Category: CBSE

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Write the name of the organism that is referred to as the ‘Terror of Bengal’.
Answer : Water hyacinth (Eichhomia Cmssipes)

Question.2. What are ‘true breeding lines’ that are used to study inheritance pattern of traits in plants ?
Answer : Breeding line having undergone with a number of repeated self pollination and shows stable trait inheritance and expression for several generations called a true breeding line.

Question.3. Name any two types of cells which acts as a cellular barriers to provide innate immunity in humans.
Answer : (i) Monocytes (natural killer)
(ii) PMNL-Neutrophils (polymorpho- nuclear leukocytes)

Question.4. Mention the type of host cells suitable for the gene guns to introduce an alien DNA.
Answer : The type of host cells suitable for the gene guns to introduce an alien DNA is plant cells.

Question.5. How is‘stratification represented in a forest ecosystem?
Answer : Stratification is the way in which different species occupying different levels or arranged in a habitat. It is a way of minimizing competition for limited but vital resources, for survival.
Vertical stratification in a forest ecosystem is represented by following strata of plantss starting from the lowest layrer :
(i) The herb layer               (ii) The shrub layer
(iii) The small tree layer   (iv) The canopy layer

Question.6. Give an example of an organism that enters ‘diapause’ and why ?
Answer : Bombyx mori (silk moth) is an insect that enters diapauses to avoid adverse environmental conditions such as drought, extreme temperature, reduced food availability, which, in turn, delays the overall development. The physiological and metabolic activities are also diminished at this particular time.

Question.7. Identify ‘a’ and ‘b’ in the figure given below representing proportionate number of major vertebrate taxa.

Answer: ‘a’ represents mammals and ‘b’ represents amphibians.

Question.8. State the cause of Accelerated Eutrophication.
Answer: Accelerated Eutrophication is the aging of a water body due to nutrient enrichment of its water. Release of nutrient-rich sewage and industrial effluents containing nutrients such as nitrogen and phosphorus causes accelerated eutrophication.

SECTION-B

Question.9. Why do algae and fungi shift to sexual mode of reproduction just before’the onset of adverse conditions ?
Answer : The fungi and algae switch to the sexual mode of reproduction during adverse conditions because sexual reproduction brings variation into the individuals. In algae and fungi, the zygote develops a thick wall that is resistant to dessication and damage. This ensures the continuity of species.

Question.10. A cross was carried out between two pea plants showing the contrasting traits of height of the plants. The result of the cross showed 50% parental characters, (i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross
Answer: carried out.

Question.11. How does the gene ‘I’ control ABO blood groups in humans ? Write the effect the gene has on the structure of red blood cells.
Answer: In humans, the ABO blood groups are controlled by a gene called gene ‘I’. Sugar polymers protrude from the surface of red blood cells and I controls the kind of sugar. It has three alleles, IA, IB and i. A person possesses any two of the three.
Table: The Genetic Basis of Blood Groups in Human Population alleles. IA and P dominate over i . But with each other, Tand P are co-dominant.

The red blood cells have sugar polymers protrude from the plasma membrane surface and it is regulated by the gene ‘I’ of ABO blood group. The alleles IA and P produce A and B types of sugar, while allele i does not produce any sugar.
OR
Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO .
(ii) Female ZW with Male ZZ
Answer : The types of sex – determination mechanisms in the following crosses are :
(1) Female XX with Male XO : In this case females has a pair of X chromosomes and males have only one X chromosomes (the O indicates absence of chromosome), so it is the case of male 0 hetrogamety eg: grasshopper (2) Female ZW with Male ZZ : This cross shows ZW type of sex determination. In this case female has one Z and one W chromosome and these chromosomes produces different kinds of gametes, so it is the case of female hetrogamety. eg: birds

Question.12. (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion.
Answer : (i) George Gamow suggested that the genetic code should be made up of a combination of three nucleotides.
(ii) He arrived at this conclusion by giving the explanation if a single nucleotide code for one amino acid, then only four amino acid could be provided. Alternatively, if two nucleotides specified one amino acid, then these could be a maximum number of 16 possible arrangements of three nucleotides code for one amino acid, then there could be 64 possible combinations. Later he suggested that every amino acid is coded by atleast one nucleotide triplet or codon.

Question.13. State the disadvantage of inbreeding among cattle. How it can be overcome ?
Answer : Continuous inbreeding among catde causes inbreeding depression, decreases the fertility and, even, the productivity of an animal. It can be overcome by applying outbreeding, in which mating is done between different breeds or individuals of the same breed with unrelated superior animals.

Question.14. Explain with the help of a suitable example the naming of a restriction endonuclease.
Answer : The nomenclature of a restriction endonuclease follows a rule as like :

1.  Ist letter of the name represents the genus of the organism from which the enzyme is derived.
2.  IInd and IIIrd letters represent the species of the organism,
   from wher it is isolated.
3. IVth letter represents the name of the strain.
4.  Last is the Roman number which represents order of isolation.
   For example, In EcoRI – Derived from E.coli, strain R. It : is the first to be isolated from the bacteria.

Question.15. State how has Agrobacterium tumifaciens been made a useful cloning vector to transfer DNA to plant cells.
Answer : Agrobacterium is a bacterium that transfers a piece of DNA to plant tissues by transferring its plasmid T-DNA to the plant genome. The plasmid T-DNA of Agrobacterium is cut with restriction endonuclease and the desired gene that has to be to transferred to a particular plant is inserted with the help of ligase enzyme. Then, this Agrobacterium plasmid is allowed to infect that
” particular plant, so that it can transfer the desired gene in to the plant genome through its T-DNA. ”

Question.16. Construct an age pyramid which reflects a stable growth status of human population.
Answer : The age pyramid that reflects a stable growth status of human population can be represented as follows :

Question.17. Apart from being a part of the food chain, predators play other important roles. Mention any two such roles supported by examples.
Answer : Predators play an important role in :

1.  Maintaining the prey population under control, this regulates intra-species competition.
   For example, if tigers are removed from a forest, spotted deer will multiply rapidly. This would result in rapid destruction of herbs and grasses in the forest and ultimately the whole forest.
2.  Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among prey species.
   For Example : the starfish pisaster is an important predator on the rocky intertidal communities of the American Pacific Coast. In a field experiment, when all the starfish were removed from the area, more than 10 species of invertebrates became extinct within a year, because of inter-specific competition.

Question.18. How are ‘sticky ends’ formed on a DNA strand ? Why are they so called ?
Answer : Sticky ends in DNA strands are produced with the help of restriction enzymes. These enzymes cut the strand of DNA a little away from the centre of the palindrome sites but between the two same bases on the opposite strands. This leaves a single stranded portions at the ends. There are overhanging stretches called ‘sticky ends’ on each. strand,

These are called sticky ends because they form hydrogen bonds with their complementary cut counterparts. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

SECTION-C

Question.19. Explain any three advantages at seeds offer to angiosperms.
Answer: Seeds offer the following advantages to angiosperms :

1. They provide nourishment and parental care to the developing embryo.
2.  They protect the embryo from harsh environmental conditions.
3. The dispersal of seeds to far-off places prevents competition among the members of the same species, thus preventing their extinction.

Question.20. Name and explain the role of the inner and middle walls of the human uterus.
Answer : The inner glandular wall of the uterus is known as endometrium.
Role : During the menstrual cycle, the endometrium wall grows into a thick, blood vessel-rich, glandular layer. It is the where embryo gets implanted. If fertilisation does not occur, the endometrium is shed during the hemorrhagic phase of the menstrual cycle.
The middle wall of the uterus is known as myometrium.
Role: It consists of smooth muscles. It brings strong contraction during delivery of the baby.

Question.21. A color-blind child is bom to a normal couple. Work out a cross to show how it is possible. Mention the sex of this child.
Answer : Color blindness is a sex-linked disease. The gene for this disorder is present on the X chromosome. The color blind child is son with genotype XCY. and sex of the child is male and carrier female.

OR
Mendel published his work on inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work.
Answer: The three reasons that delay in accepting Mendels work:

1.  Lack of communication and publicity. He published it in a journal that had limited circulation.
2.  His concept of factors (genes) as a discrete units that did not blend with each other was not accepted in the light of variations occurring continuously in nature.
3. Mathematical approach : Mendel’s approach to explain biological phenomenon with the help of mathematics was also not accepted.

Question.22. Women are often blamed for producing female children. Consequently, they are ill-treated and ostracized. How will you address this issue scientifically if you were to conduct an awareness programme to highlight the values involved ?
Answer: Women are not responsible for determination of the gender of a child. It is absolutely wrong to ill-treat a woman for giving birth to a girl child.
In human there are 22 pairs of autosomal chromosomes and one pair of sex chromosome. In human sperm (haploid) has 22 autosomes and one of the two types of sex chromosomes, i.e. either X or Y. While human females ova (haploid) have 22 autosomes and contaih only X chromosomes. The gender of a child is determined by the type of the sex chromosome (X or Y) carried by sperm that fuses with the ovum at the time of fertilization. If the fertilising sperm has an X chromosome, then the baby would be a female and if a sperm with Y chromosomefuses with the ovum, it will develop into a male child. Thus scientifically it is correct that to say that males are responsible for determination of the gender of a child. Both males and females are equally important in every respect for the balance of nature and continuity of our species and it should because of equal joy to parents. .

