Prasanna

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1

Integrate the following with respect to ‘x’:
Question 1.
(i) x¹¹
(ii) \(\frac{1}{x^{7}}\)
(iii) \(\sqrt[3]{x^{4}}\)
(iv) \(\left(x^{5}\right)^{\frac{1}{8}}\)
Solution:

Question 2.
(i) \(\frac{1}{\sin ^{2} x}\)
(ii) \(\frac{\tan x}{\cos x}\)
(iii) \(\frac{\cos x}{\sin ^{2} x}\)
(iv) \(\frac{1}{\cos ^{2} x}\)
Solution:

Question 3.
(i) 12³
(ii) \(\frac{x^{24}}{x^{25}}\)
(iii) e^x
Solution:

Question 4.
(i) (1 + x²)^-1
(ii) \(\left(1-x^{2}\right)^{-\frac{1}{2}}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1 Additional Problems

Integrate the following with respect to x.
Question 1.
(i) \(\sqrt{x^{7}}\)
(ii) (x¹⁰)^1/7
Solution:

Question 2.
(i) \(\frac{1}{x^{5}}\)
(ii) x^-1
(iii) \(\frac{1}{x^{5 / 2}}\)
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.4

Evaluate the following limits

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 17 Reproduction in Plants and Animals

Samacheer Kalvi 10th Science Reproduction in Plants and Animals Textual Evaluation Solved

I. Choose the Correct Answer.

Question 1.
The plant which propagates with the help of its leaves is ______.
(a) Onion
(b) Neem
(c) Ginger
(d) Bryophyllum.
Answer:
(d) Bryophyllum

Question 2.
Asexual reproduction takes place through budding in:
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Bacteria
Answer:
(b) Yeast

Question 3.
Syngamy results in the formation of ______.
(a) Zoospores
(b) Conidia
(c) Zygote
(d) Chlamydospores.
Answer:
(c) Zygote

Question 4.
The essential parts of a flower are:
(a) Calyx and Corolla
(b) Calyx and Androecium
(c) Corolla and Gynoecium
(d) Androecium and Gynoecium
Answer:
(d) Androecium and Gynoecium

Question 5.
Anemophilous flowers have ______.
(a) Sessile stigma
(b) Small smooth stigma
(c) Coloured flower
(d) Large feathery stigma.
Answer:
(d) Large feathery stigma.

Question 6.
Male gametes in angiosperms are formed by the division of:
(a) Generative cell
(b) Vegetative cell
(c) Microspore mother cell
(d) Microspore
Answer:
(a) Generative cell

Question 7.
What is true of gametes?
(a) They are diploid
(b) They give rise to gonads
(c) They produce hormones
(d) They are formed from gonads.
Answer:
(d) They are formed from gonads.

Question 8.
A single highly coiled tube where sperms are stored, get concentrated and mature is known as:
(a) Epididymis
(b) Vasa efferentia
(c) Vas deferens
(d) Seminiferous tubules
Answer:
(a) Epididymis

Question 9.
The large elongated cells that provide nutrition to developing sperms are ______.
(a) Primary germ cells
(b) Sertoli cells
(c) Leydig cells
(d) Spermatogonia.
Answer:
(b) Sertoli cells

Question 10.
Estrogen is secreted by:
(a) Anterior pituitary
(b) Primary follicle
(c) Graffian follicle
(d) Corpus luteum
Answer:
(c) Graffian follicle

Question 11.
Which one of the following is an IUCD?
(a) Copper – T
(b) Oral pills
(c) Diaphragm
(d) Tubectomy.
Answer:
(a) Copper – T

II. Fill in the Blanks.

Question 1.
The embryo sac in a typical dicot at the time of fertilization is ______.
Answer:
Female Gametophyte.

Question 2.
After fertilization the ovary develops into ______.
Answer:
Fruit.

Question 3.
Planaria reproduces asexually by ______.
Answer:
Regeneration.

Question 4.
Fertilization is ______ in humans.
Answer:
Internal.

Question 5.
The implantation of the embryo occurs at about ______ day of fertilization.
Answer:
6 to 7.

Question 6.
_______ is the first secretion from the mammary gland after childbirth.
Answer:
Colostrum.

Question 7.
Prolactin is a hormone produced by ______.
Answer:
Anterior Pituitary.

III. Match the following.

Question 1.

Column I	Column II
1. Fission	(a) Spirogyra
2. Budding	(b) Amoeba
3. Fragmentation	(c) Yeast

Answer:
1. (b) Amoeba
2. (c) Yeast
3. (a) Spirogyra.

Question 2.
Match the following terms with their respective meanings.

1. Parturition	(a) The duration between pregnancy and birth
2. Gestation	(b) Attachment of zygote to the endometrium
3. Ovulation	(c) Delivery of baby from a uterus
4. Implantation	(d) Release of an egg from Graafian follicle

Answer:
1. (c) Delivery of baby from a uterus
2. (a) Duration between pregnancy and birth
3. (d) Release of the egg from Graafian follicle
4. (b) Attachment of zygote to the endometrium.

IV. State whether the following statements are True or False. Correct the false statement.

Question 1.
The stalk of the ovule is called pedicle.
Answer:
False.
Correct Statement: Stalk of the ovule is called funiculus.

Question 2.
Seeds are the product of asexual reproduction.
Answer:
False.
Correct Statement: Seeds are the product of Sexual reproduction.

Question 3.
Yeast reproduces asexually by means of multiple fission.
Answer:
False.
Correct Statement: Yeast reproduces asexually by means of budding.

Question 4.
The part of the pistil which serves as a receptive structure for the pollen is called as style.
Answer:
False.
Correct Statement: The part of the pistil which serves as a receptive structure for the pollen is called as stigma.

Question 5.
Insect pollinated flowers are characterized by dry and smooth pollen.
Answer:
False.
Correct Statement: Insect pollinated flowers are characterized by larger and spiny pollen.

Question 6.
Sex organs produce gametes, which are diploid.
Answer:
False.
Correct Statement: Sex organs produce gametes, which are haploid.

Question 7.
LH is secreted by the posterior pituitary.
Answer:
False.
Correct Statement: LH is secreted by the anterior pituitary.

Question 8.
Menstrual cycle ceases during pregnancy.
Answer:
True.

Question 9.
Surgical methods of contraception prevent gamete formation.
Answer:
True.

Question 10.
The increased level of estrogen and progesterone is responsible for menstruation.
Answer:
False.
Correct Statement: The decreased level of estrogen and progesterone is responsible for menstruation.

V. Answer in a word or sentence.

Question 1.
If one pollen grain produces two male gametes, how many pollen grains are needed to fertilize 10 ovules?
Answer:
One sperm fuses with the egg and forms a diploid zygote. So 10 pollen grains are needed to fertilize 10 ovules.

Question 2.
In which part of the flower germination of pollen grains takes place?
Answer:
Germination of pollen grains takes place in the stigma of the female flower.

Question 3.
Name two organisms which reproduce through budding.
Answer:
Yeast, Hydra.

Question 4.
Mention the function of endosperm.
Answer:
Endosperm provides food to the developing embryo.

Question 5.
Name the hormone responsible for the vigorous contractions of the uterine muscles.
Answer;
Oxytocin, from the posterior pituitary, is responsible for the vigorous contractions of the uterine muscles.

Question 6.
What is the enzyme present in acrosome of sperm?
Answer:
Acrosome contain hyaluronidase an enzyme that help the sperm to enter the ovum during fertilization.

Question 7.
When is World Menstrual Hygiene Day observed?
Answer:
Every year May 28 is observed as World Menstrual Hygiene Day.

Question 8.
What is the need for contraception?
Answer:
Contraception is one of the best birth control measures to check population growth.

Question 9.
Name the part of the human female reproductive system where the following occurs.

1. Fertilization
2. Implantation

Answer:

1. Oviduct of the female genital tract.
2. Uterus

VI. Short Answer Questions.

Question 1.
What will happen if you cut Planaria into small fragments?
Answer:
Breaking of fragments of Planaria results into many fragments. Each fragment having one cell will give rise to a new Planaria, by cell division.

Question 2.
Why is vegetative propagation practiced for growing some type of plants?
Answer:
No gametic fusion is required in vegetative reproduction. In this type, new plantlets are formed from vegetative cells, buds or organ of plant. The vegetative part of plant get detached from the parent body and grows into an Independent daughter plant.

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

Binary fission	Multiple fission
1. The nucleus divides into two parts.	1. The nucleus divides into many parts.
2. It gives rise to new individuals.	2. It gives rise to many individuals
3. Cytoplasm divides after each nuclear division.	3. Cytoplasm does not divide after every nuclear division.
4. eg. Amoeba.	4. eg. Plasmodium.

Question 4.
Define Triple fusion.
Answer:
The fusion of one male gamete (n) fuses with the secondary nucleus (2n) to produce primary endosperm nucleus (3n) is called Triple fusion.

Question 5.
Write the characteristics of insect-pollinated flowers.
Answer:
Pollination with the help of insects like flies and honey bees are called Entomophily. To attract those insects, these flowers are brightly coloured, have smell and nectar.

Question 6.
Name the secondary sex organs in male.
Answer:
Secondary sex organs in male are seminiferous tubules, epididymis, sperm duct, seminal vesicles, prostrate gland, cowper’s gland and penis.

Question 7.
What is colostrum? How is milk production hormonally regulated?
Answer:
The first fluid which is released from the mammary gland after childbirth is called colostrum. Milk production from alveoli of the mammary gland is stimulated by prolactin secreted from the anterior pituitary. The ejection of milk is stimulated by the posterior pituitary hormone oxytocin.

Question 8.
How can menstrual hygiene be maintained during menstrual days?
Answer:
Menstrual hygiene to be maintained during menstrual days are:

1. Sanitary pads should be changed regularly to avoid infections due to microbes from vagina and sweat from genitals.
2. Use of warm water to clean genitals helps to get rid of menstrual cramps.
3. Wearing loose clothing rather than tight-fitting clothes will ensure the airflow around the genitals and prevent sweating.

Question 9.
How does developing embryo gets its nourishment inside the mother’s body?
Answer:
The placenta allows the exchange of food materials, diffusion of oxygen, excretion of nitrogenous wastes and elimination of carbon-di-oxide. A cord, containing blood vessels that connect the placenta with the foetus is called the umbilical cord.

Question 10.
Identify the parts A, B, C and D.

Answer:
The parts A, B, C and D are:

Question 11.
Write the events involved in the sexual reproduction of a flowering plant.
(a) Discuss the first event and write the types.
(b) Mention the advantages and the disadvantages of that event.
Answer:
(a) The first event is pollination. Pollination is the process of transfer of pollen grains from anther to stigma of a flower. The two types of pollination are self-pollination and cross pollination.

(b) Advantages of self-pollination:

1. Self-pollination is possible in certain bisexual flowers.
2. Flowers do not depend on agents for pollination.
3. There is no wastage of pollen grains.

Disadvantages of self-pollination:

1. The seeds are less in numbers.
2. The endosperm is minute. Therefore, the seeds produce weak plants.
3. New varieties of plants cannot be produced.

Advantages of cross pollination:

1. The seeds produced as a result of cross pollination, develop and germinate properly and grow into better plants, i.e., cross pollination leads to the production of new varieties.
2. More viable seeds are produced.

Disadvantages of cross-pollination:

1. Pollination may fail due to distance barrier.
2. More wastage of pollen grains.
3. It may introduce some unwanted characters.
4. Flowers depend on the external agencies for pollination.

Question 12.
Why are the human testes located outside the abdominal cavity? Name the pouch in which they are present?
Answer:
The testicles, produce sperm and testosterone. The testicles are located outside the body because the sperms develop best at a temperature, several degrees lower than normal body temperature. The pouch, is scrotum, a sac – like structure, in which the testes are present.

Question 13.
Luteal phase of the menstrual cycle is also called the secretory phase. Give reason.
Answer:
In leutal phase LH and FSH decreases, corpus luteum produces progesterone and its level increases followed by a decline progesterone also stimulates the glands in the uterus to secrete substances that maintain the endometrium and keep it from breaking down. For this reason, this phase of menstrual cycle is called secretory phase.

Question 14.
Why are family planning methods not adopted by all the people of our country?
Answer:

• Illiteracy
• Emphasis is in rural areas and not in villages.
• Door to door campaign to encourage families could not be done because of overpopulation.
• Poor economic status and poverty of most of the people in India.
• Age-Old cultural norms continue to cause, poor family planning practices, all across the country.

VII. Long Answer Questions.

Question 1.
With a neat labelled diagram describe the parts of a typical angiosperms ovule.
Answer:
The main part of the ovule is the nucellus which is enclosed by two integuments, leaving an opening called micropyle. The ovule is attached to the ovary wall by a stalk, called funiculus. The basal part is Chalaza.

