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MP Board Class 12th Biology Solutions Chapter 3 Human Reproduction

Human Reproduction NCERT Textbook Questions and Answers

Question 1.
Fill in die Blanks:

1. Humans reproduce …………………….. (asexually/ sexually)
2. Humans are …………………….. (oviparous, viviparous, ovoviviparous)
3. Fertilization is …………………….. in humans, (external / internal)
4. Male and female gametes are …………………….. (diploid / haploid)
5. Zygote is …………………….. (diploid / haploid)
6. The process of release of ovum from a mature follicle is called ……………………..
7. Ovulation is induced by a hormone called ……………………..
8. The fusion of male and female gametes is called ……………………..
9. Fertilisation takes place in ……………………..
10. Zygote divides to form …………………….. which is implanted in uterus.
11. The structure which provides vascular connection between fetus and uterus is called ……………………..

Answer:

1. Sexually
2. Viviparous
3. Internal
4. Haploid
5. Diploid
6. Ovulation
7. LH and FSH
8. Fertilisation
9. Follopian tube
10. Embryo
11. Placenta.

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:

Question 4.
Write two major functions each of testis and ovary
Answer:
Functions of testis functions:

• Production of sperms by seminiferous tubules.
• Production of male sex hormone, testosterone by leydig cells.

Functions of Ovary

• Production of ova (eggs).
• Production of female sex harmones, estrogen and progesterone.

Question 5.
Describe the structure of a seminiferous tubuls.
Answer:
Seminiferous tubules are highly coiled tubes, which are lined on the inside by :

• Male germ cells called spematogonia that undergo meiotic division to form sperm celts.
• Sertoli cells provides nutrition and molecular signals to the germ cells.
  

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis : Formation of sperms in the testis is called as spermatogenesis. It involves in the following steps :

1. Multiplication phase : In this phase, sperm cells are formed in testes.
The inner layer of seminiferous tubules of testes is formed of germinal epithelium.
Some of these cells called primary germ cells divide mitotically into spermatogo¬nia which become separated in the germi¬nal layer. Other cells of this layer serve as nutrition for the dividing cells.

2. Growth phase : In this phase, spermatogonia starts growing, absorbing nutrient substances. These large cells are called primary spermatocytes.

3. Maturation phase : It is a very important phase. Primary spermatocytes divide twice. The first division is meiotic due to which the number of chromosomes is reduced to half. In this process, primary spermatocyte divides into two halves which are known as secondary spermatocytes. The second division is mitotic and no change takes place in the number of chromosomes. Thus, from two secondary spermatocytes four spermatids are formed. In this manner from one primary spermatocyte four spermatids are formed. These spermatids change into sperm cells of spermatozoa by a process called metamorphosis.

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
The hormones involved in regulation of spermatogenesis are:

• Gonadotropin releasing hormone
• Luteinizing hormone (LH)
• Follicle stimulating hormone
• Testosteron.

Question 8.
Define spcrmiogenesis and spermiation.
Answer:
Spermiogenesis : The process involving transformation of spermatid into spermatozoa is called spermiogenesis.
Spermiation: After spermiogenesis sperm heads became embedded in the sertoli cells and are finally released from the seminiferous tubules by the process called spermiation.

Question 9.
Draw a labelled diagram of human sperm and explain its defferent parts.
Answer:
Structure of sperm: A mature sperm is a delicate microscopic, motile structure. A typi¬cal mammalian sperm consists of the following three parts:

1. Head,
2. Middle piece and
3. Tail.

(1) Head: Head is knob like terminal part of the sperm. It is composed of a large nucleus and an acrosome. At die time of entry of the sperm into egg acrosome Secretes spermlysin which dis¬solve the egg membrane and thus facilitates entry of sperm into the egg or ovum.

(2) Middle Piece : It is short and lies be¬tween head and tail. It contains two granules called the proximal and distal centrioles in front side and towards posterior side cylindrical middle part of sperm. It is considered as the power house of sperm as it contains compact mass of mitochondria, which provides energy for metabolism and movement of sperm.

(3) Tail: It is situated on posterior part of sperm. It moves with the help of axial filament. The posterior part of the tail is called as end piece and it is not covered by membrane.

Question 10.
What are the major components of seminal plasma?
Answer:
The main components of the seminal plasma are fructose, calcium ion, some enzymes and prostagladines.

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The main functions of male accessory ducts and glands are as follows :

1. Functions of accessory ducts:

(a) Rete Testis : They transport sperms from seminiferour tubule to Vas efferentia.
(b) Vas efferentia : Transports sperms to epdidymis.
(c) Epididymis : Sperms are stored here. Maturation of sperms occur.
(d) Vas deference : Transports sperms from epididymis to the urethra.

(2) Functions of glands:

(a) Prostate gland : It produces milky secretion which forms considerable part of the semen. It makes sperm motile.
(b) Bulbourethral gland : Its secretion make the penis lubricated.
(c) Seminal vesicle: It secrete mucus and watery alkaline fluid which provide energy to the sperm.

Question 12.
What is oogenesis ? Give a brief account of oogenesis. .
Answer:
Oogenesis is the process of formation of mature female gametes from primordial germ cells.

• This process is initiated during embryonic developmental stage when about two million gamete mother cells (oogonia) are formed in each foetal ovary.
• Oogonia start meiotic division which gets arrested at prophases-I stage. They are referred to as primary oocytes.
• Each of these gets covered with layers of granulose cells and are then called primary follicle.
• Many of the primary follicles degenerate from birth to puberty, leaving about 60000-80000 in each ovary at puberty.
• More layers of granulose cells and another theca layer surround it and now it is called secondary follicle. Theca layer is arranged as inner theca-intema and outer theca- extema.
• Secondary follicle transforms into tertiary follicle that has a fluid filled cavity called antrum.
• The primary oocyte within the tertiary follicle grows in size and completes its first meiotic division now.
• This is an unequal division resulting in the formation of:
  (a) A large haploid cell (that keeps the majority of nutrient rich cytoplasm) called secondary oocyte.
  (b) A tiny cell, with haploid nucleus and almost no cytoplasm called first polar
  body.
• The tertiary follicle undergoes certain changes to mature into graafian follicle.
  Secondary oocyte forms a new membrane around it, called zona pellucida.
• Under the influence of LH, the Graaffian follicle now ruptures to release secondary oocyte from the ovary by a process called ovulation.

Summary of oogenesis:

Oogonia Meiosis-I initiated -4 Primary oocyte (arrested at prophase-I) → Granu¬losa layer builds → Primary follicle → More granulose and theca layer added → Secondary follicle → Fluid filled cavity develops → Tertiary follicle → Primary oocyte completes meio- sis-I → Secondary oocyte (haploid)+ polar body → Tertiary follicle transforms to Graafian follicle → Zona pellucida builds around secondary oocyte Graafian follicle ruptures → Secondary oocyte (ovum) released.
Meiosis-II will occur only at the time of penetration of sperm into the ovum.

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:

Question 14.
Draw a labelled diagram of a graafian follice.
Answer:

Question 15.
Name the function of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae.
Answer:
The functions of the following :
(a) Corpus luteum Secretes large amount of progesterone which is essential for the maintenance of endometrium of the uterus.
(b) Endometrium is necessary for the implantation of the fertillised ovum, for contrib¬uting towards making of placenta and other events of pregnancy.
(c) Acrosome is filled with enzymes that help in dissolving the outer cover of the ovum and entry of sperm nucleus.
(d) Sperm tail facilitates motility of the sperm essential for reaching the ovum to fertilise it.
(e) Fimbriae are fmgers-like projections at the mouth of fallopian tubules that help in
collection of the ovum after ovulation.

Question 16.
Identify true/ false statement to make it true.

1. Androgens are produced by sertoli cells. (True / False)
2. Spermatozoa get nutrition from sertoli cells. (True / False)
3. Leydig cells are found in ovaiy. (True / False)
4. Leydig cells synthesise androgens. (True / False)
5. Oogenesis takes place in corpus luteum. (True / False)
6. Menstrual cycle ceases during pregnancy. (True / False)
7. Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True /False)

Answer:

1. Flase
2. True
3. False
4. True
5. False
6. True
7. True.

Question 17.
What is mentrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates is called menstrual cycle. The uterus linings becomes thick and spongy to receive fertilised egg. If the egg is not fertilised, this lining is not needed any longer so, it slowly breaks and comes out through vagina along with blood and mucous. This is called menstruation. It is repeated at an average interval of about 28/29 days.
Following hormones are regulate this cycle:

• Gonadotropin
• strogen
• Luteinizing hormone
• Follicular stimulating hormone
• Progesteron.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
The process of delivery of the foetus (Child birth) at the end of the pregnancy is called parturition. The signals for parturition originate from the fully developed foetus and the placenta, which trigger the release of oxytocin from the maternal pitutary. Oxytocin acts on the uterine muscles and induces stronger uterine contractions leading to expulsion of the baby. Relaxin hormone released by the ovary widens the vagina to facilitate birth.
Following hormones are involved in induction of parturition:

1. Cartisol
2. Estrogen
3. oxytocin.

Question 19.
In our society the women are often blamed for giving bird) to daughters. Can you explain why this is not correct?
Answer:
Women are blamed for giving birth to daughters. This is wrong because sex of the baby is determined by the sperm that can have either X or Y-chromosome. Women have only one type of chromosome (X) in all the ova.
If the sperm having X-chromosome fertillises the ovum (X), the resulting zygote (XX) will become a female.
If the sperm having Y-chromosome fertillises the ovum (X), the resulting zygote (XY) will become a male.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal?
Answer:
Only one egg is released by a human (female) ovary in a month. Only one egg is released if the mother gave birth to identical twins. Yes two or more eggs are released in case fraternal twins are born.

Question 21.
How many eggs do you think were released by the ovay of the female dog which gave birth to 6 puppies?
Answer:
Six eggs are released by the ovary of a female dog if it gave birth to six puppies.

Human Reproduction Other Important Questions and Answers

Human Reproduction Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Fertilization is related by: (CBSE PMT 2007)
(a) Realizing & gamete cells from gonad
(b) Transfer of male gametophy te to female gamete
(c) Fusion of male and female sex organs
(d) Fusion of nucleus of male gametes and nucleus of female gamete.
Answer:
(d) Fusion of nucleus of male gametes and nucleus of female gamete.

Question 2.
Cleavage is a process of fertilized egg, in which egg.
(a) Does not divide, but grows in sizes
(b) Divides continuously but does not grow in size
(c) Divided continuously and grows in size
(d) None of these.
Answer:
(b) Divides continuously but does not grow in size

Question 3.
Foetal membranes provide:
(a) Protection of embryo
(b) Nutrition to embryo
(c) Protection and nutrition to embryo
(d) None of these.
Answer:
(c) Protection and nutrition to embryo

Question 4.
Science of ageing is called:
(a) Chronology
(b) Odontology
(c) Gynecology
(d) Gerontology.
Answer:
(d) Gerontology.

Question 5.
The female counterpart of prostate gland in male (man) is:
(a) Bertholin’s gland
(b) Uterus
(c) Clitoris
(d) None of these
Answer:
(d) None of these

Question 6.
Humanis:
(a) Oviparous
(b) Viviparous
(c) Ovoviviparous
(d) None of these
Answer:
(b) Viviparous

Question 7.
Periodic cycle is:
(a) Of menstrual phase
(b) Of estrogen secretion
(c) Of fertilization
(d) None of these
Answer:
(a) Of menstrual phase

Question 8.
Ageing in mammals including man is due to:
(a) Adverse changes in environment
(b) Interaction between hereditary factors (genes) and the environment
(c) Malnutrition and stress
(d) All of these.
Answer:
(b) Interaction between hereditary factors (genes) and the environment

Question 9.
The part of the human body shows regeneration is:
(a) Spleen
(b) Kidney
(c) Brain
(d) Liver
Answer:
(d) Liver

Question 10.
Fertilization in mammals occurs in:
(a) Oviduct funnel
(b) Fallopian tubule
(c) Uterus
(d) Vagina.
Answer:
(b) Fallopian tubule

Question 11.
The capsule enclosing testes of mammal is called:
(a) Tunica albugenia
(b) Tunica membrana
(c) Tunica vaginalis
(d) Tunica vesculosa
Answer:
(a) Tunica albugenia

Question 12.
Sertoli cells are found in:
(a) Testes of Earthworm
(b) Testes of Frog
(c) Testes of Mammals
(d) Testes of Cockroach
Answer:
(a) Testes of Earthworm

Question 13.
The cavity of gastrula is known as:
(a) Blastocoel
(b) Coelome
(c) Archenteron
(d) Haemocoel
Answer:
(c) Archenteron

Question 14.
Implantation is the process in which
(a) Fertilization of egg
(b) Movemènt of egg
(c) Disappear of egg
(d) Blastosyst is formed by uterus.
Answer:
(a) Fertilization of egg

Question 15.
Seminiferous tubule is found in: ‘
(a) In testes
(b) In ovary
(c) In kidney
(d) In lungs.
Answer:
(a) In testes

2. Match the Following:

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

Answer:

1. (b)
2. (c)
3. (d)
4. (e)
5. (a).

3. Answer in One Word/Sentence:

1. How many sperms will be produced from 24 spermatocytes during spermatogenesis ?
2. Where does fertilization takes place in mammals ? (SSCE1993, CBSE 95)
3. Write the name of various phases of embryogenesis.
4. How many polar bodies are formed during the formation of one gamete?
5. Write die name of three layers of gastrula.
6. How many polar bodies are formed during the formation of ovum during oogenesis? (CBSE 1993)
7. Name the substance formed in sperms and which helps the sperms to enter into ovum. (AISB1991)
8. How many sperms and ova are formed from 100 primary spermatocytes and primary oocyte respectively ? (SSCE 1992)
9. Name the part of sperm secreting enzyme for the entrance of sperms into ovum. (SSCE 1992)
10. Name the organ where corpus luteum is formed. (SSCE 1995)
11. Write the duration of gestation period in human cases.
12. How many autosomes are found in human sperm ?
13. Where is corpus luteum formed ?
14. Name the developmental stage of man in which it is transplanted of uterus wall.(SSCE 1993)
15. Name the process of release of ovum from graafian follicle.

Answer:

1. 96,
2. Fallopian tube
3. Cleavage
4. Blastula and Gastrula Infant, 4.3
5. Ectoderm, mesoderm and endodenn
6. 2
7. Spermlysin
8. 400,100
9. Head
10. Ovary
11. 240 days
12. 22
13. Ovary
14. Blastula
15. Ovulation.

Human Reproduction Very Short Answer Type Questions

Question 1.
By which term, the process of the formation of morulla from zygote is referred?
Answer:
Cleavage.

Question 2.
What is called first menstrual stage or phase in the girls age 12-14 years?
Answer:
Menarche.

Question 3.
What is the example of spermatogenesis?
Answer:
Male gametogenesis.

Question 4.
What provides nourishment to foetus inside the uterus?
Answer:
Placenta.

Question 5.
How many germinal layers are there in gastrula stage?
Answer:
Three.

Question 6.
By which term, the middle germinal layer of gastrula is referred?
Answer:
Mesoderm.

Question 7.
After Fertilization, how many days are needed for the birth of human child?
Answer:
280 days (9 months and 10 days).

Question 8.
To produce the new organisms of their own kind is called by a term, name it
Answer:
Reproduction.

Question 9.
By which name, the reproduction happens by means of fusion of two different gametes?
Answer:
Sexual reproduction.

Question 10.
Which gametogenesis goes on till life span in human beings after attaining puberty?
Answer:
Spermatogenesis.

Question 11.
The stoppage of menstrual cycle in a 50 yrs old female is known as.
Answer:
Menopause.

Question 12.
What is pregnancy?
Answer:
The time period between fertilization to the birth of child is called pregnancy.

Question 13.
Where is carpora cavernosa found?
Answer:
Carpora cavernosa is found in penis.

Question 14.
What is the gestation period in elephant, dog and cat ?
Answer:
Elephant 641 days dog 58-68 days, cat 63 days.

Question 15.
Name the hormone that relaxes public symphysis during parturition.
Answer:
Relax in hormone.

Human Reproduction Short Answer Type Questions

Question 1.
Explain menarchy or menopause.
Answer:
Menopause: In every human female, puberty period starts from 12-13 yrs of age to 45-50 yrs of the age. During this period except pregnancy at every interval of a month during 26th day to 28th day. If pregnancy does not occur then the internal wall of the uterus secretes out mucilaginous liquid along with blood. Secretion continues for 3-4 days called menses. As it comes at definite period so, it is called menstruation cycle. At the age of 40-50 yrs. menses stop and females reach to a stage called menopause. Ability pregnancy also stops after attaining menopause.

Question 2.
Draw a well labelled diagram of the T.S. of human testis.
Answer:

Question 3.
Draw a well labelled diagram of Oogenesis.
Answer:

Question 4.
What is the position of fallopian tubule in female reproductive organs? What are their significance ?
Answer:
Fallopian tubules are a pair of small muscular tubes, one each on either side ol*the uterus. These tubules extend from the vicinity of the ovary to the ovary. Each tubule is about 10 cm in length. The free end of each tube lies near the ovary of its side. This end is funnel shaped and fimbriated. It is called ostium and infundibulum. Infundibulum opens in the abdominal cavity by means of abdominal ostium. The fallopian tubule is kept in position by a mesentery which is attached to the uterus.

Significance of Function of Fallopian Tubes:

• By their lashing movement of the cilia present in the lining of infundibulum and nearby area help in pulling the released ovum into fallopian tube.
• Passage of ovum into uterus is aided by muscular movement of fallopian tube as well as beating of cilia present in the lining layer of tube.
• Fertilization of ovum mostly takes place in the ampulla part of fallopian tube.

Human Reproduction Long Answer Type Question

Question 1.
Write differences between spermatogenesis and oogenesis
Answer:
Differences between Spermatogenesis and Oogenesis:

Question 2.
Describe the process of the formation and functions of corpus luteum.
Answer:
Corpus luteum is formed within ovary. It is a yellow body or structure which develops from ruptured Graafian follicle after the release of ovum. The cells become enlarged and conical. They are filled with a fluid called as luteum. It is developed under influence of L.H. of anterior pituitary. If fertilization do not occur it degenerates but if fertilization occurs, it remains for seven months during pregnancy and is responsible for the formation of some hormones which stimulates pregnancy and lactation.

Question 3.
Describe the structure and function of fallopian tubes.
Answer:
Fallopian tubules are a pair of small muscular tubes, one each on either side of the uterus. These tubules extend from the vicinity of the ovary to the ovary. Each tubule is about 10 cm in length. The free end of each tube lies near the ovary of its side. This end is funnel shaped and fimbriated. It is called ostium and infundibulum. Infundibulum opens in the abdominal cavity by means of abdominal ostium. The fallopian tubule is kept in position by a mesentery which is attached to the uterus.

Function of Fallopian Tubes:
By their lashing movement of the cilia present in the lining of infundi-bulum and nearby area help in pulling the released ovum into fallopian tube. ,
Passage of ovum into uterus is aided by muscular movement of fallopian tube as well as beating of cilia present in the lining layer of tube.
Fertilization of ovum mostly takes place in the ampulla part of fallopian tube.

Question 4.
In our society the women are often blamed for giving birth to daughters. Can you explain why this is not correct ?
Answer:
It is not correct to blame women for giving birth the daughters. The male sperm contains either X or Ychromosome whereas the female egg contains only x chromosomes. At the time of fertilisation when sperm carrying x chromosome combines with egg carring x chromosome of female Xygote z is formed which would be a female and when sperm with Y chromosome combines with egg containing X chromosome XY zygote is formed which would be a male. Thus scientifically sex of the baby is determined by the father and not by the mother as blamed in our society.

Question 5.
Describe the structure of ovary.
Answer:
These are a pair of female gonads or primary sex-organs lying one on each side of uterus. Ovary is attached to abdominal wall as well as uterus by means of ligaments. It is surrounded by a fold by peritoneum named mesovarium. Ovary is internally differentiated into four parts: germinal epithelium, tunica albuginea, cortex and medulla. Germinal epithelium is the outermost layer of cuboidal to flattened cells. Germinal epithelium is followed by a sheath of condensed connective tissues which is termed as tunica albuginea. It is followed by cortex. The central part of ovary contains medulla. A large number of groups of specialized cells are present in the cortex which are termed as ovarian follicles. These follicles are found in four development stages :

Incipient follicle : The central part of these follicles contain a large cell which is covered by smaller cells.
Primary follicle: These follicles are developed from incipient follicles.
Vascular follicle: It is formed from primary follicles. The oocyte of these follicle is covered by a many layered thick wall.

Graafian follicle: Mature ovarian follicle is termed as graafian follicle. It is covered by two sheaths derived from cortex. The follicle contains a single oocyte. A group of follicular cells surrounds the oocyte or ovum. It is called cumulus ovaricus. Another group produces membrane granulosa. Oocyte has two non-cellular membranes, inner vitelline membrane and outer zona peUucida.

Question 6.
Describe the process of fertilization and give its significance.
Answer:
Fertilization: Fertilization is the fusion of male and female gametes to produce a single diploid cell, called zygote. Fertilization in human female takes place in fallopian tube.

In a sexual mating or coitus the male ejaculates semen into the vaginal passage of the female using the copulatory organ, the penis. In a single coitus as many as 200 million sperms are introduced into the female genital tract but out of this huge number of sperms only one is destined to fuse with the ovum, provided the fallopian tube lodges of fully developed seqondary oocyte.Sperms travel through the vaginal passage and enter the uterus through the cervix.

They travel further through the uterus and finally enter the fallopian tube.The vaginal passage is highly acidic to prevent any bacterial infection but this acidic medium is not suitable for the survival of sperms. Many sperms die on the way. The liquid medium of the semen is alkaline which can neutralize the acidity of vagina to some extend and keep the. sperms alive and active. The sperms move with the speed of 1 -5 to 3 mm per minute.

The ovum gets surrounded by a large number of sperms but usually only one fuses with the ovum. The sperm penetrates the ovum using certain chemical substances of enzymatic nature.These chemicals are called spermlysins which are present in the acrosome. Certain receptors on the cell surface of the ovum enable the sperms to penetrate the wall of ovum. The ovum is surrounded by follicle cells. These cells are joined together by a glue like substance known as mucopolysaccharide, an acid, called hyaluronic acid. The sperm pro¬duces spermlysin, known as hyaluronidase. The over all changes in a sperm before the fertilization is called capacitation.

