CBSE Class 10 Science Chapter 8 How do Organisms Reproduce Short Answer Questions

CBSE Class 10 Science Chapter 8 How do Organisms Reproduce

Sexual Reproduction In Humans:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Sexual Reproduction In Humans

CBSE Class 10 Science Chapter 8 How do Organisms Reproduce Short Question And Answers

Question 1. What is the importance of DNA copying in reproduction?
Answer:

Reproducing organisms create new individuals that look very much like themselves. The daughter organisms and their parents look similar because their body design is similar.

  • Reproduction at its most basic level involves making copies or the blueprint of the body design. The chromosomes in the nucleus of a cell contain information for the inheritance of features from parents to the next generation in the form of DNA (Deoxyribonucleic Acid, molecules).
  • So, reproducing cells prepare copies of DNA molecules to keep one DNA copy with the parent cell and other copy/copies to pass on to the next generation/daughter cells.

Read and Learn More CBSE Class 10 Science Short Answer Questions

Question 2. Why is variation beneficial to the species but not necessary for the individual?
Answer:

A variation may not be beneficial for an individual that is not suitable for the niche. In this case, an individual or a few variants may not survive. This does not affect the existence of a species.

  • Variations are useful for the survival of species in changed environmental situations.
  • For example, if a population of reproducing organisms were suited to a particular niche (well-defined place of abode) and if the niche is drastically changed, the population could be wiped out.
  • However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive.

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

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Binary Fission From Multiple Fission

Question 4. How will an organism be benefited if it reproduces through spores?
Answer:

The nuclei in the hypha of fungus like Mucor and Rhizopus, get surrounded by a bit of cytoplasm and get a thick wall. This wall protects the spore during unfavourable conditions.

When the favourable conditions return, the spores develop into new hyphae. Thus, spores help the fungus tide over unfavourable conditions.

Question 5. Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:

In highly developed multicellular organisms, different cell types perform different specialised functions. Reproduction in such organisms is also the function of a specific cell type.

  • If the organism itself consists of many cell types, reproduction by regeneration or fragmentation cannot be achieved.
  • In multicellular organisms like Hydra and Planaria regeneration is carried out by specialised cells.
  • These cells proliferate and make large numbers of cells. From this mass of cells, different cells change to become various cell types and tissues.
  • Such specialised cells which can regenerate or develop a new individual from their body part do not exist in complex organisms.

Question 6. Why is vegetative propagation practised for growing some types of plants?
Answer:

Reasons for growing some types of plants by vegetative propagation:

  1. Plants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds.
  2. Such methods also make possible the propagation of plants such as bananas, oranges, roses and jasmine that have lost the capacity to produce seeds.
  3. All plants produced by this method are genetically similar enough to the parent plant to have its all characteristics.

Question 7. Why is DNA copying an essential part of the process of reproduction?
Answer:

Reproducing organisms create new individuals that look very much like themselves. They look similar because their body designs are similar.

  • The DNA present in chromosomes located in the nucleus of a cell contains information for the inheritance of features from parents to the next generation.
  • So, reproducing cells creates two copies of the DNA. One copy of DNA remains in the reproducing original cell. The other copy of DNA is passed to the daughter cell formed by the parent having additional cellular apparatus.

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

Pollination is the process by which the pollen from another reaches the stigma of the flower of the same species. Fertilisation is the process in which the male gamete fuses with the egg to form a zygote.

Question 9. What is the role of the seminal vesicles and the prostate gland?
Answer:

Seminal vesicles secrete a viscous fluid which forms most of the part of semen. It also lubricates the passage through which the sperms travel.

  • This fluid also protects the sperm from the acids normally present in the urethra and female reproductive duct. The prostate gland produces prostatic fluid which is discharged into the urethra through more than two dozen fine ducts.
  • There is more secretion during sexual act when they are forcefully ejected by contraction of muscular and elastic tissues.

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

The females show development of breasts, darkening of nipples, menstruation and growth of public hair.

Question 11. How does the embryo get nourishment inside the mother’s body?
Answer:

After fertilisation, the lining of the uterus thickens and is richly supplied with blood to nourish the growing embryo.

  • The embryo gets nutrition from the mother’s blood with the help of a special tissue called the placenta. This is disc-shaped tissue which is embedded in the uterine wall.
  • The placenta contains villi on the embryo’s side of the tissue and blood spaces on the mother’s side surrounding the villi. This provides a large surface area for food and oxygen to pass from the mother to the embryo and waste products from the embryo to the mother’s diseases.

Question 12. If a woman is using a copper-T, will it help protect her from sexually transmitted
Answer:

Copper-T is a contraceptive method which prevents implantation of the zygote inside the uterus. It cannot prevent a woman from sexually transmitted diseases (STDs). STDs are transmitted by contact which cannot be prevented by copper-T.

Question 13. What are the advantages of sexual reproduction over asexual reproduction? Answer:

Sexual reproduction incorporates the process of combining DNA from two different individuals during reproduction.

  • Sexual reproduction promotes variation among the offspring of the same parents and provides opportunities for new combinations of characters, thus playing an important role in the origin of new species.
  • Asexual reproduction does not require a combination of DNA from two different individuals. So, it does not create variations and does not help in creating new species and evolution.

Question 5. What are the functions performed by the testes in human beings?
Answer:

Testes are the sites for the formation of sperms. They also produce a hormone named testosterone which is responsible for secondary sex characteristics in males.

Question 14. Why does menstruation occur?
Answer:

The ovary releases one egg every month and the uterus prepares for the implantation of the zygote by thickening its walls which are richly supplied by blood vessels.

If the egg is not fertilised, the uterine lining slowly breaks down and comes out through the vagina as blood and mucus. This cycle takes place regularly every month and is known as menstruation and lasts for about two to eight days.

Question 15. Draw a labelled diagram of the longitudinal section of a flower.
Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce The Longitudinal Section Of A Flower

Question 16. What are the different methods of contraception?
Answer:

The methods used for the regulation of childbirth are:

  1. Barrier method: In this method, physical devices like condoms, cervical caps and diaphragm are used to prevent sperm from reaching upto the ovum.
  2. Chemical method: In this method, specific drugs are used by females. These drugs may be (a) oral pills or (b) vaginal pills. Oral pills mainly contain hormones and are called oral contraceptives.
  3. Intrauterine contraceptive devices (IUCDs): IUCDs like Copper-Tis are placed safely in the uterus by a skilled doctor. IUCDS prevent implantation of the fertilized ovum inside the uterus.
  4. Surgical method: This method applies to both males and females. In males, a small portion of vas deferens and in females the fallopian tube is surgically removed or ligated (tied). It is known as vasectomy in males which prevents the release of sperm from the testis. In females, it is called tubectomy.

Question 17. How are the modes of reproduction different in unicellular and multicellular organisms?
Answer:

The modes of reproduction in unicellular organisms are:

  1. Binary fission is found in unicellular animals like Amoeba, Paramecium, Euglena, etc. After attaining an optimum size the adult individual undergoes a simple division into two daughters of equal size. It consists of the division of the nucleus followed by the cleavage of cytoplasm into equal parts, each having a daughter nucleus. Binary fission can occur in any place in Amoeba.
  2. Multiple fission is also a mode of asexual reproduction found in unicellular animals or protozoans, e.g., Plasmodium, and Monocystis. The nucleus of the parent cell undergoes repeated divisions to produce several daughter nuclei. The cytoplasm cleaves around each nucleus to form many daughter cells. Ultimately under favourable conditions, the parent cell membrane ruptures and daughter individuals are released. Sometimes, the daughter cells derived from multiple fission develop resistant coverings to produce spores. The process is called sporulation.
  3. Budding: Some unicellular organisms such as yeast can put out small buds that separate and grow further.

The mode of reproduction in multicellular organisms

  1. Fragmentation: Simple multicellular organisms such as spirogyra simply break up into smaller pieces upon maturation. These pieces, known as fragments, grow into new individuals.
  2. Regeneration: In some fully differentiated organisms—an individual, such as Hydra and Planaria, if somehow cut or broken up into many pieces, many of these pieces grow into separate individuals, Regeneration is carried out by specialised cells.
  3. Budding: Organisms, such as Hydra use regenerative cells for reproduction in the process of budding. In Hydra, an outgrowth, called bud, develops at a specific site. When their buds mature, they detach from the parent body and become new independent individuals.
  4. Sexual reproduction: This mode of reproduction is common in complex multicellular organisms. In this mode of sexual reproduction, both sexes-males and females- are needed to produce a new generation. Thus, the sexual mode of reproduction involves two individuals.

Question 18. How does reproduction help in providing stability to populations of species?
Answer:

Variations in the population that help the population of a species to tide over adverse environmental conditions are induced by reproduction, especially by sexual reproduction. These variants are adapted to the changing environment of the niche. Reproduction also helps generate copies of individuals that are suited to a particular environment which is existing.

Question 19. What could be the reasons for adopting contraceptive methods?
Answer:

Reasons for adopting contraceptive methods

  1. The sexual act always has the potential to lead to pregnancy. If a woman (mother) is not in good health or mentally unprepared, it will cause adverse effects on the mother and the foetus.
  2. There are many diseases such as AIDS, gonorrhoea, and syphilis which are transmitted through sexual acts. These can be prevented by the use of contraceptives.
  3. The reproduction process causes an increase in population if adequate measures are not taken to prevent pregnancy. The increase in population in our country is increasing exponentially. This puts strains on the natural resources of the country and is a major cause of poverty in our country.

Thus, adopting contraceptives can prevent:

  1. Unwanted pregnancies.
  2. Spread of sexually transmitted diseases (STDs) and lethal diseases like AIDS and HIV.
  3. Frequent pregnancies may cause health hazards to the mother and her infant.

Question 20. List any two modes of asexual reproduction in animals. Under which mode of reproduction is vegetative propagation placed and why? List two advantages of vegetative propagation.
Answer:

Two asexual modes of reproduction in animals are:

Fission/Fragmentation/Budding/Regeneration (Any Two) Vegetative propagation is under an asexual mode of reproduction.

Reasons: Reproductive part of the plant not involved/Single parent involved.

Advantages of vegetative propagation (Any two)

  1. Plants raised can bear flowers and fruits earlier than those produced from seeds.
  2. Plants that have lost the capacity to produce seeds can be produced.
  3. All plants produced are genetically similar to the parent plant.

Question 21. State two examples of each of the following:

  1. Plants having unisexual flowers
  2. Agents of pollination
  3. Physical changes that are common to both boys and girls.

Answer:

  1. Papaya, watermelon
  2. Wind, water, air, insects (any two)
    1. Increase in height and weight. (Body becomes larger)
    2. Thinner hair appears on the legs, arms as well as on the face.
    3. Skin becomes oily and pimples develop.

Question 22. What is the placenta? State its function in human females.
Answer:

Placenta: A special tissue embedded in the uterus wall with the help of which the human embryo gets nutrition from the mother’s blood.

Functions:

  1. To provide a large surface area for glucose and O2 to pass through the mother’s blood to the embryo.
  2. To metabolic waste generated by the embryo is removed through the placenta into the mother’s blood.

Question 23. Variations are important for the survival of species over time. Justify this statement with reasons.
Answer:

Variations help the individuals to survive even after the drastic changes occur in nature. These changes may be in the niche, temperature, salinity water levels, etc.

Question 24. Mention the changes seen in the girls that depict the aspect of the sexual maturation of the body.
Answer:

  1. Growth of pubic hair. (armpits and the genital area)
  2. Breast size begins to increase with darkening of the skin of the nipples at the tips of the breast.
  3. The hips widen.
  4. Menstruation starts.

Question 25. Justify the following statements:

  1. Variations are beneficial for the species over some time.
  2. New offspring produced are similar to their parents but not identical.
  3. Binary fission is different in Amoeba and Leishmania.

Answer:

  1. Variations are beneficial for species because they give survival advantages even in adverse environmental conditions.
  2. Similar-Same DNA Not identical-Due to variation during DNA copying
  3. Amoeba-Any plane; Leishmania-definite orientation.

Question 26. Name and explain the method by which rhizopus reproduces.
Answer:

Rhizopus reproduces by spore formation.

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Spore Formation In Rhizopus

Method:

It is a method of asexual reproduction. Spores are produced in a structure called sporangium. The nucleus divides several times within the sporangium and each nucleus with a bit of cytoplasm, develops into a spore.

Question 27. What is the effect of DNA copying, which is not perfectly accurate, on the reproduction process? How does the amount of DNA remain constant though each new generation is a combination of DNA copies of two individuals?
Answer:

  1. DNA copying leads to more variations in the offspring.
  2. Sexually reproducing organisms have a special lineage of cells which have only half the amount of DNA/half the number of chromosomes as the parent cell.
  3. When such germ cells fuse, a new individual is formed with the same amount of DNA as that of a parent.

Question 28. Explain how organisms create an exact copy of themselves.
Answer:

To build the copies of DNA or the genetic material, the cells use biochemical reactions. Additional cellular apparatus along with the DNA copies are separated and so a cell divides to give rise to two almost identical cells.

Question 29. State the changes that take place in the uterus after the implantation of a young embryo. How does this embryo get nourishment inside the mother’s body?
Answer:

  1. The uterine wall thickens, i.e., richly supplied with blood.
  2. A special tissue called the placenta develops which connects the embryo to the uterine wall that provides nutrients and oxygen to it.
  3. The placenta is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue. On the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo.

Question 30.

  1. What is the location of the following:
    1. DNA in a cell
    2. Gene
  2. Expand DNA.

Answer:

    1. In the nucleus of a cell.
    2. Located on the chromosomes.
  1. Deoxyribonucleic acid

Question 31. What are chromosomes? Explain how in sexually reproducing organisms, the number of chromosomes in the progeny is maintained.
Answer:

  1. Chromosomes are thread-like structures made up of DNA found in the nucleus.
  2. The original number of chromosomes becomes half during gamete formation. Hence, when the gametes combine, the original number of chromosomes gets restored in the
    progeny.

Question 32.

  1. What are sexually transmitted diseases? Name anyone who is caused by bacteria
    and one caused by viral infection.
  2. Mention any two methods to avoid such diseases.

Answer:

  1. Diseases that spread through sexual contact are called sexually transmitted diseases.
    1. Gonorrhoea or syphilis
    2. Warts, HIV, AIDS
    1. Use of condoms on the penis.
    2. Use covering on a vagina.

Question 33. Differentiate between plumule and radicle.
Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Difference Between Plumule And Radicle

Question 34. Explain the process of regeneration in planaria. How is this process different from reproduction?
Answer:

  1. Planaria gets cut into many pieces, each piece grows into separate individuals, and specialized cells proliferate. The mass of cells gets differentiated into various cells and tissues.
  2. Regeneration is not the same as reproduction as most of the organisms would not normally depend on being cut up to be able to reproduce. (It is normally for repair of the damaged part.)

Question 35. Distinguish between the functions of the ovary and testis.
Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Difference Between Ovary And Testis

Question 36. Differentiate between the following:

  1. Placenta and uterus
  2. Unisexual and bisexual flowers
  3. Fertilization and germination

Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Difference Between Placenta And Uterus

Question 37. A part of the male reproductive system is shown below. Study the diagram and answer the questions that follow.

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Male Reproductive System

  1. Two parts have been incorrectly labelled. Identify them.
  2. State the function of the urethra.
  3. Which hormone is released by the testis?

Answer:

  1. Testis, Penis
  2. It serves as a common passage for both sperm and urine.
  3. Testosterone

Question 38. How can pregnancy be prevented surgically?
Answer:

  1. When vas deferens in males are blocked surgically, sperm transfer will be prevented.
  2. When fallopian tubes are blocked, eggs will not be able to reach the uterus.

Question 39.

  1. State, in brief, the functions of the following parts of the human female reproductive system:
    1. Ovary
    2. Fallopian tube
    3. Uterus
  2. State, in brief, the functions of the following parts of the human male reproductive system:
    1. Scrotum
    2. Testes
    3. Vas deferens

Answer:

    1. Ovary-Production of ova and sex hormones
    2. Fallopian tube-Site of fertilization
    3. Uterus-Keeps the foetus till complete development.
    1. Scrotum–Protects testes outside the abdominal cavity.
    2. Testes-Produces male gametes called sperms.
    3. Vas deferens-Delivers sperms from the testes to the urethra.

Question 40. What is the function of copper-T used by some women? What is its side effect?
Answer:

It prevents pregnancy as it prevents implantation in the uterus. It can cause side effects due to irritation of the uterus.

Question 41.

  1. How do organisms reproduce by fission?
  2. Write the names of any two organisms which reproduce by this method.
  3. Differentiate between the fission of Leishmania and Plasmodium.

Answer:

  1. Cell division takes place which leads to the creation of new individuals.
  2. Bacteria, Protozoa
  3. Leishmania-binary fission occurs in a definite orientation. Plasmodium undergoes multiple fission.

Question 42. Differentiate between binary fission and multiple fission.
Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Difference Between Binary Fission And Multiple Fission

Question 43. State the importance of chromosomal differences between sperms and eggs of humans.
Answer:

  1. Eggs always contain the same type of sex chromosomes (both X). Sperms contain X or Y sex chromosomes. Thus, sperm containing the X chromosome when combines with the X chromosome of the egg results in a female child.
  2. Whereas sperm containing the Y chromosome when combines with the X chromosome of the egg results in a male child.

Question 44. What happens to the following parts after fertilization?

  1. Ovum
  2. Ovary
  3. Ovule
  4. Sepals and petals

Answer:

  1. Forms zygote
  2. Forms fruit
  3. Forms seed
  4. Shrivel and fall off

Question 45. Differentiate between self-pollination and cross-pollination.
Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Difference Between Self Pollination And Cross Pollination

Question 46. Which kind of contraceptive method prevents STDs and how?
Answer:

  1. Barrier method
  2. Because by this method there is no direct contact between the genital organs of male and female and thus it prevents transmission of any infection.

Question 47.

  1. Identify the asexual methods of reproduction in each of the following organisms:
    1. Rose
    2. Yeast
    3. Planaria
  2. What is fragmentation? Name a multicellular organism which reproduces by this method.

Answer:

    1. Vegetative propagation by stem
    2. Budding
    3. Regeneration
  1. Fragmentation is an asexual method of reproduction in which an organism simply breaks up into smaller pieces upon maturation. These pieces or fragments grow into new individuals.

Spirogyra is a multicellular organism with a relatively simple body organisation which reproduces through this simple method of reproduction, i.e., fragmentation.

Question 48. List four points of significance of reproductive health in a society. Name any two areas related to reproductive health which have improved over the past 50 years in our country.
Answer:

Significance:

  • Prevents STDs
  • Advantages of small family
  • Less mortality among newborns
  • Reduces the cases of maternal mortality

Areas which have improved: Family planning, decrease in STD cases.

Question 49. In a bisexual flower despite the young stamens being removed artificially, the flower produces fruit. Provide a suitable explanation for the above situation.
Answer:

The pistil is intact. Cross-pollination has occurred leading to fertilisation and formation
of fruit.

Question 50. Can you consider cell division as a type of reproduction in unicellular organisms? Give one reason.
Answer: Yes, because it results in the formation of two daughter cells, that is, it results in the production of more individuals of the organism.

Question 51. What is a clone? Why do offspring formed by asexual reproduction exhibit remarkable similarity?
Answer: Clone refers to the offspring of an organism formed by the asexual method of reproduction. Since they possess exact copies of the DNA of their parent, clones exhibit remarkable similarity.

Question 52. Explain how offspring and parents of organisms reproducing sexually have the same number of chromosomes.
Answer:

Reduction division (meiosis) during gamete formation halves the chromosome number in both male and female gametes. Since these two gametes fuse during fertilisation, the original number of chromosomes (as in the parent) is restored in the offspring.

Question 53. Colonies of yeast fail to multiply in water but multiply in sugar solution. Give one reason for this.
Answer:

Sugar provides energy for sustaining all life activities in yeasts. In water, it fails to
reproduce because of inadequate energy in its cells.

Question 54. Why does bread mould grow profusely on a moist slice of bread rather than on a dry slice of bread?
Answer:

Moisture is an important factor in the growth of hyphae. Moistened bread slice offers both moisture and nutrients to the bread mould, hence it grows profusely. A dry slice of bread offers nutrients but not moisture hence hyphae fail to grow.

Question 55. Give two reasons for the appearance of variations among the progeny formed by sexual reproduction.
Answer:

  1. Sexual reproduction involves two parents with different sets of characters.
  2. The gene combinations are different in gametes.

Question 56. Would a Planaria cut vertically into two halves regenerate into two individuals? Complete the figure by indicating the regenerated regions.

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce A Planaria Cut Vertically Into Two Halves Regenerate

Answer:

Yes, the shaded part in the figures represents the regenerated halves.

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Regenerated Halves

Question 57. From the internet, gather information about the chromosome numbers of five animals and five plants. Correlate the number with the size of the organism and answer the following questions.

  1. Do larger organisms have more number of chromosomes cells?
  2. Can organisms with fewer chromosomes reproduce more easily than organisms with more chromosomes?
  3. The more the number of chromosomes/cells, the greater the DNA content. Justify.

Answer:

  1. No, there is no relationship between the size of the organism and its chromosome number.
  2. No, the process of reproduction follows a common pattern and is not dependent on the number of chromosomes.
  3. Yes, since the major component of chromosomes is DNA, if there are more chromosomes in a cell, the quantity of DNA will also be greater.

Question 58. In tobacco plants, the male gametes have twenty-four chromosomes. What is the number of chromosomes in the female gamete? What is the number of chromosomes in the zygote?
Answer:

  1. The number of chromosomes in the female gamete is 24.
  2. Number of chromosomes in zygote is 48.

Question 59. Why cannot fertilisation take place in flowers if pollination does not occur?
Answer:

  1. In a flower, fertilisation requires both male and female gametes.
  2. If pollination does not occur, male gamete is not available hence fertilisation cannot take place.

Question 60. Is the chromosome number of zygote, embryonal cells and adult of a particular organism always constant? How is the constancy maintained in these three stages?
Answer:

Yes, the constancy is maintained because cells in all these three structures undergo only mitotic divisions.

Question 61. Where is the zygote located in the flower after fertilization?
Answer: Zygote is located inside the ovule which is present in the ovary.

Question 62. Reproduction is linked to the stability of the population of a species. Justify the statement.
Answer:

In reproduction, DNA passes from one generation to the next. Copying of DNA takes place with consistency but with minor variations. This consistency leads to the stability of species.

Question 63. How are general growth and sexual maturation different from each other?
Answer:

General growth refers to different types of developmental processes in the body like an increase in height, weight gain, and changes in shape and size of the body, but sexual maturation is specific to changes reflected at puberty like cracking of voice, new hair patterns, development of breasts in females etc.

Question 64. Trace the path of sperm during ejaculation and mention the gland and its functions associated with the male reproductive system.
Answer:

Sperm comes out from the testis into the vas deferens and then passes through the urethra before ejaculation. The secretions of seminal vesicles and prostate glands provide nutrition to the sperms and also facilitate their transport.

Question 65. What changes are observed in the uterus if fertilisation does not occur?
Answer:

The thick and spongy lining of the uterus slowly breaks and comes out through the vagina as blood and mucus.

Question 66. What changes are observed in the uterus after the implantation of the young embryo?
Answer:

The uterine wall thickens and is richly supplied with blood. A special tissue called the placenta develops which connects the embryo to the uterine wall that provides nutrients and oxygen to it.

Question 67. What are the benefits of using mechanical barriers during sexual acts?
Answer:

Mechanical barriers like condoms prevent the sperm from reaching the egg. Thus it is an effective method to avoid pregnancy. It also prevents the transmission of infections during sexual acts.

Question 68. In the given figure label the parts and mention their functions.

  1. Production of egg
  2. Site of fertilisation
  3. Site of implantation
  4. Entry of the sperms
  5. Answer:

CBSE Class 10 Science Chapter 8 How Do Organisms Reproduce Female Reproductive System

  1. Ovary (production of egg)
  2. Oviduct (site of fertilisation)
  3. Uterus (site of implantation)
  4. Vagina (entry of the sperm)

Question 69. What would be the ratio of chromosome number between an egg and its zygote? How is the sperm genetically different from the egg?
Answer:

The ratio is 1: 2. Sperms contain either X or Y chromosome whereas an egg will always have an X chromosome.

Question 70. Green and red-coloured seeds are recessive and dominant traits respectively. Out of F1 and F2, in which generation will the green seed appear, if both parents are not hybrid?
Answer: F2 generation.

Question 71. What is self sterility mechanism in flowers which allows only cross-pollination?
Answer:

In self-sterility, pollen grains of the same flower are not capable of fertilising that flower, e.g. potato, tea etc. Hence, cross-pollination is the rule applied.

Question 72. Colonies of yeast fail to multiply in water but multiply in sugar solution. Give one reason for this.
Answer:

Sugar provides energy for sustaining all life activities in yeast. In water, it fails to reproduce because of inadequate energy in its cells.

Question 73. What is the main cause of variations among organisms during sexual reproduction?
Answer:

During sexual reproduction, the DNA copying mechanism is not accurate, and the resultant errors are a source of variations in populations of organisms.

Question 74. What is the advantage of reproduction through spores in the case of rhizopus?
Answer:

  1. The spores are covered by thick walls that protect them until they come into contact
    with another moist surface and can begin to grow.
  2. Spores being lighter easily get dispersed to distant places and provide better chances of survival.

Question 75. We hear and read about female foeticide, which is a wrong practice. In some families, be it rural or urban, females are tortured for giving birth to a girl child. They do not seem to understand the scientific reason behind the birth of a boy or a girl.

Answer the following questions based on the information given above.

  1. In your opinion, is the approach of society towards the mother in this regard correct? Explain the scientific reason.
  2. How can you promote awareness among people?
  3. What values will you be promoting by doing so?

Answer:

  1. In my opinion approach of society is baseless. The sex of the child is determined by the type of chromosome present in sperm (X or Y) that fuses with the ovum at the time of fertilisation.
  2. By organising poster-making, debates and seminars.
  3. Spreading social awareness against various social malaise, viz. gender, inequality female foeticide etc.

Question 76. Seeds have some food stored in the cotyledons that are helpful during the process of germination. A small seed can grow into a giant tree when allowed to grow in the right conditions/environment.

Answer the following questions based on the above passage:

  1. How is the food stored in the cotyledons helpful during the process of germination?
  2. List four values acquired by a student in the early stages of his/her growth which can
    enable him/her to achieve great heights in life.
  3. Write any two environmental conditions to be provided by the parents/family to help the child’s personality grow in the right direction.

Answer:

  1. Cotyledons contain food in the form of bio-chemical energy which is used in the process of germination.
  2. Hard work, passion for learning and knowledge, similarity, thinking skills and originality.
  3. Providing opportunities for a variety of learning experiences, motivation and freedom to experiment.

Question 77. In a certain village, the sex ratio is very low due to a large number of cases of female infanticide. A man from the same village takes his pregnant wife to the doctors to find out whether the foetus is a girl or a boy. The doctor performs a test and tells them that it is a girl.

Answer the following:

  1. Name the technique used by the doctor to determine the type of sex of the foetus.
  2. Which values are disrespected by doctors?
  3. Suggest two actions to be taken by the administration to promote desired related values.

Answer:

  1. The technique used to determine the type of sex of the foetus is called amniocentesis.
  2. Value for life, professional ethics, gender bias.
  3. Enactment of law.
    1. Awareness campaigns about the declining sex ratio and its ill effects.
    2. Education of women.

