Category: Class 12

Students can Download Tamil Nadu 12th Biology Model Question Paper 5 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 5 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.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) Sporogenous cell is epidermal
(d) Ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

Question 2.
The genotype of a plant showing the dominant phenotype can be determined by _______.
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Question 3.
An allo-hexaploidy contains ________.
(a) Six different genomes
(b) Six copies of three different genomes
(c) Two copies of three different genomes
(d) Six copies of one genome
Answer:
(b) Six copies of three different genomes

Question 4.
Match the following column I with column II


Answer:
(D) 1 – c, 2 – d, 3 – a, 4 – b

Question 5.
Assertion (A): Incubation is followed by Inoculation.
Reason (R): Explant is inoculated to media.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(b) R explains A

Question 6.
Environment of any community is called
(a) Paratope
(b) Biotope
(c) Ecotope
(d) Epitope
Answer:
(b) Biotope

Question 7.
Which of the following is not a sedimentary cycle?
(a) Nitrogen cycle
(b) Phosphorus cycle
(c) Sulphur cycle
(d) Calcium cycle
Answer:
(a) Nitrogen cycle

Question 8.
The plants which are grown in silvopasture system are
(a) Sesbania and Acacia
(b) Solanum and Crotolaria
(c) Clitoria and Begonia
(d) Teak and sandal
Answer:
(a) Sesbania and Acacia

Part – II

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

Question 9.
What are multiple alleles? Give an example.
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g: ABO blood group

Question 10.
List out the benefits of herbicide tolerant crops.
Answer:
Advantages of Herbicide Tolerant Crops:

• Weed control improves higher crop yields.
• Reduces spray of herbicide.
• Reduces competition between crop plant and weed.
• Use of low toxicity compounds which do not remain active in the soil.
• The ability to conserve soil structure and microbes.

Question 11.
Soil formation can be initiated by biological organisms. How?
Answer:
Soil formation is initiated by the biological weathering process. Biological weathering takes place when organisms like bacteria, fungi, lichens and plants help in the breakdown of rocks through the production of acids and certain chemical substances.

Question 12.
Where did Montreal Protocol was held? State its objectives.
Answer:
The International treaty called the Montreal Protocol (1987) was held in Canada on substances that deplete ozone layer and the main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the Earth’s ozone layer.

Question 13.
What is Heterosis?
Answer:
The superiority of the F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 14.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturising characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Part – III

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

Question 15.
List out the functions of tapetum.
Answer:

• It supplies nutrition to the developing microspores.
• It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
• The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
• Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Question 16.
What do you mean by Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into 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.
• Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

Question 17.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins and auxin apart from macro and micro nutrients. Most seaweed based fertilizers are made from kelp (brown algae) which grows to length of 150 metres. Liquid seaweed fertilizer is not only organic but also ecofriendly. The alginates in the seaweed that reacts with metals in the soil and form long, crosslinked polymers in the soil.

These polymers improve the crumbing in the soil, swell up when they get wet and retain moisture for a long time. They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 18.
Write a brief note on Detritus food chain.
Answer:
Detritus food chain is a type of food chain which begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems. The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.

Question 19.
Mention the name of man-made cereal. How it is developed?
Answer:

Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

1. Tetraploidy: Crosses between diploid wheat and rye.
2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye.

Hexaploidy Triticale hybrid plants demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.

Part – IV

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

Question 20.
(a) Describe the structure of a Cicer seed (dicot seed) with labelled diagram.
Answer:
Structure of a Cicer seed as an example for Dicot seed The mature seeds are attached to the fruit wall by a stalk called funiculus. The funiculus disappears leaving a scar called hilum. Below the hilum a small pore called micropyle is present. It facilitates entry of oxygen and water into the seeds during germination.

Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.

In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root.

The other end of the axis called embryonic shoot is the plumule. Embryonal axis above the level of cotyledon is called epicotyl whereas the cylindrical region between the level of cotyledon is called hypocotyl. The epicotyl terminates in plumule whereas the hypocotyl ends in radicle.

[OR]

(b) (i) Explain the three types of hyperploidy.
(ii) List out the significance of ploidy.
Answer:
(i) (a) Trisomy: Addition of single chromosome to diploid set is called Simple trisomy (2n+l). Trisomics were first reported by Blackeslee (1910) in Datura stramonium (Jimson weed). But later it was reported in Nicotiana, Pisum and Oenothera. Sometimes addition of two individual chromosome from different chromosomal pairs to normal diploid sets are called Double trisomy (2n + 1 + 1).
(b) Tetrasomy: Addition of a pair or two individual pairs of chromosomes to diploid set is called tetrasomy (2n+2) and Double tetrasomy (2n+2+2) respectively. All possible tetrasomics are available in Wheat.
(c) Pentasomy: Addition of three individual chromosome from different chromosomal pairs to normal diploid set are called pentasomy (2n+3).

(ii)

• Many polyploids are more vigorous and more adaptable than diploids.
• Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
• Autopolyploids usually have increase in fresh weight due to more water content.
• Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.
• Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Question 21.
a. (i) Define tissue culture.
(ii) Explain the basic concepts involved in plant tissue culture.
Answer:
(i) Growing plant protoplasts, cells, tissues or organs away from their natural or normal environment, under artificial condition, is known as Tissue Culture.

(ii) Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.
Totipotency: The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

[OR]

(b) What is soil profile? Explain the characters of different soil horizons.
Answer:
Soil is commonly stratified into horizons at different depth. These layers differ in their physical, chemical and biological properties. This succession of super-imposed horizons is called soil profile.

Horizon	Description
O-Horizon (Organic horizon) Humus	It consists of fresh or partially decomposed organic matter. O1 – Freshly fallen leaves, twigs, flowers and fruits. O2 – Dead plants, animals and their excreta decomposed by micro-organisms. Usually absent in agricultural and deserts.
A-Horizon (Leached horizon) Topsoil – Often rich in humus and minerals.	It consists of top soil with humus, living creatures and in-organic minerals. A1 – Dark and rich in organic matter because of mixture of organic and mineral matters. A2 – Light coloured layer with large sized mineral particles.
B-Horizon (Accumulation horizon) (Subsoil – Poor in humus, rich in minerals)	It consists of iron, aluminium and silica rich clay organic compounds.
C – Horizon (Partially weathered horizon) Weathered rock Fragments – Little or no plant or animal life.	It consists of parent materials of soil, composed of little amount of organic matters without life forms.
R – Horizon (Parent material) Bedrock	It is a parent bed rock upon which underground water is found.

Bio-Zoology [Maximum Marks: 35]

Part – I

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

Question 1.
Assertion (A): Asexual reproduction is called blastogenic reproduction.
Reason (R): It is accomplished by mitotic and meiotic divisions.
(a) A and R are correct
(b) A is correct but R is incorrect
(c) Both A and R are incorrect
(d) R is the correct explanation for A
Answer:
(b) A is correct but R is incorrect

Question 2.
The mature sperms are stored in the ________.
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Question 3.
A contraceptive pill prevents ovulation by _______.
(a) blocking fallopian tube
(b) inhibiting the release of FSH and LH
(c) stimulating the release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting the release of FSH and LH

Question 4.
Patau’s syndrome is also referred to as ________.
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 5.
Cyclosporin – A is an immunosuppressive drug produced from ________.
(a) Aspergillus niger
(b) Manascus purpureus
(c) Penicillium notatum
(d) Trichoderma polysporum
Answer:
(d) Trichoderma polysporum

Question 6.
PCR proceeds in three distinct steps governed by temperature, they are in order of ________.
(a) Denaturation, Annealing, Synthesis
(b) Synthesis, Annealing, Denaturation
(c) Annealing, Synthesis, Denaturation
(d) Denaturation, Synthesis, Annealing
Answer:
(a) Denaturation, Annealing, Synthesis

Question 7.
Match column I with column II

(a) A – 4, B – 5, C – 2, D – 3, E – 1
(b) A – 3, B – 1, C – 4, D – 2, E – 5
(c) A – 2, B – 3,C – 1, D – 5, E – 4
(d) A – 5, B – 4, C – 2, D – 3, E – 1
Answer:
(a) A – 4, B – 5, C – 2, D – 3, E – 1

Question 8.
Total number of mega biodiversity countries in the world is _______.
(a) 12
(b) 15
(c) 17
(d) 19
Answer:
(c) 17

Part – II

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

Question 9.
Why are the offsprings of oviparous animals are at a greater risk as compared to offsprings of viviparous organisms?
Answer:
Oviparous animals are egg-layers. The eggs containing embryo are laid out of their body and are highly susceptible to environmental factors (temperature, moisture etc.) and predators. Whereas, in viviparous animals, the embryo develops inside the body of female and comes out as young ones. Hence offsprings of oviparous animals are at risk compared to viviparous animal.

