Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

Question 1.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO3
(d) NaNO2
Answer:
(a) NaCN

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

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

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

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

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

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

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
Answer:
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
Solution:
For a first order reaction
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 1

Question 8.
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.24 × 10-4 mol L-1 solubility product of Ag2C2O4 is
(a) 2.42 × 10-8 mol3 L-3
(b) 2.66 × 10-12mol3 L-3
(c) 4.5 × 10-11mol3 L-3
(d) 5.619 × 10-12mol3 L-3
Answer:
(d) 5.619 × 10-12mol3 L-3
Solution:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 2

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

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

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

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
cTamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 3
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 4
(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 5
Answer:
(c) benzene diazonium chloride

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

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

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

Part – II

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

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

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 6

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

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

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

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

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

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

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

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

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

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

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

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 7
This reaction follows SN1 mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 8

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 9
(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 10
This is the reason why soaps do not work in hard water.

Part – III

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

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

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

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

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H2SO4.H2O) and di (H2SO4.H2O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 11

Question 27.
Give evidence that [CO(NH3)gCl]SO4 and [CO(NH3)5SO4]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO3 solution and BaCl2 solution. When Cl ions are the counter ions, a white precipitate will be obtained with AgNO3 solution. If SO42- ions are the counter ions, a white precipitate will be obtained with BaCl2 solution.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 12

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 13z = k [x]m [y]n …… (1)
8z = k [4x]m [y]n…… (2)
I 6z = k [4x]m [4y]n…… (3)
Dividing Eq (2) by Eq (1) we get,
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 14
1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 15
16 = 4m . 4n
16 = 42.4n
16/16 = 4n
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]m[y]n
k [x]1.5 [y]0
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 16Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 17

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

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 18

  • HPO42- and PO43- , SO3-2 and HSO3 are two conjugate acid – base pairs.
  • PO43- is the conjugate base of the acid HPO42- (or)
    HPO42- is the conjugate acid of PO43-.
  • HSO3 is the conjugate acid of SO3-2 (or) SO3-2 is the conjugate base of HSO3

(iii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 19

  • NH4+ and NH3, CO32- and HCO3 are two conjugate acid – base pairs.
  • HCO3 is the conjugate of acid CO32- (or) CO32- is the conjugate bases of HCO3
  • NH3 is the conjugate base of NH4+ (or) NH4+ is the conjugate acid of NH3.

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

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 31.
Identify A, B, C, and D
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 20
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 21
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 22

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 23

Question 33.
Differentiate soap and detergents?
Soap :

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

Detergent :

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

Part – IV

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

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

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

In dibome, the boron is sp3 hybridised. Three of the four sp3 hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 24

[OR]

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

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

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

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

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

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 25

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

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

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

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

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

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

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

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

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 26

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

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 27

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

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

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

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

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.
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Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10-3s-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of kspof two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10-15 and 6 × 10-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t1/2 = 0.693/ k
t1/2= O.693/1.54 ×10-3s -1 = 450s

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

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

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

[OR]

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

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

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

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 29
Ni(OH)2 is more soluble than AgCN.

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

2. Oxidation occurs at the anode

Pb(s) → Pb2+ (aq) + 2e …………(1)

The Pb2+ ions combine with SO42- to form PbSO4 precipitate

Pb2+(aq) + SO42- → PbS4(s) …………….(2)

3. Reduction occurs at the cathode

PbO2(s) + 4H+(aq) + 2e – → Pb2+(aq) + 2H2O ………….(3)

The Pb2+ ions also combines with SO42- ions to form sulphuric acid to form PbSO4 Precipitate

Pb2+(aq) + SO42-(aq) → PbSO4 …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb(s) + PbO2(s) + 4H+ + 2SO42-(aq) → 2PbSO4(s) + 2H2O(l)

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

6. The emf of the cell depends on the concentration of H2SO4. As the cell reaction uses SO42- ions, the concentration H2SO4 decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H2SO4 is regenerated.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 30
The above reaction is exactly the reverse of redox reaction which takes place while discharging.