Question 78. Mr Sharma was suffering from various types of diseases presently. He went for thorough health check-ups and was diagnosed to be HIV+. Soon this news spread in his neighbourhood and on account of this, he faced social isolation.

Answer the following questions based on the information given above.

  1. Do you think people’s indifference towards HIV+ people is justifiable? What kind of approach should we have toward the persons suffering from AIDS?
  2. How can one protect oneself from this disease?
  3. What values are not promoted by Mr Sharma’s neighbours?

Answer:

  1. Not. We should exhibit a humane approach/behaviour and should not treat them as isolated.
  2. One should be aware of the mode of transmission of disease, the danger of sharing needles, transfusion of contaminated blood, etc.
  3. Sympathy for fellow human beings, care, humane nature.

Question 79. List any two reasons why the Government has banned prenatal sex determination by law.
Answer:

  1. People may get female foetus absorbed.
  2. Reckless female foeticide has disturbed the male-female ratio in society.

How do Organisms Reproduce Multiple Choice Questions and Answers

Question 1. In the list of organisms given below, those that reproduce by the asexual method are

  1. Banana
  2. Dog
  3. yeast
  4. Amoeba
  1. (2) and (4)
  2. (1), (3) and (4)
  3. (1) and (4)
  4. (2), (3) and (4)

Answer: 2. (1), (3) and (4)

Question 2. In a flower, the parts that produce male and female gametes (germ cells) are

  1. Stamen and anther
  2. Filament and stigma
  3. Anther and ovary
  4. Stamen and style

Answer: 3. Anther and ovary

Question 3. Which of the following is the correct sequence of events of sexual reproduction in a flower?

  1. Pollination, fertilisation, seedling, embryo
  2. Seedling, embryo, fertilisation, pollination
  3. Pollination, fertilisation, embryo, seedling
  4. Embryo, seedling, pollination, fertilisation

Answer: 3. Pollination, fertilisation, embryo, seedling

Question 4. Offsprings formed by an asexual method of reproduction have greater similarities among themselves because

  1. Asexual reproduction involves only one parent
  2. Asexual reproduction does not involve gametes
  3. Asexual reproduction occurs before sexual reproduction
  4. Asexual reproduction occurs after sexual reproduction
  1. (1) and (2)
  2. (1) and (3)
  3. (2) and (4)
  4. (3) and (4)

Answer: 1. (1) and (2)

Question 5. Characters transmitted from parents to offspring are present in

  1. Cytoplasm
  2. Ribosome
  3. Golgi bodies
  4. Genes

Answer: 4. Genes

Question 6. Characters that are transmitted from parents to offspring during reproduction show

  1. The only similarities between parents
  2. Only variations with parents
  3. Both similarities and variations with parents
  4. Neither similarities nor variations

Answer: 3. Both similarities and variations with parents

Question 7. A feature of reproduction that is common to Amoeba, Spirogyra and Yeast is that

  1. They reproduce asexually
  2. They are all unicellular
  3. They reproduce only sexually
  4. They are all multicellular

Answer: 1. They reproduce asexually

Question 8. In Spirogyra, asexual reproduction takes place by

  1. Breaking up of filaments into smaller bits
  2. Division of a cell into two cells
  3. Division of a cell into many cells
  4. Formation of young cells from older cells.

Answer: 1. Breaking up of filaments into smaller bits

Question 9. The ability of a cell to divide into several cells during reproduction in Plasmodium is called

  1. Budding
  2. Reduction division
  3. Binary fission
  4. Multiple fission

Answer: 4. Multiple fission

Question 10. The correct sequence of reproductive stages seen in flowering plants is

  1. Gametes, zygotes, embryos, seedlings
  2. Zygote, gametes, embryo, seedling
  3. Seedling, embryo, zygote, gametes
  4. Gametes, embryos, zygotes, seedlings

Answer: 1. Gametes, zygote, embryo, seedling

Question 11. The number of chromosomes in parents and offsprings of a particular species remains constant due to

  1. Doubling of chromosomes after zygote formation
  2. Halving of chromosomes during gamete formation
  3. Doubling of chromosomes after gamete formation
  4. Halving of chromosomes after gamete formation

Answer: 2. Halving of chromosomes during gamete formation

Question 12. In Rhizopus, tubular thread-like structures bearing sporangia at their tips are called

  1. Filaments
  2. Hyphae
  3. Rhizoids
  4. Roots

Answer: 2. Hyphae

Question 13. Vegetative propagation refers to the formation of new plants from

  1. Stem, roots and flowers
  2. Stem, roots and leaves
  3. Stem, flowers and fruits
  4. Stem, leaves and flowers

Answer: 2. Stem, roots and leaves

Question 14. Factors responsible for the rapid spread of bread mould on slices of bread are

  1. A large number of spores
  2. Availability of moisture and nutrients in bread
  3. Presence of tubular branched hyphae
  4. Formation of round-shaped sporangia
  1. (1) and (3)
  2. (1) and (2)
  3. (2) and (4)
  4. (3) and (4)

Answer: 3. (2) and (4)

Question 15. The length of the pollen tube depends on the distance between

  1. Pollen grain and the upper surface of the stigma
  2. Pollen grain on the upper surface of the stigma and ovule
  3. Pollen grain in anther and the upper surface of the stigma
  4. The upper surface of the stigma and the lower part of the style

Answer: 2. Pollen grain on the upper surface of stigma and ovule

Question 16. Which of the following statements is true for flowers?

  1. Flowers are always bisexual
  2. They are the sexual reproductive organs
  3. They are produced in all groups of plants
  4. After fertilisation, they give rise to fruits
  1. (1) and (4)
  2. (2) and (3)
  3. (1) and (3)
  4. (2) and (4)

Answer: 4. (2) and (4)

Question 17. Which among the following statements is true for unisexual flowers?

  1. They possess both stamen and pistil
  2. They possess either stamen or pistil
  3. They exhibit cross-pollination
  4. Unisexual flowers possessing only stamens cannot produce fruits
  1. (1) and (4)
  2. (2), (3) and (4)
  3. (3) and (4)
  4. (1), (3) and (4)

Answer: 2. (2), (3) and (4)

Question 18. Which among the following statements is true for sexual reproduction in flowering plants?

  1. It requires two types of gametes
  2. Fertilisation is a compulsory event
  3. It always results in the formation of a zygote
  4. Offspring formed are clones
  1. (1) and (4)
  2. (1), (2) and (4)
  3. (1), (2) and (3)
  4. (1), (2) and (4)

Answer: 3. (1), (2) and (3)

Question 19. Offsprings formed as a result of sexual reproduction exhibit more variations because

  1. Sexual reproduction is a lengthy process
  2. Genetic material comes from two parents of the same species
  3. Genetic material comes from two parents of different species
  4. Genetic material comes from many parents

Answer: 2. Genetic material comes from two parents of the same species

Question 20. Reproduction is essential for living organisms to

  1. Keep the individual organism alive
  2. Fulfill their energy requirement
  3. Maintain growth
  4. Continue the species generation after generation

Answer: 4. Continue the species generation after generation

Question 21. During adolescence, several changes occur in the human body. Mark one change associated with sexual maturation in boys

  1. Loss of milk teeth
  2. Increase in height
  3. Cracking of voice
  4. Weight gain

Answer: 3. Cracking of voice

Question 22. In human females, an event that reflects the onset of the reproductive phase is

  1. Growth of body
  2. Changes in hair pattern
  3. Change in voice
  4. Menstruation

Answer: 4. Menstruation

Question 23. In human males, the testes lie in the scrotum, because it helps in the

  1. Process of mating
  2. Formation of sperm
  3. Easy transfer of gametes
  4. All the above

Answer: 2. Formation of sperm

Question 24. Which among the following are not the functions of testes at puberty?

  1. Formation of germ cells
  2. Secretion of testosterone
  3. Development of placenta
  4. Secretion of estrogen
  1. (1) and (2)
  2. (2) and (3)
  3. (3) and (4)
  4. (1) and (4)

Answer: 3. (3) and (4)

Question 25. The correct sequence of organs in the male reproductive system for the transport of sperm is

  1. Testis → vas deferens → urethra
  2. Testis → ureter → urethra
  3. Testis → urethra → ureter
  4. Testis → vas deferens → ureter

Answer: 1. Testis → vas deferens → urethra

Question 26. Which among the following diseases is not sexually transmitted?

  1. Syphilis
  2. Hepatitis
  3. HIV-AIDS
  4. Gonorrhoea

Answer: 2. Hepatitis

Question 27. Asexual reproduction takes place through budding in

  1. Amoeba
  2. yeast
  3. Plasmodium
  4. Leishmania

Answer: 2. Yeast

Question 28. Which of the following is not a part of the female reproductive system in human beings?

  1. Ovary
  2. Uterus
  3. Vas deferens
  4. Fallopian tube

Answer: 3. Vas deferens

Question 29. The anther contains

  1. Sepals
  2. Ovules
  3. Carpel
  4. Pollen grains

Answer: 4. Pollen grains

CBSE Class 10 Science Chapter 7 Control and Coordination Short Answer Questions

CBSE Class 10 Science Chapter 7 Control and Coordination

Hormones In Animals:

CBSE Class 10 Science Chapter 7 Control And Coordination Hormones In Animals

CBSE Class 10 Science Chapter 7 Control and Coordination Short Question And Answers

Question 1. What is the difference between a reflex action and walking?
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination Difference Between A Reflex Action And Walking

Question 2. What happens at the synapse between two neurons?
Answer:

At the synapse at the end of the axon of the neuron carrying information, the electrical impulse sets off the release of some chemicals, known as neurotransmitters. These chemicals cross the gap (synapse) reach the dendrites of the next neuron and start electrical impulses there.

Read and Learn More CBSE Class 10 Science Short Answer Questions

Question 3. Which part of the brain maintains posture and equilibrium of the body?
Answer: A part of the hindbrain known as the cerebellum maintains posture and equilibrium of the body.

Question 4. How do we detect the smell of an agarbatti (incense stick)?
Answer: The smell of an agarbatti is sensed by the olfactory receptor present in the nose. They send information to the olfactory lobe, located in the forebrain, which interprets the data.

Question 5. What is the role of the brain in reflex action?
Answer:

Nerves from all over the body meet in a bundle in the spinal cord on their way to the brain. Reflex arcs are formed in the spinal cord itself. So the input (stimulus) and output (response) actions are completed quickly, although the information input also reaches the brain. The brain may store the experience for the future.

Question 6. What are plant hormones?
Answer:

Plant hormones are special chemical compounds released by different parts of the plants. They help to coordinate growth, development, and responses to the environment. They are synthesized at places away from the site where they act. They diffuse to the area of action.

Question 7. How is the movement of leaves of the sensitive plant different from the movement of a shoot toward light?
Answer:

Differences between the movements of the leaves of the sensitive plant and the movement of the shoot toward light are as follows:

  • The movement of leaves of the sensitive plant is not directional to the stimulus of touch. But the movements of the shoot are directional towards the light.
  • The movement of leaves (drooping down) of sensitive plants is due to changes in the amount of water present in them. The movement of the shoot towards light is due to growth controlled by a growth hormone called auxin.

Question 8. Give an example of a plant hormone that promotes growth.
Answer:

Auxin promotes the growth of the cell, gibberellins help in the growth of the stem, and cytokinin promotes cell division (any of them).

Question 9. How do auxins promote the growth of a tendril around a support?
Answer:

The tendrils are sensitive to touch. When a part of the tendril comes in contact with a support, the growth hormone auxin diffuses from the site. This causes growth in the part of the tendril away from the support. This causes the tendril to circle the support. Thus, the tendril clings to the support.

Question 10. Design an experiment to demonstrate hydrotropism.
Answer:

Experiment to demonstrate hydrotropism: Growth movements in response to the stimulus of moisture (water) are termed hydrotropic movements (Hydrotropism).

Experiment: Place some ready-to-germinate seeds of pea/gram/beam in moist sawdust in a shallow box whose bottom has been replaced by a wire gauze.

Keep the apparatus undisturbed for 2-3 days. Keep the sawdust moist. After some time radicles will pass through the sieves. Then the radicles curve upward again to enter into the moist sawdust.

The roots are positively hydrotropic and the curvature is more than that due to the geotropic effect.

CBSE Class 10 Science Chapter 7 Control And Coordination Demostrate Hydrotropism

Question 11. How does chemical coordination take place in animals?
Answer:

In animals, chemical coordination is carried out by hormones secreted by endocrine glands. Hormones are secreted by ductless glands placed far away from the site of action of the secreted hormone.

  1. Hormones are directly secreted into the blood that carries them to different parts of the body and also to tissues/organs where they are required.
  2. The body cells have special molecules on their plasma membrane for recognition of information a hormone carries. After receiving hormonal information the organ/tissues/cells act accordingly.

Question 12. Why is the use of iodized salt advisable?
Answer:

Iodine is necessary for the synthesis of the thyroid glands’ thyroxin hormone. Thyroxin regulates carbohydrates, fat, and protein metabolism in the body to maintain the best balance for growth.

  1. The deficiency of iodine may cause a decrease in the production of thyroxine resulting in goiter disease and other metabolic disorders. The main symptom of goiter is a swollen neck.
  2. As a preventive measure, people are advised to use iodized common salt.

Question 13. How does our body respond when adrenaline is secreted into the blood?
Answer: Adrenaline is secreted by the adrenal gland directly into the blood. The target organs or the specific tissues on which it acts include the heart.

Response due to secretion of adrenaline

  1. The heart beats faster, resulting in a supply of more oxygen to our muscles.
  2. The blood to the digestive system and skin is reduced and diverted to our skeletal muscles.
  3. Breathing rate increases because of contraction of the diaphragm and the rib muscles. All these responses together enable our body to be ready to deal with the adverse situation.

Question 14. Why are some patients with diabetes treated by giving injections of insulin?
Answer:

Insulin is a hormone secreted by the pancreas. It helps in regulating blood sugar levels. It is not secreted in proper amounts, and the sugar level in the blood rises causing many harmful effects. Doctors treat diabetic patients with high blood sugar levels with insulin injections to keep blood sugar levels under control.

Question 15. What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:

Receptors are specialized tips of some nerves that are located in our sense organs such as skin, tongue, etc.

  • They detect information from the environment and pass on the information to the spinal cord and brain for proper response. If receptors do not work properly, there will be a delay in responding to a stimulus.
  • This sometimes becomes dangerous. For example, touching a hot object is an urgent and dangerous situation for us. We need to detect it and respond to it immediately.

Question 16. Draw the structure of a neuron and explain its function.
Answer:

Diagram of a neuron to show its structure:

CBSE Class 10 Science Chapter 7 Control And Coordination Structure Of Neuron

The function of neurons is to carry messages over long distances in the body of a person quickly. The message that neurons transmit in the nervous system is in the form of electrical impulses. Neurons make the whole nervous system work efficiently.

For example, the sensory neurons transmit impulses from the sensory cell towards the central nervous system and the motor neurons transmit impulses from the central nervous system towards the muscle cells for taking appropriate action.

CBSE Class 10 Science Chapter 7 Control And Coordination Neuromusular Junction

Question 17. How does phototropism occur in plants?
Answer:

The directional (or tropic) movements due to light, in plants are called phototropic movements. In this case shoot of the plant grows towards the direction of light whereas roots grow away from the light stimulus.

CBSE Class 10 Science Chapter 7 Control And Coordination Response Of The Plant To The Direction Of Light

Question 18. Why does the shoot bend toward the light?
Answer:

The growth hormone, auxin, is not able to promote growth in the plant part which is well-illuminated, from one side. So, the hormone auxin diffuses toward the shady side of the shoot.

  • The increased concentration of growth hormone auxin in the shaded part of the shoot causes an increase in the growth of cells in that region.
  • The enhanced growth of the shaded side of the shoot i.e., one side of the shoot, causes the bending of the shoot towards the source of light.

Question 19. Which signals will get disrupted in case of a spinal cord injury?
Answer:

  1. Reflex action, as information detected from the environment, will be conveyed late to the reflex arc and response to the signal may be delayed to reach the effector organ/tissues. They may lead to dangerous situations. The signal of touching a hot object, if delayed, may lead to delayed response and we may get hurt.
  2. Information detected by nerves from various parts of the body may reach the spinal cord/brain late.
  3. Message from the brain will not be conducted to various organizers: This will lead to disturbed control and coordination of the organism.

Question 20. How does chemical coordination occur in plants?
Answer:

Plants do not have a nervous system for control and coordination. They may have the following system:

  1. By making movements dependent on growth such as bending of the shoot towards the source of light. For such actions, plants produce hormones, such as auxins, cytokinins, etc.
  2. Movements that are independent of growth like shrinking or swelling as happens in touch-me-not plants.
  3. The plants also use electrical-chemical means to convey information from cell to cell, as happens in the drooping of leaves on touch.

Question 21. What is the need for a system of control and coordination in an organism?
Answer:

Living organisms make many movements, which may be due to growth, in response to changes in the environment or to procure food. Each kind of change in the environment evokes an appropriate movement in response.

In multicellular organisms, different body parts or body organs must work in unison for the survival of the individual. In other words, living organisms must use systems providing control and coordination.

Question 22. How are involuntary actions and reflex actions different from each other?
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination Involuntary Action and Reflex Action

Question 23. Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination Nervous Control And Hormonal Control

Question 24. What is the difference between how movement takes place in a sensitive plant and the movement in our legs?
Answer:

The difference between movement in a sensitive plant and the movement in our legs

CBSE Class 10 Science Chapter 7 Control And Coordination Difference Between Sensitive Plant And Our Legs

Question 25. State the functions of any three of the structural and functional units of the nervous system.
Answer:

The structural and functional unit of the nervous system is a neuron. Its functions are as follows:

  1. Cell body: Stimulus received from dendrite is changed into an impulse in the cyton.
  2. Dendrites: They receive sensation or stimulus, which may be physical, chemical, mechanical, or electrical. They pass the stimulus to cyton.
  3. Axon: It conducts impulses away from the cell body.

Question 26.

  1. Which plant hormone is present in greater concentration in the areas of rapid cell division?
  2. Give one example of a plant growth promoter and a plant growth inhibitor.

Answer:

  1. Cytokinin is present in greater concentration in the areas of rapid cell division.
  2. An example of a plant growth promoter is gibberellins and an example of a plant growth inhibitor is abscisic acid.

Question 27. Suggest proof that even unicellular organisms like amoeba respond to stimuli. Justify by giving two examples that even plants respond to stimuli.
Answer:

Amoeba moves towards food and tends to aggregate in moderately warm water. Amoeba and other protozoa avoid mechanical obstacles.

Thus, unicellular organisms respond to stimuli. Roots of plants move downward in response to gravitational force and shoots of plants move towards light. This shows that plants respond to stimuli.

Question 28.

  1. Name the part of the brain that controls:
    1. Voluntary action.
    2. Involuntary action.
  2. What is the significance of the peripheral nervous system? Name the components of this nervous system and distinguish between the origin of the two.

Answer:

    1. Voluntary action – cerebellum;
    2. Involuntary action – medulla oblongata.
  1. The communication between the central nervous system and the other parts of the body is facilitated by the peripheral nervous system. Cranial nerves arise from the brain; spinal nerves arise from the spinal cord.

Question 29. State how the concentration of auxin stimulates the cells to grow longer on the side of the shoot which is away from light.
Answer:

When a growing plant detects light, auxin synthesizes at the shoot tip to help the cell to grow longer. When light comes from one side, auxin diffuses towards the shady side of the shoot.

This concentration of auxin stimulates the cells to grow longer on the side of the shoot which is away from light and the plant appears to bend towards light.

Question 30.

  1. Explain any three directional movements in plants.
  2. How are the brain and spinal cord protected in humans?
  3. Name the master gland present in the brain.

Answer:

  1. Three directional movements in plants are:
    1. Phototropism: It is the movement of a part of a plant towards light, e.g., the movement of the stem towards light.
    2. Geotropism: The upward and downward growth of shoots and roots in response to gravity is called geotropism.
    3. Chemotropism: It is the movement of a part of a plant in response to a chemical stimulus, e.g., the growth of a pollen tube towards a chemical produced by an ovule.
  2. The brain and spinal cord are protected by a skeleton – the brain by the cranium and the spinal cord by the vertebral column.
  3. The master gland present in the brain is called the pituitary gland.

Question 31.

  1. An old man is advised by his doctor to take less sugar in his diet. Name the disease from which the man is suffering. Mention the hormone due to imbalance of which he is suffering from this disease. Which endocrine gland secretes this hormone?
  2. Name the endocrine gland which secretes growth hormone. What will be its effect on a person of:
    1. Deficiency of growth hormones?
    2. Excess secretion of growth hormone?

Answer:

  1. The man is suffering from the disease Diabetes. Insulin is the hormone that is responsible for this disease. The pancreas secretes this hormone.
  2. The gland that secretes the growth hormone is the pituitary gland.
    1. Deficiency of growth hormone causes Dwarfism.
    2. Excess secretion of growth hormone causes Gigantism in a person.

Question 32. Distinguish between estrogen and progesterone.
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination Difference Between Estrogen And Progesterone

Question 33.

  1. Identify the phytohormone used by plants while performing the following functions:
    1. Cell division in shoot tip.
    2. Inhibiting growth on the approach of unfavorable conditions.
  2. List in tabular form two differences between the movement in the ‘touch-me-not’ plant and the movement of the shoot towards the light.

Answer:

  1. Auxin,
  2. Abscisic acid.

CBSE Class 10 Science Chapter 7 Control And Coordination Difference Between Tropism And Nastic Movement

Question 34. How does chemical coordination occur in plants? Explain with the help of three examples.
Answer:

In plants, chemical coordination occurs through various phytohormones.

  1. Auxins are secreted by growing tissues. They provide growth of plants.
  2. Gibberellins cause stem elongation, seed germination, and flowering.
  3. Cytokinins are present in areas of actively dividing cells like fruits and seeds. Promote cell division.
  4. Abscisic acid inhibits growth and responds to environmental stress. (any three)

Question 35. Draw a neat diagram of the human brain and label it the following parts:

  1. Midbrain
  2. Pituitary gland

Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination Human Brain

Question 36. Illustrate with the help of a diagram, the effect of auxins in different parts of a plant.
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination The Effect Of Auxins In Different Parts Of A Plant

Question 37. List the names of hormones secreted by the following endocrine glands and mention their functions:

  1. Thyroid gland
  2. Adrenal gland
  3. Pituitary gland

Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination The Names Of Harmones

Question 38. Write one example of each of the following tropic movements:

  1. Positive phototropism
  2. Negative phototropism
  3. Positive geotropism
  4. Hydrotropism
  5. Negative geotropism
  6. Chemotropism

Answer:

  1. Positive phototropism: Movement of the stem of a plant towards light.
  2. Negative phototropism: Movement of roots away from light.
  3. Positive geotropism: Movement of roots towards gravity.
  4. Negative geotropism: Upward movement of shoots.
  5. Hydrotropism: Growth of roots of a plant towards the water in the soil.
  6. Chemotropism: Growth of pollen tubes towards ovule.

Question 39. Describe an activity to illustrate the phenomenon of phototropism and explain why this occurs.
Answer:

Activity:

Take a plant and make such an arrangement that it receives the light coming from a window as shown in the figure and observe it for a few days.

CBSE Class 10 Science Chapter 7 Control And Coordination Response Of The Plant To The Direction Of Light.

  • Now, turn the flask so that the shoots are away from light and the roots towards light. Leave it undisturbed in this condition for a few days.
  • Again observe to find the difference in the movement.
  • When growing plants detect light, auxin, synthesized at the shoot tip, helps the cells to grow longer; auxin always diffuses towards the shady side of the shoot. This concentration of auxin stimulates the cells to grow longer on the side of the shoot which is away from light. Thus, the plant appears to bend towards light.

Question 40. What is a cerebrospinal fluid? What is its function?
Answer:

The fluid which fills the space between the membranes which cover the brain is called cerebrospinal fluid.

Function: It protects the brain from mechanical shocks.

Question 41. What is a reflex arc?

  1. What are the components of the reflex arc?
  2. How do muscle cells move?

Answer:

  1. The process of detecting the signal or the input and responding to it by an output action might be completed quickly. Such a connection is commonly called a reflex arc.
  2. Stimulus → Receptors → Sensory neurons → Spinal cord → Motor neurons → Effector.
  3. Muscle cells have special proteins that change their shape and arrangement in the cell in response to electrical impulses. This leads the muscle cells to shorten.

Question 42. Write the difference between the cerebellum and cerebrum.
Answer:

CBSE Class 10 Science Chapter 7 Control And Coordination The Difference Between Cerebrum And Cerebellum

Question 43. ‘Brain and spinal cord are two vital organs of our body’. How is our body designed to protect them?
Answer:

The brain is inside a bony box contained in a fluid-filled balloon which protects it from shocks. The vertebral column protects the spinal cord. Both are protected by protective layers called meninges, protective fluid (CSF→ cerebrospinal fluid).

Question 44.

  1. How does the tendril coil around the support?
  2. Define phototropism.

Answer:

  1. Tendrils are sensitive to touch. When they come in contact with any support, the part of the tendril in contact with the object does not grow as rapidly as the part of the tendril away from the object. This causes the tendril to coil around the support.
  2. Movement of plants in response to light.

Question 45. List the sequence of events that occur when a plant is exposed to unidirectional light, leading to the bending of a growing shoot. Also, name the hormone and the type of movement.
Answer:

  1. Auxin diffuses towards the shady side.
  2. Causes elongation on one side (shady side) resulting in bending of the growing shoot.
  3. The hormone responsible for bending is auxin and the movement is phototropic.

Question 46.

  1. Name the diseases by which a person is likely to suffer due to the deficiency of:
    1. Iodine
    2. Insulin
  2. How is the timing of secretion and amount of hormone secretion regulated in the human system? Explain with an example.

Answer:

    1. Goitre
    2. Diabetes
  1. The timing and amount of hormone released are regulated by feedback mechanisms, e.g., if the sugar levels in the blood rise, they are detected by the cells of the pancreas which respond by producing more insulin. As the blood sugar level falls, insulin secretion is reduced.

Question 47.

  1. Name the hormone that is injected into a diabetic patient.
  2. Why should we use iodized salt in our diet?
  3. If iodine is insufficient in one’s diet, what might be the deficiency disease and its symptoms?

Answer:

  1. Insulin.
  2. Iodine is necessary for the thyroid gland to secrete thyroxine hormone. Thyroxine regulates carbohydrates, proteins, and fat metabolism in the body, to provide the best balance for growth.
  3. If iodine is insufficient in one’s diet, an iodine deficiency disease called goiter occurs. The symptom is a swollen neck.

Question 48.

  1. If the cerebellum is not functioning properly, what are the activities of our body that will be affected?
  2. How do muscle cells move?

Answer:

  1. If the cerebellum is not functioning properly, voluntary action of our body will be affected by maintaining the posture and balance of the body.
  2. Muscle cells move by changing their shape so that they get shortened.

Question 49. Name a hormone secreted by:

  1. Pancreas
  2. Pituitary
  3. Thyroid. Write one function of each of the hormones.

Answer:

  1. Insulin – controls blood sugar levels.
  2. The pituitary – growth hormone regulates growth and development.
  3. Thyroxine – regulates the metabolism of carbohydrates, fats, and proteins.