Question.23. (a) Name the tropical sugar cane variety grown in South India. How has it helped in improving the sugar cane quality grown in North India ?
(b) Identify ‘a’, ‘b’ and c’ in the following table :

Answer : (a) Saccharum ojficinarum variety grown in South India, which has a thicker stem and high sugar content but, it didnot grow well in Northern India. Saccharum barberi is a natively grown in Northern India. These two varieties vtere crossed to get the desirable qualities of both (high yield, higher sugar content, thicker stem and the ability to grow in Northern India).
(b) (i) Aphids (ii) Jassids and fruit borer (iii) Okra (Bhindi)

Question.24. Why are beehives kept in a crop field during flowering period ? Name any two crop fields where this is practised.
Answer: Beehives are kept in a crop field during flowering period to increase the pollination efficiency of the crop, which increases the crop yield. Also, bees collect huge amounts of nectar from the flowers of the crop in a close reach without much foraging without difficulty. This increases honey yield. Crop fields where this is practiced: Apple, sunflower and watermelon fields

Question.25. How did the process ofRNA interference help to control the nematode from infecting roots of tobacco plants ? Explain.
Answer: RNA Interference (RNAi) is a gene-silencing process that blocks the expression of genes in the parasite when it enters the host s body.

1. Meloidegyne incognitia infects roots of tobacco plants and cause a severe loss by causing reduction in yield.
2.  RNAi is a method to prevent infestation of roots of tobacco plants by a nematode Meloidegyne incognitia.
3.  It is a defense in all eukaryotic organisms.
4.  In RNAi, a complementary RNA binds to mRNA to form a double strand RNA(dsRNA) that cannot translate and blocked the expression mRNA.
5. In this process, initially nematode-specific genes (DNA) are introduced in the host plant.
6.  This introduced DNA forms both sense and antisense RNA.
7.  These two strands, being complementary to each # other so that they form dsRNA.
8. This dsRNA results in RNA interference and finally silenced the specific mRNA of nematode.
9.  Thus and the parasite cannot survive in the transgenic host expressing specific RNAi. The transgenic plant therefore got itself protected from the parasite.

Question.26. Study the graph given below and answer the questions that follow:

(i) Write the status of food and space in the curves (a) and (b).
(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population?
(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line.
Answer : (i) This curve show sample food and space for the population depicted by curve :
a : When the food and space are unlimited available
b :When the resources are limiting, the curve becomes sigmoid.
(ii) In the absence of predators, curve b would appropriately depict the prey population.
(iii) The dotted line represents the carrying capacity. The carrying capacity represents the size of population that the environment can hold by providing necessary resources.

Question.27. (i) What is primary productivity ? .Why does it vary in different types of ecosystems ?
(ii) State the relation between gross and net primary productivity.
Answer : (i) Primary productivity is the amount of biomass produced per unit area in a certain time period by plants during photosynthesis. It is expressed in terms of g/m2 or Kcal/m2.
Primary productivity depends upon the type of plant species associated with an ecosystem, photosynthetic capacity of these plants and nutrient availability. This is the reason why it varies in different types of ecosystems.
(ii) The relation between the gross and net primary productivity can be shown as :

It is the rate of production of organic matter during photosynthesis, where,
Pn = Net primary productivity (NPP)
Pg = Gross primary productivity (GPP)
R = Respiration losses.

SECTION-D

Question.28. (a) Coconut palm is monoecious, while date palm is dioecious. Why are they so called ?
(b) Draw a labelled diagram of sectional view of a mature embryo sac of an angiosperm.
Answer : (a) Cocunut palm is monoecious, while date palm is dioecious because in coconut palm both as male and female flowers are borne on the same plant, while date palm bears exclusively either male flowers or female flowers.
(b) Diagram showing sectional view of a mature embryo sac of an angiosperm

OR
(a) How is ‘oogenesis’ markedly different from ‘spermatogenesis’ with respect to the growth till puberty in the humans?
(b) Draw a sectional view of human ovary and label the different follicular stages, ovum and Corpus luteum.

Question.29. (a) Explain the process of DNA replication with the help of a schematic diagram.
(b) In which phase of the cell cycle does replication occur in
Eukaryotes ? Whast would happen if cell-division is not
followed after DNA replication.
Answer : (a) Mechanism of DNA Replication. The following
steps are involved in replication of DNA’:

1.  Origin of Replication : Replication begins at a particular region of DNA which has a particular nucleotide sequence called autonomic replicating dioxyribonucleo tides:
2. Oxynation of dioxyribonuleotides:
   
3.  Initation of Replication: Origin of replication is recognised by complex. The unwinding of DNA molecules starts at specific points called Initiation point. These are identified by specific initiator proteins.
4.  Unwinding of Helix :
   (a) The enzyme Helicase unwinds the DNA helix & unzips the double strands of DNA. This process is ATP dependent. It takes place by breaking of H-bonds.
   (b) Unwinding of DNA molecule into two strands results in the formation of DNA replication bubble which later extend as a Y-shaped structure called Replication fork.
   (c) The separated strands become stablised in this condition with the help of single strand binding proteins (SSBPs).
   (d) Due to unwinding, a supercoiling & tenstion is created, which is released by enzymes Topoisomerases I & II. Topoisomerase II of prokaryotes is also called Gyrase, which functions both as Helicase & Topoisomerases .
5. Formation of Primer Strand:
   (a)Once the primer strand formed,DNA replication occurs in 5”->3′ direction i.e during synthesis of new strand dioxyribonuleotides (dATP,dGTP,dTTP,dCTP) are added in the free 3′ OHead.
   
   
   The -release of 2 phosphate molecule & energy aid in formation of H-bonds.
   
   (b) DNA replication occurs in S phase of cell cycle in eukaryotes. If cell division is not followed aftre DNA. replication then cell enters in G0 phase and becomes permanent and specialized.
   (c) As the DNA replication proceeds on the two parental strands, synthesis of daughter parent 3’—> 5’ strand. It is called Leading Daughter strand.
   (d) Synthesis of the other daughter strand along the other parental strand, takes place in the form of short pieces because of the opposite arrangement of nucleotides. A new RNA primer is formed everytime, where new DNA strand is built in small segments. These RNA primers are removed by polymerase I & a enzymes in prokaryotes .
   (e) Since replication is continuous over one strand & discontinuous over the other, it is called semi-continuous replication.
   (f) Discontinuous pieces of the lagging strand are joined Overall Direction of Replication.

(b) DNA replication occurs in S phase of cell cycle in eukaryotes.If cell division is not followed after DNA replication then cell enters in G0 phase and becomes permanent and specialized.
OR
(a)Explain Darwinian theory of evolution with the help of one suitable example. State the two key concept of the theory.
(b) Mention any three characteristics of Neanderthal man that lived in near east and central Asia.
Answer: (a) Darwinian Theory of Evolution took place by natural selection. The number of life forms depends upon their life span and their ability to multiply. Another aspect of Darwinian Theory is natural selection, the survival of the fittest where nature selects the individuals, which are most fit to adapt to their environment. Example: Selection of the antibiotic resistant in bacteria. When a bacterial population grow on an agar plate containing antibiotic penicillin, the colonies that are sensitive to penicillin die, whereas one or few bacterial colonies that are resistant to penicillin survive. This is because these bacteria have undergone mutation results in evolution of a gene that made them resistant to penicillin drug. Hence, the advantage of an individual over the other helps in the struggle for existence.The two key concepts of the theory are :

1. Branching descent : According to this concept, various species have come into existence from a common ancestor.
2. Natural selection : According to this concept, nature selects the individuals, which are most fit to adapt to their environment.

(b) Characteristics of Neanderthal man :

1.  They possess a brain size of 1400 cc.
2.  They were short but very strong with upwards curved thigh bones. .
3.  They used the hides to protect their body and to bury the dead.

Question.30. (a) Name the technology that has helped scientists to propagate on a large scale the desired crops in a short duration. List the steps carried out to propagate the crops by the said technique.
(b) How are somatic hybrids obtained?
Answer : (a) Plant tissue culture is the technique of in vitro maintenance and growth of plant cells, tissues & organs on a suitable culture medium. The technique of tissue culture was first suggested by Gottleib Haberlandt in 1902.

