, a salt, and the solvent water. This reaction is called the neutralization reaction.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(e) Magnesium oxide was added to dilute HCl and magnesium oxide was a added to dilute NaOH.
Answer:
(i) Magnesium oxide + dil HCl.
This is a neutralization reaction. Magnesium oxide is an insoluble base, it reacts with dilute HCl to produce a soluble salt MgCl2 and water H2O.
\(\mathrm{MgO}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(n)}\)

(ii) Magnesium oxide + NaOH.
No chemical reaction takes place between magnesium oxide and sodium hydroxide.

(f) Zinc oxide was added to dilute HCl and zinc oxide was added to dilute NaOH.
Answer:

  • Zinc oxide reacts with dilute hydrochloric acid to form zinc chloride and water. It is a neutralization reaction.
    \(\mathrm{ZnO}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{ZnCl}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
  • Zinc oxide reacts with sodium hydroxide to form sodium zincate and water.
    \(\mathrm{ZnO}_{(\mathrm{s})}+2 \mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{t})}\)

(g) Dilute HCl was added to limestone.
Answer:

  • When hydrochloric acid is added to limestone, carbon dioxide is liberated. Limestone is calcium carbonate.
    CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
  • Carbon dioxide is prepared in the laboratory using these chemicals.

(h) Pieces of blue vitriol were heated in a test tube. On cooling, water was added to it.
Answer:

  • On heating, the crystalline structure of blue vitriol breaks down to form a colourless powder and water is released.
  • This water is part of the crystal structure of blue vitriol.
  • It is called water of crystallization.
  • On adding water to the white powder, a solution was formed which has the same colour as the copper sulphate salt solution.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(i) Dilute H2SO4 was taken in an electrolytic cell and electric current was passed through it.
Answer:

  • If pure water is used in the electrolytic cell, current does not flow even on putting on the switch.
  • Pure water is a bad conductor of electricity. Dilute H2SO4 is acidulated water.
  • The electrical conductivity of water increases on mixing with strong acid or base in it due to their dissociation and electrolysis of water takes place.
  • H2SO4 is fully dissociated in aqueous solution. \(\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}\)
  • H2O is a weak electrolyte and is only slightly dissociated \(\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}\)
  • During electrolysis, the hydrogen ions migrate towards the cathode and are discharged there.

[H+ ions gains electrons and are converted to hydrogen gas]
\(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}\)
Cathode reaction:
\(2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-}\)
Anode reaction:
\(2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}\)

  • For every hydrogen ion discharged at the anode, another hydrogen ion is formed at the cathode.
  • The net result is that the concentration of the sulphuric acid remains constant and electrolysis of water is overall reaction. \(2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{H}_{2}+\mathrm{O}_{2}\)
  • The volume of the hydrogen gas formed near the cathode is double that of the oxygen gas formed near the anode.

3. Classify the following oxides into three types and name the types.
CaO, MgO, CO2, SO3, Na2O, ZnO, Al2O3, Fe2O3
Answer:
There are three types of oxides : Basic oxides, Acidic oxides and Amphoteric oxides.

Basic oxides Acidic oxides Amphoteric oxides
CaO CO2 ZnO
MgO SO3 Al2O3
Na2O
Fe2O3

Generally metal oxides are basic in nature.
Exception : Al2O3 and ZnO are amphoteric oxides.
Generally non-metal oxides are acidic in nature.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

4. Explain by drawing a figure of the electronic configuration.

a. Formation of sodium chloride from sodium and chlorine.
Answer:

b. Formation of a magnesium chloride from magnesium and chlorine.
Answer:

5. Show the dissociation of the following compounds on dissolving in water, with the help of chemical equation and write whether the proportion of dissociation is small or large.
Hydrochloric acid, Sodium chloride, Potassium hydroxide, Ammonia, Acetic acid, Magnesium chloride, Copper sulphate.
Answer:
(a) Hydrochloric acid (HCl)

  • \(\mathrm{HCl}_{(g)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
  • Hydrochloric acid is a strong acid, as on dissolving in water, it dissociates almost completely and the resulting aqueous solution contains mainly H ions and the concerned acidic radical.
  • The proportion of dissociation is large.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(b) Sodium chloride (NaC1)

  • \(\mathrm{NaCl}_{(s)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-}\)
  • When an ionic compound begins to dissolve in water, the water molecules push themselves in between the positive and negative ions of the compound and separate them from each other.
  • The proportion of dissociation is large.

(c) Potassium hydroxide (KOH)

  • \(\mathrm{KOH}_{(n)} \frac{\text { Water }}{\text { dissociation }} \mathrm{K}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}\)
  • Potassium hydroxide is a strong base, as on dissolving in water, it dissociates almost completely and the resulting aqueous solution contains mainly OH+ ions and the concerned basic radical.
  • The proportion of dissociation is large.

(d) Ammonia (NH3)
(i) \(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{n}}-\frac{\text { Water }}{\text { uissociation }}>\mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{\text {(aq) }}^{-}\)
(ii) Ammonia dissolves in water to form NH4OH (ammonium hydroxide). NH4OH does not dissociate completely as it is a weak base. The aqueous solution contains a small proportion of OH ions and the concerned basic radical along with a large proportion of undissociated molecules of the base i.e. NH4OH.
(iii) The proportion of dissociation is small.

(e) Acetic acid (CH3COOH)

  • \(\mathrm{CH}_{3} \mathrm{COOH}_{(l)} \frac{\text { Water }}{\text { dissociation }} \mathrm{CH}_{3} \mathrm{COO}_{(\text {aq) }}+\mathrm{H}_{(a z)}+{ }^{+}\)
  • Acetic acid is a weak acid, on dissolving in water it does not dissociate completely, and the resulting aqueous solution contains H+ ion and the concerned acidic radical in small proportion along with large proportion of the undissociated molecules of the acid.
  • The proportion of dissociation is small.

(f) Magnesium chloride (MgCl2)

  • \(\mathrm{MgCl}_{2(\mathrm{~s})} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Mg}^{2+} \text { (aq) }+2 \mathrm{Cl}^{-} \text {(aq) }\)
  • Magnesium chloride dissolves in water and forms magnesium ions and chloride ions. When an ionic compound begins to dissolve in water, the water molecules push themselves in between the ions of the compound and separate them from each other.
  • The proportion of dissociation is large.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(g) Copper sulphate (CuSO4)

  • \(\mathrm{CuSO}_{4(s)} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4}^{2-}\)
  • When Copper sulphate is dissolved in water it forms copper ions and sulphate ions. When an ionic compound begins to dissolve in water, water molecules push themselves in between the ions of the compound and separate them from each other.
  • The proportion of dissociation is large.

6. Write down the concentration of each of the following solutions in g/L and mol/L.

a. 7.3g HCl in 100ml solution
b. 2g NaOH in 50ml solution
c. 3g CH3COOH in 100ml solution
d. 4.9g H2SO4 in 200ml solution
Answer:
To find : The concentration in g/L.
Solution:

7. Obtain a sample of rainwater. Add to it a few drops of universal indicator. Meausre its pH. Describe the nature of the sample of rainwater and explain the effect if it has on the living world.
Answer:

  • pH of rain water is 6.5 that means rain water is slightly acidic.
  • When we add universal indicator to rain water it turns orangish red, indicating pH value is between 0 to 7, which tells us that rain water is acidic in nature.
  • Most of the plants grow best when pH of soil is close to 7. If the soil is too acidic or too basic, it affects plant growth.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

8. Answer the following questions.

a. Classify the acids according to their basicity and give one example of each type.
Answer:

  • Basicity of acids : The number of H+ ions obtainable by the dissociation of one molecule of an acid is called its basicity. The acids are classified as monobasic, dibasic and tribasic acids based on the number of H+ ions present.
  • Examples of monobasic acid : HCl, HNO3, CH3COOH
  • Examples of dibasic acid: H2SO4, H2CO3
  • Examples of tribasic acid: H3BO3, H3PO4

b. What is meant by neutralization? Give two examples from everyday life of the neutralization reaction.
Answer:

  • In neutralization reaction, an acid reacts with a base to form salt and water.
  • In a neutralisation reaction the acid dissociates to form H+ ions and base dissociates to form OH ions.
  • They combine to form H2O molecules which mixes with the solvent.

Examples in daily life:

  • When people suffer from acidity, they take some antacids to neutralise the acid in their stomach.
  • If an ant stings us the pain is due to formic acid. It is neutralised by rubbing moist baking soda which is basic in nature.

c. Explain what is meant by electrolysis of water. Write the electrode reactions and explain them.
Answer:
Electrolysis of water:

  • Electrolysis of water is the decomposition of water into oxygen and hydrogen gas due to an electric current being passed through acidified water.
  • Cathode reaction:
    \(2 \mathrm{H}_{2} \mathrm{O}_{(l)}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{aq}}\)
  • Anode reaction:
    \(2 \mathrm{H}_{2} \mathrm{O}_{(l)} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+4 \mathrm{e}^{-}\)
  • It is found that the volume of gas formed near the cathode is double that of the gas formed near the anode.
  • Hydrogen gas is formed near the cathode and oxygen gas near the anode.
  • From this, it is clear that electrolysis of water has taken place and its constituent element have been released.

9. Give a reason for the following.

a. Hydronium ions are always in the form H3O+.
Answer:

  • Acids in water gives H+ ions. These H+ ions do not exist freely in water.
  • This is because H+ is a single proton, a hydrogen atom has only one proton and one electron.
  • If the electron is removed to make H+, all that is left is an extremely tiny positively charged nucleus.
  • This H ion will immediately combine with the surrounding water (H2O) molecules to form (H3O4) hydronium ion.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

b. Buttermilk spoils if kept in a copper or brass container.
Answer:

  • Buttermilk contains an organic acid called as lactic acid.
  • The lactic acid reacts with copper and brass and forms toxic compounds which are not fit for consumption.
  • They are harmful and may cause food poisoning.
  • So it is not advisable to keep buttermilk in brass or copper containers.

10. Write the chemical equations for the following activities.

(a) NaOH solution was added to HCl solution.
Answer:
When NaOH reacts with HCl, it gives NaCl and water.

(b) Zinc dust was added to dilute H2SO4.
Answer:
When zinc reacts with dilute sulphuric acid, it forms zinc sulphate and hydrogen gas is liberted.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 4

(c) Dilute nitric acid was added to calcium oxide.
Answer:
When dilute nitric acid reacts with calcium oxide, it forms calcium nitrate and water.

(e) Carbon dioxide gas was passed through KOH solution.

(f) Dilute HCl was poured on baking soda.

11. State the differences.

a. Acids and bases
Answer:

Acids Bases
(i) A substance which liberates H+ ions when dissolved in water is an acid
(ii) Blue litmus turns red in an acid.
(iii) The pH of an acid is less than 7.
(iv) Acids are sour to taste
(v) e.g. HCl, H2SO4
A substance which liberates OH ions when dissolved in water is called a base.
Red litmus turns blue in a base
The pH of a base is greater than 7.
Bases are bitter to taste,
e.g. NaOH, KOH.

b. Cation and anion
Answer:

Cations Anions
(i) Cations are ions with a net positive charge. Anions are ions with a net negative charge.
(ii) Cations are generally formed by metals. When metals donate electrons, they have excess of protons, hence they form cations. Anions are generally formed by non-metals. When non-metals accept electrons, they have excess of electrons, hence they form anions.
(iii) Cations are attracted towards the cathode which are negatively charged electrodes. Anions are attracted towards the anode which are positively charged electrodes.
(iv) e.g.: Na+, Ca2+, Mg2+, K+ etc. e.g.: O2 , S2-, Cl, Br etc.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

c. the Negative electrode and the positive electrode.

Answer:

Negative Electrode Positive Electrode
(i) Negatively charged electrodes are called as a cathode. Positively charged electrodes are called as Anode.
(ii) Positively charged cations move towards the cathode or negative electrode. Negatively charged anions move towards the anode or positive electrode.
(iii) Cathode accepts electrons from cations Anode gives electrons to anions

12. Classify aqueous solutions of the following substances according to their pH into three groups : 7, more than 7, less than 7.