The embryo sac contains seven cells and the eighth nuclei located within the nucellus. Three cells at the micropylar end form the egg apparatus and the three cells at the chalazalnte9uments end are the antipodal cells. The remaining two nuclei are called polar nuclei found in the centre. In the egg apparatus, one is the egg cell (female gamete) and the remaining two cells are the synergids.

Question 2.
What are the phases of the menstrual cycle? Indicate the changes in the ovary and uterus.
Answer:
The four phases of the menstrual cycle are:

• Menstrual or Destructive phase
• Follicular or Proliferative phase
• Ovulatory phase
• Luteal or secretory phase.

Events of the menstrual cycle and changes in ovary and changes in the uterus.

Phase	Days	Changes in Ovary	Changes in Uterus	Hormonal changes
Menstrual phase	4 – 5 days	Development of primary follicles.	Breakdown of uterine endometrial lining leads to bleeding.	The decrease in progesterone and oestrogen.
Follicular phase	6th – 13th day	Primary follicles grow to become a fully mature Graafian follicle.	Endometrium regenerates through proliferation.	FSH and Oestrogen increase.
Ovulatory phase	14th day	The Graafian follicle ruptures, and releases the ovum (egg).	Increase in endometrial thickness.	LH peak.
Luteal phase	15th – 28th day	Emptied Graafian follicle develops into corpus luteum.	The endometrium is prepared for implantation if fertilization of the egg takes place if fertilization does not occur corpus luteum degenerates, uterine wall ruptures, bleeding starts and unfertilized egg is expelled.	LH and FSH decrease, Corpus luteum produces progesterone and its level increases followed by a decline, if menstrual bleeding occurs.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
In angiosperms the pollen germinates to produce pollen tube that carries two gametes. What is the purpose of carrying two gametes when single gamete can fertilize the egg?
Answer:
In angiosperms, one sperm cell fuses with the egg cell to form the zygote, while the other fuses with the two polar nuclei that form the endosperm which nourishes the developing embryo.

Question 2.
Why the menstrual cycle does not take place before puberty and during pregnancy?
Answer:
The reproducing period of a women’s life starts and becomes functional and an increase in sex hormone production starts only in puberty. So the menstrual cycle does not take place before puberty. The release of a mature egg, maintains the lining of the uterus, during pregnancy. During pregnancy, the placenta produces progesterone. This maintains the lining of the uterus during pregnancy and it means that menstruation does not happen.

Question 3.
Read the following passage and answer the questions that follow.
Rahini and her parents were watching a television programme. An advertisement flashed on the screen which was promoting use of sanitary napkins. Rahini’s parents suddenly changed the channel, but she objected to her parents and explained the need and importance of Such advertisement.

(a) What is first menstruation called? When does it occur?
Answer:
The first menstruation is called Menarche. In human females, the menstrual cycle starts at the age of 11 – 13 years which marks the onset of puberty.

(b) List out the napkin hygiene measures taken during menstruation?
Answer:

1. The sanitary pad and tampons should be wrapped properly and discarded because they can spread infections.
2. Sanitary pad or tampon should not be flushed down the toilet.
3. Napkin incinerators are to be used properly for disposal of used napkins.

(c) Do you think that Rahini’s objection towards her parents was correct? If so, Why?
Answer:
Yes. Awareness to be created in maintaining menstrual hygeine and importance of menstrual hygeine for good health.

Samacheer Kalvi 10th Science Reproduction in Plants and Animals Additional Questions Solved

I. Fill in the blanks.

Question 1.
The reproduction process is to preserve individual species and is called ______.
Answer:
Self – perpetuation.

Question 2.
The _______ grains are produced in the anther within the pollen sac.
Answer:
Pollen.

Question 3.
The _____ contains the future plant or embryo, which develops into a seedling.
Answer:
Seed.

Question 4.
Each testis is covered with a layer of fibrous tissue called ______.
Answer:
Tunica albuginea.

Question 5.
The first fluid which is released from the mammary gland after childbirth is called ______.
Answer:
Colostrum.

Question 6.
The organs of the reproductive system are divided into ______ and ____ (accessory) sex organs.
Answer:
Primary, secondary.

Question 7.
The plasma membrane of an ovum is surrounded by inner thin _____ and an outer thick ______.
Answer:
Zona pellucida, Corona Radiata.

Question 8.
______ is the practice of healthy living and personal cleanliness.
Answer:
Hygiene.

Question 9.
Testosterone initiates the process of ______.
Answer:
Spermatogenesis.

Question 10.
The cortex of ovary is composed of a connective tissue called ______.
Answer:
Stroma.

Question 11.
In plants, the fusion of one sperm with the egg is called ______.
Answer:
Syngamy.

Question 12.
The _______ provides energy for the movement of the tail in sperm, causing sperm motility, which is essential for fertilization.
Answer:
Mitochondria.

Question 13.
Self – pollination is also known as ______.
Answer:
Autogamy

Question 14.
During spore formation, a structure called sporangium develops from the ______.
Answer:
Fungal hyphae.

II. Choose the correct pair from the following.

Question 1.
(a) Chalaza and pollination
(b) Calyx and spores
(c) Anemophily and hydrophily
(d) Stamens and Planaria.
Answer:
(c) Anemophily and hydrophily

Question 2.
(a) Oogenesis and spermatogenesis
(b) Fragmentation and Fertilization
(c) Primary follicle and fission
(d) Micropyle and Regeneration.
Answer:
(a) Oogenesis and spermatogenesis

Question 3.
(a) Scrotum and sporangium
(b) Pollination and population
(c) Spermatogenesis and seminiferous tubules
(d) Diaphragm and Blastocyst.
Answer:
(c) Spermatogenesis and seminiferous tubules

Question 4.
(a) Autogamy and Budding
(b) Funiculus and chalaza
(c) Bulbils and spirogyra
(d) Polar nuclei and pollination.
Answer:
(b) Funiculus and chalaza

Question 5.
(a) Tuberous root and granulosa cells
(b) Corolla and Graafian follicle
(c) Synergids and Regeneration
(d) The vegetative and generative cell.
Answer:
(d) The vegetative and generative cell.

III. Match the following.

Question 1.

1. Offsprings	(a) Pollination by insects
2. Anemophily	(b) Pollination by animals
3. Spermatogenesis	(c) Blood vessels
4. Umbilical cord	(d) Expulsion of the young one
5. Entomophily	(e) Seminiferous tubules
6. Parturition	(f) Sexual reproduction
7. Zoophily	(g) Pollination by wind

Answer:

1. (f) Sexual reproduction
2. (g) Pollination by wind
3. (e) Seminiferous tubules
4. (c) Blood vessels
5. (a) Pollination by insects
6. (d) Expulsion of the young one
7. (b) Pollination by animals.

IV. Choose the correct answer.

Question 1.
The male and the female gametes contain this material on the chromosomes, which transmit the character traits to the next generation.
(a) genes
(b) chromoplast
(c) septum
(d) spores
Answer:
(a) genes

Question 2.
Raphe and hilium in seed represent:
(a) Nucellus
(b) Funiculus
(c) Integument
(d) Micropyle
Answer:
(b) Funiculus

Question 3.
The corona radiata is formed of ______.
(a) Vitelline membrane
(b) Oxytocin
(c) Zygote
(d) Follicle.
Answer:
(d) Follicle.

Question 4.
Pollination is followed by:
(a) Seed formation
(b) Fragmentation
(c) Fertilization
(d) Budding
Answer:
(c) Fertilization

Question 5.
The membrane forming the surface layer of the ovum is called ______.
(a) Entomophily
(b) Exine
(c) Vitelline membrane
(d) Intine.
Answer:
(c) Vitelline membrane

V. Write True or False for the following statements. Write the correct statement for the incorrect statement.

Question 1.
Sexual reproduction involves the fusion of two diploid gametes to form a haploid individual (Zygote).
Answer:
False.
Correct Statement: Sexual reproduction involves the fusion of two haploid gametes, to form a diploid individual (Zygote).

Question 2.
Accessory sex organ in man includes the Gonads. (Testes in males and Ovaries in female).
Answer:
False.
Correct Statement: Primary sex organ in man include the Gonads. (Testes in males and Ovaries in female).

Question 3.
The connective tissue of Cortex called stroma is lined by the germinal epithelium cells called Granulosa cells.
Answer:
True.

Question 4.
The mature ovum or egg in the female is elliptical in shape and has full of yolk.
Answer:
False.
Correct Statement: The mature ovum or egg is spherical in shape and is always free of yolk.

Question 5.
The menstrual cycle ceases around 48-50 years of age and this stage is termed as menopause.
Answer:
True.

VI. Answer the following briefly.

Question 1.
What is the role of acrosome in human sperm?
Answer:
The sperm head has a like structure called acrosome. It contain hyaluronidase an enzyme that helps the sperm to enter the ovum during fertilization.

Question 2.
What are the significance of fertilization and the post-fertilization changes?
Answer:
Significance of fertilization:

• It stimulates the ovary to develop into a fruit.
• It helps in the development of new characters from two different individuals.

Post – fertilization changes:

• The ovule develops into a seed.
• The integuments of the ovule develop into the seed coat.
• The ovary enlarges and develops into a fruit.

Question 3.
What is Asexual reproduction? Explain the spore formation in Rfiizoptis with a diagram.
Answer:
Production df an offspring by a single parent without the formation and fusion of gametes is called Asexual reproduction. Asexual reproduction occurs by spore formation. In Rhizopus, during spore formation, a structure called sporangium develops from the fungal hypha. The nucleus divides several times within the sporangium and each nucleus with a small amount of cytoplasm develops into a spore. The spores are liberated and they develop into new hypha after reaching the ground or substratum.

Question 4.
What could be the reasons for adopting contraceptive methods?
Answer:
Contraception is one of the best birth control measures. A number of techniques or method have been developed to prevent pregnancies in women which leads to welfare of the family group and society.

Question 5.
Explain the following:
(a) Toilet Hygiene
(b) Napkin Hygiene
(c) Menstrual Hygiene
Answer:
(a) Toilet Hygiene:

• To reduce the bad odour and infection, the floors of the toilet should be kept clean and dry.
• Toilet flush handles, doorknobs, light switches and walls should be cleaned with disinfectants to kill harmful germs and bacteria.
• Hands should be washed with soap, before and after toilet use.

(b) Napkin Hygiene:

• To prevent infections, the sanitary pad, and tampon should be wrapped and discarded properly.
• Sanitary pad and tampon should not be flushed down the toilet.
• Napkin incinerators are to be used properly for disposal of used napkins.

(c) Menstrual Hygiene:

• Sanitary pads should be changed regularly, to avoid infections due to microbes from vagina and sweat from genitals.
• Use of warm water to clean genitals helps to get rid of menstrual cramps.
• Wear loose clothing, to have airflow around the genitals and prevent sweating.

Question 6.
Explain the structure of the following:
(a) Testes
(b) Ovary
Answer:
(a) Testes: Testes are the reproductive glands of the male, that are Oval shaped, which lie outside the abdominal cavity of a man, in a sac-like structure called Scrotum. Each testis is covered with a layer of fibrous tissue called tunica albuginea. Septa divides the testes into pyramidal lobules, in which lie seminiferous tubules, cells of Sertoli and the Leydig cells. The process of spermatogenesis takes place in the seminiferous tubules. The Sertoli cells are the supporting cells and provide nutrients to the developing sperms. The Leydig cells lie between the seminiferous tubules and secrete testosterone. It initiates the process of spermatogenesis.

(b) Ovary: Ovaries are located on either side of the lower abdomen, composed of two almond-shaped bodies, each lying near the lateral end of the fallopian tube. Each ovary has an outer cortex and an inner medulla. The cortex is composed of a network of connective tissue called stroma and is lined by the germinal epithelium. The epithelial cells called the granulosa cells to surround each ovum in the ovary together forming the primary follicle. As the egg grows larger, the follicle also enlarges and gets filled with the fluid and is called the Graafian follicle.

Question 7.
What do you know about the National Health Programme?
Answer:
To improve the reproductive health of the people, National Health Programmes such as,
(a) Family Welfare Programme:

• Maternal and Child Health Care (MCH).
• Immunization of mothers, infants and children.
• Nutritional supplement to pregnant women and children.
• Motivate couples to accept contraceptive methods and to have small family norms, which improve economic status, living status and quality of life.

(b) Reproductive and Child Health Care (RCH)

• Pregnancy and childbirth.
• Postnatal care of the mother and child.
• Importance of breastfeeding.
• Prevention of reproductive tract infections and sexually transmitted diseases.

Question 8.
What is gastrulation?
Answer:
The transformation of blastula into gastrula and the formation of primary germ layers (Ectoderm, Mesoderm and Endoderm) by rearrangement of the cells is called gastrulation.