The following steps are carried out to propagate crops by tissue culture :

1.  Preparation of suitable medium : Suitable medium, containing a carbon source, such as sucrose, and inorganic
   salts, vitamins, amino acids and growth regulators like auxin, cytokinin etc.
2. Selection of Explant : Any part of the plant, especially apical and axillary meristem can be used as explant.
3.  Incubation : Growing the explant in the test tube, under sterile conditions.
4.  Regeneration : Since explant show the property of totipotency, new plantlet can be regererated.
5. Hardening : Regenerated plants are grown in pots, to expose them to environmental conditions.
6.  Plantlet transfer: After hardening, plantlets are transferred to field.

(b) Isolated protoplasts from two different varieties of plants, each having a desirable character, can be further gronw to form a new plant. These hybrids are called somatic hybrid while the process is called somatic hybridisation.
OR
(a) Cancer is one of the most dreaded diseases of humans.
Explain ‘Contact inhibition and ‘Metastasis’ with respect to the disease. .
(b) Name the group of genes which have been identified in normal cells that could lead to cancer and How they do so?
(c) Name any two techniques which are useful to detect cancers of internal organs.
(d) Why are cancer patients often given a-interferon as part of the treatment ?
Answer : (a) Cancer is one of the most dreaded disease of humans. Normal cells show a property called contact1 inhibition, by virtue of which contact with other cells and inhibits their uncontrolled growth but cancer cells lose this property. As a result, cancer cells divide continously to give rise to a mass of cells (tumours).
Metastasis is a property of malignant tumours. Some cancer cells from tumors gets sloughed from the tumor and the reach distant sites through the blood and wherever they reach initiate the formation of new tumours by dividing actively. This property is called metastasis.
(b) Cellular oncogenes (c-onc) and proto-oncogenes are the group of genes that have been identified in normal cells. These genes when activated under certain conditions, would lead to oncogenic transformation of the cells.
(c) Techniques such as radiography, CT (Computed Tomography) and MRI (Magnetic Resonance Imaging) are useful to detect cancers of internal organs.
(d) The biological response modifiers such as a-interferoris are given to cancer patients as part of their treatment because it activates a patient s immune system and helps in destroying the tumour.

SET -II

SECTION-A

Question.4. Name the two intermediate hosts which the human liver fluke depends on to complete its life cycle so as to facilitate parasitization of its primary host.
Answer: Terrestrial snail and Fish, are two intermediate hosts on which the human liver fluke depends on to complete its life cycle so as to facilitate parasitization of its primary host.

Question.7. Mention how does DNA polymorphism arise in a population.
Answer : DNA polymorphism is a genetic variant and intro-duced in a population by mutation and genetic drift.

SECTION-B

Question.9. Name the organic materials the exine and intine of an angiosperm pollen grains are made up of. Explain the role of exine.
Answer: Exine is made of sporopollenin. Intine is made of cellulose and pectin. Sporopollenin (exine) is most resistant organic material which can with stand high temperature and acids and alkalies.

Question.13. How can healthy potato plants be obtained from a desired potato variety which is viral infected ? Explain.
Answer: Tissue culture can get us disease free potato plants from viral infected plants. The apical and axillary meristems of virus infected plant are free of virus. So meristems can be removed and their culture can give us virus free plants.

Question.15. What is Biopiracy ? State the initiative taken by the Indian Parliament towards it.
Answer : Biopiracy is defined as the use of bio-resources by mitltinational companies or other organisations without
proper authorisation from concerned country or people. Indian Parliament has declared second amendment of Indian Patents bill Act (1970) this bill considers patenting, its terns emergency provisions and research and development initiatives.

Question.18. Write the role of‘Ori’ and ‘restriction site in a cloning vector pBR322.
Answer: Role of Ori sequence and ristriction site in pBR322: Ori is a genetic sequence that acts as the initiation site for replication of DNA, when any fragment linked .to this sequence ‘ can be initiated to replicate within host cells.
Recognition site is the specific DNA sequences which contains different palindromic sequence, as recognized by respective restriction enzymes (such as EcoRI, Hind III, Pvul, BamHI etc.). Recognition sites are sequences where the restriction enzymes cut the DNA.

SECTION-C

Question.20. A cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(a) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(b) Write the conclusion you draw of the inheritance pattern of this disease.
Answer: (a)
Genotype of daughter- XX
Genotype of husband- XY

(b) Conclusion: Haemophilia is a sex-linked recessive disease which shows it is transmitted from the carrier female to the sons. From the above pedigree chart, it can be observed that the disease is being transmitted from the carrier female to her daughter (carrier) and son (affected). The carrier daughter transmits this disease to the grandson and the possibility of a female becoming a haemophilic is extremely rare. This inheritance is called crisscross inheritance.

Question.22. Draw a labelled diagram of the sectional view of a Human seminiferous tubule, (six parts to be labelled)
Answer:

SECTION-D

Question.30. Explain the ovarian and uterine events that occur during a menstrual cycle in a human female, under the influence of Pituitary and Ovarian hormones respectively.
Answer : Menstrual Cycle : Menstrual cycle is the reproductive cycle in all primates and begins at puberty (menarche). It involves cyclic changes in females reproductive tract culminating in menstruation that is flow of cast off uterine and fallopian tube lining along with blood and tissue fluid through the vagina. In human females, menstruation occurs once in 28 to 29 days. The cycle of events starting from one menstruation till the next one is called the menstrual cycle.
It consist of three phases:

1. Proliferating phases : It last for about 14 days. Lining of the uterus and fallopian tubes proliferates and its vascularization increases. A graafian follicle grows, matures and secretes oestrogen. It ruptures to release its egg (secondary oocyte) after about 14 days. The LH and FSH are at their peak in the middle before release of oocyte. This phase is also called the ovulatory phase.
2.  Secretory phase : It lasts for-about 10 days. The empty graafian follicle forms in it corpus luteum which secretes progesterone. The lining of uterus and fallopian tubes undergoes further hypertrophy. Endometrial glands of the uterus secrete a nutritive fluid for the foetus.
3.  Menstrual (Bleeding) phases : It lasts for about 4 days. If fertilization does not occur, the corpus luteum regresses, and the linning of uterus and fallopian tubes breakdown, resulting in menstrual flow. This occurs after 25 days and continues 3 to 5 days. The basal part of endometrial lining remain intact during menstruation and produces new uterine lining.
   

OR
(a) Why does endosperm development precede embryo development in angiosperm seeds ? State the role of endosperm in mature albuminous seeds.
(b) Describe with the help of three labelled diagrams the different embryonic stages that include mature embryo of dicot plants.
Answer : (a) Endosperm development precedes embryo development in angiosperm seeds because primary endosperm cell divides repeatedly and forms a triploid endosperm tissue. The cells of this tissue are filled with reserve food materials which provide nutrition to the developing embryo.
Albuminous seeds stores starch and fat to retain a part of endosperm as it is not completely used up during embryo development (e.g. wheat, maize, barley, castor, sunflower).
(b) Development of embryo :
The Zygote formed after fertilization of egg cell starts dividing and gives rise to proembryo. This proembryo further divides, forming a globular, heart-shaped and mature embryo.

Following are the steps that occur during the development of embryo:

1.  In dicots zygote elongates and divides into upper and lower cell.
2.  The lower one lying toward micropyle further divides in one direction into a row of cell called suspensor.
3.  The upper cell towards the antipodal end is called embryo cell.
4.  The first cell of the suspensor often enlarges and acts as , haustorium or absorbing organ while its terminal cell called hypophysis cell divides giving rise to the apex of the radicle.
5. The upper cell or embryo cell enlarges and divides repeatedly to form eight cells that are arranged in two tiers – epibasal (terminal) and hypobasal (near the suspensor). A typical dicot embryo consists of an embryonal axis and two cotyledons.
6.  The portion of the embryonal axis above the level of cotyledons is called epicotyl. It contains the plumule (shoot tip). The portion below the axis is called hypocotyl. It contains the radicle (root tip). The root tip is covered by the root cap.
7.  Subsequent divisions give rise to globular heart-shaped embryo.

SET -III

SECTION-A

Question.3. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments ?
Answer : Satellite DNA is separated from bulk genomic DNA by density-gradient centrifugation technique. They are separated as different peaks. The bulk DNA forms a Major peak and the other small peaks are referred to as satellite DNA.

Question.5. Name the Green House gases that contribute to total global warming.
Answer: Carbon dioxide, nitrous oxide, ozone and methane are the Greenhouse gases that are responsible for global warming.

SECTION-B

Question.12. What is gene therapy ? Name the first clinical case where it was used ?
Answer : Gene therapy is a experimental technique in which genes can be used for treating and preventing disease. A four year old girl became the first gene therapy patient on September 14, 1990 at the NIH clinical center. She was suffering from adenosine deaminase (ADA) deficiency.