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

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 31

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH3, – C2H5, – C3H7
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 32

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

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

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C6H5CONH2). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 33

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 34

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

[OR]

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

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 35

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A is a non-singular matrix such that A-1 = \(\left[\begin{array}{rr}
5 & 3 \\
-2 & -1
\end{array}\right]\), Then (AT)-1 = _________.
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 1
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 2
Answer:
(d) \(\left[\begin{array}{ll}
5 & -2 \\
3 & -1
\end{array}\right]\)

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

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

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

Question 7.
tan-1 (\(\frac{1}{4}\)) + tan-1 (\(\frac{2}{9}\)) is equal to ________.
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 3
Answer:
\(\tan ^{-1}\left(\frac{1}{2}\right)\)

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

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

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

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

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

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

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

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

Question 18.
If X is a binomial random variable with expected value 6 and variance 2.4, Then P {X = 5} is _______.
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 4
Answer:
(d) \(\left(\begin{array}{c}
10 \\
5
\end{array}\right)\left(\frac{3}{5}\right)^{5}\left(\frac{2}{5}\right)^{5}\)

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

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

Part – II

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

Question 21.
Using elementary transformation find the inverse of the matrix \(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
Answer:
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 5

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

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

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

Question 26.
In each of the following cases, determine whether the following function is homogeneous or not. If it is so, find the degree g (x, y, z) = \(\frac{\sqrt{3 x^{2}+5 y^{2}+z^{2}}}{4 x+7 y}\)
Answer:
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 6
∴ It is homogeneous function of degree 0.

Question 27.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = e-2x, y = 0, x = 0 and x = 1.
Answer:
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 7

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

Question 28.
Compute P(X = k) for the binomial distribution, B (n,p) where n = 10, p = \(\frac{1}{5}\), k = 4
Answer:
n = 10, p = \(\frac{1}{5}\), k = 4
∴ q = 1 – p = 1 – \(\frac{1}{5}=\frac{4}{5}\)
P(X = x) =nCx pxqn-x, x = 0, 1, 2, …….n.
P (X = k) = P (X = 4)
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 8

Question 29.
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 9
be any three boolean matrices of the same type. Find A ∧ B
Answer:
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 10

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

Part – III

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

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

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

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

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

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

Question 35.
If the radius of a sphere is measured as 7m with an error of 0.02 m then find the approximate error in calculating its volume.

Question 36.
Find the volume of the solid that results when the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (a > b > 0) is revolved about the minor axis.

Question 37.
Verify that the function y = e is a solution of the differential equation \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-6 y=0\)

Question 38.
Find the mean and variance of the distribution \(f(x)=\left\{\begin{array}{cc}
3 e^{-3 x}, & 0<x<\infty \\
0, & \text { elsewhere }
\end{array}\right.\)

Question 39.
Let A = {a + \(\sqrt{5}\) b : a, b ∈ Z} . Check whether the usual multiplication is a binary operation on A.

Question 40.
If \(\frac{z+3}{z-5 i}=\frac{1+4 i}{2}\) find the complex number z.

Part – IV

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

Question 41.
(a) Solve, by Cramer ’s rule, the system of equations
x1 – x2 = 3, 2x1 + 3x2 + 4x3 = 17,  x2 + 2x3 = 7
[OR]
(b) A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume.

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

Question 42.
(a) Solve the equation z3 + 8i = 0, where z ∈ C.
[OR]
(b) Solve (1 + 2ex/y)dx + 2ex/y \(\left(1-\frac{x}{y}\right)\) dy = 0

Question 43.
(a) Find the area of the region bounded between the parabola x2 =y and the curve y = |x|.
[OR]
(b) Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}\) and passing through the point (-1, 1, -1).

Question 44.
(a) Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation \(\frac{x^{2}}{30^{2}}-\frac{y^{2}}{44^{2}}=1\). The tower is 150 m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
[OR]
(b) If 2 + i and 3 – \(\sqrt{2}\) are roots of the equation
x6 – 13x5 + 62x4 – 126x3 + 65x2 + 127x – 140 = 0 find all roots.

Question 45.
(a) If u = \(\sin ^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)\) show that \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\frac{1}{2} \tan u\)
[OR]
(b) The cumulative distribution function of a discrete random variable is given by.
Tamil Nadu 12th Maths Model Question Paper 1 English Medium 11
Find (i) the probability mass function (ii) P(X < 3) and (iii) P(X ≥ 2).

