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Students can Download Tamil Nadu 12th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

General Instructions:

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.
Match the following:

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

Question 2.
Changing the codon AGC to AGA represents ___________.
(a) Mis-sense mutation
(b) Non-sense mutation
(c) Frame-shift mutation
(d) Deletion mutation
Answer:
(a) Mis-sense mutation

Question 3.
Restriction enzymes are ______.
(a) Not always required in genetic engineering
(b) Essential tools in genetic engineering ‘
(c) Nucleases that cleave DNA at specific sites
(d) Both b and c
Answer:
(d) Both b and c

Question 4.
Totipotency refers to _________.
(а) capacity to generate genetically identical plants
(b) capacity to generate a whole plant from any plant cell/explant
(c) capacity to generate hybrid protoplasts
(d) recovery to healthy plants from diseased plants
Answer:
(b) capacity to generate a whole plant from any plant cell/explant

Question 5.
In soil water available for plants is ________.
(a) gravitational water
(b) Chemically bound water
(c) Capillary water
(d) hygroscopic water
Answer:
(c) Capillary water

Question 6.
Which one is in descending order of a food chain?
(a) Producers → Secondary consumers → Primary consumers → Tertiary consumers
(b) Tertiary consumers → Primary consumers → Secondary consumers → Producers
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers
(d) Tertiary consumers → Producers → Primary consumers → Secondary consumers
Answer:
(c) Tertiary consumers → Secondary consumers → Primary consumers → Producers

Question 7.
Clean Development Mechanism (CDM) is defined is _______.
(a) Copenhagen Acord
(b) Montreal protocol
(c) Paris Agreement
(d) Kyoto protocol
Answer:
(d) Kyoto protocol

Question 8.
Pick out the odd pair.
(a) Man selection – Morphological characters
(b) Pureline selection – Repeated self pollinartion
(c) Clonal selection – Sexually propagated
(d) Natural selection – Involves nature
Answer:
(a) Man selection – Morphological characters

Part – II

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

Question 9.
Write a short note on pollen kitt.
Answer:
Pollenkitt is contributed by the tapetum and coloured yellow or orange and is chiefly made of carotenoids or flavonoids. It is an oily layer forming a thick viscous coating over pollen surface. It attracts insects and protects damage from UV radiation.

Question 10.
What is back cross?
Answer:
Back cross is a cross of F₁ hybrid with any one of the parental genotypes. The back cross is of two types; they are dominant back cross and recessive back cross. It involves the cross between the F₁ off spring with either of the two parents.

Question 11.
Productivity of profundal zone will be low. why?
Answer:
The producers of the pond ecosystem depends on phytoplankton through photosynthesis. Profundal zone lies below the limnetic zone with no effective light penetration, hence productivity rate is very low.

Question 12.
Explants for tissue culturing has to be surface sterilized. How?
Answer:
The explants are surface sterilized by first exposing the material is running tap water and then treating it in surface sterilizing agents like 0.1 % Mercuric chloride, 70% ethanol under aseptic condition inside the laminar air flow chamber.

Question 13.
Define heterosis, through which way does this condition can be maintained for generation.
Answer:
The superiority of F₁ hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, adaptability resistance to disease, pest, etc., Vegetative propogation is the best suited method to maintain hybrid vigour.

Question 14.
Discuss which wood is better for making furniture.
Answer:
Teak wood is the ideal type of wood for making household furnitures because, it is highly durable and shows great resistance against the attack of termites and fungi. Moreover it does not split or crack and is a carpenter friendly wood.

Part – III

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

Question 15.
Give the names of the scientists who rediscovered mendel’s work.
Answer:

• Hugo de Vries of Holland
• Carl Correns of Germany
• Erich von Tschermark of Austria.

Question 16.
Write the advantages and disadvantages of Bt cotton.
Answer:
The advantages of Bt cotton are:

• Yield of cotton is increased due to effective control of bollworms.
• Reduction in insecticide use in the cultivation of Bt cotton
• Potential reduction in the cost of cultivation.

Bt cotton has some limitations:

• Cost of Bt cotton seed is high.
• Effectiveness upto 120 days after that efficiency is reduced.
• Ineffective against sucking pests like jassids, aphids and whitefly.
• Affects pollinating insects and thus yield.

Question 17.
How cryopreservation works?
Answer:
Cryopreservation, also known as Cryo-conservation, is a process by which protoplasts, cells, tissues, organelles, organs, extracellular matrix, enzymes or any other biological materials are subjected to preservation by cooling to very low temperature of -196°C using liquid nitrogen. At this extreme low temperature any enzymatic or chemical activity of the biological material will be totally stopped and this leads to preservation of material in dormant status. Later these materials can be activated by bringing to room temperature slowly for any experimental work.

Question 18.
What is Co-evolution? Give examples.
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.

Examples:

• Corolla length and proboscis length of butterflies and moths (Habenaria and Moth).
• Bird’s beak shape and flower shape and size.

Question 19.
Write a short note on Chipko Movement.
Answer:
Chipko Movement:
The tribal women of Himalayas protested against the exploitation of forests in 1972. Later on it transformed into Chipko Movement by Sundarlal Bahuguna in Mandal village of Chamoli district in 1974. People protested by hugging trees together which were felled by a sports goods company. Main features of Chipko movement were,

• This movement remained non political
• It was a voluntary movement based on Gandhian thought.
• It was concerned with the ecological balance of nature
• Main aim of Chipko movement was to give a slogan of five F’s – Food, Fodder, Fuel, Fibre and Fertilizer, to make the communities self sufficient in all their basic needs.

Part – IV

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

Question 20.
(a) Name the tissue that nourishes the embryo in angiospermic seeds. Explain its types.
Answer:
Structure of ovuIe (Megasporangium)
Endosperm: The primary endosperm nucleus (PEN) divides immediately after fertilization but before the zygote starts to divide, into an endosperm. The primary endosperm nucleus is the result of triple fusion (two polar nuclei and one sperm nucleus) and thus has 3n number of chromosomes. It is a nutritive tissue and regulatory structure that nourishes the developing embryo. Depending upon the mode of development three types of endosperm are recognized in angiosperms. They are nuclear endosperm, cellular endosperm and helobial endosperm.

Nuclear endosperm: Primary Endosperm Nucleus undergoes several mitotic divisions without cell wall formation thus a free nuclear condition exists in the endosperm.
Examples: Coccinia, Capsella and Arachis.

Cellular endosperm:
Primary endosperm nucleus divides into 2 nuclei and it is immediately followed by wall formation. Subsequent divisions also follow cell wall formation.
Examples: Adoxa, Helianthus and Scoparia.

Helobial endosperm: Primary Endosperm Nucleus moves towards base of embryo sac and divides into two nuclei. Cell wall formation takes place leading to the formation of a large micropylar and small chalazal chamber. The nucleus of the micropylar chamber undergoes several free nuclear division whereas that of chalazal chamber may or may not divide. Examples : Hydrilla and Vallisneria.

Ruminate endosperm: The endosperm with irregularity and unevenness in its surface forms ruminate endosperm. Examples : Areca catechu, Passiflora and Myristica.

[OR]

(b) Describe the basic steps involved in recombinant DNA technology.
Answer:

• Isolation of a DNA fragment containing a gene of interest that needs to be cloned. This is called an insert.
• Generation of recombinant DNA (rDNA) molecule by insertion of the DNA fragment into a carrier molecule called a vector that can self-replicate within the host cell.
• Selection of the transformed host cells that is carrying the rDNA and allowing them to multiply thereby multiplying the rDNA molecule.
• The entire process thus generates either a large amount of rDNA or a large amount of protein expressed by the insert.
• Wherever vectors are not involved the desired gene is multiplied by PCR technique. The multiple copies are injected into the host cell protoplast or it is shot into the host cell protoplast by shot gun method.
  

Question 21.
(a) Derive the protocol for micro propagation of banana.
Answer:

[OR]

(b) Expand NBT and Explain how it is involved in developing new traits in plant breeding.
Answer:
New Breeding Techniques (NBT) are a collection of methods that could increase and accelerate the development of new traits in plant breeding. These techniques often involve genome editing, to modify DNA at specific locations within the plants to produce new traits in crop plants. The various methods of achieving these changes in traits include the following,

• Cutting and modifying the genome during the repair process by tools like CRISPR /Cas.
• Genome editing to introduce changes in few base pairs using a technique called Oligonucleotide-directed mutagenesis (ODM).
• Transferring a gene from an identical or closely related species (cisgenesis).
• Organising processes that alter gene activity without altering the DNA itself (epigenetic methods).

Bio-Zoology [Maximum Marks: 35]

Part – I

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

Question 1.
Assertion (A): Organisms show three phases in their life cycle.
Reason (R): Juvenile phase is a degenerative phase.
(a) A is correct. R is incorrect
(b) Both A and R are incorrect
(c) R is the correct explantation for A
(d) A is not correct but R is correct.
Answer:
(a) A is correct. R is incorrect

Question 2.
Find the wrongly matched pair
(a) Bleeding phase – fall in oestrogen and progesterone
(b) Follicular phase – rise in oestrogen
(c) Luteal phase – rise in FSH level
(d) Ovulatory phase – LH surge
Answer:
(c) Luteal phase – rise in FSH level

Question 3.
Identify the correct statements from the following.
(a) chlamydiasis is a viral disease
(b) Gonorrhoea is caused by a spirochaete bacterium, Treponema palladium
(c) The incubation period for syphilis is 2 to 14 days is males and 7 to 21 days in males.
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.
Answer:
(d) Both syphilis and gonorrhoea are easily cured with antibiotics.

Question 4.
Klinefelter’s syndrome is characterized by a karyotype of _______.
(a) XYY
(b) XO
(C) XXX
(d) XXY
Answer:
(d) XXY

Question 5.
The golden age of repetails was _______.
(a) Mesozoic era
(b) Cenozoic era
(c) Paleozoic era
(d) Proterozoic era
Answer:
(a) Mesozoic era

Question 6.
Match list I with list II

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

Question 7.
The relation ship between sucker fish and shark is ________.
(a) Competition
(b) Commensalism
(c) Predation
(d) Parasitism
Answer:
(b) Commensalism

Question 8.
Who is called as the forest man of India?
(a) Sunderlal Bahuguna
(b) M.S. Swamination
(c) Dr. V. Kurier
(d) Jadav Payeng
Answer:
(d) Jadav Payeng

Part – II

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

Question 9.
Mention the importance of the position of the testes in humans.
Answer:
The testes are positioned in such a way hanging out from the body in scrotal sac that provides optimal temperature 2°C to 3°C lower than internal body temperature for effective sperm production.

Question 10.
Type A blood should not be injected to a person having B-blood group. Why?
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 11.
Name the anticodons required to recognize the following codons. AAU, CGA, UAU, GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 12.
What is diapedesis?
Tissue damage and infection induce leakage of vascular fluid, containing chemotactic signals like serotonin, histamine and prostaglandins. They influx the phagocytic cells into the affected area. This phenomenon is called diapedesis.

Question 13.
How was Insulin obtained before the advent of rDNA technology? What were the problems encountered?
Answer:
Conventionally, Insulin was isolated and refined from the pancreas of pigs and cows to treat diabetic patients. Though it is effective, due to minor structural changes, the animal insulin caused allergic reactions in few patients.

Question 14.
How many hotspots are there is India? Name them.
Answer:
India encloses 4 biodiversity hotspots. They are

1. Himalayan
2. Indo-Burma
3. Western ghats
4. Sundalands

Part – III

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

Question 15.
Define surrogacy.
Answer:
Surrogacy is a method of assisted reproduction or agreement whereby a woman agrees to carry a pregnancy for another person, who will become the newborn child’s parent after birth. Through In Vitro Fertilization (IVF), embryos are created in a lab and are transferred into the surrogate mother’s uterus.

Question 16.
Differentiate between divergent evolution and convergent evolution with one example for each.
Answer:

Divergent Evolution	Convergent Evolution
Divergent evolution is a result of homology. eg: The wings of bird and the forelimbs of human both are homologous structures modified according to functions. In birds, it is used for flight and in humans used for writing and other purposes.	Convergent evolution is a result of analogy. eg: Root modification in sweet potato, and stem modification in potato are analogous structures both performing same function i.e., storage.

Question 17.
Under which conditions does a bacterium develops resistance against antibiotics?
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 18.
Expand (i) CFC (ii)AQI (w) PAN
Answer:
(i) CFC – Chloro fluro carbon
(ii) AQI – Air Quality Index
(iii) PAN – Peroxyacetyl nitrate .

Question 19.
Mention the number of primers required in each cycle of PCR. Write the role of primers arid DNA polymerase in PCR. Name the source organism of DNA polymerase used in PCR.
Answer:

• For each cycle of PCR two primers are required.
• Primers are the small fragments of single stranded DNA or RNA which serves as template for initiating DNA polymerization.
• DNA polymerase is an enzyme that synthesize DNA molecules by pairing the Deoxyribo Nucleotides leading to formation of new strands.
• DNA polymerase used in PCR is Taq polymerase which is isolated from a thermophilic bacteria called Thermus aquatics. Taq polymerase will remain active ever at very high temperature (80°C) and hence used in PCR amplification technique.

Part – IV

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

Question 20.
(a) Menstruation – a cyclic event occurring in every normal woman throughout her fertile period. Name the various phases of the menstruation and explain it.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases:

1. Menstrual phase
2. Follicular or proliferative phase
3. Ovulatory phase
4. Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation. These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as secretory phase. During pregnancy all events of menstrual cycle stop and there is no menstruation.

In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle.

[OR]

(b) Deoxy Ribo Nucleic Acid the life thread which acts as a genetic material for majority of living organism. Enlist the properties of DNA that makes it an ideal genetic material.
Answer:
(1) Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

(2) Stability: It should be stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided.

Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

(3) Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

(4) Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 21.
(a) Explain the structure of Immunoglobulin molecule with a suitable diagram.
Answer:
An antibody molecule is Y shaped structure that comprises of four polypeptide chains, two identical light chains (L) of molecular weight 25,000 Da (approximately 214 amino acids) and two identical heavy chains (H) of molecular weight 50,000 Da (approximately 450 amino acids). The polypeptide chains are linked together by di-sulphide (S-S) bonds. One light chain is attached to each heavy chain and two heavy chains are attached to each other to form a Y shaped structure. Hence, an antibody is represented by H2 L2. The heavy chains have a flexible hinge region at their approximate middles.

Each chain (L and H) has two terminals. They are C – terminal (Carboxyl) and amino or N-terminal. Each chain (L and H) has two regions. They have variable (V) region at one end and a much larger constant (C) region at the other end.

Antibodies responding to different antigens have very different (V) regions but their (C) regions are the same in all antibodies. In each arm of the monomer antibody, the (V) regions of the heavy and light chains combines to form an antigen – binding site shaped to ‘fit’ a specific antigenic determinant.

Consequently each antibody monomer has two such antigen – binding regions. The (C) regions that forms the stem of the antibody monomer determine the antibody class and serve common functions in all antibodies.

[OR]

(b) List out the uses of Transgenesis.
Answer:
(1) Transgenesis is a powerful tool to study gene expression and developmental processes in
higher organisms.

(2) Transgenesis helps in the improvement of genetic characters in animals. Transgenic animals serve as good models for understanding human diseases which help in the investigation of new treatments for diseases. Transgenic models exist for many human diseases such as cancer, Alzheimer’s, cystic fibrosis, rheumatoid arthritis and sickle cell anemia.

(3) Transgenic animals are used to produce proteins which are important for medical and pharmaceutical applications.

(4) Transgenic mice are used for testing the safety of vaccines.

(5) Transgenic animals are used for testing toxicity in animals that carry genes which make them sensitive to toxic substances than non-transgenic animals exposed to toxic substances and their effects are studied.

(6) Transgenesis is important for improving the quality and quantity of milk, meat, eggs and wool production in addition to testing drug resistance.

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 and score more marks in your examinations.

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

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III, and IV are to be attempted separately
4. Question numbers 1 to 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-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part-I

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

Question 1.
The metal oxide which cannot be reduced to metal by carbon is ………………….
(a) PbO
(b) Al₂O₃
(C) ZnO
(d) FeO
Answer:
(b) Al₂O₃

Question 2.
Compounds used as an eye lotion .
(a) H₃BO₃
(b) HBO₂
(C) H₂B₄O₇
(d) B₂O₃
Answer:
(a) H₃BO₃

Question 3.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) Br ₂> I₂ > F₂ > Cl₂
(b) F₂ > Cl₂ > Br₂ > I₂
(c) I₂ > Br₂ > Cl₂ > F₂
(d) Cl₂ > Br₂ > F₂ > I₂
Answer:
(d) Cl₂ > Br₂ > F₂ > I₂

Question 4.
Consider the following statements.
(i) The melting point decreases from Scandium to Vanadium in 3d series.
(ii) In 3d transition series, atomic radius decreases from Sc to V and upto copper atomic radius nearly remains the same.
(iii) As we move down in 3d transition series, atomic radius increases.
Which of the above statements is/are incorrect?
(a) i only
(b) ii only
(c) iii only
(d) i, ii and iii
Answer:
(a) i only

Question 5.
Which complex is used as an antitumor drug in cancer treatment?
(a) Ca – EDTA chelate
(b) EDTA
(c) Ti Cl₄ + Al(C₂H₅)₃
(d) Cis – Platin
Answer:
(b) EDTA

Question 6.
The number of unit cells in 8 gm of an element X ( atomic mass 40) which crystallizes in bcc
pattern is (N_A is the Avogadro number) ………..
(a) 6.023 x 10²³
(b) 6.023 x 10²²
(c) 60.23 x 10²³
(d) \(\left(\frac{6.023 \times 10^{23}}{8 \times 40}\right)\)

Hint: In bcc unit cell, 2 atoms = 1 unit cell
Number of atoms in 8g of element is Number of moles = \(\frac{8 g}{40 g, \mathrm{mol}^{-1}}\) = 0.2mol
1 mole contains 6.023 x 10²³ atoms 40g,mol
0.2 mole contains 0.2 x 6.023 x 10²³ atoms
\(\left(\frac{1 \text { unit cell }}{2 \text { atoms }}\right)\) x 0.2 x 6.023 xio23 = 6.023 x 10²² unit cells

Question 7.
Among the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is …………
Answer:

Solution:
k = Ae^-(Ea/RT)
lnk = \(\ln \mathrm{A}-\left(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\right)\left(\frac{1}{\mathrm{T}}\right)\)
this equation is in the form of a straight line equation y = c + m x
a plot of lnk vs (1/T) is a straight line with negative slope.

Question 8.
4Na+O₂ → 2Na₂O
Na₂O + H₂O → 2NaOH
In the given reaction, the oxide of sodium is …………..
(a) Acidic
(b) Basic
(c) Amphoteric
(d) Neutral
Answer:
(b) Basic
Solution:
Na₂O form NaOH so that it is basic oxide.

Question 9.
How many faradays of electricity are required for the following reaction to occur
MnO₄^– → Mn²⁺
(a) 5F
(b)3F
(e) 1F
(d) 7F
Answer:
(a) 5F
Hint: 7MnO₄^– + 5e^– → Mn²⁺4H2O
5 moles of electrons i.e., 5F charge is required.

Question 10.
Match the following:

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

Question 11.
Which of the following is not the product of dehydration of


Answer:

Question 12.
Which one of the following reaction is an example of disproportionation reaction?
(a) Aldol condensation
(b) cannizaro reaction
(c) Benzoin condensation
(d) none of these
Answer:
(b) cannizaro reaction

Question 13.
The correct order of basic strength in the case of substituted ethyl amines is ………

Answer:

Question 14.
Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence
of little hydrochloric acid, it becomes levorotatory. –
Reason: Sucrose hydrolysis gives unequal amounts of glucose and fructose. As a result of this – change in sign of rotation is observed.
(a) If both accretion and reason are true and reason is the correct explanation of assertion
(b) If both assertion and reason are true but reason is not the correct explanation of assertion
(c) If assertion is true but reason is false
(d) if both assertion and reason are false
Answer:
(a) If both accretion and reason are true and reason is the correct explanation of assertion

Question 15.
The correct structure of the drug paracetamol is ……….
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium – 9
Answer:

Part – II

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

Question 16.
What are the various steps involved in extraction of pure metals from their ores?
Answer:
The extraction of pure metals from the concentrated ores is carried out in two steps:

• Conversion of the ore into oxides of the metal of interest.
• Reduction of the metal oxides to elemental metals.

Question 17.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphoms is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphoms.

Question 18.
Give an example of coordination compound used in medicine.
Answer:
Medical uses of coordination compounds : –

• Ca-EDTA chelate, is used in the treatment of lead and radioactive poisoning. That is for removing
  lead and radiactive metal ions from the body.
• Cis-platin is used as an antitumor drug in cancer treatment.

Question 19.
Give the examples for a zero order reaction.
Answer:
Examples for a zero order reaction:
(i) Photochemical reaction between H₂ and Cl₂

(ii) Decomposition of N20 on hot platinum surface
N₂ (g) ⇌ N₂ + 1/2O₂(g)

(iii) Iodination of acetone in acid medium is zero order with respect to iodine.

Question 20.
Write the overall redox reaction which takes place in the galvanic cell,
Pt(s) | Fe²⁺(aq),Fe³⁺(aq) | MnO^–₄(aq), H+(aq),Mn²⁺ (aq) | Pt(s)
Answer:
At Anode half cell : 5Fe²⁺_(aq) → 5Fe³⁺_(aq) + 5e^–
At cathode half cell : MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H²O(l)
Overall redox reaction : 5Fe²⁺_(aq) + MnO^–_4(aq) + 8H⁺_(aq)→ 5Fe³⁺_(aq) + Mn²⁺_(aq) + 4H²O_(l)

Question 21.
In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found that all solutions containing more than 6.6 mL of AB coagulate with in 5 minutes. What is the flocculation values of AB for sol (X)?
Answer:
A minimum of 6.6mL of AB is required to coagulate the sol.
The moles of AB in the sol is, \(\frac{6.6 \times 0.01}{20}\) = 0.0033 moles
This means that a minimum of 0.033 moles or 0.0033 x 1000 = 3.3 milli moles are required
for coagulating one litre of sol.
Flocculation value of AB for X = 3.3

Question 22.
Explain Rosenmund reduction.
Answer:
Aldehydes can be prepared by the hydrogenation of acid chloride, in the presence of palladium supported by Barium sulphate. This reaction is called Rosenmund reduction.

In the above reaction, BaS04 act as a catalytic poison to palladium catalyst, so that aldehyde cannot be further reduced to alcohol.

Question 23.
‘What are the uses of aliphatic nitro compounds.
Answer:
(i) Nitromethane is used as a fuel for cars.
(ii) Chloropicrin (CCl₃NO₂) is used as an insecticide
(iii) Nitroethane is used as a fuel additive and precursor to explosive and they are good solvents for polymers, cellulose ester, synthetic rubber and dyes etc.

Question 24.
Classify the following as linear, branched or cross linked polymers
(a) Bakelite
(b) Nylon
(c) polythene
Answer:
(a) Bakelite – cross linked polymer
(b) Nylon – Linear polymer
(c) Polythene – Linear polymer

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]

Question 25.
A hydride of 2^nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). identify A, B and C.
Answer:
(i) A hydride of 2nd period alkali metal (A) is lithium hydride (LiH).
(ii) Lithium hydride (A) reacts with diborane (B) to give lithiumborohydride (C) which act as reducing agent.

Question 26.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced . container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.

Question 27.
Calculate the percentage efficiency of packing in case of body centered cubic crystal.
Answer:
Packing efficiency : In body centered cubic arrangement the spheres are touching along the leading diagonal of the cube as shown in the figure.
In ∆ABC,
AC² = AB² + BC²
In∆ACG,
AG² = AC² + CG²

∴ Volume of the sphere with radius ‘r’


Number of spheres belong to a unit cell in bec arrangement is equal to two and hence the total volume of all spheres.

i. e., 68% of the available volume is occupied.The available space is used more efficiently than in simple cubic packing.

Question 28.
H₃BO₃ accepts hydroxide ion from water as shown below
H₃BO₃ (aq) + H₂O((l) ⇌ B(OH)₄^– + H⁺
Predict the nature of H₃BO₃ using Lewis concept.
Answer:
Boric acid is also called as hydrogen borate or orthoboric acid. It is a weak mono basic Lewis acid of boron and it is written as B(OH)₃ It accepts hydroxyl (OH^–) ion from water. It does not dissociate to give hydronium (H₃O⁺) ion rather forms metaborate ion and this ions in turn give H₃O⁺ ion.
B(OH)₃ + H₂O ⇌ [B(OH)₄]^– + H₃O⁺
Hence it is considered as weak acid.

Question 29.
Explain graphical representation of chemical adsorption and physical adsorption.
Answer:
(i) Adsorption isotherms represents the variation of adsorption at constant temperature.
(ii) When amount of adsorption is plotted versus temperature at constant pressure is called adsorption isobar

(iii) In physical adsorption x/m decreases with increase in T. But in chemical adsorption x/m increases with rise in temperature and then decreases. The increase illustrate the requirement of activation of the surface for adsorption is due to the fact that formation of activated complex require certain energy. The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases.

Question 30.
Mention the uses of Glycerol.
Answer:

• Glycerol is used as a sweetening agent in confectionery and beverages.
• It is used in the manufacture of cosmetics and transparent soaps.
• It is used in making printing inks and stamp pad ink and lubricant for watches and clocks.
• It is used in the manufacture of explosive like dynamite and cordite by mixing it with china clay.

Question 31.
Give the differences between primary and secondary structure of proteins.
Answer:
Primary structure of proteins:

• Linear sequence of amino acids
• Composed of peptide bonds formed between amino acids.
• Formed during translation.
• Involved in post – translational modifications.

Secondary structure of proteins :

• Folding of the peptide chain into an a-helix and p-sheet.
• Encompasses hydrogen bonds
• Forms collagen, elastin action, myosin, and keratin – like fibres.
• Involved in forming structures such as cartilages, ligaments, skins etc.

Question 32.
Draw the structure of (i) procaine (ii) Lidocaine
Answer:
(i) Procaine

(ii) Lidocaine

Question 33.
Explain the variation in E⁰_M3+/M2+ 3d series.
Answer:
1. In transition series, as we move down from Ti to Zn, the standard reduction potential \(\mathbf{E}^{0} \mathrm{M}^{2+} / \mathrm{M}^{3+}\) value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺ .

2. \(\mathrm{E}^{0} \mathrm{m}^{2+} / \mathrm{M}\) 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⁺.

3. The standard electrode potential for the M³⁺ / M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

4. The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.

5. For Fe³⁺ /Fe²⁺ the reductionpotential is 0.77V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.

6. 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 Mn very feasible [E⁰ = +1.51 V]

Part – IV

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

Question 34.
(a) (i) Describe a method for refining nickel. (3)
(ii) Why group 18 elements are called inert gases?-Write the general electronic configuration of group 18 elements. (2)
[OR]
(b) (i) Give the uses of helium. (3)
(ii) Ni (II) compounds are more stable than Pt (II) compounds. Give reason. (2)
Answer:
(a) (i) The impure nickel is heated in a stream of carbon monoxide at around 350 K. The nickel
reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities ‘ . are left behind.
Ni (s) + 4 CO (g) → Ni(CO)₄ (g)
On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal
Ni(CO)₄(g) → Ni(s) + 4 CO (g)

(ii) The elements of group-18 have completely filled s and p orbitals, hence they are more stable and have least reactivity. Therefore group-18 elements are called inert gases. ns2np6 is the general electronic configuration of group 18 elements.

[OR]

(b) (i) Uses of helium.

• Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
• Helium is used to provide inert atmosphere in electric arc welding of metals
• Helium has lowest boiling point hence used in cryogenics (low temperature science).
• It is much less denser than air and hence used for filling air balloons

(ii) The ionisation enthalpy values can be used to predict the thermodynamic stability of their compounds.
For Nickel, I.E₁ + I.E₂ = 737 + 1753
= 2490 kJ mol-1
For Platinum, I.E₁ +I.E₂ = 864+ 1791
= 2655 kJ mol-1
Since, the energy required to form Ni²⁺ is less than that of Pt²⁺ , Ni(II) compounds are thermodynamically more stable than Pt(II) compounds.

Question 35.
(a) (i) Arrange the following in order of increasing molar conductivity (3)
1. Mg[Cr(NH₃)(Cl)₅] . 2.[Cr(NH₃)₅Cl]₃ [COF₆]₂ 3. [Cr(NH₃)₃Cl₃]
(ii) Silicon carbide is very hard. Justify this statement. (2)
[OR]
(b) (i) Write the rate law for the following reactions: (a) A reaction that is 3/2 order in x and zero order in y. (A) A reaction that is second order in NO and first order in Br₂. (3)
(ii) Classify the following species into Lewis acids and Lewis bases and show how these can act as Lewis acid / Lewis base? (2)
(a) OH^– ions (A) F^– (c) H⁺ (d) BCl₃
Answer:
(a) (i) These complexes can ionise in solution as:
Mg[Cr(NH₃)(Cl)₅] = Mg²⁺ + [Cr(NH₃)(Cl)₅]^2-
[Cr(NH₃ )₅Cl]₃ [COF₆]₂ = [Cr(NH₃)₅₆Cl]²⁺ + [COF₆]^3-
[Cr(NH₃)₃Cl₃] = does not ionize

As the number of ions in solution increases, their molar conductivity also increases.
Therefore, conductivity follows the order:
[Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl]₃ [COF < Mg[Cr(NH₃)(Cl)₅]

(ii) Silicon carbide is very hard. It is a covalent solid contains the atoms which are bound together in a three dimensional network entirely by covalent bonds. So the covalent network crystal siC is very hard and have high melting point.

[OR]

(b) (i)

(ii) (a) OH^– ions can donate an electron pair and act as Lewis base.
(b) F^– ions can donate an electron pair and act as Lewis base.
(c) H⁺ ions can accept an electron pair and act as Lewis base. .
(d) BCl₃ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.

Question 36.
(a) (z) Calculate pH of 10^-7 MHCl. (3)
(ii) Define corrosion. Give one example. (2)
[OR]
(b) What is adsorption isotherm? Explain about Freundlich adsorption isotherm. (5)
Answer:
(a) (z) If we do not consider [H₃O⁺ ] from the ionisation of H₂O, then [H₃O⁺ ] = [HCl] = 10^-7 M i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7 M). Hence, the H₃O⁺ (10^-7 M) formed due to the auto ionisation of water cannot be neglected.
So, in this case we should consider [H₃O⁺ ] from ionisation of H₂ O
[H₃O⁺ ] = 10^-7 (from HCl) + 10^-7 (from water)
= 10^-7 (1+1)
= 2 x 10^-7
pH = -log₁₀[H₃O⁺]
= log₁₀ (2 x 10^-7 )=—[log 2 + log10^-7 ]
= – log 2 -(-7). 1og₁₀¹⁰
= 7 – log 2
= 7 – 0.3010 = 6.6990
= 6.70

(ii) The redox process which causes the deterioration of metal is called corrosion.
Rusting of iron is an example of corrosion. It is an electro chemical process.

[OR]

(b) i. Adsorption isotherms represents the variation of adsorption at constant temperature. Adsorption isotherm can be studied quantitatively. ii. A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorbate at constant temperature is called adsorption isotherms.

ii. Freundlich adsorption isotherm.
According to Freundlich \(\frac{\mathbf{x}}{\mathrm{m}}=\mathrm{k} \mathrm{P}^{1 / \mathrm{n}}\)
Where x is the amount of adsorbate (or) adsorbed on ‘m’ gm of adsorbent at a pressure of R k and n are constants. Value of ‘n’ is always less than unity.

iv. This equation is applicable for adsorption of gases on solid surfaces. The Same equation becomes \(\frac{\mathbf{x}}{\mathrm{m}}=\mathrm{k} \mathrm{P}^{1 / \mathrm{n}}\) kc when used for adsorption in solutions with ‘c’ as concentration.

v. These equation quantitatively predict the effect of pressure (or concentration) on the adsorption of gases (or adsorbates) at constant temperature.

vi. Taking log on both sides of equation

vii. Hence the intercept represents the value of log k and the slope \(\frac{b}{q}\) gives \(\frac{1}{n}\)

viii. This equation explains the increase of \(\frac{x}{m}\)with increase in pressure. But experimental values shows the deviation at low pressure.

ix. Limitations:
(a) This equation is purely empirical and valid over a limited pressure range.
(b) The value of k and n also found vary with temperatures. No theoretical explanations were given.

Question 37.
(a) An organic compound (A) of molecular formula C₆H₆O gives white precipitate with bromine water. (A) on reaction with NaOH gives (B). (B) reacts with methyl iodide in presence of dry ether gives (C) of molecular formula C₇H_/8O which will not liberate H2 gas with metallic Na. (C) on reaction with acetyl chloride gives (D) and (E) of formula which are position isomers. Identify A, B, C, D & E and explain the reaction. (5)

[OR]

(A) (i) What happens when n-propyl benzene is oxidised using H⁺ / KMnO₄?(2)
(ii) Identify A, B ,and C.