Question.15. Why does Bt toxin not kill the bacterium that produces it, but kill the insect that ingests it ?
Answer : Bt toxin protein is produced by a soil bacterium called Bacillus thuringiensis in inactive prototoxin and crystalline form. The prototoxin form does not kill the bacteria. It becomes active and toxic when it is consumed by insects such as lepidopterans (armyworm), coleopterans (beedes) and dipterans (flies/mosquitoes) due to presence of alkaline pH in the gut. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause lyses and swelling, eventually killing the insect.

Question.17. Identify the following pairs as homologous of analogous organs1:
(i) Sweet potato and potato
(ii) Eye of octopus and eye of mammals ,
(iii) Thoms of Bougainvillaea and tendrils of Cucurbits
(iv) Fore limbs of bat and whale
Answer : (i) Analogous organs, (ii) Analogous organs,
(iii) Homologous organs, (iv) Homologous organs

Question.18. List the post – fertilisation events in angiosperms.
Answer : The various post-fertilisation events occurring in angiosperms are:

1. Sepals, petals and stamens of the flower dry up alid fall off.
2. The Zygote develops into an embryo.
3.  Ovules develop into the seed.
4.  The Ovary develops into the fruit.

SECTION-C

Question.21. What are Methanogens ? Name the animals they are present in and the role they play there.
Answer: Methanogens are anaerobic bacteria grow anaerobically on cellulose material that produce large amounts of methane.
Ex : Methanobacterium.
Methanogens are commonly found in the rumen of catde and help in cellulose digestion. Hence, excreta of cattle (gobar) is rich in methanogens.

Question.27. There are many animals that have become extinct in the wild but continue to be maintained in Zoological parks.
(i) What type of biodiversity conservation is observed in this case ?
(ii) Explain any other two ways which help in this type of conservation.
Answer : (i) It is an example of ex-situ conservation.
(ii) Cryopreservation and tissue culture are two ways that helps in ex-situ conservation.
Cryopreservation : The preservation of gamates of threatened species in viable and fertile conditions at sub-zero temperatures which help in preserving these cells for longer periods.
Tissue culture : Plants are propagated from a small mass of tissue, called callus.

SECTION-D

Question.29. (a) Draw a labelled diagrammatic view of human male reproductive system.
(b) Differentiate between :
(i) Vas deferens and vasa efferentia
(ii) Spermatogenesis and spermiogenesis
Answer: (a) Diagrammatic view of a human male reproductive
OR
(a) Explain the phenomenon of double fertilization.
(b) Draw a labelled diagram of a typical anatropous ovule.
Answer : (a) Phenomenon of double fertilisation :
Pollen grains gets transferred from the anther to the stigma, and then the pollen tube enters one of the synergids and released two male gametes. One gamete moves towards the egg cell and fuses to form the zygote to complete the syngamy. The other gamete fuses with the two polar nuclei and forms triploid Primary Endosperm Nucleus (PEN). This fusion of one male gamete with two polar nuclei is termed as triple fusion. Since, , two kinds of fusion—syngamy and triple fusion—take place during fertilization in a flower, the process is known as double fertilization. It is a characteristic of flowering plants.

(b) Labelled diagram of a typical anatropous ovule

CBSE Class 10 Science Practical Skills – Homology and Analogy of Plants and Animals

BASIC BUILDING CONCEPTS
The organs which have similar basic structure but perform different functions in different species are called homologous organs. For example, tendrils of pea plant and spines of Barberry. Both tendril of pea and spines of Barberry are modified leaves. So, structurally they are same. But due to different adaptations leaves in pea and Barberry are modified into different structures.

The organs which are quiet different in fundamental structure but perform same function and superficially look alike in different species are called analogous organs. For example, tendril of pea and tendril of cucurbits. The tendril of pea is modified leaf but the tendril of cucurbits is modified stem. It means both tendril of pea and cucurbits are structurally different. But in both the plants they are adapted to perform same function, i.e., to climb a support.

AIM
To study homology and analogy with the help of models/charts of animals and models/charts/ specimens of plants.

MATERIALS REQUIRED
Plant specimens such as Bougainvillea, cucurbits, potato and sweet potato. Animal models like bird, butterfly, forelimb of frog.

THEORY
Homologous organs: The organs which perform different functions in different species but have similar basic structure and similar embryonic origin are called homologous organs.
Analogous organs: The organs which are quiet different in fundamental structure and embryonic origin but perform same function and may superficially look alike in entirely different species are called analogous organs.

PROCEDURE

1. Take the specimens/models/charts and observe them carefully.
2. Record the structure and function of each specimen.
3. Draw their neat and labelled diagrams.

OBSERVATIONS
A. Tendrils of cucurbits and spines of Bougainvillea

1. Tendrils of cucurbits are modified stem which are used to climb a support.
2. Spines of Bougainvillea are modified stem which are meant for protection.
3. Both tendril of cucurbits and spines of Bougainvillea have same basic structural design but perform different functions. So, they are homologous organs.
   

B. Forelimb of frog and forelimb of bird

1. The forelimb of frog consists of upper arm, fore arm, wrist, palm and five fingers. The upper arm is having humerus, fore arm with radius and ulna, wrist with carpels, palm have metacarpels and fingers have phalanges. The forelimb of frog helps the animal to prop up the front end of the body at rest and acts as a shock absorber when the animal lands on the ground after a leap.
2. The forelimb of birds also has same basic plan of structure with an upper arm having humerus, a fore arm with radius and ulna, a wrist with carpels, a palm with metacarpels and five fingers with phalanges. The forelimb of the bird helps in flying.
3. Both forelimbs of frog and bird show similarity in basic structure but perform entirely different function. So, they are homologous organs.
   

C. Potato and sweet potato

1. In potato, eyes are present which are actually nodes. .
2. Potato is a modified stem which is adapted to store the food.
3. Sweet potato is a modified root which is meant for storage of food.
4. Both potato and sweet potato are structurally different but they perform the same function, i.e., storage of food. So, they are analogous organs.
   

D. Wings of butterfly and wings of bird.

1. The wings of butterfly are extensions of integuments which are supported by a few muscles and used for flying in the air.
2. The wings of birds are formed of limb bones covered with flesh, skin and feathers. Birds also use their wings for flying.
3. Both wings of butterfly and wings of birds are structurally different but they perform the same function. So, they are analogous organs.
   

RESULT AND DISCUSSION

1. Homologous organs have same structure but are functionally different.
2. Analogous organs have same function but are structurally different.

PRECAUTIONS
The specimens should be fresh and clean.

INTERACTIVE SESSION

Question 1.
What is the evidence of evolution?
Answer:
The fossil records, homologous organs, analogous organs, vestigial organs provide evidence for evolution.

Question 2.
What is the evidence of common ancestry?
Answer:
Homologous organs, vestigial structures, embryonic development provide evidence for common ancestry.

Question 3.
What do you understand by homology?
Answer:
Similarity in structure due to relatedness are known as homologies.

Question 4.
Give an example of homologous leaves.
Answer:
Pitcher of pitcher plant and spines of Opuntia.

Question 5.
Give an example of plant analogies.
Answer:
Tendril of cucurbits and tendril of pea.

Question 6.
Give an example of animal analogies.
Answer:
Wings of a bird and wings of a bat.

Question 7.
Give an example of animal homologies.
Answer:
Forelimb of frog, lizard, bird and human.

Question 8.
Why are thorn of Bougainvillea and tendril of Passiflora plant considered homologous organs?
Answer:
Both are structurally similar (modified stem) but perform different function.

Question 9.
How do embryological studies provide evidences for evolution?
Answer:
Embryological studies refer to study of development of an embryo of an organism from the fertilized egg to young one. This study shows that early embryos of fish, salamander, tortoise, chick, rabbit and human resemble in shape and structure. This similarity indicates that all vertebrates have evolved from a common ancestor.

PRACTICAL BASED QUESTIONS

Multiple Choice Questions/VSA (1 Mark)

Question 1.
The difference between homologous and analogous organs is that
(a) homologous organs have same design and function whereas analogous organs have different design and function.
(b) homologous organs have different design and function whereas analogous organs have same design and functions.
(c) homologous organs have same design but different functions whereas analogous organs have different design but similar function.
(d) homologous organs have different design but similar functions whereas analogous organs have same design but different functions.

Question 2.
One of the examples of analogous organs can be the wing of parrot and
(a) flipper of whale
(b) foreleg of horse
(c) front leg of frog
(d) wings of housefly

Question 3.
Which of the following pairs of organs is analogous to each other?
(a) Leaf spines and leaf tendrils
(b) Flipper of a whale and leg of a horse
(c) Wings of an insect and wings of a bat
(d) Forelimbs of a frog and human hand

Question 4.
The three long bones in your arm and in the wing of a bird are
(a) homoplastic structures
(b) homologous structures
(c) analogous structures
(d) all of the above

Question 5.
Thorns of Bougainvillea plant and tendril of cucurbits are
(a) homologous organs
(b) paralogous organs
(c) analogous organs
(d) orthologous organs

Question 6.
The similarity in bone structure in the forelimbs of many vertebrates is an example of
(a) analogy
(b) homology
(c) digitality
(d) convergence

Question 7.
An example of homologous organs is
(a) man’s arm and cat’s foreleg
(b) man’s teeth and elephant’s tusk
(c) potato and pumpkin
(d) all of the above

Question 8.
A student wants to collect analogous organs for his project. Which one of the following pairs would you suggest? [Foreign 2013]
(a) Forelimbs of frog and birds
(b) Thorn of bougainvillea and a tendril in cucurbits
(c) Wings of an insect and wings of a bird
(d) Forelimbs of horse and human arm.