Question 10.
What is “let-down reflex”?
Answer:
Oxytocin causes the “Let-Down” reflex the actual ejection of milk from the alveoli of the mammary glands. During lactation, oxytocin also stimulates the recently emptied uterus to contract, helping it to return to pre – pregnancy size.

Question 11.
In E.coli, three enzymes β- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 12.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine.the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 13.
Write the name of causative agent, infection site, mode of transmission and any two symptoms of Chikungunya.
Answer:
Causative agent – Alpha virus
Infection site – Nervous system
Mode of transmission – Aedes aegypti (Mosquito)
Symptoms – Fever, headache, joint pain and swelling.

Question 14.
Define the following terms.
(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.
Write a short note on foeto scope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

Question 16.
Under which condition does a microbe gains resistance against antibiotic?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control. Antibiotics should be used only when prescribed by a certified health professional.

When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 17.
State Fisher and Race hypothesis.
Answer:
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.

In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh⁺positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Question 18.
Extinction of Dodo bird led to the danger of Calvaria tree – Support your answer.
Answer:
An example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough homy endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 19.
Whether PCR can be done for RNA molecules? Yes or No? Explain.
Answer:
The PCR technique can also be used for amplifications of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Part – IV

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

Question 20.
(a) Define the following terms with an example
(i) Hologamy
(ii) Isogamy
(iii) Anisogamy
(iv) Merogamy
(v) Paedogamy
Answer:
(i) Hologamy: In Hologamy, the adult individuals do not produce gametes, but they themselves act as gametes and fuse to form new individuals.
E.g.: Trichonympha

(ii) Isogamy : Fusion of morphologically and physiologically similar gametes.
E.g.: Monocystis

(iii) Anisogamy : Fusion of morphologically and physiologically dissimilar gametes.
E.g.: Vertebrates.

(iv) Merogamy : Fusion of small sized morphologically different gametes (merogametes)

(v) Paedogamy : Fusion of young individuals produced immediately after the mitotic division of adult parent cell.

[OR]

(b) Write in detail about cervical cancer.
Answer:
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

1. Having multiple sexual partners
2. Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. Healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Question 21.
(a) Point out the differences between active and passive immunity.
Answer:
Active Immunity:

1. Active immunity is produced actively by host’s immune system.
2. It is produced due to contact with pathogen or by its antigen.
3. It is durable and effective in protection.
4. Immunological memory is present.
5. Booster effect on subsequent dose is possible.
6. Immunity is effective only after a short period.

Passive Immunity:

1. Passive immunity is received passively and there is no active host participation.
2. It is produced due to antibodies obtained from outside.
3. It is transient and less effective.
4. No memory.
5. Subsequent dose is less effective.
6. Immunity develops immediately.

[OR]

(b) How is the amplification of a gene sample of interest carried out using PCR?
Answer:
Denaturation, renaturation or primer annealing and synthesis or primer extension, are the three steps involved in PCR. The double stranded DNA of interest is denatured to separate into two individual strands by high temperature . This is called denaturation. Each strand is allowed to hybridize with a primer (renaturation or primer annealing). The primer template is used to synthesize DNA by using Taq – DNA polymerase.

During denaturation the reaction mixture is heated to 95°C for a short time to denature the target DNA into single strands that will act as a template for DNA synthesis. Annealing is done by rapid cooling of the mixture, allowing the primers to bind to the sequences on each of the two strands flanking the target DNA. During primer extension or synthesis the temperature of the mixture is increased to 75°C for a sufficient period of time to allow Taq DNA polymerase to extend each primer by copying the single stranded template.

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA, This process is also called DNA amplification.

Students can Download Tamil Nadu 12th Maths Model Question Paper 2 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 2 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 = \(\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]\) be such that λA^-1 = A, then λ is _______.
(a) 17
(b) 14
(c) 19
(d) 21
Answer:
(c) 19

Question 2.
If ω ≠ 1 is a cubic root of unity and (1+ ω)⁷ = A+ B ω, then (A, B) equals to _______.
(a) (1,0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)

Question 3.
The value of z – \(\bar{Z}\) is ______.
(a) 2 Im (z)
(b) 2 i Im (z)
(c) Im (z)
(d) i Im (z)
Answer:
(b) 2 i Im (z)

Question 4.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if ________.
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a < 0
Answer:
(c) a < 0

Question 5.
sin(tan^-1 x), |x| < 1 is equal to _______.

Answer:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is _____.
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Answer:
(a) (4, 7)

Question 7.
The axis of the parabola x² = – 4y is ______.
(a) y= 1
(b) x = 0
(c) y = 0
(d) x = 1
Answer:
(b) x = 0

Question 8.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=3\) are _______.
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Answer:
(d) (5, -1, 1)

Question 9.
If the vectors \(\vec{a}=3 \vec{i}+2 \vec{j}+9 \vec{k}\) and \(\vec{b}=\vec{i}+m \vec{j}+3 \vec{k}\) are parallel then m is _________.

Answer:
(b) \(\frac{2}{3}\)

Question 10.
The minimum value of the function |3 – x | + 9 is ________.
(a) 0
(b) 3
(c) 6
(d) 9
Answer:
(d) 9

Question 11.
The curve y² = x² (1 – x²) has ______.
(a) an asymptote x = -1
(b) an asymptote x = 1
(c) two asymptotes x = 1 and x = -1
(d) no asymptote
Answer:
(d) no asymptote

Question 12.
If /(x, y, z) = xy + yz + zx, then f_x – f_z is equal to _______.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 13.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is _______.
(a) 0.2%
(b) 0.4%
(c) 0.04%
(d) 0.08%
Answer:
(b) 0.4%

Question 14.
The value of \(\int_{0}^{\pi} \sin ^{4} x d x\) is _______.

Answer:
(b) \(\frac{3 \pi}{8}\)

Question 15.
\(\int_{a}^{b} f(x) d x\) is _______.

Answer:
(d) \(\int_{a}^{b} f(a+b-x) d x\)

Question 16.
The degree of the differential equation \(y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is ________.
(a) 2
(b) 3
(c) 1
(d) 4
Answer:
(c) 1

Question 17.
In finding the differential equation corresponding toy = e^mx where m is the arbitrary constant, then m is ____.
(a) \(\frac{y}{y^{\prime}}\)
(b) \(\frac{y^{\prime}}{y}\)
(c) y’
(d) y
Answer:
(b) \(\frac{y^{\prime}}{y}\)

Question 18.
Let X be random variable with probability density function f(x) = \(\left\{\begin{array}{ll}
2 / x^{3} & x \geq 1 \\
0 & x<1
\end{array}\right.\)
Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but mean does not exist
Answer:
(b) mean exists but variance does not exist

Question 19.
The random variable X has the probability density function f(x) = \(\left\{\begin{array}{cc}
a x+b, & 0<x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
and E(X) = \(\frac{7}{12}\), then a and b are respectively _______.
(a) 1 and \(\frac{1}{2}\)
(b) \(\frac{1}{2}\) and 1
(c) 2 and 1
(d) 1 and 2
Answer:
(a) 1 and \(\frac{1}{2}\)

Question 20.
A binary operation on a set S is a function from ________.
(a) S → S
(b)(S x S) → S
(c) S → (S x S)
(d) (S x S) → (S x S)
Answer:
(b)( S x S) → S

Part – II

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

Question 21.
Solve the following system of homogeneous equations.
3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
Answer:
The matrix form of the above equation is

The augmented matrix [A, B] is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……..(1)
-17y – 34z = 0 …….(2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y= \(\frac{34 t}{-17}\) = -2t
Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t = 0
3x – 4t + 7t = 0
⇒ 3x = -3t ⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t∈R

Question 22.
Show that |3z – 5 + i| = 4 represents a circle, and, find its centre and radius.
Answer:
The given equation |3z – 5 + i| = 4 can be written as

It is of the form |z – z| = r and so it represents a circle, whose center and radius are \(\left(\frac{5}{3},-\frac{1}{3}\right)\) and 4/3 respectively.

