Nirmala

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2016

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.

SECTION-A

Question.1. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Answer : It is because of Haplodiploidy which is a sex determination system in which males develop from unfertilized eggs and are haploid, and females develop from fertilized eggs and are diploid.

Question.2. Mention the role of‘genetic mother in MOET.
Answer : MOET is a programme for herd improvement to get more eggs. The genetic mother is available for another round of super ovulation in this technology.

Question.3. What is biopiracy ?
Answer : Biopiracy is the term used to refer to the use of bioresources without proper authorisation by the multinational companies and other organisations without compensatory payments to people concerned.

Question.4. Mention two advantages for preferring CNG over diesel as an automobile fuel.
Answer : Two advantages for preferring CNG over diesel are :
(1) It is a very cheap fuel.
(2) It is a greener fuel.

Question.5. Write the probable differences in eating habits of Homo habilis and Homo  erectus.
Answer : Difference in Eating habits of Homo habilis and Homo erectus are :
Homo habilis did not eat meat while Homo erectus ate meat.

SECTION-B

Question.6. A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain.
Answer : In pea plant both male and female flowers are present on the same plant which prevents autogamy but not geitonogamy which results to produce viable seeds after self pollination. In papaya plant, male and female flowers are present on different plants i.e., each plant is either male or female which prevents both autogamy and geitonogany. Thus, a single papaya plant cannot produce yiable seeds.

Question.7. Following are the features of genetic codes. What does each one indicate ?
Stop codon; Unambigous codon; Degenerate codon; Universal codon.
Answer : Stop Codon : Not code for any amino acids. Unambiguous Codon: One codon codes for only on amino acids.
Degenerate Codon : Some amino acids are coded by more than one codon.
Universal Codon : It is same for all either bacteria or human.

Question.8. Suggest four important steps to produce a disease resistant plant through conventional plant breeding technology.
Answer : Important steps to produce a disease resistant plant through conventional plant breeding technology are :

1.  Collection of variability of germplasm.
2.  Evaluation and selection of parents.
3.  Cross hybridisation among the selected parents.
4. Selection and testing of superior recombinant.

9. Name a genus of baculovirus. Why are they considered good biocontrol agents ?
Answer: Nudeopolyhedrovirus is a genus baculovirus which “* are efficient bio-control agents. They are considered to be good bio-control agents because these viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications and show no negative inpacts on plants, mammals, birds or even non-target insects.

Question.10. Explain the relationship between CFC’s and Ozone in the stratosphere.
OR
Why are sacred groves highly protected ?
Answer : Relationship between CFC’s and Ozone :
When CFC’s are released into the stratosphere, they end up being broken up by the UV light, resulting in chlorine being released. This acts like a catalyst with the ozone molecules creating chlorine monoxide and molecular oxygen. Cl atoms are not consumed in the reaction. Hence whatever CFCs are added to the stratosphere, they have permanent and continuing affects on ozone level.
OR
Sacred Groves are relic forest patches traditionally protected by communities in reverence of a deity. Sacred Groves form important repositories of forest biodiversity. It also provide vital ecosystem services to local people are the last refuges for a large number of rare and threatened plants and animal species.

SECTION-C

Question.11. (a) Name the organic material exine of the pollen grain
is made up of. How is this material advantageous to pollen grain ?
(b) Still it is observed that it does not form a continuous layer around the pollen grain. Give reason.
(c) How are ‘pollen banks’ useful ?
OR
(a) Mention the problems that are taken care of by Reproduction and Child Health Care programme.
(b) What is amniocentesis and why there is a statutory ban on it ?
Answer : (a) The hard outer layer called exine is made up of sporopollenin which is one of the most resistant organic material. It can withstand high temperature, strong acids and alkalis. It cannot be degraded by any of the known enzymes. Hence, it acts as a shield and protects the pollen grain from getting damaged.
(b) Exine does not form a continuous layer around the pollen grain. Pollen grain exine has prominent aperture called germ pore where sporopollenin is absent. Germ pores serve as an oudet for the formation of pollen tube.
(c) Pollen grains at a large can be stored for years in liquid nitrogen (— 196°C). So, after this treatment they are stored in pollen banks. Such conserved pollen grains can be used in plant breeding programs.
OR
(a) The problems which are addressed through Repro¬duction and Child Health Care Programme are :
Creating awareness among people about the various reproduction related aspects and providing facilities for building up a reproductively healthy society.
(b) A foetal sex determination test based on the chromo¬somal pattern in the amniotic fluid surrouning the developing embryo is called amino- centesis. Statutory ban on aminocentesis is imposed because this test can be used for determining the sex. of foetus which is increasing female foeticides.

Question.12. What is a test cross ? How can it decipher the heterozygosity of a plant ?
Answer : If the progenies produced by. a test cross show 50% dominant trait and 50% recessive trait, then the cross in which the genotype of an unknown dominant phenotype can be determined by crossing it with an individual homozygous recessive phenotype for that trait is called test cross.
This cross determines whether the dominant character is coming from homozygous dominant genotype or heterozygous genotype, (eg. tallness coming from Tt) when Tt is cross with it, we obtain all Tt (tall) individuals in the program. Thus, test can be used to determine the heterozygosity of the plants.

Question.13. (a) What do *Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’used in Human Genome Project (HGP). Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Answer : (a) YAC stands for Yeast Artificial Chromosomes. BAC stands for Bacterial Artificial Chromosomes-BAC and YAC are specialized cloning vectors which are used in human genome project for cloning or amplification of human DNA fragents.
(b) Less than 2% of total genome codes for proteins in humans. Around approx. 30% of gene functions are known during HGP.
(c) SNP—Single Nucleotide Polymorphism.

Question.14. Differentiate between homology and analogy. Give one example of each.
Answer :

Question.15. (a) It is generally observed that the children who had suffered from chicken-pox in their childhood may not contract the same disease in their adulthood.’ Explain giving reasons the basis of such an immunity in an individual. Name this kind of immunity. (b) What are interferons ? Mention their role.
Answer : (a) The type of immunity is Passive Acquiredimmunity. Due to the development of memory-B cells in person’s body after primary exposure to the disease, the generated antibodies help to prevent the second recurrence of the disease in adulthood.
(b) Interferon : These are proteins made and released by host cells in response to the presence of pathogens such
as viruses, bacteria, parasites etc.Role: It inhibits viral infections and stimulates the entire immune system to fight disease. It also regulates many
kinds of cell functions.

Question.16. (a) Write the two limitations of traditional breeding technique that led to promotion of micro propagation.
(b) Mention two advantages of micro propagation.
(c) Give two examples where it is commercially adopted.
Answer : (a) Two limitations of traditional breeding are :
(1) The required characteristics may not be present in the breeding population.
(2) The breeder does not know exactly what genes have been introduced to the new cultivars.
(b) Advantages of micro propagation are :
(1) Production of many plants that are clones of each other.
(2) It can be used to produce disease free plants.
(c) It is commercially adopted for :
(1) Banana, (2) Tomato

Question.17. (a) How do organic farmers control pests ? Give two examples.
(b) State the difference in their approach from that of conventional pest control methods.
Answer : (a) Organic farmers create a system where the insects are not eradicated, but are kept at manageable levels by a complex system within living and vibrant ecosystem.
Examples are :
(i) The ladybird and dragon flies are useful to get rid of aphids and mosquitoes respectively.
(ii) Bacillus thuringiensis (Bt) are used to control butterfly caterpillars.
(b) Organic farmer, works to create a system where insects that are sometimes called pests are not eradicated, but instead are kept at manageable levels by a complex system of checks and balanced within a living ecosystem, contrary to the conventional farming practices which often, use chemical methods to kill both useful and . harmful life from.

Question.18. (a) Name the selectable markets in the cloning vector pBR322 ? Mention the role they play.
(b) Why is the coding sequence of an enzyme b- galactosidase a preferred selectable marker in comparison to the ones named above ?
Answer : (a) Ampicillin, chloramphenicol are selectable markers in the cloning vector pBR322. Selectable marker, helps in identifying and eliminating non-transformants and Selectively permitting the growth of the transformants.
(b) Alternative selectable marker which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate. In this , a recombinant DNA is . inserted within the coding sequence of an enzyme, b galactosidase, which results into activation of the enzyme referred as insertional inactivation coding sequence for the enzyme b-galactosidase is preferred over antibiotic resistance genes because recombinants can be easily visualised.

Question.19. (a) Why must a cell be made ‘competent’ in biotechnology experiments ? How does calcium ion help in doing so ?
(b) State the role of ‘biolistic gun’ in biotechnology experiments.
Answer : (a) DNA being a hydrophilic molecule, can not pass through cell membranes, hence the cells should be made competent to accept the DNA molecules as competency is the ability of a cell to take up foreign DNA. Calcium ion helps in increasing the pore size in cell wall which enables the cell to take up the recombinant DNA.
(b) To introduce alien DNA into host cells, suitable for plants, cells are bombarded with high velocity micro¬particles of gold or tungsten coated with DNA molecules known as biolistic or gene gun play important role in biotechnology experiments.

Question.20. Explain enzyme-replacement therapy to treat adeno¬ sine deaminase deficiency. Mention two disadvantages of this procedure.
Answer: Adenosine deaminase (ADA) deficiency is a genetic disorder. In this disease, the gene coding for the enzyme ADA gets deleted leading to deficiency of ADA and problems in immune system. Adenosine deaminase (ADA) deficiency in patients can be treated by enzyme replacement therapy.
In this treatment, patients are regularly injected with the functional ADA enzyme.
Disadvantages of this procedure :
(i) It does not completely eradicate the disease.
(ii) Requirement of repeated doses of the enzyme
makes it expensive.

Question.21. Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.
Answer : Mycorrhizae are associations between fungi and the roots of higher plants. The type of interaction that exists in mycorrhizae is mutualism in which both fungi and plants – are dependent on each other. Fungi absorb and transport essential nutrients to the plants and in turn plants provide the fungi with other energy carbohydrates.
The interaction that exist between cattle egret and cattle is known as commensalism. In this type of interaction, one specie is benefitted whereas the other is neither benefitted nor harmed. The cattle egrets (bird) always forage close to where the cattle are grazing because the cattle, as they move, stir up and flush out from the vegetation insects that otherwise might be difficult for the egrets to find and catch. Thus, the cattle is neither benefitted nor harmed but the egret is benefitted.

Question.22. Differentiate between primary and secondary succession. Provide one example of each.
Answer :

SECTION-D

Question.23. A large number of married couples the world over are childless. It is shocking to know that in India the female partner is often blamed for the couple being childless.
(a) Why in your opinion the female partner is often blamed for such situations in India ? Mention any two values that you as a biology student can promote to check this social evil.
(b) State any two reasons responsible for the cause of infertility.
(c) Suggest a technique that can help the couple to have a child where the problem is with male partner.
Answer : (a) Due to improper educational existence and lack of moral values and also along with the orthodox male dominant society nature in India. Females are blamed for infertility issues.
As a biology student:
(1) We must provide proper biological/Sex education at such , a basic stand which can be clear to every individual.
(2) General Health awareness programme must be scheduled to persons for their health related queries.
(b) Causes of Infertility in females :
(1) Ovulation disorders
(2) Problems in the uterus or fallopian tubes Causes of Infertility in Males :
(1) Low sperm count/Low sperm mobility
(2) Genetic abnormality
(c) Artificial Insemination (AI) is a technique that can help the couple to have a child where the problem is with male partner. In this technique, the semen collected either from the husband or a healthy donor is artifically introduced into the vagina or into the uterus of the female.
ICSI (Intra Cytoplamsic Sperm Injection) is another specialized procedure to form an embryo in the lab in which a sperm is directly injected into the ovum.

SECTION-E

Question.24. (a) Explain the menstrual phase in a human female. State the levels of overian and pituitary hormones during this phase.
(b) Why is follicular phase in the menstrual cycle also referred as proliferative phase ? Explain.
(c) Explain the events that occur in a graafian follicle at the time of ovulation and thereafter.
(d) Draw a graafian follicle and label antrum and secondary oocyte.
OR
(a) As a senior biology student you have been asked to demonstrate to the students of secondary level in your school, the procedure(s) that shall ensure cross pollination in a hermaphrodite flower. List the different steps that you would suggest and provide reasons for each one of them.
(b) Draw a diagram of a section of a megasporangium of an angiosperm and label funiculus, micropyle, embryosac and nucellus.
Answer : (a) In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starting from one menstruation till the next one is called the menstrual cycle. This cycle starts with the menstrual phase, when menstrual flow occurs and it lasts for 3-5 days. The menstrual flow results due to breakdown of endometrial lining of the uterus and its blood vessels which forms liquid that comes out through vagina.
During this phase the levels of estrogen and proges- terone are low.
(b) The proliferative phase is the part of the menstrual cycle during which follicles inside the ovaries develop and mature in preparation for ovulation. The levels of FSH.
increase in the bloodstream during the proliferation phase, stimulating the maturation of follicles. Each follicle contains an ovum, or egg. Although many follicles may grow and increase in size during this phase, only one will reach full growth and release the ovum at the time of ovulation.
(c) During the mid cycle Leutinizing hormone secretes to its maximum level which induces rupture of Graafian follicle and thereby the release of ovum (ovulation). The ovulation is followed by the luteal phase during which the remaining parts of the Graafian follicle transform as the corpus luteum which secretes large amounts of progesterone which is essential for maintenance of the endometrium. Such an endometrium is necessary for implantation of the fertilised ovum and other events of pregnancy.

OR
(a) The different steps that would suggest for cross pollination
in a hermaphrodite flower are :
(i) Removal of anthers frono the flower bud before the anther dehisces using a pair of’forceps is necessary referred as emasculation.
(ii) Emasculated flowers covered with a tag of suitable size to prevent contamination of its stigma called Bagging.
(iii) When the stigma of bagged flower attains receptivity, mature pollen grains collected from anther of the male parent are dusted on the stigma, and the flowers are rebagged.

Question.25. Describe Meselson and Stahl’s experiment that was carried in 1958 on E. coli. Write the conclusion they arrived at after the experiment.
OR
(a) Describe the process of transcription in bacteria.
(b) Explain the processing the hnRNA needs to
undergo before becoming functional mRNA eukaryotes.
Answer : The experiment was performed by Meselson and Stahl. The following steps were followed in the experiment.
coli was grown in a medium containing 1SNH4 Cl the heavy isotope 15N as the sole nitrogen source. This led to the incorporation of 15N into the newly synthesised which ultimately made the DNA heavy.This heavy DNA was separated from the normal DNA by density gradient centrifugation using cesium chloride as the gradient.
The cells were then transferred into a medium with 14N as the nitorgen source. Samples were taken from this medium and the DNA was extracted.
Observation:
Since E. coli divides every 20 minutes, the DNA extracted after 20 minutes in the experiment had a hybrid density. The DNA extracted after 40 minutes had equal amounts of hybrid and light densities.
Conclusion:
DNA extracted from the culture after another generation was composed of equal amounts of this hybrid DNA and of light DNA.
This implies that the newly synthesised DNA obtained one of its strands from the parent. Thus, replication was semiconservative.

OR
(a) Transcription has three steps : Initiation, elongation and termination.
Initiation : RNA polymerase binds to promoter and initiates transcription. It associates with initiation factor and alters the specificity of RNA polymerase to initiate the transcription.
Elongation : RNA polymerase uses nucleoside triphosphate as substrate, and polymerises in a template depended fashion following the rule of complementarity and facilitates opening of the helix and continues elongation.
Termination occurs when termination factor (rho) alters the specificity of RNA polymerase.
Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination for translation of transcription.

(b) The precursor of mRNA, i.e., hnRNA, contains both introns and exons. Introns are removed and exons are joined by a process called splicing. The remaining mRNA is processed in two ways :
Capping : An unusual nucleotide (methyl gunosine triphosphate) is added to the 5′-end of hnRNA.
Tailing : Adenylate residues (200-300) are added at 3′-end in a template independent manner.
When hnRNA is full processed, it is known as mRNA, which is transported out of the nucleous.

Question.26. (a) Name the two growth models that represent population growth and draw the respective growth curves they represent.
(b) State the basis for the difference in the shape of these curves.
(c) Which one of the curves represent the human population growth at present ? Do you think such a curve is sustainable ? Give reason in support of your answer.
OR
(a) Taking an example of a small pond, explain how the four components of an ecosystem function as a unit.
(b) Name the type of food chain that exists in a pond.
Answer. (a) Two growth models are :

1.  Exponential
2. Logistic

Exponential Growth :
N : Population size b : Birth rates
d : Death rates


(b) The difference in the shape of these curves is the amount of resources available for the given population. When resources are unlimited each species realizes its innate potential to grow in number and result in a J-shaped curve in exponential growth while in logistic growth no population has unlimited resources leads to competition for resources and show S-shaped curve.
(c) Logistic growth curve represents the human population growth at present because the number of human beings are increasing rapidly but the available resources are not increasing enough. Such a curve is not sustainable because at one point the human population would reach a place where resource there would not be finite resources for every one.
OR
(a) The four components of an ecosystem are :

1.  Productivity
2.  Decomposition
3. Energy flow
4. Nutrient cycling

A pond is a shallow water body in which all the four components of in ecosystem are well exhibited. The abiotic component is the water with all the dissolved inorganic and organic substances and the rich soil deposit at the bottom of the pond. The solar input, the cycle of temperature, day- length and other climatic conditions regulate the rate of function of the entire pond. The autotrophic components include the phytoplankton, some algae and the floating sub merged and marginal plants found at the edges. The consumers are represented by the zooplancton, the free swimming and bottom’dwelling forms. This system performs all the functions of any ecosystem and of bisophere as a whole i.e. conversion of inorganic into organic material with the help of the radiant energy of the sun by the autoprophs : Consumption of the autotrophs by heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrophs. There is unidirectional movement of energy towards the higher trophic levels and its dissipation and loss as heat to the environment.
(b) Aquatic food chain exists in a pond.

CBSE Sample Papers for Class 8 Science Practice Paper 3

Test Paper III
Class VIII

Time : 2 1/2 Hrs                                                                                                     Maximum Marks : 100 

General Instructions:

1. Read the content of the question carefully and attempt them there after.
2. Follow the specified word limit, wherever mentioned.
3. Try to be brief and concise and have clarity in expression.
4. Make diagrams and figures wherever necessary.

1. Fill in the blanks:                                                                                                1 x 10 = 10               

1. The process of beating out the grains from the harvested crop plants is
   called………….
2. The yolk is surrounded by a white portion called………….
3. The growing of plants (or crops) in the fields for obtaining food is called………….
4. The flow of electrons in the wire constitutes an ………….
5. The space around a magnet in which it can exert force on magnetic objects like
   iron and steel is called its………….
6. …………is a ceramic material containing iron along with cobalt, nickel or zinc.
7. ………….are microscopic single called algae whose cell walls are made of silica.
8. ………….is a fungus commonly known as bread mould.
9. ………….is a parasitic protozoan and cause a disease called malaria.
10. The………….is a complex mixture of a large number of hydrocarbons.

2. Give one word answer of the following questions:                                1 x 10 = 10             

1. A device which is used to test the purity of milk.
2. The process of separating grain from chaff and hay with the help of wind.
3. The removal of anthers from the flowers of a plant to prevent self pollination.
4. The process of transfering the seedlings from the nursery to the main field by hand.
5. Machine used to remove honey from honey comb.
6. The rearing of bees for obtaining honey.
7. Scientist who invented the dry cell.
8. Another name of blue green algae.
9. The process by which yeast reproduces.
10. A fungi used as a food.

3. Give reasons:                                                                                                         2 x 5 = 10               

1. Virus particle not considered to be a cell. Why?
2. We use iron for making an electromagnet and not steel. Why?
3. Farmers carry out lavelling of the ploughed field. Why?
4. Weeds multiply and spread very fast. Why?
5. A person having acidity is given milk of magnesia. Why?

4.Answer the following questions:                                                                     2 x 5 = 10               

1. What js pisiculture? How are fish reared by pisiculture?
2. Give two broad groups of cattle feed. Give their constitutions.
3. What are the two types of crops based on seasons? Give two examples for both.
4. What is meant by electromagnetic induction? Name the two scientists discovered the effect of it.
5. What are mycoplasmas? Name two diseases caused by mycoplasmas.

5.Difference between the following pair with example:                       2 1/2 x 4 = 10

1. Manure and fertiliser
2. Drycell and storage cell
3. Conductors and insulators
4. Gram positive and Gram negative bacteria.

6.Define following with diagram (any four)                                           2 1/2 x 4 = 10

1. Electromagnet
2. Magnetic compass
3. Electric circuit
4. Protozoan
5. Yeast

7. Answer any two of following questions:                                                      5 x 2= 10

1. You are given two similar steel bars, one of which is a magnet. What else would you require to test which of the given steel bar is a magnet?
2. What are the various types of bacteria? Give shape and one example of each.
3. How will you find out whether an egg is
   (a) fertilised or unfertilised?
   (b) spoiled or good eggs?

8.Give the answer of the following:

1. Name two high yielding breeds/varities of the following:                       1 x 5 = 5
   (a) Cow (b) Buffalo (c) Hen (d) Wheat (e) Paddy.
2. Write the full form of the following:                                                                1 x 5 = 5
   (a) KVIC (b) DDT (c) BHC (d) TMV (e) RNA.
3. Name the types of chemical reaction:                                                              1 x 5 = 5
   
4. Name the disease causing microorganism:                                                   1 x 5 = 5
   (a) Food poisoning     (b) AIDS (c) Measles
   (d) Sleeping sickness (e) Ring worm.
5. Draw the electric symbol of the following:                                                     1 x 5 = 5
   (a) Dry cell            (b) Battery     (c) Electric bulb
   (d) Switch open   (e) Connecting wire.
6. Draw anyone labelled diagram:                                                                                               (a) Fractionating Tower (b) Dry cell

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2010

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26 questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Question.1. Name the physical quantity whose S.I. units is JC^-1. Is it a
scalar or a vector quantity?
Answer : The SI unit of electrostatic potential is JC^-1. It is a scalar quantity.

Question.2. A beam of a-particles projected along + X-axis, experiences a force due to a magnetic field along +y-axis. What is the direction of magnetic field?

Answer: The direction of magnetic field is perpendicular to x andy axis i.e. along z axis.

Question.3. Define self-inductance of a coil. Write its S.I units?
Answer : It is the phenomenon of production of induced e.m.f. in a coil when a changing current passes through it. S.I unit .is Henry (H).

Question.4. A converging lens is kept coaxially in contact with diverging
lens — both the lenses being of equal focal lengths, what is the focal length of the combination?
Answer : Let f₁ is the focal length of convex lens and  f₂ is the focal length of concave lens. Then their equivalent focal length F would be:

Question.5. Define ionization energy. What is its value for hydrogen atom?
Answer: Ionisation energy is the required energy to knock an electron out of the atom, that is energy required to take an electron from its ground state to the outermost orbit (n = ∞).
Ionisation energy of hydroge n = E_∞ – E₁
= 0 – (-13.6 eV) = 13.6 eV

Question.6. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.
Answer : The current I through the wire is related with drift velocity as :
I = nAevd
Since two wires X and Y of same diameter but different materials are joined in series.

Question.7. X-ray is the part of electromagnetic spectrum whose wavelength lies in the range of 10^-10 m. Give its one use.
Answer : The wavelength range of 10^-10m lies in x-rays. Its uses are : X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer.

Question.8. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer?
Answer: No, since energy E = hv, where h is. Planck’s constant and u is frequency that does not change when light goes from one medium to another. Hence energy does not change.