Question 50. Define reflex action. Give one example. Show with the help of a flow diagram the path of the reflex action.
Answer:

Reflex action is a sudden action in response to something in the environment. e.g., pulling out a hand from the flame if accidentally touched. The path of the reflex action is as follows:

Receptors Sensory neuron → Spinal cord → Effector → Motor neuron

Question 51. Name the hormone synthesized at the shoot tips. How does it help plants to respond to light?
Answer:

Auxin is synthesized at the shoot tips when a growing plant detects light and helps the cells to grow longer.

CBSE Class 10 Science Chapter 7 Control And Coordination Auxin Is Synthesised At The Shoot Tips

When light is coming from one side of the plant, auxin diffuses towards the shady side of the shoot. This concentration of auxin stimulates the cells to grow longer on the side of the shoot which is away from light. Thus, the plant appears to bend towards light.

Question 52.

  1. Differentiate between sensory neurons and motor neurons.
  2. How is the brain protected in our body?
  3. Name the part of the brain responsible for the precision of voluntary actions and maintaining body posture and balance of the body.

Answer:

  1. Sensory neurons carry impulses from receptors to the brain. Motor neurons carry impulses from the brain to effectors.
  2. The brain is kept inside a bony box (skull). Inside the box, the brain is contained in a fluid-filled balloon which provides further shock absorption. This fluid is known as cerebrospinal fluid (CSF).
  3. These are possible due to a part of the hindbrain called the cerebellum.

Question 53. Name and explain the functions of the hormones secreted by the pituitary gland in humans
Answer:

Hormones secreted by the pituitary gland along with their function are as follows:

  1. Growth hormone: It regulates the growth and development of bones and muscles.
  2. Trophic hormone: It regulates the secretion of hormones from other endocrine glands.
  3. Prolactin hormone: It regulates the function of mammary glands in females.
  4. Vasopressin hormone: It regulates water and electrolyte balance in the body.
  5. Oxytocin hormone: It regulates the ejection of milk during lactation.

Question 54. How do you support the statement the pancreas’ are the overall controller of the blood glucose level?
Answer:

The pancreas releases insulin which lowers blood glucose. Moreover, it also releases glucagon which increases blood glucose. Therefore, the pancreas is the overall controller of blood glucose levels.

Question 55. Name the two main organs of our central nervous system. Which one of them plays a major role in sending commands to muscles to act without involving the thinking process? Name the phenomenon involved.
Answer:

The two main organs of the CNS are the brain and spinal cord. The spinal cord plays a major role in sending commands to muscles to act without involving the thinking process. This phenomenon is called reflex action.

Question 56. Label the parts (a), (b), and and show the direction of the flow of electrical signals in the figure given below:

CBSE Class 10 Science Chapter 7 Control And Coordination The Direction Of Flow Of Electrical Signals

Answer:

  1. Sensory neuron
  2. Spinal cord
  3. Motor neuron
  4. Effector = Muscle in the arm

Question 57. Name the plant hormones responsible for the following

  1. Elongation of cells
  2. Growth of stem
  3. Promotion of cell division
  4. Falling of senescent leaves.

Answer:

  1. Auxin
  2. Gibberellin
  3. Cytokinin
  4. Abscisic acid.

Question 58. Label the endocrine glands in the figure given below:
Answer:

  1. Pineal gland
  2. Pituitary gland
  3. Thyroid
  4. Thymus.

CBSE Class 10 Science Chapter 7 Control And Coordination The Endrocrine Gland

Question 59. In figure (a), and given below, which appears more accurate and why?

CBSE Class 10 Science Chapter 7 Control And Coordination While Roots Are Positively Geotropic

Answer:

The figure is more appropriate because in a plant shoot is negatively geotropic and hence grows upward while roots are positively geotropic, so grow downward.

Question 60. Label the parts of a neuron in the figure given below:

CBSE Class 10 Science Chapter 7 Control And Coordination The Parts Of Neuron

Answer:

  1. Dendrite
  2. Cell body
  3. Axon
  4. Nerve ending.

Question 61. Match the terms of Column (A) with those of Column (B)

CBSE Class 10 Science Chapter 7 Control And Coordination The Terms Of Column A And Column B

Answer: 1-C, 2-D, 3-A, 4-B

Question 62. What is a tropic movement? Explain with an example.
Answer:

The directional growth movements of plants due to external stimuli are called tropic movements. These can be either towards the stimulus, or away from it. For example, in the case of phototropic movement, shoots respond by bending toward light while roots respond by bending away from it.

Question 63. What will happen if the intake of iodine in our diet is low?
Answer:

  1. When iodine intake is low, the release of thyroxin from the thyroid gland will be less, by which protein, carbohydrate, and fat metabolism will be affected.
  2. A person might suffer from goiter in case of iodine deficiency in the body.

Question 64. What happens at the synapse between two neurons?
Answer:

When an electrical signal reaches the axonal end of one neuron, it releases certain chemical substances that cross the synapse and move towards the dendritic end of the next neuron generating another electrical signal.

Question 65. Answer the following:

  1. Give the names of hormones responsible for the physical changes in a boy and a girl.
  2. Name the element of the halogen family required for the proper functioning of the thyroid gland.
  3. Name the stress and fear hormones.

Answer:

  1. Boy – Testosterone Girl – Oestrogen.
  2. Iodine.
  3. Adrenaline.

Question 66. What is a cerebrospinal fluid? What is its function?
Answer:

The fluid which fills the space between the membranes which cover the brain is called cerebrospinal fluid.

Function: It protects the brain from mechanical shocks.

Question 67. Define ‘nerve impulse’. Which structure in a neuron helps to conduct a nerve impulse?

  1. Towards the cell body?
  2. Away from the cell body?

Answer:

The pieces of information passing through neurons are in the form of chemical are electrical signals called nerve impulses.

  1. Nerve impulses are carried by dendrites towards the cell body.
  2. The axon is a highly specialized nerve fiber that conducts nerve impulses away from the cell body.

Question 68. ‘Brain and spinal cord are two vital organs of our body’. How is our body designed to protect them?
Answer:

The brain is inside a bony box contained in a fluid-filled balloon which protects it from shocks. The vertebral column protects the spinal cord. Both are protected by protective layers called meninges, protective fluid (CSF → cerebrospinal fluid).

Question 69. Give reasons:

  1. The pituitary is often termed as a master endocrine gland.
  2. The pancreas helps in digestion and also regulates blood sugar levels.
  3. Adrenals are known as glands of emergency.

Answer:

  1. The pituitary is often called the master endocrine gland because it controls and coordinates the secretion of all the other endocrine glands.
  2. The pancreas secretes pancreatic juice as well as a hormone called insulin. Pancreatic juice helps in digestion whereas insulin regulates blood sugar levels.
  3. Adrenalin is secreted directly into the blood and is carried to different parts of the body. It acts on the heart. As a result, the heart beats faster to supply more oxygen to our muscles. The blood to the digestive system and skin is reduced due to the contraction of muscles around.

Question 70. Answer the following:

  1. Name the endocrine gland associated with the brain.
  2. Which gland secretes digestive enzymes as well as hormones?
  3. Name the endocrine gland associated with the kidneys.
  4. Which endocrine gland is present in males but not in females?
  5. Which hormone is responsible for changes in females during puberty?
  6. Iodine is necessary for the synthesis of which hormone?

Answer:

  1. Pituitary gland
  2. Adrenal gland
  3. Oestrogen
  4. Pancreas
  5. Testes
  6. Thyroxin

Question 71. Vijay’s friend has diabetes and every day after her meal in the recess she had to check her sugar level. To monitor it at times she used to take injections also. Vijay was very fond of sweets but she never carried any sweets in her tiffin.

  1. What is the cause of diabetes?
  2. What helps in the digestion of sugars?
  3. What value of Vijay is reflected in the above act?

Answer:

  1. Deficiency of an enzyme that can digest the sugar.
  2. Insulin hormone.
  3. Sympathy, empathy, and kindness.

Question 72. Akash and Tarun planned to visit a nursery to purchase seasonal flowering plants. While walking in the nursery, Prakash saw a Mimosa pudica (sensitive plant). He asked Akash to touch the leaflets of the Mimosa plant. The leaflets of the plant closed up. Answer the following questions based on the above information:

  1. What do you infer from this activity?
  2. What would you do based on your inference?

Answer:

  1. Mimosa pudica (touch-me-not or sensitive plant) responds to the stimulus of touch by closing of leaflets.
    • Like animals, plants are also living and show movement and response to stimuli.
  2. We would organize campaigns to create awareness amongst the masses that plants are living things.

Question 73. Ram and Shyam were doing their homework when Ram’s mother called Ram. She asked Ram to bring a packet of common salt from the market. She asked him to check that the packet bears the label ‘Iodised salt’ or ‘enriched with Iodine’. Shyam also accompanied Ram to bring the packet.

  1. Ram asked Shyam, why is it important for us to have iodized salt in our diet? Ram said iodine is necessary for the formation of thyroxin hormone.
  2. Answer the following questions based on the above information:
  3. Which endocrine gland secretes thyroxin hormone?
  4. Which disease can occur in case iodine is deficient in our diet?
  5. What will you do to spread awareness among other community members about the use of iodized salt?

Answer:

  1. Thyroid gland.
  2. Goitre may occur.
  3. Organizing awareness programs among society.

Question 74. Tarushi’s mother always kept bananas in polybags and found them to be spoiled. Tarushi suggested her mother not keep the banana in a polybag.

  1. Why should we not keep bananas in polybags?
  2. Name the hormone present at the root and shoot of the plants.
  3. What value of Tarushi is found in this case?

Answer:

  1. Ethylene is released from polybag that helps in the ripening of fruits.
  2. Auxin hormone.
  3. Showed the value of a concerned and responsible child.

Control and Coordination Multiple Choice Questions

Question 1. Which of the following statements is correct about receptors?

  1. Gustatory receptors detect taste while olfactory receptors detect smell
  2. Both gustatory and olfactory receptors detect smell
  3. Auditory receptors detect smell and olfactory receptors detect taste
  4. Olfactory receptors detect taste and gustatory receptors detect smell

Answer: 1. Gustatory receptors detect taste while olfactory receptors detect smell

Question 2. The electrical impulse travels in a neuron from

  1. Dendrite → axon → axonal end → cell body
  2. Cell body → dendrite → axon → axonal end
  3. Dendrite → cell body → axon → axonal end
  4. Axonal end → axon → cell body → dendrite

Answer: 3. Dendrite → cell body → axon → axonal end

Question 3. In a synapse, a chemical signal is transmitted from

  1. The dendritic end of one neuron to the axonal end of another neuron
  2. Axon to the cell body of the same neuron
  3. Cell body to the axonal end of the same neuron
  4. Axonal end of one neuron to dendritic end of another neuron

Answer: 4. Axonal end of one neuron to dendritic end of another neuron

Question 4. In a neuron, the conversion of the electrical signal to a chemical signal occurs at /in

  1. Cell body
  2. Axonal end
  3. Dendritic end
  4. Axon

Answer: 2. Axonal end

Question 5. Which is the correct sequence of the components of a reflex arc?

  1. Receptors → Muscles → Sensory neuron → Motor neuron → Spinal cord
  2. Receptors → Motor neuron → Spinal cord → Sensory neuron → Muscle
  3. Receptors → Spinal cord → Sensory neuron → Motor neuron → Muscle
  4. Receptors → Sensory neuron → Spinal cord → Motor neuron → Muscle

Answer: 4. Receptors → Sensory neuron → Spinal cord → Motor neuron → Muscle

Question 6. Which of the following statements is true?

  1. Sudden action in response to something in the environment is called reflex action
  2. Sensory neurons carry signals from the spinal cord to muscles
  3. Motor neurons carry signals from receptors to the spinal cord
  4. The path through which signals are transmitted from a receptor to a muscle or a gland is called reflex arc
    1. 1 and 2
    2. 1 and 3
    3. 1 and 4
    4. 1, 2 and 3

Answer: 3. 1 and 4

Question 7. Which of the following statements are true about the brain?

  1. The main thinking part of the brain is the hindbrain
  2. Centers of hearing, smell, memory, sight, etc., are located in the forebrain
  3. Involuntary actions like salivation, vomiting, and blood pressure are controlled by the medulla in the hindbrain
  4. The cerebellum does not control the posture and balance of the body
    1. 1 and 2
    2. 2 and 3
    3. 1, 2 and 3
    4. 3 and 4

Answer: 3. 1, 2 and 3

Question 8. Posture and balance of the body is controlled by

  1. Cerebrum
  2. Cerebellum
  3. Medulla
  4. Pons

Answer: 2. Cerebellum

Question 9. The spinal cord originates from

  1. Cerebrum
  2. Medulla
  3. Pons
  4. Cerebellum

Answer: 2. Medulla

Question 10. The movement of the shoot towards light is

  1. Geotropism
  2. Hydrotropism
  3. Chemotropism
  4. Phototropism

Answer: 4. Phototropism

Question 11. The main function of abscisic acid in plants is to

  1. Increase the length of cells
  2. Promote cell division
  3. Inhibit growth
  4. Promote the growth of stem

Answer: 3. Inhibit growth

Question 12. Which of the following is not associated with ith growth plant slant?

  1. Auxin
  2. Gibberellins
  3. Cytokinins
  4. Abscisic acid

Answer: 4. Abscisic acid

Question 13. Iodine is necessary for the synthesis of which hormone?

  1. Adrenaline
  2. Thyroxin
  3. Auxin
  4. Insulin

Answer: 2. Thyroxin

Question 14. Choose the incorrect statement about insulin

  1. It is produced by the pancreas
  2. It regulates the growth and development of the body
  3. It regulates blood sugar level
  4. Insufficient secretion of insulin will cause diabetes

Answer: 2. It regulates the growth and development of the body

Question 15. Select the mismatched pair

  1. Adrenaline: Pituitary gland
  2. Testosterone : Testes
  3. Estrogen: Ovary
  4. Thyroxin: Thyroid gland

Answer: 1. Adrenaline: Pituitary gland

Question 16. The shape of guard cells changes due to changes in the

  1. Protein composition of cells
  2. Temperature of cells
  3. Amount of water in cells
  4. Position of the nucleus in the cells

Answer: 3. Amount of water in cells

Question 17. The growth of tendrils in pea plants is due to

  1. Effect of light
  2. Effect of gravity
  3. Rapid cell divisions in tendrillar cells that are away from the support
  4. Rapid cell divisions in tendrillar cells in contact with the support

Answer: 3. Rapid cell divisions in tendrillar cells that are away from the support

Question 18. The growth of pollen tubes towards ovules is due to

  1. Hydrotropism
  2. Chemotropism
  3. Geotropism
  4. Phototropism

Answer: 2. Chemotropism

Question 19. The movement  of sunflobywith path path path that the sun is due to

  1. Phototropism
  2. Geotropism
  3. Chemotropism
  4. Hydrotropism

Answer: 1. Phototropism

Question 20. The substance that triggers the fall of mature leaves and fruits from plants is

  1. Auxin
  2. Gibberellin
  3. Abscisic acid
  4. Cytokinin

Answer: 3. Abscisic acid

Question 21. Which of the following statements bathe the out transmission of neimpulsesulse is incorrect nerve

  1. the nerve impulse travels the rom dendritic end to the rds axonal end
  2. At the dendritic, end electrical impulses bring about the release of some chemicals which generate an electrical impulse at the axonal end of another neuron
  3. The chemicals released from the axonal end of one neuron cross the synapse and generate a similar electrical impulse in a dendrite of another neuron
  4. A neuron transmits electrical impulses not only to another neuron but also to muscle and gland cells

Answer: 2. At the dendritic end electrical impulses bring about the release of some chemicals which generate an electrical impulse at the axonal end of another neuron

Question 22. Involuntary actions in the body are controlled by

  1. Medulla in forebrain
  2. Medulla in midbrain
  3. Medulla in hindbrain
  4. Medullathe  in the spinal cord

Answer: 3. Medulla in hindbrain

Question 23. Which of the following is not an involuntary action?

  1. Vomiting
  2. Salivation
  3. Heartbeat
  4. Chewing

Answer: 4. Chewing

Question 24. When a person is suffering fa from a severe cold, he or she cannot

  1. Differentiate the taste of an apple from that of an ice cream
  2. Differentiate the smell of a perfume from that of an agarbatti
  3. Differentiate red light from green light
  4. Differentiate a hot object from a cold object

Answer: 2. Differentiate the smell of a perfume from that of an agarbatti

Question 25. What is the correct direction of the flow of electrical impulses?

CBSE Class 10 Science Chapter 7 Control And Coordination The Flow Of Electerical Impulses

Answer: 3

Question 26. Which statement is not true about thyroxin?

  1. Iron is essential for the synthesis of thyroxin
  2. It regulates carbohydrates, pro,tein, and fat metabolism in the bThe thyroidroid gland requires iodine to synthesize thyroxin
  3. Thyroxin is also called thyroid hormone

Answer: 1. Iron is essential for the synthesis of thyroxin

Question 27. Dwarfism results due to

  1. Excess secretion of thyroxin
  2. Less secretion of growth hormone
  3. Less secretion of adrenaline
  4. Excess secretion of growth hormone

Answer: 2. Less secretion of growth hormone

Question 28. Dramatic chains of body features associated with puberty are mainly because  of the secretion of

  1. Oestrogen in the rom testes and testosterone in the rom ovary
  2. Oestrogen the the rom adrenal gland and testosterone the the pituitary gland
  3. Testosterone the rom testes” estrogenogen the rom ovary
  4. Testosterone the thyroid gland estrogenogen from the pituitary gland

Answer: 3. Testosterone the testes estrogenogen from the ovary

Question 29. A doctor advised a person to take an injection of insulin because

  1. His blood pressure was low
  2. His heart was beating slowly
  3. He was suffering from fgoiteritre
  4. His sugar levels  in his blood were high

Answer: 4. His sugar levels  in his blood were high

Question 30. The hormone which increases the fertility in males is called

  1. Oestrogen
  2. Testosterone
  3. Insulin
  4. Growth hormone

Answer: 2. Testosterone

Question 31. Which of the following endocrine glands is unpaired?

  1. Adrenal
  2. Testes
  3. Pituitary
  4. Ovary

Answer: 3. Pituitary

Question 32. The junction between two neurons is called

  1. Cell junction
  2. Neuromuscular junction
  3. Neural joint
  4. Synapse

Answer: 4. Synapse

Question 33. In humans, life processes are controlled and regulated by

  1. Reproductive and endocrine systems
  2. Respiratory and nervous systems
  3. Endocrine and digestive systems
  4. Nervous and endocrine systems

Answer: 4. Nervous and endocrine systems

Question 34. Which of the following is a plant hormone?

  1. Insulin
  2. Thyroxin
  3. Oestrogen
  4. Cytokinin

Answer: Cytokinin

Question 35. The gap between two neurons is called a

  1. Dendrite
  2. Synapse
  3. Axon
  4. Impulse

Answer: Synapse

Question 36. The brain is responsible for

  1. Thinking
  2. Regulating the heartbeat
  3. Balancing the body
  4. All of the above

Answer: All of the above

Question 37. Answer the following:

  1. Which hormone is responsible for the changes noticed in females at puberty?
  2. Dwarfism results due to deficiency of which hormone?
  3. Blood sugar level rises due to deficiency of which hormone?
  4. Iodine is necessary for the synthesis of which hormone?

Answer:

  1. Oestrogen
  2. Growth hormone
  3. Insulin
  4. Thyroxin.

Question 38. Answer the following:

  1. Name the endocrine gland associated with the ith b.ain.
  2. Which gland secretes digestive enzymes as well as hormones?
  3. Name the endocrine gland associated with the ith kid. eys?
  4. Which endocrine gland is present in males but not in females?

Answer:

  1. Pituitary
  2. Pancreas
  3. Adrenal
  4. Testes.

CBSE Class 10 Science Chapter 6 Life Processes Short Answer Questions

CBSE Class 10 Science Chapter 6 Life Processes

Respiration:

CBSE Class 10 Science Chapter 6 Life Processes Respiration

CBSE Class 10 Science Chapter 6 Life Processes Short Question And Answers

Question 1. How are the lungs designed in human beings to maximise the area for the exchange of gases?
Answer:

  1. The lungs are paired, and elastic respiratory organs are placed in the thoracic cavity.
  2. The trachea divides in the thoracic cavity into bronchi-the right and the left. Each bronchus divides into very small tubes called bronchioles.
  3. Each bronchiole terminates in balloon-like structures called alveoli.
  4. Each alveolus is thin-walled and is richly supplied by blood capillaries. Alveoli also provide maximum surface area for the exchange of gases.

Read and Learn More CBSE Class 10 Science Short Answer Questions

Question 2. What are the components of the transport system in human beings? What are the functions of these components?
Answer:

The transport system, called the circulatory system in human beings, mainly consists of the heart (pumping station), blood (fluid connective tissue) and blood vessels (in which blood flows).

  1. Heart: The heart is a muscular organ which is as big as our fist. The heart is a pumping organ to push blood around the body.
    1. It receives deoxygenated blood from the body parts and pumps it to lungs for to lungs for enriching with oxygen. It receives purified. blood from the lungs and pumps it around the body.
  2. Blood: Blood is fluid connective tissue. It consists of a fluid medium known as plasma in which the cells are suspended, such as white blood cells and red blood cells.
    1. Plasma transports food, carbon dioxide and nitrogenous wastes (urea and uric acid) in dissolved form.
    2. Red blood cells transport oxygen. Blood also transports many other Von substances like hormones, salt, etc. Platelets present in the blood stop bleeding from injuries by forming blood clots.
  3. Blood vessels: Blood vessels form a closed network of tubes to reach all the tissues. Thus, the blood pushed by the heart flows in the blood vessels (arteries) and also comes back to the heart back to the heart in the blood vessels (veins).

Question 3. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:

The separation of oxygenated and deoxygenated blood from mixing allows a highly efficient supply of oxygen to the body. This is useful in animals that have energy needs such as efficient mams and birds. They constantly use energy to maintain their body temperature.

Question 4. What are the components of the transport system in highly organised plants?
Answer:

In higher plants transport system consists of vascular bundles which contain complex conducting tissues:

  1. Xylem which contains vessels tracheids, sclerenchyma (all dead tissues) and living xylem parenchyma. Vessels and tracheids form tubular structures and conduct water and dissolved minerals from roots to other parts.
  2. Phloem which contains sieve tubes, companion cells, phloem sclerenchyma and phloem parenchyma. Sieve tubes and companion cells conduct prepared food from leaves to other parts of the plant.

Question 5. How are water and minerals transported in plants? Lean abortion
Answer:

  1. The xylem conducts water and dissolved minerals through channels formed by tracheids. and vessels from roots to all parts of the plant body.
  2. The root cells actively take up ions from the soil causing steady movement of water into the root xylem. This inward flow of water pushes water upward with pressure, called root pressure. Transport of water by root pressure is more important at night, to log especially in small plants.
  3. Evaporation of water molecules from the cells of a leaf, called transpiration, creates a suction which pulls water from the xylem of roots. Thus, transpiration helps in the absorption and upward movement of water and dissolved minerals from the roots to the leaves in tall plants.

Question 6. How is food transported in plants?
Answer:

Transportation of prepared food such as carbohydrates, amino acid, etc., through the phloem from leaves to all parts of plants, is called translocation. Companion cells and sieve boron tubes conduct prepared food.

The mechanism of translocation is an active process and utilizes energy. Material like sucrose is transferred from leaf cells or the site of storage into phloem tissue. This process requires energy which is provided by ATP (adenosine triphosphate) molecules.

  • Entry of sucrose into phloem tissue causes an increase in osmotic pressure. As a result water from outside moves into the phloem. The osmotic’ pressure moves the dissolved material in the phloem to tissues which have less pressure.
  • Thus, material moves in phloem, mainly in sieve tubes, to the place of need in the plant body. For example, sugar is translocated from its storage organ, root or stem to growing buds which need energy.

Question 7. Describe the structure and functioning of nephrons.
Answer:

Structure of nephron: A nephron is the basic unit of filtration. A large number of nephrons are closely packed in the kidney.

  • Each nephron consists of a cluster of very thin-walled blood capillaries called Glomerulus. Glomerulus is associated with the cup-shaped structure present at the end of a tubule.
  • Thus, a nephron consists of a cluster of glomerulus which is a cluster of blood capillaries, a thin-walled Bowman’s capsule to collect the suig of sto filtrate and a long tubule which reabsorbs useful substances and also carries remaining urine to the collecting duct.

CBSE Class 10 Science Chapter 6 Life Processes Structure And Functioning Of Nephrons

  1. Remove excess water, wastes and other substances from your blood.
  2. Return substances like sodium, potassium or phosphorus whenever only of these substances run low in our body.

Question 8. What are the differences between the transport of materials in the xylem and phloem?
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Xylem And Phloem

Question 9. Compare the functioning of alveoli in the lungs and nephrons in the kidneys concerning their structure and functioning.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes The Functioning Of Alveoli In The Lungs And Nephrons

Question 10.

  1. What happens to the heart when muscles work harder?
  2. Which body system is directly affected when a person has heart disease?
  3. Which cells increase in number during infection?

Answer:

  1. When muscles work harder, they need more oxygen. So, the heart begins to pump blood faster to overcome that need.
  2. The circulatory system is directly affected when a person has heart disease.
  3. The number of white blood corpuscles (WBCs) increases during infection.

Question 11.

  1. Label any 4 parts in the given diagram.
  2. What are the two functions represented in this diagram?

Answer:

    1. Pulmonary artery to lungs
    2. Lung capillaries
    3. Pulmonary vein from lungs
    4. Aorta to body
    5. Capillaries in body organs
    6. Vena cava from the body.
  1. The two functions represented are:
    1. Transport of oxygen and carbon dioxide
    2. Exchange of oxygen and carbon dioxide

CBSE Class 10 Science Chapter 6 Life Processes Transport And Exchange Of Oxygen And Carbon Dioxide

Question 12. What is blood pressure? How is it measured? Give one difference between systolic pressure and diastolic pressure.
Answer:

Blood Pressure: It is the force that blood exerts against the wall of a vessel. This pressure is much greater in arteries than in veins.

  • It is measured using an instrument called a sphygmomanometer.
  • The pressure of blood inside the artery during contraction or ventricular systole is called systolic pressure and pressure in the artery during relaxation or ventricular diastole is called diastolic pressure.
  • The normal systolic pressure is about 120 mm of Hg and diastolic pressure is 80 mm of Hg.

Question 13. Draw a diagram of the human excretory system and label kidney and ureters on it.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Excretory System In Human Beings

Question 14. Explain the mechanism of gaseous exchange between tissues and blood.
Answer:

The exchange of respiratory gases, i.e., oxygen and carbon dioxide, occurs between the blood and tissues.

  • In tissues, oxygen is used up for their activities and carbon dioxide is released. The blood from the lungs has a high concentration of oxygen and a low concentration of carbon dioxide.
  • Due to this difference in concentration of oxygen and carbon dioxide, the exchange of gases takes place between tissue and blood.

Question 15. Draw a schematic diagram to show the opening and closing of the stoma.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Showing Opening And Closing Of Stoma

Question 16.

  1. Write the balanced chemical equation for the process of photosynthesis.
  2. When do the desert plants take up carbon dioxide and perform photosynthesis?

Answer:

  1. 6CO2 + 6H2O→ C6H12O6 + 6O2
  2. Desert plants take up CO2 during the night. The CO2 is stored in organic acid at night and photosynthesis takes place during day time in the presence of sunlight.