Answer:
1. An organic compound gives white precipitate with bromine water means it must be a phenol. From the molecular formula it is identified as C₆H₅OH.

2. Phenol on reaction with NaOH gives (B) as sodium phenoxide C₆H₅ONa.

3. Sodium phenoxide on reaction with methyl iodide in the pressure of dry ether undergo Williamsons synthesis and gives Anisole as (C).

4. Anisole on reaction with acetyl chloride undergoes Friedel Craft’s acetylation and yield o-methoxy acetophenone and p-methoxy acetophenone as (D) and (E).


(b)

Question 38.
(a) (i) How will you distinguish between primary secondary and tertiary alphatic amines. (3)
(ii) Convert Benzene diazonium chloride into phenol. (2)
[OR]
(b) (i) What are the functions of lipids in living organism? (3)
(ii) What is Orion? Give its preparation and use. (2)
Answer:
(a) (i)

Students can Download Tamil Nadu 12th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

General Instructions:

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.
In majority of plants, pollen is liberated at _______.
(a) 1 celled stage
(b) 2 celled stage
(c) 3 celled stage
(d) 4 celled stage
Answer:
(b) 2 celled stage

Question 2.
Which one of the following is an example for polygenic inheritance?
(a) Flower color in Mirabilis jalapa
(b) Pod shape in garden pea
(c) Production of male honey bee
(d) Skin color in humans
Answer:
(d) Skin color in humans

Question 3.
Assertion (A): Complete linkage is noticed in male species of Drosophila.
Reason (R): Completely linked genes show some crossing over.
(a) A is true, R is false
(b) Both A and R are false
(c) A is true, R is not correct explanation of A
(d) R explains A
Answer:
(a) A is true, R is false

Question 4.
Virus free germ plants are developed from _______.
(a) Organ culture
(b) Meristem culture
(c) Protoplast culture
(d) Cell suspension culture
Answer:
(b) Meristem culture

Question 5.
The unit of measuring ozone thickness is ______.
(a) Joule
(b) Kilos
(c) Dobson
(d) Watt
Answer:
(c) Dobson

Question 6.
If 1200 Joules of solar energy is trapped by producers, how much of joules of energy does the organism in the third tropic level will receive?
(a) 120 joules
(b) 12 joules
(c) 1.2 joules
(d) 0.12 joules
Answer:
(c) 1.2 joules

Question 7.
Dwarfing gene of wheat is ______.
(a) Pal 1
(b) Atomita 1
(c) Norin 10
(d) Pelita 2
Answer:
(c) Norin 10

Question 8.
Which one of the following match is correct.
(a) Palmyra – Native of Brazil
(b) Saccharum – Abundant in Kanyakumari
(c) Steveocide – Natural sweetner
(d) Palmyra sap – fermented to give ethanol
Answer:
(c) Steveocide – Natural sweetner

Part – II

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

Question 9.
Write the origin and area of cultivation of green gram and red gram.
Answer:

	Origin	Cultivation
Green gram.	Maharashtra (India).	Madhya Pradesh, Karnataka and Tamil Nadu.
Red gram.	South India.	Maharashtra, Andhra Pradesh, Madhya Pradesh, Karnataka and Gujarat.

Question 10.
State the law of independent assortment.
Answer:
When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent to the other pair of characters. Genes that are located in different chromosomes assort independently during meiosis.

Question 11.
Expand (i) PEG (ii) PHB
Answer:
(i) PEG – Poly Ethylene Glycol
(ii) PHB – Poly Hydroxy Butyrate

Question 12.
Based on the materials used, how will you classify the culture technology. Explain.
Answer:
Based on explants used culture technology are of following types:
Organ culture – Embryos, anthers, root and shoot part are used.
Meristem culture – Meristematic tissues are used.
Protoplast culture – Protoplasts are used.
Cell culture – Single cells or aggregate of cells from callus are used.

Question 13.
Give four examples of plants cultivated in commercial agroforestry.
Answer:
Casuarina, Eucalyptus, Teak, Malaivembu

Question 14.
How does an orchid ophrys ensures its pollination by bees?
Ahnswer:
The plant, Ophrys an orchid, the flower looks like a female insect to attract the male insect to get pollinated by the male insect and it is otherwise called ‘floral mimicry’.

Part – III

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

Question 15.
List out the objectives of plant breeding.
Answer:

• To increase yield, vigour and fertility of the crop.
• To increase tolerance to environmental condition, salinity, temperature and drought.
• To prevent the premature falling of buds and fruits, etc.
• To improve synchronous maturity.
• To develop resistance to pathogens and pests.
• To develop photosensitive and thermos-sensitive varieties.

Question 16.
Spindle shaped pyramid of number is noticed in forest ecosystem. Give Reasons.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T₁) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T₂) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T₄), tertiary consumers (lion) are lesser in number than the secondary consumer (T₃) (fox and snake).

Question 17.
Distinguish between mound layering and air layering.
Answer:

Mound Layering	Air Layering
In mound layering, lower flexible branch with leaves is bent to ground and a part of the stem is buried in the soil and the tip of branch is exposed above the soil. After the roots emerge from the buried stem, a cut is made in parent plant so that the buried plant grows into a new plant.	In air layering, the stem is girdled at nodal part and hormones are applied and covered with moist soil using polythene sheet. Roots emerge in these branches after 2-4 months. Such branches are removed from parent plant and grown separately.

Question 18.
Explain the sex determination mechanism in Carica papaya.
Answer:

Carica papaya, 2n = 36 (Papaya) has 17 pairs of autosomes and one pair of sex chromosomes. Male papaya plants have XY and female plants have XX. Unlike human sex chromosomes, papaya sex chromosomes look like autosomes and it is evolved from autosome. The sex chromosomes are functionally distinct because the Y chromosome carries the genes for male organ development and X bears the female organ developmental genes. In papaya sex determination is controlled by three alleles. They are m, M₁ and M₂ of a single gene.

Question 19.
Write the protocol for artificial seed preparation.
Answer:

Later these seeds are grown in vitro medium and converted into piantiets. These piantiets require a hardening period (either green house or hardening chamber) and then shifted to normal environment condition.

Part – IV

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

Question 20.
(a) Bring out the inheritance of chloroplast gene with on example.
Answer:
Chloroplast Inheritance:
It is found in 4 O’ Clock plant (Mirabilis jalapa). In this, there are two types of variegated leaves namely dark green leaved plants and pale green leaved plants. When the pollen of dark green leaved plant (male) is transferred to the stigma of pale green leaved plant (female) and pollen of pale green leaved plant is transferred to the stigma of dark green leaved plant, the F, generation of both the crosses must be identical as per Mendelian inheritance.
Chloroplast inheritance.

But in the reciprocal cross the F₁ plant differs from each other. In each cross, the F₁ plant reveals the character of the plant which is used as female plant.

This inheritance is not through nuclear gene. It is due to the chloroplast gene found in the ovum of the female plant which contributes the cytoplasm during fertilization since the male gamete contribute only the nucleus but not cytoplasm.

[OR]

(b) Explain in detail about various types of direct gene transfer method.
Answer:

(1) Chemical mediated gene transfer:
Certain chemicals like polyethylene glycol (PEG) and dextran sulphate induce DNA uptake into plant protoplasts.

(2) Microinjection: The DNA is directly injected into the nucleus using Electroporation Methods of Gene Transfer fine tipped glass needle or micro pipette to transform plant cells. The protoplasts are immobilised on a solid support (agarose on a microscopic slide) or held with a holding pipette under suction.

(3) Electroporation Methods of Gene Transfer: A pulse of high voltage is applied to protoplasts, cells or tissues which makes transient pores in the plasma membrane through which uptake of foreign DNA occurs.

(4) Liposome mediated method of Gene Transfer: Liposomes the artificial phospholipid vesicles are useful in gene transfer. The gene or DNA is transferred from liposome into vacuole of plant cells. It is carried out by encapsulated DNA into the vacuole. This technique is advantageous because the liposome protects the introduced DNA from being damaged by the acidic pH and protease enzymes present in the vacuole. Liposome and tonoplast of vacuole fusion resulted in gene transfer. This process is called lipofection.

(5) Biolistics: The foreign DNA is coated onto the surface of minute gold or tungsten particles (1-3 mm) and bombarded onto the target tissue or cells using a particle gun (also called as gene gun/micro projectile gun/shotgun). Then the bombarded cells or tissues are cultured on selected medium to regenerate plants from the transformed cells.

Question 21.
(a) Explain the steps involved in protoplast culture.
Answer:
Protoplasts are cells without a cell wall, but bounded by a cell membrane or plasma membrane. Using protoplasts, it is possible to regenerate whole plants from single cells and also develop somatic hybrids. The steps involved in protoplast culture are:

(1) Isolation of protoplast: Small bits of plant tissue like leaf tissue are used for isolation of protoplast. The leaf tissue is immersed in 0.5% Macrozyme and 2% Onozuka cellulase enzymes dissolved in 13% sorbitol or mannitol at pH 5.4. It is then incubated over-night at 25°C. After a gentle teasing of cells, protoplasts are obtained, and these are then transferred to 20% sucrose solution to retain their viability. They are then centrifuged to get pure protoplasts as different from debris of cell walls.

(2) Fusion of protoplast: It is done through the use of a suitable fusogen. This is normally PEG (Polyethylene Glycol). The isolated protoplast are incubated in 25 to 30% concentration of PEG with Ca++ ions and the protoplast shows agglutination (the formation of clumps of cells) and fusion.

(3) Culture of protoplast: MS liquid medium is used with some modification in droplet, plating or micro-drop array techniques. Protoplast viability is tested with fluorescein diacetate before the culture. The cultures are incubated in continuous light 1000-2000 lux at 25 °C. The cell wall formation occurs within 24-48 hours and the first division of new cells occurs between 2-7 days of culture.

(4) Selection of somatic hybrid cells: The fusion product of protoplasts without nucleus of different cells is called a hybrid. Following this nuclear fusion happen. This process is called somatic hybridization.

[OR]

(b) Write a note on Henna.
Answer:
Botanical name: Lawsonia inermis.
Family: Lythraceae.
Origin and Area of cultivation: It is indigenous to North Africa and South-west Asia. It is grown mostly throughout India, especially in Gujarat, Madhya Pradesh and Rajasthan.

Uses:
An orange dye ‘Henna’ is obtained from the leaves and young shoots of Lawsonia inermis. The principal colouring matter of leaves ‘lacosone’ is harmless and causes no irritation to the skin. This dye has long been used to dye skin, hair and finger nails. It is used for colouring leather, for the tails of horses and in hair-dyes.

Bio-Zoology [Maximum Marks: 35]

Part – I

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

Question 1.
Animals giving birth to young ones are ______.
(a) Oviparous
(b) Ovoviviparous
(c) Viviparous
(d) Both a and b
Answer:
(c) Viviparous

Question 2.
Messelson and Stahl’s experiment proved ________.
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA reflication.
Answer:
(d) Semi-conservative nature of DNA reflication.

Question 3.
Match column I with column II

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

Question 4.
Choose the correctly matched pair.
(a) Amphetamines – Stimulant
(b) LSD – Narcotic
(c) Heroin – Psychotropic
(d) Benzodiazepine – Pain killer
Answer:
(a) Amphetamines – Stimulant

Question 5.
The first clinical gene therapy was done for the treatment of ________.
(a) AIDS
(b) Cancer
(c) Cystic fibrosis
(d) SCID
Answer:
(d) SCID

Question 6.
Some organisms are able to maintain homeostasis by physical means ________.
(a) Conform
(b) Regulate
(c) Migrate
(d) Suspend
Answer:
(b) Regulate

Question 7.
Select the correct linear equation describing the species area relationship?
(a) log C = log S + Z log A
(b) Z log A = log S + log C
(c) log S = log C + Z log A
(d) log C = log S ± Z log C
Answer:
(b) Z log A = log S + log C

Question 8.
Oil strains in laundry can be removed using _______.
(a) Peptidane
(b) Protease
(c) Amylase
(d) Lipase
Answer:
(d) Lipase

Part – II

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

Question 9.
The unicellular organisms which reproduce by binary fission are considered immortal. Justify.
Answer:
In unicellular organisms during binary fission, the entire cell (organism) divides completely to form two daughter cells which later on develop into adult and the process goes on repeatedly during each division leading to immortality of cell (organism). Hence unicellular organisms like amoeba are ‘biologically immortal’.

Question 10.
What is colostrum? Write its significance.
Answer:
The mammary glands secrete a yellowish fluid called colostrum during the initial few days after parturition. It has less lactose than milk and almost no fat, but it contains more proteins, vitamin A and minerals. Colostrum is also rich in IgA antibodies. This helps to protect the infant’s digestive tract against bacterial infection.

Question 11.
Define haplodiploidy.
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 12.
Mention the main objections to Darwinism.
Answer:
Some objections raised against Darwinism were –

• Darwin failed to explain the mechanism of variation.
• Darwinism explains the survival of the fittest but not the arrival of the fittest.
• He focused on small fluctuating variations that are mostly non-heritable.
• He did not distinguish between somatic and germinal variations.
• He could not explain the occurrence of vestigial organs, over specialization of some organs like large tusks in extinct mammoths and over sized antlers in the extinct Irish deer, etc.

Question 13.
Differentiate between cell mediated Immunity and Antibody Mediated Immunity.
Answer:

Cell Mediated Immunity (CMI)	Antibody Mediated Immunity (AMI)
1. In CMI, pathogens are destroyed by cells without producing antibodies.	1. In AMI, pathogens are destroyed by antibodies.
2. It is carried out by T cells, Macrophages, NK cells	2. It is carried out by B cells, T helper cells, APC cells.

Question 14.
What are attenuated recombinant vaccines?
Answer:
Attenuated recombinant vaccines includes genetically modified pathogenic organisms (bacteria or viruses) that are made nonpathogenic and are used as vaccines. Such vaccines are referred to as attenuated recombinant vaccines.

Part – III

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

Question 15.
Alien species invasion is a threat to endemic species – substantiate this statement.
Answer:
Exotic species are organisms often introduced unintentionally or deliberately for commercial purpose, as biological control agents and other uses. They often become invasive and drive away the local species and is considered as the second major cause for extinction of species. Tilapia fish (Jilabi kendai) (Oreochromis mosambicus) introduced from east coast of South Africa in 1952 for its high productivity into Kerala’s inland waters, became invasive, due to which the native species such as Puntius dubius and Labeo kontius face local extinction.

Amazon sailfin catfish is responsible for destroying the fish population in the wetlands of Kolkata. The introduction of the Nile Perch, a predatory fish into Lake Victoria in East Africa led to the extinction of an ecologically unique assemblage of more than 200 nature species of cichlid fish in the lake.

Question 16.
In what way Peyang conserved the forests?
Answer:
The ‘Forest man of India’, Jadav Payeng who created 1,360 acres of dense and defiant forest was born in Arunasapori (a river island on the Brahmaputra). He had just completed his Class X exams in 1979 when he started to sow the seeds and shoots on the eroded island covered with sand and silt. Thirty-six years later he had converted the once unproductive land into a forest.

Payeng’s forest is now home to five Royal Bengal tigers, over a hundred deer, wild boar, vultures, and several species of birds. For his remarkable initiative, the Jawaharlal Nehru University invited Payeng in 2012 on Earth Day and honoured him with the title of the ‘ Forest Man of India ’.

Question 17.
Amniocentesis, the foetal sex determination test, is banned in our country. Is it necessary comment?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 18.
Write the objectives of Human Genome project.
Answer:
The main goals of Human Genome Project are as follows:

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that makeup the human DNA
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 19.
Explain the role of cry-genes in genetically modified crops.
Answer:
Bacillus thuringiensis is a soil dwelling bacterium which is commonly used as a biopesticide and contains a toxin called cry toxin. Scientists have introduced this toxin producing genes into cotton and have raised genetically engineered insect resistant cotton plants.

During sporulation Bacillus thuringiensis produces crystal proteins called Delta-endotoxin which is encoded by cry genes. Delta-endotoxins have specific activities against the insects of the orders Lepidoptera, Diptera, Coleoptera and Hymenoptera. When the insects ingest the toxin crystals their alkaline digestive tract denatures the insoluble crystals making them soluble. The cry toxin then gets inserted into the gut cell membrance and paralyzes the digestive tract. The insect then stops eating and starves to death.

Part – IV

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

Question 20.
(a) Describe the structure of human spermatozoa with a labelled diagram.
Answer:

The human sperm is a microscopic, flagellated and motile gamete. The whole body of the sperm is enveloped by plasma membrane and is composed of a head, neck and a tail. The head comprises of two parts namely acrosome and nucleus. Acrosome is a small cap like pointed structure present at the tip of the nucleus and is formed mainly from the Golgi body of the spermatid.

It contains hyaluronidase, a proteolytic enzyme, popularly known as sperm lysin which helps to penetrate the ovum during fertilization. The nucleus is flat and oval. The neck is very short and is present between the head and the middle piece. It contains the proximal centriole towards the nucleus which plays a role in the first division of the zygote and the distal centriole gives rise to the axial filament of the sperm. The middle piece possesses mitochondria spirally twisted around the axial filament called mitochondrial spiral or nebenkem.

It produces energy in the form of ATP molecules for the movement of sperms. The tail is the longest part of the sperm and is slender and tapering. It is formed of a central axial filament or axoneme and an outer protoplasmic sheath. The lashing movements of the tail push the sperm forward.

[OR]

(b) Explain the Mechanism of ‘lac’ – operon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes-permease, P-galactosidase (β-gal) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, P-galactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylase transfers acetyl group from acetyl Co A to β-galactosidase.

The lac operon consists of one regulator gene (‘i’ gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for β-galactosidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase. Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli.

In lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory gene. When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, β-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation.

Question 21.
(a) Explain stabilizing, directional and disruptive selection with examples.
Answer:
(1) Stabilising selection (centipetal selection): This type of selection operates in a stable environment. The organisms with average phenotypes survive whereas the extreme individuals from both the ends are eliminated. There is no speciation but the phenotypic stability is maintained within the population over generation. For example, measurements of sparrows that survived the storm clustered around the mean, and the sparrows that failed to survive the storm clustered around the extremes of the variation showing stabilizing selection.

(2) Directional Selection: The environment which undergoes gradual change is subjected to directional selection. This type of selection removes the individuals from one end towards the other end of phenotypic distribution. For example, size differences between male and female sparrows. Both male and female look alike externally but differ in body weight. Females show directional selection in relation to body weight.

(3) Disruptive selection: (centrifugal selection) When homogenous environment changes into heterogenous environment this type of selection is operational. The organisms of both the extreme phenotypes are selected, whereas individuals with average phenotype are eliminated. This results in splitting of the population into sub population/species.

This is a rare form of selection but leads to formation of two or more different species. It is also called adaptive radiation. (e.g:) Darwin’s finches beak size in relation to seed size inhabiting Galapagos islands. Group selection and sexual selection are other types of selection. The two major group selections are Altrusim and Kin selection.

Operation of natural selection on different traits (a) Stablishing (b) Directional and (c) Disruptive

[OR]

(b) Tabulate the various types of innate immunity and their action mechanism.
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 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-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks : 70

PART -1

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

Question 1.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be …………………… .
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

Question 2.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is …………………. .
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 NC^-1
(d) 1000 NC^-1
Hint: E = \(\frac{V}{d}=\frac{10}{1 \times 10^{-2}}\) = 1000 NC^-1
Answer:
(d) 1000 NC^-1

Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ………………… .
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

Question 4.
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer:
(a) circle

Question 5.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is ………………….. .
(a) M
(b) 2M
(c) \(\frac{M}{2}\)
(d) Zero
Answer:
(c) \(\frac{M}{2}\)

Question 6.
The current flowing in a coil varies with time as shown in ® the figure. The variation of induced emf with time would be …………….. .


Answer:

Question 7.
Which of the following electromagnetic radiation is used for viewing objects through fog?
(a) microwave
(b) gamma rays
(c) X- rays
(d) infrared
Answer:
(d) infrared

Question 8.
Stars twinkle due to …………………. .
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction

Question 9.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Hint: Only the frequency of the electromagnetic wave remains unchanged.
Answer:
(b) Frequency

Question 10.
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would …………………… .
(a) increase by 2 times
(b) decrease by 2 times
(c) decrease by 4 times
(d) increase by 4 times
Hint: At Voltage, V = 14 kV
de-Broglie wavelength of electron, λ = \(\frac{12.3}{\sqrt{14000}}\) Å = 0.104 Å
At voltage, V = 224 kV
λ’ = \(\frac{12.3}{\sqrt{224000}}\) Å = 0.026 Å
\(\frac{\lambda}{\lambda^{\prime}}=\frac{0.104}{0.0260}\) = 4 ⇒ λ = 4λ’ ⇒ λ’ = \(\frac{\lambda}{4}\)
Answer:
(c) decrease by 4 times

Question 11.
The charge of cathode rays is …………………. .
(a) positive
(b) negative
(c) neutral
(d) not defined
Answer:
(a) positive
(b) negative

Question 12.
The primary use of a zener diode is …………………… .
(a) Rectifier
(b) Amplifier
(c) Oscillator
(d) Voltage regulator
Answer:
(d) Voltage regulator

Question 13.
For a common base circuit if \(\frac{I_{C}}{I_{E}}\) = 0.98, then current gain for common emitter circuit will be ………………. .
(a) 49
(b) 98
(c) 4.9
(d) 25.5
Ir a 0 98

Answer:
(a) 49

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

Question 15.
“Sky wax” is an application of nano product in the field of …………………….. .
(a) Medicine
(b) Textile
(c) Sports
(d) Automotive industry
Answer:
(c) Sports

PART – II

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

Question 16.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 17.
A potential difference across 24 ft resistor is 12 V. What is the current through the resistor?
Answer:
V = 12 V
and R = 24 Ω
Current, I = ?
From Ohm’s law, \(I=\frac{V}{R}=\frac{12}{24}=0.5 \mathrm{A}\)

Question 18.
State Coulomb’s inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between ‘ them.
\(\overrightarrow{\mathrm{F}} \propto \frac{q_{m_{\mathrm{A}}} q_{m_{b}}}{r^{2}} \hat{r}\)

Question 19.
State Lenz’s law.
Answer:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 20.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, d = 180 – (i + r). As, i = r in reflection, we can write angle of deviation in  reflection at plane surface as. d = 180 – 2i

Question 21.
What is photoelectric effect?
Answer:
The ejection of electrons from a metal plate when illuminated by light or any other electromagnetic radiation of suitable wavelength (or frequency) is called photoelectric effect.

Question 22.
Define impact parameter.
Answer:
The impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance.

Question 23.
In a transistor connected in the common base configuration, α = 0.95, I_E = mA. Calculate the values of I_c and I_B
Answer:
α = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{E}}}\)
I_c = αI_E = 0.95 × 1 = 0.95 mA
I_E = I_B + I_c
∴ I_B = I_c – I_E = 1 – 0.95 = 0.05 mA

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

PART-III

Answer any six questions. Question No. 28 is compulsory. [6 × 3 = 18]

Question 25.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₂ is given by
\(\overrightarrow{\mathrm{F}}_{12}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{21}\)
Here r̂₂₁ is the unit vector from charge q₂ to q₁
But r̂₂₁ = -r̂₂₁

Therefore, the electrostatic force obeys Newton’s third law.

Question 26.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
\(P=\frac{d U}{d t}=\frac{1}{d t}(V \cdot d Q)=V \cdot \frac{d Q}{d t}\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I²R
P = IV = \(\left(\frac{V}{R}\right) V=\frac{V^{2}}{R}\)

Question 27.
The repulsive force between two magnetic poles in air is 9 × 10^-3 N. If the two poles are equal in strength and are separated by a distance of 10 cm, calculate the pole strength of each pole.
Answer:
The force between two poles are given by \(\overrightarrow{\mathrm{F}}=k \frac{q_{m_{\mathrm{A}}} q_{m_{\mathrm{B}}}}{r^{2}}\)
The magnitude of the force is F = \(F=k \frac{q_{m_{A}} q_{m_{B}}}{r^{2}}\)
Given : F = 9 × 10^-3 N, r = 10 cm = 10 × 10^-2 m
Therefore 9 × 10^-3 = 10^-7 × \(\frac{q_{m}^{2}}{\left(10 \times 10^{-2}\right)^{2}}\) ⇒ q_m = 30NT^-1

Question 28.
A 200 turn coil of radius 2 cm is placed co-axially within a long solenoid of 3 cm radius. If the turn density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil.
Answer:
Numbef of turns of the solenoid, N₂ = 200
Radius of the solenoid, r cm = 2 × 10^-2 m
Area of the solenoid, A = πr² = 3.14 × (2 × 10^-2)^-2 = 1.256 × 10^-3 m²
Turn density of long solenoid per cm, N₁ = 90 × 10²

= 283956.48 × 10^-8 ⇒ M = 2.84 mH

Question 29.
Compute the speed of the electromagnetic wave in a medium if the amplitude of electric and magnetic fields are 3 × 10⁴ N C^-1 and 2 × 10^-4 T, respectively.
Answer:
The amplitude of the electric field, E₀ = 3 × 10⁴ N C^-1.
The amplitude of the magnetic field, B_o = 2 × 10^-4 T. Therefore, speed of the electromagnetic wave in a medium is
\(v=\frac{3 \times 10^{4}}{2 \times 10^{-4}}=1.5 \times 10^{8} \mathrm{ms}^{-1}\)

Question 30.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of refractive index of the second medium n₂ to that of the refractive index of the first medium n₁.
\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)

Question 31.
A proton and an electron have same de Broglie wavelength. Which of them moves faster and which possesses more kinetic energy?
Answer:
We know that λ = \(\frac{h}{\sqrt{2 m K}}\)
Since proton and electron have same de Broglie wavelength, we get

Since m_e < m_p , K_p < K_e, the electron has more kinetic energy than the proton.

Since m_e < m_p ν_p < ν_e , the electron moves faster than the proton.

Question 32.
In alpha decay, why the unstable nucleus emits ⁴₂He nucleus? Why it does not emit four separate nucleons?
Answer:
After all ⁴₂He consists of two protons and two neutrons. For example, if ²³⁸₉₂U nucleus decays into ²³⁴₉₀Th by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent (²³⁸₉₂U) nucleus. This kind of process cannot occur in nature because it would violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.

Question 33.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

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

PART – IV

Answer all the questions. [5 × 1 = 5]

Question 34.
(a) Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole: Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\)
Potential at P due to charge -q = \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\)
Total potential at the point P,
V = \(\frac{1}{4 \pi \varepsilon_{0}} q\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ……… (1)
Suppose if the point P is far away from the dipole, such that r>>a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

r²₁ = r² + a² – 2ra cos θ = r²\(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)

Since the point P is very far from dipole, then r >> a. As a result the term \(\frac{a^{2}}{r^{2}}\) is very small and can be neglected. Therefore

since \(\frac{a}{r}\)<< 1, we can use binominal theorem and retain the terms up to first order r
\(\frac{1}{r_{1}}=\frac{1}{r}\left(1+\frac{a}{r} \cos \theta\right)\) …… (2)
Similarly applying the cosine law for triangle AOP,
r₂² = r² + a² – 2ra cos(180 – θ) since cos (180 – θ) = -cos θ we get
r₂² = r² + a² + 2ra cos θ
Neglecting the term \(\frac{a^{2}}{r^{2}}\) (because r >> a)

Using Binomial theorem, we get
\(\frac{1}{r_{2}}=\frac{1}{r}\left(1-a \frac{\cos \theta}{r}\right)\) ………. (3)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec{r}\) . r̂ where r̂ is the unit vector from the point O to point P.
Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) …(4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.

Special cases:
Case (1) If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}\)

Case (ii) If the point Plies on the axial line of the dipole on the side of -q. then θ = 180°, then
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{2}}\)

Case (iii) If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 34.
(b) Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance: The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the ceíl. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is
V = IR …….. (1)

Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Then V = ξ – Ir
Ir = ξ – V ……… (2)
Dividing equation (2) by equation (1), we get
\(\frac{I r}{I R}=\frac{\xi-V}{V}\)
\(r=\left|\frac{\xi-V}{V}\right| R\) ……… (3)
since ξ, V and R are known, internal resistance r can be determined

Question 35.
(a) Calculate the magnetic induction at a point on the axial line of a bar magnet.
Answer:
Magnetic field at a point along the axial line of the magnetic dipole (bar magnet): Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength q, and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance from the geometrical center O of the bar magnet
can be computed by keeping unit north pole (q_MC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb’s law of magnetism as follows:
The force of repulsion between north pole of the bar magnet and unit north pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{(r-l)^{2}} \hat{l}\) …. (1)
where r – l is the distance between north pole of the bar magnet and unit north pole at C.
The force of attraction between South Pole of the bar magnet and unit North Pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{(r+l)^{2}} \hat{l}\) ……. (2)
where r + l is the distance between South pole of the bar magnet and unit north pole at C.

From equation (1) and (2), the net force at point C is \(\vec{F}\) = \(\vec{F}\)_N + \(\vec{F}\)_s . From definition, this net force is the magnetic field due to magnetic dipole at a point C (\(\vec{F}\) = \(\vec{B}\))

Since, magnitude of magnetic dipole moment is |\(\vec{P}\)_m| = P_m = q_m .2l the magnetic field point C equation (3) can be written as
\(\vec{B}\)_axial = \(\frac{\mu_{0}}{4 \pi}\left(\frac{2 r p_{m}}{\left(r^{2}-l^{2}\right)^{2}}\right) \hat{i}\) …… (4)
If the distance between two poles in a bar magnet are small (looks like short magnet) compared to the distance between geometrical centre O of bar magnet and the location of point C i.e.,
r >> l then (r² – l²)² ≈ r⁴ ……. (5)
Therefore, using equation (5) in equation (4), we get

[OR]

Question 35.
(b) Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.
Answer:
Induction of emf by changing relative orientation of the coil with the magnetic field:
Consider a rectangular coil of N turns kept in a uniform magnetic field \(\vec{B}\) figure (a). The coil rotates in anti-clockwise direction with an angular velocity co about an axis, perpendicular to the field.

At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value Φ_m = BA (where A is the area of the coil).

In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is Φ_m cos ωt., a component of Φ_m normal to the plane of the coil (figure (b)). The component parallel to the plane (Φ_m sinωt) has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is NΦ_B = NΦ_m cos ωt. According to Faraday’s law, the emf induced at that instant is


ε = \(-\frac{d}{d t}\) (NΦ_B) = \(-\frac{d}{d t}\) (NΦ_m cos ωt)
= -NΦ_m (-sin ωt)ω = NΦ_m ω sin ωt
When the coil is rotated through 90° from initial position, sin ωt = 1. Then the maximum value of induced emf is
ε_m = NΦ_m ω = NBAω since Φ_m = BA
Therefore, the value of induced emf at that instant is then given by
ε = ε_m sin ωt
It is seen that the induced emf varies as sin ωt function of the time angle ωt. The graph between induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:
1. Electromagnetic waves are produced by any accelerated charge.

2. Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.

3. Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.

4. Electromagnetic waves travel with speed which is equal to the speed of light in vacuum or free space, c = \(\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}\) = 3 × 10⁸ ms^-1

5. The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

Question 36.
(b) Derive the equation for acceptance angle and numerical aperture, of optical fiber.
Answer:
Acceptance angle in optical fibre:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n₁ cladding n₂ and the outer medium n₃. Assume the light is incident at an angle i_a called acceptance angle i at the outer medium and core boundary at A.
The Snell’s law in the product form, equation for this refraction at the point A.

n₃ sin i_a = n₁ sin r_a …(1)
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle i_c . Snell’s law in the product form, equation for the refraction at point B is,
n₁ sin i_c = n₂ sin 90° …(2)
n₁ sin i_c = n₂ ∵ sin 90° = 1
∴ sin i_c = \(\frac{n_{2}}{n_{1}}\) …(3)
From the right angle triangle ∆ABC,
i_c = 90° – r_a
Now, equation (3) becomes, sin (90° – r_a) = \(\frac{n_{2}}{n_{1}}\)
Using trigonometry, cos r_a = \(\frac{n_{2}}{n_{1}}\) …(4)
sin r_a = \(\sqrt{1-\cos ^{2} r_{a}}\)
Substituting for cos r_a
sin r_a = \(\sqrt{1-\left(\frac{n_{2}}{n_{1}}\right)^{2}}=\sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}}}\) …… (5)
Substituting this in equation (1)
n₃ sin i_a = \(n_{1} \sqrt{\frac{n_{1}^{2}-n_{2}^{2}}{n_{1}^{2}}}=\sqrt{n_{1}^{2}-n_{2}^{2}}\) ….. (6)
On further simplification,

If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes,
i_a = sin^-1\((\sqrt{n_{1}^{2}-n_{2}^{2}})\) …… (9)

Light can have any angle of incidence from 0 to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n₃ sin i_a) is called numerical aperture NA of the optical fibre.
NA = n₃ sin i_a \((\sqrt{n_{1}^{2}-n_{2}^{2}})\) ….. (10)

If outer medium is air, then n₃ = 1. The numerical aperture NA becomes,
NA = sin i_a \(\sqrt{n_{1}^{2}-n_{2}^{2}}\) ….. (11)

Question 37.
(a) Explain the effect of potential difference on photoelectric current.
Answer:
Effect of potential difference on photoelectric current:
To study the effect of potential difference V between the electrodes on photoelectric current, the frequency and intensity of the incident light are kept constant. Initially the potential of A is kept positive with respect to C and the cathode is irradiated with the given light.