Question 23.
Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.
Answer:
Solving 3x – 2y = 1 and 4x + y = 27
Simultaneously, we get x = 5 and y = 7
∴ The point of intersection of the lines is (5, 7)
Now we have to find the equation of a circle whose centre is
(2, -3) and which passes through (5, 7)

∴ Required equation of the circle is
(x – 2)² + (y + 3)² = \((\sqrt{109})^{2}\)
⇒ x² + y² – 4x + 6y – 96 = 0

Question 24.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})=12\) on the coordinate axes.
Answer:
\(\vec{r} \cdot(6 \vec{i}+4 \vec{j}-3 \vec{k})=12\)
Compare the above equations into \(\vec{r} \cdot \vec{n}=q\) so q = 12
Let a, b, c are intercepts of x-axis, y-axis and z-axis respectively.
Clearly

x – intercept = 2; y – intercept = 3; z – intercept = -4

Question 25.
Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x² for 1 ≤ x ≤ 2.
Answer:
f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that
f'(c) = \(\frac{f(2)-f(1)}{2-1}\) = 1 – 2c
That is, 1 – 2c = -2 ⇒ c = \(\frac{3}{2}\)

Question 26.
Show that the percentage error in the n^th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.
Answer:

Question 27.
Solve the differential equation: tany \(\frac{d y}{d x}\) = cos (x + y) + cos (x -y)
Answer:
tan y \(\frac{d y}{d x}\) = cos (x + y) + cos(x – y)
tan y \(\frac{d y}{d x}\) = cos x cos y – sin x sin y + cos x cos y + sin x sin y
tan y \(\frac{d y}{d x}\) = 2 cos x cos y
seperating the variables
\(\int \frac{\tan y}{\cos y}\) dy = 2∫cos x dx ⇒ ∫sec y tan y dy = 2∫cos x dx
sec y = 2 sin x + c

Question 28.
The probability density function of X is given by \(f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Construct the truth table for the following statement. \(\neg(p \wedge \neg q)\).
Answer:

Question 30.
Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-hand rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Answer:
Here a = 1; b = 1.5; n = 5; f(x) = x²
So, the width of each subinterval is

x₀ = 1; x₁ = 1.1; x₂ = 1.2; x₃ = 1.3; x₄ = 1.4; x₅ = 1.5
The Right hand rule for Riemann sum,
S = [f(x₁) + f(x₂) + f(x₃) + f(x₄) + f(x₅)] Δx
= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)
= [8.55] (0.1)
= 0.855.

Part – III

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

Question 31.
Find a matrix A if adj (A) = \(\left[\begin{array}{ccc}
7 & 7 & -7 \\
-1 & 11 & 7 \\
11 & 5 & 7
\end{array}\right]\)

Question 32.
Obtain the Cartesian form of the locus of z = x + iy in the following case Im[(1 – i)z +1] = 0

Question 33.
If \(\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}\), show that \([\vec{a} \vec{b} \vec{c}]\) depends on neither x nor y.

Question 34.
The Taylor’s series expansion of f(x) = sin x about x = \(\frac{\pi}{2}\) is obtained by the following way.

Question 35.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing (i) the volume of the cube and (ii) the surface area of cube.

Question 36.
Evaluate \(\int_{0}^{1} \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^{2}} d x\)

Question 37.
Find the particular solution of (1 + x³) dy – x² ydx = 0 satisfying the condition y(1) = 2.

Question 38.
If X is the random variable with distribution function F(x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
x, & 0 \leq x<1 \\
1, & 1 \leq x
\end{array}\right.\)
then find (z) the Probability density function f(x)

Question 39.
Show that \(((\neg q) \wedge p) \wedge q\) is a contradiction.

Question 40.
Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) By using Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂.
[OR]
(b) \(\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^{2}}\), given that y = 2 when x = 1

Question 42.
(a) Find the real values of x and y for the equation \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i\)
[OR]
(b) Find the area between the line y = x + 1 and the curve y = x² – 1.

Question 43.
(a) Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
[OR]
(b) Evaluate: \(\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^{2} x}\)

Question 44.
(a) A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m, and the height at the edge of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately. How wide must the opening be?
[OR]
(b) Using truth table check whether the statements \(\neg(p \vee q) \vee(\neg p \wedge q)\) and \(\neg p\) are logically equivalent.

Question 45.
(a) Find the value of cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
[OR]
(b) Verify Euler’s theorem for f(x, y) = \(\frac{1}{\sqrt{x^{2}+y^{2}}}\)

Question 46.
(a) Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
[OR]
(b) If X is the random variable with probability density function f(x) given by,
\(f(x)=\left\{\begin{array}{rc}
x+1, & -1 \leq x<0 \\
-x+1, & 0 \leq x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
then find (z) the distribution function f(x) (ii) P (-0.5 ≤ X ≤ 0.5)

Question 47.
(a) Sketch the graph of the function: y = \(x \sqrt{4-x}\)
(b) The velocity v, of a parachute falling vertically satisfies the equation, \(v \frac{d v}{d x}=g\left(1-\frac{v^{2}}{k^{2}}\right)\)
where g and k are constants. If v and x are both initially zero, find v in terms of x.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 5 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 5 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.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to …………..
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Question 2.
Compound used for propellant is
(a) BN
(b) H₂B₄O₇
(c) B₂H₆
(d) Borax
Answer:
(c) B₂H₆

Question 3.
P₄O₆ reacts with cold water to give …………..
(a) H₃PO₃
(b) H₄P₂O₇
(c) HPO₃
(d) H₃PO₄
Answer:
(a) H₃PO₃

Question 4.
Which one of the following elements show high positive electrode potential?
(a) Ti²⁺
(b) Mn²⁺
(c) CO²⁺
(d) Cr²⁺
Answer:
(c) CO²⁺

Question 5.
As per IUPAC guidelines, the name of the complex [Co(en)₂(ONO)Cl]Cl is …………..
(a) chlorobisethylenediaminenitritocobalt(III) chloride
(b) chloridobis(ethane-l,2-diamine)nitro k-Ocobaltate(III) chloride
(c) chloridobis(ethane-l,2-diammine)nitrito k-Ocobalt(II) chloride
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride
Answer:
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride

Question 6.
Solid NH₃ solid CO₂ are examples of …………………..
(a) Covalent solids
(b) polar molecular solids
(c) molecular solids
(d) ionic solids
Answer:
(b) polar molecular solids

Question 7.
After 2 hours, a radioactive substance becomes\(\left(\frac{1}{16}\right)^{4}\) of original amount. Then the half life ( in min ) is ………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

Question 8.
What is the decreasing order of strength of bases?

Answer:

Question 9.
Which electrolyte is used in Leclanche cell?
(a) ZnSO₄ + CuSO₄
(6) NH₄Cl + ZnCl₂
(e) NaCl + CuSO₄
(d) MnSO₄ + MnO₂
Answer:
(6) NH₄Cl + ZnCl₂

Question 10.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………………
(a) Cataphoresis
(b) Eleætrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect

Question 11.
In the following sequence of reactions,

(a) Butanal
(b) n-butyl alcohol
(c) propan-1-ol
(d) Propanal
Answer:
(c) propan-1-ol

Question 12.
Of the following, which is the product formed when cyclohexanone undergoe’s aldol condensation followed by heating?

Answer:
(a)

Question 13.

Answer:
(a) A – 2, B – 1, C – 4, D – 3

Question 14.
Insulin, a hormone chemically is ………………..
(a) Fat
(b) Steroid
(c) Protein
(d) Carbohydrates
Answer:
(c) Protein

Question 15.
The role of phosphate in detergent powder is
(a) control pH level of the detergent water mixture
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water
(c) provide whiteness to the fabric
(d) more soluble in soft water
Answer:
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water

Part – II

Answer any six questions. Question No. 23 is compulsory. [6 x 2 = 12]

Question 16.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

• Concentration of the ore
• Extraction of crude metal
• Refining of crude metal

Question 17.
Give the uses of Borax.
Answer:

• Borax is used for the identification of coloured metal ions.
• In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
• It is also used as a flux in metallurgy and also acts as a good preservative.