Question 9.
Which of the following statements is true?
(a) Tendril of cucurbits and phylloclade of Opuntia are homologous.
(b) Tendril of cucurbits and phylloclade of Opuntia are analogous.
(c) Wings of birds and limbs of lizards are analogous.
(d) Wings of birds and wings of bat are homologous.

Question 10.
A basket of vegetables contain carrot, potato, radish and tomato. Which of them represent the correct analogous structures?
(a) Carrot and potato
(b) Carrot and tomato
(c) Radish and carrot
(d) All of the above

Question 11.
Choose the right pair of analogous organs
(a) Forelegs of rabbit and frog
(b) Spines of cacti and Euphorbia
(c) Fins of fish and forelegs of lizard
(d) Fruits of apple and orange

Question 12.
Analogous pair of a sweet potato is
(a) onion
(b) ginger
(c) potato
(d) all of the above

Question 13.
Which of the following is correct?
(a) Homologous organs are structurally same.
(b) Analogous organs are structurally same.
(c) Homologous organs are functionally same.
(d) Analogous organs are functionally not same.

Question 14.
Which of the following are the examples of homologous organs?
(a) Venus fly trap and spines of cactus
(b) Tendrils of cucurbits and Venus fly trap
(c) Spines of Barberry and fruit of lemon
(d) Venus fly trap and pitcher plant.

Question 15.
When two organs have same structures but different functions they are
(a) homologous organs
(b) analogous organs
(c) vestigial organs
(d) none of the above

ANSWER KEY

1. (c)
2. (d)
3. (c)
4. (b)
5. (a)
6. (b)
7. (a)
8. (c)
9. (a)
10. (a)
11. (b)
12. (d)
13. (a)
14. (a)
15. (a)

Short Answer Questions (2 Marks)

Question 1.
Giving an example explain what are homologous organs.
Answer:
The organs which have similar basic structures but perform different functions in different species are called homologous organs. For example, tendril of pea plant and spines of Barberry. Both tendril of pea plant and spines of Barberry are modified leaves. So, structurally they are same. But due to different adaptations, leaves in Barberry and pea are modified into different structures.

Question 2.
What are analogous organs? Give any two examples.
Answer:
The organs which are quite different in fundamental structure and embryonic origin but perform the same function and may superficially look alike in entirely different species are called analogous organs.
Examples:

1. Wings of a butterfly and wings of a bird.
2. Potato and sweet potato.

Question 3.
Out of potato, carrot, spines of cacti, spinach

1. which two are homologous organs?
2. which two are analogous organs?

Answer:

1. Spines of cacti and spinach are leaf so structurally they both are same. But they perform different functions therefore, they are homologous organs.
2. Potato is a underground stem and carrot is a root. Structurally they both are different but they are modified to perform the same function which is storage of food. Therefore, potato and carrot are analogous organs.

Question 4.
Differentiate between homologous and analogous organs.
Answer:

Question 5.
Tendrils of cucurbits and spines of bougainvillea are homologous organs. Why?
Answer:
Tendril of cucurbits are modified stem which are used to climb a support. Spines of Bougainvillea are modified stem which are meant for protection. Both the organs have same basic structural design but perform different functions. So, they are homologous organs.

Science Practical SkillsScience LabsMath LabsMath Labs with Activity

CBSE Class 10 Science Practical Skills – Focal Length of Concave Mirror and Convex Lens

BASIC BUILDING CONCEPTS
Spherical Mirror: A spherical mirror is a part of hollow sphere with one side having silver/mercury coating, further coated with paint to protect it from damage.
According to the position of silvered surface, spherical mirrors are of two types:
(a) Concave mirror: Silvered at outer surface so that reflection takes place from inner (concave) surface.
(b) Convex mirror: Silvered at inner surface so that reflection takes place from outer (convex) surface.

IMPORTANT TERMS RELATED TO SPHERICAL MIRROR:
(а) Pole: Pole is the central point ‘P’ of the reflecting spherical surface.
(b) Aperture: Aperture is the width (LL’ or MM’) of the reflecting surface from which reflection takes place actually.
(c) Centre of curvature: Centre of curvature is the geometrical centre of the hollow sphere ‘C’ from which the mirror is formed.
(d) Radius of curvature: Radius of curvature is the radius ‘R’ of the hollow sphere whose part is spherical mirror or distance between the pole and center of curvature.
(e) Principal axis: Principal axis is the straight line which joins the pole and centre of curvature.
(f) Principal focus: Principal focus is the point ‘F’ on the principal axis where a parallel beam of light parallel to the principal axis actually meets (in case of concave mirror) or appears to come from (in case of convex mirror) after reflection is called principal focus.

(g) Focal length: Focal length is the distance ‘f’ between the pole and principal focus of the spherical mirror.

You can also download Class 10 NCERT Science Solutions to help you to revise complete syllabus and score more marks in your examinations.

Mirror formula: The relation between the object distance (u), the image distance (v) and the focal length (f)of a mirror is called mirror formula and is given by

This expression is valid nd remains the same for various positions of the object with concave as well as convex spherical mirrors

SOME MORE FACTS ABOUT CONCAVE MIRROR:

1.  It reflects the sunlight leading to brighter dazzling patch of light.
2. The sharpest and brightest spot of light on the screen is the focus of a concave mirror.
3.  Using the relation R = 2f, we can calculate the radius of curvature of the mirror.
4. The nature, size and position of the image formed by a concave mirror depends on the position of object in front of the mirror.
5. A virtual, erect and enlarged image of an object is formed behind the concave mirror while the real and inverted image is formed in front of the mirror.
6.  A concave mirror is called a converging mirror because it converges the parallel beam of light rays after reflection at a point.
7. If a light emitting source is kept at the focus of a concave mirror, it spreads a powerful parallel beam of light to a larger distance.
8.  If a piece of carbon paper is kept at the focus of a concave mirror, it begins to burn with smoke initially and after that it catches fire and burns quickly.
9.  It is used to concentrate the heat at a point, so it can be used in solar heating devices.

AIM
To determine the focal length of a concave mirror by obtaining the image of a distant object.

MATERIALS REQUIRED
A concave mirror, a metre scale, a mirror holder.

THEORY

1. The nature of the image formed by a concave mirror, depends upon the position of the object in front of it.
2. When a parallel beam of light coming from a distant object, such as tree or pole is incident on the reflecting surface of a mirror, then after reflection, the rays converge at a point and this point is called principal focus of the concave mirror as shown in the figure.
3. If a screen is placed at the point of focus, a sharp, real and inverted image of the distant object is obtained.
4.  The distance between the pole and principal focus of a spherical mirror is called focal length of the mirror. It is equal to half the radius of curvature of the mirror.

PROCEDURE

1.  Select a distant object such as a tree or pole or the sun.
2.  Mount the concave mirror in a mirror holder.
3.  Adjust the concave mirror in such a way that the rays of light coming from the tree fall on its reflecting surface.
4.  Obtain a well defined and sharp image on a white wall of the laboratory by moving the mirror backward or forward.
5. Measure the distance between the wall and the concave mirror with the help of a meter scale as shown in figure (a) or figure (b).
6. This will give you approximate focal length of the mirror.
7. Repeat the experiment by selecting the different distant objects at different distances.

OBSERVATION AND CALCULATION

RESULT
The approximate focal length of the given concave mirror is………. cm, as determined.
According to the sign conventions, the focal length of a concave mirror is negative.  Therefore,
f = -………cm

PRECAUTIONS

1.  The distant object should be clearly visible.
2. The image on the wall should be well defined and sharp.
   (Student must pay special attention to get the sharp image. Obtaining a blurred image will give you an incorrect measurement of the focal length of a concave mirror).
3.  While measuring the distance, the meter scale should be kept parallel to the ground.
4.  The meter scale must be correctly positioned between the wall and center of the concave mirror. Important note: This experiment can be performed outside the laboratory by taking the image on the screen (white paper) or white painted board or outer wall of the laboratory.

INTERACTIVE SESSION

Question 1:
What are you doing?
Answer:
I am finding the focal length of a concave mirror by obtaining the image of a distant object.

Question 2:
What is spherical mirror?
Answer:
It is a part of a hollow sphere having a highly polished reflecting surface inside or outside.