Question 9.
What is implantation?
Answer:
The process of attachment of the blastocyst to the uterine wall (endometrium) is called implantation. The blastocyst (fertilized egg) reaches the uterus and gets implanted in the uterus. The fertilized egg becomes implanted in about 6 to 7 days after fertilization.

VII. Draw a labelled diagram of the following.

Question 1.
Cross Section of:
(a) Human testes
(b) Human ovary
Answer:
(a) Human testes:

(b) Human ovary:

VIII. Write the expansion for the following abbreviations.

Question 1.

1. WHO
2. RCH
3. MCH
4. STD
5. IUD
6. UTI
7. LH
8. FSH.

Answer:

1. WHO – World Health Organization
2. RCH – Reproductive and Child Health Care
3. MCH – Maternal and Child Health Care
4. STD – Sexually Transmitted Diseases
5. IUD – Intra-Uterine Device
6. UTI – Urinary Tract Infection
7. LH – Luteinizing Hormone
8. FSH – Follicle Stimulating Hormones.

IX. Answer the following in detail. Draw diagrams wherever necessary.

Question 1.
With the help of a neat labelled diagram, explain, how does fertilization take place in flowering plants.
Answer:
The mature pollen grain contains two cells, the vegetative and the generative cell. The vegetative cell contains a large nucleus. The generative cell divides mitotically and forms two male gametes.

Pollen grains reach the stigma and begin to germinate. Pollen grain forms a small tube-like structure called pollen tube, which emerges through the germ pore. The contents of the pollen grain move into the tube. The pollen tube grows through the tissues of the stigma and style and finally reaches the ovule through the micropyle.

The vegetative cell degenerates and the generative cell divides to form two sperms or male gametes. Tip of pollen tube bursts and the two sperms enter the embryo sac. One sperm fuses with the egg and this process is called Syngamy and forms a diploid zygote. The other sperm fuses with the secondary nucleus, which is called the Triple fusion, to form the primary endosperm nucleus, which is triploid in nature.

Since two types of fusion syngamy and triple fusion take place in an embryo sac, the process is termed as double fertilization. After triple fusion, primary endosperm nucleus develops into an endosperm, which provides food to the developing embryo. Later the synergids and antipodal cells degenerate.

Question 2.
With examples and with the help of a neat labelled diagram, explain the different types of vegetative reproduction of plants.
Answer:
Vegetative reproduction takes place through,
(i) Leaves:

In Bryophyllum, small plants grow at the leaf notches.

(ii) Stem:
In plants like, strawberry aerial weak stems, touch the ground and give adventitious roots and buds. The offspring becomes an independent plant, when it is detached from the parent plant.

(iii) Root: Tuberous roots are used to develop new plants, eg. Asparagus, Sweet potato.

(iv) Bulbils: In some plants, the flower bud, modifies into the globose bulb, which is called bulbils. When these bulbils fall on the ground they grow into new plants, eg. Agave.

(v) Fragmentation: The breaking of the filament into many fragments is called fragmentation. Each fragment having one cell will give rise to a new filament by cell division, eg. Spirogyra.

(vi) Fission:

The parent cell divides into two daughter cells and each cell develops into a new adult organism eg. Amoeba.

(vii) Budding:

Formation of a daughter individual from a small projection, the bud arising on the parent body is called budding, eg. Yeast.

(viii) Regeneration: The ability of the lost body parts of an individual organism to give rise to a whole new organism is called regeneration. Regeneration takes place by the specialized mass of cells, eg. Hydra and Planaria.

Question 3.
Explain the two types of pollination and write the advantages and disadvantages of the types of pollination.
Answer:
The two types of pollination are
(a) Self – pollination: The transfer of pollen grains from the anther to the stigma of the same flower or another flower borne on the same plant is called self – pollination.
eg. Hibiscus. Self – pollination is also called Autogamy.
Advantages of self – pollination:

• Self – pollination is possible in certain bisexual flowers.
• Flowers do not depend on agents for pollination.
• There is no wastage of pollen grains.

Disadvantages of self – pollination:

• The seeds are less in numbers.
• The endosperm is minute, so the seeds produce weak plants.
• New varieties of plants cannot be produced.

(b) Cross-pollination: The transfer of pollen grains from the anthers of a flower to the stigma of a flower on another plant of the same species is called Cross-pollination, eg. Apples, Grapes, Plum.
Advantages of Cross-pollination:

• The seeds produced as a result of cross-pollination, develop and germinate properly and grow into better plants.
• Cross-pollination leads to the production of new varieties.
• More viable seeds are produced.

Disadvantages of Cross-pollination:

• Pollination may fail due to the distance barrier.
• More wastage of pollen grains.
• It may introduce some unwanted characters.
• Flowers depend on the external agencies for pollination.

Question 4.
Explain the structure of Human Sperm and Ovum with a neat labelled diagram.
Answer:
1. Structure of Human sperm:
The spermatozoa consist of a head, middle piece and tail. The sperm head is elongated and formed by the condensation of the nucleus. The anterior portion has a cap-like structure called acrosome, which contains an enzyme hyaluronidase, that helps the sperm to enter the ovum during fertilization. A short neck connects the head and middle piece, which comprises the centrioles. The middle piece contains the mitochondria which provides energy for the movement of tail. It brings about sperm motility, which is essential for fertilization.

2. Structure of Ovum:
The mature ovum or egg is spherical in shape, free of yolk and contains abundant cytoplasm and the nucleus. The ovum is surrounded by three membranes. The plasma membrane is surrounded by inner thin zona pellucida and an outer thick corona radiata, which is formed of follicle cells. The membrane forming the surface layer of the ovum is called the vitelline membrane. The fluid-filled space between zona pellucida and the surface of the egg is called perivitelline space.

Question 5.
Explain the Agents of cross-pollination.
Answer:
Cross – pollination takes place through the agency of animals, insects, wind and water.
(a) Pollination by the wind: The pollination with the help of wind is called Anemophily. The Anemophilous flowers produce an enormous amount of pollen grains, which are small, smooth, dry and light in weight. This kind of pollen can blow off at a distance of more than 1,000 km. The stigmas are large, protruding and hairy to trap the pollen grains, eg. Grasses and some Cacti.

(b) Pollination by insects: Pollination with the help of insects like honey bees, flies are called entomophily. These flowers are brightly coloured, have the smell and have nectar, to attract insects. The pollen grains are larger in size, and the exine is pitted and spiny, so it can easily adhere firmly on the sticky stigma. About 80% of the pollination by insects is carried by honey bees.

(c) Pollination by water: The pollination with the help of water is called hydrophily. This takes place in aquatic plants. Pollen grains are produced in large numbers. Pollen grains float on the surface of water till they land on the stigma of female flowers, eg. Hydrilla, Vallisneria.

(d) Pollination by Animals: When pollination takes place with the help of animals, it is called Zoophily. The flowers have bright colour, size and scent to attract animals, eg. Sunbird pollinates flowers of Canna, Gladioli etc., Squirrels pollinate flowers of silk cotton tree.

Question 6.
Explain the parts of a flower, with a neat labelled diagram.
Answer:
The flower is the reproductive organ of a flowering plant. A flower is a modified shoot. A flower consists of four whorls borne on the thalamus. The whorls are from outside. The parts of a typical flower are as follows:

(a) Calyx: Calyx consists of sepals, forms the outermost whorl. They are non-essential or accessory whorls. It protects the flower bud.

(b) Corolla: Corolla consists of petals, which are modified leaves. They are often brightly coloured and in different shapes to attract pollinators. They are non-essential or accessory whorls because they do not directly take part in the reproduction.

(c) Androecium:

Androecium is the male reproductive part of a flower and is called an essential whorl, as they take part directly in reproduction. The androecium is composed of stamens. Each stamen consists of a stalk called the filament and a small bag like structure called anther at the tip. The pollen grains are produced in the anther within the pollen sac.

Pollen grains are spherical and have a two-layered wall. The hard outer layer is exine, which has apertures, called Germpore. The inner thin layer is intine, which is continuous and made up of cellulose and pectin. Mature pollen grain contain two cells, the vegetative and generative cell. Vegetative cell contains a large nucleus. The generative cell divides, to form two male
gametes.

(d) Gynoecium:
Gynoecium is the female reproductive part of a flower. It is an essential whorl because they take part directly in reproduction. Gynoecium is made up of carpels. It has (a) Ovary (b) Style (c) Stigma. Ovary contains the ovules. The main part of the ovule is the nucellus and is enclosed by two integuments, Jk. leaving an opening called as micropyle. The ovule is attached to the ovary wall, by a stalk called Funiculus.

Question 7.
Explain the steps involved from fertilization to foetal development in human.
Answer:
Fertilization in human is internal and occurs in the Oviduct of the female genital tract. It takes place in the ampulla of the fallopian tube. An oocyte is alive for about 24 hours after it is released from follicle. The sperm enters into the ovum, fuses and forms a zygote, which is a fertilized ovum. This process is called fertilization.

(a) Cleavage and formation of Blastula: The first cleavage takes place about 30 hours after fertilization. Cleavage is a series of rapid mitotic divisions of the zygote, to form many-celled blastula (Blastocyst), which comprises an outer layer of smaller cells and inner mass of larger cells.

(b) Implantation: The blastocyst (fertilized egg) reaches the uterus and gets implanted. The process of attachment of the blastocyst to the uterine wall is called Implantation. The fertilized egg, implanted in about 6 to 7 days after fertilization.

(c) Gastrulation: After implantation, the transformation of blastula into Gastrula and the formation of primary germ layers (Ectoderm, Mesoderm and Endoderm) by rearrangement of the cells takes place, which is called Gastrulation.

(d) Organogenesis: The various organs of the foetus are established from the different germ layers, during Organogenesis.

(e) Formation of Placenta: Placenta is a disc-shaped structure, attached to the Uterine wall between the developing embryo and maternal tissues. It allows the exchange of food materials, diffusion of oxygen, excretion of nitrogenous wastes and elimination of carbon-di-oxide. The cord which connects the placenta with the foetus is called the umbilical cord.

(f) Pregnancy (Gestation): Embryo attains its development in the uterus. Gestation period of human is 280 days. During pregnancy the uterus expands up to 500 times of its normal size.

(g) Parturition (Child Birth): Oxytocin from the posterior pituitary stimulates the uterine contractions and provides force to expel the baby from the uterus, causing birth. Parturition is the expulsion of young one from the mother’s Uterus at the end of gestation.

(h) Lactation: The process of milk production after childbirth from the mammary glands of the mother is called lactation. Milk production from the mammary gland is stimulated by prolactin, a hormone secreted by anterior pituitary. The ejection of milk is stimulated by posterior pituitary hormone oxytocin.

Question 8.
What is population explosion? Explain the different ways of family planning.
Answer:
The sudden and rapid rise in the size of the human population is called the Population Explosion. Contraception: Contraception is one of the best birth control measures. The devices used for contraception are called contraceptive devices. The common contraceptive methods used to prevent pregnancy are as follows:

(a) Barrier methods: This method prevents the meeting of an ovum by the sperms. The entry of sperm is prevented into the female reproductive tract by a barrier.

• Condom: Condom is made of thin rubber or latex sheath. Condom prevents deposition of sperms in the vagina. A condom protects against Sexually Transmitted Diseases (STD) like Syphilis and AIDS.
• Diaphragm (Cervical cap): Vaginal diaphragm fitting into the vagina or a cervical cap fitting over the cervix. This prevents the entry of sperms into the uterus.

(b) Hormonal Methods: Hormonal preparations are in the form of pills or tablets (contraceptive pills). These hormones stop the release of an egg from the ovary.

(c) Intra – Uterine Devices (IUDs): The intrauterine device (IUD) are contraceptive devices, inserted into the uterus. Lippe’s Loop and Copper-T, made of copper and plastic are two synthetic devices, commonly used in India. This can remain for a period of 3 years. This reduces the sperm fertilizing capacity and prevents implantation.

(d) Surgical methods: Surgical contraception or sterilization techniques are terminal methods to prevent any pregnancy. This procedure in males is called as vasectomy (ligation of vas deferens) and in females, it is called tubectomy (ligation of fallopian tube). These are methods of permanent birth control.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.5

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The value of \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{CD}}\) is ………………

Solution:
(c) \(\overrightarrow{0}\)

Question 2.
If \(\vec{a}+2 \vec{b}\) and \(3 \vec{a}+m \vec{b}\) are parallel, then the value of m is ………………
(a) 3
(b) \(\frac{1}{3}\)
(c) 6
(d) \(\frac{1}{6}\)
Solution:
(c) 6

Question 3.
The unit vector parallel to the resultant of the vectors \(\hat{i}+\hat{j}-\hat{k}\) and \(\hat{i}-2 \hat{j}+\hat{k}\) is ………………

Solution:

Question 4.
A vector \(\overrightarrow{O P}\) makes 60° and 45° with the positive direction of the x and y axes respectively. Then the angle between \(\overrightarrow{O P}\) and the z-axis is …………….
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Solution:
(b) 60°
α = 60°, β = 45°
We know cos²α + cos²β + cos²γ = 1
(i.e.,) \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}\) + cos²γ = 1
cos²γ = 1 – \(\frac{1}{4}-\frac{1}{2}=\frac{1}{4}\)
cos γ = \(\frac{1}{2}\) ⇒ y = π/3 = 60°

Question 5.
If \(\overrightarrow{B A}=3 \hat{i}+2 \hat{j}+\hat{k}\) and the position vector of B is \(\hat{i}+3 \hat{j}-\hat{k}\), then the position vector A is …………………

Solution:

Question 6.
A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to …………..

