CBSE Class 10

NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 11 – Human Eye and Colourful World solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 11 – Human Eye and Colourful World Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
What is meant by power of accommodation of the eye ?
(CBSE Sample Paper 2010, 2012, 2014, 2015, 2016, 2017)
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision ?
Answer:

Question 3.
What is the far point and near point of the human eye with normal vision ? (CBSE 2011, 2012, 2014)
Answer:
The farthest position of an object from the human eye so that its sharp image is formed on the retina is at infinite distance from the eye.
The nearest position of an object from a human eye so that its sharp image is formed on the retina is at 25 cm from the eye.

Question 4.
A student has difficulty in reading the black board while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ? (CBSE 2012)
Answer:
Near sightedness or myopia. This defect can be corrected by using a concave lens of suitable focal length.

NCERT Chapter End Exercises

Question 1.
Choose the correct option :
Human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) near sightedness
(c) accommodation
(d) far sightedness.
Answer:
(c).

Question 2.
Human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d).

Question 3.
The least distance of distinct vision for a young adult with normal vision is about (CBSE 2012, Bihar Board 2012)
(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 50 cm. (CBSE 2011)
Answer:
(c).

Question 4.
The change in focal length of an eye lens is caused by the action of
(a) pupil
(b) retina
(c) ciliary muscles
(d) iris. (CBSE 2011)
Answer:
(c).

Question 5.
A person needs a lens of power – 5.5 diopters for correcting distant vision. For correcting his near vision, he needs a lens of power +1.5 diopter. What is the focal length of the lens required for correcting
(i) distant vision and
(ii) near vision ?
Answer:

Question 6.
The far point of a myopic person is 150 cm in front the eye. What is the nature and power of the lens required to correct the problem ?
Answer:

The lens is concave lens.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.
(CBSE 2011)
Answer:
For diagram,

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm ? (CBSE 2011)
Answer:
The nearest position of an object from a normal human eye so that its sharp image is formed on retina is 25 cm. If the object is placed at a distance less than 25 cm, then the blurred image of the object is formed on retina as the focal length of eye lens cannot be decreased below a certain limit. Hence, eye cannot see it clearly.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
[CBSE (Delhi) 2008, 2011]
Answer:
The image distance remains the same in the eye because the eye has the ability to change the focal length of its lens to make the image always on the retina when the object distance increases from the eye.

Question 10.
Why do stars twinkle ? (CBSE 2011, 2012, 2015, 2016)
                                     Or
Explain with the help of a labelled diagram, the cause of twinkling of stars. (CBSE 2014)
Answer:
Twinkling of Stars:
Light emitted by distant stars (act as point sources of light) passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities. The layers close to the surface of the earth are optically denser. As we go higher and higher, the density of layers and refractive index decreases progressively. As the light from a star enters the upper­most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star (Figure 13).

Question 11.
Explain why the planets do not twinkle. (CBSE 2011, 2012, 2015)
Answer:
Planets do not twinkle:
Planets are very close to the earth as compared to the stars. The planets act as extended sources of light. So the intensity of light we receive from the planets is very large. Therefore, the variation in the brightness of the planets is not detected. Hence, planets do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning ? (CBSE 2011, 2012)
Answer:
When sunlight enters the atmosphere of the earth, the atoms and molecules of different gases present in the atmosphere absorb this light. Then these atoms and molecules of the gases re-emit light in all directions. This process is known as scattering of light. The atoms or particles scattering light are known as scatters.
The intensity of scattered light is inversely proportional to the fourth power of the wavelength of incident light, if the size of the particles (say atoms or molecules) scattering the light is less than the wavelength of the incident light.
That is, intensity of scattered light,                I ∝ 1/λ⁴.
We know, wavelength of red light is greater than the wavelength of blue or violet light. Therefore, the intensity of scattered red light is less than the intensity of the scattered blue or violet light.
The blue colour of sky, greenish blue colour of sea water, red colour of sunset and sunrise and white colour of clouds are due to the scattering of sun light by the particles present in the atmosphere of the earth.

Question 13.
Why does the sky appear dark instead of blue to an astronaut ?
[CBSE (Delhi) 2008, 2011, 2012, CBSE (Foreign) 2016]
                                                     Or
What will the colour of the sky be for an astronaut staying in international space station orbiting the earth 1 Justify your answer giving reason. (CBSE 2014)
Answer:
The blue colour of sky is due to the scattering of sunlight. The scattering of sunlight in the atmosphere is due to the presence of atoms and molecules of gases, droplets and dust particles. When the astronaut is in space, then there is no atmosphere (or atoms and molecules of gases, droplets and dust particles) around him. Therefore, sunlight does not scatter and hence sky appears dark.

Practical Skills Based Questions ( 2 Marks)

Question 1.
Draw a path of Light ray passing through a prism.
Label angle of incidence and angle of deviation in the ray diagram. (CBSE Sample Paper 2017-18)
Answer:

Question 2.
Draw the path of light ray passing through a glass prism. Label angle of incidence, angle of deviation and angle of emergence.
Answer:

Question 3.
The path of a ray of light through a glass prism is shown below :

Write the names of the angles represented by 1, 3, 4 and 5 respectively.
Answer:
1. represents angle of incidence.
3. represents angle of prism.
4. represents angle of deviation
5. represents angle of emergence

Question 4.
The path of a ray of light through a glass prism is shown below :

Name the incident ray of light, refracted ray of light and emergent ray of light.
Answer:

1. represents incident ray of light
2. represents refracted ray of light
3. represents emergent ray of light

Question 5.
The path of a ray of light passing through a glass prism is shown below :

Name the angles X, Y, Z and O
Answer:
∠X = Incident angle
∠Y = Angle of prism
∠Z = Angle of emergence
∠O = Angle of deviation

Question 6.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 on a screen as shown in figure.

Name the colours similar to the colour of

1. danger or stop signal light
2. core of hard boiled egg
3. colour of clear sky
4. solution of potassium permanganate

Answer:

1. 1 is red colour which corresponds to the colour of danger or stop signal light.
2. 3 is yellow colour which corresponds to the colour of hard boiled egg.
3. 5 is blue colour which corresponds to the colour of clear sky.
4. 7 is violet colour which corresponds to the colour of the solution of potassium permanganate.

NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

Hope given NCERT Solutions for Class 10 Science Chapter 11 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 9 – Heredity and Evolution solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 9 – Heredity and Evolution Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same species, which trait is likely to have arisen earlier ?
Answer:
In asexually reproducing population, there is no reshuffling of traits. New traits do develop due to small inaccuracies produced during DNA copying. They will be in smaller proportion than the traits already present. Therefore, trait B which exists in 60% of population must have arisen earlier than the trait A which occurs in 10% of the population.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
How does creation of variations in a species promote survival ?
Answer:
A number of different types of variations develop in a population. All of them do not have survival value. However, some of them are pre-adaptations which can be beneficial under certain environmental conditions. For example, in a heat wave most of the bacteria will die but a few having pre-adaptation or variation to tolerate heat wave will survive and multiply. Actually selection of variants by different environmental factors constitutes the basis for evolution.