Question 17. State in brief the function of:

  1. Renal artery
  2. Ureter
  3. Kidney
  4. Urinary bladder

Answer:

  1. Renal artery: It brings blood to the kidney for filtering waste.
  2. Kidney: It filters the blood and forms urine.
  3. Ureter: It passes urine from the kidneys to the urinary bladder.
  4. Urinary bladder: It stores the urine until it is released through the urethra.

Question 18. Draw a neat labelled diagram of the structure of a chloroplast.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Structure Of Chloroplast

Question 19. What will happen if:

  1. Peristaltic movements do not occur all along the oesophagus.
  2. There are no platelets in the blood.
  3. Rings of cartilage are not present in the trachea.

Answer:

  1. If peristaltic movements do not occur, the food will not be able to be pushed forward, into the stomach.
  2. In case of no platelets, blood will not clot during bleeding.
  3. In the absence of rings of cartilage, air passage will collapse.

Question 20.

  1. Name the following:
    1. The part where the air is filtered by fine hair and mucus.
    2. The part which terminates in balloon-like structures.
    3. Balloon-like structures where the exchange of gases takes place. part which separates the chest cavity from the abdominal cavity.
  2. Why is the rate of breathing in aquatic organisms much faster than in terrestrial organisms?

Answer:

    1. Nasal passage
    2. Bronchioles
    3. Alveoli
    4. Diaphragm.
  1. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that in terrestrial organisms.

Question 21. What is lymph? How is the composition of lymph different from blood plasma? What is the direction of its flow? List two functions of the lymphatic system.
Answer:

Lymph is a circulatory fluid which flows in the lymphatic capillaries which join to form large lymph vessels.

Composition of Lymph:

Lymph is colourless and contains less protein than plasma.

Lymph flows from tissues to the heart.

Functions:

  1. Lymph carries digested and absorbed fat from the intestine.
  2. It drains excess fluid from extracellular space back into the blood.

Question 22.

  1. Why does a piece of bread taste sweet when chewed for some time?
  2. Cellulose acts as roughage in man but serves as a source of nutrients in cows. Justify the statement.

Answer:

  1. The saliva present in the mouth contains an enzyme called salivary amylase which breaks down starch present in the bread into sugar. That is why, a piece of bread tastes sweet when chewed for some time.
  2. Cellulose requires time to digest. Herbivore animals like cows have longer intestines which allows cellulose to be digested whereas human beings have comparatively short intestines. Thus, it is not digested and serves merely as roughage.

Question 23. List the three kinds of blood vessels of the human circulatory system and write their functions in tabular form.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Blood Vessels And Functions

Question 24.

  1. A product is formed in our muscles due to the breakdown of glucose when there is a lack of oxygen. Name the product and also mention the effect of the build up of this product.
  2. Differentiate between fermentation in yeast and aerobic respiration based on end products formed.

Answer:

  1. Lactic acid is formed. The formation of lactic acid in the muscles causes cramps.
  2. During fermentation in yeast, ethanol is formed while in aerobic respiration, carbon dioxide and water are formed. The energy (ATP) released in the process of fermentation is also very low compared to aerobic respiration.

Question 25. Name the glands associated with the digestion of starch in the human digestive tract and mention their role. How is the required pH maintained in the stomach and small intestine?
Answer:

  1. Salivary glands salivary amylase breaks starch into sugar.
  2. Pancreas – pancreatic amylase (juice) digests starch.
  3. Intestinal glands – intestinal amylase – digests starch.
  4. Acidic pH in the stomach – due to Hydrochloric acid secreted by the stomach.
  5. Alkaline pH in small intestine – due to bile (liver)/pancreatic juice (pancreas).

Question 26. Name the two glands associated with the digestive system in humans Name their secretion also.
Answer:

  1. Gastric glands: These are present in the walls of the stomach. These release HCl, a protein-digesting enzyme called pepsin and mucus.
  2. Pancreatic enzymes: The pancreas secretes pancreatic juice which contains enzymes like trypsin for digesting proteins and lipase for breaking down emulsified fats.

Question 27. What are the final products produced after digestion of carbohydrates, proteins and fats?
Answer:

  1. Carbohydrates – glucose
  2. Proteins – amino acid
  3. Fats glycerol + fatty acid

Question 28. Name the following concerning the alimentary canal:

    1. The largest gland.
    2. The gland secretes digestive enzymes as well as hormones.
    3. The part where digested food is absorbed.
  1. What are villi? Mention their function.

Answer:

    1. Liver
    2. Pancreas
    3. Small intestine
  1. Numerous finger-like projections present in the inner lining of the intestine which increase the surface area for absorption.

Function: These absorb digested food in the small intestine and absorb water in the large intestine.

Question 29. What are the common features of the respiratory organs in aquatic and terrestrial animals?
Answer:

  • Large surface area
  • Thin-walled
  • Richly supplied with blood
  • Well protected.

Question 30. Where do the following processes take place?

  1. Filtration of blood
  2. Reabsorption of glucose and water
  3. Secretion of salts to make final filtrate
  4. Collection of the final filtrate.

Answer:

  1. Glomerulus
  2. Renal tubules
  3. Tubular cells
  4. Collecting duct

Question 31. Explain how Paramecium obtains its food.
Answer:

In Paramecium, the food is taken in at a specific spot and is moved to this spot by the movement of cilia which covers the entire surface of the cell.

Question 32. Assume that you are a veterinary surgeon and you have removed a good length of the small intestine of a bear that was suffering from an intestinal tumour. Now, would you suggest a plant-based or a meat-based diet for the bear after its recovery? Give reasons for your answer.
Answer:

I would suggest a meat-based diet. Because to digest a plant-based diet, the bear will need a longer small intestine, since the cellulose present in the plant diet requires a longer time for digestion.

Question 33.

  1. Name the following:
    1. Part in which urine is produced
    2. The part which stores the urine
    3. The part which connects 1 and 2
    4. A part from which urine is passed out.
  2. Name the factors on which the amount of water reabsorbed along the tubular part of the nephron depends.

Answer:

    1. Kidney
    2. Ureter
    3. Urinary bladder
    4. Urethra,
  1. The amount of water reabsorbed depends on how much excess water there is in the body and on how much-dissolved waste there is to be excreted.

Question 34. State two functions of stomata.
Answer:

A massive amount of gaseous exchanges takes place in leaves through stomata. Large amounts of water can be lost through stomata (transpiration).

Question 35.

  1. Name the following:
    1. The organ which stores bile
    2. The organ which produces bile
  2. Name one enzyme present in pancreatic juice and write its functions.
  3. What are peristaltic movements?

Answer:

    1. Gall bladder
    2. Liver
    1. Trypsin: Digests proteins into amino acids.
    2. Amylase: Digests carbohydrates into simple sugars.
    3. Lipase: Digests fats into fatty acids and glycerol.
  1. The lining of the alimentary canal has muscles that contract rhythmically to push food forward. These are known as peristaltic movements.

Question 36. Define transpiration. State its two functions.
Answer: The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

Functions:

  1. It helps in the absorption and upward movement of water and minerals dissolved in it from roots to the leaves.
  2. It also helps in temperature regulation.

Question 37. How would digestion of food be affected if the bile duct is completely blocked?
Answer: If the bile duct is completely blocked, bile juice will not reach the small intestine and the digestion of fats will be affected.

Question 38. What is saliva? State its role in the digestion of food.
Answer:

Saliva is a watery fluid secreted by the salivary glands in the mouth. In the mouth, food gets mixed up with saliva secreted by salivary glands. Saliva contains an enzyme called salivary amylase which breaks polysaccharide starch into disaccharide maltose (sugar).

Question 39. Distinguish between saprozoic and holozoic nutrition.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Saprozonic Nutrition And Holozoic Nutrition

Question 40. What are the final products after the digestion of carbohydrates and proteins?
Answer:

The final product produced after digestion of carbohydrates is glucose and that of proteins is amino acid.

Question 41. Explain the process of nutrition in Amoeba.
Answer:

The mode of nutrition in Amoeba is holozoic. It feeds on unicellular plants or animals. The various steps of nutrition are ingestion, digestion, assimilation and egestion.

  • When Amoeba comes in contact with food particles, they send out pseudopodia, which engulf the prey by forming a food cup, which is known as ingestion.
  • When the tips of the encircling pseudopodia touch each other, the food is encaptured into a bag called a food vacuole. Various enzymes from the cytoplasm enter the food vacuole and break them down into simple soluble molecules.
  • This food is then absorbed by the cytoplasm of Amoeba through the process of diffusion. Amoeba cell obtains energy from the absorbed food. The undigested food is thrown out from its cell membrane.

Question 42. Name the following

  1. The process in plants that links light energy with chemical energy
  2. Organisms that can prepare their food
  3. The cell organelle where photosynthesis occurs
  4. Cells that surround a stomatal pore
  5. Organisms that cannot prepare their food
  6. An enzyme secreted from gastric glands in the stomach that acts on proteins.

Answer:

  1. Photosynthesis
  2. Autotrophs
  3. Chloroplast
  4. Guard cells
  5. Heterotrophs

Question 43. “All plants give out oxygen during the day and carbon dioxide during the night”. Do you agree with this statement? Give reason.
Answer:

During this time, as the rate of photosynthesis is more than the rate of respiration, the next result is the evolution of oxygen. At night there is no photosynthesis, so, they give out carbon dioxide due to respiration.

Question 44. How do the guard cells regulate the opening and closing of stomatal pores?
Answer:

The swelling of guard cells due to the absorption of water causes the opening of stomatal pores while the shrinking of guard cells closes the pores. Opening and closing of stomata occurs due to turgor changes in guard cells.

When guard cells are turgid, the stomatal pore is open while in flaccid conditions, the stomatal aperture closes.

Question 45. Two green plants are kept separately in oxygen-free containers, one in the dark and the other in continuous light. Which one will live longer? Give reasons.
Answer:

Plants kept in continuous light will live longer because they will be able to produce the oxygen required for their respiration through the process of photosynthesis.

Question 46. If a plant is releasing carbon dioxide and taking in oxygen during the day, does it mean that there is no photosynthesis occurring? Justify your answer.
Answer:

Release of CO2 and intake of O2 gives evidence that either photosynthesis is not taking place or its rate is too low. Normally during the daytime, the rate of photosynthesis is much more than the rate of respiration. So, CO2 produced during respiration is used up for photosynthesis hence CO2 is not released.

Question 47. Why do fishes die when taken out of water?
Answer:

Fishes respire with the help of gills. Gills are richly supplied with blood capillaries and can readily absorb oxygen dissolved in water. Since fishes cannot absorb gaseous oxygen they die soon after they are taken out of water.

Question 48. Differentiate between an autotroph and a heterotroph.
Answer:

CBSE 10 Science Chapter 6 Life Processes Difference Between Autotroph And Heterotroph

Question 49. Is ‘nutrition’ a necessity for an organism? Discuss.
Answer:

Food is required for the following purposes:

  1. It provides energy for the various metabolic processes in the body.
  2. It is essential for the growth of new cells and the repair or replacement of worn-out cells.
  3. It is needed to develop resistance against various diseases.

Question 50. What would happen if green plants disappeared from Earth?
Answer:

Green plants are the sources of energy for all organisms. If all green plants disappear from the earth, all the herbivores will die due to starvation and so will the carnivores.

Question 51. Leaves of a healthy potted plant were coated with Vaseline. Will this plant remain healthy for long? Give reasons for your answer.
Answer:

This plant will not remain healthy for a long time because

  1. It will not get oxygen for respiration.
  2. It will not get carbon dioxide for photosynthesis.
  3. The upward movement of water and minerals would be hampered due to lack of transpiration.

Question 52. How does aerobic respiration differ from anaerobic respiration?
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Right Atrium And Left Atrium

Question 53. Match the words of Column with that of Column

CBSE Class 10 Science Chapter 6 Life Processes The Words Of Column A And Column B

Answer:

  1. 2
  2. 1
  3. 4
  4. 3

Question 54. Differentiate between an artery and a vein.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Artery And Vein

Question 55. What are the adaptations of leaves for photosynthesis?
Answer:

  1. Leaves provide a large surface area for maximum light absorption.
  2. Leaves are arranged at right angles to the light source in a way that causes overlapping.
  3. The extensive network of veins enables quick transport of substances to and from the mesophyll cells.
  4. Presence of numerous stomata for gaseous exchange.
  5. The chloroplasts are more in number on the upper surface of leaves.

Question 56. Why is the small intestine in herbivores longer than in carnivores?
Answer:

Digestion of cellulose takes a longer time. Hence, herbivores eating grass need a longer small intestine to allow complete digestion of cellulose. Carnivorous animals cannot digest cellulose, hence they have a shorter intestine.

Question 57. What will happen if mucus is not secreted by the gastric glands?
Answer:

Gastric glands in the stomach release hydrochloric acid, enzyme pepsin and mucus. The mucus protects the inner lining of the stomach from the action of hydrochloric acid and enzyme pepsin.

If the mucus is not released, it will lead to erosion of the inner lining of the stomach, leading to acidity and ulcers.

Question 58. What is the significance of the emulsification of fats?
Answer:

Fats are present in food in the form of large globules which makes it difficult for enzymes to act on them. Bile salts present in bile break them down mechanically into smaller globules which increases the efficiency of fat digesting enzymes.

Question 59. What causes the movement of food inside the alimentary canal?
Answer:

The wall of the alimentary canal contains muscle layers. Rhythmic contraction and relaxation of these muscles push the food forward. This is called peristalsis, which occurs all along the gut.

Question 60. Why does absorption of digested food occur mainly in the small intestine?
Answer:

Maximum absorption occurs in the small intestine because

  1. Digestion is completed in small intestine
  2. The inner lining of the small intestine is provided with villi which increases the surface area for absorption.
  3. The wall of the intestine is richly supplied with blood vessels (which take the absorbed food to every cell of the body).

Question 61. Match Group (A) with Group (B)

CBSE Class 10 Science Chapter 6 Life Processes Match Group A With Group B

Answer:

  1. D
  2. C
  3. A
  4. B

Question 62. Why is the rate of breathing in aquatic organisms much faster than in terrestrial organisms?
Answer:

Aquatic organisms like fishes obtain oxygen from water present in a dissolved state through their gills. Since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air, the rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms.

Question 63. Why is blood circulation in the human heart called double circulation?
Answer:

The blood circulation in the human heart is called double circulation because the blood passes through the heart twice in one complete cycle of the body – once through the right half in the form of deoxygenated blood and once through the left half in the form of oxygenated blood.

Question 64. What is the advantage of having a four-chambered heart?
Answer:

In a four-chambered heart, the left half is completely separated from the right half by septa. This prevents oxygenated and deoxygenated blood from mixing.

This allows a highly efficient supply of oxygenated blood to all parts of the body. This is useful in animals that have high energy needs, such as birds and mammals.

Question 65. Mention the major events during photosynthesis.
Answer:

The major events during photosynthesis are:

  1. Absorption of light energy by chlorophyll
  2. Conversion of light energy to chemical energy
  3. Splitting of H2O into H2, O2 and eTM
  4. Reduction of CO2 to carbohydrates

Question 66. In each of the following situations, what happens to the rate of photosynthesis?

  1. Cloudy days
  2. No rainfall in the area
  3. Good manuring in the area
  4. Stomata get blocked due to dust

Answer:

  1. Decreases.
  2. Decreases.
  3. Increases.
  4. Decreases.

Question 67. Name the energy currency in the living organisms. When and where is it produced?
Answer:

Adenosine triphosphate (ATP) is produced during respiration in living organisms and also during photosynthesis in plants.

Question 68. What is common for Cuscuta; ticks and leeches?
Answer:

All are parasites, they derive nutrition from plants or animals without killing them.

Question 69. Explain the role of the mouth in the digestion of food.
Answer:

  1. Food is crushed into small pieces by the teeth.
  2. It mixes with saliva and the enzyme amylase (found in saliva) breaks down starch into sugars.
  3. The tongue helps in the thorough mixing of food with saliva.

Question 70. What are the functions of gastric glands present in the wall of the stomach?
Answer:

  1. Production of pepsin enzyme that digests proteins
  2. Secretion of mucus for protection of the inner lining of the stomach.

Question 71. Match the terms in Column (A) with those in Column (B)

CBSE Class 10 Science Chapter 6 Life Processes The Terms In Column A With Those In Column B

Answer:

  1. A
  2. D
  3. B
  4. C

Question 72. Name the correct substrates for the following enzymes:

  1. Trypsin
  2. Amylase
  3. Pepsin
  4. Lipase

Answer:

  1. Protein
  2. Starch
  3. Protein
  4. Fats

Question 73. Why do veins have thin walls as compared to arteries?
Answer:

Arteries carry blood from the heart to various organs of the body under high pressure so they have thick and elastic walls. Veins collect the blood from different organs and bring it back to the heart.

The blood is no longer under pressure so the walls are thin with valves to ensure that blood flows only in one direction.

Question 74. What will happen if platelets are absent in the blood?
Answer: In the absence of platelets, the process of clotting will be affected.

Question 75. Plants have low energy needs as compared to animals. Explain.
Answer: Plants do not move. In a large plant body, there are many dead cells like sclerenchyma, as a result, it requires less energy as compared to animals.

Question 76. Why and how does water enter continuously into the root xylem?
Answer:

Cells of the root are in close contact with soil and so actively take up ions. The ion- concentration increases inside the root and hence osmotic pressure increases the movement of water from the soil into the root which occurs continuously.

Question 77. Why is transpiration important for plants?
Answer:

Transpiration is important because

  1. It helps in the absorption and upward movement of water and minerals from roots to leaves.
  2. It prevents the plant parts from heating up.

Question 78. How do the leaves of plants help in excretion?
Answer:

Many plants store waste materials in the vacuoles of mesophyll cells and epidermal cells. When old leaves fall, the waste materials are excreted along with the leaves.

Question 79. What is the functional difference between the four chambers of the heart?
Answer:

CBSE Class 10 Science Chapter 6 Life Processes Difference Between Right Atrium And Left Atrium

Question 80. Give one structural and one functional difference between an artery and a vein.
Answer:

CBSE Class 10 Science Chapter 6 Life Processes The Three Kind Of Blood Vessels Of Human Circulatory System

Question 81. Give the role of the following in plants and animals:

  1. Guard cells
  2. Hymph
  3. Bio catalyst
  4. Phloem

Answer:

  1. Stomatal opening
  2. Fat transport
  3. Speeding up the digestive process
  4. Transport of food

Question 82. Name the excretory wastes of the following:

  1. Lungs
  2. Kidney
  3. Skin
  4. Leaves.

Answer:

  1. CO2
  2. Urea
  3. Sweat
  4. O2, H2O and CO2

Question 83. After a vigorous exercise, you may experience cramps in your leg muscles. Why does this happen?
Answer:

This happens due to the build-up of lactic acid in our muscles. During heavy exercise, there is a lack of oxygen in our muscle cells and therefore, another pathway for the breakdown of pyruvate of lactic acid takes place.

Question 84. Which organ secretes a hormone when the blood sugar rises? Name the digestive enzymes released by this organ.
Answer:

The pancreas secretes the hormone insulin which regulates the blood sugar level. Digestive enzymes secreted by the pancreas are trypsin and lipase.

Question 85. Explain the assimilation of carbohydrates, proteins and fats inside the cells.
Answer:

  1. Carbohydrates: Glucose is used to release energy. Excess of glucose is brought to the liver and converted to glycogen (glycogenesis) and stored there. As the need arises the glucose is reformed from glycogen (glycogenolysis) and released into the bloodstream.
  2. Proteins: Most of the amino acids are used in protein synthesis which helps in the growth and repair of body tissues. Proteins may act as structural proteins or act as enzymes and thus control various metabolic activities of the body.
    1. Amino acids which are not used as such, are deaminated in the liver i.e., the amino group (-NH2) is removed. The amino group forms ammonia (NH3) which is highly toxic and must be removed. This is done by liver cells which convert it to harmless urea, which is then a Diagram showing absorption of food in the intestine excreted by the kidneys. Urea is formed only in the liver.
  3. Fats: Fats release energy. Excess fats are stored in the liver and the form of adipose tissue. Some fats enter into the composition of cell organelles.

Question 86. For the last few days Rakesh has started cycling long distances. But on his way back he feels cramps in his calf muscles.

  1. What can be the possible cause of it?
  2. Name the process involved and its equation.
  3. How can Rakesh have quick relief from muscle cramps?

Answer:

1. Accumulation of lactic acid in the muscles.

2. Anaerobic respiration.

CBSE Class 10 Science Chapter 6 Life Processes Anaerobic Respiration

3. By having lots of water baths and a massage.

Question 87. During sports practice in the school, many students got tired very soon and were found to be anaemic. Health and Wellness group students made a presentation on the ‘right diet for right health’ and uploaded it on school website.

  1. What causes anaemia?
  2. Name two foods that can help to overcome the problem of anaemia.
  3. What value of Health and Wellness group students is displayed in the above act?

Answer:

  1. A deficiency of haemoglobin in red blood cells leads to anaemia.
  2. Spinach and beetroot.
  3. Health and Wellness group students showed the value of group work and being responsible citizens.

Question 88. A group of students from class X were given the project on ‘respiratory diseases’. The students spread an awareness camp on how to overcome respiratory problems.

  1. What are the causes of respiratory diseases?
  2. Name two respiratory diseases.
  3. What value of students of class X is displayed in the above act?

Answer:

  1. Too small and suspended particles inhaled causing respiratory diseases.
  2. Asthma and bronchitis.
  3. Students of class X showed the value of group work, being cooperative learners and responsible citizens.

Life Processes Multiple Choice Questions

1. Which of the following statements about the autotrophs is incorrect?

  1. They synthesise carbohydrates from carbon dioxide and water in the presence of sunlight and chlorophyll
  2. They store carbohydrates in the form of starch
  3. They convert carbon dioxide and water into carbohydrates in the absence of sunlight
  4. They constitute the first trophic level in food chains

Answer: 3. They convert carbon dioxide and water into carbohydrates in the absence of sunlight

2. In which of the following groups of organisms, food material is broken down outside the body and absorbed?

  1. Mushrooms, green plants, Amoeba
  2. Yeast, mushroom, bread mould
  3. Paramecium, Amoeba, Cuscuta
  4. Cuscuta, lice, tapeworm

Answer: 2. Yeast, mushroom, bread mould

3. Select the correct statement

  1. Heterotrophs do not synthesise their food
  2. Heterotrophs utilise solar energy for photosynthesis
  3. Heterotrophs synthesise their food
  4. Heterotrophs are capable of converting carbon dioxide and water into carbohydrates

Answer: 1. Heterotrophs do not synthesise their food

4. Which is the correct sequence of parts in the human alimentary canal?

  1. Mouth → stomach → small intestine → oesophagus → large intestine
  2. Mouth → oesophagus → stomach → large intestine → small intestine
  3. Mouth → stomach → oesophagus → small intestine → large intestine
  4. Mouth → oesophagus → stomach → small intestine → large intestine

Answer: 4. Mouth → oesophagus → stomach → small intestine → large intestine

5. If salivary amylase is lacking in the saliva, which of the following events in the mouth cavity will be affected?

  1. Proteins breaking down into amino acids
  2. Starch breaking down into sugars
  3. Fats breaking down into fatty acids and glycerol
  4. Absorption of vitamins

Answer: 2. Starch breaking down into sugars

6. The inner lining of the stomach is protected by one of the following from hydrochloric acid. Choose the correct one

  1. Pepsin
  2. Mucus
  3. Salivary amylase
  4. Bile

Answer: 2. Mucus

7. Which part of the alimentary canal receives bile from the liver?

  1. Stomach
  2. Large intestine
  3. Small intestine
  4. Oesophagus

Answer: 2. Large intestine

8. A few drops of iodine solution were added to rice water. The solution turned blue-black. This indicates that rice water contains

  1. Complex proteins
  2. Simple proteins
  3. Fats
  4. Starch

Answer: 4. Starch

9. In which part of the alimentary canal food is finally digested?

  1. Stomach
  2. Mouth cavity
  3. Large intestine
  4. Small intestine

Answer: 4. Small intestine

10. Choose the function of the pancreatic juice from the following

  1. Trypsin digests proteins and lipase carbohydrates
  2. Trypsin digests emulsified fats and lipase proteins
  3. Trypsin and lipase digest fats
  4. Trypsin digests proteins and lipase-emulsified fats

Answer: 4. Trypsin digests proteins and lipase-emulsified fats

11. When air is blown from the mouth into a test tube containing lime water, the lime water turns milky due to the presence of

  1. Oxygen
  2. Carbon dioxide
  3. Nitrogen
  4. Water vapour

Answer: 2. Carbon dioxide

12. The correct sequence of anaerobic reactions in yeast is

CBSE Class 10 Science Chapter 6 Life Processes Anaerobic Reactions

Answer: 4.

13. Which of the following is most appropriate for aerobic respiration?

CBSE Class 10 Science Chapter 6 Life Processes Aerobic Respiration

Answer: 4.

14. Which of the following statement(s) is (are) true about respiration?

  1. During inhalation, ribs move inward and the diaphragm is raised
  2. In the alveoli, an exchange of gases takes place i.e., oxygen from alveolar air diffuses into blood and carbon dioxide from blood into alveolar air
  3. Haemoglobin has a greater affinity for carbon dioxide than oxygen
  4. Alveoli increase surface area for the exchange of gases
  1. 1 and 4
  2. 1 and 3
  3. 2 and 3
  4. 2 and 4

Answer: 4. 2 and 4

15. Which is the correct sequence of air passage during inhalation?

  1. Nostrils → larynx → pharynx → trachea → lungs
  2. Nasal passage → trachea → pharynx → larynx → alveoli
  3. larynx → nostrils → pharynx → lungs
  4. Nostrils → pharynx → larynx → trachea → alveoli

Answer: 4. Nostrils → pharynx → larynx → trachea → alveoli

16. During respiration exchange of gases takes place in

  1. Trachea and larynx
  2. Alveoli of lungs
  3. Alveoli and throat
  4. Throat and larynx

Answer: 2. Alveoli of lungs

17. Which of the following statement(s) is (are) true about heart?

  1. The left atrium receives oxygenated blood from different parts of the body while the right atrium receives deoxygenated blood from the lungs
  2. The left ventricle pumps oxygenated blood to different body parts while the right ventricle pumps deoxygenated blood to the lungs
  3. The left atrium transfers oxygenated blood to the right ventricle which sends it to different body parts
  4. Right atrium receives deoxygenated blood from different parts of the body while the left ventricle pumps oxygenated blood to different parts of the body
  1. 1
  2. 2
  3. 2 and 4
  4. 1 and 3

Answer: 3. 2 and 4

18. What prevents the backflow of blood inside the heart during contraction?

  1. Valves in heart
  2. Thick muscular walls of ventricles
  3. Thin walls of atria
  4. All of the above

Answer: 1. Valves in the heart

19. Single circulation i.e., blood flows through the heart only once during one cycle of passage through the body, is exhibited by

  1. Labeo, Chameleon, Salamander
  2. Hippocampus, Exocoetus, Anabas
  3. Hyla, Rana, Draco
  4. Whale, Dolphin, Turtle

Answer: 2. Hippocampus, Exocoetus, Anabas

20. In which of the following vertebrate groups/groups, the heart does not pump oxygenated blood to different parts of the body?

  1. Pisces and amphibians
  2. Amphibians and reptiles
  3. Amphibians only
  4. Pisces only

Answer: 4. Pisces only

21. Choose the correct statement that describes arteries.

  1. They have thick elastic walls, blood flows under high pressure; collect blood from different organs and bring it back to the heart
  2. They have thin walls with valves inside, blood flows under low pressure and carry blood away from the heart to various organs of the body
  3. They have thick elastic walls, blood flows under low pressure; and carry blood from the heart to various organs of the body
  4. They have thick elastic walls without valves inside, blood flows under high pressure and carries blood away from the heart to different parts of the body.