The hyaluronidase enzyme facilitates the sperm to pen¬etrate through the corona radiata (follicle cells), zona pellucida and the plasma membrane of ovum. The nucleus and the cytoplasmic components get inside the ovum, leaving the tail outside. The entry of sperm stimulates the ovum and the signal is transmitted to the egg surface incapacitating all the other cells surrounding the ovum. Nucleus of sperm move towards the nucleus of ovum and they fuses with each other to form zygote. It takes 2 \(\frac { 1 }{ 2 }\) hours to complete fertilization process. Once the ovum has been fertilized a barrier forms around it that normally prevents other sperms from entering.

Now fertilized egg reaches to the uterus and within seven days of fertilization it is transplanted to the wall of the uterus.

Significance of Fertilization
Egg become active after entry of sperm and completes its second maturation division.
Formation of fertilization membrane prevent entry of other sperms in the ovum. In human this membrane is not formed.
It restores the diploid number (2n) of chromosomes in the zygote.
It combines characters of male and female resulting in the introduction of variations. These variations make the offspring better equipped for struggle against environmental conditions to ensure the existence.
The ovum is stimulated to cleavage.
After fertilization ovum rotate inside the membrane.
Ovum do not contain centriole and obtain it from sperm during this process thus zygote continuously divides.
It is necessary for the egg to attain maturity.

Question 7.
Describe development of embryo up to the formation of three germ layer. Give the names of organs formed by three germ layer.
Or
Define cleavage. Describe the changes that occur in embryo till gastrulation.
Answer:
The term cleavage refers to a series of rapid mitotic divisions of the zygote following fertilization forming a many celled blastula. Following are the various steps of embryonic development in human up to the formation of three germ layer:

1. Formation of morula : The fertilized zygote divides. It undergoes successive quick mitotic cell divisions called cleavage. First cleavage is holoblastic, unequal and meridional. It divides the ovum completely into two unequal blastomeres. The plane of cleavage passes through animal vegetal axis, i.e., it is meridional. Large blastomere divides little earlier than the small one giving a transitional three cell stage. As a result of further cleavages, a solid mass of mulberry-shaped embryo is formed called morula.
Morula is of the size of zygote but consists of 32 cells. These cells are of two types : The outer layer of smaller cells around, inner larger cells. Within 72 hours of fertilization, morula reaches the uterus.

2. Formation of blastula : Transformation of morula into a blastula starts by the rearrangement of blastomeres. This leads to the formation of a central cavity inside the morula. The outer cells of morula absorb the nutritive fluid secreted by the uterine mjueous membrane and transform into trophoblast. The fluid absorbed by these cells collects in the central cavity called blastocoel. This cavity separates the trophoblast from the inner mass of larger cells except on one side and is termed blastocyst. The inner cell mass is pushed to one pole as a small knob. This knob gives rise to the embryo and is termed as embryonal knob.

3. Formation of gastrula : In this embryonic stage development of three germ layer occurs. In this stage morphogenic movement of cells of embryo occurs, as a result of this three germ layers are formed. A cavity develops at the centre called as archenteron which open outside through blastopore.

4. Formation of three germ layers :

Formation of endoderm : The enlargement of blastodermic vesicle is followed by the separation of few cells from the inner cell mass. These cells push out from the blastocoel to become the initial cells of the innermost layer of gastrula, the pattern of a tube within a tube. The inner tube is called primitive gut. It is differentiated into gut tract which is within the body and a yolk sac that communicates with the gut of the embryo. The remaining cells of the inner Cell mass are organized to form the embryonic disc.

Formation of mesoderm : After the formation of endoderm an increased rate of cell proliferation takes place at the caudal end of the embryonic disc. This results in the localised increase in the thickness of the disc. These cells subsequently get detached from the embryonic disc and get organized to a well demarcated mesodermal layer.
Formation of ectoderm: After the formation of endoderm and mesoderm, the remain¬ing cells of the embryonic disc arrange themselves in a layer to form ectoderm.

Fate of three germinal layer : Three germ layers differentiated at the time of gastrula- tion give rise to all the tissues and organs of the body of adult by the process of differentia¬tion and organogenesis.
Germ layers

1. Germ layers : Ectoderm
Tissue or organs of adult :

• Epidermis and its derivatives.
• Cutaneous sensory organs.
• Olfactory organs.
• Lens of eye.
• Membranous labyrinth (internal ear).
• Anterior lobe of pituitary gland.
• Lining of buccal cavity.
• Enamel of teeth.
• Endoderm
• Lining of proctodaeum.
• Brain, spinal cord, spinal and cranial nerves.

2. Germ layer : Endoderm
Tissue or organs of adult :

• Lining of alimentary canal except buccal cavity and rectum.
• Lining of larynx, trachea and lungs.
• Urinary bladder.
• Parathyroid, thyroid and thymus.
• Liver and pancreas.

3. Germ layer : Mesoderm
Tissue or organs of adult :

• Dermis.
• Vertebral column.
• Muscles of the body.
• Excretory organs.
• Reproductive tracts.
• Peritoneum.
• Circulatory system, heart, blood vessels, blood, lymph and spleen.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 16 Environmental Issues

Environmental Issues NCERT Textbook Question and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage are the waste originating from the kitchen, toilet, laundry and other sources. They contain impurities such as suspended solid (sand, salt, clay), colloidal materials (faecal matters, bacteria, plastic and cloth fiber), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia) and disease-causing microbes.

When organic wastes from the sewage enter the water bodies, they serve as a food source for microorganisms such as algae and bacteria. As a result, the population of these microorganisms in the water body increases. Here, they utilize most of the dissolved oxygen for their metabolism. This results in an increase in the levels of BOD in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate at home, school or during your trips to other places, could you very easily reduce. Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkin, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at school include waste paper, plastics, vegetable and fruit peels, food wrapping, sewage, etc. Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimized by writing on both sides of the paper and by using recycled paper.

Plastic and glass waste can also be reduced by recycling and re-using. Also, substituting plastics bags with biodegradable jute bags can reduce wastes generated at home, school or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities.

Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because microorganisms do not have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface due to greenhouse effect.

Causes of global warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane and water vapour. These gases trap solar radiation released back by the earth. Global warming is a result of industrialization, burning of fossil fuels, and deforestation.

Effects of global warming: Global warming is defined as an increase in the average temperature of the Earth’s surface. It has been observed that in the past three decades, the average temperature of the Earth has increased by 0-6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control measures for preventing global warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG, etc.
• Reforestation.
• Recycling of materials.

Question 4.
Match the items given in column A and B :

Answer:

1. (b)
2. (a)
3. (c)
4. (d)

Question 5.
Write critical notes on the following :
(a) Eutrophication.
(b) Biological magnification.
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication : It is the natural ageing process of lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilizers, and sewage from land which leads to an increased fertility of the lake. As a result it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting into algal blooms. Later, the decomposition of theses algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological magnification : The increase in concentration of harmful non- biodegradable substances into higher tropic level is called biological magnification. DDT used to protect the crops reach the soil and are absorbed by plants with water and minerals from the soil. Due to rain, these chemicals can also enter water sources and into the body of aquatic plants and animals. As a result, chemicals enter the food chain. Since, these chemicals cannot be decomposed, they keep on accumulating at each trophic level. The maximum concentration is accumulated at the top carnivore’s level.

The producers (phytoplankton) were found to have 0 04 ppm concentration of DDT. Since many, types of phytoplankton were eaten by zooplankton (consumers), the concentration of DDT in the bodies of zooplankton as found to be 0-23ppm. Small fish that feed on zooplankton accumulate more DDT in their body. Thus, large fish (top carnivore) that feed on several small fish have the highest concentration of DDT.

(c) Groundwater depletion and ways for its replenishment : The level of ground¬water has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers, etc.

As a result, the source of groundwater is depleting. This is because the amount or groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for replenishing groundwater:

• Preventing over-exploitation of groundwater.
• Optimizing water use and reducing water demand.
• Rainwater harvesting.
• Preventing deforestation and plantation of more trees.

Question 6.
Why ozone hole forms over Antarctica? How will enhanced ultraviolet radiations affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released ‘from Chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s release chlorine atoms by the action of UV rays on them. The release of chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion.
CF₂Cl₂ → CF₂Cl+ Cl
Cl + O₃ → ClO+O₂
ClO + O₂ → Cl+O₂
The formation of the ozone hole will result in an increased concentration of UV-B radiation on the Earth’s surface. UV-B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV-B cause corneal cataract in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.

The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the King of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and the workers went to bishnoi village. There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnoi’s showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

The Chipko movement was started in 1974, in die Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual you would take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:

Measures for preventing air pollution:

• Planting more trees.
• Use of clean and renewable energy sources such as CNG and Bio-fuels.
• Reducing the use of fossil fuels.
• Use of catalytic converters in automobiles.

Measures for preventing water pollution :

• Optimizing the use of water.
• Using kitchen waste water in gardening and other household purposes measures for controlling noise pollution:
  • Avoid burning crackers on Diwali,
  • Plantation of more trees. ‘

Measures for decreasing solid waste generation:

• Segregation of waste.
• Recycling and reuse of plastic and paper.
• Composting of biodegrable kitchen waste.
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes. ‘
(b) Defunct ships and e-wastes.
(c) Municipal solid wastes.
Answer:
(a) Radioactive wastes : Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionizing radiation such as gamma rays. These rays cause mutation in organisms which often results in skin cancer. At high dosage, these rays can be lethal. Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct ships and e-wastes : Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thu$,4hey contribute to solid wastes that are hazardous to health. E-wastes or electronic wastes generally include electronic goods such as computers, etc.

Such wastes are rich in metals such as copper, iron, silicon, gold, etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal solid wastes : Municipal solid wastes are generated from schools, offices, homes and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes and other disease-causing microbes. Hence, it is necessary to dispose municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi. Various steps have been taken to improve the quality of air in Delhi:

• Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unbumt particles.
• Phasing out of old vehicles.
• Use of unleaded petrol.
• Use of low-sulphur petrol and diesel.
• Use of catalytic converters.
• Application of stringent pollution-level norms for vehicles.
• Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.
• The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of Suspended Particulate Matter (SPM) and Respiratory
• Suspended Particulate Matter (RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Green house gases
(b) Catalytic converter
(c) Ultraviolet-B.
Answer:
(a) Green house gases: The greenhouse effect refers to an overall increases in the average temperature of the earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane and water vapour. When solar radiation reach the Earth, some of these radiation are absorbed. These absorbed radiation are released back into the atmosphere. These radiation are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival. However an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic converter: Catalytic converters are devices fitted in automobiles to reduce vehicular pollution. These devices contain expensive metals such as platinum, palladium and rhodium that act as catalysts. As the vehicular discharge passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas (respectively).

(c) Ultraviolet-B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light. It is a harmful radiation that comes from sunlight and penetrates through the ozone hole on to the earth’s surface. It induces many health hazards in humans. UV-B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV-B cause corneal cataract in human beings.

Environmental Issues Other Important Question and Answers

Environmental Issues Objective Type Questions

1. Choose the Correct Answer:

Question 1.
It is a greenhouse gas:
(a) H₂
(b) CO
(c) CO₂
(d) N₂
Answer:
(c) CO₂

Question 2.
Combustion of fissil fuel resulting in:
(a) SO₂ pollution
(b) NO₂ pollution
(c) N₂O pollution
(d) NO pollution.
Answer:
(a) SO₂ pollution

Question 3.
The causes of greenhouse effect is the increase in the concentration of:
(a) CO₂
(b) CO
(c) O₃
(d) Nitrogen oxide.
Answer:
(a) CO₂

Question 4.
Which of the following is responsible for the reduction of 03 from the atmosphere :
(a) CFC
(b) NO₂
(c) CO₂
(d) SO₂
Answer:
(a) CFC

Question 5.
Which of the following is highly dangerous radioactive pollutant:
(a) Strontium-90
(b) Phosphorus-32
(c) Sulphur-35
(d) Calcium-40.
Answer:
(a) Strontium-90

6. Causes of acid rain is:
(a) SO₂ pollutant
(b) CO
(c) Insecticide pollution
(d) Dust pollution.
Answer:
(a) SO₂ pollutant

Question 7.
Carbon monoxide is the chief pollutant of:
(a) Water
(b) Air
(c) Noise
(d) Soil.
Answer:
(b) Air

Question 8.
Plants are the purifier of air because:
(a) Respiration
(b) Photosynthesis
(c) Transpiration
(d) Drying.
Answer:
(a) Respiration

Question 9.
Bhopal gas tragedy of 1984 took place because methyl isocyanate related with:
(a) DDT
(b) Ammonia
(c) CO₂
(d) Water.
Answer:
(d) Water.

Question 10.
Maximum ozone layer depletion is caused by :
(a) CO₂
(b) CFC
(c) SO₂
(d) CH4.
Answer:
(b) CFC

Question 11.
Minimata disease is caused by the pollution of:
(a) Mercury and Lead
(b) Mercury and Cadmium
(c) Lead and Tin
(d) Lead and Strontium.
Answer:
(b) Mercury and Cadmium

Question 12.
Taj Mahal is affected by :
(a) Air pollution due to Mathura refinery
(b) Decomposition of marble
(c) Pollution of Yamuna
(d) All of the above.
Answer:
(a) Air pollution due to Mathura refinery

Question 13.
Which of the following is not a pollutant:
(a) Thermal energy plant
(b) Automatic vehicles
(c) Nuclear power energy plant
(d) Hydroelectric project.
Answer:
(d) Hydroelectric project.

Question 14.
Green house effect is increase in temperature of earth due to :
(a) High concentration of NO₂
(b) High concentration of SO₂
(c) High concentration of CO₂
(d) High concentration of CO.
Answer:
(c) High concentration of CO₂

Question 15.
Green house gases are:
(a) CO₂, CFC, CH_4, NO₂
(b) CO₂, O₂, N₂, NO₂, NH₃
(c) CH₄, N2, CO₂, NH₃
(d) CFC, CO_2, NH₃, H₂
Answer:
(d) CFC, CO₂, NH₃, H₂.

Question 16.
Which of the following is an indicator of water pollutant:
(a) BOD
(b) COD
(c) E. coli
(d) All of the above
Answer:
(a) BOD

Question 17.
Cause of water pollution is :
(a) 2,4-D and insecticides
(b) Smoke
(c) Automatic engines
(d) Aeroplanes.
Answer:
(a) 2,4-D and insecticides

Question 18.
Polluted water is treated by :
(a) Lichen
(b) Fungi
(c) Fern
(d) Phytoplankton
Answer:
(d) Phytoplankton

Question 19.
Which country produce maximum green house gas:
(a) India
(b) Brazil
(c) USA
(d) France.
Answer:
(c) USA

Question 20.
Which are not affected by acid rain :
(a) Lithosphere
(b) Plants
(c) Ozone layer
(d) Animals.
Answer:
(c) Ozone layer

Question 21.
Ozone hole causes:
(a) Global warming
(b) Reduction in the rate of photosynthesis
(c) More UV rays come to earth
(d) All of the above.
Answer:
(c) More UV rays come to earth

Question 22.
Highly polluted river of India is :
(a) Ganga
(b) Yamuna
(c) Gomti
(d) Hugh.
Answer:
(a) Ganga

Question 23.
Effect of pollution is forst marked on:
(a) Microorganism
(b) Vegetation
(c) Food crops
(d) None of the above
Answer:
(b) Vegetation

Question 24.
Ozone hole enhances:
(a) UV-radiations reaching earth
(b) Number of characters
(c) Skin cancer
(d) All of the above.
Answer:
(d) All of the above.

2. Fill in the Blanks :

1. …………………… is everything that is around us.
2. …………………. Nature of environmental study is
3. …………………. is the highest layer of the earth.
4. An interacting group of various species in a common location is called……………………..
5. …………………….. is a layer or a set of layers of gases surrounding a planet.
6. ……………………….. is electromagnetic radiation within a certain portion of the electromagnetic spectrum.
7. The is defined as a vertical section of the soil. ……………………….
8. …………………… water is available in plants easily.
9. ……………………. is essential for respiration of animals which are found in the soil.
10. …………………… is essential part of the earth for continuity of life.
11. …………………… gas is essential for life.
12. …………………….. fish is used to eat the larva of mosquitoes.
13. Quantity of atmospheric C02 is …………………… by forestation.
14. Soil pollution is caused by ……………………..
15. ………………….. rays are absorbed by ozone layer.

Answers:

1. Environment
2. Multidisciplinary
3. Biosphere
4. Bio-community
5. Atmosphere
6. Light
7. Soil profile
8. Capillary
9. Oxygen
10. Death
11. Oxygen
12. Gambusia
13. Balanced
14. Chemical fertilizers
15. Ultraviolet rays.

3. Match the Following :
(i)

Answer:

1. (c)
2. (a)
3. (d)
4. (b).

Answer:

1. (c)
2. (a)
3. (b)
4. (d).

Environmental Issues Very Short Answer Type Questions

Question 1.
The surroundings in which a person, animal or plant lives are called.
Answer:
Environment.

Question 2.
When two individuals can produce fertile offsprings, typically by sexual reproduction is called.
Answer:
Species.

Question 3.
What called, a biological community of interacting organisms and their physical environment?
Answer:
Ecosystem.

Question 4.
Name the lower part of the atmosphere which has found more than 90% gases.
Answer:
Troposphere,

Question 5.
Rays of light which are low wavelength from purple colour rays?
Answer:
Ultraviolet rays.

Question 6.
Light effect of flowering in plants.
Answer:
Photoperiodism.

Question 7.
Smallest particle which are found in soil.
Answer:
Clay.

Question 8.
Water which are absorbed from the atmosphere and held very tightly by the soil particles.
Answer:
Hygroscopic water.

Question 9.
Plants are one marked by short life-cycle.
Answer:
Ephemeral plant.

Question 10.
When an animal or plant resembles another creature or in animate object?
Answer:
Mimicry.

Question 11.
Expand BOD.
Answer:
Biological Oxygen Demand.

Question 12.
Name two gases which are caused air pollution.
Answer:
SO₂ and CO₂.

Question 13.
How much is hearing capacity of man?
Answer:
10 to 12 decibels.

Question 14.
Write the name of highly polluted river of India.
Ans.
Ganga.

Question 15.
What is called the increase in atmospheric temperature due to increased C02 concentration?
Answer:
Greenhouse effect.

Question 16.
Write the full form of DDT.
Answer:
Dichloro Diphenyl Trichloroethane.

Question 17.
How much quantity of CO₂ in atmosphere.
Answer:
0 03%.

Question 18.
Smoke which is caused eye irritation.
Answer:
No.

Question 19.
When the World Environment Day is celebrated?
Answer:
5th June.

Question 20.
Write the full form of PAN.
Answer:
Peroxy Acetyle Nitrate.

Question 21.
Write the full form of CFCs.
Answer:
Chlorofluourocarbons.

Question 22.
How much percent found of CO₂ in atmospheric temperature?
Answer:
60 %.

Environmental Issues Short Answer Type Questions

Question 1.
What is the definition of pollution?
Answer:
Pollution is defined as an undesirable change in physical, chemical or biological characteristics of air, land, water and soil.

Question 2.
What are the cause of air pollution?
Answer:
The causes of air pollution: The main causes of air pollutions are fossil fuels industries, factories and particulate matter are produced due to combustion of petrol, diesel, kerosene etc. in vehicles, houses and factories which pollute air.

In thermal power plants, steel and glass industries, paper and sugar mills etc. combustion of coal and furnace oil produce carbon monoxide, carbon dioxide, sulphur dioxide, ash, dust particles and some heavy metals which are released into atmosphere. Similarly various types of chemicals released from cloth mills, cement industries, asbesters industries, pesticides, industries etc. also causes air pollution.

Question 3.
What is acid rain? Write any two effects of acid rain on human beings.
Ans.
Acid rain: SO₂ gas is released into the environment by burning of fossil fuels containing sulphide from industries. e.g., Coal. In presence of moisture SO₂ reacts with water and forms the droplets of sulphurous and sulphuric acid in the environment. Likewise nitrogen oxides released from motor vehicles, burning materials and chemical industries . also react with water to produce the droplets of nitric acid. The droplets of sulphurous, sulphuric acid and nitric acids fall down on the earth with rainwater and thus, it is known as acid rain.
SO₂ + \(\frac { 1 }{ 2 }\)O₂ → SO₃
H₂O +SO₃ →H₂SO₄
H₂O + NO₂ → HNO₃
Effects on human beings: Skin diseases, irritation, metabolic diseases.

Question 4.
Write short note on pollution caused due to combustion activities.
Answer:
Combustion of fuel, such as wood, coal, natural gas for cooking and for some other purposes gives out gases like CO₂, CO, SO₂ and consumes oxygen from the air.

Question 5.
Write the effect of air pollution in plants.
Answer:
Effect of air pollution on plants :

• Increase in the concentration of SO₂ causes chlorosis of leaves in plants.
• The cells and chlorophyll of leaves are degraded and fall down.
• Vegetative and reproductive growth is inhibited.
• The development of plant is inhibited.

Question 6.
Write preventive measures to control air pollution.
Answer:
Measures for the control of air pollution :

• Industrial smokes must be filtered before releasing it into the atmosphere.
• Tree plantation should be increased and deforestation prevented.
• Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
• The use of crude fuels should be avoided and use of high quality fuels should be recommended.
• Nuclear explosions should be avoided.
• Legal control of air pollution.
• Plantation of air purifying plants.
• Development of parks and gardens in cities.

Question 7.
What is global warming? List four strategies for reducing global warming.
Answer:
Increase in the level of green house gases in the atmosphere causes the rise in global mean temperature called global warming. Four strategies for reducing global warming are:

1. Reducing deforestation.
2. Planting trees.
3. Slowing down the growth of human population.
4. Reduction in the emission of greenhouse gases.

Question 8.
Write any four measures to control green house effect
Answer:
Following important measures can be sited as the base steps towards controlling greenhouse effect:

• The aim is achieved to some extent by reducing the consumption of fossil fuels such as coal and petroleum. This can be achieved by depending more on non-conventional renewable sources of energy such as solar, wind, tidal, biogas and nuclear energies.
• Disposing of the greenhouse gases as they are formed elsewhere than in the atmosphere.
• By recovering greenhouse gases present already in the atmosphere and disposing them off elsewhere.
• Preventation of deforestation and planting of more trees.