Question 46.
(a) Prove that: \(\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)
[OR]
(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for following operation on the given set. m*n = m + n – mn ; m, n ∈ Z

Tamil Nadu 12th Maths Model Question Paper 1 English Medium

Question 47.
(a) Find the equation of the circle passing through the points (1, 1), (2, -1), and (3, 2).
[OR]
(b) Evaluate: \(\int_{0}^{\pi / 2} \frac{d x}{4+9 \cos ^{2} x}\)

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

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

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

General Instructions:

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

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between points A and C is……………………………..
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 1/4 μF
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 1
Answer:
(b) 2 μF

Question 2.
If the electric field in a region is given by \(\overrightarrow{\mathrm{E}}=5 \hat{i}+4 \hat{j}+9 \hat{k}\), then the electric flux through a surface of area 20 units lying in the
y – z plane will be …………………
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hint. The area vector
\(\overrightarrow{\mathrm{A}}=20 \hat{i} ; \overrightarrow{\mathrm{E}}=(5 \hat{i}+4 \hat{j}+9 \hat{k})\)
Flux ( φ)\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) = 5 x 20 = 100 units

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 3.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 2
(a) π Ω
(b) \(\frac{\pi}{2} \Omega\)
(c) 2π Ω
(d) \(\frac{\pi}{4} \Omega\)
Answer:
(b) \(\frac{\pi}{2} \Omega\)

Question 4.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed co. Find the ratio of its magnetic moment with angular momentum is …………..
(a) M
(b) \(\frac{3}{\pi} \mathrm{M}\)
(c) \(\frac{2}{\pi} \mathrm{M}\)
(d) \(\frac{1}{2} \mathrm{M}\)
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 3
Answer:
(b) \(\frac{3}{\pi} \mathrm{M}\)

Question 5.
A proton enters a magnetic field of flux density 1.5 Wb/m2 with a speed of 2 x 107 m/s at angle of 30° with the field. The force on the proton will be ……………….
(a) 0.24 x 10-12 N
(b) 2.4 x 10 -12 N
(c) 24 x 10-12 N
(d) 0.024 x 10-12 N
Answer:
(b) 2.4 x 10 -12 N
Hint: F = Bqv sin θ = 1.5 x 1.6 x 10-19 x 2 x 107 x sin 30°= 2.4 x 10-12 N

Question 6.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\) Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) . The power factor of the of the circuit is ……………
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(c) 1

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 7.
The inductance of a coil is proportional to…………………………………….
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 8.
The electric and magnetic fields of an electromagnetic wave are……………….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other

Question 9.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……………
(a) get shifted downwards
(b) get shifted upward
(c) will remain the same
(d) data insufficient to conclude
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 4
Answer:
(b) get shifted upward

Question 10.
The wavelength λe of an electron and λp of a photon of same energy E are related ………..
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 5
Answer:
(d) \(\lambda_{p} \propto \lambda_{e}^{2}\)

Tamil Nadu 12th Physics Model Question Paper 2 English Medium 6

Question 11.
A system consists of No nucleus at t = 0. The number of nuclei remaining after half of a half-life (that  is, at time \(t=\frac{1}{2} \mathrm{T}_{\frac{1}{2}}\)
Answer:
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 7
Hint:
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 8

Question 12.
In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called……………………………….
(a) acceptor
(b) donor
(c) intrinsic semiconductor
(d) extrinsic semiconductor
Answer:
(c) intrinsic semiconductor
Hint: Pure semiconductors are called intrinsic semiconductors.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 13.
The light emitted in an LED is due to……………………………
(a) Recombination of charge carriers
(b) Reflection of light due to lens action
(c) Amplification of light falling at the junction
Answer:
(a) Recombination of charge carriers

Question 14.
The frequency range of 3 MHz to 30 MHz is used for………………………………..
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
The materials used in Robotics are……………………..
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Part – III

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

Question 16.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 17.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
\(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)
The S.I. unit of current density. \(\frac{\mathrm{A}}{\mathrm{m}^{2}}(\text { or }) \mathrm{Am}^{-2}\)

Question 18.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to magnetising field \((\overrightarrow{\mathrm{H}})\)
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 19.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 20.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Answer:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, φ = 4 mWb = 4 x 10-3 Wb
Induction of the coil , L
\(\mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4}=2 \mathrm{H}\)