Answer: 4. They have thick elastic walls without valves inside, blood flows under high pressure and carries blood away from the heart to different parts of the body.

22. The filtration units of kidneys are called

  1. Ureter
  2. Urethra
  3. Neurons
  4. Nephrons

Answer: 4. Nephrons

23. Oxygen liberated during photosynthesis comes from

  1. Water
  2. Chlorophyll
  3. Carbon dioxide
  4. Glucose

Answer: 1. Water

24. The blood leaving the tissues becomes richer in

  1. Carbon dioxide
  2. Water
  3. Haemoglobin
  4. Oxygen

Answer: 1. Carbon dioxide

25. Which of the following is an incorrect statement?

  1. Organisms grow with time
  2. Organisms must repair and maintain their structure
  3. The movement of molecules does not take place among cells
  4. Energy is essential for life processes

Answer: 3. Movement of molecules does not take place among cells

26. The internal (cellular) energy reserve in autotrophs is

  1. Glycogen
  2. Protein
  3. Starch
  4. Fatty acid

Answer: 3. Starch

27. Which of the following equations is the summary of photosynthesis?

  1. 6CO2+ 12H2O → C6H12O6+ 6O2+ 6H2O
  2. 6CO2 + H2O + Sunlight → C6H12O6 + O2+ 6H2O
  3. 6CO2 + 12H2O + Chlorophyll + Sunlight → C6H12O6 + 6O2+ 6H2O
  4. 6CO2 + 12H2O + Chlorophyll + Sunlight → C6H12O6+6CO2+ 6H2O

Answer: 3. 6CO2 + 12H2O + Chlorophyll + Sunlight → C6H12O6 + 6O2+ 6H2O

28. Choose the event that does not occur in photosynthesis

  1. Absorption of light energy by chlorophyll
  2. Reduction of carbon dioxide to carbohydrates
  3. Oxidation of carbon to carbon dioxide
  4. Conversion of light energy to chemical energy

Answer: 3. Oxidation of carbon to carbon dioxide

29. The opening and closing of the stomatal pore depends upon

  1. Oxygen
  2. Temperature
  3. Water in guard cells
  4. The concentration of CO2 in stomata

Answer: 3. Water in guard cells

30. Choose the forms in which most plants absorb nitrogen

  1. Proteins
  2. Nitrates and Nitrites
  3. Urea
  4. Atmospheric nitrogen
  1. 1 and 2
  2. 2 and 3
  3. 3 and 4
  4. 1 and 4

Answer: 2. 2 and 3

31. Which is the first enzyme to mix with food in the digestive tract?

  1. Pepsin
  2. Cellulase
  3. Amylase
  4. Trypsin

Answer: 3. Amylase

32. Which of the following statement(s) is (are) correct?

  1. Pyruvate can be converted into ethanol and carbon dioxide by yeast
  2. Fermentation takes place in aerobic bacteria
  3. Fermentation takes place in mitochondria
  4. Fermentation is a form of anaerobic respiration
  1. 1 and 3
  2. 2 and 4
  3. 1 and 4
  4. 2 and 3

Answer: 3. 1 and 4

33. Lack of oxygen in muscles often leads to cramps among cricketers. This results due to

  1. Conversion of pyruvate to ethanol
  2. Conversion of pyruvate to glucose
  3. Non-conversion of glucose to pyruvate
  4. Conversion of pyruvate to lactic acid

Answer: 4. Conversion of pyruvate to lactic acid

34. Choose the correct path of urine in our body

  1. Kidney → ureter → urethra → urinary bladder
  2. Kidney → urinary bladder → urethra → ureter
  3. Kidney → ureters → urinary bladder → urethra
  4. Urinary bladder → kidney → ureter → urethra

Answer: 3. Kidney → ureters → urinary bladder → urethra

35. During deficiency of oxygen in tissues of human beings, pyruvic acid is converted into lactic acid in the

  1. Cytoplasm
  2. Chloroplast
  3. Mitochondria
  4. Golgi body

Answer: 1. Cytoplasm

CBSE Class 10 Science Chapter 10 Light-Reflection And Refraction Short Answer Questions

Class 10 Science Chapter Light-Reflection And Refraction

Question 1. Define the principal focus of a concave mirror.
Answer:

It is a point on the principal axis at which a beam of incident light parallel to the principal axis converges after reflection from the concave mirror.

Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:

Focal length = \(\frac{1}{2}\) x Radius of curvature

= \(\frac{1}{2}\) × 20 cm

= 10 cm

Question 3. Name a mirror that can give an erect and enlarged image of an object.
Answer: Concave mirror

Question 5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:

As the radius of curvature and focal length of a convex mirror are positive, so

R = +32 cm

and f= \(\frac{R}{2}\)

f = +16 cm

Question 6. A concave mirror produces a times magnified (enlarged) real image of an object placed 10 cm in front of it. Where is the image located?
Answer:

As the image is real, so magnification m must be negative.

∴ \(m=\frac{h^{\prime}}{h}=-\frac{v}{u}=-3\)

Or v = 3u

But u = 10 cm

∴ v = 3 x (-10)

v = -30 cm

Thus, the image is located at a distance of 30 cm from the mirror on the object side of the mirror.

Question 7. A ray of light traveling in air enters obliquely into water. Does the light ray bends towards the normal or away from the normal? Why?
Answer:

Light travels faster in the rarer mediums as air and slower in the denser mediums as water. As a ray of traveling in air enters obliquely into the water, it slows down and bends towards the normal.

Question 8. Light enters from air to glass and has an ing refractive info of dex 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3 x 108 ms-1.
Answer:

The refractive index of glass, ng = 1.50

Speed of light in vacuum, c = 3 × 108 ms-1

Speed of light in glass = \(\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{1.5}\) = 2 × 108 ms-1

Question 9. Find out, the medium having the highest optical density. Also, find the medium with the lowest optical density.
Answer:

The diamond has the highest refractive index (2.42), so it has the largest optical density. Air has the lowest refractive index (1.0003), so it has the lowest optical density.

Question 10. You are given kerosene, turpentine, and water. In which of these does the light travel fastest? Use the information given.

Refractive indices of some material media (concerning vacuum):

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Indices Of Some Material Media

Answer:

For kerosene, n = 1.44

For turpentine oil, n = 1.47

For water, n = 1.33

As water has the lowest refractive index, so light travels faster in this optically rarer medium than in kerosene and turpentine oil.

Question 11. The refractive index of a diamond is 2.42. What is the meaning of this statement?
Answer:

The refractive index of a diamond is 2.42. It indicates that the ratio of the speed of light in air
to that in diamond is 2.42.

Question 12. Define 1 dioptre of power of a lens.
Answer: One dioptre is the power of a lens whose focal length is 1 meter.

Question 13. 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:

Here, v = +50 cm [v is positive for real image]

As the real image is of the same size as the object, so

⇒ \(m=\frac{h^{\prime}}{h}=\frac{v}{u}=-1\)

Or u = -v=-50 cm

Now,\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{+50}-\frac{1}{-50}=+\frac{2}{50}=+\frac{1}{25}\)

Or f = 25 cm = 0.25 m

P = \(\frac{1}{f}=\frac{1}{+0.25 \mathrm{~m}}\)= + 4 D

Question 14. Find the power of a concave lens of focal length 2 m.
Answer:

As the focal length of a concave lens is negative, so f = − 2 m

Power,

P = \(\frac{1}{f(\text { in } \mathrm{m})}\)

⇒ \(\frac{1}{-2 \mathrm{~m}}=-0.5 \mathrm{D}\)

Question 15. Which one of the following materials cannot be used to make a lens?

  1. Water
  2. Glass
  3. Plastic
  4. Clay

Answer: 4. Clay being opaque, does not transmit light.

Question 16. 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?

  1. Between the principal focus and the center of curvature.
  2. At the center of curvature.
  3. Beyond the center of curvature.
  4. Between the pole of the mirror and its principal focus.

Answer: 4. Between the pole of the mirror and its principal focus.

Question 17. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

  1. At the principal focus of the lens.
  2. At twice the focal length.
  3. At infinity.
  4. Between the optical center of the lens and its principal focus.

Answer: 2. At twice the focal length.

Question 18. 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

  1. Both concave.
  2. Both convex.
  3. The mirror is concave and the lens is convex.
  4. The mirror is convex, but the lens is concave.

Answer: 1. Both concave (as both concave mirror and concave lens have a negative focal length).

Question 19. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

  1. Plane.
  2. Concave.
  3. Convex.
  4. Either plane or convex.

Answer: 3. The mirror is convex.

Question 20. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

  1. A convex lens of focal length 50 cm.
  2. A concave lens of focal length 50 cm.
  3. A convex lens of focal length 5 cm.
  4. A concave lens of focal length 5 cm.

Answer: 3. A convex lens of short focal length (5 cm).

Question 21. 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:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Concave Mirror

The object should be placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. Then the image will be virtual, erect, and larger than the object.

Question 22. Name the type of mirror used in the following situations:

  1. Headlights of a car
  2. Side/rear-view mirror of a vehicle
  3. Solar furnace
  4. Support your answer with reason.

Answer:

  1. Concave mirrors are used as reflectors in the headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.
  2. A convex mirror is used as a side/rear-view mirror of a vehicle because of the following reasons:
    1. A convex mirror always forms an erect, virtual, and diminished image of an object placed anywhere in front of it.
    2. A convex mirror has a wider field of view than a plane mirror of the same size.
  3. Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Question 23. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:

As shown in the figure, when the lower half of the convex lens is covered with black paper, it still forms the complete image of the object as that with the uncovered lens. However, the intensity of the image is reduced when one-half of the convex lens is covered with black paper.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction One Half Of A Convex Lens

Question 24. 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 nature of the image formed.
Answer:

A converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1:5, i.e., 1 cm represents 5 cm.

Therefore, on this scale, a 5 cm high object, an object distance of 25 cm, and a focal length of 10 cm can be represented by 1 cm high, 5 cm, and 2 cm lines respectively.

Now, we draw the ray diagram as follows:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Converging Lens Of Focal Length

  1. Draw a horizontal line to represent the principal axis of the convex lens.
  2. The Centre line is shown by DE.
  3. Mark two foci F and F” on two sides of the lens, each at a distance of 2 cm from the lens.
  4. Draw an arrow AB of height 1 cm on the left side of the lens at a distance of 5 cm from the lens.
  5. Draw a line AD parallel to the principal axis and then, allow it to pass straight through the focus (F) on the right side of the lens.
  6. Draw a line from A to C (center of the lens), which goes straight without deviation.
  7. Let the two lines starting from A meet at A’.
  8. Draw A’B’ perpendicular to the principal axis from A’.
  9. Now A’B’ represents the real, but inverted image of the object AB.
  10. Then, measure CB’ and A’B’. It is found that CB’ = 3.3 cm and A’B’ = 0.7 cm.
  11. Thus, the final position, nature, and size of the image A’ B’are:
    1. Position of image A’B’ = 3.3 cm x 5 = 16.5 cm from the lens of the opposite side.
    2. The nature of image A’ B’ is real and inverted.
    3. Height of image A’B’ = 0.7 cm x 5 = 3.5 cm, i.e., the image is smaller than the object.

Question 25. A concave lens of focal length 15 cm forms an image of 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:

Focal length, f= -15 cm [If is -ve for a concave lens]

Image distance, v = -10 cm

[Concave lens forms a virtual image on the same side as the object, so v is -ve]

As \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

∴ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\)

⇒ \(\frac{1}{-10}-\frac{1}{-15}\)

⇒ \(\frac{-3+2}{30}\)

⇒ \(-\frac{1}{30}\)

Object distance, u = −30 cm

Drawing the ray diagram: Using a scale of 1:5, we get u = -2 cm, f= -3 cm. We draw the ray diagram as follows:

  1. Draw the principal axis (a horizontal line).
  2. Draw a convex lens, keeping the principal center (C) on the principal axis.
  3. Mark points F and B on the left side of the lens at a distance of 15 cm and 30 cm. respectively.
  4. Join any point D (nearly at the top of the lens) and F by a dotted line.
  5. Draw a line AD, parallel to the principal axis.
  6. Draw a line A’B’ perpendicular to the principal axis from B’.
  7. Draw a line CA’ backward, so that it meets the line from D parallel to the principal axis at A.
  8. Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B.
  9. The AB is the position of the object. Measure distance BC. It will be found to be equal to 6 cm.

The object is placed at a distance of 6 cm x 5 = 30 cm from the lens.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Concave Lens Of Focal Length

Question 26. 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.
Answer:

Object distance, u = -10 cm

Focal length, f= +15 cm  [f is +ve for a convex mirror]

Image distance, v =?

As \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

∴ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

⇒ \(\frac{1}{+15}-\frac{1}{-10}\)

⇒ \(\frac{2+3}{30}\)

⇒ \(\frac{1}{6}\)

Image distance, v = +6 cm

As u is +ve, so a virtual, erect image is formed at a distance of 6 cm behind the mirror.

Question 27. The magnification produced by a plane mirror is +1. What does this mean?
Answer:

As m = \(\frac{h^{\prime}}{h}=-\frac{v}{u}\)

Given a plane mirror, m = +1, so

h’ h and v=-u

  1. m = 1 indicates that the size of the image is the same as a positive time sign of m indicates that a virtual image is formed behind the mirror.

Question 28. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror with a radius of curvature of 30 cm. Find the position of the image, its nature, and size.
Answer:

Object size, h = +5 cm

Object distance, u = -20 cm

The radius of curvature, R= +3.0 cm [R is +ve for a convex mirror]

Focal length, \(f=\frac{R}{2}=+15 \mathrm{~cm}\)

From the mirror formula,

∴ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

⇒ \(=\frac{1}{+15}-\frac{1}{-20}\)

⇒ \(\frac{4+3}{60}=\frac{7}{60}\)

Image distance, \(v=\frac{60}{7}=8.6 \mathrm{~cm}\)

Magnification, \(m=-\frac{v}{u}=\frac{h^{\prime}}{h}\)

Image size, \(h^{\prime}=-\frac{v h}{u}\)

⇒ \(-\frac{8.6 \times 5}{-20}\)

= 2.15 ≅ 2.2 cm

A virtual, erect image of a height of 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.

Question 29. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of a focal length of 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:

Object size, h = +7.0 cm

Object distance, u = −27 cm

Focal length, f = -18 cm

Image distance, v =?

Image size, h’ = ?

As \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

∴ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

⇒ \(\frac{1}{-18}-\frac{1}{-27}\)

⇒ \(\frac{-3+2}{54}=\frac{-1}{54}\)

Or V = – 54 cm

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image.

Magnification, \(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\)

Image size, \(h^{\prime}=-\frac{v h}{u}\)

⇒ \(\frac{(-54) \times(+7)}{(-27)}\)

= -14 cm

The image is real, inverted, and enlarged in size.

Question 30. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Answer:

Here, P = -2.0 D

⇒ \(f=\frac{1}{P}\)

⇒ \(=\frac{1}{-2.0 \mathrm{D}}\)

= − 0.5 m

The type of lens is concave.

Question 31. 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?
Answer:

Here, P = +1.5 D

⇒ \(f=\frac{1}{\mathrm{P}}\)

⇒ \(\frac{1}{+1.5 \mathrm{D}}\)

⇒ \(+\frac{10}{15} \mathrm{~m}\)

= +66.67 cm

As the focal length is positive, the prescribed lens is converging.

Question 32. Define absolute refractive index. Absolute refractive indices of medium ‘A’ and medium ‘B’ are ‘n’ and ‘n’ respectively, what is the refractive index of medium ‘B’ concerning medium ‘A’? How does the velocity of light vary with the change in the optical density of the media?
Answer:

The absolute refractive index of a medium is defined as the ratio of the velocity of light in a vacuum to the velocity of light in the medium. If c is the velocity of light in the vacuum and v in the medium, then n = c/v.

If na and nb are the absolute refractive indices of the medium A and B respectively, then

na = \(\frac{c}{v_a} \text { and } n_b=\frac{c}{v_b}\)then

bna = \(\frac{\text { Velocity of light in B }}{\text { Velocity of light in A }}=\frac{v_b}{v_a}=\frac{n_a}{n_b}\)

Question 33. What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror.
Answer:

At least two rays are required for locating the image of an object formed by a concave mirror.

A concave mirror forms a virtual image of an object when it is placed between the focus and pole of the mirror as shown:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Concave Mirror Forms A Virtual Image Of An Object

Question 34.

  1. Define the ine power of a lens and write its SI unit.
  2. 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 lens, the the if image size is equal to the object size? Also, find the power of the lens.

Answer:

  1. It is defined as the reciprocal of focal length lens cans to bend the rays of light falling on it. Its SI unit is dioptre.
  2. Given v = 50 cm, since the size of the image and that of the object are the same, therefore, the image is formed at 2f.

Hence, 2f = 50 or f= 25 cm = 0.25 m

Now, P = \(\frac{1}{f}=\frac{1}{0.25}=\frac{100}{25}\)= + 4 D

Question 35. Discuss the position and nature of the image formed by a concave mirror when the object is moved from infinity toward the pole of the mirror.
Answer:

As the object is moved from infinity towards the pole of a concave mirror the image shifts from the focus of the mirror to infinity.

When the object is at infinity, the image is formed at the principal focus or in the focal plane. As the object is shifted, further the image is formed between the principal focus centimeter of curvature, then at the centimeter of curvature, beyond the centimeter of curvature, at infi, city, and finally the image is formed behind the mirror. The size of the increasing.

Question 36. With the help of a ray diagram, show the formation of an image by a convex lens when the object is placed between F and 2F. Write two characteristics of the image so formed.
Answer: The image formed is real, I’ve,  red and diminished.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction The Image Is Real Inverted And Diminished

Question 37.

  1. Define the line radius of curvature and focal length of a spherical mirror and show it on a figure.
  2. Write the relation between the radius of curvature and the focal length of a spherical mirror.

Answer:

  1. The radius of curvature is the radius of the hollow sphere of which the reflecting surface of the spherical mirror forms a part. Focal length is the distance from the pole of the mirror to the focal point F.
  2. Relation: \(f=\frac{\mathrm{R}}{2}\)

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Focal Length Of A Spherical Mirror

Question 38. An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the opticenterntre of the lens is 20 cm, determine the position, an true, and size of the image formed using the lens formula.
Answer:

Given h+5 cm, f = – 10 cm, u = -20 cm, v = ?, h’ =?

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

∴ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{(-10)}+\frac{1}{(-20)}=\frac{-2-1}{20}=\frac{-3}{20}\)

∴ \(v=-\frac{20}{3} \mathrm{~cm}\)

Nature: Virtual, erect.

⇒ \(h^{\prime}=\frac{v}{u} \times h=\left(\frac{-20}{3} \times \frac{1}{-20} \times 5\right) \mathrm{cm}=+\frac{5}{3} \mathrm{~cm}\)

Question 39. If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror?
Answer:

Convex mirror

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Convex Mirror

Use: As rear view mirror in vehicles. Also in malls, hotels, and airports for security reasons.

Reason:

  • Forms erect image
  • Wider field of view.

Question 40. A convex lens has a focal length of 25 cm. Calculate the distance of the object from the lens if the image is to be formed on the opposite side of the lens at a distance of 75 cm from the lens. What will be the nature of the image?
Answer:

Given, f= + 25 cm, v = 75 cm, u =?

Using \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) We have

⇒ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{75}-\frac{1}{+25}=\frac{-2}{75}\)

Therefore, u = -37.5cm

The image is real, inverted, and magnified.

Question 41. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and its magnification.
Answer:

Given that O= 4.5 cm, u = − 12 cm, f= + 15 cm

Using the mirror formula \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

Or \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-12}=\frac{9}{60}[/latex

Therefore, v = 6.7 cm

The positive sign indicates that the image formed is on the other side of the mirror.

Also, [latex]m=\frac{1}{\mathrm{O}}=-\frac{v}{u} \Rightarrow \frac{1}{4.5}=-\frac{6.7}{-12}\)

Therefore, I = \(\frac{6.7 \times 4.5}{12}=+2.5 \mathrm{~cm}\)

I = + 2.5 cm

The positive sign indicates that the hat image formed is erect and virtual.

Question 42. How will you decide whether a given piece of glass is a convex lens, concave, lens, or plane glass plate?
Answer:

Hold the given piece of glass over some printed matter.

  1. If the letters appear magnified, the given piece is a convex lens.
  2. If the letters appear diminished, the given piece is a concave lens.
  3. If the letters appear to be of the same size, then it is a plane glass piece.

Question 43. A student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the flame at a distance of 15 cm from its pole.

  1. Write the type of mirror he should use.
  2. Find the linear magnification of the image produced.
  3. What is the distance between the object and its image?
  4. Draw a ray diagram to show the image formation in this case.

Answer:

  1. Concave mirror
  2. Linear magnification is given by
    1. \(m=\frac{v}{u}=\frac{-60}{-15}=-4\)
  3. 60-15 = 45 cm
  4. The diagram is as shown:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Candle Frame

Question 44. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location the the of image and magnification. Describe what happens to the image as the needle is moved farther from the mirror.
Answer:

h1 = 4.5 cm, u = – 12 cm, f = 15 cm, v = ?

Using the equation \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) we have

⇒ \(\frac{1}{15}=\frac{1}{v}+\frac{1}{-12} \text { or } \frac{1}{15}+\frac{1}{12}=\frac{1}{v}\)

Solving for u, we have v = + 6.6 cm

Now, \(m=-\frac{v}{u}=-\frac{6.6}{-12}\) = 0The imagemage will diminish.

Question 45. With the help a  of ray diagram, show that angle of incidence is equal to the angle of reflection when a ray is incident on the concave mirror.
Answer:

In the ray diagram shown below, CB is the normal and ∠i = ∠r, AB is the incident ray and BD is the reflected ray.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Ray Incident On The Concave Mirror

Question 46.

  1. State Snell’s law of refraction.
  2. When a ray of traveling in air enters obliquely into a glass slab, it is observed that the light ray emerges parallel to the incident ray but it is shifted sideways slightly. Draw a ray diagram to illustrate it.

Answer:

  1. The ratio the the of sine of the angle of incidence to the signoff the angle of refraction is constant for the light of the color and a given pair of the medium.
  2. The diagram is as shown:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Snells Law of Refraction

Question 47. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?
Answer:

Power of a lens is P = \(\frac{1}{f}\)

Since power = 2.5 dioptre

2.5 = \(\frac{1}{f}\)

or f= \(\frac{1}{f} \text { or } f=\frac{1}{2.5}\) m = 0.4 m = 40 cm

The focal length of the lens is 40 cm. Since the focal length is positive, therefore, the lens is a convex lens.

Question 48. The refractive index of dense flint glass is 1.65 and for all,ohol it is 1. concerning to air. What is the refractive index of dense flint to alcohol?
Answer:

Given \({ }_a \mu_g=1.65 \text { and }{ }_a \mu_{a l c}=1.36\)

Now, \({ }_{\text {alc }} \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_{\text {alc }}}=\frac{1.65}{1.36}=1.21\)

Question 49.

  1. Water a has a refractive of dex 1.33 and alcohol a has a refractive of dex 1.36. Which of the two mediums is optically denser? Ga’s reason’s your answer. Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol.
  2. The absolute refractive index of diamond is 2.42 and the absolute refractive index of glass is 1.50. Find the refractive index of a diamond concerning glass.

Answer:

1. Alcohol is optically denser than water. This is because use greater the refractive index more dense the liquid/material.

The diagram is as shown:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Alcohol Is Optically Dencer Than Water

2. Given and = 2.42, ang = 1.50, gnd = ?

Using the expression

⇒ \({ }_g n_d=\frac{{ }_a n_d}{{ }_a n_g}=\frac{2.42}{1.50}=1.61\)

Question 50. Show that the relative refractive index of medium concerning medium 1 can be expressed as the ratio of their absolute refractive indices.
Answer:

If n1 and n2 are the absolute refractive indices of media 1 and 2 the and speeds of light in the two media, then

⇒ \(n_1=\frac{c}{v_1} \quad \text { or } \quad v_1=\frac{c}{n_1}\)

and \(n_2=\frac{c}{v_2} \quad \text { or } \quad v_2=\frac{c}{n_2}\)

But the relative refractive index of medium concerning to medium 1 is

⇒ \(n_{21}=\frac{v_1}{v_2}=\frac{c / n_1}{c / n_2}\)

⇒ \(n_{21}=\frac{n_2}{n_1}\)

Question 51. Name the spherical mirror used as:

    1. Shaving mirror
    2. Rearview mirror in vehicles
    3. Reflector in searchlights.
  1. Write any three differences between a real and virtual image.

Answer:

    1. Concave mirror
    2. Convex mirror
    3. Convex mirror

2. Points of difference between a real and virtual image:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Difference Between Real Image And Virtual Image

Question 52.

  1. For the same angle of incidence, 45°, the angle of refraction in two transparent media; 1 and 2 is 20° and 30° respectively. Out of 1 and 2, which medium is optically denser and why?
  2. Light enters from the air to the diamond which has a refractive index of 2.42. Calculate the speed of light in a diamond, if the speed of light in air is 3.00 × 108m s-1. Answer:
  1. Medium 1 is optically denser as the angle of refraction is lesser in it, hence light bends more towards normal.
  2. n = speed of light in vacuum/speed of light in medium 2.42 = 3 x 108/v or v = 1.24 × 108 ms-1

Question 53. What is meant by the power of a lens? Give its SI unit. When two or more lenses are placed in contact, what will be their combined power?
Answer:

The power of a lens is defined as the ability of the lens to bend the rays of light incident on it. i.e., P = 1/f

The SI unit of power is called a dioptre. If P1 and P2 are the powers, then P of the combination is given by P = P1 + P2.

Question 54.

  1. What is meant by the refraction of light?
  2. If applying the Cartesian sign convention for spherical lenses the image distance obtained is negative. State the significance of the negative sign.

Answer:

  1. When light travels obliquely from one medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as the refraction of light.
  2. A negative sign of image distance means the image is virtual and erect and is formed on the same side of the object concerning the lens.

Question 55. A student-focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Candle Frame On A White Screen

  1. From the above table, find the focal length of the lens without using the lens formula.
  2. Which set of observations is incorrect and why?
  3. In which case the size of the object and image will be the same? Give a reason for your answer.

Answer:

  1. From the table of observation, it is clear that for u = 30, v = 30 cm. This means that both u and v must be equal to twice the focal length of the convex lens.
    1. Focal length of the convex lens = \(=\frac{30}{2}\) = 15 cm
  2. The observation (v) is incorrect. For this observation, u = 15f, i.e., the object is at
    the focus and the image must be formed at infinity and not 70 cm.
  3. Since m = \(\frac{v}{u}\) the size of the image will be equal to the size of the object, if v = u i.e., for the observation (iii).

Question 56.

  1. What happens to a ray of light when it travels from one medium to another having equal refractive indices?
  2. State the cause of the refraction of light.

Answer:

  1. No refraction or bending would take place. The light will travel in a straight line.
  2. The refraction occurs due to a change in the speed of light as it enters from one medium to another.

Question 57.

  1. What should be the position of the object, when a concave mirror is to be used:
    1. As a shaving mirror, and
    2. In torches producing parallel beams of light?
  2. A man standing in front of a mirror finds his image having a very small head and legs of normal size. What type of mirror are used in designing such a mirror?