Common salt, sodium acetate, hydrochloric acid, carbon dioxide, potassium bromide, calcium hydroxide, ammonium chloride, vinegar, sodium carbonate, ammonia, sulphur dioxide.
Answer:

pH = 7 pH > 7 pH < 7
(a) common salt. sodium acetate. sulphur dioxide.
(b) potassium bromide. sodium carbonate hydrochloric acid.
(c) ammonia. carbon-dioxide.
(d) calcium hydroxide. ammonium chloride.
(e) vinegar

Class 9 Science Chapter 5 Acids, Bases and Salts Intext Questions and Answers

Question 1.
How are the following substances classified into three groups with the help of litmus?
Lemon, tamarind, baking soda, buttermilk, vinegar, orange, milk, lime, tomato, milk of magnesia, water, alum.
Answer:

Basic Substance: Turns Red Litmus Blue Acidic Substance: Turns Blue Litmus Red Neutral Substance: No change in the colour of litmus
Baking Soda Lemon water
Lime Tamarind
Milk of Magnesia Buttermilk
Vinegar
Orange
Milk
Tomato
Alum

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 2.
Fill in the columns in the part of the following table:
Answer:

Question 3.
Complete the following table of the concentration of various aqueous solutions.
Answer:

Question 4.
Complete the following table of neutralization reactions and also write down the names of the acids, bases and salts in it.
Answer:

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 5.
What are the names of the following compounds? NH3, Na2O, CaO.
Answer:
NH3 : Ammonia
Na2O : Sodium oxide
CaO : Calcium oxide

Question 6.
Into which type will you classify the above compounds, acid, base or salt?
Answer:
NH3 : base
Na2O : base
CaO : base

Question 7.
ive examples of monobasic, dibasic and tribasic acids.
Answer:
Monobasic acid examples: HNO3, HCl, CH3COOH
Dibasic acid examples: H2SO4, H2CO3
Tribasic acid examples: H3BO3, H3PO4

Question 8.
Give the three types of bases with their examples.
Answer:
Types of bases:
Monoacidic base examples : NaOH, KOH, NH4OH
Diacidic base examples: Ca(OH)2, Ba(OH)2
Triacidic base examples: A(OH)3, Fe(OH)3

Question 9.
What are the colours of the following natural and synthetic indicators in acidic and basic solutions? Litmus, turmeric, jamun, methyl orange, phenolphthalein?
Answer:

Indicator Colour in Acidic Solution Colour in basic Solution
Litmus Red Blue
Turmeric Yellow Red
Jamun Red Blue-Green
Methyl orange Red Yellow
Phenolphthalein Colourless Pink

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 10.
On mixing the substances as shown here, what are the resulting mixtures formed by mixing the following substances called?
(a) Water and salt
(b) Water and sugar
(c) Water and sand
(d) Water and sawdust
Answer:
(a) Water and salt – Homogeneous mixture
(b) Water and sugar – Homogeneous mixture
(c) Water and sand – Heterogeneous mixture
(d) Water and sawdust – Heterogeneous mixture

Question 11.
What would be the definition of an acid and a base with reference to the neutralization reaction?
Answer:

  • Acid: An acid is a substance which neutralises a base to form salt and water.
  • Base: A base is a substance which neutralises an acid to form salt and water.

Answer the following questions:

Question 1.
Take aqueous solution of sodium chloride, copper sulphate, glucose, urea, dil.H2SO4 and dil.NaOH in a beaker and test the electrical conductivity of the solutions. Answer the given below questions.

(a) With which solutions did the bulb glow?
Answer:
Solutions with which bulb glows: Aqueous solution of NaCl, CuSO4, H2SO4 and NaOH.

(b) Which solutions are electrical conductors?
Answer:
Solutions which are electrical conductors:
NaCl, CuSO4, H2SO4 and NaOH.

Question 2.
During electrolysis of copper sulphate, if electric current is passed through the electrolytic cell for a long time, what change would be seen at the anode?
Answer:

  • When electricity is passed for a long time through copper sulphate solution, the following reaction is seen at the anode: Anode Reaction: Cu(s) → Cu2+(aq) + 2e
  • All the copper atoms will get converted into copper ions and get deposited on the cathode. This process continues till the copper anode exists.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Would water be a good conductor of electricity?
Answer:
Pure water is not a good conductor of electricity.

Answer the following questions:

Question 1.
(i) With which solutions did the bulb glow?
(ii) Which solutions are electrical conductors?

Answer:
(i) The bulbs glows when the wire are immersed in NaCl solution.
(ii)

Solution Results
1g Copper Sulphate Solution Bulb glow
1g Glucose Solution Bulb does not glow
1g Urea Solution Bulb does not glow
5ml dil.H2SO4 Solution Bulb glows
5ml dil. NaOH Solution Bulb glows

Question 2.
What would be the definition of an acid and a base with reference to the neutralization reaction?
Answer:

  • Acid: An acid is a substance which neutralises a base to form salt and water.
  • Base: A base is a substance which neutralises an acid to form salt and water.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Dissolve 2g salt in 500ml pure water. Take 250 ml of this solution in a 500ml capacity beaker. Connect two electrical wires to the positive and negative terminals of a power supply. Remove the insulating cladding from about 2cm portions at the other ends of the wires. These are the two electrodes. Fill two test tubes upto the brim with the prepared dilute salt solution. Invert them on the electrodes without allowing any air to enter. Start the electric current under 6V potential difference from the power supply. Observe what happens in the test tubes after some time.
(a) Did you see the gas bubbles forming near the electrodes in the test tubes?
(b) Are these gases heavier or lighter than water?
(c) Are the volumes of the gases collected over the solution in the two test tubes the same or different?
(d) Test the solutions in the two test tubes with litmus paper, what do you see?
(e) Repeat the activity by using dilute H2SO4 as well as dilute NaOH as the electrolyte.
Answer:
(a) Yes, gas bubbles can be seen forming near the electrodes in the test tube.
(b) These gases are lighter than water.

(c) The volumes of gases collected over the solution in the two test tubes are different.
(d) The solution in the cathode turns red litmus blue while solution in anode turns blue litmus red.
(e) When the above experiment is repeated with dil H2SO4 and dil NaOH, Hydrogen gas is liberated at cathodes and oxygen gas is liberated at anode.

Answer the following questions:

Question 1.
Cut a lemon into two equal parts. Take the juice of each part into two separate beakers. Pour 10 ml of drinking water in one beaker and 20 ml in the second. Stir the solutions in both the beakers and taste them. Is there any difference in the tastes of the solutions in the two beakers? What is that difference?
Answer:

  • In the above activity, the sour taste of the solutions is because of the solute, lemon juice in them.
  • The quantity of the lemon juice is the same in both the solutions. Yet their tastes are different.
  • The solution in the first beaker is more sour than the one in the second.
  • Although the quantity of the solute is the same in both the solutions, the quantity of the solvent is different.
  • Ratio of the quantity of the solute to the quantity of the resulting solution is different. This ratio is larger for the solution in the first beaker and therefore that solution tastes more sour.
  • On the other hand, the proportion of the lemon juice in the total solution in the second beaker is smaller and taste is less sour.
  • The taste of foodstuff depends upon the nature of the taste-giving ingredient and its proportions in the foodstuff.
  • Similarly, all the properties of a solution depend on the nature of the solute and solvent and also on the proportion of the solute in a solution
  • The proportions of a solute in a solution is called the concentration of the solute in the solution.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 2.
Take a big test tube. Choose a rubber stopper in which a gas tube can be fitted. Take a few pieces of magnesium ribbon in the test tube and add some dilute HCl to it. Take a lighted candle near the end of the gas tube and observe. What did you observe?

Answer:

  • Magnesium metal reacts with dilute HCl and an inflammable gas, hydrogen, is formed.
  • During this reaction, the reactive metal displaces hydrogen from the acid to release hydrogen gas.
  • At the same time, the metal is converted into basic radical which combines with the acidic radical from the acid to form the salt.
    \(\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq}) \longrightarrow} \underset{\rightarrow}{\mathrm{MgCl}_{2(\mathrm{aq})}}+\mathrm{H}_{2(\mathrm{~g})}\)

Question 3.
Take some water in a test tube and add a little red oxide (the primer used before painting iron articles) to it. Now add a small quantity of dilute HCl to it, shake the test tube and observe.
(a) Does the red oxide dissolve in water?
(b) What change takes place in the particles of red oxide on adding dilute HCl?
Answer:
(a) The chemical formula of red oxide is Fe2O3. It is insoluble in water.
(b) The water-insoluble red oxide reacts with HCl to produce a water-soluble salt FeCl3.

  • This gives a yellowish colour to the water.
  • The following chemical equation can be written for this chemical change.
    \(\mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{~s})}+6 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{FeCl}_{3(\mathrm{aq})}+3 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 4.
Fit a bent tube in a rubber cork. Take some lime water in a test tube and keep it handy. Take some baking soda in another test tube and add some lemon juice to it. Immediately fit the bent tube over it. Insert its other end into the lime water. Note down your observations of both the test tubes. Repeat the procedure using washing soda, vinegar and dilute HC1 properly. What do you see?
Answer:

  • In this activity, when limewater comes in contact with the gas released in the form of an effervescence, it turns milky.
  • This is a chemical test for carbon dioxide gas.
  • When lime water turns milky, we infer that the effervescence is of CO2.
  • This gas is produced on the reaction of acids with carbonate and bicarbonate salts of metals.
  • A precipitate of CaCO3 is produced on its reaction with the lime water Ca(OH)2.
  • This reaction can be represented by the following chemical equation.
    \(\mathrm{Ca}(\mathrm{OH})_{2(\mathrm{aq})}+\mathrm{CO}_{2(\mathrm{~g})} \longrightarrow \mathrm{CaCO}_{3(\mathrm{~s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
  • Washing soda is sodium carbonate Na2CO3. It will react same as baking soda (NaHCO3).
  • Vinegar and HCl are acids, they do not react chemically with lemon juice.

Class 9 Science Chapter 5 Acids, Bases and Salts Additional Important Questions and Answers

Select the correct option

Question 1.
…………………. acid is present in lemon.
(a) malic acid
(b) tartaric acid
(c) citric acid
(d) butyric acid
Answer:
(c) Citric acid

Question 2.
Tamarind contains …………………. acid.
(a) Lactic acid
(b) tartaric acid
(c) matlic acid
(d) butyric acid
Answer:
(b) tartaric

Question 3.
Butter milk contains …………………. acid.
(a) butyric acid
(b) Lactic acid
(c) matlic acid
(d) citric acid
Answer:
(a) Butyricacid

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 4.
If the basic radical is H+ the type of compound is ………………… .
(a) neutral
(b) base
(c) acid
(d) alkali
Answer:
(c) Acid

Question 5.
The name of compound NH3 is ………………… .
(a) nitric acid
(b) ammonium
(c) nitride
(d) ammonia
Answer:
(b) ammonia

Question 6.
The bases which are soluble in water are called as ………………… .
(a) indicators
(b) acids
(c) alkalis
(d) salts
Answer:
(c) alkalis

Question 7.
H3PO4 is a …………………. acid.
(a) monobasic
(b) tribasic
(c) tetrabasic
(d) dibasic
Answer:
(b) tribasic

Question 8.
According to pH scale pure water has a pH of ………………… .
(a) 6
(b) 7
(c) 5
(d) 8
Answer:
(b) 7

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 9.
With reference to neutralization, metallic oxides are ………………… in nature.
(a) basic
(b) neutral
(c) acidic
(d) saline
Answer:
(a) basic

Question 10.
Molecular formula of blue vitriol is ………………… .
(a) CuSO3 5H2O
(b) CuSO4 4H2O
(c) CUSO3 4H2O
(d) CUSO4 5H2O
Answer:
(d) CUSO4 5H2O

Question 11.
Molecular formula of crystalline alum is ………………… .
(a) KSO4, AISO4, 24H2O
(b) K2SO4, ALSO4, 24H2O
(c) K2SO4, AL2(SO4)3, 24H2O
(d) KSO4, Al2(SO4)3, 24H2O
Answer:
(c) K2SO4, AL2(SO4)3, 24H2O

Question 12.
Molecular formula for sodium oxide is ………………… .
(a) Na2O
(b) NaO2
(c) NaO
(d) Na2O2
Answer:
(a) Na2O

Question 13.
H2CO3 is …………………. acid.
(a) monobasic
(b) dibasic
(c) tribasic
(d) tetrabasic
Answer:
(b) dibasic

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 14.
Molecular formula of Red oxide is ………………… .
(a) Fe2O3
(b) CuO
(c) Fe3O4
(d) Na2O
Answer:
(a) Fe2O3

Question 15.
The positive terminal electrode is called as ………………… .
(a) anode
(b) cathode
(c) anion
(d) cation
Answer:
(a) anode

Question 16.
…………………. produced in stomach helps in digestion.
(a) Hydrochloric acid
(b) Oxalic acid
(c) Sulphuric acid
(d) Nitric acid?
Answer:
(a) Hydrochloric acid

Question 17.
The solution turns blue litmus red, its pH is likely to be ………………… .
(a) 7
(b) 4
(c) 14
(d) 9
Answer:
(b) 4

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 18.
An ionic compound NaCl is formed by ………………… .
(a) Na+ and Cl
(b) Na+ and Cl+
(c) Na and Cl
(d) Na and Cl+
Answer:
(a) Na+ and Cl

Question 19.
pH of strong acid is ………………… .
(a) 0
(b) 7
(c) 8
(d) 14
Answer:
(a) 0

Question 20.
HCl + NaOH → NaCl + H2O is a …………………. reaction.
(a) neutralization
(b) crystallisation
(c) electrolysis
(d) dissociation
Answer:
(a) neutralization

Question 21.
Adding water to acid is an …………………. reaction.
(a) endothermic
(b) exothermic
(c) neutralisation
(d) crystallisation
Answer:
(b) exothermic

Find the odd one out:

(a) Rose Petal, Turmeric, Phenolphthalein, Indigo.
Answer:
Phenolphthalein is odd one out as rest are natural indicators while phenolphthalein is a synthetic indicator.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

(b) Lime water, Vinegar, Acetic acid, Tartaric acid.
Answer:
Lime water is odd one out as this is basic in nature while rest are acidic.

(c) NaHCO3, HCl, H2SO4, HNO3
Answer:
NaHCO3 is odd one out as it is base while rest are acids.

(d) Oxalic acid, Nitric acid, Citric acid, acetic acid.
Answer:
Nitric acid is odd one out as others are weak acids while Nitric acid is a strong acid.

(e) Crystalline, Liquid, Gases, Solid.
Answer:
Crystalline is odd one out as this is nature of a compound while others are physical states of compounds.

(f) Ca(OH)2, Mg (OH)2, NaOH, NH4OH.
Answer:
NaOH is odd one out even though all are bases but NaOH is highly soluble in water compared to others.

Name the following:

Question 1.
Name the three types of ionic compounds.
Answer:
The three types of ionic compounds are acids, bases and salts.

Question 2.
Name the two constituents of molecule of an ionic compound.
Answer:

  • Cation (positive ion/ basic radical)
  • Anion (negative ion/acidic radical).