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 21.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 22.
How many photons of frequency 1014 Hz will make up 19.86 J of energy?
Answer:
Total energy emitted per second = Power x time
19.863 = Power x is
∴ Power 19.86 W
Number of photons =
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 9

Question 23.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 1010 decays/s

Question 24.
A transistor having α =0.99 and VBE = 0.7V, is given in the circuit. Find the value of the collector current.
Answer:
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 10
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 11

Part – III

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

Question 25.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 26.
A conductor of linear mass density 0.2 g m-1 suspended ,by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the x page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s-2 = 111.87.
Answer:
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 12
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 13

Question 27.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 28.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = (u)c

Question 29.
If the focal length is 150 cm for a glass lens, what is the power of the lens?
Answer:
Given: focal length,f = 150 cm (or) f= 1.5 m
Equation for power of lens is, P = 1/f
Substituting the values,
\(P = \frac{1}{1.5}\)= 0.067 diopter
As the power is positive, it is a converging lens.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 30.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglic wavelength of the particle is \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\) \(\text { i.e. } \lambda \propto \frac{1}{\sqrt{m}}\)
As me <<mp ,so λe >>λP
Hence protons have greater de- broglic wavelength

Question 31.
Distinguish between avalanche and zener breakdown.
Answer:

Avalanche Breakdown Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths. It occurs in junctions which are heavily doped and have narrow depletion widths.
It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision. It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
At reverse voltage above 6V breakdown is due to avalanche effect. At reverse voltage below 6V breakdown is due to zener effect.
Electric field produced is weak in nature. A strong electric field is produced
Charge carriers obtain energy from the applied potential. Zener current is independent of applied voltage.

Question 32.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Question 33.
What are black holes?
Answer:
Black holes are end stage of stars Which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Part – IV

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

Question 34.
(a) How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 14

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements
Δq1, Δq2, Δq3 ….. Δqn ……… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such change elements
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 15
Here Δ qi is the ith charge element, r is the distance of the point P from the zth charge element and \(\hat{r}_{i \mathrm{P}} \) is the unit vector from ith charge element to the point P.

However, the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit Δq → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}\)

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r}\) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\overrightarrow{\mathrm{F}}=q \overrightarrow{\mathrm{E}}\)

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = Q/L. Its unit is coulomb per meter (Cm-1). The charge present in the infinitesimal length dl is dq = λ dl
The electric field due to the line of total charge Q is given by
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 16
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 17

(b) Surface charge distribution: If the charge Q is uniformly distributed on. a surface of area A, then surface charge density (charge per unit area) is \(\lambda=\frac{Q}{L}\). Its unit is coulomb per square meter (Cm-2 ). The charge present in the infinitesimal area dA is dq = adA. The electric field due to a of total charge Q is given by
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 18

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by \(\rho=\frac{Q}{V}\) . Its unit is coulomb per cubic meter (Cm-3 ). The charge present in the infinitesimal volume element dV is dq = ρdV.
The electric field due to a volume of total charge Q is given by
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 19

[OR]

(b) Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as
V = El
As we know, the magnitude of current density
\(\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{\mathrm{V}}{l}\)
But \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\),so we write the equation as
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
By rearranging the above equations, we get
\(\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma \mathrm{A}}\right)\)
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 20
The quantity \(\frac{l}{\sigma \mathrm{A}}\) is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR …………….. (3)

Question 35.
(a) Calculate the magnetic held inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid: Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 21
In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\) = μ x (total current enclosed by Amperian loop)
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 22
Since the elemental lengths along be and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 23
Since the magnetic field outside the   solenoid is zero, the integral
\(\int_{c}^{d} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0\)
For the path along ab, the integral is
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 24
where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 25
Let NI be the current passing through the solenoid of N turns, then
Tamil Nadu 12th Physics Model Question Paper 2 .29
The number of turns per unit length is given by \(\frac{\mathrm{NI}}{\mathrm{L}}=n\) then
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
Since n is a constant for a given solenoid and μo is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

[OR]

(b) Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle: The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 27

Construction: In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working: If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil. This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.