Answer:

    1. The object should be between the pole and the focus.
    2. At the focus.
    1. Small head – convex mirror
    2. Legs of normal size – plane mirror.

Question 58. How can you identify the three types of mirrors without touching them?
Answer:

We will look into the mirror by going close to it:

  1. If an image is of the same size and erect – –plane mirror.
  2. If an image is bigger and erect – a concave mirror.
  3. If an image is of a smaller size and erect – convex mirror.

Question 59.

  1. Why does a ray of light pass through the center of curvature of a concave mirror after reflection is reflected along the same path?
  2. Why are convex mirrors commonly used as rear-view mirrors in vehicles?
  3. A pencil partly immersed in water in a glass tumbler appears to be displaced at the interface of air and water. Name the phenomenon of light responsible for it.

Answer:

  1. Because the incident ray falls on the mirror along the normal to the reflecting surface.
  2. Because they always give an erect, though diminished image. Thus, convex mirrors enable the driver to view a much larger area.
  3. Refraction of light

Question 60. An object is placed at the following distances from a concave mirror of focal length 15 cm

  1. 10 cm
  2. 20 cm
  3. 30 cm
  4. 40 cm

Which position of the object will produce:

  1. Virtual image
  2. A diminished real image
  3. An enlarged real image
  4. An image of the same size

Answer:

  1. 10 cm
  2. 40 cm
  3. 20 cm
  4. 30 cm.

Question 61. A ray of light is incident obliquely on a glass slab. Draw a ray diagram showing the path of the light ray. Mark the angle of incidence, angle of refraction, angle of emergence, and lateral displacement of the ray. Give a formula to find the refractive index of the glass slab in terms of the angle of incidence and angle of refraction.
Answer:

The ray diagram is as shown:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Ray Of Light Is Obliquely On A Glass Slab

The required formula is \({ }_a n_g=\frac{\sin i}{\sin r}\)

Question 62. Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.

  1. The object is placed between the device and its focus, image formed is enlarged and behind it.
  2. The object is placed between the focus and the device, image formed is enlarged and on the same side as that of the object.
  3. The object is placed between infinity and the device, image formed is diminished and between focus and optical center on the same side as that of the object.
  4. The object is placed between infinity and the device, and an image formed is diminished and between the pole and focus, behind it.

Answer:

  1. Concave mirror
  2. Convex lens
  3. Concave lens
  4. Convex mirror

Question 63. Why does a light ray incident on a rectangular glass slab immersed in any medium emerge parallel?
Answer:

The extent of bending of the ray of light at the opposite parallel faces of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray on a rectangular glass slab. However, the light ray is shifted sideward slightly.

When the glass slab is immersed in any medium the interface and glass medium are equal and opposite, so, the emergent ray will always be parallel to the incident ray.

Question 64. When dipped in water in a glass tumbler, a pencil appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like kerosene or turpentine? Support your answer with reason.
Answer:

A pencil partly immersed in water in a glass tumbler, appears to be displaced at the interface of air and water. The light reaching out from the portion of the pencil inside the water seems to come from a different direction, compared to the part above the water due to the refraction of light.

The pencil appears to be bent to different extents, if instead of water, liquids like kerosene or turpentine are used as their refractive indices are different which in turn produces deviation from incident ray by different extents.

Question 65. How is the refractive index of a medium related to the speed of light? Obtain an expression for the refractive index of a medium concerning another in terms of speed of light in these two media.
Answer:

Refractive index, μ \(=\frac{\text { Speed of light in vacuum, } c}{\text { Speed of light in medium, } v}\)

For medium 1, refractive index of first medium, μ1 \(=\frac{c}{v_1}\)

For medium 2, refractive index of second medium, μ2 \(\mu_2=\frac{c}{v_2}\)

∴ \(\mu_{21}=\frac{\mu_2}{\mu_1}=\frac{\frac{c}{v_2}}{\frac{c}{v_1}}=\frac{v_1}{v_2}\)

Question 66. The refractive index of diamond concerning glass is 1.6 and the absolute refractive index of glass is 1.5. Find out the absolute refractive index of a diamond.
Answer:

The refractive index of diamond concerning glass, and \(=\frac{v_g}{v}\) = 1.6, absolute refractive

Index of glass, ng= \(\frac{c}{v_g},\) and absolute refractive index of diamond, nd = \(\frac{c}{v_d}\)

Therefore, nd= \(\frac{v_g}{v_d} \times \frac{c}{v_g}\) = 1.6 x 1.5= 2.40

Question 67. A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer:

The statement is correct if the object is placed within 20 cm of the lens in the first case and between 20 cm and 40 cm in the second case.

Question 68. Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus on the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?
Answer:

Sudha should move the screen towards the lens to obtain a clear image of the building. The approximate focal length of this lens is 15 cm.

Question 69. How are the power and focal length of a lens related? You are provided with two lenses of focal lengths of 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?
Answer:

P= \(\frac{1}{f}\) P∞ \(\frac{1}{f}\) Power of a lens is inversely proportional to its focal length therefore lens has a focal length of 20 cm will provide more convergence.

Question 70. Under what condition in an arrangement of two plane mirrors, incident ray, and reflected ray will always be parallel to each other, whatever may be the angle of incidence? Show the same with the help of a diagram.
Answer:

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Two Plane Mirrors Are Always Parallel To Each Other

When two plane mirrors are placed at a right angle to each other then the incident and reflected rays will always be parallel to each other.

Question 71. Draw a ray diagram showing the path of rays of light when they enter with oblique incidence

  1. From air into water,
  2. From water into air.

Answer:

When a ray of light enters with oblique incidence from air into water, then it goes from an optical rarer medium to an optical denser medium, and the velocity of light decreases which in turn bends the incident light towards the normal. Also i > r.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction From Air Into Water

When a ray of light enters with oblique incidence from water into air, then it goes from an optical denser medium to an optical rarer medium, and the velocity of light increases which in turn bends the incident light away from the normal. Also I < r.

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction From Water Into Air

Question 72. At a small town fair, Akshay took his friend and showed him a mirror in which his image showed his upper half very fat and his lower body very thin. Akshay’s friend got upset but Akshay explained to him by showing him a similar image in the mirror.

  1. Name two mirrors used in this fair shop.
  2. Name the mirror in which the size of the image is small.
  3. What values of Akshay are reflected?

Answer:

  1. Concave and convex mirror
  2. Convex mirror
  3. Akshay showed compassion and empathy.

Light-Reflection And Refraction Multiple Choice Question And Answers

Question 1. Which of the following can make a parallel beam of light when light from a point source is incident on it?

  1. Concave mirror as well as convex lens
  2. Convex mirror as well as concave lens
  3. Two plane mirrors placed at 90° to each other
  4. Concave mirror as well as concave lens

Answer: 1. Concave mirror as well as convex lens

Question 2. A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is

  1. -30 cm
  2. -20 cm
  3. -40 cm
  4. -60 cm

Answer: 2. -20 cm

Question 3. Under which of the following conditions a concave mirror can form an image larger than the actual object?

  1. When the object is kept at a distance equal to its radius of curvature
  2. When an object is kept at a distance less than its focal length
  3. When an object is placed between the focus and center of curvature
  4. When an object is kept at a distance greater than its radius of curvature

Answer: 3. When an object is placed between the focus and center of curvature

Question 4. The figure shows a ray of light as it travels from medium A to medium B. Refractive index of medium B relative to medium A is

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Refractive Index Of The Medium B Relative To Medium A

  1. √3/√2
  2. √2/√3
  3. 1/√2
  4. √2

Answer: 1. √3/√2

Question 5. A light ray enters from medium A to medium B as shown in the figure. The refractive index of medium B relative to A will be

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Light Ray Enters From Medium A To Medium B

  1. Greater than unity
  2. Less than unity
  3. Equal to unity
  4. Zero

Answer: 2. Less than unity

Question 6. Beams of light are incident through holes A and B and emerge out of box through the holes C and D respectively as shown in the figure. Which of the following could be inside the box?

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction Beams Of Light Are Incident Through The Holes A And B

  1. A rectangular glass slab
  2. A convex lens
  3. A concave lens
  4. A prism

Answer: 1. A rectangular glass slab

Question 7. A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the figure. Which of the following could be inside the box?

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Beam Of Light Is Incident Through The Hole On Side A

  1. Concave lens
  2. Rectangular glass slab
  3. Prism
  4. Convex lens

Answer: 4. Convex lens

Question 8. Which of the following statements is true?

  1. A convex lens has 4 dioptre power having a focal length of 0.25 m
  2. A convex lens has -4 dioptre power having a focal length of 0.25 m
  3. A concave lens has 4 dioptre power having a focal length of 0.25 m
  4. A concave lens has -4 dioptre power having a focal length of 0.25 m

Answer: 1. A convex lens has 4 dioptre power having a focal length of 0.25 m

Question 9. Magnification produced by a rearview mirror fitted in vehicles

  1. Is less than one
  2. Is there more than one
  3. Is equal to one
  4. Can be more than or less than one depending upon the position of the object in front of it

Answer: 1. Is less than one

Question 10. Rays from the Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that the size of its image is equal to the size of the object?

  1. 15 cm in front of the mirror
  2. 30 cm in front of the mirror
  3. Between 15 cm and 30 cm in front of the mirror
  4. More than 30 cm in front of the mirror

Answer: 2. 30 cm in front of the mirror

Question 11. A full-length image of a distant tall building can be seen by using

  1. A concave mirror
  2. A convex mirror
  3. A plane mirror
  4. Both concave as well as plane mirror

Answer: 2. A convex mirror

Question 12. In torches, searchlights, and headlights of vehicles the bulb is placed

  1. Between the pole and the focus of the reflector
  2. Very near to the focus of the reflector
  3. Between the focus and center of curvature of the reflector
  4. At the center of the curvature of the reflector

Answer: 2. Very near to the focus of the reflector

Question 13. The laws of reflection hold good for

  1. Plane mirror only
  2. Concave mirror only
  3. Convex mirror only
  4. All mirrors irrespective of their shape

Answer: 4. All mirrors irrespective of their shape

Question 14. The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C, and D in the figure. Which one of them is correct?

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Rectangular Glass Slab

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 15. You are given water, mustard oil, glycerine, and kerosene. In which of these media a ray of light incident obliquely at the same angle would bend the most?

  1. Kerosene
  2. Water
  3. Mustard oil
  4. Glycerine

Answer: 4. Glycerine

Question 16. Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in the figure?

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction The Ray Of Light Incident On A Concave Mirror

  1. 1
  2. 2
  3. 3
  4. 4

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction The Ray Of Light Incident On A Concave Mirror.

Answer: 4. 4

Question 17. Which of the following ray diagrams is correct for the ray of light incident on a lens shown in the figure?

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction The Ray Of Light Incident On A Lens

  1. 1
  2. 2
  3. 3
  4. 4

CBSE Class 10 Science Chapter 10 Light Reflection And Refraction A Ray Of Light Incident On A Lens

Answer: 1. 1

Question 18. A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size, and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.

  1. Plane, convex, and concave
  2. Convex, concave, and plane
  3. Concave, plane, and convex
  4. Convex, plane, and concave

Answer: 3. Concave, plane, and convex

Question 19. In which of the following, the image of an object placed at infinity will be highly diminished and point-sized?

  1. Concave mirror only
  2. Convex mirror only
  3. Convex lens only
  4. Concave mirror, convex mirror, concave lens, and convex lens

Answer: 4. Concave mirror, convex mirror, concave lens, and convex lens

CBSE Solutions For Class 10 Mathematics Chapter 8 Introduction To Trigonometry

Class 10 Maths Introduction To Trigonometry

Question 1. In ΔABC, ∠B = 90° and, Sin A = \(\frac{3}{5}\) , then find all other trigonometric rotiss for ∠A.
Solution:

Sin A = \(\frac{3}{5}=\frac{perpendicular}{\text { hypo }}\)

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry Then All Other Trigonometric Ratios For Angle A

ΔABC

∠B=90°, BC=3t, AC=5k

AB2 + BC2= AC2

AB2 = AC2 – BC2

= (25k2) – (9k2)

AB2 = 16k2

AB = 4K

Read and Learn More Class 10 Maths

Cos A = \(=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}\)

tan A = \(\frac{\text {perpendicular }}{\text {base }}=\frac{B C}{A B}=\frac{3 k}{4 k}=\frac{3}{4}\)

Cosec A = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{3 k}=\frac{5}{3}\)

Sec A = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\frac{5 k}{4 k}=\frac{5}{4}\)

Cot A = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{4 k}{3 k}=\frac{4}{3}\)

Question 2. In ΔABC, ∠B = 90° and Cos A = \(\frac{9}{41}\), then find all other trigonometric ratios for ∠A.

Solution:

Cos A = \(\frac{9}{41}\)

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry Then All Other Trigonometric Ratios For Angle A.

ΔABC

∠B=90°, AB=9k, AC = 41k

AB2 + BC2 = AC2

BC2 = AC2 – AB2

BC2 = (41k) = (9k)2

BC2 = 1681k2 – 81k2

BC2 = 1600k2

BC = 40k

Sin A = \(\frac{\text { perpendicular }}{\text { hypotemuse }}=\frac{B C}{A C}=\frac{40}{41}\)

tan A = \(\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{40}{9}\)

Cosec A = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{41}{40}\)

SecA = \(\frac{\text { hypotenusc }}{\text { base }}=\frac{A C}{A B}=\frac{41}{9}\)

Cot A = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{9}{40}\)

Question 3. In ΔABC, ∠A= 90° and tan B= \(\frac{5}{6}\), then find all other trigonometric ratios for ∠B.

Solution:

ΔABC, ∠Ạ = 90°

tan B = \(\frac{\text { Perpendicular }}{\text { Base }}=\frac{5}{6}\)

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry Then All Other Trigonometric Ratios For Angle B

AC = 6K, BC=5K

AB2 = AC2+BC2

AB2 = (6k)2 + (5k)2

AB2 = 36k2 +25k2

AB2 = 61k2

AB = \(\sqrt{61 k}\)

Sin B = \(\frac{\text { Perpendicular }}{\text { hyotenuse }}=\frac{B C}{A B}=\frac{5 K}{\sqrt{61 K}}=\frac{5}{\sqrt{61}}\)

Cos B = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}=\frac{6 k}{\sqrt{61 k}}=\frac{6}{\sqrt{61}}\)

Cosec B = \(\frac{\text { hypotenuse }}{\text {Perpendicular}}=\frac{A B}{B C}=\frac{\sqrt{61}}{5}\)

Sec B = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{A B}{A C}=\frac{\sqrt{61}}{6}\)

Cot B = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{6}{5}\)

CBSE Class 10 Maths Solutions Introduction To Trigonometry

Question 4. In ΔPQR, ∠R = 90° and Cosec P =\(\frac{13}{5}\), then find all trignometric ratio for ∠Q.

Solution:

ΔPQR, ∠R=90°

Cosec P = \(=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13}{5}=\frac{P Q}{Q R}\)

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry Then All Other Trigonometric Ratios For Angle Q

PR2 + QR2 = PQ2

PQ = 13K, QR = 5k

PR2 = PQ2 – QR2

= (13k)2– (5k)2

PR2 = 169k2-25k2

PR2 = 144k2

PR = 12k

Sin Q = \(\frac{\text { perpendicular }}{\text { hypotenuesc }}=\frac{Q R}{P Q}=\frac{5 K}{13 K}=\frac{5}{13}\)

Cos Q = \(\frac{\text { base }}{\text { hypdenuse }}=\frac{P R}{P Q}=\frac{12 k}{13 k}=\frac{12}{13}\)

tan Q = \(\frac{\text { perpendicular }}{\text { base }}=\frac{Q R}{P R}=\frac{5 k}{12 k}=\frac{5}{12}\)

Sec Q = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{P Q}{P R}=\frac{13 k}{12 k}=\frac{13}{12}\)

Cot Q = \(=\frac{\text { base }}{\text { perpendicular }}=\frac{P R}{Q R}=\frac{12 k}{5 k}=\frac{12}{5}\)

Question 5. In ΔABC, ∠C = 90° and Sec B = \(\frac{5}{4}\), then find all other trigonometric ratios for ∠B.

Solution:

SecB = \(\frac{\text { hypo }}{\text { base }}=\frac{5}{4}=\frac{A B}{A C}\)

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry Then All Other Trigonometric Ratios For Angle B.

AB = 5K, AC = 4K

BC2 = AB2 – AC2

BC2 = (5k)2 = (4k)2

BC2 = 25k2 – 16k2

BC2 = 9k2

BC = 3K

Sin B = \(\frac{\text { perpendiular }}{\text { hypo }}=\frac{B C}{A C}=\frac{3 k}{4 k}=\frac{3}{4}\)

COS B = \(\frac{\text { base }}{\text { hypo }}=\frac{A B}{A C}=\frac{5 k}{4 k}=\frac{5}{4}\)

tan B = \(=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{3 k}{5 k}=\frac{3}{5}\)

Cosec B = \(\frac{\text { hypo }}{\text { base }}=\frac{A C}{B C}=\frac{4 k}{3 k}=\frac{4}{3}\)

Cot B = \(=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{5 k}{3 k}=\frac{5}{3}\)

Question 6. If tanθ = 2, then they find the value of \(\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta}\)

Solution:

tanθ = 2 = \(\frac{\sin \theta}{\cos \theta}=\frac{2}{1}\)

⇒ \(\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta}\)

⇒ \(\frac{2(2)(1)}{(1)^2-(2)^2}\)

⇒ \(\frac{4}{1-4}\)

⇒ \(\frac{-4}{3}\)

Question 7. If tanθ = \(\tan \theta=\frac{a}{b},\), then find the value of \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

Solution:

⇒ \(\tan \theta=\frac{a}{b}\)

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{a}{b}\)

⇒ \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

⇒ \(\frac{2 a-3 b}{2 a+3 b}\)

Question 8. If tanθ = \(\frac{3}{5}\), then find the Value of \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

Solution:

⇒ \(\tan \theta=\frac{3}{5}\)

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{3}{5}\)

⇒ \(\frac{3 \sin \theta+4 \cos \theta}{3 \sin \theta-4 \cos \theta}\)

⇒ \(\frac{3(3)+4(5)}{3(3)-4(5)}\)

⇒ \(\frac{9+20}{9-20}\)

⇒ \(\frac{-29}{11}\)

Question 9. Find the value of Sin 60° Cos 30° – Cos 60° sin 30°.

Solution:

Sin 60° Cos 30°- Cos 60° Sin 30°

⇒ \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

⇒ \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2\)

⇒ \(\frac{3}{4}-\frac{1}{4}\)

⇒ \(\frac{12-4}{16}\)

⇒ \(\frac{8}{16}\)

⇒ \(\frac{1}{2}\)

Question 10. Find the Value of \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

⇒ \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}\)

⇒ \(\sqrt{3}\)

Question 11. Find the value of 2sin 30° Cos30°

Solution:

2 sin 30° Cos 30°

⇒ \(2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\)

⇒ \(2\left(\frac{\sqrt{3}}{4}\right)\)

⇒ \(\frac{\sqrt{3}}{2}\)

Question 12. Show that: 5 Cos2 60° + 4 sec2 30° tan2 4s° + Cos2 90° = \(\frac{67}{12}\)

Solution:

5 Cos2 60° + 4 Sec2 30° – tan2 45° + Cos2 90°

⇒ \(5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-1+(0)^2\)

⇒ \(5\left(\frac{1}{4}\right)+4\left(\frac{4}{3}\right)-1\)

⇒ \(\frac{5}{4}+\frac{16}{3}-1\)

⇒ \(\frac{3 \times 5+4 \times 16-12}{12}\)

⇒ \(\frac{15+64-12}{12}\)

⇒ \(\frac{67}{12}\)

Question 13. Find the value of 4sin2 30° + tan2 60° + Sec2 45°

Solution:

4sin2 30°+ tan2 60°+ Sec2 45°.

⇒ \(4\left(\frac{1}{2}\right)^2+(\sqrt{3})^2+(\sqrt{2})^2\)

⇒ \(4\left(\frac{1}{4}\right)+3+2\)

= 1+3+2

= 6

Question 14. Find the Value of \(\frac{\sin 30^{\circ}}{\cos ^2 45^{\circ}}-\tan ^2 60^{\circ}+3 \cos 90^{\circ}+\sin 0^{\circ}\)

Solution:

⇒ \(\frac{\sin 30^{\circ}}{\cos ^2 45^{\circ}}-\tan ^2 60^{\circ}+3 \cos 90^{\circ}+\sin 0^{\circ}\)

⇒ \(\frac{\frac{1}{2}}{\left(\frac{1}{\sqrt{2}}\right)^2}-(\sqrt{3})^2+3(0)+0\)

⇒ \(\frac{\frac{1}{2}}{\frac{1}{2}}-3\)

⇒ 1 – 3 = -2

Question 15. Evaluate: \(\frac{\sin ^2 30^{\circ}+\sin ^2 45^{\circ}-4 \cot ^2 60^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\frac{1}{2} \tan 60^{\circ}}\)

Solution:

⇒ \(\frac{\sin ^2 30^{\circ}+\sin ^2 45^{\circ}-4 \cot ^2 60^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\frac{1}{2} \tan 60^{\circ}}\)

⇒ \(\frac{\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2-4\left(\frac{1}{\sqrt{3}}\right)^2}{2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{2}(\sqrt{3})}\)

⇒ \(\frac{\frac{1}{4}+\frac{1}{2}-4\left(\frac{1}{3}\right)}{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}\)

⇒ \(\frac{\frac{1}{4}+\frac{1}{2}-\frac{4}{3}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{1}{6}-\frac{4}{3}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{1-8}{6}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{-7}{6}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{-7}{6}}{\frac{\sqrt{3}}{2}}\)

⇒ \(\frac{-7}{12 \sqrt{3}}\)

Question 16. Show that: \(\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{1}{\frac{2}{\sqrt{3}}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\)

Question 17. Show that: Cos2 60°- Sin2 60° = -Sin30°

Solution:

Cos2 60°-sin260°= -Sin30°

⇒ \(\frac{1}{4}-\frac{3}{4}= -\frac{1}{2}\)

⇒ \(\frac{-2}{4}=\frac{-1}{2}\)

⇒ \(\frac{-1}{2}=\frac{-1}{2}\)

Question 18. Show that: tan 60° = \(\tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

⇒ \(\sqrt{3}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}\)

⇒ \(\sqrt{3}=\frac{2/ \sqrt{3}}{\frac{3-1}{3}}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{2 / 3}\)

⇒ \(\sqrt{3}=\frac{1}{1 / \sqrt{3}}\)

⇒ \(\sqrt{3}=\sqrt{3}\)

Question 19. Show that: \({Cos} 30^{\circ}=\sqrt{\frac{1+\cos 60^{\circ}}{2}}\)

Solution:

⇒ \({Cos} 30^{\circ}=\sqrt{\frac{1+\cos 60^{\circ}}{2}}\)

⇒ \(\frac{\sqrt{3}}{2}=\sqrt{\frac{1+\frac{1}{2}}{2}}\)

⇒ \(\frac{\sqrt{3}}{2}=\sqrt{\frac{2+1}{2}}\)

⇒ \(\frac{\sqrt{3}}{2}=\sqrt{\frac{\frac{2+1}{2}}{2}}\)

⇒ \(\frac{\sqrt{3}}{2}=\sqrt{\frac{\frac{3}{2}}{2}}\)

Question 20. Evaluate: \(\frac{\tan 45^{\circ}}{2 \sin 30^{\circ}-\cos 60^{\circ}}\)

Solution:

⇒ \(\frac{\tan 45^{\circ}}{2 \sin 30^{\circ}-\cos 60^{\circ}}\)

⇒ \(\frac{1}{2\left(\frac{1}{2}\right)-\frac{1}{2}}\)

⇒ \(\frac{1}{1-\frac{1}{2}}\)

⇒ \(\frac{1}{\frac{2-1}{2}}\)

⇒ \(\frac{1}{\frac{1}{2}}\)

= 2

Question 21. If A = 45°, then show that: Cos 2A = Cos2A – Sin2A

Solution:

Given A = 45°

Cos2(45°) = Cos2(45°)-Sin2(45°)

Cos 90° = \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{1}{\sqrt{2}}\right)^2\)

0 = \(\frac{1}{2}-\frac{1}{2}\)

0 = 0

Question 22. If A=30° and B=60°, then show that: Cos (A+B) = CosA CosB – Sin A Sin B

Solution:

Given A= 30°, B=60°

Cos (30°+60°) = Cos 30° Cos 60° – Sin 30° Sin60°

⇒ \(\cos 90^{\circ}=\frac{\sqrt{3}}{2}\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2}\)

0 = \(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\)

0=0

Question 23. If A = 30°, then show that: \(\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}\)

Solution:

Given A = 30°

tan 2(30°) = \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

tan 60° = \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{\frac{3-1}{3}}\)

⇒ \(\sqrt{3}=\frac{2/\sqrt{3}}{2 / 3}\)

⇒ \(\sqrt{3}=\sqrt{3}\)

Question 24. If \(\cos \theta=\frac{\sqrt{3}}{2}\), then find the value of Sin 3θ.

Solution:

⇒ \(\cos \theta=\frac{\sqrt{3}}{2}\)

Cos θ = Cos30°

θ = 30°

= Sin 3θ

= Sin 3(30°)

= Sin 90°

= 1

Question 25. If tan (A+B) = \(\sqrt{3}\) and Sin (A-B) = \(\frac{1}{2}\), then find the value of tan (2A-3B)

Solution:

tan (A+B)= \(\sqrt{3}\)

Sin (A-B)= \(\frac{1}{2}\)

tan (2A-3B)

tan (A+B)= \(\sqrt{3}\)

A+B = 60°

Sin (A-B) = Sin 30° =) A-B = 30°

Acting equations

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry The Value Of Tan 2A minus 3B Adding Equations

⇒ \(A=\frac{90^{\circ}}{2}=45^{\circ}\)

Subtracting equations

CBSE School For Class 10 Maths Chapter 8 Introduction To Trigonometry The Value Of Tan 2A minus 3B Substracting Equations

⇒ \(B=\frac{30^{\circ}}{2}\)

B = 15°

tan (2A-3B)

= tan (2(45°)-3((5°))

= tan (90°-45°)

= tan 45° = 1

Question 26. If A+B = 90° and tan A = \(\sqrt{3}\), then find the Value of B.

Solution:

A+B=90°

tan A = \(\sqrt{3}\)

tan A = tan 60°

A = 60°

60°+B=90°

B = 30°

Question 27. If A-B = 30° and Sin A = \(\frac{\sqrt{3}}{2}\), then find the value of B.