Now, the potential of A is increased and the corresponding photocurrent is noted. As the potential of A is increased, photocurrent is also increased. However a stage is reached where photocurrent reaches a saturation value (saturation current) at which all the photoelectrons from C are collected by A. This is represented by the flat portion of the graph between potential of A and photocurrent.

When a negative (retarding) potential is applied to A with respect to C, the current does not immediately drop to zero because the photoelectrons are emitted with some definite and different kinetic energies. .

The kinetic energy of some of the photoelectrons is such that they could overcome the retarding electric field and reach the electrode A.

When the negative (retarding) potential of A is gradually increased, the photocurrent starts to decrease because more and more photoelectrons are being repelled away from reaching the electrode A. The photocurrent becomes zero at a particular negative potential V₀ , called stopping or cut-off potential.

Stopping potential is that the value of the negative (retarding) potential given to the collecting electrode A which is just sufficient to stop the most energetic photoelectrons emitted and make the photocurrent zero.

At the stopping potential, even the most energetic electron is brought to rest. Therefore, the initial kinetic energy of the fastest electron (K_max) is equal to the work done by the stopping potential to stop it (eV₀).
K_max = \(\frac { 1 }{ 2 }\) mv²_maxK = eV₀ ……… (1)
where v_max is the maximum speed of the emitted photoelectron.

From equation (1),
K_max = eV₀ (in joule) (or) K_max – V₀ (in eV)

From the graph, when the intensity of the incident light alone is increased, the saturation current also increases but the value of V₀ remains constant.

Thus, for a given frequency of the incident light, the stopping potential is independent of intensity of the incident light. This also implies that the maximum kinetic energy of the photoelectrons is independent of intensity of the incident light.

[OR]

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

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

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

(ii) Determination of specific charge:
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
eV = \(\frac{1}{2}\)mV² ⇒ \(\frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1), we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ….. (2)
Substituting the values of E, B and V, the specific charge can be determined as
\(\frac{e}{m}\) = 1.7 × 10¹¹ C kg^-1

(iii) Deflection of charge only due to uniform electric field:
When the magnetic field is turned off, the deflection is only due to electric field. The deflection in vertical direction is due to the electric force.

F_e = eE …(3)
Let m be the mass of the electron and by applying Newton’s second law of motion, acceleration of the electron is
a_e = \(\frac{1}{m}\) F_e ……. (4)
Substituting equation (4) in equation (3),
a_e = \(\frac{1}{m} e E=\frac{e}{m} E\)

Let y be the deviation produced from Deviation of path by applying uniform electric field original position on the screen. Let the initial upward velocity of cathode ray be u = 0 before entering the parallel electric plates. Let t be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
t = \(\frac{1}{v}\) …(5)
Hence, the deflection y’ of cathode rays is (note: u = 0 and a_e = \(\frac{e}{m}\)E)

Therefore, the deflection y on the screen is
y ∝ y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get y = C \(\frac{1}{2} \frac{e}{m} \frac{l^{2} \mathrm{B}^{2}}{\mathrm{E}}\) ……. (7)
Rearranging equation (7) as \(\frac{e}{n}=\frac{2 y \mathrm{E}}{\mathrm{Cl}^{2} \mathrm{B}^{2}}\) …… (8)
Substituting the values on RHS, the value of specific charge is calculated as
\(\frac{e}{m}\) = 1.7 × 10¹¹Ckg^-1m

(iv) Deflection of charge only due to uniform magnetic field
Suppose that the electric field is switched off and only the magnetic field is switched on. Now the deflection occurs only due to magnetic field. The force experienced by the electron in uniform magnetic field applied perpendicular to its path is
F_m = evB (in magnitude)
Since this force provides the centripetal force, the electron beam undergoes a semicircular path. Therefore, we can equate F_m to centripetal force \(\frac{m v^{2}}{R}\)
F_m = evB = \(m \frac{v^{2}}{R}\)
where v is the velocity of electron beam at the point where it enters the magnetic field and R is the radius of the circular path traversed by the electron beam.
eB = \(m \frac{v}{\mathrm{R}} \Rightarrow \frac{e}{m}=\frac{v}{\mathrm{BR}}\) ……. (9)
Further, substituting equation (1) in equation (9), we get
\(\frac{e}{m}=\frac{E}{B^{2} R}\) …….. (10)
By knowing the values of electric field, magnetic field and the radius of circular path, the value of specific charge \(\left(\frac{e}{m}\right)\) can be calculated

Question 38.
(a) Draw the circuit diagram of a half wave rectifier and explain its working
Answer:
Half wave rectifier circuit:
The half wave rectifier circuit. The circuit consists of a transformer, a p-n junction diode and a resistor. In a half wave rectifier circuit, either a positive half or the negative half of the AC input is passed through while the other half is blocked. Only one half of the input wave reaches the output. Therefore, it is called halfwave rectifier. Here, a p-n junction diode acts as a rectifying diode.

During the positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal A becomes positive with respect to terminal B. The diode is forward biased and hence it conducts. The current flows through the load resistor R_L and the AC voltage developed across R_L constitutes the output voltage V₀ and the waveform of the diode current.

During the negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal A is negative with respect to terminal B. Now the diode is reverse biased and does not conduct and hence no current passes through R_L. The reverse saturation current in a diode is negligible. Since there is no voltage drop across R_L, the negative half cycle of ac supply is suppressed at the output.

The output of the half wave rectifier is not a steady dc voltage but a pulsating wave. This pulsating voltage is not sufficient for electronic equipments. A constant or a steady voltage is required which can be obtained with the help of filter circuits and voltage regulator circuits. Efficiency (η) is the ratio of the output dc power to the ac input power supplied to the circuit. Its value for half wave rectifier is 40.6 %

[OR]

Question 38.
(b) Fiber optic communication is gaining popularity among the various transmission media – justify.
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems.
Light has very high frequency (400 THz – 790 THz) than microwave radio systems. The fibers are made up of silica glass or silicon dioxide which is highly abundant on Earth.
Now it has been replaced with materials such as chalcogenide glasses, fluoroaluminate crystalline materials because they provide larger infrared wavelength and better transmission capability.
As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and isolation is required from electrical and electromagnetic interference.

Applications
Optical fiber system has a number of applications namely, international communication, inter¬city communication, data links, plant and traffic control and defense applications.

Merits

• Fiber cables are very thin and weight lesser than copper cables.
• This system has much larger bandwidth. This means that its information carrying capacity is larger.
• Fiber optic system is immune to electrical interferences.
• Fiber optic cables are cheaper than copper cables.

Demerits

• Fiber optic cables are more fragile when compared to copper wires.
• It is an expensive technology.

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 and score more marks in your examinations.

TN State Board 12th Physics 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. 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-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART -1

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

Question 1.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 2.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 × 10² Ω
(c) 4 × 10³ Ω
(d) 4.2 Ω
Answer:
(b) 41 × 10² Ω

Question 3.
The magnetic field at the center O of the following
(a) \(\frac{\mu_{0} \mathbf{I}}{4 r} \otimes\)
(b) \(\frac{\mu_{0} I}{4 r} \odot\)
(c) \(\frac{\mu_{0} I}{2 r} \otimes\)
(d) \(\frac{\mu_{0} I}{2 r} \odot\)

Answer:
(a) \(\frac{\mu_{0} \mathbf{I}}{4 r} \otimes\)

Question 4.
The horizontal component of earth’s magnetic field at a place is 3.6 × 10^-5T. If the angle of dip at this place is 60°, the vertical components of earth’s field at this place is
(a) 1.2 × 10^-5T
(b) 2.4 × 10^-5T
(c) 4 × 10^-5T
(d) 6.2 × 10^-5T
Hint: B_v = B_H tan δ = 3.6 × 10^-5 × tan 60°
B_v = 6.2 × 10^-5 T
Answer:
(d) 6.2 × 10^-5T

Question 5.
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{6}\)
(d) zero
Answer:
(a) \(\frac{\pi}{4}\)

Question 6.
Alternating current can be measured by
(a) moving coil galvanometer
(b) hot wire ammeter
(c) tangent galvanometer
(d) none of the above
Answer:
(b) hot wire ammeter

Question 7.
Let E = E₀ sin[10⁶ × -ωt] be the electric field of plane electromagnetic wave, the value of ω is
(a) 0.3 × 10^-14 rad s^-1
(b) 3 × 10^-14 rad s^-1
(c) 0.3 × 10¹⁴ rad s^-1
(d) 3 × 10¹⁴ rad s^-1
Answer:
(d) 3 × 10¹⁴ rad s^-1

Question 8.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are
(A) 51 and I
(b) 51 and 31
(c) 91 and I
(d) 91 and 31
Answer:
(c) 91 and I

Question 9.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation

Question 10.
Emission of electrons by the absorption of heat energy is called emission.
(a) photoelectric
(b) field
(c) thermionic
(d) secondary
Answer:
(c) thermionic

Question 11.
Proton and α – particle have the same de-Broglie wavelength. What is same for both of them?
(a) Time period
(b) Energy
(c) Frequency
(d) Momentum
Hint: λ = h/p, when wavelength λ is same, momentum p is also same.
Answer:
(d) Momentum

Question 12.
The ratio of the wavelengths for the transition from n = 2 to n = 1 in Li⁺⁺, He⁺ and H is
(a) 1:2: 3
(b) 1:4: 9
(c) 3:2:1
(d) 4: 9: 36
Answer:
(d) 4: 9: 36

Question 13.
The principle in which a solar cell operates
(a) Diffusion
(b) Recombination
(c) Photovoltaic action
(d) Carrier flow
Answer:
(c) Photovoltaic action

Question 14.
The output transducer of the communication system converts the radio signal into
(a) Sound
(b) Mechanical energy
(c) Kinetic energy
(d) None of the above
Answer:
(a) Sound

Question 15.
The alloys used for muscle wires in Robots are
(a) Shape memory alloys
(b) Gold copper alloys
(c) Gold silver alloys
(d) Two dimensional alloys
Answer:
(a) Shape memory alloys

PART – II

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

Question 16.
What is Polarisation?
Answer:
Polarisation \(\overrightarrow{\mathrm{P}}\) is defined as the total dipole moment per unit volume of the dielectric.
\(\overrightarrow{\mathrm{P}}\) = χ_e \(\overrightarrow{\mathrm{P}}\)_ext

Question 17.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\overrightarrow{\mathrm{j}}\) . \(\overrightarrow{\mathrm{A}}\)
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

Question 18.
The horizontal component and vertical components of Earth’s magnetic field at a place are 0.15 G and 0.26 G respectively. Calculate the angle of dip and resultant magnetic field.
Answer:
B_H = 0.15 G and B_v = 0.26 G
tan I = \(\frac{0.26}{0.15}\) ⇒ I = tan^-1 (1.732) = 60°
The resultant magnetic field of the Earth is
\(\mathrm{B}=\sqrt{\mathrm{B}_{\mathrm{H}}^{2}+\mathrm{B}_{\mathrm{H}}^{2}}=0.3 \mathrm{G}\)

Question 19.
State Fleming’s right hand rule.
Answer:
The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

Question 20.
The wavelength of a light is 450 nm. How much phase it will differ for a path of 3 mm?
Answer:
The wavelength is, λ = 450 nm = 450 × 10^-9 m
Path difference is, δ = 3 mm = 3 × 10^-3 m
Relation between phase difference and path difference is, Φ = \(\frac{2 \pi}{\lambda} \times \delta\)
Substituting, Φ = \(\frac{2 \pi}{450 \times 10^{-9}} \times 3 \times 10^{-3}=\frac{\pi}{75} \times 10^{6}\)
Φ = \(\frac{\pi}{75} \times 10^{6} \mathrm{rad}\)

Question 21.
How will you define threshold frequency?
Answer:
For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.

Question 22.
Calculate the number of nuclei of carbon-14 undecayed after 22,920 years if the initial number of carbon-14 atoms is 10,000. The half-life of carbon-14 is 5730 years.
Answer:
To get the time interval in terms of half-life, n = \(\frac{t}{\mathrm{T}_{1 / 2}}=\frac{22,920 \mathrm{yr}}{5730 \mathrm{yr}}=4\)
The number of nuclei remaining undecayed after 22,920 years
\(\mathrm{N}=\left(\frac{1}{2}\right)^{n} \mathrm{N}_{0}=\left(\frac{1}{2}\right)^{4} \times 10,000 \Rightarrow \mathrm{N}=625\)

Question 23.
A diode is called as a unidirectional device. Explain
Answer:
Diode is called as a unidirectional device, i.e., current flows in only one direction (anode to cathode internally) when a forward voltage is applied, the diode conducts and when reverse voltage is applied, there is no conduction. A mechanical analogy is a rat chat, which allows motion in one direction only.

Question 24.
Give the factors that are responsible for transmission impairments.
Answer:

• Attenuation
• Distortion (Harmonic)
• Noise

PART-III

Answer any six questions ¡n which Q.No. 26 ¡s compulsory. (6 × 3 = 18)

Question 25.
A sample of HO gas ¡s placed in a uniform electric field of magnitude 3 × 10⁴ N C^-1. The dipole moment of each HCI molecule is 3.4 × 10^-30 Cm. Calculate the maximum torque experienced by each HCl molecule.
Answer:
The maximum torque experienced by the dipole is when it is aligned perpendicular to the applied field.
\(\tau_{\max }\) = pE sin90° = 3.4 × 10^-30 × 3 × 10⁴Nm
\(\tau_{\max }\) =10.2 × 10^-26Nm

Question 26.
The resistance ola wire is 20 Ω . What will be new resistance, ¡fit is stretched uniformly 8 times its original length?
Answer:
R₁ = 20 Ω, R₂ = ?
Let the original length (l₁) be 1.
The new length, l₂ = 8l₁ (i.,e) l₂ =8l
The original resistance,


Though the wire is stretched, its volume is unchanged.
Initial volume = Final volume
A₁l₁ = A₂l₂ , A₁l =A₂8l
\(\frac{A_{1}}{A_{2}}=\frac{8 l}{l}=8\)
By dividing equation R₂ by equation R₁, we get

Substituting the value of \(\frac{A_{1}}{A_{2}}\) we get
\(\frac{R_{1}}{R_{2}}\) = 8 × 8 = 64 ⇒ R₂ = 64 × 20 = 1280 Ω
Hence, strecthing the length of the wire has increased its resistance.

Question 27.
State Biot-Savart’s law.
The magnitude of magnetic field \(d \vec{B}\) at a point P at a distance r from the small elemental length taken on a conductor carrying current varies

• directly as the strength of the current I
• directly as the magnitude of the length element \(d \vec{l}\)
• directly as the sine of the angle (say,0) between d\(d \vec{l}\) and r̂ .
• inversely as the square of the distance between the point P and length element \(d \vec{l}\).
  This is expressed as
  \(d \mathrm{B} \propto \frac{\mathrm{I} d l}{r^{2}} \sin \theta\)

Question 28.
Give the principle of AC generator.
Answer:
Alternators work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. The magnitude of the induced emf is given by Faraday’s law of electromagnetic induction and its direction by Fleming’s right hand rule.

Question 29.
A transformer is used to light a 140 W, 24 V bulb from a 240 V AC mains. The current in the main cable is 0.7 A. Find the efficiency of the transformer.
Answer:

Question 30.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

• The Incident light is taken from left to right (i.e. object on the left of mirror).
• All the distances are measured from the pole of the mirror (pole is taken as origin).
• The distances measured to the right of pole along the principal axis are taken as positive.
• The distances measured to the left of pole along the principal axis are taken as negative,
• Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
• Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

Question 31.
Write the relationship of de Broglie wavelength k associated with a particle of mass m in terms of its kinetic energy K.
Answer:
Kinetic energy of the particle, K = \(\frac{1}{2}\) mv² = \(\frac{p^{2}}{2 m}\)
p = \(\sqrt{2 m \mathrm{K}}\)
de-Broglie wavelength of the particle λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\)

Question 32.
What is binding energy of a nucleus? Give its expression.
Answer:
when Z protons and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding energy is released. This is called the binding energy of the nucleus (BE) and is equal to (Δm)c² .
BE = (Zm_p + Nm_n – M ) c²

Question 33.
Distinguish between avalanche and zener breakdown.
Answer:

Avalanche Breakdown	Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths.	It occurs injunctions 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.

PART – IV

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

Question 34.
(a) Calculate the electric field due to a dipole on its axial line and equatorial plane.
Answer:
Case (i) Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. line
The electric field at a point C due to +q is


Since the electric dipole moment vector \(\vec{p}\) is from -q to +q and is directed along BC, the above equation is rewritten as

where P̂ is the electric dipole moment unit vector from -q to +q.
The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec{E}\)_– \(\vec{E}\)₊ us stronger than \(\vec{E}\)_–. Therefore, the length of the \(\vec{E}\)₊ vector is drawn large than that of \(\vec{E}\)_– vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec{E}\)₊ since +q is closer to C than -q.

The direction of \(\vec{E}\)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a).
Under this limit the term(r² – a²) ≈ r⁴ Substituting this into equation, we get

If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of \(\vec{p}\).

Case (ii) Electric field due to an electric dipole at a point on the equatorial plane

Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since the point C is quite-distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E is along BC and the direction of \(\vec{E}\)₊ is along and the direction of \(\vec{E}\)_– CA. \(\vec{E}\)₊ and \(\vec{E}\)_– are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right|\) sin θ and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| sin θ\) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of \(\vec{E}\)₊ and \(\vec{E}\)_– and its direction is along -P̂

The magnitudes \(\vec{E}\)₊ and \(\vec{E}\)_– are the same and are given by


By substituting equation (1) into equation (2), we get

Question 34.
(b) Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected, A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.

Applying Kirchhoff’s current rule to junction B,
I₁ – I_G – I₃ = 0 ……. (1)
Applying Kirchhoff’s current rule to junction D,
I₂ + I_G – I₄ = 0 ……. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁p + I_GG – I₂R = 0 ……. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁p – I₃Q – I₄S – I₂R = 0 ……. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0).
Substituting I_G = O in equation, (I), (2) and (3), we get
I₁ = I₃ …… (5)
I₂ = I₄ ……… (6)
I₁p = I₂R ………. (7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₂S – I₂R= 0
I₁(P + Q) = I₂(R+S) …… (8)
Dividing equation (8) By equation (7), we get

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

Question 35.
(a) Show the time period of oscillation when a bar magnet is kept in a uniform magnetic field is T = 2π\(\sqrt{\frac{1}{p_{m} \mathrm{B}}}\). in second, where I represents moment of interia of the bar magnet, p<sub<m is the magnetic moment and is the magnetic field.
Answer:
The magnitude of deflecting torque (the torque which makes the object rotate) acting on the bar magnet which will tend to align the bar magnet parallel to the direction of the uniform magnetic field \(\overrightarrow{\mathrm{B}}\) is
\(\vec{\tau}\) = p_m B sin θ
The magnitude of restoring torque acting on the bar magnet can be written as
\(|\vec{\tau}|=\mathrm{I} \frac{d^{2} \theta}{d t^{2}}\)
Under equilibrium conditions, both magnitude of deflecting torque and restoring torque will be equal but act in the opposite directions, which means
\(\mathrm{I} \frac{d^{2} \theta}{d t^{2}}=-p_{m} \mathrm{B} \sin \theta\)
The negative sign implies that both are in opposite directions. The above equation can be written as
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{p_{m} B}{I} \sin \theta\)
This is non-linear second order homogeneous differential equation. In order to make it linear, we use small angle approximation, i.e., sin θ ≈ θ, we get
\(\frac{d^{2} \theta}{d t^{2}}=-\frac{p_{m} \mathrm{B}}{\mathrm{I}} \theta\)
This linear second order homogeneous differential equation is a Simple Harmonic differential equation. Therefore,
Comparing with Simple Harmonic Motion (SHM) differential equation \(\frac{d^{2} x}{d t^{2}}=-\omega^{2} x\)
where ω is the angular frequency of the oscillation.

where, B_H is the horizontal component of Earth’s magnetic field.

[OR]

Question 35.
(b) An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
Answer:
Energy stored in an inductor: Whenever a current is established in the circuit, the inductance opposes the growth of the current. In order to establish a current in the circuit, work is done against this opposition by some external agency. This work done is stored as magnetic potential energy. Let us assume that electrical resistance of the inductor is negligible and inductor effect alone is considered. The induced emf eat any instant t is
ℰ = \(-\mathrm{L} \frac{d i}{d t}\) …(1)
Let dW be work done in moving a charge dq in a time dt against the opposition, then
dW = -edq = -ℰdq = -ℰidi [ ∵ dq = idt]
Substituting for e value from equation (1)
= \(-\left(-\mathrm{L} \frac{d i}{d t}\right) i d t\)
dW = Lidt ……. (2)
Total work done in establishing the current i is
This work done is stored as magnetic potential energy.

W = \(\frac{1}{2}\) Li² ………. (3)
This work done is stored as magnetic potential energy.
∴ U_B = \(\frac{1}{2}\) Li²

Question 36.
(a) Using Faraday’s law of electromagnetic induction, derive an equation for motional emf.
Answer:
Motional emf from Faraday’s law: Let us consider a rectangular conducting loop of width l in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field.

When the loop is pulled with a constant velocity \(\overrightarrow{\mathrm{v}}\) to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday’s law, an electric current is induced in the loop which flows in a direction so as to oppose the pull of the loop.
Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is
Φ_B = \(\int_{A} \vec{B} \cdot d \vec{A}=B A \cos \theta\)
Here θ = 0° and cos 0° = 1
= BA
Φ_B = Blx ……… (1)

As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
\(\varepsilon=\frac{d \Phi_{\mathrm{B}}}{d t}=\frac{d}{d t}(\mathrm{B} l x)\)
Here, both B and l are constants. Therefore,
\(\varepsilon=\mathrm{B} l \frac{d x}{d t}=\mathrm{B} l v\) …… (2)
where v = \(\frac{d x}{d t}\) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.

[OR]

Question 36.
(b) Derive the mirror equation and the equation for lateral magnification. The mirror equation:
Answer:
The mirror equation establishes a relation among object distance u, image distance v and focal length/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C.
Let us consider three paraxial rays from point B on the object.
The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PB’. The third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.

As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B’PA’ . The triangles ∆BPA and ∆B’PA’ are similar. Thus, from the rule of similar triangles,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}\) …… (1)
The other set of similar triangles are, ∆DPF and ∆ BA.’ F. (PD is almost a straight vertical line)
\(\frac{A^{\prime} B^{\prime}}{P D}=\frac{A^{\prime} F}{P F}\)
As, the distances PD = AB the above equation becomes,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{A^{\prime} F}{P F}\) …….. (2)
From equations (1) and (2) we can write,
\(\frac{P A^{\prime}}{P A}=\frac{A^{\prime} F}{P F}\)

As, A’ F PA’ – PF, the above equation becomes,
\(\frac{P A^{\prime}}{P A}=\frac{P A^{\prime}-P F}{P F}\)
We can apply the sign conventions for the various distances in the above equation.
PA= -u. PA’= -v, PF = -f
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes.

The above equation (4) is called mirror equation

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.

m = \(\frac{h^{\prime}}{h}\) …… (5)
Applying proper sign conventions for equation (1), A’B’ PA’
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}\)
A’B’ = -h, AB = h, PA’ = -v, PA = -u
\(\frac{-h^{\prime}}{h}=\frac{-v}{-u}\)
On simplifying we get,
\(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\) …….(6)
Using mirror equation, we can further write the magnification as,
\(m=\frac{h^{\prime}}{h}-\frac{f-v}{f}=\frac{f}{f-u}\) ……… (7)

Question 37.
(a) Obtain Einstein’s photoelectric equation with necessary explanation.
Answer:
Einstein’s explanation of photoelectric equation:
When a photon of energy hv is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected.

In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photoelectric work function Φ₀) and the remaining energy as the kinetic energy of the ejected electron. From the law of conservation of energy,
hυ = Φ₀ + \(\frac { 1 }{ 2 }\) mv² …… (1)
where m is the mass of the electron and u its velocity

If we reduce the frequency of the incident light, the speed or kinetic energy of photo electrons is also reduced. At some frequency v0 of incident radiation, the photo electrons are ejected with almost zero kinetic energy. Then the equation (1) becomes
hυ₀ = Φ₀
where v₀ is the threshold frequency. By rewriting the equation (1), we get
hυ = hυ₀ + \(\frac { 1 }{ 2 }\) mv² …….(2)
The equation (2) is known as Einstein’s Photoelectric equation.
If the electron does not lose energy by internal collisions, then it is emitted with maximum kinetic energy K_max. Then
K_max = \(\frac { 1 }{ 2 }\) mv²_max
where υ_max is the maximum velocity of the electron ejected. The equation (1) is rearranged as follows:
K_max = hυ – Φ₀

[OR]

Question 37.
(b) Derive the energy expression for hydrogen atom using Bohr atom model.
Answer:
The energy of an electron in the nth orbit
Since the electrostatic force is a conservative force, the potential energy for the orbit is

This implies that U_n = -2 KE_n. Total energy in the «th orbit is
E_n = KE_n + U_n = KE_n – 2KE_n = -KE_n
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \frac{Z^{2}}{n^{2}}\)
For hydrogen atom (Z = 1),
E_n = \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \frac{1}{n^{2}} \text { joule }\) ……. (1)
where n stands for principal quantum number. The negative sign in equation (1) indicates that the electron is bound to the nucleus.
Substituting the values of mass and charge of an electron (m and e), permittivity of free space s0 and Planck’s constant h and expressing in terms of eW. we get
E_n = -13.6\(\frac{1}{n^{2}}\)eV
For the first orbit (ground state), the total energy of electron is E₁ = – 13.6 eV. For the second orbit (first excited state), the total energy of electron is E₂ = -3.4 eV. For the third orbit (second excited state), the total energy of electron is E₃ =1.51 eV and so on.

Notice that the energy of the first excited state is greater than the ground state, second excited state is greater than the first excited state and so on. Thus, the orbit which is closest to the nucleus (r₁) has lowest energy (minimum energy compared with other orbits). So, it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (-13.6 eV ) is used as a unit of energy called Rydberg (1 Rydberg = -13.6 eV ).

The negative value of this energy is because of the way the zero of the potential energy is defined. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.

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

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

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D₂ and reverse biases diode D₁. Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC . As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure (b). Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature.

The efficiency ( η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

[OR]

Question 38.
(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the challenges and risk factors.
(a) ICT is widely used in increasing food productivity and farm management.
(b) It helps to optimize the use of water, seeds and fertilizers etc.
(c) Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
(d) Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining
(a) ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
(b) Information and communication technology provides audio-visual warning to the trapped underground miners.
(c) It helps to connect remote sites.

Students can Download Computer Science Chapter 11 Database Concept Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 11 Database Concept

Samacheer Kalvi 12th Computer Science Database Concept Text Book Back Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
What is the acronym of DBMS?
Answer:
(a) DataBase Management Symbol
(b) Database Managing System
(c) DataBase Management System
(d) DataBasic Management System
Answer:
(c) DataBase Management System

Question 2.
A table is known as ……………………..
(a) tuple
(b) attribute
(c) relation
(d) entity
Answer:
(c) relation

Question 3.
Which database model represents parent-child relationship?
(a) Relational
(b) Network
(c) Hierarchical
(d) Object
Answer:
(c) Hierarchical

Question 4.
Relational database model was first proposed by ……………………..
(a) E F Codd
(b) E E Codd
(c) c) E F Cadd
(d) E F Codder
Answer:
(a) E F Codd

Question 5.
What type of relationship does hierarchical model represents?
(a) one-to-one
(b) one-to-many
(c) many-to-one
(d) many-to-many
Answer:
(b) one-to-many

Question 6.
Who is called Father of Relational Database from the following?
(a) Chris Date
(b) Hugh Darween
(c) Edgar Frank Codd
(d) Edgar Frank Cadd
Answer:
(c) Edgar Frank Codd

Question 7.
Which of the following is an RDBMS?
(a) Dbase
(b) Foxpro
(c) Microsoft Access
(d) SQLite
Answer:
(d) SQLite

Question 8.
What symbol is used for SELECT statement?
(a) σ
(b) π
(c) X
(d) Ω
Answer:
(a) σ

Question 9.
A tuple is also known as ………………………
(a) table
(b) row
(c) attribute
(d) field
Answer:
(b) row

Question 10.
Who developed ER model?
(a) Chen
(b) EF Codd
(c) Chend
(d) Chand
Answer:
(a) Chen

PART – II
II. Answer The Following Questions

Question 1.
Mention few examples of a database?
Answer:
Examples of popular DBMS: Dbase, FoxPro

Question 2.
List some examples of RDBMS?
Answer:
SQL server, Oracle, mysql, MariaDB, SQLite.

Question 3.
What is data consistency?
Answer:
Data Consistency
On live data, it is being continuously updated and added, maintaining the consistency of data can become a challenge. But DBMS handles it by itself. Data Consistency means that data values are the same at all instances of a database

Question 4.
What is the difference between Hierarchical and Network data model?
Answer:
Network database model is an extended form of hierarchical data model. The difference between hierarchical and Network data model is :

1. In hierarchical model, a child record has only one parent node,
2. In a Network model, a child may have many parent nodes. It represents the data in many-to-many relationships.
3. This model is easier and faster to access the data.

Question 5.
What is normalization?
Answer:

1. Normalization is a process of organizing the data in the database to avoid data redundancy and to improve data integrity.
2. Database normalization was first proposed by Dr. Edgar F Codd as an integral part of RDBMS. These rules are known as E F Codd Rules.

PART – III
III. Answer The Following Questions

Question 1.
What is the difference between Select and Project command?
Answer:
Select:
THE SELECT operation is used for selecting a subset with tuples according to a given condition. Select filters out all tuples that do not satisfy C.

Project:
The projection eliminates all attributes of the input relation but those mentioned in the projection list. The projection method defines a relation that contains a vertical subset of Relation.

Question 2.
What is the role of DBA?
Answer:
Database Administrator

1. Database Administrator or DBA is the one who manages the complete database management system.
2. DBA takes care of the security of the DBMS, managing the license keys, managing user accounts and access etc.

Question 3.
Explain Cartesian Product with a suitable example?
PRODUCT OR CARTESIAN PRODUCT (Symbol: X)
Answer:
Cross product is a way of combining two relations. The resulting relation contains, both relations being combined.
A × B means A times B, where the relation A and B have different attributes.
This type of operation is helpful to merge columns from two relations.

Question 4.
Explain Object Model with example?
Answer:
Object Model

1. Object model stores the data in the form of objects, attributes and methods, classes and Inheritance.
2. O This model handles more complex applications, such as Geographic information System (GIS), scientific experiments, engineering design and manufacturing.
3. It is used in file Management System.
4. It represents real world objects, attributes and behaviors. It provides a clear modular structure.
5. It is easy to maintain and modify the existing code.

An example of the Object model is Shape, Circle, Rectangle and Triangle are all objects in this model.

1. Circle has the attribute radius.
2. Rectangle has the attributes length and breadth.
3. Triangle has the attributes base and height.
4. The objects Circle, Rectangle and Triangle inherit from the object Shape.

Question 5.
Write a note on different types of DBMS users?
Answer:
Types of DBMS Users
(i) Database Administrator:
Database Administrator or DBA is the one who manages the complete database management system. DBA takes care of the security of the DBMS, managing the license keys, managing user accounts and access etc.

(ii) Application Programmers or Software Developers:
This user group is involved in developing and designing the parts of DBMS.

(iii) End User:
End users are the one who store, retrieve, update and delete data.

(iv) Database designers: are responsible for identifying the data to be stored in the database for choosing appropriate structures to represent and store the data.

PART – IV
IV. Answer The Following Questions

Question 1.
Explain the different types of data model?
Answer:
Types of Data Model
Following are the different types of a Data Model

1. Hierarchical Model
2. Relational Model
3. Network Database Model
4. Entity Relationship Model
5. Object Model

1. Hierarchical Model
Hierarchical model was developed by IBM as Information Management System.
In Hierarchical model, data is represented as a simple tree like structure form. This model represents a one-to-many relationship i.e. parent-child relationship. One child can have only one parent but one parent can have many children. This model is mainly used in IBM Main Frame computers.

2. Relational Model
The Relational Database model was first proposed by E.F. Codd in 1970 . Nowadays, it is the most widespread data model used for database applications around the world.
The basic structure of data in relational model is tables (relations). All the information’s related to a particular type is stored in rows of that table. Hence tables are also known as relations in a relational model. A relation key is an attribute which uniquely identifies a particular tuple (row in a relation (table)).

3. Network Model
Network database model is an extended form of hierarchical data model. The difference between hierarchical and-Network data model is :

1. In hierarchical model, a child record has only one parent node,
2. In a Network model, a child may have many parent nodes. It represents the data in many-to-many relationships.
3. This model is easier and faster to access the data.