Question.10. A spherical conducting shell of inner radius r₁ and outer radius r₂  has a charge ‘Q’, a charge‘q’ is placed at centre ofdie shell.
(a) What is the surface charge density on (i) inner surface (ii) outer surface of the shell?
(b) Wire the expression for the electric field at a point x > r₂, from the centre of the shell.
Answer : The charge (+ q) at the centre induces charge – q on the inner surface of the shell and charge + q on the outer surface of the shell

Question.11.Draw a sketch of a plane electromagnetic wave propagating along the Z-direction. Depict clearly the directions of electric and magnetic fields varying sinusoidally with z
Answer : E along x-axis, B along y-axis.or vice-versa X

Question.12. Show that the electric field at the surface of a charged conductor is given by.

where a is the surface charge density and n is unit vector normal to the surface in the outward direction?
Answer:

Question.13.Two identical loops one of copper and the other of aluminum, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the two coils, justify your answer.

Question.14. An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths?

Question.15. Write two factors justifying the need of modulating a signal. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be peak voltage of the modulating signal in order to have a modulation index of 75%.
Answer : Modulation of a signal is necessary to transmit a signal in the audio frequency range over a long distance due to
(i) Size of antenna
(ii) Effective power radiated by antenna.

Question.16 Write Einstein photoelectric equation? State clearly the three salient features observe in photoelectric effect, which can be explained on basis of above equation?
Answer : Einstein photoelectric equation :

Where ø₀ is work function. If the energy of the photon absorbed by the electron is less than the work function ø₀ of the metal, then the electron will not be emitted. Therefore, for the given metal, the thershold, frequency of light be v₀, then an amount of energy hv₀ of the photon of light will be spent in ejecting the electron out of the metal, that is, it will be equal to the work function W. Thus

This equation is called ‘Einstein Photoelectron equation. Salient features :
1. The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the * frequency of incident radiation.
2. There exists a minimum cut-off frequency v₀, for which the stopping potential is zero.
3. Photoelectric emission is instantaneous.

Question.17.Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
OR
Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A≤ 240. How do you explain the constancy of binding energy per nucleon in range 30 < A < 170 using the property that nuclear force is short-ranged?
Answer : Plot of potential energy of a pair of nucleons as a fuction of their seperation:

Conclusions:
(i) The nuclear force is much stronger than the Coulomb force acting between the charges or gravitational forces between
(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than few fermi.
(iii) For a separation greater than r₀ the force is attractive and for separations less than ro force is strongly repulsive.

Constancy of binding energy per nucleon in range 30 < A < 170 a consequence of the fact that nuclear force is short-ranged. It will be under the influence of only some of its neighbours.

Question.18. (i) Identify the logic gates marked P and Q in the given logic circuit.


(ii) Write down output at X for the inputs A = 0, B = 0 and A = 1, B = 1.
Answer : (i) The logic gate P is NAND gate and logic gate Q is OR gate.
(ii) Let output of logic gate P is Y.

Question.19. Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz to 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. Why there an upper limit to frequency of waves used in this mode?
Answer : Sky wave propagation can be used by short wave broadcast services having frequency r and from a MHz to 30 MHz.

In this mode sky waves are directed towards the sky and are reflected by the ionosphere toward the desired station on the earth.
Sky wave propagation is restricted to frequencies upto 30 MHz because ionosphere cannot reflect electro-magnetic waves having frequencies greater than 30 MHz.

Question.20.Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V When the terminals of the cell are also connected accross to a resistance of 5 Ω as shown in the circuit, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.

Answer: The internal resistance of a cell depends on nature of electrolyte as well as the concentration of the electrolyte.
Let r be the internal resistance of the cell

Question.21. A network of four capacitors each of 12 μF capacitance is connected to a 500 V supply as given in the figure. Determine (a) equivalent capacitance of the network and
(b) charge on each capacitor.

Question.22. (i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly its working.
(ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment?
OR
(i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working.
(ii)Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
Answer: (i)

Working : Telescope has an objective and eyepiece. The objective has a large focal length and much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image at infinity.
(ii) Calculation of magnifying power :


The objective forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual.
(ii) To achieve a large magnification of a small object; both the * objective and eyepiece should have small focal lengths.

Question.23. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength  450 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved
away from the slits?
Answer :

Question.24. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I₁, I₂ and I₃ in the circuit diagram shown.

Answer: (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.
∑I = 0
(b) Loop rule : The algebraic sum of changes in potential, around any closed loop, involving resistors and cells in the loop, is zero.

Question.25.(a) Write symbolically the
(b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life.
Answer:

(b) Derivation of average life :

Hence the average life period of a radioactive element is 1.44 times the half life period of the element.

Question.26. How does an unpolarised light get polarised when passed through a Polaroid? Two polaroids are set in crossed positions. A third Polaroid is placed between the two making an angle θ with the pass axis of the first Polaroid. Write the expression for the intensity of light transmitted from the second Polaroid. In what orientations will be transmitted intensity be (i) minimum and (ii) maximum?
Answer. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid, then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules.
When the polaroid is rotated in the path of plane polarized light, its intensity will vary from maximum to minimum, let Io be intensity of polarized light after passing through first polarizer P₁. Then intensity of polarized light after passing through second polarizer P₂ is given by
I = (I₀ cos² θ)
Expression for the intensity transmitted through second polaroid
I = (I₀ cos² θ) cos² (90° – θ) = I₀ (cos θ sin θ)² = I₀sin² 2θ/4
where I0 is the intensity of the polarized light after passing through the first polaroid.
(i) Intensity will be maximum when θ= 45°
(ii) Minimum when θ=0°

Question.27. An illuminated object and a screen are placed 90 cm apart.
Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.
Answer : For real image, m = v/u =2
v = 2u

Question.28. (a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance’in series in required whereas in an ammeter a shunt is used?
OR
(a) Derive an expression for the force between two long parallel current carrying conductors.
(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction?

Answer : (a) Moving coil galvanometer : It is a device used for the detection and measurement of small electric current.

Construction : It consists of a coil having a large number of turns of insulated copper wire wound around metallic frame.
A hair spring is attached to lower end of coil, the other end is attached to scale through the pointer.
Principle : Galvanometer action is based upon the principle that when electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current.
Working : The magnetic torque tends to rotate the coil. A spring provides a counter torque that balances the magnetic torque; resulting in a steady angular deflection. The deflection is indicated on the scale by a pointer attached to the spring.
If I is current flowing through coil.
B = magnetic field supposed to be uniform and parallel to coil
A = area of coil
Deflecting torque acting on the coil is
τ = NI AB sin 90° = NIBA             ………. [sin 90° = 1 ]
Due to deflecting torque, the coil rotates and suspension wire gets twisted.
(b) Importance and production of radial magnetic field :
In a radial magnetic field, magnetic torque remains maximum for all positions of the coils. It is produced due to cylindrical pole pieces and soft iron core.
Reason:
Voltmeter : This ensures that a very low current passes through the voltmeter and hence does not change (much) the original potential difference to be measured.
Ammeter : This ensures that the total resistance of the circuit
does not change much and the current flowing remains (almost) as its original value.
(c) A galvanometer can be converted into a voltmeter by connecting a high resistance in series with galvanometer to
draw a very small current. A galvanometer can be converted into an ammeter by connecting a low resistance shunt in parallel with galvanometerto draw a very large current.

Two long parallel conductors ‘a’ and ‘b’ are separated by a distance d and carry (parallel) currents  I_a and  I_b, respectively. The conductor ‘a’produces, the same magnetic field B_a at all points along the conductor ‘b’ perpendicular to the plane of the page directed downwards.

By Fleming left hand rule, this force acts on CD towards AB
force. This force will be attractive.
(b) Definition of ampere : The ampere is the value of that teady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed on meter apart in vacuum, would produce on each of these conductors a force equal to 2 x 10^-7 newton per meter of length.
(c) Magnetic field due to the straight wire AB at a perpendicular , distance r from it.

The direction of force on proton acts in plane of paper towards right. ,

Question.29.State Faraday’s law of electromagnetic induction.
Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B
from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variations of these quantities with distance 0≤ x ≤ 2b

OR
Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances?
Answer: Part I: Faradays law of induction : The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Where ∅_B is flux linked with one turn of the coil. The negative sign indicates the direction of 6 and hence the direction of current in a closed loop. If the circuit is closed a current
I=∈/R is set up in it.
Part II : Refer to following figure. The arm PQ of the rectangular conductor is moved from x = 0 outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero the situation when the arm PQ possesses substantial resistance r. Consider the situation when arm PQis pulled outwards from x=0 to x=2b and is then moved back to x = 0 with constant speed v.




OR
Step up Transformer:
Principle : When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it (mutual induction). The value of this emf depends on the number of turns in the secondary.

secondary windings. The core laminations are joined in the form of strips; in between the strips you can see that there are some narrow gaps right through the cross-section of the core. These staggered joints are said to be ‘imbricated’. Working : Both the coils have high mutual inductance. A mutual electro-motive force is induced in the transformer from the alternating flux that is setup in the laminated core, due to the coil that is connected to a source of alternating voltage. Most of the alternating flux developed by this coil is linked with the other coil and thus produces the mutual induced electro-motive force. Thus so produced electro-motive force can be explained with the help of Faradays laws of Electromagnetic Induction as
If the secondary coil circuit is closed, a current flow in it and thus electrical energy is transferred magnetically from the first to the second coil.
The alternating current supply is given to the first coil and hence it can be called as the primary winding. The energy is drawn out from the second coil and thus can be called as the secondary winding.

Large scale transmission : The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is  stepped-up (so that current is reduced and consequendy, the I²R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.

Question.30. (a) Draw the circuit arrangement for studying the input and output characteristics of an n-p-ti transistor in CE configuration. With the help of these characteristics define
(i) input resistance, (ii) current amplification factor. ‘
(b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce self-sustained oscillations.
Answer: (a)


(b) Circuit diagram :

Working : In an oscillator, we get an ac output without any external input signal. Hence, the output in an oscillator is self-sustained. To attain this, an amplifier is taken. A portion of the output power is returned back (feedback) to the input in phase with the starting power (this process is termed positive feedback).

SET II

Note : Except for the following questions, alL the remaining questions have been asked in Previous Set.

Question.1.Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom form its
(i) second permitted energy level to the first level, and
(ii)the highest permitted energy level to the first permitted level.
Answer:

Question.4. A beam of electrons projected along + x-axis, experiences a force due to a magnetic field along the +y-axis. WTiat is the direction of the magnetic field?

Answer : The direction of magnetic field is along z-axis.

Question.12. A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity V as shown in the figure. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. The magnetic field is normal to the loops.

Answer : The induced emf is expected to be constant only in the area of rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of one field region is not constant. Hence induced emf will vary accordingly.

Question.19. A network of four capacitors each of 15 pF capacitance is connected to a 500 V supply as shown in the figure. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.

Question.20. Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.0 V, when the terminals of the cell are also connected to a resistance of 3 Ω as shown in the circuit, the voltmeter reading drops to 1.5 V Find the internal resistance of the cell.

Question.22.In Youngs double slit experiment, the two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 420 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved
away from the slits?

the screen is moved away from the slits fringes become further apart as D increases.

Question.23. State KirchhofFs rules. Apply Kirchhoff’s rules to the loops ACBPA and ACBQA to write the expressions for the currents I₁, I₂ and I₃ in the network.

Question.27. The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. A beam of protons projected along +x-axis, experience a force due to a magnetic field along the —y-axis. What is the direction of the magnetic field?

Answer. The direction of megnetic field is along z-axis.

Question.8.The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of electron in this state ?
Answer:

Question.11. (i) Identify the logic gates marked P and Q in the given logic circuit.

Question.21. A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. If-the magnification is 19, find the focal length of the lens.

Question.24. Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.5 V. When the terminals of the cell ate also connected to a resistance of 5 D as shown in the circuit, the voltmeter reading drops to 2.0 V. Find the internal resistance of the cell.

Question.25.In Young’s double slit experiment, the two slits 0.20 mm apart are illuminated by monochromatic light of wavelength 600 nm. The screen is 1.0 m away from the slits.
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum.
(b) How will the fringe pattern change if the screen is moved
away from the slits?

(b) When the screen is moved away from the slits fringes become further apart as D increases.

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2012

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26 questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Question.1. When electrons drift in a metal from lower to higher potential, Does it mean that all the free electrons of the metal are moving in the same direction ?
Answer : No, when electric field is applied the electrons will have net drift from lower to higher field but locally electrons may Collide with ions and may change its direction of motion.

Question.2. The horizontal component of the earths magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator?
Answer : On the equator, the value of both angle of dip (8) and vertical component of earth’s magnetic field is zero. So, in t this case, B_v= 0.

Question.3. Show on a graph, the variation of resistivity with temperature for a typical semiconductor.
Answer: The following curve shows the variation of resistivity with temperature for a typical semiconductor.

This is because for a typical semiconductor,resistivity decreases rapidly with increasing temperature.

Question.4. Why should electrostatic field be zero inside a conductor?
Answer : Charge on conductor resides on its surface. So if we consider a Gaussian surface inside the conductor to find the electrostatic field,

Where, q = charge enclosed in Gaussian surface.
q= 0, inside the conductor, hence the electrostatic field inside the conductor is zero.

Question.5. Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum.
Answer : Both microwaves and UV rays are a part of the electromagnetic spectrum. Thus, the physical quantity that remains same for both types of radiation will be their speeds which is equal to c (3 x10⁸m/s).

Question.6. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer : A biconvex lens twill act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction
as itself.

Question.7.Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

Question.8. State de-Broglie hypothesis.
Answer : de Broglie postulated that the material particles may exhibit wave aspect. Accordingly a moving material particle behaves as wave and the wavelength associated with material particle is

Question.9. A ray of light incident on an equilateral prism (μ_g = √3 ) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.
Answer : It is given that the prism is equilateral in shape. So, all the angles are equal to 60°. Thus, the angle of

Question.10. Distinguish between‘Analog and Digital signals’,
OR
Mention the functions of any two of the following used in communication system :

1. Transducer
2.  Repeater
3. Transmitter
4. Bandpass Filter

Answer : Analog Signal: It is continuous signal, which varies continuously with variable may be time or distance etc. E.g. Voice of human
Digital Signal : It is a type of signal which has only two values high or low. In digital high means 1 and low means 0.

1.  Transducer : It is an electric device which coverts energy from one form to another form. E.g. microphone, which converts sound energy into electric energy and vice-versa.
2.  Repeater : It is an electronic device used in transmission system to regenerate the signal. It picks up a signal amplifies it and transmits it to receiver.
3.  Transmitter : Transmitter is an electronic device which is used to radiate electromagnetic waves. The purpose of the transmitter is to boost up the signal to be radiated to the required power level, so that it can travel long distances. The most familiar transmitters are mobile transmitter antennas, radio and T.V broadcasting antennas etc. ‘
4.  Bandpass filter : It is an electronic filter, which pass the certain band (range) of frequency and reject rest of all.

Question.11. A cell of emf E and internal resistance r is connected to two external resistance Ri and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance Ri only
(iii) withR₁ and R₂ in series combination
(iv) with R₁ and R₂ in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
Answer : The current relating to corresponding situations is as follows :

Question.12. The susceptibility of a-magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties.
Answer : Diamagnetic materials have negative susceptibility. So the given magnetic material is diamagnetic.
Two properties of diamagnetic material:

1. They do not obey Curie’s law.
2. They are feebly repelled by a magnet.

Question.13. Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two cods.
Answer: Magnetic field produced by the two coils at their common centre having currents I₁ and I₂, radius a₁ and a₂, number of turns N₁, N₂, are given by :

Question.14. When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used the current flows continuously. How does one explain this based on the concept of displacement current?
Answer : When an ideal capacitor is charged by dc battery, charge flows till the capacitor gets fully charged.
When an ac source is connected then conduction current

Question.15. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.
Answer : We know, that for a point charge Q

Question.16. Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by 1/2 L I².
Answer : Self inductance is the inherent inductance of a circuit, given by the ratio of the electromotive force produced in the circuit by self-induction to the rate of change of current producing it. It is also called coefficient of self-induction.

Question.17. The current in the forward bias is known to be more (∼ mA) than the current in the reverse bias (∼ μA). What is the reason, then, to operate the photodiode in reverse bias ?
Answer : The current in the forward bias is due to majority carriers where as current in the reverse bias is due to minority carriers. So current in forward bias is more (~mA) than current in reverse bias (~μA).
On illumination of photodiodes with light, the fractional change in the majority carriers would be much less than that’ in -minority carriers. It implies fractional change due to light on minority carrier dominated reverse bias current is more easily measurable
than fractional change in forward bias current. f So photodiodes are operated in reverse bias condition.

Question.18.A metallic rod of ‘L’ length is rotated with angular frequency of ‘o’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Question.19. The figure shows a series LCR circuit with L = 75 H, C = 80 μF, R = 40Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii)the current at the resonating frequency,
(iii)the rms potential drop across the inductor at resonance.

Question.20. A rectangular loop of wire of size 4 cm x 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wife are coplanar, find
(i) the torque acting on the loop and
(ii)the magnitude and direction of the force on the loop due to the current carrying wire.

Question.21. (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the n^th orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is. initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?
Answer : (a) According to Bohr’s ‘second postulate of quantization, the electron can revolve around the nucleus only in those circular orbits in which the angular momentum of the electron is integral multiple of where h is Planck’s

Question.22. In the figure a long uniform potentiometer wire AB is having .a constant potential gradient along its length. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε_1/ε₂ and (ii) position of null point for the cell ε₁.
How is the sensitivity of a potentiometer increased?

Using Kirchhoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential between A and D.

Question.23.

1.  What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
2. Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number A.

Answer:

1.  The constancy of BE/A over most of the range is saturation property of nuclear force, ‘
   In heavy nuclei: nuclear size > range of nuclear force.
   So, a nuclear sense approximately a constant number of neighbours and thus, the nuclear BE/A levels off at high ‘A’.
   This is saturation of the nuclear force.
2.  To find the density of nucleus of an atom, we have an atom with mass number let say A and let mass of the nucleus of the atom of the mass number A be m

This expression is independent of mass number A and is constant.

Question.24.Write any two factors which justify the need for modulating a signal. Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave.
Answer : Factors needed for modulating a signal:

1.  To send the signal over large distance for communication.
2. Practical size of antenna.
   

Question.25.Write Einsteins photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be explained using this equation.
Answer : Einstein’s photoelectric equation,

According to Planck’s quantum theory, light radiations consist of small packets of energy.
Einstein postulated that a photon of energy hv is absorbed by the electron of the metal surface, then the energy equal to øis used to liberate electron from the surface and rest of the energy hv —ø becomes the kinetic energy of the electron.

Sailent features observed in photoelectric effect:

•  The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation. –
•  There exists a minimum cut — off frequency V₀, for which the stopping potential is zero.
• Photoelectric emission is instantaneous.

Question.26.(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Youngs double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference isλ, is K units. Find out the intensity of light at a point where path difference is λ/3.
Answer : (a) Coherent sources have constant phase difference between them i.e., phase difference does not change with time. Hence, the intensity distribution on the screen remains constant and sustained.
(b) We know

Question.27.Use Huygens principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band?
Answer : Consider a parallel beam of monochromatic light is incident normally in a slit of width b as shown in figure. According to Huygens principle every point of slit acts as a source of secondary wavelets spreading in all directions. Screen is placed at a larger distance.
Consider a particular point P on the screen receives waves from all the secondary sources. All these waves start from different point of the slit and interference at point P to give resultant intensity.

Point P₀ is a bisector plane of the slit. At P₀, all waves are travelling equal optical path. So all wavelets are in phase thus interface constructively with each other and maximum, intensity is observed. As we move from P₀, the wave arrives with different phases and intensity is changed. Intensity at point P is given by

Where, D = Distance between screen and slit,
λ = Wavelength of the light,
b = size of slit
So with the increase in size of slit the width of central maxima decrease. Hence, double the size of the slit would result in half the width of the central maxima.

Question.28. Explain the principle of a device that can build up high voltage of the order of a few million volts.
Draw a schematic diagram and explain the working of this device.
Is there any restriction on the upper limit of the high voltage set up in this machine? Explain.
OR
(a) Define electric flux. Write its S.l. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
Answer: Van de Graff generator is the device used for building up high voltages of the order of a few million volts.
Such high voltages are used to accelerate charged particles such as electrons, protons, ions, etc.
It is based on the principle that charge given to a hallow conductor is transferred to outer surface and is distributed uniformly over it.
Construction:

It consists of large spherical conducting shell (S) supported over the insulating pillars. A long narrow belt of insulating material is wound around two pulleys P₁ and P₂,B₁ and B₂ are two sharply pointed metal combs. B₁ is called the spray comb and B₂ is called the collecting comb.
Working: The spray comb is given a positive potential by high tension source. The positive charge gets sprayed on the belt. As the belt moves and reaches the sphere, a negative charge is induced on the sharp ends of collecting comb B2 and an equal positive charge is induced on the farther end of B₂ . This positive charge shifts immediately to the outer surface of S. Due to discharging action of sharp points of B₂ , the positive charge on the belt is neutralized. The uncharged belt returns down and collects the positive charge from B₁ , which in turn is collected by B₂. This is repeated. Thus, the positive charge of S goes on accumulating. In this way, potential differences of as much as 6 or 8 million volts (with respect to the ground) can be built up. The main limiting factor on the value of high potential is the radii If the electric field just outside the sphere is sufficient for dielectric breakdown, of air, no more charge can be transferred to it.
For a conducting sphere,
Electric field just outside sphere

Question.29. Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 130 cm and an eye-piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye-piece.
OR
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
Answer : Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at the least distance of distinct vision to the angle subtended at the eye by the object lying at infinity, when seen directly.
The formula for magnifying power is,


A microscope is used to look into smaller objects like structure of cells etc. On the other hand, a telescope is used to see larger objects that are very far away like stars, planets etc.
Telescope mainly focuses on collecting the light into the objective lens, which should thus be large, where the microscope already has a focus and the rest is blurred around it. There is a big difference in their magnification factors.
For telescope the angular magnification is given by

Question.30. Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, Av, of the amplifier is give by is the current gain; ri RL is the load resistance and r, is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?
OR
(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode.
Explain is working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate (ii) NAND gate

Answer : Circuit diagram of CE transistor amplifier :
Working : If a small sinusoidal voltage is applied to the input of a CE configuration, the base current and collector current will also -have sinusoidal variations. Because the collector current drives the load, a large sinusoidal voltage Vo-will be observed at the output. The expression for voltage gain of the transistor in CE configuration is :




Working: When the diode rectifies the whole of the AC wave, it is called full wave rectifier.
The figure shows the arrangement for using diode as full wave rectifier. The alternating input signal is fed to the primary P₁ P₂ of a transformer. The output signal appears across the load resistance R_L.
During the positive half of the input signal, suppose P₁ and P₂ are negative and positive respectively. This would mean that S₁ and S₂ are positive and negative respectively. Therefore, the diode D₁is forward biased and D₂ is reverse biased. The flow of current in the load resistance AL is from A to B.
During the negative half of the input signal, S₁ and S₂ are negative and positive respectively. Therefore, the diode D₁is reverse biased and D₂ is forward biased. The flow of current in the load resistance R_L is from A to B.

SET II

Note : Except for the following questions, all the remaining . questions have been asked in Set-I 2012.

Question.1. Why must electrostatic field be normal to the surface at every point of a charged conductor?
Answer : The electrostatic field must be normal to the surface , of the conductor of every point because if the electric field is not normal to the surface of the charged conductor, there will be a component of the electric field along the surface of the conductor, which would exert a force on the charges at the surface. Due to this charge starts flowing which is not possible.

Question.6. Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed v. The conductor is carrying current I in the direction shown in the figure.

Answer: Clockwise

Question.10. Derive the expression for the self inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.
Answer : Self-Inductance of a long Solenoid : The magnetic field B at any point inside such a solenoid is constant and given by,

Question.16. Two identical circular loops, P and Q, each of radius r and carrying currents I and 21 respectively are lying in parallel planes such that they have a common axis. The direction of
current in both the loops is clockwise as seen from O which is equidistance from the both loops. Find the magnitude of the net magnetic field at point O.

Question.20. A series LCR circuit with L = 4.0, C = 100 μF, R = 60 Ω connected to a variable frequency 240 V source as shown in figure calculate:
(i) the angular frequency of the source which drives the circuit at resonance,
(ii)the current at the resonating frequency,
(iii)the rms potential drop across the inductor at resonance.