Solution:

Question 7.
The vectors \(\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}\) are ……………
(a) parallel to each other
(b) unit vectors
(c) mutually perpendicular vectors
(d) coplanar vectors
Solution:
(d) coplanar vectors

Question 8.
If ABCD is a parallelogram, then \(\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}\) is equal to ……………

Solution:

Question 9.
One of the diagonals of parallelogram ABCD with \(\vec{a}\) and \(\vec{b}\) as adjacent sides is \(\vec{a}+\vec{b}\). The other diagonal \(\overrightarrow{\mathrm{BD}}\) is ……………

Solution:

Question 10.
If \(\vec{a}\), \(\vec{b}\) are the position vectors A and B, then which one of the following points whose position vector lies on AB, is ………….

Solution:
(c) \(\frac{2 \vec{a}+\vec{b}}{3}\)

Question 11.
If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of three collinear points, then which of the following is true?

Solution:
(b) \(2 \vec{a}=\vec{b}+\vec{c}\)

Question 12.
If \(\vec{r}=\frac{9 \vec{a}+7 \vec{b}}{16}\), then the point P whose position vector \(\vec{r}\) divides the line joining the points with position vectors \(\vec{a}\) and \(\vec{b}\) in the ratio ………………
(a) 7 : 9 internally
(b) 9 : 7 internally
(c) 9 : 7 externally
(d) 7 : 9 externally
Solution:

Question 13.
If \(\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}\) is a unit vector, then the value of λ is ……………..
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{2}\)
Solution:

Question 14.
Two vertices of a triangle have position vectors \(3 \hat{i}+4 \hat{j}-4 \hat{k}\) and \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) .If the position vector of the centroid is \(\hat{i}+2 \hat{j}+3 \hat{k}\), then the position vector of the third vertex is ………………….

Solution:

Question 15.

(a) 42
(b) 12
(c) 22
(d) 32
Solution:
(c) 22

Question 16.
If \(\vec{a}\) and \(\vec{b}\) having same magnitude and angle between them is 60° and their scalar product is \(\frac{1}{2}\) then \(|\vec{a}|\) is ……………
(a) 2
(b) 3
(c) 7
(d) 1
Solution:
(d) 1

Question 17.
The value of θ ∈ (0, \(\frac{\pi}{2}\)) for which the vectors are perpendicular, is equal to …………………

Solution:

Question 18.

(a) 15
(b) 35
(c) 45
(d) 25
Solution:
(d) 25

Question 19.
Vectors \(\vec{a}\) and \(\vec{b}\) are inclined at an angle θ = 120°
is equal to ………….
(a) 225
(b) 275
(c) 325
(d) 300
Solution:
(d) 300

Question 20.
If \(\vec{a}\) and \(\vec{b}\) are two vectors of magnitude 2 and inclined at an angle 60°, then the angle between \(\vec{a}\) and \(\vec{a}+\vec{b}\) is ………………
(a) 30°
(b)60°
(c) 45 °
(d) 90°
Solution:
(a) 30°

Question 21.
If the projection of \(5\hat{i} -\hat{j}-3 \hat{k}\) on the vector \(\hat{i}+3 \hat{j}+\lambda \hat{k}\) is same as the projection of \(\hat{i}+3 \hat{j}+\lambda \hat{k}\) on \(5\hat{i}- \hat{j}-3 \hat{k}\) then λ is equal to ………………
(a) ±4
(b) ±3
(c) ±5
(d) ±1
Solution:
(c) ±5

Question 22.
If (1, 2, 4) and (2, -3λ, -3) are the initial and terminal points of the vector \(\hat{i}+5 \hat{j}-7 \hat{k}\), then the value of λ is equal to ……………..
(a) \(\frac{7}{3}\)
(b) \(-\frac{7}{3}\)
(c) \(-\frac{5}{3}\)
(d) \(\frac{5}{3}\)
Solution:

Question 23.
If the points whose position vector \(10 \hat{i}+3 \hat{j}, 12 \hat{i}-5 \hat{j}\) and \(\vec{a} \hat{i}+11 \hat{j}\) are collinear then a is equal to ………………
(a) 6
(b) 2
(c) 5
(d) 8
Solution:
(d) 8

equating \(\hat{j}\) components
⇒ -8 = 8t ⇒ t = -1
equation \(\hat{i}\) components
t(a – 10) = 2
(i.e.,) (-1) (a – 10) = 2
a – 10 = -2
a = – 2 + 10 = -8

Question 24.
If then x is equal to …………..
(a) 5
(b) 7
(c) 26
(d) 10
Solution:
(c) 26

Question 25.
If \(\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k},|\vec{b}|=5\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\), then the area of the triangle formed by these two vectors as two sides, is …………….
(a) \(\frac{7}{4}\)
(b) \(\frac{15}{4}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{17}{4}\)
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3

Question 1.
Find \(\vec{a} \cdot \vec{b}\) when

Solution:

Question 2.
Find the value of λ for which the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular, where

Solution:
(i) \(\vec{a}\) = 2î + λĵ – k̂ and \(\vec{b}\) = î – 2ĵ + 3k̂
Given \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors
∴ \(\vec{a}\) . \(\vec{b}\) = 0
(2î + λĵ – k̂ ) . (î – 2ĵ + 3k̂) = 0
(2) (1) + (λ) (- 2) + (1) (3) = 0
2 – 2λ + 3 = 0
2λ = 5
λ = \(\frac{5}{2}\)

(ii) \(\vec{a}\) = 2î + 4ĵ – k̂ and \(\vec{b}\) = 3î – 2ĵ + λk̂
Given \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors
∴ \(\vec{a}\) . \(\vec{b}\) = 0
(2î + 4ĵ – k̂) . (3î – 2ĵ + λk̂) = 0
(2) (3) + (4) (-2) + (-1) (λ) = 0
6 – 8 – λ = 0
λ = – 2

Question 3.
If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 15 and \(\vec{a} \cdot \vec{b}\) = 75\(\sqrt{2}\) , find the angle between \(\vec{a}\) and \(\vec{a}\) .
Solution:

Question 4.
Find the angle between the vectors

Solution:

Question 5.
If \(\overrightarrow{\boldsymbol{a}}, \overrightarrow{\boldsymbol{b}}, \overrightarrow{\boldsymbol{c}}\) are three vectors such that \(\vec{a}+2 \vec{b}+\vec{c}=\overrightarrow{0}\) and \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=7\) find the angle between \(\vec{a}\) and \(\vec{b}\)
Solution:

Question 6.
Show that the vectors are mutually orthogonal.
Solution:

Question 7.
Show that the vectors form a right-angled triangle.
Solution:

So, the given vectors form the sides of a right-angled triangle

Question 8.

Solution:

Question 9.
Show that the points (2, -1, 3) (4, 3, 1) and (3, 1, 2) are collinear
Solution:
Let the given points be A, B, C

Question 10.
If \(\vec{a}, \vec{b}\) are unit vectors and θ is the angle between them, show that

Solution:

Question 11.
Let \(\vec{a}, \vec{b}, \vec{c}\) be the three vectors such that \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5\) and each one of them being perpendicular to the sum of the other two, find \(|\vec{a}+\vec{b}+\vec{c}|\)
Solution:

Question 12.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(2 \hat{i}+6 \hat{j}+3 \hat{k}\)
Solution:

Question 13.
Find λ, when the projection of is 5 units.
Solution:

Question 14.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.3 Additional Problems

Question 1.
Find λ so that the vectors are perpendicular to each other.
Solution:

Question 2.

Solution:

Question 3.
If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is \(\sqrt{3}\)
Solution:

Question 4.
Show that the vectors form a right-angled triangle.
Solution:

⇒ The given vectors form the sides of a right-angled triangle.

Question 5.
Find the projection of

Solution:

Question 6.
Show that the vector \(\hat{i}+\hat{j}+\hat{k}\) is equally inclined with the coordinate axes.
Solution:

Question 7.
If \(\vec{a}, \vec{b}, \vec{c}\) are three mutually perpendicular unit vectors, then prove that \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}\)
Solution:

Question 8.
Show that the points whose positions vectors from a right-angled triangle.
Solution:

⇒ The given points form a right-angled triangle.

Question 9.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequences whose n^th terms are given below and classify them as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic progression and none of them.

Solution:


It is not a G.P. or A.P. or H.P. or A.G.P.

It is not an A.P. or G.P. or H.P. or A.G.P

It is not an A.P. or G.P. or H.P. or A.G.P.


It is a A.G.P.

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below.

Solution:
a₁ = 1 + 1 = 2 ; a₂ = 2
a₃ = 3 + 1 = 4 ; a₄ = 4
a₅ = 5 + 1 = 6 ; a₆ = 6
So, the first 6 terms are 2, 2, 4, 4, 6, 6

Solution:
a₁ = 1 ; a₂ = 2, a₃ = 3
a₄ = a₃ + a₂ + a₁ = 3 + 2 + 1 = 6 ⇒ a₄ = 6
a₅ = a₄ + a₃ + a₂ = 6 + 3 + 2 = 11 ⇒ a₅ = 11
a₆ = a₅ + a₄ + a₃ = 11 + 6 + 3 = 20 ⇒ a₆ = 20
So the first 6 terms are 1, 2, 3, 5, 8, 13.

Solution:

Question 3.
Write the n_th term of the following sequences.
Solution:
(i) 2, 2, 4, 4, 6, 6……
Solution:

Solution:
Nr: 1, 2, 3, ……t_n = n
Dr: 2, 3, 4, …..t_n = n + 1


Solution:
Nr: 1, 3, 5, 7, . . .which is an A.P. a = 1, d = 3 – 1 = 2
t_n = a + (n – 1)d
t_n = 1 + (n – 1)2 = 1 + 2n – 2 = 2n – 1.
Dr : 2, 4, 6, 8, . . .
So the n^th term is 2 + (n – 1)2 = 2 + 2n – 2 = 2n.

(iv) 6, 10, 4, 12, 2, 14, 0, 16, -2,….
Solution:
t₁ = 6 ; t₂ = 10
t₃ = 4 ; t₄ = 12
t₅ = 2 ; t₆ = 14
t₇ = 0 ; t₈ = 16
When n is odd, the sequence is 6, 4, 2, 0,…
(i.e.) a = 6 and d = 4 – 6 = -2.
So, t_n = 6 + (n – 1)(-2) = 6 – 2n + 2 = 8 – 2n
When n is even, the sequence is 10, 12, 14, 16,…
Here a = 10 and d = 12 – 10 = 2
t_n = 10 + (n – 1)2 = 10 + 2n – 2 = 2n + 8 (i.e.) 8 + 2n

Question 4.
The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an AP. Find the numbers in GP.
Solution:
The 3 numbers in a G.P. is taken as \(\frac{a}{r}\), a, ar
Their product is 5832.

6r² + 6 = 13
6r² – 13r + 6 = 0
(3r – 2)(2r – 3) = 0
r = 2/3 or 3/2

Question 5.

Solution:

Question 6.
If t_k is the k^th term of a G.P., then show that t_{n – k}, t_n, t_{n + k} also form a GP for any positive integer k.
Solution:
Let a be the first term and r be the common ratio.
We are given t_k = ar^{k – 1}
We have to Prove : t_{n – k}, t_n, t_{n + k} form a G.P.

Question 7.
If a, b, c are in geometric progression, and if \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\), then prove that x, y, z are in arithmetic progression.
Solution:
Given a, b, c are in G.P.
⇒ b² = ac
⇒ log b² = log ac
(i.e.) 2log b = log a + log c …(1)

Substituting these values in equation (1) we get 2y = x + z ⇒ x, y z are in A.P.

Question 8.
The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers.
Solution:
Let the two numbers be a and b.