Question 3.
How do Mendel’s experiments show that traits may he dominant or recessive ?
(CCE 2012, CBSE Delhi 2016, 2017)
Answer:
Mendel crossed Garden Pea plants having contrasting visible traits, e.g., tall and dwarf, violet and white flowered.
In F₁ generation there were no halfway characteristics. A cross between pure tall and pure dwarf plants yielded only tall plants in F₁ generation. There were no medium height plants. When F₁ plants were self bred, the F₂ plants were not all tall plants. Instead, both tall and dwarf plants appeared in ratio of 3 : 1. It means that the trait for dwarfness was present in F₁ generation but was not expressed while the trait for tallness expressed itself.

The trait of tallness which expresses itself in the presence of its contrasting form is called dominant. The other trait of dwarfness which is unable to express its effect in the presence of its contrasting trait is known as recessive.

Question 4.
How do Mendel’s experiments show that traits are inherited independently ? (CCE 2012, CBSE A.I. 2016, Delhi 17)
Answer:
Independent inheritance of traits is proved by employing dihybrid crosses and obtaining dihybrid ratios. Mendel crossed pure breeding tall plants having round seeds (TTRR) with pure breeding short plants having wrinkled seeds (ttrr). The plants of F₁ generation were all tall and with rounded seeds (TtRr) indicating that the characteristics of tallness and round seededness were dominant. Self breeding of F₁ yielded plants in the ratio of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled seeded and short round seeded plants are new combinations which can develop only if the traits are inherited independently. If the two traits are considered individually, F₂ ratio would be same as for monohybrid crosses, i.e., 12 tall : 4 short, 12 round seeded : 4 wripkled seeded.

TR, Tr, rR, tr x TR, Tr, rR, tr Gametes 9 tall rounded : 3 tall wrinkled : 3 short rounded : 1 short wrinkled

Question 5.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits—blood group A or O, is dominant ? Why or why not ?
Answer:
No. The information is not enough to tell whether the trait of the blood group A (I^A) or blood group
0(I⁰) is dominant. Either can be possible. Each individual carries two alleles. A recessive trait appears only when the two alleles are similar.
Possibility I: Blood Group A is Dominant and O Recessive. The trait of blood group O can appear only when both the recessive alleles occur together as in mother and daughter (I⁰I⁰). A group father should carry both the alleles of A and O (I^AI⁰).
Possibility 2: Blood Group O is Dominant and A Recessive. In this case the father should carry the alleles of A(I^AI^A) while the mother can be homozygous or heterozygous (I⁰I⁰, I⁰I^A). The daughter will have one dominant alleles of 0(I⁰I^A).
As both the possibilities can occur, the given information is unable to tell whether allele for blood group A or O is dominant.

Question 6.
How is the sex of the child determined in human beings ? (CCE 2013)
Answer:
Sex of the child is determined by the gametes that fuse to form zygote which later grows into offspring. Human females (44 + XX) produce only one type of ova (22 + X). Human males (44 + XY) form two types of sperms, androsperms (22 + Y) and gynosperms (22 + X). Both are formed in equal number. It is a chance factor whether an androsperm or a gynosperm fuses with egg to form 44 + XY or 44 + XX child. A child that obtains an X-chromosome from father will be girl and the one who inherits a Y-chromosome will be boy.

Question 7.
Draw Fig. 4.12. What are the different ways in which individuals with a particular trait may increase in a population ?
Answer:
There are three different ways in which individuals with a particular trait can increase in a population.

1. Survival Value (Natural Selection): The trait has survival value. It is picked up by natural selection. Through differential reproduction, it increases in population, e.g., green colour in beetles instead of red providing camouflage in bushes against being picked up by crows.
2. Genetic Drift: There is seasonal or accidental decline in population. The survivors have certain combination of traits which increase in number with the increase in population. The traits may not give any extra benefit to population.
3. Food: Individuals with particular trait may have extra abundance of food in their environment. They will naturally increase in number.

Question 8.
Why are traits acquired during the life time of an individual not inherited ?
(CCE 2011, 2012, 2013, CBSE A.I. 2017)
Answer:
Acquired traits are structural, functional and behavioural changes that an individual develops during its life time due to a particular environment, disease, trauma, use and disuse, conditioning or learning. The traits are not passed on to DNA of germ cells. They remain restricted to somatic cells. They are destroyed with the death of the individual. Therefore, intelligence, experiences and structural changes acquired during life time of an individual cannot pass to the progeny. Weismann (1892) cut the tails of mice for 21 generations but a tail still developed in 22nd generation.

Question 9.
Why are the small number of surviving tigers a cause of worry from the point of view of genetics ?
Answer:
A small population is always at a risk of degeneration and extinction due to

1. Excessive inbreeding that brings about inbreeding depression or degeneration,
2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
3. Lesser adaptability to changes in the environment,
4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 10.
What factors could lead to the rise of a new species ? (CCE 2012, 2014, CBSE Foreign 2017)
Answer:

1. Absence of gene flow amongst sub-populations due to the presence of physical barriers, long distance, differences in habitats, environmental and climatic conditions.
2. Accumulation of different variations in the different sub-populations of the species.
3. Natural selection of particular traits in a particular environment.
4. Genetic Drift. Separation of a small population, changes in its allele frequency, new mutations and adaptations to new habitat.
5. Reproductive Isolation. Accumulation of different variations and genetic drift result in absence of interbreeding in the previous subpopulations of a species. This results in the formation of new species. e.g., Finches of Galapogos islands.

Question 11.
Will geographical isolation be a major factor in the spéciation of a self pollinating plant species ? Why or why not ?
(CCE 2012)
Answer:
No. Geographical isolation has little role in spéciation of self pollinating plant species because there is already no gene flow among members of the species. Pea or Wheat which is self pollinated (due to pollination in bud condition) is not affected by any type of isolation. However, self pollinated plants can accumulate variations due to mutations and other factors and form new species.

Question 12.
Will geographical isolation be a major factor in the spéciation of an organism that reproduces asexually ? Why or why not ?
(CCE 2012)
Answer:
Recombination of genes is absent in asexually reproducing organisms. Therefore, variations originating in them do not get diluted but spread to all the» subsequent generations. Geographical isolation, which helps in spéciation due to formation of a separate gene pool, has no role in spéciation of asexually reproducing organisms.