Answer:

Question.22. A rectangular loop of wire of size 2 cm x 5 cm carries a steady current of 1 A. A straight long wire carrying 4 A current is kept near the loop as shown in the figure. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Question.27. Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode ofpropagationusedintheftequencyrangeabove40MHz.
Answer : Three different modes of propagation of electromagnetic waves are :

1.  Ground (surface) wave propagation.
2.  Sky waves propagation
3. Space wave propagation (LOS communication)
   For frequencies about 40 MHz, space wave propagation is being used as the ionosphere will not reflect the signals and ground transmission is not possible. In space wave the transmitter and receiver must be on the line of sight (LOS) together.
   Such wave propagation used for Television broadcast and Satellite Communication
   In this, propagation the uplink and downlink frequencies are kept different to avoid the mixing up of the Signals.
   

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II 2012.

Question.6. Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface?
Answer : We know that the electric field inside a conductor is zero, so E = 0. This is the reason why electrostatic potential is

Question.8. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing?

Answer : Using Lenz’s law we can predict the direction of induced current in the ring. Induce current oppose the cause of increase of magnetic flux in moving towards the conductor.
In metal Ring 1 induced current will be in is clockwise In metal Ring 2 induced current will be in is anticlockwise

Question.9. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its own properties.
Answer : A magnetic material having relative permeability of 800 would be classified as a ferromagnet. A few examples of such materials include iron and nickel.
Its two properties are :

1.  All ferromagnetic materials become paramagnetic when heated to a temperature above the Curie temperatue (Tc).
2.  These materials show a strong attraction towards magnetic fields and have a’tendency to become magnets themselves.

Question.16. Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and. in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O.

Now, as the current flowing in loop P is clockwise by using right hand thumbs rule the direction of the magnetic field will be towards left and as the current in loop Q is clockwise then the direction of magnetic field is towards left. So the net magnetic field at point O will be the sum of the magnetic fields due to loops P and Q.
Also, as the fields produced are at an equal distance to O, B_p = B_Q, So, net field

Question.23.A rectangular loop of wire of size 2.5 cm x 4 cm carries a steady current of 1 A. A straight wire carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.

CBSE previous Year Solved  Papers  Class 12 Physical Education Outside Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  The question paper consists of 26 questions.
2. All question are compulsory.
3. Answer to questions carrying 1 mark should be in approximately 10-20 words.
4. Answer to questions carrying 3 marks should be in approximately 30-50 words ,
5.  Answer to questions carrying 5 marks should be in approximately 75-100 words.

Question.1. Playgrounds are essential for creating sports environment. Justify your answer. **
Answer. Playgrounds are essential to create the right environment for the development of physical activities. Physical activities and sports promote good health and well being. Playgrounds are considered as labs for nurturing the talents of young individuals. (Any other relevant answer may also be considered.)

Question.2. What do you mean by “surfing” in adventure sports?
Answer . Surfing involves riding ocean waves with a surfboard. The surfer swims a ways out into the ocean lying on their stomach on the surfboard, to catch a ocean wave.

Question.3. Enlist two non-nutritive components of diet.
Answer . Non-nutritive components of diet are :

1.  Colour compounds
2.  Flavour compounds

Question.4. What does the school intend by stating that, “only such
students shall participate in the Basketball Intramurals who have not represented the school in Basketball in the past and minimum 10 substitutions shall be compulsory.”
Answer . For promoting mass participation. To explore the hidden talent of the student.

Question.5. What is “an abnormal curvature of spine at front” termed as?
Answer . Lordosis is an abnormal curvature of spine at front.

Question.6. What type of resistances can be used for developing strength among children?
Answer .

1.  Own body weight
2.  Gravitational force Static and dynamic resistance own body weight, gravitational force can be used for developing strength among children.

Question.7.Which test would you suggest for your grandmother to test lower body flexibility?
Answer: Chair sit and reach test.

Question.8.Why does involvement in regular exercise delay the onset of fatigue?
Answer . If we do regular exercise, our fitness level will be increased and it also develops endurance because of the fatigue level delays.

Question.9.What is energy?
Answer . Energy is defined as an ability or capacity of a body to perform work. Energy is denoted by the letter “E” and the “SI” unit of energy is joule (J).

Question.10. Explain intrinsic motivation.
Answer . Intrinsic motivation is internal. It occurs when people are compelled to do something out of interest, pleasure, importance and desire.

Question.11. “Pace races means, running the whole distance of a race at a constant speed”. Which are the races included in pace races?
Answer . 800 mts and above or 800 and 1500 mts

Question.12. Mention any three objectives of adventure sports.
Answer .

1.  Builds concentration : By participating in adventure sports on a regular basis, an individual gets many fascinating experiences. This activity usually helps to develop alertness and focus on attention-grabbing tips. This increases the capability to concentrate.
2.  Develops bonding with nature : The majority of the adventure sports are open-air activities giving the individuals various opportunities to increase familiarity with nature.
   Thus, they develop bonding with nature, which gives a chance to get in touch with natural environment.
3.  Builds-up courage : These sports give plentiful
   opportunities to prepare the participants for building up qualities for facing challenges in all tough situations courageously and with determination.

Question.13. Recently Sarita Devi refused to accept the bronze medal during the ceremony. The international body (AIBA) which regulates boxing has taken a stringent action against Sarita Devi and the coaches.

1.  Do you agree with the decision of Sarita Devi? Justify your answer.
2. What values do you think Sarita Devi has not shown by her behavior during the medal distribution ceremony?

Answer .

1.  I do not agree with the decision of Sarita Devi because she did not show sportsmanship/ she did not obey the AIBA rules.
   OR
   I do agree with the decision of Sarita Devi because she raised the voice against the decision of the umpire/ injustice.
2.  Sportsmanship, conventionality.

Question.14. What do you mean by “round shoulders”? Suggest any four physical activities for correcting round shoulders.
Answer . Round shoulder is a postural deformity in which the shoulders become round and sometimes they seem to be bent forward.
Physical activities for round shoulder :

1.  Sit on a chair, rest the back against it pull the shoulders backward and see upwards.
2. Hold the horizontal bar for some time
3.  Perform dhanurasana (arch formation) regularly.
4.  Perform Chakraasana (arch formation) regularly.

Question.15. Critically explain the use of dietary supplements in heavy dose for longer duration. Justify your answer with two suitable examples.
Answer .

1.  Excess calcium in diet for longer time can cause heart diseases/formation of stones in kidney.
2.  Excess iron causes siderosis (vomiting and headache)
3.  Vitamin E can cause prostates cancer

Question.16. Explain in brief “The Harvard Step Test”.
Answer . The Harvard step test is a cardiovascular fitness test. It is also called aerobic fitness test. It is used to measure the cardiovascular fitness or aerobic fitness by checking the recovery rate.
Equipment required : A gym bench or box of 20 inches high for man and 16 inches for woman, stopwatch and cadence tape.
Procedure : The athlete stand in the front of the bench or box. On the command “GO” the athlete steps up and down on the bench or box at a rate of 30 steps per minute. Stopwatch is also started’at the start of the stepping. Calculation of the scores : Calculate with the help of following formula.
“fitness index score = (100 x test duration in seconds)/ (2 x sum of heartbeat in recovery period)”.

Question.17. “Regular physical activity can delay your ageing process”
Justify your answer in light of the effect of activities on physiological changes.
Answer .

1.  Change in the nervous system : After age 25, everyone loses nerve cells. Gradually over time, this /esults in a reduced ability of nerve transmission, changing response time and coordination. The brain also shrinks in size, which does not significantly effect functioning except in the most extreme cases. These changes may also affect sleeping patterns somewhat by decreasing the length of total sleep time and REM sleep.
2.  Maintains bone density : With advancement of age, bone density decreases as well generally leading to the rupture or osteoporosis. Physical exercises assist in maintaining bone mass and prevent osteoporosis. Bone growth gets stimulated with resistance exercises. With the help of regular exercise, the aged persons can increase their bone density as depicted by several research studies.
3.  Changes in the endocrine system : The endocrine or metabolic system is responsible for changing food into energy. After age 25, everyone experiences approximately a 1% decrease per year in their metabolic rate. This overall slowing results in food being less well absorbed and utilized as well as a decrease in the overall metabolism of drugs. Consequences can include reduced stamina and reserves as well as greater susceptibility to drug toxicity.

Question.18. How does angle of projection help as a factor for athletes in games and sports?
Answer .

1.  When the height of the release is equal to the height of landing the optimum angle of release is 45 degree.
2.  When the height of release if greater than the height of landing as in a hammer throw, the optimum angle of release is less than 45 degree.
3.  When the height of release is less than the height of landing as in a bunker shot in golf, the optimum angle of release is more than 45 degree.
   

Question.19. Dynamic strength is divided into’ three parts. Write in brief about each.
Answer .

1.  Maximum Strength : In a single muscular contraction it is the ability of muscle to contract over resistance of utmost intensity of stimulus. The most excellent examples are weight lifting and throwing events in track and field.
2.  Explosive strength s It can be stated as the ability to prevail over resistance by means of high speed. It combines strength and speed abilities and based on the nature of the blend of strength and speed, the explosive strength can be sub¬divided further into start strength, power and speed strength.
3.  Strength Endurance : Similar to explosive strength, it is a result of two motor abilities as well. Under conditions of fatigue, it is the ability to work against resistance. Depending on the actuality whether the movement is static or dynamic, strength endurance can be there in form of static or dynamic strength. The strength endurance can be divided further into proper strength endurance and strength endurance depending on the kind of the blend of strength and endurance.
   20. What are the five essential elements of positive sports
   environment ?

Question.21. Draw a knock out fixture of 21 teams mentioning all the steps involved.
Answer .

Question.22. What are the various factors affecting physiological fitness? Explain.
Answer . The various physiological factors which determine the strength of an individual are stated below :

1. Size of the muscle : The strength of the muscle largely depends upon the size of the muscle. It is a well known fact that bigger and larger muscles can produce more force. The force produced by the same size of muscle in males and females is approximately the same but males are found to be stronger because they have larger muscles and bigger muscles in comparison to females. With the help of different methods of strength training such as weight training the size of the muscle can be increased and as a result of that strength is improved. So, the strength is determined by the size of the muscle.
2.  Body weight : It is also a well known fact that the individuals who are heavier are stronger than the individuals who are lighter. There is a positive correlation between body weight and strength among international weightlifters. The heavier weightlifters lift the heavier weight. So, body weight also determines the strength of an individual.
3.  Muscle composition : Each muscle consists of basically two types of muscle fibres i. e., fast twitch fibres (white fibres) and slow twitch fibres (red fibres). The fast twitch fibres are capable to contract faster and therefore, they can produce more force. On the contrary, the slow twitch fibres are not capable to contract faster but they are capable to contract for a longer duration. The muscles which have more percentage of fast twitch fibres, ca.n produce more strength. The percentage of fast twitch fibres and slow twitch fibre is genetically determined and can not be changed through training. So it can be said that the percentage of these fibres determines the strength.
   
   
   4. Intensity of the nerve impulse : A muscle is composed of a number of motor units. The total force of the muscle depends on the number of contracting motor units. Whenever, a stronger nerve impulse from central nervous system excite more number of motor units, the muscle will contract more strongly or it can said that the muscle will produce more force or strength. So, the intensity of the nerve impulse also determines the amount of strength.

Question.23. Explain the cognitive aspect of stress. Suggest any three techniques briefly to overcome stress
Answer .
(a) Inability to concentrate
(b) poor judgment .
(c) seeing only negative
(d) anxious or racing
thoughts and constant worrying (Explanation of Any Two) Techniques :

1.  Eat a healthy diet : Well-nourished bodies are better prepared to cope with stress, so, be mindful of what you eat. Start your day right with breakfast, and keep your energy up and your mind clear with balanced, nutritious meals throughout the day.
2.  Get enough sleep : Adequate sleep fuels your mind, as well as your body. Feeling tired will increase your stress because it may cause you to think irrationally.
3.  Do something you enjoy everyday : Make time for leisure activities that bring you joy, whether it be stargazing, playing the piano, or working on your bike.
4.  Keep your sense of humor : This includes the ability to laugh at you. The act of laughing helps your body fight stress in a number of ways.

Question.24. Differentiate between 1 : 1 and 1 : 2 ratio interval training, with suitable examples.
Answer : 1 : 1 means load and the rest is equal, eg. I minute exercise followed by 1 minute of rest. Similarly 1 : 2 means that the period of rest is double of the load. eg. 1 minute exercise followed by 2 minute of rest.
The slow and extensive interval training methods can be given by using 1: 1 and 1:2 Fast or intensive interval training methods can be given by using 1 : 1 and 1 : 2.
Student is required to explain the relationship between load and rest, with suitable examples from their respective games and sports.

Question.25. Vitamins are very essential for working of the body and are divided into two groups. Explain about them.
Answer :

1.  Fat soluble vitamins : The fat-soluble vitamins are those vitamins that are soluble in fat. These include vitamin A, D, E and K stored in the liver and in body fat.
   (a) Vitamin ‘A’ ’. Vitamin A (retinol) is a fat-soluble vitamin that acts as an antioxidant. The Vitamin A that we obtain from animal products is called retinoid and can be used by our body without any modification. The form of Vitamin A found in fruits and vegetables is known as carotenoids.
   (b) Vitamin ‘D’ : Vitamin D is different from other essential vitamins because our own “body can manufacture it with sunlight exposure. The main function of Vitamin D is to regulate the absorption of calcium and phosphorus in our bones and aid in cell-to-cell communication throughout the body.
   (c) Vitamin ‘E’ : Vitamin E (tocopherol) is a powerful, fat- soluble antioxidant that helps to protect cell membranes against the damage caused by free radicals, prevents the oxidation of LDL cholesterol and is also important in the formation of red blood cells (RBC), thus essential for blood coagulation.
   (d) Vitamin K : Vitamin K occurs in two forms- Vitamin K1 and Vitamin K2. Vitamin K is, known as the clotting vitamin, because without it, blood would not clot. It also helps in, B prevention of haemorrhage.
2.  Water soluble vitamins : The water-soluble vitamins are those vitamins that are soluble in water. Vitamin C and members of the Vitamin B complex are water-soluble vitamins.
   (a) Vitamin ‘B₁’ (Thiamin) : Vitamin Bj or Thiamin, helps to release energy from foods, promotes normal appetite, and is important in maintaining proper nervous system function.
   (b) Vitamin ‘B₂’ (Riboflavin) : Riboflavin or Vitamin B2, helps to release energy from foods, promotes good vision, and healthy skin. It also helps to convert the amino acid tryptophan (which makes up protein) into niacin.
   (c) Vitamin ‘B₃’ (Niacin) : Vitamin B3, or niacin, works with other b-complex vitamins to metabolize food and provide energy for the body. Vitamin B3 is involved in energy production, normal enzyme function, digestion, promoting normal appetite, healthy skin and nerves.
   (d) Vitamin ‘B₆’(Pyridoxine) : Vitamin Bg is a keyfactor in protein and glucose metabolism as well as in the formation of hemoglobin. Hemoglobin is a component of red blood cells-it carries oxygen. Vitamin Bg is also involved in keeping the lymphnodes, thymus and healthy.
   (e) Vitamin ‘B₁₂’(Cobalamin) : Vitamin B₁₂ is also known as Cobalamin, aids in the building of genetic material, production of normal red blood cells, and maintenance of the nervous system.
   (f) Biotin : Biotin is a coenzyme and a B vitamin and is also known as Vitamin H. As a supplement, biotin is sometimes used for diabetes, brittle nails, and other conditions. Biotin helps to release energy from carbohydrates and aids in the metabolism of fats, proteins and carbohydrates from food.
   (g) Vitamin ‘C’ : Vitamin C is also called Ascorbic acid,
   , which is a water-soluble vitamin and cannot be stored in the body. Most plants and animals can produce their own vitamin C but humans cannot. Vitamin C is needed for proper growth, development, and to heal wounds. It is used to make the collagen tissue for healthy teeth, gums, blood vessels and bones. .

Question.26. Weight training is one of the oldest methods for development of strength. What are its advantages and disadvantages?
Answer : Weight training refers to any activity which, involves the use of weights. Weight training is an exercise in which muscles of the body are forced to contract under tension using weights, body weight or other devices in order to stimulate growth, strength power and endurance.
Advantages of Weight Training :

1.  Improved muscle tone and strength : Weight training improves muscle strength and tone to protect joints from injury. It also helps children to maintain flexibility and balance weight management and increased muscle to fat ratio. As a child gains muscle, the body burns more kilojoules when at rest.
2. Greater Stamina : Stamina helps to work longer without tiring. People with good stamina are less likely to suffer muscle injuries, backaches and soreness. Building stamina also improves overall fitness, general helath and appearance. As a child grows stronger, he won’t get tired easily.
3. Pain management : It increases strength and flexibility, reduces joint pain, and helps combat fatigue, Exercise is vital for people with arthritis. Even moderate exercise can ease your pain and help you maintain healthy weight.

Disadvantages of Weight Training :

1.  Threat of injury : The risk of getting injured is always high while performing weight training particularly when doing weight training with no companion. When repetitions of exercises are performed, one should not be alone at that critical movement as chance of getting injured is more. Thus, to perform weight training with a companion or a supporter is always advisable or to perform under supervision of an expert to avoid the worst to happen.
2. Less flexibility : The level of flexibility gets reduced if proper exercises of flexibility are not done alongwith weight training. It will be negligible if flexibility exercises are performed persistently.

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2011

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. Name the embryonic stage that gets implanted in the uterine wall of a human female.
Answer : The blastocyst gets implanted in the uterine wall of a human female.

Question.2. State the importance of biofortification.
Answer: To increase the amount of vitamin, minerals, protein, fat, Micronutrieflt and mineral content of the plants.

Question.3. Biotechnologists refer to Agrobacterium tumifaciens as a natural genetic engineer of plants. Give reasons to support the statement.
Answer : Agrobacterium tumifaciens is a pathogen of several dicot plants. It is able to deliver a piece of its DNA, called Ti (Tumor inducing) plasmid into normal plant cells where it gets incorporated in the host DNA & replicates along with it.

Question.4. How do algal blooms affect the life in water bodies ?
Answer: Algal blooms diminish water quality and deteriorate the water quality by depleting the oxygen content resulting asphyxiation and increase BOD. It also causes fish mortality and extremely toxic to human beings and animals.

Question.5. Name the common ancestor of the great apes and man.
Answer: Ramapithecus and Dryopithecus.

Question.6. Write a difference between net primary productivity and gross productivity.
Answer : Gross Primary (GPP) productivity : It is the total amount of energy captured or he total organic matter or biomass manufactured by the producet by the process of photosynthesis per unit time per unit area.
Net Primary productivity (NPP) : It is the amount of energy or biomass stored by the produces per unit areaper unit time. It is calculated by substracting the amount of energy utilised in respiration from the GPP.

Question.7. Mention the contribution of genetic maps in human genome project.
Answer: Genetic and physical maps act as instruments for the completion of human genome project. Genetic maps were used to study polymorphism among the genes, use of RE (Restriction Endonuclease) to determine specific repetitive DNA sequence commonly present.

Question.8. Name the phase all organisms have to pass through before they can reproduce sexually.
Answer: (i) Juvenile phase (animals) or vegetative phase (plants)
(ii) Reproductive phase .

SECTION-B

Question.9. Name the emyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer : Streptokinase enzyme produced by Streptococcus bacterium. It is extremely useful in medical practice due to its ability to break down blood clots. .

Question.10. How is ‘Rosie’ considered different from a normal cow ?Explain.
Answer : The ‘Rosie’ is first, transgenic cow, Rosie, produced human protein-enriched milk. Milk contained the human gene alpha-lactalbumin and it was nutritionally a more balanced product for human babies than natural cow-milk.

Question.11. State the use of Biodiversity in modem agriculture.
Answer : Use of biodiversity in development of agriculture:-
(i) Modern agricultural technologies increases the productivity per unit area of land, and helps in the conservation and promotes farming of all wild and native varieties of plants.
(ii) Base for our agricultural food chain, development and safeguard of livestock’s etc.
(iii) The use of plant protection products is an important tool to control some of these invasive Which species and protect biodiversity, including that found on the farm.

Question.12. Write the full form of VNTR. How is VNTR different from “Probe”?
Answer : VNTR stands for Variable Number of Tandem Repeats. Probe is a small fragment of DNA or RNA used for identification of genes in biological system. VNTR is a small fragment of DNA containing tandemely repeated sequence, whose number and length vary among chromosome and’ individuals. A probe can be used for VNTR to find out the relationship between the people, to find out the criminals as to confirm parents of a child.

Question.13. Differentiate between benign and malignant tumours.
Answer : Difference between benign and malignant tumours :

Question.14. The above graph shows Species-Area relationship. Write the equation of the curve ‘a’ and explain.

Answer: curve ‘a’represent the equation S=C\({ A }^{ Z }\) curve ‘b’ represent the equation log S = log C.+ Z log A, where S= Species richness, A= Area, Z = slope of the line (regression coefficient) C = Y-intercept. The characteristic feature of curve is-
(i) Within a region richness of species increases with exploration of new areas in limit.
(ii) Straight line in the graph represent logarithmic value of species richness.
OR
Differentiate between in situ and ex situ approaches of conservation of biodiversity.
Answer: Difference between in situ and ex situ approaches of conservation of biodiversity

Question.15. The cell division involved.in gamete formation is not of the same type in different organisms. Justify.
Answer : Cell division results in the formation of gamete, heterogametic species produce male and female gametes. In Monera, Fungi, Algae and Bryophytes gametes are haploid (n), the gametes are produced by mitotic division. While organism such as pteridophytes, gymnosperms, angiosperms and animals including human beings, the parental body is diploid (2n), they undergo reduction division, to produce haploid gametes.

Question.16. Identify the type of the given ecological pyramid and give one . example each of pyramid of number and pyramid of biomass in such cases.

Answer : The given ecological pyramid is the inverted pyramid.
Inverted pyramid of biomass: Lake: Phytoplankton —> Zooplankton —> fishes.
Inverted pyramid of number: Tree —> insects —> birds.

Question.17. Describe the Lactational Amenorrhea method of birth control.
Answer : Lactational Amenorrhea method (LAM) is considered a natural method of birth control based on the fact that higher lactation around the clock decreases menstruation and ovulation does not occur. Female will not get pregnant during the first six months after parturition.

Question.18. Name the type of bioreactor shown. Write the purpose for which it is used.

Answer : The given bioreactor is the simple stirred tank bioreactor.
Its purpose is a large scale production of recombinant protein or enzymes, using microbial plants/animals/human cells. It is usually cylindrical or with a curved base to facilitate mixings of reactor content. The stirrer facilitates mixing and use available oxygen. It contains agitator system, an oxygen delivery system, a foam control system, a temperature control system, pH control system and sampling ports so that small volume of the culture can be withdrawn periodically.

SECTION – C

Question.19. Draw a labelled diagram of the reproductive system in a human female.
Answer:

Question.20. Branching descent and natural selection are the two key concepts of Darwinian Theory of Evolution. Explain each concept with the help of a suitable example.
Answer : Branching descent : Branching descent is the process of development of a new species from, single common descendant. New developed species became, geographically adapted to a new environment ultimately results in complete development of new species, e.g., Darwins Finches-varieties of Finches arose from grain eaters; Australian marsupials evolved from common marsupial.
Natural selection : Natural selection is a process in which better adapted organisms lead to better adaptation and survival while less adapted organisms gets eliminated in successive stages. Selected organisms reproduce and produce stable genetic quality to sustain the changes, e.g., white moth surviving before the industrial revolution and black moth surviving after the industrial revolution.