So, (a + b – 20)² = 4ab
(i.e.) (a + b)² + 400 – 40(a + b) = 4ab
(a + b)² – 4ab = 40(a + b) – 400
from(3) (a + b)² – 4ab = 32(a + b)
⇒ 32(a + b) = 40(a + b) – 400
(÷ by 8) 4(a + b) = 5(a + b) – 50
4a + 4b = 5a + 5b – 50
a + b = 50
a = 50 – b
Substituting a = 50 – b in (3) we get
(50 – b – b)² = 32(50)
(50 – 2b)² = 32 × 50

When b = 5, a = 50 – 5 = 45
When b = 45, a = 50 – 45 = 5
So the two numbers are 5 and 45.

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + p – q = 0 are equal, then show that p, q and r are in AP.
Solution:
The given quadratic equation is (q – r)x² + (r – p)x + (p – q) = 0
Given that the roots of the equation are equal. ∴ The discriminant is equal to zero.
(r – p)² – 4(q – r) (p – q) = 0
r² – 2rp + p² – 4 (pq – q² – rp + qr) = 0
r² – 2rp + p² – 4pq + 4q² + 4rp – 4qr = 0
r² + p² + 4q² + rp – 4pq – 4rq = 0
r² + p² + (-2q)² + 2r.p + 2p(-q) + 2(-2q)r = 0
(r + p – 2q)² = 0
r + p – 2q = 0 ⇒ 2q = p + r
∴ p, q , r are in A. P.

Question 10.
If a, b, c are respectively the p^th, q^th and r^th terms of a G.P., show that (q – r) log a + (r – p) log b + (p – q) log c = 0.
Solution:
Let the G.P. be l, lk, lk²,…
We are given t_p = a, t_q = b, t_r = c

LHS = (q – r) log a + (r – p) log b + (p – q) log c

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Additional Questions Solved

Question 1.

Solution:
Here a₁ = 1
Substituting n = 2, we obtain a₂ = a₁ + 2 = 1 + 2 = 3
Substituting n = 3, 4 and 5, we obtain respectively
a₃ = a₂ + 2 = 3 + 2 = 5, a₄ = a₃ + 2 = 5 + 2 = 7
a₅ = a₄ + 2 = 7 + 2 = 9
Thus, the first five terms are 1, 3, 5, 7 and 9.

Question 2.
Find the 18^th and 25^th term of the sequence defined by

Solution:
When n = 18 (even)
a_n = n(n + 2) = 18(18 + 2) = 18(20) = 360
When n = 25(odd)

Question 3.
Write the first six terms of the sequences given by
(i) a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
(ii) a₁ = 4, a_{n + 1} = 2na_n
Solution:
(i) Here a₁ = a₂= 1 = a_{n – 1} + a_{n – 2} (n ≥ 3)
Putting n = 3, a₃= a₂ + a₁ = 1 + 1 = 2
Putting n = 4, a₄ = a₃ + a₂ = 2 + 1 = 3
Putting n = 5, a₅ = a₄ + a₂ = 3 + 2 = 5
Putting n = 6, a₆ = a₅ + a₄ = 5 + 3 = 8
∴ First six terms of the sequence are 1, 1, 2, 3, 5, 8

(ii) Here a₁ = 4 and a_{n + 1} = 2na_n
Putting n = 1, a₂ = 2 × 1 × a₁ = 2 × 1 × 4 = 8
Putting n = 2, a₃ = 2 × 2 × a₂ = 4 × 8 = 32
Putting n = 3, a₄ = 8 × 192 = 1536
Putting n = 4, a₅ = 2 × 4 × a₄ = 8 × 192 = 1536
Putting n = 5, a₆ = 2 × 5 × a₅ = 10 × 1536 = 15360

Question 4.
An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the series.
Solution:
Let a₁ be the first term and d, be a common difference.
Here n = 21.
∴ The three middle terms are a₁₀,a₁₁, a₁₂
Now, a₁₀ + a₁₁ + a₁₂ = 129 [Given]
∴ (a₁ + 9d) + (a₁ + 10d) + (a₁ + 11d) = 129
⇒ 3a₁ + 30d = 129 ⇒ a₁ + 10d = 43 ……(i)
The last three terms are a₁₉, a₂₀, a₂₁
a₁₉, a₂₀, a₂₁ = 237 [Given]
∴ (a₁ + 18d)+(a₁ + 19d) + (a₁ + 20d) = 237
(i.e.,) 3a₁ + 57d = 237 ⇒ a₁ + 19d = 79 … (2)
Subtracting (i) from (ii), we get 9d = 36, ⇒ d = 4
∴ From (i), a₁ + 40 = 43 ⇒ a₁ = 3
Hence, the series is 3, 7, 11, 15 …….

Question 5.
Prove that the product of the 2^nd and 3^rd terms of an arithmetic progression exceeds the product of the first and fourth by twice the square of the difference between the 1^st and 2^nd.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of A.P.
Then, a₁ = a, a₂ = a + (2 – 1)d = a + d
a₃ = a + (3 – 1)d = a + 2d, a₄ = a + (4 – 1)d = a + 3d
We have to show that a₂.a₃ – a₁.a₄ = 2(a₂ – a₁)²
LHS = a₂.a₃ – a₁.a₄ = (a + d)(a + 2d) – a(a + 3d)
= a₂ + 3ad + 2d₂ – a₂ – 3ad = 2d₂
RHS = 2(a₂ – a₁)² = 2d₂
Since LHS = RHS.
Hence proved.

Question 6.
If the p^th, q^th and r^th terms of an A.P. are a, b, c respectively, prove that a(q – r) + b (r – p) + c(p – q) = 0.
Solution:
Let A be the first term and D be the common difference of A.P.
a_p = a, ∴ A + (p – 1)D = a ….. (1)
a_q = b, ∴ A + (q – 1)D = b ……. (2)
a_r = c, ∴ A + (r – 1)D = c …….. (3)
∴ a (q – r) + b (r – p) + c (p – q) = [A + (p – l) D] (q – r) + [A + (q – 1) D]
(r – p) + [A + (r – 1) D] (p – q) [Using (1), (2) and (3)]
= (q – r + r – p + p – q)A + [(p – l)(q – r) + (q – l)(r – p) + (r – l)(p – q)]D
= (0) A + (pq – pr – q + r + qr – pq – r + p + pr – p – qr + q)D
= (0)A + (0)D = 0.

Question 7.
If a, b, c are in A.P. and p is the A.M. between a and b and q is the A.M. between b and c, show that b is the A.M. between p and q.
Solution:
a, b, c are in A.P.
2b = a + c …… (1)
p is the A.M. between a and b

q is the A.M. between b and c


Hence, b is the A.M. between p and q

Question 8.
If x, y, z be respectively the p^th, q^th and r^th terms of a G.P. show that x^{q – r}, y^{r – p}, z^{p – q} = 1
Solution:
Let A be the first term and R be the common ratio of G.P.
a_p = x ⇒ x = AR^{p – 1} ……… (1)
a_q = y ⇒ x = AR^{q – 1} ……… (2)
a_r = z ⇒ x = AR^{r – 1} ……… (3)
Raising (1), (2), (3) to the powers q – r, r – p, p – q respectively, we get

Multiplying (4), (5), and (6), we get

Hence, x^{q – r}, y^{r – p}, z^{p – q} = 1

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Find the derivatives of the following functions with respect to corresponding independent variables.
Question 1.
f(x) = x – 3 sinx
Solution:
f(x) = x – 3 sin x
f'(x) = 1 – 3 cos x

Question 2.
y = sin x + cos x
Solution:
\(\frac{d y}{d x}\) = cosx + (-sinx) = cos x – sin x

Question 3.
f(x) = x sin x
Solution:
f(x) = x sin x
f'(x) = x cos x + sin x – 1
f'(x) = x cos x + sin x

Question 4.
y = cos x – 2 tan x
Solution:
\(\frac{d y}{d x}\) = -sin x = 2 (sec²x)
= – sin x – 2 sec²x

Question 5.
g(t) = t³ cos t
Solution:
g(t) = t³ cos t
g'(t) = t³ × – sin t + cos t × 3t²
g'(t) = – t³ sin t + 3t² cos t
= 3t² cos t – t³ sin t

Question 6.
g(t) = 4 sec t + tan t
Solution:
g{t) = 4 sect + tan t
g'(t) = 4(sec t tan t) + sec²t
= 4sec t tan t + sec²t

Question 7.
y = e^x sin x
Solution:
y = e^x sin x
\(\frac{d y}{d x}\) = e^x cos x + sin x e^x
\(\frac{d y}{d x}\) = e^x (sin x + cos x)

Question 8.
y = \(\frac{\tan x}{x}\)
Solution:

Question 9.
y = \(\frac{\sin x}{1+\cos x}\)
Solution:
y = \(\frac{\sin x}{1+\cos x}=\frac{u}{v}\) (say)
u = sin x v = 1 + cosx
u’ = cos x v’ = -sin x
y = \(\frac{u}{v} \Rightarrow y^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)

Question 10.
y = \(\frac{x}{\sin x+\cos x}\)
Solution:

Question 11.
y = \(\frac{\tan x-1}{\sec x}\)
Solution:

Question 12.
y = \(\frac{\sin x}{x^{2}}\)
Solution:

Question 13.
y = tan θ (sin θ + cos θ)
Solution:

Question 14.
y = cosex x. cot x
Solution:
y = u v ⇒ y’ = uv’ + vu’
u = cosec x ⇒ u’ = -cosec x cot x
v = cot x ⇒ v’ = – cosec² x
(cosec x)(-cosec²x) + cot x(-coseç x cot x)
= cosec³x – cosec x cot²x
= – cosec x (cosec²x + cot²x)
= \(-\frac{1}{\sin x}\left(\frac{1+\cos ^{2} x}{\sin ^{2} x}\right)=-\frac{\left(1+\cos ^{2} x\right)}{\sin ^{3} x}\)

Question 15.
y = x sin x cos x
Solution:

Question 16.
y = e^-x. log x
Solution:
y = e^-x logx = uv (say)
Here u = e^-x and v = log x
⇒ u’ = -e^-x and v’ = \(\frac{1}{x}\)
Now y = uv ⇒ y’ = uv’ + vu’
(i.e.) \(\frac{d y}{d x}\) = e^-x \(\left(\frac{1}{x}\right)\) + log x(-e^-x)
= e^-x(\(\frac{1}{x}\) – log x)

Question 17.
y = (x² + 5) log (1 + x)e^-3x
Solution:

Question 18.
y = sin x⁰
Solution:

Question 19.
y = log₁₀x
Solution:

Question 20.
Draw the function f'(x) if f(x) = 2x² – 5x + 3
Solution:
f(x) = 2x² – 5x + 3
f'(x) = 4x – 5
x = 0, f'(0) = 4 × 0 – 5 = – 5
x = 1, f'(0) = 4 × 1 – 5 = – 1
x = 2, f'(0) = 4 × 2 – 5 = – 3

Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2 Additional Questions

Question 1.
Find the derivation of the following functions
Question 1.
3 sin x + 4 cos x – e^x
Solution:
y = 3 sin x + 4 cos x – e^x
\(\frac{d y}{d x}\) = 3 (cos x) + 4 (- sin x) – (e^x)
= 3 cos x – 4 sin x – e^x

Question 2.
sin 5 + log₁₀x + 2 secx
Solution:
y = sin 5 + log₁₀x + 2 secx
\(\frac{d y}{d x}\) = 0 + \(\left(\frac{1}{x}\right)\) log₁₀ e + 2[sec x + tan x] = \(\frac{\log _{10} e}{x}\) + 2 sec x tan x

Question 3.
6 sin x log₁₀x + e
Solution:

Question 4.
(x⁴ – 6x³ + 7x² + 4x + 2) (x³ – 1)
Solution:
Let u = x⁴ – 6x³ + 7x² + 4x + 2 and v = x³ – 1
u’ = 4x³ – 6 (3x²) + 7 (2x) + 4 (1) + 0
= 4x³ – 18x² + 14x + 4
v’= 3x³
y = uv’ + vu’
i.e. \(\frac{d y}{d x}\) = (x⁴ – 6x³ + 7x² + 4x + 2) (3x²) + (x³ – 1) (4x³ – 18x² + 14x + 4)
= 3x⁶ – 18x⁵ + 21x⁴ + 12x³ + 6x² + 4x⁶ – 18x⁵ + 14x⁴ + 4x³ – 4x³ + 18x² – 14x – 4
= 7x⁶ – 36x⁵ + 35x⁴ + 12x³ + 24x² – 14x – 4

Question 5.
(3x² + 1)²
Solution:
y = (3x² + 1)² = (3x² + 1) (3x² + 1)
Let u = 3x² + 1 and v = 3x² + 1
∴ u’ = 3(2x) = 6x and v’ = 6x
y’ = uv’ + vu’
(i.e.,) \(\frac{d y}{d x}\) = (3x² + 1) (6x) + (3x² + 1) 6x = 12x (3x² + 1)