Question 13.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
(CBSE Foreign 2017)
Answer:

1. Closeness of species is determined by presence or absence of fundamental characteristics and correlated characters. Two species of bacteria are closely related as they possess fundamental similarities of occurrence of nucleoid (instead of nucleus), absence of membrane covered cell organelles and presence of 70 S ribosomes. Human beings are close to monkeys because they possess similar eukaryotic multicellular body with vertebrate characters, mammalian traits and primate characters.
2. These days DNA matching is undertaken to find out the degree of closeness of the species.

Question 14.
Can the wing of a butterfly and the wing of a bat be considered homologous ? Why or why not ?
Answer:
No. Wings of butterfly and bat are fundamentally different in their origin and structure. In butterfly they are integumentary outgrowths having hollow tubes. In bat they are modified fore limbs which are covered by skin. Such organs which have a different origin and basic structure but are functionally similar are called analogous organs.

Question 15.
What are fossils ? What do they tell us about the process of evolution ? (CCE 2013, CBSE Foreign 2017)
Answer:
Fossils are remains or impressions of the past organisms that are found in the rocks of the old ages. They are often called written documents of evolution because they directly indicate the presence of different types of organisms in different ages. The path of evolution is known by arranging the fossils in a proper sequence age-wise. The early fossils are of simple organisms. Later on different complex forms arose, flourished and died down. They were replaced by newer forms. Study of fossils can also indicate the evolutionary stages of organisms. For example, modern horse (Equus) arose from a fossil animal Eohippus that existed on earth 60 million years back as a 30 cm high small animal. It evolved into 60 cm high goat sized Mesohippus about 40 million years back. Mesohippus gave rise to Merychippus (16-18 million years back) that formed Pliohippus (100-120 cm high, 10 million years ago). The modern horse evolved only 0-5 million years ago from Pliohippus.

Question 16.
Why are human being who look so different from each other in terms of size, colour and looks said to belong to same species ? (CBSE A.I. 2009 C, CBSE Foreign 2017)
Answer:
Delimitation of a species is based on the presence of a common gene pool, free inbreeding and reproductive isolation. Differences in size, colour and looks are based on preponderance of specific alleles and their interactions with the environment. All human beings, despite presence of different races, belong to same species {Homo sapiens) because they share the same gene pool, can marry amongst themselves and produce fertile offspring.

Question 17.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzee have a better body design ? Why or why not ?
Answer:
A better body design is the one which has more complexity, more elaboration and more controls which gives the organism a better competitive edge over others. There is no doubt that out of the four (bacteria, spiders, fish and chimpanzee), chimpanzee has a more elaborate body design or organisation. However, since body design is meant for competitive survival in their environment, all the four organisms or for that all living organisms, have a good body design that is suited to their environment.

NCERT Chapter End Exercises

Question 1.
A Mendelian experiment consisted of breeding tall Pea plants bearing violet flowers with short Pea plants bearing white flowers. The progeny all bore violet flowers but almost half of them were short. This suggests that genetic make up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) T_tWW
(d) T_tWw.
Answer:
(c) T_tWW.

Question 2.
An example of homologous organ is
(a) Our arm and a dog’s fore leg
(b) Our teeth and elephant tusks
(c) Potato and runners of grass
(d) All the above.
Answer:
(d) All the above.

Question 3.
In evolutionary terms we have more in common with
(a) A Chinese school boy
(b) A chimpanzee
(c) A spider
(d) A bacterium.
Answer:
(a) A Chinese school boy.

Question 4.
A study found that children with light coloured eyes are likely to have parents with light coloured eyes. On this basis can we say anything about whether the light eye colour is dominant or recessive ? Why or why not ?
Answer:
No. We cannot say with certainity whether the light eye colour is dominant or recessive. But since both the parents as well as the children have light eye colour, the probability is that it is a recessive trait. A recessive trait appears only when an individual possesses both the recessive alleles. As the parents are pure for the trait, the children also possess the trait and are pure for the same. Had the light eye colour been a domination trait, the recessive dark colour trait will have the chance to segregate and appear in some of the children.

Question 5.
How are the two areas of study, evolution and classification, interlinked ? (CBSE A.I. 2016)
Answer:
Classification is based on similarities and differences amongst organisms. The more characteristics two species have in common, the more closely related they are. They must have evolved from a common ancestor. Similarly more differences mean different adaptations and divergence from common ancestor in the remote past.

Question 6.
Explain the terms analogous and homologous organs with examples. (CBSE A.I. 2008 C)
Answer:
Analogous Organs: They are organs which have similar appearance and function but are quite different in their origin, development and anatomy.
Examples: Wings of Butterfly (integumentary outgrowths) and bird (modified fore-limbs).
Homologous Organs: They are organs which have similar origin, similar development and similar internal structure but have different forms and functions.
Examples: Fore-limbs of Horse, human hand, flipper of whale, wing of bird or bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:

1. Survey the dog population in and around your locality. Find out the percentage of different colours.
2. Observe the lineages where same colour is present in both parents and offspring. There is possibility that in these lineages both the alleles of coat colour are similar.
3. Allow crossing between two lineages having different coat colours.
4. Find the colour of F₁ individuals. It is probably the dominant coat colour.
5. Cross the F₁ dogs with the one having the other probably recessive colour. Is the ratio similar to test cross, Le., 1 : 1 ?

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

1. Fossils indicate the path of evolution of different groups.
2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
3. Some fossils have characteristics intermediate between two groups,
   e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 9.
What evidence do we have for the origin of life from inanimate matter ?
Answer:
Miller and Urey (1953) assembled an apparatus which had a spark chamber (for producing lightning), a flask for boiling and a condenser. They introduced a mixture of methane, ammonia, hydrogen and water into the apparatus. The gaseous mixture was exposed to electric discharges, boiling (800°C) and condensation with the temperature kept just below 100°C. The experiment was continued for a few days. At the end of one week, 15% of carbon (from methane) had been converted into simple organic compounds of amino acids, organic acids, sugars and nitrogen bases. It clearly proved that organic compounds or building blocks of life developed from inanimate matter in the remote past when the hot earth was cooling.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually reproducing organism?
Answer:
Variations arising during sexual reproduction occur due to

1. Chance separation of homologous chromosomes during gametogenesis.
2. Crossing over between homologous chromosomes.
3. Chance coming together of chromosomes during fertilisation,
4. Errors or mutations occurring during DNA replication.