Question 18.
Differentiate primary valency and secondary valency.
Answer:
Primary Valency :

1. The primary valence of a metal ion positive in most of the cases and zero in certain cases.
2. The primary valence is always satisfied by negative ions.
3. The primary valences are non directional
4. Example: In COCl₃.6NH₃, the primary valence of cobalt is +3 Example: In CoCl3.6NH3, the secondary valence of cobalt +3

Secondary Valency :

1. The secondary valence as the coordination number.
2. The secondary valence is satisfied by negative ions, neutral molecular or positive ions.
3. The secondary valences are directional
4. Example: In COCl₃.6NH₃, the secondary valence of cobalt is 6

Question 19.
Write a note about molecular solids.
Answer:

• In molecular solids, the constituents are neutral molecules. They are held together by weak vander waals forces.
• Molecular solids are soft and they do not conduct electricity. Eg., Solid CO₂

Question 20.
What are the limitations of Arrhenius concept?
Answer:

• Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.
• This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 21.
Write a note on catalytic poison.
Answer:
Catalytic poison: Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.
For example, in the reaction, 2SO₂ + O₂ → 2SO₃ with a Pt catalyst, the poison is AS₂O₃. i. e., AS₂O₃ destroys the activity of pt. AS₂O₃ blocks the activity of the catalyst. So, the activity is lost.

Question 22.
Convert phenyl magnesium bromide to phenyl methanol (or) How would you prepare phenyl methanol from Grignard reagent?
Answer:

Question 23.
Identify compounds A,B and C in the following sequence of reactions.

Answer:

Question 24.
Why cannot Vitamin C be stored in our body?
Answer:
Vitamin C is water soluble, therefore it is readily excreted in urine and hence cannot be stored in the body.

Part – III

Answer any six questions. Question No. 29 is compulsory. [6 x 3 = 18]

Question 25.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that – contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 26.
Discuss the Commercial method to prepare Nitric acid.
[OR]
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.

4NH₃ + 5O₂ → 4NO + 6H₂O + 120 kJ
2NO + O₂ → 2NO₂

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.

6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 27.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:

• Actinoid contraction is greater from element to element than lanthanoid cintraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
• Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
• In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only. ,
• Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 28.
What are the examples of first order reaction?
Answer:
(i) Decompostion of dinitrogen pentoxide
2N₂O_5(g) → 2NO_2(g) + 1/2 O_2(g)

(ii) Decomposition of thionylchloride
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

(iii) Decompostion of H₂O₂ in aqueous solution.
H₂O_2(aq) → H₂O_(l) + 1/2 O_2(g)

(iv) Isomerisation of cyclopropane to propene

Question 29.
Complete the following reaction.

Answer:

Question 30.
How will you calculate degree of dissociation of weak electrolytes and dissociation constant using Kohlrausch’s law?
Answer:
(i) The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity in infinite dilution using the formula \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)

(ii) According to Ostwald’s dilution law \(\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{C}}{1-\alpha}\)
Substituting α value in the above equation

Question 31.
What are the characteristics of adsorption?
Answer:

• Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid. .
• Adsorption is always accompanied by decrease in free energy. When AG reaches zero, the equilibrium is attained.
• Adsorption is a spontaneous process.
• When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
  ∆G = ∆H – T ∆S where ∆G = change in free energy
  ∆H = change in enthalpy
  ∆S = change in entropy
  .’. ∆H = ∆G + T∆S
• Adsorption is exothermic and it is a quick process.
• If simultaneous adsorption and absorption take place, it is termed as ‘sorption’ and sorption of gases on metal surface is called occlusion.

Question 32.
What are the uses of cellulose?
Answer:

• Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
• Gun cotton – nitrated ester of cellulose an explosive is prepared from cellulose.
• Cellulose act as food for animals.

Question 33.
How will you prepare PHBV? Give its use?
Answer:
(i) The biodegrable polymer PHBV (Poly hydroxy butyrate-co hydroxyl valerate) is prepared by the polymerisation of monomers 3 – hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

(ii) It is used in orthopaedic devices and in controlled release of drugs.

Part – IV

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

Question 34.
(a) (i) Which type of ores can be concentrated by froth floatation method?
Give two examples for such ores. (2)
(ii) Explain the variation In E°_M3+/M2+ series. (3)
[OR]
(b) (i) Mention the uses of silicon tetrachtoride. (2)
(ii) What are all the conditions that are necessary for catenation? (3)
Answer:
(a) (i) Suiphide ores can be concentrated by froth floatation method.
e.g., (i) Copper pyrites (CuFeS₂) (ii) Zinc blende (ZnS) (iii) Galena (PbS)

(ii)

• In transition series, as we move down from Ti to Zn, the standard reduction potential E° _M3+/M2+value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺
• E°_M3+/M2+ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
• The standard electrode potential for the M³⁺ /M²⁺ half cell gives’the relative stability between M³⁺ and M²⁺.
• The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
• For Fe³⁺/Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.
• Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.5 IV]

[OR]

(b) (i) Silicon tetrachloride is used in the production of semiconducting silicon.
It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

(ii) Essential condition for catenation:

• The valency of elements is greater than or equal to two.
• Element should have an ability to bond with itself.
• The self bond must be as strong as its bond with other elements.
• Kinetic inertness of catenated compound towards other molecules.

Question 35.
(a) (i) Discuss the manufacture of chlorine. (3)
(ii) What is inert pair effect? (2)
[OR]
(b) (i) Calculate the magnetic moment of Ti³⁺ and V⁴⁺. (2)
(ii) Draw all possible geometrical isomers of the complex [CO(en)₂Cl₂]⁺ and identify the optically active isomer. (3)
Answer:
(a) (i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na⁺ and Cl^– ions are formed. Na⁺ ion reacts with OH^– ions of water and forms sodium hydroxide.

Hydrogen and chlorine are liberated as gases.

Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

(ii) In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

[OR]

(b) (i) Ti (Z = 22) Ti³⁺3d¹
V (Z = 23) V⁴⁺ 3d¹
.’. \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mu_{\mathrm{B}}\) So they are paramagnetic.

(ii) 1. Cis – [Co(en)₂Cl₂]⁺

The coordination complex [Co(en)₂Cl₂]⁺ has three isõmers two optically active cis forms and the optically inactive trans form.

Question 36.
(a) (i) Calculate the number of atoms in a fee unit cell.
(ii) How do nature of the reactant influence rate of reaction?
[OR]
(b) (i) Account for the acidic nature of HClO₄
In terms of Bronsted – Lowry theory, identify its conjugate base.
(ii) IS it possible to store copper sulphate in an iron vessel for a long time?
Given \(\mathbf{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\) and \(\mathbf{E}_{\mathbf{F e}^{2+} \mathbf{F e}}^{s}=+\mathbf{0 . 4 4 V}\)
Answer:
(a) (i) Number of atoms in a fcc unit cell,

(ii) Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Let us compare the following two reactions that we carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺ . In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60°C.

The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts. If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantaneously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

[OR]

(b) (i) HClO₄ ⇌ H⁺ + ClO₄^–

According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

Let us consider the stabilities of the conjugate bases ClO₄^– , CIO₃^– , ClO₂^– and CIO^– formed from these acid HClO₄, HClO₃ , HClO₂, HOCl respectively. These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

The charge stabilization increases in the order, ClO^– < ClO^–₂ < ClO^–₃ < CIO^–₄ . This means ClO^–₄ will have maximum stability and therefore will have minimum attraction for H⁺. Thus ClO^–₄ will be weakest base and its conjugate acid HClO₄ is the strongest acid.

ClO^–₄ is the conjugate base of the acid HClO₄.