Question 3.
Name any three acids with their molecular formula.
Answer:

  • Hydrochloric acid – HC1
  • Sulphuric acid – H2SO4
  • Nitric acid – HNO3

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 4.
Name any three bases with their molecular formula.
Answer:

  • Sodium hydroxide – NaOH
  • Potassium hydroxide – KOH
  • Calcium hydroxide – Ca(OH)2

Question 5.
Name any three salts with their molecular formula.
Answer:

  • Sodium chloride – NaCl
  • Potassium sulphate – K2 SO4
  • Calcium chloride – CaCl2

Question 6.
Name any two strong acids.
Answer:

  • Hydrochloric acid – HC1
  • Sulphuric acid – H2 SO4

Question 7.
Name any two weak acids.
Answer:

  • Acetic acid – CH3 COOH
  • Carbonic acid – H2 CO3

Question 8.
Name any two strong bases.
Answer:

  • Sodium hydroxide – NaOH
  • Potassium hydroxide – KOH

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 9.
Name a weak base.
Answer:
Ammonium hydroxide – NH4OH

Question 10.
Name any two alkalis.
Answer:

  • Sodium hydroxide – NaOH
  • Potassium hydroxide – KOH

Question 11.
Name any two acids with their basicity 1 (monobasic)
Answer:

  • Hydrochloric acid – HC1
  • Nitric acid – HNO3

Question 12.
Name any two acids with their basicity 2 (dibasic)
Answer:

  • Sulphuric acid – H2 SO4
  • Carbonic acid – H2CO3

Question 13.
Name any two acids with their basicity 3 (tribasic)
Answer:

  • Boric acid – H3BO3
  • Phosphoric acid – H3PO4

Question 14.
Name any two bases with their acidity 1 (monoacidic)
Answer:

  • Sodium hydroxide – NaOH
  • Potassium hydroxide – KOH

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 15.
Name any two bases with their acidity 2 (diacidic)
Answer:

  • Calcium hydroxide – Ca(OH)2
  • Barium hydroxide – Ba(OH)2

Question 16.
Name any two bases with their acidity 3 (triacidic)
Answer:

  • Aluminium hydroxide – Al(OH)3
  • Ferric hydroxide – Fe(OH)3

Question 17.
Name the two units to express the concentration of the solution.
Answer:

  • mass of solute in grams dissolved in one litre of the solution grams per litre, (g/L).
  • Number of moles of the solute dissolved in one litre of the solution. Molarity (M) of the solution.

Match the columns:

Question 1.

Column ‘A’ Column ‘B’
(1) HNO3
(2) H3PO4
(3) CH3COOH
(4) H2CO3
(a) Acetic acid
(b) Carbonic acid
(c) Phosphoric acid
(d) Nitric acid

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 2.

Column ‘A’ Column ‘B’
(1) NH4OH
(2) Ca(OH)2
(3) Al(OH)3
(4) Ba(OH)2
(a) Aluminium Hydroxide
(b) Barium Hydroxide
(c) Calcium Hydroxide
(d) Ammonium Hydroxide

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Question 3.

Column ‘A’ Solution Column ‘B’ pH
(1) Milk (a) 0
(2) Milk of Magnesia (b) 14
(3) 1 M HCl (c) 10.5
(4) 1 M NaOH (d) 6.5

Answer:
(1-d),
(2- c),
(3 – a),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 4.

Column ‘A’ Column’B’
(1) Urine
(2) Apples
(3) Orange
(4) Butter
(a) Butyric acid
(b) Uric acid
(c) Malic acid
(d) Citric acid

Answer:
(1 – b),
(2 – c),
(3 – d),
(4 – a)

Question 5.

Column ‘A’ Column ‘B’
(1) Crystalline blue vitriol
(2) Crystalline green vitriol
(3) Crystalline
(4) washing soda Crystalline alum
(a) FeS04-7H20
(b) K2S04-A12(S04)3.24H20
(c) CuS04-5H20
(d) Na2CO310H2O

Answer:
(1-c),
(2- a),
(3 – d),
(4 – b)

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

State whether the following statements are true or false and if false. Correct the false statement:

(1) The separation of H and CI in HCI is in absence of water.
(2) \(\mathrm{NaOH}_{(\mathrm{s})} \frac{\text { Water }}{\text { dissociation }}>\mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
(3) H2SO4 is a strong acid.
(4) NaC1 is an ionic compound.
(5) Turmeric is synthetic indicator.
(6) Metal + Dilute acid forms salt and water.
(7) Copper oxide is called red primer.
(8) Oxide of non-metal + Acid → Salt + Water.
(9) Zinc oxide reacts with sodium hydroxide to form sodium zincate.
(10) Al2O3 is an amphoteric oxide.
(11) Blue vitriol ZnSO4. 7H2O.
(12) Molecular formula for crystalline Ferrous sulphate is Fe5O4. 5H2O.
(13) NaCl in water does not conduct electricity.
(14) Phenolphthalein is colourless in base.
Answer:
(1) False. The separation of H+ and Cl in HC1 is in presence of water.
(2) True
(3) True
(4) True
(5) False. Turmeric is a natural indicator.
(6) False. Metal + Dilute acid forms salt and hydrogen gas.
(7) False. Iron oxide is called red primer.
(8) False. Oxide of izan-metal + Base → Salt + Wafer.
(9) True
(10) True
(11) False. Blue Vitriol is CuSO4. 5H2O.
(12) False. Molecular formula for crystalline Ferrous sulphate is FeSO4. 7H2O.
(13) False. NaCl in water conducts electricity.
(14) False, Phenolphthalein is colourless in acid and pink in base.

Define the following:

Question 1.
Acid
Answer:
An acid is a substance which on dissolving in water gives rise to H ion as the only cation. For example, HC1, H2 SO4, H2CO3.Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 1

Question 2.
Base
Answer:
A base is a substance which on dissolving in water gives rise to the OW ion as the only anion. For, NaOH, Ca(OH)2Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 2

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Strong Acid
Answer:
On dissolving in water, a strong acid dissociates almost completely, and the resulting aqueous solution contains mainly H ions and the concerned acidic radical. e.g. HC1, HBr, HNO3, H2SO4.

Question 4.
Weak Acid
Answer:
On dissolving in water a weak acid does not dissociate completely, and the resulting aqueous solution contains H+ ion and the concerned acidic radical in small proportion along with large proportion of the undissociated molecules of the acid, e.g., H2CO3 (Carbonic acid), CH3COOH (Acetic acid)

Question 5.
Strong Base
Answer:
On dissolving in water, a strong base dissociates almost completely and the resulting aqueous solution contains mainly OH ions and the concerned basic radicals, e.g. NaOH, KOH, Ca(OH)2, Na2O.

Question 6.
Weak Base
Answer:
On dissolving in water, a weak base does not dissociate completely and the resulting aqueous solution contains a small proportion of OH ions and the concerned basic radical along with a large proportion of undissociated molecules of the base. e.g. NH4 OH.

Question 7.
Alkali
Answer:
The bases which are highly soluble in water are called alkali, e.g. NaOH, KOH, NH3. Here, NaOH and KOH are strong alkalis while NH3 is a weak alkali.

Question 8.
Basicity of acids
Answer:
The number of H+ ions obtainable by the dissociation of one molecule of an acid is called its basicity.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 9.
Acidity of bases
Answer:
The number of OH ions obtainable by the dissociation of one molecule of a base is called its acidity.

Question 10.
Concentration of solute in the solution.
Answer:
The proportion of a solute in a solution is called the concentration of the solute in the solution.

Question 11.
Concentrated solution.
Answer:
When the concentration of a solute in its solution is high, it is a concentrated solution.

Question 12.
Dilute solution
Answer:
When the concentration of a solute in its solution is low, it is a dilute solution.

Question 13.
Neutralization
Answer:
A reaction in which an acid reacts with a base to form salt and water is called a neutalization reaction.
NaOH + HC1 → NaCl + H2O
Base + Acid → Salt + Water

Question 14.
Cathode
Answer:
The electrode connected to the negative terminal of a battery by means of a conducting wire is called a cathode.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 15.
Anode
Answer:
The electrode connected to the positive terminal of a battery by means of a conducting wire is called an anode.

Question 16.
Cations
Answer:
Cations are positively charged ions which are attracted towards negative electrode (cathode) when electricity is passed into a solution of an ionic compound.

Question 17.
Anions
Answer:
Anions are negatively charged ions which are attracted towards the positive electrode (anode) when electricity is passed into a solution of an ionic compound.

Question 18.
Electrolytic cell
Answer:
An assembly that consists of a container with an electrolyte and the electrodes dipped in it, is called an electrolytic cell.

Question 19.
Molarity of a solution
Answer:
The number of moles of the solute dissolved in one litre of the solution is called the molarity of that solution. The molarity of a solute is indicated by writing its molecular formula inside a square bracket for example [NaCl] = 1

Question 20.
Acid – base indicators
Answer:
Some natural and synthetic dyes show two different colours in acidic and basic solution, and such dyes are acid base indicators.

Explain the following chemical reactions with the help of balanced equations:

Question 1.
Magnesium reacts with dilute hydrochloric acid.
Answer:
When magnesium reacts with dilute hydrochloric acid, it forms magnesium chloride and hydrogen gas is liberated.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 3

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 2.
When copper reacts with nitric acid.
Answer:
When copper reacts with nitric acid, it forms copper nitrate and hydrogen gas is liberated.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 5

Question 3.
When ferric oxide reacts with diluted hydrochloric acid.
Answer:
When Ferric oxide reacts with diluted hydrochloric acid, it forms ferric chloride and water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 9

Question 4.
When calcium oxide reacts with dilute hydrochloric acid.
Answer:
When calcium oxide reacts with dilute hydrochloric acid, it forms calcium chloride and water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 7

Question 5.
When Magnesium oxide reacts with dilute hydrochloric acid.
Answer:
When magnesium oxide reacts with dilute hydrochloric acid, it forms magnesium chloride and water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 8

Question 6.
When zinc oxide reacts with dilute hydrochloric acid.
Answer:
When zinc oxide reacts with dilute hydrochloric acid, it forms zinc chloride and water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 57

Question 7.
When aluminium oxide reacts with hydrogen fluoride
Answer:
When aluminium oxide reacts with hydrogen fluoride, it forms Aluminium fluoride arid water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 10

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 8.
When carbon dioxide reacts with sodium hydroxide.
Answer:
When carbon dioxide reacts with Sodium hydroxide, it forms Sodium carbonate and water.

Question 9.
When carbon dioxide reacts with potassium hydroxide.
Answer:
When carbon dioxide reacts with potassium hydroxide, it forms potassium carbonate and water.

Question 10.
When sulphur trioxide reacts with sodium hydroxide.
Answer:
When sulphur trioxide reacts with sodium hydroxide, it forms sodium sulphate and water.

Question 11.
When calcium hydroxide reacts with carbon dioxide.
Answer:
When calcium hydroxide reacts with carbon dioxide, it forms calcium carbonate and water.

Question 12.
When sodium carbonate reacts with hydrochloric acid.
Answer:
When sodium carbonate reacts with hydrochloric acid, it forms sodium chloride, carbon dioxide and water.

Question 13.
When sodium carbonate reacts with sulphuric acid.
Answer:
When sodium carbonate reacts with sulphuric acid, it forms sodium sulphate, carbon dioxide and water.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 14.
When calcium carbonate reacts with nitric acid.
Answer:
When calcium carbonate reacts with nitric acid, it forms calcium nitrate, carbon dioxide and water.

Question 15.
When potassium carbonate reacts with sulphuric acid.
Answer:
When potassium carbonate reacts with sulphuric acid, it forms potassium sulphate, carbon dioxide and water.

Question 16.
When sodium bicarbonate reacts with hydrochloric acid. OR Dilute HC1 was poured on baking soda
Answer:
When sodium bicarbonate reacts with hydrochloric acid, it forms sodium chloride, carbon dioxide and water.

Question 17.
When potassium bicarbonate reacts with nitric acid.
Answer:
When potassium bicarbonate reacts with nitric acid, it forms potassium nitrate, carbon dioxide and water.

Question 18.
When sodium bicarbonate reacts with acetic acid.
Answer:
When sodium bicarbonate reacts with acetic acid, it forms sodium acetate, carbon dioxide and water.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 21

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 19.
When copper sulphate is heated.
Answer:
When copper sulphate is heated it loses it’s water of crystallization to form white anhydrous copper sulphate.
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 58

Question 20.
When Ferrous sulphate is heated.
Answer:
When ferrous sulphate is heated it loses its water of crystallization to form white anhydrous ferrous sulphate.

Q.2. (B)-3. Complete the following table.

Question 1.
Complete the following table.
Answer:

Complete the following reactions.

Question 1.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 31

Question 2.
Metal + Dilute acid → Salt + Hydrogen
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 32

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Metal oxide + Dilute acid → SaIt + Water
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 33
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 34

Question 4.
Oxide of non-metal + base → salt + water
Answer:

Question 5.
Carbonate salt of metal + dilute acid → Another salt of metal + Carbon dioxide + water
Answer:

Question 6.
Bicarbonate salt of metal + dilute acid → Another salt of metal + carbon dioxide + water
Answer:

Give scientific reasons:

Question 1.
Ionic compound NaCl has very high stability.
Answer:

  • The outermost shell of both Na+ and Cl ions is a complete octet.
  • An electronic configuration with a complete octet indicates a stable state.
  • A molecule of NaCl has Na+ and Cl ions. An ionic bond is formed between these ions.
  • The force of attraction between them is very strong as it is formed between the oppositely charged Na+ and Cl ions.
  • Therefore NaCl, an ionic compound has very high stability.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 2.
Ionic compound dissociates while forming an aqueous solution.
Answer:

  • On dissolving in water, an ionic compound forms an aqueous solution.
  • In the solid state, the oppositely charged ions in the ionic compound are sitting side by side.
  • When an ionic compound being to dissolve in water, the water molecules push themselves in between the ions of the compound and it separates them from each other, that is to say, an ionic compound dissociates while forming an aqueous solution.

Question 3.
Blue coloured copper sulphate crystals become colourless on heating.
Answer:

  • Copper sulphate crystals are blue in colour and crystalline in form due to presence of water of crystallisation.
  • Each molecule of crystalline copper sulphate contains five molecules of water of crystallisation (CuSO4.5H2O).
  • On heating, the copper sulphate crystals lose the water of crystallisation and turns into white amorphous powder called as anhydrous copper sulphate.
  • Therefore, blue coloured copper sulphate crystals become colourless on heating.

Question 4.
During electrolysis of water, a few drops of sulphuric acid are added to it.
Answer:

  • Pure water is a covalent compound and hence it is a non-electrolyte and does not conduct electricity.
  • When a few drops of sulphuric acid (H2SO4) are added to water.
  • Being a strong acid it dissociates almost completely in its solution forming H+ cations and \(\mathrm{SO}_{4}^{2-}\) anions.
  • The movement of these ions in the solution towards the respective electrodes amount to the conduction of electricity through the solution.
  • Therefore, during electrolysis of water, a few drops of sulphuric acid are added to it.

Question 5.
Glucose is a non-electrolyte.
Answer:

  • Glucose is a covalent compound
  • It does not form any ions in its aqueous solution.
  • Due to this aqueous solution of glucose does not conduct electronic current.
  • Hence, glucose is a non-electrolyte.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 6.
Pure water is a poor conductor of electricity but aqueous solution of sodium chloride conducts electricity.
Answer:

  • Pure water does not contain any free ions.
  • Sodium chloride (NaCl) is an ionic compound made up of sodium cation (Na+) and chloride anion (Cl)
  • When sodium chloride is dissolved in water , these ions dissociates in its aqueous solution. ‘
  • These ions are free to move in the solution and conduct electricity.
  • Therefore, pure water is a poor conductor of electricity but aqueous solution of sodium Chloride conducts electricity.

Question 7.
When carbon dioxide gas is passed through freshly prepared lime water, the limewater turns milky.
Answer:

  • Limewater traditionally means a weak solution of the alkali calcium hydroxide Ca(OH)2.
  • When CO2 is passed through limewater, it reacts with calcium hydroxide to form insoluble particulates (precipitate of calcium carbonate (CaCO3).
  • Calcium carbonate is weak basic salt and this gives a milky white precipitate.
    Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 38
  • Hence, lime water turns milky when CO2 gas is passed through it.

Q.3.1. Answer the following:

Question 1.
Write down chemical equations for
(a) Zinc oxide reacts with sodium hydroxide
(b) Aluminium oxide reacts with sodium hydroxide.
Answer:
(a) When zinc oxide reacts with sodium hydroxide, it forms sodium zincate and water
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 59

(b) When Aluminium oxide reacts with sodium hydroxide, it forms sodium aluminate and water.

Question 2.
Can we call Al2O3 and ZnO acidic oxides on the basis of above reactions.
Answer:

  • No, because they also react with acids to form their respective salts and water.
  • So, they show the properties of basic oxides also.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Define ‘amphoteric oxides’ and give two examples.
Answer:

  • Amphoteric oxides are those oxides which react with both adds as well as bases to form their respective salts and water.
  • Amphoteric oxides show the properties of both acidic oxides as well as basic oxides. ZnO and Al2 O3 are amphoteric oxides.?

Question 4.
Take a solution of 1g copper sulphate in 50ml water in a 100 ml capacity beaker. Use a thick plate of copper as anode and a carbon rod as cathode. Arrange the apparatus as shown in the figure and pass an electric current through the circuit for some time. Do you see any changes?
Answer:

  • A thin film of copper metal is deposited on cathode which is immersed in solution. There is no change in colour of solution
  • The electrons from cathode combines with Cu2+ ion from the solution forming Cu atoms which were then deposited on the cathode.

Q.3.5. Answer in brief:

Question 1.
What are acids, bases and salts?
Answer:

  • Compounds having H+ as the basic radical in their molecules are called Acids.
  • Compounds having OH as the acidic radical in their molecule are called Bases.
  • Ionic compounds which have a basic radical other than H+ and an acidic radical other than OH are called salts.

Question 2.
What is an ionic bond?
Answer:

  • The molecule of an ionic compound has two constituents namely cation (positive ion / basic radical) and anion (negative ion / acidic radical).
  • There is a force of attraction between these ions as they are oppositely charged, and that is called the ionic bond.
  • The force of attraction between one positive charge on a cation and one negative charge on an anion makes one ionic bond.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 3.
Give examples to show that proportions of H+ and OH ions in aqueous solution determines the properties of those solutions.
Answer:
The examples to show that proportions of H+ and OH ions in aqueous solution determines the properties of those solutions are :

  • The proportions of H+ and OH ions divides soil into the acidic, neutral and basic, types of soil.
  • It is necessary for blood, cell sap etc to have H+ and OH ions in certain definite proportions for their proper functioning.
  • Fermentation carried out with the help of micro-organisms, other biochemical processes and also many chemical processes require the proportion of H+ and OH ions to be maintained within certain limits.

Question 4.
What is pH scale?
Answer:

  1. In 1909, the Danish scientist Sorensen introduced a convenient new scale of expressing H+ ion concentration which is found to be useful in chemical and biochemical processes.
  2. It is the pH scale (pH: power of hydrogen). The pH scale extends from 0 to 14. According to this scale pure water has a pH of 7. pH 7 indicates a neutral solution. This pH is the midpoint of the scale.
  3. The pH of an acidic solution is less than 7 and?

Question 5.
Give the pH of following solutions.
Answer:

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 6.
What is universal indicator? Which is the most accurate method of measuring the pH of a solution?
Answer:

  • In the pH scale, the pH of solutions varies from 0 to 14 in accordance with the strength of the acid or base.
  • To show these variations in pH, a universal indicator is used.
  • A universal indicator shows different colours at different values of pH. A universal indicator is made by mixing several synthetic indicators in specific proportions.
  • The pH of a solution can be determined by means of a universal indicator solution or the pH paper made from it.
  • However, the most accurate method of measuring the pH of a solution is to use an electrical instrument called pH meter.
  • In this method, pH is measured by dipping electrodes into the solution.

Answer in detail:

Question 1.
Explain the Arrhenius theory of acids and bases.
Answer:
The Swedish scientist Arrhenius put forth a theory of acids and bases in the year 1887. This theory gives definitions of acids and bases as follows:
Acid : An acid is a substance which on dissolving in water gives rise to H ion as the only cation. For example, HCl, H2SO4, H2CO3

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 61

Base: Abase is a substance which on dissolving in water gives rise to the OH ion as the only anion, For example, NaOH, Ca(OH)2
Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts 51

Question 2.
Write a short note on Neutralization.
Answer:

  • Take 10 ml of dilute HCl in a beaker, go on adding dilute NaOH drop by drop and record the pH.
  • Stop adding the NaOH when the green colour appears on the pH paper, that is when the pH of solution becomes 7.
  • Both HCl and NaOH dissociate in their aqueous solutions. Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts
  • Addition of NaOH to HCl solution is like adding a large concentration of OH ions to a large concentration of H+ ions.
  • However water dissociates into H+ and OH ions to a very small extent.
  • Therefore on mixing the excess OH ions combines with excess H+ ions and forms H2O molecules which mix with solvent water.
  • This change can be represented by the ionic equation shown as follows. H+ + Cl + Na+ + OH → Na+ + Cl + H2O
  • It can be observed that Na+ and CT ions are there on both the sides. Therefore the net ionic reaction is H+ + OH → H2O
  • As NaOH solution is added drop by drop to the HCl solution, the concentration of ff goes on decreasing due to combination with added OH ions, and that is how the pH goes on increasing.
  • When enough NaOH is added to HCl, the resulting aqueous solution contains only Na+ and Cl ions, that is, NaCl, a salt, and the solvent water. The only source of H+ and OH ions in this solution is a dissociation of water.
  • Therefore, this reaction is called the Neutralization reaction. The Neutralization reaction is also represented by the following simple equation.

Question 3.
Explain the water of Crystallization.
Answer:

  • Take some crystals of blue vitriol (CuSO4. 5H2O) in a test tube. Heat the test tube on a low flame of a burner.
  • It was observed that on heating, the crystalline structure of blue vitriol broke down to form a colourless powder and water came out.
  • This water was part of the crystal structure of blue vitriol. It is called water of crystallization.
  • On adding water to the white powder a solution

    was formed which had the same colour as the solution in the first test tube.
  • From this we come to know that no chemical change has occurred in the crystals of blue vitriol due to heating.
  • Losing water on heating blue vitriol, breaking down of the crystal structure, losing blue colour and regaining blue colour on adding water are all physical changes.
  • Similarly, ferrous Sulphate crystals also contain 7 molecules of water of crystallization which are lost on heating.
  • The reaction is represented as
  • Ionic compounds are crystalline in nature. These crystals are formed as a result of definite arrangement of ions.
  • In the crystals of some compounds, water molecules are also included in this arrangement.
  • That is the water of crystallization. The water of crystallization is present in a definite proportion of the chemical formula of the compound.

Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

Question 4.
Explain the conduction of electricity through solutions of ionic compounds
Answer:

  1. Electrons conduct electricity through electrical wires, and ions conduct electricity through a liquid or a solution.
  2. Electrons leave the battery at the negative terminal, complete the electric circuit and enter the battery at the positive terminal.
  3. When there is a liquid or a solution in the circuit, two rods, wires or plates are immersed in it. These are called electrodes.
  4. Electrodes are made of conducting solid. The electrode connected to negative terminal of a battery by means of a conducting wire is called a cathode and the electrode connected to the positive terminal of a battery is called anode.
  5. We have seen that salts, strong acids and strong bases dissociates almost completely in their aqueous solutions.
  6. Therefore the aqueous solutions of all these three contain large number of cations and anions.
  7. A characteristic of liquid state is the mobility of its particles. Due to its mobility the positive charged ions of the solution are attracted to the negative electrode or cathode.
  8. On the other hand, the negative charged ions of the solution are attracted to the positive electrode or anode.
  9. The movement of ions in the solution towards the respective electrodes accounts to the conduction of electricity through the solution.
  10. From this we can understand that those liquids or solutions which contain a large number of dissociated ions conduct electricity. Maharashtra Board Class 9 Science Solutions Chapter 5 Acids, Bases and Salts

 

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 6 Classification of Plants

Class 9 Science Chapter 6 Classification of Plants Textbook Questions and Answers

1. Match the proper terms from columns A and C with the description in column B.
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 10

2. Complete the sentences by filling in the blanks and explain those statements.
(angiosperms, gymnosperms, spore, Bryophyta, thallophyta, zygote)
a. ……………….. plants have soft and fiber-like body.

b. ……………….. is called the ‘amphibian’ of the plant kingdom.
Answer:
Bryophyta plant is called the ‘amphibian’ of the plant kingdom

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

c. In pteridophytes, asexual reproduction occurs by ……………….. formation and sexual reproduction occurs by ………………..formation.
Answer:
Spore, zygote: Pteridophyta plants show alteration of generation. One generation reproduces by spore-formation and the next generation reproduces sexually by zygote formation.

d. Male and female flowers of ……………….. are borne on different sporophylls of the same plant.
Answer:
Gymnosperms bear their male and female flowers on different sporophylls of the same plant

3. Answer the following questions in your own words.

a. Write the characteristics of subkingdom Phanerogams.
Answer:

  • Plants which have special structures for reproduction and produce seeds are called Phanerogams.
  • In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food.
  • During the germination of the seed, the stored food is used for the initial growth of the embryo.
  • Depending upon whether seeds are enclosed in a fruit or not phanerogams are classified into gymnosperms and angiosperms.

b. Distinguish between monocots and dicots.
Answer:

Dicots Monocots
Seed Two cotyledons Single cotyledon
Root Well developed, primary root (Taproot) Fibrous roots
Stem Strong, hard. e.g. Banyan tree Hollow, e.g. Bamboo
False, e.g. Banana
Disc-like, e.g. Onion.
Leaf Reticulate venation Parallel venation
Flower Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous) Flowers with 3 parts or in multiples of three (trimerous).

c. Write a paragraph in your own words about the ornamental plants called ferns.
Answer:

  • Ferns belong to the group of plants called Pteridophyta.
  • They have well-developed roots, stem and leaves but do not bear flowers and fruits.
  • They have separate tissues for the conduction of food and water.
  • They reproduce with the help of spores formed along the back or posterior surface of their leaves.
  • They reproduce asexually by spore formation and sexually by zygote formation.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

d. Sketch, label and describe the Spirogyra.

e. Write the characteristics of the plants belonging to division Bryophyta.
Answer:

  • Bryophyta group of plants are called the amphibians of the plant kingdom because they grow in moist soil but need water for reproduction.
  • These plants are thalloid, multicellular and autotrophic.
  • They reproduce by spore-formation.
  • Their plant body structure is flat, ribbon-like, long, without true roots, stem and leaves.
  • Instead, they have stem-like or leaf-like parts and root-like rhizoids.
  • They do not have specific tissues for the conduction of food and water.
  • Examples: Moss (Funaria), Anthoceros, Riccia etc.?