As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil εp is almost equal and opposite to the applied voltage υp and is given by
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil εs is given by
\(\varepsilon_{\mathrm{s}}=-N_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ………………… (1)
where Np and Ns are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then  εs = υs where υs is the voltage across secondary coil.
\(v_{s}=\varepsilon_{s}=-\mathrm{N}_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ……….. (2)

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

From equations (1) and (2),
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=K\) …………………… (3)

This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υp ip= Output power υsis
where ipand is are the currents in the primary and secondary coil respectively. Therefore,
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=\frac{i_{p}}{i_{s}}\) ………………….. (4)

Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}=K\)

(i) If Ns> Np ( or K > 1), ∴ Vs > Vp and Is < Ip. This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.

(ii) If Ns < Np (or K < 1) , ∴ Vs < Vp and Is > Ip . This is step-down transformer where voltage is decreased and the current is increased.

Question 36.
(a) Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves: Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows. If the charged -particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 28

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves. Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
Ey =E0 sin (kz-ωt)
Br = B0 sin(kz – ωt) where, Eo and Bo are amplitude of oscillating electric and magnetic field,\(\hat{k} \) is a wave number, ω is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between Eo and Bo is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac{E_{0}}{B_{0}}\)

In any medium, the ratio of Eo and Bo is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
\(v=\frac{E_{0}}{B_{0}}<c\)
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

[OR]

(b) Explain about compound microscope and obtain the equation for magnification.
Answer:
Compound microscope:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium 29

Magnification of compound microscope
From the ray diagram, the linear magnification due to the objective is,
\(m_{0}=\frac{h^{\prime}}{h}\)
from the figure ,\(\tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L} \), then
\(\frac{h^{\prime}}{h}=\frac{L}{f_{0}} ; m_{o}=\frac{L}{f_{o}}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as fo and fe are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification me of the eyepiece is,
\(m_{e}=1+\frac{D}{f_{e}}\)

The total magnification m in near point focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)

If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is
\(m_{e}=\frac{D}{f_{e}}\)

The total magnification m in normal focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 37.
(a) Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:

  • This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
  • The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
  • De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
  • As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
  • Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium 30

Working:

  • The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
  • The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
  • With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.

[OR]

(b) Discuss the process of nuclear fission and its properties.
Answer:

  • When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
  • The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
  • The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.
  • Uranium undergoes fission reaction in 90 different ways. The most common fission reactions of
    Tamil Nadu 12th Physics Model Question Paper 2 .35
  • Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state. \(_{ 92 }^{ 236 }{ U }\) . But this excited state does not last longer than 10-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium 32

Question 38.
(a) Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut-off regions functions like an electronic switch that helps to. turn ON or OFF a given circuit by a small control signal.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 33

Presence of dc source at the input (saturation region):
When a high input voltage (V. = +5V) is applied, the base current (I ) increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current (Ic) increases the voltage drop across Rc .thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.

Absence of dc source at the input (cut-off region):
A low input voltage (Vin = OV), decreases the base current (IB) and in turn decreases the collector current (Ic ). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (Ic) decreases the drop across, thereby increasing the output voltage, dose to +5 V. The transistor acts as an open switch which is considered as the OFF condition.

It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

[OR]

(b) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified.
They are

  1. Amplitude modulation,
  2. Frequency modulation, and
  3. Phase modulation.

1. Amplitude Modulation (AM): If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure
(a) is the message signal or baseband signal that carries information, figure
(b) shows the high-frequency carrier signal and figure
(c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 34

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. An increase in the amplitude of the ‘ baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave.

Tamil Nadu 12th Physics Model Question Paper 2 English Medium

Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C). The increase in amplitude in the negative half cycle (B, D) reduces the frequency of the modulated wave (Figure (c)).
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 35

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.
Tamil Nadu 12th Physics Model Question Paper 2 English Medium 36

The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium

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

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

Question 1.
Wolframite ore is separated from tinstone by the process of
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer:
(d) Electromagnetic separation

Question 2.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 1
Identify A
(a) BN3
(b) B3N
(c) (BN)3
(d) BN
Answer:
(d) BN

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

Question 7.
For a first order reaction, the rate constant is 6.909 min-1.The time taken for 75% conversion
in minutes is …………………………..
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 2
Answer:
(b) \(\left(\frac{2}{3}\right) \log 2\)
Solution:
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 3