(ii) E⁰_cell = E⁰_ox + E⁰_red = 0.44 V +0.34 V = 0.78 V
These +ve E⁰_cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 37.
(a) (i) Explain the formation of water with copper catalyst by intermediate compound formation theory. (3)
(ii) O-nitro phenol is slightly soluble in water where as P-nitro phenol is more soluble. Give reason. (2)
[OR]
(b) (i) What happens when the following alkenes are subjected to reductive ozonolysis.
1. propene
2.1-Butene
3. Isobutylene (3)
(ii) What are reducing and non – reducing sugars? (2)
Answer:
(a) (i) Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as

(ii) O-nitro phenol is slightly soluble in water and more volatile due to intra molecular hydrogen bonding, whereas P-nitro phenol is more soluble in water and less volatile due to intermolecular hydrogen bonding.

[OR]

(b)

(ii) Reducing sugars: Those carbonhydrates which contain free aldehyde or ketonic group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosacchaides whether aldose or ketone are reducing sugars.

Non-reducing sugars: Cabohydrates which do not reduce Tollen’s reagent and Fehling’s solution are called non-reducing sugars. Example Sucrose. They do not have free aldehyde group.

Question 38.
(a) A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO₂ (B) on heating with liquid ammonia followed by treating with Br₂ /KOH gives (C) which on treating with NaNO₂ and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D. (5)
[OR]

(b) Explain the mechanism of cleansing action of soaps and detergents.(5)
Answer:

[OR]

(b) Mechanism of cleansing action of soaps and detergents:

• The cleansing action of both soaps and detergents from their ability to lower the surface tension of water, to emulsify oil or grease and to hold them in a suspension in water.
• This ability is due to the structure of soaps and detergents.
• In water a sodium soap dissolves to form soap anions and sodium cations. For example, the following chemical equation shows the ionisation of sodium palmitate.
  
• A soap anion consists of a long hydrocarbon chain with a carboxylate group on one end. The hydrocarbon chain, which is hydrophobic, is soluble in oils or grease. The ionic part is the carboxylate group which is hydrophilic, is soluble in water.
  
• In water, detergent dissolves to form detergent anions and sodium cations. For example the following chemical equations show the ionisation of sodium alkyl sulphate and sodium alkyl benzene sulphate.
  

6. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy stain:

• A soap or detergent anion consists of a hydrophobic part and a hydrophilic part.
• Soap or detergent reduces the surface tension of water. Therefore the surface of the cloth is wetted thoroughly.
• The hydrophobic parts of the soap or detergents anions are soluble in grease.
• The hydrophilic parts of the anions are soluble in water.
• Scrubbing or mechanical agitation helps to pull the grease away from the cloth and the
• grease is broken into smaller droplets.
• Repulsion between the droplets causes the droplets to be suspended in water, forming an emulsion.
• Thus the droplets do not coagulate or redeposit on the cloth. Rinsing washes away the droplets.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 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 4 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.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO₃
(d) NaNO₂
Answer:
(a) NaCN

Question 2.
The basic structural unit of silicates is
(a)(SiO₃)^2-
(b) (SiO₄)^2-
(c) (SiO)^–
(d) (SiO₄)^4-
Answer:
(d) (SiO₄)^4-

Question 3.
………………… is a pungent smelling gas.
(a) Ammonia
(b) Nitric acid
(c) Fluorite
(d) Sodium chloride
Answer:
(a) Ammonia

Question 4.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most.of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) i only
(b) i and ii
(c) ii and iii
(d) i and iii
Answer:
(d) i and iii

Question 5.
How many geometrical isomers are possible for [Pt (Py) (NH₃) (Br) (Cl)]?
(a) 3
(b) 4
(c) 0 3
(d) 15
Answer:
(a) 3

Question 6.
Metal excess defect is possible in……………….
(a) AgCl
(b) AgBr
(c) KCl
(d) Fes
Answer:
(c) KCl

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
Answer:
(b) The half life of a first order reaction does not depend on [A₀]; the half-life of a second order reaction does depend on [A₀].
Solution:
For a first order reaction

Question 8.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 × 10^-4 mol L^-1 solubility product of Ag₂C₂O₄ is
(a) 2.42 × 10^-8 mol³ L^-3
(b) 2.66 × 10^-12mol³ L^-3
(c) 4.5 × 10^-11mol³ L^-3
(d) 5.619 × 10^-12mol³ L^-3
Answer:
(d) 5.619 × 10^-12mol³ L^-3
Solution:

Question 9.
Assertion(A): Copper Sulphate can be stored in a Zinc vessel.
Reason (R): Zinc is less reactive than Copper.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are wrong

Question 10.
Fog is colloidal solution of……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
c
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.

(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid

Answer:
(c) benzene diazonium chloride

Question 13.
Which one of the following is the IUPAC name of CH₃ – CH₂ – CH₂ CN?
(a) Propiono nitrile
(b) Butane cyanide
(c) Isobutyro nitrile
(d) Butane nitrile
Answer:
(d) Butane nitrile

Question 14.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ……………
(a) L-Erythrose, L-Threose, L-Eiythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer:
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose

Question 15.
Tranquilisers are substances used for the treatment of ………………
(a) cancer
(b) AIDS.
(c) mental diseases
(d) blood infection
Answer:
(c) mental diseases

Part – II

Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12]

Question 16.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.

Question 18.
Write the biologically importance of coordination compounds.
Answer:
Biological important of coordination compounds:

(i) Ared blood corpuscles (RBC) is composed of heme group, which is Fe²⁺– Porphyrin complex, it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg²⁺ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO₂ and water into carbohydrates and oxygen.

(iii) Vitamin B₁₂(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is CO⁺ surrounded by Porphyrin like ligand.

(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.

Question 19.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

• Nature and state of the reactant .
• Concentration of the reactant
• Surface area of the reactant
• Temperature of the reaction
• Presence of a catalyst

Question 20.
What is meant by standard reduction potential? What is its application?
Answer:

• The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
• The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
• So higher the E° values, lesser is the tendency to undergo corrosion.

Question 21.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

• It is reversible
• It has low heat of adsorption
• It has weak van der Waals forces of attraction with adsorbent.
• It increases with increase in pressure.
• It forms multimolecular layer.

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:

This reaction follows SN¹ mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.

(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Part – III

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

Question 25.
Write the application of Iron (Fe).
Answer:

• Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
• Cast iron is used to make pipes, valves and pumps stoves etc.
• Magnets can be made of iron and its alloys and compounds.

An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H₂SO_4.H₂O) and di (H₂SO_4.H₂O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 27.
Give evidence that [CO(NH₃)gCl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO₃ solution and BaCl₂ solution. When Cl^– ions are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If SO₄^2- ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
z = k [x]^m [y]ⁿ …… (1)
8z = k [4x]^m [y]ⁿ…… (2)
I 6z = k [4x]^m [4y]ⁿ…… (3)
Dividing Eq (2) by Eq (1) we get,

1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m . 4ⁿ
16 = 4².4ⁿ
16/16 = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]^m[y]ⁿ
k [x]^1.5 [y]⁰
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Answer:

• HF and F^– , HS^– and H₂S are two conjugate acid – base pairs.
• F^– is the conjugate base of the acid HF (or) HF is the conjugate acid of F^– .
• H₂S is the conjugate acid of HS^– (or) HS^– is the conjugate base of H₂S.

(ii)

• HPO₄^2- and PO₄^3- , SO₃^-2 and HSO₃^– are two conjugate acid – base pairs.
• PO₄^3- is the conjugate base of the acid HPO₄^2- (or)
  HPO₄^2- is the conjugate acid of PO₄^3-.
• HSO₃^– is the conjugate acid of SO₃^-2 (or) SO₃^-2 is the conjugate base of HSO₃^–

(iii)

• NH₄⁺ and NH₃, CO₃^2- and HCO₃^– are two conjugate acid – base pairs.
• HCO₃^– is the conjugate of acid CO₃^2- (or) CO₃^2- is the conjugate bases of HCO₃^–
• NH₃ is the conjugate base of NH₄⁺ (or) NH₄⁺ is the conjugate acid of NH₃.

Question 30.
In the following fields, how adsorption is applied?
Answer:
(i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer:
(i) Medicine: – Drugs cure diseases by adsorption on body tissues.

(ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil.

(iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

(iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting absorbed on precipitate. It is used to indicate the end point of filtration.