Question 6.
(3 sec x – 4 cosec x) (2 sin x + 5 cos x)
Solution:
y = (3 sec x – 4 cosec x) (2 sin x + 5cos x)
Let u = 3 secx-4 cosecx and v = 2 sinx + 5 cosx
u’ = 3 (sec x tan x) – 4 (-cosec x cot x) ; v’ = 2 (cos x) + 5 (- sin x)
u’ = 3 sec x tan x + 4 cosec x cot x); v’ = 2 cos x – 5 sin x .
∴ y’ = uv’ + vu’
So \(\frac{d y}{d x}\) = (3 sec x – 4 cosec x) (2 cos x – 5 sinx) + (2 sin x + 5 cos x) (3 sec x tan x + 4 cosec x cot x) = 6 sec x cos x – 15 sec x sin x – 8 cosec x cos x + 20 cosec x sin x + 6 sinx secx tanx + 8 sinx cosecx cotx+ 15 cosx secx tanx + 20 cos x cosec x cot x
= 6 \(\frac{1}{\cos x}\) cosx – 15 \(\frac{1}{\cos x}\)sin x – 8 \(\frac{1}{\sin x}\) cos x + 20 \(\frac{1}{\sin x}\) sin x + 6 sin x \(\frac{1}{\cos x}\) tan x + 8 sin x \(\frac{1}{\sin x}\) cot x + 15 cos x \(\frac{1}{\cos x}\) tan x + 20 cos x \(\frac{1}{\sin x}\) cot x
= 6 – 15tan x – 8cot x + 20 + 6 tan²x + 8 cot x + 15 tan x + 20cot²x
= 26 + 6 tan²x + 20 cot²x

Question 7.
x² e^x sinx
Solution:
y = x² e^x sin x
Let u = x², v = e^x and w = sinx
u’ = 2x, v’ = e^x and w’ = cos x
y’ = uvw’ + vwu’ + uwv’
= (x² e^x) cos x + (e^x sin x)(2x) + (x² sin x)e^x
= x² e^x cos x + 2xe^x sin x + x² e^x sin x
= xe^x {x cos x + 2 sin x + x sin x}

Question 8.
\(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Solution:
y = \(\frac{\cos x+\log x}{x^{2}+e^{x}}\)
Let u = cos x + log x and v = x² + e^x
∴ u’ = – sin x + \(\frac{1}{x}\), v’ = 2x + e^x

Question 9.
\(\frac{\tan x+1}{\tan x-1}\)
Solution:

Question 10.
\(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Solution:
y = \(\frac{\sin x+x \cos x}{x \sin x-\cos x}\)
Let u = sinx + x cosx and v = x sin x – cos x
u’ = cos x + x( – sin x) + cos x (1)
= cos x – x sin x + cos x = 2 cos x – x sin x
v’=x (cos x) + sin x(1) – (- sin x)
= x cos x + sin x + sin x = 2 sin x + x cos x
y = \(\frac{u}{v}\) ∴ y’ = \(\frac{v u^{\prime}-u v^{\prime}}{v^{2}}\)

Students can Download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 1.
Find the area and perimeter of the following parallelograms.

Solution:
(i) Given base b = 11 cm ; height h = 3 cm
Area of the parallelogram = b × h sq. units = 11 × 3 cm²
= 33 cm²
Also perimeter of a parallelogram = Sum of 4 sides
= 11 cm + 4 cm + 11 cm + 4 cm = 30 cm
Area = 33 cm²; Perimeter = 30 cm.

(ii) Given base b = 7 cm
height h = 10 cm
Area of the parallelogram = b × h sq. units
= 7 × 10 cm² = 70 cm²
Perimeter = Sum of four sides
= 13 cm + 7 cm + 13 cm + 7 cm = 40 cm
Area = 70 cm², Perimeter = 40 cm

Question 2.
Find the missing values.

Solution:
(i) Given Base 6 = 18 cm ; Height h = 5 cm
Area of the parallelogram = b × h sq. units
= 18 × 5 cm²
= 90 cm²

(ii) Base b = 8m; Area of the parallelogram = 56 sq. m
b × h = 56
8 × A = 56
h = \(\frac{56}{8}\)
h = 7 m

(iii) Given Height h = 17 mm
Area of the parallelogram = 221 sq. mm
b × h = 221
b × 17 = 221
b = \(\frac{221}{17}\)
b = 13 m
Tabulating the results, we get

Question 3.
Suresh on a parallelogram shaped trophy in a state level chess tournament. He knows that the area of the trophy is 735 sq. cm and its base is 21 cm. What is the height of that trophy?

Solution:
Given base 6 = 21 cm
Area of parallelogram = 735 sq. cm
b × h = 735
21 × h = 735
h = \(\frac{735}{21}\)
h = 35 cm
∴ Height of the trophy = 35 cm

Question 4.
Janaki has a piece of fabric in the shape of a parallelogram. Its height is 12 m and its base is 18 m. She cuts the fabric into four equal parallelograms by cutting the parallel sides through its mid-points. Find the area of each new parallelogram.
Solution:

Area of a parallelogram = (base × height) sq. units
Base length = \(\frac{18}{2}\) = 9 m
Height = \(\frac{12}{2}\) = 6 m
Area = 9 × 6 = 54 m²
Area of each parallelogram = 54 m²

Question 5.
A ground is in the shape of parallelogram. The height of the parallelogram is 14 metres and the corresponding base is 8 metres longer than its height. Find the cost of levelling the ground at the rate of ₹ 15 per sq. m.
Solution:

Height of the parallelogram h = 14 m
Base = 8 m longer than height
= (14 + 8) m = 22 m
Area of the parallelogram = (base × height) sq. units
= (22 × 14)m² = 308 m²
Cost of levelling 1 m² = ₹ 15
Cost of levelling 308 m² = 308 × 15 = ₹ 4,620
Cost of levelling the ground = ₹ 4,620

Objective Type Questions

Question 6.
The perimeter of a parallelogram whose adjacent sides are 6 cm and 5 cm is
(i) 12 cm
(ii) 10 cm
(iii) 24 cm
(iv) 22 cm
Solution:
(iv) 22 cm
Hint:
= 2(6 + 5) = 2 × 11 = 22 cm

Question 7.
The area of parallelogram whose base 10 m and height 7 m is
(i) 70 sq.m
(ii) 35 sq.m
(iii) 7 sq.m
(iv) 10 sq.m
Solution:
(i) 70 sq. m
Hint: = base × height = 10m × 7m = 70 sq.m

Question 8.
The base of the parallelogram with area is 52 sq. cm and height 4 cm is
(i) 48 cm
(ii) 104 cm
(iii) 13 cm
(iv) 26 cm
Solution:
(iii) 13 cm
Hint:

Question 9.
What happens to the area of the parallelogram if the base is increased 2 times and the height is halved?
(i) Decreases to half
(ii) Remains the same
(iii) Increase by two times
(iv) None
Solution:
(ii) Remains the same
Hint:
Area = b × h sq. units
New base = 2 × old base
New height = \(\frac{1}{2}\) × old height
New Area = New base × New height = (2 × b)\(\frac{1}{2}\) × h = bh = old Area.

Question 10.
In a parallelogram the base is three times its height. If the height is 8 cm then the area is
(i) 64 sq. cm
(ii) 192 sq. cm
(iii) 32 sq. cm
(iv) 72 sq. cm
Solution:
(ii) 192 sq. cm
Hint: Given b = 3 × h; h = 8 cm
Area = b × h = 3h × 8 = 3 × 8 × 8 = 192 cm²

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 16 Plant and Animal Hormones

Samacheer Kalvi 10th Science Plant and Animal Hormones Textual Evaluation Solved

I. Choose the Correct Answer.

Question 1.
Gibberellins cause _____.
(a) Shortening of genetically tall plants
(b) Elongation of dwarf plants
(c) Promotion of rooting
(d) Yellowing of young leaves.
Answer:
(c) Promotion of rooting

Question 2.
The hormone which has positive effect on apical dominance is:
(a) Cytokinin
(b) Auxin
(c) Gibberellin
(d) Ethylene
Answer:
(b) Auxin

Question 3.
Which one of the following hormones is naturally not found in plants?
(a) 2, 4 – D
(b) GA3
(c) Gibberellin
(d) IAA.
Answer:
(a) 2, 4 – D

Question 4.
Avena coleoptile test was conducted by:
(a) Darwin
(b) N. Smit
(c) Paal
(d) F.W. Went
Answer:
(d) F.W. Went

Question 5.
To increase the sugar production in sugarcanes they are sprayed with _____.
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene.
Answer:
(d) Ethylene.

Question 6.
LH is secreted by:
(a) Adrenal gland
(b) Thyroid gland
(c) Anterior pituitary
(d) Hypothalamus
Answer:
(c) Anterior pituitary

Question 7.
Identify the exocrine gland _____.
(a) Pituitary gland
(b) Adrenal gland
(c) Salivary gland
(d) Thyroid gland.
Answer:
(c) Salivary gland

Question 8.
Which organ acts as both exocrine gland as well as endocrine gland?
(a) Pancreas
(b) Kidney
(c) Liver
(d) Lungs
Answer:
(a) Pancreas

Question 9.
Which one is referred to as “Master Gland”?
(a) Pineal gland
(b) Pituitary gland
(c) Thyroid gland
(d) Adrenal gland.
Answer:
(b) Pituitary gland

II. Fill in the blanks.

Question 1.
______ causes cell elongation, apical dominance and prevents abscission.
Answer:
Auxin.

Question 2.
______ is a gaseous hormone involved in abscission of organs and acceleration of fruit ripening.
Answer:
Ethylene.

Question 3.
____ causes stomatal closure.
Answer:
Abscisic acid.

Question 4.
Gibberellins induce stem elongation in _____ plants.
Answer:
Corn and Pea.

Question 5.
The hormone which has a negative effect on apical dominance is _____.
Answer:
Cytokinin.

Question 6.
Calcium metabolism of the body is controlled by _____.
Answer:
Parathormone.

Question 7.
In the Islets of Langerhans, beta cells secrete _____.
Answer:
Insulin.

Question 8.
The growth and functions of the thyroid gland are controlled by _____.
Answer:
Hormone T₃ and T₄

Question 9.
Decreased secretion of thyroid hormones in the children leads to _____.
Answer:
Cretinism.

III. Match the following.

Question 1.
(a) Match the Column I with Columns II and III.

Column I	Column II	Column III
Auxin	Gibberella fujikuroi	Abscission
Ethylene	Coconut milk	Intermodal elongation
Abscisic acid	Coleoptile tip	Apical dominance
Cytokinin	Chloroplast	Ripening
Gibberellins	Fruits	Cell division

(b) Match the following hormones with their deficiency states.

Hormones	Disorders
1. Thyroxine	(a) Acromegaly
2. Insulin	(b) Tetany
3. Parathormone	(c) Simple goitre
4. Growth hormone	(d) Diabetes insipidus
5. ADH	(e) Diabetes mellitus

Answer:
(a)

Column I	Column II	Column III
Auxin	Coleoptile tip	Apical dominance
Ethylene	Fruits	Ripening
Abscisic acid	Chloroplast	Abscission
Cytokinin	Coconut milk	Cell division
Gibberellins	Gibberella fujikuroi	Intermodal elongation

(b)
1. (c) Simple goitre
2. (e) Diabetes mellitus
3. (b) Tetany
4. (a) Acromegaly
5. (d) Diabetes insipidus

IV. State whether True or false, If false write the correct statement.

Question 1.
A plant hormone concerned with stimulation of cell division and promotion of nutrient mobilization is cytokinin.
Answer:
True.

Question 2.
Gibberellins cause parthenocarpy in tomato.
Answer:
True.

Question 3.
Ethylene retards senescence of leaves, flowers and fruits.
Answer:
False.
Correct Statement: Ethylene hastens the senescence of leaves, flowers and fruits.

Question 4.
Exophthalmic goitre is due to the over secretion of thyroxine.
Answer:
True.

Question 5.
The pituitary gland is divided into four lobes.
Answer:
False.
Correct Statement: The Pituitary gland is composed of two lobes and performs different functions.

Question 6.
Estrogen is secreted by corpus luteum.
Answer:
False.
Correct Statement: Estrogen is produced by the Graafian follicles of the ovary.

V. Assertion and Reasoning Questions

Direction: In each of the following questions a statement of assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Mark the correct statement as,
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) Both A and R are false.

Question 1.
Assertion: Application of cytokinin to marketed vegetables can keep them fresh for several days.
Reason: Cytokinins delay senescence of leaves and other organs by mobilisation of nutrients.
Answer:
(d) Both A and R are false.