Question.21. Scientists have succeeded in recovering healthy sugarcane plants from a diseased one.
(a) Name the part of the plant used as explant by the scientists.
(b) Describe the procedure the scientists followed to recover the healthy plants.
(c) Name this technology used for crop improvement.
Answer: (a) Meristematic cells, apical and axillary
(b) Meristem tissue (sugarcane) is grown to nutrient medium leads in invitro.The tissue proliferates to form undifferentiated mass/callus. This callus formed is transferred to a medium containing growth hormones like auxins and cytokinins.
(c) Tissue culture or micropropagation.

Question.22. (i) Name the enzyme that catalyses the transcription of hnRNA.
(ii) Why does the hnRNA need to undergo changes ? List the changes hnRNA undergoes and where in the cell such changes take place.
Answer: (i) RNA polymerase II.
(ii) hnRNA has non-functional introns in between the functional exons. To remove these, it undergoes changes. The changes that hnRNA undergoes includes:
(a) Capping; Methyl guanosine triphosphate is added to 5′ end.
(b) Tailing: In which poly A tail is added at 3′ end.
(c) Splicing: Through which introns are removed and exons are joined.

Question.23. (i) Write the scientific names of the two species of filarial worms causing filariasis.
(ii) How do they affect the body of infected person(s) ?
(iii) How does the disease spread ?
Answer: (i) Filariasis is caused by organism called Wuchereria, two principal species belong to this category is Wuchereria bancrofti and Wuchereria malayi.
(ii) Filarial worm remain in the body for a long time and develops chronic inflammation. They inhabit lymphatic vessels of lower limbs resulting in the swelling of lower limbs and the disease is called elephantiasis or filariasis. Genital organ also gets affected resulting in deformities’ in its shape and size.
(iii) Transmission of infection generally occurs through bite of female mosquito vectors.

Question.24. Name the genus to which baculoviruses belong. Describe, their role in the integrated pest management programmes.
Answer: Nucleopolyhedrovirus.
Role : Baculoviruses is regarded as natural pest management microorganism. They control only species-specific pest, do not affect non-target organisms or beneficial insects are conserved, they thus aid in IPM problems and there is no negative impact on plants or other animals. These viruses can attack wide range of arthropod and some other insects.

Question.25. Unambiguous, universal and degenerate are some of the terms used for the genetic code. Explain the salient features of each one of them.
Answer : This small combination of amino acids sequence was named as genetic code or codon. Sir Har Gobind Khorana developed chemical method for synthesis of RNA molecule. Some of the important characteristic feature of genetic codes is-
(1) Unambiguous — One codon codes for one amino acid,e.g. AUG(Methionine).
(2) Universal – Codon and its corresponding amino acid are the same in all organisms, eg. Bacteria to human UUU ‘ codes for phenylalanine.
(3) Degenerate – Coding of some amino acids are done by more than one sets of codon therefore they are termed as degenerate, e.g. – UUU and UUC code or phenylalanine.

Question.26. Water is very essential for life. Write any three features both for plants and animals which enable them to survive in water scarce environment.
Answer : Features which enable them to survive in water scarce environments are :
Animal adaptation
(i) Kangaroo rats of North America maintain water requirement by internal fat oxidation.
(ii) Some animals have the ability to concentrate urine so that minimum amount of water is lost during excretion.
(iii) Some have burrowing nature to minimize water loss. Plant adaptation
(i) Desert plants develop thick cuticle and deeply placed stomata to decrease rate of transpiration.
(ii) Use of CAM photosynthetic pathway helps stomata to remain inactive or closed during day time.
(iii) Some desert plants develop spikes to replace leaf so that rate of photosynthesis is done by flat stems.
OR
How do organisms cope with stressful external environmental conditions which are localised or of short duration?
Answer : The following methods are employed by organisms to cope with stressful environmental conditions :
(i) Migrate temporarily from the stressful habitat to a hospitable area.
(ii) Suspend activities
(iii) Form thick walled spores
(iv) Form dormant seed
(v) Hibernate during winter
(vi) Planktons undergo diapause

Question.27. (i) State the consequence if the electrostatic precipitator of a thermal plant fails to function.
(ii) Mention any four methods by which the vehicular air pollution can be controlled.
Answer : (i) Particulate matter is considered very harmful, major step for the removal of particulate matter is the implementation of electrostatic precipitator. They are placed near the exhaust of thermal power plant. Precipitator has an electrode wire with thousands of volts, releasing electrons which get attached to the dust particles (negatively charged). On the other side collecting plates attract charged dust particle, where scrubber cleans gases like sulphur oxide.
(ii) Automobile are major source of air pollution, PCB have suggested that particulate size 2.5 micrometers or less in diameter (PM 2.5) are responsible for causing the greatest harm to human health. Four important methods that can be implemented to reduce vehicular pollution :
(a) Use of CNG, (b) Phasing out of old vehicles,
(c) Use of unleaded petrol, (d) Use of low sulphur fuel

SECTION-D

Question.28. Give reasons why:
(i) Most zygotes in angiosperms divide only after certain amount of endosperm is formed.
(ii) Groundnut seeds are exalbuminous and castor seeds are albuminous.
(iii) Micropyle remains as a small pore in the seed coat of a seed.
(iv) Integuments of an ovule harden and the water content is highly reduced, as the seed matures.
(v) Apple and cashew are not called true fruits.
Answer:
(i) To obtain nutrition from the endosperm for the developing embryo.
(ii) The Groundnut seeds rates are albuminous because the endosperm is completely consumed. Whereas, castor seeds rates albuminous because the endosperm persists.
(iii) For the entry of water and oxygen required for germination.
(iv) To protect the embryo and keep the seed viable, until favourable conditions return for germination.
(v) In apple and cashew, ovary does not take part in fruit formation, instead thalamus contributes to fruit formation.
OR
(a) Draw a labelled diagram of L.S. of an embryo of grass (any six labels).
(b) Give reason for each of the following:
(i) Anthers of angiosperm flowers are described as dithecous.
(ii) Hybrid seeds have to be produced year after year.
Answer:
(a)

(b) (i) Each anther of angiosperm is dithecous i.e bilobed in nature with two layered protection called theca (dithecous). (ii) Use of hybrid vegetable and crop is growing exponentially in present era. They have significantly increased productivity of the plants with higher nutritive value and because progeny shows segregation and do not maintain hybrid characters.

Question.29. Describe the mechanism of pattern of inheritance of ABO blood groups in humans.
Answer: In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles IA, IB and i. Hence, referred to as multipleallelism. A person possesses any two of the three alleles. Z4 and IB dominate over i. But with each other, Z and Z4 are co-dominant.
Table: The Genetic Basis of Blood Groups in Human Population

OR
(a) Why is haemophilia generally observed in human males ? Explain the conditions under which a human female can be haemophilia
(b) A pregnant human female was advised to undergo M.T. E It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote farmed by an XX- egg fertilized by Y-carrying sperm. Why was she advised to undergo M.T P.?
Answer : (a) Haemophilia is sex linked recessive disease. It is transmitted from unaffected female carrier to a male child with haemophilia. Y has no allele for this. If male inherits Xh from the mother, he will be haemophilic (with the genotype XhY). If female inherits XhXh, one from the carrier mother and one from her haemophilic father, then she can be haemophilic.
(b) In normal human being fusion of two gamete sperm and egg results in the development of zygote, with 23 pairs of chromosomes, or 46 chromosomes. Gamete develops after meiosis containing one set of chromosome so called haploid (22 autosome) and one sex chromosome. Fusion of sperm (male) and egg (female) gamete forms diploid zygote with 22 pairs of autosome and a pair of sex chromosome. Male (sperm) possess heterogametic sex chromosome X and Y, while female have only hohaogafnetic chromosome XX. Above given problem deals with trisomic condition, nondisjunction abnormality such as Klinefelter syndrome where males have an extra X chromosome. Genotype of which is XXY or sometime it may be XXYY, XXXY. She was advised to undergo MTP since the child will have the following problems :
(i) Male with feminine traits
(ii) Gynaecomastia
(iii) Underdeveloped testes
(iv) Sterile

Question.30. (i) Describe the characterisucs a cloning vector must possess,
(ii) Why DNA cannot pass through the cell membrane ? Explain. How is a bacterial cell made “competent” to take up recombinant DNA from the medium ?
Answer: (i) Characteristics features of cloning vector must have:
(a) Presence of selectable marker genes encoding for an antibiotic resistant or genes encoding for a-galactosidase.
(b) Cloning site or recognition site for convenient insertion and removal of plasmid DNA, by the use of RE.
(c) Easy propagation and maintenance of the clone.
(d) Ori or origin of replication
(ii) DNA is a hydrophilic molecule, therefore it cannot pass through the cell membrane. Bacterial cells are made competent by:
(a) Treating bacteria with specific concentration of divalent cation results in increase efficiency of DNA to move inside bacterium cell wall.
(b) Followed by the incubation of cell with recombinant DNA on ice, then place them briefly at 42°C (heat shock), and again back on ice.
OR
If a desired gene is identified in an organism for some experiments, explain the process of the following :
(i) Cutting this desired gene at specific location .
(ii) Synthesis of multiple copies of this desired gene Answer : (i) Cutting of the desired gene at specific location is attained by the implementation of restriction enzymes (RE). Firstly, the restriction endonucleases that recognise the palindromic nucleotide sequence of the desired gene is identified. These enzymes are specialized to cut the fragment of DNA at specific locations. It cuts each of the double helix . at a specific point which is a little away from the centre of the palindromic site. The cutting site is between the same two bases on the opposite strands. This results in over¬hanging single stranded stretches which act as sticky ends, (ii) Multiple copies of the desired gene is synthesised by polymerase chain reaction (PCR) method. In this method, the desired gene is synthesised in vitro. The double stranded DNA is denatured using high temperature of 95°C arid the strands are separated. Each separated strand acts as template. Two sets of chemically synthesised oligonucleotides (primers) and DNA polymerase are being used in vitro for the multiplication of desired gene. The thermostable Taq polymerase extends the primers using nucleotides provided in the reaction mixture.

SET-II

SECTION-A

Question.3. Why is it essential to have a ‘selectable marker’ in a cloning vector ?
Answer : Selectable markers are essential to identify and eliminate non-transformants, and selectively permiting the growth of the transformants.

SECTION – B

Question.9. Why are some molecules called bioactive molecules? Give two examples of such molecules.
Answer : Bioactive molecules are those molecules which are biologically active. Because microbes like bacteria or fungi are used in their production.
e.g., Citric acid produced by a fungus Aspergillus niger.
Acetic Acid produced by a bacteria Acetobacter aceti.

Question.13. List the two types of immunity a human baby is bom with. Explain the differences between the Jyvo types.
Answer : The two types of immunity are innate and passive/ acquired immunity. Innate immunity is a non-specific type of defense that provides a barriers to the entry of antigens. Passive immunity is a pathogen-specific type of defense that develops in response to encounter with pathogen. The foetus receives antibodies through the placenta.

Question.16. Explain the response of all communities to environment over time.
Answer : Environmental factors like temperature, water, light, soil, etc., may influence the members of communities in varying degrees. Organisms in response to these factors shall try to adapt according to their capacities. This change is orderly,’sequential and parallel with the changes in the physical environment. In this process, they may also try to maintain a constant internal environment through homeostasis or migrate to a less stressful environment or may even suspend activities till favourable conditions return. These changes lead to a community is in real equilibrium with the environment called the climax community.
(i) The cow is administered with FSH to induce follicular maturation and super-ovulation to produce 6 to 8 eggs.
(ii) The animal is either mated with an elite bull or artificially inseminated.
(iii) The fertilised eggs 8-32 cells stage are recovered non- surgically and transferred to surrogate mother where they develop into an improved variety.
(iv) This technology is being used to get high milk yielding females and to increase herd size in short time in cattle.

Question.23. (a) Name the causative agent of typhoid in humans.
(b) Name the test administered to confirm the disease.
(c) How does the pathogen gain entry into the human body?
Write the diagnostic symptoms and mention the body organ that gets affected in severe cases.
Answer : (a) The causative agent of typhoid in humans is Salmonella typhi.
(b) The test administerd to confirm the disease is Widal test.
(c) The pathogen gains entry via small intestine through contaminated food and water and migrate to other organ through blood.
The symptoms include sustained high fever (30°C to 40°C), weakness, stomach pain, constipation, headache, loss of appetite’.

Question.25. (a) Name the scientist who called t-RNA an adaptor molecule.
(b) Draw a clover leaf structure of t-RNA showing the following:
(i) tyrosine attached to its amino acid site.
(ii) anticodon for this amino acid in its correct site (codon for tyrosine is UCA).
(c) What does‘the actual structure of t-RNA look like ?
Answer : (a) Francis Crick (b)

(c)the actual structure of tRNA looks like inverted L.

SECTION-C

Question.21. Describe the technology that has successfully increased the herd size of cattle in a short time to meet the increasing demands of growing human population.
Answer: Multiple ovulation embryo transfer technology (MOET) has successfully increased the herd size of cattle.

SECTION-D

Question.30. (a) With the help of diagrams show the different steps in the formation of recombinant DNA by action of restriction endonuclease enzyme EcoRI.
Name the technique that is used for separating the fragments of DNA cut by restriction endonucleases.
Answer: (a)

Steps in formation of recombinant DNA by action of restriction endonuclease enzyme EcoBl
(b) Gel electrophoresis is used for separating the fragments of DNA cut by restriction endonucleases.
OR
(a) Name the source from which insulin was extracted earlier. Why is this insulin no more in use by diabetic people ?
(b) Explain the process of synthesis of insulin by Eli Lilly Company. Name the technique used by the company.
(c) How is the insulin produced by human body different from the insulin produced by the above mentioned company?
Answer: (a) Earlier, insulin was extracted from pancreas of slaughtered catde and pig. However, this insulin use caused some patients to develop an allergic reaction to this foreign protein.
(b) Eli Lilly used the following producer for insulin synthesis :
(i) Two DNA sequences corresponding to A and B chains of insulin were prepared.
(ii) These sequences were then introduced in plasmids of E. coil.
(iii) The two insulin chains are produced separately.
(iv) The two chains are extracted and combined with creating disulphide bonds to form the assembled mature molecule of insulin.
(c) The pro-hormone (like a pro-enzyme) synthesiszed in the human body has an extra stretch of C peptide.

SET -III

SECTION-A

Question.1. What stimulates pituitary to release the hormone responsible for parturition ? Name the hormone.
Answer: The signal from the fully developed foetus and placenta or the foetal ejection reflex induces mild uterine contraction for parturition. The responsible hormone is oxytocin.

Question.2. Pollinating species of wasps show mutuajism with specific fig plants. Mention the benefits the female wasps derive from the fig trees from such an interaction.
Answer :The female wasp uses the fruit as oviposition and developing seeds for nourishing its larvae.

SECTION-B

Question.9. A relevant portion of P – chain of haemoglobin of a normal human is given below:

The codon for the sixth amino acid is GAG. The sixth codon GAG mutates to GAA as a result of mutation A’ and into GUG as a result of mutation ‘B’. Haemoglobin structure did not change as a result of mutation A’ whereas haemoglobin structure changed because of mutation ‘B’ leading to sickle shaped RBCs. Explain giving reasons how could mutation ‘B’ change the haemoglobin structure and not mutation A’.
Answer: Due to mutation A, GAG mutates to GAA. Both GAG and GAA code for glutamic acid and hence there is no change in RBCs. Whereas GUG formed due to mutation ‘B’ codes for valine and so the RBCs become sickle-shaped.

Question.10. Biopiracy should be prevented. State why and how.
Answer: Biopiracy is unauthorized exploitation of bioresources of developing or under-developed countries. There has been growing realization of the injustice, inadequate compensation and benefit sharing between developed and developing countries. Hence, it should be prevented. It can be prevented by developing laws to obtain proper authorization and by paying compensatory benefits.

Question.13. Why is tobacco smoking associated with rise in blood pressure f and emphysema (oxygen deficiency in the body) ? Explain.
Answer : Tobacco has nicotine, an alkaloid, that stimulates the release of adrenaline and noradrenaline which raise blood pressure and increase the heart rate. Smoking tobacco releases carbon monoxide which reduces the concentration of haem-bound oxygen. This causes emphysema.

Question.18. What is polyblend ? Why did the plastic manufacturers think of producing it ? Write its usefulness.
Answer : Polyblend is a fine powder of recycled modified plastic. Polyblend was produced to recycle plastic waste. It was developed by Ahmed Khan, a plastic manufacturer to solve the ever increasing problem and accumulation of waste. Polyblend can be used to lay roads that have increased road life. When blended with bitumen, it enhances the bitumens water repellent properties and increase the life of road.

SECTION-C

Question.23.

Study the diagram showing replication of HIV in humans and answer the following questions accordingly:
(i) Write the chemical nature of the coat ‘A’.
(ii) Name the enzyme ‘B’ acting on ‘X’ to produce molecule ‘C’. Name ‘C’.
(iii) Mention the name of the host cell ‘D’. the HIV attacks first when it enters into the human body.
(Iv) Name the two different cells the new viruses ‘E’ subsequendy attack.
Answer: (i) Coat A’ is made up of protein.
(ii) The enzyme ‘B’ is reverse transcriptase, ‘C’ is viral DNA.
(iii) The host cell ‘D’ is macrophage.
(iv) The new viruses ‘E’ subsequendy attack macrophages and helper T-lymphocytes.

Question.25. Answer the following questions based on Messlson and Stahl’s experiment:
(a) Write the name of the chemical substance used as a source of nitrogen in the experiment by them.
(b) Why did the scientists synthesise the light and the heavy DNA molecules in the organism used in the experiment ?
(c) How did the scientists make it possible to distinguish the heavy DNA molecule from the light DNA molecule ? Explain.
(d) Write the conclusion the scientists arrived at after completing the experiment.
Answer: (a) Ammonium chloride (NH4CI).
(b) To check if DNA replication was semi-conservative.
(c) The heavy and light DNA molecules were distinguished by centrifugation in a Cesium chloride density gradient.
(d) The scientists concluded that DNA replicates semi- conservatively.

CBSE previous Year Solved  Papers  Class 12 Physical Education Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  The question paper consists of 26 questions.
2. All question are compulsory.
3. Answer to questions carrying 1 mark should be in approximately 10-20 words.
4. Answer to questions carrying 3 marks should be in approximately 30-50 words ,
5.  Answer to questions carrying 5 marks should be in approximately 75-100 words.

PART-A

Question.1.Define active flexibility
Answer . Active flexibility focuses on improving soft tissue extensibility and increasing neuromuscular control. It includes foam rolling (self- myofascial release) as well as active – isolated stretching.

Question.2.What is a bye?
Answer . A bye refers to a dummy team that does not play in the first round but participates in the II nd round. The number of byes in a fixture is the actual difference between the teams participating in the tournament and the next highest number which is the power of 2.

Question.3. What do ’ you understand by Social environment in sports?
Answer . Environment is the surroundings or conditions in which a person, animal, or plant lives or operates. It is the natural world, as a whole or in a particular geographical area, especially as affected by human activity

Question.4. What is a correct posture?
Answer . Posture is the position of the body or body parts, the relative placement of body parts to each other, or an individual’s general carriage or bearing. Posture is the way people hold their bodies upright against the downward pull of gravity. Proper alignment of the spine is necessary for good posture. The spine is not straight; it has three curves, bending inward at the neck, curving slightly out at the upper back and in again at the lower back.

Question.5.What is pratyahar in yoga? **
Answer. Pratyahara means literally “control of ahara,” or “gaining mastery over external influences.” It is compared to a turtle withdrawing its limbs into its shell — the turtle’s shell is the mind and the senses are the limbs. The term is usually translated as “withdrawal from the senses,” but it has many meanings.

Question.6.Define a balanced diet.
Answer . Eating a balanced diet means choosing a wide variety of foods and drinks from all the food groups. It also means eating certain things in moderation, namely saturated fat, trans fat, cholesterol, refined sugar, salt and alcohol. The goal is to take in nutrients you need for health at the recommended levels.

Question.7. Explain Fartlek.
Answer .  Fartlek, which means “speed play” in Swedish, is a training method that blends continuous
training with interval training.The variable intensity and continuous nature of the exercise places stress on both the aerobic and anaerobic systems. It differs from traditional interval training in that it is unstructured; intensity and/or speed varies, as the athlete wishes.

Question.8. Define sports psychology.
Answer : Sports psychology is the study of how psychology influences sports, athletic performance, exercise, and physical activity. Some sports psychologists work with professional athletes and coaches to improve performance and increase motivation.

Question.9. Differentiate between fitness and wellness. **
Answer.  Fitness refers to the physical part. A fit person is supposed to have a good stamina, strength and flexibility. He or she maintains the fitness by doing various kinds of exercises (indoor or outdoor, freehand or with weights, aerobic or anaerobic etc).
Wellness is a more general term encompassing the whole being which is the mind and soul in addition to the body. Your body can be in a great shape but you could be in a mental mess. You could even be fine mentally but may not be spiritually oriented.
These represent various levels of well being. Mostly, people start with taking care of their bodies, then start working on the mental peace and later graduate towards spiritual progress.
In a small number of people ( like Saints and Monks), the focus on mental and spiritual wellness is very high whereas the effort for physical fitness may or may not be there. Buddha emphasised the importance of a strong and fit body for the purpose of progressing on the spiritual path and that’s why he prohibited self-mortification, prescribing the middle path.

Question.10. Mention any two essential elements of positive sports environment.
Answer. Two essential elements of positive sports environment:

1. Role of individual in improving sports environment: The individual can be a player, the captain of the team or sports captain of the school. They can motivate other students to play and participate in the games and sports by telling them the benefits of sports. An individual can” play a very effective role in the improvement of sports environment for prevention of sports related accidents. For this purpose a well planned and coordinated programme should be designed to minimise the accidents in the field of sports.
2. Prevention of Sports related accidents : In this the Physical Education Teachers and Trainers/ Coaches play very important role. They shall see that there is no obstacle in the ground. The ground is neat, clean and levelled. The training should be done in organized manner. Extra attention should be given to injury oriented sports like Cricket, Hockey, Football, Archery, Javelin Throw, Discus Throw, Shot-put etc. The Physical Education Teacher should be there to provide help at the time of injury or accident that may happen during the game.

Question.11. Write a brief about two elements of yoga. **
Answer. 
The Earth Element (Prithvi) in the body refers to our connection with the earth, the place where the raw elements for our bodies came from. We may bring more “earth” awareness into our yoga practice when we focus on grounding, building a firm base of support, connecting with the earth beneath us. Many times in our lives we become too “lofty” in our thoughts and need to come down to Earth to feel grounded and secure. In balancing postures, for instance, our stability is facilitated through deepening our connection with the Earth. Sometimes the Earth also refers to the Planet that we call home…exploring our connection with our Planet and the beauty and wonders that it holds… which we are all responsible to preserve. The Earth element is represented in the 1st/Root Chakra of the body.

The Water Element (Apah, Jala) in the body refers to the water that flows through our veins, or the circulatory system in the body. Health is only possible as the water element remains flowing in the body. While in our yoga postures (asanas) we may need to consciously increase the flow of blood into a particular area of the body. Sometimes the practice also takes on a flowing action, with one movement fluidly moving into the next. Water has almost magical properties and is essential to life (our bodies are 70% water), with incredible healing and cleansing powers. It is a universal symbol for the soul and is the element representing the 2nd/Sacral Chakra, the area where new life is generated.

Question.12. What are the two methods of flexibility development?
Answer .

1. Warm-up before stretching: The very first thing you must do before stretching is a warm up jog, run, or bike to get loose. You don’t have to run or bike that hard Or far, but it is good to do at least 20 minutes of warm up before stretching.
2.  Do dynamic stretching : Dynamic stretching includes motion and is meant to mimic and exaggerate the movements of actual exercise and daily motions.

Question.13. Differentiate between State and Trait Anxiety.
Answer. State anxiety is characterized by a state of heightened emotions that develop in response to a fear or danger of a particular situation. State anxiety can contribute to a degree of physical and mental paralysis, preventing performance- of a task or where performance is severely affected, such as
forgetting movements during a dance or gymnastic routine; to breaking in sprint or swim starts or missing relatively easy shots at goal i.e. pressure situations.
Trait anxiety refers to,a general level of Stress that is characteristic of an individual, that is, a trait related to personality. Trait anxiety varies according to how individuals have conditioned themselves to respond to and manage the stress.