Question 31.
Identify A, B, C, and D

Answer:

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:

Question 33.
Differentiate soap and detergents?
Soap :

1. Soaps are sodium or potassium salt of long chain fatty acid.
2. Soaps are made from animal (or) plant fats and oils.
3. Soaps have lesser cleansing action.
4. Soaps are bio degradable.
5. Soaps are less effective in hard water.
6. They have a tendency to form a scum in hard water.
7. Example: Sodium palmitate.

Detergent :

1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene sulphonic acid.
2. Detergents are made from petrochemicals.
3. Detergents have more cleansing action.
4. Detergents are non-bio degradable.
5. Detergents are more effective even in hard water.
6. They do not form scum with hard water.
7. Example: Sodium lauryl sulphate.

Part – IV

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

Question 34.
(a) (i) Explain the role of carbon monoxide in the purification of nickel? (2)
(ii) Describe the structure of diborane. (3)
[OR]
(b) (i) What type of hybridisation occur in 1. BrF₅. 2. BrF₃ (2)
(ii) Most of the transition metals act as catalyst. Justify this statement. (3)
Answer:
(a) (i) During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl.
Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

(ii) In diborane two BH₂ units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to
form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds.
1. e. two three centred B-H-B bonds utilise two electrons each. Hence, these bonds are three centre- two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure.

In dibome, the boron is sp³ hybridised. Three of the four sp³ hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.

[OR]

(b) (i) 1. BrF₅. is a AX₅ type. Therefore is has sp³ d² hybridisation. Hence, BrF₅ molecule has square pyramidal shape.
2. BrF₃ is a AX₃ type. Therefore it has sp³ d hybridisation. Hence, BrF₃ molecule has T-shape.

(ii) Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 35.
(a) (i) Why tetrahedral complexes do not exhibit geometrical isomerism. (2)
(ii) Explain about the importance and application of coordination complexes. (3)
[OR]
(b) Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
(a) (i) In tetrahedral geometry

• AH the four ligands are adjacent or equidistant to one another.
• The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
• It has plane of symmetry

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.

(ii) 1. Phthalo blue – a bright blue pigment is a complex of copper (II) ion and it is used in printing ink and packaging industry.

2. Purification of Nickel by Mond’s process involves formation of [Ni (CO)₄ ] which yields 99.5% pure on decomposition.

3 . EDTA is used as a chelating ligand for the separation of lanthanides, in softening of hard water and also in removing lead poisoning.

4. Coordination complexes are used in the extraction of silver and gold from their ores
by forming soluble cyano complex. These cyano complexes are reduced by zinc to
yield metals. This process is called Mac – Arthur Forrest cyanide process.

5. Some metal ions are estimated more accurately by complex fonnation. For eg., Ni present in Nickel chloride solution is estimated accurately forming an insoluble, complex called [Ni (DMG)₂].

6. Many of the complexes are used as catalyst in organic and inorganic reactions. For e.g.,
(i) Wilkinson’s Catalyst – [(PPh₃)₃, Rh Cl] is used for hydrogenation of alkenes.
(ii) Ziegler – Natta Catalyst [TiCl₄ + Al (C₂H₅) ]₃ is used in the polymerisation of ethene.

7. In photography, when the developed film is washed with sodium thio sulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodium dithio sulphate argentate (I) which can be removed easily by washing the film with water.

AgBr + 2 Na₂ S₂O₃ → Na₃ [Ag (S₂O₃)₂] + 2NaBr

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

(iii) ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer. Third layer gives different arrangement. Fourth Layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (Le.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated with the addition of more layers.

In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12.

Voids: The empty spaces between the three dimensional layers are known as voids. There are
two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void:
A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom r Layer a layer is a hole between four spheres. The spheres are arranged at the vertices Layerc of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10^-3s^-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of k_spof two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t_1/2 = 0.693/ k
t_1/2= O.693/1.54 ×10^-3s ^-1 = 450s

(ii) 1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A smaLl amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of ImI of 10% NaCI solution. Smaller the gold number, greater the protective power.

[OR]

(b) (i) 1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.

2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.

3. Acidic buffer solution : Solution containing acetic acid and sodium acetate.
Basic buffer solution : Solution containing NH₄O and NH₄Cl.

(ii)

Ni(OH)₂ is more soluble than AgCN.

Question 37.
(a) Describe about lead storage battery construction and its uses. (5)
[OR]
(b) (i) What happens when m-cresol is treated with acidic solution of sodium dichromate? (3)
(ii) Formic acid is more stronger than acetic acid. Justify this statement. (2)
(a) Lead storage battery :
1. Anode : Spongy lead
Cathode : Lead plate bearing PbO₂
Electrolyte : 38% by mass of H₂SO₄ with density 1.2 g/ml

2. Oxidation occurs at the anode

Pb_(s) → Pb²⁺ _(aq) + 2e^– …………(1)

The Pb²⁺ ions combine with SO₄^2- to form PbSO₄ precipitate

Pb²⁺_(aq) + SO₄^2- → PbS_4(s) …………….(2)

3. Reduction occurs at the cathode

PbO_2(s) + 4H⁺_(aq) + 2e^– – → Pb²⁺_(aq) + 2H₂O ………….(3)

The Pb²⁺ ions also combines with SO₄^2- ions to form sulphuric acid to form PbSO₄ Precipitate

Pb²⁺_(aq) + SO₄^2-_(aq) → PbSO₄ …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb_(s) + PbO_2(s) + 4H⁺ + 2SO₄^2-_(aq) → 2PbSO_4(s) + 2H₂O(l)

5. The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volts.

6. The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2- ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.

The above reaction is exactly the reverse of redox reaction which takes place while discharging.

8. Uses: Lead storage battery is used in automobiles, trains, inverters.

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH₃, – C₂H₅, – C₃H₇

Question 38.
(i) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating
forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’
of molecular formula C₆H₇N. Write the structures and IUPAC names of compound
A,BandC. (3)
(ii) Lactose act as reducing sugar. Justify this statement. (2)
[OR]
(b) (i) Explain the mechanism of enzyme action? (3)
(ii) Which chemical is responsible for the antiseptic properties of dettol? (2)
Answer:
(a) (i) Step-1: To find out the structure of compounds‘B’and‘C’.
1. Since compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ + KOH (i.e, ., Hoffmann bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C₆H₇N is C₆H₅NH₂ (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.

(ii) Lactose is a disaccharide and contains one galactose unit and one glucose unit.
In lactose, the β-D galactose and β-D glucose are linked by β-1, 4 – glycosidic bond.
The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

[OR]

(b) (i) Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.

(ii) (a) Chloroxylenol and (b) Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.

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.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 3 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 3 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.
Wolframite ore is separated from tinstone by the process of
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer:
(d) Electromagnetic separation

Question 2.

Identify A
(a) BN₃
(b) B₃N
(c) (BN)₃
(d) BN
Answer:
(d) BN

Question 3.
Among the following the correct order of acidity is
(a) HClO₂ < HCIO < HClO₃ < HClO₄
(b) HClO₄ < HClO₂ < HClO < HClO₃
(c) HClO₃ < HClO₄ < HClO₂ < HClO
(d) HClO < HClO₂ < HClO₃ < HClO₄
Answer:
(d) HClO < HClO₂ < HClO₃ < HClO₄

Question 4.
Which of the following transition metal is present in Vitamin B₁₂ ?
(a) Cobalt
(b) Platinum
(c) Copper
(d) Iron
Answer:
(a) Cobalt

Question 5.
A magnetic moment of 1.73BM will be shown by one among the following.
(a)TiCl₄
(b) [COCl₆]^4-
(c) [CU(NH₃)₄]²⁺
(d) [Ni(CN)₄]^2-
Answer:
(c) [CU(NH₃)₄]²⁺

Question 6.
Consider the following statements.
(,i) metallic solids possess high electrical and thermal conductivity
(ii) solid ice are soft solids under room temperature
(iii) In non polar molecular solids constituent molecules are held together by strong electrostatic forces of attraction
Which of the above statements is./ are not correct?
(a) (i) & (ii)only
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 7.
For a first order reaction, the rate constant is 6.909 min^-1.The time taken for 75% conversion
in minutes is …………………………..