School represents the parent node
Library, Office and Staff room is a child to school (parent node)
Student is a child to library, office and staff room (one to many relationship)

4. Entity Relationship Model. (ER model)
In this database model, relationship are created by dividing the object into entity and its characteristics into attributes.
It was developed by Chen in 1976. This model is useful in developing a conceptual design for the database. It is very simple and easy to design logical view of data. The developer can easily understand the system by looking at ER Model constructed. Rectangle represents the entities. E.g. Doctor and Patient

Ellipse represents the attributes E.g. D-id, D-name, P-id, P-name. Attributes describes the characteristics and each entity becomes a major part of the data stored in the database. Diamond represents the relationship in ER diagrams E.g. Doctor diagnosis the Patient

5. Object Model
Object model stores the data in the form of objects, attributes and methods, classes and Inheritance. This model handles more complex applications, such as Geographic information System (GIS), scientific experiments, engineering design and manufacturing. It is used in file Management System. It represents real world objects, attributes and behaviors. It provides a clear modular structure. It is easy to maintain and modify the existing code.

Question 2.
Explain the different types of relationship mapping?
Answer:
Types of Relationships:
Following are the types of relationships used in a database.

1. One-to-One Relationship
2. One-to-Many Relationship
3. Many-to-One Relationship
4. Many-to-Many Relationship

1. One-to-One Relationship
In One-to-One Relationship, one entity is related with only one other entity. One row in a table is linked with only one row in another table and vice versa.
For example: A student can have only one exam number.

2. One-to-Many Relationship
In One-to-Many relationship, one entity is related to many other entities. One row in a table A is linked to many rows in a table B, but one row in a table B is linked to only one row in table A. For example: One Department has many staff members.

3. Many-to-One Relationship
In Many-to-One Relationship, many entities can be related with only one in the other entity. For example: A number of staff members working in one Department. Multiple rows in staff members table is related with only one row in Department table.

4. Many-to-Many Relationship
Amany-to-many relationship occurs when multiple records in a table are associated with multiple records in another table.

Example 1: Customers and Product:
Customers can purchase various products and Products can be purchased by many customers

Example 2: Students and Courses:
A student can register for many Courses and a Course may include many students

Example 3: Books and Student:
Many Books in students.

Question 3.
Differentiate DBMS and RDBMS?
Answer:

Question 4.
Explain the different operators in Relational algebra with suitable examples?
Answer:
Relational Algebra Operations from Set Theory

1. UNION (∪)
2. INTERSECTION (∩)
3. DIFFERENCE (-)
4. CARTESIAN PRODUCT (X)

SELECT (symbol: σ)
General form σ_c ( R) with a relation R and a condition C on the attributes of R.
The SELECT operation is used for selecting a subset with tuples according to a given condition. Select filters out all tuples that do not satisfy C.

PROJECT (symbol : II)
The projection eliminates all attributes of the input relation but those mentioned in the projection list.
Example 1 using Table A
π_course (STUDENT)
Result

Course Big Data R language PythonProgramming
Note: duplicate row is removed in the result UNION (Symbol :u)
It includes all tuples that are in tables A or in B. It also eliminates duplicates. Set A Union Set B would be expressed asAuB

Example 2:
Consider the following tables

Result

SET DIFFERENCE ( Symbol: -)
The result of A – B, is a relation which includes all tuples that are in A but not in B. The attribute name of A has to match with the attribute name in B.

Example 4:
( using Table B)
Result

INTERSECTION (symbol: ∩) A∩B
Defines a relation consisting of a set of all tuple that are in both in A and B. However, A and B must be union-compatible

Example 5:
(using Table B)

PRODUCT OR CARTESIAN PRODUCT (Symbol: X)
Cross product is a way of combining two relations. The resulting relation contains, both relations being combined.
A × B means A times B, where the relation A and B have different attributes.
This type of operation is helpful to merge columns from two relations.

Cartesian product : Table A × Table B

Question 5.
Explain the characteristics of DBMS?
Characteristics of Database Management System
Answer:

Samacheer kalvi 12th Computer Science Database Concept Additional Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
…………………… are raw facts stored in a computer
(a) data
(b) Information
(c) row
(d) tuple
Answer:
(a) data

Question 2.
……………………. gives meaningful information
(a) data
(b) Information
(c) row
(d) tuple
Answer:
(b) Information

Question 3.
……………………. is a repository collection of related data
(a) data
(b) Information
(c) database
(d) tuple
Answer:
(c) database

Question 4.
……………………… is a software that allows us to create, define and manipulate databases
(a) data
(b) Information
(c) DBMS
(d) Tuple
Answer:
(c) DBMS

Question 5.
Which one of the following is not a characteristic of DBMS?
(a) Redundancy
(b) consistency
(c) Normalization
(d) Insecure
Answer:
(d) Insecure

Question 6.
Find the wrong statement about DBMS?
(a) segregation of application program
(b) Maximum data Redundancy
(c) Easy retrieval of data
(d) Reduced development time
Answer:
(b) Maximum data Redundancy

Question 7.
How many major components of DBMS are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 8.
Which one of the following is not a component of DBMS?
(a) Data
(b) Methods
(c) DataBase Access Language
(d) Modules
Answer:
(d) Modules

Question 9.
Which is the language used to write commands to access, insert, update data stored in database?
(a) DataBase Access Languages
(b) Javascript
(c) Basic
(d) Foxpro
Answer:
(a) DataBase Access Languages

Question 10.
A column is known as an ……………………….
Answer:
Attribute

Question 11.
Entire collection of related data in one table, is referred to as ……………………… or …………………………
Answer:
File or Table

Question 12.
Each table ………………………. represents a Field
Answer:
column

Question 13.
Each ……………………… in table represents a record.
Answer:
row

Question 14.
How many different types of a data model are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 15.
Hierarchical Model was developed by ……………………….
(a) Apple
(b) IBM
(c) Microsoft
(d) Macromedia
Answer:
(b) IBM

Question 16.
Which database Model is the extended form of hierarchical data Model?
(a) Network
(b) Relational
(c) Flat File
(d) Object
Answer:
(a) Network

Question 17.
1. Relational – (i) classes
2. object model – (ii) Mainframe
3. ER model – (iii) key
4. Hierarchical – (iv) Entity
(a) 1-iii, 2-i, 3-iv, 4-ii
(b) 1-i, 2-ii, 3-iii, 4-iv
(c) 1-iv, 2-iii, 3-i, 4-ii
(d) 1-iv, 2-ii, 3-i, 4-iii
Answer:
(a) 1-iii, 2-i, 3-iv, 4-ii

Question 18.
The relational model was developed in the year ……………………..
(a) 1980
(b) 1970
(c) 1965
(d) 1985
Answer:
(b) 1970

Question 19.
…………………… uniquely identifies a particular tuple in a table
Answer:
Relation key

Question 20.
Which model establishes many to many relationships?
(a) Network
(b) Relational
(c) Hierarchical
(d) Object
Answer:
(a) Network

Question 21.
ER Model Expand ………………………..
(a) Entry Relation
(b) Entity Relationship
(c) Entire Row
(d) Entity Row
Answer:
(b) Entity Relationship

Question 22.
ER Model was developed in the year …………………….
(a) 1978
(b) 1972
(c) 1976
(d) 1975
Answer:
(c) 1976

Question 23.
In ER Model, objects are said to be ……………………..
Answer:
entity

Question 24.
Find the wrongly matched pair.
(a) Entities – Rectangle
(b) Ellipse – attributes
(c) Diamond – relationship
(d) row – square
Answer:
(d) row – square

Question 25.
…………………….. represents the relationship in ER diagrams.
Answer:
Diamonds

Question 26.
GIS stands for ………………………
Answer:
Geographic Information System

Question 27.
Which model is used in file management systems?
(a) object
(b) Hierarchical
(c) Network
(d) ER
Answer:
(a) object

Question 28.
DBA means ………………….
Answer:
Database Administators

Question 29.
………………….. is one who manages the complete database management system
(a) Manager
(b) Engineer
(c) DBA
(d) Service Person
Answer:
(c) DBA

Question 30.
……………………. are the one who store, retrieve, update and delete data.
Answer:
End User

Question 31.
Choosing appropriate structures to represent and store the data are the task of ……………………..
Answer:
database designer

Question 32.
RDBMS means ………………………..
(a) Relational Database Manipulation System
(b) Relational Database Management system
(c) Rapid DataBase Management Server
Answer:
(b) Relational Database Management system

Question 33.
Pick the odd out.
Oracle, Foxpro, MariaDB, SQLite
Answer:
Foxpro

Question 34.
Find the true statement
(a) Data redundancy is exhibited by DBMS
(b) Data redundancy is not present in DBMS
Answer:
(a) Data redundancy is exhibited by DBMS

Question 35.
Find the false statement
(a) Distributed Databases supported by DBMS
(b) Distributed Databases supported by RDBMS
Answer:
(a) Distributed Databases supported by DBMS

Question 36.
……………………. Model of data storage is used in DBMS.
Answer:
Navigational

Question 37.
In which database systems, transaction management is efficient?
(a) DBMS
(b) RDBMS
(c) ERDMS
(d) DBMS
Answer:
(b) RDBMS

Question 38.
How many types of relationships are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 39.
Identify which one of the following is an example for many to one relationship?
(a) student with exam number
(b) many staff members in one department
(c) customer, products
(d) Books and students
Answer:
(b) many staff members in one department

Question 40.
…………………… is a procedural query language used to query the database tables using SQL
Answer:
Relational Algebra

Question 41.
Find the wrong pair
(a) Union U
(b) cartesian product P
(c) project n
(d) select o
Answer:
(b) cartesian product P

Question 42.
Which method defines a relation that contains a vertical subset of relation?
(a) project
(b) select
(c) difference
(d) union
Answer:
(a) project

Question 43.
Duplicate row is removed in ……………………….
(a) o
(b) π
(c) x
(d) –
Answer:
(b) π

Question 44.
…………………. is used to merge columns from two relations.
(a) σ
(b) π
(c) x
(d) –
Answer:
(c) x

Question 45.
…………………. is security from unauthorized users.
Answer:
Data Integrity

Question 46.
…………………. means duplication of data in a database.
Answer:
Redundancy.

PART- II
II. Answer The Following Questions

Question 1.
Differentiate data from Information?
Answer:
Data:
Data are raw facts stored in a computer. A data may contain any character, text, word or a number.
Example:
600006, DPI Campus, SCERT, Chennai, College Road

Information:
Information is formatted data, which allows to be utilized in a significant way. It gives meaningful information.
Example:
SCERT
College Road
DPI Campus
Chennai 600006

Question 2.
Define database?
Answer:
Database is a repository collection of related data organized in a way that data can be easily accessed, managed and updated. Database can be a software or hardware based, with one sole purpose of storing data.

Question 3.
Name the components of DBMS?
Answer:
Components of DBMS:
The Database Management System can be divided into five major components as follows:

1. Hardware
2. Software
3. Data
4. Procedures/Methods
5. Database Access Languages

Question 4.
Define Table?
Answer:
Table is the entire collection of related data in one table, referred to as a File or Table where the data is organized as row and column.

Question 5.
Define row and column?
Answer:
Each row in a table represents a record, which is a set of data for each database entry. Each table column represents a Field, which groups each piece or item of data among the records into specific categories or types of data. Eg. StuNo., StuName, StuAge, StuClass, StuSec.

Question 6.
What is meant by data model?
Answer:
Data Model

1. A data model describes how the data can be represented and accessed from a software after complete implementation
2. It is a simple abstraction of complex real world data gathering environment.
3. The main purpose of data model is to give an idea as how the final system or software will look like after development is completed..

Question 7.
Write note on relational Algebra?
Answer:
Relational Algebra is a procedural query language used to query the database tables using SQL. Relational algebra operations are performed recursively on a relation (table) to yield an output.

PART – III
III. Answer The Following Questions

Question 1.
Define DBMS?
Answer:
A DBMS is a software that allows us to create, define and manipulate database, allowing users to store, process and analyze data easily. DBMS provides us with an interface or a tool, to perform various operations to create a database, storing of data and for updating data, etc. DBMS also provides protection and security to the databases. It also maintains data consistency in case of multiple users.

Question 2.
What are the advantages of DBMS?
Answer:
Advantages of DBMS

1. Segregation of application program
2. Minimal data duplication or Data Redundancy
3. Easy retrieval of data using the Query Language
4. Reduced development time and maintenance

PART – IV
IV. Answer The Following Questions

Question 1.
Explain the Various Components of DBMS?
Answer:
Components of DBMS:
The Database Management System can be divided into five major components as follows:

1. Hardware
2. Software
3. Data
4. Procedures/Methods
5. Database Access Languages

1. Hardware:
The computer, hard disk, I/O channels for data, and any other physical component involved in storage of data

2. Software:
This main component is. a program that controls everything. The DBMS software is capable of understanding the Database Access Languages and interprets into database commands for execution.

3. Data:
It is that resource for which DBMS is designed. DBMS creation is to store and utilize data.

4. Procedures/Methods:
They are general instructions to use a database management system such as installation of DBMS, manage databases to take backups, report generation, etc.

5. DataBase Access Languages:
They are the languages used to write commands to access, insert, update and delete data stored in any database.

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 and score more marks in your examinations.

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

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 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-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part-I

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

Question 1.
Gibbs free energy change for the electrolysis process is expressed by
(a) ∆G° = – nFE°
(b) ∆G° = – nF
(c) ∆G° = – nE°
(d) ∆G° = nFE°
Answer:
(a) ∆G° = – nFE°

Question 2.
Match items in column -II with the items of column – II and assign the correct code.

Answer:
(a)
A – 2
B – 1
C – 4
D – 3

Question 3.
Shape of ClF₃ is  ……………..
(a) Linear
(b) T-shape
(c) Pyrimidal
(d) Square planar
Answer:
(b) T-shape

Question 4.
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to ……………..
(a) their magnetic behaviour
(b) their unfilled d orbitals
(c) their ability to adopt variable oxidation states
(d) their chemical reactivity
Answer:
(c) their ability to adopt variable oxidation states

Question 5.
Which one of the following pairs represents linkage isomers?
(a) [CU(NH₃)₄] [PtCl₄] and [Pt(NH₃)₄] [CuCl₄]
(b) [CO(NH₃)₅(NO₃)]SO₄ and [CO(NH₃)₅(ONO)]
(c) [CO(NH₃)₄(NCS)₂]Cl and [CO(NH₃)₄(SCN)₂]Cl
(d) both (b) and (c)
Answer:
(c) [CO(NH₃)₄(NCS)₂]Cl and [CO(NH₃)₄(SCN)₂]Cl’

Question 6.
Each atom in the comer of the cubic unit cell is shared by how many unit cells?
(a) 8
(b) 6
(c) 1
(d) 12
Answer:
(a) 8

Question 7.
The rate constant of a reaction is 5.8 × 10^-2 s^-1 . The order of the reaction is ……………..
(a) First order
(b) zero order
(c) Second order
(d) Third order
Answer:
(a) First order
Solution:
The unit of rate constant is s”1 and it indicates that the reaction is first order.

Question 8.
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by…………………
(a) [H⁺] = \(\frac{\mathbf{K}_{\mathrm{a}}[\mathrm{a} \mathrm{cid}]}{[\mathrm{salt}]}\)
(b) [H⁺] = K_a[salt]
(c) [H⁺] = K_a[acid]
(d) [H⁺] = \(\frac{\mathrm{K}_{\mathrm{a}}[\mathrm{salt}]}{[\text { acid }]}\)
According to Henderson equation

Question 9.
Kohlrausch’s law is applied to calculate ……………..
(a) molar conductance at infinite dilution of a weak electrolyte
(b) degree of dissociation of weak electrolyte
(c) solubility of a sparingly soluble salt
(d) all the above
Answer:
(d) all the above

Question 10.
For freudlich isotherm a graph of \(\log \frac{x}{\mathrm{m}}\) is plotted against log P. The slope of the line and its y – axis intercept respectively corresponds to ………
(a) 1/n , k
(b) log 1/n ,k
(c) 1/n , log k
(d) log 1/n, log k
Answer:
(c) 1/n , log k
Solution:

Question 11.
]The reaction of sodium methoxide with ethyl bromide follows
(a) SN¹ mechanism
(b) SN² mechanism
(c) E₁ reaction
(d) E₂ reaction
Answer:
(b) SN² mechanism

Question 12.
In which of the following reactions new carbon – carbon bond is not formed?
(a) Aldol condensation
(b) Friedel craft reaction
(c) Kolbe’s reaction
(d) Wolf kishner reduction
Answer:
(d) Wolf kishner reduction

Question 13.
The reagent used to convert Nitromet’nane to methyl amine is
(a) Zn/NH₄Cl
(b) Sn/HCl
(c) H₂SO₅
(d) H₂S₂O₈
Answer:
(b) Sn/HCl

Question 14.
If one strand of the DNAhas the sequence ‘ATGCTTGA’, then the sequence of complementary
strand would be
(a) TACGAACT
(b) TCCGAACT
(c) TACGTACT
(d) TACGRAGT
Answer:
(a) TACGAACT

Question 15.
The ratio between the maximum tolerated dose of a drug and a minimum curative dose is called ……………..
(a) iso electric point
(b) therapeutic index
(c) critical point
(d) iso thermal point
Answer:
(b) therapeutic index

Part – II

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

Question 16.
Describe the role of Sodium cyanide in froth floatation.
Answer:

• Sulphide ores are concentrated by the froth floatation process.
• Depressants are used to prevent certain type of particles from forming the froth.
• NaCN act as a depressant to separate ZnS from PbS.

Question 17.
Which is more stable? Fe³⁺ or Fe²⁺ – explain.
Fe (Z = 26) Fe → Fe²⁺ + 2e^–
Fe → Fe³⁺ + 3e^–
Fe²⁺ [Number of electrons 24]
Electronic configuration = [Ar]3d⁶
Fe³⁺ [Number of electrons 23]
Electronic configuration = [Ar]3d⁵
Among Fe³⁺ and Fe²⁺, Fe³⁺ is more stable due to half filled d-orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So Fe³⁺ is more stable than Fe²⁺ .

Question 18.
Write briefly about the applications of coordination compounds in volumetric analysis.
Answer:
Hardness of water is due to the presence of Ca²⁺ and Mg²⁺ions in water. EDTA forms stable complexes with Ca²⁺ and Mg²⁺ . So the total hardness of water can be estimated by simple volumetric titration of water with EDTA.

Question 19.
Classify the following solids
(a) P4 (b) Brass
(c) Diamond
(d) NaCl
(e) Iodine
Answer:
(a) P₄ – Molecular solid
(b) Brass – Metallic solid
(c) Diamond – Covalent solid
(d) NaCl – Ionic solid .
(e) Iodine – Molecular solid

Question 20.
A lab assistant prepared a solution by adding a calculated quantity of HC1 gas 25°C to get a solution with [H₃O⁺]= 4 × 10^-5 s M. Is the solution neutral (or) acidic (or) basic.
Answer:
[H₃O⁺] = 4 × 10^-55 M
pH = -log₁₀[H₃O⁺]
PH = – log₁₀ [4 × 10^-5]
pH = -log₁₀[4]-log_1o[10^-5]
pH = – 0.6020 – (-5) = – 0.6020 + 5
pH = 4.398
Therefore, the solution is acidic.

Question 21.
0.1M copper sulphate solution in which copper electrode is dipped at 25°C. Calculate the electrode potential of copper. [Given: E° _Cu²⁺|Cu = 0.341
Answer:
Given that
[Cu²⁺] = 0.1M
E° _Cu²⁺|Cu = 0.34
E_Cell = ?
Cell reaction is Cu²⁺(aq) + 2e^– → Cu (s)

Question 22.
Can we use nucelophiles such as NH3,CH₃O^– for the Nucleophilic substitution of alcohols
Answer:
Increasing order of nucelophilicity,

• Higher electron density will increase the nucelophilicity.
• Negatively charged species are almost always more nucelophiles than neutral species.
• ROe has an alkyl group attached, allowing a greater amount of polarizability. This means oxygen’s lone pairs will be more readily available to reach in \(\mathrm{RO}^{\ominus}\) than in \(\mathrm{OH}^{\ominus}\) . Hence CH₃O^– is the better nucelophile for the nucleophilic substitution of alcohols.
• NH₃ cannot act as nucelophile for the nucleophilic substitution of alcohols.

Question 23.
Aniline does not undergo Friedel – Crafts reaction. Explain.
Answer:
Aniline being a Lewis base reacts with Lewis acid AlCl₃ to form a salt.

Due to the presence of a positive charge on N-atom in the salt the group – NH₂ AlCl^– acts as a strongly deactivating group. As a result, it reduces the electron density in the benzene ring and which inhibits the electrophilic substitution reaction. Therefore aniline does not undergo Friedel – Crafts reaction.

Question 24.
What are hormones? Give examples.
Answer:
Hormone is an organic substance that is secreted by one tissue into the blood stream and induces a physiological response in other tissues. It is an inter cellular signaling molecule. Virtually every process is a complex organism is regulated by one or more hormones. Example, insulin, epinephrine, estrogen, androgen etc.

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
Write a note on zeolites.
Answer:
Zeolites:

• Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxygen in their regular three dimensional framework.
• They are hydrated sodium alumino silicates with general formula.
  Na₂O.(Al₃O₃).x(SiO₂)y(H₂O)
  (x – 2 to 10; y = 2 to 6)
• Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
• The Si and A1 atoms are tetrahederally coordinated with each other through shared oxygen atoms.
• Zeolites structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
• Zeolite crystal act as a molecular sieve. They help to remove permanent hardness of water.

Question 26.
What is Royal water? Mention its uses.
Answer:
When three parts of concentrated hydrochloric acid and one part of concentrated nitric acid are mixed, aquaregia is obtained. This is also known as Royal water. This is used for dissolving gold, platinum etc.
Au + 4H⁺ + NO₃^– + 4 Cl^– → AUCl₄^– + NO + 2H₂O
3Pt + 16H⁺ + 4NO₃^– + 18Cl^– → 3PtCl₃^2- + NO + 8H₂O

Question 27.
Ni(CN)₄]^2- is diamagnetic, while [NiCl₄]^2- is paramagnetic , explain using crystal field theory.
Answer:
(a) [Ni(CN)₄]^2-
Ni = 3d⁸ 4S²
Ni²⁺ = 3d⁸

• Nature of the complex – Low spin (Spin paired)
• Ligand filled electronic configuration of central metal ion, t⁶_2g e⁶_g
• Magnetic property : No unpaired electron (CN^– is strong filled ligand), hence it is diamagnetic.
• Magnetic moment : μ_s = 0

(b) [NiDl₄]^2-
Ni = 3d⁸ 4S²
Ni²⁺ = 3d⁸

• Nature of the complex, – high spin
• Ligand filled electronic configuration of central metal ion, t⁶_2g e²_g
• Magnetic property : Two unpaired electron (CF is weak field ligand). Hence it is paramagnetic
• Magnetic moment : μ_s= \(\sqrt{2(2+2)}\) = √8 = 2.83BM

Question 28.
Calculate the number of atoms per unit cell of bcc type.
Answer:

• In a body centered cubic unit cell, each comer is occupied by an identical particle and in addition to that one atom occupied the body centre.
• Those atoms which occupy the comers do not touch each other, however they all touch the one that occupies the body centre.
• Hence each atom is surrounded by eight nearest neighbours and coordination number is 8. An atom present at the body centre belongs to only a particular unit cell, i.e., unshared by other unit cell.

∴ number of atoms in a bcc unit cell = \(\frac{N_{c}}{8}+\frac{N_{b}}{1}=\frac{8}{8}=\frac{1}{1}\)
= 1 + = 2.

Question 29.
Identify the order for the following reactions
(i) Rusting of Iron
(ii) Radioactive disintegration of ₉₂U²³⁸
(iii) 2A + B → products ; rate = K[A]^1/2[B]²
Answer:
(i)
Theoritically order value may be more than one but practically one.
(ii) All radioactive disintegrations are first order reactions
(iii) 2 A + 3B → products
rate = K[A]^1/2[B]²
Order = \(\frac{1}{2}+2=\frac{5}{2}=2.5\)

Question 30.
Calculate the (i) hydrolysis constant, (ii) degree of hydrolysis and (iii) pH of 0.05M sodium carbonate solut
Answer:
(i) Hydrolysis constant:

Given Kw = 1 × 10^-14
c = 0.05M
pK_a = 10.26
pK = -log k_a
K_a = antilog of (-pK_a)
K_a = antilog of (-10.26)
K_a = 5.49 × 10^-11

Question 31.
Write the structure of the aldehyde, carboxylic acid and ester that yield 4 – methylpent -2-en-l-ol.
Answer:
(i) Aldehyde yield 4-methylpent-2-en-l-ol is,

(ii) Acid yield 4-methylpent-2-en-l-ol is,

(iii) Ester yield 4-methylpent-2-en-l-ol is,

The above shown compounds udergo reduction reaction to yield 4-methylpent-2-en-l-ol.

Question 32.
Predict the major product that would’ be obtained on nitration of the following compounds.

Answer:

Question 33.
What are the biological functions of nucleic acids?
Answer:

• Energy carriers (ATP)
• Components of enzyme cofactors. Example : Co enzyme A, NAD⁺, FAD
• Chemical messengers. Example : Cyclic AMP, CAMP

Part – IV

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

Question 34.
(a) (i) Explain the observations from the Ellingham diagram. (3)
(ii) Write a short note on anamolous properties of the first element of p-block. (2)
[OR]
(b) (i) Write the products formed in the reaction of concentrated nitric acid with zinc. (2)
(ii) d-block elements readily form complexes. Give reason. (3)
Answer:
(a) (i) 1. For most of the metal oxide formation, the slope is positive. It can be explained as follows.
Oxygen gas is consumed during the formation of metal oxides which results in the decrease in randomness. Hence, ∆S becomes negative and it makes the term, T∆S positive in the straight line equation.

2. The graph for the formation of carbon monoxide is a straight line with negative slope. In this case ∆S is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas. It indicates that CO is more stable at higher temperature.

3. As the temperature increases, generally ∆G value for the formation of the metal oxide become less negative and becomes zero at a particular temperature. Below this temperature, ∆G is negative and the oxide is stable and above this temperature ∆G is positive. This general trend suggests that metal oxides become less stable at higher temperature and their decomposition becomes easier.

4. There is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO. This is due to the phase transition (melting or evaporation).

(ii) In p-block elements the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour.

• Small size of the first member.
• High ionisation enthalpy and high electronegativity.
• Absence of d-orbitals in their valance shell.

The first member of the group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows diagonal relationship with silicon of group 14. The oxides of boron and silicon are similar in their acidic nature.

[OR]

(b) (i) Zinc with Cone.

(ii) 1. Transition elements (d-block elements) have a tendency to form coordination • compounds (complexes) with a species that has an ability to donate an electron pair to
form a coordinate covalent bond.

2. Transition metal ions are small and highly charged and they have vacant low energy orbitals to accept an electron pair donated by other groups. Due to these properties, transition metals form large number of complexes.

3. Examples: [Fe(CN)₆]^4- , [CO(NH₃)₆]³⁺

Question 35.
(a) (i) Write the IUPAC names for the following complexes. (2)
1. Na₂[Ni(EDTA)]
2. [CO(en)₃]₂(SO₄)₃
(ii) What is meant by piezo electricity? (3)
[OR]
(b) (i) Consider the oxidation of nitric oxide to form NO₂ (3)
2NO_(g) + O_2(g) 2NO_2(g)
(a) Express the rate of the reaction in terms of changes in the concentration of NO, O₂ and NO ₂
(b) At a particular instant, when [O₂]is decreasing at 0.2 mol L^-1s^-1 at what rate is [NO₂] increasing at that instant?
(ii) Classify the following as acid (or) base using Arrhen ¡us concept (2)
1. HNO₃ 2. Ba(OH)₂ 3. H₃PO₄ 4. CH₃COOH
Answer:
(a) (1) 1. Na₂[Ni(EDTA)] = Sodium Ethylenediaminetetraacetatonickelate (Il) (or)
Sodium 2,2′,2”,2”’-(ethane- I ,2-diyldinitrilo) tetraacetatonickelate(II)
2. [CO(en)₃]₂(SO₄)₃ = tris(ethylenediamine)cobalt(III) sulphate

(ii) Piezo electricity is the appearance of an electrical potential across the sides of a crystal. When you subject it to mechanical stress. The word piezo electricity means electricity resulting from pressure and latent heat. Even the inverse is possible which is known as inverse piezo electric effect.

(b) (i)

(ii) 1. HNO₃ : Nitric acid, dissociates to give hydrogen ions in water. .. HNOS is acid.
2. Ba(OH)₂ : Barium hydroxide, dissociates to give hydroxyl ions in water.
∴ Ba(OH)₂ is base.
3. H₃PO₄: Orthophosphoric acid, dissociates to give hydrogen ions in water.
∴ H₃PO₄ is acid.
4. CH₃COOH : Acetic acid, dissociates to give hydrogen ions in water.
∴ CH₃COOH is acid.

Question 36.
(a) Derive an expression for Nernst equation. (5)
[OR]
(b) Describe adsorption theory of catalysis. (5)
(a) Nernst equation is the one which relates the cell potential and the concentration of the specie:
involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
xA+yB ⇌ lC+mD
The reaction quotient Q is,

We know that,
∆G = ∆G°+RT In Q ……..(1)
∴ ∆G = -nFE_cell ;∆G°=-nFE°_cell
∴ equation (1) becomes
– nFE_cell = nFE°_cell + RT lnQ ………(2)
Substitute the Q value in equation (2)

This is called the Nernst equation.
At 25°C (298 K) equation (4) becomes,

[OR]

(b) Adsorption theory:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also be called as contact catalysis.
According to this theory, the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes and gives the product.
The various steps involved in a heterogeneous catalysed reaction are given as follows:

• Reactant molecules diffuse from bulk to the catalyst surface.
• The reactant molecules are adsorbed on the surface of the catalyst.
• The adsorbed reactant molecules are activated and form activated complex which is decomposed to form the products.
• The product molecules are desorbed.
• The product diffuse away from the surface of the catalyst.

Advantages of adsorption theory:
The adsorption theory explains the following

• Increase in the activity of a catalyst by increasing the surface area. Increase in the surface area of metals and metal oxides by reducing the particle size increases the rate of the reaction.
• The action of catalytic poison occurs when the poison blocks the active centres of the catalyst.
• A promoter or activator increases the number of active centres on the surfaces.

Question 37.
(a) (i) What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis. (3)
(ii) What are essential and non-essential amino acids? Give one example of each type. (2)
[OR]
(b) How are the following conversions effected (5)
1. propanal into butanone
2. Hex-3-yne into hexan-3-one
3. phenylmethanal into benzoic acid
4. phenylmethanal into benzoin
Answer:
(a)

(ii) Essential amino acids: Amino acids which are not synthesised by the human body are called essential amino acids. Example: Valine, Leucine.
Non-essential amino acids: Amino acids which are synthesised by human body are called non-essential amino acids. Example: Glycine, Aspartic acid, etc.

[OR]

(b) 1. Propanal into butanone:

2. Hex-3-yne into hexan-3-one:

3. Phenylmeihanal into benzoic acid:

4. Phenyl methanal into benzoin:

Question 38.
(a) Identify A to E in the following frequency. of reactions. (5)

[OR]
(b) (i) What are the biological importance of proteins? (3)
(ii) Name one substance which can act as both analgesic and antipyretic. (2)
(a)

[OR]

(b) (i) Proteins are the functional units of living things play vital role in all biological processes

• All biochemical reactions occur in the living systems are catalysed by the catalytic proteins called enzymes.
• Proteins such as keratin, collagen acts as structural back bones.
• Proteins are used for transporting molecules (Haemoglobin), organelles (Kinesins) in the cell and control the movement of molecules in and out of the cells (Transporters).
• Antibodies help the body to fight various diseases.
• Proteins are used as messengers to coordinate many functions. Insulin & glucagon controls the glucose level in the blood.
• Proteins act as receptors that detect presence of certain signal molecules and activate the proper response.
• Proteins are also used to store metals such as iron (Ferritin) etc.

(ii) Aspirin (acetylsalicylic acid) is a chemical substance which lowers body temperature (to normal) and also reduces body pain. Therefore it acts as both antipyretic and analgesic.

Students can Download Computer Science Chapter 12 Structured Query Language (SQL) Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 12 Structured Query Language (SQL)

Samacheer Kalvi 12th Computer Science Structured Query Language (SQL) Text Book Back Questions and Answers

PART – 1
I. Choose The Best Answer

Question 1.
Which commands provide definitions for creating table structure, deleting relations, and modifying relation schemes?
(a) DDL
(b) DML
(c) DCL
(d) DQL
Answer:
(a) DDL

Question 2.
Which command lets to change the structure of the table?
(a) SELECT
(b) ORDER BY
(c) MODIFY
(d) ALTER
Answer:
(d) ALTER

Question 3.
The command to delete a table is ……………………..
(a) DROP
(b) DELETE
(c) DELETE ALL
(d) ALTER TABLE
Answer:
(a) DROP

Question 4.
Queries can be generated using …………………….
(a) SELECT
(b) ORDER BY
(c) MODIFY
(d) ALTER
Answer:
(a) SELECT

Question 5.
The clause used to sort data in a database ……………………….
(a) SORT BY
(b) ORDER BY
(c) GROUP BY
(d) SELECT
Answer:
(b) ORDER BY

PART – II
II. Answer The Following Questions

Question 1.
Write a query that selects all students whose age is less than 18 in order wise?
Answer:
SELECT * FROM STUDENT WHERE AGE <= 18 ORDER BY NAME.