Question 9.
Write short note :

1. Bio-magnification.
2. UV rays.
3. Bio digradable pollutants.
4. Non-biodegradable.

Answer:
1. Bio-magnification : Bio-magnification, also known as bio-amplification or biological magnification is the increasing concentration of a substance, such as a toxic chemical, in the tissues of organisms at successively higher levels in a food chain.

2. UV rays : Sunlight is the chief source of ultraviolet rays. The rays of sunlight having a wavelength range from 200 to 390 nm are called as UV rays.

Effect of UV rays: They have a direct effect on living cells. DNA and other chemical substances in cells are inactivated due to UV rays. Prevention of replication of the DNA molecule and its distortion cause many ill effects.

3. Non-biodegradable: The pollutants which cannot be purified by natural methods are called non-biodegradable pollutants. Plastic products, many chemicals, DDT, long chain detergents, glass aluminium, mercury salt and other synthetic products manufactured by man come under this category. These non-biodegradable pollutants not only accumulate but often “biologically magnified” as they move in biochemical cycles and along food chains.

4. Biodegradable: The pollutants which are degraded by natural factors and decomposed by natural activities are known as biodegradable pollutants, e.g., Domestic sewage, heat. The domestic sewage can be rapidly decomposed by natural processes or in engineered system (sewage treatment plant) that enhance natures great capacity to decompose and recycle.

Question 10.
What is the effect of SOz in the Environment?
Answer:

Question 11.
Write only the sources of water and air pollution.
Answer:
1. Sources of Water pollution: Following are the chief sources of water pollution:
(a) Human sources :

• Domestic sewage,
• Industrial effluents,
• Agricultural wastes,
• Oil pollution,
• Thermal and nuclear power station.

(b) Natural sources: Water pollution takes place by natural ways like soil erosion, mixing of metallic substances, plant leaves, humus and faecal matter of animals etc.

2. Sources of Air pollution: Following are the chief sources of air pollution:
(a) Human sources:

• Combustion activities,
• Industrial activities,
• Agricultural works,
• Use of solvents,
• Activities concerned with atomic energy.

(b) Natural sources: Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

Question 12.
Write a note on the effect of water pollution on aquatic organisms.
Answer:
Effect of water pollution on aquatic plants:

• Inorganic nitrates and phosphates in excess amount stimulates excessive plant growth in lakes and reservoirs. These plants deplete the oxygen contents of the water during night. This leads to suffocation of fish and other aquatic life.
• The rapid algal growth leads to the diminishing of nutrient in the medium causing rapid decay of algal filament. This increases productivity of lake and stream water brought about by nutrient enrichment is called eutrophication.
• The number of microorganisms increases in polluted water.
• Siltation occurs in water.
• Temperature of water increases and 02 ratio is reduced.

Effect of water pollution on aquatic animals: Animal life depends on aquatic plants. Aquatic animals are affected from water pollution as follows :

• Decrease in the value of B.O.D. resulting in the number of aquantic animals.
• Animals found in freshwater and killed.
• Animal diversity is also decreased and fishes suffer from various types of diseases.
• Terrestrial animals are also affected by the use of polluted water.

Question 13.
Write the measures for control and prevention of water pollution.
Answer:
Measures for control and prevention of water pollution are:
Preventive measures:

• Use of harmful pesticides and weedicides must be stopped completely.
• Discharge of effluents into rivers, lakes and sea should be strictly prohibited without treatment.
• Oil spill should be prevented.
• Proper disposal of sewage so, that it does not find its way into water bodies.
• Preventing bathing, washing cloths, throwing dead bodies and other wastes into water source.

Curative measures:
1. Adequate waste water treatment: The domestic sewage and the industrial waste should be properly treated before its disposal into water ways.
Waste water treatment involves three steps :

(a) Primary treatment: During this treatment large objects and suspended undissolved solids are removed and converted into the sludge, a valuable fertilizer.

(b) Secondary treatment: Aeration is supplied to promote bacterial decomposition, followed by chlorination to reduce its content of bacteria.

(c) Tertiary treatment: During this phase nitrates and phosphates are removed. The treated water is then released. Sewage treatment is quite expensive and in many developing countries only the first two steps are followed.

2. Treatment of industrial effluents: Industrial effluents should be properly treated to remove the pollutants. These involve neutralization of acids and alkalies, removal of harmful chemicals, coagulation of colloidal impurities, precipitation of metallic compounds and reducing the temperature of wastes to decrease thermal pollution. Chemical oxidation can be achieved by chlorination through reaction with ozone. However, there are certain chemicals which are difficult to remove.

3. Recycling: One of the best methods of prevention and control of water pollution is the recycling of the various kinds of pollutants and wastes, e.g., Dung of cow and buffalo can be used for the production of gobar gas a cheap source of fuel and also as manure.

Question 14.
Write short notes on air pollution caused due to industries.
Answer:
Industrial pollution is caused by industrial pollutants which are released by the industries such as SO_2, CO₂ , CO, H₂S and hydrocarbons together with dust and smoke. These are produced by the burning of coal and petroleum.

The chemical industries releases HCl, Cl₂, nitrogen oxide, oxides of copper, zinc, lead, arsenic etc. Industrial pollutants causes air pollution and water pollution.

Question 15.
Write preventive measures of sound pollution.
Answer:
Preventive measures of noise pollution:

• Noise pollution can be controlled by designing quieter machines, proper lubrication and better maintenance of machines. Noise producing parts of machines may be covered by suitable insulating materials.
• Noise producing industries should be installed away from residential areas.
• Use of public addressing systems should be restricted at a fixed intensity and hours of the day.
• Stuffing of cotton plugs in the ear or use of ear-muffs and minimize the danger of occupational exposure of noise.
• Trees absorb sound vibrations and thus, reduce the noise pollution. Thus, plantation of trees along highways, establishment of parks help in reducing noise pollution.
• By use of silencer in all automobiles and engines.

Question 16.
Write the effects of noise pollution.
Answer:
Effects of Noise pollution :

• The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
• Interference with speech communication.
• Noise pollution leads to neurosis, anxiety hypertension, cardiovascular disease, hepatic stress, giddiness.
• Annoyance leading to ill-temper, bickering, mental disorientation and violent behaviour.
• The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be short into blood stream with conse¬quence rise in blood pressure. This ultimately results in tension and nervousness.
• Affects different metabolic activities.

Question 17.
How many types of pollution are there ? Explain in brief.
Answer:
Pollution are of following five types :

1. Air pollution,
2. Noise pollution,
3. Water pollution,
4. Radioactive pollution,
5. Soil pollution.

1. Air pollution: Pyrotechnics’ results in the release of harmful gases like CO₂, CO and SO₂ in the environment which produces various types of diseases in living beings.

2. Noise pollution: Pyrotechnics produces the unwanted sound dumped into the atmosphere leading to health hazards.

3. Water pollution: Pyrotechnics produces some waste materials, on reaching water it causes water pollution.

4. Radioactive pollution: Pollution created due to radioactive substance is called radioactive pollution. Atomic energy centres and bombarding also causes radioactive pollution.

5. Soil pollution: Any change in physical and chemical characteristic of soil due to addition of unwanted substance which adversely affects productivity of soil it termed as soil pollution.

Environmental Issues Long Answer Type Questions

Question 1.
Describe the diseases caused due to radioactive pollution.
Answer:
Following diseases are caused by radioactive pollution:

• Leukaemia and bone cancer: In human beings and other animals, radioactive pollution causes blood and bone cancer.
• Ageing: The reproductive capacity of organisms is decreased due to radioactive
  pollution and ultimately causes ageing.
• Epidemic disease: Radioactive pollution causes the decreased rate of antitoxin production which affects the immune system and results in the production of epidemic diseases.
• Nervous system and sensory cells become irritated.
• Radioactive pollution results skin cancer.
• Mutations also take place due to radioactive pollution.
• Thyroid cancer is also produced due to radioactive iodine.

Question 2.
Burning of crackers spread pollution in the environment. Explain.
Answer:
Explosive substances are used in crackers. Crackers are burned for celebration but it causes many harms to the living organisms. Some of the harm effects of burning of crackers are as follows :

1. Air pollution: It releases harmful gases such as CO₂, CO, SO₂ etc. into the atmosphere. Which causes respiratory diseases.

2. Noise pollution: Burning of crackers produces noise which causes noise pollution which effects our hearing capacity, causes headache, increases heartbeat and blood pressure.

3. Water pollution: Waste of burning of crackers are drained into the pond, river with rainwater or through drain which causes harm to aquatic organisms.

4. Soil pollution: Burning of crackers not only produces smoke of harmful gases but its harmful chemicals mixes with soil and pollute it, which makes the soil sterile and reduced growth of plants.

Question 3.
Write an essay on air pollution.
Answer:
Air pollution : Any change in the composition of air is called pollution. The chief reason of air pollution is the releasing of harmful gas in the environment that affectdur eyes, lungs, skins, heart and brain and’produce various disorders in them. Sources of Air polution:

1. Sources of Water pollution: Following are the chief sources of water pollution:
(a) Human sources :

• Domestic sewage,
• Industrial effluents,
• Agricultural wastes,
• Oil pollution,
• Thermal and nuclear power station.

(b) Natural sources: Water pollution takes place by natural ways like soil erosion, mixing of metallic substances, plant leaves, humus and faecal matter of animals etc.

2. Sources of Air pollution: Following are the chief sources of air pollution:
(a) Human sources:

• Combustion activities,
• Industrial activities,
• Agricultural works,
• Use of solvents,
• Activities concerned with atomic energy.

(b) Natural sources: Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

(A) Effects of air pollution :

• Air pollutant like SO₂ enter the soft tissues causing drying of the mouth, scratching throat and smarting eyes.
• Hydrocarbons and many other pollutants are responsible for causing cancer.
• Oxides of carbon, sulphur and nitrogen diffuse into the blood stream to combine with haemoglobin causing reduction in its oxygen carrying capacity. CO severely damages cardiovascular system and disturbs psychometry functions.
• P.A.N. inhibits Hill reaction and thus, decreases photosynthetic production of an ecosystem.

Control of Air pollution:  Measures for the control of air pollution :

• Industrial smokes must be filtered before releasing it into the atmosphere.
• Tree plantation should be increased and deforestation prevented.
• Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
• The use of crude fuels should be avoided and use of high quality fuels should be recommended.
• Nuclear explosions should be avoided.
• Legal control of air pollution.
• Plantation of air purifying plants.
• Development of parks and gardens in cities.

Question 4.
Name the major air pollutants and their individual effects.
Answer:
Major air pollutants and their individual effects are as follows :

1. Carbon monoxide: It is highly poisonous gas. On entering blood stream, it combines with hemoglobin to block oxygen transport function. It causes laziness, headache, disturbance of psychometry functions, decrease in visual perception, serious effects on cardiovascular system etc.

2. Sulphur dioxide: The animals and human population when exposed to SO₂, suffer from respiratory diseases. It causes chest constriction, headache, vomiting and death from respiratory ailments.

3. Hydrogen sulphide: Its rotten egg like smell cause nausea, irritation in eyes and throat.

4. Nitrogen oxide: It inhibits cilia action so that soot and dust penetrate far into the lungs and finally resulting respiratory diseases in human beings and animals. In plants, it causing chlorosis.

5. Aerosoles: These are the chemicals affecting ozone layer of the atmosphere due to which UV rays enter in the atmosphere and causing various diseases in plants and animals.

6. Ammonia : It inflame upper respiratory passage in human beings. In plants, it inhibits seed germination, destruction of chloroplast and inhibition of the growth of roots and shoots.

7. Hydrogen chloride: It affects the leaves of the plants, eyes, respiratory organs of animals and human beings.

8. Hydrocarbons: It causes yellowing of plants, drying of buds, whereas in man and other animals, mucous glands of eyes and nose are irritated.

Question 5.
Write down the methods to control soil pollution.
Answer:
Control of Soil pollution: By following methods soil pollution can be prevented:

• Government should make provisions for latrines and people should not be allowed to litter in open fields.
• Solid wastes such as tin, copper, iron, glass etc., should not be dumped into soil.
• The solid wastes should be recycled to form new materials.
• Fertilizers and pesticides should be judiciously used.
• Instead of pesticides to kill pests, biological control methods should be adopted-*
• The excretory waste of man an’d cattle should be used to prepare biogas.
• Clean and well covered dustbins should be used in villages and cities.
• Closed hollow pipes should be used to collect and discharge liquid wastes.
• To prevent soil erosion, grass and small plants should be grown.
• Soil management should be adopted.

Question 6.
What are the sources of soil pollution?
Answer:
Sources of soil pollution:

• Acid rainwater and water from mines are main sources of soil pollution.
• Mixing of debris, waste products with the soil cause soil pollution.
• By excessive use of pesticides and fertilizers, pollute the soil.
• When industrial wastes are discharged in the soil, they also pollute the soil.
• Heavy metals e.g., Cadmium, Zinc, Nickel, Arsenic etc., are mixed with the soil
  from mines. These metals are harmful for plants as well as primary and secondary consumers, in a food chain. ’
• Bones of dead animals, paper putrified flesh and food, iron, copper, mercury etc., pollute the soil.
• In our villages and rural areas, where there are no latrines, people go in fields for being at ease. From their stool also the soil is polluted.
• Insecticides like D.D.T. is very dangerous substance. When these enter the body of consumers from producers, their concentration is increased because these are non-degradable substances. Moreover, these can remain in the atmosphere for upto 15.years.

Question 7.
Explain non-ionising and ionising radiation.
Answer:
Types of Radiations on the basis of their action on cells, radiations are of two types:
1. Non-ionising radiations: These include ultraviolet rays (UV rays; 100-300 nm), these have low penetration capacity. These affect only those cells which absorb them. The cause are as following :

• Sunburn: It includes rupturing of sub-cutaneous blood capillaries, blisters, reddening of skin and injury to stratum germinativum.
• Snow blindness: It damages eyesight due to damage of corneal cells.
• Inactivation of organic bio-molecules arid formation of thymin dimer in DNA.
• However they cause cancer, tumor, skin disease, etc.

2. Ionizing radiations : These include X-rays, cosmic rays and atomic radiations. These radiations have low wavelengths but high penetration power. These damage the living cells shifting the electron from one to other bio-molecule. Their harmful effects are as follows :

• Short range effects: These may appear within few days or a few weeks after exposure. The effects include loss of hair, nails, subcutaneous bleeding, metabolic changes, change in proportion of blood cells, dead tissues or death of the organisms inliigh dose.
• Long range effects: These appear in several months or even several years after their exposure. These cause tumours, cancers, mutations, genetic deformities, shorter life span and developmental defects.

Question 8.
What are sources of noise ’ pollution? Write any four effects of noise pollution.
Answer:
Sources of noise pollution: Noise is either natural such as thunder or man-made. The common sources of noise pollution are :

• Industries such as textile mills, printing-press, engineering establishments, etc.
• Defence equipment and vehicles such as tanks, artillery and rocket launching explosions, practise firing, etc.
• Transport vehicles as trains, trucks, buses, two-wheelers, jet planes, etc. mid accessory noise produced by horns, sirens.
• Other applications as dynamite blasting, use of jack hammers, pile drivers, bulldozer, lawn mowers, etc.

Effects of noise pollution:

• The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
• Interference with speech communication.
• Noise pollution leads to neurosis, anxiety, hypertension, cardiovascular disease, hepatic stress, giddiness.
• Annoyance leading to ill-temper, bickering, mental disorientation and violent behaviour.
• The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be shot into blood stream with consequence rise in blood pressure. This ultimately results in tension and nervousness.
• Affects different metabolic activities.

Question 9.
What is Green house effect? Write any four impacts of Green house effect
Answer:
Green house effect: The Green house effect may therefore, be defined as “The progressive warming up of the earth’s surface due to blanketting effect of man-made CO₂ in the atmosphere. It means the excessive presence of those gases blocked in the infrared radiation from earth’s surface to the atmosphere leading to an increase in temperature, which in turn would make life difficult on earth to forthcoming future generations.
Four impacts of Green house effects are :

1. Effect on global climate : The climate of earth has never been free of change. Its composition, temperature and self-cleansing ability have all varied since, the planet first formed. Yet the pace in the past two countries has been remarkable. The atmosphere’s composition in particular has changed significantly faster than it has at any time in human history.

Increase in Green house gases does not increase die temperature of the earth i.e„ it is not uniform. Temperature is normal at poles and is very less at tropics. In Iceland, Greenland, Sweden, Norway, Finland, Alaska, Siberia, temperature is more, thus at poles, ice starts melting.

2. Effect of Green house Effect on forests: When the atmospheric temperature increases, only those plants can survive which can bear high temperature. Because of this, new species of plant also come into existence, Herbaceous plant can not survive in this high temperature. Plants with high wood Will increase in number. According to a survey if the a nount of CO₂ in the atmosphere is doubled, there will be a great decrease in a green biomass.

3. Effect on crops: The positive aspect of greenhouse effect will be on agriculture. Because, CO₂ is a natural fertilizer, the plants will grow larger and faster with increasing CO₂ in the atmosphere. At first sight, the abnormal growth of plants might be expected to be beneficial because the yields of major crops might increase, however chances of depletion of characters and productivity of soil are also associated with this.

4. Effect on ozone layer : During depletion, the chlorine, fluorine, or bromine molecules of CFCs and halogens are converted into reactive free radical form by photochemical reactions from their intial non-reactive sites. Oxides of nitrogen generally inactivate Cl but the lowering of stratospheric temperature changes. NO₂ into non-reactive ritric acid. Thus, Cl or F are free to react with ozone, distintegrating it into O₂ + O.

The tiny ice particles during winter favours the conversion of chlorine into chlorine monoxide, which behaves as a catalytic compound. The abundance of chloromonoxide rich air in stratosphere continues to rise. Chlorine monoxide react with nascent oxygen and thereby gets converted into chlorine. Thus, the cycle continues destroying the ozone-level.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 2 Sexual Reproduction in Flowering Plants

Sexual Reproduction in Flowering Plants NCERT Textbook Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Answer:
Development of male gametophyte takes place in anther and female gametophyte in ovary.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structure formed at the end of these two events.
Answer:
Differences between Microsporogenesis and Megasporogenesis:

Meiosis Occurs during micro and megasporogenesis microspores (Pollen grain) are formed at the end of micro sporogenesis and female gametophyte (enibro sac) are formed at the end of megasporogenesis.

Question 3.
Arrange the following terms in the correct developmental sequence:
Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male garnets.
Answer:
Sporogenous tissue → Pollen mother cell → Microspore → Tetrad → Pollen grain → Male garnets.

Question 4.
With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.
Answer:
Structure of Ovule: Each ovule consists of the following parts as visible in a longitudinal section:

• • A small stalk or funicle by which the ovule remains attached with the placenta of the
    ovary.
  • Hilum is the point at which it is attached with the ovule. In inverted ovule the funicle fuses with the main body of the ovule and is called as raphe.
  • The ovule is surrounded on all sides by two integuments but not at the apex where an aperture called micropyle is present. This end of the ovule is called as micropylar, while the end of the ovule opposite to it is called as chalazal end.
  • Embryo sac is situated inside the nucellus.
  • Towards the micropyle end of embryo sac one egg or oospore and two synergids are found, and towards the chalaza end of embryo sac three antipodal cells are found. At the centre secondary nuclei is found.
    

Question 5.
What is mean by monosporic development of a female gametophyte ?
Answer:
Out of theTour megaspores, three degenerate and only one remains funtional which develops in, to a female gametophyte or embryo sac. This is called monosporic development, i.e., when embryosac develops from one single megaspore it is called monosporic embryo sac.

Question 6.
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
Explanation: Nucleus of the functional megaspore undergoes mitosis resulting in 2-nuclei that move to two opposite poles forming 2-nucleate embryo sac.
Two more mitotic nuclear divisions result in 4-nucleate and later 8-nucleate stages of the embryo sac. so, far no cytokinesis (cytoplasmic division) has taken place.
Now cell walls start to build leading to the organisation of female gametophyte or embryo sac.

Six of the 8-nuclei are bound by cell wall and the remaining 2-celled polar nuclei lie below the egg apparatus in the large central cell.
Seven-celled stage
Three cells are grouped together at the micropylar end and constitute the egg appara¬tus, which is constituted of two synergids and one egg cell.
Three cells at the chalazal end are called antipodals. The large central cell has 2-polar nuclei.
Thus, a typical angiosperm embryo sac at maturity is 7-celled but 8-nucleated as the central cell has 2-nuclei

Question 7.
What are chasmogamous flowers? Can cross pollination occur in cleistogamous flower ? Give resion for your answer.
Answer:
Chasmogamous flowers are open flowers with exposed stamens and stigma which facilitate cross pollination.
No cross pollination occurs in cleistogamous flowers. As these flowers are closed and never open and thus no transfer of pollen from outside to stigma of the flower is possible. So there are no occurs cross pollination.

Question 8.
Mention two strategies evolved to prevent self-pollination in flowers.
Answer:
Two strategies evolved to prevent self-pollination are :
(i) Pollen release and stigma receptivity is not synchronized.
(ii) Anthers and stigma are placed at such positions that pollen doesn’t reach stigma.

Question 9.
What is self-incompatibility ? Why does self-pollination not lead to seed forma¬tion in self-incompatible species ?
Answer:
Self-incompatibility is a genetic mechanism to prevent self pollen from fertilizing the ovules by inhibiting pollen germination or pollen tube growth in the pistil.
In these cases, self-pollination does not lead to seed formation because fertilization is inhibited.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Answer:
It is the covering of female plant with butter paper or polythene to avoid their contamination from foreign pollens during breeding programme.

Question 11.
What is triple fusion? Where and how does it take place ? Name the nuclei involved in triple fusion?
Answer:
Triple fusion refers to the process of fusion of three haploid nuclei. It takes place in the embryo sac.
The 3-nuclei that fuse together are, nucleus of the male gamete and 2-polar nuclei of the central cell to produce a triploid primary endosperm nucleus.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilized ovule?
Answer:
The zygote is dormant in fertilized ovule for sometime because at this time, endosperm needs to develop. As endosperm is the source of nutrition for the developing embryo, the nature ensures the formation of enouth endosperm tissue before starting the process of embryogenesis.