Only the last method of variations is found in asexually reproducing organisms. Therefore, rate of appearance of variations is quite high in sexually reproducing organisms as compared to asexually reproducing organisms. Further, variations developed in sexually reproducing organisms are quite viable as most of them are due to reshuffling of genes. It is not so in asexually reproducing organisms. Here, most of the changes are harmful. They have a negative impact on evolution except when changing environment finds them useful. Because of the abundance and viability of variations, the rate of evolution is also high in sexually reproducing organisms.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny ?
Answer:
Most organisms are diploid. Their genetic material consists of two sets of chromosomes. Gametes carry single set of chromosomes, i.e., they are haploid. Sexual reproduction involves the formation and fusion of two types of gametes, male and female. Male gamete brings one set of chromosomes from the male parent. Female gamete also brings one set of chromosomes from the female parent. When two gametes fuse during sexual reproduction, the normal diploid chromosome complement is restored. It consists of 50% chromosomes from male parent and 50% chromosomes from female parent. Therefore, both the parents contribute equal genetic material to the offspring through formation and fusion of gametes.

Question 12.
Only variations that confer an advantage to an individual will survive in a population. Do you agree with this statement ? Why or why not ?
Answer:
No. Alongwith advantageous variations, a number of indifferent variations remain in the populations. Only the disadvantageous variations which are either lethal or extremely harmful are eliminated. All other variations persist in the population. Many of them function as preadaptations.

Selection Type Questions

Alternate Response Type Questions
(True/False (T/F), Right (√)/Wrong (x), Yes/No)

Question 1.
Mendel studied science and mathematics at the university of Amsterdam.
Question 2.
Both the parents contribute DNA equally to the offspring.
Question 3.
A factor which shows its effect in the hybrid is called recessive.
Question 4.
Sex of the child is determined by the type of ovum provided by the mother.
Question 5.
A recessive trait can also be common as blood group O.
Question 6.
Dromaesaurs were the first to fly.
Question 7.
Attached ear lobe is recessive trait.
Question 8.
Charles Darwin discovered the law of independent assortment,

Matching Type Questions

Question 9.
Match the articles given in columns A and B (single matching) :

Question 10.
Match the contents of columns I, II and III (double matching) :

Question 11.
What type of similarity, homologous (H) and analogous (A) occurs in the pairs of organs ?

Question 12.
Match stimulus with appropriate Response

Fill In the Blanks

Question 13.The term genetics was coined by ……………
Question 14.Mendel chose ………….. characters in Pea for his experiments.
Question 15. Broccoli has been developed from ……………… cabbage through artificial selection.
Question 16. ……………… Speciation occurs in geographically separated populations.
Question 17. Fossils are written documents of ………………..

Answers:

NCERT Solutions for Class 10 Science Chapter 9 – Heredity and Evolution

Hope given NCERT Solutions for Class 10 Science Chapter 9 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 10 – Light Reflection and Refraction solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 10 – Light Reflection and Refraction Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis where the parallel rays of light after reflecting from a concave mirror meet.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is focal length ? (Bihar Board 2012)
Answer:
Radius of curvature, R= 20 cm.

Question 3.
Name a mirror that can give an erect and magnified image of an object.
Answer:
A concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles ?
[CBSE (All India) 2007, 2011, 2012]
Answer:
This is because a convex mirror forms an erect and diminished (small in size) images of the objects behind the vehicle and hence the field of view behind the vehicle is increased.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = +32 cm. Therefore, f = R/2 = +32/2 = +16 cm.
Thus, focal length of the convex mirror = +16 cm.

Question 6.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?
Answer:
m – -3, But m = -v/u, so v = 3u
u = -10 cm
v = 3 (-10 cm) =-30 cm
Thus, the image is located at a distance of 30 cm to the left side of the concave mirror.

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?
Answer:
The ray of light bends towards the normal because the speed of light decreases when it goes from air (rarer medium) into water (denser medium).

Question 8.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3 x 10⁸ m s^-1 (CBSE 2011)
Answer:

Question 9.
You are given kerosene, turpentine and water. In which of these does the light travel faster ?
Answer:
We know, v = c/n. Refractive index (n) of water is 1.333, whereas refractive index of kerosene is 1.44 and that of turpentine is 1.47. As refractive index of water is least, so speed of light in water is more than in kerosene and turpentine. Hence, light travels faster in water.

Question 10.
The refractive index of diamond is 2.42. What is the meaning of this statement ? [CBSE (Delhi) 2008, 2012, 2013; Bihar Board 2012]
Answer:

Question 11.
Define 1 dioptre of power of a lens.
Answer:
Power = I/f (in m).
Power of a lens is 1 dioptre if focal length of the lens is 1 metre or 100 cm.

Question 12.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens.
Answer:

Question 13.
Find the power of concave lens of focal length 2m?
Answer:

NCERT Chapter End Exercises

Question 1.
Which one of the following materials cannot be tised to make a lens 1
(a) water
(b) glass
(c) plastic
(d) clay.
Answer:
(d). This is because clay is opaque (i.e. light cannot pass through it).

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ?
(a) between the principal focus and the centre of curvature
(b) at the centre of curvature
(c) beyond the centre of curvature
(d) between the pole of the mirror and its principal focus.
Answer:
(d).

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object ? (Bihar Board 2012)
(a) at the principal focus of the lens
(b) at twice the focal length
(c) at infinity
(d) between the optical centre of the lens and its principal focus.
Answer:
(b).

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of — 15 cm. The mirror and the lens are likely to be
(a) both are concave
(b) both are convex
(c) the mirror is concave and the lens is convex
(d) the mirror is convex but the lens is concave.
Answer:
(a).

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane only
(b) concave only
(c) convex only
(d) either plane or convex.
Answer:
(d).

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary ?
(a) a convex lens of focal length 50 cm
(b) a concave lens of focal length 50 cm
(c) a convex lens of focal length 5 cm
(d) a concave lens of focal length 5 cm.
Answer:
(c). Magnifying power of a reading glass (Convex lens) = 1/f.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case.
Answer:
A concave mirror produces an erect image if the object is placed between the pole and the focus of the concave mirror. Thus, object may be placed at any position whose distance is less than 15 cm from the concave mirror. The image is virtual and erect. The image is larger than the object. For a ray diagram, see figure 24.