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

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

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

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

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

Question 13.
Match the following:
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 4
Answer:
(a) A – 3, B – 4, C – 1, D – 2

Question 14.
Which one of the following is levorotatory?
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 5
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 6

Question 15.
Which one of the following structures represents nylon 6,6 polymer?
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 7
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 8

Part – II

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

Question 16.
Carbon monoxide is more effective reducing agent than carbon below 983 K but, above this temperature, the reverse is true -Explain.
Answer:
From the Ellingham diagram, we find that at 983 K, the curves intersect.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 9
The value of ∆G° for change of C to CO2 is less than the value of ∆G° for change of CO to CO2 Therefore, coke (C) is a better reducing agent than CO at 983K or above temperature. However below this temperature (e.g. at 673K), CO is more effective reducing agent than C.

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

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

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

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

Question 22.
Alcohol can act as Bronsted base. Prove this statement.
Answer:
Alcohols can also act as a Bronsted bases. It is due to the presence of unshared electron pairs on oxygen which make them to accept proton. So proton acceptor are Bronsted bases. i. e., alcohols are Bronsted bases.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 10

Question 23. Arrange the following in decreasing order of basic strength
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 11
Answer:
(i) Aliphatic amines are more basic than aromatic amines. Therefore CH3CH2NH2 and CH3NH2 are more basic. Among the ethylamine and methylamine, ethylamine was experienced more +1 effect than methylamine and hence ethylamine is more basic than methylamine.

(ii) Nitrogroup has a powerfol electron withdrawing group and they have both -R effect as well as -I effect. As a result, all the nitro anilines are weaker bases than aniline. In P-nitroaniline
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 12
both R effect and -I effect of the NO2 group decrease the basicity.

(iii) Therefore decreasing order of basic strength is,
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 13

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

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

Part – III

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

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

BaO2 + Mg → BaO + Mgo

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is : 852 kJ-moF1) which facilitates the reduction of Cr2O3 by aluminium power.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 14

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

(ii) On heating potash alum (A) 500 K give anhydrous potash alum (or) burnt alum (B).
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 15

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

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

Question 28.
Define half life of a reaction. Show that for a first order reaction half life is independent of initial concentration.
Answer:
Half life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value.
For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.
The rate constant for a first order reaction is given by,
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 16

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

Determination of the molar conductivity of weak electrolyte at infinite dilution:
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraush’s Law. For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCl, NaCl and CH3COONa .
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 17
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 18

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

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

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

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

Question 32.
Write the structure of all possible dipeptides which can be obtained from glycine and alanine.
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 19

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

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

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

Part – IV

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

Question 34.
(a) (i) How will you purify metals by using iodine?(3)
(ii) Boron does not react directly with hydrogen. Suggest one method to
prepare diborane from BF3(2)
[OR]
(b) (i) Write the reason for the anomalous behaviour of Nitrogen. (3)
(ii) Mn2+ is more stable than Mn4+. Why? (2)
Answer:
(a) (i) This method is based on the thermal decomposition of metal compounds which lead to the formation of pure metals. Titanium and zirconium can be purified using this method. For example, the impure titanium metal is heated in an evacuated vessel with iodine at a temperature of 550 K to form the volatile titanium tetra-iodide.( TiI4 ). The impurities are left behind, as they do not react with iodine.
Ti(s) + 2I2(s) → TiI4 (vapour)

The volatile titanium tetraiodide vapour is passed over a tungsten filament at a
temperature around I 800 K. The titanium tetraiodide is decomposed and pure titanium
is deposited on the filament. The iodine is reused.
TiI4 (vapour) ) → Ti(s) +2I2(s)

(ii) Boron does not react directly with hydrogen. However it forms a variety of hydrides
called boranes. Treatment of gaseous boron triuluoride with sodium hydride around
450 K gives diborane.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 20

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

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

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

Question 35.
(a) (i) Draw all possible stereo isomers of a complex Ca[CO(NH3)Cl(Ox)2] (3)
(ii) What is Bragg’s equation? (2)
(b) (i) What is an elementary reaction? Give the differences between order and molecularity of a reaction. (3)
(ii) In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction? (2)
Answer:
(a) (i) Possible stereo isomers of a complex Ca[CO(NH3)Cl(Ox)2]
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 21