Question 13.
Differentiate between:
(a) Hypocotyl and Epicotyl
(b) Colcoptile and Coleorhiza
(c) Integument and Testa
(d) Perisperm and Pericarp.
Answer:
(a) Differences between Hypocotyl and Epicotyl:

(b) Differences between Coleoptile and Coleorhiza:

(c) Differences Between Integement and Testa:

(d) Differences between Peris perm and Pericarp:

Question 14.
Why is apple called a false fruit? Which parts of the flower forms the fruit?
Answer:
Apple is called a false fruit because it develops from the thalamus instead of ovary (thalamus is the enlarged structure at the the base of the flower).

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Answer:
Emasculation means removal of anthers, with a forceps, from the flower bud before dehiscence.
Plant breeder employs this technique to prevent contamination of stigma with the undesired pollen. This is useful in artificial hybridisation, where desired pollen is required.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?
Answer:
Oranges, lemons, lichis could be potential fruits for inducing the parthenocarpy because seedless variety of these fruits would be much appreciated by the consumers.

Question 17.
Explain the role of tape turn in the formation of pollengrain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It produces the exine layer of the pollen grains, which is composed of the sporopollenin, the most resistant fatty, substance.
During microsporongenesis, the cells of tapetum produce various enzymes, hormones, amino acids and other nutritious material required for the development of pollen grains.

Question 18.
What is apomixis and what is its importance?
Answer:
Apomixis is the process of asexual production of seeds, without fertilisation.
The plants that grow from these seeds are identical to the mother plant.

Uses :

• It is a cost effective method for producing seeds.
• It has great use for plant breeding when specific traits of a plant have to be pre¬served.

Sexual Reproduction in Flowering Plants Other Important Questions and Answers

Sexual Reproduction in Flowering Plants Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Fertilization process is:
(a) Fusion of male gamete with egg
(b) Transference of pollengrains to stigma
(c) Fusion of polar nucleus with male gamete
(d) Formation of seed from ovule.
Answer:
(a) Fusion of male gamete with egg

Question 2.
Female gametophy te of angiosperms is:
(a) Megaspore mother cell
(b) Ovule
(c) Endosperm
(d) Nucellus.
Answer:
(c) Endosperm

Question 3.
Effect of nector demonstrate by Xenia word on:
(a) Morphological tissue
(b) Root
(c) Flower
(d) Endosperm.
Answer:
(d) Endosperm.

Question 4.
In a diploid flowering plant, no. of chromosomes are 12, then 6 chromosomes will be present in:
(a) Cotyledons cells
(b) Endospermic cells
(c) Synergids cells
(d) Leaf cells.
Answer:
(c) Synergids cells

Question 5.
Seed coats of seed after fertilization develop from:
(a) Integuments
(b) Endosperms
(c) Chalaza
(d) Ovule.
Answer:
(a) Integuments

Question 6.
Perispermis:
(a) Degenerative secondary nucleus
(b)Apomixis
(c) Peripheral part of endosperm
(d) Parthenogenesis.
Answer:
(b)Apomixis

Question 7.
Development of ovary into fruit from ovule without fertilization, the process is called:
(a) Parthenio ovule
(b) Apomixis
(c) Pathenocarpy
(d) Parthenogenesis.
Answer:
(c) Pathenocarpy

Question 8.
Mitotic cell division in an ovule takes place in:
(a) Nucellus
(b) Megaspore mother cell
(c) Macrosporangia
(d) Archisporium.
Answer:
(b) Megaspore mother cell

Question 9.
Development of sporophyte without the fusion of a gamete is called:
(a) Apomixis
(b) Apospory
(c) Apogamy
(d) Pollination.
Answer:
(c) Apogamy

Question 10.
Male gametes are found in angiospermic plants :
(a) In pollengrain
(b) In anther
(c) In ovule
(d) All of these
Answer:
(a) In pollengrain

Question 11.
The function of pollentube is:
(a) Help in pollination
(b) Protection of stigma
(c) Male gamete carrier
(d) All of these
Answer:
(c) Male gamete carrier

Question 12.
Tapetum is a part of:
(a) Male gametophyte
(b) Female gametophyte
(c) Ovary wall
(d) Anther wall.
Answer:
(d) Anther wall.

Question 13.
Cotyledons is situated in monocots:
(a) Axial
(b) Terminal
(c) Basal
(d) In Any Position
Answer:
(a) Axial

Question 14.
Which type of pollination found in sunflower:
(a) Self pollination
(b) Cross pollination
(c) All of the above
(d) None of these.
Answer:
(c) All of the above

Question 15.
Endospermic nucleus are :
(a) Haploid
(b) Diploid
(c)Triploid
(d) None of these.
Answer:
(b) Diploid

Question 16.
Double fertilization is discoverd by: :
(a) By Navaschin
(b) Leuwenhock
(c) Strasburger
(d) Hofmeister.
Answer:
(a) By Navaschin

Question 17.
The type of pollination in which genetically different pollen grains are brought to
stigma is : (AMU 2012)
(a) Xenogamy
(b) Geitonogamy
(c) Cleistogamy
(d) Dichogamy. ,
Answer:
(a) Xenogamy

Question 18.
If the number of chromosome in root cell is 14 then what will be the number of chromosome in synergid cell of in ovule of that parent (BHU2012)
(a) 14
(b) 21
(c) 7
(d) 28.
Answer:
(c)7

Question 19.
Presence of many embryo (Polyembryony) is characters tic feature of: (BHU2012)
(a) Citrus
(b) Mango
(c) Banana
(d) None of these.
Answer:
(a) Citrus

Question 20.
Sporopollenin is formed by the polymerisation of : (BHU2012)
(a) Fat and Phenol
(b) Carotenoid and Fat
(c) Fat and Ester
(d) Carotenoid and Ester.
Answer:
(a) Fat and Phenol

2. Fill in the Blanks:

1. The main function of endosperm in a seed is the storage of ………………
2. Pollination takes place by the agency of bird is called ………………
3. Apple is an example of fruit. ………………
4. Tissue on which ovules are attached is called ………………
5. ……………… types stamens are found in sunflower.
6. In plants development of fruit takes place from ………………
7. One cotyledon of maize is called ………………
8. Tetradynamous stamens are found in ………………
9. ………………is thin in endospermic seed.
10. In ovule …………….. is joined with funiculus.
11. Anther is ……………… in angiospermic plants.
12. ……………… type of pollination are found in Typha.
13. ……………… pollination is occurs in Salvia.
14. Outer wall of pollen grain is called ………………
15. Process of seedless fruit is called ………………

Answer:

1. Food material
2. Omithophilly
3. False
4. Placenta
5. Syngenecious
6. Ovary
7. Scutellum
8. Mustard
9. Cell wall
10. Hylum
11. Reproductive organ
12. Joint
13. Insect
14. Exine
15. Parthenocarpy.

3. Match the Following:

Answer:

1. (e)
2. (d)
3. (b)
4. (a)
5. (c)

4. Answer in One Word/Sentence:

1. Name the four whorls of flower.
2. Give the term for plants which produce flower and fruits many times.
3. Name the outer coat of the seed.
4. Give example of plant in which flowers are pollinated by bat.
5. Give the term used for similar calyx and corolla.
6. What parts of the plant produce seed and fruit after fertilization?
7. Name the type of pollination carried out of bat.
8. Give the term for fruit development without fertilization.

Answer:

1. Calyx, Corolla, Androecium and Gynoecium
2. Perrenial plants
3. Seed coat
4. Kadamb
5. Petalloid
6. Ovary
7. Chiropteriphilly
8. Parthenocarpy.

Sexual Reproduction in Flowering Plants Very Short Answer Type Questions

Question 1.
Development of female gametophyte occurs in which cell?
Answer:
By functional megaspore mother cell.

Question 2.
What is the other name of female gametophyte?
Answer:
Embryo sac.

Question 3.
Ovule derives nourishment from which part of the carpel?
Answer:
Ovule derives nourishment from placenta.

Question 4.
What are cleistogamous flowers ? Give an example.
Answer:
The flowers which do not open are called cleistogamous flowers, e.g. Commelina

Question 5.
What is the substance found on the exine of pollen grains?
Answer:
Sparopollenin.

Question 6.
Give the characters of wind pollinated flowers.
Answer:
White in colour, small in size and pollengrains are formed in large number.

Question 7.
What is the ploidy of angiospermic endosperm?
Answer:
Triploid.

Question 8.
Give an example of a monocotyledonous endospermic seed.
Answer:
Ricinus.

Question 9.
Give example of two false fruit.
Answer:
Apple, Jackfruit.

Question 10.
What are monocious plants?
Answer:
Plants which has both male and female flower.

Sexual Reproduction in Flowering Plants Short Answer Type Questions

Question 1.
What is parthenogenesis ?
Answer:
Parthenogenesis : It is the method of reproduction in which ovum develops into embryo and forms a new plant without fertilization. It has been observed that pollination is necessary for parthenogenesis. It stimulates the process. Ulothrix, Spirogyra and Solan- aceae and Malvaceae family plants reproduce by this method.
Sometimes in some plants ovary also develops into fruit during this process. Such fruits are called as parthenocarpic fruits.
Example: Banana, Apple, Grapes, Guava etc. shows parthenocarpic fruits.

Question 2.
What is Polyembryony?
Answer:
Polyembryony : When more than one embryo develops in one seed then this condition is called as Polyembryony. It is generally found in citrus family. It is also found in Nicotiana, conifers, rice, wheat. It occurs when fertilization occurs in all embryo sacs found in the ovule.

Question 3.
Describe structure of anther with labelled diagram.
Answer:
Structure of Anther : Transverse section of anther shows that it consists of 2 lobes which are connected by connective. Each lobe contains two Pollen sacs. Innermost layer of pollen sac is called as Tapetum.

Tapetum is a layer rich in nutritive contents which supplies food material for the developing pollen grains. At first many Pollen mother cells are formed in them which divides meiotically to form haploid pollen grains.

Question 4.
Give four contrivances for self pollination.
Answer:
Contrivances for self-pollination:
(1) Bisexuality : When male and female parts are found in same flower then possibility of self-pollination increases.

(2) Cieistogamy : In Commelina benghalensis both cleistogamous and chasmogamous flowers are produced. The former are the underground flowers and the latter are the aerial ones developed on branches. In the small, inconspicuous cleistogamous flowers the pollen are shed within the closed flowers so that self-pollination is a must. This is also observed in Impatiens, Oxalis, Viola, Portulaca etc.

(3) Homogamy : Here the stamens and carpels mature at the same time. So, there is a greater chance of self-pollination as compared to cross-pollination, e.g., Mirabilis, Argemone etc.

(4) Failure of cross pollination : In some flowers generally cross pollination occurs but if they fails to do cross pollination then self-pollination occurs.

Question 5.
What do you mean by micropropagation?
Answer:
Micropropagation : It is a modern method of reproduction. By this process thousands of new plants can be obtained from few tissues of mother plant. This method is based on tissue and cell culture technique.
In this process a small part of tissue is separated from the plant and then it is grown in nutrient medium in aseptic condition. The tissue develops to form a cluster of cells which is called as callus. This callus can be preserved for long time for multiplication. A small part of the callus is transferred to nutrient medium, where it grows into a new plant. This plant is then transferred to the field. By this process Orchids, Carnations, Chrysanthemum plants can be grown successfully.

Question 6.
What are the advantages of cross-pollination ?
Answer:
Advantages of cross-pollination:

• The weaker characteristics of the race elimi¬nated and replaced by better character of nature.
• New improved varieties cap be produced by cross-pollination. ‘
• It increases the adaptability of the offsprings.
• Seeds are more vigorous.

Question 7.
What is fertilization? Draw labelled diagram showing path of pollen tube during fertilization.
Or
What do you mean by sexual reproduction? Describe the process of fertilization in plants with the help of diagram or sexual reproduction.
Answer:
Fertilization : The process of fusion of male and female gamete is called as fertilization or sexual reproduction. As a result of this process a diploid zygote is formed.

Question 8.
Differences between self and cross pollination
Answer:
Differences between Self and Cross poltination :

Sexual Reproduction in Flowering Plants Long Answer Type Questions

Question 1.
What is self-pollination ? Give advantages and disadvantages of self pollination.
Answer:
Self-pollination : When the pollen grain of one flower are transferred to the stigma of the same flower then the process is called as self-pollination.

Advantages of self-pollination:
These are the advantages of self pollination:

• Parental characters can be preserved indefinitely in several generations.
• Self-pollination helps in maintaining pure lines for experimental hybridization.
• It is most economical method of pollination. The plants do not consume their energies in the production of
• large number of pollen grains, nectar and coloured corolla.
• It ensures seed production and flowers do not take chances of the failure of fertilization.

Disadvantages of self-pollination:
These are the disadvantages of self-pollination :

• The weaker characteristics or defects of the plant can never be eliminated from the race.
• No useful characters can be introduced in the race.
• The immunity of race towards diseases falls and ultimately it falls prey to many diseases.

Question 2.
What is double fertilization?
Or
Explain double fertilization in angiosperms.
Answer:
Double fertilization : It is found in all angiospermic plants. It was first discovered by S.N. Navaschin (1899) in Lilium and Fritillaria. When pollen tube come in contact of embryo sac then the tip of the pollen tube disintegrate and release two male gametes into the embryo sac. Out of which one male gamete (x) fuses with egg cell (x) to form a diploid zygote. This process is called as fertilization. The second male gamete (x) fuses with secondary nuclei (2x) to form a triploid body.
The act of two nuclear fusion is called as double fertilization.

Question 3.
Describe development of embryo sac or Female gametophyte in angiosperms.
Answer:
Development of embryo sac or Female gametophyte in angiosperms :
During the initial stage of development of ovule primary archesporial cell gets demarcated at the apex of the nucellus below the epidermis. It divides periclinally into a outer primary parietal cell or primary wall cell and an inner primary sporogenous cell. The later acts as a megaspore mother cell which enlarges in size and divide meiotically to form a row of four megaspores. Of the four cells the upper three cells degenerate and appear as dark caps while the lowest one functions arid is called the functional megaspore. The latter greatly enlarges and forms the embryo sac.

Question 4.
Describe development of Endosperm.
Answer:
Development of Endosperm : It develops from the triploid tissue of the fertilized embryo sac after the act of double fertilization. It is of the following three types:

1. Free nuclear endosperm.
2. Cellular type of endosperm.
3. Helobial type of endosperm.

The name refers to the type of nuclear divisions of the endosperm nucleus. If triploid nucleus divides by free nuclear division the endosperm produced contains many nuclei lying freely in it and hence it is termed as free nuclear endosperm. If the nuclear division is followed by wall formation it is called as cellular type. If endosperm is intermediate between the two types it is called as helobial type.


Question 5.
What is male gametogenesis ? Describe male gametogenesis in plants.
Or
Describe development of pollen grain.
Answer:
Male gametogenesis: The process of formation of male gametes is called as male gametogenesis or microsporogenesis.

Each anther usually consists of two lobes, connected together by a connective. Each lobe contains two sacs called as Pollen sac. In the early stage of development the anther comprises of a homogenous mass of cells limited by a well demarcated epidermis. During the course of development the anther gets four lobed. Each lobe exhibits a hypodermal lining, few cells thick with distinct nuclei. These hypodermal cells constitute the archesporium. The cell of archesporium divide periclinally, cutting of parietal cells towards the periphery and sporogenous cells within it two chambers or loculi, which are termed as pollen sacs or microsporangia.

Fig. Development of Pollen grain : (A) Formation of Pollen lobes, (B) Archesporial cell, (C) and (D) Division in Parietal cell, (E) and (F) Parietal layer : Tapetum sporogenous tissue (G), (H) and (I) Pollen mother cell, (J) Pollen grain, (K) Outer layer Degenerate, (L) Four free Pollen grains.

Besides the above changes in the wall layers the sporogenous layer comprising of primary sporogenous cells also undergo several divisions to form microspore mother cells or pollen mother cells. In the initial stage the microspore mother cells are closely packed but as the anther enlarges the pollen sac becomes spacious and the pollen mother cells assume spherical shape and get loosely arranged.

In many cases some of the sporogenous cells are nonfunctional and serve only as the nutritive material for the functional microspore mother cells. Each microspore mother cell is diploid in nature and by a single meiotic division give rise to four haploid nuclei. The four nuclei so formed are arranged tetrahedrally, and called as Tetrad. Innermost layer of pollen sac is called as Tapetum. Outer to which endothecium layer is formed. In outer most layer of pollen sac epidermis is formed. Surrounding each nuclei exine and intine are formed, now each structure is called as Pollen grain. Pollen grains are liberated during the dehiscence of anthers through stomium.

Question 6.
Differenciate between:
(i) Embryo sac and Endosperm
(ii) Seed and Ovule
Answer:
(i) Difference between Embryosac and Endosperm:

(ii) Difference between Seed and Ovule

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ Ex 10.3

प्रश्न 1.
निम्नलिखित समीकरणों को ढाल अंतःखण्ड रूप में रूपान्तरित कीजिए और उनके ढाल तथा y-अंतः खण्ड ज्ञात कीजिए :
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.
हल:
(i) x + 7y = 0
∴ y = – \(\frac{1}{7}\)x + 0
∴ ढाल = – \(\frac{1}{7}\), y-अंत: खण्ड = 0.
(ii) 6x + 3y – 5 = 0,
3y = – 6x + 5
∴ y = – 2x + \(\frac{5}{3}\)
ढाल = – 2, y-अंत: खण्ड = \(\frac{5}{3}\)
(iii) y = 0
या y = 0. x + 0.
ढाल = 0, y-अंत: खण्ड = 0

प्रश्न 2.
निम्नलिखित समीकरणों को अंतःखण्ड रूप में रूपान्तरित कीजिए और अक्षों पर इनके द्वारा काटे गए अंतःखण्ड ज्ञात कीजिए :
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0.
हल:
(i) 3x + 2y – 12 = 0
या 3x + 2y = 12
12 से दोनों पक्षों में भाग देने पर
\(\frac{x}{4}+\frac{y}{6}\) = 1
अतः अंत: खण्ड 4 तथा 6 हैं।
(ii) 4x – 3y = 6
6 से दोनो पक्षों में भाग देने पर,
\(\frac{4 x}{6}-\frac{3 y}{6}\) = 1
\(\frac{x}{\frac{3}{2}}+\frac{y}{-2}\) = 1.
अत: अंत:खण्ड \(\frac{3}{2}\) तथा – 2 हैं।
3y + 2 = 0
या 3y = – 2
y = – \(\frac{2}{3}\).
अन्त: खण्ड हेतु समीकरण का रूप:
\(\frac{x}{0}+\frac{y}{-\frac{2}{3}}\) = 1
अंत:खण्ड 0 और –\(\frac{2}{3}\) हैं।

प्रश्न 3.
निम्नलिखित समीकरणों को लम्ब रूप में रूपान्तरित कीजिए। उनकी मूल बिन्दु से लांबिक दूरियाँ और लम्ब तथा धन x-अक्ष के बीच का कोण ज्ञात कीजिए :
(i) x – \(\sqrt{3}\)y + 8 = 0,
(ii) y – 2 = 0,
(iii) x – y = 4.
हल:

(ii) y – 2 = 0 या y = 2
0·x + y·1 = 2
x cos 90° + y sin 90° = 2 [∵ cos 90° = 0, sin 90° = 1]
∴ p = 2, α = 90°.
(iii) x – y =4
\(\sqrt{2}\) से भाग देने पर
\(\frac{1}{\sqrt{2}} x+\left(-\frac{1}{\sqrt{2}}\right) y\) = 2\(\sqrt{2}\)
\(\frac{1}{\sqrt{2}}\) = cos (360° – 45°) = cos 315°
और – \(\frac{1}{\sqrt{2}}\) = sin 315०
∴ x – y = 4 का लम्ब रूप
x cos 315° + y sin 315° = 4
की तुलना x cos α + y sin α = p से करने पर,
p = 2\(\sqrt{2}\) , α = 315°.

प्रश्न 4.
बिन्दु (- 1, 1) की रेखा 12(x + 6) = 5(y – 2) से दूरी ज्ञात कीजिए।
हल:
12(x + 6) = 5(y – 2).
या 12x + 72 = 5y – 10
12x – 5y + 82 = 0

प्रश्न 5.
x-अक्ष पर बिन्दुओं को ज्ञात कीजिए जिनकी रेखा \(\frac{x}{3}+\frac{y}{4}\) = 1 से दूरियाँ 4 इकाई हैं।
हल:
दिया गया समीकरण है: \(\frac{x}{3}+\frac{y}{4}\) = 1
12 से गुणा करने पर
4x + 3y – 12 = 0 …(1)
x- अक्ष पर माना कोई बिन्दु (a, 0) हो, तो बिन्दु (a, 0) से रेखा (1) की दूरी
= \(\frac{4 a+0-12}{\sqrt{16+9}}\) = ± \(\frac{4 a-12}{5}\)
∴ ± \(\frac{4 a-12}{5}\) = 4
या ± (4a – 12) = 20
+ ve चिन्ह लेकर 4a = 32 या a = 8
x-अक्ष पर अभीष्ट बिन्दु (8, 0) है।
– ve चिन्ह लेकर, –\(\frac{4 a-12}{5}\) = 4 या – 4a + 12 = 20
4a = – 8, a = – 2
दूसरा अभीष्ट बिन्दु (- 2, 0) है।

प्रश्न 6.
समान्तर रेखाओं के बीच की दूरी ज्ञात कीजिए
(i) 15x + 8y – 34 = 0 और 15x + 8y + 31 = 0
(ii) l(x +y) + p = 0 और l(x + y) – r= 0
हल:

प्रश्न 7.
रेखा 3x – 4y + 2 = 0 के समान्तर और बिन्दु (-2, 3) से जाने वाली रेखा का समीकरण ज्ञात कीजिए।
हल:
3x – 4y + 2 = 0
या 4y = 3x +2
∴ y = \(\frac{3}{4} x+\frac{2}{4}\)
∴ रेखा की ढाल = \(\frac{3}{4}\)
दिया गया बिन्दु (- 2, 3) और ढाल m = \(\frac{3}{4}\) से जाने वाली रेखा का समीकरण
y – y₁ = m(x – x₁)
y – 3 = \(\frac{3}{4}\)(x + 2)
या 4y – 12 = 3x + 6
या 3x – 4y + 18 = 0.
दूसरी विधि : कोई भी रेखा ax + by + c = 0 के समान्तर ax + by + k = 0 के रूप में लिखी जा सकती है।
∴ 3x – 4y + 2 = 0 के समान्तर रेखा 3x – 4y + k = 0 है
यह (- 2, 3) से होकर जाती है।
∴ 3 x (- 2) – 4 x 3 + k = 0 या k = 18
अभीष्ट समान्तर रेखा का समीकरण: 3x – 4y + 18 = 0.