Question 2.
Differentiate Unique and Primary Key constraint?
Answer:
The unique constraint ensures that no two rows have the same value in the specified columns. Primary key constraint declares a field as a Primary key which helps to uniquely identify a record. The primary key is similar to unique constraint except that only one field of a table can be set as primary key.

Question 3.
Write the difference between table constraint and column constraint?
Column constraint:
Column constraint apply only to individual column.

Table constraint:
Table constraint apply to a group of one or more columns.

Question 4.
Which component of SQL lets insert values in tables and which lets to create a table?
Answer:
Insert values in tables – DML Create a table – DDL

Question 5.
What is the difference between SQL and MySQL?
Answer:
SQL-Structured Query Language is a language used for accessing databases while MySQL is a database management system, like SQL Server, Oracle, Informix, Postgres, etc. MySQL is a RDBMS.

PART – III
III. Answer The Following Questions

Question 1.
What is a constraint? Write short note on Primary key constraint?
Answer:
Constraints are used to limit the type of data that can go into a table. This ensures the accuracy and reliability of the data in the database. Constraints could be either on a column level or a table level.

Primary Key Constraint:
This constraint declares a field as a Primary key which helps to uniquely identify a record. It is similar to unique constraint except that only one field of a table can be set as primary key. The primary key does not allow NULL values and therefore a field declared as primary key must have the NOT NULL constraint.
Example showing Primary Key Constraint in the student table:
CREATE TABLE Student
(
Admno integer NOT NULL PRIMARY KEY, → Primary Key constraint
Name char(20)NOT NULL,
Gender char(l),
Age integer,
Place char(10),
);

Question 2.
Write a SQL statement to modify the student table structure by adding a new field?
Answer:
ALTER TABLE Student MODIFY Address char(25);

Question 3.
Write any three DDL commands
a. CREATE TABLE Command
Answer:
You can create a table by using the CREATE TABLE command.
CREATE TABLE Student
(Admno integer,
Name char(20), \
Gender char(1),
Age integer,
Place char(10),
);

b. ALTER COMMAND
The ALTER command is used to alter the table structure like adding a column, renaming the existing column, change the data type of any column or size of the column or delete the column from the table.
Alter table Student add address char;

c. DROP TABLE:
Drop table command is used to remove a table from the database.
DROP TABLE Student;

Question 4.
Write the use of Savepoint command with an example?
SAVEPOINT command
Answer:
The SAVEPOINT command is used to temporarily save a transaction so that you can rollback to the point whenever required. The different states of our table can be saved at anytime using different names and the rollback to that state can be done using the ROLLBACK command.
SAVEPOINT savepoint_name;
UPDATE Student SET Name = ‘Mini ’ WHERE Admno=105;
SAVEPOINT A;

Question 5.
Write a SQL statement using DISTINCT keyword?
Answer:
DISTINCT Keyword:
The DISTINCT keyword is used along with the SELECT command to eliminate duplicate rows in the table. This helps to eliminate redundant data. For Example:
SELECT DISTINCT Place FROM Student;
Will display the following data as follows :
SELECT * FROM Student;
Output
Place
Chennai
Bangalore
Delhi

PART – IV
IV. Answer The Following Questions

Question 1.
Write the different types of constraints and their functions?
Answer:
Type of Constraints
Constraints ensure database integrity, therefore known as database integrity constraints. The different types of constraints are :

(i) Unique Constraint
This constraint ensures that no two rows have the same value in the specified columns. For example UNIQUE constraint applied on Admno of student table ensures that no two students have the same admission number and the constraint can be used as:
CREATE TABLE Student
(
Admno integer NOT NULL UNIQUE, → Unique constraint
Name char (20) NOT NULL,
Gender char (1),
Age integer,
Place char (10),
);
The UNIQUE constraint can be applied only to fields that have also been declared as NOT NULL.
When two constraints are applied on a single field, it is known as multiple constraints. In the above Multiple constraints NOT NULL and UNIQUE are applied on a single field Admno, the constraints are separated by a space and at the end of the field definition a comma(,) is added. By adding these two constraints the field Admno must take some value ie. will not be NULL and should not be duplicated.

(ii) Primary Key Constraint
This constraint declares a field as a Primary key which helps to uniquely identify a record. It is similar to unique constraint except that only one field of a table can be set as primary key. The primary key does not allow NULL values and therefore a field declared as primary key must have the NOT NULL constraint.
Example showing Primary Key Constraint in the student table:
CREATE TABLE Student
(
Admno integer NOT NULL PRIMARY KEY, → Primary Key constraint
Name char(20)NOT NULL,
Gender char(I),
Age integer,
Place char(10),
);
In the above example the Admno field has been set as primary key and therefore will help us to uniquely identify a record, it is also set NOT NULL, therefore this field value cannot be empty.

(iii) DEFAULT Constraint
The DEFA ULT constraint is used to assign a default value for the field. When no value is given for the specified field having DEFAULT constraint, automatically the default value will be assigned to the field.
Example showing DEFAULT Constraint in the student table:
CREATE TABLE Student
(
Admno integer NOT NULL PRIMARY KEY,
Name char(20)NOTNULL,
Gender char(1),
Age integer DEFAULT = “17”, → Default Constraint
Place char(10),
);
In the above example the “Age” field is assigned a default value of 17, therefore when no value is entered in age by the user, it automatically assigns 17 to Age.

(iv) Check Constraint:
This constraint helps to set a limit value placed for a field. When we define a check constraint on a single column, it allows only the restricted values on that field. Example showing check constraint in the student table:
CREATE TABLE Student
(
Admno integer NOT NULL PRIMARY KEY
Name char(20)NOTNULL,
Gender char(1),
Age integer (CHECK<=19),
→ Check Constraint
Place char(10),
);
In the above example the check constraint is set to Age field where the value of Age must be less than or equal to 19.
Note
The check constraint may use relational and logical operators for condition.

(v) TABLE CONSTRAINT
When the constraint is applied to a group of fields of the table, it is known as Table constraint. The table constraint is normally given at the end of the table definition. Let us take a new table namely Studentl with the following fields Admno, Firstname, Lastname, Gender, Age, Place: CREATE TABLE Student 1
Admno integer NOT NULL,
Firstname char(20),
Lastname char(20),
Gender char(1),
Age integer,
Place char(10),
PRIMARY KEY (Firstname, Lastname) → Table constraint
);
In the above example, the two fields, Firstname and Lastname are defined as Primary key which is a Table constraint.

Question 2.
Consider the following employee table. Write SQL commands for the qtns.(i) to (v)?
Answer:

1. To display the details of all employees in descending order of pay.
2. To display all employees whose allowance is between 5000 and 7000.
3. To remove the employees who are mechanic.
4. To add a new row.
5. To display the details of all employees who are operators.

Output:

1. SELECT * FROM employee ORDER BY DESC;
2. SELECT * FROM employee WHERE ((allowance >= 5000) AND(allowance <= 7000));
3. DELETE FROM employee WHERE desig = “Mechanic”;
4. INSERT INTO employee(Empcode, Name, desig, pay, allowance) VALUES(‘M1006’, ‘RAM’, ‘Mechanic’,22000, 8000);
5. SELECT * FROM employee WHERE desig = ‘operator’;

Question 3.
What are the components of SQL? Write the commands in each? Components of SQL?
SQL commands are divided into five categories:
Answer:

a. Data Definition Language
The Data Definition Language (DDL) consist of SQL statements used to define the database structure or schema. It simply deals with descriptions of the database schema and is used to create and modify the structure of database objects in databases.
SQL commands which comes under Data Definition Language are:

b. Data Manipulation Language
A Data Manipulation Language (DML) is a computer programming language used for adding (inserting), removing (deleting), and modifying (updating) data in a database.
SQL commands which comes under Data Manipulation Language are :

c. Data Control Language:
A Data Control Language (DCL) is used for controlling privileges in the database SQL commands: GRANT, REVOKE

d. Transactional Control Language:
Transactional control language (TCL) is used to manage transactions i.e. changes made to the data in the database.
SQL commands: COMMIT, ROLLBACK, SAVEPOINT

e. Data Query Language:
The Data Query Language (DQL) have commands to query or retrieve data from the database.
SQL commands: SELECT.

Question 4.
Construct the following SQL statements in the student table –
(i) SELECT statement using GROUP BY clause.
(ii) SELECT statement using ORDER BY clause.

(i) GROUP BY clause
The GROUP BY clause is used with the SELECT statement to group the students on rows or columns having identical values or divide the table into groups. For example to know the number of male students or female students of a class, the GROUP BY clause may be used. It is mostly used in conjunction with aggregate functions to produce summary reports from the database.
The syntax for the GROUP BY clause is
SELECT <column-names> FROM <table-name> GROUP BY <column-name>HAVING condition];
To apply the above command on the student table :
SELECT Gender FROM Student GROUP BY Gender;
The following command will give the below given result:

SELECT Gender, count(*) FROM Student GROUP BY Gender;

(ii) ORDER BY clause
The ORDER RTclause in SQL is used to sort the data in either ascending or descending based on one or more columns.
1. By default ORDER BY sorts the data in ascending order.
2. We can use the keyword DESC to sort the data in descending order and the keyword ASC to sort in ascending order.
The ORDER BY clause is used as:
SELECT <column-name> [,<column-name>,….] FROM <table-name> ORDER
BY <column1>,<column2>, …ASC\ DESC;
For example :
To display the students in alphabetical order of their names, the command is used as
SELECT * FROM Student ORDER BY Name;
The above student table is arranged as follows :

Question 5.
Write a SQL statement to create a table for employee having any five fields and create a table constraint for the employee table?
Answer:
CREATE TABLE EMPLOYEE
(Empcode integer NOT NULL,
Name char(20),
desig char(20),
pay integer,
allowance integer,
PRIMARY KEY(Name, desig));

Practice Programs

Question 1.
Create a query of the student table in the following order of fields name, age, place and admno?
Answer:
CREATE TABLE Student(Name char(30), age integer, place char(30), admno integer)).

Question 2.
Create a query to display the student table with students of age more than 18 with unique city?
Answer:
SELECT * FROM student WHERE age >= 18 GROUP BY city.

Question 3.
Create a employee table with the following fields employee number, employee name, designation, date of joining and basic pay?
Answer:
CREATE TABLE employee(empNo integer, ename char(30), desig char(30), doj datetime, basic integer);

Question 4.
In the above table set the employee number as primary key and check for NULL values in any field?
Answer:
CREATE TABLE employee(empno integer NOT NULL PRIMARY KEY, ename char(30) NOT NULL, desig char(30), doj datetime, basic integer).

Question 5.
Prepare a list of all employees who are Managers?
Answer:
SELECT * FROM employee WHERE desig = ’Managers’.

Samacheer kalvi 12th Computer Science Structured Query Language (SQL) Additional Questions and Answers

PART – I
I. Choose the correct answer

Question 1.
The SQL was called as …………………….. in early 1970’s.
(a) squel
(b) sequel
(c) seqel
(d) squeal
Answer:
(b) sequel

Question 2.
SQL means ………………………..
Answer:
Structured Query Language

Question 3.
RDBMS Expansion ………………………
Answer:
Relational DataBase Management System

Question 4.
Expand ANSI ………………………
(a) American North South Institute
(b) Asian North Standard Institute
(c) American National Standard Institute
(d) Artie National Standard Institute
Answer:
(c) American National Standard Institute

Question 5.
ANSI Published SQL standard in the year ………………………..
(a) 1986
(b) 1982
(c) 1984
(d) 1989
Answer:
(a) 1986

Question 6.
The latest SQL was released in ……………………….
(a) 1987
(b) 1992
(c) 2008
(d) 2012
Answer:
(c) 2008

Question 7.
The latest SQL standard as of now is …………………….
Answer:
SQL 2008

Question 8.
Identify which is not a RDBMS package …………………….
(a) MySQL
(b) IBMDB2
(c) MS-Access
(d) Php
Answer:
(d) Php

Question 9.
A ……………………… is a collection of tables.
Answer:
database

Question 10.
CRUD means ……………………
(a) creative reasoning under development
(b) create read update delete
(c) create row update drop
(d) calculate relate update data
Answer:
(b) create read update delete

Question 11.
A …………………… is a collection of related data entries and it consist of rows and columns.
Answer:
table

Question 12.
……………………….. is the vertical entity that contains all information associated with a specific field in a table
(a) Field
(b) tuple
(c) row
(d) record
Answer:
(a) Field

Question 13.
A ………………………… is a horizontal entity in the table.
Answer:
record

Question 14.
DDL means …………………………
Answer:
Data Defnition Language

Question 15.
Match the following:
1. DDL – (i) Modify Tuples
2. Informix – (ii) Create Indexes
3. DML – (iii) MySQL
4. DCL – (iv) Grant
(a) 1-ii, 2-iii, 3-i, 4-iv
(b) 1-i, 2-ii, 3-iii, 4-iv
(c) 1-iv, 2-iii, 3-ii, 4-i
(d) 1-iv, 2-i, 3-ii, 4-iiii
Answer:
(a) 1-ii, 2-iii, 3-i, 4-iv

Question 16.
The SQL used in high level programming languages is ………………………….
Answer:
Embedded Data Manipulation Language

Question 17.
WAMP stands for ……………………….
Answer:
Windows, Apache, MySQ1 and PHP

Question 18.
To work with the databases, the command used is …………………….. database
(a) create
(b) modify
(c) use
(d) work
Answer:
(c) use

Question 19.
Which is used to serve live websites?
(a) WAMP
(b) SAMP
(c) DAMP
(d) TAMP
Answer:
(a) WAMP

Question 20.
How many components of SQL are there?
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(c) 5

Question 21.
Which among the following is not a WAMP?
(a) PHP
(b) MySQL
(c) DHTML
(d) Apache
Answer:
(c) DHTML

Question 22.
Which is used to define database structure or schema?
(a) DML
(b) DDL
(c) DCL
(d) DQL
Answer:
(b) DDL

Question 23.
Identify which is not a SQL DDL command?
(a) create
(b) delete
(c) drop
(d) truncate
Answer:
(b) delete

Question 24.
Which command changes the structure of the database?
(a) update
(b) alter
(c) change
(d) modify
Answer:
(b) alter

Question 25.
Identify which statement is given wrongly?
(a) DDL statement should specify the proper data type
(b) DDL should not identify the type of data division
(c) DDL may define the range of values
(d) DDL should define the size of the data item
Answer:
(b) DDL should not identify the type of data division

Question 26.
Identify which is wrong?
DML means
(a) Insertion
(b) Retrieval
(c) Modification
(d) alter
Answer:
(d) alter

Question 27.
How many types of DML commands are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 28.
Pick the odd out
Insert, update, alter, delete
Answer:
alter

Question 29.
Grant and Revoke commands comes under ……………………..
(a) DML
(b) DCL
(c) DQL
(d) DDL
Answer:
(b) DCL

Question 30.
…………………….. is a DQL command
(a) select
(b) commit
(c) update
(d) delete
Answer:
(a) select

Question 31.
Which one restores the database to last commit state?
(a) commit
(b) Grant
(c) rollback
(d) save point
Answer:
(c) rollback

Question 32.
Which is used to query or retrieve data from a database?
(a) DQL
(b) DML
(c) DCL
(d) DCM
Answer:
(a) DQL

Question 33.
Variable width character string is given by the data type ……………………….
(a) char
(b) varchar
(c) dec
(d) real
Answer:
(b) varchar

Question 34.
If the precision exceeds 64, then it is
(a) float
(b) real
(c) float
(d) decimal
Answer:
(c) float

Question 35.
…………………… have special meaning in SQL
(a) keywords
(b) commands
(c) clauses
(d) arguments
Answer:
(a) keywords

Question 36.
…………………… are the values given to make the clause complete
Answer:
Arguments

Question 37.
Each table must have at least ………………………. column
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 38.
Which one of the following ensures the accuracy and reliability of the data in the database?
(a) Arguments
(b) constraints
(c) column
(d) clauses
Answer:
(b) constraints

Question 39.
How many types of constraints are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 40.
The …………………….. constraint can be applied only to fields that have also been declared a NOT Null.
Answer:
unique

Question 41.
When two constraints are applied on a single field, it is known as ……………………….. constraints.
Answer:
multiple

Question 42.
Which key helps to uniquely identify the record in the table?
(a) unique
(b) primary
(c) secondary
(d) null
Answer:
(b) primary

Question 43.
Which constraint is used to assign a default value for the field?
(a) unique
(b) primary
(c) secondary
(d) default
Answer:
(d) default

Question 44.
The check constraint may use ………………….. operators for condition.
(a) relational
(b) logical
(c) both
(d) None of these
Answer:
(c) both

Question 45.
When the constraint is applied to a group of fields of the table, then it is ………………………. constraint.
(a) table
(b) column
(c) multiple
(d) primary
Answer:
(a) table

Question 46.
The …………………….. command is used to insert, delete and update rows into the table.
(a) DCL
(b) DML
(c) DTL
(d) TCL
Answer:
(b) DML

Question 47.
If the data is to be added for all columns in a table
(a) specifying column is optional
(b) specifying column is must
Answer:
(a) specifying column is optional

Question 48.
Find the wrong statement from the following delete command
(a) permanently removes one or more records
(b) removes entire row
(c) removes individual fields
(d) deletes the record
Answer:
(c) removes individual fields

Question 49.
The update command specifies the rows to be changed using the …………………….. clause.
(a) where
(b) why
(c) what
(d) how
Answer:
(a) where

Question 50.
Set keyword in update command is used to assign new data.
True / false
Answer:
True

Question 51.
Find the wrong one about alter command
(a) remove a column
(b) remove all columns
(c) rename a column
(d) delete row
Answer:
(d) delete row

Question 52.
The keyword …………………….. is used along with the select command to eliminate duplicate rows in the table.
Answer:
distinct

Question 53.
The ……………………. keyword in select command includes an upper value and a lower value.
Answer:
betweeen

Question 54.
How many types of sorting are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 55.
The default sorting order is ……………………….
Answer:
ascending

Question 56.
……………………. clause is used to filter the records.
Answer:
where

Question 57.
The ……………………… clause is used to select the group of students on rows or columns having identical values.
Answer:
group by

Question 58.
Which is to count the records?
(a) +
(b) *
(c) =
(d) /
Answer:
(b) *

PART – II
II. Answer The Following Questions

Question 1.
Write note on RDBMS?
Answer:
RDBMS stands for Relational DataBase Management System. Orqcle, MySQL, MS SQL Server, IBM DB2 and Microsoft Access are RDBMS packages. RDBMS is a type of DBMS with a row-based table structure that connects related data elements and includes functions related to Create, Read, Update and Delete operations, collectively known as CRUD.

Question 2.
What does data manipulation means?
Answer:
By Data Manipulation we mean,

1. Insertion of new information into the database
2. Retrieval of information stored in a database.
3. Deletion of information from the database.
4. Modification of data stored in the database.

Question 3.
What are the 2 types of DML?
Answer:
The DML is basically of two types:
Procedural DML – Requires a user to specify what data is needed and how to get it. Non-Procedural DML – Requires a user to specify what data is needed without specifying how to get it.

Question 4.
What is meant by data type?
Answer:
The data in a database is stored based on the kind of value stored in it. This is identified as the data type of the data or by assigning each field a data type. All the values in a given field must be of same type.

Question 5.
Write about ALL keyword in select?
Answer:
ALL Keyword
The ALL keyword retains duplicate rows. It will display every row of the table without considering duplicate entries.
SELECT ALL Place FROM Student:
The above command will display all values of place field from every row of the table without considering the duplicate entries.

PART – III
III. Answer The Following Questions

Question 1.
Write note on SQL?
Answer:

1. The Structured Query Language (SQL) is a standard programming language to access and manipulate databases.
2. SQL allows the user to create, retrieve, alter, and transfer information among databases.
3. It is a language designed for managing and accessing data in a Relational Data Base Management System (RDBMS).

Question 2.
What are the various processing skills of SQL?
Answer:
The various processing skills of SQL are :
(i) Data Definition Language (DDL) :
The SQL DDL provides commands for defining relation schemes (structure), deleting relations, creating indexes and modifying relation schemes.

(ii) Data Manipulation Language (DML) :
The SQL DML includes commands to insert, delete, and modify tuples in the database.

(iii) Embedded Data Manipulation Language :
The embedded form of SQL is used in high level programming languages.

(iv) View Defintion :
The SQL also includes commands for defining views of tables.

(v) Authorization :
The SQL includes commands for access rights to relations and views of tables.

(vi) Integrity :
The SQL provides forms for integrity checking using condition.

(vii) Transaction control :
The SQL includes commands for file transactions and control over transaction processing.

Question 3.
How to create and work with database?
Answer:
Creating Database
(i) To create a database, type the following command in the prompt:
CREATE DATABASE database_name;
For example to create a database to store the tables:
CREATE DATABASE stud;

(ii) To work with the database, type the following command.
USE DATABASE;
For example to use the stud database created, give the command
USE stud;

Question 4.
What are the functions performed by DDL?
Answer:
A DDL performs the following functions :

1. It should identify the type of data division such as data item, segment, record and database file.
2. It gives a unique name to each data item type, record type, file type and data base.
3. It should specify the proper data type.
4. It should define the size of the data item.
5. It may define the range of values that a data item may use.
6. It may specify privacy locks for preventing unauthorized data entry.

Question 5.
Name the SQL Commands under TCL. Explain?
Answer:
SQL command which come under Transfer Control Language are:

Question 6.
Write about the parts of SQL Commands?
Answer:
Keywords They have a special meaning in SQL. They are understood as instructions.
Commands They are instructions given by the user to the database also known as statements.
Clauses They begin with a keyword and consist of keyword and argument.
Arguments They are the values given to make the clause complete.

Question 7.
Write note on delete command?
Answer:
DELETE COMMAND
The DELETE command permanently removes one or more records from the table. It removes the entire row, not individual fields of the row, so no field argument is needed. The DELETE command is used as follows :
DELETE FROM table-name WHERE condition;
For example to delete the record whose admission number is 104 the command is given as follows:
DELETE FROM Student WHERE Admno=104;

The following record is deleted from the Student table.
To delete all the rows of the table, the. command is used as :
DELETE * FROM Student;
The table will be empty now and could be destroyed using the DROP command.

Question 8.
Write note on delete, truncate, drop commands?
Answer:
DELETE, TRUNCATE AND DROP statement:
The DELETE command deletes only the rows from the table based on the condition given in the where clause or deletes all the rows from the table if no condition is specified. But it does not free the space containing the table.
The TRUNCATE command is used to delete all the rows, the structure remains in the table and free the space containing the table.
The DROP command is used to remove an object from the database. If you drop a table, all the rows in the table is deleted and the table structure is removed from the database. Once a table is dropped we cannot get it back.

Question 9.
Differentiate between and not between?
Answer:
BETWEEN and NOT BETWEEN Keywords
The BETWEEN keyword defies a range of values the record must fall into to make the condition true. The range may include an upper value and a lower value between which the criteria must fall into.
SELECT Admno, Name, Age, Gender FROM Student WHERE Age BETWEEN 18 AND 19;

The NOT BETWEEN is reverse of the BETWEEN operator where the records not satisfying the condition are displayed.
SELECT Admno, Name, Age FROM Student WHERE Age NOT BETWEEN 18 AND 19;

Question 10.
Differentiate IN and NOT IN
Answer:
IN Keyword
The IN keyword is used to specify a list of values which must be matched with the record values. In other words it is used to compare a column with more than one value. It is similar to an OR condition.
For example:
SELECT Admno, Name, Place FROM Student WHERE Place IN (“Chennai, “Delhi ”);

NOT IN: The NOT IN keyword displays only those records that do not match in the list. For example:
SELECT Admno, Name, Place FROM Student WHERE Place NOT IN (“Chennai”, ‘ “Delhi”);
will display students only from places other than “Chennai” and “Delhi”.

Question 11.
Write note on NULL?
Answer:
NULL Value:
The NULL value in a field can be searched in a table using the IS NULL in the WHERE c lause. For example to list all the students whose Age contains no value, the command is used as:
SELECT * FROM Student WHERE Age IS NULL.

Question 12.
Write note on Roll Back?
Answer:
ROLLBACK command
The ROLLBACK command restores the database to the last committed state. It is used with SAVEPOINT command to jump to a particular savepoint location. The syntax for the ROLLBACK command is :
ROLL BACK TO Save point name.

Question 13.
Write note on having clause?
Answer:
HAVING clause:
The HAVING clause can be used along with GROUP BY clause in the SELECT statement to place condition on groups and can include aggregate functions on them. For example to count the number of Male and Female students belonging to Chennai.
SELECT gender, COUNT(*) FROM Student GROUP BY Gender HAVING Place = ‘Chennai’;

The above output shows the number of Male and Female students in Chennai from the table student.

PART – IV
IV. Answer The Following Questions

Question 1.
Write about data type and description?
Answer:

Question 2.
Write about DML Commands?
Answer:
DML COMMANDS
Once the schema or structure of the table is created, values can be added to the table. The DML commands consist of inserting, deleting and updating rows into the table.

(i) INSERT command
The INSERT command helps to add new data to the database or add new records to the table. Syntax:
INSERT INTO <table-name> [column-list] VALUES (values);

(a) INSER T INTO Student (Admno, Name, Gender, Age, Place)
VALUES (100, ‘Ashish ’, ‘M\ 17, ‘Chennai);

(b) INSERT INTO Student (Admno, Name, Gender, Age, Place)
VALUES (10, ‘Adarsh’, ‘M’, 18, ‘Delhi);

(c) INSERT INTO Student VALUES (102, ‘Akshith \ ‘M’, ‘17, ’ ‘Bangalore);
The above command inserts the record into the student table.
To add data to only some columns in a record by specifying the column name and their data, it can be done by:

(d) INSERT INTO Student(Admno, Name, Place) VALUES (103, ‘Ayush’, ‘Delhi’);

(e) INSERT INTO Student (Admno, Name, Place) VALUES (104, ‘Abinandh ‘Chennai); The student table will have the following data:

(ii) DELETE COMMAND
The DELETE command permanently removes one or more records from the table. It removes the entire row, not individual fields of the row, so no field argument is needed. The DELETE command is used as follows:
DELETE FROM table-name WHERE condition;
For example to delete the record whose admission number is 104 the command is given as follows:
DELETE FROM Student WHERE Admno=104;

The following record is deleted from the Student table.
To delete all the rows of the table, the command is used as :
DELETE * FROM Student;
The table will be empty now and could be destroyed using the DROP command.

(iii) UPDATE COMMAND
The UPDATE command updates some or all data values in a database. It can update one or more records in a table. The UPDATE command specifies the rows to be changed using the WHERE clause and the new data using the SET keyword. The command is used as follows: UPDATE <table-name> SET column-name = value, column-name = value,… WHERE condition;
For example to update the following fields:
UPDATE Student SET Age = 20 WHERE Place = “Bangalore ”;
The above command will change the age to 20 for those students whose place is “Bangalore”.
The table will be as updated as below:

To update multiple fields, multiple field assignment can be specified with the SET clause separated by comma. For example to update multiple fields in the Student table, the command is given as:
UPDATE Student SETAge=18, Place = ‘Chennai’ WHERE Admno = 102;

The above command modifies the record in the following way.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Chemistry 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. 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-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part-I

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

Question 1.
The following set of reactions are used in refining Zirconium

This method is known as ……………..
(a) Liquation
(b) Van Arkel process
(c) Zone refining
(d) Mond’s process
Answer:
(b) Van Arkel process

Question 2.
The stability of+1 oxidation state increases in the sequence ……………
(a) Al < Ga < In < Tl
(b) Tl < In < Ga < Al
(c) In < Tl < Ga < Al
(d) Ga < In < Al < Tl
Answer:
(a) A1 < Ga < In < Tl

Question 3.
Assertion : bond dissociation energy of fluorine is greater than chlorine gas.
Reason: chlorine has more electronic repulsion than fluorine.
(a) Both assertion and reason are true and reason is the correct explanation of assertion
(b) Both assertion and reason are true but reason is not the correct explanation of assertion
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(d) Both assertion and reason are false

Question 4.
Which of the following pair has d¹⁰ electrons?
(a) Ti³⁺ , V⁴⁺
(b) CO³⁺ , Fe²⁺
(c) Cu⁺, Zn²⁺
(d) Mn²⁺, Fe³⁺
Answer:
(c) Cu⁺, Zn²⁺

Question 5.
Which is used for the separation of lanthanides, in softening of hard water and also in removing lead poisoning?
(a) [Ni(CO)₄]
(b) EDTA
(c) [Ni(DMG)₂]
(d) TiCl₄ + Al (C₂H₅ )₃
Answer:
(b) EDTA

Question 6.
The yellow colour in NaCl crystal is due to ……………..
(a) excitation of electrons in F centers
(b) reflection of light from CF ion on the surface
(c) refraction of light from Na+ ion
(d) all of the above
Answer:
(a) excitation of electrons in F centers

Question 7.
For a reaction Rate = k[acetone] then unit of rate constant and rate of reaction respectively is
(a) (mol L^-1 s^-1),(mol^-1/2 L^1/2 s^-1)
(b) (mol^-1/2 L^-1/2 s^-1) , (mol L^-1 s^-1)
(c) (mol^-1/2 L^-1/2 s^-1), (mol L^-1 s^-1)
(d) (mol L s^-1), (mol^-1/2 L^-1/2 s)
Answer:

In this case, rate = k [Acetone]^3/2
n= 3/2
mol^1-(3/2)L^(3/2)-1s^-1 ⇒ mol^-(1/2)L^(1/2)s^-1

Question 8.
Arrange the acids
(i) H₂SO₃
(ii) H₃PO₃ and
(iii) HClO₃ in the decreasing order of acidity.

(a) (i) > (iii) > (ii)
(b) (i) > (ii) > (iii)
(c) (ii) > (iii) > (i)
(d) (iii) > (i) > (ii)
Answer:
(d) (iii) > (i) > (ii)
Acidity is directly proportional to oxidation number. As the oxidation number of S, P and Cl in H₂SO₃, H₃PO₃ and HClO₃ is +4, +3, +5 respectively. So decreasing order of acidity will be (iii) > (i) > (ii)

Question 9.
The value of cell emf of Mercury button cell is –
(a) 1.35 V
(b) – 0.76 V
(c) 0.34 V
(d) 100 V
Answer:
(a) 1.35 V

Question 10.
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K₂SO₄ (i), Na₃O₄ (ii), K₄[Fe(CN)₆] (iii) and NaCl (iv). Their coagulating power should be ……….
(a) II > I > IV > III
(b) III > II > I > IV
(c) I > II > III > IV
(d) none of these
Answer:
(b) III > II > I > IV

Question 11.
Which compound has the highest boiling point?
(a) Acetone
(b) Diethyl ether
(c) Methanol
(d) Ethanol
Answer:
(d) Ethanol

Question 12.
The reagent used to distinguish between acetaldehyde and benzaldehyde is
(a) Tollens reagent
(b) Fehling’s solution
(c) 2,4 – dinitrophenyl hydrazine
(d) semicarbazide
Answer:
(b) Fehling’s solution

Question 13.

(a) bromomethane
(b) a – bromo sodium acetate
(c) methanamine
(d) acetamide

Answer:
(c) methanamine

Question 14.
Match the following.

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

Question 15.
Regarding cross-linked or network polymers, which of the following statement is incorrect?
(a) Examples are Bakelite and melamine
(b) They are formed from bi and tri-functional monomers
(c) They contain covalent bonds between various linear polymer chains
(d) They contain strong covalent bonds in their polymer chain
Answer:
(d) They contain strong covalent bonds in their polymer chain

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 × 2 = 12]

Question 16.
Give the uses of zinc.
Answer:

• Metallic zinc is used in galvanising metals such as iron and steel structures to protect them . from rusting and corrosion.
• Zinc is also used to produce die-castings in the automobile, electrical and hardware industries.
• Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles and electrical equipment. Zinc sulphide is used in making luminous paints, fluorescent lights and x-ray screens.
• Brass an alloy of zinc is used in water valves and communication equipment as it is highly resistant to corrosion.

Question 17.
Explain why fluorine always exhibit an oxidation state of-1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibit an oxidation state of-1 and others in halogen family shows +1, +3, +5 and +7 oxidation states.

Question 18.
What is Zeigler -Natta catalyst? In which reaction it is used? Give equation.
Answer:
A mixture of TiCl₄ and trialkyl aluminium is Zeigler – Natta catalyst. It is used in the polymerization.

Question 19.
Give any three characteristics of ionic crystals.
Answer:

• Ionic solids have high melting points.
• These solids do not conduct electricity, because the ions are fixed in their lattice positions. ‘
• They do conduct electricity in molten state (or) when dissolved in water because, the ions are free to move in the molten state or solution.