Question 2.
Assertion (A): Pituitary gland is referred to as “Master gland”.
Reason (R): It controls the functioning of other endocrine glands.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

Question 3.
Assertion (A): Diabetes mellitus increases blood sugar levels.
Reason (R): Insulin decreases blood sugar levels.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

VI. Answer in a word or sentence.

Question 1.
Which hormone promotes the production of male flowers in Cucurbits?
Answer:
Gibberellins promote the production of male flowers in Cucurbits.

Question 2.
Write the name of synthetic auxin.
Answer:
2, 4 Dichloro phenoxy Acetic Acid is the synthetic hormone.

Question 3.
Which hormone induces parthenocarpy in tomatoes?
Answer:
Gibberellins are efficient in the formation of seedless fruit, inducing parthenocarpy in tomatoes.

Question 4.
What is the hormone responsible for the secretion of milk in female after child birth?
Answer:
Prolactin stimulates the secretion of milk in female after child birth.

Question 5.
Name the hormones which regulate water and mineral metabolism in man.
Answer:
Antidiuretic or vasopressin hormone regulates water and mineral metabolism in man.

Question 6.
Which hormone is secreted during emergency situation in man?
Answer:
Adrenaline and Noradrenaline is secreted during emergency situation in man.

Question 7.
Which gland secretes digestive enzymes and hormones?
Answer:
The pancreas is exocrine and endocrine in nature. The exocrine pancreas secretes pancreatic juice which plays a role in digestion. The endocrine portion (Islets of Langerhans) secrete hormones.

Question 8.
Name the endocrine glands associated with kidneys.
Answer:
Adrenal gland is associated with kidneys.

VII. Short Answer Questions.

Question 1.
What are synthetic auxins? Give examples.
Answer:
Artificially synthesized auxins, which have the properties like auxins are called as synthetic auxins, eg. 2, 4-D (2, 4-Dichlorophenoxy Acetic acid, Indole-3-Propionic Acid, alpha – Naphthalene Acetic Acid (NAA), 2, 4, 5-T (2, 4, 5-Trichlorophenoxy Acetic acid) are some of the synthetic auxins.

Question 2.
What is bolting? How can it be induced artificially?
Answer:
Rosette plant (genetic dwarfism) plant exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of a stem followed by flowering is called bolting.

Question 3.
Bring out any two physiological activities of abscisic acid.
Answer:

• During water stress and drought conditions, Abscisic acid causes stomatal closure.
• ABA induces bud dormancy towards the approach of winter in trees like birch.

Question 4.
What will you do to prevent leaf fall and fruit drop in plants? Support your answer with reason.
Answer:
Artificially synthesized auxin to be sprayed to prevent leaf fall and fruit drop as Auxin prevent the formation of abscission layer.

Question 5.
What are chemical messengers?
Answer:
A chemical messenger is any compound that serves to transmit a message. A chemical messenger refers to hormones.

Question 6.
Write the differences between endocrine and exocrine gland.
Answer:
The exocrine gland secretes juices, which plays a role in digestion. An endocrine portion is made up of endocrine glands, which secrete hormones. The pancreas is exocrine and endocrine in nature.

Question 7.
What is the role of parathormone?
Answer:
Parathormone regulates calcium and phosphorous metabolism in the body. They act on bone, kidney and intestine to maintain blood calcium levels.

Question 8.
What are the hormones secreted by the posterior lobe of the pituitary gland? Mention the tissues on which they exert their effect.
Answer:

• Vasopressin or Antidiuretic hormone.
• Oxytocin is the two hormones of the posterior lobe of the pituitary gland.

In kidney tubules, the vasopressin hormone increases the reabsorption of water. It reduces the loss of water through urine.
Oxytocin helps in the contraction of the smooth muscles of the uterus at the time of childbirth and ” milk ejection from the mammary gland after childbirth.

Question 9.
Why are thyroid hormones refered as personality hormone?
Answer:
Thyroid hormones are essential for normal, physical, mental and personality development. So, it is called as personality hormone.

Question 10.
Which hormone requires iodine for its formation? What will happen if the intake of iodine in our diet is low?
Answer:
The hormones secreted by the thyroid gland are:

• Triiodothyronine (T₃)1
• Tetraiodothyronine or Thyroxin (T₄), which need an amino acid tyrosine and Iodine for its formation.

If the intake of Iodine in our diet is low or due to the inadequate supply of iodine in our diet leads to the enlargement of the thyroid gland, which protrudes, as swelling in the neck and is called as goitre.

VIII. Long Answer Questions

Question 1.
(a) Name the gaseous plant hormone. Describe its three different actions in plants.
Answer:
Ethylene is a gaseous plant hormone.

1. Ethylene promotes the ripening of fruits. Eg: Tomato, Apple, Mango, Banana, etc.
2. Ethylene inhibits the elongation of stem and root in dicots.
3. Ethylene hastens the senescence of leaves and flowers.

(b) Which hormone is known as a stress hormone in plants? Why?
Answer:
Abscisic acid increases the tolerance of plants to various kinds of stress. So, it is also called a stress hormone. It is found in the chloroplast of plants.

Question 2.
Describe an experiment which demonstrates that growth-stimulating hormone is produced at the tip of the coleoptile?
Answer:
Went’s experiment was done by Frits warm out. This experiment demonstrates the growth-stimulating hormone Auxin is produced at the tip of the coleoptile.

He did a series of experiments in Avena Coleoptiles. In his first experiment, he removed the tips of Avena coleoptiles. The cut tips did not grow, indicating that the tips produced something essential for growth.
In his second experiment, he placed the agar blocks on the decapitated coleoptile tips. The coleoptile tips did not show any response.

In his third experiment, he placed the detached coleoptile tips on agar blocks. After an hour, the tips were discarded and placed this agar block on the decapitated coleoptile. It grew straight up, indicating that some chemical had diffused from the cut coleoptile tips into the agar block, which stimulated the growth.

From these experiments, Went concluded that, the chemical diffusing from the tip of coleoptile was responsible for the growth and he named it as “Auxin”.

Question 3.
Write the physiological effects of gibberellins.
Answer:

1. Application of gibberellins on plants stimulate extraordinary elongation of intemode. Eg: Com and Pea.
2. Treatment of rosette plants with gibberellin induces sudden shoot elongation followed by flowering. This is called bolting.
3. Gibberellins promote the production of male flowers in monoecious plants (Cucurbits).
4. Gibberellins break dormancy of potato tubers.
5. Gibberellins are efficient than. auxins in inducing the formation of seedless fruit – Parthenocarpic fruits (Development of fruits without fertilization) Eg: Tomato.

Question 4.
Where are estrogens produced? What is the role of estrogens in the human body?
Answer:
Estrogen, the female sex hormone is produced by the Graafian follicles of the ovary.
Role of estrogen:

• It brings about the changes that occur during puberty.
• It initiates the process of oogenesis.
• It stimulates the maturation of ovarian follicles in the ovary.
• It promotes the development of secondary sexual characters (breast development and high pitched voice, etc).

Question 5.
What are the conditions which occur due to lack of ADH and insulin? How are the conditions different from one another?
Answer:
ADH:
Deficiency of ADH causes a disorder called diabetes insipidus.
Deficiency of ADH reduces the reabsorption of water and causes an increase in urea output (polyuria). This deficiency disorder is called Diabetes insipidus.

IX. Higher Order Thinking Skills (HOTS) Questions

Question 1.
What would be expected to happen if,

1. Gibberellin is applied to rice seedlings?
2. A rotten fruit gets mixed with unripe fruits.
3. When cytokinin is not added to the culture medium.

Answer:

1. When gibberellins are applied, it stimulates, the extraordinary elongation of internodes.
2. The mass of undifferentiated cell is called callus. If the rotten fruits get mixed with unripe fruits, then the ethylene produced from the rotten fruits will hasten the ripening of the unripe fruits.
3. If cytokinin is not added to the culture medium, then the cell division, growth and differentiation will not be observed.

Question 2.
A plant hormone was first discovered in Japan when rice plants were suffering from Bakanae disease caused by Gibberella fujikoroi. Based on this information answer the following questions:
(a) Identify the hormone involved in this process.
Answer:
Gibberellin is involved in this process.

(b) Which property of this hormone causes the disease?
Answer:
It stimulates extraordinary elongation of intemode.

(c) Give two functions of this hormone.
Answer:

1. Gibberellin break dormancy of potato tubers.
2. Gibberellin promotes the production of male flower in monoecious plant (Cucurbits).

Question 3.
Senthil has high blood pressure, protruded eyeball and increased body temperature. Name the endocrine gland involved and hormone secretion responsible for this condition.
Answer:
The endocrine gland is Thyroid gland, and the hormones secreted are Triiodothyronine (T₃) and Tetraiodothyronine or Thyroxine (T₄). The excess secretion of the Thyroid hormone leads to Grave’s disease. The symptoms are protrusion of eyeballs (exophthalmia), increased metabolic rate, high body temperature, profuse sweating, loss of body weight and nervousness.

Question 4.
Sanjay is sitting in the exam hall. Before the start of the exam, he sweats a lot, with increased rate of heart beat. Why does this condition occur?
Answer:
It is due to secretion of Epinephrine and Norepinephrine as it produced during conditions of stress and emotion. They are called as “Emergency hormones” or flight, fright and fight hormone.

Question 5.
Susan’s father feels very tired and frequently urinates. After clinical diagnosis, he was advised to take an injection daily to maintain his blood glucose level. What would be the possible cause for this? Suggest preventive measures.
Answer:
The deficiency of insulin causes Diabetes mellitus. He has Diabetes mellitus. Increase in blood sugar level (Hyperglycemia). He was advised to take an injection, Insulin, which helps in the conversion of glucose into glycogen, which is stored in the liver. It decreases the concentration of glucose in the blood.
Preventive measures:

• Manage the weight
• Exercise regularly
• Eat a balanced healthy diet
• Limit alcohol intake
• Quit smoking
• Control the blood pressure

Textbook Activities Solved

Activity 1.
Place two or three unripe tomatoes in a brown paper bag with a ripe banana and roll the top closed. In another bag place two or three unripe tomatoes only and roll the top closed, Observe what happens to the tomatoes? Why?
Answer:
As the banana continues to ripen in the first bag, it produces ethylene gas. The gas trapped in the bag will cause tomatoes to ripen. The tomatoes remain unripe in the second bag.

Samacheer Kalvi 10th Science Plant and Animal Hormones Additional Questions Solved

I. Fill in the blanks.

Question 1.
The chemical substances produced by plants are called _____.
Answer:
Hormones.

Question 2.
Tie two lobes of thyroid glands are connected by means of a narrow band of tissue called _____.
Answer:
Isthmus.

Question 3.
The chief cells of parathyroid glands are _____.
Answer:
Parathormone.

Question 4.
The other name for adrenal gland is _____.
Answer:
Supra Renal.

Question 5.
The hormones secreted by the adrenal cortex are _____.
Answer:
Corticosteroids

Question 6.
During water stress and drought conditions _______ causes stomatal closure.
Answer:
Abscisic acid.

Question 7.
Glucagon helps the break down of ______ to glucose in the liver.
Answer:
Glycogen.

Question 8.
The adrenal medulla is composed of ______ cells.
Answer:
Chromaffin.

Question 9.
The male sex hormone _______ is responsible for the development of secondary sexual characters.
Answer:
Testosterone.

Question 10.
_______ is a gaseous plant hormone, mainly concerned with maturation and ripening of fruits.
Answer:
Ethylene.

Question 11.
The other name for pituitary gland is _____.
Answer:
Hypophysis.

Question 12.
The glandular follicles of the thyroid gland are filled with colloid material called _____.
Answer:
Thyroglobulin.

II. Match the following.

Question 1.

1. Morphogenesis	(a) Chromaffin cells
2. Bakanae disease	(b) Male sex hormone
3. Tetany	(c) Thymus
4. Testosterone	(d) Gibberella fujikuroi
5. Adrenal medulla	(e) Female sex hormone
6. Thymosin	(f) Callus in tissue culture
7. Estrogen	(g) Muscle spasm

Answer:

1. (f) Callus in tissue culture
2. (d) Gibberella fujikuroi
3. (g) Muscle spasm
4. (b) Male sex hormone
5. (a) Chromaffin cells
6. (c) Thymus
7. (e) Female sex hormone.

III. Choose the odd one out.

Question 1.
Auxins, Parthenocarpy, apical dominance, Parathormone?
Answer:
Parathormone.

Question 2.
Senescence, Dormancy, Estrogen, abscission?
Answer:
Estrogen.

Question 3.
Glucagon, Endocrine, Exocrine, Gibberellins?
Answer:
Gibberellins.

Question 4.
Norepinephrine, Isthmus, tyrosine, thyroglobulin?
Answer:
Norepinephrine.

Question 5.
Gonads, Thyroid, Cytokinin, Thymus?
Answer:
Cytokinin.

IV. Match the following endocrine glands with their location.

Question 1.