Question 8.
Name the type of mirror used in the following situations :
(a) head lights of a car
(b) side rear view mirror of a vehicle
(c) solar furnace.
Support your answer with reason. (CBSE 2012, 2013)
Answer:
(a) Concave mirror. When a bulb is placed at the focus of a concave mirror, then the beam of light from the bulb after reflection from the concave mirror goes as a parallel beam which lights up the front road.
(b) Convex mirror. Image formed by a convex mirror is erect and small in size. The field of view behind the vehicle is large.
(c) Concave mirror. Concave mirror focuses rays of light coming from the sun at its focus. So, the temperature at the focus is raised.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ?
(CBSE 2015)
Answer:
A complete image of the object is formed as shown in figure.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Image is real and inverted.
Answer:

Here, u = -25 m, f = 10 m

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. Hou> far is the object placed from the lens ? Draw the ray diagram.
Answer:

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image. (CBSE 2014)
Answer:

Since, m is positive, so the orientation of both object and image is same. Thus,image is erect and virtual.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean ? (CBSE 2011, 2014)
Answer:

It means, size of the image formed by plane mirror is equal to the size of the object. Positive sign with m tells that both object and image are erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size. (Similar CBSE 2011)
Answer:
h = 5 cm, u = -20 cm


Since h’ is positive, so image is erect and virtual.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and the nature of the image.
Answer:

Since h’ is negative, so image is inverted.

Question 16.
Find the focal length of a lens of power -2.0 D. What type of lens is this ?
Answer:

The lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ? (CBSE 2012)
Answer:

Since focal length is positive, so the lens is converging.

Practical Skills Based Questions (2 Marks)

Question 1.
A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again ? How will the magnification of the image be affected ? (CBSE 2015)
Answer:

When distance of object u increases, V must decrease so that focal length of the lens remains the same. Thus, screen on which sharp image of object is formed is moved towards the lens.
As the distance of object u increases, the size of the image decreases. Hence, magnification of the imag decreases.

As u increases, v decreases and hence m also decreases.

Question 2.
In the given incomplete ray diagram for image A’B’ of a convex lens, what is the position of object AB ? Also complete the ray of diagram? (CBSE 2015)

Answer:
The position of object AB is between F₁ and 2F₁

Question 3.
To find the image-distance for varying object-distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen.
(a) In which direction towards or away from the lens, does he move the screen to focus the object ?
(b) What happens to the size of image does it increase or decrease ?
(c) What happens when he moves the object very close to the lens ?
Answer:

As distance of object decreases, distance of image V must increase so that focal length of lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases, therefore, magnification increases. In other words, size of the image increases.
(c) When object is very close to convex lens, virtual and magnified image is formed.

Question 4.
A student performed an experiment for the image formation by a convex lens at different positions of an object. If focal length of lens is 15 cm
Answer:

Question 5.
A student performed an experiment with convex lens and found the virtual image of an object. Find :
(a) position of the object.
(b) draw ray diagram for the above situation.
(CBSE 2015)
Answer:
(a) A convex lens forms a virtual image if the object lies between the optical centre and focus- of the convex lens. That is, the distance of the object from the less must be less than the focal length of the lens.
(b)

Question 6.
A student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
(A) In which direction does he move the lens to focus the flame on the screen 1
(BJ What happens to the size of the image of the a formed on the screen 1
(C) What difference is seen in the intensity (brightness) of the image of the a screen ?
(D) What is seen on the screen when the flame is very close (at about 5 cm) to the lens ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When the flame is at about 5 cm, which is less than the focal length (10 cm) of the lens.
In this case, magnified, erect and virtual image of the flame is formed. Since, virtual image cannot be obtained on the screen, so nothing is seen on the screen.

Question 7.
A student places a candle flame at a distance of about 60 cm from a convex lens of focal length 10 cm and focuses the image of the flame on a screen. After that he gradually moves the flame towards the lens and each time focuses the image on the screen.
(A) In which direction-toward or away from the lens, does he move the screen to focus the image ?
(B) How does the size of the image change ?
(C) How does the intensity of the image change as the flame moves towards the lens ?
(D) Approximately for what distance between the flame and the lens, the, image formed on the screen is inverted and of the same size ? (CBSE 2017)
Answer:

As the distance of the candle flame (object) u decreases, distance of image u must increase so that the focal length of the lens remains the same. Therefore, the screen must be moved away from the lens.

As u decreases, v increases therefore, magnification increases, other words, size of the image increases.
(C) As the size of the image increases, so the intensity (brightness) of the image of the flame on the screen decreases.
(D) When object is placed at 2F, a real and inverted image of same size as that of the object is formed at 2F on the other side of the lens. Therefore, in this case, the distance between flame and the lens is approximately 20 cm so that the image formed on the screen is inverted and of the same size.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

Hope given NCERT Solutions for Class 10 Science Chapter 10 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

NCERT Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 8 – How do Organisms Reproduce? solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 8 – How do Organisms Reproduce? Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
What is the importance of DNA copying in reproduction ? (CCE 2011, 2015)
Answer:
DNA carries hereditary information not only for controlling cellular functions but also all the structural and functional traits of organism. It is because of the latter that single celled zygote is able to form the whole multicellular organism. During reproduction there is formation of new cells which must carry the same amount and type of hereditary information as present in the parent cell. This is accomplished by DNA copying, which occurs prior to each cell division. DNA copying is not error proof. Errors give rise to variations.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Why are variations beneficial to the species but not necessarily for the individual ? (CCE 2011, 2012)
Answer:
Many of the variations are pre-adaptations which have no immediate benefit to the individuals. However, they remain in the population. Whenever, environment undergoes a drastic change, the pre-adaptations present in some members of the population allow the latter to survive, grow and regain its former size. Therefore, it is not necessary that variations are beneficial to individuals developing them but can prove useful to the species.

Question 3.
How does binary fission differ from multiple fission ?
Answer:

Question 4.
How will an organism be benefited if it reproduces through spores ? (CCE 2011, 2012)
Answer:
Sporulation or spore formation is a method of asexual reproduction where each individual produces a number of spores. On germination each spore forms a new individual, e.g., Rhizopus.

1. All the daughters formed through spores are genetically similar.
2. Spores are a means of dispersal. They help in spreading the organism far and wide.
3. Spores can also function as a means of perennation or passage through unfavourable conditions.

Question 5.
Can you think of reasons why more complex organism cannot give rise to new individuals through regeneration.
Answer:
Regeneration is the ability of an organism to replace lost or injured parts so as to form the whole individual from an incomplete form or fragment by remodelling and growth of somatic cells through dedifferentiation, division, morphogenesis and redifferentiation. The ability for regenerative multiplication is present in simpler organisms because most of their cells can undergo dedifferentiation. However, it is limited to certain cells in complex organisms.

1. The stem cells of complex organisms can form lost tissues and organs but not the complete individual as the highly differentiated tissues and organs do not allow this.
2. In complex organisms regeneration is under neurohormonal control. Fragments do not have nervous or hormonal stimulus to grow into complete organisms.