Question 20.
How is surface area of the reactant affect the rate of the reaction?
Answer:

• In heterogeneous reactions, the surface area of the solid reactants play an important role in deciding the rate.
• For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased.
• For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO₃ as marble.

Question 21.
K_sp of Al(OH)₃ is 1 x 10^-15M. At whatpH does 1.0 × 10^-3 3M Al³⁺ precipitate on the addition of buffer of NH₄Cl and NH₄OH solution?
Answer:
Al(OH)₃ ⇌ Al³⁺ (aq) + 3OH^– (aq)
K_sp – [Al³⁺ ][OH^–]³
Al(OH)₃ precipitates when
[Al³⁺ ] [OH^– ]³ > K_sp
(1 × 10^-3 )[OH^– ]³ > 1 × 10^-15
[OH^–]^-3] > 1 × 10^-12
[OH^– ] > 1 × 10^-4 M
[OH^– ] = 1 × 10^-4 M
pOH = – log_1o[OH^–] = – log(1 × 10^-4) = 4
pH = 14 – 4 = 10
Thus, Al(OH)₃ precipitates at a pH of 10

Question 22.
Arrange the following in the increasing order of their boiling point and give a reason for your ordering: Butan – 2- ol, Butan -l-ol, 2 -methylpropan -2-ol
Answer:
Boiling points increases regularly as the molecular mass increases due to a corresponding increase in their Van der waal’s force of attraction. Among isomeric alcohols 2°-alcohols have lower boiling points than 1 “-alcohols due to a corresponding decreases in the extent of H-bonding because of steric hindrance. Thus the boiling point of Butan – 2- ol is lower than that of Butan – l-ol. Overall increasing order of boiling points is,
2-methylpropan-2-ol < Butan-2- ol < Butan – l-ol

Question 23.
What are the uses of Benzaldehyde?
Answer:

• as a flavoring agent
• in perfumes
• in dye intermediates
• as starting material for the synthesis of several other organic compounds like cinnamaldehyde, cinnamic acid, benzoyl chloride etc.

Question 24.
What are anti fertility drugs? Give examples.
Answer:
Artificially drugs are chemical substances which suppress the action of hormones that promote pregnancy. These drugs actually reduce the chances of pregnancy and act as a protection. Antifertility drugs are made up of derivatives of synthetic progesterone or a combination of derivatives of oestrogen and progesterone.
Example : Ethynylestradiol, menstranol and norethynodrel etc.

Part – III

Answer any six questions. Question No. 30 is compulsory. [6 ×3 = 18]

Question 25.
Give the uses of silicones.
Answer:
Uses of silicones:

• Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
• They are used for making water proofing clothes
• They are used as insulting material in electrical motor and other appliances
• They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

Question 26.
Complete the following reactions.
1. NaCl + MnO₂ + H₂SO₄ →
2. I₂ + S₂O₃^2-
3. P₄ + NaOH + H₂O →
Answer:
1. 4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O
2. I₂ + 2S₂O₃^2- → S₄O₆^2- + 2I^–
3. P₄ + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃↑

Question 27.
Draw and explain about the structure of chromate and dichromate ion,

Answer:

• Both chromate and dichromate ions are oxo anions of chromium and they are moderately strong oxidising agents.
• In both structures, chromium is in +6 oxidation state.
• In an aqueous solution, chromate and dichromate ions can be inter convertible, and in an alkaline solution, chromate ion is predominant, whereas dichromate ion becomes predominant in acidic solutions.

Question 28.
Why ionic crystals are hard and brittle?
Answer:
The ionic compounds are very hard and brittle. In ionic compounds the ions are rigidly held in a lattice because the positive and negative ions are strongly attracted to each other and difficult to separate. But the brittleness of a compound is now easy to shift the position of atoms or ions in a lattice. If we apply a pressure on the ionic compounds the layers shifts slightly. The same charged ions in the lattice comes closer. A repulsive forces arises between same charged ions, due to this repulsions the lattice structure breaks down chemical bonding.

Question 29.
Paracetamol is prescribed to take once in 6 hours. Justify this statement.
Answer:

• Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever
  and body pain.
• Paracetamol has a half life of 2.5 hours within the body, (i.e) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided.
• In the case of paracetamol, it is usually prescribed to take once in 6 hours.

Question 30.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0. 15 amperes. The mass of the metal deposited at the cathode is 0.783g. Calculate the equivalent mass of the metal.
Answer:
Given, I = 0.15 amperes
t = 150 mins ⇒ t = 150 x 60 sec ⇒ t = 9000 sec
Q = It ⇒ Q = 0.15 x 9000 coulombs ⇒ Q = 1350 coulombs Hence, 135 coulombs of electricity deposit is equal to \(\frac{0.783 \times 96500}{1350}=55.97\) g of metal.
∴ Hence equivalent mass of the metal is 55.97

Question 31.
What happens when
i. 2 – Nitropropane boiled with HCl
ii. Nitrobenzen electrolytic reduction in strongly acidic medium.
Answer:
i. 2 – Nitropropane boiled with HCl: 2-nitropropane upon hydrolysis with boiling HCl give a ketone (2-propanone) and nitrous oxide.

ii. Nitrobenezen electrolytic reduction in strongly acidic medium: Electrolytic reduction of nitrobenzene in weakly acidic medium gives aniline but in strongly acidic medium, it gives p-aminophenol obviously through the acid – catalysed rearrangement of the initially formed phenylhydroxylamine.

Question 32.
Write a short note on peptide bond.
Answer:

• The amino acids are linked covalently by peptide bonds.
• The carbonyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage (-CONH) between these aminoacids. This amide linkage is called peptide bond.

• The resulting compound is called a dipeptide. Because, two amino acids are inovlved for getting one peptide bond.
• If large number of amino acids combined through peptide bond, the resulting giant molecule is called a protein.
• The amino end of the peptide is known as N-terminal, while the carboxy end is called C – terminal.

Question 33.
(i) What class of drug is Ranitidine?
(ii) If water contains dissolved Ca ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
(iii) Which of the following is an antisepctic? 0.2% phenol, 1% phenol.
Answer:
(i) It is an antacid.
(ii) In this case we use synthetic detergents because it give foam with hard water.
(iii) 0.2% solution of phenol acts as antiseptic.

Part – IV

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

Question 34.
(a) (0 Explain the concentration of copper pyrites and galena ores. (3)
(ii) Out of LU(OH)₃ and La(OH)₃ which is more basic and why? (2)
(b) Explain the preparation of silicones. (5)
Answer:
(a) (i) Froth floatation: This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc blende (ZnS) etc. In this method, the metallic ore particles which are preferentially wetted by oil can be separated from gangue.

In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. A small quantity of sodium ethyl xanthate which acts as a collector is also added. A froth is generated by blowing air through this mixture.

The collector molecules attach to the ore particle and make them water repellent. As a result, ore particles, wetted by the oil, rise to the surface along with the froth. The froth is skimmed off and dried to recover the concentrated ore. The gangue particles that are preferentially wetted by water settle at the bottom.

When a sulphide ore of a metal of interest contains other metal sulphides as impurities, depressing agents such as sodium cyanide, sodium carbonate etc are used to selectively prevent other metal sulphides from coming to the froth. For example, when impurities such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the floatation property of ZnS by forming a layer of zinc complex Na₂[Zn(CN)₄] on the surface of zinc sulphide.

(ii) 1. As we move from Ce³⁺ to Lu³⁺, the basic character of Lu³⁺ ions decreases.
2. Due to the decrease in the size of Lu³⁺ ions, the ionic character of Lu – OH bond decreases, covalent character increases which results in the decrease in the basicity.
3. Hence, La(OH)₃ is more basic than Lu(OH)₃.

[OR]

(b) Generally silicones are prepared by the hydrolysis of dialkyldichlorosilanes (R₂SiCl₂ ) or diaryldichlorosilanes Ar₂SiCl₂, which are prepared by passing vapours of RCl or ArCl over silicon at 570 K with copper as a catalyst.

The hydrolysis of dialkylchloro silanes R₂SiCl₂ yields to a straight chain polymer which grown from both the sides

The hydrolysis of monoalkylchloro silanes RSiCl₃ yields to a very complex cross linked polymer. Linear silicones can be converted into cyclic or ring silicones when water molecules is removed from the terminal -OH groups.

Question 35.
(a) Bleaching action of chlorine is permanent – Justify this statement and also give the uses of chlorine. (5)
[OR]
(b) (i) Complete the following (3)


(ii) What is linkage isomerism? Explain with an example.
Answer:
(a) Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.
H₂O + Cl₂ → HCl + HOCl (Hypo chlorous acid)
HOCl → HCl + (O)

Colouring matter + Nascent oxygen → Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.

Uses of chlorine:

• Purification of drinking water
• Bleaching of cotton textiles, paper and rayon
• It is used in extraction of gold and platinum

[OR]

(b) (i)

(ii) This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms.
For examples,
[CO(NH₃)₅ONO]Cl₂
(Pentaammine nitrito cobalt (III) chloride)
O – attached. (Red in colour).
[CO(NH₃)₅NO₂]Cl₂
(Pentaammine nitro cobalt (III) chloride)
N – attached (Yellow-brown in colour).

Question 36.
(a) (i) What is the two dimensional coordination number of a molecule in square close packed layer? (2)
(ii) Derive the integrated rate law for a first order reaction? (3)
[OR]
(b) (i) Define solubility product. (2)
(ii) What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate and making the volume equal to 500 ml. (Given: Ka for acetic acid is 1.8 × 10^-5) (3)
Answer:
(a) (i) Square close packing : When the spheres of the second ” row are placed exactly above those of the first row.
This way the spheres are aligned horizontally as well as vertically. The arrangement is AAA type. Coordination number is 4.
(ii) Areaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction.
First order reaction is A → product

Rate law can be expressed as, Rate = k [A]1
Where, k is the first order rate constant

Integrate the above equation (1) between the limits of time t = O and time equal to t,
while the concentration varies from initial concentration [A₀]to [A] at the later time.

This equation (2) is in natural logarithm. To convert it into usual logarithm with base 10,
we have to multiply the term by 2.303

[OR]

(b) (i) Solubility product : It is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co-efficient in a balanced equilibrium equation.

(ii) According to Henderson — Hessalbalch equation,

Question 37.
(a) (i) Why for CH₃COOH cannot be determined experimentally? (2)
(ii) Write about the classification of organic nitro compounds. (3)
[OR]
(b) Describe about condensation methods of preparation of colloids. (OR)
Describe chemical methods of preparation of colloids. (5)
Answer:
(a) (i) Molar conductivity of weak electrolytes keeps on increasing with dilution and does not
become constant even at very large dilution.

(ii)

[OR]

(b) When the substance for colloidal particle is present as small sized particle, molecule or ion, they are brought to the colloidal dimension by condensation methods.
i. Oxidation method:- When hydroiodic acid is treated with iodic acid I sol is obtained.
HIO₃ +5HI → 3H₂O + 3I _2(sol)

ii. Reduction method:- Gold sol is prepared by reduction of auric chloride using formaldehyde.
2 AuCl₃ + 3 HCHO + 3 H₂O (sol) → 2 Au_(sol) + 6HCl + 3HCOOH

iii. Hydrolysis:- Ferric chloride is hydrolysed to get ferric hydroxide colloid
FeCl₃ + 3H₂O → Fe(OH)₃ (sol) + 3HCl

iv. Double decomposition:- When hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
As₂O₃ + 3H₂ → As ₂S₃ + 3H₂O

v. Decomposition:- When few drops of an acid is added to a dilute solution of sodium thiosulphate, sulphur colloid is produced by the decomposition of sodium thio sulphate.
S₂O₃^2- + 2H⁺ → S_(sol) + H₂O + SO₂

Question 38.
(a) (i) What is Clemmensen reduction ? Explain it. (2)
(ii) Write the structure of the major product of the aldol condensation of benzaldehyde
with acetone. (3)
[OR]
(b) (i)How will you convert nitrobenzene into (2)
1. 1,3, 5 – trinitrobenzene 2. o añd p- nitrophenol
(ii) Differentiate between Globular and fibrous proteins. (3)
Answer:
(a) (i) Aldehyde and ketones when heated with zinc amalgam and concentrated hydrochloric acid give hydrocarbons. This reaction is known as Clemmensen reduction.

(ii) Aldol condensation of benzaldehyde with acetone:

(b) (i) 1. Conversion of nitrobenzene into 1,3,5 – trinitrobenzene:

2. Conversion of nitrobenzene into o and p- nitrophenol:
Nitrobenzene heated with solid KOH at 340 K gives a low yield of a mixture of O-and P-nitrophenols.

(ii) Difference between Globular and fibrous proteins.

No.	Globular proteins	Fibrous Proteins
i	They form a-helix structure	They have (3-pleated structure.
ii	They are soluble in water:	They are insoluble in water.
iii	They are cross lined condensation polymers of acidic and basic amino acids.	They are linear condensation polymeric proteins.

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Multiple Choice Questions

Question 1.
Select the molecule which has only one ir bond.
(a) CH₃-CH=CH-CH₃
(b) CH3-CH=CH-CHO
(c) CH3-CH=CH-COOH
(d) All of these
Answer:
(a) CH₃-CH=CH-CH₃

Question 2.
In the hydrocarbon the state of hybridisation of carbon 1,2,3,4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) none of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 3.
The general formula for alkadiene is ……….
(a) C_nH_2n
(b) C_nH_2n-1
(c) C_nH_2n-2
(d) C_nH_n-2
Answer:
(c) C_nH_2n-2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept-2-ene is ……….

Answer:

Question 5.
The IUPAC name of the compound is …………

(a) 2,3 – Diemethyiheptane
(b) 3 – Methyl – 4 – ethyloctane
(c) 5 – ethyl – 6 – methyloctanc
(d) 4 – Ethyl -3 – methyloctane.
Answer:
(d) 4 – Ethyl – 3 – methyloctane.

Question 6.
Which one of the following names does not fit a real name?
(a) 3 – Methyl – 3 – hexanone
(b) 4- Methyl – 3 – hexanone
(c) 3 – Methyl – 3 – hexanol
(d) 2 – Methyl cyclo hexanone.
Answer:
(a) 3 – Methyl – 3 – hexanone

Question 7.
The TUPAC name of the compound CH3— CII = CH -C CH is ……………
(a) Pent – 4 – yn – 2 – ene
(b) Pent -3-en – 1- yne
(c) pent – 2 – en – 4 – yne
(d) Pent – 1 – yn – 3 – ene
Answer:
(b) Pent -3-en – 1- yne

Question 8.
IUPAC name of is ……….
(a) 3, 4, 4 – Trimethylheptane
(b) 2 – Ethyl – 3, 3 – dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 Butyl – 2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of is ………
(a) 2 – Hydroxypropionic acid
(b) 2, 4. 4 – Trimethylpent -3-ene
(c) Propan – 2 – ol 1 – oie acid
(d) 2, 2, 4 – Trimethylpent -2-ene
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 10.
The IUPAC name of the compound is ………
(a) 3 – Ethyl – 2- hexene
(b) 3 – Propyl – 3. hexene
(c) 4 – Ethyl – 4 – hexene
(d) 3 – Propyl -2-hexenc
Answer:
(a) 3 – Ethyl – 2- hexene

Question 11.
The IUPAC name of the compound is ……..
(a) 2 – Hydroxypropionic acid
(b) 2 – Hydroxy Propanoic acid
(c) Propan 2 – ol – 1 – oic acid
(d) 1 – Carboxyethanol.
Answer:
(b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of is ………
(a) 2 – Bromo – 3 – methylbutanoic acid
(b) 2 – methyl – 3 – bromobutanoic acid
(c) 3 – Bromo – 2 – methylbutanoic acid
(d) 3 – Bromo – 2. 3 – dimethyl propanoic acid.
Answer:
(c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is ………

Answer:

Question 14.
The number of stereoisomers of 1, 2-dihydroxycyclopentane is ……..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 15.
Which of the following is optically active?
(a) 3 – Chioropentane
(b) 2 – Chioropropane
(c) Meso – tat-tat-ic acid
(d) Glucose
Answer:
(d) Glucose

Question 16.
The isomer of ethanol is ……….
(a) acetaldehyde
(b) dimethylether
(c) acetone
(d) methyl carbinol
Answer:
(b) dimethylether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C₃H₆O?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 18.
Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) CH₂Cl₂
Answer:
(c) ethanol

Question 19.
are ………
(a) resonating structure
(b) taulomers
(c) optical isomers
(d) conformers
Answer:
(b) tautomers

Question 20.
Nitrogen detection in an organic compound is earned out by Lassaigne’s test. The blue colour formed is due to the formation of ………….
(a) Fe₃[Fe(CN)₆]₂
(b) Fe₄[Fe(CN)₆]₃
(c) Fe₄[Fe(CN)₆]₂
(d) Fe₃[Fe(CN)₆]₃
Answer:
(b) Fe₄[Fe(CN)₆]₃

Question 21.
Lassaigne’s test for the detection of nitrogen fails in ………..
(a) H₂N -CO – NH. NH₂. HCl
(b) NH₂ – NH₂. HCl
(c) C₆H₅ – NH – NH₂. HCl
(d) C₆H₅CONH₂
Answer:
(c) C₆H₅ – NH – NH₂. HCl

Question 22.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) NH₂NH₂HCl and CICH₂ – CHO
(b) NH₂CS NH₂ and CH₃ – CH2Cl
(c) NH₂CH₂ COOH and NH₂CONH₂
(d) C₆H₅NH₂ and ClCH₂ – CHO
Answer:
(d) C₆H₅NH₂ and ClCH₂ – CHO

Question 23.
Sodium nitropruside reacts with suiphide ion to give a purple colour due to the formation of ………..
(a) [Fe(CN)₅N0]^3-
(b) [Fe(NO)₅CN]⁺
(c) [Fe(CN)₅NOS]^4-
(d) [Fe(CN)₅NOS]^3-
Answer:
(c) [Fe(CN)₅NOS]^4-

Question 24.
An organic compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the compound will be close to ……….
(a) 46%
(b) 34%
(c) 3.4%
(d) 4.6%
Answer:
(b) 34%

Question 25.
A sample of 0.5g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5M H₂SO₄. The remaining acid after neutralisation by ammonia consumed 80 mL of 0.5 M NaOH, The percentage of nitrogen in the organic compound is ……….
(a) 14%
(b) 28%
(c) 42%
(d) 56%
Answer:
(b) 28%

Question 26.
In an organm compound, phosphorus is estimated as ……….
(a) Mg₂P₂O₇
(b) Mg₃(PO₄)₂
(c) H₃PO₄
(d) P₂O₅
Answer:
(a) Mg₂P₂O₇

Question 27.
Ortho and para-nitro phenol can be separated by ………….
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer:
(c) steam distillation

Question 28.
The purity of an organic compound is determined by …………
(a) Chromatography
(b) Crystallisation
(c) melting or boiling point
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by …………
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer:
(b) distillation under reduced pressure

Question 30.
Assertion: is 3-carbethoxy -2- butenoicacid.
Reason: The principal functional group gets lowest number followed by double bond (or) triple bond.
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d both the assertion and reason are false.
Answer:
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Short Answer Questions

Question 31.
Give the general characteristics of organic compounds.
Answer:

• All organic compounds are covalent compounds of carbon and are insoluble in water and soluble in organic solvents.
• They are inflammable (except CCl₄).
• They possess low boiling and melting points due to their covalent nature.
• They are characterized by functional groups.
• They exhibit isomerism.

Question 32.
Describe the classification of organic compounds based on their structure.
classification of organic compounds based on the structure
Answer:

Question 33.
Write a note on the homologous series.
Answer:

• A series of organic compounds each containing a characteristic functional group and the successive members differ from each other in molecular formula by a CH₂ group is called homologous series.
• e.g., Alkanes, Methane (CH₄) ethane (C₂H₆), Propane (C₃H₈) etc.
• Compounds of the homologous series are represented by a general formula. e.g., Alkanes: C₂H_2nAlkene: C_nH_2n
• They can be prepared by general methods.
• They show regular gradation in physical properties but have almost similar chemical properties.

Question 34.
What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(c) dimethyl ether
(d) methylamine
Answer:
1. A functional group ¡s an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.

Question 35.
Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol – C_nH_2n+1 + OH
(b) Aliphatic ketones – C_nH_2nO
(c) Aliphatic amines – C₂H_2n+1NH₂

Question 36.
Write the molecular formula of the first six members of homologous series of nitroalkanes.
Nitroalkanes:
Answer:

• CH₂NO₂ Nitromethane
• CH₂-CH₂NO₂ Nitroethane
• CH₃-CH₂-CH₂NO₂ 1- nitropropane
• CH₃-CH₂-CH₂-CH₂-NO₂ 1- nitrobutane
• CH₃-CH₂-CH₂-CH₂-CH₂-NO₂ 1 – nitropentane
• CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-NO₂ 1- nitrohexane

Question 37.
Write the molecular and possible structural formula of the first four members of homologous series of carhoxylic acids.
Answer:

Question 38.
Give the IUPAC names of the following compounds.



Answer:
(i) 2,3,5-tnmethylhexane
(ii) 2-bromo-3-methylbutane
(iii) methoxymethane
(iv) 2-hydroxybutanal
(v) buta-1,3-diene
(vi) 4-chioropent-2-yne
(vii) 1 -bromobut-2-ene
(viii) 5-oxohexanoic acid
(ix) 3-ethyl-4-ethenylheptane
(x) 2, 4, 4-trimethylpent-2-ene
(xi) 2- methyl-I -phenyipropan- I -amine
(xii) 2,2- dimethyl-4-oxopentanenitrile
(xiii) 2-ethoxypropane
(xiv) I -fluoro-4-methyl-2-nitrobenzene
(xv) 3-bromo-2-methylpentanal

Question 39.
Give the structure for the following compound.
(i) 3 – ethyl – 2 methyl -1 – pentene
(ii) 1, 3, 5 – Tnmethyl cyclohex – 1 – ene
(iii) tertiary butyl iodide
(iv) 3 – Chlorobutanal
(y) 3 – Chlorobutanol .
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2-dimethyl- 1 – chioropropane
(viii) 3 – methylbut – 1 – ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1 ,3 – diene
(xi) 1 ,5 – Dimethylcyclohexane
(xii) 2 – Chlorobut – 3 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde
Answer:

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
Detection of Nitrogen:
The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Estimation of halogens:
Answer:
carius method:
1. A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.
2. C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide

3. The precipitate AgX is filtered, washed, dried and weighed.
4. From the mass of AgX and the mass of organic compound taken, the percentage of halogens are calculated.

Question 42.
Give a brief description of the principles of:
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This is one method to purify and separate liquids present in the mixture having their boiling point close to each other. In the fractional distillation, a fractionating column is fitted with distillation flask and a condenser. A thermometer is fitted in the fractionating column near the mouth of the condenser. This will enable to record the temperature of vapour passing over the condenser.

The process of separation of the components in a liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. The process of fractional distillation is repeated, if necessary. This method finds a remarkable application in the distillation of petroleum, coal-tar and crude oil.

2. Column Chromatography:
This is the simplest chromatographic method carried out in long glass column having a stop cock near the lower end. This method involves the separation of a mixture over a column of adsorbent (Stationery phase) packed in a column. In the column a plug of cotton or glass wool is placed at the lower end of the column to support the adsorbent powder. The tube is uniformly packed with suitable adsorbent constitutes the stationary phase. (Activated aluminum oxides (alumina), Magnesium oxide, starch are also used as adsorbents).

The mixture to be separated is placed on the top of the adsorbent column. Eluent which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Different components depending upon the degree to which the components are adsorbed and complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column.

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography:
1. It is an example of partition chromatography. A strip of paper acts as an adsorbent. This method involves continues differential partitioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as the chromatographic paper is used. This paper act as a stationary phase.

2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatograrn is developed.

3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Constitutional isomers:
These isomers having the same molecular formula but differ in their bonding sequence. It is classified into 6 types:
1. Chain (or) nuclear (or) skeletal isomerism:
The phenomenon in which the isomers have similar molecular formula but differ in the nature of carbon skeleton (i.e., straight (or)
branched)
e.g., C₅H₁₂:

2. Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
e.g., C₅H₁₀:

3. Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
e.g., C₃H₆O:
(i) CH₃-CH₂-CH₂-CH=CH₂ propanal (Aldehyde group)

4. Metamerism:
This isomerism anses due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to either side of the same functional group and having same molecular formula.
e.g., C₄H₁₀O:

5. Tautomerism:
It is an isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus generally hydrogen.
e.g., C₂H₄O:

6. Ring chain isomerism:
It is an isomerism in which compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures.
e.g., C₂H₆:

Question 45.
Describe optical isomerism with a suitable example.
Answer:
Compounds having same physical and chemical property but differ only in the rotation of plane of the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Example:
Some organic compounds such as glucose have the ability to rotate the plane of the plane-polarized light and there called are said to be optically active compounds and this property of a compound is called optical activity. The optical isomer, which rotates the plane of the plane polarised light to the right or in clockwise direction is said to be dextrorotary (dexter means right) denoted by the sign (+), whereas the compound which rotates to the left or anticlockwise is said to be leavorotatory (leavues means left) denoted by sign (-). Dextrorotatory compounds are represented as ‘d’ or by sign (+) and laevorotatory compounds are represented as ‘l’ or by sign (-).

Question 46.
Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
Answer:
2-butene: Geometrical isomerism : CH₃  -CH = CH – CH₃

1. Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

2. In 2-butene, the carbon-carbon double bond is sp2 hybridised. The carbon-carbon double bond consists of a a bond and a it bond. The presence of it bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.

4. These two compounds are termed as geometrical isomers and are termed as cis and trans form.

5. The cis-isomer is the one in which two similar groups arc on the same side otthe double bond. The trans-isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-Irans isomerism.

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Answer:
Weight of organic compound = 0.30 g
Weight of carbon-dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.54 g of water contain = \(\frac {2}{18}\) × 0.54 g of hydrogen
% of hydrogen = \(\frac {2}{18}\) × \(\frac {0.54}{0.30}\) × 100 = \(\frac {2}{18}\) × \(\frac {54}{0.3}\)
% of H = 0.111 × 180 = 19.888 ≈ 20%

Percentage of carbon:
44 g of CO₂ contains 12 g of carbon
0.88 g of CO₂ contains = \(\frac {12}{44}\) × 0.88 g of carbon
% of carbon = \(\frac {12}{44}\) × \(\frac {0.88}{0.30}\) × 100 = \(\frac {12}{44}\) × \(\frac {88}{0.3}\) = \(\frac {24}{0.3}\)
% of carbon = 80 % .

Question 48.
The ammonia evolved form 0.20 g of an organic compound by kjeldahl method neutralised 15m1 of N/20 Sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
weight of organic compound = 0.20 g
Normality of acid = \(\frac{\mathrm{N}}{20}\)
Volume of standard acid neitralized by ammonia = 15 ml
1000 ml of N ammonia contains = 14 g of nitrogen
15 ml of ammonia of normality \(\frac{\mathrm{N}}{20}\) contains nitrogen = \(\frac{14 \times 15 \times 1}{1000 \times 20}\)

0.20 g of compound contains nitrogen = \(\frac{14 \times 15}{1000 \times 20}\)

100 g of compound contains nitrogen = \(\frac{14 \times 15 \times 100}{1000 \times 20 \times 0.20}\) = 5.25 g
Percentage of nitrogen = 5.25 %

Question 49.
0.32 g of an organic compound. after heating with fuming nitric acid and barium nitrate crystals is a scaled tube gave 0.466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Answer:
Weight of organic compound = 0.32 g
Weight of BaSO₄ formed = 0.466 g
233 g of BaSO₄ contains = 32 g of sulphate
0.466 g of l3aSO₄ contain = \(\frac {32}{233}\) x \(\frac {0.466}{2.32}\) x 100
= \(\frac {32}{233}\) x \(\frac {46.6}{0.32}\) = 19.999 g of sulphur
% of sulphur = 20 %

Question 50.
024 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer:
Weight of organic compound = 0.24 g
Weight of silver chloride = 0.287 g
143.5 g of AgCl contains = 35.5 g of Cl
0.287 g of AgCl contains = \(\frac {35.5}{143.5}\) x 0.287 g of Cl
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.24}\) x 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15°C and 760 mm Hg pressure. Calculate the percentage of nitrogen ¡n the compound.
Answer:
Weight of organic compound = 0.35 g
Volume of moist nitrogen (V₁) = 20.7 ml = 20.7 x 10^-3 L
Temperature = T₁ = 15°C = 273 + 15°C = 288K
Pressure of moist nitrogen P₁ = 760 mmHg

V₀ = 19.62 x 10^-3L
Percentage of nitrogen = \(\frac {28}{22.4}\) x \(\frac{\mathrm{V}_{0}}{\mathrm{W}}\) x 100
= \(\frac {28}{22.4}\) x \(\frac{19.62 \times 10^{-3}}{0.35}\) x 100
= \(\frac {28}{22.4}\) x = \(\frac {19.62}{0.35}\) x 10^-1
= 56.05 x 10^-3 x 100 = 7.007%
Percentage of nitrogen = 7.007%

In Text Questions – Evaluate Yourself

Question 1.
Give two examples for each of the following type of organic compounds.
1. non-benzonoid aromatic
2. aromatic heterocyclic
3.  alicycic
4. aliphatic open chain.
Answer:
1. Non benzenoid aromatic compounds

2. Aromatic heterocyclic compounds

3. Alicyclic compounds

4. Aliphatic open chain compounds

• CH₃-CH₂-CH₂-CH₂-CH₃ n-pentane
• CH₃-CH₂-CH₂OH 1-propanol

Question 2.
Write structural formula for the following compounds
1. Cyclohexa-1, 4-diene
2. Ethynykyclohexane

Question 3.
Write structural formula for the following compounds
1. m – dinitrobenzene
2. p-dichlorobenzene
3. 1, 3, S- Trimethytbeuzene
Answer:

Question 4.
Write all the possible isomers of molecular formula C₄H₁₀O and identify the isomerisms found in them.
Answer:
C₄H₁₀O isomers:

Question 5.
0.2346 g of an organic compound containing C, H & O, o comhution giweb 0.2754 g of H₂O and 0.4488 g CO₂. Calculate the % composition of C, H & O in the organic compound.
Answer:
Weight of organic substance (w) = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO₂ (y) = 0.4488 g
Percentage of carbon = \(\frac {12}{44}\)  x  \(\frac {y}{w}\) x 100
= \(\frac {12}{44}\) – \(\frac {0.4488}{0.2346}\) x 100 = 52.17%
Percentage of hydrogen = \(\frac {2}{18}\) x \(\frac {y}{w}\) x 100
= \(\frac {2}{18}\) x \(\frac {0.2754}{0.2346}\) x 100 = 13.04%
Percentage of oxygen = [100- (52.17 +13.04)] = 100 – 65.21 = 34.79%

Question 6.
0.16 g of an organic compound was heated in a carlus tube and H₂SO₄ acid formed was precipitated with BaCl₄. The mass of BaSO₄ was 0.35 g. Find the percentage of sulphur.
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of Sulphur = \(\frac {32}{233}\) x \(\frac {x}{w}\) x 100
= \(\frac {32}{233}\) x \(\frac {0.35}{0.16}\) x 100 = 30.04%

Question 7.
0.185 g of an organic compound when treated with Conc. HNO₃ and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
Answer:
Weight of organic substance (w) 0.185 g ;
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\frac {80}{188}\) x \(\frac {x}{w}\) x 100 = \(\frac {80}{188}\) x \(\frac {0.32}{0.185}\) x 100 = 73.6%

Question 8.
0.40 g of an iodo-substituted organic compound gave 0.235 g of Agi by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127).
Answer:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
127 x 127 0.235
Percentage of iodine = \(\frac {127}{235}\) x \(\frac {x}{w}\) x 100
= \(\frac {x}{w}\) x \(\frac {0.235}{0.40}\) x 100 = 31.75%

Question 9.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound.
Answer:
Weight of organic compound = 0.33g ;
Weight of Mg,P,07 = 0.397g
222 g of Mg₂P₂O₇ contains 62 g of phosphorous.
0.397 g of Mg₂P₂O₇ will contain \(\frac {62}{222}\) x 0.397 g of P.
0.33 g of organic compound contains \(\frac {62}{222}\) x 0.397 g of P
100 g of organic compound will contain \(\frac {62}{222}\) x \(\frac {0.397}{0.33}\) x 100
= \(\frac {2,461.4}{73.26}\) = 33.59 %
Percentage of phosphorous = 33.59 %

Question 10.
0.3 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralise 30 mL of 0.1N H₂SO₄. Calculate the percentage of nitrogen in the compound.
Answer:
Weight of organic compound (w) = 0.3 g
Strength of sulphuric acid used (N) = 0.1 N
Volume of sulphuric acid used (V) = 30 mL
30 ml of 0.1 N sulphuric acid 30 ml of 0.1 N ammonia
Percentage of nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times w}\right)\) x 100
= \(\left(\frac{14 \times 0.1 \times 30}{1000 \times 0.3}\right)\) x 100 = 14%

Example Problems

Question 1.
Classify the following compounds based on the structure
1. CH≡C-CH₂-C≡CH
2. CH₃-CH₂-CH₂-CH₂-CH₃


Answer:

1. Unsaturated open chain compound
2. Saturated open chain compound
3. Aromatic benzenoid compound
4. Alicyclic compound

Question 2.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g
Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac {2}{18}\) x \(\frac {0.039}{0.26}\) of H
% of hydrogen = \(\frac {0.039}{0.26}\) x \(\frac {2}{18}\) x 100 = 1.66%
Percentage of carbon:
44 g of CO₂ contains 12 g of C
0.245 g of CO₂ contains = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) g of C
% of Carbon = \(\frac {12}{44}\) x \(\frac {0.245}{0.26}\) x 100 = 25.69%

Question 3.
In an estimation of sulphur by Carius method, 0.2 175 g of the substance gave 0.5825 g of BaSO₄, calculate the percentage composition of S ¡n the compound.
Answer:
Weight of organic compound = 0.2 175 g
Weight of BaSO₄ = 0.5825 g
233 g of BaSO₄ contains = 32 g of S
0.5825 g of BaSO₄ contains = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) g of S
Percentage of S = \(\frac {32}{233}\) x \(\frac {0.5825}{0.2175}\) x 100 = 36.78%

Question 4.
0.284 g of an organic substance gave 0.287 g AgCl in Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Answer:
Weight of the organic substance = 0.284 g
Weight of AgCl = 0.287 g
143.5 g of AgCl contains 35.5 g of chlorine
0.287 g of AgCl Contains = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\)
% of chlorine = \(\frac {35.5}{143.5}\) x \(\frac {0.287}{0.284}\) x 100 = 24.98%

Question 5.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of phosphorous ¡n the compound
Answer:
Weight of an organic compound = 0.24 g
Weight of Mg₂P₂O₇ = 0.66 g
222 g of Mg₂P₂O₇ contains = 62 g of P
o. 66 g contains = \(\frac {62}{222}\) x 0.66 g of P
Percentage of P = \(\frac {62}{222}\) x \(\frac {0.66}{0.24}\) x 100 = 76.80%

Question 6.
0.1688 g when analysed by the Dumas method yield 31.7 mL of moist nitrogen measured at 14°C and 758mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C =12 mm of Hg).
Answer:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V₁) = 31.7 ml. = 31.7 x 10^-3 L
Temperature (T₁) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C = 12 mm of Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{0} V_{0}}{T_{0}}\)
V₀ = \(\frac{746 \times 31.7 \times 10^{-3}}{287} \times \frac{273}{760}\)
V₀ = 29.58 x 10^-3 L
Percentage of nitrogen:

= 21.90 %

Question 7.
0.6 g of an organic compound was Kjeldhalised and NH₃ evolved was absorbed into 50 mL of semi-normal solution of H₂SO₄. The residual acid solution ws diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac {N}{2}\) NaOH solution for complete neutralisation. Calculate the % of N in the compound.
Answer:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric acid = \(\frac {35 x 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with ammonia = V₁ mL
Strength of H₂SO₄ = 0.5 N
Volume of the diluted H₂SO₄ = 150 mL
Strength of the diluted sulphuric acid = 0.0875 N
V₁ = \(\frac {150 x 0.087}{0.5}\) = 26.25 mL
Volume of H₂SO₂consumed by ammonia = 50 – 26.25 = 23.75 mL
23.75 mL of 0.5 N H₂SO₄ = 23.75 mL of 0.5N NH₃
The amount of Nitrogen present in the 0.6 g of organic compound
= \(\frac{14 \mathrm{g}}{1000 \mathrm{mL} \times 1 \mathrm{N}}\) x 23.75 x 0.5 N = 0.166 g
Percentage of Nitrogen \(\frac {0.166}{0.6}\) x 100 = 27.66%

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry Additional Questions Solved

I. Choose the correct answer.
Question 1.
Statement 1. The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation.
Statement 2. The high strength of C-C bond is responsible for its catenation property.
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.
(b) Statement 1 & 2 arc correct but statement 215 not the correct explanation of statement 1.
(c) Statement 1 is correct but statement 2 is wrong.
(d) Statement 1 is wrong but statement 2 is correct.
Answer:
(a) Statement 1 & 2 are correct and statement 2 is the correct explanation of statement 1.