Question 3:
Which surface in your mirror is reflecting surface, inner or outer?
Answer:
Outer surface is silvered and regular reflection takes place from the inner surface of the mirror.

Question 4:
Is the image formed by a concave mirror real or virtual?
Answer:
Image formed by the concave mirror depends on the position of the object. When the object is kept between pole and its focus, a virtual image is obtained, whereas a real image is obtained for all other positions of the object.

Question 5:
What is the difference between a real and virtual image?
Answer:
A real image is obtained on the screen, whereas a virtual image cannot be obtained on the screen. Also, when the rays of light after reflection actually meet, a real image is formed. If the rays of light appear to meet after reflection, a virtual image is formed.

Question 6:
What is focal length of a concave mirror?
Answer:
The distance between the pole and the focus of a mirror is called focal length of the concave mirror.

Question 7:
What is radius of curvature?
Answer:
The radius of the hollow sphere, from which a mirror is formed, is called its radius of curvature.

Question 8:
Is there any relationship between radius of curvature and focal length of the mirror?
Answer:
Yes, the radius of curvature is twice the focal length.

Question 9:How will you get more accurate value of focal length?
Answer:
By obtaining the sharp image of the Sun.

Question 10:
State the laws of reflection of light.
Answer:

1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane.

Question 11:
Suppose two spherical mirrors with reflecting surface towards your face are lying on the table. How will you distinguish between them just by looking?
Answer:
The mirror which forms the virtual, erect and magnified image of our face is concave. If it produces diminished image then it is convex mirror.

Question 12:
What are the uses of a concave mirror?
Answer:
It is used as a reflector in ophthalmoscope, telescope, headlights of motor vehicles, shaving or make up mirror and also reflector in searchlight and an electric room heater.

NCERT LAB MANUAL QUESTIONS

Question 1:
How will you distinguish between a concave and a convex mirror?
Answer:

Question 2:
To determine the focal length of a concave mirror, a student focusses a classroom window, a distant tree and the sun on the screen with the help of a concave mirror. In which case will the student get more accurate value of focal length?
Answer:
In case of the sun student will get more accurate value of focal length.

Question 3:
What will be the nature of the image formed by a concave mirror for a distant object?
Answer:
The image formed by a concave mirror for a distant object is real and inverted.

Question 4:
In reflector type solar cookers, special concave (parabolic) mirrors are used. In such cookers, what should be the preferable position of food vessel for cooking?
Answer:
The preferable position of food vessel should be at focus of the concave mirror.

Question 5:
What type of mirror is used in a torch? Give reasons.
Answer:
Concave spherical or parabolic mirror is used because when the bulb (source) is kept at the focus of a mirror, parallel beam of light is obtained which travels a large distance.

Question 6:
What type of mirror is used as shaving mirror or in vanity boxes?
Answer:
Concave mirror because when the object is placed between its focus and pole, the magnified, erect and virtual image of the object will be formed.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
In the set-up shown below, a clear image of a distant object is obtained on the screen. The focal length of the concave mirror is

(a)11.4 cm
(b)9.4 cm
(c)9.8 cm
(d)9.9 cm

Question 2:
Four students A, B, C and D carried out measure-ment of focal length of a concave mirror as shown in the following four diagrams.

The best result will be obtained by student
(a) A
(b) B
(c) C
(d) D

Question 3:
A student determines the focal length of a device X, by focussing the image of a far off object on the screen positioned as shown in the figure below.

The device X is a
(a) convex lens
(b) concave lens
(c) convex mirror
(d) concave mirror

Question 4:
To find the focal length of a concave mirror Rahul focuses a distant object with his mirror. The chosen object should be
(a) a tree
(b) a building
(c) a window
(d) the sun

Question 5:
While doing an experiment, a student found that if the object moves from infinity towards pole of a mirror, the image also moves from pole to infinity. The mirror must be
(a) parabolic
(b) concave
(c) convex
(d) all of these

Question 6:
A student selected the sun as a distant object to find the focal length of a concave mirror. His teacher advised him to use a screen of wood or hard cardboard. It is because
(a) white paper is not available in the lab.
(b) white paper is costlier than other papers
(c) white paper may start burning when sun rays converge on it.
(d) white paper scatters the light rays falling on it.

Question 7:
The image of a distant object is obtained on a screen by using a concave mirror. The focal length of the mirror can be determined by measuring the distance between
(a) the object and the mirror
(b) the object and the screen
(c) the mirror and the screen
(d) the mirror and the screen as well as that between the object and the screen

Question 8:
Out of the following, the best way to do the experiment on finding the focal length of a concave mirror by obtaining the image of a distant object, is to
(a) hold the mirror in hand and keep the screen in a stand kept behind the mirror
(b) hold the mirror in a stand and hold the screen in hand, with the screen in front of the mirror.
(c) keep both the mirror and screen in suitable stands with the screen put in front of the mirror.
(d) keep both the mirror and the screen in suitable stands with the screen put behind the mirror.

Question 9:
A student obtains a blurred image of an object on a screen by using a concave mirror. In order to obtain a sharp image of the same object on the screen, he will have to shift the mirror .
(a) to a position very far away from the screen
(b) little away from the screen
(c) towards the screen
(d) either towards or away from the screen depending upon the position of the object

Question 10:
A student focused the image of a distant object using a device X on a white screen S as shown in the figure. If the distance of the screen from the device is 30 cm, select the correct statement about the device X.

(a) The device X is a concave mirror of focal length 15 cm.
(b) The device X is a concave mirror of focal length 30 cm
(c) The device X is a concave mirror of radius of curvature 30 cm.
(d) The device X is a convex mirror of focal length 30 cm

Question 11:
A student obtained a sharp image of grills of a window on a white screen using a concave mirror. His teacher remarked that for getting better value of focal length of the mirror he should focus a distant object (preferably the sun) on the screen. What should the student do for this purpose?
(a) Move the mirror and the screen towards the object
(b) Move the mirror and the screen away from the object
(c) Move the screen slightly away from the mirror
(d) Move the mirror slightly towards the screen

Question 12:
A student has to do the experiment, on finding the focal length of a given concave mirror by using a distant object. Out of the following ‘set-ups’ (A, B, C, D) available to her –
(A) a screen, a mirror holder and a scale
(B) a mirror holder, a screen holder and a scale
(C) a screen holder and a scale
(D) a mirror holder and a screen holder the ‘set-up’ that is likely to give her the best result, is the ‘set-up’ labelled as
(a) A
(b) B
(c) C
(d) D

Question 13:
A student recalls the rules to draw ray diagrams with the help of a concave mirror.
(i) The ray of light parallel to the principal axis will converge at the focus after reflection.
(ii) The ray of light passing through the centre of curvature will retrace its path after reflec-tion.
(iii) The ray of light falling at the pole gets reflected at a different angle on the other side of principal axis.
The correct statement(s) is/are
(a) (i) only
(b) (i) and (ii) only
(c) (ii) and (iii) only
(d) (i) and (iii) only

Question 14:
Which of the following pictures depicts the correct image of a distant object by using a concave mirror?

Question 15:

In the above set-up, the focal length of the concave mirror is 4.0 cm. The marks on the scale on which the screen be placed to obtain a sharp image will be
(a) at 8 cm
(b) at 2 cm
(c) at 6 cm
(d) at 1 cm

Short Answer Questions

Question 1:
State the condition when a concave mirror forms an image

1.  larger than the actual object.
2.  smaller than the actual object.

Question 2:
Why is a concave mirror of large focal length often used as a shaving mirror?

Question 3:
What happens to the ray of light after reflection from a concave mirror, if it

1. is parallel to the principal axis?
2.  passes through the center of curvature?

Question 4:
A student takes a mirror which is depressed at the centre and mounts it on a mirror stand. An erect and enlarged image of his face is formed. He places the mirror on a stand along a metre scale at 10 cm mark. In front of this mirror, he
mounts a white screen and moves it back and forth along the metre scale till a highly sharp, well-defined image of a distant building is formed on the screen at 25.5 cm mark.

1.  Name the mirror and find its focal length.
2. Why does the student get sharp image of the distant building at 25.5 cm mark?

Question 5:

1. A ray of fight passing through the center
   of curvature of a concave mirror is incident on its reflecting surface. What is the angle of incidence and angle of reflection of this ray?
2. The radius of curvature of concave mirror is 42 cm. What is its focal length?

ANSWER KEY
Multiple Choice Questions/VSA –

1. (c)
2. (a)
3. (d)
4. (d)
5. (b)
6. (c)
7. (c)
8. (c)
9. (d)
10. (b)
11. (d)
12. (a)
13. (b)
14. (c)
15. (d)

Short Answer Questions

1.

1.  When object is placed anywhere between
   pole and centre of curvature.
2.  When object is placed anywhere between infinity and centre of curvature.