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

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

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

[OR]

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

Differences between order and molecularity:

Order of a reaction :

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

Molecularity of a reaction :

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

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 22

Question 36.
(a) (i) Derive the relation between pH and pOH (3)
(ii) Give three uses of emulsions.(2)
[OR]
(b) How would you measure the conductivity of ionic solutions? (5)
Answer:
(a) (i) pH = -log1o [H3O+] …………..(1)
pOH = -log1o [OH] ……..(2)
Adding equations (1) and (2),
pH + pOH = (-log1o[H3O+]) + (-log1o [OH])
= -[(log10tHO+]) + (logio [OfT])]
pH + pOH = -log10[H3O+] [OH ]
[H3O+] [OH] = Kw
∴ pH + pOH = – log1oKw
pH + pOH = pKw [pkw = -log1oKw]

At 25°C, the ionic product of water Kw =1 x 10-14.
pKw = -log1o10-14 = 14log1o10= 14
pKw = 14
∴ pH + pOH = 14 at 25°C.

(ii)

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

[OR]

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

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

A wheatstone bridge is constituted using known resistances P,Q, a variable resistance S and conductivity cell. An AC source (550 Hz to 5 KHz) is connected between the junctions A and C.
A suitable detector is connected between the junctions B and D.
The variable resistance S is adjusted until the bridge is balanced and in this conditions, there is no current flow through the detector.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 23
Under balanced condition
P/Q = R/S
∴ R = P/Q x S …………(1)
The resistance of the electrolytic solution (R) is calculated from the known resistance values P, Q and the measured S value using the equation (1).

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

Question 37.
(a) (1) What is metamerism? Give the structure and ¡UPAC name of metamers of 2-methoxy propane (2)
(ii) Explain the following reactions.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 24(2)
(OR]
(b) An organic compound (A) of molecular formula C7H8O on oxidation with alkaline KMnO4 gives (B) of formula C7H6O.
(B) on reaction with Cl2 in the presence of catalyst FeCl3 gives
(C) of formula C7H5OCl. (B) on reaction with Cl in the absence
of catalyst gives C7H5OCl. Identify A,B,C,D and explain the reaction involved.
Answer:
(a) (i) Metamerism: It is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 25
Ethoxy ethane and 1-methoxy propane are metamers of 2-methoxy propane.

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 26

(b)
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 27

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

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

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

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

(ii)

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

[OR]

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

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

2. Phenol reacts with methanal to form ortho or para hydroxyl methyl phenols which on further reaction with phenol gives linear polymer called novolac. Novolac on further healing with formaldehyde undergoes cross linkages to form bakelite.
Tamil Nadu 12th Chemistry Model Question Paper 3 English Medium - 28

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Students can Download Tamil Nadu 12th Biology Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Biology Model Question Paper 5 English Medium

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

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

Question 1.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) Sporogenous cell is epidermal
(d) Ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

Question 4.
Match the following column I with column II
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 1
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 2
Answer:
(D) 1 – c, 2 – d, 3 – a, 4 – b

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

Part – II

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

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

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

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

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

Part – III

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 16.
What do you mean by Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into embryoids.

Applications:

  • Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
  • Somatic embryoids can be used for the production of synthetic seeds.
  • Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

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

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

Question 18.
Write a brief note on Detritus food chain.
Answer:
Detritus food chain is a type of food chain which begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems. The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 3

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 19.
Mention the name of man-made cereal. How it is developed?
Answer:
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 4
Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

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

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

Part – IV

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

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

Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 5
In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root.

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

[OR]

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

(ii)

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

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

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

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

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

[OR]

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

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

Bio-Zoology [Maximum Marks: 35]

Part – I

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

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

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

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

Question 7.
Match column I with column II
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 6
(a) A – 4, B – 5, C – 2, D – 3, E – 1
(b) A – 3, B – 1, C – 4, D – 2, E – 5
(c) A – 2, B – 3,C – 1, D – 5, E – 4
(d) A – 5, B – 4, C – 2, D – 3, E – 1
Answer:
(a) A – 4, B – 5, C – 2, D – 3, E – 1

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

Part – II

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

Part – III

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

Question 15.
Write a short note on foeto scope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 17.
State Fisher and Race hypothesis.
Answer:
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 7
In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

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

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

Part – IV

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

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

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

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

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

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

[OR]

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

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

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

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

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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

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

Passive Immunity:

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

[OR]

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

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

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA, This process is also called DNA amplification.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 8

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

Question 4.
The correct order of increasing oxidizing power in the series
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 1
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 2
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 3

Hint: VO2+ < Cr2O72- < MnO4 greater the oxidation state, higher is the oxidising power.