Question 2.
Generally, organic compounds are
a) Amorphous
b) Complexes
c) Covalent
d) Electrovalent
Answer:
c) Covalent

Question 3.
Which of the following is an example of an organic reaction?
(a) Rusting of iron
(b) Combustion of magnesium
(c) Biochemical reactions
(d) All the above
Answer:
(c) Biochemical reactions

Question 4.
The first carbon compound prepared from its elements is
a) Urea
b) Acetic acid
c) Methane
d) benzene
Answer:
b) Acetic acid

Question 5.
Which of the following is an example of the non-benzenoid aromatic compound?
(a) Tolucnc
(b) Phenol
(c) Benzyl alcohol
(cl) azulene
Answer:
(d) azulene

Question 6.
The first organic compound synthesized in the laboratory from an inorganic compound is
a) NH₄NCO
b) NH₂ – CO – NH₂
c) CH₃COOH
d) CH₄
Answer:
b) NH₂ – CO – NH₂

Question 7.
Which of the following is an example of carbocyclic alicyclic compound?
(a) Pyrrole
(b) Thiophene
(c) Cyclopropane
(d) Phenol
Answer:
(c) Cyclopropane

Question 8.
Which one of the following is the functional group of ketone?
(a)-CHO

(c) -O-
(d)-OH
Answer:

Question 9.
Which one of the following indicates isothiocyanate functional group?
(a) -NC
(b) -NCS
(c) -SCN
(d) -NCO
Answer:
(b) -NCS

Question 10.
n – Butane and isobutane are a pair of
a) chain isomers
b) position isomers
c) metamers
d) functional isomers
Answer:
a) chain isomers

Question 11.
Which structure ¡s named as 3-chlorocyclobut-1-ene?

Answer:

Question 12.
Which one of the following is called 2-cyclobutyipropanal?

Answer:

Question 13.
Which one of the following is called cyclopentyl benzene’s?

Answer:

Question 14.
Which one of the following is commonly called mesitylene?

Answer:

Question 15.
The number of possible monochloro benzenes is
a) 1
b) 3
c) 5
d) 6
Answer:
a) 1

Question 16.
Which of the following pair are called functional isomers?
(a) methyl propyl ether and diethyl ether
(b) 2-pentanone & 3-pentanone
(c) propanoic acid and methyl acetate
(d) I -butanol and 2-butanol
Answer:
(c) propanoic acid and methyl acetate

Question 17.
Which of the following does not show optical isomerism’?
(a) Glucose
(b) Tartane acid
(c) Lactic acid
(d) Methane
Answer:
(d) Methane

Question 18.
Which metal is used to prepare Lassaigne’s extract?
(a) Copper
(b) Sodium
(c) Aluminium
(d) Iron
Answer:
(b) Sodium

Question 19.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) Butan – 1 – ol
c) 2 – Methylpropan – 2 – ol
d) Butanone
Answer:
d) Butanone

Question 20.
Which one of the following is called feme ferrocyanide?
(a) Na₄[Fe(CN)₆]
(b) Na₄[Fe(CN)₆]₃
(c) Fe₄[Fe(CN)₆]
(d) Fe₄[Fe(CN)₆]₃
Answer:
(d) Fe₄[Fe(CN)₆]₃

Question 21.
What is the colour formed in Lassaigne’s test of an organic compound containing N and S?
(a) Prussian blue colour
(b) Blood red colour
(c) Purple colour
(d) Canary yellow colour
Answer:
(b) Blood red colour

Question 22.
Which one of the following is the formula of sodium nitroprusside?
(a) Na₄[Fe(CN)₅N0₅]
(b) Na₄[Fe(CN)₅SON]
(c) Na₄[Fe(CN)₆]
(d) Fe₄[Fe(CN)₆]₃
Answer;
(a) Na₄[Fe(CN)₅N0₅]

Question 23.
According to Huckel’s rule, a compound is said to be aromatic if’ it contains
a) 4n bonds
b) 4n atoms
c) (4n + 2) atoms
d) (4n + 2) π electrons
Answer:
d) (4n + 2) π electrons

Question 24.
Which one of the following solutions are added to Lassaigne’s extract to identify halogens?
(a) Acetic acid + Lead acetate
(b) dil HNO₃ + AgNO₃
(c) Fe(OH)₂ + FeCl₃
(d) Na₂CO₃ + KNO₃
Answer:
(b) dil HNO₃ + AgNO₃

Question 25.
Which is a saturated compound?
a) alkanes
b) alkenes
c) alkynes
d) cyclo alkenes
Answer:
a) alkanes

Question 26.
Which one of the following test is used to detect phosphorous in an organic compound?
(a) Silver nitrate test
(b) Copper oxide test
(c) Ammonium molybdate test
(d) Lassaigne’s test
Answer:
(c) Ammonium molybdate test

Question 27.
Identify the colour formed in the test kr phosphorous using ammonium molybciate.
(a) Crimson red colour
(b) Deep violet colour
(c) Prussian blue colour
(d) Canary yellow colour
Answer:
(d) Canary yellow colour

Question 28.
Which of the following will absorb CO₂?
(a) Conc. H₂SO₄
(b) KOH
(c) HCl
(d) Copper
Answer:
(b) KOH

Question 29.
IUPAC name of ester is
a) Alkoxy alkane
b) Alkyl alkanoate
c) Alkanoyl halide
d) Alkanoic anhydride
Answer:
b) Alkyl alkanoate

Question 30.
Which method is used to estimate sulphur?
(a) Lassaigne’s test
(b) Oxide test
(c) Carius method
(d) Kjedahl’s method
Answer:
(c) Cari us method

Question 31.
Which method is used to estimate nitrogen?
(a) Dumas method and Kjeldahl’s method
(b) Carius method & Oxide method
(c) Lassaignes test & Copper oxide test
(d) Ammonium molybdate test & Silver nitrate test
Answer:
(a) Dumas method and Kjeldahl’s method

Question 32.
IUPAC name of CH₂OH – CH₂OH is
a) 1, 2 – dihydroxy ethane
b) ethylene glycol
c) ethane – 1, 2 – diol
d) ethane – 1, 2 – dial
Answer:
c) ethane – 1, 2 – diol

Question 33.
Which of the following is used to decolourise the organic compounds?
(a) Chlorine
(b) Bleaching powder
(c) Animal charcoal
(d) Iodine
Answer:
(c) Animal charcoal

Question 34.
Which method is used to extract essential oils from plants and flowers?
(a) Crystallization
(b) Sublimation
(c) Steam distillation
(d) Differential extraction
Answer:
(c) Steam distillation

Question 35.
Which of the following is used as adsorbent?
(a) silica gel and alumina
(b) glass wool and cotton
(c) glass plate and paper
(d) glucose and fructose
Answer:
(a) silica gel and alumina

Question 36.
Which of the following compounds gives prussian blue colour in Lassaigne’s test?
(a) CH₄ and CH₃OH
(b) CH₃NH₂ and CH₃NO₂
(c) CH₃Cl and CHCl₃
(d) CH₃CHO and CH₃COCH₃
Answer:
(b) CH₃NH₂ and CH₃NO₂

Question 37.
Which of the following is the functional isomer of methyl acetate?
a) Ethyl acetate
b) Propanoic acid
c) Ethyl formate
d) Propanone
Answer:
b) Propanoic acid

Question 38.
Which one of the following is not used as air adsorbent in chromatography?
(a) Alumina
(b) Silica gel
(c) Magnesia
(d) Sucrose
Answer:
(d) Sucrose

Question 39.
The IUPAC name of ……..
(a) 2-methyl butanal
(b) butan-2-aldehyde
(c) 2-ethyipropanal
(d) 3-methyl isobutraldehyde
Answer:
(c) 2-ethyipropanal

Question 40.
Which of the following compounds will exhibit cis-trans isomerism?
Answer:
(a) 2-Buiene
(b) 2-Butyne
(c) 1-Butene
(d) 2-Butanol
Answer:
(a) 2-Butene

Question 41.
Which of the following sodium fusion extract of organic compound gives brilliant violet colour with sodium nitroprusside solution?
(a) Urea
(b) Thiourea
(c) Benzoic acid
(d) Aniline
Answer:
(b) Thiourea

Question 42.
The IUPAC name of Cinnamaldehyde is
a) 3 – Phenyl prop – 2 – enal
b) 1 – Phenyl – prop – 1 – enal
c) 1 – Phenyl – prop – 2 – enal
d) 3 – Phenyl – prop – 1 – enal
Answer:
a) 3 – Phenyl prop – 2 – enal

Question 43.
In which of the following, functional group isomerism is not possible?
(a) Alcohols
(b) Aldehydes
(c) Alkyl halides
(d) Cyanides
Answer:
(c) Alkyl halides

Question 44.
Which one of the following is used as a column in the separation of pigments of chlorophyll by chromatography technique?
(a) Petroleum ether
(b) CaCO₃
(c) Activated charcoal
(d) Ethanoic acid
Answer:
(b) CaCO₃

Question 45.
Which one of the following compound does not give Prussian blue colour in Lassaigne’s test?
(a) C₆H₅NH₂

(c) C₆H₅CONH₂
(d) C₆H₅COCl
Answer:
(d) C₆H₅COCl

Question 46.
Which one of the following shows geometrical isomerism?
(a) n-Butane
(b) 1-butene
(c) 2-butene
(d) butyne
Answer:
(c) 2-butene

Question 47.
Which one of the following shows functional group isomerism?
(a) Ethene
(b) Acetone
(c) Ethane
(a) Propane
Answer:
(b) Acetone

Question 48.
Which of the following pair gives curdy white precipitate and yellow precipitate respectivety in their Lassaignes test?
(a) C₂H₅I and C₂H₅Br
(b) C₂H₅NO₂ and C₂H₅NH₂
(c) C₆H₅Cl and CH₃
(d) CH₄ and CH₃OH
Answer:
(c) C₆H₅Cl and CH₃

II. Match the following.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

III. Fill in the blanks.

Question 1.
The property catenation is possible in ……..
Answer:
Carbon.

Question 2.
Acetic acid was synthesised by ………
Answer:
Kolbe

Question 3.
Methane was synthesised in laboratory by ………
Answer:
Berthiot
CH₃

Question 4.
is an example of ………
Answer:
aromatic benzenoid compound

Question 5.
2-butene is an example of compound.
Answer:
unsaturated open chain

Question 6.
The IUPAC name of is …….
Answer:
2, 2, 5-trimethyl heptane

Question 7.
The IUPAC name of is ……..
Answer:
3, 4-diethyl, 4-methylheptane

Question 8.
The name of is ……….
Answer:
1-cyclobutyl-2-cyclopropylethene

Question 9.
The name of
Answer:
2-(cyclobut-2-en-1-yl)-propanal

Question 10.
The name ofis …………
Answer:
2-cyclopentyipropanal

Question 11.
Esopentane and neopentane are the examples for ………
Answer:
chain isomerism

Question 12.
are called ……….
Answer:
functional isomers

Question 13.
Copper oxide test is used to detect ………..
Answer:
Carbon & Hydrogen

Question 14.
The formula of feme suiphocyanide is ………
Answer:
Fe(CNS)₃

Question 15.
In Lassaigne’s test for halogens, if colour of the precipitate is curdy white, the halogen present is ……….
Answer:
chlorine

Question 16.
The formula of ammonium phospho molybate is ……….
Answer:
(NH₄)₃.PO₄. MO₃

Question 17.
Silver nitrate test is used to detect the presence of ……….
Answer:
Halogens

Question 18.
During the estimation of carbon and hydrogen, presence of nitrogen can be avoided by using ………
Answer:
a spiral of copper

Question 19.
In Carius method, the sulphur in an organic compound is oxidised to ……….
Answer:
H₂SO₄

Question 20.
The method used to estimate nitrogen in foods and fertilisers is ………
Answer:
Kjeidahl’s method

Question 21.
The mixture of diethyl ether and ethanol can he purified by ………
Answer:
simple distillation

Question 22.
The method used to purify petroleum. coal-tar and crude oil is ………
Answer:
fractional distillation

Question 23.
The method used in the manufacture of aniline and turpentine is ……….
Answer:
steam distilation

Question 24.
The mixture of exhanol and water are separated by ………..
Answer:
azeotropic distillation

Question 25.
The different coloured constituents of chlorophyll are separated by ………
Answer:
chromatography

Question 26.
The large number of organic compounds is due to of carbon ………..
Answer:
catenation

Question 27.
The IUPAC name of the compound shown below is:

Answer:
1, 1-dichloropropane

Question 28.
name of this compound is ………..
Answer:
2-chloro-3-ethyl- 1, 4-pentadiene

Question 29.
Carboxylic acids are isomenc with ……….
Answer:
esters

Question 30.
Alcohols are isomeric with ……….
Answer:
ethers

Question 31.
The correct IUPAC name for the following structure is ……….

Answer:
5-hex-i -en-3-ol

Question 32.
The Prussian blue colour confirms the presence of nitrogen in an organic compound is due to the formation of ………
Answer:
Fe₄[Fe(CN)₆]I

Question 33.
The principle involved in paper chromatography is ……….
Answer:
partition

Question 34.
Steam distillation is used for the extraction of ……..
Answer:
essential oils

Question 35.
In chromatography, if the stationary phase is solid, the basis is ……….
Answer:
adsorption

Question 36.
In chromatography, if the stationary phase is liquid, the basis is ………
Answer:
partition

Question 37.
The isomer of acetaldehyde is ……….
Answer:
acetone

Question 38.
The general formula of alkyne is ……….
Answer:
C_nH_2n-2

Question 39.
The IUPAC name of (CH₃)₂CH-CH₂-CH(CH₃)₂-CH(CH₃)₂ is ……..
Answer:
2, 3, 5-trimethyl hexane

IV. Choose the odd one out.

Question 1.
(a) THF
(b) Pyridine
(c) Phenol
(d) Thiophene
Answer:
(c) Phenol. ft is a homocyclic compound whereas others are heterocyclic compounds.

Question 2.

Answer:

Question 3.
(a) Azulene
(b) Propane
(c) Butane
(d) Ethene
Answer:
(a) Azulene. It is a non-benzenoid aromatic homocyclic compound whereas others are aliphatic compounds.

Question 4.
(a) Dyes
(b) Polymers
(c) Cosmetics
(d) Common salt
Answer:
(d) Common salt. It is an inorganic compound whereas others are organic compound.

Question 5.
(a) Renzene
(b) Water
(c) Ether
(d) Chloroform
Answer:
(b) Water. It is a polar solvent whereas others are non-polar solvents.

V Choose the correct pair.

Question 1.
(a) Benzene: Aliphatic compound
(b) Propane: Aromatic compound
(c) Pyridine: Heterocyclic compound
(d) Cyclohexane: Polycyclic compound
(c) Pyridine: Heterocyclic compound

Question 2.
(a)-OH : Ketone
(b)-CHO : Carboxylic acid

(d)-NO₂ : Amine
Answer:

Question 3.
(a) Organic compounds: inflammable
(b) Organic compounds: ionic compound
(c) Organic compounds: high boiling point and high melting point
(d) Organic compounds: soluble in water
Answer:
(a) Organic compounds: inflammable

Question 4.
(a) C_nH_2n+2 : C₂H₄
(b) C_nH_2n : C₃H₆
(c) C_nH_2n-2 : C₂H₆
(d) C_nH_2n+2 : C₃H₄
Answer:
(b) C_nH_2n : C₃H₆

Question 5.

Answer:

Question 6.
(a) n-pentane and iso pentane : position isomerism
(b) neopentane and n-pentane : chain isomerism
(c) propanal and propanone : position isomerism
(d) propañoic acid and methyl acetate : chain isomerism
Answer:
(b) neopentane and n-pentane : chain isomerism

VI. Choose The incorrect pair.

Question 1.
(a) Dumas method : Estimation of nitrogen
(b) Kjeldahls method : Estimation of nitrogen
(c) Carius method : Estimation of halogens
(d) Dumas method : Estimation of sulphur
Answer:
(d) Dumas method : Estimation of sulphur

Question 2.
(a)-CHO : Aldehyde
(b)-COOH : Carboxylic acid
(c)-NH₂ : Nitro group
(d)-O- : Ether
Answer:
(c)-NH₂ : Nitro group

Question 3.
(a) Benzene and nitro benzene : Distillation
(b) Coal tar and crude oil : Fractional distillation
(c) Aniline and turpentine : Steam distillation
(d) Naphthalcne and benzoic acid : Crystallization
Answer:
(d) Naphthalene and benzoic acid: Crystallization

Question 4.
(a) BaSO₄ : White colour precipitate
(b) Ag₂S : Black colour precipitate
(c) Fe(CNS)₃ : Prussian blue colour
(d) PbS : Black colour precipitate
Answer:
(c) Fe(CNS)₃ : Prussian blue colour

Question 5.
(a) propanal and propropane : Functional isomerism
(b) Nitrite fòrrn and nitro form : Tautomerism
(c) Pent- 1-ene and pcnt-2-ene : Chain isomerism
(d) Propanoic acid and methyl acetate: Functional isomerism
Answer:
(c) Pent-1-ene and pent-2-ene : Chain isomerism

VII. Assertion & Reason.

Question 1.
Assertion (A) : Carbon cannot form ionic bond.
Reason (R) : It is not possible for the carbon to form either C4t or C ions, as it requires large amount of energy.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct hut R is not the correct explanation of A.
(c) A is.correct but R is wrong.
(d) A is wrong but Ris correct.
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a dîftèrence of more than 30°C in their boiling points can be separated by simple distillation.
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.
Answer:
(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question 3.
Assertion (A) : Pent-1 -ene and pent-2-ene are position isomers.
Reason (R) : Position isomers ditTer in the position of functional group or substituent.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(c) Assertion is correct but reason is wrong.
(d) Assertion is wrong but reason is correct.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) All organic compounds are ionic compounds.
(b) All organic compounds have high boiling point and high melting point.
(c) Many of the organic compounds are inflammable.
(d) Organic compounds are mostly soluble in water.
Answer:
(c) Many of the organic compounds are inflammable.

Question 2.
(a) Propane is heterocyclic compound.
(b) Azulene is a non benzenoid and aromatic homocyclic compound.
(c) Pyridine is a homocyc lic compound.
(d) Cyclopropane is an aromatic compound.
Answer:
(b) Azulene is a non henzenoid and aromatic homocyclic compound.

Question 3.
(a) CH≡CH-CH₂-C≡CH is a saturated open chain compound.
(b) CH₃-CH₂-CH₂-CH₂-CH₃ is an aromatic benzenoid compound.
is an aromatic benzenoid compound.
Answer:
is an aromatic benzenoid compound.

Question 4.
(a) Organic compounds are covalent and generally insoluble in water.
(b) Organic compounds are ionic but generally soluble in water.
(c) Organic compounds non-inflammable
(d) Organic compounds do not show catenation.
Answer:
(a) Organic compounds are covalent and generally insoluble in water.

Question 5.
(a) Fe₄[Fe(CN)₆] is prussian blue precipitate.
(b) Ag₂S is a white precipitate.
(c) PbS is a blood red colour precipitate.
(d) BaSO4 is a black colour precipitate.
Answer:
(a) Fe₄[Fe(CN)₆] is prussian blue precipitate.

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 2 Mark Question and answers

Write brief answer to the following questions:

Question 1.
What is catenation?
Answer:
The tendency of an atom to form a chain of bonds with the atoms of the same element is called catenation. The high strength of C-C bond is responsible for its catenation property.

Question 2.
Explain the following terms in IUPAC system of nomenclature. of organic compounds.
(i) Root word
(ii) prefix
(iii) suffix
Answer:
(i) Root word:
Root word denotes the number of carbon atoms in the longest continuous chain in molecules.
(ii) prefix:
Prefix denotes the group(s) attached to the main chain which is placed before the root.
(iii) suffix:
Suffix denotes the functional group and is paced after the root word.

Question 3.
What is meant by functional group? Give two example.
Answer:
A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way. irrespective of organic molecule in which it is present. The reaction of an organic compound takes place at the functional group.
e.g.. Alcohol -OH group
Ether -O- group

Question 4.
Classify the following compounds based on the structure.
(i) CH₂=CH-CH=CH₂
(ii) CH₃-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃

Answer:
(i) Unsaturated open chain compound
(ii) Saturated open chain compound
(iii) Aromatic benzenoid compound
(iv) Alicylic compound

Question 5.
Give one example for each of the following type of organic compounds.

1. Non-benzeoid
2. Aromatic heterocclic
3. Alicyclic
4. Aliphatic open chain

Answer:
1. Non-benzenoid aromatic compound

2. Alicyclic compound

3. Aromatic heterocyclic Compound

4. Atiphatic open chain compound

Question 6.
Give two examples for each of the following type of organic compounds.

1. Aromatic homocyclic compound
2. Aromatic heterocyclic compound

Answer:
1. Aromatic homocyclic compound

2. Aromatic heterocyclic compound

Question 7.
Write the functional group of the following compounds

1. Aldehyde
2. Ester
3. Ether
4. alcohol

Answer:

Question 8.
Write the functional group of –

1. cyanide
2. socyanide
3. cyanate
4. isocyanate

Answer:

1. cyanide -CN
2. isocyanide -NC
3. cyanate -CNO
4. isocyanate -NCO

Question 9.
Write the functional group of –

1. thiocyanate
2. isothiocyanate
3. thiols
4. thioether

Answer:

1. Thio cyanale -SCN
2. Isothiocyanate -NCS
3. thiols -SH
4. thicethers -S-

Question 10.
Write the IUPAC names of the following compounds.

Answer:

1. 3-methylpentane
2. 2, 2, 5-trimethylheptane

Question 11.
Write the IUPAC names of the following compounds.

Answer:

1. 3-ethyl-2-methyl pentane
2. 2-methylbutanal

Question 12.
What are the IUPAC names of the following compounds?


Answer:

1. 2-ethyl-but-4-ene-oic acid
2. 2, 2-dimethyl-hexanoic acid

Question 13.
Write the IUPAC names of the following compounds.

Answer:

1. 2-ethyl-3-propyl pentane-dioic acid
2. 3-methy-1 hexane

Question 14.
Predict the IUPAC names of the following compounds

Answer:

1. cyclobutane
2. cyclopentane
3. cyclobutene
4. cyclo octane

Question 15.

Answer:

1. Ethyl cyclobutane
2. Methylcyclohexane
3. Cyclohexanol

Question 16.
Write the ¡UPAC names of the following compounds.


Answer:

1. 2-cyclobutyl propanol
2. 3-cyclohexyl pentan-2-one

Question 17.
Write the structural formula for the following compounds.
(i) cyclohexa- 1, 3-diene
(ii) methyl cyclopentane
Answer:

Question 18.
Draw the structures of –
(i) o-xylene
(ii) m-xylene
(iii) p-xylene
Answer:

Question 19.
Draw the structure of –
(i) Mesitylene
(ii) 1,2, 3-trimethyl benzene
Answer:

Question 20.
Write the structure of –
(i) p-dinitrobenzene
(ii) o-dichlorobenzene
Answer:

Question 21.
Draw the structure of –
(i) 2-cyclopentyl propanal
(ii) 2-(cyclo-but-cnyl) propanal
Answer:

Question 22.

Answer:

1. N, N-dimethylbenzene amine
2. N-ethyl-N-methylpropan-I-amine

Question 23.
Draw the structure of 4-hydroxy-3(2-hydroxy ethyl) hexanoic acid.
Answer:

Question 24.
Explain about dash line structure with a suitable example.
Answer:
The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a lewis structure. A single line or dash represents a single covalent bond.
e.g., n- propanal:

Question 25.
What ¡s meant by condensed structure? Explain with an example.
Answer:
The bond line structure can be further abbreviated by omitting all these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1, 3-butadiene.

Question 26.
What are bond-line structures? Give one example.
Answer:
The condensed structural formula is simplified in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bond are shown in zig-zag fashion. The only atoms are specially written are oxygen, chlorine, nitrogen etc. Example, Ten. butyl chloride
condensed structure

bond line structure

Question 27.
Define isomerism. Give example.
Answer:
Isomerism represents the existence of two or more compounds with the same molecular formula but different structure and properties. Compounds exhibiting this isomerism are called isomers.
e.g., C₂H₆O:

1. CH₃-CH₂OH Ethanol
2. CH₃-O-CH₃ Methoxyrnethane

Question 28.
Write the possible isomers for the formula C₅H₁₀ with their name and type of isomerism present in it.
Answer:
C₅H₁₀:

1. CH₃-CH₂-CH₂-CH = CH₂ (Pent-1-ene)
2. CH₃-CH₂-CH=CH-CH₃ (Pent-2-ene)
   The type of isomerism present above is position isomerism.

Question 29.
Write the possible isomers for the formula C₅H₁₀O with their name indicating position isomerism.
Answer:


Question 30.
Draw the functional isomers for the formula C₃H₆O₂ with their names.
Answer:
C₃H₆O₂:

1. CH₃-CH₂-COOH (Propanoic acid)
2. CH₃-COOCH₂ (Methyl acetate)

Question 31.
What is metamerism? Give an example.
Answer:
Metamerism is one kind of structural isomerism which arises due to the unequal distribution of carbon atoms on either side of the functional group or diffèrent alkyl groups attached to the either side of the same functional group and having the same molecular formula.
e.g., C₄H₁₀O:

Question 32.
What is tautomerism?
Answer:
It is a special type of functional isomerism in which single compound exists in two readily interconvertible structures that differ markedly in the relative position of atleast one atomic hydrogen. The two different structures are known as tautomers.

Question 33.
What ¡s meant by dyad system? Explain with example.
Answer:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.

Question 34.
What is triad system? Give example.
Answer:
1. In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1. 3-migration of hydrogen atom from one polyvalent atom to other with in the molecule.
2. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.

Question 35.
What is enolisation? What is labile form?
Answer:
Enolisation is a process in which keto form is converted into eno! form. Both tautomeric forms are equally stable. The less stable form is known as labile form.

Question 36.
Give the structures of Nitro-aci tautomerism.
Answer:

Question 37.
Explain ring chain isomerism with the formula C₄H₈.
Answer:
In ring chain isomerism, compounds have the same molecular formula but differ in terms of bonding of carbon atoms to form open chain and cyclic structures.
C₄H₈:

Question 38.
Define stereo-isomerism.
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

Question 39.
Define geometrical isomerism with an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.

Question 40.
Trans isomer is more stable than cis isomer. Justify this statement.
Answer:
Trans isomer is more stable than cis isomer. This is because in the cis isomer, the bulky groups are on the same side of the double bond. The steric repulsion of the groups makes the cis isomers less stable than the trans isomers in which bulky groups are on the opposite side.

Question 41.
Draw the cis, trans isomeric structures of 1, 3-butadiene.
Answer:

Question 42.
What are the condition for optical isomerism (or) enantiomerism.
Answer:
1. A carbon atom whose tetravalency in satisfied by four different substituents (atoms (or) groups) is called asymmetric carbon (or) chiral carbon. The optical isomer should have one or more chiral carbon to show optical activity.

2. The molecule possessing chiral carbon atom and is non-superimposable its own mirror image is said to be chiral ntolecule and the property is called chirality or dissymmetry.

Question 43.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

• A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
• The tube is heated till reaction ceases arid become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
• The contents of the dish is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 44.
What Is the need for purification of organic compounds?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds, they must be in the pure state. So organic compounds must be purified.

Question 45.
Define sublimation. Give two examples.
Answer:
The process of conversion of solid to vapour without melting or heating and on cooling the vapours getting back solids, such phenomenon is known as sublimation.
e.g., Naphthalene, Camphor.

Question 46.
Explain the process of chromatography in chlorophyll.
Answer:
The separation of different coloured constituents of chlorophyll is done by chromatography by M.S. Tswelt. He achieved it by passing a petroleum ether solution of chlorophyll present in leaves through a column of CaCO3 firmly packed into a narrow glass tube. Different components of the pigments got separated and lands to form zones of different colours.

Question 47.
Draw the first six members of the carboxylic acid homologous series.
Answer:

1. HCOOH
2. CH₃COOH
3. CH₃-CH₂-COOH
4. CH₃-CH₂-CH₂-COOH
5. CH₃-CH₂-CH₂-CH₂COOH
6. CH₃-CH₂-CH₂-CH₂-CH₂-COOH

Question 48.
Give the condensed formula and bond line formula of 2, 2,4- trimethylpentane.
Answer:
2, 2, 4 –  irimethylpentane
(CH₃)₃CCH₂CH(CH₃)₂-Condensed formula

Question 49.
Differentiate between the principle of estimation of nitrogen in an organic compound by

1. Dumas method
2. Kjeldahl’s method.

Answer:
1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO₂ where free nitrogen, CO₂ and H₂O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with conc. H₂SO₄, a little amount of potassium sulphate and a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 50.
Explain the principle of paper chromatography.
Answer:
This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified whose solution is prepared in a suitable solvent.

Question 51.
Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer:
Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na₂S, NaX and Na₃PO₄. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Question 52.
Name a suitable technique of separation of the components from a mixture of calcium, sulphate and camphor.
Answer:
Sublimation. Because camphor can sublime whereas CaSO₄ does not.