Question.14. Explain the role of games and sports as means of fitness development. **
Answer.
Healthy mind can be found only in a healthy body. In a weak body there cannot be a healthy and active mind. And for a healthy body physical exercise is a must. Without physical exercise, our body will grow weak, lethargic and dull. The aim of education is the all round development of a personality. It cannot afford to neglect the physical aspect of a student. Development of mind and body are equally important in any good education.

Want of proper physical exercise in the form of games and sports develops many mental problems. Mere intellectual attainment is not enough. Good health and sound body are also a must to face the challenges of life. Therefore, games and sports are an integral part of school education. Education will remain incomplete without physical training and exercise. “All work and no play makes Jack a dull boy” is a famous saying. After studies some kid of physical exercise, games and sports are necessary. The refresh body and mind and provide recreation. A game of football or wallyball in the open air is very refreshing. A game of hockey or a match of badminton will help a student regain his lost mental and physical energy. There will be greater intake of oxygen, better blood circulation and digestion because of these. In the open, where games are played there is fresh air, openness and presence of nature. They have a very healthy influence on the players. Running, Jumping, kicking, swimming etc. provide vigorous exercise to our limbs and organs of the body. They provide us physical fitness, courage endurance, cooperation and team spirit. The players are more disciplined and fit than others. Sports and games along with education prepare us to stand up and face the challenges of life.

Games and sports are a valuable form of education. They develop our skills and abilities to the maximum. They teach discipline, obedience and cooperation. Every game has it own rules and regulations. They are binding on the players. All players have to follow them. There is penalty on their violation. One can never win a match without following the laws of the game. He has to abide by the judgement of the refree. It teaches a player how important are laws. It makes clear how important it is to follow the rules and regulations games and sports help us in producing very disciplined citizens, leaders and professionals. They teach how to cooperate with one another and achieve success. It is a playground or gymnasium where team spirit, cooperation and endurance can be taught best, games teach players how to ignore individual interests for the sake of greater interests of the team and society. This teaching of sacrifice is of great social and national interest. Games also teach fair play and faith in equality and justice. They enable us to take defeat and victory in a cheerful spirit.

Games also allow an outlet to our suppressed energy. It helps us a lot in remaining peaceful and non-violent. When our energy is suppressed we become irritative, short-tempered, violent hooliganism and acts of lawlessness. Games also provide us the best use of leisure time.
Famous players and sports persons bring credit for themselves and the country. They are famous and popular and work as country’s cultural ambassadors. They strengthen international relations. They also develop and promote patriotism and national integration. But games are a means and not an end in themselves. They should not be practised at the cost of studies. Excess of everything is bad. They should be played and enjoyed only in spare time. They are real boon and blessing if done properly and wisely.

India needs good and great players and sports persons. They are in great demand to participate in national and international events. India’s record in this respect has been very poor. The boys and girls should be caught at the very young age and trained in different games and sports. There should be no school without a proper playground attached to it. Mere mental education is of no use without physical education. They should go hand-in-hand as integral parts of an education. They are complementary to each other.

Question.15. What is the role of a spectators in creating a positive sports environment? Explain.
Answer : Positive sports environment is the condition and circumstances which are favourable and beneficial for the sportspersons who perform sports activities. Behaviour and attitude of the spectators towards coaches, players and officials should be positive. They should not pass any negative comment towards players, referee, umpire and any other game officials. They should not indulge them in any type of violence. They should try to motivate the players so that they may put up better performance. In this way the spectators can play a vital role for creating positive sports environment.

Question.16. What is the role of yoga in sports? Explain.
Answer. Yoga focuses on harmony between mind and body. To achieve this, yoga uses movements, breath, posture, relaxation and mediation in order to establish a healthy, lively and balanced approach to life. Yoga plays an important role in games and sports also. Yoga improves near about all physical fitness and wellness components required by sportsman.
Paschimottanasans: Paschimottanasans literally translated as “intense stretch of the west”. A yoga position where one sits on the floor with legs flat on the floor, straight ahead. Lift spine long, hinge from the hips instead of the waist. Learn forward without bending your knees. Focus on bringing chest forward, not on bringing the head to the floor.
A similar frontbend is uttanasana which is a standing front bend. Some consider Paschimottanasans to be a safer stretch since gravity is less of a factor than active flexibility in achieving flexibility in the furthest reaches of the stretch.  It is more passive in its initial stages, making it a good transition between the two forms. The arms can also more easily support the upper body in this vulnerable position, and can be used both to move further into or move out of the stretch.
Unlike Uttanasana it is also much easier to move the legs, rotating them inwards or outwards, abducting or adducting them at the hip, flexing or extending the knees, or enacting plantar or dorsi flexion of the ankle. These variations can be performed either as a combined stretch, to change emphasis on different tissues, or simply to takes one’s mind off the hamstrings and lower back being stretched. They can be rhythmically to aid in relaxation.
Dhanurasana : Dhanurasana or Bow pose is a yoga posein which the practitioner lies on their bolly, grabs their feet, and lifts the legs into the shape of a bow. Dhanurasana is also called Urdva Chakrasana. Improoerly executed, this pose could compromise the knees and the spine. Because the spine is placed out of alignment and pressure is applied, this pose can lead to ruptured disks. Any backward extension of the spine should be done exclusively using the back muscles and should never be forced. In addition the spine should always be extended fully from hip bones to the head before moving into any pose. It is always appropriate in any yoga class to skip a dangerous pose in favour of another. In this case, locust pose is appropriate.
Halasana : Halasana or the plow pose, is yoga pose or asana, in which the practitioner lies on the floor, lifts their legs, and then places them behind the head. Practitioners should be advised that this pose can put significant strain on the cervical spine. This pose can cause an injury if not performed properly. It is recommended practitioners conult with a qualified yoga instructor before attemting this or other advanced pose.

Question.17. Explain any two methods for speed development.
Answer: Speed means the velocity with which an individual can execute his movements in other words, it means the capacity of moving a body part or the whole body with the greatest possible velocity. For example, the movements of a smashers arm in volley ball, has the maximum speed or velocity at the time of smashing the volly ball. The following methods are usually adopted for the development of speed in sprinting events.

1.  Acceleration runs : These are usually adopted to develop speed, specially in attaining maximum speed from stationary position. It should be kept in mind that the technique of any event should be learnt in the beginning. Only then, We should switch over to acceleration runs.
2.  Pace Races : It means, running the whole distance of a race at a constant speed. In pace races, an athlete runs the uniform speed. Generally 800 meters and above races are inculded in pace races.

Question.18. Prepare a fixture for 21 teams on a knock out basis.
Answer:

Fixture of 21 teams on knockout basis

Question.19. Suggest five exercises as corrective measures for Round Shoulders.
Answer : These stretches, yoga poses and exercises are very important to work into your regular training program for improved posture and to combat rounded shoulders.

1.  Back bound hand pose
2. Shoulder squeeze
3. Cow face pose
4. Baby cobra pose
5.  Bridge pose

Question.20. Explain any five essential elements of diet.
Answer : Some sources state that sixteen chemical elements are required to support human biochemical processes by serving structural and functional roles. However, as many as 26 elements in total (including the common hydrogen, carbon, nitrogen and oxygen) are suggested to be used by mammals, as a result of studies of biochemical, special uptake, and metabolic handling studies. However, many of these additional elements have no well-defined biochemical function known at present. Most of the known and suggested dietary elements are of relatively low atomic weight, and are reasonably common on land, or at least, common in the ocean (iodine, sodium).
5 most essential elements are :

1.  Calcium
2. Iron
3. Iodine
4. Vitamins
5.  Carbohydrates

Question.21. Explain in detail the developmental characteristics of childhood. **
Answer. Child Development Ages & Stages. … Ages and Stages is a term used to broadly outline key periods in the human development time line. During each stage growth and development occur in the primary developmental domains including physical, intellectual, language and social – emotional.

Question.22. Explain any 5 essential elements of diet. **
Answer. The human body needs a long list of nutrients every day. The essentials, though, are called “macronutrients,” and your body needs them to stay healthy and perform optimally. They include the following five:

1.Carbohydrates
Main function: Provide energy
“Carbohydrates are the body’s main energy source and the brain’s only source of fuel,” says Kate Patton, MEd, RD, a registered dietitian at the Cleveland Clinic in Ohio. Your body breaks carbohydrates down into glucose, which cells require to create energy.
Get more: The best sources of carbohydrates are whole grains and foods made from those grains, such as whole-wheat bread, bulgur, barley, oatmeal, brown rice, and cornmeal. Limit your intake of sugar and refined grains (including white pasta, white rice, and white breads), the U.S. Department of Agriculture (USDA) recommends.

2. Protein
Main function: Build and repair tissue
“Protein is another important source of energy for the body,” Solomon says. Protein consists of amino acids that act as the body’s main building blocks for tissues, such as muscle, skin, bone, and hair. Proteins also assist in many reactions in the body, including the production of enzymes (the catalysts that keep all body processes running smoothly), hormones, and antibodies, Solomon explains.
Get more: The best protein sources are lean meats, poultry and seafood, beans and peas, nuts and seeds, eggs, and soy products, according to the USDA.

3. Fats
Main function: Provide backup energy
“Your body uses fats for energy when carbohydrates aren’t available,” Patton says. “You also need fats as insulation, to help your body absorb fat-soluble vitamins, and to protect your organs.”
Get more: Fats come in both liquid and solid forms. The USDA notes that the best sources of healthful fats are the liquid monounsaturated and polyunsaturated fats found in olive oil, canola oil, sunflower oil, soybean oil, corn oil, nuts, seeds, and avocados, as well as fatty fish rich in omega-3 fatty acids. Limit foods high in unhealthy saturated fats (red meat, cheese, butter, and ice cream) and trans fats (processed products that contain partially hydrogenated oil), which increase your risk for disease.

4. Vitamins and Minerals
Main function: Maintain optimal health
“You need vitamins and minerals for numerous physiological functions that help you survive,” Patton says. They’re essential for normal growth and development, and each one plays a unique role in helping to maintain optimal health. For example, calcium and vitamin D are necessary for healthy bones, and the B vitamins help support the nervous system, explains Tricia L. Psota, PhD, RDN, president-elect of the DC Metro Area Dietetic Association.
Get more: Vitamins and minerals come from a variety of foods, including fruits and vegetables, dairy products, and lean protein sources. “Eat a selection of colorful fruits and vegetables every day, and vary the types of proteins you eat,” Solomon says.

5. Water
Main function: Enables vital bodily functions
You’ve probably heard that you can live for weeks without food but only days without water. That’s because water is the most important essential nutrient. It is involved in many of your body’s vital functions, and it distributes other essential nutrients to your cells.
Get more: The Institute of Medicine recommends that men consume about 125 ounces of water a day and women 91 ounces per day. About 20 percent can come from foods, and the remaining 80 percent should come from drinking water — about 12 cups a day for men and 8.8 cups for women.

PART – B

Question.23. Write a brief about 4 fundamental skills from any game/ sport of your choice. **

Question.24. Write a short note on SGFI. **

Question.25. Draw a diagram of field/table/court from any game/sport of your choice.**

Question.26. Mention any 4 latest rules from any game/sport of your choice.**

Question.27. Briefly explain the historical development from any game/ sport of your choice.**

 

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2012

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. Why is banana considered a good example of parthenocarpy ? 
Answer: Banana is considered a good example of parthenocarpy because formation of fruit in banana occurs without fertilization (parthenocarpy), i.e., there is no formation of seeds.

Question.2. State two different roles of spleen in the human body.
Answer : The roles of spleen in the human body is that the spleen is the secondary lymphoid organ that produce lymphocytes and the red pulp of spleen removes the old or damaged red blood cells from the body.

Question.3. A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer : The dominant traits are : Axial, violet flower.

Question.4. Why is it desirable to use unleaded petrol in vehicles fitted with catalytic converters ?
Answer: It is desirable to use unleaded petrol in vehicles fitted with catalytic converters because lead in petrol inactivates the catalysts which convert harmful pollutants (CO, unburnt hydrocarbons, nitric oxide) to lesser harmful pollutants (C02,H2o,N2).

Question.5. Where is acrosome present in humans ? Write its function.
Answer: The anterior portion of the sperm head in human beings is covered by a cap-like structure called acrosome. Function of Acrosome :
(i) Acrosome is filled with hydrolytic enzyme-Hyaluronidase that aids in the entry of sperm into the ovum.

Question.6. Write the name of the following :
(a) The most common species of bees suitable for apiculture
(b) An improved breed of chicken
Answer : (a) Apis indica is the most common species of bees for apiculture.
(b) Leghorn is an improved breed of chicken.

Question.7. Comment on the similarity between the wing of a cockroach
and the wing of a bird. What do you infer from the above, with reference to evolution ?
Answer : The wing of a cockroach and the wing of a bird are not similar anatomically i.e., not similar in structure but similar in function. Thus we infer that these organs are analogous which has resulted in convergent evolution.

Question.8. Mention the role of cyanobacteria as a biofertiliser.
Answer: The role of cyanobacteria as a biofertiliser : Cyanobacteria (Anabaena and- Nostoc) are free-living in the root nodules of leguminous plants and they fix atmospheric nitrogen. They act as a bio-fertilizers especially in paddy fields.

SECTION-B

Question.9. (a) Draw a neat labelled diagram of a nudeosome.
(b) Mention what enables histones to acquire a positive charge.
Answer: (a)

(b) Depend on the abundance of basic amino acid residues of lysines and arginines with charged side chain.

Question.10. State one advantage and one disadvantage of cleistogamy.
Answer: Advantage of cleistogamy: Self-pollination is assured, seed production is also assured in the absence of pollinators., Disadvantage of cleistogamy: In cleistogamous flowers due to self-pollination least variations observed.

Question.11. (a) Where do the signals for parturition originate from in humans ?
(b) Why is it important to feed the newborn babies on colostrum ?
Answer : (a) The signals for parturition in humans originate from the fully developed foetus and the placenta which include mild uterine contraction.
(b) The colostrum or first milk is important to feed the new born babies because it contains antibodies (IgA), to provide passive immunity to the baby.

Question.12. (a) A recombinant vector with a gene of interest inserted
within the gene of a-galactosidase enzyme, is introduced into a bacterium. Explain the method that would help in selection of recombinant colonies from non-recombinant ones.
(b) Why is this method of selection referred to as “insertional inactivation” ?
Answer: (a) Bacteria is grown in a medium with chromOgenic substrate, blue coloured colonies with no recombinations and colonies with no blue colour show presence of recombinants.
Chromogenic substrate used to identify recombinants and non-recombinants
(b) Gene for the enzyme is inactivated by insertion which is referred to as insertional inactivation.

Question.13. Explain brood parasitism with the help of an example.
Answer: Brood parasitism involves the use of host individuals of the same or different species to raise the young of the brood parasite. This relieves the parasitic parent from the investment of rassing young or building nests, producing offspring etc.
eg : The cowbird family is the bird family in North America is an brood parasite (because are not capable of building a nest). While most brood parasites have eggs that mimic the hosts eggs the cowbird is again different.

Question.14. Give reasons for the following :
(a) The human testes are located outside the abdominal cavity.
(b) Some organisms like honey-bees are called parthenogenetic animals.
Answer : (a) To maintain the temperature (2-2.5°C) lower than the normal internal body temperature,, which is essential for spermatogenesis.
(b) The phenomenon of development of female gamete directly into an individual without fertilization is called parthenogenesis. Example : The drones/males develop from unfertilised eggs of honey bees.

Question.15. Name the plant source of ganja. How does it affect the body of the abuser ?
Answer : Plant source : Cannabis Sativalhemp plant.
Affect: It damages cardio-vascular system of the body
OR
Name the two special types of lymphocytes in humans. How do they differ in their roles in immune response ?
Answer: B lymphocytes, T lymphocytes.
B-cells produce pathogen specific antibodies called humoral immune response.
T-cells help the B-cells to produce antibodies and are responsible for direct cell mediated immunity.

Question.16. (a) Mention the cause and the body system affected by ADA deficiency in humans.
(b) Name the vector used for transferring ADA-DNA into the recipient cells in humans. Name the recipient cells.
Answer : (a) The body system affected by ADA deficiency in humans is immune system. ADA deficiency is caused due to lack the gene coding for adenosine deaminase.
(b) A retroviral vector is used to transfer ADA-DNA into recipient cells. The recipient cells are lymphocytes.

Question.17. How did Ahmed Khan, plastic sacks manufacturer from Bangalore, solve the ever – increasing problem of accumulating plastic waste ?
Answer : Ahmed Khan, a plastic sacks manufacturer solve the ever increasing problem and accumulation of waste. Polyblend fine powder of recycled modified plastic can be used to lay roads that will increased road life. When blended with bitumen, it enhances the bitumen’s water repellent properties and increase the life of road.

Question.18. Name the bacterium that causes typhoid. Mention two diagnostic symptoms. How is this disease transmitted to others ?
Answer : Bacterium: Salmonella typhi.
Diagnostic Symptom: Constipation, stomach pain, headache, weakness, loss of appetite, high fever.
The disease is transmitted through contaminated food and water.

SECTION-C

Question.19. (a) Explain the phenomena of multiple allelism and codominance taking ABO blood group as an example.
(b) What is the phenotype of the following:
(i) \({ I }^{ A }\)i (ii) ii
Answer: Multiple allelism : (a) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles, \({ I }^{ A }\), \({ I }^{ B }\) and i, hence, referred to as multiple allelism. Co-dominance : If \({ I }^{ A }\) and \({ I }^{ B }\) both are present in an individual, and they both are expressed because of the phenomenon of co-dominance.
(b) (i) Phenotype of \({ I }^{ A }\) / : A blood group.
(ii) Phenotype of i i: O blood group,

Question.20. How does industrial melanism support Darwins theory of Natural Selection ? Explain.
Answer: In England, before industrial revolution the environment was unpolluted. The white-winged moths were more and lichens on the barks of trees were pale. The white-winged moths could easily camouflage, while the darkwinged were spotted out by the birds for food. Hence, they could not survive. After industrial revolution the lichens became dark (due to soot deposit). This favoured the dark-winged moths while the white-winged were picked by birds. The population of the former which was naturally selected increased.

Question.21. (a) What is the programme called that is involved in
improving success rate of production of desired hybrid and herd size of catde ?
(b) Explain the method used for carrying this programme for cows.
Answer: (a) Multiple ovulation embryo transfer Technology/ MOET is used for improving success rate of production of desired’hybrid and herd size of cattle.
(b) Methods used for carrying this programme for cows :

1.  The cow is administered with FSH to induce follicular maturation and super – ovulation to produce 6 to 8 eggs.
2.  The animal is either mated with an elite bull or artificially inseminated.
3.  The fertilised eggs 8-32 cells stage are recovered non- surgically and transferred to surrogate mother where they develop into an improved variety.

Question.22. Explain the function of each of the following :
(a) Coleorhiza (b) Umbilical cord (c) Germ pores
Answer: The function of each of the following:
(a) Goleorhiza : Protects the radical of (monocot) embryo.
(b) Umbilical cord : Transports nutrients and respiratory gases and metabolic wastes to and from mother and foetus!
(c) Germ pores : Allow germination of pollen grain and formation of pollen tubes.

Question.23. How is the amplification of a gene sample of interest carried out using Polymerase Chain Reaction (PCR) ?
Answer: Polymerase chain reaction (PCR) is a method in which the desired gene is synthesised in vitro in following steps:

1.  Denaturation : The double-stranded DNA is denatured by applying high temperature of 95°C for 15 seconds. Each separated single stranded strand now acts as template for DNA synthesis.
2.  Annealing: Two sets of primers are added which anneal to the 3′ end of each separated strand. Primers act as initiators of replication.
3.  Extension : DNA polymefase extends the primers by adding nucleotides complementary to the template provided in the reaction. A thermostable DNA polymerase (Taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.
   

Question.24. Trace the life-cycle of malarial parasite in the human body when bitten by an infected female Anopheles.
Answer : Plasmodium requires two hosts to complete its life cycle.When female Anopheles mosquito bites a healthy person. Sporozoite of Plasmodium gets into human blood through the bite of female Anopheles mosquito. The parasite multiply in liver cells and finally burst in liver cells and released in blood, then they get into red blood cells, where they further multiply asexually and burst in RBCs also and released toxic substance haemozoin (associated with fever and chills). After a while they change into gametocytes, which are picked up by the mosquitoes and the entire cycle occurs again.

Question.25. List the salient features of double helix structure of DNA.
Answer: The salient features of Double Helix structure of DNA:

1.  It is made of two polynucleotido chains where backbone is sugar phosphats constituted and bases project inside.
2.  There is complementary base pairing between the two strands of DNA. The amount of adenine is equal to thymine and the amount of guanine is equal to cytosine.
3.  The two strands are coiled in right-handed fashion and are anti-parallel in orientation. One chain has a 5’—>3′ polarity while the other has 3’—>5′ polarity.
4.  The diameter of the strand is always constant due to pairing of purine and pyrimidine, i. e., adenine is complementary
   to thymine while guanine is complementary to cytosine.
5.  The distance between the base pairs in a helix is 0.34 nm and a complete turn contains approximately ten base pairs. The pitch of the helix is 3.4 nm and the two strands are right-handed coiled.

OR
How are the structural genes activated in the lac operon in E. coli ?
Answer: The structural genes activated in the lac operonin E. coil in the following manner :
Lactose consists of the genes lac z, y and a. Lactose acts as the inducer that binds with repressor protein and frees the operator gene. RNA polymerase freely moves over the structural genes, transcribing lac mRNA, which in turn – produces the enzymes responsible for the digestion of lactose.

Question.26. Alien species are highly invasive and are a threat to indigenous species. Substantiate this statement with any three examples.
Answer : The three examples of the above statement are :

1.  Nile perch introduced into Lake Victoria in East Africa led to the extinction of Cichlid fish.
2. Invasive plants like Parthenium /Lantana /Eichhomia caused environmental damage and posed a threat to indigenous species.
3.  Introduction of African catfish (Clarias gariepinus) to aquaculture is a threat to indigenous Indian catfishes.

Question.27. (a) Tobacco plants are damaged severely when infested with
Meloidegyne incognitia. Name and explain the strategy that is adopted to stop this infestation.
(b) Name the vector used for introducing the nematode specific gene in tobacco plant.
Answer: (a) The infestation jyas prevented by strategy which was based on RNA interference or RNAi or gene silencing. During this process nematode specific gene is introduced into host plant (using Agrobacterium) which produce dsRNA. This specific mRNA of the nematode silenced and parasite dies.
(b) Agrobacterium tumifaciens vector are used for introducing the nematode specific gene in tobacco plant.

SECTION – D

Question.28. (a) Taking one example each of habitat loss and fragmentation, explain how are the two responsible for biodiversity loss.
(b) Explain two different ways of biodiversity conservation.
Answer : (a) Habitat loss and fragmentation are responsible for biodiversity loss are :

1. Habitat loss : The Amazon rainforest (called the “lungs of the planet”) is being cut and cleared for cultivation of soya beans and for conversion into grasslands for raising beef catde.
2. Fragmentation : When large-sized habitats are broken or fragmented due to human setdements, building of roads, digging of canals, etc., the population of animals requiring large territories and some animals with migratory habitats declines.