4. Sketch and label the figures of the following plants and explain them into brief.
Marchantia, Funaria, Fern, Spirogyra.

Question 1.
Spirogyra.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 5

  • Spirogyra belongs to the division thallophyta. They are called as algae.
  • It grows mainly in water.
  • It does not have specific parts like root-stem- leaves-flowers but are autotrophic due to the presence of chlorophyll.
  • The plant body of Spirogyra is soft and fibre-like.
  • It has spirally arranged chloroplasts in its cell.

Question 2.
Funaria and Marchantia (Bryophyta)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 6

  • These plants are called ‘amphibians’ of the plant kingdom because they grow mostly in soil and need water for reproduction.
  • They do not have specific tissues for the conduction of food and water.
  • The plant body is fiat, ribbon-like long, without true roots, stem and leaves
  • Instead, they have stem-like or leaf-like parts and root like rhizoids.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Fern (Pteridophyta):
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 7

  • They have well-developed roots, stem and leaves for the conduction of food and water.
  • They do not bear flowers and fruits.
  • They reproduce with the help of spores present along the back or posterior surface of the leaves.

5. Collect a monocot and dicot plant available in your area. Observe the plants carefully and describe them in scientific language.

6. Which criteria are used for the classification of plants? Explain with reasons.
Answer:
Criteria for classification of plants:

  • If plants do not bear flowers, fruits and seeds, they are non-seed bearing plants. If they bear flowers, fruit and seeds, they are seed-bearing plants.
  • Presence or absence of conducting tissues- Plants such as pteridophytes, gymnosperms and angiosperms which possess conducting tissues are included in vascular plants whereas thallophytes and bryophytes which do not possess conducting tissues are included under non-vascular plants.
  • Depending upon whether the seeds are enclosed in fruit or not, plants are classified as gymnosperms (naked-seeds) and angiosperms (seeds covered by fruit) Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
  • Depending upon the number of cotyledons in seeds, plants are classified into dicotyledons and monocotyledons

Class 9 Science Chapter 6 Classification of Plants Intext Questions and Answers

Can you recall?

Question 1.
How have living organisms been classified?
Answer:
(i) Organisms have been classified based on the following:

  • Cell structure
  • Body Organisation
  • Mode of nutrition
  • Reproduction

(ii) Organisms are also classified at kingdom level and groups and subgroups.

Activity-based questions

Question 1.
You may have seen a lush green soft carpet on old walls, bricks and rocks in the rainy season. Scrape it gently with a small ruler, observe it under a magnifying lens and discuss.
Answer:

  • It shows considerable tissue complexity and is differentiated into two main parts: a root and a shoot.
  • They have a variety of specialized tissues within these two regions of the body.
  • Same kind of cells are seen throughout the whole body except reproductive cells.

Question 2.
You may have seen ferns among the ornamental plants in a garden. Take a leaf of a fully grown fern and observe it carefully.
Answer:

  • New leaves typically expand by the unrolling in a tight spiral manner.
  • The anatomy of fern leaves can either be simple or highly divided.
  • They show the presence of spores formed along the back or posterior surface of their leaves.

Question 3.
Observe all garden plants like Cycas, Christmas tree, Hibiscus, Lily, etc. and compare them. Note the similarities and differences among them. Which differences did you notice in gymnosperms and angiosperms?
Answer:
Cycas and Christmas tree are gymnosperms, whereas Hibiscus and lily are angiosperms.

  1. Similarities: These plants have special structures for reproduction and produce seeds. During the germination of the seed, the stored food is used for the initial growth of the embryo.
  2. Differences: In gymnosperms, reproductive organs have cones whereas in angiosperms reproductive organs have flowers.
  3. In gymnosperms, seeds are without natural coverings whereas in angiosperms seeds are enclosed in natural coverings called fruits.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
Soak the seeds of corns, beans, groundnut, tamarind, mango, wheat, etc. in water for 8 to 10 hrs. After they are soaked, check each seed to see whether it divides into two equal halves or not and categorize them accordingly.
Answer:
Monocots: com, wheat (it cannot be divided into equal halves)
Dicots: beans, groundnut, tamarind and mango (it can be divided into two equal halves)

Class 9 Science Chapter 6 Classification of Plants Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
The five-kingdom classification was proposed b7
(a) Robert Whittaker
(b) Robert Hooke
(c) Eichler
(d) Louis Pasteur
Answer:
(a) Robert Whittaker

Question 2.
In 1883, classified plants into two sub-kingdoms.
(a) Robert Whittaker
(b) Alexander Fleming
(c) Eichler
(d) Robert Hooke
Answer:
(c) Eichler

Question 3.
Ulothrix, ulva, sargassum belong to
(a) Bryophyta
(b) Thallophyta
(c) Pteridophyta
(d) Gymnosperms
Answer:
(b) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 4.
is a bryophyte.
(a) Ulva
(b) Nephrolepis
(c) Funaria
(d) Equisetum
Answer:
(c) Funaria

Question 5.
In the seeds are naked.
(a) Pteridophyta
(b) Angiosperms
(c) Gymnosperms
(d) Bryophyta
Answer:
(c) Gymnosperms

Question 6.
In the flowers are reproductive organs.
(a) Angiosperms
(b) Gymnosperms
(c) Pteridophyta
(d) Bryophyta
Answer:
(a) Angiosperms

Question 7.
In the flowers are tetramerous or pentamerous.
(a) Monocotyledons
(b) Dicotyledons
(c) Gymnosperms
(d) Pteridophyta
Answer:
(b) Dicotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 8.
In monocotyledonous plants, the stem is
(a) hollow
(b) false
(c) disc-like
(d) all of these.
Answer:
(d) all of these

Question 9.
Lycopodium belongs to
(a) Thallophyta
(b) Bryophyta
(c) Gymnosperms
(d) Pteridophyta
Answer:
(d) Pteridophyta

Question 10.
Leaves of show reticulate venation.
(a) Bamboo
(b) Banana
(c) Onion
(d) Banyan
Answer:
(d) Banyan

Question 11.
Various types of fungi like yeasts and moulds are included in the group
(a) Thallophyta
(b) Halophyte
(c) Xenophyta
(d) Angiosperms
Answer:
(a) Thallophyta

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 12.
Bryophytes have a root-like structure called
(a) Nodes
(b) Rhizoids
(c) Nodules
(d) Aerenchyma
Answer:
(b) Rhizoids

Question 13.
reproduce with the help of spores formed along the back or posterior surface of their leaves.
(a) Halophyta
(b) Pteridophyta
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Pteridophyta

Question 14.
In ……………………….., the reproductive organs cannot be seen.
(a) Pteridophyta
(b) Cryptogams
(c) Thallophyta
(d) Angiosperms
Answer:
(b) Cryptogams

Question 15.
are mostly evergreen, perennial and woody.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 16.
Gymnosperms bear male and female flowers on different of the same plant.
(a) Branches
(b) Roots
(c) Sporophylls
(d) Flowers
Answer:
(c) Sporophylls

Question 17.
In the seeds are not enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(c) Gymnosperms

Question 18.
In the seeds are enclosed by fruit.
(a) Pteridophyta
(b) Thallophyta
(c) Gymnosperms
(d) Angiosperms
Answer:
(d) Angiosperms

Question 19.
The plants whose seeds cannot be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(d) Monocotyledons

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 20
The plants whose seeds can be divided into equal parts are called
(a) Algae
(b) Fungus
(c) Dicotyledons
(d) Monocotyledons
Answer:
(c) Dicotyledons

Find the odd one out:

Question 1.
Ulothrix, Ulva, Nephrolepis, Sargassum
Answer:
Nephrolepis: It belongs to division Pteridophyta whereas the others belong to division thallophyta.

Question 2.
Funaria, Marchantia, Anthoceros, Spirogyra
Answer:
Spirogyra:

Question 3.
Marsilea, Pteris, Lycopodium, Riccia
Answer:
Riccia:

Question 4.
Cycas, Mango, Apple, Banyan
Answer:
Cycas:

Question 5.
Onion, Papaya, Wheat, Green peas
Answer:
Green peas:

Complete the analogy:

(1) Spirogyra : Thallophyta : : Riccia :
(2) Moss : Bryophyta : : Selaginella :
(3) Nephrolepis : Pteridophyta :: Ulothrix :
(4) Pteridophyta : Roots :: Bryophyta :
(5) Gymnosperms : naked seeds : : Angiosperms :
(6) Dicotyledon : Reticulate venation : : Monocotyledon:
(7) Bamboo stem: Hollow:: Onion Stem:
(8) Monocotylendon : Tap root:: Dicotyledon :
Answer:
(1) Bryophyta
(2) Pteridophyta
(3) Thallophyta
(4) Rhizoids
(5) Covered seeds
(6) Parallel venation
(7) Disc like
(8) Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Difference between:

Question 1.
Thallophyta and Bryophyta
Answer:

Thallophyta Bryophyta
These plants grow mainly in water They grow in moist soil but need water for reproduction

Question 2.
Gymnosperms and Angiosperms
Answer:

Gymnosperms Angiosperms
No natural covering on seeds Seeds are formed in fruits

Question 3.
Algae and Moss
Answer:

Algae Moss
These plants mainly grow in water. These plants need water for reproduction.

State whether the following statements are true or false. Correct the false statements:

(1) Thallophyta are called as the amphibians of the plant kingdom.
(2) Fungi like yeasts and moulds are included in division bryophyta.
(3) Moss (Funaria) belongs to division bryophyta.
(4) Bryophyta have specific tissues for conduction of food and water.
(5) Plants belonging to Thallophyta group are only unicellular.
(6) Pteridophytes have well developed roots, stems and leaves.
(7) Pteridophytes reproduce with the help of spores formed along the back or posterior surface of their leaves.
(8) Nephrolepis belongs to division Pteridophyta.
(9) Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into monocots and dicots.
(10) Gymnosperms are mostly evergreen, perennial and woody.
(11) Gymnosperms bear male and female flowers on different sporophylls of different plants.
(12) In Angiosperms, the seeds are covered by fruits.
(13) Dicotyledonous plants show reticulate venation.
(14) Moncotyledonous plants have trimerous flowers.
(15) In dicotyledonous plants, the stem is strong and hard.
Answer:
(1) False. Thallophyta plants grow mainly in water.
(2) False. Fungi like yeasts and moulds are included in division thallophyta.
(3) True
(4) False. Bryophyta do not have specialised tissuesfor conduction of food and water.
(5) False. Plants belonging to thallophyta group may be unicellular or multicellular.
(6) True
(7) True Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants
(8) True
(9) False. Depending whether seeds are enclosed in. a fruit or not, angiosperms are classified into monocots and dicots.
(10) True
(11) False. Gymnosperms bear male and female flowers on different sporophylls of the same plant.
(12) True
(13) False. Dicotyledonous plants show parallel venation.
(14) True
(15) True.

Give name

Question 1.
What are ornamental plants are called?
Answer:
Ferns

Question 2.
Plants with two cotyledons are called.
Answer:
Dicots

Question 3.
Plants with single cotyledon are called.
Answer:
Monocots

Question 4.
Type of venation showed by hibiscus plant leaves
Answer:
Reticulate venation

Question 5.
Type of venation showed by lily plant leaves
Answer:
Parallel venation

One line answers

Question 1.
Which plants are mostly evergreen, perennial and woody?
Answer:
Gymnosperms are mostly evergreen, perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Which type of venation showed by dicot plants?
Answer:
Leaves of dicot plants show reticulated venation.

Question 3.
Which type of venation showed by monocot plants?
Answer:
Leaves of monocot plants show parallel venation

Question 4.
How are angiosperms classified into monocot and dicot?
Answer:
Depending whether seeds and enclosed in fruit or not, angiosperms are classified into monocot and dicot

Question 5.
In which division are fungi like moulds and yeast classified?
Answer:
Fungi like moulds and yeast classified in division thallophyta.

Question 6.
Plants belonging to which group may be unicellular or multicellular?
Answer:
Plants belonging to thallophyta group may be unicellular or multicellular

Give scientific reason

Question 1.
Thallophyta plants have thin and fibre like body
Answer:
Thallophyta: These plants grow mainly in water i.e. fresh water as well as in saline water, therefore they usually have a soft and fibre-like (filamentous) body.

Question 2.
Bryophyta plants are called the amphibian plants.
Answer:
Bryophyta: They grow in moist soil but need water for reproduction. Therefore, they are called ‘amphibians of plant kingdom’.

Question 3.
Gymnosperms bear their male and female flowers on different sporophylls of the same plant
Answer:
Gymnosperms: As these plants do not take the assistance of pollinators i.e. vectors, the male and female flowers are present on the different sporophyll of the same plant for successful fertilisation.

Write note on

Question 1.
August W. Eichler
Answer:
In 1883, Eichler, a botanist, classified the Kingdom Plantae into two subkingdoms. As a result, two subkingdoms, cryptogams and phanerogams were considered for plant classification.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Thallophyta
Answer:
These plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 3.
Bryophyta
Answer:
This group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 4.
Pteridophyta
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 5.
Phanerogams
Answer:
Plants which have special structures for reproduction and produce seeds are called phanerogams. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 6.
Gymnosperms
Answer:
Gymnosperms are mostly evergreen, perennial and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 7.
Angiosperms
Answer:
The flowers these plants bear are their reproductive organs. Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Complete the flow chart.

Question 1.
Living Organisms
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 1
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 2

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Kingdom: Plantae
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 3
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 4

Distinguish between:

Question 1.
Bryophyta and Pteridophyta:
Answer:

Bryophyta Pteridophyta
Bryophytes grow in soil but need water for reproduction. Pteridophytes grow in soil.
Plant body is without specific parts like true roots, stem and leaves. Plant body is differentiated into true roots, stem and leaves.
Conducting tissues for food and water absent. Conducting tissues for food and water present.
Examples: Moss (Funaria), Marchantia, Anthoceros, etc. Examples: Nephrolepis, Marsilea, Pteris, Adiantum, Lycopodium etc.