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

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

Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 4

Question 7.
In a homogeneous reaction
A → B + C + D, the initial pressure was P0 and after time t it was P. Expression for rate constant in terms of P0 , P and t will be ……..
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 5
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 6
Solution:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 7
Question 8.
The solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is
(a) 4x3
(b) 108x5
(c) 27x4
(d) 9x
Answer:
(a) 4x3
Solution:
Mg(OH)2 Mg2+(x) + 2OH ((2x)2)
Ksp = 4x3

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0. lmol of chlorine gas using a current of 3 A is ……………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Hint: m = ZIt (mass of 1 mole of Cl2 gas = 71)
Answer:
(b) 107.2 minutes
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 8

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

Question 11.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 9
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 10
Answer:
(a) A- 2, B – 3, C – 4, D – 1

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

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

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

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

Part – II

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

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

Question 17.
Complete the following reactions:
(a) B(OH)3 + NH3
(b) Na2B4O7 + H2SO4 + H2O
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 11

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

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

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

Question 21.
Write a note about Helmoholtz double layer.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This layer attracts the oppositely charged ions in the medium and hence the boundary separating the two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges, they cannot come close and condense. Hence this helps to explain the stability of the colloid.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 12

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

Question 23.
What happens when
Oxidation of acetoneoxime with trifluoroperoxy acetic acid.
Answer:
Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 13

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

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

Part – III

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

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

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

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 14

Question 26.
How will you prepare phosphine and explain the purification of phosphine?
Phosphine’is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 15

Phosphine is freed from phosphine dihydride (P2H4) by passing through a freezing mixture. The dihydride condenses while phosphine does not.
Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 16
Phosphine is prepared in pure form by heating phosphorous acid.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 17
A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda S0lution.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 18

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

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

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

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

Anisotropy :

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

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

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

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

Question 31.
1mole of Hlis allowed to react with t-butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 19
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 20

Question 32.
How will you prove the presence of aldehyde group in glucose?
Answer:
(i) Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen’s reagent) and alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver and Fehling’s solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 21

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

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

Part – IV

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

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

(ii) The alunite the alum stone is the naturally occurring form and it is K2SO4 . Al2(S04)3.4Al(0H)3. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of – potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 22 (2) (2)

(b)(i) Reaction between chlorine with cold NaoH
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 23
Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

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

Question 35.
(a) (i) Discuss the ortho and pyro silicates. (2)
(ii) Compare lanthanides and actinides. (3)
[OR]
(b) (i) Ni2+ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction. (3)
Cu+, Zn2+, Sc3+, Ti4+ are colourless. Prove this statement. (2)
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO4]4- tetrahedral units are called ortho silicates or neso silicates.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 24
Examples: Phenacite – Be2SiO4(Be2+ ions are tetrahedrally surrounded by O2- ions), Olivine – (Fe/Mg)2 SiO4 (‘Fe2+ and Mg2+ cations are octahedrally surrounded by O
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 25

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

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

(ii)
Lanthanoids

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

Actinoids

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

[OR]

(b) (i) Ni2+ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called [Ni(DMG)2]
Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of [Ni(DMG)2] complex.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 26

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

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

(OR)

(b) (i) Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of Kh for that reaction. (3)
(ii) Identify the Lewis acid and the Lewis base in the following reactions. (2)
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 27
Answer:
(a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water

NaOH(aq) + CH3COOH(aq) ⇌ CH3COONa(aq) + H2O

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

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

[OR]

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

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

4.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 28
such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 29
Kh value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law Kh = h2C and \(\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}}\)
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 30

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

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

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

[OR]

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

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

4. The relationship between molar conductance and equivalent conductance is
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 31

(ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 32

2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 33

3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions.
Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 34

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium

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

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

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

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

5. Structure of α – helix

Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium - 35

[OR]

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

Thermoplastic :

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

Thermosetting

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