प्रश्न 8.
रेखा x – 7y + 5 = 0 पर लम्ब और x-अन्तः खण्ड 3 वाली रेखा का समीकरण ज्ञात कीजिए।
हल:
∵ x-अंत:खण्ड = 3
∴ रेखा A(3, 0) से होकर जाती है।
रेखा PQ : x – 7y + 5 = 0
या 7y = x +5
या y = \(\frac{1}{7}\) x + \(\frac{5}{7}\)

इसलिए PQ की ढाल = \(\frac{1}{7}\)
∵ PQ ⊥ AB
∴ A से होकर जाने वाली रेखा AB की ढाल = – 7
∴ बिन्दु (3, 0) से रेखा AB का समीकरण,
y – 0 = – 7(x – 3).
= – 7x + 21
या 7x + y – 21 = 0.
दूसरी विधि : ax + by + c = 0 की लम्ब कोई रेखा bx – ay + k = 0
∴ x – 7y + 5 = 0 की लम्ब रेखा 7x + y + k = 0
यह रेखा (3, 0) से होकर जाती है।
∴ 7 x 3 + 0 + k = 0, अर्थात् k = – 21
∴ अभीष्ट रेखा का समीकरण 7x + y – 21 = 0.

प्रश्न 9.
रेखाओं \(\sqrt{3}\)x + y =1 और x +\(\sqrt{3}\)y =1 के बीच का कोण ज्ञात कीजिए।
हल:

θ = 30° = \(\frac{\pi}{6}\) रेडियन। .

प्रश्न 10.
बिन्दुओं (h, 3) और (4, 1) से जाने वाली रेखा, रेखा 7x – 9y – 19 = 0 को समकोण पर प्रतिच्छेद करती है। का मान ज्ञात कीजिए।

माना रेखा AB बिन्दु A(h, 3), B(4, 1) से जाने वाली रेखा की ढाल,

चूँकि दोनों रेखाएँ एक-दूसरे को समकोण पर प्रतिच्छेद करती हैं, ∴ m₁,m₂ = – 1
\(\frac{2}{h-4} \times \frac{7}{9}\) = – 1
14 = – 9(h – 4) = – 9h + 36
∴ 9h = 36 – 14 = 22
h = \(\frac{22}{9}\)

प्रश्न 11.
सिद्ध कीजिए कि बिन्दु (x₁, y₁) से जाने वाली और रेखा Ax + By + C = 0 के समान्तर रेखा का समीकरण
A(x – x₁) + B(y – y₁) = 0 है।
हल:
रेखा Ax + By + C = 0
या y= – \(\frac{A}{B}\)x – \(\frac{C}{B}\)
रेखा की ढाल = – \(\frac{A}{B}\)
∴ समान्तर रेखा की ढाल = – \(\frac{A}{B}\)

प्रश्न 12.
बिन्दु (2, 3) से जाने वाली दो रेखाएँ परस्पर 60° के कोण पर प्रतिच्छेद करती हैं। यदि एक रेखा की ढाल 2 है तो दूसरी रेखा का समीकरण ज्ञात कीजिए।
हल:
माना दूसरी रेखा की ढाल m है।
दोनों रेखाओं के बीच कोण
tan θ = \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)
जहाँ θ = 60°, m₁ = m और m₂ = 2
∴ tan 60 = ± \(\frac{m-2}{1+2 m}\) = \(\sqrt{3}\)

प्रश्न 13.
बिन्दुओं (3, 4) और (- 1, 2) को मिलाने वाली रेखाखण्ड के लम्ब समद्विभाजक रेखा का समीकरण ज्ञात कीजिए।
हल:
माना बिन्दुओं A(3, 4) और B(- 1, 2) को मिलाने वाले रेखाखण्ड का मध्य बिन्दु

रेखा CD बिन्दु D से होकर जाती है
∴ रेखा CD का समीकरण
y – 3 = – 2(x – 1)
= – 2x + 2
∴ 2x + y – 5 = 0.

प्रश्न 14.
बिन्दु (- 1, 3) से रेखा 3x – 4y – 16 = 0 पर डाले गए लम्बपाद के निर्देशांक ज्ञात कीजिए।
हल:
मान लीजिए रेखा AB का समीकरण, 3x – 4y – 16 = 0 …(i)
या y = \(\frac{3}{4}\)x – 4
रेखा AB की ढाल = \(\frac{3}{4}\)
बिन्दु C(- 1, 3) से AB पर डाला गया लम्ब CD है
∴ AB ⊥ CD.

अतः रेखा CD का समीकरण,
y – y₁ = m(x – x₁)
y – 3 = \(\frac{-4}{3}\)(x + 1)
या 3y – 9 = – 4x – 4
या 4x + 3y – 5 = 0 …(ii)
समी (i) को 3 से और (ii) को 4 से गुणा करने पर,
9x – 12y = 48
16x + 12y = 20
इनको जोड़ने पर
25x = 68 या x = \(\frac{68}{25}\)
x का मान (i) में रखने पर,

प्रश्न 15.
मूल बिन्दु से रेखा y = mx + c पर डाला गया लम्ब रेखा से बिन्दु (-1, 2) पर मिलता है। m और … c के मान ज्ञात कीजिए।
हल:
माना रेखा AB का समीकरण, y = mx + c
रेखा AB की ढाल = m

O से रेखा AB पर लम्ब OC डाला गया है जो बिन्दु C(- 1, 2) पर मिलता है।
∴ लम्ब रेखा OC की ढाल = –\(\frac{1}{m}\)
अब रेखा OC का समीकरण,
y – 0 = –\(\frac{1}{m}\)(x – 0)
या x + my = 0
OC की प्रवणता = \(\frac{2-0}{-1-0}\) = – 2
∴ लम्ब रेखा OC की ढाल = –\(\frac{1}{m}\)
बिन्दु C (- 1, 2) निम्न रेखा पर स्थित है :
y = mx + c
⇒ 2 = – m + c
m = \(\frac{1}{2}\) रखने पर,
2 = – \(\frac{1}{2}\) + c
∴ c = 2+ \(\frac{1}{2}\) = \(\frac{5}{2}\)
अतः m = \(\frac{1}{2}\), c = \(\frac{5}{2}\)

प्रश्न 16.
यदि p और q क्रमशः मूल बिन्दु से रेखाओं x cos θ – y sin θ = k cos 2θ और x sec θ +y cosec θ = k पर लम्ब की लंबाइयाँ हैं तो सिद्ध कीजिए कि
p₂ + 4q₂ = k₂.
हल:
मूल बिन्दु (0, 0) से x cos θ – y sin θ = k cos 2θ की दूरी,

समीकरण (1) और (2) को वर्ग करके जोड़ने पर,
k₂ = p₂ + 4q₂
अतः p₂ + 4q₂ = K₂.

प्रश्न 17.
शीर्षों A(2, 3), B(4, – 1) और C(1, 2) वाले त्रिभुज ABC के शीर्ष A से उसकी सम्मुख भुजा पर लम्ब डाला गया है। लम्ब की लम्बाई तथा समीकरण ज्ञात कीजिए।
हल:
मान लीजिए AM रेखा BC पर लंब डाला गया है
(i) रेखा BC की ढाल

रेखा AM बिन्दु A से जाती है और ढाल = 1 है।
∴ AM का समीकरण
y – y₁ = m(x – x₁)
y – 3 = 1. (x – 2)
या x – y + 1 = 0
(ii) बिन्दु B(4, – 1) और C(1, 2) से होकर जाने वाली रेखा BC का समीकरण

प्रश्न 18.
यदि p मूल बिन्दु से उस रेखा पर डाले गए लम्ब की लम्बाई हो जिसस पर अक्षों पर कटे अंत: खण्ड a और b हों, तो दिखाइए कि \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).
हल:
उस रेखा का समीकरण, जिसकी अक्षों पर कटे अंत:खण्ड a और b हों,
\(\frac{x}{a}+\frac{y}{b}\) = 1 (अंत:खण्ड समीकरण)
मूल बिन्दु (0, 0) बसे इस रेखा पर डाले गए लम्ब की लम्बाई

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 1 Reproduction in Organisms

Reproduction in Organisms NCERT Textbook Questions and Answers

Question 1.
Why is reproduction essential (neccessary) for organisms?
Answer:
An organism gives rise to young ones by reproduction. The offspring grow, mature and in turn produce new offspring. Thus, there is a cycle of birth, growth and death. Reproduction enables the continuity of the species, generation after generation. So, therefore reproduction is essential.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual mode of reproduction is better because it is biparental reproduction and introduces variation among offsprings and their parants due to crossing over and recombination during gamete formation by meiosis.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
In sexual reproduction, the offspring is morphologically and genetically identical to the parent and to each other. Hence, it is called clone.

Question 4.
Offsprings formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Offspring formed due to sexual reproduction have better chance of survival because:

• The offspring consists its hybrid characters which may adapt better with the different environment.
• Genetic variations are introduced among the offspring’s, which increases the biological tolerance.
• Sexual reproduction occurs in adverse condition in lower plant kingdom, so sexual spores survive in adverse condition.
• Sexual reproduction may not always show better chances of survival because the offspring may be inferior to the parents.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:
The progenies have similar genetic make up and are exact copies of their parents in asexual reproduction but the progenies have different genetic make up and different from each other and dissimilar to the parent in sexual reproduction.

Variation is absent in asexual reproduction but it is a common phenomenon of sexual reproduction. In asexual reproduction, variation may occur due to mutation whereas varia¬tion occurs due to mutation, crossing over and recombination in sexual reproduction.

Question 6.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction aslo considered as a type of asexual reproduction?
Answer:
Differences between Asexual and Sexual reproduction:

Vegetative reproduction is considered as a type of asexual reproduction because :

• It is uniparental reproduction.
• There is no involvement of gametes or sex cells.
• Cell division and no reductional division takes place.
• Vegetative propagules are somatic cells.

Question 7.
What is vegetative propagation ? Give two suitable examples.
Answer:
In plants, the vegetative propagules (runner, rhizome, sucker, etc.) are capable of producing new offsprings by the process called vegetative propagation. As the formation of these vegetative propagules does not involve both the parents, the process involved is asexual.
Examples:

• Adventitious buds in the notches along the leaf margins of bryophyllum grow to form new plants.
• Potato tuber having buds when grown, develops into a new plant.

Question 8.
Define:
(a) Juvenile phase
(b) Reproductive phase
(c) Senescent phase.
Answer:
(a) Juvenile phase : It is the pre-reproductive in which all organisms require a certain growth and maturity in the life before reproducing sexually.
(b) Reproductive phase: Reproductive phase is the phase in the life cycle, where an organism possess all the capacity and potential to reproduce sexually. It is the end of juvenile phase or vegetative phase.
(c) Senescent phase : It is the post reproductive phase in the life cycle where an organism slowly losses the rate of metabolism, reproductive potential and show deterioration of the physiological activity of the body.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Higher organisms have resorted to sexual reproduction to :

• Get over the unfavourable condition
• Restore high gene pool in a population
• Restore vigour and vitality of the race and .
• Get proper parental care ‘
• Introduce variation to enable better adaptive capacity.

Question 10.
Explain, why meiosis and gametogenesis are always interlinked.
Answer:
Gametogenesis (formation of male and female gametes) is associated with reduction in chromosome number thus, the gamete formed contains half chromosome set of the parental cell. So, gametogenesis is interlinked with meiosis because in meiosis reduction of chromosome number from diploid set (2n) to haploid set (n) takes place.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
Answer:

Question 12.
Define external fertilization mention its disadvantages.
Answer:
The fusion of compatible gametes (Male and female) outside the body of an organism is called external fertilization, e.g., in Frog.
Disadvantages of external fertilization:

• It requires a medium for fusion of gametes.
• The young ones are often exposed to the predators.
• After fertilization, offsprings are produce large in number but no parental care is provided

Question 13.
Differentiate between a Zoospore and Zygote.
Answer:
Differences between Monera and Protista:

Question 14.
Differentiate between Gametogenesis and Embryogenesis.
Answer:
Differences between Gametogenesis and Embryogenesis:

Question 15.
Describe the post-fertilization changes in a flower.
Answer:
The post-fertilization changes that take place in a flower are as follows :

• The formation of zygote which later develops into an embryo and primary endosperm cell which develops into endosperm takes place.
• While the sepals, petals and stamens are shed, the pistil remains intact.
• The fertilized ovule develops into seeds.
• The ovary matures into a fruit that later develops a thick protective wall, called pericarp.
• Seeds after dispersal germinate under favourable conditions which later develop into a new plant.

Question 16.
What is a bisexual flower ? Collect five bisexual flowers from your neigh bourhood and with the help of your teacher find our their common and scientific names.
Answer:

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that bears unisexual flowers ?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow colored petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure. Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
In viviparous animals, the young one develops inside the body of the female organism. As a result of this, the young one gets better production and nourishment for proper development. In case of oviparous animals, they lay eggs and the young ones develop inside the calcareous shell, outside the body of the female. So, the young ones are not effectively protected and nourished and are vulnerable, to predators so, they are at a greater risk as compared to the offsprings of the viviparous animals.

Reproduction in Organisms Other Important Questions and Answers

Reproduction in Organisms Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Callose wall is found: (CBSE PMT 2007)
(a) In male gamete
(b) In ovum
(c) In pollen grain
Answer:
(d) In megaspore mother cell.

Question 2.
Name the plant in which new plant arise from the notches of leaves through vegetative propagation: (AFMC2012)
(a) Asparagus
(b) Chrysanthemum
(c) Agave
(d) Bryophyllum.
Answer:
(d) Bryophyllum.

Question 3.
Which type of vegetative propagation is occurs in banana: (AMU 2012)
(a) From tuber
(b) From rhizome
(c) From bulb
(d) From stolen.
Answer:
(b) From rhizome

Question 4.
Which of the following is viviparous:
(a) Tortoise
(b) Bony fish
(c) Hummingbird
(d) Whale
Answer:
(d) Whale

Question 5.
Which of the following is characteristics of the ‘clone’:
(a) Similar to ancestors in genetic character
(b) A group of plants developed in vegetative propagation
(c) Plants developed from same parents
(d) All of the above.
Answer:
(d) All of the above.

Question 6.
Organism which give birth to young ones:
(a) Viviparous
(b) Amphibians
(c) Oviparous
(d) Coelomate.
Answer:
(a) Viviparous

Question 7.
Set of chromosome found in zygote:
(a) X
(b) 2X
(c) 3X
(d) 4X.
Answer:
(b) 2X

Question 8.
The plant which can be obtained same as parent plant:
(a) Through seeds
(b) Through fruits
(c) By cutting the stem
(d) By hybridization.
Answer:
(c) By cutting the stem

Question 9.
Vegetative propagation in potato is:
(a) By rhizome
(b) By bulb
(c) By stem
(d) By tuber.
Answer:
(d) By tuber.

Question 10.
Which of the following is oviparous animal:
(a) Bat
(b) Whale
(c) Penguin
(d) Amoeba.
Answer:
(c) Penguin

2. Fill in the Blanks:

1. Reproductive organs in plants are developed in ……………….. phase.
2. ……………….. nucleus are present in pollen tube of angiosperms.
3. The ……………….. are not form in Rose and Banana, therefore, reproduction occurs through ………………..
4. Reproductive organs of animals are differentiate in ……………….. stage.
5. Eichomia propagate through ………………. in water.
6. Rate of multiplication is greater ……………….. reproduction than ……………….. reproduction.

Answer:

1. Adolescence
2. Three
3. Seeds, Vegetative propagation
4. Embryo
5. Offset
6. Sexual, Asexual.

3. Match the Following

Answer:

1. (e)
2. (d)
3. (f)
4. (a)
5. (c)
6. (b)

Answer in One Word/Sentence:

1. What we called uniparental structure formed from asexual reproduction which is genetically and morphologically similar.
2. Write the name of an algae reproduce by means of zoospores.
3. Animal reproduce asexually.
4. What we called spores having flagella.
5. In Bryophyllum what kind of bud reproduce vegetatively ?

Answer:

1. Clone
2. Chlamydomonas
3. Spongilla
4. Zoo-spores
5. Adventitious bud.

Reproduction in Organisms Very Short Answer Type Questions

Question 1.
What is meant by life span ?
Answer:
The period from birth to the natural death of an organism is called life span.

Question 2.
Give definition of fertilization.
Answer:
Fertilization is a process in which male and female gamete fused to form the zygote.

Question 3.
What is clone ?
Answer:
Morphologically and genetically similar individuals who are produced by single parent is called clone.

Question 4.
Give the name of an organism in which asexual reproduction occurs through conidia.
Answer:
Penicillum.

Question 5.
In which plant vegetative propagation occurs through rhizome ?
Answer:
In Ginger, Termeric.

Question 6.
Give the name of an organism in which transverse binary fission occurs.
Answer:Paramoecium.

Question 7.
What is aging ?
Answer:
When humans are not capable for reproduction is known as aging.

Question 8.
Give two examples of Hermaphrodite plants.
Answer:
Cucurbita and Coconut are hermaphrodite plant.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ Ex 10.2

प्रश्न 1 से 8 तक रेखा का समीकरण ज्ञात कीजिए जो दिए गए प्रतिबंधों को संतुष्ट करता है।
प्रश्न 1.
x- अक्ष और y- अक्ष के समीकरण लिखिए।
हल:
x- अक्ष का समीकरण y = 0.
तथा y- अक्ष का समीकरण x = 0.

प्रश्न 2.
ढाल \(\frac{1}{2}\) और बिन्दु (- 4, 3) से जाने वाली।
हल:
ढाल m = \(\frac{1}{2}\) , बिन्दु (- 4, 3)
अभीष्ट रेखा का समीकरण
y – y₁ = m(x – x₁)
y – 3 = \(\frac{1}{2}\) = (x + 4)
या 2y – 6 = x + 4
∴ x – 2y + 10 = 0.

प्रश्न 3.
बिन्दु (0, 0) से जाने वाली और ढाल m वाली।
हल:
दिया है : बिन्दु (0, 0), ढाल = m
ढाल m, तथा (x₁, y₁) से जाने वाली रेखा का समीकरण
y – y₁ = m(x – x₁)
∴ y – 0 = m(x – 0)
अतः अभीष्ट समीकरण y = mx.

प्रश्न 4.
बिन्दु (2, 2\(\sqrt{3}\)) से जाने वाली और x-अक्ष से 75° के कोण पर झुकी हुई।
हल:
चूँकि रेखा x-अक्ष के साथ 75° पर झुकी हुई है, तब रेखा की ढाल
m = tan 75° = tan (45° + 30°)

या (2 + \(\sqrt{3}\))x – y + 2\(\sqrt{3}\) – 4 – 2\(\sqrt{3}\) = 0
अतः (2 + \(\sqrt{3}\))x – y – 4 = 0.

प्रश्न 5.
मूल बिन्दु के बाईं ओर x- अक्ष को 3 इकाई की दूरी पर प्रतिच्छेद करने तथा ढाल – 2 वाली।
हल:
मूल बिन्दु से बाईं ओर 3 इकाई की दूरी पर स्थित बिन्दु (- 3, 0) होगा तथा ढाल m = – 2
m तथा (x₁, y₁) के द्वारा रेखा का समीकरण,
y – y₁ = m(x – x₁)
वहाँ x₁ = – 3 तथा y₁ = 0 रखने पर,
y – 0 = – 2(x + 3)
या y = – 2x – 6
या 2x + y + 6 = 0.

प्रश्न 6.
मूल बिन्दु से ऊपर y-अक्ष को 2 इकाई की दूरी पर प्रतिच्छेद करने वाली और x-अक्ष की धन दिशा के साथ 30° का कोण बनाने वाली।
हल:
मूल बिन्दु से y-अक्ष पर 2 इकाई की दूरी पर स्थित बिन्दु (0, 2) होगा। x-अक्ष की धन दिशा के साथ रेखा 30° का कोण बनाती है।
∴ m = tan 30° = \(\frac{1}{\sqrt{3}}\)
रेखा का समीकरण,
y – y₁ = m (x – x₁)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – 0)
या \(\sqrt{3}\)y – 2\(\sqrt{3}\) = x
या x – \(\sqrt{3}\)y + 2\(\sqrt{3}\) = 0.

प्रश्न 7.
बिन्दुओं (-1, 1) और (2 – 4) से जाते हुए।
हल:

या 3y – 3 = – 5x – 5
अतः 5x + 3y + 2 = 0.

प्रश्न 8.
उस रेखा का समीकरण ज्ञात कीजिए जिसकी मूल बिन्दु से लांबिक दूरी 5 इकाई और लंब धन x-अक्ष से 30° का कोण बनाती है।
हल:
हम जानते हैं कि लंब रूप में रेखा AB का समीकरण,
x cos ω + y sinω = P
यहाँ पर दिया है: ω = 30°, तथा p = 5

∴ रेखा AB का समीकरण,
x cos 30 + y sin 30 = 5
x.\(\frac{\sqrt{3}}{2}\) + y.\(\frac{1}{2}\) = 5
∴ \(\sqrt{3}\)x + y = 10.

प्रश्न 9.
∆PQR के शीर्ष P(2, 1), Q(- 2, 3) और R(4, 5) हैं। शीर्ष R से जाने वाली माध्यिका का समीकरण ज्ञात कीजिए।
हल:
PQ का मध्य बिन्दु अर्थात् m (0, 2) है।
∴ दो बिन्दुओं से जाने वाली रेखा का समीकरण,

अब बिन्दुओं R (4, 5) तथा M(0, 2) से जाने वाली रेखा का समीकरण,
y – 5 = \(\frac{2-5}{0-4}\) = (x – 4)
या 4(y – 5) = 3 (x – 4)
या 3x – 4y + 8 = 0
अतः माध्यिका RM का समीकरण 3x – 4y + 8 = 0.