Question 2.
Angiosperms and Gymnosperms.
Answer:

Angiosperms Gymnosperms
(i) In Angiosperms, the stems have branches. (i) In Gymnosperms, the stems are without branches.
(ii) Reproductive organs are flowers. (ii) Reprodcutive organs are cones.
(iii) Seeds are enclosed in natural coverings, i.e., fruits. (iii) Seeds are not enclosed in natural coverings.
(iv) Examples: Mango, Bamboo, etc. (iv) Examples: Cycas, Picea etc.

Question 3.
Cryptogams and Phanerogams.
Answer:

Cryptogams Phanerogams
(iii) Their reproductive organs are hidden. (iii) Their reproductive organs are exposed.
(iii) They reproduce by forming spores. (iii) They reproduce by forming seeds.
(iii) They are less evolved plants. (iii) They are highly evolved plants.
(iv) They are divided into Thallophyta, (iv) They are divided into Gymnosperms and
Bryophyta, Pteridophyta. Angiosperms.

Distinguish between:

Question 1.
Thallophyta

Answer:
Spirogvra, Ulothrix, Ulva, Sargassum

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
Bryophyta
Answer:
Moss (Funaria), Marchantia, Anthoceros, Riccia

Question 3.
Pteridophyta
Answer:
Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium

Question 4.
Gymnosperms
Answer:
Cycas, Picca (Christmas tree), Thuja (Morpankhi), Pinus (Deodar)

Question 5.
Angiosperms
Answer:
Tamarind, Mango, Apple, Lemon

Question 6.
Monocot plants
Answer:
Bamboo, bananas, corn, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

Question 7.
Dicot plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

Observe the figure and answer the questions

1. Dicot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 8

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Well developed, primary root (Tap root)

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 4 or 5 parts or in their multiples (tetramerous or pentamerous)

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Reticulate Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Strong and hard

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Two cotyledons

Question 6.
Give example of the following types of plants
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

2. Monocot Plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 9

Question 1.
What are the characteristics of the above plants in terms of root system?
Answer:
Fibrous roots

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
What are the characteristics of the above plants in terms of flowers?
Answer:
Flowers with 3 parts or in multiples of three (trimerous).

Question 3.
What are the characteristics of the above plants in terms of leaf venations?
Answer:
Parallel Venation

Question 4.
What are the characteristics of the above plants in terms of type of stem?
Answer:
Hollow, False or Disc-like

Question 5.
What are the characteristics of the above plants in terms of seed?
Answer:
Single cotyledons

Question 6.
Give example of the following types of plants
Answer:
Bamboo, bananas, com, daffodils, garlic, ginger, grass, lilies, onions, orchids, rice, sugarcane, tulips, and wheat

3. Spirogyra
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 11

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division I Thallophyta.

Question 2.
Where does this plant grow?
Answer:
These plants grow mainly in water.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Are these types of plants unicellular or multicellular?
Answer:
They may be unicellular or multicellular and microscopic or large.

Question 4.
Are these types of plant autotropic?
Answer:
They are autotrophic due to the presence of chlorophyll but types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Question 5.
Do these plants have a root-stem-leaves-flowers system?
Answer:
They do not have specific parts like root-stem- leaves-flowers.

Question 6.
How is the body of these types of plants?
Answer:
These plants usually have a soft and fibre-like body.

4. Funaria
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 12

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division II Bryophyta.

Question 2.
Where does this plant grow?
Answer:
They grow in moist soil but need water for reproduction.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
What are these group of plants called in the plant kingdom?
Answer:
This group of plants is called the ‘amphibians’ of the plant kingdom.

Question 4.
Are these types of plant autotropic?
Answer:
They reproduce by spore formation.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves.

Question 6.
What do these plants have instead of roots?
Answer:
They have root like rhizoids.

5. Fern
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 13

Question 1.
Which division of plants does this plant come? under?
Answer:
This plant come under Division III Pteridophy ta.

Question 2.
Where does this plant grow?
Answer:
They grow in soil.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How do these plants reproduce?
Answer:
These plants reproduce asexually by spore- formation and sexually by zygote formation.

Question 4.
Do these plants produce flowers and fruits?
Answer:
They do not bear flowers and fruits.

Question 5.
Do these plants have root-stem-leaves-flowers system?
Answer:
Plants from this group have well developed roots, stem and leaves and separate tissues for conduction of food and water.

Question 6.
Where are the spores formed in the plants body?
Answer:
The spores formed along the back or posterior surface of their leaves.

6. Cycas
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 14

Question 1.
Which division of plants does this plant come under?
Answer:
This plant come under Division III Phanerogams Division I Gymnosperms.

Question 2.
Explain structure of these types of plants?
Answer:
Gymnosperms are mostly evergreen perennial and woody.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
How is stem and leaves of these types of plants?
Answer:
Their stems are without branches and the leaves form a crown.

Question 4.
Where are the male and female flowers located?
Answer:
These plants bear male and female flowers on different sporophylls of the same plant.?

Question 6.
Give some examples of these types of plants?
Answer:
Rose, sunflower, grapes, strawberries, tomatoes, peas, peanuts and potatoes

7. Monocot and Dicot plants
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 15

Question 1.
Which division of plants does this plant come under?
Answer:
ThisplantcomeunderDivisionlllPhanerogams Division II Angiosperms

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
How are the seeds of these types of planis?
Answer:
The seeds are formed within fruits thus these seeds are covered

Question 3.
How can we classify the plants according to their seeds in this division?
Answer:
The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Question 4.
How the venations are present on the leaves of these types of plants?
Answer:
These plants bear parallel or reticulated venations on the leaves.

Question 5.
How is the root system of these types of plants?
Answer:
The root systems of these types of plant are tap roots or fibrouš roots.

Complete the paragraph

Question 1.
Thallophyta plants grow mainly in …………….. . This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of …………….., is called algae. Algae show great diversity. They may be unicellular or …………….., and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a …………….. and fibre-like body. Various types of …………….. like yeasts and moulds which do not have …………….. are also included in this group.
Answer:
Thallophyta plants grow mainly in water. This group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll, is called algae. Algae show great diversity. They may be unicellular or multicellular, and microscopic or large. Examples of algae are Spirogyra, Ulothrix, Ulva, Sargassum, etc. Some of these are found in fresh water while some are found in saline water. These plants usually have a soft and fibre-like body. Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 2.
…………….. group of plants is called the amphibians’ of the plant kingdom because they grow in moist soil but need …………….. for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by …………….. formation. The structure of the plant body of bryophytes is flat, ribbon?like long, without true …………….., stem and leaves. Instead, they have stem-like or leaf?like parts and root-like ……………. . They do not have specific …………….. for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.
Answer:
Bryophyta group of plants is called the ‘amphibians’ of the plant kingdom because they grow in moist soil but need water for reproduction. These plants are thalloid, multicellular and autotrophic. They reproduce by spore formation. The structure of the plant body of bryophytes is flat, ribbon-like long, without true roots, stem and leaves. Instead, they have stem-like or leaf-like parts and root-like rhizoids. They do not have specific tissues for conduction of food and water. Examples are Moss (Funaria), Marchantia, Anthoceros, Riccia, etc.

Question 3.
Plants from Pteridophyta group have well developed roots, stem and leaves and separate …………….. for conduction of food and water. But,
they do not bear …………….. and ……………… They reproduce with the help of …………….. formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce …………….. by spore-formation and sexually by …………….. formation. They have a well-developed conducting system.
Answer:
Plants from Pteridophyta group have well developed roots, stem and leaves and separate tissues for conduction of food and water. But, they do not bear flowers and fruits. They reproduce with the help of spores formed along the back or posterior surface of their leaves. Examples are ferns like Nephrolepis, Marsilea, Pteris, Adiantum, Equisetum, Selaginella, Lycopodium, etc. These plants reproduce asexually by spore-formation and sexually by zygote formation. They have a well-developed conducting system.

Question 4.
Phanerogams plants which have special structures for …………….. and produce …………….. In these plants, after the process of reproduction, seeds are formed which contain the …………….. and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in …………….. a or not, phanerogams are classified into …………….. and ……………. .
Answer:
Phanerogams plants which have special structures for reproduction and produce seeds. In these plants, after the process of reproduction, seeds are formed which contain the embryo and stored food. During germination of the seed, the stored food is used for the initial growth of the embryo. Depending upon whether seeds are enclosed in a fruit or not, phanerogams are classified into gymnosperms and angiosperms.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 5.
Gymnosperms are mostly …………….., perennial and woody. Their stems are without …………….. The leaves form a …………….. These plants bear male and female flowers on different …………….. of the same plant …………….. of these plants do not have natural coverings, i.e. these plants do not form …………….. and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.
Answer:
Gymnosperms are mostly evergreen, perennial • and woody. Their stems are without branches. The leaves form a crown. These plants bear male and female flowers on different sporophylls of the same plant. Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms. (gymnos: naked, sperms: seeds). Examples Cycas, Picea (Christmas tree), Thuja (Morpankhi), Pinus (Deodar), etc.

Question 6.
The flowers of Angiosperms plants bear are their …………….. organs Flowers develop into …………….. and seeds are formed within …………….. . Thus, these seeds are ……………..; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called …………….. plants and those whose seeds cannot be divided into equal parts are called …………….. plants.
Answer:
The flowers of Angiosperms plants bear are their reproductive orgAnswer: Flowers develop into fruits and seeds are formed within fruits. Thus, these seeds are covered; hence, they are called angiosperms (angios: cover, sperms: seeds). The plants whose seeds can be divided into two equal halves or dicotyledons are called dicotyledonous plants and those whose seeds cannot be divided into equal parts are called monocotyledonous plants.

Answer the questions in detail:

Question 1.
Write the characteristics of Thallophyta.
Answer:

  • Thallophyta plants grow mainly in water.
  • The group of plants, which do not have specific parts like root-stem-leaves-flowers but are autotrophic due to the presence of chlorophyll are called algae.
  • Algae show great diversity They may be unicellular or multicellular and microscopic or large.
  • Some of these are found in freshwater while some are found in saline water.
  • Various types of fungi like yeasts and moulds which do not have chlorophyll are also included in this group.
  • Examples: Spirogyra, Ulothrix, Ulva, etc.

Question 2.
Write the characteristics of Gymnosperms.
Answer:

  • Gymnosperms are mostly evergreen, perennial and woody.
  • Their stems are without branches.
  • The leaves form a crown.
  • These plants bear male and female flowers on different sporophylls of the same plant.
  • Seeds of these plants do not have natural coverings, i.e. these plants do not form fruits and are therefore called gymnosperms (gmnos: naked, sperms: seeds)
  • Examples: Cycas, Picea (christmas tree), Thuja, Pinus (deodar), etc.

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Make concept diagram

Question 1.
Plant classification
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 16

Question 2.
Taxonomy of carnivorous 1ant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 17

Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants

Question 3.
Taxonomy of mango plant
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 6 Classification of Plants 18

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 9 Science Solutions Chapter 3 Current Electricity

Class 9 Science Chapter 3 Current Electricity Textbook Questions and Answers

1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 26
A. By which method are the appliances connected?
Answer:
Appliances are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. What must be the potential difference across individual appliances?
Answer:
The potential difference across all appliances is same in parallel connection.

C. Will the current passing through each appliance be the same? Justify your answer.
Answer:
No, as every appliance has a different load (resistance), the current flowing through each appliance will be different.

D. Why are the domestic appliances connected in this way?
Answer:
The domestic appliances are connected in parallel as the potential difference remains same.

E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer:
No, the other devices will not stop working as the current flowing through them is along different paths.

2. The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 1
Which law can you prove with the help of the above circuit?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 2
(b) This circuit can be used to prove Ohm’s law.
(c) V = 1R is the expression of Ohm’s law

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.

A. Which method should he use to connect the bulbs?
B. What are the characteristics of this way of connecting the bulbs depending on the answer of A above?
C. What will be the effective resistance in the above circuit?

4. The following table shows current in Amperes and potential difference in Volts.

a. Find the average resistance.
b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
c. Which law will the graph prove? Explain the law.

5. Match the pairs

‘A’ Group – ‘B’ Group
1. Free electrons – a. V/ R
2. Current – b. Increases the resistance in the circuit
3. Resistivity – c. Weakly attached
4. Resistances in series – d. VA/LI

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

6. The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?

7. Resistances R1, R2, R3 and R4 are connected as shown in the figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 3

a. Both S1 and S2 are closed.
b. Both S1 and S2 are open.
c. S1 is closed but S2 is open.
Answer:
(a) When both S1 and S2 are dosed, the effective resistance of the circuit decreases and hence, current will increase.
(b) When both S1 and S2 are open, the effective resistance of the rircuit increases and hence, current will decrease.
(c) When S2 is closed and S2 is open, the effective resistance of the tircuit decreases and hence current will increase. [Current will be more than case (b) but less than in case (a)]

8. Three resistances x1, x2 and x3 are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting x1, x2 and x3 are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)

a. Current I flows through x1, x2 and x3
b. x is larger than x1, x2 and x3
c. x is smaller than x1, x2 and x3
d. The potential difference across x1, x2and x3 is the same
e. x = x1 + x2 + x3
\(\text { f. } x=\frac{1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}}\)

9. Solve the following problems.

A. The resistance of a 1m long nichrome wire is 6Ω. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2Ω)
Answer:
The resistance of 70cm wire will be 4.2 Ω

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

B. When two resistors are connected in series, their effective resistance is 80Ω. When they are connected in parallel, their effective resistance is 20Ω. What are the values of the two resistances? (Answer : 40Ω, 40Ω)
Answer:
The values of the two resistances R1 and R2 are 40Ω and 40Ω.