Answer:
(b) \(\left(\frac{2}{3}\right) \log 2\)
Solution:

Question 8.
The pH of 10^-5 M KOH solution will be ……………….
(a) 9
(b) 5
(c) 19
(d) none of these
Answer:
(a) 9

Solution:
KOH → K⁺ + OH^–
10^-5M 10^-5M 10^-5M
[OH^–]= 10^-5M
pH = 14 – pOH
pH = 14 – (-log [OH^–])
= 14 + log [OH^– ] = 14 + log10^-5
= 14 – 5 = 9

Question 9.
The Lead storage battery is used in
(a) pacemakers
(b) automobiles
(c) electronic watches
(d) flash light
Answer:
(b) automobiles

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
Hint: coagulating power ± \(\frac{1}{\text { coagulation value }}\)
Answer:
(a) III > II > I

Question 11.
Oxygen atom in ether is
(a) very active
(b) replacable
(c) comparatively inert
(d) less active
Answer:
(c) comparatively inert

Question 12.
During nucleophilic addition reaction, the hybridisation of carbon changes from
(a) sp² to sp³
(b) sp³ to sp²
(c) sp to sp³
(d) dsp² to sp³
Answer:
(a) sp² to sp³

Question 13.
Match the following:

Answer:
(a) A – 3, B – 4, C – 1, D – 2

Question 14.
Which one of the following is levorotatory?

Answer:

Question 15.
Which one of the following structures represents nylon 6,6 polymer?

Answer:

Part – II

Answer any six questions. Question No. 23 is compulsory. [6 × 2 = 12]

Question 16.
Carbon monoxide is more effective reducing agent than carbon below 983 K but, above this temperature, the reverse is true -Explain.
Answer:
From the Ellingham diagram, we find that at 983 K, the curves intersect.

The value of ∆G° for change of C to CO₂ is less than the value of ∆G° for change of CO to CO₂ Therefore, coke (C) is a better reducing agent than CO at 983K or above temperature. However below this temperature (e.g. at 673K), CO is more effective reducing agent than C.

Question 17.
Mention the uses of the potash alum.
Answer:

• It is used for purification of water
• It is also used for water proofing and textiles
• It is used in dyeing, paper and leather tanning industries
• It is employed as a styptic agent to arrest bleeding.

Question 18.
Why Gd³⁺ is colourless?
Answer:
Gd – Electronic Configuration : [Xe] 4f⁷5d¹ 6s²
Gd³⁺ Electronic Configuration : [Xe] 4f⁷
In Gd³⁺ , no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.

Question 19.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free to move but remain held together by strong electrostatic forces of attraction, so they cannot conduct electricity in the solid state.

Question 20.
What are Lewis acids and bases? Give two example for each.
Answer:
I. Lewis acids:
(i) Lewis acid is a species that accepts an electron pair.
(ii) Lewis acid is a positive ion (or) an electron deficient molecule.
(iii) Example, Fe²⁺, CO₂, BF₃ , SiF₄ etc…

II. Lewis bases:
(i) Lewis base is a species that donates an electron pair.
(ii) Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons.
(iii) Example, NH₃, F^–, CH₂= CH₂ CaO etc….

Question 21.
NH₃, CO₂ are readily adsorbed where as H₂, N₂ are slowly adsorbed. Give reason.
Answer:

• The nature of adsorbate can influence the adsorption. Gases like NH, C02 are easily liquefiable as have greater Van der Waals forces of attraction and hence readily adsorbed due to high critical temperature.
• But permanent gases like H₂. N₂ can not be easily liquefied and having low critical temperature and adsorbed slowly.

Question 22.
Alcohol can act as Bronsted base. Prove this statement.
Answer:
Alcohols can also act as a Bronsted bases. It is due to the presence of unshared electron pairs on oxygen which make them to accept proton. So proton acceptor are Bronsted bases. i. e., alcohols are Bronsted bases.

Question 23. Arrange the following in decreasing order of basic strength

Answer:
(i) Aliphatic amines are more basic than aromatic amines. Therefore CH₃CH₂NH₂ and CH₃NH₂ are more basic. Among the ethylamine and methylamine, ethylamine was experienced more +1 effect than methylamine and hence ethylamine is more basic than methylamine.

(ii) Nitrogroup has a powerfol electron withdrawing group and they have both -R effect as well as -I effect. As a result, all the nitro anilines are weaker bases than aniline. In P-nitroaniline

both R effect and -I effect of the NO₂ group decrease the basicity.

(iii) Therefore decreasing order of basic strength is,

Question 24.
How are biopolymers more beneficial than synthetic polymers?
Answer:
Durability of synthetic polymers is advantageous, however it presents a serious waste disposable problem. In renewal of the disposable problem, biodegradable polymers are useful to us.

Biopolymers are safe in use. They disintegrate by themselves in biological system during a certain period of time by enzymatic hydrolysis and to some extent by oxidation and hence, are biodegradable. As a result, they do not cause any pollution.

Part – III

Answer any six questions. Question No. 26 is compulsory. [6 × 3 = 18]

Question 25.
Explain Aluminothermic process.
Answer:
Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process. In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide) is used.

BaO₂ + Mg → BaO + Mgo

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is : 852 kJ-moF1) which facilitates the reduction of Cr₂O₃ by aluminium power.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with BaCl₂ and gives a red colour compound with alizarin. Identify A and B.
Answer:
(i) A double salt which contains fourth period alkali metal (A) is potash alum
K₂SO₄ Al₂ (SO₄ )₃ .24H₂O

(ii) On heating potash alum (A) 500 K give anhydrous potash alum (or) burnt alum (B).

(iii) Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion.
Sulphate ion reacts with BaCl2 to form white precipitate of Barium Sulphate:
(SO₄ )^2- + BaCl₂ → BaSO₄ + 2Cl^–
Aluminium ion reacts with alizarin solution to give a red colour compound.

Question 27.
[Ti (H₂O)₆ ]²⁺ is purple in colour. Prove this statement.
Answer:
(i) In [Ti (H₂O)₆ ]²⁺, the central metal ion is Ti³⁺ which has d¹ configuration. This single electron occupies one of the t_tgorbitals in the octahedral aqua ligand field.

(ii) When white light falls on this complex, the d electron absorbs light and promotes itself to eg level.

(iii) The spectral data show the absorption maximum is at 20000 cm^-1 corresponding to the crystal field splitting energy (∆₀) 239.7 kJ mol^-1. The transmitted colour associated with this absorption is purple and hence the complex appears in purple in colour.

Question 28.
Define half life of a reaction. Show that for a first order reaction half life is independent of initial concentration.
Answer:
Half life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value.
For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.
The rate constant for a first order reaction is given by,

Question 29.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law: It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution:
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraush’s Law. For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH₃COONa .

Question 30.
Explain the following observations.
(a) Lyophilic colloid is more stable than lyophobic colloid.
(b) Coagulation takes place when sodium chloride solution added to a colloidal solution of ferric hydroxide.
(c) Sky appears blue in colour.
AnsweR:
(a) A lyophilic sol is stable due to the charge and the hydration of the sol particles. Such a sol can only be coagulatd by removing the water and adding solvents like alcohol, acetone, etc. and then an electrolyte. On the other hand, a lyophobic sol is stable due to charge only and hence it can be easily coagulated by adding small amount of an electrolyte.

(b) The colloidal particles get precipitated, i.e., ferric hydroxide is precipitated.

(c) The atmospheric particles of colloidal range scatter blue component of the white sunlight preferentially. That is why the sky appears blue.

Question 31.
Mention the uses of formic acid?
Answer:
Formic acid. It is used
(i) for the dehydration of hides.
(ii) as a coagulating agent for rubber latex
(iii) in medicine for treatment of gout
(iv) as an antiseptic in the preservation of fruit juice

Question 32.
Write the structure of all possible dipeptides which can be obtained from glycine and alanine.
Answer:

Therefore two dipeptides structures are possible from glycine and alanine. They are (i) glycyl alanine and (ii) Alanyl glycine

Question 33.
How the tranquilizers work in body?
Answer:

• They are neurologically active drugs.
• Tranquilizer acts on the central nervous system by blocking the neurotransitter dopamine in the brain.
• This drug is used for treatment of stress anxiety, depression, sleep disorders and severe mental diseases like schizophrenia.