(b) The two different ways of biodiversity conservation are :

1.  Ex situ
2.  In situ conservation.

Ex situ conservation : In this conservation threatened organism are taken out from the natural habitat and placed in special setting with care and protected, eg., zoological park, botanical garden, wild safari.
In situ conservation: In this conservation threatened organisms are conserved in their natural habitat, e.g, national , parks, biosphere reserves.
OR
(a) What depletes ozone in the stratosphere? How does this affect human life?
(b) Explain biomagnification of DDT in an aquatic food chain. How does it affect the bird population?
Answer: (a) Chlorofluorocarbons (CFCs) released from the refrigerators air conditioners deplete ozene in the stratosphere. Ozone acts as a shield and protects the earth from, the harmful UV rays of the sun.
Effect on Human life : Chlorofluorocarbons depletes ozone layer causing UV rays to reach to earth which damages DNA causing mutation, skin cancer, inflammation of cornea, cataract, aging of skin, snow blindness.
(b) If DDT leaches from the agricultural field, it gets into the water body (the concentration is 0.0003 ppm) and enters the food chain:
zooplanktons (0.04 ppm) —> small fish (0.05 ppm) —» large fish (2 ppm) —» any fish eating bird (5 ppm). Concentration of DDT increases along the food chain, reaching a high level in the top carnivore bird.
Effect on Bird population : DDT concentration disturbs Ca++ metabolism, egg shells become thin, premature breaking resulting in decline in bird population.

Question.29. The following is the illustration of the sequence of ovarian events “a” to “i” in a human female:

(a) Identify the figure that illustrates corpus luteum and name the pituitary hormone that influences its formation.
(b) Specify the endocrine function of corpus luteum. How
does it influence the uterus ? Why is it essential ?
(c) What is the difference between “d” and “e” ?
(d) Draw a neat labelled sketch of Graafian follicle.
Answer: (a) Corpus luteum is illustrated by ‘g’ and the hormone influencing its formation is luteininsing hormone (LH).
(b) Corpus luteum produces the hormone progesterone, causes proliferation of the endometrium which gets highly vascularised. It is essential for the implantation of the fertilized ovum and maintains the same during pregnancy.
(c) “d” is the developing tertiary follicle, “e” is the Graafian follicle.

OR
(a) Why is fertilisation in an angiosperm referred to as double fertilisation ? Mention the ploidy of the cells involved.
(b) Draw a neat labelled sketch ofL.S. of an endospermous monocot seed.
Answer: (a) Fertilisation of haploid egg cell by one haploid male gamete to form diploid zygote is called syngamy. Fertilisation of two (diploid) polar nuclei by the other haploid male gamete to form triploid primary endosperm nucleus is called triple fusion.
(b) L.S. of an endospermic monocot seed.

Question.30. Describe Frederick Griffiths experiment on Streptococcus pneumoniae. Discuss the conclusion he arrived at.
Answer : Frederick Griffiths experiment Transforming Principle conducted by him in 1928.

1. Frederick Griffith (1928) conducted experiments with Streptococcuspneumonia (bacterium causing pneumonia).
2.  He observed two strains of this bacterium—one forming smooth shiny colonies (S-type) with capsule, while other forming rough colonies (R-type) without capsule.
3.  When live S-type cells were injected into mice, they died due to pneumonia.
4.  When live R-type cells were injected into mice, they survived and he arrived at this conclusion
5. When heat-killed S-type cells were injected into mice, they survived and there were no symptoms of pnuemonia.
6. When, heat-killed S-type cells were mixed with live R-type cells and injected into mice, they died due to unexpected symptoms of pneumonia.
7. He concluded that heat-killed S-type bacteria caused a transformation of the R-type bacteria into S-type bacteria but he was not able to understand the cause of this bacterial transformation.

OR
(a) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross up to F2 generation.
(b) State the laws of inheritance that can be derived from such a cross.
(c) How is the phenotypic ratio of F2 generation different in a dihybrid cross ?
Answer : (a) A monohybrid cross taking seed coat colour as a trait in Pisum sativum, for e.g if pea plant with yellow seed coat is crossed with pea plant having green seed coat then in the Fj generation all the plants are yellow seeds.

F2 Phenotypic ratio = 3:1
F2 Genotypic ratio =1:2:1
The law of inheritance can be derived from such a cross :
(b) (i) Law of Dominance: Factor occurs in pairs. In a contrasting pair of factors one member of the pair dominates (dominant) the other (recessive).
(ii) Law of Segregation : Factors or allele of pair segregate from each other such that gamete receives only one of the two factors. Paired condition is restored at the time of zygote’ formation.
(c) Phenotypic ratio of F2 generation in monohybrid cross is 3 :1 whereas in a dihybrid cross the phenotypic ratio is 9 : 3 : 3 : 1.

SET-II

SECTION – A

Question.1. How do the pollen grains of Vallisneria protect themselves ?
Answer : The pollen grains of Vallisneria have mucilaginous covering to prevent them from getting wet.

Question.2. Name the respective pattern of inheritance where Ft phenotype.
(a) does not resemble either of the two parents and is in between the two.
(b) resembles only one of the two parents.
Answer : (a) The respective pattern of inheritance where phenotype not resemble either of the two parents and is in
between the two known as Incomplete dominance.
(b) resembles only one of the two parents Law of Dominance.

Question.5. How is the entry of only one sperm and not many ensured into an ovum during fertilisation in humans ?
Answer : During fertilisation in humans: At the sperm head there is an enzyme to dissolve the follicles of ovum and facilitate entry of the sperm nucleus for fertilization and help the sperm enter into the cytoplasm of the ovum.

Question.7. State the significance of Coelacanth in evolution.
Answer : It is an ancestor of amphibians. The latest analysis shows that the genes of modern coelacanths’can-themselves be considered living fossils.

SECTION – B

Question.12. Name the source organism that possesses Taq polymerase.
What is so special about the function of this enzyme ?
Answer: Thermus aquaticus.
The enzyme can tolerate high temperature and is thermostable. It does not get denatured during PCR at high temperature.

Question.13. State one advantage and one disadvantage of cleistogamy.
Answer : Advantage : Cleistogamy flowers produce assured seed set even in the absence of pollinators to increase genetic variations. Disadvantage: The disadvantage of Cleistogamy is the offspring produced have limited genetic diversity.

Question.15. Name the source of cyclosporin-A. How does this bioactive molecule function in our body ?
Answer: The source of cyclosporin-A Trichoderma polysporum.
Bioactive molecule function in our body: It is used as an immuno-suppressant agent in organ transplant patient.
OR
(a) Name the group of viruses responsible for causing AIDS in humans. Why are these viruses so named ?
(b) List any two ways of transmission of HIY infection in humans, other than sexual contact.
Answer : (a) The group of viruses responsible for causing AIDS in humans iis Retrovirus. These are named so because they (have RNA g;enome) have reverse transcriptase enzyme which carries on die processes RNA —> DNA —» RNA.
(b) (i) Transfusion , of infected blood.
(ii) Sharing infect ed syringes and needles. .
(iii) Children bor n to HIV mother through placents.

Question.17. Name any two or ganisms that are responsible for ringworms in humans. Mention two diagnostic symptoms. Name the specific parts of the human body where these organisms thrive and explain why.
Answer : Micro sporum and Trichophyton are two organisms that are responsible for ringworms in humans.
Symptoms : Dry and scaly lesion on skin, nails, scalp, intense itching These thrive in body groin, between toes, thrive better in heat, moisture, perspiration.

SECTION – C

Question.19. Differentiate between perisperm and endosperm giving one example of each.
Answer:

Question.25. (a) List any three ways of measuring population density of a habitat.
(b) Mention the essential information that can. be obtained by studying the population density of an organism.
Answer : There are different ways of measuring population density of a habitat are as follows :

1.  Quadrat method: This method involves the use of square of particular dimensions to measure, no. of organisms.
2. Direct observation : This method is used for counting  of organism.
3. Indirect method : This method is used for the number fish caught per trap gives the measure of their total t density in a given water bod y.

(b) The population density of an organism provides us the status of habitat, whether competition for survival exists or not, whether population is increas ing or declining, natality, mortality, emigration, immigration.

SECTION – D

Question.28. (a) Explain the significance of ecological pyramids with the help of an example.
(b) Why are the pyramids referred to as1 upright’ or ‘inverted’ ?
Answer : (a) Ecological pyramid expresses the relationship between the organisms at differen t trophic levels with reference to their number, energy and biomass.

Upright Pyramid : In this pyramid producers are more in number and in biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. The Pyramid of energy is always upright as only 10% energy is transferred from one trophic level to the next.
Inverted pyramid: It shows less number/biomass of producers when compared to primary consumers. For e.g. large number of insects feeding on a big tree give inverted pyramid of number.
OR
(a) Explain giving reasons why the tourists visiting Rohtang Pass or Mansarovar are advised to resume normal ‘active life’ only after a few days of reaching there.
(b) It is impossible to find small animals in the polar regions. Give reasons.
Answer: (a) The tourists visiting Rohtang Pass or Mansarovar are advised to resume normal active life’ only after a few days of reaching there because initially the person suffers from altitude sickness, nausea, fatigue and heart palpitation because of low oxygen availability and low atmospheric pressure. Gradually the body increases RBC production, decreasing binding capacity of Hb and increases the breathing rate to get acclimatised. •
(b) Small animals are rarely found in polar regions because small birds have larger surface area relative to their volume, so they lose heat much faster spend more energy to generate body heat. They have to expand much energy to generate body heat through metabolism.

SET -III

SECTION-A

Question.1.Identify the figure given below and the part Labelled “A”.

Answer:The figure is of blastula/blastocyst.
A—Trophob’last.

Question.2. How do interferons protect us ?
Answer : Interferons protect uninfected cells from further viral infections, by creating cytokine barriers.

Question.3. Name the interaction between a whale and the bar nacles growing on its back?
Answer : Commensal ism is the interaction between a wdiale and the barnacles grow ing on its back.

Question.7.In a dihybrid cross, when would the proportion of parental . gene combinations be much higher than non-parental types, as experimentally shown by Morgan and his group ?
Answer: When the genes are present on a same chromosome or incomplete linkage is present then the proportion of parental gene combinations be much higher than non-parental types,* according to Morgan and his group.

SECTION—B

Question.12. Name the cells that act as HIV factory in humans when infected by HIV. Explain the events that occur in the infected cell.
Answer : Macrophages/Helper T-cells act as HIV factory in humans.
The events occur in the infected cell:

1.  HIV attached to CD-4 cell with GP-120 to CD-4 protein.
2.  Reverse transcription by reverse transcriptase, enzyme.
3.  Its DNA attached with host cell DNA.
4.  Multiplication of HIV.
5.  Lysis of infected cell.

Question.15. Name and explain the two types of immune responses in humans.
Answer: The two types of immunity are active immunity and passive immunity.
Active immunity: Immunity developed in the host body due to production of antibodies in response to antigens which of low intensity and produce memory cells.
Passive immunity : When ready-made antibodies are direcdy given to protect the body against foreign agents which is of very high intensity.

Question.16. How does the study of different parts of a flower help in identifying in wind as its pollinating agent ?
Answer: Pollination in which wind distributed the pollens is called anemophily.
Wind pollinated flowers have light and non sticky pollen grains, well exposed stamens to disperse pollen easily, large and often feathery stigma for easy trapping of pollen, single ovule in each ovary, numerous flowers packed into an inflorescence.

SECTION-C

Question.22. Explain how do the following act as contraceptives :
(a) CuT (b) “Saheli”
Answer: (a) CuT release Cu ions when inserted into the uterus which suppress sperm motility, lowers the fertilising capacity of sperms. These devices inserted by doctors as experts in the uterus through vagina
(b) Oral contraceptive for the female! contain a non-steroidal preperation. It is once a week pill with very few side effects and high contraceptive value inhibit ovulation, implantation, as well as alter the quality of cervical mucus to prevent or retard the entry of sperms.

Question.26. Name and explain the evolutionary concept represented in the illustration given below:

Answer : The illustration represents adaptive radiation or divergent evolution.
It is the example of adaptive radiation in placental animals of Australia.

1.  A variety of placental mammals have evolved which appear similar to a corresponding marsupial.
2.  When more than one adaptive radiation appeal; to have occurred in an isolated geographical area, and two or more
   groups of unrelated animals come to resemble each other for similar mode of life or habitat, it is called convergent evolution.

OR
(a) Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
(b) State the functions of the following in a prokaryote :
(i) tRNA (ii)rRNA
Answer: (a) The father never passes on the gene for haemophilia to his sons because haemophilia is a sex-linked recessive disease and the defective.Gene is present on X chromosome only and not on Y chromosome.
(b) Function of t-RNA in prokaryotes –
(i) tRNA reads the genetic codes, carries amino acids to the site of protein synthesis and act as an adap tor molecule.
(ii) rRNA plays structural and catalytic role during translation .

SECTION – D

Question.30. (a) A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing aixial, violet flowers, produced axial, violet flower and axtial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny, (b) Name and state the law that can be derived from this cross and not from a monohybrid cross.
Answer: (a)

(b) This cross is based on Law of Independent Assortment. This law states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of character.
OR
(a) Describe the process of synthesis of fully functional mRNA in a eukaryotic cell.
(b) How is this process of mRNA synthesis different from that in prokaryotes?
Answer : (a) The process of synthesis of fully functional mRNA in a eukaryoic cell.

1. The primery transcrips are non-functional, containing both the coding region, exon, and non-coding reigion, irron, in RNA and are called heterogenous RNA or hnRNA.
2.  The hnRNA undergoes two additional process called cappint and tailing.
3.  In cappint, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′-end of hnRNA.
4. In tailing, adenylate rersidues (about 200-300) are added at 3–end in a template independent manner.
5.  Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.
   

(b)In prokaryotes, there is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. In bac teria, mRNA does not require any processing as it does not have any introns.

CBSE previous Year Solved  Papers  Class 12 Physical Education Outside Delhi 2014

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  The question paper consists of 26 questions.
2. All question are compulsory.
3. Answer to questions carrying 1 mark should be in approximately 10-20 words.
4. Answer to questions carrying 3 marks should be in approximately 30-50 words ,
5.  Answer to questions carrying 5 marks should be in approximately 75-100 words.

PART-A

Question.1. What do you understand by Recreation?**
Answer. Recreation is an activity of leisure, leisure being discretionary time. The “need to do something for recreation” is an essential element of human biology and psychology. Recreational activities are often done for enjoyment, amusement, or pleasure and are considered to be “fun”.

Question.2. Explain the objectives for Intramurals.
Answer . Objectives of Intramurals :

1.  To provide opportunities to actively engage in programs and activities promoting an enhanced quality of life.
2. To provide opportunities for personal development through leadership, diversity, and teamwork.
3. To provide a forum for an experiential education for students, faculty, and staff. ‘

Question.3. Explain sport environment.
Answer . Sports environment is the conditions and circumstances which are favourable and beneficial for the sports persons who perform sports activities.

Question.4. Define flat foot.
Answer . A common and usually painless condition, flatfeet may occur when the arches don’t develop during childhood. In other cases, flatfeet may develop after an injury or from the simple wear-and-tear stresses of age.

Question.5. What is Swadhyaya in Yogic Niyajmas? **
Answer. The term svadhyaya literally means ‘one’s own reading’ or ‘self-study’, it is the fourth of the Niyamas of Patanjali’s Yoga Sutras, and has the potential to deepen our Yoga practice way beyond the mat.

Question.6. What are vitamins?
Answer . Vitamins are compounds found in certain food which are vital to our health and longevity. Our body needs them for growth, function, energy, tissue repair and waste removal. There are two categories of vitamins: water-soluble and fat-soluble vitamins. Water-soluble vitamins are vitamins B and C, and fat-soluble vitamins, are vitamins A, D, E and K.

Question.7. Define acceleration runs.
Answer .  Acceleration runs is a special kind of training in which running speed is gradually increased from jogging to striding and finally to sprinting at maximum speed. Each component is about 50 meter long. Its progressive nature reduces the risk of muscles injuries.

Question.8. What is goal setting?
Answer . Goal setting involves establishing specific, measurable, achievable, realistic and time-targeted (S.M.A.R.T) goals. The goal setting theory suggests that an effective tool for making progress is to ensure that participants in a group with a common goal are clearly aware of what is expected from them.

Question.9. Explain any two factors affecting wellness.**
Answer. Factors Affecting Physical Fitness
Beth has heard that kids today are less physically fit than ever, but why is this? Both long term and short term physical fitness are affected by a great number of factors, such as:

• Injury – Any serious injury will have an adverse effect on a student’s physical fitness. When the body takes time to heal, it takes away opportunities to move and build strength, stamina and coordination.
• Lack of physical activity – When Beth’s students show up at the gym, she can tell they’ve been spending more time in front of screens than they do on their feet. Increasing school demands, families with both parents working, neighborhoods with fewer parks, and cuts in school recess all mean that kids spend less time outside and moving, which results in overall lower fitness.
• Poor diet/nutrition – The amount of fast food and quick options kids can grab has increased over the years. They seem to be eating more junk and fewer whole foods, which means they don’t get as much nutrition as they could. This results in an overall lower level of physical fitness. Beth knows its super important to teach her students about making healthy food choices.
• Poor and inconsistent rest/sleep – Although it may seem contrary, rest is just as important to physical fitness as physical activity. The body needs to replenish and rebuild when resting.
• Dehydration – Although there is no expert consensus definition of dehydration, being dehydrated will decrease both physical and mental performance. The body needs water to survive and thrive.

Question.10. ” Explain any two types of causative factors related to accidents in sports. **
Answer.
Causes:
Sports injuries are most commonly caused by poor training methods; structural abnormalities; weakness in muscles, tendons, ligaments; and unsafe exercising environments. The most common cause of injury is poor training. For example, muscles need 48 hours to recover after a workout. Increasing exercise intensity too quickly and not stopping when pain develops while exercising also causes injury.

Everyone’s bone architecture is a little different, and almost all of us have one or two weak points where the arrangement of bone and muscle leaves us prone to injury. Common predisposing factor in injuries to the ankles, legs, knees, and hips include:

• uneven leg length
• excessive pronation (flat feet)
• cavus foot (over-high arches)
• bowlegged or knock-knee alignment

Uneven leg length may lead to awkward running and increases the chance of injury, but many people with equal-length legs suffer the same effects by running on tilted running tracks or along the side of a road that is higher in the centre. The hip of the leg that strikes the higher surface will suffer more strain.

Pronation is the inward rolling of the foot after the heel strikes the ground, before the weight is shifted forward to the ball of the foot. By rolling inwards, the foot spreads the shock of impact with the ground. If it rolls too easily, however, it can place uneven stress on muscles and ligaments higher in the leg.

While an overly flexible ankle and foot can cause excessive pronation, a too-rigid ankle will cause the effects of caves foot. Although the arch of the foot itself may be normal, it appears very high because the foot doesn’t flatten inwards when weight is placed on it. Such feet are poor shock absorbers and increase the risk of fractures higher in the legs.
Bowlegs or knock knees add extra stress through knees and ankles over time, and may make ankle sprains more likely.

Other structural conditions that make sports injuries more common include:

• lumbar lordosis: forward curve in the lower spine
• patella alta: a kneecap that’s higher than usual
• high Q angle: kneecap displaced to one side, as with knock knees

Having some muscles that are very strong and others that are weak can lead to injury. If your quadriceps (front thigh muscles) are very strong, it can increase the risk of a stretched or torn hamstring (rear thigh muscle). Tight iliotibial bands may be the cause of knee pain for many athletes in running sports.

Overuse injuries are caused by repeated, microscopic injuries to a part of the body. Many long distance runners experience overuse injuries even after years of running. For road runners, the surface is hard and sometimes uneven, and the running movements are repetitive. In addition, there are usually both up- and downhill elements, and these increase the stress on tendons and muscles in the lower leg. You will more likely develop running injuries if you wear the wrong shoes or sneakers. You should use footwear that doesn’t allow side-to-side movement of the heel, and that adequately cushions the foot.

People who play request sports tend to injure their upper body. The need to firmly grasp the request and the shock of impact with the ball can cause various injuries to the tendons of the wrist and elbow, such as “tennis elbow,” which may extend into the muscles of the forearm. In addition, the human arm really isn’t designed to handle strenuous activity above the head. Tennis is a leading cause of rotator cuff (shoulder joint) tendinitis. This is potentially one of the most difficult sports injuries. If you continue to play tennis when you have a sore shoulder, the rotator cuff tendons can fray or tear and may require surgery.

Question.11. Mention any four points to show the importance of yoga. **
Answer.
A foundation for healthy living. By closely observing the lifestyles and needs of people in the west Swami Vishnudevananda synthesized the ancient wisdom of yoga into five basic principles that can easily be incorporated into one’s lifestyle and provide a solid foundation for healthy living. It is around these four principles that the teachings are based.

1. Proper Exercise (Asana)
2. Proper Breathing (Pranayama)
3. Proper Relaxation (Savasana)
4. Proper Diet (Vegetarian)

Proper Exercise (Asana)
Acts as a lubricating routine for the joints and muscles and other parts of the body by increasing circulation and flexibility. The asanas not only produce physical benefits, but are also mental exercises in concentration and meditation, promoting optimum health. Our physical body is meant to move and exercise. If our lifestyle does not provide natural motion of muscles and joints, then disease and great discomfort will ensue with time. Proper exercise should be pleasant to the practitioner while beneficial to the body, mind and spiritual life.
There are numerous modern physical culture systems designed to develop the muscles through mechanical movements and exercises. As Yoga regards the body as a vehicle for the soul on its journey towards perfection, Yogic physical exercises are designed to develop not only the body. They also broaden the mental faculties and the spiritual capacities.
The Yogic physical exercises are called Asanas, a term which means steady pose. This is because the Yoga Asana (or posture) is meant to be held for some time. However this is quite an advanced practice. Initially, our concern is simply to increase body flexibility.
The body is as young as it is flexible. Yoga exercises focus on the health of the spine, its strength and flexibility. The spinal column houses the all-important nervous system, the telegraphic system of the body. By maintaining the spine’s flexibility and strength through exercise, circulation is increased and the nerves are ensured their supply of nutrients and oxygen.The Asanas also affect the internal organs and the endocrine system (glands and hormones)

Proper Breathing (Pranayama)
Connects the body to the solar plexus, where tremendous potential energy is stored. Through specific breathing techniques this energy is released for physical and mental rejuvenation. By far the most important thing about good breathing is the Prana, or subtle energy of the vital breath. Control of the Prana leads to control of the mind. Breathing exercises are called Pranayamas, which means to control the Prana.The two main Pranayamas practiced are Kapalabhati and Anuloma Viloma

Proper Relaxation (Savasana)
Yoga teaches three levels of relaxation – physical, mental and spiritual.
Long before the invention of cars, planes, telephones, computers, freeways and other modern triggers of stress, the Rishis (sages or seers) and Yogis of yore devised very powerful techniques of deep relaxation. As a matter of fact, many modern stress-management and relaxation methods borrow heavily from this tradition. By relaxing deeply all the muscles the Yogi can thoroughly rejuvenate his nervous system and attain a deep sense of inner peace.When the body and the mind are constantly overworked, their natural efficiency to perform work diminishes. Modern social life, food, work and even the so-called entertainment, such as disco dancing, make it difficult for modern people to relax. Many have even forgotten that rest and relaxation are nature’s way of recharging. Even while trying to rest, the average person expends a lot of physical and mental energy through tension. Much of the body’s energy is wasted uselessly.More of our energy is spent in keeping the muscles in continual readiness for work than in the actual useful work done. In order to regulate and balance the work of the body and mind, it is best to learn to economize the energy produced by our body. This may be done by learning to relax.It may be remembered that in the course of one day, our body usually produce all the substances and energy necessary for the next day. But it often happens that all these substances and energy may be consumed within a few minutes by bad moods, anger, injury or intense irritation. The process of eruption and repression of violent emotions often grows into a regular habit. The result is disastrous, not only for the body, but also for the mind.During complete relaxation, there is practically no energy or “Prana” being consumed, althouth a little is keeping the body in normal condition while the remaining portion is being stored and conserved.In order to achieve perfect relaxation, three methods are used by yogis: “Physical”, “Mental”, and “Spiritual” relaxation. Relaxation is not complete until the person reaches that stage of spiritual relaxation, which only advanced spiritual aspirants know.