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current? (Answer : 1.4 A)
Answer:
Given: Electric charge (Q) = 420 C
Time (t) = 5 min = 5 x 60
= 300 sec.
To find: Electric current (1) = ?
Formula:
\(I=\frac{Q}{t}\)
Solution:
\(I=\frac{Q}{t}\)
The current in the circuit is 1.4 A.

Class 9 Science Chapter 3 Current Electricity Intext Questions and Answers

Can you recall?

Question 1.
You must have seen a waterfall. Which way does the water flow?
Answer:
Water flows from a certain height of a mountain towards the ground.

Question 2.
Material: Copper and aluminium wires, glass rod, rubber.
Make connection as shown in figure 3.8. First connect a copper wire between points A and B and measure the current in the circuit. Then in place of the copper wire, connect the aluminium wire, glass rod, rubber, etc one at a time and measure the current each time. Compare the values of the current in different cases.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 4
Also take different metal strips (Iron, Copper, Zinc, and Aluminium) and connect it in slot AB. Now observe the difference in the resistance using Ohm meter.
Answer:
When copper and aluminium wires are connected to the circuit, current flows through it, as both are good conductors of electricity. When glass rod or rubber was connected to the circuit, current does not flow through it, as both are bad conductors of electricity.

Copper displays lowest resistance while the resistance increases with aluminium, zinc and iron respectively.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Set up the experiment as shown in figure. Then remove the clamp from the rubber tube.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 27

(a) What happens when the clamp is removed?
Answer:
When the clamp is removed, water flows from higher level to lower level.

(b) Does the water stop flowing? Why?
Answer:
Yes, the water stops flowing. This happens when the level of water becomes equal in both the bottles, i.e., there is no difference in the water levels.

(c) What will you do to keep the water flowing for a longer duration?
Answer:
The difference in the water level has to be maintained till that time. The difference must never be zero.

Question 4.
Point out the mistakes in the figure below:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 28
Answer:
A: Wire is broken at the negative terminal. Bulb will not glow as the circuit is incomplete.
B: Wire is disconnected at the negative terminal. Bulb will not glow as the circuit is incomplete.
C: The circuit is complete. Therefore, bulb will glow.
D: Rubber is a bad conductor of electricity. Hence, it will not allow current to flow and the bulb will not glow.

Question 5.
Why are the bulbs in Figures B, C and D not lighting up?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 29
Answer:

  • In B, the blue wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In C, the red wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
  • In D, both wires are connected to the same terminal. Hence, there is no potential difference and current does not flow. Therefore, bulb will not light up.

Class 9 Science Chapter 3 Current Electricity Additional Important Questions and Answers

Choose and write the correct option:

Question 1.
1mA = …………… A.
(a) 103
(b) 10-3
(c) 106
(d) 10-6
Answer:
(a) 103

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
To increase the effective resistance in a circuit the resistors are connected in ………….. .
(a) Series
(b) Parallel
(c) Both ways
(d) None of these
Answer:
(a) series

Question 3.
1 kilowatt hr = …………… joules.
(a) 4.6 x 106
(b) 3.6 x 106
(c) 30.6 x 106
(d) 3.6 x 1O5
Answer:
(b) 3.6 x 106

Question 4.
The voltage difference in India between the live and neutral wires is about ………….. .
(a) 110 V
(b) 220 V
(c) 440 V
(d) 60 V
Answer:
(b) 220 V

Question 5.
Resistivity is the specific property of a ………….. .
(a) Area of cross-section
(b) Temperature
(c) Length
(d) Material
Answer:
(d) material

Question 6.
If a P.D. of 12 V is applied across a 3Ω resistor then the current passing through it is ………….. .
(a) 36 A
(b) 4 A
(c) 0.25 A
(d) 15 A
Answer:
(b) 4 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 7.
In order to measure the electric current flowing through a circuit, we connect …………… with the circuit.
(a) a voltmeter in parallel
(b) a voltmeter in series
(c) an ammeter in parallel
(d) an ammeter in series
Answer:
(d) an ammeter in series

Question 8.
P and Q are two wires of same length and different cross-sectional areas and made of same material. Name the property which is same for both the wires.
(a) Resistivity
(b) Resistance
(c) Current
(d) Both (a) and (b)
Answer:
(a) Resistivity

Question 9.
The following is true for identical bulbs connected in parallel.
(a) All bulbs glow with unequal brightness.
(b) If one bulb is non-functional, all will stop working.
(c) All bulbs glow with equal brightness.
(d) Bulbs function for longer time.
Answer:
(c) All bulbs glow with equal brightness

Question 10.
The …………… wire is either yellow or green in colour.
(a) Live
(b) Neutral
(c) Earth
(d) Fuse
Answer:
(c) earth

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
A current flows through a circuit due to the difference in …………… between two points in the conductor.
(a) Gravity
(b) Potential
(c) Resistance
(d) Fuse
Answer:
(b) potential

Question 12.
…………… is the amount of charge flowing through a particular cross sectional area in unit time.
(a) Electric current
(b) Ampere
(c) Volt
(d) Force
Answer:
(a) Electric current

Question 13.
The flow of …………… constitutes the electric current in a wire.
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) electrons

Question 14.
The conventional direction of flow of current is from …………… terminal to …………… terminal.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(c) positive, negative

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 15.
Current stops flowing when potential difference between two ends of a wire becomes ………….. .
(a) Zero
(b) Positive
(c) Negative
(d) Higher
Answer:
(a) zero

Question 16.
Resistances are connected in …………… so as to pass the same current through them.
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(a) series

Question 17.
To decrease the effective resistance in a circuit, the resistances are connected in ………….. .
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer:
(b) parallel

Question 18.
1μV = …………… V
(a) 102
(b) 10-6
(c) 106
(d) 103
Answer:
(b) 10-6

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 19.
Good conductors contain a large number of ………….. .
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer:
(c) free electrons

Question 20.
Electrons flow from …………… terminal to …………… terminal in a conductor when a potential difference is applied.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer:
(a) negative, positive

Find the odd one out:

Question 1.
Voltmeter, Ammeter, Galvanometer, Thermometer
Answer:
Thermometer

Question 2.
Rubber, Silver, Copper, Gold
Answer:
Rubber

Question 3.
Wood, Glass, Steel, Rubber
Answer:
Steel

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
Graphite, Diamond, Fullerenes, Coal
Answer:
Fullerenes

Distinguish between:

Question 1.
Voltmeter and Ammeter
Answer:

Voltmeter Ammeter
(i) It is an instrument used to measure the potential difference between two terminals of a cell. (i) It is an instrument to measure the electric current flowing through a circuit.
(ii) It is connected in parallel with the cell. (ii) It is connected in series with the cell.
(iii) It has a very high resistance. (iii) It has a very low resistance.
(iv) Voltmeter has range of volts. (iv) Ammeter has range of amps.

Question 2.
Ohmic conductors and Non-Ohmic conductors
Answer:

Conductors Insulators
(i) Substances which have very low electrical resistances are called conductors. (i) Substances which have extremely high electrical resistances are called Insulators.
(ii) They contain a large number of free electrons. (ii) They contain practically no free electrons.
(iii) Conductors are mostly metals. (iii) Insulators are mostly non metals.
(iv) Conductor example iron, copper. (iv) Insulator example rubber, plastic.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 3.
Conductors and Insulators
Answer:

Resistance Resistivity
(i) The hindrance to the flow of electrons is called resistance. (i) Resistivity is the specific property of the material of a conductor.
(ii) The S.I. unit of resistance is ohm (Q). (ii) The S.I. unit of resistivity is ohm-metre (Q – m).
(iii) It depends on temperature, area of cross-section, length of conductor and material of the conductor. (iii) It depends on material of the conductor.
(iv) Resistance can be changed as it depends of external factor as well. (iv) Resistivity cannot be changed as it depends of internal factors.

Question 4.
Resistance in Series and Resistance in Parallel
Answer:

Resistance in Series Resistance in Parallel
(i) Effective resistance of the resistors is equal to the sum of their individual resistances. (i) Inverse of the effective resistance is equal to the sum of the inverse of individual resistances.
(ii) The same current flows through each resistor. (ii) The total current flowing through the circuit is the sum of the currents flowing through individual resistors.
(iii) The effective resistance is larger than each of the individual resistances. (iii) The effective resistance of resistors connected in parallel is less than the least resistance of individual resistors.
(iv) This arrangement is used to increase the resistance in a circuit. (iv) This arrangement is used to decrease the resistance in a circuit.

Question 5.
Answer:

Electric current Potential difference
(i) The flow of electric charge per unit time is called electric current. (i) The difference in potential between the positive and negative terminal of a cell is the potential difference of that cell.
(ii) The S.I. unit of electric current is ampere. (ii) The S.I. unit of potential difference volt.
(iii) Ammeter is used to measure electric current. (iii) Voltmeter is used to measure electric current.
(iv) Current is represented by: \(\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}\) (iv) Potential difference is represented by: \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make pair:

Question 1.
Copper : Conductor :: Rubber : ……………….
Answer:
Insulator

Question 2.
Aluminium : ………………. :: Indium oxide : Super Insulator
Answer:
Super conductor

Question 3.
Parallel Connection : \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\) :: Series Connection : ……………….
Answer:
Rs = R1 + R2

(4) Electric Current : ………………. :: Electric charge : Coulomb
Answer:
Ampere

(5) Electric resistance : Ohm :: Potential difference : ……………….
Answer:
Volt

State whether the following statements are true or false. Correct the false statements:

(1) The SI unit of charge is volt.
(2) Voltmeter is always connected in series with the device.
(3) The conventional direction of flow of current is from positive terminal to negative terminal.
(4) Silver and copper are good conductors.
(5) Resistivity of pure metals is more than alloys.
(6) Resistance in series arrangement is used to decrease resistance of circuit.
(7) A conducting wire offers less resistance to flow of electrons.
(8) Charges are measured in ampere.
(9) The unit of potential difference is ampere.
(10) Resistance of a conductor is inversely proportional to the length of the conductor.
(11) Ammeter is connected in parallel to the cell to measure current.
(12) Fuse is made of wire having high melting point.
Answer:
(1) False. The SI unit of charge is coulomb.
(2) False. Voltmeter is ahvays connected in parallel with the device.
(3) True
(4) True
(5) False. Resistivity of pure metals is less than alloys.
(6) False. Resistance in series arrangement is used to increase resistance of circuit.
(7) True
(8) False. Charges are measured in coulomb.
(9) False. The unit of potential difference is volt.
(10) False. Resistance of a conductor is directly proportional to the length of the conductor.
(11) False. Ammeter is connected in series to the cell to measure current.
(12) False. Fuse is made of wire having low melting point.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Answer the following in one sentence:

Question 1.
Which is the unit used to measure large voltages?
Answer:
Kilovolts and Megavolts are the units used to measure large voltages.

Question 2.
What is the SI unit of potential difference?
Answer:
The SI unit of potential difference is volt (V).

Question 3.
What is lightning?
Answer:
Lightning is the electric discharge travelling from clouds at high potential to earth’s surface which is at zero potential.

Question 4.
What is the unit of resistivity.
Answer:
The unit of resistivity is ohm metre (Qm).

Question 5.
Which substances are called conductors of electricity?
Answer:
Those substances which have very low electrical resistance are called conductors of electricity.

Question 6.
What is Earth wire?
Answer:
Earth wire is generally yellow or green colour, it is connected to a metal plate buried deep underground near the house and is for safety purpose.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write formula:
(1) E1ectriccurrent \(=\frac{Q}{t}\)
(2) Electric charge = It
(3) Potential difference = IR
(4) Electric resistance \(=\frac{V}{I}\)
(5) Current \(=\frac{V}{R}\)
(6) Resistivity \(=\frac{RA}{L}\)

Give scientific reasons:

Question 1.
Free electrons are required for conduction of electricity.
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to the nucleus.
  • These are called free electrons. These electrons can easily move from one part of a conductor to its other parts. The negative charge of the electrons also gets transferred as a result of this motion.
  • The free electrons in a conductor are the carriers of negative charge. Hence, free electrons are required for conduction of electricity.

Question 2.
Wood and glass are good insulators.
Answer:

  • Those substances which have infinitely high electrical resistance are called insulators.
  • Wood and glass have high resistance and negligible free electrons for conduction of electricity.
  • Hence, wood and glass are good insulators.

Question 3.
Connecting wires in a circuit are made of copper and aluminium.
Answer:

  • Copper and aluminum are good conductors of electricity.
  • They have low electrical resistance and large number of free electrons.
  • As they are malleable and ductile, they can be drawn into thin wires. Hence, connecting wires in a circuit are made of copper or aluminum.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
A thick wire has a low resistance.
Answer:

  • The resistance (R) of a wire is inversely proportional to the cross-sectional area (A) of a wire. i.e., R i
  • Thus, greater is the cross-sectional area of a conductor (wire), lower is its resistance. Hence, a thick wire has a low resistance.

Question 5.
A series combination of resistances is used to increase the resistance of a circuit.
Answer:

  • When resistances are connected in series, the effective resistance of the resistors is equal to the sum of their individual resistances. Rs = R1 + R2 ………….. Rn
  • The effective resistance is larger than each of the individual resistances. Hence, This arrangement is used to increase the resistance in a circuit.