Solution:

A-B = 30°

Sin A = \(\frac{\sqrt{3}}{2}\)

Sin A = Sin 60°

A = 60°

60°-B = 30°

B = 30°

Question 28. \(\frac{1}{1+\tan ^2 \theta}+\frac{1}{1+\cos ^2 \theta}=1\)

Solution:

⇒ \(\frac{1}{1+\tan ^2 \theta}+\frac{1}{1+\cot ^2 \theta}\)

⇒ \(\frac{1}{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}+\frac{1}{1+\frac{\cos ^2 \theta}{\sin ^2 \theta}}\)

⇒ \(\frac{1}{\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}}+\frac{1}{\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta}}\)

⇒ \(\frac{\cos ^2 \theta+\sin ^2 \theta}{1}\)

= 1 = R.H.S

Question 29. \(\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}=1\)

Solution:

⇒ \(\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}\)

⇒ \(\sin ^2 \theta+\frac{1}{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)

⇒ \(\sin ^2 \theta+\frac{\cos ^2 \theta}{1}\)

⇒ \(\sin ^2 \theta+\cos ^2 \theta\)

= 1 = R.H.S

Question 30. \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-{cosec}^2 \theta=0\)

Solution:

⇒ \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-{cosec}^2 \theta \Rightarrow \text { L.H.S }\)

⇒ \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}\)

⇒ \(\frac{\sin ^2 \theta+\cos ^2 \theta-1}{\sin ^2 \theta}\)

⇒ \(\frac{1-1}{\sin ^2 \theta}=0=\text { R.H.S }\)

Question 31. \(\frac{1+\tan ^2 \theta}{{cosec}^2 \theta}=\tan ^2 \theta\)

Solution:

⇒ \(\text { L.H.S }=\frac{1+\tan ^2 \theta}{{cosec}^2 \theta}\)

⇒ \(\frac{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}{\frac{1}{\sin ^2 \theta}}\)

⇒ \(\frac{\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}}{\frac{1}{\sin ^2 \theta}}\)

⇒ \(=\frac{1}{\cos ^2 \theta} \times \frac{\sin ^2 \theta}{1}\)

⇒ \(\tan ^2 \theta=\text { R.HS }\)

Question 32. Prove that: Cosec A-Cot A = \(\frac{1}{{cosec} A+\cot A}\)

Solution:

L.H.S = CosecA – Cota = \(({cosec} A-\cot A) \cdot \frac{({cosec} A+\cot A)}{({cosec} A+\cot A)}\)

⇒ \(\frac{{cosec}^2 A-\cot ^2 A}{{cosec} A+\cot A}\)

⇒ \(\frac{1}{{cosec} A+\cot A}\)

= R.H.S

Question 33. Prove that \(\frac{\sec A+1}{\tan A}=\frac{\tan A}{{Sec} A-1}\)

Solution:

L.H.S = \(\frac{\sec A+1}{\tan A}=\frac{\sec A+1}{\tan A} \times \frac{\sec A-1}{\sec A-1}\)

⇒ \(\frac{\sec ^2 A-1}{\tan A(\sec A-1)}=\frac{\tan ^2 A}{\tan A(\sec A-1)}\)

⇒ \(\frac{\tan A}{(\sec A-1)}\)

= R.H.S

Question 34. Check whether the equation \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\sec \phi+1}{\sec \phi-1}\) is an identity or not?

Solution:

L.H.S = \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}\)

⇒ \(\frac{\frac{\sin \phi}{\cos \phi}+\sin \phi}{\frac{\sin \phi}{\cos \phi}-\sin \phi}\)

⇒ \(\frac{\sin \phi \sec \phi+\sin \phi}{\sin \phi \sec \phi-\sin \phi}\)

⇒ \(\frac{\sin \phi(\sec \phi+1)}{\sin \phi(\sec \phi-1)}\)

⇒ \(\frac{\sec \phi+1}{\sec \phi-1}\)

= R.H.S

Question 35. If \(\cos ^2 30^{\circ}+\cos ^2 45^{\circ}+\cos ^2 60^{\circ}=x\), then find the value of “x”.

Solution:

⇒ \(\cos ^2 30^{\circ}+\cos ^2 45^{\circ}+\cos ^2 60^{\circ}=x\)

⇒ \(x=\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2\)

⇒ \(x=\frac{3}{4}+\frac{1}{2}+\frac{1}{4}\)

⇒ \(x=\frac{6+4+2}{8}\)

⇒ \(x=\frac{12}{8}\)

⇒ \(x=\frac{3}{2}\)

Question 36. Without using trigonometric tables, evaluate:

1. \(\frac{\sin 11^{\circ}}{\cos 79^{\circ}}\)

Solution:

⇒ \(\frac{\sin 11^{\circ}}{\cos 79^{\circ}}=\frac{\sin 11^{\circ}}{\left.\cos \left(90^{\circ}-79\right)^{\circ}\right)}=\frac{\sin 11^{\circ}}{\sin 11^{\circ}}=1\)

2. \(\frac{\sec 15^{\circ}}{{cosec} 75^{\circ}}\)

Solution:

⇒ \(\frac{\sec 15^{\circ}}{{cosec} 75^{\circ}}=\frac{\sec 15^{\circ}}{{cosec}\left(90^{\circ}-75^{\circ}\right)}=\frac{\sec 15^{\circ}}{{cosec} 15^{\circ}}=1\)

3. \(\frac{\tan 54^{\circ}}{\cot 36^{\circ}}\)

Solution:

⇒ \(\frac{\tan 54^{\circ}}{\cot 36^{\circ}}=\frac{\tan 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}=\frac{\tan 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}=\frac{\tan 54^{\circ}}{\tan 54{ }^{\circ}}=1\)

Question 37. Evaluate: tan42° – Cot 48°

Solution:

tan 42°-Cot 48° = tan 42°- Cot (90°- 42°)

tan42°-tan 42°

= 0

Question 38. Evaluate: Sex 36° – Cosec 54°

Solution:

Sec 36°- Cosec 54° = Sec 36° – Cosec (90°-36°)

= Sec 36° – Sec 36°

= 0

Question 39. Prove that: Sin 42°Cos 48°+ Sin 48° Cos42° = 1

Solution:

L.H.S= Sin 42° Cos48° + Sin48° Cos 42°

= Sin 42°cos (90°-42°)+ Sin(90° – 42°) Cos 42°

= Sin 42° Sin 42° + Cos42° Cos42°

Sin2 42° + Cos2 42°

= 1

Question 40. If Sin 3A = Cos(A-26°) where 3A is an acute angle, then find the value of A.

Solution:

Given that,

Sin 3A = Cos (A-26°)

Cos (90°-3A) = Cos (A-26°)

90°-3A = A-26°

-4A = -116°

A = 29°

CBSE Solutions For Class 10 Mathematics Chapter 9 Some Applications of Trigonometry

Class 10 Maths Some Applications of Trigonometry

Question 1. The length of the Shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height. Show that the angle of elevation of the Sun is 60°.

Solution:

Let PQ be a vertical pole whose height is h. Its Shadow is OQ whose height is \(\frac{h}{\sqrt{3}}\)

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Length Of The Shadow A Vertical Pole

Let the angle of elevation of the Sun is ∠POQ = 0

In APOQ,

⇒ \(\tan \theta=\frac{P Q}{O Q}=\frac{h}{h / \sqrt{3}}=\sqrt{3}=\tan 60^{\circ}\)

θ = 60°

∴ The angle of elevation of Sun = 60°

Read and Learn More Class 10 Maths

Question 2. If a tower 30m high, casts a shadow \(10 \sqrt{3} \mathrm{~m}\) long on the ground, then what is the angle of elevation of the Sun?

Solution:

It is given that AB = 30m be the tower and BC = \(10 \sqrt{3} \mathrm{~m}\) m be its shadow on the ground.

Let θ be the angle of elevation.

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Angle Of Elevation Of The Sun

In a right triangle,

tan θ = \(\frac{AB}{BC}\)

⇒ \(\frac{30}{10 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\)

= tan 60°

θ = 60°

∴ Hence, the angle of elevation θ = 60°

Question 3. A ladder 15 meters long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

Solution:

Let PR be a ladder of length 15m and QR, a wall of height h.

Given that ∠PRQ = 60°

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry A Ladder 15 Meters Long Just Reaches The Top Of A Vertical Wall

In ΔPQR,

Cos 60° = \(\frac{h}{PR}\) = \(\frac{1}{2}\) = \(\frac{h}{15}\)

⇒ h = \(\frac{15}{2}\)m

∴ Height of the wall = \(\frac{15}{2} m\)

Question 4. A Circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the Pole, if the angle made by the rope with the ground level is 30°.

Solution:

In ΔABC,

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Angle Made By The Rope With The Ground Level Is 30 Degrees

Sin 30°= \(\frac{A B}{A C}\)

⇒ \(\frac{1}{2}=\frac{A B}{20}\)

AB = 10

∴ Height of pole = 10m

Question 5. A tree breaks due to stom and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.

Solution:

Let the part CD of the tree BD broken in air and touches the ground at point A.

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Distance Between The Foot Of The Tree And The Height Of The Tree

According to the problem,

AB = 8M

and ∠BAC = 30°

In ΔABC,

tan 30° = \(\frac{BC}{AB}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{B C}{8}\)

⇒ \(B=\frac{8}{\sqrt{3}} \mathrm{~m}\)

and Cos 30° = \(\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{A C}\)

AC = \(\frac{16}{\sqrt{3}} m\)

CD = \(\frac{16}{\sqrt{3}} m\) (AC = CD)

Now, the height of tree = BC + CD

⇒ \(\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8 \sqrt{3} \mathrm{~m}\)

CBSE Class 10 Maths Solutions Some Applications Of Trigonometry

Question 6. The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is 30°: Find the height of the tower.

Solution:

Let AB be the tower.

The angle of elevation of the top of the tower from Point C, 30m away from A is 30°

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Angle Of Elevation Of The Top Of A Towe From A Point On The Ground

∴ In ABAC,

tan 30°= \(\frac{A B}{A C}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{A C}\)

AB = \(\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m}\)

∴ Height of the tower = \(10 \sqrt{3} \mathrm{~m}\)

Question 7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building and 60° respectively. Find the height of the tower.

Solution:

Let, CD be the height of the transmission tower.

Here, the height of the building

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Height Of The Transmission Tower

BC = 20m

In ΔABC,

tan 45° = \(\frac{B C}{A B}\)

1 = \(\frac{20}{A B}\)

AB = 20m

In ΔABD,

tan 60° = \(\frac{B D}{A B} \Rightarrow \sqrt{3}=\frac{B D}{20}\)

⇒ \(B D=20 \sqrt{3} \mathrm{~m}\)

⇒ \(B C+C D=20 \sqrt{3}\)

⇒ \(20+C D=20 \sqrt{3}\)

⇒ \(C D=20(\sqrt{3}-1) m\)

∴ Height of transmission tower = \(20(\sqrt{3}-1) m\)

Question 8. The Shadow of a tower Standing on a level plane is found to be much longer when the Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Solution:

Let AB be a tower of height ‘h’ meters and BD and BC be its shadows when the angles of elevation of the sun are 30° and 60° respectively.

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Shadow Of A Tower Standing On A Level Plane And Find The Height Of THe Tower

∴ ∠ADB =30°, ∠ACB = 60° and CD = 50m

Let BC = X meters.

In ΔABC

tan 60° = \(\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x}\)

⇒ \(x=\frac{h}{\sqrt{3}}\)

In ΔABD

tan 30° = \(\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50}\)

⇒ \(\sqrt{3} h=x+50 \Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50\)

2h= \(50 \sqrt{3}\)

h = \(25 \sqrt{3}\)

∴ Height of the tower = \(25 \sqrt{3}\)m

Question 9. The angle of elevation of the top of a tower from a point on the ground is 30: After walking nom towards the tower, the angle of elevation becomes 60° Find the height of the tower.

Solution:

Let AB be a tower of height ‘h’ meters. From points D and c on the ground, the angle of elevation of top A of the tower is 30° and 60° respectively.

Given that CD = 40m

let BC = x meters

CBSE School For Class 10 Maths Chapter 9 Some Application Of Trigonometry The Angle Of Elevation Tower From Point On Ground 30 Degrees After The Angle Of Elevation Becomes 60 Degrees

In ΔABC

tan 60° = \(\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x}\)

⇒ \(x=\frac{h}{\sqrt{3}}\)

In ΔABD

tan 30° =\(\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40+x}\)

⇒ \(\sqrt{3} h=40+x\)

⇒ \(\sqrt{3} h=40+\frac{h}{\sqrt{3}}\) [from (1)]

⇒ \(3 h=40 \sqrt{3}+h\)

⇒ \(2 h=40 \sqrt{3}\)

⇒ \(h=20 \sqrt{3}\)

∴ Height of the tower = \(20 \sqrt{3} \mathrm{~m}\) m

CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

CBSE Class Class 10 Maths Volume And Surface Area Of Solids

Question 1. A tent of cloth is Cylindrical upto I’m height and Conical above it of the Same radius of base. If the diameter of the tent is 6m and the slant height of the Conical part is 5m, find the cloth required to make this tent.
Solution:

Diameter of base 2r = 6m

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Tent Of Cloth Is Cylindrical Up To Height And Conical Of The Sum Radius Of Base

r = \(\frac{6}{2}\) = 3m

Height of Cylindrical path h = 1m

The slant height of Conical part l = 5m

Cloth required in tent = 2πrh +πrl

= πr(2h+1)

⇒ \(\frac{22}{7}\) x 3 (2×1 +5)

⇒ \(\frac{22}{7}\) x 3(10)

= 66m2

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CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Question 2. The volume and Surface area of a Solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Solution:

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3}\) = 3πr2

⇒ \(\frac{2}{3}\) r = 3

2r = 9

Hence, the diameter of the hemisphere = 9 units.

Question 3. 2 Cubes each of volume 64 cm3 are joined end to end. Find the Surface area of the resulting Cuboid.
Solution:

Given,

volume of cube = 64 cm3

(Side)3 = 64

(Side)3 = 43

Side = 4cm

Side of cube = 4cm

A Cuboid is formed by joining two Cubes together.

∴ For Cuboid

length l = 4+4=8cm,

breadth b = 4cm

height b = 4cm

Now, the total surface area of the cuboid.

= 2(l.b+b.h+l.h)

2(8×4+ 4×4+8×4)

= 2(32+16+32)

2(80)

= 160 cm2

Question 4. From a Solid Cylinder whose height is 2.4cm and diameter is 1.4 cm, a Conical Cavity of the Same height and diameter is hollowed out. Find the total Surface area of the remaining Solid to the nearest Cm2.
Solution:

Diameter of Cylinder 2r = 1.4cm

r = 0.7cm

∴ Radius of Cylinder = radius of cone = r= 0.70m

Height of Cylinder = height of cone

h = 2.40m

If the slant height of a cone is l, then

l2 = h2+r2 = (2.4)2 + (0.7)2

5.76 +0.49 = 6.25

1 = \(\sqrt{6.25}\) = 2.5cm

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Solid Cylinder Whose Height And Diameter The Total Surface Area Of The Remaining Solid To The Nearest Cm

The surface area of the remaining solid = area of the base of the cylinder + curved Surface of the cylinder + curved surface of the cone

πr2 + 2πrh + πrl

=πr (r+2h+1)

= \(\frac{22}{7}\) × 0.7 (0.7+2×2.4+2.5)

= \(\frac{22}{7}\) × 0.7 × (5.86) 223×0.7

= \(\frac{22}{7}\): × 4.102

= 17.6cm2

Question 5. The radius and height of a solid right Circular Cone are in the ratio of 5:12. If its volume is 314 cm3, find its total Surface area. [Take π = 3.14]
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Radius And Height Of A Solid Right circular Cone

Let the radius of Cone = 5x

∴ Height of Cone =12x

l2 = (5x)2+ (12x)2

l2 = 25x2 +144x2

l2 = 169x2

l = \(\sqrt{169 x^2}\)

l = 13x

It is given that volume = 314 cm2

∴ \(\frac{1}{3}\)π(5x)2 (12x) = 314

⇒ \(\frac{1}{3}\) × 3.14 x 25×12 × x3 = 314

⇒  x3 = \(\frac{314 \times 3}{3.14 \times 25 \times 12}\) = 1

∴ x = 1cm

∴ Radius r = 5×1 = 5cm

Height h = 12×1 = 12cm

and slant height 1 = 13×1 = 13cm

Now, total surface area of Cone = πr(l+r) = 3.14 × 5(13+5) = 3.14×5×18

= 282.60cm2

Hence, the total surface area of cone is 282.60cm2

Question 6. The Curved Surface area of a Cone of height 8m is 188.4m2. Find the volume of Cone.
Solution:

πrl = 188.4

⇒ rl = \(\frac{188.4}{3.14}\) = 60

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids The Curved Surface Area Of A Cone

r2l2 = 3600

r2 (h2 +r2)=3600

r2 (64+r2)=3600

r4+64r2 = 3600 = 0

(r2 +100) (r2 – 36)=0

∴ r2 = -100 or r2=36

r = 6

(r2 = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Question 7. A Conical tent is required to accommodate 157 persons, each person must have 2m2 of space on the ground and 15m3 of air to breathe. Find the height of the tent” Also Calculate the slant height.
Solution:

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Conical Tent Is Required To Accommodate Persons And Also Calculate The Slant Height

1 person needs 2m2 of Space.

∴ 157 Persons needs 2 × 157 m2 of Space on the ground.

πr2 = 2 × 157

∴ r2 = \(\frac{2 \times 157}{3.14}\)

r2 = 100

r = 10m

Also, 1 person needs 15m3 of air.

∴ 157 Persons need 15×157 m2 of air.

∴ \(\frac{1}{3}\) πr2h = 15×157

⇒ h = \(\frac{15 \times 157 \times 3}{3.14 \times 100}\) = 22.5m

∴ l2 = h2 + r2 = (225)2 + (10)2 = 606.25

∴ l = \(\sqrt{606.25}\) = 24.62m

Question 8. Three Cubes of metal whose edges are in the ratio 3:4:5 are melted down into a single cube whose diagonal is \(12 \sqrt{3}\) cm. Find the edges of the three cubes.
Solution:

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x3, 64x3, and 125x3 in cm3 respectively.

Now, the Sum of the Volumes of these three Cubes = 27x3 + 64x3+125x3

216x3 cm3

Let the edge of the new cube be a cm.

∴ Diagonal of new Cube = \(a \sqrt{3}\) cm

∴ \(a \sqrt{3}=12 \sqrt{3}\)

⇒ a = 12

∴ Volume of new Cube = (12)3 = 1728 cm3

Now by the given Condition

216x3 = 1728

x3 = 8

x = 2

∴ Edge of 1 Cube = 3×2 = 6cm

Edge of 2 Cube = 4×2 = 8 Cm

Edge of 3 Cube = 5×2 = 10cm

Question 9. A Solid is in the shape of a Cone Standing on a hemisphere with both their radii being equal to Icm and the height of the cone is equal to its radius. Find the volume of the Solid in terms of π.
Solution:

Rodius of hemisphere = radius of Cone = r = 1 cm

Height of Cone h = radius of Cone = 1 cm

Volume of hemisphere = \(\frac{2}{3}\) πr3

volume of Come = \(\frac{1}{3}\) πr2h

∴ Volume of Solid = Volume of hemisphere + volume of Come

⇒ \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

⇒ \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

⇒ \(\frac{2}{3} \pi+\frac{1}{3} \pi\)

= π cm3

Question 10. A granary is in the shape of a Cuboid of a size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags Can be stored in the granary?
Solution:

The Size of the granary is 8m x 6m x 3m.

∴ volume of granary =8×6×3 = 144m3

the volume of one bag of grain = 0.65m3

∴ The number of bags that can be stored in the granary

= \(\frac{\text { volume of granary }}{\text { volume of each bag }}\)

= \(\frac{144}{0.65}\)

= 221.54 or 221 bags.

Question 11. A cylindrical bucket 28cm in diameter and 72cm high is full of water. the water is emptied into a rectangular tank 66 cm long and 28cm wide. Find the height of the water level in the tank.
solution:

Let the height of the water level in the tank = xm, then according to problem

πr2h = l×b×x

Or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

Or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}\)

x = 24 Cm

Question 12. A Solid Spherical ball of Iron with a radius of 6cm is melted and recast into three Solid Spherical balls. The radii of the two balls are 3cm and 4cm respectively, determining the diameter of the third ball.
Solution:

Let the radius of the third ball = r cm

∴ The volume of three balls formed = volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 +64 +r3 = 216

⇒ r3 = 125, i.e., r = 5cm

The diameter of the third ball = 2×5cm = 10cm

Question 13. A Semicircle of radius 17.5cm is rotated about its diameter. Find the Curved Surface of So generated Solid.
Solution:

The Solid generated by a circle rotated about its diameter is a Sphere

Now, a radius of Sphere r = 17.5 cm

and its Curved Surface = 4π2

= \(4 \times \frac{22}{7} \times 17.5 \times 17.5\)

= 3850 cm2

Question 14. A sphere of radius 6cm is melted and recast into a Cone of height 6cm. Find the radius of the Cone.
Solution:

Radius of Sphere = 6cm

∴ Volume of Sphere = \(\frac{4}{3} \pi(6)^3\) = 288π cm3

Let radius of Cone = r

Height of Cone = 6cm

∴ Volume of Cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of Cone = volume of a sphere

2πr2 = 288π

r2 = 144

r = 12

Therefore, a radius of Cone = 12cm

Question 15. A metallic Cylinder of diameter 16 cm and height 9cm is melted and recast Into Sphere of diameter 6cm. How many such Spheres can be formed?
Solution:

For the Cylinder,

Radius = \(\frac{16}{2}\) = 8cm

Height = 9cm

∴ Volume of Cylinder = π (8)2(9) = 576 π cm3

Diameter of Sphere = 6cm

∴ Radius of Sphere = \(\frac{6}{2}\) = 3 cm

Now, Volume of one Sphere = \(\frac{4}{3} \pi(3)^3\) = 36π cm3

∴ Number of Spheres formed = \(=\frac{\text { volume of Cylinder }}{\text { volume of one Sphere }}\)

⇒ \(\frac{576 \pi}{36 \pi}\)

= 16

Question 16. The volume of a sphere is 288π Cm3, 27 Small Spheres Can be formed with this Sphere. Find the radius of the Small Sphere.
Solution:

Volume of 27 Small Spheres = Volume of One big Sphere = 288π

⇒ Volume of 1 Small Sphere = \(\frac{288}{27} \pi\)

⇒ \(\frac{4}{3} \pi r^3=\frac{22}{3} \pi\)

⇒ r3 = 8

⇒ r = 2cm

Question 17. A metallic Sphere of radius 4.2cm is melted and recast into the Shape of a Cylinder of radius 6cm, Find the height of the cylinder.
Solution:

Radius of Sphere R = 4.2cm

Radius of Cylinder r = 6cm

Let the height of the Cylinder = h

Now, the volume of the cylinder = Volume of the Sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ h = \(\frac{4 R^3}{3 r^2}\)

⇒ \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 сm

∴ Height of Cylinder = 2.744cm

Question 18. Metallic Spheres of radii 6cm, 8cm, and 10 cm, respectively, are melted to form a Single Solid Sphere. Find the radius of the resulting Sphere.
Solution:

Let r1 = 6cm r2 =8cm and r3 = 10cm

let the radius of a bigger Solid Sphere = R

The volume of bigger Solid Volume = Sum of volumes of three given Spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

⇒ \(R^3=6^3+8^3+10^3\)

⇒ R3 = 216 +512 +1000

⇒ R3 = 1728

⇒ R2 = 123

⇒ R = 12cm

∴ Radius of new Solid Sphere = 12 cm

Question 19. A 20m deep well with a diameter 7m is dug and the earth from digging is evenly Spread out to form a platform 22m of the platform.
Solution:

Radius of well, r = \(\frac{7}{2} m\)

and depth h = 20m

Let the height of the platform be H meter.

∴ The volume of platform = Volume of well

⇒ 22 x 14 x H = πr2h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ H = \(\frac{7 \times 5}{14}\)

⇒ H = \(\frac{35}{14}\)

⇒ H = 2.5mn

Height of platform = 2.5m

Question 20. A Cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.
Solution:

We Can Solve this using Similarity

Let r and h be the radius and height of a Cone OAB

CBSESE School For Class 10 Maths Chapter 10 Volume And Surface Area Of Solids A Cone Is Divided Into Two Parts By Drawing A Plane Through The Mid Point Of Its Axis

Let OE = \(\frac{h}{2}\)

As OED and OFB are Similar

∴ \(\frac{OE}{O F}=\frac{ED}{F B}\)

⇒ \(\frac{h / 2}{h}=\frac{ED}{r}\)

⇒ ED = \(\frac{r}{2}\)

Now volume of Cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}\)

⇒ \(\frac{\pi r^2 h}{24}\)

and volume of Cone DAB = \(\frac{1}{3} \times \pi x r^2 \times h\) = \(\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { volume of Part } O C D}{\text { volume of Part } C D A B}\)

⇒ \(\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{8-1}{24}}\)

⇒ \(\frac{1}{7}\)

Question 21. A well of diameter 3m is dug lum deep. The earth taken out of it has been Spread evenly all around it in the Shape of a Circular ring of width um to form an embankment. Find the height of the embankment.
Solution:

Diameter of well 2r = 3m

r = \(\frac{2}{3}\) = 1.5m

and depth h = 14m

∴ Volume of earth taken out from well = πr2h

⇒ \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5+4 = 5.5m

∴ Area of the ring of platform = π(R2– r2)

⇒ \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

⇒ \(\frac{22}{7}[30.25-2.25]\)

⇒ \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

let the height of the embankment = H

∴ 88 x H=99

⇒ \(H=\frac{99}{88}=\frac{9}{8}\)

= 1.125m

∴ Height of embankment = 1.125m

CBSE Solutions For Class 10 Mathematics Chapter 14 Statistics

CBSE Solutions For Class 10 Mathematics Chapter 14 Statistics

Question 1. Find the mean by the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean By Direct Method

Now,

mean \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}=\frac{1100}{50}=22\)

Question 2. Find the mean using the Direct Method:

CBSE School For Class 10 Maths Chapter 14 Statistics CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method.

Read and Learn More Class 10 Maths

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Find The Mean Using Direct Method

Now, mean \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}=\frac{13,200}{50}=264\)

Question 3. The mean of the following frequency distribution is 25. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 25

Now, Mean

⇒ \(\bar{x}=\frac{\sum f_{i x i}}{\sum f_i}\)

⇒ \(\bar{x}=\frac{1230+15 p}{42+p}\)

⇒ \(25=\frac{1230+15 P}{42+P}\)

⇒ 1050 + 25P = 1230 + 15P

⇒ 25P – 15P = 1230 + 15P

⇒ 10P = 180

⇒ P = \(\frac{180}{10}\)

P = 18

CBSE Class 10 Maths Solutions Statistics

Question 4. The mean of the following distribution is 54. Find the value of P using the direct method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54.