2. While shaving, we keep our face between the pole and focus of a concave mirror to obtain an erect and enlarged image of the face. This enlarged image of the face helps us in having a better and a closer shave.

3.

1. After reflection, the ray of light passes
   through the focus of a concave mirror in this case.
2.  In this case, the ray of light gets reflected back along the same path, i.e. through centre of curvature.

4.

1. The mirror is a concave mirror. The focal
   length is the difference between mirror and screen, so, f = 25.5 cm-10 cm = 15.5 cm
2. The rays of light coming from a distant object are parallel to principal axis and meets at the focus of concave mirror where the screen is placed. Thus, a sharp image of the distant building is formed.

5.

1. The ray of light passing through the centre
   of curvature of a concave mirror strikes the mirror normally and gets reflected back along the same path; hence the angle of incidence and the angle of reflection will be equal to zero.
2.  f = R/2 = 42/2 = 21 cm

 

Experiment  11(b)

BASIC BUILDING CONCEPTS
Lens: A homogeneous transparent material or medium bounded by two surfaces of different or same radii of curvature is called lens.
The curved surface of the lens may be either convex or concave and the other surface may be either plane or spherical.

TYPES OF LENS:
(а) Double convex lens: If both the refracting surfaces of the lens are convex, then the lens is said to be double convex lens or simply convex lens. It is thicker at the middle and thinner at the edges.
(б) Double concave lens: If both the refracting surfaces of the lens are concave, then the lens is said to be double concave lens or simply concave lens. It is thinner at the middle and thicker at the edges.

IMPORTANT TERMS ASSOCIATED WITH LENS:
(а) Optical centre: The central point ‘O’ on the principal axis of the lens, through which incident ray of light passes undeviated with negligible lateral displacement is called optical centre of the lens.
(b) Center of curvature: The center of the sphere, whose part is a spherical surface of the lens is called center of curvature. Since the lens has two spherical surfaces so two different center of curvatures C1 and C2 lie on either side of the surface.
(c) Radius of curvature: The radius of the sphere from which the spherical surface of lens is made, is called radius of curvature of lens. So each surface of lens have separate radii of curvature R1 and R2 respectively.
(d) Principal axis of lens: A line joining the centre of curvatures and C2 of the two spherical surfaces is the principal axis of the lens.

For the refraction by spherical lens, we assume
(i) Lens to be a thin lens having small thickness.
(ii) Aperture of the thin lens is small, much less than its radius of curvature.
(e) Principal focus:
(i) Convex lens: When a parallel beam of light is incident on one of the spherical surface of a convex lens, the rays after refraction through it, meet at a fixed point on the principal axis. This point is called the principal focus of the convex lens.
Same effect can also be obtained from the other surface also. So the convex lens has two focus Fx and F2 one on each side of the lens.

(ii)Concave lens: When a parallel beam of light is incident on either of the spherical surface of a concave lens, after refraction through it, they appear to come from a fixed point on the principal axis. This point is called the principal focus of a concave lens.
Fig. Principal Focus of a Concave Lens

(f) Focal length: The distance of the principal focus from the optical centre of the spherical lens is called the focal length T of the lens.
If the lens (both spherical surfaces) is surrounded by the same medium having equal refractive indices, then the two principal focal lengths of the lens will be equal, i.e., OFx = OF, = f ,
The focal length of the lens depends upon the following factors:
(i) Radius of curvature of both the refracting surfaces and
(ii) Refractive index of the material of the lens.
Lens Formula: The relation between object distance (u), image distance (v) and focal length (f) of lens is
given by

This expression is valid for all the various positions of the object with a spherical lens including the virtual
image formed by concave lens or a convex lens.

SOME MORE FACTS ABOUT CONVEX LENS

1.  It converges the incident parallel beam of light on the other side of the lens. So it is called convergent lens also.
2.  A ray of light passing through the first principal focus (in a convex lens), or appearing to meet at it (in a concave lens) emerges parallel to the principal axis after refraction.
3. The image formed by the convex lens may be real or virtual, inverted or erect, diminished, same size or magnified, i.e., enlarged depending on the various positions of the object.
4.  Focal length for convex lens is positive while it is negative for concave lens.
5. The linear magnification produced by a convex lens can be less than one, equal to one or greater than one and it has no unit.
6. In case of convex lens, magnification is positive for virtual image and negative for real image. In case of concave lens magnification is always positive because virtual image is always formed.
7.  The lens behaves as a plane glass plate only when it is kept in a medium whose refractive index is equal to that of lens.
8. The degree of convergence or divergence of light rays incident on any refracting surface of the lens is called the power of a lens.
9. A lens of large focal length bends the light rays less after refraction through a small angle by focussing them away from the optical center.
10. . Magnification and sharpness of the image formed by a lens can be increased by combining a number of lenses. Such a lens system is commonly used in various optical instruments such as cameras, telescope, microscope etc.

AIM
To determine the focal length of a convex lens by obtaining the image of a distant object.

MATERIALS REQUIRED
A convex lens, lens holder, white screen such as wall or white painted board, meter scale, distant object.

THEORY

1. 1. The rays coming from the distant object such as tree/sun/electric bulb or tall building travelling a large distance, can be considered as a parallel beam of light.

1. These rays after refraction through the convex lens converge at a point called focus.
2.  The separation between the lens and the screen placed at the focus gives the approximate focal length of a convex lens.
3. The image obtained on the screen kept at principal focus is real, inverted and highly diminished i.e. much smaller in size than the object.

PROCEDURE

1. Select a far off object such as tree/windows of other building/any other far away object in the laboratory/ Sun.
2. Mount the lens in a lens holder.
3. Keep the lens in a vertical position throughout the experiment.
4. Place the lens towards a selected distant object.
5. Keep a screen on the other side of the lens.
6. Move the lens forward or backward to get a sharp, real and inverted image on the screen.
7.  Measure the distance between lens and screen at this position of the lens.
8.  Repeat this procedure by changing the distant object.
9. The distance between the screen (when sharp image is obtained) and convex lens give approximate focal length of the convex lens.

OBSERVATION AND CALCULATION

RESULT
The approximate focal length of a convex lens = …………cm.

PRECAUTIONS

1. The lens holder along with lens should be kept vertical throughout the experiment.
2.  Screen should be neat and clean to get clear image. It must be white.
3.  While measuring the distance, metre scale should be kept horizontal and it must be parallel to the ground.
4.  Distance should be measured only when well defined sharp image of the distant object is obtained.
5. Lens and screen should be at same level.
6. The lens used in the experiment must be clean.

Important Note: The procedure to find the approximate focal length of a convex lens does’ not apply to the concave lens because concave lens always forms a virtual and erect image for any position of the object. The virtual image cannot be taken on the screen.

INTERACTIVE SESSION

Question 1:
What are you doing?
Answer:
I am finding the focal length of a convex lens by forming the image of a far off object on a screen.

Question 2:
What type of image will you get?
Answer:
Real, inverted and diminished i.e. much smaller in size to that of the object.

Question 3:
Which other lens produces a diminished image?
Answer:
Concave lens produces a diminished image of an object held between optical centre and infinity.

Question 4:
What do you mean by infinity?
Answer:
When the object is at sufficiently large distance from the lens/mirror as compared to the focal length of the lens/mirror, the object is said to be at infinity.

Question 5:
Which optical phenomena is represented by this experiment?
Answer:
Refraction.

Question 6:
What is refraction?
Answer:
When the light rays travelling obliquely from one medium to another, a change in the direction of propagation in the second medium is observed. This phenomena is known as refraction of light.

Question 7:
It means light does not travel in the same direction in all media.
Answer:
Yes, it is true.

Question 8:
Suppose a lemon is kept in water in a bowl and viewed from outside. Will it appear larger or smaller than its actual size and why?
Answer:
It appears larger than its actual size because of refraction.

Question 9:
In what terms will you express the extent of the change in direction that takes place in a given pair of media?
Answer:
It is expressed in terms of refractive index.

Question 10:
Which important physical quantity can be linked with refractive index?
Answer:
Speed of light in different media.

Question 11:
How?
Answer:
The refractive index of second medium with respect to first medium is given by the ratio of the speed of light in media first to the speed of light in media second.

Question 12:
What is the relationship between angle of incidence and angle of refraction?
Answer:
The ratio of the sine of the incident angle to the sine of refracted angle is a constant and is equal to the refractive index of second medium with respect to first medium i.e.

Question 13:
What will be the position of the object to get the enlarged image on the same side of the lens as the object?
Answer:
Between focus F and optical centre O.

Question 14:
How the image is formed by the lens?
Answer:
Lens forms the image by refracting light. .

Question 15:
On what factor, the ability of a lens to converge the light rays depends?
Answer:
It depends on its focal length.