Question 6.
Why is vegetative propagation practised for growing some types of organisms ?
Answer:
Vegetative propagation is practised in a number of horticulturally and economically important plants because it is advantageous.
Advantages:

1. Seedless Plants. Vegetative propagation is the only known method of multiplication of seedless plants, g., Banana, Sugarcane, Pineapple, Jasmine, some varieties of Orange, Rose.
2. Uniform Yield. Seeds and fruits are of uniform quality, size, taste and aroma.
3. Genetic Uniformity. Vegetative propagation gives a genetically uniform population.
4. Good Qualities. Good qualities of a variety can be maintained indefinitely.
5. Survival Rate. Survival rate of the daughters is nearly 100% while in case of seed grown plants, it is 10%.
6. Quicker Method. Vegetatively reproduced plants bear flowers and fruits earlier than the plants raised through seeds. Potato requires only three months for forming a new crop if raised from tubers. It takes 15 months if raised from seeds.
7. Introduction in New Areas. In areas where seed germination fails to form mature plants, vegetative reproduction can help in establishing the plants.

Question 7.
Why is DNA copying an essential part of the process of reproduction ?
Answer:
Cell multiplication is essential for reproduction either as a means of multiplication in unicellular organisms or as a means of development of multicellular body from a single celled zygote. Cell multiplication cannot occur without DNA replication or DNA copying because each new cell must carry the full DNA complement.

Question 8.
How is the process of pollination different from fertilization ?
Answer:

Question 9.
What is the role of seminal vesicles and the prostate gland ? (CGE 2011, 2012)
Answer:
Seminal Vesicles: They secrete 60-70% of semen plasma that is alkaline and viscous having fructose (for nourishing the sperms), fibrinogen, proteins and prostaglandins. Prostaglandins cause movements in the genital tract of the female. Sperms are also activated by secretion of seminal vesicles.
Prostate Gland: It produces 20-30% of semen plasma. The secretion is alkaline and viscous. It has clotting enzyme and chemical essential for sperm activity.

Question 10.
What are the changes seen in girls at the time of puberty ?
Answer:

1. Breast size begins to increase. There is darkening of skin of nipples below the tips of breasts.
2. Menarche or beginning of menstruation.
3. Broadening of pelvis,
4. Deposition of fat in face, buttocks and thighs,
5. Increased vasculature of skin and hence increased warmth of skin,
6. Rounding of body ccthtours.
7. High pitched voice,
8. Slow growth of ovaries, fallopian tubes, uterus, vagina, enlargement of labia, etc.

Question 11.
How does the embryo get nourishment inside the mother’s body ? (CCE 2012)
Answer:
Embryo gets nourishment from mother’s body with the help of placenta through a cord called umbilical cord. Placenta contains many finger-like villi from the chorion covering of the embryo. They occur in contact with blood sinuses of the mother present in the endometrial lining of uterus. All nutrients (glucose, amino acids, vitamins, etc.) diffuse from mother’s blood into villi and from there to embryo through the umbilical cord.

Question 12.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases ?
Answer:
No, sexually transmitted diseases occur due to fluid to fluid contact that takes place in the vagina.

NCERT Chapter End Exercises

Question 1.
Asexual reproduction takes place through budding in
(A) Amoeba
(B) Yeast
(C) Plasmodium
(D) Leishmania.
Answer:
(B).

Question 2.
Which of the following is not a part of the female reproductive system in human beings.
(A) Ovary
(B) Uterus
(C) Vas deferens
(D) Fallopian tube.
Answer:
(C).

Question 3.
The anther contains
(A) Sepals
(B) Ovules
(C) Carpel
(D) Pollen grains.
Answer:
(D).

Question 4.
What are the advantages of sexual reproduction over asexual reproduction ?
Answer:
Asexual reproduction is monoparental, with no gametes, no meiosis and very little variations. Sexual reproduction is generally biparental involving fusion of gametes, meiosis and lot of variations.

Question 5.
What are the functions performed by testis in human beings ? (CCE 2011)
Answer:

1. Formation of sperms from germinal cells found in seminiferous tubules.
2. Secretion of hormone testosterone by Leydig cells. Testosterone induces secondary sexual characters at puberty. It helps in maintenance and functioning of secondary sex organs.

Question 6.
Why does menstruation occur ?
Answer:
Menstruation occurs in response to low level of estrogen and progesterone hormones which causes constriction of blood vessels in uterine wall, stoppage of nourishment to overgrown endometrium that sloughs off, passing out broken mucosal membrane, blood and mucus.

Question 7.
Draw a labelled diagram of L.S. flower.
Answer:

Question 8.
What are the different methods of contraception ?
Answer:

1. Mechanical Barriers like condoms, cervical cap, diaphragm.
2. Oral Contraceptives or oral pills like Mala D, Saheli
3. Intrauterine Contraceptive Devices (IUCD) like loop, bow, Cu-T.
4. Surgical Methods like vasectomy in males and tubectomy in females.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms ?
Answer:

Question 10.
How does reproduction help in providing stability to population of species ?
Answer:

1. Replication of DNA.
2. Growth and differentiation of cellular machinery.
3. Cell division. It is mode of reproduction in single celled organisms,
4. Development of special reproductive structures and formation of new individuals.
5. Continued replication of DNA, growth and cell division, formation of tissues, organs, etc. and maturation into a multicellular organism.

Question 11.
What could be the reasons for adopting contraceptive methods ?
Answer:

1. Enjoying a good reproductive health.
2. Protecting from sexually transmitted diseases.
3. Restricting the number of children.
4. Spacing the birth of children so as to properly look after them, provide them proper education without depleting the resources of the family.
5. Controlling population.

Selection Type Questions

Alternate Response Type Questions
(True/False (T/F), Right(√)/Wrong (x), Yes/No)

Question 1.
Basic event in reproduction is creation of DNA copy.
Question 2.
Plasmodium multiplies by binary fission.
Question 3.
Bryophyllum propagates through spore formation.
Question 4.
Copper-T is a contraceptive device used by women.
Question 5.
Hibiscus has unisexual flowers.
Question 6.
At the time of birth a girl baby has thousands of immature eggs.
Question 7.
Ovulation occurs in reproductively active females roughly in the middle of menstrual cycle.
Question 8.
Sperms mature at a temperature higher than that of human body.

Matching Type Questions

Question 9.
Match the articles given in columns A and B (single matching)

Question 10.
Match the contents of columns I, II and III (double matching) :

Question 11.
List the type of reproduction (A-asexual, V-vegetative, S-sexual) in the following organisms (Key or check list items)

Question 12.
Match each stimulus with appropriate response :

Fill In the Blanks

Question 13. ………………….. help in survival of the species in changing environment.
Question 14. ………………….. is common method of multiplication of Yeast and Hydra.
Question 15. Reproductive tissues begin to …………………….. when rate of general body growth begins to ………………… .
Question 16. Fallopian tubes are cut and ligated in ………………… .
Question 17. Bartholin’s glands are components of ……………………. reproductive system.