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Of The Following Distribution Is 54

Now, Mean \(\bar{x}=54\)

⇒ \(\bar{x}=\frac{\sum f_{i x_i}}{\sum f_i}\)

54 = \(\frac{2070+70 p}{41+P}\)

2214+54p=2070+70P

70P-54P = 2214-2070

16P = 144

P = \(\frac{144}{16}\)

P = 9

Question 5. Find the mean from the following table using the Short Cut Method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Short Cut Method

Now, Mean a = 22.5

Mean \(\bar{x}=a+\frac{\sum f_{i d i}}{\sum f_i}\)

⇒ \(\bar{x}=22.5+\frac{0}{67}\)

⇒ \(\bar{x}=22.5\)

Question 6. Find the mean from the following table using the step. deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics Step Deviation Method

Now, a = 37.5, h = 5

Mean \(\bar{x}=a+\frac{\sum f_{i u_i}}{\sum f_i} \times h\)

⇒ \(\bar{x}=37.5+\frac{-46}{50} \times 5\)

⇒ \(\bar{x}=37.5-\frac{46}{10}\)

⇒ \(\bar{x}=\frac{375-46}{10}\)

⇒ \(\bar{x}=\frac{329}{10}\)

⇒ \(\bar{x}=32.9\)

Question 7. Find the mean from the following table using the step deviation method:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean From The Following Table Using Step Deviation Method

Now, a = 27.5, h = 5

mean \(\bar{x}=a+\frac{\sum f_i u_i}{\sum f_i} \times h\)

⇒ \(\bar{x}=27.5+\frac{-20}{40} \times 5\)

⇒ \(\bar{x}=\frac{220-20}{8}\)

⇒ \(\bar{x}=\frac{200}{8}\)

⇒ \(\bar{x}=25\)

Question 8. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 8

Here, N = 58

∴ For Median class \(\frac{N}{2}=\frac{58}{2}=29\)

∴ Median class = 10-13

Here l1 = 10, l2 = 13

⇒ i = 13-10 = 3

f = 0, C.f = 29

∴ Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(10+\frac{29-29}{0} \times 3\)

M = \(10+\frac{0}{0} \times 3\)

M = 10

Question 9. Find the Median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data 9

Here, N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

=25

∴ Median Class = 20-30

Now, l1 = 20, l2 = 30

i = 30-20 = 10

f = 12, C = 15

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(20+\frac{(25-15)}{12} \times 10\)

M = \(20+\frac{10}{6} \times 5\)

M = \(\frac{120+50}{6}\)

M = \(\frac{170}{6}\)

M = 28.33

Question 10. Find the median from the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median From The Following Data

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

∴ The median class is 15-20

Now, l1 = 15, l2 = 20, i = 20-15 = 5, f = 15, C = 20

and Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(15+\frac{25-20}{15} \times 5\)

M = \(\frac{45+5}{3}\)

M = \(\frac{50}{3}\)

M = 16.67

Question 11. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median For The Following Frequency Distribution

Now, N=340

⇒ \(\frac{N}{2}=\frac{340}{2}=170\)

∴ Median class = 39.5-46.5

∴ l1 = 39.5, l2 = 46.5, i= 46.5-39.5 = 7, f = 102, C = 199

Median M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

M = \(39.5+\frac{(170-199)}{102} \times 7\)

M = \(39.5-\frac{29}{102} \times 7\)

M = \(39.5-\frac{203}{102}\)

M = \(\frac{4029-203}{102}\)

M = \(\frac{3826}{102}\)

M = 36.5

Question 12. Find the Median for the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution

Here N = 123

⇒ \(\frac{N}{2}=\frac{123}{2}=61.5\)

∴ Median Class = 20,5-25.5

l1 = 20.5, l2 = 25.5, i = 25.5-20.5 = 5, f = 24, C = 65

Median M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{F} \times i\)

M = \(20.5+\frac{(61.5-65)}{24} \times 5\)

M = \(20.5-\frac{3.5}{24} \times 5\)

M = \(\frac{492-17.5}{24}\)

M = \(\frac{474.5}{24}\)

M = 20.1

Question 13. If the Median of the following frequency distribution is 32.5. Find the value of F.

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Median Of The Following Frequency Distribution Is 325

Here N = 34+ P

⇒ \(\frac{N}{2}=\frac{34+P}{2}\)

Median = 325 ⇒ Median class=30-40

∴ 11 = 30, l2 =40, i = 40-30=10

f = 12, C = 17

M = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{f} \times i\)

32.5 = \(30+\frac{\left(\frac{34+p}{2}-17\right)}{12} \times 10\)

32.5-30 = \(\frac{10}{12}\left(\frac{34+P}{2}-17\right)\)

2.5 = \(\frac{5}{6}\left(\frac{34+p-34}{2}\right)\)

2.5 = \(\frac{5 P}{12}\)

30 = 5P

P = \(\frac{30}{5}\)

P = 6

Question 14. Determine the median for the following income distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean For The Following Income Distribution

Here N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}=50\)

∴ Medion class = 300-400

∴ l1 = 300, l2 = 400, i= 400-300 = 100, f = 30, C = 33

M = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(300+\frac{50-33}{30} \times 100\)

M = \(300+\frac{17}{30} \times 100\)

M = \(\frac{9000+1700}{30}\)

M = \(\frac{10700}{30}\)

M = 356.67

Question 15. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data.

Solution:

Clearly, the modal class is 60-80 as it has the maximum frequency

∴ l = 60, f1 = 12, f0 = 10, f2 = 6, h = 20

Mode M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(60+\frac{12-10}{2(12)-10-6} \times 20\)

M = \(60+\frac{2}{24-16} \times 20\)

M = \(60+\frac{2}{8} \times 20\)

M = \(\frac{480+40}{8}\)

M = \(\frac{520}{8}\)

M = 65

Question 16. Given below is the frequency distribution of the heights of players in a school;

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Frequency Distribution Of The Heights Of Players In A School

Clearly, the modal class is 165.5-1685 as it has a maximum frequency

∴ l = 165.5, f1 = 142, f0 =118, f2 = 127, h=3

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(165.5+\frac{142-118}{2(142)-118-127} \times 3\)

M = \(165.5+\frac{24}{284-245} \times 3\)

M = \(165.5+\frac{24}{39} \times 3\)

M = \(\frac{6454.5+72}{39}\)

M = \(\frac{6526.5}{39}\)

M = 167.35

Question 17. Find the mode of the following frequency distribution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Frequency Distribution

Solution:

Clearly, the modal class is 50-60, as it has the maximum frequency

∴ l = 50, f1 = 11, f0 = 9, f2 = 6, h = 10

Mode = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(50+\frac{11-9}{2(11)-9-6} \times 10\)

M = \(50+\frac{2}{22-15} \times 10\)

M = \(50+\frac{2}{7} \times 10\)

M = \(\frac{350+20}{7}\)

M = \(\frac{370}{7}\)

M = 52.86

Question 18. The following distribution represents the height of 160 students in a class;

CBSE School For Class 10 Maths Chapter 14 Statistics The Height Of 160 Students Of A Class

Solution:

Clearly, the modal class is 155-160 as it has the maximum

∴ l = 155, f1 = 38, f0 =30, f2 =24, h=5

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(155+\frac{38-30}{2(38)-30-24} \times 5\)

M = \(155+\frac{8}{76-54} \times 5\)

M = \(155+\frac{40}{22}\)

M = \(\frac{3410+40}{22}\)

M = \(\frac{3450}{22}\)

M = 156.82

Question 19. The following table gives the weekly wage of workers in a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory.

Find (1) the mean (2) the modal class (3) the mode

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Weekly Wage Of Workers In A Factory

1. Mean:

Mean = \(\frac{\sum f_{i x i}}{\sum f_i}=\frac{5520}{80}\)

= 69

2. Modal class = 55-60

3. Mode: clearly, the Modal class is 55-60 it has the maximum frequency

l = 55, f1 = 20, f0 = 5, f2 = 10, h = 5

M = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(55+\frac{20-5}{2(20)-5-10} \times 5\)

M = \(55+\frac{15}{40-15} \times 5\)

M = \(55+\frac{15}{25} \times 5\)

M = \(\frac{1375+75}{25}\)

M = \(\frac{1450}{25}\)

M = 58

Question 20. The mode of the following Series is 36. Find the missing frequency in it:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Series Is 36 Find The Missing Frequency

Solution:

Clearly, 30-40 is the modal class as mode 36 lies in this class

Here l = 30, f1 = 16, f0 = x (Say), f2 = 12 and h = 10 and mode 36

Mode (M) = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

36 = \(30+\frac{16-x}{2(16)-x-12} \times 10\)

36 = \(30+\frac{16-x}{32-x-12} \times 10\)

36 = \(30+\frac{16-x}{20-x} \times 10\)

36 – 30 = \(\frac{16-x}{20-x} \times 10\)

6(20-x) = 10(16-x)

120-6x = 160-10x

10x-6x = 160-120

4x = 40

x = \(\frac{40}{4}\)

x = 10

Question 21. Compute the mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Data

Clearly, Modal class 29.5 – 29.5 as it has the maximum frequency

∴ l = 19.5, f1 = 23, f0 = 16, f2 = 15, h = 10

M = \(\ell+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

M = \(19.5+\frac{23-16}{2(23)-16-15} \times 10\)

M = \(19.5+\frac{7}{46-31} \times 10\)

M = \(\frac{292.5+70}{15}\)

M = 24.17

Question 22. Find the mean, Median, and mode of the following data:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mode Of The Following Frequency Distribution

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Mean Median And Mode Of The Following Data

let assumed mean A = 70, h = 20, Ef = 50, and Σfu = -19

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=70+\left[20 \times \frac{-19}{50}\right]\)

⇒ \(\bar{x}=70+[20 x-0.38]\)

⇒ \(\bar{x}=70-7.6\)

⇒ \(\bar{x}=62.4\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}\) = 25

Cumulative frequency just greater than 25 is 36 and the Corresponding class is 60-80.

∴ l1 = 60, f = 12, l2 = 80, C = 24, i = 80-60 = 20

Now, median (M) = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(60+\frac{(25-24)}{12} \times 20\)

M = \(60+\frac{1}{12} \times 20\)

M = \(\frac{720+20}{12}\)

M = \(\frac{740}{12}\)

M = 61.66

Mode = 3(Median)-2(Mean)

Mode = 3(61.66)-2(62.4)

= 184.98-124.8

M = 60.18

Question 23. 100 Surnames were randomly picked from a local directory and the distribution of a number of letters of the English alphabet in the Surname was obtained as follows:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics 100 Surnames Were Randomly Picked From A Local Directony And Letter Of The English Alphabet

let assumed mean A = 11.5, h = 3, Σf = 100, Σfu = -106

Mean \(\bar{x}=A+\left[h \times \frac{\sum(f u)}{\sum f}\right]\)

⇒ \(\bar{x}=11.5+\left[3 \times \frac{-106}{100}\right]\)

⇒ \(\bar{x}=11.5+[3 \times-1.06]\)

⇒ \(\bar{x}=11.5-3.18\)

⇒ \(\bar{x}=8.32\)

Here, N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}\)

= 50

Cumulative frequency just greater than 50 is 76 and the corresponding class is 10-13

∴ l1 = 7, f = 40, l2 = 10, C = 36, i = 10-7 = 3

Median (M) = \(\ell+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

M = \(7+\frac{50-36}{40} \times 3\)

M = \(7+\frac{14}{40} \times 3\)

M = \(\frac{280+42}{40}\)

M = \(\frac{322}{40}\)

M = 8.05

Mode = 3(Median)-2(Mean)

= 3(8.05)-2(8-32)

= 24.15-16.64

M = 7.51

Question 24. The following table gives the daily income of such workers of a factory:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory.

Find the mean, mode, and median of the above data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Income Of 50 Workers Of A Factory

let assumed mean A = 150, h = 20, Σf = 50, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_4}{\sum P}\right]\)

⇒ \(\bar{x}=150+\left[20 \times \frac{-12}{50}\right]\)

⇒ \(\bar{x}=150+[20 x-0.24]\)

⇒ \(\bar{x}=150-4.8\)

⇒ \(\bar{x}=145.2\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 36 and the Corresponding is 120-140

∴ l1 = 120, f = 14, l2 = 140, C = 12, i = 140-120 = 20

Now, Median = \(l_1+\frac{\left(\frac{N}{2}- C\right)}{f} \times i\)

⇒ \(120+\frac{(25-12)}{14} \times 20\)

⇒ \(120+\frac{13}{14} \times 20\)

⇒ \(\frac{1680+260}{14}\)

⇒ \(\frac{1940}{14}\)

M = 138.57

Mode = 3(Median) – 2 (Mean)

M = 3(138.57)-2((45.2)

= 415.71-290.4

= 125.31

Question 25. A Survey regarding the heights (in cm) of so girls in a class was conducted and the following data was obtained:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data.

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics A Survey Regarding The Height In Cm Of 50 Girls Of A Class Was Conducted And The Following Data

Let assumed Mean A=145, h=10, Σf=50, and Σfu=24

Mean \(\bar{x}=A+\left[h \times \frac{\sum f_u}{\sum f}\right]\)

⇒ \(\bar{x}=145+\left[10 \times \frac{24}{50}\right]\)

⇒ \(\bar{x}=145+[10 \times 0.48]\)

⇒ \(\bar{x}=145+4.8\)

⇒ \(\bar{x}=149.8 \mathrm{~cm}\)

Here N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency just greater than 25 is 42 and the Corresponding is 150-160

l1 = 150, f = 20, l2 = 160, C = 22, i = 160-150 = 10

Now, Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

⇒ \(150+\frac{(25-22)}{20} \times 10\)

⇒ \(150+\frac{30}{20}\)

⇒ \(\frac{3000+30}{20}\)

⇒ \(\frac{3030}{20}\)

= 151.5 cm

Mode = 3(Median)-2(Mean)

Mode = 3(151.5)-2(149.8)

= 454.5-299.6

= 154.9

Question 26. The table below shows the daily expenditure on food of 30 households in a locality:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality

Solution:

CBSE School For Class 10 Maths Chapter 14 Statistics The Daily Expenditure On Food Of 30 House Holds In A Locality.

Let assumed mean A=225, h = 50, Σf=30, Σfu = -12

Mean \(\bar{x}=A+\left[h \times \frac{\sum f u}{\sum f}\right]\)

⇒ \(\bar{x}=225+\left[50 \times \frac{-12}{30}\right]\)

⇒ \(\bar{x}=225+[50 \times -0.4]\)

⇒ \(\bar{x}=225-20\)

⇒ \(\bar{x}=205\)

Here, N = 30

⇒ \(\frac{N}{2}=\frac{30}{2}=15\)

Median Class = 200-250

l1 = 200, f = 12, l2 = 250, C = 13, i = 250-200 = 50.

Median (M) = \(l_1+\frac{\left(\frac{N}{2}-c\right)}{F} \times i\)

M = \(200+\frac{15-13}{12} \times 50\)

M = \(\frac{2400+100}{12}\)

M = \(\frac{2500}{12}\)

M = 208.33

CBSE Solutions For Class 10 Mathematics Chapter 15 Probability

CBSE Solutions For Class 10 Mathematics Chapter 15 Probability

Question 1. In any situation that has only two possible outcomes, each outcome will have Probability. Find whether it is true or false.
Solution: False.

The probability of each outcome will be \(\frac{1}{2}\), only when the two outcomes are equally likely.

Question 2. A Marble is chosen at random from 6 marbles numbered 1 to 6. Find the Probability of getting a marble having number 2 and 6 on it.
solution:

The favourable Case is to get a marble on which both numbers 2 and are written. But there is no such marble.

So, N(E) = 0 and n(S) = 6

∴ Required probability = \(\frac{n(E)}{n(S)}=\frac{0}{6}=0\)

Question 3. A marble is chosen at random from 6 marbles numbered Ito 6. Find the Probability of getting a marble having number 2 or 6 on it.
Solution:

Here N(E) = 2

and n(S) = 6

∴ Required Probability = \(\frac{2}{6}=\frac{1}{3}\)

Read and Learn More Class 10 Maths

CBSE Class 10 Maths Solutions Probability

Question 4. It is given that in a group of 3 Students, the probability of 2 students not having the Same birthday is 0.992. What is the probability that the 2 students have the Same birthday.
Solution:

The probability of 2 students not having the Same birthday = 0.992

∴ Probability of 2 students having the same birthday

= 1-0.992

= 0.008

Question 5. A bag Contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

  1. Red?
  2. Not ved?

Solution:

Total balls = 3+5 = -8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favorable outcomes of drawing a batt at random red ball =3

∴ The probability of drawing a red ball

⇒ \(\frac{\text { Favourable outcomes of drawing a red ball }}{\text { Total possible outcomes }}\)

⇒ \(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1- probability that the ball drawn

⇒ \(1-\frac{3}{8}=\frac{5}{8}\)

Question 6. A die is thrown once. Find the probability of getting:

  1. A prime number
  2. A number lying between 2 and 6
  3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2,3,5}=3

∴ Probability of getting prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3, 4, 5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1,3,5}=3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

Question 7. 12 defective pens are accidentally mixed with 132 good to just look at a pen and tell whether or not it is taken out at random from this lot. Determine the probability Out is a good one.
Solution:

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 +12 = 144

Total Favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}=\frac{11}{12}\)

Question 8. A child has a die whose Six faces show the letters as

CBSE School For Class 10 Maths Chapter 15 Probability A Child Has A Die Whose Six Fases Show The Letters

given below; The die is thrown once. what is the probability of getting

  1. A?
  2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting \(A=\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 9. A box Contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

  1. A two-digit number,
  2. A number divisible by 5.

Solution:

We have, n(S) = 90

1. Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10,11,12,13,14, …….., 90

∴ n(A) = 81

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

2. let B be the event of getting ་་ a number divisible by 5″.

∴ Favourable Cases are 10, 15, 20, 25, 30, …… 90.

Let there be n in numbers.

∴ Tn = 90

∴ 10+ (n-1)5 = 90

⇒ (n-1)5=80

⇒ n-1 = 16

⇒ n = 17

∴ n(B) = 17

P(B) = \(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 10. It is known that a box of 600 electric bulbs Contains 12 defective bulbs. One bulb is taken out at random from this box, what is the probability that it is a non-defective bulb?
Solution:

The number of non-defective bulbs in the box = 600-12=588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)

=0.98

Important Questions for Class 12 Chemistry Chapter 10 Biomolecules

Biomolecules

Question 1. Glucose or succrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:

  • A glucose molecule contains five -OH groups while a sucrose molecule contains eight -OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water.
  • Hence, they are soluble in water.
  • But, cyclohexane and benzene do not contain -OH groups. Hence, they cannot undergo II- bonding with water and as a result, they are insoluble in water.

Question 2. What are the expected products of the hydrolysis of lactose?
Answer:

Lactose is composed of β-D galactose and β-D glucose. Tims, on hydrolysis, gives β-D galactose and β-D glucose.

Biomolecules Lactose

Biomolecules Lactose And D Glucose

Question 3. How do you explain the absence of an aldehyde group in the pentaacctate of D-glucose?
Answer:

D-glucose reacts with hydroxylamine (NH2, OH) to form an oxime because of the presence of an aldehydic (-CHO) group of carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2, OH to give an oxime.

Biomolecules Glucose And Oxime

But pentaacctate of D-glucose does not react with NH2, OH. This is because pentaacctate does not form an open chain structure.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer:

Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept, thus giving rise to a dipolar ion known as a zwitter ion.

Biomolecules Zwitter Ion

  • Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. However, halo-acids do not exhibit such dipolar behaviour.
  • For this reason, the melting points and the solubility of amino acids in water are higher than those of the corresponding halo acids.

Question 5. Where does the water present in the egg go after boiling the egg?
Answer:

When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg. the water present in it is absorbed by the coagulated protein through H-bonding.

CBSE Class 12 Chemistry Chapter 10 Biomolecules Question And Answers

Question 6. Why cannot vitamin C be stored in our bodies?
Answer: Vitamin C cannot be stored in our body because it is water-soluble. As a result, it is readily excreted in the urine.

Question 7. What products would be formed when a nucleotide from DMA containing thymine is hydrolysed?
Answer: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine. β-D-2- deoxyribose and phosphoric acid are obtained as products.

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:

  • A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine.
  • Therefore, on hydrolysis of DNA. the quantity of adenine produced is equal of that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
  • But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.

Question 9. What are monosaccharides?
Answer:

  • Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified based on a number of carbon atoms and the functional group present in them.
  • Monosaccharides containing an aldehyde group are known as aldose and those containing a keto group are known as ketose.
  • Monosaccharides are further classified as triose, tetrose, pentose, Itexose, and heptose according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3  carbon atoms is called aldotriose.

Question 10. What are reducing sugars?
Answer: Reducing sugars are carbohydrates that reduce l’ehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.

Question 11. Write two main functions of carbohydrates in plants.
Answer:

Two main functions of carbohydrates in plants are:

  1. Polysaccharides such as starch serve as storage molecules.
  2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 12. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose. maltose, galactose, fructose and lactose
Answer:

  1. Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
  2. Disaccharides: Maltose, lactose

Question 13. What do you understand by the term glycosidic linkage?
Answer:

  • Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
  • For example, in a sucrose molecule, two monosaccharide units, α- glucose and β- fructose, are joined together by a glycosidic linkage.

Biomolecules Surcose

Question 14. What is glycogen? How is it different from starch?
Answer:

Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components – amylose (15-20%) and amylopectin (80-85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.

Question 15. What are the hydrolysis products of (1) sucrose and (2) lactose?
Answer: On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

Biomolecules The Hydrolysis Production Of Sucrose And Lactose 1

The hydrolysis of lactose gives β-D-galactose and β-D-glucose.

Biomolecules The Hydrolysis Production Of Sucrose And Lactose 2

Question 16. What is the basic structural difference between starch and cellulose?
Answer:

Starch consists of two components – amylose and atm lopectin. Amylose is a long linear chain of α-D-(+)-glucose units joined by C1– C4 glycosidic linkage (α-link).

Biomolecules Basic Steructural Difference Between Starch And Cellulose 1

Amylopectin is a branched-chain polymer of α—D—glucose units, in which the chain is formed by C1 – C4 glycosidic linkage and the branching occurs by C1 – C6, glycosidic linkage.

Biomolecules Basic Steructural Difference Between Starch And Cellulose 2

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1 – C4 glycosidic linkage (β-link).

Biomolecules Basic Steructural Difference Between Starch And Cellulose 3

Question 17. What happens when D-glucose is treated with the following reagents?

  1. HI
  2. Bromine water
  3. HNO3

Answer:

When D-glucose is heated with III for a long time, n-hexane is formed.

Biomolecules D Glucose Is Treated With HI

When D-glucose is treated with Br2 water, D-gluonic acid is produced.

Biomolecules D Glucose Is Treated With Bromine Water

On being treated with HNO3, D-glucose gets oxidised to give saeeliaric acid.

Biomolecules D Glucose Is Treated With HNO3

Question 18. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

  1. Aldehydes give 2, 4-DNP test. Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
  2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free-CHO group is absent from glucose.
  3. Glucose exists in two crystalline forms -α and β. The α-form (m.p. 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p.=423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 19. What are essential and non-essential amino acids? Give two examples of each type.
Answer:

Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food. For example: valine and leucine Non-essential amino acids are also required by the human body, but they can be synthesised in the body.

For example: Glycine, and Alanine

  1. Peptide linkage
  2. Primary structure
  3. Denaturation

Question 20. Define the following as related to proteins

  1. Peptide linkage
  2. Primary structure
  3. Denaturation

Answer:

Peptide linkage: The amide for pied between the -COOH group of one molecule of an amino acid and the -NH2, group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.

Biomolecules Peptide Linkage

Primary structure: The primary structure of a protein refers to the specific sequence in which various amino acids are present in it. i.e. the sequence of linkage between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

Denaturation: In a biological system, a protein is found to have a unique 3- dimensional structure and a unique biological activity. In such a situation, the protein is called native protein.

  • However, when the native protein is subjected to physical changes such as changes in temperature or chemical changes such as changes in pH. its H-bonds are disturbed.
  • This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation.
  • During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.  One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 21. What are the common types of secondary structures of proteins?
Answer:

There are two common types of secondary .structure of proteins:

  1. α- helix structure
  2. β-pleated sheet structure

α- Helix Structure: In this structure, the Nil group of an amino acid residue forms H-bonds with the group of the adjacent turn of the right-handed screw (α-helix).

β-pleated Sheet Structure: This structure is called so because it looks like the pleated folds of draper). In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Biomolecules Alpha And Beta Plated Steructure

Question 22. What type of bonding helps in stabilising the a -helix structure of proteins?
Answer: The H-bonds formed between the -Ml group of each amino acid residue and the Biomolecules Amino Acid group of
the adjacent turns of the α-helix help in stabilising the helix.

Question 23. Differentiate between globular and fibrous proteins.
Answer:

Biomolecules Fibrous Protein And Globular Protein

Question 24. How do you explain the amphoteric behaviour of amino acids?
Answer: In an aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as a zwitter ion.

Biomolecules Zwitter Ion.

Therefore, in Zwitter ionic form, the amino acid can act both as an acid and as a base.

Biomolecules Zwitter Ion Forms The Amino Acid

Thus, amino acids show amphoteric behaviour.

Question 25. What are enzymes?
Answer:

  • Enzymes are proteins that catalyse biological reactions. They are very specific and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after a particular substrate or class of substrate and sometimes after a particular reaction.
  • For example, enzymes used to catalyse the hydrolysis of maltose into glucose are named as maltase.

Biomolecules Maltose And Glucose

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named oxidoreductase enzymes, the name of an enzyme ends with ’-ase’.

Question 26. What is the effect of denaturation on the structure of proteins?
Answer:

  • As a result of denaturation. globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation.
  • secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 27. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:

Biomolecules Vitamins Classified

Based on their solubility in water or fat. vitamins are classified into two groups.

  1. Fat-soluble Vitamins: Vitamins that are soluble in fat and oils, but not in ‘water, belong to this group.
    • For example: Vitamins A, D, E, and K
  2. Water-soluble Vitamins: Vitamins that are soluble in water belong to this group,
    • For example: B group vitamins (B1, B2, B6, B12, etc.) and vitamin C

However, biotin or vitamin H is neither soluble in water nor fat. Vitamin K is responsible for the coagulation of blood.

Question 28. Why are vitamin A and vitamin C essential to us? Give their important sources,
Answer:

  • The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums).
  • The source of vitamin A is fish liver oil, carrots, butter, and milk. The sources of vitamin C are citrus fruits. amla, and green leafy vegetables.

Question 29. What are nucleic acids? Mention their two important functions.
Answer:

Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids – deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.

Two main functions of nucleic acids are:

  1. DNA is responsible for the transmission of inherent characteristics from one generation to the next. This process of transmission is called heredity.
  2. Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell, even though the proteins are synthesised by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.

Question 30. What is the difference between a nucleoside and a nucleotide?
Answer:

A nucleoside is formed by the attachment of a base to the 1′ position of the sugar.

Nucleoside = Sugar + Base

On the other hand, all three basic components of nucleic acids (i.e. pentose sugar, phosphoric acid, and base) are present in a nucleotide.

Biomolecules Nucleoside And Nucleotide

Question 31. The two strands in DNA are not identical bill are complementary. Explain.
Answer:

In the helical structure of DNA. the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosin forms hydrogen bonds with guanine, while adenine forms hydrogen bonds with thymine. As a result, the two strands are complementary to each other.

Question 32. Write the important structure and functional difference between DNA and RNA.
Answer: The structural differences between DNA and RNA are as follows:

Biomolecules Structural Difference Between DNA And RNA

The functional differences between DNA and RNA are as follows:

Biomolecules Functional Difference Between DNA And RNA

Question 33. What are the different types of RNA found in the cell?
Answer:

  1. Messenger RNA (m-RNA)
  2. Ribosomal RNA (rRNA)
  3. Transfer RNA (t-RNA)

Question 34. What are hormones and write to its functions?
Answer:

Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly into the bloodstream which transports them to the site of action.

In terms of chemical nature, some of these are steroids, for example., estrogens and androgens: some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine.

  1. Insulin keeps the blood glucose level within a narrow limit. Hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.
  2. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play a role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of the amino acid tyrosine.
  3. Steroid hormones are produced by the adrenal cortex and gonads (testes in males and ovaries in females).
  4. Glucocorticoids control carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.
  5. Testosterone is the major sex hormone produced in males. It is responsible for the development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It’s responsible for the development of secondary female characteristics and participates in the control of the menstrual cycle.
  6. Progesterone is responsible for preparing the uterus for the implantation of a fertilised egg.