NCERT LAB MANUAL QUESTIONS

Question 1:
How will you distinguish between a convex and concave lens?
Answer:

1.  Image formed by the convex lens may be real or virtual depending upon the position of object while concave lens always forms a virtual image wherever the object may be.
2. Convex lens converges the parallel beam of light incident on it after refraction while concave lens diverges the same parallel beam of light which incident on it.
3. The image formed by the convex lens may be equal in size, smaller or larger than the object while concave lens always forms diminished/smaller image of the object.

Question 2:
To determine the focal length of a convex lens, a student focusses a classroom window, a distant tree and the sun on the screen. In which case will the student is closer to accurate value of focal length?
Answer:
Sun

Question 3:
What is the nature of an image formed by a thin convex lens for a distant object? What change do you expect if the lens were rather thick?
Answer:
The image formed by a thin convex lens for a distant object will be real and inverted. A decrease in focal length will be observed when the lens becomes thicker.

Question 4:
You are provided with two convex lenses of same aperture and different thickness. Which one of them will be of shorter focal length?
Answer:
Thicker convex lens. .

Question 5:
If we cover one half of the convex lens while focussing a distant object, in what way will it affect the image formed?
Ans. A full size of image will still be obtained but only the intensity or brightness of image will reduce.

Question 6:
Can this method be used to find the approximate focal length of a concave lens?
Answer:
No, because concave lens always forms a virtual image wherever the object may be.

Question 7:
Which type of lens is used by the watch makers while repairing fine parts of a wrist watch?
Answer:
Convex lens because when the fine parts of a wrist watch are kept between its optical center and focus, magnified, virtual and erect images are formed.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA (1 Mark)

Question 1:
Parallel rays, from a distant tree, incident on the device X, form its distinct image on a screen as shown. The diagram, correctly showing the image of the tree on the screen, is diagram

(a) A and the device X is a convex lens
(b) A and the device X is a concave mirror
(c) B and the device X is a convex lens
(d) B and the device X is a concave mirror

Question 2:
In an experiment to determine the focal length of a convex lens, a student obtained a sharp and inverted image of a distant tree on the screen behind the lens. She then removed the screen and looked through the lens in the direction of object. She will see
(a) an inverted image of the tree at the focus of the lens.
(b) no image as the screen has been removed.
(c) a blurred image on the wall of the laboratory
(d) an erect image of tree on the lens.

Question 3:
While performing an experiment on determining
the focal length of a convex lens, a student obtains  a sharp and inverted image of the laboratory ; window grill on the screen and measures the distance d between the screen and the lens. She then repeats the experiment and takes a distant tree as the object in the second case. In order to  get a sharp image on the screen, she will now need to move the screen
(a) slightly nearer to the lens
(b) slightly farther away from the lens
(c) very close to the lens
(d) very far away from the lens

Question 4:
If you are to determine the focal length of a convex lens, you should have
(a) a convex lens and a screen.
(b) a convex lens and a lens holder.
(c) a lens holder, a screen holder and a scale.
(d) a convex lens, a screen, holders for them and a scale.

Question 5:
After performing the experiment to determine focal length of a convex lens by focussing a distant object, a teacher asked Asha to draw a j ray diagram of her experiment and show where did she place the screen for getting sharp image. The figure drawn by her is given below

The point at which she had placed the screen during her experiment was
(a) at A
(b) at B
(c) at C
(d) at D

Question 6:
If the upper half portion of the lens is covered with black paper, the effect on image is
(a) upper half of the image will be absent
(b) the brightness of the image will be reduced
(c) size of the image depends on covered area of lens
(d)) there will be no effect

Question 7:
Which of the following can be used to find the focal length of a convex lens?
(a) Object at a distance of 10 cm for a lens of focal length 30 cm
(b) Light from a laboratory window
(c) Light from the sun
(d) Light from a tree next to the window

Question 8:
A sharp image of a distant object is obtained on a screen by using a convex lens. In order to determine the focal length of the lens, you need to measure the distance between the
(a) lens and the object
(b) lens and the screen
(c) object and the screen
(d) lens and the screen and also object and the screen

Question 9:
A student obtained a sharp image of the grill
of a window on a screen, using a convex lens. For getting better results, the teacher suggested focussing of a distant tree instead of the grill. In which direction should the lens be moved for this purpose?
(a) Away from the screen
(b) Very far away from the screen
(c) Behind the screen {d) Towards the screen

Question 10:
For the experiment on finding the focal length of
a convex lens by obtaining the image of a distant object, a laboratory assistant keeps the following apparatus on the table. [Foreign 2008]
A. A stand with a metal needle fixed in it
B. A stand with the given convex lens fitted in it
C. A thick cardboard white screen
D. A metre scale
A student can do the experiment by using the
apparatus listed as
(a) A, B, C, D
(b) B, C, D, A
(c) C, D, A, B
(d) D, A, B, C

Question 11:
Three students measured the focal length of a convex lens using parallel rays from a distant object. All of them measured the distance as given
Student A: distance between lens and screen
Student B: distance between object and screen
Student C: distance between object and lens Which student will get the correct focal length of lens?
(a) Student A only
(b) Student B and C both
(c) Student A and B both
(d) Student C only

Question 12:
Out of the following objects which one would you prefer to determine the focal length of a given convex lens by focusing its image on a screen?
(a) A burning candle placed on the far end of a laboratory table
(b) Grills of the laboratory window
(c)A tall tree visible from the laboratory window
(d)Sun rays entering the laboratory through its window.

Question 12:
Given below are few steps (not in proper sequence) followed in the determination of focal length of a given convex lens by obtaining a sharp image of a distant object:
(i) Measure the distance between the lens and screen.
(ii) Adjust the position of the lens to form a sharp image.
(iii) Select a suitable distant object.
(iv) Hold the lens between the object and the screen with its faces parallel to the screen.

Question 14:
A student focussed the image of a distant object using a device ‘X’ on a white screen ‘S’ as shown in the figure. If the distance of the screen from the device is 40 cm, select the correct statement about the device.

(a) The device X is a convex lens of focal length 20 cm.
(b) The device X is a concave mirror of focal length 40 cm.
(c) The device X is a concave mirror of radius of curvature 40 cm.
(d) The device X is a convex lens of focal length 40 cm.

Short Answer Quetions

Question 1:
A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations as given below:

(a) What is the focal length of a given convex lens?
(b) Which set of observations is incorrect and why?

Question 2:
With the help of a diagram, differentiate between a converging and diverging lens. State one use of a convex lens.

Question 3:
A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm on the optical bench. She asks the students to suggest the position of the screen on the optical bench so that a distinct image of a distant tree is obtained on it. What should be the correct position of screen as suggested by the students and why?

Question 4:
A point-sized image of a distant object is formed at the principal focus of a convex lens. If the object starts moving towards the lens, how will the position of the image change? Illustrate it with suitable diagrams.

Question 5:
Explain how a converging lens acts as a magnifying glass. For this purpose, why a lens having a short focal length is chosen instead of lens having a long focal length?

ANSWER KEY
Multiple Choice Questions/VSA

1. (a)
2. (a)
3. (a)
4. (d)
5. (a)
6. (b)
7. (c)
8. (b)
9. (d)
10. (b)
11. (a)
12. (d)
13. (b)
14. (d)

Short Answer Questions

1.

1. When the position of the object is at 2F, the
   image formed at 2F on the end other side of the lens is real, inverted and of same size as that of the object. Therefore,
   object distance = image distance
   2f = 30 cm
   f = 15 cm
   or
2. For the last observation, we find that the object lies between the optical center and focus of the lens, so virtual image is formed and does not fall on the screen. Hence last observation is incorrect.

2. Converging lens: When a parallel beam of light is incident on one of the spherical surfaces of a convex lens, the rays after refraction through it meet at the principal focus on the principal axis. Since the convex lens converges the parallel beam of light, so it is called converging lens.

Diverging lens: When a parallel beam of light is incident on one of the spherical surfaces of a concave lens, the rays after refraction through it appear to come from the principal focus on the principal axis. Since the concave lens diverges the parallel beam of light, so it is called diverging lens.

One use of a convex lens: It is used as magnifying glass.

3. Distant object can be considered to be at infinity. The rays coming from the distant tree are parallel to each other and therefore, its image is formed at the principal focus of the convex lens. Hence, the screen should be placed at 20.5 + 15 = 35.5 cm mark on the optical bench.

4. When an object starts moving from infinity towards the lens and reaches its focus, the image starts moving from its principal focus on the other side of the lens towards infinity. When an object lies in between focus and optical center, the image is formed on the same side of the lens
Object lies between the focus and optical center of the lens

5. To use a converging lens or convex lens as a magnifying glass, the object should be placed within its focus to get a magnified and erect image of the object as shown in the figure below:

A convex lens of short focal length has a greater magnifying power in comparison to a convex lens of large focal length. Hence, the object will appear much larger when seen through a convex lens having short focal length.

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