Answers:

NCERT Solutions for Class 10 Science Chapter 8 – How do Organisms Reproduce?

Hope given NCERT Solutions for Class 10 Science Chapter 8 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

In this chapter, students will come to know about early attempts at the classification of elements, Newlands Law Of Octaves, Mendeleev’s Periodic Table, the modern periodic table, metallic and non-metallic properties. At the end students are supposed to attempt question and answer exercise, multiple choice questions and group activity.

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Chemistry) Chapter 5 – Periodic Classification of Elements solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 5 – Periodic Classification of Elements Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

Periodic Classification of Elements Class 10 Science In Text Book Questions

Question 1.
Did Doberiener’s triads also exist in the columns of Newlands’ octaves ? Compare and find out.
Answer:
Yes, some of the Doberiener’s triads did exist in the columns of Newlands’ octaves. For example,

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
What were the limitations of Doberiener’s triads ?
Answer:
Doberiener was in a position to identify three triads. It could not apply to all the elements known at that time. Therefore, the classification was not so useful.

Question 3.
What were the limitations of Nawlands’ Law of Octaves ?
Answer:

1. Actually this classification was successful only upto the element calcium. After that, every eighth element did not possess the same properties as by the element lying above it in the same group. For example, the elements cobalt (Co) and nickel (Ni) placed below chlorine had different properties. Same was the case with copper (Cu) placed after potassium in the same group.
2. Newland committed another mistake. He placed two elements in the same slot in a particular group. For example, Co and Ni in the first group after chlorine. Similarly, elements cerium (Ce) and lanthanum (La) were placed after yitterium (Y) in the same group. Newland could not offer any explanation for such an arrangement.
3. Newland somehow thought that only 56 elements existed in nature and no more elements were likely to be discovered. But this belief ultimately proved to be wrong.
4. When noble gas elements were discovered at a later stage, their inclusion in the table disturbed the entire arrangement.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of the elements : K, C, Al, Si, Ba.
Answer:
Oxygen is a member of group VTA in Mendeleev’s periodic table. Its valency is 2. Similarly, the valencies of all the elements listed can be predicted from their respective groups. This can help in writing the formulae of their oxides.

1. Potassium (K) is a member of group IA. Its valency is 1. Therefore, the formula of its oxide is K₂O.
2. Carbon (C) is a member of group IVA. Its valency is 4. Therefore, the formula of its oxide is C₂O₄ or CO₂.
3. Aluminium (Al) belongs to groups IIIA and its valency is 3. The formula of the oxide of the element is Al₂O₃.
4. Silicon (Si) is present in group IVA after carbon. Its valency is also 4. The formula of its oxide is Si₂O₄ or SiO₂
5. Barium (Ba) belongs to group IIA and the valency of the element is 2. The formula of the oxide of the element is Ba₂O₂ or BaO.

Question 5.
Besides gallium which two other elements have since been discovered that fill the gaps left by Mendeleev in creating his periodic table ?
Answer:
Two other elements are scandium (Sc) and germanium (Ge). In their gaps, the elements with names Eka-boron and Eka-silicon were placed.

Question 6.
What was the criteria used by Mendeleev in creating his periodic table ?
Answer:
Mendeleev used atomic masses of the elements as the criteria for creating his periodic table. In this table, the elements were arranged in order of increasing atomic masses.

Question 7.
Why do you think that the noble gases should be placed in a separate group ?
Answer:
In the Mendeleev’s periodic table, the elements have been arranged in the different groups on the basis of valency. For example, the elements placed in group I (IA and IB) have Valency equal to one. Same is the
case with the elements placed in other groups. Since the noble gas elements He, Ne, Ar, Kr, Xe, and Rn have zero valency, they could not find a place in Mendeleev’s periodic table. These have been placed in a separate group called zero group in the periodic table. Please note that the noble gas elements were not a part of the Mendeleev’s periodic table. They were added later on.

Question 8.
How could the Modern Periodic table remove various anomalies of Mendeleev’s periodic table ?
Answer:
We have already studied that the Mendeleev’s periodic table is based on the atomic masses of the elements whereas Modern Periodic Table takes into account their atomic numbers. Since the properties of the elements are linked with the electronic configuration of their atoms (i.e., atomic number), this means that Modern Periodic Table is better than the Mendeleev’s Periodic Table. The important advantages are listed.

1. The position of the elements in the periodic table are linked with their electronic configuration.
2. Each group is an independent group and the idea of sub-groups has been discarded.
3. One position for all the isotopes of an element is justified since the isotopes have the same atomic number. For example, the three isotopes of the element hydrogen e., protium, deuterium and tritium have atomic number one (Z = 1). Similarly, two isotopes of chlorine i.e. Cl-35 and Cl-37 are placed in the same slot since they have same atomic number (Z = 17).
4. The positions of certain elements which were earlier misfits in the Mendeleev’s periodic table  are now justified because it is based on   atomic numbers of the elements.  For example, Ar precedes K because its atomic number (18) is less than that of K (19).
5. It is quite easy to remember and reproduce.

Question 9.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice ?
Answer:
Magnesium (Mg) belongs to group 2 known as Alkaline Earth Family. The two other elements belonging to the same group are calcium (Ca) and strontium (Sr). The basis of choice is the electronic distribution in the valence shell of these elements. All of them have two electrons each.

Question 10.
Name :
(a) three elements that have a single electron in their outermost shells.
(b) three elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Lithium, sodium, potassium (Alkali metals present in group 1)
(b) Beryllium, magnesium, calcium (Alkaline earth metals present in group 2)
(c) Helium, neon, argon (Noble gases present in group 18).

Question 11.
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements ?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common ?
Answer:
(a) The atoms of all these elements have one electron each in their valence shell. That is why, these elements are placed in the group 1 known as alkali metal group. The electronic configurations of these elements are given :

All the three elements evolve hydrogen gas on reacting with water
2 Li + 2 H₂O ———-> 2 LiOH + H₂
2 Na + 2 H₂O ————> 2 NaOH+ H₂
2 K + 2 H₂O ————> 2 KOH + H₂
Apart from this, all the elements happen to be the first elements of their respective periods. For example,

• Second period starts from lithium (Li)
• Third period starts from sodium (Na)
• Fourth period starts from potassium (K).

(b) Both these elements have completely filled shells. Helium (Z = 2) has two electrons in the only shell (K shell). The other element neon (Z = 10) has both K and L shells fully filled (2, 8). Because of the filled shells, the atoms of these elements do not have any desire to take part in chemical combination and they have been placed together in the same group known as group 18 or zero group.

Question 12.
In the modern periodic table, which are the metals among the first ten elements ?
Answer:
Metals among the first ten elements are lithium (Li) and beryllium (Be) . These are placed towards the left of the table.