प्रश्न 10.
(- 3, 5) से होकर जाने वाली और बिन्दु (2, 5) और (- 3, 6) से जाने वाली रेखा पर लंब रेखा का समीकरण ज्ञात कीजिए।
हल:
बिन्दु A(2, 5) और B(- 3, 6) से होकर जाने वाली रेखा का ढाल

यदि PL बिन्दु P(- 3, 5) से AB पर लम्ब डाला गया हो तो उसकी ढाल m₂ मान लीजिए।
रेखाएँ PL और AB परस्पर लम्ब हैं।
यदि PL की ढाल × AB की ढाल = – 1
अर्थात m₂ × \(\left(-\frac{1}{5}\right)\) = – 1
∴ m₂ = 5
PL की ढाल 5 है और P(- 3, 5) से होकर जाती है तो PL का समीकरण,
y – y₁ = m₂(x – x₁)
या y – 5 = 5 (x + 3)
∴ 5x – y + 20 = 0.

प्रश्न 11.
एक रेखा (1, 0) तथा (2, 3) बिन्दुओं को मिलाने वाली रेखाखंड पर लम्ब है तथा उसको 1 : n के अनुपात में विभाजित करती है। रेखा का समीकरण ज्ञात कीजिए।
हल:
रेखा AB बिन्दु A(1, 0) तथा B(2, 3) से होकर जाती है।
∴ AB की ढाल = \(\frac{3-0}{2-1}=\frac{3}{1}\)
PQ ⊥ AB
AB की ढाल = \(\frac{3}{1}\)

∴ PQ की ढाल, m = \(-\frac{1}{\frac{3}{1}}=-\frac{1}{3}\)
PQ रेखा AB को C पर प्रतिछेदन करती है।
साथ ही बिन्दु C रेखाखंड AB को 1 : n के अनुपात में बांटता है।

या (n + 1)x + 3(n + 1)y = n + 2 + 9 = n + 11
या (n+ 1) x + 3 (n + 1)y = n + 11.

प्रश्न 12.
एक रेखा का समीकरण ज्ञात कीजिए जो निर्देशांक अक्षों से समान अंत:खण्ड काटती है और बिन्दु (2, 3) से जाती है।
हल:
(i) रेखा AB बिन्दु P(2, 3) से होकर जाती है और निर्देशांक अक्षों पर समान अंत: खंड बनाती है।
OA = OB
∠BAO = 45°,
∠BAX = 135°

AB की ढाल, m = tan 135° = – 1
रेखा का समीकरण, y – y₁ = m(x – x₁)
जहाँ x₁ = 2, Y₁ = 3 तथा m = – 1
y – 3 = – (x – 2)
या x + y – 5 = 0
या x + y = 5.

प्रश्न 13.
बिन्दु (2, 2) से जाने वाली रेखा का समीकरण ज्ञात कीजिए जिसके द्वारा अक्षों से कटे अंतःखंडों का योग 9 है।
हल:
मान लीजिए P(2, 2) से होकर जाने वाली रेखा से अक्षों पर बने अंतः खंड a तथा b हैं।

अंतः खंड रूप में रेखा का समीकरण
\(\frac{x}{a}+\frac{y}{b}\) = 1
यह रेखा P(2, 2) से होकर जाती है।
∴ \(\frac{2}{a}+\frac{2}{b}\) = 1
दिया है कि अंतः खंडों का योग 9 है।
∴ a + b = 9
b = 9 – a
b का मान (1) में रखने पर,
\(\frac{2}{a}+\frac{2}{9-a}\) = 1
या 2(9 – a) + 2a = a (9 – a)
18 – 2a + 2a = 9a – a₂
या a₂ – 9a + 18 = 0
या (a – 6)(a – 3) = 0
a = 6, 3
b = 3, 6
जब a = 6 तथा b = 3 हो, तो रेखा का अभीष्ट समीकरण
\(\frac{x}{6}+\frac{y}{3}\) = 1
या 3x + 6y = 18
या x + 2y = 6.
जब a = 3 तथा b = 6 हो, तब रेखा का अभीष्ट समीकरण,
\(\frac{x}{3}+\frac{y}{6}\) = 1
या 6x + 3y = 18
या 2x + y = 6

प्रश्न 14.
बिन्दु (0, 2) से जाने वाली और धन x-अक्ष से \(\frac{2 \pi}{3}\) के कोण बनाने वाली रेखा का समीकरण ज्ञात कीजिए। इसके समांतर और y-अक्ष को मूल बिन्दु से 2 इकाई नीचे की दूरी पर प्रतिच्छेद करती हुई रेखा का समीकरण भी ज्ञात करो।
हल:
माना एक रेखा PQ बिन्दु P(0, 2) से होकर जाती है और धन x-अक्ष के साथ \(\frac{2 \pi}{3}\) का कोण बनाती है।
∴ PQ की ढाल = tan\(\frac{2 \pi}{3}\)
= – \(\sqrt{3}\)

∴ रेखा PQ का समीकरण, y – y₁ = m(x – x₁ )
y – 2 = – \(\sqrt{3}\) (x – 0)
या \(\sqrt{3}\)x + y – 2 = 0.
दूसरी रेखा RS रेखा PQ के समांतर है
∴ RS का ढाल = –\(\sqrt{3}\)
यह रेखा (0, – 2) से होकर जाती है।
रेखा RS का समीकरण, y – y₁ = m(x – x₁)
y + 2 = –\(\sqrt{3}\) (x – 0)
\(\sqrt{3}\)x + y + 2 = 0.

प्रश्न 15.
मूल बिन्दु से किसी रेखा पर डाला गया लम्ब रेखा से बिन्दु (- 2, 9) पर मिलता है। रेखा का समीकरण ज्ञात कीजिए।
हल:
मान लीजिए रेखा AB पर मूल बिन्दु से डाला गया लम्ब AB पर मिलता है।
OP की ढाल = – \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{9-0}{-2-0}=-\frac{9}{2}\)
परन्तु AB ⊥ OP

अब AB की ढाल \(\frac{2}{9}\) है और P(- 2, 9) से होकर जाती है।
∴ AB का समीकरण
y – y₁ = m(x – x₁)
अर्थात् y – 9 = \(\frac{2}{9}\) = (x + 2) या
या 9y – 81 = 2x +4
या 2x – 9y + 85 = 0.

प्रश्न 16.
तांबे की छड़ की लम्बाई L (सेमी में) सेल्सियस ताप C का रैखिक फलन है। एक प्रयोग में यदि L= 124.942, जब C = 20 और L= 125.134 जब C = 110 हो, तो L को C के पदों में व्यक्त कीजिए।
हल:
L ताप C का रैखिक फलन है।
(20, 124.942), (110, 125.134) इसका रैखिक फलन है। इन दो बिन्दुओं से संतुष्ट फलन

प्रश्न 17.
किसी दूध भण्डार का स्वामी प्रति सप्ताह 980 लीटर दूध, 14 रू प्रति लीटर के भाव से और 1220 लीटर दूध 16 रू प्रति लीटर के भाव से बेच सकता है। विक्रय मूल्य तथा मांग के मध्य के संबंध को रैखिक मानते हुए ज्ञात कीजिए कि प्रति सप्ताह वह कितना दूध 17 रू प्रति लीटर के भाव से बेच सकता है?
हल:
दूध के भाव और मात्रा में रैखिक सम्बन्ध है। यह रेखा दो बिन्दुओं (14, 980), (16, 1220) से होकर जाती है।
इससे प्राप्त रेखा का समीकरण,
y – y₁ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) (x – x₁)
y – 980 = \(\frac{1220-980}{16-14}\)(x – 14)
= \(\frac{240}{2}\)(x – 14)
= 120(x – 14)
y = 980 + 120 (x – 14)
जब x का मान 17 है तो y का मान नीचे दिया गया है।
y = 980 + 120(17 – 14)
= 980 + 120 × 3
= 980 + 360
= 1340
अतः 17 रू प्रति लीटर भाव का 1340 लीटर दूध बिकेगा।

प्रश्न 18.
अक्षों के बीच रेखाखंड का मध्य बिंदु P(a, b) है। दिखाइए कि रेखा का समीकरण \(\frac{x}{a}+\frac{y}{b}\) = 2 हैं।
P(a, b)
हल:
माना रेखा AB अक्षों पर p और q अंतःखंड बनते हैं।
∴ बिन्दु A और B के क्रमशः निर्देशांक (p, 0) और (0, q) हैं।

प्रश्न 19.
अक्षों के बीच रेखाखण्ड को बिन्दु R(h, k), 1 : 2 के अनुपात में विभक्त करता है। रेखा का समीकरण ज्ञात कीजिए।
हल:
अक्षों के बीच रेखाखंड AB को R(h, k) AR : RB = 1 : 2 के अनुपात में विभक्त करता है।
मान लीजिए अक्षों पर अंत:खण्ड OA = a और OB = b है।

a और b के मान रखने पर,
\(\frac{x}{\frac{3 h}{2}}+\frac{y}{3 k}\) = 1
या \(\frac{2 x}{h}+\frac{y}{k}\) = 3
या 2kx + hy = 3hk.

प्रश्न 20.
रेखा के समीकरण की संकल्पना का प्रयोग करते हुए सिद्ध कीजिए कि तीन बिन्दु (3, 0), (-2,-2) और (8, 2) संरेख हैं।
हल:
बिन्दु A(3, 0), B(- 2, – 2) से होकर जाने वाली रेखा का समीकरण

बिन्दु C(8, 2) इस रेखा पर पड़ता है तब इसके निर्देशांक इस समीकरण को संतुष्ट करेंगे।
∴ 2 × 8 – 5 × 2 – 6 = 0
या 16 – 16 = 0
अतः दिए हुए बिन्दु A, B, C सरेख हैं।

MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ Ex 10.1

प्रश्न 1.
कार्तीय तल में एक चतुर्भुज खींचिए जिसके शीर्ष (- 4, 5), (0, 7), (5, – 5) और (- 4, – 2) हैं। इसका क्षेत्रफल भी ज्ञात कीजिए।
हल:
दिए गए बिन्दुओं (- 4, 5), (0, 7), (5, -5) और (- 4, – 2) क्रमशः A, B, C, D द्वारा दर्शाया गया है। चतुर्भुज ABCD को दो भागों में बाँटा गया है। जो ∆ABD तथा ∆BDC के रूप में हैं।

∆ABD के शीर्ष A(- 4, 5), B(0, 7), D(- 4, – 2) हैं।
∴ ∆ABD का क्षेत्रफल = \(\frac{1}{2}\)|[ – 4(7 + 2) +0(- 2 – 5)+ (- 4)(5 – 7)]|
= \(\frac{1}{2}\)|[-36+8]|= \(\frac{1}{2}\) x 28
= 14 वर्ग इकाई
∆BDC के शीर्ष B(0, 7), D(-4, – 2), C ( 5, – 5) हैं।
∆BDC का क्षेत्रफल = \(\frac{1}{2}\)[0(- 2 + 5) – 4( – 5 – 7) + 5(7 + 2)]
= \(\frac{1}{2}\) [48 +45] = = \(\frac{1}{2}\) × 93
= 46.5 वर्ग इकाई
∴ चतुर्भुज ABCD का क्षेत्रफल = ∆ABD का क्षेत्रफल + ∆BDC का क्षेत्रफल
= 14+ 46.5
= 60.5 वर्ग इकाई।

प्रश्न 2.
2a भुजा के समबाहु त्रिभुज का आधार y-अक्ष के अनुदिश इस प्रकार है कि आधार का मध्य बिन्दु मूल बिन्दु पर है। त्रिभुज के शीर्ष ज्ञात कीजिए।
हल:
माना ∆ABC की भुजा BC, y- अक्ष के अनुदिश है जिसका मध्य बिन्दु मूल बिन्दु O है।
⇒ B और C के शीर्ष बिन्दु (0, a) और (0, – a) हैं।
बिन्दु A, x- अक्ष पर है, AB = 2a, OB = a

समकोण त्रिभुज OAB में,
OA² = AB² – OB² = (2a)² – a²
= 4a² – a² = 3a²
∴ OA = \(\sqrt{3}\)a
∴ A के निर्देशांक (\(\sqrt{3}\)a,0) हैं।
अतः AABC के निर्देशांक (\(\sqrt{3}\)a,0), (0, a), (0 – a) हैं।

प्रश्न 3.
P(x₁,y₁) और Q(x₂, Y₂) के बीच की दूरी ज्ञात कीजिए जब :
(i) PQ,y- अक्ष के समांतर है,
(ii) PQ, x- अक्ष के समांतर है।
हल:
(i) जब कोई रेखा y-अक्ष के समांतर होती है तो उस पर जितने भी बिन्दु होंगे उनके x- निर्देशांक बराबर होते हैं अर्थात् X₁ = X₂.

(ii) जब कोई रेखा x-अक्ष के समांतर होती है तो उसके प्रत्येक बिन्दु का y- निर्देशांक बराबर होता है।
अर्थात् y₁ = Y₂

प्रश्न 4.
x- अक्ष पर एक बिन्दु ज्ञात कीजिए जो (7, 6) और (3, 4) बिन्दुओं से समान दूरी पर है।
हल:
मान लीजिए x- अक्ष पर बिन्दु A(a, 0), बिन्दु B(7, 6) और C(3, 4) से समान दूरी पर है।

अर्थात् AB = AC
या AB₂ = AC₂
या (a – 7)₂ + (0 – 6)₂ = (a – 3)₂ + (0 – 4)₂
∴ a₂ – 14a + 49 + 36 = a₂ – 6a + 9+ 16
– 14a + 6a = 25 – 85
= – 60
या – 8a = – 60
या a = \(\frac{60}{8}=\frac{15}{2}\)
अतः बिन्दु 4 के निर्देशांक \(\left(\frac{15}{2}, 0\right)\) है।

प्रश्न 5.
रेखा की ढाल ज्ञात कीजिए जो मूल बिन्दु और P(0, -4) तथा B(8, 0) बिन्दुओं को मिलाने वाले रेखाखंड के मध्य बिन्दु से जाती है।
हल:
बिन्दु P(0, – 4) और B(8, 0) को मिलाने वाले रेखाखंड का मध्य बिन्दु

⇒ PB का मध्य बिन्दु M के निर्देशांक (4, -2) है।
मूल बिन्दु 0 के निर्देशांक (0, 0) हैं।
∴ OM की ढाल = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\).

प्रश्न 6.
पाइथागोरस प्रमेय के प्रयोग बिना दिखलाइए कि बिन्दु (4, 4), (3, 5) और (- 1, – 1) एक समकोण त्रिभुज के शीर्ष हैं।
हल:
माना दिए गए बिन्दु A(4, 4), B(3, 5) और C(- 1, – 1) हैं, तब

प्रश्न 7.
उस रेखा का समीकरण ज्ञात कीजिए जो y- अक्ष की धन दिशा से वामावर्त्त मापा गया 30° का कोण बनाती है।
हल:
माना रेखा OP, y- अक्ष से वामावर्त्त 30° का कोण बनाती है।
∴ x- अक्ष की धन दिशा से 90° + 30° = 120° का कोण बनाती है।
⇒ रेखा OP की ढाल = tan 120 = – \(\sqrt{3}\)

यह रेखा मूल बिन्दु (0, 0) से होकर जाती है। रेखा का बिन्दु ढाल रूप है
y – y₁ = m(x – x₁)
∴ OP का समीकरण y – 0 = – \(\sqrt{3}\) (x – 0)
या y = – \(\sqrt{3}\)x.

प्रश्न 8.
x का वह मान ज्ञात कीजिए जिसके लिए बिन्दु (x,- 1), (2, 1) और (4, 5) सरेख हैं।
हल:
मान लीजिए बिन्दु A (x, – 1), B(2, 1), C(4, 5) संरेख हैं यदि
AB की ढाल = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1+1}{2-x}=\frac{2}{2-x}\) …(1)
BC की ढाल = \(\frac{5-1}{4-2}=\frac{4}{2}\) = 2 …(2)
∴ समीकरण (1) और (2) से,
\(\frac{2}{2-x}\) = 2
या 1 = 2 – x .
x = 1.

प्रश्न 9.
दूरी सूत्र का प्रयोग किए बिना दिखलाइए कि बिन्दु (-2,-1), (4,0), (3, 3) और (-3, 2) एक समांतर चतुर्भुज के शीर्ष हैं।
हल:
मान लीजिए एक चतुर्भुज के शीर्ष A(- 2, – 1), B(4, 0), C(3, 3), तथा D(- 3, 2) हैं।

अर्थात् BC || AD
अतः AB || DC, BC || AD
अतः ABCD एक मांस चतुर्भुज है।

प्रश्न 10.
x- अक्ष और (3, – 1) और (4, – 2) बिन्दुओं को मिलाने वाली रेखा के बीच का कोण ज्ञात कीजिए।
हल:
माना 4(3, – 1), B(4, – 2) को मिलाने वाली रेखा AB की ढाल = \(\frac{-2+1}{4-3}=\frac{-1}{1}\) = – 1
यदि x- अक्ष और AB के बीच θ कोण हो, तो
tan θ = – 1 = tan 135°
θ = 135°.

प्रश्न 11.
एक रेखा की ढाल दूसरी रेखा की ढाल का दुगुना है। यदि दोनों के बीच के कोण की स्पर्शज्या (tangent) \(\frac{1}{3}\) है तो रेखाओं की ढाल ज्ञात कीजिए।
हल:
माना रेखाओं की ढाल m₁, m₂ हों, तब
∴ m₁ = 2m₂ यदि दोनों रेखाओं के बीच कोण हो, तो
tan θ = \(\frac{1}{3}\)

– ve चिन्ह लेने पर, \(1+2 m_{2}^{2}=-3 m_{2}\) या \(2 m_{2}^{2}+3 m_{2}\) +1 = 0
या (m₂ + 1) (2m₂ + 1) = 0 अर्थात् m₂ = – 1, – \(\frac{1}{2}\)
∴ रेखा की ढाल – 2, – 1, तथा – 1, – \(\frac{1}{2}\)

प्रश्न 12.
एक रेखा (x₁, y₁) और (h, k) से जाती है। यदि रेखा की ढाल m है तो दिखाइए
k – y₁ = m(h – x₁).
हल:
माना रेखा AB बिन्दु A(x₁, y₁) और B(h, k) से गुजरती हो, तब
∴ AB की ढाल = \(\frac{k-y_{1}}{h-x_{1}}\) = m
अर्थात् k – y₁ = m(h – x₁)

प्रश्न 13.
यदि तीन बिन्दु (h, 0), (a, b) और (0, k) एक रेखा पर हैं तो दिखाइए कि \(\frac{a}{h}+\frac{b}{k}\) = 1
हल:
मान लीजिए बिन्दु A (h, 0), B(a, b), तथा C(0, k) एक रेखा पर हों, तब

या (a – h) (k – b) = – ab
या ak – ab – hk + hb = – ab
∴ ak + hb = hk
hk से भाग देने पर, \(\frac{a}{h}+\frac{b}{k}\) = 1

प्रश्न 14.
जनसंख्या और वर्ष के निम्नलिखित लेखाचित्र पर विचार कीजिए। (देखिए आकृति में) रेखा AB की ढाल ज्ञात कीजिए और इसके प्रयोग से बताइए कि वर्ष 2010 में जनसंख्या कितनी होगी ?
हल:
दी गयी आकृति में रेखा AB बिन्दु A(1985, 92) और B(1995,97) से होकर जाती है।
∴ AB की ढाल = \(\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}\)
मान लीजिए सन् 2010 में जनसंख्या y₁ करोड़ होगी जो बिन्दु P(2010, y₁), AB पर पड़ता है।

∴ ABP सरेखीय हैं।
या AB की ढाल = BP की ढाल
\(\frac{1}{2}=\frac{y_{1}-97}{2010-1995}=\frac{y_{1}-97}{15}\)
∴ 2(y₁ – 97) = 15
2y₁ = 15 + 2 × 97
= 15 + 194 = 209
∴ y₁ = \(\frac{209}{2}\) = 104.5
सन् 2010 में जनसंख्या 104.5 करोड़ होगी।

MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 9 अनुक्रम तथा श्रेणी विविध प्रश्नावली

प्रश्न 1.
दर्शाइए कि किसी समांतर श्रेढ़ी के (m + n) वें तथा (m – n) पदों का योग m वें पद का दुगुना है।
हल:
मान लीजिए समांतर श्रेढ़ी का पहला पद a और सार्व अंतर d है।

प्रश्न 2.
यदि किसी समांतर श्रेढ़ी की तीन संख्याओं का योग 24 है तथा उनका गुणनफल 440 है तो संख्याएँ ज्ञात कीजिए।
हल:
मान लीजिए समांतर श्रेढ़ी की तीन संख्याएँ a – d, a और a + d हैं।
तीनों संख्याओं का योग = (a – d) + a + (a + d) = 24
∴ 3a = 24 या a = 8
तीन संख्याओं का गुणनफल = (a – d). a .(a + a)
= a (a² – d²)
= 8(64 – d²) [∵ a = 8]
या 8(64 – d²) = 440
या 64 – d² = 55
d² = 64 – 55 = 9 या d = 3
अतः अभीष्ट संख्याएँ 5, 8, 11.

प्रश्न 3.
माना कि किसी समांतर श्रेढ़ी के n, 2n तथा 3n पदों का योगफल क्रमशः S₁, S₂ तथा S₃ हैं, तो दिखाइए कि S₃ = 3(S₂ – S₁).
हल:
मान लीजिए समांतर श्रेढ़ी का पहला पद a और सार्व अंतर d है।

प्रश्न 4.
200 और 400 के मध्य आने वाली उन सभी संख्याओं का योगफल ज्ञात कीजिए जो 7 से विभाजित हों।
हल:
200 से 400 के मध्य आने वाली संख्याएँ 203, 210, 217,…….., 399
मान लीजिए 399, n वाँ पद है।
∴ 399 = a + (n – 1).7
= 203 + 7 (n – 1)
या 399 – 203 = 196 = 7(n – 1)
∴ n – 1 = \(\frac{196}{7}\) = 28 या n = 29
∴ 203 + 210 + 217 +……+ 399
= \(\frac{29}{2}\)[203 + 399] [∵ S = \(\frac{n}{2}\)(a + l)]
= \(\frac{29}{2}\)(602) = 29 × 301
= 8729.