Proper Diet (Vegetarian) is eating with conscious awareness. A yogi takes food which has the most positive effect on the body and mind and the least negative effect on the environment. Besides being responsible for building our physical body, the foods we eat profoundly affect our mind. For maximum body-mind efficiency and complete spiritual awareness, Yoga advocates a lacto-vegetarian diet. This is an integral part of the Yogic lifestyle.
The yogic diet is a vegetarian one, consisting of pure, simple, natural foods which are easily digested and promote health. Simple meals aid the digestion and assimilation of foods. Nutritional requirements fall under five categories: protein, carbohydrates, minerals, fats and vitamins. One should have a certain knowledge of dietetics in order to balance the diet. Eating foods first-hand from nature, grown in fertile soil (preferably organic, free from chemicals and pesticides) will help ensure a better supply of these nutritional needs. Processing, refining and overcooking destroy much food value.
There is a cycle in nature known as the “food cycle” or “food chain”. The Sun is the source of energy for all life on our planet; it nourishes the plants (the top of the food chain) which are then eaten by animals (vegetarian), which are then eaten by other animals (carnivores). The food at the top of the food chain, being directly nourished by the Sun, has the greatest life promoting properties. The food value of animal flesh is termed as “second-hand” source of nutrition, and is inferior in nature. All natural foods (fruits, vegetables, seeds, nuts and grains) have, in varying quantities, different proportions of these essential nutrients. As source of protein, these are easily assimilated by the body. However, second-hand sources are often more difficult to digest and are of less value to the body’s metabolism. Many people worry about whether they are getting enough protein, but neglect other factors. The quality of the protein is more important than the quantity alone. Dairy products, legumes, nuts and seeds provide the vegetarian with an adequate supply of protein. The high protein requirement still being used by many Health Departments is based on antiquated data and has been scientifically disproved many times in the laboratory. A healthy motto is: “Eat to live, not live to eat”. It is best if we understand that the purpose of eating is to supply our being with the lifeforce,or Prana, the vital life energy. So the greatest nutritional plan for the Yoga student is the simple diet of natural fresh foods.
However, the true Yogic diet is actually even more selective than this. The Yogi is concerned with the subtle effect that food has on his mind and astral body. He therefore avoids foods which are overly stimulating, preferring those which render the mind calm and the intellect sharp. One who seriously takes to the path of Yoga would avoid ingesting meats, fish, eggs, onions, garlic, coffee, tea (except herbal), alcohol and drugs.
Any change in diet should be made gradually. Start by substituting larger portions of vegetables, grains, seeds and nuts until finally all flesh products have been completely eliminated from the diet.The Yogic diet will help you attain a high standard of health, keen intellect and serenity of mind.

Question.12. Explain any two methods for flexibility development.
Answer:

1.  Warm-up before stretching: The very first thing you must do before stretching is a warm up jog, run, or bike to get loose. You don’t have to run or bike that hard Or far, but it is good to do at least 20 minutes of warm up before stretching.
2.  Do dynamic stretching : Dynamic stretching includes motion and is meant to mimic and exaggerate the movements of actual exercise and daily motions.

Question.13. Explain ethics in sports. **
Answer.
To understand the role ethics plays in sport and competition, it is important to make a distinction between gamesmanship and sportsmanship.
Gamesmanship is built on the principle that winning is everything. Athletes and coaches are encouraged to bend the rules wherever possible in order to gain a competitive advantage over an opponent, and to pay less attention to the safety and welfare of the competition. Some of the key tenants of gamesmanship are:
Winning is everything
It’s only cheating if you get caught
It is the referee’s job to catch wrongdoing, and the athletes and coaches have no inherent responsibility to follow the rules
The ends always justify the means
Some examples of gamesmanship are:
Faking a foul or injury
Attempting to get a head start in a race
Tampering with equipment, such as corking a baseball bat in order to hit the ball farther
Covert personal fouls, such as grabbing a player underwater during a water polo match
Inflicting pain on an opponent with the intention of knocking him or her out of the game, like the Saint’s bounty scandal
The use of performance-enhancing drugs
Taunting or intimidating an opponent
A coach lying about an athlete’s grades in order to keep him or her eligible to play
All of these examples place greater emphasis on the outcome of the game than on the manner in which it is played.
A more ethical approach to athletics is sportsmanship. Under a sportsmanship model, healthy competition is seen as a means of cultivating personal honor, virtue, and character. It contributes to a community of respect and trust between competitors and in society. The goal in sportsmanship is not simply to win, but to pursue victory with honor by giving one’s best effort.
Ethics in sport requires four key virtues: fairness, integrity, responsibility, and respect.
Fairness:
All athletes and coaches must follow established rules and guidelines of their respective sport.
Teams that seek an unfair competitive advantage over their opponent create an uneven playing field which violates the integrity of the sport.
Athletes and coaches are not discriminated against or excluded from participating in a sport based on their race, gender, or sexual orientation.
Referees must apply the rules equally to both teams and cannot show bias or personal interest in the outcome.
Integrity:
Similar to fairness, in that any athlete who seeks to gain an advantage over his or her opponent by means of a skill that the game itself was not designed to test demonstrates a lack of personal integrity and violates the integrity of the game. For example, when a player fakes being injured or fouled in soccer, he or she is not acting in a sportsmanlike manner because the game of soccer is not designed to measure an athlete’s ability to flop. Faking is a way of intentionally deceiving an official into making a bad call, which only hurts the credibility of the officiating and ultimately undermines the integrity of the game.
Responsibility:
To be sportsmanlike requires players and coaches to take responsibility for their performance, as well as their actions on the field. This includes their emotions.
Many times athletes and coaches will make excuses as to why they lost the game. The most popular excuse is to blame the officiating. The honorable thing to do instead is to focus only on the aspects of the game that you can control, i.e. your performance, and to question yourself about where you could have done better.
Responsibility requires that players and coaches be up to date on the rules and regulations governing their sport.
Responsibility demands that players and coaches conduct themselves in an honorable way off the field, as well  as on it.
Respect:
All athletes should show respect for team mates, opponents, coaches, and officials.
All coaches should show respect for their players, opponents, and officials.
All fans, especially parents, should show respect for other fans, as well as both teams and officials.
The sportsmanship model is built on the idea that sport both demonstrates and encourages character development, which then influences the moral character of the broader community. How we each compete in sports can have an effect on our personal moral and ethical behaviour outside of the competition.

Some argue for a “bracketed morality” within sports. This approach holds that sport and competition are set apart from real life, and occupy a realm where ethics and moral codes do not apply. Instead, some argue, sports serves as an outlet for our primal aggression and a selfish need for recognition and respect gained through the conquering of an opponent. In this view, aggression and victory are the only virtues. For example, a football player may be described as mean and nasty on the field, but kind and gentle in everyday life. His violent disposition on the field is not wrong because when he is playing the game he is part of an amoral reality that is dictated only by the principle of winning.

An ethical approach to sport rejects this bracketed morality and honors the game and one’s opponent through tough but fair play. This means understanding the rules and their importance in encouraging respect for your opponent, which pushes you to be your best.

Question.14. “Games and sports are the best means for attaining fitness.” Justify. **
Answer. It is known that fitness and wellness makes an individual physically fit,mentally stable and helps becoming a good citizen. Fitness helps individual achieve satisfactory level of strength, endurance and flexibility. It further improves the confidence and energy level. One feels more energetic and fresh for the whole day. It also leads to sound sleep followed by more relaxed body which leading to mental satisfaction and social stability. Body becomes more resistant to general ailments. Fitness improves efficiency of heart and lungs by improving cardio-respiratory fitness. It helps in maintaining normal blood pressure of the body. In nutshell, we can say – fitness and wellness helps to achieve the aim of physical education i.e. – “All round development of personality of the individual”.
Principles of fitness:-

1. Before taking any fitness programme one go for medical check-up.
2. One can take-up the fitness program at any age.
3. Fitness program must be scientifically chalked out depending on age,sex and ability.
4. Programme must be simple to complex.

Question.15. Elucidate the role of media for improvement of positive sports environment.
Answer : The media, whether it is print media or electronic media, plays a significant and effective role for creating positive sport environment. Media is providing its valuable contribution for the encouragement and promotion of sports and games. Electronic media is the source by which millions of people watch the various World cups and Olympic games. There are many sports channels are available such as Ten sports, NEO sports, Star sports that provide the facility of watching live and recorded programs of games and sports. Thus media is promoting the games and sports.
The media encourages people to develop reasonable interests in engaging in sports. Either for fun, excitement, recreation, physical fitness or health care. In the homes, countless number of youths involves themselves in sporting activities. This has contributed to the remarkable improvement of fitness among the youths. Another role that the media play in sports is the advocacy of unity among diverse races, religions, ethinic groups, language, colour, peoples, idiosyncrasies and world¬views. Media not only publicise the games and sports but it also gives publicity to sportspersons which is a motivating source for them.

Question.16. Explain any three techniques of meditation. **
Answer. 1. BUDDHIST MEDITATION
Zen Meditation (Zazen)
Zazen means “seated Zen”, or “seated meditation”, in Japanese. It has its roots in the Chinese Zen Buddhism (Ch’an) tradition, tracing back to Indian monk Bodhidharma (6th century CE). In the West, its most popular forms comes from Dogen Zenji (1200~1253), the founder of Soto Zen movement in Japan. Similar modalities are practiced in the Rinzai school of Zen, in Japan and Korea.

2.HINDU MEDITATION (Vedic & Yogic)
Mantra Meditation (OM Meditation)
A mantra is a syllable or word, usually without any particular meaning, that is repeated for the purpose of focusing your mind. It is not an affirmation used to convince yourself of something.
Some meditation teachers insist that both the choice of word, and its correct pronunciation, is very important, due to the “vibration” associated to the sound and meaning, and that for this reason an initiation into it is essential. Others say that the mantra itself is only a tool to focus the mind, and the chosen word is completely irrelevant.
Mantras are used in Hindu traditions, Buddhist traditions (especially Tibetan and “Pure Land” Buddhism), as well as in Jainism, Sikhism and Daoism (Taoism). Some people call mantra meditation “om meditation”, but that is just one of the mantras that can be used. A more devotion oriented practice of mantras is called japa, and consists of repeating sacred sounds (name of God) with love.

3.CHINESE MEDITATION
Daoism is a Chinese philosophy and religion, dating back to Lao Tzu (or Laozi). It emphasizes living in harmony with Nature, or Tao, and it’s main text is the Tao Te Ching, dating back to 6th century B.C. Later on some lineages of Taoism were also influenced by Buddhist meditation practices brought from India, especially on the 8th century C.E..
The chief characteristic of this type of meditation is the generation, transformation, and circulation of inner energy. The purpose is to quieten the body and mind, unify body and spirit, find inner peace, and harmonize with the Tao. Some styles of Taoist Meditation are specifically focused on improving health and giving longevity.

Question.17. Explain any three principles of training in brief.
Answer: Three principles of training:

1.  Principle of overload : The overload principle is a basic sports fitness training concept. It means that in order to improve, athletes must continually work harder as they their bodies adjust to existing workouts. Overloading also plays a role in skill learning.
2. Principle of Specificity : The principle of specificity states that the more specific a training activity is to a given sport—muscle group, work load, velocity and pattern of movement, body posture, and range of motion—the more it will contribute to increasing performance in that sport. .
3. Principle of Individualization : This could also be called the snowflake principle, since it highlights that no two climbers—or their optimal conditioning program-are the same. The best training program for a person will target his/ her specific weaknesses, address past or present injuries, provide sufficient time for recovery, and be structured to provide the greatest output for the available training input.

Question.18. Suggest the formation of various committees for systematic and smooth conduct of sports day in your school.
Answer :

1. Select members who are excited about the project : If you’ve been working with the same committee for years, they may be growing bored or frustrated with the annual undertaking.
2.  Clearly define tasks : Progress on your event will come to a halt if your committee members don’t know what to do next. At the end of each planning session, members should know their next steps and the deadlines for completing each task.
3.  Plan ahead : Send committee members the meeting schedule several weeks in advance so they can clear the time to attend meetings.
4.  Establish goals and expectations : When someone joins your committee, communicate clearly how much time they should set aside to complete their tasks.

Question.19. Explain in detail about any five advantages of correct posture.
Answer: Posture is the position you maintain while standing, sitting or lying down. You have good posture when your position creates the least amount of strain on supporting muscles and ligaments when you move or perform weight-bearing activity. It has many advantages. Some of them are :

1.  Better for Your Body : Good posture and back support are essential for avoiding back and neck pain. Good posture also prevents muscle aches and muscle fatigue. It keeps your bones and joints in proper alignment so you use your muscles more efficiendy, preventing strain and overuse.
2. Future Health : Proper posture reduces abnormal wear and tear on joint surfaces, which can lead to arthritis. It also reduces stress on ligaments that connect spinal joints. Good posture helps you avoid developing an abnormal permanent position, which can cause spinal disk problems and constricted blood vessels and nerves. Good posture also protects spinal joints from injury and deformity.
3. Breathe Right: Good posture helps to open the airways and ensure proper breathing. Proper breathing allows enhanc-ed oxygen flow in the cardiopulmonary system. The blood then carries sufficient oxygen to the nervous system, organs and other tissues, so they function effectively.
4. Looking Good : Maintaining good posture does wonders * for your appearance. Proper posture can help you make a good
   first impression, and appear more attractive and-confident.
5.  Good posture : Good posture prevents fatigue because muscles are being used more efficiently, allowing the body to use less energy.

Question.20. What is the role of various elements of diet on performance
of an athlete?
Answer: Eating a balanced diet means choosing a wide variety of foods and drinks from all the food groups. It also means eating certain things in moderation, namely saturated fat, trans fat, cholesterol, refined sugar, salt and alcohol. The goal is to take in nutrients you need for health at the recommended levels.

1. Most of the energy used by the body is provided by the carbohydrate and the fats.
2. Proteins build the body and perform the function of repair of damage tissues.
3. Minerals an’d vitamins regulate the functions of the body; they are needed for important chemical reactions taking place in the body.
4. Water is present in all tissues of the body and plays an important role in various life processes. For example, digestion, excretion and transport of important materials within the body. It also helps in cellular reactions.

Question.21. What is endurance? Explain the various methods for its development.
Answer : Endurance refers to the body’s ability to continue using muscular strength and endure repeated contractions for an extended period of time. It is essential in exercise and when doing heavy tasks as it allows the muscles to perform for long periods of time without becoming tired. There are various methods of its development. Some are as follows :

1.  Continuous Training : Continuous training is a type of sports training that involves activity of moderate intensity with a duration of-more 15 minutes with resting intervals. It is the most common type of training and is for maintaining general health and well being. Generally, this type of training is used to prepare the body for sustained workouts such as marathons and triathlons, but can also be effective for more casual athletes. It allows the body to work from its aerobic energy stores to improve overall fitness and endurance. Chief
   -benefits of continuous training include fat burning, muscle building, and increasing maximum aerobic potential.
2.  Interval training: Interval training involves periods of hard work followed by a timed period of rest, repeated several times in one training session. The periods of hard work are called high intensity activity. Rest can be active (walking, jogging etc) An example of interval training is 10 fast runs over 40 metres, with a two minute rest between each run. Variables to consider during interval training are Distance/duration of activity, Intensity of activity, duration of rest, activity during rest, number of sets, and frequency of training. By varying any of the variables athletes can be progressively overloaded. This form of training also increases fitness levels for people involved in exercise.
3. Fartlek training : Fartlek, which means “speed play” in Swedish, is a training method that blends continuous’ training with interval training. The variable intensity and continuous nature of the exercise. places stress on both the aerobic and anaerobic systems. It differs from traditional interval training in that it is unstructured; intensity and/or speed vary, as the athlete wishes. Fartlek training can be used to improve both the aerobic and anaerobic systems by mixing moderate activity with bursts of speed.

PART-B

Question.22. Write about any four tournaments of the game/sport of your choice.

Question.23. Explain any 6 terminologies from the game/sport of your choice. **

Question.24. Explain any five latest rules from the game/sport of your choice. **

Question.25. Explain Arjuna Award. **

Question.26. Write about achievements of any three important sports personalities from the game/sport of your choice. **

Question.27. Explain any five common soft tissue injuries in the game/ sport of your choice. **

CBSE Previous Year Solved  Papers  Class 12 Physical Education Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  The question paper consists of 26 questions.
2. All question are compulsory.
3. Answer to questions carrying 1 mark should be in approximately 10-20 words.
4. Answer to questions carrying 3 marks should be in approximately 30-50 words ,
5.  Answer to questions carrying 5 marks should be in approximately 75-100 words.

Question.1. Suggest any four ways through which women participation in sport across age group can be enhanced.
Answer :

1. Modification in Legislation.
2. Better coverage of women sports
3. Improvement in fitness and wellness movements –
4. Educating Women

Question.2. Trekking is a long adventurous journey undertaken on foot in areas where common means of transport are generally not available. Name any four important materials required that should be carried along.
Answer :

1. First Aid Box
2.  Sleeping Bags
3.  Pair of Shoes And Socks
4.  Rope

Question.3. Enlist two objectives of Intramurals.
Answer :

1. Develops moral and ethical values of the students.
2.  Provide opportunity to the maximum number of students to participate in sports.

Question.4. Enlist two sources for calcium and iron separately
Answer : Calcium sources : Cheese, Milk, Orange Juice, Eggs, Yogurt. Iron sources : Liver, meat, orange juice, egg.

Question.5. Explain correct sitting posture.
Answer : When we sit in a chair, our hips should be as far as back in the chair as possible. Head, spinal column, Shoulder and hips should be in straight line and erect. Legs should touch the ground and not in hanged position. Thighs should be in horizontal position.

Question.6. Calculate the physical fitness index using short formula for a 12 year old boy having completed Harvard Step Test for a duration of 3 minutes and a pulse rate of 54 beats for 1 to 1.5 minute.
Answer : The athlete’s fitness index score is calculated with the help of following formula. ‘
Fitness index score = (100 x test duration in seconds) divided by (2 x sum of heart beats in recovery period). (100 x 180 Sec)/ 2 x 54 = 500/3 = 166.66

Question.7. Your grandmother feels she has reduced her upper body flexibility and therefore she wants to test herself. Which test would you suggest her?
Answer : Back scratch test for upper body flexibility.

Question.8. Explain the term Hypertrophy of muscles.
Answer : Increase in size of the muscle fiber due to regular exercises or Hypertrophyis enlargement of heart due to regular exercises which is called “Athletic heart”.

Question.9. What do you understand by linear movement?
Answer : Linear movement refers to any movement along a straight line in one direction.

Question.10. Explain the term “realistic” in goal setting principles.
Answer : Realistic goads are achievable goals.

Question.11. Suggest any two isometric exercises for shoulder region.
Answer :

1.  Pushing against the wall.
2.  Holding push up position.

Question.12. What safety measures children should be taught while participating in trekking?
Answer :

1.  Avoid trekking during bad weather conditions.
2. To prevent insect bite do wear full sleeves shirts and full pants.
3.  Wear proper footwear so that you don’t slip while trekking.
4.  Don’t eat leaves, flowers etc. while trekking, they may be poisonous.

Question.13. Briefly explain the functions and resources of three fat soluble vitamins.
Answer : Fat Soluble vitamins are A, D, E, K.
Functions :

1.  Vitamin ‘D’ : The main function of vitamin D is to regulate the absorption of calcium and phosphorus in our bones and aid in cell-to-cell communication throughout the body.
2. Vitamin ‘E’ : Vitamin E (tocopherol) is a powerful, fat-
   soluble antioxidant that helps to protect cell membranes against the damage caused by free radicals, prevents the oxidation of LDL cholesterol and is also important in the formation of red blood cells (RBC), thus essential for blood coagulation.
3. Vitamin ‘K’ : Vitamin K occurs in two forms-Vitamin K] and Vitamin K2. Vitamin K is known as the clotting vitamin, because without it, blood would not clot. It also helps in prevention of haemorrhage.
   Sources of Vitamin A : Ghee, milk, curd, egg yolk, fish, tomato, papaya, spinach, carrot, pumpkin etc.
   Sources of Vitamin D : Egg yolk, fish, sunlight. Vegetables, cod liver oil, milk, cream, butter, tomato,
   carrot etc.
   Sources of Vitamin E : Green vegetables, kidney, liver, heart cotton seed, sprouts seeds, coconut oil.
   Sources of Vitamin K : Cauliflower, spinach, cabbage, tomatoes, wheat, egg and meat etc.

Question.14. Neeti along with her father was regular at district park in early morning. She realized that most of the children are obese. She along with her few classmates wanted to help those children. She discussed with her physical education teacher and the principal of the school. School decided to organize awareness rally for the neighbourhood.

1.  How obesity can be prevented? Give two ways.
2. Give any two disadvantages of obesity.
3.  What values are shown by Neeti and her
   classmates?

Answer :

1. Obesity can be prevented by :
   1. Avoid fast food, fatty food and over eating
   2. Regular exercise/ physical activity
2. Disadvantages of obesity :
   1. More chances of injury
   2. More disease/ physical health problems
3. Values shown:
   1. Good moral character
   2. Self-discipline
   3. Decisiveness
   4. Logical and decision maker

Question.15. Briefly explain the six physical benefits of exercise to children.
Answer :

1.  Exercise helps in healthy growth and development : Encouraging healthy lifestyles in children and adolescents is important for when they grow older. Participating in organized sports and games is not only of great fun but is very essential for healthy growth and development.
2.  Exercises improve self-esteem : Self-esteem can play a great role in how children feel about themselves and also how much they enjoy things or worry about things. Exercise reduces depressive symptoms and improves self-esteem in children.
3.  Exercise reduces blood sugar level : Lowering blood sugar level is not only the health issue of adults but also for children as they are also becoming trapped with high sugar level. Exercise is helpful in preventing sugar from accumulating in the blood by triggering muscles to receive more glucose from blood and utilize it for energy.
4.  Exercise helps in motor development : Exercise helps in increasing the motor development of infants and children at a very rapid pace which in due course improves in building fine movements later on in life. The movements of muscles become efficient and smooth by doing exercise in premature age; their movements turn out to be attractive.
5.  Exercise strengthens the heart : Exercise helps improve heart health, and can even reverse some heart disease risk factors. A child who exercises often, has the lowest risk for heart disease, but any amount of exercise is beneficial.
6.  Exercise makes stronger bones, muscles and joints : Exercise is vital for strong muscles, bones and joints. Exercise may help to children lower their risk of chronic pain related to muscles, bones and joints in the future. Exercise increases bone density which helps to make bones stronger.

Question.16. Explain the procedure for conducting Kraus- Weber test for measuring minimum muscular strength.
Answer : Kraus -Weber Test. This test consists of six items. It is commonly known as the Kraus-Weber Tests. These tests are supposed to measure the minimum muscular fitness of an individual. Infact, they measure a level of strength and flexibility of certain key muscle groups below which the functioning of whole body as a healthy individual seems to be endangered. These tests are graded on a pass-fail basis. But partial movements on each test can be scored from 0 to 10.
Six tests measures minimum muscular fitness of an individual
Test No.1. The subject lies down in supine position i.e., flat on his back and hands behind his neck. The examiner holds his feet to keep him on the ground. The subject is asked to perform on sit- up. If he performs one sit- up, he passes this test. If he cannot raise his shoulders from Ehe table or ground, his score remains zero.
Test No.2. The lying position for this test remains same i.e., in supine position except that his knees are bent and ankles
remains in touch with his buttocks. He is asked to perform one sit- up. If he is able to perform full sit- up, he passes this test. If he is unable to raise his shoulders from the table or ground, he gets zero.
Test No. 3. Subject lies in supine’position i.e., lies flat,on his back with his hands behind the neck. He is asked to raise his feet 10 inches from the ground. His knees should be straight. The examiner counts to 10 seconds. He passes this test if he holds that position for ten seconds. Scoring from 0-10 depends on the number of seconds he holds the appropriate position.
Test no. 4. Subject lies in prone position i.e., on his stomach with a pillow under his lower abdomen and his hands behind his neck. The examiner holds his feet down. The subject is asked to raise his chest, head and shoulders, while the examiner counts to 10 seconds. He passes the test if he is able to hold the exact position up to 10 seconds. Scoring from 0-10 depends on the number of seconds he holds the exact position.
Test No. 5. The subject’s position remains the same, but the examiner holds his chest down. The subject is asked to raise his feet. His knees should be straight. The examiner counts to 10 seconds. Scoring from 0-10 depends on the number of seconds he holds the position.
Test No.6. It is also known as floor- touch test. It measures the flexibility of trunk. The subject stands erect, bare footed, hands at sides and feet together. He is asked to lean down slowly to touch the floor with fingertips for 3 seconds. In this test bouncing or jerking is not allowed. The examiner holds his knees in order to prevent any bend; if it occurs. Scoring from 0-10 depends on the number of seconds he holds the position.

Question.17. Maintaining physical activities for a longer period, brings desirous changes in circulatory system. Justify your answer by highlighting three benefits of exercise.
Answer :

1. Heart size increases : The size of heart and strength of the cardiac muscles increases due to regular exercises as to the maximum extent the left ventricle adapts. The walls of the heart develop into stronger and thicker as shown in recent studies and the thickness of myocardial wall increases as well.
2.  Resting heart rate decreases : The resting heart rate decreases due to regular exercises. After duration of 10 week training programme, the resting heart rate may reduce up to 10 beats per minute from the normal of 72 beats per minute. The heart becomes more efficient due to regular exercises. In highly conditioned athletes the resting heart rate decreases to 30 beats/minute..
3.  Stroke volume increases at resting conditions : The stroke volume increases at resting conditions due to regular exercises. The stroke volume at rest remains up to 50-70 ml/ beat in untrained individuals; in trained individuals it ranges from 70-90 ml/beat and in the elite endurance athletes it ranges from 90-110 ml/beat.

Question.18. What is the difference between linear and angular ‘ motion? Explain through example.
Answer : Linear motion is any motion that moves along a straight line in one direction. The direction can either be horizontal, vertical or inclined direction. Example, approach run.
Angular motion is rotatory motion, it occurs when all points on a body or object move in a circular path about the same fixed central line or axis. A child swings and rotations in i hammer throw are the best example.

Question.19. What do you understand by relative strength? Explain ? the importance of body weight in determining relative
strength.
Answer: Relative strength is strength in relation to your body
weight. Relative strength have a determining importance in sports in which the athlete shifts his body in space without any additional external weight. (H/J and L/J) as well as in sports in which he has to restrict his own weight within the framework of weight division (e.g. boxing, wrestling, weight f lifting etc.) e.g. if 1 RM (repetition maximum) is 50 kgs and body wt. is 50 kg and if 1 RM is 70 kg and body wt. is 50 kg, in second category the relative strength is more.

Question.21.Being sports captain of the school, prepare five important committees with their responsibilities to conduct one day run for health race.
Answer :

1.  Publicity committee : The committee for publicity informs and announces the sports events, dates and venues in advance to the institutions, departments and public through newspapers, media, e-mail and website etc. Their main responsibility is to advertise the sports events,
2.  Boarding and lodging committee : The main responsibility of this committee is ensuring and making : needful arrangements for providing accommodation and
   serving food to all the officials and sports persons.
3.  Transportation committee : This committee is mainly responsible for providing the facilities regarding transportation of various teams to the venue of sports events or to the place of boarding and lodging. Their main duty is to make arrangements of transportation.
4.  Grounds and equipment’s committee : This committee comes under the technical committee which mainly deals with the technical area of ground marking and layouts of track arid field etc. This committee also makes necessary arrangements of equipment’s related to the game and athletic meet.
5. Official committee : This committee selects various officials such as referees, judges, and various qualified officials as per requirements of the particular sport.

Question.22. What are the important functions of our skeletal system?
Answer :
The adult human skeletal system consists of 206 bones, as well as a network of tendons, ligaments and cartilage that connects them. The skeletal system performs vital functions — support, movement, protection, blood cell production, calcium storage and endocrine regulation — that enable us to survive.

Question.23. Explain Sheldon’s classification of personality and explain
its importance in physical education and sports.
Answer :

1.  Endomorphic : An Endomorphic somatotype is also known as a viscerotonic. The characteristic traits of this somatotype usually includes being relaxed, tolerarit, comfortable, and sociable. Psychologically, they are also fun- loving, good humored, even-tempered, and they love food and affection. The Endomorph is physically “round”. They have wide hips and narrow shoulders that give a pear-shape They tend to have a lot of extra fat on their body and on their arms and thighs. They have skinny ankles and wrists
   that make the rest of their body look even bigger.
2. Ectomorphic : An ectomorph is the complete opposite of the Endomorph. Physically, they have narrow shoulders, thin legs and arms, little fat on the body, a narrow face and a narrow chest. They may eat just as much as the endomorph but never seem to gain any weight. They always stay skinny. Personality wise, they tend to be self- conscious, socially anxious, artistic, thoughtful, quiet, and private. They always keep to themselves and are a afraid to branch out.
3.  Mesomorphic : The mesomorph is in between the endomorph and thin ectomorph. They  have an attractive and desirable body. Physically, they tend to have a large head and broad shoulders with a narrow waist. They have a strong muscular body and strong arms and legs and little fat on the body. They work for the body they have so that they could have an attractive body. Psychologically, the mesomorph is adventurous and courageous. They are not afraid to break out and do new things with new people. They are assertive.

Question.24. What is movement speed? Explain the methods to develop speed endurance.
Answer : Movement speed is the time taken between the initiation of movement and the completion of the movement. It depends upon techniques, explosive strength, flexibility and coordinative abilities. It plays a vital role in boxing, gymnastics, swimming; throws and jumps etc. Where the minimum time is taken to complete the movement.
To develop the speed endurance we will have to work more on pace races because pace races means running the whole distance at a constant speed.Generally, 800 meters and above races are included in pace races. As a matter of fact, an athlete’s can run a distance of 300 meters at full speed but, in longer races such as 800 meters or above races he must conserve his energy by reducing his speed. For example, if there is a runner of 800 meter race his best time is 1 minute 40 second, so, he should run first 400 m in 49 seconds and next 400m in 51 seconds.

Question.25. Diet for sportspersons are important. What should be the aims of preparing diet for sportsperson?
Answer : Aims of preparing diet for sports person :

1. Maintaining body weight and body composition desired for that specific sport
2. Maintaining adequate pool of nutrient levels in the body
3. Adopting healthy nutritional practices during training and competition.
4. Carrying on with healthy nutritional practices during off season as well i.e when competition are not taking place.

Question.26. “Involvement in physical activities for longer period of time with moderate intensity can improve the quality of life.” Justify your answer.
Answer :

1.  Exercise helps in healthy growth and develop-ment : Exercise is an important part of keeping children healthy. Encouraging- healthy lifestyles in children and adolescents is important for when they grow older. Participating in organized sports and games is not only of great fun but is very essential for healthy growth and’ development.
2.  Exercises improve self-esteem : Exercise is necessary for your physical and mental health. Self-esteem can play a great role in how children feel about themselves and also how much they enjoy things or worry about things. Exercise reduces depressive symptSms and improves self-esteem in children.
3. Enhances flexibility : The stiffness of joints decreases due to exercising, in a way improving the flexibility. The elasticity of tendons, figments and joint capsules improves due to regular exercise.
4. Lessens stress and tension : Regular exercise has a distinctive capability to slow down the depression process by-reducing stress and tension. Actually, regular exercise lessens the levels of bodys stress hormones like adrenaline! and cortisol. The body’s natural painkillers and mood elevators i.e. the endorphins are produced due to regular exercise. These benefits of work out facilitate in delaying the process of ageing.
5. Connect with others : Spend time with positive people who enhance your life. A strong support system will buffer you from the negative effects of stress.
6. Keep your sense of humor : This includes the ability to laugh at you. The act of laughing helps your body fight stress in a number of ways.

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2009

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26
   questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Question.1. What is the electrostatics potential due to an electric dipole at an equatorial point?
Answer : The electric potential due to an electric dipole at an equatorial point is zero.

Question.2. Name the EM waves used for studying crystal structure of solids. What is its frequency range?
Answer : X-rays are used to study crystal structure of solids. The frequency range is 1 nm to 10^-3 nm.

Question.3. An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is direction of the magnetic field?
Answer : The magnetic field is given to be in the direction in which the electron moves. The force is given by

Question.4. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled?

Question.5. Two thin lenses of power + 6D and -2D are in contact. What is the focal length of the combination.

Question.6. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Question.7. Two nuclei have mass numbers in the ratio 1: 8. What is the ratio of their nuclear radii?

Question.8. Give the logic symbol of NOR gate.

Question.9. Draw 3 equipotential surfaces corresponding to a field that uniformly increase in magnitude but remains constant along z-direction. How are these surfaces different from that of a constant electric field along z-direction?
Answer : For constant electric field in z-axis equipotential surfaces will be plane parallel to xy- planes .

If field that increases in magnitude, the equipotential surfaces will be planes parallel to XY plane But, as the field increases such planes will get closer.

Question.10. Define electric flux. Write its S.I. unit A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?
Answer: Electric flux is the total number of electric field lines crossing an area. Its SI units in Nm /C. The electric flux through a spherical surface of radius R for a charge q enclosed by the surface is If radius is reduced to half, the electric flux remains the same.

Question.11. Define refractive index of a transparent medium. A ray of light passes through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.
Answer : Refractive Index : The ratio of velocity of light in vacuum to the velocity of light in medium is called absolute refractive index of the medium.
Graph : The plot of angle of deviation versus angle of 0 incidence for a triangular prism is shown below :

Question.12. Calculate the current drawn from the battery in the given network.

Question.13. Answer the following questions:
(a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Why?
(b)The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer :
(a) X-rays from celestial objects cannot reach the earth’s surface. So for X-ray astronomy to be possible the satellite has to be present in space. However, visible light and lower wavelengths of light emitted by celestial objects reach the surface of the earth. So these objects can be observed with optical and radio telescopes.
(b) The ozone layer is crucial for human survival as they help to block UV radiations and other high frequency harmful radiations and prevent them from reaching the earth’s surface.

Question.15. Define the term ‘linearly polarised light’.
When does the intensity of transmitted light become maximum, when a polaroid sheet is rotated between two crossed polaroids?
Answer: ‘Linearly polarized light’ is an electromagnetic wave, in which the vibrations of electric field are restricted to a single plane. ‘
The intensity of transmitted light is given by I =I_m cos² θ
The intensity of transmitted light is maximum, when θ =0° or 180°, or the polarizing axis of the two polaroids is parallel to each other.

Question.16. A wire of 15 Ω resistance is gradually stretched to double
its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery.
Answer : When any resistor is stretched to double its original length, the new resistance becomes four times of original resistance.

Question.17. (a)a nucleus in its ground state is always less than the total mass of its constituents — neutrons and protons. Explain.
(b) Plot of graph showing the variation of potential energy of a pair of nucleons as a function of their separation.
Answer :
(a) The mass of a nucleus in its ground state is always less than the total mass of its constituents because some mass is converted into energy, in accordance with the equation E = mc². This difference in mass is called the mass defect and the energy corresponding to the mass defect is the binding energy. This
is the energy that has to be supplied*to the nucleus to break it up into its constituents..
(b) The graph is

Question.18. Write the function of (i) Transducer and (ii) Repeater in the context of communication system.
OR
Write two factors justifying the need of modulation for transmission of a signal.
Answer :

1. Transducer converts energy from one form of another either at the input or at the output. For example sound signals are converted to electrical signals so that they can be transmitted through the communication channel.
2. Repeater stations receive the signal, amplify it and then transmit it. So they are a combination of a receiver and , transmitter. This helps to increase the range of transmission of signals.

OR
Two factors justifying the need of modulation for transmission of signal are :

1. Modulation helps to increase the frequency of the signal. This helps to transmit it over larger distances. This is because the power radiated by an antenna is proportional to 1/λ₂ The power radiated by the antenna increase for high frequencies.
2.  The size of the antenna required is proportional to (wavelength/4). So if waves of large wavelength are transmitted then the size of the antenna required is impractical. By increasing the frequency of the signal by modulation, its wavelength is decreased. So now shorter and more practical size antennas can be built.

Question.19. A positive point charge (+ q) is kept in the vicinity of an uncharged conducting plate. Sketch electric field lines originating from the point on to the surface of the plate. Derive the expression for the electric field at the surface of a charged conductor.
OR
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.
Answer : Representation of electric field :
The electric field due to a positive charge (+q) is represented as :

Electric field due to a point charge : Consider a point charge +q placed at the origin O of the coordinate frame. Let qo be any test charge placed at P.

Question.20. (i) State the principle of working of a meter bridge.
(ii) In a meter bridge balance point is found at a distance l1 t with resistance R.and S as shown in the figure.

Question.21.

1. State Faradays law of electromagnetic induction.
2.  A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 x 10^-4 T and the dip angle is 30°?

Answer:

1.  Faraday’s law of electromagnetic induction states that the induced emf is proportional to the rate of change of magnetic flux.
2.  The induced emf will be given by E = Blv where the magnetic field B is perpendicular to the length l. In the question the earth’s magnetic field is given and the angle of dip is 30  degrees, so the magnetic field perpendicular to the direction of the plane is B sin 30°. ’
   

Question.22. In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on. the interference fringes if the’ monochromatic source is replaced by a source of white light?
Answer:

When the monochromatic source is replaced by a source of white light, the fringe width is changed.

Question.23. Draw a schematic arrangement of the Geiger-Marsden experiment. How did the scattering of a-particles by a-thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly.
Answer: Diagram of the Geiger Marsden experiment.

The alpha particles that are incident head-on with the gold nucleus experience a very large force of repulsion and undergo maximum deflection. Equating the kinetic energy of the incident alpha particle with the potential energy of the  alpha particle and gold nucleus, the sum of the approximate radius of the gold nucleus and alpha particles, can be found.

Question.24. Distinguish between sky wave and space wave propagation. Give a brief description with the help of suitable diagrams indicating how these waves are propagated.
Answer : The sky waves are reflected from the ionosphere and received by a receiver. Space waves penetrate the ionosphere  and are intercepted by a satellite. They can also be used for line of sight communication.
This mode of propagation is used by short-wave broadcast service.
The space waves are the radiowaves of very high frequency (i.e. between 54 MHz to 4.2 MHz). The space waves can travel through atmosphere from transmitter antenna to- receiver antenna either directly or after reflection from ground in the earth’s troposphere region. That is why space wave propagation in also known as tropbspherical propagation. The space waves travel in straight line from transmitting
antenna to receiving antenna. Therefore, the space waves are used for line of sight communication such as television
broadcast, microwave link and satellite communication.

Question.26. Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw a input and out waveforms of the single. Write the expression for its voltage gain.
Answer. Circuit diagrame of a common emitter amplofire using an n-p-n- transisstor.

Question.27. Draw a plot showing the variation of binding energy per nucleon versus the mass number A. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
Answer: Binding energy curve : The variation of average B.E per nucleon with mass number A, shown in figure below.

A graph between the average binding energy per nucleon and the mass number A of different nuclei is a curve, called ‘binding energy curve’.

1.  For nuclei having mass number A=50 to A=80 are almost stable.
2.  For nucler having mass number above 80, the average BE/nucleon decreases slowly and drops to about 7.6 MeV for uranium (A = 238). This lower value of binding energy per nucleon fails to overcome the Coulombian repulsion.
3.  For nuclei having mass number below 50 also, the average BE/nucleon decreases and below 20, it decreases sharply, e.g. : For heavy hydrogen (H²), it in only about 1.1 MeV. This shows that the nuclei having mass number below 20 are comparatively less stable.
4. Below A = 50, the curve does not fall continuously, but has subsidiary peaks at 8o¹⁶, 6C¹² and  2He⁴. This shows that these nuclei are more stable than their immediate neighbours.
5.  Binding energy per nucleon is small for both light and heavy nuclei. When light nuclei fuse to form a heavy nucleus, high value of B.E is released in Nuclear Fusion. When a heavy nucleus splits into light nuclei, high value of B.E is released in Nuclear fission.

Question.28. Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.

1.  Show that time period of ions in a cyclotron is
   independent of both the speed and radius of circular path. .
2. What is resonance condition? How is it used to accelerate the charged particles?

OR
(a) Two straight long parallel conductors carry currents Ii and I2 in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them,
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.

1.  What is the direction ,of the magnetic moment of the current loop?
2.  When is the torque acting on the loop (A) maximum, (B) zero?

Answer : Diagram of Cyclotron

Charged particles are introduced between the dees. An alternating voltage applied between the dees accelerates, these particles by an electric field. A magnetic field that is perpendicular to the plane of the dees exerts a force of the particles that is given by F = q (V x B). This cause the particles to follow a circular trajectory. As’ they reach the dees the polarity is reversed and the particles are once more accelerated. This continues and highly energetic beams of charged particles are obtained. Magnetic field is perpendicular to dees.

Hence, the time period is independent of speed and radius of circular path.
(ii) Resonance condition : When the angular frequency of the rotating charged particle and the angular frequency of the
alternating voltage applied across the dees of the cyclotron match. The charged particles are only accelerated by the electric field. The magnetic field only keeps it moving along a circular tra.ck. The acceleration happens when the charged particle crosses the gap between the two dees. At this instant the field between the dees has to be reversed so that the electric field can accelerate the charged particle.

Question.29. (a) What are eddy currents ? Write their two applications.
(b) Figure shows a rectangular conducting loop PQSR in which arm RS of length ‘l’ is movable. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. The arm RS is moved with a uniform speed V.

Deduce an expression for
(i) The emf iriduced across the arm ‘RS’,
(ii) The external force required to move the arm, and
(iii) The power dissipated as heat.
OR
(a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy”. Justify this statement.
(b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.
Answer : (a) When a bulk piece of conductor is subjected to changing magnetic flux, the induced current developed in it is called eddy current.
Applications of eddy currents :

1.  Magnetic brakes in trains.
2. Electromagnetic damping.
3.  Induction furnaces.
4. Electric power meter.

(b) (i) The emf induced across the arm B lv
On account of the presence of the magnetic field, there will be a force on the arm RS. This force I (lx B), is directed outwards in the direction opposite to the velocity of the arm RS. The magnitude of this force is,

(a) Lenz’s Law : An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.
Lenz’s law is shown with the negative sign in Faraday’s law of induction :

Lenz’s law states that the current induced in a circuit due to a change or a motion in a magnetic field is so directed as to oppose the change in flux or to exert a mechanical force opposing the motion.
Explanation of conservation of energy : When a magnet is moved near a current carrying coil, the direction of the induced current opposes the motion of the magnet. When the north pole of the magnet is moved towards the coil, the induced current flows in a direction so that near face of the coil acts as a magnetic north pole. The repulsion between two poles opposes the motion of the magnet towards the coil. Similarly, when the north pole of the magnet is from the coil, the direction of the induced current is such as to make the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil. In either case, therefore, work has to be done in moving the magnet. This mechanical work appears as electrical energy in the coil. Thus Lenz’s law is in accordance with the principle of conservation of energy.
(b) Mutual Inductance of two long coaxial solenoids : Consider the following fig. which shows two long co-axial solenoids each of length l. Let the radius of the inner solenoid S₁be r₁ and the number of turns per unit length be n₁.

Question.30. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object, (ii) Write three distinct advantages of a reflecting type telescope
(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system.
OR
(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror.
(b) The near point of a hypermetropic person is 50 cm
from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

(ii)Advantage of reflecting type telescope over a refracting
type telescope are :

1.  Lenses suffer from chromatic aberrations that are not there in mirrors.
2.  Lenses also have spherical aberration; a parabolic mirror will be free of spherical aberration.
3.  It is easier to support large mirrors as the back surface is nonreflecting, but a lens needs support around its rim.

(a) Mirror forumula for concave Mirror :


(b) The near point of the eye of this person is at 50 cm. The required lens must have a focal length such that the virtual image of the book placed at 25 cm is formed at 50 cm. Then image can be focussed by the eye.

SET II

Note: Except for the following questions, all the remaining questions have been asked in Set-I, 2009.

Question.1. What is work done in moving a test charge q through a distance of 1 cm along the equatorial axis of electric dipole?
Answer : Since potential difference for equipotential surface
ΔV = 0
Work done in moving a positive test charge q through a distance 1 cm is
w= qΔV = q x 0= 0

Question.5. Two thin lenses of power +4 D and — 2D are in contact. What is focal length of combination?

Question.6. Give the logic symbol of NAND gate?

Question.7.Two nuclei have mass number in the ratio 8 the ratio of their nuclear radii?

Question.8.The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?

Question.9.

1. State the principle on which the working of an optical fiber is based.
2.  What are the necessary conditions for this phenomenon
   to occur?

Answer :

1. The phenomenon of reflection of light when the light travelling in a denser medium strike* the interface separating the denser medium and the rarer medium at an angle greater than the critical angle is called total internal reflection.
2.  Conditions :
   (a) Light must travel from a denser medium to rarer medium.
   (b) The angle of incidence in the denser medium must be greater than the critical angle for the two media in contact.

Question.10. (i) State the law that gives the polarity of the induced emf. (ii) A 15.0μF capacitor is connected to 220 V, 50 Hz source.
Find the capacitive reactance and the rms current.
Answer :
(i) Lenz’s law : According to this law direction of induced emf Or current in a circuit is such that it opposes the cause or change that produces it.

Question.23. Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively.
OR
(a) A charge +Q is placed on a huge spherical conducting shell
of radius R. Another small conducting sphere of radius r carrying charge is introduced inside the laige shell and is placed at its centre. Find the potential difference between two points, one lying on the sphere and the other on the shell.
(b) How would the charge between the two flow if they are
connected by a conducting wire? Name the device which works on this fact.
Answer : Electric field due to a uniformly charged infinite plane sheet: Suppose a thin non-conducting uniform surface

Electric field intensity E on the either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.
Let P be any point at a distance r from the sheet. Let the small area elements are perpendicular on the surface of the imagined cylinder, so electric flux is zero. are parallel on the two cylindrical edges P and Q, which contributes electric flux.
.^.. Electric flux over the edges P and Q of the cylinder is




(b) When both spheres are connected by a conducting wire then charge q on smaller sphere wifi flow onto the large sphere having charge Q. ,
Van de Graff generator works on this fact that charge to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

SET III

Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-II, 2009.

1.Define the term ‘potential energy’ of charge q’ at a distance ‘r’ in an external electric field.
Answer: Potential energy of a single charge ‘q’ ara distance Y in an external field = q.V (r)

Question.4.The stopping potential in an experiment on photoelectric effect is 2V What is the maximum kinetic energy of the photoelectrons emitted?

Question.5.Two thin lenses of power + 5D and —2.5 D are in contact: What is the focal length of the combination?
Answer :

Question.6.How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen in halved?

Question.7. Give the logic symbol of AND gate.
:

Question.8.Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of the their nuclear radii?

Question.11. (i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.

Question.16. A wire of 20 Ω resistance is gradually stretched” to double its original length. It is then cut into two equal parts. These parts are then connected in. parallel across a 4.0 volt battery. Find the current drawn from the battery.
Answer: When any resistor is stretched to double its original length. The new resistance becomes four times of its original resistance.

Question.25. Explain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. Draw its I-V characteristics.
Answer : Zener diode is fabricated such that both the p-type and the n-type are highly doped. This makes the depletion region thin. When an electric field is applied, a high electric field appears across the thin depletion region. When the electric field becomes very high, it knocks off electrons from the host atoms to create a large number of electrons. This results a large value of current inside the circuit.

Zener diode has a sharp breakdown voltage and this property of zener is used for voltage regulation.

Question.27. Define the activity of a radionuclide. Write its S.I. unit. Give a plot of the activity of a radioactive species versus time.
How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8th of its initial value?
Answer : The total decay rate R of one or more radionuclide is called the activity of that sample. The S.I. unit for activity is becquerel.
1 becquerel = 1 Bq = 1 decay per second
Graph: The graph between the activity of a radioactive species and time is :