Question 6.
A parallel combination of resistances decreases the effective resistance of the circuit.
Answer:

  • In a parallel combination, the inverse of the effective resistance is equal to the sum of the inverses of individual resistances. \(\frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}} \ldots \ldots \cdot \frac{1}{\mathrm{R}_{\mathrm{n}}}\)
  • The effective resistance of resistors connected in parallel is less than the individual resistors.
  • Due to this, any addition of an individual resistance in parallel combination will decrease the overall resistance of the circuit. Hence, a parallel combination of resistance decreases the effective resistance of the circuit.

Question 7.
Lightning occurs from sky to earth.
Answer:

  • Lightning is the electric discharge travelling from clouds at high potential to the earth’s surface, which is at zero potential.
  • The earth is always at lower potential as compared to the clouds.
  • Hence, lightning occurs from sky to earth.

Question 8.
In streetlights, bulbs are connected in parallel.
Answer:

  1. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows through the other paths, and the rest of the bulbs light up.
  2. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually. Hence, streetlights are connected in parallel.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Numerical:
Numericals based on the formula: (1) Q= It (2) W= VQ

Question 1.
A current of 0.4 A flows through a conductor for 5 minutes. How much charge would have passed through the conductor?
Answer:
Given: Current (I) = 0.4 A
Time (t) = 5 min = 5 x 60 = 300 s
To find: Charge (Q) =?
Formula: Q = 1 x
Solution: Q = 0.4 x 300
Q= 120 C.
Charge passing through the conductor is 120

Question 2.
Find the amount of work done if 3 C of charge is moved through a potential difference of 9 V.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 5
The work done is 27 joule.

Question 3.
The resistance of the filament of a bulb is 1000Ω. It is drawing a current from a source of 230 V. How much current is flowing through it?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 6
The current flowing through the filament of bulb is 0.23 A.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 4.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 7
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 8
The resistivity of the wire is 4.71 x 10-5 Qm.

Question 5.
A current of 0.24 A flows through a conductor when a potential difference of 24 V is applied between its two ends. What is its resistance?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 9
The resistance of a conductor is 100Ω.

Question 6.
If three resistors 15Ω, 3Ω and 4Ω each are connected in series, what is the effective resistance in the circuit?
Answer:
Given:
R1 =15Ω
R2 = 3Ω
R3 = 4Ω
Effective resistance in series (Rs) = ?
Rs = R1 + R2 + R3
Rs = 15 + 3 + 4
Rs = 22Ω
The effective resistance in the circuit is 22Ω.

Question 7.
Three resistances 15Ω, 20Ω and 10Ω are connected in parallel. Find the effective resistance of the circuit.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 10
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 11
The effective resistance of the circuit is 4.615 Ω. It is less than the least of the three i.e., 10Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Write a note on the following:

Question 1.
Electric current
Answer:
An electric current is the flow of electrons through a conductor. Quantitatively, current
(I) is defined as the charge passing through a conductor in unit time.
\(T=\frac{Q}{t}\)

Question 2.
1 ampere
Answer:
One ampere current is said to flow in a conductor if one coulomb charge flows through it every second.
\(1 \mathrm{~A}=\frac{1 \mathrm{C}}{1 \mathrm{~s}}\)

Question 3.
1 volt
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
\(1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)

Question 4.
Potential Difference
Answer:
The amount of work done to carry a unit positive charge from point A to point B is called the electric potential difference between the two points.
\(V=\frac{W}{Q}\)

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Conductor
Answer:
Those substances which have very low resistance are called conductors. Current can flow easily through such materials.

Question 6.
Insulators
Answer:
Those substances which have extremely high resistance and through which current cannot flow are called insulators.

Question 7.
1 ohm
Answer:
If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.
\(\frac{1 \text { Volt }}{1 \text { Ampere }}=1 \mathrm{Ohm}\)

Question 8.
Potential
Answer:
The level of electric charge present is known as potential.

Question 9.
Ohm’s Law
Answer:
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
V = IR

Question 10.
Superconductors
Answer:
The resistance of some conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Such conductors are called superconductors.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 11.
Non-ohmic conductors
Answer:
Conductors which do not obey Ohm’s law are called non-ohmic conductors.

Complete the flow charts:

(1) Protection from Electricity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 12

(2) Resistance
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 13

(3) Resistivity
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 14

Write properties/characteristics/advantages of the following:

Superconductors
Answer:
The resistance of these conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Aluminium is an example of Super Conductor. Superconductors can be used in space missions to increase/ boost the signal strength. They are also used i in the data fibres to increase the speed of data transfer.

Give explanations of the given statements:

Question 1.
Safety precautions are to be taken while using electricity.
Answer:

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing’ through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • If not, then you should use a wooden pole to push the person away from the electric wire.

Question 2.
In a domestic circuit colour code is followed while setting up electrical wiring.
Answer:

  • The electricity in our homes is brought through the main conducting cable either from the electric pole or from underground cables.
  • Usually, there are three wires in the cable.
    (a) Live wire which brings in the current. It has a red or brown insulation.
    (b) Neutral wire through which the current returns. It is blue or black.
    (c) Earth wire is of yellow or green colour. This is connected to a metal plate buried deep underground near the house and is for safety purposes.
  • In India, the voltage difference between the live and neutral wires is about 220 V.
  • Live and neutral wires are connected to the electric meter through a fuse.
  • They are connected through a main switch, to all the conducting wires inside the home so as to provide electricity to every room.
  • In each separate circuit, various electrical appliances are connected between the live and neutral wires.
  • The different appliances are connected in parallel and the potential difference across every appliance is the same.

Question 3.
Fuse used in electrical circuit can save electrical objects from damage.
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1 A, 2 A, 3 A, 4 A, 5 A and 10 A are used.

Question 4.
Bulbs arranged in parallel glow brighter than bulbs arranged in series.
Answer:

  • The amount of light given out by bulbs in parallel combination will be more than that in series combination.
  • In parallel combination the resistance of the overall circuit decreases whereas in series it increases, so the current flowing through the bulbs in parallel circuit is more.
  • Due to this, intensity of light given out by bulbs in parallel combination is more than the bulbs in series combination.

Complete the following table:
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 15

Solve the numerical:

Question 1.
The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 16
The resistivity of the wire is 4.71 x 10-5 Ωm.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Determine the current that will flow when a potential difference of 33 V is applied between two ends of an appliance having a resistance of 110 Ω. If the same current is to flow through an appliance having a resistance of 500 Ω, how much potential difference should be applied across its two ends?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 17
The current is 0.3 A and potential difference to be applied is 150 V.

Question 3.
Determine the resistance of a copper wire having a length of 1 km and diameter of 0.5 mm.
Answer:
Given: Resistivity of copper (p)
= 1.7 x 10-8 Ω m
Converting all measures into metres.
Length of wire (L) = 1 km
= 1000 m = 103 m
Diameter of wire (d) = 0.5 mm
= 0.5 x 10-3m
To find: Resistance of wire (R) = ?
Formula:
\(R=\rho \frac{L}{A}\)
Solution:
If d is the diameter of the wire then, its area of cross-section
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 18
The resistance of a copper wire is 85Ω and area of cross section is 0.2 x 10-6 m2

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 5.
Two resistors having resistance of 16 and 14 are connected in series. If a potential difference of 18 V Is applied across them, calculate the current flowing through the circuit and the potential difference across each individual resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 19
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 20
The current in the circuit is 0.6 A and potential across 16 Ω retor is 9.6 volt and 14 Ω resistor is 8.4 voIt.

Question 6.
If the resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to battery of 12 V, find the effective resistance in the circuit. Calculate the total current and current in each resistor.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 21
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 22
(a) The total current is 4 A and current in each resistor is 2.4 A, 1.2 A and 0.4 A respectively.
(b) The effective resistance in the circuit is 3Ω.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Complete the diagram and answer the questions:

Question 1.
The following figure shows the symbols for components used in the accompanying electrical circuit.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 23
(a) Place them at proper places and complete the circuit.
(b) Which law can you prove with the help of the above circuit?
(c) State expression for Ohm’s law
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 24
(b) This circuit can be used to prove Ohm’s law.
(c) V = IR is the expression of Ohm’s law

Question 3.
Explain with the help of a diagram, what are free electrons and how they move through the conductor?
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 25
Answer:

  • Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to nucleus.
  • These are called free electrons.
  • These electrons can easily move from one part of a conductor to its other parts.

Complete the paragraph:

Question 1.
If resistors are connected in series,
Answer:
The same current flows through each resistor. The effective resistance of the resistors is equal to the sum of their individual resistances. The potential difference between the two extremes of the arrangement is equal to the sum of the potential differences across individual resistors. The effective resistance is larger than each of the individual resistances. This arrangement is used to increase the resistance in a circuit. This type of connection is used in electrical heating equipment like geysers, iron, and hair dryers.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
If a number of resistors are connected in parallel,
Answer:
The inverse of the effective resistance is equal to the sum of the inverses of individual resistances. The current flowing through an individual resistor is proportional to the r inverse of its resistance and the total current flowing through the circuit is the sum of the currents flowing through individual resistors. The potential difference across all r resistors is the same. The effective resistance of resistors connected in parallel is less than r the least resistance of individual resistors.

This arrangement is used to reduce the resistance in a circuit. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows, through the other paths, and the rest of the bulbs light up. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually.

Read the paragraph and answer the questions.

Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket. Before cleaning an electrical appliance it should be switched off and its plug removed from the socket. One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. As rubber is an insulator, it prevents | the current from flowing through our body, thereby protecting it. If a person gets an electric shock, you should not touch that person. You should switch off the main switch and if the switch is too far or you do not know where it is located, then you should remove the plug from the socket if possible. If not, then you should use a wooden pole to push the person away from the electric wire.

(i) Why should the electrical sockets be fitted at a certain height?
Answer:
Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside.

(ii) Why plug wires should not be pulled out while removing any electrical device?
Answer:
Plug wires should not be pulled out while removing any electrical device as it may cause the wire to break causing short circuit which can lead to fire or death.

(iii) Why should a person wear footwear with rubber soles while handling electrical appliances. .
Answer:
As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. Hence a person should wear footwear with rubber soles while handling electrical appliances.

(iv) Saee is touching an electrical button socket with wet hands what will you advise her and why?
Answer:
We will advise her to dry her hands before touching any electrical sockets or devices as water on the hands can cause an electrical short circuit producing shock to the person touching it.

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

(v) Sneha is getting an electrical shock what will you do the save her life?
Answer:
We should switch off the main switch and if the switch is too far or we do not know where it is located, then we should remove the plug from the socket if possible. If not, then we should use a wooden pole to push the person away from the electric wire.

(vi) Give a title to the above passage.
Answer:
Precautions to be taken while using electricity

Answer the questions in details:

Question 1.
Find the expression (i.e., derive the expression) for the resistors connected in series.
Answer:
Expression for the resistance connected in series:
(i) Let R1, R2 and R3 be three resistances connected in series between C and D.
(ii) Let Rs be the effective resistance in circuit and V1, V2 and V3 be the potential difference across R1, R2 and R3 respectively.
(iii) Let the potential difference across CD be V.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 30
(iv) In series combination.
v = v1 + v2 + v3 ……………………(i)
By using Ohm’s law
V = IRs
∴ V1= IR1, V2 = IR2 and V3 = 1R3
Substituting these values in equation (j) we get
IRs = IR1 + IR2 + IR3
∴ Rs = R1 + R2 + R3
For ‘n’ number of resistors conneded in series we get
Rs = R1 + R2 + R3 + …………………. + Rn

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Question 2.
Find the expression (i.e., derive the expression) for the resistors connected in parallel.
Answer:
Expression for the resistance connected in parallel.
(i) Let R1, R2 and R3 be the three resistances connected in parallel combination between points C and D and let R be their effective resistance.
(ii) Let I1, I2 and I3 be the currents flowing through resistances R1, R2 and R3 respectively. Let I be the current flowing through the circuit and V be the potential difference of the cell.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 31
(iii) For parallel combination of resistances,
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 32
(iv) Substituting the values of (I, I1, I2 and I3) in equation (i) we get
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 33

Question 3.
Find the expression for resistivity of a material.
Answer:
(i) At a given temperature, the resistance (R) of a conductor depends on its length (L), area of cross-section (A) and the material it is made of. If the resistance of a conductor is R, then
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 34
(ii) p is the constant of proportionality and is called the resistivity of the material.
(iii) The unit of resistivity in SI units is Ohm metre (Ω m).
(iv) Resistivity is a specific property of a material and different materials have different resistivity. ’

Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity

Make the concept diagram and explain:

Question 1.
Make the concept diagram of an electrical circuit and explain the working of a fuse.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 35
Answer:

  • Fuse wire is used to protect domestic appliances.
  • It is made of a mixture of substances and has a specific melting point.
  • It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. The circuit gets broken and the flow of current stops, thus protecting the appliance.
  • This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1A, 2A, 3A, 4A, 5A, and lO Aareused.

(2) Show motion of electrons in an circuit and explain precautions while using an electrical device.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 37

  • Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
  • Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
  • One’s hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity
  • As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
  • If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
  • 1f not, then you should use a wooden pole to push the person away from the electric wire.

Q.4.4.Complete the incomplete figure and give an explanation:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 38
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 39
If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.

I α V
I = kV (k = constant of proportionality)
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 40

This is known as Ohm’s law.
We can obtain the SI unit of resistance from the above formula, Potential difference and current are measured in Volts and Amperes respectively. The unit o resistance is called Ohm. It is indicated by the symbol Ω.
Maharashtra Board Class 9 Science Solutions Chapter 3 Current Electricity 41
The resistance of one Ohm : If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm.