Search Results for:

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 70

Part – I

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

Question 1.
Which one of the following ore is best concentrated by froath – floatation method?
(a) Magnetite
(b) Haematite
(c) Galena
(d) Cassiterite
Answer:
(c) Galena

Question 2.
Which compound is used as flux in metallurgy?
(a) Boric acid
(b) Borax
(c) Diborane
(d) BF₃
Answer:
(b) Borax

Question 3.
The shape of XeOF₄ is
(a) T Shaped
(b) Pyramidal
(c) Square planar
(d) Square pyramidal
Answer:
(d) Square pyramidal

Question 4.
How many moles of acidified KMnO₄ required to oxidise one mole of oxalic acid?
(a) 5
(b) 0.6
(c) 1.5
(s) 0.4
Answer:
(b) 0.6

Question 5.
The type of isomerism exhibited by [Pt(NH₃)₂ Cl₂] ?
(a) coordination isomerism
(b) linkage isomerism
(c) optical isomerism
(d) geometrical isomerism
Answer:
(d) geometrical isomerism

Question 6.
The fraction of the total volume occupied by the atoms in a fcc is

Answer:
(a) \(\frac{\pi \sqrt{2}}{6}\)

Question 7.
The half life period of a radioactive element is 140 days. After 280 days 1g of element will be
reduced to which amount of the following?
(a) 1/4
(b) 1/16
(c) 1/8
(d) 1/2
Answer:
(a) 1/4
Solution:

Question 8.
Which is not a Lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F
Answer:
(a) BF₃

Question 9.
During electrolysis of molten copper chloride, the time required to produce 0.2 mole of chlorine gas using a current of 2A is ………..
(a) 32.66 min
(b) 321.66 min
(c) 378 min
(d) 260 min
Solution:
m = ZIt (mass of 1 mole of Cl2 gas = 71) m
t = \(\frac{\mathrm{m}}{\mathrm{Zl}}\) (mass of 0.2 mole of Cl2 gas = 14.2 g)

Question 10.
Smoke is a colloidal solution of …………
(a) Solid in gas
(b) Gas in gas
(c) Liquid in gas
(d) Gas in liquid
Answer:
(c) Liquid in gas

Question 11.
Iso propyl benzene on oxidation in presence of air and dilute acid gives
(a) C₆H₅COOH
(b) C₆H₅COCH₃
(c) C₆H₅COC₆H₅
(d) C₆H₅OH
Answer:
(d) C₆H₅OH

Question 12.
But-2 ene on ozonolyis followed by subsequent cleavage with Zn and water gives …………….
(a) ethanal
(b) Propanal
(c) Propanone
(d) Methanal
Answer:
(a) ethanal

Question 13.
3(i)
This reaction is known as
(a) Friedal- craft’s reaction
(b) HVZ reaction
(c) Schotten – Baumann reaction
(d) Cannizaro reaction
Answer:
(c) Schotten – Baumann reaction

Question 14.
The pyrimidine bases present in DNA are
(a) Cytosine and Adenine
(b) Cytosine and Guanine
(c) Cytosine and Thiamine
(d) Cytosine and Uracil
Answer:
(c) Cytosine and Thiamine

Question 15.
Nylon is an example of
(a) Polyamide
(b) Polythene
(c) Polyester
(d) Polysaccharide
Answer:
(a) Polyamide

Part – II

Answer any six questions. Question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Write a test to identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H₂SO₄, an ester triethyl borate is formed. The Vapour of this ester bums with a green edged flame and this reaction is used to identify the presence of borate.

Question 17.
How is pure phosphine prepared from phosphorous acid ?
Answer:
Phosphine is prepared in pure form by heating phosphorous acid.

Question 18.
What are ionisation isomers? Explain with an example.
Answer:
1. Ionisation isomerism arises when an ionisable counter ion (simple ion) itself can act as a ligand.

2. The exchange of such counter ions with one or more ligands in the coordination entity will result in ionisation isomers. These isomers will give different ions in solution.

3. For example, consider the coordination compound [Pt (en)₂ Cl₂]Br₂. In this compound, both Br^– and Cl^– have the ability to act as a ligand and the exchange of these two ions result in a different isomer [Pt(en)₂Br₂]Cl₂. In solution, the first compound gives Br^– ions while the later gives Cl ions and hence these compounds are called ionization isomers.

Question 19.
What is pseudo first order reaction? Give one example.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction.
Let us consider the acid hydrolysis of an ester,

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., concentration of water remains almost a constant.
Now we can define k [H₂O] = k’
.’. The above rate equation becomes, Rate = k’ [CH₃COOCH₃]
Thus it follows first order kinetics.

Question 20.
State Faraday’s second law of electrolysis.
Answer:
When the same quantity of charge is passed through the solutions of different electrolytes, the , amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

Question 21.
How will you convert glycerol into acrolein?
Answer:

 

Question 22.
Give any four differences between DNA and RNA.
DNA

1. It is mainly present in nucleus, mitochondria and chloroplast
2. It contains deoxyribose sugar
3. It’s life time is high
4. It is stable and not hydrolysed easily by alkalis
5. It can replicate itself

RNA

1. It is mainly present in cytoplasm, nucleolus and ribosomes
2. It contains ribose sugar
3. It is Short lived
4. It is unstable and hydrolyzed easily by alkalis
5. It cannot replicate itself. It is formed from DNA.

Question 23.
Write short notes on Antioxidants.
Answer:

• Antioxidants are substances which retard the oxidative deteriotations of food. Food containing fats and oils is easily oxidised and turn rancid.
• To prevent the oxidation of fats and oils, chemical BHT (butyl hydroxy toluene), BHA (butylated hydroxy anisole) are added as antioxidants.
• These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils there by stop the chain reaction of oxidation of food.
• Sulphur dioxide, sulphites are also used as antioxidant and also act as antimicrobial agents and
  enzyme inhibitors.

Question 24.
50ml of 0.05M HNO₃ is added to 50ml of 0.025M KOH. Calculate the pH of the resultant solution.
Answer:
Number of moles of HNO₃ = 0.05 × 50 × 10^-3 = 2.5 × 10^-3
Number of moles of KOH = 0.025 × 50 × 10^-3 = 1.25 × 10^-3
Number of moles of HNC₃ after mixing = 2.5 × 10^-3 – 1.5 × 10 ^-3 = 1.25 × 10^-3

After mixing, total volume = 100 ml = 100 × 10^-3 L

pH = – log [H⁺]
pH = -log(l.25 × 10^-2) = 2 – 0.0969
= 1.9031

Part – III

Answer any six questions. Question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Explain the electro metallurgy of aluminium.
Answer:
Electrochemical extraction of aluminium -Hall-Herold process: In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as a anode. A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber. About 10% calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K. The chemical reactions involved in this process are as follows:

Ionisation of alumina : Al₂O₃ → 2Al³⁺ + 3O^2-
Reaction at cathode : 2Al³⁺ (melt) + 6e^– → 2Al _(l)
Reaction at anode : 6O^2- (melt) → 3O₂ + 12e^–
Since carbon acts as anode the following reaction also takes place on it.
C (s) + O^2- (melt) → CO + 2e^– ; C_(s) + 2O^2- (melt) → CO₂ + 4e^–

Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction can be written as follows:

4Al³⁺ (melt) + 6O^2- (melt) + 3C_(s) → 4Al_(l) + 3CO_2(g)

Question 26.
Give the uses of helium.
Answer:

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

Question 27.
Explain chromyl chloride Test.
Answer:
(i) When potassium dichromate is heated with any chloride salt in the presence of Conc.H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved. This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis.

(ii) The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained. Cr02Cl2 + 4NaOH -► Na2Cr04

Question 28.
A face centred cubic solid of an element (atomic mass 60 g mol ) has a cube edge of 4A. Calculate its density.
Answer:
For FCC unit cell n 4
Edge length(a) = 4A = 4 × 10^-8cm
Mass (M) = 60 g mol^-1
Dcnsity(ρ) = ?

Question 29.
Describe the construction of Daniel cell and write its cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell: The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.
Joining the half cell : The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCl, Na₂SO₄ etc., The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.

When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:- The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zr²⁺ ions and the electrons. The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (- ve).
Zn(s) → Zn²⁺+(aq) + 2e^–
(loss of electron-oxidation)

Cathodic reduction:
As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).

cu²⁺ 2e^– cu_(s)( gain of electron – reduction)

Salt bridge: -The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn²⁺ in solution, i.e., the solution contains more number of Zn²⁺ ions as compared to SO^2-₄ and hence the solution in the anodic compartment would become positively charged. Similarly, the solution in the cathodic compartment would become negatively charged as the Cu²⁺ ions are reduced to copper i.e., the cathodic solution contain more number of SO^2-₄ ions compared to Cu²⁺ .

Completion of circuit:- Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes :- As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted into Zn²⁺ or the entire Cu²⁺ ions are converted into mettalic copper.
Daniel cell is represents as

Question 30.
Write short notes on (i) Negative catalyst (ii) Phase transfer catalyst
Answer:
(i) Negative catalyst:
(i) In certain reactions, presence of certain substances decreases the rate of the reaction. Such
substances are called negative catalyst and the process is called negative catalysis.

(ii) In oxidation of chloroform, ethanol decreases the rate of the reaction and ethanol act as negative catalyst.

(iii) Decomposition of H202 rate is decreased by glycerol and it act as negative catalyst

(ii) Phase transfer catalyst:

(i) Consider the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immisible.

(ii) As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy. For such situation, a third solvent is added which is miscible with both. So, the phase boundary is eliminated the reactants freely mix and react fast.

(iii) But for large scale preparation of any product, use of a third solvent is not convenient as it . may be expensive. For such problems, phase transfer catalysis provides a simple solution, which avoids the use of solvents.

(iv) It directs the use of a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

(v) Example : Substitution of Cl^– and CN^– in the following reaction.

R Cl = 1- chloro octane
R CN = 1- cyano octane
(vi) By direct heating of two phase mixture of organic 1- chloro octane with aqueous sodium cyanide for several days, 1- cyano octane is not obtained. However, if a small amount of quartemary ammonium salt like tetra alkyl ammonium cation which has hydrophobic and hydrophilic ends, transports CN- from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reactions with 1- chloro octane as shown below

(vii) So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Explain the mechanism of Aldol condensation of acetaldehyde.
Answer:
In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α – hydrogen add together to give β- hydroxyl aldehyde (aldol) or β – hydroxyl ketone (ketol). The reaction is called aldol condensation reaction.

Mechanism
The mechanism of aldol condensation Of acetaldehyde takes place in three steps.
Step 1: The carbanion is formed as the a – hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion.

Step 3: The alkoxide ion formed is protonated by water to form aldol.

The aldol rapidly undergoes dehydration on heating with acid to form a, p unsaturated aldehyde.

Question 32.
Explain the preparation of Nylon – 6,6 and Buna- S.
Answer:
Nylon 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene diamine. With the elimination of water to form amide bonds.

Buna – S is prepared by the polymerisation of buta -1,3 – diene and styrene in the ratio of 3:1 in the presence of sodium.

Question 33.
Identify A to C in the following sequence.

Part – IV

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

Question 34.
(a) (i) Explain how gold ore is leached by cyanide process
(ii) Explain the classification of lnosilicates

[OR]

(b) (i) What are interhalogen compounds? Give examples
(ii) Explain the preparation of KMn04.
Answer:
(a) (i) 4Au (s) + 8CN^– (aq) + 2H₂O(aq) + O₂(g)→ 4Au(CN)₂]^–(aq) + 4OH^–
2[Au(CN)₂]^–(aq) + Zn (s) → 2Au (s) + [Zn(CN)₄]^-2(aq)
In the first reaction Au changes into Au+, i.e. its oxidation takes place. In the second reaction: Au⁺ → Au° (i.e. ) reduction takes place.

(ii) Ino silicones: Silicates which contain V number of silicate units liked by sharing two or more oxygen atoms are called inosilicates. They are further classified as chain silicates and double chain silicates.

Chain silicates (or pyroxenes): These silicates contain [(SiO₃)_n]^2n- ions formed by linking ‘ri number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.

Example: Spodumene – LiAl(SiO₃)₂ Double chain silicates (or amphiboles): These silicates contains [Si₄O₁₁] _n^6n-. ions. In these silicates there are two different types of tetrahedra : (i) Those sharing 3 vertices (ii) those sharing only 2 vertices.

Examples:
Asbestos: These are fibrous and noncombustible silicates. Therefore they are used for thermal insulation material, brake linings, construction material and filters. Asbestos being carcinogenic silicates, their applications are restricted.

[OR]

(b) (i) Each halogen combines with other halogens to form a series of compounds called interhalogen compounds, e.g., ClF₃, IF₅, IF₇
(ii) Potassium permanganate is prepared from pyrolusite (MnO₂) ore. The preparation involves the following steps.

Conversion of MnO₂ to potassium manganate:
Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO₃ A green coloured potassium manganate is formed.

Oxidation of potassium manganate to potassium permanganate:
Potassium manganate can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.

Chemical oxidation: In this method potassium manganate is treated with ozone (O₃) or chlorine to get potassium permanganate.

Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.
K₂MnO₄ ⇌ 2K⁺ + MnO₄^2-
H₂O ⇌ H⁺ + OH^–

Manganate ions are converted into permanganate ions at anode.

The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

Question 35.
(a) (i) Explain [Fe(CN)₆]^3- is paramagnetic, using Crystal Field theory
(ii) What is schottky detect?

[OR]

(A) (0 Derive Henderson – Hassel balch equation
(ii) What is kohlraush’s law?
Answer:
(a) (i)

(ii) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.Ionic solids in which the cation and anion are of almost of similar size show schottky defect. Example: NaCl.

Presence of large number of schottky defects in a crystal, lowers its density.
For example, the theoretical density of vanadium monoxide (VO) calculated using the edge length of the unit cell is 6.5 g cm^-3 , but the actual experimental density is 5.6 g cm^-3 .It indicates that there is approximately 14% Schottky defect in VO crystal. Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

[OR]

(b) (i) Henderson – Hasselbalch equation:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution, i.e.,

2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.

4. Taking logarithm on both sides of the equation

5. Reverse the sign on both sides

Equation (6) & (7) are called Henderson – Hasselbalch equations.

(ii) Kohlrauselrs 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.

Question 36.
(a) (i) Explain Intermediate compound formation theory.
(ii) Write short notes on ultra filtration.

[OR]

(b) How the following conversions are effected?
(i) Phenol → Salicylaldehvde
(ii) Phenol → Phenolphthalein
(iii) Glycol → 1.4dioxane
Answer:
(a) (i) The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation.. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the
catalyst is regenerated.
Consider the reactions:
A+B → AB (1)
A+C →AC (intermediate) (2)
C is the catalyst
AC+B → AB+C (3)

Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.

(ii) Ultrafiltration:

• The pores of ordinary filter papers permit the passage of colloidal solutions. In ultrafilteration, the membranes are made by using collodion, cellophane or visiking.
• When a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.
• This process is quickened by application of pressure. The separation of sol particles from electrolyte by filtration through an ultrafilter is called ultrafiltration.
• Collodian is 4% solution of nitrocellulose in a mixture of alcohol and water.

[OR]

(b) (i) Phenol → Salicylaldehyde:

(ii) Phenol → Phenolphthalcin:
On heating phenol with phthalic anhydride in presence of con.H2S04, phenolphthalein
is obtained.

(iii) Glycol →1,4 dioxane:
When distilled with Conc.112S04, glycol forms 1, 4-dioxane

Question 37.
(a) Write short notes on
(i) Mustard oil reactions
(ii) Carbylamine reaction
(iii) Gabriel pathalimide synthesis

[OR]

(b) Explain the structure of Fructose.
Answer:
(a) (i) Mustard oil reactions:
When primary amines are treated with carbon disulphide (CS₂), N – alkyldithio carbonic acid is formed which on subsequent treatment with HgCl₂, give an alkyl isothiocyanate.

(ii) Carbylamine reaction:
Aliphatic (or) aromatic primary amines react with chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test is used to identify the primary amines.

(iii) Gabriel pathalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.

[OR]

Structure of Fructose:

• Elemental analysis and molecular weight determination of fructose show that it has the molecular formula C₆H₁₂O₆.
• Fructose on reduction with HI and red phosphorus gives a mixture of n – hexane (major product) and 2 – iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.
  
• Fructose reacts with NH₂OH and HCN. It shows the presence of carbonyl groups in the molecules of fructose.
• Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of five hydroxyl groups in a fructose molecule.
• Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde (-CHO) group.
• Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C-2. This confirms the presence of keto group.
  

On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose.

This shows that a keto group is present in C-2. It also shows the presence of 10 alcoholic groups at C- 1 and C- 6. From the above reaction the structure of fructose is

Question 38.
(a) (i) A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
(ii) K_sp of Ag₂CrO₄ is 1.1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.1M K₂CrO₄?
[OR]
(b) Compound A of molecular formula C₇H₆O reduces Tollen’s reagent when A reacts with 50% NaOH gives compound B of molecular formula C₇H₈O and C of molecular formula C₇H₅O₇Na. Compound C on treatment with dil.HCl gives compound D of molecular formula C₇H₆O₂. When D is heated with sodalime gives compound E.
Identify A,B,C,D & E. Write the corresponding equations.
Answer:
(a) (i) 1. For the first order reaction k =\(\frac{2.303}{t} \log \frac{a}{(a-x)}\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60.
k = (2.303/ 50) log (100/ 60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1
2. t = ?, when x = 80%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.010216 min^-1
t = (2.303/0.010216) log (100/20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

(ii)

[OR]

(b)

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

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

Instructions:

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

Time: 3 Hours
Max Marks: 70

PART -1

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

Question 1.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C

Question 2.
What is the value of resistance of the following resistor?

(a) 100 k Ω
(b) 10 kΩ
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Question 3.
A flow of 10⁷ electrons per second in a conduction wire constitutes a
(a) 1.6 × 10^-26A
(b) 1.6 × 10¹²A
(c) 1.6 × 10^-12A
(d) 1.6 × 10²⁶A
Hint:

Answer:
(c) 1.6 x 10^-12A

Question 4.
The force experienced by a particle having mass m and charge q accelerated through a potential difference V when it is kept under perpendicular magnetic field B is ……………….. .
(a) \(\sqrt{\frac{2 q^{3} \mathrm{BV}}{m}}\)
(b) \(\sqrt{\frac{q^{3} \mathrm{B}^{2} V}{2 m}}\)
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)
(d) \(\sqrt{\frac{2 q^{3} B V}{m^{3}}}\)
Answer:
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)

Question 5.
A magnetic needle is kept in a non-uniform magnetic field. It experiences
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque

Question 6.
The instantaneous values of alternating current and voltage in a circuit are i = \(\frac{1}{\sqrt{2}}\) sin( 100πt) A and v = \(\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right) \mathrm{V}\) The average power in watts consumed in the circuit is
(a) \(\frac{1}{4}\)
(b) \(\frac{\sqrt{3}}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{8}\)
Answer:
(d) \(\frac{1}{8}\)

Question 7.
Which of the following is an electromagnetic wave’?
(a) α – rays
(b) β – rays
(c) γ – rays
(d) all of them
Answer:
(c) γ – rays

Question 8.
When light is incident on a soap film of thickness 5 × 10^-5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be.
(a) 1.22
(b) 1.33
(c) 1.51
(d) 1.83
Hint. The condition for constructive interference, (for reflection)

Answer:
(b) 1.33

Question 9.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Hint. Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.
Answer:
(d) violet

Question 10.
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
(a) \(\sqrt{\frac{h v_{0}}{m}}\)
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)
(c) \(2 \sqrt{\frac{h v_{0}}{m}}\)
(d) \(\sqrt{\frac{h v_{0}}{2 m}}\)

Hint: From Einstein’s photoelectric equation
K_max = hυ – hυ₀ [υ = 4υ₀]
\(\frac { 1 }{ 2 }\) mV²_max = 4hυ₀ – hυ₀
V²_max = \(\frac{6 h v_{0}}{m}\)
V_max = \(\sqrt{\frac{6 h v_{0}}{m}}\)
Answer:
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)

Question 11.
The number of photo-electrons emitted for light of a frequency u (higher than the threshold frequency υ₀) is proportional to
(a) Threshold frequency (υ₀)
(b) Intensity of light
(c) Frequency of light (υ)
(d) υ – υ₀
Hint: Photoelectric current of Intensity of incident light
Answer:
(b) Intensity of light

Question 12.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
(a) 1
(b) 2
(c) 3
(d) 4
Hint: The ionisation energy of a hydrogen atom is, IE = \(\frac{13.6 z^{2}}{n^{2}}\)

Answer:
(c) 3

Question 13.
The given electrical network is equivalent to ………….. .
(а) AND gate
(b) OR gate
(c) NOR gate
(d) NOT gate

Answer:
(c) NOR gate

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.
Which one of the following is the natural nanomaterial?
(a) Peacock feather
(b) peacock beak
(c) Grain of sand
(d) Skin of the wale
Answer:
(a) Peacock feather

PART – II

Answer any six questions in which Q. No 21 is compulsory. [6 × 2 = 12]

Question 16.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:
Charge produced in each object q = 50 nC
q = 50 × 10^-9C
Charge of electron (e) = 1.6 × 10^-19C
Number of electron transferred, n = \(\frac{q}{e}=\frac{50 \times 10^{-9}}{1.6 \times 10^{-19}}\) = ,n =31.25 × 10^-9 × 10¹⁹
n = 31.25 × 1o¹⁰ electrons

Question 17.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 18.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
\(\overrightarrow{\mathrm{P}}\) = q_m\(\overrightarrow{\mathrm{d}}\)

Question 19.
What is meant by ‘Wattful current’?
Answer:
The component of current (I_RMS cos Φ) which is in phase with the voltage is called active component. The power consumed by this current = V_RMSI_RMS cos Φ . So that it is also known as ‘WattfuT current.

Question 20.
What are electromagnetic waves?
Answer:
Electromagnetic waves are non-mechanical waves which move with speed equals to the speed of light (in vacuum).

Question 21.
Two light sources have intensity of light as 10. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I₀.
The resultant intensity is, I = 4 I₀ cos² (φ/2)
Resultant intensity when, Φ = π / 3, is I = 4I₀ cos² (π / 6)
I = 4I₀(√3/2)² = 3I₀

Question 22.
State de Broglie hypothesis.
Answer:
De Broglie hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves.

Question 23.
Calculate the energy equivalent of 1 atomic mass unit.
Answer:
We take, m = 1 amu = 1.66 × 10²⁷ kg
c = 3 × 10⁸ ms^-1
Then, E = mc² = 1.66 × 10²⁷ × (3 × 1o⁸)²⁷ J

E ≈ 981 MeV
∴ 1 amu = 931 MeV

Question 24.
hat do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

PART III

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

Question 25.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 × 10¹¹ N m² kg^-2. The value of the constant k in Coulomb law is k = 9 × 10⁹ N m² C^-2.
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Question 26.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.
Answer:
Resistance of a nichrome wire at 0°C, R₀ = 10 Ω
Temperature co-efficient of resistance, α = 0.004/°C
Resistance at boiling point of water, R_T = ?
Temperature of boiling point of water, T = 100 °C
R_T = R₀ ( 1 + α T) = 10[1 + (0.004 × 100)]
R_T = 10(1 + 0.4) = 10 × 1.4
R_T = 14Ω
As the temperature increases the resistance of the wire also increases.

 

Question 27.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic held) inside the specimen \(\overrightarrow{\mathrm{B}}\) is equal to the sum of the magnetic field \(\overrightarrow{\mathrm{B}}\)₀ produced in vacuum due to the magnetising field and the magnetic field \(\overrightarrow{\mathrm{B}}\)_m due to the induced magnetisation of the substance.

Question 28.
An inductor blocks AC but it allows DC. Why? and How?
Answer:
An inductor L is a closely wound helical coil. The steady DC current flowing through L produces uniform magnetic field around it and the magnetic flux linked remains constant. Therefore there is no self-induction and self-induced emf (back emf). Since inductor behaves like a resistor, DC flows through an inductor.

The AC flowing through L produces time-varying magnetic field which in turn induces self- induced emf (back emf)- This back emf, according to Lenz’s law, opposes any change in the current. Since AC varies both in magnitude and direction, its flow is opposed in L. For an ideal inductor of zero ohmic resistance, the back emf is equal and opposite to the applied emf. Therefore L blocks AC.

Question 29.
Derive the relation between/and R for a spherical mirror.
Answer:
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection.
If MP is the perpendicular from M on the principal axis, then from the geometry, The angles

∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
tan i = \(\frac{P M}{P C}\) and tan 2i = \(\frac{P M}{P F}\)
As the angles are small, tan i ≈ i, i = \(\frac{P M}{P C}\) and 2i = \(\frac{P M}{P F}\)
Simplifying further, 2 \(\frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}}\) ; 2PF = PC
PF is focal length f and PC is the radius of curvature R.
2f = R (or) f = \(\frac{R}{2}\)
f = \(\frac{R}{2}\) is the relation between/and R.

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

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

Question 31.
Show that nuclear density is almost constant for nuclei with Z > 10.
Answer:

The expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z >10) have the same density and it is an important characteristics of the nuclei.

Question 32.
Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 pH. The frequency band is from 500 kHz to 1500 kHz.
Answer:
Resonant frequency f₀ = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
On simplifying we get C = \(\frac{1}{4 \pi^{2} f_{0}^{2} L}\)
When frequency is equal to 500 kHz

When frequency is equal to 1500 kHz

There fore, the capacitor range is 75 – 676 pF

Question 33.
Distinguish between wireline and wireless communication.
Answer:

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

PART – IV

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

Question 34.
(a) Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction: A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor. Two comb shaped metallic conductors E and D are ‘ fixed near the pulleys.
The comb D is maintained at a positive potential of 10⁴ V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working: Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 V which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

[OR]

Question 34.
(b) Explain the equivalent resistance of a series resistor network.
Answer:
Resistors in series: When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R₁, R₂ and R₃ connected in series.


The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V₁, V₂ and V₃ be the potential difference (voltage) across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR₁ V₂ = IR₂ and V₃ = IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃ ….. (1)
V = I(R₁ + R₂ + R₃)
V = I.R_s ….. (2)
where R_s is the equivalent resistance,
R_s = R₁ + R₂ + R₃ ….. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).
Note: The value of equivalent resistance in series connection will be greater than each individual resistance.

Question 35.
(a) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O. Consider an element of length dl of the wire at a distance l from point O and \(\vec{r}\) be the vector joining the element dl with the point P. Let θ be the angle between dl and \(\vec{r}\) Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} \vec{l}}{4 \pi r^{2}} \sin \theta\) (unit vector perpendicular to \(d \vec{l}\) and \(\vec{r}\) )-(1)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors \(d \vec{l}\) and \(\vec{r}\) (let it be n̂). The net magnetic field can be determined by integrating equation with proper limits.
\(\vec{B}\) = ∫d\(\vec{B}\)
From the figure, in a right angle triangle PAO,

tan (π – θ) = \(\frac{a}{l}\)
l = \(-\frac{a}{\tan \theta}\) (since tan (π – θ) = – tan θ) ⇒ \(\frac{1}{\tan \theta}\) = cot θ
l = -a cot θ and r = a cosecθ

Differentiating,
dl = a cosec²θ dθ

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating \(d \vec{B}\) by varying the angle from θ = φ₁ to θ = φ₂ is

For a an infinitely long straight wire, l = 0 and 2 = , the magnetic field is
\(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n}\) …… (3)
Note that here n̂ represents the unit vector from the point O to P.

[OR]

Question 35.
(b) Show that the mutual inductance between a pair of coils is same (M₁₂ = M₂₁).
Answer:
Mutual induction: When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.

Consider two coils which are placed close to each other. If an electric current q is sent through coil i₁ the magnetic field produced by it is also linked with coil 2.
Let Φ₂₁ be the magnetic flux linked with each turn of the coil 2 of N₂ turns due to coil 1, then the total flux linked with coil 2 (N₂Φ₂₁) is proportional to the current i₁ in the coil 1.
N₂Φ₂₁ ∝ i₁
N₂Φ₂₁ = M₂₁i₁ or M₂₁ = \(\frac{N_{2} \Phi_{21}}{i_{1}}\)
The constant of proportionality M₂₁ is the mutual inductance of the coil 2 with respect to coil 1. It is also called as coefficient of mutual induction. If i₁ = 1 A, then M₂₁ = N₂Φ₂₁.
Therefore, the mutual inductance M₂₁ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
When the current i₁ changes with time, an emf ξ₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ξ₂₁ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current with respect to time. If \(\frac{d i_{1}}{d t}\) = 1 As^-1, then M₂₁ = -ξ₂.
Mutual inductance M₂₁ is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil I is l As^-1.
Similarly, if an electric current i₂ through coil 2 changes with time, then emf ξ₁ is induced in coil 1. Therefore, N

where M₁₂ is the mutual inductance of the coil I with respect to coil 2. It can be shown that for a given pair of coils, the mutual inductance is same. i.e., M₂₁ = M₁₂ = M.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.

Question 36.
(a) Derive the equation for refraction at single spherical surface. Equation for refraction at single spherical surface:
Answer:
Let us consider two transparent media having refractive indices n₁ and n₂ are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n₁. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.

Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n₂ > n₁, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed. Snell’s law in product form for the refraction at the point N could be written as,
n₁ sin i = n₂ sin r …… (1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n₁i = n₂r ……… (2)
Let the angles
∠NOP = α ∠NCP = β ∠NIP = γ
tan α = \(\frac{P N}{P O}\) tan β = \(\frac{P N}{P C}\) tan γ = \(\frac{P N}{P I}\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac{P N}{P O}\) ; β = \(\frac{P N}{P C}\) ; γ = \(\frac{P N}{P I}\) …… (3)
For the triangle, ∆ONC,
i = α + β ….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ …… (5)
Substituting for i and r from equations (4) and (5) in the equation (2).
n₁ (α + β) = n₂ (β – γ)
Rearranging,
n₁α + n₂γ = (n₂ – n₁)β
Substituting for α, β and γ from equation (3)
\(n_{1}\left(\frac{\mathrm{PN}}{\mathrm{PO}}\right)+n_{2}\left(\frac{\mathrm{PN}}{\mathrm{PI}}\right)=\left(n_{2}-n_{1}\right)\left(\frac{\mathrm{PN}}{\mathrm{PC}}\right)\)
Further simplifying by cancelling method
\(\frac{n_{1}}{\mathrm{PO}}+\frac{n_{2}}{\mathrm{PI}}=\frac{n_{2}-n_{1}}{\mathrm{PC}}\) ….. (6)
Following sign conventions, PO = -υ , PI = +ν and PC = +R in equation (6),
\(\frac{n_{1}}{-u}+\frac{n_{2}}{v}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\)
After rearranging, finally we get,
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\) ……… (7)
Equation (7) gives the relation among the object distance υ, image distance ν , refractive indices of the two media (nl and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n₁ = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac{n}{v}-\frac{1}{u}=\frac{(n-1)}{\mathrm{R}}\) ……… (8)

[OR]

Question 36.
(b) What do you mean by electron emission? Explain briefly various methods of electron emission.
Answer:
Electron emission:

• Free electrons possess some kinetic energy and this energy is different for different electrons. The kinetic energy of the free electrons is not sufficient to overcome the surface barrier.
• Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface.
• The liberation of electrons from any surface of a substance is called electron emission. There are mainly four types of electron emission which are given below.

(i) Thermionic emission:
When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known as thermionic emission.

The intensity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.

Examples: cathode ray tubes, electron microscopes, X-ray tubes etc.

(ii) Field emission:
Electric field emission occurs when a very strong electric field is applied across the metal. This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.

Field emission
Examples: Field emission scanning electron microscopes, Field-emission display etc.

(iii) Photo electric emission:
When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the.free electrons. Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place. The number of electrons emitted depends on the intensity of the incident radiation.

Examples: Photo diodes, photo electric cells etc.

(iv) Secondary emission:
When a beam of fast moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface. Thus the free electrons get sufficient kinetic energy so that the secondary emission of electron occurs.

Examples: Image intensifires, photo multiplier tubes etc.

Question 37.
(a) Discuss the spectral series of hydrogen atom.
Answer:
The spectral lines of hydrogen are grouped in separate series. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit. These series are named as Lyman series. Balmer series, Paschen series, Brackett series, Pfund series, etc. The wavelengths of these spectral lines perfectly agree with the equation derived from Bohr atom model.
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=\bar{v}\) …… (1)
where ν̂ is known as wave number which is inverse of wavelength, R is known as Rydberg constant whose value is 1.09737 × 10⁷ m^-1 and m and n are positive integers such that m > n. The various spectral series are discussed below:

(a) Lyman series:
Put n = 1 and m = 2,3,4 in equation (I). The wave number or wavelength of spectral lines of Lyman series which lies in ultra-violet region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\)

(b) Balmer series:
Put n = 2 and m = 3,4,5 in equation (I). The wave number or wavelength of spectral lines of Balmer series which lies in visible region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{m^{2}}\right)\)

(c) Paschen series:
Put n = 3 and m = 4,5,6 in equation (I). The wave number or wavelength of spectral lines of Paschen series which lies in infra-red region (near IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{m^{2}}\right)\)

(d) Brackett series:
Put n = 4 and m = 5,6,7 in equation (I). The wave number or wavelength of spectral lines of Brackett series which lies in infra-red region (middle IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{m^{2}}\right)\)

(e) Pfund series:
Put n = 5 and m = 6,7,8 in equation (I). The wave number or wavelength of spectral lines of Pfund series which lies in infra-red region (far IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{5^{2}}-\frac{1}{m^{2}}\right)\)

[OR]

Question 37.
(b) State and prove De Morgan’s First and Second theorems.
Answer:
De Morgan’s First Theorem:
The first theorem states that the complement of the sum of two logical inputs is equal to the product of its complements.

Proof:
The Boolean equation for NOR gate is Y = \(\overline{\mathrm{A}+\mathrm{B}}\) . The Boolean equation for a bubbled AND gate is Y = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) Both cases generate same outputs for same inputs. It can be verified using the following truth table.

From the above truth table, we can conclude \(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\). Thus De Morgan’s First Theorem is proved. It also says that a NOR gate is equal to a bubbled AND gate. The corresponding logic circuit diagram is shown in figure.

De Morgan’s Second Theorem:
The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.

Proof
The Boolean equation for NAND gate is Y = \(\overline{\mathrm{AB}}\)
The Boolean equation for bubbled OR gate is Y = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\) . A and B are the inputs and Y is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table.

From the above truth table we can conclude \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\). Thus De Morgan’s Second Theorem is proved. It also says, a NAND gate is equal to a bubbled OR gate. The corresponding logic circuit diagram is shown in figure.

Question 38.
(a) 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
(i) Amplitude modulation,
(ii) Frequency modulation, and
(iii) Phase modulation.

(i) 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.
Amplitude Modulation

(ii) Frequency Modulation (FM):
The frequency of the camer 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. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction
(A, C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (figure (c)).

(iii) Phase Modulation (PM):
The instantaneous amplitude of the baseband signal modifies the phase of the camer signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency. The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase
lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

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

[OR]

Question 38.
(b) Elaborate any two types of Robots with relevant examples.
Answer:
(i) Human Robot:
Certain robots are made to resemble humans in appearance and replicate the human activities like walking, lifting, and sensing, etc.

• Power conversion unit: Robots are powered by batteries, solar power, and hydraulics.
• Actuators: Converts energy into movement. The majority of the actuators produce rotational or linear motion.
• Electric motors: They are used to actuate the parts of the robots like wheels, arms, fingers, legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
• Pneumatic Air Muscles: They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. They contract almost 40% when the air is sucked inside them.
• Muscle wires: They are thin strands of wire made of shape memory alloys. They can contract by 5% when electric current is passed through them.
• Piezo Motors and Ultrasonic Motors: Basically, we use it for industrial robots.
• Sensors: Generally used in task environments as it provides information of real-time knowledge.
• Robot locomotion: Provides the types of movements to a robot. The different types are (a) Legged (b) Wheeled (c) Combination of Legged and Wheeled Locomotion (d) Tracked slip/skid

(ii) Industrial Robots:
Six main types of industrial robots:

1. Cartesian
2. SCARA (Selective Compliance Assembly Robot Arm)
3. Cylindrical
4. Delta
5. Polar
6. Vertically articulated

Six-axis robots are ideal for:

• Arc Welding
• Spot Welding
• Material Handling
• Machine Tending
• Other Applications

Students can Download Computer Science Chapter 13 Python and CSV Files Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Science Solutions Chapter 13 Python and CSV Files

Samacheer Kalvi 12th Computer Science Python and CSV Files Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
A CSV file is also known as a
(a) Flat File
(b) 3D File
(c) String File
(d) Random File
Answer:
(a) Flat File

Question 2.
The expansion of CRLF is ………………………
(a) Control Return and Line Feed
(b) Carriage Return and Form Feed
(c) Control Router and Line Feed
(d) Carriage Return and Line Feed
Answer:
(d) Carriage Return and Line Feed

Question 3.
Which of the following module is provided by Python to do several operations on the CSV files?
(a) py
(b) xls
(c) csv
(d) os
Answer:
(c) csv

Question 4.
Which of the following mode is used when dealing with non-text files like image or exe files?
(a) Text mode
(b) Binary mode
(c) xls mode
(d) csv mode
Answer:

Question 5.
The command used to skip a row in a CSV file is ……………………….
(a) next( )
(b) skip( )
(c) omit( )
(d) bounce( )
Answer:
(b) skip( )

Question 6.
Which of the following is a string used to terminate lines produced by writer( )method of csv module?
(a) Line Terminator
(b) Enter key
(c) Form feed
(d) Data Terminator
Answer:
(a) Line Terminator

Question 7.
What is the output of the following program?
import csv
d=csv.reader(open(‘c:\PYPRG\chl3\city.csv’))
next(d)
for row in d:
print(row)
if the file called “city.csv” contain the following details
chennai,mylapore
mumbai,andheri
(a) chennai,mylapore
(b) mumbai,andheri
(c) chennai, mumbai
(d) chennai,mylapore,mumbai,andheri
Answer:
(b) mumbai,andheri

Question 8.
Which of the following creates an object which maps data to a dictionary?
(a) listreader( )
(b) reader( )
(c) tuplereader( )
(d) DictReader( )
Answer:
(d) DictReader( )

Question 9.
Making some changes in the data of the existing file or adding more data is called
(a) Editing
(b) Appending
(c) Modification
(d) Alteration
Answer:
(c) Modification

Question 10.
What will be written inside the file test.csv using the following program import csv
D = [[‘Exam’],[‘Quarterly’],[‘Halfyearly’]]
csv.register_dialect(‘M’,lineterminator = ‘\n’)
with open(‘c:\pyprg\chl3\line2.csv’, ‘w’) as f:
wr = csv.writer(f,dialect=’M’)
wr.writerows(D)
f.close( )
(a) Exam Quarterly Halfyearly
(b) Exam Quarterly Halfyearly
(c) Q H
(d) Exam, Quarterly, Halfearly
Answer:
(d) Exam, Quarterly, Halfearly

PART – II
II. Answer The Following Questions

Question 1.
What is CSV File?
Answer:
A CSV file is a human readable text file where each line has a number of fields , separated by commas or some other delimiter. A CSV file is also known as a Flat File. Files in the CSV format can be imported to and exported from programs that store data in tables, such as Microsoft Excel or OpenOfficeCalc.

Question 2.
Mention the two ways to read a CSV file using Python?
Answer:
Read a CSV File Using Python
There are two ways to read a CSV file.

1. Use the csv module’s reader function
2. Use the DictReader class.

Question 3.
Mention the default modes of the File?
Answer:
You can specify the mode while opening a file. In mode, you can specify whether you want to read ‘r’, write ‘w’ or append ‘a’ to the file. You can also specify “text or binary” in which the file is to be opened.
The default is reading in text mode. In this mode, while reading from the file the data w’ould . be in the format of strings.

Question 4.
What is use of next( ) function?
Answer:
# skipping the first row(heading)
Example: next( reader)

Question 5.
How will you sort more than one column from a csv file? Give an example statement?
Answer:
To sort by more than one column you can use itemgetter with multiple indices: operator .itemgetter (1,2).
#using operator module for sorting multiple columns
sortedlist = sorted (data, key=operator.itemgetter(1))

PART – III
III. Answer The Following Questions

Question 1.
Write a note on open( ) function of python. What is the difference between the two methods?
Answer:
Python has a built-in function open() to open a file. This function returns a file object, also called a handle, as it is used to read or modify the file accordingly.
For Example
>>> f = openf’sample.txt”) bopen file in current directory andf is file object
>>> f = open(‘c:\ \pyprg\ \chl3sample5.csv’) #specifyingfull path
You can specify the mode while opening a file. In mode, you can specify whether you want to read ‘r’, write ‘w’ or append ‘a’ to the file, you can also specify “text or binary” in which the file is to be opened.
The default is reading in text mode. In this mode, while reading from the file the data would be in the format of strings.
On the other hand, binary mode returns bytes and this is the mode to be used when dealing with non-text files like image or exe files.
f = open(“test.txt”) # since no mode is specified the default mode it is used
#perform file operations
f.close( )
The above method is not entirely safe. If an exception occurs when you are performing some operation with the file, the code exits without closing the file. The best way to do this is using the “with” statement. This ensures that the file is closed when the block inside with is exited. You need not to explicitly call the close() method. It is done internally.

Question 2.
Write a Python program to modify an existing file?
Answer:
import csv
row= [‘3’, ‘Meena’, ‘Bangalore’]
with open(‘student.csv’, ‘r’) as readFile:
reader= csv.reader(readFile)
lines =list(reader) # list( ) – to store each row of data as a list
lines[3] =row
with open(‘student.csv’, ‘w’) as writeFile:
# returns the writer object which converts the user data with delimiter
writer= csv.writer(writeFile)
#writerows( )method writes multiple rows to a csv file
writer, writerows(lines)
readFile.close( )
writeFile.close( )
When we Open the student.csv file with text editor, then it will show:

Question 3.
Write a Python program to read a CSV file with default delimiter comma (,)?
Answer:
CSV file with default delimiter comma(,)
The following program read a file called “sample l.csv” with default delimiter cpmma(,) and print row by row.
#importing csv
import csv
#opening the csv fde which is in different location with read mode
with open(‘c:\ \pyprg\\sample l.csv’, V) as F:
#other way to open the file is f = (‘c:\\pyprg\\sample l.csv’, ‘r’)
reader = csv.reader(F)
Sprinting each line of the Data row by row print(row)
F.close( )
OUTPUT
[‘SNO’, ‘NAME’, ‘CITY’]
[12101’,’RAM’, ‘CHENNAI’]
[‘12102′,’LAVANYA’,’TIRUCHY’]
[‘12103′,’LAKSHMAN’,’MADURA’]

Question 4.
What is the difference between the write mode and append mode?
Answer:
Append mode write the value of row after the last line of the “student.csv file:”
The ‘w’ write mode creates a new file. If the file is already existing ‘w’ mode over writs it. Where as ‘a’ append mode add the data at the end of the file if the file already exists otherwise creates a new one.

Question 5.
What is the difference between reader( ) and DictReader( ) function?
Answer:
Reading CSV File Into A Dictionary:
To read a CSV file into a dictionary can be done by using DictReader class of csv module which works similar to the reader( ) class but creates an object which maps data to a dictionary. The keys are given by the fieldnames as parameter. DictReader works by reading the first line of the CSV and using each comma separated value in this line as a dictionary key.

The columns in each subsequent row then behave like dictionary values and can be accessed with the appropriate key (i.e. fieldname). The main difference between the csv.reader( ) and DictReader( ) is in simple terms csv. reader and csv.writer work with list/tuple, while csv.DictReader and csv.DictWriter work ‘ with dictionary. csv.DictReader and csv.DictWriter take additional argument fieldnames that are used as dictionary keys.

PART – IV
IV. Answer The Following Questions

Question 1.
Differentiate Excel file and CSV file?
Answer:
The difference between Comma-Separated Values (CSV) and eXceL Sheets(XLS) file formats is
Answer:
Excel:

1. Excel is a binary file that holds information about all the worksheets in a file, including both content and formatting
2. XLS files can only be read by applications that have been especially written to read their format, and can only be written in the same way.
3. Excel is a spreadsheet that saves files into its own proprietary format viz. xls or xlsx
4. Excel consumes more memory while importing data

CSV:

1. CSV format is a plain text format with a series of values separated by commas.
2. CSV can be opened with any text editor in Windows like notepad, MS Excel, OpenOffice, etc.
3. CSV is a format for saving tabular information into a delimited text file with extension .csv
4. Importing CSV files can be much faster, and it also consumes less memory

Question 2.
Tabulate the different mode with its meaning?
Python File Modes:
Answer:

Question 3.
Write the different methods to read a File in Python?
Answer:
Read a CSV File Using Python
There are two ways to read a CSV file.

1. Use the csv module’s reader function
2. Use the DictReader class.

CSV Module’s Reader Function:
You can read the contents of CSV file with the help of csv.reader( )method. The reader function is designed to take each line of the file and make a list of all columns. Then, you just choose the column you want the variable data for. Using this method one can read data from csv files of different formats like quotes (” “),pipe (|) and comma(,).
The syntax for csv.reader( ) is
csv. reader (fileobject, delimiter,fmtparams)
where
file object:- passes the path and the mode of the file
delimiter:- an optional parameter containing the standard dilects like, I etc can be omitted
fmtparams:- optional parameter which help to override the default values of the dialects like skipinitialspace,quoting etc. Can be omitted

CSV file with default delimiter comma(,)
The following program read a file called “sample l.csv” with default delimiter comma(,) and print row by row.
#importing csv
import csv
#opening the csv file which is in different location with read mode
with open(‘c:\ \pyprg\\sample l.csv’, ‘r’) as F:
#other way to open the file is f= (‘c:\\pyprg\\sample l.csv’, ’r)
reader = csv.reader(F)
#printing each line of the Data row by row
print( row)
F.close( )
OUTPUT
[‘SNO’, ‘NAME’, ‘CITY’]
[‘12101′,’RAM’,’CHENNAI’]
[‘12102’, ‘LAVANYA’, ‘TIRUCHY’]
[‘12103’, ‘LAKSHMAN’, ‘MADURAI’]
CSV files- data with Spaces at the beginning
Consider the following file “sample 2.csv” containing the following data when opened through notepad

The following program read the file through Python using “csv.reader( )”.
import csv
csv.register_dialect(‘myDialect’,delimiter = ‘,’,skipimtialspace=True)
F=open(‘c:\ \pyprg\ \sample 2.csv’,’r’)
reader= csv.reader(F, dialect=’myDialect’)
for row in reader:
print( row)
F.close( )
OUTPUT
[‘Topic 1’, ‘Topic 2’, ‘Topic 3′]
[!one’, ‘two’, ‘three’]
[‘Example 1’, ‘Example 2’, ‘Example 3’]
As you can see in “sample 2.csv” there are spaces after the delimiter due to which the output is also displayed with spaces.
These whitespaces can be removed, by registering new dialects using csv.register_dialect( ) class of csv module. A dialect describes the format of the csv file that is to be read. In dialects the parameter “skipinitialspace” is used for removing whitespaces after the delimiter.

CSV File-Data With Quotes
You can read the csv file with quotes, by registering new dialects using csv.register_dialect( ) class of csv module.
Here, we have quotes.csv file with following data.
SNO,Quotes
(a) “The secret to getting ahead is getting started.”
(b) “Excellence is a continuous process and not an accident.”
(c) “Work hard dream big never give up and believe yourself.”
(d) “Failure is the opportunity to begin again more intelligently.”
(e) “The successful warrior is the average man, with laser-like focus.”
The following Program read “quotes.csv” file, where delimiter is comma (,)but the quotes are within quotes (“”). import csv
csv.register_dialect(‘myDialect’,delimiter = ‘,’,quoting=csv.QUOTE_ALL,
skipinitialspace=True)
f=open(’c:\\pyprg\ \quotes.csv’,’r’)
reader= csv.reader(f, dialect-myDialect’)
for row in reader:
print (row)
OUTPUT
[‘SNO’, ’Quotes’]
[(a), ’The secret to getting ahead is getting started.’]
[(b), ’Excellence is a continuous process and not an accident.’]
[(c), ’Work hard dream big never give up and believe yourself.’]
[(d), ’Failure is the opportunity to begin again more intelligently.’]
[(e), ’The successful warrior is the average man, with laser-like focus. ’]
In the above program, register a dialect with name myDialect. Then, we used csv. QUOTE_ ALL to display all the characters after double quotes.
CSV files with Custom Delimiters
You can read CSVfile having custom delimiter by registering a new dialect with the help of csv.register_dialect( ).
In the following file called “sample 4.csv”,each column is separated with | (Pipe symbol)

The following program read the file “sample4.csv” with user defined delimiter “|”
import csv
csv.register_dialect(‘myDialect’, delimiter ‘|’)
with open(‘c:\\pyprg\\sample 4.csv’, ‘r’) as f:
reader= csv.reader(f, dialect-myDialect’)
for row in reader:
print(row)
f.close( )
OUTPUT
[‘RollNo’, ‘Name’, ‘City’]
[‘12101’, ‘Arun’, ‘Chennai’]
[‘12102’, ‘Meena’, ‘Kovai’]
[T21031,’Ram’,’Nellai’]

Reading CSV File Into A Dictionary:
To read a CSV file into a dictionary can be done by using DictReader class of csv module which works similar to the reader() class but creates an object which maps data to a dictionary. The keys are given by the fieldnames as parameter. DictReader works by reading the first line of the CSV and using each comma separated value in this line as a dictionary key. The columns in each subsequent row then behave like dictionary values and can be accessed with the appropriate key (i.e. fieldname).

If the first row of your CSV does not contain your column names, you can pass a fieldnames parameter into the DictReader’s constructor to assign the dictionary keys manually. The main difference between the csv.reader( ) and DictReader( ) is in simple terms csv. reader and csv.writer work with list/tuple, while csv.DictReader and csv.DictWriter work with dictionary. csv.DictReader and csv.DictWriter take additional argument fieldnames that are used as dictionary keys.

For Example:
Reading “sample 8.csv” file into a dictionary
import csv
filename= ‘c:\ \pyprg\ \sample 8.csv’
input_file =csv.DictReader(open(filename;’r’))
for row in input_file:
print(dict(row))
print (data(row)) #dict( ) to print data
OUTPUT
{‘ItemName ‘^Keyboard’, ‘Quantity’: ’48’}
{‘ItemName VMonitor’, ‘Quantity’: ’52’}
{‘ItemName VMouse’, ‘Quantity’: ’20’}
In the above program, DictReader( ) is used to read “sample 8.csv” file and map into a dictionary. Then, the function dict( ) is used to print the data in dictionary format without order. Remove the dict( ) function from the above program and use print(row). Check you are getting
the following output
OrderedDict([(‘ItemName Keyboard’), (‘Quantity’, ’48’)])
OrderedDict([(‘ltemName ‘,’Monitor’), (‘Quantity’, ’52’)])
OrderedDict([(‘ItemName ‘,’Mouse’), (‘Quantity’, ’20’)])

Question 4.
Write a Python program to write a CSV File with custom quotes?
CSV File with quote characters
Answer:
You can write the CSV file with custom quote characters, by registering new dialects using csv.register_dialect( ) class of csv module,
import csv
csvData = [[‘SNO’,Items’], [‘l’,’Pen’], [‘2′,’Book’], [‘3′,’Pencil’]]
csv.register_dialect(‘myDialect’, delimiter = ‘|’,quotechar = “”,
quoting=csv.QUOTE_ALL)
with open(‘c:\\pyprg\ \chl3\\quote.csv’, ‘w’) as csvFile:
writer= csv.writer(csvFile, dialect-myDialect’)
writer, write rows(csvData)
print(“writing completed”)
csvFile.close( )
When you open the “quote.csv” file in notepad, we get following output:

In the above program, myDialect uses pipe (|) as delimiter and quotechar as doublequote “” to write inside the file.

Question 5.
Write the rules to be followed to format the data in a CSV file?
Answer:
Rules to be followed to format data in a CSV file:
(i) Each record (row of data) is to be located on a separate line, delimited by a line break by pressing enter key. For example:
xxx,yyy
denotes enter Key to be pressed

(ii) The last record in the file may or may not have an ending line break. For example:
PPP, qqq
yyy,xxx

(iii) There may be an optional header line appearing as the first line of the file with the same format as normal record lines. The header will contain names corresponding to the fields in the file and should contain the same number of fields as the records in the rest of the file. For example: field_ name 1,field_name 2,field_name_3 aaa,bbb,ccc
zzz,yyy,xxx CRFF( Carriage Return and Line feed)

(iv) Within the header and each record, there may be one or more fields, separated by commas. Spaces are considered part of a field and should not be ignored. The last field in the record must not be followed by a comma. For example: Red, Blue

(v) Each field may or may not be enclosed in double quotes. If fields are not enclosed with double quotes, then double quotes may not appear inside the fields. For example: “Red”,”Blue”,”Green” #Field data with double quotes
Black,White,Yellow #Field data without double quotes

(vi) Fields containing line breaks (CRLF), double quotes, and commas should be enclosed in double-quotes. For example:
Red, “;:Blue CRLF #comma itself is a field value, so it is enclosed with double quotes Red, Blue, Green

(vii) If double-quotes are used to enclose fields, then a double-quote appearing inside a field must be preceded with another double quote. For example:
“Red,” “Blue1: “Green”, #since double quotes is a field value it is enclosed with another double quotes,, White

Practice Programs

Question 1.
Write a Python program to read the following Namelist.csv file and sort the data in alphabetically order of names in a list and display the output

import csv.operator
data = csv.reader(open(‘c:\\PYPRG\\NameList.scv’))
next(data)
sorted list = sorted(data, key = operator.itemgetter(1))
for row in sorted list:
print(row)
Output:

Question 2.
Write a Python program to accept the name and five subjects mark of 5 students. Find the total and store all the details of the students in a CSV file?
Answer:
import csv
csvData = [[‘student’, ‘ml Vm2′,’m3′,’m4′,’m5′,’total’],
[‘Ram’, ’90’,’90’,”90′,”90′,”90′,”450′],
[‘Hari’,’100′,’100′,’100′,’10’,’90’,’490′],
[‘Sai’, ’90’,’90’,’ 100′,’ 100′,’ 100′,’480′],
[‘Viji’, ’ 100′,’90’,’90’,’90’,’ 100′,’470′],
[‘Raja’, ’80’,’’80’,”80′,’ 100′,’ 100′,’440′]]
with open(‘c:\\pyprg\\chl3\\st.csv’,’w’)as CF:
writer = csv.writer(CF) writer. writerows(csvData)
CF.close( )
Output:

Samacheer kalvi 12th Computer Science Python and CSV Files Additional Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
CSV means ……………………… files
(a) common server values
(b) comma separated values
(c) correct separator values
(d) constructor separated value
Answer:
(b) comma separated values

Question 2.
The file extension to save excel files are
(a) xls Civdsx
(b) XL
(c) exc or XL
(d) XL or xlx
Answer:
(a) xls Civdsx

Question 3.
csv files cannot be opened with ………………………..
(a) notepad
(b) MS Excel
(c) open office
(d) html
Answer:
(d) html

Question 4.
Identify the wrong statement
(a) Excel is a binary file
(b) csv is a plain text
(c) Excel is a plain text
(d) csv has tabular information
Answer:
(c) Excel is a plain text

Question 5.
Identify the statement which is correct.
(a) csv consumes less memory and faster
(b) Excel consumes less memory and slower
Answer:
(a) csv consumes less memory and faster

Question 6.
Find the wrong statement.
(a) csv files can be opened with any text editor
(b) Excel files can be opened with any text editor
Answer:
(b) Excel files can be opened with any text editor

Question 7.
…………………… file is used to store tabular data such as spreadsheet or database.
Answer:
csv

Question 8.
How will you open a new file in Notepad?
(a) File → New
(b) Ctrl + N
(c) both a and b
(d) shift + N
Answer:
(c) both a and b

Question 9.
If the fields of data in csv file has commas, then it should be given with …………………..
(a) ,
(b) ”
(c) ‘
(d) :
Answer:
(b) ”

Question 10.
Any field containing a newline as part of its data should be given in ……………………..
(a) quotes
(b) double colon
(c) colon
(d) double quotes
Answer:
(d) double quotes

Question 11.
If the fields contains double quotes as part of the data, the internal quotation marks need to be
(a) same
(b) quarter
(c) doubled
(d) tripled
Answer:
(c) doubled

Question 12.
The line // white indicates
(a) the first two fields of the row are empty
(b) It can be deleted
(c) comma not necessary
(d) only one field is there and , can be deleted
Answer:
(a) the first two fields of the row are empty

Question 13.
Find the correct statement
(I) The last record in the file may or may not have an ending line break
(II) Header is must with same format as record lines.
(a) (I) is true, (II) is False
(b) (I) is False, (II) – True
(c) (I), (II) – both are true
(d) (I), (II) – both are false
Answer:
(a) (I) is true, (II) is False

Question 14.
Identify the wrong statement from the following.
(a) Each field may or may not be enclosed in double quotes.
(b) If the fields are not enclosed with double quotes, then double quotes may not appear inside the fields
(c) Fields containing line breaks, double quotes and commas should be enclosed in single quotes.
(d) the last field in the record must not be followed by a comma
Answer:
(c) Fields containing line breaks, double quotes and commas should be enclosed in single quotes.

Question 15.
There are ……………………. ways to read a csv file.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 16.
In open command, file name can be represented in ………………………
(a) ” ”
(b) ”
(c) $
(d) both a & b
Answer:
(d) both a & b

Question 17.
open( ) returns a file called ………………………… which is used to read or modify the file accordingly.
Answer:
handle

Question 18.
The default reading mode is ………………….. mode.
Answer:
text

Question 19.
The default mode when you open a file is
(a) r
(b) w
(c) x
(d) a
Answer:
(a) r

Question 20.
In text mode, while reading from the file the data would be in the format of ……………………..
(a) int
(b) float
(c) char
(d) strings
Answer:
(d) strings

Question 21.
What will happen when you open a file for writing and file already exists there?
(a) creates a new file
(b) truncates the file
(c) overwrite the file
(d) append the contents
Answer:
(b) truncates the file

Question 22.
To open the file updating data, click ……………………
(a) a
(b) b
(c) t
(d) +
Answer:
(d) +

Question 23.
…………………… opens a file for exclusive creation.
(a) r
(b) w
(c) x
(d) +
Answer:
(c) x

Question 24.
………………….. opens the file for read and write in binary mode.
(a) r
(b) b
(c) x + b
(d) r + b
Answer:
(d) r + b

Question 25.
Python has a ……………………… collector to clean up unreferenced objects.
Answer:
garbage

Question 26.
closing a file will free up the resources that were tied with the file and is done by ……………………. method.
(a) Exit
(b) close
(c) Quit
(d) None of these
Answer:
(b) close

Question 27.
Which format is not allowed to read data from cav files?
(a) quotes
(b) pipe
(c) comma
(d) Asterisk
Answer:
(d) Asterisk

Question 28.
How many arguments are there in csv.reader( ) functions?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 29.
Which one of the following cannot be omitted in cav. reader?
(a) file object
(b) delimiter
(c) fmtparams
(d) space
Answer:
(a) file object

Question 30.
A ……………………. describes the format of the csv file that is to be read.
Answer:
dialect

Question 31.
In dialects, the parameter …………………… is used for remaining whitespaces after the delimiter.
Answer:
skipintialspace

Question 32.
By default, what will be the value of skipinitial space?
(a) True
(b) False
(c) 0
(d) 1
Answer:
(b) False

Question 33.
A ………………….. is a class of csv module which helps to define paramters for reading and writing csv.
Answer:
dialect

Question 34.
We can register for new dialects using class of csv module.
Answer:
csv.register-dialects

Question 35.
Which of the following is used to display all the characters after double quotes.
(a) Quote
(b) Quote-all
(c) double quotes
(d) single quotes
Answer:
(b) Quote-all

Question 36.
Which one of the following is used to add the elements in the list.
(a) add
(b) insert
(c) append
(d) update
Answer:
(c) append

Question 37.
Which one of the following cannot be used as a column separator?
(a) delimiter
(b) pipe
(c) comma
(d) #
Answer:
(d) #

Question 38.
List literals are written using ………………………..
(a) [ ]
(b) ( )
(c) { }
(d) <>
Answer:
(a) [ ]

Question 39.
An ordered sequence of elements which are mutable or changeable are called ……………………..
(a) object
(b) tuple
(c) list
(d) dictionary
Answer:
(c) list

Question 40.
…………………….. command arranges a list value in ascending order.
Answer:
sort( )

Question 41.
……………………… is used to arrange a list in descending order.
Answer:
sort(reverse)

Question 42.
To sort by more than one column, we can use ………………………. with multiple indices.
Answer:
itemgetter

Question 43.
To sort second column which option have to be selected?
(a) itemgetter(0)
(b) itemgetter(1)
(c) itemgetter(2)
(d) itemgetter(3)
Answer:
(b) itemgetter(1)

Question 44.
(I) csv.writer work with list/tuple
(II) csv.Dictwriter work with dictionary
(III) csv.DictReader work with list/tuple/dictionary.
(a) (I),(II) – True (III) – False
(b) (I) – True (II), (III) – False
(c) (I),(II),(III) – True
(d) (I),(II),(III) – False
Answer:
(a) (I),(II) – True (III) – False

Question 45.
The additoinal argument fieldnames that are used with csv.DictReader and csv. Dictwriter are called as …………………………
Answer:
dictionary keys

Question 46.
Which function is used to print the data in dictionary format without order?
(a) dictionary
(b) print( )
(c) dict( )
(d) dictprint( )
Answer:
(c) dict( )

Question 47.
Which is a dictionary subclass that saves the order in whcih its contents are added?
(a) orderedDist
(b) SortDist
(c) DistSort
(d) Sorting
Answer:
(a) orderedDist

Question 48.
……………………. is used to remove the ordered Diet.
Answer:
Dist( )

Question 49.
…………………. method writes a row of data into the specified file.
Answer:
writerow( )

Question 50.
The number of parameters in csv.writer( ) …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 51.
Identify the wrong statement.
(a) The writerow( ) writes one row at a time
(b) The writerows( ) writes all the data at once
(c) No such writerows( ) function in csv
Answer:
(c) No such writerows( ) function in csv

Question 52.
Identify the true statement.
(a) writerow( ) takes 1 dimensional data
(b) writerows( ) takes 2 dimensional data
(c) both are true
(d) both are false
Answer:
(c) both are true

Question 53.
By default, csv files open automatically in ……………………..
Answer:
Excel.

PART – II
II. Answer The Following Questions

Question 1.
What is the syntax for csv.reader( )?
Answer:
The syntax for csv.reader( ) is
where csv.reader(fileobject,delimiter,fmtparams)
file object – passes the path and the mode of the file
delimiter – an optional parameter containing the standard dilects like, | etc can be omitted
fmtparams – optional parameter which help to override the default values of the dialects like skipinitialspace,quoting etc. Can be omitted

PART – III
III. Answer The Following Questions

Question 1.
How will you create a csv file?
Answer:
Creating CSV Normal File
To create a CSV file in Notepad, First open a new file using
File → New or Ctrl +N.
Then enter the data you want the file to contain, separating each value with a comma and each row with a new line.
For example consider the following details
Topic 1,Topic 2,Topic 3
one,two,three
Example 1,Example 2,Example 3
Save this content in a file with the extension .csv . You can then open the same using Microsoft Excel or any other spreadsheet program. Here we have opened using Microsoft Excel. It would create a table of data similar to the following:

Question 2.
How will you create csv file that contains comma with data?
Answer:

To retain the commas in “Address” column, you can enclose the fields in quotation marks.
For example:
RollNo, Name, Address
12101, Nivetha, “Mylapore, Chennai”
12102, Lavanya, “Adyar, Chennai”
12103, Ram, “Gopalapuram, Chennai”
As you can see, only the fields that contain commas are enclosed in quotes. If you open this in MS Excel, It looks like as follows

Question 3.
How will you create csv file that contains double quotes with data?
Answer:
Creating CSV File That contains Double Quotes With Data
If your fields contain double-quotes as part of their data, the internal quotation marks need to be doubled so that they can be interpreted correctly. For Example, given the following data:
Answer:

It should be written in csv file as RollNo, Name, FavoriteSports, Address
12101,Nivetha””” Cricket’”‘”, FootballMylapore chennai 12102, Lavanya,””” Basketball Cricket “””, Adyar chennai 12103, Ram,””” Soccer”””,”” Hockey”””, Gopalapuram chennai
The output will be

Question 4.
How will you create csv file using MS-Excel?
Answer:
Create A CSV File Using Microsoft Excel:
To create a CSV file using Microsoft Excel, launch Excel and then open the file you want to save in CSV format. For example, below is the data contained in our sample Excel worksheet:

Item Name, Cost-Rs, Quantity, Profit
Keyboard,480, 12,1152 ,
Monitor, 5200, 10, 10400
Mouse,200,50,2000
Total Profit = 13552

Question 5.
What are the different formats to create csv files?
Answer:

1. CSV file – data with default delimiter comma (,)
2. CSV file – data with Space at the beginning
3. CSV file – data with quotes
4. CSV file – data with custom Delimiters

Question 6.
Write a program create csv files data with sapces at the beginning?
Answer:
import csv
csv.register_dialect(‘myDialect’,delimiter = 7,skipinitialspace=True)
F=open(‘c:\ \pyprg\ \sample2.csv’,’r’)
reader= csv.reader(F, dialect=’myDialect’)
for row in reader:
print( row)
F.close( )
OUTPUT
[‘Topic 1’, ‘Topic2’, ‘Topic3’]
[‘one’, ‘two’, ‘three’]
[‘Example 1’, ‘Example2’, ‘Example3’]

Question 7.
Write a program to read csv files with custom delimiters?
Answer:
import csv
csv.register_dialect(‘myDialect’, delimiter ‘|’)
with open(‘c:\\pyprg\\sample4.csv’, ‘r’) as
f: reader= csv.reader(f, dialect=’myDialect’)
for row in reader: print(row)
F.close( )
Output
[‘RollNo’, ‘Name’, ‘City’]
[‘12101’, ‘Arun’, ‘Chennai’]
[‘12102’, ‘Meena’, ‘Kovai’]
[‘12103’, ‘Ram’, ‘Nellai’]

Question 8.
Give the syntax for csv.writer( )
Answer:
The syntax for csv.writer( ) is
csv. writerffileobject, delimiter,fmtparams)
where
fileobject : passes the path and the mode of the file.
delimiter : an optional parameter containing the standard dilects like , | etc can be omitted.
fmtparams : optional parameter which help to override the default values of the dialects like skipinitialspace,quoting etc. can be omitted.

Question 9.
Give the program to add new row to the csv file?
Answer:
import csv
row= [‘6’, ‘Sajini’Madurai’]
with open(‘student.csv’, ‘a’) as CF: # append mode to add data at the end
writer= csv.writer(CF)
writer.writerow(row) # writerow( ) method write a single row of data in file
CF.close( )
Output

Question 10.
Give a program for csv file with a line terminator?
import csv
Answer:
Data= [[‘Fruit’, ‘Quantity’], [‘Apple’, ‘5’], [‘Banana’, ‘7’], [‘Mango’, ‘8’]] csv.register_dialect(‘myDialect’, delimiter= ‘|’, lineterminator = ‘\n’) with open(‘c:\\pyprg\ \chl3\\line.csv’, ‘w’) as f:
writer= csv.writer(f, dialect=’myDialect’) writer.writerows(Data)
f.close( )
Output

PART – IV
IV. Answer The Following Questions

Question 1.
Write a program to read a specific column in a file?
Answer:
Read a specific column In a File:
To get the specific columns like only Item Name and profit for the “sampleS.csv” file. Then you have to do the following:
import csv
#opening the csv file which is in different location with read mode
f=open(“c:\ \pyprg\ \chl3sample5.csv”,’r’)
#reading the File with the help of csv.reader( )
readFile=csv.reader(f)
#printing the selected column for col in readFile :
print col[0],col[3]
f.close( )

Question 2.
Write a python program to read a csv file and store it in a list?
Answer:
Read A CSV File And Store It In A List
In this topic you are going to read a CSV file and the contents of the file will be stored as a list. The syntax for storing in the List is
list = [ ] # Start as the empty list
list.append(element) # Use append( ) to add elements
For example all the row values of “sample.csv” file is stored in a list using the following
program
import csv
# other way of declaring the filename
inFile= ‘c:\ \pyprg\\sample.csv’
F=open(inFile,’r’)
reader= csv.reader(F)
# declaring array
arrayValue = [ ]
# displaying the content of the list
for row in reader:
array Value, appendfrow)
print(row)
F.close( )

Question 3.
Write a python program to read a csv file and store a column value in a list for sorting?
Answer:
python file
F=open(inFile;’r’)
# reading the File with the help of csv.reader( )
reader = csv.reader(F)
# skipping the first row(heading) next( reader)
# declaring a list array Value = [ ]
a= int(input (“Enter the column number 1 to 3:-“))
# sorting a particular column-cost for row in reader:
array Value. append(row[a]) array Value.sorif) for row in array Value: print (row)
F.close( )
OUTPUT
Enter the column number 1 to 3:- 2
50
12
10

Question 4.
Write a python program to get data at runtime and write it in a csv file?
Answer:
import csv
with open(‘c:\ \pyprg\ \chl3\\dynamicfile.csv’, ‘w’) as f:
w = csv. writer (f)
ans=’y’
while (ans= =’y’):
name= input(“Name?: “)
date = input(“Date of birth: “)
place = input(“Place: “) w.writerow([name, date, place])
ans=input(“Do you want to enter more y/n?: “)
F=open(‘c:\ \pyprg\ \chl3\\dynamiefde.csv,’r’)
reader = csv.reader(F)
for row in reader:
print(row)
F.close( )
OUTPUT
Name?: Nivethitha
Date of birth: 12/12/2001
Place: Chennai
Do you want to enter more y/n?: y
Name?: Leena
Date of birth: 15/10/2001
Place: Nagercoil
Do you want to enter more y/n?: y
Name?: Padma
Date of birth: 18/08/2001
Place: Kumbakonam
Do you want to enter more y/n?: n
[‘Nivethitha’, ’12/12/2001′, ‘Chennai’]
[ ]
[’Leena’, ’15/10/2001′, ‘Nagercoil’]
[ ]
[‘Padma’, ’18/08/2001′, ‘Kumbakonam’]

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

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

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. 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 (QAB)^-1 = \(\left[\begin{array}{cc}
12 & -17 \\
-19 & 27
\end{array}\right]\) and A^-1 = \(\left[\begin{array}{cc}
1 & -1 \\
-2 & 3
\end{array}\right]\) then B^-1 = ……………

Answer:
(a) \(\left[\begin{array}{cc}
2 & -5 \\
-3 & 8
\end{array}\right]\)

Question 2.
If z is a complex number such that z ∈ C\R and \(z+\frac{1}{z}\) ∈ R, then |z| is ……………
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 3.
Let A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\) and 4B = \(\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & x \\
-1 & 1 & 3
\end{array}\right]\). If B is the inverse of A, then the value of x is ……………
(a) 2
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 4.
If sin^-1x + sin^-1y = \(\frac{2 \pi}{3}\) then cos^-1x + cos^-1y is equal to ……………
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) π
Answer:
(b) \(\frac{\pi}{3}\)

Question 5.
The eccentricity of the ellipse (x – 3)² + (y – 4)² = \(\frac{y^{2}}{9}\) is ……………
(a) √3/2
(b) 1/3
(c) 3√2
(d) 1/√3
Answer:
(b) 1/3

Question 6.
The directrix of the parabola y² = 4x is ……………
(a) y = -1
(b) x = -1
(c) y = 1
(d) x = 1
Answer:
(b) x = -1

Question 7.
The angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-3 \hat{k})+t(2 \hat{i}+\hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+\hat{j})+4=0\) is ……………
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Answer:
(c) 45°

Question 8.
The d.c.s. of a vector whose direction ratios are 2, 3, -6 are

Answer:
(a) \(\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)\)

Question 9.
A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. Find the rate of change of the balloon’s angle of elevation in radian per second when the balloon is 30 metres above the ground.
(a) 1/25 radians/sec
(b) 4/25 radians/sec
(c) 1/5 radians/sec
(d) 1/3 radians/sec
Answer:
(b) 4/25 radians/sec

Question 10.
The asymptote to the curve y²(2 +x) = x² (6 – x) is
(a) x = 2
(b) x = -2
(c) x = 6
(d) x = -6
Answer:
(b) x = -2

Question 11.
The change in the surface area S = 6x² of a cube when the edge length varies from x₀ to x₀ + dx is
(a) 12x₀ + dx
(b) 12x₀dx
(c) 6x₀dx
(d) 6x₀ + dx
Answer:
(b) 12x₀dx

Question 12.
The differential of y if y = sin 2x is
(a) 2 cos 2x
(b) 2 cos 2x.dx
(c) -2 cos lx.dx
(d) cos2x.dx
Answer:
(b) 2 cos 2x.dx

Question 13.
The value of \(\int_{-4}^{4}\left[\tan ^{-1}\left(\frac{x^{2}}{x^{4}+1}\right)+\tan ^{-1}\left(\frac{x^{4}+1}{x^{2}}\right)\right] d x\) is
(a) π
(b) 2π
(c) 3π
(d) 4π
Answer:
(d) 4π

Question 14.
\(\begin{aligned}
&\int^{b} f(x) d x=\\
&a
\end{aligned}\) ………

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

Question 15.
The integrating factor of the differential equation \(\frac{d y}{d x}+y=\frac{1+y}{x}\) is
(a) \(\frac{x}{e^{\lambda}}\)
(b) \(\frac{e^{x}}{x}\)
(c) λe^x
(d) e^x

Question 16.
The differential equation representing the family of curves y = A cos (x + B), where A and B are parameters, is

Answer:
(b)\(\frac{d^{2} y}{d x^{2}}+y=0\)

Question 17.
A rod of length 21 is broken into two pieces at random. The probability density function of
the shorter of the two pieces is \(f(x)=\left\{\begin{array}{ll}
\frac{1}{l}, & 0< x, \mathrm{x}=0, & l \leq x<2 l
\end{array}\right.\)
The mean and variance of the shorter of the two pieces are respectively ,

Answer:

Question 18.
A computer salesperson knows from his past experience that he sells computers to one in every twenty customers who enter the showroom. What is the probability that he will sell a computer to exactly two of the next three customer?

Answer:
(a) \(\frac{57}{20^{3}}\)

Question 19.
The operation * defined by a * b = \(\frac{a b}{7}\) is not a binary operation on
(a) Q⁺
(b) Z
(c) R
(d) C

Question 20.
If X is a discrete random variable then P(X >a) =
(a) P(X < a)
(b) 1 – P(X < a)
(c) 1 – P(X < a)
(D) = 0

Part – II

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

Question 21.
State the reason for \(\cos ^{-1}\left[\cos \left(-\frac{\pi}{6}\right)\right] \neq-\frac{\pi}{6}\)
Answer:
We know cos(-θ) = cos θ

Question 22.
Find the equations of the tangent and normal to the circle x² + y² = 25 at P(-3, 4).
Answer:
Equation of tangent to the circle at P (x₁, y₁) is xx₁ +yy₁ = a².
That is, x(-3) + y(4) = 25
-3x + 4y = 25
Equation of normal is xy₁ – yx₁ = 0
That is, 4x + 3y = 0.

Question 23.
Determine whether the three vectors \(2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+2 \hat{k}\) and \(3 \hat{i}+\hat{j}+3 \hat{k}\) are coplanar.
Answer:

Question 24.
Find the intervals of monotonicities and hence find the local extremum for the following
functions f(x) = \(\frac{x}{x-5}\)
Answer:
f(x) = \(\frac{x}{x-5}\)


(when x ≠ 5)
f(x) is strictly decreasing on (-∞, 5) and (5, ∞)
And there is no local extremum

Question 25.
Find the linear approximation for f(x) = [altex]\sqrt{1+x}[/latex] , x ≥ -1, at x₀ = 3. Use the linear approximation to estimate f(3.2).
Answer:
We know from that L(x) = f(x₀) + f'(x₀)(x – x₀). We have x₀ = 3, ∆x = 0.2 and hence

Actually, if we use a calculator to calculate we get \(\sqrt{4.2}=2.04939\)

Question 26.
Evaluate \(\int_{0}^{2 \pi} \sin ^{4} x \cos ^{3} d x\)
Answer:
f(x) = sin⁴x cos³x
f(2π – x) = sin⁴(2π – x) cos³ (2π- x)
= sin⁴x cos³x = f(x)
\(2\int_{0}^{\pi} \sin ^{4} x \cos ^{3} d x\)
Again f(x) = sin⁴x cos³x
f(π – x) = sin⁴(π – x) cos³ (π – x)
f(π – x) = – sin⁴(π -x) cos³(π -x) = -f(x) = 0.

Question 27.
Solve the differential equation \(\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Answer:

Question 28.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
\(f(x)=\left\{\begin{array}{lr}
k & 200 \leq x \leq 600 \\
0 & \text { otherwise }
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Let * be defined on R by (a*b) = a + b + ab – 7. Is * binary on R ? If so, find \(3*\left(\frac{-7}{15}\right)\)
Answer:
a*b = a + b + ab – 7
Now when a, b ∈ R, then ab ∈ R also a + b ∈ R.
So, a + b + ab ∈ R.
We know – 7 ∈ R.
So, a + b + ab – 7 ∈ R.
(i.e.) a * b ∈ R

Question 30.
Examine for the rational roots of x⁸ – 3x + 1 = 0.
Answer:
x⁸ – 3x + 1 = 0……..(1)
Here a_n = – 1, a₀ = 1
If p/q is a rational root of (1)
Then q is a factor a_n, p is a factor of aQ
The possible values of p and q are ±1. Among the possible values 1,-1, [(p,q) = 1]
None of them satisfy the equation (1)
.’. The above equation has no rational roots.

Part – III

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

Question 31.
Solve the following system of linear equations by matrix inversion method:
2x + 5y = -2, x + 2y = -3
Answer:

Question 32.
If (x₁ + iy₁)(x₂ + iy₂)(x₃ + iy₃)…………(x_n +iy_n) = a + ib , show that
(i) (x₁² + y₁²)(x₂² + y₂²)(x₃² + y₃²)…….(x_n² + y_n²) = a² + b²
(ii)

Answer:

Question 33.
Find the sum of squares of roots of the equation 2x⁴ – 5x³ + 6x² – 3 = 0.
Answer:
The given equation is 2x⁴ – 8x³ + 6x² – 3 = 0.

Let the roots be α β γ δ
α + β + γ + δ = -b = 4
α β + βγ + γδ + aδ + aγ + βδ ) = c = 3
α βγ + βγδ + γαδ = -d = 0
αβγδ???? = -3/2
To Find α² + β² + γ² + δ ² = ( α + β + γ + δ)² – 2 (αβ + βγ + γδ + αδ + αγ +βδ)
= (4)² – 2(3) = 16 – 6 = 10

Question 34.
Find the non-parametric form of vector equation, and Cartesian equation of the plane
passing through the point (0, 1, -5) and parallel to the straight lines .
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+s(2 \hat{i}+3 \hat{j}+6 \hat{k})\) and \(\hat{r}=(\hat{i}-3 \hat{j}+5 \hat{k})+t(\hat{i}+\hat{j}-\hat{k})\)
Answer:
We observe that the required plane is parallel to the vectors \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{c}=\hat{i}+\hat{j}-\hat{k}\) and passing through the point (0, 1, —5) with position vector \(\vec{a}\). We observe that \(\vec{b}\) is not parallel to \(\vec{c}\)
Then the vector equation of the plane in non-parametric form is given by

If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as -9x + 8y – z = 13 or 9x – 8y + z +13 = 0.

Question 35.
Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = x³ – 5x² – 3x in [1,3]
Answer:
f(x) = x³ – 5x² – 3x
a = 1, b = 3
f(a) = f(1)= 1 – 5 – 3 = -7
f(b) = f (3) = 21 – 5(9) – 3(3)
= 27 – 45 – 9 = -27

f(x) = x³ – 5x²– 3x
f(x) = 3² – 10x – 3
f'(c) = 3c² – 10c – 3 ………(2)
From (1) and (2),
3c² – 10c – 3
3c² – 10c – 3 + 10 = 0
3c² – 10c + 7 =0
3c² – 3c – 7c + 7 = 0

So, Lagrange’s mean value theorem is true with c = 7/3

Question 36.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. (255)^1/4
Answer:

Question 37.
Find the area of the region bounded by the line 6x + 5y = 30, x – axis and the lines x = – 1 and x = 3.
Answer:
The region is sketched in Fig. It lies above the x – axis.
Hence, the required area is given by

Question 38.
A six sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Answer:
Six sided die marked ‘2’ on one face, ‘3’ on two faces and ‘4’ on three faces.
When it is thrown twice, we get 36 sample points.
‘X’ denotes sum of the face numbers and the possible values of ‘X’ are 4, 5, 6, 7 and 8
For X = 4, the sample point is (2, 2)
For X = 5, the sample points are (2, 3), (3, 2)
For X = 6, the sample points are (3, 3), (2, 4), (4, 2)
For X = 7, the sample points are (3, 4), (4, 3)
For X = 8, the sample point is (4, 4)

Question 39.
Check whether the statement p —>• (q —► p) is a tautology or a contradiction without using the truth table.
Answer:

Question 40.
Solve: \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x \sqrt{1-x^{2}}\)
Answer:

Part – IV

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

Question 41.
(a) Solve the following system of homogeneous equation.
3x + 2y + 7z – 0, 4x – 3y – 2z – 0, 5x + 9y + 23z = 0,
Answer:
The matrix form of the above equations is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0
-17y – 34z = 0
Taking z = t in (2) we get -17y – 34t = 0

Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t= 0
3x – 4f + 7f = 0 ⇒ 3x = -3f ⇒ x = -t
So the solution is x = -t; y = -2t; and z=t, t∈R [OR]

(b) Two cards are drawn with replacement from a well shuffled deck of 52 cards. Find the mean and variance for the number of aces.
Answer:
n (S) = 52
Number of aces = n (A) = 4

Let X denotes number of aces when two cards are drawn.

a) Show thatis purely imaginary.

Question 42.
(a) Show that \(\left(\frac{19+9 i}{5-3 i}\right)^{15}-\left(\frac{8+i}{1+2 i}\right)^{15}\) is purely imaginary.
Answer:

[OR]

(b) (i) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right) \) a,b ∈ Q . Examine the closure, commutative, and associative properties satisfied by * on Q
(ii) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right) \) a,b ∈ Q
existence of identity and the existence of inverse for the operation * on Q
Answer:
(i) 1. Closure property:
Let a, b ∈ Q.
a*b = \(\left(\frac{a+b}{2}\right) \) ∈ Q (∵ a, b, 2 ∈ Q)
So, closure property is satisfied.

2. Commutative property:
Let a,b ∈ Q.
a*b = \(\left(\frac{a+b}{2}\right) \) …..(1)
b*a = \(\left(\frac{a+b}{2}\right) \) ……(2)
(1) = (2) ⇒ Now a * b = b * a
⇒ Commutative property is satisfied.

3. Associative property:
Let a, b,c ∈ Q.
To prove associative property we have to prove that a * (b * c) = (a * b) * c

So, associative property is not satisfied.

(ii) a* b = \(\left(\frac{a+b}{2}\right) \)
Let e ∈ Q be the identity element. Then a * e = a
To find e : a * e = a
(i.e.) \(\left(\frac{a+b}{2}\right) \) = a ⇒a + e = 2a ⇒ e = 2a- a
(i.e.) the identity element e = a which is not possible.
So, the identity element does not exist and so inverse does not exist.

Question 43.
(a) Solve (v – 3) (x – 6) (x – 1) (x + 2) + 54 = 0.
Answer:
(x – 3) (x – 1) (x – 6) (x + 2) + 54 = 0
(x²– 4x + 3) (x² – 4x – 12) + 54 = 0
Put x² – 4x = y
(y + 3)(y – 12) + 54 = 0
y²– 9y – 36 + 54 = 0
y² – 9y + 18 = 0
(y – 3) (y – 6) = 0
(y – 3) = 0
x² – 4x – 3 = 0


[OR]

(b) Find the local extrema of the function f(x) = 4x⁶ – 6x⁴
Answer:
Differentiating with respect to x, we get
f (x) = 24x⁵ – 24x³
= 24x³ (x² – 1)
= 24x³ (x + 1) (x – 1)
f’ (x) = 0 ⇒ x = -1, 0, 1. Hence the critical
numbers are x = -1, 0, 1
Now, f'”(x) = 120x⁴ – 72x² = 24x² (5x² – 3)
⇒ f” (-1)= 48, f” (0) = 0, f”(1) = 48.
As f” (-1) and f”(1) are positive by the second derivative test, the function f(x) has local minimum. But at x = 0, f ” (0) = 0. That is the second derivative test does not give any information about local extrema at x = 0 . Therefore, we need to go back to the first derivative test. The intervals of monotonicity is tabulated in the table.

Question 44.
(a) Prove that \(\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}\)
Answer:

(b) If v(x, y) = \(\log \left(\frac{x^{2}+y^{2}}{x+y}\right)\), prove that \(x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=1\)
Answer:

‘ f ‘ is a homogeneous function of degree 1. By Euler’s theorem, we have

Question 45.
(a) Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x² + 7y² = 14 .
Answer:
2x² + 7y² = 14

The equation of tangent to the above ellipse will be of the form

Here the tangents are drawn from the point (5, 2)

Squaring on both sides we get
(2 – 5m)² = 7m² + 2
25m² + 4 – 20m – 7m² – 2 = 0
18m² – 20m + 2 = 0
(= by 2) => 9m² – 10m + 1 = 0
(9m – 1) (m – 1) = 0
m = 1 (or) m = 1/9
When m = 1, the equation of tangent is
y = x + 3 or x-y + 3 = 0
When m = 1/9 the equation of tangent is 9

[OR]

(b) Find the coordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Answer:

Direction of the normal plane (1, 2, 3) JP(4, 3,2)
d.c.s of the PQ is\(\frac{x_{1}-4}{1}=\frac{y_{1}-3}{2}=\frac{z_{1}-2}{3}=k\)
x₁k + 4, y₁ = 2k + 3, z₁ = 3k + 2
This passes through the plane x + 2y + 3z = 2
k + 4 + 2(2 k + 3) + 3(3 k + 2) = 2
k+ 4 + 4k + 6 + 9k + 6 = 2
14k = 2 – 16
14k = -14
k = -1
∴ The coordinate of the foot of the perpendicular is (3, 1, -1)
∴ Length of the perpendicular to the plane is

Question 46.
(a) Solve (x + y)²\(\frac{d y}{d x}\) = a² dx
Answer:
Put x+y = z. Differentiating with respect to x we get

(b) A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
Answer:

Question 47.
(a) Let w(x, y’ z) = \(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}(x, y, z)\) ≠ (0, 0, 0). Show that \(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}=0\)
Answer:

(b) Evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{e^{-\tan x}}{\cos ^{6} \pi x} d x\)
Answer:

Students can Download Tamil Nadu 12th Commerce Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Commerce Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Commerce 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. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part – I

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

Question 1.
With a wider span, there will be ………… hierarchical levels.
(a) more
(b) less
(c) multiple
(d) additional
Answer:
(b) less

Question 2.
Spot market is a market when the delivery of the financial instrument and payment of cash occurs.
(a) Immediately
(b) In the future
(c) Uncertain
(d) After one mouth
Answer:
(a) Immediately

Question 3.
How many times a security can be sold in a secondary market?
(a) only one time
(b) two time
(c) three times
(d) multiple times
Answer:
(d) multiple times

Question 4.
Money market provides ……………
(a) medium term funds
(b) short term funds
(c) long term funds
(d) shares
Answer:
(b) short term funds

Question 5.
Jobbers transact in a stock exchange.
(a) For their Clients
(b) For their Own Transactions
(c) For other Brokers
(d) For other Members
Answer:
(b) For their Own Transactions

Question 6.
Registering and controlling the functioning of collective investment schemes as ………..
(a) Mutual funds
(b) Listing
(c) Rematerialisation
(d) Dematerialisation
Answer:
(d) Dematerialisation

Question 7.
Human resource management determines the …… relationship.
(a) internal, external
(b) employer, employee
(c) owner, servant
(d) principle, agent
Answer:
(b) employer, employee

Question 8.
Advertisement is a ………. source of recruitment.
(a) internal
(b) external
(c) agent
(d) outsourcing
Answer:
(b) external

Question 9.
Selection is usually considered as a,process.
(a) positive
(b) negative
(c) natural
(d) none of these
Answer:
(b) negative

Question 10.
Improves skill levels of employees to ensure better job performance.
(a) Training
(b) Selection
(c) Recruitment
(d) Performance appraisal
Answer:
(d) Performance appraisal

Question 11.
Which one of the market deals in the purchase and sale of shares and debentures?
(a) stock exchange market
(b) Manufactured goods market
(c) local market
(d) family market
Answer:
(a) stock exchange market

Question 12.
In the following variables which one is not the variable of marketing mix?
(a) place variable
(b) product variable
(c) program variable
(d) price variable
Answer:
(c) program variable

Question 13.
Social marketing deals with
(a) Society
(b) Social clan
(c) Social change
(d) Social evil
Answer:
(c) Social change

Question 14.
The main objective of all business enterprises is …………….
(a) Providing service
(b) Providing better standard of life
(c) Providing necessities to the society
(d) Earn profit
Answer:
(d) Earn profit

Question 15.
As the consumer is having the rights, they are also having
(a) measures
(b) Promotion
(c) responsibilities
(d) duties
Answer:
(c) responsibilities

Question 16.
Which of the following is not a consumer right summed up by John.F.Kennedy.
(a) Right to safety
(b) Right to choose
(c) Right to consume
(d) Right to be informed
Answer:
(a) Right to safety

Question 17.
The chairman of the District Forum is
(a) District Judge
(b) High court Judge
(c) Supreme court Judge
(d) None of the above
Answer:
(a) District Judge

Question 18
ownership makes bold management decisions due to their strong foundation in the intimation level.
(a) Private
(b) Public
(c) Corporate
(d) MNC’s
Answer:
(a) Private

Question 19.
The property in the goods means the
(a) Possession of goods
(b) Custody of goods
(c) Ownership of goods
(d) Both (a) and (b)
Answer:
(c) Ownership of goods

Question 20.
Number of parties is a bill of exchange are
(a) 2
(b) 6
(c) 3
(d) 4
Answer:
(c) 3

Part – II

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

Question 21.
What are the objectives of MBO?
Answer:
Management by objectives is intended primarily:

• To measure and judge performance.
• To relate individual performance to organisational goals.
• To clarify both the job to be done and the expectations of accomplishment.
• To foster the increasing competence and growth of the subordinates.

Question 22.
Write a note on OTCEI.
Answer:
The OTCEI was set up by a premier financial institution to allow the trading of securities across the electronic counters throughout the country.

Question 23.
Define Stock Exchange.
Answer:
According to Husband and Dockerary, “Stock exchanges are privately organized markets which are used to facilitate trading in securities.”

Question 24.
Give the meaning of Human Resource.
Answer:
In an organisation, the human resource are the employees who are inevitable for the survival t and success of the enterprise.

Question 25.
What is service marketing?
Answer:
Service marketing denotes the processing of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism, professional services, repairs etc.

Question 26.
Write short notes on: “Right to be informed.”
Answer:
Consumers should be given all the relevant facts about the product. This implies that the manufacturer and the dealer are expected to disclose all the material facts relevant and relating to the product.

Question 27.
What is GST?
Answer:
GST is the indirect tax levied on goods and services across the country. It is a comprehensive, multi-stage, destination-based tax that is levied on every value addition.

Question 28.
What is a contract of sale of goods?
Answer:
Contract of sale of goods is a contract whereby the seller transfers or agrees to transfer the property (ownership) of the goods to the buyer for a price.

Question 29.
What is the other name of business entrepreneur?
Answer:
The term entrepreneur means the person who takes steps for commencing the business. So he is otherwise called as organiser or proprietor.

Question 30.
Define Director.
Answer:
The Companies Act 2013 section 2 (34) defines a director appointed to the board of a Company is “A Person who is appointed or elected member of the Board of Directors of a company and has the responsibility of determining and implementing policies along with others in the board”.

Part – III

Answer any seven in which question No. 31 is compulsory. [7 x 3 = 21]

Question 31.
What are the process involved in MBO?
Answer:

• Defining Organisational Objectives
• Goals of Each Section
• Fixing Key Result Areas
• Setting subordinate objectives or targets
• Matching Resources with objective
• Periodical Review meetings
• Appraisal of Activities
• Reappraisal of objectives

Question 32.
Differentiate spot market from future market.
Answer:

Spot Market	Future Market
Spot market is otherwise called cash market. It is a market where the delivery of the financial instrument and payment of cash occurs immediately, i.e. settlement is completed immediately.	Future market is otherwise called forward market. It is a market where the delivery of asset and payment of cash takes place at a predetermined time frame in future.

Question 33.
What are the documents required for a Demat account?
Answer:
For opening a demat account, we submit proof of identity and address along with a passport size photo and the account opening form.

• Documents for Identity: Pan card, Voters ID, Passport, Driver’s License, IT Returns, Electricity and Telephone Bills are the Identity documents.
• Documents for Address: Ration card, Passport, Voter’s ID card, Driving License, Telephone Bills, Electricity bills are the address documents.

Question 34.
Mention any three Role of Marketer.
Answer:
(1) Instigator: As an instigator, marketer keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever changing market.

(2) Integrator: Marketer plays a role of integrator in the sense that he collects feedback or vital inputs from channel members and consumers.

(3) Implemented Marketer plays a role of implementer when he/she actually converts marketing opportunities into marketable product.

Question 35.
Discuss the objectives E-Marketing.
Answer:
The following are the objectives of E-Marketing:

• Expansion of market share
• Reduction of distribution and promotional expenses
• Achieving higher brand awareness
• Strengthening database

Question 36.
What do you understand by “Right to redressal”?
Answer:
The complaints and protests are not just to be heard: but the aggrieved party is to be granted compensation within a reasonable time period . There should be prompt settlement of complaints and claims lodged by the aggrieved customers.

Question 37.
Explain the concept of Privatisation.
Answer:
Privatisation means permitting the private sector to set up industries which were previously reserved for the public, sector. Under this policy many Public Sector Units (PSUs) were sold to private sector. The main reason for privatisation was that PSUs were running in losses due to mismanagement and political interference.

Question 38.
What are the characteristics of a bill of exchange?
Answer:
Characteristics of a Bill of Exchange:

• A bill of exchange is a document in writing.
• The document must contain an order to pay.
• The order must be unconditional.
• The instrument must be signed by the person who draws it.
• The name of the person on whom the bill is drawn must be specified in the bill itself.

Question 39.
Distinguish between Entrepreneur and Manager.

Basis of difference	Entrepreneur	Manager
Motive	The very motive of an entrepreneur is to start a venture by setting of an entity.	The very motive of manager is to render service in an entity setup for execution of venture.
Status	Entrepreneur is owner of the entity.	Manager is a salaried employee in the entity set up for carrying on the venture.
Risk Bearing	Entrepreneur bears the eventual risk and uncertainty in operating the enterprise.	Manager doesn’t bear any risk in the venture, where the venture is unsuccessful he/she simply. quits the enterprise.

Question 40.
When are alternative directors appointed?
Answer:
Alternate director is appointed by the Board of Directors, as a substitute to a director who may be absent from India, for a period which is not less than three months.

Part – IV

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

Question 41
(a) Explain the various functions of management.
Answer:

• Planning:- Planning is the primary function of management. Planning is a constructive reviewing of future needs so that present actions can be adjusted in view of the established goal.
• Organising:- Organising is the process of establishing harmonious relationship among the members of an organisation and the creation of network of relationship among them.
• Staffing:- Staffing function comprises the activities of selection and placement of competent personnel.
• Directing:- Directing denotes motivating, leading, guiding and communicating with subordinates on an ongoing basis in order to accomplish pre-set goals.
• Controlling:- Controlling is performed to evaluate the performance of employees and deciding increments and promotion decisions.
• Co-ordination:- Co-ordination is the synchronization of the actions of all individuals, working in the enterprise in different capacities.
• Motivating:- The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers.

Subsidiary Functions:

• Innovation:- Innovation includes developing new material, new products, new techniques in production, new package, new design of a product and cost reduction.
• Representation:- A manager has to act as representative of a company. It is the duty of every manager to have good relation with others.
• Decision-making:- Every employee of an organisation has to take a number of decisions every day. Decision-making helps in the smooth functioning of an organisation.
• Communication:- Communication is the transmission of human thoughts, views or opinions from one person to another person. Communication helps the regulation of job and co-ordinates the activities.

[OR]

(b) Compare the concept of social marketing with service marketing.
Answer:
Social marketing: Social marketing is a new marketing tool.
It is the systematic application of marketing philosophy to achieve social good.
The primary aim of social marketing is ‘social good’ such as anti-tobacco, anti-drug, anti-pollution, anti-dowry, road safety, protection of girl child.

• Service marketing: A service is any activity or benefit that one party can offer to another which is essentially intangible.
• Service marketing is a specialized branch of marketing.
• Service marketing denotes the process of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism and repairs.

Question 42
(a) Explain the various disadvantages of MBO.
Answer:

• MBO fails to explain the philosophy; most of the executives do not know how MBO works? what is MBO? and why is MBO necessary? and how participants can benefit by MBO.
• MBO is a time consuming process. Much time is needed by senior people for framing the MBO. Next, it leads to heavy expenditure and also requires heavy paper work.
• MBO emphasises only on short-term objectives and does not consider the long-term objectives.
• The status of subordinates is necessary for proper objectives setting. But, this is not possible in the process of MBO.
• MBO is rigid one. Objectives should be changed according to the changed circumstances, external or internal. If it is not done, the planned results cannot be obtained.

[OR]
(b) Explain the duties of consumers.
Answer:
Apart from rights, there are certain duties imposed on the consumer. The following are the duties of consumers.

• Buying Quality Products at Reasonable Price: It is the duty of a consumer to purchase a product after gaining a thorough knowledge of its price, quality and other terms and conditions.
• Ensure the Weights and Measurement before Purchase: The consumer should ensure that he/she is getting the product of exact weight and. measure.
• Reading the Label Carefully: It is the duty of the consumer to read the label of the product thoroughly.
• Beware of False and Attractive Advertisements: It is the prime duty of the consumer about the genuineness of the advertisement, before purchasing the product.
• Ensuring the Receipt of Cash Bill: It is a legitimate duty of consumers to get the cash receipt and warranty card supplied along with the bill.

Question 43.
(a) Explain the Instruments of Money Market.
Answer:
(i) Treasury Bills: Treasury bills are very popular and enjoy a higher degree of liquidity since they are issued by the Government. A Treasury bill is nothing but a promissory note issued for a specified period stated therein. The Government promises to pay the specified amount mentioned therein to the bearer of the instrument on the due date. The period does not exceed a period of one year.

(ii) Certificate of Deposits: Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

(iii) Commercial Bills: The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. It possesses the advantages like self-liquidating in nature, recourse to two parties, knowing exact date of transactions, transparency of transactions etc.

[OR]

(b) Explain the overall performance of State Commission.
The State Commission is to be appointed by the State Government. The person who is a judge or retired judge of high court is the president of the state commission.

Performances:

• The compensation of the value should not exceed Rs.20 lakhs and below Rs.1 crore.
• The state commission has the power to call for the records and pass orders in the district forum.
• To furnish the information which is required for the purpose of the Act to any officer.

Question 44.
(a) Explain the powers of SEBI.
Answer:
The various powers of a SEBI are explained below:

• Powers Relating to Stock Exchanges and Intermediaries: SEBI has wide powers to get the information from the stock exchange and intermediaries regarding their business transactions for inspection.
• Power to Impose Monetary Penalties: SEBI has power to impose monetary penalties on capital market intermediaries for violations.
• Power to Initiate Actions in Functions Assigned
• Power to Regulate Insider Trading: SEBI has power to regulate insider trading or can regulate the functions of merchant banker.
• Powers Under Securities Contracts Act: For the regulation of stock exchange, the Ministry of Finance issued a notification, for delegating several of its powers under the securities contract Act.

[OR]

(b) Explain the impact of LPG on Indian Economy.
Impact of Liberalization:

• Liberalization has opened up new business opportunities abroad and increased foreign direct investment.
• New market for various goods came into existence and resulted not only in urban but also in rural development.
• It became very easy to obtain loans from banks for business expansion.

Impact of Privatization:

• Privatization has a positive impact on the financial growth by decreasing the deficits and debts.
• Increase in the efficiency of government undertakings.
• Provide better goods and services to the consumers.

Impact of Globalization:

• Multinational corporations can manufacture, buy and sell goods worldwide.
• Globalization has led to a boom in consumer products market.
• The advent of foreign companies and growth in economy has led to job creation.

Question 45
(a) Describe the significance of External source of recruitment.
Answer:
External sources of recruitment include sources that lie outside the organisation. This provides a wider collection of potential employees with the necessary skillset, especially for managerial and technical positions. Existing employees can recommend suitable candidates, which may lead to a higher level of teamwork and synchronisation among employees. Hiring new employees can lead to the introduction of new blood and thus the introduction of a new set of skills and ideas. External sources of recruitment offer jobs to unskilled, semi-skilled and skilled workers.

[OR]

(b) Discuss in detail the features of a cheque.
Answer:
A cheque is a negotiable instrument drawn on a particular banker.
Features:
(i) Instrument in Writings
A cheque or a bill or a promissory note must be an instrument in writing. Though the law does not prohibit a cheque being written in pencil, bankers never accept it because of risks involved. Alternation is quite easy but detection is impossible in such cases.

(ii) Unconditional Orders
The instrument must contain an order to pay money. It is not necessary that the word ‘order’ or its equivalent must be used to make the document a cheque. It does not cease to be a cheque just because the world ‘please’ is used before the word pay. Further the order must be unconditional.

(iii) Drawn on a Specified Banker Only
The cheque is always drawn on a specified banker. A cheque vitally differs from a bill in this respect as latter can be drawn on any person including a banker. The customer of a banker can draw the cheque only on the particular branch of the bank where he has an account.

(iv) A Certain Sum of Money Only
The order must be for payment of only money. If the banker is asked to deliver securities, the document cannot be called a cheque. Further, the sum of money must be certain.

(v) Payee to be Certain
The cheque must be made payable to a certain person or to the order of a certain person . or to the bearer of the instrument. The word, person includes corporate bodies, local authorities, associations, holders of the office of an institution etc.

(vi) Signed by the Drawer
The cheque is to be signed by the drawer. Further, it should tally with specimen signature furnished to the bank at the time of opening the account

Question 46.
(a) What are the differences between on the job training and off the job training?
Answer:

	Basis for comparison	On the Job Training	Off the Job Training
1.	Meaning	The employee learns the job in the actual work environment.	The training of employees is done outside the actual work place.
2.	Cost	It is cheapest to carry out.	It is costly due to the expenses like separate training room, specialist, etc.
3.	Suitable for	Suitable for manufacturing related jobs.	It is suitable for managerial jobs.
4.	Approach	Practical approach	Theoretical approach
5.	Carried out	Provided by the experienced employee	Provided by the experts
6.	Methods	Coaching, mentoring, apprenticeship, job rotation	Seminar, lectures, vestibule, field trip, e-leaming

[OR]

(b) What are the various kinds of Debentures?
Answer:
Debenture is a document issued by the company for acknowledging the loan from the public. Debentures are classified into different categories on the basis of:

• Convertibility of the Instrument
• Security of the Instrument;
• Redemption ability; and
• Registration of Instrument.

On the basis of convertibility:

• Non-Convertible Debentures: These instruments cannot be converted into equity shares.
• Partly Convertible Debentures: A part of these instruments are converted into equity shares.
• Fully Convertible Debentures: These are fully convertible into equity shares.
• Optionally Convertible Debentures: The investor can have the option to either convert the debentures at a price decided by the issuer or agreed upon at the time of issue.

On the basis of Security:

• Secured Debentures: These instruments are secured by a charge on the fixed assets of the issuer company.
• Unsecured Debentures: These instruments are unsecured against the assets.

On the basis of Redeemability:

• Redeemable Debentures: It refers to the debentures which will be redeemed in the future.
• Irredeemable Debentures: It is a debenture, in which no specific time is specified by the companies to pay back the money.

On the basis of Registration:

• Registered Debentures: These are issued in the name of a particular person, who is registered by the company.
• Bearer Debentures: These are issued to the bearer and are negotiable instruments, and are transferred by mere delivery.

Question 41.
(a) Why the marketing is important to the society and individual firm? Explain.
Answer:
Importance of Marketing:
To the Society:

• Marketing is a connecting link between the consumer and the producer.
• Marketing helps in increasing the living standard of people.
• Marketing helps to increase the nation’s income.
• Marketing process increases employment opportunities.
• Marketing helps to maintain economic stability and rapid development in various countries.

To the Individual Firms:

• Marketing generates revenue to firms.
• Marketing gives information to the top management for taking overall decisions on production.
• Marketing and innovation are the two basic functions of all businesses.

[OR]

(b) State the qualification of directors.
Qualifications of Director: As regards to the qualification of directors, there is no direct provision in the Companies Act, 2013. In general, a director shall possess appropriate skills, experience and knowledge in the fields of finance, law, management, sales, marketing, research and other disciplines related to the business.

The following are the qualifications:

• A director must be a person of sound mind.
• A director must hold share qualification if the article of association provides such.
• A director must be an individual.
• A director should be a solvent person.
• A director should not be convicted by the Court for any offence, etc.

Students can Download Samacheer Kalvi 10th Tamil Model Question Paper 1 Pdf, Samacheer Kalvi 10th Tamil Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamil Nadu Samacheer Kalvi 10th Tamil Model Question Paper 1

நேரம்: 3.00 மணி
மதிப்பெண்கள் : 100

(குறிப்புகள்:

• இவ்வினாத்தாள் ஐந்து பகுதிகளைக் கொண்டது. அனைத்து பகுதிகளுக்கும் விடையளிக்க – வேண்டும். தேவையான இடங்களில் உள் தேர்வு வினாக்கள் கொடுக்கப்பட்டுள்ளது. காக
• பகுதி I, II, III, IV மற்றும் Vல் உள்ள அனைத்து வினாக்களுக்குத் தனித்தனியே விடையளிக்க வேண்டும்.
• வினா எண். 1 முதல் 15 வரை பகுதி-1ல் தேர்வு செய்யும் வினாக்கள் தரப்பட்டுள்ளன. ஒவ்வொரு வினாவிற்கும் ஒரு மதிப்பெண். சரியான விடையைத் தேர்ந்தெடுத்து குறியீட்டுடன் எழுதவும்.
• வினா எண் 16 முதல் 28 வரை பகுதி-IIல் இரண்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன: ஏதேனும் 9 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 29 முதல் 37 வரை பகுதி-IIIல் மூன்று மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. –
  ஏதேனும் 6 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 38 முதல் 42 வரை பகுதி-IVல் ஐந்து மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. ஏதேனும் 5 வினாக்களுக்கு மட்டும் விடையளிக்கவும்.
• வினா எண் 43 முதல் 45 வரை பகுதி-Vல் எட்டு மதிப்பெண் வினாக்கள் தரப்பட்டுள்ளன. அனைத்து வினாவிற்கும் விடையளிக்கவும்.

பகுதி – 1 (மதிப்பெண்கள்: 15) 

(i) அனைத்து வினாக்களுக்கும் விடையளிக்கவும்.
(ii) கொடுக்கப்பட்ட நான்கு விடைகளில் சரியான விடையினைத் தேர்ந்தெடுத்துக் குறியீட்டுடன் விடையினையும் சேர்த்து எழுதுக. [15 x 1 = 15]

(குறிப்பு: விடைகள் தடித்த எழுத்தில் உள்ளன.)

Question 1.
”சாகும் போதும் தமிழ் படித்துச் சாகவேண்டும் என்றன் சாம்பலும் தமிழ் மணந்து வேகவேண்டும்” என்று பாடியவர் யார்?
(அ) பாரதிதாசன்
(ஆ) பெருஞ்சித்திரனார்
(இ) சச்சிதானந்தன்
(ஈ) ஆறுமுகநாவலர்
Answer:
(இ) சச்சிதானந்தன்

Question 2.
‘சிந்துக்குத் தந்தை என்று பாராட்டப்பட்டவர்……….
(அ) பாரதியார்
(ஆ) பாரதிதாசன்
(இ) சுரதா
(ஈ) கவிமணி
Answer:
(அ) பாரதியார்

Question 3.
ஒரு தொடர்மொழியில் இரண்டு சொற்கள் இருந்து அவற்றின் இடையே சொல்லோ உருபோ இன்றி பொருள் தருவது……………… எனப்படும்.
(அ) தொகை நிலைத் தொடர்
(ஆ) பெயரெச்சத் தொடர்
(இ) வினையெச்சத் தொடர்
(ஈ) விளித் தொடர்
Answer:
(அ) தொகை நிலைத் தொடர்

Question 4.
மொழிபெயர்த்தல் என்னும் தொடரைத் தொல்காப்பியர் கையாண்ட இடம் ……………
(அ) பெயரியல்
(ஆ) வினையியல்
(இ) மரபியல்
(ஈ) உயிரியல்
Answer:
(இ) மரபியல்

Question 5.
புகழ்ந்தால் என்னுடல் புல்லரிக் காது இகழ்ந்தால் என்மனம் இறந்து விடாது’ – இவ்வடிகளில் அமைந்துள்ள முரண் சொல்………
அ) என்னுடல், என் மனம்
(ஆ) புல்லரிக்காது இறந்துவிடாது
(இ) புகழ்ந்தால், இகழ்ந்தால்
(ஈ) புகழ்ந்தால் என் மனம்
Answer:
(இ) புகழ்ந்தால், இகழ்ந்தால்

Question 6.
உயிரினங்களில் மனிதரை உயர்த்திக்காட்டுவது அவர்களின்
(அ) சிந்தனை ஆற்றல்
(ஆ) செல்வம்
(இ) வாழ்நாள்
(ஈ) ஆற்றல்
Answer:
(அ) சிந்தனை ஆற்றல்

Question 7.
சாலச் சிறந்தது’ என்பது …….. தொடராகும்.
(அ) பெயரெச்சத்தொடர்
(ஆ) உரிச்சொல்தொடர்
(இ) விளித்தொடர்
(ஈ) வினையெச்சத்தொடர்
Answer:
(ஆ) உரிச்சொல்தொடர்

Question 8.
நன்மொழி என்பதன் இலக்கணக் குறிப்பு ….
(அ) வியங்கோள் வினைமுற்று
(ஆ) பண்புத்தொகை
(இ) வினையெச்சம்
(ஈ) உம்மைத்தொகை
Answer:
(ஆ) பண்புத்தொகை

Question 9.
அறிஞருக்கு நூல், அறிஞரது நூல் ஆகிய சொற்றொடர்களில் பொருளை வேறுபடுத்தக் காரணமாக அமைவது
(அ) வேற்றுமை உருபு
(ஆ) எழுவாய்
(இ) உவம உருபு
(ஈ) உரிச்சொல்
Answer:
(அ) வேற்றுமை உருபு

Question 10.
விருந்தினரைப் பேணுவதற்குப் பொருள் தேவைப்பட்டதால் தன் கருங்கோட்டுச் சீறியாழைப் பயணம் வைத்து விருந்தளித்தான் என்கிறது புறநானூறு – இதன் விருந்து போற்றிய நிலை……….
(அ) நிலத்திற்கேற்ற விருந்து
(ஆ) இன்மையிலும் விருந்து
(இ) இரவிலும் விருந்து
(ஈ) உற்றாரின் விருந்து
Answer:
(ஆ) இன்மையிலும் விருந்து

Question 11.
முல்லைப்பாட்டை இயற்றியவர்………… ஆவார்.
(அ) பேயனார்
(ஆ) ஓதலாந்தையார்
(இ) நப்பூதனார்
(ஈ) நக்கண்ணையார்
Answer:
(இ) நப்பூதனார்

பாடலைப் படித்துப் பின்வரும் வினாக்களுக்கு (12, 13, 14, 15) விடை தருக.

இவைசரி யென்றால் இயம்புவதென் தொழில்
இவைதவ றாயின் எதிர்ப்பதென் வேலை!
ஆக்கல் அளித்தல் அழித்தல் இம் மூன்றும்
அவனும் யானுமே அறிந்தவை! அறிக!

Question 12.
இப்பாடலின் ஆசிரியர் யார்?
(அ) கண்ண தாசன்
(ஆ) பாரதிதாசன்
(இ) ஜெயகாந்தன்
(ஈ) பாரதியார்
Answer:
(அ) கண்ண தாசன்

Question 13.
முத்தொழில்கள் யாவை?
(அ) அறம், பொருள், இன்பம்
(ஆ) இயல், இசை, நாடகம்
(இ) ஆக்கல், அழித்தல், காத்தல்
(ஈ) ஆடல், பாடல், ஓடுதல்
Answer:
(இ) ஆக்கல், அழித்தல், காத்தல்

Question 14.
இப்பாடல் இடம் பெற்ற அடி எதுகைகளை எழுதுக.
(அ) இவை சரி, இவை தவறு
(ஆ) சரி, தவறு
(இ) இவை, இயம்பு
(ஈ) மூன்றும், ஆறும்
Answer:
(அ) இவை சரி, இவை தவறு

Question 15.
இப்பாடல் இடம் பெற்ற நூல் எது?
(அ) அன்னை மொழியே
(ஆ) காசிக்காண்டம்
(இ) முல்லைப்பாட்டு
(ஈ) காலக்கணிதம்
Answer:
(ஈ) காலக்கணிதம்

பகுதி – II (மதிப்பெண்கள்: 18)

பிரிவு – 1

எவையேனும் நான்கு வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்க.
21 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [4×2 = 8]

Question 16.
விடைக்கேற்ற வினா அமைக்க.
Answer:
(அ) “ஒரு மொழியில் உணர்த்தப்பட்டதை வேறொரு மொழியில் வெளியிடுவது மொழி பெயர்ப்பு ” என்கிறார் மணவை முஸ்தபா.
(ஆ) இறைவனுக்கு உகந்த கலையாகப் பரதநாட்டியம் கருதப்படுகிறது.

விடை: அ) மொழிபெயர்ப்பு பற்றி மணவை முஸ்தபா கருத்து யாது?
ஆ) பரத நாட்டியம் எவ்வகைக் கலையாகக் கருதப்படுகிறது?

Question 17.
மழைமேகம் எவ்வாறு பெருமழை பெய்கிறது?
Answer:
மழைமேகம் ஒலிக்கும் கடலின் குளிர்நீரைப் பருகிப் பெருந்தோற்றம் கொண்டு வலமாய் எழுந்து மலையைச் சூழ்ந்து வேகத்துடன் பெருமழை பொழிகிறது.

Question 18.
மருத்துவத்தில் மருந்துடன் அன்பும் நம்பிக்கையும் ஆற்றும் பாங்கினை எழுதுக.
Answer:

1. மருத்துவர் உடலில் ஏற்பட்ட புண்ணைக் கத்தியால் அறுத்துச் சுட்டாலும் அது நன்மைக்கே என்று உணர்ந்து நோயாளி அவரை நேசிப்பார்.
2. வித்துவக்கோட்டில் எழுந்தருளியிருக்கும் அன்னையே! அதுபோன்று நீ உனது நாக்கை வைகை மாதிரி வினாத்தாள் விளையாட்டால்.
3. நீங்காத துன்பத்தை எனக்குத் தந்தாலும் உன் அடியவனாகிய நான் உன் அருளையே எப்பொழுதும் எதிர்பார்த்து வாழ்கின்றேன்.

Question 19.
அமிலமழை குறிப்பு வரைக.
Answer:
கந்தக டை ஆக்சைடு, நைட்ரஜன் டை ஆக்சைடு ஆகியவை மழை பெய்யும் போது நீரில் கரைந்துவிடுவதால் அமிலமழை பெய்கிறது.

Question 20.
தானியம் ஏதும் இல்லாத நிலையில் விதைக்காக வைத்திருந்த தினையை உரலில் இட்டுக் குற்றியெடுத்து விருந்தினருக்கு விருந்தளித்தாள் தலைவி’ என்பது இலக்கியச் செய்தி விருந்தோம்பலுக்குச் செல்வம் மட்டுமே இன்றியமையாத ஒன்றா? உங்கள் கருத்தைக் குறிப்பிடுக.
Answer:
விருந்தளிக்க செல்வம் மட்டும் இருந்தால் போதாது. விருந்து கொடுக்க வேண்டும் என்ற எண்ணமும் வேண்டும். பணம் இல்லை என்றாலும், எண்ணம் இருந்தால் தன்னிடம் இல்லை என்றாலும், பிறரிடம் கடன் பெற்றாவது விருந்து அளிக்கப்படும். அதற்குப் பணம் முக்கியம் இல்லை. எண்ணம் தான் முக்கியம்.

Question 21.
குன்றேறி ‘ எனத் தொடங்கும் திருக்குறளை எழுதுக.
Answer:
குன்றேறி யானைப்போர் கண்டற்றால் தன்கைத்தொன் றுண்டாகச் செய்வான் வினை.

பிரிவு – 2

எவையேனும் ஐந்து வினாக்களுக்கு மட்டும் குறுகிய விடையளிக்கவும். [5×2 = 10]

Question 22.
வினைமுற்றை வினையாலணையும் பெயராக மாற்றுக. பொது அறிவு நூல்களைத் தேடிப் படித்தார். போட்டித் தேர்வில் வென்றார்.
Answer:
பொது அறிவு நூல்களைத் தேடிப் படித்தவர் போட்டித் தேர்வில் வென்றார்.

Question 23.
தொடரில் விடுபட்ட வண்ணங்களை உங்கள் எண்ணங்களால் நிரப்புக. (வெள்ளை , பச்சை , கருக்கத், சிவந்தது)
Answer:

• வானம் கருக்கத் தொடங்குகிறது.
• அனைவரின் பாராட்டுகளால், வெட்கத்தில் பாடகர் முகம் சிவந்தது.
• வெள்ளை மனம் உள்ளவரை அப்பாவி என்கிறோம்.
• கண்ணுக்குக் குளுமையாக இருக்கும் பச்சை புல்வெளிகள்

Question 24.
இருசொற்களைப் பயன்படுத்தி ஒரு சொல் அமைக்க இயற்கை – செயற்கை
Answer:
பாதை தெரியாத இயற்கைக் காடுகளில் பயணிக்கச் செயற்கைக் கருவிகள் யன்படுகின்றன.

Question 25.
கலைச்சொற்கள் தருக.
Answer:

• Cosmic rays – விண்வெளிக் கதிர்கள்
• Bio technology – உயிரித் தொழில்நுட்பம்

Question 26.
பார் – வேர்ச்சொல்லைக் கொண்டு எழுவாய்த் தொடர், வினையெச்சத் தொடர் ஆகியவை எழுதுக.
Answer:

Question 27.
நிறுத்தக் குறிகளை இடுக. மென்பொருள்கள் கவிதைகள் கதைகள் விதவிதமான எழுத்து நடைகள் போன்றவற்றைக் கற்றுக் கொண்டு மனிதர்களுடன் போட்டியிட்டாலும் வியப்பதற்கில்லை.
Answer:
மென்பொருள்கள் கவிதைகள், கதைகள், விதவிதமான எழுத்து நடைகள் போன்றவற்றைக் கற்றுக் கொண்டு மனிதர்களுடன் போட்டியிட்டாலும் வியப்பதற்கில்லை.

Question 28.
இகழ்தல், நண்பா – இலக்கணக் குறிப்பு தருக.
Answer:
இகழ்தல் – தொழிற்பெயர்
நண்பா – விளித்தொடர்

பகுதி – III (மதிப்பெண்க ள்: 18)

பிரிவு -1

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [2 x 3 = 6]

Question 29.
மனிதர்களின் மூளையைப் போன்றது. செயற்கை நுண்ணறிவு கொண்ட கணினியின் மென்பொருள். மனிதனைப் போலவே பேச, எழுத, சிந்திக்க இத்தொழில்நுட்பம் மேம்படுத்தப்படுகிறது. இதனால் மனிதகுலத்துக்கு ஏற்படுகிற நன்மைகளைப் பற்றி அறிவியல் இதழ் ஒன்றுக்கு எதிர்காலத் தொழில்நுட்பம் என்ற தலைப்பில் எழுதுக.
Answer:
வேலை வாய்ப்புகளில் கணிசமான மாற்றங்களைச் செயற்கை நுண்ணறிவு கொண்டுவரப்போகிறது. எதிர்காலத்தில் ரோபோ’ விடம் குழந்தையை ஒப்படைத்துவிட்டு நிம்மதியாக அலுவலகம் செல்லும் பெற்றோர்களை நாம் பார்க்கப்போகிறோம். வயதானவர்களுக்கு உதவிகள் செய்தும், அவர்களுக்கு உற்ற தோழனாய்ப் பேச்சுக் கொடுத்தும் பேணும் ரோபோக்களை நாம் பார்க்கப்போகிறோம்! செயற்கை நுண்ணறிவுள்ள ரோபோக்களால், மனிதன் செய்ய இயலாத, அலுப்புத் தட்டக்கூடிய, கடினமான செயல்களைச் செய்யமுடியும்; மனித முயற்சியில் உயிராபத்தை விளைவிக்கக் கூடிய செயல்களைச் செய்யமுடியும்!

புதிய வணிக வாய்ப்புகளைச் செயற்கை நுண்ணறிவு நல்குகிறது. பெருநிறுவனங்கள் தங்கள் பொருள்களை உற்பத்தி செய்யவும் சந்தைப்படுத்தவும் செயற்கை நுண்ணறிவைப் பயன்படுத்துகின்றன. விடுதிகளில், வங்கிகளில், அலுவலகங்களில் தற்போது மனிதர் அளிக்கும் சேவைகளை ரோபோக்கள் அளிக்கும். மேலும், நம்முடன் உரையாடுவது, ஆலோசனை வழங்குவது, பயண ஏற்பாடு செய்து தருவது, தண்ணீர் கொண்டு வந்து தருவது, உடன் வந்திருக்கும் குழந்தைகளுக்கு வேடிக்கை காட்டுவது எனப் பலவற்றைச் செய்யும்.

எதிர்காலத்தில் நாம் பயணிக்கும் ஊர்திகளைச் செயற்கை நுண்ணறிவைக் கொண்டு இயக்கவேண்டியிருக்கும். இத்தகைய ஊர்திகள் ஏற்படுத்தும் விபத்துகள் குறையும், போக்குவரத்து நெரிசல் இருக்காது. அதன்மூலம் பயண நேரம் குறையும், எரிபொருள் மிச்சப்படும். இத்தகைய மென்பொருள்கள் கவிதைகள், கதைகள், விதவிதமான எழுத்து நடைகள் போன்றவற்றைக் கற்றுக்கொண்டு மனிதர்களுடன் போட்டியிட்டாலும் வியப்பதற்கில்லை! கல்வித் துறையில் இத்தொழில் நுட்பத்தைப் பலவிதங்களில் பயன்படுத்தும் சாத்தியக்கூறுகள் இருக்கின்றன.

Question 30.
ஜெயகாந்தன் சிறுகதை மன்னன்’ என எவ்வாறு பெயர் பெற்றார்?
Answer:

• ஜெயகாந்தன் பேசி, “எதற்காக எழுதுகிறேன்?” என்ற தலைப்பில் கட்டுரையாகத் தொகுக்கப்பட்ட பகுதியும், ‘யுகசந்தி’ என்ற தொகுப்பில் இடம்பெற்றுள்ள ‘தர்க்கத்திற்கு அப்பால்’ என்னும் சிறுகதையும் பாடப்பகுதியில் இடம்பெற்றுள்ளன.
• தான் வாழ்ந்த காலத்தில் சிக்கல்கள் பலவற்றை ஆராய், எடுத்துச் சொல்ல, தன் பார்வைக்கு உட்பட்ட தீர்ப்பைச் சொல்ல அவர் மேற்கொண்ட நடவடிக்கையே படைப்பு.
• அவருடைய படைப்புகள் உணர்ச்சி சார்ந்த எதிர்வினைகளாக இருக்கின்றன. இதுவே அவருக்குச் சிறுகதை மன்னன் என்ற பட்டத்தைத் தேடித்தந்தது.
• இவர் குறும்புதினங்களையும், புதினங்களையும், கட்டுரைகளையும், கவிதைகளையும் படைத்துள்ளார். தன் கதைகளைத் திரைப்படமாக இயக்கியிருக்கிறார்.
• தலைசிறந்த உரத்த சிந்தனைப் பேச்சாளராகவும் திகழ்ந்தார்; சாகித்திய அகாதெமி விருதையும். ஞானபீட விருதையும் பெற்ற இவருடைய கதைகள் பிறமொழிகளில் மொழிபெயர்க்கப்பட்டுள்ளன.

Question 31.
உரைப்பத்தியைப் படித்து வினாக்களுக்கு விடை தருக.
Answer:
கொடையின் சிறப்பால் வள்ளல் எழுவர் போற்றப்படுவது, பழந்தமிழர் கொடை மாட்சியைப் புலப்படுத்துகிறது. எழுவரின் கொடைப் பெருமை சிறுபாணாற்றுப்படையிலும், பெருஞ்சித்திரனார் பாடலிலும் பதிவு செய்யப்பட்டிருப்பது குறிப்பிடத்தக்கது. ஆற்றுப்படை இலக்கியங்கள், கொடை இலக்கியங்களாகவே உள்ளன. பதிற்றுப்பத்து சேர அரசர்களின் கொடைப் பதிவாகவே உள்ளது.

(அ) வள்ளல்களின் எண்ணிக்கை யாது?
கடையேழு வள்ளல்கள்

(ஆ) எழுவரின் கொடைச் சிறப்பைக் கூறும் நூல்கள் எவை?
Answer:

1. சிறுபாணாற்றுப்படை
2. பெருஞ்சித்திரனார் பாடல்

(இ) பதிற்றுப்பத்து எந்த அரசரைப் பற்றிக் கூறுகிறது?
Answer:
சேர அரசர்

பிரிவு – 2

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க.
34 ஆவது வினாவிற்குக் கட்டாயமாக விடையளிக்க வேண்டும். [2 x 3 = 6]

Question 32.
வைத்தியநாதபுரி முருகன் குழந்தையாக அணிந்திருக்கும் அணிகலன்களுடன் செங்கீரை ஆடிய நயத்தை விளக்குக.
Answer:

• திருவடியில் அணிந்த சிறு செம்பொன்னால் ஆன கிண்கிணிகளோடு சிலம்புகளும் சேர்ந்து ஆடட்டும்.
• இடையில் அரைஞாண் மணியோடு ஒளிவீசுகின்ற அரைவடங்கள் ஆடட்டும். பசும்பொன் என ஒளிரும் தொந்தியுடன் சிறுவயிறு சரிந்தாடட்டும்.
• பட்டம் கட்டிய நெற்றியில் விளங்குகின்ற பொட்டின் வட்டி வடிவான சுட்டி பதிந்தாடட்டும். கம்பிகளால் உருவான குண்டலங்களும் காதின் குழைகளும் அசைந்தாடட்டும்.
• உச்சிக் கொண்டையும் அதில் சுற்றிக் கட்டப்பட்டுள்ள ஒளியுள்ள முத்துக்களோடு ஆடட்டும்.
• தொன்மையான வைத்தியநாதபுரியில் எழுந்தருளிய முருகனே! செங்கீரை ஆடி அருள்க!
• இவற்றுடன் அழகிய பவளம் போன்ற திருமேனியும் ஆட, செங்கீரை ஆடுக.

Question 33.
தமிழழகனார் தமிழையும் கடலையும் இரட்டுறமொழியும் பாங்கினை விளக்குக.
Answer:
தமிழ் :

• தமிழ், இயல், இசை, நாடகம் என முத்தமிழாய் வளர்ந்தது; முதல் இடை கடை ஆகிய முச்சங்கங்களால் வளர்க்கப்பட்டது.
• ஐம்பெருங்காப்பியங்களை அணிகலன்களாகப் பெற்றது. சங்கப் பலகையில் அமர்ந்திருந்த சங்கப்புலவர்களால் காக்கப்பட்டது.

கடல் :

• கடல், முத்தினையும் அமிழ்தினையும் தருகிறது.
• வெண்சங்கு, சலஞ்சலம், பாஞ்சசன்யம் ஆகிய மூன்று வகையான சங்குகளைத் தருகிறது. மிகுதியான வணிகக் கப்பல்கள் செல்லும்படி இருக்கிறது.
• தன் அலையால் சங்கினைத் தடுத்து நிறுத்திக் காக்கிறது.

Question 34.
அடிபிறழாமல் எழுதுக.
(அ) “சிறுதாம்பு ” எனத் தொடங்கும் முல்லைப்பாட்டு பாடல்.
Answer:
சிறுதாம்பு தொடுத்த பசலைக் கன்றின்
உறுதுயர் அலமரல் நோக்கி, ஆய்மகள்
நடுங்கு சுவல் அசைத்த கையள், ” கைய
கொடுங்கோற் கோவலர் பின்நின்று உய்த்தர
இன்னே வருகுவர், தாயர்” என்போள்
நன்னர் நன்மொழி கேட்டனம் – நப்பூதனார்

(அல்லது)

(ஆ) “வாளால் அறுத்து” எனத் தொடங்கும் பெருமாள் திருமொழி பாடல்.
Answer:
வாளால் அறுத்துச் சுடினும் மருத்துவன்பால்
மாளாத காதல் நோயாளன் போல் மாயத்தால்
மீளாத் துயர்தரினும் வித்துவக் கோட்டம்மா! நீ
ஆளா உனதருளே பார்ப்பன் அடியேனே. – குலசேகராழ்வார்

பிரிவு – 3

எவையேனும் இரண்டு வினாக்களுக்கு மட்டும் சுருக்கமாக விடையளிக்க. [2 x 3 = 6]

Question 35.
ஆசிரியப்பாவின் பொது இலக்கணத்தை எழுதுக.
Answer:

• அகவல் ஓசை பெற்று வரும்.
• ஈரசைச் சீர் மிகுதியாகவும், காய்ச்சீர் குறைவாகவும் பயின்று வரும்.
• ஆசிரியத் தளை மிகுதியாகவும் வெண்டளை, கலித்தளை ஆகியவை விரவியும் வரும்.
• மூன்று அடி முதல் எழுதுபவர் மனநிலைக்கேற்க அமையும்.
• ஏகாரத்தில் முடித்தல் சிறப்பு.

Question 36.
‘ஐயத்தின் நீங்கித் தெளிந்தார்க்கு வையத்தின் வானம் நணியது உடைத்து’ – இக்குறட்பாவினை அலகிட்டு வாய்பாடு தருக.
Answer:

Question 37.
நிரல்நிறைப் பொருள்கோள் விவரி.
Answer:
ஒரு செய்யுளில் சொற்கள் முறை பிறழாமல் நிரல்நிறையாக (வரிசையாக அமைந்து வருவது ‘நிரல்நிறைப் பொருள்கோள் ‘ ஆகும். இது முறை நிரல்நிறைப் பொருள்கோள், எதிர் நிரல்நிறைப் பொருள்கோள் என இருவகைப்படும்.

(அ) முறை நிரல்நிறைப் பொருள்கோள்:
செய்யுளில் எழுவாயாக அமையும் பெயர்ச்சொற்களை அல்லது வினைச்சொற்களை வரிசையாக நிறுத்தி, அவை ஏற்கும் பயனிலைகளையும் அவ்வரிசைப்படியே நிறுத்திப் பொருள் கொள்ளுதல் முறை நிரல்நிறைப் பொருள்கோள் ‘ ஆகும்.

(எ.கா.) அன்பும் அறனும் உடைத்தாயின் இல்வாழ்க்கை
பண்பும் பயனும் அது’

இக்குறளில் பண்பு பயன் என்ற இரு சொற்களை வரிசைப்படுத்தி, அவற்றிற்குரிய விளைவுகளாக அன்பு, அறன் என்று வரிசைப்படுத்தி உள்ளார். அவற்றை இல்வாழ்க்கையின் பண்பு, அன்பு என்றும் அதன் பயன், அறன் என்றும் பொருள் கொள்ள வேண்டும். எனவே, அன்புக்குப் பண்பும் அறத்துக்குப் பயனும் பயனிலைகளாக – நிரல்நிறையாக – நிறுத்திப் பொருள் கொள்வதால், இப்பாடல் முறை நிரல்நிறைப் பொருள்கோள்’ எனப்படும்.

(ஆ) எதிர் நிரல்நிறைப் பொருள்கோள்:
செய்யுளில் எழுவாய்களை வரிசைப்படுத்தி அவை ஏற்கும் பயனிலைகளை எதிர் எதிராகக் கொண்டு பொருள் கொள்ளுதல் எதிர் நிரல்நிறைப் பொருள்கோள்’ ஆகும்.

(எ.கா) விலங்கொடு மக்கள் அனையர் இலங்குநூல்
கற்றாரோடு ஏனையவர்’

இக்குறளில் முதல் அடியில் விலங்கு, மக்கள் என்று எழுவாய்களை வரிசைப்படுத்திவிட்டு, அடுத்த அடியில் பயனிலைகளாகக் கற்றார், கல்லாதார் (ஏனையவர்) என வரிசைப்படுத்தி உள்ளார். அவற்றைக் கற்றார் மக்கள் என்றும், கல்லாத ஏனையவர் விலங்குகள் என்றும் எதிர் எதிராகக் கொண்டு பொருள் கொள்ள வேண்டும். எனவே, இக்குறள் ‘எதிர் நிரல்நிறைப் பொருள்கோள் ‘ ஆகும்.

பகுதி – IV (மதிப்பெண்கள்: 25)

அனைத்து வினாக்களுக்கும் விடையளிக்க. [5 x 5 = 25]

Question 38.
(அ) ஆற்றுப்படுத்தல் என்பது அன்றைக்குப் புலவர்களையும் கலைஞர்களையும் வள்ளல்களை நோக்கி நெறிப்படுத்துவதாக இருந்தது. அது இன்றைய நிலையில் ஒரு வழிகாட்டுதலாக மாறியிருப்பதை விளக்குக.
Answer:

• ஆற்றுப்படுத்துதல் என்பது வள்ளலை நாடி எதிர்வருபவர்களை அழைத்து யாம் இவ்விடத்தைச் சென்று இன்னவெல்லாம் பெற்று வருகின்றோம்.
• நீயும் அந்த வள்ளலிடம் சென்று வளம்பெற்று வாழ்வாயாக என்று கூறுதல் ஆற்றுப்படை ஆகும்.
• ஆற்றுப்படுத்துதல் என்பது இன்றைய நிலையில் ஒரு வழிகாட்டுதலாக இருக்கிறது.
• தன்னிடம் இல்லை என்றோ அல்லது தெரியாது என்றோ யார் வந்தாலும் அவர்களுக்கு – வழிகாட்டுதலாகவும் இருக்கிறது.
• அவர்களுக்கு அறிவுரை கூறி அவர்களை வழிகாட்டுகின்றனர். அன்றைய ஆற்றுப்படுத்துதல் இன்றைய வழிகாட்டுதலாக மாறியுள்ளது.
• இது ஒவ்வொரு நிலையிலும் மாற்றம் அடைந்துள்ளது. உதவி தேவைப்படுபவர்களுக்கு பெரும் உதவியாக இருந்து வருகிறது. இதுவே இன்றைய ஆற்றுப்படுத்துதல் ஆகும்.

(அல்ல து)

(ஆ) இறைவன், புலவர் இடைக்காடன் குரலுக்குச் செவிசாய்த்த நிகழ்வை நயத்துடன் எழுதுக.
Answer:
இடைக்காடனார் இறைவனை வணங்குதல் :

• இடைக்காடனார் இறைவன் திருமுன் விழுந்து வணங்கி எழுந்து, “தமிழறியும் பெருமானே!
• அடியார்க்கு நல்நிதி போன்றவனே ! திருஆலவாயிலில் உறையும் இறைவனே ! அழகிய வேப்பமலர் மாலையை அணிந்த பாண்டியன்.
• பொருட்செல்வத்தோடு கல்விச் செல்வமும் மிக உடையவன் எனக் கூறக் கேட்டு.
• அவன் முன் சொற்சுவை நிரம்பிய கவிதை பாடினேன். அவனோ சிறிதேனும் சுவைத்துத் தலை அசைக்காமல் புலமையை அவமதித்தான்” என்றார்.

இடைக்காடனாரின் சினம்:

• இடைக்காடனார் இறைவனிடம், “பாண்டியன் என்னை இகழவில்லை, சொல்லின் வடிவாக உன் இடப்புறம் வீற்றிருக்கும் பார்வதி தேவியையும்.
• சொல்லின் பொருளாக விளங்கும் உன்னையுமே அவமதித்தான்” என்று சினத்துடன் கூறிச் சென்றார்.
• அவரது சொல் வேற்படைபோல் இறைவனின் திருச்செவியின் சென்று தைத்தது.

இறைவன் இலிங்க வடிவை மறைத்தல்:

1. கோவிலை விட்டு வெளியேறிய இடைக்காடனாருக்கும் அவர் நண்பராகிய கபிலருக்கும் மனமகிழ்ச்சி உண்டாக்க நினைத்தார். இறைவன் ஞானமயமாகிய தம்முடைய இலிங்க வடிவத்தை மறைத்து உமாதேவியாரோடும் திருக்கோவிலைவிட்டு வெளியேறி நேர் வடக்கே வையை ஆற்றின் தென் பக்கத்தே ஒரு திருக்கோவிலை ஆக்கி அங்கு சென்று இருந்தார். பாண்டிய மன்னனின் வேண்டுதல்:

2. இறைவனே, என்னால், என் படைகளால், என் பகைவரால், கள்வரால், காட்டில் உள்ள விலங்குகளால் இத்தமிழ்நாட்டில் தங்களுக்கு இடையூறு ஏற்பட்டதா? மறையவர் நல் ஒழுக்கத்தில் குறைந்தனரோ? தவமும் தருமமும் சுருங்கியதோ? இல்லறமும் துறவறமும் தத்தம் நெறியில் இருந்து தவறினவோ?

எமது தந்தையே யான் அறியேன்” என்று வேண்டினான் பாண்டிய மன்னன். இறைவன் மன்னனிடம், “சிறந்த குளிர்ந்த வயல்கள் சூழ்ந்த கடம்பவனத்தை வி ஒருபோதும் நீங்கமாட்டோம். இடைக்காடனார் பாடலை இகழ்ந்த குற்றம் தவிர வேறு குற்றம் உன்னிடம் இல்லை.

• இடைக்காடனார் மீது கொண்ட அன்பினால் இவ்வாறு இங்கு வந்தோம்” என்றார். தமாகா லைலாகாத

Question 39.
(அ) உனது பகுதியில் நூலகம் அமைத்துத் தர வேண்டி விண்ணப்பம் ஒன்று வரைக.
Answer:

சென்னை,
29.10.2019

அனுப்புநர்
ச.தருண்,
15, காந்தி தெரு,
சென்னை – 600088.

பெறுநர்
மாநகராட்சி அலுவலர்
மாநகராட்சி அலுவலகம்,
சென்னை – 600 088.

மதிப்பிற்குரிய ஐயா,

பொருள்: குடியிருக்கும் பகுதியில் நூலக வசதி வேண்டுதல்’.

வணக்கம். நான் மேற்கண்ட முகவரியில் சில ஆண்டுகளாக வசித்து வருகின்றேன். இன்றைய உலகில் பலவகையான புத்தகங்களைப் படித்து அறிவை வளர்க்க வேண்டும். இங்கு நூலகம் இல்லாததால் மாணவர்கள் மிகவும் சிரமப்படுகிறார்கள். வேறு நூலகத்திற்குச் சென்று வருவதற்குள் காலதாமதம் ஏற்படுகின்றது. இதனால் மாணவர்கள் வீட்டில் சரியாகப் படிக்க முடியவில்லை . மதிப்பெண் எடுக்க முடியவில்லை . கல்லூரி மாணவர்களும் அவதிப்படுகிறார்கள். எனவே, எங்கள் பகுதிக்கு ஒரு நூலகம் அமைத்துத் தர வேண்டி மிகப் பணிவுடன் கேட்டுக்கொள்கிறேன்.

நன்றி.

இப்படிக்கு,
தங்கள் உண்மையுள்ள,
(ச. தருண்

இடம் : சென்னை
நாள்: 29.10.2019
உறைமேல் முகவரி

பெறுநர்
மாநகராட்சி அலுவலர்,
மாநகராட்சி அலுவலகம்,
சென்னை – 600 088.

அல்லது)

(ஆ) பள்ளித்திடலில் கிடைத்த பணப்பையை உரியவரிடம் ஒப்படைத்ததையும் அதற்குப் பாராட்டுப் பெற்றதையும் பற்றி வெளியூரில் இருக்கும் உறவினர் ஒருவருக்குக் கடிதம் எழுதுக.

4.4.2019
காஞ்சிபுரம்.

– அன்புள்ள மாமா,
நான் இங்கு நலம் அது போல் வீட்டில் உள்ள அனைவரும் நலமாக இருக்கிறார்கள். உங்கள் வீட்டில் அனைவரும் நலம் என நம்புகிறேன். நடந்து முடிந்த தேர்வுகளில் நான் நல்ல மதிப்பெண் பெற்றுள்ளேன். கடந்த வாரம் எங்கள் பள்ளியில் ஆண்டு விழா நடைபெற்றது. அதில் நான் நடனம் ஒன்றில் பங்கு கொண்டேன். ஆண்டுவிழாவிற்கு வந்திருந்த ஒருவர் பணப்பையைத் தவற விட்டு விட்டார். அதை நான் எடுத்துத் தலைமையாசிரியரிடம் ஒப்படைத்தேன்.

தலைமையாசிரியர் ஒலிபெருக்கியில் சொல்லி அந்தப் பணப்பையை உரியவரிடம் ஒப்படைத்தார் அந்த பணப்பைக்குச் சொந்தக்காரர் என்னை மேடைக்கு அழைத்து பொன்னாடை அணிவித்துப் பாராட்டினார், சிறு உதவி செய்ததற்குப் பெரிய அளவில் மரியாதை செய்தார். என் தம்பி தங்கைகளை அழைத்துக் கொண்டு காஞ்சிபுரம் வர வேண்டும் என்றார். இந்த மகிழ்ச்சியான நிகழ்வை உங்களுடன் பகிர்ந்துகொள்ள விரும்பினேன். பகிர்ந்து கொண்டேன். சுபம் நிறைக.

இங்ஙனம்,
தங்கள் அன்பு,
யாழினி.

உறைமேல் முகவரி
பெறுநர் மங்களம் நகர்,
சூளைமேடு,
சென்னை – 600 004.

Question 40.
படம் உணர்த்தும் கருத்தை நயமுற ஐந்து வரிகளில் எழுதுக.
Answer:

விருப்புடன் செய்க ஈகை
வெறுப்பு வேண்டாம் தம்பி
ஒரு பிறவியில் செய்க நன்மை
ஏழு பிறவியிலும் தவிர்க்க தீமை!
வாழ்த்தும் வயது இருப்பின் வாழ்த்தட்டும்
இல்லையெனில்
வணங்கட்டும் உங்களின் செய்கையை!

Question 41.
வங்கிக் கணக்கிலிருந்து பணம் எடுக்கும் படிவம் நிரப்புக.
Answer:

Question 42.
(அ) தொலைக்காட்சி நிகழ்வுகளையே பார்த்துக்கொண்டிருக்கும் தம்பி; திறன் பேசியிலேயே விளையாடிக்கொண்டிருக்கும் தங்கை, காணொலி விளையாட்டுகளில் மூழ்கியிருக்கும் தோழன், எப்போதும் சமூக ஊடகங்களில் இயங்கியபடி இருக்கும் தோழி. இவர்கள் எந்நேரமும் நடப்புலகில் இருக்காமல் கற்பனை உலகில் மிதப்பவர்களாக இருக்கிறார்கள் ! இவர்களை நெறிப்படுத்தி நடைமுறை உலகில் செயல்படவைக்க நீங்கள் செய்யும் முயற்சிகளைப் பட்டியல் இடுக.
Answer:
J

(அல்ல து)

(ஆ) மொழிபெயர்க்க.
Answer:
Malar : Devi, switch off the lights when you leave the room.
Devi : Yeah. We have to save electricity…
Malar : Our nation spends a lot of electricity for lighting up our streets in the night.
Devi : Who knows? In future, our country may launch artificial moons to light our night time sky!
Malar : I have read some other countries are going to launch these types of illumination satellites near future.
Devi : Superb news! If we launch artificial moons, they can assist in disaster relief by beaming light on areas that lost power!

விடை
மலர் : தேவி, அறையை விட்டு வெளி செல்லும் போது மின்விளக்குகளை அணைத்துவிட்டு செல்ல வேண்டும்.
தேவி : ஆமாம்! நாம் மின்சாரத்தைச் சேமிக்க வேண்டும்.
மலர் : இரவு நேரங்களில் மட்டும் தெருவிளக்குகளில் நம்நாட்டில் அதிக அளவு மின்சாரம் பயன்படுகிறது.
தேவி : யாருக்குத் தெரியும்? இனி வரும் காலங்களில் நமது நாட்டில் செயற்கை நிலாக்களை உருவாக்கி இரவு நேரங்களில் ஆகாயத்தில் வெளிச்சங்களை உருவாக்கலாமோ!
மலர் : நாடுகளில் வெளிச்சத்தை உண்டாக்கும் ஒரு சில செயற்கை கோள்கள் அமைத்திருப்பதாக நான் படித்திருக்கிறேன்.
தேவி : சிறப்பான செய்தி ! பேரிடர் காலங்களில் இந்த செயற்கை நிலாக்களைக் கொண்டு மின்சாரம் துண்டிக்கப்பட்ட இடங்களில் இதையே பயன்படுத்தலாம்.

பகுதி – V(மதிப்பெண்கள் : 24)

அனைத்து வினாக்களுக்கும் விரிவாக விடையளிக்க. [3 x 8 = 24]

Question 43.
(அ) மொழிப் பெயர்ப்பின் பயன்கள் யாவை?
Answer:
மொழிபெயர்ப்பின் துணை:

• இது மொழிபெயர்ப்புக் காலம், காலையில் எழுந்தவுடன் நாளிதழ்ப்படிப்பு, மொழிபெயர்ப்பு மூலமே நமக்குச் சாத்தியமாகிறது.
• இரவு தொலைக்காட்சியில் காணும், கேட்கும் செய்திகளும் மொழிபெயர்ப்பு மூலமே கிடைக்கின்றன.
• நம் பணிகள் பலவற்றிலும் மொழிபெயர்ப்பின் துணை இருந்து கொண்டே இருக்கிறது.

மொழிகளின் பங்கு :

1. இன்றைய வளரும் நாடுகளில் அறிவியலை உருவாக்க – அரசியலை உருவாக்க – பொருளியலை உருவாக்க – சமூகவியலை உருவாக்க – இலக்கியத்தை உருவாக்க மொழிபெயர்ப்பே உதவுகிறது.

2. மொழிபெயர்ப்பு, மனிதர்களையும், நாடுகளையும், காலங்களையும் இணைக்கிற நெடுஞ்சாலையாக இருக்கிறது. காலத்தால் இடத்தால் மொழியால் பிரிக்கப்பட்ட மானுடத்தை இணைக்கிறது;

3. கடந்த காலத்தை எதிர்காலத்துடன் இணைக்கும் அது மனித வாழ்வின் ஒரு பகுதியாகவே இருக்கிறது.

4. பல மொழிகளிலும் காணப்படும் சிறப்புக்கூறுகளை எல்லாம் ஒருங்கு சேர்த்து அனைவருக்கும் பொதுமையாக்குகிறது.

5. ஜெர்மனியில் ஓர் ஆண்டில் பிற மொழிகளிலிருந்து 5000 நூல்கள் வரை மொழிபெயர்க்கப்படுகின்றன.

6. புள்ளிவிவரப்படி அதிகமான தமிழ் நூல்கள் பிறமொழிகளில் மொழிபெயர்க்கப்பட்டுள்ளன.

7. அவ்வரிசையில் முதலிடம் ஆங்கிலம், இரண்டாமிடம் மலையாளம், அதைத் தொடர்ந்து அடுத்தடுத்து நிலைகளில் முறையே தெலுங்கு, இந்தி, கன்னடம், வடமொழி, ரஷ்யமொழி, வங்கமொழி, மராத்தி மொழி போன்றவை இடம்பெறுகின்றன.

அறிவியல் கண்டுபிடிப்புக்களும், இலக்கியப் படைப்புகளும் :

• மொழிபெயர்ப்பினால் புதிய சொற்கள் உருவாகி மொழிவளம் ஏற்படுகிறது.
• பிற இனத்தவரின் பண்பாடு, நாகரிகம், பழக்கவழக்கம் போன்றவற்றை அறியமுடிகிறது.
• அதிலிருந்து நல்லனவற்றை நாம் பெற்றுக்கொள்ள முடிகிறது. பிறமொழி இலக்கிய அறிவு கிடைக்கிறது. அதன்மூலம் நம் இலக்கியத்தை வளப்படுத்த முடிகிறது.
• உலகப்புகழ் பெற்ற அறிவியல் கண்டுபிடிப்புகளையும் இலக்கியப் படைப்புகளையும் அறிவதற்கு வாய்ப்பு ஏற்படுகிறது.
• கருத்துப் பகிர்வைத் தருவதால் மொழிபெயர்ப்பைப் பயன்கலை என்று குறிப்பிடுவார்கள்.
• மொழிபெயர்ப்பு மூலம் ஒரு நாட்டின் வரலாற்றிலும், இலக்கியத்திலும், பண்பாட்டிலும் வலிமையான தாக்கத்தை ஏற்படுத்த முடியும்.

(அல்ல து)

(ஆ) நாட்டு விழாக்கள் – விடுதலைப் போராட்ட வரலாறு – நாட்டின் முன்னேற்றத்தில் மாணவர் பங்கு – குறிப்புகளைக் கொண்டு ஒரு பக்க அளவில் மாணவப் பருவமும் நாட்டுப் பற்றும்’ என்ற தலைப்பில் மேடை உரை எழுதுக.
மாணவப் பருவமும் நாட்டுப் பற்றும்

முன்னுரை:
இந்தியா நிலப்பரப்பில் உலகில் ஏழாவது பெரிய நாடு மக்கள் தொகையில் இரண்டாவது பெரிய நாடு, இந்தியாவின் முதல் தலைமை அமைச்சர் ரோஜாவின் ராஜா, சமாதானப்புறா ஜவஹர்லால் நேரு இந்தியாவைப் பொருளாதாரத்தில் முன்னேற்ற வேண்டும் என்று ஐந்தாண்டு திட்டங்களைத் தீட்டினார். வளர்ச்சிப் பாதையில் சென்றுக் கொண்டிருக்கிற இந்தியாவில் இன்றைய மாணவர்கள் நாளைய மன்னர்கள், ஆகையால் மாணவர்களுக்குச் சமுதாயத் தொண்டு ஆற்ற வேண்டிய கடமைகள் உள்ளன.

விடுதலைக்கு முன்:
மாண்பு என்றால் மாட்சி, பெருமை, சிறப்பு என்று பொருள் அதனால் மாணாக்கர் என்று அழைக்கப்பட்ட சொல். மாணவர் என்று மருவிவிட்டது. இந்தியாவின் விடுதலைப் போராட்டத்திற்கு மகாத்மா காந்தி அடிகள் தலைமையேற்றார். ஒத்துழையாமை இயக்கத்தை அறிவித்தார். மாணவர்கள் கல்லூரியை விட்டு விலகினர். போராட்டங்களில் ஈடுபட்டனர். சிறை சென்றனர். இறுதியில் விடுதலை கிடைத்தது. உலகெங்கும் மாணவர்கள் கலந்து கொண்டு நடத்திய அறப்போராட்டங்கள் தோற்றதாக
வரலாறு இல்லை.

நாட்டின் முன்னேற்றம்:
இந்தியப் பேரரசு எத்துணை திட்டங்களைத் தீட்டிப் பொருள் உற்பத்தி செய்தாலும் வளர்ச்சிப் பாதையில் சென்றாலும் சாதிமத வேற்றுமை, மக்கள் பெருக்கம், அறியாமை, தீண்டாமை, பதுக்கல், கடத்தல் ஆகியவைகள் வளர்ச்சிக்குத் தடைக்கற்களாக உள்ளன. இத்தகைய விலங்குகளால் மக்கள் சிறைப்படுத்தப்பட்டுள்ளனர்.

சமுதாயத் தொண்டில் மாணவனின் பங்கு :
சாலைப் போக்குவரத்தைச் சீர் செய்ய காவலர்களுடன் சேர்ந்து பணியாற்ற வேண்டும். சாலையில் விபத்துகளினால் பாதிக்கப்பட்டவரை உடனே மருத்துவமனையில் கொண்டு போய்ச் சேர்க்க வேண்டும். உடல் ஊனமுற்றவர்களுக்கு மாணவர்கள் வழிகாட்டிகளாக விளங்கலாம்.

சாலையில் கிடக்கும் கண்ணாடித் துண்டுகள், ஆணிகள், வாழைப்பழத் தோல்கள், புகைந்து கொண்டிருக்கும் வெண்சுருட்டுத் துண்டுகள் போன்றவற்றை மாணவர்கள் வெட்கம் பாராமல் எடுத்துச் சென்று மக்கள் நடமாட்டம் இல்லாத இடத்தில் போட வேண்டும். மாணவர்கள் கண்மூடி வழக்கமான மூடப்பழக்க வழக்கங்களை மண்மூடிப் போகச் செய்தல் வேண்டும்.

முடிவுரை:
”ஒல்லும் வகையான் அறவினை ஓவாதே
செல்லும் வாயெல்லாம் செயல்”

என்றார் வள்ளுவர் தன்னால் இயலும் தொண்டுகளை எங்கெங்கு செய்ய இயலுமோ அங்கெல்லாம் செய்தல் வேண்டும் என்பது இக்குறளின் பொருள். தொட்டில் பழக்கம் சுடுகாடு மட்டும் என்பது பழமொழி இவைகளை எல்லாம் மாணவர்கள் பசுமையாக உள்ளத்தில் கொண்டு சமுதாயத் தொண்டு ஆற்றிட வேண்டும்.

Question 44.
(அ) அறிவியலாளர் ஸ்டீபன் ஹாக்கிங்குடன் விண்வெளிப் பயணம் என்னும் தலைப்பில் கற்பனைக் கதை ஒன்று எழுதுக.
Answer:
அறிவியலின் வளர்ச்சி மனிதனின் அறிவை விரிவாக்குகிறது. ஐயங்களை நீக்குகிறது. பழைய தவறான புரிதல்களை நீக்குகிறது. எண்ணங்களை மாற்றுகிறது அறிவியலால் ஒருகாலத்தில் நிறுவப்பட்டிருந்த கருத்து பின்னால் மறுக்கப்படுவதும் நேர்கிறது. மீண்டும் புதிய தடங்களைப் பதித்துப் புதிய பாதையிலே அறிவியல் இயங்குகிறது.

இயற்கையின் மர்ம முடிச்சுகளை அவிழ்க்கும் அறிவியல் சிந்தனை, போற்றுதலுக்குரியதாக இருக்கிறது. அதிலும் தன்னால் எந்த இயக்கமும் மேற்கொள்ள இயலாத நிலையிலும் அறிவியலின் இயங்கும் தன்மையை அறிந்து புது உண்மைகளைச் சொன்ன ஒருவரை உலகம் போற்றுவதில் வியப்பில்லை.

அவருடன் விண்வெளி பயணம் என்பது மிகவும் சுவாரசியமான ஒன்று. நானும் அவரும் விண்வெளியில் சென்று கொண்டு இருக்கும் போது நமது பால்வீதியில் கோடிக்கணக்கான விண்மீன்கள் ஒளிர்கின்றன. அவற்றுள் நம் ஞாயிறும் ஒன்று. ஒரு விண்மீனின் ஆயுள் கால முடிவில் உள்நோக்கிய ஈர்ப்பு விசை கூடுகிறது. அதனால் விண்மீன் சுருங்கத் தொடங்குகிறது. விண்மீன் சுருங்கச் சுருங்க அதன் ஈர்ப்பாற்றல் உயர்ந்து கொண்டே சென்று அளவற்றதாகிறது.

சில நேரங்களில் உண்மைப் புனைவை விடவும் வியப்பூட்டுவதாக அமைந்துவிடுகிறது. அப்படி ஓர் உண்மைதான் கருந்துளைகள் பற்றியதும், புனைவு இலக்கியம் படைப்பவர்களது கற்பனைகளையெல்லாம் மிஞ்சுவதாகவே கருந்துளைகள் பற்றிய உண்மைகள் உள்ளன. அதனை அறிவியல் உலகம் மிக மெதுவாகவே புரிந்துகொள்ள முயல்கிறது என்று ஸ்டிஃபன் ஹாக்கிங் கூறினார்.

(அல்லது)

(ஆ) அனுமான் ஆட்டத்தைக் கூறுக.
Answer:

• திடீரென்று மேளமும் நாதசுரமும் துரித கதியில் ஒலிக்கத் தொடங்கின.
• எதற்கென்று தெரியாமல் கூட்டம் திகைத்துப் பந்தலை நோக்குகையில் பெருங்குரல் எழுப்பியபடி அனுமார் பந்தல் கால் வழியாகக் கீழே குதித்தார்.
• அனுமார் வாலில் பெரிய தீப்பந்தம். ஜ்வாலை புகைவிட்டுக் கொண்டு எரிந்தது. கூட்டம் தானாகவே பின்னால் நகர்ந்தது.
• அனுமார் கால்களைத் தரையில் பதித்து உடம்பை ஒரு குலுக்குக் குலுக்கினார்.
• தீயின் ஜ்வாலை மடிந்து அலை பாய்ந்தது. கைகளைத் தரையில் ஊன்றி அனுமார் கரணமடித்தார்.
• சுருண்ட வால் இவன் பக்கமாக வந்து விழுந்தது.
• கூட்டம் அச்சத்தோடு கத்தியபடி அலைக்கழிந்தது.
• அனுமார் பெரிதாகச் சிரித்துக்கொண்டு நின்றார். அனுமார் நின்றதும் கூட்டம் கொஞ்சம் அமைதியுற்றது.
• முன்நோக்கி நகர்ந்து வந்தது. அனுமார் நேசப்பான்மையோடு சிரித்து வாலை மேலே தூக்கிச் சுற்றினார்.
• தீ வட்டமாகச் சுழன்றது. வேகம் கூடக்கூட, கூட்டம் இன்னும் முன்னால் நகர்ந்து வந்தது.
• இவன் நெருங்கி அனுமார் பக்கம் சென்றான்.
• அனுமார் இன்னொரு பாய்ச்சல் பாய்ந்து வேகமாக ஆட ஆரம்பித்தார். வர வர ஆட்டம் துரிதகதிக்குச் சென்றது.
• பதுங்கியும் பாய்ந்தும் ஆடினார்.
• ஆட ஆட, புழுதி புகை போல எழுந்தது. கழுத்துமணி அறுந்து கீழே விழுந்தது.
• ஒன்றையும் பொருட்படுத்தாமல் ஆட்டத்தில் தன்னை இழந்தவராக ஆடினார்.
• மேளமும், நாதசுரமும் அவர் ஆட்டத்தோடு இணைந்து செல்ல முடியவில்லை. தடுமாறிவிட்டது.
• மேல் மூச்சு வாங்க அனுமார் ஆட்டத்தை நிறுத்தினார். மேளமும் நாதசுரமும் நின்றன.
• அயர்ச்சியோடு மேளக்காரன் தோளிலிருந்து தவுலை இறக்கிக் கீழே வைத்தான்.
• ஆட்டம் முடிந்தது. தீர்மானமாகியது போல எஞ்சி இருந்த கூட்டமும் அவசர அவசரமாகக் கலைய ஆரம்பித்தது.

Question 45.
(அ) நூலகம் காட்டும் அறிவு’ – என்னும் தலைப்பில் கட்டுரை ஒன்று தருக. முன்னுரை:
Answer:
“வாழ்க்கை என்றொரு புத்தகம்
பக்கங்கள் எத்தனை யார் அறிவார்?”

எனும் வினாவால் வாழ்க்கையே புத்தகம்தான் என எடுத்தியம்பும் வல்லிக்கண்ணனின் பார்வை வீச்சு சிறப்புடையதாகும். வாழ்க்கையையே புத்தக நோக்கினில் பார்த்ததற்கும், வாழ்க்கையில் பள்ளிப் புத்தகம் தவிர வேறு புத்தகங்களைப் பார்த்ததேயில்லை என்பதற்கும் எத்தனை வேறுபாடு. இங்குதான் நூலகத்தை மறந்த நிலை என்பது வெளிப்படுகிறது.

அறிஞர்களை உருவாக்கும் நூலகம்:
அடுத்தவரைப் பற்றி பேசிப்பேசி நாட்களை ஓட்டியும், போட்டியும், வஞ்சமும் நிறைந்த உலகில் ஒரு நிமிடம் நூலகத்தை நோக்கிப் பயணத்தைத் திருப்புங்கள். அயர்வுகளைத் தீர்க்கும் அருமருந்து அங்குதான் உள்ளது. பல்வேறு அறிவியலறிஞர்களும், அறிஞர்களும் இங்கிருந்து தான் வெளிப்படுகின்றனர்.

எல்லா நூலையும் நாம் விலை கொடுத்து வாங்கிக் கற்க முடியாது. ஆனால் எல்லா நூல்களின் இருப்பிடமான நூல் நிலையம் சென்றால் அங்கிருந்து நாம் பல நூல்களைக் கற்கலாம் அல்லவா?

நூலகத்தினை பயன்படுத்தும் விதம்:
நூலகத்தினை எவ்வாறெல்லாம் பயன்படுத்துகின்றனர் எனில், பத்திரிகை படிக்க வருபவர் சிலர் ; விளையாட்டுச் செய்திகளை விருப்பமுடன் படிப்பவர் பலர்; இதழ்களில் அட்டைப்படங்களைக் காண வருபவர்கள் சிலர்; திரையுலகை தரிசிக்க வருபவர் பலர் என எண்ணற்ற முகங்களை வழி நடத்துவது இந்நூல் நிலையங்களாகும்.

நூல்களைக்கூட படித்திட வாங்கிச்சென்று, வேண்டிய பக்கங்களைக் கிழித்து எடுத்துத் திருப்பித் தருபவர் உண்டு. நூலினில் பல படங்களை வரைந்து வைத்தல், சில பெயர்களை எழுதுதல் என எண்ணற்ற சிறு செயல்கள் செய்து தமது சிறுமையை வெளிப்படுத்துபவர் உளர்.

அறிவுக்களஞ்சியம் :
பிறமொழி அறிவு வளர்ந்திட உதவும் நூல்கள் உதவியால், பிற மொழியாளரிடம் பேசும் அளவிற்கு தம்மை உயர்த்திக் கொண்டவர் உண்டு. மொழிகளைப் பற்றி நூல்கள் மட்டுமல்லாமல் அறிவியல்,பழங்கால வரலாறுகள், கதைகள், நாவல்கள், கவிதை நூல்கள், கட்டுரைத் தொகுப்புகள், சிறு தொழில் கற்றிட உதவும் நூல்கள், சமையல் குறிப்புகள் எனப் பல்வேறு பிரிவுகளில் அமைந்து அறிவுக் களஞ்சியமாய்த் திகழ்கிறது.

பல்வேறு நூலகங்களின் பெயர்கள் :
நூலகங்கள் இல்லாத இடங்களில்லை. பள்ளிகள், கல்லூரிகள், பல்கலைக் கழகங்கள் முதலிய இடங்களில் உள்ளது. ஊர்கள் தோறும், மாவட்டங்கள் தோறும் நூலகங்கள் உண்டு. மாநிலத்தின் தலைமையிடத்திலும் நூலகம் உண்டு. சென்னையில் மாநில மைய நூலகமான கன்னிமாரா நூலகம்’ அமைந்துள்ளது.

தேவநேயப் பாவாணர் நூலகம், மறைமலையடிகள் நூலகம், சரஸ்வதி மஹால் நூலகம், தஞ்சை தமிழ் பல்கலைக்கழக நூலகம், சாது சேஷய்யா ஓரியண்டல் நூலகம், வ.உ.சி. நூலகம், கவிமணி நூலகம் என்பன போன்ற பல நூலகங்கள் மாவட்டங்கள் வாரியாக அமைக்கப்பட்டுள்ளன. தற்போது வள்ளுவரின் பெயரில் நூலகங்கள் ஏற்படுத்தப்பட்டு வருகின்றன.

அறிவுச் சுரங்கமாய் நூலகம்:
நூலகங்களில் நூல்கள் பெற வேண்டுமானால், நூலக உறுப்பினராகிக் கொண்டு அதன் பிறகு நூலை எடுத்துக் கொள்ளலாம். மனிதன் அறிவுச் சுரங்கமாய் விளங்க நூலகமே முக்கியக் காரணம். பள்ளிகளில் மாணவர்கள் பேச்சுப் போட்டியில், கட்டுரைப் போட்டியில் பரிசுகளைப் பெற்றிட நூலகமும் ஒரு காரணமே.

எவரொருவர் அறிவின் பிறப்பிடமாகத் திகழ்கிறாரோ அவரைத் துன்பம் நெருங்குவதில்லை. அவரது அறிவுத் திறனால் துன்பம் வராமல் காக்கப்படுகிறது. இதையே வள்ளுவர்,

“எதிரதாக் காக்கும் அறிவினார்க்கு இல்லை
அதிர வருவதோர் நோய் ” – எனும்

குறள் மூலம் அறிவுறுத்துகிறார். இவ்வாறு ஒருவர் அறிவின் சுடராய்த் திகழ நூலகம் மிக முக்கியமான வழிகாட்டியாகும்.

முடிவுரை:
இவ்வாறு நூலகமானது ஒரு மனிதனுக்கு அறிவு, சிந்திக்கும் ஆற்றல் வழங்குவதோடு தகுதியுடையவராய் எழச் செய்யும் அற்புத மருந்தாகும். மாணவப் பருவத்திலேயே நூலகத்தினைப் பயன்படுத்துதல் இன்றியமையாதது.

எங்கே கிளம்பி விட்டீர்கள், நூலக உறுப்பினர் ஆகத்தானே!

(அல்ல து)

(ஆ) குறிப்புகளைப் பயன்படுத்திக் கட்டுரை ஒன்று தருக.
முன்னுரை – வரதட்சணை – பண்டைய பெண்களின் பெருமை – ”திருமணம் என்பது ஆயிரம் காலத்துப் பயிர்” – பலியாகும் பெண்கள் – ” மாமியார் உடைத்தால் மண்சட்டி திருமணம் – வேண்டாம் வரதட்சணை – “பெண்கள் நாட்டின் கண்கள்’ – முடிவுரை.

வேண்டாம் வரதட்சணை
குறிப்புச் சட்டகம்

முன்னுரை
பொருளுரை: வரதட்சணைக் கொடுமை
பலியாகும் பெண்கள்
திருமணம்
வேண்டாம் வரதட்சணை
முடிவுரை –

முன்னுரை
வரதட்சணை என்பது, மணமான பெண் தன் தாய் வீட்டிலிருந்து கொண்டு வரும் சீதனம், பெண்களுக்கு மேலும் பொறுப்புகளையும், உறவுகளையும் சேர்க்கக் கூடியது. அவளது திருமண வாழ்க்கையாகும். இது திருமணத்தின் தனித்தன்மை என்று நாம் குறிப்பிடலாம். பெண்கள் குடும்பத் தலைவி என்ற பொறுப்பை ஏற்று, செயல்பட்டு தன் வாழ்க்கையில் முன்னேறுவது திருமணத்திற்குப் பின் என்றும் நாம் கூறலாம்.

‘திருமணம் என்பது ஆயிரம் காலத்துப் பயிர்’ என்பார்கள். அக்காலங்களில் பெண்கள் வேலைக்குச் செல்லாத காரணத்தால், வீட்டில் எல்லா வேலைகளையும் சுமந்து தன் கணவனது சொற்படி நடந்து அவனது பொருளாதார மேம்பாட்டிற்காக சிறு அன்பளிப்பாக இச்சீதனம் வழங்கப்பட்டது.

வரதட்சணைக் கொடுமை:
திருமணம் சொர்க்கத்தில் நிச்சயிக்கப்படுகிறது என்பார்கள். ஆனால், திருமணம் வரதட்சணையால் தான் நிச்சயிக்கப்படுகிறது. அக்காலத்திலிருந்தே மணமகனின் வீட்டார். பெண் வீட்டாரிடம் வரதட்சணைக் கேட்டு வருகின்றனர். பெண் வீட்டாரோ, அதை எப்படிக் கொடுக்கலாம் என்ற வருத்தத்தில் இருப்பார்கள்.

பலியாகும் பெண்கள் :
திருமணம் முடிந்து சில நாட்கள் செல்ல, மாமியாரும், அப்பெண்ணுடைய கணவரும் அவளைக் கொடுமைப்படுத்துவார்கள். “மாமியார் உடைந்தால் மண்சட்டி, மருமகள் உடைத்தால் அது பொன் ஓடாக மாறுவதேன்?” அவர்களோ, தொடர்ந்து கொடுமைப்படுத்திக் கொண்டே இருப்பர். இதனால் ஏராளமான பெண்கள் தற்கொலை முயற்சியில் ஈடுபடுகிறார்கள்.

திருமணம்:
திருமணம் என்பதொரு புனிதமானச் செயல், அப்படிப்பட்ட செயலை வரதட்சணை என்னும் பயங்கர உருவத்தால் நாம் பாவச் செயலாக்கக்கூடாது. வரதட்சணை கேட்டாலும் அவரவர் தகுதிக்கு ஏற்றபடி கேட்கவேண்டும். இக்காலத்தில் வரதட்சணை கேட்பவர்கள் மீது வழக்குத் தொடரலாம்.

முன், சீதனம் என்று அழைக்கப்பட்டது. பின் கட்டாயமாக்கப்பட்டு வரதட்சணை என்று அழைக்கப்படுகிறது. முன், அன்பளிப்பாக வழங்கப்பட்டது. இப்போது பல கொடுமைகளைப் புரியும் வழக்கமாக்கப்பட்டுள்ளது. இதனால் பெண்களுக்கு நேரும் கொடுமைகள் பற்பல.

“உன் விழிப்பில்தான் நாட்டின் பிழைப்பே உள்ளது”

வேண்டாம் வரதட்சணை :
பெண்கள் இல்லையேல் நாட்டின் வளர்ச்சி இல்லை. இத்தகைய பெண்களைக் கொடுமைப்படுத்தி பணம் பறிப்பது வேதனைக்குரிய விஷயம்.

முடிவுரை :
வீட்டிற்கு வரும் பெண் வரதட்சணையோடு வராமல், பாசம், பண்பு, நல்ல குணம், நட்பு போன்ற நற்குணங்கள் பலவற்றைக் கொண்டு வந்தால், புன்னகையும் பொன் நகையாகும். ஓர் ஆணுக்குக் கல்வியளிப்பது அவருக்கு மட்டுமே சிறப்பு. ஆனால், ஒரு பெண்ணுக்குக் கல்வியளித்தல் என்பது, அது அக்குடும்பத்திற்கே கல்வியளிப்பது போன்றதாகும். வரதட்சணைக்குப் பதிலாகப் படித்த பெண்ணாக, கல்வி கேள்விகளில் சிறந்தவளாக்குதல் வீட்டிற்கும், ஏன் நாட்டிற்கும் பெருமை. எனவே, வரதட்சணையை ஒழிக்க முயன்று நமது நாட்டினை ஒரு பூஞ்சோலையாக மாற்றிக் காட்டுவோம்.

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

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

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

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

Question 1.
Match the following:

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

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

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

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

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

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

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

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

Part – II

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

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

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

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

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

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

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

Part – III

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

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

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

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

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

Bt cotton has some limitations:

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

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

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

Examples:

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

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

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

Part – IV

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

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

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

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

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

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

[OR]

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

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

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

[OR]

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

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

Bio-Zoology [Maximum Marks: 35]

Part – I

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

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

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

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

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

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

Question 6.
Match list I with list II

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

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

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

Part – II

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

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

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

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

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

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

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

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

Part – III

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

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

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

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

Question 17.
Under which conditions does a bacterium develops resistance against antibiotics?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control.

Antibiotics should be used only when prescribed by a certified health professional. When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Question 18.
Expand (i) CFC (ii)AQI (w) PAN
Answer:
(i) CFC – Chloro fluro carbon
(ii) AQI – Air Quality Index
(iii) PAN – Peroxyacetyl nitrate .

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

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

Part – IV

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

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

Menstrual cycle comprises of the following phases:

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

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

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

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

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

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

[OR]

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

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

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

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

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

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

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

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

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

[OR]

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

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

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

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

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

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

Students can Download Bio Botany Chapter 8 Environmental Issues Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 8 Environmental Issues

Samacheer Kalvi 12th Bio Botany Environmental Issues Text Book Back Questions and Answers

Question 1.
Which of the following would most likely help to slow down the greenhouse effect?
(a) Converting tropical forests into grazing land for cattle.
(b) Ensuring that all excess paper packaging is buried to ashes.
(c) Redesigning landfill dumps to allow methane to be collected.
(d) Promoting the use of private rather than public transport.
Answer:
(d) Promoting the use of private rather than public transport.

Question 2.
With respect to Eichhornia
Statement A: It drains off oxygen from water and is seen growing in standing water.
Statement B: It is an indigenous species of our country.
(a) Statement A is correct and Statement B is wrong.
(b) Both Statements A and B are correct.
(c) Statement A is correct and Statement B is wrong.
(d) Both statements A and B are wrong.
Answer:
(a) Statement A is correct and Statement B is wrong.

Question 3.
Find the wrongly matched pair.
(a) Endemism – Species confined to a region and not found anywhere else.
(b) Hotspots – Western ghats
(c) Ex-situ Conservation – Zoological parks
(d) Sacred groves – Saintri hills of Rajasthan
(e) Alien sp Of India – Water hyacinth
Answer:
(d) Sacred groves – Saintri hills of Rajasthan

Question 4.
Depletion of which gas in the atmosphere can lead to an increased incidence of skin cancer?
(a) Ammonia
(b) Methane
(c) Nitrous oxide
(d) Ozone
Answer:
(d) Ozone

Question 5.
One green house gas contributes 14% of total global warming and another contributes 6%. These are respectively identified as ________
(a) N₂O and CO₂
(b) CFCs and N₂O
(c) CH₄ and CO₂
(d) CH₄ and CFCS
Answer:
(b) CFCs and N₂O

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

Question 7.
Deforestation means ______
(a) growing plants and trees in an area where there is no forest
(b) growing plants and trees in an area where the forest is removed
(c) growing plants and trees in a pond
(d) removal of plants and trees
Answer:
(d) removal of plants and trees

Question 8.
Deforestation does not lead to _______
(a) Quick nutrient cycling
(b) Soil erosion
(c) alternation of local weather conditions
(d) Destruction of natural habitat weather conditions
Answer:
(a) Quick nutrient cycling

Question 9.
The unit for measuring ozone thickness ______
(a) Joule
(b) Kilos
(c) Dobson
(d) Watt
Answer:
(c) Dobson

Question 10.
People’s movement for the protection of environment in Sirsi of Karnataka is ________
(a) Chipko movement
(b) Amirtha Devi Bishwas movement
(c) Appiko movement
(d) None of the above
Answer:
(c) Appiko movement

Question 11.
The plants which are grown in silivpasture system are ________
(a) Sesbania and Acacia
(b) Solenum and Crotalaria
(c) Clitoria and Begonia
(d) Teak and sandal
Answer:
(a) Sesbania and Acacia

Question 12.
What is ozone hole?
Answer:

• The ozone shield is being damaged by the chlora fluoro carbon widely used in refrigeration, aerosols, chemicals used and cleaners in many industries.
• The decline in the thickness of the ozone layer over the restricted area is called ozone hole.

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

Question 14.
Expand CCS.
Answer:
CCS – Carbon Capture and Storage.

• Carbon capture and storage is a technology of capturing carbon dioxide and inject it deep into the underground rocks at a depth of 1 km (or) more.
• It is an approach to mitigate global warming.

Example:

• It is capturing CO₂ released from industries and power plants.
• Such as declining oil fields, gas fields saline aquifers and unmineable coal have been suggested as storage sites.
• Various safe sites have been selected for permanent storage.
• liquid storage in ocean and solid storage by reduction of CO₂ with metal oxide to produce stable carbonates.
• It is also known as Geological sequestration.

Question 15.
Hpw do forests help in maintaining the climate?
Answer:
Forests play a major role in regulating the CO₂ level in the atmosphere. As the result global warming in highly reduced.

Question 16.
How do sacred groves help in the conservation of biodiversity?
Answer:

• Sacred groves are the patches (or) grove of cultivated trees.
• which are community protected and are based on strong religious belief system.
• These groves provide a number of ecosystem services to the neighbourhood like protecting watershed, fodder, medicinal plants and microclimate control,
• These groves can conserve water and prevent soil nutrient loss,
• It has several Ayurvedic medicinal plants that are not to be found in the forest are abundant in the sacred groves.
• Further rare, endangered, threatened and endemic species are concentrated in sacred grove.
• SGS are with rich biodiversity and harbour and many rare species of plants and animals.
• Each grove is an abode of deity mostly village god (or) Goddesses like Aiyanar (or) Amman (eg) – 448 groves were documented throughout Tamilnadu

Question 17.
Which one gas is most abundant out of the four commonest greenhouse gases? Discuss the effect of this gas on the growth of plants?
Answer:
Carbondioxide is the most abundant greenhouse gas. Increase in CO₂ level in the air decreases the uptake of nitrogen components leading to protein deficiency and chlorophyll formation.

Question 18.
Suggest a solution to water crisis and explain its advantages.
Answer:

• Rainwater harvesting is the solution to the water crisis.
• Promotes adequacy of underground water and water conservation.
• Mitigates the effect of drought.
• Reduces soil erosion as surface run-off is reduced.
• Reduces flood hazards.
• Improves groundwater quality and water table / decreases salinity.
• Storing water underground is an eco – friendly measure and a part of sustainable water storage strategy for local communities. Rain water harvesting – RWH (solution to water crisis – A ecological problem)
• Rainwater harvesting is the accumulation and storage of rain water for reuse in – site rather than allowing it to run off.
• Rainwater can be collected from rivers, roof tops and the water collected is directed to a deep pit.
• The water percolates and gets stored in the pit RWH is a sustainable water management practice implemented not only in urban area but also in agricultural fields.
• Which is an important economical cost-effective method for the culture.

Question 19.
Explain afforestation with case studies.
Answer:
Afforestation is planting of trees where there was no previous tree coverage and the conversion of non-forested lands into forests by planting suitable trees to retrieve the vegetation. Example: Slopes of dams afforesed to reduce water run-off, erosion and siltation. It can also provide a range of environmental sendees including carbon sequestration, water retention. The Man who Single Handedly Created a Dense Forest

Jadav “Molai” Payeng (bom 1963) is an environmental activist has single-handedly planted a forest in the middle of a barren wasteland. This Forest Man of India has transformed the world’s largest river island, Majuli, located on one of India’s major rivers, the Brahmaputra, into a dense forest, home to rhinos, deers, elephants, tigers and birds. And today his forest is larger than Central Park.

Former vice-chancellor of Jawahar Lai Nehru University, Sudhir Kumar Sopory named Jadav Payeng as Forest Man of India, in the month of October 2013. He was honoured at the Indian Institute of Forest Management during their annual event Coalescence. In 2015, he was honoured with Padma Shri, the fourth highest civilian award in India. He received honorary doctorate degree from Assam Agricultural University and Kaziranga University for his contributions.

Question 20.
What are the effects of deforestation and the benefits of agroforestry?
Answer:
The conversion of a forested area into a non-forested area is known as deforestation.

Effects of deforestation:

• Burning of forest wood releases stored carbon, a negative impact just opposite of carbon sequestration.
• Trees and plants bind the soil particles. The removal of forest cover increases soil erosion and decreases soil fertility. Deforestation in dry areas leads to the formation of deserts.
• The amount of runoff water increases soil erosion and also creates flash flooding, thus reducing moisture and humidity.
• The alteration of local precipitation patterns leading to drought conditions in many regions. It triggers adverse climatic conditions and alters water cycle in the ecosystem.
• It decreases the bio-diversity significantly as their habitats are disturbed and disruption of natural cycles.
• Loss of livelihood for forest dwellers and rural people.
• Loss of life support resources, fuel, medicinal herbs and wild edible fruits.

Benefits of agroforestry:

• It is an answer to the problem of soil and water conservation and also to stabilise the soil (salinity and water table) reduce landslide and water run-off problem.
• Nutrient cycling between species improves and organic matter is maintained.
• Trees provide micro climate for crops and maintain O₂, – C0₂ balanced, atmospheric temperature and relative humidity.
• Suitable for dry land where rainfall is minimum and hence it is a good system for alternate land use pattern.
• Multipurpose tree varieties like Acacia are used for wood pulp, tanning, paper and firewood industries.
• Agro-forestry is recommended for the Farm Forestry for the extension of forests, mixed forestry, shelter belts and linear strip plantation.

Samacheer Kalvi 12th Bio Botany Environmental Issues Additional Questions and Answers

1 – Mark Questions

Question 1.
Which is not a greenhouse gas?
(a) CO₂
(b) N₂O
(c) O₃
(d) CFC
Answer:
(C) O₃

Question 2.
Identify the incorrect statement with regard to Global warming.
(a) Leads to species enrichment
(b) Decreases irrigation
(c) Increases vector population
(d) Frequent heat waves
Answer:
(a) Leads to species enrichment

Question 3.
__________ is the unit of measurement of total ozone.
Answer:
Dobson unit

Question 4.
The total ozone layer over the earth surface is __________
(a) 30 DU
(b) 300 DU
(c) 3000 DU
(d) 0.3 DU
Answer:
(b) 300 DU

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

Question 6.
Which is not a beneficial aspect of Agroforestry?
(a) Nutrient cycling is improved
(b) Balance in O₂ – CO₂ composition
(c) Suitable for wetland where rainfall is maximum
(d) Reduces water run-off problem
Answer:
(c) Suitable for wetland where rainfall is maximum

Question 7.
The production of woody plants combined with pasture is referred to system.
Answer:
Silvopasture

Question 8.
Assertion (A): CO₂ is a main cause for global warming
Reason (R) : Greenhouse gases trap the radiant heat from sun
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(c) R explains A.

Question 9.
Assertion (A): Ozone acts as a natural sun block.
Reason (R): UV rays reaching the earth are deviated from earth.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(c) R explains A.

Question 10.
Assertion (A): Social forestry refers to management of forests and afforestation on barren lands.
Reason (R): Afforestation involves the cutting of trees.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(a) A is correct R is incorrect.

Question 11.
Assertion (A): Prosopis juliflora is native to Afganisthan.
Reason (R): Alien species refers to non-native species.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(b) A is incorrect R is correct.

Question 12.
Assertion (A): In zoological parks, the animals are maintained in their natural habitat.
Reason (R): Ex-situ conservation refers to protecting species in their natural habitat.
(a) A is correct R is incorrect.
(b) A is incorrect R is correct.
(c) R explains A.
(d) Both A and R are incorrect.
Answer:
(d) Both A and R are incorrect.

Question 13.
Which is not reduced by deforestation?
(a) Amount of habitat
(b) Amount of animal population
(c) Amount of biodiversity
(d) Amount of agricultural land
Answer:
(d) Amount of agricultural land

Question 14.
Identify the potent cause for deforestation.
(a) Agriculture
(b) Soil erosion
(c) Afforestation
(d) Forest fire
Answer:
(a) Agriculture

Question 15.
Total number of forestry extension centres in Tamil Nadu is __________
(a) 16
(b) 32
(c) 18
(d) 51
Answer:
(b) 32

Question 16.
Who is celebrated as Forest Man of India?
(a) Anand Mohan Chakrabarthy
(b) Dr. M.S Swaminathan
(c) Jadav Molai Payeng
(d) Choudary Ram Dhan
Answer:
(c) Jadav Molai Payeng

Question 17.
Invasive species __________
(a) alter the soil system
(b) are more adapted
(c) are fast growing
(d) all the above
Answer:
(d) all the above

Question 18.
Pick out the odd one out __________
(a) Biosphere reserve
(b) National Park
(c) Wild life sanctuaries
(d) Botanical gardens
Answer:
(d) Botanical gardens

Question 19.
Which is not true with respect to prosopis juliflora?
(a) Invasive species native to Mexico
(b) Arrest wind erosion
(c) Absorb hazardous chemical from soil
(d) Decreases O₂ content of water bodies
Answer:
(d) Decreases O₂ content of water bodies

Question 20.
How many numbers of scared grooves were documented in tamil nadu?
(a) 484
(b) 844
(c) 488
(d) 448
Answer:
(c) 488

Question 21.
Match the following:

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

Question 22.
Process of heating biomass in low oxygen environment is called as ______
Answer:
Pyrolysis

Question 23.
Biochar is __________
(i) a kind of char coal used as a soil amendment
(ii) a potent way of sequestring carbon
(iii) made from biomass via pyrolysis
(iv) a notable solid, rich in carbon.
(a) (i) and (iii) is correct
(b) (ii) and (iv) is correct
(c) (i) and (ii) is correct
(d) all the above is correct.
Answer:
(d) all the above is correct.

Question 24.
Which is not a true statement regarding rain water harvesting?
(a) Mitigates groundwater quality
(b) Reduces soil erosion
(c) Decreases soil salinity
(d) No wastage of land for storing
Answer:
(a) Mitigaes groundwater quality

Question 25.
EIA stands for __________
(a) Ecological Information Analysis
(b) Environmental Information Assessment
(c) Environmental Impact Analysis
(d) Environmental Impact Assessment
Answer:
(d) Environmental Impact Assessment

Question 26.
__________ is the 100th Satellite launched to watch border surveillance.
(a) GSAT-6A
(b) SCAT SAT-I
(c) INSAT 3DR
(d) CARTOSAT-2
Answer:
(d) CARTOSAT-2

Question 27.
The ozone layer of __________ is called bad ozone.
(a) Stratosphere
(b) Mesosphere
(c) Troposphere
(d) Exosphere
Answer:
(c) Troposhere

Question 28.
When does World Ozone Day is observed?
(a) June 17th
(b) December 1st
(c) October 12th
(d) September 16th
Answer:
(d) September 16th

Question 29.
Clean Development Mechanism (CDM) is defined in __________
(a) Copenhagen Acord
(b) Montreal Protocol
(c) Paris Agreement
(d) Kyoto Protocol
Answer:
(d) Kyoto Protocol

Question 30.
__________ is a plant species which acts as an indicator of Nitrate pollution.
(a) Petunia
(b) Lichens
(c) Gladiolus
(d) Pinus
Answer:
(a) Petunia

Question 31.
Identify the plant species that is not used as a live fence.
(a) Sesbania grandiflora
(b) Acacia species
(c) Petunia species
(d) Erythrina species
Answer:
(c) Petunia species

Question 32.
Match the following:

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

2 – Mark Questions

Question 1.
What are greenhouse gases? Give an example.
Answer:
The gases that capture heat are called Green House Gases.
E.g: Carbon dioxide (CO₂)

Question 2.
Name any four green house gases.
Answer:
CO₂, CH₄, N₂O and CFC

Question 3.
Define global warming.
Answer:
The increase in mean global temperature due to increased concentration of green house gases is called global warming.

Question 4.
Methane is one of a potent green house gas. Point out few sources from where methane is generated.
Answer:
Paddy field, fossil fuel production, bacteria in water bodies, cattle rearing, non-wetland soils and forest & wild fires.

Question 5.
List out the anthropogenic sources of Nitrous oxide.
Answer:
Man-made sources include nylon and nitric acid production, use of fertilizers in agriculture, manures cars with catalytic converter and burning of organic matter.

Question 6.
What is Dobson unit?
Answer:
Dobson Unit is the unit of measurement for total ozone. One DU is the number of molecules of ozone that would be required to create a layer of pure ozone 0.01 millimetre thick at a temperature of 0° C and a pressure of 1 atmosphere.

Question 7.
What do you mean by good ozone and bad ozone?
Answer:

1. The ozone layer of troposphere is called bad ozone.
2. The ozone layer of stratosphere is called good ozone.

Question 8.
What is ozone layer? Why it is essential?
Answer:
Ozone layer is a region of Earth’s stratosphere that absorbs most of the Sim’s ultra violet radiation. The ozone layer is also called as the ozone shield and it acts as a protective shield, cutting the ultra-violet radiation emitted by the sun.

Question 9.
Define ozone hole. Name any one potent chemical that is responsible for the effect.
Answer:
The decline in the thickness of the ozone layer over restricted area is called Ozone hole. Chlorofluorocarbons (CFC) damages the ozone layer to a great extent.

Question 10.
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 11.
What is meant by biological indicators? Give example.
Answer:
The plant species or plant community acts as a measure of environmental conditions, it is referred as biological indicators or phytoindicators or plant indicators.
E.g: Lichens as SO₂ pollution indicator.

Question 12.
Name the plant species which act as indicators for Nitrate pollution & Flouride pollution.
Answer:

1. Petunia species acts as indicators for Nitrate pollution.
2. Gladiolus species acts as indicators for Flouride pollution.

Question 13.
Define the term Agroforestry. Name any two major tree species cultivated in Agroforestry.
Answer:
Agroforestry is an integration of trees, crops and livestock on the name plot of land. The commercial Agroforestry trees are casuarina and Eucalyptus.

Question 14.
What is silvopasture system? How it helps economy?
Answer:
The production of woody plants combined with pasture is referred to silvopasture system. The trees and shrubs may be used primarily to produce fodder for livestock or they may be grown for timber, fuel wood and fruit or to improve the soil.

Question 15.
Mention the name of any four plant species that are widely used as live fences in agricultural practicer.
Answer:

1. Gliricidia sepium
2. Sesbaniagrandiflora
3. Erythrina species
4. Acacia species

Question 16.
Define social forestry.
Answer:
Social forestry refers to the management of forests and afforestation on barren lands with the purpose of helping the environmental, social and rural development and benefits.

Question 17.
Compare Deforestation and Afforestation
Answer:

1. Deforestation: The conversion of forested area into a non- forested area is known as deforestation.
2. Afforestation: Afforestation is planting of trees where there was no previous tree coverage and the conversion of non-forested lands into forests by planting suitable trees to retrieve the vegetation.

Question 18.
What is invasive species?
Answer:
A non-native species to the ecosystem or country under consideration that spreads naturally, interferes with the biology and existence of native species, poses a serious threat to the ecosystem and causes economic loss

Question 19.
Point out the adverse effect caused by invasive species.
Answer:
Invasive species are fast growing and are more adapted. They alter the soil system by changing litter quality thereby affecting the soil community, soil fauna and the ecosystem processes. It has a negative impact on decomposition in the soils by causing stress to the neighbouring native species.

Question 20.
Distinguish between In-situ conservation and Ex-situ conservation.
Answer:
In-situ conservation:

1. Conservation of species in their natural habitat.
2. E.g: Biosphere reserves

Ex-situ conservation:

1. Conservation of species outside their natural habitat.
2. E.g: Zoological parks

Question 21.
Mention any two historical community level conservation movements held for the protection of environment.
Answer:
Chipko Movement and Appiko Movement

Question 22.
Give the names of any four famous sacred grooves in Tamil Nadu.
Answer:

1. Banagudi Shola
2. Thirukurungudi
3. Udaiyankudikadu
4. Sittannavasal

Question 23.
What does the term ‘Endemic’ refers to?
Answer:
Any species found restricted to a specified geographical area is referred to as ENDEMIC.

Question 24.
Name any two endemic trees of Peninsular India.
Answer:

1. Agasthiyamalaia pauciflora
2. Harawickia binata

Question 25.
What is carbon sequestration?
Answer:
Carbon sequestration is the process of capturing and storing CO₂ which reduces the amount of CO₂ in the atmosphere with a goal of reducing global climate change.

Question 26.
Carbon sequestration occurs naturally by plants and oceans. Name any four microalgal species involved in the process.
Answer:

1. Chlorella
2. Scenedesmus
3. Chroococcus
4. Chlamydomonas

Question 27.
Explain the term Carbon sink.
Answer:
Any system having the capacity to accumulate more atmospheric carbon during a given time interval than releasing CO₂.
Example: forest, soil, ocean are natural sinks. Landfills are artificial sinks.

Question 28.
How EIA is beneficial to a society?
The benefits of EIA to society
Answer:

1. A healthier environment
2. Maintenance of biodiversity
3. Decreased resource usage
4. Reduction in gas emission and environment damage

Question 29.
What is GIS?
Answer:
Geographic Information System (GIS) is a computer system for capturing, storing, checking and displaying data related to positions on Earth’s surface. Also to manipulate, analyse, manage and present spacial or geographic data.

3 – Mark Questions

Question 30.
List out the effects of global warming.
Answer:

1. Rise in global temperature which causes sea levels to rise as polar ice caps and glaciers begin to melt causing submergence of many coastal cities in many parts of the world.
2. There will be a drastic change in weather patterns bringing more floods or droughts in some areas.
3. Biological diversity may get modified, some species ranges get redefined. Tropics and sub-tropics may face the problem of decreased food production.

Question 31.
Mention any three sources of carbon dioxide emission.
Answer:

1. Coal based power plants, by the burning of fossil fuels for electricity generation.
2. Combustion of fuels in the engines of automobiles, commercial vehicles and air planes contribute the most of global warming.
3. Agricultural practices like stubble burning result in emission of CO₂.

Question 32.
What are the adverse effects of global warming on plants?
Answer:

1. Low agricultural productivity in tropics
2. Frequent heat waves (Weeds, pests, fungi need warmer temperature)
3. Increase of vectors and epidemics
4. Strong storms and intense flood damage
5. Water crisis and decreased irrigation
6. Change in flowering seasons and pollinators
7. Change in Species distributional ranges
8. Species extinction

Question 33.
Suggest few ways to overcome global warming.
Answer:

1. Increasing the vegetation cover, grow more trees
2. Reducing the use of fossil fuels and green house gases
3. Developing alternate renewable sources of energy
4. Minimising uses of nitrogeneous fertilizers, and aerosols.

Question 34.
Ozone acts a a natural sun screen – Justify.
Answer:
Ozone depletion in the stratosphere results in more UV radiations especially UV B radiations (shortwaves). UV B radiation destroys biomolecules (skin ageing) and damages living tissues. UV – C is the most damaging type of UV radiation, but it is completely filtered by the atmosphere (ozone layer). UV – a contribute 95% of UV radiation which causes tanning burning of skin and enhancing skin cancer. Hence the uniform ozone layer is critical for the wellbeing of life on earth.

Question 35.
Give a detailed account on clean development mechanism.
Answer:
Clean Development Mechanism (CDM) is defined in the Kyoto protocol (2007) which provides project based mechanisms with two objectives to prevent dangerous climate change and to reduce green house gas emissions. CDM projects helps the countries to reduce or limit emission and stimulate sustainable development.

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

Question 36.
What are the major activities carried out by forestry extension centres?
Answer:

1. Training on tree growing methods
2. Publicity and propaganda regarding tree growing
3. Formation of demonstration plots
4. Raising and supply of seedlings on subsidy
5. Awareness creation among school children and youth about the importance of forests through training and camps.

Question 37.
Prosopis juliflora, though an alien invasive species to India, it is beneficial to certain extent. Give reason.
Answer:
Prosopis juliflora is used to arrest wind erosion and stabilize sand dunes on coastal and desert areas. It can absorb hazardous chemicals from soil and it is the main source of charcoal.

Question 38.
Write a short note on Chipko Movement.
Chipko Movement:
Answer:
The tribal women of Himalayas protested against the exploitation of forests in 1972. Later on it transformed into Chipko Movement by Sundarlal Bahuguna in Mandal village of Chamoli district in 1974. People protested by hugging trees together which were felled by a sports goods company. Main features of Chipko movement were,

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

Question 39.
Give a brief account on In-situ conservation.
Answer:
In-situble conservation means conservation and management of genetic resources in their natural habitats. Here the plant or animal species are protected within the existing habitat. Forest trees, medicinal and aromatic plants under threat are conserved by this method. This is carried out by the community or by the State conservation which include wildlife, National park and Biosphere reserve. The ecologically unique and biodiversity rich regions are legally protected as wildlife sanctuaries, National parks and Biosphere reserves. Megamalai, Sathyamangalam wildlife, Guindy and Periyar National park, and Western ghats, Nilgiris, Agasthyamalai and Gulf of Mannar are the biosphere reserves of Tamil Nadu.

Question 40.
What do you mean by Biochar? How it helps the environment?
Answer:
Biochar is a long term method to store carbon. To increase plants ability to store more carbon, plants are partly burnt such as crop waste, waste woods to become carbon rich slow decomposing substances of material called Biochar. It is a kind of charcoal used as a soil amendment. Biochar is a stable solid, rich in carbon and can endure in soil for thousands of years. Like most charcoal, biochar is made from biomass via pyrolysis. (Heating biomas in low oxygen environment) which arrests wood from complete burning.

Biochar thus has the potential to help mitigate climate change via carbon sequestration. Independently, biochar when added to soil can increase soil fertility of acidic soils, increase agricultural productivity, and provide protection against some foliar and soil borne diseases. It is a good method of preventing waste woods and logs getting decayed instead we can convert them into biochar thus converting them to carbon storage material.

Question 41.
Explain the role of lakes in an ecosystem.
Answer:
Lakes as a storage of rain water provides drinking water, improves ground water level and preserve the fresh water bio-diversity and habitat of the area where in occurs.

In terms of services lakes offer sustainable solutions to key issues of water management and climatic influences and benefits like nutrient retention, influencing local rainfall, removal of pollutants, phosphorous and nitrogen and carbon sequestration.

5 – Mark Questions

Question 42.
List out the major effects of Ozone depletion.
Answer:
The main ozone depletion effects are:

1. Increases the incidence of cataract, throat and lung irritation and aggravation of asthma or emphysema, skin cancer and diminishing the functioning of immune system in human beings.
2. Juvenile mortality of animals.
3. Increased incidence of mutations.
4. In plants, photosynthetic chemicals will be affected and therefore photosynthesis will be inhibited. Decreased photosynthesis will result in increased atmospheric CO₂ resulting in global warming and also shortage of food leading for food crisis.
5. Increase in temperature changes the climate and rainfall pattern which may result in flood / drought, sea water rise, imbalance in ecosystems affecting flora and fauna.

Question 43.
What are the objectives of Afforestation programme?
Answer:
Afforestation Objectives:

1. To increase forest cover, planting more trees, increases CO₂ production and air quality.
2. Rehabilitation of degraded forests to increase carbon fixation and reducing CO₂ from atmosphere.
3. Raising bamboo plantations.
4. Mixed plantations of minor forest produce and medicinal plants.
5. Regeneration of indigenous herbs / shrubs.
6. Awareness creation, monitoring and evaluation.
7. To increase the level and availability of water table or ground water and also to reduce nitrogen leaching in soil and nitrogen contamination of drinking water, thus making it pure not polluted with nitrogen.
8. Nature aided artificial regeneration.

Question 44.
How Eichhornia crassiper spoils the Indian ecosystem?
Answer:
Eichhornia crassipes is an invasive weed native to South America. It was introduced as aquatic ornamental plant, which grows faster throughout the year. Its widespread growth is a major cause of biodiversity loss worldwide. It affects the growth of phytoplanktons and finally changing the aquatic ecosystem.

It also decreases the oxygen content of the waterbodies which leads to eutrophication. It poses a threat to human health because it creates a breeding habitat for disease causing mosquitoes (particularly Anopheles) and snails with its free floating dense roots and semi submerged leaves. It also blocks sunlight entering deep and the waer ways hampering agriculture, fisheries, recreation and hydropower.

Question 45.
Write a comparative note on Carbon Foot Print (CFP).
Answer:
Every human activity leaves a mark just like our footprint. This Carbon foot print is the total amount of green house gases produced by human activities such as agriculture, industries, deforestation, waste disposal, buring fossil fuels directly or indirectly. It can be measured for an individual, family, organisation like industries, state level or national level. It is usually estimated and expressed in equivalent tons of CO₂ per year.

The burning of fossil fuels releases CO₂ and other green house gases. In turn these emissions trap solar energy and thus increase the global temperature resulting in ice melting, submerging of low lying areas and inbalance in nature like cyclones, tsunamis and extreme weather conditions.
To reduce the carbon foot print we can follow some practices like

1. Eating indigenous fruits and products
2. Reduce use of your electronic devices
3. Reduce travelling
4. Do not buy fast and preserved, processed, packed foods
5. Plant a garden
6. Less consumption of meat and sea food. Poultry requires little space, nutrients and less pollution comparing cattle farming
7. reduce use of Laptops (when used for 8 hours, it releases nearly 2 kg. of CO₂ annually)
8. Line dry your clothes.
   (Example: If you buy imported fruit like kiwi, indirectly it increases CFP. How? The fruit has travelled a long distance in shipping or airliner thus emitting tons of CO₂)

Question 46.
Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants.
Specific uses

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

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Name the movement launched by the people of Mandal village to protect the trees by hugging them.
Answer:
Chipko movement.

Question 2.
Recently in January 2019, our Tamil Nadu Government had imposed a ban on using 14 different products of plastic origin. Mention any six plastic products that you know.
Answer:

1. Water Packets
2. Plastic straws
3. Plastic carry bags
4. Thermocol cups
5. Plastic coated paper plates
6. Plastic flags.

Question 3.
Due to plastic ban scheme in our state, people are gradually switching over to other optionals for daily activities. As a biology student, suggest few eco-friendly alternatives for this issue.
Answer:

1. Cloth / Jute bags
2. Plantain leave, palmyra plates as plates
3. Metal cups
4. Earthen pots or ceramic wares
5. Paper made flags

Question 4.
Deforestation is creating various adverse effect on environment. Enlist the consequences of deforestation.
Answer:

1. Increased global warming.
2. Loss of livelihood for forest dwellers.
3. Loss of bio-diversity
4. Decrease in soil fertility.
5. Decline in annual rainfall.

Students can Download Bio Botany Chapter 6 Principles of Ecology Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Bio Botany Solutions Chapter 6 Principles of Ecology

Samacheer Kalvi 12th Bio Botany Principles of Ecology Text Book Back Questions and Answers

Question 1.
Arrange the correct sequence of ecological hierarchy starting from lower to higher level.
(a) Individual organism → Population Landscape → Ecosystem
(b) Landscape → Ecosystem → Biome → Biosphere
(c) community → Ecosystem → Landscape → Biome
(d) Population → organism → Biome → Landscape
Answer:
(a) Individual organism → Population Landscape → Ecosystem

Question 2.
Ecology is the study of an individual species is called
(i) Community ecology
(ii) Autecology
(iii) Species ecology
(iv) Synecology
(a) i only
(b) ii only
(c) i and iv only
(d) ii and iii only
Answer:
(b) ii only

Question 3.
A specific place in an ecosystem, where an organism lives and performs its functions is
(a) habitat
(b) niche
(c) landscape
(d) biome
Answer:
(b) niche

Question 4.
Read the given statements and select the correct option.
(i) Hydrophytes possess aerenchyma to support themselves in water.
(ii) Seeds of Viscum are positively photoblastic as they germinate only in presence of light.
(iii) Hygroscopic water is the only soil water available to roots of plant growing in soil as it is present inside the micropores.
(iv) High temperature reduces use of water and solute absorption by roots.
(a) i, ii, and iii only
(b) ii, iii and iv
(c) ii and iii only
(d) i and ii only
Answer:
(d) i and ii only

Question 5.
Which of the given plant produces cardiac glycosides?
(a) Calotropis
(b) Acacia
(c) Nepenthes
(d) Utricularia
Answer:
(a) Calotropis

Question 6.
Read the given statements and select the correct option.
(i) Loamy soil is best suited for plant growth as it contains a mixture of silt, sand and clay.
(ii) The process of humification is slow in case of organic remains containing a large amount of lignin and cellulose.
(iii) Capillary water is the only water available to plant roots as it is present inside the micropores.
(iv) Leaves of shade plant have more total chlorophyll per reaction centre, low ratio of chi a and chi b are usually thinner leaves.
(a) i, ii and iii only
(b) ii, iii and iv only
(c) i, ii and iv only
(d) ii and iii only
Answer:
(d) ii and iii only

Question 7.
Read the given statements and select the correct option.
Statement A: Cattle do not graze on weeds of Calotropis.
Statement B: Calotropis have thorns and spines, as defense against herbivores.
(a) Both statements A and B are incorrect.
(b) Statement A is correct but statement B is incorrect.
(c) Both statements A and B are correct but statement B is not the correct explanation of statement A.
(d) Both statements A and B are correct and statement B is the correct explanation of statement A.
Answer:
(b) Statement A is correct but statement B is incorrect.

Question 8.
In soil water available for plants is
(a) gravitational water
(b) chemically bound water
(c) capillary water
(d) hygroscopic water
Answer:
(c) capillary water

Question 9.
Read the following statements and fill up the blanks with correct option.

1. Total soil water content in soil is called ______
2. Soil water not available to plants is called ______
3. Soil water available to plants is called ______

Answer:
(a) Holard, Echard and Cheresard

Question 10.
Column I represent the size of the soil particles components. Which of the following is correct match for the Column I and Column II

Answer:
(c) i, ii, i and iv

Question 11.
The plant of this group are adapted to live partly in water and partly above substratum and free from water
(a) Xerophytes
(b) Mesophytes
(c) Hydrophytes
(d) Halophytes
Answer:
(b) Mesophytes

Question 12.
Identify the A, B, C and D in the given table:


Answer:
(a) (+) Parasitism (-) Amensalism

Question 13.
Ophrys an orchid resembling the female of an insect so as to able to get pollinated is due to phenomenon of
(a) Myrmecophily
(b) Ecological equivalents
(c) Mimicry
(d) None of these
Answer:
(c) Mimicry

Question 14.
A free living nitrogen fixing cyanobacterium which can also form symbiotic association with the water fem Azolla
(a) Nostoc
(b) Anabaena
(c) Chlorella
(d) Rhizobium
Answer:
(b) Anabaena

Question 15.
Pedogenesis refers to
(a) Fossils
(b) Water
(c) Population
(d) Soil
Answer:
(d) Soil

Question 16.
Mycorrhiza promotes plant growth by
(a) Serving as a plant growth regulators
(b) Absorbing inorganic ions from soil
(c) Helping the plant in utilizing atmospheric nitrogen
(d) Protecting the plant from infection
Answer:
(d) Protecting the plant from infection

Question 17.
Which of the following plant has a non-succulent xerophytic and thick leathery leaves with waxy coating?
(a) Bryophyllum
(b) Ruscus
(c) Nerium
(d) Calotropis
Answer:
(d) Calotropis

Question 18.
In a fresh water environment like pond, rooted autotrophs are
(a) Nymphaea and typha
(b) Ceratophyllum and Utricularia
(c) Wolffia and pistia
(d) Azolla and lemna
Answer:
(a) Nymphaea and typha

Question 19.
Match the following and choose the correct combination from the options given below:

Answer:
(d) iv, iii, ii, v and i

Question 20.
Strong, sharp spines that get attached to animal’s feet are found in the fruits of
(a) Argemone
(b) Ecballium
(c) Heraitier
(d) Crossandra
Answer:
(a) Argemone

Question 21.
Sticky glands of Boerhaavia and Cleome support
(a) Anemochory
(b) Zoochory
(c) Autochory
(d) Hydrochory
Answer:
(b) Zoochory

Question 22.
Define ecology.
Answer:

• “The study of living organisms, both plants, and animals, in their natural habitats or homes” – Reiter (1885)
• “Ecology is the study of the reciprocal relationship between living organisms and their environment”. – Earnest Haeckel (1889)

Question 23.
What is the ecological hierarchy?
Name the levels of ecological hierarchy.
Answer:
The interaction of organisms with their environment results in the establishment of a grouping of organisms which is called ecological hierarchy.

Question 24.
What are ecological equivalents? Give one example.
Answer:
Taxonomically different species occupying similar habitats (Niches) in different geographical regions are called Ecological equivalents.

Examples:
Certain species of epiphytic orchids of Western Chats of India differ from the epiphytic orchids of South America. But they are epiphytes.

Question 25.
Distinguish habitat and niche.
Answer:
Habitat:

1. A specific physical space occupied by an organism (species).
2. Same habitat may be shared by many organisms (species).
3. Habitat specificity is exhibited by organism.

Niche:

1. a functional space occupied by an organism in the same eco-system
2. A single niche is occupied by a single species
3. Organisms may change their niche with time and season

Question 26.
Why are some organisms called eurythermal and some others as stenohaline?
Answer:
Eurythermal:
Organisms which can tolerate a wide range of temperature fluctuations are called Eurythermal.
Eg : Zostera (a marine angiosperm) and Artemisia tridentata.

Stenohaline:
Organisms which can withstand only a small range of salinity are called stenohaline.
Eg : Plants of estuaries.

Question 27.
‘Green algae are not likely to be found in the deepest strata of the ocean’. Give at least one reason.
Answer:
As the name indicates, green algae possess photosynthetic pigments which use light as an energy source for survival and they are not found in the deepest sea since there is a lack of light.

Question 28
What is Phytoremediation?
Answer:
Some plants can also be used to remove cadmium from the contaminated soil, this is known as phytoremediation.
Example: Rice and Eichhornia (water hyacinth) tolerate cadmium by binding it to their proteins.

Question 29.
What is the Albedo effect and write their effects?
Answer:
Gases let out to atmosphere causes climatic change. Emission of dust and aerosols from industries, automobiles, forest fire,  and DMS (dimethyl sulphur) play an important role in disturbing the temperature level of any region. Aerosols with small particles is reflecting the solar radiation entering the atmosphere. This is known as Albedo effect.

Question 30.
The organic horizon is generally absent from agricultural soils because tilling, e.g., plowing, buries organic matter. Why is an organic horizon generally absent in desert soils?
Answer:

• Organic horizon consist of fallen leaves twigs, flowers and fruits, dead plants, decomposers, animals and their excreta
• Usually it is absent in agricultural and desert land, because
• Desert soil is driest soil. It consist of mostly sandy particle (90-95%). Its water holding capacity is very low.
• It does not have any plants, animals and decomposers to enrich the soil with organic horizon.
• Desert soil is not enough to support a large diverse plant and animal community because of low organic matter and lack of water holding capacity organic horizon generally absent.

Question 31.
Soil formation can be initiated by biological organisms. Explain how?
Answer:
Soil formation is initiated by the biological weathering process. Biological weathering takes place when organisms like bacteria, fungi, lichens and plants help in the breakdown of rocks through the production of acids and certain chemical substances.

Question 32.
Sandy soil is not suitable for cultivation. Explain why?
Answer:

• Sandy soil loses water at a high rate leading to a low level of water retention that is unsuitable for plant growth.
• Sandy soils are less with humus, minerals bacteria, fungi nematodes, insects, earthworm which is helpful for plants growth.
• Soil productivity and soil fertility of sandy soil will be very low.
• The pH value of the soil solution determines the availability of plant nutrients. The pH value of the sandy soil will be very less.

Question 33.
Describe the mutual relationship between the fig and wasp and comment on the phenomenon that operates in this relationship.
Answer:
Wasps present in the figs is an example of mutualism where both the interacting species are benefitted from the obligate association. Wasps acts as a pollinating agents for figs intum larvae of wasps are nourished and protected by fruits of figs.

Question 34.
Lichen is considered as a good example of obligate mutualism. Explain.
Answer:
Lichen is a mutual association of an alga and a fungus, where the algal partner nourishes and provides shelter to fungi, intum fungi confers protection from bacterial infections and also maintains moisture for algal growth.

Question 35.
What is mutualism? Mention any two example where the organisms involved are commercially exploited in modern agriculture.
Answer:

1. Mutualism is an interaction between two species of organisms in which both are benefitted from the obligate
2. association. Nitrogen fixing cyanobacteria associated with Azolla (a fern) and Rhizobium found in the root nodules of leguminous plant are used in the field of agriculture to increase the soil fertility.

Question 36.
List any two adaptive features evolved in parasites enabling them to live successfully on their host?
Answer:
Holoparasites:
The organisms which are dependent upon the host plants for their entire nutrition are called Holoparasites. They are also called total parasites.

Examples:

• Cuscuta is a total stem parasite of the host plant Acacia, Duranta, and many other plants. Cuscuta even gets flower inducing hormone from its host plant.
• Balanophora, Orobanche, and rafflesia are the total root parasites found on higher plants.

Hemiparasites:
The organisms which derive only water and minerals from their host plant while synthesizing their own food by photosynthesis are called Hemiparasites. They are also called partial parasites.
Examples:
Viscum and Loranthus are partial stem parasites.

Question 37.
Mention any two significant roles predation plays in nature.
Answer:
Predation maintains the stability of the food chain in an ecosystem. The population of the insects and small animals is in control due to predation or else it may lead to overgrazing and browsing thereby altering the vegetation.

Question 38.
How does an orchid Ophrys ensure its pollination by bees?
Answer:

• Mimicry is the phenomenon in which a living organism modifies its form, appearance structure (or) behaviour and looks like another living organism as a self defence and increases the chance of its survival.
• Floral mimicry is usually inviting pollinators.

Example:
The plant, Ophrvs an orchid, the flower looks like a female insect to attract the male insect to get pollinated bv the male insect and it is otherwise called ‘floral mimicry1.

Question 39.
Water is very essential for life. Write any three features for plants which enable them to survive in a water scarce environment.
Answer:

1. Presence of a highly developed root system to absorb water.
2. Stems and leaves are covered with waxing coating or dense hairs to avoid transpirational loss.
3. Modified leaves generally leathery and shiny to reflect light and heat.

Question 40.
Why do submerged plants receive weak illumination than exposed floating plants in a lake?
Answer:
Submerged plants like Vallisneria receive dim illumination because the majority of the light are reflected back by the water surface whereas, the floating hydrophytes receive and absorb maximum light as they are on the water surface.

Question 41.
What is vivipary? Name a plant group which exhibits vivipary.
Answer:
Viviparity is the phenomenon, where the seeds germinate and then starts developing to some extent before they detach from the parent plant body. In-plant, it is noticed in Halophytes like Rhizophora.

Question 42.
What is thermal stratification? Mention their types.
Answer:
Thermal Stratification:
It is usually found in aquatic habitats. The change in the temperature profile with increasing depth in a water body is called thermal stratification. There are three kinds of thermal stratification.

1. Epilimnion – The upper layer of warmer water.
2. Metalimnion – The middle layer with a zone of a gradual decrease in temperature.
3. Hypolimnion- The bottom layer of colder water.

Question 43.
How is rhytidome act as the structural defence by plants against fire?
Answer:
Rhytidome is the structural defense by plants against fire. The outer bark of trees which extends to the last formed periderm is called Rhytidome. It is composed of multiple layers of suberized periderm, cortical and phloem tissues. It protects the stem against fire, water loss, invasion of insects and prevents infections by microorganisms.

Question 44.
What is myrmecophily?
Answer:

• Sometimes, ants take their shelter on some trees such as Mango, Litchi, Jamun, Acacia etc.
• These ants act as bodyguards of the plants against any disturbing agent and the plants in turn provide food and shelter to these ants.
• These ants act as bodyguards of the plants against any disturbing agent and the plants, in turn, provide food and shelter to these ants.

Question 45.
What is a seed ball?
Answer:
Seed ball is an ancient Japanese technique of encasing seeds in a mixture of clay and soil humus (also in cow dung) and scattering them on to the suitable ground, not planting trees manually. This method is suitable for barren and degraded lands for tree regeneration and vegetation before the monsoon period where the suitable dispersal agents become rare.

Question 46.
How is anemochory differ from zoochory?
Answer:
Anemochory:

1. Anemochory refers to the seed dispersal by wind.
2. Anemochory seeds are very minute and may have wings or feathery appendages for dispersal.
3. E.g: Orchids

Zoochory:

1. Zoochory refers to the seed dispersal by animals.
2. Zoochory seeds and fruits are very fleshy and succulent and sticky they may have hooks to adhere to the body of animals.
3. E.g: Mango

Question 47.
What is co-evolution?
Answer:

• The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms.
• This type of evolution is called co-evolution. It is a kind of co-adaptation and mutual change among interactive species.

Example:
Corolla length and probosci’s length of butterflies and moths (Habenaria and Moth).

Question 48.
Explain Raunkiaer classification in the world’s vegetation based on the temperature.
Answer:
Raunkiaer classified the world’s vegetation into the following four types. They are megatherms, mesotherms, microtherms and hekistotherms.

Question 49.
List out the effects of fire on plants.
Answer:
Effects of fire:

1. Fire has a direct lethal effect on plants.
2. Burning scars are the suitable places for the entry of parasitic fungi and insects.
3. It brings out the alteration of light, rainfall, nutrient cycle, fertility of soil, pH, soil flora and fauna.
4. Some fungi which grow in soil of burnt areas are called pyrophilous.
5. Example: Pyronema confluence.

Question 50.
What is the soil profile? Explain the characters of different soil horizons.
Answer:
Soil is commonly stratified into horizons at different depths. These layers differ in their physical, chemical, and biological properties. This succession of super-imposed horizons is called a soil profile.

Question 51.
Give an account of various types of parasitism with examples.
Answer:
Parasitism:
It is an interaction between two different species in which the smaller partner (parasite) obtains food from the larger partner (host or plant). So the parasitic species is benefited while the host species is harmed. Based on the host-parasite relationship, parasitism is classified into two types they are holoparasite and hemiparasite.

Holoparasites: The organisms which are dependent upon the host plants for their entire nutrition are called Holoparasites. They are also called total parasites.

Examples:

• Cuscuta is a total stem parasite of the host plant Acacia, Duranta and many other plants. Cuscuta even gets flower inducing hormone from its host plant.
• Balanophora, orobanche and Refflesia are the total root parasites found on higher plants.

Hemiparasites:
The organisms which derive only water and minerals from their host plant while synthesizing their own food by photosynthesis are called Hemiparasites. They are also called partial parasites.

Examples:

• Viscum and Loranthus are partial stem parasites.
• Santalum (Sandal Wood) is a partial root parasite.
• The parasitic plants produce the haustorial roots inside the host plant to absorb nutrients from the vascular tissues of host plants.

Question 52.
Explain different types of hydrophytes with examples.
Answer:
Hydrophytes
The plants which are living in water or wet places are called hydrophytes. According to their relation to water and air, they are subdivided into the following categories:

1. Free-floating hydrophytes
2. Rooted-floating hydrophytes
3. Submerged floating hydrophytes
4. Rooted- submerged hydrophytes
5. Amphibious hydrophytes.

1. Free-floating hydrophytes: These plants float freely on the surface of the water. They remain in contact with water and air, but not with soil. Examples: Eichhornia, Pistia, and Wolffia (smallest flowering plant).

2. Rooted floating hydrophytes: In these plants, the roots are fixed in the mud, but their leaves and flowers are floating on the surface of the water. These plants are in contact with soil, water and air. Examples: Nelumbo, Nymphaea, Potomogeton and Marsilea.

3. Submerged floating hydrophytes: These plants are completely submerged in water and not in contact with soil and air. Examples: Ceratophyllum and Utricularia.

4. Rooted-submerged hydrophytes: These plants are completely submerged in water and rooted in soil and not in contact with air. Examples: Hydrilla, Vallisneria and Isoetes.

5. Amphibious hydrophytes (Rooted emergent hydrophytes): These plants are adapted to both aquatic and terrestrial modes of life. They grow in shallow water. Examples: Ranunculus, Typha and Sagittaria.

Question 53.
Enumerate the anatomical adaptations of xerophytes.
Answer:

1. Presence of multilayered epidermis with heavy cuticle to prevent water loss due to transpiration.
2. The hypodermis is well developed with sclerenchymatous tissues.
3. Sunken shaped stomata are present only in the lower epidermis with hairs in the sunken pits.
4. Scotoactive type of stomata found in succulent plants.
5. Vascular bundles are well developed with several layered bundle sheath.
6. The mesophyll is well-differentiated into palisade and spongy parenchyma.
7. In succulents, the stem possesses a water storage region.

Question 54.
List out any five morphological adaptations of halophytes.
Answer:
Morphological adaptations

1. The temperate halophytes are herbaceous but the tropical halophytes are mostly bushy.
2. In addition to the normal roots, many stilt roots are developed.
3. A special type of negatively geotropic roots called pneumatophores with pneumatophores to get sufficient aeration are also present. They are called breathing roots.
   Example: Avicennia.
4. Presence of thick cuticle on the aerial parts of the plant body.
5. Leaves are thick, entire, succulent and glossy. Some species are aphyllous (without leaves).

Question 55.
What are the advantages of seed dispersal?
Answer:
Advantages of seed dispersal:

• Seeds escape from mortality near the parent plants due to predation by animals or getting diseases and also avoiding competition.
• Dispersal also gives a chance to occupy favourable sites for growth.
• It is an important process in the movement of plant genes, particularly this is the only method available for self-fertilized flowers and maternally transmitted genes in outcrossing plants.
• Seed dispersal by animals helps in the conservation of many species even in human-altered ecosystems.
• Understanding of fruits and seed dispersal acts as a key for proper functioning and establishment of many ecosystems from deserts to evergreen forests and also for the maintenance of biodiversity conservation and restoration of ecosystems.

Question 56.
Describe dispersal of fruit and seeds by animals.
Answer:
Birds and mammals, including human beings, play an efficient and important role in the dispersal of fruit and seeds. They have the following devices.

1. Hooked fruit: The surface of the fruit or seeds have hooks (Xanthium), barbs (Andropogon), spines (Aristida) by means of which they adhere to the body of animals or clothes of human beings and get dispersed.
2. Sticky fruits and seeds:
   • Some fruits have sticky glandular hairs by which they adhere to the fur of grazing animals. Example: Boerhaavia and Cleome.
   • Some fruits have viscid layer which adhere to the beak of the bird which eat them and when they rub them on to the branch of the tree, they disperse and germinate.
     Example: Cordia and Alangium.
3. Fleshy fruits: Some fleshy fruits with conspicuous colours are dispersed by human beings to distant places after consumption.

Samacheer Kalvi 12th Bio Botany Principles of Ecology Additional Questions and Answers

1 – Mark Questions

Question 1.
Who is called the father of Modem Ecology?
Answer:
Eugene P. Odum

Question 2.
Autoecology deals with the study of ________
(a) Community
(b) Population
(c) Individual species
(d) Niche of species
Answer:
(c) Individual species

Question 3.
The environment of any community is called
(a) Paratope
(d) Biotope
(c) Opitope
(d) Biotope
Answer:
(d) Biotope

Question 4.
Match Coloumn I with Column II

Answer:
(a) – iii
(b) – i
(c) – iv
(d) – ii

Question 5.
The study of soil is called as __________
(a) Lithotripsy
(b) Lithosphere
(c) Pedology
(d) Pedology analysis
Answer:
(c) Pedology

Question 6.
Identify the indicators of fire.
(a) Pucinia
(b) Pyricularia
(c) Pyronema
Answer:
(c) Pyronema

Question 7.
The surface features of earth are called __________
Answer:
Topography

Question 8.
Amensalism is called as __________
Answer:
Antibiosis

Question 9.
is the transition zone between two ecosystems.
Answer:
Ecotone

Question 10.
Match the type of species interaction with correct combination. Interaction Type Combination

Answer:
(a) – iii (b) – i (c) – iv (d) – ii

Question 11.
Wasps is the fruits of fig is an example for ________ type of species interaction.
Answer:
Mutualism

Question 12.
Statement 1: Latitudes represent distance from the equator.
Statement 2: Height above the seal level from longitude.
(a) Statement 1 is correct. Statement 2 is incorrect.
(b) Statement 1 is incorrect. Statement 2 is correct.
(c) Both the statements are correct.
(d) Both the statements are incorrect.
Answer:
(a) Statement 1 is correct. Statement 2 is incorrect.

Question 13.
Statement 1: Holoparsites depend totally on other organisms for nutrition.
Statement 2: Dumta is holoparasite.
(a) Statement 1 is correct. Statement 2 is incorrect.
(b) Statement 1 is incorrect. Statement 2 is correct.
(c) Both the statements are correct.
(d) Both the statements are incorrect.
Answer:
(c) Both the statements are correct.

Question 14.
Statement 1: Ephemerals are drought evaders.
Statement 2: They are not true xerophytes.
(a) Statement 1 is correct. Statement 2 is incorrect.
(b) Statement 1 is incorrect. Statement 2 is correct.
(c) Both the statements are correct.
(d) Both the statements are incorrect.
Answer:
(c) Both the statements are correct.

Question 15.
Assertion (A) : Plains and valleys are rich in vegetation
Reason (R): Slow drain of surface water and better water retention is noticed.
(a) A is true R is false
(b) R explains A
(c) A and R are false
(d) A and R are true. But R doesnot explains A
Answer:
(b) R explains A

Question 16.
Utricularia is a _______
(a) Rooted floating hydrophyte
(b) Submerged floating hydrophyte
(c) Rooted submerged hydrophyte
(d) Amphibious hydrophyte
Answer:
(b) Submerged floating hydrophyte

Question 17.
Earth day is observed on
(a) April 22nd
(b) March 21st
(c) July 07th
(d) September 16th
Answer:
(a) April 22nd

Question 18.
Plants in sandy soils are commonly called as _______
Answer:
Psammophytes

2 – Mark Questions

Question 1.
How Earnest Haeckel defined ecology?
Answer:
Earnest Haeckel defined “Ecology is the study of the reciprocal relationship between living organisms and their environment.”

Question 2.
What is ecological hierarchy?
Answer:
The interaction of organisms with their environment results in the establishment of grouping of organisms which is called ecological hierarchy.

Question 3.
Sequentially arrange the different units of ecological hierarchy.
Answer:

Question 4.
Define

1. Autecology
2. Synecology.

Answer:

1. Autecology is the ecology of an individual species and is also called species ecology.
2. Synecology is the ecology of a population or community with one or more species and also called as community ecology.

Question 5.
What is Niche?
Answer:
An ecological niche refers to an organism’s place in the biotic environment and its functional role in an ecosystem.

Question 6.
What are ecological factors?
Answer:
The environment (surrounding) includes physical, chemical and biological components. When a component surrounding an organism affects the life of an organism, it becomes a factor. All such factors together are called environmental factors or ecological factors.

Question 7.
Name the climatic factors that affect plant life.
Answer:
Light, Temperature, Water, Wind and Fire.

Question 8.
Name any four physiological processes in plants, where the light plays a crucial role.
Answer:

1. Photosynthesis
2. Transpiration
3. Seed germination
4. Flowering

Question 9.
Heliophytes differ from Sciophytes. How?
Answer:
Heliophytes – Light loving plants. Example: Angiosperms.
Sciophytes – Shade loving plants. Example: Bryophytes and Pteridophytes.

Question 10.
Based on temperature prevalence, Raunkiaer classified world’s vegetation into four types. Name them.
Answer:
Raunkiaer classified the world’s vegetation into the following four types. They are megatherms, mesotherms, microtherms and hekistotherms.

Question 11.
Distinguish between evergreen forests and sclerophyllous forests.
Answer:
Evergreen Forests: Found where heavy rainfall occurs throughout the year.
Sclerophyllous Forests: Found where heavy rainfall occurs during winter and low rainfall during summer.

Question 12.
What does the term ‘Timber line’ refers to?
Answer:
Timber line is an imaginary line in a mountain or higher areas of land that marks the level above which trees do not grow. The altitudinal limit of normal tree growth is about 3000 to 4000 m.

Question 13.
Compare Euryhaline organisms with stenohaline organisms.
Answer:

1. Euryhaline: Organisms which can live in water with wide range of salinity. Examples: Marine algae and marina angiosperms.
2. Stenohaline: Organisms which can withstand only small range of salinity. Example: Plants of estuaries.

Question 14.
Write the composition of gases in atmosphere.
Answer:
Nitrogen -78% , Oxygen -21%, Carbon-di-oxide -0.03%, Argon and other gases – 0.93%.

Question 15.
What is Albedo effect?
Answer:
Gases let out to atmosphere causes climatic change. Emission of dust and aerosols from industries, automobiles, forest fire, S02 and DMS (dimethyl sulphur) play an important role in disturbing the temperature level of any region. Aerosols with small particles is reflecting the solar radiation entering the atmosphere. This is known as Albedo effect.

Question 16.
Point any two adverse effects of fire in an environment.
Answer:

1. Fire has a direct lethal effect on plants.
2. It brings out the alteration of light, rainfall, nutrient cycle, fertility of soil, pH, soil flora and fauna

Question 17.
Pyronema confluens is the indicator of fire – comment.
Answer:
Pyronema confluens is a fungus which grow on the soil of burnt or fire disturbed areas. Hence it is called as indicator of fire.

Question 18.
What are edaphic factors?
Answer:
Edaphic factors are the abiotic factors related to soil, include the physical and chemical composition of the soil formed in a particular area.

Question 19.
Name the study that deals with soil factors. Also mention the optimal soil pH for crop cultivation.
Answer:
The study of soil is called pedology. The best pH of soil for crop cultivation is 5.5 to 6.8.

Question 20.
Define soil profile.
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.

Question 21.
Given below are few types of plants. Mention their habitats.

1. Halophytes
2. Chasmophytes
3. Cryophytes
4. Psammophytes.

Answer:

1. Halophytes: Plants living in saline soils.
2. Chasmophytes: Plants living in rocky crevices.
3. Cryophytes: Plants living in ice surface.
4. Psammophytes: Plants living in sandy soils.

Question 22.
Mention any four topographic factors that affect vegetation.
Answer:

1. Latitude
2. Altitude
3. Direction of mountain and
4. Steepness of mountain.

Question 23.
How the steepness of mountain affects the vegetation?
Answer:
The steepness of the mountain or hill allows the rain to run off. As a result the loss of water causes water deficit and quick erosion of the top soil resulting in poor vegetation. On the other hand, the plains and valley are rich in vegetation due to the slow drain of surface water and better retention of water in the soil.

Question 24.
Name any two positive interactions with an example for each.
Answer:

1. Mutualism – E.g: Lichen
2. Commensalism – E.g: Orchids.

Question 25.
Define mutualism with an example.
Answer:
Mutualism is an interaction between two species of organisms in which both are benefitted from the obligate association. E.g: Lichens (alga with fungus).

Question 26.
What is the principle of commensalism?
Answer:
Commensalism is an interaction between two organisms in which one is benefitted and the other is neither benefitted nor harmed. The species that derives benefit is called the commensal, while the other species is called the host.

Question 27.
Specify the type of interactions between the given pair of species.

1. Spanish moss and Oak tree
2. Cuscuta and Acacia
3. Nepenthes and Ants
4. Alga and fungus

Answer:

1. Commensalism
2. Parasitism
3. Predation
4. Mutualism

Question 28.
Explain the concept of proto co-operation.
Answer:
Proto Co-operation is an interaction between organisms of different species in which both organisms benefit but neither is dependent on the relationship. Example: Soil bacteria / fungi and plants growing in the soil.

Question 29.
What are Holoparasites? Give example.
Answer:
The organisms which are dependent upon the host plants for their entire nutrition are called Holoparasites. They are also called total parasites.
Example: Cuscuta is a total stem parasite of the host plant Acacia.

Question 30.
What are hemiparasites? Give example.
Answer:
Hemiparasites is the organisms which derive only water and minerals from their host plant while synthesizing their own food by photosynthesis are called Hemiparasites. They are also called partial parasites.
Example: Viscum.

Question 31.
Cite an example for partial stem parasite and partial root parasite.
Answer:

1. Partial stem parasite – E.g: Loranthus.
2. Partial root parasite – E.g: Santalum.

Question 32.
Define Amensalism.
Answer:
Amensalism is an interspecific interaction in which one species is inhibited while the other species is neither benefitted nor harmed. The inhibition is achieved by the secretion of certain chemicals called allelopathic substances. Amensalism is also called antibiosis.

Question 33.
Point out any two morphological adaptations noticed in the roots of hydrophytes.
Answer:

1. Roots are totally absent in Wolffia and Salvinia or poorly developed in Hydrilla or well developed in Ranunculus.
2. The root caps are replaced by root pockets.
   Example: Eichhomia.

Question 34.
What are hygrophytes? Give example.
Answer:
The plants which can grow in moist damp and shady places are called hygrophytes. Examples: Habenaria (Orchid) and Mosses (Bryophytes), etc.

Question 35.
What are trichophyllous plants? Give example.
Answer:
In xerophytic plants, with the leaves and stem are covered with hairs are called trichophyllous plants.
Example: Cucurbits (Melothria and Mukia)

Question 36.
Give an example for following type of adaptations.

1. Phyllode
2. Cladode

Answer:

1. Phyllode – E.g: Acacia.
2. Cladode – E.g: Asparagus.

Question 37.
Write a brief note on pneumatophores. Give an example.
Answer:
Pneumatophores are the special type of negatively geotropic roots developed by halophytes. It possess pneumathodes to get sufficient aeration. They are also called breathing roots.
Example: Avicennia.

3 – Mark Questions

Question 38.
Differentiate habitat from niche.
Answer:
Habitat:

1. A specific physical space occupied by an organism (species).
2. Same habitat may be shared by many organisms (species).
3. Habitat specificity is exhibited by organism.

Niche:

1. A functional space occupied by an organism in the same eco-system.
2. A single niche is occupied by a single species.
3. Organisms may change their niche with time and season.

Question 39.
What is thermal stratification? Explain its types.
Answer:
Thermal stratification is usually found in aquatic habitat. The change in the temperature profile with increasing depth in a water body is called thermal stratification. There are three kinds of thermal stratifications.

1. Epilimnion – The upper layer of warmer water.
2. Metalimnion – The middle layer with a zone of gradual decrease in temperature.
3. Hypolimnion – The bottom layer of colder water.

Question 40.
What are the adverse effects of temperature on plant?
Answer:

1. Temperature affects the enzymatic action of all the bio-chemical reactions in a plant body.
2. Low temperature with high humidity can spread diseases to plants.
3. The varying temperature with moisture determines the distribution of the vegetation types.

Question 41.
Explain briefly about the three types of fire.
Answer:

1. Ground fire – Which is flameless and subterranean.
2. Surface fire – Which consumes the herbs and shrubs.
3. Crown fire – Which bums the forest canopy.

Question 42.
Classify soil based on its formation.
Answer:
Based on soil formation (pedogenesis), the soils are divided into:

1. Residual soils -These are soils formed by weathering and pedogenesis of the rock.
2. Transported soils – These are transported by various agencies.

Question 43.
Loamy soil is ideal for crop cultivation – Justify.
Answer:
Loamy soil is ideal soil for cultivation, since it consists of 70% sand and 30% clay or silt or both. It ensures good retention and proper drainage of water. The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 44.
Direction of mountain determines the richness of vegetation – Justify.
Answer:
North and south faces of mountain or hill possess different types of flora and fauna because they differ in their humidity, rainfall, light intensity, light duration and temperature regions. The two faces of the mountain or hill receive different amount of solar radiation, wind action and rain. Of these two faces, the windward region possesses good vegetation due to heavy rains and die leeward region possesses poor vegetation due to rain shadows (rain deficit).

Question 45.
What are epiphytes? Explain their characteristic features.
Answer:
The plants which are found growing on other plants without harming them are called epiphytes. They are commonly found in tropical rain forest The epiphytic higher plant (Orchids) gets its nutrients and water from the atmosphere with the help of their hygroscopic roots which contain special type of spongy tissue called Velamen. So it prepares its own food and does not depend on the host. They use the host plant only for support and does not harm.

Question 46.
Discuss on predator – prey interaction with example.
Answer:
Predation: It is an interaction between two species, one of which captures, kills and eats up the other. The species which kills is called a predator and the species which is killed is called a prey. The predator is benefitted while the prey is harmed. Many herbivores are predators. Cattles, Camels and Goats etc., frequently browse on the tender shoots of herbs, shrubs and trees.

Question 47.
Give an account of Mimicry.
Answer:
Mimicry is a phenomenon in which living organism modifies its form, appearance, structure or behavior and looks like another living organism as a self defence and increases the chance of their survival. Floral mimicry is for usually inviting pollinators but animal mimicry is often protective. Mimicry is a result of evolutionary significance due to shape and sudden heritable mutation and preservation of natural selection.

Question 48.
Mention any two species that exhibits protective mimicry.
Answer:

1. Carausium morosus (Stick insect)
2. PhyIlium frondosum (Leaf insect).

Question 49.
What is co-evolution? Explain with example.
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.
Examples:

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

Question 50.
How physical dryness differ from physiological dryness?
Answer:

1. Physical dryness: In these habitats, soil has a little amount of water due to the inability of the soil to hold water because of low rainfall.
2. Physiological dryness: In these habitats, water is sufficiently present but plants are unable to absorb it because of the absence of capillary spaces.
   Example: Plants in salty and acidic soil.

Question 51.
Point out the Anatomical adaptations exhibited by the Halophytes.
Answer:
Anatomical adaptations:

1. Epidermal cells of stem is heavily cutinized, almost squarish and are filled with oil and tannins.
2. Star’ shaped sclereids and ‘H’ shaped heavy thickened spicules that provide mechanical strength to cortex are present in the stem.
3. The leaves may be dorsiventral or isobilateral with salt secreting glands.

5-Mark Question

Question 52.
Explain various edaphic factors that affect vegetation.
Answer:
The important edaphic factors which affect vegetation are as follows:

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

Question 53.
What does competition refers to? Classify and describe it.
Answer:
1.Competition: It is an interaction between two organisms or species in which both the organisms or species are harmed. Competition is the severest in population that has irregular distribution. Competition is classified into intraspecific and interspecific.

2. Intraspecific competition: It is an interaction between individuals of the same species. This competition is very severe because all the members of species have similar requirements of food, habitat and pollination etc., and they also have similar adaptations to fulfill their needs.

3. Interspecific competition: It is an interaction between individuals of different species. In grassland, many species of grasses grow well as there is little competition when enough nutrients and water is available. During drought Shortage of water occurs. A life and death competition starts among the different species of grass lands.

Survival in both these competitions is determined by the quantity of nutrients, availability of water and migration to new areas. Different species of herbivores, larvae and grass hopper competing for fodder or forage plants. Trees, shrubs and herbs in a forest struggle for sunlight, water and nutrients and also for pollination and dispersal of fruits and seeds, The Utricularia (Bladderwort) competes with tiny fishes for small crustaceans and insects.

Question 54.
Point out any five morphological adaptations of epiphytes.
Answer:
Morphological adaptations:

1. Root system is extensively developed. These roots may be of two types. They are Clinging roots and Aerial roots. Clinging roots fix the epiphytes firmly on the surface of the supporting objects. Aerial roots are green coloured roots which may hang downwardly and absorb moisture from the atmosphere with the help of a spongy tissue called velamen.
2. Stem of some epiphytes are succulent and develop pseudo bulb or tuber.
3. Generally the leaves are lesser in number and may be fleshy and leathery.
4. Myrmecophily is a common occurrence in the epiphytic vegetation to prevent the predators.
5. The fruits and seeds are very small and usually dispersed by wind, insects and birds.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
Being a tropical country, India is the largest producer of delicious mangoes. These mango tree don’t grow in temperate countries. Give reason.
Answer:
Mango tree show very narrow range of thermal tolerance. Hence they cannot grow in temperate countries.

Question 2.
What is cause of flag forms in trees?
Answer:
Unidirectional wind stimulates the development of flag forms in trees.

Question 3.
In the picture given below, A and B represents the two different biomes. What does the letter C denotes? What will be its impact on the organisms in C Explain with example.
Answer:

The letter C denotes Ecotone – a transition zone between two different ecosystem. Those species found in the ecotone will have impact of environment of two habitats.
E.g. Owl in the ecotone area between forest and grassland.

Question 4.
Observe the tabular column and complete it using proper terms.

Answer:
A = +
B = Commensalism
C = –
D = Parasitism
E = –

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

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 70

Part-I

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

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

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

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

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

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

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

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

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

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

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

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

Question 10.
Match the following:

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

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


Answer:

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

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

Answer:

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

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

Part – II

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Part – III

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

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

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

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

∴ Volume of the sphere with radius ‘r’


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

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

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

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

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

Question 30.
Mention the uses of Glycerol.
Answer:

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

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

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

Secondary structure of proteins :

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

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

(ii) Lidocaine

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

2. \(\mathrm{E}^{0} \mathrm{m}^{2+} / \mathrm{M}\) value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.

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

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

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

6. Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn very feasible [E⁰ = +1.51 V]

Part – IV

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

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

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

[OR]

(b) (i) Uses of helium.

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

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

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

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

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

[OR]

(b) (i)

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

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

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

[OR]

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

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

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

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

vi. Taking log on both sides of equation

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

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

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

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

[OR]

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

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

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

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

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


(b)

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