1. Pituitary gland	(a) Female sex gland
2. Thyroid gland	(b) Male sex gland
3. Parathyroid	(c) Above the kidney
4. Islets of Langerhans	(d) The posterior surface of the thyroid lobe
5. Adrenal gland	(e) The upper part of the chest, the lower end of the trachea
6. Testes	(f) Base of midbrain
7. Ovary	(g) Pancreas
8. Thymus	(h) Trachea

Answer:

1. (f) Base of midbrain
2. (h) Trachea
3. (d) The posterior surface of the thyroid lobe
4. (g) Pancreas
5. (c) Above the kidney
6. (b) Male sex gland
7. (a) Female sex gland
8. (e) The upper part of the chest, the lower end of Trachea.

V. Write ‘True’ or ‘False’ for the following statements. Write the correct statement for false:

Question 1.
Auxins, cytokinins and gibberellins inhibit plant growth, while abscisic acid and Ethylene promote plant growth.
Answer:
False.
Correct Statement: Auxins, cytokinins and gibberellins promote plant growth, while abscisic acid and ethylene inhibit plant growth.

Question 2.
Glucagon helps in the breakdown of glycogen to glucose in the liver.
Answer:
True.

Question 3.
The thymus is partly an endocrine gland and partly a lymphoid gland.
Answer:
True.

Question 4.
Leydig cells from the cells of a female. Gonads located in the pelvic cavity.
Answer:
False.
Correct Statement: Leydig cells from the endocrine part of the testes.

Question 5.
Cretinism is caused due to increased secretion of the thyroid hormones in children.
Answer:
False.
Correct Statement: Cretinism is caused due to decreased secretion of the thyroid hormones in children.

Question 6.
Cytokinin promotes the growth of lateral buds even in the presence of apical buds.
Answer:
True.

VI. Answer the following in a word or with a sentence.

Question 1.
Where are Auxins produced?
Answer:
Auxins are produced at the tip of stems and roots from where they migrate to the zone of elongation.

Question 2.
What is Richmond Lang effect?
Answer:
Delaying the process of ageing in plants with the application of cytokinin is called Richmond Lang effect.

Question 3.
What is Richmond Lang effect?
Answer:
Application of cytokinin delays the process of ageing in plants. This is called Richmond Lang effect.

Question 4.
Give the uses of progesterone.
Answer:
Progesterone maintains pregnancy and regulates the menstrual cycle.

Question 5.
What are the secretions of alpha and beta cells of Islets of Langerhans?
Answer:
Alpha cells secrete glucagon and beta cells secrete Insulin.

Question 6.
What is the main function of Glucogen.
Answer:
Glucogon converts excess amount of glycogen stored under the muscle and liver to glucose. Thus raising the blood glucose level.

Question 7.
Name the two types of sex glands.
Answer:
Testes and Ovaries.

Question 8.
Specify the symptoms of acromegaly.
Answer:
Acromegaly leads to abnormal enlargement of head, face, hand and feet.

Question 9.
Name the three layers of the adrenal cortex.
Answer:

• Zona glomerulosa
• Zona fasciculata
• Zona reticularis

Question 10.
What is Endocrinology?
Answer:
The branch of biology which deals with the study of the endocrine glands is called Endocrinology.

VII. Answer the following briefly.

Question 1.
Name the types of plant hormones.
Answer:

• Auxins
• Cytokinins
• Gibberellins
• Abscisic acid
• Ethylene.

Question 2.
Growth hormone is important for normal growth. Explain.
Answer:
Growth hormone promotes the development and enlargement of all tissues •of the body. It stimulates the growth of muscles, cartilage and long bones. It controls cell metabolism.

Question 3.
What are Ductless glands? Why are they called so?
Answer:
Endocrine glands are called ductless glands because their secretions are diffused into the bloodstream, and are carried to the different parts of the body. They do not have specific ducts to carry the hormones.

Question 4.
Why pancreas is called as dual gland?
Answer:
The exocrine part of the pancreas produces pancreatic juice. The endocrine part produce islets of Langerhans, consists of two cells namely alpha cells that produce a hormone called glucagon and Beta cells that produce insulin. So, it is called as dual gland.

Question 5.
Write any three physiological effects of cytokinins.
Answer:

• Cytokinins induce cell division.
• Cytokinins promote the growth of lateral buds even in the presence of apical buds.
• Cytokinesis causes cell enlargement.

Question 6
Which gland is a link between endocrine and lymphoid gland.
Answer:
Thymus is partly an endocrine gland and partly a lymphoid gland. It is located in the upper part of the chest covering the lower end of trachea. Thymosin is the hormone secreted by the thymus.

Question 7.
What are the functions of thyroid hormones? Write any three points.
Answer:

• Production of energy by maintaining the Basal Metabolic Rate (BMR) of the body.
• Helps to maintain normal body temperature.
• Influences the activity of the Central Nervous System.

Question 8.
What are plant hormones?
Answer:
Plant hormones are organic molecules that are produced at extremely low concentration in plants. These molecules control morphological, physiological and biochemical responses.

Question 9.
Explain any three functions of Testosterone.
Answer:

• It influences the process of spermatogenesis.
• It stimulates protein synthesis and controls muscular growth.
• Responsible for the development of secondary sexual characters (distribution of hair on body and face and deep voice pattern, etc.).

Question 10.
Write a short note on the thymus gland.
Answer:
The thymus gland is partly an endocrine gland and partly a lymphoid gland. It is located in the upper part of the chest covering the lower end of the trachea. Thymosin is the hormone secreted by Thymus.
Functions:

• It has a stimulatory effect on the immune function.
• It stimulates the production and differentiation of lymphocytes.

VIII. Draw a labelled diagram for the following.

Question 1.
Hormonal interaction in plant growth and development.
Answer:

Question 2.
Pancreas
Answer:

Question 3.
Adrenal gland
Answer:

IX. Answer the following in detail.

Question 1.
With a neat labelled diagram, explain the pituitary gland and the types of hormones.
Answer:

The pituitary gland is a pea-shaped compact mass of cells located at the base of the midbrain. As it regulates and controls the other endocrine glands, it is called “Master gland”

1. Hormones secreted by anterior pituitary:
(a) Growth Hormone (GH): It promotes the development and enlargement of all tissues. It stimulates the growth of muscles, cartilage and long bones. It controls cell metabolism. Decreased secretion of growth hormone leads to Dwarfism in children characterised by stunted growth, delayed skeletal formation and mental disability. Oversecretion of growth hormone leads to gigantism in children. Characterised by abnormal enlargement of head, face, hands and feet.

(b) Thyroid Stimulating Hormone (TSH) – a growth of thyroid gland.

(c) Adrenocorticotrophic hormone (ACTH) – Protein synthesis in the adrenal cortex.

(d) Gonadotrophic hormones (GTH) – for the normal development of Gonads.
The other two hormones are Follicle Stimulating Hormone (FSH); Luteinizing Hormones (LH)

(e) Prolactin (PRL) Initiates the development of mammary glands during pregnancy and production of milk after childbirth.

2. Hormones secreted by the posterior lobe:
(a) Vasopressin or Antidiuretic hormone (ADH) – It reduces the loss of water through urine. Deficiency of ADH reduces the reabsorption of water and causes an increase in urine output (polyuria). This deficiency disorder is called Diabetes insipidus.

(b) Oxytocin – It helps in the contraction of the smooth muscles of the uterus at the time of childbirth and milk ejection from the mammary gland after childbirth.

Question 2.
With a neat labelled diagram, explain thyroid gland, functions of thyroid hormones and the thyroid dysfunction.
Answer:
The thyroid gland is composed of two distinct lobes lying one on either side of the trachea. The two lobes are connected by means of a narrow band of tissue called as the isthmus, the gland is composed of glandular follicles and lined by cuboidal epithelium. The follicles are filled with colloid material called thyroglobulin. An amino acid tyrosine and iodine are involved in the formation of thyroid hormone.
The hormones secreted by the thyroid gland are:

1. Triiodothyronine (T₃)
2. Tetraiodothyronine or Thyroxine (T₄)
   

Functions of Thyroid hormone:

• Production of energy by maintaining the Basal Metabolic Rate (BMR) of the body.
• Helps to maintain normal body temperature.
• Influences the activity of the Central Nervous System.
• Controls growth of the body and bone formation.
• Essential for normal physical, mental and personality development. So it is called personality hormone.
• Regulates cell metabolism.

When the thyroid gland fails to secrete the normal level of hormone, the condition is called thyroid dysfunction. It leads to the following conditions:
1. Hypothyroidism: It is caused due to the decreased secretion of the thyroid hormones.

• Goitre: Goitre is caused due to the inadequate supply of iodine in our diet. It leads to the enlargement of the thyroid gland, protruded, marked swelling in the neck and is called goitre.
• Cretinism: It is caused due to decreased secretion of thyroid hormone in children. The conditions are stunted growth, mental defect, lack of skeletal development and deformed bones. They are called as cretins.
• Myxoedema: It is caused by the deficiency of thyroid hormone in adults. They are mentally sluggish, increase in body weight, puffiness of the face, hand etc.

2. Hyperthyroidism: It is caused due to the excess secretion of the thyroid hormone, which leads to Grave’s disease. The symptoms are protrusion of the eyeballs (Exopthalmia), increased metabolic rate, high body temperature, sweating, loss of body weight and nervousness.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Differentiate the following
Question 1.
y = (x² + 4x + 6)⁵
Solution:
y = (x² + 4x + 6)⁵
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 5(x² + 4x + 6)^5-1 × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 4x + 6)
= 5 (x² + 4x + 6) (2x + 4)
= 10 (x + 2) (x² + 4x + 6)⁴

Question 2.
y = tan 3x
Solution:
y = tan 3x
put u = 3x
\(\frac{d u}{d x}\) = 3
Now y = tan u
⇒ \(\frac{d u}{d x}\) = sec² u
So \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\) = (sec² u) (3)
= 3 sec² 3x

Question 3.
y = cos (tan x)
Solution:
y = cos (tan x)
[ y = f(g(x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f'(g(x)) . g'(x)]
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (tan x) × \(\frac{\mathrm{d}}{\mathrm{d} x}\) (tan x)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sin (tan x) × sec²x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = – sec² . sin (tan x)

Question 4.
y = \(\sqrt[3]{1+x^{3}}\)
Solution:

Question 5.
y = \(e^{\sqrt{x}}\)
Solution:

Question 6.
y = sin (e^x)
Solution:
y = sin (e^x)
Let u = e^x

Question 7.
F(x) = (x³ + 4x)⁷
Solution:
F(x) = (x³ + 4x)⁷
Put u = x³ + 4x

Question 8.
h(t) = \(\left(t-\frac{1}{t}\right)^{\frac{3}{2}}\)
Solution:

Question 9.
f(t) = \(\sqrt[3]{1+\tan t}\)
Solution:

Question 10.
y = cos (a³ + x³)
Solution:

Question 11.
y = e^-mx
Solution:

Question 12.
y = 4 sec 5x
Solution:

Question 13.
y = (2x – 5)⁴ (8x² – 5)^-3
Solution:

= \(\frac{8(2 x-5)^{3}}{\left(8 x^{2}-5\right)^{4}}\) (-4x² + 30x – 5)

Question 14.
y = (x² + 1) \(\sqrt[3]{x^{2}+2}\)
Solution:

Question 15.
y = xe^-x2
Solution:
y = x e^-x2
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = x . e^-x2 (- 2x) + e^-x2 . 1 = – 2x² e^-x2 + e^-x2
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = e^-x2 (1 – 2x²)

Question 16.
s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)
Solution:

Question 17.
f(x) = \(\frac{x}{\sqrt{7-3 x}}\)
Solution:

Question 18.
y = tan (cos x)
Solution:
y = tan (cos x)

Question 19.
y = \(\frac{\sin ^{2} x}{\cos x}\)
Solution:

Question 20.
y = \(5^{-\frac{1}{x}}\)
Solution:

Question 21.
y = \(\sqrt{1+2 \tan x}\)
Solution:

Question 22.
y = sin³x + cos³x
Solution:
y = sin³x + cos³x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 3 sin²x cos x + 3 cos² × – sin x
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = 3 sin x cos x (sin x – cos x)

Question 23.
y = sin² (cos kx)
Solution:

Question 24.
y = (1 + cos²x)⁶
Solution:

Question 25.
y =\(\frac{e^{3 x}}{1+e^{x}}\)
Solution:

Question 26.
y = \(\sqrt{x+\sqrt{x}}\)
Solution:

Question 27.
y = e^{x cos x}
Solution:

Question 28.
y = \(\sqrt{x+\sqrt{x+\sqrt{x}}}\)
Solution:

Question 29.
y = \(\sin (\tan (\sqrt{\sin x}))\)
Solution:

Question 30.
y = sin^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution: