Category: Class 12

Students can Download Tamil Nadu 12th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 3 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identity the mismatched pair regarding the anther walls.
(a) Epidermal layer – Protective in function
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 2.
Extra nuclear inheritance is a consequence of presence of genes in _______.
(a) Mitochondria and chloroplasts
(b) Endoplasmic reticulum and mitochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(b) Endoplasmic reticulum and mitochondria

Question 3.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(c) 9 cM

Question 4.
Plasmids are _______.
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 5.
Solar energy used by green plants for photosynthesis is only ________.
(a) 2 – 8%
(b) 2 – 10%
(c) 3 – 10%
(d) 2- 9%
Answer:
(b) 2 – 10%

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is _____.
(a) over hunting and poaching
(b) green house effect
(c) competition and predation
(d) habitat destruction
Answer:
(d) habitat destruction

Question 7.
Assertion (A): Genetic variation provides the raw material for selection.
Reason (R): Genetic variations are differences in genotypes of the individuals.
(a) (A) is right and (R) is wrong
(b) (A) is wrong and (R) is right
(c) Both (A) and (R) are right
(d) Both (A) and (R) are wrong
Answer:
(c) Both (A) and (R) are right

Question 8.
Groundnut is native of ______.
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1 ,
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 10.
What are the materials used to grow microorganism like Spirulina.
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Question 11.
Compare Redifferentiation with Dedifferentiation.
Answer:
Redifferentiation:
A process by which an already differentiated cell undergo further differentiation to form another type of cell.

Dedifferentiation:
A process of reversion of cells (differentiated cells) to meristematic cells leading to formation of callus.

Question 12.
Loamy soil is ideal for crop cultivation – Justify.
Answer:
Loamy soil is ideal soil for cultivation, since it consists of 70% sand and 30% clay or silt or both. It ensures good retention and proper drainage of water. The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 13.
Mention any four environmental benefits of Rain Water Harvesting.
Answer:

• Promotes adequacy of underground water and water conservation.
• Mitigates the effect of drought.
• Improves groundwater quality and water table / decreases salinity.
• Reduces soil erosion as surface run-off water is reduced.

Question 14.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
How the flowers of salvia are adopted for mellitophily?
Answer:
Pollination in Salvia (Lever mechanism): The flower of Salvia is adapted for Bee pollination. The flower is protandrous and the corolla is bilabiate with 2 stamens. A lever mechanism helps in pollination. Each anther has an upper fertile lobe and lower sterile lobe which is separated by a long connective which helps the anthers to swing freely. When a bee visits a flower, it sits on the lower lip which acts as a platform.

It enters the flower to suck the nectar by pushing its head into the corolla. During the entry of the bee into the flower the body strikes against the sterile end of the connective. This makes the fertile part of the stamen to descend and strike at the back of the bee. The pollen gets deposited on the back of the bee. When it visits another flower, the pollen gets rubbed against the stigma and completes the act of pollination in Salvia.

Question 16.
What is the difference between mis-sense mutation and non-sense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
Give an account on germplasm conservation. .
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank and DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 18.
How are microbial innoculants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil. They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose.

They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers. .

Question 19.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Find out the molecular explanation for the wrinkled pea seeds used by Mendel.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in r allele. In the homozygous mutant form of the gene (rr) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas. The wrinkled seed accumulates more sucrose and high water content.

Hence the osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.

[OR]

(b) Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants. .

Specific uses:

• Helps predicting favourable climate, for the study of spreading of disease and controlling it.
• Mapping of forest fire and species distribution.
• Tracking the patterns of urban area development and the changes in Farmland or forests over several years.
• Mapping ocean bottom and its resources.

Question 21.
(a) Explain various edaphic factors that affect vegetation.
Answer:
The important edaphic factors which affect vegetation are as follows:

1. Soil moisture: Plants absorbs rain water and moisture directly from the air.
2. Soil water: Soil water is more important than any other ecological factors affecting the distribution of plants. Rain is the main source of soil water. Capillary water held between pore spaces of soil particles and angles between them is the most important form of water available to the plants.
3. Soil reactions: Soil may be acidic or alkaline or neutral in their reaction. pH value of the soil solution determines the availability of plant nutrients. The best pH range of the soil for cultivation of crop plants is 5.5 to 6.8.
4. Soil nutrients: Soil fertility and productivity is the ability of soil to provide all essential plant nutrients such as minerals and organic nutrients in the form of ions.
5. Soil temperature: Soil temperature of an area plays an important role in determining the geographical distribution of plants. Low temperature reduces use of water and solute absorption by roots.
6. Soil atmosphere: The spaces left between soil particles are called pore spaces which contains oxygen and carbon-di-oxide.
7. Soil organisms: Many organisms existing in the soil like bacteria, fungi, algae, protozoans, nematodes, insects and earthworms, etc., are called soil organisms.

[OR]

(b) Describe the procedure involved in Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme β-galactosidase and contains several recognition sites for restriction enzyme.

β-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- β-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated. Therefore, no-blue colour will develop (white) because p-galactosidase is not synthesized due to inactivation of lacZ.

Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify the proper sequence.
(a) juvenile phase, senescent phase, vegetative phase
(b) juvenile phase, maturity phase, senescent phase
(c) vegetative phase, maturity phase, juvenile phase
(d) senescent phase, juvenile phase, vegetative phase
Answer:
(b) juvenile phase, maturity phase, senescent phase

Question 2.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 3.
Darwin’s finches are an excellent example of _______.
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) geographical isolation
Answer:
(c) adaptive radiation

Question 4.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogenes
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 5.
Which of the following microorganism is used for the production of citric acid in industries?
(a) Lactobacillus bulgaris
(b) Penicillium citrinum
(c) Aspergillus niger
(d) Rhizopus nigricans
Answer:
(c) Aspergillus niger

Question 6.
The interaction in nature, where one gets benefit on the expense of other is __________.
(a) Predation
(b) Mutualism
(c) Amensalism
(d) Commensalism
Answer:
(d) Commensalism

Question 7.
Which of the following region has maximum bio-diversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 8.
The thickness of stratospheric ozone layer is measured in ____________.
(a) Sieverts units
(b) Melson units
(c) Dobson units
(d) Beaufort Scale
Answer:
(c) Dobson units

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What is parthenogenesis? Give two examples from animals.
Answer:
Development of an egg into a complete individual without fertilization is known as parthenogenesis. It was first discovered by Charles Bonnet in 1745.
E.g. Honey bees, Aphis.

Question 10.
Mention the production site and action site of following hormones.
(a) GnRH (b) Relaxin
Answer:

Hormone	Production Site	Action Site
GnRH	Hypothalamus	Pituitary gland
Relaxin	Placenta	Pelvic joints and cervix

Question 11.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 12.
State Van’t Hoff’s rule.
Answer:
Van’t Hoff’s rule states that with the increase of every 10°C, the rate of metabolic activity is doubled or the reaction rate is halved with the decrease of 10°C.

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 14.
List any four adverse effect of noise.
Answer:

1. High blood pressure
2. Stress related ailments
3. Sleep disruption
4. Hearing impairment

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Ban- body.

Question 16.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

• The structural gene codes for proteins, rRNA and tRNA required by the cell.
• Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
• The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 17.
Explain the principles of Lamarckian theory.
Answer:

• The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
• The theory of inheritance of acquired characters – Characters that are developed during the life time of an organism are called acquired characters and these are then inherited.

Question 18.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid.
Answer:

Question 19.
ELISA is a technique based on the principles of antigen-antibody reactions. Can this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative result.

[Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of Phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) The following is the illustration of the sequence of ovarian events (a-i) in a human female.

(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) A- Primordial follicle; B- Primary follicle; C- Secondary follicle; D-Tertiary follicle; E- Mature graafian follicle; F- Ovulation (release of egg); G- Empty Graafian follicle; H- Corpus luteum; I – Corpus albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle
H – Corpus luteum
During development of ovum, the primary follicle gets surrounded by many layers of granular cells and forms a new layer called secondary follicle.

Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

[OR]

(b) Write short notes on the following.
(i) Brewer’s yeast
(ii) Ideonella sakaiensis
(iii) Microbial fuel cells
Answer:
(i) Brewer’s yeast – Saccharomyces cerevisiae is a widely used fungal species in preparation and softening of bakery products like dough.

(ii) Ideonella sakaiensis is a bacterium is used to recycle PET plastics. The enzyme PETase and MHETase in the bacterium breakdown the PET plastics into terephthalic acid and ethylene glycol.

(iii) A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around.

A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.

Question 21.
(a) Write the salient features of Human Genome Project.
Answer:

• Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
• An average gene consists of3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
• Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
• The chromosomal organization of human genes shows diversity. .
• There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes, whereas chromosome ‘Y’ has 231 genes.
• Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

[OR]

(b) Tropical regions are rich in biodiversity. Why?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

• Warm tropical regions between the tropic of Cancer and Capricorn on either side of equator possess congenial habitats for living organisms.
• Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
• The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
• The average rainfall is often more than 200 mm per year.
• Climate, seasons, temperature, humidity, photo periods are more or less stable and
  encourage both variety and numbers. .
• Rich resource and nutrient availability.

Students can Download Tamil Nadu 12th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 2 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 2.
pBR 322, BR stands for ________.
(a) Plasmid Bacterial Recombination
(b) Plasmid Bacterial Replication
(c) Plasmid Boliver and Rodriguez
(d) Plasmid Baltimore and Rodriguez
Answer:
(c) Plasmid Boliver and Rodriguez

Question 3.
Which of the given plant produces cardiac glycosides?
(a) Calotropis
(b) Acacia
(c) Nepenthes
(d) Utricularia
Answer:
(a) Calotropis

Question 4.
Methane is _______ times as effective as CO₂ at trapping heat.
(a) 5
(b) 10
(c) 20
(d) 100
Answer:
(c) 20

Question 5.
Statement 1: Arachis hypogea belongs to Fabaceae
Statement 2: It is a native of Brazil.
(a) Statement 1 is correct and Statement 2 is incorrect
(b) Statement 1 is incorrect and Statement 2 is correct
(c) Both the Statements are correct
(d) Both the Statements are incorrect
Answer:
(c) Both the Statements are correct

Question 6.
Which of the following scientist developed world’s first cotton hybrid?
(a) Dr. B.P. Pal
(b) C.T. Patel
(c) Dr. K. Ramiah
(d) N.G.P. Rao
Answer:
(b) C.T. Patel

Question 7.
Ecosystem is the structural and functional unit of ecology. This statement was given by _______.
(a) Tansley
(b) Odum
(c) Charles Elton
(d) Edwin
Answer:
(b) Odum

Question 8.
Name the scientist(s) who rediscovered the Mendelian work? .
(i) Hugo de Vries (ii) Carl Correns (iii) Tschermak (iv) T.H. Morgan
(a) (i), (ii)and (iv) (b) (i), (ii)and (iii) (iii) ii, iii, iv (d) i, iii and iv
Answer:
(b) (i), (ii)and (iii)

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Detached leaf of Bryophyllum produces new plants. How?
Answer:
In Bryophyllum, the leaf is succulent and notched on its margin. Adventious buds develop at these notches and are called epiphyllous buds. They develop into new plants forming a root system and become independent plants when the leaf gets decayed.

Question 10.
Crossing over occurs only in germinal cells. Yes or No? Support your answer.
Answer:
No. Though crossing over is a common process in germinal cells rarely it also occurs in somatic cells during mitosis. Such crossing over is called mitotic crossing over or somatic crossing over.

Question 11.
Why do we use cryoprotectants in conservation process? Name any two cryoprotectant.
Answer:
Cryoprotectants are the protective agents that are used to protect the cells or tissues from the stress of freezing temperature. E.g: Sucrose, dimethyl sulphoxide.

Question 12.
What is Eltonian pyramid?
Answer:
Eltonian pyramid or Ecological pyramid is a graphic representation of the trophic structure and function at successive trophic levels of an ecosystem.

Question 13.
What do you understand by the term good ozone and bad ozone?
Answer:
The ozone layer of troposphere is called bad ozone.
The ozone layer of stratosphere is called good ozone because the layer acts as an shield for absorbing UV rays coming from sun which is harmful for living organisms causing ONA damage.

Question 14.
A person got irritation while applying chemical dye. What would be your suggestion for alternative?
Answer:
If a grey haired person is allergic on using chemical dyes then he can go for natural dyes like Henna. Henna is an organic dye obtained from leaves and young shoots of Lawsonia inermis. The principal colouring matter is Tacosone’ which is harmless and causes no irritation on skin.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Differentiate between incomplete dominance and codominance.
Answer:
Incomplete Dominance:

• In incomplete dominance, neither of the allele is not completely dominant to another allele rather combine and produce new trait.
• New phenotype is formed due to character blending (not alleles)
• Example : Pink flowers of Mirabilis Jalapa

Co-dominance:

• In co-dominance, both the alleles in heterozygote are dominant and the traits are equally expressed (joint expression)
• No formation of new phenotype rather both dominant traits are expressed, conjointly.
• Example : Red and white flowers of camellia.

Question 16.
What are the features that a vector must possess to facilitate cloning?
Answer:
The following are the features that are required to facilitate cloning into a vector.

• Origin of replication (ori): This is a sequence from where replication starts and piece of DNA when linked to this sequence can be made to replicate within the host cells.
• Selectable marker: In addition to ori the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.
• Cloning sites: In order to link the alien DNA, the vector needs to have very few, preferably single, recognition sites for the commonly used restriction enzymes.

Question 17.
Define somatic embryogenesis? Give any two of its applications.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids.

Applications:

• Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
• Somatic embryoids can be used for the production of synthetic seeds.

Question 18.
What is TSM? How it is classified and what does it focuses on?
Answer:
TSM stands for Traditional Systems of Medicines India has a rich medicinal heritage. A number of Traditional Systems of Medicine (TSM) are practiced in India some of which come from outside India. TSM in India can be broadly classified into institutionalized or documented and non-institutionalized or oral traditions. Institutionalized Indian systems include Siddha and Ayurveda which are practiced for about two thousand years.

These systems have prescribed texts in which the symptoms, disease diagnosis, drugs to cure, preparation of drugs, dosage and diet regimes, daily and seasonal regimens. Non-institutional systems, whereas, do not have such records and or practiced by rural and tribal peoples across India. The knowledge is mostly held in oral form. The TSM focus on healthy lifestyle and healthy diet for maintaining good health and disease reversal

Question 19.
How do sacred groves help in the conservation of biodiversity?
Answer:
These are the patches or grove of cultivated trees which are community protected and are based on strong religious belief systems which usually have a significant religious connotation for protecting community. Each grove is an abode of a deity mostly village God Or Goddesses like Aiyanar or Amman. 448 grooves were documented throughout Tamil Nadu, of which 6 groves (Banagudi shola, Thirukurungudi and Udaiyankudikadu, Sittannnavasal, Puthupet and Devadanam) were taken up for detailed floristic and faunistic studies.

These groves provide a number of ecosystem services to the neighbourhood like protecting watershed, fodder, medicinal plants and micro climate control.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Draw a flow chart depicting the various types of ploidy.
Answer:

[OR]

(b) Mention the application of biotechnology.
Answer:

• Biotechnology is one of the most important applied interdisciplinary sciences of the 21st century. It is the trusted area that enables us to find the beneficial way of life.
• Biotechnology has wide applications in various sectors like agriculture, medicine, environment and commercial industries.
• This science has an invaluable outcome like transgenic varieties of plants e.g. transgenic cotton (Bt-cotton), rice, tomato, tobacco, cauliflower, potato and banana.
• The development of transgenics as pesticide resistant, stress resistant and disease resistant varieties of agricultural crops is the immense outcome of biotechnology.
• The synthesis of human insulin and blood protein in E.coli and utilized for insulin deficiency disorder in human is a breakthrough in biotech industries in medicine.
• The synthesis of vaccines, enzymes, antibiotics, dairy products and beverages are the products of biotech industries.
• Biochip based biological computer is one of the successes of biotechnology.
• Genetic engineering involves genetic manipulation, tissue culture involves aseptic cultivation of totipotent plant cell into plant clones under controlled atmospheric conditions.
• Single cell protein from Spirulina is utilized in food industries.
• Production of secondary metabolites, biofertilizers, biopesticides and enzymes.
• Biomass energy, biofuel, bioremediation and phytoremediation for environmental biotechnology.

Question 21.
(a) Differentiate Primary succession and Secondary succession.
Answer:
Primary succession:

• Developing in an barren area.
• Initiated due to a biological or any other external factors.
• No soil, while primary succession starts
• Pioneer species come from outside environment.
• It takes more time to complete.

Secondary succession:

• Developing in disturbed area.
• Starts due to external factors only.
• It starts where soil covers is already present.
• Pioneer species develop from existing environment.
• It takes comparatively less time to complete.

(b) Enumerate the characters of anemophilous flowers
Answer:
Anemophilous plants have the following characteristic features:

• The flowers are produced in pendulous, catkin-like or spike inflorescence.
• The axis of inflorescence elongates so that the flowers are brought well above the leaves.
• The perianth is absent or highly reduced.
• The flowers are small, inconspicuous, colourless, not scented, do not secrete nectar.
• The stamens are numerous, filaments are long, exerted and versatile.
• Anthers produce enormous quantities of pollen grains compared to number of ovules available for pollination. They are minute, light and dry so that they can be carried to long distances by wind.
• In some plants anthers burst violently and release the pollen into the air. Example: Urtica.
• Stigmas are comparatively large, protruding, sometimes branched and feathery, adapted to catch the pollen grains. Generally single ovule is present.
• Plant produces flowers before the new leaves appear, so the pollen can be carried without hindrance of leaves.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only?
(a) Syphilis, gonorrhoea and candidiasis
(b) Syphilis, chlamydiasis and gonorrhoea
(c) Syphilis, gonorrhoea and trichomoniasis
(d) Syphilis, trichomoniasis and pediculosis
Answer:
(b) Syphilis, chlamydiasis and gonorrhoea

Question 2.
ABO blood group in man is controlled by __________.
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Hershey and Chase experiment with bacteriophage showed that ____________.
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 4.
Modern man belongs to which period?
(a) Quaternary
(b) Cretaceous
(c) Silurian
(d) Cambrian
Answer:
(a) Quaternary

Question 5.
Allergy involves _______.
(a) IgE
(b) IgG
(c) IgA
(d) IgM
Answer:
(a) IgE

Question 6.
Recombinant Factor VIII is produced in the ______ of the Chinese Hamster.
(a) Liver cells
(b) blood cells
(c) ovarian cells
(d) brain cells
Answer:
(c) ovarian cells

Question 7.
Match with correct pair

(a) A – ii, B – iv, C – i, D – iii
(b) A – ii, B – iii, C – i, D – iv
(d) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – ii, C – i, D – iii
Answer:
(b) A – ii, B – iii, C – i, D – iv

Question 8.
Assertion (A): Snake is a stenotherm.
Reason (R): Organism can tolerate narrow temperature fluctuations.
(a) Both A and R are correct R explain A
(b) A is correct R is incorrect
(c) A and R are correct, R doesn’t explains A
(d) Both A and R are incorrect
Answer:
(a) Both A and R are correct R explain A

PART – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Mention the differences between spermiogenesis and spermatogenesis.
Answer:
Spermiogenesis:
Transformation of spermatids into mature sperm.

Spermatogenesis:
Spermatogenesis is the sequence of events in the seminiferous tubules of testes that produces male gametes, the sperms.

Question 10.
Draw and label a gemmule of sponge.
Answer:

Question 11.
State Wiener’s Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R², R⁰, R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 12.
Why t-RNA is called an adapter molecule?
Answer:
The transfer RNA (t-RNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule. This term was postulated by Francis Crick.

Question 13.
Why Red list is prepared periodically?
Answer:
The purpose of preparation of Red List are:

• To create awareness on the degree of threat to biodiversity
• Identification and documentation of species at high risk of extinction
• Provide global index on declining biodiversity .
• Preparing conservation priorities and help in conservation of action
• Information on international agreements on conservation of biological diversity

Question 14.
Write notes on (a) Eutrophication (b) Algal Bloom
Answer:
Eutrophication refers to the nutrient enrichment in water bodies leading to lack of oxygen and will end up in the death of aquatic organisms.
Algal Bloom is an excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Explain Foetal-ejection reflex.
Answer:
As the pregnancy progresses, increase in the oestrogen concentration promotes uterine contractions. These uterine contractions facilitate moulding of the foetus and downward movement of the foetus. The descent of the foetus causes dilation of cervix of the uterus and vaginal canal resulting in a neurohumoral reflex called Foetal ejection reflex or Ferguson reflex.

This initiates the secretion of oxytocin from the neurohypophysis which in turn brings about the powerful contraction of the uterine muscles and leads to the expulsion of the baby through the birth canal.

Question 16.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 17.
State immunological surveillance theory.
Answer:
The concept of immunological surveillance postulates that the primary function of the immune system is to “seek and destroy” malignant cells that arise by somatic mutation. The efficiency of the surveillance mechanism reduces either as a result of ageing or due to congenital or acquired immuno deficiencies, leads to increased incidence of cancer. Thus, if immunological surveillance is effective, cancer should not occur. The development of tumour represents a lapse in surveillance.

Question 18.
Write a short note on Pathaneer.
Answer:
In some parts of South India, a traditional drink called pathaneer is obtained from fermenting sap of palms and coconut trees. A common source is tapping of unopened spadices of coconut. It is a refreshing drink, which on boiling produces jaggery or palm sugar. When pathaneer is left undisturbed for few hours it gets fermented to form toddy with the help of naturally occurring yeast, to form a beverage that contains 4 percent alcohol. After 24 hours toddy becomes unpalatable and is used for the production of vinegar.

Question 19.
Elucidate the methodology of ELISA test.
Answer:
During diagnosis the sample suspected to contain the antigen is immobilized on the surface of an ELISA plate. The antibody specific to this antigen is added and allowed to react with the immobilized antigen. The anti-antibody is linked to an appropriate enzyme like peroxidase.

The unreacted anti-antibody is washed away and the substrate of the enzyme (hydrogen peroxidase) is added with certain reagents such as 4-chloronaphthol. The activity of the enzyme yields a coloured product indicating the presence of the antigen.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) What are IUD’s? Explain its way of functioning. Also describe their types.
Answer:
Intrauterine Devices (IUDs) are inserted by medical experts in the uterus through the vagina. These devices are available as copper releasing IUDs, hormone releasing IUDs and nonmedicated IUDs. IUDs increase phagocytosis of sperm within the uterus. IUDs are the ideal contraceptives for females who want to delay pregnancy. It is one of the popular methods of contraception in India and has a success rate of 95 to 99%.

Copper releasing IUDs differ from each other by the amount of copper. Copper IUDs such as Cu T-380 A, Nova T, Cu 7, Cu T 380 Ag, Multiload 375, etc. release free copper and copper salts into the uterus and suppress sperm motility. They can remain in the uterus for five to ten years.

Hormone-releasing IUDs such as Progestasert and LNG – 20 are often called as intrauterine systems (IUS). They increase the viscosity of the cervical mucus and thereby prevent sperms from entering the cervix.

Non-medicated IUDs are made of plastic or stainless steel. Lippes loop is a double S-shaped plastic device.

[OR]

(b) Explain how Urey – Miller’s experiment supports the origin of life?
Answer:

Urey and Miller (1953), paved way for understanding the possible synthesis of organic compounds that led to the appearance of living organisms is depicted in the Figure In their experiment, a mixture of gases was allowed to circulate over electric discharge from an tungsten electrode.

A small flask was kept boiling and the steam emanating from it was made to mix with the mixture of gases (ammonia, methane and hydrogen) in the large chamber that was connected to the boiling water. The steam condensed to form water which ran down the ‘U’ Diagrammatic representation of Urey-Miller’s experiment tube. Experiment was conducted continuously for a week and the liquid was analysed. Glycine, alanine, beta alanine and aspartic acid were identified.

Thus Miller’s experiments had an insight as to the possibility of abiogenetic synthesis of large amount of variety of organic compounds in nature from a mixture of sample gases in which the only source of carbon was methane. Later in similar experiments, formation of all types of amino acids, and nitrogen bases were noticed.

Question 21.
(a) Discuss any two Allosomal abnormalities in human beings.
Answer:
Allosomal abnormalities in human beings:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities.
Several sex chromosomal abnormalities have been detected. E.g. Klinefelter’s syndrome and
Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males):
This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47, XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females):
This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45 ,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome:

[OR]

(b) Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
1. Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than the large population due to inbreeding depression (less adaptivity and variation)

2. Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow water marine invertebrates disappeared.

3. Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, over exploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction.

The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Students can Download Tamil Nadu 12th Tamil Model Question Paper 4 Pdf, Tamil Nadu 12th Tamil Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Tamil Model Question Paper 4

நேரம்: 2.30 மணி 
மதிப்பெண்கள் : 90

குறிப்புகள்:

• இவ்வினாத்தாள் ஐந்து பகுதிகளைக் கொண்டது. அனைத்து பகுதிகளுக்கும் விடையளிக்க வேண்டும். தேவையான இடங்களில் உள் தேர்வு வினாக்கள் கொடுக்கப்பட்டுள்ளது.
• பகுதி I, II, III, IV மற்றும் Vல் உள்ள அனைத்து வினாக்களுக்கும் தனித்தனியே விடையளிக்க வேண்டும்.
• வினா எண் 1 முதல் 14 வரை பகுதி-1ல் தேர்வு செய்யும் வினாக்கள் தரப்பட்டுள்ளன.
  ஒவ்வொரு வினாவிற்கும் ஒரு மதிப்பெண். சரியான விடையைத் தேர்ந்தெடுத்துக் குறியீட்டுடன் விடையினையும் சேர்த்து எழுதவும்.
• வினா எண் 15 முதல் 30 வரை பகுதி-பால் இரண்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன.
• வினா எண் 31 முதல் 43 வரை பகுதி-IIIல் நான்கு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன.
• வினா எண் 44 முதல் 46 வரை பகுதி-IVல் ஆறு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன.அனைத்து வினாவிற்கும் விடையளிக்கவும்.
• வினா எண் 47-ல் பகுதி-Vல் மனப்பாடப்பகுதி தரப்பட்டுள்ளன.

பகுதி – I

அனைத்து வினாக்களுக்கும் விடை தருக. [14 × 1 = 14]

(விடைகள் தடித்த எழுத்தில் தரப்பட்டுள்ளன)

Question 1.
சாபவிமோசனம், அகலிகை கதைகளில் தொன்மங்களைப் பயன்படுத்திய எழுத்தாளர்…………………
(அ) டாக்டர் உதயமூர்த்தி
(ஆ) கல்கி
(இ) புதுமைப்பித்தன்
(ஈ) புலமைப்பித்தன்
Answer:
(இ) புதுமைப்பித்தன்

Question 2.
‘முருகு உறழ் முன்பொடு’ என்ற தொன்மம் விளக்கும் பாடல் …………… நூலில் இடம் பெற்றுள்ளது. (அ) நற்றிணை
(ஆ) நல்ல குறுந்தொகை
(இ) நன்னூல்
(ஈ) தண்டியலங்காரம்
Answer:
(அ) நற்றிணை

Question 3.
பல சிற்றுறுப்புகளின் தொகுதியாக உள்ள பேருறுப்பைக் குறிப்பது ………………….
(அ) காதை
(ஆ) சருக்கம்
(இ) காண்டம்
(ஈ) படலம்
Answer:
(இ) காண்டம்

Question 4.
வள்ளலார் சமரச சன்மார்க்க சபையை நிறுவிய இடம். ………….
(அ) வண்டலூர்
(ஆ) வடக்கூர்
(இ) வண்டியூர்
(ஈ) வடலூர்
Answer:
(ஈ) வடலூர்

Question 5.
…………… ஆண்டின் மக்கள் தொகைக் கணக்கெடுப்பின் படி சென்னை நகரின் மக்கட்தொகை 1900
ஆகும்.
(அ) 1646
(ஆ) 1746
(இ) 1846
(ஈ) 1846
Answer:
(அ) 1646

Question 6.
ஓதற்பிரிவிற்கு உரிய காலம் ………… ஆண்டுகள்.
(அ) இரண்டு
(ஆ) மூன்று
(இ) நான்கு
(ஈ) ஐந்து
Answer:
(ஆ) மூன்று

Question 7.
கவிஞர் நகுலனின் இயற்பெயர் ……….
(அ) பாலசந்திரன்
(ஆ) இந்திரன்
(இ) சாமிக்கண்ணு
(ஈ) துரைச்சாமி
Answer:
(ஈ) துரைச்சாமி

Question 8.
………….. நூலுக்கு இளம்பூரணர்தான் முழுமையான உரை தந்துள்ளார்.
(அ) தொல்காப்பியம்
(ஆ) நன்னூல்
(இ) மாறனலங்காரம்
(ஈ) தண்டியலங்காரம்
Answer:
(அ) தொல்காப்பியம்

Question 9.
தமிழ்ச் சமூகம், பண்பாடு சார்ந்த மானிடவியல் ஆய்வுகளை முன்னெடுத்து………………… வருபவர் ….
(அ) தாராபாரதி
(ஆ) பக்தவத்சல பாரதி
(இ) பாரதி சண்முகம்
(ஈ) சுத்தானந்த பாரதி
Answer:
(ஆ) பக்தவத்சல பாரதி

Question 10.
‘கூடி வாழ்தல்’ என்ற பொருளில் வருவது……………………
(அ) குடும்பு
ஆ) கடும்பு
(இ) குடம்பை
(ஈ) கடம்பை
Answer:
(அ) குடும்பு

Question 11.
தேடலை விரிவாக்குவது என்பர்…………………
(அ) புகழ்
(ஆ) செல்வம்
(இ) கல்வி
(ஈ) உறவு
Answer:
(இ) கல்வி

Question 12.
அகநானூற்றின் களிற்றியானை நிரையில் உள்ள பாடலின் எண்ணிக்கை ………….. ஆகும்.
(அ) 100
(ஆ) 120
(இ) 180
(ஈ) 110
Answer:
(ஆ) 120

Question 13.
புனவன் என்பது……………………….. ஐக் குறித்தது.
(அ) மீனவன்
(ஆ) பாடகன்
(இ) நூலகன்
(ஈ) கானவன்
Answer:
(ஈ) கானவன்

Question 14.
ஓட்டுநர் உரிமம் இன்றி ஊர்தியை இயக்கினால் சிறைத்தண்டனையோ அல்லது……… அபராதமோ
விதிக்கப்படும்.
(அ) ரூ. 2000
(ஆ) ரூ. 3000
(இ) ரூ. 40000
(ஈ) ரூ. 5000
Answer:
(ஈ) ரூ. 5000

பகுதி – II

இரண்டு அல்லது மூன்று வரிகளில் விடை தருக.[12 x 2 = 24]

பிரிவு – 1

எவையேனும் மூன்றனுக்கு விடை தருக.

Question 15.
வாகைத் திணை என்றால் என்ன?
Answer:
வெற்றி பெற்ற அரசனும் அவனது வீரர்களும் வெற்றியின் அடையாளமாக வாகைப்பூவைச் சூடி
வெற்றியைக் கொண்டாடுவது வாகைத் திணையாகும்.

Question 16.
நும் பாடப்பகுதியில் கொடுக்கப்பட்டுள்ள அகநானூற்று பாடலில் உள்ளுறைப் பொருளை எழுதுக.
Answer:

• வண்டியை இழுக்கும் எருதுகளின் துன்பத்தைத் தந்தை போக்கியது.
• தலைவியைக் கண்டதனால் எனக்கேற்பட்ட துன்பத்தை நீ போக்குதற்கு உரியவன் என்று தலைவன் பாங்கனிடம் உள்ளுறுத்துக் கூறினான்.
• எருதைத் தலைவனுக்கும் தந்தையைப் பாங்கனுக்கும் உப்பின் எடையால் எருது வருந்தும் நிலையைக் காதல் வருத்தத்திற்கும் உள்ளுறையாக வைத்துப் பாடல் புனையப்பட்டுள்ளது.

Question 17.
முகம் முகவரியற்றுப் போனதற்கு, சுகந்தி சுப்ரமணியன் கூறும் காரணத்தை எழுதுக.
Answer:
பல முகங்களோடு முகம் காணும் போது எனது முகம் காணவில்லை. எனக்குள்ளே என்னைத் தொலைத்த பின் எனது முகம் முகவரியற்றுப் போனது இன்னும் என்னைத் தேடிக்
கொண்டிருக்கிறேன்.

Question 18.
மொழியின் இயல்பு வழக்குகளை கலையின் வழக்குகளாக மாற்றுபவை எவை?
Answer:
மொழியின் இயல்பு வழக்குகளைக் கலையியல் வழக்குகளாக மாற்றுபவை.

• உவமம்
• உருவகம்
• எச்சம்
• குறிப்பு
• உள்ளுறை
• இறைச்சி

பிரிவு – 2

எவையேனும் இரண்டனுக்கு விடை தருக

Question 19.
வரியெழுத்தின் உறுப்புகள் யாவை?
Answer:

• புள்ளி
• கால்
• கொம்பு
• விலங்கு

முதலியவை வரியெழுத்தின் உறுப்புகள் ஆகும்.

Question 20.
“வெள்ளையர் நகரம்” “கருப்பர் நகரம்”- விளக்குக.
Answer:

• செயின்ட் ஜார்ஜ் கோட்டைக்கு உள்ளே வீடுகள் இருந்த பகுதி ‘வெள்ளையர் நகரம்’ என்று அழைக்கப்பட்டது.
• கோட்டைக்குள் இருப்பவர்களின் தேவைகளை நிறைவேற்றும் பணியாளர்கள் வணிகர்கள் போன்றோருக்காக வெளியே அமைத்த குடியிருப்புகள் உள்ள பகுதி ‘கருப்பர் நகரம்’ என அழைக்கப்பட்டது.

Question 21.
முப்பரிமாணக் கலை என்றால் என்ன?
Answer:

• திரைப்படத்தில், நடிப்பவரை முன் பின் மேல் என்று பல கோணங்களில் படப்பிடிப்புக் கருவியால் இடம் மாற்றி மாற்றிப் படம் பிடித்துத் திரையில் காட்ட முடியும்.
• திரைப்படத்தை முப்பரிமாணக் கலை என வகைப்படுத்துகிறோம்.
• ஒருவன் ஓடிவருவதைக் காட்டிவிட்டு அதைப் பார்ப்பவன் ஏன் இப்படி ஓடி வருகிறீர்கள் என்று
  கேட்கவேண்டியதில்லை. என்னாச்சு? என்று கேட்டால் போதும்.
• இது காட்சிமொழியின் தன்மை. உணர்ச்சிகளைக் காண்பிக்க முகத்தை மட்டுமே காண்பித்தால் போதும்.

பிரிவு – 3

எவையேனும் ஏழனுக்கு விடை தருக.

Question 22.
ஏதேனும் ஒன்றனுக்கு மட்டும் பகுபத உறுப்பிலக்கணம் தருக.
(அ) களைந்து (ஆ) வந்தனர்
Answer:
(அ) களைந்து = களை + த் (ந்) + த் + உ
களை – பகுதி
த் – சந்தி ; ந் – விகாரம்
த் – இறந்தகால இடைநிலை
உ – வினையெச்ச விகுதி

(ஆ) வந்தனர் = வா (வந்) + த் + அன் + அர்
வா – பகுதி ; (வந்) – விகாரம்
த் – இறந்தகால இடைநிலை
அன் – சாரியை
அர் – பல்பால் வினைமுற்று விகுதி

Question 23.
ஏதேனும் ஒன்றனுக்கு மட்டும் இலக்கணக்குறிப்பு தருக.
(அ) கருந்தடம்
(ஆ) உன்ன லிர்
Answer:
(அ) பண்புத்தொகை
(ஆ) முன்னிலைப் பன்மை வினைமுற்று

Question 24.
மரபுப்பிழை நீக்குக.
வண்டுகள் பாட குயில்கள் கத்த மயில்கள் கூவின.
Answer:
வண்டுகள் முரல் குயில்கள் கூவிட மயில்கள் அகவின.

Question 25.
ஏதேனும் ஒன்றனுக்குப் பிரித்துப் புணர்ச்சி விதி தருக.
அ) நன்மொழி
(ஆ) பத்துப்பாட்டு
Answer:
(அ) நன்மொழி = நன்மை + மொழி
விதி : ஈறுபோதல்

(ஆ) பத்துப்பாட்டு = பத்து + பாட்டு
விதி : இயல்பினும் விதியினும் நின்ற உயிர்முன் கசதப மிகும்.

Question 26.
தனித்தமிழில் எழுதுக.
விவாகத்திற்கு பந்து மித்திரர்களுடன் வருக.
Answer:
திருமணத்திற்கு உறவினர்களுடனும், நண்பர்களுடனும் வருக.

Question 27.
மயங்கொலிப் பிழையின்றி பொருள் வேறுபட ஒரே தொடரில் விடையளி.
உன் — உண்
Answer:
உணவு இடைவேளையில் காலம் தாழ்த்தாமல் உன் சாப்பாட்டைத் தவறாமல் உண்.

Question 28.
கொச்சை நீக்கிச் சரியாக எழுது.
இன்னக்கி சாயுங்காலம் தங்கச்சி வரும்.
Answer:
இன்று மாலை தங்கை வருவாள்.

Question 29.
தனித்தமிழில் எழுதுக.
போலீஸ் ஸ்டேஷன் போய் கம்ளைண்ட் கொடு.
Answer:
காவல் நிலையம் சென்று புகார் கொடு.

Question 30.
விடைக்கேற்ற வினா எழுதுக.
பொதுத்தேர்வுக்கான வினாக்கள் பாடப்பகுதியின் உட்பகுதியிலிருந்துதான் வருகின்றன.
Answer:
பொதுத்தேர்வுக்கான வினாக்கள் பாடப்பகுதியின் எப்பகுதியிலிருந்து வருகின்றன?

பகுதி – III

ஐந்து அல்லது ஆறு வரிகளில் விடை தருக. [7 x 4 = 28]

பிரிவு – 1

எவையேனும் இரண்டனுக்கு மட்டும் விடை தருக.

Question 31.
‘செம்பரிதி மலைமேட்டில் தலையைச் சாய்ப்பான் செந்நிறத்துப் பூக்காடாம் வான மெல்லாம்’ தொடர் வெளிப்படுத்தும் காட்சி நயத்தை விளக்குக.
Answer:

• செம்பரிதி மலைமேட்டில் தலையைச் சாய்ப்பான் – செம்மையான சூரியன் அதாவது மாலைப் பொழுதில் தோன்றும் சிவப்பு நிற சூரியன் மலைகளின் மேடு அதாவது மலையின் உச்சியில் சென்று மறைந்து போவான்.
• செந்நிறத்துப் பூக்காடாம் வானமெல்லாம் – செம்மையான நிறம் கொண்ட பூக்கள் போல அத்தருணத்தில் வானம் எல்லாம் சிவப்பு வண்ணமாய் அந்த மாலைப் பொழுதில் நிறம் மாறி நிற்கும்.

Question 32.
சுரதா குறிப்பு வரைக.
Answer:

• உவமைக் கவிஞர் என்று சிறப்பிக்கப்படும் சுரதாவின் இயற்பெயர் இராசகோபாலன்.
• அப்பெயரைப் பாரதிதாசன்மீது கொண்ட பற்றுதலால் சுப்புரத்தினதாசன் என்று மாற்றி, அதன் சுருக்கமான சுரதா என்னும் பெயரில் மரபுக் கவிதைகள் எழுதினார்.
• முழுக்க முழுக்கக் கவிதைகளையே கொண்ட காவியம் என்ற இதழை நடத்தியதோடு இலக்கியம், விண்மீன், ஊர்வலம் போன்ற இலக்கிய ஏடுகளையும் நடத்தியுள்ளார்.
• தேன்மழை, துறைமுகம், மங்கையர்க்கரசி, அமுதும் தேனும் உள்ளிட்ட பல நூல்களைப் படைத்துள்ளார்.
• இவர் தமிழக அரசின் கலைமாமணி விருது, பாரதிதாசன் விருது, தஞ்சைத் தமிழ்ப் பல்கலைக்கழக்தின் இராசராசன் விருது உள்ளிட்ட பல விருதுகளைப் பெற்றவர்.

Question 33.
சூதும் கள்ளும் கேடு தரும் – திருக்குறள் வழி விவரிக்க.
Answer:
“களித்தானைக் காரணம் காட்டுதல் கீழ்நீர்க்
குளித்தானைத் தீத்துரீஇ யற்று”

கள்ளுண்டு மயங்கியவனிடம் நல்லன சொல்லித் திருத்த முயல்வது, நீரில் மூழ்கிய ஒருவரைத் தீப்பந்தம் கொண்டு தேடுவது போன்றது.

“சிறுமை பலசெய்து சீர்அழிக்கும் சூதின்
வறுமை தருவது ஒன்று இல்”

ஒருவருக்குத் துன்பம் பலவற்றையும் உண்டாக்கி அவருடைய புகழையும் கெடுக்கின்ற சூதைப்போல் வறுமை தருவது வேறொன்றும் இல்லை.

Question 34.
தலைக்கோல் என்றால் என்ன? அவற்றின் சிறப்பு யாது?
Answer:

• அரங்கேற்றத்தை வெற்றிகரமாக முடிக்கும் ஆடல் மகளுக்குத் தலைக்கோல் அளித்துச் சிறப்பிப்பர்.
• தலைக்கோல் என்பது, பெரும்புகழ் கொண்ட பகை மன்னனுடன் நிகழ்த்திய போரில், தோற்றுப் புறங்காட்டிய அவனிடமிருந்து பறிக்கப்பட்ட அழகுமிக்க வெண்கொற்றக் குடையின் காம்பில் செய்யப்படுவது.
• அக்காம்பின் கணுக்கள் முழுவதும் நவமணிகளை இழைத்து அக்கணுக்களுக்கு இடைப்பட்ட பகுதிகளில் சாம்பூந்தம் எனும் பொன் தகட்டை வலம்புரியாகவும், இடம்புரியாகவும் சுற்றிக்கட்டி அதனை ஒரு கோலாக்குவர்.
• வெண்கொற்றக் குடையுடன் உலகாளும் மன்னனின் அரண்மனையில் அதனை வைத்து இந்திரன் மகன் சயந்தன் என அக்கோலை நினைத்து மந்திர விதியாலே வழிபாடு செய்வர்.
• அத்தலைக்கோலைப் புண்ணிய நதிகளிலிருந்து பொற்குடங்களில் முகந்து வந்த நன்னீரால் நீராட்டுவர்.
• மாலைகளை அணிவித்துப் பொருத்தமான ஒரு நல்ல நாளிலே பொன்னாலான பூணினையும் முகப்படாத்தையும் கொண்டிருக்கிற பட்டத்து யானையின் கையில் வாழ்த்தித் தருவர்.
• முரசுகள் முழங்கப் பல்வேறு வாத்தியங்கள் ஒலிக்க அரசரும் அவரின் ஐம்பெருங்குழுவினரும் சூழ்ந்து வரப்பட்டத்து யானை, தேரை வலம் வந்து அதன் மேல் உள்ள கவிஞனிடம் அத்தலைக்கோலைக் கொடுக்கும்.
• அனைவரும் ஊர்வலமாக அரங்கிற்கு வந்தபின், அத்தலைக்கோலைக் கவிஞன் ஆடலரங்கில்
  வைப்பான்.
• மாதவியின் ஆடலரங்கில் தலைக்கோல் வைக்கப்பட்டது.

பிரிவு – 2

எவையேனும் இரண்டனுக்கு மட்டும் விடை தருக.

Question 35.
மயிலை சீனி. வேங்கடசாமியின் படைப்புகள் குறித்து எழுதுக.
Answer:

• ஆசிரியர் பயிற்சி பெற்றுத் தொடக்கப் பள்ளியில் 25 ஆண்டுகள் ஆசிரியராகப் பணியாற்றினார். பாடம் கற்பிக்கும் நேரத்தைத் தவிர மற்ற நேரங்களில் அவர் வளாகம் சாராத ஆய்வாளராகத் தம்மை உருவாக்கிக் கொண்டார்.
• கட்டுரையோ, நூலோ எழுதும் முன் தரவுகளைச் சேகரித்துத் தமது ஆய்வுக்கு ஏற்றதை ஒப்பிட்டு ஐயமிருப்பின் மற்றவர்களைக் கேட்டுத் தெளிந்த பிறகே வெளியிடுவார்.
• மயிலை சீனி. ஒரு தமிழ்த் தேனீ. அறிவின் வாயில்களை நோக்கியே அவர் கால்கள் நடந்தன. நூலகங்களே அவரது தாயகங்களாயின. அறிவை விரிவு செய்து அல்லும் பகலும் ஆய்வில் மூழ்கிக் கருத்து முத்துகளைத் தமிழ்நாட்டிற்கு வழங்கினார்.
• புதிய செய்தி தருதல், புது விளக்கமளித்தல், இருண்ட பக்கங்களுக்கு ஒளியூட்டுதல், தவறுகளை மறுத்து உண்மையை எடுத்துரைத்தல் என்பவற்றை ஆய்வு அணுகு முறைகளாகக் கொண்டார்.

மயிலை சீனி. வேங்கடசாமியின் படைப்புகள்:

• விபுலானந்த அடிகள்…. இதழ்களில் வெளியாகின.
• பௌத்தமும் தமிழும், சமணமும் தமிழும்
• களப்பிரர் ஆட்சியில் தமிழகம்
• தமிழர் வளர்த்த அழகுக் கலைகள்
• தமிழ்நாட்டு வரலாறு
• சாசனச் செய்யுள் மஞ்சரி
• மறைந்து போன தமிழ் நூல்கள்

Question 36.
பெருங்காப்பியத்தில் இடம் பெற வேண்டிய சிறப்புகள் யாவை?
Answer:

• வாழ்த்துதல், தெய்வத்தை வணங்குதல், வருபொருள் உரைத்தல் என்ற மூன்றில் ஒன்றினைத் தொடக்கத்தில் பெற்று வரும். அவற்றுள் இரண்டோ மூன்றோ வரலாம்.
• அறம், பொருள், இன்பம், வீடு ஆகிய நான்கும் பெருங்காப்பியத்தின் திரண்ட பொருளாக அமைந்திருக்க வேண்டும். எனினும் இவற்றுள் பாவிகத்திற்கு ஏற்றவண்ணம் ஒன்றும் பலவும் சிறப்பாக அமைந்திருக்க வேண்டும்.
• தன்னிகர் இல்லாத் தன்மை உடையவனைக் காப்பியத் தலைவனாகக் கொண்டிருத்தல் வேண்டும்.
• மலை (1), கடல் (2), நாடு (3), நகர் (4), சிறுபொழுது (5-10), பெரும்பொழுது (11-16), கதிரவன் தோற்றம் (17), சந்திரனின் தோற்றம் (18) ஆகிய பதினெட்டு உறுப்புகளும் இயற்கை வருணனைகளாக அமைதல் வேண்டும்.
• திருமணம் புரிதல், மக்களைப் பெற்றெடுத்தல், முடிசூடல் முதலான நிகழ்வுகளைக் கொண்டிருக்க வேண்டும்.
• அமைச்சர்களுடன் கலந்துரையாடல், தூது செல்லல், போர்ப் புரிய படைகள் அணிவகுத்தல், போர்நிகழ்ச்சி, வெற்றி பெறுதல் போன்ற நிகழ்வுகளும் இடம் பெறுதல் வேண்டும்.
• சந்தி எனப்படும் கதைப்போக்கு (தொடக்கம், வளர்ச்சி, விளைவு, முடிவு என்பவை) வரிசைப்படி அமைந்திருக்க வேண்டும்.
• அமைப்பு முறையில் பெருங்காப்பிய உட்பிரிவுகளுள், சருக்கம், இலம்பகம், பரிச்சேதம் என்ற பெயர்களில் ஒன்றைப் பெற்றிருத்தல் வேண்டும்.
• எண்வகைச் சுவையும் மெய்ப்பாட்டுக் குறிப்புகளும் கேட்போர் விரும்பும் வண்ணம் அமைக்கப்பட்டிருக்க வேண்டும்.

Question 37.
மாணக்கர் அக்காலத்தில் சுவடிகளை எவ்வாறு அமைத்தனர்?
Answer:

• இளம்பிள்ளைகளுக்கு உபாத்தியாயர் ஓலையை வாரி ஒழுங்காக நறுக்கித் துளையிட்டுக் கயிறு கோர்த்துத் தருவார்.
• ஒரு துளையிடுவதும் இரண்டு துளையிடுவதும் உண்டு. மற்ற பிள்ளைகள் தாங்களே செய்து
  கொள்ளுவார்கள். பனையேடு, சீதாள பத்திரம் முதலியவற்றில் எழுதுவது வழக்கம்.
• மேலே சட்டமாகப் பனைமட்டையின் காம்பை நறுக்கிக் கோர்ப்பார்கள். மரச்சட்டங்களையும் அமைப்பார்கள்.
• செப்புத் தகட்டாலும் சட்டஞ் செய்து கோர்ப்பார்கள். அந்தச் சட்டங்களின் மேல் வர்ண மையினாற் பல வகையான சித்திரங்கள் எழுதுவதுண்டு.
• இரட்டைத் துளையுள்ள ஏடுகளில் ஒரு துளையில் செப்புக் கம்பி அல்லது மூங்கிற் குச்சியைச் – செருகிக் கட்டுவார்கள். அதற்கு நாராசம் என்று பெயர்.
• சுவடியைக் கோக்கும் கயிற்றின் ஒரு தலைப்பில் டையாக, பனையோலையை ஈர்க்கோடு கிளிமூக்குப் போலக் கத்தரித்து அமைப்பார்கள். அதற்குக் கிளிமூக்கென்று பெயர்.
• இப்போது அச்சுப் புத்தகங்களின் அளவில் எவ்வளவு வேறுபாடுகள் உண்டோ அவ்வளவு பனையோலைச் சுவடிகளிலும் உண்டு.

Question 38.
தென்னிந்திய சினிமாத் தொழில் வளர காரணமானவர் யாவர்?
Answer:

• படங்காட்டுதல் மூலம்தான் முதன்முதலாகத் தென்னிந்திய சினிமாத் தொழில் தோன்றியது.
• மனைவியின் வைரமாலையை விற்று சாமிக்கண்ணு வின்சென்ட், பிரெஞ்சுக்கார் டுபான் என்பவரிடமிருந்து 2500 ரூபாய்க்கு ஒரு புரொஜக்டரையும் சில துண்டுப்படங்களையும் வாங்கினார். • திருச்சியில் ஒரு கூடாரத்தில் படங்காட்ட ஆரம்பித்த அவர், பின்னர் திருவனந்தபுரம், மதுரை
  நகர்களில் முகாமிட்டு, மதராசுக்கு வந்து காட்சிகள் நடத்தினார்.
• அங்கிருந்து வடக்கே சென்று பெஷாவர், லாகூர் பின்னர் லக்னோ நகரங்களில் படக்காட்சிகள் நடத்திவிட்டு 1909 இல் மதராஸ் திரும்பினார்.
• அங்கே எஸ்பிளனேட்டில் (இன்றைய பாரிஸ் அருகே) கூடாரம் போட்டுச் சலனப்படங்களைத் திரையிட்டார்.
• சென்னையிலிருக்கும் போது சினிமாத்தொழிலை இங்கு நிறுவ ஒரு முக்கியமான அடியெடுத்து வைத்தார்.
• புரொஜக்டர்களை இறக்குமதி செய்து விற்க ஆரம்பித்தார். இதனால் புதிய திரையரங்குகள் வர ஏதுவாயிற்று.

பிரிவு – 3

எவையேனும் மூன்றனுக்கு விடை தருக

Question 39.
குறிஞ்சித்திணை அல்லது நெய்தல்திணையை விவரி.
Answer:
குறிஞ்சித்திணை :
அகத்திணைகள் ஐந்து, அவற்றுள் ஒன்று குறிஞ்சித்திணை.

முதற்பொருள் :
நீலம் – மலையும் மலை சார்ந்த இடமும்
பொழுது – சிறுபொழுது – யாமம்
பெரும்பொழுது – கூதிர், முன்பனி

கருப்பொருள்:
தெய்வம் – முருகன்
மக்கள் – சிலம்பன், வெற்பன், பொருப்பன், கொடிச்சி, குறத்தி, குறவர், குறத்தியர் கானவர்
பறவை – கிளி, மயில்
விலங்கு – புலி, கரடி, யானை, சிங்கம்
பூ – காந்தள், குறிஞ்சி, வேங்கை
தொழில் – தேனெடுத்தல், கிழங்கு அகழ்தல், வெறியாடல், தினைகாத்தல்
உணவு – தினை, மலைநெல், மூங்கிலரிசி
ஊர் – சிறுகுடி

உரிப்பொருள்:
புணர்தலும் புணர்தல் நிமித்தமும்

சான்று:
ஒழுகுநீ ராரல் பார்க்கும்
குருகு முண்டுதான் மணந்த ஞான்றே

விளக்கம்:
தலைவன் நெடுங்காலம் தலைவியைத் திருமணம் செய்து கொள்ளாமல் காலந்தாழ்த்த அதனால் வருந்திய தலைவி “தலைவர் கருணைகொண்டு என்னைத் திருமணம் செய்து கொண்டாலன்றி எனக்கு உதவியாகச் சான்று கூறுவார் வேறு ஒருவரும் இலர்” என்று தோழியிடம் கூறியது.

(அல்லது)

நெய்தல் திணை:
Answer:
முதற்பொருள்
நிலம் – கடலும், கடல் சார்ந்த இடமும்.
பொழுது – சிறுபொழுது – எற்பாடு
பெரும்பொழுது – முன்பனி, பின்பனி, இளவேனில், முதுவேனில். கூதிர்

கருப்பொருள்
தெய்வம் – வருணன்
மக்கள் – பரதவர், பரத்தியர், நுளையர்
பறவை – நீர்க்காக்கை
விலங்கு – சுறா
ஊர் – பட்டினம், பாக்கம்
நீர் – உவர்நீர்க்கேணி, உவர்க்கழி
உணவு – மீனும், உப்பும் விற்றால் பெறும் பொருள்
தொழில் – மீன் பிடித்தல், உப்பு விற்றல்

உரிப்பொருள்
‘இரங்கலும், இரங்கல் நிமித்தமும்’

சான்று:
இறவுப்புறத் தன்ன பிணர்படு தடவுமுதற்
சுறவுக்கோட் டன்ன முள்ளிலைத் தாழை

என்ற நற்றிணைப் பாடல் நெய்தல் திணைக்குச் சான்றாகும்.

துறை:
இது புணர்ந்து நீங்கிய தலைவனைத் தோழி வரைவு கடாயது.

விளக்கம்:
பகற்குறியில் தலைவன் தலைவியைச் சந்தித்துச் செல்கின்றான். அப்பொழுது தோழி தலைவனை நோக்கி “இவ்வாறு செல்வாயானால் மீண்டும் நீ வருவதற்குள் தலைவி இறந்து விடுவாள். ஆதலால் அதற்கு ஏற்றது செய்”, என வரைவு (மணஞ்செய்து கொள்வது) தோன்றக்
கூறுவது வரைவு கடாதலாகும்.

Question 40.
உருவக அணி அல்லது பிறிது மொழிதல் அணியை சான்றுடன் விவரி.
Answer:
உருவக அணி:
அணி விளக்கம்:
உவமானமும், உவமேயமும் வேறு வேறு பொருள் எனத் தோன்றாமல் ஒன்றுபோல் காட்டி, உவமானத்தின் தன்மை முழுவதும் உவமேயத்தில் மறைந்து நிற்கும்படிக் கூறுவது ‘உருவக அணி’ எனப்படும்.

(எ.கா.) முகத்தாமரை

விளக்கம்
முகமானது தாமரையாக உருவகப்படுத்தப்பட்டுள்ளது. நாம் கூறும் பொருள் ‘உவமேயம்’ எனப்படும். ஒப்புமையாகக் காட்டும் பொருள் ‘உவமை’ எனப்படும்.

(அல்லது)

பிறிது மொழிதல் அணி:
Answer:
அணி விளக்கம் :
புலவர் தாம் கூறக் கருதியதை வெளிப்படையாகக் கூறாமல், அதனோடு தொடர்புடைய வேறொன்றின் மீது ஏற்றிக் கூறி, அதன்மூலம் தாம் கருதியதைப் பெற வைத்தல் ‘பிறிது மொழிதல் அணி’யாகும்.

(எ.கா.) பீலிபெய் சாகாடும் அச்சிறும் அப்பண்டம்
சால மிகுத்துப் பெயின்.

விளக்கம்:
மயிலிறகு மென்மையானது என்றாலும், அதனையே அளவுக்கு மேல் வண்டியில் ஏற்றினால், வண்டியானது பளு தாங்காமல் அச்சு முறிந்துவிடும் என்பது இக்குறளின் பொருள். ஆனால், வள்ளுவர் இக்கருத்தை உணர்த்த இந்தக் குறளைக் கூறவில்லை.

எதிரிகள் வலிமையற்றவர்களாக இருந்தாலும், அவர்கள் எண்ணிக்கையில் அதிகமாக இருந்து, வலிமையுடைய வனைத் தாக்கினால் அவன் நிலைகுலைந்து போவான். இக்கருத்தை வலியுறுத்தவே வள்ளுவர் இக்குறளை எழுதியுள்ளார். எனவே, இது ‘பிறிது மொழிதல் அணி’ ஆகும். (பிறிது – வேறொன்று; மொழிதல் – கூறுதல்))

Question 41.
கீழ்க்காணும் பாடலைப் படித்தறிந்து ஏதேனும் 5 நயங்களை மட்டும் எழுதுக.
Answer:
தாயென அன்பு செய்து
தந்தைபோல் பரிந்து, சொந்தச்
சேயென அணைத்துப் பேசி
செவ்விய அறிவு கூறி
தூயநன் னடத்தை கற்கத்
துணையென நடந்து காட்டும்
ஆயநற் குணமுள் ளோனே
ஆசானென் றழைக்கத் தக்கோன். (- நாமக்கல் கவிஞர்)

ஆசிரியர் குறிப்பு:

இயற்பெயர் : வெ. இராமலிங்கம்
பிறப்பு : 29 அக்டோபர், 1888
ஊர் : மோகனூர் – நாமக்கல் மாவட்டம்
சிறப்பு பெயர் : காந்தியக் கவிஞர்
படைப்பு : மலைக்கள்ளன், நாமக்கல் கவிஞர் பாடல்கள்
இறப்பு : 24 ஆகஸ்டு 1972)

திரண்ட கருத்து:
ஒரு நல்ல ஆசிரியன் என்பவன் தாயைப் போல அன்பு செய்தும், தந்தையைப் போல பரிவுடனும், பெற்ற குழந்தையைப் போல அணைத்து பேசியும் நல்ல அறிவுரைகளை கூறுதல் வேண்டும். தூய நன்னடத்தைகளைக் கற்க துணையைப் போல நடந்துகாட்ட வேண்டும். இவ்வகையான தூய்மையான நல்ல குணங்கள் உள்ளவனே நல்ல ஆசான் ஆவார்.

மையக் கருத்து:
ஒரு நல்ல ஆசான் தாய், தந்தை, குழந்தை, துணையை போல இருக்க வேண்டும் என கூறுகிறார்.

மோனை:
மோனை – சீர்தோறும் முதல் எழுத்து ஒன்றி வருவது
அன்பு – அறிவு
ஆயநற்குணம் – ஆசானெள்

எதுகை:
அடிதோறும் இரண்டாவது எழுத்து ஒன்றி வருவது அடி எதுகை.
தாயென – சேசெயன
தூயறள் – ஆயநள்

அணி:
உவமையணி

Question 42.
தமிழாக்கம் தருக.
1. Walls have ears.
2. Wealth is best known by want.
3. While in rome, be a roman.
4. Think every – body alike.
Answer:
1. பகலில் பக்கம் பார்த்துப் பேசு.
2. நிழலின் அருமை வெயிலில் தெரியும்.
3. ஊருடன் ஒத்துவாழ்.
4 தன்னைப்போலப் பிறரை நினை.

Question 43.
பின்வரும் தலைப்புகளில் ஏதேனும் ஒன்றில் 8 வரிகளில் கவிதை புனைக.
Answer:

வான்மழை :

வரவேற்பேன் வரவேற்பேன்
வான்மழையே நீ வந்துவிடு
உன்னை அழைக்கத்தான்
உவந்துநான் மரம் வளர்த்தேன்
மாமயில்கள் அகவி ஆடுவதும்
மாமழை உன்னை வரவேற்கத்தான்
வாடிடும் பயிர்கள் காக்க
வாஞ்சையுடன் நீ வருவாயே!

பகுதி – IV

பின்வரும் வினாக்களுக்கு இரு பக்கங்களுக்கு மிகாமல் விடை தருக.[3 x 6 = 18]

Question 44.
(அ) நெடுநல்வாடையில் நக்கீரர் காட்டும் மழைக்கால வருணனையைச் சொல்லில் வடிக்க.
Answer:

• “ஐப்பசி அடை மழை! கார்த்திகை கனமழை!” என்பது சொலவடை, ஓராண்டை ஆறு பருவங்களாக வகைப்படுத்திய பழந்தமிழர் ஜப்பசி, கார்த்திகை மாதங்களைக் கூதிர்ப்பருவம் என்று அழைத்தனர்.
• பருவ மாற்றங்களால் உயிரனங்களின் இயல்பு வாழ்க்கை , மாற்றம் பெறுகிறது.
• முல்லை நில மக்கள், பறவைகள், விலங்குகள் இவற்றின் வாழ்வில் மழையும் குளிரும் ஏற்படுத்தும் மாற்றத்தைச் சங்க இலக்கியம் பதிவு செய்துள்ளது.
• தான் தங்கியிருந்த மலையை வலப்பக்கமாகச் சூழ்ந்து எழுந்த மேகமானது உலகம் குளிருமாறு புதிய மழையைப் பொழிந்தது.
• தாழ்வான பகுதிகளில் பெருகிய வெள்ளத்தை வெறுத்த, வளைந்த கோலையுடைய ஆயர் எருமை, பசு, ஆடு ஆகிய நிரைகளை வேறு மேடான நிலங்களில் மேய விட்டனர்.
• தாம் பழகிய நிலத்தை விட்டுப் பெயரும் நிலையால் வருத்தம் அடைந்தனர்.
• அவர்கள் தலையில் சூடியிருந்த நீண்ட இதழ்களையுடைய காந்தள் மாலை கசங்கியது.
• பலருடன் சேர்ந்து கொள்ளி நெருப்பினால் கைகளுக்குச் சூடேற்றிய போதிலும் அவர்களது பற்கள் நடுங்கின.
• விலங்குகள் குளிர்மிகுதியால் மேய்ச்சலை மறந்தன. குரங்குகள் நடுங்கின.
• மரங்களில் தங்கியிருந்த பறவைகள் நிலத்தில் வீழ்ந்தன.
• பசுக்கள் பாலுண்ண வந்த கன்றுகளைத் தவிர்த்தன. மலையையே குளிரச் செய்வது போன்றிருந்தது அக்குளிர்கால நள்ளிரவு என மழைக்காலத்தை வருணிக்கிறார் நக்கீரர்.

(அல்லது)

(ஆ) நாட்டியக் கலைஞருக்குத் தமிழர் உரிய சிறப்பு அளித்தனர் என்பதைச் சிலப்பதிகாரப் பாடற்பகுதி கொண்டு நிறுவுக.
Answer:
முன்னுரை:
சிலப்பதிகாரத்திலுள்ள அரங்கேற்று காதையின் ஒருபகுதி பாடமாக வைக்கப்பட்டுள்ளது. தமிழரின் கலை, நாகரிகம், பண்பாடு முதலானவற்றை உள்ளடக்கிய கருவூலமாகச் சிலப்பதிகாரம் திகழ்கிறது. அரசகுடி அல்லாதவர்களைக் காப்பியத்தின் தலைமக்களாக வைத்துப் பாடியதால் இது ‘குடிமக்கள் காப்பியம்’ எனப்படுகிறது. புகார், மதுரை, வஞ்சிக் காண்டங்கள் முறையே சோழ, பாண்டிய, சேர மன்னர்களைப் பற்றியவை என்பதால் ‘மூவேந்தர் காப்பியம்’ எனவும் அழைக்கப்படுகிறது.

மாதவியின் நாட்டியப் பயிற்சி:
மாதவி, அழகிய தோள்களை உடையவள்; தேனும் தாதுவும் நிறைந்த பூக்களை அணிந்த சுருண்ட கூந்தலை உடையவள். ஆடல், பாடல், அழகு என்னும் இம்மூன்றில் ஒன்றும் குறைபடாமல் (ஐந்தாண்டில் ஆடல் கற்பதற்கான சடங்குகளைச் செய்து) ஏழு ஆண்டுவரை ஆடல் கலையைப் பயின்றாள். அவள் தனது பன்னிரண்டாவது வயதில் ஆடலை அரங்கேற்ற விரும்பினாள், (வீரக் கழல் பூண்ட சோழ மன்னனது அரசவைக்கு வந்தாள் அவளுடன் ஆடல், ஆசான், இசை ஆசான், கவிஞன், தண்ணுமை ஆசான், குழல் ஆசான், யாழ் ஆசான் ஆகியோரும் வந்திருந்தனர்.)

நாட்டிய அரங்கின் அமைப்பு:
திறம்படக் கற்றுணர்ந்த சிற்ப நூலாசிரியரால் சொல்லப்பட்ட இயல்புகளிலிருந்து மாறுபடாத நன்னிலத்தை, ஆடல் அரங்கிற்காகத் தேர்ந்தெடுத்தனர். பொதிகை மலை போன்ற மலைகளிலே நீண்டு வளர்ந்த மூங்கில்களிலே, ஒன்றுக்கொன்று இடையே ஒரு சாண் அளவு கணுக்களைக் கொண்ட மூங்கிலைக் கொண்டு வந்தனர். நூல்களில் கூறப்பட்ட முறையாலே அரங்கம் அமைந்தனர். தம் கைப்பெருவிரலில் இருபத்து நான்கு அளவினைக் கொண்டதாக அம்மூங்கிலை வெட்டினர். அதை அரங்கம் அமைக்கும் கோலாகக் கொண்டு அதில் ஏழுகோல் அகலமும் எட்டுக்கோல் நீளமும் ஒரு கோல் உயரமும் உடையதாக நாட்டிய அரங்கத்தை அமைத்தனர்.

தலைக்கோல் :
அரங்கேற்றத்தை வெற்றிகரமாக முடிக்கும் ஆடல்மகளுக்குத் தலைக்கோல் அளித்துச் சிறப்பிப்பர். தலைக்கோல் என்பது. பெரும்புகழ் கொண்ட பகை மன்னனுடன் நிகழ்த்திய போரில், தோற்றுப் புறங்காட்டிய அவனிடமிருந்து பறிக்கப்பட்ட அழகுமிக்க வெண்கொற்றக் குடையின் காம்பில் செய்யப்படுவது. அக்காம்பின் கணுக்கள் முழுவதும் நவமணிகளை இழைத்து அக்கணுக்களுக்கு இடைப்பட்ட பகுதிகளில் சாம்பூந்தம் எனும் பொன் தகட்டை வலம்புரியாகவும், இடம்புரியாகவும் சுற்றிக்கட்டி அதனை ஒரு கோலாக்குவர். வெண்கொற்றக் குடையுடன் உலகாளும் மன்னனின் அரண்மனையில் அதனை வைத்து இந்திரன் மகன் சயந்தன் என அக்கோலை நினைத்து மந்திர விதியாலே வழிபாடு செய்வர். அத்தலைக்கோலைப் புண்ணிய நதிகளிலிருந்து பொற்குடங்களில் முகந்து வந்த நன்னீரால் நீராட்டுவர்.

மாலைகளை அணிவித்துப் பொருத்தமான ஒரு நல்ல நாளிலே பொன்னாலான பூணினையும் முகபடாத்தையும் கொண்டிருக்கிற பட்டத்து யானையின் கையில் வாழ்த்தித் தருவர். முரசுகள் முழங்கப் பல்வேறு வாத்தியங்கள் ஒலிக்க அரசரும் அவரின் ஐம்பெருங்குழுவினரும் சூழ்ந்து வரப்பட்டத்து யானை, தேரை வலம் வந்து அதன் மேல் உள்ள கவிஞனிடம் அத்தலைக்கோலைக் கொடுக்கும். அனைவரும் ஊர்வலமாக அரங்கிற்கு வந்தபின், அத்தலைக்கோலைக் கவிஞன் ஆடலரங்கில் வைப்பான். மாதவியின் ஆடலரங்கில் தலைக்கோல் வைக்கப்பட்டது.

மாதவி மன்னனிடம் பரிசு பெறுதல்:
பொன்னால் செய்யப்பட்ட ஒரு பூங்கொடி வந்து நடனமாடியது போல மாதவி அரங்கில் தோன்றி நாட்டிய நூலில் சொல்லப்பட்ட முறைமை தவறாது பாவம், அபிநயம் இவற்றைச் சரியாகக் கடைப்பிடித்து அனைவரும் கண்டுகளிக்கும்படி அழகுற ஆடினாள். மாதவி கூத்துக்கு உரிய இயல்பினிலிருந்து சற்றும் வழுவாது ஆடினாள். ஆடலைக் கண்டு அகமகிழ்ந்த மன்னனிடமிருந்து ‘தலைக்கோலி’ என்னும் பட்டமும் பெற்றாள். அரங்கேற்றம் செய்யும் நாடகக் கணிகையர்க்குப் ‘பரிசு இவ்வளவு ‘ என நூல் விதித்த முறைப்படி ‘ஆயிரத்தெட்டுக் கழஞ்சுப் பொன் மாலையை’ மன்னனிடமிருந்து பரிசாகப் பெற்றாள்.

மாதவியின் நாட்டியம்:
அரசன் முதலானோர் யாவரும் தத்தம் தகுதிக்கு ஏற்ப அமைக்கப்பட்ட இருக்கையில் அமர்ந்தனர். அதனருகே இசைக் கருவிகளை வாசிப்போர், நிற்க வேண்டிய முறைப்படி அவரவர்க்கு உரிய இடத்தில் நின்றனர். அரங்கேற்றம் செய்ய வேண்டிய நாடகக் கணிகையாகிய மாதவி அரங்கில் வலக்காலை முன்வைத்து ஏறி, பொருமுக எழினிக்கு நிலையிடனான வலத்தூண் அருகே போய் நிற்க வேண்டியது மரபு என்பதால் அங்குப் போய் நின்றாள். ஆடலில் தேர்ச்சிபெற்று அரங்கேறிய தோரியமகளிரும் தொன்றுதொட்டு வரும் முறைப்படி ஒருமுக எழினிக்கு நிலையிடனான இடப்பக்கத்தூணின் அருகே போய் நின்றனர்.

முடிவுரை:
தமிழர்கள் வீரத்திற்கு முக்கியத்துவம் கொடுத்து தன் வெற்றியை போற்றி தலைக்கோலை பாதுகாத்தனர். அத்தகைய தலைக்கோலையே நாட்டியத்திற்காக பரிசு வழங்குகையில் அவர்கள்
நாட்டியக் கலைக்கு எத்தகைய முக்கியத்துவம் அளித்தனர் என்பதனை உணர்ந்து கொள்ள முடிகிறது.

Question 45.
(அ) மயிலையார் ஓர் “ஆராய்ச்சிப் பேரறிஞர்” என்னும் கூற்றினைச் சான்றுகளுடன்
கட்டுரைக்க.
Answer:
முன்னுரை:

• ஒவ்வொரு தேசிய இனமும் தன்னுடைய கடந்தகால வரலாற்றை அறிந்திருந்தால் மட்டுமே ‘ எதிர்கால இலக்குகளை அடைய முடியும்.
• அத்தகைய இன வரலாற்றை இலக்கியம், தொல்லியல், கல்வெட்டு, பண்பாடு ஆகியவற்றின் அடிப்படையில் தரவுகளைத் தேடி எழுதிய ஆளுமைகளில் முக்கியமானவர் மயிலை சீனி. வேங்கடசாமி.
• இதழ் ஆசிரியராக வாழ்க்கையைத் தொடங்கிய அவர், பள்ளி ஆசிரியராகப் பணியாற்றியபடியே எழுதி முடித்த புத்தகங்களே, தமிழர் தம் பழம்பெருமையை உணர, உதவும் புதையலாக விளங்குகிறது.

தொடக்ககால ஆய்வுகள்:

• 1934 இல் தெ. பொ. மீனாட்சி சுந்தரனார் சிந்தாதிரிப்பேட்டை உயர்நிலைப் பள்ளியில் நடத்திய தமிழ் ஆராய்ச்சி மாநாட்டில் ச.த. சற்குணர், கிறித்துவமும் தமிழும் என்ற பொருள் குறித்து உரையாற்றினார்.
• அவ்வுரையைக் கேட்டுப் பெற்ற ஆர்வத்தினால் ‘கிறித்துவமும் தமிழும்’ என்னும் நூலை மயிலையார் எழுதினார். இதுவே அவருடைய முதல் நூலாகும்.
• தமிழின் தொல் இலக்கியம், தொல் கலை, தொல்லெழுத்து ஆகியவற்றில் கொண்ட ஈடுபாடு
  காரணமாக, ‘பௌத்தமும் தமிழும்’, ‘சமணமும் தமிழும்’ ஆகிய நூல்களை அவர் இயற்றினார்.
• சமயம், மானுடவியல், தமிழக வரலாறு, தொல்பொருள் ஆய்வு, கலை வரலாறு, மொழியாய்வு முதலான பல துறைகளிலும் இவர் கவனம் செலுத்தினார்.
• குறிப்பாக, கல்வெட்டு ஆய்வில் இவருக்குப் பயிற்சி அதிகம். தமிழ் எழுத்தியலின் வளர்ச்சியை நன்குணர்ந்தவர். வட்டெழுத்து, கோலெழுத்து. தமிழ் பிராமி ஆகியவற்றில் புலமை பெற்றிருந்த காரணத்தால் சாசனங்களை அவரால் எளிதாக வாசிக்க முடிந்தது.
• வரலாறு, இலக்கியம், கலையியல், சமயம் எனப் பல துறைகளிலும் நூல்களை எழுதியிருப்பது வேங்கடசாமியின் பன்முக அறிவை விளக்குகிறது.

வரலாற்று ஆய்வு :

• மகேந்திரவர்மன், நரசிம்மவர்மன் ஆகியோர் குறித்த நூல்களைத் தொடர்ந்து மூன்றாம்
  நந்திவர்மன் என்னும் பல்லவ மன்னனைப் பற்றியும் அவர் எழுதினார்.
• தமிழில் அம்மன்னனைப் பற்றி எழுதப்பட்ட முதல் நூல் என்னும் பெருமையைப் பெற்றது.
• சங்க கால மூவேந்தர்கள், கொங்குநாட்டு மன்னர்கள், துளு நாட்டு மன்னர்கள், களப்பிரர் மற்றும் இலங்கை குறித்த வரலாறு ஆகியவற்றையும் எழுதியுள்ளார்.
• சங்க காலத்துக்குப் பிற்பட்ட காலத்தில் தமிழகத்தைக் களப்பிரர்கள் என்ற மன்னர்கள் ஆட்சி புரிந்தனர். இக்காலம் தமிழர்களின் வரலாற்றில் இருண்ட காலம் என்று மரபுவழிப்பட்ட வரலாற்று ஆசிரியர்கள் குறித்தனர்.
• இம்மன்னர்கள் குறித்த ஆய்வை வேங்கடசாமி விரிவாக ஆராய்ந்து களப்பிரர் குறித்த புதிய கண்ணோட்டத்தைக் ‘களப்பிரர் ஆட்சியில் தமிழகம்’ என்னும் நூல் மூலம் வெளிப்படுத்தினார்.

கலையியல் ஆய்வு:

• கலையியல் சார்ந்து தமிழில் வெளியான பல நூல்களுக்கும் வேங்கடசாமியே வழிகாட்டியாக விளங்கினார்.
• தமிழரது கலைத்திறனை எடுத்தோதும் ‘தமிழர் வளர்த்த அழகுக்கலைகள்’ என்னும் நூல், கவின் கலைகள் குறித்துத் தமிழில் வெளிவந்த முழுமையான முதல் நூல் ஆகும்.
• இந்நூல் தமிழக அரசின் முதற்பரிசைப் பெற்றது. இறைவன் ஆடிய எழுவகைத் தாண்டவம். நுண்கலைகள், இசைவாணர் கதைகள் ஆகியவை கலை பற்றிய இவருடைய பிற நூல்கள் ஆகும். • தமது நூல்களின் படங்களைத் தானே வரைந்து வெளியிட்டது இவரது கலைத்திறனுக்குச் சான்றாகும்.

கல்வெட்டு ஆய்வுகள்:

• சங்ககாலத் தமிழக வரலாற்றில் சில செய்திகள், பழங்காலத் தமிழர் வணிகம், களப்பிரர் ஆட்சியில் தமிழகம், கொங்கு நாட்டு வரலாறு, தமிழ்நாட்டு வரலாறு – சங்க காலம் (அரசியல்) ஆகிய நூல்களையும் எழுதினார்.
• ஆய்வுலகில் மயிலை சீனி. வேங்கடசாமியின் குறிப்பிடத்தக்க பங்களிப்பைக் கூற வேண்டுமானால் தமிழக வரலாற்றினைப் பல கோணங்களில் மீட்டுருவாக்கம் செய்ததைக் குறிப்பிடலாம். தமிழ்நாட்டு வரலாறு என்னும் நூல் இவருடைய மீட்டுருவாக்க முயற்சிக்குச் சரியான சான்றாகும்.
• தமிழியலுக்குத் தேவையான பல்வேறு ஆவணங்களையும் தொகுத்து ஆய்வு செய்யும் பணியை இவர் மேற்கொண்டார். இப்பணியின் விளைவாக, சாசனச் செய்யுள் மஞ்சரி, மறைந்துபோன தமிழ்நூல்கள் ஆகிய நூல்களை எழுதினார்.

பன்மொழிப் புலமை:

• தமிழ் ஆய்வு மரபில், சொல் ஆய்வுகளை மேற்கொள்ளாத ஆய்வாளர் இல்லை. நீண்ட வரலாறு கொண்ட மொழியின் சொற்களில் இவ்வகையான ஆய்வுக்குரிய ஏதுக்கள் மிகுதியாகும். வேங்கடசாமி தொடர்ச்சியாக இத்தகைய சொல்லாய்வுப் பணியை மேற்கொண்டு வந்துள்ளார். ‘செந்தமிழ்ச் செல்வி’ என்னும் இதழில் அவர் எழுதிய சொல்லாய்வுக் கட்டுரைகள் ‘அஞ்சிறைத் தும்பி’ என்ற தொகுப்பாக வெளியிடப்பட்டது.
• மகேந்திரவர்மன் இயற்றிய ‘மத்த விலாசம்’ என்ற நாடக நூலை ஆங்கிலம் வழியாகத் தமிழாக்கியுள்ளார்.
• தமிழ்ப் பண்பாட்டிலும் தமிழர் வரலாற்றிலும் ஆழ்ந்த அக்கறை கொண்ட மயிலை சீனி. அவர்கள் தமிழர் பண்பாட்டின் தனித்தன்மையை நிறுவுவதில் உறுதியான பார்வை கொண்டவர்.

ஆராய்ச்சிப் பேரறிஞர்:

• மயிலை சீனியாரால் பல ஆய்வுகள் தமிழுலகுக்கு முதன்முதலாக வழங்கப்பட்டவை. இவரது ஆய்வுகள் அறிஞருக்கு மட்டுமன்றிப் பொதுமக்களுக்கும் அறிவு விருந்தோம்பியவை.
• பல ஆய்வுகள் கிளைவிடுவதற்கு அடிமரமாக இருந்தவை. இவரது ஆய்வுகள், வேண்டாத நூலிது என்றோ நூலில் வேண்டாத பகுதி என்றோ ஒதுக்க முடியாத வகையில் இவரது எழுத்தாளுமை திகழ்ந்தது.

முடிவுரை:

• தமிழக வரலாற்றுக் கழகத்திலும் தமிழகப் புலவர் குழுவிலும் உறுப்பினராக இருந்து பெருந்தொண்டாற்றிய வேங்கடசாமிக்கு 1962இல் தமிழ் எழுத்தாளர் சங்கம் பாராட்டிக் கேடயம் வழங்கியது.
• மதுரைக் காமராசர் பல்கலைக்கழகம் தமிழ்ப் பேரவைச் செம்மல்’ என்ற விருதினை அளித்தது.
• தமது ஓயாத தேடலினாலும் கடுமையான உழைப்பாலும் அரிய ஆய்வு முடிவுகளைக் கொணர்ந்த இம்மாமனிதருக்கு அறிஞர்கள் கூடிச் சென்னை கோகலே மண்டபத்தில் மணிவிழா எடுத்து ஆராய்ச்சிப் பேரறிஞர்’ என்ற பட்டத்தை வழங்கினார்.

(அல்லது)

(ஆ) குடும்பம் என்னும் சிறிய அமைப்பிலிருந்தே மனித சமூகம் என்னும் பரந்த அமைப்பு
கட்டமைக்கப்படுகிறது – எவ்வாறு? விளக்குக.
Answer:

• குடும்பம் எனும் சிறிய அமைப்பிலிருந்தே மனித சமூகம் எனும் பரந்த அமைப்பு கட்டமைக்கப்படுகிறது. குடும்பம் தொடங்கிக் குலம், கூட்டம், பெருங்குழு, சமூகம் என்ற அமைப்புவரை விரிவு பெறுகிறது.
• குடும்பமே மனித சமூகத்தின் அடிப்படை அலகாக உள்ளது. வாழுங்காலம் முழுவதும் தொடர்ந்து வேறு எந்த நிறுவனமும் இந்த அளவுக்கு மனிதனைச் சமூகவயப்படுத்தும் பணியைச் செய்ததில்லை.

குடும்பம்:
குடும்பம் எனும் அமைப்பு ஏற்படுவதற்கு அடிப்படை திருமணமே, குடும்பம், திருமணம் இரண்டும் ஒன்றையொன்று சார்ந்தே செயல்படுகின்றன – நாணயத்தின் இரண்டு பக்கங்களைப் போல, இன்று நாம் வழங்கும் ‘திருமணம்’, குடும்பம் ஆகிய இரண்டு சொற்களும் தொல்காப்பியத்திலும் சங்க இலக்கியங்களிலும் எங்கும் இடம்பெறவில்லை . குடும்பம் எனும் சொல் முதன் முதலில் திருக்குறளில்தான் (1029) வருகிறது.

வாழிடம்:
மருதத்திணைப் பாடல் ஒன்றில் மகளிர் ‘தம்மனை’, ‘நும்மனை’ என மனைவியின் இல்லத்தையும் கணவனின் இல்லத்தையும் பிரித்துப் பேசும் போக்கினைக் காண முடிகிறது. இன்னும் சில இடங்களில் தற்காலிகத் தங்குமிடம் புக்கில் எனவும், திருமணத்திற்குப்பின் கணவனும் மனைவியும் பெற்றோரிடமிருந்து பிரிந்து, தனியாக வாழுமிடம் ‘தன்மனை’ எனவும் வழங்கப்பெற்றுள்ளன.

மணந்தகம்:
குடும்பமும் உயிரிகளைப் போன்றே தோன்றுகிறது; வளர்கிறது, பல கட்டங்களைக் கடக்கிறது. அதன் வாழ்க்கைச் சுழற்சியில் பல வடிவங்களில் நிலைமாற்றம் பெறுகிறது. இத்தகைய நீண்ட பாதையில் குடும்பத்தின் தொடக்கம் திருமணமே. மணம்புரிந்த கணவனும் மனைவியும் சேர்ந்து இல்லற வாழ்வில் ஈடுபடக்கூடிய தொடக்கக் கட்டமே ‘மணந்தகம்’ எனப்படுகிறது. முதல் குழந்தை பிறக்கும்வரை உள்ள காலகட்டத்தை இந்நிலை குறிக்கிறது. தனிக்குடும்ப உருவாக்கத்தின் தொடக்க நிலையாக இது அமைகிறது.

தாய்வழிக் குடும்பம்:
சங்ககாலத்தில் கண சமூகத்துக்குத் தாயே தலைமை ஏற்றிருந்தாள். தாய்வழியாகவே குலத்தொடர்ச்சி குறிக்கப்பட்டது. பதிற்றுப்பத்து கூறும் சேரநாட்டு மருமக்கள் தாய முறை இதற்குச் சிறந்த எடுத்துக்காட்டாகும்.

சிறுவர்தாயே பேரிற் பெண்டே
செம்முது பெண்டின் காதலஞ்சிறா அன்
வானரைக் கூந்தல் முதியோள் சிறுவன்
முளரிமருங்கின் முதியோள் சிறுவன்
என்மகள் ஒருத்தியும் பிறள்மகன் ஒருவனும்

முதலான தொடர்களில் ‘ இவளது மகன்’ என்றே கூறப்பட்டது. இவனது மகன் எனக் கூறப்படவில்லை என்பது நோக்கத்தக்கது. இவை அனைத்தும் சங்ககாலத்தில் காணப்பட்ட தாய்வழிச் சமூகத்தின் நிலையைக் காட்டுகின்றன.

சங்ககாலத்தில் பெண் திருமணம் செய்த பின்னரும் தன் இல்லத்திலேயே தொடர்ந்து வாழ்க்கை நடத்தும் தாய முறை இருந்துள்ளது. திருமணத்திற்குப்பின் மனைவியின் இல்லத்துக்குச் சென்று கணவன் வாழ்வதே நடைமுறையாக இருந்துள்ளது.

தந்தைவழிக் குடும்பம்:
மனித குலத்தில் ஆதியில் தோன்றி வளர்ந்த தாய்வழி முறையானது தமிழர்களிடம் இருந்ததைச் சங்க இலக்கியங்களின் வழி அறிய முடிந்தாலும், சங்க காலத்திலேயே ஆண் மையச் சமூக முறை வலுவாக வேர் ஊன்றிப் பரவலாகி விட்டதையும் காணமுடிகிறது.

ஆண் மையச் சமூகத்தில் பெண் திருமணத்திற்குப்பின் தன் கணவனுடைய தந்தையகத்தில் வாழ வேண்டும். மணமானபின் தலைவன் தலைவியை அவனுடைய இல்லத்திற்கு அழைத்து வந்தபோது அவனுடைய தாய் அவளுக்குச் சிலம்புகழி நோன்பு செய்திருக்கிறாள்.

தனிக்குடும்பம் :
தனிக்குடும்பம் தோன்றுவதற்கான தொடக்கநிலைக் குடும்பங்கள் பற்றிச் சங்க இலக்கியங்கள் மிகுதியாகப் பேசியிருக்கின்றன. இளமகவுநிலைக் குடும்பங்களின் காட்சிகளை ஐங்குறுநூறு தெளிவுபடுத்துகிறது. “மறியிடைப் படுத்த மான்பிணை போல்” மகனை நடுவணாகக்கொண்டு தலைவனும் தலைவியும் வாழ்ந்திருக்கின்றனர். தாய், தந்தை, குழந்தை மூவருமுள்ள தனிக்குடும்பம் மிகவும் நெருக்கமானது என்பதால் இது தொடக்கநிலை / எளிய நெருக்கமான குடும்பம் எனப்படும்.

விரிந்த குடும்பம்:
சங்க காலத்தில் தனிக்குடும்ப அமைப்பு விரிவுபெற்று இவர்களுடன் பெற்றோர் ஒருவரின் தந்தையும் உடன் வாழும் “விரிந்த குடும்ப முறையையும் காண முடிகிறது. கணவன், மனைவி, மகன் ஆகியோருடன் தந்தை சேர்த்து வாழ்ந்த நேர்வழி விரிந்த குடும்ப முறையை ஒக்கூர் மாசாத்தியாரின் புறநானூற்றுப் பாடல் கூறுகிறது.

இவ்வாறு குடும்பம் என்னும் சிறிய அமைப்பு மனித சமூகம் என்னும் பரந்த அமைப்பாக கட்டமைக்கப்படுகிறது.

Question 46.
(அ) பொறுப்புணர்ச்சியின்றி இருந்த ஊரைத் தன் பொறுப்புணர்வால் மாற்றிய மருதனின்
பண்பு நலத்தை விவரிக்க.
Answer:
மருதன் கண்ட காட்சி:
“ஐப்பசி மாதம் அடைமழை காலம்” கண்ணுக்கெட்டிய தூரம் வரை பச்சைப் பசேல் என்று விரிந்து கிடந்த வயல்வெளிகள். வயல்வெளியெங்கும் நடவு முடிந்து ஒரு வாரம் பத்து நாளான பச்சை பிடிக்கத் தொடங்கியிருந்த இளம்பயிர். நான்கு நாள் அடைமழையில் எல்லா வாய்க்கால்களும் பொங்கி வழிந்து வரப்பு எது, வயல் எது என்று அடையாளம் தெரியாமல் இப்பொழுதோ சற்றுப் பொறுத்தோ நீருக்குள் மூழ்கிவிடும் ஆபத்தில் இருந்தது.

மருதனின் சிந்தனைகள்:
எந்த மழையின் உதவியில் நட்டார்களோ, அதே மழையின் ‘அபரிமித அன்பினால்’ இப்போது பயிர் தெப்பலாடுகிறது. ஒரு நாள் மூழ்கினால் போதும். முழுவதும் அழுகிவிடும். மறுபடி புதிதாக நாற்றுவிட்டு…. புதிய சாகுபடிதான். அதற்கு யாரால் முடியும்…? இதற்கே அங்கே வாங்கி, இங்கே பிடுங்கி என்று ஏகப்பட்ட அல்லாடல்கள். மறுபடியும் என்றால் …. தரிசுதான். சோற்றுக்கு லாட்டரிதான். வேறு என்ன செய்ய? என்ன செய்யலாம் என்று மருதனுக்கும் ஆயிரம் யோசனைகள்.

மருதனுக்கு தோன்றிய யோசனை:
கரைவழியே நடந்தான். உபரித் தண்ணீர் வடிய வேண்டிய வடிகால் மதகை எட்டிப் பார்த்தான். மதகின் கீழ்க்குமிழி மட்டுமல்ல. ஊரைச்சுற்றி வளைந்து ஓடிவரும் மூன்றுமைல் நீள வடிவாய்க்கால் முழுவதுமே சுவர் வைத்துத் தடுத்ததைப் போல் காடாய் மண்டிக் கிடந்த நெய்வேலி காட்டாமணக்குச் செடிகளின் அசுரத்தனமான ஆக்கிரமிப்பு. பயிர்கள் மூழ்காமல் மொத்தத் கிராமமும் தப்பித்துக்கொள்ள வழி கண்டுபிடித்து விட்ட சந்தோஷம் மருதனுக்கு. இந்த பேய்ச் செடிகளைப் பிடுங்கி எறிந்தால் போதும். ஒரே நாளில் உபரி நீர் முழுவதும் வடிந்துவிடும்.

மருதனுக்கு ஏற்பட்ட சந்தேகம்:
சரி இவ்வளவு நீளம் மண்டிக்கிடக்கும் செடிகளை அரித்து எறிவது லேசான காரியமா? இந்த மலைப்பிற்கும் ஒரு சில நிமிட யோசனைக்குப் பிறகு வழி தெரிந்தது. உற்சாகமாக நடக்கத் தொடங்கினான்.

மருதன் மாரியப்பனிடம் கூறினான்:
மருதனைக் கண்ட மாரியப்பன் ஊர்க்காரங்க எல்லோரும் ஒண்ணு சேந்தோம்னு வச்சுக்க. ஆளுக்கொரு செடின்னாகூட ஒரே நாள்லே வாய்க்காலும் சுத்தமாயிடும். தண்ணியும் கடகடன்னு வடிஞ்சிடும். இப்படிச் சொன்ன மருதனை ஏற இறங்கப் பார்த்தான். பார்த்ததோடு சரி. காதில் கேட்காதவாறு மாரி அவன் வேலையைச் செய்ய ஆரம்பித்தான் பொறுமையிழந்த மருதன் “ஏண்டா மாரி, நான் சொன்னது உங்கிட்டதான். நீ சாஞ்சுகிட்டிருந்த பனைமரத்துக்கிட்டேயில்லை.” “தெரியுது … ஏதாவது நடக்கிற காரியமா இருந்தா பதில் சொல்லலாம். நீயோ போகாத ஊருக்கு வழி கேக்கிற …. நானென்ன சொல்ல முடியும்?” “ச்சே… நீயெல்லாம் ஒரு மனுஷன் ….. முதமுத உங்கட்ட வந்து கேட்டேன் பாரு… என்னைச் சொல்லணும்..” கோபம் மாறாமல் கீழே இறங்கினான் மருதன்.

மருதன் காளியப்பனின் உதவியை நாடினான்:
வடக்கேயிருக்கும் எட்டூரு தண்ணியும் நம்மூரு வழியாத்தானே வடிஞ்சாகனும். மேற்கொண்டு மழை பேயணும் கூட அவசியமில்லை…… ராத்திரிக்குள்ளே எல்லாத் தண்ணியும் இங்கே வந்திறங்கிடுச்சின்னா…. அவ்வளவுதான் … இப்பவே எல்லாப் பயிரும் தோகையாடுது. எல்லாமே அப்புறம் தண்ணிக்குள்ளதான். ராத்திரி நம்ம ஊர்க்காரர்களை ஒண்ணு கூட்டி ஒரு வார்த்தை சொன்னீங்கன்னா போதும். கிழவரின் புருவம் ஏறி இறங்கியது. ஏன்டா மருதா … உனக்கு விவரம் தெரிஞ்சு நம்ப ஊரு பயலுவ எந்த நல்ல காரியத்துக்காகவாவது ஒண்ணு கூடியிருக்கானுவளா? மூலைக்கு ஒருத்தனா முறுக்கிக்கிட்டுல்லே போவானுங்க.

சொல்ற விதத்திலே சொன்னா எல்லாருமே கேப்பாங்க …. அதிலும் உங்க சொல்லுக்க மதிப்பு ஜாஸ்தி. யோசிக்காதீங்க பெரியப்பா …. ஒருநாள் தாமதிச்சாலும் ஊரே பாழாப் போயிடும்…… மருதனின் கவலையும், பதைப்பும் கிழவரை என்னவோ செய்தது. இருந்தாலும் கண்மூடி யோசித்தார். நாளைக்குக் காலையிலேயே பலபலன்னு விடியறப்ப வானமா தேவியிலே கட்டிக் கொடுத்திருக்கிற எம்மக வீட்லே இருந்தாகணும். குடும்பத்தோட வில் வண்டியிலே போறோம். அங்கே பேத்திக்குத் தலை சுத்துறாங்க திரும்ப வர மூணு நாளாகும். கிழவரின் சாதுரியம் மருதனுக்குப் புரிந்துவிட்டது.

மருதன் பட்டதாரி பிரேம்குமாரை சந்தித்தல் :
பிரேம்குமார் கிராமத்தின் முதல் பட்டதாரி. “நாகூர்பிச்சை” என்று அப்பா, அம்மா வைத்த பெயரை ”பிரேம்குமார்” என்று மாற்றி வைத்துக்கொண்டு “மன்றம் ” அது இது வென்று என்னவென்னவோ சதா சர்வகாலமும் செய்து கொண்டிருப்பவன். நாம நினைக்கிற காரியத்துக்கு இவன்தான் பொருத்தமானவன். முகம் மலர பிரேம்குமாரை வழி மறித்தான். “என்னண்ணே …” சிரித்தபடி பிரேம்குமார். கடகடவென்று எல்லாவற்றையும் சொல்லி முடித்தான் மருதன். ஆனால் அவன் எதிர்பார்த்தபடி, பிரேமிடமிருந்து சட்டென்று எந்தப் பதிலும் வரவில்லை. சிறிது நேர யோசிப்புக்குப் பிறகு மருதனை ஏறிட்டான்.

மருதண்ணே… நீங்க சொல்ற வேலையைச் செய்றதுக்குன்னே பிடபின்யூ டின்னு கவர்மெண்ட்லே ஒரு டிபார்ட்மெண்ட் இருக்கு. நாளைக் காலையிலே அவங்களைப் பார்த்து ஒரு ‘பெட்டிஷன்’ கொடுத்தீங்கன்னா செஞ்சுட்டுப் போறாங்க என்று கூறிவிட்டு மருதனின் பதிலை எதிர்பார்க்காமல் அவ்ன் போய்க்கொண்டே இருந்தான்.

மருதனின் புலம்பல்:
மருதனால் சும்மா இருக்க முடியவில்லை. ஒருத்தர் பாக்கி இல்லாமல் ஊர்க்காரர்களிடம் சொல்லிச் சொல்லிப் புலம்பினான். அத்தனை பேரும் அவன் சொன்னதை ஒப்புக்கொண்டார்கள். ஆனால் அதென்ன அதிசயமோ தெரியவில்லை. ஒவ்வொருவருக்கும் ஒவ்வொரு வேலை இருந்தது. அதுவும் தள்ளிப்போட முடியாத அவசர வேலை.

மருதன் தன் மனைவி அல்லியிடம் கூறுதல்.
தன் மனைவி அல்லியிடம் மருதன் நடந்ததைக் கூறினான். அதனைக் கேட்ட அல்லி இந்த ஊர்ல இருக்கிற மொத்தம் அறுபது வேலி நிலத்திலே நமக்குன்னு ஒரு “சக்கரைக்குழி” நிலம் கூட இல்லே. எந்த நிலம் எப்படிப்போனா நமக்கென்ன… நமக்குன்னு சொந்தம் கொண்டாட இருக்கிறது நம்ம கையும் காலும்தான். இந்த ஊரு இல்லேன்னா … இன்னொரு ஊரு …. வேலையைப் பாப்பியா ….. ஆவேசமாய்க் கொட்டி முழக்கிவிட்டு உள்ளே போனாள்.

மருதனின் முயற்சி:
தளும்புகின்ற வடி வாய்க்காலில் ஜில்லென்ற இடுப்பளவு தண்ணீரில் தன்னந்தனியே நின்றபடி மண்டிக்கிடந்த காட்டாமணக்குச் செடிகளை “சரக் சரக்” கென்று அறுத்து மேலே எறிந்து கொண்டிருந்தான் மருதன். அப்படியே திகைத்துப் போய் நின்றுவிட்டாள் அல்லி. அவளையறியாமலேயே புடவையை வரிந்து கட்டிக்கொண்டு வாய்க்காலுக்குள் இறங்கிவிட்டாள். “நீ சொல்றது” நிஜம்தான் மாமா. ஊரு நல்லா இருந்தாதான் நாமளும் நல்லாயிருக்கலாம். அதுக்காக இவ்ளோ நீளமான வாய்க்காலை நீயும் நானும் மட்டுமே சுத்தம் பண்ணிட முடியுமா? ஆற்றாமையுடன் கேட்டவளைத் திரும்பிப் பார்க்காமலே பதில் தந்தான். “முதல்லே நம்மாலே முடிஞ்சதை நாம செய்வோம்…”!

மருதன் முயற்சியால் ஏற்பட்ட மாற்றம்:
மருதன் மற்றும் அல்லி செய்து கொண்டிருந்த வேலையைக் கண்டு மாரியும் வேலையில் ஈடுபட்டான். இவர்கள் மூவரையும் கண்ட காளியப்பன் வண்டியில் இருந்து இறங்கி அவரும் செய்தார். இதனை வண்டிக்காரர் மூலம் அறிந்த ஊர்மக்கள் ஒவ்வொருவரும் வந்து செய்ய ஆரம்பித்தனர். “ஊர் கூடித் தேர் இழுக்கும் போதும்“ வேர்வடத்தைப் பிடிக்கும் முதல் கரமாக இருந்தது மருதனின் கரம்.

முடிவுரை:
“ஆக்கமும் அழிவும் நம்மாலே” என்னும் பழமொழிக்கு இணங்க மருதனின் பொறுப்புணர்வால் அவனுக்கும் ஊருதிடும் நன்மை ஏற்பட்டது. மருதனின் பண்பு பாராட்டிற்கு உரியது.

(அல்லது)

(ஆ) ‘கிராமங்கள் தங்கள் முகவரியை இழந்து வருகின்றன’ – தலைக்குளம் கதையின்றி உங்கள் கருத்தை விவரிக்க.
Answer:
குளத்துக்கரை விநாயகரும், அரசமரமும், சுத்தமான காற்றும், காதிற்கினிய குயில் ஓசையும் கோடிக்கணக்கான பணம் கொடுத்தாலும் கிடைக்காத சந்தோசம் தரும் இடமாக கிராமங்கள் திகழ்கின்றன.

எந்தத் தொழில் முறை மாறினாலும் அழிந்தாலும் உலகம் இயங்குவது பாதிக்காது. ஆனால் உயிர் கொடுக்கும் உழவுத்தொழில் அழிந்தால் உலகம் இருக்கும். ஆனால் உயிர்கள் இருக்காது. அப்படிப்பட்ட உழவுத்தொழில் செய்யும் கிராமங்கள் நகரை நோக்கித் திரும்பிச் செல்வது காலத்தின் கட்டாயம் என்றாலும், கஷ்டகாலம் நம்மை நோக்கி வருகிறது என்பதும் நம் அறிய வேண்டியதும் கட்டாயம். நம் கிராமங்கள் அழிந்து வருவதை நாம் கண்கூடாகப் பார்த்து வருகின்றோம்.

நகரத்திற்கு வரும் கிராமத்தினர் பட்டியலில் இந்திய அளவில் தமிழகம் 3 வது இடத்தில் உள்ள மொத்த மக்கள் தொகையில் 14.5% பேர் கிராமங்களை காலி செய்துவிட்டு நகருக்குள் வந்துவிட்டனர். மொத்தமுள்ள மக்களின் 48.40% பேர் நகரங்களில் வசிப்பது உணவு உற்பத்திக்கு விடப்பட்ட அபாய எச்சரிக்கை. ஏற்கனவே கேரளா, கர்நாடகா, ஆந்திராவிடம் தண்ணீருக்காகப் போராடிவரும் நாம், இனிவரும் காலங்களில் உணவுக்கு அண்டை மாநிலத்தை நம்பி இருக்க வேண்டிய காலம் வெகு தொலைவில் இல்லை.

கிராமங்களுக்கு என சில அடையாளங்கள் உண்டு. வயல்வெளி, திண்ணைவீடுகள், மரத்தடி கோயில், குளம் இவற்றுடன் மண் மணக்கும் விளையாட்டுகள் – கபடி தமிழகக் கிராமங்களுக்கே உரிய அடையாளம். ஒரு காலத்தில் ஊருக்கு ஊர் கேட்ட சடு … குடு சப்தங்களை இப்போது கேட்க முடியவில்லை. எங்கே செல்கிறோம் நாம் என்றே புரியவில்லை.

“ஆற்றங்கரை ஓரம் அமைந்த வீடுகள்
பச்சை கம்பளம் விரித்தாற் போல
பசுமை நிறைந்த வயல் வெளிகள்
குடும்பத்தோடு அகம் மகிழும்
திண்ணை அமர்வு”

என அனைத்தையும் தொலைத்துவிட்டு ஆன்ராய்ட் அலைபேசியின் தொடு திரையில் அழகிய படங்களைத் தேடுதலால் மட்டுமே தொலைந்த நம் கிராமத்தைத் திருப்பி விட முடியாது. இயன்ற வரை இயற்கையைப் போற்றி வாழ்வோம்.
கிராமத்தை அழிவிலிருந்து மீட்போம் !

பகுதி – V

அடிமாறாமல் செய்யுள் வடிவில் எழுதுக.

Question 47.
(அ) சுரதாவின் விண்வேறு என்று துவங்கும் பாடலை எழுதுக.[1 x 4 = 4]
Answer:
விண்வேறு; விண்வெளியில் இயங்கு கின்ற
வெண்மதியும் செங்கதிரும் முகிலும் வேறு;
மண்வேறு; மண்ணோடு கலந்தி ருக்கும்
மணல்வேறு; பனித்துளியும் மழையும் வேறு;
புண்வேறு; வீரர்களின் விழுப்புண் வேறு;
புகழ்வேறு; செல்வாக்கு வேறு;
காணும் கண்வேறு; கல்விக்கண் வேறு; கற்றார்
கவிநடையும் உரைநடையும் வேறு வேறு. (சுரதா)

(ஆ) ‘சினம்’ என முடியும் குறள் வெண்பாவை எழுதுக.[1 x 2 = 2]
Answer:
தன்னைத்தான் காக்கின் சினம்காக்க; காவாக்கால்
தன்னையே கொல்லும் சினம்.( – திருவள்ளுவர்)

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 2 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts An internal choice of questions is provided wherever applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences
6. Question numbers 25 to 33 in Part III are three-marks questions. These are to be answered in about three to five short sentences
7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Blistered copper is .
(a) 98% pure copper
(b) 96% pure copper
(c) 97% pure copper
(d) 88% pure copper
Answer:
(a) 98% pure copper

Question 2.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) Mg₂SiO₄ is an orthosilicate
(c) SiO₄ is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is not aluminosilicate

Question 3.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent
Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Answer:
(c) (i) and (iii)

Question 4.
The correct order of increasing oxidizing power in the series


Answer:

Hint: VO₂⁺ < Cr₂O₇^2- < MnO₄^– greater the oxidation state, higher is the oxidising power.

Question 5.
The geometry possible in [Fe F₆]^4- and [COF₊₆]^4- is ……………
(a) Trigonal bipyramidal
(b) Square planar
(c) Octahedral
(d) Tetrahedral
Answer:
(c) Octahedral

Question 6.
The number of carbon atoms per unit cell of diamond is
(a) 8
(b) 6
(c) 1
(d) 4
Answer:
(a) 8

Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.

Question 7.
In a homogeneous reaction
A → B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀ , P and t will be ……..

Answer:

Solution:

Question 8.
The solubility of an aqueous solution of Mg(OH)₂ be x then its K_sp is
(a) 4x³
(b) 108x⁵
(c) 27x⁴
(d) 9x
Answer:
(a) 4x³
Solution:
Mg(OH)₂ Mg²⁺(x) + 2OH^– ((2x)²)
K_sp = 4x³

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0. lmol of chlorine gas using a current of 3 A is ……………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Hint: m = ZIt (mass of 1 mole of Cl₂ gas = 71)
Answer:
(b) 107.2 minutes

Question 10:
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS₂S₃ are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO₄) = 0.22
The correct order of their coagulating power is
(a) III > II > I
(b) I > II > III
(c) I > III > II
(d) II > III> I
coagulating power ± \(\frac{1}{\text { coagulation value }}\)
Answer:
(a) III > II > I

Question 10.
Assertion : Coagulation power of Al³⁺ is more than Na⁺ .
Reason : greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)

Question 11.


Answer:
(a) A- 2, B – 3, C – 4, D – 1

Question 12.
Which of the following represents the correct order of acidity in the given compounds (
(a) FCH₂COOH > CH₃COOH > BrCH₂COOH > ClCH₃COOH
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH
(c) CH₃COOH > ClCH₂COOH > FCH₂COOH > Br – CH₂COOH
(d) Cl CH₂COOH > CH₃COOH > BrCH₂COOH > ICH₂COOH
Hint. -I effect increases the acidity. If electronegativity is high, -I effect is also high.
Answer:
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH

Question 13.
1 -nitrobutane and 2-methyl-1 -nitropropane are belong to
(a) position isomerism
(b) functional isomerism
(c) Tautomerism
(d) chain isomerism
Answer:
(d) chain isomerism

Question 14.
Haemoglobin is
(a) an enzyme
(b) a globular protein
(c) a vitamin
(d) carbohydrate
Answer:
(b) a globular protein

Question 15.
An example of antifertility drug is
(a) novestrol
(b) seldane
(c) salvarsan
(d) Chloramphenicol
Answer:
(a) novestrol

Part – II

Answer any six questions. Question No. 20 is compulsory. [6 x 2 = 12]

Question 16.
Name the method used for the refining of (i) Nickel (ii) Zirconium
Answer:
(i) Mond’s process
(ii) Van Arkel’s method

Question 17.
Complete the following reactions:
(a) B(OH)₃ + NH₃ →
(b) Na₂B₄O₇ + H₂SO₄ + H₂O
Answer:

Question 18.
KMnO₄ does not act as oxidising agent in the presence of HC1. Why?
Answer:
HCl cannot be used for making acidified KMnO₄ as oxidising agent, since it reacts with KMnO₄ as follows.
2MnO₄^– + 10 Cl^– + 16H⁺ → 2Mn²⁺ + 5Cl₂ + 8H₂O

Question 19.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled? ,
Answer:
(i) \(\frac{d x}{d t}=k[\mathrm{A}]^{1}[\mathrm{B}]^{2}\)
(ii) If concentration of ‘B ’ is tripled, then the rate will become 9 times.
(iii) When concentration of both A and B are doubled, then the rate will become 8 times.

Question 20.
Calculate the pH of 0.04 M HNO₃ Solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = -log[H₃O⁺]
= – log(0.04)
= – log(4 x 10^-2)
= 2 -log 4 = 2-0.6021
= 1.3979= 1.40

Question 21.
Write a note about Helmoholtz double layer.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This layer attracts the oppositely charged ions in the medium and hence the boundary separating the two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges, they cannot come close and condense. Hence this helps to explain the stability of the colloid.

Question 22.
Why HCOOH does not give Hell-Volhard Zelinsky reaction but CH₃3COOH does?
Answer:
CH₃COOH contains α – hydrogen atom and hence gives HVZ reaction but HCOOH does not contain an α – hydrogen atom and hence does not give HVZ reaction.

Question 23.
What happens when
Oxidation of acetoneoxime with trifluoroperoxy acetic acid.
Answer:
Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.

Question 24.
What are food preservatives?
Answer:
Food preservatives are chemical substances are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by growth of micro organisms.
Examples:

• Acetic acid is used mainly as a preservative for preparation of pickles.
• Sodium meta sulphite is used as preservative for fresh vegetables and fruits.
• Sodium benzoate is used as preservative for juices.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 x 3 = 18]

Question 25.
Explain the principle of electrolytic refining with an example.
Answer:
The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing aqueous solution of the salts of the metal of interest. The rods of impure metal are used as anode and thin strips of pure metal are used as cathode. The metal of interest dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Let us understand this process by considering electrolytic refining of silver as an example.
Cathode: Pure silver
Anode: Impure silver rods

Electrolyte: Acidified aqueous solution of silver nitrate.
When a current is passed through the electrodes the following reactions will take place
Reaction at anode: 2Ag(s) → Ag⁺ (aq) + le^–
Reaction at cathode: Ag⁺(aq)+ le^– → Ag (s)

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

Question 26.
How will you prepare phosphine and explain the purification of phosphine?
Phosphine’is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

Phosphine is freed from phosphine dihydride (P2H4) by passing through a freezing mixture. The dihydride condenses while phosphine does not.
Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda S0lution.

Question 27.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
(i) In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
(a) Lanthanoids have general electronic configuration [Xe] 4f2 14 5d° 1 6s2
(b) The common oxidation state of lanthanoids is +3
(c) All these elements have similar physical and chemical properties.

(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties.

(iii) If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.

(iv) Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.

Question 28.
Distinguish between isotropy and anisotropy?
Answer:
Isotropy:

1. Isotropy means uniformity in all directions.
2. Isotropy means having identical values of physical properties such as refractive index, electrical conductance in all directions.
3. Isotropy is the property of amorphous solids.

Anisotropy :

1. Anisotropy means non-uniformity in all directions.
2. Anisotropy is the property which depends on the direction of measurement. They show different values of physical properties when measured along different directions.
3. Anisotropy is the properly of crystalline solids.

Question 29.
What are the merits and limitations of the intermediate compound theory?
Answer:
Merits:
(i) The specificity of a catalyst
(ii) The increase in the rate of the reaction with increase in the concentration of catalyst.

Limitations:
(i) This theory fails to explain the action of catalytic poison and promoters. .
(ii) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 30.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection.
Al, Zn and Mg are used as sacrificial anodes.

Question 31.
1mole of Hlis allowed to react with t-butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:

Question 32.
How will you prove the presence of aldehyde group in glucose?
Answer:
(i) Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen’s reagent) and alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver and Fehling’s solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.

Question 33.
Explain about anaesthetics with their types.
(i) Local anaesthetics: It causes loss of sensation in the area in which it is applied without losing consciousness. They block pain perception that is transmitted via peripheral nerve fibres to the brain.
Example: Procaine, Li do caine.
They are often used during minor surgical procedures.

(ii) General anaesthetics: They cause a controlled and reversible loss of consciousness by affecting central nervous system.
Example: Propofol, Isoflurane.
They are often used for major surgical procedures.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) CO is more stable at higher temperature. Why? (2)
(ii) How will you prepare potash alum? (3)
[OR]
(b) (i) Give the balanced equation for the reaction between chlorine with cold NaOH. (3)
(ii) Nitrogen exists as diatomic molecule and Phosphrus as P₄ Why? (2)
Answer:
(a) (i) In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case AS is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

(ii) The alunite the alum stone is the naturally occurring form and it is K₂SO₄ . Al₂(S0₄)₃.4Al(0H)₃. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of – potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.
(2)

(b)(i) Reaction between chlorine with cold NaoH

Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

(ii) Nitrogen has a triple bond between its two atoms because of its small size and high electronegativity. Phosphorus P₄ has single bond, that is why it is tetra-atomic.

Question 35.
(a) (i) Discuss the ortho and pyro silicates. (2)
(ii) Compare lanthanides and actinides. (3)
[OR]
(b) (i) Ni²⁺ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction. (3)
Cu⁺, Zn²⁺, Sc³⁺, Ti⁴⁺ are colourless. Prove this statement. (2)
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or neso silicates.

Examples: Phenacite – Be₂SiO₄(Be²⁺ ions are tetrahedrally surrounded by O^2- ions), Olivine – (Fe/Mg)₂ SiO₄ (‘Fe²⁺ and Mg²⁺ cations are octahedrally surrounded by O

Pyro silicates: Silicates which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates. They are formed by joining two [SiO₄]^4-

tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while joining). Example: Thortveitite – Sc₂Si₂O₇

(ii)
Lanthanoids

1. Differentiating electron enters in 4f orbital
2. Binding energy of 4f orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.

Actinoids

1. Differentiating electron enters in 5f orbital
2. Binding energy of 5f orbitals are lower
3. They show greater tendency to form complexes
4. Most of the actinoids are coloured.
5. E.g : U³⁺ (red), U⁴⁺ (green), UO₂²⁺ (yellow)
6. They do form oxo cations such as UO₂²⁺ NpO₂²⁺ etc.
7. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7.

[OR]

(b) (i) Ni²⁺ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called [Ni(DMG)₂]
Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of [Ni(DMG)₂] complex.

(ii) 1. Cu⁺, Zn²⁺ have d10 configuration and Sc³⁺, Ti⁴⁺have d¹ configuration.
2. d-d transition is not possible in the above complexes. So they are colourless.

Question 36.
(a) i)0 What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure? (2)
(ii) Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation
\(\log \mathbf{k}=\log \mathbf{A} \frac{\mathbf{E}_{\mathbf{a}}}{2.303 \mathbf{R}}\left(\frac{1}{\mathbf{T}}\right)\)

Where E_a is the activation energy. When a graph is plotted for logk Vs \(\frac{1}{T}\) a straight line
with a slope of- 4000K Is obtained. Calculate the activation energy. (3)

(OR)

(b) (i) Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h for that reaction. (3)
(ii) Identify the Lewis acid and the Lewis base in the following reactions. (2)

Answer:
(a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water

NaOH_(aq) + CH₃COOH_(aq) ⇌ CH₃COONa_(aq) + H₂O

2. Coordination number of atoms in a bcc structure is 8

(ii) logk = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{T}}\right)\)
y = c + mx
m = \(-\frac{E_{a}}{2.303 R}\)
E_a =—2.303Rm
E_a = – 2.303 x 8.314 J K^-1 mol^-1 x (— 4000 K)
E_a = 76,589 J mol^-1
E_a =76589kJmol^-1

[OR]

(b) (i) 1. Coordination number: The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. In aqueous solution, CH₃COONa is completely dissociated as follows.
CH₃COONa_(aq), →  CH₃COO^–_(aq) + Na⁺_(aq)

3. CH₃COO^– is a conjugate base of the weak acid CHCOOH and it has a tendency to react with H⁺ from water to produce unionised acid. But there is no such tendency
for Na⁺ to react with OH^–

4.

such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law K_h = h²C and \(\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}}\)

Question 37.
(a) (i) What are lyophilic and lyophobic sols? Give one example of each type. Why are
Answer:
(a) (i) Lyophilic Sols: Colloidal sols directly formed by mixing substances like gums, gelatin, starch, rubber, etc. with a suitable liquid (The dispersion medium) are lyophilic sols.
An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated.

Lyophobic sols: These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.
Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

(ii) 1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons.
2. In the reaction 2SO₂ + O₂ → 2SO₃ with Pt catalyst, the catalyst poison is AS₂O₃

[OR]

(b) (i) 1. Substances like AgCl, PbS04 are sparingly soluble in water. The solubility product can be determined using conductivity experiments.
2. Let us consider AgCl as an example
Agci_(s)) ⇌ Ag⁺ + Cl^–
K_Sp = [Ag ⁺] [Cl^–]

3. Let the concentration of [Ag⁺] be ‘C’ mol L^-1
If [Ag⁺] = C, then [Cl^–] is also equal to C mol L^-1.
.’. K_sp = C.C
K_sp = C²

4. The relationship between molar conductance and equivalent conductance is

(ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.

2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver.

3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions.

Question 38.
(a) (i) What are the uses of nitrobenzene? (2)
(ii) Write a note on formation of -helix. (3)
[OR]
(b) (i) Define TFM value. (2)
(ii) Differentiate thermoplastic and thermosetting. (3)
Answer:
(a) (i) 1. Nitro benzene is used to produce lubricating oils in motors and machinery.
2. It is used in the manufacture of dyes, drugs, pesticides, synthelic rubber, aniline and explosives like TNT, TNB.
(ii) 1. In the a-helix sub-structure, the aminoacids are arranged in a right handed helical (spiral) structure and are stabilised by the hydrogen bond between the carbonyl oxygen one aminoacid (nth residue) with amino hydrogen of the fifth residue (n+4th residue).

2. The side chains of the residues protrude outside of the helix. Each turn of an a-helix contains about 3.6 residues and is about 5.4 A long.

3. The amino acid pro line produces a line in the helical structure and often called as a helical breaker due to its rigid cyclic structure.

4. Many fibrous proteins such as a-Keratin in hair, nails, wool, skin and myosin in muscles have a-helix structure. Stretching property of human hair is due to the helical structure of a-keratin in hair.

5. Structure of α – helix

[OR]

(b) (i) 1. The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the total amount of fatty matter that can be separated from a sample after spliting with mineral acids. Higher the TFM value in the soap, better is its quality.
2. As per BIS standards, Grade I soaps should have 76% minimum TFM value (ii) Difference between thermoplastic and thermosetting:

Thermoplastic :

1. They soften on heating and harden on cooling, and they can be remoulded.
2. They consists of linear long chain polymers and low molecular weights polymers.
3. All the polymer chains are held together by weak Vanderwaals forces.
4. They are weak, soft and less brittle.
5. They are formed by addition polymerisation.
6. They are soluble in organic solvents.
7.  Example: PVC, polythene, polystrene etc.

Thermosetting

1. They do not soften on heating and they cannot be remoulded.
2. The consist of three dimensional network structure and high molecular weight polymers.
3. All the polymer chains are linked by strong covalent.
4. They are strong, hard and more brittle.
5. They are formed by condensation polymerisation.
6. They are insoluble in organic solvents.
7. Example: Bakelite, melamine etc.

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 4 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

Answer any six questions in which Q.No. 28 is compulsory. [6 x 3 = 18]

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication	Wireless communication
It is a point-to-point communication.	It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres.	It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

Answer all the questions.  [5 x 5 = 25]

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁, is included in the secondary circuit and the balancing length l₁, is found by adjusting the jockey for zero deflection. Then the second cells ξ₂, is included in the circuit and the balancing length l₂, is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have
ξ₁ = Irl₁ ……………….. (1)
ξ₂ = Irl₂ ……………….. (2)
By dividing equation (1) by (2)
\(\frac{\xi_{1}}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle : When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction : The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

(b) Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

(2) Eddy current brake : This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

(4) Electro magnetic damping : The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer
• For a given cathode, the magnitude of
  the current depends on the intensity to incident radiation and
  the potential difference between anode and cathode.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then eE = eBv
\(v=\frac{E}{B}\) …………….. (1)

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.

Students can Download Tamil Nadu 12th Biology Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 4 English Medium

General Instructions:

1. 1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
   2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
   3. All questions of Part I, II, III and IV are to be attempted separately.
   4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
   5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
   6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
   7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
A Plant called ‘X’ possesses small flower with reduced perianth and versatile anther. The probable agent for pollination would be _________.
(a) water
(b) air
(c) butterflies
(d) beetles
Answer:
(b) air

Question 2.
“Gametes are never hybrid”. This is a statement of _______.
(a) Law of dominance
(b) Law of independent assortment
(c) Law of segregation
(d) Law of random fertilization
Answer:
(c) Law of segregation

Question 3.
In which techniques Ethidium Bromide is used?
(a) Southern Blotting techniques
(b) Western Blotting techniques
(c) Polymerase Chain Reaction
(d) Agarose Gel Electrophoresis
Answer:
(d) Agarose Gel Electrophoresis

Question 4.
Which of the following statement is correct?
(a) Agar is not extracted from marine algae such as seaweeds
(b) Callus undergoes differentiation and produces somatic embryoids
(c) Surface sterilization of explants is done by using mercuric bromide
(d) pH of the culture medium is 5.0 to 6.0
Answer:
(d) pH of the culture medium is 5.0 to 6.0

Question 5.
The term pedogenesis is related to _______.
(a) Fossils
(b) Water
(c) Population
(d) Soil
Answer:
(d) Soil

Question 6.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
(a) Ammonia
(b) Methane
(c) Nitrous oxide
(d) Ozone
Answer:
(d) Ozone

Question 7.
A wheat variety, Atlas 66 which has been used as a donor for improving cultivated wheat is rich in _______.
(a) iron
(b) carbohydrates
(c) proteins
(d) vitamins
Answer:
(c) proteins

Question 8.
Statement A: Coffee contains caffeine.
Statement B: Drinking coffee enhances cancer.
(a) A is correct, B is wrong
(b) A and B – Both are correct
(c) A is wrong, B is correct
(d) A and B – Both are wrong
Answer:
(b) A and B – Both are correct

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Draw and label the structure of a typical pollen grain.
Answer:

Question 10.
What is test cross? Why it is done?
Answer:
Test cross is crossing an individual of unknown genotype with a homozygous recessive. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character.

Question 11.
You are working in a biotechnology lab with a bacterium namely E.coli. How will you cut the nucleotide sequence? Explain it.
Answer:
The DNA nucleotide sequence can be cut using Restriction endonucleases (RE). Restriction

Question 12.
What is ecological hierarchy? Name the levels of ecological hierarchy.
Answer:
The interaction of organisms with their environment results in the establishment of grouping of organisms which is called ecological hierarchy.

Question 13.
Mutagens are the substances that induces mutation. Name any two physical and chemical mutagens.
Answer:
UV short waves, X-rays – Physical mutagens.
Nitromethyl, Urea – Chemical mutagens.

Question 14.
What is pseudo cereal? Give an example.
Answer:
The term pseudo-cereal is used to describe foods that are prepared and eaten as a whole grain, but are botanical outliers from grasses. Example: quinoa. It is actually a seed from the Chenopodium quinoa plant, belongs to the family Amaranthaceae.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
“Endothecium is associated with dehiscence of anther” Justify the statement.
Answer:
The inner tangential wall develops bands (sometimes radial walls also) of α cellulose (sometimes also slightly lignified). The cells are hygroscopic. The cells along the junction of the two sporangia of an anther lobe lack these thickenings. This region is called stomium. This region along with the hygroscopic nature of endothecium helps in the dehiscence of anther at maturity.

Question 16.
What is gene mapping? Write its uses.
Answer:
The diagrammatic representation of position of genes and related distances between the adjacent genes is called genetic mapping. It is directly proportional to the frequency of recombination between them. It is also called as linkage map.

Uses of genetic mapping:

• It is used to determine gene order, identify the locus of a gene and calculate the distances between genes.
• They are useful in predicting results of dihybrid and trihybrid crosses.
• It allows the geneticists to understand the overall genetic complexity of particular organism.

Question 17.
How synthetic seeds are developed?
Answer:
Artificial seeds or synthetic seeds (synseeds) are produced by using embryoids (somatic embryos) obtained through in vitro culture. They may even be derived from single cells from any part of the plant that later divide to form cell mass containing dense cytoplasm, large nucleus, starch grains, proteins, and oils, etc. To prepare the artificial seeds different inert materials are used for coating the somatic embryoids like agrose and sodium alginate.

Question 18.
Discuss the three zones of a lentic ecosystem.
Answer:
There are three zones, littoral, limnetic and profundal. The littoral zone, which is closest to the shore with shallow water region, allows easy penetration of light. It is warm and occupied by rooted plant species. The limnetic zone refers the open water of the pond with an effective penetration of light and domination of planktons.

The deeper region of a pond below the limnetic zone is called profundal zone with no effective light penetration and predominance of heterotrophs. The bottom zone of a pond is termed benthic and is occupied by a community of organisms called benthos (usually decomposers).

Question 19.
Write a short note on clean development mechanism.
Answer:
Clean Development Mechanism (CDM) is defined in the Kyoto protocol (2007) which provides project based mechanisms with two objectives to prevent dangerous climate change and to reduce green house gas emissions. CDM projects helps the countries to reduce or limit emission and stimulate sustainable development.

An example for CDM project activity, is replacement of conventional electrification projects with solar panels or other energy efficient boilers. Such projects can earn Certified Emission Reduction (CER) with credits / scores, each equivalent to one tonne of CO₂, which can be counted towards meeting Kyoto targets.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Describe dominant epistasis with an example.
Answer:

Dominant Epistasis – It is a gene interaction in which two alleles of a gene at one locus interfere and suppress or mask the phenotypic expression of a different pair of alleles of another gene at another locus. The gene that suppresses or masks the phenotypic expression of a gene at another locus is known as epistatic. The gene whose expression is interfered by non-allelic genes and prevents from exhibiting its character is known as hypostatic. When both the genes are present together, the phenotype is determined by the epistatic gene and not by the hypostatic gene.

In the summer squash the fruit colour locus has a dominant allele ‘W’ for white colour and a recessive allele ‘w’ for coloured fruit. ‘W’ allele is dominant that masks the expression of any colour. In another locus hypostatic allele ‘G’ is for yellow fruit and its recessive allele ‘g’ for green fruit. In the first locus the white is dominant to colour where as in the second locus yellow is dominant to green. When the white fruit with genotype WWgg is crossed with yellow fruit with genotype wwGG, the F₁ plants have white fruit and are heterozygous (WwGg). When F₁ heterozygous plants are crossed they give rise to F₂ with the phenotypic ratio of 12 white : 3 yellow : 1 green.

Since W is epistatic to the alleles ‘G’ and ‘g’ the white which is dominant, masks the effect of yellow or green. Homozygous recessive ww genotypes only can give the coloured fruits (4/16). Double recessive ‘wwgg’ will give green fruit (1/16). The Plants having only ‘G’ in its genotype (wwGg or wwGG) will give the yellow fruit(3/16).

[OR]

(b) Point out the significance of plant succession.
Answer:
Significance of Plant Succession:

• Succession is a dynamic process. Hence an ecologist can access and study the seral stages of a plant community found in a particular area.
• The knowledge of ecological succession helps to understand the controlled growth of one or more species in a forest.
• Utilizing the knowledge of succession, even dams can be protected by preventing siltation.
• It gives information about the techniques to be used during reforestation and afforestation.
• It helps in the maintenance of pastures.
• Plant succession helps to maintain species diversity in an ecosystem.
• Patterns of diversity during succession are influenced by resource availability and disturbance by various factors.
• Primary succession involves the colonization of habitat of an area devoid of life.
• Secondary succession involves the re-establishment of a plant community in disturbed area or habitat.
• Forests and vegetation that we come across all over the world are the result of plant succession.

Question 21.
(a) Compare the various types of Blotting techniques.
Answer:

[OR]

(b) Explain different types of hybridization.
Answer:
Types of Hybridization:
According to the relationship between plants, the hybridization is divided into.
1. Intravarietal hybridization – The cross between the plants of same variety. Such crosses are Useful only in the self-pollinated crops.

2.. Intervarietal hybridization – The cross between the plants belonging to two different varieties of the same species and is also known as intraspecific hybridization. This technique has been the basis of improving self-pollinated as well as cross pollinated crops.

3. Interspecific hybridization – The cross between the plants belonging to different species belonging to the same genus is also called intragenic hybridization. It is commonly used for transferring the genes of disease, insect, pest and drought resistance from one species to another.
Example: Gossypium hirsutum x Gossypium arboreum – Deviraj.

4. Intergeneric hybridization – The crosses are made between the plants belonging to two different genera. The disadvantages are hybrid sterility, time consuming and expensive procedure. Example: Raphanobrassica and Triticale.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Select the correct production site and action site of Relaxin.
(a) Hypothalamus and pituitary gland
(b) Pituitary gland and Pelvic joints and cervix
(c) Placenta and pelvic joint and cervix
(d) Hypothalamus and placenta
Answer:
(c) Placenta and pelvic joint and cervix

Question 2.
Fusion of young individuals produced immediately after the mitotic division of adult parent cell is called _______.
(a) Merogamy
(b) Anisogamy
(c) Hologamy
(d) Paedogamy
Answer:
(d) Paedogamy

Question 3.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number ______.
(a) 20
(b) 21
(c) 23
(d) 19
Answer:
(b) 21

Question 4.
DNA finger printing techniques was developed by _______.
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
(d) Hershey and Chase
Answer:
(b) Alec Jeffreys

Question 5.
Identify the correct sequence of periods from oldest to youngest
(a) Cambrian → Permian → Devonian → Silurian → Ordovician
(b) Permian → Silurian → Devonian → Ordovician → Cambrian
(c) Permian → Devonian → Silurian → Cambrian → Ordovician
(d) Cambrian → Ordovician → Silurian → Devonian → Permian
Answer:
(d) Cambrian → Ordovician → Silurian → Devonian → Permian

Question 6.
Spread of cancerous cells to distant sites is termed as _________.
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 7.
Assertion (A): Streptomycin is an antibiotic.
Reason (R): Antibiotic are microbial chemicals inhibits the growth of pathogenic microbe.
(a) A is right R is wrong
(b) R explains A
(c) A and R are wrong
(d) A is wrong but R is right
Answer:
(b) R explains A

Question 8.
World Ozone Day was observed on _________.
(a) September 16^th
(b) October 12^th
(c) December 1^th
(d) August 18^th
Answer:
(a) September 16^th

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Question 10.
State any two unique features of ELISA test.
Answer:
ELISA is highly sensitive and can detect antigen even in nanograms.
ELISA test does not require radioisotopes or radiation counting apparatus.

Question 11.
Define Anaphylaxis.
Answer:
Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation

Question 12.
Who is Cro-Magnon?
Answer:
Cro-Magnon was one of the most talked forms of modem human found from the rocks of Cro-Magnon, France and is considered as the ancestor of modem Europeans. They were not only adapted to various environmental conditions, but were also known for their cave paintings, figures on floors and walls.

Question 13.
What is S – D sequence?
Answer:
The 5′ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 14.
Expand (a) GIFT (b) ICSI
Answer:
GIFT – Gamete Intra – Fallopian Transfer
ICSI – Intra-cytoplasmic sperm injection

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Write a short note on encystment in amoeba.
Answer:
During unfavorable conditions (increase or decrease in temperature and scarcity of food) Amoeba withdraws its pseudopodia and secretes a three-layered, protective, chitinous cyst wall around it and becomes inactive. This phenomenon is called encystment. When conditions become favourable, the encysted Amoeba divides by multiple fission and produces many minute amoebae called pseudopodiospore or amoebulae.

The cyst wall absorbs water and breaks off liberating the young pseudopodiospores, each with a fine pseudopodia. They feed and grow rapidly to lead an independent life.

Question 16.
Draw a schematic representation of human oogenesis.
Answer:

Question 17.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

• Sex-education in school and public forums.
• Promoting the uses of contraception.
• Compulsory sterilization for mentally retarded and criminals.
• Egg donation.
• Artificial insemination by donors.
• Prenatal diagnosis of genetic disorders and performing MTP.
• Gene therapy.
• Cloning.
• Egg/sperm donation of healthy individuals.

Question 18.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.
1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.

2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 19.
What is Q₁₀ value? How it is calculated?
Answer:
The effect of temperature on the rate of reaction is expressed in terms of temperature coefficient or Q₁₀ value. The Q₁₀ values are estimated taking the ratio between the rate of reaction at X°C and rate of reaction at (X-10°C).

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Differentiate between r-selected species and k-selected species.
Answer:
r – selected species:

• Smaller sized organisms
• Produce many offspring
• Mature early
• Short life expectancy
• Each individual reproduces only once or few times in their life time
• Only few reach adulthood
• Unstable environment, density independent

k – seleced species:

• Larger sized organisms
• Produce few offspring
• Late maturity with extended parental care
• Long life expectancy
• Can reproduce more than once in lifetime
• Most individuals reach maximum life span
• Stable environment, density dependent

[OR]

(b) Give a detailed account on ethanol production by microbes and the uses of ethanol.
Answer:
Ethanol production:
Saccharomyces cerevisiae (Yeast) is the major product of ethanol.

Since ethanol is used for industrial, laboratory and fuel proposes, it is called as industrial alcohol.

Organism used: Saccharomyces cerevisiae, bacteria like Zymomonas mobilis and Sarcina ventriculi.
Substances used: Molasses, Com, Potatoes, wood waste.

Process of ethanol production:
Step-1: Milling of fees stock.
Step-2: Adding fungal (Aspergillus) amylase to break down starch into sugar.
Step-3: Yeast is added to convert sugar into ethanol.
Step-4: Distillation yield 96% concentrated ethanol.

Uses of Ethanol:
Ethanol and bio-diesel are the two commonly used first generation bio-fuels.
Ethanol is used as fuel, mainly as bio-fuel additive for gasoline.

Question 21.
(a) How DNA is packed in an eukaryotic cell?
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (H1) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an H1 molecule. Chromatin lacking H1 has a beads-on-a-string appearance in which DNA enters and leaves the nucleosomes at random places. H1 of one nucleosome can interact with H1 of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chromatin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucleosome, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different H1 molecules. DNA is a solenoid and packed about 40 folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of proteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In a typical nucleus, some regions of chromatin are loosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

[OR]

(b) Explain Oparin – Haldane hypothesis on evolution.
Answer:
According to the theory of chemical evolution primitive organisms in the primordial environment of the Earth evolved spontaneously from inorganic substances and physical forces such as lightning, UV radiations, volcanic activities, etc. Oparin (1924) suggested that the organic compounds could have undergone a series of reactions leading to more complex molecules. He proposed that the molecules formed colloidal aggregates or ‘coacervates’ in an aqueous environment.

The coacervates were able to absorb and assimilate organic compounds from the environment. Haldane (1929) proposed that the primordial sea served as a vast chemical laboratory powered by solar energy. The atmosphere was oxygen free and the combination of CO₂, NH₃ and UV radiations gave rise to organic compounds. The sea became a ‘hot’ dilute soup containing large populations of organic monomers and polymers.

They envisaged that groups of monomers and polymers acquired lipid membranes and further developed into the first living cell. Haldane coined the term prebiotic soup and this became the powerful symbol of the Oparin-Haldane view on the origin of life (1924-1929). Oparin and Haldane independently suggested that if the primitive ‘ atmosphere was reducing and if there was appropriate supply of energy such as lightning or UV light then a wide range of organic compounds can be synthesized.

Students can Download Tamil Nadu 12th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 1 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is …………….
(a) 80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 80 x 10^-17 J

Question 2.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Question 3.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at  R/2 distance between two coils is …………..

Answer:

Question 4.
An electron moves on a straight line path XY as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?

(a) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil

Question 5.
The resistance of an ideal ammeter is ……………
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 6.
In a step-down transformer the input voltage is 22 kV and the output voltage is 550 V. The ratio of the number of turns in the secondary to that in the primary is ……………
(a) 1 : 20
(b) 20 : 1
(c) 1 : 40
(d) 40 : 1
Answer:
(c) 1 : 40

Question 7.
If the magnetic monopole exists, then which of the Maxwell’s equation to be modified?

Answer
\(\oint \overrightarrow { { E } } \cdot d\overrightarrow { { A } } =0\)

Question 8.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint. Let d₁ = 5 cm and d₂ = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d₁n + d₂n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 9.
When a ray of light enters a glass slab from air
(a) its wavelength decreases
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint: \(Wavelength,\lambda =\frac { { Velocity } }{ { Frequency } } =\frac { u }{ v } \)
When light travels from air to glass, frequency ν remains unchanged, velocity u decreases and hence wave length λ also decreses

Question 10.
A particle of mass 3 x 10^-6 g has the same wavelength as an electron moving with a velocity
6 x 10⁶ m s^-1. The velocity of the particle is…………………………………………
(a) 1.82 x 10^-18 ms^-1
(b) 9 x 10^-2 ms^-1
(c) 3 x 10^-31 ms^-1
(d) 1.82 x 10¹⁵ ms^-1
Answer:
(d) 1.82 x 10¹⁵ ms^-1
Hint: de – Broglie wavelength of electron

Question 11.
If the nuclear radius of ²⁷A1 is 3.6 fermi, the approximate unclear radius of ⁶⁴Cu is
(a) 2.4
(b) 1.2
(c) 4.8
(d) 3.6
Answer:
(c) 4.8
Hint :

Question 12.
Energy of characteristic X-ray is a consequence of ………….
(a) energy of projectile electron
(b) thermal energy of target
(c) transition in target atoms
(d) none of the above
Answer:
(c) transition in target atoms

Question 13.
The specific characteristic of a common emitter amplifier is ……………
(a) High input resistance
(b) Low power gain
(c) Signal phase reversal
(d) Low current gain
Answer:
(c) Signal phase reversal

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called ………………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as ……………..
(a) Bulk material
(b) Nanomaterial
(c) Soft material
(d) Magnetic material
Answer:
(b) Nanomaterial

Part – II

Answer any six questions in which Q. No 17 is compulsory. [6 x 2 = 12]

Question 16.
Write a short note on the superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\(\overrightarrow{\mathrm{F}}_{1}^{\mathrm{tot}}=\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{13}+\overrightarrow{\mathrm{F}}_{14}+\cdots+\overrightarrow{\mathrm{F}}_{1 n}\)

Question 17.
If an electric field of magnitude 570 N C^-1, is applied in the copper wire, find the acceleration experienced by the electron.
Answer:

Question 18.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed loop is p0 times net current enclosed by the loop.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\)

Question 19.
Write down the equation for a sinusoidal voltage of §0 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.
Answer:

Question 20.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.

Question 21.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 22.
How many photons per second emanate from a 50 mW laser of 640 nm?
Answer:

Question 23.
An ideal diode and a 5 Ω resistor are connected in series with a 15 V power supply as shown in figure below. Calculate the current that flows through the diode.
Answer:
The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage.Therefore, current that flows through the diode can be calculated using Ohm’s law.

V = IR
\(I=\frac{V}{R}=\frac{15}{5}=3 \mathrm{A}\)

Question 24.
What do you mean by Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

Part – III

Answer any six questions in which Q.No. 26 is compulsory. [6 x 3 = 18]

Question 25.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 26.
If the resistance of coil is 3 Ω at 20°C and a = 0.004/°C then determine its resistance at 100°C.
Answer:
R₀ =3Ω = 100°C, T₀ = 20°C
α = 0.004/°C, R_T = ?
Rr= R₀(1+∝ (T-T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
⇒ R₁₀₀ = 3(1 + 0.32)
R₁₀₀= 3(1.32)
⇒ R₁₀₀ = 3.96 Ω

Question 27.
State ‘Tangent Law’.
Answer:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Question 28.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

Question 29.
One type of transparent glass has refractive index 1.5. What is the speed of light through this glass?
Answer:

Question 30.
What is a photo cell? Mention the different types of photocells.
Answer:
Photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:

• Photo emissive cell
• Photo voltaic cell
• Photo conductive cell

Question 31.
Half lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. Initially, the samples have equal number of nuclei. Calculate the ratio of decayed numbers of A and B nuclei after 80 minutes.
Answer:
80 minutes = 4 half lives of A = 2 half live of B
Let the initial number of nuclei in each sample be N.

Question 32.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Intrinsic	Extrinsic
These are pure semiconducting tetravalent crystals.	These are semiconducting tetravalent crystals doped with impurity atoms group III (or) V
Their electrical conductivity is low.	Their electrical conductivity is high.
There is no permitted energy state between valence and conduction band.	There is no permitted energy state of the impurity atom between valence and conduction band.
Their electrical conductivity depends on temperature.	Their electrical conductivity depends on temperature as well as dopant concentration.

Question 33.
Distinguish between Nanoscience and Nanotechnology.
Answer:

Nanoscience	Nanotechnology
Nanoscience is the science of objects with typical sizes of 1-100 nm. Nano means one- billionth of a metre that is 10-9 m. If matter is divided into such small objects the mechanical, electrical, optical, magnetic and other properties change.	Nanotechnology is a technology involv­ing the design, production, characteriza­tion, and applications of nano structured materials.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Obtain the expression for electric field due to an uniformly charged spherical shell. Electric field due to a uniformly charged
Answer:
spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a) At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially
outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,

The electric field \(\overrightarrow { { E } } { and }\quad d\overrightarrow { { A } } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\overrightarrow { { E } } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But = total area of Gaussian surface  = 4πr² Substituting this value in equation (2)

 

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\overrightarrow{\mathrm{E}}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}} \hat{r}\) ………. (4)

Case (c) At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law.

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes
E = 0    (r < R)  ………….. (6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

[OR]

(b) Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:

An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff’s current rule to junction B.
I₁ – I_G – I₃ = 0   ………….. (1)
Applying Kirchhoff’s current rule to junction D.
I₂ + I_G – I₄ = 0   ………….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁P + I_GG – I₂R = 0  ………….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁P + I₃Q – I₄S – I₂R = 0  ………….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B through galvanometer (I_Q = 0). Substituting I_G = 0 in equation, (1), (2) and (3), we get
I₁ = I₃  ………………….. (5)
I₁ = I₄ ……………………. (6)
I₁P =I₂R ……………….(7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₁S – I₁R = 0
I₁( P + Q) = I₂( P + S) …………………(8)
Dividing equation (7)

This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance  (fourth one) can be determined.

Question 35.
(a) Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 21. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follows:

The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{i}+\mathrm{F}_{\mathrm{N}} \sin \theta \hat{j}\) ……………. (1)
Where \(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{r^{\prime 2}}\) . The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is

From equation (1) and equation (2), the net force at point C is \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\mathrm{F}_{\mathrm{S}}\). This net force is equal to the magnetic field at the point C.

If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r>> l, then,

Therefore , using equation (7) in equation (6) ,we get
\(\overrightarrow{\mathrm{B}}_{\text {equatorial }}=-\frac{\mu_{0}}{4 \pi} \frac{p_{m}}{r^{3}} \hat{i}\)
Since in genaral , the magnetic filed at equatorial points is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatoral and the direction of B_axial and B_equatoral are opposite.

[OR]

Question 35.
(b) Give the advantage of AC in long distance power transmission with an example.
Answer:
Advantages of AC in long distance power transmission: Electric power is produced in a large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places. Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission.

But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i²R) in the transmission lines which are hundreds of kilometer long. This power loss can be tackled either by reducing current I or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases die cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent. At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer.

Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.

Illustration: An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case (i) : P = 2 MW; R = 40 Ω; V = 10 kV
Power, P= VI
Power loss = Heat produced = I²R = (200)² x 40 = 1.6 x 10⁶ W
% of power loss = \(\frac{1.6 \times 10^{6}}{2 \times 10^{6}} \times 100 \%\) = 0.8 x 100% = 80%

Case (ii): P = 2 MW; R = 40 Ω; V = 100 kV
∴ Current, \(I=\frac{P}{V}=\frac{2 \times 10^{6}}{100 \times 10^{3}}=20 \mathrm{A}\)
Power loss = Heat produced = I²R = (20)² x 40 = 0.016 x 10⁶ W
% of power loss = \(\frac{0.016 \times 10^{6}}{2 \times 10^{6}} \times 100 \%=\) = 0.008 x 100% = 0.8%

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[OR]

(b) Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n₂ sin i = n₂ sin r for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin 90° ………………. (1)
n₁ sin i_c = n₂
∵ sin 90° = 1
sin i_c = \(\frac{n_{2}}{n_{1}}\) ………………….(2)
From the right angle triangle ΔABC,

Question 37.
(a) Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Failures of classical wave theory:
Answer:
From Maxwell’s theory, light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of photoelectric effect using wave picture of light.

When light is incident on the target, there is a continuous supply of energy to the electrons. According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, Intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.

According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.

Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount of time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than 10^-9 s after the surface is illuminated) which could not be explained by wave theory.

[or]

(b) Obtain the law of radioactivity.
Answer:
Law of radioactive decay:
At any instant t, the number of decays per unit time, called rate of decay \(\left(\frac{d \mathrm{N}}{d t}\right) \) to the number of nuclei at the same instant.
\(\frac{d \mathrm{N}}{d t} \propto \mathrm{N}\)

By introducing a proportionality constant, the relation can be written as
\(\frac{d \mathrm{N}}{d t}=-\lambda \mathrm{N}\) ……………………. (1)

Here proportionality constant X is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. By rewriting the equation (1), we get
dN = -XNdt …………… (2)

Here dN represents the number of nuclei decaying in the time interval dt. Let us assume that at time t = 0 s, the number of nuclei present in the radioactive sample is N_o. By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t.
From equation (2), we get

N = N_o e^-λt …………… (4)
[Note: e^lnx = e^y ⇒ x = e^y ]
Equation (4) is called the law of radioactive decay. Here N denotes the number of undecayed nuelei present at any time t and N₀ denotes the number of nuclei at initial time t = 0. Note that the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite. Equation (4) is plotted.

We can also define another useful quantity called activity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as R = \(R=\left| \frac { dN }{ dt } \right| \)

Note : That activity R is a positive quantity. From equation (4), we get.

The equation (6) is also equivalent to radioactive law of decay. Here R₀ is the activity of the sample at t = 0 and R is the activity of the sample at any time t. From equation (6), activity also shows exponential decay behavior. The activity R also can be expressed in terms of number of undecayed atoms present at any time t. From equation (6), since N = N₀ e ^-λt we write
R = λN    ………………… (7)
Equation (4) implies that the activity at any time t is equal to the product of decay constant and number of undecayed nuclei at the same time t. Since N decreases over time, R also decreases.

Question 38.
(a) Explain the construction and working of a full wave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transformer, and a load resistor (R_L). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂. Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D, and reverse biases diode D_r Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC. As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature. The efficiency (r|) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

[OR]

(b) Explain the three modes of propagation of electromagnetic waves through space. Propagation of electromagnetic waves:
Answer:
The electromagnetic wave transmitted by the transmitter travels in three different modes to reach the receiver-according to its frequency range:
(i) Ground wave propagation (or) surface wave propagation (nearly 2 kHz to 2 MHz)
(ii) Sky wave propagation (or) ionospheric propagation (nearly 3 MHz to 30 MHz)
(iii) Space wave propagation (nearly 30 MHz to 400 GHz)

(i) Ground wave propagation
If the electromagnetic waves transmitted by the transmitter glide over the surface of the earth to reach the receiver, then the propagation is called ground wave propagation. The corresponding waves are called ground waves or surface waves.

 

Increasing distance: The attenuation of the signal depends on

• power of the transmitter
• frequency of the transmitter, and
• condition of the earth surface.

Absorption of energy by the Earth: When the transmitted signal in the form of EM wave is in contact with the Earth, it induces charges in the Earth and constitutes a current. Due to this, the earth behaves like a leaky capacitor which leads to the attenuation of the wave.

Tilting of the wave: As the wave progresses, the wavefront starts gradually tilting according to the curvature of the Earth. This increase in the tilt decreases the electric field strength of the wave. Finally, at some distance, the surface wave dies out due to energy loss.

The frequency of the ground waves is mostly less than 2 MHz as high frequency waves undergo more absorption of energy at the earth’s atmosphere. The medium wave signals received during the day time use surface wave propagation.

It is mainly used in local broadcasting, radio navigation, for ship-to-ship, ship-to-shore communication and mobile communication.

(ii) Sky Wave Propagation:
The mode of propagation in which the electromagnetic waves radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth is called sky wave propagation or ionospheric propagation. The corresponding waves are called sky waves.

The frequency range of EM waves in this mode of propagation is 3 to 30 MHz. EM waves of frequency more than 30 MHz can easily penetrate through the ionosphere and does not undergo reflection. It is used for short wave broadcast services. Medium and high frequencies are for long-distance radio communication. Extremely long distance communication, is also possible as the radio waves can undergo multiple reflections between the earth and the ionosphere. A single reflection helps the radio waves to travel a distance of approximately 4000 km.

Ionosphere acts as a reflecting surface. It is at a distance of approximately 50 km and spreads up to 400 km above the Earth surface. Due to the absorption of ultraviolet rays, cosmic ray, and other high energy radiations like a, (3 rays from sun, the air molecules in the ionosphere get ionized. This produces charged ions and these ions provide a reflecting medium for the reflection of radio waves or communication waves back to earth within the permitted frequency range. The phenomenon of bending the radio waves back to earth is nothing but the total internal reflection.

(iii) Space wave propagation:
The process of seeding and receiving information signal through space is called space wave communication. The electromagnetic waves of very high frequencies above 30 MHz are called as space waves. These waves travel in a straight line from the transmitter to the receiver. Hence, it is used for a line of sight communication (LOS).

For high frequencies, the transmission towers must be high enough so that the transmitted and received signals (direct waves) will not encounter the curvature of the earth and hence travel with less attenuation and loss of signal strength. Certain waves reach the receiver after getting reflected from the ground.

Students can Download Tamil Nadu 12th Commerce Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Commerce Model Question Paper 4 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part -I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
Drucker stresses jobs of management.
(a) three
(b) two
(c) four
(d) Six
Answer:
(a) three

Question 2.
Choose the odd one out?
(a) Directing
(b) Motivating
(c) Innovation
(d) Controlling
Answer:
(c) Innovation

Question 3.
…………… harmonises the goal of an individual with the organisation’s goal.
(a) MBO
(b) MBE
(c) MBA
(d) MBM
Answer:
(a) MBO

Question 4.
Indian organised financial system consist of sectors.
(a) Two
(b) Four
(c) Three
(d) Six
Answer:
(b) Four

Question 5.
Capital market otherwise called as …………..
(a) Securities market
(b) Share market
(c) Debenture market
(d) Money market
Answer:
(a) Securities market

Question 6.
……………………. deals with the financial assets and securities whose maturity period does not exceed one year.
(a) Money market
(b) Capital market
(c) Stock exchange
(d) Government bonds
Answer:
(a) Money market

Question 7.
The consumer is to be protected against any ……………..
(a) unfair practices of trade
(b) family functions
(c) profit making firm
(d) loss in business
Answer:
(a) unfair practices of trade

Question 8.
The National Commission should have members.
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(d) 6

Question 9.
………………. is the latest outcome of liberalization.
(a) Privatisation
(b) Globalisation
(c) Foreign collaboration
(d) None of these
Answer:
(c) Foreign collaboration

Question 10.
A bill of exchange drawn on a specified banker is ……………
(a) promissory note
(b) cheque
(c) hundi
(d) share
Answer:
(b) cheque

Question 11.
The entrepreneurs classified on the basis of types of business are …………….
(i) Industrial entrepreneur
(ii) Technical entrepreneur
(iii) Professional entrepreneur
(iv) Business entrepreneur
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iv)

Question 12.
Board meetings to be conducted minimum times in a year.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 13.
A public company shall have a minimum of ……………. directors.
(a) Twelve
(b) Seven
(c) Three
(d) Two
Answer:
(c) Three

Question 14.
Which of the following types of shares are issued by a company to rains capital from the existing shareholders?
(a) Equity shares
(b) Rights shares
(c) Preference shares
(d) Bonus shares
Answer:
(b) Rights shares

Question 15
should cover aspects like sources of finance, technical know-how, source of labours and raw material, market potential and profitability.
(a) Technical Report
(b) Finance Report
(c) Project Report
(d) Progress Report
Answer:
(c) Project Report

Question 16.
Find the odd one out is context of Trading Entrepreneur.
(a) Selling
(b) Commission
(c) Buying
(d) Manufacturing
Answer:
(d) Manufacturing

Question 17.
Which of the below is not classified into managerial function.
(a) Planning
(b) Marketing
(c) Organising
(d) Controlling
Answer:
(c) Organising

Question 18.
Section 6 of Negotiable Instruments Act 1881 deals with ……………
(a) Promissory Note
(b) Bills of exchange
(c) Cheque
(d) None of the above
Answer:
(c) Cheque

Question 19.
……………… results from the removal of barriers between national economics to encourage the flow of goods, services, capital and labour.
(a) privatisation
(b) Liberalisation
(c) Globalisation
(d) Foreign Trade
Answer:
(c) Globalisation

Question 20.
The state commission can entertain complaints where the value of the goods or services and the compensation, if any claimed exceed.
(a) ₹ 2 lakhs but does not exceed ₹ 5 lakhs
(b) ₹ 20 lakhs but does not exceed ₹ 1 crore
(c) ₹3 lakhs but does not exceed ₹ 5 lakhs
(d) ₹ 4 lakhs but does not exceed ₹ 20 lakhs
Answer:
(d) ₹ 4 lakhs but does not exceed ₹ 20 lakhs

PART – II

Answer any seven in which Question No.30 is compulsory. [7 x 2 = 14]

Question 21.
State the meaning of Authority.
Answer:
Authority means the right of a superior to give the order to his subordinates. This is the issue of commands followed responsibility for their consequences.

Question 22.
How is price decided in a secondary market?
Answer:
Financial markets allow for the determination of the price of the traded financial asset through the interaction of buyers and sellers. They provide a signal for the allocation of funds in the economy, based on the demand and supply, through the mechanism called price discovery processes.

Question 23.
What is meant by Remiser?
Answer:
Remiser is an agent of a member of a stock exchange. He obtains business for his principal i.e. the member.

Question 24.
What is Human Resource Management ?
Answer:
Human Resource Management is a function of management concerned with hiring, motivating and maintaining people in an organisation. It focuses on people in an organisation.

Question 25.
Define Marketer.
Answer:
“A person whose duties include the identification of the goods and services desired by a set of consumers, as well as the marketing of those goods and services on behalf of a company”.

Question 26.
What is Caveat Emptor?
Answer:
‘Caveat emptor’ is a Latin term that means “let the buyer beware.” Similar to the phrase “sold as is,” this term means that the buyer assumes the risk that a product fails to meet expectations or have defects.

Question 27.
Expand VUCA.
Answer:
VUCA – volatility, uncertainty, complexity and ambiguity.

Question 28.
What is meant by goods?
Answer:
The term goods mean every kind of movable property other than actionable claim and money.

Question 29.
List down the managerial functions of entrepreneurs.
Answer:

• Planning
• Organising
• Directing
• Controlling
• Coordination

Question 30.
What do you mean by Equity Share?
Answer:
The share of a company which do not have any preferential rights with regard to dividend and repayment of share capital at the time of liquidation of a company, is called as equity share or ordinary share.

Part – III

Answer any seven in which question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
What are the principles of Taylor?
Answer:
Principles of scientific management propounded by Taylor are:

• Science, Not Rule of Thumb
• Harmony, Not Discord
• Mental Revolution
• Cooperation, Not individualism
• Development of each and every person to his or her greatest efficiency and prosperity.

Question 32.
Write a note on Secondary Market.
Answer:
Secondary Market is the market for securities that are already issued. Stock Exchange is an important institution in the secondary market.

Question 33.
Explain National Stock Market System. (NSMS)
Answer:
National stock market system was advocated by the High Powered Group. It is headed by Shri. Pherwani (popularly known as Pherwani Committee). At present the National Stock Market comprises the following:

• National Stock Exchange of India Limited (NSE)
• Stock Holding Corporation of India Limited (SHCIL)
• Securities Trading Corporation of India (STCI)

Question 34.
Marketer is an innovator? Do you agree?
Answer:
Marketer seeks to distinguish his products/services by adding additional features or functionalities to the existing product, modifying the pricing structure, introducing new delivery pattern, creating new business models, introducing change in production process and so on.

Question 35.
What is meant by artificial scarcity?
Answer:
There are certain situations where the shop-keepers put up the board ‘No Stock” in front of their shops, even though there is plenty of stock in the store. In such situations consumers who are desperate to buy such goods have to pay hefty price to buy those goods and thus earning more profit unconscientiously.

Question 36.
State the framework of Corporate Governance in India.
Answer:
The Indian Corporate Governance framework requires listed companies

• To have independent directors on the board; At least one third of the directors have to be independent directors.
• To have at least one independent woman director,
• To disclose all deals and payments to related parties.
• To disclose details of managerial compensation.

Question 37.
Discuss the implied conditions and warranties in sale of goods contract.
Answer:
In every contract of sale, there are certain expressed and implied conditions and warranties. The term implied conditions means conditions which can be inferred from or guessed from the context of the contract.Following are the implied conditions:

• Conditions as to Title
• Conditions as to Description
• Sale by Sample
• Conditions as to Quality or Fitness
• Conditions as to Merchantability
• Condition as to Wholesomeness
• Condition Implied by Trade Usage Following are the implied warranties:
• Quiet Possession
• Free from Any Encumbrances
• Warranty in the case of Dangerous Goods

Question 38.
List the commercial functions of Entrepreneur and explain them shortly.
Answer:
Commercial Functions of Entrepreneur:
Production or Manufacturing: Under production function, entrepreneur has to take decision relating to selection of factory site, design and layout, type of products to be manufactured, research and development, product design, etc.

Marketing: Entrepreneur has to carry out following functions pertaining to marketing aspect namely consumer research, product planning and development, standardisation, packaging, pricing, warehousing, distribution, promotion etc.

Accounting: Entrepreneur has to arrange to prepare trading and profit and loss account in order to know the profit or loss incurred out of operation of the business and prepare balance sheet to know the financial status of business at a particular day.

Finance: In the sphere of financial function, an entrepreneur has to take decisions like choosing the right type of financing, framipg the best dividend policy, acquiring of funds, efficiently managing fixed and current assets, maximising shareholders wealth and investing of funds efficiently and effectively.

Human Resource Management: Entrepreneur has to estimate the manpower needs of the enterprise and accordingly decide the size of manpower required for various slots of organisational structure.

Question 39.
What is issue of shares at discount? What conditions should be fulfilled?
Answer:
When the shares are issued at a price below the face value they are said to be issued at a discount. For example, a share having the face value of Rs 10 is issued at Rs 8. The companies act 2013, prohibits the issue of shares at discount (Section 53), except sweat Equity share.

Question 40.
What do you understand by ‘Poll’?
Answer:
Poll means tendering or offering vote by ballot to a specially appointed officer, called the polling officer. Under the Companies Act, poll means exercising voting right in proportion to shareholder’s contribution to the paid-up capital of a limited company having a share capital.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41.
(a) Discuss the implications of span of management.
Answer:
The Span of Management has two implications:
(i) Influences the complexities of the individual manager’s job.
(ii) Determine the shape or configuration of the Organisation.
The span of management is related to the horizontal levels of the organisation structure. There is a wide and a narrow span of management. With the wider, span, there will be less hierarchical levels, and thus, the organisational structure would be flatter. Whereas, with the narrow span, the hierarchical levels increases, hence the organisational structure would be tall.

• Both these organisational structures have their advantages and the disadvantages. But however the tall organisational structure imposes more challenges.
• Since the span is narrow, which means less number of subordinates under one superior, requires more managers to be employed in the organisation. Thus, it would be very expensive in terms of the salaries to be paid to each senior.
• With more levels in the hierarchy, the communication suffers drastically. It takes a lot of time to reach the appropriate points, and hence the actions get delayed.
• Lack of coordination and control because the operating staff is far away from the top management.

[OR]

(b) What are the functions of Financial Markets?
Answer:
Intermediary Functions
(i) Transfer of Resources: Financial markets facilitate the transfer of real economic resource from lenders to ultimate borrowers.

(ii) Enhancing Income: Financial markets allow lenders earn interest/dividend on their surplus
investible funds and thus contributing to the enhancement of the individual and the national income.

(iiI) Productive Usage: Financial markets allow for the productive use of the funds borrowed and thus enhancing the income and the gross national production.

(iv) Capital Formation: Financial markets provide a channel through which new savings flow to aid capital formation of a country.

(v) Price Determination: Financial markets allow for the determination of the price of the traded financial asset through the interaction of buyers and sellers.

(vi) Sale Mechanism: Financial market provides a mechanism for selling of a financial
asset by an investor so as to offer the benefits of marketability and liquidity of such assets.

(vii) Information: The activities of the participants in the financial market result in the generation and the consequent dissemination of information to the various segments of the markets, so as to reduce the cost of transaction of financial assets.

Financial Functions

• Providing the borrowers with funds so as to enable them to carry out their investment plans.
• Providing the lenders with earning assets so as to enable them to earn wealth by deploying the assets in productive ventures.
• Providing liquidity in the market so as to facilitate trading of funds

Question 42.
(a) Explain the various types of New Financial Institutions.
Answer:
(i) Venture Fund Institutions: Venture capital financing is a form of equity financing designed especially for funding new and innovative project ideas. Venture capital funds bring into force the hi-technology projects which are converted into commercial  production.

(ii) Mutual Funds: Financial institutions that provide facilities for channeling savings of small investors into avenues of productive investments are called ‘Mutual Funds’.

(iii) Factoring Institutions: “Factoring” is an arrangement whereby a financial institution provides financial accommodation on the basis of assignment/sale of account receivables.

(iv) Over the Counter Exchange of India (OTCEI): The OTCEI was set up by a premier
financial institution to allow the trading of securities across the electronic counters throughout the country.

(v) National Stock Exchange of India Limited (NSEI): NSEI was established in 1992 to function as a model stock exchange. The Exchange aims at providing the advantage of nation-wide electronic screen based “scripless” and “floorless” trading system in securities.

(vi) National Clearance and Depository System (NCDS): Under the scripless trading system, settlement of transactions relating to securities takes place through a book entry.

(vii) National Securities Depositories Limited: The NSDL was set up in the year 1996 for achieving a time bound dematerialization as well as rematerialization of shares.

(viii) Stock Holding Corporation of India Limited (SHCIL): Stock Holding Corporation of India Limited (SHCIL) aims at serving as a central securities depository in respect of transactions on stock exchanges. The Corporation also takes up the administration of clearing functions at a national level.

[OR]

(b) What are the features of Government Securities?
Answer:
(i) Agencies: Government securities are issued by agencies such as Central Government, State Governments, semi-government authorities like local Government authorities.
(ii) RBI Special Role: RBI takes a special and an active role in the purchase and sale of these securities as part of its monetary management exercise.
(iii) Nature of Securities: Securities offer a safe avenue of investment through guaranteed payment of interest and repayment of principal by the Government.
(iv) Liquidity Profile: The liquidity profile of gilt-edged securities varies. Accordingly liquidity profile of securities issued by Central Government is high.
(v) Tax Rebate: A striking feature of these securities is that they offer wide-range of tax incentives to investors.

(vi) Market: As each sale and purchase has to be negotiated separately, the Gilt-Edged Market is an Over-The-Counter Market.
(vii) Forms: The securities of Central and State Government take such forms as inscribed stock or stock certificate, promissory note and bearer bond.
(viii) Participants: The participants in Government securities market include the Government sector comprising Central and State Governments
(ix) Trading: Small and less active, banks and corporate holders who purchase and sell Government securities on the stock exchanges participate in trading.
(x) Issue Mechanism: The Public Debt Office (PDO) of the RBI undertakes to issue government securities.
(xi) Issue opening: A notification for the issue of the securities is made a few days before the public subscription is open.
(xii) Grooming Gradual: It is the acquisition of securities nearing maturity through the stock exchanges by the RBI.
(xiii) Switching: It is the purchase of one security against the sale of another security carried out by the RBI in the secondary market as part of its open market operations.
(xiv) Auctioning: A method of trading whereby merchants bid against one another and where the securities are sold to the highest bidder.

Question 43.
(a) Distinguish between Stock Exchange and Commodity Exchange.
Differences between Stock Exchange and Commodity Exchange.

	Stock Exchange	Commodity Exchange
(0	Meaning: Stock Exchange is an organised market for purchase and sale of industrial and financial securities.	A commodity exchange is an exchange where commodities are purchased and sold. E.g.: Gold, Silver, Rice, Wheat
(ii)	Function: Providing easy marketability.	Offering hedging or price insurance services.
(Hi)	Object: It is facilitating capital formation.	It will facilitate goods flow through risk reduction.
(iv)	Participants: Investors and Speculators are participating in stock exchange.	Producers, dealers, traders are involved in the commodity exchange.
(v)	Articles traded: Industrial securities such as stocks and bonds and government securities.	Only durable and graded goods are traded in this.

[OR]

(b) Differentiate HR from HRM.

S.No.	Human Resource (HR)	Human Resource Management (HRM)
1.	Meaning: Human resource, i.e., the employees create other sources.	Managing the human resource is known as human resource management.
2.	Human resource exhibits innovation and creativity.	Human resource management has universal relevance.
3.	Human resource alone can think, act and analyse.	The accomplishment of organisational goal is made possible through human resource in an organisation.
4.	Human resources are movable.	Human resource management is intangible function.
5.	Human resource can work as team.	Human resource management focuses on the development of manpower through training . and development programmes.

Question 44.
(a) Explain the important methods of interview.
Answer:
Interview means a face to face interaction between the interviewer and interviewee.
Interview may be of various types:-

• Preliminary Interview: It is conducted to know the general suitability of the candidates who have applied for the job.
• Structured Interview: In this method, a series of questions is to be asked by the interviewer. The questions may be pre-prepared.
• In depth Interview: This interview is conducted to test the level of knowledge of the interviewee in a particular field.
• Panel Interview: Where a group of people interview the candidate. The panel usually comprises the chair person, subject expert, psychological experts and so on.
• Stress Interview: This type of interview is conducted to test the temperament and emotional balance of the candidate.
• Online Interview: Due to tremendous growth in information and communication technology, interviews are conducted by means of internet via Skype, Google duo, Whatsapp.

[OR]

(b) What is your contribution to promote the market in the modern society?
Answer:
Market is a place where buyers and sellers gather for purchase and sale.
Market may be of Local market, national market and international or Global market.
To develop and promote the market the following are needed:

• Eligible and satisfied and customer is needed.
• Quality and durable goods are to be marketed.
• Recent trends like E-marketing, online marketing are to be encouraged.
• After sales service are to be provided to durable goods.
• Customers are to be financed for buying costly articles.
• New Innovations and marketing research are to be introduced to develop the market.

Question 45.
(a) Discuss any two new methods of marketing with its advantages.
Answer:
Rural Marketing: Rural marketing is a process of developing pricing, promoting and distributing rural specific goods and serv ices leading to exchange with rural customers.There is inflow of goods into rural markets for production and consumption and there is also outflow of products to urban areas.
The rural to urban flow consists of agricultural products like rice, wheat and sugar etc.

Service Marketing: Service marketing is a special branch of marketing. It denotes the processing of selling service goods like telecommunication, banking, insurance, healthcare, tourism and professional services. The service products are mostly intangible. The unique feature of services marketing warrant different strategies compared with the marketing of physical goods.

[OR]

(b) Write about five important consumer legislations.
Answer:
To protect the consumers from the unfair traders, the government passed various legislative Acts – They are follows:

• The Indian Contract Act, 1982 was passed to bind the people on the promise made in the contract.
•  The Essential Commodities Act, 1955 protects the consumers against artificial shortages created by the sellers by hoarding the goods.
• The Prevention of Food Adulteration Act, 1954 checks the adulteration of food articles and ensures purity of goods supplied.
•  Weights and Measures Act, 1958 protects the consumer against malpractices of underweight dr under measurement.

Question 46.
(a) Explain the term District Forum and explain the functions of District Forum.
Answer:
As per the Consumer Protection Act 1986, a district forum is established in each and every district to solve the problems in the concerned district. Present or Retired district judge is the president of District forum

Functions:

• The complaints relating to the district can be solved by district forum.
• Compensation that can be claimed is less than Rs. 20 lakhs.
• Sometimes the district forum also may pass orders against traders.

[OR]

(b) Distinguish between Condition and Warranty.

Basis for Difference	Condition	Warranty
1. Meaning	It is a stipulation which is essential to the main purpose of the contract of sale.	It is a stipulation which is collateral to the main purpose of contract.
2. Significance	Condition is necessary to the contract that the breaking of which cancels out the contract.	The violation of warranty will not revoke the contract.
3. Transfer of Ownership	Ownership on goods cannot be transferred without fulfilling the conditions.	Ownership on goods can be transferred on the buyer without fulfilling the warranty.
4. Remedy	In case of breach of contract, the affected party can cancel the contract and claim damages.	In the case of breach of warranty, the affected party cannot cancel the contract but can claim damages only.
5. Treatment	Breach of condition may be treated as breach of warranty.	Breach of warranty cannot be treated as breach of condition.

Question 47.
(a) Explain the different kinds of endorsements.
Answer:
When the person signs on the back of the instrument to transfer his interest, it is known as endorsement. The endorsement are of various types:

(i) Blank or general endorsement : When the endorser puts his mere signature on the back of an instrument without mentioning the name of the person to whom the endorsement is made, it is called Blank Endorsement

(ii) Endorsement in full or special endorsement: If the endorser, in addition to his signature, mentions the name of the person to whom it is endorsed, is known as endorsement in full or special endorsement.

(iii) Conditional endorsement: When the endorser of a negotiable instrument makes his liability dependent upon the happening of an event which may or may not happen, it is called conditional endorsement.

(iv) Restrictive endorsement : When an endorsement restricts or prohibits further negotiability of the instrument, it is, called Restrictive Endorsement.

(v) Partial Endorsement :Where the endorsement seeks to transfer only a part of the amount payable under the instrument, the endorsement is called Partial Endorsement.

[OR]

(b) Write the difference between Share Certificate and Share Warrant.
Answer:

S.No.	Share Certificate	Share Warrant
1.	A share certificate is an instrument in writing for the legal proof of the ownership.	A share warrant is a negotiable instrument, issued against fully paid up shares.
2.	Every company must issue share certificate to its shareholders.	There is no compulsion of the issue of share warrants by the company.
3.	Normally, the holder of the share certificate is to be the member of the company.	Generally, the holder of the share warrant is not the member of the company.
4.	The share certificate is issued by the company within three months of the allotment of shares.	Shares warrant can be issued only when the shares are fully paid up.
5.	No need to authorized in the Articles of Association to issue share certificate.	The issue of a share warrant must be authorized in the Articles of Association of the company.

 

Students can Download Tamil Nadu 12th Maths Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 1 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If A is a non-singular matrix such that A^-1 = \(\left[\begin{array}{rr}
5 & 3 \\
-2 & -1
\end{array}\right]\), Then (A^T)^-1 = _________.


Answer:
(d) \(\left[\begin{array}{ll}
5 & -2 \\
3 & -1
\end{array}\right]\)

Question 2.
If Δ ≠ 0 then the system is ________.
(a) Consistent and has unique solution
(b) Consistent and has infinitely many solutions
(c) Inconsistent
(d) Either consistent or inconsistent
Answer:
(a) Consistent and has unique solution

Question 3.
The solution of the equation |z| – z = 1 + 2i is _______.
(a) \(\frac{3}{2}\) – 2i
(b) \(-\frac{3}{2}\) + 2i
(c) 2 – \(\frac{3}{2}\) i
(d) 2 + \(\frac{3}{2}\) i
Answer:
(a) \(\frac{3}{2}\) – 2i

Question 4.
The value of e^iθ + e^-iθ is _________.
(a) 2 cos θ
(b) cos θ
(c) 2 sin θ
(d) sin θ
Answer:
(a) 2 cos θ

Question 5.
The polynomial x³ – kx² + 9x has three real zeros if and only if, k satisfies __________.
(a) |k| ≤ 6
(b) k = 0
(c) |k| > 6
(d) |k| ≥ 6
Answer:
(d) |k| ≥ 6

Question 6.
The domain of the function defined by f (x) = sin^-1 \(\sqrt{x-1}\) is ________.
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d)[-1, 0]
Answer:
(a) [1, 2]

Question 7.
tan^-1 (\(\frac{1}{4}\)) + tan^-1 (\(\frac{2}{9}\)) is equal to ________.

Answer:
\(\tan ^{-1}\left(\frac{1}{2}\right)\)

Question 8.
8. The equation of the latus rectum of y² = 4x is _______.
(a) x = 1
(b) y = 1
(c) x = 4
(d) y = -1
Answer:
(a) x = 1

Question 9.
The circle passing through (1, -2) and touching the axis of x at (3, 0) passing through the point _______.
(a) (-5, 2)
(b) (2, -5)
(c) (5, -2)
(d) (-2, 5)
Answer:
(c) (5, -2)

Question 10.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(\frac{1}{5}\), then the value of λ is _______.
(a) 2\(\sqrt{3}\)
(b) 3\(\sqrt{2}\)
(c) 0
(d) 1
Answer:
(a) 2\(\sqrt{3}\)

Question 11.
The tangent to the curve y² – xy + 9 = 0 is vertical when ________.
(a) y = 0
(b) y = ± \(\sqrt{3}\)
(c) y = \(\frac{1}{2}\)
(d) y = ± \(\sqrt{3}\)
Answer:
(b) y = ± \(\sqrt{3}\)

Question 12.
The volume of a sphere is increasing in volume at the rate of 3π cm³/sec. The rate of change of its radius when radius \(\frac{1}{2}\) cm _______.
(a) 3 cm/s
(b) 2 cm/s
(c) 1 cm/s
(d) \(\frac{1}{2}\) cm/s
Answer:
(a) 3 cm/s

Question 13.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is _______.
(a) 0.4 cu.cm
(b) 0.45 cu.cm
(c) 2 cu.cm
(d) 4.8 cu.cm
Answer:
(d) 4.8 cu.cm

Question 14.
If v (x, y) = log (e^x + e^y ), then \(\frac{\partial v}{\partial x}+\frac{\partial v}{\partial y}\) is equal to _____.
(a) e^x + e^y
(b) \(\frac{1}{e^{x}+e^{y}}\)
(c) 2
(d) 1
Answer:
(d) 1

Question 15.
The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \cos x d x\) is _______.
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{2}\)
(c) 0
(d) \(\frac{2}{3}\)
Answer:
(d) \(\frac{2}{3}\)

Question 16.
The general solution of the differential equation log \(\left(\frac{d y}{d x}\right)\) = x + y is ______.
(a) e^x + e^y = c
(b) e^x + e^-y = c
(c) e^–x + e^y = c
(d) e^–x + e^-y = c
Answer:
(b) e^x + e^-y = c

Question 17.
The order and degree of the differential equation \(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 3}+x^{1 / 4}=0\) are respectively.
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Answer:
(a) 2, 3

Question 18.
If X is a binomial random variable with expected value 6 and variance 2.4, Then P {X = 5} is _______.

Answer:
(d) \(\left(\begin{array}{c}
10 \\
5
\end{array}\right)\left(\frac{3}{5}\right)^{5}\left(\frac{2}{5}\right)^{5}\)

Question 19.
A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is ______.
(a) 6
(b) 4
(c) 3
(d) 2
Answer:
(d) 2

Question 20.
If a*b = \(\sqrt{a^{2}+b^{2}}\) on the real numbers then * is ________.
(a) commutative but not associative
(b) associative but not commutative
(c) both commutative and associative
(d) neither commutative nor associative
Answer:
(c) both commutative and associative

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Using elementary transformation find the inverse of the matrix \(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
Answer:

Question 22.
Evaluate the zw if z = 5 – 2i and w = -1 + 3i
Answer:
zw = (5 – 2i) (-1 + 3i) = -5 + 15i + 2i – 6i² = -5 + 17i + 6 = 1 + 17i

Question 23.
Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.
Answer:
Given roots is (3 + 2i), the other root is (3 – 2i); Since imaginary roots occur in with real co-efficient occurring conjugate pairs.
x² – x(S.O.R) + P.O.R = 0 ⇒ x² – x(6) + (9 + 4) = 0
x² – 6x + 13 = 0

Question 24.
Is cos^-1 (-x) = π – cos^-1 x true? Justify your answer.
Answer:
Let θ = cos^-1 (-x)
⇒ cos θ = -x ⇒ -cosθ = x
i.e. cos(π – θ) = x
⇒ π – θ = cos^-1 x ⇒ π – cos^-1 x = θ
i.e. π – cos^-1 x = cos^-1(-x)

Question 25.
Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x – axis for the following functions: f(x) = x² – x, x ∈ [0, 1]
Answer:
Tangent is parallel to x axis. So \(\frac{d y}{d x}=0\)
f (x) = x² -x
f’ (x) = 2x – 1
f'(x) = 0 ⇒ 2x – 1 = 0 ⇒ x = \(\frac{1}{2}\) ∈[0, 1]

Question 26.
In each of the following cases, determine whether the following function is homogeneous or not. If it is so, find the degree g (x, y, z) = \(\frac{\sqrt{3 x^{2}+5 y^{2}+z^{2}}}{4 x+7 y}\)
Answer:

∴ It is homogeneous function of degree 0.

Question 27.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = e^-2x, y = 0, x = 0 and x = 1.
Answer:

Question 28.
Compute P(X = k) for the binomial distribution, B (n,p) where n = 10, p = \(\frac{1}{5}\), k = 4
Answer:
n = 10, p = \(\frac{1}{5}\), k = 4
∴ q = 1 – p = 1 – \(\frac{1}{5}=\frac{4}{5}\)
P(X = x) =nC_x p^xq^n-x, x = 0, 1, 2, …….n.
P (X = k) = P (X = 4)

Question 29.

be any three boolean matrices of the same type. Find A ∧ B
Answer:

Question 30.
The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Answer:
Slope of the tangent is the reciprocal of four times the ordinate
i.e., \(\frac{d y}{d x}=\frac{1}{4 y}\)
4∫y dy = ∫ dx
4\(\frac{y^{2}}{2}\) = x + c ⇒ 2y² = x + c
Passes through (2, 5)
∴ c = 50 – 2 = 48
Equation of the curve is 2y² = x + 48

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was ₹19,800 per month at the end of the first month after 3 years of service and ₹23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method to solve the problem.)

Question 32.
If the equations x² + px + q = 0 and x² + p’x + q’ = 0 have a common root, show that it must be equal to \(\frac{p q^{\prime}-p^{\prime} q}{q-q^{\prime}} \text { or } \frac{q-q^{\prime}}{p^{\prime}-p}\)

Question 33.
Find the value of tan^-1 (-1) + \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)

Question 34.
A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of s =16t² in t seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?

Question 35.
If the radius of a sphere is measured as 7m with an error of 0.02 m then find the approximate error in calculating its volume.

Question 36.
Find the volume of the solid that results when the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (a > b > 0) is revolved about the minor axis.

Question 37.
Verify that the function y = e is a solution of the differential equation \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0\)

Question 38.
Find the mean and variance of the distribution \(f(x)=\left\{\begin{array}{cc}
3 e^{-3 x}, & 0<x<\infty \\
0, & \text { elsewhere }
\end{array}\right.\)

Question 39.
Let A = {a + \(\sqrt{5}\) b : a, b ∈ Z} . Check whether the usual multiplication is a binary operation on A.

Question 40.
If \(\frac{z+3}{z-5 i}=\frac{1+4 i}{2}\) find the complex number z.

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) Solve, by Cramer ’s rule, the system of equations
x₁ – x₂ = 3, 2x₁ + 3x₂ + 4x₃ = 17,  x₂ + 2x₃ = 7
[OR]
(b) A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume.

Question 42.
(a) Solve the equation z³ + 8i = 0, where z ∈ C.
[OR]
(b) Solve (1 + 2e^x/y)dx + 2e^x/y \(\left(1-\frac{x}{y}\right)\) dy = 0

Question 43.
(a) Find the area of the region bounded between the parabola x² =y and the curve y = |x|.
[OR]
(b) Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}\) and passing through the point (-1, 1, -1).

Question 44.
(a) Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation \(\frac{x^{2}}{30^{2}}-\frac{y^{2}}{44^{2}}=1\). The tower is 150 m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
[OR]
(b) If 2 + i and 3 – \(\sqrt{2}\) are roots of the equation
x⁶ – 13x⁵ + 62x⁴ – 126x³ + 65x² + 127x – 140 = 0 find all roots.

Question 45.
(a) If u = \(\sin ^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)\) show that \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\frac{1}{2} \tan u\)
[OR]
(b) The cumulative distribution function of a discrete random variable is given by.

Find (i) the probability mass function (ii) P(X < 3) and (iii) P(X ≥ 2).

Question 46.
(a) Prove that: \(\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)
[OR]
(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for following operation on the given set. m*n = m + n – mn ; m, n ∈ Z

Question 47.
(a) Find the equation of the circle passing through the points (1, 1), (2, -1), and (3, 2).
[OR]
(b) Evaluate: \(\int_{0}^{\pi / 2} \frac{d x}{4+9 \cos ^{2} x}\)

Students can Download Tamil Nadu 12th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 2 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
  These are to be answered by choosing the most suitable answer from the given four
  alternatives and writing the option code and the corresponding answer
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between points A and C is……………………………..
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 1/4 μF

Answer:
(b) 2 μF

Question 2.
If the electric field in a region is given by \(\overrightarrow{\mathrm{E}}=5 \hat{i}+4 \hat{j}+9 \hat{k}\), then the electric flux through a surface of area 20 units lying in the
y – z plane will be …………………
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hint. The area vector
\(\overrightarrow{\mathrm{A}}=20 \hat{i} ; \overrightarrow{\mathrm{E}}=(5 \hat{i}+4 \hat{j}+9 \hat{k})\)
Flux ( φ)\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) = 5 x 20 = 100 units

Question 3.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a) π Ω
(b) \(\frac{\pi}{2} \Omega\)
(c) 2π Ω
(d) \(\frac{\pi}{4} \Omega\)
Answer:
(b) \(\frac{\pi}{2} \Omega\)

Question 4.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed co. Find the ratio of its magnetic moment with angular momentum is …………..
(a) M
(b) \(\frac{3}{\pi} \mathrm{M}\)
(c) \(\frac{2}{\pi} \mathrm{M}\)
(d) \(\frac{1}{2} \mathrm{M}\)

Answer:
(b) \(\frac{3}{\pi} \mathrm{M}\)

Question 5.
A proton enters a magnetic field of flux density 1.5 Wb/m² with a speed of 2 x 10⁷ m/s at angle of 30° with the field. The force on the proton will be ……………….
(a) 0.24 x 10^-12 N
(b) 2.4 x 10 ^-12 N
(c) 24 x 10^-12 N
(d) 0.024 x 10^-12 N
Answer:
(b) 2.4 x 10 ^-12 N
Hint: F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

Question 6.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\) Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) . The power factor of the of the circuit is ……………
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(c) 1

Question 7.
The inductance of a coil is proportional to…………………………………….
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 8.
The electric and magnetic fields of an electromagnetic wave are……………….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other

Question 9.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……………
(a) get shifted downwards
(b) get shifted upward
(c) will remain the same
(d) data insufficient to conclude

Answer:
(b) get shifted upward

Question 10.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related ………..

Answer:
(d) \(\lambda_{p} \propto \lambda_{e}^{2}\)

Question 11.
A system consists of N_o nucleus at t = 0. The number of nuclei remaining after half of a half-life (that  is, at time \(t=\frac{1}{2} \mathrm{T}_{\frac{1}{2}}\)
Answer:

Hint:

Question 12.
In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called……………………………….
(a) acceptor
(b) donor
(c) intrinsic semiconductor
(d) extrinsic semiconductor
Answer:
(c) intrinsic semiconductor
Hint: Pure semiconductors are called intrinsic semiconductors.

Question 13.
The light emitted in an LED is due to……………………………
(a) Recombination of charge carriers
(b) Reflection of light due to lens action
(c) Amplification of light falling at the junction
Answer:
(a) Recombination of charge carriers

Question 14.
The frequency range of 3 MHz to 30 MHz is used for………………………………..
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
The materials used in Robotics are……………………..
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Part – III

Answer any six questions. Question No. 20 is compulsory.   [6×2 = 12]

Question 16.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 17.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
\(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)
The S.I. unit of current density. \(\frac{\mathrm{A}}{\mathrm{m}^{2}}(\text { or }) \mathrm{Am}^{-2}\)

Question 18.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to magnetising field \((\overrightarrow{\mathrm{H}})\)
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 19.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 20.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Answer:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, φ = 4 mWb = 4 x 10^-3 Wb
Induction of the coil , L
\(\mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4}=2 \mathrm{H}\)

Question 21.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 22.
How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer:
Total energy emitted per second = Power x time
19.863 = Power x is
∴ Power 19.86 W
Number of photons =

Question 23.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 10¹⁰ decays/s

Question 24.
A transistor having α =0.99 and V_BE = 0.7V, is given in the circuit. Find the value of the collector current.
Answer:

Part – III

Answer any six questions. Question No. 26 is compulsory.   [6 x 3 = 18]

Question 25.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 26.
A conductor of linear mass density 0.2 g m^-1 suspended ,by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the x page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s^-2 = 111.87.
Answer:

Question 27.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 28.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = (u)c

Question 29.
If the focal length is 150 cm for a glass lens, what is the power of the lens?
Answer:
Given: focal length,f = 150 cm (or) f= 1.5 m
Equation for power of lens is, P = 1/f
Substituting the values,
\(P = \frac{1}{1.5}\)= 0.067 diopter
As the power is positive, it is a converging lens.

Question 30.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglic wavelength of the particle is \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\) \(\text { i.e. } \lambda \propto \frac{1}{\sqrt{m}}\)
As m_e <<m_p ,so λ_e >>λ_P
Hence protons have greater de- broglic wavelength

Question 31.
Distinguish between avalanche and zener breakdown.
Answer:

Avalanche Breakdown	Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths.	It occurs in junctions which are heavily doped and have narrow depletion widths.
It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision.	It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
At reverse voltage above 6V breakdown is due to avalanche effect.	At reverse voltage below 6V breakdown is due to zener effect.
Electric field produced is weak in nature.	A strong electric field is produced
Charge carriers obtain energy from the applied potential.	Zener current is independent of applied voltage.

Question 32.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Question 33.
What are black holes?
Answer:
Black holes are end stage of stars Which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Part – IV

Answer all the questions.  [5 x 5 = 25]

Question 34.
(a) How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements
Δq₁, Δq₂, Δq₃ ….. Δq_n ……… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such change elements

Here Δ qi is the i^th charge element, r is the distance of the point P from the z^th charge element and \(\hat{r}_{i \mathrm{P}} \) is the unit vector from ith charge element to the point P.

However, the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit Δq → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}\)

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r}\) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\overrightarrow{\mathrm{F}}=q \overrightarrow{\mathrm{E}}\)

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = Q/L. Its unit is coulomb per meter (Cm^-1). The charge present in the infinitesimal length dl is dq = λ dl
The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on. a surface of area A, then surface charge density (charge per unit area) is \(\lambda=\frac{Q}{L}\). Its unit is coulomb per square meter (Cm^-2 ). The charge present in the infinitesimal area dA is dq = adA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by \(\rho=\frac{Q}{V}\) . Its unit is coulomb per cubic meter (Cm^-3 ). The charge present in the infinitesimal volume element dV is dq = ρdV.
The electric field due to a volume of total charge Q is given by

[OR]

(b) Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as
V = El
As we know, the magnitude of current density
\(\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{\mathrm{V}}{l}\)
But \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\),so we write the equation as
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
By rearranging the above equations, we get
\(\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma \mathrm{A}}\right)\)

The quantity \(\frac{l}{\sigma \mathrm{A}}\) is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR …………….. (3)

Question 35.
(a) Calculate the magnetic held inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid: Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\) = μ₀  x (total current enclosed by Amperian loop)

Since the elemental lengths along be and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals.

Since the magnetic field outside the   solenoid is zero, the integral
\(\int_{c}^{d} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0\)
For the path along ab, the integral is

where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is

Let NI be the current passing through the solenoid of N turns, then

The number of turns per unit length is given by \(\frac{\mathrm{NI}}{\mathrm{L}}=n\) then
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
Since n is a constant for a given solenoid and μ_o is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

[OR]

(b) Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle: The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction: In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working: If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil. This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.

As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil ε_p is almost equal and opposite to the applied voltage υ_p and is given by
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by
\(\varepsilon_{\mathrm{s}}=-N_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ………………… (1)
where N_p and N_s are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then  ε_s = υ_s where υ_s is the voltage across secondary coil.
\(v_{s}=\varepsilon_{s}=-\mathrm{N}_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ……….. (2)

From equations (1) and (2),
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=K\) …………………… (3)

This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υ_p i_p= Output power υ_si_s
where i_pand i_s are the currents in the primary and secondary coil respectively. Therefore,
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=\frac{i_{p}}{i_{s}}\) ………………….. (4)

Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}=K\)

(i) If N_s> N_p ( or K > 1), ∴ V_s > V_p and I_s < I_p. This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.

(ii) If N_s < N_p (or K < 1) , ∴ V_s < V_p and I_s > I_p . This is step-down transformer where voltage is decreased and the current is increased.

Question 36.
(a) Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves: Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows. If the charged -particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves. Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
E_y =E₀ sin (kz-ωt)
B_r = B₀ sin(kz – ωt) where, E_o and B_o are amplitude of oscillating electric and magnetic field,\(\hat{k} \) is a wave number, ω is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between E_o and B_o is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac{E_{0}}{B_{0}}\)

In any medium, the ratio of E_o and B_o is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
\(v=\frac{E_{0}}{B_{0}}<c\)
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

[OR]

(b) Explain about compound microscope and obtain the equation for magnification.
Answer:
Compound microscope:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Magnification of compound microscope
From the ray diagram, the linear magnification due to the objective is,
\(m_{0}=\frac{h^{\prime}}{h}\)
from the figure ,\(\tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L} \), then
\(\frac{h^{\prime}}{h}=\frac{L}{f_{0}} ; m_{o}=\frac{L}{f_{o}}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f_o and f_e are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification m_e of the eyepiece is,
\(m_{e}=1+\frac{D}{f_{e}}\)

The total magnification m in near point focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)

If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is
\(m_{e}=\frac{D}{f_{e}}\)

The total magnification m in normal focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 37.
(a) Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:

• This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
• The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
• De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
• As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
• Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:

• The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
• The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
• With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.

[OR]

(b) Discuss the process of nuclear fission and its properties.
Answer:

• When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
• The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
• The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.
• Uranium undergoes fission reaction in 90 different ways. The most common fission reactions of
  
• Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state. \(_{ 92 }^{ 236 }{ U }\) . But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 38.
(a) Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut-off regions functions like an electronic switch that helps to. turn ON or OFF a given circuit by a small control signal.

Presence of dc source at the input (saturation region):
When a high input voltage (V. = +5V) is applied, the base current (I ) increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current (I_c) increases the voltage drop across R_c .thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.

Absence of dc source at the input (cut-off region):
A low input voltage (V_in = OV), decreases the base current (I_B) and in turn decreases the collector current (I_c ). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (I_c) decreases the drop across, thereby increasing the output voltage, dose to +5 V. The transistor acts as an open switch which is considered as the OFF condition.

It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry

[OR]

(b) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified.
They are

1. Amplitude modulation,
2. Frequency modulation, and
3. Phase modulation.

1. Amplitude Modulation (AM): If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure
(a) is the message signal or baseband signal that carries information, figure
(b) shows the high-frequency carrier signal and figure
(c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. An increase in the amplitude of the ‘ baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave.

Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C). The increase in amplitude in the negative half cycle (B, D) reduces the frequency of the modulated wave (Figure (c)).

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.

The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.