Part – IV

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

Question 34.
(a) (i) How will you purify metals by using iodine?(3)
(ii) Boron does not react directly with hydrogen. Suggest one method to
prepare diborane from BF₃(2)
[OR]
(b) (i) Write the reason for the anomalous behaviour of Nitrogen. (3)
(ii) Mn²⁺ is more stable than Mn⁴⁺. Why? (2)
Answer:
(a) (i) This method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Titanium and zirconium can be purified using this method. For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide.( TiI₄ ). The impurities are left behind, as they do not react with iodine.
Ti_(s) + 2I_2(s) → TiI₄ (vapour)

The volatile titanium tetraiodide vapour is passed over a tungsten filament at a
temperature around I 800 K. The titanium tetraiodide is decomposed and pure titanium
is deposited on the filament. The iodine is reused.
TiI₄ (vapour) ) → Ti_(s) +2I_2(s)

(ii) Boron does not react directly with hydrogen. However it forms a variety of hydrides
called boranes. Treatment of gaseous boron triuluoride with sodium hydride around
450 K gives diborane.

(b) (i) 1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N₂ has a unique ability to form pπ- pπ multiple bond whereas the heavier members of this group (15) do not form pπ- pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ- dπ bond due to the absence of d-orbitals whereas other elements can.

(ii) The relative stability of different oxidation states of 3d metals is correlated with the extra stability of half-filled and fully filled electronic configurations.
Example: Mn²⁺ (3d⁵) is more stable than Mn⁴⁺ (3d³)

Question 35.
(a) (i) Draw all possible stereo isomers of a complex Ca[CO(NH₃)Cl(Ox)₂] (3)
(ii) What is Bragg’s equation? (2)
(b) (i) What is an elementary reaction? Give the differences between order and molecularity of a reaction. (3)
(ii) In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction? (2)
Answer:
(a) (i) Possible stereo isomers of a complex Ca[CO(NH₃)Cl(Ox)₂]

(ii) 1. X-ray diffraction analysis is the most powerful tool for the determination of crystal structure.

2. The interplanar distance (d) between two successive planes of atoms can be calculated using the following equation form the X-ray diffraction data 2d sin θ = nλ. The equation is known as Bragg’s equation
Where λ = wavelength of X-ray; d = Interplanar distance
θ = The angle of diffraction n = order of reflection
By knowing the values of λ, λ and n, we can calculate the value of d
\(d=\frac{n \lambda}{2 \sin \theta}\)

Using these values, the edge of the unit cell can be calculated.

[OR]

(b) (i) Elementary reaction: Each and every single step in a reaction mechanism is called an elementary reaction.

Differences between order and molecularity:

Order of a reaction :

• It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
• It can be zero (or) fractional (or) integer.
• It is assigned for a overall reaction.

Molecularity of a reaction :

• It is the total number of reactant species that are involved in an elementary step.
• It is always a whole number, cannot be zero or a fractional number.
• It is assigned for each elementary step of mechanism.

(ii)

Question 36.
(a) (i) Derive the relation between pH and pOH (3)
(ii) Give three uses of emulsions.(2)
[OR]
(b) How would you measure the conductivity of ionic solutions? (5)
Answer:
(a) (i) pH = -log_1o [H₃O⁺] …………..(1)
pOH = -log_1o [OH^–] ……..(2)
Adding equations (1) and (2),
pH + pOH = (-log_1o[H₃O⁺]) + (-log_1o [OH^–])
= -[(log10tHO+]) + (logio [OfT])]
pH + pOH = -log₁₀[H₃O⁺] [OH^– ]
[H₃O⁺] [OH^–] = K_w
∴ pH + pOH = – log_1oK_w
pH + pOH = pKw [pk_w = -log_1oK_w]

At 25°C, the ionic product of water K_w =1 x 10^-14.
pK_w = -log_1o10^-14 = 14log_1o10= 14
pK_w = 14
∴ pH + pOH = 14 at 25°C.

(ii)

1. The cleansing action of soap is due to emulsions.
2. It is used in the preparation of vanishing cream.
3. It is used in the preparation of cold liver oil.

[OR]

(b) The conductivity of an electrolytic solution is determined by using a wheatstone bridge arrangement in which one resistance is replaced by a conductivity cell filled with the electrolytic solution of unknown conductivity.

In the measurement of specific resistance of a metallic wire, a DC power supply is used. Here AC is used for this measurement to prevent electrolysis. Because DC current through the conductivity cell leads to the electrolysis of the solution taken in the cell.

A wheatstone bridge is constituted using known resistances P,Q, a variable resistance S and conductivity cell. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C.
A suitable detector is connected between the junctions B and D.
The variable resistance S is adjusted until the bridge is balanced and in this conditions, there is no current flow through the detector.

Under balanced condition
P/Q = R/S
∴ R = P/Q x S …………(1)
The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured S value using the equation (1).

Specific conductance (or) conductivity of an electrolyte can be calculated from the resistance value (R) using the following expression
\(\kappa=\frac{1}{R}\left[\frac{l}{A}\right]\)

The value of cell constant \(\frac{l}{A}\) is usually provided by the cell manufacturer. Alternatively the cell constant may be determined using KC1 solution whose concentration and specific conductance are known.

Question 37.
(a) (1) What is metamerism? Give the structure and ¡UPAC name of metamers of 2-methoxy propane (2)
(ii) Explain the following reactions.
(2)
(OR]
(b) An organic compound (A) of molecular formula C₇H₈O on oxidation with alkaline KMnO₄ gives (B) of formula C₇H₆O.
(B) on reaction with Cl₂ in the presence of catalyst FeCl₃ gives
(C) of formula C₇H₅OCl. (B) on reaction with Cl in the absence
of catalyst gives C₇H₅OCl. Identify A,B,C,D and explain the reaction involved.
Answer:
(a) (i) Metamerism: It is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.

Ethoxy ethane and 1-methoxy propane are metamers of 2-methoxy propane.

(ii)

(b)

Question 38.
(a) (0 Account for the following:
1. Primary amines (R – NH2)have higher boiling point than tertiary amines (R₃N).
2. Aniline does not undergo Friedel-Crafts reaction.
3. (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. (3)
(H) Name two fat soluble vitamins, their sources and the diseases caused due to
their deficiency in diet. (2)
[OR]
(b) (i) Why ranitidine is a better antacid than magnesium hydroxide? (2)
(ii) What is bakelite? How is it prepared? Give its uses. (3)

(a) (i) 1. Due to maximum intermolecular hydrogen bonding in primary amines (due to presence of more number of H-atoms), primary amines have higher boiling point in comparison to tertiary amines.

2. Aniline does not undergo Friedel-Crafts reaction due to acid-base reaction. Aniline and a Lewis Acid / Protic Acid, which is used in Friedel-crafts reaction.

3. In (CH₃)₃N there is maximum steric hindrance and least solvation but in (CH₃)₂NH the solvation is more group; di-methyl amine is still a stronger base than trimethyl amine.

(ii)

Vitamins	Sources Deficiency	Diseases
Vitamin A	Fish, liver oil, carrot	Night blindness
Vitamin D	Sunlight, milk, egg yolk	Rickets and osteomalacia

[OR]

(b) (i) To treat acidity, weak base such as magnesium hydroxide is used. But this weak base make the stomach alkaline and trigger the production of much acid. This treatment only relieves the symptoms and does not control the cause. But ranitine stimulate the secretion of HC1 by activating the receptor in the stomach wall which binds the receptor and inactivate them. So ranitine is a better antacid than magnesium hydroxide.

(ii) 1. Bakelite is a thermo setting plastic. It is prepared from the monomers such as phenol and formaldehyde. The condensation polymerisation take place in the presence of acid or base catalyst.

2. Phenol reacts with methanal to form ortho or para hydroxyl methyl phenols which on further reaction with phenol gives linear polymer called novolac. Novolac on further healing with formaldehyde undergoes cross linkages to form bakelite.

• Novolac is used in paints ‘Soft bakelites are used in making glue for binding laminated
  wooden planks and in varnishes
• Hard bakelites are used to prepare combs, pens.

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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.

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