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristics ?
Answer:
Before identifying the metallic character from the list of the elements, we must remember two points :

• The metallic character of an element is related to the electron releasing tendency of its atoms. Greater the tendency, more will be the metallic character.
• In general, metallic character of the elements increases down the group and decreases along a period.
  With the help of the Modern Periodic Table, let us identify the group and period to which these elements

Since the metallic character increases down a group and decreases along a period, the obvious choice is between two elements. These are Be (beryllium) present in group 2 and Ga (gallium) present in group 13. Now, the size of Ga is very big as compared to that of Be. Actually, the element Ga has three shells since it belongs to period 4 while the element Be has only two shells as it is a member of period 2. This means that the element Ge has a greater electron releasing tendency of its atom as compared to the element Be. Therefore, Ge has more metallic character, rather maximum metallic character among the elements that are listed.

Periodic Classification of Elements Class 10 Science NCERT End Exercise

Question 1.
Which of the following statements is not correct about the trends while going from left to right across the periodic table ?
(a) The elements become less metallic in nature
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c). The atoms lose their valence electrons with difficulty and not easily. This is on account of the reason that

• nuclear charge increases from left to the right since the atomic number of the elements gradually increases.
• with the increase in nuclear charge, the force binding the electrons with the nucleus increases. Therefore, the atoms lose their valence electrons with difficulty.

Question 2.
Element X forms a chloride with the formula XCI2 which is a solid with high melting point. X would most likely to be in the same group of the periodic table as :
(a) Na
(b) Mg
(c) Al
(d) Si.
Answer:
(b). The formula of the chloride of the element is XCl₂. This means that the valency of the element X is 2 since chlorine is monovalent. The element with valency 2 is expected to be present in group 2 to which magnesium (Mg) belongs.

Question 3.
Which element has :
(a) two shells, both of which are completely filled with electrons ?
(b) the electronic configuration 2, 8, 2 ?
(c) a total of three shells with four electrons in the valence shell ?
(d) a total of two shells with three electrons in the valence shell ?
(e) twice as many electrons in the second shell as in the first shell ?
Answer:
(a) The elements with completely filled shells are noble gas elements and they belong to group 18. Since the element has two shells ; it must be present in second period and is neon (Ne) with electronic configuration 2, 8.
(b) The electronic configuration suggests that the element belongs to third period and second group. It is therefore, magnesium (Mg).
(c) The element with three shells is present in third- period. Since it* has four electrons in the valence
shell, it must belong to group 14 and is silicon (Si) with electronic configuration 2, 8, 4.
(d) The element with two shells is expected to be present in the second period. With three electrons in
the valence shell, it must belong to group 13 and is boron (B) with electronic configuration 2, 3.
(e) The element has only two shells. The first shell can have a maximum of two electrons. The second shell has four electrons which is twice the number of electrons present in the first shell. Therefore, the electronic configuration of element is 2, 4. It is carbon with atomic number (Z) equal to 6.

Question 4.
(a) Which property do all elements in the same column of the periodic table as boron have in common ?
(b) Which property do all elements in the same column of the periodic table as fluorine have in common ?
Answer:
(a) The element boron (B) is the first member of group (also called column) 13. It has three electrons in
the valence shell (2, 3). The other elements included in the same column are aluminium (Al),
gallium (Ga), indium (In) and thalium (Tl). They too have three electrons in the valence shell of their atoms. Just like boron, these elements also show a valency of 3 in their compounds.
(b) The element fluorine (F) is the first member of group (also called column) 17. It has seven electrons in the valence shell (2, 7). The other members present in the same group known as halogen family are chlorine (Cl), bromine (Br), iodine (I) and astatine (At). They have also seven electrons in the valence shell of their atoms. Like fluorine, they all show a valency of 1 in their compounds.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element ?
(b) To which of the following elements would it be chemically similar ? (Atomic numbers are given in parentheses).
N(7), F(9), P(15), Ar(18).
Answer:
(a) The atomic number of the element is 17 (2 + 8 + 7 = 17).
(b) It would be chemically similar with fluorine (F) which has also 7 electrons in valence shell (2, 7)

Question 6.
The position of three elements A, B and C in the periodic table are shown below :

(a) State whether A is metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B ?
(d) Which type of ion, cation or anion will be formed by the element A ?
Answer:
(a) Group 17 represents halogen family. All the elements included in the family are Therefore, element A is a non-metal.
(b) Reactivity of non-metals is generally due to the electron accepting tendency of their atoms. Down the group, the atomic size increases. Therefore, the attraction of the nucleus for the outside electrons decreases. This means that down the group of non-metals, reactivity decreases. Thus, the element C is less reactive than the element A.
(c) Atomic size of the elements decreases along a period. The elements B and C are present in the same period. Since C is placed after B, the size of the element C is less than that of B.
(d) The elements A, as pointed out earlier is a non-metal which belongs to group 17. It has seven valence electrons (2, 8, 7). In order to have the configuration of the nearest noble gas element, it will take up one electron and change to anion i.e., A^– ion.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write their electronic configuration. Which of these will be more electronegative and why ? (CBSE 2012)
Answer:
The electronic configurations of the two elements are :
Nitrogen (Z = 7) 2, 5 ;
Phosphorus (Z = 15) 2, 8, 5
Since the size of nitrogen is small as compared to phosphorus, it has a greater tendency to take up electrons. It is therefore, more electronegative than phosphorus.

Question 8.
How does the electronic configuration of an atom relate to its position in the modern periodic table ?
(CBSE Delhi 2011)
Answer:
The modern periodic table is based upon atomic numbers of the elements. Since electronic configurations of the elements depend upon their atomic numbers, this means that the periodic table is based on the electronic configurations of the elements. For example, all the alkali metals have one electron each in their valence shell. These are placed in group 1. Similarly, the alkaline earth metals with two electrons in their valence shell are placed in group 2 and so on.

Question 9.
In the modern periodic table, calcium (Z = 20) is surrounded by the elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium ?
(CBSE All India 2011)
Answer:
Only those elements are placed in the same group in which the gaps of atomic numbers are : 8, 8, 18, 18, 32. If we look at the atomic numbers of the elements that are listed, it becomes clear that the elements with atomic numbers 12, 20 (Ca), 38 fit into this pattern. They are placed in the same group and have also similar physical and chemical properties.

Question 10.
Compare and contrast the arrangement of elements in Mendeleevs periodic table and the modern periodic table.
Answer:
The main points of distinction between Mendeleevs periodic table and Modern periodic table are as follows :

NCERT Solutions for Class 10 Science Chapter 5 – Periodic Classification of Elements

Hope given NCERT Solutions for Class 10 Science Chapter 5 helpful to you.