Question 53.
Explain, why an organic liquid vapourises at a temperature below its boiling point on steam distillation?
Answer:
It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

Question 54.
Will CCl₄give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
No. CCl₄ is a completely non-polar covalent compound whereas AgNO₃ is ionic in nature. Therefore they are not expected to react and thus a white ppt. of silver chloride will not be formed.

Question 55.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
CO₂ is acidic in nature and therefore it reacts with the strong base KOH to form K₂CO₃:
2KOH + CO₂ – K₂CO₂ + H₂O

Question 56.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
Sulphur sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test.
Pb(OCOCH₃)₂ + H₂SO₄ → PbSO₄-+ 2CH₃COOH

Question 57.
Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur?
Answer:
On Ilising with sodium metal the elements present in an organic compound are converted into sodium salts which are water soluble which can be filtered and detected by the respective tests.

Question 58.
Under what conditions can the process of team distillation is used?
Answer:
Steam distillation is used to purify the Liquids which are steam volatile and not miscible with water.

Question 59.

Answer:
1, 2-dichloropropane

Question 60.
Write bond-line formulas for: Isopropyl alcohol, 2, 3-dimethvlbutanal, Heptan-4-one.
Answer:

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 3-Mark Questions

Question 1.
Write the functional group of the following compounds:
(i) carboxylic acid
(ii) Acid anhydride
(iii) Acyichioride
(iv) Amide
(v) imines
(vi) Nitroso compound
Answer:

Question 2.
What are the general molecular formula and functional group of the following compounds?
Answer:
(i) Ilydrazines
(ii) Hydrazo compound
(iii) Imide
(iv) Phenols
(v) Amine
(vi) Nitroalkane
Answer:

Question 3.
Write the tUPAC names of the following compounds.

Answer:
(i) Pentan-2a1
(ii) Pentan-(2-ene-2-propyl)- i -oic acid
(iii) 4-methyl-i -cyanohexane

Question 4.

Answer:
(I) Hex-4-ene-2-oI
(ii) 3-ethyl-5-methyl heptane

Question 5.
Draw the structure of
(i) 1-ethyl-2-methyl cyclopentane
(ii) 1-ethyl-2, 3-dimethyl cyclohexane
(iii) 5-ethyl-2-methylcyclohcx- I -ene
Answer:

Question 6.
Draw the structures of:
(i) 2-cyclobutyl propane
(ii) 2-cyctopropyl butane
(iii) chiorocyclo but-2-eue
Answer:

Question 7.
Give the IUPAC name of the following compounds:

Answer:
(i) 2-(cyclo but-2-en-1-yl)-propanal
(ii) 4-(cyclopent-3-en- 1-yl )-3-methylbutanoic acid
(iii) 3-(3-nitro cyclopenryl)-prop-2-enoic acid

Question 8.
Write the IUPAC names of the following compounds

Answer:
(I) 2-(2-hydroxypropyl) cyclohexan- 1-01
(ii) CyclopentyÍ benzene
(iii) Cyclohexane carboxyl Ic acid

Question 9.
Draw the structure
(i) 1-(cyclo bytyl)-2 (cylopropyl) ethane
(ii) 2-carbamyl cyclobutane-1-carboxylic acid
Answer:

Question 10.
Draw the structures of:
(i) Bromohenzene
(ii) 1, 2-dichlorobenzene
(iii) 1-chloro-3-methvlbenzene
Answer:

Question 11.
Draw the structures of –
(i) Benzvl chloride
(ii) Benzal dichloride
(iii) Benzotrichloride
Answer:

Question 12.
Write the IUPAC names of the following compounds.

Answer:
(i) 3-methylpentane
(ii) 2, 2. 5-trimethylheptane
(iii) 2-methylbutanal

Question 13.
Write the IUPAC names of the following compounds.

Answer:
(i) 3-ethyl-2-methylpentane
(ii) 2-ethyl but-3-enoic acid
(iii) 2-forrnyl-2-methylheptanoic acid

Question 14.
Draw the structures of
(i) 3-methylpentanal
(ii) 5-hydroxy 2,2-dimethyl heptanoic acid
(iii) 2-ethyl-4-propy Ipentane-d ioic acid
Answer:

Question 15.

Answer:
(i) 3-methylhexane
(ii) 2-methylbutanal
(iii) 2-ethylbut-3-enoic acid

Question 16.
Give the IUPAC name Of –

Answer:
(i) 4-methyl/hexanenitrite
(ii) 2-methyl but-3-en-amide
(iii) I-Iex-4-en-2-ol

Question 17.
Draw the structures of –
(i) 3-ethyl-5-methylheptane
(ii) 3-ethyl-2-methylhexane
(iii) 2, 4-dimethylpent-2-ene
Answer:

Question 18.
Draw the structures of:
(i) 3-methylhcpta 1, 3, 5-triene
(ii) pent-1-yne
(iii) 2-methylpropan-2-oI
Answer:

Question 19.
Give the IUPAC name of the following compounds.

Answer:
(i) 4-mcthylpcntan-1-ol
(ii) 2, 2-dimethylpropan-1-ol
(iii) Propanoic acid

Question 20.
Draw the structural formula of:
(i) 4-methylpent-3-en-2rone
(ii) pent-1-yne-3-one
Answer:

Question 21.
Write the IUPAC names of the following compounds.
CH₃-CH₂-CH₂-NH-CH₃

Answer:
(i) N-methylpropan- 1 -amine
(ii) N-rnethylpropan-2-amine
(iii) N, N-dimethylpropan- 1-amine

Question 22.
Draw the structurai formula of the following compounds.
(i) N-cthyl-N-methylpropan-1-amine
(ii) N, N-dimethyl benzenamine
Answer:

Question 23.
Draw the complete structural formula, condensed structure and bond line structure of
(i) n-propanol
(ii) 1, 3-butadiene.
Answer:

Question 24.
Draw the dash line structure, condensed structure and bond line structure of 1, 3-dimethyl cyclopentane.
Answer:

Question 25.
What is wedge formula? Explain with suitable example.
Answer:
1. The simplest convention is solid and dashed wedge formula in which 3-D image of a molecule can be perceived from two dimensional picture.

2. In this representation, a tetrahedral molecule with four atoms or groups a, b, e and d bonded to it can be represented by wedge formula as follows.

3. A solid wedge
or a heavy line is used to indicate a bond projecting above the plane of the paper and dashed wedge
or a dashed line is used to depict the bond below the plane. The bonds lying in the plane of the paper are shown by normal lines.

Question 26.
Draw the fisher projection formula for tartaric acid.
Answer:

Question 27.
Explain the advantage of sawhorse projection formula over the fisher projection formula with an example.
Answer:
1. The fisher projection fonnula inadequately portrays the spatial relationship between ligands attached to the atoms. The sawhorse projection attempts to clarify the relative location of the groups.

2. In sawhorse projection formula, the bond between two carbon atoms ¡s drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back.

Question 28.
Explain about the Newmann projection formula with an example.
Answer:
1. In this method, the molecules are viewed from the front along the carbon-carbon bond axis.

2. The two carbon atom forming the G bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point where as the carbon lying farther from the eye is represented by the origin of the circle.

3. Therefore the C-H bonds of the front carbon are depicted from the circle while the C H bonds of the back carbon are drawn from the circumferance of the circle with an angle of 120 to each other.

Question 29.
Write the possible isomers for the formula C₅H₁₂ with their names and structures.
Answer:

(i), (ii) and (iii) are chain isomers.

Question 30.
What are the possible isomers for the formula C₄H₉Cl? Give their structures and IUPAC names.
Answer:
C₄H9Cl:

Question 31.
Write the metamers for the formula C₅H₁₀O with their IUPAC names.
Answer:

Question 32.
Explain about the geometrical isomerism possible in oximes.
Answer:
1. Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here syn and anti are used instead of cis and trans respectively.

2. In the syn isomer the H atom of a doubly bonded carbon and OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.

3. for e.g.,

Question 33.
What are enantiomers?
Answer:
1. An optically active substance may exist in two or more isomeric forms which have same physical and chemical properties but diflèr in terms of direction of rotation of plane polarised light, such optical isomers which rotate the plane polarised light with equal angle but in opposite directions are known as enantiomers and (he phenomenon is known as enantiomerism.

2. Isomers which are non-super impossible mirror images of each other are called enantiomers.

Question 34.
How would you detect sulphur?
Answer:
1. To a portion of the Lassaigne’s extract, freshly prepared sodium nitroprusside solution is added. If deep violet or purple colour is formed, the presence of sulphur is confirmed.

2. To another portion of Lassaignes extract, acetic acid and lead acetate solution are added. If black precipitate is formed, sulphur presence is confirmed.

Question 35.
Explain about the oxidation test for sulphur.
Answer:
1. Oxidation test:
The organic substances are fused with a mixture of KNO₃ and Na₂CO₃. The sulphur if present is oxidised to sulphate.
Na₂CO₃ + S + 3(O) Na₂SO₄ + CO₂

2. The fused mass is extracted with water, acidified with HCl and the BaCl₂ solution is added to it. A while precipitate indicates the presence of sulphur.

Question 36.
How would you detect the halogens in an organic compound?
Answer:

• To a portion of the Lassaigne’s filtrate. dii. HNO₃ is added, warmed gently and AgNO₃ solution is added.
• Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
• Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
• Appearance of yellow precipitate insoluble in ammonia solution indicates the presence of iodine.
  

Question 37.
Why nitric acid is added in the Lassaigne’s test for halogens?
Answer:
1. if N (or) S is present in the organic compound along with the halogen, we might obtain Na₂S and NaCN in the solution which interfere with the detection of the halogen in the AgNO₃ test.

2. Therefore we boil the Lassiagne’s extract with HNO₃ which decomposes NaCN and Na₂ S as follows:

Question 38.
Explain about the test for phosphorous in an organic compound.
Answer:

• A solid organic compound is strongly heated with a mixture of Na₂CO₃ and KNO₃. Phosphorous present in the compound is oxidised to sodium phosphate.
• The residue is extracted with water and boiled with cone. HNO₃. A solution of ammonium molyixiate is added to this solution.
• A canary yellow precipitate shows the presence of phosphorous.

Question 39.
Explain about principle and reactions involved in carius method of estimation of sulphur. Carius method:
Answer:
1. Principle:
A known mass of the organic substance is heated strongly with fuming HNO₃. C and H get oxidized to CO₂. and H₂O while sulphur is oxidised to sulphuric acid as follows.

2. The resulting solution is treated with excess of BaCl₂ solution, H₂SO₄ present in the solution is converted into BaSO₄. From the mass of BaSO4, the percentage of sulphur can be calculated.

Question 40.
Explain about the procedure and calculation behind the carius method of estimation of sulphur.
Answer:
Carius method:
(I) Procedure:
A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and bcatcd for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape. The carius tube is broken and the content collected in a beaker. Excess of BaCl, is added to the beaker. BaSO₄ furmed is converted to BaSO₄ (white ppt.) The precipitate is filtered. washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(ii) Calculation:
Mass of organic compound = Wg
Mass of BaSO₄ formed = r g
233 g of BaSO₄ contains 32 g of sulphur
∴ x g of BaSO₄ contain \(\frac {32}{233}\) x x g of sulphur

Question 41.
What is Homologous Series? Give a suitable example.
Answer:
Homologous series:
It is a series of compounds in which the adjacent members differ by a – CH₂ unit. Individual members of such series are called homologues, and the phenomenon is called homology. All the members of such a series of alkane have the general formula C_nH_{2n + 2}. Few members of this family are
CH₄ – Methane
C₂H₆ – Ethane
C₃H₈ – Propane
C₄H₁₀ – Butane
C₅H₁₂ – Pentane

Question 42.
Explain the copper oxide test for the detection of carbon and hydrogen present in the given organic compound.
Answer:
Copper oxide test:
The organic substance is mixed intimately with about three times its weight of dry copper oxide by grinding. The mixture is then placed in a hard glass test tube fitted with a bent delivery tube. The other end of which is dipping into lime water in another test tube. The mixture is heated strongly and the following reaction take place.
C + 2CuO → CO₂ + 2CuO
2H + CuO → H₂O + Cu

Thus if carbon is present, it is oxidized to CO₂ which turns lime water millcy. If hydrogen is also present, it will be oxidized to water which condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water collected in the bulb is separated on anhydrous CuSO₄ which turn blue. This confirms the presence of C and H in the compound. If however, H is not present water droplet is not obtained in the bulb.

Question 43.
Explain differential extraction.
Answer:

• The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction.
• When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel.
• The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl₃). The organic solvent immiscible with water will form a separate layer and the contents are shaken gently.
• The solute being more soluble in the organic solvent is transferred to it.
• The solvent layer is then separated by opening the tap of separating funnel and the substance is recovered.

Question 44.
Explain about the principle involved in chromatography. Give its types.
Answer:

1. The principle behind chromatography is selective distribution of the mixture of organic substances between two phases-a stationary phase and a moving phase. The stationary phase can be a solid or liquid while the moving phase is a liquid or a gas.
2. if the stationary phase is solid, the basis is adsorption and when it is a liquid, the basis is partition.
3. Chromatography is defined as technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.
4. The various methods of chromatography are:
   • Column chromatography (CC)
   • Thin layer chromatography (TLC)
   • Paper chromatography (PC)
   • Gas liquid chromatography (GLC)
   • Ion exchange chromatography

Question 45.
Describe about adsorption chromatography.
Answer:

• The principle involved is different compounds are adsorbed on an adsorbent to different degree.
• Silica gel and alumina are the commonly used adsorbent. The components of the mixture move by varying distances over the stationary phase.

Question 46.
What are hybridisation states of each carbon atom ¡n the following compounds?
Answer:

Question 47.
Give the IUPAC names of the following compounds.

Answer:
(a) Propylbenzene
(b) Methylpentanenitrile
(c) 2, 5-Dimethyllieptanc
(d) 3-Bromo-3-chloroheptane
(e) 3-Chioropropanal
(f) 2, 2-Dichloroethanol

Question 48.
Write the formulas for the first five members of each homologous series beginning with the following compound. CH₃COCH₃
Answer:
CH₃COCH₃
CH₃COCH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃CO(CH₂)₄CH₃

Question 49.
Write the formulas for the first five members of each homologous series beginning with the following compound: H-CH=CH2
Answer:
H-CH=CH₂
CH₃CH=CH₂
CH₃CH₂CH=CH₂
CH₃CH₂CH₂CH=CH₂
CH₃CH₂CH₂CH₂CH=CH₂

Question 50.
Identify the functional groups ¡n the following compounds.
Answer:

Question 51.
What is the difference between distillation, distillation under reduced pressure and steam distillation?
Answer:
Distillation:
it is used in case of volatiLe liquid mixed with a non-volatile impurities.

Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.

Steam distillation:
it is used to purify steam volatile liquids associated with water immiscible impurities.

Question 52.
(a) What is Lassaigne’s extract? Will NaCN give a positive Lassaigne’s test for nitrogen?
(b) Which colour will appear in the Lassaigne’s test if the compound contains both nitrogen and sulphur.
(c) Why is Lassaigne’s extract prepared in distilled water? Can we detect oxygen iii a compound by Lassaigne’s test?
Answer:
(a) When organic compounds is fused with sodium metal and then extracted by water it is called Lassaigne’s extract. Yes.
(b) Blood red colour.
(c) Lassaigne’s extract is prepared in distilled water since ta water contains Cl’ ions. No, oxygen cannot be detected by Lassaigne’s test.

Question 53.
0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer:
Mass of the compound = 0.3780
Mass of silver chloride 0.5740 g

Question 54.
In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.
Answer:
Percentage of sulphur

Samacheer Kalvi 11th Chemistry Fundamentals of Organic Chemistry 5 marks Questions and Answers

Question 1.
Explain about the steps involved in naming an organic compound as per IUPAC nomenclature.
Answer:
The following steps should be followed for naming an organic compound as per IUPAC nomenclature.

• Choose the longest carbon chain (Root word). Consider all other groups attached to this chain as substituents.
• Numbering of the longest carbon chain.
• Naming the substituents (prefixes) or (suffixes).
• Arrange the substituents in the alphabetical order.
• Write the name of the compound as below.

Prefix + Root word Primary suffix + r Secondary suffix

Question 2.
How will you detect the presence of carbon and hydrogen in an organic compound?
Answer:
Copper oxide test:
1. The organic substance is mixed with three times its weight of dry copper oxide by grinding. The mixture is placed in a hard glass test rube fitted with a bent delivery tube. The other end of which is dipping into lime water in an another test tube. The mixture is heated strongly.

2. 2CuO + C → CO₂ + 2Cu
CuO + 2H → H₂O + Cu

3. Thus if carbon is present, it is oxidized to CO₂ which turns lime water milky. If hydrogen is also present, it will be oxidized to water and condenses in small droplets on the cooler wall of the test tube and inside the bulb. Water is collected on white anhydrous CuSO₄ which turns blue.

4. This confirms the presence of C and H in the compound.

Question 3.
Explain lassaigne’s test for detection of nitrogen ¡n an organic compound.
Answer:
I step:
Preparation of sodium fusion extract:
A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is heated. When it melts to a shining globule, a pinch of organic compound is added to it. The tube is then heated till the reaction ceases and becomes red hot.

Then the test tube is plunged in about 50 ml of distilled water taken in a china dish and break the bottom of the tube by striking against the dish. The contents of the dish is boiled for about 10 minutes and then filtered. This filtrate is known as lassaigne’s extract (or) sodium fusion extract.

II step :
Test for Nitrogen:
If Nitrogen is present, it gets converted to sodium cyanide which reacts with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl and gives a Prussian blue colour (or) green coloured precipitate, it confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the action of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate.

Reactions involved:
Na + C + N → NaCN
FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄


if both N & S are present. a blood red colour is obtained due to the following reactions

Question 4.
Explain about the estimation of carbon and hydrogen.
1. Principle:
A known weigIt of organic substance is brunt in excess of oxygen and the carbon and hydrogen present in it are oxidised to CO₂ and H₂O respectively.

The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the organic substance are calculated.

2. Description of the apparatus:

(a) The oxygen supply (b) combustion tube (c) Absorption tube

Oxygen supply:
To remove the moisture from oxygen. it is allowed to bubble through sulphuric acid and then passed through a U-tube containing sodalime to remove CO₂. The oxygen gas free from moisture and CO₂ enters the combustion tube.

Combustion tube:
A hard glass tube open at both ends used for the combustion. It contains (i) an oxidized copper gauze to prevent the backward diffusion of the products of combustion (ii) a porcelain boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidized copper gauze placed towards the end of the combustion tube. The combustion tube is heated by a gas burner.

Absorption apparatus:
The combustion products containing moisture and CO₂ are then passed through the absorption apparatus which consists of(i) a weighed U-tube packed with pumice soaked in conc. H₂SO₄ to absorb water (ii) a set of bulbs containing a Strong solution of KOH to absorb CO₂ and finally (iii) a guard tube filled with anhydrous CaCl₂ to prevent the entry of moisture from atmosphere.

3. Procedure:
The combustion tube is heated strongly to dry its content. It is then cooled and connected to absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till all the substance in the boat is burnt away. This takes about 2 hours. Finally a strong current of oxygen is passed. Then the U-tube and potash bulbs are then
detached and increase in weight of each of them is determined.

4. Calculation:
Weight of organic substance = W g
Increase in weight of H₂O = x g
Increase in weight of CO₂ = y g
18g of H₂O contains 2 g of hydrogen
∴ x g of H₂O contain \(\frac {1}{2}\) × x g of hydrogen.
∴ Percentage of hydrogen = (\(\frac {2}{18}\) x \(\frac {x}{w}\) x 100
44g of CO₂ contains 12 g of carbon
∴ y g of CO₂ contain \(\frac {12}{44}\) × y g of carbon
∴ Percentage of carbon = (\(\frac {12}{44}\) × \(\frac {12}{44}\) × 100)%

Question 5.
Explain about the estimation of halogens by carius method.
Answer:
Carius method:
A known mass of the substance is taken along with fuming HNO₃ and AgNO₃ taken in a clean carius tube. The open end of the carius tube is sealed and placed in a iron tube for 5 hours in a range at 530 to 540 K. Then the tube is allowed to cool and a small hole is inade in the tube to allow the gases to excape. The tube is broken and the precipitate is filtered, washed, dried and weighed. From the mass of AgX produced percentage of halogen in the organic compound is calculated.

Calculation:
Weight of the organic compound = Wg
Weight of AgCl = a g
143.5 g AgCl contains 35.5g of Cl
a g of AgCl contain \(\frac {35}{143.5}\) x 9 g of Cl
Wg of organic compound contains \(\frac {35}{143.5}\) × 9 g of Cl
% of chlorine = (\(\frac {35}{143.5}\) × \(\frac {a}{w}\) × 100)%
Weight of silver bromide = b g
188 g of AgBr contains 80 g of Br
b g of AgBr contain = \(\frac {80}{188}\) × b g of Br
% of Bromine = (\(\frac {80}{188}\) × \(\frac {b}{w}\) × 100)
Weight of silver iodide = c g
235 g of AgI contains 127 g of I
c g of AgI contain = \(\frac {80}{188}\) × c g of I
% of Iodine = (\(\frac {80}{188}\) × \(\frac {c}{w}\) × 100)

Question 6.
How will you estimate phosphorous in an organic compound?
Answer:
Carius method:
Procedure:
A known mass of organic compound (wg) containing phosphorus is heated with fuming HNO₃ in a sealed tube where C is converted into CO₂ and H to H₂O. Phosphorous present in the organic compound is oxidised to phosphoric acid which is precipitated as ammonium phospho molybdate by heating with conc. HNO₃ and by adding ammonium molybdate.

The precipitate of ammonium phrospho molybdate is filtered, washed, dried and weighed.

Calculation:
Mass of organic compound = Wg
Mass of ammonium phospho molybdate = x g
Molar mass of ammonium phospho molybdate 1877 g
1877 g of ammonium phospho molybdate contains 31 g of phosphorous
x g of ammonium phospho molybdate contain = \(\frac {31}{1877}\) × x g of phosphorous
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {x}{w}\) × 100)% of phosphorous
In an alternate method, phosphoric acid is precipitated as magnesium-ammoniumphosphate by adding magnesia mixture. The ppt. is washed dried and ignited to get magnesium pyrophosphate which is washed, dried and weighed.

Weight of magnesium pyrophosphate = y g
Molar mass of magnesium pyrophosphate = 222 g
222 g of magnesium pyrophosphate contains 62 g of P
y g of magnesium pyrophosphate contain = \(\frac {62}{222}\) × y g of P
% of phosphorous = (\(\frac {31}{1877}\) × \(\frac {y}{w}\) × 100)%

Question 7.
Explain Dumas method of estimation of nitrogen.
Answer:
Dumas method:
Principle:
This method is based on the fact that nitrogeneous compound when heated with cupric oxide in an atmosphere of CO₂ yields free nitrogen.
Traces of nitrogen are reduced to elemental nitrogen by passing over heated copper spiral.
Description of the apparatus:

CO₂ Generator:
CO₂ needed in this process’is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube (or) by the action of dil. HCl on marble in a kipps apparatus. The gas is passed through the combustion tube after dried by bubbling through cone. H₂SO₄.

Combustion tube:
The combustion tube is heated in a furnace is charged with (a) A roll of oxidised copper gauze to prevent the back diffusion of products of combustion and to heat the organic substance mixed with CuO by radiation (b) a weighed amount of organic substance mixed with excess of CuO (c) a layer of CitO packed in about 2/3 length of the tube and kept in position by loose asbestos plug on either side and (d) a reduced copper piral which reduces any oxides of nitrogen formed during combustion of nitrogen.

Schiff’s nitromctc:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO₂. It is estimated by passing nitro meter when CO₂ is absorbed by KOH and the nitrogen gas gets collected in the upper part of the graduated tube.

Calculation:
Weight of the substance taken Wg
Volume of nitrogen = V₁L
Room temperature = T₁K
Atmospheric pressure = p mm Hg
Aqueous tension al room temperature P’ nun of Hg
Pressure of dry nitrogen = P—P’ = P’₁ mm Fig
P₀, V₀ and T₀ be the pressure, volume and temperature respectively of dry nitrogen at S.T.P.
Then, \(\frac{P_{0} V_{0}}{T_{0}}=\frac{P_{1} V_{1}}{T_{1}}\)
V₀ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}}\)
V₀ = \(\frac{P_{1} V_{1}}{T_{1}} \times \frac{273 \mathrm{K}}{760 \mathrm{mm} \mathrm{Hg}}\)
22.4 L of N₂ at STP weigh 28 g of N₂
V₀L of N₂ at STP weigh \(\frac {28}{22.4}\) x V₀
W g of organic compound contain \(\frac {28}{22.4}\) x V₀ g of N₂
100 g of organic contain \(\frac {28}{22.4}\) x \(\frac{V_{o}}{w}\) x 100 = % of Nitrogen

Question 8.
Explain Kjeldahl’s method.
Answer:
Principle:
This method is based on the fact that an organic compound containing nitrogen is heated with cone. H₂SO₄. the nitrogen is conerted Lo ammonium sulphate. The resultant liquid is heated with excess of alkali and then liberated ammonia gas is absorbed in excess of standard acid. The amount of ammonia (nitrogen) is determined by finding the amount of acid neutralised by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance 0.3 to 0.5g is placed in a special long necked Kjeldahl flask made of pyrex glass. About 25 ml of cone. H₂SO₄ together with a little K₂SO₄ and CuSO₄ [catalyst] are added to il. the flask is loosely stoppered by a glass bulb and heated gently in an inclined position.

The heating is continued till the brown colour of the liquid disappears leaving the content clear as before. At this point all the nitrogen is converted to ammonium sulphate. The kjeldahl flask is cooled aiid its contents are diluted with distilled water and carefully transferred into a I litre round bottom flask. An excess NaOH is poured down the side of the flask and it is filled with a kjeldhals trap and a water condenser.

The lower end of the condenser dips in a measured volume of excess of \(\frac {N}{20}\) H₂SO₄ solution. The liquid in the round bottom flask is heated and liberated ammonia is distilled to sulphuric acid. When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = Wg
Volume of H₂SO₄ required for the complete neutralisation of evolved NH₃ = V ml
Strength H₂SO₄ used to neutralise NH₃ = N.
Let the volume and strength of NH₃ formed are V₁ and N₁ respectively.
V₁N₁ = VN
The amount of nitrogen present in W g of organic compound = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times w}}\)
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times w} \times 100\) = \(\frac {1.4NV}{w}\) %

Question 9.
How will you estimate carbon and hydrogen present in the given organic compound?
Answer:
Both carbon and hydrogen are estimated by the same method. A known weight of the organic substance is burnt in excess of oxygen and the carbon and hydrogen present in it are oxidized to carbon dioxide and water, respectively.
C_XH_Y + O₂ + XCO₂ + \(\frac{\mathrm{y}}{2}\)H₂O

The weight of carbon dioxide and water thus formed are determined and the amount of carbon and hydrogen in the original substance is calculated. The apparatus employed for the purpose consists of three units (1) oxygen supply (2) combustion tube (3) absorption apparatus.

Oxygen supply:
To remove the moisture from oxygen it is allowed to bubble through sulphuric acid contained in a drech gel bottle and then passed through a U-tube charged with soda lime to remove CO₂. The oxygen gas free from moisture and carbon dioxide enters the combustion tube.

Combustion tube:
A hard glass tube open at both end is used for the combustion of the organic substance. It is filled with
(i) a role of oxidized copper gauze to prevent the backward diffusion of the product of combustion
(ii) a porcelain boat containing a known weight of the organic substance
(iii) coarse copper oxide packed in about 2/3 of the entire length of the tube, and kept in position by loose asbestos plugs on either side and
(iv) a roll of oxidized copper gauze placed towards the end of the combustion tube to prevent any vapours of the organic substance having the tube unoxidized. The combustion tube is enclosed in a furnace, and heated by a gas burner.

Absorption Apparatus:
The products of combustion containing moisture and carbon dioxide are then passed through the absorption apparatus which consists of
(i) a weighed U-tube packed with pumice soaked in Conc. H₂SO₄ to absorb water
(ii) a set of bulbs containing strong solution of KOH to absorb CO₂ and finally
(iii) a guard tube filled with anhydrous CaCl₂ to prevent the entry of moisture from atmosphere.

Procedure:
To start with, before loading it with the boat, the combustion tube is detached from the absorption unit. The tube is heated strongly to dry its content and CO₂ present in it is removed by passing a current of pure, dry oxygen through it. It is then cooled slightly and connected to the absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced.

The tube is again heated strongly till the substance in the boat is burnt away. This takes about 2 hours. Finally, a strong current of oxygen is passed through the combustion tube to sweep away any traces of carbon dioxide or moisture which may have been left in it. The U-tube and the potash bulbs are then detached and the increase in weight of each of them is determined.

Calculation:
Weight of the organic substance taken = w g
Increase in weight of H₂O = x g
Increase in weight of CO₂ = y g
18 g of H₂O contain 2 g of hydrogen
∴ x g of H₂O contain = \(\left(\frac{2}{18} \times \frac{x}{w}\right)\)
% of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\) %

44 g of CO₂ contains = 12 g of carbon
∴ y g of CO₂ contain = \(\left(\frac{12}{44} \times \frac{y}{w}\right)\) g of carbon
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\) %

Question 10.
Explain steam distillation (or) How is essential oils are recovered from plants and flowers.
Steam distillation:
This method is applicable for solids and liquids. If the compound to be steam distilled and it should not decompose at any steam temperature should have a fairly high vapour pressure at 273 K, it should be insoluble in water and the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round-bottomed flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam.

The mixture in the flask is heated and then a current of steam passed into it. The vapours of the compound mix up with the steam and escape into the condenser. The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers also in the manufacture of aniline and turpentine oil.

Question 11.
Explain about azeotropic distillation.
Answer:
The mixture of liquids that cannot be separated by fractional distillation can be purified by azeotropic distillation. The mixture are called azeotropes. These azeotropes are constant boiling mixture which distill as a single component at a fixed temperature for example ethanol and water in the ratio of 95.87: 4.13. In this method, the presence of a third component C₆H₆, CCl₄, ether, glycol glycerol which act as dehydrating agent depress the partial pressure of one component of azeotropic mixture and raises the boiling point of that component and thus the other component will distil over. Substance like C₆H₆, CCl₄ have low b.pt. and reduce the partial vapour pressure of alcohol more than that of water while substance like glycerol and glycol have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 12.
Explain about thin layer chromatography.
Answer:
A sheet of glass is coated with a thin layer of adsorbent (cellulose, silica gel (or) Alumina). This sheet of glass is called chromplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluant (solvent) .

The eluant is drawn up the adsorbent layer by capillary action. The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retention factor (R_f) value.

The spots of coloured compounds are visible on TLC plate due to their original colour. The colourless compounds are viewed under Uy light or in another method using Iodine crystals or by using appropriate reagent.

Question 13.
An organic compound contains 69% carbon and 4.9% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.
Answer:
Step I.
Calculation of mass of CO₂ produced
Mass of compound = 0.20 g
Percentage of carbon = 69 g

Mass of CO₂ formed = \(\frac{69 \times 44 \times(0.20 \mathrm{g})}{12 \times 100}\) = 0.506g

Step II.
Calculation of mass of H₂O produced
Mass of compound = 0.20 g
Percentage of hydrogen = 4.8%

Mass of H₂O formed = \(\frac{4.8 \times 18 \times(0.20 \mathrm{g})}{2 \times 100}\) = 0.0864 g

Question 14.
0.50 g of an organic compound was Kjeldahlished. The ammonia eolved was passed in 50cm³ of IN H₂SO₂. The residual acid required 60 cm³ of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.
Answer:
Step 1.
Calculation of volume of unused acid
Volume of NaOli solution required = 60 cm³
Normality of NaOH solution = \(\frac {1}{2}\) N
Normality of H₂SO₄ solution = \(\frac {1}{N}\)
Volume of unused acid can be calculated by applying normality equation

1 x V = \(\frac {1}{2}\) x 60 = 30cm³

Step II.
Calculation of volume of acid used
Volume of acid added = 50 cm³
Volume of unused acid = 30 cm³
Volume of acid used = (50 – 30) = 20 cm³

Step III.
Calculation of percentage of nitrogen
mass of compound = 0.05 g
Volume of acid used = 20 cm³
Normality of acid used = 1 N

percentage of nitrogen = \(\frac{1.4 \times 20 \times 1}{0.50}=56 \%\)

Question 15.
In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm3 of N₂ collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen ¡n the compound. (‘apour pressure of water at 300 K is 15 mm Hg)
Answer:
P₁ = 715 – 15 = 700mm Hg, P₁ = 760 mm Hg
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
T₁ = 300 K, T₂ = 273K, V₁ = 50 cm³, V₂ = ?
V₂ = \(\frac{700 \times 50 \times 273}{300 \times 760}\) = 41.9 cm³
% of N = \(\frac {28}{22400}\) x 41.9 x \(\frac {100}{W}\) = 17.46%