प्रश्न 5.
1 से 100 तक आने वाले उन सभी पूर्णांकों का योगफल ज्ञात कीजिए जो 2 या 5 से विभाजित हों।
हल:
2 से विभाजित होने वाले पूर्णांक 2, 4, 6,…., 100
इनकी कुल संख्या = 50
5 से विभाजित होने वाले पूर्णांक 5, 10, 15, 20,……100
इनकी कुल संख्या = 20
2 और 5 दोनों से विभाजित होने वाले पूर्णांक 10, 20, 30,…., 100
इनकी कुल संख्या = 10
1 से 100 तक आने वाले पूर्णांक जो 2 या 5 से विभाजित हों, तब
= (2 + 4 + 6 + ……50 पदों तक) + (5 + 10 + 15 +…… 20 पदों तक) – (10 + 20 + 30 +……10 पदों तक)
= \(\frac{50}{2}\)[4 + (50 – 1). 2] + \(\frac{20}{2}\)[10 + (20 – 1).5] – \(\frac{10}{2}\)[20 + (10 – 1). 10]
= \(\frac{50 \times 102}{2}\) + 10 x 105 – 5 x 110
= 2250 + 1050 – 550
= 3050.

प्रश्न 6.
दो अंकों की उन सभी संख्याओं का योगफल ज्ञात कीजिए, जिनको 4 से विभाजित करने पर शेषफल 1 हो।
हल:
दो अंको की वे संख्याएँ जो 4 से विभाजित करने पर 1 शेष रहता है 13, 17, 21,….., 97
मान लीजिए n पद हों, तब n वाँ पद,
97 = 13 + (n – 1). 4
∴ 84 = (n – 1) × 4
∴ n = 22
∴ 13 + 17 + 21 +…..+ 97 = \(\frac{22}{2}\)[26 + (22 – 1).4]
= 11 × (26 + 84)
= 11 × 110
= 1210.

प्रश्न 7.
सभी x, y ϵ N के लिए f(x + y) = f(x).f(y) को संतुष्ट करता हुआ f एक ऐसा फलन है कि f(1) = 3 एवं \(\sum_{x=1}^{n}\)f(x) = 120 तो n का मान ज्ञात करो।
हल:
f(1) = 3, f(2) = f(1 + 1) = f(1) .f(1) = 3.3 = 9
f(3) = f(1 + 2) = f(1). f(2) = 3.9 = 27
f(4) = f(1 + 3) = f(1). (3) = 3 . 27 = 81
इस प्रकार f(1) + f(2) + f(3) +……, n पदों तक
= 3 + 9 + 27 + 81 + ……., n पदों तक = 120
⇒ \(\frac{3\left(3^{n}-1\right)}{3-1}\) = 120
या 3(3ⁿ – 1) = 120 × 2 = 240
3ⁿ – 1 = \(\frac{240}{3}\) = 80
या 3ⁿ = 81 = 3⁴
अतः n = 4.

प्रश्न 8.
गुणोत्तर श्रेढ़ी के कुछ पदों का योग 315 है, उसका प्रथम पद तथा सार्व अनुपात क्रमशः 5 और 2 हैं।
अंतिम पद तथा पदों की संख्या ज्ञात करो।
हल:
दी हुई गुणोत्तर श्रेणी
5 + 10 + 20 + 40 +…….
n पदों का योग = \(\frac{5\left(2^{n}-1\right)}{2-1}\) = 315
∴ 2ⁿ – 1 = 63
या 2ⁿ = 64 = 2⁶
n = 6
6 वाँ पद = 5 × 2⁶ – 1
= 5.2⁵
= 5 × 32 = 160.

प्रश्न 9.
किसी गुणोत्तर श्रेढ़ी का प्रथम पद 1 है। तीसरे एवं पाँचवें पदों का योग 90 हो, तो गुणोत्तर श्रेढ़ी का सार्व अनुपात ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेढ़ी का सार्व अनुपात r है।
तीसरा पद = ar² = 1.r² = r²
पाँचवाँ पद = ar⁴ = r⁴
तीसरे और पाँचवें पद का योग = r² + r⁴ = 90
r⁴ + r² – 90 = 0
या (r² + 10)(r² – 9) = 0
∴ r² = – 10 मान्य नहीं है।
∴ r² – 9 = 0, r² = 9
∴ r = ± 3.

प्रश्न 10.
किसी गुणोत्तर श्रेढ़ी के तीन पदों का योग 56 है। यदि हम क्रम से इन संख्याओं में से 1, 7, 21 घटाएँ तो हमें एक समांतर श्रेढ़ी प्राप्त होती है। संख्याएँ ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेढ़ी की तीन संख्याएँ a, ar, ar² हैं।
तीनों पदों का योग = a + ar + ar² = 56 …..(1)
इन संख्याओं में से 1, 7, 21 घटाने पर संख्याएँ
ar – 1, ar – 7, ar² – 21 समांतर श्रेढ़ी में हैं।
∴ 2(ar – 7) = (a – 1) + (ar² – 21)
या 2ar – 14 = ar² + a – 22
ar² – 2ar + a = 22 – 14 = 8 ….(2)
समी. (1) को (2) से भाग देने पर
= \(\frac{a\left(1+r+r^{2}\right)}{a\left(1-2 r+r^{2}\right)}\) = \(\frac{56}{8}\) = 7
या 7(1 – 2r + r²) = 1 + r + r²
6r² – 15r + 6 = 0
2r² – 5r + 2 = 0
या (r – 2) (2r – 1) = 0 या r = 2, \(\frac{1}{2}\)
समी (1) में r = 2 रखने पर,
a(1 + 2 + 4) = 56 या a = \(\frac{56}{7}\) = 8
इस प्रकार तीन संख्याएँ हैं: 8, 16, 32.
पुन: समी (1) में r = \(\frac{1}{2}\) रखने से,
a\(\left(1+\frac{1}{2}+\frac{1}{4}\right)\) = 56
a = \(\frac{56 \times 4}{7}\) = 32
∴ तीन संख्याएँ 32, 16, 8.
अतः अभीष्ट संख्याएं 8, 16, 32 हैं।

प्रश्न 11.
किसी गुणोत्तर श्रेढ़ी के पदों की संख्या सम है। यदि उसके सभी पदों का योगफल, विषम स्थान पर रखे पदों के योगफल का 5 गुना है, तो सार्व अनुपात ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेढ़ी का पहला पद = a सार्व अनुपात = r और पदों की संख्या = 2n
सभी पदों का योगफल = \(\frac{a\left(r^{2 n}-1\right)}{r-1}\)
विषम स्थानों पर रखे पद a, ar², ar⁴, …. n पदों तक
इनका योग = a + ar² + ar² +……n पदों तक
= \(\frac{a\left[\left(r^{2}\right)^{n}-1\right]}{r^{2}-1}=\frac{a\left(r^{2 n}-1\right)}{r^{2}-1}\)
दिया है :
गुणोत्तर श्रेढ़ी के 2n पदों का योगफल = 5 × [विषम स्थानों पर स्थित पदों का योगफल]

प्रश्न 12.
एक समांतर श्रेढ़ी के प्रथम चार पदों का योगफल 56 है। अंतिम चार पदों का योगफल 112 है। यदि इसका प्रथम पद 11 है, तो पदों की संख्या ज्ञात कीजिए।
हल:
मान लीजिए समांतर श्रेणी
a + (a + d) + (a + 2a) +……+ l जबकि l अंतिम पद n वाँ पद है।
प्रथम 4 पदों का योगफल = \(\frac{4}{2}\)[2a + (4 – 1) d]
= 2[22 + 3d] [∵ a = 11]
दिया है: 2[22 + 3d) = 56
⇒ 3d + 22 = 28 या d = 2
अंतिम पद = a + (n – 1) d = 11 + (n – 1).2
= 2n + 9
अंतिम चार पद 2n + 9, 2n + 7, 2n + 5, 2n + 3
इनका योगफल = \(\frac{4}{2}\)[2(2n + 9) + (4 – 1). (- 2)]
= 2[4n + 18 – 6]
= 2[4n + 12]
दिया है : 2(4n + 12) = 112
∴ 4n + 12 = 56
4n = 56 – 12 = 44
∴ n = 11.

प्रश्न 13.
यदि \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\) (x ≠ 0) हो, तो दिखाइए कि a, b, c, d गुणोत्तर श्रेढ़ी में है।
हल:

अतः a, b, c, d गुणोत्तर श्रेढ़ी में है।

प्रश्न 14.
किसी गुणोत्तर श्रेढ़ी में S,n पदों का योग, P उनका गुणनफल तथा R उनके व्युत्क्रमों का योग हो तो सिद्ध कीजिए कि P²Rⁿ = Sⁿ.
हल:
मान लीजिए गुणोत्तर श्रेढ़ी a + ar + ar² +….. + ar^{n – 1}
इन n पदों का गुणनफल, P = a. ar . ar²….. ar^{n – 1}
= aⁿ. r^{1 + 2 +…+ (n – 1)}
= \(a^{n} r \frac{n(n-1)}{2}\)

अतः P²Rⁿ = Sⁿ.

प्रश्न 15.
किसी समांतर श्रेढ़ी का p वाँ, धूवाँ, वाँ पद क्रमशः a, b, c हैं, तो सिद्ध कीजिए
(q – r)a + (r – p)b + (p – q) c = 0.
हल:
मान लीजिए समांतर श्रेणी
A + (A + d) + (A + 2d) +…. है।
p वाँ पद = A + (p – 1) d = a ….(1)
q वाँ पद = A + (q – 1) d = b ….(2)
r वाँ पद = A + (r – 1) d=c …..(3)
समी (2) में से समी (3) को, समी (3) में से समी (1) को, समी (1) में से समी (2) को घटाने पर
(q – r)d = b – c ….(4)
(r – p)d = c – a …(5)
(p – q)d = a – b ….(6)
समीकरण (4), (5) तथा (6) को क्रमशः a, b तथा c से गुणा करके जोड़ने पर,
a(q – r)d + b(r – p)d + c(p – d)d
= a(b – c) + b(c – a) + c(a – b)
= ab – ac + bc – ba + ca – bc
= 0
दोनों पक्षों में d से भाग देने पर,
(q – r)a + (r – p)b + (p – q)c = 0.

प्रश्न 16.
यदि \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) समांतर श्रेड़ी में हैं, तो सिद्ध करो कि a, b, c समांतर भेट्टी में हैं।
हल:

प्रश्न 17.
यदि a, b, c, d गुणोत्तर श्रेढ़ी में हैं, तो सिद्ध कीजिए कि \(\left(a^{n}+b^{n}\right),\left(b^{n}+c^{n}\right),\left(c^{n}+d^{n}\right)\) गुणोत्तर श्रेढ़ी में हैं।
हल:
a, b, c, d गुणोत्तर श्रेढ़ी में हैं।
मान लीजिए सार्व अनुपात r है।

प्रश्न 18.
यदि x² – 3x + p = 0 के मूल a तथा b हैं तथा x² – 12x + q = 0 के मूल c तथा d हैं, जहाँ a, b, c, d गुणोत्तर श्रेढ़ी के रूप में हैं। सिद्ध कीजिए कि
(q+ p) : (q – p) = 17 : 15.
हल:
यदि समीकरण Ax² + Bx + C = 0 के मूल a , B हैं, तो
α + β = \(\frac{-B}{A}\), αβ = \(\frac{C}{A}\)
दिया है कि x² – 3x + p = 0 के मूल a, b हैं
∴ a+ b = 3, ab = p …..(1)
इसी प्रकार x² – 12x + q = 0 के मूल c, d हैं
∴ c + d = 12. cd = q …..(2)
अब a, b, c, d गुणोत्तर श्रेढ़ी में हैं, जिसका मान लीजिए r सार्व अनुपात है।
∴ b = ar, c = ar², d= ar³
a + b = 3, a + ar = 3 …(3)
c + d = 12 या ar² + ar³ = 12 …(4)
समी (3) को (4) से भाग देने पर,

प्रश्न 19.
दो धनात्मक संख्याओं a और b के बीच समांतर माध्य तथा गुणोत्तर माध्य का अनुपात m : n है। दर्शाइए कि
a : b = \((m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})\).
हल:
a और b के बीच समांतर माध्य = \(\frac{a+b}{2}\)
a और b के बीच गुणोत्तर माध्य = \(\sqrt{a b}\)
दोनों माध्यों का अनुपात m : n

प्रश्न 20.
यदि a, b, c समांतर श्रेढ़ी में हैं; b, c, d गुणोत्तर श्रेढ़ी में हैं तथा \(\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\) समांतर श्रेढ़ी में हैं, तो सिद्ध कीजिए कि a, c, e गुणोत्तर श्रेढ़ी में हैं।
हल:
a, b, c समांतर श्रेढ़ी में हैं ∴ \(\frac{a+c}{2}\) = b …(1)
b, c, d, गुणोत्तर श्रेढ़ी में हैं, ∴ bd = c² …(2)
\(\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\) समांतर श्रेढ़ी में हैं, ∴ \(\frac{2}{d}\) = \(\frac{1}{c}+\frac{1}{e}\)
⇒ d = \(\frac{2 c e}{c+e}\) …..(3)
b और d का मान (1) और (3) से लेकर (2) में रखने पर

प्रश्न 21.
निम्नलिखित श्रेढ़ियों के n पदों का योग ज्ञात कीजिए :
(i) 5 + 55 + 555+ ……
(ii) 0.6 + 0.66 + 0.666 +…..
हल:
(i) S = 5 + 55 + 555 +…..n पदों तक

(ii) S = 0.6 + 0.66 + 0.666 +….n पदों तक

प्रश्न 22.
श्रेढ़ी का 20वाँ पद ज्ञात कीजिए :
2 × 4 + 4 × 6 + 6 × 8 +…..+ n पदों तक
हल:
2, 4, 6,….. का 20 वाँ पद = 2n = 2 × 20 = 40.
4, 6, 8….. का 20 वाँ पद = 4 + 19 × 2 = 4 + 38 = 42
∴ 2 × 4 + 4 × 6 + 6 × 8+…… का 20 वाँ पद
= 40 × 42 = 1680.

प्रश्न 23.
श्रेणी 3 + 7 + 13 + 21 + 31 +….. के n पदों का योगफल ज्ञात कीजिए।
हल:

= n² + n – 2 +3
= n² + n + 1
∴ दी हुई श्रेणी का योग
= Σn² + Σn + n
= \(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}+n\)
= \(\frac{n}{6}\)[(n + 1)(2n + 1) + 3(n + 1) + 6]
= \(\frac{n}{6}\)[2n² + 6n + 10]
= \(\frac{n}{3}\)[n² + 3n + 5].

प्रश्न 24.
यदि S₁, S₂, S₃, क्रमशः प्रथम n प्राकृत संख्याओं का योग, उनके वर्गों का योग तथा घनों का योग है, तो सिद्ध कीजिए कि
\(9 S_{2}^{2}=S_{3}\left(1+8 S_{1}\right)\)
हल:
S₁ = n प्राकृत संख्याओं का योग
= 1 + 2 + 3 +…..n पदों तक
= \(\frac{n(n+1)}{2}\) …..(1)
S₂ = n प्राकृत संख्याओं के वर्गों का योग
= 1² + 2² + 3² +…..+ n²
= \(\frac{n(n+1)(2 n+1)}{6}\) …..(2)
S₃ = n प्राकृत संख्याओं के घनों का योग
= 1³ + 2³ + 3³ +…..+ n³
= \(\frac{n^{2}(n+1)^{2}}{4}\)

प्रश्न 25.
निम्नलिखित श्रेणियों के n पदों का योग ज्ञात कीजिए :
\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots\)
हल:
अंश में दी हुई संख्याएँ 1³, 1³ + 2³, 1³ + 2³ + 3³, …..
n वाँ पद = 1³ + 2³ + 3³ +…..+ n³
= \(\frac{n^{2}(n+1)^{2}}{4}\)
हर में दी हुई संख्याएँ 1, (1 + 3), (1 + 3 + 5), ……
n वाँ पद = 1 + 3 + 5 +……n पदों तक

प्रश्न 26.
दशाइए कि \(\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots . .+n(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots .+n^{2}(n+1)}\) = \(\frac{3 n+5}{3 n+1}\).
हल:

प्रश्न 27.
कोई किसान एक पुराने ट्रैक्टर को 12000 रू. में खरीदता है। वह 6000 रु. नकद भुगतान करता है और शेष राशि को 500 रू की वार्षिक किस्त के अतिरिक्त उस धन पर जिसका भुगतान न किया गया हो 12% वार्षिक ब्याज भी देता है। किसान को ट्रैक्टर की कुल कितनी कीमत देनी पड़ेगी?
हल:
पुराने ट्रैक्टर का मूल्य = 12000 रू
नकद भुगतान = 6000 रू
शेष = 12000 – 6000 = 6000 रू
एक किस्त का भुगतान = 500 रू
कुल किस्तें = \(\frac{6000}{12}\) = 12
P मूलधन पर 12% प्रतिवर्ष की दर से 1 वर्ष का ब्याज
= \(\frac{P \times 12 \times 1}{100}=\frac{3}{25} P\)

कुल भुगतान = (12000 + 4680) रू
= 16680 रू।

प्रश्न 28.
शमशाद अली 22000 रू में एक स्कूटर खरीदता है। वह 4000 रू नकद देता है और शेष राशि को 1000 रू वार्षिक किस्त के अतिरिक्त उस धन पर जिसका भुगतान न किया गया हो 10% वार्षिक ब्याज भी देता है। उसे स्कूटर के लिए कुल कितनी राशि चुकानी पड़ेगी?
हल:
स्कूटर की कीमत = 22000 रू
नकद भुगतान = 4000 रू
शेष = 22000 – 4000 = 18000 रू
एक किस्त की राशि = 1000 रू
∴ कुल किस्तें = \(\frac{18000}{1000}\) = 18
P मूलधन पर एक वर्ष का 10% प्रति वर्ष की दर से ब्याज
= \(\frac{P \times 10 \times 1}{100}\) = \(\frac{P}{10}\)
किस्त देने के बाद शेष राशि जिस पर एक वर्ष का ब्याज लगना है,
= 18000, 17000, 16000,….., 1000
कुल ब्याज की राशि
= \(\frac{1}{10}\)(18000 + 17000 + 16000 +…..+ 18 पदों तक)
= \(\frac{1}{10} \times \frac{18}{2}\)[2 × 18000 – (18 – 1) × 1000]
= \(\frac{9}{10}\)[36000 – 17000)
= \(\frac{9 \times 19000}{10}\) = 17100 रू
कुल किश्तों की राशि = 18000 रू
नकद = 4000 रू
कुल भुगतान = (18000 + 17000) + 4000 रू
= 39,100 रू।

प्रश्न 29.
एक व्यक्ति अपने चार मित्रों को पत्र लिखता है। वह प्रत्येक को उसकी नकल करके चार दूसरे व्यक्तियों को भेजने का निर्देश देता है, तथा जिनसे यह भी करने को कहता है कि प्रत्येक पत्र प्राप्त करने वाला व्यक्ति इस श्रृंखला को जारी रखे। यह कल्पना करके कि श्रंखला न टूटे तो 8वें पत्रों के समूह भेजे जाने तक कितना डाक खर्च होगा जबकि एक पत्र का डाक खर्च 50 पैसे है।
हल:
पहला व्यक्ति चार पत्र लिखता है। पत्र प्राप्त करने वाले 4 व्यक्ति फिर चार-चार पत्र लिखते हैं। इस प्रकार श्रृंखला बढ़ती चली जाती है।
हर अवसर पर पत्रों की संख्याएँ 4, 16, 24…… 8 पदों तक
कुल पत्रों की संख्या = 4 + 16 + 64 + ……………8 पदों तक
= \(\frac{4\left(4^{8}-1\right)}{4-1}\) = \(\frac{4}{3}\)(65536 – 1)
= \(\frac{4}{3}\) × 65535 = 87380
एक पत्र का डाक खर्च = 50 पै. = \(\frac{1}{2}\)रू
कुल डाक खर्च = 87380 x \(\frac{1}{2}\)
= 43690 रू

प्रश्न 30.
एक आदमी ने एक बैंक में 10000 रूपये 5% वार्षिक साधारण ब्याज पर जमा किया। जब से रकम बैंक में जमा की गई तब से, 15वें वर्ष में उसके खाते में कितनी रकम हो गई तथा 20 वर्षों बाद कल कितनी रकम हो गयी, ज्ञात कीजिए।
हल:
बैंक में जमा की गई राशि = 10000 रू
ब्याज की दर = 5% प्रति वर्ष
एक वर्ष बाद ब्याज = \(\frac{10000 \times 5 \times 1}{100}\) = 500रू
इस प्रकार हर वर्ष उसे 500 रू ब्याज के मिलेंगे।
1 वर्ष, 2 वर्ष, 3 वर्ष,…….बाद ब्याज की राशि
500, 1000, 1500, ….
15 वें वर्ष में ब्याज = (n – 1) × 500 = (15 – 1) x 500
= 14 × 500
= 7000 रू
मूलधन = 10000 रू
उसके खाते में 15वें वर्ष में = 10000 + 7000
= 17000 रू होंगें
20 वर्ष का ब्याज = 20 × 500
= 10000 रू
मूलधन = 10000 रू
20 वर्ष बाद बैंक में कुल जमा राशि = 10000 + 10000 = 20000 रू।

प्रश्न 31.
एक निर्माता घोषित करता है कि उस की मशीन जिसका मूल्य 15625 रूपये है, हर वर्ष 20% की दर से उसका अवमूल्यन होता है। 5 वर्ष के बाद मशीन का अनुमानित मूल्य ज्ञात कीजिए।
हल:
यदि किसी मशीन का r% की दर से अवमूल्यन हो रहा है n वर्ष बाद मशीन का मूल्य \(P\left(1-\frac{r}{100}\right)^{n}\) होगा।
प्रारभ में मशीन का मूल्य P रूपये है।
यहां पर P = 15625, r = 20% प्रति वर्ष, n = 5 वर्ष
∴ उस मशीन का 5 वर्ष बाद का मूल्य
= 15625 \(\left(1-\frac{20}{100}\right)^{5}\)
= 15625 \(\left(\frac{4}{5}\right)^{5}\)
= 15625 x (.8),sup>5 = 5120 रू।

प्रश्न 32.
किसी कार्य को कुछ दिनों में पूरा करने के लिए 150 कर्मचारी लगाए गए। दूसरे दिन 4 कर्मचारियों ने काम छोड़ दिया, तीसरे दिन चार और कर्मचारियों ने काम छोड़ दिया तथा इस प्रकार अन्य। अब कार्य पूरा करने में 8 दिन अधिक लगते हैं, तो दिनों की संख्या ज्ञात कीजिए, जिनमें कार्य पूरा किया गया।
हल:
150 कर्मचारी उस कार्य को n दिनों में समाप्त करते हैं
150 कर्मचारियों का 1 दिन का काम = \(\frac{1}{n}\)
1 कर्मचारी का 1 दिन का काम = \(\frac{1}{150n}\)
पहले दिन 150 कर्मचारी 1 दिन में \(\frac{150}{150 n}\) कार्य करते हैं
दूसरे दिन 146 कर्मचारी 1 दिन में \(\frac{146}{150 n}\) कार्य करते हैं
तीसरे दिन 142 कर्मचारी 1 दिन में \(\frac{146}{150 n}\) कार्य करते हैं
वह काम n + 8 दिन में पूरा हुआ
∴ \(\frac{150}{150 n}+\frac{146}{150 n}+\frac{142}{150 n}\) +…….(n + 8) पदों तक = 1
या \(\frac{1}{150 n}\)[150 + 146 + 142 +…n + 8) पदों तक] = 1
या \(\frac{n+8}{2(150 n)}\)[2 x 150 +(n + 8 – 1) x (- 4)] = 1
(n + 8)[300 – 4(n + 7)] = 300n
या (n + 8)(- 4n + 272) = 300n
या (n + 8)(n – 68) = – 75n
या n² – 60n – 544 = – 75n
या n² + 15n – 544 = 0
या (n + 32)(n – 17) = 0
n ≠ – 32 या n = 17
कुल समय = n + 8 दिन
= 17 + 8 = 25 दिन।

MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 9 अनुक्रम तथा श्रेणी Ex 9.4

प्रश्न 1 से 7 तक प्रत्येक श्रेणी के n पदों का योग ज्ञात कीजिए :
प्रश्न 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ….
हल:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +……
प्रत्येक पद के दो गुणनखण्ड हैं।
पहले गुणनखंडों से बनी श्रेढ़ी 1, 2, 3, 4……
∴ n वाँ पद = n
दूसरे गुणनखंडों से बनी श्रेढ़ी 2, 3, 4, 5……
n वाँ पद = (n + 1)
1 × 2 + 2 × 3 + 3 × 4+…. का n वाँ पद = n(n + 1) = n² + n

प्रश्न 2.
1 × 2 × 3 + 2 × 3 × 4+ 3 × 4 × 5 +…..
हल:
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +……
पहले गुणनखंडों की श्रेढ़ी 1, 2, 3, 4, …..n
n वाँ पद = n
दूसरे गुणनखंडों की श्रेढ़ी 2, 3, 4, 5,….
n वाँ पद = (n + 1)
तीसरे गुणनखंडों की श्रेढ़ी. 3, 4, 5….
n वाँ पद = (n + 2)
∴ 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +…… का n वाँ पद
= n(n + 1)(n + 2) = n(n² + 3n + 2)
=n³+ 3n² + 2n

प्रश्न 3.
3 × 1² + 5 × 2² + 7 × 3² +……
हल:
3 × 1² + 5 × 2² + 7 × 3² +…..
पहले गुणनखंड 3, 5, 7,….. का n वाँ पद = 3 + (n – 1). 2 = 2n + 1
दूसरे गुणनखंड 1², 2², 3²….. का nवाँ पद = n²
∴ 3 × 1² + 5 × 2² + 7 × 3² +…… का गवाँ पद
= (2n + 1) n² = 2n³ + n²

प्रश्न 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) +…….
हल:
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) +…….
1, 2, 3,….. का गवाँ पद = n
2, 3, 4,……का n वाँ पद = (n + 1)

प्रश्न 5.
5² + 6² + 7² +….+ 20².
हल:
n वें पद वाली इस श्रेणी में,
(n + 4)² = n² + 8n + 16
S_n = ΣT_n = Σn² + 8 Σn + (16 + 16 +……n. पदों तक)
= \(\frac{n(n+1)(2 n+1)}{6}\) + 8 × \(\frac{n(n+1)}{2}\) + 16n

प्रश्न 6.
3 × 8 + 6 × 11 + 9 × 14 +…..
हल:
3 × 8 + 6 × 11 + 9 × 14 +….
3, 6, 9 का n वाँ पद = 3n
8, 11, 14,…..का n वाँ पद = 8 + (n – 1). 3 = 3n + 5
∴ 3 × 8 + 6 × 11 + 9 × 14 +……का nवाँ पद = 3n(3n + 5)
= 3 (3n² + 5n)
दी हुई श्रेढ़ी के n पदों का योगफल

प्रश्न 7.
1² + (1² + 2²) + (1² + 2² + 3²) +…..
हल:

दी हुई श्रेढ़ी के n पदों का योगफल

प्रश्न 8 से 10 तक प्रत्येक श्रेणी के n पदों का योग ज्ञात कीजिए जिसका nवाँ पद दिया है :
प्रश्न 8.
n(n + 1)(n + 4).
हल:
Tⁿ = n(n + 1)(n + 4) = n(n² + 5n + 4)
= n³ + 5n² + 4n
दी हुई श्रेढ़ी के n पदों का योग = Σn³ + 5Σn² + 4Σn
= \(\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}\)
= \(\frac{n(n+1)}{12}\)[3n(n + 1) + 10(2n + 1) + 24]
= \(\frac{n(n+1)}{12}\)[3n² + 3n + 20n + 10 + 24]
= \(\frac{n(n+1)}{12}\)[3n² + 23n + 34]

प्रश्न 9.
n² + 2ⁿ.
हल:
दी हुई श्रेढ़ी के n पदों का योग
= Σn² + Σ2ⁿ

प्रश्न 10.
(2n – 1).
हल:
Tⁿ = (2n – 1)² = 4n² – 4n + 1
दी हुई श्रेढ़ी के n पदों का योग

MP Board Class 11th Maths Solutions

MP Board Class 11th Maths Solutions Chapter 9 अनुक्रम तथा श्रेणी Ex 9.3

प्रश्न 1.
गुणोत्तर श्रेणी \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8} \dots\) का 20वाँ तथा nवाँ पद ज्ञात कीजिए।
हल:
गुणोत्तर श्रेणी का पहला पद, a = \(\frac{5}{2}\)
दूसरा पद = \(\frac{5}{4}\), सार्व अनुपात = \(\frac{1}{2}\)
n वाँ पद = \(a r^{n-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}\) = \(\frac{5}{2^{n}}\).
n = 20 रखने पर,
20 वाँ पद = \(\frac{5}{2^{20}}\)

प्रश्न 2.
उस गुणोत्तर श्रेणी का 12वाँ पद ज्ञात कीजिए, जिसका 8वाँ पद 192 तथा सार्व अनुपात 2 है।
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद = a
सार्व अनुपात = 2

प्रश्न 3.
किसी गुणोत्तर श्रेणी का 5वाँ, 8वाँ तथा 11 वाँ पद क्रमशः p, q तथा s हैं, तो दिखाइए कि q² = ps.
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद = a
सार्व तथा अनुपात =r
5वाँ पद = ar^{5 – 1} = ar⁴ = p
8वाँ पद = ar^{8 – 1} = ar⁷ = q
11वाँ पद = ar^{11 – 1}= ar¹⁰ = s
बायाँ पक्ष = q² = (ar⁷)² = a² . r¹⁴
दायाँ पक्ष = ps = ar⁴ ar¹⁰= a² . r¹⁴
अतः q² = ps.

प्रश्न 4.
किसी गुणोत्तर श्रेणी का चौथा पद उसके दूसरे पद का वर्ग है तथा प्रथम पद – 3 है, तो 7 वाँ पद ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद, a = – 3
तथा सार्व-अनुपात = r
चौथा पद = ar^{4 – 1} = ar³ = – 3r³
दूसरा पद = ar = – 3r
दिया है : चौथा पद = (दूसरे पद)²
⇒ – 3r³ = (-3r)² = 9r²
r= – 3
7वाँ पद = \(a r^{7-1}=a r^{6}=(-3)(-3)^{6}\)
= (- 3)⁷ = – 2187.

प्रश्न 5.
अनुक्रमों का कौन सा पद :
(a) 2, 2\(\sqrt{2}\), 4, … ; 128 है ?
(b) \(\sqrt{3}\), 3, 3, …. ; 729 है ?
(c) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}\), ….; 19683 है?
हल:
(a) गुणोत्तर श्रेणी का पहला व दूसरा पद क्रमशः 2 और 2\(\sqrt{2}\)

∴ \(\frac{n-1}{2}\) = 6, n – 1 = 12 या n = 13.

प्रश्न 6.
x के किस मान के लिए संख्याएँ –\(\frac{2}{7}\), x, – \(\frac{7}{2}\) गुणोत्तर श्रेणी में हैं ?
हल:
संख्याएँ a, b और c गुणोत्तर श्रेणी में है यदि b² = ac
∴ –\(\frac{2}{7}\), x, – \(\frac{7}{2}\) गुणोत्तर श्रेणी में हैं
\(x^{2}=\left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right)\) = 1
x = ± 1.

प्रश्न 7 से 10 तक प्रत्येक गुणोत्तर श्रेणी का योगफल निर्दिष्ट पदों तक ज्ञात कीजिए।
प्रश्न 7.
0.15, 0.015, 0.0015,…..20 पदों तक।
हल:
गुणोत्तर श्रेणी 0.15, 0.015, 0.0015
पहला पद, a = 0.15
सार्व अनुपात, r = \(\frac{0.015}{0.15}\) = 0.1

प्रश्न 8.
\(\sqrt{7}, \sqrt{21}, 3 \sqrt{7}\),…..n पदों तक।
हल:
गुणोत्तर श्रेणी \(\sqrt{7}, \sqrt{21}, 3 \sqrt{7}\), …….
पहला पद, a = \(\sqrt{7}\) , सार्व अनुपात, r = \(\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\)

प्रश्न 9.
1, – a, a², – a³,…. पदों तक (यदि a ≠ – 1).
हल:
गुणोत्तर श्रेणी 1, – a, a, ², – a³,…..
पहला पद, a = 1, सार्व अनुपात, r = \(\frac{-a}{1}\) = – a

प्रश्न 10.
x³, x⁵, x⁷, …..n पदों तक (यदि x ≠ ± 1).
हल:
गुणोत्तर श्रेणी x³, x⁵, x⁷, …..

प्रश्न 11.
मान ज्ञात कीजिए \(\sum_{k=1}^{11}\left(2+3^{k}\right)\).
हल:

प्रश्न 12.
एक गुणोत्तर श्रेणी के तीन पदों का योगफल \(\frac{39}{10}\) है तथा उनका गुणनफल 1 है। सार्व अनुपात तथा पदों को ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेणी के तीन पद \(\frac{a}{r}\), a तथा ar हैं।

प्रश्न 13.
गुणोत्तर श्रेणी 3,3², 3³,… के कितने पद आवश्यक हैं ताकि उनका योगफल 120 हो जाए।
हल:
मान लो गुणोत्तर श्रेणी के कुल पद = n

या 3(3ⁿ – 1) = 120 × 2 = 240
3 से भाग देने पर
3ⁿ – 1 = \(\frac{240}{3}\) = 80
या 3ⁿ = 80 + 1 = 81 = 3⁴
अत:
n = 4.

प्रश्न 14.
किसी गुणोत्तर श्रेणी के प्रथम तीन पदों का योगफल 16 है तथा अगले 3 पदों का योग 128 है तो गुणोत्तरं श्रेणी का प्रथम पद, सार्व अनुपात तथा n पदों का योगफल ज्ञात कीजिए।
हल:
मान लीजिए गुणोत्तर श्रेणी a, ar, ar²,…. है।
पहला पद = a, सार्व अनुपात = r
तीन पदों का योगफल = \(\frac{a\left(1-r^{3}\right)}{1-r}\) = 16 …(1)
चौथा पद = a × r^{n – 1} = ar^{4 – 1} = ar³
अगले तीन पदों का योगफल = \(\frac{a r^{3}\left(1-r^{3}\right)}{1-r}\) = 128

प्रश्न 15.
एक गुणोत्तर श्रेणी का प्रथम पद a = 729 तथा 7वाँ पद 64 है, तो S⁷ ज्ञात कीजिए।
हल:
गुणोत्तर श्रेणी का पहला पद, a = 729
मान लीजिए सार्व अनुपात = r

प्रश्न 16.
एक गुणोत्तर श्रेणी को ज्ञात कीजिए, जिसके प्रथम दो पदों का योगफल – 4 है तथा 5 वाँ पद तृतीय पद का 4 गुना है।
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद = a
सार्व अनुपात = r
पहले दो पदों का योग = a + ar = – 4 ……(1)
5 वाँ पद = ar⁴, तीसरा पद = ar²
5 वाँ पद = 4 × तीसरा पद
ar⁴ = 4 × ar²
∴ r² = 4 या r = ± 2
समी (1) में r = 2 रखने पर
a (1 + 2) = – 4
∴ a = – latex]\frac{4}{3}[/latex]
∴ गुणोत्तर श्रेणी – 5, 3…. है
और जब r = – 2, ∴ a (1 – 2) = – 4, या a = 4
गुणोत्तर श्रेणी है: 4, – 8, 16, – 32,….

प्रश्न 17.
यदि किसी गुणोत्तर का 4वाँ, 10वाँ तथा 16वाँ पद क्रमशः x, y तथा z हैं, तो सिद्ध कीजिए कि x, y, z गुणोत्तर श्रेणी में हैं।
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद = a,
सार्व अनुपात =r

प्रश्न 18.
अनुक्रम 8, 88, 888, …. के n पदों का योग ज्ञात कीजिए।
हल:
मान लीजिए S = 8 + 88 + 888 + … पदों तक
= 8 [1 + 11 + 111 + … n पदों तक]
= \(\frac{8}{9}\)[9 + 99 + 999 +…. पदों तक]

प्रश्न 19.
अनुक्रम 2, 4, 8, 16, 32, तथा 128, 32, 8, 2, \(\frac{1}{2}\) के संगत पदों के गुणनफल से बने अनुक्रम का योगफल ज्ञात कीजिए।
हल:
अनुक्रम 2, 4, 8, 16, 32 तथा 128, 32, 8, 2,\(\frac{1}{2}\) के संगत पदों के गुणनफल 2 × 128, 4 × 32, 8 × 8, 16 × 2, 32 × \(\frac{1}{2}\) या 256, 128, 64, 32, 16.
गुणोत्तर श्रेणी का पहला पद, a = 256

प्रश्न 20.
दिखाइए कि अनुक्रम a, ar, ar²,… ar^{n – 1} तथा A, AR, Ar²,…. AR^{n – 1} के संगत पदों के गुणनफल से बना अनुक्रम गुणोत्तर श्रेणी होती है तथा सार्व अनुपात ज्ञात कीजिए। .
हल:
%अनुक्रम a, ar, ar²,….ar^{n – 1} तथा A, AR, AR²,… AR^{n – 1} के संगत पदों के गुणनफल से बना अनुक्रम
या aA, arAR, ar². AR², ….
या aA, aArR, aAr² R², ….
स्पष्ट है कि यह पद गुणोत्तर श्रेणी में है।
इसका पहला पद = aA
सार्व अनुपात = \(\frac{a A r R}{a A}\) = rR.

प्रश्न 21.
ऐसे चार पद ज्ञात कीजिए जो गुणोत्तर श्रेणी में हो, जिसका तीसरा पद प्रथम पद से 9 अधिक हो, तथा दूसरा पद चौथे पद से 18 अधिक हो।
हल:
मान लीजिए गुणोत्तर श्रेणी a, ar, ar², ar³,… है
तीसरा पद = ar², प्रथम पद = a
∴ ar² – a = 9 …(1)
दूसरा पद = ar, चौथा पद = ar³
ar – ar³ = 18 …(2)
समी (1) को (2) से भाग देने पर,

प्रश्न 22.
यदि किसी गुणोत्तर श्रेणी का p वाँ, q वाँ तथा वा पद क्रमशः a, b, तथा c हो, तो सिद्ध कीजिए कि \(a^{q-r} \cdot b^{r-p}-c^{p-q}\) = 1.
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद A और सार्व अनुपात R है
p वाँ पद = AR^{p – 1} = a ….(1)
q वाँ पद = AR^{q – 1} = b ….(2)
r वाँ पद = AR^{r – 1} = c …..(3)
समी. (1) की q – 7, समी (2) की r – p, समी (3) की p – q घात का प्रयोग करने पर,

प्रश्न 23.
यदि किसी गुणोत्तर श्रेणी का प्रथम तथा nवाँ पद a तथा b हैं, एवं P, n पदों का गुणनफल हो, तो सिद्ध कीजिए कि P² = (ab)ⁿ.
हल:
मान लो गुणोत्तर श्रेणी का सार्व अनुपात है।
पहला पद = a, n वाँ पद = ar ^{n – 1} = b
P = n पदों का गुणनफल
= a. ar. ar². ar³ ….ar^{n – 1}
= a ⁿ. r ^{1 + 2 + 3 +…+ (n – 1)} = \(a^{n} r^{\frac{n(n-1)}{2}}\)

प्रश्न 24.
दिखाइए कि एक गुणोत्तर श्रेणी के प्रथम n पदों का योगफल तथा (n + 1) वें पद से (2n)वें पद तक के पदों के योगफल का अनुपात में है।
हल:
मान लीजिए गुणोत्तर श्रेणी का पहला पद a और सार्व अनुपात = \(\frac{1}{r^{n}}\) हों, तब

प्रश्न 25.
यदि a, b, c तथा d गुणोत्तर श्रेणी में हैं तो दिखाइए कि \(\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}\).
हल:
मान लीजिए गुणोत्तर श्रेणी का सार्व अनुपात 7 है।

प्रश्न 26.
ऐसी दो संख्याएँ ज्ञात कीजिए जिनको 3 और 81 के बीच रखने पर प्राप्त अनुक्रम एक गुणोत्तर श्रेणी बन जाए।
हल:
मान लीजिए G₁, G₂ ऐसी दो संख्याएँ हैं जिससे 3, G₁, G₂, 81 गुणोत्तर श्रेणी बनाते हैं।
यह कुल चार पद हैं। यदि r सार्व अनुपात हो तो
∴ 81 = 3.r^{4 – 1} = 3 . r³
⇒ r=3
G₁ = 3r = 3 . 3 = 9
G₂ = 3r² = 3.3² = 27
अतः संख्याएँ 9 और 27 हैं।

प्रश्न 27.
n का मान ज्ञात कीजिए ताकि \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\), a तथा b के बीच गुणोत्तर माध्य हो।
हल:
a और b के बीच गुणोत्तर माध्य = \(\sqrt{a b}\)

या \(\left(\frac{a}{b}\right)^{n+\frac{1}{2}}\) = 1 = \(\left(\frac{a}{b}\right)^{0}\)
⇒ n+ \(\frac{1}{2}\) = 0 या n = – \(\frac{1}{2}\).

प्रश्न 28.
दो संख्याओं का योगफल उनके गुणोत्तर माध्य का 6 गुना है तो दिखाइए कि संख्याएँ (3 + 2\(\sqrt{2}\)) : (3 – 2\(\sqrt{2}\)) के अनुपात में हैं। .
हल:
मान लीजिए संख्याएँ a और b हों, तब

प्रश्न 29.
यदि A तथा G दो धनात्मक संख्याओं के बीच क्रमशः समांतर तथा गुणोत्तर माध्य हों, तो सिद्ध करो कि संख्याएँ \(\mathbf{A} \neq \sqrt{(A+G)(A-G)}\) हैं।
हल:
मान लीजिए संख्याएँ a और b हैं।

प्रश्न 30.
किसी कल्चर में बैक्टीरिया की संख्या प्रत्येक घण्टे के पश्चात् दुगुनी हो जाती है। यदि प्रारंभ में उसमें 30 बैक्टीरिया उपस्थित थे, तो बैक्टीरिया की संख्या दूसरे, चौथे तथा nवें घण्टों बाद क्या होगी ?
हल:
प्रारम्भ में बैक्टीरिया की संख्या a = 30
प्रत्येक घण्टे बाद बैक्टीरिया की संख्या दुगुनी हो जाती है
∴ सार्व अनुपात = 2.
दूसरे घण्टे बाद बैक्टीरिया संख्या = ar² = 30 × 2² = 120
चौथे घण्टे बाद बैक्टीरिया संख्या = ar⁴ = 30 × 2⁴ = 480
n वें घण्टे बाद बैक्टीरिया संख्या = arⁿ = 30 × 2ⁿ.

प्रश्न 31.
500 रुपए धनराशि 10% वार्षिक चक्रवृद्धि ब्याज पर 10 वर्षों बाद क्या हो जाएगी, ज्ञात कीजिए ?
हल:
माना A मिश्रधन, P मूलधन, r% प्रतिवर्ष ब्याज की दर तथा n वर्ष का समय हो, तो
A = \(P\left(1+\frac{r}{100}\right)^{n}\)
दिया है: P = 500, r = 10%, n = 10 वर्ष
A = 500 \(\left(1+\frac{10}{100}\right)\)
= 500 × (1.1)¹⁰.

प्रश्न 32.
यदि किसी द्विघात समीकरण के मूलों के समांतर माध्य एवं गुणोत्तर माध्य क्रमशः 8 तथा 5 हैं, तो द्विघातीय समीकरण ज्ञात कीजिए।
हल:
मान लीजिए द्विघात समीकरण के मूल α और β हों, तब
\(\frac{\alpha+\beta}{2}\) = 8 ∴ α + β = 16
तथा \(\sqrt{\alpha \beta}\) = 5 ∴ αβ = 25
∴ द्विघातीय समीकरण.
x ² – (α + β) x + αβ = 0
⇒ x ² – 16x + 25 = 0..

MP Board Class 11th Maths Solutions